Physical Chemistry for the JEE and Other Engineering Entrance Examinations 9788131787618, 9789332516366

Table of contents :
Cover
Contents
Preface
Chapter 1: Structure of Atom
1.1 Introduction
1.2 Atomic Theory
1.3 Sub-Atomic Particles
1.3.1 Discovery of Electron
1.3.2 Charge on the Electron
1.3.3 Discovery of Proton
1.3.4 Discovery of Neutron
1.4 Atomic Models
1.4.1 Thomson Model of Atom
1.4.2 Rutherford’s Nuclear Model of Atom
1.5 Atomic Number
1.5.1 Isobars, Isotopes and Isotones
1.6 Developments Leading to the Bohr Model of Atom
1.6.1 Nature of Light and Electromagnetic Radiation
1.6.2 Quantum Theory of Radiation
1.6.3 Photoelectric Effect
1.6.4 Compton Effect
1.6.5 Dual Nature of Electromagnetic Radiations
1.6.6 Atomic Spectra
1.6.7 Types of Spectra
1.7 Bohr’s Model of the Atom
1.7.1 Bohr’s Theory of the Hydrogen Atom
1.7.2 Origin of Spectral Lines and the Hydrogen Spectrum
1.7.3 Limitations of the Bohr’s Model
1. 8 Waves and Particles
1.8.1 Dual Nature of Matter
1.8.2 Heisenberg’s Uncertainty Principle
1.8.3 Significance of Uncertainty Principle
1.8.4 Quantum Mechanical Model of Atom
1.8.5 Schrodinger Wave Equation
1.8.6 The Meaning of Wave Function
1.9 Quantum Numbers
1.10 Shapes of Orbitals
1.10.1 Boundary Surface Diagrams
1.10.2 Energies of Orbitals
1.11 Filling of Orbitals
1.11.1 Electronic Configuration of Atoms
1.11.2 Relative Stabilities of Electronic Configurations
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 2: Basic Concepts of Chemistry
2.1 Introduction
2.2 Submicroscopic Models
2.3 Mixture and Pure Substances
2.4 Physical Properties of Matter
2.4.1 The States of Matter
2.4.2 Chemical Change
2.4.3 The Elements
2.4.4 Compounds and Mixtures
2.4.5 Difference between Physical and Chemical Change
2.5. Properties of Matter and their Measurement
2.5.1 Physical Measurements
2.5.2. The International System of Units (SI Units)
2.5.3 S1 Base Units and S1 Prefixes
2.5.4 Derived Units
2.5.5 Dimensional Analysis
2.6 Significant Figures
2.6.1 Number of Significant Figures
2.6.2 Significant Figures in Calculations
2.6. 3 Exact Numbers
2.7 Laws of Chemical Combinations
2.7.1 Law of Conservation of Mass
2.7.2 Law of Definite Proportions
2.7.3 Law of Equivalents or Law of Reciprocal Proportions
2.7.4 Dalton’s Atomic Theory
2.7.5 Law of Multiple Proportions
2.7.6 Gay-Lussac's is Law of Volumes
2.7.7 Avogadro’s Hypothesis
2.8 Atomic Weights
2.8.1 The Atomic Weight Scale
2.9 Percentage Composition and Formula
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 3: The States of Matter
3.1 Introduction
3.2 Inter-Molecular Forces
3.2.1 Dipole-Dipole Interaction
3.2.2 Ion-Dipole Interactions
3.2.3 Ion-Induced Dipole Interaction and Dipole-Induced Dipole Interaction
3.2.4 Instantaneous Dipole-Induced Dipole Interaction
3.2.5 Hydrogen Bond
3.2.6 Intermolecular Forces vs Thermal Interactions
3.3 The Gaseous State
3.3.1 Mass (m)
3.3.2 Volume (V)
3.3.3 Pressure
3.3.4 Temperature (T)
3.4 The Gas Laws
3.4.1 Boyle’s Law
3.4.2 Charles – Gay Lussac’s Law
3.4.3 Gay Lussac’s Law
3.4.4 Avogadro’s Hypothesis
3.4.5 Ideal Gas Equation
3.4.6 Gas Density
3.4.7 Graham’s Law of Diffusion
3.4.8 Dalton's Law of Partial Pressures
3.5 The Kinetic Theory of Gases
3.5.1 Kinetic Gas Equation
3.5.2 Derivation of Kinetic Gas Equation
3.5.3 Deduction of Gas Laws
3.6 Distribution of Molecular Velocities
3.7 Collision Properties
3.7.1 Calculation of Mean Free Path
3.7.2 Mean Free Path
3.8 Real Gases: Deviation From Ideal Gas Behaviour
3.8.1 Van der Waals Equation
3.8.2 Applicability of Van der Waals Equation to Real Gases
3.8.3 Compressibility Factor
3.9 The Heat Capacities of Gases
3.9.1 Molar Heat at Constant Volume
3.9.2 Molar Heat at Constant Pressure
3.9.3 Joule-Thomson coefficient
3.10 Liquefaction
3.10.1 Law of Corresponding States
3.10.2 Limitations of Van der Waals Equation
3.10.3 Vapour Pressure
3.10.4 Vapour Pressure of Salt Hydrates
3.10.5 Surface Tension
3.10.6 Viscosity
3.10.7 Factors Affecting Viscocity of Liquids
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 4: Solid State
4.1 Introduction
4.2 Crystalline and Amorphous Solids
4.2 Classification of Crystalline Solids
4.2.1 Molecular Solids
4.2.2 Ionic Solids
4.2.3 Covalent or Network Solids
4.2.4 Metallic Solids
4.3 Allotropy and Polymorphism
4.3.1 Enantiotropy
4.3.2 Monotropy
4.3.3 Dynamic Allotropy
4.4 Isomorphism
4.5 The Vapour Pressure and Melting Points of Solids
4.6 Space Lattice and Unit Cell
4.6.1 Primitive and Unit Cell
4.6.2 The Seven Crystal Systems
4.6.3 Contribution of Lattice Points to the Unit Cell
4.7 Packing of Equal Spheres
4.7.1 Interstitial Sites or Interstitial Voids
4.7.2 Radios Ratio of Tetrahedral Void
4.7.3 Radius Ratio of Octahedral Void
4.7.4 Radius Ratio of Triangular Void
4.7.5 Radius Ratio of Cubic Void
4.7.6 Coordination Number
4.7.7 Locating Tetrahedral and Octahedral Voids in Cubic Close Packing
4.8 Efficiency of Packing
4.8.1 Relationship Between the Nearest Neighbour Distance (d) and Radius of Atom (r) and Edge of Unit Cell (a)
4.9 Calculations Involving Unit Cell Dimensions
4.10 Metal Crystals
4.11 Radius Ratio and Structure of Ionic Compounds
4.11.1 Structure of Ionic Compoundsof AB Type
4.12 Crystallography
4.12.1 The Law of Constancy of Interfacial Angles
4.12.2 Elements of Symmetry
4.12.3 Law of Rational Indices
4.12.4 X-rays and Internal Structure of Crystal
4.12.5 Determination of Crystal Structure
4.12.6 Structure of Crystals
4.13 Imperfections in Solids
4.13.1 Point Defects
4.14 Properties of Solids
4.14.1 Electrical Properties
4.14.2 Electrical Conductivity in Metals
4.14.3 Electrical Conductivity in Semi Conductors
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 5: Solutions
5.1 Introduction
5.2 Types of Solutions
5.3 Methods For Expressing The Concentration of A Solution
5.4 Types of Binary Solutions
5.5 Solubility
5.5.1 The Solubility of Solids in Liquids
5.5.2 Cause of Solubility of Solids in Liquids
5.6 Solubility of Gases in Liquids Henry’s Law
5.6.1 Other Forms of Henry’s Law
5.6.2 Characteristics of Henry’s Law Constant KH
5.6.3 Limitations of Henry’s Law
5.6.4 Application of Henry’s Law
5.7 Solution of Liquids in Liquids
5.7.1 Cause of Miscibility and Immiscibility
5.7.2 Distillation of Binary Solutionsof Liquid in Liquid
5.7.3 Fractional Distillation
5.7.4 Fractional Distillation of Solutions Showing Large Positive Deviation from Raoult’s Law
5.7.5 Fractional Distillation of Solutions Showing Larger Negative Deviations from Raoult’s law
5.7.6 Azeotropic Mixtures or Constant Boiling Mixtures
5.8 Colligative Properties
5.8.1 Vapour Pressure of Solutions of Solids in Liquids
5.8.2 Types of Colligative Properties
5.9 Lowering of Vapour Pressure
5.9.1 Raoult’s Law
5.9.2 Lowering of Vapour Pressure –A Colligative Property
5.9.3 Limitations of Raoult’s Law
5.9 4 Derivation of Raoult’s Law
5.9.5 Determination of Molecular Weight or Molecular Mass from Lowering of Vapour Pressure
5.9.6 Determination of the Lowering of Vapour Pressure
5.10 Osmosis and Osmotic Pressure
5.10.1 Demonstration of Osmosis and Osmotic Pressure
5.10.2 Berkeley and Hartely’s Method
5.10.3 Van’t Hoff Theory of Dilute Solutions The Laws of Osmotic Pressure
5.10.4 Van’t Hoff – Avogadro’s Law for Solutions
5.10.5 Osmotic Pressure: A Colligative Property
5.10.6 Relation Between Osmotic Pressure and Vapour Pressure Lowering
5.10.7 Determination of Molecular Weight or Molecular Mass from Osmotic Pressure
5.10.8 Usefulness and Limitations of Van’t Hoff’s Theory of Dilute Solutions
5.10.9 Osmotic Pressure of Mixture of Two Solutions
5.10.10 Reverse Osmosis
5.10.11 Silicate Gardens
5.10.12 Biological Importance of Osmosis
5.11 Elevation in Boiling Point
5.11.1 Determination of Molecular Weight or Molecular Mass from Elevation in Boiling Point
5.11.2 Thermodynamic Derivation
5.11.3 Boiling Point Elevation – A Colligative Property
5.11.4 Relation Between Elevation of B.P.and Relative Lowering of Vapour Pressure
5.11.5 Determination of Boiling Point Elevation
5.12 Depression of Freezing Point
5.12.1 Determination of Molecular Weight or Molecular Mass of a Solute from Depression in F.Pt
5.12.2 Relation Between Depression in F.Pt and Lowering of Vapour Pressure
5.12.3 Relation Between Depression in F.Pt and Osmotic Pressure
5.12.4 Freezing Point Depression – A Coligative Property
5.12.5 Determination of Freezing Point Depression
5.12.6 Application of Depression of Freezing Point
5.13 Abnormal Molecular Weights
5.13.1 Dissociation
5.13.2 Association
5.13.3 Abnormal Molecular Weights
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 6: Thermodynamics
6.1 Introduction
6.2 Thermodynamic Systems
6.2.1 Types of Systems
6.2.2 Thermodynamic Variables
6.2.3 Thermodynamic Equilibrium
6.2.4 State of a System
6.2.5 Homogeneous and Heterogeneous Systems
6.2.6 State Functions
6.2.7 Methods of Studying Thermodynamics
6.2.8 Types of Process
6.2.9 Path
6.3 Heat and Work
6.4 Zeroth Law
6.5 Internal Energy
6.6 The First Law of Thermodynamics
6.6.1 Applications of First Law of Thermodynamics
6.6.2. Work Done in Isothermal and Reversible Expansion of an Ideal Gas
6.6.3 Work Done in an Isothermal Irreversible Expansion of an Ideal Gas
6.6.4 Adiabatic Expansion of Ideal Gas
6.6.5 Work Done in the Irreversible Expansion of an Ideal Gas in Adiabatic Process
6.6.6 Comparison Between Isothermal and Adiabatic Expansion of an Ideal Gas
6.7 Enthalpy
6.7.1 Enthalpy and Standard States
6.8 Second Law of Thermodynamics
6.8.1 The Decrease in Enthalpy is not a Criterion but a Contributor for Spontaneity
6.8.2 Entropy and Spontaneity
6.8.3 The Concept of Entropy
6.8.4 Changes Occurring in an Isolated System
6.8.5 Quantitative Aspects of Entropy
6.8.6 Entropy as a State Function
6.8.7 Entropy Change Spontaneity and Equilibrium: Second law of Thermodynamics
6.8.8 Entropy and Equilibrium State
6.8.9 Entropy Change in Reversible Process
6.8.10 Entropy Change in Irreversible Processes
6.8.11 Entropy Change for Ideal Gases
6.8.12 Entropy Change During Phase Transition
6.8.13 Standard Entropies
6.8.14 Entropy Changes of a Reaction
6.9 Gibbs Energy
6.9.1 Standard Gibbs Energies
6.9.2 The Gibbs Energy of Formation of an Element in its Standard State is Zero
6.9.3 Gibbs Energy Changes Under Non-Standard Conditions
6.9.4 Predicting Spontaneity of a Process
6.9.5 Qualitative Treatment of Gibbs Energy
6.9.6 Equilibrium and Gibbs Energy
6.9.7 Equilibrium and Equilibrium Constants
6.9.8 Coupled Reactions
6.9.9 Gibbs Energy Change and Non-Mechanical Work
6.9.10 Variation of Gibbs Energy with Temperature and Pressure
6.9.11 Clapeyron Equation
6.9.12 Clausius–Clapeyron Equation
6.9.13 Application of Clapeyron’s Classius Equation for Liquid ←→ Vapours Equilibrium
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 7: Thermochemistry
7.1 Introduction
7.2 Energy Stored in Atoms and Molecules
7.2.1 Measuring Heats of Reaction
7.2.2 Thermochemical Equations
7.3. Standard Enthalpies
7.3.1 Standard Enthalpy of an Element
7.3.2 Standard Heats of Formation
7.3.3 Enthalpy Changes in Chemical Reactions
7.3.4 Variation of Heat of Reaction with Temperature: Kirchhoff’s Equation
7.3.5 Heat of Combustion
7.3.6 Enthalpy of Phase Transitions
7.3.7 Enthalpy Changes in Solution
7.4. Laws of Thermochemistry
7.4.1 Enthalpy of Atomization
7.4.2 Bond Enthalpy (ΔbondHO)
7.4.3 Lattice Energies
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 8: Chemical Kinetics
8.1 Introduction
8.2 Rate of Reaction
8.2.1 Reactions Involving Different Stoichiometric Coefficients of Reactants and Products
8.2.2 Average Rate and Instantaneous Rate
8.2.3 Units of Rate of Reaction
8.2.4 Determination of the Rate of Reaction
8.3 Factors which influence the Rate of Reactions
8.4 Rate Laws
8.4.1 Characteristics of Rate Constant
8.4.2 Differences Between Rate of Reaction and Rate Constant
8.5 Rate Law Expression
8.6 Order of Reaction
8.6.1 Pseudo Chemical Reactions
8.7 Molecularity of a Reaction
8.7.1 Why the Reactions of Higher Order are Rare
8.7.2 Mechanism and Rate Law
8.7.3 How to Assign Mechanism to a Reaction
8.8 Integrated Rate Equations
8.8.1 Zero Order Reaction
8.8.2 First Order Reaction
8.8.3 Some Typical First Order Reactions
8.8.4 Applications of the First Order Rate Law Equation
8.8.5 Second Order Reactions
8.8.6 Example of Second Order Rate Equation
8.8.7 Third Order Reactions
8.8.9 Complications in the Determination of Order of Reaction: Complex Reactions
8.9 Effect of Temperature on Rate of Reaction
8.9.1 Activated Molecules and Temperature
8.10 Theories of Reaction Rates
8.10.1 Collision Theory of Reaction Rate
8.10.2 Catalysts and Activation Energy
8.10.3 Transition State Theory
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 9: Chemical Equilibrium
9.1 Introduction
9.1.1 Reversibility of Reactions
9.1.2 Equilibrium in Physical Processes
9.2.1 Solid–Liquid Equilibrium
9.2.2 Liquid–Vapour Equilibrium
9.2.3 Solid–Vapour Equilibrium
9.2.4 Equilibrium Involving Dissolution of Solid or Gases in Liquids
9.2.5 General Characteristics of Equilibrium Involving Physical Process
9.3 Chemical Equilibria
9.3.1 Characteristics of Chemical Equilibrium
9.3.2 Limitations of the Equation for Chemical Equilibrium
9.3.3 Types of Chemical Equilibria
9.4 Law of Mass Action–Equilibrium Constant
9.4.1 Application of Law of Mass Action
9.4.2 Characteristics of Equilibrium Constant
9.4.3 Factors Influencing Equilibrium Constant
9.4.4 Units of Kc and Kp
9.4.5 Relationship Between Kc and Kp
9.4.6 Change in the Values of Kc and Kp with the Change in the Form of Chemical Equation
9.4.7 Heterogeneous Chemical Equilibria
9.5 Aplications of Equilibrium Constants
9.5.1 Predicting the Extent of Reaction
9.5.2 Predicting the Direction of the Reaction
9.6 Temperature Dependenceof Equilibrium Constant
9.7 Calculation of Equilibrium Concentrations and Equilibrium Presures
9.8 Effect of Addition of an Inert Gas
9.8.1 Effect of Addition of Reactants to the Reaction
9.8.2 Dissociation of Dinitrogen Tetroxide
9.8.3 Calculation of Degree of Dissociation from Density Measurements
9.9 Homogeneous Chemical Equilibrium in Liquid State
9.10 Relationship Between Equilibrium Constant K, Reaction Quotient Q and Gibbs Energy G
9.11 Le Chatelier’s Principle
9.11.1 Application to physical Equilibrium
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 10: Ionic Equilibrium
10.1 Introduction
10.2 Arrhenius Theory of Ionization
10.2.1 Ostwald’s Dilution Law
10.2.2 Factors Affecting the Degree of Dissociation
10.2.3. Limitations of Arrhenius Theory
10.3 Acids, Bases and Salts
10.3.1 Arrhenius Theory
10.3.2 The Bronsted-Lowry Theory
10.3.3 Lewis Theory
10.3.4 Influence of Solvents on Acid Strength
10.3.5 Alkalis, Acids and Amphoteric Hydroxides
10.3.6 Determination of Relative Strengths of Acids
10.3.7 Factors Influencing the Strength of an Acid
10.3.8 Acid–Base Strength and the Molecular Structure
10.4 Dissociation of Weak Acids and Weak Bases
10.4.1 Dissociation Constants of Polybasic Acids
10.4.2 Dissociation of Weak Base
10.5 Ionization Constant of Water and its Ionic Product
10.5.1 The pH Scale
10.5.2 Relation Between Ka and Kb
10.5.3 Common Ion Effect in the Ionization of Acids and Bases
10.6 Acid–Base Neutralization —Salts
10.6.1 Hydrolysis of Salts and the pH of Their Solutions
10.7 Buffer Solution
10.7.1 pH Values of Buffer Mixtures
10.8 Acid – Base Indicators
10.8.1 Theory of Indicators
10.8.2 Selection of Indicators in Acid-Base Titrations
10.9 Solubility Product
10.9.1 Common Ion Effect
10.9.2 Applications of Solubility Product
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 11: Redox Reactions
11.1 Introduction
11.2 Redox Reactions Involving Electron Transfer and Bond Breaking
11.2.1 Electron Transfer Involving Essentially Covalent Molecules
11.2.3 Competitive Electron Transfer Reactions
11.3 Oxidation Numbers
11.3.1 Oxidation Numbers of Elements in Covalent Compounds
11.3.2 Rules for Assigning Oxidation Number to an Atom
11.3.3 Average Oxidation Numbers
11.3.4 Oxidation State
11.3.5 Distinction Between Oxidation Number and Valency
11.3.6 Redox Reactions in Terms of Oxidation Number
11.3.7 Oxidation Number and Naming of Compounds
11.4 Types of Redox Reactions
11.4.1 Balancing of Redox Reactions
11.4.2 Balancing of Redox Reactions by Oxidation Number Method
11.4.3 Balancing of Redox Reactions by Ion – Electron Method
11.5 Redox Reactions as the Basis For Titrations
11.5.1 Limitations of Concepts of Oxidation Number
11.6 Redox Reactions and Electrode Processes
11.6.1 Electrode Potential
11.6.2 The Standard Hydrogen Electrode
11.6.3 Standard Electrode Potentials
11.6.4 Reference Electrodes
11.6.5 Factors Affecting the Values of Standard Electrode Potentials
11.6.6 Salt Bridge and its Function
11.6.7 Electrochemical Series
11.6.8 Application of Electrochemical Series
11.7 Importance of the Redox Reactions in Human Activity
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 12: Electrochemistry
12.1 Introduction
12.2 Electrochemical Cells
12.2.1 Representation of an Electrochemical Cell
12.2.2 Electrochemical Changes: Electrolytic Cells and Galvanic Cells
12.2.3 Types of Electrodes
12.2.4 Electromotive Force of the Cell
12.2.5 Reversible and Irreversible Cells
12.3 Electrical Energy
12.3.1 Standard Free Energies of Half-Cell Reactions
12.3.2 Dependence of Redox Potential on Ionic Concentration and on Temperature
12.3.3 Nernst Equation for Single Electrode Reaction
12.3.4 Equilibrium Constant from Nernst Equation
12.3.5 Electrochemical Cell and Gibbs Energy of the Reaction
12.3.6 Relationship Between Electrical Energy and Enthalpy Change of Cell Reaction
12.4 Concentration Cells
12.5 Conductance
12.5.1 Conductance of Electrolytes
12.5.2 Factors Affecting Electrical Conductivity of Electrolytic Solutions
12.5.3 Molar Conductivity or Molar Conductance
12.5.4 Equivalent Conductance
12.5.5 Measurement of the Conductance of Solutions
12.5.6 Factors Affecting Variation of Molar Conductance
12.6 Kohlrausch’s Law
12.6.1 Applications of Kohlrausch's Law
12.6.2 Conductometric Titrations
12.7 Electrolysis
12.8 Overvoltage
12.8.1 Cathode Products
12.8.2 Anode Products
12.8.3 Mercury Cathodes
12.9 Faraday’s Laws of Electrolysis
12.9.1 Faraday's First Law of Electrolysis
12.9.2 Faraday’s Second Law of Electrolysis
12.9.3 Application of Electrolysis
12.10 Batteries
12.10.1 Primary Cells
12.10.2 Secondary Cells
12.10.3 Fuel Cells
12.11 Corrosion
12.11.1 Hydrogen Evolution Type
12.11.2 Differential Oxygenation Corrosion
12.11.3 Passivity of Metals
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 13: Stoichiometry
13.1 Atomic Weights and Equivalent Weights
13.1.1 Determination of Equivalents
13.1.2 Determination of Atomic Weights (Chemical Methods)
13.1.3 Physical Methods of Determining Atomic Masses
13.2 Molecular Weights and Formulae of Gases
13.3 Methods for Determination of Density
13.4 Eudiometry or Gas Analysis
13.4.1 Abnormal Vapour Densities
13.5 Balanced Chemical Equation
13.5.1 Information Conveyed by Chemical Equation
13.5.2 Balancing of Chemical Equation
13.6 Numerical Calculation Based on Chemical Equations
13.7 Titrimetric Method of Analysis
13.7.1 Acid-Base Titrations
13.7.2 Redox Titrations
13.7.3 Precipitation Titrations
13.7.4 Complexometric Titrations
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 14: Surface Chemistry
14 A Adsorption
14.1 Introduction
14.2 Enthalpy of Adsorption
14.3 Types of Adsorption
14.4 Factors Affecting the Adsorption of Gas by Solid
14.5 Adsorption Isotherms
14.5.1 Freundlich Isotherm
14.5.2 Adsorption from Solution
14.5.3 Langmuir Adsorption Isotherm- Langmuir Adsorption Equation
14.6 Applications of Adsorption
Key Points
14 B Catalysis
14.7 Introduction
14.7.1 Promoters
14.7.2 Catalytic Poisons or Anticatalysts
14.7.3 Autocatalyst
14.7.4 Induced Catalysis
14.8 Types of Catalysis
14.8.1 Homogeneous Catalytic Reactions
14.8.2 Heterogeneous Catalytic Reactions
14.9 Characteristics of Catalyst
14.10 Theories of Catalysis
14.10.1 Intermediate Compound Formation Theory
14.10.2 Adsorption Theory
14.10.3 Activity and Selectivity of Heterogeneous Catalysis
14.10.4 Shape Selective Catalysis by Zeolites
14.11 Enzyme Catalysis
14.11.1 Characteristics of Enzyme Catalysis
14.11.2 Mechanism of Enzyme Catalysis
Key Points
14 C Colloids
14.12 Introduction
14.13 Colloidal State –An Intermediate State
14.13.1 Classification of Colloids
14.13.2 Classification Based on Physical State of Dispersed Phase and Dispersion Medium
14.13.3 Classification Based Upon Appearance
14.13.4 Classification Based on Interaction of Phases
14.13.5 Classification Based on Type of Particles of Dispersed Phase
14.13.6 Classification Based on Charge
14.14 Preparation of Colloids
14.15 Purification of Sols
14.16 Properties of Colloidal Solutions
14.16.1 Physical Properties
14.16.2 Colligative Properties
14.16.3 Optical Properties
14.16.4 Kinetic Properties
14.16.5 Electrical Properties
14.16.6 Origin of Charge
14.16.7 Electrical Double Layer
14.16.8 Electrosmosis
14.17 Stability of Colloids
14.17.1 Coagulation
14.17.2 Hardy–Schulz Law
14.17.3 Coagulation of Lyophilic Sols
14.17.5 Gold Number
14.18 Emulsions
14.18.1 Theories of Emulsification
14.19 Associated Colloids
14.20 Gels
14.21 Applications of Colloids
Key Points
Practice Exercise
Answer Keys
Hints and Solutions
Chapter 15: Nuclear Chemistry
15.1 Discovery of Natural Radioactivity
15.1.1 Radioactive Radiations
15.1.2 Nuclear Stability
15.1.3 Measurement of Radioactivity
15.1.4 Units of Radioactivity
15.2 Nuclides
15.2.1 Types of Radioactive Decay
15.2.2 Disintegration Theory
15.2.3 Soddy–Fajan–Russel’s Group Displacement Law
15.3 Rate of Disintegration
15.3.1 Half-Life Period
15.3.2 Average Life or Mean Life Period
15.3.3 Radioactive Equilibrium
15.3.4 Parallel Path Decay
15.3.5 Maximum Yield of Daughter Nuclide
15.4 Radioactive Disintegration Series
15.5 Theories of Nuclear Stability
15.5.1 Neutron / Proton (n/p) Ratio
15.5.2 Mass Defect–Binding Energy
15.5.3 Mass Defect and Packing Fraction
15.6 Nuclear Reactions
15.6.1 Nuclear Reactions Versus Chemical Reactions
15.6.2 Artificially Induced Nuclear Reactions
15.6.3 Types of Nuclear Reactions
15.6.4 Artificial or Induced Radioactivity
15.6.5 Cause of Artificial Radioactivity – Bohr Theory of Compound Nucleons
15.6.6 Nuclear Fission
15.6.7 Nuclear Fusion
15.6.8 Plutonium and the Actinides
15.7 Radioactive Isotopes as Tracers
15.8 Aplications of Radioactive Isotopes
15.8.1 Carbon Dating
15.8.2. The Age of the Earth
15.8.3. Applications in Industry
15.8.4 Application in Agriculture
15.8.5 Application in Medicine
15.8.6 Therapeutic Uses
15.8.7 Applications in Biochemistrys
15.8.8 Applications of Radioisotopes in Chemistry
Answer Keys
Practice Exercise
Answer Keys
Hints and Solutions

Citation preview

Physical Chemistry for the JEE and Other Engineering Entrance Examinations

K. Rama Rao S.V.V. Satyanarayana

Delhi Ÿ Chennai

Copyright © 2013 Dorling Kindersley (India) Pvt. Ltd. Licensees of Pearson Education in South Asia No part of this eBook may be used or reproduced in any manner whatsoever without the publisher’s prior written consent. This eBook may or may not include all assets that were part of the print version. The publisher reserves the right to remove any material in this eBook at any time. ISBN 9788131787618 eISBN 9789332516366 Head Office: A-8(A), Sector 62, Knowledge Boulevard, 7th Floor, NOIDA 201 309, India Registered Office: 11 Local Shopping Centre, Panchsheel Park, New Delhi 110 017, India

Contents Preface

Chapter 1. Chapter 2. Chapter 3. Chapter 4. Chapter 5. Chapter 6. Chapter 7. Chapter 8. Chapter 9. Chapter 10. Chapter 11. Chapter 12. Chapter 13. Chapter 14. Chapter 15.

iv

StruCture of atom BaSiC ConCeptS of ChemiStry the StateS of matter Solid State SolutionS thermodynamiCS thermoChemiStry ChemiCal KinetiCS ChemiCal equiliBrium ioniC equiliBrium redox reaCtionS eleCtroChemiStry StoiChiometry SurfaCe ChemiStry nuClear ChemiStry

1.1–1.86 2.1–2.38 3.1–3.114 4.1–4.90 5.1–5.96 6.1–6.80 7.1–7.46 8.1–8.98 9.1–9.84 10.1–10.112 11.1–11.48 12.1–12.108 13.1–13.86 14.1–14.70 15.1–15.64

Preface Physical Chemistry is the coordination of Physics and Chemistry, of which Oswald laid the foundation. Without the knowledge of those physical methods, which raised Chemistry from being a mere collection of facts to a science with a rational basis, it is simply impossible to study Chemistry, as Robert Bunsen pointed out, “A chemist who is no physicist is almost valueless.” Although certain branches of Chemistry, such as Organic Chemistry continue to retain traces of individuality, it is not possible to study one branch of chemistry in isolation. (It must be noted here that recent studies conducted on organometallic compounds has significantly blurred the line between Organic Chemistry and Inorganic Chemistry.) It is true that the study of neither Organic Chemistry nor Inorganic Chemistry is possible without a clear understanding of Physical Chemistry. This book, targeted at students of classes XI and XII preparing for various competitive examinations, presents a logical and modern approach to the basic concepts of Physical Chemistry. Although the historical approach is often believed to be the best way of inculcating a research outlook—ideally, the basis of all science teaching—we have purposely retained only those ideas and obsolete experimental procedures that provide a historical introduction or establish the experimental basis for current theories. Physical Chemistry involves the theoretical interpretation of experimental observations. However, all too frequently, students are unaware of the methods used in making these observations. So, throughout the book, sections have been devoted to brief descriptions of the apparatus employed and discussions of the fundamental principles concerned in the measurement of various physical properties. This will help students appreciate the logic of the various interpretations. From experience, we know that combining the theory with its numerical application enables students to grasp the subject more easily. Hence, many solved examples and problems for practice have been included in the text, with a clear emphasis on the questions that have appeared in several competitive examinations. The intent is to give students a clear idea of the level of numerical questions that appear in competitive examinations. Solutions to numerical questions are attached as an aid to the understanding of the principles of Physical Chemistry. Physical Chemistry naturally starts with the basic concepts such as development of structure of matter, element, compound, mixtures, development of chemical combinations. These are dealt with in the chapter ‘Basic Concepts of Chemistry’. The study of atom is one of the most outstanding advances in Physical Chemistry that brought about great changes in the presentation of facts of Inorganic as well as Organic Chemistry. All the topics usually dealt with under the heading of Physical Chemistry will be found in this book. It has been made up-to-date as much as possible by the inclusion of recent developments in the theories of several works.

K. Rama Rao S. V. V. Satyanarayana

CHAPTER

1 Structure of Atom

T

he universe is a concourse of atoms. Marcus Aurelius

1.1 iNTRODUCTiON The introduction of atomic theory by John Dalton early in the nineteenth century marks the inception of a modern era in chemical thinking. The virtue of Dalton’s theory was not that it was new or original, for theories of atoms are older than the science of chemistry, but that it represented the first attempt to place the corpuscular concept of matter upon a quantitative basis. The theory of the atomic constitution of matter dates back at least 2,500 years to the scholars of ancient Greece and early Indian philosophers who were of the view that atoms are fundamental building blocks of matter. According to them, the continued subdivision of matter would ultimately yield atoms which would not be further divisible. The word ‘atom’ has been derived from the Greek word ‘a-tomio’ which means ‘uncuttable’ or non-divisible. Thus, we might say that as far as atomic theory is concerned Dalton added nothing new. He simply displayed a unique ability to crystallize and correlate the nebulous notions of the atomic constitution prevalent during the early nineteenth century into a few simple quantitative concepts.

1.2 ATOmiC TheORy The essentials of Dalton’s atomic theory may be summarized in the following postulates: 1. All matter is composed of very small particles called atoms. 2. Atoms are indestructible. They cannot be subdivided, created or destroyed. 3. Atoms of the same element are similar to one another and equal in weight. 4. Atoms of different elements have different properties and different weights. 5. Chemical combination results from the union of atoms in simple numerical proportions.

John Dalton was born in England in 1766. His family was poor, and his formal education stopped when he was eleven years old. He became a school teacher. He was color blind. His appearance and manners were awkward, he spoke with difficulty in public. As an experiment he was clumsy and slow. He had few, if any outward marks of genius. From 1808, Dalton published his celebrated New Systems of Chemical Philosophy in a series of publications, in which he developed his conception of atoms as the fundamental building blocks of all matter. It ranks among two greatest of all monuments to human intelligence. No scientific discovery in history has had a more profund affect on the development of knowledge. Dalton died in 1844. His stature as one of the greatest scientists of all time continues to grow. Thus, to Dalton, the atoms were solid, hard, impenetrable particles as well as separate, unalterable individuals. Dalton’s ideas of the structure of matter were borne out of considerable amount of subsequent experimental evidence as to the relative masses of substances entering into chemical combination. Among the experimental results and relationship supporting this atomic theory were Gay-Lussac’s law of combination of gases by volume, Dalton’s law of multiple proportions, Avagadro’s hypothesis that equal volumes of gases under the same conditions contain the same number of molecules, Faraday’s laws relating to electrolysis and Berzelius painstaking determination of atomic weights.

modern Atomic Theory Dalton’s atomic theory assumed that the atoms of elements were indivisible and that no particles smaller than atoms

1.2

Structure of Atom

exist. As a result of brilliant era in experimental physics which began towards the end of the nineteenth century extended into the 1930s paved the way to our present modern atomic theory. These refinements established that atoms can be divisible into sub-atomic particles, i.e., electrons, protons and neutrons—a concept very different from that to Dalton. The major problems before the scientists at that time were (i) How the sub-atomic particles are arranged within the atom and why the atoms are stable? (ii) Why the atoms of one element differ from the atoms of another elements in their physical and chemical properties? (iii) How and why the different atoms combine to form molecules? (iv) What is the origin and nature of the characteristics of electromagnetic radiation absorbed or emitted by atoms?

1.3.1 Discovery of electron The term ‘electron’ was given to the smallest particle that could carry a negative charge equal in magnitude to the charge necessary to deposit one atom of a 1-valent element by Stoney in 1891. In 1879, Crookes discovered that when a high voltage is applied to a gas at low pressure streams of particles, which could communicate momentum, moved from the cathode to the anode. It did not seem to matter what gas was used and there was strong evidence to suppose that the particles were common to all elements in a very high vacuum they could not be detected. The cathode ray discharge tube is shown in Fig 1.1.

1.3 SUb-ATOmiC PARTiCleS We know that the atom is composed of three basic subatomic particles namely the electron, the proton and the neutron. The characteristics of these particles are given in Table 1.1.

Table 1.1 The three main sub-atomic particles Particle

Symbol

Electron

e

Proton

p

Neutron

n

Mass

Charge

1/1837 of H-atom or 9.109 ×10–28 g or 9.1×10–31 kg 1.008 amu or 1.672×10–24 g or 1.672×10–27 kg (1 unit) 1.0086 amu or 1.675×10–24 g or 1.675×10–27 kg (1 unit)

– 4.8 × 10–10 esu or –1.602 × 10–19 coulombs (–1 unit) + 4.8 × 10–10 esu or + 1.602 × 10–19 coulombs (+ 1 unit) No charge

It is now known that many more sub-atomic particles exist, e.g., the positron, the neutrino, the meson, the hyperon etc, but in chemistry only those listed in Table 1.1 generally need to be considered. The discovery of these particles and the way in which the structure of atom was worked out are discussed in this chapter.

Fig 1.1 Cathode ray discharge tube The properties of these cathode rays are given below: (i) When a solid metal object is placed in a discharge tube in their path, a sharp shadow is cast on the end of the discharge tube, showing that they travel in straight lines. (ii) They can be deflected by magnetic and electric fields, the direction of deflection showing that they are negatively charged. (iii) A freely moving paddle wheel, placed in their path, is set in motion showing that they possess momentum, i.e., particle nature. (iv) They cause many substances to fluoresce, e.g., the familiar zinc sulphide coated television tube. (v) They can penetrate thin sheets of metal. J.J. Thomson (1897) extended these experiments and determined the velocity of these particles and their charge/ mass ratio as follows. The particles from the cathode were made to pass through a slit in the anode and then through a second slit. They then passed between two aluminium plates spaced about 5 cm apart and eventually fell onto the end of the tube, producing a well-defined spot. The position of the spot was noted and the magnetic field was then switched on, causing the electron beam to move in a circular arc while under the influence of this field (Fig 1.2.).

Structure of Atom

Thomson proposed that the amount of deviation of the particles from their path in the presence of electrical and magnetic field depends on (i) Greater the magnitude of the charge on the particle, greater is the interaction with the electric and magnetic fields, and thus greater is the deflection. (ii) Lighter the particle, greater the deflection (iii) The deflection of electrons from its original path increase in the voltage across the electrodes, or the strength of the magnetic field. By careful and quantitative determination of the magnetic and electric fields on the motion of the cathode rays, Thomson was able to determine the value of charge to mass ratio as e = 1.758820 × 1011 C kg -1 me

Fig 1.3 Millikan’s apparatus for determining the value of the electronic charge

(1.1)

me is the mass of the electron in kg and e is the magnitude of the charge on the electron in Coulomb.

me =

Thomson’s experiments show electrons to be negatively charged particles. Evidence that electrons were discrete particles was obtained by Millikan by his well known oil drop experiment during the years 1910-14. By a series of very careful experiments Millikan was able to determine the value – electronic charge, and the mass. Millikan found the charge on the electron to be –1.6 × 10–19 C. The present-day accepted value for the charge on the electron is 1.602 × 10–19 C. When this value for ‘e’ is compared with the most modern value of e/m, the mass of the electron can be calculated.

Cathode (–)

1.6022 × 10-19 C e = e / me 1.758820 × 1011 C kg -1 = 9.1094 × 10–31 kg

1.3.2 Charge on the electron

Anode (+)

1.3

Gas at low pressure

(1.3)

Small droplets of oil from an atomiser are blown into a still thermostated airspace between parallel plates, and the rate of fall of one of these droplets under gravity is observed, from which its weight can be calculated. The airspace is now ionized with an X-ray beam, enabling the droplets to pick up charge by collision with the ionized air molecules. By applying a potential of several thousand volts across the parallel metal plates, the oil droplet can either be speeded up or made to rise, depending upon the direction of the electric field. Since, the speed of the droplet can be related to its weight, the magnitude of the electric field, and the charge it picks up, the value of the charge can be determined.

+

Spot of light when top plate is positive

Spot of light when plate is not charged Deflecting plates

(1.2)

– Fluorescent screen

Fig 1.2 Thomson’s apparatus for determining e/m for the electron

1.4

Structure of Atom

1.3.3 discovery of Proton If the conduction of electricity, through gases is due to particles, which are similar to those involved during electrolysis, it was to be expected that positive as well as negative ones should be involved, and that they would be drawn to the cathode. By using a discharge tube containing a perforated cathode, Goldstein (1886) had observed the formation of rays (shown to the right of the cathode in Fig 1.4.).

noticed a particle of great penetrating power which was unaffected by magnetic and electric fields. It was found to have approximately the same mass as the proton (hydrogen ion). The reaction is represented as 9 4

1 Be + 42 He → 12 6 C +0 n

Where the superscript refers to the atomic mass and the subscript refers to the atomic number (the number of protons in the nucleus). Notice that a new element, carbon, emerges from this reaction.

1.4 Atomic models

Fig 1.4

J.J. Thomson (1910) measured their charge/mass ratio from which he was able to deduce that the particles were positive ions, formed by the loss of electrons from the residual gas in the discharge tube. The proton is the smallest positively charged particle equal in magnitude to that on the electron and is formed from the hydrogen atom by the loss of an electron.

The discovery that atoms contained electrons caused some consternation. Left to themselves, atoms were known to be electrically neutral. So, the negative charge of the electrons had to be balanced by an equal amount of positive charge. The puzzle was to work out how the two types of charges were arranged. To explain this, different atomic models were proposed. Two models proposed by J.J. Thomson and Earnest Rutherford are discussed here though they cannot explain about the stability of atoms.

1.4.1 thomson model of Atom “A theory is a fool and not a creed.” J.J. Thomson Sir Joseph John Thomson 1856-1940 Thomson’s researches on the discharge of electricity through gases led to the discovery of the electron and isotopes.

H → H+ + e– Unlike cathode rays, the characteristics of posi tively charged particles depend upon the nature of the gas present in the cathode ray tube. These are positively charged ions. The charge to mass ratio of these parti cles is found to be dependent upon the gas from which these originate. Some of the positively charged particles carry a multiple of the fundamental unit of electrical charge. The behaviour of these particles in the magnetic or electric field is opposite to that observed for electron or cathode rays.

In 1898, Sir J.J. Thomson proposed that the electrons are embedded in a ball of positive charge (Fig 1.5). This model of the atom was given the name plum pudding or raisin pudding or watermelon. According to this model

1.3.4 discovery of Neutron The neutron proved to be a very elusive particles to track down and its existence, predicted by Rutherford in 1920, was first noticed by Chadwick in 1932. Chadwick was bombarding the element beryllium with α-particles and

Fig 1.5 The Thomson model of atom. The positive charge was imagined as being spread over the entire atom and the electrons were put in this background.

Structure of Atom

we can assume that just like the seeds of a watermelon are embedded within the reddish juicy material, the electrons are embedded in a ball of positive charge. It is important to note that in Thomson’s model, the mass of the atom is uniformly distributed over the atom. Though this model could explain the overall neutrality of the atom, but could not explain the results of later experiments.

the discovery of X-rays and Radioactivity In 1895, Rontgen noticed when electrons strike a material in the cathode ray tubes, produces a penetrating radiation emitted from the discharge tubes, and it appeared to originate from the anode. The radiation had the following properties: (i) It blackened the wrapped photographic film. (ii) It ionized gases, so allowing them to conduct electricity. (iii) It made certain substances fluoresce, e.g., zinc sulphide. Furthermore the radiation was shown to carry no charge since it could not be deflected by magnetic or electric fields. Since Rontgen did not know the nature of the radiation, he named them as X-rays and the name is still carried on. The true nature of X-rays was not discovered until 1912, when it became apparent that its properties could be explained by assuming to be wave like in character, i.e., similar to light but is much smaller wavelength. It is now known that X-rays are produced whenever fastmoving electrons are stopped in their tracks by impinging on a target, the excess energy appearing mainly in the form of X-radiation. The year after Rontgen discovered X-rays, Henry Becquerel observed that uranium salts emitted a radiation with properties similar to those possessed by X-rays. The Curies followed up this work and discov ered that the ore pitchblend was more radioactive than purified uranium oxide; this suggested that something more intensely radioactive than uranium was responsible for this increased activity and eventually the Curies succeeded in isolating two new elements called polonium and radium, which were responsible for this increased activity. In 1889, Becquerel reported that the radiation from the element radium could be deflected by a magnetic field and in the same year, Rutherford noticed that the radiation from uranium was composed of at least two distinct types. Subsequently, it was shown that the radiation from both sources contained three distinct components and are named as α – β and γ-rays. Rutherford found that α-rays consist of high energy particles carrying two units of

1.5

positive charge and four units of atomic mass. He concluded that α-particles are helium nuclei as when α-particles combined with two electrons yielded helium gas. β-rays are negatively charged particles similar to electrons. The γ-rays are high energy radiations like X-rays, are neutral in nature and do not consist of particles. As regards penetrating power, α-particles have the least followed by β-rays (100 times that of α-particles) and γ-rays (1000 times that of α-particles).

1.4.2 Rutherford’s Nuclear model of Atom ernest Rutherford 1871-1937 Rutherford was born in New Zealand in 1871. He was educated at the university of New Zealand and at Cambridge university. He taught at McGill University, and at the university of Manchester. In 1919, he became director of the celebrated Cavendish laboratory at Cambridge. In 1908, he received Nobel prize in chemistry. Rutherford made many of the basic discoveries in the field of radioactivity. With Bohr and others, he elaborated a theory of atomic structure. In 1919, he produced the first artificial transmutation of an element (that of nitrogen into oxygen). For many years. he was a vigorous leader in laying the foundations of the greatest developments in atomic science, which he did not live to see. He died in 1937.

In 1911, Earnest (later Lord) Rutherford demonstrated a classic experiment for testing the Thomson’s model. Rutherford, Geiger and Harsden studied in detail the effect of bombarding a thin gold foil by high speed α-particle (positively charged helium particles). A thin parallel beam of α-particles was directed onto a thin strip of gold and the subsequent path of the particles was determined. Since the α-particles were very energetic, it was thought they would go right through the metal foils. To their surprise they observed some unexpected results which are summarized as follows: (i) It was observed that 99% of the alpha particles passed straight through the foil and struck the screen at the centre. (ii) A few of the alpha particles deflected from their original path through varying angles. (iii) Hardly one out of the 20, 000 α-particles was bounced back on its path. The results of Rutherford’s experiment is represented more explicitly in Fig 1.6.

1.6

Structure of Atom ++ ++

(a) One layer of metal atoms each with nucleus. –

– ++ ++

Alpha particles

–

+ Nucleus ++

–

– –

On the basis of above observations and conclusions, Rutherford proposed the nuclear model of atom as follows: (i) Rutherford suggested that an atom has a centre of the nucleus, in which positive charge and mass are concentrated and he called this centre as the nucleus. The quantitative results of scattering experiments such as Rutherford’s indicate that the nucleus of an atom has a diameter of approximately one to six Fermis (1 Fermi = 10–10 m) and atoms have diameters about 100000 times as great as the size of the nucleus, i.e., of the order 10–5 m. (ii) The atom as a whole is largely an empty space and the nucleus is located at the centre of the atom. (iii) The nucleus is surrounded by the electrons which are revolving round the nucleus in closed paths like the planets. (iv) Electrons and the nucleus are held together by electrostatic forces of attraction. Rutherford, therefore, contemplated the dynamical stability and imagined the electrons to be whirling about the nucleus, similar to the planetary motion.

– ++

++

(b) One atom of the metal with nucleus. Fig 1.6 Results of Rutherford’s experiment.

Rutherford’s Explanation Rutherford pointed out that his results are not in agreement with Thomson’s model. He explained the above results by drawing the following conclusions: (i) As most of the alpha particles passed very nearly straight through the foil, it means that the atom is extra ordinarily hollow with a lot of empty space inside. (ii) The only way to account for the large deflection is to say that the positive charge and mass in the gold foil are concentrated in very small regions. Although most of the alpha particles can go through without any deflection, occasionally some of them which come closer to the region of positive charge, they repel each other, and the repulsion may be big enough to cause the α-particles to undergo large deviations from its original path. (iii) Due to the rigid nature of the nucleus, some a-particles on colliding with the positive charge, turn back on their original paths.

Drawbacks of the Rutherford model Following objections were raised against the Rutherford’s model, and were reported: (i) It did not explain how the protons could be closepacked to give a stable nucleus. (ii) When an electron revolves round the nucleus, it will radiate out energy, resulting in the loss of energy. This loss of energy will make the electron to move slowly and consequently, it will be moving in a spiral path and ultimately falling onto the nucleus. So, the atom should be unstable, but the atom is stable.

Fig 1.7 An electron that is accelerating, radiates energy. As it loses energy, it spirals onto the nucleus. (iii) If an electron starts losing energy continuously, the observed spectrum would be continuous and have broad bands merging into one another. But most of the atoms give line spectra. Thus, Rutherford’s model failed to explain the origin of line spectra.

Structure of Atom

1.5 ATOmiC NUmbeR In 1913, H.G.J. Moseley, a young English Physicist and one of Rutherford’s brilliant students, examined the Xray spectra of 38 elements. When Moseley bombarded different elements with cathode rays, the X-rays were generated which had different frequencies. He suggested that the frequency of X-rays produced in this manner was related to the charge presented on the nucleus of an atom of the element used as anode. When he took the frequencies of any particular line in all elements such as Ka or Kb lines, then the frequencies were related to each other by the equation. v == a (Z–b)

(1.3)

Where, v is the frequency of any particular Ka or Kb etc, line. Z is the atomic number of the element and ‘a’ and ‘b’ are constants for any particular type of line. A plot of v vs Z gives a straight line showing the validity of the above equation. This is shown in Fig 1.8.

Fig 1.8 number.

Variation of x – rays frequency with atomic

However, no such relationship was obtained when the frequency was plotted against the atomic mass. Moseley, further found that the nuclear charge increases by one unit in passing from one element to the next element arranged by Mendeleef in the order of increasing atomic weight. The number of unit positive charges carried by the nucleus of an atom is termed as the atomic number of the element. Since, the atom as a whole is neutral, the atomic number is equal to the number of positive charges present in the nucleus. Atomic number = Number of protons present in the nucleus = Number of electrons present outside the nucleus of the same atom. While the positive charge of the nucleus is due to protons, the mass of the nucleus is due to protons and

1.7

neutrons. The total number of protons and neutrons is called the mass number. The particles present in the nucleus are called nucleons. Mass number (A) = Number of protons (Z) + Number of neutrons (n)

1.5.1 isobars, isotopes and isotones In modern methods, the symbols of elements are written n A as Z [ X ]x , where the left hand superscript A is the mass number, left hand subscript Z is the atomic number, right hand superscript n is the number of charges and right hand subscript is the number of atoms. Different atoms having same mass number but with different atomic numbers are called isobars, e.g., 14 6 C and 14 N . The atoms having same atomic number but with dif7 ferent mass numbers are called isotopes. The difference between isotopes is due to the difference in the number of neutrons present in the nucleus, e.g., 11 H, 12 D and 13T are isotopes of hydrogen namely protium, deuterium and tritium respectively. Isotopes exhibit similar chemical properties be cause they depend on the number of protons in the nucleus. Neutrons present in the nucleus show very little effect on the chemical properties of an element. So, all the isotopes of a given element show same chemical behavior. Sometimes atoms of different elements contain same number of neutrons. Such atoms are called 14 isotones. For example, 13 6 C and 7 N . Isotones differ in both atomic number and mass numbers but the difference in atomic number and mass number is the same.

1.6 DevelOPmeNTS leADiNg TO The bOhR mODel OF ATOm During the period of development of new models to improve Rutherford’s model of atom, two new concepts played a major role. They are (i) Dual behaviour of electromagnetic radiation This means that light has both particle like and wave like properties. (ii) Atomic spectra The experimental results regarding atomic spectra of atoms can only be explained by assuming quantized (fixed) electronic energy levels in atoms. Let us briefly discuss about these concepts before studying a new model proposed by Niels Bohr known as Bohr Model of Atom.

1.8

Structure of Atom

1.6.1 Nature of light and electromagnetic Radiation

Characteristics of the Wave motion 1. Wavelength (l): It is the distance between two nearest crests or troughs. It is denoted by the Greek letter Lambda l and is expressed in Angstrom units (Å). It is also expressed as micron meter (m), mill micron meter (mm), nanometer (nm), picometer (pm) etc. These units are related to SI unit (m) as 1 Å = 10–10 m; 1 m = 10–6 m, 1 mm = 10–9 m, 1 nm = 10–9 m; 1 pm = 10–12 m

Fig 1.9 Diffraction of the waves while passing through a slit

2. Frequency (n): The number of waves passing through a given point in a unit time is known as its frequency. It is denoted by the Greek letter ν (nu). The frequency is inversely proportional to the wavelength. Its unit in cycles per second (cps) of Hertz (Hz); 1 cps = 1 Hz. A cycle is said to be completed when a wave consisting crest and trough passes through a point. 3. Velocity (c): The distance travelled by the wave in one second is called the velocity of wave. It is equal to product of wavelength and frequency of the wave. Thus, c = vl or

Fig 1.10 Propagation of wave

A radiation is a mode of transference of energy of different forms. Light, X-rays and g-radiations are examples of the radiant energy. The earliest view of light, due to Newton, regarded light as made up of particles (commonly termed as corpuscles of light). The particle nature of light explained some of the experimental facts such as reflection and refraction of light. However, it failed to explain the phenomenon of interference and diffraction. The corpuscular theory was therefore, discarded and Huygens proposed a wave-like character of light. With the help of wave theory of light, Huygens explained the phenomena of interference and diffraction. We know that when a stone is thrown into water of a quiet pond, on the surface of water waves are produced. The waves originate from the centre of the disturbance and propagate in the form of up and down movements. The point of maximum upward displacement is called the crest and the point of maximum downward displacement is called trough. Thus, waves may be considered as disturbance which originate from some vibrating source and travel outwards as a continuous sequence of alternating crests and troughs as shown in Fig 1.10.

v=

c l

(1.4)

4. Amplitude: It is the height of crest or depth of wave’s trough and is generally expressed by the letter ‘a’. The amplitude of the wave determines the intensity or brightness of radiation. 5. Wave number (v ) : It is equal to the reciprocal of wavelength. In other words, it is defined as the number of wavelengths per centimeter. It is denoted by v and is expressed in cm–1. Thus, 1 (1.5) v= l c c But v = or l = l v v=

l c

(1.6)

electromagnetic Radiation In 1873 Maxwell proposed that light and other forms of radiant energy propagate through space in the form of waves. These waves have electric and magnetic fields associated with them and are, therefore, called electromagnetic radiations and electromagnetic waves. The important characteristics of electromagnetic radiations are (i) These consist of electric and magnetic fields that oscillate in the directions perpendicular to the direction in which the wave is travelling as shown in

Structure of Atom

Fig 1.11. The two field components have the same wavelength and frequency. (ii) All electromagnetic waves travel with the same speed. In vacuum, the speed of all types of electromagnetic radiation is 3. 00 × 108 ms–1. This speed is called the velocity of light. (iii) These electromagnetic radiations do not require any medium for propagation. For ex ample, light reaches us from the sun through empty space.

Fig 1.11 Electric and magnetic fields association with an electromagnetic wave electromagnetic Spectrum The different electromagnetic radiations have different wavelengths. The visible light in the presence of which our eyes can see, contains radiations having wavelength

1.9

between 3800 Å to 7600 Å. The different colours in the visible light corresponds to radiations of different wavelengths. In addition to visible light there are so many other electromagnetic radiations, such as X-rays, ultraviolet rays, infra-red rays, microwaves and radiowaves. The arrangement of different types of electromagnetic radiations in the order of increasing wavelengths (or decreasing frequencies) is known as electromagnetic spectrum. Different regions of electromagnetic spectrum are identified by different names. The complete electromagnetic spectrum is shown in Fig 1.12. The various types of electromagnetic radiations have different energies and are being used for different purposes as listed in Table 1.2. Table 1.2 waves

Some applications of electromagnetic

Name

Frequency

Wavelength

g-rays X-rays

21

10 – 10 1017 – 1019

10 10–10

Ultraviolet Visible Infrared Microwave Radio frequency

1015 – 1016 1013 – 1014 1012 – 1013 109 – 1011 105 – 108

10–7 10–6 10–4 10–2 102

20

Cosmic rays

Fig 1.12 Complete electromagnetic spectrum

–12

Uses Cancer treatment Medical “pictures”, material testing Germicidal lamps Illumination Heating Cooking, radar Signal transmission

1.10

Structure of Atom

Solved Problem 1

1.6.2 Quantum Theory of Radiation

Calculate and compare the energies of two radiations, one with a wavelength of 8000 Å and the other with 4000 Å. Solution: According to Planck’s equation

When an object is heated, its colour gradually changes. For example, a black coal becomes red, orange, blue and finally ‘white hot’ with increasing temperature. This shows that red radiation is most intense at a particular temperature, blue radiation is most intense at another temperature and so on. It indicates that the intensities of radiations of different wavelengths emitted by a hot body depend upon the temperature. The curves representing the distribution of radiation from a black body at different temperatures are shown in Fig 1.13.

E = hv = h

c l

h = 6.60 × 10–34 J s; c = 3 × 1010 ms–1 Also l1 = 8000 Å = 8000 × 10–10 m and l2 = 4000 Å = 4000 × 10–10 m E1 =

6.62 × 10-34 J s × 3 × 1010 m s -1 8000 × 10-10 m

= 2.475 × 10–19 J and E2 =

6.62 × 10-34 J s × 3 × 1010 m s -1 4000 × 10-10 m

= 4.95 × 10–19 J E1 2.475 × 10-19 J = E2 4.95 × 10-19 J or

E1 1 = E2 2

or 2E1 = E2 Fig 1.13 Energy distribution for black body radiation at different absolute temperatures

Solved Problem 2 A radio station is broadcasting a programme at 100 MHz frequency. If the distance between the radio station and the receiver set is 300 Km, how long would it take the signal to reach the receiver set from the radio station? Also calculate wavelength and wave number of these radio waves. Solution: All electromagnetic waves travel in vacuum or in air with the same speed of 3 × 10–8 m s–1 Time taken =

Distance 300 × 1000 m = = 1× 10-3 s Velocity 3 × 108 m s -1

Calculate of wavelength (l), c = vl l=

c 3 × 108 m s -1 = 3m = 100 × 106 s -1 v

Calculation of wave number ( v ) v =

1 1 = = 0.33 m -1 l 3m

A black body is one which will absorb completely the incident radiation of wavelength. The ideal black body does not reflect any energy, but it does radiate energy. Although no actual body is perfectly black, it is possible to postulate a condition where radiation is completely enclosed in a space surrounded by thick walls at a constant temperature. Thus, we may imagine a cavity inside a solid body containing a small hole in one wall whereby the investigator may study the enclosed radiation without altering its character. When a beam of light is passed through the hole, it would be absorbed by the inner walls of the enclosure and their temperature increases. As soon as the beam was removed, since the enclosed kept at a constant temperature the excess energy emerges through the hole in the form light. This gives a distinctive spectrum, usually referred to as the black body spectrum. If the energy emitted is plotted against its frequency or wavelength where n = c / l, the curve shows a maximum, as indicated in Fig 1.13. The graph shows that the energy emitted is

Structure of Atom

1.11

greatest at the middle wavelengths in the spectrum and least at the highest and lowest frequencies. If curves are plotted for a series of temperatures, as given in Fig 1.13. It is found that the maximum moves towards shorter wavelengths as the temperature rises.

max Planck 1858-1947 A German physicist. He received his Ph. D. in theoretical physics from university of Munich in 1879. Max Planck was well known for his Quantum theory which won the Nobel Prize in physics in 1918. He also made significant contributions in thermodynamics and other areas of Physics. The shapes of the curves could not be explained on the basis of wave theory of radiation. Max Planck in 1900 resolved this discrepancy by postulating the assumption that the black body radiates energy — not continuously but discontinuously in the form of energy packets called quanta. The general quantum theory of electromagnetic radiations can be stated as (i) When atoms or molecules absorb or emit the radiant energy, they do so in separate units of waves. These waves are called quanta or photons. (ii) The energy of a quantum or photon is given by E = hv

(1.7)

where, v is the frequency of the emitted radiation and h is the Planck’s constant. (iii) An atom or molecule emits or absorbs either one quantum of energy (hv) or any whole number multiples of this unit. This theory provided the basis for explaining the photoelectric effect, atomic spectra and also helped in understanding the modern concepts of atomic and molecular structure.

Quantized energy (a)

Continuous change of energy (b)

Fig 1.14

1.6.3 Photoelectric effect When a beam of light falls on a clean metal plate in vacuum, the plate emits electrons. This effect was discovered by Hertz in 1887, and is known as the photoelectric effect. The metal surface emits electrons by the action of light, that can be demonstrated using negatively charged gold leaf electroscope (Fig 1.15). As the light from carbon arc falls on the metal plate, the diversion of leaves is reduced slowly. This shows that electrons come out of all metal plates. Extensive studies have revealed the following facts: (i) Photoelectric effect is instantaneous, i.e., as soon as the light rays fall on the metal surface, electrons are ejected. (ii) Only light of a certain characteristic frequency is required for expulsion of electrons from a particular metal surface.

Quantization of energy The restriction of any property to discrete values is called quantization. A quantity cannot vary continuously to have any arbitrary values but can change only discontinuously to have some specific values. For example, a ball moves down a staircase (Fig 1.14a) then the energy of the ball changes discontinuously and it can have only certain definite values of energy corresponding to the energies of various steps. Energy of the ball in this case is quantized. On the other hand, if the ball moves down a ramp (Fig 1.14 b), then the energy of the ball changes continuously and the ball can have any value of energy corresponding to any point on the ramp. Energy in this case in not quantized.

Fig 1.15 (a) Einstein’s explanation of photoelectric effect, and (b) Experimental device for photoelectric effect.

1.12

Structure of Atom

(iii) The number of electrons emitted are directly proportional to the intensity of the light. (iv) The gas surrounding the metal plate has no effect on photoelectric transmission. (v) Kinetic energy of photoelectron depends upon the nature of metal. (vi) For the same metal kinetic energy of photoelectrons varies directly with the frequency of light. If the frequency is decreased below a certain value (Threshold frequency), no electrons are ejected at all. The classical wave theory of light was completely inadequate to interpret all these facts. Einstein applied the quantum theory of radiation to explain this phenomenon. An electron in a metal is found by a certain amount of energy. The light of any frequency is not able to cause the emission of electrons from the metal surface. There is certain minimum frequency called the threshold frequency which can just cause the ejection of electrons. Suppose the threshold frequency of the light required to eject electrons from the metal is v0. When photon of light of this frequency strikes the metal surface, it imparts its entire energy (hv0) to the electron. This enables the electrons to break away from the atom by overcoming the attractive influence of the nucleus. Thus, each photon can eject one electron. If the frequency of the light is less than n0, the electrons are not ejected. If the frequency of the light (say v) is more than v0, more energy is supplied to the electron. The remaining energy, which will be the energy hv imparted by incident photon and the energy used up, i.e., hv0, would be given as kinetic energy to the emitted electron. Hence, 1 hv = hv0 + mv 2 2

(1.8)

1 2 where mv is the kinetic energy of the emitted 2 electron and hv0 is called the work function. 1 2 mv = h(v – v0) 2

or

(1.9)

This equation is known as Einstein’s photoelectric equation. If the frequency of light is less than the threshold frequency, there will be no ejection of electrons. Values of photoelectric work functions of some metals are given below. Metal

Li

Na

K

Mg

Cu

Ni

W (eV)

2.42

2.3

2.25

3.7

4.8

4.3

Solved Problem 3 Calculate the energy in one photon of yellow light of wavelength 589 nm.

Solution: The energy E, of one photon E = hn =

hc l

Substituting the values E=

(6.626 × 10-34 Js) (3.00 × 108 m s -1 ) = 3.375 × 10-19 J 5890 × 10-10 m

Solved Problem 4 When a radiation of wavelength 200 nm falls on a metal surface, the work function of the metal being 5 eV, the electron is ejected. What is the velocity of the electron? Solution: Work function 5 eV = 5 × 1.602 × 10–19 J = 8.01 × 10–19 J (\1eV = 1.602 × 10–19 J) Work function = hv0 v0 = v=

Work function 8.01× 10-19 J = = 1.21× 10-15 S-1 h 6.626 × 10-34 J c 3.0 × 108 m s -1 = l 200 × 10-9 m

= 1.5 × 1015 s -1 V2 =

1 2 mv = h ( v - v0 ) 2

2h 2 × 6.626 × 10-34 ( v - v0 ) = (1.5 - 1.21) ×1015 m 9.11× 10-31 kg

V = 0.649 × 106 m s -1 Solved Problem 5 A photon of wavelength 253.7 nm strikes a copper plate, the work function of the metal being 4.65. eV. Calculate the energy of photon and kinetic energy of the emitted photoelectron. Solution: (i) Energy of the photon v=

c 3 × 108 m s -1 = = 11.824 × 1014 s -1 -9 l 253.7 × 10 m

E = hv = ( 6.626 × 10-34 Js )(11.824 × 1014 s -1 ) = 7.83 × 10–19 J Since, 1 eV = 1.602 × 10–19 J 7.83 × 10-19 J \E = = 4.887 eV 1.602 × 10-19 J (ii) Kinetic energy of the photon Work function = 4.65 eV KE = hv – Work function = 4.887 eV – 4.65 eV = 0.237 eV

Structure of Atom

1.6.4 Compton effect According to classical electromagnetic wave theory, monochromatic light falling upon matter should be scattered without change in frequency. But when X-rays are impinged on matter of low atomic weight, X-rays of slightly longer wavelength than those of impinging beam were produced. In other words when a photon collides with an electron, it may leave with a lower frequency, and the electron thereby acquires a greater velocity. In the collision the energy of the photon is reduced and the energy of the electron is increased. This is called Compton effect and can be explained as a result of the impact between two bodies, the photon and the electron and it is of course additional proof of the corpuscular nature of light. This impact is elastic, and both kinetic energy and momentum are conserved. After the encounter each body has a different energy and a different momentum from the values it had before the contact, but the sum of the energies and the sum of the momenta are unchanged.

1.6.5 Dual Nature of electromagnetic Radiations As described, photoelectric effect and Compton effect could be explained considering that electromagnetic radiations consisting of small packets of energy called quanta or quantum (or single photon). These packets of energy can be treated as particles, on the other hand, radiations exhibit phenomena of interference and diffraction which indicate that they possess wave nature. So, it may be concluded that electromagnetic radiation possesses the dual nature, i.e., particle nature as well as wave nature. Einstein (1905) even calculated the mass of the photon associated with a radiation of frequency v as given below: The energy E of the photon is given as E = hv Albert einstein 1879-1955 Born in Germany but later shifted to America. He was regarded as one of the two great Physicists the world has known, the other being lssac Newton. He is well known for his three research papers on special relativity, Brownian motion and photoelectric effect. These research papers were published in 1905 while he was working as a technical assistant in a Swiss patent office in Berne. He received the Nobel Prize in Physics in 1921 for his explanation of the photoelectric effect. His work has influenced the development of physics in significant manner. “To him who is discoverer in the field (or science), the products of his imagination appear so necessary and natural that he regards them, and would like to have them regarded by others, not as creations of thought but as given realities”— Albert Einstein

1.13

Also according to Einstein’s mass energy equation E = mc2 where m is the mass of photon. From the above two equations, we get hv = mc2

(1.10)

hv (1.11) c2 h c (1.12) m= 2⋅ c l h (1.13) m= cl Thus, the mass of the photon can be calculated. m=

1.6.6 Atomic Spectra The velocity of light depends upon the nature of the medium through which it passes. Because of this reason while a beam of light passing from one medium to another, deviates or refracts from its original path, the radiations with shorter wavelength deviate more than the one with a longer wavelength. If a ray of sunlight or any white light is allowed to pass through a prism, it splits up into a continuous band of seven colors from red to violet as observed in a rainbow. This phenomenon is known as dispersion and the pattern of colors is called a spectrum. The red color light radiation having longest wavelength will be deviated least while the radiation of violet color having shortest wavelength will be deviated more. The spectrum of white light that we can see ranges from violet at 7.50 × 1014 Hz to red at 4 × 1014 Hz. In this spectrum one color merges into another color adjacent to it viz violet merges into blue, blue into green and so on. When electromagnetic radiation interacts with matter, atoms and molecules may absorb energy and reach to a higher energy state. While coming to the ground state (more stable lower energy state) the atoms and molecules emit radiations in various regions of the electromagnetic spectrum.

1.6.7 Types of Spectra Spectra are of two types: (i) Emission spectra, and (ii) Absorption spectra (i) Emission spectra When energy is supplied to a sample by heating or irradiating it, the atoms or molecules or ions present in the sample absorb energy and the atoms, molecules or ions having higher energy are said to be excited. While coming to the ground (normal) state the excited species emit the absorbed energy. The spectrum of such emitted radiation is called the emission spectrum. Emission spectra are further classified according to its appearance as continuous, line and band spectra.

1.14

Structure of Atom

Film Excited sample

Prism Incresing wavelength

(ii) Absorption spectra It is produced when the light from a source emitting a continuous spectrum is first passed through an absorbing substance and then recorded after passing through a prism in a spectroscope. In that spectrum, it will be found that certain colours are missing which leave dark lines or bands at their places. This type of spectrum is called the absorption spectrum. Similar to emission spectra, absorption spectra are also of three types:

Emission spectrum

Absorbing sample

Film

Source of white light

Prism Increasing wavelength

Red

Yellow Orange

Green

Indigo Blue

Absorption spectrum

Violet

Fig 1.16 (a) Continuous spectrum When the source emitting light is an incandescent solid, liquid or gas at a high pressure, the spectrum so obtained is continuous. In other words, this type of spectrum is obtained whenever matter in the bulk is heated. For example, hot iron and hot charcoal gives the continuous spectra. (b) Line spectrum Line spectrum is obtained when the light emitting substance is in the atomic state. Hence, it is also called as the atomic spectrum. Line spectrum consists of discrete wavelengths extended throughout the spectrum and are generally obtained from the light sources like mercury, sodium, neon discharge tube, etc. (c) Band spectrum This type of spectrum arises when the emitter in the molecular state is excited. Each molecule emits bands which are characteristics of the molecule concerned and that is why we call this as molecular spectrum also. The sources of band spectrum are (i) carbon with a metallic salt in its core, (ii) vacuum tube, etc. In the emission spectra bright lines on black background will appear.

Visible spectrum

Prism Slit Source of white light

Emission spectrum of Barium

Continuous emission spectrum of white light/sun

Fig 1.18 Absorption spectrum of Barium

Fig 1.17 Illustrating Emission and Absorption Spectra It may be noted that the dark lines in absorption spectra appear exactly at the same place where the coloured lines appear in the emission spectra.

Emission and absorption spectra

(a) Continuous absorption spectrum This type of spectrum arises when the absorbing material absorbs a continuous range of wavelengths. An interesting example is the one in which red glass absorbs all colors except red and hence, a continuous absorption spectrum will be obtained. (b) Line absorption spectrum In this type, sharp dark lines will be observed when the

Structure of Atom

Brackett Series

0

20,000

hydrogen Spectrum As explained above, each element emits its own characteristic line spectrum which is different from that of any other element. Since, hydrogen contains only one electron, its spectrum is the simplest to analyze. To get the spectrum of hydrogen, the gas is enclosed in a discharge tube under low pressure and electric discharge is passed through it. The hydrogen molecules dissociate into atoms and get excited by absorbing the energy: On the principle that what goes up must come down, sooner or later the atoms must loose their energy and fall back to the ground state, by loosing the energy as light. The light that is given out can be measured in a spectrometer and the pattern recorded on a photographic paper. The pattern is called hydrogen spectrum. The hydrogen spectrum obtained consists of a series of lines in the visible, ultraviolet and infrared regions. These have been grouped into five series in Fig 1.19 which are named after their discoverers. These are given in Table 1.3 with the year when discovered. The Balmer series happened to be the first series of lines in the hydrogen spectrum. This was because the lines were in the visible part of the spectrum and therefore the easiest to observe. Balmer (1885) discovered a relationship between the wave number and the position of the line in the series.

Lyman Series

Balmer Series

Pfund Series Paschen Series

absorbing substance is a vapour or a gas. The spectrum obtained from sun gives Fraunhofer absorption lines corresponding to vapours of different elements which are supposed to be present on the surface of the sun. (c) Band absorption spectrum When the absorption spectrum is in the form of dark bands, this is known as band absorption spectrum An interesting example is that of an aqueous solution of KMnO4 giving five absorption bands in the green region. The pattern of lines in the spectrum of an element is characteristics of that element and is different from those of all other elements. In other words, each element gives a unique spectrum irrespective of even the form in which it is present. For example, we always get two important lines 589 nm and 589.6 nm in the spectrum of sodium whatever may be its source. It is for this reason that the line spectra are also regarded as the fingerprints of atoms. Since atoms of different elements give characteristic sets of lines of definite frequencies, emission spectra can be used in chemical analysis to identify and estimate the elements present in a sample. The elements rubidium, caesium, thallium, gallium and scandium were discovered when their minerals were analyzed by spectroscopic methods. The element helium was discovered in the Sun by spectroscopic method.

Violet δ

α /cm–1

40,000

60,000

80,000

3500

Blue Blue-green γ β

4000

4500 5000 5500 6000 Wavelength in angstroms

100,000 (b) The Balmer-series spectrum of hydrogen.

(a) Relative location of the Lyman, Balmer, Paschen, Brackett and Pfund series of hydrogen spectrum.

Fig 1.19

1.15

Hydrogen spectrum

Table 1.3 Series

Region of spectrum

Lyman (1914) Balmer (1885) Paschen (1908) Brackett (1922) Pfund (1924)

Ultraviolet Visible Infrared Infrared Infrared

Value of n

Values of m

1 2 3 4 5

2, 3, 4, ... 3, 4, 5, ... 4, 5, 6, ... 5, 6, 7, ... 6, 7, 8, ...

Red α

6500

7000

1.16

Structure of Atom

The relation is

Solved Problem 7

1 1   1 = v = RH  2 - 2  (1.14) l 2 m  where, RH is the Rydberg constant and is equal to 109677 cm–1. v is the wave number and m is an integer having values 3, 4, 5, 6, etc. Ritz (1908) gave a generalization known as Ritz Combination principle which was nicely applicable to hydrogen spectrum. According to this principle, the wave number in any line in a series can be represented as a difference of two square terms, one of which is consistent and the other varies throughout the series. Thus, 1  1 v = RH  2 - 2  (1.15) m  n where, RH is Rydberg constant. All the lines appearing in the hydrogen spectrum are governed by the above equation. For, 1  1 Lyman series v = RH  2 - 2  ; m = 2,3, 4 1 m 

(1.16)

1  1 Paschen series v = RH  2 - 2  ; m = 4,5, 6 3 m  

(1.17)

1  1 Brackett series v = RH  2 - 2  ; m = 5, 6, 7 4 m 

(1.18)

1  1 v = RH  2 - 2  ; m = 6, 7,8 5 m 

(1.19)

Pfund series

Solved Problem 6 An electronic transition from M shell (m = 3) to K shell (n = 1) takes place in a hydrogen atom. Find the wave number and the wavelength of radiation emitted (R = 1,09,677 cm–1). Solution: v=

1 1  1 = RH  2 - 2  , n = 1 and m = 3 l m  n 8 1 1  = 1, 09, 677  2 - 2  = 109677 × 9 1 3 

v = 97, 491 cm -1 Wavelength, l=

 1 1 = = 1.026 × 10-5 cm = 1026 A v 97491

In hydrogen atom, an electron jumps from fourth orbit to first orbit. Find the wave number, wavelength and the energy associated with the emitted radiation. Solution: v=

1 15 1 1  = RH  2 - 2  cm -1 = 109677 × cm -1 = 102822 cm -1 16 l 1 4 

1.7 bOhR’S mODel OF The ATOm Neils bohr 1885–1962 Born in Denmark in 1885. He received Ph. D. from the university of Copenhagen in 1911. He then spent a year with J.J. Thomson and Earnest Rutherford in England. In 1913, he returned to Copenhagen and in 1920 he was named as Director of the Institute of Theoretical Physics. Bohr’s theory of atomic structure, which was presented in 1913, laid a broad foundation for the great atomic progress of recent years. Second only in importance to his celebrated theory are the facts that it was in Bohr’s laboratory that the implications of nuclear fission were first predicted and that he obtained an understanding of nuclear stability that contributed greatly to the spectacular development of atomic energy. After the first world war, Bohr worked energies for peaceful uses of atomic energy. He received the first Atoms for Peace award in 1957. Bohr was awarded the Nobel Prize in Physics in 1922. “...... The very word “experiment” refers to a situation where we can tell others what we have done and what we have learned” — Neils Bohr. “Our experiments are questions that we put to Nature.” — Neils Bohr. In order to explain the line spectra of hydrogen and the overcome to objections leveled against Rutherford’s model of the atom, Neils Bohr (1913) proposed his quantum mechanical structure of the atom. He made use of quantum theory, according to which energy is lost or gained not gradually, but in bundles or quanta. The concept of nucleus explained by Rutherford was retained in Bohr’s model. Through this model does not meet the modern quantum mechanics, it is still in use to rationalize many points in the atomic structure and spectra. The important postulates of the Bohr Theory are 1. Electrons move in certain fixed orbits associated with a definite amount of energy. The energy of an electron moving in an orbit remains constant as long as it stays in the same orbit called stationary state or orbit.

Structure of Atom

Hence, a certain fixed amount of energy is associated with each electron in a particular orbit. Thus, stationary orbits are also known as energy levels or energy shells. Bohr gave number 1, 2, 3, 4 etc, (starting from the nucleus to these energy levels). The various energy levels are now designated as K, L, M, N etc, and are termed as principal quantum numbers. 2. Energy is emitted or absorbed only when an electron jumps from one orbit to another, i.e., form one energy level (or principal quantum number) to another. Since each level is associated with a definite amount of energy, the farther the energy level from the nucleus, the greater is the energy associated with it.

1.17

Based on the above postulates, Bohr calculated the radii of the various orbits and the energies associated with the electrons present in those orbits. The frequencies of the spectral lines determined experimentally by Lyman, Balmer and others are in excellent agreement with those calculated by Bohr’s theoretical equations.

1.7.1 Bohr’s Theory of the Hydrogen Atom Bohr pictured the hydrogen atom as a system consisting of a single electron with a charge designed as e, rotating in a circular orbit of radius r about the nucleus of charge Ze with a velocity v.

M NOP

Tangential velocity of the revolving electron

Fig 1.20 Bohr's orbit like representation of various energy levels 3. Permissible orbits are those for which the angular momentum is an integral multiple of (h/2π). The electrons move in such orbits without any loss of energy. Thus, angular momentum me vr =

nh 2π

(1.20)

where, n is an integer 1, 2, 3 ....... me is mass of electron and r is the radius of orbit. The n values corresponds to the principal quantum number. 4. When an electron gets sufficient energy from outside, an electron from an inner orbit of lower energy state E1 moves to an outer orbit of higher energy state E2. The excited state lasts for about 10–8 s. The difference of energy E2 –E1 is thus radiated out. During the emission or absorption of radiant energy to Planck’s, Einstein equation E = hv is obeyed. Thus the frequency of emitted radiation is given by hv = ∆E = E2 – E1

(1.21)

Calculation of velocity V: The angular momentum of the electron is defined as the product of the velocity of the electron in its orbit, its mass, and the radius of its orbit. The product is symbolized as mvr. According to Bohr’s postulates, the angular momentum are whole multiples of h / 2π. Therefore, according to this quantum restriction, angular momentum may be restricted as vr = mvr

nh 2π

∴ The velocity of the electron in an orbit, v=

nh 2πmr

(1.22)

Radius of an Orbit According to Coulomb’s law, the electrostatic force of attraction Fe between the charges may be evaluated mathematically as Ze 2 Fee = 2 (1.23) r By definition, the magnitude of the centrifugal force Fc for an electron of mass m, with a velocity in its orbit of V and with an orbit radius r is given as mv 2 Fc = (1.24) r

1.18

Structure of Atom

In order that the electron’s orbit may remain stable, it is necessary to assume that the two forces, electrostatic and centrifugal, are equal and opposed to each other. From this assumption may be deduced the following mathematical relationship mv 2 Ze 2 = 2 (1.25) r r and upon solving for v=

Ze 2 mr

(1.26)

Equations (1.26) and (1.22) may be equated since they are equal to each other

En = -

(1.27)

Solved Problem 8 Calculate the (i) radius of first orbit (ii) velocity of electron in the first orbit, and (iii) energy of first orbit of hydrogen atom. Solution: n2 h2 (i) Radius of an orbit is given by r = 2 4p mZe 2 Since, n = 1 and Z for hydrogen = 1 r=

h2 4p2 me 2

( 6.626 ×10

and upon solving for r, we obtain n2 h2 4p2 mZe 2

r=

r= (1.28)

Consequently, the radius r0 of the smallest orbit (first orbit) for the hydrogen atom is h2 (1.29) r0 = 2 2 4p me

By definition the kinetic energy of body is equal to

1 2 mv . 2

The total energy E of the electron is the sum of its kinetic and potential energies. If the potential energy of an electron is taken as zero when it is at an infinite distance from the nucleus, the value at a distance r is given by –Ze2 / r. This value may be obtained by integrating Eq. (1.24) between the limits of r and infinity. The negative sign indicates that work must be performed on the electron to transfer it to infinity. Therefore, the total energy of the electron is E=

mv 2 Ze 2 2 r

(1.30)

From Eq (1.25) we may obtain 2

ergs )

2

4 × p2 × ( 9.109 × 10-28 g )( 4.803 × 10-10 esu )

2

(ii) Velocity of electron in the first orbit of hydrogen nh V= 2pmr 1× 6.626 × 10-27 ergs

2 × p× ( 9.109 × 10-28 g )( 0.529 × 10-8 cm )

= 2.188 × 108 cm–1 (iii) Energy of electron En =

2π2 me 4 Z2 × 2 h2 n

Since, n = 1, and Z = 1 Energy of electron in the first orbit 2p2 me 4 E= h2 E=

2p2 ( 9.109 × 10-28 g )( 4.803 × 10-10 esu )

( 6.626 ×10

-27

ergs )

4

2

= 2.179 × 10–11 erg / atom

2

mv Ze 2 2r

(1.31)

and upon combining Eqs (1.30) and (1.31) the following expression may be derived E=-

-27

= 0.529 × 10–8 cm = 0.529 Å

V=

energy of the electron

(1.33)

where, En is the total energy of the electron in the orbit designated by the quantum number n.

2

Ze nh = mr 2pmr

2p2 me 4 Z 2 × 2 h2 n

Ze 2 2r

(1.32)

Now if we substitute the value r from Eq. (1.28) into (1.32) the total energy of the electron may be stated as

The energy of an electron is usually expressed in Kcal or KJ mol–1 or electron volt (eV) One erg/molecule = 1.44 × 1013 Kcal / mol = 6.22 × 1013 KJ / mol \ E1 = (–2.179 × 10–11) × (1.44 × 1013) = –313.77 Kcal / mol = –1312.19 KJ / mol Now 1eV = 1.602 × 10–12 ergs \ E1 =

2.179 × 10-11 erg = -13.6 eV /atom 1.602 × 10-12

Structure of Atom

If the term wave number is used for frequency v v= c

1.7.2 Origin of Spectral lines and the hydrogen Spectrum As described by Bohr, energy is radiated when an electron moves from one orbit of a quantum number of n2 to an orbit of quantum number n1, where n1 is the inner orbit. Energy is absorbed if the electron moves in opposite direction, i.e., from inner orbit to outer orbit. The energy DE which is emitted may be represented as a difference in energy of the two electronic states and can be indicated by the equation DE = E2 - E1 = hv

(1.34)

where, E2 is the energy of orbit n2 and E1 is the energy of inner orbit n1. Spectral lines are produced by the radiation of photons, and the position of the lines on the spectral scale is determined by the frequency, or frequencies, of the photons emitted. Transitions to the innermost level n1 from orbits n2, n3, n4 etc, gives rise to the first, second, third etc, lines of the Lyman series in the UV region. Transitions from the outer most energy levels gives rise to spectral lines of higher frequencies. Transition of electron to n2 level from outer level gives Balmer series of lines in the visible region. Similar transitions from higher orbits to the third orbit (n = 3) produces Paschen series in the infrared region. Other series of lines have been discovered for similar shifts in the far infrared region. A sketch indicating the transitions which produces these spectral lines is given in Fig 1.21. From the Eq. (1.34) the difference between the energies of an electron in the two orbits n2 and n1 may be indicated as 2p 2 Z 2 e 4 m  1 1  DE = En2 - En1 = hv =  2 - 2  (1.35) h  n1 n2 

2p 2 Z 2 e 4 m  1 1  (1.36)  2- 2 3 hc  n1 n2  All the terms in the fraction 2p2Z2 e4 m/h3c are constant. For hydrogen atom Z is 1. If we evaluate the fraction from the constant terms, the following value is obtained: v=

2p2 12 e 4 m = 109, 700 cm -1 = R h3 c

(1.37)

where, R is the Rydberg constant. Moreover, when we substitute R in Eq. 1.36  1 1  v = R 2 - 2  (1.38)  n1 n2  Equation 1.38 is identical with the general equation for the wave number of any spectral line within any of the spectral series of the hydrogen atom as given in Eq. (1.15).

1.7.3 limitations of the bohr’s model Bohr’s atomic model was a very good improvement over Rutherford’s nuclear model and enabled in calculation of radii energies of the permissible orbits in the hydrogen atom. The calculated values were in good agreement with the experimental values. Bohr’s theory could also explain the hydrogen spectra successfully and also the spectra of hydrogen like atoms (He+, Li2+, Be3+, etc). The theory was, therefore, largely accepted and Bohr was awarded Nobel prize in recognition of

Paschen series

Brackett series

Balmer series

1.19

Pfund series

Lyman series

Fig 1.21 Origin of emission spectrum of hydrogen atom

1.20

Structure of Atom

this work. However, Bohr’s model could not explain the following points: (i) Bohr’s model could not explain the spectra of atoms containing more than one electron. Bohr’s model could not explain hydrogen spectrum obtained using high resolution spectroscopes. Each spectral line, on high resolution was found to consist of two closely spaced lines. (ii) It was observed that in the presence of a magnetic field each spectral line gets splitted up into closely spaced lines. This phenomenon, known as Zeeman effect, could not be explained by the Bohr’s model. Similarly, the splitting of spectral line under the effect of applied electric field, known as Stark effect, could not be explained by the Bohr’s model. (iii) Bohr’s model could not explain the spectra of atom of elements other than hydrogen. (iv) Bohr’s model could not explain the ability of atoms to form molecules and the geometry and shapes of molecule.

1. 8 WAveS AND PARTiCleS Because of the limitations of Bohr’s atomic model several scientists tried to develop a more subtle and general model for atoms. During that period two important proposals were contributed significantly for the modern quantum mechanical model of the atom. They are (i) Dual nature, i.e., wave as well as particle nature of matter. (ii) Heisenberg’s uncertainty principle.

1.8.1 Dual Nature of matter According to Maxwell’s concept, light, radiation consists of waves while Planck’s quantum theory considers photons as particles. Thus, light, which consists of electromagnetic radiation, is both a wave and particle. Based on this analogy in 1924, the Frenchman, Prince Louis de Broglie published an exceedingly complicated account of the wave-particle duality. de Broglie stated that any form of matter such as electron, proton, atom or molecules, etc, has a dual character. The waves predicted by de Broglie are known as matter waves. These waves are quite different from electromagnetic waves in following two respects: (i) Matter waves cannot radiate through empty space like the electromagnetic waves. (ii) Speed of matter waves is different from that of electromagnetic waves.

louis de broglie 1892–1987 He was a French physicist. At the beginning he studied history but while working on radio communications in first world war as an assignment, he developed interest in science. He received his Dr Sc. from the university of Paris in 1924. He war professor of theoretical physics at the university of Paris from 1932 to 1962. He was awarded Nobel Prize in physics in 1929. By making use of the Einstein’s (E = mc2) and Planck’s quantum theory, (E = hv) de Broglie deduced a fundamental relation called the de Broglie equation: h l= (1.39) mv This equation gives the relationship between the wavelength of the moving particle and its mass. In the Eq. (1.39), l is the wavelength of the wave associated with an electron of mass ‘m’ moving with velocity v. Eq. (1.39) can be written as h (1.40) mv = l 1 (1.41) mv ∝ l 1 (1.42) p∝ l In the E.q. (1.42) ‘p’ represents the momentum mv of particle; and l corresponds to the wave character of matter and p its particle character. Thus, the momentum (p) of a moving particle is inversely proportional to the wavelength of the waves, associated with it. The revolutionary postulate of de Broglie received direct experimental verification in 1927 by Davisson and Germer, G.P. Thomson and later by Stern. They showed that heavier particles (H2, He etc.) showed diffraction patterns when reflected from the surface of crystals. Particularly Davisson and Germer found that electron beam was diffracted when striken on a single crystal of nickel, which proves the wave like character of electrons. de brolie’s Relationship and bohr’s Theory Application of de Broglie’s relationship to a moving electron around a nucleus puts some restrictions on the size of the orbits. It means that electron is not a mass particle moving in a circular path but instead a standing wave train (non-energy, radiating motion) extending around the nucleus in the circular path as shown in Fig 1.22 (a) and 1.22 (b). For the wave to remain continually in phase, the circumference of the orbit should be an integral multiple of wavelength l,

Structure of Atom

1.21

meaning of y and Its Significance

Fig 1.22 Diagrammatic representation of electron orbits one of which is (a) in phase; and the other (b) out of phase. 2pr = nl (1.43) where, r is the radius of the orbit, and n is a whole number. From Eq. (1.40) we know h (1.44) l= mv Substituting the value of l in Eq. (1.43), we get nh mv nh mvr = 2p

y is the wave function or the amplitude of the wave. The value of amplitude increases and reaches the maximum which is indicated by peak in the curve. This is shown by the upward arrow in the figure. The value of the amplitude decreases after reaching the maximum value. This is shown by the downward arrow. Above the X-axis the amplitude is shown as +ve, along the X-axis it is zero and below the X-axis it is negative. The intensity of light is proportional to the square of amplitude (or A2). Therefore, A2 can be taken as a measure of the intensity of light since light is considered to consist of photons (corpuscular theory), the density of photons is considered to be proportional to A2. Thus, as far as light waves are concerned A2 indicates the density of photons in space or the intensity of light.

2pr = or

Fig 1.23 (1.45)

which is the same as Bohr’s postulate for angular moment of electron. From the E.q. (1.45) it can be known that electrons can move only is such orbits for which the angular momentum must be an integral multiple of h/2π. If the circumference is bigger or smaller than the value as given by in Fig 1.22 b, the electron wave will be out of phase. Thus, de Broglie relation provides a theoretical basis for the Bohr’s postulate for angular momentum. In Fig 1.22 (a) the wave is in phase continually. In the summer of 1927 physicists from all over the world arrived in Brussels at the Solway congress. At this congress de Broglie’s concept on the relationship between waves and particles was totally rejected. For many years to come, a complete different representation of this relationship led the way. It was Heisenberg and Schrodinger who supported and strongly represented the concept of de Broglie at another congress and got it accepted. de Broglie’s equation is true for material particles of all sizes and dimensions. However, in the case of small micro objects like electrons, the wave character is of significance only. In the case of large macro-objects the wave character is negligible and cannot be measured properly. Thus, de Broglie equation is more useful for small particles.

This concept can be extended even to the y functions moving in the atom since electron moving with high speed is associated with a wave characteristics. Therefore, y2 in the case of electron wave denotes the probability of finding an electron in the space around the nucleus or the electron density around the nucleus. If y2 is maximum, the probability of finding an electron is also maximum. Solved Problem 9 An electron with mass of 9.1 × 10–31 kg is moving with a velocity of 103 m/s. Calculate its kinetic energy and wavelength (Planck’s constant h = 6.6 × 10–34 kg m2 s–1). Solution: Kinetic energy KE 2 1 1 = mv 2 = × ( 9.1× 10-31 kg )(103 m s -1 ) 2 2 = 4.55 × 10–7 J

h 6.6 × 10-34 kg m 2 s -1 = mv 9.1× 10-31 kg × 103 ms -1 = 7.25 × 10–7 m

Wavelength l =

Solved Problem 10 A moving electron has 2.8 × 10–25 J of kinetic energy. Calculate its wavelength (mass of electron = 9.1 × 10–31 kg).

1.22

Structure of Atom

Solution: 1 Kinetic energy = mv 2 2 2 Kinetic energy × V2 = m =

2 × 2.8 × 10-25 2 -2 m s 9.1× 10-31

2 × 2.8 × 10-25 2 -2 = m s 9.1× 10-31 2 2

= 61538.61 m s \ v = 784.46 m s–1 By de Broglie’s equation, we have l=

h 6.6 × 10-34 kgm 2 s -2s = × mv 9.1× 10-31 × 784.46 kgm s -1 = 9.2455 × 10–7 m

1.8.2 heisenberg’s Uncertainty Principle Werner heisenberg (1901–1976) He received his Ph. D. in physics from the university of Munich in 1923. Later he worked with Max Born at Gottingen and with Niels Bohr at Copenhagen. He spent about 14 years (1927 – 1941) as professor of physics at the university of Leipzig. He lead the German team working for atomic bomb during second world war. Once the war ended he became the director of Max Planck Institute for physics in Gottingen. He was awarded Nobel Prize in physics 1932. “The exact sciences also start from the assumption that in the end, it will always be possible to understand nature, even in every new field of experience but we make no priori assumptions about the meaning of the word understand — W. Heisenberg According to classical mechanics, a moving electron is considered to be a particle. Therefore, its position and momentum could be determined with a desired accuracy. On the other hand de Broglie also considered a moving electron to be wave-like. Therefore, it becomes impossible to locate the exact position of the electron on the wave because it is extending throughout a region of space. So, the following fundamental question arises: “If an electron is exhibiting the dual nature, i.e., wave and particle, is it possible to know the exact location of the electron in space at some given instant.” The answer to the above question was given by Heisenberg in 1927 who stated that

“For a subatomic object like electron, it is impossible to simultaneously determine its position and velocity at any given instant to an arbitary degree of precision.” Heisenberg gave mathematical relationship for the uncertainty principle by relating the uncertainty in position (Dx) with uncertainty in momentum (Dp) as Dp × Dx ≥

h 4p

(1.46)

The ≥ in the Eq. (1.46) means that the product of Dx and Dp can be either greater than or equal to but would be never smaller than h / 4p. But h / 4p is constant. Therefore, it follows from Eq. (1.46) that smaller the uncertainty in locating the exact position (Dx), greater will be the uncertainty in locating the exact momentum (Dp) of the particle and vice versa. As Dp is equal to mDv the Eq. (1.46) is equivalent to saying that position and velocity cannot be simultaneously determined to an arbitrary precision. However, in our daily life, these principles have no significance. This is because, we come across only large objects. The position and velocity of these objects can be determined accurately because in these cases the changes that occur due to the impact of light are negligible. The microscopic objects suffer a change in position or velocity as a result of the impact of light. For example, to observe an electron, we have to illuminate it with light or electromagnetic radiation. The light must have a wavelength smaller than the wavelength of electron. When the photon of such light strikes the electron, the energy of the electron changes. In this process, no doubt, we shall be able to calculate the position of the electron, but we would know very little about the velocity of the electron after the collision.

1.8.3 Significance of Uncertainty Principle The most important consequence of the Heisenberg uncertainty principle is that it rules out existence of definite paths or trajectories of electrons and other similar particles. The trajectory of an object is determined by its location and velocity at various moments. At any particular instant if we know the position of a body and also its velocity and the forces acting on it at that instant we can predict at what position it will present at a particular time. This indicates that the position of an object, and its velocity completely determine its trajectory. Because, it is not possible simultaneously to determine the position and velocity of an electron at any given instant precisely it is not possible to talk of the trajectory of an electron. The effect of Heisenberg uncertainty principle is significant only for motion of microscopic objects and is negligible for that of macroscopic objects. This can be seen from these illustrated examples.

Structure of Atom

1.23

Solved Problem 11

Concept of Probability

Calculate the uncertainty (Dv) in the velocity of a cricket ball of mass 1 kg, if certainty (Dx) in its position is of the order of 1 Å. Solution: In the above example. h ∆V = 4π∆ x m But h = 6.6 × 10–34 kg m2 s–2 s: Dx = 1Å = 10–10 m and m = 1 kg 6.6 × 10-34 kg m 2 s -1 s DV = × -10 m kg 4 × p× 10 × 1

The consequences of the uncertainty principle are far reaching and probability takes the place of exactness in velocity (which is related to kinetic energy) of an electron. The Bohr concept of the atom, which regards the electrons as rotating in definite orbits around the nucleus, must be abandoned and should be replaced by a theory which considers probability of finding the electrons in a particular region of space. This means that, it is possible to state the probabilities of the electron to be various distances with respect to the nucleus. In the same way, probable values of velocity can also be given. It should be clearly understood, however, that a knowledge of probability for an electron ‘moves’ from one location to another.

7 × 6.6 × 10-34 m s -1 4 × 22 × 10-10 = 5.25 × 10–25 m s–1

=

Solved Problem 12 A microscope using suitable photons is employed to locate an electron in an atom within a distance of 0.1 Å. What is the uncertainty involved in the measurement of its velocity? Solution: h Dv = 4pDx m But h = 6.626 × 10-34 kg m 2 s -2s; Dx = 0.1Å = 1 × 10–11 m and m = 9.11 × 10–31 kg \ Dv =

6.626 × 10 -34 kg m 2 s -2 s × m kg 4 × p × 10 -11 × 9.11 × 10 -31

= 0.579 × 107 m s–1 = 5.79 × 106 m s–1 Why bohr’s model was a Failure? Bohr considered the electron as a charged particle moving in well defined circular orbits around the nucleus. So, the position and the velocity of the electron can be known exactly at the same time. Uncertainty principle shows that it is impossible to measure these variables simultaneously because the wave character of electron was not considered in the Bohr model. Calculation of the trajectory of an electron in an atom or molecule is, therefore, a futile exercise. We can now appreciate the major fault of the Bohr model. In calculating electron orbits precisely, Bohr was violating this fundamental requirement and hence his theory was only partially successful. So, a model which can account the wave-particle duality of matter and be consistent with Heisenberg uncertainty principle was in quest. This came with the advent of quantum machanics.

1.8.4 Quantum mechanical model of Atom Classical mechanics based on Newton’s laws of motion successfully describes the motion of all macroscopic objects since the uncertainties in position and velocity are small enough to be neglected. However, it fails in the case of microscopic objects like electrons, atoms, molecules etc. So, while considering the motion of microscopic objects, the concept of dual behaviour of matter and the uncertainty principle are taken into account. The branch of science that takes into account this dual behaviour of matter is called quantum mechanics. As far as the other dynamical variables of an electron are concerned, we can show that the uncertainty principle leads to the following result. The total energy of an electron in an atom or molecule has a well-defined (sharp) value. The probability distribution as well as the sharp values can be calculated from a function designated as y (x) and called the wave function or the psi function. y (x) is obtained by solving the Schrodinger equation which is the fundamental equation in quantum mechanics in the same manner that Newton’s equation is fundamental in classical mechanics. important Features of the Quantum mechanical model of Atom Quantum mechanical model of the atom is the outcome of the application of Schrodinger wave equation to atoms. Its important features are • Quantization of the energy of electron in atoms, i.e., it can have certain discrete values. • The allowed solutions of Schrodinger wave equation tell about the existence of quantized electronic energy levels for electrons in atoms having wave like properties.

1.24

•

•

Structure of Atom

As per Heisenberg’s uncertainty principle, both the position and velocity of an electron cannot be determined simultaneously and exactly. So, the path of the electron can never be determined accurately. Because of this reason the concept of probability was introduced for finding the electron at different points in an atom. The wave function y for an electron in an atom is the atomic orbital. Whenever we say about electron by a wave function it occupies that orbital. Because several wave functions are possible for an electron, there also as many atomic orbitals. These one electron orbital wave functions or orbital form the basis for electronic structure of atoms. Every orbital possess certain energy and can accommodate only two electrons. In atoms having several electrons, the electrons are filled in the order of increasing energy. In the multielectron atoms each electron has an orbital wave function characteristic of the orbital it occupies. All the information about the electron in an atom is stored in its orbital wave function y and quantum mechanics makes it possible to extract this information out of y. The probability of finding the electron at a point within an atom is proportional to y2 at that point. Though y is sometimes negative, y2 is always positive and is known as the probability density. The values of y2 predicts the different points in a region within an atom at which the electron will be most probably found.

1.8.5 Schrodinger Wave equation

Schrodinger published his ideas in January 1926. This date represents one of the mile-stones in the history of Chemistry. His work formed the basis of all our present ideas on how atoms bond together. The heart of his method was his prediction of the equation that governed the behaviour of electrons, and the method of solving it. Schrodinger’s equation is ∂ 2 ψ ∂ 2 ψ ∂ 2 ψ 8π2 m + + + 2 ( E −U)ψ = 0 ∂x 2 ∂y 2 ∂z 2 h

(1.43)

where, m = Mass of electron E = Total energy of the electron (kinetic energy + potential energy) U = Potential energy h = Planck’s constant y = Wave function This equation applies to stationary waves as it does not have the time dependence part of the wave function. Schrodinger won the Nobel prize in physics in 1933. The solution of this equation are very complex and you will learn them for different systems in higher classes. For a system (such as an atomic or a molecule whose energy does not change with time), the Schrodinger equation is written as Ĥy = Ey, where Ĥ is a mathematical operator called Hamiltonian. Schrodinger gave a recipe of constructing this operator from the expression for the total energy of the system. The total energy of the system takes into account the kinetic energies of all the sub-atomic particles (electrons, nuclei) attractive potential between the electrons and nuclei and repulsive potential among the electrons and nuclei individually. Solutions of this equation gives E and y.

erwin Schrödinger 1887–1966 Erwin Schödinger was born in Austria. He received his Ph. D. in theoretical physics from the university of Vienna in 1910. At the request of Max Planck, Schrödinger became his successor at the university of Berlin in 1927. Because of his opposition to Hitler and Nazi policies, he left Berlin and returned to Austria and in 1936, when Austria was occupied by Germany, he was forcibly removed from his professorship. Then he moved to Dublin, Ireland. He shared the 1933 Nobel Prize for physics with P.A.N. Dirac. During 1920 there was a great deal of interest in waveparticle duality and de Broglie’s matter waves. It was the Austrian physicist Erwin Schrödinger who invented a method of showing how the properties of waves could be used to explain the behaviour of electrons in atoms.

1.8.6 The meaning of Wave Function The wave function could only be used to provide information about the probability of finding the electron in a given region of space around the nucleus. Max Born, a German physicist proposed that we must give up ideas of the electron orbiting the nucleus at a precise distance. In this respect Bohr was wrong in thinking that the electron in the ground state of the hydrogen atom was always to be found at a distance a0 form the nucleus. Rather, it was only most probably to be found at this distance. The electron had a smaller probability of being found at a variety of other distances as well. It is important to realise that we should not try to talk about finding the electron at a given point. The reason for this is that there is an infinite number of points around the nucleus. So, the probability of finding the electron at any one of these points is infinitely small, i.e., zero. Then

Structure of Atom

to predict the position of electron, we can imagine taking out a series of photographs of an atom to give us an instantaneous picture of the whereabouts of the electron. If we combine all pictures we would end up with a picture as shown in Fig 1.24. The separate dots have overlapped to give regions in which the density of dots is very high and regions where the density of dots is much lower. In the high density regions, we say that there is a high probability density. The maximum in the density comes at exactly the same distance a0, as Bohr predicted in his work. The circulated symmetry of the probability density is clear. However, we should remember that atoms exist in three dimensions, so really the diagram should be shown as a sphere.

a0

Fig 1.24 Electron density It is easier to draw circles rather than spheres, so usually we draw the density diagram as a circle. Also, it is common practice not to include the shading and to agree that when a circle is drawn, it provides a boundary surface within which, say, there is a 95% probability of finding the electron. When Schrodinger equation is solved for hydrogen atom, the solution gives the possible energy levels the electron can occupy and the corresponding wave function(s) y of the electron associated with each energy level. Accepted solutions to the wave functions are called Eigen wave functions. The probability of finding an electron at a point in space whose coordinates are x, y and z is given by y2 (x, y, z). This three dimensional region obtained where the probability to find the electron is about 95%, is called an orbital. Because atomic behavior is so unlike ordinary experience, it is very difficult to get used to and it appear peculiar and mysterious to everyone, both to the novice and to the experienced scientist. Even the experts do not understand it the way they would like to, and it is perfectly reasonable that they should not, because all of direct human experience and of human intution applies to large objects. We know how large objects will act, but things on a small scale just do not act that way. So, we have to learn about them in a sort of abstract or imaginative fashion and not by connection with our direct experience. Feynman Lectures on Physics Vol. l Chapter 37.

1.25

1.9 QUANTUm NUmbeRS The mathematical solution of three dimensional Schrodinger wave equation gives three values of E for acceptable values of y. These values are related to one another through whole numbers. These values are termed as Quantum numbers and represented with n, l, m — called principal, azimuthal and magnetic quantum numbers, respectively. These three quantum numbers together with the fourth called spin quantum number describe fully the location and energy of an electron. Thus, quantum numbers are the numbers which determine the energy of electron, the angular momentum, shape of the electron orbital, the orientation of the orbital and spin of the electron. Thus, each quantum number is associated with a particular characteristic of the electron. These quantum numbers are discussed below briefly. 1. Principal Quantum Number (n). The number allotted to Bohr’s original stationary states, visualized as circular orbits is called the principal quantum number. The innermost orbit, i.e., that nearest to the nucleus has a principal quantum number 1, the second orbit has a quantum number of 2, and so on. So, the principal quantum number denoted by n have value 1, 2, 3, 4, ..... Alternatively, letters are used to characterize the orbits, K, L, M, N, ... for 1, 2, 3, 4, .... The choice of letters originates from Mosely’s work on the X-ray spectra of the elements. He called group of lines in the spectra the K, L, M, N, ... groups. The number of electrons in an atom which can have the same principal quantum number is limited and is given by 2n2 where n is the principal quantum number concerned. Thus, Principal quantum number (n) Letter designation

1 K

2 L

3 M

4 N

Maximum number of electrons

2

8

18

32

This is the most important quantum number as it determines to a large extent the energy of an electron. It also determines the average distance of an electron from the nucleus. As the value of n increases, the electron gets farther away from the nucleus and its energy increases. The higher the value of n, the higher is the electronic energy. For hydrogen and hydrogen like species, the energy and size of the orbital are determined by the principal quantum number alone. The principle quantum number also identifies the number of allowed orbitals within a shell. With the increase in the value of ‘n’ the number of allowed orbitals increases and are given by n2. All the orbitals of a given value of ‘n’ constitute a single shell of atom.

1.26

Structure of Atom

2. Azimuthal Quantum Number (l). It is also known as orbital angular momentum or subsidiary quantum number. It defines the three dimensional shape of the sub-shell. For each value of the principal quantum number there are several closely associated orbitals, so that the principal quantum number represents a group or shell of orbits. In any one shell, having the same principal quantum number the various subsidiary orbits are denoted as s, p, d, f, ... sub-shells. The letters originates from the sharp, principal, diffuse and fundamental series of the lines in spectra. The number of sub-shells or sub-levels in a principal shell is equal to the value of n. The values of subshells is represented with l. l can have values ranging from 0 to n – 1, i.e., for a given value of n the possible value of l can be 0, 1, 2......(n–l). For example when n = 1 value of l is only 0. For n = 2 the possible value of l can be 0 and 1. For n = 3 the possible values are 0, 1 and 2. Sub-shells corresponding to different values are as follows: value for l Notation for subshell

0 s

1 p

2 d

3 f

4 g

5…….. h ...

The permissible values for ‘l’ for a given principal quantum number and the corresponding sub-shell notation are as follows. l values 0 1 2 3

1 1s

2 2s 2p

3 3s 3p 3d

4 4s 4p 4d 4f

5 5s 5p 5d 5f

6 6s 6p

7 7s 7p

3. Magnetic Quantum Number (ml). This quantum number which is denoted by ml refers to the different orientations of the electron cloud in a particular subshell. These different orientations are called orbitals. The number of orbitals in a particular sub-shell within a principal energy level is given by the number of values allowed to ml which in turn depends upon on the value of l. The possible values of ml range from + l through 0 to – l, thus making a total of (2l + 1) values. Thus, in a subshell, the number of orbitals is equal to (2l + 1). For l = 0 (i.e., s-subshell) ml can have only one value ml = 0. It means that s-subshell has only one orbital. For l = 1 (i.e., p-subshell) ml can have three values +1, 0 and –1. This implies that p-subshell has three orbitals. For l = 2 (i.e., d-subshell) ml can have five values + 2, + 1, 0, and –2. It means that d-subshell has five orbitals. For l = 3 (i.e., f-subshell) ml can have seven values + 3, + 2, + 1, 0, –1, –2 and –3. It means that f-subshell has seven orbitals. The number of orbitals in various types of subshells are as given below. Sub-shell Value of l No. of orbitals (2l + 1)

s 0 1

p 1 3

d 2 5

f 3 7

g 4 9

The relationship between the principal quantum number (n), angular momentum quantum number (l) and magnetic quantum number (ml) is summed up in Table 1.4.

Table 1.4 Relationship Amongst the Value of n, l and ml Number of orbitals

Principal quantum number (n)

Possible values of (l)

Designation of sub-shell

Possible values of ml

In a given sub-shell

K

1

0

1s

0

1

L

2

0 1 0 1 2 0 1 2 3

2s 2p 3s 3p 3d 4s 4p 4d 4f

0 + 1, 0, – 1 0 + 1, 0, –1 + 2, + 1, 0, – 1, –2 0 + 1, 0, –1 + 2, +1, 0, –1, –2 + 3, + 2, + 1, 0, – 1, – 2, – 3

1 3 1 3 5 1 3 5 7

Energy level

M

N

3

4

In a given energy level 1 4 9

16

Structure of Atom

4. Spin Quantum Number (ms). This quantum number which is denoted by ms does not follow from the wave mechanical treatment. Electron spin was first postulated in 1925 by Uhlenbeck and Goudsmit to account for the splitting of many single spectral lines into double lines when examined under a spectroscope of high resolving power. The electron in its motion about the nucleus also rotates or spins about its own axis. In other words, an electron has, besides charge and mass, an intrinsic spin angular quantum number. Spin angular momentum of the electron –a vector quantity, can have two orientations relative to a chosen axis. The spin quantum number can have only two values which are +

1 1 1 and - . The + value indicates the 2 2 2

clockwise spin, generally represented by an arrow upwards, i.e., ↑ and the other indicates anti-clockwise spin, generally represented by an arrow downwards, i.e., ↓. The electrons that have different ms values (one +

1 1 and the other - ) 2 2

are said to have opposite spins. An orbital cannot hold more than two electrons and these two electrons should have opposite spins.

Orbit and Orbital Orbit and orbitals are different terms. Bohr proposed an orbit as a circular path around the nucleus in which the electrons moves. As per Heisenberg’s uncertainty principle the precise description is not possible. So, its physical existence cannot be demonstrated experimentally. An orbital is a three dimensional region calculated from the allowed solutions of the Schrodinger’s equation. It is quantum mechanical concept and refers to one electron wave function y in an atom. It is characterised by three quantum numbers (n, l and ml) and its value depends upon the coordinates of the electron. y has no physical meaning but its square y2 at any point in an atom gives the value of probability density at that point. Probability density (y2) is the probability per unit volume and the electron in that volume. y2 varies from one region to another region in the space but its value can be assumed to be constant within a small volume. The total probability of finding the electron in a given volume can be calculated by the sum of all products of y2 and the corresponding volume elements. It is thus possible to get the probable distribution of an electron in an orbital.

1.27

1.10 ShAPeS OF ORbiTAlS The nature of the Schrodinger equation is such that the wave function y may be regarded as an amplitude function. This corresponds, on a three-dimensional scale, to the function which expresses the amplitude of vibration of a plucked string. As it is the square of the amplitude of the vibrating string which measures the intensity of the wave involved, so y2 measures the probability of an electron existing at a point, meaning therefore, that the chance of finding an electron at that point is zero. A high value of y2 at a point means that there is a high chance of finding an electron at the point. The value of the wave function y, for an electron in an atom is dependent, in general, both on the radial distance, r, of the electron from the nucleus of the atom and on its angular direction away from the nucleus. The nucleus, itself, however, cannot provide any sense of direction until a set of arbitrary chosen axes are superimposed. If Cartesian axes are chosen as in Fig 1.25 (a) then a point in space, P, around the nucleus, N, can be defined in terms of x–, y– and z-coordinates. Such axes do not, however, have any absolute directional significance until an external magnetic field is applied. The axes then have a definite direction in relation to the direction of the magnetic field. Alternatively, polar coordinates in Fig 1.25 (b) can be used.

Fig 1.25 The position of a point P relative to point N can be expressed in terms of (a) Cartesian coordinates x, y, z or (b) Polar coordinates r, θ and f. The mathematical relationship between y and r and the angular direction can be established accurately for a single electron in a hydrogen like atom. For more complicated atoms containing more than one electron, the corresponding relationships can only be established approximately because of the difficulty involved in solving the mathematical equations.

1.28

Structure of Atom

Fig 1.26 Plots of y n, l versus r for electron belonging to different orbitals The differences between s–, p– and d-orbitals depend on the different ways in which y and / or y2 vary with r and with the angular direction, as explained in the following sections: The probability of finding an electron in a given volume of space is represented by radial probability distribution curves. These curves indicate how the probability of finding an electron varies with the radial distance from the nucleus without any reference to its direction from the nucleus. The radial wave function is generally written as yn, l. Let us now plot the function yn, l against r for electrons belonging to different orbitals and try to correlate them with the probability density around a point at a distance r from the nucleus. The plots are as shown in Fig 1.26. From these plots, it is clear that y n, l cannot be related with probability density around any point at a distance r from the nucleus because (i) y n, l is maximum at r = 0. If y n, l represents probability density the electron will have maximum probability of occurring at the nucleus. This cannot be true as the actual probability of finding an electron at the nucleus is zero. (ii) y n, l has both positive as well as negative values (if probability curves for 2s and 3s electrons). However, probability density cannot have a negative value. Objection No. (ii) can be removed by considering yn, l to be related to probability density instead of y n, l. Now even if y n, l is negative at some space its square has to be positive. The plots of y n, l versus r for electrons belonging to different orbitals are as shown in Fig 1.28. As can be seen the curves do not exhibit negative value for the radial wave functions at any distance. But even y n, l cannot be related with probability density because yn, l is maximum at r = 0. This means that the probability of finding electron is maximum on the nucleus which again cannot be true (as already explained). Thus, neither y n, l nor y2 n, l

can be directly related with probability of finding electron at a point which is at a distance r from the nucleus. Now let us consider the space around the nucleus (taken at the centre) to be divided into a large number of thin concentric spherical shells of thickness dr. Consider one of these shells with the inner radius r (Fig 1.27).

r O r + dr

Fig 1.27 Division of space around the nucleus into small spherical shells of very small thickness The volume of such a shell viz, dV, will be given by dV = (4/3) p (r + dr)3 – (4/3) pr3

(1.44)

4 3 =   p[r p ++dr3 + 3r2dr + 3rdr2 – r3] 3

(1.45)

Neglecting very small terms dr2 and dr3, we get dV = 4p2 dr

(1.46)

The probability of finding electron within the small radial shell of thickness dr around the nucleus, called radial probability will therefore be given by r dV = yn, l × dV = 4pr2 dr y2n, l

(1.47) (1.48)

The radial probability distribution between r = 0 to r = r will be given by the summation of all the probability distributions for concentric radial shells from r = 0 to r = r, i.e., by

∫

r =r

r =0

4pr 2 dr y n2 ,l

(1.49)

Structure of Atom

1s

ψ 21,0

2s

ψ 22,0

r

1.29

3s

ψ 23,0

r

r

Fig 1.28 Plots of y n, l. versus r for electrons belonging to different orbitals In other words, the radical probability distribution* of electron may be obtained by plotting the function 4pr2 y2 n, l against r, its distance from the nucleus. Such graphs are radial probability distribution curves. The radial probability distribution curves for 2s, 3s, 3p and 3d electrons are shown in Fig 1.29. It may be noted that for 1s orbital, the probability density is maximum at the nucleus and it decreases sharply as we move away from it. On the other hand, for 2s orbital the probability density first decreases sharply to zero and again

starts increasing. After reaching a small maxima it decreases again and approaches zero as the value of r increases further. The region where this probability density function reduces to zero is called nodal surfaces or nodes. In general it has been found that ns-orbital has (n – 1) nodes, that is, number of nodes increases with increase of principal quantum number n. In other words, number of nodes for 2s orbital is one, two for 3s and so on. These probability density variations can be visualized in terms of charge cloud diagrams (Fig 1.30). In these

3s

4πr 2 R 2,2 0

4πr 2 R 3,2 2

3p

r

r

3d 2 4πr 2 R 3,1

2 4πr 2 R 3,1

2s

r

r

Fig 1.29 Radial probability distribution curves for 2s, 3s, 3p and 3d electrons.

1.30

Structure of Atom Z

Z

Y

–X

Z

Y

X –X

–Y

X –X

–Y

–Z

1s orbital

Y

Radial node –Z

X

–Y

2s orbital

Radial node –Z

3s orbital

Fig 1.30 The shapes of various s-orbitals diagrams, the density of the dots in the region represents electron probability density in that region.

1.10.1 boundary Surface Diagrams s-orbitals The shapes of the orbitals in which constant probability density for different orbitals can be represented with boundary surface diagram. The value of probability density y2 is constant in this boundary surface drawn in space. In principle, many such boundary surfaces may be possible. However, for a given orbital, only that boundary surface diagram of constant probability density is taken to be good representation of the shape of the orbital which encloses a region or volume in which the probability of finding the electron is very high, say 90%, but cannot be 100% because always there will be some value, however small it may be, at any finite distance from the nucleus. Boundary surface diagram for an ‘s’ orbital is actually a sphere centred on the nucleus. In two dimensions, this sphere looks like a circle. The probability of finding the electron at any given distance is equal in all directions. It is also observed that the size of the s-orbital increases in n, that is, 4s > 3s> 2s> 1s and the electron is located further away from the nucleus as the principal quantum number increases. p-orbitals The probability density diagrams for p orbitals are very different to those of s-orbitals. If you look back at quantum numbers you will find that we said that the magnetic quantum number can tell us the number of orbitals of a given type. The result is that for s-orbitals the magnetic quantum number, ml, has only one value. This means that there is only one variety of s-orbitals that is the variety

we have already met; the spherically symmetric ones. For p-orbitals there are three possible values of m (+ 1, 0, –1) — the three types of p-orbitals are called px, py and pz. The boundary surface diagrams are known as radial probability distribution curves or radial charge density curves or simply as radial distribution curves. Such curves truly depict the variation of probability density of electronic charge with respect to r (distance of charge from the nucleus.) Radial probability at a distance r is the probability of finding electron at all points in space which are at a distance r from the nucleus and the radial probability distribution is the graph of these probabilities as a function of r. The radial probability distribution curves for 1s and 2p electrons are as shown in Fig 1.31 2 The radial probability function 4pr2 ψ n , l written 2 2 for the sake of simplicity as 4pr y is evidently the product of two factors. While the probability factor y2 decreases as r increases. This gives rise to curves of the type shown in Fig 1.31. At r = 0, though the factor y2 is maximum, (Fig 1.28) the factor 4 pr2 is zero. Similarly, when r is very large, the factor 4pr2 is no doubt very large but the probability factor y2 is negligible so that the radial probability is exceedingly small as shown in Fig 1.32. Here unlike s-orbitals, the boundary surface diagrams are not spherical. Instead each p-orbital consists of two sections called lobes that are one either side of the plane that passes through the nucleus. The probability density function is zero on the plane where the two lobes touch each other. This plane is called nodal plane. The nodal plane for px orbital is YZ-plane, for py orbital, the nodal plane is XZ plane, and for pz orbital nodal plane is XY plane. It should be understood, however, that there is no simple relation between the values of ml (–1, 0 and + 1) and the x, y and z directions. The size, shape and energy of the three orbitals are identical.

Structure of Atom

1.31

4πr 2 R1,2 0

zero and infinite distance, as the distance increases. The number of nodes are given by n – 2, that is, number of radial nodes is 1 for 3p-orbital, two for 4p-orbital three and so on. The shapes of 2px, 3px and 4px orbits with radial nodes are shown in Fig 1.33.

− +

+

−

+

−

+

−

4πr 2 R 22,1

Fig 1.33 The exact shapes of 2px, 3px and 4px orbitals showing the radial nodes.

Fig 1.31 Radial probability distribution curves for 1s and 2p electrons

Fig 1.32 The shapes and boundary surface diagrams for 2px, 2py and 2pz orbitals Like s-orbitals, p-orbitals increase in size and energy with increase in the principal quantum number and hence the order of the energy and size of various p-orbitals is 4p > 3p> 2p.Further, like s-orbitals, the probability density functions for p-orbital also pass through value zero, besides at

d-orbitals The orbitals having l value as 2 are known as d-orbitals. The minimum value of principal quantum number must be 3 to have l value 2 as the value of l cannot be greater than n – 1. There are five d-orbitals corresponding to ml values (–2, –1, 0, + 1 and + 2) for l = 2. Their boundary surface diagrams are shown in Fig 1.34. The five d-orbitals are designed as d xy , d yz , d xz , d x2 - y 2 and d z 2 The orbital d z 2 has a dumbell shaped curve symmetric structure about the z-axis having a ring-like collar in XY plane. The dumbell shaped part of the curve has a positive geometric sign (because whatever be the sign of z positive or negative, its square is always positive), ring the collar in the XY plane has a negative geometric sign. The orbital d xy has a double dumbell shape. The quantity XY will be positive when both x and y are positive or when both are negative. However, XY will be negative when either of x or y is negative. Thus, the sign of the curve is positive in first and third quadrants while it is negative in second and fourth quadrants. The shape of d xz , d yz orbirals can be explained in a similar manner. The orbital d x2 - y 2 is also double dumbell shaped but its lobes lie on X and Y axes. The signs of the lobes on Xaxis will always be positive ( whatever be the sign of X, X2 will always be positive) whereas the sign of the lobes on Y axis will always be negative ( whatever be the sign of Y, – Y2 will always be negative). The exact shapes of d orbitals are obtained by taking into consideration the total wave function. Accordingly 3d x2 - y 2 orbital would be similar in shape to 4d x2 - y 2 orbital or 5d x2 - y 2 orbital except for the fact that 1. A 5d orbital would have two, a 4d orbital have one and a 3d orbital have no radial node.

1.32

Structure of Atom

dx

2

− y2

orbital

3dx2−y2 Fig 1.34

3dz2

Shapes and boundary surface diagrams for 3dxy ,3dyz ,3dxz ,3dx2 – y2 and 3dz2 orbitals. The nodal planes for

dx2 – y2 are shown. Similarly there will be two nodal planes for each dxy ,dyz and dxz orbitals.

2. A 5d orbital would occupy more space than a 4d orbital and a 4d orbital would occupy more space than a 3d orbital as shown in Fig 1.35. Every d-orbital also have two nodal planes passing through the origin and bisecting the XY plane containing Z axis. These are called angular nodes and are given by l.

For any orbital (i) The number of angular nodes (nodal planes) is equal to l value. (ii) The number of radial nodes (nodal surfaces or nodes) is equal to (n–l–1). (iii) The total number of radial nodes and angular nodes for any orbital is equal to (n – l – 1) + l = n – 1. f-orbitals

Fig 1.35 The 3dx2 -y2 , 4dx2 - y2 and 5dx2 - y2 orbitals in three dimensional space showing radial nodes. z

z 3 nodal of planes of f-orbital

z

+

– +

+ –

x x

–

+

x x dz2

fz3 (fx3 and fy3 are similar)

+

+

y

fz (x2–y2) fx (y2–z2) and fy (z2–x2) are similar

fxyz orbital

Fig 1.36 Shapes of f-orbitals

The orbitals having l value as 3 are known as f-orbitals. The minimum value of principal quantum number must be 4 to have l value 3 as the value of l cannot be greater than n – 1. There are seven f-orbitals corresponding to ml values (–3, –2, –1, 0, + 1, + 2, + 3) for l = 3.Their boundary surface diagrams are shown in Fig 1.36.

1.10.2 energies of Orbitals In the hydrogen atom the energy of an electron depends mainly on the principal quantum number. So, the energies of all the sub shells in a principal quantum number are equal. For example, 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f Through the shapes of different sub shells are different, the energy of the electron is the same when present

Structure of Atom

in any one of the sub shells belonging to the same principal quantum number. The orbitals having same energy are called degenerate orbitals. In hydrogen atom when electron is present in 1s orbital, the hydrogen atom is considered in most stable condition and it is called the ground state. This is because the electron is nearer to the nucleus. But when electron is present in 2s, 2p or higher orbitals in the hydrogen atom, it is said to be in the excited state. In multielectron atoms, unlike the hydrogen atom, the energy of an electron depends both on its principal quantum number (shell), and the azimuthal quantum number (sub shell). This means that different sub shells viz., s, p, d, f, ... in a principal quantum number will have different energies, because of the mutual repulsion between the electrons. In hydrogen atom there is only attractive force between nucleus and electron but no repulsive forces. The stability of an electron in a multielectron atom is because the total attractive interactions are more than the repulsive interactions experienced by every electron with other electrons. The attractive force of nucleus on the electrons increases with increase in the positive charge (Ze) of nucleus. The attractive power of nucleus on the electrons will be less due to screening effect of inner electrons. The net positive charge ex perienced by the electron from the nucleus is known as effective nuclear charge (Z*). 7 6

Principal quantum number

5 7p 7s 6d 5f

4

3

2

1

50 Atomic number

75

1.11 FilliNg OF ORbiTAlS The distribution of electrons in various orbitals is known as electronic configuration. Having derived the energy level sequence, it is now a simple matter to write the electronic configurations of atoms by making use of Aufbau principal, Pauli’s exclusion principle and the Hund’s rule of maximum multiplicity.

3d 3p 3s

1. Aufbau Principle

1s 25

The attractive and repulsive interactions depend upon the shell and shape of the orbital in which the electron is present. The spherical s orbital can shield the electrons from the nuclear attraction more effectively than the p-orbital with dumbell shape. This is because of the spherical nature of the s-orbital can shield nucleus from all directions equally but the electron in a p-orbital say px can shield the nucleus in only one direction, i.e., x-direction. Similarly because of difference in their shape and more diffused character the d-orbitals have less shielding power than p-orbitals. Further because of their different shapes the electron in spherical orbital spends more time close to the nucleus when compared to p-orbital and the p-electron spends more time in the vicinity of nucleus when compared to d-electron. For this reason the effective nuclear charge experienced by different sub shell decreases with increase in the azimuthal quantum number (l). From this we can easily understand that s-electron is strongly attracted by the nucleus than the p-electron which in turn will be strongly attracted than the d-electron and so on. Thus, the different sub-shells belonging to same principal quantum number have different energies. However, in hydrogen atom, these have the same energy. In multielectron atoms, the dependence of energies of the sub shells on ‘n’ and ‘l’ are quite complicated but in simple way is that of combined value of n and l. Lower the value n + l for a sub shell, lower is its energy. If the two sub-shells have the same n + l value, the orbital having lower n value have the lower energy. The energies of sub-shells in the same shell decrease with increase in the atomic number (Z*). For example, the energy of 2s orbital of hydrogen atom is greater than that of 2s orbital of lithium and that of lithium is greater than that of sodium and so on, i.e., E2s (H) > E2s (Li) > E2s (Na) > E2s (K).

6p 6s 5d 5p 5s 4f 4d 4p 4s

2p 2s

1

1.33

100

Fig 1.37 The relative energies of the atomic orbitals as a function of atomic number

Aufbau is a German term meaning “building up”. This principle is utilized to deduce the electronic structure of polyelectron atoms by building them up, by filling up of orbitals with electrons. The Aufbau principle states that “in the ground state of the atoms, the orbitals are filled in order of their increasing energies.” In other words, in the ground state of atom, the orbital with a lower energy is filled up first before the filling of the orbital with a higher energy commences.

1.34

Structure of Atom

The increasing order of energy of various orbitals is 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, ... The above sequence of energy levels can be easily remembered with the help of the graphical representation shown in Fig 1.38. 2. Pauli’s exclusion Principle The four quantum numbers define completely the position of electron in an atom. It is thus possible to identify an electron in an atom completely by stating the values of its four quantum numbers. Wolfgang Pauli, an Austrian scientist, put forward an ingenious principle which controls the number of electrons to be filled in various orbitals, and hence, it is named as the exclusion principle. It states that no. two electrons in an atom can have the same set of four quantum numbers Thus, in the same atom any two electrons may have three quantum numbers identical but not the fourth which must be different. This means that the two electrons can have the same value of three quantum numbers, n, l and ml but must have the opposite spin quantum numbers. This restricts that only two electrons may exist in the same orbital and these electrons must have opposite spins. Pauli’s exclusion principle helps in calculating the number of electrons to be present in any sub shell. For example, the 1s sub-shell comprises one orbital and thus the maximum number of electrons present in 1s sub shell can be two because of the following two possibilities. 1 n = 1, l = 0, m = 0, s = + 2

Fig 1.38 The order in which the atomic orbitals are used in building up the electron configuration of many-electron atoms. The orbitals are used in sequence, from the bottom in accordance with the Aufbau principle, Hund’s rule and Pauli’s exclusion principle.

1 2 The p-sub-shell which is having 3 orbitals (px, py and pz) can accommodate 6 electrons, the d-sub shell which having 5 orbitals can accommodate 10 electrons and the f-sub shell having 7 orbitals can accommodate 14 electrons. This can be summed up as the maximum number of electrons in the shell with principal quantum number n is equal to 2n2. n = 1, l = 0, m = 0, ss == -

Here a question arises? Why only two electrons can present in one orbital? It can be answered as follows. We may recall from our knowledge of elementary physics that the motion of an electric charge creates a magnetic field. In an orbital the electron spin may be clockwise or anti-clockwise as shown in Fig 1.39. This results in the formation of two magnets. Placing of two electrons together in the same orbital results in considerable repulsion due to same negative charge. By overcoming the repulsive forces and to keep the two electrons together in the same orbital some energy is required, called pairing energy. If the two electrons also have the same spin there will be magnetic repulsion also, causing the requirement of more pairing energy. But, if the two electrons are spinning in opposite directions and forming a pair in an orbital, mutually cancel their magnetic moments. Now, we can easily understand that for an electron if pairing energy is less than the energy of next higher energy level the electron will be paired in the orbital. But if the pairing energy is greater than the energy of next higher energy level the electron goes to the next higher energy level.

4b

Structure of Atom

1.35

(iii) Hund’s rule of maximum multiplicity. The electronic configuration of different atoms can be represented in two ways: (i) nlx or sa pb dc ...... notation (ii) Orbital diagram method

Fig 1.39 No two electrons in an atom can have the same set of four quantum numbers 3. hund’s Rule of maximum multiplicity We know that the orbitals belonging to same sub shell have same energy and are called degenerate orbitals. Hund’s rule states that the electron pairing in degenerate orbitals of a given sub shell will not take place unless all the available orbitals of a given shell contains one electron each. Since there are three p, five d and seven f orbitals, pairing of electrons will start in p, d and f orbitals with the entry of 4th, 6th and 8th electrons, respectively. It is also found that half-filled and fully filled degenerate set of orbitals acquire extra stability.

1.11.1 Electronic Configuration of Atoms The filling of electrons in different orbitals in an atom are governed by the above three rules viz., (i) Aufbau principle, (ii) Pauli’s exclusion principle, and

In the nlx method, n represents the principle quantum number, l represents the azimuthal quantum number (sub shell) and x represents the number of electron in that sub shell. In sa pb dc ... notation the sub shell is depicted with a superscript like a, b, c, ... etc, along with the principal quantum number before the respective sub shell. In the orbital diagram notation each orbital of the sub shell is represented by a box and the electron is represented by an arrow (↑) for positive spin and an arrow (↓) for negative spin. The advantage of the second notation over the first notation is that it represents all the four quantum numbers. Based on these notations, the electronic configurations of atoms of various elements of the Periodic Table in their ground state are as given in Table 1.5. Hydrogen has only one electron (Z = 1). This electron will enter the lowest energy orbital which is 1s. This, the solitary electron of hydrogen atom, will occupy the 1s orbital. The electronic configuration of hydrogen atoms is, therefore, represented as 1s1. The next element, helium, has two electrons (Z = 2). One of these electrons occupies the 1s orbital as in the case of the hydrogen atom. The second electron can also enter this orbital so as to fill it completely. The electronic configuration of helium is therefore, represented as 1s2. The two electrons occupying this orbital will have the

Table 1.5 Electronic Configuration of First 10 Elements 1

Hydrogen

1s1

2

Helium

1s2

3

Lithium

1s2 2s1

4

Beryllium

1s2 2s2

5

Boron

1s2 2s2 2 p1x

6

Carbon

1s2 2s2 2 p1x 2 p1y

7

Nitrogen

1s2 2s2 2 p1x 2 p1y 2 p1z

8

Oxygen

1s2 2s2 2 px2 2 p1y 2 p1z

9

Fluorine

1s2 2s2 2 px2 2 p y2 2 p1z

Neon

1s2 2s2 2 px2 2 p y2 2 pz2

10

1.36

Structure of Atom

opposite spins as can be seen from the following orbital diagram:

The next element is lithium. It has three electrons (Z = 3). The third electron of lithium (Li) is not allowed into the 1s orbital because of the Pauli exclusion prin ciple. If third electron enters into 1s orbital its spin will become similar with one of the two electrons due to which the charge repulsion and magnetic repulsion will become more. So more pairing energy will be required. This pairing energy will be more than the energy of the next higher energy level 2 s. So, the electron goes into 2s. Hence, the electronic configuration of Li is 1s2 2s1. The 2s orbital can accommodate one more electron. The configuration of beryllium (Be) atom is therefore 1s2 2s2. The next element boron (Z = 5) has five electrons. The 2s orbital has already been filled up. But the principal quantum number is now 2, there are three p-orbitals still available. The fifth electron, therefore, enters one of these p-orbitals. The electronic configuration of born is 1s2 2s2 2p1. The next five elements, namely carbon (Z = 6), nitrogen (Z = 7), Oxygen (Z = 8), fluorine (Z = 9) and neon (Z = 10) of this period get their p-orbitals successively filled up. Their electronic configurations are given Table 1.5. It may be noted that in carbon, the arrangement of electrons in p-orbitals is 2 p1x , 2 p1y , and not the 2 px2 . This is because to pair the electron in 2 px pairing energy is required but to enter into the 2 p y orbital having energy equal (degenerate) to 2 px pairing energy is not required. So, the electron goes into 2 p y . Similarly in nitrogen, the arrangement is 2 p1x 2 p1y 2 p1z and not 2 px2 2 p1y . This is in accordance with Hund’s rule of maximum multiplicity. In oxygen the last electron has two options. One to pair in one of three p-orbitals or to go the next higher energy level 3s. Since the pairing energy is less than the energy required to go to 3s orbital, pairing of electrons starts in oxygen and continues till all the three 2p orbitals are filled up. The principal quantum 3 (M-shell) begins with sodium (Z = 11). the eleventh electron will evidently, enter, into the 3s orbitals, the electronic configuration being 1s2 2s2 2p6 3s1. The next seven elements namely magnesium (Z = 12) aluminum (Z = 13), silicon (Z = 14), phosphorus (Z = 15), sulphur (Z = 16), chlorine (Z = 17), and argon (Z = 18) follow the sequence described above.

Consider now the electronic configuration of the next element which is potassium ( Z = 19). The additional electron (i.e., nineteenth electron) instead of entering 3d orbitals, all of which are lying unoccupied so far, goes into the 4s orbital. This is because 4s orbital is at lower energy level than 3d orbitals. The electronic configuration of potassium may be represented as 1s2 2s2 2p6 3s2 3p6 4s1. The additional electron in calcium (Z = 20) follows the same course and goes into the same, i.e., 4s orbitals, to fill it up completely. So, the electronic configuration of calcium is 1s2 2s2 2p6 3s2 3p6 4s2. The electrons in the completely filled shells are known as core electrons and the electrons that are added to the electronic shell with the highest principal quantum number are called the valence electrons. For example, the electrons in Ne are the core electrons and the electrons in the highest principal quantum number from Na to Ar are the valence electrons. The next ten elements (Z = 21 to 30) which follow calcium are scandium (Sc), titanium (Ti), vanadium (V), chromium (Cr), manganese (Mn), iron (Fe), cobalt (Co), nickel (Ni), copper (Cu) and zinc (Zn). In these elements, addition of electrons takes place in the inner 3d orbitals, while the outer 4s orbitals remains fully occupied. The electronic configuration of scandium (Z = 21), the first element in this series, is 1s2 2s2 2p6 3s2 3p6 3d1 4s2 while that (Z = 30) of the last element of the series is 1s2 2s2 2p6 3s2 3p6 3d10 4s2. Thus, while filling of 3d orbitals begins with scandium, it ends up with zinc. Anomalous configurations of chromium and copper It may be noted that the configurations of chromium (Z = 24) and copper (Z = 29) do not follow the general trend. The electronic configurations of Cr and Cu are expected to be as follows: and

Cr 1s2 2s2 2p6 3s2 3p6 3d4 4s2, Cu 1s2 2s2 2p6 3s2 3p6 3d9 4s2

but actually their configuration are Cr 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 5 4 s1 and Cu 1s 2 2 s 2 2 p 6 3s 2 3 p 6 3d 10 4 s1 These anamolies are attributed to the fact that exactly half-filled and completely filled orbitals (i.e., p3, p6, d5, d10, f 7 and f 14) are more stable because they possess the lower energy. The cause of this extra stability has been discussed in detail in the subsequent sections. Thus, to acquire an increased stability, one of the electrons from 4s orbital goes into the nearby 3d orbital so that 3d orbitals become exactly half-filled in the case of chromium and completely filled up in the case of copper.

Structure of Atom

After completely filling the 3d orbitals, the filling up of 4p orbitals starts at gallium (Ga) and is complete at krypton (Kr). In the next 18 elements from rubidium (Rb) to xenon (Xe) the pattern of filling the 5s, 4d and 5p orbitals are similar to that 4s, 3d and 4p orbitals as discussed above. Then comes the turn of the 6s orbital. In caesium (Cs) and barium (Ba), this orbital contains one and two electrons respectively. After the filling up of 6s sub shell with barium (Z = 56) the next electron in lanthanum (Z = 57) goes to 5d sub shell and not the 4f sub shell. The subsequent electrons in the succeeding elements of the lanthanide series (from Ce to Lu Z = 58 to 71) however enters the 4f sub shell by following the normal sequence of energies. The filling up of 5d sub shell recommences only after the 4f sub shell is completely filled up with Lu (Z = 71). For the sake of simplicity, we do not always write the complete electronic configuration. Instead, the configuration of the noble gas preceding the valence shell is not given; it is represented by its symbol in square bracket. For example, for Na (Z = 11), we may write [Ne] 3s1, and for Ca (Z = 20), we may write [Ar] 4s2. Thus, the configurations of La, Ce and Pr would be as follow: La (Z = 57) : [Xe] 4f 0 5d1 6s2 Ce (Z = 58) : [Xe] 4f 1 5d1 6s2 Pr (Z = 59) : [Xe] 4f 2 5d1 6s2 Similarly, after filling up of 7s sub shell with radium (Z = 88) the next electron in actinium (Z = 89) goes to 6d sub shell and not to 5f sub shell. The subsequent electrons in the succeeding elements of the actinides (from Th to Lr Z = 90 to 103), however, enter the 5f sub shell. Thus, the configurations of Ac, Th and Pa would be as follows: Ac (Z = 89) : [Rn] 6f 1 7s2 Th (Z = 90) : [Rn] 5f 1 6d1 7s2 Pa (Z = 91) : [Rn] 5f 2 6d1 7s2 The electronic configurations of the known elements (as determined by spectroscopic methods are given in Table 1.6).

1.11.2 Relative Stabilities of electronic Configurations Before we discuss the relative stabilities of various electronic configurations, we should clearly understand the difference between electronic configuration and electronic arrangement. For this, let us consider the nitrogen atom. Its outer electronic configuration is 2s2 2p3. From this we get information that three electrons are present in a set of three 2p orbitals. This configuration gives us no information about the manner in which these three electrons are arranged in three 2p orbitals.

1.37

It can be easily shown through permutations and combinations that there are as many as 20 different ways of placing these three electrons in the three 2p orbitals. In other words, the electronic configuration 2p3 can have 20 different electronic arrangements as shown in Table 1.7 These different electronic arrangements would have different energies inspite of the fact that they belong to the same electronic configuration. (Some of the electronic arrangements may have the same energy due to similar inter electronic repulsions). Therefore, when we talk of the stability of the electronic configuration, we are ipso facto concerned with the stability of the electronic arrangement of that configuration. Of all the electronic arrangements permitted by a given configuration the most stable configuration can be predicted based on the following two factors: 1. Symmetrical distribution of charge 2. Hund’s rule of maximum multiplicity To explain these two factors we shall consider here the different electronic arrangements of p3 configuration as shown in Table 1.7. Symmetrical Distribution of Charge In order to understand the effect of the symmetrical distribution of charge in orbitals on the stability of electronic arrangements, let us recall the shapes of the various sets of orbitals one by one. The s-orbital is spherical in shape which implies that the electronic charge would be distributed uniformly in all directions whether there is one or two electrons present in the orbital. Of the three p-orbitals, the px is symmetrical along the X-axis while the orbitals py and pz are symmetrical along the Y- and Z-axes respectively. If one electron is present in px -orbital, i.e., for p1x configuration, the electronic charge would be mainly concentrated in the X-direction only. Likewise the electronic charge would be mainly concentrated in the Y-direction of p1y configuration. Similarly, in p1x p1y configuration, the electronic charge would be mainly concentrated in the xy plane. In a similar manner, the electronic charge would be concentrated more along the X-direction in the configuration px2 p1y p1z and more in the plane xy in the configuration px2 p y2 p1z . It is evident that in all the p configurations mentioned above, the charge is not evenly spread along all the directions. In other words, the distribution of charge is nonuniform or unsymmetrical. On the other hand, the distribution of charge in the configurations p1x p1y p1z and px2 p y2 pz2 would be uniform or symmetrical in all directions. Configurations with even or uniform or symmetrical distribution of charge in all directions would evidently be associated with lower energy and hence higher stability than the configurations with unsymmetrical distribution of electronic charge.

1.38

Structure of Atom

Table 1.6 Electronic Configuration of the Elements Element

Z

1s

2s

2p

3s

3p

H He Li Be B C N O F Ne Na Mg Al Si P S Cl Ar K Ca Sc Ti V Cr * Mn Fe Co Ni Cu* Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb * Mo* Tc Ru* Rh* Pd* Ag* Cd* In

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

3d

4s

4p

1 2 3 5 5 6 7 8 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

1 2 2 2 2 1 2 2 2 2 1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

1 1 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6

4d

1 2 4 5 5 7 8 10 10 10 10

4f

5s

5p

5d

5f

6s

6p

6d

7s

1 2 2 2 1 1 2 1 1 1 2 2

1 Table Continues

Structure of Atom

1.39

Table 1.6 Electronic Configuration of the Elements–continued Element

Z

1s

2s

2p

3s

3p

3d

4s

4p

4d

Sn Sb Te l Xe Cs Ba La * Ce* Pr Nd Pm Sm Eu Gd * Tb Dy Ho Er Tm Yb Lu Hf Ta W Re Os Ir Pt* Au Hg Tl Pb Bl Po At Rn Fr Ra Ac Th Pa U Np Pu Am Cm Bk Cf

50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96 97 98

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

4f

5s

5p

2 3 4 5 6 7 7 9 10 11 12 13 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14 14

2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 3 4 5 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6 6

5d

1

1

1 2 3 4 5 6 7 9 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10 10

5f

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

2 3 4 6 7 7 8 10

6s

1 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2

6p

6d

7s

3 4 5 6 6 1 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2 6 2 Table Continues

1.40

Structure of Atom

Table 1.6 Electronic Configuration of the Elements–continued Element

Z

1s

2s

2p

3s

3p

3d

4s

4p

4d

4f

5s

5p

5d

5f

6s

6p

Es Fm Md No Lr Rf Db Sg Bh Hs Mt Ds Rg **

99 100 101 102 103 104 105 106 107 108 109 110 111

2 2 2 2 2 2 2 2 2 2 2 2 2

2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6

2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6

10 10 10 10 10 10 10 10 10 10 10 10 10

2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6

10 10 10 10 10 10 10 10 10 10 10 10 10

14 14 14 14 14 14 14 14 14 14 14 14 14

2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6

10 10 10 10 10 10 10 10 10 10 10 10 10

11 12 13 14 14 10 11 12 13 14 14 14 14

2 2 2 2 2 2 2 2 2 2 2 2 2

6 6 6 6 6 6 6 6 6 6 6 6 6

6d

7s

1 2 3 4 5 6 7 8 9 10

2 2 2 2 2 2 2 2 2 2 2 2 1

* Elements with exceptional electronic configurations. * * Elements with atomic number 112 and above have been reported but not yet fully authenticated and named.

Table 1.7 Arrangements of Three Electrons in Three p-Orbitals S. No.

px

py

pz

1 2 3 4 5 6 7

↑ ↑ ↑ ↓ ↓ ↓ ↓

↑ ↑ ↓ ↑ ↓ ↓ ↑

↑ ↓ ↑ ↑ ↓ ↑ ↓

8 9 10

↑ ↑↓ ↑↓

↓ ↑

↓

11 12 13 14 15 16 17 18 19 20

↑↓ ↑ ↑ ↑ ↑↓

↑↓ ↑↓ ↓

↑↓

↓ ↑↓ ↓

↓ ↓

↑ ↑ ↑↓

↑↓

↓ ↑↓ ↑↓

To sum up: a symmetrical distribution of electronic charge leads to a decrease in energy and hence an increase in the stability of the system. It can be easily seen that among 20 electronic arrangements of p3 configurations of Table 1.7 1 to 8 arrangements would have a uniform or symmetrical distribution of charge in space. Therefore, these configurations would be more stable than all other p3 configurations. Now a question arises that among the 1 to 8 arrangements of p3 configurations which will be more stable ? This can be explained based on Hund’s rule of maximum multiplicity. Then what is spin multiplicity? The spin multiplicity is the value of 2S + 1 where S is the resultant spin quantum number. The relation between the number of unpaired electrons, the resultant spin quantum number S and multiplicity is given in Table 1.8. Table 1.8 Spin Multiplicity No. of unpaired electrons

S

Multiplicity

Name of state

0

0

1

Singlet

1

1 2

2

Doublet

2

1

3

Triplet

3

3 2

4

Quartet

4

2

5

Quintet

Structure of Atom

Maximum spin multiplicity is possible only when all the degenerate orbitals occupy with electrons of parallel spin. In the Table 1.7 the 1 and 5 electronic arrangements have maximum number of unpaired electrons with parallel spin and thus have maximum multiplicity. So, these two (1 and 5) electronic arrangements are more stable than the other electronic arrangements. This is because of lowering of energy in those arrangements. Then again another question arises as to why such arrangements have less energy? The answer for this question can be ex plained involving the contribution of exchange energy as discussed below. By advanced mathematical treatment, it can be shown that if on exchanging the position in space of two electrons with parallel spins, there is no change in the electronic arrangement, it would lead to decrease in energy. The pair of electrons is called the exchange pair. The larger the number of exchange pairs of electrons, the greater would be the decrease of energy. The energy decrease per exchange pair of electrons is termed as columbic exchange energy and is represented with E. it would evidently, carry a negative sign. In the electronic arrangements 1 and 5 of p3 configurations in Table 1.7, three exchange pairs are possible (1, 2; 1, 3; and 2, 3). So the lowering of energy due to columbic exchange energy will be – 3E. From 1-8 electronic arrangements of p3 configurations (Table 1.7) except 1and 5 there is only one exchange pair since one electron is with anti-clockwise spin. So, the decrease in energy due to columbic exchange energy is only – E. In the remaining electronic arrangements 9 to 20 (Table 1.7) two electrons are paired in one of the p orbitals which requires pairing energy represented by P. Obviously, the pairing energy is positive. Since the pairing energy has positive sign, while the columbic exchange energy have negative sign, pairing energy will destabilize the atom while exchange energy will give stability to the atom. Thus, the overall stability of a system would be decided by the aggregate of exchange energy (E) and pairing energy (P). In the case of electronic

1.41

arrangements 9 to 20 (Table 1.7) the net energy change will be – E + P, and is the least stable. The electronic arrangements 1 and 5 (Table 1.7) are most stable because of symmetrical distribution of charge and more number of columbic exchange energies. Similarly, the stability of anomalous electronic configuration of chromium can be explained.

The 3d 4 4s2 configuration has only 10 exchange pairs ( C2 pairs) whereas the 3d5 4s1 has 15 exchange pairs (6C2 pairs) because the 3d4 4s2 has 5 electrons with parallel spins while 3d5 4s1 arrangement has 6 electrons with parallel spins. Hence, in 3d5 4s1 arrangement the decrease in energy is –15E while in 3d4 4s2, is –10E. Further, in 3d4 4s2 configuration, there is one pair of electrons in 4s orbital which requires the pairing energy. So, the net decrease in energy of 3d4 4s2 configuration is –10E + P. Also, the 3d4 4s2 configuration has no symmetrical distribution of charge. So, the 3d5 4s1 configuration for chromium is more stable than 3d4 4s2 configuration. It can thus be concluded that electronic arrangements with exactly half-filled or completely filled degenerate orbitals would be more stable than any other electronic arrangement. 5

Fig 1.40 Possible exchange pairs for a d5 configuration

1.42

Structure of Atom

Key POiNTS Sub-Atomic Particles

Rutherford’s Atomic model

•

•

•

The fundamental particles of every atom are (i) electron, (ii) proton, and (iii) neutron The only element whose atoms do not contain a fundamental particle (neutron) is hydrogen.

• •

electron • •

• • • • • • •

Discovered by JJ Thomson in 1897. Experiment in a discharge tube: cathode rays are formed in a discharge tube voltage (10, 000 V) and low pressure. Electron is the lightest fundamental particle. Charge on electron is equal to –4.8 × 10–10 esu or –1.602 × 10–19 coulombs. Mass of one electron is equal to 1/1837 of hydrogen atom or 9.109 × 10–28 g or 9.1 × 10–31 kg. Mass of one mole of electrons is equal to 0.55 mg. Energy of electron is not influenced by any magnetic field. Specific charge decreases with increase in velocity of electron due to increase in mass. e/m ratio is constant for the cathode ray.

• •

•

Atomic Number • •

Proton • • • • • • •

Discovered by Goldstein in canal rays experiment. Properties of the canal rays are characteristics of the gas taken in the discharge tube. The canal rays contain positively charged ions of gas present in the discharge tube. The smallest +ve charged particle is proton. The mass of proton 1.008 u or 1.672 × 10–24 g or 1.672 × 10–27 kg. The charge of proton is equal to 4.8 × 10–10 esu or 1.602 × 10–19 coulombs. Proton is H+ ion.

• • • • •

Neutron was discovered by Chadwick. Neutron is emitted when lighter elements like beryllium and boron are bombarded with α-particles. Neutron has no charge. Mass of neutron is 1.00866 u or 1.675 × 10–24 g or 1.675 × 10–27 kg. Neutron is the heaviest fundamental particle of the atom.

The number of electrons or protons in an atom is equal to its atomic number. Moseley discovered the relation between the frequencies of the characteristic X-rays (v) of an element and its atomic number (z). v = a ( z - b ) a and b are constants.

• • • • • •

Neutron

Rutherford’s experiment proved the presence of nucleus. The total mass of the atom is concentrated at the centre of the spherical atom known as nucleus. The positive charge of the nucleus is counter-balanced by the negatively charged electrons revolving round the nucleus. Rutherford model is also known as the Planetary model. As per the laws of electrodynamics a moving electron (charged particle) should continuously loose energy by emission of radiation. Consequently it should fall into the nucleus. If the electron loses energy continuously, the atomic spectrum should be band spectrum but the atomic spectra are line spectra.

• •

Sum of the number of protons and neutrons in an atom is called its mass number. Mass number is the corrected atomic weight to a whole number. Mass number A = Z + N where Z is atomic number and N is the number of neutrons. Atoms of an element which differ in their mass but have the same atomic number are called isotopes. Isotopes have same number of protons but differ in the number of neutrons present in them. Different atoms having same mass number but with different atomic numbers are called isobars. Atoms of different elements containing the same number of neutrons are called isotones. Isotones differ in both atomic numbers and mass number but difference in the atomic number and mass number is the same.

Structure of Atom

Developments leading to the bohr’s model of Atom, Nature of light

•

• •

•

• •

•

•

• • • • • • •

•

All radiant energy propagates in the form of waves. The radiant energy is in the form of electromagnetic waves. The radiations are associated with electric and magnetic field perpendicular to one another. On the propagation of an electromagnetic radiation there is only propagation of wave but not that of the medium. Wavelength is the distance between two successive crests or between two successive troughs of waves. (i) It is denoted by l. (ii) It is measured in Å (Angstroms) or nm (nanometers) 1Å = 10–8 cm or 1 nm = 10–9 m. Frequency is the number of waves per second passing at a given point. It is denoted by v. Units of frequency are Hertz or cycles s–1. Velocity of light is the distance travelled by one wave in one second. Light or all electromagnetic radiations travel in vacuum or air with the same velocity. Velocity of light c = 3 × 108 m s–1. Wavelength is inversely proportional to the frequency of the wave. Velocity of light = Frequency × Wavelength or c = νλ. Wave number is the number of waves spread in one centimeter, denoted by v . Wave number v is the reciprocal of wavelength. 1 c v = , but v = , \ v = c v l l Units of wave number are cm–1 or m–1.

•

• • • • •

• •

•

• • • •

•

Corpuscular theory of light was proposed by Newton. According to corpuscular theory of Newton, light is propagated in the form of small invisible particles. Quantum Theory was proposed by Max Planck and was extended by Einstein. A hot vibrating body does not emit or absorb energy continuously but emits or absorbs energy discontinuously in the form of small energy packets called quanta. The energy of radiation (E) is proportional to its frequency (ν). E∝v

(or)

E = hv

h is a constant known as Planck’s constant. Its value is 6.6256 × 10–34 Js or kg m2 s–1 or 3.6253 × 10–27 erg second or g cm2.

A hollow sphere coated inside with platinum black and having a small hole in its wall acts as a near black body. Black body is a perfect absorber and perfect emitter of radiant energy. At a given temperature the intensity of radiation increases with wavelength, reaches a maximum and then decreases. As the temperature increases the intensity of the radiation will be more towards the lower wavelengths. Planck’s quantum theory explains only the black body radiations. Einstein extended the quantum theory to all types of electromagnetic radiations. Einstein called the energy packets of electromagnetic radiations as photons. The emission of electrons from metal surface when exposed to radiation of suitable wavelength is known as photoelectric effect. The photoelectric effect is readily exhibited by alkali metals like K and Cs. When photon strikes the metal, its energy is ab sorbed by the electron and emission of electrons takes place. A part of the energy of the photon is used up to escape the electron from the attractive forces and the remaining energy is used in increasing the kinetic energy of the electron. hv = W + KE

•

Quantum Theory of Radiation

1.43

•

•

hv = Energy of photon; W = Energy required to overcome the attractive forces on an electron in the metal; KE = Kinetic Energy of the emitter electron. A photon exists independently until it is absorbed by an another body. A body can absorb light energy in the form of pho tons and goes to the excited state E2 from the ground state E1. Emission of energy will be accompanied during the transition from excited state to ground state. E2 – E1 = DE = hv

Types of Spectra •

• •

The arrangement obtained by splitting of electromagnetic radiation into its component wave lengths when passed through a prism is called a spectrum. When white light is passed through the prism, it gives a continuous spectrum of seven colours (VIBGYOR). In a continuous spectrum each colour fades into the next colour as in a rainbow.

1.44

•

•

• • •

• • • • •

Structure of Atom

The spectrum of incandescent white light obtained by heating a solid to very high temperature is a continuous spectrum. The spectrum formed by the emission of energy in the form of light radiation is called an emission spectrum. Emission spectrum consists of bright lines or bands on a dark background. Absorption spectrum is just opposite to emission spectrum. The dark lines in the absorption spectrum and bright lines in a the emission spectrum of a particular substance appear at the same places (same wave length). The apparatus used to record the spectrum is called the spectrometer or spectrograph. Emission spectrum is due to the emission of light by the excited atoms or molecules. Absorption spectrum is due to the excitation of atoms or molecules by absorbing the energy. Each element has its own characteristic line spectrum by which it can be identified. Line spectra are given by atoms so known as atomic spectra and band spectra are given by molecules so known as molecular spectra.

hydrogen Spectrum • • • •

• • •

• • • • •

•

1 1  2 - 2  n1 n2  R is Rydberg’s constant and its value is 109677 cm–1 for hydrogen atom. In Balmer series first line is called as Hα, second line is Hβ, third line as Hγ, fourth line as Hδ and so on. The wavelengths of Hα, Hβ, Hγ and Hδ are obtained by substitution series in hydrogen spectrum are 1 = RH l

Name of series

n1

n2

Lyman series Balmer series Paschen series Brackett series Pfund series

1 2 3 4 5

2, 3, 4, 5, 6,........ 3, 4, 5, 6,........ 4, 5, 6,........ 5, 6, 7,........ 6, 7 ......

Spectral Region Ultraviolet Visible Near infrared Infrared Far infrared

The spectral lines get closer and closer as we move from n2 = 3 to 4 to 5 to 6 etc. If n2 is taken as ∝ the limiting wavelength of the lines in any series can be obtained. From Lyman series to Pfund series, wavelength increases but frequency and energy decrease. Maximum difference in energy for hydrogen atom is found when n1 = 1 and n2 = 2 or higher value. RH value 109677 cm–1 is valid only for lines in the hydrogen spectrum. For a spectral line of any one electron species like He+ or Li2+, the value of R is equal to 109677 z2 cm–1 where z is the atomic number. Number of spectral lines when an electron is coming to ground state or to the orbit n1 from excited state in orbit n2 can be calculated as. n ( n + 1) 2

, where n = n2 – n1.

bohr’s model of the Atom • •

•

Of all the spectra, hydrogen spectrum is the simplest spectrum. Hydrogen spectrum consists of a groups of lines classified into various series. Only Balmer series is the visible series in hydrogen spectrum. Wavelength of a spectral line in hydrogen spectrum can be calculated by using Rydberg’s equation. v=

•

• • •

• • •

The basis for the Bohr’s model of atom is the Planck’s quantum theory and hydrogen spectrum. Electrons revolve round the nucleus in concentric circular orbits which are represented by K, L, M, N, O, P..... or 1, 2, 3, 4, 5, 6..... etc. Each orbit is associated with some energy. So they are known as energy levels or stationary orbits. The energy of the orbit increases with increase in the distance form the nucleus. The energy of an electron moving in an orbit remains constant. The electron can move from one orbit to another orbit only when they lose or gain the energy difference of the orbitals. ΔE = E2 – E1 = hv An electron while moving from higher energy level to the lower energy level emits energy in the form of light. The angular momentum (mvr) is quantized. It is an h where ‘h’ is the Planck’s integral multiple of 2p constant. mvr =

•

The force of attraction between the nucleus and the electron =

•

nh 2p

e2 . r2

The centrifugal force gained by the electron due the revolving round the nucleus = -

mv 2 . r

Structure of Atom

•

Bohr’s equation to calculate the radius of an orbit is 2

nh 4 p 2 2e 2 m

•

For hydrogen, since z = 1, the radius of orbit is equal to 0.529 × n2 Å. Radius of 1st orbit = 0.529 Å, radius of 2nd orbit = 2.116 Å etc.

•

r= •

•

Kinetic energy of the electron due to motion =

e . r

•

Potential energy of the electron due to position = -

•

Total energy of the electron E = Kinetic energy + potential energy 2

=

2

1 2 e e e mv - = r 2 2r r

= -

• • • • •

1 1 v = RH  2 - 2   n1 n2 

 e 2  = mv   r 

The RH value =

•

Bohr’s theory could explain the spectra of atoms and ions like He+, Li2+. Be3+ having one electron only. Bohr’s theory correlates the velocity of light, electronic mass, Planck’s constant and electronic charge. The energy of the orbit, radius of the orbit and the value of Rydberg’s constant RH calculated from Bohr’s theory are in good agreement with the experimental value. Bohr’s theory could not explain the spectra of atoms having electrons more than one. Bohr’s theory could not explain the wave nature of electron established by de Broglie. It could not explain Zeeman effect and Stark effect. Splitting of spectral lines in a strong magnetic field is called as the Zeeman effect. Splitting of spectral lines in strong electric field is called the Stark effect. The further splitting of ordinary spectrum when taken by super-spectrometer is called the fine spectrum.

• •

•

Bohr’s equation for calculating the energy of an orbit 2π2 me 4 z 2 × 2 h2 n

The energy of an electron in an atom is negative and inversely proportional to the square of the n value (n is the number of orbit). As the electron moves away from the nucleus the kinetic energy decreases and potential energy increases. The potential energy of an electron is zero when the electron is at infinite distance from the nucleus. As the n value of the orbit decreases, the size and energy of the orbit decreases. The kinetic energy of an electron (positive value) is greater than the potential energy (negative value). The energy of an electron in the hydrogen atom. 313.6 Kcal mol–1 n2 1312 = - 2 KJ mol–1 n

• • • • •

Quantum Numbers • •

En = −

2.18 × 10−11 ergs atom–1 =− n2 13.6 = - 2 eV atom–1 n •

To explain fine spectrum each electron in an atom is assigned with a set of four quantum numbers. Quantum numbers explain the orbital concept of atomic model i.e., size, shape, orientation and spin of the electron.

Principal Quantum Number • • • •

Bohr’s equation to calculate the velocity of an electron in hydrogen atom,

•

nh 2pe 2 z = × h n 2pmr

•

Vn =

2p2 me 4 = 109699 cm–1. ch3

•

2

e2 2r

En = − •

2

The velocity of electron in first orbit of hydrogen is 2.188 ms–1. The energy emitted or absorbed when an electron moves from one orbit to another orbit can be calculated from DE = hv = hcv . v can be calculated by using Rydberg’s-Ritz equation.

1 2 mv . 2 2

•

•

2

1.45

It was proposed by Bohr, and is denoted by n. It will have any integer value except zero. It gives the size of the orbit and hence energy of orbit. As the value of n increases the size and energy of the orbit increases. It also represents the distance of the electron from the nucleus. The number of electrons that can be present in an orbit is equal to 2n2.

1.46

Structure of Atom

Azimuthal Quantum Number • • • •

• • • • • • • • • •

It was proposed by Sommerfeld denoted by l. It gives the shape of an orbital. ‘l’ values depend upon the ‘n’ values and are equal to ‘n - l’ i.e., 0, 1, 2, 3,..... Depending upon the ‘l’ values the subshell in which the electron present can be known. If l = 0, the electron belongs to s-sub-shell. If l = 1, the electron belongs to p-sub-shell. If l = 2, the electron belongs to d-sub-shell. If l = 3, the electron belongs to f-sub-shell. The number of electrons that can be present in any sub-shell is equal to 2(2l + 1) or 4l + 2. The number of sub-shells present in an orbit is equal to the value of principal quantum number ‘n’ value. Number of subs-shells in K shell (n = 1) = 1 (i.e., 1s) Number of sub-shells in L shell ((n = 2)) = 2 (i.e., 2s and 2p) Number of sub-shells in M shell (n = 3) = 3(i.e., 3s, 3p and 3d) Number of sub shells in N shell (n = 4) = 4(i.e., 4s, 4p, 4d and 4f) s-orbital is spherically symmetrical in shape. p-orbital is dumb-bell in shape. d-orbital is double dumb-bell in shape. f- orbital is four fold dumb-bell in shape.

magnetic Quantum Number • • • • • • • • •

It was proposed by Lande to account for the Zeeman effect. It is denoted by ml. It gives the orientation of the orbital. The values of ‘ml’ depends upon ‘l’ and are equal to (2l + 1) ranging from –l passing through ‘0’ to + l. s-sub-shell will have only one value ‘0’. So contain only one orbital i.e., s-orbital. p-sub-shell will have three orbitals having ‘ml’ value –1, 0, +1. These are px, py and pz. d-sub-shell will have five orbitals having ‘ml’ values –2, –1, 0, +1, +2. f-sub-shell will have seven orbitals having ‘ml’ values –3, –2, –1, 0, +1, +2, +3. The + and – signs indicate only change in direction but the magnetic quantum number ‘ml’ as such is neither positive nor negative.

Spin Quantum Number •

It was proposed by Uhlenbeck and Goudshmit and is denoted by ‘s’.

• • •

•

•

Electron moving in an orbital can spin on its own axis. The spin of the electron may be clockwise or anti-clockwise. 1 The clockwise spin is denoted by + or ↑ and 2 1 anti-clockwise spin is denoted by - or ↓. 2 The difference between the two spin quantum numbers is 1 which is equal to the difference between successive quantum numbers. Modern theory of atomic structure was proposed on the basis of quantum mechanics or wave mechanics proposed independently by de Broglie, E. Schrodinger and W. Heisenberg.

Waves and Particles • •

Wave theory of moving particles was proposed by de Broglie. Wavelength of a moving particle can be calculated by the relation l =

•

•

•

h h = mv p

where l is wavelength of a moving particle h is Planck’s constant, m is mass of the particle, v is velocity of particle, and p is the momentum of particle. The dimensions of the Planck’s constant depend upon the dimensions of momentum. de Broglie derived the wave nature of a moving particle from Einstein’s mass energy equation (E = mc2) and Planck’s quantum theory (E = hv). While the electron wave is moving in an orbit if the two ends meet to give regular series of crests and troughs, the electron wave is said to be in phase. For the electron to be in phase it is necessary that the circumference of the Bohr’s orbit (= 2πr) must be equal to the whole number of the wavelength (l) of electron wave nl = 2πr.

heisenberg’s Uncertainty Principle •

•

It is impossible to know exactly both position and momentum of an electron or any other smaller particle simultaneously and accurately. Mathematically the uncertainty principle can be expressed as Dx ⋅ Dp ≥ (Where n = 1, 2, 3, 4.....)

h np

Structure of Atom

•

For an electron revolving round the nucleus in an atom the value of n is nearly 4. Dx ⋅ Dp ≥

Hence,

h 4p

Schrodinger Wave equation •

• •

The consequence of Heisenberg uncertainty principle is that since the exact position of an electron in an atom cannot be predicated, only probability of finding an electron in space around the nucleus can be predicted. Schrodinger wave equation gives the probability of finding the electron in space around the nucleus. Schrodinger wave equation is 2

2

2

Where y is the wave function, m is the mass of electron, E is the total energy of electron, U is the potential energy of the electron. Schrodinger wave equation indicates the variation of the y value along x, y and z axes. y is the amplitude of the wave function. y2 denotes the particle density when applied to the particles.

•

•

• • • •

•

•

Probability Concept • • • • •

p-orbital has both nodal regions equal to n – 2 and nodal planes equal to the value of l, i.e., = 1 The nodal plane for px orbital is yz; for py is xz and for pz is xy. Each d-orbital has nodal regions equal to n – 3 and nodal planes equal to the value of l i.e., 2. For any orbital the number of nodal regions is equal to n – l –1 and nodal planes equal to l. The nodal regions are known as radial nodes and the nodal planes are known as angular nodes.

Filling of Orbitals Pauli’s exclusion Principle • •

No two electrons in the same atom can have the same values for all the four quantum numbers. The two electrons present in a given orbital may have the same values of n, l and ml but they must have opposite spin, i.e., either +

•

1 1 or 2 2

Pauli’s principle helps the determination of maximum number of electrons that can be placed in an orbital of sub-shell and a main shell.

Pauli’s principle is followed only while writing the electronic configuration of atoms in their ground state.

Aufbau Principle

2

∂ ψ ∂ ψ ∂ ψ 8π m + + + 2 ( E −U)ψ = 0 ∂x 2 ∂y 2 ∂z 2 h

• •

•

1.47

The electron configurations of successive elements differ only in the last electron which is known as differentiating electron. According to aufbau principle “the electrons tend to occupy orbitals of minimum energy” in the ground state of atom. The energies of different orbitals can be calculated from n + l values. In case two orbitals have same n + l values, the electron goes into the orbital whose n values is less. The sequence of filling different orbitals by electrons be known form Moeller diagram. (Fig 1.38). The energy of 4s orbital is less than that of 3d orbital in elements having atomic number up to 20, but in elements having atomic number above 20, the energy of 3d orbital is less than 4s. The energy of 4f orbital has more energy than 6s in the elements having atomic number up to 57, but in the elements having atomic number around 90 the energy of 4f is less than 5s orbital. Electronic configurations are written in nlx method where n is the principal quantum number. l is the azimuthal quantum number and x is the number of electrons present in it.

hund’s Rule • •

• • •

The orbitals in a sub-shell having equal energy are called degenerate orbitals. According to Hund’s rule all the available orbitals are singly filled first with electrons of parallel spin and then only pairing of electrons starts. The half-filled and completely filled degenerate orbitals in atoms provide the stability. When two electrons are paired with opposite spins columbic repulsion energy is lowered. When two electrons are paired with parallel spins they remain farther apart due to columbic repulsion.

Stability of Atoms •

On exchanging the position in space of two electrons with parallel spins, if there is no change in the electronic arrangement it leads to decrease in energy. This pair is called the exchange pair.

1.48

• •

•

•

Structure of Atom

The decrease in energy per exchange pair of electrons is referred as Columbic exchange energy. As the number of exchange pairs increases the stability of an atom increases by the lowering of exchange energy E for each pair. For chromium with electronic configuration 3d 5 4s1 has five unpaired electron in d-orbitals and one electron in 4s orbital having parallel spin which gives 15 exchange pairs (6C2), so average lowering energy is 15E. For chromium with electronic configuration 3d4 4s2 has only four unpaired electrons in d-orbitals and which form 10 exchange pairs ( 5C2), so average lowering of energies is 10E.Hence 3d5 4s1 configuration rather than 3d4 4s2 configuration is stable for chromium.

magnetic Properties of Atoms • • • • • •

Atoms which contain unpaired electrons exhibit paramagnetism. Paramagnetic substances are attracted by the external magnetic field. Atoms in which all the electrons are paired exhibit diamagnetism. Diamagnetic substances are repelled by the external magnetic field. The paramagnetic moments of atoms depend upon the number of unpaired electrons. The paramagnetic moment can be calculated by using the formula m = n(n + 2) BM where n is the number of unpaired electrons.

Structure of Atom

1.49

PRACTiCe eXeRCiSe multiple Choice Questions with Only One Answer level i 1. Which statement does not form part of Bohr’s model of the hydrogen atom. (a) energy of the electrons in the orbit is quantized. (b) the electron in the orbit nearest the nucleus is in the lowest energy. (c) electrons revolve in different orbits around the nucleus. (d) the position and velocity of the electrons in the orbit cannot be determined simultaneously. 2. If S1 be the specific charge (e/m) of cathode rays and S2 be that of positive rays then which is true. (a) S1 = S2 (b) S1 S2 (d) either of these 3. The mass of an electron is m, its charge e and it is accelerated from rest through a potential difference V. The kinetic energy of the electron in joules will be (a) V (b) eV (c) MeV (d) none 4. In the above question, the velocity acquired by the electron will be (a)

(V / m)

(b)

(eV / m)

(c)

(2eV / m)

(d) none

5. For an electron in a hydrogen atom, the wave function, ψ is proportional to exp (r/ao), where ao is the Bohr’s radius. What is the ratio of the probability density the electron at the nucleus to the probability density at ao (a) e (b) e2 1 (d) 0 (c) 2 e 6. Photoelectric effect is the phenomenon in which (a) photons come out of a metal when it is hit by a beam of electrons. (b) photons come out of the nucleus of an atom under the action of an electric field. (c) electrons come out of a metal with a constant velocity which depends on the frequency and intensity of incident light wave. (d) electrons come out of a metal with different velocities not greater than a certain value which depends only on the frequency of the incident light wave not on its intensity. 7. In photoelectric effect, the photo-current (a) increases with increase of frequency of incident photon.

8.

9.

10.

11.

12.

13.

14.

(b) decreases with increases of frequency of incident photon. (c) does not depend on the frequency of photon but depends only on the intensity of incident light. (d) depends both on the frequency of the incident photon. Increase in the frequency of the incident radiations increases the (a) rate of emission of photo electrons (b) work function (c) kinetic energy of photo electrons (d) threshold frequency A photo sensitive metal is not emitting photo electrons when irradiated. It will do so when threshold is crossed. To cross the threshold we need to increase. (a) intensity (b) frequency (c) wavelength (d) none If E1, E2 and E3 represent respectively the kinetic energies of an electron, an alpha particle and a proton each having same de Broglie wavelength then. (a) E1 > E3 > E2 (b) E2 > E3 > E1 (c) E1 > E2 > E3 (d) E1=E2=E3 A surface ejects electrons when hitted by green light but not when hitted by yellow light. Will electrons be ejected if the surface is hitted by red light. (a) yes (b) no (c) yes, if the red beam is quite intense (d) yes, if the red beam continues to fall upon the surface for a long time. The angular speed of the electron in the nth orbit of Bohr hydrogen is. (a) Directly proportional to n (b) inversely proportional to n (c) inversely proportional to n2 (d) inversely proportional to n3 The minimum energy required to excite a hydrogen atom from its ground state is. (a) 3.4 eV (b) 13.6 eV (c) –13.6 eV (d) 10.2 eV The binding energy of the electron in the lowest orbit of the hydrogen atomis 13.6 eV. The energies required in eV to remove an electron from three lowest orbits of the hydrogen atom are (a) 13.6, 6.8, 8.4 eV (b) 13.6,10.2,3,4 eV (c) 13.6,27.2, 40.8 eV (d) 13.6,3.4,1.5 eV

1.50

Structure of Atom

15. The frequency of first line of Balmer series in hydrogen atom is n0. The frequency of corresponding line emitted by singly ionized helium atom is. (a) 2ν0 (b) 4ν0 ν0 ν (d) 0 (c) 2 4 16. The orbital angular momentum of an electron in 2sorbital is. h (a) (b) zero 4p h h (c) (d) 2 2p 2p 17. Which one represents an impossible arrangement. n 3 4 3 5

(a) (b) (c) (d)

l 2 0 2 3

m –2 0 –3 0

s ½ ½ ½ ½

18. Two electrons in the same orbital may be identified with (a) n (b) l (c) m (d) s 19. Photoelectric emission is observed from a surface for frequencies ν1 and ν2 of the incident radiation (ν1 > ν2). If the maximum kinetic energies of the photoelectrons in the two cases are in the ratio 1:k then the threshold frequency v0 is given by. ν − ν1 kν1 − ν 2 (a) 2 (b) k −1 k −1 (c)

kν 2 − ν1 k −1

(d)

ν 2 − ν1 k

20. The ratio of the different in energy of electron between the first and second Bohr’s orbits to that between second and third Bohr’s orbit is. 1 27 (a) (b) 3 5 (c)

9 4

(d)

4 9

21. The wavelength of a moving body of mass 0.1 mg is 3.31 × 10–29 m. The kinetic energy of the body in J would be (a) 2.0 × 10–6 (b) 1.0 × 10–3 –3 (c) 4.0 × 10 (d) 2.0 × 10–4 22. If the speed of electron in the Bohr’s first orbit of hydrogen atom is x, the speed of the electron in the third orbit is x x (b) (a) 9 3 (c) 3x

(d) 9x

23. If each hydrogen atom is excited by giving 8.4 eV of energy, then the number of spectral lines emitted is equal to. (a) none (b) two (c) three (d) four 24. When 4f- level of an atom is completely filled with electrons, the next electron will enter (a) 5 s (b) 6 s (c) 5d (d) 5p 25. The electron identified by quantum numbers n and 1, (i) n = 4, l = 1 (ii) n = 4, l = 0 (iii) n = 3, l = 2 (iv) n = 3, l = 1 can be placed in order of increasing energy from the lowest to highest (a) iv < ii < iii < i (b) ii < iv < i < iii (c) i < iii < ii < iv (d) iii < i < iv < ii 26. The energy of an electron in the Bohr’s orbit of H atom is –13.6 eV. The possible energy value (s) of the excited state (s) for electrons in Bohr’s orbits of hydrogen is (are) (a) –3.4 eV (b) –4.2 eV (c) –6.8 eV (d) +6.8 eV 27. The number of d-electrons in Fe+2 (at no of Fe = 26) is not equal to that of the (a) p-electrons in Ne (at. no = 10) (b) s-electrons in Mg (at. no = 12) (c) d-electrons in Fe (d) p-electrons in Cl– (at. no = 17) 28. Which have the same number of s-electrons as the d-electrons in Fe2+ (a) Li (b) Na (c) N (d) P 29. If ‘RH’ is the Rydberg constant, then the energy of an electron in the ground state of hydrogen atom is 1 R C (a) H (b) RH ch h hc (c) (d) - RH hc RH 30. When atoms are bombared with α- particles, only a few in a million of the α- particles suffer deflections while others pass through undeflected. This is because (a) the force of attraction on the α-particles by the oppositely charged electrons is not sufficient (b) the nucleus occupies much smaller volume compared to the volume of the atom (c) the force of repulsion on the fast moving α-particles small (d) the effect in the nucleus do not have any effect on the α-particles 31. The nucleus and an atom can be assumed to be spherical. The radius of the nucleus of mass number A is given by 1.25 × 10-13 × A1/3 cm . The atomic radius of

Structure of Atom

32.

33.

34.

35.

36.

37.

38.

39.

40.

41.

atom is one A0. If the mass number is 64, the fraction of the atomic volume that is occupied by the nucleus is. (a) 1.0 × 10–3 (b) 5.0 × 10-5 -2 (c) 2.5 × 10 (d) 1.25 × 10-13 The nucleus of an atom is located at x = y = z = 0. If the probability of finding an s-orbital electron in a tiny volume around x = a, y = z = 0 is 1× 10-5 . What is the probability of finding the electron in the same sized volume around x = z = 0, y = a? (a) 1× 10-5 (b) 1× 10-5 x a -5 2 (c) 1× 10 x a (d) 1× 10-5 x a–1 What is the probability at the second site if the electron were in a pz orbital for data in (Q. No. 32) above? (a) 1× 10-5 (b) 2 × 10-5 -5 (c) 4 × 10 (d) 0 1  1 If there are three possible values  - , 0, +  for the 2  2 spin quantum number ms, then the electronic config of K(19) will be (a) 1s 2 2 s 3 2 p 9 3s 3 3 p1 (b) 1s 2 2 s 2 2 p 6 3s 2 4 s1 (c) 1s 2 2 s 2 2 p 9 3s 2 4 p 4 (d) None is correct If the Aufbau rule is not followed in filling of suborbitals then block of the element will change in (a) K (19) (b) Sc (21) (c) V(23) (d) Ni (28) If Hund’s rule is not followed, magnetic moment of Fe2+, Mn+ and Cr all having 24 electrons will be in order (a) Fe 2 + < Mn + < Cr (b) Fe 2 + = Cr < Mn + 2+ + (c) Fe = Mn < Cr (d) Mn + = Cr < Fe 2 + Following ions will be be coloured if Aufbau rule is not followed. (a) Cu2+ (b) Fe2+ 3+ (c) Sc (d) 1,2 true The wavelength is equal to the distance travelled by the electron in one second, then (a) l = h / n (b) l = h / m (c) l = h / p (d) l = h / m Magnitude of the charge on the helium ion is. (a) 4.8 × 10–10 esu (b) 2.4 × 1010 esu (c) 9.6 × 10–10 esu (d) 1.6 × 10–10 esu A photon was absorbed by a hydrogen atom in its ground state and the electron was promoted to the fifth orbit, then the excited atom returned to its ground state, visible and other quanta were emitted, other quanta are (a) 2  →1 (b) 5  →2 (c) 3  (d) 4  →1 →1 Consider the following ions (i) Ni 2 + (ii) Co 2 + 2+ (iii) Cr (iv) Fe3+ (atomic numbers: Cr = 24, Fe = 26, Co = 27, Ni = 28)

42.

43.

44.

45.

46.

47.

48.

1.51

The correct sequence of the increasing order of the number of unpaired electrons in these ions is (a) a, b, c, d (b) d, b, c, a (c) a, c, b, d (d) c, d, b, a The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of Be3+ is (a) 1:4 (b) 1:8 (c) 1:16 (d) 16:1 Bohr model can explain spectrum of (a) the hydrogen atom only (b) the hydrogen molecule only (c) an atom or iron having one electron only (d) the sodium atom only Of the following radiation with maximum wavelength is. (a) UV (b) radio wave (c) X-ray (d) IR Zeeman effect explains splitting of lines in (a) magnetic field (b) electric field (c) both (d) none In presence of magnetic field d-suborbit is. (a) 5-fold degenerate (b) 3-fold degenerate (c) 7-fold degenerate (d) 2-fold degenerate Size of the nucleus is (a) 10–13 cm (b) 10–10 cm –1 (c) 10 mm (d) all correct The orbit and orbital angular momentum of an electron are

3h 3 h and . respectively. The number of radial 2p 2 p

and angular nodes for the orbital in which the electron is present are respectively (a) 0,2 (b) 2,0 (c) 1,2 (d) 2,2 49. Each orbital has a nodal plane. Which of the following statements about nodal planes are not true? (a) a plane on which there is zero probability that the electron will be found (b) a plane on which there is maximum probability that the electron will be found (c) both (d) none 50. The radial distribution curve of 2s sublevel consists of x nodes, x is (a) 1 (b) 3 (c) 2 (d) 0 51. Magnetic moments of V (z = 23), Cr (z = 24), Mn (z = 25) are x, y, z Hence (a) x = y = z (b) x < y < z (c) x < z < y (d) z < y < x

1.52

Structure of Atom

52. Uncertainty in position and momentum are equal. Uncertainty in velocity is. (b) h / 2p (a) h / p (c)

1 h 2m p

(d) none

53. If Aufbau principle is not used, 19 th electron in Sc (z = 21) will have (a) n = 3, l = 0 (b) n = 3, l = 1 (c) n = 3, l = 2 (d) n = 4, l = 0 54. Number of electrons that F(z=9) has in p-orbitals, is equal to (a) number of electrons in s-orbitals in Na (11e) (b) number of electrons in d-orbitals in Fe3+ (23e) (c) number of electrons in d-orbitals in Mn (25e) (d) 1,2,3 are true 55. If each orbital can hold a maximum of 3 electrons, the number of elements in 4th period of periodic table (long form) is. (a) 48 (b) 54 (c) 27 (d) 36 56. If the radius of first Bohr orbit is x, then de Broglie wavelength of electron in 3rd orbit is nearly (a) 2p x (b) 6p x (c) 9x (d) x 3 57. The series of lines present in the visible region of the hydrogen spectrum is (a) Balmer (b) Lyman (c) Paschen (d) Brackett 58. Which orbital gives an electron a greater probability of being found close to the nucleus (a) 3p (b) 3d (c) 3s (d) equal 59. Hund’s rule deals with the distribution of electrons in (a) quantum shell (b) an orbit (c) an orbital (d) degenerate orbitals 60. s-orbital is spherically symmetrical hence (a) it is directional independent (b) angular dependent (c) both correct (d) both incorrect 61. In centro-symmetrical system, the orbital angular momentum, a measure of the momentum of a particle travelling around the nucleus, is quantized. Its magnitude is. (a)

l (l + 1)

h 2p

(b)

l (l - 1)

h 2p

(c)

s ( s + 1)

h 2p

(d)

s ( s - 1)

h 2p

62. If travelling at equal speeds, the longest wavelength of the following matter is that of (a) electron (b) proton (c) neutron (d) alpha particle. 63. The line spectra are the characteristics of (a) atoms in the excited state (b) molecules in the excited state (c) atoms in ground state (d) molecules in ground state 64. In the Bohr model of the hydrogen atom, the ratio of the kinetic energy to the total energy of the electron in quantum state is. (a) 1 (b) 2 (c) –1 (d) –2 65. The number of waves made by a Bohr electron in an orbit of maximum magnetic quantum number +2 (a) 3 (b) 4 (c) 2 (d) 1 66. The distance of closest approach of an α-particle fired towards a nucleus with momentum p is r. What will be the distance of closest approach when the momentum of the α-particle is 2p? (a) 2r (b) 4r (c) r/2 (d) r/4 67. The radius of Bohr’s first orbit in H atom is 0.53 mm. the radius of second orbit in He+ would be (a) 0.0265 mm (b) 0.0530 mm (c) 0.1060 mm (d) 0.2120 mm 68. The ionization potential of hydrogen atom is 13.6 eV. The energy required to remove an electron from the n = 2 state of hydrogen atom is. (a) 27.2 eV (b) 13.6 eV (c) 6.8 eV (d) 3.4 eV 69. The value of charge on the oil droplets experimentally observed were –1.6 × 10–19 and –4 × 10–19 coulomb. The value of the electronic charge, indicated by these results is. (a) 1.6 × 10–19 (b) –2.4 × 10–19 –19 (c) –0.4 × 10 (d) –0.8 × 10–19 70. The ratio of energy of a photon of 2000 A0 wave length radiation to that of 4000 A0 radiation is (a)

1 4

(b)

1 2

(c) 2 (d) 4 71. Atomic weight of Ne is 20.2. Ne is a mixture of 20 Ne and 22Ne. Relative abundance of heavier isotope is. (a) 90 (b) 20 (c) 40 (d) 10

Structure of Atom

72. The velocity of electron in the hydrogen atom is 2.2 × 106 m / s . The de Broglie wavelength for this electron is. (a) 33nm (b) 45.6nm (c) 23.3nm (d) 0.33nm 73. What is the energy in joule of a photon of light with wavelength 4.0 × 103 nm (a) 7.5 × 10–20 (b) 5.0 × 10–20 –10 (c) 2.0 × 10 (d) 2.5 × 10–10 74. The maximum wavelength of light that can excite an electron from first to third orbit of hydrogen atom is. (a) 487nm (b) 170nm (c) 103nm (d) 17nm 75. The specific charge of a proton is 9.6 × 107 C kg–1, then for an α-particles it will be (a) 2.4 × 107 C kg–1 (b) 4.8 × 107 C kg–1 (c) 19.2 × 107 C kg–1 (d) 38.4 × 107 C kg–1 76. The work function for a metal is 4 eV. To emit a photo electron of zero velocity from the surface of the metal, the wavelength of incident light should be (a) 2700 A0 (b) 1700 A0 0 (c) 5900 A (d) 3100 A0 77. Ultraviolet light of 6.2 eV falls on aluminum surface (work function = 4.2 eV ). The kinetic energy (in joule) of the fastest electron emitted is approximately (a) 3 × 10–21 (b) 3 × 10–19 –17 (c) 3 × 10 (d) 3 × 10–15 78. The threshold wavelength for photoelectric effect on sodium is 5000 A0. Its work function is. (a) 4 × 10–19 J (b) 1 J (c) 2 × 10–19 J (d) 3 × 10–10 J 79. Photons of energy 6 eV are incident on a potassium surface of work function 2.1 eV. what is the stopping potential? (a) –6V (b) –2.1 V (c) –3.9 V (d) –8.1 V 80. In an atom two electrons move around the nucleus in circular orbits of radii R and 4R. The ratio of the time taken by them to complete one revolution is. (a) 1:4 (b) 4:1 (c) 1:8 (d) 8:7 81. In the series limit of wavelength of the Lyman series for the hydrogen atom is 912 A0, then the series limit of wavelength for the Balmer series of the hydrogen atom is. (a) 912 A0 (b) 912×2A 0 0 (c) 912×4A (d) 912/2A0

1.53

82. The difference in angular momentum associated with the electron in two successive orbits of hydrogen atom is h h (a) (b) p 2p h h (d) (n - 1) (c) 2p 2 83. Ionization potential of hydrogen atom is 13.6 eV. H y drogen atom in the ground state are excited by monochromatic light of energy 12.1 eV. The spectral lines emitted by hydrogen according to Bohr’s theory will be (a) one (b) two (c) three (d) four 84. Energy levels A,B,C of a certain atom corresponding to increasing values of energy, i.e; EA < EB < EC. If λ1, λ2 and λ3 are the wavelengths of radiations corresponding to the transitions C to B, B to A and C to A respectively, which of the following statement is correct λ

C 1

λ

B λ

2

3

A (a) λ3 = λ1 + λ2

85.

86.

87.

88.

(b) λ3 =

l1 l 2 l1 + l 2

(c) λ1+λ2+λ3 = 0 (d) l 32 = l12 + l 22 The ratio of the radius of the orbit for the electron orbiting the hydrogen nucleus to that of an electron orbiting a deuterium nucleus is. (a) 1:1 (b) 1:2 (c) 2:1 (d) 1:3 Suppose 10-17 J of light energy is needed by the interior of human eye to see an object. The photons of green light (λ = 550 nm) needed to see that object are (a) 27 (b) 28 (c) 29 (d) 30 A photon of 300 nm is absorbed by a gas and then reemits two photons. One reemitted photon has wavelength 496 nm, the wavelength of second reemitted photon is. (a) 757 (b) 857 (c) 957 (d) 657 The shortest λ for the Lyman series is ……. (given (RH = 109678 cm-1) (a) 912 A0 (b) 700 A0 0 (c) 600 A (d) 811 A0

1.54

Structure of Atom

89. The longest λ for the Lyman series is…….. (given (RH = 109678 cm-1) (a) 1215 A0 (b) 1315 A0 0 (c) 1415 A (d) 1515 A0 90. The λ for Hα line of Balmer series is 6500 A0. Thus λ for Hβ line of Balmer series is. (a) 4864 A0 (b) 4914 A0 0 (c) 5014 A (d) 4714 A0 91. The momentum of particle of wavelength 0.33 nm is…..kg m sec-1 (a) 2 × 10–24 (b) 2 × 10–12 –6 (c) 2 × 10 (d) 2 × 10–48 92. An oil drop has charge 6.39 × 10–19 C. The total number of electrons on oil drop are (a) 1 (b) 2 (c) 3 (d) 4

multiple Choice Questions with Only One Answer level ii 1. Select the correct statement. h for the electron (a) Orbital angular momentum is p having n = 5 and lower value of the azimuthal quantum number. (b) If n = 3, l = 0, m = 0 for the last valence shell electron, then the possible atomic number may be 12 or 13. (c) Total spin of electrons for the atom (Z = 25) Mn is ±

7 . 5

(d) Magnetic moment due to spin for noble gas is zero. 2. Identify the correct statement (s) (i) The radial probability distribution curves for 1s, 2s, 2p and 3d are identical. (ii) The number of nodal planes for 2p, 3d and 4p orbitals is the same. (iii) Theoretically calculated spin only magnetic moment is same for Ti2+ and Ni2+. (iv) A ‘p’ orbital can accommodate a maximum of six electrons. (a) I and II are correct. (b) II and III are correct. (c) III and IV are correct. (d) Only IV is correct. 3. The magnetic moment of two ions Mx+ and My+ of the element M(Z = 26) is found to be 5.916 BM.

If x > y, then which of the following statement is correct? (a) My+ is more stable than Mx+. (b) My+ is less stable than Mx+. (c) Both are equal stable. (d) Cannot be predicted exactly. 4. If l0 is threshold wavelength for photo electronic emission, l is the wavelength of light falling on the surface of metal, and m, mass of electron, then de Broglie wavelength 1of the emitted electron is 1 2 2  h(ll 0 )   h (l 0 - l )  (a)  2mc(l - l)  (b)   0    2mcll 0  1

1

 h (l - l 0 )  2  hll 0 )  2 (c)  (d)     2mc   2mcll 0  5. From the following observation of Mr. Gupta, Mr. Agarwal and Mr. Maheshwari, predict the type of orbital. Mr. Gupta: The angular function of the orbital intersect the three axes at the origin only. Mr. Agarwal: XY plane acts as a nodal plane. Mr. Maheshwari: The graph of radial probability vs ‘r’ intersects the radial axis at three separate regions, points including origin. (a) 5dyz (b) 4pz (c) 6dyz (d) 6dxy 6. The probability of finding the electron is maximum for d x2 - y 2 orbital (a) along x-axis only (b) along y-axis only (c) along x, y-axis only (d) between x, y axes 7. Which one of the following statement is not correct? (a) Rydberg’s constant and wave number have same units. (b) Lyman series of hydrogen spectrum occurs in the UV region. (c) The angular momentum of the electron into the ground state hydrogen atom is equal to

h . 2p

(d) The radius of first Bohr orbit of hydrogen atom is 2.116 × 10–8 cm. 8. Which one of the following statement is correct? (a) 2s orbital is spherical with two nodal planes. (b) The de Broglie wavelength (l) of a particle of mv . h (c) The principal quantum number n indicates the shape of an orbital. (d) The electronic configuration of phosphorus is given by Ne 3s 2 3 p1x p1y p1z . mass m and velocity v is equal to

Structure of Atom

9. What is true regarding the length of photographic film occupied by various series in hydrogen emission spectrum? (L1: Length of Lyman series; L2: Length of Balmer series; L3 Length of Paschen series) (a) L1 = L2 = L3 (b) L1 < L2 < L3 (c) L1 < L2 > L3 (d) L1 > L2 > L3 10. Photons having energy equivalent to binding energy of 4th state of He+ ion are used the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of 4V, then the minimum value of de Broglie’s wavelength associated with the electron is (a) 1.1 Å (b) 9.15 Å (c) 5 Å (d) 11 Å 11. If the angular momentum of an electron changes from 2h h to during a transition in a hydrogen atom, it p p

12.

13.

14.

15.

16.

17.

18.

results in the formation of spectral lines in (a) UV region (b) visible region (c) IR region (d) Far IR region Proton, alpha particles are accelerated with same potential. The de Broglie wavelength ratio is (b) 1: 2 2 (a) 2 2 :1 (c) 4 : 1 (d) 1 : 4 A stationary He+ ion emitted a photon corresponding to the first line of the Lyman series. The photon liberates a photo electron from a stationary hydrogen atom in ground state. The kinetic energy of the electron is (a) 10.2 eV (b) 13.6 eV (c) 20.4 eV (d) 27.2 eV The uncertainty in position is of the order of 1 Å. The uncertainty in velocity of a cricket ball (a) 1.7 K × 10–23 m/s (b) 1.7 × 10–24 m/s (c) 3.4 × 10–23 m/s (d) 3.4 × 10–24 m/s A standing wave in a string 35 cm long has a total of six nodes including ends. Hence, the wavelength is (a) 14 cm (b) 5.82 cm (c) 7 cm (d) 17.5 cm The m value for an electron in an atom is equal to the number of m values for l = 1. The electron may be present in (a) 3d x2 - y 2 (b) 5fxyz (c) 2px (d) 3 S An electron travels with velocity of v ms–1. For a proton to have same de Broglie wavelength, the velocity will be approximately (a) v ms–1 (b) 1840 v ms–1 –1 (c) v/1840 ms (d) 1840 v ms–1 The dye Acriflavine, when dissolved in water, has its maximum light absorption at 4530 Å and the maximum fluorescence emission at 5080 Å. The number

1.55

of fluorescence quanta is, on the average 53% of the number of quanta absorbed. Using the wavelengths of maximum absorption and emission, what percentage of absorbed energy is emitted as fluorescence? (a) 100% (b) 47% (c) 36% (d) 12% 19. A gaseous particle ‘X’ has a proton (atomic) number ‘n’ and a charge of +1. An another gaseous particle ‘Y’ has a proton (atomic number) of (n + 1) and is isoelectronic with X. Which of the following statements correctly describes X and Y? (a) X has larger radius than Y. (b) X requires more energy than Y when a further electron is removed from each particle. (c) X releases more energy than Y when an electron is added to each particle. (d) X and Y both have the same electro negativity. 20. When an electron is transited from 2E to E energy level, the wavelength of photon produced is l1. If electronic transition involves 4/3 E to E level, the wavelength of resultant photon is l2. Which of the following is a correct relation? (a) l 2 = l1

(b) l 2 = 3l1

3 4 l1 (d) l 2 = l1 4 2 21. If a certain metal was irradiated by using two different light radiations of frequency ‘x’ and ‘2x’ the kinetic energies of the ejected electrons are ‘y’ and ‘3y’ respectively. The threshold frequency of the metal will be (c) l 2 =

(a)

x 3

(b)

x 2

3x 2x (d) 2 3 22. During the excitation of an electron, it travels at a distance of nearly 0.7935 nm in the hydrogen atom. The maximum number of spectral lines formed during the deexcitation is (a) 1 (b) 3 (c) 6 (d) 10 23. The ratio between the wave numbers of spectral lines of Balmer series of hydrogen having the least and the highest wavelength values is (a) 5 : 27 (b) 27 : 5 (c) 5 : 9 (d) 9 : 5 24. The number of nodal planes, radial nodes and peaks in the radial probability curve of 5d orbital are respectively (a) 5, 2, 3 (b) 4, 3, 2 (c) 2, 3, 2 (d) 2, 2, 3 (c)

1.56

Structure of Atom

25. The number of revolutions made by electron in Bohr’s 2nd orbit of hydrogen atom in one second is (a) 6.55 × 1015 (b) 8.2 × 1014 15 (c) 1.64 × 10 (d) 2.62 × 1016 26. When an excited state of H-atom emits a photon of wavelength l and returns to the ground state, the principal quantum number of the excited state is given by (a)

(lR - 1) lR

(b)

lR (lR - 1)

(c)

lR (lR - 1)

(d)

lR (lR + 1)

27. What is the maximum number of electrons in an atom that can have the quantum numbers n = 4, ml = +1? (a) 4 (b) 15 (c) 3 (d) 6 28. The wave function (y) of 2s is given by 1/ 2

 1   r  - r / 2 a0 (y ) 2 s = . At r = r0,   2 - e a0  2 2l  a0   radial node is formed. Thus, for 2s, r0 in terms of a0 is (a) r0 = a0 (b) r0 = 2a0 (c) r0 = a0/2 (d) r0 = 4a0 29. Which of the following statements in relation to the hydrogen atom is correct? (a) 3s, 3p and 3d orbitals all have the same energy. (b) 3s and 3p orbitals are of lower energy than 3d orbital. (c) 3p orbital is lower in energy than 3d orbital. (d) 3s orbital is lower in energy than 3p orbital. 30. Choose the wrong statement. (a) The shape of an atomic orbital depends upon the azimuthal quantum number. (b) The orientation of an atomic orbital depends upon the magnetic quantum number. (c) The energy of an electron in an atomic orbital of multi-electron atom depends upon the principal number only. (d) The number of degenerate atomic orbital of one type depends upon the value of azimuthal and magnetic quantum number only. 31. The ground state electronic configurations of the elements, U, V, W, X and Y (the symbols do not have any chemical significance) are as follows: U 1s2 2s2 2p3 V 1s2 2s2 2p6 3s1 W 1s2 2s2 2p6 3s2 3p2 X 1s2 2s2 2p6 3s2 3p6 3d5 4s2 Y 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 1

Determine which sequence of elements that satisfy the following statements? (i) Element-forms a carbonate which is not decomposed by heating. (ii) Element-is most likely to form coloured ionic compounds. (iii) Element-has largest atomic radius. (iv) Element-forms only acidic oxide. (a) VWYU (b) VXYW (c) VWYX (d) VXWU 32. If the quantum numbers are defined as follows, n = 1, 2, 3...., l = 0, 1, 2,...(n – 1), m = –l in integral steps to +l, how many electrons could be fitted in Ist shell? (a) 2 (b) 8 (c) 10 (d) 18 33. If electrons were to fill up progressively with orbit saturation (neglecting Aufbau rule) and each orbital were to accommodate the three electrons, instead of two; which of the following would NOT hold correct for the new electronic arrangement in zirconium atom (Z = 40)? (a) The number of p-electrons would be double the number of s-electrons. (b) Spin quantum number would become super flows. (c) Zirconium would continue to belong to the 4th period (in new Periodic Table) (d) Zirconium would show retention of block. 34. If the above radial probability curve indicates 2s orbital, the distance between the peak points x, y is

(a) 2.07 Å (b) 1.59 Å (c) 0.53 Å (d) 2.12 Å 35. An electron is continuously accelerated in a vacuum tube by applying a potential difference. If the de Broglie’s wavelength is decrease by 10% the change in the kinetic energy of the electron is nearly (a) decreased by 11% (b) increased by 23.4% (c) increased by 10% (d) increased by 11.1%

Structure of Atom

36. The angular momentum of electron in Li2+ was found to be (7h/11). The distance of the electron from nucleus is (a) 1.69 Å (b) 0.68 Å (c) 2.82 Å (d) 8.19 Å 37. The frequency ‘v’ of certain line of the Lyman series of the atomic spectrum of hydrogen satisfies the following conditions (i) It is the sum of the frequencies of another Lyman and a Balmer line (ii) It is the sum of the frequencies of a certain line, a Balmer line and a Paschen line (iii) It is the sum of the frequencies of a Lyman and a Paschen line but no Brackett line To which transition does v correspond? (a) n2 = 3 to n1 = 1 (b) n2 = 3 to n1 = 2 (c) n2 = 2 to n1 = 1 (d) n2 = 4 to n1 = 1 38. A hydrogen atom sample in the ground state is excited by monochromatic light radiation of wavelength l Å. The resulting spectrum consists of maximum 15 different lines. What is l? (RH = 109677 cm–1) (a) 937.3 Å (b) 1025 Å (c) 1236 Å (d) None of these 39. The ratio between time periods taken by electron in Bohr’s 2nd and 3rd orbits for each revolution is (a) 9 : 4 (b) 4 : 9 (c) 8 : 27 (d) 27 : 8 40. The velocity of an electron in nth orbit of hydrogen atom bears the ratio 1 : 411 to the velocity of light. The number of coloured lines formed when electron jumps from (n + 3) state to ground state, is (a) 4 (b) 3 (c) 5 (d) 6 41. The de Broglie wavelength (l) of electron is related to its accelerating potential (V) by the equation 12.27 (a) l  = ( A) V 12.27 (b) l ( nm ) = V 150 (c) l  = ( A) V 150 (d) l ( nm ) = V 42. Hydrogen-like ion has the wavelength difference between the first lines of Balmer and Lyman series equal to 33.4 nm. The atomic number of that ion is equal to (a) 4 (b) 3 (c) 2 (d) 5

1.57

43. When a certain metal was irradiated with light of frequency 3.2 × 1016 Hz, the photo electrons emitted has twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light of frequency 2.5 × 1016 Hz. The threshold frequency of the metal is ........1015 Hz. (a) 18 (b) 20 (c) 22 (d) 16 44. What is the wavelength of the photon emitted by a hydrogen atom when an electron makes transition from n = 2 to n = 1? Given that ionization potential is 13.786 V. (Answer expressed in Angstrom units) (a) 1200 (b) 900 (c) 800 (d) 1000 45. The ratio of the energy of the electron in ground state of hydrogen to the electron in first excited state of Be3+ is (a) 1 : 4 (b) 1 : 8 (c) 1 : 16 (d) 16 : 1 46. The configuration of Cr atom is 3d5 4s1 but not 3d4 4s2 due to reason R1 and the configuration of Cu atom 3d10 4s1 but not 3d9 4s2 due to Reason R2. R1 and R2 are (a) R1: The exchange energy of 3d5 4s1 is greater than that of 3d4 4s2. R2: The exchange energy of 3d10 4s1 is greater than that of 3d9 4s2. (b) R1: 3d5 4s1 and 3d4 4s2 have same exchange energy but 3d5 4s1 is spherically symmetrical. R2: 3d10 4s1 is also spherically symmetrical. (c) R1: 3d5 4s1 has a greater exchange energy than 3d4 4s2. R2: 3d10 4s1 has a spherical symmetry. (d) R1: 3d5 4s1 has greater energy than 3d4 4s2. R1: 3d10 4s1 has a greater energy than 3d9 4s2. 47. Maximum value (n + 1 + m) for unpaired electrons in second excited state of chlorine 17Cl is (a) 28 (b) 25 (c) 20 (d) None of these 48. If r be the radius of first Bohr’s orbit for hydrogen atom, the de Broglie wavelength in the nth orbit of hydrogen atom is given by (a) 2p / n (b) 2pr ⋅ n (c) 2pr / n 2 (d) 2prn 2 49. When two electrons are present in two degenerate orbitals of an atom, the energy is lower if their spin is parallel. The statement is based upon (a) Pauli’s exclusion principle (b) Aufbau principle (c) Hund’s rule (d) Bohr’s rule 50. If the nitrogen atom had electronic configuration 1s7, it would have energy lower than that of the normal

1.58

Structure of Atom

ground state configuration 1s22s22p3, because the electrons would be closer to the nucleus. yet 1s7 is not observed because it violates. (a) Heisenberg uncertainly principle (b) Hund’s rule (c) Pauli exclusion principle (d) Bohr’s postulate of stationary orbits 51. If RH be the Rydberg constant, then the energy of an electron in the ground state of hydrogen atom is RH C hC (b) RH h 1 (c) (d) - RH hC RH Ch 52. When electronic transition occurs from higher energy state to a lower energy state with energy difference equal to DE eV, the wavelength of the line emitted is approximately equal to (a)

(a)

12400 × 10-10 13397 × 10-10 m (b) m DE DE

(c)

14322 × 10-10 12387 × 10-10 m (d) m DE DE

53. In Mosley’s equation given constants a = b = 1 then the frequency is 100s–1. The element will be (a) K (b) Na (c) Rb (d) Cs 54. Which of the pair of orbitals have electronic density along the axis? (a) dxydyz (b) d x2 - y 2 d z 2 (c) dxydxz (d) d xy d z 2 55. If azimuthal quantum number could have value of n also in addition to normal value, then E. C. of Cr (z = 24) would have been (a) 1s2 2s2 2p6 3s2 3p6 4s2 3d4 (b) 1s2 2s2 2p6 3s2 3p6 4s1 3d5 (c) 1s2 1p6 2s2 2p6 3s2 2d6 (d) 1s2 1p6 2s2 2p6 2d6 56. In radial probability distribution curve of 2s orbital, the distance between the two peaks is (a) 0.53 (b) 2.1 (c) 1.1 (d) 1.57 57. If n and l are respectively principal and azimuthal quantum numbers, then the expression for calculating the total number of electrons in any energy level is l = n -1

(a)

∑ 2(2l + 1)

l =n

(b)

l =0

l =1

l = n -1

(c)

∑ 2(2l + 1) l =1

∑ 2(2l + 1) l =n

(d)

∑ 2(2l + 1) l =0

58. Consider the following statements concerning the sub shells: (i) The maximum value of m for g sub shell is 4 (ii) The maximum value number of unpaired electrons in g sub shell is 9 (iii) The lowest value for the shell having g sub shell is 5 Select correct statements (a) ii, i only (b) ii, iii only (c) i, iii only (d) i, ii, iii 59. If a 1 g body is travelling along the X-axis at 100 cm s–1 within 1 cm s–1 than 4 uncertainty in its position is (a) 2.64 × 10–30 m (b) 5.28 × 10–30 m (c) 1.32 × 10–30 m (d) 0.66 × 10–30 m 60. The photoelectric emission requires a threshold frequency V0 for a certain metal l1 = 2000 Å and l2 = 1660 Å produce electrons with kinetic energy KE1 and KE2 of KE2 = 2KE1. The threshold frequency V0 is (a) 1.15 × 1015 s–1 (b) 1.15 × 1014 s–1 (c) 1.19 × 1015 s–1 (d) 1.19 × 1014 s–1 61. Calculate the wavelength emitted during the transition of electrons in between two levels of Li2+ ion whose sum is 4 and the difference is 2. (a) 1.14 × 10–6 cm (b) 1.14 × 10–5 cm (c) 2.12 × 10–6 cm (d) 2.14 × 10–5 cm 62. A certain dye absorbs light of l = 4700 Å and then fluorescence a light X Å. Assuming that 47% of absorbed light is reemitted as fluorescence, the ratio of quanta out to the number of quanta absorbed is 0.5. The value of X is (a) 4000 Å (b) 5000 Å (c) 5500 Å (d) 6000 Å 63. The bond dissociation energy of H—H bond is 400 KJ/mol. The energy required to excite 0.04 moles of H2 gas to the first excited state is (a) 94.6 KJ (b) 47.3 KJ (c) 70.9 KJ (d) 80.4 KJ 64. The angular momentum of an electron in a Bohr’s orbit of hydrogen atom is 4.217 × 10–34 kg-m2/s. The number of specified lines formed in visible region when electron falls from this level is (a) 4 (b) 3 (c) 2 (d) 6 65. What transition in the hydrogen spectrum would have the same wavelength as the transition n = 4, n = 2 in He+ ion? (a) 4 → 2 (b) 3 → 1 (c) 3 → 2 (d) 2 → 1 66. If the radius of first Bohr orbit is X1, then the de Broglie wavelength of electron in 3rd orbit is nearly (a) 2 π x (b) 6 π x (c) 9 x (d) x/3

Structure of Atom

67. The angular speed of the electron in nth orbit of Bohr hydrogen atom is (a) directly proportional to n (b) inversely proportional to n (c) inversely proportional to n2 (d) inversely proportional to n3 68. In hydrogen atom, the electronic motion is represented as follows. The number of revolutions made by that electron in one second is equal to

75. Imagine an atom made up of a proton and a hypothetical particle of double the mass of the electron but having the same charge as the electron. Apply the Bohr atomic model and consider possible transitions of this hypothetical particle to the first excited level. The largest wavelength of the photon that will be emitted has wavelength λ equal to (R is the Rydberg constant.) 9 36 (a) (b) 5R 5R (c)

(a) 6.5 × 1015 (b) 2 × 1014 13 (c) 2 × 10 (d) 1.01 × 1014 69. The length of the minor axis corresponding to the elliptical path for which n = 4 and k = 2. (a) 8.464 Å (b) 16.92 Å (c) 2.116 Å (d) 4.23 Å 70. In which quantum level does the electron jumps in He+ ion to ground state if it is given an energy corresponding to 99% of the ionisation potential of He+ ion. (a) 4 (b) 6 (c) 8 (d) 10 71. Which among the following is correct of 5B in normal state? (a)

: Against Hund’s rule

(b)

: Against Aufbau principle as well as Hund’s rule

(c)

: Violation of Pauli’s exclusion principle and not Hund’s rule

(d)

: Against Aufbau principle

72. How many times does light travel faster in vacuum than an electron in Bohr’s first orbit of hydrogen atom? (a) 13.7 times (b) 67 times (c) 137 times (d) 97 times 73. When a hydrogen atom emits a photon of energy 12.1 eV. The orbit angular momentum changes by (a) 1.05 × 10–34 J-s (b) 2.11 × 10–34 J-s (c) 3.16 × 10–34 J-s (d) 4.22 × 10–34 J-s 74. If kinetic energy of a particle is doubled; de Broglie wavelength becomes (a) 2 times (b) 4 times (c) 2 times (d) 1 / 2 times

1.59

18 5R

(d)

27 5R

76. When a certain metal was irradiated with light having a frequency of 3.0 × 1016 s–1, the photo electrons emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light having a frequency of 2.0 × 1016 s–1. The threshold frequency of the metal (in 1016 s–1) is (a) 1 (b) 2 (c) 3 (d) 4 77. Given that ionisation potential of hydrogen atom is 2.0 × 10–18 J, Planck’s constant is 6.0 × 10–34 J-s. The frequency Hb line in Balmer series is in the values of (1012 Hz) is (a) 625 (b) 840 (c) 925 (d) 520 78. The radius of the orbit in hydrogen atom is 0.8464 nm. The velocity of electron in this orbit (in the order of 103 m/s), is (a) 547 (b) 637 (c) 342 (d) 832 79. The calculated magnetic moment of Cr2+ is (a) 1.73 BM (b) 2.8 BM (c) 4.8 BM (d) 5.9 BM 80. The number of d electrons in Fe2+ is not equal to that of (a) p-electrons in Ne (b) s-elelctrons in Mg (c) p-electrons in Cl– (d) p-electrons in Mg 81. de Broglie wavelength of the Bohr’s first orbit is l. The circumference of the Bohr fourth orbit is (a) 4 λ (b) 8 λ (c) 16 λ (d) 2 λ 82. What hydrogen like ion has the wavelength difference between the first line of Balmer and Lyman series equal to 133 nm: (a) He+ (b) Li2+ 3+ (c) Be (d) B4+

1.60

Structure of Atom

83. The shortest wavelength of hydrogen atom in Lyman series is X. The longest wavelength in Balmer series of He+ is? 9x 36 x (b) (a) 5 5 (c)

x 4

(d)

will be (n1 is not necessarily ground state) [Assume for this atom, no spectral series shows overlaps with other series in the emission spectrum]

5x 9

84. Pick out the correct option. Where T stands for True and F stands for False. (i) The energy of the 3d orbital is high compared to the 4s orbital in hydrogen atom. (ii) The electron density in xy plane in d x2 - y 2 orbital is zero. (iii) 24th electron in Cr goes to 3d orbital. (iv) The three quantum numbers were clearly explained in terms of Schrodinger wave equation. (a) TTFF (b) TFTT (c) TFTF (d) FFTT 85. Who modified Bohr's theory by introducing elliptical orbits for electron path? (a) Hund (b) Thomson (c) Rutherford (d) Sommerfeld 86. The velocity of electron in a certain Bohr orbit of hydrogen atom bears the ratio 1 : 275 to the velocity of light. So the number of orbits is (a) 1 (b) 2 (c) 3 (d) 4 87. If there were three possibilities of electron spin, K (19) would be placed in (a) s-block (b) p-block (c) d-block (d) f-block 88. Consider the following plots for 2s-orbitals

x, y and z are respectively, (a) y, y2 and 4pr2 y2 (b) y2, y and 4pr2 y2 (c) 4pr2 y2 and y2 and y (d) y2, 4pr2 y2 and y 89. For a hypothetical H-like atom which follows Bohr’s model, some spectral lines were observed as shown. If it is known that line ‘E’ belongs to the visible region, then the lines possibly belonging to ultraviolet region

90.

91.

92.

93.

94.

(a) B and D (b) D only (c) C only (d) A only In any sub shell, the maximum number of electrons having same value of spin quantum number is (a) l (l + 1) (b) l +2 (c) 2l +1 (d) 4l +2 Given the wave function of an orbital of hydrogen  1   e - r / a0 , the most probable atom; y1.0 =  3/ 2  pa  0   distance of the electron present in the given orbital from the nucleus is (a) a0 × e–2r (b) a0 (c) (3/2)a0 (d) 2a0 Given the correct order of initials T (True) or F (False) for following statements: (i) Maximum kinetic energy a photoelectron emitted from a metal [work function = 2 eV] by a photon of wavelength 310 nm is 3.2 × 10–19 J. (ii) The ratio of magnitude of total energy; kinetic energy: Potential energy for any orbit is 1 : 1 : 2. (iii) Number of maxima for the curve 4pr2 R2 (r) vs are three for the 5px orbital. (iv) The ratio of de Broglie wavelength of a ‘H’ atom, ‘He’ atom and CH4 molecule moving with equal kinetic energies 4 : 2 : 1. (a) FFTT (b) TTFT (c) TFTF (d) FTTT The difference of nth and (n + 1)th Bohr’s radius of Hatoms is equal to (n – 1)th Bohr’s radius. Hence, the value of ‘n’ is (a) 1 (b) 2 (c) 3 (d) 4 Which orbital is represented by complete wave function y420? (a) 4d z 2 (b) 3d x2 - y 2 (c) 4px (d) 4dyz

Structure of Atom

1.61

95. The wave function of the atomic orbital of an H-like species is given as y2s =

1 4 2p

Z 1/ 2 (2 - Zr )e - Zr / 2

The radius for nodal surface for He+ ion in Å is (a) 1.5 Å (b) 1 Å (c) 2 Å (d) 4 Å 96. For the photoelectric effect, the maximum kinetic energy EK of the emitted photo electrons is plotted against the frequency v of the incident photons as shown in figure. The slope of the curve gives

(a) charge of the electron (b) work function of the metal (c) Planck’s constant (d) ratio of the Planck’s constant to electronic charge 97. de Broglie wavelength of two particles A and B are  1  plotted against   ; where V is the potential on the  V  particles. Which of the following relation is correct about the mass or the particles?

(a) mA = mB (b) mA > mB (c) mA < mB (d) mA ≤ mB 98. Which of the following curves represent the speed of the electron of hydrogen atom as a function of principal quantum number ‘n’?

(a) 1 (b) 2 (c) 3 (d) 4 99. In the graph between v and Z for the Mosley’s equation, v = a (Z – b), the intercept OX is 1 on v axis.

What is the frequency v when the atomic number Z is 52? (a) 7.14 s–1 (b) 7 s–1 –1 (c) 2401 s (d) 2601 s–1

multiple Choice Questions with One or more Than One Answer 1. Which of the following can be concluded correctly from the solution of Schrodinger equation? (a) For all orbitals, the radial wave-function (y) approaches to value zero as r (distance from nucleus) (b) The radial probability density (y)2 for 1s orbital is maximum at nucleus. (c) The radial distribution function (4pr2, y2) for 1s orbital is minimum at nucleus. (d) In plot of radial wave function (y), y changes its sign at the nodes. 2. When photons of energy 4.25 eV strike the surface of a metal A, the ejected photo electrons have maximum kinetic energy TA (expressed in eV) and de Broglie wavelength lA. The maximum kinetic energy of the photo electrons liberated from another metal ‘B’ by photons of energy 4.20 eV is TB = TA – 1.5 (expressed in eV). If the de Broglie wavelength of these photo electrons is lB = 2lA, then (a) work function of A is 2.25 eV. (b) work function of B is 4.20 eV. (c) TA = 2 eV. (d) TB = 2.50 eV.

1.62

Structure of Atom

3. In a photoelectric experiment, the stopping potential is plotted against

1 of incident radiation for two l

different metals, the curve is like as shown in figure. Predict which of the following statements are correct?

(a) Slope of the curves for both the metals is 1.212 × 10–6 Jm C–1 (b) When an electromagnetic radiation of wavelength 100 nm strikes the two metals separately, the stopping potential for metal I is 12.18 V. (c) The stopping potential for metal I is more than that of metal II, if both the metals are exposed to electromagnetic radiations of wavelength 200 nm, separately. (d) The stopping potential for metal II is more than that of metal I, if both the metals are exposed to electromagnetic radiation of wavelength 200 nm, separately. 4. Which of the following are the correct statements? (a) Light waves were considered electromagnetic in nature. (b) In vacuum, all types of electromagnetic radiation regardless of their wavelength, travel at the same speed, i.e., 3 × 108 ms–1. (c) The ideal body which emits and absorbs all frequencies, is called a black body. (d) Quantum is also called photon. 5. Which of the following pairs have identical values of magnetic moment? (a) Zn2+, Cu+ (b) Co2+, Ni2+ 4+ 2+ (c) Mn , Co (d) Mg2+, Sc+ 6. Which of the following statements are (is) correct? (a) Electrons in motion behaves as if they were waves. (b) s-orbital is non directional. (c) An orbital can accommodate a maximum of two electrons with parallel spins. (d) The energies of the various sub shells in the same shell are in the order s > p > d > f.

7. The nucleus of an atom is located at x = y = z = 0. If the probability of finding an ‘s’ orbital electron in a tiny volume around x = a, y = z = 0 is 1.0 × 10–5, choose the correct statement (s) regarding the probability of e– (a) The probability of finding the electron in the same-sized volume around x = z = 0, y = a is the same i.e., 1× 10–5. (b) The probability of finding the electron in the same-sized volume around x = y = 0, z = a is zero. (c) The probability at the second site if the electron were in a pz orbital is zero. (d) The probability at the second site if the electron were in a pz orbital is the same i.e., 1.0 × 10–5. 8. Choose the correct statement (s) regarding BohrSommerfeld’s model. (a) All paths around the nucleus are elliptical. (b) When an electron revolves around the nucleus following circular path, only the angle of rotation is changed. (c) When an electron revolves around the nucleus following elliptical path, both the angle of rotation and the distance from the nucleus are changed. (d) For an elliptical path, k 0 mean that the relatively greater electrostatic repulsion between two electrons in the same orbital within a sub shell as compared with occupancy of two orbitals having different ml values. (b) Electrons in singly occupied orbitals tend to have their spins in the same direction so as to maximize the net magnetic moment. (c) In the ground state, an atom adopts a configuration with the greatest number of unpaired electrons. (d) Electrons in different orbitals with parallel spins show mutual attraction, so the electron-nucleus interaction is improved. Which of the following statement is/are correct? (a) For all values of ‘n’ the p-orbitals have the same shape, but the overall size increases as ‘n’ increases, for a given atom. (b) The fact that there is a particular direction, along which each p-orbital has maximum electron-

16.

17.

18.

19.

1.63

density, plays an important role in determining molecular geometries. (c) The charge cloud of a single electron in 2px atomic orbital consists of two lobes of electrondensity. (d) The very fact that an e– is present in both the lobes at equal times, and the fact that in order to move from one lobe to another, it must cross the node, implies, there is at least some probability of finding an e– in the node. In a hydrogen-like sample electron is in 2nd excited state, the binding energy of 4th state of this sample is 13.6 eV, then (a) A 24.17 eV photon can set free the electron from the second excited state of this sample. (b) 3 different types of photons will be observed if electrons make transition up to ground state from the second excited state. (c) If 23 eV photon is used, then KE of the ejected electron is 1 eV. (d) 2nd line of Balmer series of this sample has same energy value as 1st excitation energy of H-atoms. Identify the correct statement (s) among the following. (a) If Aufbau rule is not followed, then chromium will still remain in d-block. (b) If there are three possible values for spin quantum numbers, then chromium would have been lying in s-block (c) If Hund’s rule of maximum multiplicity is not followed, then the magnetic moment of chromium would be zero. (d) If Hund’s rule of maximum multiplicity is not followed, then Cr3+ (aq) ion would be colorless. The energy of an electron in the first Bohr orbit of H-atom –13.6 eV; then which of the following statement (s)/are correct for He+? (a) The energy of electron in second Bohr orbit is –13.6 eV. (b) The KE of electron in the first orbit is 54.46 eV (c) The KE of electron in second orbit is 13.6 eV. (d) The speed of electron in the second orbit is 2.19 × 106 m/s. If the electron of the hydrogen atom is replaced by another particle of same charge but of the double mass, then (a) radii of different shells will increase. (b) energy gap between two levels will become double. (c) ionization energy of the atom will be double. (d) speed of new particle in a shell will be lesser than the speed of electron in the same shell.

1.64

Structure of Atom

20. Which of the following is/are correct about the radial probability curves? (a) 3d z2 has three angular nodes. (b) The number of angular nodes are l. (c) The number of radial nodes is equal to n – l –1. (d) The number of maximum peaks in 2s-orbitals are two. 21. Select the correct statements about the wave function y. (a) y must be real. (b) y must be single values, continuous. (c) y has no physical significance. (d) y2 gives the probability of finding the electrons.

Comprehensive Type Questions Passage i Imagine a universe in which the four quantum numbers can have the same possible values as in our universe except the angular momentum quantum number ‘l’ can have integral values of 0, 1, 2,..., n instead of 0, 1, 2,.....(n – 1) and magnetic quantum number ‘m’ can have integral values of – (l + 1) to 0 to (l + 1) instead of –l to 0 to +l. 1. Electronic configuration of the element with atomic number 25, based on the above assumption is (a) 1s2 1p6 2s2 1d10 2p5 (b) 1s2 1p10 2s2 1d11 (c) 1s3 1p9 2s3 2p9 3s1 (d) 1s6 1p10 2s6 2p3 2. Based on above assumption magnetic moment of the element with atomic number 18, is (a) 24 BM (b) 8 BM (c) Zero BM (d) 15 BM 3. Based on assumption after completion of 3s orbital electron enters into .... orbital (a) 3p (b) 2d (c) 4s (d) 1d 1 1 4. If spin quantum numbers are – , 0, + based on the 2 2 above assumption electronic configuration of the element with atomic number 30 is (a) 1s3 2s3 2p9 3s3 3p9 4s3 (b) 1s3 1p15 2s3 1d9 (c) 1s9 1p15 2s6 (d) 1s3 1p9 2s3 2p9 1d6 Passage ii Imagine an atom made up of a proton and a hypothetical particle of half the mass of the electron but having the same charge as the electron. Bohr’s model is applicable and all

transition of this hypothetical particle from higher level to ground state is possible. 1. The largest wavelength of photon that will be emitted when hypothetical particle jumps to first excited level is 18 72 (a) (b) 5R 5R (c)

36 5R

(d)

8 R

2. The velocity of this particle in second Bohr orbit is (a) 2.188 × 106 m/s (b) 1.094 × 106 m/s (c) 0.547 × 106 m/s (d) 4.376 × 106 m/s 3. The radius of third Bohr orbit is (a) 9.522 Å (b) 4.761 Å (c) 2.380 Å (d) 1.587 Å 4. The kinetic energy of the hypothetical particle in first orbit is (a) –6.8 eV (b) 6.8 eV (c) 27.2 eV (d) –27.2 eV Passage iii A single electron atom has nuclear charge +Ze. Here, Z is the atomic number and e is charge. It requires 47.2 eV to excite the electron from second Bohr orbit to third Bohr orbit. 1. Atomic number of the element is (a) 2 (b) 3 (c) 4 (d) 5 2. The wavelength required to remove electron from the first Bohr orbit in above data, is (a) 0.22 × 10–7 cm (b) 1.0 × 10–7 cm (c) 5.6 × 10–7 cm (d) 3.7 × 10–7 cm 3. The kinetic energy of the electron in first Bohr orbit in above data, is (a) 1.962 × 10–10 ergs (b) 3.48 × 10–10 ergs (c) 0.872 × 10–10 ergs (d) 5.45 × 10–10 ergs Passage iv A gas of identical H-like atom has some atoms in the lowest energy level A and some atoms in a particular upper (excited) energy level B and there are no atoms in any other energy level. The atoms of the gas make transition to a higher energy level by absorbing monochromatic light of

Structure of Atom

photon of energy 2.55 eV. Subsequently, the atoms emit radiation of only six different photon energies. Some of the emitted photons, have energy 2.55 eV. Some have more and some have less than 2.55 eV. 1. The initially excited level B is (a) 2nd orbit (b) 3rd orbit th (c) 4 orbit (d) 5th orbit 2. The ionization energy of the gaseous atom is (a) 13.6 eV (b) 14.4 eV (c) 27.2 eV (d) 54.4 eV 3. The minimum energy of the emitted photon is (a) 2.7 eV (b) 2.0 eV (c) 0.66 eV (d) 0.54 eV Passage v In the photoelectric effect, an incoming photon brings a definite quantity of energy, hv. It collides with an electron close to the surface of the metal target, and transfers its energy to it. The difference between the work function, f, and the energy hv appears as the kinetic energy of the ejected electron. The minimum potential at which the plate photoelectric potential. If V0 is the stopping potential, then eV0 = hv. – hv0 current becomes zero is called stoping.

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Passage vi According to Planck’s quantum theory of radiation, the radiant energy is emitted or absorbed discontinuously in the form of quanta. The energy of a quantum is given by E = hv, where ‘v’ is the frequency of radiation. Moreover, a body can emit or absorb energy only in an integral multiples of quanta. 1. Planck’s quantum theory of radiation favors the following nature of electromagnetic radiation. (a) Particle (b) Wave nature (c) Dual nature (d) None of these 2. The energy associated with one photon of wavelength 3000 Å is (a) 5.42 × 10–19 J (b) 6.63 × 10–19 J –19 (c) 3.73 × 10 J (d) 3.73 × 10–19 J 3. Which of the following is not the correct representation of Einstein’s photoelectric equation? 1 2 (a) mv = h ( v + v0 ) 2 (b) hv = hv0 + (c) hv = W + (d)

1 2 mv 2

1 2 mv 2

1 2 mv = h (v - v0 ) 2

Passage vii

1. Light of wavelength 4000 Å is incident on a metal whose work function is 2 eV. Then the maximum possible kinetic energy of the photoelectron is (a) 3.09 eV (b) 1.9 eV (c) 1.09 eV (d) None of these 2. The threshold frequency for emitting photoelectrons from a metal surface is 5 × 1014 s–1. Which of the following could be the frequency of radiation to produce photoelectrons having twice the kinetic energy of those produced by the radiation of frequency 1015 s–1? (a) 2.15 × 1014 s–1 (b) 3.70 × 1013 s–1 (c) 4.11 × 1014 s–1 (d) 15 × 1014 s–1 3. If one photon has 25 eV energy and the work function of the material is 7 eV, then value of the stopping potential will be (a) 32 V (b) 18 V (c) 3.3 V (d) Zero

The properties of electrons indicate that they have a dual nature, i.e., they behave both as particle and as wave. de Broglie suggested that every object which possesses mass and velocity behaves both as a particle and as a wave. According to de Broglie, the wavelength l of a particle of mass ‘m’, moving with a velocity ‘v’ is given by l = h/mv. The dual nature of electron further gets support by Heisenberg’s uncertainty principle which states that it is not possible to determine simultaneously and accurately the position and momentum of a moving electron or some other microscopic particle. h , where Dx and According to Heisenberg DxDp ≥ 4p Dp respectively represent the uncertainties involved in the simultaneous determination of position and momentum of a moving particle. 1. The de Broglie wavelength of a cricket ball of mass 0.2 kg moving with a velocity of 30 ms–1 would be (a) 1.8 × 10–34 m (b) 3.6 × 10–34 m (c) 2.2 × 10–34 m (d) 1.1 × 10–34 m

1.66

Structure of Atom

2. An electron can be located with an error equal to 10–10 m. What is the uncertainty in its momentum? (a) 5.28 × 10–25 kg ms–1 (b) 5.5 × 10–26 kg ms–1 (c) 3.34 × 10–14 kg ms–1 (d) 6.63 × 10–27 kg ms–1 3. If the uncertainty in the position and momentum for an electron are equal, then the uncertainty in its velocity is h h (a) (b) 2p p (c)

1 m

h 2p

(d)

1 2m

h p

Passage viii In 1908, W. Ritz introduced his combination principle, which is virtually a generalization of Balmer formula. In its simplest form the principle states that the wave number of any spectral line may be represented as the combination of two terms, one of which is constant and the other variable throughout each spectral series. v=

R R ; x and y integers for H-atom. The later on x2 y 2

excited state can make a transition to the second excited state by emitting two photons of energy 4.25 eV and 5.95 eV respectively. 1. The value of n is (a) 8 (b) 6 (c) 5 (d) 4 2. The atomic number Z is (a) 1 (b) 2 (c) 3 (d) 4 Passage X A formula analogous to the Rydberg formula applies to the series of spectral lines which arise from transitions from higher energy levels to the lower energy level of the hydrogen atom. A muonic hydrogen ion is like a hydrogen atom in which the electron is replaced by a heavier particle the muon. The mass of the muon is 207 times the mass of an electron while charge is the double that the electron (assume charge of proton is same). 1. Radius of first Bohr orbit of muonic hydrogen ion is (a)

0.529 Å 207

(b)

(c)

0.529 Å 828

(d) 0.529 × 828 Å

theories have modified the above principle as for H-atom.  1 1  v = RH  2 -− 2   n1 n 2  RH is Rydberg number = 109,677.76 cm–1 and n1 and n2 are lower and higher energy levels of the H-atom. 1. What transition H-spectrum would have the same l as the Balmer transition n = 4 to n = 2 of He+ spectrum? (a) 2 → 1 (b) 3 → 1 (c) 4 → 1 (d) 5 → 1 2. The wavelength of radiation of the longest transition in Lyman series of He+ ion is (a) 22.8 nm (b) 32.2 nm (c) 40 nm (d) 48 nm 3. If the shortest l of H-atom in Balmer series is l1, the longest l in Lyman series is (a) 0.14 l1 (b) 0.33 l1 (c) 0.44 l1 (d) 0.66 l1 Passage iX A hydrogen-like atom (atomic number Z) is in a higher excited state of quantum number ‘H’. This excited atom can make a transition to the first excited state by successively emitting two photons of energies 10.2 eV and 17.00 eV respectively. Alternatively, the atom from the same

0.529 Å 414

2. Ionization energy of muonic hydrogen ion is (a) 13.6 × 207 eV (b) 13.6 × 414 eV (c) 13.6 × 828 eV (d) 13.6 × 3312 eV 3. Distance between 1st and 3rd Bohr orbit of muonic hydrogen ion will be 0.529 0.529 (a) ×8Å ×8Å (b) 207 414 0.529 0.529 (c) ×5Å (d) ×8Å 207 828

matching Type Questions 1. Match the Column I with Column II. Column I (Chemical Prop)

Column II (Metals)

(a) Distance of maximum probability of 1s electron (b) Radial node of 2s electron is

(p) 1.1 Å (q) 1.06 Å

(c) Radius of 2nd orbit in He+ ion is

(r) 1.59 Å

(d) Radius of 3rd orbit in Li2+ ion is

(s) 0.53 Å

Structure of Atom

2. Match the following:

7. Match the following.

Emission spectrum Absorption spectrum Band spectrum Line spectrum

Column II (p) (q) (r) (s)

Cold nitrogen gas Excited atomic hydrogen Very hot carbon dioxide Neon gas at room temperature

3. Match the following: Column I (a) Magnetic moment of an

Column I (a) Violation of Aufbaus

rule (b) Violation of Pauli’s Exclusion Principle (c) Violation of Hund’s rule (d) Violation of above three principles

(p) Principle QN (q) Azimuthal QN (r) Magnetic QN (s) Spin QN

tum of electron

Column I

(b) Nitrogen atom (c) Li2+ ion (d) Helium atom

(q) ↑↑

↑ ↑ ↓

(r)

↑

↑↑ ↑↑ ↑

(s)

↑

↑↓ ↑

8. Match the following: Column I (a) (b) (c) (d)

Column II (p) Principal quantum

number (q) Azimuthal quantum number (r) Exchange energy (s) Symmetry

Column II Stark effect Schrodinger’s equation Bohr’s atomic model Pauli’s exclusivity (t) Aufbau sequencing

n l m s

(p) (q) (r) (s)

4. Match the following:

(a) Hydrogen atom

(p) ↑↓ ↑↑ ↑

Column II

atom or ion (b) Radial node (c) Hund’s rule (d) Orbital angular momen-

Column Ii

↑

Column I (a) (b) (c) (d)

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9. Match the following: Column I

Column II

(a) 2s (b) 2p

(p) Sum of (n + l) is 3 (q) Total number of radial

(c) 3s

(r) no. of peaks in radial

(d) 5f

(s) Spherical symmetry

nodes = 1 probability graph = 2

5. Match the following: Column I

of orbital around nucleus

Column II

(a) Magnetic moment

(p) Principal quantum

(b) Splitting of spectral

(q) Azimuthal quantum

number lines in magnetic field

number

(c) Number of spherical

(r) Magnetic quantum

nodes (d) Fine lines spectra

number (s) Spin quantum number

6. n → orbit no; Z → at no; rn, z → radius Vn, z → velocity; Tn, z → Time period of revolution Kn, z → kinetic energy of the electron. Column I (of single electron species) (a) (b) (c) (d)

r2,1 : r1,2 V1,3 : V3,1 T1,2 : T2,1 K1,2 : K2,1

Column II (Ratio) (p) (q) (r) (s)

9:1

8:1 16 : 1 1 : 32

10. Match the following Column I with Column II. Column I (a) For transition: n = 5 to

n=2 (b) For transition: n = 5 to n=3 (c) For transition: n = 4 to n=2 (d) For transition: n = 4 to n=1

Column II (p) Only two Balmer lines

are obtained (q) Total three spectral

lines are obtained (r) At least one Lyman line

is obtained (s) Hb for Balmer series

and Ha for Paschen series both are obtained

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Structure of Atom

Assertion (A) and Reason (R) Type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): The wave functions of p-type (l =1) with the same value of ‘n’ but with different values of ‘m’ will have the same radial distribution function. Reason (R): Radial distribution function depends only on the values ‘n’ and ‘l’ and not on the value of ‘m’. 2. Assertion (A): According to Bohr IE2 of He is 54.4 eV/atom. Z eff2 Reason (R): IE of atom is given by 13.6 2 eV/atom n 3. Assertion (A): An electron can never be found in the nucleus. Reason (R): Velocity of the electron wave is less than the velocity of light. 4. Assertion (A): No two electrons of an atom can have the same set of all four quantum numbers. Reason (R): An orbital can accommodate a maximum of two electrons. 5. Assertion (A): 3dxy, 4dxy, 5dxy.......... all are having the same shape i.e., the double dumb bell. Reason (R): The size of orbitals are independent of the principal quantum number. eratio of anode rays is different for 6. Assertion (A): m gasses.

7.

8.

9.

10.

Reason (R): Proton is the fundamental particle present in gases. Assertion (A): y2 measures the electron probability density at a point in an atom. Reason (R): y and y2 vary as a function of the three coordinates ‘r’ (radial part), θ and f (angular part). Assertion (A): Two orbitals with same value of l but different values of ml will have same number of angular nodes. Reason (R): Angular wave function of an orbital depends upon its l and ml values. Assertion (A): Aufbau principle is violated in writing electronic configuration of Pd. Reason (R): Pd is diamagnetic in nature. Assertion (A): Spectral lines would not be seen for a 2px – 2py transition. Reason (R): p-orbitals are degenerate orbitals.

11. Assertion (A): Energy of orbitals in hydrogen atom is 1s < 2s = 2p < 3s = 3p = 3d < 4s = 4p = 4d = 4f Reason (R): The energy of orbitals in hydrogen atom is determined by using principal quantum number. 12. Assertion (A): Angular momentum of an electron in an orbit is quantized. Reason (R): The electron wave propagating around the nucleus remains in the same phase, 2pr = nl. 13. Assertion (A): Fine lines are observed in spectra if an atom is placed in a strong magnetic field. Reason (R): Degenerate orbitals split in the presence of a magnetic field. 14. Assertion (A): Spin quantum number can have the value +1/2 or –1/2. Reason (R): Sign here signifies the wave function. 15. Assertion (A): The transition of electrons n3 → n2 in the atom will emit greater energy than n4 → n3. Reason (R): n3 and n2 are closer to nucleus than n4.

integer Type Questions 1. In which quantum level does the electron jumps in He+ ion to ground state if it is given an energy corresponding to 99% of the ionization potential of He+ ion? 2. When a certain metal was irradiated with light having a frequency of 3.0 × 1016 s–1, the photo electron emitted had twice the kinetic energy as did photoelectrons emitted when the same metal was irradiated with light having a frequency of 2.0 × 1016 s–1. The threshold frequency of the metal (in 1016 s–1) is ______________. 3. Given ionization potential of hydrogen atom is 2.0 × 10–18 J, Planck’s constant is 6.0 × 10–34 J-s. The frequency of Hb line in Balmer series has the values of (1012 Hz) is ______________. 4. The radius of the orbit in hydrogen atom is 0.8464 nm. The velocity of electron in this orbit (in the order of 103 m/s) is ______________. 5. The velocity of electron in a certain Bohr’s orbit of hydrogen atom bears in the ratio 1 : 275 to the velocity of light. The qunatum number (n) of the orbit is ___________. 6. The wavelength of a line in Balmer series of hydrogen atom is 4814 Å. In which orbit deexitation take place? 7. The de Broglie wave length of an electron in a certain orbit of hydrogen atom is 13.3 Å. So the number of waves present in an orbit is ______________. 8. Give the number of angular nodes of orbital belong to the subshell for which minimum value of magnetic quantum number is –2.

Structure of Atom

9. What will be the number of lines corresponding to Lyman series when an electron is in 3rd excited state of hydrogen atom? 10. In a single hydrogen atom, the electron is excited to its 6th orbit. The maximum no. of distinct lines possible, when it comes to the ground state is ______________. 11. For the hydrogen atom, En =

1 RH . n2

Where, RH = 2.178 × 10–18 J. Assuming that the electrons in other shells exert no effect, find the minimum no. of Z (atomic no.) at which a transition from the second energy level to the first results in the emission of an X-ray. Given that the lowest energy rays have l = 4 × 10–8 m. 12. Find the quantum number ‘n’ corresponding to the excited state of He+ ion if on transition to the ground state that ion emits two photons in succession with wavelengths 1078.5 and 30.4 nm. 13. Energy required to stop the ejection of electrons from Cu-plate 0.27 eV. Calculate the work function (in eV) when radiation of l = 235 nm strikes down the plate. 14. What is the degeneracy of the level of the hydrogen atom that has the energy

- RH ? 9

15. The number of visible lines when an electron returns from the 5th orbit to ground state in the hydrogen spectrum is ______________. 16. A photon of wavelength 5000 Å strikes a metal surface, the work function of the metal being 0.475 eV. The kinetic energy of the emitted photo electron is ______________ eV. 17. The de Broglie wave length of on electron in a certain Bohr’s orbit H-atom is 6.64 AO.The quantum number of orbit is. 18. In a nonconventional basis, there are three allowed value of spin quantum numbers, then how many more elements can be accommodated in the second period as compared to convertional periodic table 19. The number of revolutions made by electron in 1 sec in H –atom in its orbit is twice of the number of revolutions made by electron in 1 sec in the 2nd orbit of He+ ion, then on is 20. An electron in a hydrogen atom in its ground state has 1-5 times as much energy as the minimum required for its escape from the atom. The velocity of electron in the scientific notation is x × 10 y m / sec . Then the value of Y is ______________. 21. In a collection of H-atoms, all the electrons tend to flow to n = 5 to ground level finally (directly or indirectly) without emitting any line in Balmer series the possible different radiations are.

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Previous years’ iiT Questions 1. The electrons, identified by quantum numbers n and l, (1999) (i) n = 4, l = 1, (ii) n = 4, l = 0, (iii) n = 3, l = 2 and (iv) n = 3, l = 1 can be placed in order of increasing energy, from the lowest to highest as: (a) (iv) < (ii) < (iii) < (i) (b) (ii) < (iv) < (i) < (iii) (c) (i) < (iii) < (ii) < (iv) (d) (iii) < (i) < (iv) < (ii) 2. The number of nodal planes in a px orbital is: (2000S) (a) one (b) two (c) three (d) zero 3. The electronic configuration of an element is ls2, 2s22p6, 3s23p63d5, 4s1. This represents its: (2000S) (a) excited state (b) ground state (c) cationic form (d) anionic form 4. The wavelength associated with a golf ball weighing 200 g and moving at a speed of 5 m/h is of the order: (2001S) (a) 10–10 m (b) 10–20 m (c) 10–30 m (d) 10–40 m 5. The quantum numbers +1/2 and –1/2 for the electron spin represent: (2001S) (a) rotation of the electron in clockwise and anticlockwise direction respectively (b) rotation of the electron in anticlockwise and clockwise direction respectively (c) magnetic moment of the electron pointing up and down respectively (d) two quantum mechanical spin states which have no classical analogue 6. Rutherford’s experiment, which established the nuclear model of the atom, used a beam of: (2002S) (a) β-particles, which impinged on a metal foil and got absorbed. (b) γ-rays, which impinged on a metal foil and ejected electrons. (c) helium atoms, which impinged on a metal foil and got scattered. (d) helium nuclei, which impinged on a metal foil and got scattered. 7. If the nitrogen atom has electronic configuration ls7, it would have energy lower than that of the normal ground state configuration ls22s22p3, because the

1.70

Structure of Atom

electrons would be closer to the nucleus. Yet ls7 is not observed because it violates: (2002S) (a) Heisenberg uncertainty principle (b) Hund’s rule (c) Pauli exclusion principle (d) Bohr postulate of stationary orbits 8. The radius of which of the following orbit is same as that of first Bohr’s orbit of hydrogen atom? (2004S) (a) He+ (n=2) (b) Li2+ (n=3) (c) Li2+ (n=2) (d) Be3+ (n=2) 9. The number of radial nodes of 3s and 2p – orbitals are respectively: (2005S) (a) 2,0 (b) 0.2 (c) 1,2 (d) 2,1 10. The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is [a0 is Bohr radius] (2012) h2 h2 (a) (b) 16p 2 ma02 4p 2 ma02 (c)

h2 32p 2 ma02

h2 64p 2 ma02

(d)

11. Match the following: (2006) Column I

Column II

(a) Vn/Kn = ? (b) If radius of nth orbital ∝ En

(p) 0 (q) –1

x; x = ? (c) Angular momentum in

(r) –2

lowest orbital (d)

1 ∝ zy; y = ? rn

(s) 1

12. Match the entries in Column-I with the correctly related quantum number(s) in column-II (2008) Column I (a) Orbital angular momen-

tum of the electron in a hydrogen like atomic orbital.

Column II (p) Principle quantum number

(b) A hydrogen-like one

(q) Azimuthal quantum electron wave function number obeying Pauli principle. (c) Shape, size and orienta- (r) Magnetic quantum tion of hydrogen like number atomic orbitals. (d) Probability density of (s) Electron spin quantum electron at the nucleus in number hydrogen-like atom.

Comprehensive Type Questions The hydrogen-like species Li2+ is in spherically symmetrical state S1, with one radial node. Upon absorbing light the ion undergoes transition to a state S2. The state S2 has one radial node and its energy is equal to the ground state energy of the hydrogen atom. (2010) Answer the following questions: 13. The state S1 is: (a) 1s (b) 2s (c) 2p (d) 3s 14. Energy of the state S1 in units of the hydrogen atom ground state energy is: (a) 0.75 (b) 1.50 (c) 2.25 (d) 4.50 15. The orbital angular momentum quantum number of the state S2 is: (a) 0 (b) 1 (c) 2 (d) 3

integer Type Questions Answer to the following question in a single digit integer, ranging from 0 to 9. 16. The maximum number of electrons that can have principle quantum number, n =3 and spin quantum numbers ms = –

1 is ______. 2

(2011) 17. The work function of some metals is listed below. The number of metals which will show photoelectric effect when light of 300 nm wavelength falls on the metal is ______. (2011) Metal Li eV

Na

K

Mg Cu

Ag

Fe

Pt

2.4 2.3

2.2

3.7

4.3

4.7

6.3 4.75

4.8

W

Structure of Atom

1.71

ANSWeR KeyS multiple Choice Questions with Only One Answer level i 1. d 2. c 3. b 4. c 5. c 6. d 7. c 8. c 9. b 10. a 11. b 12. d 13. d 14. d 15. b 16. b 17. c 18. d 19. b

20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38.

b d b a c a a d d d b d a d d a b d d

39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57.

c a a a c b a d a a b a b c c d c b a

58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76.

c d a a a a c a d c d d c d d b c b d

77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92.

b a c c c b c b a b a a a c a d

21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40.

b c d d b c d b a c b d c a b c d a c a

41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

a a a a a c b b c c d a b b c b a d a c

61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80.

a b a c d b d d a d c c b d c a a a c c

1. 2. 3. 4. 5. 6. 7.

a,b,c,d a,c a,b,c a,b,c,d a,c a,b a,c

8. 9. 10. 11. 12. 13. 14.

b,c,d b,c,d b,c,d c,d b,c a,c,d a,b,c

Passage i 1. d

2. b

3. b

4. c

2. b

3. a

4. b

2. d

3. d

2. a

3. c

2. d

3. b

2. b

3. a

2. a

3. d

Passage ii 1. b

1. d

81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99.

c b a d d b b b d c b b d a b c b a d

15. 16. 17. 18. 19. 20. 21.

Comprehensive Type Questions

Passage iii

multiple Choice Questions with Only One Answer level ii 1. d 2. b 3. b 4. a 5. a 6. c 7. d 8. d 9. b 10. c 11. b 12. a 13. d 14. d 15. a 16. b 17. c 18. b 19. a 20. b

multiple Choice Questions with One or more Than One Answer

Passage iv 1. a Passage v 1. c Passage vi 1. a Passage vii 1. d

Passage viii 1. a

2. a

3. b

Passage iX 1. b

2. c

Passage X 1. b

2. c

3. b

a,b,c a,b a,c a,b,c,d b,c b,c,d a,b,c,d

1.72

Structure of Atom

matching Type Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

(a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

s rs qs p qs q rs qsrt qrs s

(b) (b) (b) (b) (b) (b) (b) (b) (b) (b)

p ps pq pqrs r p pqr qst p q

(c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

integer Type Questions q ps pqrs p pq s pqr pqs ps pqs

(d) (d) (d) (d) (d) (d) (d) (d) (d) (d)

r qs qr pq q r r s qr prs

Assertion (A) and Reason (R) Type Questions 1. a 2. a 3. a

4. b 5. c 6. b

7. b 8. b 9. b

10. a 11. a 12. a

13. c 14. c 15. b

1. 2. 3. 4. 5.

10 1 625 547 2

6. 7. 8. 9. 10.

4 4 2 3 5

11. 12. 13. 14. 15.

2 5 5 9 3

16. 17. 18. 19. 20.

2 2 4 1 6

21. 6

Previous years’ iiT Questions 1. 2. 11. 12. 13.

a 3. a 4. (a) r (a) qr 2 14.

b 5. d 7. c 9. a c 6. d 8. d 10. c (b) q (c) p (d) s (b) pqrs (c) pqr (d) pq 3 15. 2 16. 9 17. 4

Structure of Atom

1.73

hiNTS AND SOlUTiONS multiple Choice Questions with Only One Answer level i 2. Mass of electron is less when compared to mass of positive rays r

ø = e ao 5. y Probability density means y2 10. λë =

2KEm

h 4p

Dp =

\DV =

V r

16. Orbital angular momentum =

l (l +1)

h 2p

h v1 = h vo + x v1 - vo 1 = v2 - vo k vo =

kv1 - v2 k -1

-13.6 -13.6 E2 - E1 1 = 27 = 4 20. 5 E3 - E2 -13.6 -13.6 9 4 21. l =

h 6.625 × 10

3.31 × 10 -29 =

-34

2( KE ) × 0.1 × 10 -5 -4

\ KE = 2 × 10 J = 1.25×10-13×(64)1/3 =5×10-13cm 4 π(5×10-13 )3 3 =1.25×10-13 Fraction of volume = 4 -8 3 π (10 ) 3

31. Radius of nucleus

32. The probability of finding election in all directions are same

h ml h \l = m

56. 2πr = nλ 2πx = λ In third orbit 2π(9x) = 3λ \ λ = 6πx

mv = p p v= m P2 1 \ mV 2 = 2 2m 2 2 p Ze \ = 2m r \ If momentum is 2 P

2 KEm

38. V = l

1 h m 4p

66. When reaches nucleus kinetic energy converted to potential energy

h v2 = hvo + kx

\l =

52. Dχ = Dp

h

12. Angular speed =

19.

48. n = 3, l = 2 So orbital is 3d

4 P 2 Ze 2 = 2m r1 \4

Ze 2 Ze 2 = r r1

\r1 =

r 4

80 The time taken by one revolution = \ Ratio = \1: 8

2p r V

2p r 2p (4r ) : -8 2.188 × 10 2.188 × 158 / 2

1.74

Structure of Atom

multiple Choice Questions with Only One Answer level ii 1. Orbital angular momentum =

l (l + 1)

h . For n 2p

= 5l values are 0, 1, 2, 3, 4 and these cannot satisfy the given condition. 2. (i) The no. of peaks in a radial probability curve is equal to n – l. So, no. of peaks in 1s and 2s are not equal. (ii) No. of nodal planes equals to l, so it is same for 2p, 3p and 4p orbitals. (iii) Ti2+ and Ni2+ both have same no. of unpaired e– and have same magnetic moments. (iv) Any orbital of any sub shell can accommodate only two e–. +x 3. M and M+y both have same magnetic moment 5.916 which means that both have same no. (5) of unpaired e–. It is possible only if x and y are +3 and +1. Electron configuration corresponding to these two oxidation states are M+x → 3s2 3p6 3d5 M+y → 4s1 3d6 So, M+x is more stable due to the half-filled configuration. 4. According to photoelectric effect eq. hc hc + E where E = KE = λ λ0

 l -l  E = hc  0   ll 0  de Broglie’s wavelength h , where M = mass of e– and E = KE 2mE or

h  l -l  2mhc  0   ll 0 

=

1/ 2

6. Along x, y axis the shape of orbital is as shown below:

h 2 (ll 0 ) 2mhc(l 0 - l)

  hll 0 =    2mhc(l 0 - l)  5. From the observations no of nodal regions for that orbital n – l – 1 = 2. xy is a nodal plane. So orbital is dyz

8. (i) No. of nodal planes = l = 0 (ii) de Broglie’s wavelength l =

h h = mv p

(iii) The principal quantum no. n indicates the size and energy of orbit. (iv) p has electronic configuration 3s2 3p3. 9. The length of a series depends on the difference between wavelength of successive lines. It is more in Lyman series. 13.6 2 × 2 = 3.4 eV 42 Work function = 1.4 eV The additional energy supplied due to acceleration is 4 eV. So, net resultant KE = (3.4 – 1.4) + 4 = 6 eV

10. Energy given to the e– =

12.27

de Broglie’s wavelength = 11. Angular momentum of e– =

V

Å=

12.27 6

5Å

nh 2p

So the transition occurs from 4 → 2; that means the transition lies in Balmer series and will be observed in the visible region. 12. de Broglie’s wavelength =

h 2meV

where m is mass of particle, e is charge and V is stopping potential. h k : l :l = So, p a 2m × e × 0 2 × 4m × 2e × v p

or l p : l a =

1

p

1 =2 2 :1 2 4 :

13. Energy of emitted photon 13.6   13.6 =  2 × 4 - 2 × 4  eV 2  1  = 3 × 13.6 eV \ KE of e– = [3 × 13.6 – 13.6] = 27.2 eV

Structure of Atom

14. DX × DV = \ DV =

21. From the photoelectric effect h(x) = hv0 + y h(2x) = hv0 + 3y from Eq. (1) → y = hx – hv0 ⇒ h (2x) = hv0 + 3[h(x) – h(v0)] x ⇒ v0 = 2

h 4pm -34

6.625 × 10 1 × 4 × 3.142 × 0.1505 10-10

@ 3.4 × 10-24 m/s 15. Six nodes are as shown:

So, it has 2.5 waves, so wavelength =

35 = 14 cm. 2.5

16. No. of m values = 2l + 1 So, no. of m values for l = 1 is 3 This value is possible for m in case of f and higher sub shells only. 17. To have the same de Broglie’s wavelength they must have the same momentum. ⇒ Pp = Pe ⇒ mp × vp = me ve ⇒ 1837 me × vp = me ve So,

vp =

ve 1837

18. Let us suppose that ‘na’ is the no. of quanta absorbed and ‘ne’ is the no. of quanta emitted. na and ne were related as ne = 0.53 na hc So, the absorbed energy Ea = na × 4530 Emitted energy Ee = ne × ⇒

hc 5080

19. From the data no. of e– in both cases is (n – 1). So, x has +1 charge and y has +2 charge. 20. Energy involved in transition E1 = 2E – E = E = hc . l

4E E –E= = 1. Energy involved in transition E1 = 3 3 E1 hc / l1 l E \ = = ⇒ 2 =3 l1 E2 E / 3 hc / l 2

(1) (2)

22. From the data the difference between radii of ground state and excited state orbit radius ⇒ rn – r1 = 0.7935 × 10–9 m ⇒ rn – (0.0529 × 10–9 m) = 0.7935 × 10–9 m ⇒ rn = 0.8464 × 10–9 m = 0.0529 × 10–9 × n2 ⇒ n=4 So, the maximum spectral lines formed =

(n2 - n1 )(n2 - n1 + 1) =6 2

23. For Balmer series n1 = 2 and for the least wavelength n2 = ∞ and for the highest wavelength n2 = 3. So, wave no. for spectral line with least wavelength, 1  R 1 v1 = RH  2 - 2  = H 4 ∞  2 wave no. for spectral line highest wavelength 1  5R 1 v2 = RH  2 - 2  = H 36 3  2 So, v1 : v2 =

RH 5 RH : =9:5 4 36

24. No. of nodal planes = l = 2 No. of Radial nodes = n – l – 1 = 5 – 2 – 1 = 2 No. of peaks = n – l = 5 – 2 = 3 25. No. of revolutions made by one e– per sec

ne hc 4530 0.53na ⋅ × na 5080 na hc

= 0.4726

1.75

=

velocity of e - z 2 @ 3 × 6.66 × 1015 2prn n

26. From Rydberg’s equation 1 1 1  = R 2 - 2  l 1 n  ⇒

lR 1 1 ⇒ n2 = = 1lR - 1 lR n2

28. At radial node i.e., at r = r0, y2 = 0. 2  1  1 1/ 2  r0  - r0 / 2 a0   =0 ⇒   2 - e a0   2 2p  a0    r ⇒ 2 - 0 = 0 ⇒ r0 2a0 a0

1.76

Structure of Atom

29. In case of unielectron systems energy can be determined only from n values i.e., orbital of same shell have same energy. 30. The energy of an e– in a multielectronic atom can be determined from (n + l) value. 31. (i) Carbonates with thermal stability can be produced by alkali metals. So, — V. (ii) Colored compounds will be formed by transition elements. So, — X. (iii) Largest atomic radius is possible with maximum no. of shells. So, — Y. (iv) Acidic oxides will be formed by non-metals. So, — W. 32. No. of l values possible for first shell = 3 (0, 1, 2) So, the total possible no. of m values are nm = 9(1 + 3 + 5) [No. of m values for given l =2l +1] So, the no. of e– that can be fitted = 2 × n.18e–. 33. According to the given rules Zr (40) has electronic configuration as follows; 1s3 2s3 2p9 3s3 3p9 3d13 35. l1 is the wavelength corresponding to initial kinetic energy E1. l2 is the wavelength corresponding to final kinetic energy E2 and l2 = 0.9l1. l1 =

⇒

hc 2mE1

: l2 =

l1 1 = = l 2 0.9

hc 2mE2

E2 E1

⇒ E2 = 1.23456 E1 So, E2 is 23.456% is more than that of E1. nh 7 h = 36. Angular momentum = 2p 11 ⇒ =n=4 0.529 × n 2 Radius of nth orbit rn = 2 ⇒ rn = 2.82 Å 37. The data indicates that the transition is only possible with Lyman, Balmer, Paschen series that means transition to 4th orbit is not possible which means that transition from 4th orbit occurs.

38. Maximum no. of spectral lines possible = Σ (n – 1) = 15 ⇒ n=6 So, from Rydberg’s equation v=

1 1 1  = RH  2 - 2  l 1 6 

l=

36 × 912 Å 35

 1  = 912 Å    RH 

= 938.05 n3 Z2 2 T1 Z 2 n13 = × ⇒ T2 Z12 n23 [z is the same] 23 ⇒ 3 = 8 : 27 3 40. Velocity of an e– in nth orbit

39. Time period ∝

2.17 × 106 m/s n From the data 2.17 × 106 3 × 108 = n 4 ⇒ n=3 No. of lines in Balmer series is 6 – 2 = 4 =

41. l =

h 2meV0

42. Wavelength of first line in Lyman series l1 =

1 3 1 ⋅ × RH 4 Z 2

Wavelength corresponding to Balmer series. l2 =

1 36 1 × × RH 5 Z 2

⇒

l 2 - l1 =

⇒

91.2 ×

⇒

Z@4

1 Z2

1 1 ⋅ RH Z 2

 36 3   5 - 4  

 129   20  = 33.4  

43. From the data h (3.2 × 1016) = hv0 + 2E h (2.5 × 1016) = hv0 + E By solving Eqs. (i) and (ii) v0 = 18 × 1015 Hz

(i) (ii)

Structure of Atom

44.

1.77

hc = E2 – E1 l 6.625 × 10-34 × 3 × 108 ⇒ l

53. Moseley’s equation is v = a (Z – b). So, from the data 100 = 1 (Z – 1) ⇒ Z = 11

 -13.786  -13.786   -  = 1.602 × 10–19  4 1   

55. By the given condition, each primary shell should consist of (n + 1) subshells. So, configuration will be 1s2 1p6 2s2 2p6 3s2 2d6

= l = 1.199 × 10–7 m = 1200 Å –

th

45. Energy of e in n orbit of an H like system, En =

-13.6 × Z 2 eV/atom n2

n -1

I = ∑ 2(2l + 1) i =0

E Z 2 n2 So, 1 = 12 × 22 E2 Z 2 n1 =

57. No. of e– corresponding for a given l value equals to 2 (2l + 1). For a given n value the possible ‘l’ values are 0 to n – 1. So, total no. of e– in a given orbit

58. l value for g sub shell is 4. (i) Maximum no. of m values = 2l + 1. So, for g = 9 (ii) Maximum no. of unpaired e–1 = 2l + 1. So, for g = 9. (iii) The minimum principal quantum no. of gshell = 5.

1 4 × = 1: 4 16 1

59. Dx ⋅ Dv = 47. Electronic configuration in second excited state = 1s2 2s2 2p6 3s2 3p3 3d2 So, max (n + l + m) value for this configuration = [[(3 + 1 +(–1)] + (3 + 1 + 0) + (3 + 1 + 1)] + (3 + 2 + 2) + (3 + 2 + 1)] = 25 48. In any orbit the no. of waves by an e– equals to its principal quantum no. and total length of all the waves must equal n circumference of the orbit. ⇒ nl = 2prn ⇒ nl = 2pr (n2) –

51. Energy of e in ground state En =

-2p2 Z 2 me 4 n2 n2

RH = Rydberg constant =

⇒

2p2 nZ 2 e 4 Ch3

6.625 × 10-27 1 5.272 × 10-28 × = cm 4 × 3.142 × 1 2 2

@ 2.64 × 10–30 m 60. From the data hc = hv0 + E1 l1 hc = hv0 + 2E1 l2

61. From the data n1 + n2 = 4 and n1 – n2 = 2 ⇒ n1 = 1 and n2 = 3 So, from Rydberg’s constant 1 1 1 = RH ⋅ Z 2  2 - 2  l  n1 n2 

hc l

l=

Dx =

By substitution and on solving 1/4 and 3/4 ⇒ v0 = 1.19 × 1015 s–1

So, energy of e– in ground state configuration –RH × ch 52. DE =

⇒

h 4pm

hc 6.625 × 10-34 × 3 × 108 = DE ( DEe.V . ) × 1.602 × 10-19

= 1.2406 × 10–10 m

⇒

1 1 1  = 109677 × 9  2 - 2  l 1 3 

⇒ l = 114 Å = 1.14 × 10–6 cm

1.78

Structure of Atom 2

62. If na is the number of absorbed quanta and ne is the no. of emitted quanta, then

2

ne = 0.5 na Emitted energy Ee is 47% of the absorbed energy Ea hc ne ⋅ X = 0.47 So, hc na 4700 ⇒

4700 ne × = 0.47 X na

⇒

X =

4700 × 0.5 = 5000 Å 0.47

63. Before the excitation the molecule should undergo the homolytic cleavage. So, energy required for dissociation Ed = 0.04 × 400 = 16 KJ. On dissociation the sample consists of 0.08 moles of H gas atoms. Energy required for excitation  -13.6  -13.6   - = 0.08 × 6.023 × 1023    1   4 = 4.914768 × 1023 eV =

4.914768 × 1023 × 1.602 × 10-19 KJ 103

= 78.734 KJ So, total energy required = 16 + 78.734 = 94.7345 KJ –

64. Angular momentum of e mvr =

or

1 2  2  = n 2 ⇒ n1 = 1   1 2

66. de Broglie’s wavelength in nth orbit is nl = 2prn ⇒ nl = 2pr1n [for hydrogen] ⇒ l = 2pr1n = 2pr (x)3 ⇒ l = 6px 67. If w is angular velocity and r is radius of orbit, v is velocity of e–. v = rw v 1 ⇒ w= ∞ 3 r n 68. We know that the no. of waves made by an e– in an orbit equals to its principle quantum number ‘n’. So, from the figure n = 4 No. of revolutions made by an e– per sec (or) Frequency =

=

Z2 × 6.66 × 1015 s–1 n3

1 × 6.66 × 1015 = 1× 1014 s -1 (4)3

69. In case of an elliptical orbit, we know that length of major axis n = length of minor axis k And length of major axis is double the radius of that circular orbit

nh 2p

nh = 4.217 × 10-34 kg.m2/s 2p

⇒ n = 3.999 @ 4 The no. of spectral lines in visible region = n – 2 = 4 – 2=2 65. From the data 1 1 RH × 4 ×  2 - 2  = RH 2 4   By comparison

2  1  4 =n     2 ⇒ n2 = 2

1 1  2 - 2  n1 n2 

⇒ ⇒

2 × 16 × 0.529 4 = l 2 l = 8.464 Å

70. Ionization potential in H-like system means, it is the energy of e– in ground state. E So, En = (0.01) E1 = 21 n ⇒ n = 10. 73. DE = 12.1 = En2 - En1

Structure of Atom

-13.6 - (-13.6) n2

So, 12.1 = ⇒ n=3 Therefore, Dn =

change

in

angular

momentum

=

Magnetic moment m = =

@ 2.11 × 10–34 J s. 74. If E is KE of a particle, de Broglie wavelength h 2mE

So, if KE doubles, l becomes

1 2

times

81. de Broglie e’s wavelength ln = 2pr = nl So,

1 1  2 - 2  n1 n2  – As e has double the mass in hypothetical system. Rydberg’s constant in the equality will be equal to 2RH.

1 1 1 = 2 R  2 - 2  [transition is to the first l 2 3   excited level i.e., to 2n orbit] 18 5R

83. Shortest wavelength is possible for any series if n2 = ∞. Longest wavelength is possible for any series if n2 = n1 + 1. So, for hydrogen 11 1  1 = RH  2 - 2  Xx 1 ∞ 

E E So, hv = 21 - 21 n2 n1

⇒ RH =

[Here, E1 is –ve of ionization Energy]

@

2.18 × 106 m/s n

2pZe 2 nh

⇒ Velocity of e–1 in 4th orbit = 545 × 103 m/s 79. Electronic configuration of Cr and Cr2+ are as given

1 x

For helium 1 1 1 = RH ⋅ Z 2  2 - 2  l 2 3   From Eqs (i) and (ii), 9x l= 5

 -2 × 10-18  -2 × 10-18   ⇒ 6.000 × 10-34 × v =  -  4    16 ⇒ v = 6.25 × 1014 = 625 × 1012 Hz

Velocity of e– in nth orbit =

2pr4 4

⇒ Circumference of the fourth orbit = 4l4 = 16l1

77. For, Hb line of Balmer series n1 = 2 and n2 = 4.

78. Radius of nth orbit rn = 0.529 × n2 Å ⇒ 8.464 Å = 0.529 × n2 ⇒ n=4

l1 2pr1 = l 4 2pr1 × 4 l 4 = 4l1 =

So,

l=

4(4 + 2) = 4.89 BM

80. Electronic configuration of Fe2+ = 1s2 2s2 2p6 3s2 3p6 3d6 So, no. of d e– in Fe2+ = 6 Configuration of Ne — 1s2 2s2 2p6 ” ” Mg — 1s2 2s2 2p6 3s2 ” ” Cl– — 1s2 2s2 2p6 3s2 3p6

1 = RH 75. l

⇒

n(n + 2)

-34

2p

l=

below: Cr → 1s2 2s2 2p6 3s2 3p6 4s1 3d5 Cr+2 → 1s2 2s2 2p6 3s2 3p6 3d4 So, no. of unpaired e– = 4.

h 2p

( 3 - 1) × 6.625 ×10

1.79

(i)

(ii)

84. (a) In H-like systems, energy of electron in an orbital increases with increase in ‘n’. So, energy of 3d is less than energy of 4s. (b) In d x2 - y 2 orbitals cloud is along x and y-axes. So, it lies in xy plane. (c) Electronic configuration of Cr — 1s2 2s2 2p6 3s2

1.80

Structure of Atom

3p6 4s1 3d5 (d) Except the spin quantum no. the remaining Quantum number are obtained from Schrodinger equation. 86. From the data =

Velocity of e -1 1 = Velocity of light 275

⇒ V = 1090909 m/s Velocity of e– in nth orbit Vn =

1:

93. From the data r(n + 1) – rn = r(n – 1) ⇒ [(0.529) (n + 1)2] – [0.529 × n2] = (0.529)(n – 1)2 ⇒ n2 + 2n + 1 – n2 = n2 – 2n + 1 ⇒ n2 = 4n ⇒ n = 4 94. Orbital can be represented in terms of wave function as ‘ynlm’ and we know that in general, 0 value for m can be assigned for orbital along with z axis.

2.18 × 106 m/s ⇒ n

n=2

87. If three spin quantum no. were possible, then each orbital can accommodate 3e–. So, electronic configuration of K is 1s3 2s3 2p9 3s3 3p1.

95. At nodal surface, y2 value must be equal to zero. By equating y2= 0 we will get the expresion as (2 – Zr) = 0 ⇒

89. For the line in ultraviolet region, the energy must be more than that corresponding to E. Energy of A > Energy of E

⇒ v = v0 ×

91. To find the distance at which finding probability d y = 0. dr 2

⇒ e - r / a0 = 0 ⇒ – In (r / a0 ) = – In 1 ⇒ r = a0

=

98. Velocity of e– =

= (2 × 10–19) (ii) Total energy : Kinetic energy : Potential energy

1 : –1 : 2 If magnitude is only considered, ratio, 1 : 1 : 2. (iii) No. of maximum (or) no. of peaks =n–l=5–4=1 (iv) In terms of KE (E) de Broglie’s wavelength h 1 1

:

1 4

:

2meV

1 16

2.18 × 106 m/s n

So, graphic representation must be a hyperbola curve. 99.

-mv 2 mv 2 mv 2 :+ : 2r r 2r

So, lH : lHe : l CH4 =

h

from the curve, we can say that at any point lB > lA, means that mB < mA.

hc – Work function l

2mE

1 ⋅ KE h

97. de Broglie’ wavelength l =

6.625 × 10-34 × 3 × 108 – (2 × 1.602 × 10–19) 310 × 10-9

l=

2 2 = =1Å Z 2

or KE = hv – hv0 y = mx – c So, slope of the curve m = h.

2

of e– is maximum, we have to count

r=

96. Equation for photoelectric effect hv = hv0 + KE

90. In the given sub shell, half of the max. no. of e– in sub shell will have the same spin quantum number.

92. (i) KE =

1 1 : =4:2:1 2 4

v = a (z – b) v = az – ab a = slope of the curve = tan 45° = 1 ab = Intercept on the axis = 1 ⇒ v = 52 – 1 ⇒ v = 2601 s–1

multiple Choice Questions with One or more Than One Answer 2. (a, c) From the data we can write the expression as

Structure of Atom

follows: 4.25 = WA + TA 4.20 = WB + TB TB = TA – 1.5 h 2h = 2mTB 2mTA

(i) (ii) (iii)

⇒ TA = 4TB (iv) So, from above equations TB = 0.5 eV TA = 2 eV WA = 2.25 eV WB = 4.15 eV 3. hv = hv0 + KE and stopping potential is equal to K.E. ⇒

hc hc = + v0 l l0 v0 =

hc 1 - hc ⋅ l l0

v0( eV ) =

y = mx – c 6.625 × 10-34 × 3 × 108 1.602 × 10-19

Work function for metal = 0.24 eV hc = w + v0 l ⇒

v0 =

-34

6.625 × 10 × 3 × 10 100 × 10-9

8

–0.24 × 1.602 × 10–19 = 1.949 × 10–18 J = 12.1 eV The work function for metal II is more than metal I. So, if both metals were exposed to same light, KE i.e., stopping potential for metal II is less than metal I. 5. (a, c) For the same magnetic moment ions (a) atoms should contain same no. of unpaired e–. 16. BE = Ionization energy of H-like system =

2.

3. 4.

Based on the given conditions, No. of sub shells = n + 1, and No. of orbitals in a sub shell = 2l + 3 That means 1st orbit consists of s and p sub shells with 3 and 5 orbitals and 2nd orbit consists of s, p and d sub shells with 3, 5 and 9 orbitals, respectively. EC with atomic no. 25 1s2 1p10 2s6 2p3 EC corresponding to atomic No. 18 is 1s6 1p10 2s2, so it has two unpaired e– which causes a magnetic moment of 8 BM. After 3s, 3p and 2d have the same (n + l) values. So, because of lower n value e– enters into 2d. If three spin quantum no. were possible each orbital can accommodate 3 electrons, So, configuration of atom with atomic no. is 1s9 1p15 2s6

Passage ii

hc 1 hc 1 ⋅ - ⋅ e l e l0

So, slope = m =

1.

13.6 13.6 × Z 2 = 2 × Z 2 = 13.6 4 n2

Because of half of the mass of –ve particle in the hypothetical system. Energy, velocity, radius, expression were as follows: -2p2 Z 2 me 4 E , So Ehypothetical = E= 2 2 2 nh V=

2pZe 2 , So Vhypothetical = V nh

r=

n2 h2 , So rh = 2r 4p2 me 2

Rydberg’s constant, R =

So,

=

1 1 1 = Rn  2 - 2  l 3  2

R 9 - 4 72 R ⇒ l= 2  4 × 9  5

2. Vn = V =

=

Comprehensive Type Questions

2p2 mZ 2 e 4 R ; So, Rh = 2 ch3

1. Longest wavelength is possible if transition occurs from the immediate orbit. So, the given transition is 3 → 2.

⇒ Z=4

Passage i

1.81

2.188 × 106 m/s n

2.188 × 106 = 1.094 × 106 m/s 2

3. r = rn × n2 = (2r) n2 = 0.529 × 2 × 9 = 9.522 Å 4. We know that KE and total energy are with

1.82

Structure of Atom

opposite sign and are of equal magnitude. So, KEn = – En =

E 13.6 = 6.8 eV/atom = 2 2

3. Stopping potential is the –ve potential that must be applied on collection plate to stop 1 photoelectrons. It will be equal to KE of e– in (eV.) So, V0 = 25 – 7 = 18 eV.

Passage iii Passage vi

1. We know that E3 – E2 =

-13.6 × 22  -13.6  - ×Z2  9  4 

1 1 So, 47.2 = 13.6 × 2  -  ⇒ Z = 5 4 9 2.

hc = (–13.6 × 25 × 1.602 × 10–19) l

⇒ l = 3.6489 × 10–9 m 3. KE = – E = – (13.6 × 1.602 × 10–12) ergs = 5.4468 × 10–9 ergs Passage iv From the data A is the 1st orbit and B is nth orbit. After absorption of 2.55 eV, e–1 excites to n1 orbit. The possible no. of spectral lines for de-excitation is Σ (n1 – 1) = 6. Which mass that n1 = 4 and n may be 2 (or) 3. From the given data En1 - En = 2.55 -13.6  -13.6  × Z2 –  2  n2 = 2.55 i.e., 16  n  The above expression is possible only if n = 2 and Z = 1. 3. Minimum energy is possible due to transition of e– from 4th orbit to 3rd orbit. E=-

13.6  -13.6  -  = 0.6611 V 4  9 

Passage v hc 1. = W + KE l

hc 6.625 × 10-34 × 3 × 102 = 6.625 × 10–19 J = l 3000 × 10-80

Passage vii 1. de Broglie wavelength of a cricket ball =

h 6.625 × 10-34 = = 1.1 × 10–34 m mV 0.2 × 30

2. DP =

-34

8

h 6.625 × 10-34 = 5.272 × 10–2 = 4p ⋅ Dx 4p× 10-10

3. Dx = DP = m × DV So, according to Heisenberg’s principle h Dx × DP = m2 (DV)2 = 4p 1 h ⇒ DV = 2m 2p Passage viii 2. Longest wavelength corresponding to the transition (n + 1) → n, so in case of Lyman series the transition is 2 → 1. So,

1 1 1 = RH ⋅ Z 2  2 - 2  l  n1 n2 

1 = RH l ⇒l =

 6.625 × 10 × 3 × 10  1 × eV  -10 -19  4000 × 10 1.602 × 10   ⇒ 2 + KE ⇒ KE = 1.10 eV

⇒

2. E =

1 1  ⋅ -  1 a 

1 912 = Å = 22.8 nm 4 RH 4

3. Shortest wavelength corresponds to the transition of ∞ → n. So,

1 1  R 1 = RH  2 - 2  = H l1 2 ∞  4

For the longest wavelength 2. KE1 = hv1 – hv0 = h (10 × 1014 – 5 × 1014) = h (5 × 1014) In second case KE2 = 2KE1 ⇒ h (10 × 1014) = hv1 – 5 × 1014 × h ⇒ v1 = 15 × 1014 s–1

1  1 1  3R = RH  2 - 2  = H l 4 1 2 

Structure of Atom

⇒

l=

l1 = 0.33 l1 3

Passage iX The data can be represented in the form of a diagram as follows:

1.83

r1 n12 Z 2 = × r2 n22 Z1 V1 Z1 n2 = × V2 Z 2 n1 T1 n13 Z 22 = × T2 n23 Z12 ⇒

K1 n22 = K 2 n12

Assertion (A) and Reason (R) Type Questions

So,

En – E2 = (10.2 + 17)

⇒

-13.6  -13.6  × Z 2 -  2  Z 2 = 27.2 n2  2 

Similarly, -13.6  -13.6  × Z 2 -  2  Z 2 = 10.2 n2  3  By solving Eqs. (i) and (ii), n = 6, Z = 3 Passage X 1. Radius r =

n2 h2 , So for hypothetical, it can be 4p2 mZe 2

modified as rm =

n2 h2 0.529 = Å 414 4p2 (207 m) (2e)

2. Ionization energy is the energy of e– in first orbit with +ve sign. E=

-2p2 Z 2 me 4 for hypothetical n2 h2 Em =

-2p2 Z 2 (207 m) (2e) n2 h2

= 828 × 13.6 eV 0.529 2 2 0.529 3 - 1  = 3. r3 – r1 = ×8 Å 414  414

matching Type Questions 2. Note: Molecules at very high temperature are monoatomic gases that gives a line-spectrum. 6. We know that

1. Peaks in radial distribution curve = n – l radial nodes = n – l – 1 Which means that the radial probability curves depend only on ‘n’ and ‘l’ and not on ‘m’. 3. An electron can never be found in the nucleus and this can be explained by uncertainty principle. Any object can not move with velocity of light. 4. Reason is the consequence of assertion and it can not explain assertion. 5. The shape of orbital is independent of principal quantum number. 6. Anode rays are +ve ions of different gases and they have different charges and masses. So, e/m ratio for anode rays depends on nature of gas. 7. y2 is the square of amplitude of e– wave and we know that I ∝ a2 where I is intensity and a is amplitude. 8. The no. of angular nodes = l and angular wave functions depend on ‘I’ and m 9. Diamagnetic nature is due to pairing of e– 10. In the absence of electric and magnetic fields px, py and pz orbitals will have same energy and transition in between px and py can not give spectral line. 11. In case of H like systems, energy of e– can be determined only from ‘n’ values and all the sub shells with same ‘n’ value will have same energy. 12. The electron can revolve in a particular orbit if it’s wave is in phase only. For the He, condition is nl = 2pr which can be modified as mvr =

nh . 2p

14. + (or) – (ve) sign indicates that electrons are with opposite spins only. 15. As the orbit moves towards the nucleus, the energy difference between successive orbits increases.

integer Type Questions 1. After the absorption of energy, energy of e– = –0.01 E -E So –0.01 E = 2 n

1.84

Structure of Atom

1 1   = 6.625 × 10–34 × 3 × 108  + -9 -9  108.5 10 30.4 10 × ×   ⇒ n=5

⇒ n = 10 2. 1 3. Hb line is due to the transition of e– from 4 to 2 -2 × 10-18  -2 × 10-18  -  16 4   v = 6.25 × 1014 or 625 4. Radius of orbit rn = 0.529 × n2 = 0.8464 ⇒ n=4 So, hvz =

Velocity of e– =

2.188 × 106 m/s n

= 547.00 m/s 5. Velocity of e– 2.188 × 106 3 × 108 = n 275 ⇒ n=2 6.

1 1 1 = RH  2 - 2  l n  2 1 1 1 1  = 4814 912  4 n 2  ⇒ n=4 ⇒

7. If l is de Broglie’s wavelength and n is principal quantum no, nl = 2prn = 2p × 0.529 × n2 .. ⇒ l = 2p × 0.529 × n ⇒ n=4 8. Minimum value of magnetic quantum no. = –l So, ‘l’ for the given orbital is 2. No. of angular nodes = l = 2 9. 3rd excited state means 4th orbit. So, spectral lines in Lyman series = n – 1 = 4 – 1 = 3 10. Max no. of spectral lines = Σ (n2 –n1) (6 – 1) = 5 11.

hc = En2 - En1 l hc 1 1  = Z 2  -  × 2.178 × 10–8 l 1 4  ⇒ Minimum possible number for Z = 2

12. En2 - En1 = ⇒

hc hc + l1 l 2

 13.6  -13.6  2  2 × 1.602 × 10-19  - 2 -    1   n 

13.

hc = w + eV0 l

⇒ w = 5 eV 14. Degeneracy means the total no. of degenerate orbitals. In H-like system all the orbitals present in a particular orbit will have the same energy. So, no. of degenerate orbitals (d) degeneracy n2 = 9 15. No. of visible lines means, lines in Balmer series, =n–2=5–2=3 16. 16.2 17. l = 2pr1n 6.64 =2 ⇒ n= 2p× 0.529 18. In second period 2s, 2p orbital are to be filled with three e– in each orbital. So, total no. of elements in second will be 12. 19. No. of revolutions (a) frequency f∝

z2 n3

So,

f1 Z12 n23 = × f 2 Z 22 n13

⇒

2 1 23 ⇒n=1 = × 1 4 n2

20. From the data KE of e– = 13.6 × 0.5 eV 1 ⇒ mv 2 = 13.6 × 0.5 × 1.602 × 10–19 2 ⇒ v = 1.5473 × 106 So, the correct answer is 6. 21. (6)

Previous years’ iiT Questions 1. Smaller is the value of (n+l), lesser is the energy of sub shell. When (n+l) values are same then energy of sub shell depends on the principal quantum number ‘n’; smaller will be the value of ‘n’ lesser is the energy of sub shell.

Structure of Atom

(iv) n = 3, l = 1 < (ii) n = 4, l = 0 < (iii) n = 3, l = 2 < (i) n = 4, l = 1 2. px orbital has only one nodal plane i.e, yz-plane 3. 1s2, 2s22p6; 3s23p63d5, 4s1 is ground state electronic configuration of Cr24 4. l =

Vn =−2 Kn rn ∝ En−1 In the lowest orbit (1s), the angular momentum h 2π = 0 ( for l = 0) = l (l + 1)

h 6.626 × 10-34 = mv 0.2 × 5 / 3600

rn =

6. Beam of α-particles (nuclei of helium) are used in the Rutherford’s gold leaf experiment 7. An orbital can accommodate maximum two electrons with opposite spin and no two electrons in the atom can have same values of all four quantum numbers. 8. r =

n2 × 0.529 Å z

For Be3+, n = 2, z = 4 \ r = 0.529 Å 9. Number of radial nodes = (n-l-1) Number of radial nodes for 3s = (3-0-1) = 2 Number of radial nodes for 2p = (2-1-1) = 0 10.

2p 2 e 4 m h 2 e 4 m p 2 m 4 = = e 4h 2 2h 2 2h 2 h2 ao = 2 2 4h e m h4 \ e4 = 16p 4 m 2 a 2 h2 \ KE = 32p ma 2 KE =

matching Type Questions 11. (a-r) (b-q) (c-p) (d-s) − KZe 2 ; 2r KZe 2 Z2 K n ( Kinetic energy ) = − 2 × 2.18 × 10−18 J = 2r n KZe 2 Vn = potential energy = − r n2 rn = Radius of Orbit = × 0.5529 Å Z V E n ( Total energy ) =

1.85

n2 × 0.529 Å Z

1 ∝ Z y i.e , ∝ Z 1 i.e., y = 1 rn

12. Orbital angular momentum =

l (l + 1)

h cos θ 2p

i.e., it depends on azimuthal quantum number, magnetic quantum number Shape, size and orientation of hydrogen like atomic orbitals is determined by principal, azimuthal and magnetic quantum numbers Probability density of electron at the nucleus is determined by the principal and azimuthal quantum number Comprehensive Type Questions 13. 2s orbital is spherically symmetrical and it has one radial node. Number of radial nodes of 2s orbital = n-l-1=2-0-1=1 Z2 32 9 × 13.6 eV = - 2 × 13.6 eV = - × 13.6 eV 2 4 n 2 12 EH = - 2 × 13.6 eV 1 ES1 9 \ = = 2.25 EH 4

14.. ES1 = -

15. ES2 = EH Z2 × 13.6 eV = - 13.6 eV n2 Z 2 = n2

-

Z = n = 3 for Li 2 + Thus,S2 will be 3p because it has one radial node For 3 p, n = 3, l = 1 Number of radial node = n - l - 1 = 3 - 1 - 1 = 1

1.86

Structure of Atom

integer Type Questions 16. Total number of electrons with (n = 3) = 2n2 = 18 Out of these 18 electrons, 9 electrons will have 1  anticlockwise spin  ms = -  2 

17. Photoelectrons are ejected only when the energy of absorbed quantum is greater than the threshold energy or work function (Φ) of metal. hc 6.626 × 10 -34 × 3 × 108 Absorbed Energy, E = = l 300 × 10 -9 6.626 × 10 -19 = 6.626 × 10 -19 J = eV = 4.17eV 1.6 × 1019 Thus, four metals, i.e., Li, Na, K, Mg will show photoelectron emission.

CHAPTER

2 Basic Concepts of Chemistry

I

t cannot be said too often that science is not mathematics but reasoning; not equipment but inquiry. JR Platt

2.1 introduction Chemistry is the science that deals with the structure and composition of matter and the transformations it undergoes. It is a field of study that has engaged the attention of a large number of people over a rather long period of time, both because of the complexity and great variety of matter, and the infinite detail with which the chemist wishes to understand these problems. Chemistry tries to answer such questions as these: what is the make-up or composition of this particular kind of water? Why does it have certain qualities and not others? Will it undergo changes into new, and perhaps more interesting kind of matter? Can it be produced through the transformation of other, more plentiful kinds of matter? Above all, the chemist is interested in the structure of matter. He believes that each kind of matter, if it could be examined with sufficient magnification, would be found to have characteristic structure, and that if enough were known about the detailed structure for different kinds of matter, all the individual qualities of matter could be understood. This belief must, in a large measure, rely on faith because the kind of structure that is envisaged is best described as submicroscopic: it is so fine that even the most powerful of microscopes will not render it visible. Nevertheless, it has been possible to derive a remarkably detailed picture of the structure of matter by a variety of indirect means. A chemist trying to learn about the structure of matter is not unlike a detective on the scene of a crime. He obtains a clue from this set of observations, and another one from that; then he puts these clues together and builds up a circumstantial case. There is one important difference, however, whereas the detective’s reconstruction of the crime involves real persons who can be questioned, and objects that can be observed, the chemist’s picture of

the structure of matter can never be verified by direct recourse to the human senses. His case must remain forever circumstantial. The chemist does the best he can to elucidate the structure of matter within the framework set by these limitations. He tries to make his circumstantial case so airtight, to arrive at a unified kind of evidence, that no one but a confirmed could entertain any serious doubt about the picture. This is an ambiguous goal. Clues must be used wherever they are found. While much of the work of chemists on the information concerning the structure of matter comes from the work of chemists on the composition and reactions of individual kinds of matter, an equally important share comes from the work of physicists, whose business is to discover and such aspects as are common to all matter. Thus, physics includes such studies as the motion of objects under the influence of a force, the behaviour of matter after electrification or magnetization, or the flow of heat from one position of matter to another where there is a difference in temperature. The two sciences, chemistry and physics, complement one another. The former is concerned with the specific characteristics of individual kinds of matter, and the latter with general properties shared by all matter. It is not physicists working on different aspects of the same general problem.

2.2 SubmicroScopic modelS No matter how strong the case may be in favour of a cer tain picture of the structure of matter, it must be remem bered that the submicroscopic world and everything in it are still only creations of the human mind. This is neces sary so, for the objects that populate the submicroscopic

2.2

Basic Concepts of Chemistry

world are much too small to be sensed directly. Why, then, should chemists consider what would seem to be a purely imaginary realm? After all, isn’t chemistry reputed to be an experimental science, standing four squares on direct observation? And aren’t there enough mysteries to be solved right here in the macroscopic (Greek “macro” means large and “scopic” means to view) world – the world of direct observation–to keep an army of chemists for a thousand years? The answer to these questions is that chemists are not satisfied with merely observing nature: they want to understand and explain their observations. To do this, the submicroscopic world is indispensable. If the examination has been through and imagina tive, we may decide that the failure to obtain direct per ceptual evidence is not due to lack of skill in observation, but rather to the limitations set by the human senses. In other words, the structural features we have been hop ing to observe must be too small to be perceived. When faced with such a conclusion, the physical scientist will usually resort to explanation by means of a submicro scopic model. To the physical scientist, a model is one of the two things: it may be a picture he draws of the world as it would appear if our senses were superhuman (unlimited in quality or in range, large or small) or the model may be purely mathematical, and so abstract and general in scope that a pictorialization of it is not possible. The picture models are relatively easy to understand and use as all of us have been conditioned since childhood to accept as “real” a great many things that we have seen only in the form of pictures or toys. Fortunately, most of chemistry can be explained by means of picture models, and we shall use them often. However, the picture models often prove to be inadequate because nature is more complex than any model that is simple enough to be visualized. Moreover, a picture model, like a photograph, focuses attention on a particular limited aspect of nature and may not be as general as we would like. A mathematical model, which consists, is not subject to limitations of this kind furthermore, manipulation of the equations frequently brings to light new relationships that were not envisaged by the inventors of the later, but only when the picture models seen to be inadequate.

2.3 mixture and pure SubStanceS The matter we encounter every day is almost a complex mixture. The things that are most familiar to us and which we might expect to be easy to classify and study are of such complexity as to make the analytical chemist shudder. The air we breathe is a mixture of at least five

substances; gasoline is a complex mixture of more than a dozen substances. Portland cement consists of five or more substances. Steel, though consisting largely of iron, owes its properties to the mixture of a number of other substances during its manufacture. Virtually all living or once-living forms of matter are complicated almost beyond description; it would be a hopeless job to count all different substances present in the human body. There are few cases where the matter that we encounter in ordinary experience is “pure”, that is, consists of only a single substance. Distilled water is a pure sub stance; so is the copper that we use in electrical wiring, the granulated sugar that we add to our coffee; baking soda, and some moth flakes. But one is hard put to extend this list much further. Even such a pure-looking substance as ordinary table salt consists not only of sodium chloride crystals, but also of small amounts of additives that delay clumping. Because of the great complexity of ordinary mat ter, chemistry as a science had a rather slow start. It was only after the necessity for studying the behaviour of ‘unmixed’ substances had been recognized, and meth ods for separating the mixtures in ordinary use into their components had been perfected, that the development of chemistry became rapid. The wonderfully simple and straightforward laws of chemical combination and com position could be discovered only after chemists began to deal with pure materials. This procedure—the separation of complex mixture into their components which are then studied individually— illustrates a general strategy of the laboratory sciences. Complex, seemingly unsolved, problems can often be solved by breaking them down into a number of smaller problems, each amenable to solution.

2.4 phySical propertieS of matter There are several materials which contain different phases. We know just from their appearance that we are dealing with mixtures of substances. Matter with these characteristics is referred to as heterogeneous (Greek “hetero” means of other kinds) E.g., a mixture of iron fillings and sand. There are, however, a good many materials that are perfectly uniform throughout, that is, any one portion is just like every other portion. In many cases, these materials consist of more than one substance e.g., air, gasoline and steel. Materials that are uniform throughout are referred as homogeneous (Greek “homo” = “of the same kind or nature”). When a homogeneous substance consists of more than one substance, it is called a solution. A solution may be gaseous, liquid or solid.

Basic Concepts of Chemistry 2.3

Matter

Heterogeneous

Homogeneous

Solutions (Homogeneous mixtures)

Pure substances

Matter

Pure substances Mixtures

Elements

Compounds

Homogeneous mixtures

Heterogeneous mixtures

Pure substances

Solutions (Homogeneous mixtures)

2.4.1 the States of matter The preceding classification of matter on the basis of its purity compliments the common classification of matter on the basis of its state which may be solid, liquid or gas eous. Water can exist in all three states as ice, as liquid water and steam. Water is one of the few substances with which the average person is familiar in all three of its states. The reason is simple: the temperature range over which transitions among the various states take place is quite narrow (100° on the Centigrade scale) and the melting point and boiling point are not too far from room temperature. Most other common substances are familiar to us in only one state because the transition temperatures are far from room temperature. An example is the substance iron, familiar only in solids. However, at a temperature of 1535°C, iron becomes liquid, and at 3000°C and one atmosphere pressure, it is converted to gas (a vapour). On

the other hand, oxygen is best known as gas. However, at a temperature of –183°C and one atmosphere pressure it becomes a liquid, and with further cooling it solidifies at –218.4°C. The three states of matter can be operationally de fined in terms of their macroscopic characteristic as follows: The solid state is characterized by 1. High density, usually in the range from 1 to 10 grams per cubic centimetre 2. Rigid shape 3. Slight compressibility 4. Only slight expansion at atmospheric pressure as the temperature is raised (usually less than 0.01% per degree centigrade) The liquid state is characterized by 1. High density, but usually somewhat less than that of the corresponding solid;

2.4

Basic Concepts of Chemistry

2. Lack of definite shape (liquids conform to the shape of the containing vessel) 3. Slight compressibility, but usually somewhat greater than that of the corresponding solid 4. Only slight expansion at atmospheric pressure as the temperature is raised (usually about 0.1% per degree Centigrade) The gaseous state is characterized by 1. Low density 2. Indefinite shape and volume (gases fill completely any container into which they are placed) 3. High compressibility 4. Considerable expansion (about 0.3% per degree Centigrade) when heated at constant pressure

2.4.2 chemical change In order to formulate a satisfactory sub microscopic model of chemical change, we must first become familiar with its macroscopic attributes. Let us, therefore, con sider a few examples of substances undergoing chemical change. The rusting of iron is a familiar transformation. The strong, ductile metal is slowly transformed into brittle, easily crumbled, red powder of little mechanical strength. This red powder is a new substance in no way like the parent metal. For example, the magnetic property commonly associated with iron no longer exists in this new material, its density is less, and it is no longer a good conductor for electricity. The process of combustion in which coal, wood, or oil is consumed is another dramatic chemical change. The products of combustion of wood in no way resemble the original wood, these being largely a gas (carbon dioxide), water and ash. The conversion of the food we eat to living tissue, the growth of plants, the souring of milk and the decay of vegetation are all examples of chemical changes being carried on by living organisms. The cooking of an egg, the conversion of highboiling crude oils into gasoline, the production of metal lic aluminum from the stony looking bauxite, the hard ening of mortar, the evolution of gas that occurs when baking powder is moistened, these are all further exam ples of the literally millions of chemical changes known to chemists. The characteristics which all chemical changes (also called chemical reactions) have in common are: 1. The substances that are present initially (the reactants) disappear. 2. One or several new substances (the products) appear as the reaction proceeds and the reactants disappear.

The properties of the products are recognizably different from those of the reactants. 3. Energy in the form of heat, light or electricity is released or absorbed in the course of the chemical change.

2.4.3 the elements Man has always shown a desire to provide simple explanations for complex events. The earliest writings and legends reveal a preoccupation with a uniform, underlying reality. Thus the Greek philosopher, Empedocles, proclaimed in the fifth century B.C. that the universe is composed of a multitude of combinations of the four “elements” earth, air, fire and water. A century earlier, Thales, another Greek, had pro posed the theory that there is only one element, water, and that the multiplicity of the universe consists sim ply of various other substances, championed as being the ultimate building blocks of the universe, but none was universally accepted. All the various elements that were espoused had one serious shortcoming: the experiments that the various theories suggested did not lead to the developments that the various theories suggested or to the development of a fruitful science of chemistry. It was not until 1661 that the British scientist, Robert Boyle, in his book the Sceptical Chemist, proposed a definition of an element that has withstood the test of time and experiment and with slight modification, is still in use today. In Boyle’s own words: “And to prevent mistakes I must advertise you that I now mean by Elements…… certain primitive and simple, or perfecting unmingled bodies; which not being made of any other bodies, or of one another, are the ingradients of which all those called perfectly mixt bodies are immediately compounded, and into which they are ultimately resolved”. In other words, Boyle is saying that there are certain substances (Boyle specifies no number) that can be thought of as being of ultimate simplicity, that is, they cannot be broken down into yet simpler substance, and it is these that, in various combinations, make up all of the complex substance (mixt bodies) of the universe. Note that Boyle gives us a test, or experiment, the results of which can be used to determine whether or not a given substance is an element. If the substance cannot be broken down into yet simpler substance by any means known to the chemist, it must be classified as an element.

Basic Concepts of Chemistry 2.5

The reader will note that Boyle’s definition is an operational definition since a substance is classified as an element solely on the basis of experimental operations.

2.4.4 compounds and mixtures Let us now consider Boyle’s “mixt bodies” or compounds as we shall call them. A compound is a pure substance composed of two or more elements joined in chemical combination in a definite proportion by weight. Thus a compound is not merely a mixture of elements. As a result of the reaction, the original properties of the elements have been lost and replaced by an entirely new set of properties that are characteristic of the compound. While reacting to form a compound, the elements have entered into a state of firm chemical combination, they are held together in the compound by chemical bonds that are so strong that the compound remains intact even when it is subjected to purification and separation procedures. Let us consider a few examples of compound in the light of this definition. Water consists of 11.2% by weight of hydrogen and 88.8% by weight of oxygen in chemical combination. That is, hydrogen has reacted with oxygen to produce the new substance water. Carbon dioxide is a compound consisting of 27.3% by weight of carbon and 72.7% by weight of oxygen in chemical combination. Sodium chloride is a compound consisting of 39.3% sodium and 60.7% chloride in chemical combination. Sulphuric acid is a compound consisting of 32.70% sulphur, 65.26% oxygen and 2.04% hydrogen, in chemical combination. The fact that the combination of a compound has definite and fixed value gives us a criterion for distinguishing of a mixtures is not fixed but can be varied arbitrarily. In other words, we can vary the percentage composition in any way we like and still have a product that is correctly labelled of the contents present in it with their percentages. In all chemical reactions, the elements unite to form only one compound. When more than one compound can be formed, the situation is naturally more complicated, but the composition of each of these compounds is still fixed and definite.

2.4.5 difference between physical and chemical change You may wonder how a given change is known to be a chemical change in the sense already discussed, rather than a physical change. In a physical change, a sub stance is heated, or cooked, or compressed, or other wise exposed to some new external conditions without, however, losing its chemical identity. When the original

external conditions are restored, the substance regains its original physical properties. Water is still the same chemical substance, “water” whether in the liquid, solid or gaseous state, the main difference is the temperature at which these states are stable. On the other hand, in a chemical change the initial substances disappear and new substances with entirely different physical and chemical properties appear in its place.

2.5. propertieS of matter and their meaSurement Every substance has its own characteristic properties. They are categorized into two types: (i) physical properties and (ii) chemical properties. Physical properties are those which can be measured or observed without changing the identity or the composition of the substance. Properties like colour, odour, melting point, density etc include the physical properties. Chemical properties on the other hand are the properties whose measurement or observation require a chemical change to occur. The examples of chemical properties are characteristic reactions of different substances such as acidity, basicity, combustibility etc.

2.5.1 physical measurements Chemists characterise and identify substances by their particular properties. To determine many of these properties requires physical measurements. In a modern chemical laboratory, measurements often are complex, but many experiments begin with simple measurements of mass, volume, time and so forth. Measurement is the comparison of a physical quantity to be measured with a unit of measurement – that is, with a fixed standard measurement. On a centimetre scale, the centimetre unit is the standard comparison. A steel rod that measures 9.12 times the centimetre unit has a length of 9.12 centimetres. To record the measurement, one must be careful to give both the measured number (9.12) and the unit (centimetres).

2.5.2.the international System of units (Si units) The earlier measurements were probably based on the human body (the length in foot, etc). Later, fixed stand ards were developed, but these varied from place to place. Each country or government (and often each trade) adopted its own units. As science become more quan titative in the 17th and 18th centuries scientists found the lack of standard units was a problem. They began to seek a simple, international system of measurement.

2.6

Basic Concepts of Chemistry

Iron rod

1

2

3

4

5

6

7

8

9

10

11

12

Centimetres fig 2.1 Precision measurement with centimetre ruler

In 1971, a study committee of the French academy of science devised such a system, called metric system. It became the official system of measurement for France and was soon used by scientists throughout the world.

2.5.3 S1 Base Units and S1 Prefixes The international system of units (in French Le systeme international d’ units – abbreviated as SI) was established by the 11th general conference on weights and measurement (CGPM from conference general des poids at measures). The CGPM is an inter-governmental treaty organisation created by a diplomatic treaty known as Metre Convention, which was signed in Paris in 1875. In 1960, the general conference of weights and measures adopted the international system of units. This system has seven base units, SI units from which all others can be derived. Table 2.1 lists these base units and the symbols used to represent them. One advantage of any metric system is that it is decimal system. In SI, a larger or smaller unit for physical quantity is indicated by SI prefix, which is a prefix used in the international system to indicate a power of ten. For example, the base unit and length in SI is the metre and 10–2 metre is called a centimetre. Thus 2.54 centimetres equals 2.54 × 10–2 metre. The SI prefixes are given in Table 2.2. table 2.1 SI base units Quantity

Unit

Symbol

Length Mass Time Temperature Amount of substance Electric Current Luminous intensity

Metre Kilogram Second Kelvin Mole Ampere Candela

m Kg s k mol A cd

table 2.2 SI prefixes Multiple

Prefix

Symbol

1018 1015 1012 109 106 103 102 10 10–1 10–2 10–3 10–6 10–9 10–12 10–15 10–18

exa peta tera giga mega kilo hecto deka deci centi milli micro nano pico femto atto

E P T G M k h da d c m m n p f a

Length: The metre (m) is SI base unit of length. By combining it with one of the SI prefixes, one can get a unit of appropriate size for any length measurement. For the very small lengths used in chemistry, the nano metre (nm: 10–12m) is an acceptable SI unit. A non-SI unit of length traditionally used by chemists is the angstrom (Å), which equals 10–10 m. The metre was originally defined in terms of standard platinum iridium bar kept at sèvres France. In 1983, the metre was defined as the distance travelled by light in a vacuum in 1/299, 792.458 seconds. Mass and Weight: In unsophisticated discussions, the terms “weight” and “mass” are often used inter-changeably, but it should be noted that, in reality, these terms mean

Basic Concepts of Chemistry 2.7

different things. The mass of an object is a measure of the amount of matter it contains, the weight is a measure of the force with which the object is attracted to the earth. The weight is always proportional to the mass and the units have been chosen so that a one gram mass has a weight of one gram at sea level. Since the same names are used for the units of mass and of weight, and since the quantities are numerically equal (at sea level) they are often used interchangeably. It should be noted, however, that an object is moved away from the earth the gravitational attraction and therefore the weight diminishes. The mass however remains constant. The mass of a substance can be determined very accurately in the laboratory by using an analytical balance (Fig. 2.2). The kilogram is the SI base unit of mass. This is an unusual base unit in that it contains a prefix. In form ing other SI mass units, prefixes are added to the word gram (g) to give a unit such as milligram (mg; 1 mg = 10–3 g) The present standard of mass is the platinum iridium kilogram mass kept at the International Bureau of Weights and Measures in Sèvers France. Time: The second (s) is the SI base unit of time combining this unit with prefixes such as milli-, micro-, nano-, and pico-, one can create units appropriate for measuring

very rapid events. For example, the time it takes to add two 10 - digit numbers on a modern high-speed computer is roughly a nano second. The time required for the fastest chemical processes is picoseconds. When you measure time much longer than a few hundred seconds, you revert to minutes and hours, an obvious exception to the prefix base format of the international system. Atomic clock is the seventh generation of atomic clocks unveiled in 1993 at the national institute of standards and technology (NIST) in Boulder, Colorado. Circuitry in the clock counts the oscillations in the microwave radiation absorbed by caesium atoms. After 9,192,631,770 oscillations the clock registers the passage of one second. This clock is so precise that it loses or gains no more than one bil lionth of a second per day (one second in more than 3 million years).

Temperature: Temperature is difficult to define precisely but we all have an initiative idea of what we mean by it. It is a measure of “hotness”. A hot object placed next to a cold one becomes less hot, while the cold object becomes hotter. Energy passes from a hot object to a cold one, and the quantity of heat passed between the objects depends on the difference in temperature between the two. Therefore, temperature and heat are different, but related concepts.

Rider

Beam

Pan

Pointer and scale

Pan release Levelling scrow

Beam release fig 2.2 Analytical balance

2.8

Basic Concepts of Chemistry

In scientific work, the Centigrade scale of temperature is used on this scale, the melting point of ice is defined as 0° and the boiling point of water at a pressure of 1 atm as 100°. The interval between melting and boiling points is therefore divided into one hundred equal units, each called one degree Centigrade (1°C). In practice, temperatures are measured by means of a substance, some convenient property of which varies with temperature. The most commonly used thermometre is made of liquid mercury sealed in a glass tube. An increase in temperature causes an elongation of the mercury column. The familiar household thermometre employs the Fahrenheit scale. On this scale, ice melts at 32° and water boils at 212°. Normal body temperature is 98.6° on the Fahrenheit scale or 37.0° on the Centigrade scale. The SI base unit of temperature is the Kelvin (k), a unit on an absolute temperature scale. On any absolute scale, the lowest temperature that can be attained theoretically is zero. The Celsius and the Kelvin scales have equal size unit but 0°C is equivalent to 273.15 K. Thus it

373 K

100°C

310 K

37°C

293 K

25°C

273 K

0°C

Kelvin

is easy to convert from one scale to the other, using the formula K = °C + 273.15 A temperature of 25°C (about room temperature) equals to 298 k. Note: The degree sign (°) is not used with the Kelvin scale and the unit of temperature is simply the Kelvin. The Fahrenheit scale is at present the common temperature scale in the United States. Fig 2.3 compares the Kelvin, Celsius and Fahrenheit scales. As the figure shows 0°C is the same as 32°F (both exact) and 100°C corresponds to 212°F (both exact). Therefore there are 212−32 = 180 Fahrenheit degrees in the range of 100 Celsius degrees. That is there are 1.8 Fahrenheit degrees for every Celsius degree. Knowing this that 0°C equals to 32°F, we can derive a formula to convert degree Celsius to degree Fahrenheit. °F = (1.8 × °C) + 32 By rearranging this, we can obtain formula for converting Degrees Fahrenheit to Celsius.

Celsius

°C =

Boiling point of water

Human body temperature Room temperature

Freezing point of water

212°F

98.6°F 77°F

32°F

Fahrenheit

fig 2.3 Thermometres with different temperature scales

°F − 32 1.8

Basic Concepts of Chemistry 2.9

2.5.4 derived units Once base units have been defined, for a system of measurement, you can derive other units from them. One can do this by using the base units in equations that defined other physical quantities. For example, area is defined as length times length. Therefore SI unit of area = (SI unit of length) x (SI unit of length). From this, one can see that the SI unit of area is metre x metre or m2. Similarly, speed is defined as the rate of change of distance with time, that is speed = distance/ time. So SI unit of distance SI unit of speed = SI unit of time

The SI unit of speed is metres per second (that is, metres divided by second). The unit is symbolised m/s or m.s.–1. The unit of speed is an example of an SI derived unit, which is derived by combining SI base units. Some derived units are given in Table 2.3. Volume: Volume is defined as length cubed and has the SI unit of cubic metre (m3). This is too large a unit for normal laboratory work, so we use either cubic decimetres (dm3) or cubic centimetres (cm3, also written cc). Traditionally, chemists have used the litre (L) (approximately one quart). In fact, most laboratory glassware is calibrated in litres or millilitres (1000 mL = 1 L) because 1 dm equals 10 cm a cubic decimetre, or one litre, equals (10 cm)3 = 1000 cm3. Therefore, a millilitre equals a cubic centimetre. In summary 1 L = 1 dm3 and 1 mL = cm3

table 2.3 Derived units Quantity

Definition of Quantity

SI unit

Area Volume Density Speed Accelaration Force Pressure Energy

Length squared Length cubed Mass per unit volume Distance travelled per unit Speed changed per unit time Mass times acceleration of object Force unit area Force times distance travelled

m2 m3 kg m–3 ms–1 ms–2 kg.m.s–2( = newton N) kg m–1 s–2( = Pascal, Pa) kg.m2s–2( = joule J)

Volume: 1000 cm3; 1000 mL; 1 dm3; 1 L.

1 cm 10 cm = 1 dm

Volume:1 cm3 1 mL 1 cm

fig 2.4 Different units to express the volumes

2.10

Basic Concepts of Chemistry

Density: The density of an object is its mass per unit volume. You can express this as m d= V Where d is the density, m is the mass, and V is the volume. Suppose an object has a mass of 15.0 g and volume

now equals 1 and no units are associated with it. Because it is always possible to multiply any quantity by 1 without changing that quantity. You can multiply the previous expressions for the volume by the factor 12/103 cm3 without changing the actual volume. You are changing only the way you express this volume.

15.0 g = 1.50 of 10.0 cm3. Substituting you find that d = 10.0 cm3 g cm–3. The density of the object is 1.50 g cm–3 (or 1.50 g/ 3 cm ). Density is characteristic of a substance and can be used for identifying it. The metals likely to be mixed with gold, such as silver or copper, have lower densities than gold. Therefore an adultered (impure) gold can be expected to be far less dense than pure gold.

V = 5.003 cm3 × 103 cm3 = 125 × 10–3 L = 0.125 L

2.5.5 dimensional analysis In performing numerical calculations with physical quantities, it is good practice to enter each quantity as a number with its associated unit. Both the number and the units are then carried through the indicated algebric operations. The advantages of this are two-fold. 1. The units for the answer will come out of the calculations. 2. If one makes an error in arranging factors in the calculation (for example, if one uses the wrong formula), this will became apparent because the final units will be nonsense. Often while calculating, there is a need to convert units from one system to other. The method used to accomplish this is called factor label method or unit factor method or dimensional analysis. It is the method of calculation in which one carries along the units for quantities. As an illustration, suppose one wants to find the volume V of a cube, given ‘s’ the length of the side of cube. Because V = s 3, if s = 5.00, one can find that V = (5.00 cm) 3 = 125 cm3. There is no guesswork about the unit of volume here, it is cubic centimetre (cm3). Suppose, however, if one wishes to express the volume in litres (L), a metric unit that equals 103 cubic centimetres one can write equality as 1 L = 103 cm3 If both sides of the equality are divided by the right hand quantity, we get 1L 103 cm3 = 3 3 =1 3 3 10 cm 10 cm

Observe that the units are treated in the same way as algebric quantities. Note too that the right-hand side

1L

The ratio 1 L/10 3 cm is called conversion factor’ because it is a factor equal to 1 that converts a quantity expressed in one unit to quantity expressed in another unit. Note that the numbers in this conversion factor are exact, because 1 L equals exactly 1000 cm 3. Such exact conversion factors do not affect the number of significant figures (explained later) in an arithmetic result. In the previous calculation, the quantity 5.00 cm (the measured length of the side) does determine or limit the number of significant figures. The conversion-factor method can be used to convert any unit to another unit provided a conversion equation exists between the two units. Relationships between certain US units and metric units are given in Table 2.4. table 2.4 Relationship of some US and metric units Length

Mass

Volume

1 in = 2.54 cm (exact) 1 yd = 0.9144 m (exact)

1 lb = 04536 kg 1 lb = 16 oz (exact) 1 oz = 28.35 g

1 qt = 0.9464 L 4 qt = 1 gal

1 mi = 1.609 km 1 mi = 5280 ft (exact)

One can use these to convert between US and metric units. Suppose you wish to convert 0.547 lb to grams. From table 1 lb (note that 1 lb = 0.4536 kg or 1 lb = 453.69) the conversion factor from pounds to grams is 453.6 g/lb. Therefore, 0.547 lb ×

453.6 g = 248 g. 1 lb

Solved problem 1 Nitrogen gas is the major component of air. A sample of nitrogen gas in a glass bulb weighs 243 mg. What is this mass in SI base units of mass (kilograms)? Solution: Since 1 mg = 10–3 g 243 mg ×

10 −3 g = 2.43 × 10–1 g 1 mg

Basic Concepts of Chemistry 2.11

1 kg = 103 g 1 kg 2.43 × 10–1g × 3 = 2.43 × 10–4 kg 10 g It can be written in single step 243 mg ×

10 −3 g 1 Kg = 2.43 × 10–4 kg × 1 mg 103 g

Solved problem 2 The oceans contain approximately 1.35 × 109 km3 of water. What is this volume in oceans? Solution: 3 3  103 m   1 dm  9 3 × 1.35 × 10 km ×     1 km   10 −1 m  = 1.35 × 1021 dm3. ∴ 1 cubic decimetre is equal to a litre, the volume of the ocean is 1.35 × 1021 L. Solved problem 3 How many centimetres are there in 6.51 miles? Use the exact definitions 1mi = 5280ft : 1ft = 12 in, and 1 in = 2.54 cm. Solution: 1=

5280 ft 12 in 2.54 cm 1= 1= 1 mi 1 ft 1 in

rod as being somewhere between 9.11 cm and 9.13 cm. Thus you record the length of the rod as being somewhere between 9.11 cm and 9.13 cm. The spread of values indicates the precision with which a measurement can be made by this centimetre ruler. Measurements that are of high precision are usually accurate. It is possible, however, to have a systematic error in a measurement. Suppose that, in calibrating a ruler, the first centimetre is made too small by 0.1 cm. Then although the measurement of length on this ruler is still precise to 0.01 cm, it is accurate to only 0.1 cm This indicates the precision of a measured number (or result of calculations on measured numbers) and we often use the concept of significant figures. Significant figures are those digits in a measured number (or results of a calculation with measured numbers). That includes all certain digits plus a final one having some uncertainty. When you measured the rod, you obtained the values 9.12 cm, 9.11 cm and 9.13 cm. You could report the result as the average, 9.12 cm. The first two digits (9.1) are certain, the next digit (2) is estimated, so it has some uncertainty. It would be incorrect to write 9.120 cm for the length of the rod. This would say that the last digit (0) has some uncertainty but that the other digits (9.12) are certain which is not true.

Then 6.51 mi ×

5280 ft 12 in 2.54 cm × × 1 mi 1 ft 1 in = 1.05 × 106 cm

2.6 Significant figureS If you repeat a particular measurement, the result may not be obtained precisely, because each measurement is subject to experimental error. The measured values vary slightly from one another. Suppose you perform a series of identical measurement of quantity. The term precision refers to the closeness of the set of values obtained from identical measurements of a quantity. Accuracy is a related term; it refers to the closeness of a single measurement to its true value. To illustrate the idea of precision, consider a simple measuring device, the centimetre ruler. In Fig. 2.1 a steel rod has been placed near a ruler subdivided into tenths of a centimetre. You can see that the rod measures just over 9.1 cm. With care, it is possible to estimate by eye to hundredths of a centimetre. Here you might give the measurement at 9.12 cm. Suppose you measure the length of this

2.6.1 Number of Significant Figures Number of significant figures refers to the number of digits reported for the value of a measured or calculated quantity, indicating the precision of the value. Thus, there are three significant figures in 9.12 cm, whereas 9.123 has four. To count the number of significant figures in a given measured quantity, you observe the following rules. 1. All digits are significant except zeroes at the beginning of the number and possibly terminal zeroes (one or more zeroes at the end of a number) Thus, 9.12 cm, 0.912 cm, and 0.00912 cm all contain three significant figures. 2. Terminal zeroes ending at the right of the decimal point are significant. Each of the following has three significant figures: 9.00 cm, 9.10 cm, 90.0 cm. 3. Terminal Zeroes in a number without an explicit decimal point may or may not be significant. If someone gives a measurement as 900 cm, you do not know whether one, two, or three significant figures are intended. If the persons writes 900.0 cm (note

2.12

Basic Concepts of Chemistry

the decimal point), so the zeroes are significant. More generally, you can remove any uncertainty in cases by expressing the measurement in scientific notation. Scientific Notation is the representation of a number in the form A × 10 n where A is a number with a single none zero digit to the left of the decimal point and n is an integer or whole number. In scientific notation, the measurement 900 cm precise to two significant figures is written 9.0 × 102 cm. If precise to three significant figures, it is written as 9.00 × 102 cm. Scientific notation is also convenient for expressing very large or very small quantities. It is much easier (and simplifies calculations) to write the speed of light as 3.00 × 108 (rather than as 300,000,000 metres per second)

2.6.2 Significant Figures in Calculations Measurements are often used in calculations. How do you determine the correct number of significant figures to report for the answer to a calculation? The following are two rules that we use: 1. Multiplication and division: When multiplying and dividing measured quantities, give as many significant figures in the answer as there are in the measurement with the least number of significant figures. 2. Addition and subtraction: When adding or subtracting measured quantities, give the same number of decimal places in the answer as there are in the measurement with the least number of decimal places. Let us see how you apply these rules. Suppose you have a substance believed to be cisplatin and in an ef fort to establish its density, you measure its solubility (the amount that dissolves in a given quantity of water) you find that 0.0634 gram of the substance dissolves in 25.31 grams of water. The amount dissolving in 100.0 grams is 100.0 grams of water ×

0.0634 gram cisplatin 25.31 grams of water

The value 0.2504938 for the numerical part of the answer (100.0 × 0.0634 ÷ 25.31) It would be incorrect to give this number as the final answer. The measurement 0.0634 gram has the least number of significant figures (three). Therefore, report the answer to three significant figures – that is 0.250 gram. Now consider the addition of 184.2 grams and 2.324 grams. The resultant value is 186.524. But because the quantity 184.2 grams has the least number of decimal places one, whereas 2.324 grams has three – the answer is 186.5 grams.

2.6. 3 exact numbers So far we have discussed only numbers that involves uncertainties. However, you will also encounter exact numbers. An exact number is a number that arises when you count items or some times when you define a unit. For example, there are 9 coins in a bottle i.e., exactly 9 but not 8.9 or 9.1. Also when you say there are 12 inches in a foot, you mean exactly 12. Similarly, the inch is defined to be exactly 2.54 centimetres. The conventions of significant figures do not apply to exact numbers. Thus the 2.54 in the expression “1 inch equals 2.54 centimetres” should not be interpreted as a measured number with three significant figures. In effect 2.54 has an infinite number of significant figures, but of course it would be impossible to write out an infinite number of digits. You should make a mental note of any numbers in is in a calculation that is exact, because they have no effect on the number of significant figures in a calculation. The number of significant figures in a calculation result depends only on the numbers of significant figures in quantities having uncertainties. For example, suppose you want the total mass of 9 coins when each coin has a mass of 3.0 grams. The calculation is 3.0 grams ω 9 = 27 grams. You report answer to significant figures because 3.0 grams has two significant figures. The number 9 is exact and does not determine the number of significant figures Rounding In reporting the solubility of a substance in 100 grams of water as 0.250 gram, the resultant value is rounded in the example discussed in significant figures in calculations. Rounding is the procedure of dropping non-significant digits in a calculation result and adjusting the last digit reported. The general procedure is as follows: 1. If the digit is 5 or greater, add 1 to the last digit to be retained and drop all digits farther to the right. Thus rounding 1.2151 to three significant figures gives 1.22. 2. If the digit is less than 5, simply drop it and all digits farther to the right and rounding 1.2143 to three significant figures gives 1.21. In doing a calculation of two or more steps, it is desirable to retain non-significant digits for intermediate answers. This ensures that accumulated small errors from rounding do not appear in the final result. If you use a calculator, you can simply enter numbers one after the other, performing each arithmatic operation and rounding only the final answer. To keep track of the correct number of significant figures, you may want to record intermediate answers with a line under the last significant figure, as shown in the solved problem 4 in part (d).

Basic Concepts of Chemistry 2.13

(b) 5.41−0.398

system of nomenclature had developed. What we know as oxygen, for example, was called ‘de phlogisticated air. The simpler nomenclature of today was devised by Lavoisier. During the quantitative work of the second half of the 18th century, many interesting facts had come to light which led to the development of laws of chemical combinations.

(d) 4.18−58.16 × (3.38−3.01)

2.7.1 law of conservation of mass

Solved problem 4 Calculate and round the answers to the correct number of significant figures (units of measurement have been omitted). 2.568 × 5.8 4.186 (c) 3.38−3.01

(a)

Solution: (a) The factor 5.8 has the lowest significant figures; therefore the answer should be reported to two significant figures. Round the answer to 3.6 (b) The number with the least number of decimal places is 5.41 therefore, round the answer to two decimal places, to 5.01 (c) The answer is 0.37 note that one significant figure is lost on subtraction (d) First do the subtraction within parentheses, underlining the least significant digits. 4.18−58.16 × (3.38−3.01) = 4.18−58.16 × 0.37 Following the convention, do the multiplication before the subtraction. 4.18−58.16 × 0.37 = 4.18 – 21.591 = –17.3392 The final answer is –17.

2.7 lawS of chemical combinationS Upto the 18th century, chemistry was primarily qualitative, and changes in weight were regarded as of little importance. It was not until the latter half of the 18th century as a result of the work of Black (on carbonates and oxides), Cavendish (on the composition of water) and Lavoisier’s quantitative work began to assume its essential place in the study of science. Priestley’s preparation of oxygen in 1774 and the subsequent use of this discovery by Lavoisier in his famous experiments on the combustion of mercury, completed the overthrow of erroneous phlogiston theory and gave us our present-day explanation of combustion (combustion was supposed to consist in the escape of an ‘element’ or ‘principle’ to which the name of phlogiston was given. When a metal burned in air, phlogiston escaped into air to leave the metal calx (i.e., oxide). Combustion continued until the air was ‘saturated’ with phlogiston, thus

By weighing substances before and after chemical change, Antoine Lavoisier (1743−1794), a French chemist, demonstrated the law of conservation of mass, which states that The total mass remains constant during a chemical change (chemical reaction). In a series of experiments, Laviosier applied the law of conservation of mass to clarify the phenomenon of burning, or combustion. He showed that when a material burns, a component of air (which he called oxygen) combines’ chemically with the material. For example, when the liquid metal mercury is heated in air, it burns or combines with oxygen to give a red-orange substance, whose modern name is mercury (II) oxide. We can represent the chemical change as follows: Mercury + Oxygen → Mercury (II) oxide The principle of conservation of mass or indestructibility of matter that in a chemical change, matter is neither created nor destroyed is the basis of all our chemistry. It makes possible, for example, our shorthand method of representing chemical reactions by means of the equations of the mathematician. Its discovery stands out as one of the greatest achievements of the 18th century.

2.7.2 Law of Definite Proportions Joseph Proust, a French chemist, proposed this law. He stated that Whenever two or more elements combine to form a given compound, they do so in a definite proportion by weight Another way of stating this law is Whenever two samples of a pure substance are identical in all physical properties, their chemical composition is also same. Proust determined the composition of two samples of cupric carbonate of one of which was natural origin and the other was a synthetic one.

Metal – phlogiston = calx Because in some cases, the calx could be reconverted to the metal by heating with charcoal, it was suggested that charcoal was very rich in phlogiston, but there were other important results. Under the phlogistonists, a complicated

Natural sample Synthetic sample

% of Cu

% of O2

% of C

51.35 51.35

9.74 9.74

38.91 38.91

2.14

Basic Concepts of Chemistry

This law, sometimes referred to as the law of definite proportions or the law of definite composition, is true regardless of the source of the samples, which may be obtained from a natural source or be prepared by a chemist in his laboratory, and regardless of the method of analysis that is used. Thus, the chemical composition of a given pure substance is just as definite and characteristic property of the substance as any of the physical properties, and it may be used in the same way to identify the substance. The French man Berthelot attacked the view of constancy of composition, maintaining that the affinity of attraction between elements affected their combining proportions and that variable rather than constant composition was the rule. Oxides of certain metals were cited as examples and we now know that some of these were actually different compounds. There was too much confusion as to the exact nature of a compound and solutions, and alloys were regarded by some chemists as compounds. In 1801, a controversy on two subjects started in France and continued until 1807. Berthelot was opposed by his country man Proust, who was extremely careful to prepare specimens of a high degree of purity. Proust emphasised compounds. In his work on oxides, incidentally, he was the first to explain the constitution of hydroxides which had a different proportion of oxygen than is present in the normal compound. While Proust may be regarded as the discoverer of the law of constant composition, other factors than his analyses contributed largely to the general acceptance of the law.

2.7.3 law of equivalents or law of reciprocal proportions The Swedish chemist Berzelius published (in 1812) a series of results on the investigations of the combining relations of several metals with the non-metals oxygen and sulphur. This work and that of Richter led to the law of equivalents or reciprocal proportions, which for elements may be stated as follows: The weights of two elements A and B which combine with or replace a fixed weight of a third element C, are those same weights of A and B which combine with or replace each other. To take a case in illustration: 12 units of weight of magnesium and 1 unit of hydrogen combine respectively with 8 units of oxygen; and 12 units of magnesium displace 1 unit of hydrogen from, for example, a dilute acid.

2.7.4 dalton’s atomic theory Although the origin of idea that matter is composed of small indivisible particles called ‘a – tomio’ (meaning – indivisible), dates back to the time of Democritus, a Greek

philosopher (460-370 BC). It again started emerging as a result of several studies which led to laws mentioned above. In 1808, John Dalton published a new system of chemical of philosophy in which he proposed the following: 1. Elements consist of extremely small particles called atoms which remain indivisible during a chemical change 2. Atoms of the same element are alike and have the same mass 3. There are thus as many different kinds of atoms as kinds of elements 4. Chemical combinations consist in the union between atoms to form groups of atoms (which later were called molecules) Note: Points (2) and (3) are now known to be unique because of the existence of isotopes. John Dalton was born in England in 1766. His family was poor and his formal education stopped when he was 11 years old. He became a school teacher. He was colour blind, his appearance and manners were awkward, he spoke with difficulty in public. As an experimenter he was clumsy and slow; he had few, if any, outward marks of genius. Dalton’s atomic theory was presented in 1808. It ranks among the greatest of all monuments to human intelligence. At that time John Dalton was secretary of the Philosophical Society of Manchester. He was interested in meteorology from an early age, and this probably led him to a consideration of the physical constitution of gases. It was this interest perhaps rather than the quantitative work of the continent (of much of which he may have been ignorant) that led him to the formulation of his theory. His interest in chemical combination came from his analysis of the two hydrocarbons, methane and ethylene, which showed that in these two compounds the respective weights of carbon combining with a given weight of hydrogen were in the proportion of 1:2. He found a similar regularity among the oxides of nitrogen. Dalton thus discovered (in 1804) the law of multiple proportions. It was no doubt this law (not published until 1808) led Dalton to his conception of the atom as the unit of chemical combination. His theory was first published in 1808, but the substance of it had already been given previously in lectures. Dalton’s atomic theory gave an explanation to the law of constant composition which settled not only the French controversy, but also the explanation of the law of multiple proportions, which Dalton published later. He had withheld previous publication for until the law of constant composition was accepted, as there was little chance that the law of multiple proportions would be recognized.

Basic Concepts of Chemistry 2.15

2.7.5 law of multiple proportions The law of multiple proportions states that when two elements, A and B, combine to form two different compounds, the weight of B in the two compounds combined with a fixed weight of A, are in the ratio of small whole numbers, see table 2.5. The data in this table is based on the work of J. J. Berzelius; Within limits that can reasonably be ascribed to experimental error, the weights of the second element in combination with 100 g of the first element are always in the ratio of small whole numbers. Berzelius was a magnificent experimentalist. laws,theories and hypotheses If a process or object can be perceived directly by the human senses, a statement about it that has been verified by these senses is called a fact. Consequently, verified observations of physical or chemical properties, or macroscopic processes involving matter are called facts. A natural law is a generalization based upon a large body of facts related to one phenomenon and process. It make the form of a concise statement, or it may be a mathematical equation. In either case, it summaries all previous observations and predicts that, under the same circumstances, the same observations will be made in the future. For example, the law of definite composition is the simple statement that any two samples of the same pure substance always contain the same elements in the same proportion by weight, the composition of substances and it predicts that the

same sort of behaviour will be found in all future investigations. However, since scientists are frequently redetermining the composition of known substances and are preparing new substances, some of which may be radically different from those known today, it is not impossible that some day something will turn up that is not consistent with this law. If this happens, the law will have to be modified, or restated, or perhaps even abandoned. Thus, the very quality of prediction which makes a law useful to the scientist also makes it vulnerable, for a single inconsistent observation may necessitate revision. Laws are formulated in the belief that there is a fundamental orderliness to nature so that, under the same set of circumstances, the same results will be observed again and again. Even when an observation is inconsistent with an existing law, the scientist is not likely to lose his faith in the ‘lawfullness” of Nature. He will search for a new, more general law that is consistent also with the inconsistent observation, or he will restate the law so that it no longer applies to this observation. For example, there was a time when the law of conservation of mass was thought to apply to all the changes that matter undergoes. Then it was found that this law fails when it is applied to nuclear reactions. As a result a more general law of conservation of mass-energy has replaced it; the new law applied to both chemical and nuclear reactions. And the law of conservation of mass has been restated so that it applies to both chemical reactions, for which it is still valid. As has been ported out, scientists are not satisfied only to observe facts and formulate laws, but they

table 2.5 Examples of multiple proportions Ratio

Nearest ratio of small whole numbers

Brown Oxide Yellow Oxide

Lead and Oxygen 100 g : 15.6 g 100 g : 7.8 g

2.00 : 1

2:1

Sulphur Trioxide Sulphur Dioxide

Sulphur and Oxygen 100 g : 146.427 g 100 g : 97.83 g

1.497 : 1

3:2

Black Oxide Red Oxide

Copper and Oxygen 100 g : 25 g 100 g : 12.5 g

2.00 : 1

2:1

Ferric Oxide Ferrous Oxide

Iron and Oxygen 100 g : 44.25 g 100 g : 29.6 g

1.495 : 1

3:2

2.16

Basic Concepts of Chemistry

try also to explain the facts and the laws. Because they are unable to perceive the innermost workings of matter directly, they invent possible explanations involving sub-microscopic particles and events. These explanations by means of sub-microscopic models are caused theories. To be acceptable, a theory must be able to explain a substantial body of facts; it will not do to invent a special theory to explain each individual fact. Furthermore a theory must be clear and simple, so that when it is applied to a new situation it will readily lead to an unambiguous prediction; it is not satisfactory to populate the sub-microscopic world with unpredictable figures like demons or elves those behaviour in a new situation is anybody’s guess. A theory in chemistry can never become a law because one can never hope to observe directly the particles and processes which the theory postulates. For example, even though there is now a tremendous amount of evidence in support of the atomic theory of matter, the theory can never become a law simply because we can never hope to perceive atoms directly through the human senses. Sometimes we makes guess about the behaviour of matter, which are later capable of being verified by direct observations. For such guesses we have reserved the term hypothesis. When verified, a hypothesis may become either a fact or a law. For example, a speculation about the chemical and physical properties of a yet unprepared compound is a hypothesis. After the compound has been prepared, if its measured properties agree with our expectation, the hypothesis becomes a fact. Bertrand Russell has given an illuminating account of the stages in the gradual maturing of such a hypothesis into an accepted theory. “Every physical theory which survives goes through three stages. In the first stage, it is a matter of controversy among specialists; in the second stage, the specialists are agreed that it is the theory which best fits the available evidence, though it may well hereafter be found incompatible with new evidence; in the third stage it is thought very unlikely that any new evidence will do more than somewhat modify it.”

2.7.6 gay-lussac's is law of Volumes This was the law as proposed by Gay-Lussac in 1808. The law states that When gases unite to form new compounds, the volumes of the reactants and those of the gaseous products are in the ratio of small whole numbers.

One of the reactions Gay-Lussac studied was the combination of hydrogen with oxygen to form steam. Since these relationships apply to gaseous reactions, it is obvious that the reaction must be carried out at a temperature greater than 100°C to prevent the steam from condensing to form liquid water. Gay-Lussac found that when the reaction take place, two volumes of hydrogen gas (at 100°C and 1 atm pressure) reacted with one volume of oxygen gas (at 100°C and 1 atm pressure) to produce two volumes of steam (at 100°C and 1 atm pressure). Similarly, two volumes of carbon monoxide gas reacted with one volume of oxygen gas to produce two volumes of carbon dioxide gas.

+ Hydrogen gas 2 volumes

Oxygen gas 1 volume

Steam 2 volumes

+ Carbon monoxide 2 volumes

Oxygen gas 1 volume

Carbon dioxide 2 volumes

Gay-Lussac’s discovery of integer ratio in volume relationship is actually the law of definite proportion by volume. The law of definite proportions, stated earlier, was with respect to mass. Gay-Lussac’s law was explained properly by the work of Avogadro in 1811.

2.7.7 avogadro’s hypothesis Another chemist, an Italian named Amedeo Avogadro, saw in the law of combining volumes an important clue to the molecular formulae of the reactants and products. He suggested that the problem could be solved by making the following two assumptions: 1. Equal volumes of all gases under the same conditions of temperature and pressure contain the same number of molecules. 2. Molecules of certain elements (oxygen, hydrogen, nitrogen, etc,.) are not indivisible as had been supposed by Dalton but instead consist of two identical atoms that are easily separated when the element reacts to form a compound. Armed with these two assumptions it becomes easy to deduce, for example, the correct molecular formula of water. Recall that two volumes of hydrogen react with one volume of oxygen to form two volumes of steam. Therefore, according to Avogadro’s first assumption, twice as many hydrogen

Basic Concepts of Chemistry 2.17

molecules as oxygen molecules are required to form a given amount of water. Also the number of water molecules that are formed is equal to the number of hydrogen molecules that are used up because the volumes of steam and hydrogen are equal. To complete the determination of molecular formula of water, we must make use of Avogadro’s second postulate of each oxygen molecule produces two water molecules, then each water molecule must contain one oxygen atom. The same line of reasoning, when applied to the relation between the volumes of hydrogen and of water (2:2) leads to the conclusion that each water molecule contains two hydrogen atoms. Thus, the molecular formula of water must be H2O. Dalton objected vigorously to the second postulate as mentioned before. Dalton and others believed at that time oxygen and hydrogen containing two atoms did not exist. Avogadro’s proposal was published in the French Journal de Physidue. In spite of being correct, it did not gain much support. About 50 years later, in 1860, the first international conference on chemistry was held in Karlsruhe in Germany to resolve various ideas. At the meeting, Stanislao Cannizaro presented a sketch of a course of chemical philosophy which emphasized the importance of Avogadro’s work.

2.8 atomic weightS Recall that Dalton’s atomic theory was based mostly on weight relationships and that it assumed a characteristic weight for the atoms of each element. The assignment of atomic weights to the individual elements was a central problem of the theory and could not be solved uniquely until molecular formulae became known. With acceptance of Avogadro’s hypothesis, molecular formulae became available and therefore it became possible to develop an accurate scale of atomic weights. First let us see how a knowledge of the molecular formula and the composition of a compound enables us to calculate the relative weights of the constituent atoms. For example, the molecular formula of water has been found to be H2O, and the composition is 88.8 weight per cent (wt.%) of oxygen, and 11.2 weight per cent of hydrogen. The calculation then proceeds as follows: Wt % of hydrogen = Wt of hydrogen in 100 g of water = No. of hydrogen atom in 100 g of water × Wt. of 1 hydrogen atom (1) Wt % of oxygen = Wt of oxygen in 100 g of water = No. of oxygen atoms in 100 g of water × Wt of 1 oxygen atom (2)

Dividing (1) by (2) gives Wt % of hydrogen = Wt % of oxygen No. of hydrogen atoms in100 g in of water × No. of oxygen atoms in100 g of water Wt of 1 hydrogen atom Wt of 1 oxygen atom

(3)

The quantity Wt of 1 hydrogen atom Wt of 1 oxygen atom is precisely the ratio of atomic weights we are looking for. It can be calculated because the percentage composition is known, and because the quantity No. of hydrogen atoms in 100 g of water No. of oxygen atoms in 100 g of water Must equal the quantity No. of hydrogen atoms in 1 molecule No. of oxygen atoms in 1 molecule The latter is equal to 2/1 because the molecular formula is H2O. Thus 11.2 2 Wt of 1 hydrogen atom = × 88.8 1 Wt of 1 oxygen atom

(4)

Solving equation (4), We find the desired ratio of the weights of the atoms to be

11.2 , or very nearly 2(88.8)

1 16

. In

other words, the weight of one hydrogen atom is very nearly, one-sixteenth the weight of an oxygen atom.

2.8.1 the atomic weight Scale By a series of calculations of this type, involving compounds whose composition and molecular formula are known, it becomes possible to relate the weights of all the atoms to each other. Thus we are able to set up an atomic weight scale, that is a table that lists the weights of all the elements relative is some common standard. Until quite recently the standard universally chosen was the element oxygen. Oxygen was chosen for practical reasons because most of the other elements form compounds with it. The weights of the other elements were then compared to oxygen. In this comparison, oxygen is assigned the atomic weight of exactly 16. This assignment makes the atomic weight of the lightest element hydrogen, very nearly 1.0 and therefore avoids the use of atomic weights less than one.

2.18

Basic Concepts of Chemistry

In 1961, the atomic weight standard was changed by the International Union of Pure and Applied Chemists and the mass of carbon – 12 isotope is taken as the standard. Here carbon – 12 is one of the isotopes of carbon and represented as 12C. In this system, 12C is assigned a mass of exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard. One atomic mass unit is defined as a mass exactly equal to one-twelfth the mass of one carbon −12 atom And 1 amu = 1.66056 × 10–24 Mass of an atom of hydrogen = 1.6736 × 10

Thus, in terms of amu, the mass of hydrogen atom 1.6736 × 10−24 g = 1.0078 amu = 1.008 amu. 1.66056 × 10−24 g Thus, the mass of oxygen – 16 (16O) atom would now become 15.995 amu. Nowadays ‘amu’ has been replaced by ‘U’ which is known as unified mass. When we use atomic masses of elements in calculations, we actually use average atomic masses of elements.

=

average atomic mass Many naturally occurring elements consist of more than one isotope. So the atomic weights determined are the average atomic mass of that element. This can be illustrated by taking the carbon as an example in the following manner.

12

C C 14 C 13

2(12.01U)+ 4(1.008U) = 24.02U + 4.032U = 28.052U Similarly, molecular weight or molecular masses of other molecules can be calculated. formula masses

–24

Isotope

mass of ethane which contains two carbon atoms and six hydrogen atoms can be obtained as follows: Molecular mass of ethane molecule (C2H6) is

Relative Abundance %

Atomic mass (amu)

98.892 1.108 2 × 10–10

12 13.00335

14.00317

The average atomic mass of carbon is (0.98892)(12U) + (0.01108) (13.00335U) + (2 × 10–12 (14.00317U) = 12.011U In a similar manner, atomic masses of other elements can be calculated. The atomic masses of the elements given in the periodic table for different elements are their average atomic masses. molecular mass Molecular weight or molecular mass of a molecule of a substance is the sum of the atomic weights or atomic masses of atoms present in the molecules. For example, molecular

Substances like sodium chloride do not have exact independent molecules as their constituent units. In such cases, only the simple ratio of constituent ions is found and then a formula is written. Therefore, for such compounds we get formula masses but not molecular masses. For sodium chloride, the formula is NaCl and formula mass is 58.5U. mole concept The basic assumption in Dalton’s atomic theory is that when atoms combine to form molecules, they do so in the ratio of small whole numbers. In carrying out a reaction, it is therefore convenient to weigh the elements in such quantities that the numbers of atoms that are weighed are in the same ratio as in the final compound. Thus if we wish to prepare water, we would like to be able to weigh twice as many hydrogen atoms as oxygen atoms. It would obviously not do to weigh 2 g of hydrogen and 1 g of oxygen, since the weight of a hydrogen atom is not the same as that of an oxygen atom. To ensure that we are weighing the correct numbers of each, the atomic weights of the elements must be taken into consideration. It is therefore convenient to introduce the concept of the gram atomic weight. The gram atomic weight is that quantity of the element, taken in grams, that is numerically equal to the atomic weight of that element. The gram atomic weight is such a convenient unit because of the following reason: One gram atomic weight of any element contains exactly the same number of atoms as one gram atomic weight of any other element. A gram molecular weight (or mole) of a pure substance is defined in an analogous manner: One gram molecular weight of any substance is that weight of the substance, in grams, that is numerically equal to the molecular weight of the substance.

Basic Concepts of Chemistry 2.19

It can be shown that one gram molecular weight of any substance contains exactly the same number of molecules as one gram molecular weight of any other substance. The definition of gram molecular weight leads to another unit that is very useful in chemical calculations. This unit is the gram molecular volume.

It may be emphasized that the mole of a substance always contains the same number of entities, no matter what the substance may be. In order to determine this number precisely, the mass of a carbon – 12 atom was determined by a mass spectrometre and found to be equal to 1.992648 × 10–23 kg. Knowing that one mole of carbon weighs 12 g, the number of atoms in it is equal to

The gram molecular volume is the volume occupied by one gram molecular weight of the given gaseous element or compound under the standard conditions of 0°C and one atmospheric pressure (STP).

12 g mol−1 12 C = 6.0021367 × 1023 atom 1.992648 × 10−23 g 12 C atom −1 mol–1

Thus, one gram molecular volume of oxygen is the volume occupied by 32.00 g of oxygen gas at 0°C and one atmospheric pressure; this volume is found to be 22.4 litres (22,400 cc). Similarly, one gram molecular volume of carbon dioxide is the volume occupied by 44.0 g of carbon dioxide gas at 0°C and 1 atmospheric pressure, and this volume turns out also to be 22.4 litres. In fact, the gram molecular volume of all gaseous substance is the same, namely 22.4 litres. This should come as no surprise if we consider Avogadro’s first hypothesis, that equal volumes of gases under the same conditions contain an equal number of molecules. If we take equal numbers of carbon dioxide and oxygen molecules, the ratio of weights of the two samples must be the same as the ratio of their molecular weights. Conversely, if we take amounts of two gaseous substances in ratio to their molecular weights, their volumes must be equal. A further corollary exists if we take one gram atomic weight or molecular weight of any element or compound, whether solid, liquid or gaseous. It will contain a specific number of atoms or molecules, which will be same for all substances. This number is called Avogadro’s number. It is enormously important whenever we want to study the properties of individual molecules. There are several rather indirect methods for determining Avogadro’s number. All of these methods give concordant answers, namely 6.02 × 1023 atoms in any gram atomic weight, or molecules in any gram molecule weight. This number represents an unimaginably large number of particles. Thus if all the molecules in one gram molecular weight of water (18 g) were enlarged to the size of sand grains, they would form a pile of sand over one mile high and one mile on each side. In SI system, mole (symbol mol) was introduced as the seventh base quantity for the amount of a substance. One mole is the amount of a substance that contains as many particles or entities as there are atoms in exactly 12 g (or 0.012 kg) of the 12C isotope.

This number of entities in 1mol is named as Avogadro Constant denoted by NA in honour of Amedeo Avogadro. When one knows the actual number of molecules in a given sample of material, it becomes easy to calculate the actual weight of a single molecule of different substances. For example, if we take 18.0 g of water, which is one gram molecular weight, we have 6.02 × 1023 molecules. Therefore, one molecule will weigh 18 ÷ (6.02 × 1023) g which is equal to about 3 × 10–23 g. With this information one can also compute the dimensions of a molecule. For example, since 18.0 g of water containing 6.02 × 1023 molecules occupies 18.0 cubic centimetres, each molecule occupies approximately 18.0 ÷ (6.02 × 1023) cc, or 3 × 10–23 cc. Let us assume that each water molecule is a cube whose volume is equal to 3 × 10–23 cc. Then the length of the side of this cube, which is computed from the formula 3

Length =

volume

is found to be 3.1 × 10 cm. Thus, the length of a water molecule must be very close to 3.1 × 10–8 cm. It is a wonder that we are unable to see or weigh an individual molecule. –8

relationship between molecular mass and Vapour density The vapour density of any gas is the ratio of the densities of the gas and hydrogen under similar conditions of temperature and pressure. Vapour density (VD) =

=

Density of gas Denisty of hydrogen

Mass of certain volume of the gas Mass of the same volume off hydrogen at the same temp and pressure

If n molecules are present in the given volume of a gas and hydrogen under similar conditions of temperature and pressure.

2.20

Basic Concepts of Chemistry

VD =

=

=

Mass of n molecules of gas Mass of n molecules of hydrogen

Mass of 1 molecule of gas Mass of 1 molecule of hydrogen

Mol mass (since mol mass of hydrogen = 2) 2 Hence 2 × V.D = Mol mass

This formula can be used for the determination of molecular mass of volatile substances from vapour density mainly either by Victor Meyer method or by Dumas method. Solved problem 5 A piece of copper weights 0.635 g. How many atoms of copper does it contain? (C.E.E. Bihar, 1992) Solution: Gram atomic mass of copper = 63.5 g No. of moles of copper in 0.635 g =

0.635 = 0.01 63.5

No. of Cu atoms in 1 mole = 6.02 × 1023 No. of Cu atoms in 0.01 mole = 0.01 × 6.02 × 1023 = 6.02 × 1021 problems for practice 1. How many years it would take to spend one Avogadro’s number of rupees at a rate of 10 lakh of rupees in one second? 2. One atom of an element x weights 6.644 × 1023 g. Calculate the number of gram atoms in 40 kg of it. 3. From 200 mg CO2, 1021 molecules are removed. How many moles of CO2 are left? 4. Calculate the number of SO 2− 4 ions in 100 mL of 0.001 MH2SO4 solutions. 5. The density of helium at NTP is 0.1784 g/L. What is the weight of 1 mole of it? 6. Calculate the number of oxygen atoms and weight, in 50 g of CaCO3. 7. Calculate the total number of electrons present in 1.6 g of CH4. 8. Calculate the charge on 1 g ion of N3– in coulombs. 9. Equal masses of oxygen, hydrogen and methane are taken in a container in identical conditions. Find the ratio of the volumes of the gases.

10. A plant virus is found to consist of uniform cylindrical particles of 150 A° in diametre and 5000 A° long. The specific volume of the virus is 0.75 cm3/g. If the virus is considered to be a single particle, find its molecular weight. IIT (1999) 11. Vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27°C. Calculate the number of moles of NO2 in 100 g mixture. (MLNR 1993) 12. Calculate number of oxalic acid molecules in 100 mL of 0.02 N oxalic acid. (Roorke 1992) 13. Calculate the weight of iron which will be converted into its oxide by the action of 18 g of steam. (UPSEAT 1996) 14. Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO3). 15. An alloy has Fe, Co and Mo equal to 71%, 12% and 17%, respectively. How many cobalt atoms are there in a cylinder of radius 2.50 cm and a length of 10.0 cm? The density of alloy is 8.20 g/mL. Atomic weight of cobalt = 58.9. 16. How many grams of CaWO4 would contain the same mass of tungsten that is present in 569 g of FeWO4 (At wt of W = 184) 17. It has been estimated that 93% of all atoms in the entire universe are hydrogen and that vast majority of those remaining are helium. Based on only these two elements, estimates the mass percentage composition of the universe. 18. Molecular weight of hemoglobin is about 65,000 g/ mol. Hemoglobin contains 0.35% Fe by mass. How many iron atoms are there in hemoglobin molecule? 19. The density of mercury is 13.69 g/cc. Calculate approximately the diametre of an atom of mercury, assuming that each atom is occupying a cube of edge length equal to the diametre of the mercury atom. 20. A polystyrene, having the formula Br3C6H3(C3H8)n was prepared by heating styrene with tribromobenzoyl peroxide in the absence of air. If it was found to contain 10.46% bromine by weight, find the value of n. 21. If the components of air are N 2, 78%; O2, 21%; Ar, 0.9%, and CO2, 0.1% by volume, what would be the molecular weight of air? 22. At room temperature, the density of water is 1.0 g/L and the density of ethanol is 0.789 g/mL. What volume of ethanol contains the same number of molecules as are present in 175 mL of water?

Basic Concepts of Chemistry 2.21

2.9 percentage compoSition and formula From a knowledge of chemical composition by weight and the atomic weights of the elements involved, it is possible to find the formula of a chemical compound. This important class of problems will be illustrated by several examples. In 25.0 grams of calcium sulphide there is 13.9 grams of calcium and 11.1 grams sulphur. With this information and the atomic weights of calcium and sulphur, we shall proceed to find the formula of calcium sulphide. To do this it is necessary to find the ratio of gram – atoms of calcium to gram-atoms of sulphur The atomic weight of calcium is 40.1 (rounded off to the first decimal for ease in making the calculations). From this it follows that 40.1 grams Ca would be 1 gram-atom Ca. Hence 13.9 grams Ca would be

13.9 grams = 0.347 gram-atom Ca 40.1 grams gram-atom −1 Similarly, as the atomic weight of sulphur is 32.1, 11.1 grams of sulphur would be

11.1 grams = 0.347 gram-atom S 32.1 gram gram-atom −1 It appears, therefore, that in this particular sample of calcium sulphide 0.347 gram-atoms Ca is combined with 0.347 gram-atom of S. In other words, the ratio of gramatoms, and hence of atoms is 1:1 and the formula of calcium sulphide must be written as CaS. The arithmetical operations shown above may conveniently be summarized as follows.

Symbol Ca S

Parts by weight

Atomic weight

13.9 11.1

40.1 35.1

Gram-atoms 13.9/40.1 = 0.347 11.1/32.1 = 0.347

The ratio 0.347 to 0.347 is equal to the ratio 1:1, hence the formula is CaS. Now we shall find the formula by a slightly more complicated compound, sodium carbonate (washing soda). A 10 grams sample of sodium carbonate contains 4.34 grams of sodium, 1.13 grams of carbon, and 4.52 grams of oxygen. Proceeding as before

Symbol Na C O

Parts by weight 4.34 1.13 4.52

Atomic weight Gram-atoms 23.0 12.0 16.0

4.34/23.0 = 0.189 1.13/12.0 = 0.094 4.52/16.0 = 0.283

The ratio of numbers is 0.189 Na to 0.094 C to 0.283 O. It is not clear what the ratio would be in whole numbers, but these whole numbers are readily found by dividing each of the numbers by the smallest one of these three i.e., by 0.094. Then 0.189/0.094 = 2 gram-atom Na 0.094/0.094 = 1 gram-atom C 0.283/0.094 = 3 gram-atom O The formula of sodium carbonate is therefore Na2CO3. The reader will notice that 0.283 divided by 0.094 is not exactly 3, nearer 3.01. But 3.01 is nearer to 3. So we conclude that the small discrepancy is due to experimental error. One more example of formula finding will be given. This is for the compound Ferric (iron) oxide. This sub stance contains 70.1 per cent iron and 30.0 per cent oxy gen. Expression of the composition in percentage means simply that 100 part weight of ferric oxide contains 70.1 part of iron and 30.0 parts of oxygen. The very slight deviation of the total from 100 percent is not unusual in experimentally measured composition. Symbol Fe O

Parts by weight

Atomic weight

Gram-atoms

70.1 30.0

55.8 16.0

70.1/55.8 = 1.26 30.0/16.0 = 1.88

Dividing as usual, by smaller one of the two quanti ties we have 1.26/1.26 = 1 gram-atom of Fe and 1.88/1.26 = 1.49 gram-atom of O. It still is not clear what the formula should be in whole-number subscripts unit. We multiply both quantities by 2, then 2 gram-atoms of Fe is combined with 3 gram-atoms of O. The formula is Fe 2O3. But if we calculate the formula for compounds such as acetylene or benzene. We get the formula as CH, which do not represent the actual formula of these compounds. Such a formula which represents the simplest ratio of atom in a compound is called empirical formula. Empirical formula of a compound is the simplest formula showing the relative number of atoms of different elements in one molecule of the compound. Molecular formula represents the actual number of atoms of different elements present in one molecule of the compound. For certain compounds, the molecular formula and the empirical formula may be one and same as explained in the examples. The molecular formula of a compound may be same as empirical formula or whole number multiple of it. Thus the molecular formula = (empirical formula) n where n is an integer 1, 2, 3,….etc.

2.22

Basic Concepts of Chemistry

Example Compound Ethene Ethyne Benzene Glucose

Empirical formula

n

Molecular formula

CH2 CH CH CH2O

2 2 6 6

C2H4 C2H2 C6H6 C6H12O6

Step 2 Dividing the percentage composition by the respective atomic weight of the elements 10.06 0.84 89.10 = 0.84; = 0.84; = 2.51 12 1 35.5

Step 3 Dividing each value in Step 2 by the smallest number among them to get simple atomic ratio 0.84 0.84 2.57 = 1, = 1, =3 0.84 0.84 0.84

Since molecular formula = (empirical formula)n Molecular Weight = Empirical formula weight × n n=

Molecular weight Empirical formula weight

Step 4 Ratio of the atoms present in the molecule C:H: Cl = 1:1:3 ∴ The empirical formula of the compound is CHCl3 Solved problem 7

From the discussion made so for, we can summarise the different steps involved in determination of molecular formula of a compound from the percentage composition by weight of each element present in a compound. Step 1 The percentage of each element is divided by its atomic weight. It gives atomic ratio of the element present in the compound. Step 2 Divide each number obtained for the respective elements in Step 1 by the smallest number among those numbers so as to get the simplest ratio. Step 3 If any number obtained in Step 2 is not a whole number then multiply all the numbers by a suitable integer to get whole number ratio. This ratio will be the simplest ratio of the atoms of different elements present in the compound. Step 4 Write the empirical formula using the symbols of various elements present written side by side with their respective whole number ratio as a subscript to the lower right hand corner of the symbol. Step 5 Calculate the empirical formula mass and divide the molecular mass with empirical formula mass to get the value of n, i.e., simple multiple of empirical formula. Step 6 Multiply the empirical formula with n to get the molecular formula.

A carbon compound contains 12.8% carbon, 2.1% hydrogen, 85.1% bromine. The molecular weight of the compound is 187.9. Calculate the molecular formula. Solution: Step 1 Per cent composition of the elements present in the compound Carbon

Hydrogen

Bromine

12.8

2.1

85.1

Step 2 Dividing with the respective atomic weights 12.8 2.1 85.1 12 1 80 1.067 2.1 1.067 Step 3 Dividing by the smallest number to get simple atomic ratio 1.067 21 1.067 1.067 1.067 1.067 1 2 1

Step 4 Empirical formula is CH2Br Step 5 Empirical formula weight = 12 + (2 × 1) + 80 = 94 n=

Solved problem 6 Chemical analysis of a carbon compound gave the following percentage composition by weight of the elements present, carbon = 10.06%, hydrogen = 0.84%, chlorine = 89.10%. Calculate the empirical formula of the compound. Solution: Step 1 Percentage of the elements Carbon

Hydrogen

Chlorine

10.06

0.84

89.10

187.9 =2 94

The molecular formula

= (empirical formula)n = (CH2Br)2 = C2H4Br2

problems for practice 23. A carbon compound on analysis gave the following percentage composition: C, 14.5%; H, 1.8%, Cl, 64.46%, O, 19.24%. Calculate the empirical formula.

Basic Concepts of Chemistry 2.23

24. 0.200 g of an organic compound gave on combustion, 0.147 g of carbon dioxide and 0.12 g of water and 74.6 c.c of nitrogen gas at STP. Calculate the empirical formula of the compound. 25. The empirical formula of a compound is CH2O. Its molecular weight is 90. Calculate the molecular formula of the compound. 26. 0.188 g of an organic compound having an empiri cal formula CH2Br displaced 24.2 cc of air at 14°C and 752 mm pressure. Calculate the molecular formula of the compound (aqueous tension at 14°C is 12 mm). 27. Four grams of copper chloride on analysis was found to contain 1.890 g of copper (Cu) and 2.110 g of chlorine. What is the empirical formula of copper chloride? 28. Buteric acid contains C, H and O. A 4.24 mg sample of buteric acid is completely burned. It gives 8.45 mg of CO2 and 3.46 mg of H2O. The molecular mass of buteric acid is 88U. What is the molecular formula of buteric acid?

29. A compound on analysis gave the following percentage composition: Na = 14.31%, S = 9.97%, H = 6.22%; O = 69.5%. Calculate the molecular formula of the compound on the assumption that all the hydrogen in the compound is present in combination with oxygen as water of crystallisation. The molecular mass of the compound is 322 (At masses: Na = 23 S = 32, H = 1 and O = 16) 30. A welding fuel gas contains carbon and hydrogen only. Burning a small sample of it gives 3.38 g of carbon dioxide, 0.690 g of water and no other products. A volume of 10L (measured at S.T.P) of this welding gas is found to weigh 11.6 g. Calculate (i) empirical formula, (ii) molar mass of the gas and (iii) molecular formula. 31. A crystalline salt on being rendered anhydrous loses 45.6% of its weight. The composition of anhydrous salt is: Al = 10.5%, K = 15.1%, S = 24.8% and O = 49.6%. Calculate the formula of anhydrous salt and crystalline salt (at mass of Al = 27, K = 39, S = 32, O = 16).

2.24

Basic Concepts of Chemistry

Key pointS

• • •

•

A substance which cannot be broken down into yet simpler substances by any means is known as element. A compound is a pure substance composed of two or more elements joined in chemical combination into a definite proportion by weight. The mass of an object is a measure of the amount of matter it contains while the weight is a measure of the force with which the object is attracted to the earth by gravitational force. Weight is always proportional to mass. Mass and weight of a substance are numerically equal at sea level. If the object is moved away from the earth, the gravitational attraction decreases and therefore the weight diminishes, but the mass remains constant.

physical Quantities and their measurement

• • • • • • • •

•

• •

Properties of matter like mass, length, time, temperature which can be quantified and expressed in numerals with suitable units are called physical quantities. Mass of matter can be determined by measuring resistance to movement. The size of the object is measured in terms of length, area and volume. Length refers to one dimension, area to two dimensions and volume to three dimensions of space. Time is another term of measurement used to know how long it takes for a process or a chemical transformation takes place. Temperature is physical quantity used for measuring the degree of hotness and coldness. Temperature determines the heat flow from a body at a higher temperature to a body at lower temperature. The physical quantities which are derived with the help of two or more physical quantities are called derived physical quantities e.g., density is a derived physical quantity, derived from mass and volume. The three aspects of the numerical result of measurement are (i) Numerical magnitude (ii) Precision (iii) Indicator of scale The physical quantities which depend on the size of the sample like mass, weight, volume, pressure, etc., are called extensive properties. The physical quantities which do not depend on the size of the sample like temperature, density, refractive

•

index, surface tension viscosity, etc., are called intensive properties. The physical quantities like energy, work, heat, electric charge etc., are based on SI base units.

Measurement and Significant Figures

• • • • •

•

Precision is the closeness of the set of values obtained from identical measurements of a quantity. Accuracy is the closeness of single measurement to its true value. The digits in a properly recorded measurement are known as significant figures (or) significant figures are the meaningful digits in a measured or calculated quantity. Greater the number of significant figures in a reported result, smaller is the uncertainty and greater the precision. Rules for reporting the number of significant figures: (i) All non-zero digits are significant, starting with the first digit that is not zero. (ii) Zeroes between non-zero digits are significant. (iii) Zeroes to the right of significant figures are significant for a number less than 1. (iv) When a number ends in zeroes that are not to the right of a decimal point, the zeroes may not be significant. Calculation involving significant figures: (i) While carrying out addition or subtraction of a number of terms, the result should be reported to the same number of decimal places as that of the term with the least number of decimal places. (ii) In the calculation involving multiplication or division, the number of significant figures in the final answer should not be greater than the number of significant figures in the least precise factor. (iii) The numbers that come from direct count of objects or that result from definitions are called exact numbers and are considered to possess an infinite number of significant figures.

matter

• • •

Any thing which has mass and occupies space is matter. Matter exists in three states viz., solid, liquid and gaseous state. Gaseous state of matter at very high temperatures containing gaseous ions and free electrons is referred as the plasma state.

Basic Concepts of Chemistry 2.25

•

• • •

•

•

An element is pure substance that can be neither decomposed into nor built from simpler substance or ordinary physical or chemical means. An element consists of only one kind of atoms. Molecules are identifiable units of matter consisting of two or more atoms of the same element or of different elements combined in a definite ratio. Atomicity is the number of atoms in a molecule. Compound is a substance formed by the union of two or more elements in a fixed proportion by weight and into which it may be decomposed by suitable chemical methods. Mixture is a material obtained as a result of mixing two or more substances (elements or compounds) in any proportion so that the compounds do not lose their identity. A mixture may be homogeneous or heterogeneous. A homogeneous mixture has the same composition throughout. A heterogeneous mixture may have two or more phases each with its own set of properties.

laws of chemical combination

•

•

•

•

•

•

Law of conservation of mass: It was proposed by Lavoisier. Total amount of matter in the universe is unaltered whatever the change take place in its distribution (or) the sum of the weights of reacting substances are equal to the sum of the weights of products (or) matter can neither be created nor destroyed. Law of definite proportions: Proposed by Proust, verified by Strass and Richards. When two elements combine to form a compound they combine in definite proportion by weight (or) a chemical compound has same compositions, by whatever method is prepared. Law of multiple proportions: Proposed by Dalton, proved by Berzelius and Strass. When two elements combine to form more than one compound, the different weights of one of the elements combining with a fixed weight of the other are in proportions of whole number. Law of reciprocal proportions: Proposed by Ritcher. The weights of two or more substances which separately react chemically with identical weights of third element are also the weights in which they combine with each other or multiples of them. Gay Lussac’s law of gaseous volumes: Volumes of gases used and produced in a chemical reaction stand in a simple integral ratio when these volumes are measured under the same conditions of temperature and pressure. Berzelius hypothesis: Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of atoms.

•

• • • • • • • •

Avogadro’s hypothesis: Equal volumes of all gases under the same conditions of temperature and pressure contain equal number of molecules (or) the volumes of a gas at fixed pressure and temperature is proportional to the number of moles (or molecules of gas present). Mathematically, Avogadro’s law is V ∝ n. Dalton’s Atomic Theory: The observations of law of multiple proportions led to formulation of the atomic theory: An atom is the smallest indivisible particle of an element that can take part in a chemical change. All atoms of a given element are identical both in mass and chemical properties. Atoms of different elements have different masses and different chemical properties. Compounds are formed by the combination of different atoms in the ratio of small whole numbers. Chemical reaction involves only combination, separation and rearrangement of atoms. Atoms are neither created nor destroyed in the course of an ordinary chemical reaction. The two modifications made in the hypotheses of John Dalton are (i) An atom is divisible and destructive (ii) All atoms of an element are not identical in mass.

mole concept

• • • • • • • • •

Atomic weight scale is a table that lists the weights of all the elements relative to some common standard. The mass of C-12 is taken as standard and its mass is assigned exactly 12 atomic mass unit (amu) and masses of all other atoms are given relative to this standard. One amu is defined as a mass exactly equal to onetwelfth the mass of one carbon – 12 atom 1 amu = 1.66056 × 10–24 g Many naturally occurring elements consist of more than one isotope. For such elements, the atomic weights determined are the average atomic weights of isotopes Molecular weight or molecular mass of a molecule of a substance is the sum of the atomic weights or atomic masses of atoms present in the molecule. For ionic compounds which do not contain molecules, formula weights are calculated as indicated in the formula of ionic compound. The atomic weight expressed in grams is known as gram atomic weight. Molecular weight expressed in grams is known as gram molecular weight or gram mole. At STP one gram mole of any gas occupies 22.414 litre or 22414 cc of volume. It is known as gram molar volume.

2.26

• •

•

Basic Concepts of Chemistry

One gram mole of any substance will contain 6.022 × 1023 molecules. The number of atoms in one gram atom of an element or the number of molecules present in one gram mole of substance i.e., 6.022 × 1023 is known as Avogadro’s number denoted by N. The mass of 1 mole of a substance in gram is known as molar mass. Gram molecular weight = weight of one molecule in g × 6.022 × 1023

calculation of empirical and molecular formula

• • •

∴ wt. of one molecule in grams = Molecular wt molecular wt = N 6.022 × 1023 Wt of one atom in grams = Atomic wt = Atomic wt23 N 6.022 × 10 • The gaseous substances at NTP will have 6.022 × 1023 molecules in 22.4 litres or 22,400 cc whose weight is equal to their molecular weight. • Reciprocal of Avogadro’s number is known as Avogram Avogram =

• •

1 1 = = 1.67 × 10–23 N 6.022 × 1023

The number of molecules present in one cc of gas at STP is known as Loschmidt number. Its value is 2.617 × 1019 The mole is the amount of substance containing the same number of atoms in exactly 12 grams of 12C isotope. Weight in grams W = Gram molecular weight MW

•

Number of moles’ n =

•

Weight of a substance containing a definite number of moles = Number of moles × gram molecular weight W = n × MW Number of molecules present in a given substance

•

•

=

Weight in grams × Avogadro’s number Gram molecular weight

=

W ×N MW

Number of atoms = Weight × Avogadro’s number Gram atomic weight

• • •

Empirical formula of a compound gives the simplest ratio of the number of atoms of different elements present in one molecule of a compound. Empirical formula will not give the actual number of atoms of different elements present in one molecule of the compound. Calculation of the empirical formula involves the following steps: 1. The percentage composition by weight of each element in the compound should be determined. 2. The percentage of each element should be divided by its atomic weight to get the relative number of atoms of each element. 3. The numbers obtained in the above step (2) should be divided with smallest value to get the simplest ratio. 4. If the numbers obtained in the above step (3) are not whole numbers they should be multiplied by a suitable integer to get whole number ratio. 5. The ratio obtained in the above step (4) gives the empirical formula. Molecular formula represents the actual number of atoms of different elements present in one molecule of the compound. For certain compounds, the empirical formula and molecular formula may be same. Molecular formula = Empirical formula × n n=

•

•

Molecular Weight Empirical formula weight

Molecular weight of a substance can be determined by different methods such as i) vapour density method, ii) methods based on colligative properties. 2 × vapour density = Molecular weight The molecular weights of volatile compounds are determined by Victor Meyer’s method.

Basic Concepts of Chemistry 2.27

practice exerciSe multiple choice Questions with only one answer level i 1.

C OOH |

(C HOH) n + AgNO3 ( excess ) → Silver Salt → Ag |

COOH

2.

3.

4.

5.

6.

7.

8.

9.

0.02 mole

10.8 g

The molecular weight of CH3 (CH2)n (CH3) considering OH group also acting as acidic groups is (a) 72 (b) 44 (c) 106 (d) 96 The vapour density of completely dissociated NH4Cl would be (a) Half that of NH4Cl(g) (b) Less than half that of NH4Cl (c) More than half that of NH4Cl (d) Depends on the amount of NH4Cl taken at start A sample of clay was partially dried and then contained 50% silica, 10% water. The original clay contained 19% water. The percentage of silica present in the sample is (a) 45% (b) 47.3% (c) 52% (d) 59.1% If 1021 molecules are removed from 200 mg of CO2, then the number of moles of CO2 left is (a) 2.88 × 10–3 (b) 4.54 × 10–3 –3 (c) 1.66 × 10 (d) 1.66 × 10–2 The number of molecules of glucose present in 1 mL of 0.1 M solution is (a) 6 × 1022 (b) 6 × 1019 18 (c) 6 × 10 (d) 6 × 1021 2− The number of C2 O 4 ions in 100 mL of 0.2 M aqueous Na2C2O4 is (assume complete ionisation) (a) 0.62 × 1022 (b) 1.204 × 1022 22 (c) 1.86 × 10 (d) none The number of electrons present in 63 mg of HNO 3 is (a) 1.92 × 1023 (b) 1.92 × 1021 22 (c) 1.92 × 10 (d) none Which of the following contains the maximum number of atoms? (a) 10 g CaCO3 (b) 4 g of hydrogen (c) 9 g of NH4NO3 (d) 1.8 g of C6H12O6 How many significant figures are in 1.00 × 103? (a) 3 (b) 2 (c) 1 (d) 4

10. If the percentage of water of crystallization in MgSO4. xH2O is 13%, what is the value of ‘x’? (a) 1 (b) 4 (c) 5 (d) 7 [Relative atomic masses H = 1, O = 16, Mg = 24, S = 32] 11. The weight of magnesium which has same number of atoms as present in 8 g of methane is (a) 48 g (b) 120 g (c) 60 g (d) 56 g 12. What is the volume (in lit) of oxygen at STP required for complete combustion of 32 g of CH4? (a) 44.8 (b) 89.6 (c) 22.4 (d) 179.2 13. 4 g of hydrocarbon on complete combustion gave 12.571 g of CO2 and 5.143 grams of H2O. What is the empirical formula of hydrocarbon? (a) CH (b) CH2 (c) CH3 (d) C2H3 14. The sodium salt of an acid dye contains 7% of sodium. What is the minimum molar mass of the dye? (a) 336.5 (b) 286.5 (c) 300.6 (d) 306.5 15. X forms an oxide X2O3, 0.36 grams of X forms 0.56 grams of X2O3. So the atomic weight of X is (a) 36 (b) 565 (c) 28 (d) 43.2 16. Study the following table: Compound

Weight of compound taken

I. CO2 (44) 4.4 g II. NO2 (46) 2.3 g III. H2O2 (34) 6.8 g IV. SO2 (64) 1.6 g Which two compounds have least weight of oxygen? (a) II and IV (b) I and III (c) I and II (d) III and IV 17. An element X is found to combine with oxygen to form X4O6. If 8.40 g of this element combines with 6.50 g of oxygen, the atomic weight of the element in grams is (a) 24.0 (b) 31.0 (c) 50.4 (d) 118.7 18. 0.078 grams of hydrocarbon occupy 22.414 mL volume at STP. The empirical formula is CH. The molecular formula is (a) C2H2 (b) C6H6 (c) C8H8 (d) C4H4

2.28

Basic Concepts of Chemistry

19. Using scientific notation 1000 metre in centimetres upto two significant figures is expressed as (a) 1.0 × 104 (b) 1 × 105 5 (c) 1.0 × 10 (d) 1.00 × 103 20. Which of the following has three significant figures? (a) 0.52 (b) 543.00 (c) 126.0 (d) 0.0600 21. Among the following pairs, law of multiple proportions is illustrated by (a) H2S and SO2 (b) BeO and BeCl2 (c) NH3 and NO2 (d) N2O and NO 22. Irrespective of the source, pure samples of water always yield 88.89% mass of oxygen and 11.11% mass of hydrogen. This is explained by the law of (a) Conservation of mass (b) Definite proportion (c) Gay-Lussac (d) Constant Volume 23. Which of the following statements is true regarding rounding off the digits? (a) If the digit coming after the desired number of significant figures is more than 5, the preceding digit is increased by one. (b) If the digit coming after the desired number of significant figures is less than 5, the preceding digit is not changed. (c) If the digit coming after the desired number of significant figures is 5, the preceding digit is increased by one if it is odd and not changed if it is even. (d) All of the above. 24. Four 1 litre flasks are separately filled with the gases hydrogen, helium, oxygen and ozone at the same room temperature and pressure. The ratio of total number of atoms of these gases present in the different flasks would be (a) 1:1 : 1:1 (b) 1:2 : 2:3 (c) 2:1 : 2:3 (d) 3:2 : 2:1 25. The empirical formula of an organic compound containing carbon and hydrogen is CH2. The mass of one litre of this organic gas is exactly equal to that of one litre of N2. Therefore, the molecular formula of the organic gas is (a) C2H4 (b) C3H6 (c) C6H12 (d) C4H8 26. If two compounds have the same empirical formula but different molecular formula, they must have (a) Different percentage composition (b) Different molecular weights (c) Same viscosity (d) Same vapour density

multiple choice Question with one or more than one answer 1. 2 moles of CaCO3 present in 1 litre aqueous solution has a density equal to 1.2 g/cc equal to (a) 2 M solution of CaCO3 (b) 4 N solution of CaCO3 (c) Solution with mole fraction of CaCO3 is 0.0347 (d) Solution which contains 55.5 moles of water 2. The vapour density of a mixture containing NO2 and N2O4 is 38.3 at 27°C. Which of the following is correct for 100 moles of mixture? (a) Moles of NO2 in mixture are 33.48 (b) Moles of N2O4 in mixture are 66.52 (c) Weight of NO2 in mixture is 1540 g (d) Weight of N2O4 in mixture is 1540 g 3. Select the correct statements: (a) At STP, volume occupied by one mole liquid water is 22.4 lit. (b) Volume occupied by 1 g H2 gas at STP is equal to volume occupied by 2 g He at STP. (c) 1 g of CH4 real gas occupies 1.4 lit volume at STP. (d) SO2Cl2 reacts with H2O to give mixture of H2SO4, HCl. Aqueous solution of 1 mole SO2Cl2 will be neutralized by 2 mole of Ca(OH)2. 4. In which of the following number; all zeroes are significant? (a) 4,00004 (b) 0.0060 (c) 20.000 (d) 0.800 5. Which of the following statements are true? (a) One gram atom of carbon contains Avogadro’s number of atoms. (b) One mole oxygen gas contains Avogadro’s number of molecules. (c) One mole of the hydrogen molecules contains Avogadro’s number of atoms. (d) One mole of electrons stands for 6.023 × 1023 electrons. 6. Which pair of species have same percentage of carbon? (a) CH3COOH (b) C6H12O6 (c) C12H22O11 (d) C2H5OH

comprehensive type Questions passage i The total number of non-zero digits in a number is called the significant figure (SF). It is equal to the number of digits written, including the last one, even through its value is uncertain. While calculating the figure, one should remember the following rules: First is that all digits are significant except

Basic Concepts of Chemistry 2.29

zero at the beginning of the number and the second is that the zeroes to the right of the decimal points are significant. It may be noted that zero at the end of a number and without the decimal point may not be significant. Any number can be easily transformed into scientific notation by shifting the decimal point in the number to get a new number. If the decimal point is moved to the right, we should multiply the number by 10n. Where n is equal to the number of places moved and if the decimal point is moved to the right, we should multiply the number by 10–n. 1. Given the numbers 141 cm, 0.141 cm, 0.0141 cm the number of significant figure for the three numbers are (a) 3, 4 and 5 respectively (b) 3, 3 and 3 respectively (c) 3, 3 and 4 respectively (d) 3, 4 and 4 respectively 2. In which of the following number all zeroes are significant? (a) 0.0003 (b) 0.0040 (c) 40.000 (d) 0.700 3. 0.000345 is written in the three significant figure. Which scientific notation is correct? (a) 3.45 × 10–4 (b) 345.0 × 10–6 –4 (c) 34.5 × 10 (d) 3.450 × 10–4

matching type Questions 1. Match the following Column I (a) 78 g/L solution of aluminium hydroxide (b) 100 mL solution of sulphuric acid containing 9.8 g of it (c) 9.8% by mass solution of orthophosphoric acid (density = 1 g/cc) (d) 9.8% (W/V) solution of sulphuric acid

Column II (p) 3.0 N (q) 2.0 N

(r) 1.11 M

(s) 1 M

2. Match the following Column I (a) One gram molecule of oxygen gas (b) Gram equivalent volume of H2 (c) 44 g of CO2 (d) 18 g of water

Column II (p) 11.2 L at STP (q) One mole of O2 (r) 22.4 L at STP (s) 3 N atoms

assertion (a) and reason (r) type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): For a reaction between 10 L H2(g), 10 L O2(g), only 5 L of liquid water is obtained. Reason (R): For any reaction, the volume of gaseous reactants and products involved in it are in the same ratio as their stoichiometric coefficients. 2. Assertion (A): Boron has relative atomic mass 10.81. Reason (R): Boron has two isotopes 105 B and 115 B . Their relative abundance is 19% and 81%. 3. Assertion (A): The percentage of nitrogen in urea is 46.6%. Reason (R): Urea is covalent compound. 4. Assertion: The number of significant figure in 106000 is three. Reason: In 106000, all the zeroes are significant. 5. Assertion: A number 143.72 can be written as 1.4372 × 102 in scientific notation. Reason: In scientific notation, as number is generally expressed in the form of N × 10n, where N is a number between 1.00…and 9.999…and n is exponent. 6. Assertion: Nitrogen (28 parts) forms five stable oxides with oxygen containing 16,32,48,64 and 80 parts respectively. This data illustrates the law of multiple proportions. Reason: According to law of multiple proportions, the relative amounts of an element combining with some fixed amount of second element in a series of compounds are the ratios of simple whole numbers. 7. Assertion: Atomicity of oxygen is 2. Reason: Number of atoms present in a molecule is called its atomicity. 8. Assertion: Number of moles of H2 in 0.224 litre of hydrogen is 0.01 mole. Reason: 22.4 Litres of H2 at STP contain 6.023 × 1023 molecules. 9. Assertion: One mole of NaCl contains 6.023 × 1023 molecules of sodium chloride. Reason: 58.5 g of NaCl also contains 6.023 × 1023 molecules of NaCl.

2.30

Basic Concepts of Chemistry

10. Assertion: Element and compound are the examples of pure substances. Reason: Element contains one kind of atoms while compounds contain more than one kind of atoms of elements. 11. Assertion: One mole of SO2 contains double the number of molecules present in one mole of O2. Reason: Molecular weight of SO2 is double to that of O2.

previous years’ iit Questions 1. How many moles of electrons weigh one kilogram? (a) 6.023 × 1023 1 (b) × 1031 9.108 (c)

6.023 × 1054 9.108

(d)

1 × 108 9.108 × 6.023

integer type Questions 1. Crystalline oxalic acid (Mol wt = 126) is subjected to prolonged isotopic exchange treatment with D2O and recrystallised. What is the maximum increase in the molecular weight of the oxalic acid? 2. Ozone is getting converted into oxygen in a chamber. The average molecular weight of the mixture is 40. If 10x% of ozone is decomposed, what is ‘x’?

(2002)

2. Which has maximum number of atoms? (a) 24 g of C (12) (b) 56 g of Fe (56) (c) 27 g of Al (27) (d) 108 g of Ag (108) (2003) 3. A student performs a titration with different burettes and finds titre values of 25.2 mL, 25 .25 mL, and 25.0 mL. the number of significant figures in the average titre value is: (2010)

Basic Concepts of Chemistry 2.31

anSwer KeyS multiple choice Questions with only one answer level i 1. a 2. a 3. a 4. a 5. b 6. b

7. 8. 9. 10. 11. 12.

c b a a c b

13. 14. 15. 16. 17. 18.

b d d b b b

19. 20. 21. 22. 23. 24.

c d d b d c

25. a 26. b

multiple choice Questions with one or more than one answer 1. a, b, c, d 2. a, b, c

3. b, d 4. a, c

5. a, b, d 6. a, b

assertion (a) and reason (r) type Questions 1. d 2. a 3. b

4. c 5. a 6. a

7. a 8. b 9. a

10. a 11. d

comprehensive type Questions passage i 1. b

2. c

3. a

matching type Questions 1. (a) ps 2. (a) qr

(b) qs (b) p

(c) pqs (c) qrs

integer type Questions 1. 6

2. 4

previous years’ iit Questions 1. b

2. a

3. c

(d) qs (d) s

2.32

Basic Concepts of Chemistry

hintS and SolutionS hints to problems for practice 6

1. No. of rupees spent in one second = 10 No. of rupees spent in one year = 106 × 60 × 60 × 24 × 365 Avogadro’s number of rupees will be spent in = 6.02 × 1023 106 × 60 × 60 × 24 × 365 = 19.089 years 2. Wt of 1 mole of atoms of X = 6.644 × 10–23 × 6.022 × 1023 = 40 g ∴ Atomic mass of X = 40 40 × 1000 No. of moles (or gram atoms) of X = 40 = 1000 3. Total no. of moles of CO2 removed =

0.2 = 0.00454 44

8. Each N3– ion carries 3 × 1.602 × 10–19 coulombs (∴ Charge on the electron = 1.602 × 10–19 coulombs) ∴ 6.022 × 1023 ion carries a charge of 3 × 1.602 × 10–19 × 6.022 × 1023 = 2.894 × 105 coulombs 9. Each gas has a mass x g. Therefore O 2: Weight x x No. of moles 32 Volume ratio ∴ O2: H2: CH4 = 1:16:2

22 150 150 × × 7 2 2 −16 −8 × 10 × 5000 × 10

10. Volume of virus = πr 2 l =

= 0.884 × 10–16 cm3

21

10 No. of moles of CO2 removed = 6 022 . × 1023 = 0.00166 No. of moles of CO2 left = 0.00454 – 0.00166 = 0.00288 4. No. of moles = Molarity × volume in litres = 0.001 × 0.1 = 0.0001 1 mole of H2SO4 contains 1 mole of SO42– ions 0.0001 mole of H2SO4 contains 0.0001 mole of SO42– ions ∴ No. of sulphate ions = 0.0001 × 6.022 × 1023 = 6.022 × 1019 5. Weight of 1 mole = density × gram molar volume = 0.1784 × 22.4 =4g 50 6. No. of moles of CaCO3 = = 0.5 (M wt of CaCO3 100 = 100) Each molecule of CaCO3 contains 3 atoms of O Or 1 mole of CaCO3 contains 3 moles of O atoms Or 0.5 mole of CaCO3 contains 1.5 moles of O atoms No. of atoms of O = 1.5 × 6.022 × 1023 = 9.033 × 1023 Wt of O atoms = 1.5 × 16 = 24 g. 7. No. of moles of CH4 =

1.6 = 0.1 16

No. of molecules of CH4 in 1.6 g = 0.1 × 6.022 × 1023 = 6.022 × 1022 molecules One molecule of CH4 contains 10 electrons (6 from C and 4 from 4 H atoms) ∴ 6.022 × 1022 molecules of CH4 contain 10 × 6.022 × 1022 = 6.022 × 1023 electrons

H2: CH4 x x x x 2 16

Wt of one virus =

0.884 × 10−16 = 1.178 × 10−16 g 0.75

Mol Wt of virus = 1.178 × 10–16 × 6.022 × 1023 = 7.095 × 107 11. Mol Wt of mixture of NO2 and N2O4 = 38.3 × 2 = 76.6 Let ‘a’ g of NO2 present in 100 g mixture ∴ Mole of NO2 + Mole of N2O4 = mole of mixture a 100 − a 100 + = 46 92 76.6 a = 20.10 g Mole of NO2 in mixture =

20.1 = 0.437 46

20.1 (∵ Valency Factor 12. Normality = 0.02; Molarity = 46 = 2)

0.02 100 (∵ mole × No of moles of oxalic acid = 2 1000 = M × V inl.) No. of moles of oxalic acid = 0.001 × 6.022 × 1023 = 6.022 × 1020 13. The reaction between iron and steam is 3Fe + 4H2O → Fe3O4 + 4H2 Mole ratio of reaction suggests Mole of Fe 3 = Mole of H 2 O 4

Basic Concepts of Chemistry 2.33

0.35 65000 ∴ 1 mole of hemoglobin contains = × 56 100 4.06 moles of Fe Thus 1 molecule of hemoglobin contains four iron atoms.

18 3 3 × = 18 4 4 3 Wt of iron = × 56 = 42 g 4 Mole of Fe =

14. 100 kg sample of impure CaCO3 has 95 kg CaCO3 95 × 200 = 190 kg ∴ 200 kg impure sample contains 100 CaCO3 CaCO3 decomposes as CaCO3 → CaO + CO2 100 g 54 g 44 g 100 kg CaCO3 gives = 56 kg of CaO 190 kg CaCO3 gives CaO =

95 × 200 = 106.4 kg 100

15. Wt of alloy cylinder = V × d = πr2 h × d 22 = × (2.5)2 × 10 × 8.2 = 1610.7 g 7 1610.7 × 12 = 193.3 g Wt of Co in alloy = 100 ∴ No. of Cobalt atoms in 193.3 g

6.022 × 1023 × 193.3 = 58.9 = 19.8 × 1023 16. Let the mass of CaWO4 be x g. Mass of W in x g of CaWO4 = FeWO4 = mass of w in 569 g of FeWO4 Moles of W in CaWO4 × At wt of W = moles of W in FeWO4 × at wt of W As both CaWO4 and FeWO4 contain 1 atom of W each ∴ Moles of CaWO4 × At wt of W = moles of FeWO4 × At wt of W X 569 × 184 = × 184 288 304 X = 539.059 17. In every 100 atoms, 93 are H atoms and 7 are He atoms Mass of H = 93 × 1 = 93 g Mass of He = 7 × 4 = 28 g 93 × 100 = 76.86% ∴ Per cent mass of H = 93 + 28 ∴ Per cent mass of He = 23.14% 18. No. of moles of hemoglobin =

100 65000

No. of moles of iron in 100 g of hemoglobin = 0.35 56

19. The density of mercury is 13.6 g/cc. Suppose the length of the side of the cube is x cm, i.e., the diametre of one Hg atom ∴Volume occupied by 1 Hg atom = x3 cc Mass of one Hg atom = 13.6 × x3 g Mass of one Hg atom = At.wt 200 = g Avogadro const 6.022 × 1023 200 6.022 × 1023 200 x3 = = 2.44 × 10–23 13.6 × 6.022 × 1023 x = 2.9 × 10–8 cm 13.6 × x3 =

20. Let the wt of polystyrene be 100 g. 10.46 No. of moles of Br in 100 g of polystyrene = 79.9 = 0.1309 From the formula of polystyrene No. of moles of Br = 3 × moles of Br3C6H3 (C3H8)n Or 0.1309 = 3 ×

W 3 × 100 = MW 314.7 × 44 n

∴ n = 44.9 ≈ 45 21. The volume ratio of the gases will be the same as their mole ratio M Wt of air = 78 × 28 + 21× 32 + 0.9 × 40 + 0.1× 44 78 + 21 + 0.9 + 0.1 = 28.964 22. Let the volume of ethanol containing the same number of molecules as are present in 175 mL of H2O be V mL as given Moles of C2H5OH in V mL = moles of H2O in 175 mL Wt of C2 H 5 OH Wt of H 2 O Now = = M wt of C2 H 5 OH M wt of H 2 O 0.789 × V 1.0 × 175 = = 46 18 V = 566.82 mL or

2.34

Basic Concepts of Chemistry

23. Step

C

H

Cl

O

1. % of element

14.5

1.8

64.46

19.24

2. Dividing with at wts

14.5 = 1.21 12

1.8 = 1.8 1

64.46 = 1.81 35.5

1.2 =1 1.21

3. Simplest ratio

1.21 =1 1.21

1.8 = 1.5 1.21

1.81 = 1.5 1.21

1.2 =1 1.21

4. Multiplying with a suitable integer

1×2=2

1.5 × 2 = 3

1.5 × 2 = 3

1×2=2

Empirical formula = C2H3Cl3O2

24. Step

C

H

N

0.147 × 12 100 × 44 0.20

0.12 × 2 100 × 18 0.200

74.6cc × 289 100 × 22400 0.200

100−(20.04 + 6.66 46.03)

= 20.014

= 6.66

= 46.03

= 26.67

1. % of element

O

2. Dividing with at mass

20.014 = 1.67 12

6.66 = 6.66 1

46.03 = 3.33 14

26.67 = 1.66 16

3. Simplest ratio

1.67 =1 1.67

6.66 =4 1.67

3.33 =2 1.67

1.66 =1 1.67

Empirical formula = CH4N2O

25. Empirical formula weight of CH2O = 12 + 2 + 16 = 30 M wt 90 n= = =3 empirical formula weight 30 The molecular formula (CH2O)3 = C3H6O3 26. Weight of the carbon compound W = 0.188 g Volume of air displaced V = 24.2 cc Pressure of moist air P = 752 mm Pressure of dry air (P1) = (752−12) = 740 mm Temperature (T1) = 14 + 273 = 287 K Volume of dry air displaced under STP V2 = ? Where P2 = 760 mm; T2 = 273 K P1V1 T2 74 mm × 24.2 cc × V2 = T × P = 287 k 1 2 273 k = 22.42 cc 760 mm Volume of vapour at STP = 22.42 cc The weight of 22.42 cc of the vapour at STP = 0.188 g Weight of 22,400 cc of the vapour at STP

0.188 g × 22400 cc 22.42 cc = 187.9 g ∴ Gram molecular weight of the compound = 187.9 g Empirical formula CH2 Br Empirical formula weight (12 + 2 + 80) = 94 Molecular weight 187.9 = =2 Empirical formula weight 94 The molecular formula = (empirical formula)2 (CH2 Br)2 = C2H4 Br2 27. Step

Cu

Cl

1. % element

1.89 × 100 = 47.3 4.00

2.11 × 100 = 52.7 4.00

2. Dividing with atomic mass

47.3 = 0.74 63.5

52.7 = 1.48 35.5

3. Simplest ratio

0.74 =1 0.74

1.48 =2 0.74

The empirical formula of the compound is CuCl2.

Basic Concepts of Chemistry 2.35

28. Step 1. % Element

2. Dividing with atomic mass 3. Simplest ratio

C

H

O

12 8.45 × × 100 = 54.3% 44 4.24

2 3.46 × × 100 = 9% 18 4.24

36.7%

54.3 = 4.52 12

9.0 = 8.93 1.008

36.7 = 2.29 16

4.52 =2 2.29

8.93 =4 2.29

2.29 =1 2.29

Empirical formula of buteric acid is C2H4O Empirical formula Wt = (2 × 12) + (4 × 1) + (1 × 16) = 44 Molecular mass 88 n= = =2 Emperical formula wt 44 Molecular formula = (Empirical formula)2 = (C2H4O)2 = C4H8O2 29. Step

Na

S

H

O

1. % Element

14.31

9.97

6.22

69.50

2. Dividing with atomic mass

14.31 = 0.62 23

9.97 = 0.31 32

6.22 = 6.22 1

69.50 = 4.34 16

3. Simplest ratio

0.62 =2 0.31

0.31 =1 0.31

6.22 = 20 0.31

4.34 = 14 0.31

The empirical formula of the compound = Na2SH20O14 Empirical formula wt = 322 Molecular mass 322 = =1 Now n = Empirical formula mass 322 Molecular formula Na2SH20O14 Since all hydrogens are present as water molecules 20 hydrogens must be present as 10H2O molecules. The remaining oxygen atoms are present with the molecule. Molecular formula of the compound = Na2SO4. 10H2O 30. Step

C

H

1. Mass of element

12 × 3.38 = 0.92 14

2 × 0.69 = 0.077 18

2. % Element

0.922 × 100 = 92.2% 0.922 × 0.077

0.077 × 100 = 7.7% 0.922 × 0.77

3. Dividing with atomic mass

92.2 = 7.7 12

7.7 = 7.7 1

4. Simplest ratio

7.7 =1 7.7

7.7 =1 7.7

Empirical formula = CH

2.36

Basic Concepts of Chemistry

(ii) Calculation of molar mass Wt of 10 L of gas at STP = 11.6 g 11.6 × 22.4 = 26 g mol–1 Wt of 22.4 L of gas at STP = 10 (iii) Calculation of molecular formula Empirical formula mass = 12+1 = 13 Molecular mass = 26 Molecular mass 26 Now n = = =2 Empirical formula mass 13 Molecular formula = (Empirical formula)2 = (CH)2 Molecular formula = C2H2 31. Step

K

Al

S

O

1. % Element

15.1

10.5

24.8

49.6

2. Dividing with atomic mass

15.1 39 = 0.38

10.5 27 = 0.38

24.8 32 = 0.775

49.6 16 = 3.1

3. Simplest ratio

0.38 =1 0.38

0.38 =1 0.38

0.775 =2 0.38

3.1 =8 0.38

∴ The empirical formula of compound = KAlS2O8 Empirical formula mass of the compound = 39 + 27 + 64 + 128 = 258 100 g of salt loses 45.6 g of water So 100 g salt contains 100–45.6 g = 54.4 g of anhydrous salt 54.4 g of anhydrous salt lost 45.6 g of water 258 g of anhydrous salt lost? 258 × 45.6 = 216.2 g 54.4 216.2 No of molecules of water of crystallisation = 18 = 12 The simplest formula of crystalline salt = KAlS2O8 12H2O

hints to multiple choice Questions with only one answer level i 1. 0.02 mole silver salt gives 0.1 moles of silver So no. of OH groups are 5. Then compound is CH3(CH2)3 CH3. 2. NH4Cl → NH3 + HCl 17 + 36.5 So MW = 2

3. Clay contains: x % silica, 81–x% Impurities, 19 % H2O After partial drying 50% silica 40% Impurities 10 % H2O X 81 − X = So = 50 40 X = 45% 4. Initial no. of moles of CO2 =

200 × 10−3 = 4.545 × 10–3 44

No. of moles of CO2 removed is = 1.66 × 10–3 No. of moles of CO2 remaining is = 2.88 × 10–3 5. No. of moles of C6H12O6 = 10–4 So no. of molecules of C6H12O6 = 6 × 1019 = 0.02 6. No. of moles of C2 O 2− 4 No. of ions = 1.204 × 1022 7. No. of moles = 10–3 Each molecule contains = 32 electrons No. of electrons = 10–3 × 6.023 × 1023 × 32 = 1.92 × 1022 8. (a) no. of atoms in CaCO3 =

10 × No × 5 100

(b) no. of atoms of hydrogen =

4 × No × 2 2

(c) no. of atoms in NH4 NO3 =

9 × No × 9 80

(d) no. of atoms in C6H12O6 =

1.8 × No × 24 180

10. 120 g MgSO4 contain 18 x g of H2O 87 g of MgSO4 contain 13 g of H2O 120 18x = 87 13 x=1 8 × No × 5 11. No of atoms in 8 g CH4 is 16 Weight of Mg = x ∴

x 8 ×N= × No × 5 24 16

X = 60 g 12. CH4 + 2O2 → CO2 + 2H2O(l) 2 moles CH4 require 4 moles of oxygen

Basic Concepts of Chemistry 2.37

13. Moles of CO2:H2O = 0.285: 0.285 C:H = 1:2 14. 7 g sodium, 93 g anion present in sodium salt of dye

23 × 93 g anion 7 Mw of acid dye = 1+305.5 = 306.5

23 g sodium reat with

15. 0.36 g of X combines with 0.2 g of Oxygen 0.36 × 48 = 86.4 g Wt of x react with 48 g of oxygen is 0.2 Aw of x = 43.2 16. (a) (b) (c) (d)

4.4 g CO2 contain 3.2 g of oxygen 2.3 g NO2 contain 1.6 g of oxygen 6.8 g H2O2 contain 6.4 g oxygen 1.6 g SO2 contain 0.8 g of oxygen

17. 8.4 g of X combines with 6.5 g of oxygen 96 g of oxygen combines with

96 × 8.4 = 124 gm of X 6.5

Aw = 31 18. MW of compound is 78 EW of compound is 13 So molecular formula is C6H6 24. P, V, T are same. So number of molecules are same.

previous years’ iit Questions 1. Weight of one electron = 9.108 × 10–31 kg No. of electrons in 1 kg =

1 9.108 × 10−31

2. No. of Moles of carbon 24 = 2.0 12 No. of atoms = 2 × 6.023 × 1023 atoms Rest all elements have one mole each = 6.023 × 1023 atoms 3. Titration values are 25.2, 25.25 and 25.0. The lowest number of significant figures in titration = 3 Average titre value = 25.2 + 25.25 + 25.0 3 =

75.45 = 25.15 3 Since the lowest value of significant figures is 3. The answer is 25.1 =

This page is intentionally left blank.

CHAPTER

3 The States of Matter

N

othing in nature is more ancient than motion, and the volumes that philosophers have compiled about it are neither few nor small, yet have I discovered that there are many things of interest about it, that have hitherto been unperceived. Galileo

3.1 IntroductIon When the properties of a material are studied, it is not the properties of single molecules that are involved, but the properties of aggregates of a large number of molecules. The behaviour of matter in the bulk depends not only on the nature of individual molecules, but also on the effect they have on each other, as well as the conditions (e.g., temperature, pressure etc.) under which they are studied. All matter may exist in three states—solid, liquid and gas. A solid is rigid and has definite shape. A liquid can flow, and so alter its shape, but like a solid, it has a definite bounding surface. A gas can also flow but differs from both liquid and solid in that it has no surface, so that it will completely fill any space available to it. Another characteristic property of a gas is that it changes in pressure causing marked volume changes. Solid and liquid, on the other hand, are almost incompressible. The conditions of temperature and pressure decide the state in which a material will exist, and by a suitable change in the conditions, it can be transformed from one state to another. Supposing the pressure to be kept same (normal atmospheric pressure, for example), an increase in temperature may cause a solid to melt and thus become a liquid, while at a still higher temperature, the liquid will boil and become vapour (i.e., a., gas). The term vapour is used to describe a material in the gaseous state when it can be condensed to the liquid form by increasing the pressure at the same temperature. A vapor is therefore a gas below its critical temperature. Thus, water is a solid (ice) below 0°C, melts at the latter temperature to a liquid, and if the temperature is raised to 100°C, it boils and becomes a gas (water, vapour or steam). In the process transition from solid → liquid →

gas, there is a continual intake of energy by the material which may be separated into the following steps: I. Raising the temperature of the solid to the melting point: The amount of heat that has to be supplied depends on the heat capacity of the material (the heat capacity of a system is the quantity of heat required to raise the temperature by one Kelvin). II. Melting the solid: During the transition, the temperature remains constant, but the solid takes up heat as it melts (the enthalpy of fusion). III. Raising the temperature of the liquid to the boiling point: Heat must be supplied according to the heat capacity of the liquid. IV. Boiling the liquid: The temperature remains constant during the transition liquid → gas, but the enthalpy of vaporization is taken up. The whole process can be reversed by cooling the material, i.e., extracting energy from it. When a vapour condenses, or a liquid solidifies, the heat evolved is exactly equal to the amount of heat absorbed during boiling or melting. It follows that when a material is in any one state (either solid, liquid or gas), its temperature rises when it absorbs energy, and falls when it loses energy. In other words, temperature is a measure of the amount of energy possessed by the material. We must now consider what forms this energy may take. The quantity of energy involved is usually much smaller than the quantum of the electronic energy levels (it is only at very high temperature that the electrons in atoms are moved to higher levels as shown by the spectra emitted by some atoms when strongly heated in a flame). There are, however, various forms of mechanical energy that a molecule may possess. If the molecules can move about, they will have

3.2

States of Matter

kinetic energy of motion; if they can rotate they will have rotational energy; and lastly the atoms in a molecule may vibrate to give vibrational energy. Fig 3.1 shows rotation and vibration in a diatomic molecule. Since it is the movement of mass that matters, only the nuclei of the two atoms are shown. These three forms of energy have one factor in common: they all involve kinetic energy, which is the energy possessed by bodies in motion. This approach to an explanation of the physical behaviour of matter is therefore called the kinetic theory, from the Greek ‘kinesis’ (motion). A solid has a rigid shape, which suggests that the molecules are not able to move about: it is, in fact, known that they can only vibrate. Liquids and gases, on the other hand, move readily and also show the property of diffusion, the latter phenomenon can best be explained by a simple illustration of a gas tap left open for a few seconds. A smell of gas is quickly noticed some distance away showing that the gas molecules have dispersed and mixed with the surrounding molecules of air. This rapid diffusion can only be explained by assuming that the molecules of a gas are in constant and rapid motion. Diffusion also occurs in liquids, but much more slowly because the closer packing of the molecules hinders motion. The first direct evidence that molecules are in constant motion was obtained by Robert Brown in 1827, when he observed that grains of pollen suspended in water were in a continual state of agitation (The Brownian movement) owing to bombardment by water molecules.

3.2 Inter-molecular forces The two main similarities between solids and liquids are that they are almost incompressible and that both

have a definite surface. The first point suggests that the molecules are practically touching, so that it is impossible to force them much closer together even when a high pressure is applied. It may seem strange at first sight that the molecule, with its rather nebulous clouds of electrons, should behave in this way like a hard and solid object. The reason is that the outer (valence) electrons are all in filled orbitals as a result of electronpairing during bond formation. These filled orbitals cannot overlap with the orbitals of other molecules, so that when two molecules approach each other closely, strong repulsive force is encountered. For the purposes of the kinetic theory, we can regard molecules as solid objects of definite size. The existence of a surface in solids and liquids indicates that there are strong forces holding the molecules together, so that they cannot easily break away (some manage to escape in the form of vapour; this will be disregarded for the time being). These are the forces of inter-molecular attraction, sometime called Van der Waals force, which operate between any molecules that approach within a certain distance of each other. The various types of interactions between molecules which are responsible for inter-molecular forces are discussed here.

3.2.1 dipole-dipole Interaction When a substance contains molecules with permanent dipole moments, they interact with one another. The negative end of one molecule attracts the positive end of another molecule or vice versa (Fig 3.2).

Vibration

Rotation fig 3.1 Rotation and vibration of a diatomic molecule

States of Matter

δ

δ

+

–

–

+

δ

V

δ

+

(a)

–

More charge density towards chlorine (a) δ + δ

–

Cl

H

Cl

H

–

Cl

H +

3.3

(b)

fig 3.2 (a) Distribution of an electron cloud in HCl molecule (b) Dipole-dipole interaction between two HCl molecules

3.2.2 Ion-dipole Interactions Ion-dipole interactions are similar to ion –ion interactions except that they are more sensitive to distance and tend to

+

+

–

–

–

+

+

be somewhat weaker since the charges (q+,q–) comprising the dipole are usually considerably less than a full electronic charge. Ion-dipole interaction attracts an ion (either a cation or anion) and a polar molecule to each other. The charge density on the cation is higher than on anions, therefore a cation interacts more strongly with dipoles than does an anion having same charge but with bigger size. Ion–dipole interactions are important in solutions of ionic compounds in polar solvents where solvated species such as (Na(H2O)x]+ and [F(H2O)y]– (for solution of NaF in H2O) exist. Thus, hydration of different ions is an example of iondipole interaction. (Fig 3.3)

–

The force of attraction is electrostatic in nature. The dipole-dipole interactions depend upon the distance and orientation of the two dipoles. The orientation of molecular dipoles is such that the ends of similar charge are as far apart as possible. A favorable orientation of two dipoles results in attractive dipole-dipole interactions. The dipole-dipole interaction is one of the factors which determines the melting and boiling points of polar substance. A substance with no dipole moment will have a lower melting point or boiling point than a polar molecule, if molecular weight and shape are same.

A–

M+

– –

+

–

+

–

+

+

fig 3.3 Ion-dipole interaction

3.4

States of Matter

3.2.3 Ion-Induced dipole Interaction and dipole-Induced dipole Interaction If a charged particle such as anion or a molecule with permanent dipole approaches a neutral molecule, the charge of the ion or the dipole of the polar molecule interacts with the electron cloud of the neutral molecule or atom causing a distortion in the electron cloud of the neutral atom or molecule. As a result of this distortion, there is a charge separation in the second molecule and it also behaves as a dipole. This molecule also behaves as a dipole. This molecule is then said to have an induced dipole (Fig 3.4). The attraction will be then between opposite poles. The polarization of the neutral species will depend upon its inherent polarizability and on the polarizing field afforded by the charged ion Z±.

3.2.4 Instantaneous dipole-Induced dipole Interaction Molecules and atoms are neutral in the sense that the electrons are uniformly distributed around the nuclei, i.e., the centre of negative charges coincides with the centre of positive charge. If the electron cloud at any instant is even slightly displaced relative to the positive centre, a small instant-aneous dipole is developed. This instantaneous dipole can interact with a nearly neutral atom or molecule

δ –

δ + A

and cause instantaneous induced dipole in second molecule or atom. A schematic representation of attraction is shown in the Fig 3.5. Another way of looking at this phenomenon is to consider the electrons in two or more “non-polar” molecules as synchronizing their movements (at least partially) to minimize electron–electron repulsion and maximize electron–nucleus attraction. Such attractions are extremely short ranged and weak, as are dipole–induced dipole forces. The attraction between induced dipoles is called London dispersion forces. The dipole – dipole, dipole – induced dipole and dispersion forces are collectively called Van der Waals forces. One important factor that determines the magnitude of Van der Waals forces is the relative polarisability of the electrons of the atoms involved. In the halogen family, for example, polarisability increases in the order F1 (q) Law of corresponding state

  

3   (3VR − 1) = 8TR VR2  



RT 

aV

(r) Critical state

ab

c (d) V 3 +  b + V2 + =0 Pc Pc Pc  

(s) Shows negative deviation, Z 1

gas in high pressure region (c) Very large volume conditions (d) For H2 gas at 0°C in all pressure region

(s) TC = 80 K

9. Match the following Column I with Column II. Column I

Column II

(a) (b) (c) (d)

(p) Temperature (q) Pressure (r) Attractive forces (s) Molecular mass

Diffusion of gases Density Critical temperature Kinetic energy

10. Match the following Column I with Column II. Column I

Column II

(a) Helium gas (P = 200 atom, T = 273 K) (b) Helium gas (P = 0 atom) (c) H2(P = 200 atom, T = 273 K) (d) CH4 (P = 1 atom, T = 273 K)

(p) P(V – nb) = nRT (q) Compressibility factor = 1 (r) Repulsive forces dominate (s) Compressibility factor = 1

3.87

assertion (a) and reason (r) type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): NH3 is more absorbed on activated coconut charcoal than H2 Reason (R): Critical temperature for NH3 is more than that of H2 2. Assertion (A): Van der Waals constant ‘a’ is larger for N2 than NH3 Reason (R): There is H-bonding in NH3 3. Assertion (A): The measurement of the pressure of dry gas collected over water is based upon Boyle’s law. Reason (R): Volume of a given mass of gas is directly proportional to its absolute temperature at given pressure 4. Assertion (A): 1/4th of the gas is expelled if air present in an open vessel is heated from 27°C to 127°C. Reason (R): Rate of diffusion of gas is directly proportional to the square root of its molecular mass at constant temperature and pressure 5. Assertion (A): Compressibility factor (Z) for nonideal gases is always greater than 2 Reason (R): Non-ideal gases always exert less pressure than expected. 6. Assertion (A): At constant temperature, if pressure on the gas is doubled, density is also doubled. Reason (R): At constant temperature, molecular masses of a gas are directly proportional to the density and inversely proportional to pressure 7. Assertion (A): Z > 1 repulsive forces dominate the behaviour of a particular gas Reason (R): With increase of pressure, Z decreases continuously 8. Assertion (A): The value of Boyle temperature TB =

a for Van der Waal’s gases Rb

Reason (R): At Boyles temperature TB, real gases behave ideally over a large range of pressure

3.88

States of Matter

9. Assertion (A): In low speed region, fraction of molecules having a particular speed range increases with increase in temperature Reason (R): Increase in temperature of a gas system decreases the fraction of molecules possessing most probable speed 10. Assertion (A): SO2 gas is easily liquefied while H2 is not Reason (R): SO2 has low critical temperature while H2 has high critical temperature 11. Assertion (A): If ‘a’ is Van der Waal’s constant, aI2 > aCl2 Reason (R): Dipole attractions among the above gaseous molecules is proportional to molecular weight 12. Assertion (A): Gas with lower molecular weight will effuse or diffuse faster at constant temperature and pressure Reason (R): Kinetic energy of any gas depends upon its mass temperature 13. Assertion (A): Greater the value of Van der Waal’s constant ‘a’, greater is the liquefaction of gas Reason (R): ‘a’ indirectly measures the magnitude of attractive forces between the molecules. 14. Assertion (A): Critical temperature is the temperature at which the real gas exhibits ideal behaviour for considerable range of pressure Reason (R): At critical point the densities of a substance in a gaseous and liquid states are same 15. Assertion (A): Aqueous tension depends on temperature Reason (R): Aqueous tension is the force experienced by water molecule on its surface

4. A jar contains a gas and a few drops of water at Tk. The pressure in the jar is 830 mm of Hg. The temperature of the jar is reduced by 1%. The vapor pressure of water at two temperatures is 30 mm and 25 mm Hg. The new pressure in the jar is expressed in mm of Hg. Answer is divided by 90.77. Express the integer. 5. A compound exists in the gaseous state both as monomer (A) and dimer (A2). The molecular weight of A is 48. In an experiment 96 gm of the compound was confined in a vessel of volume 33.6 litres and heated to 273°C. The extent of dimer is 50% by weight. Then the pressure of mixture is 6. A spherical balloon of volume 5 lit is to be filled up with H2 at NTP from a cylinder of 6 lit volume containing the gas at 6 atm at 0°C. The no. of balloons that can be filled up is 7. Two flasks A and B have equal volumes. A is maintained at 300 K and B at 600 k. while A contains H2 gas, B has an equal mass of CH4 gas. Assuming ideal behaviour for the both gases, find the ratio of (uav)A : (uav)B. 8. A cylindrical container with movable piston initially had 3.0 mole of a gas at 8.0 atm pressure and a volume of 5.0 litre. If piston is moved to create a volume 10.0 litre, while simultaneously withdrawing 1.5 mole of gas, calculate the final pressure 9. A graph is plotted between PVm on y – axis, P-along x-axis, Vm is molar volume of the gas. The gas is real gas at 49 K. find the intercept on the y – axis as P → O at 49 K (R = 0.082 lit atom. K–1 mol–1.) 10. The volume to be excluded due to only two molecules of a gas in collision with a fixed point of impact is 0.09 Litre (NA = Avogadro’s number). If the value NA

Integer type Questions 1. The compressibility factor for 1 mole of a Van der Waals gas at 27°C and 82.1 atoms is found to be 0.333. Assuming that the volume of gas molecule is negligible, so Van der Waal’s constant is expressed in the units of 10–2 lit 2 atm mol2. Express the answer in terms of nearby integer 2. At 27°C, two balloons of equal volume are filled with 32 kg O2, 2 kg of H2 at a pressure 2 atm. The O2 balloon leaks to a pressure of ½ atms in 24 min. The time in min for H2 balloon to reach a pressure of ½ atom is 3. A mixture containing 1.12 lit D2 and 2.24 lit H2 at NTP is taken in a bulb and connected to the fully evacuated bulb with a stop cock. The stop cock is opened for some time and then closed. The first bulb contains 0.1 gm of D2. The weight of H2 present in second bulb is expressed in terms of 10–2/2 gm.

11.

12.

13. 14.

of ‘a’ is 3.6 atom L2mol–2, then the value of Boyle’s temperature is 10x K. What is the value of ‘x’? (R = 0.08 L atom K–1m–1) The mass of gas molecule ‘Y’ is twice the mass of gas molecule ‘Z’. The rms speed of ‘Y’ is twice the rms speed of ‘Z’ and their respective pressures are Pv and Pz. if two samples of ‘Y’ and ‘Z’ contain same no. of molecules, what will be the ratio of respective pressures of two samples in separate containers of equal volume? Find out the mass of CH4 effused from a mixture containing 16 g each of CH4 and He, if 8 g of He is effused within the same time. Ratio of excluded volume and actual molecular volume of real gas is equal to……. Urms of CH4 at Tk is 6 times of Ump of SO2 at T1k. the temperature of CH4 gas is …………times of SO2

States of Matter

15. Boyle’s temperature of various gases is given below Gas 1 2 3 4 TB(k) 117 23 498 406 Which gas can be liquefied most easily? 16. If the slope of Z (compressibility) v/s pressure curve is constant (slope = π/492 atm–1) at a particular temperature 300 K and very high pressure, then calculate the diametre of the molecules in Å. (Given N = 6 × 1023, R = 0.0821 atom lit mol–1 K–1) 17. Two moles of a gas are confined to a 5 litre flask at 27°C, calculate its pressure using Van der Waal’s equation. For ammonia a = 6.25 atom lit–2 mol–2 and b = 0.037 lit mol.

V(L)

3.89

(28.6 L,373 K)

(b) (22.4 L, 273 K) T(K) V(L)

(30.6 L,373 K)

(c) (22.4 L, 273 K) T(K) V(L)

Previous years’ IIt Questions

(d)

1. The rms velocity of hydrogen is 7 times the rms velocity of nitrogen. If T is the temperature of the gas, then (2000S) (a) T(H2) = T(N2) (b) T(H2) > T(N2) (c) T(H2) < T(N2) (d) T(H2) = 7 T(N2) 2. The compressibility of a gas is less than unity at STP. Therefore, (2000S) (a) Vm > 22.4 litres (b) Vm < 22.4 litres (c) Vm = 22.4 litres (d) Vm = 44.8 litres 3. At 100°C and I atm, if density of liquid water is 1.0 g cm3 and that of water vapour is 0.0006 g cm3, then the volume occupied by water molecules in 1 litre of steam at that temperature is (2000S) (a) 6 cm3 (b) 60 cm3 (c) 0.6 cm3 (d) 0.06 cm3 4. The root mean square velocity of an ideal gas at constant pressure varies with density (d) as (2001S) (a) d2 (b) d (d) 1 d (c) d 5. Which of the following volume (V) temperature (T) plots represents the behaviour of one mole of an ideal gas at one atmospheric pressure? V(L)

(38.8 L,373 K)

(a) (22.4 L, 273 K) T(K)

(22.4 L, 273 K) (14.2 L,373 K) T(K)

6. When the temperature is increased, surface tension of water (2002S) (a) Increases (b) Decreases (c) Remains constant (d) Show irregular behaviour 7. Positive deviation from ideal behaviour takes place because of (2003S) (a) Molecular interaction between atoms and PV/ nRT>1 (b) Molecular interaction between atoms and PV/ nRT1 (d) Finite size of atoms and PV/nRT octahedral > tetrahedral > trigonal. Thus relatively small cations occupy tetrahedral holes and larger cations occupy the octahedral holes. If cations are too large to fit into octahedral holes, the anions may occupy the larger cubic holes made possible by the more open spacing. This can be explained basing on the following basic principles. (i) Ions of opposite charge try to get as close together as possible; but (ii) Ions of the same charge will avoid coming in contact with each other.

B-

A+ B

B-

B-

(a)

A+

r++r45° r-

-

B-

r++r-

B-

-

B

B-

B-

(b)

30°

A+ B-

B-

r-

(c)

fig 4.41 (a) Showing the limiting r+/r– value for an octahedral arrangement (b) The unstable octahedral arrangement r+/r– becomes less than 0.414 (C) The limiting r+/r– value for A+

4.24

Solid State

In this way, the attractions between the oppositely charged ions will be maximized while the repulsions between ions of the same charge will be minimized. Consider, for example, a structure in which a central A+ ion is surrounded, octahedrally by six B– ions the coordination number will be 6. Fig 4.41 shows the general arrangement in such a structure but indicates only the nuclei of the ions concerned and not their actual relative sizes. Fig 4.41 shows the possible relative sizes of A+ and B– in the limiting case. There will be B– ions both directly above and below the central A+ ion but these have been omitted for the sake of clarity. In this limiting case, the B– ions are in contact with each other as well as with A+. Such a state of affairs can only exist when r–/(r++r–) is equal to cos 45°, i.e., when r+/r– is equal to 0.414. If different ions with larger r– and/or smaller r+ values, are concerned, a state of octahedral arrangement would be much less likely, for as shown in Fig 4.41 b, the B– ions would be in contact with each other, and repelling each other without being in contact with, and being attracted by A+. Such ions would rearrange into tetrahedral structure with a coordination number of 4. In simple language, the cation has become too small, or the anion too large for the centred cation to be surrounded by six anions; instead it is surrounded by four. If the radius ratio r+/r- becomes smaller still, the central cation may only be able to be surrounded by three anions in a coplanar, triangular arrangement with a coordination number of 3. Such an arrangement is shown in Fig 4.41c, from which it can be seen that the limiting radius ratio for such structure occurs when r–(r++r–) is equal to cos 30°, i.e., when r+/r– is equal to 0.155. If cations are too large to fit into the octahedral holes, the anions may occupy the larger cubic holes made possible by the more open spacing. By similar arguments and calculations the limiting radius ratio value for various coordination number can be summarized in Table 4.3 table 4.3 Radius ratio rules Limiting radius ratio r+/r

Coordination number

Structure

0-0.155

2

Linear

0.155-0.225

3

Triangular

0.225-0.414

4

Tetrahedral

0.414-0.732

6

Octahedral

0.732-1

8

Cubic

Co-ordination numbers of 5, 7, 9, 10 and 11 do not occur because of the impossibility of balancing the electrical charges of the ions concerned. When the radius ratio becomes

r −> r +

0.4

r+ r−
0.732 (c) Coordination number increases from 6 to 8 Unstable Coordination number decreases from 6 to 4

fig 4.42 Change in coordination number with the change in r+/r– value. equal to 1, ions of the same size are making up the crystal. Such a state of affairs is to be found in crystals of metals. In a given compound, the fraction of octahedral or tetrahedral voids that are occupied depends upon the chemical formula of the compound. Normally the anions are larger than cations and the larger anions occupy the lattice points while the cations occupy the interstitial voids occupied. From the knowledge of the packed structure and the voids occupied, we can have an idea about the structure of simple ionic compounds. For example, consider a compound of general formula AB in which B ions form a close packed lattice, there are two possibilities. (i) As the number of octahedral voids are equals to the number of B– ions that form a close packed lattice, all octahedral voids will be occupied by A+ ions. So the number of A+ ions and B– ions are equal, e.g., NaCl has this type of structure in which Cl– ions form a close packed structure and Na+ ions occupy all the octahedral sites. This structure is known as rock salt structure. (ii) There will be two tetrahedral voids per atom in a close packed structure, i.e., two tetrahedral sites available for every B– ion.To form the compound AB, only one half of the tetrahedral voids will be occupied. Zinc blend (ZnS) has this type of structure, in which S2– ions form cubic close packed lattice and Zn2+ions occupy one half of the tetrahedral site. This structure is known as zinc blend structure (iii) In the compounds of the type A 2B in which B – ions adopt cubic close packed lattice, then all the tet rahedral sites will be occupied by A + ions. Since

Solid State

there are two tetrahedral sites per atom, and all the sites are occupied, there will be two A + ions for each B – ion. This structure is known as antifluorite structure. (iv) Though the size of A+ ions is smaller than B– ions, sometimes they may adopt cubic close packed structure and B– ions occupy all the tetrahedral sites then the formula of the compound is AB2. This structure is known as fluorite structure.

4.11.1 structure of Ionic compounds of aB type

Thus, the number of NaCl units per unit cell = 4 The sodium chloride structure is also called as rock–salt structure. Alkali metal halides (except CsCl, CsBr and CsI) and oxides of alkaline–earth metals have this type of structure. (v) The ratio of (rNa+/rCl–) is 0.525 which exceeds the value 0.414 required for precise packing. So the arrangement of Cl– ions slightly opens up to accommodate sodium ions. Thus Cl– ions do not touch each other and from an expanded face centred lattice. Packing Efficiency in Rock–Salt Structure

1. Sodium chloride (i) In sodium chloride, chloride ions occupy all the corners and face centres of cubic close packed arrangement, i.e., face-centred cubic. (ii) The Na+ ions occupy all the octahedral sites Cl- ion octahedrally surrounded by six Na+ ions

4.25

ClNa+

Na+ ion octahedrally surrounded by six Cl- ions

fig 4.43 Rock salt structure (iii) We know that there are six octahedral holes around each Cl– ion, each Cl– ion is surrounded by 6 Na+ ions. Similarly each Na+ ion is surrounded by 6 Cl– ions (iv) Thus, in NaCl the coordination number of Cl– ion is 6 and that of Na+ ion is also 6. This is called 6:6 coordination and the ratio of Na+ and Cl– ions in this structure is 1:1 and the formula of the compound is Na+ Cl–. This can be calculated as follows Number of Na+ ions = 1 12 (at edge centres) × + 1 (at body centre)× 4 1=4 Number of Cl– ions = 1 1 8 (at corners) × + 6 (at face centres) × = 4 2 8

Let the edge length of rock- salt type crystal is 'a' Edge length a = rc + ra Face diagonal = 2a In the diagonal three anions are in touch each other ∴ Face diagonal = 2a = 4 ra 4 × Mw Noa 3

Density of face centred cubic = (Each unit cell contains 4 ionic pairs) Packing fraction = 3

3

4  a  a  4  4× π + 4 × π  0.414   3  2 2 3  2 2 3 a = 0.79 2. Zinc Sulphide Zinc sulphide occurs in two different forms (i) Zinc blende and (ii) Wurtzite (i) Zinc blende: The unit cell representation of zinc blende structure is shown in Fig 4.44. The unit cell of the zinc blende structure consists of S2– ion in a face centred cubic arrangement. The Zn2+ ions occupy half the tetrahedral voids. f

f f

S2-

f

f

2+

Zn

f

fig 4.44 Zinc blende structure. There are fourteen sulphide ions at the corners or faces (f) of the unit cell. Four zinc ions are bonded tetrahedrally to some of the sulphide ions.

4.26

Solid State

Since there are two tetrahedral voids available for each S2– ions and only half these voids are occupied by Zn2+ ion there is one Zn 2+ for every S2– ion. Thus the compound has formula ZnS. In this structure each Zn 2+ ion is surrounded by four sulphide ions in a a tetrahedral manner. Similarly, each S 2– ions is also surrounded by four Zn2+ ions in tetrahedral manner. Since there are 4 Zn 2+ and 4 S 2– ions per unit cell, there are 4 ZnS units per unit cell. The actual ratio of (rZn2+/ rS2–) is 0.40 which is a higher value than required for exact fitting of Zn2+ ions in the tetrahedral voids in the close packing of S 2– ions i.e., 0.225 for r+/ r–. The bond between zinc and sulphur has appreciable covalent character because (i) due to small difference in electronegativity and (ii) due to more polarising power of Zn2+ ion with pseudo inert gas configuratio and polarisability of bigger S2– (electronegativity of Zn =1.6 and that of S= 2.5). This structure is also known as sphalarite structure and is similar to that of diamond. In diamond all the positions are occupied by carbon atom. Other examples for sphalarite structure are CuCl, CuBr, CuI and AgI.

≡ Sulphide ions ≡ Zinc ions Formula, ZnS

fig: 4.45 Structure of wurtzite. Hexagonal structure.

= Cs = Cl

Packing Efficiency in Zinc Blende Structure Since zinc blende structure is face centred cubic, each unit cell consist of 4 ionic pairs. Let the edge length = a In the fcc structure the face diagonal consists of three anions in touch each other ∴ Face diagonal =

Density of fcc =

2a = 4 ra

4 × Mw Noa 3

Packing fraction = 3

4  4  a  a  4× π + 4 × π  0.225   3  3  2 2 2 2 3 a = 0.748 = 0.75

3

(ii) Wurtzite: Wurtzite has a hexagonal structure as shown in Fig 4.45. It has an overall shape of a triangular prism which may have shape ob tained by joining six trigonal prisms to make hexagonal structure, Sulphide ions occupy the corners of the triangular face. Four zinc ions make up the corners of tetrahedron centred on another sulphide ion at a central point in the

fig: 4.46 Caseium chloride structure showing (right) the cubical arrangement of eight caesium ions around one chloride ion.

cell. The coordination number of each ion is 4. 3. Caesium chloride In the structure of caesium chloride, the Cl– ions are at the corners of a cube whereas Cs+ ion is at the centre of the cube or vice versa. This structure can be assumed as that being made of two interlocking cubic arrangements of the two ions. This structure has 8:8 coordination i.e., each Cs+ is surrounded by eight Cl– ions and each Cl– ion is surrounded by eight Cs+ ions. As the rCs+ /rCl– value 0.93 is slightly greater than the cubic void ratio 0.732 the Cl– ions slightly open up to accommodate Cs+ ions. structure of Ionic compounds of the type aB2 The ionic crystals containing the cations and anions in the ratio of 1:2 or 2:1 are popularly known as AB 2 type or A2B Type. The common example of AB 2 type structure is calcium fluoride (CaF2) called fluorite structure and that of A2B type structure is called antifluorite structures.

Solid State

4.27

F− ion surrounded by 4 Ca2+ ions F

≡ Calcium ions

Ca2+ ≡ Fluoride ions

fig 4.47 Calcium fluoride structure Ca2+ ions are at the corners and faces of large cube inside there is small cube with F– ions at the corners.

4. Calcium fluoride: The unit cell of calcium fluoride is shown in Fig 4.47 The calcium ions are arranged in face centred cubic structure i.e., the Ca2+ ions are present at all the corners and at the centre of each face of the cube. The fluoride ions occupy all the tetrahedral voids. There are two tetrahedral voids per atom in a cubic closed packed lattice. So there are two tetrahedral voids for every Ca+2 ion. Since F– ions occupy all the tetrahedral sites, there will be two F– ions for Ca2+ ion. Thus, the formula of the compound is CaF2. In this structure each F– ion is surrounded by four Ca2+ ions while each Ca+2 ions is surrounded by eight F– ions. Thus, the coordination numbers of Ca2+ and F– ions are 8:4. In calcium fluoride unit cell has four calcium ions and eight fluoride ions as explained below: Number of Calcium ions =

Na+ ion surrounded by 4O2- ions

Na+ O2

O2- ion surrounded by 8 Na+ ions

fig 4.48 Structure of sodium oxide

1 1 8(at corner) × +6 (at face centres) × = 4 2 8 Number of fluoride ions = 8(within the body) × 1 = 8 Thus, the number of CaF2 units per unit cell is 4. This structure is also known fluorite structures. Other ionic compounds that have fluorite structure are SrF2, BaF2 BaCl2 and CdF2 5. Sodium oxide Na2O has the structure opposite to CaF2. The O2– ions constitute a cubic close packing type of lattice and the Na+ ions occupy all the tetrahedral voids i.e., the Na+ ions occupy the positions of fluoride ions and larger O2– ions occupy the positions of calcium ions of fluoride structure. So it is known as antifluorite structure. In this case coordination number of Na+ ions is 4 and that of O2– ions is 8. Thus, Na2O

has 4:8 coordination. This structure is known as antifluorite. structure. Several oxides and sulphides such as Li2O, K2O, Rb2O and Rb2S have the antifluorite structure. 6. Rutile structure: This is a form of titanium dioxide. The unit cell has tetragonal, rather than cubic struc ture as shown in Fig 4.49 titanium ions are at the corners and at the centre of cube. The titanium ion at the centre is surrounded by six oxide ions. When the unit cells are build up one on another, every titanium ion is surrounded by six oxide ions and every oxide has three titanium ions as their nearest neighbours. From this ratio one titanium has two oxide ions as in its formula unit.

4.28

Solid State

≡ Titanium(IV) ions ≡ Oxide ions

Formula TiO2

fig 4.49 Structure of rutile. The titanium ion at the centre of the larger cube also at the centre of a six oxide ions

structures of some solid crystals 7. Perovskite structure: It is a mineral with formula CaTiO3. The Ca2+ ions occupy the corners of the cube, the O2– ions occupy the face centres of the cube and the titanium ion Ti4+ lies at the centre of the cube.

≡ Titanium(iv) ≡ Calcium ions ≡ Oxide ions Formula, CaTiO3

fig 4.50 Structure of perovskite. Ca2+ ions are at the corners, O2– at the faces and Ti4+ ion at the centre.

8. Diamond: The structure of diamond is based on the face centred cubic structure but with extra carbon atom inside the cube. The arrangement around each carbon is tetrahedral. The coordination number of each carbon is 4.

f f f f f f

fig: 4.51 The face centred cubic structure of diamond. All the bonds are not shown. The atoms marked f are on the six faces of the cube. Bonds into neighbouring unit cell are not shown.

9. Packing in oxides of iron (i) Iron oxide (FeO): Iron oxide (FeO) has rock salt (NaCl) structure i.e., face centred cubic structure in which Fe 2+ ions should occupy the octahedral voids while O 2– ions adopt face centred cubic arrangement. This is the ideal arrangement. However, this oxide is always non – stoichiometric i.e., thus composition of Fe2+ and O 2– ions is not 1:1. It has been found to generally Fe 0.95 O (wustite). This composition can be obtained if a small number of Fe 2+ ions are replaced by Fe 3+ two thirds as many Fe2+ ions in octahedral sites. (ii) Magnetite (Fe 3O 4): The composition of Magnetite may be considered as FeO. Fe 2O 3. If 2/3 of Fe 2+ ions in FeO are con verted into Fe 3+ ions (for every three Fe 2+ ions two Fe 3+ ions to balance charge) we get this composition (FeO·Fe 2O 3). In Fe 3O 4 the oxide ions are arranged in cubic close packing arrangement. Fe 2+ ions occupy oc tahedral sites and the Fe 3+ ion are equally distributed between octahedral and tetra hedral voids. This structure is known as inverse spinal structure. Fe 3O 4 is the load stone used by ancient travellers to find the direction. Another example of this structure is MgFe 2O 4 in which Fe 2+ ions of Fe 3O 4 are replaced by Mg 2+ ions. 10. Normal spinel structure: A spinel is an important class of oxides consisting of two types of metal ions with the oxide ions. The oxide ions are ar ranged in ccp with Mg 2+ ions occupying tetrahedral voids and Al 3+ ions in a set of octahedral voids. This type of structure is known as spinel structure. The general formula of the compounds adopting spinel structure is AB 2O 4 e.g., Ferrites such as ZnFe2O 4. In this case the oxide ions are arranged in cubic close packing, the divalent cations (Mg 2+ or Fe2+ or Zn 2+) are in tetrahedral sites and trivalent ions (Al 3+ or Fe 3+) are in octahedral sites. In normal spinel structure one-eighth of the tetrahedral holes are occupied by divalent metal ions and one-half of the octahedral holes are occupied by trivalent met al ions. These are very important magnetic mate rial and are used in telephones and memory-loops in computers.

Solid State

practIce exercIse I unit cell edge length-atomic radius

Because it is bcc arrangement No. of atoms in the unit cell Z = 2 Atomic mass of the element = 50

solved problem 1 Mass of an atom = A metal crystallizes in a simple cubic unit cell. The length of the edge of the unit cell is 0.622 nm. What is the radius of each atom of the metal? Solution: For simple cubic Radius =

a 0.622 = = 0.311 nm 2 2

problems for practice 1. Iron crystallizes in a body centred cubic structure calculate the radius of Fe atom if edge length of unit cell is 286 pm. 2. Gold (atomic radius = 0.144 nm) crystallizes in a face centred unit cell. What is the length of the side of the cell? 3. Xenon crystallizes in the face centred cubic lattice and the edge of the cell is 620 pm. What is the nearest neighbour distance and radius of xenon atom? 4. The radius of an atom is 200 pm. If it crystallizes as a face centred cubic lattice, what is the length of the side of the unit cell? 5. Potassium crystallizes in body centred cubic lattice. What is the approximate number of unit cell in 4.0 g of potassium? Atomic mass of potassium is 39. 6. Metallic gold crystallizes in the face centred cubic lattice. What is the approximate number of unit cell in 2.0 g of gold? Atomic mass of gold is 197. 7. Aluminum crystallizes in a cubic close structure. Its metallic radius is 125 pm. (a) What is the length of side of the unit cell? (b) How many unit cells are there in the 1.00 cm3 of aluminum?

II unit cell dimensions and density solved problem 2 A compound having bcc geometry has atomic mass 50. Calculate the density of the unit cell, if its edge length is 290 pm. Solution: Edge length (a) = 290pm = 290×10–10 cm Volume of unit cell = (290×10–10 cm)3 = 24.39×10–24 cm3

4.29

50 6.02 × 1023

Mass of unit cell = 2×Mass of an atom =

2 × 50 100 = 23 6.02 × 10 6.02 × 1023

Density =

=

Mass of unit cell Volume of unit cell

100 6.02 × 10 × 24.39 × 10−24 23

= 6.819 cm–3 problems for practice 8. An element crystallizes in a structure having a fcc unit cell of an edge length 200pm. Calculate its density if 200 g of this element contains 24 × 1023 atoms. 9. A metal (atomic mass = 50) has a body centred cubic crystal structure. The density of the metal is 5.96 g cm–3. Find the volume of the unit cell. 10 Metallic gold crystallizes in the face centred cubic lattice. The edge length of cube unit cell a = 4.070A°. Calculate the closest distance between gold atoms and the density of gold. Atomic mass of Au = 197 amu.

III lattice type from density and unit cell dimensions solved problem 3 The density of chromium metal is 7.2 g cm–3. If the unit cell is cubic with edge length of 289 pm determine the type of unit cell (simple body centred or face centred). Atomic mass of Cr = 52 amu. No = 6.02×1023 mol–1. Solution: Length of the edge = 289pm = 289×10–12 = 289×10–10 Volume of unit cell = (289×10–10cm)3 = 24.14×10–24 cm3 Atomic mass Mass of atom = Avogadro Number 52 = 8.63 × 10−23 g 6.02 × 1023 Mass of unit cell = Density × Volume of unit cell = 7.2×24.14×10–24 = 1.738×10–22 g =

4.30

Solid State

Mass of the unit cell No. of atoms in unit cell = Mass of an atom 1.738 × 10−22 =2 8.63 × 10−23 Since the unit cell contains 2 atoms it is body centred cubic (bcc). =

IV atomic mass and number of atoms from density and unit cell dimensions solved problem 4 An element crystallizes into a structure which may be described by a cubic type of unit cell having one atom on each corner of the cube and two atoms on one of its diagonals. If the volume of this unit cell is 24×10–24 cm3 and density ‘d’ is 7.2 g cm–3. Calculate the number of atoms present in 200 g of the element. Solution: In a simple cubic unit cell x=1 Volume of unit cell a3 = 24×10–24 cm–3 d = 7.2 g cm–3 d=

M×Z No × a 3

7.2 =

200 × 1 No × 24 × 10−24

16. An element with molar mass 2.7×10–2 kg mol–1 forms cubic unit cell with edge length 405 pm. If its density is 27×103 kg/m3 what is the nature of the cubic unit cell? 17. Calculate the density of silver which crystallizes in the fcc cubic structure. The distance between the nearest silver atoms in this is 287 pm. 18. Copper crystallizes into a fcc lattice with edge length 3.61×10–8 cm. Show that the calculated density is in agreement with its measured values of 8.92 cm–3. 19. The unit cell of an element of atomic mass and density 108, 10.5 g cm–3 is a cube with edge length of 409 pm. Find the structure of the crystal lattice (simple cube, fcc or bcc). Avogadro’s number is 6.023×1023.

V the atomic radius from atomic mass and density solved problem 5 Niobium crystallizes in body centred cubic structure. If density is 8.55 g cm3, calculate atomic radius of niobium, given its atomic mass 93 g mol–1. Solution: Density = 8.55 g cm–3 Let the length of the edge = a cm Volume of unit cell = a3 cm3 Mass of an atom =

∴ No = 11.57×1023 atoms

=

problems for practice

11. Copper crystallizes in a fcc arrangement with cell edge of 3.608×10–8cm. The density of copper has been found to be 8.92 cm–3. Calculate the atomic (molar) mass of copper. 12. An element has a body centred cubic (bcc) structure with a cell edge of 288 pm. The density of the element is 7.2 g cm–3. How many atoms are present in 208 g of the element. 13. An element A crystallizes in fcc structure 200 g of this element has 4.12×1024 atoms. The density of A is 7.2 g cm–3. Calculate the edge length of the unit. 14. Chromium metal crystallizes with a body centred cubic lattice. The length of unit cell is found it be 287 pm. Calculate the atomic radius, the number of atoms per unit cell and density of chromium (Atomic, mass of Cr = 52 g mol–1 and No = 6.02 ×1023 mol–1). 15. Silver crystallizes in fcc lattice. If edge length of the unit cell is 4.077×10–8 cm and density is 10.5 g cm–3 calculate the atomic mass of silver.

Atomic mass 6.022 × 1023 93 6.022 × 1023

No. of atoms per unit cell = 2 (bcc) Mass of unit cell =

2 × 93 6.022 × 1023

Volume of unit cell = =

Mass Density

2 × 93 = 36.12 × 10 −24 cm3 6.022 × 1023 × 8.55

a3 = 36.12×10–24 cm3 Edge length a=(36.12×10–24)1/3 = 3.306×10–8 cm = 3.306×10–10 m Now radius in body centred cubic r = 3 × 3.306 × 10−10 = 1.431× 10−10 m 4 = 0.143 nm

=

3 a 4

Solid State

VI Packing Efficiency

Edge length, a = 400 pm

solved problem 6

Interionic distance =

Calcium crystallizes in a face centred cubic unit cell. Calculate the packing fraction for the calcium unit cell. Solution: In fcc lattice 1 1 Z =× × 8 +× × 6 = 1 + 3 = 4 8 2 Atomic radius r =

2 ×a 4

 2a  4 Volume of 4 atoms = 4 × 4 3 πr 3 = 4 × × π   3  4 

( )

=

2πa 3 6

Packing fraction =

=

3

2πa 3 3 a 6

2π 2 × 3.14 + = 0.74 6 6

problems for practice 20. A metal crystallizes into two cubic phases, face centred cubic (fcc) and body centred cubic (bcc) whose unit cell lengths are 3.5 and 3.0 A0, respectively. Calculate the ratio of densities of fcc and bcc. 21. The density of solid argon is 1.65 g/ml at –223°C. If the argon atom is assumed to be sphere of radius 1.54×10–8 cm what percentage of solid argon is apparently empty space? 22. Compute the packing factor for sphere occupying (i) a body centred cubic and (ii) a simple cubic structure when closest neighbours in both cases are in contact.

VII calculation of Ionic radii from edge length

CsCl has bcc arrangement and its unit cell edge length is 400 pm. Calculate the interionic distance in CsCl. Solution: CsCl has body centred cubic structure Interionic distance =

3 ×400 2

= 1.732 ×200 = 346.4 pm

problems for practice 23. CsCl forms a body centred cubic lattice. Caesium and chloride ions are in contact along the body diagonal of the cell. The length of the side of the unit cell is 412 pm and Cl– ion has radius of 181 pm, calculate the radius of Cs+ ion. 24. A solid AB has CsCl type of structure. The edge length of the unit cell is 4.04 A0. What is the distance of closest approach between A+ and B– ? 25. A unit cube length for LiCl (NaCl structure) is 5.4 A0. Assuming anion – anion contact, calculate the ionic radius of chloride ion.

VIII unit cell dimensions and density solved problem 8 An element having atomic mass 60 has face centred cubic unit cell. The edge length of the unit cell is 400 pm. Find out the density of the element. Solution: Unit cell edge length = 400 pm = 400×10–10 cm Volume of unit cell = (400×10–10cm)3 = 64×10–24cm3 Mass of the unit cell = Number of atoms in the unit cell × Mass of each atom 1 Number of atoms in the fcc unit cell = (8 × ) + 8 1 (6 × ) = 4 2 Atomic mass 60 = Mass of one atom = Avogadro's No 6.03 × 1023 ∴Mass of the unit cell = 4 ×

Density of the unit cell =

solved problem 7

3 a 2

4.31

=

60 6.03 × 1023

Mass of unit cell Volume of unit cell

4 × 60 1 × = 6.2 g cm −3 6.03 × 1023 64 × 10−24

4.32

Solid State

problems for practice

No = –3

26. An element (density 6.8 g cm ) occurs in bcc structure with cell edge of 290 pm. Calculate the number of atoms present in 200 g of the element. 27. Caesium chloride crystallizes as a body centred cubic lattice and has a density 4.0 g cm–3. Calculate the length of the edge of the unit cell of caesium chloride crystal. Molar mass of CsCl = 168.5 g mol–1 NA = 6.02×1023 mol–1. 28. Calculate the distance between Na+ and Cl– ions in NaCl crystal if the density is 2.165 g cm–3 [Molar mass of NaCl = 58.5 g mol–1 NA= 6.02×1023 mol–1].

Ix avogadro constant from density and unit cell dimensions solved problem 9 Calculate the value of Avogadro number from the following data. Density of NaCl = 2.165 g cm–3, distance between + Na and Cl– ions in NaCl crystal is 281 pm. Solution: Sodium chloride has face centred cubic structure. Therefore, the number of formula units or molecules per unit cell = 4 Mass of the unit cell = No of molecules per unit cell × mass of a molecule Mass of each molecule =

Molar mass Avogadro number

Let Avogadro number = No Molar mass = 23+35.5 = 58.5 ∴ Mass of each molecule =

58 : 5 No

58.5 No Mass of unit cell Density of unit cell = Volume of unit cell Mass of unit cell = 4 ×

Since distance between Na+ and Cl– ions is 281 pm, the length of edge is double the distance between Na+ and Cl– ions Edge of unit cell = 2×281 pm = 562 pm Or = 562× 10–12 m = 562 pm Volume of unit cell = (562×10–10)3 cm3 = 17.75×10–23 cm3 2.165 g cm–3 =

4 × 58.5 No(17.75 × 10−23 cm −3 )

4 × 58.5 2.165 × 17.75 × 10−23

= 6.09 × 1023 problems for practice 29. Calculate the value of the Avogadro constant from the inter nuclear distance of adjacent ions in NaCl, 0.282 nm and the density of the solid NaCl, 2.17×10 3 kg/m 3. A unit cell contains 4 NaCl formula units. 30. The density of a particular crystal LiF is 2.65 g/ cc. X-ray analysis shows that Li+ and F– ions are arranged in a cubic array at a spacing of 2.01 A°. From these data calculate the apparent Avogadro constant. 31. CsCl has body centred cubic lattice with the length of a side of a unit cell 412.1 pm and aluminium is face centred cubic lattice with length of the side of unit cell 405 pm. Which of the two has larger density (Atomic mass Cs = 132.9, Al = 29.9).

x radius ratio-Voids-structure-coordination number solved problem 10 In a cubic close packed structure of mixed oxides, the lattice is made up of oxide ions, one-eight of tetrahedral voids are occupied by divalent ions (A2+) while one-half of octahedral voids are occupied by trivalent ions (B3+) what is the formula of the oxide? Solution: In a close packed arrangement, there is one octahedral and two tetrahedral voids corresponding to each atom consituting the lattice. Therefore, if Number of oxide ions (O2–) per unit cell = 1 Number of tetrahedral voids per oxide ion in lattice = 1×2 = 2 Number of divalent (A2+) ions =

1 1 ×2 = 8 4

Number of octahedral voids per oxide ion in lattice = 1×1 = 1 Number of trivalent (B3+) ions = 1×

1 1 = 2 2

Formula of the compound = A1/4B1/2O Whole number formula = AB2O4

Solid State

problems for practice 32. Zinc sulphide crystalises with zinc ions one-half of the tetrahedral holes in a closest-packed array of sulphide ions. What is the formula of zinc sulphide? 33. Aluminium oxide crystalises with aluminium ion in two thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminium oxide? 34. Lithium forms a bcc lattice. If the lattice constant is 3.5 × 10–10 and the experimental density is 5.3 × 102 kg m–3. Calculate the percentage occupancy of Li metal (Li = 7). 35. A compound alloy of gold and copper crystallizes in a cube lattice in which the gold atoms occupy the lattice points at the corners of the cube and the coppers atoms occupy the centres of each of the cube faces. Determine the formula. 36. The coordination numbers of Ba2+ in BaF2 is 8. What must be the coordination number of F? 37. The ionic radii of Ca2+ and O2– are 99 pm and 140 pm respectively. What type of void Ca2+ ion is likely to occupy? What is the most probable coordination number of Ca2+ ion? 38. Predict the coordination number of ions M+ and Y– in MY. Ionic radii of M+ and Y– are 97 and 251 pm respectively. 39. A solid AB has the NaCl structure of the radius of the cation is 100 Pm. What is the maximum value of radius of anion? 40. KF has rock – salt structure. What is the distance between K+ and F– in KF, if its density is 2.48 g cm–3? 41. If three elements P, Q and R crystallizes in a cubic solid lattice with P atoms at the corners, Q atoms at the cube centre and R atoms at the centre of edges, then write the formula of the compound. 42. When atoms are placed at the corners of all 12 edg es of a cube, how many atoms are present per unit cell? 43. A unit cell consists of a cube in which there are A atoms at the corners and B atoms at the face centres and A atoms are missing from 2 corners in each unit cell. What is the simplest formula of the compound? 44. A unit cell consists of a cube in which there are anions (Y) at each corner and cations (X) at the centre of alternate faces of the unit cell. What is the simplest formula of compound? 45. A solid is made of two elements X and Y. Atoms of X are in bcc arrangement and Y atoms occupy all the octahedral sites and alternate tetrahedral sites. What is the formula of the compound?

4.33

4.12 crystallography So far we have discussed what is a solid, what are the possible arrangements of particles in solids and what are the different type crystal structures. Now we will see how to determine the structure of a crystal. The branch of science which deals with the geometrical properties and structure of crystals and crystalline is called crystallography. Geometric crystallography is concerned with the outward spatial arrangement of crystal plane and the geometric shape of crystals and is based on three funda mental laws, namely (a) the law of constancy of interfa cial angles (b) the laws of symmetry and (c) the law of rationality of indices.

4.12.1 the law of constancy of Interfacial angles (a) An interfacial angle is defined as the angle between lines drawn normal to two faces. The law of con stancy of interfacial angles first proposed by Nicolus Sten in 1669 states that for a given substance the corresponding faces or planes forming the external surface of crystal intersect a definite angle called interfacial angle and remains always constant no matter how the faces develop. This is illustrated in Fig 4.52.

fig 4.52 Interfacial angles in a crystal

Although the external shape is different but the interfacial angles are same. This constancy of interfacial angles introduces enormous simplification in crystallography in the study of crystal structure. (b) The laws of symmetry: The internal structure of a crystal is due to a definite arrangement of its component atoms, molecules or ions and this arrangement repeat itself over and over in a definite space, like wall paper in two dimensions. This characteristic is expressed by saying that crystal is a symmetrical structure. The internal symmetry corresponds to the external symmetry

4.34

Solid State

of the form of the ideal crystal. There are three possible types of symmetry. (i) Plane of symmetry: It is an imaginary plane which passes through the centre of a crystal and divides it into two equal portions such that one part is the exact mirror image of the other. (ii) Axis of symmetry: An axis of symmetry is a line about which the crystal may be rotated such that it presents the same appearance more than once during the complete revolution. A n-fold (or n-gonal) axis of symmetry is an axis such that when an ideal crystal is rotated around it the crystal occupies the same position in space n-times in complete turn of 360°. The axes are shown in the Fig 4.53.

2-FOLD

4-FOLD

(a)

(b)

3-FOLD

(c)

6-FOLD

(d)

fig 4.53 Various axes of symmetry

It the same appearance is repeated 2 times during complete rotation of 360°, the crystal is known to have 2 – fold axis of symmetry. Similarly, if same appearance is repeated 3-times during complete rotation of crystal, the crystal is known to have 3-fold axis of symmetry. Axes 2-, 3-, 4-, and 6- fold symmetry are also called diagonal, trigonal, tetragonal and hexagonal axes respectively or diad, triad, tetrad and hexad respectively when the crystal comes to occupy the same position in space 2, 3, 4 and 6- times in a complete rotation. The angle through which the crystal will have to be rotated to get same appearance will be 360/n i.e., 180°,120°, 90°, and 60° respectively for 2-, 3-, 4-and 6- fold symmetry. However the five fold axis of symmetry does not occur. (iii) Centre of symmetry: It is a point in the body of the crystal such that a line drawn through it intersects the opposite faces at equal distances in both directions. Only one centre of symmetry is possible for any crystal.

4.12.2 elements of symmetry The total number of planes, axes and centre of symmetries possessed by a crystal is known aselements of symmetry. It should be noted that a crystal may have one or more number of plane or axis of symmetry, but it has only one centre of symmetry. Some crystals have no plane of symmetry, others have no axis of symmetry and some have no element of symmetry at all. Now we calculate the total elements of symmetry in a simple cubic crystal such as NaCl crystal. This crystal has 23 elements of symmetry as is clear from the Fig 4.54 These elements of symmetry are. (i) Rectangular plane of symmetry: This is shown in Fig. 4.54. There can be two more such planes each of which will be at right angles to the plane already shown. Thus there are three rectangular planes of symmetry. (ii) Diagonal plane of symmetry: The plane passing diagonally through the cube is shown in the Fig. 4.54. There can be a total of six such planes as a little imagination will show. (iii) Axis of four–fold symmetry: One of four fold axis is shown in Fig. 4.54. There are three axes mutually perpendicular to each other. (iv) Axis of three–fold symmetry: One such axis is shown in the Fig 4.54. There can be a total of four such three fold axes. (v) Axis of two fold–symmetry: One such axis emerging from opposite edges is shown in the Fig 4.54. There are evidently six such axes of two fold symmetry. (vi) Centre of symmetry: There is only one centre of symmetry lying at the centre of the cube.

SQUARE PLANAR PLANE OF SYMMETRY

DIAGONAL PLANE OF SYMMETRY AXIS OF FOUR FOLD SYMMETRY

CENTRE OF SYMMETRY AXIS OF THREE FOLD SYMMETRY

AXIS OF TWO FOLD SYMMETRY

fig 4.54 Elements of symmetry

Solid State

Thus, the number of symmetry elements of various types is. (a) Plane of symmetry = 3 + 6 = 9 elements (b) Axes of symmetry = 3 + 4 + 6 = 13 elements (c) Centre of symmetry = 1 element.

4.12.3 law of rational Indices Crystals are bounded by a number of surfaces which are generally perfect flat. These surfaces are known as faces. All the faces corresponding to a crystal are said to constitute a form. A simple form of a crystal is entirely made up of like faces, while a crystal which consists of two or more simple forms is said to be a combination. The principal planes of symmetry of a crystal which contain atoms are called the plane of crystal. Let us consider three axes OX, OY and OZ represent the three crystallographic axes with ‘O’ as origin (Fig 4.55) ABC is a unit plane which cuts three axes at ABC respectively giving unit intercepts OA = a, OB = b, and OC = c. The lengths a, b, and c may or may not be equal but their ratio OA : OB ; OC = a : b : c is constant. The ratio of unit intercepts is called axial ratios. The law of rational indices was first proposed by Havy. The law simply states that the intercepts of the planes of a crystal on a suitable set of axes can be expressed by small multiples of unit distances. The indices used in practice to denote the direction of a plane of a crystal are the Miller indices. Miller indices of a plane intercept on the various crystallographic axes. For calculating Miller indices, a reference plane, known as parametral plane, is selected having intercepts a, b and c along x, y and z axes. Then, the

intercepts of the unknown plane are given with respect to a, b and c of the parametral plane. Thus, the Miller indices are h=

a intercept of the plane along x-axis

k=

b intercept of the plane along y-axis

l=

c intercept of the plane along z-axis

In order to make an idea of Miller indices more clear let us consider a cube having three rectangular axes OX, OY and OZ. The face LRTS cuts one of these axis (OX axis) at a point S and is parallel to the other OY and OZ. Supposing the sides of the cube of unit length, the intercept made by this plane on the three axes are in the ratio 1/1:1/∝: 1/∝ (Miller indices) d (hkl) =

a 2

h + k 2 + l2

Where a is the length of the cube side while h, k and l are the Miller indices of the plane. The spacings of the three planes (100), (110) and (111) of simple cubic lattice can be calculated.

d(100) =

d (110) =

Z

a 2

1 +0+0

=a

a 2

2

1 +1 + 0 d(111) =

K

4.35

a 2

2

2

1 +1 +1

=

=

a 2 a 3

The ratio is d(100) : d(110) : d(111) :1:

C c a

O

1 2

:

1 3

= 1 : 0.707 : 0.577

Similarly dhkl ratios of face-centred cubic and body centred cubic can be calculated. For face - centred cubic.

b B

A L

M

Y

d(100) : d(110) : d(111) = 1:

1 2

:

2 3

= 1 : 0.707 : 1.154

For body - centred cubic

X

d(100) : d(110) : d(111) = fig 4.55 Crystallographic axes

1 1 1 : : 2 2 2 3

4.36

Solid State Z M

The distances between the parallel planes in crys tal are designated as d hkl. For different cubic lattic es these inter planar spacings are given by general formula.

T R

P

4.12.4 x-rays and Internal structure of crystal O S N

X

L

Y

Z

(100)

Y

(110)

X

(112)

(111)

fig 4.56 Miller indices of planes in cubic lattice

= 1: 2 :

1 3

= 1:1.414:0.577 Hence the Miller indices of the face LRTS will be 1:0:0. This face is known as cubic face and is usually expressed as (100) face. The face MRLN cuts the axes OY and OZ and is parallel to the axis OX. Hence the resulting face is called diagonal face and is expressed as (110) face. Similarly, the face NSM cuts all faces with equal intercepts and its indices will therefore 1:1:1. This face is called cube diagonal face and is denoted as (111) face. Miller indices of the planes in a cubic crystal are given in the Fig 4.56.

The conclusion that X-rays were light rays of very short wave length and the realization that a crystal consisted of regular array of planes of atoms prompted Max Von Lane to suggest that the crystal should behave as a diffraction grating for X-rays. If the wavelength of X-rays might be the same order of magnitude as the spacing between the ion or atoms of the crystal, one should expect to observe a diffraction effect. William H. Bragg and his son William L. Bragg followed up on Von Lanes work and gave a simpler explanation of the diffraction of X-rays by crystals. According to them X-rays behave as if they are reflected by planes of atoms in a crystal and when a beam of X-rays is allowed to fall on a crystal, each atom there in acts as a source of scattered radiation of the same wave length. They developed a useful relationship between the wavelength of X-rays and the spacing between the lattice planes. As the atoms in a crystal are orderly arranged, a beam of X-rays is reflected from a plane of atoms similar to the reflection of light wave from a plane mirror. Let us consider a beam of monochromatic X-rays incident on a set parallel and equidistant planes called lattice planes or Bragg planes, in the crystal structure. The beam of X-rays striking the atoms which constitute these planes will be diffracted in such a manner as to cause either interference with or reinforcement of the beam diffracted from the first or outer plane, and the whole beam will behave as if it had been reflected from the surface of the crystal as shown in Fig 4.57. Two waves AB and DG are incident on planes 1 and 2 respectively at an angle θ. These waves are in phase when these waves are reflected from planes 1 and 2 respectively. The wave 2 travels a greater distance than wave 1. Drawing perpendiculars from point B on DE and EF as BG and BH respectively, the path difference of the two waves equal to GE+EH can be calculated. The reflected waves BC and HF must be in phase so that the two waves reinforce each other. This is possible when the path difference is either one whole wavelength λ or any integral multiple of it i.e., nλ i.e., GE + EH = nλ From Fig 4.57, ∆BGE and ∆BHE are congruent ∴ GE = HE

Solid State

A

λ λ

D

B

θ

θ

C

4.37

WAVE 1 e WAVE 2

F PLANE 1

H

G

PLANE 2 d

E

fig 4.57 X-rays diffraction from crystals

GE GE = = Sin θ BE d ∴ d Sin 𝜽 + d sin 𝜽 = nλ or 2d sin 𝜽 = nλ

This simple equation is known as Bragg’s equation. If θ, n and λ are known, d can be calculated.

4.12.5 determination of crystal structure

1. Bragg’s method The apparatus used by Bragg for the investigation of crystals is shown in the Fig 4.58. X-rays are gener ated from the X-ray tube A and made as far as possible

A + R

X-RAY TUBE TURN TABLE

TARGET

SAMPLE CRYSTAL

S

FOCUSSING SLIT

L CRYSTAL PLANE

fig 4.58 X-ray spectrometre (rotating crystal technique)

monochromatic by passing through suitable absorbing screens S, after which they pass through lead slit L to obtain a fine beam. The beam strikes the centre of a giv en face of crystal. The crystal itself is placed just at the centre of a rotating table coaxially with the table and the crystal is a rotating arm carrying an ionization cham ber. The turn tables as well as the arm are provided with circular scales so that the angles through which that are turned can be obtained directly. The X-rays filtered and collimated by the slits S and L are allowed to fall on the crystal and the reflected rays are passed into the ioniza tion chamber which is connected with an electrometre. The chamber C filled with SO 2 gas or methyl bromide vapour and the extent of ionization is measured with the help of electrometre attached to the ionization chamber. The crystal face and the ionization chamber are rotated about as their common axis, the chamber being rotated through twice the angle through which the crystal is turned. In this manner the angle between the face and the ionization chamber and the face and the incident beam are kept the same. As the glancing angle is changed by turning the crystal face slowly, the strength of ionization is recorded from the electrometre. These are plotted θ vs λ and the maxima are obtained from the graph. These maxima should be governed by the equation. 2d sin 𝜽 = nλ This method has successfully been applied in the study of NaCl and KCl crystals. The method is however, time consuming and it is used mainly to obtain accurate intensity data. 2. The powder method The Bragg's method of obtaining inter planar spacing’s has the disadvantage that mounting the crystal on a pre axis is time consuming. The simplest method of obtaining inter

4.38

Solid State

MONOCHROMATIC CRYSTAL SHIELD

CAMERA

POWDERED CRYSTALS IN GLASS CAPILLARY

FILM

X-RAY TUBE

fig 4.59 Crystal structure determination by powder method.

planar spacing is the Debye – Scherrer powder method. A sample of the crystal is ground to a powder and placed in a thin walled glass tube which is mounted in the X-ray beam. Since many crystals are present, all having different orientations, there are some which will be oriented as to satisfy the Bragg’s relation for given set of planes. Another group will be oriented so that the Bragg’s relation is satisfied for another set of planes and so on. The experimental set up is shown in the Fig 4.59. In this method a cone of reflected radiation is produced which is recorded on the film as curved lines. The distance between the lines on the film is measured accurately. After obtaining the powder pattern the lines are indexed. The distance x of each line from the central spot is measured, usually by halving the distance between the two replications on either side of the centre. If r is the radius of the film, then its circumference 2πr corresponds to a scattering angle of 360°. Then x 2θ = 2πr 360 From the Bragg angle 𝜽 the inter planar spacing, d can be calculated. Using the densitometre, a device for measuring the light transmission of the developed film, the intensity or the lines can be measured.

4.12.6 structure of crystals 1 Sodium chloride (NaCl): In case of NaCl the first order peaks are formed in a particular experiment to

occur at 5°18', 7°31', 4°36' for reflection from (100), (110) and (111) type planes, respectively. The inter planar distances are then in the ratio of the reciprocals of the sin θ values i.e., 2d sin 𝜽 = n λ or Hence d100 : d110 : d111 = =

d=

nλ 2sin θ

d∝

1 sin θ

1 1 1 : : sin 5°18' sin7°31' sin4°36'

1 1 1 : : 0.0924 0.1307 0.0801

= 1:0.707:1.154 =1:

1 2

:

2 3

This is the ratio predicted for face centred cubic structure. Hence in crystals of rock salt the structural units are arranged in a face centred cubic lattice. Fig 4.60 (a) shows the structure of NaCl which consists of a face centred array of Na+ ions and an interpenetrating face centred array of Cl- ions. The (100) and (110) planes contain an equal number of both kinds of ions but the (111) planes consists of either all Na+ or Cl– ions. Now if X-rays are scattered from the (111) plane, the scattered rays from successive Na+ planes are exactly in phase, the rays scattered from the inter cleaved Cl– planes are retarded by half a wavelength

Solid State

4.39

100

Cl-

110 Na+

111 0

5°

10°

15°

20°

25°

θ (a)

(b)

fig 4.60 Space lattice of sodium chloride (b) intensity vs 𝜽 for Bragg scattering from (100) (110), and (111) type of KCl and therefore exactly out of phase. The first order (111) deflection is therefore weak, the second strong, the third again weak and fourth again strong. Such a weakening of reflection on odd orders indicate that reflections are taking place at dissimilar planes. This alteration of these intensities is due to alter nate planes of chlorine atoms separated by the layers of planes of sodium ions. The odd order reflections which are weakened, indicate that mid way between two lattice planes of the same kind, there may appear one of a dif ferent type. 2. Potassium chloride (KCl). When sylvine (KCl) crystals were similarly investigated the intensity- order of reflection diagram was obtained as shown in the Fig 4.60 (b) The first order spectrum from (100), (110) and (111) planes of KCl was observed 5°13', 7°18, and 9°3' respectively. This yield the ratio d100 : d110 : d111 : =

1 1 1 : : sin 5°13 ' sin 7°18 ' sin 9°3 '

= 1:

1

= 1:

2 3 : 2 3

2

:

1 3

= 1:0.704:0.575 The ratio is very close to that expected to exist between spacing along a simple cubic lattice. Hence KCl crystal has a simple cubic lattice.

The first order reflection from (100) plane of NaCl and KCl takes place at 5°18' and 5°13' respectively. Now according to equation 1 ∝sin θ d Interplanar distance is inversely proportional to sin 𝜽 . Hence d100 (KCl) sin 5°18 ' = = 1.11 d100 ( NaCl) sin 5°13 '

Since the volume of a small cubic lattice is (d100) 3 taking into consideration the idea that. NaCl and KCl have the same crystal structure the ratio of the molecular volume of the two salts is given by. (1.11)3 = 1.37 The value is in good agreement with the experimental value of 1.39. This suggests that both NaCl and KCl have same type of space lattice. According to Bragg, both NaCl and KCl actually have the same structure i.e., they have face centred cubic lattice as shown above in which Na or K atoms and chlorine atoms are arranged as shown in Fig. 4.60. According to Bragg, the X-ray scattering factor for an atom is equal to the number of extra nuclear electrons at small glancing angles. The atomic numbers of potassium (19) and chlorine (17) are very close to each other. More

4.40

Solid State

over potassium and chloride ions in KCl have equal number of electrons. If we assume that all the atoms are identical it means that face centred arrangement of NaCl become simple cubic arrangement for KCl. In (111) face of KCl the atomic numbers of both K and Cl are so close that the reflected intensities are nearly same and the odd orders are completely cut - off and hence KCl appears to have a simple cubic lattice. But in the case of NaCl, the atomic number of sodium and chlorine are quite different. As a result of this difference, their scattering factors are also different and hence give rize to a face centred cubic lattice. Thus, both NaCl and KCl may be considered to have the same structure.

As there is only one ion each elementary cube in the system either sodium or chlorine. M = ρd 3 2N Here M = 58 g = 58×10–3 kg N = 6.023 × 1023, ρ = 2.16 × 103 kg m–3  M  d=  2 Nρ 

1

3

  58 × 10−3 = 23 3  2 6 023 10 2 16 10 . . × × × ×  

solved problem 11 A beam of X-rays of wave length 0.154 nm produces blackening of the photographic plate at an angle 11°27' from 100 plane for first order maxima. Calculate the distance between the successive planes of molecules in the crystal. Solution: λ = 0.154nm 𝜽 =11°27' n=1 2d sin 𝜽 = n λ 2d 11°27’ = 1 × 1.54×10–10 d=

=

1.54 × 10 −10 2 sin 11°27 ' 1.54 × 10 −10 2 × 0.1985

= 3.88×10–10 m solved problem 12 Find the lattice spacing in a cubic crystal of NaCl having a density of 2.16×103 kg m–3. Assume the molecular weight of sodium chloride to be 58. Solution: If M is the molecular weight of NaCl and N the Avogadro’s number, since there are 2N ions or diffracting centres i.e., N ions of Na+ and N ions of Cl– in one g molecule, the mass associated with each ion is M 2N

1

3

=2.814×10–10 m.

problems for practice 46. Calculate the distance between 111 planes in a crystal of Ca. Also calculate the distance between 222 planes. Which planes are closer (a = 0.556 nm)? 47. The angle of diffraction 2θ for a first order nature was bound to be 27°8' using X-rays of wavelength 2.29 A0. Calculate the distance between two diffracted planes. 48. Calculate the wavelength of X-rays which produces a diffraction angle 2θ equal to 16.800 for a crystal. Assume first order diffraction with inter particle distance in crystal of 0.2 nm. 49. The first order reflection of a beam of X-rays of wavelength 1.54 A 0 from the (100) plane of a crystal of the simple cubic type occurs at an angle of 11.29 0 Calculate the length of the unit cell (sin 11.29 0 = 0.1991) 50. Metallic gold (at. wt. =197) crystallizes in a bcc unit cell. The second order reflection (n = 2) of X-rays for the planar that makes up the tops and bottoms of the unit cell at θ = 22.20°. The wavelength of the X-rays is 1.5A°. What is the density of the metallic gold? 51. When an electron in an excited molybdenum atom falls from the L to the K shell an X-ray is emitted. These X-rays are diffracted at an angle of 7.75° by planes with a separation of 2.64A°. What is the difference in energy between the K shell and L shell in molybdenum assuming a first order dif fraction?

Solid State

4.13 ImperfectIons In solIds Even though in crystalline solids the particles are ar ranged in an orderly manner, there are certain imper fections. Generally a solid consists of large number of small crystals. These small crystals have defects in them. The imperfections in solid occur when crystalli zation process is carried at fast or moderate rate. Single crystals are formed when the crystallization process is carried at extremely slow rate. Such crystals are also not free from defects. Any deviation from completely ordered arrangement in a crystal constitutes a disorder or a defect or imperfection. These imperfections will not only change the properties of crystals such as electrical conductivity and mechanical strength but also give rise to new properties. The defects that occur in crystalline solids are broadly classified into two types. 1. Point defects: These are due to the irregularities or deviations from ideal arrangement around a point or an atom in a crystalline substance. These are also called as atomic imperfections and are due to missing or misplaced ions. 2. Line defects: These are the irregularities or deviations from ideal arrangement in entire rows of lattice point. These are called Lattice imperfections and extend along lines (line defects) or planes (plane defects). Line defects are also called dislocations These irregularities are collectively called as crystal defects.

4.41

on temperature. Basically these are of two types: Vacancy defects and interstitial defects. (i) Vacancy defect: When some constituent particles (atoms or molecules) are absent in the lattice sites and create vacant sites, the vacancy defects arise in crystals. Due to these defects the density of a substances decrease. This defect can also develop when a substance is heated.

Vacancy defects fig 4.61 Vacancy defects (ii) Interstitial defect: When some constituent particles (atoms or molecules) occupy an interstitial site (extra to lattice points) the crystal is said to have interstitial defect. This defect increases the density of the substance due to the presence extra atom or molecules in interstitial sites.

4.13.1 point defects The point defects in solids are again two types a) stoichiometric point defects b) Non-stoichiometric point defects. (A) Stoichiometric point defects: The compounds in which the number of positive and negative ions are exactly in the ratio indicated by the chemi cal formula are called stoichiometric compounds. The defects that do not disturb the stoichiometry (the ratio of number of positive and negative ions) are called stoichiometric defects. Stoichiometric compounds obey the law of constant composition that the same chemical compound always contains the same composition by weight. At one time these are called Daltonide compound in contrast to Berthollide or nonstoichiometric compounds where the chemical composition of a compound was variable, not constant. The stoichiometric de fects are also called intrinsic or thermodynamic defects because the number of these defects depend

Interstitial defects fig 4.62 Interstitial defects

Vacancy and interstitial defects as explained above can be shown by non-ionic solids. Ionic solids must always maintain electrical neutrality i.e., the defects should not disturb the stoichiometry (the ratio of numbers of positive and negative ions or charges). These are of following types. 1. Frenkel defect: This defect was discovered by a Russian scientist Frenkel in 1926. It arises when an ion

4.42

Solid State

is missing from its normal position and occupies an interstitial site between the lattice points. Generally the smaller ion (usually cation) is dislocated from its normal site to an interstitial site. It creates a vacancy defect at its original site and an interstitial defect at its new location. This defect is also called as dislocation defects. Cation in Interstitial Site

Cation vacancy A+

B-

A+

A+ A+

B-

B-

A+

B-

A+

B-

A+

B-

A+

B-

A+

B-

A+

B-

fig 4.63 A Frenkel defect Frenkel defect generally occurs in compounds in which (i) coordination number is low and (ii) anions are much larger in size than the cations. In pure alkali metal halides which contain larger cations, Frenkel defects are not common because the larger cations cannot get into interstitial positions. These defects are mainly found in silver halides, zinc sulphide in which the small sized cations Ag+ and Zn2+ can occupy the interstitial sites. 2. Schottky defect: This defect was discovered by German scientist Schottky in 1930. It arises due to missing of ions from their normal sites. It is basically a vacancy defect in ionic solids. In order to maintain electrical neutrality, the number of missing cations and anions are equal. Cation Vacancy

A+

B-

B-

Anion Vacancy

A+

B-

A+

B-

A+

B-

A+

B-

A+

B-

A+

B-

A+ A+

B-

fig 4.64 The Schottky defect

Schottky defect occurs mainly in highly ionic compounds where the positive and negative ions are of similar size and hence the coordination number is high (usually 8 or for example, NaCl, KCl, KBr and CsCl ionic solids have Schottky defects. It has been observed that in NaCl there are about 10 6 Schottky pairs per cm 3 at room temperature. In one cm 3 there are about 10 22 ions and this means that there will be one Schottky defect per 10 16 ion in NaCl. Similar to simple vacancy defect, Schottky defect also causes decrease in the density of the substance. In certain ionic solids such as AgBr, both Frenkel and Schottky defects occur.

consequences of schottky and frenkel defects Schottky and Frenkel defects causes several interesting changes in the properties of ionic crystals. 1. Because of the presence of Frenkel and Schottky de fects the electrical conductivity of crystal increases. When an electric field is applied a nearby ion moves from its lattice point to occupy a hole. This results in creating a new hole and another nearby ion moves into it and so on. This process continues and a hole, there by moves from one end to another end. Thus, it conducts electriccity across the whole of crystal. 2. In Schottky defects due to the presence of holes in crystal, its density decreases. 3. The presence of holes also causes the decrease in lattice energy resulting in the decrease of stability of crystal. If large numbers of holes are present the lattice may also collapse. 4. The closeness of similar charges in Frenkel defects tends to increase the dielectric constant of the crystals. differences Between schottky and frenkel defects Schottky defect

Frenkel defect

It is due to the missing of ions It is due to dislocation of ions from their normal crystal sites. from their normal site to an interstitial site. Due to this defect density of crystal decreases.

It does not affect the density of the crystal.

This defect is present in the ionic solids having high coordination number and in which cations and anions are of equal sizes, e.g., NaCl, CsCl

This defect generally occurs, in ionic solids having low coordination number and in which anions are larger in size than cations, e.g., AgCl, ZnS

Solid State

B. Impurity defects: Impurity defects in ionic solids may be created by introducing impurity ions. If the impurity ions have different valence state than that of the host ions vacancies are created. For example if molten NaCl containing a little amount of SrCl2 is crystallised some of the sites of Na+ ions are occupied by Sr2+ ions. Each Sr2+ ion replaces two Na+ ions. It occupies the site of one ion and it other site remains vacant. The cationic vacancies thus produced are equal in number to that of Sr2+ ions. Another similar example is the solid solution of CdCl2 and AgCl.

Na+

Cl-

Cl-

Na+

Sr2+

Cl-

Na+

Cl-

Cl-

Na+

Cl-

Na+

Cl-

Cl-

4.43

maintained either by having extra electrons in the structure or changing the charge on some of the metal ion. This makes the structure imperfect in some way, i.e., it contains defects which are in addition to the normal thermodynamic defects already discussed. 1. Metal excess defects These defects occur due to the presence of excess positive ions. They may occur in the following different ways. (i) Anion vacancies: In this type of defect nega tive ion may be missing from its lattice site leaving a hole which is occupied by an elec tron there by maintaining the electrical neu trality. Evidently there is an excess of positive ions although the crystal as a whole is electri cally neutral. This defect is rather similar to a Schottky defect in that there are holes and not interstitial ions. This type of defect is formed by crystals which would be expected to form Schottky defects. For example, when alkali metal halide crystals are heated in the atmos phere of corresponding metal vapours anion vacancies are produced. The Cl – ions diffuse to the surface of the crystal and combine with sodium atoms to form NaCl.

Na+

Fig 4.65 Impurity defect by introducing cation vacancy in NaCl by substitution on of Na+ by Sr2+

Sometimes the impure ions may occupy interstitial sites in a crystal. For example in the non stoichiometric cuprous oxide (Cu2O) in which the copper to oxygen ratio is slightly less than 2:1 the two Cu+ ions are replaced by one Cu2+ ion due to which some of the positions to be occupied become vacant. (B) Non–stoichiometric defects Non stoichiometric or Berthollide compound exist over a range of chemical composition. The ratio of the number of atoms of one kind to the number of atoms of the other kind does not correspond exactly to the ideal whole number ratio implied by the formula. Such compounds do not obey the law of constant proportions. There are many examples of these compounds particularly in the oxides and sulphides of transition elements. Thus, in FeO, FeS or CuS the ratio of Fe: O, Fe: S or Cu: S differs from that indicated by the ideal chemical formula. If the ratio of atoms is not exactly 1:1 in the above cases, there must be either an excess of metal ions or a deficiency of metal ions (e.g., Fe0.84O - Fe0.94O, Fe0.9S). Electrical neutrality is

A+

B-

A+

B-

B-

A+

B-

A+

A+

e-

A+

B-

B-

A+

B-

A+

Electron remains trapped in anion vacancy

fig : 4.66 Metal excess defect due to anion vacancy (F- centre) The electrons released from sodium atoms during the formation of sodium chloride diffuse into the crystal and occupy the anionic sites. As a result the crystal now has an excess of sodium. The electrons trapped in anionic sites are referred as F-centres (from the German word Farben Zenter for colour centre). These give very interesting colours. For example the excess sodium in NaCl make the crystal yellow, the excess of potassium in KCl makes

4.44

Solid State

the crystal appear violet and the excess of lithium in LiCl makes the crystal appear pink. These colours are due to the excitation of the electrons present in anionic vacancies when they absorb energy from the visible light falling on the crystals. (II) Excess cations occupying interstitial sites: Metal excess defects also occur when an extra positive ion occupies an interstitial position in the lattice and electrical neutrality is maintained by the inclusion of an interstitial electron. This defect may be visualized as the loss of non metal atoms which leave their electrons behind. The excess metal ion and electron both occupy the interstitial positions. This type of defect is found in crystals which are likely to develop Frenkel defect. For example, zinc oxide on heating at high temperatures loses oxygen reversibly. So there is excess of zinc in the crystal and its formula become Zn1+x O. The excess zinc ions move into the interstitial sites and the electrons to neighboring interstitial sites. These electrons give rise to enhanced electrical conductivity. Also because of these electrons, zinc oxide is yellow in colour when hot but becomes white when cold since it absorb oxygen again.

A B+-

A+

B-

B-

A+ B-

A+

A+

Be-

Cation and elecron in interstitial sites

fects the number of free electrons is also small in number. So their conductivities are very small. Because of this reason these are also called semi conductors. Since the electrical conductivity is due to normal electron conduction, these are called n-type semi conductors. (b) The crystals with metal excess defects are generally coloured as explained already. 2. Metal deficiency defect These contain less number of positive ions than negative ions. (i) Cation vacancies: In some cases, the positive ions may be missing from their lattice sites. The extra negative charge may be balanced by some nearby metal ion acquiring two positive charges instead of one. This type of defect is possible in metal which show variable oxidation states. The common examples of compounds having this defect are ferrous oxide, ferrous sulphide, nickel oxide etc. For example, in the case of iron py rites (FeS) two out of three ferrous ions in lattice may be converted into Fe 3+ state and third Fe 2+ ion may be missing from its lattice site. Now the crystal contains both Fe 2+ and Fe 3+ ions and gives rise to exchange of electrons from Fe 2+ ion to Fe3+ ion due to which Fe 2+ ion converts into Fe3+ ion and Fe 3+ ion converts into Fe 2+. Because of this the crystals have metallic luster and the iron pyrites is nick – named as fool’s gold as it shines like gold. Due to exchange of electrons these substances become conductors.

A+

B-

A+

B-

A–

B-

A+

B-

A+

B–

A–

B+ Cation vacancy

fig 4.67 Metal excess defect due to extra cation If this type of defect oxide is heated in dioxygen then cooled to room temperature, its conductivity decreases. This is because the dioxygen oxidises some of the interstitial ions and these subsequently remove interstitial electrons which reduces the conductivity.

B+

B–

A+

A+

B–

A2+

B–

B–

A+

B–

A+

Metal acquiring higher charge

fIg 4.68 Metal deficient defect due to cation due to cation vacancy

consequences of metal excess defects (a) The crystals with metal excess defects conduct electricity due to the presence of free electrons as they migrate. Since there are only a small number of de-

(ii) Extra anions occupying interstitial sites: In principle it might be possible to have an extra negative ion in an interstitial position and to

Solid State

balance the charges by means of an extra charge on an adjacent metal ion. However, since negative ions are usually large, it would be difficult to fit them into interstitial positions. No example of crystals containing such negative interstitial ions are known

Anion occupying interstitial site A+

B–

A+

4.45

53. What fraction (n/N) of the lattice sites are vacant at 298 K for a crystal for which the energy required to make a defect is 1 ev. (1 ev = 1.602 × 10–19J) 54. Analysis shows that nickel oxide has formula Ni0.98O. What fractions of the nickel exist as Ni2+ and Ni3+ ions? 55. The edge length of NaCl crystal is 0.5628 nm and the observed density is 2.165 × 103 kg m–3. (Molecular weight of NaCl is 58.44 × 10–3 kg). Calculate the fraction of sites unoccupied.

B–

B–

4.14 propertIes of solIds

B–

A+

B–

A+

A+

B–

A2+

B–

B–

A+

B–

A+

Metal acquiring higher charge

fig 4.69 Metal deficient defect due to extra anion. Crystals with metal deficient defects are semiconductors. The conductivity is due to the movement of electrons from one ion to another. For example, when an electron moves as explained from Fe2+ to Fe 3+, the Fe2+ changes to Fe3+. It is also called movement of positive hole and the substances are called p-type Semi-conductors. solved problem 13 If NaCl is doped with 10–3 mol % of SrCl2 what is the concentration of cation vacancies. Solution: On doping NaCl by SrCl2 each Sr2+ ion replaces 2Na+ ions from NaCl crystal causing one cation vacancy. Thus No. of moles of cation vacancy in 100 mL of NaCl = 10–3 No. of moles of cation vacancy in 1 mol = 10–3/100 = –5 10 mol Total cation vacancies = 10–5 × No =10–5×6.023×023 = 6.02×108 problems for practice 52. The composition of a sample of wustite is Fe 0.93O. What percentage of iron is present in the form of Fe (III)?

The properties of solids are very much related to their composition and structure. The properties of solids such as electrical and magnetic properties are utilised in the devices such as transistors, computers, tele phones etc. Some important properties of solids are discussed here.

4.14.1 electrical properties On the basis of electrical conductivity, solids can be classified into three types. (i) Conductors (ii) Semi-conductors (iii) Insulators The conductivity of a solid is one of its characteristic properties and throws light on the internal structure and bonding in the solid. 1. Conductors: The solid which allow the passage of electric current are called conductors. These are of two types: (a) Metallic conductors: These allow the electricity to pass through them without under going any chemical change. e.g., copper, silver etc. Electrical conductivity of metals is very high and is of the order of 107Ω–1 m–1. In metallic conductors, the conductance is due to the movement of electrons under the influence of an applied electric potential. The streams of electrons constitute the current. (b) Electrolytic conductors: These allow the electricity to pass through them by under going chemical change. The ionic solids do not conduct electricity in solid state but conduct electricity in their molten state or in their aqueous solution. In these conditions the ions of the electrolyte are free and conduct electric current due to the movement of ions. The ionic solids with defects may conduct electricity to a small extent because the migration becomes possible due to the presence of vacancies or interstitial sites.

4.46

Solid State

2. Semi conductors: The solids whose conductivity lies between those of typical metallic conductors and insulators are called semiconductors. They have conductivity in the range 10–6 to 104 Ω m–1 3. Insulators: The solids which do not allow the passage of electric current through them are called insulators e.g., wood, sulphur, rubber, plastic etc. The conductivity range of insulators is 10–10 to 10–20 Ω–1 m–1. The magnitude of electrical conductivity strongly depends on the number of electrons available to participate in the conduction process. Particularly in metals the conductivity depends on the number of valence electrons present per atom. In order to understand the bonding and properties of a metallic element in a crystal, one should know what happens when a number of metal atoms interact.

4.14.2 electrical conductivity in metals The electrical conductivity of metals can be explained basing on band theory of the bonding in metals. We start with considering the sodium metal. If two sodium atoms come close to each other, the 3s orbital of one sodium atom can combine with 3s orbital in the second sodium atom to form two molecular orbitals (MOs) of which one is bonding MO with lower energy and the other is antibonding MO with higher energy. If a third sodium atom also come close to these two sodium atoms three MOs are formed. In a similar way if n sodium atoms approach together, n MOs can be formed. The building up of such MOs can be represented as in Fig 4.70. 3s Energy band

Empty orbitals

3s

Energy

Filled orbitals

Na

Na2

Na3

Na6

Na20

Nan

fig 4.70 Formation of 3s energy bands in sodium metal.

Depending upon different types of atomic orbitals which overlap, different energy bands gives a pictorial representation of the energy levels in metallic crystal. The arrangement of electrons in the different energy bands determines the characteristics of a metal. The energy bands formed from different atomic orbitals may overlap or be separated from each other. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called the conduction band. The energy gap between the top of the valence band and the bottom of the conduction band is called, energy gap or forbidden zone, e.g., in the case of metals, the valence band may be half filled or there may be an overlapping between the valence and conduction bands. This makes it possible for the electrons to go to vacant bands and hence is responsible for electrical conductivity. In sodium metal 1s atomic orbitals form 1s band, 2s atomic orbitals form 2s band, 2p atomic orbitals form 2p band and 3s atomic orbitals form 3s band. 1s, 2s and 2p bands are completely filled with electrons while 3s band is half filled since ‘N’ orbitals contain ‘N’ electrons but each orbital can accomodate 2 electrons. The presence of electrons in the half filled 3s band leads to high electrical conductance as a number of unoccupied energy levels are available in the valence band itself. The electronic configuration of magnesium is 1s2 2s2 6 2 2p 3s . Unlike sodium, 3s band is completely filled by the electrons as every 3s orbital of Mg contribute 2 electrons. However, in Mg it has been found that the 3p energy band (formed by unoccupied 3p AOs of the Mg atoms) overlaps the 3s band. There is therefore, no energy gap between the valence band and the conduction band and Mg is an excellent conductor.

4.14.3 electrical conductivity in semi conductors In the case of insulators the energy gap is very large and therefore the vacant conduction band is not available to the electrons of the completely filled valence band. Several solids have electrical properties intermediate between those of metals and insulators. These are called semi conductors. The energy gap in these substances in very small and increase in temperature gives thermal energy for some of the electrons in the valence band and move into conduction band. The energy gap value in KJ mol–1 for carbon group elements are diamond (511), silicon (111) and germanium (63). Diamond with large band gap is an insulator. Silicon and germanium have fairly small band gaps and are called semi conductors. Their electrical conductivity is substantially greater than that of insulator but far less that of conductors. The conduction by pure substances such as

Solid State

4.47

Conduction band Empty band

Empty band

Energy

Forbidden zone (Large energy gap)

Small energy gap

Filled band Partially filled band

Overlapping bands Metal (a)

Insulator (b)

Semi conductor (c)

fig 4.71 Difference in (a) metals (b) insulators and (c) semi conductors. In each case an unshared area represents a conduction band. silicon and germanium is called intrinsic conduction and these pure substances exhibiting electrical conductivity are called intrinsic semi conductors. The conductivity of pure silicon and germanium is very low at room temperature. The conductivity of silicon and germanium can be increased by adding appropriate amount of suitable impurity. The process is called doping. Doping can be done with an impurity which is electron rich or electron deficit as compared to the intrinsic semi conductor. Such impurities introduce electronic defects in them.

conduction far above the possible by intrinsic conduc tion. Since the current is carried by excess electrons, it is n- type semi conduction.

Ge

Ge

Ge

Ge

Ge

Ge

Ge

Ge

doping With electron rich Impurities The electron rich impurities are also known as donor impurities. When silicon is doped with small amount of group 15 elements such as P, As or Sb its electrical conductivity increases sharply. Silicon or germanium are first obtained in extremely pure condition by zone refining. Some atoms with five outer electrons such as arsenic ‘As’ are deliberately added to the silicon crystal. This process is called ‘doping’ the crystal. A minute proportion of silicon atoms are randomly replaced by arsenic atoms with five electrons in their outer shell. Only four of the outer electrons on each arsenic atom are required to form bands in the lattice. At absolute zero or low temperature the fifth electron is localized on the arsenic atom. However, at normal temperatures some of these fifth electrons on arsenic are excited into the conduction band, where they can carry current quite readily. This is extrinsic conduction and it increases the amount of semi

-

e Ge

Ge

As

Ge

Ge

Ge

Ge

Ge

Surplus mobile electron

fig 4.72 Arsenic doped germanium semi conductor

Electron Deficient Impurities These are also known as acceptor impurities. If a crystal of pure silicon may be doped with some atoms with only three outer electrons, such as indium (In). Each indium atom uses three outer electrons to form three bonds in the

4.48

Solid State

Ge

Ge

Ge

Ge

Ge

Ge

In

Ge

Ge

Ge

Ge

Ge

Positive hole (no electron)

+

Ge

Ge

Ge

p-type semi conductor and the other an n-type semi conductor. In the middle there will be a boundary re gion where the two sides meet, which is p-n junction. Such junctions are the important part of modern semi conductor devices.

applications of n-type and p-type semi conductors

Ge

fig 4.73 Indium doped germanium semi conductor

lattice, but they are unable to form four bonds to complete the covalent structure. One bond is incomplete and the site normally occupied by the missing electron called a ‘Positive hole’ At absolute zero or low temperatures, the positive holes are localized around the indium atoms. However, at normal temperatures a valence electron on an adjacent silicon atom may gain sufficient energy to move into the hole. This forms a new positive hole on the germanium atom. The positive hole seems to have moved in the opposite direction to the electron. By a series of ‘hops’ the positive hole can migrate across the crystal. This is equivalent to moving an electron in the opposite direction, and thus current is carried. Since the current is carried by the migration of positive centres, this is p-type semi conduction. If a single crystal is doped with indium at one end, and with arsenic at the other end, then one part is a

Various combinations of n-type and p-type semi conductors are used for making electronic components. Rectifiers: A rectifier will only allow current from an outside source to flow through it in one direction. This is invaluable in converting alternating current Ac into direct current Dc and it is common to use a square of four diodes in a circuit to do this. A diode is simply a transistor with two zones, one p-type and the other n-type with p-n junction in between them. Suppose that a positive voltage is applied to the p-type region and a negative (or more negative) voltage applied to the n- type region. In the p-type region posi tive holes will migrate towards the p-n junction. In the ntype region electrons will migrate towards the junction. At the junction, the two destroy each other. In another way it can be explained that at the junction the migrating electrons from the n- type region move into the vacant holes in the valence band of the p-type region. The mi gration of electrons and holes can continue indefinitely and a current will flow for as long as the external voltage is applied. If voltage is reversed, the p-type region becomes negative while the n-type region becomes positive. In the p-type region positive holes move away from the junction and in the n-type region electrons migrate away from the junction. At the junction there are neither positive holes nor electrons, so no current can flow.

n

p A

diode

Positive holes p

n

p

n

electrons conducts fig 4.74 A n-p junction as a rectifier

does not conduct

Solid State

An n-p-n transistor

A p-n-p transistor

0V

n

p

-0.2 V

n

Emitter

Base

p

n

Base (b)

Collector

p

Emitter

Base

p

n

4.49

0V

Collector

p

-2 V Emitter (e)

Collector (c)

-2V

-0.2 V fig 4.75 n- p-n and p-n–p transistors

Photovoltaic cell: If a p- n is junction is irradiated with light provided that the energy of the light pho tons exceeds the band gap, then some bonds will break, giving electrons and positive holes and these electrons are promoted from the valence band to the conduction band. The extra electrons in the conduction band make the n- type region more negative whilst in the p- type region the electrons are trapped by some positive holes. If the two regions are connected in an external circuit, then electrons can flow from the p- type to the n- type region. Such device acts as a battery that can generate electricity from light which harness solar energy. Transistors: Transistors are made by sandwiching a layer of one type of semi conductor between two layers of the other type of semi conductor Different voltages must be applied to the three regions transistor to make it work. The npn and pnp type of transistors are used to detect or amplify radio or audio signals. Now, a large variety of solid materials have been prepared by combination of elements of group -13 and 15 or 12 and 16 which have an average valence electrons of 4 as in germanium and silicon. Typical examples of group 13-15 compounds are AlP, GaAs or InSb, Gallium arsenide (GaAs) semi conductors have gained much response and have revolutionized the design of semi conductor devices. The examples of group 12-16 compounds are ZnS, CdS, CdSe and HgTe. In these compounds, the bonds are not perfectly covalent and the ionic character depends upon the electronegative differences between the two elements. It is interesting to know that transition metal oxides show marked differences in electrical properties. TiO, VO, CrO2and ReO3 behave like typical metals. Rhenium oxide, ReO3 behave metallic, like copper in conductivity and

appearance. Certain other oxides like VO, VO2, and TiO3 show metallic or insulating properties depending upon temperature. magnetic properties We know that an electric current travelling in a loop of wire produces magnetic character. The flow of electric current is nothing but the flow of electrons. Similar to this the movement of electrons in an atom also produces magnetic character. The magnetic character of electrons is due to their spin angular momentum and also due to orbital angular momentum. The magnetic properties of any individual atom or ion will result from the combination of these two properties, i.e., the spin moment and orbital moment of the electron. Types of magnetism: When a substance is placed in a magnetic field (between the poles of magnets) the magnetic field produced by the electrons in that substance interacts with the externally applied magnetic field. It is interesting to note that when the various substances are placed between the poles of a magnet (i.e., in a magnetic field) they do not behave similarly, i.e., they show different behaviours which are known as magnetic behaviours. These are classified as diamagnetism, paramagnetism, ferromagnetism, anti ferromagnetism and ferri magnetism. Diamagnetism: If the magnetic lines of force are repelled by the substance, the field (B) in the substance is lesser than the applied field (H) (i.e., B H) are called paramagnetic substances. This phenomenon is known as paramagnetism. A paramagnetic substance tends to set itself with its length parallel to the magnetic field. Thus paramagnetic substances are attracted into the magnetic field Paramagnetism is due to the presence of unpaired electrons in the atom, or ions or molecules of the substance. The greater the number of unpaired electrons greater will be paramagnetism shown by the substance. In substances containing one or more unpaired electrons, the magnetic fields caused by the unpaired electrons are not mutually cancelled, since each of the unpaired electrons has equal magnetic moment and thus some permanent and definite value of resultant magnetic moment is obtained. Thus resultant magnetic moment which is combination of spin and orbital magnetic moments induced by the externally applied magnetic field. Such a substance therefore instead of experiencing repulsion like diamagnetic substances, experiences attraction on a magnetic field i. e it shows paramagnetic behaviour, e.g., O2 Cu2+, Fe3+ Cr3+ etc. Ferromagnetism: It is a special case of paramagnetism. In ferromagnetic substances the field strength (B) is very much greater than H. (i. e B>>H). In ferromagnetic substances the lines of force attracted very strongly and the substances may even be magnetized permanently. In solid, state, the metal ions of ferromagnetic substances are grouped together into small regions called domains. Thus each domain acts as a tiny magnet. In an unmagnetised piece of ferromagnetic substance the domains or randomly oriented and their magnetic moment get cancelled. When the substance is placed in a magnetic field all the domains

get oriented in the direction of the magnetic field and a strong magnetic effect is produced. Thus ordering of the domains persists even when the magnetic field is removed and the ferromagnetic substance becomes a permanent magnet. Examples for ferromagnetic substances are metals like iron, cobalt, nickel, gadolinium compounds like CrO2

(a) Ferromagnet

(b) Anti-ferromagnet

(c) Ferrimagnet fig 4.76 Alignment of magnet moments in a) ferromagnet b) Anti ferromagnet c) ferrimagnet Anti ferromagnetism: Certain substances like MnO have domain structure similar to ferromagnetic substances but their domains are oppositely oriented in a compensatory way so as to give zero net magnetic moment due to cancellation of the individual magnetic substances. These are called anti ferromagnetic substances. Ferrimagnetism: When the magnetic moments of the domains in the substance are aligned in parallel and anti parallel directions in unequal numbers resulting in net magnetic moment. This phenomenon is known as ferrimagnetism. These are weakly attracted by magnetic field as compared to ferromagnetic substances. Fe 3O4 (magnetic) and ferrites like MgFe 2O4 and ZnFe 2O4 are examples of such substances. These substances lose their ferrimagnetic character on heating and become paramagnetic. It may be noted that all the magnetically ordered solids (ferromagnetic, anti- ferromagnetic and ferrimagnetic) transform to paramagnetic state at a higher temperature due to randomization of their spins. e.g., (i) V2O3 transforms form anti-ferromagnetic state to paramagnetic state at 150 K (ii) Fe3O4 becomes paramagnetic from ferromagnetic at 850 K.

Solid State

dielectric properties A dielectric is a substance in which an electric field gives rise to no net flow of electric charge. This is because that in insulators the electrons are closely bound to individual atoms or ions and therefore they do not generally migrate under the applied electric field. However, due to shift in charges, dipoles are created which result into polarization. The alignment of these dipoles in different ways i.e., compensatory way (zero dipole) or non compensatory way (net dipole) impart certain characteristic properties to solids. Depending upon the alignment of the dipoles, the crystals have very interesting properties (i) Piezoelectricity: The dipoles may align in such a way so that there is net dipole moment in the crystals. Crystals of this type exhibit Piezoelectricity or pressure electricity. When such a crystal is deformed by mechanical stress, electricity is produced due to displacement of ions or conversely when an electric field is applied in such crystals, there will be mechanical strain due to atomic displacements This is sometimes called inverse piezoelectric effect. Thus the piezoelectric crystal acts as mechanical – electrical energy into mechanical strains or vice versa. The piezoelectric crystals are used in record players as pick ups in which they produce electric signals by the application of pressure (i.e., mechanical strain) Stress +++++++++++ +

+++++++++++ +

+

+

+

+

+

+

+

+

+

+

+

+

V

Stress a

b

fig 4.77 (a) Dipoles with in a piezeoelectrical solid (b) Voltage is generated when solid is subjected to mechanical stress. These are also used in microphones, ultrasonic generators, stain gauges, sonar detectors, etc. The common example of piezoelectric materials are titanates of barium and lead zirconate (PbZrO3) ammonium dihydrogen phosphate (NH4H2PO4), quartz etc.

4.51

(ii) Ferro electricity: In certain piezoelectric solids in which dipoles are spontaneously aligned in a particular direction, even in the absence of electric field. The direction of polarization can be changed by applying an electric field. This phenomenon is known as Ferro electricity. The Common examples are barium titanate (BaTiO3), Sodium potassium tartrate (Rochelle salt), potassium dihydrogen phosphate (KH2PO4) etc. (iii) Pyro electricity: Certain piezoelectric crystals, when heated acquire electric charges on opposite faces. This phenomenon is called pyroelectricity (iv) Anti-ferro electricity: If the alternate dipoles are in opposite direction, then the net dipole moment will be zero and the crystal is called anti ferroelectric. The common example is lead zirconate (PbZrO3) super conductivity Metals are good conductors of electricity and their conductivity increases as the temperature is decreased. In 1911 the Dutch scientist Heike Kamerlingh Onnes discovered that metals such as Hg and Pb become super conductors at temperature near absolute zero. A super conductor has zero or almost zero electrical resistance. It can therefore carry an electric current without losing energy and in principal the current can flow for ever. There is a critical temperature Tc at which the resistance drops sharply and super conduction occurs. Later Meissner and Ochsemfeld found that some super conducting materials will not permit a magnetic field to penetrate their bulk. This is now called Meissner effect and gives rise to levitation. Levitation occurs when objects float on air. This can be achieved by the mutual repulsion between a permanent magnet and a super conductor. A super conductor also expels all internal magnetic fields (arising from unpaired electrons), so super conductors are diamagnetic. Most metals have transition temperature (critical temperature) between 2K-5K. The highest temperature at which super conductivity was known till recently was 23 K for alloys of niobium. Research is going on to find materials which behave as a super conductors at higher temperatures. Since 1987 many complex metal oxides have been studied and found to possess super-conductivity at somewhat higher temperatures. Some of them are as follows. YBa2Cu3O7 90 K Bi2Ca2Sr2Cu3O10 105 K Ti2Ca2Ba2Cu3O10 125 K Super conducting materials have great technologi cal applications. They can be used in electronics, in building magnets, in power transmission and in levita tion transportation (trains which move in air without rails).

4.52

Solid State

key poInts solids liqrids and gases • •

• • •

• •

• •

• •

• •

•

The cooling of liquid molecules leads to the formation of solids. In the solid state, inter molecular attractive forces are extremely strong and as a result the molecular motion is completely stopped. In solids, the molecules assume fixed positions and their motion is restricted to just vibration. Unlike gases and liquids, the solids possess definite volume and shape. The constituent particles in solid are closely packed and this leads to the properties like incompressibility, rigidity, mechanical strength, slow diffusion and non - fluidity. Like liquids solids also evaporate and hence exhibit vapour pressure. The molecules at the surface of a solid possess more kinetic energy and breaks away from the surface to enter into vapour state. The process whereby a solid directly converts into vapour is called sublimation. Substances like ammonium chloride, iodine, camphor, solid carbon dioxide sublime at ordinary temperature and pressure. Snow sublimes when the surrounding temperature is less than 0oC. The white cloud like exhaust fumes of high flying jets contain water vapour convert directly into microcrystalline ice slowly converts into water vapour without passing through liquid state. The property of sublimation is used in the freeze drying of substances containing water. Freeze drying is a process of cooling the substances containing water to below –10oC at 1 atm, where the water molecules freeze to ice. This when subjected to evacuation, water sublimes leaving behind the non volatile components. Instant coffee, tea, powdered milk and many medicines are dried by freeze drying method.

•

•

•

•

Classification of Crystalline Solids •

•

types of solids • • •

•

Solids are of two types: crystalline solids and amorphous solids. In crystalline solids the constituent particles are arranged in an orderly manner, e.g., NaCl, ZnS etc. The solids which do not possess crystalline form are called amorphous solids, e.g., glass, rubber, many plastics, red phosphorous and amorphous sulphur. Amorphous solids are regarded as super cooled liquids or pseudo solids because they do not melt sharply

at a definite temperature but tend to soften on heating and gradually change into a viscous liquid. Isotropy is that, the physical properties such as refractive index, thermal and electrical conductivity and solubility are same in different directions. Amorphous solids are isotropic. Anisotropy is that, the physical properties as above are in different directions. Crystalline solids are different anisotropic isotropic. Amorphous solids are characterised by limited in compressibility, limited rigidity, no definite geometric shape, no sharp melting point and isotropy. Crystalline solids are characterised by incompressibility, rigidity, mechanical strength, definite geometric shape, sharp melting point and anisotropy.

•

•

The crystalline solids are four types (i) covalent solid (ii) ionic solid (iii) molecular solid and (iv) metallic solid. Depending on the type of attractive forces between molecules the molecular solids are again categorized into three types. (i) Non polar molecular solids in which atoms such as Ar, Kr, Xe etc or molecules of covalent compounds are held together by weak Van der Waals forces. These solids have low m pts and the particles are widely separated than in close packed ionic or metallic lattices. (ii) Polar molecular solids in which molecules are held together by relatively stronger dipoledipole-dipole interactions. The solids are soft and non-conductors of electricity. Their m. pts are higher than those of non-polar molecular solids, e.g., solid CO2. (iii) Hydrogen bonded molecular solids are those in which molecules participate in hydrogen bonding. In crystals of benzoic acid, hydrogen bonds cause the association into dimers which are then held together by Van der Waals forces. These are also non conductors of electricity. In ionic solids: the structural units are positive and negative ions, each ion being surrounded by a definite number a oppositely charged ions known as coordination number. In ionic solids, ions are held together by strong electrostatic attractive forces and they have high m. pts. They are hard and brittle. Since the ions in a solid are not free to move they are non conductors of electric-

Solid State

•

•

• •

ity in solid state but in molten state and in aqueous solution they ionises and become good conductors of electricity. In Covalent or Network solid the atoms are connected to one another by covalent linkages forming giant network. E.g., diamond, graphite, carborrundum quartz. In metallic solids the units are positive metallic ions surrounded by a sea of mobile electrons, each electron belonging to a number of positive ions and each positive ion belonging to a number of electrons. The binding force in metallic solids is metallic bond. Metallic solids exhibit metallic lusture, high electrical and thermal conductivity due to free moving electrons.

allotropy and polymorphism •

•

•

•

Allotropy is the phenomenon of existence of a substance (element or compound) in two or more forms having different physical and chemical properties. Enantiotropy is that type of allotropy where each allotropy will keep indefinitely, i.e., is stable over a given temperature range. e.g., Sulphur, tin, ammonium chloride, red and yellow forms of mercuric iodide. If one allotropy is stable under normal conditions but the other allotrope is unstable or metastable, then it is known as monotropic allotropy. Examples of monotropy are white and violet phosphorous, diamond and graphite, oxygen and ozone; calcite and aragonite. Under normal conditions the metastable form changes to the stable form but never in the reverse direc tion. The phenomenon in which one allotrope changes into the other at exactly the same rate as the reverse occurs is known as dynamic allotropy. Both allotropes are stable over a wide range of conditions. In liquid state Sλ and Sµ exhibit dynamic allotropy.

Vapour pressure and melting point of solids •

•

•

•

When two different chemical substances have the same crystalline form they are said to be isomorphous. Isomorphism is exhibited by the solids in which the packing in crystal lattice is same which depends on the nature of the forces holding the units of struc ture together and on the relative sizes and shapes of units. Ionic compounds exhibit isomorphism when the relative sizes of the ions are same and have similar shape. e.g., NaCl and KCl are isomorphous. K2SO4, K2 SeO4; KMnO4 and BaSO4; all alums are isomorphous.

Solids have a vapour pressure but is very small due to strong forces which have to be overcome to vapourise. The temperature and pressure at which the three states solid, liquid and vapour of a substance are in equilibrium is called triple point.

space lattice and unit cell •

•

•

•

•

• •

•

•

Isomorphism •

4.53

A space lattice is a regular arrangement of constituent particles (atoms, ions or molecules) of a crystalline solid in three dimensional space. The positions which are occupied by the atoms, ions or molecules in the crystal are called lattice points or lattice sites. The smallest repetitive unit of a crystal lattice which is used to describe the lattice is called the unit cell. Crystal possess the same symmetry as their constituent unit cells. Primitive unit cells are drawn with lattice points at all corners and each primitive cell contain the equivalent of one atom. When a primitive cell contains one or more constituent particles present at positions other than corners in addition to those at corners, it is called centred unit cell. In the simple unit cell the particles are present only at the corners of the unit cell. In the body centred unit cell there is one particle present at the centre of the unit cell in addition to the particles at the corners of the unit cell. In a face centred unit cell there is one particle present on the centre of each face in addition to the particles at the corners of the unit cell. In the end centred unit cell there is one particle in the centre of two opposite faces in addition to the particles at the corners of the unit cell.

crystal systems • •

• •

There are seven crystal systems arising due to different symmetry of the crystal lattices. A crystal system is characterized by the dimensions of a unit cell along the three axes (a, b, c) and the size of angles (a, β, γ) between the three axes. The various types of unit cells possible are given in Table 4.2 and Fig 4.14. The total number of three dimensional lattices are 14 which are known as Bravis lattices.

4.54

Solid State

number of atoms per unit cell • •

•

•

1 = 1 particle per unit cell 8

•

8× •

In a body centred cubic unit cell particles are located at the centre of the cell as well as at the corners. Therefore the number of atoms per unit cell in body centred cubic unit cell is 1 8 (at corners) × + 1 (at body centre) × 1 = 2 particles 8

•

•

In a simple or primitive cubic lattice the lattice points are located only at the corners. In different cubic unit cells there are mainly four kinds of lattice points. The four types of lattice points and the contribution of each particle at the lattice point to the unit cell are. (i) A particle in the body of the unit cell belong to that unit cell only and counts 1. (ii) A particle on a face is shared by two unit cells and contributes ½ to the unit cell. (iii) A particle on the edge is shared by four unit cells and contributes ¼ to the unit cell. (iv) A particle at a corner is shared by eight cells that share the corner and so contributes 1/8 to the unit cell. In a simple or primitive cubic lattice the lattice points are located at the corners of each unit cell and can contribute only 1/8 of each particle at the corner to the unit cell shared by 8 unit cells in space lattice. So a simple cubic unit cell has

In a face centred cubic unit cell atoms are found at the centre of the six faces of the cell as well as at each of the eight corners. The number of particles per unit cell in fcc is

•

• •

•

the atoms in the first layer. This results in AB AB AB…………..pattern. Metals that crystallize in the hcp structure are Be, Mg, Ti, Zn, and Cd. In cubic close packing (ccp) the atoms in third lay er are not immediately above the atoms in the first layer. This type of stacking of layers results in ABC ABC … pattern. Metals that crystallize in the ccp structure are Al, Cu, Ag, Au, Pt, Ni. In hcp and ccp structures the coordination number, i.e., the number of surrounding atoms in contact with an atom is 12. It is impossible to pack identical spheres (atoms) together with coordination number more than 12.

Interstitial sites or Interstitial Voids •

• • •

•

1 1 6 (at centre of each face) × + 8 (at corners) × = 4 2 8 particles

In close packed arrangement of spherical particles (atoms/ions/molecules) two type of voids or holes are created. The void created when six spherical particles are in contact with each other is called octahedral void or octahedral hole. The void created when four spherical particles are in contact with each other is called tetrahedral void or tetrahedral hole. If tetrahedral voids of second layer are covered, hexagonal close packed (hcp) structure results. If octahedral voids of second layer are covered, cubic close packed (ccp) structure results. In a close packed structure of N atoms there are 2N tetrahedral voids and N octahedral voids. The octahedral voids are larger than tetrahedral voids. The radius ratio of different voids is 0.225 for tetrahedral void, 0.414 for octahedral void, 0.155 for triangular void and 0.732 for cubic void.

locating tetrahedral and octahedral Voids in cubic close packing packing of equal spheres (structure of metals) • • •

In metals the atoms in a close-packed layers are arranged in a regular hexagon. In metals the close-packed layers of atoms can be staked in two ways. In hexagonal close packing (hcp). After arrang ing the atoms of second layer in the depressions of the first layer, the atoms of the third layer are arranged so that they will be immediately above

•

•

•

In cubic close packing there are eight tetrahedral voids two on each body diagonal at one fourth distance from each end. In cubic close packing or face centred cubic lattice. There is one octahedral hole at the cube centre and 12 octahedral voids on the centres of the 12 edges of the cube. The octahedral hole at centre of the cube surrounded the atoms belonging to the same cube. The octahedral hole at the edge centre is surrounded by three atoms

Solid State

•

belonging to the same unit cell (2 on the corners and 1 on the face centre) and three belonging to two adjacent unit cells. Each octahedral hole on the edge centre is being shared by four unit cells. The number of octahedral voids in cubic close packing per unit cell are 4.

Here Z is the number of atoms present in one unit cell and m is the mass of a single atom. m=

Density d =

The atomic radius r is expressed in terms of edge length a. r=

r=

r= •

a in the case of simple cubic lattice 2 a 2 2

in the case of face centred cubic lattice

3 a in the case of body centred cubic lattice 4

Ionic radii of an ion r in terms of edge length in different types of ionic crystals can be expressed as. rc + ra = a in the case of simple cubic lattice rc + ra =

a for cubic lattice of NaCl type 2

3 a for a body centred lattice of CsCl type 2 Volume occupied in fcc or ccp arrangement is 74.06%. Volume occupied in body centred cubic (bcc) arrangement is 68%. Packing efficiency of simple cubic lattice is 52.4%. Packing efficiency of hexagonal close packing is 74%.

• •

• •

•

•

•

•

•

•

calculations Involving unit cell dimensions •

If the edge length of the unit cell is 'a' pm i.e., a×10–10 cm Volume of the unit cell = a3× 10–30 cm3

Density of the unit cell =

Mass of unit cell Volume of unit cell

•

•

• •

Mass of the unit cell = (No. of atoms or formula units per unit cell) × (Mass of one atom or one formula unit) =Z×m

ZM g Cm–3 N A a 3 × 10−30

radius ratio: structure of Ionic compounds

rc + ra =

• •

M where M is gram molecular weight and NA is NA

Avogadro’s number.

Efficiency of Packing •

4.55

•

In ionic solids every ion is surrounded by a number of oppositely charged ions. The number of oppositely charged ions surrounding an ion is called as the coordination number of that particular ion. The arrangement of ions in a crystal and the coordination number of an ion depends upon ratio of the ions or atoms surrounded it. In the case of ionic solids the ratio of the radius of the cation to the radius of an anion is called limiting radius ratio. In simple ionic crystals, anions are normally larg er than cations and arranged in a closest-packed array. Being smaller in size cations occupy the voids. If all the octahedral voids are occupied by cations, the number of cations is equal to the number of anions. If all the tetrahedral voids are occupied by cations, the number of cations is twice the number of anions. Relatively small cations occupy the tetrahedral holes while larger cations occupy the octahedral holes. If the cation is too larger to fit into the octahe dral hole the anions make larger cubic holes for cations. Coordination numbers of 5, 7, 9 10 and 11 do not occur because of the impossibility of balancing the electrical charges. When the radius ratio becomes equal to 1, ions of the same size are making up the crystal. This is found in the crystals of metals. When all the octahedral voids are occupied by the cations the crystal will get the rock-salt structure. When half the tetrahedral voids are occupied by cations the crystal will get the zinc-blende or sphalarite structure. If all the tetrahedral voids are occupied by anions while cation adopt cubic close packed structure the

4.56

•

• •

Solid State

crystal will get the fluorite structure and the formula of the compound will be AB2 as the tetrahedral voids are double to the ions making cubic close packed structure. If anions adopt cubic close packing and cations occupy all the tetrahedral voids the crystal will get the antifluorite structure and the formula of the compound will be A2 B. Packing efficiency of rock-salt structure is 79% and that of zinc blend structure is 75%. The type of hole occupied can be determined from the radius ratio.

Radius ratio

Coordination number

•

•

a2x type Ionic solid Anti fluorite Structure •

• Structure

< 0.155

2

linear

0.155-0.225

3

Triangular

0.225-0.414

4

Tetrahedral

0.414-0.732

6

Octahedral

0.732-1

8

Cubic

•

rutile structure

structure of Ionic compounds sodium chloride structure

•

• •

In NaCl each Na+ or Cl– ion is surrounded by six (coordination number) other oppositely charged ions octahedrally. NaCl structure is regarded as cubic close packed array of Cl– ions with Na+ ions occupying all the octahedral holes. The unit cell of NaCl has 4 Na+ ions and 4 Cl– ions and thus 4 formula units per unit cell. The ionic solids having NaCl type crystal structure are HgO, CaO, SrO, NaBr, KCl.

•

• •

• •

Zinc blende or Sphalarite (ZnS) Structure •

•

The radius ratio 0.4 of Zn2+ to S2– in zinc blende suggest a tetrahedral arrangement for ZnS.

ay2 type Ionic solid

•

fluorite (calcium fluoride) structure • •

In fluorite each Ca2+ ion is surrounded by eight F– ions giving bcc arrangement. Each F– ion in CaF2 is surrounded by four Ca2+ ions tetrahedrally.

In anti fluorite structure anions form cubic close packed arrangement and cations occupy tetrahedral holes. In anti fluorite structure coordination number of cation is 4 and that of anion is 8. Ionic solids having anti fluorite structure are Li2O, Na2O, K2O and Rb2O.

structures of some solid crystals

•

•

Since the number of F– ions are double the number of Ca2+ ions the coordination number of Ca2+ and F– are 8 and 4 respectively. Ionic solids having fluorite structure are UO2, ThO2, SrF2

• •

Rutile is a form of titanium dioxide. Its unit cell has a tetragonal structure in which titanium ions are at the corners and at the centre of the cube. The titanium ion at the centre is surrounded by six oxide ions. Every titanium ion is surrounded by six oxide ions and every oxide ion has three titanium ions as their nearest neighbours. The ratio of titanium and oxide ions is 1:2 and the formula of rutile is TiO2. In ZnS each Zn2+ ion is surrounded by four S2– ions and each S2– ion is tetrahedrally surrounded by four Zn2+ ions. In ZnS the coordination number of Zn2+ and S2– is 4:4 arrangement. ZnS is related to cubic close packed structure and Zn2+ ions occupy tetrahedral holes. Since there are twice as many tetrahedral holes as there are S 2– ions it follows that to obtain a formula ZnS, only half of the tetrahedral holes are occupied by Zn 2+ (that is every alternate tetrahedral site is occupied. ZnS has face centred cubic structure in which S2– ions occupy the lattice points and Zn2+ ions are at one fourth of the distance along each body diagonal. Each unit cell of ZnS consists of four Zn2+ and four S2– ions making 4 ZnS formula units. Ionic solids having ZnS crystal structure are BeO, CuCl, CuBr etc.

Solid State

Wurtzite structure •

•

Wurtzite has hexagonal structure. Sulphide ions occupy the triangular faces. Four zinc ions make up the corners of tetrahedron centred on another sulphide ion at a central point in the cell. The coordination number of each is 4.

caesium chloride structure •

• •

The radius ratio of CsCl is 0.93 indicates that CsCl has bcc type arrangement in which each Cs+ ion is surrounded by 8 Cl– ions and vice-versa and the coordination number of CsCl is 8:8. In CsCl if the ions at the corners are Cl– ions there will be Cs+ ion at the body centred position or vice-versa The ionic solids having CsCl structure are CsBr, CsI, TlBr.

perovskite structure • •

Perovskite is a mineral with formula CaTiO3. The Ca2+ ions occupy the corners of the cube, the O2– ions occupy the face centres of the cube and the titanium ion lies at the centre of the cube.

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•

•

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•

structure of magnetite (fe3o4) • •

•

•

The Fe3O4 contain Fe3+ and Fe2+ ions in the ratio 2:1 and considering the composition is FeO.Fe2O3. In Fe3O4 oxide ions are arranged in ccp and Fe 2+ ions occupy octahedral voids while Fe 3+ ions are equally distributed between octahedral and tetrahedral holes. MgFe2O4 also has the structure similar to magnetite in which Mg2+ ions are present in place of Fe2+ ion in Fe3O4. Magnetite has inverse spinel structure.

•

• •

•

normal spinel structure • •

•

Spinel is a mineral MgAl2O4. In spinel oxide ions are arranged in ccp with Mg2+ ions occupying tetrahedral voids and Al3+ ions occupy octahedral voids. Ferrites such as ZnFe2O4 also possess spinel structure.

crystallography •

The branch of science which deals with the geometrical properties and structure of crystals and crystalline substances is called crystallography.

• •

4.57

The law of constancy of interfacial angles states that for a given substance the corresponding faces or planes forming the external surface of crystal intersect a definite angle called interfacial angle and remains always constant no matter how the faces develop. Plane of symmetry is an imaginary plane which passes through the centre of a crystal and divide it into two equal portions, such that one part is the exact mirror image of the other. Axis of symmetry is a line about a crystal may be rotated such that it presents same appearance more than once during the complete rotation. A n-fold (or n-gonal) axis of symmetry is an axis such that when an ideal crystal is rotated around it, the crystal occupies the same position in space n-times in a complete rotation of 360°. Centre of symmetry is a point in the body of the crystal such that a line drawn through it intersects the opposite faces at equal distances in both directions. Only one centre of symmetry is possible for any crystal. The total number of planes, axes and centre of symmetry possessed by a crystal is known as elements of symmetry. A cube has 3 rectangular planes of symmetry, 6 diagonal planes of symmetry. 3 four fold axis of symmetry 4 three fold axis of symmetry. 6 two fold axis of symmetry and one centre of symmetry, thus the total of symmetry elements is 23. The law of rational indices states that the intercepts of the planes of a crystal on a suitable set of axes can be expressed by small multiples of unit distances. The indices used in practice to denote the direction of a plane of a crystal are the Miller indices. For calculating Miller indices, a reference plane, known as parametral plane is selected having intercepts a, b and c along x, y and z axes. The intercepts of the unknown plane are given with respect to a, b and c of the parametral plane as. h=

a intercept of the plane along x-axis

k=

b intercept of the plane along y-axis

l=

c intercept of the plane along z-axis

The distances between the parallel planes in crystals are designated as dhkl. For different cubic lattices the interplanar spacings are given by the general formula.

4.58

Solid State

Imperfections in solids

a

d(hkl) =

•

h 2 + k 2 + l2

Where a is the length of the cube side while h, k and l are the Miller indices of the plane. • The spacings of three planes (100), (110) and (111) of simple cubic lattice can be calculated d(100) =

d(110) =

d(111) = •

a 1 +0+0 a 12 + 12 + 0 a 2

2

•

a 2

3

The ratio of three planes for a simple cube is 1 2

:

•

a

=

2

1 +1 +1

d(100) : d(110) : d(111) =1 :

•

=

• •

=a

2

•

• •

1 3

= 1 : 0.707 : 0.57 •

The dhkl ratio of face centred cubic and body centred cubic are

•

d(100) : d(110) : d(111) (for face-centred cube)

•

=1:

1 2

:

2 3

= 1 : 0.707 : 1.154 •

d(100) : d(110) : d(111) for bcc =

=1:

2:

1 3

1 1 1 : : 2 2 2 3

= 1: 1.414 : 0.577. •

determination of structure of solids By x-ray diffraction • •

•

When x- rays are incident on a crystal face, they are reflected by the atoms in different planes. Bragg’s equation to calculate the distance between the repeating planes of particles in crystals from the reflected x-rays is. n λ = 2d sinθ Where n is an integer like 1, 2, 3 and represents order of reflection, λ is the wavelength of the x-rays used and d is the distance between the repeating planes.

• •

• • •

The imperfections or defects which arise due to irregularity in the arrangement of atoms or ions in a crystal are called atomic imperfections. The atomic defects caused by missing ions are called point defects. Point defects are two types (i) stoichiometric defects and (ii) Non-stoichiometric defects. A Schottky defect consists of a pair of holes in the crystalline lattice due to the absence of one positive ion and one negative ion. In the case of Schottky defect the density of solid decrease. Schottky defect occurs mainly in the ionic compounds which contain smaller ions of similar size which have high coordination number, e.g., NaCl, CsCl, KCl and KBr. Frenkel defect is created when an ion occupies an interstitial site instead occupying its correct lattice site. The small cations (compared to anions) occupy the interstitial positions causing Frenkel defect. Frenkel defect is favoured when there is large difference in the size between cation and anion and having low coordination number 4 or 6. AgCl, AgBr and AgI show Frenkel defect. In these solids Ag+ ion occupy the interstitial sites. There is no change in the density of the solid due to Frenkel defect but the lattice may be distorted and show increase in the unit cell dimensions due to the presence of ions in interstitial positions. In non-stoichiometric defects the ratio of positive ion to negative ion is not exactly one. The metal excess defect is due to the absence of a negative ion from its lattice site leaving a hole which is occupied by an electron, thereby maintaining the electrical balance. NaCl/Na, KCl/K shows metal excess defect. The nonstoichometric NaCl/Na is yellow in colour. In metal excess defect the anion site occupied by an electron is called F-centre. The solids having F-centres have colour and the intensity of the colour increases with increase in the number of F-centres. Solids containing F-centres are paramagnetic since the electrons occupying the vacant sites are unpaired. When materials with F-centres are irradiated with light become photo conductors. Metal excess defects also occurs when an extra positive ion occupies an interstitial position in the lattice and to maintain electrical neutrality one electron is included in an interstitial position, e.g., ZnO, CdO, Cr2O3 and Fe2O3.

Solid State

• •

•

•

• •

•

•

The solids with metal excess defect contain free electrons and behave as n-type semiconductor. ZnO is white at low temperature but yellow at high temperature because it loses oxygen reversibly at high temperature and forms a non-stoichiometric defect with metal excess. Metal deficiency defect is due to the absence of a positive ion from its lattice point and the charge can be balanced by an adjacent metal ion having an extra positive charge, e.g., FeO, NiO, FeS and CuI. If an Fe2+ ion is missing from its lattice site in FeO then there must be two Fe3+ ions some where in the lattice to balance the electrical charges. Impurity defects in ionic solids may be created by introducing impurity ions. Impurity defect is created when the impurity ions have different valance state than that of the host ions. When molten NaCl is crystallized containing little SrCl2 some of the sites of Na + ions are occupied by Sr2+ ions, each Sr 2+ ion replaces two Na + ions occupying the site of one ion and the other site remain vacant. Crystals with metal deficiency defects are p-type semiconductors.

properties of solids

electrical conductivity in metals •

•

• • • •

•

•

•

electrical properties •

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Good conductors are those which allow the maximum current to flow throw them and their conductivity is of the order 108 ohm–1 cm–1. Metallic conductors allow the current to pass through them without undergoing any chemical change. In metallic conductors the conductance is due to the movement of electrons under the influence of an applied electric potential. The stream of electrons constitute the current. Electrolytic conductors allow the electricity to pass through them by undergoing chemical change. The conductivity of electrolytic conductors is due to the movement of ions in their molten state or in their aqueous solutions. Semiconductors are the solids whose conductivity lies between those of typical metallic conductors and insulators. Their conductivity range is 10–6 to 104 ohm–1 cm–1. Insulators are those which do not allow electricity to flow through them. Their conductivity order is 10–22 ohm –1 cm–1.

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According to molecular orbital (MO) theory in a metal crystal the orbitals of valence shell of all atoms combine to form a large number of MOs around all the atoms in that metal crystal. Since large number of MOs having almost equal energy are formed. They are very close to each other and form as band of MOs. Depending upon different types of atomic orbitals which ovelap different energy bands are formed. The arrangement of electrons in the different energy bands determines the characteristics of a metal. The energy bands formed from different atomic orbital may overlap or be separated from each other. The highest occupied energy band is called the valence band while the lowest unoccupied energy band is called conduction band. In the case of metals, the valence band may be half filled or there may be an overlapping between the valence band and conduction bands which makes it possible for the electrons to go into vacant bands and hence responsible for electrical conductivity. In the case of insulators the energy gap is very large and therefore the vacant conduction band is not available to the electrons of the completely filled valence band. In semiconductors the energy gap is intermediate between those of metals and insulators. The increase in temperature gives thermal energy for some electrons in valence band to move into the conduction band and hence their electrical conductivity increases with increase in temperature. Semiconductors are perfect insulators at absolute zero. Silicon and germanium which crystallize in diamond type network lattice are semiconductors. At room temperature the conductivity of silicon and germanium is extremely low but at higher temperature the bonds begin to break down ejecting the electrons and hence conductivity increases. The conduction introduced in the crystal without adding an external substance is called intrinsic conduction. Doping is a process of mixing pure silicon or germanium with an impurity. Doping enhances the conductivity and the products are called extrinsic semi conductors. n- type semi conductors (n- stands for negative) are obtained due to metal excess defect or by adding trace amounts of V or 15th group elements (P, As) to pure silicon or germanium.

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Solid State

When P or As is added to silicon or germanium some of the Si or Ge atoms in the crystal are replaced by P or As atoms and 4 out 5 electrons of P or As atom will be used for bonding with Si or Ge atoms while the fifth electron serve to conduct electricity. p-type semiconductors (p-stands for positive) are obtained due to metal deficiency defect or by doping with impurity atoms containing less electrons (i.e., atoms of III or 13 th group). When B, Ga or In is added to silicon or germanium some of the Si or Ge Atoms in the crystal are replaced by B, Ga or In atoms and only three valencies of Si or Ge are satisfied leaving an electron at Si or Ge because they have one electron less. Due to the shortage of electrons when Si or Ge is doped with B, Ga or In electron vacancies com monly known as positive holes rises in p- type semi conductors. When electric field is applied in p- type semiconductors flow of current takes place due to migratio of positive holes due to the movement of electrons from adjacent site into positive hole. Unlike metals the conductivity of semiconductors increases with increases in temperature because the weakly bound extra electron or positive hole become free by the increased temperature and can be used for conduction. A super conductor is that whose electrical resistance is zero. The electrical resistance of a material usually becomes zero near the absolute zero. The temperature at which a substance starts behaving like a super conductor is called transition temperature.

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magnetic properties • • •

Diamagnetic solids contain paired electrons (↑↓) and repell the external magnetic field. Paramagnetic solids contain unpaired electrons and are attracted into the applied magnetic field. In ferromagnetic solids there occurs magnetic interactions between the neighbouring centres (domains) and the electrons in these centres interact in parallel direction (↑↑↑↑↑). This interaction leads to an increase in magnetic moment. Iron, cobalt and nickel are ferromagnetic substances.

In antiferromagnetic solids, there occurs magnetic interaction between the neighbouring centres and the electrons in these centres interact in anti paral lel (↑↓ ↑↓ ↑↓) direction which leads to a decrease in magnetic moment, e.g., [Cu(CH 3COO) 2H 2O] 2; VO(CH 3COO) 2, MnO, MnO 2, Mn 2O 3. In ferrimagnetic solids there occurs magnetic interactions between the neighbouring centres and the electrons in these centrees interact in such a way which leads to the presence of uncompensated spins (↑↑ ↓↑ ↑↓) in the opposite direction resulting some magnetic moment e.g., magnetite (Fe3O4); Ferrite MFeO4 (Where M = Mg2+, Cu2+, Zn2+ etc).

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A dielectric substance is that which may not allow electric current through it but charges are induced on its faces on the application of electric field. When electric field is applied displacement of charges takes place and dipoles are created which results in polarization. Crystals in which dipoles may align to produce a net dipolement are called piezoelectrics. When piezoelectric crystals are subjected to pres sure or mechanical stress electricity is produced due to displacement of ions and this is known as piezoelectricity. In some piezoelectric crystals the dipoles are spon taneously aligned in a definite direction even in the absence of electric field and are called ferroelectric substances and this phenomenon is called ferroelectricity. Potassium hydrogen phosphate (KH2PO4), barium titanate (BaTiO3) and sodium potassium tartrate (NaKC4H4O6. 4H2O Rochelle salt) are ferroelectric substances. The crystals in which alternate dipoles are in opposite direction and have net dipole moment equal to zero are called antiferroelectric substances, e.g., lead zirconate (PbZrO2). The polar crystals which attain charges on opposite faces and produce a small electric current on heating are called pyroelectric substances and this phenomenon is called pyroelectricity.

Solid State

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practIce exercIse multiple choice questions with only one answer level I 1. A crystal plane intercepts the three crystallographic axes at a, 1/2 b and 3/2 c where a, b and c are the unit lengths along X, Y and Z axes, respectively. The miller indices of this plane will be (a) 1:2:0.67 (b) 1:0.5:1.5 (c) 3:6:2 (d) 2:1:3 2. The unit cell of highest symmetry is (a) Cubic (B) Triclinic (c) Hexagonal (d) Monoclinic 3. The unit cell of lowest symmetry is (a) Cubic (b) Triclinic (c) Hexagonal (d) Monoclinic 4. The number of planes of symmetry in cubic crystal is (a) 1 (b) 9 (c) 13 (d) 23 5. The number of axis of symmetry in a cubic crystal is (a) 1 (b) 9 (c) 13 (d) 23 6. A match box exhibits ________ geometry. (a) Cubic (b) Orthorhombic (c) Triclinic (d) Monoclinic 7. If the three inter-axial angles defining the unit cell are all equal in magnitude, the crystal can not belong to the ________ system. (a) Orthorhombic (b) Hexagonal (c) Tetragonal (d) Cubic 8. The axial angles in triclinic crystal system are (a) α = β = γ = 90° (b) α = γ = 90°, β ≠ 90° (c) α ≠ β ≠ γ ≠ 90° (d) α = β = γ ≠ 90° 9. TiO2 is well known example of (a) Triclinic system (b) Tetragonal system (c) Monoclinic system (d) Cubic system 10. For a certain crystal, the unit cell axial lengths are found to be a = 5.62 Å, b = 7.41 Å and c = 10.13 Å. The three coordinate axes are mutually perpendicular. The crystal system to which the crystal belongs is (a) Tetragonal (b) Orthorhombic (c) Monoclinic (d) Cubic 11. Most crystals show good cleavage because their atoms, ions or molecules are (a) Weakly bonded together (b) Strongly bonded together (c) Spherically symmetrical (d) Arranged in planes

12. In a crystal, the atoms are located at the position of (a) Zero potential energy (b) Infinite potential energy (c) Minimum potential energy (d) Maximum potential energy 13. A big RED spherical balloon (radius = 6a) is filled up with gas. On this balloon six small GREEN_ spherical balloons (radius = a) are stuck on the Surface in a specific manner. As RED balloon is slowly deflated, a point comes when all these six GREEN balloons touch and green balloons arrange themselves in a 3-D closed packing arrangement. At that stage the radius of the RED balloon would have reduced by approximately (a) 14.5 times (b) 1.414 times (c) 6.0 times (d) 2.42 times 14. An alloy of copper, silver and gold is found to have copper constituting the fcc lattice. If silver atoms occupy the edge centres and gold is present at body centre, the alloy has a formula (a) Cu4Ag2Au (b) Cu4Ag4Au (c) Cu4Ag3Au (d) CuAgAu 15. An element occurring in the bcc structure has 12.08 × 1023 unit cells. The total number of atoms of the element in these cells will be (a) 24.16 × 1023 (b) 36.18 × 1023 23 (c) 6.04 × 10 (d) 12.08 × 1023 16. Copper metal has a face-centred cubic structure with the unit cell length equal to 0.361 nm. The apparent radius of a copper ions is (a) 0.128 nm (b) 1.42 nm (c) 3.22 nm (d) 4.22 nm 17. The rank of a cubic unit cell is 4. The type of cell as. (a) Body centred (b) Face centred (c) Primitive (d) None of these 18. A solid PQ has rock salt type structure in which Q atoms are at the corners of the unit cell. If the body centred atoms in all unit cells are missing, the resulting stoichiomtery will be (a) PQ (b) PQ2 (c) P3Q4 (d) P4Q3 19. Gold crystallises in fcc lattice with edge length 4.07Å. The closest distance between gold atoms is (a) 2.035 Å (b) 8.140 Å (c) 2.878 Å (d) 1.357 Å

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Solid State

20. Packing fraction in sample cubic lattice is (a) 1/6π (c)

3 π 8

(b)

2 π 6

(d) None of these

21. The distance between two nearest neighbors in simple cubic lattice of axial length, l, is 3 l (a) l (b) 2 2 (c) l (d) none 2 22. An element (Atomic weight = 100) having bcc structure has unit cell edge length 400 pm. The density of this element will be (a) 5.188 gm/mL (b) 16.37 gm/mL (c) 7.29 gm/mL (d) 2.14 gm/mL 23. A metallic element exists as bcc lattice. Each edge of the unit cell is 2.88 Å. The density of the metal is 7.20 g cm–3. How many unit cells will be present in 100 g of the metal? (a) 6.85×102 (b) 5.82×1023 5 (c) 4.37×10 (d) 2.12×106 24. The unit cell of a metallic element of atomic mass 108 and density 10.5 g/cm3 is a cube with edge length of 409 pm. The structure of the crystal lattice is (a) fcc (b) bcc (c) hcp (d) Simple cubic 25. Packing fraction in body centred cubic lattice is (a) 1/6π (c)

3 π 8

(b)

2 π 6

(d) None of these

26. α form of iron exists in bcc form and g form of iron exists in fcc structure, Assuming that the distance between the nearest neighbors is the same in the two forms, the ratio of the density of g form to that of α form is (Atomic weight of Fe = 56) (a) 1.089 (b) 1.25 (c) 0.89 (d) 2.2 27. The most malleable metals (Cu, Ag, Au) have close – packing of the type (a) Hexagonal close-packing (b) Cubic close-packing (c) Body-centred cubic packing (d) Simple cubic 28. The anions (A) form hexagonal closest packing and atoms (C) occupy only 2/3 of octahedral voids in it then the general formula of the compound is (a) CA (b) A2 (c) C2A3 (d) C3A2

29. The number of tetrahedral and octahedral voids in hexagonal primitive unit cell are. (a) 8,4 (b) 2,1 (c) 12,6 (d) 6,12 30. Which type of hole is smaller in close packing? (a) Tetrahedral (b) Octahedral (c) Cubic (d) Square 31. The number of octahedral voids per unit body centred cubic cell is: (a) 1.0 (b) 2.0 (c) 3 (d) 4.0 32. The co-ordination number of a metal crystallizing in a hexagonal closed structure is (a) 12 (b) 4 (c) 8 (d) 6 33. For an octahedral arrangement of ions around each other in a structure, the range of radius ratio value should be (a) 0.155-0.255 (b) 0.255-0.414 (c) 0.414-0.732 (d) 0.732-1.000 34. In a face centred cubic arrangement of metallic atoms, what is the relative ratio of sizes of tetrahedral and octahedral voids? (a) 0.543 (b) 0.732 (b) 0.414 (d) 0.637 35. The inter-metallic compound LiAg crystallises in cubic lattice in which both Li and Ag have co-ordination number of eight. The class of crystal is: (a) Simple cubic (b) Body centred cubic (c) Face centred cubic (d) None of these 36. Transition metal, when they form interstitial com pounds, the non-metal (H,B,C) are accommodated in (a) Voids or holes in cubic-packed structure (b) Tetrahedral voids (c) Octahedral voids (d) All of these 37. In a body centred cubic packing, the nearest neighbors lie along the (a) Edges of the cube (b) Face diagonal (c) Line joining the two opposite corners of the face (d) Cube diagonal 38. A mineral having the formula AB2 crystallizes in the cubic close packed lattice, with the A atoms occupying the lattice points. What is the co-ordination number of the B Atoms? (a) 4 (b) 6 (c) 8 (d) 12 39. Copper has a face centred cubic lattice with a unit cell edge length of 0.361 nm. What is the size of the largest

Solid State

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atom that could be fit into octahedral holes of the lattice without disturbing the lattice? (a) 0.09 nm (b) 0.187 nm (c) 0.106 nm (d) 0.053 nm What is the size of the largest atom that could be fit into tetrahedral holes of the lattice without disturbing the lattice (a) 0.029 nm (b) 0.053 nm (c) 0.09 nm (d) 0.058 nm The ionic radii of Rb+ and I– are 1.46 and 2.16 Å. The most probable type of structure exhibited by it is (a) CsCl type (b) NaCl type (c) ZnS type (d) CaF2 type For an ionic crystal of general formula AX and co-ordination number 6, the value of radius ratio will be (a) Greater than 0.73 (b) In between 0.73 and 0.41 (c) In between 0.41 and 0.22 (d) Less than 0.22 In calcium fluoride structure, the co-ordination numbers of calcium and fluoride ions are (a) 8 and 4 (b) 6 and 8 (c) 4 and 4 (d) 4 and 8 The radius of Ag+ ion 126 pm while of I– ion is 216 pm. The Co-ordination number of Ag in AgI is (a) 2 (b) 4 (c) 6 (d) 8 A binary solid (A+ B–) has a rock salt structure. If the edge length is 400 pm and radius of cation is 75 pm, the radius of anion is (a) 100 pm (b) 125 pm (c) 250 pm (d) 325 pm Each unit cell of sodium chloride consists of 13 Na atoms and (a) 13 Cl atoms (b) 14 Cl atoms (c) 8 Cl atoms (d) 6 Cl atoms Solid AB has a rock salt type structure. If the radius of the cation is 200 pm, what is the maximum possible radius of the anion? (a) 483 pm (b) 273 pm (c) 888 pm (d) 573 pm Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two third of the octahedral voids in the closest packed array of oxide ions. What is the formula of aluminum oxide? (a) Al2O3 (b) AlO2 (c) Al3O4 (d) Al3O2 What is the formula of the magnetic oxide of cobalt, used in recording tapes, that crystallizes with cobalt atoms occupying one-eight of the tetrahedral holes and

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one half of the octahedral holes in a closest packed array of oxide ions? (a) Co2O3 (b) Co5O8 (c) CoO (d) Co3O4 A compound AB crystallizes as body centred cubic lattice with a unit cell side distance of 387 pm. The distance between the oppositely charged ions in the lattice is (a) 335.15 pm (b) 387 pm (c) 193.5 pm (d) 580.5 pm A solid contains An+ and Bm– ions. The structure of solid is fcc for Bm– ions and An+ ions are present in one fourth of tetrahedral voids as well as in one fourth of octahedral voids. What is the simplest formula of solid? (a) A3B4 (b) A4B3 (c) AB2 (d) A2B Caesium chloride on heating to 760 K changes into (a) CsCl (g) (b) NaCl Structure (c) antifluorite structure (d) ZnS structure BaO has a rock salt type structure. When subjected to high pressure, the ratio of the co-ordination number of Ba2+ ion to O2– ion changes to (a) 4:8 (b) 8:4 (c) 8:8 (d) 4:4 An ionic solid is hexagonal close packing of Q2– ions and Px+ ions are in half of the tetrahedral voids. The value of x should be. (a) 1 (b) 2 (c) 4 (d) ½ In a compound, oxide ions are arranged in cubic close packing arrangement. Cation A occupy one-sixth of the tetrahedral voids and cations B occupy one-third of the octahedral voids. The formula of the compound is (a) AB2O4 (b) A2BO2 (c) ABO3 (d) ABO4 In a compound XY 2O4 the oxide ions are arranged in cubic close packing arrangement and cation X are present in octahedral voids. Cations Y are equally distributed between octahedral and tetrahedral voids. The fraction of the octahedral voids occupied is (a) 1/2 (b) 1/4 (c) 1/6 (d) 1/8 A solid X melts slightly above 273 K and is a poor conductor of heat and electricity. To which of the following categories does it belong (a) Ionic solid (b) Covalent solid (c) Metallic (d) Molecular

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Solid State

58. Which of the following is an example of metallic crystal solid? (a) C (b) Si (c) W (c) AgCl 59. Which of the following is an example of covalent crystal solid? (a) SiO2 (b) Al (c) Ar (d) NaF 60. The particles would be stationary in a lattice only at (a) 273 K (b) 0 K (c) 298 K (d) 373 K 61. Solids in which the dipoles may align themselves in an ordered manner so that there is a net dipole moment, exhibit (a) Pyro-electricity (b) Piezo-eletricity (c) Ferro-electricity (d) Anti ferro-electricity 62. Which of the following is a ferromagnetic substance? (a) Fe2O3 (b) Cr2O3 (c) Fe3O4 (d) CrO2 63. In cubic closest packed structure of mixed oxides, the lattice is made up of oxide ions, 20% of tetrahedral voids are occupied by divalent X2+ ions 50% of the octahedral voids are occupied by trivalent Y3+ ions. The formula of the oxide is: (a) X2YO4 (b) X4Y5O10 (c) X5Y4O10 (d) XY2O4 64. Which of the following acts as a superconductor at 2.2 K? (a) He (b) Cu (c) K (d) Mg 65. Addition of arsenic in small amount to pure germanium will result in the formation of (a) n-type semiconductor (b) Germanium arsenide (c) p-type semiconductor (d) A super conducting alloy 66. CaO and NaCl have the same crystal structure and nearly the same ionic radii. If ‘u’ is the lattice energy of NaCl, the lattice energy of CaO is very nearly (a) 2u (b) u (c) 4u (d) u/2 67. The presence of excess sodium in sodium chloride makes the crystal appearance yellow. This is due to presence of (a) Schottky defect (b) Frenkel defect (c) F-centres (d) Interstitial defect

68. When AgCl crystal is doped with CdCl2, then it produces a (a) Schottky defect. (b) Frenkel defect. (c) p-type semiconductor. (d) Interstitial defect. 69. Superconductors are substances which. (a) Conducts electricity at low temperatures. (b) Conduct electricity at high temperature. (c) Offers very high resistance to the flow of current. (d) Offers no resistance to the flow of current. 70. Particles of quartz are packed by (a) Electrical attraction forces. (b) Covalent bond forces. (c) Van der Waal’s forces. (d) H – bond forces. 71. Ionic solids are characterized by (a) Good conductivity in solid state. (b) High vapour pressure. (c) Low melting point. (d) Solubility in polar solvents. 72. Packing fraction in face centered cubic lattice is 1 2 (b) (a) π π 6 6 (c)

3 π 8

(d) none of these

73. If the volume occupied in the crystal by an unit cell of NaCl is 47×10–24 ml, calculate the volume of a crystal weighing 1.0 gm. (a) 2.13×1022 mL (b) 8.03×10–25 mL (c) 0.48 mL (d) 0.12 mL 74. The density of KCl is 1.9893 g cm–3 and length of a side unit cell is 6.29 Å as determined by X-rays diffraction. The value of Avogadro’s number calculated from these data is. (a) 6.017×1023 (b) 6.023×1023 (c) 6.03×1023 (d) 6.017×1019 75. Lithium borohydride crystallizes in an orthorhombic system with 4 molecules per unit cell. The unit cell dimensions are a = 6.8 A° , b = 4.4 A° and c = 7.2 A°. The density of crystals is. (a) 0.6708 g cm–3 (b) 1.6708 g cm–3 (c) 2.6708 g cm–3 (d) None 76. A certain solid metal of atomic weight 107.88 crystallizing in the cubic system has a density of 10.53 gm

Solid State

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per cm3. The side length of the unit cell = 4.0774 Å. Identify the lattice type. (a) Simple cubic (b) Face centered cubic (c) Body centered cubic (d) fcc with lattice vacancies A packing consists of a base of spheres, followed by a second layer where each sphere rests in the hollow at the junction of four spheres below it and the third layer then rests on these in an arrangement which corresponds exactly to that in the first layer. This packing is known as. (a) Hexagonal close packing. (b) Cubic close packing. (c) Square close packing. (d) Body centered cubic packing. The low density of alkali metals is due to (a) Their body centered structure in which about 32% of the available space is unfilled. (b) Their hexagonal close packed structure in which about 74% of the available space is unfilled. (c) Their cubic close packed structure in which about 74% of the available space is unfilled. (d) Their body centered cubic structure in which about 47% of the available space is unfilled. An ionic crystalline solid, MX3, has a cubic unit cell. Which of the following arrangement of the ions is consistent with the stoichiometry of compound? (a) M3+ ions at the corners and X– ions at the face centers. (b) M3+ ions at the corners and X– ions at the body centers. (c) X– ions at the corners and M3+ ions at the face centers. (d) X- ions at the corners and M3+ ions at body centers. In which of following crystals alternate tetrahedral voids are occupied? (a) NaCl (b) ZnS (c) CaF2 (d) Na2O CaBr has been structure with edge length 4.3. The shortest inter ionic distance in between Cs + and Br- is. (a) 3.72 (b) 1.86 (c) 7.44 (d) 4.3 A compound containing Zn, Al and S crystallizes with a closest packed array of sulphide ions. Zinc ions are found in one-eighth of the tetrahedral holes and aluminum ions in one half of the octahedral holes.

4.65

What is the empirical formula of the compound? (a) ZnAlS (b) ZnAl2S4 (c) Zn2Al2S3 (d) Zn3AlS4 83. In a face centred cubic lattice, atom A occupies the corner positions and atom B occupies the face centre positions. If one atom of B is missing from one of the face centre points, the formula of the compound is: (a) A2B3 (b) A2B5 (c) A2B (d) AB2

multiple choice questions with only one answer level II 1. In an ionic solid anions occupy the corners and face centres of cubic cell. Whereas the cation occupy all the octahedral holes as well as half of the tetrahedral voids. The effective number of all type of cations in the unit cell is (a) 4 (b) 6 (c) 8 (d) 9 2. In a fcc arrangement of atoms. What is the relative ratio sizes of tetrahedral and octahedral voids. (a) 0.543 (b) 0.732 (c) 0414 (d) 0.637 3. In compound XY2O4 oxide ions are arranged in CCP and cations X are present in octahedral voids. Cations Y are equally distributed between octahedral and tetra hedral voids. The fraction of the octahedral voids occupied is (a) 1/2 (b) 1/4 (c) 1/8 (d) 1/6 4. In Fe0.93 O the percentage of Fe+3 ions present in the compound is (a) 14% (b) 14.8% (c) 15.2% (d) 15.8% 5. The addition of CaCl2 crystal to KCl crystal (a) Lowers the density of the KCl crystal (b) Increases the density of the KCl crystal (c) Does not effect the density of KCl crystal (d) Increase the Frenkel defect of the crystal 6. The no. of cation vacancies if NaCl is dopped with 10–3 % mole of SrCl2 is (a) 3.01×1019 (b) 3.01×1018 18 (c) 6.02×10 (d) 6.02×1019 7. A metal crystallizes in fcc, bcc form whose edge lengths are 3.6A0 and 3.0A0 respectively. The ratio of densities of fcc and bcc is (a) 1.157 (b) 1.51 (c) 1.71 (d) 2.0

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Solid State

8. Iron crystallizes in bcc, and fcc forms. The distance between two nearest neighbours is same. The ratio of the density of fcc form to that of bcc is (a) 1.42 (b) 1.71 (c) 2.12 (d) 1.08 9. Lithium selenide can be described as a closest packed array of selenide ions with lithium ions in all of the tetrahedral holes, formula of lithium selenide is: (a) Li2Se (b) LiSe (c) LiSe2 (d) Li3Se 10. The diametre of marble is 10 mm. They are to be placed such their centres are lying in a square bond by four lines each of length 40 mm. The arrangement of marbles in places so that maximum number of marbles can be placed inside the area. The no. of marbles per unit area per cm2 is (a) 1 (b) 1.115 (c) 1.215 (d) 1.125 11. The edge length of LiCl is 4.0A0. It is in fcc structure so radius of anion is (a) 2A0 (b) 1.414A0 0 (c) 1.732A (d) 1.61A0 12. An element crystallizes in a ‘bcc’ lattice. Nearest neighbors and nearest neighbors of the element are respectively. (a) 8, 8 (b) 8,6 (c) 6,8 (d) 6,6 13. Zinc blende structure is obtained with Zn2+ occupies (a) All tetrahedral sites (b) Half tetrahedral sites (c) All octahedral sites (d) Half octahedral sites 14. In a non-stoichiometric sample of ferrous oxide with NaCl structure, the ratio of Fe3+ to Fe2+ was found to be 0.15. The fraction of octahedral sites remain as vacancies is (a) 0.0913 (b) 0.0387 (c) 0.0613 (d) 0.052 15. A metal has face centred cubic structure with a unit cell edge length of 5.64. What is the size (radius) of the largest atom which could fit into the interstices of the metal lattice without distorting it? (a) 0424Å (b) 0.828Å (c) 0.564Å (d) 0.212Å 16. In an ABC, ABC packing of Giant Buckies (C-240); Bucky babies – Chotu (C-32) and Motu (C-44) are doped in tetrahedral and octahedral voids respectively. In the unit cell thus created all balls along any one body diagonal are removed and Bucky Onions (C-240 containing C-60 suspended inside it, with their centres coinciding) are fitted at either extremes. The number of C atoms finally present in the unit are

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24.

25.

(a) 960 (b) 1284 (c) 1299 (d) 1392 An element X (atomic weight = 24 gm/mol) forms a face centred cubic lattice. If the edge length of the lattice is 4×10–8 cm and the observed density is 2.40 × 103 kg/m3 then the percentage occupancy of lattice points by element X is : (use NA=6×1023) (a) 96 (b) 98 (c) 99.9 (d) None of the above Solid AB has NaCl structure. If the radius of cation is 100 pm. The radius of anion is (a) 222 pm (b) 444 pm (c) 120.7 pm (d) 241.5 pm Number of formula units in unit cells of MgO (rock salt) ZnS (Zinc blende) and Pt (fcc) respectively are: (a) 4, 3, 2 (b) 4, 3, 4 (c) 4, 4, 4 (d) 4, 3, 1 CsCl arranged in bcc structure edge length is 4.0 A 0. The distance between neighbouring Cs + and Cl– Ions is (a) 2.82 A0 (b) 2.0 A0 0 (c) 3.46 A (d) 3.0 A0 In a spiral structure, oxide ions are cubical closest packed whereas 1/8 of tetrahedral voids are occu pied by A 2+ cations and 1/2 of octahedral voids are occupied by B +3 cations. The formula of compound as (a) A2B2O4 (b) AB2O4 (c) A2B4O4 (d) A4B2O4 Number of unit cells in 58.5 gm of NaCl is nearly (a) 6×1020 (b) 3×1022 23 (c) 1.5×10 (d) 0.5×1024 Potassium crystallizes in body centred cubic lattice with a unit cell length a = 5.2A0. The distance between nearest and next nearest neighbours are (a) 4.5 A0 and 9.01 A0 (b) 4.5 A0 and 5.2 A0 (c) 3.7 A0 and 9.01 A0 (d) 9.01A0and 4.5 A0 Titanium crystallizes in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially, by allowing these elements to occupy hole in the host lattice. Hydrogen occupies tetrahedral holes while carbon occupies octahedral holes. The formulae of hybride and carbide respectively are (a) TiH2, TiC (b) TiH4, TiC2 (c) TiH, TiC2 (d) TiH, TiC Addition of CdCl2 to AgCl yields solid solutions, where the divalent ions Cd2+ occupy the Ag+ sites. Which one of the following statements is true? (a) The number of cationic vacancies is equal to one half of number of divalent ions added.

Solid State

26.

27.

28.

29.

30.

31.

32.

33.

(b) The number of cationic vacancies is equal to that of divalent ions added. (c) No cationic or anionic vacancies are created. (d) The number of anionic vacancies is equal to number of divalent ions added. In the crystalline state of an ionic compound MX of a metal M. having cations and anions of radius rM and rX respectively the co-ordination number of M will be 4 if the value of rX/rM is between (a) 1 and 1.4 (b) 1.4 and 2.4 (c) 2.4 and 4.6 (d) 4.6 and 5.6 The Van der Waals constant ‘b’ for N 2 and H2 has the values 0.0399 lit/mole and 0.0266 lit/mole. The density of solid N 2 is 1 gm/cm 3. Assuming the molecules in the solid to be close packed with the same percentage void. The density of solid H 2 would be in gm/cm3. (a) 0.628 (b) 0.107 (c) 1.466 (d) 0.071 The edge length of unit cell of metal having molecular weight 75 gm/mol. is 5Å which crystallizes in cubic lattice. If the density is 2 gm/cc then find the radius of metal atoms. (N A = 6 × 1023) Give the answer in pm. (a) 217 (b) 187 (c) 257 (d) 165 Total volume of atoms present in a face-centred cubic unit cell of a metal is (r is atomic radius) (a) 24/3 πr3 (b) 12/3 πr3 3 (c) 16/3 πr (d) 20/3 πr3 KCl crystallizes in the same type of lattice as does NaCl. Given that rNa+/rcl– = 0.55 and rNa+/rK+ = 0.74, calculate the ratio of the side of the until cell for KCl to that for NaCl. (a) 1.122 (b) 1.224 (c) 1.414. 4).0.732 A particular solid is very hard and has a very high melting point. In solid state it is non conductor and its melt is a conductor of electricity. Classify the solid. (a) Metallic (b) Molecular (c) Network (d) Ionic The energy gap between conduction band and valence band is zero in (a) Sodium (b) Magnesium (c) Silicon (d) Germanium An alloy of copper, silver and gold is found to have copper forming the simple cubic close packed lattice. If the silver atoms occupy the face centres and gold is present at the body centre, then the formula of the alloy will be (a) Cu4Ag4Au (b) Cu4Ag2Au (c) Cu Ag Au (d) CuAg3Au

4.67

34. A binary solid (A+ B–) has a zinc blende structure with B– ions constituting the lattice and A+ ions occupying 25% tetrahedral holes. The formula of solid is (a) AB (b) A2B (c) AB2 (d) AB4 35. Give the correct order of initials T (true) or F (false) for following statements. I. In an anti-fluorite structure anions form fcc lattice and cations occupy all tetrahedral voids. II. If the radius of cations and anions are 0.2Å and 0.95Å then co-ordination number of cation in the crystal is 4. III. Cation is transferred from a lattice site to an interstitial position in Frenkel defect. IV. Density of crystal always increases due to substitutional impurity defect. (a) TFFF (b) FTTF (c) TFFT (d) TFTF 36. A certain solid mixed oxide crystallizing in the cubic system contains cations M1 and M2 and the oxide ion O2–. Each M1 ion is surrounded by 12 equidistant nearest neighbour oxide ions. If the oxide ions occupy face centres of the cubic unit cell, where are the M1 ions situated? (a) At the centre of the until cell (b) At the corners of the cube (c) At the edge centres (d) Occupying half the number of edge centres 37. In a hypothetical solid C atoms are found to form cubical close packed lattice, A atoms occupy all tetrahedral voids B atoms occupy all octahedral voids. A and B atoms are of appropriate size, so that there is no distortion in ccp lattice of C atoms. Now if a plane as shown in the following figure is cut, then the cross section of this plane will look like

C

(a)

B C A B BA B C

B

C

4.68

Solid State

(b)

C

C

C

B

B

B

C

C

C

C (c)

(d)

B

C A A

B

A A

C B

C

C

C

C

C

C

A A B B A A C C C

B

38. If NaCl is doped with 10–4 mol % SrCl2, the concentratio of cation vacancies will be (NA= 6.02 × 1023 mol–1) (a) 6.02×1016 mol–1 (b) 6.02×1017 mol–1 (c) 6.02×1014 mol–1 (d) 6.02×1015 mol–1 39. In case of an ionic crystal, the experimental density is found to be higher that the theoretical density. Hence the defect could be (a) Frenkel (b) Interstitial (c) Schottky (d) None of these 40. In a face-centred cubic arrangement of metallic atoms, what is the relative ratio of the sizes of tetrahedral and octahedral voids? (a) 0.543 (b) 0.732 (c) 0.414 (d) 0.637 41. The fraction of total volume occupied by the atoms present in a simple cube is π π (a) (b) 4 2 3 2 (c) π/4 (d) π/6 42. Barium titanate crystallizes in pevovskite structure which is cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying centres of the unit cell. If Ti 4+ ion is described as occupying the holes in BaO lattice, the type of hole and fraction of these holes occupied by these ions are (a) 100% of octahedral holes (b) 25% of tetrahedral holes (c) 25% of octahedral holes (d) 25% of cubic holes.

43. A metal M crystallized in both fcc lattice and in bcc lattice. If the densities of these forms are in 2:1 ratio the corresponding cubic edges are in the ratio (a) 1:2 (b) 2:1 (c) 1:4 (d) 1:1 44. An ion that leaves its regular site and occupies a position in the space between the lattice sites is called (a) Frenkel defect (b) Schottky defect (c) Impurity defect (d) Vacancy defect 45. In cubic packed structure of mixed oxides, the lattice is made up of oxide ions, one-fifth of tetrahedral voids are occupied by (X)m+ ions, while half of the octahedral voids are occupied by (Y)n+ then the formula of the oxide is (a) XY2O4 (b) X2YO4 (c) X4Y5O10 (d) X5Y4O10 46. In an ionic solid, anions occupy the corners and face centres of a cubic unit cell whereas, the cations occupy all the octahedral voids as well as half of the tetrahedral voids. The effective number of all types of cations in the unit cell are (a) 4 (b) 6 (c) 8 (d) 10 47. In an fcc crystal, which of the following shaded planes contains the following type of arrangement of atoms?

(a)

(b)

(c)

(d)

48. Crystal is made of particles A and B. A forms fcc packing and B occupies all the octahedral voids. If all the particles along the plane as shown in the fig. are removed, then, the formula of the crystal would be: A B

Solid State

(a) AB (b) A5B7 (c) A7B5 (d) None of these 49. In a CCP lattice of X and Y, X atoms are present at the corners while Y atoms are at face centres. Then the formula of the compound would be if one of the X atoms from a corner is replaced by Z atoms (also monovalent)? (a) X7Y24Z2 (b) X7Y24Z (c) X24Y7Z (d) XY24Z 50. The following diagram shows arrangement of lattice point with a=b=c and α=β=γ=900. Choose the correct options:

4.69

53. In an fcc unit cell, atoms are numbered as shown below. The atoms not touching each other are (Atom numbered 3 is face centre of front face) 1 4 3 2

(a) 3 and 4 (c) 1 and 2

(b) 1 and 3 (d) 2 and 4

54. Marbles of diametre 2 cm are to be placed either inside or upon an equilateral triangle (edge length 4 cm) drawn on a floor. The maximum number of marbles that can be accommodated is/are: (a) 1 (b) 2 (c) 3 (d) 6

multiple choice questions with one or more than one answer (a) The arrangement is SC with each lattice point surrounded by 6 nearest neighbours (b) The arrangement is SC with each lattice point surrounded by 8 nearest neighbours (c) The arrangement is fcc with each lattice point surrounded by 12 nearest neighbours (d) The arrangement is bcc with each lattice point surrounded by 8 nearest neighbours 51. Spinel is an important class of oxides consisting of two types of metal ions with the oxide ions arranged in CCP pattern. The normal spinel has 1/8 of the tetrahedral holes occupied by one type of metal ion and ½ of the octahedral holes occupied by another type of metal ion. Such a spinel is formed by Zn 2+, Al3+ and O2– with Zn2+in the tetrahedral holes. If CCP arrangement of oxide ions remains undistorted in the presence of all the cations, formula of spinel and fraction of the packing fraction of crystal are respectively. (a) ZnAl2O4, 77% (b) ZnAl2O4, 74% (c) Zn2AlO4, 74% (d) Zn3Al2O6, 74% 52. Which of the following statements is correct about the conduction of electricity in pure crystal of silicon at room temperature? (a) The conduction is due to electrons present in fully occupied lowest energy states (b) The conduction is due to only some electrons capable of leaving the bonds at room temperature (c) The conduction is only due to the holes formed from the release of electrons (d) The conduction is due to the movement of both the electrons released and holes formed

1. Which of the following compounds are used as pick up in the record players? (a) Lead zirconate (b) Barium titanate (c) Rochelle salt (c) Quartz 2. Which of the following statements are correct? (a) The lattice positions left vacant by anion and occupied by electrons are called F-centres. (b) The presence of F-centres make lattice electrically neutral and diamagnetic. (c) The presence of F-centres give colours to the crystal. (d) F-centres contains unpaired electrons. 3. Frenkel defect is notified in (a) AgBr (b) ZnS (c) AgI (d) NaCl 4. Which of the following are correct with respect to zinc blende structure (a) Zn+2 ions are present at corners and at the centres of each face. (b) Only alternate tetrahedral holes occupied by Zn+2 ions. (c) The coordination number of Zn+2 and S–2 is 4 each. (d) The number of ZnS units is a unit cell is 4. 5. Which of the following statements are correct? (a) The co-ordination number of each type of ion in CsCl is 8. (b) A metal that crystallises in bcc structure has coordination number 12. (c) A unit cell of ionic crystal shares some of its ions with other unit cells. (d) The length of the unit cell in NaCl is 552 pm (rNa+ = 95 pm, rCl– = 181 pm).

4.70

Solid State

6. The unit cell which are also primitive are (a) Simple cubic (b) Rhombohedral (c) bcc (d) Triclinic 7. Identify the correct statement among the following (a) CsCl is fcc structure. (b) The unit cell of KCl is fcc. (c) The hydrated copper ion in CuSO4·5H2O is three coordinate with a tetrahedral disposition of H2O molecules. (d) Both ccp and hcp have the same % of void space in the unit cell. 8. Identity the correct statements among the following (a) The cation-anion radius ratio given on insight in to the coordination number of ion in purely ionic structures, but cannot be applied to covalent compound. (b) The ceasium halide structure with 8 coordination number. (c) The hcp, ccp structures are the closed packed structures with minimum void space. (d) The colour of PbI2 (solid) is due to its structural defects. 9. Radius ratio (rc /ra) calculations (a) Assume 100% ionic nature of crystal is apply (b) Indicate the coordination number 8, 6 etc. (c) Are only approximate since many factors such as polarizabilities of ions are ignored. (d) Can decide the crystal structure of ionic solids . 10. In which of the following solids, cations occupy tetrahedral voids? (a) NaCl (b) CsBr (c) ZnS (d) Na2O 11. Crystalline solids have (a) Sharp melting point (b) Anisotropic character (c) The character of super cooled liquid (d) Smooth cooling curve 12. Which of the following informations are correct about CsCl? (a) It has body centred cubic unit cell. (b) coordination number of both Cs+ and Cl– ions are eight. (c) C/– ions are present at body centre. (d) Cs+ ions are present at body centre. 13. Select the correct statements among following. (a) Nearest neighbor distance in NaCl =

a 2

(b) Nearest neighbor distance in CaF2 = a (c) Nearest neighbor distance in Na 2 O = a

14.

15.

16.

17.

18.

19.

20.

21. 3/4

3/4

(d) Nearest neighbor distance in CsCl = a 3 / 2 (a = edge length of unit cell) In the unit cell of NaCl, which of the following statements are correct? (a) Na+ ions have six Cl– ions in its nearest neighborhood. (b) Cl– ions have six Na+ ions in its nearest neighborhood. (c) Second nearest neighbor of Na+ ion are twelve Na+ ions. (d) NaCl has 68% of occupied space. In face centred cubic structure, the octahedral voids are located at (a) Edge centres (b) Body centres (c) Face centres (d) Corners Which of the following is/are correct about the point defects (a) In Frenkel defect, the dielectric constant of solid increases. (b) In Schottky defect, the density of solid decreases. (c) In Frenkel defect, the density of solid decreases. (d) In Schottky defect, the dielectric constant of solid increases. The number of nearest neighbors with which a given sphere of packing is in contact, is called co-ordination number. The correct statements about co-ordination number are (a) Coordination number decreases on heating. (b) Coordination number increases on applying pressure. (c) CsCl acquires NaCl type structure on heating. (d) Coordination number of octahedral site is 6. In which of the following type of solids, the coordination number and packing efficiency are same: (a) bcc type solids (b) simple cubic type solid (c) hcp type solids (d) fcc type solid Sodium oxide Na2O is a crystalline solid. It has: (a) Antiflourite structure (b) Na+ ions are present at body diagonals (c) Na+ ions are present at tetrahedral voids (d) O2– ions are present at octahedral voids In a AB unit crystal of NaCl type assuming Na+ forming fcc (a) The nearest neighbour of A+ is 6B–ion (b) The nearest neighbour of B– is 6A+ ion (c) The second neighbour of A+ is 12A+ (d) The packing fraction of AB crystal is 3π / 8 Which of the following is/are a Bravis lattice? (a) Face centred orthorhombic (b) Face centred cubic (c) Body centred tetragonal (d) End centred monoclinic

Solid State

22. Which of the following have two Bravais lattices? (a) Cubic (b) Orthorhombic (c) Tetragonal (d) Monoclinic 23. Which of the following have fcc packing? (a) NaCl (b) NaF (c) KCl (d) CsCl 24. Which of the following statements are correct? (a) ccp structure has 3-layers. (b) In ccp structure first and fourth layers are repeated. (c) In hcp structure first and third layers are repeated. (d) In an hcp structure first and third layers are repeated. 25. Which of the following statements are correct? (a) Na has bcc unit lattice. (b) Mg has hcp unit lattice. (c) CsCl has bcc unit lattice. (d) In CsBr, Br– surrounds Cs+ which is present in corner from the body centre of eight cubes. 26. Which of the following have layered lattices? (a) B (b) Graphite (c) NaF (d) CsCl 27. Which one of the following contains only one Bravais lattice: (a) Hexagonal (b) Monoclinic (c) Rhombohedral (d) Triclinic 28. Stacking (a pile) of square close packed layer give rise to (a) bcc structure (b) fcc structure (c) sc structure (d) None of these 29. Packing density in identical solid spheres is 74% in (a) sc (b) fcc (c) hcp (d) bcc 30. The correct statement(s) regarding defects in solids is (are) (a) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion. (b) Frenkel defect is a dislocation defect. (c) Trapping of an electron in the lattice leads to the formation of F-centre. (d) Schottky defects have no effect on the physical properties of solids. 31. Which are amorphous solids? (a) NaCl (b) CaF2 (c) Glass (d) Plastics 32. Which of the following have z = 4 (Z is rank of unit cell) (a) ZnS (b) Na2O (c) CaF2 (d) B2O3

4.71

33. Which of the following statement (s) is/are true? (a) Conductivity of semiconductors increases with increase in temperature (b) Pure ionic solids are insulators (c) NaCl is diamagnetic substance (d) TiO2 is paramagnetic substance 34. Which of the following crystals do not crystallize in hcp structures? (a) Na (b) Be (c) Ca (d) Ba 35. In a body centred unit lattice of A2 type (a) the edge length is equal to 4 R

3

2

(b) the edge length is equal to 2R + 2R2 (c) the edge length is equal to 2 D nearest neigbouring distances

36.

37.

38.

39.

3 where d =

(d) the square of the edge length is equal to 16R2 where (R) is radius of atom Which of the following statements are correct regarding hcp (a) The coordination number is 12 (b) There are 3 spheres inside this cube (c) The number of particles per unit lattice is 6 (d) There are 5 spheres present per unit lattice The composition of sample of wustite is Fe0.93O1.00. Which of the following statements are correct related to this compound? (a) The percentage of Fe (III) by mass is 11.5% (b) The ratio of Fe (III) to Fe (II) ion is 0.17 (c) The percentage of Fe (II) ion 11.5% (d) The amount of Fe3+ is 7.84 g In a CsCl type crystal (a) Cs+ forms a simple cubic lattice, Cl– forms a simple cubic lattice (b) Cl– occupies body centre of Cs+ (c) Cs+ occupies body centre of Cl– (d) It is impossible for Cl– to occupy body centre of Cs+ because the body centre void of Cs+ is smaller than the Cl– ion size Which of the following statements are correct? (a) A NaCl type AB crystal lattice can be interpreted to be made up of two individual fcc type unit lattice of A+ and B– fused together in such a manner that the corner of one unit lattice becomes the edge centre of the other (b) In a face centred unit lattice, the body centre is an octahedral void (c) In an SCC lattice, there can be no octahedral void (d) In an SCC lattice, the body centre is the octahedral void

4.72

Solid State

40. A metallic lattice crystallizes in such a manner that the atoms are arranged in a hexagonal manner in one layer and the repetition is ABABAB……of the hexagonal layers. (a) The rank of the unit lattice is 6 (b) The rank of the unit lattice is 12   3 (c) The volume of the unit lattice is  6 × × 4r 2  . 4   4r 2 3 (d) The percentage of vacant space=26 41. The ZnS zinc blende structure is cubic. The unit cell may be described as a face-centred sulfide ion lattice with zinc ions in the tetrahedral voids. Choose the correct statement(s) among the following: (a) The nearest neighbours of Zn2+ are four (b) The angle made by the lines connecting any Zn2+ to any two of its neighbours 109028’. (c) The nearest neighbours of S–2 are four (d) The nearest neighbours of S–2 are eight 42. Gold has fcc structure. Choose the correct statement (s) among the following: (a) The closest distance between impurity atom and a gold atom, if the impurity atom occupies a tetrahedral hole is

3a (a = edge length). 4

(b) The closest distance between an impurity atom and a gold atom if the impurity atom occupies on octahedral hole is a/2. (c) The impurity in octahedral hole has more nearest neighbours to interact with than the one in tetrahedral hole. (d) Number of octahedral holes is more than that of tetrahedral holes. 43. BaTiO3 crystallises in the perovskite structure. This structure may be described as cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying the face centres and titanium ions occupying the holes (a) Ti+4 occupies octahedral void. (b) Ti+4 occupies the octahedral void which is present at the body centre. (c) Ti+4 occupies the octahedral void which is present along the edge of the unit cell. (d) Ti+4 occupies either type of the octahedral voids. 44. Some following statements about ThO2 structure (Fluorite). The correct statements are (a) In flourite, each Th+4 ion is surrounded by eitht O–2 ions in a bcc angagement. (b) The co-ordination number of both the ions is not same. (c) O–2 ion occupies tetrahedral voids (d) Flourite structure is found when the radius ratio is 0.73 or more.

45. If the radius of Na+ is 95 pm and that of Cl– ion is 181 pm then (a) co-ordination no. of Na+ is 6 (b) co-ordination no. of Na+ is 8 (c) length of the unit cell is 552 pm (d) length of the unit cell is 380 pm 46. Pick up the false statement. (a) In the fluorite structure (CaF2), the Ca 2+ ions are located at the lattice points and the fluoride ions fill all the tetrahedral holes in the ccp crystal. (b) In the antifluorite structure (Li2O, Rb2S) the cations are located at the lattice points and anions fill the tetrahedral holes in the ccp structure. (c) The radius of a metal atom is taken as half the nearest metal-metal distance in metallic crystal. (d) One tetrahedral void per atom is present in hep structure. 47. The hcp and ccp structure for a given element would be expected to gave: (a) The same co-ordination number (b) the same density (c) The same packing fraction (d) Same number of atoms 48. In a AB unit cell (Rock salt type) assuming A+ forming fcc (a) The nearest neighbour of A+ is 6B– ion. (b) The nearest neighbour of B– is 6A+ ion. (c) The second neighbour of A+ is 12A. (d) The packing fraction of AB crystal is 0.79.  3  − 1 49. In the fluorite structure if the radius ratio is   2  how many ions does each cation touch? (a) 4 anions (b) 12 cations (c) 8 anions (d) No cations 50. Select the correct statement about fcc (ABCAB….) structure. (a) Distance between nearest octahedral void and tetrahedral void is 3a / 4 . (b) Distance between two nearest octahedral void a 2 (c) Distance between two nearest tetrahedral void 3a

4

 2 (d) Distance between layer A and B is  2r  3 

Solid State

4.73

comprehensive type questions

passage III

passage I

Potassium crystallizes in bcc lattice. Edge length is 5.20 A

An element crystallizes into a structures in atoms occupies corners and two atoms are present in one of body diagonal. The volume of unit is 25×10–24 cm3 and density is 6.0 gm/ cm3 1. The atomic weight of element is (a) 30 (b) 45 (c) 60 (d) 75 2. The number of atoms present in 180 gm of element is: (a) 3.6×1024 atoms (b) 2.4×1024 atoms (c) 1.8×1024atoms (d) 1.44×1024 atoms 3. The number of unit cells present in 180 gm of element is (a) 1.2×1024 (b) 0.8×1024 24 (c) 0.6×10 (d) 0.48×1024 passage II Atoms A occupies, corners, face centres and atoms B occupies octahedral voids. Then the structure is

is A,

is B

1. If atoms present along one axis is removed then the formula of the compound is (a) A4B3 (b) A3B4 (c) A2B (d) AB2 2. The co-ordination number of A, B is (a) 4,6 (b) 6,4 (c) 4,4 (d) 6,6 3. If edge length of unit cell is 400 pm then the size of B atoms is (a) 31.8 pm (b) 113.2 pm (c) 58.6 pm (d) 28.3 pm 4. If AB is ionic solid, A ion left its site and electrons are left in this site. The effect is (a) Scholtky (b) Frenkel (c) F-centres (d) All defects 5. If B atoms present in tetrahedral voids instead of octahedral voids then co-ordination numbers of A,B are (a) 8,4 (b) 4.8 (c) 4,4 (d) 6,6

5.20 A 1. What is the distance between nearest neighbours (a) 4.50 A (b) 5.20 A (c) 9.00 A (d) 10.4 A 2. What is the distance between next nearest neighbours (a) 4.50 A (b) 5.20 A (c) 9.00 A (d) 10.4 A 3. How many nearest neighbours does each K atoms have (a) 12 (b) 6 (c) 4 (d) 8 4. How many next nearest neighbours does each K atom have (a) 12 (b) 6 (c) 4 (d) 8 5. The packing fraction value of this crystal is (a) 0.72 (b) 0.54 (c) 0.68 (d) 0.52 passage IV By X-ray studies, the packing of atoms in a crystal of gold is found to be in layers such that starting from any layer, every fourth layer is found to be exactly identical. The density of gold is found to be 19.4 g/cm3 and its atomic mass is 197 amu 1. The coordination number of gold atom in the crystal is (a) 4 (b) 6 (c) 8 (d) 12 2. The approximate number of unit cells present in 1 g gold is (a) 3.06×1021 (b) 1.53×1021 20 (c) 3.82×10 (d) 7.64×1020 3. The length of the edge of the cell will be (a) 407 pm (b) 189 pm (c) 814 pm (d) 204 pm

4.74

Solid State

passage V

passage VII

When a metallic element combines with a non-metallic element, generally an electrovalent substance is formed. Metal atom loses electron or electrons to from cations. Non metal atom gains electrons to from anion. The oppositely charged ions are held together in an ionic solid by electrostatic attraction forces. An example of formation of electrovalent substance:

The density of crystalline CsCl is 4.0 gm/cm3.(Cs = 132.5, Cl = 35.5) 1. What is the smallest Cs-to-Cs internuclear distance, if it is equal to the length of the side of a cube corresponding to the volume of one CsCl ion pair? (a) 412 Å (b) 4.12 Å (c) 10.4 Å (d) 2.6 Å 2. What is the smallest Cs-to-Cl internuclear distance in the crystal, assuming each Cs+ ion to be located in the centre of a cube with Cl– ions at each corner of the cube? (a) 7.17 Å (b) 1.78 Å (c) 3.57 Å (d) 2.91 Å

K(s)+1/2Br2(l)→KBr(s) 1. During melting of potassium bromide (a) Ions are formed (b) Ions are separated (c) Ions are made volatile (d) Molecules are separated 2. In the rock-salt structure of potassium bromide – (a) Br is in octahedral arrangement, K+ is in octahedral arrangement – (b) Br is in octahedral arrangement K+ is cubic arrangement – (c) Br is in cubic arrangement K+ is in octahedral arrangement – (d) Br is in cubic arrangement K + is in cubic arrangement. 3. Potassium bromide in its native state does not conduct electricity because (a) It has no electrons (b) It has no ions (c) Positive charge of cation is nullified by negative charge of anions (d) Ions does not move in the systematic lattice of the solid.

passage VIII Potassium crystallizes in a body-centred cubic lattice, with a unit cell length a = 5.20 Å 1. What is the distance between nearest neighbours? (a) 5.20 Å (b) 4.5 Å (c) 9.01 Å (d) 2.60 Å 2. What is the distance between next nearest neighbours? (a) 4.50 Å (b) 5.40 Å (c) 9.01 Å (d) 5.20 Å 3. How many nearest neighbours does each K atom have? (a) 8 (b) 6 (c) 4 (d) 12 4. How many next nearest neighbours does each K atom have? (a) 8 (b) 6 (c) 4 (d) 12

passage VI

passage Ix

A cubic unit cell contains manganese ions at the corners and fluoride ions at the centre of each edge. 1. What is the empirical formula? (a) MnF (b) MnF2 (c) MnF3 (d) MnF4 2. What is the co-ordination number of the Mn ion? (a) 4 (b) 6 (c) 8 (d) 12 3. What is the edge length of the cell if the radius of Mn ion is 0.65 Å and F-ion is 1.36 Å? (a) 2.01Å (b) 4.02Å (c) 5.68Å (d) 2.89Å 4. What is the density of the compound? (Mn = 55, F = 19) (a) 2.86 gm/cm3 (b) 5.72 gm/cm3 3 (c) 11.45 gm/cm (d) 8.59 gm/cm3

Titanium crystallizes in a face centred cubic lattice. It reacts with carbon or hydrogen interstitially, by allowing atoms of these elements to occupy holes in the host lattice. Hydrogen occupies tetrahedral holes, but carbon occupies octahedral holes. 1. Predict the formulae of titanium hydride and titanium carbide formed by saturating the titanium lattice with either ‘foreign’ element. (a) TiH2, TiC (b) TiH4, TiC (c) TiH4, TiL4 (4) TiH, TiL4 2. What is the maximum ratio of “foreign” atom radius to host atom radius that can be tolerated in a tetrahedral hole without causing a strain in the host lattice? (a) 0.225 (b) 0.155 (c) 0.414 (d) 0.732

Solid State

3. What is the maximum allowable radius ratio in an octahedral hole? (a) 0.155 (b) 0.225 (c) 0.414 (d) 0.732 4. Account for the fact that hydrogen occupies tetrahedral holes while carbon occupies octahedral holes. (a) rH = rC (b) rH < rC (c) rH > rC (d) Hydrogen is more reactive than carbon

(r)

(d)

(s)

2. Match the following Column I with Column II Column I (Crystal system)

passage x KCl crystallizes in the same type of lattice as does NaCl (Rock salt). Given that rNa+/rCl– = 0.5 and rNa+/rK+ = 0.7. 1. What is the ratio of the side of the cell for KCl to that for NaCl? (a) 1.143 (b) 2.57 (c) 2.4 (d) 1.2 2. What is the ratio of density of NaCl to that of KCl? (a) 1.49 (b) 1.17 (c) 1.90 (d) 1.143 passage xI BaTiO3 crystallizes in the perovskite structure. This structure may be described as a cubic lattice, with barium ions occupying the corners of the unit cell, oxide ions occupying the face centres and titanium ions occupying the centres of the unit cells. 1. If titanium is described as occupying holes in BaO lattice, what type of holes does it occupy? (a) Tetrahedral (b) Octahedral (c) Cubic (d) Triangular 2. What fraction of the holes of the type does it occupy? (a) 0.25 (b) 0.50 (c) 1.00 (d) 0.75

matching type questions 1. Match the following Column I with Column II Column I (Arrangement of the atoms/ions)

(c)

Column II (Planes in fcc lattice)

(a)

(p)

(b)

(q)

4.75

Column II (Edge length and angle)

(a) Monoclinic

(p) a≠b≠c

(b) Hexagonal (c) Rhombohedral (d) Triclinic

(q) a=b≠c (r) a=b=c (s) α ≠ β ≠ γ ≠ 90°

3. Match the following Column I with Column II Column I

Column II

(a) Rock salt structure

(p) co-ordination number of cation is 4

(b) Zinc blend structure

(q)

(c) Anti fluorite structure

(r) co-ordination number of cation and anion are same (s) Distance between two

(d) bcc Structure

3a +r+ + r– 2

nearest anions is

a 2

4. Match the following Column I with Column II Column I

Column II

(a) (b) (c) (d)

(p) (q) (r) (s)

C (solid) CH3OH Water (solid) HNO3

Covalent solid Molecular solid Hydrogen bonding Dipole-dipole interactions

5. Match the crystal system/unit cells mentioned in Column I with their characteristic features mentiioned in Column II Column I

Column II

(a) Zinc Blende

(p) Anion and Cation have similar co-ordination number (b) Rock Salt (q) Voids along one face plane are not occupied (c) Fluorite (r) All the voids along one of the body diagonal line are not occupied 4 × molar mass (d) Anti fluorite (s) d  gm3  =  cm 

volume of unit cell × Avogardo's number

4.76

Solid State

6. Match the following Column I with Column II Column I (a) NaCl (b) ZnS (c) Li2O (d) CsBr

10. Match the following Column I with Column II

Column II (p) Nearest no. of anions to a cation = 8 (q) Nearest no. of cations to an anion = 6 (r) Nearest no. of anions to an anion = 12 (s) Nearest no of cations to an anions = 8

7. Match the following Column I with Column II Column I (a) NaCl (b) CsCl (c) ZnS (d) CaF2

Column II +

–

(p) Body diagonal is 2r + 2r (q) Coordination no of cation is 8 (r) Alternate tetrahedrons are filled (s) No.of formula units per unit cell is 4 (t) Diamond follows the same structure

8. Match the following Column I with Column II Column I (Arrangement in unit cell, radius ratio in higher limit)

Column II (Coordination number of cation: Anion)

(a) Cations in CCP and anions (p) wwwin alternate unit tetrahedral voids (b) Cations in simple cubic and (q) anions in the body centre (c) Anions in CCP and cations (r) in all tetrahedral voids

Ratio of number of cation to anion in one cell is 1:1

Ratio of coordination number of cation to anion is 1:1 Ratio of number of cation to anion just touching each other is not 1 (d) Cations in CCP and anions (s) Number of next neighbours in all octahedral voids of ion is greater than 10

9. Match the following Column I with Column II Column I (a) Rock salt structure (b) Anti fluonite structure (c) Zinc belende structure (d) Corundum structure

Column II (p) % void volume 24 to 26 % (q) % of packing fraction ranging from 75 to 80 (r) Four anions and eight cations per unit cell (s) Six anions and four cations per unit cell

Column I (Packing in metallic crystals)

Column II

(a) (b) (c) (d)

(p) (q) (r) (s)

Simple cubic B.C.C. F.C.C. H.C.P

a = 2r Stacking sequence ABABA co-ordination number =12 Packing fraction = 0.74

11. Match the following Column I Crystal type

Column II Column III Cell Coordination Column IV constant number void space

(a) (b) (c) (d)

(p) (q) (r) (s)

Simple cubic bcc fcc or CCP hcp

1 2 4 6

(w) (x) (y) (z)

12 8 6 4

(K) (L) (M) (N)

26% 32% 48% 74%

12. Match the following Column I with Column II Column I (a) (b) (c) (d)

Triclinic Cubic Tetragonal Rhombohedral

Column II (p) (q) (r) (s)

one 4 fold axis a=b=c a≠b≠c α = β = γ = 90°

assertion (a) and reason (r) type questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion. b. If both assertion and reason are true but reason is not the correct explanation of assertion. c. If assertion is true but reason is false. d. If assertion is false but reason is true. 1. Assertion (A): In close packing of spheres, a tetrahedral void is surrounded by four spheres whereas an octahedral void is surrounded by six spheres. Reason (R): A tetrahedral void has a tetrahedral shape whereas an octahedral void has an octahedral shape. 2. Assertion (A): Due to Frenkel defect there is no effect on the density of the crystaline solid. Reason (R): In Frenkel defect no cation or anion leaves the crystal. 3. Assertion (A): 8 : 8 coordination of CsCl at low temperature changes to 6:6 coordination at high temperature.

Solid State

4.

5

6

7.

8.

Reason (R): Temperature influences the structure of solids. Assertion (A): Solids having more F-centres posses intense colors. Reason (R): ZnO exist yellow color at high temperature due to metal deficiency defect. Assertion (A): Space or crystal lattice differ in symmetry of the arrangement of points. Reason (R): nλ = 2d sinθ, is known as Bragg’s equation. Assertion (A): In any ionic solid (MX) with Schottky defects the number of positive and negative ions are same. Reason (R): Equal number of cation and anion vacancies are present. Assertion (A): Electrical conductivity of semiconductors increases with increasing temperature. Reason (R): With increase in temperature, large number of electrons from the valence bond can jump to the conduction bond. Assertion(A): Solids having more F-centres posses intense colors. Reason(R): ZnO exist yellow color at high temperature due to metal deficiency defect.

Integer type questions 1. PbS has NaCl structure. Its density is 12.7 gm/cm3. Molar mass if PbS is 239. Calculate inter ionic distance in pm units, answer devided by 50 and express the number. 2. Chromium (Aw = 52) crystallizes with bcc lattice. The length of the unit cell edge is found to be 290 pm. The value of density in gm/cm3 is__________. 3. Find ratio of body diagonal length to that of nearest distance between centres of tetrahedral and octahedral voids in fcc system. 4. In a FeO crystal lattice, 4% of cations sites are vacant and electroneutrality is maintained by oxidation of some Fe(II) into Fe(III). The number of Fe(III) ions present per hundred of O–2 ions are_________. 5. The density of ionic solid B+A–(Crystallizes in rock salt type) is 4 g/cc and the value of face diagonal of unit cell is 600 2 pm. If 2.96 of an ionic solid B+A– (salt of WB +S.A.) dissolved in water to make one liter solution, then its pH is (T = 298 K, Kb for BOH is 10–4, NA = 6×1023). 6. An ionic substance AB has NaCl like structure in which ions A occupy corners of the cubic unit cell. If all the face centred ions along one of axes are removed. The total no of ions A & B in one unit cell will be_________.

4.77

7. Density of lithium atom is 0.53 g/cm 3. The edge length of Li is 3.5A 0. The number of lithium atoms in a unit cell will be _____(Atomic mass of lithium is 6.94). 8. The concentration of cation vacancies if NaCl is doped with 10–3 mole percentage of SrCl2 is 6.023 ×10(x+y). If x is 10 they y is____. 9. MX is a crystalline ionic compound. X– Ions make the ccp structure and M+ ions occupy the octahedral voids. The total number of voids left unoccupied in 0.5 mol. If the compound is a×1023. What is 'a'? 10. A crystal is made up of atoms X, Y, Z. Atoms X are in fcc packing. Y occupies all octahydral voids and Z occupies all tetrahydral voids. If all the atoms along two body diagonals are removed. The ratio of sum of effective number of atoms of Y and Z to the effective number of atoms of X is a:1 The value of a is______. 11. No. if atoms present in fcc unit cell of cubic system is 12. A cubic is made up of two elements X and Y. X is present at corners and Y present at body centre. The Co. No. of X is________. 13. In ABC, ABC packing no. of tetrahedral voids present per unit cell is________. 14. How many number of effective Na+ ions, present in unit cell of Rock salt structure, is__________.

previous years’ IIt questions 1. In a solid ‘AB’ having the NaCl structure, ‘A’ atoms occupy the corners of the cubic unit cell. If all the face–centred atoms along one of the axes are removed, then the resultant stoichiometry of the solid is (2001S) (a) AB2 (b) A2B (c) A4B3 (d) A3B4 2. A substance AX BY crystallizes in a face centred cubic (fcc) lattice in which atoms ‘A’ occupy each corner of the cube and atoms ‘B’ occupy the centres of each face of the cube. Identify the correct composition of the substance AXBY (2002S) (a) AB3 (b) A4B3 (c) A3B (d) Composition cannot be specified. 3. In which of the following crystal alternate tetrahedral voids are occupied (2005) (a) NaCl (b) ZnS (c) CaF2 (d) Na2O

4.78

Solid State

4. The packing efficiency of two – dimensional square unit cell shown below is

L

(2010) (a) 39.27% (c) 74.05%

(b) 68.02% (d) 78.54%

5. Read the following statement and explanation and answer as per the options given below: Assertion: In any ionic solid [MX] with Schottky defects, the number of positive and negative ions are same. Reason: Equal number of cation and anion vacancies are present. (2001S) (a) If both assertion and reason are CORRECT, and reason is the CORRECT explanation of the assertion. (b) If both assertion and reason are CORRECT, but reason is NOT the CORRECT explanation of the assertion. (c) If assertion is CORRECT, but reason is INCORRECT. (d) If assertion is INCORRECT, but reason is CORRECT. 6. The correct statement (s) regarding defects in solids is (are). (a) Frenkel defect is usually favoured by a very small difference in the sizes of cation and anion. (b) Frenkel defect is a dislocation defect. (c) Trapping of an electron in the lattice leads to the formation of F-centre. (d) Schottky defects have no effect on the physical properties of solids. (2009) 7. Passage VI In hexagonal systems of crystals, a frequently encountered arrangement of atoms is described as a hexagonal prism. Here, the top and bottom of the cell are regular hexagons and three atoms sandwiched in between them.

A space-filling model of this structure, called hexagonal close-packed (hcp), is constituted of a sphere on a flat surface surrounded in the same plane by six identical spheres as closely as possible. Three spheres are then placed over the first layer so that they touch each other and represent the second layer. Each one of these three spheres touches three spheres of the bottom layer. Finally, the second layer is covered with a third layer that is identical to the bottom layer in relative position. Assume radius of every sphere to be ’r’ (i) The number of atoms on this hcp unit cell is (a) 4 (b) 6 (c) 12 (d) 17 (ii) The volume of this hcp unit cell is (a) 24 2r 3

(b) 16 2r 3

(c) 12 2r 3

(d)

64r 3 3 3

(iii) The empty space in this hcp unit cell is (a) 74% (b) 47.6% (c) 32% (d) 26% 8. Integer Questions Silver (atomic weight = 108 g mol–1) has a density of 10.5 g cm–3. The number of silver atoms on a surface of area 10–12 m2 can be expressed in scientific notation as y×10x. The value of x is: (2010) 9. Match the crystal system/unit cells mentioned in column 1 with their characteristic features mentioned in column II. Indicate your answer by darkening the appropriate bubbles of the 4×4 matrix given in the ORS. 2007 Column I (a) Simple cubic and facecentred cubic parametres (b) Cubic and rhombohedral (c) Cubic and tetragonal

Column II (p) have these cell (q) are two crystal systems

(r) have only two crystallographic angles of 90° (d) Hexagonal and monoclinic (s) belong to same crystal system

Solid State

4.79

ansWer keys multiple choice questions with only one answer level I 1. d 2. a 3. b 4. b 5. c 6. b 7. b 8. c 9. b 10. b 11. d 12. c 13. a 14. c 15. a 16. a 17. b

18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

c c a a a d a c a b c c a c a c a

35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51.

b d d a d a b b d b b b a a d a a

52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68.

b c b c a b c a b d d b a a b c d

69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83.

d b d b d c d b d c a b a b b

25. 26. 27. 28. 29. 30. 31. 32. 33.

a, b, c, d a, b c, d a, c b, c b, c c, d a, b, c a, b, c

34. 35. 36. 37. 38. 39. 40. 41. 42.

a, c, d a, c a, b, c b, d a, b, c a, b, c a, c, d a, b, c a, b, c

43. 44. 45. 46. 47. 48. 49. 50.

a, b b, c a, c b, d a, b, c a, b, c, d b, c a, b, d

comprehensive type questions passage I 1. a

2. a

3. a

2. d

3. c

4. c

5. a

2. b

3. d

4. b

5. c

2. d

3. a

2. a

3. d

2. b

3. b

4. a

2. b

3. a

4. b

2. a

3. c

4. b

passage II 1. b passage III

multiple choice questions with only one answer level II 1. c 2. a 3. a 4. c 5. a 6. c 7. a 8.d 9. a 10. d 11. b

12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

b b c b c a d c c b c

23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

b a b c b a c a d a d

34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44.

c d b c b b a d c d a

45. 46. 47. 48. 49. 50. 51. 52. 53. 54.

c c a a b a a b c b

1. a passage IV 1. d passage V 1. b passage VI 1. c passage VII

multiple choice questions with one or more than one answer 1. 2. 3. 4. 5. 6. 7. 8.

b, c a, c, d a, b, c b, c, d a, c, d a, b, d b, d a, c, d

9. 10. 11. 12. 13. 14. 15. 16.

a, b, c, d c, d a, b a, b, c, d a, b, c, d a, b, c a, b a, b

17. 18. 19. 20. 21. 22. 23. 24.

a, b, c, d c, d a, c a, b, c a, b, c, d c, d a, b, c a, b, d

1. b

2. c

passage VIII 1. b passage Ix 1. a

4.80

Solid State

assertion (a) and reason (r) type questions

passage x 1. a

2. b

1. c 2. a 3. a

passage xI 1. b

2. a

(a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

s p rs p psq qr s pqs q p py, m r.

7. a 8. c

Integer type questions

matching type questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

4. c 5. b 6. a

(b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b)

r q prs rs prs r pq pq pqr q qxl qs

(c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

p r ps qrs qs rs rst rs pq rs rwk ps

(d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d)

q ps qr rs qs ps qs pqs qs pqrs swk q

1. 5 2. 7 3. 4

4. 8 5. 6 6. 7

7. 2 8. 8 9. 6

10. 2 11. 4 12. 8

previous years’ IIt questions 1. d 3. b 2. a 4. d 7. Passage (i) b (ii) a (iii) d 8. 7 9. (a) ps (b) pq

5. a 6. b, c

(c) q

(d) qr

13. 8 14. 4

Solid State

4.81

hInts and solutIons hints to problems for practice

2 ×N 197 Unit cell contain = 4 atom ( fcc unit cell contain four atoms) 2 N × No of unit cell = 197 4 =

1. Edge length a = 286 pm For body centred cubic, radius of atom r=

3 ×a 4

3 1.732 × 286 = × 286 = = 123.8 pm 4 4 2. For face centred unit cell radius of atom r= or ∴

a 2 2

= 1.52×1021 7. (a) In cubic close packed structure, the face diagonal of the unit cell is equal to four times of the atomic radius Face diagonal = 4×r = 4 × 125 pm = 50 pm 2 × edge length

But face diagonal =

a = 2 2×r r = 0.144nm a = 2×1.414×0.144 = 0.407nm

∴Edge length = =

3. Edge length = 620 pm

Face diagonal 2 500 2

= 354 pm

(b) Volume of unit cell = a3 = (3.54×10–8cm)3

a = 4r Radius of xenon = 219.2pm Nearest neighbours distance = 2r = 2×219.2 = 438.4 pm

No. of unit cells in 1.00 cm3=

4. For face centred cubic unit cell 2 a = 4r a = 2 2×r = 622.16 pm 5. In a bcc unit cells there are eight atoms at corners of the cube and one atom at the body centre

8. Length of edge = 200 pm = 200×10–12 m. = 200×10–10 cm. Volume of unit cell = (200×10–10 cm)3 = 8×10–24 cm3 Mass of unit cell = No. of atoms in unit cell × Mass of each atom Since the element has fcc arrangement, the number of atoms per unit cell = 4 200 24 × 1023

 1 ∴ Number of atoms per unit cell =  8 ×  + (1× 1) = 2  8

Mass of an atom =

4 Number of atoms in 4.0 g of potassium = × 6.02 × 1023 39

∴ Mass of unit cell =

Number of unit cells in 4.0 g of potassium =

Density of unit cell =

4 6.02 × 1023 × 39 2

=

=3.09×1022 6. No of gold atoms =

wt ×N At Wt

1.00 =2.26×1022 (3.54 × 10−8 )3

9. Density =

200 × 4 = 33.3 × 10−23 g 24 × 1023 Mass of unit cell Volume of unit cell 33.3 × 10−23 = 41.6 cm −3 8 × 10−24 cm3

Mass of unit cell Density of unit cell

4.82

Solid State

Mass of unit cell = Mass of an atom × No. of atoms per unit cell

No. of atoms per unit cell in fcc arrangement = 4

Atomic mass Mass of atom = Avogadro number 50 = 6.023 × 1023

Mass of unit Cell =

Density =

No. of atoms per unit cell = 2(bcc)

A ×4 6.023 × 10−23

Mass of unit cell Volume of unit cell

2 × 50 Mass of unit cell = 6.023 × 1023

8.92 =

A×4 6.023 × 1023 (3.608 × 10−8 )3

2 × 50 6.023 × 1023 × V

Or A =

8.92 × 6.023 × 1023 × (3.608 × 10−8 )3 4

5.96 =

V=

2 × 50 6.023 × 1023 × 5.96

= 27.8×10–24cm3 10. In a face-centred cubic cell radius2=×

2a a = 4

∴ The closest distance between two atoms = diametre = 2× =

4.07 2

= 63.1 g mol–1 Atomic mass of copper = 63.1 12. Edge length of the unit cell = 288 pm = 288×10–10 cm Volume of the unit cell = (288×10–10)3 cm3 Mass of element = 208 g Let N is the number of atoms present in 208 g of element Mass of an atom =

2a a = 4 2

Since the structure is bcc, the number of atoms present in the a unit cell = 2

A° = 2.878A°

Mass of unit cell = 2 ×

Number of atoms in a face-centred unit cell 1 1 = (8 × ) (6 × )=4 8 2

Density. =

Mass of 4 atoms per unit cell = 4 × 197 amu = 4×197× (1.66 × 10–24) g = 1.308 ×10–21 g Volume of the unit cell = a3 = (4.07×10–8)3 cc

7.2 =

∴ density of gold =

1.308 × 10 −21 = 19.40 g / cc. (4.07 × 10 −8 )3

11. Edge length of the unit cell = 3.608×10–8 cm Volume of the unit cell = (3.608×10–8)3cm3 Density of the unit cell = 8.92 g cm–3 Mass of an atom =

=

208 N

A 6.023 × 10−23

Atomic mass Avogadro number

N=

208 N

Mass of unit cell Volume of unit cell

2 × 208 N(288 × 10−10 )3

2 × 208 7.2(288 × 10 −10 )3

= 2.42×1024 13. Let length of each edge = a cm Volume of unit cell = a3cm3 Density = 7.2 g cm–3 Mass of each atom =

200 = 48.54 × 10−24 g 4.12 × 1024

No. of atoms per unit cell = 4 (fcc) Mass of unit cell = 4×48.54 × 10–24 = 194.16×10–24 g ∴ Density =

Mass Volume

Solid State

7.2 = 3 or a =

194.16 × 10−24 a3

194.16 × 10−24 = 26.97 × 10−24 cm3 7.2

Edge length =

512 2

Mass of unit cell =

3a = 4r 3 a = 124.1 pm 4 bcc unit cell contain 2 atoms in a unit cell Radius of atom =

∴d = d=

2 × 52 = 7.3 g cm −3 6.023 × 1023 × (287 × 10−10 )3

ZM No a 3

10.5 =

Density =

4M 6.023 × 10 × (4.077 × 10−8 )3

4 × 63.5 g = 4.22 × 10–22 g 6.02 × 1023

4.22 × 10−22 g Mass of unit cell = 47.4 × 10−24 cm 2 Volume of unit cell

= 8.9 g cm–3 19. density d =

10.5 = 23

= 362 pm = 362 × 10–24 cm

Volume of the unit cell = (362 × 10–10) = 47.7×10–24 cm3 In fcc unit cell there are 4 atoms per unit cell

or a = 2.999×10–8cm = 299.9 pm 14.

4.83

ZM No a 3

Z(108) 6.023 × 10 × (4.09 × 10 −8 )3 23

Z=4 So the unit cell is face centre cubic

M = 107 16. d =

20. Unit cell length of fcc = 3.5 A°= 3.5×10–8 cm Unit cell length of bcc = 3.0A°= 3.0×10–8 cm

ZM No a 3

2.7 × 103 =

z × 2.7 × 10−2 (6.023 × 1023 ) (4.05 × 10−10 )3

density of fcc =

Mz1 No × (a1 )3

density of bcc =

Mz 2 No × (a 2 )3

Z=4 So the unit cell face centred cubic ∴

d fcc z1 (a 2 )3 = × d bcc z 2 (a1 )3

=

4 × (3.0 × 10 −18 )3 (Z1 = 4 (fcc); Z2 = 2(bcc) 2 × (3.5 × 10 −8 )3

17. Distance between nearest silver atom = 287 pm 2 a = 2x = a d=

= 2 x 405.8 pm ZM No a 3

4 × 108 (6.023 × 1023 )(4.058 × 10−8 )3

= 1.259 21. Density of argon d = 1.65 g ml–1 Radius = 1.54×10–8cm Volume of one atom of argon = (4/3) πr3 No. of atoms in 1.65 g or one mL =

= 10.79 g/cc.

1.65 ×6.02 ×1023 40

Total volume of all atoms of argon in solid state 18. In face centred cubic arrangement face diagonal is four times the radius of atom. Face diagonal = 4×128 = 512 pm Face diagonal =

2 × edge length

= (4/3) πr3×

1.65 ×6.02 ×1023 40

= (4/3)× 3.14 ×(1.54×10–8)3× = 0.38 cm3

1.65 ×6.02 ×1023 40

4.84

Solid State

Volume of solid argon = 1 cm3 ∴% of empty space =

(1 − 0.38) ×100 = 62% 1

4r = 2 × 5.14 2 × 5.14 = 1.82A 0 4

r=

22. (i) In bcc 1 Z= ( × 8) +1 = 2 8

Body diagonal =

3 × a = 4r

4 4  3a  3 Volume of atoms = 2 × πr 3 = 2 × π   = 0.68a 3 3  4  Volume of unit cell = a3 Packing factor = 0.68 (a3/ a3) = 0.68

26. In fcc structure there are two atoms per unit cell. Let M be the gram molecular mass of the element Mass per unit cell = No of atoms in unit cell × Mass per atom = 2×

M 6.023 × 1023

Volume of the unit cell = (290 ×10–10cm)3 = 24.4 × 10–24 cm3

(ii) In a simple cubic structure, atoms per unit cell Density =

1 = ×8 = 1 8 r=

Density given = 6.8 cm–3

a 2

6.8 = 3

Volume of atom V =

4 3 4 a πr = π   = 0.524 a3 3 2 3

Volume of unit cell = a3 Packing factor = 0.524 (a3 / a3) = 0.524 23. Body diagonal

Mass of unit cell Volume of unit cell

3 a = 712.16

2× M 6.023 × 1023 × 24.4 × 10−24

6.8 × 6.023 × 1023 × 24.4 × 10−24 = 50 g 2 Thus gram molar mass of the element is 50 g ∴50 g of the element contain 6.023 ×1023 atoms M=

200 g of the element would contain

2rc+ 2ra = 712.16 2rc + 2(181) = 712.16 rc= 175.38 pm

=

6.023 × 1023 × 200 = 24.09×1023 atom 50

27. No of atoms per bcc unit cell is 2. 24. In CsCl structure the distance of closest approach between cation and anion is equal to half the body diagonal of the cube Body diagonal = =

3 × edge length

3 × 4.04 A 0 = 7.0A 0

1 The distance of closest approach =× of the body 2 diagonal 1 =× ×7.0A0 = 3.5A0 2 25. In NaCl structure, anions have fcc arrangement. In such an unit cell face diagonal is four times the radius of anion Face diagonal = 2 a = 2 × 5.14 A 0 Face diagonal = 4r

d=

ZM No × a 3

4=

2 × 168.5 6.023 × 1023 × a 3

∴a3 = 139×10–24 cm3 Edge length a = 512 pm

28. d =

ZM No × a 3

2.165 =

4 × 58.5 6.023 × 1023 × a 3

a3 = 179 × 10–24 cm3 a = 562 pm 2(rc + ra) = edge length rc + ra = 281 pm.

Solid State

29. To calculate the Avogadro constant, i.e., the no. of NaCl formula units per mole, let us take 1 mole of NaCl

33. Since there is one octahedral hole per anion (O2–) and only 2/3 of these holes are occupied, the ratio of Al and O should be

mass of 1 mole Volume of 1 mole of NaCl = density 58.5 × 10−3 = 2.7 × 10−5 m3 3 2 . 17 10 × =

2 : 1 , i.e., 2:3 3

Thus the formula is Al2O3

34. Density =

Volume of 1 unit cell = a3= (0.564× 10–9)3 = 1.79×10–28 m3 ∴ Number of unit cells per mole =

ZM ZM = NV N(a 3 )

For a bcc lattice Z = 2 and given that a =3.50×10–10 m and M =7× 10–3 kg mol–1

2.7 × 10 −5 = 1.51 × 1023 1.79 × 10 −28

dCal =

Since 1 unit cell of NaCl has 4 NaCl formula units Avogadro’s constant (formula unit per mole) = 4 × (1.51×1023) =6.04 ×1023

2 × (7 × 10−3 ) 6.023 × 1023 × (3.5 × 10−10 )3

= 5.42 × 102 kg m–3 ∴ Percentage occupancy =

30. Mol wt of LiCl = 25.937 g (LiF = 6.939 + 18.998) wt of 1 mole 25.937 = Vol of 1 mole = wt per cc (density) 2.65 = 9.78 cc. Since this volume is supposed to be of a cube. The length of each edge of the cube = 3 9.78 = 2.138 cm. 2.138 No. of ions present in one edge = 2.01× 10−8 = 1.063×108 No. of ions (Li+ and F–) in the cube = (1.063×108)3 = 1.201 ×1024 No of LiF molecules per mole, i.e., Avogadro’s constant =

=

1. × 168.4 6.023 × 1023 × (4.12 × 10−8 )3

= 3.99 g cm–3 Density of aluminium =

4. × 27 6.023 × 1023 × (4.05 × 10−8 )3

= 2.69 g cm–3 So calcium chloride has larger density.

Thus the formula is ZnS.

d cal

×100

5.3 × 102 × 100 5.42 × 102

35. One-eighth of each corner atom (Au) and one-half of each face centred atom (Cu) are contained within the unit cell of the compound 1 Thus number of Au atoms per unit cell =8 × = 1 8 1 and no. of Cu atoms per unit cell = 6 × = 3 2 The formula of the compound is AuCl3. 36. The coordination number and the charges of ions always balance out to give charge neutrality. The coordination number of Ba2+ indicates that each Ba2+ ion is surrounded by 8F– ions. In order to balance 8 negative charges of F– ions 4 Ba2+ ions are required. Hence Coordination number of F– is 4.

37.

rc 99 = = 0.7 ra 170 Ca2+ ion occupies the octahedral void. Coordination number is 4.

32. Since there are two tetrahedral holes per each anion (S2–ion) and one- half of these holes are occupied by zinc ions there must be 1 × × 2 =1 zinc ion per S2– ion 2

d expt

= 97.78%

1.201× 1024 = 6.01× 1023 2

31. Density of CsCl =

4.85

38.

rc 97 = = 0.386 ra 251 ∴ Coordination number = 4

4.86

Solid State

39. rc = 0.414 ra 100 ∴ra = = 241.5 pm 0.414

a

46. d =

(h

d111 = 40. d =

ZM Noa 3

2.48 =

4 × 58 6.023 × 1023 × a 3

0.556

= 0.321 nm

2

(1 + 12 + 12 ) 0.556

2 + 22 + 22 2

= 0.161 nm

The separation of 222 planes is half as that of the 111 planes. 47. nλ= 2d sin θ

537 = 268.5 pm 2

41. P atoms are at corners So no. of P atoms = 8 ×

1 =1 8

No. of Q atom = 1 1 No. of R atoms =× ×12 = 3 4

Given n= 1; λ= 2.29×10–10; θ=

42. Corners of edges means corners of the unit cell. 1 =1 8

1 6 3 43. No. of atoms of A = 6 × = = 8 8 4 1 No. of atoms of B = 6 × = 3 2

∴1 × 2.29 ×10–10 = 2 × d × sin 13°34' d = 4.88 ×10–10 m

48. nλ = 2d sin θ

Given n =1, d =0.2 × 10–9m, θ = λ=

16.80 = 8.40° 20

2d sin θ 2 × 0.2 × 10−9 × sin 8.4° = n 1

49. nλ = 2d sinθ 1 ×(1.54 ×10–8) = 2 × d × sin 11.29 d = 3.86 × 10–8cm Further d =

a

(h

2

3.86 × 10−8 =

or

+ k 2 + l2 ) a 2

1 + 02 + 02

a = 3.68×10–8 cm

Formula = A3/4B3≈AB4 1 44. No. of anions (Y) =8 × =1 8 No. of cations (X) = 3 ×

27°8′ = 13°34′ 2

= 5.84 × 10–11 m

So the formula = PQR3

So No. of atoms = 8 ×

+ k 2 + l2 )

and d222 =

a3 = 155 × 10–24 cm3 Edge length = 537 pm a = 2 (rc + ra) rc + ra=

2

1 3 = 2 2

Formula = X3/2Y ≈ X3Y2 45. No. of atoms of X = 4 No. of atoms of Y = 4 + 4 = 8 Formula = X4Y8 ≈ XY2

50. 2d sin θ = nλ ; n = 2; λ= 1.54 A°= 1.54×10–8 cm a=

nλ = 2 × 1.54 × 10–8 2sin θ

Now d=

197 × 2 Mz = = 19.42 g cm −3 3 23 6.02 × 10 × (3.23 × 10 −8 )3 Noa

Solid State

51. 2d sinθ = nλ 2a sin θ 2 × 2.64 × 10 −10 × sin 7.75 = n 1 = 0.712 ×10–10 m λ=

E=

hc 6.62 × 10 −34 × 3 × 108 = = 27.89 × 10 −16 J λ 0.172 × 10 −10

55. Edge length = 0.5628 nm Observed density = 2.165×103 kg m–3 Molecular weight = 58.44 ×10–3 kg ∴ Volume of unit cell = (0.5628×10–9m)3 There are 4 Na+ and 4 Cl– ions per unit cell ∴ Mass per unit cell =

52. It is clear from the valency of Fe (II) and Fe (III) and three Fe (II) ions will be replaced by 2 Fe (III) ion causing a loss of one iron ion. Total loss of iron from one mole of FeO = 1 – 0.93 =0.07 ∴Total Fe (III) present in one mole of FeO =2 ×0.07 = 0.14 Total number of Fe (II) and Fe (III) present in one molecule of FeO = 0.93 ∴% of Fe (III) =

0.14 × 100 = 15.05% 0.93

ln or log

=

58.44 × 10−3 × 4 = 3.883 × 10−25 kg 23 (6.023 × 10 )

Density (Theoritical) = 3.883 × 10 −25 kg = 2.178 × 103 kg −3 m (0.5628 × 10 −9 m)3 Fraction of sites unoccupied = (2.178 − 2.165) × 103 = 5.96 × 10−3 2.178 × 103

multiple choice questions with only one answer level I 1. Weiss Indices = (1, 1/2, 3/2) Miller Indices = (2 : 1 : 3)

n = e–(Ev/ KT) 53. N Ev n =− N KT

13. In final stage radius of red balloon is 0.414 a Radius reduced by = 6/0.414 =14.5

n Ev =− N 2.303 KT

14. Cu atoms: 6 × 1/2 + 8 × 1/8 = 4 Silver atom = 12 × 1/4 = 3 Gold atom = 1 So Cu4 Ag3 Au

1.602 × 10−19 8.314 × 1023 ) × 298 2.303 × ( 6.022

Solving for

15. bcc contains 2 atoms So atoms: 12.08 × 1023 × 2

n = 1.24× 10–17. N

16.

54. From the given formula it is clear that for every 100 oxide ions there are only 98 nickel ions. Suppose out of 98 nickel ions x exist as Ni2+ and the remaining (98-x) exist as Ni3+. Total positive charge on 98 nickel ions = x×2 +(98-x)3 Total negative charge on 100 oxide ions = 100× 2 Due to electrical neutrality. x ×2 + (98-x)3 = 100×2 2x + 294–3x = 200 x = 94 ∴ Fraction of nickel as Ni2+ =

94 × 100 = 96% 98

Fraction of nickel as Ni3+ = (100-96)% = 4%

4.87

2 a = 4r

17. Rank means no. of atoms 19.

2 a = 2x

20. Pf = 4/3 π(a/2)3/a3 24. d = ZM/Noa3 10.5 =

Z(108) 6.023 × 10 × (40.9 × 10 −12 )3 23

3

4  3a  25. Pf = 2 π  / a3  3  4 

4.88

Solid State

26. In α form ∴a =

multiple choice questions with only one answer

3a = 2 x

2x 3

In γ form

2a = 2 x

a = 2x dγ 4M = dα No 2 x

(

)

3

2M  2x  No    3

3

= 1.089 34. 0.225 r / 0.414 r = 0.547 39.

level I 4. Assume No. of Fe+3 ion = x No. of Fe+2 ion are = 0.93x 32+(0.93–x) 2 = 2 n = 0.14 0.14 × 100 percentage of Fe+3 = 0.93 6. Sr+2 replaces 2Na+ ions and occupy one position. So No. of vacancies =10–5 per mole ∴ 10–5 × 6.023×1023 = 6.023×1018 7.

2 a = 4r

41. rc/ra = 0.675

4M d1 No(3.6)3 = 2M d2 No(3)3

10.

44. rc/ra = 0.583 45. a = 2 rc +2 ra 47. rc = 0.414 ra 50.

3 a = 2(rc + rc )

54. Formula = PQ 55. No. of oxide ion = 4 No. of cation A = 8/6 No. of cation B = 4/3 Formula = A8/6 B4/3 O4 56. No. of oxide ion = 4 X present in one octahedral of void Y present in one octahedral, one tetrahedral void 57. Compound = H2O 63. X8/5 Y2 O4 ∴ X8 Y10 O2O = X4 Y5 O10 73. No. of moles = 1/58.5 1 No. of NaCl units = ×N 58.5 = 1022 No. of unit cells = 1022/4 Volume = 0.12 mL

11. 2a = 4r 14. Two Fe+3 Ions replaces 3 Fe+2 ions and occupy two sites. One position is vacant Fe+3/Fe+2 =15/100 No. of vacancies=7.5/122.5 15. The largest – atom that can fit in the void is 0.414 r 16. Packing is fcc. If one body diagonal is removed then Giant Buckies one 3+3/4 Motu are = 3 Chotu are = 6 Corners are replaced by C–30 No. of C–300 = 1/4 Total no of atoms 3(240) + 3/4 (240) + 1/4 (300) + 3×44 + 6 × 32 4 × 24 × 10 −3 17. dthe = 6.023 × 1023 × (4 × 10 −10 )3 = 2.49 × 103 kg/m3 Percentage occupancy = 2.40×103/2.49×103 = 96% 18. rc = 0.414ra 20.

3a = 2 x

22. No. of unit cells = 58.5 6.023 × 1023 × = 1.5 × 1023 58.5 4

Solid State

27. b = 4V No b So V = 4 No

passage VI d H2

Density = mass/volume d N 2 2 0.0399 d H2 = × 28 0.0266

2 b H2 = 28 b N2

28. d = ZM/Noa3 z = 2 (bcc) 3a = 4r 30. Assume ra- = x rNa+ = 0.53x rK+ = 0.743x rKCl 0.743x + x = = 1.122 rNaCl 0.55x + x 48. No. of A ions removed is = 1 No. of B ions removed is = 1 So formula is AB 49. No. of X atom = 7/8 No. of Y atom = 3 No. of Z atom = 1/8 54.

3. Edge length a = 2rc+ 2ra 4. d =

ZM Noa 3

(1)(112) 6.023 × 1023 × (4.02 × 10 −8 )3 = 2.86 gm/ cc passage.VII: 1. d =

1(168) =4 (6.023 × 1023 ) × (a 3 )

a3= 69.7×10–24cm3 a = 4.12 A0 2. 3a = 2 ( rc + ra ) rc+ ra = 3.56A0 passage VIII 3a = 2 x x = 4.5 A0 2. Next nearest neighbors means edge length

1.

passage x 1. Assume rCl– =x

comprehensive type questions passage I 1. d =

3M ZM = Noa 3 6 × 1023 × 25 × 10 −24

2. mole = 180/30 = 6 No. of atom = 6×6×1023 3.6 × 1024 3. No. of unit cell = 3

rNa + x

= 0.5, rNa+ = 0.5x

rK+ = 0.5x/0.7 = 0.714x a KCl 2(0.714 x + x ) = = 1.142 a NaCl 2(0.5x + x ) 4 × 58.5 d NaCl N(1.5x )3 2. = 1.17 = 4 × 74.5 d KCl N(1.714 x )3

passage III

Integer type questions 1.

3 a = 2x 1. d =

2. Next nearest means edge length passage IV 1. It is fcc

ZM Noa 3

12.7 =

(4)(239) (6.023 × 1023 )(a 3 )

Inter ionic distance =

a 2

4.89

4.90

Solid State

ZM 2. d = Noa 3 5.

2 a = 600 2 a = 600 pm ZM d= Noa 3 MW = 129.6 [M] = 0.0228 M pH = 1/2(pKw–pKb–log C)

8. Sr+2 replaces 2Na+ ions and occupies one site. One site is vaccant. So No. of Sr+2 Ions = 6.023×1023×10–5 No. of vaccant sites = 6.023×1023 9. Tetrahedral voids are vaccant So No. of vacant tetrahydral voids = (0.5) × 6.023×1023×2 = 6 ×1023 10. No. of X atom = 3+4×1/8 = 3+ 1/2 = 7/2 No. of Y atom = 12×1/4 = 3 No. of Z atom = 4 Y+Z: X 7 : 7/2 2:1

CHAPTER

5 Solutions

S 5.1 inTroducTion A true solution may be defined as a ‘molecularly’ homogeneous mixture of two or more substances (a colloidal solution which shows heterogeneity is not a true solution). This means that the molecules of the two components (solvent and solute) are distributed uniformly throughout the whole of the solution. Another way of putting it is to say that a solution is a one–phase system, since the solution, being a molecular mixture, has no surfaces separating any one part from another. If, however the solution is in equilibrium with second phase (e.g., excess of solid or a gas) then the system would be heterogeneous. The substances making up the solution are called components of the solution. Depending upon the total number of components present in the solution, it is called the binary solution (two components), ternary solution (three components), quaternary solution (four components). In this unit only binary solutions are considered. The component which is present in excess in a binary solution is called solvent while the other component is called solute. In other

olutions fill oceans, solutions are running in our veins….. Savante Arrhenius

words, a solute is a substance that dissolves and a solvent is a substance in which dissolution takes place. For example, if sodium chloride is dissolved in water, sodium chloride is solute and water is solvent.

5.2 Types of soluTions Depending upon the physical states of the solute and solvent, the solutions can be classified into the following types. Among the different types of solutions given in Table 5.1, the liquid solutions, i.e., solid in liquid, liquid in liquid and gas in liquid are very common. In all these three types of solutions, the solvent is liquid. The solution in which water is the solvent are called aqueous solutions, while those in which water is not the solvent are called nonaqueous solutions. Some common non-aqueous solvents are ether, benzene, carbon tetrachloride etc. In the process of forming a solution, the two components may mix with each other to different extent. These

Table 5.1 Types of solutions Type of Solution

Solute

Solvent

Examples

1)

Gaseous solutions

2)

Liquid solutions

3)

Solid solutions

Gas Liquid Solid Gas Liquid Solid Gas Liquid Solid

Gas Gas Gas Liquid Liquid Liquid Solid Solid Solid

Mixture of N2 and O2 gases, air Chloroform vapours mixed with nitrogen gas, water vapor in air Camphor vapours in nitrogen gas, dust and smoke particles in air Oxygen dissolved in water, soda water Ethanol dissolved in water Sugar or salt in water Solution of hydrogen in palladium (occlusion) Amalgams of mercury with metals Copper dissolved in gold (All alloys)

5.2

Solutions

may be even completely miscible or completely immiscible. The extent to which one material (solute) will dissolve in another (solvent) is governed by the following factors: (a) The attractive forces between the solute and solvent molecules compared with the attractions that the solute (or solvent) molecules have for each other. (b) Solvation of the solute, i.e., the attachment of molecules of the solvent to solute molecules or ions, either by electrical attraction or chemical bonding. The effect of unlike inter molecular forces can be approached by considering whether, say pentane (a liquid hydrocarbon, with non-polar molecules and therefore small attractive forces) will mix with water, where the molecules are polar and the intermolecular forces are also increased by the formation of hydrogen bonds. It will be difficult in the first place, for pentane molecules to force their way between the strongly attracted water molecules. It will also be difficult for the water molecules to pass into the pentane, since they cannot easily break away from the adjacent water molecules. Very little mixing will take place – in other words, neither liquid is more than very slightly soluble in the other. This is an extreme case of differing intermolecular forces, and where there is less difference, the mutual solubility will correspondingly increase. To take the other extreme, if two liquids have almost identical attractive forces, there will be nothing to stop the molecules of the liquids from mixing completely. It is more difficult for a solid to dissolve in a liquid than for one liquid to dissolve in (i.e., mix with) another liquid. This is due to the much stronger forces holding the molecules together in the solid, which have to overcome before the molecules can pass into the liquid phase and mix with the molecules of the solvent. High solubility therefore requires strong attractive forces between solvent and solute molecules. The strongest forces of this kind are those between ions and polar solvent molecules, where the electrical attraction is so strong that the ions, when in solution, will have molecules of solvent attached to them: this is known as solvation. Since electrical forces are inversely proportional to the square of the distance, a small ion will be more solvated than a large one because the solvent molecules can approach closer to the charge. The water molecule is highly polar and so water is an excellent solvent for ionic substances; the solvation which takes place in this case being referred to as hydration. Fig 5.1 shows how the ions of sodium chloride hydrate in solution; the number of water molecules attached to an ion is not a fixed quantity, but the illustration shows the average number.

Water dipoles Ionic crystal

-

+

-

+

+

-

+

-

-

+

-

+

+

-

+

-

Hydration of ionic crystal

-

+

Hydrated ions fig 5.1 Hydration of ions

If both solvent and solute consist of polar molecules, there is also strong electrical attraction, in this case between the two dipoles. This is likely to give good solubility, and solvation can again occur. Solvation of molecules is often increased by the formation of hydrogen bonds. It follows that polar solvents (and particularly water) are good solvents for polar molecules and for ionic substances, whereas non-polar solvents will dissolve non polar molecules, but will be poor solvents for polar molecules or ions. This may be expressed shortly as like dissolves like.

5.3 MeThods for expressing The concenTraTion of a soluTion The composition of a solution can be described by expressing its concentration. The latter can be expressed either qualitatively or quantitatively. In qualitative manner, we can say a solution may be dilute (i.e., relatively very small quantity of solute). Quantitatively the concentration of a solution may be defined as the amount of sol ute present in the given quantity by of the solution. Any solution of which the strength is known is a standard

Solutions

solution. If the solubility of a substance is known, its saturated solution is thus a standard solution. Other standard solutions are more frequently used. The concentration is usually expressed in any one of the following ways: 1. Mass percentage: The mass percentage of a component in a given solution is the mass of the component per 100 g of the solution Mass percent of a component Mass of the component in the solution = × 100 Total mass of the solution If WB be the mass of solute (B) and WA be the mass of solvent (A), then WB Mass percentage of B = ×100 WA + WB For example, a 5% (aqueous) solution of sugar by mass means that 100 g of solution contain 5 g of sugar. Mass percentage can also be expressed as mass percent of solute

=

Mass of solute Volume of solution × Density of solution

Mass percentage of solute = mass fraction × 100 solved problem 1 If 11 g of oxalic acid are dissolved in 500 mL of solution (density =1.1 g mL–1), what is the mass % of oxalic acid in solution? Solution: 500 mL of solution contain 11 g of oxalic acid Density of solution = 1.1 g mL–1 Mass of solution = (500 mL) × (1.1 g mL–1) = 550 g Mass of oxalic acid = 11 g 11 Mass % of oxalic acid = × 100 = 2% 550

5.3

Volume % of a component =

Volume of the component × 100 Total volume of solution

3. Percent mass by volume (W/V): It is the mass of solute dissolved in 100 mL of the solution. Percent of solute mass by volume PPM =

Mass of solute ×10 × 100 Mass of solution

This method is generally used in medicines and pharmacy. It is mass by volume (W/V) percentage. 4. Parts per million (ppm): When a solute is present in very minute amounts (trace amounts) the concentration is expressed in parts per million. It is the parts of a component per million parts of the solution. PPM =

Mass of solute ×106 Mass of solution

Atmospheric pollution in cities due to harmful gases is generally expressed in ppm though in this case the values refer to volumes rather than masses. For example, the concentration of SO2 in Delhi has been found to be as high as 10 ppm. This means that 10 cm3 of SO2 is present in 106 cm3(or 103 L) of air. The hardness of water, the pollutants in water is often expressed in terms of Mg mL–1 or ppm. solved problem 2 Sea water contains 5.8 × 10–3 g of dissolved oxygen per kilogram Express the concentration of oxygen in sea water per million. Solution: Mass of solute Parts per million ( ppm ) = × 106 Mass of solution =−

5.8 × 10 −3 × 106 = 5.8 ppm 103

problem for practice

problem for practice

1. Calculate percentage composition in terms of a solution obtained by mixing 300 g 25% solution of NH4NO3 with 400 g of a 40% solution of solute

2. In an industrial town the pollution of SO2 in air has been reported as 10 ppm by volume. Calculate the volume of SO2 per litre of air.

2. Volume percentage (V/V) In the case of a solution of liquid in liquid, the concentration can be expressed conveniently in volume percentage. It is defined as the volume of the component per 100 parts by volume of the solution.

5. Mole fraction: It is generally denoted by c and subscript used on the right hand side of c denotes the component.

5.4

Solutions

It is the ratio between the number of moles of a component of the solution to the total number of moles of all components of the solution. Let ‘A’ is the solute and ‘B’ is the solvent in a binary solution, the number of moles of A be ns and that of B be no in the solution. Total number of moles in the solution= (ns+no) The mole fraction of solute A is given by no. of moles of A c= total no. of moles in the solution ns nA = = ns + no n A + n B Similarly, mole fraction of the solvent B in a solution no nB χB = = ns + no n A + n B The sum of the mole fractions of all the components is equal to unity. ∴ cA + cB = 1

G / wA (a / w B ) + G / w A

χB =

a / wB (a / w B ) + (G / w A )

and

For a solution containing n1 n2, n3….. etc moles of the various components, the mole fraction of ith component can be written as

ci =

solved problem 3 A solution has 25% ethanol, 25% acetic acid and 50% water by mass calculate the mole fraction of acetic acid. Solution: nB Mole fraction of acetic acid c B = nA + nB + nC 25 / 60 25 / 46 + 25 / 60 + 50 / 18 0.416 = = 0.111 0.513 + 0.416 + 2.777 =

problems for practice

If no = a/WB and ns = G/WA - where a and G are the weights of the solvent and the solute respectively, WB and WA are the corresponding gram molecular weights. Then the mole fractions of A and B are as follows χA =

 G WB  n ns × becomes = s =   n o  WA a  ns + no G a Since n s = and n o = WA WA χA =

ni n1 + n 2 + .....

The sum of mole fractions of all the components in solution is always equal to one. If the mole fraction of one component of a binary solution is known, that the other can be calculated. For example, the mole fraction cA is related to cB as cA= 1– cB or cB= 1–cA It may be noted that the mole fraction is independent of temperature. In the case of dilute binary solution, the weight of the solute (G) is very small compared to that of the solvent (a). Under these conditions the number of moles of the solute (ns) can be neglected in comparison with the number of moles of the solvent (no). Then the mole fraction of the solute

3. A solution to be used as hand lotion, is prepared by dissolving 3.20 g of methanol and 9.20 g of glycerol (C3H8O3) in 90 g of water. Calculate the mole fraction of glycerol. (Mol mass of glycerol = 92, methanol = 32) 4. 12% (W/V) aqueous solution has a density of 1.2 g/ mL. What are the mole fractions of the components. (Given mol wt of solute as 40) 5. The mole fraction of benzene in a solution with toluene is 0.50. Calculate the mass percent of benzene in the solution.

6. Molarity (M): Molarity of a solution is defined as the number of gram moles of the solute present in one litre of the solution or millimole of the solute present in mL of solution. It is represented by M. Mathematically Moles of solute Volume of solution in litres Mass of solute in grams = Gram molecular weight of solute × Vol of solution (L)

Molarity ( M ) =

Moles of solute (nB) = ∴M =

Mass of solute in grams (WB ) Molar mass of solute M B

nB WB or or V( L ) M B × V( L )

WB (g ) × 1000 M B × V( mL )

Solutions

A solution having molarity equal to unity is called molar solution. Such solution contains one mole of solute per litre solution. The solutions having molarity equal to 0.5 M, 0.1 M and 0.01 M are called semi molar, decimolar and centimolar solutions respectively. Molarity is expressed in units of mol L–1 or mol dm–3. It may be noted that both normality as well as molarity of a solution change with change in temperature. The units of molarity are moles per litre (mol L–1) or moles per cubic decimeter (mol dm–3). Demal unit: Demal is another unit for expressing the concentration of the solution. It is represented by D and expresses the molarity of the solution at or 0°C. For example 2.5 D solution implies the presence of 2.5 mole of the solute in one litre solution at 0°C. solved problem 4 What mass of CaBr2 is needed to prepare 150 mL of a 3.5 M solution (Atomic mass of Ca = 40, Br = 80)? Solution: Weight Number of millimoles = M × 10−3 = Molarity ×Volume (in mL) Molecular weight of CaBr2 = 40+(80×2)=200 weight of CaBr2 needed = 3.5 × 150 × 200 × 103 = 105 g solved problem 5 Determine the molarity of a 40% solution of HCl which has a density of 1.2 g mL–1. Solution: 1000 % Strength × Molarity = Sp. gravity × 100 Mol. wt 40 1000 = 1.2 × × = 13.15 M 100 36.5 problems for practice 6. Find the molality of H2SO4 solution whose sp gravity is 1.98 and contains 95 g of H2SO4 per 100 mL (molecular mass of H2SO4 = 98). 7. 3.645 g of HCl are dissolved in water to give 1 M solution. What is the volume of the solution in mL? 8. Calculate the molarity of 10.6% (W/V) Na2CO3 solution. 9. Calculate the molarity of pure water (density of water = 1 g mL–1).

8. Normality (N): The normality of a given solution (N) is defined as the number of gram equivalent weights of the solute present in one litre of solution.

5.5

If n gram equivalent weights of the solute are dissolved in ‘V’ litres of the solution, then the normality of the solution n G. eq wt ⋅ v lit Weight of the solute WB but n = Gram equivalent weight of the solute N=

1   WB 1000   W = × N= B ⋅    GEW V in lit   GEW V in mL  If one gram equivalent weight of solute is present in one litre or 1000 mL of the solvent the solution is known as 1 normal solution. The normality of a solution can be calculated if the equivalent weight of the solute is known. The equivalent weights of different substances can be calculated as follows. (i) Equivalent weight of an acid E=

Molecular weight or formula weight of the acid Basicity of the acid

The number of displaceable hydrogen atoms in a molecule of the acid is known as basicity of the acid. For example, the number of displaceable hydrogen atoms in HCl, HNO3, CH3 COOH is one, in H2SO4, H2C2O4 is two and in H3PO4 is three. (ii) Equivalent weight of a base E=

Molecular weight or formula weight of the base acidity of the base

The number of hydroxyl groups present in a molecule of the base is known as acidity of the base. For example, the acidity (number of hydroxyl groups) in NaOH, KOH is one and in Ba (OH)2 is two. (iii) Equivalent weight of a salt E=

Formula weight of the salt Total no. of +Ve charges on cation or V− e charges on anion given by the formula unit of salt

For example, (a) Sodium chloride (NaCl) Cation Na+; total no. of cations = total no. of the +Ve charges per formula unit = 1 Anion Cl–; total No. of anions = total no. of –Ve charges per formula unit = 1 Formula weight of NaCl 58.5 ∴ E= = = 58.5 1 1 (b) Sodium carbonate (Na2CO3) Cation Na+; total No. of cations per formula unit = 2 Total No. of positive charges on cations given by Na2CO3 = 2 Anion CO32–: Total number of anions per formula unit = 1

5.6

Solutions

Total no. of negative charges of anions per formula unit = 2 ∴ E=

Formula weight of Na 2 CO3 106 = = 53.0 2 2

(iv) Equivalent weight of an oxidant or a Reductant The equivalent weight of an oxidant or a reductant depends on the redox reaction. Thus the equivalent weight of the oxidants and reductants changes with the nature of the reaction. The number of moles of electrons transferred to 1 mole of oxidant or released from 1 mole of reductant indicates the equivalent weight of the oxidant or reductant respectively. Thus equivalent weight of an oxidant or reductant E=

Molecular wt of oxidant or reductant No. of electrons involved in redox reaction

solved problem 6 One litre solution of N/2 HCl is heated in a beaker. It was observed that when the volume of the solution is reduced to 600 mL, 3.25 g of HCl is lost. Calculate the normality of the new solution. Solution: 1 L of 1 N HCl solution contains = 1 gram equivalent of HCl = 36.5 g 1 L of N/2 HCl solution contains = 36.5/2 = 18.25 g of HCl Wt of HCl lost on heating = 3.25 g Wt of HCl left after heating = 18.25–3.25 = 15.0 g 15.0 ( Eq mass = 36.5) 36.5 Gram equivalents of solute Normality = × 1000 Volume of solution in mL

Gram equivalents of HCl left =

=

15.0 × 1000 = 0.685 N 36.5 × 600

9 Molality (m): The molality of a given solution (m) is defined as the number of moles of the solute present per kg (or 1000 g) of the solvent. If ‘n’ moles of the solute is dissolved in ‘a’ Kg of the solvent, the molality of the solution m= but n =

no. of moles of solute (n) no. of Kg of the solvent (a)

weight of the solute (G) gram mol. wt of the solute (GMW)

  G  G 1 1000  ∴m =  × = ×    GMW ' a ' in Kg   GMW ' a ' in g  If ‘n’ = 1 and ‘a’ = 1 kg, then the molality of the solute is equal to 1.0. The solution is known as 1 molal solution or 1 m solution. Similarly when 0.1 moles of solute is dissolved in 1 kg or 1000 g of the solvent, the resulting solution is known as decimolal or 0.1m solution. Since the quantities of the solute and the solvent are expressed in weights, the molality does not change with the change in temperature, i.e., molality is independent of temperature. solved problem 7 Calculate the molality of a 13% (W/W) solution of H2SO4. Solution: 13% (W/W) solution means 13 g of the solute are present in 100 g of solution ∴ Weight of the solvent in 100 g of solution = (100– 13) = 87 g G 1000 × Molality of the solution = GMW ' a ' in g =

13 1000 × = 1.53 m 98 87

problems for practice 10. The density of 8.653% (W/V) Na 2CO 3 solution is 1.018 g mL –1. What is the molality of the solution? 11. 100 mL of alcohol is made upto a litre with distilled water. If the density of the alcohol solution is 0.789 g/ mL. What is the molality of the alcohol solution? 12. Calculate the molality of 1 litre solution of 93% H2SO4 (mass/volume). The density of the solution is 1.4 g/mL. 13. An aqueous solution of certain substance has been labelled as 1 M. Find mole fraction of solvent and solute.

10. Formality (F): Formality is the number of formula masses of the solute dissolved per litre of the solution. It is represented by F. Formality =

Number of formula masses of solute Volume of the solution in litre

The term formality is used to express the concentrations of ionic substances. The ionic compounds such as NaCl, KNO3, CuSO4 etc. do not exist as discrete molecules. In such cases, we do not use the term mole for expressing the concentration. The sum of the atomic masses of various atoms constituting the formula of the ionic compound is called gram formula mass instead of mass.

Solutions

some important formulae and relationships Some very significant formulae about different modes of concentrations of solution are given below. 1. Relationship between molarity and normality The normality (N) and (M) of a solution are related as. Molarity × Mol mass (solute) = Normality × Eq mass (solute) 2. Relationship between molarity (M) and mass percentage (%) % mass × d × 10 Molarity = Molar mass (solute) % mass × d × 10 Normality = Eq. wt (solute) 3. Relationship between molarity (M) and molality (m) m= or

1000 M (1000 × d) − (M × MW (solute) 1 MW ( solute) d = + M m 103

4. Dilution formula: If the solution of some substance is diluted by adding solvent from volume V1 to volume V2, then initial molarity M1 and final molarity M2 are related as M 1V 1 = M 2V 2

Similarly N1V1 = N2V2

5. Molarity of a mixture: If V1 mL of a solution of molarity M1 is mixed with V2 mL of another solution of same substance with molarity M2, then molarity of the resulting mixture solution (M) can be obtained from the formula M= Similarly N =

M1V1 + M 2 V2 V1 + V2

N1V1 + N 2 V2 V1 + V2

6. Relationship between molality (m) and mole fraction of solute (xB) m × (MWA ) cB = 1000 + m × MWA Also m =

1000 c B c A × MWA

7. Relationship between molarity (M) and mole fraction of solute (χB) cB =

M × MWA M (MWA − MWB ) + 1000 d

Also M =

5.7

1000 d c B c × MWA + c B × MWB

In the above expressions, the subscript A refers to solvent and B refers to solute comparison Between the strengths of Molar and Molal solutions For most of the aqueous solutions, 1 M solutions have higher concentrations than corresponding 1 m solutions. It is so because in 1m solutions, 1 mol of solute is present in 1000 g, i.e., 1000 cm3 (d  1) of water. But in 1 M solution 1 mol of solute is present in 1000 cm3 of solution which contain less than 1000 cm3 of solvent. However in case of non-aqueous solutions, 1 M solution can be less, more or even has same concentration as that of 1 m solution depending on density of solution. The molarity (M) and molality (m) of the solution are related as 1 M m m= or = d − (M × MWB × 10 −3 ) M d − (M × MWB × 10 −3 ) (i) Conditions for m = M or m/M = 1 1 ∴ 1= (d − M ) × MWB × 10 −3 or (d–M) × MWB×10–3 = 1 or d = 1 + (M × MWB×10–3)...(i) Now for dilute solution, the factor M × MWB × 10–3 being very small can be neglected in comparison to 1. Hence d = 1 m (ii) Conditions for m > M or >1 M Substituting in (i) we get d–(M×MWB×10–3) > 1 or d > 1 m >< 1 (iii) When m < M or M Substituting in (i) we get d–(M-MWB × 10–3) < 1 or d < 1

problems for practice 14. What is the mole fraction of the solute in 2.5 m aqueous solution? 15. Commercially available sample of sulphuric acid is 15% H2SO4 by weight (density = 1.10 g mL–1). Calculate (i) molarity (ii) normality (ii) molality of the solution. 16. An antifreeze solution is prepared from 222.6 g of ethylene glycol [C2H4(OH)2] and 200 g of water, calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of solution?

5.8

Solutions

17. Calculate the volume of 80% H2SO4 by weight (density =1.80 g mL–1) required to prepare IL of 0.2 MH2SO4. 18. A sample of drinking water was found to be severely contaminated with chloroform (CHCl3), supposed to be carcinogen. The level of contamination was 15 ppm (by mass) (i) Express this in percent by mass. (ii) Determine the molality of chloroform in the water sample. 19. How many mL of sulphuric acid of density 1.84 g mL–1 containing 95.6 mass % of H2SO4 should be added to one litre of 40 mass % solution of H2SO4 of density 1.31 g mL–1 in order to prepare 50 mass % solution of sulphuric acid of density 1.40 g mL–1? 20. Two solutions of ethanol marked as A and B have been labeled as 25% ethanol by mass and 25% ethanol by volume respectively. If density of ethanol is 0.789 g mL–1, and that of solution is 0.968 g mL–1, then point out which solution has larger molarity? 21. You are provided with one litre each of 0.5 M NaOH and 0.25 M NaOH solution. What maximum volume of 0.30 M NaOH can be obtained from these solutions without using water at all? 22. Show that for an aqueous solution of solute x with molar mass Wg mol–1, the molarity ‘M’ and molality ‘m’ are related as d/M = (1/m) – (W/1000) (d being the density of the solution).

5.4 Types of Binary soluTions We have already discussed about the different types of binary solutions depending on the physical states of solute and solvents. The three main types of solutions are (i) solid solutions (ii) liquid solutions and (iii) gaseous solutions. 1. Solid Solutions Solutions in which solute and solvents are solids are referred to as solid solutions. These are generally formed by mixing of solid components. Alloys are very familiar examples of solid solutions. If the lattice sites in the host solid (the solvent) are occupied by the solute particles, the solutions formed are known as substitutional solutions. This type of solutions are formed when the particle sizes of both solvent and solute are similar. Alloys like bronze, brass, monel metal etc are common examples of substitutional solutions. The solutions formed by the occupation of smaller solute particles in the interstitial voids or interstitial sites in a crystal lattice of the host component (the solvent) are referred to as interstitial solutions. Cast iron, steel, tungsten carbide (WC) are common examples of interstitial solutions. In WC, the tungsten atoms adopt fcc lattice while carbon atoms occupy interstitial octahedral sites. Tungsten carbide is extremely hard substance which is used in making cutting and grinding tools.

Solvent Solute (a)

Solvent Solute (b) fig 5.2 Solid solutions (a) Substitution solutions (b) Interstitial solution 2. Liquid Solutions The solutions which exist in liquid state are known as liquid solutions. In these solutions, solvent is liquid while solute can be solid, liquid or gas. (a) Solutions of solids in liquids (b) Solutions of liquids in liquids (c) Solutions of gases in liquids. (a) Solutions of solids in liquids The binary solution of solid in liquid are of solids in liquid type solutions. Solid is referred to as solute and liquid is solvent. Solid solutes may be ionic solids or molecular solids. Ionic solids ionize in water and conduct electric current. So they are referred to as electrolytes. Molecular solids do not ionize but they dissolve in water because of their ability to form hydrogen bonds with water molecules. The molecular solids are referred to as non electrolytic solutes. (b) Solutions of liquids in liquids The liquid in liquid solutions are of the following types (i) The two components may be almost immiscible: If one liquid is polar and another liquid is non polar, they are immiscible. For example, benzene and water. (ii) The miscibility of the components may be partial: If the inter molecular attractions of one liquid is different from inter molecular attraction of the other, there may be a partial miscibility of the two liquids. For example, ether and water. (iii) The two components may be completely miscible: In this case the two liquids are of the same nature, i.e., they are either polar (like alcohol and water) or non – polar (like benzene and hexane).

Solutions

(c) Solutions of gases in liquids Most of the gases are soluble in water and also in some other liquids to different extent. Solutions of gases in water are very common in our daily routine. The familiar examples are carbonated beverages i.e., solution of CO2 in water, formaline (an aqueous solutions of formaldehyde, and variety of household cleaners which contain aqueous solutions of ammonia. All natural waters also contain dissolved O2, N2 and traces of other gases.

5.5 soluBiliTy Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. It depends upon the nature of solute and solvent as well as temperature and pressure.

5.5.1 The solubility of solids in liquids When a solid is shaken up with a liquid at a constant temperature, after a time, a point is reached when no more solid will dissolve. The solution is then said to be saturated. We have here an example of a state of equilibrium, where the rate at which solid is dissolving is equal to the rate at which material crystallizes out again from the solution. The equilibrium is not affected by the amount of solid present, since if a small area of surface is considered, the rate at which molecules escape from the surface depends on the number of molecules in that area of surface, which remains constant so that Rate of dissolution = Constant While the rate at which molecules return to the portion of the surface of the solid is proportional to the concentration of those molecules in the solutions (C), hence Rate of recrystallisation = constant × C At equilibrium, Rate of dissolution = Rate of crystallization C = Constant The concentration of solution is therefore constant as long as an excess of solid is present. It follows that a saturated solution is a solution that is in equilibrium with an excess of solid at a given temperature. The concentration of a saturated solution represents what is called the solubility of a material in a particular solvent. It is expressed as the molality, i.e., the number of moles of solute dissolved in 1 kg of solvent to form a saturated solution. In the absence of solid particles, it is possible for a solution to be super saturated, i.e., to contain a higher concentration of the solute than does a saturated solution.

5.9

Such solutions may be prepared by making a solution that is nearly saturated (but entirely free from solid) at a certain temperature and then cooling it. Since the solubilities of most substances decrease as the temperature is reduced, the solution will now be super saturated, but it is unlike that solid will form at once, since it is difficult for crystals to start growing unless there is solid nucleus on which the growth can start. If a small crystal of the solute is added, solid will at once crystallize out and equilibrium will be established. This is known as seeding the solution.The presence of dust particles or small particles of glass caused by scratching the walls of the container with glass rod will have the same result in relieving super saturation. Very small particles of solid have a greater solubility than the normal value, a fact first discovered by Ostwald. This has some importance in gravimetric analysis because precipitates, when first formed, frequently contain extremely small particles which can pass through a filter paper (e.g., the precipitate of barium sulphate, in the determination of sulphate). If the precipitate is allowed to stand for a time before filtering, preferably hot, the smallest particles will slowly redissolve, the solid being redeposited on the larger paticles. The process is sometimes called granulation of the precipitate. Gravimetric measurements involving the precipitation of so called insoluble compounds are actually based on the fact that these substances are only sparingly soluble, so that the amount of material remaining in solution is negligibly small. The same principle is used in qualitative analysis when metals are precipitated as chlorides, sulphides hydroxides, carbonates etc.

5.5.2 cause of solubility of solids in liquids In accordance with the general principles expressed above, water is a good solvent for ionic compounds (inorganic and organic salts) and for organic molecular solids which are strongly polar and particularly those with which the water molecules can form hydrogen bonds. There are two reasons why water is good solvent for ionic crystals. The first is that water is a good insulator. If water molecules come between ions at the edge of a crystal, it weakens the attraction between them. This helps them to float away from the crystal. Also, when the ions are in water, they are surrounded by a layer of water molecules. This layer insulates the ions from one another and prevents them joining together into a crystal again. Other liquids can act as insulator as well. We can compare how effective they are by observing the values of their relative permittivity (Another name for relative permittivity is dielectric constant). The higher the value, the better the insulator.

5.10

Solutions

Table 5.2 The relative permittivities (dielectric constants) of some liquids

Liquid

Formula

*Relative permittivity (Dielectric constant)

Water Ethanol Ethanoic acid Benzene Tetrachloro methane

H 2O C2H5OH CH3COOH C6H6 CCl4

80.1 25.7 6.2 2.3 2.3

If we concentrate on the particles in the solid, there is an increase in entropy.

*Relative permittivities has no units

Add ions

The second reason why water is a good solvent is that water is a polar molecule with a positive end that can attract negative ions and a negative end that can attract positive ions.

H

δ +

H

δ + O

OH

δ +H δ -

δ Na+

ADD WATER

δ + δ δ +H O H

Cl–

δ + δ OH δ +H δ +

fig 5.3 Hydration of ions It is the ability of water to gather round both negative and positive ions that makes it such a good solvent for ionic crystals. The layer of water molecules that surrounds the ion is called the hydration sphere. When sodium chloride is put into water, the ions on the outside of the crystal feel two competing influences. (i) The attractions of the oppositely charged ions in the crystal, which tend to stop the outside ions escaping; and (ii) The attractions between the surrounding water molecules and the outer ions, which tend to pull the ions off the crystal. One measure of the attractions between the ions in the crystal is the lattice energy, ΔHL and how strongly ions are attracted by water molecules is the hydration energy ΔHh. Approximately, it is the enthalpy change when one mole of an ion in the gaseous state dissolves in water. The energy change when one mole of sodium chloride dissolves in water is ΔHsol ΔHsol + ΔHL = ΔHh or ΔHsol = ΔHh – ΔHL

-

+

+

+ -

-

fig 5.4 Contributions to the entropy change when solid dissolve in water The hydration energies of the ions and the lattice energy are exothermic changes, so they have negative sign. If ΔHh is more negative than ΔHL, then ΔHsol will be negative and the crystal should dissolve exothermically. For sodium chloride, we have ΔHL = –781 KJ mol–1 ΔHh = ΔHh (Na+) + ΔHh (Cl–) = –390 KJ mol–1 –384 KJ mol–1 = –774 KJ mol–1 So ΔHsol = –774 KJ mol–1 – (–781 KJ mol–1) = +7 KJ –1 mol (a) Effect of entropy change on solubility From the above results it can be seen that sodium chloride dissolves endothermically in water. Not only sodium chloride but many ionic solids dissolve endothermically. The difficulty this leaves us in is that it appears that ionic crystals dissolve even though it is energetically unfavorable. This is where we turn to thermodynamics for the explanation. According to free energy change equation. ΔG = ΔH–TΔS Where ΔG is free energy change, ΔH is enthalpy change and ΔS is entropy change. Though ΔH is positive, the ΔG may be negative, if the entropy change is large. When crystals dissolve endothermically, it is the entropy change that is responsible for the change taking place. The increase in entropy occurs because there is greater spread of the ions and molecules among the available energy levels, i.e., the number of complex ions increases. However, it should be noted that when an ion leaves a crystal lattice it becomes surrounded by hydration sphere. The water molecules in the sphere are less free to move

Solutions

than they were in the bulk of the water. Indeed, for a small, highly charged ion the hydration sphere may consist of several layers of water molecules. Because of this reason there may be reduction in entropy.

Add ions

+

+ -

fig 5.5 of ions

-

+

-

Decrease in entropy of water due to hydration

Thus whether there is overall increase or decrease in entropy will depend on the following factors. (i) Change in the number of complex ions* in the crystal when ions leave. (ii) Change in the number of complex ions open to individual ions owing to their release into the liquid; and (iii) Change in the number of complex ions owing to the formation of hydration spheres. Calculating the precise change in the total number of complex ions is a fearsome task; fortunately we need not to attempt to it. (b) Effect of sizes of ions in a crystal on solubility When a crystal is made of small highly charged ions the lattice energy is too large, but the hydration energies of the ions are not large enough to overcome the lattice energy, so the substance is either insoluble or only very slightly soluble. The heat of solution is such a large positive value, even if the entropy change is positive, the free energy change will remain positive. So the crystal will not dissolve. Table 5.3 Factors influencing the solubilities of ionic crystals. Factor

Implication

Lattice energy large

Ions held together very strongly. Crystal likely to be insoluble.

Lattice energy small Ions large and/or small charge

Ions held together weakly. Crystal likely to be soluble.

Ions small and/or high charge

Lattice energy likely to be small Entropy change likely to be positive ΔGsol will be negative. Crystal should be soluble. Lattice energy likely to be very large. Entropy change likely to be negative ΔGsol will be positive. Crystal should be insoluble.

* For complex ions refer to Thermodynamics.

5.11

The effect of ionic sizes on the solubility can be studied by comparing the solubilities of nitrates and chlorides. The general rule is that, even if the chloride is insoluble, the nitrate will be soluble. Indeed, there is a rule in chemistry that says “all nitrates are soluble”. The main reason is that the nitrate ion is larger than the chloride ions. This has two effects. First, the lattice energy of a metal nitrate is usually smaller than that of the chloride (due to bigger size of nitrate ion). Secondly, the larger size of the nitrate ion results in the decrease of entropy of the water around it is less than it is around the smaller chloride ion. (Fewer water molecules would be held in the hydration sphere of larger nitrate ion). Both these effects work together to make it more likely that the nitrate will dissolve. An extreme example is the great insolubility of silver chloride, and the great solubility of silver nitrate. However, the nature of the bonding in silver chloride is a complicating factor here. (c) Solubility of covalent substances in water It is not only ionic substances that dissolve in water. Many covalent substances dissolve also. These are of two types: One type react with water rather than dissolve in it. Such reaction is called a hydrolysis reaction. Hydrogen chloride under goes hydrolysis. HCl(g)+H2O(l) → H3O+(aq)+Cl–(aq) The second type are usually polar and able to hydrogen bond to water molecules. Organic compounds that dissolves in water are often like this. Typical examples are molecules containing the groups –O–H (alcohols, aldehydes, carbohydrates, phenols, carboxylic acids) and >N–H (amines, amides). The solution will consist of a homogeneous mixture of water molecules and hydrated solute molecules or ions. Ethanol acts as solvent for the same classes of compounds, but the solubility of salts is lower than in water. We have already seen in the beginning “like dissolves like”. Now, we shall try to explain why, for example, the organic solid naphthalene will not dissolve in water, even though it will dissolve in the organic liquid benzene. Naphthalene is not polar, so the molecules in the crystal are held together by Van der Waals forces. These forces are also responsible for holding benzene molecules together. Van der Waals forces are relatively weak, so it does not require a great deal of energy to move one naphthalene molecule apart from another. Essentially, when naphthalene dissolves in benzene, it is due to Van der Waals forces in the solution, we should expect only a small enthalpy change in such a case which is what happens. It is rare to find a large enthalpy of solution for an organic solid dissolving in an organic liquid. On the other hand, the entropy change when the solid dissolves should be positive, just as it is for an ionic solid dissolving in water. Indeed, it is the entropy change that drives the process along.

Solutions Solubility /g per 100 g water 180 nit rat e

160 140

tas siu m

(d) Effect of temperature on solubility If volumes are measured for water before and after an ionic solid is dissolved in it, it would often be found that the volume decrease slightly. This is due to the water molecules clinging tightly to the ions as they float away from the crystals. The effective volume of water molecules in hydration sphere is less than water molecules free to roam through the solution. However, it is not always the case that the volume decreases. For example, crystals that contain water of crystallization, e.g., CuSO4. 5H2O release water molecules. In the case of these crystals, 1 mol of the solid would release 5 mol of water, i.e., 90 cm3.

120

Po

5.12

ead

100

L

80 60

Sodium sulphate

40

Table 5.4 Solubilities of solids in water at 25°C Substance NaCl NaBr NaI MgSO4 CaSO4

Solubility/g per 100 g water

0.615

35.98

0.919

94.57

1.23

184.25

1.83×10

–1

22.01

4.66×10

–3

0.63

–5

SrSO4

7.11×10

0.1

BaSO4

9.43×10–7

2.2×10–4

NaNO3

1.08

91.69

BaNO3

3.91×10

–2

10.22

The solubilities of ionic solids often, but not always, increases with temperature, some times considerably. Applying Le Chatelier’s principle, as the tem perature is raised the system will tend to oppose the change, i.e., to absorb heat. Since the result is that more material dissolves, it follows that the process of solution is usually accompanied by an absorption of heat. This is due the energy required to pull the molecules or ions away from the crystal lattice (Lattice energy). Where solvation occurs however, energy is also liberated during the process. The result of this can be seen by comparing the effect of temperature on the solubilities of various inorganic salts in water (Fig 5-6). When it will be seen that the solubilitiess of certain salts vary very little with temperature (in some cases, e.g., calcium citrate the solubility may even decrease as the temperature is rised). This is because the energy of hydration almost balances the lattice energy, so that the overall energy changes is nearly zero. This may occur in the case of anhydrous salts, e.g., NaCl.

Sodium chloride (v) hlorate ium c Potass

20 0

Solubility/mol per 100 g water

e rat

nit

10

20

30

40

50

60

70

80 90 100 Temperature /0C

fig 5.6 Variation of solubilities of some salts with temperature On the other hand, when salts are hydrated already in the solid state (e.g., Na2SO4.10H2O) the energy of hydration is found to be small and the solubility increases with rise in temperature. Sodium sulphate is a very good example of this behavior, because at 32°C the hydrate changes to the anhydrous form, and the solubility of the latter actually decreases with temperature. The solubility curve has a break at 32°C which is the transition temperature for the equilibrium. Na2SO4.10H2O ⇌ Na2SO4 +10H2O

(e) Effect of pressure on solubility Pressure does not have any significant effect on solubility of solids in liquids. It is so because solids and liquids are highly incompressible and practically remain unaffected by changes in pressure. From the discussion made so far it may be concluded (i) If the value of ∆Hsol < 0; i.e., the solution process is exothermic, then by Le – chatlier’s principle, the solubility of such a solute with decrease with the rise in temperature, e.g., CuSO4, Na2SO4. (ii) If the value of ∆Hsol > 0, i.e., solution process is endothermic then, solubility of such a solute will increase with rise in temperature, e.g., KNO3, NaNO3, hydrated salts etc. (iii) If the value of ∆Hsol ≈ 0; the solution process is neither exothermic nor endothermic. Solubility of such solids does not increase or decrease regularly or continuously with rise in temperature. This happens if a substance under goes a change from one polymeric form to another at a particular temperature.

Solutions

Where there is no reaction as for example between oxygen, nitrogen and hydrogen with water, the solubility is low. In all cases however, the solubility varies from solvent to solvent. Oxygen and nitrogen, for example, are more soluble in alcohol than in water. There is usually greater solubility when there is similarity in chemical character between solute and solvent. For instance, the gaseous hydrocarbons are more soluble in the liquid hydrocarbons than in water. 2. Temperature: In contrast to the solution of a solid, the solution of a gas in a liquid solvent is, in general, an exothermic process. Hence, provided the pressure is kept constant, the solubility of a gas decreases with rise of temperature (unless required dry, nitrous oxide is usually collected in the laboratory over hot water because of its appreciable solubility in cold). This means that, in general, a gas may be expelled from its solution in a liquid by boiling. There are exceptional cases, e.g., a dilute solution of hydrogen chloride in water actually increases in strength; on boiling, to a maximum at the concentration of the constant boiling mixture. Hydrogen chloride would be evolved from a saturated solution on boiling until, again, the concentration of the constant boiling mixture is reached. 3. Pressure: Provided the temperature is kept constant, the mass of gas dissolving in a given amount of liquid is directly proportional to the pressure of the gas above the liquid at equilibrium. This is Henry’s law.

or

m∝P m = KP

yg

en

where K is the proportionality constant. The magnitude of K depends on the nature of gas, nature of the solvent, temperature and the units of pressure. According to the above equation when solubility of the gas is plotted against the equilibrium pressure at a given temperature, a straight line passing through the origin is obtained. Fig 5.7 shows the variation of solubility of some gases against equilibrium pressure. The straight line graphs show the validity of Henry’s law.

Ox

The solubility of a gas in liquid is usually measured in terms of absorption coefficient. This is defined as the volume of gas measured at STP which will dissolve in unit volume of solvent at a particular temperature and under a pressure of 1 atmosphere. For example, the absorption coefficient of oxygen in water at 15°C is 0.034. This means that one litre of water at 15°C will dissolve 0.034 litre, i.e., 34 mL of oxygen at STP under a pressure of 1 atmosphere. The solubility of a gas in a liquid is affected mainly by the following three factors. 1. The chemical nature of the gas and the liquid: If there should be a chemical reaction between the gas and the solvent, then the absorption coefficient is high. Common examples are hydrogen chloride, ammonia and sulphur dioxide in water HCl+H2O⇄H3O++Cl– NH3+H2O⇄NH4++OH– SO2+3H2O⇄2H3O++SO32–

If ‘m’ is the mass of the gas dissolved in a unit volume of the solvent and ‘p’ is the pressure of the gas in equilibrium with the solution, then

Solubility

5.6 soluBiliTy of gases in liquids henry’s law

5.13

tr Ni

en

og

m

liu He

Equilibrium pressure p

fig 5.7 Variation of solubilities of certain gases with equilibrium pressure. If the mole fraction of the gas in the solution is used as measure of its solubility, then Mole faction of the gas in the solution is proportional to the partial pressure of the gas in the solution, i.e., χ∝P or χ = K′P 1 or P = KH c (KH = ) K' Where KH is called Henry’s law constant. Thus Henry’s law may alternatively be stated as The pressure of a gas over a solution in which the gas is dissolved is proportional to the mole fraction of the gas dissolved in the solution. When a mixture of a number of gases is brought in contact with a solvent, each constituent gas dissolves in proportion to its own partial pressure but not to their total pressure. Therefore Henry’s law is applied to each gas independent of the presence of other gas.

5.14

Solutions

The value of KH can be calculated by plotting the graph of PB/χB versus χ and extrapolating to χB = 0. Different gases have different KH values at the same temperature. This suggest that KH is a function of the nature of the gas. It is obvious from the equation P = KH χ, that higher the value of KH at a given pressure, the lower is the solubility of the gas in the liquid. It can be seen from the Table 5.5 that KH values for both N2 and O2 increase with increase of temperature indicating that the solubility of gases increases with decrease in temperature. Table 5.5 Values of Henry’s law constants for certain gases in water Gas

Temperature (K)

He H2 N2 N2 O2 O2 Ar CO2 CH4

293 293 293 303 293 303 298 298 298

Table 5.6 at 298 K

K=

Where m = K′; Cs, K′ being Constant At Constant temperature T Cs KRT = = Z (constant) Cg K' Hence at a certain temperature, the ratio of the molar concentrations of the gas in solution and the gas phase is constant. This is another expression for the Henry’s Law. 2. Now suppose m gm of a gas has a volume V at constant temperature and pressure and let M be the molecular weight of the gas then

KH (K bar) 144.97 69.16 76.48 88.84 34.86 46.82 40.3 1.83×10–5 0.413

KH values of gases in different solvents

Gas

Solvent

KH (K bar)

H2 H2 CO2 CO2 CH4 CH4

Water Benzene Water Benzene Water Benzene

71.18 3.67 1.67 0.11 41.85

0.56

It follows too, therefore if the pressure above a solution of a gas in a liquid is reduced, gas comes out of solution. Gas also comes out of solution if some inert gas, e.g., nitrogen is bubbled through the solution, for as the nitrogen escapes at the surface, it carries away with it, gas from the solution phase there by reducing the concentration.

5.6.1 other forms of henry’s law Henry’s law can also be expressed in a number of other ways also 1. Cs and Cg represent the molar concentrations of the gas in the solution and in gas phase respectively, then P = Cg RT and m ∝ Cs. Thus

m K ' Cs = P Cg RT

m= Now K =

M. PV RT

m M PV MV = = P P RT RT

At constant T, we have K RT V= = Z" M In other words, Henry’s law can also be defined as follows. The volume of a gas dissolved in a solvent at a given temperature is independent of the pressure. The above law may be made clear from the following consideration. Suppose m grams of a gas having volume ‘V’cc are dissolved in a given volume of solvent at the equilibrium pressure P of the gas. At constant temperature T, if the pressure is allowed to increase to, say 2 P, the mass of the gas dissolved in the same volume of the solvent will be 2m which corresponds to a volume of 2V at pressure P. But the volume of gas passing into the solution at 2 P will be 1/2 × 2V = V. This is because of the fact that V is inversely proportion to P. Thus the volume of the gas dissolved will remain the same before and after inspite of the increase in pressure.

5.6.2 characteristics of henry’s law constant Kh 1. The units of Henry’s law constant are same as those of pressure, i.e., torr or K bar. 2. Different gases have different values of KH. 3. The KH values of a gas is different in different solvents and it increases with increase in temperature. 4. The knowledge of KH value of a gas can help us in calculating the solubility of gas in a solvent at a given temperature. 5. Partial pressure of a gas is a linear function of its mole fraction.

Solutions

5.15

Table 5.7 Absorption coefficients of gases at 293 K Solvent

H2

He

N2

O2

CO

CO2

H2S

NH3

HCl

Water Benzene Acetone Ethanol

0.017 0.066 0.065 0.080

0.009 0.018 0.030 0.028

0.015 0.104 0.129 0.130

0.028 0.163 0.207 0.143

0.025 0.153 0.198 0.177

0.88 – 65 3.0

2.68 – – –

710 – – –

442 – –

5.6.3 limitations of henry’s law

-

Henry’s law is valid only under the following conditions 1. The pressure of the gas should not be too high. Like other gas laws Henry’s law is not obeyed strictly under high pressure. 2. The temperature is not too low. 3. The gas should not undergo any chemical reaction with the solvent. 4. The gas law is not obeyed by highly soluble gases which react chemically with the solvent to give another kind of particle. In a non-ionizing solvent such as benzene, the solubility of hydrogen chloride would obey Henry’s law.

When deep sea divers use compressed air for breathing the compressed air is much more soluble in blood and other body fluids than air at normal pressures. Most of the oxygen component of the dissolved air is used in metabolism while nitrogen remains as such. When the diver returns to the surface quickly the excess N2 (dissolved in blood) comes out quickly as tiny bubbles which block blood flow in blood vessels. This results in severe pain in joints and muscles. It is known as decompression sickness or bends and the affected person may faint,suffer deafness, paralysis or even die. To avoid this, the divers must return to the surface slowly or spend considerable time in decompression chamber where the pressure is gradually lowered.

5.6.4 application of henry’s law

solved problem 8

Henry’s law finds many applications in industry and helps to explain several biological phenomena some important applications are 1. To increase the solubility of CO2 in soft drinks and soda water, the bottles are sealed under high pressure. 2. In lungs, where the oxygen is present in air with high partial pressure, oxygen combines with haemoglobin to form oxyhaemoglobin. In tissues, the partial pressure of oxygen is low and therefore oxyhaemoglobin releases oxygen which is used for cellular activities. 3. At higher altitudes, the partial pressure of oxygen is less than that at the ground level. This leads to low concentration of oxygen in blood and tissues of the people living at higher altitudes or those of mountaineers. This results in becoming weaker and reduces the thinking capacity of mountaineers. This is called anoxia. 4. Scuba divers (deep sea divers) carrying breathing kit containing compressed air suffer from the painful effects called bends when they come out of the surface. This is due to the release of gases dissolved in their blood while breathing air at high pressure under water. To minimize the painful effects of deep sea divers during decompression, oxygen diluted with less soluble helium gas is used as breathing gas.

Hydrogen is shaken with 160 cm3 of water at 20°C and a pressure of 66.6 K Pa. What mass of hydrogen will be dissolved? (Absorption coefficient of hydrogen is 0.017). Solution: From the absorption coefficient 1cm3 of water will therefore dissolve 0.017 cm3 of hydrogen at STP when the pressure is 101.3 K pa. For 160 cm3 of water Volume of H2 dissolved at 101.3 pKa = 0.017 ×160 cm3 Applying Henry’s law Volume of H2 dissolved at 66.6 pKa = 0.017 × 160 ×

66.6 101.3

= 1.79cm3 (at STP) 1.79 Mass of H2 dissolved = 2 × 22400 = 1.61 × 10–4 g solved problem 9 Suppose air contains 79 per cent nitrogen and 21 per cent oxygen by volume. The absorption coefficients of nitrogen and oxygen at 15°C are 0.018 and 0.034 respectively. Calculate the composition by volume of the mixture of gases

5.16

Solutions

boiled from water which was saturated with the air of the above composition at 15°C. Solution: Let the pressure of air be P 79 P Partial pressure of N2 (PN2) = 100 21 P Partial pressure of O2 (PO2) = 100 Suppose a volume V of water is saturated with air Then the volume of N2 dissolving = V × 0.018 ×

79 P 1 × 100 P

0.018 × 79 ×V 100 1.422 V = 100 =

Volume of oxygen dissolving = V×0.034 × =

21 P 1 × 100 P

0.034 × 21 ×V 100

∴ Total volume of gas dissolved  0.018 × 79 0.034 × 21 =V  +   100 100  = ∴

2.136 V 100

2.136 1.422 V of nitrogen V of gas contain 100 100

∴ 100 V of gas contain

100 × 100 1.422 × V = 66.6 2.136 V 100

∴ Composition by volume expelled air is 66.6 percent N2 and 33% percent O2. problems for practice 23. If N2 gas is bubbled through water at 293 K, how many milli moles of N2 gas would dissolve in 1 litre of water. Assume that N2 exerts a partial pressure of 0.987 bar The KH for N2 at 293 K is 76.48 bar. 24. The Henry’s law constant for oxygen dissolved in water is 4.34 × 104 atm at 25°C. If the partial pressure of oxygen in air is 0.2 atm under atmospheric conditions, calculate the concentration (in moles per litre) of dissolved oxygen in water in equilibrium with air at 25°C. 25. Henry’s law constant of CO2 in water is 1.67×108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water packed under 2.5 atm CO2 pressure at 298 K.

26. H2S a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 mol kg–1, calculate Henry’s constant.

5.7 soluTion of liquids in liquids As already discussed 5.4(2b) the solutions of liquids in liquids are of there type viz (i) completely miscible (ii) partially miscible and (iii) immiscible solutions. Here we confine to the discussion of the completely miscible binary liquid solutions. The completely miscible liquids form two types of solutions viz, (i) ideal solutions and (2) non-ideal solutions. 1. Ideal solutions An ideal gas is one in which there is no attraction between molecules and no change in internal energy when the volume is changed. In the case of liquids, ideal solution is one in which the interaction between components of the solution is the same as the interaction between the molecules of each component. Thus a solution composed of two components A and B will be ideal if the interaction between A and B molecules is same as the interaction between A molecules or B molecules. If this is considered to be true no heat will be absorbed or evolved as a result of mixing of A and B and vapour pressure of such liquids can be calculated by averaging the properties of pure liquids. The volume of the solution is exactly equal to the sum of the volume of the liquids which are mixed together. For example, if 100 mL of water is added to 200 mL of water, the volume of the mixture is 300 mL, because there is no change in the interaction of molecules as a result of mixing. Similarly, 100 mL of methanol + 100 mL of ethanol will give a total volume equals to 200 mL and no heat will be evolved as a result of mixing. This is because of the fact that the molecules of ethanol and methanol are very much alike. But if 100 mL of sulphuric acid is mixed with 100 mL of water the final volume of the mixture will be 182 mL and a considerable amount of heat is evolved. As a result of chemical interaction between water and sulphuric acid, the latter ionizes and hence in the solution of sulphuric acid and water have different properties than they had as pure liquids. Such type of solutions in which the properties of dissolved liquids are different from those of the liquids in the pure state and which are formed by evolution or absorption of heat are called non-ideal solutions. From the above discussion, it may be concluded that an ideal solution is that (i) For which DVmix = 0, i.e., there is no volume change during mixing. (ii) For which ΔHmix = 0, i.e., no heat is evolved or absorbed during mixing of the liquids components. (iii) Which obeys Raoult’s law at all concentrations and at all temperatures.

Solutions

The French chemist Francois Marte Raoult (1886) studied the vapour pressures of number of binary solutions of volatile liquids at constant temperature and proposed a generalization known as Raoult’s law. According to this law “Partial pressure of a component (say liquid A) in solution is proportional to the mole fraction.” If all the components in solutions behave like ideal gases, then the total pressure of the solution is equal to the sum of the partial pressures of the individual components. Suppose a binary solution consists of two volatile liquids A and B. If PA and PB are the partial vapour pressures of the liquids and χA and χB are their mole fractions in solution then, PA = PA0 χA PB = PB0 χB PTotal = PA + PB PA0

PB0

where and are the vapour pressure of the two volatile components A and B respectively. According to Dalton’s Law of partial pressures, the total pressure (Ptotal) over the solution phase in the container will be the sum of the partial pressures of the components of the solution and is given as Ptotal = PA+PB Substituting the values of PA and PB we get Ptotal =

χAP0A+χB P0B

= (1- xB) P0A+ χBP0B =

P0A +

(P0B -P0A)χB

Following conclusions can be drawn from the above equation: (i) Total vapour pressure can be drawn from the equation. (ii) Total vapour pressure over the solution can be related to the mole fraction of any one component. (iii) Depending on the vapour pressures of the pure components A and B, total vapour pressure over the solution decreases or increases with the increase of the mole fraction of component A. On plotting the partial pressure PA and PB against their mole fractions we should get a graph as shown in Fig 5.8 Dotted lines denote vapour pressure-mole fraction curves of the components A and B, while the thick line denotes the total pressure of the solution. The dotted lines pass through the points and respectively when χA and χB equal to unity. Similarly, the thick line of Ptotal Vs χB is also linear. The

minimum value of Ptotal is PA0 and the maximum value is PB0 assuming that component A is less volatile than component B, i.e., PA0 < PB0 . The composition in vapour phase in equilibrium with the solution is determined by the partial pressure of the components. If YA and YB are the mole fractions of the components A and B in the vapour phase, using Dalton’s law of partial pressure, PA = YA. Ptotal PB =YB. Ptotal In general Pi = Yi. Ptotal If a solution obeys Raoult’s law, then the vapour pressure at any point on the curves should be the same as calculated from Raoult’s law equation. There are only a few binary miscible liquids which obey Raoult’s law through out the complete range of concentration and thus form ideal solutions. Some of them are (i) ethylene bromide and ethylene chloride (ii) benzene and ethylene chloride (iii) chlorobenzene and bromobenzene (iv) stannic chloride and ethylene dibromide and propylene dibromide.

VAPOUR PRESSURE

raoult’s law

5.17

P0B

N UTIO

OL OF S B OF RE U SS RE

RE ESSU L PR

TOTA

P

LP

0 A

IA RT PA

PART I

AL P

A

χ A= 1 χ B= 0

(PURE A)

RESS

URE

OF A

B MOLE FRACTION

χ A= 0 χ B= 1 (PURE B)

fig 5.8 Vapour pressure of an ideal solution 2. Real and non-ideal solutions Most of the solutions, however show appreciable deviations from the ideal behavior and are called real or non-ideal solutions. A non-ideal solution is that solution (i) which does not obey Raoult’s law (ii) for which ΔVmix is not zero (iii) for which ΔHmix is not zero A solution obtained on mixing two liquids will be nonideal if the solute-solvent interactions are weaker or stronger than the solute-solute and solvent-solvent interactions. (a) Non-ideal solutions showing positive deviations from Raoult’s Law. Consider that a solution is

5.18

Solutions

formed by mixing two liquids A and B. Let the mixing liquids be such that “A-B Interactions are weaker than A-A and B-B interactions”. As a result the escaping tendency of molecules of liquids A and B from the solution becomes more than that of pure liquids A and B. Hence, total vapour pressure of this solution becomes more than that corresponding vapour pressure expected for an ideal solution. Such solutions are said to have positive deviations from Raoult’s law. (i) PA > PA0 χA

(ii)

PB > PB0 χB

The positive deviations have been shown in Fig 5.9. The dotted lines indicate the ideal behaviour, while the thick lines indicate the actual behaviour. Some important examples of the solutions showing positive deviations are (i) acetone and ethyl alcohol (ii) acetone and carbon disulphide (iii) ethyl alcohol and water (iv) carbon tetrachloride and toluene.

VAPOUR PRESSURE

+ PB P = PA

P0B

PB P0A

P

A

A χ A=1 χ B=0 (PURE A)

MOLE FRACTION

χ

B

=0 χ B=1 (PURE B) A

(b) Non-ideal solutions showing negative deviations from Raoult’s law. Consider that a solution is formed by mixing two liquids A and B. Let the mixing liquids be such that “A-B interactions are stronger than A-A and B-B interactions” As a result the escaping tendency of molecules of liquids A and B from the solution becomes less than that of pure liquids A and B. Hence, total vapour pressure of this solution becomes less than the corresponding vapour pressure for an ideal solution. Such solutions are said to have negative deviations from Raoult’s law. Mathematically. (ii) PB> n, so n can be neglected in the denominator So

(P − Ps ) n = P N

(1)

5.9.2 lowering of Vapour pressure – a colligative property It is evident from the above equation (i) that the lowering of vapour pressure of a solvent depends upon the number of moles (and hence the number of molecules) of the solute dissolved a given amount of the solvent (and not upon the

5.26

Solutions

nature of the solute dissolved in the given amount of the solvent). Hence lowering of vapour pressure is a colligative property.

=1

In case of pure solvent n = 0 and the fraction N (n+N) ∴The vapour pressure P = K By replacing K in equation (2) by P, we have

5.9.3 limitations of raoult’s law Raoult’s law is applicable in case of very dilute solutions which behave like ideal solutions. The solutions deviate from ideal behavior as their concentrations are raised. The highest concentration of a solution to which the Raoult’s law is applicable, is taken as the ideality limit, i.e., only those solutions that obey Raoult’s low are ideal nature. Raoult’s law is valid for the solutions of non-volatile and non electrolytic solutes. Thus lowering of vapour pressure can only be used for the determination of molecular weight of a non-volatile and non-electrolytic substances, because this method depends upon Raoult’s law. Solutes which dissociate or associate in a particular solution do not obey the Raoult’s law. Because in case of dissociation, the number of particles in solution gets increased and thus the lowering of vapour pressure is also increased because it is proportional to the number of effective particles in solution. In case of association of the solute the number of effective particles decreases and hence the lowering of vapour pressure also decreases.

5.9 4 derivation of raoult’s law Let P and Ps are the vapour pressure of pure solvent and solution respectively and suppose the solution is obtained by dissolving n moles of a non-volatile solute in N moles of solvent. The number of gm mole of the solution = n + N ∴ Mole fraction of the solute =

n n+N

Mole fraction of the solvent in solution =

N n+N

The vapour pressure of the solution is therefore proportional to both of them, i.e., Ps ∝

n (n + N)

and

Ps ∝

N (n + N)

Since for very dilute solution N>>n, the value n/ (N+n) is very small and so the first proportionality may be neglected Hence,

or Ps =

Ps ∝

N (n + N)

K.N where K is constant (n + N)

(2)

or

 N  Ps = P   n + N  Ps N = P (n + N)

Subtracting both the values from one p   N  1−  S  = 1−   n + N  P   or

(P − Ps ) n + N − N = P n+N

or

P − Ps n = P n+N

(3)

This is Raoult’s Law.

5.9.5 determination of Molecular weight or Molecular Mass from lowering of Vapour pressure Molecular weight of non-volatile, non electrolytic solutes can be measured by vapour pressure lowering of their dilute solutions as follows. Suppose, w gm of a solute of molecular weight m, dissolved in W gm of a solvent of molecular weight M, lowers the vapour pressure from P to Ps. Then n = (w/m) and N = (W/M) Substituting these values in Raoult’s law equation (3) P − Ps n w /m = = P N + n W /M + w /m For very dilute solution, the term w/m in denominator is negligible in comparison to W/M and so can be neglected. Then we P − Ps w /m wM = = P W /M Wm Thus by knowing the relative lowering of vapour pressure, weight of solute and solvent and molecular weight (molecular mass ) of solvent, the molecular weight or molecular mass of solute can be calculated.

Solutions

5.9.6 determination of the lowering of Vapour pressure The determination of lowering of vapour pressure of a solvent can be done accurately by the following method. Walker and Ostwald’s Method: This is a dynamic method and was proposed by Walker and Ostwald. In this method, a current of dry air is bubbled successively through two sets of bulbs containing respectively (i) the solution (vapour pressure Ps) and (ii) the pure solvent (vapour pressure P) and finally through a reagent which can absorb the vapour of the solvent. If the solvent is water, the reagent used is usually anhydrous calcium chloride. The whole assembly is show in Fig 5.22. Glass bulbs A contain solution prepared by dissolving a known amount of non volatile solute in definite amount of solvent. Glass bulbs B contain weighed amount of pure solvent. A weighed amount of anhydrous calcium chloride is taken in the ‘U’ tube at the end. The glass bulbs A and B are thermostated. Now air is bubbled gradually through the bulbs A and B (so that it gets saturated with the vapour in each bulb and then through anhydrous calcium chloride tubes). On passing through the solution the air takes up an amount of vapour, proportional to vapour pressure of the solution (Ps). On passing through the solvent B, the moist air takes up further amount of vapour, which is proportional to the difference in vapour pressure of the pure solvent and solution, i.e., (P-Ps). Thus the loss in weight of solution (bulb A) ∝ Ps Loss in weight of solvent (bulb B) ∝ P–Ps Loss in weight of solution and solvent (bulbs A+B ∝ P) Ps +P–Ps ∝ P At the end of the experiment, calcium chloride tubes are weighed. The gain in weight by the calcium chloride tubes should be equal to the loss in weight of solution and solvent. So that

Thus the relative lowering of vapour pressure and consequently the molecular weight of the solute can be calculated from knowledge of the ratio of the loss in weight of the solvent to the gain in weight of CaCl2 tubes and using the following relation. P − Ps wM = P mW solved problem 12 Dry air was drawn in succession through a series of bulbs containing 4.5 gms of substance ‘A’ in 54 gms of ethyl alcohol and through similar set of bulbs containing pure alcohol at the same constant temperature. The loss of weight in the first series of bulbs was 1.5 and in second series was 0.3 gms. Calculate the molecular mass of A. Solution: Loss in weight of solution ∝ Ps = 1.5 gms Loss in weight of alcohol ∝ P-Ps = 0.3 gms Total loss ∝ [Ps+(P-Ps)] = 1.5 + 0.3 = –1.8 gms Loss in weight of alcohol P − Ps 0.3 = = Total loss P 1.8 Suppose m be the mol. mass of A mol mass of alcohol, i.e., M = 46 gm Wt of substance A, i.e., w = 4.5 gm Substituting these values in equation (P-Ps)/P = wM/ mW 0.3 4.5 × 46 = 1.8 m × 54 m=

1.8 × 4.5 × 4.6 = 23 0.3 × 54

P − Ps Loss in weight of solvent = Gain in weight of CaCl2 tubes P PURE DRY AIR

A

A SOLUTION

A

B

5.27

B

B

PURE SOLVENT

fig 5.22 Walker and Oswald’s method

CaCl2 TUBE

5.28

Solutions

problems for practice 29. The vapour pressure of pure benzene at a certain temperature is 640 mm Hg. A non volatile non electrolyte solid weighing 2.175 g is added to 39 g of benzene. The vapour pressure of the solution is 600 mm Hg. What is the molecular weight of the solid substance? (IIT1990) 30. A solution containing 30 g of a non-volatile solute in exactly 90 g of water has a vapour pressure of 21.85 mm of Hg at 25°C. Further 18 g of water is then added to the solution; the new vapour pressure becomes 22.15 mm of Hg at 25°C. Calculate (a) molecular mass of the solute and (b) vapour pressure of water at 25°C. (M.L.N.R 1990) 31. What mass of non volatile solute (urea) needs to be dissolved in 100 g of water in order to decrease the vapour pressure of water by 25% what will be the molality of solution? (IIT 1993) 32. When a solution of sugar is prepared in water at 25°C the vapour pressure of water (17.51 mm) is decreased by 0.0614 mm. Calculate (i) relative lowering in vapour pressure (ii) vapour pressure of the solution and (iii) mole fraction of water and sugar.

When a concentrated solution of a salt, say copper sulphate solution, is separated from pure solvent by a parchment paper or animal bladder acting as a membrane, the movement of the particles of the solute from the concentrated solution into the solvent is prevented but the flow of solvent into the concentrated solution is still continued. Such a membrane which allows free passage to the solvent molecules but not to the solute molecules to pass through it, is called semipermeable membrane. This phenomenon of spontaneous flow of pure solvent through a semipermeable membrane into a solution is known as osmosis. In other words, the phenomenon of passage of pure solvent through a semipermeable membrane from a solution of low concentration to one of higher concentration is termed as osmosis.

h Water Sugar solution

(Initial state)

5.10 osMosis and osMoTic pressure Diffusion of Solute: It is well known fact that gas can distribute itself uniformly through out the whole space provided, is due to the property of diffusion. Solute in a solution also possess the property of diffusion. They have a tendency to diffuse from a higher concentration to a lower concentration so as to bring about a uniform concentration through out. The diffusion in liquids takes place much more slowly than in a gas; because of the fact that molecules of a liquid are more closely crowded together. Thus if a concentrated solution of potassium permanganate is placed at the bottom of a beaker and a layer of water or that of a dilute solution of potassium permanganate is carefully poured over it, the coloured potassium permanganate can be seen diffusing up and after some time the boundary will become indistinct and finally disappears when there will be one homogeneous solution of uniform concentration throughout. The process of diffusion is due to (i) motion of solute molecule from a concentrated solution to the less concentrated solution and (ii) motion of solvent molecules into the concentrated solution. Thus diffusion may be defined as “the phenomenon of the movement of solute particles from a more concentrated to a less concentrated solution, in order to bring a uniform concentration throughout the bulk of the solution.”

Semipermeable membrance

Water (Final state)

fig 5.23 Osmosis and the osmotic pressure in thistle funnel due to hydrostatic pressure of solution in the stem of thistle funnel The flow of the pure solvent through the semipermeable membrane will continue till equilibrium is attained when the hydrostatic pressure of the column of the solution is just sufficient to prevent the entry of more solvent into the solution through semipermeable membrane. This hydrostatic pressure, produced by the process of osmosis is called the osmotic pressure of the solution. The first experiment on osmosis was done by Nollet (1748) but extensive study of the phenomenon was conducted not earlier than the middle of 19th century. A piece of parchment is tightly stretched across the belly of thistle funnel. The funnel is filled with a concentrated solution of sugar and lowered into a beaker of water. The level of liquid gradually rises in the stem of the thistle funnel and after some time, the further rising of the solution stops. When a semipermeable membrane is interposed between a solution and water (solvent) the solute molecules bombard the membrane and set up a pressure. The pressure exerted by the dissolved substance as a result of osmosis is known as osmotic pressure.

Solutions

5.29

5.10.1 demonstration of osmosis and osmotic pressure A battery pot with copper ferrocyanide semipermeable membrane deposited in its walls is filled with a solution of cane sugar and tightly closed with a rubber piston. The pot is now allowed to stand in an outer large beaker containing the solvent water as shown in Fig 5.24.

SOLVENT

SOLUTION

SEMIPERMEABLE MEMBRANE SOLVENT

INITIAL LEVEL

SOLUTION

fig 5.25 Osmosis through vapour pressure

fig 5.24 Measurement of osmotic pressure As a result of osmosis, the water will have a tendency to flow into the pot through the semipermeable membrane and causes the level of the solution in glass tube to rise. The tendency can be opposed by applying pressure on the solution by placing increasing weights on the piston. After placing proper weights on the piston, a condition of equilibrium is reached, when the tendency of the water to enter the pot through semipermeable membrane is just sufficient to prevent the entry of more solvent into the solution as a result of the counter pressure from above. The osmotic pressure of a solution may thus be defined as the excess pressure which must be applied to a solution in order to prevent the flow of solvent into it through a semipermeable membrane. Osmotic pressure concept can also be explained on the basis of fugacity or escaping tendency. Take two beakersone containing a solution of a nonvolatile solute and the other the pure solvent. Place these beakers inside a closed vessel at constant temperature. It will be observed that water (solvent) would slowly vapourize while the solution would gradually be diluted. Thus the volume of the solution increases and that of solvent decreases. This indicates that the escaping tendency of solvent molecules is greater in the case of pure solvent than that of the solution. The flow of solvent molecules from the pure solvent to the solution through the process

of vapourization takes place because of the difference in the escaping tendencies or vapour pressure of the two liquids. This process by which the solvent molecules continue to escape (due to high vapour pressure of the solvent) and the solution continues to condense vapours above it (due to low vapour pressure of the solution) is similar to osmosis where solution air interface acts as a semipermeable membrane. When the vapour pressure of the solution becomes equal to the vapour pressure of the solvents, a condition of equilibrium is reached at which osmosis will not takes place further. This condition of equilibrium may be attained by applying pressure externally on the solution or withdrawing pressure from the solvent. Thus osmotic pressure may also be defined as the excess pressure which must be applied to a given solution in order to increase its vapour pressure until it becomes equal to that of the solvent or the pressure withdrawn externally from the pure solvent in order to decrease its vapour pressure until it becomes equal to the solution.

5.10.2 Berkeley and hartely’s Method It is the best method for the measurement of osmotic pressure. The method is based on applying sufficient pressure on the solution side in order to check osmosis. It consists of a porous pot open at both the ends. The porous pot has copper ferrocyanide membrane fused in its wall. It is sealed into an outer bronze cylinder filled with a solution under examination and water is taken in the porous pot. The bronze cylinder is fitted with a piston on which weights can be placed to exert pressure on the solution.

5.30

Solutions

P T

SOLUTION

SOLVENT

a solution in inversely proportional to the volume (V) containing one mole of the solution. p∝

SEMIPERMEABLE MEMBRANE fig 5.26 Berkeley and Hartley apparatus Water tends to pass through the porous pot towards the solution. Now pressure is applied with the help of the piston so that the water level in tube becomes stationary. The external pressure applied on the piston is equal to the osmotic pressure (p) and can be read on the manometer. Osmotic pressure up to 150 atmospheres has been measured accurately by this method. Table 5.9 Osmotic pressure of sucrose at 303 K Sucrose/Kg water

Osmotic pressure (atmospheres)

220 gm 370 gm 420 gm 1430 gm

15.48 29.72 74.94 148.46

1 (at constant temperature) V

(1)

The relationship is similar to Boyle’s law for gases and known as Boyle’s Van’t Hoff Law for dilute solution. This generalisation is given from the data of osmotic pressure of aqueous solution of sucrose obtained by Berkeley and Hartely. 2. Effect of temperature on osmotic pressure: Van’t Hoff showed that osmotic pressure of a dilute solution is directly proportional to the absolute temperature at a given temperatures. Thus π∝T

(2)

This is similar to Charles’s law applied for gases and is known as Charles’-Van’t Hoff Law for dilute solutions. We know that liquids are incompressible. Hence the volume of the solution remains constant as required by the law. 3. Van’t Hoff equation for dilute solutions: From the above two laws, we see that p∝

1 (at constant temperature) V

(3)

and π∝T (at constant concentration or volume containing one mole of solute). Combining the two laws, we get π ∝ T/V

(4)

πV ∝T

(5)

advantages of this method (i) It is less time consuming. The equilibrium is established quickly and hence results are obtained in a short time. (ii) The concentration of the solution does not change during the determination of osmotic pressure. (iii) There is no strain on the manometer and the danger of the bursting of semipermeable membrane is eliminated.

5.10.3 Van’t hoff Theory of dilute solutions The laws of osmotic pressure Van’t Hoff, a Dutch physical chemist in 1885 proposed that a substance in solution behaves exactly like a gas and the osmotic pressure of a dilute solution is equal to the pressure which the solute would exert if it were a gas at the same temperature and occupying the same volume as occupied by the solution. 1. Effect of concentration on osmotic pressure: For substances in dilute solution, the osmotic pressure is proportional to the concentration at a constant temperature. In other words, the osmotic pressure (π) of

or

or πV = RT where R is molar solution constant.

(6)

This equation is exactly similar to the one obtained for one mole of ideal gases. It has been found that the value of the product πV in case of sucrose solution is 22.6 litre atmospheres at 273 K. If the result is divided by 273 the value of R is obtained R=

pV 22.6 litre atm = T 273

= 0.0828 litre atm degree–1 mole–1 Thus the value R (solution constant) is found to agree with the value of gas constant (R = 0.0821 litre atm degree–1 mole–1) if V litres of the solution contain n moles of the solute The Vant-Hoff equation becomes πV = nRT

(7)

Solutions

5.10.4 Van’t hoff – avogadro’s law for solutions Like the gas equation which incorporates Boyle’s law, Charles law and Avogadro’s law, the dilute solutions also obey Avogadro’s law. It states as follows “Equimolar solutions of different solutes exert equal osmotic pressure under identical conditions of temperature.” Solutions which have the same osmotic pressure at the same temperature, i.e., solutions which produce no flow of solvent through the semipermeable membrane are said to be isotonic solutions. When two solutions having the same osmotic pressure are separated with one another through a semipermeable membrane, there will be no transfer of solvent from one solution to another. As will be seen from the Van’t Hoff’s theory isotonic solutions must have the same molar concentration. That is w1 w2 = m1V1 m 2 V2

(8)

Where m1 and m2 are the molecular weights of solutes in the two solutions and w1 and w2 are the weights of the two solutes (in gms) in V1 and V2 litre respectively. If a solution has more osmotic pressure than some other solution is called hypertonic. On the other hand, a solution having less osmotic pressure than the other solution is called hypotonic. Thus a hypertonic solution will be more concentrated with respect to other solutions and a hypotonic solution will be less concentrated with respect to other solutions. It is quite interesting to note that 0.91% solution of sodium chloride (known as saline water) is isotonic with human blood corpuscles. In this solution, the corpuscles neither swell nor shrink. Therefore, the medicines are mixed with saline water before being injected into the veins. However a 5% NaCl solution is hypertonic solution and when red blood cells are placed in this solution, water comes out of the cells and they shrink. On the other hand, when red blood cells are placed in distilled water (hypotonic solution) water flows into the cells and they swell or burst. When dehydrated fruits and vegetables are placed in water, they slowly swell and return to original form. This is because the cell walls of fruits and vegetable are semipermeable. The liquid inside the cell in the dried state is in the form of highly concentrated (hypertonic) solution. When such dried fruits are placed in water (hypotonic solution) osmosis starts. Due to this water molecules enter the fruit and vegetables and they return to the original form.

5.31

Similarly if two eggs are (after removing the hard shell by dissolving in dilute HCl) placed on in distilled water (hypotonic solution) and another in saturated sodium chloride (hypertonic solution), after a few minutes, it will be observed that the egg placed in water swells where as the other placed in salt solution shrinks. The skin of the egg acts as a semipermeable membrane. In the first case, pure water enter into the egg due to osmosis and hence the egg swells. On the other hand, the water comes out of the egg kept in saturated sodium chloride so it shrinks. This is because osmosis occurs from higher concentration of solvent to lower concentration of solvent.

5.10.5 osmotic pressure: a colligative property It is evident from Van’t Hoff equation (7) that πV = nRT (for dilute solutions) Here V is the volume in which n molecules of a solute are dissolved. Thus  n p =   RT  v or

π = CRT

(9)

where C is molar concentration. According to equation (9) the osmotic pressure π of a solution at any given temperature T depends on the number of moles present per unit volume, i.e., upon the molar concentration and quite independent of the nature of the solute. Hence osmotic pressure of a solution is a colligative property.

5.10.6 relation Between osmotic pressure and Vapour pressure lowering Suppose an aqueous solution present in an inverted thistle funnel closed by a semipermeable membrane at the bottom is placed in pure water contained in a beaker. The whole apparatus is then enclosed in an evacuated bell jar which is evacuated. Due to osmosis water level in the stem of thistle funnel rises to a height h. At equilibrium the hydrostatic pressure due to the liquid column is equal to the osmotic pressure of the solution. Thus π = hrg

(10)

Where r is the density of the solution approximately equal to that of pure solvent if the solution is dilute and g is the acceleration due to gravity. Let P the vapour pressure of the solvent at its level in the outer vessel Ps be the vapour pressure of the solution at the above level in the thistle funnel. The difference between them is equal to the pressure of the vapour column height h.

5.32

Solutions

5.10.7 determination of Molecular weight or Molecular Mass from osmotic pressure

ps

h

According to Van’t Hoff, dilute solutions behave like gases and obey all gas laws, including the following general equation.

VAPOUR

πV = nRT p SOLUTION SEMIPERMEABLE MEMBRANE SOLVENT

fig 5.27 Vapour pressure lowering and osmotic pressure That is P–Ps = hdg

(11)

Where π = osmotic pressure, V = volume in litres containing one gm mole of the electrolyte, R = solution constant, i.e., 0.0821 litre atmospheric and T = absolute temperature. Suppose w gms of the solute be dissolved in a litre of solution so as to get a sufficiently dilute solution and its osmotic pressure be measured at a known temperature by a suitable method. Let m be the molecular weight of the solute. So the amount present in V litres must be one gm mole (molecular weight in gms). Thus we have n=w/m. Putting this value in above equation we have

Where d is the density of the solvent vapour. Putting the value of h from (10) in (11), we have

P − Ps =

p d × dg = p rg r

or

Let M be the molecular weight of solvent vapour and V the volume occupied by one gm mole of the solvent vapour, then RT PV = RT or V = P But the density of the vapour

Substituting the value of d in equation (12), we have  p   MP  (P − Ps ) =   .    r   RT  or

P − Ps Mp = P RTr

(13)

(P − Ps ) = constant × p P (P − Ps ) ∝p P

w RT m

(16)

m=

w RT p

(17)

From this equation molecular weight m can be calculated. Here it should be noted that W gms of solute are present in one litre. ∴ m gms of the solute will be present in V=

(14) Thus relative lowering of vapour pressure is proportional to the osmotic pressure.

w litres m

m w

Substituting the value of V in equation πV = RT (For one mole, i.e., n = 1 in equation 15) We have p ×

For a given solvent at a given temperature, M, r and T are constant; thus equation (9) becomes

or

pV =

or

M M MP = = d= V (RT /r) RT

(15)

or

m = RT w m=

Or Molecular weight m =

R×T×W p

(18)

0.0821 × T × W  R = 0.0821 p

Thus knowing the absolute temperature T, weight of the solute (w) and osmotic pressure of the solutions (π) the molecular weight of the solute (m) can be calculated. Both equations (17) and (18) have been utilized in solving the problems based on molecular weight determination.

Solutions

5.10.8 usefulness and limitations of Van’t hoff ’s Theory of dilute solutions Usefulness: Van’t Hoff has successfully explained the behavior of dilute solutions. The equation πV = RT can be used for the determination of molecular weight of non electrolyte and non-volatile substances. The molecular weight of haemoglobin was first determined by osmotic pressure measurements. Limitations: The general equation for dilute solutions πV = RT holds good for dilute solutions of non-electrolytes at low temperatures only. Certain type of deviation occurs when (i) Solute is an electrolyte (due to dissociation). (ii) Concentration of solute is high. (iii) Temperature is high (above 25°C).

5.10.9 osmotic pressure of Mixture of Two solutions Case I: If two solutions of same substance having different osmotic pressure π1 and π2 are mixed, osmotic pressure of the resultant solutions can be calculated as π1V1 + π2V2 = πR(V1+V2) Where V1 and V2 are the volumes of two solutions and πR is the resultant osmotic pressure. Case II: If n1 and n2 are the number of moles of two different solutes present in V1 and V2 volumes respectively, the osmotic pressure of the mixture can be calculated as n1 i1 RT n 2 i 2 RT p = p1 + p 2 = + (V1 + V2 ) (V1 + V2 ) p=

(n1 i1 + n 2 i 2 ) RT (V1 + V2 )

Here i1 and i2 are Van’t Hoff factor for two solutes.

5.10.10 reverse osmosis Suppose a solution is separated from a pure solvent through semipermeable membrane. If external pressure applied on the solution is more than the osmotic pressure, the phenomenon of reverse osmosis will take place and the solvent will start following from the solution to the pure solvent. Reverse osmosis is used in desalination of sea water. A schematic set up for the process is shown in Fig 5.28. A tube fitted with a semipermeable membrane at one end is lowered deep into the sea. The fresh water will start following into the tube, if hydrostatic pressure due to the column of sea water outside the tube is greater than the osmotic pressure.

5.33

Piston Pressure > ̟

Salt water

Fresh water water outlet

SPM

fig 5.28 Reverse osmosis occurs by applying more pressure solutions than its osmotic pressure.

A variety of polymer membranes are available for this purpose. The pressure required for the reverse osmosis is quite high. A workable porous membrane is a film of cellulose acetate is permeable to water but impermeable to impurities and ions present in sea water. Now a days many countries are using desalination plants to meet their drinking water requirements.

5.10.11 silicate gardens When small crystals of various inorganic salts, such as copper sulphate, nickel sulphate, alum etc are added to a 5% solution of sodium silicate the so called “silicate gardens or chemical gardens” are produced because of the fact that metal ions diffuse out of a crystal and form precipitate of metal silicate. These precipitates act as semipermeable membrane round the crystal. The process of osmosis takes place and water flows from a di lute sodium silicate solution to the stronger salt solution. The membrane, therefore, bursts at its weaker points, so as to liberate more metal ions and these ions further give rise to new membrane. This process continues and as a result growths looking like gardens form around the crystal.

5.10.12 Biological importance of osmosis Plants absorb water from soil. Plants and animal bodies are made up with large number of cells. Each cell is made with living cytoplasmic walls which act as semipermeable membranes. The cells contains a fluid called cell sap. When the cell of the roots of a plant come in contact with soil water or dilute solution (hypotonic), the water enters into the cells through the cell wall due to osmosis. The pressure developed inside the cell due to inflow of water is referred to as turgor. The water thus absorbed is further transported to the other parts of the plant. If a

5.34

Solutions

cell come in contact with hypertonic solution water will diffuse out of the cell fluid and partial collapse of the cell will take place. This phenomenon is known as explasmolysis or crenation. The process of rupture or bursting of a cell by a hypotonic solution is called plasmolysis. The rupture of red blood cell when kept in hypertonic solution is called hemolysis. Salt present in mango pickles removes the water from raw mangoes and preserve from spoiling. People consuming more salt and excessive salty food suffer from edema which is swelling and puffiness produced because of retention of water in the tissue cells and intercellular space. The preservation of meat against bacterial action is done by salting it. Similarly fruits are preserved against bacteria by adding sugar. The bacteria on salted meat or candid fruits loses water through osmosis shrivels and ultimately dies. solved problem 13 Calculate the molecular weight of cellulose acetate if its 0.2% (wt/vol) solution in acetone (sp gr 0.8) shows an osmotic rise of 23.1 mm against pure acetone at 27°C. Solution: 0.2 percent solution means 0.2 g of cellulose acetate dissolved in 100mL of solution Osmotic pressure = 2.31cm of acetone P = 2.31× P=

0.80 cm of Hg = 0.136 cm of Hg 13.6

0.136 atm (1 atm = 76 cm of Hg) 76

Suppose M is the Mol wt of cellulose acetate 0.2 , V = 100 mL = 0.1 litre, R = 0.0821 Lit M atm/K/mole And T = 273 + 27 = 300 K n=

Now P = ∴

n RT V 0.136 0.2 /M = × 0.0821 × 300 76 0.1 M = 27,500

solved problem 14 At 10°C the osmotic pressure of urea solution is 500 mm. The solution is diluted and the temperature is raised to 25°C, when the osmotic pressure is found to be 105.3 m. Determine the extent of dilution.

Solution: Suppose V1 litres of the solution contains n moles of the solute at 10°C which was diluted to V2 litres at 25°C. Thus we have

And

500 n = × 0.0821 × 283 760 V1

(1)

105.3 n = × 0.0821 × 298 760 V2

(2)

Dividing (1) by (2) we get V2 =5 V1 Thus the solution was diluted 5 times. problems for practice 34. A 6.84% solution of cane-sugar was found to be isotonic with 11.8% solution of raffinose. Calculate the mole wt of raffinose (mol wt. of cane sugar = 342). 35. Calculate the amount of urea dissolved per litre if aqueous solution is isotonic with 10% cane sugar solution molecular weight of urea = 60. 36. A solution is prepared by dissolving 1.08 g of human serum albumin, a protein obtained from blood plasma in 50 cm3 of aqueous solution. The solution has an osmotic pressure of 5.85 mm Hg at 298 K. (i) What is molar mass of albumin? (ii) What is the height of water column placed in solution? d (H2O) = 1 gm cm–3 37. The osmotic pressure of blood is 8.21 atm at 37°C. How much glucose should be used per litre for an intravenous injection that is isotonic with blood. 38. Calculate the osmotic pressure of a solution obtained by mixing 100 mL of 3.4 per cent solution of urea (mol mass = 60) and 100 mL of 1.6 percent solution of cane sugar (mol mass = 342) at 293 K (R = 0.0832 bar K–1 mol–1).

5.11 eleVaTion in Boiling poinT Boiling point of a liquid may be defined as the temperature at which its vapour pressure becomes equal to the atmospheric pressure, i.e., 760 mm. We know that the vapour pressure of a liquid is lowered when a non volatile solute is added to it. Hence the vapour pressure of solution will becomes equal to that of pure solvent at the higher temperature so that it may start boiling. In other words, the boiling point of a solvent is always elevated, when a non-volatile solute is added to it.

Solutions

5.35

From (3) we have A

D

E

DP ∝

VAPOUR PRESSURE

P0

For a given solvent, M is constant. Thus

B P1

DP ∝

NT VE L O

S

N

P2

O TI

LU

SO

C

1 N

IO UT

T0

T

T2

DP ∝ DTb ∝

fig 5.29. Vapour pressure – temperature curves of solvent and solutions 1 and 2

or DTb = K b or

Fig 5.29 shows the changes in vapour pressure with temperature for pure solvent, and two solutions (1 and 2) of different concentrations. It is well evident that dissolution of non volatile substance in a solvent causes an increase in boiling point. This rise or elevation in boiling point depends upon the amount of solute dissolved in the solution. Near boiling point these curves are almost parallel for the solution is nearly equal to that for the pure solvent. This portion can therefore be regarded as having two similar triangles ABD and ACE. Then from geometry, AE AC = AD AB

T2 − Tb P − P2 = T1 − Tb P 0 − P1 0

( DT ) 2 ( DP) 2 = ( DT )1 ( DP)1

(1)

Where Tb = B.Pt of pure solvent T1 = B.Pt of solution 1 T2 = B.Pt of solution 2 P0 = Vapour pressure of pure solvent P1 = Vapour pressure of solution 1 P2 = Vapour pressure of solution 2.

∆b ∝ ∆P

or

(2)

That is, elevation in boiling point is proportional to the lowering of vapour pressure. But according to Raoult’s law DP ∝

DP ∝

(4)

w mW

TEMPERATURE °C

From the graph

w wM

From (2) and (4) we have

2

L

SO

or

Mw mW

n N

w /m w /m n (because = W /M N W /M

(3)

Here w grams of solute (molecular weight m) are dissolved in W gms of the solvent (molecular weight M).

w mW

DTb =

Kb w × 1000 × 1000 mW

DTb =

K b × w ×1000 mW

(5)

The constant, Kb is known as the boiling point constant for the solvent. When one mole of the solute is dissolved in 1000 gm of a solvent. The constant is called molal elevation constant Kb or ebullioscopic constant. ∆Tb = Kb× molality of solution

(6)

Molal elevation constant is thus the elevation in boiling point, when one mole of a non volatile solute is dissolved in 1000 gms of the solvent i.e., m = 1. The units of Kb are K kg mol–1. The value of Kb depends only upon the solvent and is independent of the nature of solute and concentration of the solution. Sometimes, the value of Kb is given for 100 g of the solvent. In such event this constant is ten times the value of the molal elevation constant. Table 5.10 Molal elevation constants of certain solvents Solvent Water Benzene CCl4 CS2 Chloroform Ethyl alcohol Ether Acetic acid Camphor Nitro benzene

B.Pt (K)

Kb (K kg mol–1)

373.0 353.3 350.0 319.4 334.4 351.5 307.8 391.3 481.3 483.8

0.52 2.53 5.03 2.34 3.63 1.20 2.02 3.07 5.95 5.24

5.36

Solutions

5.11.1 determination of Molecular weight or Molecular Mass from elevation in Boiling point Suppose the weight of the solute = w gms Weight of the solvent = W gms Molecular weight of solute = m gms The molality of the solution =

ln

w 1000 × m W

Now ln

w ×1000 mW

= When=

The vapour pressure curves of a pure solvent and a dilute solution in the boiling point range can be shown in the Fig 5.30.

VAPOUR PRESSURE

p2

P2 − P1 L v DT = × P2 R T12 P2 − P1 n = (For dilute solutions) P2 N Thus equation (10) becomes n L v ∆T = × N R T12

I

UT

ON

TEMPETATURE T1

(11)

(12)

Now let w gms of solute (molecular weight m) are dissolved in W gms of the solvent (molecular weight M), then

L SO

n w/m = N W/M

T2

fig 5.30. Change in vapour pressure with temperature for solvent and solution. At temperature T1 the vapour pressure P2 of the solvent is equal to the atmospheric pressure, hence evidently T1 is the boiling point of the pure solvent. The vapour pressure of the solution corresponding to the temperature T1 is P1 which is less than the atmospheric pressure. Hence solution will not boil at T1, but will boil at temperature T2 when the pressure of the solution becomes equal to P2. Thus Clausius Clapeyron equation may be represented as

P2 L v  T2 − T1  (8) =   P1 R  T1 T2  This equation can be applied to solutions. From Fig 5.30, it is evident that T2–T1 represents the elevation ln

2  n  RT ∆T =   1  N  Lv

or

S

(10)

But according to Raoult’s law,

NT

VE OL

P2 − P1 P2

P2 − P1 is small the equation (9) becomes P2

5.11.2 Thermodynamic derivation

p1

(9)

 (P − P )  P2 P = − ln 1 = − ln 1 − 2 1  P1 P2 P2  

(7)

From this relation, the molecular weight or molecular mass of the solute can be calculated.

ATMOSPHERE

P2 L v DT = × P1 R T12

In this equation Lv is latent heat of vaporization of one mole of the solvent and so known as molar heat of vaporization. T1 is the boiling point of pure solvent.

Putting it in equation (6) we have DTb = K b

in boiling point ∆T. For very dilute solutions T1 is not for apart from T2 and equation (8) may be reduced to the form (without any serious error)

Hence equation (12) becomes DT =

w × M RT12 RT12 w × = m × W Lv l v Wm,

(13)

Where lv (= Lv/M) is the heat of vaporization per gm of the solvent (i.e., w = 1000 gm) the equation (13) becomes ∆T =

RT12 l v × 1000

(14)

Now ∆T represents the molal elevation constant Kb of the solvent. Hence ∆T = K b =

0.002 × T12 RT12 = ( R ≈ 2) l v × 1000 lv

(15)

Solutions

5.11.3 Boiling point elevation – a colligative property Since ∆Tb ∝

n N

(16)

That is boiling point elevation ∆Tb in the case of a dilute solution is directly proportional to the number of moles of the solute dissolved in a given amount of the solvent and does not depend upon the nature of the solute. Hence equimolar amounts of different solutes (e.g., 6 gms of urea or 18 gms of glucose or 34.2 gms of sucrose) when dissolved in the same amount of solvent (water) raise the boiling point to the same extent. This is because of the fact that 6 gms of urea are equivalent to 1/10th of molecular weight of urea which is 60. Similarly 18 gm of glucose are 1/10th of molecular weight of glucose (180) and 34.2 gms of sucrose are 1/10th of molecular weight of sucrose (342). Hence the boiling point elevation is another important colligative property.

5.37

This method eleminates fluctuations of temperature as well as error arising due to super heating. The apparatus consists of a graduated boiling tube, which contains the liquid under examination. An inverted funnel tube placed in the boiling tube collects the bubbles rising from a few fragments of porous pot in the liquid. As the solvent boils, the bubbles of vapour, which form under the funnel rise up in the tube carry with them liquid, which is projected against the stem of the thermometer, held a little above the liquid. In this way, the bulb is covered with a thin layer of boiling liquid. The constant temperature is soon obtained which gives the boiling point of the pure solvent a weighted amount of the solute is added and another reading is taken. The volume of the solution is noted and the mass of solvent is calculated from the density of the solvent assuming the volume occupied by the solute to be negligible.

5.11.4 relation Between elevation of B.p. and relative lowering of Vapour pressure We know that ∆Tb =

1000 × K b × w m× W

We also know that number of moles of solute weight in gm w = = n= Mol.wt m W Similarly number of moles of solvent N = M Or W = N.M  w Putting the values of n  =  and W (=N.M) in the  m above equation, we get 1000 × K b n ∆Tb = × M N Now according to Raoult’s law, we know that P − Ps n = P N 1000 × K b P − Ps (17) Hence ∆Tb = × M P This is the required relation.

5.11.5 determination of Boiling point elevation Though there are certain methods for the determination of, boiling point elevation the better method, Cottrell’s method, is given here.

INVERTED FUNNEL TURE

A

FRAGMENTS OF POROUS POT

fig 5.31. Cottrell's boiling point apparatus

solved problem 15 The latent heat of vaporization of carbon disulphide is 86 cal and its boiling point is 46°C. A solution of 1.5 gm of benzoic acid in 60 gms of carbon disulphide boils at 47.5°C. Calculate apparent mol. wt of benzoic acid. Solution: T = 273 + 46°C = 319 K. Let Kb be the molal elevation constant. Then Kb =

0.00 2T 2 0.002 × 319 × 319 = = 2.366 Lv 86

Solutions

Now let m be the apparent mol. wt. of benzoic acid ∆T= 47.5–46 = 1.5°C, w = 1.5 gm, W = 60 gm, Kb = 2.366 Substituting these values in equation, 1000 × K b × w 1000 × 2.366 × 1.5 m= = = 39.4 DT × W 1.5 × 60 problems for practice 38. If 30 g of a solute of molecular weight 154 is dissolved in 250 g of benzene, what will be the boiling point of the resulting solution under atmospheric pressure. The molal boiling point elevation constant for benzene is 2.61°C m–1 and the boiling point of pure benzene is 80.1°C. 39. A solution containing 0.5216 g of naphthalene (mole. wt = 128.16) in 50 mL of CCl4 shows boiling point elevation of 0.402° while a solution of 0.6216 g of an unknown solute in the same weight of solvent gave a boiling – point elevation of 0.647°. Find the molecular mass of the unknown solute. 40. The b.Pt of a solution of 5 g of sulphur in 100 g of carbon disulphide is 0.476 K above the pure solvent. Calculate the molecular formula of sulphur in this solvent. The b. pt. of pure carbon disulphide is 319.3 K and molal elevation constant for CS2 is 2.44 K kg mol–1. 41. Molal elevation constant for benzene is 2.52 K/m. A solution of some organic substance in benzene boils at 0.126°C higher than benzene. What is the molality of the solution?

NT

VAPOUR PRESSURE

5.38

S

A

0

p

IO UT

N1

L

SO

P p1

D

B C

p2

N IO

2

T LU SO

E

T2 T1 Tt TEMPERATURE

fig 5.32 Vapour pressure curves of the pure solvent and solutions 1 and 2 In the Fig 5.32 points A, B and C represents the freezing points of pure solvent and the two solutions (1 and 2) respectively. It is also evident that the freezing point of the solution is lower than that of the solvent and that the freezing point of the solution decreases as the concentration of the solute increases. In the case of very dilute solutions, the curves AC, BD and CE would almost be linear and hence the triangles ABD and ACE are similar. Hence, AD BD = AE CE

(1)

From the graph, P0 − P1 T T = f− 1 P0 − P2 Tf − T2

5.12 depression of freezing poinT Freezing point is the temperature at what both solid and liquid states of a substance have the same vapour pressure. It is a well known fact that the vapour pressure of a solvent is lowered as a result of the addition of a non volatile solute. In otherwords, freezing point of the solution is always lower than that of the pure solvent. Hence the freezing point of the solvent is lowered by the addition of a non volatile solute to it. Fig 5.32 represents the vapour pressure curves of the pure solvent, solutions of different concentration (1 and 2) and of the solid solvent. The intersection of vapour pressure curve of the solution with the vapour pressure curve of the pure solid solvent is the freezing point of the solution and should be the temperature at which the first crystals of the solvent appear as a result of cooling the solution.

VE OL

or

∆P1 (∆Tf )1 = ∆P2 (∆Tf ) 2

∴ ∆P ∝ ∆Tf But from Raoult’s law n  n DP ∝ ∴ P0 − Ps ∝  N  N

or ∆P ∝

(2) (3) (4)

w/m W/M

wM mW Where w gms of the non-volatile solute (molecular weight m) are dissolved in W gms of pure solvent (molecular weight M). For a dilute solution and for a given solvent M is constant. Hence, or ∆P ∝

Solutions

∆P ∝ ∆Τf ∝

w mW

(5)

Table 5.11 Molal freezing point depression constant (Kf) for some solvents Solvent

or ∆Tf = K f

w mW

(6)

The constant, Kf is known as freezing point constant for the solvent. When one mole of the solute is dissolved in 1000 gms of a solvent, this constant is “known as molal depression constant or molal cryoscopic constant and defined as the depression in freezing point which may theoretically be produced by dissolving one mole of the solute in 1000 gms of the solvent. Therefore DTf = K f

5.39

Freezing Point K

Kf (K m–1)

273 155.7 278.6 209.6 250.5 164.2 156.9 279.5 290.0

1.86 1.99 5.12 4.70 31.8 3.83 1.79 20.0 3.9

Water Ethanol Benzene Chloroform Carbon terrachloride Carbon disulphide Ether Cyclohexane Acetic acid

K w × 1000 w = f . mW 1000 mW

w ×1000 ∆Tf = K f ⋅ mW = Kf × molality of solution

(7) (8)

5.12.1 determination of Molecular weight or Molecular Mass of a solute from depression in f.pt Let the weight of the solute = w gms Weight of the solvent = W gms Molecular weight of the solute = m ∴ Molality of the solution =

w 100 × m W

5.12.2 relation Between depression in f.pt and lowering of Vapour pressure The relation between these quantities is exactly similar to that derived in the case of elevation of boiling point (5.11.4) because expression for depression in F.P is similar to that of elevation in boiling point. Thus

∆Tf =

1000 × K f P − Ps × M P

5.12.3 relation Between depression in f.pt and osmotic pressure This relation is also similar to that derived in case of elevation of boiling point

Putting this value in equation (8) ∆Tf = K f ×

or m =

w 1000 × m W

1000. K f . w DTf W

From this equation, the molecular weight or molecular mass of the solute can be determined. The molal depression constant may also be derived thermodynamically using Clausius Clapeyron equation  to the liquid ↽ ⇀  vapour equilibrium in solution as derived for molal elevation constant (5.11.2). The thermodynamic relationship for calculating Kf of a particular solvent is Kf =

0.02 Tf2 Lf

∆Tf =

1000 × K f π × d RT

5.12.4 freezing point depression – a coligative property It is evident from the equation (4) that

∆Tf ∝

n N

In other words, depression in freezing point of dilute solution is directly proportional to the number of moles of the solute dissolved in a given amount of the solvent and quite independent of the nature of the solute. Hence freezing point depression is a colligative property.

5.40

Solutions

5.12.5 determination of freezing point depression Rast Method: Rast (1922) has devised a simple method for determining the molecular weights of solutes soluble in camphor which has a very high molecular depression. The depression in freezing point of one mole of a solute dissolved in 1000 gm of camphor has been found to be 37.7°C where as the same amount of solute when dissolved in 1000 gms of water gives the freezing point depression of only 1.8%. In other words, the depression is so great in this case that it can easily be determined by making use of ordinary thermometer. A small amount of powdered camphor acting as solvent is first introduced into a thin walled capillary tube and its melting point is determined in the usual way. A known amount solute (few milligrams) whose molecular weight is to be determined is then mixed to a known amount of camphor (10 to 15 times that of solute) and the two are mixed intimately by fusing. After cooling the melt is powdered and a small amount is taken in a capillary tube and the melting point is determined as before. This gives the freezing point of the solution of the substance in camphor. The molecular weight of the substance can then be calculated by the depression of freezing point or melting point so obtained.

5.12.6 application of depression of freezing point (a) Eutectics: As discussed so far we know that freezing point of a liquid is depressed if it contains another substance dissolved in it. The mixture with the lowest melting point is called eutectic mixture. A liquid with the eutectic composition freezes at a single temperature, without previously depositing solid. A solid with the eutectic composition melts, without change of composition, at the lowest temperature of any mixture. One technological important eutectic is solder, which has mass composition of about 67 percent tin and 33 percent lead and melts at 183°C. The eutectic formed by 23 percent NaCl and 77 percent water by mass melts at –21.1°C. Such mixtures are called freezing mixtures. (b) Antifreeze solutions: Water is used in radiators of vehicles. If the vehicle is to be used in places where the temperature is less than zero then water would freeze in radiators. To avoid this problem, certain substances are used in radiators so that water does not freeze at low temperature in radiators. These are called antifreeze solution. Ethylene glycol in water is commonly, used in car radiators which lowers the freezing point of water. Freezing point can be lowered to the desired extent by changing the amount of ethylene glycol.

(c) Clearing of ice from roads in the hills: Salts such as NaCl or CaCl2 are scattered on icy roads at higher altitudes. This helps in melting the ice as long as the outdoor temperature is above the lowest freezing point of salt water mixture. NaCl can melt ice at temperature as low as –21°C. CaCl2 is effective in melting the ice at a temperature as low as –55°C. solved problem 16 An aqueous solution containing 2 gms of a solute dissolved in 100 gm of water freezes at –0.5°C. Calculate the mol. wt of solute. Molar heat of fusion of ice at 0°C is 1.44 cal and R = 2 cal. Solution: Molar heat of fusion = Lf = 1.44 cal ∴ Heat of fusion of ice/gm (Lf) = Now K f = m=

1.44 = 80 cal. 18

RTf 2 2 × 273 × 273 = 1.86° = 1000 L f 1000 × 80

K f × 1000 × w 1.86 × 1000 × 2 = = 74.4 DTf × w 0.5 × 1000

solved problem 17 An aqueous solution contains 6% and 12% of urea and glucose (by mass) respectively. Find freezing point of the solution kf for water is 1.86 K kg mol–1. (Molecular mass of urea = 60, glucose = 180) Solution: We know ∆Tf = Kf ×m = Kf × no. of mole of solute per kg of solvent 1 1000  6 12  1000 = 1.86 ×  + × = 1.86 × ×  6 82  60 180  82 = 3.78 K Freezing point of solution T1 = T0–∆Tf = 273–3.78 = 269.22 K problems for practice 42 What weight of glycerol would have to be added in 500 g of water in order to lower its freezing point by 7.5°C if Kf for water is 1.86 K kg mol–1? 43. To elements A and B form compounds having molecular formulae AB2 and AB4 when dissolved in 20.0 g of benzene 1.0 g of AB2 lowers the freezing point by 2.3°C. Where as 1.0 g of AB4 lowers the freezing point

Solutions

44.

45.

46.

47.

48.

by 1.3°C. The molal depression constant for benzene in 1000 g is 5.1 Calculate the atomic masses of A and B. Ethylene glycol CH2OH.CH2OH, the major component of permanent antifreeze, effectively depresses the freezing point of water in an automobile radiator. What minimum weight of ethylene glycol must be mixed with 6 gallons of water to protect it from freezing at –24°C. (1 gallon = 3.785 lit Kf = 1.86). An aqueous solution of urea had a freezing point of –0.52°C predict the osmotic pressure of the same solution at 37°C. Assume that the molar concentration and the molality are numerically equal (Kf = 1.86). A solution of urea in water has a boiling point of 100 18°C. Calculate the freezing point of the same solution Molal constants for water Kf and Kb are 1.86 and 0.512 respectively. 1000 gm of 1 m aqueous solution of sucrose is cooled and maintained at –3.534°C. Find how much ice will separate out kf (H2O) = 1.86 K kg mol–1. Kf for water is 1.86°C m–1. What is the molality of the solution which freezes at –0.192°C? Assuming no change in the solute by raising the temperature at what temperature will the solution boil (Kb for H2O = 0.515°C m–1).

5.41

associate to form a bigger molecule. In such case the number of solute particles decreases and consequently the colligative properties, such as osmotic pressure, depression in freezing point, elevation in boiling point are much smaller than the calculated on the basis of single molecule. The observed molecular weight in such case is much higher than the normal molecular weight. The molecular weight of acetic acid (CH3COOH), for example, in benzene as determined by freezing point depression has been found to be 118 instead of 60. It is clear from this example that CH3 COOH exists largely as (CH3 COOH)2 in benzene. The elevation of boiling point or the depression in freezing point of a liquid caused by dissolving a solute in it is proportional to the lowering of vapour pressure and thus to the osmotic pressure of the solution. But osmotic pressure of a solution depends solely on the number of particles of the solute in a given volume of the solution. Thus, ∆T ∝ osmotic pressure π ∝ number of solute particles 1000 Kw 1 where m = But molecular wt. m ∝ DT W ∆T Molecular weight ∝

1 No. of particles of the solute

or molecular mass.

5.13 aBnorMal Molecular weighTs

5.13.3 abnormal Molecular weights

Since colligative properties does not depend upon the weight but solely upon the number of solute particles associated with a given quantity of solvent, in some cases, where the solutes undergo association or dissociation, abnormal results are obtained.

In the case of association, the number of effective particles decreases, while in case of dissociation, the number of these particles increases. This causes the molecular weight to be abnormal. In the case of dissociation, the observed molecular weight is lesser while in case of association, the observed molecular weight is higher than the normal molecular weight.

5.13.1 dissociation Soluble salts, acids and bases have a tendency to ionize in solution, i.e., the molecules break down into positively and negatively charged ions. One normal molecule gives rise to two or more ions as a result the number of soluble particles increase than the normal number. For example, sodium chloride in aqueous solution exist almost entirely as Na+ and Cl– ions. In such cases, because the number of effective particles increases, the colligative properties such as osmotic pressure, depression in freezing point, elevation in boiling point, are much higher than those calculated on the basis of undissociated single molecule.

5.13.2 association There are many organic solutes which undergo association in non aqueous solution i.e., two or more molecules

Observed Mol. wt Normal M ol. wt Normal no. of solute particles = No. of solute particles after dissociation (or) associaation Van’t hoff factor In order to explain such abnormal cases Van’t Hoff introduced a factor ‘i’ known as Van’t Hoff factor. It is defined as i= or

Observed osomotic pressure Calculated osmotic pressure

in general, i=

Observed colligative property Calculated colligative propperty

5.42

Solutions

Since colligative properties vary inversely as the molecular mass or molecular weight of the solute, it follows i=

Calculated (normal) Mol. wt or molecular mass Observed Moll. wt or molecular mass

Degree of Association: Degree of association may be defined as the fraction of the total number of molecules which combine to form bigger molecules. Suppose if one mole of a solute dissolved in a given volume of a solvent two molecules combine to form a complex molecule and let x be the degree of association.  ⇀  (B)2 2B ↽ The number of unassociated molecules = 1–x Number of associated molecules = x/2 Total number of effective molecules x x = 1− x +   = 1−   2 2   Observed osmotic effect 1 − ( x 2) ∴i = = Calculated osmotic effect 1 and i =

1 − ( x 2) Normal Mol. wt = Observed Mol. wt 1

If n molecules associate to form complex molecule, then total number of effective molecules = 1–x+(x/n) x 1− x +   n ∴i = 1 Thus the degree of association x and Van’t Hoff factor i can be calculated if the molecular weight, and n, the number of simple molecules which combine to give one associated molecule are known. Degree of Dissociation: The degree of dissociation may be defined as the fraction of total number of molecules of electrolyte present as ions in the solution. Suppose one gm mole of NaCl is dissolved in water and x be the degree of dissociation. Then  NaCl ↽ ⇀  Na+ + Cl– 1–x x x Total number of effective particles = 1–x + x + x = 1+x Hence i =

or

i=

Observed osmotic effect 1 + x = Normal osmatic effect 1 Normal molecular weight 1 + x = Observed molecular weight 1

in case of chemical reaction If a chemical reaction occurs between the species present in a solution the Van’t Hoff factor is not applicable. For example if iodine is added to a solution of potassium iodide solution the iodine combines with potassium iodide forming potassium triiodide. KI + I2 →KI3 If x moles of I2 is added to one litre of 0.1 molar solution of potassium iodide, then KI I

+ −

0.1

I 2 K → I3

+ I 2 • → I3− x

0

0.1 − y x − y

(

Initial concentration

Y Final concentration

∴ DTf = K f m K + + m I− + m I2 + m I− 3

)

DTf = K f (0.1) + (0.1 − y ) + ( x − y ) + y 

By using this equation the ∆Tf can be calculated. solved problem 18 Phenol (C6H5OH) associates in water to double molecules. A solution of 1.25 gms of phenol in 50 gms of water lowered the freezing point by 0.3°C. Calculate the degree of association of phenol. (Kf for water = 1.86° mole) Solution: ∆T = 0.3°C, Kf = 1.86, W = 50 gms Hence observed mol. wt of phenol m=

1000 × 1.86 × 1.25 = 15.5 0.3 × 50

Normal (Calculated) molecular weight of phenol from formula C6H5OH = 94 Since phenol associates to give biomolecules in solution, let x be the degree of association of phenol.  ⇀  (C6H5OH)2 2C6H5OH ↽ 1 0 1–x x/2 No. of particles before association = 1 No. of particles after association = 1–x + (x/2) = 1–(x/2) Since Normal Mol. wt No. of Particles after association = Observed Mol. wt No. of Particles before association 94 1 − ( x 2 ) 155 or x = = 0.824 = 155 1 94 × 2 Hence degree of association of phenol = 82.4%

Solutions

solved problem 19

problems for practice

The boiling point of a solution of 5 gms of BaCl2 (mol wt. 208) in 100 gms of water is 100. 3°C. Calculate the degree of dissociation, if the molal elevation constant for water is 0.52. Solution: Let m be the observed mol. wt. of BaCl2 ∆T = 100.3–100 = 0.3°C, w = 5 gms, W = 100 gms Kf = 0.52 m=

1000 × K b × w 1000 × 0.52 × 5 = = 86.6 DT × W 0.3 × 1000

Now, let x be the degree of dissociation of BaCl2 BaCl2 = Ba2+ + Cl– + Cl– 1 0 0 0 (before dissociation) 1–x x x x (after dissociation) No. of particles after dissociation. = 1–x + x + x + x = 1 + 2x Now, Normal mol wt No. of particles after dissociation = Observed mol wt No. of particles before dissociation 208 1 + 2 x or x = 0.70 = 86.6 1 Hence degree of dissociation of BaCl2 = 70% solved problem 20 The osmotic pressure of decinormal solution of KCl was found to be 4.6 atms at 20°C. Calculate the degree of dissociation of KCl. Solution: We know πv = RT V = 10 litres, R = 0.0821 lit atm and T = 20+273 = 293 K Thus calculated osmotic pressure π is given by π=

RT 0.0821× 293 = = 2.40 atmospheres V 10

But the observed osmotic pressure is 4.60 atmospheres Now suppose the degree of dissociation of KCl = x So KCl ⇌ K+ + Cl– 1 0 0 (before dissociation) 1–x x x (after dissociation) No. of particles after dissociation = 1–x + x + x = 1+ x Now

5.43

Observed OP No. of particles after dissociation = Calculated OP No. of particles before dissociation or

4.60 1 + x = 2.40 1

or x = 0.916 or degree of dissociation is 91.6%

49. How much amount of NaCl should be added to 100 gms of water so that its freezing point is depressed by 2 K. For water Kf = 1.86 K kg mol–1. 50. The degree of dissociation of Ca(NO3)2 in a dilute solution containing 16 g of the salt per 200 g of water at 100°C is 70%. If the vapour pressure of water is 760 mm, calculate the vapour pressure of solution. (IIT 1991) 51. Calculate the normal – boiling point of a sample of sea water found to contain 3.5% of NaCl and 0.13% MgCl2 by mass. The normal boiling point of water is 100°C and Kf (water) = 0.51 K kg mol–1. Assume that both the salts are completely ionized. 52. Calculate the boiling point of a solution containing 0.61 g of benzoic acid in 50 g of CS2 (l) assuming 84% dimerisation of the acid. The boiling point and Kb of CS2 are 46.2°C and 2.3 K kg mol–1 respectively. 53. x g of a non electrolytic compound (molar mass = 200) is dissolved in 1.0 litre of 0.05 M NaCl solution. The osmotic pressure of this solution is found to be 4.92 atm at 27°C. Calculate the value of ‘x’. Assume complete dissociation of NaCl and ideal behaviour of this solution. (Rookie 1998) 54. A 1.2% solution of NaCl is isotonic with 7.2% solution of glucose. Calculate the Van’t Hoff factor of NaCl. (M.L.N.R 1997) 55. 1.4 g of acetone dissolved in 100 g of benzene gave a solution which freezes at 277.12 K. Pure benzene freezes at 278.4. 2.8 g of solid (A) dissolved in 100 g of benzene gave a solution which froze at 277.76 K. Calculate the molecular mass of (A). (Roorkee 2000) 56. To 500 cm3 of water 3.0×10–3 kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? Kf and density of water are 1.86 K kg mol–1 and 0.997 g cm–3 respectively. (IIT 2000) 57. A sample of a drug C21H23O5N (Mol wt = 369) was mixed with lactose C12H22O11 (Mol wt 342) was analyzed by osmotic pressure measurement to determine the amount of sugar present. If 100 mL of solution containing 1.0 g of the drug – sugar mixture has an osmotic pressure of 5.27 mm Hg at 25°C. What is the percent sugar present? 58. A 0.025 M solution of monobasic acid had a freezing point of –0.06°C. Calculate Ka for the acid Kf (H2O) = 1.86. 59. 1.1 g of CoCl3. 6NH3 (mol wt =267) was dissolved in 100 g of H2O. The freezing point of the solution was –0.29°C. How many moles of solute particles exist

5.44

Solutions

in solution for each mole of solute introduced? Kf for H2O = 1.80°C? 60. The freezing point of an aqueous solution of KCN containing 0.1892 mole kg–1 was –0.704°C. On adding 0.095 mole of Hg (CN)2 the freezing point of the solution was –0.53°C. Assuming the complex is formed according to the equation Hg (CN)2 + m CN– → Hg (CN) mm +− 2 and also Hg (CN)2 is the limiting reactant, calculate m. 61. Mercuric iodide is insoluble in water but dissolves in potassium iodide solution to form complex anion. A solution of 3.32 grams of potassium iodide in 100 grams of water begins to freeze at –0.745°C. A saturated solution of mercuric iodide in the above potassium iodide solution contains 4.54 grams of mercuric

iodide per 100 grams of water and it begins to freeze at –0.558°C. Give reasoned deduction of the formula of the complex anion from these data. Mention any simplifying assumptions which you make (K = 39.0, I = 127, Hg = 200.6. 62. 0.148 gm of a weak acid titrated with N/10 solution of sodium hydroxide, required 20.0 cc of the alkali for neutralization 1.0 gm of the same acid, dissolved in 100 gm of water depressed the freezing point by 0.5°C. 1.00 gm of the acid dissolved in 100 gm of benzene depressed the freezing point by 0.35°C (The freezing point constants for water and benzene per 1000 gm are 1.85° and 5.12° respectively. What deductions can be made from the above information?

Solutions

5.45

Key poinTs •

• • •

• • •

• • • •

•

A homogeneous mixture of two or more substances whose composition can be varied with certain units is known as true solution e.g., salt in water, sugar in water, air, alloys etc. Homogeneous means the composition is same through out. A solution containing only two components is called binary solution. The component present in greater amount than any of all the other components in a solution is called solvent. The other components present in small in a solution are called solutes. Formation of a solution is a physical change not a chemical change. The solution containing non-volatile solute in vola tile solvent can be separated by evaporation or distillation. Depending on the nature of the solvent the solutions are different types. If water is solvent the solution is called aqueous solution. If alcohol is solvent the solution is called alcoholic solution. The solutions in which solvent is other than water are called as non-aqueous solution. CHCl3, CCl4, C6H6, ethers etc are used as non aqueous solvents. A solution in which the amount of the solute is very low is called as dilute solution.

•

•

Methods for expressing the concentration of a solution •

•

•

Types of solutions • •

•

•

•

Solutions are three types basing on the physical state of solvent. Gaseous solutions are those in which solvent is gas, solute may be gas or liquid or solid. e.g., (i) Gas in gas: air, mixture of O2 and N2 (ii) liquid in gas moisture in air (iii) solid in gas : camphor in air. Liquid solutions are those in which solvent is liquid while solute may be gas or liquid or solid. e.g., (i) Gas in liquid: aerated water, soda water (ii) liquid in liquid: alcohol in water (iii) solid in liquid: salt in water, sugar in water. Solid solutions are those in which solvent is solid and solute may be gas or liquid or solid e.g., (i) Gas in liquid: Occlusion of H2 in palladium (ii) Liquid in solid: amalgams (liquid Hg in Zn) (iii) solid in solid. Alloys like brass, bronze etc. Alloys are solid solutions.

The extent to which one material (solid) will dissolve in another (solvent) is governed by the following factors. (a) The attractive forces between the solute and solvent molecules compared with the attractions that the solute (or solvent) molecules have for each other. (b) Solvation of the solute i.e., the attachment of molecules of the solvent to solute molecules or ions, either by electrical attraction or chemical bonding. It is more difficult for a solid to dissolve in a liquid than for one liquid to dissolve in (i.e., mix with) another liquid because the much stronger forces holding the molecules together in the solid, have to overcome before the molecules can pass into the liquid phase and mix with the molecules of the solvent.

Concentration of a solution may be defined as the amount of solute present in the given quantity of the solution. Any solution of which, the strength is known is called as standard solution. If the solubility of a substance is known its saturated solution is thus a standard solution. Mass Percentage of a component in a given solution is the mass of the component per 100 g of the solution. Mass percent of a component =

or = •

Mass of solute Volume of solution × D ensity of solution

Volume Percentage is the volume of the component per 100 parts of the solution. Volume percent of a component =

•

Mass of the component in the solution ×100 Total mass of the sollution

Volume of the component × 100 Total volume of solution

Percent mass by volume (W/N) is the mass of solute dissolved in 100 mL of the solution. Percent of solute mass by volume =

Mass of solute × 100 Volume of solution

5.46

•

Solutions

Parts per million (ppm) is the parts of component per million parts of the solution. This method is used to express the concentration of trace amounts in a solution.

•

V1 M1 V2 M 2 = n1 n2

Mass of solute ppm = × 106 Mass of solution •

•

• •

•

Mole fraction is the ratio of the number of moles of one component to the total number of moles (solute and solvent) in a binary solution. n2 Where n1 and Mole fraction of a solute χ A = n1 + n 2 n2 are the number of moles of solvent and respectively. W No. of mole of a substance = M Mole fraction of a substance A in a solvent is given by ( w A / MA ) χ= ( w A / MA ) + ( w B / MB ) For a dilute solution the number of moles of solute in the denominator can be neglected. Then χA =

• • •

•

• •

•

where V1, M1, and n1 are volume, morality and number of moles of one substance, while V2, M2, and n2 are volume molority and number of moles of other substance participating in the reaction respectively. If density (d) and per cent by weight of a substance in a solution are known, then the morality M = d × per cent ×

• • •

w A MB × M A WB

In a solution, the sum of mole fraction of all components = 1 Molarity (M) is the number of moles of solute present in one litre of solution Number of moles of solute Molarity M = Volume of solution in litres

• •

•

10 M. wt

No. of moles present in a solution = Molarity × vol. in litres. No. of mille moles present in a solution = Molarity × vol in cc Molarity of a mixture of solutions of different concentrations of some substance can be calculated by M=

V1 M1 + V2 M 2 + ..........Vn M n Total Volume

Units of molarity are moles litre–1. Molarity is temperature dependent. Normality is the number of gram equivalents of solute present in one litre of solution, represented by N Normality N =

No. of gram equivalents of solute Vol. of solution in litrres

or

M=

Weight of the solute ( W ) Mol. wt (MW ) × Volume in litres (V)

or N =

or

M =

W W × 1000 or Mw × V in lit Mw × V in cc

Weight of the solute W Equivalent Weight V (EW ) × ol. in liters (V)

or N =

W W × 1000 or N = EW × V in litre EW × V in cc

Solutions having morality equal to 1 M, 0.1 M and 0.01 M are called molar, decimolar and centimolar solutions respectively. Weight of a substance present in 1 litre = Morality × M. wt. If the solution of M1 of volume V1 is diluted to volume V2 its molarity M2 can be calculated by

•

• •

V1 M1 = V2 M2 •

When the solution of two different substances react together then.

The volume to be added to a dilute solution of volume V1 of morality M1 to a concentration of M2 is given by Vol of water to be added = Volume after dilution (V2) – Volume before dilution (V1) Volume of water to be added =

V1M1 − V1 M2

Solutions having normality equal to 1 N, 0.1 N and 0.01 N are called normal, decinormal and centinormal solutions. Wt. of the substance present in 1 litre = Normality × E.W. Wt of the substance present in a given solution = Vol. in litre × Normality × Equivalent weight or W = VNE

•

•

If the solution of N1 of volume V1 is diluted to a volume V2 then its normality N2 can be calculated by V1N1 = V2N2. The volume of water to be added to a dilute solution of volume V1 of normality N1 to a concentration N2 is given by

Solutions

Volume of water to be added = Vol. after dilution–Vol. before dilution or V2–V1 Volume of water to be added = •

•

•

•

•

V1 N1 − V1 N2

When the solution of two different substances react with each other then V1N1 = V2N2 where V1 and N1 are the volume and normality of one substance while V2 and N2 are volume and normality of other substance. Pressure does not affect the solubility of solids in liquids because solids and liquids are highly incompressible and are unaffected by changes in principle. If the value of ∆Hsol < 0; i.e., the solution process is endothermic, then by Le Chatelier’s principle, the solubility of such a solute will decrease with rise in temperature e.g., CuSO4, Na2SO4. If the value of ∆Hsol > 0; i.e., solution process is endothermic then solubility of such a solute will increase with rise in temperature e.g., KNO3, NaNO3, hydrated salts etc. If the value of ∆H sol ≈ 0 the solution process is neither exothermic nor endothermic. Solubility of such solids does not increase or decrease with change in temperature.

•

•

• •

•

•

• •

•

solubility of gases in liquids •

•

•

• •

•

•

•

The absorption coefficient of a gas is the volume of a gas measured at STP which will dissolve in the unit volume of solvent at a particular temperature and under a pressure of 1 atmosphere. The solubility of gases which react with the solvent is more but the solubility of gases which do not react with solvent (e.g., O2, N2 etc with water) is less. If gases and solvents have similar chemical characters, e.g., hydrocarbon gases in hydrocarbon liquids, the solubility of a gas is more. Solubility of gas in a liquid increases with decrease in temperature. The gases dissolved in liquid can be expelled by boiling the liquids but the gases which react with solvent cannot be expelled completely and they form azeotropic mixtures, e.g., HCl in water. Solubility of a substance is its maximum amount that can be dissolved in a specified amount of solvent at a specified temperature. Solubility of a substance depends upon the nature of solute and solvent as well as temperature and pressure. A saturated solution is a solution that is in equilibrium with an excess of solid at a given temperature.

5.47

A solution that contain a high concentration of the solution than does a saturated solution is known as super saturated solution. If a small crystal of solute is added to a super satu rated solution, the solid will at once crystallize out and equilibrium will be established. This is known as seeding. More the dielectric constant of a solvent more is the solubility of ionic compounds in it. Some covalent solids dissolve in water due to hydrolysis while some other covalent solids dissolve in water due to hydrogen bonding. Ionic compounds dissolve in polar solvents while covalent compounds dissolve in non-polar solvents and this is known as like dissolves like The solubilities of ionic solids often but not always increase with temperature, because energy is required to pull the molecules or ions away from the crystal lattice (lattice energy). Energy is also liberated due to solvation of the solute particles. The solubilities of certain solids do not vary much with temperature because the solvation energy almost balances the lattice energy e.g., Calcium citrate. Solubility of hydrated salts increases with increase in temperature since the hydration energy is small. Equivalent of a salt Formula Weight of the salt Total charge on the cation or thee anion of the salt

•

Equivalent weight of an oxidant of reductant. =

•

•

Formula Weight No. of electrons transfered in the reaction per mole

Molality is the no. of moles of solute present in 1 Kg (1000 g) of solvent. Molality is represented by m. Units of molality are mole/Kg No. of moles of solute Molality = Wt. of solvent in Kg or m =

W M.w × wt of solvent in Kg

or m =

W × 1000 M.w × wt of solvent in gms

•

Formality is the number of formula masses of the solute dissolved per litre of the solution and is represented by F

•

F=

No. of formula masses of solute Vol. o f the solution in liitre

5.48

•

•

Solutions

Temperature has no effect on molality and mole fractions since the quantities of solute and solvent are expressed in weights. Molarity and normality changes with temperature since the volume of solvent changes with temperature.

•

m ∝ p or m = kp (k is constant) •

Types of Binary solutions •

•

•

Solutions are mainly three types (i) solid solution (ii) liquid solutions and (iii) gaseous solutions. Among these the liquid solutions are mainly discussed here. Liquid solutions of mainly three type’s solutions of (i) solids in liquids (ii) liquids in liquids and (iii) gases in liquids. The weight of a substance that can react with a particular volume (V) of a solution of another substance having normality (N) and equivalent weight (E) can be calculated by W = VNE

•

• • •

If the density (d) and per cent by weight of a substance in a solution are known, then its normality can be calculated by 10 N = d × per cent × Ew No. of gram equivalents presents in a solution = Normality × vol. in litres (N × V in litres) No. of milli equivalents present in a solution = Normality × volume in cc (N × V in cc) Normality of a mixture of solutions of different concentrations of some substance N=

•

V1 N1 + V2 V2 + .......Vn N n Total volume

Relation between molarity (M) and normality (N) of any solution are related as. Molecular Weight Molarity × = Normality Equivalent Weight Molarity =

Equivalent weight of an acid Molecular wt or Formula wt of an acid = Basicity of the acid

•

Basicity of the acid is the number of replaceable hydrogens (by metal ions) in a molecule of the acid. Equivalent weight of a base

•

=

•

Mol. wt or Formula wt of an base Acidity of the base

k value depends on the nature of the gas, nature of the solvent, temperature and the units of pressure. Mole fraction of the gas in the solution is proportional to the partial pressure of the gas in the solution χ ∝ p or χ = k΄p or p = KH χ (KH = 1/K΄) KH is called Henry’s law constant

•

•

•

The pressure of a gas over a solution in which the gas is dissolved is proportional to the mole fraction of the gas dissolved in the solution. When a mixture of gases is in contact with a solvent, the solubility of each gas in the liquid is in the proportion to its own partial pressures. Henry’s law can also be stated as at a certain temperature the ratio of the molar concentrations of the gas in solution and the gas phase is constant, i.e., the volume of a gas m=

•

MPV RT

The volume of a gas dissolved in a solvent at a given temperature is independent of the pressure.

solutions of liquids in liquids •

•

•

Normality × Equivalent Weight Molecular Weight

•

Henry’s law states that the mass of gas dissolving in a given amount of liquid is directly proportional to the pressure of the gas above the liquid at equilibrium.

•

•

•

An ideal solution is one in which the attraction between components of the solution is the same as the interaction between the molecules of each component. Neither heat is absorbed nor evolved during the formation of ideal solution and the volume of the solution is equal to the sum of the volumes of the component liquids. The vapour pressure of such solutions can be calculated by averaging the properties of the liquids. The solutions in which properties of dissolved liquids are different from those of the liquids in the pure state and which are formed by evolution or absorption of heat are called non-ideal solution. Raoult’s law states that partial pressure of a component (say liquid A) in solution is proportional to the mole fraction. If all the components in solutions behave like ideal gases, then the total pressure of the solution is equal to the sum of the partial pressure of the individual components. Ptotal = c A PA0 + c B P 0 Where P0A and P0B are the vapour pressures of pure solvents A and B and χA and χB are mole fractions of the A and B in solution.

Solutions

•

The composition of vapour of an ideal solution can be determined by the partial pressures of the components. If YA and YB are the mole fractions of the components A and B in the vapour phase can be calculated using Dalton’s law of partial pressures

•

PA = YA × Ptotal PB = YB × Ptotal •

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•

•

•

•

•

•

• •

•

•

A non – ideal solution is that solution (i) which does not obey Raoult’s law (ii) for which ∆Vmix is not zero and (iii) for which ∆Hmix is not zero. In non – ideal solutions the solute – solvent interactions are weaker or stronger than the solute – solute and solvent – solvent interactions. The non – ideal solutions in which solute – sol vent interactions are weaker than solute – solute or solvent – solvent show positive deviations from Raoult’s law. The total vapour pressure of the solution which shows positive deviation is more than the corresponding vapour pressure expected for an ideal solution. The reasons for the positive deviations are (i) ∆H mixing is positive because energy is required for breaking the hydrogen bonds (ii) ∆V mixing is positive because molecules are held together less tightly. The non – ideal solutions in which solute – solvent interactions are stronger than solute – solute and solvent – solvent interactions show negative deviation from Raoult’s law. The total vapour pressure of the solution which shows negative deviation is less than the corresponding vapour pressure expected for an ideal solution. The reasons for the negative deviations are (i) ∆H mix is negative because energy is released due to increase in attractive forces (ii) ∆V mixing is negative because the molecules in solution are held together more tightly. Two liquids will mix if the ∆Gmix is negative. Some liquids mix endothermically for which the enthalpy change and entropy of mixing are working together to make ∆Gmix strongly negative. If the enthalpy and entropy changes combined make ∆Gmix negative when only a small amount of one liquid mixes with the other, the two liquids are partially miscible. Here the small increase in ∆Smix is sufficient to overcome the unfavorable ∆Hmix. By increasing the temperature it is possible for T∆Smix to overcome ∆Hmix and thus the partially miscible liquids can become completely miscible and the temperature at which the partially miscible liquids become completely miscible is known as upper consulate temperature.

•

•

•

•

• •

•

5.49

Separation of a mixture of miscible liquids by distillation methods depends on. (i) A liquid boils when its vapour pressure equals to atmospheric pressure. (ii) The higher the vapour pressure of a liquid the more volatile is the liquid. (iii) A liquid with a high vapour pressure will boil at low temperature than a liquid with lower vapour pressure. Depending upon the nature of change of vapour pressure of a solution with composition mixtures (binary solutions) have been divided into three types. In the type I solutions vapour pressure increases regularly with composition of the mixture. These solutions do not show any maximum or minimum. For these solutions the value of vapour pressure is intermediate between those of pure components. In type – II solutions the vapour pressure – composition curve shows a minimum for certain temperature and the solution will boil at this lowest temperature. In type – III solutions the vapour pressure – composition curve shows a maximum for certain temperature and the solution corresponding to the composition boils at that highest temperature. The different components in type –I solutions can be separated by fractional distillation. A mixture of liquids which like a pure component boils at a constant temperature and distills over completely at the same temperature without any change in composition is known as constant boiling mixture or azeotropic mixture. Solution of the type – II which show large positive deviations from Raoult’s law will form minimum boiling azeotrope while the solutions of the type – III which show large negative deviations from Raoult’s law will form maximum boiling azeotrope.

colligative properties of dilute solutions •

•

•

•

A solution where the amount of the solute present is very much less than that of the solvent is a dilute solution. When a non-volatile substance is dissolved in a volatile solvent the vapour pressure decreases than the pure solvent due to the decrease in the number of solvent molecules at the surface of the liquid. The decrease in the vapour pressure of a liquid when a nonvolatile solute is dissolved in it is called the lowering of vapour pressure. At any given temperature the vapour pressure of a solution is less than that of pure solvent.

5.50

•

• •

Solutions

The ratio of the lowering of vapour pressure showing (P -Ps) to the vapour pressure of the pure solvent (P) is known as the relative lowering of vapour pressure. Lowering of vapour pressure is (P – Ps) while the relative lowering of vapour pressure is (P – Ps)/P. Raoult’s law states that the relative lowering of vapour pressure of a dilute solution of a non-volatile solute is equal to the mole fraction of the solute.

•

P − Ps n = Where n and N are the moles of P n+N

•

= • •

•

solute and solvent in a solution respectively. Lowering of vapour pressure is directly proportional to the amount of non – volatile solid dissolved. Relative lowering of vapour pressure has no units but the lowering of vapour pressure has units of pressure. For dilute solutions the number of moles of solute (n) in the denominator can be neglected.

DTb =

• •

•

P − Ps w M P − Ps n = Or = × P m W P N

•

•

•

•

•

Where w and W are the weights of the solute and solvents and m and M are the molecular weights of solute and solvent respectively. Raoult’s law is applicable (i) only for dilute solution (ii) when the solution is ideal i.e., no molecular interaction exists between the solute and solvent molecules (iii) when the solute is nonvolatile and (iv) when the solute molecules do not undergo association or dissociation. The properties of dilute solutions which depend on the number of particles (ions or molecules) of the solute dissolved in the solution are called colligative properties. Lowering of vapour pressure, elevation of boiling point, depression in freezing point and osmotic pressure are colligative properties. Since the vapour pressure of a solution is less than the vapour pressure of pure solvent, the solution boils at a higher temperature than the boiling point of pure solvent and is known as elevation of boiling point. Elevation of boiling point is directly proportional to the molal concentration of the solute in a solution. ∆Tb ∝ m or ∆Tb = Kb m Where ∆Tb is elevation in boiling point, m (molality) the number of moles of solute in 1 Kg of solvent and proportionality constant Kb is called boiling point elevation constant or molal boiling point constant or ebullioscopic constant.

The molecular weight of solute (M) can be calculated from the elevation in boiling (∆Tb) by using the expression

w and W are the weights of solute and solvents respectively. Since the vapour pressure of a solution is less than the vapour pressure of pure solvent, the solution freezes at a lesser temperature than the freezing point of pure solvent and is known as depression in freezing point. Depression in freezing point ∆Tb is directly proportional to molality ‘m’ of solution. The proportionality constant Kf is known as the freezing point depression constant or molal freezing point depression constant or cryoscopic constant. The molecular weight of a solute M can be calculated from the depression in freezing point (∆Tf) by using the expression ∆Tf =

•

• •

•

•

K × 1000 × w K b × 1000 × w or M = b M×W DTb × W

K × 1000 × w K f × 1000 × w or M = f ∆Tf × W M×W

The values of boiling point elevation constant and the freezing point depression constant depend on the nature of solvent and can be ascertained from the following relations Kb =

RTb2 1000 × D Vap /H M solvent

Kf =

RTf2 1000 × D fus H / M solvent

R and Msolvent stands for gas constant and molecular weight of solvent respectively and Tb and Tf denote the boiling and freezing points of the solvent respectively in Kelvin, ∆vap H, ∆fusH represents change in enthalpies for the vapourization of solvent and fusion of the pure solid solvent respectively. The mixture with lowest melting point is called eutectic mixture. Methanol, ethylene glycol, glycerol etc. will be mixed with water which is taken in radiators of automobile engines so that the water will not freezes at below 0°C. Salts such as NaCl and CaCl2 are scattered on icy roads so that ice melts at low temperatures and clears the roads NaCl makes the ice to melt at –21°C and CaCl2 makes the ice to melt at –55°C. Semi permeable membrane is that which allows only solvent to pass through it but not solute particles.

Solutions

•

•

•

•

•

•

•

• •

•

•

Some animal membranes, parchment paper, cellophane paper are some natural semipermeable membranes. Certain membranes prepared from some inorganic precipitates also act as semipermeable membranes e.g., copper ferrocyanide precipitate film deposited in the pores of a porous container acts as semipermeable membrane. The process of solvent flowing into the solution when the solvent and the solution are separated by a semipermeable membrane is called osmosis. The inflow of solvent from a dilute solution into the concentrated solution of the solute when the two solutions are separated by a semipermeable membrane is also called osmosis. Osmosis occurs as a result of some pressure exerted by solute molecules and that pressure is called osmotic pressure. The hydrostatic pressure developed on the aqueous dilute solution at equilibrium state due to inflow of water when the solution is separated from the water by a semipermeable membrane is called osmotic pressure (or) The pressure required to be applied on the solution to just stop the osmosis is also called as osmotic pressure. The phenomenon of passing solvent from the solution through semipermeable membrane by applying a high pressure greater than osmotic pressure is knowns as reverse osmosis. Desalination of sea water is carried by using reverse osmosis. Van’t Hoff proposed that a non-volatile solute present in a dilute solution behave like a gas and this is known as Van’t Hoff’s theory of dilute solutions. According to Van’t Hoff, all laws that applicable to gases are also applicable to dilute solutions and these laws are called Van’t Hoff’s laws. According to Van’t Hoff at constant temperature the osmotic pressure (π) of a dilute solution of concentration (C) is directly proportional to the concentration π∝C

V∝

1 C

1 = or πV = k; k proportionality constant (similar V

to Boyle’s law)

•

• •

The osmotic pressure (π) of a solution of constant concentration (C) is directly proportional to the temperature in the Kelvin scale (T) π ∝ T or π = ЌT. Ќ = proportionality constant (similar to charles law) Osmotic pressure (π) of a dilute solution is related to concentration (C) and absolute temperature (T) as follows π = C RT (R is constant equal to gas constant) ∴ πV = RT If n moles of solute are present in V litres we have πv = nRT The molecular weight (M) of a solute in a solution can be calculated by using osmotic pressure from the equation wRT where w is the weight of solute, V is the M= nV volume of solution.

• • • •

•

• •

• •

C = concentration mole/lit

The volume (V) of a solution is inversely proportional to concentration

p∝

•

5.51

•

Solutions of same osmotic pressure at a given temperature are called isotonic solutions. Blood is isotonic with saline i.e., 0.9% w/V NaCl solution. Plants take up water from the soil through the roots by osmosis. If the osmotic pressure of the contents inside a living cell is more than that of the contents surrounding it, outside water enter into the cell due to which it bursts. It is called explasmolysis or crenation. The rupture of red blood cell is called haemolysis. If the osmotic pressure of the contents of a living cell is less than that of the contents surrounding outside, the contents from the cell come out of the cell and the cell collapses. This is known as plasmolysis. The raw mango in pickle loses water due to osmosis when placed in salt solution. The molar mass of a solute can be determined by using the experimental results of any one of the colligative property. The lowering of vapour pressure can be determined experimentally by ostwald’s dynamic method. In Ostwald’s dynamic method when air is passed through the solution, the loss in weight of the solution is proportional to the vapour pressure of the solution Ps. When air, already saturated with vapour of solution, is passed through pure solvent, it again saturated with some more solvent vapour because the vapour pressure of pure solvent is greater than solution. The loss in weight of pure solvent is greater than solution. The loss in weight of pure solvent is proportional to lowering of vapour pressured (P–Ps).

5.52

•

•

Solutions

If the air saturated with the solvent vapour is passed over an absorber present in a U-tube, the weight of U-tube increases which is proportional to the vapour pressure of pure solvent P. The molecular mass can be calculated by substituting P–Ps and P in equation =

•

•

• •

• •

•

•

(P − Ps ) w M = × P m W

The elevation in the boiling point of a dilute solution can be determined experimentally by Cottrell’s method in which temperature difference is measured by using Beckman thermometer. The depression in freezing point can be experimentally determined by Rast’s method generally used for solid solutions i.e., a solid solute and solid solvent. Rast’s method is based on that pure solid have high melting point than the impure solid. The difference in the melting points of pure solid and impure solid is equal to the depression in freezing point of the solid solvent. Osmotic pressure can be experimentally determined by Berkeley – Hartley method. In Berkeley – Hartley method the pressure required to apply on the solution to just prevent the osmosis when the solution and the solvent are separated by a semipermeable membrane is equal to the osmotic pressure. In Berkeley – Hartley method the copper ferrocyanide precipitated in the pores of a porous tube act as semipermeable membrane. The external pressure applied on solution to prevent the osmosis is numerically equal to this osmotic pressure of the solution.

abnormal Molecular weights •

•

•

•

number of particles decreases and their colligative properties also decreases. The ratio of the observed colligative property and calculated colligatve property is called Van’t Hoff factor ‘i’

The molecular weights determined by using colligative properties of substances which associate or dissociate will be abnormal and are called abnormal molecular weights. Ionic substances like NaCl, BaCl 2, AlCl3 etc ionse in solutions. Their colligative properties are high due to increase in number of particles when compared with solutions of non electrolytes like glucose having equimolar mass. When molecules of substances like acetic acid in benzene associate as dimers, trimers or polymers the

i= or i =

Observed colligative property Calculated colligative propeerty Calculated molecular mass Observed molecular mass

•

The value of Van’t Hoff factor i is greater than i for ionic substances while it has lower value than i for associated substances. Calculated molar mass Observed molar mass Normal no. of solute particles = No. of solute particles dissociation or assocciation

•

Degree of association is the fraction of the total number of molecules which combine to form bigger molecules. e.g., 2B ⇌ (B)2

If x be the degree of association when one mole of solute is dissolved in a given volume of solvent x x No. of unassociated molecules = 1 − x + = 1 − 2 2 Experimental colligative property (1 − x / 2) = i= 1 Calculated oc lligative property

(1 − x / 2) Normal Mol. wt = Abnormal Mol. wt 1 Degree of dissociation is the fraction of the total number of an electroyte present as ions in the solution. E.g., if one mole of NaCl is dissolved in water and the degree of dissociation is x i=

•

NaCl → Na+ + Cl– 1–x x x Total number of particles = 1–x + x + x = 1 + x Experimental colligative Property 1+x = i= 1 Calculated colligative Property i=

Normal Mol. wt (1 + x ) = Abnormal Mol. wt 1

Solutions

5.53

pracTice exercise Multiple choice questions with only one answer level i 1. An azeotropic mixture of two liquids boils at a lower temperature than either of them when (a) it is saturated. (b) it does not deviate from Raoult’ s law. (c) it shows negative deviation from Raoult’ s law. (d) it shows positive deviation from Raoult’s law. 2. A solution of a substance containing 1.05 g per 100 mL was found to be isotonic with 3% glucose solution. The molecular weight of the substance is (a) 63 (b) 630 (c) 6.3 (d) 31.5 3. In cold countries ethylene glycol is added to water in the radiators of cars in winter season. It results in (a) lowering boiling point (b) reducing viscosity (d) reducing specific heat (d) lowering freezing point 4. Beckmann thermometer measures (a) boiling point of the solution. (b) freezing point of the solution. (c) any temperature. (d) elevation in boiling point or depression in freezing point. 5. When HgI2 is mixed with aqueous solution of KI, then (a) freezing point decrease (b) freezing point does not change (c) freezing point increase (d) boiling point remains unchanges 6. The freezing point of 0.05 molal solution of a nonelectrolyte in water is______. Kf = 1.86 K kg mol–1 (a) –1.86°C (b) –0.93°C (c) –0.093°C (d) 0.93°C 7. Which of the following 0.1 M aqueous solutions will have lowest freezing point (a) potassium sulphate (b) sodium chloride (c) urea (d) glucose 8. The V.P. of a dilute solution of a non-volatile solute is P and the V.P. of pure solvent is P0, the lowering of the V.P. is (a) +Ve (b) –Ve (c) P/P0 (d) P0/P

9. The Van’t Hoff factor NaCl assuming 100% dissociation is (a) 1/2 (b) 2 (c) 1 (d) 3 10. The solution in which the blood cells retain their normal form are with regard to the another (a) Isotonic (b) Hypertonic (c) Hypotonic (d) None of these 11. Two solution A and B are separated through a SPM. A has high O.P than B. The osmosis will occur from (a) A to B (b) B to A (c) Both (d) None of these 12. When an ideal binary solution is in equilibrium with its vapour, molar ratio of the two components in the solution and in the vapour phase is (a) same (b) different (c) may or may not be same depending upon volatile nature of the two components (d) None of these 13. The lowering of vapour pressure of 0.1 M aqueous solution of NaCl, CuSO4 and K2SO4 are (a) all equal (b) in the ratio 1:1:1.5 (c) in the ratio 3:2:1 (d) in the ratio 1.5:1:2.5 14. The boiling point of C6H6, C6H5NH2, CH3OH and C6H5NO2, are 80°C, 184°C, 65°C and 212°C respectively.Which has higher vapour pressure at room temperature (a) C6H6 (b) CH3OH (c) C6H5NH2 (d) C6H5NO2 15. The correct relationship between the b.pt of very dilute solution of AlCl3(t1) and CaCl2 (t2) having same molar concentration is (a) t1 > t2 (b) t2 >t1 (c) t1 = t2 (d) None of these 16. The freezing point of equimolal aqueous solutions will be highest for (a) C6H5NH3Cl (b) Ca(NO3)2 (c) La(NO3)3 (d) C6H12O6 17. The relationship between osmotic pressure at 273 K when 10 gm glucose (P1) 10 g urea (P2) and 10 g sucrose (P3) are dissolved in 250 mL of water is (a) P1>P2>P3 (b) P3>P1>P2 (c) P2>P1>P3 (d) P2>P3>P1 18. In dilute solution the depression in f.pt. is directly proportional to (a) normality (b) molality (c) molarity (d) mole fraction

5.54

Solutions

19. A maxima or minima obtained in the temperature composition curve of a mixture of two liquids indicates? (a) An azeotropic mixture (b) An eutectic formation (c) That the liquids are immiscible with one another. (d) That the liquids are partially miscible at the maximum. 20. Two solvents A and B have Kf values 1.86 and 2.79 K mole –1 Kg respectively. A given amount of a substance when dissolved in 500 g of A, it completely dimerizes and when same amount of substance is dissolved in 500 g of B, the solute undergoes trimerization. What will be ratio of observed lowering of freezing point in two cases for A and B. (a) 3:2 (b) 1:1 (c) 2:3 (d) 3:4 21. 12.2 gm of benzoic acid (M=122) in 100 gm H2O has elevation of boiling point of 0.27° Kb = 0.54° K kg/ mol. If there is 100% polymerization number of molecules of benzoic acid in associated state is (a) 1 (b) 2 (c) 3 (d) 4 22. Based on the given diagram, which of the following statement regarding the solutions of two miscible volatile liquids are correct? C

H G D

F A

E 1 0

XA XB

0 1

B

(1) Plots AD and BC show that Raoult’s law is obeyed for the solution in which B is a solvent and A is the solute and as well for that in which A is solvent and B is solute. (2) Plot CD shows that Dalton’s law of partial pressures is observed by the binary solution of components A and B. (3) EF + BG = GH, and AC and BD correspond to the vapour pressures of the pure solvents A and B respectively. Select the correct answer using the codes given below. (a) 1 and 2 (b) 2 and 3 (c) 1 and 3 (d) 1,2 and 3 23. For an ideal liquid solution, the plot of 1/XA versus 1/ YA (Where XA and YA are the mole fractions of A in

liquid and vapour phase) is linear with slope and intercept equal to (a)

P B0 − P A0 P A0 , 0 P B0 PB

(b)

P B0 − P A0 P B0 , 0 P B0 PA

(c)

P A0 P B0 − P A0 , P B0 P B0

(d)

P B0 P B0 − P A0 , P A0 P B0

24. Which of the following plots does not represent the behavior of an ideal binary liquid solution? (a) Plot of PA versus XA (mole fraction A in liquid phase) is linear. (b) Plot of PB versus XB is linear. (c) Plot of Ptotal versus XA (or XB) is linear. (d) Plot of Ptotal versus XA is non linear. 25. Which of the following plots correctly represents the behavior of an ideal binary liquid solution? (a) Plot of XA (mole fraction of A in liquid phase) versus YA (mole fraction of A in vapour phase) is linear. (b) Plot of XA versus YB is linear. (c) Plot of 1/XA versus 1/YA is linear. (d) Plot of 1/XA versus 1/YB is linear. 26. For an ideal binary liquid solution With P A0 > P B0 , Which of the following relationship between XA (mole fraction of A in liquid phase ) and YA(mole fraction A in vapour phase ) is correctly represented? (a) XA = YB (b) XA > YA (c) XA < YA (d) XA and YA cannot be correlated with each other. 27. A binary liquid solution of chloroform and acetone is prepared. Which of the following statement correctly represent the behavior of this liquid solution? (a) The solution formed is an ideal solution. (b) The solution formed is non ideal solution with positive deviations from Raoult’s law. (c) The solution formed is non ideal solution with negative deviations from Raoult’s law. (d) Chloroform exhibits positive deviation wheras acetone exhibits negative deviation from Raoult’s law. 28. A semipermeable membrane used in the measurement of osmotic pressure of a solution allows the passage of (a) solvent molecules through it. (b) solute molecules through it. (c) both solvent and solute molecules through it. (d) either solvent or solute and not both through it. 29. In which mode of expression the concentration of solution remains independent of temperature (a) Molarity (b) Normality (c) Formality (d) Molality

Solutions

30. If 0.50 mole of BaCl2 is mixed with 0.20 mole of Na3PO4, the maximum number of moles of Ba3 (PO4)2 that can be formed is (a) 0.1 (b) 0.2 (c) 0.5 (d) 0.7 31. 30 mL of solution is neutralized by 15 mL of 0.2 N base. The strength of the acid solution is (a) 0.1 N (b) 0.15 N (c) 0.3 N (d) 0.4 N 32. A 500 g tooth paste sample has 0.2 g fluoride concentration. What is the concentration of F in terms of ppm level? (a) 250 (b) 200 (c) 400 (d) 1000 33. Increasing the temperature of an aqueous solution will cause (a) decrease in molality (b) decrease in molarity (c) decrease in mole fraction (d) decrease in % W/W 34. Which of the following solutions has the highest normality (a) 1.8 g of KOH/1 lit (b) N-Phosphoric acid (c) 6 g of NaOH/100 mL (d) 0.5 m H2SO4 35. Hydrochloric acid solutions A and B have concentration of 0.5 N and 0.1 N respectively. The volumes of solutions A and B required to make 2 litres of 0.2 NHCl are (a) 0.5 lit of A + l.5 lit of B (b) 1.5 lit of A + 0.5 lit of B (c) 1.0 lit of A + l.0 lit of B (d) 0.75 lit of A + l.25 lit of B 36. The molarity of pure water is (a) 55.6 (b) 50 (c) 100 (d) 18 37. The mole fraction of water in 20% aqueous solution of H2O2 is (a) 77/68 (b) 68/77 (c) 20/80 (d) 80/20 38. Volume of 0.1 M K2Cr2O7 required to oxidize 35 mL of 0.5 M FeSO4 solution is (a) 29 mL (b) 87 mL (c) 175 mL (d) 145 mL 39. Which one is a colligative property ? (a) Boiling point (b) Vapour pressure (c) Osmotic pressure (d) Freezing point 40. Which of the following is not a colligative property? (a) Osmotic pressure (b) Elevation of b.p (c) Vapour pressure (d) Depression of f.p

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41. 100 mL of a liquid A was mixed with 25 mL of a liquid B to give a non–ideal solution of A-B mixture having positive deviations. The volume of this mixture would be (a) 75 mL (b) 125 mL (c) just more than 125 mL (d) close to 125 mL but not exceeding 125 mL 42. Which of the following pairs shows a positive deviation from Raoult’s law? (a) Water–Hydrochloric acid (b) Water–Nitric acid (c) Acetone–Chloroform (d) Benzene–methanol 43. Which one of the following solution would produce maximum elevation in B.P (a) 0.1 M Glucose (b) 0.2 M Sucrose (c) 0.1 M BaCl2 (d) 0.1 M MgSO4 44. Elevation in boiling point was 0.52°C When 6 gm of a compound X was dissolved in 100 gm of water. Molecular weight of X is (Kb of water is 5.2 K per 100 g of water) (a) 120 (b) 60 (c) 600 (d) 180 45. The latent heat of vapourisation of water is 9700 cal/ mole and if the b.p. is 100°C. The ebullioscopic constant of water is (a) 0.513 K (b) 1.026 K (c) 10.26 K (d) 1.832 K 46. The molal freezing point constant for water is 1.86 K Kg/mole. If 342 gm of cane sugar (C12H22O11) is dissolved in 1000 g of water, the solution will freeze at (a) –1.86°C (b) 1.86°C (c) –3.92°C (d) 2.42°C 47. Which has the minimum freezing point (a) one molal NaCl solution (b) one molal KCl solution (c) one molal CaCl2 solution (d) one molal urea solution 48. The freezing point of 1 molal NaCl solution assuming NaCl to be 100% dissociated in water is Kf = 1.86 K kg mol–1 (a) –1.86°C (b) –3.72°C (d) +1.86°C (d) +3.72°C 49. If 0.01 M solution each of urea, common salt and Na2SO4 are taken, the ratio of depression of freezing point is (a) 1:1:1 (b) 1:2:1 (c) 1:2:3 (d) 2:2:3 50. What is the molality of a solution of a certain solute in a solvent if there is a freezing point depression of 0.184°C and the freezing point constant is 18.4. (a) 0.01 (b) 1 (c) 0.001 (d) 100

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51. The vapour pressure of a liquid in a closed container depends upon (a) amount of liquid (b) surface area of the container (c) temperature (d) None of the above 52. The temperature at which the vapour pressure is equal to the external pressure is called the (a) critical temperature (b) boiling point (c) normal point (d) saturation point 53. What would happen if a thin slice of sugar beet is placed in a concentrated solution of NaCl? (a) Sugar beet will lose water from its cells. (b) Sugar beet will absorb water from solution. (c) Sugar beet will neither absorb nor lose water. (d) Sugar beet will dissolve in solution. 54. The freezing point of the water is depressed by 0.037°C in a 0.01 moles NaCl solution. The freezing point of a 0.02 mol solution of sucrose in °C is (a) –0.0370 (b) –0.0185 (c) –0.0740 (d) –0.1850 55. The osmotic pressure of solution increases if (a) temperature is decreased. (b) solution constant is increased. (c) number of solute molecules is increased. (d) volume is increased. 56. Osmotic pressure of sugar solution at 24°C is 2.5 atmosphere. The concentration of the solution in gm mole per litre is (a) 10.25 (b) 1.025 (c) 102.5 (d) 0.1025 57. Solutions with same osmotic pressure are called (a) Hypertonic (b) Hypotonic (c) Isotonic (d) Normal 58. Which of the following correctly expresses the Van’t Hoff factor? (a) Calculated osmotic pressure/Observed osmotic pressure (b) Observed molecular weight/Calculated molecular weight (c) Calculated boiling point/Observed boiling point (d) Observed colligative property/Calculated colligative property 59. Van’t Hoff factor for an electrolyte is (a) >1 (b) KCl>CH3COOH>sucrose. (d) Raoult’s law states that the vapour pressure of a compound over a solutions is proportional to its mole fraction?

10. The normal boiling point of toluene is 110.7°C, and its boiling elevation constant is 3.32 K kg mol–1. The enthalpy of vaporization of toluene is nearly (a) 17.0 KJ mol–1 (b) 34.0 KJ mol–1 –1 (c) 51.0 KJ mol (d) 68.0 KJ mol–1 11. The strength (gpl) of H2SO4 is numerically same as the volume strength of 1.212% (w/v) H2O2 solution. If the 200 mL of H2SO4 solution is diluted by 300 mL of water. What is its final strength (gpl) (a) 1.6 (b) 2 gm/lit (c) 1.8 (d) 1.4 12. Given

13.

14.

15.

16.

PA0 5 = For an ideal mixture of liquids A and PB0 3

B in which XA = 0.5. How many repeated distillations are required to get a small quantity of the distillate containing at least 80% of A? (a) 1 (b) 2 (c) 3 (d) 4 Three solutions are prepared by adding w gm of A into 1 kg of water and w gm of B into another 1 kg water and w gm of C in another 1 kg of water (A, B, C are non-electrolytes). Dry air is passed from these solutions in sequence (A-B-C). The loss in weight of solution A was found to be 2 gm while solution B gained 0.5 gm and solution C lost 1 gm. Then the relation between molar masses of A, B and C is (a) MA: MB: MC = 4:3:5 1 1 1 (b) MA: MB:MC = : : 4 3 5 (c) MC> MA>MB (d) MB>MA>MC The incorrect statements among the following. (a) The relative lowering of vapour pressure of a solution is independent of temperature. (b) The vapour pressure of a solution increase linearly with increase in temperature. (c) The lowering of vapour pressure of a solution is directly proportional to mole fraction of solute. (d) The lowering of vapour pressure of a solution is equal to mole fraction of solute. An aqueous solution contains 5% by mass of urea and 10% by mass of sucrose. If mol of depression constant of water is 1.86 K kg mol–1. The brezing point of solution as (a) –1.43°C (b) –2.43°C (c) –3.43°C (d) –4.43°C Depression of freezing point of 0.01 molal aq CH3COOH solution is 0.02046°C 1 molal urea solution freezes at –186°C. Assuming molality equal to molarity, pH of CH3COOH solution is (a) 2 (b) 3 (c) 3.2 (d) 4.2

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17. When 200 g of 20% solution were cooled, part of the solute precipitated. The concentration of the solution becomes 12% the mass of the precipitated substance is (a) 16.16 (b) 17.17 (c) 18.18 (d) 19.19 18. How many m moles of sucrose should be dissolved in 500 g of water so as to get a solution which has a difference of 103.57°C between boiling point and freezing point (a) 500 mmoles (b) 900mmoles (c) 750 mmoles (d) 650mmoles 19. The immiscible liquid system, 1-bromobutane and water, distills at 95°C. At this temperature the vapour pressures are in the ratio 1:5. What is the ratio of the masses distilled? (a) 1:1 (b) 7.5:1 (c) 1.51: 1 (d) 2:1 20. Calculate the amount of ice that will separate out on cooling a solution containing 50 g of ethylene glycol in 200 g water to –9.3°C. [Kf for water = 1.86 K kg mol–1] (a) 39 (b) 42 (c) 45 (d) 48 21. 100 mL of H2SO4 solution having molarity 1 M & density 1.5 gm/mL is mixed with 400 mL of water. Resulting molarity of H2SO4 solution (density = 1.25 gm/mL) is (a) 4.4 M (b) 0.145 M (c) 0.227 M (d) None 22. Which of the following represents correctly, the changes in thermodynamic properties during the formation of 1 mol of an ideal binary solution

(a) J mol-1

+

∆G-

0

T∆S-

-

∆Gmole fraction

+ (b) J mol-1

0 -

∆G∆HT∆Smole fraction

(c) J mol-1

0 -

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T∆S∆H∆Gmole fraction

+ (d) J mol-1 0 -

T∆S∆G∆Gmole fraction

23. A complex of iron and cyanide ions is 100% ionized at 1 molal. If its elevation in boiling point is 2.08° then the complex is (given: Kb = 0.25°C mol–1 kg) (a) Fe4III[FeII(CN)6]3 (b) K3[FeIII(CN)6] (c) K4[FeII(CN)6] (d) None 24. Two liquids A and B have vapour pressure in the ratio PA0 :PB0 = 1:2 at a certain temperature suppose that we have an ideal solution of A and B in the mole fraction ratio A:B = 1:2, then mole percent of A in the vapour in equilibrium with the solution at the given temperature is (a) 20 (b) 25 (c) 35 (d) 75 25. Certain mass say x g of urea dissolved in 500 g of water and cooled upto –0.5°C, whereby 128 gm of ice separates out from the solution. If cryoscopic constant for water be 1.86 K kg mol–1, the value of ‘x’ will be (a) 6 g (b) 8 g (c) 12 g (d) 15 g 26. Acetic acid exists in benzene solution in the dimeric from. In an actual experiment Van’t Hoff factor was found to be 0.52 then the degree of association of acetic acid is: (a) 0.48 (b) 0.88 (c) 0.52 (d) 0.96 27. The amount of ice that will separate out on cooling a solution containing 50 gm. of ethylene glycol in 200 gm of water to –9.3°C is (Kf = 1.86 K mol–1 (a) 38.71 gm (b) 38.71 mg (c) 42 gm (d) 42 mg 28. The molality of the urea solution which decrease the vapour pressure of pure water by 25% is (a) 1.85 m (b) 18.5 m (c) 1.388 m (d) 13.88 m

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29. 3 gm of acetic acid is added to 500 cm3 water. Acetic acid 23% dissociated into ions Kf of H2O is 2.0 K kg mol–1. The freezing point of the solution is (a) 0.246°C (b) –0.246°C (c) 0.123°C (d) –0.123°C 30. The phenomenon in which cells are swelled up and then burst if placed in hypotonic solution is called (a) Plasmolysis (b) Haemolysis (c) Exosmosis (d) None of these 31. 0.61 gm of benzoic acid is dissolved in 46 gm of carbon disulphide 80% dimerzation of acid take place Kb of CS2 is 2.3 K kg mol–1. Boiling point of CS2 is 46.2°C. The boiling point of solution is (a) 46.40°C (b) 46.25°C (c) 46.30°C (d) 46.35°C 32. A particular electrolyte MXNy was dissolved in water. The degree of dissociation of salt in water was found to be 60%. The observed molar mass (obtained by measuring the Van’t Hoff’s factor) can be corrected with the calculated molar mass using (a)

M 0.6 + 0.4( x + y)

(b) M [ 0.6 + 0.4( x + y) ] (c)

M 0.4 + 0.6( x + y)

(d) [0.4+0.6(x+y)]M 33. The concentration of copper (II) ions in an unknown solution is determined by comparing the absorbance of a diluted sample of the unknown with Beer’s Law plot at a constant wave length for several standard solutions. When a 2.0 mL sample of the unknown copper (II) ion solution is diluted to 5.0 mL, its absorbance is 0.29. what is concentration of copper (II) ions in the unknown? (a) 0.0062 M (b) 0.012 M (c) 0.031 M (d) 0.078 M 0.40 0.30 0.20

0.10

0

0

0.010

0.020

0.030

Concentration, M

0.040

34. A 5.25% solution of a substance is isotonic with a 1.5% solution of urea (molar mass = 60 g mol–1 in the same solvent. if the densities of both the solutions are assumed to be equal to 1.0 g cm–3, molar mass of the substance will be (a) 210.0 g mol–1 (b) 90.0 g mol–1 –1 (c) 115.0 g mol (d) 105.0 g mol–1 35. When dry air is passed through 0.026 M solution of MgCl2 and then through 0.072 M urea solution there is no loss in mass of urea solution. Hence the degree of dissociation of MgCl2 is (a) 0.75 (b) 0.39 (c) 0.885 (d) 0.92 36. A solution containing 0.1 g of non-volatile organic substance P (molecular weight 100) in 100 g of benzene raises the boiling point of benzene by 0.2°C, while a solution containing 0.1 g of another nonvolatile substance ‘Q’ in the same amount of benzene raise the boiling point of benzene by 0.4°C. what is the ratio of molecular weights of P and Q? (a) 1:2 (b) 2:1 (c) 1:4 (d) 4:1 37. Select the incorrect choice from th following: (a) ∆Smix for all types of solutions (ideal and nonideal solutions) is positive (b) ∆Gmix of ideal solutions is always negative (c) A non ideal solution having + Ve deviation shows ∆H = +Ve (d) Azerotrope contains only one component formed as a result of definite union of the two liquids and the composition of azoetrope does not change with pressure (consider binary liquids) 38. If 100 mL of 1 N sulphuric acid were mixed with 100 mL of 1 N sodium hydroxide, the solution will be (a) acidic (b) basic (c) neutral (d) slightly acidic 39. The mineral atacamite (A) is [CuCl2·xCu(OH)2]. 45.05 mL of 0.5089 M HCl were required to react completely with 1.6320 g of a A. Hence ‘x’ is (mole. Wt. of A = 427) (a) 2 (b) 3 (c) 4 (d) 1 40. A complex is represented as CoCl3 XNH3. Its 0.1 molal solution in water shows ∆Tf = 0.6°C Kf for H2O is 2.0 K molality–1. Assuming 100% ionization of complex and co-ordination number of cobalt is six. the formula of complex is (a) [Co(NH3)4Cl] (b) [Co(NH3)5Cl]Cl2 (c) [Co(NH3)3Cl3] (d) Both 1 and 2

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41. The osmotic pressure of 0.1 m monobasic acid if its pH is 2 at 30°C is (a) 2.95 atm (b) 3.20 atm (c) 2.7 atm (d) 2.83 atm 42. 1 gm of monobasic acid in 100 gm of water lowers the freezing point by 0.20. If 2 gm of same acid requires N 150 mL of alkali for completes neutralization. Kf 10

43.

44.

45.

46.

47.

48.

for H2O is 2.0 K kg mol–1. The degree of dissociation of weak acid is (a) 33.3% (b) 20% (c) 66.6% (d) 40% An aqueous solution of glucose boils at 100.01°C. The molal elevation constant for water is 0.5 K kg mol–1. The no. of glucose molecules in 100 gm of water is (a) 1.2x1022 (b) 2.4x1022 21 (c) 1.2x10 (d) 2.4x1021 The freezing point of an aqueous solution having mole fraction water 0.8 is. (Kf for H2O is 2 K kg mol–1) (a) –55.6°C (b) –13.8°C (c) –41.6°C (d) –27.8°C The freezing point of solution containing 50 mL of ethylene glycol in 50 gm of water is found to –30°C Kf for water is 2 K kg mol–1. Density of ethylene glycol is (a) 0.46 gm/mL (b) 0.31 gm/mL (c) 0.69 gm/mL (d) 0.93 gm/mL 0.1 M Solution of glucose was found to be isotonic with a solution of X in 100 gm water. The relative lowering in vapour pressure of solution of X in water is (a) 0.18 (b) 0.018 (c) 0.0018 (d) 0.00018 The vapour pressure of a solvent decreased by 10 mm of Hg when a non-volatile solute was added to the solvent. The mole fraction of solute in solution is 0.2. What would be mole fraction of the solvent if decrease in vapour pressure is 20 m of Hg (a) 0.2 (b) 0.4 (c) 0.6 (d) 0.8 FeCl3 on reaction with K4[Fe(CN)6] in aqueous solution gives blue colour. These are separated by a semi permeable membrane AB as shown. Due to osmosis there is 0.1 M K4Fe(CN)6

49. The expression relation molality (m) and mole fraction of solute (x2) in a solution is (a) x 2 =

mM1 1 + mM1

(b) x 2 =

1 − mM1 mM1

(c) x 2 =

mM1 1 − mM1

(d) x 2 =

1 + mM1 mM1

where M1 is the molar mass of solvent 50. Assume that incremental amounts of a volatile solute are added to a solvent so that the mole fraction of solute varies from 0 to 1. A plot of the partial pressure of the solute vs. mole fraction of solute will: (a) Be linear with slope equal to the vapour pressure of the pure solute. (b) Be linear with slope equal to the vapour pressure of the pure solvent. (c) Be linear with slope equal to the sum of the vapour pressures of pure solvent and pure solute. (d) Vary exponentially with the mole fraction of the solute 51. The molecular mass of glucose is 180 g/mol and that of sucrose is 342 g/mol. Assuming ideal behavior, a plot of freezing point vs molal concentration of sucrose would: (a) Be linear with a slope equal to that for a similar plot for glucose. (b) Be linear with a slope equal to twice that for a similar plot for glucose. (c) Be linear with a slope equal to one half that for a similar plot for glucose. (d) Be nonlinear with rate of curvature twice that for a similar plot for glucose 52. Which of the following graphs below explains the time dependence of rates of evaporation and condensation for a liquid evaporating in a closed container roe (a) Rate roc

0.01 M FeCl;

time Side X

Side Y SPM

(a) (b) (c) (d)

Blue colour formation in side X Blue colour formation in side Y Blue colour formation in both of the sides X and Y No blue colour formation

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roe (b) roc

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Solutions

roc

(c)

roe

roe (d)

roc

53. PA = (235y–125xy) mm of Hg. PA is partial pressure of A, x is mole fraction of B in liquid phase in the mixture of two liquids A and B, y is the mole fraction of A in vapour phase then P0B in mm of Hg is (a) 235 (b) 0 (c) 110 (d) 125 54. The temperature at which molality of pure water is equal to its molarity is (a) 273 K (b) 298 K (c) 277 K (d) None of these 55. The Van’t Hoff factor for 0.1 M Ba(NO3)2 solution is 2.74. The degree of dissociation is (a) 91.3% (b) 87% (c) 100% (d) 74% 56 Which will form maximum boiling point azeotrope? (a) HNO3 + H2O solution (b) C2H5OH+H2O solution (c) C6H6+C6H5CH3 solution (d) None of these 57. Relative decrease in vapour pressure of an aqueous solution containing 2 mol [Cu(NH3)3Cl]Cl in 3 mol H2O is 0.50. On reaction with AgNO3, this solution will form (a) 1 mol AgCl (b) 0.25 mol AgCl (c) 2 mol AgCl (d) 0.4 mol AgCl 58. The solubility of common salt is 36.0 in 100 g of water at 20°C. If systems, I, II and III contain 40.0, 36.0 and 20.0 g of the salt added to 100.0 g of water in each case, the vapour pressures would be in the order (a) IIII (c) I=II>III (d) I=II PB0 ;

Temperature

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T2

0.0 0.2 0.4 0.6 0.81 Mole fraction Graph(II)

Select the correct options; (a) Graph(I) represents maximum boiling point and minimum vapour pressure. (b) Graph(II) represents minimum boiling point and maximum vapour pressure. (c) Graph(I) represents minimum boiling point and maximum vapour pressure. (d) Graph(II) represents maximum boiling point and minimum vapour pressure. 22. Consider the following Solutions; (I) 1M aqueous glucose (II) 1M aqueous NaCl (III) 1M C6H5COOH in C6H6 (IV) 1 M (NH4)3PO4 (a) All are isotonic solution (b) III is hypotonic to I, II, IV (c) I, II, IV are hypertonic to III (d) IV is hypertonic to I, II, IV 23. Which of the following increase with increase in temperature? (a) Vapour pressure (b) Osmotic pressure

Solutions

24.

25.

26.

27.

28.

29.

30.

(c) Boiling point (d) Freezing point Identify the correct statement (a) 1 m NaCl has higher freezing point than 1m glucose solutions (b) 1 m NaCl solution has same boiling point as 1 m KCl solution (c) Molecular weight of NaCl will be less than 58.5 in water because it undergoes dissociation (d) i > 1 when solution undergoes association The depression of freezing point experiment, it is found that. (a) The vapour pressure of solution is less than that of solvent. (b) The vapour pressure of solution is more than that of solvent. (c) Only solute molecule solidify at the freezing point. (d) Only solvent molecule solidify at the freezing point. The lowering in vapour pressure when a solute is added. (a) Will depend upon the nature of solute. (b) Will be the same irrespective of solvent taken. (c) Will be different when different solvents are taken. (d) Will depend on the temperature. A substance effloresces. (a) Due to anhydrous layer on top. (b) When its vapour pressure is greater than that of water vapour in air. (c) When its vapour pressure is less than that of water vapour in air. (d) When its crystalline substance has water of crystallization. The vapour pressure of a liquid depends on (a) Nature of liquid. (b) Temperature. (c) Atmospheric pressure. (d) Volume of space above the liquid. Which of the following plots represents an ideal binary mixture? (a) Plot of PTotal vs XB is linear (XB is mole fraction of liquid ‘B’) (b) Plot of PTotal vs YA is linear (YA = mole fraction of A in vapour phase) (c) Plot of 1/PTotal versus YA is linear (d) Plot of 1/PTotal versus YB is non-linear Degree of ionization depends upon (a) Nature of solvent (b) Nature of electrolyte (c) Temperature (d) Pressure

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31. Identify the correct statements: (a) Gases have higher value of Tc (critical temperature) have higher solubility. (b) The solubility of gas decreases with increase in temperature because KH (Henry’s law constant). increases with increase in temperature (c) The solubility of gas in liquid is directly proportional to pressure. (d) The solubility of gas in liquid is inversely proportional to pressure. 32. The depression in freezing point of a substance depends upon: (a) The nature of solute (b) The nature of solvent (c) Molality of solution (d) Independent of nature of solvent 33. Which of the following statements are true? (a) After dissolving the outer shell of egg in acetic acid when it is put in concentrated NaCl solution the egg shrinks. (b) After dissolving the outer shell of egg in acetic acid when it is put in NaCl solution the egg enlarge in size. (c) After dissolving the outer shell of egg in acetic acid when it is put in distilled H2O it swells. (d) After dissolving the outer shell of egg in acetic acid when it is put in distilled H2O it shrinks. 34. Which of the following statements is/are correct? (a) The vapour pressure of a liquid mixture of two volatile liquids is always less than that in its pure state. (b) The vapour pressure of a liquid gets lowered to a greater extent than expected when an electrolyte which is non volatile is added. (c) The vapour pressure of a liquid gets raised when a non volatile liquid which associates is added. (d) The vapour pressure of a liquid gets lowered to a lesser extent than expected when a non-volatile substance which associates is added. 35. Which of the following do not depend on temperature? (a) Molarity (b) Molality (c) Mole fraction (d) Normality 36. Which of the following statements is/are correct about azeotropic mixture? (a) Azeotropic mixtures are non-ideal solution (b) The components of azeotrapic mixture cannot be separated by fractional distillation

5.66

37.

38.

39.

40.

41.

42.

43.

44.

Solutions

(c) Azeotropes obey Raoult’s law (d) Solution with positive deviation from Raoult’s law, forms minimum boiling azeotrope Solution of two liquids A and B showing negative deviation from Raoult’s, will show; (a) ΔHmix< 0 (b) ΔVmix< 0 (c) DP < PA0 X A +PB0 X B (d) ΔSmix< 0 Which of the following are not colligative properties? (a) Relative lowering of vapour pressure (b) Surface tension (c) Osmosis (d) Elevation in boiling point Which of the following is correct for an ideal solution? (a) ΔGmix = 0 and ΔSmix>0 (b) ΔHmix = 0 and ΔVmix = 0 (c) ΔVmix = 0 and ΔSmix > 0 (d) ΔHmix = 0 and ΔSmix > 0 5.3% (w/v) Na2CO3 Solution and 6.3% (w/v) H2C2O4.2H20 Solution have same: (a) molality (b) molarity (c) normality (d) molefraction Which of the following are correct about the binary homogeneous liquid mixture? (a) H2O and C2H5OH ΔHsol>0, ΔVSO1>0 (b) C6H6 and C6H5CH3 ΔHsol= 0, ΔVSO = 0 (c) CH3COCH3 and CHCl3 ΔHsol < 0, ΔVSO1 0, ΔVSO1 1 at normal dilution.

(c) i increase more rapidly with dilution and attain a limiting value of (x+y) at infinite dilution. (d) The increases in (i) with dilution is due to increase in molality of solution with dilution. 45. For a dilute solution having molality m of a given solute in a solvent of mol.wt. M, b.pt. Tb and heat of

46.

47.

48.

49.

 ∂T  is equal to: vaporization per mole DH;  b   ∂m  m →0 (a) Molal elevation constant of solvent (b) Where M in kg; Δvap H and R in Jmo/–1 (c) Where M in kg; Δvap S and R in Jmo/–1 (d) Where M in g; R and Δvap H expressed in same unit of heat Which facts are true when we use Van’t Hoff equation P = CST for osmotic pressure p of dilute solutions? (a) The equation is identical to that of ideal gas equation. (b) The solute particles in solution are analogous to the gas molecules and the solvent is analogous to the empty space between the gas molecules. (c) Solute molecules are dispersed in the solvent in a similar way as the gas molecules are dispersed in empty space. (d) The equation is not identical to that of ideal gas equation. Which of the following are correct about mixture of two miscible liquids forming azeotropes; (a) Has the lowest temperature at which the two liquids can exist in the liquid state. (b) Has the lowest temperature attainable at which the two liquids forms eutectic mixture. (c) eutectic point has found application to attain low melting point solids which are eutectic mixtures. (d) Composition of azeotrope and their b.pt vary with the external pressure. Which statements are correct about antifreeze mixtures to use to melt ice or snow on roads? (a) Antifreeze mixture of CaCl2+water(f.pt-50°C) is preferred over KCl+water (f.pt-10°C). (b) The low f.pt of CaCl2+water is due to the formation of CaCl2.6H2O at low temperature. (c) The use of antifreeze for salt solutions cause major problems of corrosion of steel car bodies and reinforcement bars in concrete road structures. (d) More is the amount of salt spreaded on road, easier is melting of ice. According to Henry’s law, the partial pressure of gas (P΄g) is directly proportional to mole fraction of gas in dissolved state, i.e., P΄gas = KH.Xgas where KH is Henry’s constant. Which are correct?

Solutions

(a) KH is characteristic constant for a given gassolvent system. (b) Higher is the value of KH, lower is solubility of gas for a given partial pressure of gas. (c) KH has temperature dependence. (d) KH increase with temperature. 50. Pick out the correct statements (a) The ratio of vapour pressure over solution phase on mixing two immiscible liquids is equal to ratio of their moles in vapour phase. (b) The ratio of vapour pressure over solution phase on mixing two immiscible liquids is equal to ratio of their moles in liquid phase. (c) The ratio of vapour pressure over solution phase on mixing two miscible liquids is equal to ratio of the product of their vapour pressure and their mole fraction in liquid phase. (d) The ratio of vapour pressure over solution phase on mixing two immiscible liquids is equal to the ratio of the product of their vapour pressure and their moles fraction in liquid phase.

comprehensive Type questions passage i Consider following law for the solution of a gas in a solvent and answer the questions at the end of it. “The solubility of a gas in a liquid is directly proportional to the pressure of the gas.” 1. Select the correct statement. (a) NH3 is soluble in water due to hydrogen bonding as well as due to formation of ions. (b) Gases which can be liquefied easily are more soluble in water than the gases which cannot be liquefied. (c) Both of the above are incorrect. (d) Both of the above are correct. 2. An ionic compound that attracts atmospheric water so strongly that a hydrate is formed is said to be (a) dilute (b) hygroscopic (c) immiscible (d) miscible 3. The solubility of gases in liquids (a) increases with increase in pressure and temperature. (b) decreases with increase in pressure and temperature. (c) increases with increase in pressure and decreases in temperature. (d) decreases with increase in pressure and increases in temperature. 4. The solubility of N2(g) in water exposed to the atmosphere, when the partial pressure is 593 mm is 5.3 × 10–4 M. Its solubility at 760 mm and at the same

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temperature is (a) 4.1×10–4 M (b) 6.8×10–4 M (c) 1500 M (d) 2400 M 5. The Henry’s law constant for the solubility of N2 gas in water at 298 K is 1.0 × 105 atm. The mole fraction of N2 in air is 0.8. The number of moles of N2 from air dissolved in 10 moles of water at 298 K and 5 atm pressure is. (a) 4.0×10–4 (b) 4.0×10–5 –4 (c) 5.0×10 (d) 4.0×10–6 passage ii Following statement is considered to define colligative properties: Answer the questions at the end of it. Properties whose values depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. 1. Acetic acid in benzene solution forms dimer due to intermolecular H-bonding. For this case Van’t Hoff factor is (a) i = 1 (b) i > 1 (c) i < 1 (d) inclusive 2. If pKa = –log Ka = 4, and Ka = Cx2 then Van’t Hoff factor for weak monobasic acid when C = 0.01 M is (a) 1.01 (b) 1.02 (c) 1.10 (d) 1.20 3. An aqueous solution of 0.01 M CH3COOH has Van’t Hoff factor of 1.01. If pH = –log [H+], pH of .01 M CH3COOH solution would be (a) 2 (b) 3 (c) 4 (d) 5 4. In which case Van’t Hoff factor is maximum? (a) KCl, 50% ionized (b) K2SO4 40% ionized (c) SnCl4, 20% ionized (d) FeCl3, 30% ionized 5. A complex containing K+, Pt (IV) and CI– is 100% ionized i = 3. Thus, complex is (a) K2 [PtCl4] (b) K2 [PtCl6] (c) K3 [PtCl5] (d) K[PtCl3] passage iii Consider following figure and answer the questions at the end of it. Figure explains elevation in boiling point when a nonvolatile solute is added to a solvent. Variation of vapour pressure with temperature and showing elevation in boiling point

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Solutions

Vapour pressure

1atm Atmospheric pressure P0

A

B

P solvent P ∆ TD

T0

T

Temperature

1. Given that DTb is the elevation in boiling point of the the solvent in the a solution of molality ‘m’ then

3.

4.

5.

 ∆Tb   m  is equal to   (a) Kb (molal elevation constant) (b) Lv (latent heat of vaporization) (c) DS (entropy change) (d) x(mole fraction of solute) Elevation in b.p. of an aqueous urea solution is 52° (Kb = 0.52° mol–1 kg). Hence mole fraction of urea in this solution is (a) 0.982 (b) 0.0567 (c) 0.943 (d) 0.018 A complex of iron and cyanide ions is 100% ionized at 1 m (molal). If its elevation in b.p is 2.08° (Kb 0.52°C mol–1 kg) then complex is (a) K3[Fe(CN)6] (b) Fe(CN)2 (c) K4[Fe(CN)6] (d) Fe(CN)4 Select correct statement: (a) Heats of vapourization for a pure solvent and for a solution are similar because similar intermolecular forces between solvent molecules must be overcome in both cases. (b) Entropy change between solution and vopour is smaller than the entropy change between pure solvent and vapour. (c) Boiling point of the solution is larger than that of the pure solvent. (d) All are correct statements. Consider following terms: (I) mKb (II) mKbi (III)

DTb i

Ratio of DTb/Kb of 6%, AB2 and 9% A2B (AB2 and A2B both are non- electrolytes ) is 1 mol/kg in both cases. Hence atomic masses of A and B are respectively. (a) 60,90 (b) 40,40 (c) 44.5,9.5 (d) 10,40

passage iV

P solution

2.

6.

(IV) Kb

Term which can be expressed in degree (temperature) are (a) III,IV (b) I,II (c) I,II,III (d) I,III

Question at the end are based on following phase diagram for pure solvent and solution for depression in freezing point. Freezing point of a liquid is defined as that temperature at which it is in equilibrium with its solid state.

°C

1. Freezing point of the following system is: Liquid solvent ⇌ solid solvent (a)

∆H − ∆G ∆S

(b)

DH DS

(c)

DG DS

(d)

DS DH

2. Freezing point of a solution is smaller than that of a solvent. It is due to (a) DH of solution and solvent is almost identical since intermolecular forces between solvent molecules are involved. (b) DS of the solution is larger than that for the solvent. (c) DS of the solution is smaller than that of the solvent. (d) DH of solution is much higher than of but DS of solution is smaller than that of the solvent. 3. Select correct statement. (a) Solution has more molecular randomness than a pure solvent has, the entropy change between solution and solid is larger than the entropy change between pure solvent and solid.

Solutions

passage V Read the following fact and answer question at the end of it. Osmosis, like all colligative properties, results from an increase in entropy as pure solvent passed through the membrane and mixes with the solution. 1. Desalinating of sea water is now done using. (a) Reverse osmosis (b) Osmosis (c) Filtration (d) Evaporation 2. Red blood cells are placed in a solution and neither hemolytic nor crenation occurs. Therefore the solution is: (a) Hypertonic (b) Hypotonic (c) Isotonic (d) Isotropic 3. The passage of a solvent across a semipereable membrane because of concentration difference is called (a) dialysis (b) hemolysis (c) hydration (d) osmosis 4. Match the term with definitions Term

Defination

Hemolysis

A

Crenation

B

Hygroscopic

C

Hypertonic

D

Isotonic

E

The shrinking of a red blood cell when placed in a solution of greater osmolarity than the cell itself Refers to two solutions of same osmolarity Refers to a solution having greater osmolarity than 0.30 osmol, the osmolarity of normal red blood cells An ionic compound which attracts water molecules from the atmosphere The swelling and bursting of a red blood cell when placed in a solution of lower osmolarity than the cell itself

passage Vi Read the following statement and answer the question at the end of it. The partial vapour pressure of any volatile component of an ideal solution is equal to the vapour pressure of the pure component multiplied by the mole fraction of that component in the solution. 1. At a given temperature total vapour pressure in Torr of a mixture of of volatile components A and B is given by P = 120–75 XB hence vapour pressure of pure A and B respectively (a) 120,75 (b) 120,195 (c) 120,45 (d) 75,45 2. A mixture contains 1 mol. volatile A (P0A = 100 mm Hg) and 3 mol volatile liquid B (PB0 = 80 mm Hg) if solution behave ideally, total vapour pressure of the distillate is approximately: (a) 85 mm Hg (b) 86 mm Hg (c) 90 mm Hg (d) 92 mm Hg 3. Moles of K2SO4 to be dissolved in 12 mol. Water to lower its vapour pressure by 10 mm Hg at a temperature the vapour pressure of pure water is 50 mm Hg is (a) 3 mol (b) 2 mol (c) 1mol (d) 0.5 mol passage Vii Answer the question (given below) which are based on the following. Vapour pressure plots of benzene-toluene mixtures at 20°C solution of benzene and toluene are ideal. Raoult’s law is valid for both components over entire range of concentration.

V.P(mm/Hg)

(b) Heat of fusion of solution and solvent are similar since similar forces of intermolecular force are involved. (c) Sugar containing solution freezes at a lower temperature than water. (d) All are correct statements. 4. 60 g of urea is dissolved in 1100 g solution. To keep DT/K1 as mol/kg water separated in the form of ice is (a) 40 g (b) 60 g (c) 100 g (d) 200 g

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80 70 60 50 40 30 20 10 0

0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 Molefraction of benzene

1. An ideal solution consisting of two components A and B (such as benzene and toluene) is one in that (a) The intermolecular attractions A…..A,B…..B and A…….B are equal (b) DHmix =0, DVmix=0 (c) Both of the above conditions are followed (d) None of the above conditions is followed 2. There is deviation from ideal behaviour, if mixture contains: (a) n-hexane and n-heptane (b) chlorobenzene and bromobenzene (c) o-xylene and p-xylene (d) acetone and chloroform

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Solutions

passage Viii Answer the question (given below) which are based on the following diagram. P 0A V.P P

P0B

PA

PB X4 X3

0 1

1 0

Vapour pressure diagram for real solution of two liquids A and B that exhibit a negative deviation from Raoult’s law. The dashed lines represent the plots for ideal solutions 1. Solution contain components A and B shows this type of deviation from ideal behavior when (a) (b) (c)

(d)

Attraction A…B larger than average of A..A, B…B attraction as in (a) smaller than average of A… A, B… B attraction as in (c)

DHmix DVmix b.p +ve +ve Larger than expected –ve +ve

–ve As in (a) +ve Smaller than expected

+ve

+ve As in (c)

2. This type of deviation is also expected in the following (a) Ethanol and cyclohexane (b) Ethyl bromide and ethyl chloride (c) Benzonitrile and ethyl cyanide (d) Diethyl ether and chloroform passage ix Answer the question (given below) which are based on the following diagram. Vapour pressure diagram for real solution of two liquids A and B that exhibit a positive deviation from Raoult’s law. The vapour pressure of both A and B are greater than predicted by Raoult’s law. The dashed lines represented the plots for ideal solutions. P 0A

P V.P. P0B

PA PB PB

1. Consider some facts about the above phase diagram

(A) This is observed when A … B attractions are greater than average of A … A and B … B attraction. (B) DHmix = +ve, ∆Vmix = +ve (C) Boiling point is smaller than expected such that vapourization is increased. (D) It forms azeotropic mixture. Select correct facts: (a) A, B, C (b) B, C, D (c) A, C, D (d) A, B, C, D 2. Total vapour pressure of mixture of 1 mol. of volatile component A(PA0 = 100 mm Hg) and 3 mol. volatile component B(PB0 = 60 mm Hg) is 75 mm. For such case. (a) There is positive deviation from Raoult’s law. (b) Boiling point has been lowered. (c) Force of attraction between A and B is smaller than that between A and A or between B and B. (d) All the above statements are correct. passage x Read the following sentence and answer the questions at the end of it. Benzoic acid dimerises in benzene but ionizes in aqueous solution. O O H O COH C C O H O O

O

COH + H2O

CO + H3O

1. Equimolal solutions of (A) benzoic acid in benzene and (B) in aqueous solution are taken. Thus (a) Van’t Hoff factor of (A) > (B) (b) Van’t Hoff factor of (A) < (B) (c) Van’t Hoff factor of (A) = (B) (d) Dimer formation or ionization is not possible. 2. In the following equilibrium N 2 O 4 (g )  2 NO 2 NO2 is 50% of the total volume. Hence, degree of dissociation (x) and Van’t Hoff factor (i) respectively are (a) 0.5,1.5 (b) 0.25, 1.25 (c) 0.33, 1.33 (d) 0.66, 1.66 3. Which has maximum freezing point? (a) 6 g urea solution in 100 g H2O (b) 6 g acetic acid solution in 100 g H2O (c) 6 g sodium chloride in 100 g H2O (d) All have equal freezing point. 4. Aluminium phosphate is 100% ionized in 0.01 mol. aqueous solution. Hence ∆Tb/Kb is (a) 0.01 (b) 0.015 (c) 0.0175 (d) 0.02

Solutions

passage xi Vapour pressure of C6H6 and C7H8 mixture at 50°C are given by P = 180 XB + 90 in mm, where XB is mole fraction of C6H6. 936 gm of benzene and 736 gm of toluene are mixed. The vapours are formed. The vapours are removed and condensed into liquid and again brought to the temperature of 50°C Answer the following questions: 1. Vapour pressure of benzene, and toluene are (a) 180.90 mm (b) 90,180 mm (c) 90,270 mm (d) 270,90 mm 2. Vapour pressure of first liquid mixture is (a) 198 mm (b) 162 mm (c) 126 mm (d) 108 mm 3. Mole fraction of benzene in vopour phase of firstliquid (a) 0.818 (b) 0.182 (c) 0.671 (d) 0.330 4. Vapour pressure of condensed liquid is (a) 122.7 mm (b) 172.1 mm (c) 237.2 mm (d) 297.1 mm 5. Mole fraction of benzene in vapour phase of condensed liquid is (a) 0.072 (b) 0.931 (c) 0.831 (d) 0.17 passage xii A solution containing 0.1 mol of a non volatile solute and 0.9 mol of benzene is cooled out, until some benzene is frozen out. The solution is then decanted off from the solid and warmed upto 353 K where its vapour pressure was found to be 680 torrs. The normal freezing point and normal boiling point of benzene are 278.5 K and 353 K respectively and molal depression constant of benzene is equal to 5 K kg mol-1 Assume ideal behaviour. 1. Mass of benzene frozen out during cooling is: (a) 31.2 gm (b) 70.2 gm (c) 39 gm (d) 7.8 gm 2. The temperature to which the solution was cooled originally is (a) 275 K (b) 275.48 K (c) 239.48 K (d) 262.48 K 3. Molar enthalpy of fusion of benzene is (a) 31 cal/mol (b) 2.42 kcal/mol (c) 48 cal/mol (d) K cal/mol passage xiii The vapour Pressure of two pure liquids A and B form an ideal solution are 300 torr and 800 torr respectively at a temperature ‘T’. A mixture of the vapour of A and B for

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which the mole fraction of A is 0.25 is slowly compressed at a temperature ‘T’ 1. The composition of the first-drop of the condensate is (a) XA = 0.47 (c) XA = 0.53 (c) XA = 0.37 (d) XA = 0.63 2. The composition of the solution whose normal boiling point is ‘T’ (a) XA = 0.92 (c) XA = 0.08 (c) XA = 011 (d) XA = 0.89 3. The pressure when only the last bubble of vapour remains (a) 564 torr (c) 760 torr (c) 765 torr (d) 675 torr passage xiV The freezing point of 0.02 mole fraction solution of acetic acid in benzene is 298.95 K. Acetic acid exists partly as a dimer. Freezing point of benzene is 300 K and its heat of fusion is 36 cal/gm. R = 2 cal/deg/mol. 1. The molality of the solution is (a) 0.130 m (b) 0.195 m (c) 0.391 m (d) 0.261 m 2. The value of molal cryoscopy constant is (a) 10 K kg mol–1 (b) 7.5 K kg mol–1 –1 (c) 5 K kg mol (d) 2.5 K kg mol–1 3. The percentage dimerisation of acid is (a) 80% (b) 60% (c) 40% (d) 20% passage xV Properties, whose value depend only on the concentration of solute particles in solution and not on the identity of the solute are called colligative properties. There may be change in number of moles of solute due to ionization or association hence these properties are also affected. Number of moles of the product is related to degree of ionization or association by Van’t Hoff factor ‘i’ Given by i = [1+(n-1) a] for dissociation &  1   i = 1 +  − 1 α  for association  n   Where n is the number of products (ions or molecules) obtained per mole of the reactant. A dilute solution contains ‘t’ moles of solute X in 1 Kg of solvent with molal elevation constant Kb the solute dimerises in the solution according to the following equation. The degree of association is a [Assume M = m]

2X ⇌ X2

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Solutions

1. The Van’t Hoff factor will be [if we start with one mole of X] (a) i = 1–2a (b) i = 1–a/2 (c) i = 1 + a/2 (d) i = 1 + a 2. The colligative properties observed will be: (a) DPobs>DPactual DTb, obs>DTbactual DTf, obs>DTfactual (b) DPobs = DPactual DTb, obs = DTb actual DTf, obs = DTfactual (c) DTb, obs The vapour pressure. So it is positive devotion passage x  2. N2O4 ↽ ⇀  2NO2 1–a 2a 2α = 0.5 1+ α a = 0.33 i = 1+ a = 1.33

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Solutions

3. ΔTf1 = Kf

6 1000 × = Kf 60 100

ΔTf2 = Kf

6 1000 = (>1) = > Kf × 60 100

ΔTf3 = Kf

6 1000 × 58.3 100

X ''B PB0 X1B = X ''T PT0 X1B (270)(0.818) = = 13.55 (90)(0.181) 5.

X ''B = 13.55 1 − X ''B

(2)

ΔTf less freezing point is more 4.

∆Tb = im Kb i value of AlPO4 = 2 ∆Tb = 2 (0.01) = 0.02 Kb

X’’B = 13.55-13.55X’’B X’’B = 0.931 passage xii 0 1. P − P = n 2 P n1 760 − 608 0.1 = 608 n1

passage xi

n1 = 0.25 Wt of benzene is solution = 31.2 gm Initial wt of benzene = 70.2 gm Ice = 70.2–31.2 = 39 gm

1. P = 180XB + 90 If XB = 1 then P = PB0 PB0 = 270, PT0 = 90 2. = nB = nT

936 = 12 78

2. DTf = K f

736 = 8 92

278.5 − Tf = 5(

Vapour pressure of 1st Liquid is PM = PB0 XB + PT0 X2  123   8  = 270   + 90    20   20  = 198 mm 3.

Kf =

1 X= B

9 = 0.818 11

4. Vapour pressure of condensed liquid = P0BX1B +P0TX1T = 270 (0.818)+90 (0.181) PM = 237.2mm

RT 2 M 1000DH

1.987(278.5) 2 × 78 1000 DH DH = 2.42 Kcal/ mole 5=

270 ×

2x1B = 9–9X1B

0.1 × 1000 ) 31.2

Tf = 262.481 K

X1B P 0 × B = X1T PT0 × T 3 5=9 = 2 2 90 × 5 X1B 9 = 1 1 − XB 2

w2 1000 × MW2 w1

passage xiii 1.

X1A PA0 X B = X1B PA0 X B 0.25 300 X A = 0.75 800 X B XA = 0.888 XB XA = 0.888 (1–XA) XA = 0.472

Solutions

2. Normal boiling point means VP = 760 mm 760 = 300XA + 800(1–XA) XA= 0.08 3. Last bubble remain means total vapour converted to liquid PM = 300(0.25) +800(0.75) PM = 675 Torr passage xiV n2 1. X 2 = = 0.02 n1 + n 2 n2 1 = n1 49 n2 1000 1 × MW1 × = 1000 49 w1 m=

1000 = 0.261 m 49 × 78

2. K f =

RT 2 M 1000DH

2(300) 2 = = 5K kg mol−1 1000 × 36 3. ΔTf = Kf i m 300–298.95 = 5 I (0.261) I = 0.804 CH3COOH → (CH3COOH)2 a 1–a 2 i = 1–

a 2

0.804 = 1–

a 2

a = 40% passage xV  1. 2X ↽ ⇀  X2 1–x a/2 α 1− 2 2 α i = 1− 2 3. ΔTb = Kb i m

 a DTb = K b 1 −  t  2 Kba t 2

DTb = K b t − (K b t − DTb ) 2 Kb t ∴a = 2 −

=a

2DTb Kb t

 2X ↽ ⇀  X2 t(1–a) t a/2 ta K= 2  t (1 − a ) 2  K=

K b (K b t − DTb ) [2 DTb − K b t ]2

integer Type questions 0 1. P − P = n 2 P n1

25 W2 18 = × , 75 60 90 100 =5 20

w = 100 gm

2. DTt = Kt m 9.3 = 1.86 ×

62 1000 × 62 w

ice = 205–w = 5 gm 3. DTf = K f

W2 100 × MW2 W1

(0.48) 100 × 10 MW MW = 120 2 = 50

120 =6 20 0 4. P − P = n 2 P n1

20 X 114 = × 80 36 114 x = 9 gm

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Solutions

5. DTf = Kf i m DTf = Kf mE×P X 1000 1.28 = 5.12 × MW 100 MWthe = 40x X 1000 1.4 = 1.86 × MW 100 MWExp = 13.28x i=

MWThe 40 x = =3 MWExp 13.28x

6. DTf = Kf i m 0.558 = 1.86i(0.1) i=3 So formula = [Co(NH3)xCl]Cl2 So cordination number is 6 So x = 5 7. Π = iCST  0.1 = (1 + 2)(0.712)   (0.0821)(300)  2 =3 8. DTf = Kfm 0.02046 = 1.86(1 + x)×(0.01) n = 0.1 H+ = 10–3 m pH = 3 9. DTf = Ktm X 1000 0.2 = K f 10 x Kf = 2 × 10–3 2 × 10−3 1000 0.25 = × x w ∴w = 8 gm ice = 10–8 = 2 gm 10. Van’t Hoff factor of NaCl, NaHSO4 is same 11. p ∝ n 733 Wnitrobenzene 18 = × WH2 O 27 121 W1 = 182.49 W2 W2 = 5.4 × 10−3 W1 12. Isotonic means same osmotic pressure 17.4 1000 5.85 1000 (1 + 2d )ST = × × 2 × ST × 174 100 58.5 100 1+2a=2 a = 0.5

13. PM = PA0 x A + PB0 x B 760 = 320 x A + 800 (1 − x A ) x = 0.08 14. DTf = Kf i m 0.00558 = 1.86i(0.001) i=3 No. of Ions: 3 RT 2 1000 L 8.3 × (350) 2 = 1000(340) = 3 K kg/mol

15. K f =

16. DTf = Kfm 0 − Tf = 1.8

P 1000 × 60 500

Tt = –0.06P Tb − 100 = 0.5

P 1000 × 60 500

Tb = 100 + 0.016p difference = 100 + 0.016P + 0.06P ∴P=6 17. DTf = Kfm 7.9 0.44 = 0.52 × 10 MW MW = 93 93 =3 31 18. Π = CST  x 5.49 =  + (0.06)2 0.0821 × 30   90 n=

x = 9 gm previous years’ iiT questions 2. DTb = Kb i m 13.49 = 0.52 3 134.4 = 0.0156 3. DTb = Kb i m 20 1000 2 = 1.72 i × 172 50 i = 0.5 4. P = K

n2 n1 + n 2

Solutions

n 4 = 1.0 × 105 × 2 n2 = 4 × 10–4 10 5. Van’t Hoff Factor i = 4 DTf = Kf i m 0 − Tb = 1.86 (4)

0.1 1000 × 329 100

Tf = –2.3 × 10–2 6. Wt of solution = 1000 + 120 = 1120 gm 1120 Vol of solution = = 973.9 mL 1.15 60 1000 M= × 120 973.9 = 2.05 M passage 7. Mole factor of C2 H5OH = 0.9 Mole factor of H2O = 0.1 cH2O = 0.1 n2 1 = n1 + n 2 10

n2 1 = n1 9 n2 1000 mw1 × 1000 = W1 9 M=

1000 = 2.415 9 × 46

DTf = Kf m = 2(2.415) = 4.83 K T0–Tf = 4.83 Tf = 155.7–4.83 = 150.9 k 8. Here C2H5OH is solvent H2O is solute P0 − P n 2 = P n1

40 − P 0.1 = P 0.9 360 =10p P = 36 9. Here water is solvent, C2H5OH is solute DTb = Kb m XC2H5OH = 0.1 m = 6.17 DTb = (0.52)(6.17) = 3.2 Tb = 376.2 K

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CHAPTER

6 Thermodynamics

T

here is no such thing as absolute truth and absolute falsehood. The scientific mind should never recognize the perfect truth or the perfect falsehood of any supposed theory or observation. It should carefully weigh the chances of truth and error and grade each in its proper position along the time joining absolute truth and absolute error. Henry A. Rowland

6.1 inTroducTion The study of chemical or physical process can be approached in two different ways. In the first case, the process may be treated in terms of atoms and molecules of the system, i.e., on the basis of the kinetic theory, which in the second way the process may be studied from a consideration of energy change involved. The second type approach is called thermodynamics approach. This science is concerned with the chemical and physical process which involve the conversion of one form of energy into another. These energy changes takes place because of the rearrangement of atoms in reactants to form products. Sometimes the energy changes associated with the chemical reactions are more important than the products of the reaction. For example, the fuels like methane, cooking gas or coal burns in air with liberation of energy stored in their molecules. The chemical energy may also be used to do mechanical work when a fuel burns in an engine or to provide electrical energy through a galvanic cell like dry cell. There are other forms of all machinery; electrical energy, radiant energy, chemical energy and nuclear energy. These different forms of energy may be transformed from one into another. The branch of science which deals with these energy transformations is referred to as thermodynamics. Thermodynamics mainly deals with the basic laws governing heat, followed later by statistical mechanics, which explains the laws of thermodynamics in terms of atoms, their movement etc. Broadly speaking, thermodynamics is the physics of heat as studied on macroscopic scale. This branch of science deals with the determination of the relationship among the various properties of materials without knowing their internal structure. Atoms are not considered and one merely deals with matter in bulk, i.e., objects of say, about 1 cm in size. There is certainly some connection between heat

energy and atoms, but now we only give stress on that thermodynamics is formal and quantitative study of a lot of facts known to us from practical and microscopic experience (like that when mechanical work is done, heat is produced). Thermodynamics basically revolves around a few laws. There are four of them, of which only two namely the First law and Second law are very common. The odd name Zeroth law was coined because the law was discovered after the first law but had to be put ahead of it.

6.2 Thermodynamic SySTemS Before learning about the laws of thermodynamics, there are a few things that we ought to know. We start with thermodynamic systems, which could be a cylinder with a piston and containing some gas or a mixture of two chemicals or something else like that. In all cases, the system is macroscopic, in the sense mentioned earlier. In thermodynamics, systems are treated in a formal way as in Fig 6.1. One does not bother too much about details but certainly notes that the system has a boundary which separates it from the surroundings. If the system is a drop of liquid, the boundary would be the surface of the liquid itself. For convenience we shall think of this as a wall. Thus “A specified part of the universe which is under observation is called the system.” “The remaining portion of the universe which is not a part of the system us called surroundings.” The system and the surroundings are separated by real or imaginary boundaries. The boundary also defines the limits of the system. The system and the surroundings can interact across the boundary.

6.2

Thermodynamics

if it can exchange both energy and matter with the surroundings, e.g., evaporation of water from a beaker presents an open system. Here vapours of water (matter) go into the atmosphere and heat (energy) required is absorbed by water from surrounding. In the photosynthesis, plants take up carbon dioxide, water (matter) and sunlight (energy) in the presence of chlorophyll and produce carbohydrates. In this process oxygen (matter) is transferred to the surroundings. Hence, plants constitute an open system. 2. Closed System: A system which can exchange energy but not matter with surroundings is called a closed system. E.g., hot coffee in a stainless steel flask is an example of closed system because energy can be gained or lost (through the steel walls) but not the matter. Similarly, chemical reactions carried out in closed containers are the example of closed system. 3. Isolated Systems: A system which can neither exchange matter nor energy with the surroundings is called an isolated system. For example, take some water in an insulated vessel and put a small piece of sodium metal in it. An exothermic reaction takes place. Neither hydrogen gas nor heat (energy) is transferred to the surroundings.

Surroundings

SYSTEM

Boundary

Fig 6.1 Schematic representation of a system with surroundings and boundary walls

6.2.1 Types of Systems The system can be acted on by external force which could be mechanical, electrical, magnetic and so on. Also, heat could be added to the system or removed from the system. All this depends of course on the nature of walls. The walls which allow the transmission of heat through them into or out of the system are called diathermal walls of the system. Those walls which do not permit any heat through them into or out of the system are called adiabatic walls of the system. Considering the interaction between system and its surrounding, a system may be divided into three classes: 1. Open System: A system is said to be an open system

There is no perfectly isolated system but a system which is completely sealed to prevent inflow or outflow of the matter and is thermally insulated to prevent flow of heat can be considered as an isolated system.

6.2.2 Thermodynamic Variables Consider a gas contained in a box. In this case, we would be interested in the pressure, temperature and volume of

Surroundings Matter

Matter

Matter

(Heat-Work)

Energy

System

Energy

Energy (a) Open System

(b) Closed System Fig 6.2 Open, closed and isolated systems

(c) Isolated System

Thermodynamics 6.3

the gas. Quantities such as these are referred to as thermodynamic variables. For example, if the system is a wire that is being pulled, the variables would include the length, the tension and so on. All the variables are macroscopic in nature and fall into two categories: intensive and extensive. In a thermodynamic system, in equilibrium, the intensive variables have the same values everywhere in the system. Pressure is an example of an intensive variable and in a gas in equilibrium, the pressure is the same everywhere within the gas. Extensive variables on the other hand, vary with the size of the system – mass is a good example. Table 6.1 gives a partial list of variables usually encountered. The list in Table 6.1 is based on common sense and can be drawn up without reference to the thermodynamics as such. When heat is considered as a form of energy, new members namely the intensive variable temperature (T) and the corresponding extensive variable entropy (S) should be added to the list in table 6.1. You can find that to every intensive variable, there is an extensive variable which is a sort of counterpart. Together they are referred to as a conjugate pair. The product of every conjugate pair has the dimension of work. It occasionally happens that the system under study has unknown conjugate variables. For example, while studying the thermodynamics of super fluid helium, physicists discovered the need for such additional variables. To allow for such possibilities one now includes in the above list generalize (intensive) force variables xi and their conjugate counter parts xi which are generalized (extensive) displacement variables. Thus the extensive and intensive Properties can be defined as follows Extensive Property of a system may be defined as any property whose magnitude depends upon the quantity of material present. Internal energy, enthalpy , entropy, Gibbs energy, heat, total mass, total volume, total energy etc. are the well known examples of extensive properties of the system. Intensive Property of a system is defined as property whose value is independent of the material or materials in

a system. Pressure, density, temperature, specific heat, volume per mole, energy per mole, morality normality, pH, gas constant R, viscosity, surface tension dipole moment, standard e.m.f . refractive index b.pt, m.pt, dielectric constant etc are the examples of intensive properties of the system.

6.2.3 Thermodynamic equilibrium Equilibrium is a familiar concept in chemistry. Dishes of different shapes are shown in Fig 6.3. On each dish a marble is placed. If the marble in Fig 6.3(a) is displaced slightly, it would rattle about for a while and then come to rest in its original place. This is an example of stable equilibrium. A small disturbance to a system in such a state does not produce any drastic change, and the system eventually returns to its original configuration. On the other hand, if the delicately poised marble in Fig 6.3(b) is displaced even slightly, it would roll over with no question of getting back. This is an example of unstable equilibrium. Compared to this, the marble in Fig 6.3(c) will roll over only if given a big enough push; it is said to be in a state of meta stable equilibrium. The marble on the flat plate in Fig 6.3(d) is equally comfortable wherever it is and is said to be in state of neutral equilibrium. We shall be mostly concerned with systems in stable equilibrium. An example of such a system is shown in Fig 6.4 where we have a cylinder with a piston and containing some gas at a certain temperature. The downward force due to the weight of the piston just balance the upward forces due to pressure and the system is in equilibrium if the piston is slightly depressed and released, it would spring back and come to rest in its original position rather like the marble in Fig 6.3(a). The gas can be slowly compressed by gently adding sand to the pan. To make the gas expand, sand is slowly removed from the pan. By placing the cylinder in contact with heat reservoir, one can either input heat into the system or heat from it while changes are being made. From now on, equilibrium would mean

Table 6.1 A partial list of thermodynamics variables: Intensive Variable

Extensive Variable

Wire Liquid film Magnetic Material

Pressure Tension Surface tension Magnetic field

Dielectric

Electric field

Volume Length Surface area Magnetic dipolemoment Electric dipolemoment

System Fluid (gas or liquid)

(a)

(c)

(b)

(d)

Fig 6.3 Illustration of the concept of mechanical equilibrium

6.4

Thermodynamics

thermodynamic equilibrium. It means that if a small external force is applied to the system and released, the system would come back to the thermodynamic state it was in originally, i.e., the value of all the extensive and intensive variables would recover. In thermodynamics, we always consider slow changes such as those produced when sand is gradually added to the pan in Fig 6.4 Besides being made infinitely slowly (at least in principle), the changes must be such that by reversing the process (e.g., by removing sand), we are able to get back to the original state. Such a process is called reversible. There are three different conditions which are to be fulfilled for the attainment of complete equilibrium. 1. Mechanical Equilibrium: A system is said to be in mechanical equilibrium when there is no unbalanced force existing between different parts of the system or between the system and the surroundings as explained above.

2. Thermal Equilibrium: A system is said to be in thermal equilibrium if the temperature is same throughout the whole system including the surroundings. 3. Chemical Equilibrium: A system is said to be in chemical equilibrium if the composition of the systems remain fixed and definite. In other words, mechanical equilibrium implies the uniformity of pressure (as explained in Fig 6.4), thermal equilibrium to the uniformity of temperature and the chemical equilibrium to the uniformity of chemical composition. A system may be in mechanical and thermal equilibrium without being in chemical equilibrium. For example, a mixture of hydrogen and iodine at the same temperature and pressure may continue to react slowly for days, until chemical equilibrium is attained. Similarity, even after attainment of chemical equilibrium and mechanical equilibrium, a system may allow thermal changes to take place or continue. A system is said to be in thermodynamic equilibrium if it satisfies all the above three equilibrium. When a system is in thermodynamic equilibrium, its properties have definite magnitudes.

6.2.4 State of a System Sand

A system to be in a given state must have definite values assigned to its properties, such as temperature, pressure, volume, composition, refractive index, viscosity etc. which are associated with a system but, the first four namely the temperature, pressure, volume and composition are of fundamental importance because these variables are quite sufficient to completely define a system thermodynamically. These variables are usually known as thermodynamic properties or parameters or state variables. For example, in a homogeneous system the composition is fixed and temperature, pressure and volume of a system are interrelated with one another in the form of equation of state. Thus a state variable is one that has a definite value when the state of system is specified. For example, in the gas equitation PV = RT; Gas

Fig 6.4 Illustration of thermodynamic equilibrium attained isothermally

if two of three variables (P, V and T) are known, the third can be easily calculated as R is gas constant. The two variables usually specified are temperature and pressure and are called independent variables. The third variable i.e., the volume depends upon the temperature and pressure and is said to be dependent variable. If the variables are found to change in a given system, then the system must also be changing in its state. This type of system will therefore, be in unstable condition. In

Thermodynamics 6.5

a system left itself the variables have a tendency to be uniform throughout the system and when uniformity in the magnitudes of variables is acquired, a steady condition is reached and the system is said to attain equilibrium. Some notable about the thermodynamic features are (i) Variation in one or more macroscopic properties brings a change in the state of the system, when other macroscopic properties attain new values. The macroscopic properties are thus, called state variables or state functions. (ii) Initial state refers to the starting state of system in equilibrium. After interaction with surroundings (involving exchange of matter or energy or both) the system attains another equilibrium state which is referred to as a final state (iii) Thermodynamic state of the system must not be confused with physical state or phase. (iv) A system is said to be in thermodynamic equilibrium state if its macroscopic properties do not change with time.

6.2.5 homogeneous and heterogeneous Systems Homogeneous System: A system is said to be homogeneous if it is uniform throughout. For example, a pure solid or a liquid or a gas or a mixture of gases or a solution of a solid in a liquid or a mixture of two or more completely miscible liquids are all homogeneous systems. In other words, a homogeneous system consists of only one phase. Heterogeneous system: A system is said to be heterogeneous if it is not uniform throughout. A heterogeneous system consists of two or more phases, which are separated from each other by definite bounding surfaces. However, each phase in itself is uniform throughout. Some common examples of heterogeneous system are (i) a mixture of two or more solids (ii) a mixture of two or more immiscible liquids (ii) a liquid in contact with its vapours.

6.2.6 State Functions The thermodynamic properties whose values depend only upon the initial and final states of the system and are independent of matter of the manner as to how the change is brought about, are called state functions. The concept of state function can be easily understood from the following analogy. If we consider ‘h’ as the height between the top and bottom of the mountain, then 'h' is the independent of the path followed in reaching the top of the mountain. Here, the parameter ‘h’ is analogous to state function. In thermodynamics, some common state functions are internal

energy (U), enthalpy (H), entropy (S), Gibbs energy (G) pressure (P), temperature (T), volume (V) etc. It may be noted that two very important thermodynamic parameters namely heat (q) and work (w) are not the state functions because they are path dependent.

6.2.7 methods of Studying Thermodynamics There are two general methods of studying thermodynamics. 1. By using cyclic processes, in which the exchange of heat, mechanical work, electrical work, etc are considered directly. 2. By using thermodynamic functions, such as internal energy enthalpy, entropy, free energy etc. which enable the deduction of heat changes and work in a given system in a somewhat more efficient way In modern thermodynamics, the interest has, however, shifted to the thermodynamic functions. Limitations of Thermodynamics: there are a few limitations of thermodynamics. These are 1. Laws of thermodynamics are not applicable to microscopic systems. 2. The mass energy transformations are not being considered except as a carrier of energy. 3. Thermodynamics is concerned with the initial and final states of the system and not with the process by which it was brought into that state, nor with the rates at which those processes occurred. Thus time element in any transformation is not considered in thermodynamics.

6.2.8 Types of Process A thermodynamic process is the path or operation by which a system changes from one state to another. There are different types of processes as described below. 1. Adiabatic Process: If a process is carried out under such conditions that no exchange of heat takes place between the system and surroundings, the process is known as adiabatic process. For example, sudden bursting of a cycle tube is an adiabatic process. In the case of an exothermic process, the heat evolved remains in the system and therefore, the temperature of the system rises while in the case of endothermic process, the heat absorbed is supplied by the system itself and hence the temperature of the system falls. 2. Isothermal Process: A process is said to be isothermal if it takes place at constant temperature when a gas is compressed suddenly, some heat is evolved. But if the compression is slow and heat evolved at once removed so that the temperature remains constant, the change is isothermal. On the other hand, if the gas is suddenly expanded,

6.6

Thermodynamics

work is done by the gas and some heat is absorbed. If now heat is supplied from an outside source, so that temperature remains constant, the change is isothermal. In adiabatic process, the temperature gets altered because the system is not in a position to exchange heat with the surroundings. In an isothermal process, the temperature of the system remains constant during each stage and the system can exchange heat with the surroundings. 3. Isobaric Process: If the pressure of the system remains constant during each step of the change in the state of the system, the process is called isobaric process. Consider a mixture of H2 and O2 in the ratio 2:1 contained in a cylinder fitted with a weightless and frictionless piston. When the electric discharge is passed through the mixture the volume of the system decreases. 2H 2 + O 2 → 2H 2 O

Hence, piston will be moved down so that the pressure of the system remains constant. 4. Isochoric Process: A process is said to be isochoric if the volume of the system remains constant during each step of the process again consider a reaction. N2O4(g)→2NO2(g) contained in a cylinder, fitted with frictionless and weightless piston in dissociation, the volume of the system increases. Thus in order to keep the volume fixed, weights are to be kept over the piston. 5. Cyclic Process: A process is said to be a cyclic process if a system after completing a series of changes returns to its original state.

6.2.9 Path Sequence of steps starting from the initial state to the final state of the systems is called the path. The path of a cyclic process is called a cycle. The final state of the system can be attained either directly or in stages. Accordingly, there are two types of paths of the systems. Consequently, there are two different types of process; these are 1. Reversible Process: A process is said to be reversible process if it is carried out infinitesimally slowly so that the driving force is only infinitesimally greater than the opposing force. A reversible process proceeds through a succession of equilibrium steps each of which is an equilibrium state. A reversible process must be carried out in such a way that enough time is allowed between each change for the properties to come to a constant value. Reversible process are ideal and cannot be realized in practice. This is because a reversible process requires infinite time for completion. Neverthless the concept of reversibility is highly useful and serves infinitesimal as a means to study many processes which are of immense important to chemists. A reversible process can be reversed at any stage by increasing the opposing force by an infinitesimal amount. 2. Irreversible process: A process is said to be irreversible if it is not carried out infinitesimally slow (instead it is carried out rapidly) so that the system does not get chance to attain equilibrium. An irreversible equilibrium cannot be reversed without the help of an external energy and without changing the properties of the surroundings.

Table 6.2 Reversible and irreversible processes Reversible process 1 It occurs at a very slow speed and involves a series of equilibrium states. 2 The opposing force and the driving force are nearly equal.

3 Work obtainable in a reversible expansion process is maximum. 4 It cannot be realized in actual practice and is only theoretical. 5 The reversible process occurs in an infinite number of steps, infinite time is required for its completion.

6 All changes occurring in the direct process can be reversed in the reverse process.

Irreversible process It takes place rapidly and does not involve a series of equilibrium states. The opposing force and driving force differ widely. Work obtainable is always less than the work obtainable in a reversible process. Most of the processes occurring in nature and laboratory are irreversible. Irreversible process does not involve many steps and gets completed in short time. In an irreversible process the change taking place in direct process cannot be reversed.

Thermodynamics 6.7

This can be easily understood from the illustration given in 6.2.3. The gas can be slowly compressed by gently adding sand particles one by one to the pan. To make the gas expand sand particles are removed slowly one by one from the pan. In this process the compression or expansion of the gas takes place infinitesimally slowly, i.e., in the thermodynamically reversible manner. When the pressure on the piston is suddenly changed by increase or decrease of weight in a large amount the change in the volumes of gas takes place rapidly and will be irreversible.

6.3 heaT and Work Before going to learn about the different laws of thermodynamics, we should know what is heat and temperature? Temperature is the degree of hotness of a body. Without heat we would all be dead. Living in the plains of India, as most of us do, we do not appreciate how essential heat is for life. All we have to do up in the Himalayas to realize how one cannot survive without heat. Even though we ourselves might be quiet indifferent to this question, our ancients were not. This becomes clear from the way they worshipped the sun. The source of all the heat energy we receive on the surface of the earth. The story of heat is more or less, as follows: In the seventeenth century there was some vague speculation about what heat was but in the eighteenth and the early part of the nineteenth century speculation gave way to careful experiments. Slowly it become clear that heat was a form of energy and that it could be converted back, into heat. Therefore, the study of heat became quantitative and various laws governing heat energy were formulated. Knowing that heat is a form of energy is fine but what really is this heat? If two identical balls of iron, one at 50°C and the other at 100°C, what is the difference between them? Quite clearly, one of them is hotter than the other but what exactly does that mean? If matter is made up of atoms then what do the atoms in the hotter substance do as compared to those in a substance which is at a lower temperature? The answer for question raised above received serious attention during the latter part of the nineteenth century and even today. To put it all crispy, first came thermodynamics. which deals with the basic laws governing heat, followed later by statistical mechanics which explains the laws of thermodynamics in terms of atoms and their movements etc. The first clue was provided by Count Rumford. Rumford was an interesting character. Born in America in 1753, he was named Benjamin Thomson. During the American struggle for independence he sided the British. When he knew that the British were losing, he fled to England. At the age of thirty-one he retired from the British army and went to Bavaria, now in Germany, to join the army there. The

king of Bavaria made him a Duke whereupon Thompson assumed the name Count Rumford. Rumford was very much interested in guns and explosives, and in fact it was a scholarly paper on explosive which he published, which elected as a Fellow of the Royal Society. By the way, while living in England, Rumford founded the royal institution, later made famous by Faraday. While supervising the manufacture of gun barrels for the Bavarian army, Rumford discovered that friction produces heat. In a paper sent to the royal society he drew attention to the fact that heat and work were connected– friction involves work–but was unable to express that relationship quantitatively. The relation between heat and work was first established by James Prescott Joule of England in 1847, using the apparatus shown in Fig 6.5 When the weights move down, the paddles rotate and heat the water. So as Rumford had said, work in fact produces heat. But Joule could go further than Rumford. Knowing the masses and the amount they moved downwards under the influence of gravity, he could calculate W, the amount of mechanical work done. On the other hand, by measuring the rise in temperature of the water, he could determine Q, the amount of heat produced. The falling weights turn the paddles which then stir the water and heat it. Knowing W, the work done and Q the heat liberated, Joule made the daring suggestion that W = Q. Today we would write that relationship as W = J.Q.

.

.

Fig 6.5 Joules arrangement for measuring the mechanical equivalent

6.8

Thermodynamics

Here J is a constant, now known as the mechanical equivalent of heat. In CGS units the value of J is 4.18 × 107 ergs/calorie. In CGS system mechanical work is measured in units of erg heat is measured in units of calorie . What all means is that 4.18 × 107 ergs of work must be done to produce one calorie of heat. In thermodynamics work may be defined as any quality that flows across the boundary of a system during a change in its state and is completely convertible into the lifting of a weight in the surroundings. Work is done in various ways. For example, 1. Electric work: When current flows in an electric circuit, electrical work is said to be done and is given by the product of the e.m.f (in volts) and quantity of electricity that flows Q (coulombs). The product Volt × Coulombs gives the Joule. 2. Gravitational work: If a body is moved through a certain height against the gravitational force, the work is said to be gravitational work and is equal to ‘mgh’ where m is the mass of the body, h is the height in cm against the gravitational field of acceleration g cm/sec2. 3. Mechanical work: Mechanical work is done when there is a change in the volume of a system. For Example mechanical work done at a constant pressure is given by W = P (V2–V1) = PΔV Where V2 and V1 are the volumes in the final and initial state of the system at a constant pressure P. If ΔV is +Ve the value of W is positive and the system, say a gas expands in the process. The work in this case is done by the system on its surroundings. If ΔV is –Ve, W will have a negative sign and the gas undergoes contraction. In this work is done by the surroundings on the system. In IUPAC system work done by the system is –Ve in expansion process and work done on the system is + Ve in contraction process.

A

B

6.4 ZeroTh LaW The Zeroth law is a very simple one based on experience. Suppose if a hot iron-rod is plunged into a bucket of water, there would then be a hiss accompanied by the formation of some steam and also some bubbles. If the rod is pulled out after a while, we would find that was no longer hot because heat from the rod has passed into the water in the bucket. The heat flow from the rod stops when the rod and water are at the same temperature. The zeroth law essentially summarizes all this experience. The famous German physicist Arnold Somerfeld states as follows “There exists a property called temperature. Equality of temperature is a condition for thermal equilibrium between two systems or between two parts of a single system.” A corollary is that heat flow form one system to another is possible when the two systems are not at the same temperature. The zeroth law sometimes stated differently as below “If two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other.” The common use of thermometer in comparing the temperature of any two or more systems is based on zeroth law. For example, if we want to compare temperature of two bodies A and B , then thermometer (analogous to body C) is allowed to come into thermal equilibrium first with body A and then with body B by placing it in contact with A and B turn by turn. The temperature readings on thermometer give the comparative ideas of degree of hotness of two bodies A and B. It may be noted that, the thermometer is a very small body as compared to A and B.

A

C

B

C

C is in thermal equilibrium with A as well as with B

A and B are also in thermal equilibrium with each other

Fig 6.6 Illustration of zeroth law of thermodynamics

Thermodynamics 6.9

Hence, exchange of energy between A and C or between B and C is insignificant and individual energies of body A and B remain unchanged during temperature measurement. Temperature Scales How temperature is measured? Usually, we use the mercury thermometer to measure temperature but this is not the only type of what is called the thermometric property of the substance used in the thermometer. In the early days one arbitrary took some temperature as zero temperature and measured the change in the property with respect to that zero temperature. Thus, Fahrenheit chose the zero of temperature as that of mixture of ice and salt, this being the lowest temperature he could go to. Celsius on the other hand, took the melting of ice to be zero of temperature. After defining zero temperature one must also define the unit of temperature or the interval which we call as a degree. In the centigrade scale, now called Celsius scale, the temperature interval between the melting point of ice and the boiling point of water into 100 equal parts and calls the each part as a degree (Celsius).There is a similar scheme in the Fahrenheit scale. The Celsius and Fahrenheit temperature scales are arbitrary. Anybody can take any two points, not necessarily melting of ice and condensing steam and define temperature scale. A universal or unique scale is developed as follows. Consider a constant volume thermometer. A fixed quantity ‘Q’ of some gas is made to always occupy a fixed volume ‘V’, no matter what the temperature ‘T’ is. From the pressure reading ‘P’ the temperature is deduced using PV = RT. However this applies only to perfect gas. Real gases like oxygen, nitrogen etc do not obey this law. The

temperature corresponding to the triple point of water is taken as standard temperature. The triple point of water is a fixed temperature and pressure (0.01°C and 4.58 torr) at which all the three physical states of water ice, water and water vapour: exist in equilibrium. Results obtained by measuring a fixed temperature (like that of melting ice) by using constant–volume thermometer but by varying the quantity Q of a gas in it the different curves are for different gases. Observe the same Q of the different gases lead to different readings. This is because real gases do not obey ideal gas equation (PV = RT). However, in the dilute gas limit the readings of all the thermometers converge to the same value, which is also the correct reading. From the equation PV = RT, one can notice that when T = 0, the pressure P is also zero. With zero temperature fixed in this manner one has the perfect gas scale (pgs). The temperature at triple point of water (0.01°C) is taken as 273.16. With this choice melting point of ice is equal to 273.15. But on the Celsius scale we read this temperature as 0°. Hence 0°C corresponds to 273.15°C on pgs. in turn this implies that 0°C on pgs corresponds to –273.15°C. This pgs is known as the thermodynamic scale or absolute scale the unit of temperature in the absolute scale is defined by taking the triple point as 273.16 and is called degree Kelvin in honour of Lord Kelvin who made important contribution to this subject. Temperatures on the pgs are denoted by k (degree Kelvin). T°C = Tk–273.15 Tk = 273.15 + t°C 273.15 + t°C

V Fixed O2

T ice(Q)

Air N2

237.16 K 237.15 K

He

Triple point Melting point of ice

0.01°C 0.0°C

32°F

H2 Correct value (a)

P(Q)

0K

-273.15°C

-459.69°F

(b)

Fig 6.7 (a) Results obtained by measuring a fixed temperature (like that of melting ice) by using a constant–volume thermometer but by varying the quantity Q of gas in it. (b) Comparison of Kelvin, Celsius and Fahrenheit temperature.

6.10

Thermodynamics

In short the temperature scales can be illustrated as 1. The thermodynamic scale is the absolute scale for measuring temperatures. 2. The thermodynamic scale is equivalent to the perfect gas scale (pgs). 3. The zero of the both scales is the same and is called absolute zero of temperature.

6.5 inTernaL energy Every system has within itself a definite quantity of energy called the internal energy or intrinsic energy and is represented by the symbol U or E. It is made up of kinetic energy and potential energy of the constituent particles. The kinetic energy arises due to motion of its particles and includes their translational energy, rotational energy, vibration energy etc. The potential energy arises from different types of interactions between the particles and includes electronic energy, energy due to molecular interactions, nuclear energy etc. The sum of all forms of energies stored in atoms of molecules is internal energy. Internal energy is an extensive property and is also a state function. Its value depends upon state of the substance but does not depend upon how that state is achieved. For example, CO2 can be obtained by various methods such as by heating calcium carbonate or by burning coal. However, one mole of CO2 at S.T.P is associated with a definite amount of internal energy which does not depend upon the source from which it is obtained. Different substances have different internal energies depending upon the nature of the constituting atoms, bonds and other conditions of temperature, pressure etc. For example the internal energy of 1 mole of CO2 will be different from the internal energy of 1 mole of SO2 even under similar conditions of temperature and pressure. Further, the internal energy of 1 mole of water at 300 K is different than that of one mole of water at 310 K under same atmospheric pressure. The exact magnitude of the internal energy cannot be determined, because of the fact that chemical nature includes various indeterminate factors, such as translational, rotational and vibration movements of the molecules, the nature of individual atoms, the arrangement and number of electrons. However, we can determine the change in internal energy (ΔU) of the system when it undergoes a change from minimal state (Ui) to final state (Uf). For example UA and UB are the internal energies in states of A and B respectively. Then the difference between the internal energies in the two states will DU = U A − U B

For chemical reactions the change in internal energy may be considered as the difference between the internal energies of products and that of the reactions, i.e.,

DU = DU products − U reactants U p −Ur Where Up is the internal energy of the products Ur is the internal energy of reactants and ΔU gives the change in internal energy. The significant features of U are 1. Internal energy depends upon the quantity of substance in the system and hence, it is an extensive property. 2. Change in internal energy (ΔU) represents the heat evolved or absorbed in a reaction at constant temperature and constant volume. 3. In a process involving liberation of energy Up0 or sign of ΔU is positive. 5. For isothermal process involving ideal gas, T is constant. Hence ΔU = 0 for both reversible and irreversible process. The change in internal energy in a chemical reaction can be measured using bomb calorimeter.

6.6 The FirST LaW oF ThermodynamicS Basically, the first law of thermodynamics is a law of conservation of energy. Let us say there is a system in which some work is done, increasing the internal energy U of the system by an amount ΔU from conservation of energy we have ΔW = ΔU Suppose instead of doing work on the system we just added an amount ΔQ of heat energy which raises the internal energy. In this case we would have ΔQ = ΔU If heat ΔQ is supplied and work ΔW is done on the system then the increase ΔU in internal energy is given by ΔU = ΔW + ΔR From the above observations, the first law of thermodynamics can be stated in any one of the following ways: 1. Energy can neither be created nor destroyed by any physical or chemical change but may be converted from one form into another.

Thermodynamics 6.11

2. The sum of all forms of energies in an isolated system is constant. 3. When one form of energy disappears, exactly the same amount of energy appears in some other form. This however, has been partly modified since it is now known the energy can be produced by the destruction of mass and is given by the Einstein relation E = mc2 Thus in the modified form the law states that the total mass and energy of an isolated system remains constant though these are inter convertible. In the beginning we have taken the work is done on the system but if work is done by system, the internal energy of system decreases. In this case work is taken as negative (–W). Now if Q is the amount of heat added to the system and W is the work done by the system then change in internal energy becomes ΔU = Q + (–W) = Q-W The relationship between internal energy, work and heat is a mathematical statement of first law of thermodynamics.

Sign convention for heat and Work It is very important to understand the sign convention for Q and W The signs of W and Q are related to the internal energy change. When energy has been supplied to the system as work and heat, W and Q are positive. The internal energy in such a case increases. On the other hand, if energy of the system is decreased by removing some heat or by the work done by the system, W and Q are negative. In such case the internal energy of the system decreases. The conventions can be illustrated as • If heat is absorbed by a system Q is positive (as there is increase in energy of the system) but if heat is evolved by a system Q is given negative sign (since there is loss in energy of the system). For example if we supply 25 KJ of heat energy to the system, we write Q = + 25 KJ, but if same amount of heat is given out to the surroundings, we write Q = –25 KJ Heat absorbed by the system = Q positive Heat absorbed by the system = Q negative Work done on the system = W positive Work done on the system = W negative •

Energy is a state function and ΔU depends only on the initial and final states whereas Q and W are not state functions. The values of Q and W depends upon the way in which the change is carried out. However,

•

whatever may be the process ΔU is always equal to Q + W and therefore Q + W is also a state function. For an ideal gas undergoing an isothermal change the internal energy of the system does not change, i.e., ΔU = 0, then 0 = Q + W or Q = –W

•

This means that heat absorbed by the system is equal to the negative of the work done on the system.

6.6.1 applications of First Law of Thermodynamics Many chemical reactions involve the generation of gas capable of doing mechanical work or the generation of heat. These changes must be quantified and relate them to the changes in internal energy. Work (W): It is another mode if transference of energy. Work is said to be performed if the point of application or force is displaced in the direction of the force. It is equal to the force multiplied by the displacement (distance through which the force acts). There are two main types of work which we generally come across. These are (i) electrical work and (ii) mechanical work. Electrical work is important in systems where reaction takes place between ions, whereas mechanical work performed when a system changes its volume in the presence of external pressure. Mechanical work is important specially in the systems that contain gases. This is also known as pressure–volume work.

calculation of mechanical Work (a) Work done by the system: Consider the combustion of petrol in an automobile engine 31 C10H22(l) + O2(g) → 10CO2(g) + 11H2O(l) 2 + Mechanical work Consider that the reaction takes place in a gas cylinder having frictionless and weightless piston. Having area of cross-section equal to A. Let the external pressure acting on the piston is slightly less than pressure of the gas inside the cylinder Pex. If the external pressure acting on the piston then the piston moves upward till the pressure inside the cylinder becomes equal to Pex as shown in Fig 6.8. During the expansion suppose the piston moves a very small distance l then change in volume is given by Volume change = l × A = ∆V = (Vf -Vi) Force We know that pressure = area

6.12

Thermodynamics

Pex

l

l

Pex

Pex

Area = Pex V Vi

Pressure, P

Pressure, P

Pex

Vf Volume, V

(a)

Area = Pex V Vf

Vi Volume, V

(b)

Fig 6.8 (a) Irreversible expansion against the constant pressure Pex (b) Irreversible compression against the constant pressure Pex ∴ Force = Pressure x Area So pressure on the piston F = Pex. A If W is the work done by the system due to movement of the piston then W = Force x distance = Pex. A.l = Pex (–∆V) = –Pex (Vf-Vi) Since there is increase in volume the work is done by the system. In the expansion of gas, work is done by the system. Here (Vf-Vi) will be positive and positive multiplied a by negative will become positive. Hence the sign obtained for the work will be negative. The negative sign in the above expression is required to obtain conventional sign for W. (b) Work done on the system: consider the following reaction in a cylinder 1 O2 (g) → CO2 (g) CO (g) + 2 Due to lesser number of moles of products than the reactants, there will be decrease in volume. The piston moves down. Hence work is done by the surroundings on the system. During this compression suppose piston moves a distance l and is cross sectional area of the piston is A. Then volume change = l xA = ∆V = (Vf-Vi) Force ∴ Pressure = area Force on the piston = Pex A

If W is the work done on the system by movement of the piston then W = force x distance = Pex. Al = Pex (–∆v) = –Pex (Vf-Vi) Again the negative sign of this expression is required to obtain conventional sign for w which will be positive. This indicates during compression of gas, work is done on the system. Since (Vf-Vi) is negative and negative multiplied by negative will be positive. Thus the sign obtained for the work will be positive.

6.6.2. Work done in isothermal and reversible expansion of an ideal gas In an isothermal expansion heat is allowed to flow into or out of the system so that the temperature remains constant. If the gas is allowed to expand at constant temperature, the pressure will decrease during expansion. Thus if the gas is ideal, its expansion at constant temperature will be regarded as isothermal expansion and the latter will not be accompanied by any change in internal energy. Under these conditions ∆U = O But according to first law of thermodynamics ∆U = Q-W O = Q-W Or Q = W

Thermodynamics 6.13

Let us consider the work done when a perfect gas is allowed to expand in a cylinder provided with a weightless piston. Due to expansion of the gas, the piston is raised and under such conditions the maximum work is obtained when the external pressure on the piston is less than the pressure of the gas by an infinitesimally small amount. Now, we assume further that the pressure changes in infinite number of steps in such a way that it always remains infinitesimally smaller than the pressure of gas at each step (conditions very close to reversibility). If there is small increase in volume dV at each step the net work done by the gas for the finite change of volume Vi to Vf is called reversible work (Wrev) and is given summation of all Pex dV terms . It can be written mathematically as Vf Vi

Wrev = ∫ Pex dV

vf

vf

vf

vi

vi

vi

V f = − ∫ Pex dV = − ∫ ( Pin ± dP )dV = − ∫ Pin dV ± dPdV The product (dP × dV) can be neglected, as it is product of two extremely small quantities

∫

vf

vi

Pin dV

Now, the internal pressure of the gas, Pin can be expressed in terms of its volume RT For one mol ideal gas, Pin = V (For n mol of ideal gas Pin = Wrev = − ∫

vf

vi

Wrev = Wmax It may also be noted that if Pex = 0; then W is also 0 this means that no work is done during free expansion of ideal gas. Electrical Work is significant in electrochemical cells, where redox reactions occurs and the change is transferred through the external circuit. If Q is the charge flowing through the conductor and E is the potential difference across the conductor then electrical work is given by the expression Wele = EQ Electrical work is also known as non PV-work or nonexpansion work.

Here at each stage Pex = Pin – dP . It may be noted that in case of compression of a gas under similar conditions Pex at every stage is given as Pex = Pin + dP . In general we can write Pex = Pin − dP The expression for reversible work can be written as

Wrev =

work the external pressure has to be infinitesimally smaller than the internal pressure of the gas. These conditions are nearly close to reversibility. Thus we can write that for a given change of volume

nRT ) V

Vf Vf RT dV = –RT ln = –2.303 RT log V Vi Vi

For n moles of gas the expression for reversible work under isothermal condition is given by Vf Wrev = –2.303 nRT log Vi Pf = 2.303 nRT log Pi Now, for a given change in volume (∆V), W can be maximum if Pex is maximum. But for expansion Pex to be smaller than Pin. This means that for getting maximum

Solved Problem 1 A gas cylinder of 5 L capacity containing 4 Kg of He gas at 27°C developed leakage leading to the escape of the gas into atmosphere. If atmospheric pressure is 1.0 atm. calculate the work done by the gas assuming ideal behaviour Solution: 4 × 103 = 103 moles No of moles of gas n = 4 Initial volume V1 = 5 L, T = 27 + 273 = 300 k Final Volume V2 =

nRT 103 ×0.0821×300 = = 24630 L P 1

∆V = V2-V1 = 24630 – 5 = 24625 L WPV = –P∆V = –1 × 24625 L – atm = –24625 × 101.3 J = –24625 × 101.3 × 10–3 KJ = –2.494 × 103 KJ Solved Problem 2 Calculate the maximum work done when three moles of an ideal gas are allowed to expand isothermally and reversibly from a volume of 1 litre to volume of 10 litre at 27°C. Solution: We know V W= -nRT×2.303 log10 f Vi 10 = –3 × 0.0821 × 300 × 2.303 log 1 = –170.17 litre atmospheres = –170.17 × 101.3 = 17017 ×101.3 ×10–3 KJ = –17.238 KJ

6.14

Thermodynamics

Problems for Practice 2.5 mol of ideal gas at 2 atm and 27°C expands isothermally to 2.5 times of its original volume against the external pressure of 1 atm calculate work done. b. If the same gas expands isothermally in a reversible manner, then what will be the value of W. Calculate the work performed when two moles of hydrogen expand isothermally and reversibly at 25°C from 15 to 50 litres. One mole of a liquid is converted into its vapours at its boiling point (353.2 K) by supplying heat. The vapour expands against the pressure of 1 atm. The heat of vaporization of the liquid is 395 J g–1. Calculate Q, W, and ∆U for the process if molecular mass of the liquid is 78. A sample of gas is compressed by an average pressure of 0.5 atm so as to decrease its volume from 400 cm3 to 200 cm3. During the process 8.0 J of heat flows out to surroundings. Calculate the change internal energy of the system. A system has internal energy U1. If 600 J of heat is supplied to it and at the same time it does 450 J of work. Calculate the value of U2. Calculate W and internal energy change for the conversion of 1 mole of steam at a temperature of 100°C and at a pressure of 1 atmosphere. Latent heat of vaporization of water is 9720 cal/ mole. A vessel contains 10 moles of an ideal gas 298 K and 10 atmospheres, what will be the work done if the gas is allowed to expand isothermally (i) into a vacuum (ii) under reversible conditions till the value of final pressure is 2 atmosphere (iii) against a constant opposing pressure of 2 atm. Calculate the minimum work necessary to compress 64 g of O2 from 10 to 5 litre at 300 K. How much heat is evolved in this process? A gas expands from 3 dm3 to 5 dm3 against a constant pressure of 3 atm. The work done during expansion is used to heat 10 mole of water of temperature 290 K. Calculate final temperature of water. Specific heat of water = 4.184 J g–1 K–1. Work done in expansion of an ideal gas from 4 litres to 6 litre against a constant external pressure of 1.5 atm was used to heat up 1 mole of water at 291 K. If specific heat of water is 4.18 J g–1 K–1. What is the final temperature of water?

1. a.

2.

3.

4.

5.

6.

7.

8.

9.

10.

6.6.3 Work done in an isothermal irreversible expansion of an ideal gas The irreversible expansion of an ideal gas is of two types. Type I: Expansion in Vacuum (External pressure is zero) In this case Pex = 0 W ∫ dW = ∫ PdV = 0 Hence Q = W = 0 ∆U = 0 Type II: Intermediate Expansion: In this case, work is done against a constant pressure which has a value in the range. Q < Pex ≤ P2 Thus Q = W = Or

∫

V2 V1

Pex dV = Pex (V2-V1)

Q = W = Pex (V2-V1) = Pex ∆V

 nRT nRT  − = Pex  P1   P2  P = nRT 1 − 2  P1

  ; if P2 = Pex

It may be noted that the work done on the system due to compression in an irreversible process will be different in magnitude and sign, than the work of irreversible expansion.

 nRT nRT  − (–Wirr) = P1 (V2–V1) = P1  P1   P2 P  = nRT  1 − 1  P2  Or –Wirr = –nRT

 P1  1 −   P2 

 P  Wirr = nRT 1 − 1   P2  We know that in compression V2 < V1, work done will be negative. As P1 and P2 are different the two works are also different. Value of ∆U: As the gas is ideal and temperature is constant ∆U = U2 –U1 = 0 The heat change, Q is expressed as Q = ∆U + W = 0 + W

Thermodynamics 6.15

 P  Q = W = nRT 1 − 2   P1  The following conclusions can be drawn. (i) The work done in an irreversible process is less than the work done in the reversible process. (ii) Heat absorbed in an irreversible expansion of an ideal gas is less than heat absorbed during the reversible process. (iii) Work done during compression is negative but in IUPAC system it is + ve. (iv) The magnitude of work involved in an irreversible compression will be more than that in a reversible process. (v) Work done in isobaric process W = –Pex (V2–V1) = –Pex ∆V. (vi) Work done in isochoric process is zero. Solved Problem 3 Five moles of an ideal gas at 27°C are allowed to expand isothermally from initial pressure of 1.01325 × 106 Nm–2 to a final pressure of 4 atm against a constant pressure 1.01325 × 105 Nm–2. Calculate W, Q and ∆U. Solution: As the gas expands against a constant external pressure (Pex = 1 atm) Therefore  nRT nRT  − W = Pex (V2 – V1) = Pex  P1   p2 P P  = nRT  ex − ex   P2 P1  Using the given data n = 5; P1 = 1.01325 × 106 Nm–2, P2 = 4 atm W = 4 × 1.01325 × 105 Nm–2 Pex = 1.01325 × 105 Nm–2, T = 300 K ∴ W = 5 ×8.314 JK–1 mol–1 × 300 K  1.01325 × 105 1.01325 × 105  × − 5 6  4 × 1.01325 × 10 1.01325 × 10 

3 1 1  = 5 × 8.314 × 200  −  = 12471 × J = 1870.6 J 20  4 10  For isothermal expansion ∆U = 0 Q = W = 1870.6 J

6.6.4 adiabatic expansion of ideal gas An adiabatic change is a change in which the system is thermally insulated so that no heat enters or leaves the system.

Because in an adiabatic process, the temperature changes as the system is not allowed to exchange heat with its surrounding. There will be a fall in temperature in an adiabatic expansion of a gas, while there will be a rise in temperature in an adiabatic compression. In an adiabatic process, since no heat is absorbed or evolved by the system, we have Q = 0 and thus from first law of thermodynamics ∆U = W. But in expansion, work is done on the surrounding by the system. In expansion, therefore the temperature of the system will fall and there will be a decrease in internal energy. Suppose in the expansion of the gas the volume increase from V to +dV and temperature decrease from T to T –dT. Work done by the expansion of gas will be PdV Thus dU = W = PdV  dV  Since dU = –CVdT = –PdV = RT   (for one mole  V  of the gas) dV CV dT Or = V R T Integrating the above equation between the limits V1 and T1 and V2 and T2 we get V2

CV dV =– R V V1

∫

∫

T2

T1

dT T

T CV Or ln = ln 1 T2 R . But CP–CV = R CP − CV V2 T2 ln = ln CV V1 T2

 CP   V T − 1 ln 2 = 1   V1 T2  CV   V2 T1 CP Or γ-1 ln = ln (putting = γ) V1 T2 Cv

 V2  Or ln  V   

g -1

= ln

1

 V2  Or  V    1

g -1

=

T1 T2

T1 T2

Further, since P1V1 = RT1 and P2V2 = RT2, it follows.

P1V1g = P2 V1g or PVg = Constant. Moreover P1(1−g ) × T1g = P21−g T2g + g

(1-g )  T1   P2   T  =   2  P1 

6.16

Thermodynamics

Solved Problem 4 A given mass of a perfect gas at 0°C is compressed suddenly to a pressure 20 times the initial value. Taking g = 1.42. Calculates the final temperature of the gas. Solution: We know that g

 T1   20P1   T  =  P  1 2

  P = (Cv + R) T2 = T1  R 2 + C V    P1   = Cv T2 = T1  R P2 + C V  P  1    P2  CV + R   P1 T2 =   CV  

1-1.42

Here T1 = 0.273 = 273 K, T2 = ? P1 = initial pressure P2 = 20 P1 1.42 1-1.4.2  273   20P1  =   T   P1  2 Or T2 = 662.2 K T2 = 662.2 – 273 = 389.2°C

  T  1  

Enthalpy change can be calculated as follows ∆H = nCP (T2 – T1) Substituting of T2 from the above expression, ∆H can be calculated.

6.6.5 Work done in the irreversible expansion of an ideal gas in adiabatic Process

6.6.6 comparison Between isothermal and adiabatic expansion of an ideal gas

The work done against constant pressure is given as

In an isothermal process, temperature of the system remains constant where as in adiabatic process, temperature has to be changed. The pressure-volume relation for reversible isothermal process is given by Boyle’s law

δW = Pex × dV Thus w =

∫

V2

V1

PdV

= Pex (V2 – V1) = Pex

 nRT2 nRT1   P − P  2 1

If external pressure (Pex) is equal to the final pressure P2 then  nRT2 nRT1  − W = P2  P1   P2 T P  = nRT1  2 − 2   T1 P1  Here, T2 = Final temperature in adiabatic irreversible expansion. It is different than in the adiabatic reversible expansion. Internal energy change: For adiabatic change Q=0 ∴ ∆ U = –W = –P2 (V2 – V1) = nCV (T2 –T1)  nRT1 nRT2  − = P2  P2   P1 P  or Cv (T2– T1) = RT1  2  –RT2  P1 

Temperature being constant PV = Constant Also pressure-volume relation for a reversible adiabatic process is given by. PVγ = Constant We know that Cp is always greater than Cv. Hence the CP ratio of C = γ (γ is greater than unity) V

STATE II

STATE I ISOTHE

RMAL

E=0

AD IA

BA TIC

T=0 q=0

Fig 6.9 Comparison between Isothermal and Adiabatic expansion

Thermodynamics 6.17

Clearly, the increase in volume for a given decrease of pressure will be less in adiabatic expansion than that in isothermal expansion. Hence P-V curve will be steeper for an adiabatic than for isothermal process starting at the same point. The smaller change in volume in adiabatic process than in isothermal process can be explained by the fact that in adiabatic process, expansion takes place at the expense of internal energy. Due to this, internal energy becomes less resulting in the fall in temperature. This cooling effect causes decrease in volume. But in isothermal expansion heat is absorbed to make up for expansion work done by the gas and hence the temperature remains constant. As the work of expansion is equal to the area under the curve, it is clear from the figure that work done by an ideal gas is greater in isothermal expansion than that in adiabatic expansion. Solved Problem 5 A gas originally at 1.10 atm and 298 K underwent a reversible adiabatic expansion to 1.00 atm and 287 K. What is the molar heat capacity of the gas? Solution: For reversible adiabatic expansion in terms of pressure and temperature  T2     t1   287     298 

CP /R

=

P2 P1

=

1.00 1.10

CP / R

CP 100  287  log   = log 2 110  298  −CP × 0.01633 = −0.04139 2 ∴ Cp = 5 Cal mol−1 Problems for Practice 5   11. Five moles of an ideal gas  CV = R  are allowed  2 

to expand adiabatically and reversibly at 300 K from a pressure of 10 atmosphere to a pressure of 1 atmosphere. Calculate the final temperature of the gas and the amount of work done by the gas . 12. A bulb of 60 watt is switched on in a room of dimensions 5×4×3 m2. What will be the increase in temperature and 1 atm is 0.71 J g–1 K–1 and heat capacity of four walls and the roof is 50 × 103 J K −1 (density of air = 1.22 × 10 −6 Kg mL−1

6.7 enThaLPy The change in internal energy gives the heat change accompanying a chemical reaction at constant volume. However, in laboratory, most of the chemical reactions are carried out at constant pressure rather than at constant volume. For example, all systems open to atmosphere are obviously at constant atmospheric pressure. When a chemical reaction takes place in an open atmosphere, there may be change in volume but the pressure remains constant, i.e., atmospheric pressure. To study the heat changes for reactions at constant pressure and at constant temperature, a new term called enthalpy has been introduced. We have already studied that energy change occurring during the reaction at constant temperature and constant volume is given by internal energy change. There are two types of energy changes in chemical reactions: exothermic and endothermic. Here more precisely these are referred to as heat changes. There is a convention to show when reactants lose energy in a reaction. We show the heat change as negative number. The heat change that occurs when reactions takes place under the ordinary conditions of an open vessel where the air pressure is constant is called enthalpy change of the reaction, that is an Enthalpy change is Heat change that Take Place at Constant Pressure The symbol for enthalpy is H and a change in enthalpy is shown the Greek capital letter delta, ∆, in front of the H like as ∆H. For example, the heat liberated in the reaction when zinc metal react with copper (II) sulphate solution is about 210 KJ mol–1. This can be represented as follows: Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s); H = –210 KJ mol– The equation ∆U = Q + W may be written as ∆U = QP-P∆V at constant pressure where QP is heat absorbed by the system and –P∆V represent expansion work done by the system. If the initial and final states are represented with the subscripts 1 and 2, the above equation can be written as

U 2 − U1 = Q P – P ( V2 − V1 ) Rearranging the above equation

Q P = (U 2 + PV2 ) – (U1 + PV2 ) Now, the enthalpy can be defined as H = U + PV Therefore, QP = H2 – H1 = ΔH Although Q is a path dependent function, H is a state function because it depends on U, P and V, all of which are state functions. Therefore, ∆H is independent of path. So QP is also independent of path.

6.18

Thermodynamics

For finite changes at constant pressure, we can write the equation H = U + PV as ΔH = ΔU + Δ(PV) Since P is constant, we can write ΔH = ΔU + PΔV At constant volume ΔH = ΔU + VΔP This implies that whenever heat is absorbed by the system at constant pressure, we are actually measuring the changes in the enthalpy. ∆H is negative for exothermic reactions which evolve heat during the reaction and ∆H is positive for endothermic reactions which absorb heat from surroundings. At constant volume (ΔV = 0); ΔU = Qv. Therefore the equation ΔH = ΔU becomes ΔH = ΔU = Qv Since solids and/or liquids do not show any signifi cant volume changes on heating; the difference between ∆H and ∆U is also not significant for system consist ing solids and/ or liquids. However, when gases are involved, the difference between ∆H and ∆U is significant. For example, if V1 is the total volume of the gaseous reactants, V2 is the total volume of gaseous products, n1 is the number of moles of the gaseous reactant and n2 is the number of moles of the gaseous products, all at constant pressure and temperature, then using the ideal gas law we write And Thus Or Or

PV1 = n1RT PV2 = n2RT PV2 – PV1 = n2RT = (n2 –n1) RT P (V2 –V1) = (n2 –n1)RT PΔV = ∆ng RT

Here Δng refers to the difference in the number of moles of gaseous products and gaseous reactants. Substituting the value of P∆V from equation

smaller ∆H at a higher pressure than at a lower pres sure; some show the opposite behaviour. In order to make comparisons of ∆H values one must, know about the pressure used. The convention is to choose a pressure of 1 atmosphere (in non SI units) 101.325 KPa in SI units as the standard pressure. Values of enthalpy changes that are measured at the standard pressure (or corrected to apply to this condition) are called Standard enthalpy changes. Standard enthalpy changes are given a special symbol H . Similarly, enthalpy values can change with temperature and it is accepted when ever possible enthalpy change should refer to the change taking place at 25°C i.e., 298 K. When this is done we have yet another symbol: ∆H (298 K) For example for the zinc and copper (II) sulphate reaction to be precise we should write O

O

Zn (S) + Cu2+ (aq) →Zn2 + (aq) + Cu (s) DH (298 K = –210 KJ mol–1)

However, it becomes tedious to keep writing the temperature. So we assume that enthalpies refer to 298 K, so the extra information in bracket will be omitted. The enthalpy changes for formation of different number of moles of products are also shown in the reaction. For example, 1 3 N2(g) + H2(g)→ NH3 (g) ∆H = –46.0 KJ mol–1 2 2 O

N2 (g) + 3H2 (g) → 2NH3 (l)

∆H = –92.0 KJ mol–1 O

The first equation gives the standard enthalpy change for the production of 1 mol of ammonia. The second gives the standard enthalpy change for production of 2 mol of ammonia. The units of ∆H are KJ mol–1 and always assume that ∆H value by the side of the reaction refers to the quantises shown in the reaction. In some books these value are also called as molar standard enthalpy changes and given the symbol ∆H m. However we are not using the subscript, m on the other hand, we shall use subscripts to emphasise, special types of enthalpy changes; for example, ∆H C will stand for the standard enthalpy change of combustion of a compound. The enthalpy can be summarized as • An enthalpy change DH is a heat change that takes place at constant pressure. • Enthalpy is a state function, i.e., DH does not depend on the history of the chemicals taking part in a reaction. • A standard enthalpy change DH refers to 101.325 K Pa (1 atm) and normally refers to 298 K. • Endothermic reaction: DH is positive. • Exothermic reaction: DH is negative. O

O

O

O

∆H = ∆U + P∆V using P∆V = ∆ng RT we get ∆H = ∆U + ∆ng RT This equation is useful calculating DH and DU and vice versa.

6.7.1 enthalpy and Standard States Enthalpy values can be measured for many reactions and we can calculate the enthalpy changes for the reactions that do not occur. Experimentally, it was found that ∆H changes with pressure. Some reactions have a

O

O

Thermodynamics 6.19

6.8 Second LaW oF ThermodynamicS From a study of first law of thermodynamics, it is evident that (1) various forms of energy can be converted into heat and also into one another, (2) when one form of energy disappears, an equivalent amount of another must appear. However, the first law is still incomplete, because this law does not tell us under what conditions and to what extent it is possible to bring about conversion of one form energy into the other. Moreover, it also fails to explain the direction in which the process of transformation would occur. First law of thermodynamics puts no restrictions on the directions of heat flow. However, the flow of heat is unidirectional from higher temperature to lower temperature. Infact, all naturally occurring processes whether chemical or physical will tend to proceed spontaneously in one direction only. For example, a gas expanding to fill the available volume, burning carbon in dioxygen giving carbon dioxide. But heat will not flow from colder body to hot body on its own, the gas in a container will not spontaneously contract into one corner or carbon dioxide will not form carbon and dioxygen spontaneously. Several spontaneously occurring change show unidirectional change. Then what is the driving force of spontaneously occurring changes? Then what are spontaneous, and non-spontaneous processes? 1. Spontaneous processes (Natural processes): Spontaneous process is that process which has the urge or tendency to undergo a physical or chemical change under certain given conditions. Spontaneous process can occur of their own or acquire energy for their initiation. All naturally occurring processes are spontaneous and irreversible. Some examples are given below. (i) Water flows from a higher level to the lower level. We cannot reverse the direction of flow without some external aid. (ii) In case of a bar metal having different temperatures at two ends heat flows spontaneously from the hot end to the cold end. This process can also not be reversed. Thus heat flows from a higher temperature to the lower temperature. A metal bar having uniform temperature can never become hot at one end. (iii) When two solutions of different concentrations are brought into contact diffusion of solute starts form a more concentrated solution to a less concentrated solution. A solution of uniform concentration cannot be made more concentrated at one end as compared to the other end. (iv) Electricity also flows form a higher potential to the lower potential. The direction of flow of electric current be reversed only by applying an external field in the opposite direction.

From the above examples it is clear that all the naturally occurring process proceed spontaneously (without external aid) and are thermodynamically irreversible in character and will lead to equilibrium. They are unidirectional and proceed at definite and measurable rates. 2. Non – spontaneous or unnatural process: it is reverse of spontaneous process for example (i) flow of current from a lower potential to the higher potential and so on. On analysis of these process, we find that for any spontaneous processes, it is possible to devise a mechanism by which we can obtain useful work from it. For example (i) falling water can turn a wheel (ii) an expanding gas can push a piston and so on. It is clear that a spontaneous process occurs by itself. When a gas expands into vacuum, no useful work is done because energy is lost and not harnessed. Also we note that a non-spontaneous process can be made to occur only by supplying energy from outside to the system. The decomposition of water by the passage of electric current is a non- spontaneous process. 2H2O(l)  → 2H2(g) + O2(g) Electricity

The process continues as long as the electric current is maintained. As soon as the supply of electricity is cut off, the decomposition of water stops. From this, we conclude that a non-spontaneous process is that which has no tendency to occur or which is forbidden and is made to occur only if energy from outside is continuously supplied.

6.8.1 The decrease in enthalpy is not a criterion but a contributor for Spontaneity From the study of a wide range of chemical reactions, it has been found that an energy change is invariably involved. Since reactions which proceed with the liberation of heat (exothermic reactions) are generally those which take place most readily, it is tempting to suggest that the tendency for chemicals to react in order to evolve energy is the factor which decides whether a particular reaction is possible. However, many reactions are endothermic and must therefore extract energy from surroundings, e.g., thermal decomposition reactions which generally takes place more readily at higher temperatures, and the dissolution of many salts in water. Furthermore many reactions are easily reversed and thus, if exothermic in one direction, must be endothermic to the same extent in the other direction (law of conservation of energy). Consider the Haber process: N2 + 3H2 → 2NH3

6.20

Thermodynamics

By the application of high pressure and a moderately high temperature in the presence of a suitable catalyst, the reaction can be made to proceed from left to right at a convenient rate, particularly if the ammonia is liquefied and removed. For every mole of ammonia that is formed, 46.0 KJ of heat are evolved. 1 3 N2 + H2 →2NH3 ∆H = –46.0 KJ mol–1 2 2 Ammonia can be made to decompose into nitrogen and hydrogen and this reaction is endothermic to the extent of 46.0 KJ for every mol of ammonia decomposed. NH3 →

1 3 N2 + H2 H2 DH = + 46.0 KJ mol–1 2 2

Therefore, it becomes obvious that while decrease in enthalpy may be a contributory factor for spontaneity, but it is not true for all cases.

6.8.2 entropy and Spontaneity It is clear from the example discussed above that at least one more factor, other than a tendency for a chemical system to assume a lower energy state by transferring some energy to the surroundings, is involved; otherwise no endothermic reactions would be possible. It is interesting to note that in the breakdown of the ammonia molecule into nitrogen and hydrogen there is an increase in the number of molecules, i.e., the more ordered ammonia molecule gives simpler molecules of nitrogen and hydrogen. In fact every endothermic reaction is accompanied by some decrease in order, whether this be in the actual chemical system itself or in surroundings. Order is associated with molecular structure and solids are more ordered than the liquids into which they pass on melting, since the atoms or ions which vibrate about fixed positions in the crystal lattice become more mobile in the liquid phase. Similarly, liquids are more ordered than gases into which they pass on boiling since gaseous molecules can move about at random. Polyatomic gases are likewise more ordered than the simpler ones into which they may decompose under suitable conditions. It is now known that the possibility of a chemical reaction taking place under conditions which allow a transfer of energy to and from the surroundings, but do not allow a transfer of matter (closed system conditions) is governed by two factors. (i) A tendency for the chemical system to move to minimum energy, e.g., evolve heat to the surroundings. The energy lost by the chemical system is, of course, equal to the energy gained by the surroundings (law of conservation of energy).

(ii) A tendency for the chemical system to move in such a direction which results in a decrease in order, i.e., an increase in disorder. A new thermodynamic term Entropy is introduced as a measure of order/disorder and there is reason to believe that at absolute zero the entropy of a crystalline solid is zero. At this temperature, a crystalline solid posses perfect order. An increase in entropy is equated with an increase in disorder. Whether a reaction is possible is determined by interplay of these two effects. However, it is important to realize that it is not possible to predict how fast a particular reaction will take place, even though both factors (i) and (ii) may be favourable. For instance, carbon shows no signs of reacting with oxygen at room temperature to give carbon monoxide, even though it can be shown that the reaction should be very exothermic at this temperature and should result in an increase in disorder, i.e., one extra mole of gas is produced for each mole of oxygen. 2C + O → 2CO 2 1 mole

2 moles

Carbon and oxygen are said to be thermodynamically unstable with respect to carbon monoxide, the reaction being opposed by another effect (activation energy barrier).

6.8.3 The concept of entropy The interaction of chemical substances involves the breaking of bonds (an endothermic process) and the forming of new ones (exothermic process). The enthalpy of the chemical system will be lowered if the process of making bonds to give product molecules is more exothermic than the process of breaking bonds in the reactants is endothermic, provided, of course, that the energy so released is allowed to flow out of the system into the surroundings (by the law of conservation of energy there is no overall change in energy since the energy lost by the chemical system exactly matches that gained by the surroundings). Thus consider the synthesis of hydrogen chloride from hydrogen and chlorine, which involves the breaking of H-H and Cl-Cl bonds and the making of H-Cl bonds irrespective of the actual detailed mechanism involved in this reaction. H2(g) →2H(g)

θ = 435.9 KJ mol–1 DH 298

Cl2(g) → 2Cl(g)

θ = 435.9 KJ mol–1 DH 298

O

O

θ 2H(g) + 2Cl(g) → 2HCl(g) DH 298 = –(2 × 431) KJ O

Overall process H2(g) → Cl2(g) → 2HCl

θ = –184.3 KJ DH 298 O

Thermodynamics 6.21

The reaction is exothermic to the extent of 184.3 KJ for every 2 moles of hydrogen chloride produced. The tendency for change to occur in such a manner that order is lowered (or disorder is increased) is often at conflict with the move towards minimum energy (with the consequent flow of energy into the surroundings) but in one system at least, i.e., an isolated system, a move towards greater disorder become all- important criterion for change to occur.

6.8.4 changes occurring in an isolated System An isolated system is not one in which neither energy nor matter can enter or leave. The universe is considered to be one, and the nearest approach to an isolated system in the laboratory is a closed Dewar container. Consider an isolated system which is divided into two equal volumes by a partition and which contain helium in one compartment and neon in the other at the same temperature and pressure. When the partition is removed the gases diffuse throughout all the available space and become completely mixed. Since noble gases have been specified there can be little or no interaction between their atoms; furthermore, if interaction did take place to slight extent there could not be any overall energy change since the system is an isolated one. However, the molecules of each gas can now move throughout twice the original volume or, in other words, the gases are now in a more disordered or random state. Disorder can be measured quantitatively and is given the same ‘entropy’. An increase in disorder is an increase in entropy and this is denoted as ∆S which is positive; conversely, when a decrease in entropy occurs ∆S is negative. The increase in entropy which occurs when gases mix without interaction is given by the mathematical equation. ∆S = Constant × loge

Product of final volumes of each gas Product of initial volume of each gas

The constant is the gas constant ‘R’ when 1 mole of each gas is involved; thus 1 mole of helium and 1 mole of neon, initially occupying a volume V1 each will eventually occupy a volume 2V1 each after mixing and the increase in entropy will be 2V × 2V1 ∆S = R loge 1 = R loge 4 J mol–1 deg–1 V1 × V1 Note that units of entropy are joules per degree, i.e., thermal units are involved even though in this example no heat transfer has taken place. What are the chances that helium and neon do not mix? Certainly on energetic grounds alone this is not impossible,

but statistically the process becomes more and more unlikely with increasing number of molecules. Consider two molecules of helium labelled He1 and He2 and two neon molecules labelled Ne1 and Ne2 mixed. In the system of total volume 2V1 for ‘unmixing’ to occur He1 must exchange with He2. He1

He2

He1

Ne1

Ne1

Ne2

He2

Ne2

Total volume 2V1, mixed

Total volume 2V2, unmixed

Statistically the probability of unmixing is an unlikely event for a mixture of helium and neon. There are six different ways in which two molecules of helium and two molecules of neon can be distributed evenly in the compartments A and B and these are shown below. Compartment A He1 He1 He1 He2 He2 Ne1

Compartment B

He2 Ne1 Ne2 Ne1 Ne2 Ne2

Ne1 Ne2 Ne1 Ne2 Ne1 He1

Ne2 He2 He2 He1 He1 He2

In only one of these arrangements do both helium molecules appear in compartment A and the probability of the system returning to its original state is 1/6. In four of the arrangements a helium molecule is paired with one of neon and the probability of the system remaining completely mixed is 4/6 or 2/3. Consider now four molecules of helium and four molecules of neon ‘mixed’ in a similar system; the probability of finding all four molecules of helium in compartment A and all four molecules of neon in compartment B is now 1/70. For eight molecules of each kind the probability reduces to 1/12870. The number of ways in which n molecules can be distributed so that m appears in one compartment (n-m) appear in the other is given by the formula

n! m !(n − m)! Where n! means factorial n and is n (n-1) (n-2) (n-3) 1….. In the examples we have been considering m = n/2 and (n-m) has also been n/2. If we now consider the system involved 1 mole of helium and 1 mole of neon, i.e., Avogadro’s number or approximately 6 × 1023 molecules of each kind, then the

Thermodynamics

probability of ‘unmixing’ occurring is infinitesimally small. The most probable state is one of complete mixing (greater disorder): thus the most probable state of system is one of high entropy. Consider now the dissolution of ammonium nitrate in water, carried out in an isolated system on dissolving, the highly ordered structure of the ammonium nitrate is broken down and dispersed throughout the liquid, and there is an overall increase in entropy, but the energy is unchanged since the system is an isolated one. It is instructive to examine this system in more detail as follows: the breakdown of the solid ammonium nitrate to form a solution results in an increase in entropy; but the process is endothermic and this results in the air above the solution being cooled down. The air molecules move more slowly and hence there is a decrease in the entropy of the air. A decrease in entropy also occurs owing to the water molecules being more ordered when hydration of ammonium nitrate occurs, the ammonium and nitrate ions binding water molecules. However, there is an overall increase in entropy and dissolution of ammonium nitrate ceases when no further increase in entropy can occur. In the above example, there has been no overall change of energy since the system is an isolated one, but, since the process is endothermic, thermal energy of the air has been transformed into chemical energy of the hydrated ammonium and nitrate ions. Here, and in chemical reactions in general, high entropy is associated with the maximum number of ways in which the energy of the system can be distributed. Quantum theory explains energy is held in a series of definite energy levels, for a given energy range there are more translation energy levels than rotational energy levels. Similarly there are more rotational energy levels than vibration energy levels . This means that there are far more ways of distributing a given energy among the various translation energy levels than among either rotational or vibration energy levels. Thus in a purely qualitative sense, we can visualise the increase in entropy (the spread of energy) that occurs when ammonium nitrate dissolved in water in the following manner. 1. The solid structure of ammonium nitrate is broken down and the hydrated ions are dispersed in solution. Energy of vibration is replaced by energy of translation and there is a very large increases in the number of energy levels now available. The energy therefore increases. 2. The air is cooled down and this means that there is less energy to be distributed among various energy levels

(This applies to translational , rotational and vibration energy levels but the first is the more important) the entropy there decreases. 3. Water molecules are bound to the ammonium and nitrate ions i.e., restricted, and there is a decrease in entropy. However, the conversion of vibration energy of the ammonium nitrate into translational energy of the hydrated ions results in a large increase in entropy and this completely over shadows the decrease in entropy that occurs in stages 2 and 3. The overall entropy therefore increases. In order to visualize the increase in the number of ways in which the energy of a system can be distributed when a change occurs, let us consider a sample but purely hypothetical situation. Suppose we have 3 molecules of a perfect gas with a total energy of 2 units which can be distributed between the three molecules.

Increasing energy

6.22

2 1

0 3 possible arrangements

3 possible arrangements

(Total number of arrangements = 6)

One molecule can be given two units of energy and the other two molecules will now have zero energy, there are three ways of arranging this by exchanging the molecules alternatively two molecules can be given one unit of energy each, the remaining molecules having zero energy there are also three ways of distributing the available two units of energy. Each arrangement is called a complexion or ensemble thus at a higher temperature there is a greater number of complexions or ensembles than at a lower temperature. When the number of complexions or ensembles increases we say that entropy increases. When the number of complexions or ensembles decreases we say that entropy decreases. Suppose we also have three molecules of a perfect gas with a total of four units of energy there are fifteen ways in which this energy can be distributed between the three molecules.

Increasing energy

Thermodynamics 6.23 4

6.8.5 Quantitative aspects of entropy

3

So far entropy has been rather loosely equated with order/ disorder, an increase in disorder meaning an increase in the number of ways in which the energy of a system can be distributed amongst its energy levels and corresponding to an increase in entropy. We shall now consider the entropy change which results from a reversible and isothermal expansion of a gas, and define the entropy change in precise terms. Consider an ideal gas initially at pressure P can be supplied with heat and allowed to expand. If the gas is opposed by a pressure, only infinitesimally less than the gas pressure at all stages of the expansion, then the maximum amount of work is extracted from the expanding gas; this is denoted by the symbol Wrev and the conditions are said to be reversible ones. By increasing the opposing pressure very slightly the gas can be made to contract. Since the gas is considered to be ideal, in that intermolecular attractions do not exist, and since the expansion is considered to be carried out under isothermal conditions, i.e., the temperature of the gas remaining constant, all the heat put into the gas converted into work of expansion. For a small volume change dV, the work done PdV, and for a volume change from V1 to V2, the work done is Wrev where

2 1 0

3 ways

6 ways

3 ways

3 ways

Total number of arrangements = 15

If the three molecules having a total energy of two units and the three molecules having a total of four units of energy are placed in separate halves of an isolated system and separated from each other by a perfectly conducting partition, experience tells us that there will be transfer of thermal energy from one to the other. The three molecules having a total energy of two units will now gain one unit of energy at the expense of the three molecules initially having a total of four units of energy. It can easily be shown that the number of ways in which three units of energy can be distributed between three molecules is ten, and since we have two lots of such molecules the total number of possible distributions of the total energy is 10 × 10 or 100. Before the gases were brought into thermal contact with each other there were 6 × 15 or 90 ways of distributing this same amount of energy. Energy transfer takes place from one to the other because this results in increase in the number of ways of distributing the energy. If the thermally conducting partition is now removed, there is further increase in the number of ways of distributing the total energy. The problem now is to determine how many arrangements there are for distributing a total of six units of energy between six molecules. There are in fact, 462 possible ways and this means that mixing of the gases will take place when the partition is removed. We can summaries the situation as follows changes take place in an isolated system in a direction such that when equilibrium is achieved, the number of ways in which the total energy of the system can be dis tributed is a maximum (the entropy of the system is a maximum). Thus entropy is a measure of the number of ways that energy can be shared out among molecules. The greater the number of ways (complexions or ensembles), the greater is the entropy and vice versa. The entropy can be calculated from Boltzmann’s formula S = K lnW, where K is the Boltzmann’s constant 1.38 × 10–23 JK–1 and lnW means the natural logarithm of the number of complexions or ensembles W.

Wrev =

∫

V2

V1

PdV

For 1 mole of an ideal gas PV = RT, therefore Wrev =

∫

V2

V1

V RT dV = RT loge 2 dv V V1

The amount of heat put into the gas, Qrev equals the work done Wrev. Therefore Qrev = Wrev = RT loge

Or

V2 V1

Qrev V = R loge 2 T V1

The quantity Qrev/T is defined to be the entropy change (an increase since heat is absorbed). Thus ∆S =

Q rev V = Rlog e 2 (for 1 mole of ideal gas) T V1

If one mole of ideal gas had expanded from an initial volume V1 into a vacuum so that the final volume was V2, then ofcourse no heat would have been absorbed. Nevertheless, the entropy change is still given by the expression Qrev/T1 since the final state of the gas is the same irrespective of whether it expands reversibly and isothermally or into a vacuum where it does no work.

6.24

Thermodynamics

∆STotal = ∆Ssystem + ∆Ssurr > O When a system is in equilibrium, the entropy is maximum, and the change in entropy ∆S = O. We can say that entropy for a spontaneous process increases till it reaches maximum and at equilibrium the change in entropy is zero. Since entropy is a state property, we can calculate the change in entropy of a reversible process by ∆Ssys =

Qrev T

We find that both for reversible and irreversible expansion for an ideal gas under isothermal conditions ∆U = 0,But ∆Stotal, i.e., ∆Ssys + ∆Ssurr is not zero for irreversible process. Thus ∆U does not discriminate between reversible and irreversible process, whereas ∆S does.

6.8.7 entropy change Spontaneity and equilibrium: Second law of Thermodynamics The second law of thermodynamics introduces the concept of entropy and its relation with spontaneous process. In an isolated system such as mixing of gases, there is no exchange of energy or matter between the system and surroundings. But due to increase in randomness, there is increase in entropy. Thus we can say that for a spontaneous process in an isolated system, the change in entropy is positive, i.e., ∆S>O. However, if the system is not isolated, we have to take into account the entropy changes of the system and the surroundings. Then the total energy change (∆STotal) will be equal to the sum of the change in entropy of the system (∆Ssys) and the change in entropy of the surroundings (∆Ssurr), i.e., ∆Stotal = ∆Ssys + ∆Ssurr

6.8.6 entropy as a State Function Consider a cylinder fitted with frictionless and weightless piston which contains a gas and is in contact with large heat reservoir. During isothermal and reversible expansion of the gas from volume V1 to V2, let there be absorption of heat, q at temperature T Qrev , since ∴ Change in entropy of the system ∆Ssys = T an equivalent amount of heat will be lost by the reservoir will be ∆Sres =

Qrev T

Total change in entropy ∆S1 = ∆Ssys + ∆Sres =

Qrev  −Qrev  + =O T  T 

If we now compress the gas isothermally from a volume V2 to V1 heat given by the system is Qrev ∴ DS sys =

For a spontaneous process, ∆Stotal must be positive, i.e., ∆STotal = ∆Ssys + ∆Ssurr>O But system and surroundings constitute universe for thermodynamic point of view so that for spontaneous change, ∆Suniverse >O. Second Law of Thermodynamics Second law of thermodynamics is based on the above discussions and states that The entropy of the universe always increases in the course of every spontaneous (natural) change. Thus combining the first and second laws of thermodynamics the combined statement becomes The energy of the universe is conserved where as the entropy of the universe always increase in any natural process.

Q −Qrev and .DS res = rev T T

Total change in entropy ∆S2=–

Qrev Qrev + =O T T

Table 6.3 Entropy change for the transformation H2O(l) → H2O(s) at 1 atm pressure Temperature

Total change in entropy for the complete cycle ∆S1 + ∆S2 = O At the end of the cycle, the entropy of the system is the same it had initially. Therefore entropy is a state. function.

°C –1 0 1

K 272 273 274

∆Ssys –1

∆Ssurr –1

(JK mol ) –21.85 –21.99 –22.13

–1

∆Stotal –1

(JK mol ) 21.93 21.99 22.05

(JK–1mol–1) 0.08 0 –0.08

Thermodynamics 6.25

6.8.8 entropy and equilibrium State During a spontaneous process, the entropy of the system goes on increasing. When the system reaches the equilibrium state, the entropy of the system become maximum. The mathematical condition for entropy (S) to be maximum is that the change in entropy (∆S) is, i.e., ∆S = 0 (at equilibrium) This can be illustrated by applying the entropy criterion in the conversion of water to ice at 1 atm pressure and three different temperatures. (i) At 272 K ∆STotal = ∆Ssys + ∆Ssurr = 21.85 + 21.99 = + 0.08 JK–1 mol–1 Since ∆STotal is positive at 272 K, the freezing of liquid water to ice is spontaneous. H 2 O(l) → H 2 O (s ) Spontaneous at 272 K (ii) At 273 K ∆STotal = ∆Ssys + ∆Ssurr = –21.99 + 21.99 = 0 Since ∆Stotal is zero, the process is at equilibrium i.e., neither freezing nor melting is spontaneous. At this temperature, water and ice are in equilibrium and no net change is observed. H2O(l)→H2O(s) Equilibrium at 273 K (iii) At 274 K ∆Stotal = ∆Ssys + ∆Ssurr = –22.13 + 22.05 = 0.08 JK–1 mol–1 Since ∆Stotal is negative, the freezing of water is not spontaneous at 274 K. H2O(l)→H2O(s). Not spontaneous at 274 K. But the reverse process ∆Stotal will be positive, i.e., ∆Stotal = + 0.08 JK–1, i.e., melting of ice is spontaneous at 274 K. H2O(s)→H2O(l) Spontaneous at 274 K Thus it can be said that ∆STotal is a criterion for spontaneity of a change.

6.8.9 entropy change in reversible Process In an isothermal reversible process if the amount of heat ‘Q’ is absorbed from the surrounding at a temperature T, the increase in the entropy of the system will be Q ΔSsys = + T On the other hand, surroundings lose the same amount of heat at the same temperature. The decrease in the entropy of the surroundings will be Q ΔSsys = T

[Total change in the entropy = entropy change in the system + entropy change in surroundings] ΔS total = ΔSsys + Δ Ssur Q Q =0 − T T When the reversible process is adiabatic there will be no heat exchange between system and surroundings, i.e., Q = O.

=

∴ ΔSsys = 0 ΔSsurr = 0 Δ Stotal = Δ S sys + Δ Ssurr = 0

6.8.10 entropy change in irreversible Processes Consider a system at higher temperature T1 and its surroundings at lower temperature T2. ‘Q’ amount of heat goes irreversibly from system to surroundings. Q ∴ ΔSsys = – T1 ΔSsurr = +

Q T2

ΔSprocess = ΔSsys + ΔSsurr

 T − T2  Q1 Q2 + = Q 1  T1 T2  T1 T2  But T1>T2 ∴ T1–T2 = + ve or ΔSprocess > 0 Hence, entropy increase in the an irreversible process. =–

6.8.11 entropy change for ideal gases Change in entropy for an ideal gas under different conditions can be calculated as follows. (a) When change from initial state (1) to final state (2)

T  V  ΔS = 2.303 nCv log10  2  + 2.303 nR log10 2  V   T1  1 (When T and V are variables) T  p  ΔS = 2.303 nCp log10  2  + 2.303 nR log10  1   T1   P2  (when T and P are variables) (b) Entropy change for isothermal process

V  ΔS = 2.303 nR log10  2   V1  p  Δ = 2.303 nR log10  1   P2 

6.26

Thermodynamics

(c) Entropy change for isobaric process (at constant pressure)

T  ΔS = 2.303 nCp log10  2   T1  V  ΔS = 2.303 nCp log10 2  V  1 (d) Entropy change for isochoric process (at constant volume) T  Δ S = 2.303 nCV log10  2   T1  ΔS = 2.303 nCV log10

 p2   P 

• • •

( DS = 0)

•

•

In a reversible process DS therefore

•

ΔS/mol = –2.303 R [χ1 log10 χ1 + χ2 log10 χ2] Entropy change in adiabatic expansion will be zero ΔSsys = 0. Entropy change in adiabatic reversible process ΔSSys= 0; ΔSSurr = 0; ΔS Total = 0 Adiabatic irreversible process = 0; ΔSsys >0; ΔSsurr = 0; ΔSTotal >0

Summary of the characteristics of entropy • •

• •

•

•

Entropy is measure of disorder in a system. Entropy is a state function and depends on the state variables such as T,P,V and n which govern the state of system. Entropy is extensive property. Absolute value of entropy cannot be determined easily, but its value depends on mass of the substance present in the systems. Change in entropy during the process is given as Q DS = ( S f − Si ) = rev T This value does not depend on the path adopted but depends on the final and initial states of the systems. Units of entropy are JK–1 mol–1.

or DS

universe

= 0 and

For adiabatic process,

DS sys = DS surr = DSTotal = 0 In a spontaneous irreversible process, ΔStotal or ΔSuniverse > 0

•

i.e., in spontaneous processes, there is always increase in entropy of the universe. Entropy is the unavailable energy of the system Entropy =

  n n2 Δ S/mol = –2.303 R  1 log10 x1 + log10 x2  n1 + n2   n1 + n2

total

DS sys = −DS surr

(e) Entropy change in mixing of ideal gases: If n1 mole of gas ‘A’ and n2 mole of gas B are mixed , then total entropy change is ΔS = –2.303 R[n1 log10 χ1 + n2 log10 χ2] χ1 and χ2 are mole fractions of gases A and B n1 n2 ; x2 = ; n1 + n2 n1 + n2

In the natural spontaneous process entropy of the universe is increasing.

DSuniverse > 0

1

i.e., x1 =

Entropy increases in going from solid to liquid and from liquid to gas. The entropy change for a cyclic process is zero. The entropy change in the equilibrium state is zero.

•

unavaible energy Temperature in K

Entropy is a function of probability of the thermodynamic state. Since we know that both the entropy and thermodynamic probability increase simultaneously in a process, hence the state of equilibrium is the state of maximum probability.

6.8.12 entropy change during Phase Transition Phase transition refers to the change of one state (solid, liquid or gas) to another. Such a change occurs at a definite temperature. For example, conversion of solid to liquid occurs at the melting point, while change of liquid into vapours occurs at the boiling point. The entropy change taking place during such transitions can be calculated as Q Δtrans S = rev T Where Q is the heat evolved or absorbed during transition and T is the temperature. For the transformation of one mole of a substance at constant pressure Qrev is equal to molar enthalpy change for that transformation i.e., Qrev = Δtrans H. Let us consider certain examples 1. Entropy of Fusion: The entropy change when one mole of a solid substance changes into liquid form at its melting point is known as entropy of fusion

Thermodynamics 6.27

Given ΔHv = 42.4 × 103 J mol–1 Tb = 78.4° = 351.4 K

 H2O(l) H2O(s)  The change in entropy is given by O

O

O

S water − Sice = ∆ fus S =

∆ fus H

ΔSV =

O

42.4 × 103 = 120.66 JK–1 mol–1 351.4

Tf

Where Δfus H is the enthalpy of fusion and Tf is the fusion temperature. For example, the standard enthalpy of fusion Δfus H at 273 K and 1 bar pressure for water is 6.0 KJ mol–1. O

O

H2O(s) →H2O (l) Δfus H = 6.0 KJ mol–1 O

∆ fus H Qrev = ∴ Δfus S = T T 3 −1 6.0 × 10 Jmol = 273 = 21.98 J K–1 mol–1

O

O

2. Entropy of vapourization: It may be defined as the entropy change when 1 mole of a liquid changes into vapours at its boiling point. The entropy of vapourization for a liquid at its boiling point is ∆ H Δvap S = vap Tb Where Δvap H is the standard enthalpy of vapourization at 373 K (at 1 far) is 40.79 KJ mol–1.

Problems for Practice 13. 5 moles of an ideal gas expand reversibly from a volume of 8 dm3 at a temperature of 27°C. Calculate the change in entropy. 14. Calculate the entropy of mixing of one mole of oxygen gas and 2 moles of hydrogen gas, assuming that no chemical reaction occurs and the gas mixture behaves ideally. 15. Calculate entropy change accompanying the transfer of 10.46 KJ of heat from a body A at 300°C to a body B at 77°C. 16. Calculate entropy change when 2 moles of an ideal gas are allowed to expand isothermally at 293 K from a pressure of 10 atm to a pressure of 2 atm. 17. Consider the reaction for the dissolution of ammonium nitrate.

O

O

NH4 NO3(s) → NH4+ (aq) + NO3–(aq)

O

H2 O(l)  → H2 O (v) Δ Vap H = 40.79 KJ mol–1 O

O

∴ ∆ vap S =

∆VaP H T

ΔH = + 28.1 KJ mol–1, ΔS = 108.7 JK–1 mol–1 Calculate the change in entropy of the surroundings and predict whether the reaction is spontaneous or not at 25°C.

O

40.79 × 103 Jmol −1 = 109.356 J K–1mol–1 373K 3. Entropy of sublimation: Sublimation is a process of direct conversion of solid to vapour. Entropy of sublimation may be defined as the entropy change when 1 mole of solid changes into vapour at a particular temperature. ∆ H Δsub S = sub T Where Δsub H is the standard enthalpy of sublimation at the temperature T (ΔsubH = ∆ fus H + ∆ vapH ) O

6.8.13 Standard entropies We know that the entropy of solid is less than that of a liquid and that the number of complexions or ensembles increases. This suggest that if the entropy of a substance is to be reduced to a minimum, the substance should be a solid and its temperature should be as low as possible. The lowest possible temperature is absolute zero, 0 K a perfect crystal at 0 K will have the lowest entropy possible it can be stated as

O

O

O

O

Solved Problem 6 Ethanol boils at 78.4°C, the enthalpy of vaporization of ethanol is 42.4 KJ mol–1. Calculate the entropy of vaporization of ethane. Solution: ΔSV =

DHV Tb

The entropy of a perfect crystal at 0.K is zero. This statement is also called as the third law of thermodynamics; for example, perfect crystals of hydrogen, of sodium or of iron are all assumed to have zero entropy at 0 K the zero level entropy is quite unlike the definition of zero level of enthalpy. For example, the enthalpies of formation of these elements are zero at 298 K and 101.325 K pa. It is impossible to measure entropy changes right down to 0 K because absolute zero cannot be reached. However, we can get to with in a few hundredths of a degree

6.28

Thermodynamics

Table 6.4 Standard entropies of some substances O

O

Substance

S (JK mol–1)

Substance

S (JK mol–1)

H2(g)

+ 136.6

C(s) (graphite)

+ 5.7

N2(g)

+ 191.4

S(s) (rhombic)

+ 31.9

O2(g)

+ 204.9

Na(s)

+ 51

Cl2(g)

+ 223

Zn(s)

+ 41.4

I2(s)

+ 116.1

Cu(s)

+ 33.3

H2O(l)

+ 70.0

NaCl(s)

+ 72.4

H2O(g)

+ 188.7

+ 26.8

CO2(g)

+ 213.6

MgO(s) CuSO4.5H2O(s)

+ 305.4

NO(g)

+ 210.5

Fe3O4(s)

+ 146.4

C6H6(l)

+ 172.8

C2H5OH(l)

+ 160.7

–1

–1

CH4(g)

+ 186.2

CH3COOH(l)

+ 159.8

C2H4(g)

+ 219.5

CHCl3(l)

+ 201.8

H + (aq)

0.0

Cl–(aq)

+ 56.6

2+

–

(aq)

-98.6

Br (aq)

+ 82.4

Zn2 + (aq)

-106.4

SO42–(aq)

+ 20.1

Cu

hydrogen ion in solution is zero, just as we define its enthalpy of formation as zero.

6.8.14 entropy changes of a reaction Entropy changes can be calculated in much the same ways as enthalpy changes. For example, in the reaction 1 H2(g)+ O2 (g)→H2 O(l) 2 The entropy change is S

O

product

–S

O

reactants

1  S H2O (l) – [S– H2O(g)] –  S O 2 (g)  2  1 –1 –1 = + 70.0 – 130.6– (204.9)J K mol 2 O

O

O

= –102.45 JK–1 mol–1 Notice that the entropy change is negative, which we should expect given that a liquid is made from two gases. 1 1 N2 (g) + O2(g)→NO(g) 2 2 Entropy change is S

of absolute zero. Another difficultly is that perfect crystals are hard to come by. Perfect means just that: no defects in the arrangement of the atoms, no mixtures of isotopes of any of the atoms, and so on. The entropy of a substance at temperature just above 0 K can be measured. Estimates are made for the temperature very close to 0 K. The measurement need not –concern us; we can concentrate on the results. They are given the symbol S and are known as standard entropies (They can also be called third law entropies). Some standard entropies are given in table 6.4 The values show that the entropies of gases tend to be greater than those of liquids, which in turn are greater than those of solids. There are other factors at work though. One is that, if you compare solids, the standard entropy increases as the mass of the substance increases, this is because of the energy of contribution to entropy. Heavier atoms, ions, or molecules usually have greater number of energy levels available to them. Therefore, a greater number of complexions or ensembles can occur; hence entropy increase . Other complications appear when there are different types of bonding especially liquids that are hydrogen bonded tend to have lower entropies than similar liquids that have no hydrogen bonding (hydrogen bonds restrict the motion of molecules). The entropies of ions in solution are calculated using the same convention as for the enthalpy of formation of ions in solution. We define the standard entropy of

O

O

1

1

1 1 ( ) − )1−SN−( N2g(2 ( g2)−) 1−) −SO2 2( Og2 ( g ) ) ( NO(g) 2 2 O

O

2

= + 210.5–

2

1 1 (191.4) – (204.9) JK–1 mol–1 2 2

= + 12.35 JK–1 mol–1 This reaction takes place with slight increase in entropy even though there is no overall change in volume. (There is a total of one mole of gas at the start and one mole of gas at the end).The change in entropy is the result of the gases having different ranges of rotational and vibrational energy levels open to them. Solved Problem 7 Calculate the standard entropy change associated with the following reaction at 298 K. P4(s) + 5O2(g) → P4O10(s) At 298 K S for P4 = 41.1 JK–1 mol, O2 = 205.0 JK–1 and P4O10 = 231.0 JK–1 mol–1. Solution: O

DS = O

∑ ∆S

O

Products

− ∑ ∆S Reactants O

DS = [DS P4O10 – (DS P4+5DS O2)] O

O

O

O

= 231 – (41.1 + 5 × 205.0) = 231 –1066.1 = –835.1 JK–1 mol–1

Thermodynamics 6.29

Solved Problem 8 The rusting of iron occurs as 4Fe(s) + 3O2(g) → 2Fe2O3(s) The enthalpy of formation of Fe2O3(s) is –824.2 KJ mol–1 and entropy change for the reaction is –549 JK–1 mol–1. Calculate ΔSsurr and predict whether rusting of iron is spontaneous or not at 298 K. Solution: For the reaction 4F(s) + 3O2(g) → 2Fe2O3(g) DrH = 2DfH(Fe22O33)) = 2 × −824.2 = 6 = −1648.4 KJ The release of energy by the system to the surroundings will increase the entropy of the surroundings as DS surr = −

DH sys

T 1648.4 =− = 5531 JK −1 298 DStotal = DSsyst + DSsurr = −549 + 5531 = +4982 JK −1mol −1

Since there is increase in entropy (ΔSTotal = +ve), rusting of iron is spontaneous process. misconcept about the entropy and disorder Water can exist in two different states: solid ice and liquid. We know that ice is held together by hydrogen bonds between the water molecules and that the large amount of hydrogen bonding in (liquid) water is responsible for its high boiling point. If we heat a cube of ice at 0°C the structure of the solid is disrupted. The ice melts but if the heat supplied very slowly the temperature does not change the water remaining at 0°C. We know that some hydrogen bonds have been made. However, the main change is the state of the molecules. There is a change from solid to liquid, so there will be a corresponding

(a)

(b)

Fig 6.10 Mixing if the two gases increases the entropy

increase in the number of translational energy levels and in the number of complexions. Therefore there is also an increase in entropy. However, in some other books, the authors may explained that there is an increase in entropy because there is an increase in disorder as explained below. In ice the molecules are arranged in an orderly way, with the molecules in fixed position in the crystal. In water the molecules are free to move around and they become mixed up. Therefore the arrangement of the mol ecules in water is much more disorder than it is in ice. This increase in disorder is the cause of the increase in entropy. A similar explanation for the increase in entropy when different gases mix. Suppose we had two gases A and B, in separate insulated containers joined by a tap. Then we open the tap, the gases will, of their own accord, mix this will happen even with ideal gases for which there is no heat change in mixing . Again there has been an increase in the disorder of the systems. If we were to take a sample of the gas from the lefthand container at the start of the experiment, we would be certain to pick out a molecule of A after mixing on average. We would have equal chances of picking out a molecule of A or a molecule of B. The probability of retrieving a molecule of A has reduced from 1.0 before mixing to 0.5 after mixing. This change in probabilities reflects the increase in disorder this increase in disorder represents the increase in entropy. The problems with these explanations is that they are incorrect: Entropy is not directly related to the mixing up or increase in disorder of molecules as they are distributed in space. For the change ice to water, entropy increases owing to the change in the number of complexions following the increase in translational energy levels. For the mixing of different gases, it is the increase in volume open to their molecules and the rise in the number of complexions that this causes that are responsible for entropy increasing. When all things being equal, if there is a change solid to liquid or liquid to gas, there will be an increase in entropy. The important point to realize is that it is not sensible to explain these changes in terms of ‘order’ and ‘disorder’ of arrangements of molecules in space. It is the change in number of complexions that we must consider to explain changes in entropy. To calculate precise change in number of complexions is not an easy task and one that we shall not attempt. But we can use a rule of thumb method to help us to predict whether entropy increase or decrease in a chemical change. Entropy increases when a change occurs from solid to liquid to gas while it decreases for the change from gas to liquid and liquid to solid.

6.30

Thermodynamics

6.9 giBBS energy Consider the reaction between zinc and a molar solution of hydrochloric acid

Let us now define a free energy change ∆G so that a decrease in free energy, –∆G, occurs, when the maximum useful work is done, then –∆G = Wuseful

Zn + 2H+ (molar) → Zn2+ + H2 This reaction can take place in a calorimeter at constant pressure in which there is a heat change and work done against the atmosphere by the expanding hydrogen gas. Alternatively the reaction can be carried out reversibly in an electrochemical cell, when the maximum amount of work is done Wrev. In this example, the maximum work Wrev is the sum of the electrical work denoted as Wuseful and the work done by the hydrogen against the atmosphere, the heat involved under these conditions is Qrev. Since the chemical system attains the same final state irrespective of whether the reaction takes place in a constant pressure calorimeter or in an electrochemical cell, the change in internal energy, ∆U, is the same in both cases (internal energy changes are additive like enthalpy changes and represent changes in the total amount of energy stored in chemical systems). Only differences in internal energies can be measured since there is no way of determining the absolute internal energy associated with an element. If a chemical system absorbs heat Q, and does work W the change in internal energy, ∆U, is given by the equation:

Or ∆G = ∆H–T∆S

(3)

This equation is called Gibbs Helmholtz equation and is very useful in predicting the spontaneity of a process Units of ∆G ∆G has the units of energy (i.e., KJ mol–1 ar J mol–1) because both ∆H and T∆S are energy terms ∆G (J mol–1) = ∆H (J mol–1) - T∆S (KJ K–1 mol–1). It may be noted that in earlier literature G was called Gibbs free energy because J. Willard Gibbs, introduced this term but in accordance with IUPAC recommendations, it is called Gibbs energy. We shall use the term Gibbs energy in the present unit in accordance with IUPAC recommendations.

The above expression is a statement of the law of conservation of energy. When zinc is reacted with molar hydrochloric acid under constant pressure, the heat involved is the enthalpy change, ∆H, and the work done is that due to opposing the external pressure P∆V, Therefore, under these conditions

Gibbs energy of system is defined as the maximum amount of energy available to a system during a process that can be converted into useful work. If ∆G is negative for a chemical reaction, then this reaction can be made to perform useful work, i.e., the reaction is feasible, however if ∆G is positive, no useful work can be obtained from the system, i.e., the reaction is not feasible. It is important to realize that equation (3) applies to reactions carried out at constant temperature and pressure. When reactions are done under standard conditions, i.e., at 1 atmosphere pressure and specified temperature, the Gibbs energy change, ∆G is given by the expression

∆U = ∆H–P∆V

∆G = ∆H – T∆S

Similarly when zinc reacts reversibly with molar hydrochloric acid in an electrochemical cell, the heat involved is Qrev and the work done is Wrev; the latter work comprises electrical work, Wuseful, and work done against the atmosphere, P∆V, therefore under these conditions:

When zinc reacts with molar hydrochloric acid 152.4 KJ of heat are evolved for every mole of hydrogen produced. The maximum useful work done if the reaction is carried out reversibly is 147.2 KJ for every mole of hydrogen produced. Since both values refer to standard conditions at 25°C (298 K). ∆H 298 = –152.4 KJ mol–1, ∆G 298 = –147.2 KJ mol –1 Therefore,

Q = ∆U + W or ∆U = Q–W

O

∆U = Q rev – Wrev = Qrev – (Wuseful + P∆V)

O

(1)

Equating equations (1) and (2) have ∆H–P∆V = T∆S – (W useful + P∆V) Or ∆H = T∆S – Wuseful

O

O

O

–147.2 = –152.4 - T∆S

The entropy change ∆S is equal to Q rev therefore ∆U = T∆S – (Wuseful + P∆V)

O

(2)

O

298

+147.2 − 152.4 = –0.01745 KJ mol–1 deg–1 298 Since the units of entropy are J mol–1 deg–1, the entropy change in this reaction is –17.45 J mol–1 deg–1. Since the value carries a negative sign, this amounts to a decrease of entropy in the chemical system.

Thermodynamics 6.31

Only a part of the energy released in a reaction can be used to do work; the rest is involved in an entropy change. The Gibbs energy change is the amount of energy available to do work. Spontaneous reactions can do work. Spontaneous reactions have negative Gibbs energy changes. Non-Spontaneous reactions cannot do work. Non-Spontaneous reactions have positive Gibbs energy changes.

The Gibbs Helmholtz equation can be written way by dividing the equation through by the temperature and rearranged as DG DH = + DS T T

Each term represents an entropy change. The equation says that Entropy change in the universe = entropy change in the surroundings + entropy change in the system

6.9.1 Standard gibbs energies Gibbs energy is a thermodynamic functions of state. Therefore, we can use values of standard Gibbs energies in calculations in much the same way as we have used standard enthalpies. Some standard Gibbs energies of formation for a range of compounds are given in Table 6.6.

6.9.2 The gibbs energy of Formation of an element in its Standard State is Zero We can also use Gibbs energies of formation of ions in aqueous solution (Given in Table 6.6). Here, the standard state is defined to be a solution that contain 1 mol of the ion in 1 kg of water. This type of solution is called 1 molar solution. For many purposes we can assume that a 1 molal solution is the same as a 1 molar (1 mol dm–3) solution of H+ as defined as zero. Using these values, we can calculate the Gibbs energy change of a wide variety of reactions. If the ∆G value of a reaction is negative the reaction takes place spontaneously at 298 K and that it could do work on the surroundings. On the other hand, if ∆G value of a reaction is positive , the reaction is not spontaneous. The reaction will not occur unless we do work on it. O

O

Table 6.5 Standard Gibbs energies of formation of some compounds O

Compound

∆Gf KJ mol–1

Solid Substances NaCl –384.0 KCl –408.3 CuSO4.5H2O –1879.9 Li2CO3 –1132.6 Na2CO3 –1047.7 CaCO3 –1128.8 BaCO3 –1138.8 MgO –569.4 Na2O –376.6 Al2O3 –1582.4 SiO2 –856.0 BeO –581.6 Fe3O4 –1014.2 CuO –127.2 CaO –604.2 Ca(OH)2 –896.6 NH4Cl –203.0 P(White) 0.0 P(Red) –12.0 PbO2 –219.0 ZnO –318.2

O

∆Gf KJ mol–1

Compound

Liquid Substances H2O –237.2 CH3OH –166.2 C2H5OH –174.8 C6H6 –124.5 CH3COOH –392.0 H2O2 –120.4 HCl(aq) –131.3 H2SO4 –690.0 Gaseous substances Br2 –3.1 CH4 –50.79 C2H6 –229.0 C3H8 270.0 H2O –228.6 NH3 –16.7 CO –137.3 CO2 –394.4 NO + 86.6

Table 6.6 Standard Gibbs energies of formation for some ions in solution O

O

Ion

∆Gf KJ mol–1

Ion

∆Gf KJ mol–1

H + (aq) Cl – (aq) Br – (aq) Cr2O72– (aq)

0.0 –131.2 –103.9 –1257.2

Cu2 + (aq) Zn2 + (aq) Fe3 + (aq) MnO–4 (aq)

+ 65.0 197.1 –9.7

–425.0

6.9.3 gibbs energy changes under non-Standard conditions If a reaction does not work at room temperature, we try it at high temperatures by heating. For example, coal will not burn at room temperature but on heating by keeping the coal in a flame the reaction takes place immediately. In general, there are two reasons why increasing the temperature has an effect on reactions. First, it increases the rate of a reaction. Secondly, a change in temperature can alter the thermodynamics of the reaction. This is because of the presence of the temperature term T in the equation ∆G = ∆H-T∆S.

6.32

Thermodynamics

Zinc carbonate can be kept in bottles at room temperature for years without signs of it decomposing. If you work out the enthalpy, entropy and Gibbs energy changes for the reaction where the carbonate gives off carbon dioxide you will find the following values. ZnCO3 (s) → ZnO (s) + CO2(g) ∆H = + 71 KJ mol–1 ∆S = + 175.1 JK–1 mol–1 ∆G = + 99.9 KJ mol–1 O

O

In short, we can say (a)If ∆G is negative the process is spontaneous. (b) If ∆G is Zero, the process is in equilibrium state. There is no net reaction in either direction. (c) if ∆G is positive , the process does not occur in the forward direction. It may, however, go in the reverse direction.

O

O

The fact that ∆G is positive at 298 K indicates that the reaction under standard conditions is not spontaneous, and does not occur because it is thermodynamically impossible. Zinc carbonate decomposes by heating at a temperature of about 573 K (300°C). Using the Gibbs – Helmholtz equation ∆G = ∆H - T∆S we get ∆G = +71 KJ mol–1 – 573 K × 175.1 × 10–3 KJ K–1 mol–1 = –29.33 KJ mol–1 The negative value for ∆G indicates at 573 K the reaction is possible. The increase in temperature has caused the reaction to become spontaneous. This change has been brought the large entropy change in the reaction. Such reaction which takes place mainly because of the increase in entropy is called entropy driven reaction.

Thus if a change is to be spontaneous in the forward direction, ∆G must be negative. ∆G would be negative under the following conditions. 1. Both the energy and the entropy factor are favourable and may have any magnitude, i.e., ∆H is negative and T∆S is positive. 2. When energy factor favours the change (∆H = –ve) but entropy factor opposes the change (T∆S = –ve), then the magnitude of ∆H should be more than that of T∆S. ∆H (–Ve) > T∆S (–Ve) 3. When energy factor is not favouring (∆H = +ve), but entropy favours the change (T∆S = + ve) then the magnitude of T∆S should be more than that of the energy factor. T∆S (+ve) > ∆H ( +ve)

6.9.4 Predicting Spontaneity of a Process The Gibbs energy change ∆G provides the overall criterion for the feasibility of chemical process. It gives the balance of the two tendencies, ∆H and T∆S. As seen above even if the reaction is not favoured energetically (endothermic reactions) the driving force is provided by favourable change in entropy. It can be shown that only those processes are spontaneous or feasible in which free energy undergoes a decrease i.e., ∆G is negative. By calculating ∆H and T∆S and hence ∆G we can easily predict whether reaction is spontaneous or not. 1. When energy and entropy factors are favourable i.e., ∆H is negative and T∆S is positive, Then ∆G must be negative i.e ∆G = (–) – (+) = –ve. Thus ∆G is negative for a spontaneous process. 2. If both the tendencies oppose, i.e ∆H is positive and T∆S is negative then ∆G is positive. i.e, ∆G = ( + ) – (–) = +ve Thus, ∆G is positive for a non-spontaneous process. 3. If both the tendencies are equal and opposite for example, if ∆H is positive (unfavourable) and T∆S is also positive (favourable) or vice-versa and both are of equal magnitude, then ∆G becomes zero. At this stage, the process does not proceed in either direction and the reaction is said to be in equilibrium state.

It should be noticed that while calculating ∆G Value at different temperatures, generally, we use the standard values of enthalpy and entropy changes at 298 K in our equation ∆G = ∆H-T∆S. But both ∆H and ∆S also change with temperature but in most cases the predictions made by using the standard values agree with the result obtained by using more accurate values. For this reason, in practice, we can often use the standard values at 298 K without too much error. However, in some cases it is necessary to be accurate especially when reactions take place on the scale used in industry. Table 6.7 Spontaneous and non-spontaneous reaction ∆H

∆S –T∆S

–Ve

+Ve

–Ve –Ve

+Ve +Ve

+Ve –Ve

∆G

–Ve +Ve

–Ve at all T Spontaneous at all T –Ve at low T Spontaneous at low T +Ve at high T Non Spontaneous at high T –Ve –Ve at high T Spontaneous at high T Entropy driven +Ve at low T Non spontaneous at

+Ve

+Ve at all T

low Non spontaneous at all T

Thermodynamics 6.33

6.9.5 Qualitative Treatment of gibbs energy

Solved Problem 9

Even without the help of tabulated data of different terms, the equation

For the reaction Ag2O(s)→2Ag(s)+

∆G = ∆H – T∆S Can be used to correlate a diverse range of physical and chemical behaviour. (a) Changes occurring at low temperatures. At lower and lower temperature, the factor T∆S becomes less and less important and at sufficiently low temperatures the feasibility of change occurring is controlled by the sign of ∆H, i.e., for ∆G to be negative, ∆H must also be negative. Thus exothermic reactions are the rule. Furthermore, for purely physical process such as changes of state, lowering of temperature results in the condensation of gases and the solidification of liquids, i.e, the creation of order in the closed system for this to occur ∆H must be negative than T∆S, and gases and liquids evolve a large amount of heat when a change of state occurs. (b) Changes occurring at higher temperatures. Exothermic reactions which produce disorder, for example explosive reactions are highly favoured since ∆H is negative and ∆S is positive, resulting in ∆G being very negative. As the temperature is raised, more and more endothermic reactions become possible and these are all reactions which result in an increase in disorder. An interesting endothermic reaction which readily takes place at room temperature is that between hydrated cobalt (II) chloride and thionyl chloride. CoCl2.6H2O Solid

+ 6SOCl2 → CoCl2 Solid Solid

+ 6SO2 gas

+

12HCl gas

In the above reaction, there is a large increase in entropy as the gases sulphur dioxide and hydrogen chloride are released. At extremely high temperature all chemical bonds are broken and endothermic reactions are the rule. (c) Changes occurring with little or no heat change.

∆H = 30.5 g KJ mol–1 and ∆S = 0.066 KJ mol–1 at 1 atm Calculate the temperature at which ∆G is equal to zero. Also predict the direction of the reaction at (i) this temperature and (ii) below this temperature Solution: ∆G = ∆H-T∆S 0 = 30.56–T × 0.066 30.56 ∴ T= = 463 K 0.066 ∆G = 0 at 463 K, non-spontaneous below this temperature Problems for Practice 18. ∆H and ∆S for Br2(l) + Cl2(g)→2BrCl(g) are + 29.39 KJ mol–1 and 104.0 JK–1 respectively. Above what temperature will this reaction become spontaneous? 19. Two moles of an ideal gas is allowed to expand reversibly and isothermally (300 K) from a pressure of 1 atm to a pressure of 0.1 atm (a) what is the change in Gibbs energy (2) what would be the change in Gibbs energy if the process occurred irreversibly. 20. For a reaction process, ∆H = 178 KJ and ∆S = 160 JK–1 what is the minimum temperature at which the process will be spontaneous? Assume that ∆H and ∆S do not vary with temperature. 21. Metallic mercury is obtained by roasting mercury (II) sulphide in a limited amount of air. Estimate the temperature range in which the standard reaction is product favoured. HgS(s) + O2(g)→Hg(l) + SO2(g) DH1θ = −238.6 KJ mol −1 and DSθ = +36.7 JK −1mol −1 O

O

22. Estimate the temperature range for which the following standard reaction is product – favoured SiO2 (s) + 2C (s) + 2Cl2 (g) → SiCl4 (g) + 2CO (g) ∆H = + 32.9 KJ Mol–1 and ∆S = 226.5 JK–1 mol –1 O

The diffusion of gases is an example of this change. The process is possible since ∆S is positive, i.e., a great er freedom of movement. When ∆H = 0, i.e., for the diffusion of the noble gases, the free energy equation becomes ∆G = –T∆S.

1 O2(g) 2

O

23. Predict whether it is possible or not to reduce magnesium oxide using carbon at 298 K according to the reaction MgO (s) + C(S) → Mg (s) + CO (g) ∆H = + 32.9 KJ mol–1 and ∆S = 197.67 JK–1 mol–1 If not at what temperature the reaction becomes spontaneous. O

O

6.34

Thermodynamics

24. Calculate the standard Gibbs free energy change from the free energies of formation data for the following reaction. C6 H 6 ( l) +

dG = dU + PdV + VdP–TdS

15 O 2 ( g ) → 6CO 2 ( g ) + 3H 2 O ( g ) 2

Given that ∆fG [C6H6 (l)] = 172.8 KJ mol–1 ∆fG [CO2 (g)] = –394.4 KJ mol–1 O

∆fG [H2O (g)] = –228.6 KJ mol–1 O

25. Consider the reaction 4NH3 (g) + 5O2 (g) → 4NO (g) + 6H2O (l) ∆G = –1010.5 KJ Calculate ∆fG [NO(g)] if ∆fG [NH3 (g)] = –16.6 KJ mol–1 And ∆fG [H2O(l)] = –237.2 KJ mol–1 O

O

dU = Q rev – PdV Or dU = TdS –PdV (Since Q rev = TdS) Substituting for dU in equation (1), we have dG = VdP (This equation applies at constant temperature) consider 1 mole of an ideal gas initially at pressure P1 changing to pressure P2; the change in Gibbs energy can be represented as a change from G1 to G2. Thus,

G 2 − G1 =

O

O

26. Calculate ∆rG for the following reactions using ∆fG values and predict which reactions are spontaneous. (a) Ca(s) + Cl2 (g) → CaCl2 (s)

(1)

For a reversible change in which the only work done is that against an opposing pressure, we have

O

O

changes in U, P, V and S. Using the usual calculus notation this can be represented as

∫

p2 p1

VdP

O

p2

P1

1 (b) HgO(s) → Hg(l) + O2 (g) 2

(c) NH3 (g) + 2O2 (g) → HNO3(l) + H2O(l) ∆fG Values (KJ mol–1) are CaCl2 (s) = –748.1, HgO(s) = –58.84; NH3(g) = –16.45 HNO3(l) = –80.71; H2O(l) = –237.13 O

RT dp dP (since PV = RT) P

∫

=

= RT loge

P2 P1

For a pressure change from P1 = 1 to P, the change in Gibbs energy becomes G–G RT loge P(Where G is the standard Gibbs energy) Similarly, for n moles of ideal gas the free energy change for a pressure change from 1 atmosphere to a pressure P is n(G–G (2) − G )) = nRT log e P = RT log e P n O

O

O

6.9.6 equilibrium and gibbs energy Effect of Pressure: When reactions between gases take place there is often a change of pressure and it is necessary to examine how Gibbs energy changes are influenced by changes in pressure ∆G = ∆H – T∆S Where ∆G represents the change G1 to G2, ∆H represents the change H1 to H2 etc. Thus, (G2–G1) = (H2–H1) – T (S2–S1) Or G2 = H2–TS2 or G = H–TS (subscripts omitted) similarly the equation ∆H = ∆U + P∆V can be expressed in the form

Let us now consider a specific gaseous reversible reaction, e.g., The equilibrium set up between dinitrogen tetroxide and nitrogen dioxide   → 2NO2(g) N2O4(g) ←  From equation (2) it can be seen that the free energy of n moles of ideal gas is nG = nG + RT loge Pn O

When dinitrogen tetroxide is converted into nitrogen dioxide, one mole of former gas gives 2 moles of the latter gas. Thus, 2GNO2 = 2GNO2 + RT loge P2NO2 O

(H2–H1) = (U2–U1) + P (V2–V1)

O

Or H2 – U2 + PV2 or H = U + PV (subscripts omitted) Combining the equations G = H–TS and H = U + PV we have G = U + PV–TS Suppose now there is an infinite similar change in free energy at constant temperature, due to similar infinitesimal

GN2O4 = GN2O4 + RT loge PN2O4 When PNO2 and PN2 O4 are the pressures of NO2 and N2O4 respectively in the gas mixture. The Gibbs energy change in this reaction ΔG is 2 2G NO − G N O = 2G NO − G N O + RT log e PNO − RT log e PN2 O4 2 O

2

2

4

O

2

2

4

Thermodynamics 6.35

If ΔG is negative the reaction will proceed, the partial pressure of NO2 will increase and the partial pressure of N2O4 will decrease. A stage is eventually reached when ΔG = 0, i.e., The system is in equilibrium; under these conditions the partial pressure of NO2 and N2O4 are their equilibrium pressure represented as P'NO2 and P'N2O4 respectively At equilibrium ΔG = 0 ∴ΔG = –RT loge O

At equilibrium ΔG = 0

′ 2 PNO PN′ 2 O4

Or ΔG = –RT loge Kp where Kp is the equilibration constant in terms of partial pressures Equation (3) is applicable to all gaseous reactions provided that the gases behave ideally; thus for the reaction O

aA + bB + ….⇌ yY + zZ + ΔG = ΔG + Rt loge O

y Y a A

(3)

z Z b B

We Know that if ΔG is negative the reaction is spontaneous but if the ΔG is positive the reaction is not spontaneous. Instead the reverse reaction would be spontaneous, and zinc ions would react with copper to give zinc and copper (II) ions when ΔG = 0 neither the forward reaction nor the reverse reaction if favoured. This is what happens when equilibrium is achieved.

It should be realized that the Gibbs energy change here is not standard Gibbs energy change of the reaction. ΔG is the difference in free energies of the reactants and products corrected for non-standard conditions. In a reaction that comes to equilibrium the Gibbs energy decreases to minimum rather than become zero. When equilibrium is achieved in the Daniel cell   Zn 2 + (aq )    0 = DG + RT ln  2+  Cu (aq )     Zn 2 + (aq )    Or DG = –RT ln  2+  Cu (aq )   O

P P P P

(4)

O

effect of concentration on gibbs energy It turns out that equation (4) is applicable to reactions that takes place in solution, under these conditions the concentrations of the substances present are expressed in moles per unit volume. Consider the reaction takes place in Daniel cell. The reactions is Zn(s) + Cu2 + (aq)→Zn2 + (aq) + Cu(s) We assume that the Gibbs energy of a solid does not change even though the concentration of solution in contact with it might change. Also, in this case we are dealing with elements, for which their standard Gibbs energies of formation are defined to be zero under non-standard conditions, the Gibbs energy of the ions is given by G Zn 2+ = Zn 2+ + RT ln [Zn2+(aq)] G Cu 2+ = G Cu 2+ + RT ln[Cu 2+ (aq)] The Gibbs energy change of the reaction is

2+

2+

O

For the Daniel cell we have   Zn 2 + (aq )   ΔG = –RT ln   2 +   Cu (aq )   The ratio of the concentration of the ions in this expression is called equilibrium constant for the reaction. The equilibrium constant can be written as Kc So O

ΔG = –RT ln Kc O

 Zn 2 + (aq )  Where Kc =  2 + Cu (aq )  The subscript of Kc is to show that the equilibrium constant is written using concentrations. We can also apply our condition for equilibrium to reactions that involve gases. For example, N2(g) + O2(g) → 2NO(g)

G Zn 2+ − G Cu 2+ O

6.9.7 equilibrium and equilibrium constants

= G Zn 2+ + RT ln[ Zn (aq )] − {G Cu 2+ + RT ln[Cu (aq )]}

The Gibbs energies of the gases are G o =G o +RTlnPo O

2

2

2

O

O

O

2+

2+

= G Zn 2+ − GCu2+ + RT ln[ Zn (aq )] − RT ln[Cu (aq )] = GZn 2+ − G Cu 2+ O

O

 [ Zn 2 + (aq )]  + RT ln   2+  [Cu (aq )]  O

O

O

Where ∆G = G Zn 2+ − G Cu 2+

G N =G N +RTlnPN 2

2

2

O

G NO =G NO +RTlnPNO The Gibbs energy change for the reaction is ∆G = 2G No − G N2 − G O2

6.36

Thermodynamics

O

O

O

= 2G NO + 2RT ln PNO − G N2 − RT ln PN2 − G O2 − RT ln PO2 O

O

O

= (2G NO − G NO − G O2 ) + 2RT ln PNO − RT ln PN2 − RT ln PO2 O

= ∆G = −RT ln

2 PNO PN .PO 2

Here the Gibbs energy of a solid is assumed to be independent of the pressure, so there are no correction terms for CaO or CaCO3. Then we have ∆G = G CaO + G CO2 − G CaCO3 O

O

O

O

O

= G CaO + G CaO + RT ln PCO2 − G CaCO3

2

At Equilibrium ΔG = O. So ΔG = − RT ln O

O

= (G CaO + G CO2 − G CaCO3 ) + RT ln PCO2 2 PNo PN2 ⋅ PO2

At equilibrium, DG = DG + RT ln PCO2 O

DG = – RT ln PCO2 O

In this case; the equilibrium constant is 2 PNO Kp = PN ⋅ PO 2

In this case the equilibrium constant is very simple KP = PCO2

2

Here the subscript used in Kp is to show that pressure rather than concentrations are used. Another example which involves a mixture of solids and gases is CaCO3(s) → CaO(s) + CO2 (g) We have

From this it can be understood that through there is a mixture of CaCO3, CaO and CO2 then the extent of the reaction can be gauged by measuring the pressure of CO2, often called the dissociation pressure of calcium carbonate. In fact the relation between standard Gibbs energy change and the equilibrium is the same for any reaction At Equilibrium ΔG = –RT lnK However, the way the equilibrium constant itself is written changes from reaction to reaction. Table 6.8 contain some examples which show the relation between equilibrium constant, Gibbs energy value and extent of reaction. O

O

G CO2 = G CO2 + RT ln PCO2 O

G CaO = G CaO O

G CaCO3 = G CaCO3

Table 6.8 Some reactions and their equilibrium constant expression Reaction

Equilibrium Constant Expression

 Zn2+ (aq) + Cu(s) Zn(s) + Cu (aq)  2+

+

2+

Ag (aq) + Fe

 Ag(s) + Fe3+ (aq) (aq) 

 Zn 2 + (aq )  KC = Cu 2 + (aq )   Fe3+ (aq )  KC =  Ag + (aq )   Fe 2 + (aq ) 

 Zn2+(aq) + 2OH– (aq) Zn(OH)2(s) 

KC = [Zn2 + (aq) ] [OH– (aq)]2

 2NO(g) N2(g) + O2(g) 

KP =

2 PNO PN2 ⋅ PO2 2 PNH 3

 2NH3(g) N2(g) + 3H2 (g) 

KP =

 CaO(s) + CO2(g) CaCO3(s) 

KP = PCO2

PN2 ⋅ PH2

Thermodynamics 6.37

Table 6.9 Gibbs energy values and position of equilibrium ΔG KJ mol–1 O

Equilibrium Constant 17

–100 –50 –10 –5

3.4 × 10 5.8 × 108 56.6 7.5

0

1.0

+5

0.1

+ 10 + 50 + 100

1.8 × 10–2 1.7 × 10–9 3 × 10–18

At Equilibrium Completely products Almost all products More products than reactants slightly more products than reactants Equal amounts of products and reactants Slightly more reactants than products More reactants than products Almost all reactants

Completely reactants

Solved Problem 10 Calculate the equilibrium constant Ksp for the reaction

 Ag + (aq) + Cl–(aq) AgCl(s)  Using data Δf G [AgCl(s)] = –1097 KJ, ∆f G [Aq + (aq)] = 77.1 KJ and ∆f G [Cl–(aq)] = –131.2 KJ O

O

O

Solution: ΔG for the reaction is calculated as ∆G = 77.1 + (–131.2) – (–109.7) = 55.6 KJ We have ∆G = –2.303 RT log K ∆G 55.6 × 103 =– log K = = −9.75 2.303 RT 2.303 × 8.314 × 298 O

O

O

O

K = Ksp = 1.8 × 10–10 Problems for Practice 27. For the water gas reaction

The equilibrium constants given in Table 6.10 are for  B for which an imaginary reaction A  K = [B]/[A] Completely products or reactants in the Table 6.10 means that if the reaction is carried out there would be so little of the reactants (or products) present that would not think of the reaction as being an equilibrium reaction at all. Burning magnesium in air is an example Briefly G < 0 G=0 G>0

implies K >1 implies K = 1 implies K < 1

 CO(g) + H2(g) C(s) + H2O(g)  The standard Gibbbs energy for the reaction at 1000 K is –8.1 KJ mol–1. Calculate the equilibrium constant. 28. Calculate equilibrium constant for the reaction at 400 K.

 2NO(g) + Cl2(g) 2NOCl (g)  ∆H = 77.2 KJ mol–1 and ∆S = 122 JK–1 mol–1 at 400 K O

29. Using the following data calculate the value of equilibrium constant for the following reaction at 298 K.

 C6H6(g) 3HC = CH(g) 

A general reaction can be written as

 cC + dD aA + bB  In the above reaction a moles of A react with b moles of B to give c moles of compound C and d moles of compound D. The equilibrium constant is given by KC =

[C]C [D]d [A]a [B]b

KP =

PC C PD d PA a PB b

Kc or Kp is used depending on whether the compound is in solution or gases. One important thing to remember about using equilibrium constants is that solids do not appear in formulae. In the equilibrium constant expressions the concentrations are divided by 1 mol dm–3 and pressures by the standard pressure. This means that as we have defined them the equilibrium constants have no units.

O

Acetylene

Benzene

Assuming ideal behaviour ∆fG [HC = CH(g)] = 2.09 × 105 J mol–1 ∆fG [C6H6(g)] = 1.24 × 105 J mol–1 R = 8.314 JK–1 mol–1 Can the reaction be recommended for the synthesis of benzene? 30. The equilibrium constant at 25°C for the process. O O

 Co3+(aq) + 6NH3(aq) ↽ ⇀  [Co(NH3)6]3+(aq) is 2.0 × 107 Calculate the value of ∆G at 25°C. O

(R = 8.314 JK–1 mol–1) In which direction is the reaction spontaneous when the reactants and products are under standard conditions? 31. It is planned to carry out the reaction CaCO3 (s)  CaO (s) + CO 2 (g)

6.38

Thermodynamics

At 1273 K and 1 bar pressure DrH = 176 KJ mol–1 and DrS = 157.2 KJ mole–1 O

(a) Is the reaction spontaneous at this temperature and pressure? (b) Calculate the value of (i) Kp at 1273 K for the reaction, and (ii) Partial pressure of CO2 at equilibrium.

6.9.8 coupled reactions For a reaction to take place spontaneously ΔrG must be negative. For certain reactions ΔrG is not negative and thus they are not spontaneous. However, these reactions can be made spontaneous by carrying these reactions coupling with some other reactions having very large negative Gibbs energy values so that the resultant Gibbs energy of the two combined reactions become negative. These reactions are called coupled reactions. For example, the decomposition of iron oxide Fe2O3 into iron as 2Fe2O3 (s)→4Fe (s) + 3O2(g)

(1)

Gibbs energy for this reaction ΔrG indicates that the reaction is not spontaneous. These reactions can be made spontaneous by coupling with a reactions having large negative ΔrG For example,

This reaction is coupled with various other necessary reactions in biological systems which are otherwise not spontaneous. For example, the biosynthesis of sucrose from glucose and fructose has a ∆G of + 23 KJ mol–1 By coupling this reaction with hydrolysis of ATP, the reaction becomes spontaneous. O

Glucose + Fructose + ATP → Sucrose + ADP ∆rG = –8 KJ mol–1 O

Therefore, ATP is regarded as centre of all activities of the cell.

6.9.9 gibbs energy change and non-mechanical Work We have already seen that ∆G is a measure of the spontaneity of a chemical reaction. Gibbs energy change is also related to useful work, other than pressure – volume work that can be obtained from the system. It can be shown that Gibbs energy of a process is equal to maximum possible work that can be derived from the process as given below. From the first law of thermodynamics, we know

O

O

2CO(g) + O2(g)→2CO2(g) ΔrG for the reaction is –514.4 KJ mol–1 O

(2)

In equation (1) 3 mol of O2 is involved so that we can write equation (2) as 6CO(g) + 3O2 (g)→6CO2 (g) ΔrG = 3 × (–514.4) = –1543.2 KJ mol–1

(3)

O

Let us combine equation (1) and equation (3) we get 2Fe2O3(s)→4Fe(s) + 3O2(g)ΔrG = + 1487 KJ mol–1 6 CO(g) + 3O2(g)→6CO2(g)1ΔrG = –1543.2 KJ mol–1 O

O

Or 2Fe2O3(s) + 6CO(g)→4Fe(s) + 6CO2(g) ΔrG = –56.2 KJ mol–1 O

The negative value of ΔrG indicates that iron (iii) oxide can be reduced spontaneously to iron with carbon monoxide. During the metallurgy of iron, this reaction occurs at the top part of the blast furnace where iron oxide is reduced to iron. In our body, many biological reactions occur which involve increase in Gibbs energy (ΔG = +Ve). Therefore these reactions may not be spontaneous. Adenosine triphosphate (ATP) is energy rich molecule. When ATP is hydrolyzed in the presence of an enzyme it gives adenosine diphosphate (ADP) and a phosphate ion as O

ATP + H2O →ADP + Phosphate ion ∆rG = –31 KJ mol–1 O

∆U = Q–W Where Q is the heat absorbed by the system, ∆U is the change in the internal energy and W is the work done by the system. The work may be expansion work as well as non-expansion work. The non-expansion work can be used for useful effects, for example electrical work. This is also called non-pressure-volume work or useful work. The work due to expansion at constant temperature known as pressure volume work is given by P∆V. Thus ∆U = Q – Wexpansion – Wnonexpansion = Q – P∆V – Wnonexpansion Or Q = ∆U + P∆V + Wnonexpansion Now ∆U + P∆V = ∆H Or Q = ∆H + Wnonexpansion.

(1)

(2)

For a process carried out reversibly at constant temperature

Qrev T Or Q rev = T∆S Substituting in Eq (2) T∆S = ∆H + W non-expansion Or ∆H – T∆S = –W non-expansion But ∆H–T∆S = ∆G so that ∆G = –W non-expansion Or –∆G = W non-expansion

∆S =

(3)

(4)

As already discussed that the non–expansion work may be regarded as useful work. Thus the decrease in

Thermodynamics 6.39

Gibbs energy of a system during a process is a measure of the maximum useful work done during the change. Therefore, Gibbs energy change, G of a system is a measure of its capacity of doing useful work. The greater the Gibbs energy change, the greater is the amount of work that can be obtained from the process. In the case of electrochemical cells, Gibbs energy change, ∆G is related to the electrical work done in the cell. If E is the e.m.f of the cell and n mole of electrons are involved, the electrical work done will be ∆r G = –nFE cell Here F is Faraday constant = 96500 coulombs If reactants and products are in their standard states then O

∆r G = –nFE

O

Where E is the standard cell potential.

(7) (8)

Suppose that the free energy of the system in the initial state = G1 and the free energy of the system in final state = G2. When the system has suffered an appreciate change in pressure at constant temperature. From relation (7), the free energy change ∆G is given by

∫

P2

VdP

P1

(∆G)T = RT n ∫

dG = dU + PdV + VdP –TdS – SdT (2) The first law equation for an infinitesimal change is written as dq = dU + dw or dq = dU + PdV (dW = PdV, pressure – volume work due to expansion) Also for a reversible process,

(9)

P2 P1

(∆G)T = nRT ln

dP P P2 P1

Or (∆G)T = nRT ln

V1 V2

Here V1 = volume of the system in the initial state V2 = Volume of the system in the final state. Solved Problem 11

dS =

(3) (4)

This equation gives change of free energy when a system undergoes, reversibly, a change of pressure as well as change of temperature. At constant pressure dp = 0 dG = –SdT (5)

At two different states

Then dG = VdP  ∂G  Or  =V  ∂P  T

(1)

Differentiating the above relation

 ∂G  = –S Or   ∂T  p

Also when temperature is kept constant dT = 0

Here P1 = pressure in the initial state P2 = pressure of the system in the final state We know that for n moles of an ideal gas PV = nRT nRT Or V = P Subtracting the value of V in relation (9)

6.9.10 Variation of gibbs energy with Temperature and Pressure

dq T or dq = TdS or TdS = dq + PdV Combining (2) and (3) we get dG = VdP – SdT

 ∂DG  ∴  = –∆S  ∂T  p

∆G = G2–G1 =

O

We know that G = H–TS Also H = U + PV Therefore G = U + PV – TS

 ∂G1   ∂G2    = –S1 and   = –S2 ∂T p ∂T  p

(6)

Calculate the free energy change, which occur when 2 moles of an ideal gas expand reversibly and isothermally at 300 K from initial state 4 litres to 40 litres. Solution: Given that n = 2, T = 300 K, initial vol V1 = 4 Litres Final vol V2 = 40 Litres, R = 1.987 Cal K–1 mol–1 Substituting the values in the equation (∆G)T = n RT ln

= 2.303 n RT log

V1 V2 V1 V2

6.40

Thermodynamics

= 2.303 × 2 × 1.987 × 300 × log

4 40

= –2745.6 cal

When the two are in equilibrium dG(l) = dG(g) From (1) and (2) V(l) dp – S(l) dT = V(g) dP – S(g) dT

S ( g ) − S (l ) dp = V ( g ) − V (l ) dT

Solved Problem 12 Calculate the standard Gibbs energy change for the reaction Or

Zn + Cu (aq) → Cu + Zn (aq) E = 1.10 V 2+

2+

O

Solution: We know that O

O

∆G = –nFE Zn + Cu 2+ → Cu + Zn2 + n = 2, E = 1.10 V, F = 96500 C ∴ ∆G = –2 × (96500 C) × (1.10 V) = –212300 J = –212.3 KJ ( CV = J) O

DS (vapour ) dp = DV (vapour ) dT

Where ∆S and ∆V denote the changes of entropy and volume respectively Also, we know that at equilibrium dG = O ∴ ∆G(vap) = ∆H(vap) – T∆S(vap) = O

O

∆S(vap) =

DH ( vap ) T

(4)

Substituting the value of ∆S(vap) in equation (3) we get

DH ( vap ) dP = T DV( vap ) dT

Solved Problem 13 Calculate at 25°C the equilibrium constant for the reaction Cu(s) + 2 Ag + (aq) → Cu 2+ (aq) + Ag (s) At 25°C, E cell = 0.47 V; R = 8.314 JK–1; F = 96500 C O

Solution: O

E cell =

0.059 log Kc 2 E cell × 2 Or log Kc = 0.059 E cell = 0.47 v, n = 2 0.47 × 2 ∴ log Kc = 0.059 Or Kc = 8.5 × 1015

This equation is known as Clapeyron equation and gives the rate of change of vapour pressure with temperature. It is general equation and can be applied to any equilibrium between phases. For sublimation and fusion process. Clapeyron equation can be written as

DH ( s ) dP = (for sublimation) T (Vg − Vs ) dT

O

O

6.9.11 clapeyron equation Clapeyron derived an equation which finds extensive application in one component – two phase system. For examples, liquid and vapour, solid and liquid etc. Consider a liquid in equilibrium with its vapours at temperature T and pressure P. Let the temperature change from T to T + dT and the pressure changes from P to P + dP at equilibrium dT and dP are infinitesimal changes in temperature and pressure respectively. We know That dG = Vdp – SdT For liquid dG(l) = V(l) dp – S(l) dT (1) For vapours of the liquid dG(g) = V(g) dP- S(g) dT (2)

And

DH f dP = dT T V(l ) − V( s )

(

)

(for fusion)

6.9.12 clausius–clapeyron equation Clapeyron equation (4) can be put in a convenient form which is applicable to vaporization and sublimation equilibrium in which one of the two phases is gaseous. ∆V(vap) = V(g) – V(l) We know that the volume occupied by a liquid in the gaseous form is much large. Hence, V(l) can be neglected in comparison to V(g) RT ∆V(vap) = V(g) P Substituting the values of V(vap) in relation (4) DH ( vap ) .P DH ( vap ) . × P dP = = dT TRT RT 2 DH ( vap ) dT dP Or = . 2 R P T

Thermodynamics 6.41

Integrating the above equation between the limits P1 and P2. Corresponding to the temperatures T1 and T2, we get

∫

DH ( vap ) dP = P R

P2 P1

ln

T2

∫

T1

dT T2

DH ( vap )  T2 − T1  P2 =   R  T1 T2  P1

DHV  T2 − T1  P2 =   P1 2.303R  T1 T2  This equation is called Clausius – Clapeyron equation. It is an integrated form of the equation and finds extensive application for the measurement of vapour pressure over various liquids and solids. log

6.9.13 application of clapeyron’s classius  → Vapours equilibrium equation for Liquid ←  Following are some important applications of Classius – Clapeyron equation. 1. Calculation of molar heat of vaporizations: The molar heat of vaporizations of a liquid can be calculated if its vapour pressure at two different temperatures are known. Solved Problem 14 Vapour pressure of a liquid at 85°C and 96°C and 557 mm and 645 mm respectively. Calculate the molar heat of vaporization of the liquid between 85°C and 96°C. Solution: The integrated form of Clausius – Clapeyron equation log

DH c  T2 − T1  P2 =   2.303R  T1 T2  P1

Here P1 = 557 mm, P2 = 645 mm, T1 = 85°C ≡ 358 k T2 = 96°C = 369 K, R = 8.314 Joules per degree per mole substituting the values we get log

DHV 645  369 − 358  = 2.303 × 8.314  369 × 358  557 ∆Hv 1.1 × = = 19.148 132102 log 1.1579 = 0.0000043 ∆Hv 0.0636 = 0.0000043 ∆Hv ∆Hv = 11790.47 Joules per mole = 11.79 KJ per mole

2. Effect of temperature on vapour pressure of a liquid This equation helps in calculating the vapour pressure of liquid at any temperature if its vapour pressure at a particular temperature is known. Solved Problem 15 The vapour pressure of water at 100°C is 760 mm. What will be the vapour pressure at 86°C? The latent heat of vaporization of water in this range is 41.27 KJ per mole. Solution: We know that the integrated form of Clapeyron – Classius equation is

DHV  T2 − T1  P2 =   2.303R  T1 T2  P1 Here P1 = 760 mm T1 = 100°C = 373 K T2 = 86°C = 359 K log

P2 = ? Substituting the values in the equation log

P2 41270  359 − 373  = 760 2.303 × 8.314  359 × 373 

log P2 –log 760 =

41270 × ( −14) 8.314 × 2.303 × 373 × 359

= –1.8735 log P2 –2.8808 = –0.2253 log P2 = 2.6555 P2 = 452.4 mm 3. Effect of pressure on boiling point: This equation helps in calculating the boiling point of a liquid at any pressure if the value of boiling point at some pressure is known. Solved Problem 16 A liquid boils at 45.5°C at one atmospheric pressure. At what temperature will it boil at 735 mm pressure, given that the latent heat of vaporization of the liquid is 327.2 per gram? (molecular weight of the liquid is 67). Solution: The integrated form of Clapeyron – Classius equation is

DH v  T2 − T1  P2 =   2.303R  T1 T2  P1 here P1 = 760 mm, P2 = 735 mm T1 = 45.5°C = 318.5 K , T2 = ? (Boiling Point at 735 mm) DHV = 327.2 joiles/gram = 327.2 × 67 joules per mole log

6.42

Thermodynamics

Substituting the values we get log

T − 318.5 327.2 × 67 735 = × 2 2.303 × 8.314 318.5T2 760 –0.0154 =

3.594(T2 − 318.5) T2

–0.0154 T2 = 3.594 T2 –1144.69 3.6094 T2 = 1144.69 T2 = 317.14 K = 44.14°C   Vapour ⇀ 4. Classius – Clapeyron Equation for Solid ↽ Equilibria The integrated form of the Clausius – Clapeyron equation for solid vapour equilibrium can be written as P ∆H(s)  T2 − T1  log 2 =   P1 2.303 R  T1 T2 

Here ΔH(s) = molar heat of sublimation of the substance. Note: The integrated form of classius – clapeyron DH (s ) dP could be possible only = dT T(V( g ) − V(s ) ) as the molar – volume of the substances in the gaseous state is very much greater than that in the solid state Thus (Vg–Vs) can be taken as Vg  → liq5. Classius – Clapeyron equation for solid ←  uid equilibria  → The Classius – Clapeyron equation for solid ←  liquid equilibrium is equation

∆H f dP = dT T (V1 − Vs ) Where ΔHf is the mole heat of fusion of ice. This equation can not be easily integrated. The reason is that Vs cannot be ignored as compared to Vl. Moreover the laws of liquid state are not so simple as those for gaseous state.

Thermodynamics 6.43

key PoinTS •

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The energy changes associated with physical and chemical process may be used to do mechanical work or to provide electrical energy or radiant energy or chemical energy or nuclear energy. The different forms of energy may be transformed from one into the other and the branch of science which deals with those energy transformations is referred to as thermodynamics. Thermodynamics means flow of heat and it deals with quantitative relationships between heat and other forms of energy in physico-chemical transformations. Thermodynamics mainly deals with basic laws governing heat, followed latter by statistical mechanics which explains the laws of thermodynamics. Thermodynamics is based on four experimental laws namely zeroth, the first, the second and the third laws. Laws of thermodynamics are based on experimental facts but not on theoretical facts. Thermodynamic laws provide the criteria for predicting the feasibility of a reaction and the energy transformations that occur in the processes. Laws of thermodynamics do not give any information regarding the rates of reactions or process.

Thermodynamic Variables •

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Thermodynamic equilibrium •

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Types of Systems • •

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In thermodynamics, the entire universe is considered as two parts, system and surroundings. A system is defined as any specified portion of matter in the universe, which is chosen for thermodynamic studies. The part of the universe other than the system which can exchange energy and matter with system is called surroundings. If the walls of a system allow the transmission of heat through them into or out of the system are called diathermal walls and those walls of a system which do not permit any heat through them into or out of the system are called adiabatic walls. A system which can exchange energy and matter with the surroundings is known as open system e.g., ice in an open beaker. A system which cannot exchange energy or matter with the surroundings is known as an isolated system e.g., ice in a thermos flask.

Quantities such as pressure, temperature and volume which define a system are called thermodynamic variables, and these are one of two categories intensive and extensive. The properties of the system which depend on the amount of the substance are called extensive properties e.g., mass, volume, energy, internal energy, free energy, molar entropy, molar enthalpy and heat capacity. The properties of substance which depend on the nature of the substance and is independent of the amount of the substance are called intensive properties e.g., pressure, temperature, density, viscosity, specific heat, surface tension, refractive index, boiling point, freezing point. The operation which brings about the changes from one state to another state is thermodynamic process.

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A system is said to be in mechanical equilibrium when there is no unbalanced force existing between different parts of the system or between the system and the surroundings. A system is said to be in thermal equilibrium if the temperature is same throughout the whole system including the surroundings. A system is said to be in a chemical equilibrium if the composition of the system remain constant and definite. Mechanical equilibrium implies the uniformity of pressure, thermal equilibrium to the uniformity of temperature and chemical equilibrium to the uniformity of chemical composition. A system may be in mechanical and thermal equilibrium, without being in chemical equilibrium. A system is said to be in thermodynamic equilibrium if it satisfies all the three equilibria viz, mechanical, thermal and chemical.

State Variables •

The variable properties such as temperature, pressure, volume and composition required for completely defining a system thermodynamically are known as thermodynamic properties or parameters or state variables.

6.44

• •

Thermodynamics

A state variable is one that has a definite value when the state of system is specified. In the gas equation PV = RT, if two variables are known, the third can be easily calculated as R is gas constant. The two variables usually specified are temperature and pressure and are called independent variables. The third variable i.e., volume depends on temperature and pressure and is said to be dependent variable.

homogeneous and heterogeneous Systems •

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A system is said to be homogeneous if it is uniform throughout. A homogeneous system consists only one phase. A system is said to be heterogeneous if it is not uniform throughout. A heterogeneous system consists of two or more phases which are separated from each other by definite boundaries. However, each phase itself is uniform throughout. The thermodynamic properties whose values de pend only upon the initial and final states of the system and are independent of the manner as how the change occur are called state functions, e.g., internal energy (U), enthalpy (H), entropy (S), Gibbs energy (G), pressure (P), temperature (T), volume (V) etc. The parameters which depend on the path into which system is brought are called path functions e.g., heat (q) and work (W). Thermodynamic laws are not applicable to microscopic systems. The mass energy transformations are not being considered except as a carrier of energy.

Types of Processes •

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The process in which the system does not exchange heat (q) with the surroundings is known as adiabatic process. In adiabatic processes in the case of an exothermic process, the heat evolved remains in the system and therefore the temperature of the system rises while in endothermic process the heat absorbed is supplied by the system and hence the temperature of the system decreases. A process in which the temperature ‘T’ of the system remains constant during each step of the process is known as isothermal process. For isothermal process dT = 0. A process in which the pressure (P) of the system remains constant during each step of the process is known as an isobaric process. For an isobaric process dP = 0.

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A process in which the volume of the system re mains constant during each step of the process is known as isochoric process. For an isochoric process dV = 0 A process is said to be a cyclic process if a system after completing a series of changes returns to its original state. Sequence of steps starting from the initial to the final state of the system is called path and the path of a cyclic process is called a cycle. A process which is carried out infinitesimally slowly so that the driving force is infinitesimally greater than the opposing force is known as reversible process. A process which takes place suddenly and spontaneously without occurring infinitesimally slowly is known as irreversible process. In thermodynamics work may be defined as any quantity that flows across the boundary of a system during a change in its state and is completely convertible into lifting of a weight in the surroundings.

Zeroth Law •

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Zeroth law states that equality of temperature is a condition for thermal equilibrium between two systems or between two parts of a single system. If two systems are separately in thermal equilibrium with a third, then they must also be in thermal equilibrium with each other.

internal energy •

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Every system is associated with definite amount of energy known as internal energy or intrinsic energy denoted by U. Internal energy of different substances is different and depends on the state conditions. The internal energy is the sum of all energies i.e., kinetic and potential energies of the particles making up the substance. The kinetic energy includes the translational, vibrational and rotational energies of the molecules The potential energy comprises attraction between atoms in a molecule (chemical bonds) attraction between molecules, the arrangement of electrons outside the nucleus and the arrangement of protons and neutrons inside the nucleus. Internal energy depends only on the initial and final states of the system but not on the path through which the change was brought.

Thermodynamics 6.45

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Change in internal energy is a state function but not path function. Heat (Q) and work (W) are path functions Internal energy of a system depends on the state of the substance governed by temperature, pressure, volume and concentration which are known as state variables. The magnitude of internal energy (U) of a substance cannot be determined experimentally but the change in internal energy (ΔU) due to change in volume pressure, temperature, chemical composition can be determined experimentally. The energy change can be measured in the form of heat exchanged with surroundings and the work done with the change in volume against pressure. If the volume and temperature kept constant the change in internal energy (U) is equal to the heat exchanged with surroundings ΔUv = Q (because W = O) In a chemical reaction the change in internal energies (ΔU) when reactants convert into products can be determined by measuring heat liberated or absorbed when the reaction is carried at constant temperature and volume. For a chemical reaction A→B; the ΔU can be written as ΔU = Up – Ur, if ΔU is negative the reaction is exothermic while the positive ΔU indicates the endothermic reaction.

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The work done in isothermal and reversible expansion of an ideal gas is given by the equation V Wrev = –2.303 nRT log f Vi Pf Pi Vi and Pi are the initial volume and pressure while Vf and Pf are the final volume and pressure of the gas. Work done in an isothermal irreversible expansion of an ideal gas is given by = 2.303 nRT log

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 P  W = nRT 1 − 2   P1  The work done in an irreversible process is less than the work done in the reversible process. Heat absorbed in an irreversible expansion of an ideal gas is less than heat absorbed during the reversible process. Work done during compression is negative. The magnitude of work involved in an irreversible compression will be more than that in a reversible process. In adiabatic expansion of ideal gas, since heat is not allowed into the system the work done is due to the decrease in internal energy (ΔU = w) thus causing decrease in temperature. Thus P1Vγ1 = P2Vγ2 or PVγ = Constantant γ

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First law of thermodynamics explains the heat changes accompanying chemical reactions. First law of thermodynamics states that energy can neither be created nor destroyed but energy can be converted from one form to the other. First law of thermodynamics rules out “the possibility of construction of a perpetual motion machine of first kind. A perpetual motion machine is one which operating in cycles, can produce work without any expenditure of energy on it. According to first law of thermodynamics the total energy of the system and surroundings is constant or conserved. Mathematically first law of thermodynamics can be represented as Q = ΔU + W where q is the amount of heat absorbed by the system from surroundings, ΔU is the increase in the internal energy of the system and W is the work done by the system. For infinitesimally small changes q = dU + W.

1− γ

 T1   P2    =   T2   P1 

First Law of Thermodynamics •

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Work done in the irreversible expansion of an ideal gas in adiabatic process is given by T P  W = nRT1  2 − 2   T1 P1  Internal energy change for adiabatic process is  nRT1 nRT2  ∆U = P2  −  P2   P1   P2    CV + R     P1   ∴ T2 =  T  1  CV     Enthalpy change in adiabatic irreversible expansion of an ideal gas is given by ∆H = nCP ( T2 − T1 ) . In an isothermal process, temperature of the system remains constant where as in adiabatic process temperature has to be changed.

6.46

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Thermodynamics

The pressure-volume relation for reversible isothermal process is given by the Boyle’s law PV = constant (at constant T). The pressure-volume relation for reversible adiabatic process is given by PV g = Constant. The increase in volume for a given decrease of pressure will be less in adiabatic expansion than –that in isothermal expansion. The smaller change in volume in adiabatic process than in isothermal process is due to decrease .in temperature with the decrease in internal energy, whereas in isothermal expansion heat is absorbed from surroundings to make up for expansion work done by the gas and hence the temperature remain constant. The work done by an ideal gas is greater in isothermal expansion than that in adiabatic expansion.

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heat capacity or enthalpy h •

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If the reaction is carried between solids and liquids, pressure has no effect. Hence these reactions are considered as carried at constant volume. If the substance in a reaction are gases the reaction may be carried at two different conditions i) at constant volume and ii) at constant pressure. Since atmospheric pressure is practically constant, chemical changes in open containers can be considered as taking place at constant pressure but not at constant volume. The amount of heat exchanged with the surroundings for a reaction at constant pressure and temperature differs from that exchanged at constant volume and temperature. If a gas absorbs heat at constant volume all the heat content is used to rise the internal energy of gas. If a gas absorbs heat at constant pressure some heat content is used to rise the internal energy of gas and the remaining heat content is used to do some work by the expansion of gas (ΔU + W). Since work done W is equal to PΔV; ΔU + W can be written as ΔU + PΔV. The energy change at constant pressure and temperature is called the enthalpy change denoted by ΔH The total heat content of a substance or a system at constant pressure and temperature is called enthalpy. Enthalpy of a system is equal to the sum of internal energy (U) and the product of pressure (P) and volume (V) of the system, i.e., PV The energy change taking place at constant pressure and constant temperature is called enthalpy change (ΔH) Enthalpy is a state function but not path function and it is an extensive property.

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Enthalpy (H) of a substance cannot be determined experimentally but the change in enthalpy (ΔH) can be calculated or determined. For a chemical reaction taking place at constant pressure and temperature the enthalpy change (ΔH) is the difference in the heat contents of the products (Hp) and reactants (HR); ΔH = Hp-HR. For a cyclic chemical change ΔH = 0 since the initial and final states are same. At constant volume since ΔV = 0; ΔH = ΔU. ΔH is the change in heat content at constant pressure and constant temperature while ΔU is the change in heat content at constant volume and constant pressure. The energy change taking place at constant volume and constant temperature is called internal energy change. For gaseous reactions ΔH and ΔU are related as DH = DU + nRT where n = (Total number of molecules of gaseous products) – (Total number of molecules of gaseous reactants); R = gas constant; T = absolute temperature. Heat capacity of a substance is the amount of heat required to rise its temperature through one degree. Specific heat is the amount of heat required to rise the temperature of one gram of the substance through 1°C. At constant volume the total heat capacity is utilized for the change of internal energy (ΔU) of the system with temperature. Cv =  ∂E   ∂T   V

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At constant pressure a part of the heat capacity is utilized for changing the internal energy and the remaining part is utilized for the work.

Cp CP = DU + PDV PDV = W i.e., Work done •

The relation between CP and CV is Cp-Cv = R CP =γ CV The enthalpy change measured at standard pressure are called standard enthalpy change and represented by ΔH also known as molar standard enthalpy changes ΔH m For special type of enthalpy changes to emphasize subscripts are used e.g., for enthalpy change of combustion of a compound is denoted as ΔH c . Standard enthalpy change ΔH refers to 101.325 pa (1 atm) and normally refers to 298 K. Or

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Thermodynamics 6.47

Second Law of Thermodynamics •

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First law of thermodynamics failed in (i) Predicting whether a process is spontaneous or not in the specified direction. (ii) If a transformation occurs, what fraction of one form of energy is converted into another form of energy in this transformation. The process which take place on its own accord without the aid of an external agency is called spontaneous process. All naturally occurring process are spontaneous and thermodynamically irreversible. All spontaneous process proceed till equilibrium is achieved. Examples of spontaneous processes are (i) Water flows from higher level to the lower level but cannot be reversed the direction of flow without some external aid. (ii) Heat can flow from hotter body to colder body and it cannot be reversed. (iii) If two solutions of different concentrations are mixed, it becomes homogeneous due to diffusion of solute from more concentrated solution to a less concentrated solution. (iv) Electricity flow from higher potential to lower potential. Its direction can be reversed by applying external potential in opposite direction. A non-spontaneous process is that which has no tendency to occur or which is forbidden and made to occur only if energy from outside is continuously supplied. The reverse direction of all spontaneous process is non-spontaneous processes.

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entropy and Spontaneity Entropy means transformation and denoted by S. Entropy is a measure of disorder or randomness in a system.

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Units of entropy change is J/K or JK–1 The increase in entropy which occurs when gases mix without interaction is given by mathematical equation ΔS = R loge

2V1 × 2V1 V1 × V1

In which R is gas constant when 1 mole of each gas involved and 1 mole of each of two gases occupying a volume V1 initially but after mixing a volume of 2V1. • Increase in entropy is associated with maximum number of ways in which the energy of the system can be distributed.

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According to quantum theory, energy is held in a series of definite energy levels; for a given energy range there are more translational energy levels than rotational energy levels, there are more rotational energy levels than vibrational energy levels. Thus there are several ways of distributing a given energy. The various translational energy levels are more than among either rotational or vibration energy levels. When ammonium nitrate is dissolved in water (i) The solid breaks up and hydrated ions are dispersed in solution. So energy of vibration is replaced by energy of translation and thus increase in the number of energy levels resulting in increase of entropy. (ii) The process of dissolution of ammonium nitrate is endothermic due to which air in the surroundings is cooled down resulting in the decrease of translation, rotational and vibration energy levels thus entropy decreases. (iii) Due to hydration of ions the translational and rotational motion of water molecules is restricted. So entropy decreases. The conversion of vibration energy of NH4NO3 into translational energy of the hydrated ions results in large increase on entropy causing overall increase in entropy. Changes take place in an isolated system in a direction such that when equilibrium is achieved, the number of ways in which the total energy of the system can be distributed is a maximum i.e., the entropy of the system is maximum. Entropy is a measure of the number of ways that energy can be shared out among molecules. Each arrangement for the distribution of energy between different molecules is complexion or ensemble. With increase in temperature the number of complexions increases. More the number of complexions or ensembles more of the increase in entropy. Quantitatively the entropy change ΔS is given by qrev mathematical expression ΔS = where qrev is the T heat absorbed when the process is carried reversibly and isothermally, i.e., at constant T. Entropy is a state function and is independent of path. It is an extensive property. Second law of thermodynamics states that The entropy of the universe always increase in the course of every spontaneous (natural) change. (or) The energy of the universe is conserved where as the entropy of the universe always increase in any natural process. (or)

6.48

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Thermodynamics

Heat cannot flow from a colder body to hotter body on its own. (or) Heat cannot be converted into work completely without causing some permanent changes in the system involved or in surroundings. (or) All spontaneous processes are thermodynamically irreversible and entropy of the system increases in all spontaneous process. (or) It is impossible to construct a machine working in cycles which transforms heat from lower temperature region to higher temperature region without the intervention of any external agency. A perpetual motion machine is that which can transfer heat from lower temperature to higher temperature on its own. For a spontaneous process in an isolated system the change in entropy (ΔS) is positive. If the system is not isolated the total entropy change i.e., ΔStotal the sum of the entropy change of the system (DSsystem) and the entropy change in surroundings (ΔSsurroundings) must be positive.

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The entropy change when one mole of a liquid changes into vapour at its boiling point is the entropy of vapourization ∆ Vap H ∆ VapS = (T is boiling Point) T 40.79 KJ (∆ Vap H For water = 40.79KJ mol−1 ) ∆ Vap S = 373 = 109.356 JK–1 mol–1 O

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The standard entropy change Δr S for a reaction can be determined by subtracting the standard entropies of reactants from the standard entropies of products. ∆ r Sm = ∑ VP Sm (Products) − ∑ Vr Sm (reactants ) O

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(Vp and Vr are stoichiometric coefficients, Sm is standard molar entropy). The entropy per unit amount of a substance in its standard state at the specified temperature is standard molar entropy denoted by S m. The standard entropy of a substance or ion is also called as absolute entropy. O

ΔStotal = ΔSsytem + ΔSsurroundings >0 • • •

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When an isolated system is in equilibrium the entropy is maximum. The mathmatical condition for entropy (S) to be maximum is that the change in entropy (ΔS) is zero. In some exothermic reaction the entropy of the system may decrease but the heat liberated may increase the enropy in surroundings leading to the total entropy change will become positive and in such situation the reaction will be spontaneous. When magnesium burns in oxygen, conversion of oxygen into oxide results in decrease of entropy but the heat liberated will increase the entropy of surroundings. In endothermic reactions entropy of surroundings decrease due to the flow of heat from surroundings into system, but the entropy of the system increases so that overall entropy increase is positive, the reaction is spontaneous.

gibbs energy and Spontaneity •

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entropy change during Phase Transformations •

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The entropy change when one mole of a solid substance changes into liquid form at its melting point is known as entropy of fusion. For the conversion of solid to liquid the heat required is latent heat equal to standard enthalpy of fusion ΔfusH at constant temperature and pressure. For water Δfus = 6.0 KJ mol–1 q rev ∆ fus H 6.0 × 1000Jmol−1 = = = 219.95mol−1K −1 T 273K 273

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Gibbs energy is the amount of energy available from a system which can be put to useful work at constant temperature and pressure. Gibbs energy is an extensive property. Gibbs energy or Gibbs function G has been defined as G = H-TS. The change in Gibbs energy for the system ΔGsystem at constant temperature is ΔGsystem = ΔHsystem –TΔSsystem. Only a part of the energy released in a reaction can be used to do work, the rest is involved in the entropy change. The Gibbs energy change (ΔG) is the amount of energy available to do work. Spontaneous reactions can do work. Spontaneous reactions have negative Gibbs energy change. Non-spontaneous reactions cannot do work. Nonspontaneous reactions have positive Gibbs energy change. Gibbs-Helmholtz equation is

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∆G ∆H =− + ∆S T T

Each term in the above equation represents entropy change. The equation says Entropy change in the universe = Entropy change in the surroundings + Entropy change in the system. The Gibbs energy of an element in its standard state is zero.

Thermodynamics 6.49

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The reactions which take place at high temperature are mainly due to increase in entropy and are generally termed as entropy driven reactions. If ΔG is negative the process is spontaneous. If ΔG is zero the process is in equilibrium state and there is no net reaction in either direction. If ΔG is positive the process does not occur in the forward direction, however, it may take place in the reverse direction. ΔG will be negative under the following conditions (i) Both the energy and the entropy factors are favourable and may have any magnitude i.e., ΔH is negative and TΔS is positive. (ii) When energy factor favours the change (ΔH = –Ve) but entropy factor opposes the change (TΔS = –Ve). Then the magnitude of ΔH should be more than TΔS. (iii) When energy factor is not favouring change (ΔH = +Ve) but entropy factor favours the change (TΔS = +Ve). Then the magnitude of TΔS should be more than that of energy factor. (Refer Table 6.8)

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smaller than 1. And the reaction is unlikely to form much product. In case of exothermic reactions ΔH is large and negative and ΔG is also likely to be large and negative. Hence the value of K will be much larger than 1 and reaction can go to near completion.

gibbs energy and Work •

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The Gibbs energy change for a process taking place reversibly at constant T and P is equal to the maximum possible useful work. ΔG = –Wmax In the case of galvanic cells Gibbs energy change ΔG is related to electrical work done by the cell

DG = − nFE Where E is the emf of the cell, n is the number of electrons involved and F is Faraday. • If the reactants and products are in their standard state ΔG = –nFE O

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gibbs energy change and equilibrium constant •

Third Law of Thermodynamics

The Gibbs energy change of reaction ΔG is related to the composition of the reaction mixture and to standard reaction Gibbs energy change ΔG by the equation ΔG = ΔG + RT ln Q. Where Q is the reaction quotient, R is the gas constant with the value 8.314 JK–1 mol–1 At Equilibrium Q = equilibrium constant At equilibrium ΔG = 0

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∴ 0 = ∆G + 2.303 RT log k p or∆G = −2.303 RT log K p •

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The third law of thermodynamics states that the entropy of a pure and perfectly crystalline substance at absolute zero is zero. The entropy of a solution and super cooled liquid is not zero at 0 K. For example glycerol and glass exhibit positive entropy at 0 K. Third law of thermodynamics is also referred as Nernst heat theorm. Unlike first and second laws the third law does not lead to any new concept of thermodynamics but imposes a limitation on the volume of entropy.

Gibbs energy change for isothermal expansion of an ideal gas is given by ( DG )T = 2.303nRT log

P2 V = 2.303nRT log 1 P1 V2

Where V1 = initial volume, V2 = final volume, P1 = initial pressure and P2 = final pressure O

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S1 = ∫ 0

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One of the most important applications of third law is the calculation of entropy changes in chemical reactions aA + bB + ……  → l L + mM + ……

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∆G = ∆H − T∆S = −RT ln K

ΔS is given by O

Since ΔG depends on the value of ΔH for highly exothermic reactions the value of ΔH may be large and positive and hence the value of K will be much

ΔS = lSl + mSM + ... − aSA + bSB + ....

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6.50

Thermodynamics

coupled reactions •

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The reactions in which ΔG is positive cannot take place spontaneously but they can be made spontaneous by coupling with some other reactions having very large negative Gibbs energy value so that the resulting Gibbs energy of the two combined reactions become negative. Most of the metallurgical processes and biological reactions involves coupled reactions e.g., The hydrolysis of adenosine diphosphate (ADT) and a phosphate is coupled with various other necessary reactions in biological systems which are otherwise not spontaneous.

gibbs energy change and non-mechanical Work •

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The work due to expansion at constant temperature is known as pressure-volume work and is given by PΔV. The Work that can be obtained from the system by non-expansion work such as electrical work is nonmechanical work.

Thermodynamics 6.51

PracTice exerciSe multiple choice Questions with only one answer Level i 1. A thermodynamic quantity is that (a) which is used in thermochemistry (b) which obeys all the laws of thermodynamics (c) quantity, whose value depends only on the state of the system (d) quantity, which is used in measuring thermal change 2. Maximum work can a gas do, if it is allow to expand isothermally against (a) vacuum (b) high pressure of surroundings (c) low pressure of surroundings (d) atmospheric pressure 3. The work done in an adiabatic change of gas depends on change in (a) volume (b) pressure (c) temperature (d) none of these 4. If W1, W2, W3 and W4 are the magnitudes of work done in isothermal, adiabatic, isobaric and isochoric reversible expansion of an ideal gas, the correct order (for expansion) will be (a) W1> W2> W3>W4 (b) W3> W2> W1 >W4 (c) W3>W2>W4>W1 (d) W3, W1>W2>W4 5. The volume of a system becomes twice its original volume on the absorption of 300 cal of heat. The work done on the surrounding was found to be 200 cal. What is ∆E for the system? (a) 500 cal (b) 300 cal (c) 100 cal (d) –500 cal 6. In a change from state A to state B (a) q depends only on the initial and final states (b) W depends only on the initial and final states (c) ∆E depends only on the initial and final states (d) ∆E depends upon the path adopted by A to change into B 7. The internal energy change when a system goes from state A to B is 40 KJ/mol. If the system goes from A to B by a reversible path and returns to state A by an irreversible path, what would be the net change in internal energy? (a) 40 KJ (b) >40 KJ (c) < 40 KJ (d) Zero 8. The internal energy of an ideal gas increases during an isothermal process when the gas is (a) expanded by adding more molecules to it (b) expanded by adding more heat to it (c) expanded against zero pressure (d) compressed by doing work on it

9. One mole of an ideal gas at 300 K is expanded isothermally from an initial volume of 1 L to 10 L. The ∆E for this process is (a) 163.7 cal (b) zero (c) 1381.1 cal (d) 9 L–atm 10. Temperature of one mole of gas is increased by one degree at constant pressure. Work done is (a) R (b) 2 R (c) R/2 (d) 3 R 11. Five moles of an ideal gas expand, isothermally and reversibly from an initial pressure of 100 atm to a final pressure of 1 atm at 27°C. The work done by the gas is (a) 3455 cal (b) 6909 cal (c) 0 (d) 13818 cal 12. For two moles of an ideal gas, (a) Cv – Cp = –2 R (b) Cp – Cv = 0 (c) Cp – Cv = R (d) Cp – Cv = R/2 13. If one mole of a mono atomic gas (γ = 5/3),is mixed with one mole of a diatomic gas (γ = 7/5), the value of γ for the mixture is (a) 1 (b) 1.5 (c) 2 (d) 3.07 14. How many times a diatomic gas should be expanded adiabatically so as to reduce the rms velocity to half? (a) 8 (b) 16 (c) 32 (d) 64 15. Work done by a sample of an ideal gas in a process A is double the work done in another process B. The temperature rises through the same amount in the two processes, if CA and CB be the molar heat capacities for the two processes (a) CA = CB (b) CA > CB (c) CA < CB (d) both, undefined 16. One mole of oxygen is heated from 0°C, at constant pressure, till its volume increased by 10 %. The specific heat of oxygen, under these conditions, is 0.22 cal/g-K. The amount of heat required (in J) is (a) 32 × 0.22 × 27.3 × 4.2 (b) 16 × 0.22 × 27.3 × 4.2 (c)

32 × 0.22 × 27.3 4.2

16 × 0.22 × 27.3 4.2 17. 743 J of heat energy is needed to raise the temperature of 5 moles of an ideal gas by 2 K at constant pressure. How much heat energy is needed to raise the temperature of the same mass of the gas by 2 K at constant volume? (a) 826 J (b) 743 J (c) 660 J (d) 600 J

(d)

6.52

Thermodynamics

18. If all the following gases are in monoatomic form, which has greater entropy? (a) H (b) N (c) O (d) Cl 19. What is the entropy change when 3.6 g of liquid water is completely converted into vapours at 373 K? The molar heat of vaporization is 40.85 KJ/mol. (a) 218.9 J/K (b) 2.189 J/K (c) 21.89 J/K (d) 0.2189 J/K 20. Given the following entropy values (in J/K–mol) at 298 K and 1 atm H2(g) = 130.6, Cl2(g) = 223.0 and HCl(g) = 186.7. the entropy change (in J/K–mol) for the reaction: H2(g) + Cl2(g) → 2 HCl(g), is (a) + 540.3 (b) + 727.0 (c) –166.9 (d) + 19.8 21. According to second law of thermodynamics, heat is partly converted into useful work and part of it (a) become electrical energy (b) is always wasted (c) increases the weight of the body (d) become K.E. 22. The change that does not increase entropy (a) evaporation of liquid (b) condensation (c) sublimation (d) melting of solid 23. Ammonium chloride when dissolved in water leads to cooling sensation. The dissolution of NH4Cl at constant temperature is accompanied by (a) increase in entropy (b) decrease in entropy (c) no change in entropy (d) no change in enthalpy 24. The entropy change in the fusion of one mole of a solid melting at 300 K (latent heat of fusion, 2930 J/mol) is (a) 9.77 J/K –mol (b) 10.73 J/K–mol (c) 2930 J/K – mol (d) 108.5 J/K –mol 25. When one mole of an ideal gas is compressed to half of its initial volume and heated to twice its temperature, the change in entropy is (a) Cv ln2 (b) Cpln2 (c) R ln2 (d) (cv – R) ln2 26. When a substance is heated, its entropy increases. The increase will be maximum at (a) 0°C (b) the melting point (c) the boiling point (d) 100°C 27. An isolated system comprises the liquid in equilibrium with vapours. At this stage, the molar entropy of the vapour is (a) less than that of liquid (b) more than that of liquid (c) equal to zero (d) equal to that of liquid

28. The efficiency of the reversible heat engine is ηr and that of irreversible heat engine is ηi. Which of the following relation is correct? (a) ηr = ni (b) ηr >ni (c) ηr < ni (d) ηi may be less than, greater than or equal to ηr, depending on the gas. 29. Choose the substance which has higher possible entropy (per mole) at a given temperature. (a) solid carbondioxide (b) nitrogen gas at 1 atm (c) nitrogen gas at 0.01 atm (d) nitrogen gas at 0.00001 atm 30. For a system in equilibrium, ∆G = 0, under conditions of constant (a) temperature and pressure (b) temperature and volume (c) pressure and volume (d) energy and volume 31. One mole of ice is converted into water at 273 K. The entropies of H2O (s) and H2O (l) are 38.20 and 60.01 J/K –mol, respectively. The enthalpy change for the conversion is (a) 59.54 J/mol (b) 5954 J/mol (c) 95.4 J/mol (d) 320.6 J/mol 32. The driving force of a chemical reaction is most closely related to the concept of (a) heat of formation (b) heat of reaction (c) entropy (d) free energy 33. The solubility of NaCl in water at 298 K is about 6 moles per litre. Suppose you add 1 mole of NaCl to a litre of water. For the reaction: NaCl + H2O → salt solution, (a) ΔG > 0, ΔG > 0 (b) ΔG < 0, ΔS > 0 (c) ΔG > 0, ΔS < 0– (d) ΔG < 0, ΔS < 0 34. The enthalpy and entropy change for a chemical reaction are –2.5 × 103 cal and 7.4 cal K–1 respectively. Predict that nature of reaction at 298 K is (a) Spontaneous (b) Reversible (c) Irreversible (d) Non-spontaneous 35. The factor ΔG values is important in metallurgy. T h e ΔG values for the following reactions at 800°C are given as: S2 (s) + 2O2(g)→2SO2(g); ∆G = –544 KJ 2Zn (s) + S2(s)→2ZnS (s); ∆G = –293 KJ 2Zn (s) + O2(g) →2ZnO(s); ∆G = –480 KJ The ΔG for the reaction: 2ZnS (s) + 3O2(g) →2ZnO(s) + 2SO2(g) will be (a) –357 KJ (b) –731 KJ (c) –773 KJ (d) –229 KJ

Thermodynamics 6.53

36. The enthalpy of formation steadily changes from –1789 Kcal/mol to –49.82 Kcal/mol for CH4 to 4.14 Kcal/mol for C8H18. Why? (a) As the number of carbon atoms increases the number of possible isomers increases. This reverses the expected trend ΔG values. (b) The increase in the number of C–C bonds in relation to the number of C–H bonds modifies the trend of ΔG values in relation to ΔH values. (c) In the formation of CnH2n + 2 from n carbon atoms and (n + 1) hydrogen molecules there is a large decrease in entropy. This is reflected in the ∆G values. (d) No simple reason possible. 37. What is the free energy change, ΔG, when 1.0 mole of water at 100°C and 1 atm pressure is converted in to steam at 100°C and 1 atm pressure? (a) 540 cal (b) –9800 cal (c) 800 cal (d) 0 cal 38. Pick out the correct statement from among the following: (a) In the process of boiling of a liquid work is done on the system (b) When a liquid boils at its normal boiling point, Q = ΔH (c) Joule-Thomson effect is an isochoric process (d) Molar heat of fusion is an extensive property 39. During an adiabatic process the pressure of a gas is found to be proportional to the cube of its absolute temperature. The ratio Cp/Cv = γ for the gas it is (a) 3/2 (b) 7/2 (c) 5/3 (d) 9/765

5   40. One mole of an ideal gas  Cv , m = R at 300 K,  2  5 atm is expanded adiabatically to a final pressure of 2 atm against a constant pressure of 2 atm. Final temperature is (a) 270 K (b) 273 K (c) 248.5 K (d) 200 K 41. The entropy change involved in the isothermal reversible expansion of 2 mole of an ideal gas from a volume of 10 dm3 to a volume of 100 dm3 at 27°C is: (a) 32.3 J mol–1 K–1 (b) 42.3 J mol–1 K–1 (c) 38.3 J mol–1 K–1 (d) 35.8 J mol–1 K–1 42. Assuming ∆H and S do not change with temperature, calculate the boiling point of liquid A using the thermodynamic data given below Thermodynamic data A (liq) A (gas) H (KJ/mol) –130 –100 S(J/K/mol) 100 200 (a) 300 K (b) 130 K (c) 150 K (d) 50 K O

O

43. The Haber’s process for the production of ammonia  2NH3(g) involves the equilibrium N2(g) + 3H2(g)  Assuming that (∆H and ∆S for the reaction does not change with temperature, which of the statements is true (∆H = –95 KJ and ∆S = –190 J/K) (a) Ammonia dissociates spontaneously below 500 K (b) Ammonia dissociates spontaneously above 500 K (c) Ammonia dissociates at all temperatures (d) Ammonia does not dissociate at any temperature 44. Which one of the following pairs represents the intensive properties? (a) Specific heat and temperature (b) Entropy and density (c) Enthalpy and mole fraction (d) Heat and temperature O

O

O

O

multiple choice Questions with only one answer Level ii 1. The latent heat of vaporization of water at 20°C is 10.5 Kcal mol–1 and standard heat of formation of liquid water is – 68.3 Kcal. The enthalpy change of the reaction: H2(g) + O2(g) → H2O(g) is, therefore. (a) –57.8 Kcal (b) –78.8 Kcal (c) 78.8 Kcal (d) –47.3 Kcal.  2. For SR ↽ ⇀  SM, ΔG = 76.5 ΔH = 276.5 joule at 300 K. Equilibrium temperature will be (a) 298 K (b) 414.5 K (c) 596 K (d) 207.25 K 3. The work done in an open vessel when 112 g of Fe reacts with dil. HCl to from FeCl 2 at 27°C. Will be (a) 1200 cal (b) 600 cal (c) 300 cal (d) 200 cal 4. One mole of ideal gas is allowed to expand reversibly and adiabatically from a temperature of 27°C .If the work done during the process is 3 KJ. The final temperature will be equal to (CV = 20 KJ– 1) (a) 150 K (b) 100 K (c) 26.85 K (d) 207.25 K 5. One mole of ice is converted into water at 273 K. The entropies of H2O(s) and H2O(l) are 38.20 and 60.01 J mol–1 K–1 respectively. The enthalpy change for the conversion is: (a) 59.54 J mol–1 (b) 5978 J mol–1 –1 (c) 595.4 J mol (d) 320.6 J mol–1

6.54

Thermodynamics

6. 3N2O (g) + 2NH3 (g) → 4N2(g) + 3H2O(g) ΔH = 879.6 KJ What is ΔH f for N2O in KJ/mol–1? Heat of information NH3 –45.9 KJ mol–1 H2O 241.8 KJ mol–1 (a) 246 (b) 82 (c) –82 (d) –246 7. What is the entropy change (in KJ–1 mol–1) when one mole of ice is converted into water at 0°C. The enthalpy for the conversion of ice to liquid water is 6 KJ/ mol at 0°C (a) 20.13 (b) 2.013 (c) 2.198 (d) 21.98 8. A heat engine absorbs heat q1 from a source at temperature T1 and q2 from a source at temperature T2. Work done is found to be J (q1 + q2). This is in accordance with (a) First law of thermodynamics (b) Second law of thermodynamics (c) Joules equivalent law (d) None of the above 9. 100 Joules of heat flow from a body of 50 kg at 27°C into a large Cu-block 100 kg at –23°C. What is the total change in entropy (a) 0.67 J/K (b) 15.7 J/K (c) 0.067 J/K (d) 6.7 J/K 10. Consider the following reactions at 1100°C (I) 2C + O2 →2CO, ΔG = –460 KJ/mol (II) 2Zn + O2→ 2ZnO, ΔG = –360 KJ/mol Based on these, select correct alternate (a) Zinc can be oxidized by CO (b) Zinc oxide can be reduced by carbon (d) Both are correct (d) None is correct O

1 O2(g) → CO2(g) ΔH and 2

11. For the reaction CO(g) +

ΔS are 283 KJ and –87 JK–1 respectively. It was intended to carry out this reaction at 1000, 1500, 3000 and 3500 K. At which of these temperature would this reaction be thermodynamically spontaneous? (a) 1500 and 3500 K (b) 3000 and 3500 K (c) 1000.1500 and 3000 K (d) 1500, 3000 and 3500 K 12. What should be the ΔH f of OH–ion, if the standard enthalpy of formation of liquid water is –68.31 Kcal and ΔH f (H + ) = 0? Given the standard enthalpy of neutralization of strong base is –13 Kcal equivalent–1] (a) –16.54 Kcal (b) –54.61 Kcal (c) 65.41 Kcal (d) 61.54 Kcal O

O

13. The change in entropy of the system when 2 moles of an ideal diatomic gas is heated from 400 K to 800 K under constant pressure is (a) 3 R ln2 (b) 7R ln2 (c) 5R ln2 (d) R ln2 14. A gas expands adiabatically such that its temperature, 1

T∞

15.

V

the value of γ of the gas will be:

(a) 1.30 (b) 1.50 (c) 1.70 (d) 2 The vapour pressure of water at 310 K is 25 torr. If the standard state pressure is defined I bar (750 torr). The ΔG for the process H2O(l) → H2O (g) at 310 K Assuming ΔH and ΔS remains constant with pressure for liquid (Given: log 3 = 0.47) (a) –8.76 K joule (b) + 8.76 K joule (c) –4.38 K joule (d) + 4.38 K joule An ideal gas is allowed to expand under adiabatic reversibly conditions. What is zero for such a process? (a) ΔG = 0 (b) ΔT = 0 (c) ΔS = 0 (d) none of these  B is The value of log10 K for a reaction A  (Given: ΔHr 298 K = –54.07 KJ mol–1, ΔSr 298 K = 10 JK–1 and R = 8.314 JK–1 mol–1, 2.303 × 8.314 × 298 = 5705 (a) 5 (b) 10 (c) 95 (d) 100 One mole of a an ideal monatomic gas expands isothermally against constant external pressure of 1 atm from initial volume of 1 L to a state where its final pressure becomes equal to external pressure. If initial temperature of gas is 300 K then total entropy change of the system in the above process is [R = 0.082 L atm mol–1 K–1 = 8.3 J mol–1 K–1] (a) 0 (b) R ln (24.6) O

16.

17.

O

18.

O

(c) R ln (2490) 3 R ln (24.6) 2 19. When an ideal gas undergoes unrestrained expansion, no cooling occurs because the molecules (a) are above the inversion temperature (b) exert no attractive force on each other (c) do work equal to the loss in kinetic energy (d) collide without loss of energy (d)

Thermodynamics 6.55

20. Consider the modes of transformations of a gas from state A to state B as shown in the following P-V diagram. Which one of the following is true A

25. 2 mole of an ideal gas at 27°C temperature is expanded reversibly from 2 lit to 20 lit. Find entropy change (R = 2cal/mol/ K) (a) 92.1 (b) 0 (c) 4 (d) 9.2  Cu + Zn2 + 26. If for the cell reaction Zn + Cu2 +  –1 Entropy change ΔS is 96.53 mol K–1 then temperature coefficient of the e.m. f. of a cell is: (a) 5 × 10–4 VK–1 (b) 1 × 10–3 VK–1 –3 –1 (c) 2 × 10 VK (d) 9.65 × 10–4 VK–1 27. ∆CP for a reaction is given by 2.0 + 0.2 T cal/°C its enthalpy of reaction at 10 K is –14.2 cal, its enthalpy of reaction at 100K in Kcal will be (a) 13.21 (b) –16.02 (c) 15.3 (d) 7.08 28. Which of the following statement true for ideal gas  ∂H   ∂E  (a)   =R  –  ∂ T   P  ∂T  P O

P C

B V

21.

22.

23.

24.

(a) ΔH = q along A→C (b) ΔS is same along both A→B and A→C→B (c) W is same along both A→B→C and A→C→B (d) W > O along both A→B and A→C Predict in which of the following , entropy increases/ decreases (i) A liquid crystallizes into a solid (ii) Temperature of a crystalline solid is raised from 0 to 115 K (iii) 2NaHCO3(s) →Na2CO3(g) + H2O(g) + CO2(g) (iv) H2(g) →2H(g) (a) Increase in all (b) Decrease in (i) and (ii) (c) Decreases in (i) only (d) Increase in (i), (iv)  2Fe(S) + 3CO2 (1 gm Fe2O3(s) + 3CO (1 g atom)  atom), ΔG25°C = –10 KJ The ΔG for the reaction  Fe2O3(S) + 3CO(20g 2Fe(S) + 3CO2 (2g atom)  atom) at the temperature 25°C is equal to (a) + 17.11 KJ (b) + 7.11 KJ (c) + 27.11 KJ (d) + 5.70 KJ S1: thermodynamic equilibrium in an isolated system is reached when the system’s entropy is maximized S2: Heat in internal energy of system is equal to zero in a cyclic process S3: Heat given at constant pressure is equal to change in enthalpy of closed system when only PV work is considered S4: In a closed system, capable doing only PV work, the constant temperature and pressure equilibrium condition is the minimization of the Gibbs function (a) TTTT (b) FTTF (c) FTFT (d) TTFT The standard heat of combustion of Al is –837.8 KJ mol–1 at 25°C. If Al reacts with O2 at 25°C, Which of the following releases 1250 KJ of heat? (a) The reaction of 0.624 mol of Al (b) The formation of 0.6224 mol of Al2O3 (c) The reaction of 0.312 mol of Al (d) The formation of 1.50 mol of Al2O3

 ∂H   ∂E  (b)    >  ∂T  P  ∂T  P  ∂E  (c)   =0  ∂V T (d) All of these 29. CO (g) + NO2(g) →CO2(g) + NO(g) The standard enthalpies of formation of gaseous CO, NO, CO2 and NO2 are – 110.5, 90.2 –393.5 and 33.2 KJ mol–1 respectively. Energy of activation for the forward reaction is 465.2 KJ. Calculate energy of activation for backward reaction (a) 239.2 KJ (b) 691.2 KJ (c) 226 KJ (d) 303 KJ 30. What is ∆r G( KJ/ mole) for synthesis of ammonia at 298 K at following sets of partial pressure: N2 (g) + 3H2(g)⇌2NH3 (g); ∆rG = –33 KJ/ mole [Take R = 8.3J/K mole, log 2 = 0.3; log 3 = 0.48] Gas N2 H2 NH3 Pressure (atm) 1 3 0.02 (a) + 6.5 (b) –6.5 (c) + 60.5 (d) –60.5 31. A sample of an ideal gas with initial pressure ’P’ and volume ‘V’ is taken through an isothermal process during which entropy change is found to be ∆S. The work done by the gas is PV DS (a) nR O

(b) nR∆S (c) PV PDS (d) nRV

6.56

Thermodynamics

32. If DG298 for the reaction: 2H2(g)(1 atm) + O2(g) (1 atm)  → 2H2O (g) (1atm) is –240 KJ, what is DG298 for the reaction 1 → H2(g, 4 atm) + O2(g,0.25 atm) H2O(g, 0.2 atm)  2 (a) 245.7 KJ (b) 239.3 KJ (c) 125.70 KJ (d) –245.7 KJ 33. Molar heat capacity of CD O at constant pressure is 10/ cal/ K/ mol at 100 K. The entropy change when 3.2 g of CD O vapour are heated from 1000 K to 2000 K at constant pressure is (a) 0.15 cal/K (b) 0.32 cal/ K (c) 0.46 cal/K (d) 0.69 cal/ K 2

2

3   34. 0.5 mole each of two ideal gases A  CV = R  and B 2   5    CV = R  are taken in a container and expanded 2   reversibly and adiabatically from V = 1 litre to V = 4 litre starting from temperature T = 300 K. ∆H for the process (in cal/mol) is (a) –500 cal (b) –900 cal (c) –450 cal (d) –600 cal 35. The efficiency of a fuel cell is given by (a)

DG DS

(b)

DG DH

(c)

DS DG

(d)

DH DG

36. Which of the following reaction is spontaneous? (a) Endothermic reaction with positive entropy change and high temperature (b) Endothermic reaction with negative entropy change and low temperature (c) Exothermic reaction with positive entropy change and high temperature (d) Exothermic reaction with negative entropy change and high temperature 37. P-V plot for two gases during adiabatic processes are given. Plot A and B correspond respectively To :

P

B

A

V

(a) He and O2 (c) He and Ar

(b) O2 and He (d) O2 and F2

38. PbO2 →PbO, ∆G298< 0 SnO2→SnO, ∆G298>0 Most probable oxidation state of Pb and Sn will be (a) Pb4 + , Sn4 + (b) Pb4 + , Sn2 + 2+ 2+ (c) Pb , Sn (d) Pb2 + , Sn4 + 39. One mole of a mono- atomic gas behaving as per PV ≅ nRT at 27°C is subjected to reversible isoentropic compression until final temperature reached 327°C. If the initial pressure was 1.0 atm then the value of InP(final) (In2 = 0.7) is (a) 1.75 atm (b) 0.176 atm (c) 0.0395 atm (d) 2.0 atm 40. The decomposition of limestone CaCO3 (s) → CaO (s) + CO2 (g) is non- spontaneous at 298 K. The ∆H and ∆S values for the reaction are 176.0 KJ and 160J K–1 respectively. At what temperature the decomposition becomes spontaneous? (a) At 1000 K (b) Above 827°C (c) At 500°C (d) Between 500°C and 600°C 41. Assuming that water vapour is an ideal gas, the internal energy change (ΔU) when 1 mol of water is vapourised at 1 bar pressure and 100°C, (given : molar enthalpy of Vapourization of water at 1 bar and 373 K = 41 KJ mol–1 K–1 ) will be (a) 41.00 KJ mol–1 (b) 4.100 KJ mol–1 (c) 3.1904 KJmol–1 (d) 37.9904 KJ mol–1 42. ΔH and ΔS for a reaction are + 30.558 KJ Mol–1 and 0.0666 KJ mol–1 at 1 atm pressure. The temperature at which free energy is equal to zero and the nature the reaction below this temperature are (a) 483 K, spontaneous (b) 443 K, non- spontaneous (c) 443 K, spontaneous (d) 463 K, non- spontaneous 43. Only gases remains after 12 g of carbon is treated with 49.26 lit of air 300 K and 2 atm pressure (assume 20% by volume O2N2 and 1% CO2) ΔHf for CO and CO2 are –26 Kcal, –94 Kcal The amount of heat evolved under constant pressure is (a) –96.8 Kcal (b) –66.8 Kcal (c) –23.4 Kcal (d) –33.4 Kcal 44. 1 mole of an ideal gas at 27°C is subjected to expand reversibly ten times of its initial volume. The entropy change is (a) 19.1 J/K mol (b) 29.1 J/K mol (c) 39.1 J/K mol (d) 219.1 J/K mol 45. 0.05 mole of argon at 27°C expand reversibly and adiabatically from 1.25 lit to 2.50 lit Cv for argon is 11.686 J/K mol. The enthalpy change in process is (a) 55 J (b) –55 J (c) 110 J (d) –110 J O

O

Thermodynamics 6.57

multiple choice Questions with one or more Than one answer 1. Which of the following relation(s) represent(s) adiabatic process? (a) log

V T1 = (1–γ) 2 V1 T2

(b) log

T1 V = (1–γ) log 2 T2 V1

(c) log

T1 V  = (1–γ)log  2  T2  V1 

(d) log

T2 V = (1–γ) log 2 T1 V1

2. Identify the intensive quantities from the following. (a) Refractive index (b) Temperature (c) Volume (d) Enthalpy 3. Any process will be spontaneous at constant pressure and temperature when (a) ΔSSystem = + ve (b) ΔSuniv. = + ve (c) ΔGSys = –ve (d) ΔGSys = + ve 4. 10 mol of an ideal gas (CP/CV = 1.5) undergo a change of state described by the following P-V diagram Which of the following statement is (are) true

A (600K,10atm)

P

iso

th

er

ia b ad

m

t

a

46. One mole of ideal gas is allowed to expanded reversibly and adiabatically from a temperature of 27°C If the work done by the gas in the process is 3 KJ, the final temperature will be equal to (Cv = 20 J/ K mol) (a) 100 K (b) 450 K (c) 150 K (d) 400 K 47. For the reaction at 300 K A (g) + B(g) →C(g) ΔE = –3 Kcal; ΔS = –10.0 cal/ K. Value of ΔG is (a) 6000 cal (b) –6600 cal (c) –6000 cal (d) None 48. The molar volume of ice and water were 0.0180 litres respectively. When ice (1 mole) melts at 0°C and 1 atm,1.44 Kcal of heat in absorbed value of ΔE is………cal (a) 1440.039 (b) 1448.86 (c) 844.23 (d) 14.0196 49. The molar heat capacities at constant pressure (assumed constant with respect to temperature ) of A, B and C are in the ratio 1.5:3.0: 2.5. If enthalpy change for the exothermic reaction A + 2B →3C at 300 K and 310 K are ΔH1 and ΔH2 respectively then (a) ΔH1>ΔH2 (b) ΔH1< ΔH2 (c) ΔH1 = ΔH2 (d) If T2 > T1 then ΔH> ΔH1&if T2< T1 then ΔH2 < ΔH1 50. An ideal gas is allowed to expand both reversibly and irreversibly in an isolated system. If T1 is initial temperature and Tf is the final temperature, which of the following statement is correct (a) (Tf)irr > (Tf)rev (b) Tf > T1 for reversible process but Tf = T1 for irreversible process (c) (Tf)rev = (Tf)irr (d) (Tf) = T1 for reversible process 51. In which of the following conditions a chemical reaction cannot occur (a) dH and dS increases and TdS> dS (b) dH and dS decreases and dH> TdS (c) dH increases and dS increases (d) dH decreases and dS increase 52. The efficiency of that engine is maximum when (a) Temperature of source > temperature of sink (b) Temperature of sink > temperature of source (c) Temperature difference of source and sink is minimum (d) Temperature difference of source and sink is maximum 53. 5 mole H2O(l) at 373 K and 1 atm is converted into H2O (g) at 373 K and 5 atm. ΔG for this process is [Given: R = 2 Cal /K –mol] (a) Zero (b) 1865 ln 5 cal (c) 3730 ln 5 cal (d) 3730 ln 5 cal

B

C (300K) V

(a) ΔSAB = ΔSCB (b) WABC < 0 (c) P(C) = 1.25 atm (d) P(B) = 5 atm 5. Which of the following statement is (are) true? (a) Work done in reversible isothermal expansion is always more than work done in irreversible isothermal expansion in magnitude (b) ∆G = 0 at equilibrium (c) ∆G and ∆STotal are always zero at equilibrium (d) The net enthalpy of products is less than net enthalpy of reactants in an exothermic reaction O

6.58

Thermodynamics

6. Which of the following is (are) intensive thermodynamic properties. (a) Volume (b) Kinetic energy (c) Specific conductance (d) Standard emf 7. Which of the following is (are) not true regarding internal energy of a system? (a) It essentially changes during change of state (b) Its precise and accurate measuring is impractical (c)  ∫ dE = 0 where E stands for internal energy (d) It will have different values at the same final state if reached through different paths but form the same initial state 8. Which of the following regarding the said process is (are) correct? (a) Expansion of an ideal gas against vacuum is always irreversible (b) A spontaneous process is always irreversible (c) In a reversible thermodynamic process, system always remains in equilibrium with surroundings (d) If a system containing ideal gas in a piston undergoes isothermal expansion from a given initial state to the same final volume, the surroundings loses more heat if expansion carried out irreversibly rather reversibly 9. Two moles of a monatomic ideal gas (CV = 1.5 R) initially at 400 K in an isolated, 1.0 L, piston is allowed to expand against a constant pressure of a 1.0 atm till the final volume reaches to 10 L which of the following conclusion regarding the above change (s) is (are) true? (a) The final temperature of the gas is 363.45 K (b) If the same process were carried out to the same final volume but under reversible final temperature would have been less than 363.45 K (c) In the above, process, the initial and final temperatures and volumes are related

11. Which of the following statement (s) regarding entropy change, when a system undergo charge of state is (are) correct? (a) In reversible change, ∆Suniv = 0 (b) In an irreversible change ∆Suniv>0 (c) In an adiabatic process, ∆SSyst = 0 T  (d) In an isochoric process, ∆Ssyst = nCV ln  2   T1  12. Which of the following thermodynamic relation (s) is/ are correct? (a) In a cyclic process,  ∫ dS = 0 (b) ∆G = –T∆Suniv  dG  (c)   = –S  dP  P O

13.

14.

15.

16.

g −1

 T1   V2   =   T2   V1  (d) Entropy change of a system (∆SSyst) is zero 10. Which of the following is (are) true regarding the molar heat capacities of ideal gases? (a) It is same for ideal gases independent of temperature (b) It increases with the atomicity of molecule (c) It remains constant in lower temperature range but decreases with rise in temperature range if the gas is polyatomic (d) In case of polyatomic gases, it requires more heat for same temperature rise when the temperature is very high compared to one when the temperature is low

17.

18.

(d) ∆G = ∆G + RT ln Q, Q = Reaction quotient In which of the following processes, entropy of the system is increasing? (a) NH3(g) + HCl(g) →NH4Cl(s) (b) N2(g) + 3H2(g)→2NH3(g) (c) COCl2(g) →CO(g) + Cl2(g) (d) H2O(V) + C(s) →H2(g) + CO(g) According to 2nd law of thermodynamics, the (a) Energy of universe is constant (b) Entropy of universe continuously increasing (c) Mass and energy are intercornvertibles (d) All spontaneous processes are thermodynamically irreversible Which of the following thermodynamic relation (s) is/ are correct (a) G = H–TS (b) ∆E = q + W (c) ∆H = ∆E + ∆ngRT (d) H = E + PV For which of the following processes ∆E = 0? (a) A cyclic process (b) An isothermal process (c) An isochoric process (d) Adiabatic process Which of the following is (are) state functions (s)? (a) Enthalpy (b) Heat (c) Entropy (d) Gibbs free energy One mole of ideal gas in an adiabatic, piston fitted cylinder, is allowed to undergo expansion against a constant 1.0 atm which of the following statements must be true for this process? (a) Gas must be present initially at more than 1 atm (b) Expansion occurs spontaneous (c) Enthalpy change of gas is zero (d) Entropy change for the universe is zero

Thermodynamics 6.59

19. If an ideal gas in a piston fitted cylinder is allowed to expand isothermally against vacuum, then (a) Expansion occur adiabatically (b) ∆SSys, ∆SSurr, ∆S univ are all greater than zero (c) ∆GW1>W4> W3 (c) It is given that, in a process for an ideal gas dW = 0 and dq = –ve. Then the process is accompanied by derease in internal energy (d) For a system in equilibrium ∆G = – RT ln KP O

25. Identify the correct statement (s) (a) Relation of ∆G ang ∆G in a reaction system is ∆G = ∆G – RT ln Q where, Q is the reaction quotient O

O

P  (b) For an ideal gas ∆G = nRT ln  2  for n moles at  P1  constant temperature

 dDG  (c)   =V  dp  T (d) The entropy change during isothermal reversible 1

V n expansion of an ideal gas is R ln  2  where, n  V1  is the no. moles of the gas 26. For an ideal gas undergoing isothermal irreversible expansion (a) ∆U = 0 (b) ∆H = 0 (c) ∆S = 0 (d) W = 0 27. Which of the following are correct at 298 K (a) ∆G f element = 0 (b) ∆H f element = 0 (c) ∆S f element = 0 (d) ∆G f compound = 0 28. The correct expression (s) for an adiabatic process is/are O O

O

O

(a)

V  T2 =  1 T1  V2 

γ -1

γ -1

29.

30.

31.

32.

T  γ P (b) 2 =  1  P1  T2  g 2 (c) P1V1 = P2V2 g–1 (d) P1V1 During an adiabatic reversible expansion of an ideal gas (a) Internal energy of the system decreases (b) Temperature of the system decreases (c) The value of g changes (d) Pressure increases Which is an irreversible process? (a) Mixing of two gases (b) Evaporation of water at 373 K and 1 atm in a closed system (c) Dissolution of NaCl in water (d) H2O (s) at –4°C During the Joule Thomson effect (a) A gas is allowed to expand adiabatically from high pressure region to a low pressure region (b) A gas is allowed to expand adiabatically at constant pressure (c) ∆H = 0 for ideal gas (d) ∆U = 0 for ideal gas Which of the following are thermodynamically stable? (a) C(diamond) (b) C(graphite) (c) P4(white) (d) P4(black)

6.60

Thermodynamics

33. When a solid melts there will be (a) An increase in enthalpy (b) A decrease in free energy (c) No change in enthalpy (d) A decrease in internal energy 34. ∆S is positive for the change (a) mixing of two gases (b) boiling of liquid (c) melting of solid (d) freezing of water 35. During an adiabatic expansion of an ideal gas (a) Internal energy of the system decreases (b) Temperature of the system decreases (c) The value of γ changes (d) pressure increases 36. Which of the following relationships are correct δ H  δ U  (a)   = + ve  −  δ T p  δ T v  δV  (b)   = 0 for ideal gas  δT  P R  dV  (c)   = for one mole of ideal gas  dT  P P δ H  = 0 for an ideal gas (d)   δ P  P

37. Selecte the correct statements: (a) The standard heat of formation of Cl2(l) at 298 K is taken as zero (b) Heat of reaction involving ideal gas increases with increases of pressure (c) ∆H°f of O2(g) is zero (d) ∆H°f of HCl(g) is different from ∆H°f of HCl(aq) 38. The enthalpy change for a reaction depends upon the (a) Physical states of reactants and products (b) Use of different reactants for the same product (c) Nature of intermediate steps (d) Difference in initial and final temperature of product 39. Which of the following quantities is/are state functions (a) Temperature (b) Entropy (c) Enthalpy (d) Work 40. Which of the following statements is/are correct about internal energy (a) The absolute value of internal energy cannot be determined (b) For an adiabatic process ∆E ≠0 (c) The measurement of heat change during a reaction by bomb calorimeter is equal to change in internal energy (d) Internal energy is an extensive property

41. Choose the correct statements: (a) Temperature, enthalpy and entropy are state functions (b) For reversible and irreversible both isothermal expansion of an ideal gas, change in internal energy and enthalpy is zero (c) For reaction in which ∆ng = 0 entropy change is not always zero (d) The entropy change associated with reversible isothermal expansion of an ideal gas is equal to P2 P2

2.303n R log10

comprehensive Type Questions Passage i One mole of an ideal gas undergoes reversible isothermal expansion from an initial volume V1 to an final volume 10V1 and does 10 KJ of work. The initial pressure is 1 × 107 Pascal’s 1. The temperature of the gas is (a) 422 K (b) 472 K (c) 522 K (d) 572 K 2. The volume V1 is equal to (a) 3.34 × 10–4m3 (b) 3.94 × 10–4m3 (c) 4.34 × 10–4m3 (d) 4.94 × 10–4m3 Passage ii 1 H2(g) + O2(g) → H2O(l) = –242 KJ mol–1 2 A reaction like the above is called combustion reaction . combustion reactions are exothermic. The energy produced in the combustion of fuels like hydrogen, methane and carbon monoxide is directly converted into electricity in the fuel cells. Fuel cells are not only more efficient, but also better contributors for pollution prevention. Such a fuel cell was used by Neil Armstrong on the moon, through Saturn VI rocket. 1. The calorific value of hydrogen in the conventional units per gram of fuel is (a) 484000 J (b) 242000 J (c) 121000 J (d) 60500 J 2. Equation representing the cathodic reaction of fuel cell is 4 e− (a) 2H2O(l) + O2(g) → 4OH − (aq) −2e− → 2H2O(aq) (b) H2(g) + 2OH– (aq)  −

4e (c) 4OH–(aq) − → O2(g) + 2H2O(l) −

2e (d) CO(g) + 2OH–(aq) − → CO2(g) + H2O(l)

Thermodynamics 6.61

3. All except one of the following can be attributed as reasons for an unfrequented utilisation of hydrogen in the fuel cell. The ‘odd’ one is (a) transportation of hydrogen gas is dangerous (b) efficiency of hydrogen fuel cell is a maximum of 70% only (c) hydrogen is almost not available in the free state (d) being the lightest gas, risks may be involved in storing hydrogen Passage iii Water boiled under a pressure of 1 atm. When an electric current of 0.5 amp from 12 V supply is passed for 300 sec through a resistance in thermal contact with it, it is found that 0.9 g of water is vaporized. 1. The molar enthalpy change is (a) 72 KJ/mol (b) 46 KJ/mol (c) 54 KJ/mol (d) 36 KJ/mol 2. The molar internal energy change is (a) 68.9 KJ/mol (b) 42.9 KJ/mol (c) 50.9 KJ/mol (d) 32.9 KJ/mol 3. C (graphite) +

1 O2(g)→CO2(g) ∆H = –26.4 K Cal 2

at 300 K, 1 atm. The internal energy change is (a) –26.7 Kcal (b) –26.1 Kcal (c) –30.0 Kcal (d) –25.8 Kcal Passage iV Two moles of an ideal gas at 1 atm and 27°C undergoes the following process (i) heat is absorbed at constant volume till the pressure is doubled (ii) Isothermal and reversible till the pressure is 0.8 atm (iii) Adiabatic compression till the initial stage is reached (Cv = 40 J mole–1 K–1) 1. The internal energy change (KJ) in the 1st step of the cycle and that in the whole cycle are respectively (a) 80.0 and zero (b) 600 and 25 (c) 24.0 and zero (d) None of these 2. The volume change in the last step is nearly (a) 74 dm3 (b) 4 dm3 3 (c) 123 dm (d) 50 dm3 3. The change in internal energy (KJ) in the 2nd step and that the 3rd i.e., final step are respectively (a) 9.2 and zero (b) zero and –24.0 (c) zero and –9.2 (d) 5.0 and –25.0

Passage V The extraction of metals from their oxides using carbon at high temperature involves number of important points. At a given pressure, a reaction is spontaneous if the Gibbs energy change is negative. (∆G = ∆H – T∆S). Consider MO(s)→M +

1 O2(g). The entropy change for this reaction 2

is positive. On increasing the temperature this generally causes ∆G to become more negative. The variation of ∆G for C(s) + O2(g)→CO2(g) with increasing temperature is insignificant while for C(s) + 1 O2(g)→CO(g), ∆G again falls with increase in 2 temperature. G is a state function. ∆G for formation for FeO(s), CaO(s) and CrO(s) are –500 KJ/mole–1300 KJ/mole and –700 KJ mole at 25°C. while for oxidation of C(gr) it remains practically –400 KJ/ mole and starts becoming more and more negative after 700 K. 1. Which of the following is theoretically reducible by C(s) at room temperature and 1 bar pressure (a) FeO (b) CaO (c) CrO (d) None of these 2. As temperature is increased (a) FeO(s), CaO(s) and CrO(s) become better oxidizing agents (b) FeO(s), CaO(s) become better oxidishing agents while CrO(s) becomes worse oxidizing agent. (c) FeO(s) becomes better reducing agent while CaO(s) and CrO(s) become worse reducing agents (d) All FeO(s), CaO(s) and CrO(s) become worse oxidizing agents 3. As temperature is increased (a) Carbon (s) maintains its reducing properties till 700 K and then the reducing properties fall (b) Carbon (s) maintains its oxidizing nature till 700 K and then oxidizing power increases. (c) Carbon (s) maintains its reducing tendency till 700 K and then reducing tendency increases (d) none of these O

Passage Vi A person inhales 640 g of O2 per day. If all O2 is used for converting sugar into CO2 and H2O. The heat of cumbstion of sugar is –5600 KJ mol–1 1. How much sucrose is consumed in the body in one day (a) 1140 g (b) 712.5 g (c) 855 g (d) 570 g

6.62

Thermodynamics

2. The amount of heat evolved in above problem is (a) 18666 KJ (b) 11666 KJ (c) 14000 KJ (d) 9333 KJ

5. The total internal energy change in this process is (a) 0 (b) 6.13 lit – atm (c) 6.13 lit – atm (d) –33.87 lit –atm

Passage Vii

Passage ix

A sample of sucrose weighing 0.1265 g is burnt in a calorimeter.After the reaction it was found that 2.082 KJ heat is evolved. The temperature increased in this experiment is 1.743°C ∆Hf of CO2 is –393.5 KJ/mol, ∆Hf of H2O(l) is –285.3 KJ/mol. 1. The heat of combustion of sucrose is (a) –6529 KJ (b) 6529 KJ (c) 5629 KJ (d) –5629 KJ 2. The heat of formation of sucrose (a) 3622 KJ (b) –3622 KJ (c) –2236 KJ (d) 2236 KJ 3. The heat capacity of the calorimeter and contents is (a) 1194 J/K (b) –1194 J/K (c) –4194 J/K (d) 4194 J/K

For an ideal gas, an illustration of three different paths, A, (B + C) and (D + E) from an initial state P1, V1, T1 to a final state P2, , V2, T1 is shown in the given figure,

Passage Viii Two moles of a mono atomic gas undergoes the following changes 1atm

A

B

0.5atm C 20 lit

40 lit

AB is isobaric BC is isochoric process CA is isothermal compression. Answer the following questions based on the diagram 1. What will be the total work involved in this process (a) –6.13 lit – atm (b) –6.13 lit – atm (c) 33.87 lit – atm (d) –33.87 lit – atm 2. Total entropy, enthalpy change for the overall process is (a) + Ve, + Ve (b) + Ve, 0 (c) 0, + Ve (d) 0,0 3. Entropy change in isothermal compression process is (a) –11.52 J (b) 11.52 J (c) 23.04 J (d) –23.04 J 4. The initial temperature of the gas is (a) 24.3 K (b) 243 K (c) 12.1 K (d) 121 K

P1 ,V1 , T1

D

P1 , V2 , T3

A

E

P

P2 ,V2 , T1

B

C P3 , V2 , T2

Path (A) reprsents a reversible isothermal expansion from P1, V1 to P2,V2.Path (B + C) represents a reversible adiabatic expansion (B + C) from P1, V1 to P3, V2, T2 followed by reversible heating the gas at constant volume (C) from P3, V2, T2 to P2, V2, T1. Path (D + E) repsents a reversible expansion at constant pressure P1 (D) from P1, V1, T1 to P1, V2, T3 followed by reversible cooling at constant volume V2 (E) from P1, V2, T3 to P2, V2, T1 1. What is qrev for path (A)? V (a) zero (b) –nR ln 2 V1 (c) nR1 ln

V1 V2

(d) nRT1 ln

V2 V1

2. What is ΔS for path (D + E)? T2

(a) zero

(b)

∫

T1

(c) –nR ln

V2 V1

Cv (T ) , dT T

(d) nR ln

V2 V1

Passage x A lot of thermodynamic properties are associated with a rubber band based on its structure. If you quickly stretch the rubber band and then press it against your lips, you will feel a slight warning effect. Now reverse the process. Stretch a rubber band and hold it in position for a few seconds. The quickly release the tension and press the rubber band against your lips. This time you will feel a slight

Thermodynamics 6.63

cooling effect. A thermodynamic analysis of these two experiments, given us a good insight of molecular structure of rubber. 1. Which form of rubber band has higher degree of disorder? (a) Stretched form (b) Normal form (c) Both forms has equal extent of disorder (d) It depends upon material of which the rubber is made up of 2. Let the rubber band be stretched for some time and then strain be released to return the band in to previous normal form. For the entire process: (a) ΔSrubber band S(liquid)>S(solid) (c) S(vapour) ΔUadiabatic

Thermodynamics 6.69

15. For an ideal gas, consider only P-V work in going from an initial state X to the final state Z. The final state Z can be reached by either of the two paths shown in the figure. Which of the following choice(s) is/are correct? [take ∆S as change in entropy and W as work done] [2012]

integer Type Questions 18. One mole of an ideal gas is taken from a to b along two paths denoted by the solid and the dashed lines as shown in the graph below. If the work done along the solid line path is WS and that the dotted line path is Wd, then the integer closest to the ratio Wd/Ws is:_____ (2010) 4.5

P (atmosphere)

4.0

X

Y

3.5 3.0 2.5 2.0

Z

1.5 1.0

V (liter)

0.5

(a) ΔSx→z = ΔSx→y + ΔSy→z (b) Wx→z = Wx→y + Wy→z (c) Wx→y→z = Wx→y (d) ΔSx→y→z = ΔSx→y 16. STATEMENT 1: There is a natural asymmetry between converting work to heat and converting heat to work STATEMENT 2: No process is possible in which the sole result is the absorption of heat from a reservoir and its complete conversion into work (2008) (a) Statement – 1 is True, statement – 2 is True; statement 0 2 is correct explanation for statement – 1 (b) Statement – 1 is True statement –2 is True; statement – 2 is NOT a correct explanation for statement – 1. (c) Statement – 1 is True, statement – 2 is False (d) Statement – 1 is False, statement –2 is True 17. Match the transformations in column I with appropriate options in column II Column I (a) CO2 ( s ) → CO2 ( g )

Column II (p) Phase

(c) 2 H → H 2 ( g )

transition (q) allotropic change (r) ∆H is positive

(d) P( white , solid ) → P( red , solid )

(s) ∆S is positive

(b) CaCO3 ( s ) → CaO ( s ) + CO2 ( g )

(2011)

0.0

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0

4.5

5.0

5.5

6.0

19. In a constant volume calorimeter, 3.5 g of a gas with mol.wt 28 was burnt in excess oxygen 298.0 K. The temperature of the calorimeter was found to increase from 298.0 K to 298.45 K due to the combustion process. Given that the numerical value for the enthalpy of combustion gas in KJ mol–1 is: (2009)

6.70

Thermodynamics

anSWer keyS multiple choice Questions with only one answer Level i 1. c 2. b 3. c 4. d 5. c 6. c 7. d 8. a 9. b

10. 11. 12. 13. 14. 15. 16. 17. 18.

a d c b c b a c a

19. 20. 21. 22. 23. 24. 25. 26. 27.

c d b b a a d c b

28. 29. 30. 31. 32. 33. 34. 35. 36.

b d a b d b a b c

37. 38. 39. 40. 41. 42. 43. 44.

d b a c c a b a

comprehensive Type Questions Passage i 1. c

2. c

Passage ii 1. c

2. a

3. b

2. d

3. a

2. a

3. b

2. a

3. c

Passage iii 1. d Passage iV

multiple choice Questions with only one answer Level ii 1. a 2. b 3. a 4. a 5. b 6. b 7. d 8. c 9. c 10. b 11. c

Passage V 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22.

b b b b c b b b b c c

23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33.

d d d a a d b c a c d

34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44.

b b c b d a b d d b a

45. 46. 47. 48. 49. 50. 51. 52. 53.

d b b a c a b d c

a,b,d a,b b,c a,b,c a,c,d c,d a,d a,b,c a,b b,d a,b,d a,b,d c,d b,d

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

a,b,c,d a,b a,c,d a,b a,c,d b,c a,b,d a,b,c a,b,c a,c,d b,c a,b a,b a,c

29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41.

1. d Passage Vi 1. d

2. d

Passage Vii 1. d

2. c

3. a

Passage Viii 1. a

multiple choice Questions with one or more Than one answer 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14.

1. c

a,b a,c,d a,c,d b,c a,b a,b,c a,b a,c,d c,d a,b,d a,b,c a,b,c,d a,b,c,d

2. d

3. a

4. d

5. a

Passage ix 1. d

2. d

Passage x 1. b

2. a

3. c

2. d

3. c

4. c

5. c 6. d

7. a 8. c

Passage xi 1. a

Passage xii 1. c 2. a

3. a 4. b

Passage xiii 1. b

2. b

3. a

9. a 10. a

Thermodynamics 6.71

matching Type Questions 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11.

(a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

rs ps q qr s q q t pqrs qs qr

(b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b)

ps qs qs qr prs s s r t qs pq

(c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

integer Type Questions rs qr pq p q pq pr ps pqt rt rt

(d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d)

q ps qr s r qt pr q rs pqrt rs

assertion (a) and reason (r) Type Questions 1. c 2. d 3. a

4. d 5. b 6. c

7. a 8. c 9. d

10. a 11. d 12. a

13. a

1. 2 2. 8

3. 5 4. 1

5. 0 6. 7

7. 3 8. 1

9. 4 10. 0

Previous years’ iiT Questions 1. 2. 3. 4. 17. 18.

c 5. a 6. c 7. b 8. (a) prs 2 19.

c c b a 9

9. b 10. a 11. b 12. a,b (b) rs (c) t

13. 14. 15. 16. (d)

a,d a,d a,c b pqt

6.72

Thermodynamics

hinTS and SoLuTionS hints to Problems for Practice 1. (a) W = –P (V2 –V1) Now initial volume (V1) = nRT 2.5 × 0.0821 × 300 = P 2

= 30.75 L Final volume (V2) = 30.75 × 2.5 = 76.87 L Now W = –1 (76.87 – 30.75) = –46.37 L atm = –46.37 × 101.325 J = 4698.44 J (b) For isothermal reversible expansion of ideal gas W = –2.303 nRT log

V2 V1

 76.87  = –2.303 × 2.5 × 8.314 × 300 log   30.75  = –5714. 15 J 2. We know that V W = –nRT ln 2 V1 = –2.303 nRT log10

V2 V1

50 = – (2.303 × 2 × 8.314 × 298) log 15 = – 5923 J 3. Let the heat supplied to the liquid for conversion of one mole of the liquid into vapours be Q Calculation of Q Q = Heat of vaporizations per g × mol mass of liquid = (395 J g–1) (78 g mol–1) = 30810 J mol–1 Calculation of W W = P∆V = P (Vv – Vl) = PVv (Vv ≥ Vl) = RT = 8.314 J mol–1 × 353.2 = 2936.5 J mol–1 Calculation of ∆U = Q–W = 30810 – 2936.5 = 27873.5 J mol–1. 4. Here ∆V = 200 – 400 = –200 cm3 = –0.2 litre External Pressure P = 0.50 atm = 0.50 × –0.2 = 0.1 litre – atm 1 litre atm = 101.3 J W = 0.1 × 101.3 = 10.13 J Q = –8.0 J U =Q+W = –8.0 + 10.13 = 2.13 J

5. Q = 600 J W = –450 J U2 = ? U2 = U 1 + Q + W = U1 + 600 + (–450) = U1 + 150 J 6. P = 1 atm = 1.013 × 106 dynes/cm2 V1 = Volume of 1 mole of water at 100°C = 18 mL V2 = Volume of 1 mole of steam at 100°C =

373 × 22400 = 30606ml 5 mL (Charles law) 273

Now, we have W = P (V2 –V1) = 1.013 × 106 × (30605 – 18) ergs =

1.013 × 106 × 30587 Calories 4.18 × 107

= 741 Calories Again We have ΔU = Q–W = 9720–741 = 8979 calories 7. (i) Expansion of a gas in vacuum corresponds to free expansion hence the work done W=O (ii) For reversible expansion P  W = 2.303 nRT log  1   P2  10 = 2.303 × 10 × 2 × log 2 = 9594.4 calories (iii) W = P(V2V1) From general gas equation PV = nRT 10 × RT 10 × RT V1 = ;V2 = 10 2

10 × RT 10 × RT  W = 2  −  10   2 = 2 × 4RT = 2 × 4 × 2 × 298 = 4768 calories Q = W = 831.8 calories 8. W = –2.303 nRT log = –2.303 ×

V2 V1

64 5 × 2 × 300 × log 32 10

= 831.8 calories

Thermodynamics 6.73

9. In this case work is against constant pressure P and thus irreversible W = –PΔV = –3 × 2(ΔV = 5–3 = 2 dm3 2litre) = –6 litre atm. = –6 × 101.3 J = –607.85 Now thus work is used up in heating water W = n × C × ΔT 607.8 = 10 × 4.184 × 18 × ΔT [∵C = 4.184 jg–1 ∴ = 4.184 × 18 J mole–1] ΔT = 0.81 ∴ Final temperature of water = T1 + ΔT = 290 + 0.81 = 290.81 K 10. since work is done against constant pressure P and this is irreversible W = –P × ΔV = –1.5 × 2(∵ΔV = 6–4 = 2 litre) = –3 litre atm = –3 × 101.3 J = –303.9 J Now this work is used up in heating 1 mole of water W = n × C × ΔT 303.9 = 1 × 4.184 × 18 × ΔT ΔT = 4.035 ∴Final temperature of water = T1 + ΔT = 291 + 4.035 = 295.035 K 11. From the relationship CP–CV = R 7 5 R +R= R 2 2 7 Cp 2 R = = 1.4 g= Cv 5 R 2 For an adiabatic expansion of an ideal gas

CP =

g −1

 P2   P 

g

P  =  2  P1  1 Given P1> 10 atm ; P2 = 1 atm; T1 = 200 K, r = 1.4 1.4 −1

1    10 

1.4

T  = 2   300 

0.4/1.4

 1 T2 = 300 ×    10  = 155.3 K W = nCV(T1–T2) = 5 × (5/2) × 1.987 × (300–155.3) = 5 × (5/2) × 1.987 × 144.7 3593.5 calories = 15.033 K

12. Heat evolved by 60 watt bulb in 20 minutes = 60 × 60 × 20 = 7200 J 1J 1.watt = sec Volume of air in room = 5 × 4 × 3 = 60 m3 = 60 × 106 cm3 Mass of air in room = 1.22 × 10–6 × 60 × 606 kg = 73.02 kg ∴ Heat given by bulb = Heat taken by (roof + walls) room + heat taken by air 7200 J = 50 × 103(∆T) × 0.71 × 73.2 × 103 × ∆T ∆T = 0.07 K 13. DS = 2.303nR log

V2 V1

80 DS = 2.303 × 5 × 8.314 × log 8 = 95.736 × 1 –1 = 95.736 JK 14. n1 = 1 mole , n2 = 2 moles

Mole fraction of O2 χ1 =

n1 2 = = 0.33 n1 + n 2 1 + 2

And mole fraction of H2, χ2 =

n1 2 = = 0.67 n1 + n 2 1 + Z

D Smix = –2.303R(n2 × log χ1 + n2 log χ2) = –2.303 × 8.314 × [1 × log 0.33 + 2 + log0.67] = 15.88 JK–1 15. We have DS =

=

Q rev (T2 − T1 ) T1T2

10.46 ×103 (573 - 350) = 11.63JK -1 573× 350

16. We have DS = 2.303nR log

P1 P2

Here n = 2, R = 8.314 JK–1 mol–1 T = 293 K;P1 = 10 atm, P2 = 2 atm.

 10  DS = 2.303 × 2 × 8.314 × log    2 = 38.294 × log 5 –1 –1 = 26.76 JK mol 17. The reaction is endothermic , DH = 28.1 KJ mol–1 D Entropy change in surroundings DSsurr =

=−

DH sys T

28.1×103 298

6.74

Thermodynamics

= –94.3 JK–1 mol–1 DSsys = 108.7 JK–1 mol–1 DSTotal = DSsys + DSsurr = 108.7–94.3 = 14.4 JK–1 mol–1 There is increase in total entropy. So ammonium nitrate dissolves spontaneously

O

O

O

O

O

O

∆rG = +491.18KJ mol ∆rS = 197.67 ×10-3 KJ mol1- K -1 ;T = 298 K O

18. DG = DH–TDS If D G = 0; DH–TDS = 0 Or DH = TDS T=

O

∆G becomes negative and the reaction becomes product-favored. ∆H +32.9 = 145.25 K T= = +0.2265 ∆S 23. We know ∆rG = ∆rH - T∆ r S

∴∆rG = 491.18 KJ.mol–1 –[298 K × (197.67 × 10–3 KJ.mol–1 K–1)] = 491.18 KJ mol–1–58.9 KJ mol–1 = 432.28 KJ mol–1 Since DrG is positive at 298 K the reaction is not possible. For the feasibility of the reaction we calculate T at which ∆rG becomes zero ∆rG = ∆rH –T∆rS = 0 O

DH 29.37 × 103 = = 282.4K DS 104.0

O

Hence above 282.4 K the reaction becomes spontaneous 19. We have D G = 2.303nRT log P2 P1 Here n = 2; R = 8.314 JK–1 mol–1; T = 300 K P1 = 1 atm , P2 = 0.1 atm. 0.1 (a) DG = 2.303 × 2 × 8.314 × 300 1.0 = 11488.2 × log0.1 = 11.488 KJ mol–1 (b) Since the initial and final state are same, the Gibbs energy change of process which occurred irreversibly will be same DGirr = –11.488 KJ mol–1

O

O

O

∴T = =

∆rH ∆rS

O

O

O

491.18KJ.mol-1 197.67 ×10-3 KJ.mol-1K -1

= 2484.8 k So reaction is spontaneous above 24.84.8 K (or 2211.8°C) 24. ∆G = ∑ ∆G (products) – ∑ ∆fG (reactants) ={6DfG [CO2(g)] + 3∆fG [H2O(g)]} 15 {DfG [C6 H6(i)] + ∆f G [O2(g)]} 2 = {[6 × (–394.4)] + [3 × (–228.6)]}–[172.8 + 0]– = –3225 KJ mol–1 25. for the reaction 4NH3(g) + 5O2 (g) → 4NO(g) +6H2O (l) DG = –1010.5 KJ.mol–1 DG = 4DfG [NO(g)] + 6DfG [H2O(l)]– {4DfG [NH3(g)] + 5DfG [O2(g)]} DG = –1010.5 KJ.mol–1, DfG [H2O(l)] = –237.2 KJ.mol–1 DfG [NH3 (l)] = –16.6 KJ.mol–1 and DG [O2(g)]= 0 4DfG [NO(g)]= – 1010.5 + 1423.2 = 66.4 = 346.3 KJ mol–1 26. ∆rG can be calculate by using the relation ∆rG = ∑∆frG (products)–∑ ∆frG (reactants) (a) Ca(s) + Cl2(g)→CaCl2(s) ∆rG = ∆ fG [CaCl2(s)]–{ ∆fG [Ca(s)] + ∆fG [Cl2(g)]} = –748.1–(0 + 0) = –748.1 KJ mol–1 Reaction is spontaneous. O

O

O

O

20. When the process is at equilibrium, DG = 0 ∴D H = TDS (∵) DG = DH–TDS) or T =

DH 178000 = = 1112.5K DS 160

Thus the process will become spontaneous above 1112.5 K 21. HgS(s) + O2(g) → Hg(l) + SO2(g) ∆H = –238.6 KJ.mole–1 and DS = + 36.7 JK–1 mole–1 Assume that ∆ H and ∆S values do not depend on temperature as ∆H is negative and ∆S is positive using the equation O

O

O

O

O

O

∆G = ∆H = T∆S

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

O

∆G will be be negative at all temperature and so the reaction is product-favored at all temperatures. In this problem, both the factors (∆H ) and (∆S ) are favourable to spontaneity. 22. Since ∆S is positive the spontaneous reaction is favourable. But the ∆H is positive. So the spontaneity is not favourable. So the reaction becomes product favoured above certain temperature This can be calculated by using ∆G = ∆H -T∆S at which the O

O

O

O

O

O

O

O

O

O

O

O

O

O

Thermodynamics 6.75

(b) Hg(s)→Hg(l) +

1 O2 (g) 2

O

1 ∆rG = ∆ fG [Hg(l)] + ∆fG [O2(g)]– ∆fG 2 [HgO(s)] O

O

O

O

= (0 + 0)–(–25.84) = 58.84 Reaction is not spontaneous (c) NH3(g) + 2O2(g)→HNO3(l) + H2O(l) ∆rG = ∆fG [HNO3(l)] + ∆fG [H2O(l)] {∆fG [NH3(g)] + 2∆fG [O2(g)]} = 80.71 + (–237.13)–{–16.45 + 0} = 80.71–237.13 + 16.45 = –301.39 KJ mol–1 Reaction is spontaneous 27. We know ∆rG = 2.303 RT log K −∆ r G Or log K = 2.303 RT ∆rG = 8.1 KJ mol–1, T = 1000 K; R = 8.314 × 10–3 KJ mol–1 K–1 −8.1 ∴ log K = 2.303× 8.314 ×10 −3 ×1000 = 0.423 ∴ K = 2.64 28. ∆G = ∆H –T∆S ∆H = 77200 J mol–1, ∆S = 122 JK–1, mol–1,T = 400 K ∆G = 77200–(400 × 122) = 77200–48800 = 28400 JK–mol–1 Now ∆G = –RT in K = –2.303 RT log K 28400 = –2.303 × 8.314 × 400 log K 28400 Or log K = = −3.708 2.303× 8.314 × 400 K = 1.96 × 10–4 29. ∆G for the reaction ∆G = [C6H6(g)]–3[∆fG (HC = CH)] = 1.24 × 105–3 × 2.09 × 105 = –5.03 × 105 Now ∆G = –2.303RT log K DG = 88.155 Or log K = 2.303RT O

O

O

O

= –12.023.4 KJ Since ∆G is negative the reaction is spontaneous in the forward direction. 31. (a) ∆rG for the reaction at 1273 K ∆rG = 176 KJ mol–1∆rS = 157.2 J mol–1 and T = 1273 K ∴ ∆rG = 176 KJ mol–1–1273 × 157.2 × 10–3 KJ mol–1 Since ∆rG is negative, the reaction is spontaneous. (b) (i) ∆rG = 2.303RT log KP O

O

O

O

O

O

O

O

Or log Kp =

∆rG = –24.10 KJ mol–1; R = 8.314 × 10–3 KJ mol–1; T = 1273 K O

O

∴ log Kp =

O

O

O

O

O

O

O

O O

O

multiple choice Questions with only one answer Level i 11. W = nRT ln

14. C1 = C2

T1 V1 T1 V1

38

O

g–1 g.4

16.

T1 g–1

4

= T2V2 T = 1 V2g.4 4

V2 = 32 V1

O

O

T1 T1 4 ∴ = T2 T2 1

T1 = 4T2, T2 =

−5.03×10 = 88.155 2.303× 8.314 × 298

O

T1 T2

2C1 = C1

5

∴K = 1.43 × 10 Since the value of ∆G is negative the reaction is feasible and may be recommended for making benzene. 30. ∆G is reated to equilibrium constant K as ∆G = –2.303 RT log K ∆G = ? K = 2.0 × 107, T = 25 + 273–298 K ∆G = –2.303 × 8.314 × 298 × log 2 × 107 = –12023.4 J

P1 P2

100 = 5 × 2.303 × 1.987 × 300 = log 1 = 13818 cal

O

=

24.10 2.303× 8.314 ×10-3 ×1273

= 0.9887 ∴ Kp = 9.74 (ii) For the reaction Kp = Pco2 ∴ Pco2 = 9.74 bar

O

O

∆r G 2.303 RT

T V2 = 1 T2 V1 110 T2 = 100 273

6.76

Thermodynamics

T2 = 300.3 K q = mSdT = 32 × 0.22 × 4.2 × (300.3 –273) 17. ΔH = ΔE + ΔnRT 743 = ΔE + 5(8.314) (2) 19. ΔS =

DH 40850 = 109.5 J/mol = T 373

the entropy change into when 3.6 g liquid converted to steam is 109.5 × 3.6 = 21.89 J/K 18

multiple choice Questions with only one answer Level ii 5. At 273 K it is in equilibrium. So ΔG = 0 DH T= DS ∴ ΔH = TΔS = 273 (60.1–38.2) = 5978 J 6. He of formation is enthalpy of the compound 3N2O + 2NH3 →4N2 + 3H2O Δ H = 3Hh2o + 4Hn2– 3Hn2o – 2Hnh3 879.6 = 3(–241.8) –3× –2(–45.9) ∴ HN O = 82 KJ/ mol 2

20.

H 2 + Cl2  → 2 HCl ∆S = 2 S HCl − S H 2 − SCl 2 = 2 (186.7) –130.6 – 223 ΔS = 19.8 J/K / mol

DH 2930 = 24. ΔS = = 9.77 J/K/mol T 300 31. H2O (s) ⇋ H2O (l) ΔS = SH2O (l) →SH2O (s) ΔS =

DH T

V2 V1

= 2.303× 2×8.314 log 10 = 38.3 J/K/ mol 42. A (liquid) ⇌ A (gas) ΔH = –100–(–130) = 30 KJ ΔS = 200–100 = 100 J/K/mol T=

∆H 6000 = = 21.98 J / K / mol T 273

9. ΔSTotal = ΔSsys + ΔSsurr =−

100 100 + = 0.067 J / K 300 250

10. 2ZnO + 2C → 2Zn + 2CO2 12. H2(g)

1 O2(g) → H2O(l) 2

H+(aq) + OH–(aq) → H2O(l)

ΔH = TΔS = (273) (21.81) = 5954 J/mol 34. ΔG = DH –TΔS = –2.5 × 103 – (298×7.4) ΔG = –ve 35. 2ZnS + 3O2→2ZnO + 2SO2 ΔG = + 293 –480 –544 ΔG = –731 KJ 40. nCvdT = –PdV  1RT 1R300  5 l R (T – 300) = –2   5   2 2 T = 248.5 K 41. ΔS = nR ln

7. ∆S =

DH = 300 K DS

ΔG = –100 KJ ΔH = –68.3 K. cal ΔH = –13.7 K. cal

1 H2(g) → H+(g) + e- → ΔH = 0 K.cal 2 1 1 H2(g) + O2(g) + e- → OHΔH = ? 2 2 13. ΔS = nCp ln

T2 T1

 7  800 = 2  R  ln  2  400 0.5 14. TV = Constant 15. ΔG = –2.303 RT log kp O

= –2.303 × 8.314 × 310 log

25 = 8.76 KJ 750

17. ΔG = ΔH – TΔS = –54.07 –298 (10–2) = –57.05 KJ ΔG = –2.303 RT log Kp ∴ log Kp = 10. O

O

O

O

18. Final Pressure P =

nRT V

1 × 0.0821 × 300 = 24.6 atm ΔS = nR ln

P2 P1

Thermodynamics 6.77 3 22. ΔG = ΔG + RT lnQ  20  = 10000 + (8.314 × 298) ln   = 27.11 KJ  2 3 24. 2Al + O2 →Al2O3 ΔH = –(837.8)2 = –1674.4 KJ 2 O

39. Adiabatic reversible process 40. At equilibrium ΔG = 0 Equilibrium temperature T =

V2 25. ΔS = nR ln V = 9.2 cal 1

So reaction is spontaneous at 1000k

∂E 96.53 ∴ = = 5 × 10–4 V/K ∂T 2 × 96500 27. ΔH = nCP dT 100

10

ΔH = ΔV + Δn RT ΔV = 38 KJ/ mol. 43. Moles of air =

∴ H = –13.27 Kcal 29. Heat of reaction ΔH = HCO2 + HNO –HCO – Hno2 = –226 KJ Activation energy of forward reaction = 465.2 KJ Activation energy of backward reaction = 465.2 + 226 = 691.2 KJ

C

CO +

2 PNH 3

O

Work T

PVDS NR 1 32. H2O → H2 + O2 2 Work =

ΔG = 120 KJ

1 O2 → CO 2 0.8 0.3

– 1

ΔH = –26 Kcal. I. mole F. mole

1 O2 →CO2 ΔH = –68 Kcal 2

1 0.3 – mole 0.4 – 0.3 mole Heat evolved is = 26 + 68(0.6) = 66.8 Kcal

PN2 × PH3 2

ΔG = ΔG + 2.303 RT log Q = 60.5 KJ

+

I mole –

∴ Q = 1.48 × 10-5

31. ΔS =

2 × 49.26 =4 0.08241× 300

Moles of air = 0.8

(2 + 0.2T ) dT

30. Reaction quotient =

Δn = 1

  H2O (g) ⇀ 41. H2O (l) ↽

∂E 26. ΔS = nF ∂T

∆H = 5∫

DH = 1000 K DS

44. ΔS = nRln

V2 V1

47. ΔH = ΔU + ΔnRT = –3000 + (–1) (1.987) (300) = –3596 cal ΔG = ΔH–TΔS = –3596– 300(–10) = –6596 cal 48. dq = dV + PdV

O

1440 = ΔE + 1(0.018– 0.0196)

4 × (0.25)1/ 2 = 10 Reaction quotient = 0.2

∴ ΔE = 1440.039

ΔG = ΔG + 2.303 RT log Q = 125.7 KJ O

T2 33. ΔS = nCp ln = 0.693 K.cal/ K T1 34. TVg–1 = Constant 300 × (1)0.53 = T×(4)0.53 T = 150 K ΔH = nCpdT = –900 cal/ mol

49.

DH 2 − DH1 = ΔCp T2 − T1

2 0.0821

comprehensive Type Questions Passage i 1. W = 2.303 nRT log

v2 v1

104 = 2.303(8.314)T log 10 nRT P 1× 0.0821× 522 = 100 = 0.434 lit

2. V =

Passage ii 1. Colorific value = = 121005

242000 2

nRT 2. Initial volume of gas is = = 2(0.0821)300 P = 49.26 lit Volume of gas in 2nd step is P1V1 = P2V2 (2)(49.26) = 0.8 V V = 123.15 lit So volume change 74 lit 3. Second step is isothermal ΔU = 0 nRT (T1 − T2 ) w= γ −1 2 × 8.314(300) w= 0.278 ΔU = – w ΔU = 24 KJ Passage Vi 1. C12 H22 O11 +12O2 → 12CO2 + 11H2O

Passage iii 1. The amount heat = V it = 1800 J (1800)18 Molar enthalpy = 0.9 = 36 KJ/mol 2. H2O(l) ⇌ H2O(g) Δn = 1 ΔH = ΔE + ΔnRT DE = 36 ×1010 –1(8.314)373

= 32.9 KJ / mol 3. ∆H = ∆U + ∆nRT 26.4 × 103 = ∆U + (−0.5)(2)(300) = 26.7 K Cal Passage iV 1. Final Temperature is T P1 P2 = T1 T2 2P P = 300 T2 T = 600 K At constant volume work done is zero ∴ dq = dU dU = nCvdT = 2(40)(300)=24 KJ In whole cycle ΔU = 0 because U is a state function

Mole of O2 used per day =

640 = 20 32

Mole of C12 H22O11 used per day = 20 12 20 Wt of sucrose = × 342 = 570 gm 12 Passage Vii 1. The heat evolved is = 2.082 KJ per 0.1265 g So heat of combustion is = 5629 kg 2. C12 H22 O11 +12O2 → 12CO2 + 11H2O ∆H = 12 Hco2 + 11Hh2O – Hc12h22o11 –1529 = 12 (–393.5) + 11 (–28.53)–x x = –2236 KJ 3. q = msdT 2082 ms = = 1194 J / K 1.743 Passage Viii 1. Initial temperature of the gas is PV 1 × 20 T= = 121.8 K = nR 2 ( 0.084) In isobaric process work = PdV = 1(20) = 20 lit V In isothermal process work = nRT ln 2 V1 20 2.303 × 2 × 0.0821 × 121.8. log 40 = –13.81 lit Total work = 20 – 13.81 = 6.18 lit

Thermodynamics 6.79

2. In cyclic process all state functions are zero 3. ∆S = 2.303 nRT..log

20 40

3. ΔS = 2.303nC log

R 100 log 0.66 10 = 6.9 cal / K

1

= 2.303 × 2 × 8.314.log

T2 T1

= 2.303 ×

20 40

= −11.485

integer Type Questions Passage ix 1. In isothermal process ΔU = O 2. Entropy is state function So entropy change in D + E path = A Path Passage xi 1. In state 1 PV 1× 22.44 T1 = = = 273 k nRT 1× 0.021 In state 2 1× 44.88 T2 = = 546 K 0.0821 0.5 × 44.88 T3 = = 273K 0.0821 2. Enthalpy is a state function 3. ΔU = nCvd T 3 = 1 × × 8.314 × 273 2 = 3.40 × 103 J 4. ΔU = nCVd T Passage xii 6. ΔSSys =

+q −q , DS surr = T T

7. ΔSSys =

−q +q , DS surr = T T

Passage xiii 1. ΔS = nC ln

T2 T1

R 600  R = 1 × 2.303  + .log  0.33 2  300 = 4.83 cal

1. dq = dE –dw dq = – dw V dq = 2.303 nRT log 2 V1 10 ∴ q = 2.303 × 2 × 2 × log lit – atm 2 q = 652 J 2. dq = nCvd T

R  R +  2 = 8 Cal  0.66 2  3.

4. 1 ncdT dS = T −2 10 TdT = T ∆S = 10–2 (300–200) = 1

6. ∆S = nCv ln

=

T2 P × nR ln 1 T1 P2

3 1000 R ln + Rln 100 = 7 100 2

8. C = aT3 0.42 a= 3 = 10 aT 3 dT dS = ∫ T ΔS = 1 9. ΔH = nCp d T 5 = 1 × × 8.314 × 500 2 = 104 J 10. ΔS =

T=

DH T

1.435 × 103 = 273K 5.26

6.80

Thermodynamics

Previous years’ iiT Questions 1. In a reversible process both forward and opposing forces an equal in equilibrium 2. Work is a path function 3. ∆H = ∆V + ∆PV = 30 + (4 × 5 – 2 × 3) = 44 lit – atm DH 30 × 1000 4. TB = = DS 75

O

= –54.07 – 298 O

10. 11. 18.

= 400 K 5. ∆H = nCpdT dT = 0; So∆H = 0 6. nCvdT = –PdV 3 (T – T ) = –1 (2–1) 1 R 2 2 2 T2 = T– 3 × 0.084 1×

7. AD→SB = AD→SC + AD→SD + DD→SB = 50 + 30–20 = 60esu 8. PV–1 = 1 C=

R R R R + = + = 2R 1 − ( −1) g − 1 1− ∝ 5 −1 3

O

9. ∆G = ∆H – T∆S

19.

O

10 = −57.05 KJ 1000

∆G = –2.303 R T log K –57.05 = –2.303 × 8.314 × 10–3 × 298 log K log K = 10 In vaporization ∆S = + ve Enthalpy of heat of formation of elements is zero work done in dotted lines W = 4 × 3/2 + 1 × 1 + (0.5× 2.5) = 8.25 lit-atm Work done in dark lines W = 2.303 RT log V2/V1 =2.303 × 4× 0.5 log 5.5/0.5 = 4.8 lit – atm. Ratio ≈ 2. The evolved heat = 2.5 × 0.45 = 1.125 KJ Heat of combustion = 1.125 × 28/3.5= 9 KJ / mol.

CHAPTER

7 Thermochemistry

I

f you have had your attention directed to the novelties in thought in your own life time, you will have observed that almost all really new ideas have a certain aspect of foolishness when they are first produced. Alfred North Whitehead

7.1 IntroductIon During the course of most chemical reactions, energy in one form or another whether heat, light or electrical, is taken in or given out. Those chemical reactions in which heat is given out are said to be exothermic and those in which heat is absorbed are endothermic. The study of this heat change - the subject of thermochemistry is important that the calorific value of a fuel or a food should be known. A knowledge of the heat change involved in a chemical process is sometimes essential for plant control. In some cases, the yield of product is affected too. And as we shall see, the total energy change involved in the synthesis of a compound from its constituent elements explains some of its chemical behaviour, particularly its stability. Except in giving calorific value, the unit in which the heat is measured is the gram- calorie or the kilogram-calorie, the latter unit being written either with a capital C, i.e., as calorie (= 1000 calories) or as ‘kg. cal’ or ‘Kcal’. In any chemical reaction (e.g., where A + B react to give C + D), we may consider the heat change in two distinct ways. Suppose the reaction is carried out in calorimeter, and a rise in temperature is observed. Clearly, heat has been given out, and we may write

A + B → C + D + Q Kcal Where Q is the heat of reaction, given out by the reaction and gained by the surroundings. If a fall in temperature is observed in the calorimeter, then we write A + B → C + D - Q Kcal

Because heat has been lost by the surroundings. But we may equally well consider the heat change as heat lost or gained by the reacting species. If heat is given out, this implies a loss of heat content and is usually designated by

H and is sometimes called the enthalpy (as discussed in 6.7). So we write for an exothermic reaction. A + B → C + D; ∆H = - Q Δ Indicating the change in heat content or enthalpy and negative sign a loss of heat from the system. Similarly, if heat is absorbed by the system (an endothermic reaction), then we write A + B → C + D; ∆H = + Q The ‘Q’ convention is sometimes used when we are dealing with purely thermal energy changes, as in experimental thermochemistry; the ΔH convention is more commonly used when we have to consider all types of energy changes in a reacting system, e.g., work done by a gas in expanding; thermodynamics deals with this field of chemistry.

7.2 EnErgy StorEd In AtomS And molEculES We have seen that chemical reactions are exothermic or endothermic depending upon whether they are accompanied by evolution or absorption of heat. But if we look at a chemical reaction, we find that it is simply a reaction in which reactants change into products, i.e., one form of molecules undergo change into molecules of another form. Now the question arises: how energy is evolved or absorbed in chemical reaction? In order to answer this question, it is necessary to believe that matter is always associated with a certain amount of energy which can not be observed even though it is present in all the substances. Energy changes appearing in the form of work, heat, light, etc, can, however, be observed when matter

7.2

Thermochemistry

reacts. Energy present in the matter may be regarded to be stored in molecules which are smallest particles capable of free existence and always supposed to be present in matter. The energy in matter therefore lies in the position and motion of molecules, and subatomic particles (electrons, protons, neutrons, etc). The total energy possessed by a molecule is called the internal or intrinsic energy of the molecules. The exact magnitude of this energy cannot be determined. It is, however, possible to determine the changes in internal energy that take place when a molecule undergoes a chemical reaction or any other change of state. The energy change that occurs in a chemical reaction is usually known as chemical energy and is largely due to the change in potential energy. That results due to the breaking of bonds in reactants and formation of new bonds in products. The heat change that takes place at constant volume is the change in internal energy while the heat change that takes place at constant pressure is the enthalpy change. The change in internal energy is represented by ΔU while the change in enthalpy is represented by ΔH.

7.2.1 measuring Heats of reaction For studying reactions in solution, the calorimeter consists essentially of an insulated vessel in which the reaction is carried out. An unsilvered vacuum flask may be used to provide thermal insulation (Fig 7.1) while an outer jacket reduces temperature fluctuations in the air surrounding the flask. An efficient stirrer must be provided so that the reactants are properly mixed and the temperature is kept uniform throughout the solution. For accurate work, the thermometer must be capable of reading to 0.001 degree. The reaction will probably involve mixing two solutions

or adding a solid to a solution and it is essential that the temperature of both reactants is known accurately before the reaction starts. The best method of ensuring this is to have them both inside the calorimeter, one contained in a small separate vessel, so that they can attain the same temperature before they are mixed. Errors due to heat losses can be eliminated by using an adiabatic calorimeter (Greek “a-diabatos,” which means not to be crossed). Here the calorimeter vessel is surrounded by a jacket containing water, the temperature of which is automatically kept the same as the temperature of the liquid inside the vessel. As the reacting system and its surroundings are always at the same temperature, no heat passes in either direction. For heats of combustion, the bomb calorimeter (fig 7.2) is used. This consists of a strong metal (generally steel) vessel capable of withstanding high pressures. A weighed amount of the compound is placed in the crucible, with a wire (connected to external electrodes) for starting the reaction. The top is screwed firmly on, and the bomb is then filled with oxygen under pressure and placed in a container filled with water. Combustion is effected by fusing the wire, which ignites a cotton thread touching the compound and the temperature rise of the calorimeter plus the surrounding water is measured. Corrections for heat losses are applied unless the instrument is an adiabatic one. Firing lead Stirrer

Oxygen inlet valve Ignition wire and cotton

Stirrer Vacuum flask

Fig 7.1 Calorimeter for measuring of reaction in solution

Fig 7.2 Bomb calorimeter In order to calculate the quantity of heat that corresponds to a measured change in temperature in a calorimeter, it is necessary to know the heat capacity of the system, i.e., the heat required to raise the temperature by 1 degree. The quantity of heat liberated by the reaction can then be found from the equation: Amount of heat liberated = (temperature rise) × (heat capacity)

Thermochemistry

By knowing the heat capacity of calorimeter and also the rise in temperature, the heat of combustion at constant volume can be calculated by using the expression: ∆U = - Z × ∆T ×

M W

Where Z = Heat capacity of calorimeter system ΔT = Rise in temperature M = Molecular mass of substance W = Mass of substance taken Measurements in an open calorimeter (Fig 7.3) determine the amount of heat liberated or absorbed when known amounts of substances react at constant pressure.

7.3

thermochemical equation. It may be noted that in thermochemical equations, fractional coefficients are also commonly used contrary to our usual practice for balancing the chemical equations. A thermochemical equation can be written in two ways: (a) By writing the heat evolved or absorbed as a term in the equation. 2SO 2 ( g ) + O2 ( g ) → 2SO3 ( g ) + 694.6 KJ (b) By using ΔH notation, i.e., writing ΔH = -ve for exothermic and ΔH = +ve for endothermic reactions, as 2SO 2 ( g ) + O 2 ( g ) → 2SO3 ( g ) ; ∆H = -694.6 KJ Some Important conventions about thermochemical Equation

Thermometer

Foamed polystyrene cup

Reaction mixture

1. For exothermic reactions, ΔH is –ve and for endothermic reaction, ΔH is +ve. 2. Unless otherwise mentioned in a given reaction, the ΔH values correspond to the standard states of the substances. The standard state is taken as pure form of the substance at 1 atmospheric pressure. The modern definition of standard state replaces 1 atm by 1 bar where 1 bar = 105 pa this has been used in this text. 3. While writing the thermochemical equation for a reaction, the physical states of the reactants and the products must be mentioned. The physical states are represented by the symbol’s, l, g and aq for solid, liquid, gaseous and aqueous states, respectively. 4. It is essential because the physical states of the reactants and products cause appreciable difference in the value of ΔH. For example, when one mole of liquid water is formed from gaseous hydrogen and gaseous oxygen, the heat evolved is equal to 285.8 KJ.

Fig 7.3 Calorimeter for measuring enthalpy changes (atmospheric pressure)

1 H 2 ( g ) + O 2 ( g ) → H 2 O ( l) ; ∆ r H = -285.8 KJ 2

The calorimeter is immersed in an insulated bath fitted with stirrer and thermometer. The temperature of the bath is recorded in the beginning and after the end of the reaction and the change in temperature is observed. Knowing the heat capacity of water bath and calorimeter and also the change in temperature, the heat absorbed or evolved in the reaction can be calculated. This gives the enthalpy change (ΔH) of the reaction.

On the other hand, when one mole of water in the gaseous state is formed from the gaseous oxygen and hydrogen, the heat evolved is equal to 241.8 KJ.

7.2.2 thermochemical Equations A chemical balanced equation, which gives the heat change (evolved or absorbed, ΔH) during the reaction, is called the

1 H 2 ( g ) + O 2 ( g ) → H 2 O ( g ) ; ∆ r H = -241.8 KJ 2 5. The coefficients of the substances in a thermochemical equation represent the number of moles of each substance involved in the reaction. The ΔH values correspond to these coefficients. When the coefficients in the chemical equation are multiplied or divided by a factor, ΔH value must also be

7.4

Thermochemistry

multiplied or divided by the same factor. For example, in the equation 1 O2(g) → H2O(g); ΔrH = –241.8 KJ 2 If coefficients are multiplied by 2, we write the equation as H2(g) +

From this, we can understand that the MgO is 601.7 KJ mol–1 below the zero level, i.e., 601.7 KJ mol –1 lesser energy than magnesium and oxygen. Notice that 1 fraction   is used in the equation, so that it shows  2 the formation of 1 mol of MgO.

2H2(g) + O2(g) → 2H2O(g); ΔrH = –483.6 KJ 6. When the chemical equation is reversed, the sign of ΔH value is also changed but the magnitude remains the same. For example, N2(g) + 3H2(g) → 2NH3(g); ΔrH = – 91.8 KJ 2NH3(g) → N2(g) + 3H2(g); ΔrH = + 91.8 KJ Thus a reaction which is exothermic in one direction will be endothermic in reverse direction. 7. The value of ΔrH usually shows very small change with increase in temperature (provided there are no phase changes when the temperature is increased). 8. According to IUPAC (1981 recommendations), the different types of thermodynamic property changes such as enthalpy are expressed in such a way that the symbol denoting the process is written as a subscript before the symbol of property. For example, enthalpy of reaction as ΔrH, enthalpy of vapourization as Δvap H, enthalpy of formation as ΔfH, enthalpy of combustion as ΔCH. The standard state is represented by superscripting O for H i.e., ΔrH , ΔfH etc. O

7.3.2 Standard Heats of Formation The heat liberated in the reaction of magnesium and oxygen during the formation of MgO is called standard heat of formation. In general the definition of this quantity is: The standard heat of formation is the enthalpy change when one mole of the substance is made from its elements in their standard states. The heats of formation for certain substances are given in Table 7.1. Notice that heats of formation may be either endothermic or exothermic. It is useful to show a standard heat of formation by using a subscript ‘f’ on the ∆H symbol unless otherwise stated. All values refer to a temperature of 298 K. Thus, in the case of magnesium oxide, the standard heat of formation is ∆Hf = –601.7 KJ mol–1. In Table 7.1, C(s) refers to graphite. The value of SiO2 is only approximate. The compounds which are formed from their elements with the evolution of heat (i.e., with negative heat of formation) are called exothermic compounds whereas the compounds which are formed from their elements with the absorption of heat (i.e., with positive heat of formation) are called endothermic compounds. O

O

O

7.3.3 Enthalpy changes in chemical reactions 7.3. StAndArd EntHAlpIES 7.3.1 Standard Enthalpy of an Element To define standard enthalpy, we must choose a zero level of enthalpy. We do this in the following way. First, the state of an element as it appears at 101.325 K pa is called its standard state. Usually 298 K (25°C) is chosen as standard temperature. For example, the standard state of hydrogen is a gas, water is liquid and sodium is solid. Then the definition for standard enthalpy of an element is: The enthalpy content of an element in its standard state is zero. The following example illustrates how we can use this definition. Suppose 1 mole of magnesium is burnt in oxygen and the heat liberated in this process measured, is about –601.7 KJ mol–1 Mg(s) +

1 O2(g) → MgO(s); ΔH = –601.7 KJ mol–1 2 O

The amount of heat evolved or absorbed in a reaction is known as heat of reaction. The amount of heat evolved or absorbed at constant temperature and pressure is called enthalpy change. The heat change during a chemical reaction depends upon the amount of substance (number of moles that has reacted). Thus, enthalpy of reaction may be defined as the enthalpy change (amount of heat evolved or absorbed) in a reaction when the number of moles of reactions react completely to give the products as given by the balanced chemical equation. Let us consider a chemical reaction having two reactants A and B in the initial state and suppose HA and HB are the enthalpies of A and B. Now suppose the system undergoes a chemical change at a constant temperature and pressure forming the products C and D finally and the enthalpy of the products is HC and HD. The reaction and change in enthalpy can be represented as A+B→C+D And ∆H = (HC + HD) – (HA + HB) (1) Or and ∆H = HP – HR

Thermochemistry

–393.5

Similarly, ∆H will be negative when the enthalpy of reactants is greater than the enthalpy of products, i.e., HR>HP. In other words heat will be evolved during the course of the reaction and hence the reaction will be exothermic. The heat of reaction at constant volume and temperature is defined as the difference in the internal energies of the products and that of the reactants, the quantities of the reactants and the products being the same as given in the chemical reaction. Thus, heat of reaction at constant volume, i.e., ∆U for a reaction of the type

–285.9

A+B→C+D

–277.7

is equal to the internal energy of the products minus internal energy of the reactants. Thus,

table 7.1 Some standard heats of formation

Si(s) + O2(g) → SiO2(s)

1 O2(g) → CaO(s) 2 1 Na(s) + Cl2(g) → NaCl(s) 2 Ca(s) +

C(s) + O2(g) → CO2(g)

1 O2(g) → H2O(l) 2 1 2C(s) + 3H2(g) + O2(g) → C2H5OH(l) 2 1 1 H2(g) + F2(g) → HF(g) 2 2 1 H2(g) + O2(g) → H2O(g) 2 1 C(s) +2H2(g) + O2(g) → CH3OH(l) 2

H2(g) +

8C(s) + 4H2(g) → C8H8(l) C(s) +

3 O (g) → CO(g) 2 2

∆Hf /KJ mol–1 φ

Reaction

7.5

–910.0 –635.5 –411.0

∆U = (UC + UD) – (UA +UB) –271.1 –241.8 –238.9 –224.4 –110.5

C(s) + 2H2(g)→ CH4(g)

–74.8

3 1 N2(g) + H2(g) → NH3 2 2

–46.0

6C(s) + 3H2(g) → C6H6(l) N2(g) + 2H2(g) → N2H4(l) 2C(s) + 2H2(g) → C2H4(g) 2C(s) + H2(g) → C2H2(g)

+49.0 +50.6 +52.3 +226.8

where ∆H is called the heat of reaction and is a measure of enthalpy change at constant temperature and pressure. Thus the heat of reaction of a chemical change is defined as the difference between enthalpies of the products and reactants of the chemical reaction and is designated by ∆H, when the number of molecules of the reactants indicated by the chemical reaction have completely reacted. It is well evident from the equation that ∆H will be positive when heat content of the products (HP) is greater than that of reactants (HR). But according to the first law of thermodynamics, energy cannot be created. Thus, the excess energy will come from the surroundings. In other words, there will be absorption of heat and so the reaction will be endothermic.

Since the reaction is taking place at constant temperature and constant volume, we have W = 0. From the first law of thermodynamics: ∆U = Q – W Or ∆U = QV (Since W = 0 at constant T and V) where QV is the heat absorbed at constant volume. Thus, ∆U = Uproducts – Ureactants = UP – UR It is clear from this equation that if internal energy of products is greater than that of the reactants, i.e., UP>UR then energy will be absorbed from the surrounding and hence the reaction will be endothermic. In this, ∆U will be positive. In other words ∆U is +ve for all endothermic reactions taking place at constant temperature. If internal energy of reactants is greater than that of products, i.e., UR > U P, then energy will be evolved and given out to the surroundings and the reaction will be exothermic. ∆U will be –ve in this case. Hence, all exothermic reactions taking place at constant temperature give ∆U as –ve since UR> UP. Relation between ∆U and ∆H The quantities ∆H, ∆U and P∆V are related as ∆H = ∆U +P∆V

(1)

where ∆V is the increase in volume at constant pressure. Suppose n1 is the number of moles of gaseous reactants and n2 is the number of moles of gaseous products, then increase in number of moles = (n2–n1) = ∆n. If volume occupied by one mole is V, then volume occupied by ∆n moles will be V∆n, assuming gas laws to be applicable.

7.6

Thermochemistry

standard states is called the standard enthalpy change of that reaction.

Work done by the system at constant pressure P = P × ∆nV. But PV = RT \ Work done = P∆nV = RT∆n Substituting this value in the equation (1) ∆H = ∆U + ∆nRT

(2)

Now we know that ∆U = QV and ∆H = QP . Substituting these values in equation (2), we get QP = QV + ∆nRT If ∆n is 0, ∆H is equal to ∆U. If ∆n is negative, ∆H is less than ∆U. If ∆n is positive, ∆H is greater than ∆U. The value of ∆V is negligible in reactions involving solids and liquids. Hence, for such reactions ∆H = ∆U or QP = QV In other words, heat of reaction at constant pressure is equal to heat of reaction at constant volume. Factors which influence enthalpy change (∆H) of a reaction. 1. Physical state of the reactants and the products: The ∆H of a reaction depends upon the physical states of reactants and products. For example, when hydrogen and oxygen gases combine to give liquid water, the enthalpy of reaction is different than that when they combine to form gaseous water, etc. H2(g) +

1 O2(g) → H2O(l); ∆rH = –285.8 KJ 2

H2(g) +

1 O2(g) → H2O(g); ∆rH = –241.8 KJ 2

2. Quantities of reactants: The enthalpy change (or amount of heat evolved or absorbed) depends upon the amount of reactants. For example, the heat of combustion of 2 moles of carbon is double than the heat of combustion of 1 mole of carbon. 3. Allotropic modification: The enthalpy change for different allotropic forms of the same substance is different. For example, C (Diamond) + O2(g) → CO2(g); ∆rH = –395.4 KJ C (Graphite) + O2(g) → CO2(g); ∆rH = –393.5 KJ 4. Temperature: The enthalpy of reaction depends upon the temperature of reactants and products. 5. Conditions of constant pressure or volume: The enthalpy of reaction depends upon the conditions of constant pressure or volume. The enthalpy change for a reaction when all the participating substances (elements and compounds) are in their

7.3.4 Variation of Heat of reaction with temperature: Kirchhoff ’s Equation The heat change of any physical or chemical process varies with temperature. Making use of the first law of thermodynamics, Kirchhoff (1858) deduced certain mathematical expressions, which define the variation of heat of reaction with temperature. Consider a hypothetical reaction at constant pressure. Areactants → Bproducts Let the heat change be ∆H1 when the reaction proceeds at temperature T1. Now the product B1 formed at T1 is brought to temperature T2. Let the amount of heat absorbed be CP (T2 – T1), where CP = mean molar capacity of the products between T1 to T2 at constant pressure A (T1)

∆H1

B1 (T1)

C′P (T2 – T1)

Cp(T2 – T1)

A (T2)

∆H2

B (T2)

Thus, the net heat change in the whole process = ∆H1 + CP (T2 – T1)…….(1) In the second experiment, let the temperature of the reactants A be raised from T1 to T2 Thus, the heat absorbed = C′P (T2 – T1) where C′P means molar heat capacity of reactants A between T1 and T2 at constant pressure. Now the reaction occurs at temperature T2 and let the heat of reaction be ∆H2. Thus, net heat change during the whole process = ∆H2 + C′P (T2 – T1)……….(2) Thus, we see that reactants A at temperature T1 have changed to product B at temperature T2. According to first law of thermodynamics, the two energy changes should be equal. Hence, from (1) and (2) ∆H1 + CP (T2 – T1) = ∆H2 + C′P (T2 – T1) Or ∆H2 – ∆H1 = C′P (T2 – T1) – CP (T2 – T1) = (T2 – T1) ( C′P – CP) = (T2 – T1) ∆CP Or ∆CP =

∆H 2 - ∆H1 T2 - T1

It is called Kirchhoff’s Equation.

Thermochemistry

The change in the heat of reaction (at constant pressure) with temperature depends upon the magnitude and the sign of ∆CP. Here ∆CP = difference in the heat capacities of products and reactants at constant pressure. Hence, the change in enthalpy of reaction at constant pressure per degree change of temperature is equal to the difference in the heat capacities of the products and the reactants at constant pressure.

Substituting these values in equation (1) we get  ∂ ( ∆H )    P = (CP)B – (CP)A = ∆CP  ∂T   ∂ ( ∆H )    P = ∆CP ∂ T   As ∆H is the heat of reaction at constant pressure, we can write the above expression as d ( ∆H )

= ∆CP (2) dT This equation (2) is valid only for a small difference in temperature. For a large difference in temperature, we have

Kirchhoff ’s Equation for a reaction at constant Volume For a reaction taking place at constant valume, we have ∆U 2 - ∆U1 = C V′ - C V = ∆C V T2 - T1

∆H 2

∫

Here ∆CV = difference in the heat capacities of the products and the reactants at constant volume. Hence, we say that “the change in heat of reaction at constant volume per degree change of temperature is given by the difference in the heat capacities of the products and the reactants at constant volume.” It is clear from the above discussion that in case ∆CP or ∆CV is zero, the heat of reaction involving solids only do not appreciably change with temperature as the molar capacities equal to one another. mathematical derivation of Kirchhoff ’s Equation Consider a process in which reactant A is converted into product

Let HA and HB be the enthalpies of the reactants and the products at the same temperature and pressure. Then enthalpy change for the above process is ∆H = H B – H A. Differentiating this reaction w.r.t temperature keeping the pressure constant  ∂ ( ∆H )   ∂H B   ∂H A   ∂T  =  ∂T  -  ∂T  P P P

( )

 ∂ HB  \   = (CP)B  ∂T  P

( )

 ∂ HA and   ∂T

 P

= (CP)A

T2

d ( ∆H ) =

∫ ∆C dT P

∆H1

T1

Or ∆H2 – ∆H1 = ∆CP (T2 – T1) Or

∆H 2 - ∆H1 = ∆CP T2 - T1

(3)

This reaction is applicable if ∆CP is independent of temperature. Here ∆H1 = heat of reaction at temperature T1 and ∆H2 = heat of reaction at temperature T2. CP is not independent of temperature and its variation with temperature is given by the relation CP = α + βT + γ T2 +…….. Here, α, β, γ are constants. Thus we write ∆CP = ∆α + (∆β) T + (∆γ)T2 +………… where ∆CP is variation with temperature. Substituting the value of ∆CP in equation (3), we get

A→B

∂H  But we know that CP =   ∂T 

7.7

(1)

P

T1

∆H2 = ∆H1 +

∫ ∆C dT P

T2

Or ∆H2 = ∆H1 + ∆αT+

1 1 ∆βT2 + ∆γT3 3 2

Applications of Kirchhoff ’s Equation Kirchhoff’s equation can be used to calculate the heat of reaction at particular temperature, provided heat of reaction at some other temperature and heat capacities of products and reactants are known. The molar heat capacities of the reactants and products are same, i.e., when ∆CP = 0 then the heat of reaction is independent of temperature. The molar heat capacities of solids are approximately equal to each other and hence the heat of reaction involving solids only, are almost independent of the temperature.

7.8

Thermochemistry

Solved problem 1 A system absorbs 470 J of heat and does work equivalent to 200 J on its surroundings. Calculate the change in internal energy. Solution: Heat absorbed by the systems = 470 J Or Q = 470 J Work done by the systems = 200 J Or W = –200 J According to first law of thermodynamics, ∆U = Q + W ∆U = 470 – 200 = 270 J

problems for practice 1. A gas absorbs 120 J of heat and expands against the external pressure of 1.10 atm from a volume of 0.5 L to 2.0 L. What is the change in internal energy (1 L atm = 101.3 J).

5. 0.562 g of graphite kept in a bomb colorimeter in ex cess of oxygen at 298 K and 1 atm pressure was burnt according to the reaction C(graphite) + O2(g) → CO2(g) During the reaction, the temperature rises from 298 K to 298.89 K. If the heat capacity of the calorimeter and its content is 20.7 KJ/K, what is the enthalpy change for the reaction at 298 K and 1 atm? 6. When 1 mole of ice melts at 0°C and at constant pressure of 1 atm 1440 calories of heat are absorbed by the system. The molar volumes of ice and water are 0.0196 and 0.0180 litres respectively. Calculate ∆H and ∆U. 7. 1 mole of ice at 0°C and 4.6 mm pressure is converted to water vapour at a constant temperature and pressure. Find ∆H and ∆U, if the latent heat of fusion of ice is 80 cal/g and latent heat of vapourization of liquid water at 0°C is 596 cal/g and the volume of ice in comparison to that of water (vapour) is neglected.

Solved problem 2 The heat of combustion of gaseous methane (CH4) at constant volume is measured in a bomb calorimeter at 298 K and is found to be – 885.4 KJ mol–1. Find the value of enthalpy change: Solution: Combustion of methane gives CO2(g) and H2O(l) CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) ∆U = – 885.4 KJ mol–1 = –885400 J mol–1 ∆ng = 1 – (1+2) = –2 mol T = 298 K; R = 8.314 J mol–1 K–1 Now ∆H = ∆U + ∆ng RT = – 885400 + [(–2 mol) × (8.314 J mol–1 K–1) × (298 K)] = –885400 – 4955 = –890355 = 890.355 KJ

problems for practice 2. The enthalpy change (∆H) for the reaction N2(g) + 3 H2(g) → 2NH3(g) is – 92.38 KJ at 298 K. What is ∆U at 298 K? 3. Calculate the difference between heat of reaction at constant pressure and constant volume for the reaction at 25°C in KJ 2C6H6(l) + 15O2(g) → 12CO2(g) + 6 H2O(l). 4. A chemist whilst studying the properties of gaseous C2Cl2F2, a chlorofluorocarbon refrigerant, cooled a 1.25 g sample at constant atmospheric pressure of 1.0 atm from 320 to 293 K. During cooling ∆H and ∆U for the chlorofluorocarbon for this process. For C2Cl2F2, Cp = 80.7/mol–1 K–1.

7.3.5 Heat of combustion The standard heat of combustion is the enthalpy change when one mole of the substance is completely burnt in oxygen. In other words, it is the change in heat content when one gram mole of the substance is completely oxidized. Combustion reaction is always exothermic, thus the heat of combustion, ∆H is always negative. C(s) + O2(g) → CO2(g); ∆H c = –395.5 KJ S(rhombic) + O2(g) → SO2(g); ∆H c = –297.5 KJ CH4(g) + 2O2(g) → CO2(g) + 2H2O(l); ∆H c = –890.3 KJ O

O

O

The heat of combustion depends upon the physical state of substance. The heatof combustion of hydrogen to form liquid water is – 68.4 Kcal (or 286.2 KJ) at ordinary temperature. Actually water is formed first in the form of vapours (gas) and then condenses to form liquid water. The heat of combustion, therefore, also includes latent heat of vapourization, which is 540 × 18 = 9.72 Kcals (40.63 KJ). If the combustion is carried out at or above the boiling of water which will therefore – 68.4 + 9.72 = –58.68 Kcals (–245.5 KJ.) When the substance whose heat of combustion to be measured shows allotropy, the particular allotropic form used gives its heat of combustion. Efficiency of Fuel and Foods The energy released during the combustion of foods and fuels is a measure of their efficiency. Their

Thermochemistry

efficiency is measured in terms of calorific value. Calorific value is: The amount of heat produced in calories (or joules) when one gram of a substance (food or fuel) is completely burnt. The calorific value is usually expressed in Kcal per gram or Kilo joules per gram (1 Kcal = 4.184 KJ). For example, when methane burns, 890.3 KJ mol–1 of energy is liberated.

Solution: The fuel efficiency can be predicted from the amount of heat evolved for every gram of fuel consumed. (i) The combustion of methane is as follows: CH4(g) + 2O2(g) → CO2(g) + 2 H2O(l); ∆CH = 890.3 KJ mol–1 Molar mass of CH4 = 16

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l); ∆CH = –890.3 KJ

890.3 = 55.64 KJ g–1 16 (ii) The combustion of ethane is as follows:

\ Heat produced per gram =

Therefore, calorific value of methane is 890.3 Heat produced per gram = = 55.6 KJ 16 (Molar mass of CH4 = 16) Similarly, octane, a component of gasoline, burns as

C2 H6(g) +

7 O2(g) → 2CO2(g) + 3H2 O(l); 2

∆CH = –1559.7 KJ mol–1 Molar mass of C2H6 = 30

25 C8H18(l) + O2(g) → 8CO2(g) + 9H2O(l); 2

Heat produced per gram =

∆CH = –5460 KJ 5460 = 47.9 KJ 114 (Molar mass of octane = 114)

\ Calorific value of octane =

1559.7 = 51.99 KJ g –1 30

Thus methane has greater fuel efficiency than ethane.

problems of practice

Calorific value of some important foods and fuels are given in Table 7.2. Among the fuels, hydrogen has highest calorific value. But because of safety reason and technical problems it is not commonly used as domestic or industrial fuel. Among food items, fats and carbohydrates are the principal sources of energy which have high calorific value. From the calorific value of different foods it is possible to calculate the food requirement of an average person for balanced diet. On an average, an adult requires 8000 to 12000 KJ of energy per day.

8. In a gobar gas plant, gobar gas is obtained by bacterial fermentation of animal refuse. It is found to contain mainly methane and its heat of combustion given by the equation CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) is 809 KJ mol–1. How much gobar gas would have to be produced per day for a small village community of 100 families, if we assume that each family has to be supplied 20,000 KJ of energy per day to meet all its needs. The methane content in gobar gas is 80% by weight. 9. (a) A cylinder of gas supplied by a company is assumed to contain 14 kg butane. If a normal family requires 20,000 KJ of energy per day for cooking, how long will the cylinder last?

Solved problem 3 The enthalpies of combustion of CH4 and C2H6 are –890.3 and –1559.7 KJ mol–1 respectively. Which of the two has greater efficiency of fuel per gram? table 7.2 Calorific value of some common fuels and foods Fuel Charcoal Wood Dung Cake Kerosene Fuel Oil Hydrogen Methane Butane LPG

7.9

Cal Value KJ/g

Food

33 17 6-8 48 45 150 55 55

Butter Ghee Milk Curd Honey Rice Egg Meat

Calorific Value KJ/g 30.4 37.6 3.2 2.5 13.3 14.5 7.3 12

7.10

10.

11.

12.

13.

Thermochemistry

(b) If the air supplied to the burner is insufficient, a portion of gas escapes without combustion. Assuming that 25% of gas is wasted due to this inefficiency, how long will the cylinder last? Heat of combustion of butane is 2658 KJ mol–1. A gas mixture of 3.67 litres of ethylene and methane on complete combustion at 25°C produces 6.11 litres of CO2. Find out the amount of heat evolved on burning one litre of the gas mixture. The heat of combustion of ethylene and methane are – 1423 and – 891 KJ mol–1 at 25°C (IIT 1991) A person inhales 640 g of O 2 per day. If all the O2 is used for converting sugar into CO2 and H2O, how much sucrose (C12H22O11) is consumed in the body in one day and what is the heat evolved? ∆H (combustion of sucrose) = –5645 KJ mol–1. An athlete is given 100 g of glucose (C6H12O6) of energy equivalent to 1560 KJ. He utilises 50% of this gained energy in the event. In order to avoid storage of energy in the body, calculate the weight of water he would need to perspire. The enthalpy of evaporation of water is 44 KJ mol–1. Ethylene undergoes combustion according to the thermochemical equation C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l) ∆H = –337 Kcal O

Assuming 70% efficiency, how much water at 20°C can be converted into steam at 100°C by the com bustion of 103 litres of C2H4 gas at NTP? 14. An intimate mixture of ferric oxide (Fe2O3) and aluminium (Al) is used as solid rocket fuel. Calculate fuel value per gram and fuel value per cc of the mixture. Heat of formation and densities are as follows: ∆fH (Al2O3) = 399 Kcal mol–1 ∆fH (Fe2O3) = 199 Kcal mol–1 Density of Fe2O3 = 5.2 g/cc Density of Al = 2.7 g/cc 15. The chemical reactions taking place in an oven are (i) C(s) + O2(g) → CO2(g) + 394 KJ (ii) C(s) +

1 O2(g) → CO(g) + 111 KJ 2

Coal used in oven has 80% carbon by mass. Due to insufficient supply of oxygen, only 60% of carbon gets converted into CO2 and 40% to CO. Calculate the amount of heat generated by burning 10 kg of coal. 16. The calorific value of food is determined by measuring the heat produced by burning a weighed sample of food at constant volume in a bomb calorimeter. The heat released by 1 g (3 × 10–3 moles) sample of

chicken fat was found to be 10,000 calories at 310 K under constant pressure. The reaction may be represented as: C20H32O2(s) + 27O2(g) → 20CO2(g) + 16H2O(l) Given R = 1.987 cal mol–1 17. When glucose is oxidised in the body, 40% of the energy evolved in the reaction is available for muscular activity. How much energy is obtained from the oxidation of 4.0 g of glucose?

7.3.6 Enthalpy of phase transitions Energy is required to convert a solid into liquid and a liquid into gas. To break the inter molecular forces in the solid and in the liquid state, energy is required. The change in the physical state of a substance is called phase transition. (i) Enthalpy of Fusion The enthalpy change when 1 mole of a solid substance is converted into its liquid state at its melting point is called enthalpy of fusion. It is also called molar enthalpy of fusion, e.g., when one mole of ice melts at 0°C (or 273 K) the heat absorbed is 6.0 KJ. This corresponds to the process H2O(s) → H2O(l); ∆Fus H = 6.0 KJ mol–1 Ice water O

Here ∆fus H = is standard enthalpy of fusion. If water freezes, then the process is reversed and an equal amount of heat is evolved. Thus, O

∆Freeze H = –∆Fus H \ ∆Freeze H = –6.0 KJ mol–1 O

O

O

Similarly, the enthalpy of fusion of ethyl alcohol at 156 K is +4.8 KJ mol–1 is C2H5OH(s) → C2H5OH(l) at 156 K ∆Fus H = +4.8 KJ mol–1 O

Enthalpy of fusion may also expressed as ∆Fus H = H O

O

liquid

—H

O

solid

Melting of solid is endothermic so enthalpy of fusion is always positive. Enthalpies of fusion are useful in predicting the nature of solid and the magnitude of forces acting between the particles constituting the solid (atoms, molecules or ions). Molecular solids such as O2, H2S, CCl4 etc., have low enthalpies of fusion while ionic solids have high enthalpies of fusion.

Thermochemistry

7.11

Solution: Heat of sublimation of iodine is

molecular Solids ∆FusH of O2 at 55 K is + 0.45 KJ mol–1 ∆FusH of H2S at 188 K is + 2.0 KJ mol–1 ∆FusH of CCl4 at 250 K is + 2.5 KJ mol–1 O

O

O

Ionic Solids ∆FusH of NaCl at 1074 K is + 29.0 KJ mol–1 ∆FusH of MgCl2 at 1260 K is + 43.0 KJ mol–1 O

O

(ii) Enthalpy of Vapourization

H2O(l) → H2O(g) at 373 K; ∆vap H = 40.79 KJ mol–1 O

More the value of enthalpy of vapourization, more the attraction of forces between molecules. (iii) Enthalpy of Sublimation Sublimation is a process in which a solid converts into gas directly without converting into liquid. The enthalpy of sublimation can be defined as the enthalpy change when one mole of a solid is directly converted into its gaseous state at a temperature below its melting point. It is called enthalpy of sublimation or molar enthalpy of sublimation. E.g., for one mole of solid iodine sublimation, the heat change is 62.4 KJ and is represented as I2(s) → I2(g); ∆sub H = + 62.4 KJ. O

The enthalpy of sublimation is equal to the sum of the enthalpies of fusion and vapourization of a substance. ∆sub H = ∆sub H + ∆vap H O

∆CP (Cal/g) = CP of products – CP of reactants = 0.031 – 0.055 = –0.024 Cal/g Now ∆H2 – ∆H1 = CP (T2 – T1) ∆H2 – 24 = –0.024 × (513 – 473) ∆H2 = 24 – 0.96 = 23.04 Cal/g problems for practice

When a liquid is allowed to evaporate, it absorbs heat from the surroundings. In other words, evaporation of a liquid is accompanied by increase in enthalpy. The enthalpy of vapourization may be defined as the enthalpy change when 1 mole of a liquid is converted into gaseous state at its boiling point. It is called enthalpy of vapourization or molar enthalpy of vapourization E.g., when one mole of water is vapourized into gas, 40.6 KJ heat absorbed at 100°C (373 K) is

O

I2(s) → I2(g); ∆H = 24 Cal/g at 473 K.

18. Calculate enthalpy change when 2.38 g of carbon monoxide (CO) vaporizes at its normal boiling point. ∆vapH of CO is 6.04 KJ mol–1. 19. A swimmer coming out from the pool is covered with a thin film of water weighing about 80 g. How much heat must be supplied to evaporate this water at 298 K ∆vapH of water at 373 K = 40.8 KJ mol –1 calculate the internal energy of change of vapourization of water at 100°C. Assume ∆vapH value to be same at 298 K. 20. Standard enthalpy of vapourization of benzene at the boiling point is 30.8 KJ mol–1. For how long would a 100 W electric heater have to operate in order to vaporize a 100 g sample at that temperature? 21. When 1 g of liquid naphthalene (C10H8) solidifies, 149 J of heat is evolved. Calculate the heat of fusion of naphthalene. O

O

O

7.3.7 Enthalpy changes in Solution (i) Heat of neutralization Heat of neutralization of an acid (or base) at a given temperature is defined as the heat change when one gram equivalent of an acid (or base) in dilute aqueous solution is neutralized by one gram equivalent of a dilute aqueous solution of a base (or acid) e.g., HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l) ∆H = –57.10 KJ

O

Solved problem 4 The specific heats of iodine vapour and solid are 0.031 and 0.055 Cal/g respectively. If heat of sublimation of iodine is 24 Cal/g at 473 K, calculate the values of heats of sublimation of iodine at 513 K.

HCl(aq) + LiOH(aq) → LiCl(aq) + H2O(l) ∆H = –57.10 KJ It is interesting to note that enthalpy of neutralization of all strong acids and strong bases is always equal to 57.1 KJ mol–1 irrespective of the nature of acid or the base. This can be explained on the basis of Arrhenius theory of electrolytic dissociation. According to this theory, strong acids and strong bases are completely ionised in aqueous

7.12

Thermochemistry

solutions. The neutralization reactions between strong acids and strong bases in aqueous solutions involve simply the combination of one mole of H+ ions (from an acid) and 1 mole of OH– ions (from a base) to form unionised water molecules. For example, the reaction between hydrochloric acid and sodium hydroxide solution. The neutralization can be represented as H+(aq) + Cl–(aq) + Na+(aq) + OH–(aq) → Na+(aq) + Cl–(aq) + H2O(l) ∆H = –57.1 KJ Omitting the common ions (spectator ions):

Solved problem 5 The heat of neutralization of acetic acid and sodium hydroxide is –50.6 KJ eq–1. Find the heat of dissociation of CH3COOH if the heat of neutralization of a strong acid and strong base is –55.9 KJ eq–1. Solution: ∆H (neutralization) = ∆H (ionization of CH3COOH) + ∆H (H+ + OH–) \ ∆H (ionization of CH3COOH) = –50.6 – (–55.9) = 5.3 KJ eq–1 [∆H ionization of NaOH) = 0 as NaOH is strong base]

H+(aq) + OH–(aq) → H2O(l); ∆H = –57.1 KJ So the heat of neutralization of strong acid by a strong base in diluted solution is almost constant i.e., 57.1 KJ. Thus enthalpy of neutralization may be defined as enthalpy change when one mole of H+(aq) from an acid is completely neutralized by a base or one mole of OH–(aq) from a base is completely neutralized by an acid. Neutralization of weak acid, weak bases: If the acid or base is weak, the heat of neutralization will be different from –57.10 KJ, because the reaction will now involve ionization of weak acid or weak base as well. For example, neutralization of a weak acid, say acetic acid by strong base say, NaOH,  involves both neutralization process (H+ + OH– ↽ ⇀  H2O) and the ionization of weak acid as well.  ⇀  CH3COO– + H+ CH3COOH ↽ Both neutralization as well as ionization of the weak acid proceeds side by side till acetic acid is completely neutralized. The average value of neutralization is –57.1 KJ and the heat of neutralization of CH3COOH by NaOH has been found to be –55.9 KJ. This deviation is due to the fact that some heat is utilized for the dissociation of weak acid into H+ ion. ∆H = –55.9 – (–57.1) = 1.2 KJ This heat (1.2 KJ) is called the heat of ionization or heat of dissociation of acetic acid. Similarly, the neutralization of a strong acid (HCl) by a weak base (NH4OH(aq) which takes place as HCl(aq) + NH4 OH(aq) → NH4Cl(aq) + H2O(l) ∆H = –51.1 KJ This reaction is accompanied by  ⇀  NH+4 + OH– NH4OH ↽ Thus heat of dissociation or ionization of NH4OH is (57.1 – 51.5 = 5.6 KJ). From the discussion made so far it also clear that the heat of dissociation of water is 57.1 KJ.  ⇀  H+ + OH–; ∆H = 57.1 KJ H2O ↽

Solved problem 6 What would be the heat released (i) An aqueous solution containing 0.5 mole of HCl is mixed with an aqueous solution containing 0.3 mole of NaOH. (ii) 200 cm3 of 0.1 M H2SO4 is mixed with 150 cm3 of 0.2 M KOH. Solution: (i) 0.5 mole of HCl = 0.5 mole of H+ ions 0.3 mole of NaOH = 0.3 mole of OH– ions 0.3 moles of OH– ions would react with 0.3 moles of H+ ions to give 0.3 mole H2O \ The heat liberated = 57.1 × 0.3 = 17.1 KJ (ii) 200 cm3 of 0.1 M H2SO4 = 0.1 × 200 = 0.02 mole H2SO4 1000 150 cm3 of 0.2 M KOH =

0.2 × 150 = 0.03 mole KOH 1000

0.02 mole H2SO4 = 2 × 0.02 mole H+ ions = 0.04 mole H+ ions 0.03 mole KOH = 0.03 mole OH– ions 0.03 mole OH– ions would react with 0.03 mole H+ ion to form 0.03 mole H2O. \ The heat liberated = 57.1 × 0.03 = 1.7 KJ (ii) Enthalpy of Solution When a solute is dissolved in a solvent, heat is either evolved or absorbed. Thus dissolution of a solute in a solvent is accompanied by heat change or enthalpy change ∆H of the system. For example, sodium hydroxide dissolves exothermically while sodium nitrate dissolves endothermically. The heat of solution may be defined as The enthalpy change that takes place when one mole of a substance dissolves in a solvent to give an infinitely dilute solution.

Thermochemistry

Or The change in heat content when one gram mole of the substance is dissolved in so much of water that further dilution with water produces no further change in heat content. If heat is absorbed, ∆H is positive and the solution becomes cooler. If heat is evolved, ∆H is negative and the solution becomes warmer. KCl (s) + 200 H2O(l) → KCl (200 H2O) ∆H = + 18.58 KJ The above equation indicates that when one mole of KCl(s) is dissolved in 200 moles of water, the heat absorbed (∆H is positive) is 18.58 KJ. This is known as heat of solution of KCl. It should be noted that heat change per mole of a solute varies with concentration and so internal heat of solution is sometimes used when heat of solution is expressed in terms of concentration. For example, 1. 2. 3. 4. 5.

HCl(g) + 10 H2O → HCl (10 H2O); ∆H = –69.45 KJ HCl(g) + 25 H2O → HCl (25 H2O); ∆H = –72.26 KJ HCl(g) + 40 H2O → HCl (40 H2O); ∆H = –73.01 KJ HCl(g) + 200 H2O → HCl (200 H2O); ∆H = –74.18 KJ HCl(g) + ∞ H2O → HCl (∞H2O); ∆H = –75.14 KJ

The value of ∆H shows the general dependence of the heat of solution on the amount of solvent. As more and more solvent is used, the heat of solution approaches a limiting value, the value in the infinite dilute solution. For HCl this limiting value is –75.14 KJ. When gases like HCl, HBr and NH3 are dissolved in water, the dissolution is accompanied by evolution of heat and ∆H is negative. If there is no chemical interaction ∆H is positive or in rare cases, slightly negative. Hydrated salts and salts which do not form hydrates, (KCl, KNO3, and NaCl, etc) when dissolved in water give a positive value of ∆H. Thus ∆H is positive for salts which do not form hydrates because heat is absorbed when they are dissolved in water. Anhydrous salts (CaCl2, CuSO4 etc) which when they readily combine with water to form hydrates, dissolve with the evalution of heat, i.e., ΔH is negative. The difference in the behaviour of hydrated and anhydrous salt is because of the fact that heat is evolved as heat of hydration in the formation of hydrates. (iii) Heat of dilution If we subtract the equation (1) from (2) given in heat of solution for HCl, we get HCl (10 aq) + 15 aq → HCl (25 H2O); ∆H = ∆H2 –∆H1 = 72.26 – 69.45 = 2.81 KJ This value of ∆H = –2.81 KJ is called the heat of dilution, the heat withdrawn from the surroundings when

7.13

additional solvent is added to solution. The heat of dilution can be defined as The change in enthalpy when a solution containing 1 mole of solute is diluted from one concentration to another. The heat of dilution of a solution depends on the original concentration of the solution and on the amount of solvent added. (iv) Enthalpy of Hydration The enthalpy of a hydrated salt is the change in heat content, when 1 mole of an anhydrous substance combines with requisite number of molecules of water to form the hydrate. The heat of hydration of salt can be calculated if the heats of solution of the anhydrous salt and the hydrated forms are known. For example, the thermochemical equations 1. CuSO4(s) + 800 H2O(l) → CuSO4 (800 H2O); ∆H = –15.89 Kcal (–66.50 KJ) 2. CuSO4.5H2O + 795 H2O(l) → CuSO4 (800 H2O); ∆H = +2.80 Kcal (11.71 KJ) indicate that heats of solution of anhydrous and hydrated copper sulphate are –66.50 and 11.71 KJ respectively Subtracting (2) from (1), we have CuSO4 + 5H2O(l) → CuSO4·5H2O; ∆H = –78.21 KJ (–18.69 Kcal) The heat of hydration of CuSO4·5H2O is – 18.69 Kcal or – 78.21 KJ. Solved problem 7 At 18°C the heat of solution of anhydrous CuSO4 in large volume of water is –15.90 Kcal per mole while that of CuSO4·5H2O is 2.75 Kcal per mole. What is the heat of hydration of CuSO4? Solution: From the data given (i) CuSO4(s) + aq → CuSO4(aq); ∆H = –15.9 Kcal (ii) CuSO4 .5H2O(s) + aq → CuSO4(aq); ∆H = + 2.75 Kcal The heat of hydration of CuSO4 ∆H has to be calculated according to the equation. CuSO4(s)(aq) → CuSO4.5H2O ∆H = –15.9 – 2.75 = –18.65 Kcal

(v) Heat of transition Heat of transition is defined as the change in enthalpy when one mole (or one gram atom, in case the molecular formula is not decided) of the substance transits from one

7.14

Thermochemistry

allotropic form to another. For example, thermochemical equation S(rhombic) → S(monoclinic); ∆H = 334.72 indicates that 80 Kcal or 334.72 KJ is the change in heat content when 32 g of rhombic sulphur change into 32 g of monoclinic sulphur. Other Heats of Reaction: There are many other types of heats of reaction in addition to those mentioned so far. Examples are heats of hydrogenation, polymerization etc. Some organic compounds contain double bonds. The simplest example is ethene C2H4. Such compounds are said to be unsaturated; they contain less than the maximum amount of hydrogen. Ethene, for example, can be converted into ethane, C2H6. The reaction on adding hydrogen to a double bond is known as hydrogenation. The heat change in hydrogenation reaction is the enthalpy of hydrogenation. ∆H H. It is defined as The enthalpy of hydrogenation is the heat change when one mole of an unsaturated compound reacts with hydrogen and is completely changed into the corresponding saturated compound. For example, O

C2H4(g) + H2(g) → C2H6(g); ∆H H = –132 KJ mol O

–1

dependent of the particular manner in which the reaction occurs. It depends only on the initial and final states of the system. Making use of these facts, Hess (1840) enunciated a law, which states that the amount of heat evolved or absorbed in a chemical change is the same whether the process takes place in one or several steps. For example, CO2 can be prepared by two methods 1. C(s) + O2(g) → CO2(g); ∆H = –395.5 KJ 1 2. C(s) + O2(g) → CO(g); ∆H = –113.5 KJ 2 1 CO(g) + O2(g) → CO2(g); ∆H = –282.0 KJ 2 C(s) + O2(g) → CO2(g); ∆H = –395.5 KJ From these two methods, the heat evolved is same –395.5 KJ. Similarly, NaCl may be prepared by two methods: 1 1. Na (s) + H2O(l) → NaOH (s) + H2; ∆H = –140.88 2 1 1 H2(g) + Cl2(g) → HCl(g); ∆H = –92.3 2 2 HCl(g) + NaOH (s) → NaCl(s) + H2O(l); ∆H = –177.82 1 Cl2(g) → NaCl(s); ∆H = –411.0 KJ 2 1 1 2. H2(g) + Cl2(g) → HCl(g); ∆H = –92.3 2 2 1 Na(s) + HCl(g) → NaCl(s) + H2(g); ∆H = –318.7 2 1 Na(s) + Cl2(g) → NaCl (s); ∆H = –411.0 2

Na(s) +

7.4. lAwS oF tHErmocHEmIStry There are two important laws of thermochemistry:

1.the lavoisier and laplace law This law was proposed by Lavoisier and Laplace and states that the amount of heat required to decompose a compound into its elements is equal to heat of formation of that compound from its elements. In other words, the heat evolved (or absorbed) in a particular reaction is equal to that absorbed (or evolved) when the reaction is reversed. An important consequence of this law is that thermochemical equation can be reversed only by changing the sign of the heat evolved or absorbed. For example, the equation C (graphite) + O2(g) → CO2(g); ∆H = –395 KJ may be written as CO2(g) → C (graphite) + O2(g); ∆H = + 395.5 KJ This law can be extended to reactions of all types.

2. Hess’s law of constant Heat Summation We know that U and H are functions of state of the system only, and the same must be true of ∆U and ∆H. Thus the heat evolved or absorbed in a given reaction must be in-

Again it is clear that heat evolved in both the processes is same.

theoretical proof of Hess’s law Suppose a system undergoes change from state A to state B via intermediate states C and D as follows: A → C + x KJ C → D + y KJ D → B + z KJ The total heat liberated in this process = x + y + z KJ. Now suppose the same system undergoes from states A to B via state E. Then, A → E + a KJ E → B + b KJ According to Hess’s law, (a + b) = (x + y + z) KJ. Now suppose (a + b) is greater than (x + y + z), if we pass from A

Thermochemistry

to B via E, (a + b) KJ of heat are gained. But reversing the process if we pass from B to A via D and C, (x + y + z) KJ of heat will be absorbed. In other words, in this cycle, there is a net gain of (a + b) – (x + y + z) KJ of heat. If this process is repeated several times, an indefinite amount of heat can be produced without doing any work or using any other form of energy. But this is against the law of conservation of energy. Hence, (a + b) must be equal to (x + y + z). This law can be used to calculate all sorts of heat changes. It can be used to determine heat changes for reactions that cannot be carried out in the laboratory. For any reaction the overall enthalpy change is given by: Enthalpy change of reaction = [Enthalpies of products – Enthalpies of reactants] If the substance taking part in the reaction and the products are all in their standard states, then we can write this equation in terms of the heats of formation of the reactants and products: Enthalpy change of reaction = [Heats of formation of products – Heats of formation of reactants]

Enthalpy of Impossible reactions The heats of formation of several compounds may be given in chemistry data books. These values are not measured directly by experiment. Many of these values are calculated values using Hess’s law. For example, heats of formation of hydrocarbons cannot be measured directly because hydrogen gas will not combine directly with carbon; yet the heat of formation of methane is quoted as –74.9 KJ mol–1. C(s) + 2H2(g) → CH4(g); ∆H f (CH4) = –74.9 KJ mol–1 O

This value is obtained by using Hess’s law. Reactions for which the enthalpy changes can be measured are the heats of formation of carbon dioxide and water and the heat of combustion of methane. C(s) + O2(g) → CO2(g); ∆H f (CO2) = –393. KJ mol–1 O

1 O2(g) → H2O(l); ∆H f (H2O) = –285.9 KJ mol–1 2 CH4(g) + 2O2(g) → CO2(g) + 2H2O(l); ∆H C (CH4) = –890.4 KJ mol–1

H2(g) +

O

O

The above three reactions have to be written by rearranging so that we get the heat formation of methane. The third reaction is written so that heats of formation are obtained. ∆H C (CH4) = ∆H f (CO2) + 2 ∆H f (H2O) – ∆H f (CH4) O

O

O

O

∆H f (O2) is zero. Substituting these values for the enthalpy changes, O

7.15

–890.4 KJ mol–1 = –393.5 KJ mol–1 + 2 × (–285.9 KJ mol–1) – ∆H f (CH4) = –965.3 KJ mol–1 – ∆H f (CH4) So ∆H f (CH4) = –74.9 KJ mol–1 O

O

O

This method can be applied to many other examples.

Solved problem 8 The heats of combustion of ammonia and hydrogen are 9.06 and 68.9 Kcal respectively. Calculate the heat of formation of ammonia. Solution: 3 1 3 (i) NH3(g) + O2(g) → N2(g) + H2O; 4 2 2 ∆H = –9.06 Kcal (ii) H2(g) +

1 O2(g) → H2O(g); ∆H = –68.9 Kcal. 2

Since combustion reactions are exothermic, ∆H values are given negative signs. ∆H of the following reaction is to be calculated: 1 3 (iii) N2(g) + H2(g) → NH3(g); ∆H = ? 2 2 3 Multiply the equation (ii) with to equate the 2 number of moles of H2 in equations (ii) and (iii) and then subtract equation (i) from equation (ii) 1 3 3 O2(g) → H2O(g)] × ; ∆H = –68.9 Kcal × 2 2 2 3 3 3 H2(g) + O2(g) → H2O(g); ∆H = –103.35 2 4 2

[H2(g) +

To subtract equation (i) from equation (ii); reverse the equation (i) and then add. 3 3 3 H2(g) + O2(g) → H2O(g); ∆H = –103.35 2 4 2 1 3 3 N2(g) + H2O(g) → NH3(g) + O2(g); 2 2 4 ∆H = + 9.06 Kcal 1 3 N2(g) + H2(g) → NH3(g) + ∆H = –94.29 Kcal 2 2 Solved problem 9 Given the following standard heats of reaction at constant pressure: (i) Heat of formation of water = –68.3 Kcal (ii) Heat of combustion of acetylene = –310.6 Kcal

7.16

Thermochemistry

(iii) Heat of combustion of ethylene = –337.2 Kcal. Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25°C). Solution: The given data is 1 1. H2(g) + O2 → H2O(l); ∆H = –68.3 Kcal 2 1 2. C2H2(g) + 2 O2(g) → 2CO2(g) + H2O(l); 2 ∆H = –310.6 Kcal 3. C2H4(g) + 3O2(g) → 2CO2(g) + 2H2O(l); ∆H = –337.2 Kcal ∆E of the following equation is to be calculated: 4. C2H2(g) + H2(g) → C2H4(g); ∆E = ? Sum up the equations (i) and (ii) and subtract equation (iii) 1 H2(g) + O2(g) → H2O(l); ∆H = –68.3 Kcal 2 1 C2H2(g) + 2 O2(g) → 2CO2(g); ∆H = –310.6 Kcal 2 2CO2(g) + 2H2O(l) → C2H4(g) + 3O2(g); ∆H = – +337.2 Kcal C2H2(g) + H2(g) → C2H4(g); ∆H = –41.7 Kcal ∆H is the heat of reaction at constant pressure. Heat at constant volume can be calculated by using the following equation: ∆H = ∆E + ∆ngRT ∆ng = moles of gaseous products –moles of gaseous reactants = 1 –2 = –1 R = 0.002 Kcal/k/mole T = 273 + 25 = 298 K ∆E = –41.7 – (–1 × 0.062 × 298) = –41.104 Kcal.

problems for practice 22. Calculate enthalpy of formation of ethane at 25°C if the enthalpies of combustion of carbon, hydrogen and ethane are 94.14, 68.47 and 373.3 Kcal respectively. 23. The molar heat of combustion of C2H2(g), C(graphite) and H2(g) are 310.62 Kcal, 94.05 Kcal and 68.32 Kcal respectively. Calculate the standard heat of formation of C2H2(g). 24. Calculate the heat of the following reaction at constant pressure: F2 O(g) + H2 O → O2(g) + 2HF(g) The heat of formation of F2O(g), H2O(g) and HF(g) are 5.5 Kcal , 57.8 Kcal and 64.2 Kcal respectively.

25. Calculate the heat of formation of anhydrous Al2Cl6 from the following data: (i) 2Al (s) + 6HCl(aq) → Al2Cl6(aq) + 3H2(g); ∆H = –244 Kcal (ii) H2(g) + Cl2(g) → HCl(g); ∆H = –44.5 Kcal (iii) HCl(g) + aq → HCl(aq); ∆H = –17.5 Kcal (iv) Al2Cl6(s) + aq → Al2Cl6(aq); ∆H = –153.7 Kcal 26. Heat of combustion of ethyl alcohol is 325 Kcal and that of acetic acid is 209.5 Kcal. Calculate the heat evolved in the following reaction: C2H5OH + O2 → CH3COOH + H2O 27. The standard heat of formation at 298 K for CCl4(g), H2O(g), CO2(g) and HCl(g) is –25.5, –57.8, –94.1 and –22.1 Kcal per mol respectively. Calculate ∆H at 298 K for the reaction. CCl4(g) + 2H2O(g) → CO2(g) + 4HCl(g) 28. The standard molar heats of formation of ethane, carbon dioxide and liquid water are –21.1,94.1 and –68.3 Kcal respectively. Calculate the standard molar heat of combustion of ethane. 29. The standard enthalpy of combustion at 25°C of hydrogen, cyclohexene (C6H10) and cyclohexane (C6H12) is –241, –3800 and –3920 KJ/mole respectively. Calculate the heat of hydrogenation of cyclohexene. (IIT 1989) 30. The heat of reaction for O

N2(g) + 3H2(g) → 2NH3(g) at 27°C is –91.94 KJ. What will be its value at 47°C? The molar heat capacities at constant pressure and 27°C for N2, H2 and NH3 are 28.45, 28.32 and 37.07 joule respectively. 31. Calculate ∆H at 358 K for the reaction Fe2O3(s) + 3H2(g) → 2Fe(s) + 3H2O(l) Given that, ∆H = –33.29 KJ mole–1 and Cp for Fe2O3(s), Fe(s), H2O(l) and H2(g) are 103.8, 25.1, 75.3 and 28.8 JK–1 mole–1 32. Only CO and CO2 gases remain after 15.5 g of carbon is treated with 25 litres of air at 25°C and 5.5 atm pressure. Assume air of composition O2–19%, N2–80% and CO2–1% (by volume). Calculate the heat evolved under constant pressure given. C+O2 → CO2; ∆H = –94.05 Kcal/mole C+

1 O2 → CO; ∆H = –26.41 Kcal/mole 2

7.4.1 Enthalpy of Atomization If any substance is given sufficient energy, the bonds holding the atoms together will break and the atoms will separate from

Thermochemistry

one another. The energy involved in the process converting the substance completely into a gas of atoms is called the atomization energy or the heat of atomization ∆H at this can be defined as: the heat of atomization is the enthalpy change when one mole of a substance in its standard state is completely changed into atoms in the gaseous state. For example: H2(g) → 2H(g); ∆H at = + 436 KJ mol–1 HCl(g) → H(g) + Cl(g); ∆H at = + 431 KJ mol–1 Fe(s) → Fe(g); ∆H at = + 417.7 KJ mol–1 Na(s) → Na(g); ∆H at = + 108.4 KJ mol–1 It should be noticed all these values are positive, i.e., the process is that endothermic. We have to supply energy to break bonds. For iron and sodium the process of atomization is a little different from those of hydrogen and hydrogen chloride. The solid metals which already contain atoms are converted into a gas of atoms. Also, the solids are directly converted into gases. This change of solid to gas directly is called sublimation and hence the heats of atomization of solid elements are sometimes called heats of sublimation. For example: Na(s) → Na(g); ∆H sub = + 108.4 KJ mol–1

7.17

methane, the thermochemical equation for its atomization reaction is as follows:

O

CH4(g) → C(g) + 4H(g); ∆H

O

at

= + 1663 KJ mol–1

In methane, the four C–H bonds are identical in bond length and energy. However, the energies required to break the individual C–H bonds in each successive step differ.

O

CH4(g) → CH3 + H(g); ∆H d = + 426.0. KJ mol–1 CH3(g) → CH2 + H(g); ∆H d = + 439.0. KJ mol–1 CH2(g) → CH + H(g); ∆H d = + 451. KJ mol–1 CH(g) → CH + H(g); ∆H d = + 347.0. KJ mol–1 O

O

O

O

O

O

O

7.4.2 Bond Enthalpy (∆bondH ) O

The atomization energy of hydrogen and hydrogen chloride can be given alternative names. In these cases the individual covalent bonds are broken. Alternatively, these energies are also called bond enthalpy or bond dissociation enthalpy ∆H d. Often, the bond dissociation energies are simply called bond energies. The greater the bond energy, the stronger the bond. In the case of polyatomic molecules, bond dissociation enthalpy is different for different bonds within the same molecule. For example, for a polyatomic molecule like O

O

Clearly, the bond energy of a C–H bond depends on the order in which the particular hydrogen atom is lost from the molecule. A similar situation exists for all molecules with more than two atoms: the strengths of bonds depend on the order in which they are broken. The reason for this is that as soon as one atom is lost, the electrons that remain change their energies. Indeed, the shape of the fragment that remains can be very different to the original molecule. For example, the CH3(g) fragment is planar whereas the methane molecule from which it came is tetrahedral. So in the case of polyatomic molecules, the average bond energies are considered. In the case of methane, we determine the total energy required to break all the four hydrogen atoms from the carbon atom. This should be the sum of the individual bond energies above. CH4(g) → C(g) + 4H(g); ∆H at = + 1663 KJ mol–1 O

We then take an average of this value + 1663/4 KJ mol–1 or + 416 KJ mol–1 and call this the average bond energy. It should be noted that the average value does not necessarily match the energy of any one of the individual bonds. We put up with this as a matter of convenience. From now on, we shall use ∆H d for average bond energy. Some typical values are given in Table 7.3. O

table 7.3 Some average single bond enthalpies in KJ mol–1 H

C

N

O

F

Si

P

S

Cl

Br

I

436

414 347

389 293 159

464 351 201 138

569 439 272 184 159

293 289 368 540 176

318 264 209 351 490 213 213

339 259 327 226 230 213

431 330 201 205 255 360 331 251 243

368 276 243 197 289 272 213 218 192

297 238 201 213 213 209 180 157

H C N O F Si P S Cl Br 1

7.18

Thermochemistry

table 7.4 Some average multiple bond enthalpies in KJ mol–1 C=C C≡C C=N C=N C≡N

611 837 615 741 1070

N=N N≡N O=O

418 946 498

Solved problem 10 The bond dissociation energy of gaseous H2, Cl2 and HCl are 104, 58 and 103 Kcal/mole. Calculate the enthalpy of HCl gas. (IIT 1985) Solution: Bond energy of 1 mole of H-H bond = 104 Kcal Bond energy of 1 mole of Cl-Cl bond = 58 Kcal Energy of formation of 2 moles of H – Cl bond = 2 × 103 Kcal Thus ∆H for the reaction H2(g) + Cl2 → 2HCl(g) Is 104 + 58 – 206 = –44 Kcal Since H2 + Cl2 → 2HCl; ∆H = –44 Kcal 1 1 Then for H2 + Cl2 → HCl; ∆H = –22 Kcal 2 2

problems for practice 33. The standard molar enthalpies of formation of cyclohexane (l) and benzene (l) at 25°C are – 156 and + 49 KJ mol –1 respectively. The standard enthalpy of hydrogenation of cyclohexene (l) at 25°C is –119 KJ mol –1. Use the data to estimate the magnitude of the resonance energy of benzene. (IIT 1996) 34. Using the data (all values are in Kcal per mole at 25°C) given below, calculate the bond energy of C–C and C–H bonds. ∆H Combustion (ethane) = –372 ∆H Combustion (propane) = –350 ∆H for C(graphite) → C(g) = 172 Bond energy of H – H = 104 ∆H f of H2O(l) = –68 ∆H f of CO2(g) = –94 (IIT 1990) 35. Find bond energy of C–H bond from the following data: (i) CH4(g) + 2O2(g) → 2H2O(g) + CO2(g); ∆H = –803.3 KJ (ii) C(graphite) → C(g); ∆H = + 711.3 KJ (iii) 2H2(g) → 4H(g); ∆H = + 866.1 KJ

(iv) 2H2O(g) → 2H2(g) + O2(g); ∆H = + 485.3 KJ (v) CO2(g) → C(graphite) + O2; ∆H = + 393.3 KJ 36. Calculate ∆H of the reaction H |

H − C − Cl → C(g) + 2H(g) + 2Cl(g) |

Cl The average bond energies of C–H bond and C–Cl bond are 415.0 KJ mol–1 and 326.0 KJ mol–1 respectively. 37. Using bond energy data, calculate heat of formation of isoprene. 5C(s) + 4H2(g) → H2C = CH3 – CH = CH2 Given C–H = 98.8 Kcal, C – C = 83 Kcal C = C = 147 Kcal and C(s) → C(g) = 171 Kcal 38. Calculate the resonance energy of N2O from the following data: N ≡ N = 946 KJ mole–1 N = N = 418 KJ mole–1 O = O = 498 KJ mole–1 N = O = 607 KJ mole–1 ∆H f of N2O = 82.0 KJ mole–1 39. ∆H for the reaction H H O

|

|

H–C ≡ N(g) + 2H2(g) → H − C − N − H |

H is –150 KJ. Calculate the bond energy of C ≡ N bond (given bond energies of C–H = 414 KJ mol–1; H–H = 435 KJ mol–1; C–N = 293 KJ mol–1; N–H = 389 KJ mol–1) (CBSE 1998) 40. Calculate resonance energy in CH3COOH from the following data if the observed heat of formation of CH3COOH is –439.7 KJ. Heat of Atomization (KJ) C = 716.7 H = 218.0 O = 249.1

O O

Bond Energy (KJ) C – H = 413 C – C = 348 C = O = 732 C – O = 351 O – H = 463

O

O O

41. Compute the heat of formation of liquid methyl alcohol in kilojoules per mole using the following data. The heat of vapourization of liquid methyl alcohol = 38 KJ/mole. The heats of formation of gaseous atoms from the elements in their standard states; H = 218 KJ/ mole; C, 715 KJ/mole; O, 249 KJ/mole. Average bond energies:

C – H = 415 KJ/mole C – O = 356 KJ/mole O – H = 463 KJ/mole

Thermochemistry

42. Calculate the bond energy of C ≡ C in C2H2 from the following data: 1 (i) C2H2(g) + 2 O2(g) → 2CO2(g) + H2O(g); 2 ∆H = –310 Kcal (ii) C(s) + O2(g) → CO2(g); ∆H = –94 Kcal

The amount of energy required to break one mole of ionic crystal and to separate the constituent ions from their position in the crystal to infinite distance but with positive sign to ∆U.

1 O2(g) → H2O(g); ∆H = –68 Kcal 2 Bond energy of C–H bonds = 99 Kcal Heat of atomization of C = 171 Kcal Heat of atomization of H = 52 Kcal

Born–Haber cycle

O

43. The enthalpies for the following reactions (∆H ) at 25°C are given. 1 1 H2(g) + O2(g) → OH–(g); ∆H = 10.06 Kcal 2 2 H2(g) → 2H(g) ∆H = 104.18 Kcal O2(g) → 2O(g) ∆H = 118.32 Kcal Calculate O–H bond energy in the hydroxyl radical. O

O O

44. Bond energies of F2 and Cl2 are respectively 36.6 and 58.0 Kcal per mole. If the heat liberated in the reaction F2+Cl2 → 2ClF is 26.6 Kcal, calculate the bond energy of Cl–F bond.

7.4.3 lattice Energies Many substances are composed of ions, the arrangement of ions in three dimensional array is the lattice. The lattice energy is a measure of the energetic stability of the crystal. The lattice energy ∆HLE is defined as The amount of energy liberated when one mole of gaseous cations and one mole of gaseous anions are brought O

+ INa Cl(g)

+SNa

∆H1

+DCl

together from infinite distance apart to their equilibrium position in the crystal lattice at 0 K i.e., the change in internal energy ∆U Or

(iii) H2(g) +

Na(g)

In 1919, Born and Haber devised a cycle for calculation of the lattice energies by making use of Hess’s Law. The best way of understanding the cycle is to see it at work in a particular case. Here sodium chloride is chosen. The heat of formation of sodium chloride i.e., the heat of reaction 1 Na(s) + Cl2(g) → NaCl(s); ∆Hf NaCl 2 can be calculated by considering all the following energy factors into account. (i) The ionization energy of sodium INa (ii) The energy required to dissociate solid sodium into free atoms which is equal to the heat of sublimation SNa (iii) The dissociation energy of gaseous chlorine DCl (iv) The crystal lattice energy of sodium chloride UNaCl All the above energy factors are interrelated in the Born–Haber cycle which can be summarized in the Fig 7.4.

∆H3 – ECl

∆H4

+ Na(g)

∆H2

∆H5 Na(s) + 1/2 Cl2

7.19

– ∆Hf NaCl

– Cl(g)

– UNaCl

NaCl(s) ∆H Fig 7.4 Born–Haber cycle for calculation of lattice energy of NaCl

Thermochemistry

∆H f NaCl 1 NaCl(s); Na(s) + Cl 2 (g) 2 ∆H ∆H = –410 J KJ mol–1 The heat liberated in this reaction is called heat of formation of NaCl. Since one mole of NaCl is from the constituent elements in their standard state, the heat involved in this reaction is –410.5 KJ Mol–1. This reaction can be assumed to take place in several steps, which are as following: (i) The heat absorbed in the conversion of solid sodium to gaseous sodium is equal to its sublimation energy (+S). The sublimation energy of sodium is 108.7 KJ mol–1.

∆H = ∆H1 + ∆H2 + ∆H3 + ∆H4 + ∆H5 ∆HfNaCl = +SNa + INa +

1 DCl2 – ECl + UNaCl 2

Substituting the experimental values of energy changes in various steps and solving for U, we get U = –769.97 KJ Mol–1 The energy changes in the formation of sodium chloride can be shown as enthalpy diagram for the cycle. Na+(g) + Cl(g)

O

1/2∆ Hat

-348.6 KJ mol-1

The thermochemical equation for formation of solid sodium chloride from its constituent element can be written as

+119.55 KJ mol-1

7.20

Na+(g) + 1/2Cl2(g)

1 DCl2 1 Cl2 (g) 2 Cl(g); 2 ∆H 3 ∆H3 =

1 DCl2 = + 119.55 KJ mol–1 2

(iv) Chlorine atoms convert into chloride ions by gaining an electron due to which energy liberated is equal to its electron affinity Cl(g) + e–

eg Cl ∆H 4

Cl(g); ∆H 4 = − eg Cl = −361.60 KJ mol–1

(v) The gaseous sodium and chloride ions come close together to from solid crystalline NaCl. The energy released in this process is called crystal lattice energy (–U). Crystal lattice energy cannot be measured directly. So it can be calculated by using Born–Haber cycle. Applying Hess’s law of constant heat summation to the above sequence of reactions, the heat of formation of NaCl(s) is given by

+108. KJ mol-1

(iii) Chlorine molecules should break up into chlorine atoms for which the energy absorbed is equal to its bond energy. To break half mole of Cl2 molecules into atoms, the energy absorbed is equal to half its bond energy i.e., +239/2 = + 119.55 KJ

∆i H

O

O

Na+(g) + Cl-(g)

Na(g) + 1/2Cl2(g)

O

∆sub H

∆

Na(s) + 1/2Cl2(g)

- 410. KJ mol-1

(ii) Gaseous sodium atoms ionise absorbing an energy equal to its ionization energy of 492.82 KJ mol–1 +INa Na(g) Na + (g); ∆H2 = 492.82 KJ mol -1 ∆H2

+492.82 KJ mol-1

Na(s) → Na(g); ∆H1 = –108.7 KJ mol–1

∆eg H

∆f H

O

NaCl(s)

Fig 7.5 Enthalpy diagram for lattice enthalpy of NaCl We can calculate lattice energy for different ionic compounds by using the formula. –(Lattice energy) = –(heat of formation ) + heat of atomization + ionization energies + electron affinities.

utility of lattice Engines First the more negative the value, the greater is the energetic stability of the lattice with respect to it being broken up into separate ions. However, we should be careful not to make the mistake of thinking that the lattice energy tells us whether the substance really is held together by ionic compound. The values only give information about how much energy is released if the crystal were made from gaseous ions. To some extent it is possible to check the degree of ionic character in a crystal by performing a calculation. The method of calculation was developed

Thermochemistry

by Born, Lande and Mayer. The details are too complicated but the idea behind the method is not so hard to understand. A crystal is assumed to be made up by perfectly spherical ions. Once the geometry of the crystal has been determined, the distance between the ions are known, the energy of attraction between all oppositely charged ions and the energy of repulsion between ions of the same charge are calculated by using Born–Lande equation. ∆HLE = 1.389 × 105 O

Mz+ z 1 (1 - ) KJ mol -1 r n

Here r is the shortest distance between oppositely charged ions, measured in picometres (1 picometre, 1 pm =10–12m); z+ is the number of charges on the positive ions and z– is the number of charges on negative ions. n is the factor that takes account of the repulsion that occurs, when any two ions come close together by the electron clouds of the two ions. M is called as the Made-lung constant. It usually but not always has values in the region of 1 to 5. Born-Mayer equation makes more sophisticated allowance for the repulsions between the electron clouds of any type of two ions. By including another variable ρ(rho), the Born-Mayer equation. ∆HLE = 1.389 × 105 O

Mz+ z r

 ρ -1  1 -  KJ mol r

The result obtained for calculating the lattice energies calculated for compounds like NaCl are in good agreement with the values calculated using Born-Haber cycle indicating that sodium chloride is ionic in nature. On the other hand, some substances show significant difference between the lattice energies calculated by the Born-Lande or Born–Mayer equation and Born–Haber cycle. Particularly, the well known examples are the silver halides. table 7.5 Comparison of the lattice energy of the silver halides Halide

Lattice Energy KJ mol–1 Born–Haber cycle

AgF AgCl AgBr AgI

–943 –899 –877 –867

Born–Mayer equation –925 –833 –808 –774

be made of separate individual ions. It seems that there is a significant covalent character in the solid. Silver ion (Ag+) with pseudo electronic configuration have made more polarizing power while the bigger iodide ions has more polarizability. So according to Fajan’s rules, Ag+ ion polarizes the I– ion producing a certain amount of covalent character.

I– Ag+

I–

Ag+

Ag+

I–

Fig 7.6 Polarization in silver iodide So it is not correct in making a hard and fast generalisation about the bonding in substances such as silver iodide. For example along with other ionic compounds, silver iodide has high melting point (558°C) and it does not conduct electricity when molten. Solved problem 11 Calculate the lattice enthalpy of KCl from the following data at standard states (1 bar): Enthalpy of sublimation of potassium = 89 KJ mol–1 Enthalpy of dissociation of chlorine = 244 KJ mol–1 Ionization enthalpy of potassium = 425 KJ mol–1 Electron gain enthalpy of chlorine = –355 KJ mol–1 Enthalpy of formation of KCl = –438 KJ mol–1 Solution: The enthalpy changes for the process are K(s) → K(g); ∆H 1 = 89 KJ mol–1 O

1 Cl2(g) → Cl(g); ∆H 2 = 122 KJ mol–1 2 O

K(g) → K+(g) + e–(g); ∆H 3 = 425 KJ mol–1 Cl(g) + e–(g) → Cl–; ∆H 4 = –355 KJ mol–1 K+(g) + Cl–(g) → KCl (s); ∆H 5 = –438 KJ mol–1 O

The differences between the value increase from the fluoride to the iodide. Especially the discrepancy is so large for silver iodide. This suggests that it is unlikely to

7.21

O

O

7.22

Thermochemistry

According to Hess’s law ∆fH = ∆H 1 + ∆H 2 + ∆H 3 + ∆H 4 + U0 O

O

O

O

O

1 × 244) –355 + U0 –438 = 281 + U0 2 Or U0 = –438 – 281 = –719 KJ mol–1 Lattice energy of KCl = 719 KJ mol–1

–438 = +89 + 425 + (

problems for practice 45. Calculate the lattice enthalpy of CaCl2 given that the enthalpy of (i) Sublimation of Ca = 121 KJ mol–1 (ii) Dissociation of Cl2 to Cl = 242.8 KJ mol–1 (iii) Ionisation of Ca to Ca2+ = 2422 KJ mol–1

(iv) Electron gain enthalpy for Cl to Cl– = –355 KJ mol–1 (v) ∆fH overall = –795 KJ mol–1 O

46. The thermochemical data for the formation of argon chloride (ArCl) are given below 1 Ar(g) + Cl2(g) → ArCl(g) 2 (i) Enthalpy of dissociation of chlorine = –243.0 KJ (ii) Electron gain enthalpy of chlorine = –348.3 KJ (iii) Ionization enthalpy of argon = 1526.3 KJ (iv) Lattice enthalpy of argon chloride = –703.0 KJ Predict whether the formation of argon chloride is favourable or unfavourable.

Thermochemistry

7.23

KEy poIntS Exothermic and Endothermic reaction •

•

• •

•

The chemical reactions accompanied by liberation of heat are called exothermic reactions. In exothermic reactions, the enthalpy change ∆H is negative. The chemical reactions accompanied by the absorp tion of heat are called endothermic reactions. In endothermic reactions, the enthalpy change ∆H is positive. The energy change that occurs in a chemical reaction is usually known as chemical energy. The energy change in a chemical reaction is mainly due to the change in potential energy that results due to the breaking of bonds in reactants and formation of new bonds in products. The heat change that take place at constant volume is the change in internal energy, (∆U) while the heat change that take place at constant pressure is the enthalpy change (∆H).

  m (t - t ) W =  2 2 3 - m1  (t t )   3 1 W is water equivalent of calorimeter, m1 mass of water at lower temperature t1, m2 is the mass of water at high temperature by mixing the two water samples of m1 and m2 at t1 and t2. Heat liberated = (w + volume of reaction mixture) × rise in temperature

thermochemical Equation •

•

•

measuring Heats of reaction •

•

•

•

•

The technique of measuring heat of reactions is called calorimetric and the instrument used is called calorimeter. For combustion reactions, bomb calorimeter is used. Bomb is made with non-oxidizable material. In bomb calorimeter, combustion reactions are carried out at constant volume. The heat liberated during the combustion reactions in bomb calorimeter is calculated using the formula M ∆U = –Z∆T × W Z is heat capacity of calorimeter system, ∆T is rise in temperature, M is molecular mass of substance, W is mass of substance taken. Measurements in an open calorimeter determine: the amount of heat liberated or absorbed when known amounts of substance react at constant pressure. Since the heat liberated in bomb calorimeter is at constant volume qv it can be converted into constant pressure by using the equation qp = qv + ∆nRT

•

n is the change in the number of gaseous molecules in the combustion reaction. Heat liberated during neutralization, dilution, phase, transition, etc., is measured in second type of calorimeter where the rise in temperature of water is measured in calorimeter or Dewar flask.

•

Thermochemical equation is the equation in which the heat change accompanying the chemical reaction is also specified numerically. Unless otherwise mentioned, ∆H values are given for standard state of substance i.e., when reactions occur at 298 K (25°C) and standard atmospheric pressure (1 atm). The most stable physical state of a substance at 298 K and 1 atmospheric pressure is called its standard state. For indicating physical state of each substance in a chemical equation, designation such as ‘g’ for gas, ‘l’ for liquid, ‘s’ for solid and ‘aq’ for aqueous solution are given along with chemical formula of reactants and products e.g., C graphite + O2 (g) → CO2 (g); ∆H = -393.5 KJ HCl (aq) + NaOH(aq) → NaCl (aq) + H2 O(l); ∆H = -57.3 KJ

Enthalpies of different processes • •

•

•

•

The enthalpy content of an element in its standard state is zero. The standard heat of formation is the enthalpy change when one mole of the substance is made from its element in their standard states. Compounds formed from their elements with evolution of heat are called exothermic compounds whereas the compounds formed from their elements with absorption of heat are called endothermic compounds. The enthalpy of a reaction is the enthalpy change (amount of heat evolved or absorbed) in a reaction when the number of moles of reactants react completely to give the products as given by the balanced chemical equation. The enthalpy of formation of compound is the standard enthalpy of the compound since the standard enthalpies of the element from which it is formed is zero.

7.24

• •

•

Thermochemistry

Exothermic substances are stable while endothermic substances are unstable. In endothermic reactions, the sum of the internal energies of the products is greater than the sum of the internal energies of reactants, while in exothermic reactions, the sum of the internal energies of products is less than the sum of the internal energies of reactants. The change in enthalpy of reaction at constant pressure per degree change of temperature is equal to the difference in the heat capacities of the products and the reactants at constant pressure and is given by Kirchoff’s equation ∆Cp =

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∆H 2 - ∆H1 T2 - T1

Where ∆Cp is difference in the heat capacities of products and reactants at constant pressure. The change in heat of reaction at constant volume per degree change of temperature is given by the difference in the heat capacities of the products and the reactants at constant volume. The standard heat of combustion is the enthalpy change when one mole of the substance is completely burnt in oxygen. Calorific value of fuels and foods can be determined by heat of combustion. Calorific value is the amount of heat produced in calories (or Joules) when one gram of a substance (food or fuel) is completely burnt. The enthalpy change when 1 mole of a solid substance is converted into its liquid state at its melting point is called enthalpy of fusion, also called molar enthalpy of fusion. The enthalpy change when 1 mole of a liquid is converted into gaseous state at its boiling point is called enthalpy of vapourization or molar enthalpy of vapourization. The enthalpy change when one mole of a solid is directly converted into its gaseous state at a temperature below its melting point is called enthalpy of sublimation or molar enthalpy of sublimation. Heat of neutralization of an acid (or base) at a given temperature is the heat change when one gram equivalent of an acid (or base) in dilute aqueous solution is neutralised by one gram equivalent of a dilute aqueous solution of a base (or acid). The enthalpy of neutralization of all strong acids and strong bases is always equal to 57.1 KJ mol–1 irrespective of the nature of acid or the base. The enthalpy of neutralization is nearly the heat of formation of water from H+ of an acid and OH– of a base H+(aq) + OH–(aq) → H2O(l); ∆H = –57.1 KJ

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The enthalpy of neutralization of a strong acid and strong base is always constant because they ionise completely. The enthalpy of neutralization involving a weak acid or weak base is less than 57.1 KJ since some energy is used to dissociate the weak electrolytes. If dilute acids and bases are used, the rise in temperature is less because the volume is more, the heat will be distributed. The amount of heat required for one mole of a simple molecule in the gaseous state to break into its constituent atoms is called enthalpy of atomization. The enthalpy change in the formation of an ion at unit activity (or concentration) from its elements in aqueous solution is called enthalpy of ionization. Since absolute enthalpies of ionization cannot be determined, they are calculated by taking the enthalpy of H+(aq) at 298 K as zero. Enthalpy of solution is the quantity of heat liberated or absorbed when one mole of solute is dissolved completely in large excess of the solvent, So that further dilution will not produce any heat change. Enthalpy of solution depends on the concentration of solute. The enthalpy change per mole of solute when it is dissolved in a pure solvent in the required volume to give a solution of specified concentration is called integral heat of solution. Enthalpy of dilution is the enthalpy change when a solution containing one mole of solute is diluted from a known concentration to another concentration. Enthalpy of hydration is the enthalpy change when 1 mole of an anhydrous substance combines with requisite number of molecules of water to form the hydrate. The heat of hydration is the difference in the heats of solutions of the anhydrous salt and the hydrated salt. Heat of transition is the change in enthalpy when one mole (or one gram atom, in case the molecular formula is not decided) of the substance transits from one allotropic form to another. The enthalpy of hydrogenation is the heat change when one mole of an unsaturated compound reacts with hydrogen and is completely changed into the corresponding saturated compound.

laws of thermochemistry •

Lavoisier and Laplace law states that the amount of heat required to decompose a compound into its elements is equal to heat of formation of that compound from its elements.

Thermochemistry

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Hess’s Law states that the heat of reaction is same whether the reaction take place in one step or in several steps (or) the enthalpy change involved in a reaction is the same whether the reaction take place in one single step or in several steps (or) the amount of heat evolved or absorbed in a chemical reaction or process is the same whether the reaction take place in a single or several steps. Mathematically, Hess’s Law can be represented for a reaction A gives D in a single step or in three steps:

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Single step A → D + Q Three steps A → B + q1 B → C + q2 C → D + q3 Q = q1 + q2 + q3

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According to Hess’s Law, the heat of reaction depends only on the initial reactants and final products but not the intermediate products formed. The heat of reaction is also independent of the path or manner by which a reaction is brought. Using Hess’s Law, the heat of formation of intermediate compounds which cannot be determined experimentally can be calculated. The heat of formation of compounds like CO, benzene, etc., which cannot be determined directly can be calculated by using Hess’s Law.

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7.25

Heat changes of slow reactions can be calculated using Hess’s law. The heat of transition of α– sulphur to b– sulphur which is slow can be calculated from their heats of combustion. Heats of formation or enthalpy of formation of certain compounds such as C2H4 which require drastic conditions can be calculated using Hess’s Law. Determination of lattice energy of ionic compounds using Born–Haber cycle is based on Hess’s Law. Using Hess’s Law, bond energies in some compounds can be calculated. Bond enthalpy or bond dissociation enthalpy is the amount of energy required to break the covalent bonds in one mole of diatomic molecules such as H2, Cl2, HCl etc. In the case of polyatomic molecules, bond dissociation enthalpy is the average bond energies. The amount of energy liberated when one mole of gaseous cations and one mole of gaseous anions are brought together from infinite distance apart to their equilibrium position in the crystal lattice at 0 K, i.e., the change in internal energy ∆U or the amount of energy required to break one mole of ionic crystal and to separate the constituent ions from their position in the crystal to infinite distance but with positive sign to ∆U.

7.26

Thermochemistry

prActIcE ExErcISE multiple choice Questions with only one Answer level I 1. The standard heat of combustion of solid boron is equal to (a) 1 ∆H f (B2O3) 2 O

(b) ∆H f (B2O2) O

C + O2 → CO2; ∆H = –XKJ O

2CO + O2 → 2CO2; ∆H = –YKJ The enthalpy of formation of carbon monoxide will be (a) (2X – Y)/2 (b) (y – 2X)/2 (c) 2X – Y (d) Y – 2X Which of the following salts shall cause more cooling when one mole of salt is dissolved in the same amount of water? (Integral heat of solution at 298 K is given for each solute) (a) KNO3; ∆H = 35.4 KJ/mol (b) NaCl; ∆H = 5.35 KJ/mol (c) KOH; ∆H = –55.6 KJ/mol (d) HBr; ∆H = –83.3 KJ/mol Equal volumes of one molar hydrochloric acid and one molar sulphuric acid are neutralized completely by dilute NaOH solution and X Kcal and Y Kcal of heat are liberated, respectively. Which of the following is true? (a) X = Y (b) X = Y/2 (c) X = 2Y (d) None of these The heat evolved in the combustion of glucose, C6H12O6 is – 680 Kcal/mol. The weight of CO2 produced, when 170 Kcal of heat is evolved in the combustion of glucose is (a) 264 g (b) 66 g (c) 11 g (d) 44 g One gram sample of NH4NO3 is burnt in a bomb calorimeter. The temperature of the calorimeter increases by 6.12 K. The heat capacity of the calorimeter system is 1.23 KJ/deg. What is the molar heat of combustion for NH4NO3? (a) 7.53 KJ/mol (b) 813 KJ/mol (c) –602 KJ/mol (d) 602.2 KJ/mol C12H22O11(s) + 12O2(g) → 12CO2(g) + 11H2O(l); ∆H = –5.65 × 103 KJ. Complete combustion of 1 kg sucrose is done. Heat evolved will be (a) 1.65 × 104 KJ (b) 5.25 × 103 KJ 5 (c) 7.38 × 10 KJ (d) 3.51 × 105 KJ O

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From the data, the enthalpy change for the reaction is  2 → CH3 – CH3 + CH2 = CH2 2CH3 – CH

O

(d) - 1 ∆H f (B2O3) 2

O

(c) –∆H f (B2O3) 2. Given that

8. The data below refers to gas phase reaction at constant pressure at 25°C –1   CH3 – CH3 → CH 3 - CH 2 + H ; ∆H1 = + 420 KJ mol –1   CH3 – CH2 → CH2 = CH2 + H ; ∆H2 = +168 KJ mol

(a) +250 KJ (b) +588 KJ (c) –252 KJ (d) –588 KJ 9. Study the following thermochemical data: S+O2 → SO2; ∆H = –298.2 KJ SO2+ 1 O2 → SO3; ∆H = –98.2 KJ 2 SO3+ H2O → H2SO4; ∆H = –130.2 KJ H2 + 1 O2 → H2O; ∆H = –287.3 KJ 2

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The enthalpy of formation of H2SO4 at 298 K will be (a) –433 KJ (b) –650.3 KJ (c) +320.5 KJ (d) –813.5 KJ The lattice energy of solid NaCl is 180 Kcal/mol. The dissolution of the solid in water, in the form of ions is endothermic to the extent of 1 Kcal/mol. If the solvation energies of Na+ and Cl– ions are in the ratio 6:5, what is the enthalpy of hydration of solution of sodium ion? (a) –85.6 Kcal/mol (b) –97.6 Kcal/mol (c) 82.6 Kcal/mol (d) 100 Kcal/mol Heat of neutralization of oxalic acid is –25.4 Kcal/mol using strong base, NaOH. Hence enthalpy in process H2C2O4 → 2H+ + C2O42– is (a) 2.0 Kcal (b) –11.8 Kcal (c) 1.0 Kcal (d) –1.0 Kcal The standard heat of combustion of propane is –2220.1 KJ/mol. The standard heat of vapourization of liquid water is 44.0 KJ/mole. What is the ∆H of C3H8(g) + 5O2 → 3CO2(g) + 4H2O(g)? (a) –2220.1 KJ (b) –2044.1 KJ (c) –2396.1 KJ (d) –2176.1 KJ A geyser operating on LPG (liquefied petroleum gas) heats water flowing at the rate of 3.0 litres per minute, from 27°C to 77°C. If the heat of combustion of LPG is 40000 J/g, how much fuel in g is consumed per minute? (a) 15.25 (b) 15.50 (c) 15.75 (d) 16.00 O

13.

Thermochemistry

14. Benzene burns in oxygen according to the following reaction: C6H6(l) + 7 1 O2(g) → H2O(l) + 6CO2(g) 2 If the standard heats of formation (∆H ) of C6H6(l), H2O(l) and CO2(g) are 11.7, –68.3 and –94 Kcal/mole, respectively. What is the amount of heat that will liberate by burning 1 kg of benzene? (a) 10007 Kcal (b) 780.6 Kcal (c) 4683 Kcal (d) 1000 Kcal Under identical conditions, how many mL of 1M–KOH and 2M–H 2SO4 solutions are required to produce a resulting volume of 100 mL with the highest rise in temperature? (a) 80, 20 (b) 20, 80 (c) 60, 40 (d) 50, 50 The heat capacity of bomb calorimeter is 500 J/°C. A 2°C rise in temperature has been observed on the combustion of 0.1 g of methane. What is the value of ∆E per mole of methane? (a) 1 KJ (b) 160 KJ (c) –160 KJ (d) –1 KJ Heat evolved in the reaction H2 + Cl2 → 2HCl is 182 KJ. Bond energies of H–H = 430 KJ/mole Cl–Cl = 242 KJ/mole. The H–Cl bond energy is (a) 763 KJ/mole (b) 245 KJ/mole (c) 336 KJ/mol (d) 427 KJ/mole The heat of hydrogenation of benzene is –51.0 Kcal/mole. If heat of hydrogenation of cyclohexa diene and cyclohexene is –58 Kcal/mole and –29 Kcal/mole respectively, what is the resonance en ergy of benzene? (a) 29 Kcal/mole (b) 36 Kcal/mol (c) 58 Kcal/mol (d) 7 Kcal/mol ∆H f for NF3 is –113 KJ/mol. Bond energy for the N–F bond is 273.5 KJ/mol. Calculate the bond energies of N2 and F2 if their magnitudes are in the ratio 6:1. (a) 822.6, 137.1 KJ/mole (b) 979.8, 163.3 KJ/mol (c) 943.32, 157.22 KJ/mol (d) 762.6, 127.1 KJ/mol AB, A2 and B2 are diatomic molecules. If the bond enthalpies of A2, AB and B2 are in the ratio 2:2:1 and enthalpy of formation AB from A2 and B2 is – 100 KJ mole–1, what is the bond energy of A2? (a) 200 KJ mol–1 (b) 100 KJ mol–1 (c) 300 KJ mol–1 (d) 400 KJ mol–1 O

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7.27

21. Among the following, for which reaction heat of reaction represent bond energy of HCl? (a) HCl(g) → H(g) + Cl(g) (b) 2HCl(g) → H2(g) + Cl2(g) (c) HCl(g) → 1 2 H2(g) + 1 2 Cl2(g) (d) HCl(g) → H+(g) + Cl–(g) 22.

H +H−H

H H H H

H The bond energies at 25°C are (1) C–C; 346 KJ mol–1 (2) C–H; 413 KJ mol–1 (3) H–H; 437 KJ mol–1 (4) C = C; 611 KJ mol–1 From this data, calculate the value of ∆H at 25°C for the above reaction (a) –289 KJ mol–1 (b) –124 KJ mol–1 (c) 124 KJ mol–1 (d) 289 KJ mol–1 23. Which of the following statement is incorrect about internal energy? (a) The absolute value of internal energy cannot be determined. (b) The internal energy of one mole of a substance is same at any temperature and pressure. (c) The measurement of heat change during a reaction by bomb calorimeter is equal to the internal energy change. (d) Internal energy is an extensive property. 24. Use the following data to calculate the enthalpy of hydration for caesium iodide and caesium hydroxide, respectively Compound

Lattice energy

∆H solution

CsI + 604 KJ/mol 33 KJ/mol CsOH + 724 KJ/mol –72 KJ/mol (a) –571 KJ/mol and –796 KJ/mol (b) 637 KJ/mol and 652 KJ/mol (c) –637/mol and –652 KJ/mol (d) 571 KJ/mol and 796 KJ mol 25. The difference between heats of reaction at constant pressure and constant volume for the reaction 2C6H6(l)+15O2(g)→12CO2(g)+H2O(l) at 298 K in KJ (a) –7.43 (b) +3.72 (c) –3.72 (d) +7.43 26. Molar heat capacity of water in equilibrium with ice at constant pressure is (a) zero (b) infinity (c) 40.45 KJ/K – mol (d) 75.48 J/K– mol

7.28

Thermochemistry

27. Standard molar enthalpy of formation of CO2 is equal to (a) Zero (b) The standard molar enthalpy of combustion of gaseous carbon (c) The sum of standard molar enthalpies of formation of CO and O2 (d) The standard molar enthalpy of combustion of carbon (graphite) 28. The standard enthalpy of atomization of PCl3(g) is 195 Kcal/mol. What will be standard enthalpy of atomization of PCl5(g), if the bond dissociation energies of axial P–Cl bonds in PCl 5(g) are 10% lesser and the bond dissociation energies of equatorial P–Cl bonds in PCl5(g) are 10% higher than the bond dissociation energies of P–Cl bonds in PCl 3(g) (a) 195 Kcal/mol (b) 325 Kcal/mol (c) 331.5 Kcal/mol (d) 318.5 Kcal/cal

multiple choice Questions with only one Answer level II 1. The bond energies of C ≡ C, C–H, H–H and C = C are 198, 98, 103 and 145 Kcal, respectively. The enthalpy change of the reaction HC ≡ CH + H2 → C2H4 is (a) 48 Kcal (b) 96 Kcal (c) –40 Kcal (d) –152 Kcal 2. The heat evolved in the combustion of 112 litre of water gas at NTP (equal volume of CO and H2) is H2(g) + ½ O2(g) = H2O(l); ∆H = –241.8 KJ CO(g) + ½ O2(g) = CO2(g); ∆H = –283 KJ (a) 241.8 KJ (b) 283 KJ (c) 1312 KJ (d) 1586 KJ 3. 1.2 g of carbon is burnt completely in oxygen (limited supply) to produce CO and CO2. This mixture of gases if treated with solid I2O5 (to know the amount of CO produced), the liberated iodine required 120 mL of 0.1 M hypo solution for completed titration. The % of carbon converted into CO is (a) 60% (b) 100% (c) 50% (d) 30% 4. The standard enthalpy of formation (∆fH ) at 298 K for methane, CH4(g), is –74.8 KJ mol–1. The additional information required to determine the average energy for C–H bond formation would be (a) Latent heat of vapourization of methane (b) The first four ionization energies of carbon and electron gain enthalpy of hydrogen (c) The dissociation energy of hydrogen molecule H2 (d) The dissociation energy of H2 and enthalpy of sublimation of carbon O

5. The enthalpy of combustion of methane gas in terms of given data bond energy (KJ/mol) eC - H , e O - H , x1 x2

eC = O , e O = O x3 x4

Resonance energy of CO2 (g) = Y KJ/mol ∆Hvapourization H2O, (l) = Z KJ/mol (a) 4X1 + 2X4 – 2X3 – 4X2 + Y+Z (b) 4X1 + 2X4 – 2X3 – 4X2 + Y – 2Z (c) 4X1 + 2X4 – 2X3 – 4X2 – Y + 2Z (d) 4X1 + 2X4 – 2X3 – 4X2 – Y – 2Z 6. Select the option in which heat evolved is maximum ∆fH (CO2 g) = –75 Kcal/mol ∆fH (CO, g) = –25 Kcal/mol The product will be CO if excess amount of carbon and CO2 are present. If excess of O2 is present what are the products? (a) 10 moles of carbon and 4.5 moles of O2 (b) 24 g of carbon and 64 g O2 (c) 4 moles of carbon and 3.5 moles of O2 (d) 30 g of carbon and 80 g of O2 O O

7. S(s) + 3 O2(g) → SO3(g) + 2x Kcal 2 1 SO2(g) + O2(g) → SO3(g) + y Kcal 2 Find out the heat of formation of SO2(g). (a) (y – 2x) (b) (2x + y) (c) (x + y) (d) 2x/y 8. Read the following statements I and II carefully and select the right option. (I) The solution of CaCl2·6H2O in a large volume of water is endothermic to the extent of 3.5 Kcal/mol. If ∆H = –23.2 Kcal for the reaction CaCl2(s) + 6H2O(l) → CaCl2.6H2O(s) (II) For the reaction 2Cl(g) → Cl2(g); the signs of ∆H and ∆S are negative (a) Statement I and II both are wrong (b) Both are correct (c) Only I is correct (d) Only II is correct 9. At 300 K, the standard enthalpies of formation of C6H5COOH(s), CO2(g) and H2O(l) are –408, –393 and –286 KJ mol–1 respectively. The heat of combustion of benzoic acid at constant pressure and constant volume are respectively (R = 8.31 J mol–1) (a) –3201 KJ mol–1 and –3199.7535 KJ mol–1 (b) +3201 KJ mol–1 and +3199.7535 KJ mol–1 (c) –2103 KJ mol–1 and –3799.3175 KJ mol–1 (d) +2103 KJ mol–1 and +399.3175 KJ mol–1 10. The heat of hydrogenation for 3– methylbutene and 2– pentene are –30 Kcal/mol and –28 Kcal/mol, respectively. The heats of combustion of 2–methylbutane and pentane

Thermochemistry

are –784 Kcal/mol and –782 Kcal/mol, respectively. All the values are given under same conditions?

(a) 2 (b) –4 (c) 4 11. C + ½ O2 → CO; ∆H = –110 KJ

(d) –2 19.

C(s) + H2O(g) → CO(g)+ H2(g); ∆H = 132 KJ

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Calculate the mole composition of the mixture of steam and oxygen on being passed over coke at 1000°C, keeping temperature constant. (a) 1 : 0.4 (b) 0.4 : 1 (c) 1 : 0.6 (d) 0.6 : 1 The heat of solution of anhydrous CuSO4 is –15.9 Kcal and that of CuSO4 is –15.9 Kcal. The heat of hydration of CuSO4 is (a) –18.7 Kcal (b) 18.7 Kcal (c) –13.1 Kcal (d) 13.1 Kcal The standard enthalpy of combustion of H2, C6H10 and C6H12 are –240, –3800, –3920 KJ/mol The heat of hydrogenation of cyclohexene to cyclohexane is (a) –100 KJ (b) 100 KJ (c) –120 Kcal (d) 120 KJ HA+ NaOH → NaA + H2O; ∆H = –12 Kcal HB + NaOH → NaB + H2O; ∆H = –11 Kcal Equimolar solution of which acid has higher pH? (a) HA (b) HB (c) Both have same pH (d) Data insufficient In the reaction CS2(l) + 3O2(g) → CO2(g) + 2SO2(g) ∆H = –265 Kcal. The enthalpies of formation of CO2 and SO2 are both negative and are in the ratio 4:3. The enthalpy of formation of CS2 is + 26 Kcal/mol. Calculate the enthalpy of formation of SO2 (a) –90 Kcal/mol (b) –52 Kcal/mol (c) –78 Kcal/mol (d) –71.7 Kcal/mol If the bond dissociation energies of XY, X2 and Y2 (all diatomic molecules) are in the ratio 1: 1: 0.5 and ∆Hf for the formation of XY is –200 KJ mol–1, the bond dissociation energy of X2 will be (a) 100 KJ mol–1 (b) 200 KJ mol–1 –1 (c) 300 KJ mol (d) 800 KJ mol–1 When 1 gm equivalent of strong acid reacts with strong base heat released is 13.5 Kcal. When 1 gm equivalent H2A is completely neutralised against strong base 13 Kcal heat is released. When 1 gm equivalent B(OH)2 is completely neutralised against strong acid 10 Kcal heat is released. Calculate enthalpy change. When 1 gm mole H2A is completely neutralised by B(OH)2 (a) –19 Kcal (b) –27 Kcal (c) –10 Kcal (d) –20 Kcal If enthalpy of hydrogenation of C6H6(l) into C6H12(l) is –2058 KJ and resonance energy of C6H6(l) is

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7.29

–152 KJ/mol then enthalpy of hydrogenation of 1.4–cyclohexadiene (l) is (Assume ∆vapH of C6H6(l). C6H8(l) and C6H12(l) are equal) (a) –535.5 KJ/mol (b) –238 KJ/mol (c) –357 KJ/mol (d) None of these (∆H–∆U) in the formation of carbon monoxide from its elements at 298 K is (a) 1238.78 J/mol (b) –2477.57 J/mol (c) –357 KJ/mol (d) None of these Both the lattice energy (∆U0) and hydration enthalpy (∆Hh) of a binary salt are negative quantities. However, If ∆U0 is more than ∆Hh then (a) The salt will not dissolve in water (b) Salt will dissolve in water (c) Dissolution of salt in water is exothermic (d) Dissolution of salt in water is endothermic The internal energy change when a system goes from state A to B is 40 KJ/mole. If the system goes from A to B by a reversible path and returns to state A by an irreversible path what would be the net change in internal energy? (a) 40 KJ (b) >40 KJ (c) ∆U 4. Heat of neutralization of weak acid with strong base is –50 KJ and weak acid, weak base is –41.4 KJ. then (a) Heat of ionization of weak acid is 7.4 KJ/mole (b) Heat of strong acid and strong base neutralization is –57.4 KJ (c) Heat of ionization of weak base is 16 KJ/mole (d) Heat of ionization of weak base is 8.6 KJ/mole 5. Which the following process is/are expected to be exothermic in nature? 1 (a) Na(S)+ Cl2(g) →NaCl(s) 2 (b) Conversion of gas into liquid (c) Mixing of two liquids to form a non-ideal solution with negative deviation (d) Dissolution of glucose in water O O

comprehensive type Questions passage I Chemists’s seemingly insatiable need to measure heat has led to a broad spectrum of measurement method. Beginning students are familiar with sample “coffee cup” style calorimeters that rely on the known specific heat of water to infer heat values from measurement of temperature change. A student uses the calorimeter shown in Figure.1 below to determine the enthalpy of solution, ∆Hsolution, of KOH:

Thermometer Sample

O

Insulation Coffee-cup Calorimeter Figure 1 She finds that when a 5.6 g sample of solid KOH is added to 100 mL of water, the temperature in the solution rises from 20°C to 37°C 1. If the specific heat of the solution of KOH in water is assumed to be 4.18 J/mL–K, what is approxi mate value of ∆Hsolution, for KOH? (Assume that the calorimeter is thermally insulated from its surroundings) (a) –54.34 KJ/mol (b) –5.434 KJ/mol (c) 54.34 KJ/mol (d) 5.434 KJ/mol

7.32

Thermochemistry

2. Which of the following change in the KOH solubility experiment would be the least likely to change the measured value of ∆T? (a) Doubling the amount of water used (b) Doubling the amount of KOH used (c) Doubling the amount of water and doubling the amount of KOH used (d) Substituting an equal mass of NaOH for the KOH sample passage ll Read the following passage and answer the questions at the end (a) A student heated a sample of a metal weighing 32.6 g to 99.83°C and put it into 100.0 g of water at 23.62°C in a calorimeter. The final temperature was 24.41°C. The student calculated the specific heat

of the metal, but neglected to use the heat capacity of the calorimeter. The specific heat of water is 4.184 J/g °C. What was his answer? The metal was known to be chromium molybdenum or tungsten. By comparing the value of the specific heat to those metals (Cr, 0.460; Mo, 0.025; W, 0.135 J/g °C), the student identified the metal. (b) A student at the next laboratory bench did the same experiment, obtained the same data, and used the heat capacity of the calorimeter in his calculations. The heat capacity of the calorimeter was 410 J/°C. 1. Metal identified by first student was (a) Cr (b) Mo (c) W (d) None of these 2 Metal identified by student in the next laboratory was (a) Cr (b) Mo (c) W (d) None of these

passage lll Study the following experiment and answer the questions given at the end. 27.0

26.0

Before Mixing

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A coffee-cup calorimeter is used to determine the heat of reaction for acid-base neutralization CH3COOH(aq) + NaOH(aq) → CH3COONa(aq) + H2O(l) When we add 25.00 mL of 0.700 M NaOH to 25.00 mL of 0.700 M CH3 COOH at a given temperature as shown taken in the calorimeter is = 27.8 J/g°C Specific heat of solution = 4.18 J/g°C Density of the solution = 1.00 g/mL

1

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1. What is enthalpy of the above experiment? (1) 475 J (2) +475 J (3) –950 J (4) +950 J 2. Heat of neutralization of CH3COOH(aq) by NaOH(aq) based on this experiment is (1) –54 KJ (2) –27 KJ (3) +54 KJ (4) +27 KJ

Thermochemistry

passage IV Read the following passage based on enthalpy change in the formation of solution. Answer the questions at the end. The enthalpy is the process HCl + nH2O → HCl in n moles of H2O where n is the number of moles of water, is called the integral heat of solution. When n is large enough the continued addition of water does not increase the heat of solution, one simply writes HCl + aq → HCl(aq) The enthalpy for this process is the limiting value for the integral heat of solution. The enthalpy for the process HCl in n moles of H2O + mH2O → HCl in (n+m) moles of H2O is called the integral heat of dilution. These quantities are indicated in the figure. Another quantity of interest is the differential heat of solution, defined as the slope of the enthalpy curve. The heat of solution depends on the composition of the solution as shown in the figure. 0

Mole of HCl=1

Δ HKJ

-20

Interal heat o solution HCl+2H2O → HCI(2H2O) Intergral heat of dilution HCl in 2H2O + 3H2O → HCl IN 5H2O

-40

-60

Differential heal of soution -80

10

20

30

Mole of H2O

40

50

1. Integral heat of solution for the following step is HCl + 5H2O → HCl (5H2O) (a) –12 KJ (b) –44 KJ (c) +60 KJ (d) –60 KJ 2. Which step will give integral heat of dilution? (a) HCl + 2H2O → HCl (2H2O) (b) HCl (2H2O) + 3H2O → HCl (5H2O) (c) HCl + aq → HCl(aq) (d) All of these passage V 1 g of a weak monobasic acid (MW = 100) when dissolved in 100 g of liquid A (MW = 100) increases the boiling point from 80°C to 81°C at 760 mm of Hg. The vapour pressure of liquid. At 90°C is 850 mm of Hg. Degree of dissociation of the monobasic acid is 0.189 at 90°C. Enthalpy of

7.33

neutralization of a strong acid with strong base = –57.3 KJ/ equivalent 1. ∆Hv for the liquid A is (assuming it remains constant within the temperature range considered ) (a) 11.99 KJ mol–1 (b) –11.99 KJ mol–1 (c) 119.9 KJ mol–1 (d) –119.9 KJ mol–1 2. Degree of dissociation of the monobasic acid at 80°C will be (a) 0.2 (b) 0.158 (c) 0.1158 (d) 0.52 3 Enthalpy of neutralization of weak acid with a strong base (a) –33.9 KJ mol (b) –11.99 KJ mol (c) 15.07 KJ mol (d) –15.07 KJ mol

Assertion (A) and reason (r) type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): A reaction which is spontaneous and accompanied by decrease of randomness must be exothermic. Reason (R): All exothermic reactions are accompanied by decrease of randomness. 2. Assertion (A): Enthalpy of neutralizations is always exothermic. Reason (R): Neutralization involves reaction between an acid and a base. 3. Assertion (A): The enthalpy of formation of H2O(l) is greater than that of H2O(g). Reason (R): Entropy change is negative for condensation reaction H2O(g) → H2O(l). 4. Assertion (A): Enthalpy of graphite is lower than that of diamond. Reason (R): Entropy of graphite is greater than that of diamond. 5. Assertion (A): In the following reaction C(s) + O2(g) → CO2(g), ∆H = ∆U–RT Reason (R): ∆H = ∆U + ∆ng RT. 6. Assertion (A): As temperature increases, heat of reaction also increases for exothermic as well as for endothermic reactions. Reason (R): ∆H2(at T2) = ∆H1(at T1) + ∆CP (T2 – T1).

7.34

Thermochemistry

Integer type Questions 1. From the following data, calculate the enthalpy change for the combustion of cyclopropane at 298 K. The enthalpies of formation of CO2(g), H2O (l) and propane (g) are –393.5, –285.8 and 20.10 KJ mol–1, respectively. The enthalpy of isomerisation of cyclopropane to propane is –33.0 KJ mol–1. Fill the number in answer key after dividing with 1000. 2. 2.3 g of an organic compound of molecular mass 46 is subjected to combustion in presence of excess of oxygen in a constant volume calorimeter, the thermal capacity of which is 2.0 KJ/K. in case the temperature of the calorimeter increases from 27°C to 27.225°C in the combustion process, find the heat of combustion at constant volume of the compound in KJ. 3. The standard molar enthalpies of formation of IF5(g) and IF3(g) are –470 KJ and –847 KJ, respective ly. Valence shell electron-pair repulsion theory predicts that IF5(g) is square pyramidal in shape in which I-F bonds are equivalent while IF3(g)is T-shaped (based on trigonal-bipyramidal geometry) in which all I-F bonds are of different lengths. It is observed that the axial I-F bonds in IF3 are equivalent to the I-F bonds in IF5. Calculate the equatorial I-F bond strength (in KJ/Mol) in IF 3. Some other informations given is I2(s) → I2(g); ∆H = 62 KJ F2(g) → 2F(g); ∆H = 155 KJ I2(s) → 2I(g); ∆H = 149 KJ Divide the answer with 34 and report in the key 4. One g of a compound (C6H10O5)n on combustion gave only CO2(g) and H2O(l) along with 17.49 KJ/g of energy. Given: ∆fH of H2O(l) = –285.85 KJ mol–1; ∆fH of CO2(g) = –293.7 KJ mol–1; ∆fH of (C6H10O5)n = –2148.42 KJ mol–1. The value of ‘n’ is __________. 5. At 500 kbar pressure density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain tempera ture ‘T’. [∆H – ∆U] (in KJ/mole ) for the conversion of mole graphite to 1 mole of diamond at 500 Kbar pressure. [Given: 1 bar = 105 N l m2] is 10x. X value is __________. 6. Calculate the enthalpy of ionization of week acid (H2A → 2H + A2–) in Kcal/mol, if enthalpies of neutratization of HCl and H2A by strong base are 14 Kcal/eq and 11 Kcal/eq respectively.

7. Consider the following reactions (a) Au(OH)3 + 4HCl → HAuCl4 + 3H2O; ∆H = –28 Kcal (b) Au(OH)3 + 4HBr → HAuBr4 + 3H2O; ∆H = –36.8 Kcal If 1 mole of HAuBr4 was mixed with 4 moles of HCl, 0.44 Kcal heat was absorbed. The fraction of HAuBr4 was converted into HAuCl4 is 0.0 x and the percentage conversion is x . Then ‘x’ is 8. If the bond enthalpies of A2, AB and B2 are in the ratio 1:1: 0.5 and the enthalpy of formation of AB from A2 and B2 is –100 KJ mol –1, the bond enthalpy of A2 is 50Y. Y will be….. 1  9. For the reaction, Ag2O(s) ↽ ⇀  2Ag (s) + O2(g) –1 2 ∆H, ∆S, T are 40.63 KJ mol –1 108.8 JK mol = and 373.4 K respectively. Free energy change ∆G of the reaction will be 10. Standard Gibbs free energy change ∆G for a reaction is zero. The value of equilibrium constant of the reaction will be O

previous years’ IIt Questions O

1. ∆H for CO2, C(g) and H2O(g) are –393.5 –110.5 and –241.8 KJ/mol respectively. The standard enthalpy hence (in KJ) for the reaction CO2(g) + H2(g) → CO(g) + H2O(g) is (2000) (a) 524.1 (b) 41.2 (c) –262.5 (d) –41.2 2. Which of the reaction defines ∆H f (2000) (a) C(Diamond)+O2(g)→CO2(g) O

(b)

1 1 H2 (g) + F2 (g) → HF(g) 2 2

(c) N2(g)+3H2(g) → 2NH3(g) 1 (d) CO(g)+ O2 (g) → CO2 (g) 2 provided, calculate the multiple bonds 3. Using the data energy (KJ mol–1) of a C≡C bond in C2H2. Take the bond energy of a C-H bond as 350 KJ mol–1 (2012) 2C(s) + H2(g) → C2H2(g) ∆H = 225 KJ mol–1 2C(s) → 2C(g) ∆H = 1410 KJ mol–1 H2(g) → 2H(g) ∆H = 330 KJ mol–1 (a) 1165 (b) 837 (c) 865 (d) 815

Thermochemistry

7.35

AnSwEr KEyS multiple choice Questions with only one Answer level I 1. a 2. b 3. a 4. b 5. b 6. c

7. 8. 9. 10. 11. 12.

a c d b a b

13. 14. 15. 16. 17. 18.

c a a c b b

19. 20. 21. 22. 23. 24.

c a a b b a

25. 26. 27. 28.

a b d d

multiple choice Questions with only one Answer level II 1. 2. 3. 4. 5. 6. 7. 8.

c c d d d c a b

9. 10. 11. 12. 13. 14. 15. 16.

a b c a c b d d

17. 18. 19. 20. 21. 22. 23. 24.

a b a d d d c c

25. 26. 27. 28. 29. 30. 31. 32.

a c b c a b b c

33. 34. 35. 36. 37. 38. 39.

d b c a a b b

comprehensive type Questions passage I 1. a

2. c

passage II 1. c

2. b

passage III 1. c

2. a

passage IV 1. d

2. a

passage V 1. a

2. b

3. a

Assertion (A) and reason (r) type Questions 1. c

2. b

3. a

4. b

5. d

6. d

Integer type Questions multiple choice Questions with one or more than one Answer 1. c, d 2. b, c, d 3. a, b, c, d

4. a, b, d 5. a, b, c 6. a, b, c

7. a, d 8. b, c, d

1. 2 2. 9

3. 8 4. 6

5. 2 6. 6

7. 5 8. 8

previous years’ IIt Questions 1. b

2. b

3. d

9. 4 10. 1

7.36

Thermochemistry

HIntS And SolutIonS Hints to problems for practice 1. Work of expansion W = –P∆V ∆V = 2.0 – 0.5 = 1.5 L; P = 1.1 atm P∆V = –1.10 × 1.5 = –1.650 L atm = –1.650 × 101.3 (1 L atm = 101.3 J) = –167.1 J Since work is done by the system w = –167.1 J Heat absorbed by the system = 120 J or Q = + 120 J Now ∆U = Q +W = 120 J + (–167.1) = –47.1 J 2. ∆H and ∆U are related as ∆H = ∆U + ∆ng RT N2(g) + 3H2(g) → 2NH3(g) ∆ng = 2 – (1 + 3) = –2 mol, T = 298 K ∆H = –92.38 KJ = –92380 J, R = 8.314 JK–1 mol–1 –92380 = ∆U + [(–2 mol) × (8.314 J mol–1 K–1) × (298 K)] –92380 = ∆U – 4955 ∆U = –92380 + 4955 = –87425 J = –87.425 KJ 3. Heat of reaction at constant pressure (∆H) and heat of reaction at constant volume (∆U) are related as ∆H = = ∆U + ∆ngRT ∆H = = ∆U + ∆ngRT ∆ngRT = 12–15 = –3; R = 8.314 RJ K–1 mol–1, T = 298 k ∆H – ∆U = –3 × 8.314 × 10–3 × 298 = –7.43 KJ 4. Fall in temperature = 320–293 = 27 K Molar mass of C2Cl2F2 = (2 × 12) + (2 × 35.5) + (2 × 19) = 133 Cp = 80.7 J mol–1 K–1 80.7 = 0.6068 J g–1 K–1 Or = 133 Heat evolved from 1.25 g of sample on being cooled Qp = m × Cp × ∆T = 1.25 g × 0.6068 JK–1 g–1 × 27 K = –20.48 J ∆H = –20.48 J Now ∆H = ∆U + P∆V P∆V = 1 atm ×

248 - 274 L. 1000

= –0.026 L atm = –0.026 × 101.32 J = –2.63 J –20.48 = ∆U – 2.63 ∆U = –20.48 + 2.63 = –17.84 J

5. Quantity of heat absorbed by the calorimeter. Q = CV × ∆T ∆T = 298.89–298 = 0.89 K, CV = 20.7 KJ/K Q = 20.7 × 0.89 = 18.423 KJ Quantity of heat produced = –18.423 KJ Heat produced by burning 0.562 g of graphite = –18.423 KJ Heat produced by burning 1 mol (12 g) of graphite -18.428 × 12 0.562 = –393.4 KJ Enthalpy of combustion of graphite, ∆H = –393.4 KJ mol–1 6. Since ∆H = Qp = Heat absorbed by the system at constant pressure = 1440 calories. In the equation ∆H = ∆U + P∆V P∆V = 76 × 13.6 × 981 (18–19.6) ergs 76 × 13.6 × 981 × 1.6 Calories = 4.18 × 107 = 0.039 calories (p = 1 atm = 76 × 13.6 × 981 dynes cm–2, V2 = 18 cc, V1 = 19.6 cc) Since P∆V is very small compared to ∆U, P∆V can be neglected Thus ∆H = ∆U = 1440 calories 7. Latent heat of fusion of ice per mole = 80 × 18 = 1440 cal. Latent heat of vapourization of liquid water per mole = 596 × 18 = 10728 cal. Total heat absorbed by 1 mole (18 g) of ice in being converted to 1 mole of water vapour = 1440 + 10728 = 12168 calories Since the conversion took place at a constant pressure, QP = ∆H = 12168 calories Now we have ∆U = ∆H – P∆V = QP – P∆V As the volume of ice is to be neglected, V1 = 0 and V2 = Volume of 1 mole of water vapour at 0°C and 4.6 mm Hg pressure. =

= 22400 ×

760 = 3717000 mL 4.6

∆V = (V2 – V1) = 3717000 mL P = 4.6 mm = 0.46 × 13.6 × 981 = 6137 dynes cm–2 P∆V = 6137 × 3.717000 ergs

Thermochemistry

=

6137 × 3717000 cal 4.8 × 107

= 545 cal ∆U = 12168–545 = 11623 calories 8. Energy consumption of family per day = 20,000 KJ Total number of families = 100 Energy consumption of 100 families per day = 100 × 20,000 = 2 × 106 KJ The consumption of methane present in gobar gas may be written as CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) + 809 KJ According to the equation 809 KJ of energy is evolved = 16 × 2 × 106 g 809 = 3.96 × 104 g =39.6 kg Since methane content of gobar gas is 80% by weight, therefore weight of gobar gas needed = 39.6 × 100 = 49.5 kg 80 9. (a) Molecular formula of butane = C4H10 Molecular mass of butane = 58 1 mole or 58 g of butane on complete combustion give heat = 2658 \ 14 × 103 g of butane on complete combustion gives heat 2658 × 14 × 103 = = 641586 58 The family needs 20,000 KJ of heat per day 20,000 KJ of heat is used for cooking by a fam ily in one day 641586 KJ of heat will be used for cooking a family =

641586 = 32 days 20000

The cylinder will last 32 days (b) 25 percent of the gas is wasted due to inefficiency. This means that only 75% of butane gets combusted Therefore, the energy produced by 75% combustion of butane =

641586 × 75 = 481190 KJ 100

\ The number of days the cylinder will last 481190 = 24 days 20000 10. C2H4 + CH4+ O2 → xL (3.67–x)L x mol (3.67– x) mol

CO2 +H2O 6.11 lit 6.11 mol

7.37

Or 2 × (x+1) × (3.67–x) = 1 × 6.11; x = 2.44 lit Thus the volume of C2H4 = 2.44 lit and volume of CH4 = 1.23 lit 2.44 \ Volume of C2H4 in 1 litre mixture = 3.67 = 0.665 lit and volume of CH4 in a 1 lit mixture = 1–0.665 =0.335 lit. Now thermochemical reactions for C2H4 and CH4 are C2H4 + 3O2 → 2CO2 +2H2O; ∆H = –1423 KJ CH4 + 2O2 → CO2 + 2H2O; ∆H = –891 KJ As ∆H value given is at 25°C i,e 298 K let us first calculate the volume occupied by one mole of any gas at 25°C (supposing pressure as 1 atm) Volume per mole at 25°C =

298 × 22.4 = 24.45 lit. 273

Thus heat evolved in the combustion of 0.665 lit of C2H4 = -

1423 × 0.665 = -38.70 KJ 24.45

And heat evolved in the combustion of 0.335 lit of CH4 891 =× 0.335 = -12.20 KJ 24.45 \ Total heat evolved in the combustion of 1 litre of the mixture = –38.70 + (–12.20) = –50.90 KJ 11. Moles of O2 inhaled by a person in one day 640 = = 20 32 Given that C12H22O11 + 12O2 → 12CO2+ 11H2O; ∆H = –5645 KJ Thus, 12 moles of O2 consume 1 mole sucrose 342 \ 20 moles of O2 consume × 20, i.e., 570 g of 12 sucrose Further 5645 342 g (1 mole) of sucrose liberate × 570 342 = 9408.34 KJ 12. Energy remained in the body of the athlete after the event 1560 = = 780 KJ 2 \ Weight of water to be evaporated by 780 KJ of energy 18 × 780 = 319.1 g 44

7.38

Thermochemistry

13. Mole of C2H4 = 1000 = 44.6 22.4 Total heat evolved = 44.6 × 337 = 1.5 × 104 Kcal \ Useful heat = 0.7 × 1.5 × 104 = 1.05 × 104 Kcal Now water evaporation takes place in two stages H2O (l, 20°C) → H2O (l, 100°C); ∆H = 80 Kcal kg–1 And H2O (l, 100°C) → H2O (g, 100°); ∆H = 540 Kcal kg–1 \ H2O (l, 20°C) → H2O (g, 100°C), ∆H = 620 Kcal kg–1 \ Weight of water converted to steam amount of heat available = heat required per kg 1.05 × 10 = 16.9 kg 620 4

=

14. Given that (i) 2 Al + 3 2 O2 → Al2O3; ∆H = –399 Kcal (ii) 2Fe + 3 2 O2 → Fe2O3; ∆H = –199 Kcal Subtracting equation (ii) from equation (i) we get 2 Al + Fe2O3 → Al2O3+ 2Fe; ∆H = –399–(–199) = –200 Kcal Now 2 moles of Al weigh (2 × 27), i.e., 54 g and 1 mole of Fe2O3 weigh 160 g Total weight of the mixture = 54 + 160 = 214 g \ Fuel value of the mixture =

200 = 0.9345 Kcal 214

\ Further, volume of 2 moles Al =

54 = 20 cc 2.7

160 And volume of 1 mole of Fe2O3 = = 30.76 cc 5.2 \ Total volume of the mixture = 50.76 cc 200 \ Fuel value per CC of the mixture = 50.76 = 3.94 Kcal/cc 15. Total carbon content in 10 kg of coal 80 = × 10 = 8 kg 100 The amount of carbon which gets converted to CO2 60 × 8 = 4.8 kg 100 C(s) +O2(g) → CO2(g) + 396 KJ Heat produced burning 12 g (1 mol) of carbon = 394 KJ =

\ Heat produced by burning (4.8 × 103 g) of carbon 394 = × 4.8 × 103 KJ = 157.6 × 103 KJ 12 The amount of carbon which gets converted into CO 40 = × 8 = 3.2 kg 100 1 C (s) + O2(g) → CO(g) + 111 KJ 2 Heat produced from 12 g of carbon = 111 KJ Heat produced from 3.2 × 103 g of carbon =

111 × 3.2 × 103 = 29.6 × 103 KJ 12

Net heat produced = 157.6 × 103 + 29.6 × 103 = 187.2 × 103 KJ 16 C20H32O2(s) + 27O2(g) → 20CO2(g) + 16H2O(l) ∆H = ∆U + ∆nRT This equation is for one mole of chicken fat We have to calculate for 3 × 10–3 moles ∆n = 20 – 27 = 7 × 3 × 10–3 ∆H = –10,000 + 7 × 3 × 10–3 × 1.987 × 310 = –10000–13.0 = –10013 = –10.013 Kcal 17. C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) ∆H = –2817 KJ 1 mole of glucose (180 g) gives 2817 KJ of energy 2817 × 4 = 62.6 KJ \ 4 g of glucose gives = 180 Amount of available energy for muscular activity =

62.6 × 40 = 25.04 KJ (∵ ∆H 40% of the energy 100

available) 18. ∆vapH of CO = +6.04 KJ mol–1 Molar mass of CO2 = 28 g mol–1 Energy required for vaporizing 28 g of CO = 6.04 KJ Energy required for vapoursing 2.38 g of CO = O

6.04 × 2.38 = 0.5134 or 513.4 J 28 19. Molar mass of water = 18 g mol–1 No. of moles of water in 80 g = 80 mol 18 Heat required for evaporating 1 mol of water = 40.18 KJ 80 Heat required for evaporation of mol of water 18 40.8 × 80 = 181.15 KJ = 18 For evaporation of 1 mol of water H2O(l) → H2O(g)

Thermochemistry

∆ng = 1–0 = 1 \ ∆ vap U = ∆ vap H - ∆nRT = 40.8–1 × 8.314 × 10–3 × 373 = –37.69 KJ mol–1 20. Molar mass of benzene (C6H6) = 78 g mol–1 Moles of benzene in 100 g sample =

100 = 1.282 mol 78

Heat required for evaporating 1 mol of benzene = 30.8 KJ Heat required for evaporating 1.282 mol of benzene = 30.8 × 1.282 =39.485 KJ Now Power = Energy or W = 1 JS–1 Time Thus time =

Energy 329.485 × 103 (J) = Power 100(JS-1 )

394.85 = 6.58 min 60 21. Molecular mass of naphthalene C10H8 =128 For the solidification reaction = 394.85 sec or

C10H8(l) → C10H8(s) Heat evolved when 1 g of naphthalene solidifies = 149 J Heat evolved when 128 g of naphthalene solidify = 149 × 128 =19072 J i.e. C10H8(l) → C10H8(s); ∆fusH = –19072 J For the fusion reaction C10H8(s) → C10 H8(l) This reaction is the reverse of the above solidification reaction so that ∆fusH = –∆solidH Thus ∆fusH = 19072 J or 19.072 KJ 22. Data given is (i) C(s) + O2(g) → CO2(g); ∆H = –94.14 Kcal 1 (ii) H2(g) + O2(g) → H2O(l); ∆H = –68.47 Kcal 2 1 (iii) C2H6 + 3 O2(g) → 2CO2(g) + 3H2O(l); 2 ∆H = –373.3 Kcal ∆H of the following equation has to be calculated O

O

O

O

(iv) 2C (s) + 3H2(g) → C2 H6(g); ∆H = ? Multiply the equation (i) with 2 and eqn (ii) with 3 and then subtract eqn (iii) from the sum of these eqns. [C(s) + O2(g) → CO2(g); ∆H = –94.4 Kcal] × 2 1 [H2(g) + O2(g) → H2O(l); ∆H = –68.47 Kcal] × 3 2

7.39

To subtract reverse the eqn (iii) and add to the above eqns 2C(s) +2O2(g) → 2CO2(g); ∆H = –188.28 Kcal 3 O2(g) → 3H2O(l); ∆H = –205.41 Kcal 2 1 2CO2(g) + 3H2 O(l) → C6H6(g) + 3 O2(g); 2 ∆H = +373.3 Kcal 3H2(g) +

2C (s) +3H2(g) → C2H6(g); ∆H = –20.39 Kcal 23. The data given is (i) C(s) +O2(g) → CO2(g); ∆H = –94.05 Kcal (ii) H2(g) + 1 O2(g) → H2O(g); ∆H = –68.32 Kcal 2 1 (iii) C2H2(g) + 2 O2(g) → 2CO2(g) + H2O(g); 2 ∆H = –310.62 Kcal To calculate the ∆H of the equation 2C (s) + H2(g) → C2H2(g) Multiply the equation (i) with 2 and then sum up with equation (ii) and (iii) after reversing the equation (iii) 2C(s) + 2O2(g) → 2CO2(g); ∆H = –188.1 Kcal 1 H2(g) + O2(g) → H2O(g); ∆H = –68.32 Kcal 2 1 2CO2(g) + H2O(g) → C2H2(g) + 2 O2 (g) ; 2 ∆H = +310.62 Kcal 2C(s) + H2(g) → C2H2(g); ∆H = 54.2 Kcal 24. The data given is 1 (i) F2(g)+ O2 (g) → F2 O(g); ∆H = –5.5 Kcal 2 (ii) H2(g) (iii)

1 O2 (g) → H2 O(g); ∆H = –57.8 Kcal 2

1 1 H2 (g) + F2 (g) → HF(g); ∆H = –64.2 Kcal 2 2

to get the heat of the reaction F2O(g) + H2O(g) → 2HF(g) + O2(g) Reverse the equations (i) and (ii) and add to equation (iii) after multiplying it with 2 1 F2 O(g) → F2 (g) + O2 (g); ∆H = -5.5 Kcal 2 1 H2 O(g) → H2 (g) + O2 (g); ∆H = +57.8 Kcal 2 H2 (g) + F2 (g) → 2HF(g); ∆H = -128.4 Kcal F2(g)+H2O → 2HF(g); ∆H = –76.1 Kcal

7.40

Thermochemistry

25. The given data is (i) 2Al(s) + 6HCl(aq) → Al2Cl6(aq) + 3H2; ∆H = –244 Kcal (ii) H2(g) + Cl2(g) → 2HCl(g); ∆H = –44 Kcal (iii) HCl(g) + aq → HCl(aq); ∆H = –17.5 Kcal (iv) Al2Cl6(aq) → Al2Cl6(aq); ∆H = –153.7 Kcal.

Multiply the equation (ii) with 2 and the equation (iv) with 4, subtract equation (i) and (ii) (after multiplying with 2) from the sum of equation (iii) and equation (iv) (after multiplying with 4) CCl4(g) → C(s) + 2Cl2; ∆H = +25.5 Kcal 2H2O(g) → 2H2(g) + O2(g); ∆H = +115.6 Kcal C(s) + O2(g) → CO2(g); ∆H = –94.1 Kcal 2H2(g) + 2Cl2(g) → 4HCl(g); ∆H = –88.4 Kcal CCl4(g) + 2H2(g) → C(g)4HCl(g); ∆H = –41.4 Kcal 28. The given data is (i) 2C(s)+ 3H2(g) → C2H6(g); ∆H = –21.1 Kcal (ii) C(s)+O2(g) → CO2(g); ∆H = –94.1 Kcal 1 (iii) H2(g)+ O2(g) → H2O(g); ∆H = –68.3 Kcal 2 1 To calculate the ∆H of the equation C2H6 + 3 O2(g) 2 → 2CO2(g) + 3H2O(g) Multiply the equation (ii) with 2 and the equation (iii) with 3 and then subtract the equation (i) from the sum of the equation ii and iii (after multiplying with 2 and 3 respectively) C2H6(g) → 2C(s) + 3H2(g); ∆H = +21.1 Kcal 2C(s) + 2O2(g) → 2CO2(g); ∆H = –188.2 Kcal 3 3H2(g) + O2 → 3H2O(g); ∆H = –204.9 Kcal 2 O

O

O

We have to calculate ∆H of the equation (v) 2Al(s) + 3Cl2(g) → Al2Cl6(s); ∆H =? Multiplying the equation (ii) with 3, reverse the equation (iv) and then add these reactions to equation (i) 2Al(s) + 6HCl(aq) → Al2Cl6(s) + 3H2(g); ∆H = –244 Kcal 3H2(g) + 3Cl2(g) → 6HCl(g); ∆H = –132 Kcal Al2Cl6(aq) → Al2Cl6(s) +(aq); ∆H = +153.7 Kcal (vi) 2Al(s)+3Cl2(g) → Al2Cl6(s) + 6HCl(g) + aq; ∆H = –222.3 Kcal Multiply equation (iii) with 6 and then add to equation (vi) 2Al(s) + 3Cl2(g) + 6HCl(aq) → Al2Cl6(s) + 6HCl(s) + aq; ∆H = –222.3 Kcal 6HCl(s) + 3Cl2(g) +(aq) → HCl(aq); ∆H = –105.0 Kcal 2Al(s) + 3Cl2(g) → Al2Cl6(s); ∆H = –327.3 Kcal 26. The data given is (i) C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g); ∆H = –325 Kcal (ii) CH3COOH(l) +2O2(g) → 2CO2(g) + 2H2O(g); ∆H = –209.5 Kcal We have to calculate the heat liberated in the equation C2H5OH + O2 → CH3COOH + H2O Subtracting the equation (ii) from equation (i) we get the required equation i.e., reverse the reaction (ii) and add to the reaction (i). C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(g); ∆H = –325 Kcal 2CO2(g) + 2H2O(g) → CH3COOH(l) + 2O2(g); ∆H = +209.5 Kcal C2H5OH(l) + O2(g) → CH3COOH(l) + H2O(l); ∆H = –115.5 Kcal

O

O

O

O

O

O

O

O

O

1 C2H6(g) + 3 O2(g) → 2CO2(g) + 3H2O(g); 2 ∆H = –372.0 Kcal O

29. The given data is 1 O2 → H2O(l); ∆H = –241 Kcal 2 1 (ii) C6H10(g) + 8 O2(g) → 6CO2(g)+ 5H2O(l); 2 O

(i) H2(g)+

∆H = –3800 Kcal (iii) C6H12(g) + 9O2(g) → 6CO2(g)+ 6H2O(l); ∆H = –3920 Kcal To calculate the ∆H of the reaction C6H10(g) + H2(g) → C6H12(g) Subtract the equation (iii) from the sum of the equations (i) and (ii) 1 H2(g)+ O2(g) → H2O(l); ∆H =–241 KJ 2 1 C6H10(g)+ 8 O2(g) → 6CO2(g)+5H2O(l); 2 ∆H = –3800 KJ 6CO2(g) + 6H2O(l) → C6H12(g) + 9O2(g); ∆H = +3920 KJ O

O

O

27. The data given is (i) C(s) + 2Cl2(g) → CCl4(g); ∆H = –25.5 Kcal 1 (ii) H2(g)+ O2(g) → H2O(g); ∆H = –57.8 Kcal 2 (iii) C(s)+O2(g) → CO2(g); ∆H = –94.1 Kcal O

O

O

O

O

1 1 (iv) H2(g)+ Cl2(g) → HCl(g); ∆H = –22.1 Kcal 2 2 O

C6H10(g) + H2(g) → C6H12(g); ∆H = –121 KJ O

Thermochemistry

30. Given N2(g) + 3H2(g) → 2NH3(g) ∆H = –91.94 KJ at 27°C ∆Cp = Cp of product –Cp of reactant = [2 × Cp of NH3(g)]–[Cp of N2(g) + 3 × Cp of H2(g)] = (2 × 37.07)–(28.45 + 3 × 28.32) = –39.27 Joule Also ∆H2 – ∆H1 = ∆Cp(T2 – T2) ∆H2 – (–91940) = –39.27 (320 – 300) ∆H2 = –92725.4 Joule = –92.725 KJ 31. ∆Cp = (2 × 25.1) + (3 × 75.3) –[103.8 + (3 × 28.8)] = 85.9 JK–1mole–1 According to Kirchoff’s Law ∆H2 - ∆H1 = ∆C p T2 - T1 O

∆H2 - (-33290) = 85.9 358 - 298 ∆H358 = –28136 J mole–1 –1 = –28.136 KJ mole 15.5 = 1.292 32. Moles of C = 12 Moles of O2 =

PV 5.5 × (0.19 × 25) = = 1.068 RT 0.0821 × 298

(ii) C3H8(g) +5O2(g) → 3CO2(g) + 4H2O(l); ∆H = –530 1 (iii) H2(g) + O2(g) → H2O(l); ∆H = –68 2 (iv) C(s) + O2(g) → CO2(g); ∆H = –94 O

O

O

First calculate the heat of formation of C2H6 and C3H8 2C (s) + 3H2(g) → C2H6(g); ∆H = ? 3C(s) + 4H2(g) → C3 H8(g); ∆H = ? O O

To get the heat of formation of C2H6 subtract the equation (i) from the sum of equation (iii) × 3 and equation (iv) × 2 1 2CO2(g) + 3H2O(l) → C2 H6(g)+ 3 O2 (g); ∆H = +372 2 3H2(g) + 3 O2(g) → 3H2O(l); ∆H = –204 2 2C (s) + 2O2(g) → 2CO2(g); ∆H = –188 O

2C (s) + 3H2(g) → C2H6(g); ∆H = –20 O

1 O2  → CO 2

1.292–x moles

As O2 is completely consumed = Number of moles of oxygen after reaction. = Moles of oxygen in CO2+moles of oxygen in CO Or x = 0.844 Moles of CO2 = 0.844 Moles of CO = 1.292–0.844 = 0.448 Total heat evolved = 0.844 (–94.05) + 0.448 (–26.41) = –91.2 Kcal 33. Enthalpy of formation of 3 carbons – carbon double bonds → ∆Hf = ∆Hf = –156 – (+ 49) KJ = –205 KJ Given that + H2 →

O

O

2 →CO x.moles + O2  x.moles

1.292–x moles

34. The given data is 1 (i) C2H6(g) + 3 O2(g) → 2CO2(g) + 3H2O(l), 2 ∆H = –372

O

C

C+

7.41

Similarly the heat of formation of C3 H8 can be calculated by subtracting equation (ii) from the sum of equation (iii) × 4 and equation (iv) × 3 3CO2(g) + 4H2O(l) → C3H8(g) + 5O2(g); ∆H = +530 4H2(g) + 2 O2(g) → 4H2O(l) ∆H = –272 3C(s) + 3O2(g) → 3CO2(g) ∆H = –282 O

O

O

3C(s) + 4H2(g) + 4H2(g) → C3H8(g) ∆H = –24 O

Calculation of bond energy We have 2C (s) +3H2(g) → C2H6(g); ∆H = –20 For reactants Heat of atomization of 2 moles of atoms = 2 × 172 Heat of dissociation of 3 moles of H–H bonds = 3 × 104 For products Let heat of atomization of 1 mole of C–C bonds = x Let heat of formation of 6 moles of C–H bonds = 6 Y On adding we get ∆H of formation of C2H6 (2 × 172) + (3 × 107) + x + 6Y = –20 or x + 6Y = 676 (1) Also we have 3C (s) + 4H2(g) → C3H8(g); ∆H = –24 For reactants Heat of atomization of 3 moles of C atoms = 3 × 172 Heat of dissociation of 4 moles of H–H bonds = 4 × 104 O

O

; ∆H = –119 KJ

Theoretical enthalpy of formation of 3 double bonds in benzene ring = 3 × (–119) KJ = – 357 KJ \ Resonance energy of benzene = –357 – (–205) KJ = –152 KJ mole–1

O

7.42

Thermochemistry

For products Let heat formation of 2 moles of C–C bonds = 2X Let heat of formation of 8 moles of C–H bonds = 8Y On adding, we get 3 × 172 + 4 × 104 +2X + 8Y = –24 Or 2X + 84 = –956 (2) From (1) and (2) we get x = –82 and Y = –99 \ C–C bond energy = 82 Kcal C–H bond energy = 99 Kcal 35. The required reaction is CH4(g) → C (gas) + 4H; ∆H = ? On adding the equations from (i) to (v) we get CH4(g) → C(g) + 4H(g); ∆H = 1652.7 KJ 1652.7 Average value of the C–H bond energy = 4 = 413.2 KJ 36. ∆H = Sum of bond energies of reactants – sum of bond energies of products = 2C–H + 2 C–Cl – No bond = (2 × 415) + (2 × 326) = 1482 KJ 37. ∆H is the sum of bond energies of reactants – sum of bond energies of products = [5 × (C(s) → C(g)) + 4 × (H–H) ]– [2 × (C–C) + 2 × (C = C) + 8 × (C–H)] = [(5 × 171) + (4 × 104)]– [(2 × 83) + (2 × 147) + (8 × 98.8) = 1271 – 1250.4 = 20.6 Kcal 38. Given 1 N2(g) + O2(g) → N2O(g); ∆exp = 82 KJ 2 1   ∆HCal = 1 × (N ≡ N)+ × (O = O) 2   - [1 × (N = N)+1 × (N = O)]  1  = 946 +  × 498  - [ 418 + 607 ]   2  = 1195 - 1025 = 170 KJ Resonance energy of ∆Hf N2Oexp – ∆HfN2Ocalc = 82–170 = –88 KJ mol–1 39. H–C≡N(g) + 2H2(g) → CH3NH2(g) ∆H = Bond energy of reactant – bond energy of products = (∆HC–H + ∆HC≡N + 2∆HH–H) – [3∆HC–H + ∆HC–N + 2 ∆HN–H] –150 = [414 +∆HC≡N + (2 × 435)] – [(3 × 414) + 293+ (2 × 389)] –150 = (1284 + ∆HC≡N) – (2313) Or ∆HC≡N = 879 KJ mol–1 O

40. For ∆Hf of CH3COOH 2C(s) + 2H2(g) +O2(g) → CH3COOH(l) ∆H = [2∆HC(s) → C(g) +4∆H(g) + ∆HO(g)]– [3∆H C–H + ∆H C–C + ∆H C–O + ∆H C=O + ∆H O–H] = 2803.4 – 3132.8 = –329.4 (–329.4) = 110.3 KJ Mole–1 41. C(g) +2H2(g) +

1 O2(g) → CH3OH(l); ∆H = ? 2

For reactants Heat of atomization of 1 mole of C = 715 KJ Heat of atomization of 4 moles of H = 4 × 218 KJ Heat of atomization of 1 mole of O = 249 KJ For product Heat of formation of 3 moles of C–H bonds = –3 × 415 KJ Heat of formation of 1 mole of C–O bonds = –356 KJ Heat of formation of 1 mole of O–H bonds = –463 KJ Heat of condensation of 1 mole of CH3OH to liquid = –38 KJ on adding we get ∆H of formation of CH3OH(l) ∆H = –266 KJ mole–1 42. First calculate the heat of formation from given data To get 2C(s) + H2(g) → C2H2(g); ∆H = ? Multiply the equation (ii) with 2 and then add to equation (iii). From this sum subtract equation (i) 2C(s) + 2O2(g) → 2 CO2(g); ∆H = –188 Kcal 1 H2(g) + O2(g) → H2O(g); ∆H = –68 Kcal 2 1 2CO2(g) + H2 O(g) → C2 H2(g) + 2 O2(g); ∆H = 2 +310 Kcal 2C(s) + H2(g) → C2H2(g); ∆H = 54 Kcal Now the heat changes for reactant Heat of atomization of 2 moles of C=2 × 171 Kcal Heat of atomization of 2 moles of H=2 × 52 Kcal And heat change for the product (H–C≡C–H) Heat of formation of 2 moles C–H bonds = –2 × 99 Kcal Heat of formation of 1 mole of C≡C bonds = x Summing up, we get heat of formation of C2H2 (2 × 171) + (2 × 52) – (2 × 99) + x = 54 x = –194 Kcal Hence bond energy of C≡C bond in C2H2 is + 194 Kcal 43 The given data is H2(g) + O2(g) → 2OH(g); ∆H = 20.12 Kcal For reactants Bond energy for 1 mole of H–H bonds = 104.18 Kcal Bond energy for 1 mole of O–O bonds = 118.32 Kcal

Thermochemistry

And for product Energy of formation of 2 moles of O–H bond = 2 × x Where x is the energy of formation of 1 mole of O–H bonds Adding up all the energy change, we get heat of the above reaction (= 20.12 Kcal) i.e., 104.18 + 118.32 + 2 × x = 20.12 X = –101.19 Kcal Hence bond energy of O–H bond = + 101.9 Kcal per mole 44. Given that F2 + Cl2 → 2ClF; ∆H = –26.6 Kcal For reactants Bond energy of 1 mole of F–F bonds = 36.6 Kcal Bond energy of 1 mole of Cl–Cl bonds = 58.0 Kcal For Product Energy of formation of 2 moles of Cl–F bonds = 2 × x. (Where x is the bond formation energy of Cl–F bonds in Kcal mole–1) Adding all the heat changes, we get ∆H of the given reaction, i.e., 36.6 + 58 + 2 × x = –26.6 x = –60.6 Kcal Thus the bond energy of Cl–F bonds is +60.6 Kcal per mole. 45. The enthalpy of formation of CaCl2 according to the reaction Ca(s) + Cl2(g) → CaCl2 (s); ∆fH = –795 KJ mol–1 ∆H 1 for Ca(s) → Ca(g); = +121 KJ mol–1 ∆H 2 for Ca(g) → Ca2+(g); + 2e–(g) = +2422 KJ mol–1 ∆H 3 for Cl2(g) → 2Cl(g); = +242.8 KJ mol–1 ∆H 4 for 2Cl(g) → 2Cl–(g); = 2 × (–355) KJ mol–1 Ca2+(g) + 2Cl–(g) → CaCl2(s); ∆H = U0 = ? According to Hess’s law ∆fH = ∆H 1 + ∆H 2 + ∆H 3 + ∆H 4 + U0 –795 = 121 + 2422 + 242.8 – 710 + U0 –795 = 2075.8 + U0 U0 = –2075.8 – 795.0 = –2870.8 KJ mol–1 Lattice energy of CaCl2 = 2870.8 KJ mol–1 47. The formation of argon chloride may be represented by the Born-Haber Cycle. O

O O O

7.43

243 = 121.5 KJ 2 ∆H3 = –348.3 KJ U0 = –703.0 KJ ∆fH = 1526 + 121.5 – 348.6 –703.0 = 596.5 KJ Thus the formation of hypothetical argon chloride is endothermic and therefore, it is unfavourable ∆H2 =

multiple choice Questions with only one Answer level I 5. C6 H12 O6 + 6O2 → 6CO2 + 6H2O When 680 Kcal is released then 6 mole CO2 is obtained When 170 Kcal is released, the mole of CO2 formed = (170)6 = 1.5 680 So wt of CO2 (1.5) 44 =66 gm 6. The amount of heat liberated (1.23) (6.12) = 7.52 KJ -(7.52)80 Heat of combustion = 1 = –602.2 KJ 10. ∆H = Lattice energy + Hydration energy 1 = 180 + x Hydration energy = –179 Enthalpy of hydration of Na+ ion is 6 (–179) = –97.6 Kcal 11

O

O

O

O

O

O

O

∆H3 +

Ar (g)

Cl(g)

Cl-(g) Ar+(g) uo

∆H2

∆H1 Ar(g)

+

1/2 Cl2(g)

∆f H

The heat of formation ∆fH may be expressed as ∆fH = ∆H1 + ∆H2 + ∆H3 + U0 ∆H1 = 1526.3 KJ

ArCl(g)

11. H2C2O4 → 2H++ C2O4–2 ∆H = (13.7) 2–25.4 12. C3H8 + 5O2 → 3CO2 + 4H2 O(g) ∆H = –2220.1 + 4 × 44 = –2044.1 KJ 13. The amount of heat evolved is = (3000) (4.18) (77–27) = 627 KJ The amount of fuel 627 = = 15.75 g 40 14. C6H6 + 15 O2 → 3H2O + 6CO2 2 ∆Hcomb = 3 (–68.3) +6 (–94) – 11.7 = –780.6 Kcal The heat evolved by burning 1 kg Benzene is = 780.6 × 1000 = 10000 Kcal 78

7.44

Thermochemistry

16. q = mST = (500) 2 = –1000 J -1000 × 16 So ∆E = = –160 KJ 0. 1 17. ∆H = eH–H+eCl–Cl –2eH–Cl 182 = 430 +242 – 2X X = 245 KJ/mol 19.

1 3 N2 + F2 → NF3 2 2 1 3 ∆H = e N ≡ N + eF-F - 3e N-F 2 2 1 3 -113 = (6X) + × -3(273.5) 2 2 X = 157.22 KJ / mol

22. ∆H = eC=C + eH–H – eC–C –2 eC–H = 611 + 437 – 346 – 2 (413) = –124 KJ/mol 195 = 65 Kcal/mol 28. Bond enthalpy of P–Cl in PCl5 is = 3 Bond energy of axial P–Cl in PCl5 = 58.5 Kcal/mol Bond energy of eqatorial P–Cl in PCl5 = 71.5 Kcal/ mol Enthalpy of atomization of PCl5 = 3(58.5) +2 (71.5) = 318.5 Kcal/mol

multiple choice Questions with only one Answer level II 1. HC≡CH +H2 → C2 H4 ∆H = eC ≡ C + eH-H - eC=C - 2eC - H = 198 + 103 + (–145)–2 (98) = –40 Kcal 2. The gas contains 2.5 mole of H2, 2.5 moles of CO Heat evolved is (241.8) 2.5 + 283(2.5) = 1312 KJ 3. I2 + 2 Na2S2O3 → 2 NaI + Na2S4O6 (120)(0.1) 1 Moles of I2 = × = 0.006 1000 2 I2O5 + 5CO → I2 + 5 CO2 So moles of CO in the mixture is = (0.006) 5 = 0.03 Moles of Carbon = 0.1 Moles of CO + CO2 = 0.1 % of carbon converted into CO is = 30% 5. CH4 + 2O2 → CO2 + H2O ∆H = 4eC-H + 2eO=O - 2eC=O - 4eO-H –Resonance energy of CO2 –∆Hvap H2O(l)

7. S + O2 → SO2 = –2x + y 9. C6H5COOH +

15 O2 → 7CO2 + 3H2O(l) 2

\ ∆H = 7(–393) +3 (–286) – (–408) = –3201 KJ ∆H = ∆E + ∆nRT \ ∆E = –3199.75 KJ + H2 ∆H = –36 Kcal 10. + H2 ∆H = –28 Kcal + 8O2 5CO2 + 6 H2O ∆H = –784 Kcal + 8O2 5CO2 + 6 H2O ∆H = –782 Kcal According to Hess’s law ∆H = – 4 Kcal 11. Evolved heat = absorbed heat 132 x = (110 y)2 x 220 = y 32 x:y = 1:0.6 12. CuSO4 + H2O → CuSO4(aq); ∆H = –15.9 Kcal CuSO4·5H2O + H2O → CuSO4(aq); ∆H = 2.8 Kcal CuSO4 + 5H2O → CuSO4·5H2O; ∆H = –18.7 Kcal 1 13. H2 + O2 → H2O; ∆H = –240 KJ 2 17 C6H10 + O2 → 6CO2 + 5H2O; ∆H = –3800 KJ 2 C6H12 + 9O2 → 6CO2 + 6H2O; ∆H = –392.0 KJ C6H10 +H2 → C6H12; ∆H = –120 KJ 15. CS2 + 3O2 → CO2 + 2SO2; ΔH = –265. Kcal ∆H = HCO2 + 2 HSO2 – HCS2 \ HSO2 = –71.7 Kcal 16.

1 1 X 2 + Y2 → XY 2 2 1 1  a –200 = (a)+   - a 2 2  2 a = 800 KJ/mol

19. C + 1 O2 → CO 2 1 (8.314)298 = 1238 J 2 24. C + O2 → CO2 ∆H = 718 + 498 → (339)2 = 538 KJ The resonance energy = ∆Hexp – ∆Hthe = –393–538 = –931 KJ 25. The amount of heat evolved = (100)4(0.03) = 12 cal The amount of heat evolved for complete oxygenation ∆H – ∆U =

 60, 000  = 12 cal of one mole of hemoglobin is = 12   6  ∆H for one g mole of oxgen is 120 = 30 Kcal 4

Thermochemistry

26. The amount of heat evolved is = 1000(37) = 37 Kcal 37 Calorific value = = 7.4 Kcal/gm 27. C2H6(g) + 7 O2(g) 5→ 2CO2(g) + 3H2O(l) 2 ∆H = 2X3 + 3X2 – X4 ∆H2O(l) → H2O(g)

∆S =

∆H T

7 O2(g) → 2CO2(g) + 3H2O(g) 2 \ ∆H = 2X3 + 3X2 – X4 + 3X1 T1

comprehensive type Questions passage I 1. The amount of heat evolved is (100)(4.18)13 = 5.437 KJ 5.437 ∆Hsolution = = 54.34 KJ mol 0. 1

C2H6(g) +

7 28. C2H6 + O2 → 2CO2 + 3H2O 2 ∆H = 2 (–94.1) +3 (–68.3) – (–21.1) = –372 Kcal 30. CH3COOH + NaOH → CH3COONa + H2O \ ∆H = –57.3 + 2.1 = –55.2 KJ 15 31. C6H5 COOH + O2 → 7CO2 + 3H2O(l) 2 ∆H = ∆U + ∆nRT 15   ∆H–∆U =  7 -  (8.314) (300) = –1247 KJ  2 32. HgO → Hg +

1

2

O2 ; ∆H = 90 KJ/mol

When 45 KJ heat absorbs half moles of Hg is formed. 33. (Cp–Cv) M = R 1.987 M= = 40 0.05

35.

+ H2 → +3H2 →

passage II 1. Heat loss by metal = heat absorbed by water 32.6 (5) 99.83–24.41 = 100 × 4.18 × (24.41–23.62) S = 0.134 J/g°C The value equal to W value 2. (32.6)(S)75.42 = (100 × 4.18 + 410) 0.79 S = 0.266 The values equal to Mo value. passage III 1. Initial temperature is = 22.5°C First temperature is = 26.5°C The enthalpy of the solution is = (50 × 4.18 + 27.8) (26.5 – 22.5) = 947.2 J Heat evolved in this process is equal to –947.2 J (25)(0.7) 2. Equivalent of acid = = 0.0175 1000 Heat of neutralization = 947.2/0.175 = 54 KJ

34. (C6H10O5)n + O2 → 6n CO2 + 5nH2O \ ∆H = (15) (162n) = –2430n KJ ∆H = 6n HCO2 + 5n HH2O – H(C6H10O5)n \ H(C6H10O5)n = –770 n KJ enthalpy in formation 1 g starch is -770n = –4.75 KJ 162n ; ΔH = –119 KJ ; ΔH = (–119)3 KJ

\ ∆Hp–Hr –357 = –156 – Hr Hr = 201 KJ Resonance energy = 49–201= –152 KJ

7.45

passage V 1. 2.303 log

p2 ∆H  1 1  = R  T1 T2  p1

2.303 log 850 = ∆H  1 - 1  76 8.314  353 363  ∆H = 11.99 KJ/mol 2. Kb = =

RT2 M 1000 ∆H

8.314 × (353)2 × 100 1000 × 11.99 × 103

= 8.64 KKg mol–1 ∆ Tb = Kb (1 + α)m 1 1000 1 = 8.64 (1+ α) × 100 100 1 + α = 1.158 α = 0.158

7.46

Thermochemistry

3. Equilibrium constant at 80°C H+ + A cα cα

 HA ↽ ⇀  c(1 - α)

Cα 2 (0.1)(0.158) = 1- α 0.842 = 0.01876 Kb at 90°C is

Kb =

=

(0.1)(0.189) = 0.0233 0.811

2.303 log

K b1 K b2

=

1 ∆H  1 R  T1 T2 

∆H = 23.4 KJ HA + NaOH → NaA + H2O ∆H = –573 + 23.4 = –33.9 KJ

Integer type Questions 1. ∆ → CH3 CH = CH2 –33 = 20.1–Hcyclopropane Hcyclopropane = 53.1 KJ 9 ∆ + O2 → 3CO2 + 3H2 O 2 ∆H = 3(–393.5) + 3 (–285.8) –53.1 = –1985 KJ 2. The amount of heat evolved = msdT = 2(0.225) = 0.45 Heat of combustion of constant volume ∆E = 0.45 × 46 = 9 KJ / mol 0.23 3.

Bond energy of I–F is = 268 KJ

1 5 I2 (s) + F2 (g ) → IF5 (g ) 2 2 1 1 5 ∆H = e + eI - I + eF - F - 5eI -FF 1 2 I2 (s )→ 2 ( g ) 2 2 1 1 5 -847 = (62) + (149) + (155) - 5eI - f 2 2 2

1 3 I2 + F2 → IF3 ; ∆H = -470 KJ 2 2 1 3 1 -470 = (62) + (149) + (155 - 22.68 - eI - F 2 2 2 Equation bond energy = 272 KJ 5. C(Graphite) → C (Diamond) ∆H–∆U = P∆V = 2 6. H2A + 2NaOH → Na2A + H2O; ∆H = –22 Kcal \ H2 A → 2H + + A 2 - = 28 - 22 = 6 Kcal 7. HAuBr4 + HCl → HAuCl4 + 4HBr; ∆H = 8.8 Kcal When 0.44 Kcal heat was absorbed then 0.05 mole of HAuBr4 converted to AuCl4. 8.

1 1 A 2 + B2 → AB 2 2 1 1 ∆H = e A - A + eB - B - e A - B 2 2 x 0. 5 x-x -100 = + 2 2 \ x = 400 KJ

9. ∆G = ∆H – T∆S 10. ∆H = –2.303 RT log Kp O

previous years’ IIt Questions 1. The reaction CO2 + H2 → CO+H2O ∆H = HCo + HH2O – HCO2 = –110.5 + (–241.8) – C (–393.5) = 41.2 KJ 2. One more product obtained from elements 3. 2C (s) + H2(g) → C2H2(g) ∆H = 1410 + 330 – 2 × 350 – eC=C 225 = 1410 + 330 – 700 – eC ≡ C eC = C = 815 KJ

CHAPTER

8 Chemical Kinetics

I

wish that my room had a floor. I don’t so much care for a door, but this crawling around without touching the ground is getting to be quite a bore. Glett Burgess

8.1 IntroductIon Thermodynamics explains that whether a reaction should, or should not, take place. This can be worked out by finding the free energy change. If the change is negative, then the reaction can occur; it is a spontaneous reaction. If the change is positive, the reaction cannot take place. Thus thermodynamics shows that reactions in which there is a large evolution of heat and an increase in disorder are highly likely to occur; yet glyceryl trinitrate shows no tendency to decompose if it is treated with due respect; and coal shows no signs of reacting with the oxygen in the air at room temperature. However, glyceryl trinitrate decomposes with spectacular success if it is dropped and once the combustion of coal is underway the reaction is self sustaining. In these two cases, some energy is needed to initiate reaction even though, the reactants are thermodynamically unstable with respect to their products. Further, thermodynamics does not tell how fast a spontaneous reaction will be. For example, at room temperature, the free energy change for diamond reacting with oxygen to make carbon dioxide is negative, yet they are not burning away from the jewellery in which they are fixed. On the other hand, the reaction between white phosphorous and oxygen is also spontaneous and takes place very easily. In fact, white phosphorous has to be kept under water to keep oxygen away from it. If it is put in air, it ignites. A question of importance which does not touch is “how rapidly and what mechanism does a reaction take place”? Thermodynamics only considers (1) The energy relations between reactants and products of reactions. (2) It does not indicate the stages through which reactants may have to pass to reach the final product (3) it does not indicate the rate at which equilibrium is attained. Chemical kinetics, however compliments thermodynamics by supplying information about the rate of approach

to equilibrium and about the mechanism responsible for the change of reactants to products. The aim of chemical kinetics is to predict the rates and mechanism of chemical reactions. It may also be defined as the branch of physical chemistry which concerns itself with the study of the velocity of chemical reactions and with the elucidation of the mechanisms by which they proceed. It should, however, be noted that all reactions cannot be studied readily by chemical kinetics. Different reactions vary widely in velocity. At room temperature, some reactions are very fast, for example ionic reactions in solution often take place practically instantaneous, thus silver chloride precipitates as fast as silver and chloride ions are mixed. Some reactions on the other hand are so slow that it is not possible to detect any change at all. For example, hydrogen and oxygen will react to from water, but if the two gases are mixed at ordinary temperatures, nothing appears to happen because the reaction is extremely slow. Between these two extremes are many reactions which take place at measurable speed, so that it may take some time to obtain a good yield of the products; this is often the case, for example in organic preparations. The study of the reaction rates is also known as reaction kinetics or chemical kinetics. To study the reaction kinetics one will make many measurements of the speed of the reactions. Once the measurements are made, the next stage will be to try to explain the observations. The explanation is called the mechanism of the reaction. Among other things, the mechanism helps us imagine how new bonds are made and old ones are broken. It should be noted that in slow reactions, a large number of bonds have to be broken in reactant molecules and a large number of new bonds have to be formed in the product molecules. In case of simple ionic reactions, no bonds are to be broken. These reactions are therefore very fast: From now on we shall normally assume that the reactions we discuss are all spontaneous, i.e., thermodynamics

8.2

Chemical Kinetics

say that the reactions can take place but we have to explain why the reactions are fast or why they are slow.

8.2 rate of reactIon Chemical reactions seldom take place in accordance with the stoichiometric equation and generally, reaction proceeds through a number of stages in which one step — the slowest, controls the rate at which reactants are consumed and products are formed. However, at this stage, we shall limit our discussion to the reaction that takes place in one stage. The quantity of a reactant species consumed or the quantity of a product species formed in unit time in a chemical reaction is called the rate of that reaction. Consider a general reaction A→B With the passage of time the concentration of A goes on decreasing whereas the concentration of B goes on increasing. Graphically, the change in concentration with the passage of time is shown in Fig 8.1.

t duc

concentration

Pro

If [A]1 and [B]1 are the concentrations of A and B respectively at time t1 and [A]2 and [B]2 are their concentrations at time t2, then Δt = t2 – t1 Δ [A] = [A]2 – [A]1 Δ [B] = [B]2 – [B]1 where Δ[A] and Δ[B] are the changes in concentration of A and B during time interval ΔT. Then rate of reaction may be expressed as ∆ [A] Rate of reaction = − (7.1) ∆t Or Rate of reaction =

∆[B] ∆t

(7.2)

Where Δ[A] gives the decrease in concentration of A and Δ[B] represents increase in concentration of B. The square brackets around the substances are used to express the molar concentration (mol/ litre). It may be noted that in the case of concentration of reactants, minus sign is used. This implies that the concentration of reactants is decreasing with time. The above rate is also called average rate of the reaction. It has the dimension of concentration/ time. The units of concentrations and time are chosen according to the convenience. In the above example, the stoichiometric coefficients for the reactants and products are same. Therefore the rate at which the concentration of A decreases will be the same as the rate at which the concentration of B increases. Therefore, the rates expressed by Eq (7.1) and (7.2) are same.

8.2.1 reactions Involving different Stoichiometric Coefficients of Reactants and Products

reacta

nt

Consider the reaction A + B → 2C

Time

fig 8.1 Change in concentration of reactant and product with time The rate of the reaction may be expressed in either of the following two ways: (i) The rate of disappearance or decrease in concentration of A (reactants) Rate of reaction =

Decrease in concentration of A Time taken

(ii) The rate of increase in concentration of B (products) Rate of reaction =

Increase in concentration of B Time taken

In this case, one mole of A reacts with one mole of B to form 2 moles of C. This means that the rates of disappearance of A and B are same but the rate of appearance of C must be twice the rate of disappearance of A and B. Thus = 2 × Rate of disappearance of B = Rate of appearance of C To get unique value of the reaction, rate (independent) of the concentration terms chosen, we divide the rate of reaction defined with any of the reactants or products by the stoichiometric coefficient of that reactant or product involved in the reaction. Thus for the above reaction Rate of reaction = −

∆[A] ∆[B] 1 ∆[C] =− = ∆t ∆t 2 ∆t

Chemical Kinetics

Some other examples are as follows:  C + D (i) A + B 

Rate = −

∆ [A] ∆t

= -

∆[C ] ∆[ D] ∆[B] = = ∆t ∆t ∆t

 2C (ii) A + 2B 

Rate = -

1 ∆[B] 1 ∆[C] ∆[A] == ∆t 2 ∆t 2 ∆t

 3C + 2D (iii) 2A + 3B 

D[D] 1 ∆ [A] 1∆ 1 ∆[C ] ∆ 1 ∆ [A] = Rate = - − = - − = 3 ∆t 2∆t∆t 2 ∆t 3 ∆t  4NO2 (g) + O2 (g) (iv) 2N2O5 (g) 

1 ∆[N 2 O5 ] 1 ∆[NO 2 ] ∆[O 2 ] = = Rate = 2 ∆t 4 ∆t ∆t  2NH3(g) (v) N2 (g) + 3H2(g) 

Rate = -

∆[N 2 ] 1 ∆[H 2 ] 1 ∆[NH 3 ] == ∆t 3 ∆t 2 ∆t

change. This is due to the fact that in the case of chemical reactions, the rate depends upon the concentration of the reactants keeps on decreasing. Thus the rate of reaction may not be constant in the time interval which we measure. Thus the rate cannot be determined simply by dividing the total change in concentration by the time taken as the case of mechanical speed on the other hand, the rate of reaction may be expressed at a particular moment of time, known as instantaneous. Thus the instantaneous rate may by defined as the change of concentration of any one of the reactants or products at a given time. For instantaneous rate, the time interval ∆t is made as small as possible so that the rate of reaction remains almost constant during the time interval. Mathematically, instantaneous rates may be represented by -

∆ [A] d[A] d[B] ∆[B] or instead of or respectively. ∆t dt ∆t dt

Here, d[A] or d[B] represents infinitesimally small interval of time dt. Thus average rate approaches the instantaneous  ∆[ A]  rate as ∆t becomes smaller and smaller i.e., = -   ∆t  ∆t → 0 d[B]  ∆[B]   d[A]  and  = = -  ∆t  ∆t → 0  dt  dt Thus R inst = -

(vi) For any general reaction like  rR + sS pP + qQ  1 ∆[P] 1 ∆[Q] 1 ∆[R] 1 ∆[S] == = Rate = p ∆t q ∆t r ∆t s ∆t

8.2.2 Average Rate and Instantaneous Rate The rate measured over a long time interval is called average rate and the measure for an infinitesimally small time interval is called instantaneous rate. The rate expressions given so far give the average rate reaction over the time interval Δt. Rate =

Change in concentration ∆x = ∆t Time interval

This concept is similar to mechanical speed. For example, the average speed of a car can be easily calculated by knowing the distance travelled by it in a given time. However, during the journey, the speed keeps on changing depending upon the conditions of the road, weather, traffic density etc. The speed of the car may vary 10 km/h to 100 km/h. For a chemical reaction also, the average rate depends on the values of t2 and t1 chosen. Similar to speed of the car, the instantaneous rates of chemical reaction also

8.3

d[A] d[B] = dt dt

In general, if dx represent very small (infinitesimally small) change in concentration of any species during the very small (infinitesimally small change of time dt, the rate of reaction may be expressed as dx Rate of reaction = dt For a general reaction  rR + sS pP + qQ 

Rate = −

1 d[P] 1 d[Q] 1 d[R] 1 d[S] =− = = p dt q dt r dt s dt

8.2.3 units of rate of reaction The units of rate of reaction are concentration time–1. As concentration of substance is expressed in mol L–1 and the time is expressed in seconds or minutes or hours the units for reaction rate therefore are mole litre–1 sec–1 or moles litre–1 min–1 or moles litre–1 hr–1. For gaseous reactions, if the concentration of reactants and products are given in terms of partial pressure, then the units of rate of reaction will be atm min–1 or atm sec–1 or atm hr–1 (pressure is expressed in atmospheres).

8.4

Chemical Kinetics

The partial pressure of any gaseous component can be calculated by any one of the following methods.

reaction mixture so as to stop the reaction or decrease its rate considerably. (iii) Plot a graph by taking molar concentration along Y- axis and corresponding time along X-axis.

(i) From ideal gas equation PV = nRT nRT w RT dRT = = or P = = CRT V V M M

Calculation of average rate: The average rate of reaction is determined by noting the concentration of reactants at two different times (Fig: 8.2) x1 and x2 are the concentrations. It is clear that the concentration of reactant decreases with time making (x2-x1) a negative quantity. Thus the minus sign makes the rate as positive quantity. This gives the average rate of reaction in time interval. Calculation of instantaneous rate of reaction: The instantaneous rate gives the rate at any specific instant of time during a chemical reaction. It can be determined first by plotting a graph by taking molar concentration along y-axis and corresponding time along x-axis. The instantaneous rate can be determined by drawing a tangent to the curve at a point corresponding to the given time.

(ii) When mole fraction is given then by using the expression P = P total × mole fraction of gas in question Interconversion of units. The partial pressure can also be converted into units of concentration i.e., moles per litre by using the following equation PV = nRT or P =

P n RT = CRT or C = RT V

8.2.4 determination of the rate of reaction In order to determine the rate of a reaction: (i) Select a reactant or product species whose concentration is measurable such as change in volume, pressure, refractive index, pH, electrical conductivity or thermal conductivity etc. The change in this property is measured as a function of time. (ii) Note down the concentration of that species at regular interval of time. It is done by withdrawing a small amount of reaction mixture freezing it and then analysing it quantitatively. Freezing is done by cooling the

Consider a general reaction A+B→C+D If it is convenient to measure the concentration of A, note down the concentration of A after regular intervals of time (Say 5 minutes) plot a graph from the collected data (Fig 8.3).

Concentration of reactant

[R]Ө

rav =

−Δ [R]

[R1]

dt

=

− [R2] - [R1] (t2 - t1)

∆[R] [R2]

d[R]

rinst =

−d[R] = − slope dt

∆t

dt

t1

t2

t

Time

fig 8.2 Determination of instantaneous and average rate of reaction following the concentration of reactant

Chemical Kinetics

rinst =

d[P] = slope dt

8.5

[P]

d[P] Concentration of product

dt [P2] Δ [P] rav [P1]

=

Δ [P] Δ t

=

[P1 - P2] (t2 - t1)

Δ t

t2

t1

t Time

fig 8.3 Determination of instantaneous and average rate of a reaction following the concentration of product

Suppose, we want to find the rate of these general reactions after ten minutes (t = 10 mts). Locate the point P on the curve by drawing a perpendicular to the x-axis at a point B and y-axis at a point A. OA ∆x 0.09 = = Slope of the tangent = tan θ = = OB ∆t 20 –3 –1 –1 4.5 × 10 moles litre minute This gives the rate of the reaction after the time 10 minutes. Here OA tells the change in concentration, say ∆x and OB tells the change in time say ∆t. ∴ The rate of reaction at anytime (t) =

∆x ∆t

Consider that the change in concentration of the reactant species in an infinitesimally small time (dt) = dx Hence rate of reaction =

dx dt.

(i) If the rate is expressed in terms of the change in concentration of the reactants, there will be a decrease in concentration with time Thus the rate of reaction = -

dx dt.

-dx represents the decrease in concentration in time dt. (ii) If the rate is expressed in terms of the change in concentration of any one of the products, there will be increase in concentration with time.

∴ Rate of reaction = +

dx dt

+ dx represents the increase in concentration with time dt Solved Problem 1 The concentrations of a reactant A at different times are given below: t (s)

[A] (Mol L–1)

0 5 10 20 30

160 × 10–3 80 × 10–3 40 × 10–3 10 × 10–3 2.5 × 10–3

Calculate the average rate of reaction A × B during different intervals of time. Solution: The average rate of reaction can be calculated as rav = −

[A]2 − [A]1 . ∆t

8.6

Chemical Kinetics

[A]1 160

[A]2 80

t1 0

t2

rav (mol L–1 s–1) -3

5

-

(80 - 160) × 10 5-0

(40 - 80) × 10 -3 = 8 × 10 -3 10 - 5

80

40

5

10

-

40

10

10

20

-

10

2.5

20

30

= 16 × 10 -3

(10 - 40) × 10 -3 = 3 × 10 -3 10 (2.5 - 10) × 10 -3 = 0.75 × 10-3 30 - 20

Problems for Practice 1. Ammonia and oxygen react at high temperature as 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(g) In an experiment rate of formation of NO is 3.6 × 10–3 Mol L–1 S–1. Calculate (i) rate of disappearance of ammonia (ii) rate of formation of water. 2. The decomposition of N2O5 in CCl4 solution at 318 K has been studied by monitoring the concentration of N2O5 in the solution. Initially, the concentration of N2O5 is 2.33 M and after 184 minutes. It is reduced to 2.08 M the reaction takes place according to the equation. 2N2O5 → 4NO2 + O2 Calculate the average rate of this reaction in terms of hours, minutes and seconds. What is the rate of production of NO2 during this period? 3. The following reaction was carried out in water: Cl2 + 2I– → 2Cl– + I2 The initial concentration of 2I– was 0.50 mol l–1 and concentration after 10 minutes was 0.46 mol–1. Calculate the rate of disappearance of I– and rate of appearance of iodine. 4. The oxidation of iodide ion by peroxydisulphate ion is given as 3I– + S2 O82- → I–3 + 2 SO 24 ∆[S2 O82 - ] = 1.5 × 10 -3 mol L–1S–1 for a given ∆t interval then what is the value of - ∆[ I ] for the ∆t same interval? (ii) Also calculate the average rate of formation of SO 2− 4 for the same interval. (i) If -

8.3 factorS whIch Influence the rate of reactIonS Various factors that influence the rate of reaction are as follows. 1. Temperature: An increase in temperature leads almost invariably to an increase in the rate of reaction. In fact for many reactions, an increase of 10°C in temperature may increase the rate of the reaction two fold or more. 2. Concentration: With the exception of certain (Zero) order reactions, upon which concentration is without effect, an increase in the initial concentration of reactants also results in the acceleration of rate of reaction. 3. Nature of reactants and the products: Rate of reactions are influenced by the nature of reactants and products. A chemical reaction involves the breaking of old bonds and formation of new bonds. The reactivity of a substance can, therefore, be related to the ease with which the specific bonds involved. For example, the oxidation of nitric oxide to nitrogen dioxide takes place fairly rapidly while oxidation of carbon monoxide to carbon dioxide takes place slowly. 2NO + O2 → 2NO2 2CO + O2 → 2CO2

(Fast) (Slow)

In these two reactions, the reacting species appear to be very similar to each other; still they differ in reaction rates. 4. Presence of catalyst: Many reactions are influenced by the presence of substance which posses the ability of increasing or decreasing the rates of such reactions. Substances possessing this property of affecting the rates of chemical reactions are called catalysts while the reactions thus affected are designated as catalysed reactions. For example, a mixture of H2 and O2 does not react at room temperature. However, in the presence of a catalyst such as finely divided platinum, the reaction becomes quite vigorous. 5. Surface area: The larger the surface area of the reactants, the faster the rate of reaction. It has been found that if one of the reactant is solid, then the rate of the reaction depends upon the state of sub-division of the solid. This fact is supported by the general observation that finely divided solids react faster than massive substances. For example, a log of wood burns slowly but if it is cut into small wooden chips, the burning takes place rapidly. This is due to the fact that the total surface area of the smaller particles is greater than that of large particles and this permits more molecules of the reactants to come in contact and form products.

Chemical Kinetics

6. Pressure: Pressure has a very little effect on rates of reactions involving solids and liquids. For reactions involving gases, however, an increase in pressure increases the concentration by bringing molecules close together and hence increases the rate of reaction. 7. Exposure to radiation: Certain reactions are not influenced by light, while some reactions called photochemical reactions are greatly stimulated when light of suitable wavelength is allowed to pass through the reaction mixture. For example, reaction between hydrogen and chlorine or methane and chlorine takes place very slowly in the absence of light. However, in the presence of light these reactions takes place very rapidly.

8.7

8.4.1 characteristics of rate constant (i) The value of rate constant gives an idea about the speed of a reaction, i.e., greater is the value of rate constant faster is the reaction. In simple words, it is a measure of rate of reaction. (ii) Each reaction has a definite value of the rate constant at a particular temperature. (iii) The value of rate constant depends on the temperature i.e., changes with change in temperature. (iv) The value of rate constant is independent of the concentration of reacting species. (v) The units of rate constant depend on the order of reaction.

8.4 rate lawS Based on quantitative observations, Guldberg and Wage in 1867 investigated the effect of concentration or mass on the rate of chemical reaction and put forward a law known as law of mass action. According to this law, “The rate at which a substance reacts is directly proportional to its active mass and the rate at which a chemical reaction proceeds is proportional to the product of the active masses of the reacting substances”. The term active mass used in the above statement implies molecular concentration. In the case of gases and solutions, it means the number of gram molecules present per litre. The active mass of solids taken as unity. The active mass is usually expressed by enclosing the symbol or the formula of substance in square brackets. For example, [HCl] = 1 means 36.5 g of HCl per litre. Consider the following reversible reaction taking place at constant temperature.

 C + D A + B  The active masses or molar concentrations of A, B, C and D are [A], [B], [C] and [D] respectively. According to law of mass action, the rate of the reaction may be written as Rate ∝ [A] [B] = K [A] [B] where K is constant of proportionality and called rate constant. The rate constant is also called as velocity constant or specific rate constant and is a measure of the reaction. Now if the concentration of each of the reactants involved in the reaction is unity i.e., [A] = [B] = 1 then substituting these values in the above expression , we get Rate of reaction = K × 1 × 1 = K Thus the rate constant of a reaction at a given temperature may be defined as rate of the reaction when the concentration of each of the reactants is unity.

8.4.2 differences Between rate of reaction and Rate Constant The important difference between the rate of the reaction and rate constant of the reaction are as follows: Rate of Reaction

Rate Constant of Reaction

1. It is the speed at which the reactants are converted into the products at any moment of time. 2. It depends upon the concentration of reactant species at that moment of time. 3. It generally decreases with the progress of reaction.

1. It is constant of proportionality in the law expression.

2. It refers to the rate of reaction at the specific point when concentration of every reacting species is constant. 3. It is constant and does not depend on the progress of the reaction.

8.5 RATE LAw ExPRESSIoN Certain reactions proceed through more than one step. Let us now understand that which of these steps can be used to write the rate expression. Out of various steps of reaction, the slowest step will decide the rate of overall reaction because the reaction cannot take place faster than the slowest step. Thus the slowest step of the complex reaction is called the rate controlling step or rate determining step. A few examples are discussed to illustrate the above point more clearly. 1. Thermal decomposition of nitrous oxide 2N2O → 2N2 + O2 Experiments show that the rate of this reaction depends on the single power concentration of N2O. thus

8.8

Chemical Kinetics

a possible mechanism of the reaction may be proposed as Step I

→N O N2O  N2 + O

Step II

O2 N2 O + O → N N2 + O

Slow

fast

2N2O

 → 2N2 + O2 over all reaction

The rate law expression for this reaction can be written as Rate = K [N2O] The above postulated mechanism is consistent with the rate law expression. 2. Thermal decomposition of dinitrogen pentoxide 2N2O5  → 4NO2 + O2 The rate of above reaction is found to be dependent on single power of concentration of N2O5. Thus the possible mechanism for the reaction is Step I

→ NO NO2 + NO3 N2O5 

Step II

N2O5 + NO3 → 33NO2 + O2

on which the rate of reaction depends. It may be noted that the values of x, y and z are determined experimentally and may or may not be equal to a, b and c coefficients in the reaction. The above expression is the rate law or rate equation. It may be defined as the mathematical expression which denotes the experimentally observed rate of reaction in terms of the concentrations of the reacting species which influence the rate of the reaction. Thus it may be noted that the rate law gives the experimentally observed rate which depends on the concentration of reactants. The power of concentrations of reactants in rate law expression may or may not be same as the coefficient of reaction in balanced chemical equation. On the other hand, the law of mass action give the expression on the basis of overall balanced equation for the reaction. For example, for the reaction 2N2O5(g) →4NO2(g) + O2(g) Rate law expression

slow

fast

Rate = K [N2O5] Rate expression according to law of mass action Rate = K [N2O5]

The rate law expression for this reaction is Rate = K [N2O5]

Another example is the reaction between NO2 and F2 to from NO2 F

3. Reaction between nitric oxide and hydrogen 2NO + 2H2  → N2 + 2H2O Experiments shows that the rate of this reaction depends upon square of concentration of NO and single power of concentration of H2. Consistent with this data, the logical mechanism for the reaction may be written as Step I

Slow 2NO + H2  → N 2 ×+ H2O2

Step II

H2 O2 + H2  → 22H2O

The rate law expression for this reaction is Rate = K [NO]2 [H2] Consider a general reaction

2NO2 (g) + F2 (g) → 2NO2F(g) Experiments show that the rate of the reaction is proportional to the product of the concentrations of the nitrogen dioxide and fluorine. So the rate of the reaction, may be written as. Rate = K [NO2][F2] Rate expression according to law of mass action 2

Rate = K  NO 2   F2  It may be noted that generally the rate law expressions are not simple. These may differ even for the same reaction depending upon the conditions of the reaction; for example, for the reaction

aA + bB + cC  → products Where A, B and C are the reactants and a,b and c are stoichiometric coefficients of the balanced equation. From the kinetic study of the reaction, the dependence of concentration of reactants on the rate of reaction have been found to be Rate = [A]x [B]y [C]z Where x y and z are constant numbers or the powers of the concentrations of the reactants, A, B and C respectively

H2 (g) + I2 (g) → 2HI (g) Rate = K [H2] [I2] However, for a similar type of reaction H2 (g) + Br2 (g) →2HBr (g) The rate expression is 1

Rate =

Ka [ H 2 ][ Br2 ] 2

1 + Kb {[ HBr ] [ Br2 ]}

Chemical Kinetics

Here Ka and Kb are constants which depend upon the temperature of the reaction. It can be seen that in the absence of HBr here, the initial rate of reaction may be written as Rate = Ka  H 2   Br2 

1

2

The rate expression becomes more complicated especially when a number of parallel reaction take place simultaneously. It may have more than one term on the right hand side.

8.6 order of reactIon It is an important parameter for every chemical reaction, It is always determined experimentally and cannot be written from the balanced chemical equation. It refers to the number of reacting particles whose concentration terms determine the reaction rate. Moreover, the concentration dependence of various reactants, as we have seen, is given by the rate law. Thus the order of a reaction is defined as The sum of the powers to which the concentration terms are raised in the rate law expression. The power of the concentration of particular reactant in the rate law is called the order of the reaction with respect to that reactant Thus, if the rate of a reaction aA + bB + cC → products is given by the rate law as Rate = -

dx = K [ A]x [ B ] y [C ]z dt

Then the order of the reaction, n is n=x+y+z Where x, y and z are the order with respect to individual reactants and overall order of the reaction is sum of these exponents i.e., x + y + z When n = 1 the reaction is said to be first order reaction, if n = 2, it is said to be second order reaction and so on. Some common examples of reactions of different orders

1. Reactions of first order (i) Decomposition of H2O2 in the presence of I– Ions 1 IH2O2 → H2O + O2 2 Rate = K[H2O2] (ii) Decomposition of nitrogen pentoxide (N2O5) 1 N2O5 → 2NO2 (g) + O2 2 Rate = K[N2O5]

8.9

(iii) Decomposition of ammonium nitrite in aqueous solution NH4NO2 (aq) → N2(g) + 2H2O(l) Rate = K[NH4NO2] (iv) Dehydrohalogenation of ethyl chloride C2H5Cl(g) → C2H4(g) + HCl(g) Rate = K [C2 H5 Cl] 2. Reactions of second order (i) Oxidation of nitric oxide to nitrogen dioxide by ozone NO + O3 → NO2 + O2 Rate = K [NO] [O3] (ii) Reaction between H2 and I2 to give HI H2 + I2 → 2 HI Rate = K [H2] [I2] (iii) Decomposition of nitrogen dioxide 2NO2 → 2NO + O2 Rate = K [NO2]2 3. Reactions of third order (i) Reaction between nitric oxide and oxygen 2NO + O2 → 2NO2 Rate = K [NO]2[O2] (ii) Reaction between nitric oxide and Cl2 2NO + Cl2 → 2NOCl Rate = K [NO]2 [Cl2] 4. Reactions of fractional order (i) Formation of carbonyl chloride (Phosgene) CO + Cl2 → COCl2 1

Rate = K [ CO ] [ Cl2 ] 2 order = 2.5 (ii) Decomposition of carbonyl chloride COCl2 → CO + Cl2 2

3

order = 1.5 Rate = K [COCl ] 2 (iii) Decomposition of acetaldehyde CH3 CHO → CH4 + CO 3

Rate = K CH 3 CHO  2 order = 1.5 (iv) Reaction of chlorine with chloroform to convert into carbon tetrachloride CHCl3 (g) + Cl2 (g) → CCl4 + HCl 1

Rate = [ CHCl3 ]  Cl2  2 order = 1.5 5. Zero order reactions: The reaction in which the rate of reaction does not depend on the concentrations of the reactants are known as zero order reactions. A number of zero order reactions are known. Some examples are given below: (i) The decomposition of ammonia at the surface of metals like gold, platinum etc, is a zero order reaction. Pt 2NH3  → N2 + 3H2

8.10

Chemical Kinetics

It has been observed that the rate of the reaction is independent of the concentration of ammonia i.e., dx = K [ NH 3 ]0 or rate = K. Rate = dt Order = 0 Many photochemical reactions (e.g., formation of HCl from H2 and Cl2) are also examples of zero order reactions. H2 + Cl2  → 2HCl hv

Rate = K[H2]0 [Cl2]0 order = 0 Decomposition reactions of N2O, HI,PH3 etc, in the presence of metal are zero order reactions. N 2O → N 2 + 1 O 2 2 Rate = K[N2O]0 order = 0 2HI → H2 + I2 Rate = K [HI]0 order = 0  → 2 P + 3H 2 2PH3 high pressure Mo

Rate = K[PH3]0 order = 0 units of reaction rate constants (i) Units of rate constant for zero order reaction For zero order reaction, rate may be expressed as Rate = K[A]0 (or) = K mol L-1 =K s or K = mol L–1 S–1 (ii) Units of rate constant for first order reaction The rate of first order reaction may be expressed as Rate = K [A] mol L-1 = K (mol L–1) s K = s–1 (iii) Units of rate constant for second order reaction Rate = K [A]2 -1

mol L s K=

= K (mol L–1)2

1 = L mol–2 s–1 mol 2 s

(iv) Units of rate constant for third order reaction The rate expression for third order reaction is Rate = K [A]3 mol L-1 = K (mol L–1)3 s

K=

1 = L2 mol–2S–1 mol 2 L-2 s

In general, the units for rate constant for the reaction of nth order are Rate = K [A]n mol L-1 = K (mol L–1)n s

or K = mol 1–n Ln–1 s–1 Units of Rate Constants for Gaseous Reactions In case of the gaseous reactions, the concentrations are expressed in terms of pressure in the units of atmosphere. Therefore, the rate has the units of atm per second. Thus, the unit of different rate constant would be: (i) Zero order reaction: atm–1 (ii) First order reaction: s–1 (iii) second order reaction: atm–1 s–1 (iv) Third order reaction: atm–2 s–1 so on In general, for the gaseous reaction of the nth order, the units of rate constant are (atm)1–n s–1.

8.6.1 Pseudo Chemical Reactions Some reactions are first order each with respect to two different reactants i.e., A + B → Products Rate = K [A] [B] However, if one of the reactants is present in high concentration (solvent) then there is very little change in its concentration. In other words, the concentration of that reactant remains practically constant during the reaction. For example, if [A] = 0.01 M and that of solvent water [B] = 55.5 M, the concentration of B changes only from 55.50 to 55.49 M even after the completion of the reaction. Under such conditions, we may write Rate = K0 [A] where K0 = K [B] The reaction, therefore, behaves as a first order reaction in A. Such reactions are called pseudo first order reactions For example; consider the following reaction:

  CH3COO C2H5 + H2O   CH3COOH + C2H5OH Acid

The molecularity of the reaction is two because it involves two reacting species, namely ethyl acetate and water. However, the concentration of ethyl acetate changes during the reaction while water is present in such a large excess that its concentration remains practically unchanged.

Chemical Kinetics

Therefore, the rate of the reaction depends only on the concentration of ethyl acetate and hence the order of the reaction is one. Thus the reaction appears to be second order but follows the first order kinetics. Such reactions which appear to be higher order but actually follow lower order kinetics are called pseudo chemical reactions. Hence the above acid catalysed ester hydrolysis reaction is called pseudo first order reaction Another example is the hydrolysis of cane sugar to give glucose and fructose. C12H22O11 + H2O  → C6H12O6 + C6H12O6 Sugar water glucose fructose In this reaction also, molecularity is two while the order is one. Solved Problem 2

→ 2C (a) A + 3B  Obeys the rate equation Rate = K [A]1/2 [B] 3/2 What is the order of the reaction? (b) The reaction A + B → C has zero order. Write rate equation. Solution: (a) Rate equation Rate = K [A]1/2 [B] 3/2 Order of reaction = ½ + 3/2 = 2 (b) For the zero order reaction A+B→C Rate equation Rate = K [A]0 [B]0 Solved Problem 3 (a) If concentrations of A and B are expressed in terms of mol dm3 and time in min, calculate the units for the rate constant for the following reaction. A + B → AB (b) What are the units for a zero order reaction? (Concentrations are expressed in mol L–1 and time in seconds) Solution: (a) For the reaction A + B → AB Rate = K [A] [B] The rate is change in concentration with time. The time and concentration are expressed in min and mol dm–3 respectively mol dm -3 = K (mol dm–3)(mol dm–3) min

Or K = (mol dm–3)–1 (min–1) = mol–1 dm3 min–1

8.11

Problems for Practice 5. State the order with respect to each reactant and overall order for the following reactions. (i) C2H5Cl(g) → C2H4 (g) + HCl(g) Rate = K[C2H5Cl] – – (ii) H2O2 + 3I + 2H + → 2H2O + I 3 (aq) – Rate = K[H2O2] [I ] (iii) CH3CHO(g) → CH4(g) + CO(g) Rate = K[CH3CHO]3/2 (iv) 2NOBr(g) → 2NO(g) + Br2(g) Rate = K[NOBr]2 (v) 3NO(g) → N2O(g) + NO2(g) Rate = K[NO]2 (vi) CHCl3 + Cl2 (g) → CCl4 (g) + HCl (g) Rate = K[CHCl3] [Cl2]1/2 What are the dimensions of rate constant in each case?

8.7 MolecularIty of a reactIon According to collision theory for a chemical reaction to occur, the reacting molecules must collide with each other and according to kinetic molecular theory “The rate of a reaction is directly proportional to the number of collisions taking place between the reacting molecules”. The number of reacting species (Molecules, atoms or ions) which collide simultaneously to bring about a chemical reaction is called molecularity of a reaction. As we have seen that reactions can be classified as first, second, third or nth order, depending upon the influence exerted by concentration on the rate of reaction, they can also be classified in term of molecularity as unimolecular, bimolecular, termolecular, depending upon the number of molecules involved in the collision that lead to the chemical reaction. Examples (i) Dissociation of PCl5 involves single species which undergoes the change to form the products. Hence it is unimolecular reaction. PCl5 → PCl3 + Cl2 (ii) Dissociation of hydrogen iodide is a bimolecular reaction because two molecules collide to bring about the reaction. So it is bimolecular reaction. 2HI → H2 + I2 (iii) The reaction of nitric oxide and oxygen is a termolecular reaction because it involves collision of three reacting molecules. 2NO (g) + O2 (g) → 2NO2 (g)

8.12

Chemical Kinetics

8.7.1 why the Reactions of Higher order are rare It is observed that the molecularity more than three is rare or uncommon. It is so because the occurrence of such reactions will involve the collision of four or more molecules simultaneously. But the chances for colliding more than three different species simultaneously, each possessing energy equal to or greater than the threshold energy are rare. Consider the reaction MnO4– + 5Fe2+ + 8H+ → Mn2+ + 4H2O + 5 Fe3+ The above stoichiometric equation give only an approximate picture since we cannot imagine that in these equations the fourteen species simultaneously react together. This would involve an initial step with 14 collisions, which is not possible.

8.7.2 Mechanism and Rate Law Now the question arises that how is it possible to explain the mechanism of reactions involving six, seven or even more molecules. In order to account for such reaction, Vant Hoff proposed that the order of a complex reaction is given by the order of the slowest step in the sequence of various steps involved in that reaction. Such type of reactions which takes place through a sequence of two or more consecutive steps are called complex reactions. The detailed description of various steps by which reactants change into products is called mechanism of the reaction. The steps which contribute to the overall reactions are called elementary processes. In case of multi-step reactions, since each elementary step involves different types of reactions and each step will occur with its own distinctive rate. If one step takes place slowly than others, it will definitely control the overall reaction rate. This means that all the steps have to wait for the occurrence of this step and the rate of the reaction cannot be less than the rate of this slowest step. Once this slowest step has occurred, the other steps will take place to form the products. Thus the rate of reaction is The rate of the reaction which is determined by the slowest step in the sequence. The slowest step is called rate determining step in the proposed mechanism. This can be explained by considering the reaction between NO2 and F2 to form NO2F as an example. The experimental observations reveal that the rate of the reaction is proportional to the product of the concentrations of nitrogen peroxide and fluorine. This indicates that the rate determining step in the mechanism of this reaction

must be the reaction between NO2 and F2 only. Keeping this in mind, a mechanism of this reaction may be suggested as NO2 + F2 → NO2F + F Slow step NO2 + F → NO2F Fast step 2NO2 + F2 → 2NO2F Net reaction Even though this reaction proceeds through two steps, the rate of the overall reaction is determined by the first step which is the slowest step. Every step in this reaction is called an elementary process. All elementary processes are always either unimolecular or bimolecular. Rarely, they may be termolecular. The sequence of steps through which a reaction is supposed to take place is called reaction mechanism. A reaction mechanism is only a tentative proposal which is consistent with the available evidence. It can be modified when some new experimental facts come to light. Following are the characteristics of an elementary step: (i) It may be real or imaginary (ii) It involves three or less than three reactant molecules or ions (iii) Each elementary step is represented by a balanced chemical equation.

8.7.3 how to assign Mechanism to a reaction It has been studied above that the slowest step of a complex reaction determines its rate and order. Therefore whatever be the other steps, the mechanism must involve one-key step which is the slowest step. Once this step (slowest) is ascertained from rate law, all other steps are more predictions. Sometimes more than one mechanism can be advanced for the same reactions as illustrated in Fig.8.4. Consider reaction R → P. This reaction has two mechanisms (1) and (2). In the mechanism (1) there are two steps S1 and S2 followed by the slowest step S3. Mechanism (2) involves three steps viz., S4, S5 and S6 before the step S3. Step S3 is common in both the reactions. It is thus the rate determining step. Hence, for assigning mechanism, proceed as follows: 1. Experimentally, find out rate law of that reaction. 2. From the rate law, write the slowest step. Rate law tells us the molecularity and order of the slowest step. 3. Predict the other fast steps in such a way so that on adding all these steps, we get the original equation for the complex reaction. Accordingly, the experimentally observed rate of the reaction for the reaction between NO2 and F2 is given by the expression.

Chemical Kinetics

Mechanism(1)

Reactants (R)

S1

8.13

Mechanism(2) S4 S5

S2

S6

Intermediate products

(slowest) S3

S3(slowest) Final products P

fig 8.4 Reaction mechanism depends upon the slowest step Solved Problem 4 dx = K [NO2] [F2] dt This is the rate law for the reaction

Rate = -

Table 8.1 Comparison of molecularity and order of a reaction Molecularity (A) i ii

(B) i

ii iii

iv v

Common Points Its value rarely exceeds three. It depends upon the condition of reaction. Points of difference Molecularity is the number of molecules or ions taking part in a single step chemical reaction It is always a whole number It is purely a theoretical value determined from the balanced single step reaction It is never zero Molecularity of a complex reaction has no meaning. It is expressed for each elementary step.

Order

(i) (ii)

Its value rarely exceeds three. It depends upon the conditions of reaction.

(i)

It is the sum of exponents to which the concentration terms are raised in the rate equation (ii) It may be a whole number or a fraction. (iii) It is an experimental value from rate equation

From the following mechanism of a complex reaction, find out the order of reaction, molecularity of each step and rate law. A+B→M M+B→N+L (slow) N+L+B→C A + 3B → C. Solution: Rate law can be predicted from the slowest step as R = K [M] [B] Therefore, order of reaction = n = 1 + 1 = 2 Molecularity of first step = 2 Molecularity of second step = 2 Molecularity of third step = 3 Dimensions of rate constant mol L-1 = K (mol L–1) (mol L–1) s

Or K = mol–1 = mol–1 L s–1 Dimension of Rate constants for the reactions in problems for practice 5 (p. 8.11) (iii) Rate = K [CH3CHO]3/2 Order w. r. t CH3CHO = 3/2 over all order = 3/2 Dimensions of rate constant mol L-1 K (mol L–1) 3/2 s

Or K = mol–1/2 L1/2 s–1 (iv) It may be zero (v) Order is same for the whole reaction, no matter it is simple or a complex one.

(iv) Rate = K [NOBr] 2 Order w.r.t [NOBr] = 2 Over all order = 2 Dimensions of rate constant mol L-1 K (mol L–1)2 s

Or K = mol–1 L s–1

8.14

Chemical Kinetics

(v) Rate = K [NO]2 Order w.r.t NO = 2 Overall order = 2 Dimensions of rate constant mol L-1 K (mol L–1)2 s

(vi) Rate = K [CHCl3] [Cl2]1/2 Order w.r.t CHCl3 = 1 order w.r.t Cl2 = ½ Overall order = 1 +

1 = 3/2 2

Dimensions of rate constant. mol L–1 = K (mol L–1) 3/2 or K = mol–1/2 L1/2 s–1

8.8 Integrated rate equatIonS We have learnt already that the concentration dependence of rate is called differential rate equation. It is not always convenient to measure the instantaneous rate because it is measured by determination of slope of the tangent at points in concentration versus time graph. This makes it difficult to determine the rate law and hence the order of the reaction. To avoid this difficulty, we can integrate the rate equation and obtain integrated rate equation which gives a relation between directly measured experimental quantities i.e., concentrations at different times. For different reactions of different orders, the integrated rate equations are also different. The instantaneous rate of a reaction is given by differential rate law equation. For example, for a general reaction

dx = k dt Integrating on both sides we get

∫ dx = K ∫ dt Or x = Kt + c where c = constant of integration When t = 0, x = 0 ∴C=0 Hence x = Kt According to this relation (for zero order reaction), amount of substance reacted is proportional to the time. (a) Units of Zero Order Rate Constant (K) For zero order reaction, x = Kt Or K =

x mole L-1 = = mole L–1 S–1 t Time in sec

As concentration term is involved in the unit of K, the numerical value of rate constant K depends on the unit in which concentrations are expressed. (b) Half Life Period of Zero Order Reaction The expression for the half life period can be written as below: When x =

a ; t = t0.5 2

Where a = initial concentration Substituting the values of x and t in the zero order reaction x = Kt ∴

a a = K t0.5 or t0.5 = 2k 2

(c) Evolution of K A plot of x against ‘b’ gives a straight line passing through the origin. The slope of this straight line is K.

aA + bB → products

dx = K [A]a [B]b dt The differential form of rate law can be converted into integral form of rate law by simple mathematics (calculus). The integrated rate equations for different order reactions can be derived as follows.

8.8.1 Zero order Reaction A reaction whose rate is independent of the molar concentrations of reactants is called zero order reaction. The rates of such reactions remain uniform throughout. Thus rate = constant dx Order = constant = K dt Where K = rate constant for zero order reaction.

concentration (mol lit-1) product

the differential rate law equation is

Time t (sec)

fig 8.5 Determination K for zero order reaction

Chemical Kinetics

8.8.2 First order Reaction

Examples of Zero order Reactions 1. Photochemical reactions: These reactions take place in the presence of sunlight. Hydrogen and chlorine react in presence of sunlight to form hydrogen chloride. Light H2(g) + Cl2(g)  → 2 HCl ( g )

For example, the decomposition of ammonia on the surface of tungsten or platinum is first order when the pressure (or concentration) of NH3 in the system is low. However, at high concentration when the surface of the catalyst becomes completely covered with layer of NH3 molecules, the reaction becomes zero order reaction. In general the rate of this reaction is given as

K1 [ NH 3 ] 1 + K 2 [ NH 3 ]

A + B (excess)  → Products E.g, +

H C12H22O11 + H2O  → C6H12O6 + C6H12O6 Sucrose Glucose Fructose In both cases, the reaction rate is given by dx d[A] == K[A] Rate = dt dt d[A] = K [A] dt

(1)

The reaction shows infinitesimally small change of concentration in infinitesimally small time. Time value of rate constant can be determined by integration. 1. Derivation: Let the initial concentration of A be C0 and the concentration after time t seconds be C Substituting the value of concentration in (1) we get dC = Kc dt dC = Kdt C Integrating both sides dC -∫ = K ∫ dt dt - ln C = Kt + I

(2)

Where I is integration constant (1)

Where K1 and K2 are constants. When [NH3] is very low, K2 [NH3] can be neglected in comparison to unity so that the above equation becomes Rate = K1[NH3]

The reactions whose rate is determined by the change of only one concentration term are known as first order reactions. Such reactions can be represented by following general types: A  → Products

This reaction is zero order only when HCl formed is removed spontaneously by dissolving in water. If HCl is not removed spontaneously, concentrations of H2 and Cl2 change with time due to which the rate of reaction will not remain constant. 2. Some heterogeneous reactions: Zero order reactions generally takes place in heterogeneous system. In such systems, the reactant is adsorbed on the surface of a solid catalyst, where it is converted into product. The fraction of the surface of the catalyst covered by the reactant is proportional to the concentration of the reactant at low values. The rate of reaction is first order. However, after certain concentration limit of the reactant, the surface of the catalyst becomes completely covered. Further increase in the concentration of the reactant, the reaction rate does not change: in other words, the rate becomes independent of the concentration and therefore becomes zero order reaction.

Rate =

8.15

(2)

This reaction is first order in ammonia At higher concentration of ammonia, we may neglect 1 as compared to [NH3] so that equation 1 becomes

K1 [ NH 3 ] K1 = =K K 2 [ NH 3 ] K 2 This means that rate becomes constant and independent of [NH3] and therefore, becomes zero order reaction. Similarly, the decomposition of hydrogen iodide on the surface of gold is also zero order reaction.

When t = 0, C = C0, we get 0 = –ln C0 + I Or I = ln C0 Substituting the value of I in equation (2) Kt = ln C0 – ln C C C K = ln 0 = 2.303 log 0 C C C 2.303 K= log 0 t C

(3)

This is the equation for the first order reaction. Here K is called the first order rate constant Suppose the initial concentration of the reactant (when t = 0) is ‘a’ mol L–1 and after time ‘t’ seconds the concentration reduces to (a-x) moles L–1 of the reactant. It means that x moles have reacted. When C0 = a and c = a-x, the equation becomes

Chemical Kinetics

K or K1 =

2.303 a log t a−x

(4)

The equation (3) may also be written in the exponential form as C ln 0 = K1 t C C Or ln = - K1 t C0 C = e - K1t C0 ∴ C = C0. e–K1t Characteristics of First order Reactions

Change in concentration Time

= mol litre–1 s–1 If the concentration is expressed in mol litre–1 and time in second. for a first order reaction rate = K [A] Or mol litre–1 s–1 = K mol litre–1 Or K = s–1 Hence the units of first order rate constant are s –1 (t–1) 2. Half-change time or half-life period of a first order reaction is independent of the initial concentration. According to the first order rate equation a 2.303 log (5) K= t a-x Where a is the initial concentration of the reactants when t = 0 and (a-x) is the concentration at time t.

a = when t = t0.5 (half According to definition x = 2 change time) a 2.303 log ∴K= t 0.5 a-a 2 2.303 Or K = log 2 t0.5

(

2.303 log 2 K 2.303 Or t0.5 = × 0.3010 K 0.693 Or t0.5 = K

2.303 a′ log (a ′ - x ′) t 2.303 na = log n(a - x) t K=

=

1. The units of first order rate constant K are independent of the units in which the concentration is measured. This is because the quantity a/(a-x) is dimension less Rate of reaction =

According to the equation (6), the half change time for a first reaction is found to be constant and is independent of the initial concentration the reactant. 3. The change in concentration units does not change the value of rate constant Let the value of concentration in new units be n times those of the value in initial units then a΄ = an x΄ = nx and (a΄- x΄) = n (a-x) Where a΄, x΄, (a΄- x΄) are equations in new units

)

a 2.303 log t a-x

Which is the same as the value in the initial units Evolution of K (i) The equation (5) can be used for the evolution of K. Starting with a known concentration of the reactants (a-x) or c at different time intervals (t) may be noted. By substituting these values of (a-x) for the corresponding value of time t in equation (5) the value of K can be calculated. For a first order reaction the value of K obtained from equation (5) will almost constant (ii) The value of K can also be calculated graphically. The equation (5) on rearrangement may be expressed as t=

a 2.303 log K (a - x)

Slope = –

K 2.303

log ( a - x )

8.16

Or t0.5 =

Time (t)

(6)

fig 8.6 Plot showing relation between log (a-x) vs time (t) for first order reaction

Chemical Kinetics

Or t =

(a-x) value at different intervals of time are found. The values are then substituted in the first order equation. The value of K, will be found to be the same in each case.

C 2.303 log O K C

2.303 [log a – log (a-x)] K Kt + log a Or log a-x = 2.303

8.17

∴T=

Solved Problem 5 (7)

This is the equation of a straight line (Y = mx + C). Thus on plotting log (a-x) (or log C) against time ‘t’ the graph obtained should be straight line (Fig 8.6) The intercept on The Y- axis would be ‘a’ and the slope of the line would

-K  be   i.e., 2.303  

K Slope = – 2.303

From which K can be calculated (iii) The value of K can also be calculated for the reactions where the initial concentration is not known. The integration rate law for a first order reaction is t=

The catalysed decomposition of hydrogen peroxide in aqueous solution which is of first order is followed by titrating equal volume of samples of solution with potassium permanganate at stated times as follows: Time (in minute): 0 mL of KMnO4: 37

(8)

(9)

Subtraction of equation (7) from (6) we get (t2 - t1) =

2.303 2.303 log (a - x1) – log (a-x2) K K

Or (t2 - t1) =

25 12.3

45 5.0

Solution: It is clear from the data, that the titre value at zero time may be taken as initial concentration i.e., a = 37.0 At different intervals, the value of (a - x) are given Substituting the values in the first order rate equation a 2.303 K= log t a-x

(i) K =

2.303 37.0 log = 4.339 × 10-2 min -1 5 29.8

(ii) K =

2.303 37.0 log = 4.339 × 10-2 min -1 15 29.8

Let (a-x1) and (a-x2) be the reactant concentrations at the time interval t1 and t2 from the start then

2.303 2.303 t1 = log a log (a-x1) K K

15 19.6

Each time we get

2.303 2.303 log a log (a-x) K K

2.303 2.303 t2 = log a log (a-x2) K K

5 29.8

(a - x1 ) 2.303 log K (a - x2 )

8.8.3 Some Typical First order Reactions 1. Decomposition of Hydrogen Peroxide: In the decomposition of hydrogen peroxide, colloidal platinum acts as a catalyst. The progress of reaction can be measured by noting the volume of oxygen evolved after different intervals. 1 → H2O + O2 H2O2  2 The progress of the reaction can be studied by withdrawing small volumes of hydrogen peroxide after different intervals and then titrating it against standard potassium permanganate solution. If the molar concentration of hydrogen peroxide is ‘a’ to start with (i.e., at t = 0) then

(iii) K = 2.303 log 37.0 = 4.407 ×10-2 min–1 12.3 25 (iv) K = 2.303 log 37.0 = 4.447 ×10-2 min–1 5.0 45 The average value of K (rate constant) is thus, equal to 4.335 × 10–2 min–1 2. Decomposition of ammonium nitrite in aqueous solution: When aqueous solution of ammonium nitrite is warmed, nitrogen is evolved.

→ N2 + 2H2O NH4 NO2  The progress of reaction can be noted by measuring the volume of nitrogen gas evolved after different intervals of time. It may be noted that the volume of nitrogen evolved is a measure of the quantity of ammonium nitrite decomposed. The total volume of nitrogen (v∝) evolved at the end of the reaction (infinite time t∝) measures the total molar concentration of ammonium nitrite or the initial concentration of ammonium nitrite i.e., ‘a’ represents (a-x) values at different intervals t. Thus the first order rate equation is written as K=

2.303 V∝ log t V∝ - Vt

8.18

Chemical Kinetics

Or (a-x) ∝ (V∝ - Vt )

Solved Problem 6 From the following data for the decomposition of ammonium nitrite aqueous solution, show that the decomposition of ammonium nitrite is of the first order. Time (minutes): 10 Volume of N2 (mL) 6.25

15 9.0

20 11.4

25 13.65

∝ 35.05

Solution: Here a = V∝ = 35.05 The first order equation is K = 2.303 log V∝ V∝ - Vt t Substituting the given values in each set in the above equation:

35.05 –1 (i) K = 2.303 log = 0.019652 min 10 35.05 - 6.25 35.05 –1 (ii) K = 2.303 log = 0.01976 min 15 35.05 - 9.0 2.303 35.05 –1 (iii) K = log = 0.01964 min 20 35.05 - 11.4 –1 35.05 (iv) K = 2.303 log = 0.01971 min 25 35.05 - 13.65

Since the value of K is nearly the same in each case, we say that the given data corresponds to first order reaction. 3. Hydrolysis of methyl acetate in aqueous solution: the hydrolysis of methyl acetate with water is written as

 CH3 COOH + CH3OH CH3COOCH3 + H2O  It is an example of pseudo first order reaction. The reaction is too slow but its rate can be increased by adding a few drops of the mineral acid. Since water is present in large excess, its molar concentration is assumed to be constant. Thus, the progress of reaction can be noted by noting the change in concentration of the acid after different intervals of time. For this, small volume of the solutions are withdrawn and then titrated against standard acid (and of course the mineral acid acting as catalyst the volume of mineral acid remains the same throughout the reaction) produced due to hydrolysis. To start with (at zero time), the volume of alkali (V0) used is equivalent to mineral acid alone. If after time t, V, is the volume of standard alkali consumed then Vt - V∝ is a measure of acetic acid formed (or the ester decomposed). If V∝ is the volume of alkali consumed at the end of the reaction, then V∝ - Vo is proportional to ‘a’. Form This

(

)

(a-x) ∝ V∝ - Vo - (Vt – Vo)

Thus, the first order rate equation is reduced to V - Vo 2.303 log ∝ K1 = V∝ - Vt t Solved Problem 7 1 mL of methyl acetate was added to a flask containing 20 mL of N/20 HCl maintained at a temperature of 85°C. 2 mL of the reaction mixture was withdrawn at different intervals and titrated with a standard alkali, calculate the specific reaction rate: Time (in minutes): mL of alkali:

0 75 119 19.24 24.20 26.60

180 ∝ 29.32 42.03

Solution: Applying the first order rate equation. K1 =

V - V0 2.303 log ∝ t V∝ - Vt

Here V∝ - Vo = 42.03 - 19.24 = 22.79 Substituting the values in the above relation. (i) K1 =

2.303 22.79 log = 0.00395 min–1 75 42.03 - 24.20

(ii) K1 =

2.303 22.79 log = 0.00321 min–1 119 42.03 - 26.60

(iii) K1 =

2.303 22.79 log = 0.00316 min–1 180 42.03 - 29.30

Hence the mean value of K = 0.00344 min–1 4. Hydrolysis or inversion of cane sugar (Sucrose): The inversion of cane sugar in presence of mineral acid as catalyst is another example of pseudo unimolecular reaction. C12H22O11 + H2O  → C6H12O6 + C6H12O6 Glucose Fructose The course of the reaction can be followed by measuring the angle of rotation of plane polarized light in polarimeter. Sucrose rotates the plane in one direction where as the mixture of glucose and fructose rotates the plane in the opposite direction. Let r0 = angle of rotation in the beginning when t = 0 Let r∝ = rotation when reaction is complete

rt = rotation at time t ∴ ro - r∝ = a; rt - r∝ = ( a - x)

Chemical Kinetics

Thus, the first order rate equation is reduced to r -r 2.303 log o ∝ K= t rt - r∝ Solved Problem 8 The optical rotation of sucrose in 0.9 N HCl at different time intervals is given in the following data: Time in minutes 0 7.18 18 27.05 ∝ Rotation (in degrees) + 24.09 + 21.4 + 17.7 + 15 -10.74 Show that the reaction is of the first order. Solution: Here r0 - r∝ = 24.09 - (-10.74) = 34.83 Substituting the values in the first order rate equation K=

r -r 2.303 log o ∝ we get t rt - r∝

(i) K =

2.303 34.83 log 7.18 21.4 - (-10.74)

2.303 34.83 = 0.0047 min–1 log 7.18 32.14

=

34.83 2.303 log 17.7 (-10.74) 18 2.303 34.83 = log = 0.0046 min–1 18 28.44 2.303 34.83 (iii) K = log 27.05 15 - (-10.74) (ii) K =

=

KI + HCl NCOCH3 +

NHCOCH3 + HCl + I2

2HI

N - chloroacetanilide NHCOCH3

+

HI

Cl

The progress of reaction can be studied from time to time by adding excess of KI solution to a known volume of the reaction mixture. The amount of I2 is studied by titrating with standard solution of sodium thiosulphate. The amount of sodium thiosulphate used will correspond to the concentration of unchanged N-chloroacetanilide. At time t = 0, volume of thiosulphate (V0) will correspond to the initial conc of N–chloroacetanilide a ∝ V0 Similarly, after time t, Vt will be proportional to the amount of N-chloracetanlide (a-x) left (a-x) ∝ Vt Substituting various values in the first order equation

V 2.303 log 0 t Vt

From the following data of the isomeric change of N-chloroacetanilide into p-chloracetanilide in acidic medium, show that the isomeric change is a reaction of first order. Time in hrs: 0 1 2 3 Vol of hypo used in mL: 49.3 35.6 25.75 18.50 Solution: The first order rate constant for the conversion of N-chloroacetanilide into p-chloroacetanilide is given by K=

NHCOCH3 H+

N-Chloroacetanilide

No reaction

Solved Problem 9

We see that the values of K are approximately constant and the reaction is of the first order. 5. Reactions involving isomeric change: Conversion of N-chloroacetanilide into p-chloroacetanilide. It is an example of the first order reaction.

NCOCH3

KCl + HI

Cl

K=

34.83 2.303 log = 0.0048 min–1 25.74 15

Cl

8.19

Cl p-Chloroacetanilide

This reaction takes place in the acidic medium. The acid (H+ ions) acts as a catalyst. When N-Chloroacetanilide is treated with KI solution in acidic medium, iodine is liberated. On the other hand, p-chloroacetanilide does not liberate iodine with KI.

V 2.303 log 0 t Vt

In this problem V0 = 49.3 at zero time. Substituting the values, we get 2.303 49.3 = 3.25 × 10–1 hr–1 log 1 35.6 2.303 49.3 (ii) K = = 3.25 × 10–1 hr–1 log 2 25.75 (i) K =

(iii) K =

2.303 49.3 = 3.27 × 10–1 hr–1 log 3 18.50

As the value of K are fairly constant, the given isomeric reaction is of the first order.

8.20

Chemical Kinetics

6. Decomposition of benzene diazonium chloride (C6H5N=NCl) in water: Benzene diazonium chloride decomposes in water according to the following reaction: C6H5N = NCl + H2O → C6H5OH + N2 + HCl Benzene diazonium chloride The progress of the reaction can be studied by measuring the volume of nitrogen produced by the decomposition of benzene diazonium chloride at different intervals of time. The volume of nitrogen liberated is proportional to the decrease in the concentration of benzene diazonium chloride. Let V∝ be the volume of nitrogen measured at the end of the reaction. This corresponds to the initial concentration. If Vt is the volume of nitrogen at the end of the time t, then

Solved Problem 11 Show that the time taken for the completion of any fraction of a first order reaction is independent of initial concentration of reactant. Solution: Suppose ‘a’ is the initial concentration of a reactant a In general let x = n Where n >1 and time taken for this change = t1/n a Then a-x = an  1  n - 1 = a 1 -  = a   n  n  Substituting in equation t = 2.303 log a k (a - x) 2.303 a log t1/n = k a (n - 1) / n 2.303 n log = k n -1

(a-x) ∝ (V∝ - Vt) Substituting the values in the first order equation K=

V∝ 2.303 log = t V∝ − Vt

Solved Problem 10 Decomposition of benzene diazonium chloride was studied at a constant temperature by measuring the volume of N2 gas evolved at different intervals of time. Using the following data show that the reaction is of the first order. Time in minutes: 0 20 50 70 ∝ Vol. of N2 measured in mL: 0 10 25 33 162 Solution: When t = 20, (Vα - Vt) = 162 – 10 = 152 (i) K =

2.303 162 log = 3.22 × 10–3 min–1 20 162 − 10

Since right hand side does not contain concentration term, time taken for the completion of any fraction of a first order reaction is independent of initial concentration of the reactant. Solved Problem 12 A First Order reaction has a specific reaction rate of 10–3 sec–1. Calculate half life period for this reaction. Solution: Half-life period for first order reaction is given by K0.5 =

(ii) K =

(iii) K =

2.303 162 log = 3.36 × 10–3 min–1 50 162 − 25 2.303 162 log = 3.26 × 10–3 min–1 70 162 − 33

The fairly constant values of K. indicate that the reaction is of the first order. Half-Life Period The half-life period for the first order reaction is independent of the initial concentration of reactant and can be calculated by using the equation t0.5 =

0.693 k

0.6931 k

K = 10–3 sec–1 T0.5 =

0.6931 = 693 sec 10 - 3

Problems for Practice 6.

3 th of a reaction of first order was completed in 4

10 minutes. Calculate the half life period of this reaction. 7. An acid solution of sucrose was hydrolysed to the extent of 57% after 66 minutes. Assuming the reaction of first order, calculate the time taken for 75% hydrolysis. 8. Show that the time taken for 99 % of a first order reaction to take place is twice the time required for 90% of the reaction.

Chemical Kinetics

= 0.13026 Taking anti-logs, we get

8.8.4 Applications of the First order Rate Law Equation The first order rate law equation can be used to solve problems concerning population growth, radioactive disintegration and compound interest etc. (i) Rate law and population growth. Let us suppose that: Present population = PC Population after time t (in years) = P Percentage rate of growth of population per year = r

Rearranging

dp = rp dt

(1)

dp = r.dt p

(2)

p Or t = 2.303 log p0 r The time after which the population will get doubled p = 2 is given by p0

t 0.5 =

2.303 0.693 log 2 = r r

= 1.3498 × 800 million = 1087.8 million

(ii) Population in 2100 p 0.025 × (2100 - 1988) log = 2.303 p0 = 1.2158 p = 16.436 p0 P = 16.436 × P0 = 16.436 × 800 million = 13148.8 million (ii) Rate law and Radioactive Disintegration Radioactive nucleus disintegrate to give α, β and γ rays. The radioactive disintegration also follows first order kinetics. Radioactivity of a radioactive material is measured with Geiger Muller (G-M) counter. Let us suppose that N0 = Initial number of radioactive nuclides. N = Number of radioactive nuclides left after time t Instantaneous rate of disintegration = rate of radioactive decay dn = λN (1) dt Rearranging equation (1) we get dN = λ.dt (2) N Integrating equation (2) we get λt = ln

No N

No N Here λ = Disintegration constant Half-life period of radioactive disintegration = 2.303.log

Solved Problem 13 The population of India in 1988 was 800 million. What will be the population in 2000 and 2100 if there is no change in present growth rate which is 2.5% (25 per thousand) per year? Solution: The population growth follows the first order rate law 2.303 log

p = 1.3498 p0

Taking anti-logs

Integrating equation (2) In P = rt + integration constant When t = 0, P = P0 = initial population (3) Integration constant = P0 Substituting the value of integration constants in equation, we get ln P = rt + ln P0 p Or ln = rt p0 p Or 2.303 log = rt p0

i.e., when

8.21

p = rt p0 –1

Now r = 2.5% per year = 0.025 year (i) Population in 2000 p 0.025 × (2000 - 1988) log = 2.303 p0

0.693 λ On plotting log N versus t, a straight line is obtained t 0.5 =

λ . Thus the value of 2.303 can be calculated from the line.

whose slope is equal to -

λ = -2.303 × slope

8.22

Chemical Kinetics

(iii) Rate law and compound interest: Compounding of money also follows the first order rate law. If P0 is the initial principal money invested and p is the money after ‘t’ years, then the instantaneous rate is directly proportional to the money, i.e., dp = rp dt

(1)

Where r is the proportionality constant rearranging equation ………(1) dp = rt (1) p Integrating equation (2) gives ln p = rt + integration constant When t = 0, p = p0 ∴ ln p0 = integration constant. p Or ln = rt p0 Or 2.303 log

(3)

2.303 p log r p0

The time after which money will become double of its initial i.e., when t2 =

p = 2 is given by p0

2.303 0.693 log.2 = r r

How long will it take to double the money at a fixed annual rate of interest of 12%? Solution:

=

x

dx = ( a − x )2 0

12 = 0.12 100

1 1  1   a − x  = Kt...or... a −x − a = kt  0 1 x or K = (3) t a(a −x) If the initial concentration of reactant is C0 and at any instant of time t it is Ct then a = C0 and a-x = Ct, we get 1 1 1  K=  -  (4) t  C t C0  Case II: If the two reacting substances are different, the reaction is represented as A + B  → Products dx = K [A] [B] dt If a and b represent the initial concentration of A and B respectively and x is the amount of each reactant after time t1 we get dx = K (a-x) (b-x) (5) dt Upon integration of the concentration term 0 to x and time from 0 to t, we get X t dx (6) ∫0 (a − x)(b − x) = K ∫0 dt x

0.693 0.693 = = 5.775years r 0.12

∫

O

 → Products dx = K[A]02 dt

t

b( a - x ) 1 ln = Kt..........( a (b - x) ( a - b)

The rate of second order reaction depends upon the two concentration terms. Case I: If the reacting substances are the same, the general reaction is written as ∴ Rate of reaction -

1 (a - x) - (b - x) dx = K ∫ dt (a - b) (a - x)(b - x) O

1 [-ln (b-x) + ln (a-x)]0x = Kt ( a - b)

8.8.5 Second order Reactions

2A

1

∫ Kdt

-

Solved Problem 14

r=

∫ 0

2

p = rt p0 t=

If ‘a’ is the initial concentration of A and x is the amount consumed after time ‘t’ the rate equation becomes d(a - x) = K(a - x) 2 dt dx or = K(a - x) 2 dt dx (2) = Kdt (a - x) 2 Integrating between the limits of time from 0 to t and concentration from 0 to x1 we get

K=

2.303 b( a - x ) log t ( a - b) a (b - x)

(7) (8) (9) (10)

It can also be written as (1) K=

C (B) Ct(A) 2.303 . log 0 C0 (A) Ct(B) t[C0 (A) − C0 (B)]

(11)

Chemical Kinetics

Some important conclusions drawn from the above rate expressions 1. Time taken for completion of some fraction of change is inversely proportional to the concentration. At half change; X = 0.5a t0.5 =

1 0.5a 1 . = ...............( (12) K a (a - 0.5a ) Ka

2. The rate constant is not independent of units of concentration. It can be seen that there is one more concentration term in the denominator than in the numerator. So K is expressed in conc–1 time–1 units. If time is in seconds and concentration is expressed in moles per litre, then K is expressed in lit mol–1 sec–1. 3. Second order reaction confirms to the first order when one of the reactant is in large excess. When ‘a’ is in large excess x and b can be neglected compared to a. ba 1 ln K= (14) t.a a (b - x) Since a is constant Ka = K ' =

2.303 b log t b-x

(15)

This equation is identical with first order equation. Important characteristics of second order reactions 1. Time taken for completion of same fraction of change is inversely proportional to concentration At half change x = 0.5a ∴ From the above equation we have 1 0.5a 1 t0.5 = = K a (a − 005 .5a ) Ka 1 Or t0.5 ∝ (16) a 2. The rate constant is not independent of units of concentration. 1 x We know that K = . (17) t a(a - x) It can be seen from the above equation that there is one more concentration term in the denominator than in the numerator. Therefore the units of K will be conc–1 time–1 units. If time is expressed in seconds and concentration in moles per litre (dm3) then K=

mol(dm3 )-1 = dm3 mol–1 s–1 s.mol2 (dm3 )-2 –1 –1

(18)

Or K = litre mol s Or K = litre mol–1 min–1 (when time is expressed in minutes)

8.23

3. The value of K changes with change in concentration units. If the value of concentration in new units is n times the value in initial units, the value of rate constant is

1 of the original value. n

Let a΄ = n a, x΄ = nx and a΄- x΄ = n (a-x) where a΄, x΄ and (a΄-x΄) are molar concentration in new units. 1  x′  1  nx  = (19) K' =   a ′t  (a ′ - x ′ )  nat  na - nx 

1  nx  (20) nat  n(a - x)  1 which is of the original value of rate constant. n 4. The second order reaction follows first order kinetics if one of the reactants is in large excess. Evolution of K (i) Integration Method: The value of second order rate constants can be calculated by using the second order rate equation depending upon the type of reaction. =

K=

1 x ( when reacting substances are same) (21) t a(a − x)

And K =

b( a - x ) 2.303 log t ( a - b) a (b - x)

(when reacting

substances are different)

(22)

All that we have to do is to determine at different intervals of time and then knowing the value of initial concentration, rate constant K can be evaluated as illustrated in numerical problems for second order reactions (ii) Graphical Method: The value of K can also be obtained by a graphical method. The equation to be used is 1 1 Kt = − C t C0 1 1 = Κt + Ct C0 This equation is like Y = mx + C i.e., the equation of Or

straight line. Hence a plot of

1 versus time will be a Ct

straight line (Fig 8.7). If the reaction is of the second order, the slope of the line will directly give the value t+ of K and intercept on concentration axis will be t

1 . C0

For the reactions involving two different reactants having different initial concentrations, equation (11) can be used. It may be expressed as

8.24

Chemical Kinetics

C (A) C0 (B) K[C0 (A) − C0 (B)] t = log t 2.303 Ct (B) C0 (A) or

C (B) K[C0 (A) − C0 (B)] C (A) + log 0 t = log t C0 (A) 2.303 Ct (B)

= log

Ct (A) C (A) − log. 0 C0 (B) Ct (B)

C (A) C (A) K[C0 (A) − C0 (B)] t + log 0 = or log t 2.303 C0 (B) Ct (B) This is again an equation of straight line. Hence plot of log

Ct (A) versus t should be straight line if the Ct (B)

reaction is second order reaction. The slope of the line would K[C0 ( A ) − C0 (B)] 2.303 From which the value of K can be evaluated The intercept on the concentration axis would be equal to log C0 (A) . C0 (B)

8.8.6 Example of Second order Rate Equation 1. Hydrolysis of Ester: Out of many examples available for kinetically second order reaction, a typical example is the hydrolysis of ethyl acetate with sodium hydroxide solution.

→ CH3 COO Na + C2H5OH CH3 COO C2H5 + NaOH  A known volume of standard sodium hydroxide solution is added to the known volume of ethyl acetate. Let ‘a’ and ‘b’ be the initial concentrations of ethyl acetate and sodium hydroxide respectively. The progress of the reaction is noted by withdrawing known small volume of the mixture and titrating it with standard acid. As hydrolysis proceeds, the concentration of sodium hydroxide will go on decreasing. Let x be the concentration of sodium hydroxide at time’t; then (a-x) is the amount of the unreacted ester and (b-x) is the amount of unreacted sodium hydroxide. The value of (a-x), (b-x) obtained at different intervals are then substituted in the second order rate equation. The constant value of K each time tells that the reaction is of the second order reaction. Solved Problem 15 A gram mole of ethyl acetate was hydrolyzed with a gram mol of NaOH when the concentration fell according to the following data. Show that reaction is of second order and also calculate the value of K. Time (mts) (a-x)

1/a

(a)

Time (t)

log Ct (A) Ct (B) Time (t)

fig 8.7 (a) Plot of

1 vs t for second order reaction of a

2A → P and (b) Plot log Ct (A) Ct(B) for second reaction A+B→P

4 5.3

6 4.58

10 3.5

15 2.74

20 2.22

Solution: Here a = 8.04; therefore, 5.3, 4.58, 3.5, etc. are the values of (a-x) at different time intervals for the second order rate equation. K=

(b)

0 8.04

1 x . litre mol–1 min–1 at (a - x)

Substituting the value, we get 1 8.04 - 5.3 (i) K = × [ x = a - (a - x)] 4 × 8.04 5.3 1 2.74 = × = 1.615 × 10 -2 4 × 8.04 5.30 1 8.4 - 458 (ii) × 6 × 8.04 4.58 1 3.46 × = 1.571 × 10 -2 6 × 8.04 4.58 (iii) K =

1 8.04 - 3.5 × 10 × 8.04 3.5

Chemical Kinetics

1 4.54 × = 1.621 × 10 -2 8.04 3.5 1 8.04 - 2.74 (iv) K × 15 × 8.04 2.74

=

K=

1 x × t Pi ( Pi - x)

Solved Problem 17 A second order reaction in which initial concentration of both the reactants is same is 25% complete in 20 min utes how long will it take for the reactions to complete to 75% Solution: For a second order reaction?

1 5.30 × = 1.611× 10−2 = 15 × 8.04 2.74 1 8.04 - 2.22 × 20 × 8.04 2.22 1 5.82 = × = 1.635 × 10 -2 20 × 8.04 2.22

(v) K

The value of K in each case is constant. Thus, the given data represents second order reaction.

1 x K= × t a(a - x) For the first case

25 ;aa = 0.25aa 100

Solved Problem 16

X=

In the hydrolysis of ethyl acetate by sodium hydroxide using equivalent concentrations, the progress of reaction was examined by titrating 25 mL of the reaction mixture at regular intervals against standard acid. The following figures give the value of acid used by the unchanged alkali.

t = 20 minutes 1 0.25a K= ⋅ 20 a (a − 0.25a )

Time (mts) mL of acid (a-x)

0 16

5 10.24

15 6.13

25 4.32

35 3.41

Show that the reaction is of the second order solution. Solution: K=

Time

(a-x)

x

5 15 25 35

10.24 6.13 4.32 3.41

5.76 9.87 11.68 12.59

1 x . at a - x

0.0070 0.0067 0.0068 0.0066

The value of K is fairly constant which shows that the reaction is of second order. 2. Thermal decomposition of acetaldehyde 2 CH3 CHO

→

2CH4 + 2CO

The reaction is studied by observing the increase in pressure with the help of manometer. Let P, be the initial pressure of acetaldehyde. If after time t, the pressure fall by x, the pressure of CH4 and CO will increase by X. the total pressure P in time t is P = PCH CHO + PCH + PCO 3

4

= (Pi – x) + x + x = Pi + x Substituting the value in second order rate equation 1 t K= × we get t a(a - x)

8.25

K=

1 0.25 1 ⋅ = 20 0.75a 60a

For the second case X=

75 a 100

t=? K=

=

1 0.75a t a (a - 0.75a )

1 0.75a 3 = t 0.25a at

Since K is constant for a given reaction.

1 3 = 60a at t = 180 min Solved Problem 18 In a second order reaction, half-life period is 60 minutes when the initial concentrations is 0.02 mole per litre. Calculate the value of specific reaction rate. Solution: For a second order reaction

1 x K . t a(a - x) According to the given data

8.26

Chemical Kinetics

A = 0.02 mole liter–1 t = 60 minutes X = 50 × a = 0.5a 100 ∴ Substituting the value in the second order rate equation, we get K=

=

1 0.5a . 60 a (a - 0.5a )

0.5 1 1 × = 0.5a 60 60a

1 K= = 0.833 litre mol–1 min –1 60 × 0.02 Solved Problem 19 Decomposition of a gas is of second order. It takes 50 minutes for 50% of the gas to be decomposed when the initial concentration is 1.5 × 10–2 mole/litre. Calculate the specific reaction rate. Solution: For second order reaction 1 x K= . t a(a - x) According to the given data a = 1.5 × 10–2 mole/litre t = 50 minutes 50 × a = 0.5a X= 100

If ‘a’ is the initial concentration of A and x is the amount of ‘a’ consumed by any instant of time t, then concentration of A at any instant of time t = (a-x) mole litre–1. According to the law of mass action Rate of reaction =

dx ∝ [ A]3 dt

dx ∝ ( a - x )3 dt Or

dx = K ( a - x )3 dt

Where k is rate constant for third order reaction rearranging equation (1), we get

dx = K dt ( a - x )3 Integrating the above equation, we get dx

∫ (d - x)

3

= K ∫ dt

(a - x) 2 (-1) = Kt + C (-2) (a - x) -2 = Kt + C Or 2

When t = 0, x = 0 Putting the values in equation (2) we get

K = 1 . 0.5a 50 a (a - 0.5a)

1 1 = Kt + 2 2(a - x) 2a 2

K=

0.5 1 1 × = 0.5a 50 50a

Kt =

1 = 1.33 litre mol–1 min–1 50 × 1.5 × 10-2 Kt =

8.8.7 Third order Reactions A reaction is said to be of the third order if the reaction depends upon three concentration terms. The general equation for such reaction is 3A  → Products

(2)

Where C is constant of integration

Substituting the value in the second order rate equation, we get

=

(1)

1 1 1 1 1 - 2 =  - 2 2 2 2(a - x) 2a 2  (a - x) a 

1  a 2 - (a - x) 2  1  a 2 - a 2 - x 2 + 2ax   =   2  a 2 (a - x) 2  2  a 2 (a - x 2 ) 2 

1  2ax - x 2  Kt =  2  2  a (a - x) 2  K=

1  x(2a - x)    2t  a 2 (a - x) 2 

Chemical Kinetics

8.27

Table 8.2 Summary of order, rate equation, half-life and units of different order reactions S.No

Order Rate law

Half-Life equation

Rate equation

1

Zero

dx = K0 dt

K0 =

x t

2

First

dx = K1 (a - x) dt

K1 =

2.303 a log t a-x

3

Second

dx = K 2 (a - x) 2 dt

x 1 K2 = . t a(a - x)

4

Second

dx = K 2 (a - x)(b - x) dt

K2 =

5

Third

dx = K 3 ( a - x )3 dt

K3K=3

6

n

dx = K n (a - x) n dt

Kn =

Mole litre–1 Sec–1 t0.5 is independent of initial concentration a

2.303 b( a - x ) log t ( a - b) a (b - x)

1 1 1 -   2t  (a - x) 2 a 2  1  1 1    t (n - 1)  (a - x) n -1 a n -1 

1 a

Litre mole–1 sec–1

t0.5 ∝

1 a

Litre mole–1 sec–1

t0.5 ∝

1 a2

Mole–2 litre2 sec–1

t0.5 ∝

1 a n -1

Mole1–n litren–1 sec–1

Solved Problem 20

A balanced chemical equation in all cases does not give the correct picture of reaction. Rate law and hence order of a reaction are always determined experimentally. Some of the important methods used for the determination of order of a reaction are described below. 1. Integration Method (Method of Trial and Error): This is the most direct and simplest method of determining the order of reaction. In this method known amounts of reactants are mixed and the progress of the reaction is determined by analysing the reaction mixture from time to time. The data thus obtained are then substituting in the kinetic equation of first, second and third orders. Order of the reaction is then known from that equation which gives constant values of the rate constant.

For the reaction

rate equation

0

K=

First

2.303 a K= log t a-x

Second

K=

Third

x t

1 x . t a(a - x) 1 x(2a - x) K= . 2t a 2 (a - x) 2

Sec–1

t0.5 ∝

Methods For Determining the order of Reaction

Order

Units

CH3COO CH3 + NaOH  → CH3 COO Na + CH3OH Following data were obtained Time in minutes 0 [CH3 COOCH3] = 0.01 [NaOH]

5 0.00634

10 15 0.00464 0.00363

Find the order of reaction. Solution: We substitute the data in the first order rate equation first and if this does not give a constant value of K, the same data is inserted in second order rate equation and so on a = 0.01 Time

a-x

x

K=

2.303 a 1 x log K= . t t a (a - x ) a-x

5

0.00634

0.00364

= 0.091

11.48

10

0.00464

0.00536

= 0.076

11.55

15

0.00363

0.00637

= 0.067

11.69

It is seen from the above calculations that the first order rate equation does not give constant value K but the second order rate equation gives a fairly constant value of K. Hence the reaction is a second order reaction.

8.28

Chemical Kinetics

2. Graphical Method: In this method, the value of x are dx plotted against t (Fig 8.8) and rate of the reaction, dt at any time is determined from the slope, since dx = tan θ dt

The various values of

dx are plotted against (a-x). dt

Zero order

1st order

Log [A]

Conc [A]

2nd order

3rd order 1 [A]2

1 [A]

dx versus dt

(a-x)2, it is a second order reaction. The reaction will be dx a third order reaction if graph between and (a-x)3 is a dt straight line. Alternatively, (i) if the plot of (a-x) verses t is a straight line, the reaction is of the first order. (ii) The reaction is of the second order if the graph of 1 versus t is a straight line. (a - x ) 2

(iii) The reaction is of the third order if the graph of 1 versus t is a straight line. (a - x ) 2 t

t

If a straight line is obtained by plotting

3. Fractional Change Method: It has already been shown that time tf required to complete a definite fraction (say one half) of the reaction is (i) Directly proportional to initial concentration for zero order reactions. (ii) Independent of initial concentration for first order reactions i.e., t ∝

1 . a0

(iii) Inversely proportional to the initial, concentration t

t

Conc

Conc 3rd order

Rate

Rate

2nd order

of second order reaction i.e., tf ∝

(iv) Inversely proportional to the square of the initial

where n is the order of the reaction. If a1 and a 2 are the initial concentrations in two different experiments and t1 and t 2 are corresponding times for a definite fraction of the reaction to be completed then, 1 ( t f )1 ∝ n -1 (1) a1 (t f )2 ∝

2

(Conc) fig 8.8 method

1 . a1

concentration of third order reaction i.e., tf ∝ 1 . a2 1 In general, we can say that tf ∝ n −1 . a

Rate

Rate

(a) Plots of rate Vs Concentration [Rate = K (Conc.) n] 1st order Zero order

3

(Conc)

(b) plots for integrated rate equations Determination of order of reaction by graphical

1 a

n -1 2

(2)

Dividing Equation (1) by (2) ( t f )1  a 2  = ( t f ) 2  a1 

n -1

(3)

Chemical Kinetics

Taking logarithm of equation (3) we get  tf n = 1 +  log 1 tf  2 

  a2     log     a1   

Thus starting with two initial concentrations and finding (say t0.05)in each case of reaction can be determined Alternatively, if t0.5 is the time for half change and ‘a’ is the initial concentration, then (i) For first order reaction, t0.5 = constant (ii) For second order reaction, t0.5 × a = constant (iii) For third order reaction, t0.5 × a2 = constant (iv) For nth order reaction, t0.5 × an–1 = constant Solved Problem 21 The half life of a chemical reaction at a particular concentration is 50 minutes. When the concentration is doubled, the half life becomes 100 minutes. Find out the order of the reaction. Solution: On doubling the concentration, the value of t0.5 becomes 2 times the original value or t0.5 is directly proportional to the initial concentration. Hence reaction is a zero order reaction. Solved Problem 22 The half-life period for the thermal decomposition of phosphine at three different pressures are given below Pressure in mm Hg 707 79 37.5 Half-life period in seconds 84 84 83 Solution: Since the half-periods are the same irrespective of the initial concentration i.e, pressure, the reaction is of the first order Solved Problem 23 In the thermal decomposition of a gaseous substance, the time taken for the decomposition of half of the reactants was 105 minutes when the initial pressure was 750 mm and 950 minutes when the initial pressure was 250 mm find the order of reaction. Solution: (t0.5)1 = 105 minutes P1 = 750 mm (t0.5)2 = 950 minutes P2 = 250 mm For reaction of nth order

(t ) (t )

0.5 1

0.5 2

a  =  2  a1 

n -1

P  = 2  P1 

n -1

or

105  250  =  950  750 

or log

8.29

n −1

105  250  = log   950  750 

n −1

 250  = (n − 1) log    750  105 log 950 = −0.9565 or n − 1 = 250 −0.47774 log 750 or n − 1 = 2.00 n = 2.00 + 1.00 = 3.00 Hence the reaction is of third order. 4. Vant Hoff Differential Method: it has already been discussed that dx = K(a - x) For first order reaction dt dx = K(a - x) 2 For second order reaction dt dx = K(a - x)3 For third order reaction dt In general dx = K(a - x) n For nth order reaction dt Putting (a-x) = C, the concentration at any instant, in above equation dC1 (1) = KC n dt For two different concentrations C1 and C2 of the reactants We have dC (2) - 1 = KC1n dt dC - 2 = KC n2 (3) dt -

Taking logs  -dC1  = log K + n logC1 log   dt 

(4)

 -dC2  = log K + n logC2 log   dt 

(5)

8.30

Chemical Kinetics

Subtracting equation (5) from (4)

Now dividing (1) by (2) we get

 dC   dC  n (log C1-log C2) = log  − 1  − log  − 2   dt   dt 

r 1 =  a  b =     8r 8  2a   2b 

 dC   dC  log  − 1  − log  − 2  dt    dt  Or n = log C1 − log C2

1  1  1 =  =   2 8  2

n

n

Thus for determining the order of the reaction, we need the values of the rates of reaction

dC1 dC and 2 dt dt

at two different concentrations C1 and C2 respectively. Those can be obtained from the slopes of C vs, t plots. Substitution of these values for two different initial concentrations, n can be calculated.

mA + nB → products

For a reaction between gaseous chlorine and nitric oxide 2NO + Cl2 → 2NOCl it is found that doubling the concentration of both reactants increase the rate eight times but doubling the chlorine concentration alone doubles the rate. What is the order of reaction with respect to nitric oxide and chlorine? Solution: Suppose the initial concentration of NO is ‘a’ moles/ litre and the concentration of Cl2 is b moles/litre and the order of the reaction with respect to NO is n and with respect to Cl2 is m. Then initially, the rate of reaction will be given by (1)

Now when the concentrations of both reactants is doubled, the rate will be expressed as 8r = K (2a)n (2b)m

(2)

In the second case, when the concentration of only Cl2 is doubled We have 2r = K (a)n (2b)m Dividing equation (1) by equation (3) we have (a) n (b) m r 1 = = 2r 2 (a) n (2b) m 1  1 Or =   2 2

m

m = 1 or the order w.r.t Cl2 is 1

m

Putting n = 1 in the above equation, we get n = 2 or the order w.r.t NO = 2 Thus the total order of the reaction is 2 + 1 = 3 5. Ostwald’s Isolation Method: This method is employed in determining the order of complicated reactions by isolating one of the reactants. This is achieved by taking all but one of the reactants in a very large excess so that their active masses remain constant. Variation of concentration of this reactant only, permits a direct determination of the order of the reaction. Consider for example,

Solved Problem 24

R = K (a)n (b)m

m

(3)

By taking A in large excess, the determination of the order of reaction from the concentration change of B will give the value of n. Similarly, by taking B in large excess, value of m can be determined. The total order of reaction will be (m + n).

8.8.9 Complications in the Determination of order of Reaction: Complex Reactions The study of chemical kinetics gets highly complicated due to the occurrence of various reactions as described below. 1. Consecutive Reactions: Those reactions in which the products obtained in the first stage react with each other or with the original reactant to give rise to new products are known as consecutive reactions. The rate and order of such reactions are determined by that reaction which is the slowest. Few examples of consecutive reactions are given below: (a) Oxidation of hydroiodic acid by hydrogen peroxide in aqueous solution.

 → 2H2O + I2 + → 2H2O + I2 Or H2O2 + 2H + 2I– 

H2O2 + 2HI

The rate of this reaction is given by Rate =

dx = K [H2O2] [I–] dt

Evidently, the reaction is of second order. The reaction has been found to proceed in the following three stages. H2O2 + I– → H2O + IO– (slow) H + + IO– → HIO (fast)

Chemical Kinetics

HIO + H + + I–→ H2O + I2 (fast) The first stage is slowest and hence determines the rate of the reaction. (b) Reduction of hydrogen peroxide by hydrobromic acid in aqueous solution. H2O2 + 2H + 2Br → 2H2O + Br2 The rate of the reaction is given by +

Evidently, the reaction is of third order. The following mechanism has been proposed. (1) H2O2 + H+ + Br– → HOBr + H2O (slow) (2) HOBr + H+ + Br– → H2O + Br2 (fast) 2. Reversible Reactions: Those reactions in which the products of a chemical change together to original reactants are known as reversible reactions. These are also called counter or opposing reactions. In other words, a reaction in which the direct and reverse reactions occur simultaneously is known as reversible reaction. The reversible change occurs under the same conditions as the direct change and this causes a serious disturbance in the measurement of the reaction rate. For example, consider the following reversible reaction K

3  ⇀ 2NO + O2 ↽   2NO2 K 2

According to Bodenstein and Linder (1992), the rate of the forward reaction is third order below 290°C and the reaction proceeds without complications. Above this temperature, the rate of dissociation becomes noticeable and leads to a decrease in the rate of disappearance of NO and O2. 3. Side Reactions or Parallel Reactions: The study of the kinetics of a reaction becomes more complicated when two or more reactions takes place simultaneously reactions. The reactions forming the largest amount of product are generally known as the main reaction, while the others forming smaller amounts of products are known as side reactions. Such reactions are common both in organic as well as inorganic chemistry. For example, in the nitration of phenol both o- and pnitrophenols are formed simultaneously by the parallel reactions. Suppose a and b are the initial Concentrations of phenol and nitric acid, and X is the amount OH HNO3

+

K2

HNO3

NO2 p - nitrophenol

of these which reacts then at any instant the rates of formation of o- and p- nitro phenols are given.

dx d [ Br2 ] = K [H2O2] [H + ][Br–] = dt dt

+

OH

OH

–

OH K1

8.31

NO2 o - nitrophenol

d (o - nitrophenol ) = K1 (a – x) (b – x) dt

(1)

d ( p - nitrophenol ) = K1 (a – x) (b – x) dt

(2)

Hence the rate of disappearance of phenol and nitric acid i.e., those of reactants is given by = (K1 + K2) (a-x) (b-x) Moreover, from equations (1) and (2) K1 (a - x)(b - x) K1 d (o - nitrophenol ) = = K 2 (a - x)(b - x) K2 d ( p - nitrophenol ) Thus, the sum of the rate constants can be calculated by determining the rate of disappearance dx/dt of the reactants and the ratio of constants K1/K2 can be calculated from the rate of formation of o-nitrophenol and p-nitro phenol respectively. Other example C6H5Cl + HCl Chloro benzene

(Main reaction)

C6H6 + Cl2 C6H6Cl6 (Side reaction) Benzene hexachloride

CH2 = CH2 (Main reaction) Ethylene CH3CH2Cl + KOH CH3CH2OH (Side reaction) Ethyl alcohol

4. Surface Reactions: There are many homogeneous gaseous reaction such as thermal decomposition of phosphine, arsine and vapours of hydrogen peroxide which take place exclusively on the walls of the containing vessel. The rates of such reactions are determined by the number of gaseous molecules colliding against the walls per unit time and hence of the gas. 5. Chain Reactions: There are various reactions, for example, the combination of hydrogen and chlorine

8.32

Chemical Kinetics

gases to form hydrogen chloride, in which the first stage is the formation of chlorine atoms. These atoms may be produced by thermal or photochemical dissociation of the molecules; for example Cl2 + light (hn) → 2Cl These atoms may be produced by introducing a small quantity of the sodium vapour, when the reaction Na + Cl2 → NaCl + Cl takes place and the chlorine atoms thus formed can set up the hydrogen, chlorine chain. If atomic chlorine is introduced in a suitable manner into a mixture of chlorine and hydrogen generally by exposure to light, the following two reactions occur, both being very rapid. Cl + H2 → HCl + H (a) H + Cl2 → HCl + Cl (b) While the primary stage is Cl2 + hn → 2Cl Cl atom which is responsible for the reaction (a) is regenerated in reaction (b). This chlorine atom again reacts with a molecule of hydrogen giving a molecule of HCl and the hydrogen atom i.e., reaction (a) This hydrogen atom reacts with a molecule of chlorine to regenerate an atom of chlorine with a molecule of HCl. This process is repeated several times and a chain is set up due to the dissociation of chlorine molecule into atoms. A reaction like the above, proceeding in a series of successive reactions initiated by a suitable primary process is called a chain reaction. The reaction goes on and on until the chain terminates in one of the steps involved. In the above reaction, removal of hydrogen or chlorine atoms by any of the following reactions will result in the termination of chain. 2H → H2 2Cl → Cl2 H + Cl →HCl These reactions can occur at the walls of the containing vessel and do not take place readily in the gas phase. Combination of hydrogen and bromine to form HBr is another example of chain reaction.

8.9 EFFECT oF TEMPERATURE oN RATE of reactIon Rate of reaction increase with increase in temperature. In most cases, the rate of chemical reaction is largely increased as the temperature rises. Then why so much increase in rate with increase in temperature?

A chemical reaction takes place as a result of collisions between the reacting molecules which are in random motion. Since the rise in temperature enhances the speed of the molecules, this perhaps, responisble to increase the rate of reaction. But this is not all because the kinetic molecular theory fails miserably of reaction rate with 10°C rise in temperature. We know that according to kinetic molecular theory, the number of molecular collision is proportional to the square root of the absolute temperature of the gas i.e., Z∝ T. Where Z is the number of collision per second at TK. Let the number of collision at 27°C or 300 K = Z300 and the number of collision at 37°C or 310 K = Z310

∴

Z310 310 = = 1.106 Z300 300

Thus Z310 = 1.016 × Z300 It shows that with rise of 10°C in temperature, the rate of reaction is only slightly increased. But actually it has been found that the rate is almost double by a rise of 10°C in temperature. Arrhenius Equation We know from kinetic theory that the kinetic energy of a gas is distributed within the gas in such a manner that some molecules possess high energy and some low energy The number of gaseous molecules, n having an energy greater than a value E is related to the total number of molecules no by the Maxwell distribution law n = n0e–Ea/ RT Where R is gas constant and T is the absolute temperature. According to Arrhenius, reaction takes place between activated molecules and the rate of reaction is directly proportional to the number of such activated molecules. He suggested that in order to react, a molecule or a pair of molecules in collision must possess a certain minimum energy called threshold energy. For a reaction which proceeds slowly at room temperature, this threshold energy is much greater than average kinetic energy of the molecules. Before a molecule can react, it must acquire extra energy either by collisions or by absorption of radiation such as infrared, ultraviolet or other radiation. For an activation energy E, the number of molecules with energy greater than E, will be n Thus rate of reaction as well as rate constant of reaction will be proportional to n.

8.33

Log K

No of molecules

Chemical Kinetics

Velocity of molecules Fig 8.9 Distribution of molecular velocities 1 T

Thus Rate of reaction K = A.e–Ea/RT (2) Where A is called the Arrhenius factor or frequency factor, the quantity e–Ea/RT is called Boltzmann factor. The equation K = Ae–Ea/RT is known as Arrhenius equation. On taking logarithm ln K = ln A -

or log K = log A-

Ea RT

Ea (2.303RT )

Fig 8.10 Variation of log K with

(i) The intercept of this line of equal to log A and (ii) Slope = -Ea /2.303 R Thus knowing the value of slope and gas constant R, activation energy can be calculated as follows: Ea = –2.303 R × Slope

(3)

Where A is a constant and this expression can be written as d[lnK] Ea d[log.K] Ea = ;or = dt dt RT 2 2.303RT 2 If log10 K is plotted against 1/T, a straight line is obtained and the energy of activation Ea can be deduced from the slopes of the lines. An alternative method of evaluating Ea is to take the rate constant K1 and K2 at two different temperature T1 and T2. Then K2 Ea  T2 - T1  (4) =   K1 2.303R  T1T2  Thus the value of Ea can readily be calculated if rate constant K 1 and K 2 at temperatures T1 and T2 are known. The equation (3) is of the straight line. When log K is

Solved Problem 25 The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation for such a reaction. Solution: Given that K2 =4 K1 Also T1 = 293 K and T2 = 313 K Applying the reaction K2 Ea  T2 - T1  =   K1 2.303R  T1T2  Substituting the values in the above equation

log

log

plotted against

1 we get a straight line. T

1 T

log 4 =

Ea  313 - 293  2.303R × 8.314  293 × 313 

20 Ea 2.303 × 8.314 × 293 × 313 0.6020 × 2.303 × 8.314 × 293 × 313 ∴ Activation energy Ea = 20

∴ 0.620 =

= 52854.55 J mol–1 Hence activation energy = 52.85 KJ mol–1

8.34

Chemical Kinetics

Solved Problem 26

8.9.1 Activated Molecules and Temperature

The rate constants for the decomposition of N2O5 at various temperatures are t°C Rate Constant (K) (s–1) 0 7.87 × 10–7 20 1.70 × 10–5 40 2.57 × 10–4 60 1.78 × 10–3 80 2.14 × 10–2

Increase in temperature increases the number of activated molecules to a much greater extent than the number having average kinetic energy. From kinetic theory of gases, we know that average kinetic energy of a gas at constant temperature T is equal to 3/2 RT. At 1000 K, the average kinetic energy of the molecules will be

1 Plot log10 K against and calculate the energy of T activation. Predict what the reaction rate would be at 303 K, 323 K and 343 K. Solution: From the given data, log K is calculated for value of K. Ks–1 7.87×10–7 1.7×10–5 2.57×10–4 1.78×10–3 2.14×10–2 log K -6.104 -4.7696 -3.5901 -2.7496 -1.6696

1 On plotting log K versus the slope of the line is T found to be -5300. Ea We know that slope = 2.303R

∴ Activation energy Ea = -2.303 slope = -2.303 × 8.314 × (-5300) = 101479.85 Jk–1 mol–1 = 101.5 KJ mol–1 K–1 Calculating rate constant at 303 K For the given data K1 = 7.87 × 10–7 T1 = 273 K K2 = ? T2 = 303 K Applying the reaction log

K2 Ea (T2 - T1 ) = K1 2.303RT1T2

Or log

K2 = 7.87 × 10 -7

101479.85(303 - 273) = 1.9222. 2.303 × 8.314 × 273 × 303 Taking anti-log K2 = 83.31 7.87 ×10-7 ∴ K2 = 7.87 × 10–7 × 83.31 = 655.6 × 10–7

(3 × 2 × 1000) = 3000 cal per mole (R = 2) 2 Similarly, at 2000 K, it will be (3 × 2 × 2000) = 6000 cal per mole (since R = 2 cal/ 2

degree/ mole) In other words, the average kinetic energy of the gas molecules is doubled by doubling the absolute temperature. Now the fraction of molecules having energy greater than, e.g., 20 Kcal at 1000 K is given by e–Ea/Rt = e–20000/2×1000 = 0.00458 per cent But the fraction with energy greater than 20 Kcals (20,000 cal) at 2000 K is given by e–Ea/Rt = e–20000/2×2000 = 0.678 per cent Hence, a change of temperature which only doubles the number of molecules having an average kinetic energy increase the fraction of activated molecules more than 100 times. It clearly indicates that why temperature has such a great effect on rate of reaction. The less the energy of activation for a reaction the more easily that reaction will take place. For example, for a reaction with energy of activation of 20 Kcal per mole the rate of reaction will be quite good even at ordinary temperatures, while for an activation energy of 40 Kcals per mole, the reaction will proceed at a reasonable rate provided the temperature is raised to about 400-500°C. We know that rate constant and temperature are related as (d ln K ) E = dT RT 2

Similarly, equilibrium constant and temperature are related as (d ln K ) ∆H = dT RT 2

This similarity might be expected because of the fact that the equilibrium constant is a ratio of two rate constants and the heat of reaction ΔH is related to the activation energies of the forward and backward reactions.

Chemical Kinetics

8.10.1 Collision Theory of Reaction Rate According to this theory, chemical reaction takes place only by the collision between molecules. If the number of collisions between the reactant molecules per unit volume per second is Z and the reaction takes place at collision, the rate of reaction will be equal to Z. However, this does not happen. The number of collisions between any two reacting species at any given concentration and temperature can be calculated from the kinetic theory of gases. The value of the rate constant derived from the number of collisions has been found to be much higher in most of the reactions than their experimental value. In other words, all collisions in a reaction are not effective collisions between the molecules; only very few are effective which lead to reactions. Arrhenius suggested that there is certain minimum energy which the colliding molecules must acquire before they are capable of reacting. This minimum energy is called threshold energy and may be defined as the minimum energy that a molecules or pair of molecules must possess in order to react in a collision. Most of the molecules, however, have much less kinetic energy than the threshold energy. The excess energy that the reactant molecules having less energy than the threshold energy must acquire some extra energy in order to react or participate in a reaction to yield products. This extra energy is known as energy of activation. Thus Activation Energy = Threshold Energy –Energy actually possessed by the molecules Before a molecule can react, it must acquire extra energy either by collisions or by absorption of energy. The concept of energy of activation may be made clear by using the Fig 8.11. Suppose EA represents the average energy of reactants, EC that of products and EB the minimum energy which the reactants must possess in order to react.

Ea2 C

Reactants

Products

Energy of activation

fig 8.11 Energy of activation

Reactants Heat of Reaction Products Reaction coordinate

fig 8.12 Exothermic reaction

For the process A→C, the energy of activation EA absorbed by the reactants A in order to become activated and react EA1 = EB-EA.

EA2 = EB – EC where EC is the energy of products thus EA1-EA2 = ΔE = EC-EA.

Δ E

A

Activated State

EB is the energy in the activated state and EA, the corresponding energy of reactants. Similarly, the energy of activation EA2, for the reverse process C→A is given by

Activated complex B

Ea1

The molecules in the state B are said to be in an activated or transition state. Now since molecules must be activated before the reaction (i.e., the reactant molecules have to acquire a certain amount of the energy before they can react to yield the product). The reaction must take place not directly from reactants A to products C, but along the path ABC. In other words, there is an energy barrier that has to be crossed by the reactants before they can yield products.

Energy

8.10 THEoRIES oF REACTIoN RATES

8.35

Where ΔE represents the energy difference between products and reactants or heat of reaction at constant volume. It is evident from this concept that in moving from the activated complex state to the products, the molecules give up not only the energy absorbed on activation from A to B but also ΔE corresponding to the difference in energy levels of C to A.

8.36

Chemical Kinetics

Suppose the activation energy for forward reactions X→Y be Ex and that of the reverse reaction y→x be Ey. The heat of reaction Q is the difference between these two energies Q = Ex-Ey The rate of forward reaction = K1 [x] The rate of backward reaction = K2 [y]

Activated Complex

Energy

Activated energy Heat of reaction

The equilibrium constant K =

K1 K2

Thus for forward reaction d ln K1 Ex = dT RT 2

Reactants

Reaction coordinate

Thus

fig 8.13 Endothermic reaction Reactions generally proceed more readily when the products have a lower energy content than the reactants, the difference being the free energy of activation. In order for a reaction to proceed, energy has to be supplied to the reactants in order to carry them to the top of barrier. The amount of the work to get the reactants up to the top of the barrier is called energy of activation or free energy of activation ΔG. The energy of activation is the energy that the reactants in the original state must absorb in order to get activated and react. Highly endothermic reactions are unlikely to take place readily at low temperatures. The energy of activation has to be greater than the heat of reaction so that the activation energy for any endothermic reaction with a high heat of reaction must be high. The paths followed by an exothermic reaction are given in fig. 8.14. Consider, a general reversible reaction

 Y.; ΔH = -Q calories (heat evolved). X  The reaction path can be represented as in Fig 8.14.

X

Ex

Ey

Energy

Q

Y Reaction coordinate

fig 8.14 Activation energies of forward and backward reactions

d [ln K ] d [ln K1 ] d [ln K 2 ]  K1  = −   K = dT dT dT K2  

= Ex - Ey = 1  E - E  x y RT 2 RT 2 RT 2  =- Q RT 2 Using ΔH instead of –Q, we have

d ln K ∆H = dT RT 2 Which on integration gives ln K = C- ∆H RT Where C is a Constant and ΔH does not alter with temperature. Solved Problem 27 For the dissociation of gaseous HI, the energy of activation is 44.3 Kcal. Calculate the energy of activation for the reverse reaction. Given ΔH for the formation of 1 mole of HI from H2 and I2 gases is -1.35 Kcal. Solution:  H ( g ) + 1 I ( g ); E = 44.3 Kcals 2HI(g)  2 2 2 1 1  HI(g); ΔH = -1.35 Kcals H 2 + I 2  2 2

 2HI; ΔH = 2 × 1.35 = 2.70 Now H2(g) + I2(g)  Kcals But ΔH = E-E´ So that E´- ΔH = 44.3-2.70 = 41.6 Kcals

Chemical Kinetics

Solved Problem 28 For the reaction of hydrogen and iodine, the rate constant is 2.45 × 10–4 litre moles–1 sec–1 at 302°C and 0.950 at 508°C. Calculate the energy of activation and the frequency factor of the reaction. What is the value of rate constant at 400°C? Solution: We know that  K1  Ea  T2 - T1   K  = 2.303R  T T  2  2 1  In this problem K = 2.45 × 10–4, K2 = 0.950, T1 = 302 + 273 = 575 k T2 = 508 + 278 = 781 k

log10

∴ log

0.95 Ea  781 - 575  = -4 2.303 × 1.987  781 × 575  2.45 × 10

∴ Ea = 35700 cals/ mole Now we know K = A e –Ea/Rt ∴ 2.45 × 10–4 = A.e–35700/1.987 × 575 A = 9.0 × 109 Now Ea = 35700, A = 9.0 × 109 and T = 400 + 273 = 673 K ∴ Rate constant K at 400°C = A e –Ea/Rt A = 9.0 × 109 e–35700/1.987 × 673 K = 3.17 × 10–2 Solved Problem 29 In the decomposition of acetaldehyde in gaseous phase, various values of log10 K were plotted against 1/T and a straight line with a slope of -9920 was obtained. Similarly, a slope of -5070 was obtained in case of the decomposition of acetone dicarboxylic acid. Calculate the energy of activation in the cases given R = 2.0. Solution: We know that Ea Slope = 2.303R Ea = 2.303 × 2 Thus for the decomposition of acetaldehyde - 9920 = Ea 4.606 Ea = 45632 cal/ mole Similarly, for the decomposition of acetone dicarboxylic acid - 5070 = Ea 4.606 Ea = 23322 cal/mole

8.37

Orientation Factor: Another important factor is that the colliding molecules should have proper orientation so that the old bonds may break and new bonds are formed the reaction between hydrogen and iodine takes place only when those molecules possessing threshold or more energy collide with proper orientation. The molecules which do not have proper orientation at the time of collision do not form product molecules as shown in Fig 8.15. Derivation of Rate Constant: Out of Z colliding molecules, if q is the fraction of the molecules which are in activated state, then the rate constant (K) of the bimolecular reaction in terms of collision theory is K = Z.q

(1)

From MaxWells’s distribution law of molecular velocities, it is shown that in a gas containing n molecules per c.c at temperature T the number of molecules (‘n’) which possess’ activation energy, Ea is given by n' = n.e–Ea/RT n ′ = e–Ea/RT = q n K = Ze–Ea/RT

(2)

Thus, the rate of reaction is equal to the number of activated collisions in unit time

dx = Ze–Ea/RT dt

(3)

Since the reaction involves two molecules of the same kind, the relationship between rate of reaction and concentration is given by

dx = Kn2 dt

(4)

Where K is the specific rate and n is the concentration of reactants expressed as the number of molecules per mL from equation (3) and (4). K=

Z –Ea/RT e n2

(5)

From kinetic theory of gases, the value of Z in case of two molecules of the same as in the reaction whereby gaseous hydrogen iodide decomposed thermally is given by πRT  Z = 4n2s2   M 

1/ 2

(6)

Where M is the gram molecular weight and σ is the molecular collision diameter 1/2

Thus K = 4s2  π RT  e.- Ea / RT  M 

8.38

Chemical Kinetics

Products Activated complex Reactant molecules

No product formation

Reactants fig 8.15 Orientation of collisions

and.K = Z .e

Where 6.024 ×1023 is the Avogadro’s number. Substituting M = 127.9 and 1 A° = 10–8 cm in equation (7) Z comes out to be equal to 9.90 × 1010

- Ea / RT

1/2

Where Z = 4σ2  π RT   M 

= Collision number

(7)

Now, consider the thermal decomposition of gaseous hydrogen iodide, which is found by experiment to have a second order rate constant of 3.6 × 10–7 litre mol–1 sec–1 at 556 K.

 H2 + I2 2HI  Hydrogen iodide has a collision diameter of 3.5 A°. The molecular weight M = 127.9 and Ea the energy of activation = 44000 calories Thus e Ea / RT = e -44000/1.987×556 = 5.10 ×10-18 The number of molecules per cc 23 n = 6.024 ×10 = 6.024 × 1020 103

Thus K = Z. e

- Ea / RT

K = 9.9 ×1010 × 5.1 × 10–18 = 5.1 × 10–7 moles per litre per second. Which is in good agreement with experimental value of 3.6 × 10–7 Although collision theory of reaction rate is quite satisfactory when applied to a number of bimolecular reactions, both in gas phase and in solution, however, certain reactions, both in gaseous phase and in solution were found to proceed more readily than calculated. These include certain reactions of the first order and some ionic reactions in solution. In order to account for these delviations, the equation (7) was modified as K = P.Z .e - Ea / RT Where P is a steric or probability factor. The value of P ranges from unity (in this case the reaction obeys collision theory) to about 10–9 (in this case the

Chemical Kinetics

reaction occurs far more slowly than required by the collision theory). P may be regarded as the ratio of the observed rate constant and that calculated by the simple collision equation. P is correction term which tells us that only those collisions in which the molecules are in correct orientation result in a reaction.

8.10.2 Catalysts and Activation Energy Positive catalysts increase the rate of a reaction without the necessity of increasing the temperature. They do this by providing a different route for the reaction to take place i.e., the mechanism of the reaction is different in the presence of the catalyst. The key thing is that the new mechanism has a lower activation energy than the uncatalysed reaction with the decrease in the activation energy, many more reactant particles are able to surmount the new energy barrier and turn into products. A negative catalyst increases the activation energy of reaction by forming a new intermediate of higher energy, i.e., by changing the reaction mechanism. Due to increased activation energy, some active molecules become inactive, therefore, rate of reaction decreases.

8.39

2. Molecules are considered to be rigid spheres which are not applicable in the case of more complex types of molecules. 3. Reaction may also occur by a chain mechanism, causing a high P factor. 4. In surface catalysed reactions, the rate of reaction does not depend upon the number of collisions in the gas phase. Hence collision theory is not applicable in such type of reactions. 5. Collision theory omits from consideration an entropy of activation which cannot be disregarded. 6. The values of P do not correlate with the structures and properties of the reacting molecules. 7. The rate of reaction is usually very slow compared to the number of collisions occurring per cm. 8. Collision theory is also not very successful in the case of reversible reactions.

8.10.3 Transition State Theory This theory has been developed by Eyring. According to this theory the reactant molecules form an activated complex, prior to the formation of the products, which remains in thermodynamic equilibrium with the reactants. For example, K K  Reactants ↽ → products ⇀  Activated complex  ∓

Reasons For The Failure of Collision Theory 1. Large molecules require a correct orientation before they can react. They are also required to have necessary activation energy.

The activated complex possesses the properties of an ordinary molecule, but it has a transient existence. The energy of the activated complex is higher than both reactants and products. It decomposes at a definite rate to form the

Activation Energy Without Catalyst

Energy

Activation Energy with Catalyst via new reaction path

products

Reactants

Reaction coordinate fig 8.16 Course of reaction in the absence of and in presence of catalyst

8.40

Chemical Kinetics

products. The rate of the decomposition of the activated complex and the concentration of the activated complex determine the rate of the reaction. On the basis of these ideas, the expression for rate constant K is given as

Transistion State

Reactants

RT is a universal constant for Nh

given temperature, the specific reaction rate at a constant temperature depends on the free energy change of the activation process. The magnitude of free energy change being related to enthalpy change and entropy change (GibbsHelmholtz equation), the probability factor introduced in the collision theory can be explained in terms of entropy. Transition state theory concentrates on what happens to the reactants when they are about to change into products. This change takes place at the top of the energy barrier, so this is called the transition state region (Fig 8.17). When a reactant molecule or combination of molecules is changing into products, some of its old bonds are breaking and new bonds are being made. When the reactants are in this state of change they make up the activated complex. The activated complex does not always change into products; it can equally well change back into the reactants. One example is the reaction between a hydrogen atom and a hydrogen molecule. The transition state is linear with the two bonds some where between being made and being broken (Fig 8-17). If the energy possessed by the transition state is concentrated in bond A, then it changes back into reactants; if it concentrates in bond B, then it makes products. It can happen that the activated complex changes into an intermediate that is more energetically stable than the activated complex. The intermediate may last for sufficiently long time to allow it to be isolated but this is rate. More often, it is possible to spot an intermediate by virtue of its spectrum. The intermediate may change into another activated complex before being converted into products +

CH 3 CH CH 3 The energy profile for the reaction is shown in Fig 8.18. It is the aim of transition theory to explain the rates of reactions by working out the nature of the activated complex, how it is made from the reactants and how it changes into products.

H

H-H

H-H H Products

Course of Reaction fig 8.17 The reaction between a hydrogen atom and a H2 molecule

Intermediate + CH3CH CH3

Energy

changes, and the factor

H------H----H

RT ‡ K Nh

Where R = gas constant, T = absolute temperature, N = Avogadro’s number, h = Planck’s constant and K‡ = equilibrium constant between the activated complex and the reactants. Since equilibrium constant is related to free energy

Bond B

Energy

K=

Bond A

CH3.CH = CH2 + HCl

Cl CH3CH-CH3

Course of reaction

fig 8.18 The intermediate formed stabilised and remain at lower energy level than the activated complex. It is meta stable state. Problems for Practice 9. The rate constant is numerically the same for three reactions of first, second and third orders respectively, the unit of concentration being in moles per litre. Which reaction should be the fastest and is this true for all ranges of concentrations?

Chemical Kinetics

10. A first order gas reaction has K = 1.5 × 10–6 per second at 200°C. If the reaction is allowed to run 10 hours, what percentage of the initial concentration would have changed in the product? What is the half-life of this reaction? (IIT 1987) 11. Radioactive decay is a first order process. Radioactive carbon in wood sample decays with a half life of 5770 years. What is the rate constant (in year –1) for the decay? What fraction would remain after 11540 years? 12. While studying the decomposition of N2O5 it is observed that a plot of logarithm of its partial pressure versus time is linear. What kinetic parameters can be obtained from this observation? (IIT 1987) 13. The rate of first order reaction is 0.04 mole /L/ s at 10 minutes and 0.03 moles /L/S at 20 minutes after initiation. Find the half-life of the reaction. (IIT 2001) 14. The vapour pressure of the miscible liquids A and B are 300 and 500 mm Hg respectively. In a flask, 10 moles of A is mixed with 12 moles of B. However, as soon as B is added, A starts polymerizing into completely insoluble solid. The polymerization follows the first order kinetics. After 100 minutes, 0.525 mole of a solute is dissolved which arrests the polymerization completely. The final vapour pressure of the solution is 400 mm of Hg. Estimate the rate constant of the polymerization reaction. Assume negligible volume change on mixing and polymerization, and ideal behaviour for the final solution. (IIT 2001) 15. The rate constant for an isomerisation reaction A→B –3

is 4.5 × 10 minute –1. If the initial concentration of a is 1M. Calculate the rate of the reaction after 1 hour. (IIT 1999) 16. At 380°C the half life period for the first order decomposition of H2O2 is 360 minute. The energy of activation of the reaction is 200 KJ mole–1. Calculate the time required for 75 % decomposition at 450°C. (IIT 1995) 17. The rate constant for the first order decomposition of a certain reaction is given by the equation ln K (sec–1) = 14.34 -

1.25 × 10 -4 T

Calculate (a) the energy of activation (b) the rate constant at 500 K (c) At what temperature will its half-life period be 256 minute? (IIT July 1997)

8.41

18. A hydrogenation reaction is carried out at 500 K. If the same reaction is carried out in presence of a catalyst at the same rate the temperature required is 400 K. Calculate the activation energy of the reac tion if the catalyst lowers the activation energy barrier by 20 KJ mol –1. (IIT 2000) 19. A first order reaction is 50 % completed in 30 minutes at 27°C and in 10 minutes at 47°C. Calculate the reaction rate constant at 27°C and the energy of activation of the reaction in KJ mol–1. (IIT 1988) 20. In Arrhenius equation for a certain reaction, the values of A and Ea (activation energy) are 4 × 1013 sec–1 and 98.6 KJ mol–1 respectively. If the reaction is of first order, at what temperature will its half-life period be 10 minutes? (IIT 1990) 21. The decomposition of N2O5 according to the equation N2O5 (g) → 4NO2 (g) + O2 (g) is a first order reaction. After 30 minutes from the start of decomposition in a closed vessel, the total pressure developed is found to be 284.4 mm Hg and on completion, the total pressure is 584.5 mm Hg. Calculate the rate constant of the reaction. (IIT 1991) 22. The gas phase decomposition of dimethyl ether follows first order kinetics: CH3OCH3 (g) → CH4 (g) + H2(g) + CO(g) The reaction is carried out in a constant volume container at 500°C and has a half-life of 14.5 minutes initially only dimethyl ether is present at a pressure of 0.40 atm. What is the total pressure after 12 minutes? Assume ideal gas behaviour. (IIT 1993) 23. From the following data for the reaction between A and B [A]

[B] –1

Expt (mol L ) (mol L–1) 1 2 3

2.5 × 10–4 5.0 × 10–4 1.0 × 10–3

3.0 × 10–5 6.0 ×10–5 6.0 × 10–5

Initial Rate (mole L–1 S–1) 300 K 5.0 × 10–4 4.0× 10–3 1.6 × 10–2

320 K 2.0 × 10–3 -

Calculate the following: (i) The order of the reaction with respect to A and with respect to B (ii) The rate constant at 300 K (iii) The energy of activation (iv) The preexponential factor (IIT 1994)

8.42

Chemical Kinetics

24. The half life of first order decomposition of nitramide is 2.1 hours at 15°C

NH 2 NO2 (aq )  → N 2 O( g ) + H 2 O(l ) If 6.2 g of NH2NO2 is allowed to decompose calculate (i) time taken for NH2NO2 to decompose 99% and (ii) the volume of dry N2O produced at this point, measured at STP. (ROORKE 1994) 25. Catalytic decomposition of nitrous oxide by gold at 900°C at an initial pressure of 200 mm was 50% in 53 minutes and 73% in 100 minutes. (a) What is the order of reaction? (b) How much it will decompose in 100 minutes at the same temperature but at an initial pressure of 600 mm? (ROORKE 1990) 26. The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is a first order reaction. (i) After 55 seconds at 400 K the pressure of Cl2O7 falls from 0.062 to 0.044 atm. Calculate the rate constant. (ii) Calculate the pressure of Cl2O7 after 100 seconds of decomposition at this temperature (ROORKE 1989) 27. Thermal decomposition of a compound is of the first order. If 50% of a sample of the compound is decomposed in 120 minutes. How long will it take for 90% of the compound to decompose? (ROORKE 1988) 28. For a reaction at 800°C 2NO + 2H2 → N2 + 2H2O

Calculate the rate constant and rate of formation of AB when [A2] = 0.01 [B] = 0.2 30. Which reaction will have the greater temperature dependence for rate constant one with a small value E or one with large value of E? 31. Bicyclohexane was found to undergo two parallel first order rearrangements. At 730 K, the first order rate constant for the formation of cyclohexene was measured as 1.26 × 10–4 s–1 and for the formation of methyl cyclopentene the rate constant was 3.8 × 10–5 s–1. What is the percentage distribution of the rearrangement products? 32. The complexion of Fe2 + with the chelating agent dipyridyl has been studied kinetically in both forward and reverse directions. Fe2+ + 3 dipy → [Fe(dipy)3]2+ Rate of forward reaction is (1.45 × 1013)[Fe2 + ][dipy] Rate of backward reaction is (1.22 × 10–4)[Fe(dipy)3]2+ Find the stability constant for the complex. 33. The approach to the following equilibrium was ob2– served kinetically from both directions PtCl4 + H2O – – ⇌ [P[(H2O)Cl3]+ Cl at 25°C at was found that d[PtCl4 ] = (3.9 × 10−5 )(PtCl4 2 − ) dt −(2.1× 10−3 )[Pt (H 2 O)Cl3− ][Cl− ] −

Calculate the equilibrium constant for the complexation of the fourth Cl– by Pt (II). 34. For the reaction 1 [Cr (H 2 O) 4 Cl2 ]+ (aq ) K →[Cr (H 2 O)5 Cl]2 + (aq ) 2 K →[Cr (H 2 O)6 ]3+ (aq )

The following data were obtained:

1 2 3

[NO]×10–4 mole/Litre

[H2] × 10–4 mole/litre

d[NO]/dt × 10–4 mole litre–1mm–1

1.5 1.5 0.5

4.0 2.0 2.0

4.4 2.2 0.24

What is the order of this reaction with respect to NO and H2? 29. The following data were obtained for a gaseous reaction A2 + 2B → 2AB

1 2 3

-

K1 = 1.78 × 10–3 s–1. and K2 = 5.8 × 10–5 s–1 for the initial concentration of [Cr(H2O)4Cl2]+ is 0.0174 mole. 1 litre at 0°C/Calculate the value of t at which the concentration of [Cr (H2O)5Cl]2+ is maximum. 35. Some PH3(g)is introduced into a flask at 600°C containing an inert gas. PH3 proceeds to decompose into P(g) and H2(g) and the reaction goes to completion. The total pressure is given below as a function of time. Find the order of the reaction and calculate the rate constant for the reaction. 4PH3 → P4 + 6H2

d [A2 ]

[A2] Mol L-1

[B] Mol L-1

Mole L-1 min-1

0.10 0.10 0.20

0.01 0.04 0.01

0.072 0.288 0.144

dt

Time (s): P mm (Hg):

0 262.4

60 272.9

12 275.51

α 276.4

36. Copper–64 emits a β particle. The half-life is 12.8 hour At the time you received a sample of this radioactive

Chemical Kinetics

isotope, it had a certain initial activity (Disintegration / min). To do the experiment you have in mind you have calculated that the activity must not go below 4% of the initial value. Calculate the required time for completion of the experiment. 37. In a first order reaction, the concentrations of reactants 10 and 20 minutes after the beginning of reaction corresponds to 13.8 and 8.25 (arbitrary units). Calculate

8.43

the initial concentration of the reactants and the velocity constant. Ni → 38. At 25°C for the reaction vegetable oil + H2  vegetable ghee, pressure of H2 reduces from 2 atm to 1.2 atm in 50 minute. What is the rate of reaction in term of change of (i) pressure per minute and (ii) molarity per second?

8.44

Chemical Kinetics

KEY PoINTS • • • •

•

Chemical kinetics deals with the study of rates and mechanisms of chemical and biochemical reactions. Chemical kinetic deals with the rates of reactions and factors that influence the rate of reaction. Reactions that takes place violently and explosively are called explosive reactions. Reactions whose rates are very high and cannot be measured by ordinary methods are called as fast reactions. Reactions which proceeds in a measurable time are called slow or moderately slow reactions.

•

•

• • •

The rate of reaction is defined as the change in molar concentration or product per unit time. The rate of a chemical reaction at any instant is the decrease in concentration of the reactant (s) or the increases in concentration of the product (s) in unit time at that instant during the progress of reaction. The rate at which the reaction is proceeding can be followed by measuring the concentration of either a reactant or a product. Any of the reactants or any of the products may be utilized to specify the rate. Since the concentration of reactants decrease with time d (reactant) is negative and product is positive. If dc represents the amount of the reactant decreased after a short interval of time dt rate = -

•

dc dt

The rate can be made equivalent by dividing the rate expression by the stoichiometric coefficient present in the balanced equation for a general reaction such as pP + qQ ⇌ rR + sS rate = −

•

dc dt

If dc represents the amount of the product changed during a small interval of time dt then the rate of reaction is represented by dx . dt rate = -

•

• • •

•

•

rate of reaction •

•

• •

-

•

∆x x 2 - x1 = where x1 and x2 are the concentrations ∆t t 2 - t1

in a time interval dt. dx Instantaneous rate of reaction is determined by -

dt

Where dx represent the decrease in concentration in a time interval dt.

Factors Influencing the Rate of Reaction Nature of Reactant • •

•

•

1 d[P] 1 d[Q] 1 d[R ] 1 d[S] =− = = p dt q dt r dt s dt

At any time, the rate of reaction depends on the concentration of the reactants at that instant.

The rate a reaction goes on decreasing with time progressing as the concentration of reactants decreases. The concentration of the reactants in a reaction varies exponentially but not linearly. Any reaction does not take places with uniform rate. The slope of the curve drawn between concentration of the reactant (c) and time (t) of a reaction gives the rate of reaction. The rate measured over a long time interval is called average rate and the rate measured for an infinites mally small time interval is called instantaneous rate. The change in concentration of any one of the reactants or products at a given time is known as instantaneous rate. The units of rate of reaction are moles litre–1 sec–1 or moles litre–1 min–1 mole litre–1 hr–1. Average rate of reaction is determined by

•

The rate of reaction depends on the nature of the reactants. During reaction between covalent molecules, the existing bonds in a molecules breaks up and new bonds will be formed. Ionic reactions takes place much faster because there is no breaking of bonds in reactants. E.g., Oxidation of Fe2+ to Fe3+ with permanganate is fast. Reaction between NaCl and AgNO3 in solution takes place very fast. The rate of reaction involving covalent molecules is slow because the bonds in reactant molecules should break and new bonds in products must form e.g., Oxidation of oxalate ion by permanganate is slow. The reaction between ethyl alcohol and acetic acid occurs slowly. If the number of covalent bonds to be broken in a re actant molecule and formation in product molecules

Chemical Kinetics

are more, the reaction is slow when compared to the reaction involving lesser number of bond breaking and bond formation (or) reaction involving lesser bond rearrangements are rapid at room temperature than those involve greater bond rearrangement. effect of concentration of reactants • • • • •

The rate of zero order reaction does not depend on the concentration of reactants. Except for zero order reactions, the rate of reaction depends on the concentration of the reactants. The rate of reaction is directly proportional to the (concentration of the reactants)n or Cn For gaseous reactions the rate of reaction is directly proportional to (pressure of the reactants)n. The dependence of rate of eaction on the concentration of reactants can be mathematically expressed as

-

dc = KC m dt

Effect of Temperature • • •

The rate of reaction increases with increase in temperature. Generally, rise of 10°C in temperature doubles the specific rate of reaction. The ratio of the specific rates measured at temperatures that differ by 10°C is called temperature coefficient.

Rate Constant, order of Reaction and Rate Law •

•

• •

The rate of reaction is influenced by catalyst. In some reactions the rate of reaction is directly proportional to the concentration of the catalyst e.g.,. In acid catalysed hydrolysis of ester reaction, the rate depends on the concentration of the acid. Catalyst is involved in the reaction but not consumed in the reaction. Catalyst increases the rate of reaction by changing the path of the reaction i.,e making the reaction to proceed in the path of lower activation energy.

•

Certain reactions called photochemical reactions are greatly stimulated in the presence of light of suitable wavelength e.g., the reaction between hydrogen and chlorine or methane and chlorine takes place slowly in the absence of light but takes place very rapidly in the presence of light.

K is known as specific rate constant or rate per unit concentration of the reactants. Units of rate constant are mole1–n litren–1 sec–1.

Molecularity of the Reaction •

•

•

•

•

Several chemical reactions take place in a sequence of steps and the overall rate of reaction is governed by the slowest step. In certain cases the slowest or rate determining step may involve the formation of an unstable intermediate from the reactant molecules. The total number of reactant molecules taking part in the slowest step may involve the formation of an unstable intermediate. The total number of reactant molecules taking part in the slowest step or limiting step in the formation of intermediate species is known as molecularity of the reaction. A unimolecular reaction is a step in which a single molecule spontaneously undergoes reaction  PCl3 + Cl2 Molecularity = 1 PCl5 

•

A bimolecular reaction requires two molecules to come together. 2HI H2 + I2 Molecularity = 2

•

A termolecular reaction is one in which three molecules interact in a single step.  2SO3 Molecularity = 3 2SO2 + O2 

•

Exposure to Radiations •

The mathematical expression of the rate of reaction on concentration terms of reactants is known as rate expression or rate equation or rate law. For the reaction A + B → Products; the rate equation is rate ∝ [A] [B] Or rate = K[A] [B]

effect of catalyst • •

8.45

Reactions of molecularity more than three are rare because in such reactions, collision of 4 or more molecules simultaneously is necessary. Chances for colliding more than three different species simultaneously each possessing energy equal to or greater than the threshold energy are rare.

order of Reaction •

The order of a reaction is the total number of molecules whose concentrations change during the chemical

8.46

Chemical Kinetics

→ 2Cl2(g) + O2(g) (iii) 2Cl2O (g)  rate = K [Cl2O]2; order = 2 → CH3 COONa + (iv) CH3COOC2H5 + NaOH  C2H5OH rate = [CH3COOC2H5][NaOH]; order = 2

reaction or the sum of the powers of the concentration terms in the rate equation e.g., → 2NO2 2NO + O2  Rate = K [NO]2 [O2] Order of reaction with respect to NO = 2 Order of reaction with respect to O2 = 1 The overall order of reaction is 2 + 1 = 3

Examples of third order reactions: → 2NO2 (i) 2NO (g) + O2 (g)  rate = K[NO]2[O2]; order = 3 → 2NOCl (ii) 2NO (g) + Cl2 (g)  rate = K [NO]2[Cl2]; order = 3 (iii) 2Fe Cl3(aq) + SnCl2(aq) → 2FeCl2(aq) + SnCl4(aq) Rate = K [FeCl3]2[SnCl2]; order = 3

Examples of first order reactions: 1 → 2NO2(g) + O2(g) (i) N2O5 (g)  rate = K [N2O5]; order = 1 2 → SO2 (g) + Cl2 (g) (ii) SO2Cl2 (g)  rate = K [SO2Cl2]; order = 1 1 → H2O (l) + O2 (g) (iii) H2O2 (g)  rate = [H O ]; order = 1 2 2

•

The order of reaction may be fraction or zero or an integration but molecularity is always an integer, but never zero. The ratio of the order of reaction to that of molecularity of reaction will always be 1 or less than 1 Order of reactions can be determined experimentally but molecularity will be elucidated from the mechanism of the reaction The order of reaction is equal to the sum of the indices of the concentration terms in the rate equation while molecularity is the number atoms or ions or molecules in any elementary reaction of the mechanism of the reaction

2

→ 2H2O (l) + N2(g) (iv) NH4NO2 (aq)  rate = K [NH4NO2]; order = 1

•

H 2O (v) CH3COOC2H5  → CH3COOH + C2H5OH H+

•

rate = K [CH3COOC2H5]; order = 1 • Examples of second order reactions: → 3O2 Rate = K [O3]2; order = 2 (i) 2O3 (g)  → 2H2(g) + O2(g) (ii) 2N2O (g)  rate K [N2O]2; order = 2

Integrated Rate Equations S.NO Order

Rate law

Rate equation

dx = K0 dt

1

Zero order

2

First order

3

Second order

4

Second order

5

Third order

dx = K 3 ( a - x )3 dt

6

nth order

dx = K n (a - x) n dt

dx = K1 (a - x) dt dx = K 2 (a - x) 2 dt dx = K 2 (a - x)(b - x) dt

K0 =

x t

2.303 a log t a-x 1 x K2 = . t a(a - x) 2.303 b( a - x ) log K2 = t ( a - b) a (b - x) K1 =

t0.5 is independent of initial concentration 1 a 1 t0.5 ∝ a t0.5 ∝

1 1 1 - 2  2 2t  (a - x) a 

t0.5 ∝

1 a2

1  1 1  - n -1   n -1 t (n - 1)  (a - x) a 

t0.5 ∝

1 a n -1

KK 3= 3

Kn =

Half-Life equation

Chemical Kinetics

8.47

In these reactions, molecularity is two while the order is one. So reactions are called pseudo unimolecular reactions.

Examples of Fractional order → COCl2 (i) CO(g) + Cl2 (g)  1

rate = K[CO]2 [ Cl2 ] 2 order = 2.5

Methods for Determination of order of reaction

→ CO(g) + Cl2(g) (ii) COCl2(g) 

•

3

rate = K [ COCl2 ] 2 Order = 1.5

→ CH4(g) + CO(g) (iii) CH3CHO(g)  3

rate = K [ CH 3 CHO ] 2 order = 1.5

→ CCl4(g) + HCl(g) (iv) CHCl3(g) + Cl2(g)  rate = K[CHCl3] [ Cl2 ]

1

2

•

(i) Zero Order (R  → P°) x = Kt or

Order = 1.5

Zero order Reactions •

K=

Reactions in which the rate of reaction is independent of the concentration of the reacting substances is called zero order reaction. For zero order reaction x = Kt or K =

•

Decomposition of NH3 on metal surface such as gold and molybdenum is zero order reaction. Many photochemical reactions (e.g., formation of HCl from H2 and Cl2) are zero order reactions. Decomposition of N2O, HI, and PH3 in the presence of metal catalysts are zero order reactions.

•

K=

K = MolL–1 s–1 K = s–1 K = L mol–1S–1 K = L2mol–2 s–1 K = mol1–nLn–1 s–1

2.303 [ R ]0 a 2.303 log log = t a-x t [ R ]t

(iii) Second Order (2R  → P) K=

1 x 1 [ R ]0 - [ R ]t . . = [ R ]t at a - x [ R ]0

(iv) Second Order (R1 + R2  → P)

units of rate constants Zero order reaction First order reaction Second order reaction third order reaction nth order reaction

x a - (a - x) [ R ]0 - [ R ] = = t t t

(ii) First Order (R  → P)

x . t

•

•

In the trial and error method or integrated form of rate equation method, the concentration of reactants at zero time (a) and the concentrations at various intervals of time (a-x) are measured and substituted in the rate equations of different orders. The order corresponding to the equation which gives a constant K value is taken as the order of reaction. The rate equations for different orders are

or or or or or

atm–1 s–1 atm–1 s–1 atm–2 s–1 atm1–n s–1

Pseudo Chemical Reactions The reactions which appear to be of higher order but actually follow lower order kinetics are called pseudo chemical reactions. E.g., acid catalysed ester hydrolysis and hydrolysis of cane sugar to give glucose and fructose are pseudo first order reactions.

K=

•

2.303 b( a - x ) log t ( a - b) a (b - x)

In graphical method (i) If x vs t is a straight line parallel to time axis, it is zero order. (ii) If log

a vs t is straight line with positive a-x

slope passing through the origin, it is first order. (iii) If

a vs t is a straight line with positive a(a - x)

+

H → CH3COOH + C2H5OH CH3COOC2H5 + H2O  +

H C12H22O11 + H2O  → C6H12O6 + C6H12O6

slope passing through the origin, it is second order → P. for the reaction 2A 

8.48

Chemical Kinetics

(iv) If

second order reaction since the reaction proceed in the following steps:

b( a - x ) vs t is a straight line with positive a (b - x)

H2O2 + I–  → H2O + IO– (slow)

slope passing through the origin, it is second order

•

t0.5  a ′  =  t0.5  a 

• •

•

•

H + + IO–  → HIO (fast)

for the reaction A + B  →P In the half-life method, the order of reaction can be determined by using the equation

HIO + I– + H–  → H2O + I2 (fast) The first stage is slowest and hence determines the rate of the reaction.

n -1

Where t0.5 and t0.5 are the half lives with initial concentrations a and a´ and n is the order of reaction. The half-life of the first order reactions is independent of the initial concentration of the reactant The half-life of the second order reactions are inversely proportional to the initial concentration of the reactant. The product of the half-life and initial concentration of the reactant is constant The half-life of the third order reactions are inversely proportional to the square of the initial concentration of the reactant. The product of the half-life and the square of initial concentration of the reactant is constant. In the Vant Hoff differential method, the rate of a reaction at two different concentrations are determined from which the order can be determined.

Reversible Reactions •

2NO + O2

•

•

1

Side Reactions or Parallel Reactions •

•

•

The reactions in which the products formed in the first stage may react with each other or with the original reactant to give new products are known as consecutive reaction. The rate and order of consecutive reactions are determined by the reaction which is the slowest. E.g., reductions of H2O2 by HI and HBr is evidently

If two or more reactions take place simultaneously or side by side, these are called parallel or side or simultaneous reactions. The reaction in which largest amount of product formed is called main reaction and other reactions are called side reactions. E.g., in the nitration of phenol both o- and p – nitro phenols are formed simultaneously by parallel reactions. In parallel reaction, the rate constants of different side reaction will be determined from the rate of disappearance dx/dt of the reactants and the ratio of the constants K1/K2 can be calculated from the rate of formation of products.

Surface Reactions •

Consecutive Reactions •

K3     2NO2 K

The forward reaction is third order below 290°C but above 290°C the rate of disappearance of NO and O2 decreases due to their formation by the decomposition of NO2.

 dc   dc  log  - 1  - log  - 2   dt   dt  n= log C1 - log C2 Where n is the order of reaction, C1 and C2 are two initial concentrations From two different initial concentrations C1 and C2 the rates are determined as the slope of the curve from C vs t graph and from the data n is calculated. In the Ostwald’s isolation method, the order with respect to each reactant is determined by taking the concentration of other reactants in large excess and the order of reaction is determined by summing up the orders of all reactants in that reaction.

The reversible reactions are also called as counter or opposing reactions. The backward reaction causes a serious disturbance in the measurement of the reaction rate. E.g., in the reaction

Certain gaseous reactions such as thermal decomposition of phosphine, arsine and H2O2 vapours takes place exclusively on the walls of the vessel and are called surface reactions. The rates of these reactions are determined by the number of gaseous molecules colliding on the walls per unit time and hence the gas.

chain reactions •

The reactions proceeding in a series of successive reactions initiated by a suitable primary process are called chain reaction e.g., formation of HCl from H2 and Cl2, chlorination of alkanes.

Chemical Kinetics

theories of reaction rates • •

• • •

•

• • •

• •

Collision theory was proposed by a Swedish chemist Savante Arrhenius. A reaction between two substances is possible only when the atom or molecules of two substances collide with each other. When collision occurs between molecules, the old bonds are broken and new bonds are formed. All the collisions do not lead to the formation of the products. If all collisions of the reactant molecules lead to the formation of products, the rate of reaction at any given temperature is proportional to the number of collisions in unit time at that temperature. The minimum energy which must be associated with reactant molecules so that their mutual collision result in a chemical reaction is called threshold energy. The collisions which yield the product are called effective collisions or fruitful collisions or activated collisions. The molecules having energy less than threshold energy are called normal molecules. The difference in energy between the threshold energy and the energy of the normal colliding molecules is known as activations energy. Activations energy = Threshold energy—Energy of normal colliding molecules. Greater the activation energy lower is the rate of reaction The temperature dependence of a reaction rate can be represented by Arrhenius equation.

K = A.e - Ea / RT

•

Log K vs

•

• •

•

• •

• •

• •

•

The pre exponential factor A is called as frequency factor and Ea is the energy of activation. The units of Ea are J/mol or Kcal/mol. The rate constants at two different temperatures are related by log

•

•

K2 Ea T2 - T1 = K1 2.303R T1T2

1 gives a linear graph with negative slope. T

The reactant molecules collide with each other to cross over an energy barrier existing between the reactants and products. If the value of the difference in the internal energies of reactants and product is positive, the reaction is exothermic and if it is negative, the reaction is endothermic.

8.49

In some reactions, the catalyst decreases the activation energy so that more number of molecules can participate in the reaction. A catalyst carries the reaction through lower activation energy hence the rate of reaction increases. If the temperature is raised, the kinetic energy of the molecules increases which causes increases in (i) number of collisions (ii) number of molecules having higher energy than threshold energy. For every 10°C rise in temperature, the increase in kinetic energy is about 3.3%. So the increase in number of collisions is about 3.3 i.e., 1.8%. Hence the rate of reaction must increase only by about 1.8%. For every 10°C rise in temperature, the rate of reaction increases by 100% i.e., two times. If the rate of reaction is doubled for every rise of 10 K temperature, the rate of reaction increased for rise of temperature from 30°C to 80°C is 32 times. The activation energy does not depend on the concentration The ratio of the rate constants at two different temperatures (preferably 35°C and 25°C) is known as temperature coefficient. If the activation energy is zero then all the collisions will be fruitful and the reaction is 100% complete. The colliding molecules should have proper orientation so that the old bonds may break and new bonds are formed. The molecules which do not have proper orientation at the time of collision do not form product molecules even though they have more than threshold energy. Transition state theory was developed by Eyring: According to transition state theory, the reactant molecules before they convert into products form an activated complex which remains in thermodynamic equilibrium with the reactants. ±

K  Reactants ↽ → Products ⇀  Activated complex 

• •

The activated complex possesses the properties of an ordinary molecule but it exists only for small time. The energy of activated complex is higher than the reactants and products. The rate of decomposition of the activated complex determine the rate of reaction. The constant K is K=

RT ∗ K Nh

Where R = gas constant, T = absolute temperature, N = Avogadro number, h = Planck’s constant and K* = equilibrium constant

8.50

Chemical Kinetics

PRACTICE ExERCISE Multiple Choice Questions with only one answer level I 1. The rate of chemical reaction (Except Zero Order) (a) Decreases from moment to moment (b) remains control throughout (c) depends upon order of reaction (d) none 2. Radioactive decay follows ____order kinetics. (a) 0 (b) I (c) II (d) III 3. A certain reaction proceeds in a sequence of three elementary steps with the rate constants k1, k2 and k3. If the observed rate constant, kobs of the reaction is expressed as kobs = k3 [

k1 1/2 ] ; the observed energy of k2

activation of the reaction is (a) E3 +

1  E1    2  E2  1/2

4.

5.

6.

7.

8.

 E1 - E3  (b) E2 +    2 

 E1  E - E2 (c) E3   (d) E3 + 1 2  E2  The ratio of the rate constant of a reaction at any temperature T to the rate constant T →∝ is equal to (a) Energy of the activation of the reaction (b) Fraction of molecules in the activated state (c) Average life of the reaction (d) Pre-exponential factor in the Arrhenius equation For a reaction of II order kinetics, its t0.5 is (a) ∝a (b) ∝a1 2 (c) ∝a (d) ∝a–1 If ‘a’ is the initial conc. of a substance which reacts according to zero order kinetic and K is rate constant, the time for the reaction to go to completion is (a) a/K (b) 2 K/a (c) K/a (d) 2 K/a In a reaction, the rate is K[A][B]2/3, the order of reaction is (a) 1 (b) 2 (c) 5/3 (d) Zero Which one does not influence the rate of reaction? (a) Nature of reactant (b) Conc. of reactant (c) Temperature (d) Molecularity

9. For the reaction A + B → products, it is found that the order of A is 2 and of B is 3 in the rate expression when conc. of both is doubled. The rate will increase by (a) 10 (b) 6 (c) 32 (d) 16 10. For the reaction A →B, it is found that the rate of reaction increase by 8 times when the conc. of A is doubled. The reaction is of ……..order. (a) 1 (b) 2 (c) 3 (d) 16 11. The rate at which a substance react is proportional to its (a) equivalent weights (b) molecular weights (c) number of moles (d) number of moles per litre 12. The rate constant of a reaction has same units as the rate of reaction. The rate is of (a) zero order (b) first order (c) second order (d) none of these 13. Increases in the concentration of the reactants leads to the change in (a) heat of reaction (b) activation energy (c) collision frequency (d) none of these 14. For a molecular collision to be effective, it should satisfy the following condition: (a) It should involve the molecules having a certain minimum amount of energy (b) It should involve only ionic compounds (c) It should involve molecules having a certain minimum amount of energy and a proper orientation (d) It should involve two or three molecules 15. An endothermic reaction A → B has an activation energy as x KJ mole of A. If energy change of the reaction is y KJ, the activation energy of the reverse reactions is (a) –x (b) x-y (c) x + y (d) y-x 16. The reactions of high molecularity are rare because (a) Many body collisions have a low probability (b) Many body collisions are not favoured energetically (c) Activation energy of many body collisions is very high (d) Activation energy of many body collisions is very low

Chemical Kinetics

17. The activation energy of exothermic reaction A→ B is 20 Kcal. The heat of reaction is –50 K.cal. The activation energy for the reaction B → A will be (a) 20 Kcals (b) 30 Kcals (c) 70 Kcals (d) 50 Kcals 18. For a chemical reaction 2x + y → z the rate of appearance of Z is 0.05 mol L–1 per min. The rate of disappearance of x will be (a) 0.05 mol L–1 per hour (b) 0.005 mol L–1 per min (c) 0.1 mol L–1 min–1 (d) 0.025 mol L–1 per min 19. The activation energy of a reaction is zero. The rate constant of the reaction (a) increases with increase of temperature (b) decreases with increase of temperature (c) decreases with decrease of temperature (d) is nearly independent of the temperature 20. Order of reaction is (a) always equal to its molecularity (b) always a fraction (c) always an integral number (d) the sum of the exponents in rate law 21. The specific rate of a first order reaction depends on (a) The concentration of the reactants (b) the concentration of the products (c) Temperature (d) time 22. For the first order reaction, half-life is 14s. The time required for the initial concentration to reduce to 1/8th of its value is (a) 28 s (b) 42 s (c) (14)3 s (d) (14)2 s 23. The rate law for the reaction RCl + NaOH (aq) → ROH + NaCl is given by, Rate = K [RCl] the rate of the reaction will be (a) Doubled on doubling the concentration of sodium hydroxide (b) Halved on reducing the concentration of alkyl halide to one half (c) Decreased on increasing the temperature of the reaction (d) Unaffected by increasing the temperature of the reaction 24. For a given rate law 2A + B → C + D, the active mass of B is kept constant but that of A is tripled. The rate of reaction will (a) Decrease by 3 times (b) increase by 9 times (c) Increase by 3 times (d) depends on rate law 25. Taking the reaction A + 2B → products to be of second order which of the following is the rate law expression for the reaction?

(a)

dx dx = K [A] [B] (b) = K [A] [B]2 dt dt

(c)

dx dx = K [A]2 [B] (d) = K [A] [B]2 dt dt

8.51

26. Consider a gaseous reaction, the rate of which is given by K [X] [Y]. The volume of the reaction vessel containing these gases is suddenly reduced to 1/4th of the initial volume. The rate of the reaction as compared with original rate is (a) 1/16 times (b) 16 times (c) 1/8 times (d) 8 times 27. For a reaction, a plot of log (a-x) versus time (t) is a straight line with slope equal to K/2.303, the reaction is of (a) zero order (b) first order (c) second order (d) third order 28. The rate constant of a reaction is 5.2 × 10–1 minutes-1 the order of the reaction is (a) One (b) zero (c) two (d) three 29. The first order rate constant for decomposition of N2O5 is 6.93 × 10–4 sec-1. What is half change time for decomposition? (a) 102 sec (b) 103 sec 4 (c) 10 sec (d) 10 sec 30. The half-life period of a first order reaction is 15 minutes. The amount of substances left after one hour will be (a) One half (b) one fourth (c) one eight (d) one sixteenth 31. For a reaction A + 2B → C + D, the following data, were obtained Initial conc. (mol. L–1) [A]

Initial conc. (mol L–1) [B]

Initial rate of formation of D (mol L–1 min–1)

1

0.1

0.1

6.0 × 10–3

2

0.3

0.2

7.2 × 10–2

3 4

0.3 0.4

0.4 0.1

2.88 × 10–1 2.4 × 10–2

expt.

The correct rate law expression will be (a) Rate = K [A] [B] (b) Rate = K [A] [B]2 (c) Rate = K [A]2 [B]2 (d) Rate = P [A]2 [B] 32. The rate of the reaction A + B + C → products is given by r =

-d [ A] = K [A]1/2[B]1/2[C]1/4 the order of dt

reaction is (a) 1 (c) ½

(b) 2 (d) 13/12

8.52

Chemical Kinetics

33. In a reaction, the rate was found to be independent of conc of the reactants. The reaction is of (a) 1st order (b) 2nd order (c) order 1.5 (d) zero 34. In a multi-step reaction, the overall rate of reaction is (a) equal to the rate of slowest step (b) equal to the rate of fastest step (c) equal to the average rate of various steps (d) equal to the rate of last step 35. Which of the following expression is correct for second order reactions (C0 refers to initial concentration of reactant)? (a) t1/2 ∝ C0 (b) t1/2 ∝ C–1 –2 (c) t1/2 ∝ C0 (d) t1/2 ∝ C00 36. The rate of reaction 2NO + Cl2 ⇌ 2NOCl becomes doubled when the concentration of Cl2 is doubled however, when the concentration of both the reactants are doubled the rate becomes eight times. What is order of the reaction? (a) first (b) second (c) third (d) zero 37. The rate constant of a reaction is 1.2 × 10–2 mol–2 lit2 sec–1. The order reaction is (a) Zero (b) 1 (c) 2 (d) 3 38. For the reaction A + 2B + C → D + 2E the rate of formation of D is found to be (A) doubled when [A] is doubled keeping [B] and [C] constant. (B) doubled when [C] is doubled keeping [A] and [B] constant. (C) The same when [B] is doubled keeping [A] and [C] constant. Which one is the rate equation for the reaction? (a) Rate = K [A] [B] [C] (b) Rate = K [A]° [B][C] (c) Rate = K [A] [B]° [C] (d) Rate = K [A] [B][C]° 39. The concentration of a reactant in a solution falls (i) from 0.2 to 0.1 M in 2 h (ii) from 0.2 to 0.05 M in 4 h. The order of the hydrolysis of the reactant is (a) zero (b) two (c) one (d) half 40. Two gases A and B are filled in a container, the experimental rate law for the reaction between them, has been found to be rate = K [A] [B] predict the effect on the rate of the reaction when pressure is doubled (a) the rate is doubled (b) the rate becomes four times (c) the rate becomes eight times (d) none of the above

41. The rate of the first order reaction X → products is 7.5 × 10–4 mol lit–1 s–1. When the concentration of x is 0.5 mole L–1. The rate constant in seconds is (a) 3.75 × 10–4 S–1 (b) 2.5 × 10–4 S–1 (c) 1.5 × 10–3 S–1 (d) 3.0 × 10–4 S–1 42. For the reaction 2A + B → D, -d [ A] = K[A]2[B]. dt -d [ B ] The expression for will be dt (a) K[A]2 [B] (b) 2 K[A]2 [B] 1 (c) K [2A]2 [B] (d) K [A]2 [B] 2 43. A hypothetical reaction A2 + B2 → 2AB follows the mechanism as given below: A2 ⇌ A + A (fast); A + B2  → AB + B (slow)

A + B  → AB (fast)

44.

45.

46.

47.

The order of the overall reaction is (a) 2 (b) 1 1 (c) 1 (d) 0 2 If concentration of reactants is increased by X the rate constant k becomes (a) ek/X (b) k/X (c) k (d) X/E An endothermic reaction A→B has an activation energy of 10 Kcals/mole and heat of the reaction is 5 Kcals/mole. The activation energy of the reaction B → A is (a) 20 Kcal/mole (b) 5 Kcal/mole (c) 10 Kcal/mole (d) Zero The decomposition of a substance ‘R’ takes place according to first order kinetics. Its initial concentration is reduced to 1/8th in 24s. The rate constant of the reaction is (a) 1/24 S–1 (b) 0.69/16 S–1 (c) ln 2/8 S–1 (d) 1/8 S–1 The half life period of a first order reaction K =

2.303 a is log t a-x

(a) (b) (c) (d)

directly proportional to ‘a’ inversely proportional to ‘a’ independent of ‘a’ proportional to (a-x)

Chemical Kinetics

48. The incorrect order indicated against the rate of reaction A + B → C is Rate Order (a)

d [C ] = K [ A] dt

1

(b)

d [C ] = K [ A][ B] dt

2

(c)

d [ A] = K [ A][ B]0 dt

2

(d)

d [ A] = K [ A] dt

1

49. A first order reaction has specific rate constant of 2 min–1. The half-life of this reaction will be (a) 1.653 min (b) 0.347 min (c) 2 min (d) 0.5 min 50. For a first order reaction we have k = 100 sec–1. The time for completion of 50% reaction is (a) 1 m sec (b) 4 m sec (c) 7 m sec (d) 10 m sec 51. For the reaction A + 2B  → C, the rate of reaction at a given instant can be represented by (a) +

d[A] 1 d[B] d[C] =+ =+ dt 2 dt dt

(b)

d[A] 1 d[B] d[C] =+ =dt 2 dt dt

(c) –

d[A] 1 d[B] d[C] ==+ dt 2 dt dt

(d) +

55. If the half-time for a particular reaction is found to be constant and independent of the initial concentration of the reactions then reaction is (a) first order (b) zero order (c) second order (d) none of these 56. If initial concentration of reactants in certain reaction is doubled, the half life period of the reaction doubles, the order of a reaction is (a) zero (b) first (c) second (d) third 57. The rate of reaction A + B → products is given by the equation r = K [A][B]. If B is taken in large excess, the order of the reaction would be (a) 2 (b) 1 (c) 0 (d) unpredictable 58. In the reaction, A + 2B → C + 2D, the initial rate

-d [ A] at t = 0 was found to be 2.6 × 10–2 m sec–1. dt What is the value of

59.

60.

61.

1 d [ B] d [ A] d [C ] =+ =+ dt 2 dt dt

52. In the reaction A + B → AB, if the concentration of ‘A’ is doubled, the rate of reaction will (a) be doubled (b) be decreased to one half (c) increase four times (d) remain unaffected 53. The amount of 53D128 (t1/2 – 25 minutes) left after 50 minutes will be (a) 1/4 (b) ½ (c) 1/3 (d) none of these 54. The rate constant is given by the equation k = P. Ze–Ea/RT which factor should register a decrease for the reaction to proceed more rapidly? (a) T (b) Z (c) P (d) Ea–

8.53

62.

63.

-d [ B] at t = 0 in m sec–1? dt

(a) 2.6 × 10–2 (b) 5.2 × 10–2 –1 (c) 1.0 × 10 (d) 6.5 × 10–3 The conversion of molecules A to B follows second order kinetics. Doubling the concentration of A will increase the rate of formation of B by (a) A factor of 2 (b) a factor of 4 (c) Factor of ½ (d) a factor of ¼ Of the concentration of a reactant A is doubled and the rate of its reaction increases by a factor of 2, the order of reaction with respect to A is (a) First (b) zero (c) second (d) A factor of ¼ If initial concentration is tripled, the time for half reaction is also tripled, the order of reaction is (a) Zero (b) first (c) second (d) third The hydrolysis of methyl formate in acid solution has rate expression rate = K [HCOOCH3][H + ] the balanced equation being HCOOCH3 + H2O → HCOOH + CH3OH. The rate law contains [H+ ] though the balanced equation does not contain [H+ ] because (a) More for convenience to express the rate law (b) H+ ion is a catalyst (c) H+ is an important constituent of any reaction (d) All acids contain H + ions In a catalytic conversion of N2 to NH3 by Haber’s process, the rate of change in the concentration of NH3 per time is 40 × 10–3 mol 1–1. If there is no side reaction, the rate of the change as expressed in term of hydrogen is (a) 60 × 10–3 mol 1–1 s–1 (b) 1200 mol 1–1 s–1 (c) 20 × 10–3 mol 1–1 s–1 (d) 10.3 × 10–3 mol 1–1 s–1

8.54

Chemical Kinetics

64. In which of the following cases does the reaction go to farthest to completion? (a) k = 10 (b) k = 1 (c) k = 103 (d) k = 10–2 65. For an exothermic reaction, the energy of activation of the reactant is (a) equal to the energy of activation of products (b) less than the energy of activation of products (c) greater than the energy of activation of products (d) sometimes greater and sometimes less than that of the products → products, it was found that 66. In a reaction x + y  (i) on doubling the concentration of x, the rate doubled (ii) on doubling the concentration of y, the rate of the reaction increased four times, the overall order of the reactant is (a) 3 (b) 2 (c) 1 (d) 0 67. The hydrolysis of ester in alkaline medium is a (a) 1st order reaction with molecularity 1 (b) 2nd order reaction with molecularity 2 (c) 1st order reaction with molecularity 2 (d) 2nd order reaction with molecularity 1 68. For the chemical change A → B, It is found that the rate of reaction doubles when the concentration is increased four times. The order of A for this reaction is (a) Two (b) zero (c) one (d) half 69. For reaction 4A + B → 2C + 2D which of the following statements is not correct? (a) The rate of disappearance of B is one fourth of the rate of disappearance of A. (b) The rate of appearance of C is one half of the rate of disappearance of B. (c) The rate of formation of D is one half the rate of consumption of A. (d) The rate of formation C and D are equal. 70. The rate of certain reaction at different times are as follows: Time (min) Rate (moles litre–1 min–1)

0 10 20 30 2.8 × 10–2 2.8 × 10–2 2.81 × 10–2 2.79 × 10–2

The order of the reaction is (a) One (b) Two (c) Three (d) Zero 71. The reaction, 2A + 2B  → products, the following initial rates were obtained at various initial concentrations [A] 0.1 M 0.2 M 0.2 M

[B] 0.2 M 0.2 M 0.1 M

Rate (mol1–1 sec–1 0.46 1.84 0.92

The rate law for the reaction is (a) Rate = K[A]2[B]0 (b) Rate = K[A][B] (c) Rate = K[A]2[B] (d) Rate = K[A][B]2 72. A substance A decomposed in solution following the first order kinetics. Flask I contains 1 L of 1 M solution of A and flask II contains 100 mL of 0.6 M solution. After 8 h. the concentration of A in flask I becomes 0.25 M, what will be time for concentration of 'A' in flask II to become 0.3 M? (a) 0.4 h (b) 2.4 h (c) 4.0 h (d) unpredictable as rate constant is not given 73. The rate a chemical reaction doubles for every 10°C rise in temperature. If the temperature is increased by 60°C, the rate of reaction increase by about (a) 20 times (b) 32 times (c) 64 times (d) 128 times → 2NH3 74. The rate of the reaction N2(g) + 3H2(g)  was measured as +

1 d [ NH 3 ] = 2 ×10-4 mol L–1 sec–1. 2 dt

The rates of the reaction expressed in terms of N2 and H2 are (mol L–1 sec–1)

(mol L–1 sec–1)

(a) 1 × 10–4 1 × 10–4 –4 (b) 3 × 10 1 × 10–4 –4 (b) 1 × 10 1 × 10–4 –4 (d) 2 × 10 6 × 10–4 75. The rate of a gaseous reaction is equal to K[A][B]. the volume of the reaction vessel containing these gases is suddenly reduced to one-fourth the initial volume. The rate of the reaction would be (a) 1/6 (b) 16/1 (c) 1/8 (d) 8/1 76. Diazonium salt decomposes as C6H5N+2Cl  → C6H5Cl + N2. At 0°C, the evolution of N2 becomes two times faster when the initial concentration of the salt is doubled. Therefore it is (a) a first order reaction (b) a second order reaction (c) independent of the initial concentration of the salt (d) a zero order reaction 77. The conversion of A  → B follows second order kinetics. Doubling the concentration of A will increase the rate of reaction by a factor of (a) 2 (b) ½ (c) 4 (d) ¼

Chemical Kinetics

8.55

K1

(a)

2 K1 [ NO2 ]2 K2

(b) 2 K1 [NO2]2 – 2K2 [N2 O4] (c) 2K1 [NO2]2 (d) (2K1-K2) [NO2] 79. The given reaction 2FeCl3 + SnCl2  → 2FeCl2 + SnCl4 is an example of (a) 1 Order (b) 2nd Order (c) 3rd Order (d) Zero Order → C, it is found that dou80. For the reaction A + B  bling the concentration of A increases the rate 4 times and doubling the concentration of B doubles the reaction. What is the order of the reaction? (a) 3/2 (b) 4 (c) 1 (d) 3 81. A first order reaction is half-completed in 45 minutes. How long does it need for 99.9% of the reaction to be completed? (a) 20 hours (b) 10 hours 1 (c) 7 hours 2 (d) 5 hours

→ B, the rate of reaction increase 82. In a reaction: A  two times on increasing the concentration of the reactants four times, then the order of the reaction is (a) 0 (b) 2 (c) ½ (d) 4 83. Milk turns sour at 40°C three times as faster at as 0°C. hence Ea (activation energy) of turning of milk sour is (a)

2.303 × 2 × 313 × 273 log 3 cal 40

(b)

2.303 × 2 × 313 × 273 log (1/3) cal 40

(c)

2.303 × 2 × 40 log 3 cal 273 × 313

(d)

2.303 × 2 × 40 log (1/3) cal 273 × 313

84. A graph between log (T50), and log (conc.) for nth order reaction is a straight line. Reaction of this nature is completed 50% in 10 minutes when conc. is 2 mol L–1. This is decomposed 50% in t minutes at 4 mol L–1 n and t are

log(conc) (a) 0.20 min (c) 1, 20 min

(b) 1, 10 min (d) 0, 5 min

85. Graph between log k and

1 (k is rate constant (s–1) T

and T the temperature (K) is a straight line with OX = 5,  1  θ = Tan–1   . Hence Ea will be  2.303 

log k

of disappearance of NO2 is equal to

log t0.5

 ⇀ 78. In the reversible reaction, 2NO2 ↽   N2O4 the rate K2

θ

1/T

(a) 2.303 × 2 cal

(b)

2 cal 2.303

(c) 2 cal (d) none of these 86. Half life (T1) of the first order reaction and half life (T2) of the second order reaction are equal. Hence ratio of the rate at the start of the reaction 1     T2 = rate constant × initial constant   (a) 1 (b) 2 (c) 0.693 (d) 1.44 87. Rate constant k of a reaction is 0.0693 min–1. Starting with 20 mol L–1, rate of the reaction after 10 min will be (a) 0.693 mol L–1 min–1 (b) 1.386 mol L–1 min–1 (c) 0.0693 mol L–1 min–1 (d) 6.93 mol L–1 min–1

8.56

Chemical Kinetics

88. The rate of a chemical reaction generally increases rapidly even for small temperature increase because of a rapid increase in the (a) Collision frequency (b) Fraction of molecules with energies in excess of the activation energy (c) Activation energy (d) Average kinetic energy of molecules 89. Rate constant of a reaction k is 3.0 × 10–4 s–1, energy of activation Ea = 104.4 KJ mol–1 and Arrhenius constant A is 6.0 × 1014 s–1 at 298 K. The value of rate → ∞ is constant k as a T  (a) 2.0×1018 s–1 (b) 6.0×1014 s–1 (c) infinite (d) 3.6×1030 s–1 90. A tangent drawn on the curve obtained by plotting concentration of product (mol L–1) of a first order reaction Vs time (min) at the point corresponding to time 20 minute makes an angle to 30° with concentration axis. Hence rate of formation of products after 20 minutes will be (a) 0.580 mol L–1 min–1 (b) 1.723 mol L–1 min–1 (c) 0.290 mol L–1 min–1 (d) 0.866 mol L–1 min–1 91. For reaction 3A  → products, it is found that the rate of reaction increases 4-fold when concentration is increased 16 times keeping the temperature constant. The order of reaction is (a) 2 (b) 1 (c) 3 (d) 0.5 92. The reaction; 2O3  → 3O2, is assigned the following mechanism:  O2 + O I. O3  II. O3 + O  → 2O2 The rate law of the reaction will therefore be (a) r ∝ [O3]2 [O2] (b) r ∝ [O3]2 [O2]–1 (c) r ∝ [O3] (d) r ∝ [O3] [O2]–2 → C(g) + D(g) is 93. The reaction A (g) + 2B(g)  an elementary process. In an experiment, the pressure of A and B are PA = 0.60 and PB = 0.80 atm. When PC = 0.2 atm the rate of reaction relative to initial rate is (a) 1/48 (b) 1/24 (c) 9/16 (d) 1/6 94. A radioactive element has a half life period of 140 days. How much of it will remain after 1120 days? slow

(a)

1 31

(b)

1 256

(c)

1 512

(d)

1 128

95. The rate law for the reaction RCl + NaOH (aq)  → ROH + NaCl is given by rate = k [RCl]. The rate of the reaction will be (a) Unaffected by increasing temperature of the reaction. (b) Doubled on doubling the concentration of NaOH. (c) Halved on reducing the concentration of NaOH to one half. (d) Halved on reducing the concentration of RCl to one half. 96. For a given reaction of first order, it takes 20 minutes for the concentration to drop from 1.0 to 0.6 mol litre–1 will be. The time required for the concentration to drop from 0.6 mol litre–1 to 0.36 mol litre (a) more than 20 minutes (b) less than 20 minutes (c) Equal to 20 minutes (d) infinity 97. A catalyst lowers the activation energy of the forward reaction by 20 KJ mol–1. It also change energy of the backward reaction by an amount (a) Equal to that of forward reaction (b) Equal to twice that of the forward reaction (c) Which is determined only by the average energy of products (d) Which is determined by the average energy of products relative to that of reactants 98. The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to (a) (1/2) g (b) (1/8) g (c) (1/8) g (d) (1/16) g 99. For a second order reaction of the type rate = k[A]2, the plot of 1/[A]t versus t is linear with a (a) Positive slope and zero intercept (b) Positive slope and non zero intercept (c) Negative slope and zero intercept (d) Negative slope and zero intercept 100. The decomposition of Cl2O7 at 400 K in the gas phase to Cl2 and O2 is a first order reaction. (i) After 55 seconds at 400 K the pressure of Cl2O7 falls from 0.062 to 0.044 atm. The rate constant and pressure of Cl2O7 after 100 sec of decomposition at this temperature are (a) 5.2 × 10–4 sec–1; 0.05 atm (b) 6.2 × 10–3 sec–1; 0.033 atm (c) 5.8 × 10–3 sec–1; 0.44 atm (d) 4.6 × 10–3 sec–1; 0.005 atm

Chemical Kinetics

Multiple Choice Questions with only one answer

8.57

coordinate for the above reaction is (a)

K1 K2 X  → A + B and X  →C + D If 50% of the reaction of X was completed when 96% of the reaction of Y was completed, the ratio of their rate constants (K2/K1) is (a) 4.06 (b) 0.215 (c) 1.1 (d) 4.65 4. Consider this gas phase reaction

AB+I A+B IAB

reaction coordinate

P.E.

(b)

A+P IAB

reaction coordinate

(c)

AB+I

K1 K2   A + B  → AB + I  →P+ A  IAB  Fast

If K1 is much smaller than K2. The most suitable qualitative plot potential energy (PE) versus reaction

A+P

A+B

The reaction is found experimentally to follow this rate law

IAB

reaction coordinate

(d)

P.E.

Based on this information, what conclusions can be drawn about this proposed mechanism? Step 1 Cl2 (g) ⇌2Cl (g) Step 2 Cl (g) + CHCl3 (g) → HCl (g) + CCl3 (g) Step 3 Cl (g) + CCl3 (g) → CCl4 (g) (a) Step 1 is the rate-determining step (b) Step 2 is rate-determining step (c) Step 3 is the rate-determining (d) The rate-determining step cannot be identified 5. A first order reaction is 50% completed in 20 minutes at 27°C and in 5 min at 47°C the energy of activation of the reaction is (a) 43.85 KJ/mol (b) 55.14 KJ/mol (c) 11.97 KJ/mol (d) 6.65 KJ/mol 6. The following mechanism has been proposed for the exothermic catalysed complex reaction:

AB+I AB+1

A+B

Cl2 (g) + CHCl3 (g) → HCl(g) + CCl4(g)

Rate = k [CHCl3] [Cl2]1/2

A+P

P.E.

1. The activation energies of two reactions are 18 KJ mol–1 and 4.0 KJ mol–1 respectively at 300 K. Assuming the pre-exponential factor to be the same for both reactions, the ratio of their rate constants is (a) 3.66 × 10–8 (b) 3.66 × 10–3 –2 (c) 6.312 × 10 (d) 6.31 × 10–10 2. Reaction A + B → C + D follows’ following law: rate = k[A]+1/2 [B]+1/2. Starting with initial conc. of one of A and B each, what is the time taken for amount of A to become 0.25 moles? Given k = 2.31 × 10–3 sec–1 (a) 300 sec (b) 600 sec (c) 900 sec (d) none of these 3. Consider the following first order competing reactions:

P.E.

level II

AB+I

A+P

A+B IAB

reaction coordinate

7. For the 1 order reaction 2A(g)→3B(g), t1/2 = 12 min, initial pressure exerted by A is 640 mm of Hg. The pressure of the reaction mixture after the time period of 36 min will be [given anti-log (0.9) = 7.94] (a) 560 mm Hg (b) 680 mm Hg (c) 920 mm Hg (d) 600 mm Hg 8. In a certain reaction, 10% of the reactant decomposes in one hour, 20% in two hours, 30% in three hours and so on. The dimensions of the rate constant is (a) Hour–1 (b) mol litre–1 sec–1 –1 –1 (c) litre mol sec (d) mol sec–1 st

8.58

Chemical Kinetics

9. Rate of reaction can be expressed by Arrhenius equation is k = Ae–E/RT. In this equation, E represents (a) The activation energy below which colliding molecules will not react. (b) The total energy of the reacting molecules at a temperature T. (c) The fraction of molecules with energy greater than the activation energy of the reaction. (d) The energy above which all the colliding molecules will react. 10. The rate constant of a first order reaction at 27°C is 10–3 min–1. The temperature coefficient of this reaction is 2. What is the rate constant (in min–1) at 17°C for this reaction? (a) 10–3 (b) 5 × 10–4 –3 (c) 2 × 10 (d) 10–2 11. In a hypothetical reaction A(aq)⇌2B(aq) + C(aq) (1st order decomposition)

A is optically active (dextrorotatory) while B and C are optically inactive but B takes part in titration reaction (fast reaction) with H2O. Hence the progress of reaction can be monitored by measuring rotation of plane polarized light or by measuring volume of H2O consumed in titration with a fixed volume of reaction mixture. In an experiment the optical rotation was found to be θ = 30° at t = 20 min and θ = 15°C at t = 50 min (time is measured from start of the reaction) if the progress would have been monitored by titration method, volume of H2O2 consumed at t = 30 min (from start) is 30 mL then volume of H2O2 consumed at t = 90 min be will be (a) 60 mL (b) 45 mL (c) 52.5 mL (d) 90 mL 12. For reaction A→B, The rate constant k1 = A1e–Ea /RT and for the reaction P → Q, the rate constant K2 = A2 e–Ea /RT. If A1 = 10 8; A2 = 10 10 and Ea1 = 600 cal/mol, Ea2 = 1200 cal/mol, then the temperature at which K1 = K2 is (R = 2 cal/K-mol) (a) 600 K (b) 300 × 4.606 K 1

2

(c)

300 K 4.606

(d)

4.606 K 600

13. A gaseous compound ‘A’ reacts by three independent first order process (as shown in Figure) with rate constant 2 × 10–3, 3 × 10–3 and 1.93 × 10–3 sec–1 respectively for products B,C and D respectively. If initially, pure ‘A’ was taken in a closed container with P = 2

atm, find the partial pressure of ‘B’ (in atm) after 100 sec from start of experiment.

B (g) K1 A(g)

K2

C (g)

K3 D (g) (a) 0.288 (b) 0.287 (c) 0.432 (d) 0.99 14. For a zero order reaction, linear plot was obtained for [A] vs, t the slope of the line is equal to (a) K0 (b) -K0 0.639 K0 (d) (c) K 0 2.303 B

K1

15. Consider the reaction A

K2

. The rate C

constant for two parallel reactions were found to be 10 –2 dm3/mol/sec (K1) and 4 × 10 –2 dm3/mol/sec (K2). If the corresponding energies of activation of the parallel reactions are 80 and 100 KJ respectively. What is the apparent (net) energy of activation of given reaction? (a) 96 KJ/mol (b) 120 KJ/mol (c) 100 KJ/mol (d) 140 KJ/mol 16. The reaction [Co(NH3)5 Br]2+ + H2O → [Co(NH3)5 (H2O)]3+ + Br– is followed by measuring a property of the solution known as the optical density of the solution which may be taken to be linearly related to be concentration of the reactant. The values of optical density are 0.80, 0.35 and 0.20 at the end of 20 minutes, 40 minutes and infinite time after of the reaction which is first order. Calculate the rate constant (units: min–1). (a) 0.693 (b) 0.0693 (c) 1.386 (d) 0.1386 17. A and B are two different chemical species undergoing first order decomposition with rate constants KA and KB which are in the ratio 3:2 in magnitude. If the initial concentration of A and B are in the ratio (C0) A: (Co) B :: 3:2, what would be the ratio of concentration CA, CB after three half lives of A? (a) 1:1 (b) 3:4 (c) 1:2 (d) 2:1

Chemical Kinetics

18. The reaction 2SO2 (g) + O2 (g) → 2SO3 (g) is carried out in a 1 dm3 vessel and 2 dm3 vessel separately. The ratio of the reaction velocities will be (a) 1:8 (b) 3:4 (c) 4:1 (d) 8:1 19. A graph is drawn between log t1/2 and log a is as shown below. Slope of this graph is -1 and intercept is 6.5228. The ratio of the reaction velocities will be (a) 0.5228 × 10–6 mole/lit/sec (b) 3 × 10–6 lit/mol/sec (c) 5.228 × 10–6 lit/mol/sec (d) 5.228 × 10–6 mole/lit/sec

8.59

40 mL then volume of H2O2 consumed at t = 60 min will be: (a) 60 mL (b) 75 mL (c) 52.5 mL (d) 90 mL 22. The reaction I – + OCl–→Cl– + OI– follows the rate law d[OI –]/dt = K'[I –][OCl–], but K' proves to be a function of hydroxide ion concentration. For hydroxide ion concentrations of 1M, 0.5 M, and 0.25 M, K' is equal to 61, 120 and 230 L mol –1 S–1 respectively. And the mechanism of this reaction is OCl– + H2O ⇌ HOCl + OH– rapid HOCl + I–→HOI + Cl– slow HOl + OH–→H2O + Ol– rapid

log t 1/2

The order of the reaction with respect to hydroxide ion concentration is (a) 1 (b) 0 (c) -1 (d) 2 n+

Mn 23. Tl+ + 2Ce 4+  → Tl3+ + 2Ce3+ The experimentally determined rate law is:

Rate = K [Ce4 + ][Mn2+ ]

log a

With the following proposed mechanism involving ions of manganese: Step 1 : Ce 4 + + Mn 2 + → Ce3+ + Mn3+

–

20. A certain reactant XO3 is getting converted to X2O7 in solution. The rate constant of this reaction is measured by titrating a volume of the solution with a reducing – agent which reacts only with XO3 and X2O7. In this process of reduction, both the compounds are converted to X–. At t = 0, The volume of the reagent consumed is 30 mL and at t = 9.212 min, the volume used up is – 36 mL, the rate constant of the conversion of XO3 to st X2O7 assuming reaction is of 1 order (Given that ln 10 = 2.303, log 2 = 0.30) (a) 0.2 min–1 (b) 0.02 min–1 –1 (c) 0.01 min (d) 0.1 min–1 21. In a hypothetical reaction A (aq) ⇌ 2B(aq) + C(aq) (1st order decomposition) ‘A’ is optically active (dextrorotatory) while ‘B’ and ‘C’ are optically inactive but ‘B’ takes part in a titration reaction (fast reaction) with H2O2. Hence the progress can be monitored by measuring rotation of plane of polarized light or by measuring volume of H2O2 consumed in titration. In an experiment, the optical rotation was found to be θ = 40° at t = 20 min and θ = 10° at t = 50 min. from start of the reaction. If the progress would have been monitored by titration method, volume of H2O2 consumed at t = 5 min (from start) is

Step 2 : Ce 4 + + Mn3+ → Ce3+ + Mn 4 + Step 3 :Tl + + Mn 4 + → Tl 3+ + Mn 2 + Which ion of manganese is the catalyst according to the information provided above? (a) Mn4 + (b) Mn3+ 2+ (c) Mn (d) All of the above K1 K2  ⇀ →C 24. For A ↽   2 B, 3B  K -1

K1 = 1 × 10–4mol/l-sec, K-1 = 2 × 10–4 l/mol = sec K2 = 3 × 10–4 sec, K-1,d[B]/dt equals to (a) 2K1-2K-1[B]2-3K2[B] (b) 2K1 [A]-2K-1[B]2-3K2[B] (c) 2K1 [A]-2K-1[B]2-3K2[B]3 (d) K1–K-1 [B] 2-K2 [B] K 25. A reaction 2A + B  → C + D is first order with respect to A and second order with respect to B. Initial concentration (t = 0) of A is C0 while B is 2C0 if at t = 30 minutes the concentration of C is C0/4 then rate expression at t = 30 minutes is (a) R = 7C03 K / 16

(b)

R = 27C03 K / 32

(c) R = 247C03 K / 64

(d)

R = 49C03 K / 32

8.60

Chemical Kinetics

26. The activation energy for the decomposition of H2O2 is 76 KJ/mole at room temperature and the decomposition is slow. When a catalyst is added, rate of decomposition of H2O2 increases and the activation energy decrease to 57 KJ/mole. The rate constant increase after adding catalyst approximately by a factor of (a) 500 (b) 1000 (c) 1500 (d) 2000 27. Nitramide O2NNH2 decomposes slowly in aqueous solution according to the equation O2NNH2→N2O + H2O The experimental rate law is dN 2 O = K[O2 N.NH 2 ] / [H + ] dt Which of the following mechanism seems most appropriate? K1 (a) O2 N.NH 2  → N 2 O + H 2 O(slow) K

2 +  (b) O2 N.NH 2 + H + ↽ ⇀  O2 N.NH 3 (fast)

K3 O2 NH 3+  → N 2 O + H 3 O+ (slow) K4 +  (c) O3 N.NH 2 ↽ ⇀  O2 N.NH + H (fast) K5 O3 N.NH -  → N 2 O + OH - (slow)

H + OH → H 2 O(Fast) +

K6

-

K7 → N 2 O + H + + OH - (slow) (d) O2 N.NH 2 

28. An organic compound A decomposes following two parallel first order reactions: B

K1 A

K2

K1 1 = ; K1 = 1.09× 10 -1 hr -1 K2 9

C

If an experiment is start with A for one hour, the ratio of A to B is (a) 0.1 (b) 0.15 (c) 0.2 (d) 0.25 29. 2P (g)→ 4Q(g) + R(g) + S(l)is a first order reaction. After 23 minutes from start of decomposition in a closed vessel the total pressure developed is found to be 317 mm of Hg. After a long period the total pressure observed to be 617 mm Hg. Vapour pressure of S(l) at this temperature is 17 mm Hg. log (1.2) = 0.079. The value of the constant is (a) 3.95 × 10–3 (b) 3.95 × 10–2 (c) 7.9 × 10–2 (d) 7.9 × 10–3

30. Decomposition of dimethyl ether is a first order reaction CH3OCH3 (g) → CH4(g) + H2(g) + CO(g) Half life period is 14.5 minutes. Initially the pressure of ether is 0.4 atm. The total pressure of system after 29 Minutes is (a) 0.1 atm (b) 0.3 atm (c) 0.7 atm (d) 1 atm 31. In a reaction carried out at 500 K 0.001% of the total number of collision are effective. The energy of activation of the reaction is approximately (a) 15.8 Kcal mol–1 (b) 11.5 Kcal mol–1 (c) 12.8 Kcal mol–1 (d) Zero 32. The catalyst decrease the Ea from 100 KJ mol–1 to 80 KJ mole–1. At what temperature the rate of reaction in the absence of catalyst at 500 K will be equal to rate reaction in presence of catalyst (a) 400 K (b) 200 K (c) 625 k (d) None of these 33. The thermal decomposition of acetaldehyde: CH3CHO→CH4 + CO, has rate constant of 1.8 × 10–3 mol–1/2 min–1 at a given temperature. How would

d[CH 3 CHO] dt

change if concentration of

acetaldehyde is doubled keeping the temperature constant? (a) Will increase by 2.828 times (b) Will increase by 11.313 times (c) Will not charge (d) Will increase by 4 times 34. A reaction 2A + B → C + 2D is first order with respect to A and second order with respect to B. Initial concentration of A is ‘x’ while that of B is ‘2x’. Initial concentration of C is zero. If at t = 30 minutes, the concentration of C is x/4, the rate expression at t = 30 minutes is: (a) R = 27x3 k/64 (b) R = 7x3 k/32 (c) R = 247x3 k/64 (d) R = 49x3 k/32 35. 50 bacteria are placed in a flask containing nutrients for the bacteria so that they multiply. The production of bacteria is first order. After 30 minutes, the number of bacteria is 100. The rate constant of this process is: (a) 1.15 × 10–1 min–1 (b) 1.15 × 10–2 min–1 (c) 2.3 × 10–1 min–1 (d) 2.3 × 10–2 min–1

Chemical Kinetics

(a) I → II (c) III → II

36. Consider the reaction:

k1

8.61

(b) II → III (d) III → IV

(B)

(A) (C)

The rate constants for two parallel reactions were found to be 10–2 s–1 and 4 × 10–2 s–1. If the corresponding energies of activation of the parallel reactions are 100 and 120 KJ mol–1 respectively, what is the net energy of activation (Ea) of A? (a) 100 KJ mol–1 (b) 120 KJ mol–1 (c) 116 KJ mol–1 (d) 220 KJ mol–1 37. The reaction of hydrogen and iodine monochloride is given as:

Energy

k2

III

I

II IV

H2(g) + 2ICl(g) → 2HCl(g) + I2(g) This is of first order with respect to H2(g) and ICl(g), following mechanisms were proposed. Mechanism A: H2(g) + ICl(g) → HCl(g) + HI(g); slow Mechanism B: H2(g) + ICl(g) → HCl(g) + HI(g); slow HI(g) + ICl(g) → HCl(g) + I2(g); fast

38.

39.

40.

41.

Which of the above mechanism (s) can be consistent with the given information about the reaction? (a) A and B both (b) neither A nor B (c) A only (d) B only For the decomposition of H2O2 that follows 1st order kinetics, the half-life period is 20 min. If the reaction is started with 32 volumes of H2O2 the time taken for the H2O2 to come to one volume is (a) 80 min (b) 100 min (c) 120 min (d) 40 min If 60% of a first order reaction was completed in 60 minute, 50% of the same reaction would be completed in approximately (a) 45 minutes (b) 60 minutes (c) 40 minutes (d) 50 minutes (log 4 = 0.60, 5 = 0.69) The energies of activation for forward and reverse  2AB are 180 KJ mol –1 and 200 KJ for A2 + B2  –1 mol . The enthalpy change of the reaction (A2 + B2 → 2AB) in the presence of a catalyst will be (in KJ mol –1) (a) 20 (b) 300 (c) 120 (d) 280 According to the reaction profile given, which reaction step is rate determining in the forward direction?

Reaction progress 42. A gaseous compound ‘A’ decomposes to give B (g), C(g) and D (g) with rate constants K1 = 2 × 10–3 sec–3 K2 = 3×10–3sec–1 and K3 = 1.93 × 10–3 sec–1 respectively. If initially pure ‘A’ was taken in a closed container with P-2 atm, the partial pressure of ‘B’ (atm) at 100 second after the start of the reaction is (a) 0.144 (b) 0.576 (c) 0.288 (d) None of these 43. A “diffusion-controlled” reaction is a reaction in which all collisions between the reacting species lead to products (these reactions are called “diffusion – controlled” because the rate is controlled only by how fast the reaction molecules can “diffuse” together). In aqueous solution at 25°C, the reaction of a strong acid, H3O +, with a strong base, OH– , is an example of this type of reaction. The rate constant K, for this reaction is 1.4 × 1011 M–1 s–1 H3O + (aq) + OH– (aq)–> 2H2O (I) Which of the following statements is more likely TRUE? (a) Increasing the temperature would have no effect on the rate of this reaction. (b) The activation energy for this reaction must be very large. (c) The rate of this reaction would be independent of the concentration of H3O + and OH– (d) The rate constant for this would be different if the reaction were carried out in a more viscous solvent than water. 44. At 350 K, a particular second-order reaction, consisting of a single reactant, A has a rate constant equal to 4.5 10–3 M–1 s–1. If the initial concentration of A is

8.62

Chemical Kinetics

0.80 M, how many half-lives are required for the concentration of A to become equal to 0.10 M? (a) 7 Half-life (b) 2 Half-life (c) 3 Half-life (d) 4 Half-life 45. Under the same reaction condition, initial concentration of 1.386 mol dm–3 of a substance becomes half in 40 seconds and 20 seconds through first order and  K1  zero order kinetics respectively. Ratio  K  of the  0 constants for order (K1) and zero order (K0) of the reaction is (a) 0.5 mol–1 dm3 (b) 1.0 mol dm–3 –3 (c) 1.5 mol d (d) 2.0 mol–1 dm3 46. The bromination of acetone that occurs in the solution is represented by this equation CH3COCH3(aq) + Br2(aq) → CH3COCH2Br(aq) + H+(aq) + Br–(aq) These kinetic data were obtained for the given reaction concentrations Initial concentration, M

Initial rate, disappearances of Br2Ms–1

[H+ ] 0.05 0.05 0.10 0.20

[CH3COCH3] [Br2] 0.30 0.05 0.30 0.10 0.30 0.10 0.40 0.05

5.7×10–5 5.7×10–5 1.2×10–4 3.1×10–4

Based on these data, the rate equation is (a) Rate = k[CH3COCH3][Br2][H + ]2 (b) Rate = k[CH3COCH3][Br2][H + ] (c) Rate = k[CH3COCH3][H + ] (d) Rate = k[CH3COCH3][Br2] → N2O(g) 47. Consider the reaction NH2NO2(g)  H2O(g). Concentration of nitramide as a function is shown below. Which line represents the correct tangent to the graph at the time (t = 0)? 0.4

∆ 48. Ethylene is produced by C4H8  → 2C2H4. The rate –3 –1 constant is 2.303 × 10 sec . What time will be taken so that the molar ratio of ethylene to cyclobutane in reaction mixture is one (a) 1.76 × 102 sec (b) 1.76 × 103 sec (c) 3.01 × 102 sec (d) 3.01 × 103 sec 49. When a catalyst is introduced in a first order reaction, rate increases by 4000 times at 27°C. The change in activation energy in the presence of catalyst is (a) 4.94 Kcal (b) 2.49 Kcal (c) 9.82 Kcal (d) 7.41 Kcal 50. A solid ‘P’ kept in a vessel containing argon at 1 atm at 27°C was heated. This resulted in entire sublimation of ‘P’ and the total pressure was found to be 2 atm at 127°C. On further heating to 327°C, gaseous ‘P’ further dissociated as per the reaction

2P(g)  → Q(g) + R(g) Final pressure in the vessel will be (a) 2 atm (b) 3 atm (c) 3.33 atm (d) 4 atm 51. For an exothermic chemical process occurring in two steps as follows: (i) A + B  → X (slow) (ii) X  → A + B (fast) The progresses of reaction can be best described by (a)

P.E. X AB A+B Reaction coordinate

(b)

conc reactant 0.2

P.E.

0.1 0.05

(a) I (c) III

I

II III

30

60

IV

120

time (min)

A+B 173

(b) II (d) IV

240

360

Reaction coordinate

AB

Chemical Kinetics

(c)

8.63

A0

Conc.

P.E. x A+B AB

time

Reaction coordinate

(d)

(a)

X AB

 ln 3   t (t≥0°C)  10 

log T50

450

log a (a) 0 (b) 1 (c) 2 (d) 3 dc ac 53. The rate expression for a reaction is – = dt 1 + bc where a, b > 0. The half of the reaction is (a)

bc 1 ln 2 + 0 a 2a

(b) a ln 2 + (c)

2bc0 a

bc 1 ln 2 + 0 a 2a

2bc0 (d) a ln 2 + 2a 54. At the point of intersection of two curves shown for → nB, the concentration of B is givthe reaction A  en by the 1st order reaction

ln kt = ln K0 + 

The value of temperature coefficient is (a) 4 (b) 3 (c) 10 (d) 2 56. Rate law of a given equation of nth order A  →  dx  product is given by   = K[A]n A straight line is  dt   dx  inclined at 45° when a graph between log   and  dt  log [A] is plotted. Hence ‘n’ is (a) 0 (b) 1 (c) 2 (d) 0.5 57. The variation of concentration of A within two experiments starting with two different initial concentrations of A is in the following graph. The reaction is repre→ B (aq). What is the rate of resented as A (aq)  action (M/min) when concentration of A in aqueous solution was 1.8 M?

Concentration(M)

52. Following is the graph between log T50 and log a. a = initial concentration for given reaction at 27°C. Hence order is (T50 = half life)

A0 n -1

nA0 n +1

A+B Reaction coordinate

(b)

 n -1  (d)   A0  n +1 55. For a certain reaction, the variation of the rate constant with temperature is given by the equation

(c)

P.E.

nA0 2

1.7 1.6 1.5 1.4 1.3 1.2 1.1 1 0.9 0.8 0.7 0.6

Experiment-1 Experiment-2

5

10

15

20

time (min.)

(a) 0.080 M min–1 (b) 0.036 M min–1 (c) 0.1296 M min–1 (d) 1 M min–1

8.64

Chemical Kinetics

58. Which of the following will have the slowest rate of reaction with HCl? (a) Marble chips at 40°C (b) Powdered marble at 25°C (c) Marble chips at 25°C (d) Powdered marble at 40°C 59. A gaseous reactant. ‘A’ decomposes to produce gases ‘B’ and ‘C’ in a parallel reactions. Both by first order as follows: K1=2×10-3 min-1

B(g)

A(g)

K2=3×10-2 min-1

60.

61.

62.

63.

C(g)

If the decomposition is carried out in a sealed flask, partial pressure of ‘B’ after very long time was found to be 100 mm Hg determine the time when partial pressure of A is 100 mm Hg. (a) 50 min (b) 87.5 min (c) 12.5 min (d) 75 min The rate of a reaction in pressure of catalyst at temperature 27°C may be made equal to the rate of reaction in absence of catalyst but for this we have to raise the temperature up to 47°C and activation energy in the absence of catalyst is 9.6 KJ/mole. The activation energy in the presence of catalyst is (a) 10.2 KJ/mol (b) 0.6 KJ/mol (c) 9 KJ/mol (d) 8 KJ/mol In a reaction carried out at 27°C, 0.001% of the reactant molecules are in activated state. The rate constant of this reaction at 27°C is 3 × 10–3 sec. Arhenius constant at 37°C is (a) 30 sec (b) 300 sec (c) 60 sec (d) 600 sec Decomposition of 22.4 vol of H2O2 is first order reaction. After 6 hours, 10 mL of this solution is diluted to 100 mL. 10 mL of this diluted solution is titrated with 20 mL of 0.02 mL 0.02 M KMnO4 in acid medium. Thus rate constants of H2O2 is (a) 0.23 hr–1 (b) 0.172 hr–1 –1 (c) 0.115 hr (d) 0.057 hr–1 The inversion of sugar in the presence of acid follows 1st order. The parameter readings are ∞ Time in minutes 0 23 Rotation in degrees + 45 + 15 -15 Rate constant of reaction in min–1 (a) 3 × 10–1 (b) 3 × 10–2 –1 (c) 4.7 × 10 (d) 4.7 × 10–2

64. Hydrolysis of methyl acetate with dil HCl is first order kinetics. Titre values are determined in different time intervals by using NaOH. They are Time in minutes 0 23 ∝ Titra value 20 40 50 Velocity constant of hydrolysis is (a) 3.0 × 10–1 min–1 (b) 3.0 × 10–2 min–1 (c) 4.7 × 10–1 min–1 (d) 4.7 × 10–2 min–1 65. For the decomposition of H2O2(aq) it was found that Vo2 (t = 15 min) was 100 mL (at 0°C and 1 atm) while Vo2 (maximum) was 200 mL (at 0°C, 2 atm). If the reaction had been followed by titration method and if VKMnO4 (t = 0) had been 40 mL what would VKMnO4 (t = 15 min) be? (a) 30 mL (b) 25 mL (c) 20 mL (d) 15 mL 66. The first order reaction sucrose → glucose + fructose take place at 308 K in 0.5 N HCl. At time zero, the initial rotation of the mixture is 34°. After 10 minutes, the total rotation is 30°. If the rotation of sucrose per mole is 85° that of glucose is 74° and fructose is -89°. The rate constant of the reaction is (a) 2.303 log

10 min–1 9

(b) 2.303 × 10–1 log

10 min–1 9

(c) 2.303 × 10–2 log

10 min–1 9

10 min–1 9 67. A + B → products. The rate is first order with respect to A and second order with respect to B. If 1 mole each of A and B are introduced into a 1 lit vessel and the initial rate was 10 –2 mol/ lit/ sec. The rate of reaction when half of the reactant used in the reaction is (a) 10–2 mol/ lit/ sec (b) 1.2 × 10–2 mol/ lit/ sec (c) 10–3 mol/lit/sec (d) 1.2 × 10–3 mol/ lit/ sec 68. The rate of first order reaction is 0.04 mol/lit/sec at 10 min and 0.03 mol/lit/ sec at 20 min after initiation. The rate constant of the reaction is (a) 2.87 × 10–2 min–1 (b) 2.87 × 10–3 min–1 (c) 1.438 × 10–2 min–1 (d) 1.438 × 10–3 min–1

(d) 2.303 × 10–3 log

Chemical Kinetics

69. A certain reaction A → B follows the given concentration (Molarity) –time graph. Which of the following statement is true?

The mol% of CH4 in reaction mixture excluding CH3CHO would

0.5

(a)

50K1 K1 + K 2

0.4

(b)

100K1 K1 + K 2

0.3

(c)

150K1 K1 + K 2

[A] 0.2

(d)

200K1 K1 + K 2

0.1

0

20

40

60

80

100

Time (sec) (a) The reaction is second order with respect to A. (b) The rate for this reaction at 40 second will be approximately 3.465 × 10–3 ms–1. (c) The rate for this reaction at 80 second will be 1.75 × 10–3 ms–1. (d) The [B] will be 0.25 M at t = 60 second 70.

k1

A

2k1

3B

4C

1.5k1 5D

All reactions are of 1st order at time t = t1 (t1>0) [B] = α Therefore at [C] Time t = t2 (where t2 ≥ t1) [C] =β [D] Which of the following is correct? (a) α > β (b) α = β (c) αβ = 0.4 (d) α + β = 0.4 71. CH3CHO(g) undergoes decomposition at very high temperature according to first order parallel reaction K1 CH3CHO (g)  → CH4(g) + CO (g) K2 CH3CHO (g)  → CH2CO (g) + H2(g)

8.65

72. Catalytic decomposition of an oxide by gold at 900°C. at an initial pressure of 200 mm was 50% in 23.03 minute and 92% in 84.1 minute. Then the velocity constant is (a) 0.2 mol–1L–1 min–1 (b) 0.0620 min–1 (c) 0.03010 min–1 (d) 0.3010 mol–1 L–1 min–1 73. Identify the true version for the following two first order reactions: (A) X  → Y k = 1010 e–500/T (B) W  → Z k = 1012 e–1000/T (a) Reaction 1 has higher activation energy barrier than reaction 2. (b) The temperature at which both reactions will have identical ‘K’ is above 100°C. (c) At 1010 degree centigrade reaction 1 will proceed faster than reaction 2. (d) Reaction 1 involves higher percentage of effective collisions than reaction 2 at 100°C.

Multiple Choice Questions with one or More than one answer 1. A+B → C+D is stoichiometrically balanced reaction. The initial rate of the reaction is doubled if the initial concentration of A is doubled, but is quadrupled if the initial concentration of B is doubled. Select the correct statement(s). (a) The reaction is first order in B and second order in A. (b) The reaction is first order in A and second order in B. (c) The reaction cannot be a single–step reaction. (d) The overall order of the reaction is 3.

8.66

Chemical Kinetics

2. A certain reaction A → B follows the given concentration (molarity) time graph. Which of the following statements are correct? 0.5

Molarity

0.4 0.3 0.2 0.1

0

20

40

60 80 Time/Second

100

5. Some of the following statements are given below. Choose the correct statements: (a) A collision in proper orientation between reactant molecules must occur with a certain minimum energy, known as threshold energy, it is effective in forming the products. (b) Greater the value of energy of activation, smaller the value of rate constant (c) Greater the value of activation energy, greater the value of rate constant. (d) At a lower temperature, an increase in temperature causes more change in the value of rate constant than that at higher temperature. 6. Zn + 2 H + → Zn 2 + + H 2 Half-life period is independent of concentration of zinc at constant pH. For the constant concentration of zinc, rate becomes 100 times when pH decreases from 3 to 2. Hence

dx = K [Zn]0[H+]2 dt dx (b) = K[Zn]1[H+]2 dt (a)

(c) Rate is not effected if concentration of zinc is made four times and that of H+ ion is halved. (d) Rate becomes four times if concentration of H+ is doubled and concentration of zinc is doubled. 7. Attainment of the equilibrium A(g) ⇌ B(g) + 2C(g). Initial concentration of B and C are not equal to zero find the correct options:

conc. (mol/lit)

(a) The reaction is first order with respect to A. (b) The rate of this reaction at 40 sec. will be equal to 3.5 × 10–3M/sec. (c) The rate of this reaction at 80 sec. will be equal to 1.75 × 10–3M/sec. (d) The [B] will be 0.25 M at 60 sec 3. Select the correct statements: (a) Molecularity of a reaction may include the number of product molecules take part in the reaction (b) Larger the value of Ea, greater is the effect on the value of ‘K’ for a given temperature range. (c) At lower temperature, increase in temperature causes more changes in the value of ‘K’ than higher temperature. (d) If concentration of catalyst appear in rate law that it becomes pseudo first order reaction rate = k [reactant]1 [catalyst]1. 4. According to collision theory, not all collisions between molecules lead to reaction. Which of the following statements provide reasons why is this so? (a) The average energy of the two colliding molecules is less than some minimum amount of energy required for the reaction. (b) Molecules cannot react with each other unless a catalyst is present. (c) Molecules that are improperly oriented during collision will not react. (d) Molecules in different states of matter cannot react with each other.

18

C

12

A B

10

time

5 sec

(a) At t = 5 sec. equilibrium has been reached and Kc = 270 (b) At t = 5 sec. equilibrium has been reached and % dissociation of A is 40% (c) Initial conc. of B and C is 2 molar each. (d) Initial conc. of A is 20 molar. 8. Consider the following case of competing 1st order

K1 reactions A

K2

C after the start of the reaction

D

Chemical Kinetics

t = 0 with only A, the [C] is equal to [D] at all times. The time in which all three concentrations will be equal is given by 1 (a) t = ln 3 2k1 (b) t =

1 ln 3 2k2

(c) t =

1 ln 3 3k1

1 ln 3 3k2 9. Thallium (I) is oxidized by Ce(IV) as follows: Tl+ + 2Ce+4 → Tl+3 + 2Ce+3 Following are the elementary steps involved in the above reaction: (a) Ce+4 + Mn+2 → Ce+3 + Mn+3 (b) Ce+4 + Mn+3 → Ce+3 + Mn+4 (c) Tl+ + Mn+4 → Tl+3 + Mn+2 If rate = k [Ce+4][Mn+2]; the catalyst, intermediate and rate determining step are respectively: (a) Mn+2, Mn+3, 1 (b) Mn+2, Mn+3, 2 +2 +4 (c) Mn , Mn , 1 (d) Tl+1, Mn+2, 1 10. Which of the following statements are correct? (a) Increase in concentration of reactant increases the rate of zero order reaction. (b) Rate constant K is equal to collision frequency A1 if Ea = 0 (c) Rate constant k is equal to collision frequency A1 if Ea = ∝ 1 (d) Log K Vs is a straight line. (d) t =

T

11. Which of the following is correct? (a) In a first order reaction, time taken for completion of 95% of reaction is constant at a given temperature, at any concentration of reactant. (b) In a zero order reaction time taken for completion of reaction is [A]0/K (here [A]0 – initial conc. of reactant, K = rate constant). (c) Rate equation of every elementary reaction is given by law of mass action. (d) With the increase of temperature, rate of exothermic reaction decreases. 12. A (aq) + B(aq) → C(aq). This reaction is an exothermic reaction following second order kinetics. To increase the rate of the reaction, which of the following can be done? (a) decrease of temperature of the reaction (b) increase of temperature of the reaction (c) increase of concentration of the reactants (d) addition of water to the reaction vessel

8.67

13. For a first order reaction, which of the following are true (K = rate constant, t = time of reaction)? (a) The degree of dissociation of the reactant is (1–e–kt) (b) The pre-exponential factor in Arrhenius equation has the dimensions of t–1. (c) t94% = 3t50% (nearly) (d) A plot of reciprocal of concentration of reactant vs. time gives a straight line. 14. Decomposition of 3A(g) → 2B(g) + 2C(g) follows 1st order kinetics. Initially, only A is present in the container. Pressure developed after 20 min and infinite time are 3.5 and 4 atm respectively. Which one is correct? (b) t50% = 20 min (b) t75% = 40 min (c) t99% = 64/3 min (d) t87.5% = 60 min 15. For a hypothetical elementary reaction, 2B k1 k1 1 A where k = 2 2 k2 2C Initially only 2 moles of A are present. The correct relation among no. of moles of A, B and C at the end of 75% reaction are: (a) (nB) = 2(nA) (b) (nC) = 2(nB) (c) (nB +nC) = 3 1 (nC) 4 16. The acid hydrolysis of ester is (a) 1st order reaction (b) bimolecular reaction (c) pseudo unimolecular reaction (d) none 17. The reaction CH3COOC2H5 + NaOH → CH3COONa + C2H5OH is (a) Bimolecular reaction (b) II order reaction (c) III order reaction, (d) none 18. For a general reaction aA+bB →cC+dD the rate of reaction may be give as (d) (nA) =

(a) r = -

dC A 1 × = kC ACBb dt a

(b) r = -

dCB 1 × = kC Aa CBb dt b

(c) r = -

dCC 1 × = kC Aa CBb dt c

(d) r = -

dCD 1 × = kC Aa CBb dt d

8.68

Chemical Kinetics

Comprehensive Type Questions Passage I Oxidation of metals is generally a slow electrochemical reaction involving many steps. These steps involve electron transfer reactions. A particular type of oxidation involve overall first order kinetics with respect to fraction of unoxidised metal (1–f) surface thickness relative to maximum thickness (T) of oxidized surface, when metal surface is exposed to air for considerable period of time Rate law: 0

df = k(l - f ) , where f = x/T, dt 200 hrs t

-3 ln(1-f) X = thickness of oxide film at time‘t’ and T = thickness of oxide film at t = ∞ 1. The time for thickness to grow 50% of ‘T’ is (a) 23.1 hrs (b) 46.2 hrs (c) 100 hrs (d) 92.4 hrs 2. The exponential variation of ‘f’ with t (hrs) is given by

(a) [l- e (c) e

-3t /200

-3t /200

]

(b) e (d) e

-3/200

-1

3t /200

Passage II The half life for nth order reaction n A  → products can be calculated as, -d[A] d[A] Rate = = K n [A]n or - ∫ = K n ∫ dt : dt [A]n t1/ 2

2n -1 - 1 K n (n - 1)a 0n -1

where a0 is initial concentration of A 1 So, t1/2 ∝ n -1 a0

t 1. For a reaction of nth order, the ratio of 1/ 2 is a t 3/ 4 function of (a) a0 (b) kn (c) n (d) t 2. The following data was obtained for a chemical reaction a0(m/l) 0.5 1.1 2.48 t1/2(seconds) 4280 885 174 The order of the reaction is (a) 1 (b) 2 (c) 3 (d) 4

3. Which of this statement is true? (a) t1/2 increases with increase in temperature (b) t1/2 decreases with increases in temperature (c) t1/2 is independent of temperature (d) t1/2 is the time taken when half the actual time is taken for the chemical reaction to complete Passage III Chemical Kinetics of peroxydisulphate ion

The peroxydisulphate ion is one of the strongest oxidants that are known, although the oxidation reaction is relatively slow. Peroxydisulphate ions are able to oxidize all, expect fluoride, to halogens. The initial rate (r0) of the iodine formation according to S2 O -82 + 2I–  → 2 SO -42 + I2 was determined as a function of the initial concentration (Co) of the reactants at 25°C. C0 ( S2 O -82 ), M C0(I–), M r0[× 10–8 mol L–1 s–1] 0.0001 0.010 1.1 0.0002 0.010 2.2 0.0002 0.005 1.1 1. Oxidation states of oxygens in peroxydisulphate ion is/are (a) -2 only (b) -2 and -1 1 (c) -1 only (d) –2 and 2 2. The rate constant of the above reaction is (a) 1.1×10–3 L. mol–1 s–1 (b) 0.011 L. mol–1 s–1 (c) 0.015 L. mol–1 s–1 (d) 1.1 L. mol–1 s–1 3. Iodine reacts with thiosulphate ions (S2O2– 3 ) forming iodide ions rapidly. Assuming that there is an excess of thiosulphate ions relative to the peroxydisulphate ions and the ions in the solution, rate law for the reaction –2 S2O 8 + 2I–  → 2 SO–42 + I2 at 25°C would be –2 (a) r = k [S2O 8 ][I–]2 where K is not equal to the one you found the previous question (b) r = k [ S2 O -82 ] where k is equal to the one you found in the previous question (c) r = k [ S2 O -82 ] where k is not equal to the one you found in the previous question (d) r = k [ I–] where k is not equal to the one you found in the previous question Passage IV From the following data for the reaction between A and B [A] mol/lit [B] mol/lit Initial rate mol/lit/sec. at 2.5×10–4 3.0×10–5 5.0×10–4 2.0×10–3 –4 –5 –3 5.0×10 6.0×10 4.0×10 1.0×10–3 6.0×10–5 1.6×10–2

Chemical Kinetics

1. The order of reaction when A is taken excess (a) 0 (b) 1 (c) 2 (d) 3 2. Rate constant at 300 K is (a) 2.67 × 108 (b) 5.34 × 108 8 (c) 4.00 × 10 (d) 6.21 × 108 3. Activation energy of the reaction in this temperature range (a) 2.64 × 104 cal/mol (b) 2.64 × 103 cal/mol (c) 1.32 × 103 cal/mol (d) 1.32 × 104 cal/mol Passage V The decomposition of NH3 is as follows 2NH3  → N2 + 3H2 if the rate obeys

-dNH 3 K1 [ NH 3 ] = Then K1, K2 are constants dt 1 + K 2 [ NH 3 ] 1. If NH3 is very less the order of reaction is (a) 0 (b) 1 (c) 2 (d) 3 2. If NH3 is very high the order of reaction is (a) 0 (b) 1 (c) 2 (d) 3

8.69

Passage VII For the following first order gaseous reaction k1 A(g)

3B(g)

k2 2C(g)

The initial pressure in a rigid container of capacity V lit is 1 atm. Pressure at time t = 10 seconds is 1.6 atm and after infinite time it becomes 2.2 atm 1. Partial pressure of gas C after infinite time will be (a) 0.6 atm (b) 0.5 atm (c) 1.5 atm (d) 1.6 atm 2. The value of ratio of

k1 is k2

(a) 3:2 (b) 1:4 (c) 4:1 (d) 2:3 3. The value of rate constant K1 is (a) ln 2 (b) ln 2 50 25 (c) 2 ln 2 25

(d) None

Passage VIII

 n B with time is presented in the The reaction A  figure:

Passage VI For the following first order gaseous reaction A  →P

Threshold energy of the uncatalysed reaction at 500 K is 140 KJ. The reaction is carried out in the presence of a catalyst 300 K in such a way that the rate of the catalysed reaction becomes equal to that of the uncatalysed reaction at 500 K. 1. ΔH of the reaction is (a) -60 KJ (b) +60 KJ (c) -20 KJ (d) + 20 KJ 2. Activation energies of uncatalysed forward and backward reactions respectively are (a) 100, 200 (b) 90, 130 (c) 100, 120 (d) 110, 125 3. The percentages of reacting molecules crossing over the energy barrier in the catalysed and uncatalysed forward reactions respectively are (a) 10–11 and 10–11 (b) 10–5 and 10–3 (c) 10–3 and 10–5 (d) 3% and 5%

Conc. mol lit-1

Average energy of A = 40 KJ Average energy of P = 20 KJ

B 0.5 A

0.3 0.1 1

3

5

7

Time (hour)

1. The value of ‘n’ is (a) 1 (b) 2 (c) 3 (d) 4 2. The equilibrium constant is (a) 1.2 (b) 1.8 (c) 2.0 (d) 2.4 3. The initial rate of disappearance of A is in mol/lit/ hour (a) 0.025 (b) 0.05 (c) 0.075 (d) 0.1

8.70

Chemical Kinetics

Passage Ix The hydrolysis of ethyl acetate in aqueous solution is first order with respect to ethyl acetate. Upon varying the pH of the solution the first order rate constant varies as follows: pH 3 2 1 1.1 11 110 K in 10–4 sec–1 1. The order of reaction with respect to H+ is (a) 1 (b) 2 (c) 3 (d) 4 2. The value of rate constant in above problem is (a) 1.1 × 10–1 lit/mol/sec (b) 1.1 × 10–2 lit/mol/sec (c) 1.1 × 10–3 lit/mol/sec (d) 1.1 × 10–4 lit/mol/sec Passage x A mixture of two different substances A and B undergo simultaneous first order reactions to produce the same product C as A

K1 = 1×10-2×2.303 min-1

C

The partial t1/2 of A along path I and path II are 173.25 min and 346.5 min, respectively. The energies of activation of the reaction along path I, path II and path III are 40 KJ mol –1, 60 KJ mol –1 and 80 KJ mol –1 respectively. 1. The per cent distribution of C in the product mixture B, C and D at any time is equal to (a) 20 (b) 60 (c) 80 (d) 40 2. The initial rate of consumption of A and the sum of the initial rate of formation of B, C and D are respectively, taking [A] =0.25 M, equal to (a) 2.0 × 10–3 mol L–1 min–1 and 2.5 × 10–3 mole L–1 min–1 (b) 2.0 × 10–3 mol L–1 min–1 and 2.0 × 10–3 mole L–1 min–1 (c) 2.5 × 10–3 mol L–1 min–1 and 3.0 × 10–3 mole L–1 min–1 (d) 4.0 × 10–3 mol L–1 min–1 and 2.0 × 10–3 mole L–1 min–1 3. The overall energy of activation of A along all the three parallel path is equal to (a) 52 KJ mol–1 (b) 60 KJ mol–1 –1 (c) 55 KJ mol (d) 80 KJ mol–1

B

Passage xII

K2 = 2×10-2×2.303 min-1

After 30 min (t = 30 min) from the beginning, mole percentage of C was found to be 60 (take log10 2 = 0.3 and mole percentage = mole fraction ×100) 1. Mole percentage of B at t = 0 (a) 37.5 (b) 40 (c) 45 (d) 60 2. Mole percentage of A at t = 60 min (a) 37.5 (b) 40 (c) 45 (d) 15 3. Mole percentage of B at t = 60 min (a) 37.5 (b) 40 (c) 45 (d) 2.5

Following facts are taken to express rate of the reaction in terms of stoichiometric coefficients. Answer the question at the end of it For the reaction aA+bB→cC+dD dx 1 d[C ] 1 d[ D] 1 d [ A] 1 d[ B] ===+ =+ dt a dt b dt c dt d dt 1. For the reaction in alkaline aqueous solution, 3BrO–→BrO3–+2Br– The value of rate constant at 80° C in the law for −

Passage xI ‘A’ is a substance that converts into B, C and D by three first order parallel paths simultaneously according to the following stoichiometric reactions. Path I B

∆ [ BrO − ] ∆t

Rate constants when the rate law is written for ∆ [ BrO3− ] ∆ [ Br − ] and are ∆t ∆t K, in terms of

A

t1/2 = 86.625 min.

Path II

Path III

2C

D

was found to be 0.054 mol–1 s–1.

(a) (b) (c) (d)

∆ [ BrO3- ] ∆

0.018 L mol–1 s–1 0.018 L mol–1 s–1 0.162 L mol–1 s–1 0.162 L mol–1 s–1

In terms of

∆ [ Br - ] ∆t

0.036 L mol–1 s–1 0.027 L mol–1 s–1 0.108 L mol–1 s–1 0.036 L mol–1 s–1

Chemical Kinetics

2. Rate of formation of SO3 in the following reaction 2SO2+O2→2SO3

–1

is 100 g min hence rate of disappearance of O2 is (a) 50 g min–1 (b) 100 g min–1 –1 (c) 200 g min (d) 20 g min–1 3. A reaction follows the given concentration time graph. The rate for this reaction at 20 seconds will be

Then relation between K1K2 and K3 is (a) 1.5 K1 = 3 K2 = K3 (b) 2K1 = K2 = 3 K3 (c) K1 = K2 = K3 (d) K1 = 3K2 = 2K3 Passage xIII Consider the following statement and answer the questions at the end of it. The rate at which a substance reacts is proportional to its active mass and the rate at which a reaction proceeds is proportional to product of active masses of the reacting substances. 1. For the complex reaction: +  Ag + + 2 NH 3 ↽ ⇀  [ Ag ( NH 3 )2 ]

0.4 0.3 0.2 0.1

0

20

40

60 80 Time/Second

100

(a) 4 × 10–3 Ms–1 (b) 8 × 10–3 Ms–1 –3 –1 (c) 2 × 10 Ms (d) 0.693 × 10–3 Ms–1 4. In the following reaction xA→yB d[ B]   d [ A]  + 0.3 = log  log  dt   dt   where –ve sign indicates rate of disappearance of the reactant. Thus, x:y is (a) 1:2 (b) 2:1 (c) 3:1 (d) 3:10 5. For a gaseous reaction, the rate is often expressed in  dn   dp   dC  terms of   instead of  and   , where   dt   dt   dt  C is the concentration and n is the number of mole. Hence relation between expressions is dC 1  dn  1  dP  =  = (a)   dt V  dt  RT  dt  dC  dn   dp  (b) = =  dt  dt   dt  dC  dn  V  dP  (c) = =   dt  dt  RT  dt  (d) None of these 6. For the reaction 2NH3→N2+3H2 d[NH 3 ] d[N 2 ] = K1 [NH 3 ]; dt dt d[H 2 ] = K 2 [NH 3 ] =K 2 [NH 3 ]; dt

 dx  2 7 2 -2 -1 +   = 2 × 10 L mol s [Ag ] [NH 3 ] dt -1 × 10 -2 s -1[Ag(NH 3 ) 2+ ] Hence ratio of constants of the forward and backward reaction is: (a) 2×107 L2 mol–2 (b) 2×109 L2 mol–2 (c) 1×10–2 L2 mol–2 (d) 0.5×10–9 L2 mol–2 2. In the reaction  2NH3 N2+3H2   dx  2 -3 3 -3 -1 2   = 1 × 10 M [ N 2 ][ H 2 ] - 1 × 10 M [ NH 3 ] dt

[ N2 ] [ H 2 ] 3  dx  = 10 -5 M 2 ; if   is  dt  [ NH 3 ] 2 (a) 0 (b) 1 × 10–5 –5 (c) 1 × 10 (d) 1 × 10–3 3. In the following graphical representation for the reaction A→B, there are two types of regions:

[B]

Concentration

0.5

Molarity

8.71

[A]

I:

II:

(a) I and II both represent kinetic region at different time intervals.

Chemical Kinetics

(b) I and II both represent equilibrium region at different time interval I represents kinetic region while II represents equilibrium region. (c) I represents kinetic while II represents equilibrium region. (d) I represents equilibrium while II represents kinetic region. 4. At a given temperature K1 = K2 for the reaction  C+D A+B 

 dx  If   =K1 [A][B]-K 2 [C][D]  dt  In which set of the concentration reaction ceases? [A] [B] [C] [D] (a) 0.1 M 0.2 M 0.3 M 0.4 M (b) 0.4 M 0.25 M 0.2 M 0.5 M (c) 0.2 M 0.2 M 0.3 M 0.2 M (d) 0.2 M 0.2 M 0.4 M 0.2 M

Mechanism I:

H2(g)+2ICl(g) → HCl(g)+I2(g)

slow Mechanism II: H2(g)+ICl(g)  → HCl(g)+HI(g)

(a) I only (b) II only (c) both I and II (d) neither I and II 3. For the following reaction (CH3)3CCl+H2O→(CH3)3COH+HCl  dx    = K [(CH 3 ).CCl], Hence, rate-determining dt step is (a) (CH3)3CCl→(CH3)3C⊕+Cl– (b) (CH3)3CCl+H2O→(CH3)3COH+HCl (c) (CH3)3C⊕+H2O→(CH3)3COH+H+ (d) H⊕+Cl–→HCl Passage xV

Passage xIV A+2B→product  dx  x y If rate law can be written as   = k [ A ] [ B ] dt Then x is said to be order w.r.t A and y is said to be order w.r.t B. If a reaction involves more than one step, the overall reaction is obtained by adding these elementary steps. In such cases, molecularity can’t be decided by overall reaction on the basis of its stoichiometric and rate equation, however, the order of an elementary step can be predicted from its molecularity. In fact, the order of an elementary step is always equal to its molecularity. The slowest step is the overall rate determining step and given order of the reaction. The following reaction is first order in A and first order in B. For the reaction

A+B → Product. Rate = k[A][B]

A B

I

This reaction is first-order in H2(g) and also first-order in ICl(g). Which of these of proposed mechanism can be consistent with the given information about this reaction?

II

1. Relative rate of this reaction in vessel I and II of equal volume is (a) 1:1 (b) 1:2 (b) 2:1 (d) 1:4 2. The reaction of hydrogen and iodine monochloride is represented by the equation H2 (g) +2ICl (g)→2HCl(g)+I2(g)

Following property is for the given order of a reaction. Based on this, answer the question given at the end of it Time of undergoing a definite fraction of a reactant is independent of the concentration. 1. For such reactions (as above), concentration of the reactant after two average lives (also called natural life time) is reduced to (a) 0.25 (b) 1/e (c) 1/e2 (d) 0.75 2. Which represents above-type reaction out of I,II and III?

log(a-x)

8.72

O

 a  log   a-x

O

I time

II time

T50

O

III a(conc.)

(a) I,II and III (b) I and III (c) II and III (d) I and III 3. In the following reaction A→ Product, I and II are two different sets of above type reaction: X is equal to I:

II: t=0

(a) 10 min (c) 30 min

t = 2 min

t=0

(b) 5 min (d) 2 min

t = x min

Chemical Kinetics

4. Rate constant of reaction is 0.0693 min–1. If we start with 20 mol L–1 it is reduced to 2.5 mol.L–1 in (a) 10 min (b) 20 min (c) 30 min (d) 40 min

Transition states are shown by (a) IR3 [A]>1 R3>R2>R1 10. For first order reaction K=

2.303 a log t a-x

2.303 100 log . Or t = 0.0127 100 - 75

1.5 × 10–6 =

2.303 × 0.6021 = 109.18. = 0.0127

Log

=

a 2.303 log K a - 0.99a

99 × a   ∴= 100 = 0.99a   

2.303 × 2.0 K Or t0.5 4.606 ...........(1) K a 2.303 log Similarlyt0.90= K a - 0.9a

=

Percentage of reactant remained =

90 × a   ∴ x = 100 = 0.9a   

2.303 .........(2) K

Dividing (1) by (2)we get 2.303 =2 K

Hence time taken for 99% of first reaction to take place is twice the time taken to complete 90% of the reaction.

a-x × 100 a

= 0.9473 ×100 = 94.73% Percentage of the initial concentration changed to product = 100-94.73 = 5.27% Further, t0.5 =

a 2.303 2.303 log log10 K 0.1a K

t 0.99 4.606 = K t 0.9

a 1.5 × 10-6 × 10 × 60 × 60 = a-x 2.303

a-x = 0.9473 a

a 2.303 2.303 log = log100 K 0.01a K

Or t0.9 =

2.303 a log 10 × 60 × 60 a-x

= 0.02344 a-x Or log = -0.002344 = 1.9765 a

8. We know that For first order reaction 2.303 a K= log . t a-x t0.99 =

8.83

0.6932 0.6932 = = 0.462 ×106s. k 1.5 × 10-6

0.693 11. Disintegration constant K = t0.5 (or) rate constant

K=

0.693 = 0.00012yr -1 5770

For disintegration process a initial concentration  → A Product a – I. conc a-x x conc. after 11540 years We know that K=

2.303 a log t a-x

8.84

Chemical Kinetics

Vapour pressure of the mixture of solvents =

2.303 a log 0.00012 = K = 11540 a-x

χ Α PAo + χΒ PBo

0.00012 × 11540 a = 0.6013 = 2.303 a-x Taking antilog

P0 =

log

Vapour pressure of the solution, P = 400 (After the addition of 0.525 mole of solute) Applying Raoult’s law

a = 3.993 a-x Fraction of A remained =

1 1 a-x = = 3.993 4 a

12. The reaction is first order. 13. Let the concentration of the reactant after 10 min and 20 min be C1 and C2 respectively Rate after 10 min = KC1 = 0.04 × 60 And rate after 20 min = KC2 = 0.03 × 60 C 4 ∴ 1 = C2 3 Supposing the reaction starting after 10 minutes C 2.303 2.303 4 log 1 = log = 0.02878 10 C2 10 3

T0.5 =

0.693 0.693 = = 24.086 min. k 0.02878

0.525 0.525 + (x + 12)

(4)

15. A → B a 0 (a-x) x K = 2.303 log a t a-x

initial concentrations Concentrations after time t.

2.303 1 log (t = 1 hr = 60 min) 60 (1 - x)

4.5 ×10-3 × 60 1 = 1- x 2.303 1-x = 0.7634 Rate = K (1-x) = 4.5 × 10–3 (0.7634) = 3.4353 × 10–3 moles litre–1 minute–1 Log

B ( l )(12 mole )  → An(s) Polymerisation

After 100 min: x moles Rate law: Rate = - 1 d [ A] = k[ A] n dt 1 dx Or = k ( x) n dt

2.303 a log On integration: nk s = K = t x 1

(3)

Solving equations (2), (3) and (4) we get X = 2.84 Substituting x = 2.84; t = 100 m and a = 10 in equation (1) 2.303 10 K= log = 1.2 × 10-2 min–1 100 2.84

4.5 × 10–3 =

14. 10 moles nA(l)

(2)

p0 - p = mole fraction of the solute p0 =

K=

12 x × 500 × 300 + x + 12 x + 12

16. K1 at 653 K = (1)

[In this problem, n is not given. As n, i.e., number of molecules of polymerizing to give one molecule, is a constant for this reaction and so the multiplication of n with the rate constant K1 gives another constant k1 which may also be called rate constant, though not according to its definition] Now calculate x to get the value of K from equation (1) When 0.525 mole of a solute is added, polymerization stops and x moles of 'a' remain. Just before the addition of the solute Moles of the solvent A = x Moles of the solvent B = 12

0.693 0.693 = = 1.925 × 10 -3 min t0.5 360

We know that Log

k2 Ea  T2 - T1  =   k1 2.303R  T1 T2 

Log

200 × 103  723 - 653  k2 = 1.925 × 10 -3 2.303 × 8.314  653 × 723 

K2 = 0.068 min–1 t=

2.303 a 2.303 100 log log = = 20.39 minute 25 k2 a - x 0.068

Chemical Kinetics

17. (a) Given ln k(sec–1) = 14.34- 1.25 ×10 T From Arrhenius equation Ea ln K = ln A RT On comparing these two equations

-4

Ea = 1.25 × 104 R Ea = 1.25 × 104×R = 1.25 × 104 × 1.987 = 24.83 Kcal mol–1 (b) k at 500 K 4 ln K = 14.34 - 1.25 ×10 500 K = 2.35× 10–5 sec –1 0.693 0.693 (c) K = = = 4.51 × 10 -5 sec -1 t0.5 256 × 60 t0.5 4.51 × 10–5 = 14.34 T = 513 K.

1.25 ×104 T

18. Let Ea = activation energy in presence of catalyst And E'a = activation energy in the absence of catalyst Then K = Ae–Ea/RT K1 = Ae–E'a /(R×500) in presence of catalyst K2 = Ae–E'a /(R×400) in absence of catalyst K2 = Ae–E'a /(R×400) in absence of catalyst Given the two rates are same i.e., r1 = r2 K 1 = K2 ∴ e–Ea/(R×500) = e–Ea/(R×400) Ea E ′a = Or R× 500 R× 400 or

Ea E ′a (  Ea - Ea′ = 20 ) = 500 400

∴Ea = 400 KJ mol–1 19. For first order reaction K = At 27°C K27°C =

0.693 t0.5

0.693 = 0.231 min -1 30

0.693 min -1 10 Now by using the following equation

8.85

K1 - Ea  T2 - T1  =   K 2 2.303 R  T1 ,T2  - Ea 0.231  320 - 300  = log10 0.693 2.303 × 8.314  300 × 320  Ea 20 - log10 0.3333 = × 19.1471 96000 19.1471 × 96000 Ea = × log 0.3333 20 = -91906 × ( -0.4772 )

log10

= 43857 J .mol -1 = 43.857 KJ .mol -1 20. According to Arrhenius equation K = Ae–Ea/RT Ea Or log K = Log ART Ea Or 2.303 log10 K = 2.303 log10 ART 0.693 For a first order t0.5 = K 0.693 -1 So K = Sec .(t0.5 = 10min = 600sec) 600 = 1.1×10–3sec–1 98.6 × 10 3 Hence log (1.1×10–3) = log(4×1013)2.303× 8.314× t T = 310.95 K 21. 2 moles of N2O5 on decomposition given, 4 moles of NO2 and 1 mole of O2. So the total pressure after completion corresponds to 5 moles and initial pressure to 2 moles. Initial pressure of N2O5 P0 =

2 ×584.5=233.8 mm Hg 5

after 30 minutes, the total pressure = 284.5 mm Hg 2N2O5 → 4NO2 + O2 P0-2P ∴ P0 + 3P = 284.3 Or 3P = 284.5-233.8 = 50.7 mm Hg 50.7 Or P = =16.9 mm Hg 3 Pressure of N2O5 after 30 minutes = 233.8-(2×16.9) = 200 mm Hg 2.303 233.8 K= log = 5.2× 10 -3 min -1 30 200.0

At 47°C K47°C =

22. K =

0.693 0.693 =. = 0.047793min -1 14.5 t0.5

Let the pressure of diethyl ether after 12 minutes be P atm. Applying first order equation K=

P 2.303 log O t P

8.86

Chemical Kinetics

log

0.4 0.047793× 12 = 0.2490 = P 2.303

0.4 = 1.7743 P 0.4 Or P = = 0.2254 atm 1.7743 Decrease in pressure x = 0.4- 0.2254 = 0.1746 atm → CH4(g) + H2(g) + CO(g) CH3COCH3 (g)  P0–x x x x Total pressure = P0+2x = 0.4 + 20.1746 = 0.7492 23. Let the rate law be Rate = K[A]x[B]y From Expt (1) 5.0 × 10–4 = K[2.5 × 10–4+ ]x[3.0 × 10–5]y (1) From Expt (2) 4.0 × 10–3 = K[5.0 × 10–4]x[6.0 × 10–5]y (2) 4.0 × 10-3 -1 = 2 x.2 = 8 Dividing (2) by (1) 5.0 × 10-4 From expt (3) 1.6 × 10–2 = K[1 × 10–3]x[6.0 × 10–5]y (3) 1.6 × 10-2 x Dividing (3) by (2) =2 =4 4.0 × 10-3 Or x = 2 and Y = 1 Hence order w.r.t. A is 2nd and order w.r.t B is 1st (II) Rate = [A]2 [B] From Expt (1) 5 × 10–4 = K [2.5 × 10–4]2[3.0 × 10–5] 5 × 10-4 Or K = = 2.67 × 108 L2 mol–2 s–1 [2.5 × 10-4 ]2 [3.0 × 10-5 ] (ii) From Arrhenius equation Ea 2.0 × 10-3 20 = × log 5.0 × 10-4 2.303 × 8.314 300 × 320 Ea =

2.303 × 8.314 × 300 × 320 × log 4 20

= 55.333 KJ mol–1 (iv) Applying log K = Log A -

Ea 2.303RT

A 55.333 = 9.633 = K 2.303×8.314×300 A Or = 4.29 × 109 K

log

A = 4.29 × 109 × 2.67 × 108 = 1.145 × 1018

0.693 0.693 = = 0.33hr -1 t0.5 2.1 Applying the first order rate equation a 2.303 K= log a-x t 2.303 100 t= log .033 100 - 99 = 13.96 hr

24. (i) K =

(ii) No. of moles of NH2NO2 decomposed = 0.99 × 6.2 = 0.099 62 No. of moles N2O formed = 0.099 Volume of N2O at STP = 0.099 × 22400 mL = 2217.6 mL

25. (a) Using first order kinetic equation and substituting with given values 2.303 200 (i) K = log = 0.0131min -1 53 200 - 100 (ii) K =

2.303 200 log = 0.0131min -1 100 200 - 146

As the values of k are same in both cases, the reaction is of first order. (b) Since the required time for the completion of same fraction in a first order reaction is independent of initial concentration; the percentage decomposition in 100 minutes when the initial pressure is 600 mm will also be 73 %. 26. (i) As pressure concentration K=

2.303 Pi (initial pressure) log t Pt (pressure after time t)

2.303 0.062 log = 6.2 × 10-3 s -1 55 0.044 = (ii) Again applying the first order equation 2.303 Pi (initial pressure) K= log t Pt (pressure after time t) 2.303 0.062 log 100 Pt 6.2×10-3 ×100 Or = log 0.62 - log(Pt) 2.303

6.2 × 10–3 =

Or log (Pt) = log 0.062- 0.2692 = 2.7924-0.2692 Pt = 0.033 atmosphere Pressure after 100 sec = 0.033 atm

Chemical Kinetics

27. Half life of reaction = 120 min For first order reaction 0.693 0.693 K= = = 5.77 × 10-3 min -1 t0.5 120 Applying first order equation 2.303 a t= log K a-x

-.

=

2.303 100 log -3 10 5.77 × 10

2.303 = 399 min 5.77 × 10-3

28. From the data (i) and (ii) we see that when the concentration of H2 is halved, the rate is also halved at constant concentration of NO. Hence the reaction is of first order with respect to H2. Let us now consider the data (II) and (III) to determine the order with respect to NO as [H2] is constant The rate law for the above reaction is 1 d[NO] Rate = - × = K[NO]m [H 2 ]1 2 dt Where m is the order with respect to NO d[NO] Or = 2K[NO]m [H 2 ] dt Substituting data (ii) and (iii) we get 2.2 × 10–4 = 2 K (1.5 × 10–4)m (2 × 10–3) 0.24 × 10–4 = 2 K (0.5 × 10–4)m (2 × 10–3) Dividing (1) by (2)

d[A 2 ] = K [A2] [B] dt

Substituting the data from (1) 0.072 = K(0.1) (0.1) K = 72 L mol–1 min–1 1 d[AB] = K[A 2 ] Rate of formation of AB (per mole) = 2 dt d[AB] Rate of formation of AB = = 2 × 72 × 0.01 dt –1 –1 = 0.288 mole L min

If a = 100, x = 90 or (a-x) = 10 Then t=

8.87

30. We have Ae–Ea/Rt The temperature dependence of K can be found by taking the derivative of the above equation with respect to T. Thus dK = Ae - Ea RT . E 2 = KE2 dt RT RT Thus the temperature dependence will be greater for reactions with large value of E. K1

31. Bicyclohexane

K2

Percentage of cyclohexene = = (1)

2.2 (1.5 × 10-4 ) m = = 3m 0.24 (0.5 × 10-4 ) m 220 = 3m 24 Taking log log 220- log 24 = m log 3 2.3424-1.3802 = m × 0.4771 = 0.9622 = 0.4771 0.9622 Or m = ≈2 0.4771

Or

29. From (1) and (2), we see that [A2] is constant and when [B] increases 4 times, the rate also increases 4 time i.e., rate K [B] Again from (1) and (3) we see [B] is constant and doubling the concentration of A2 increase the rate by two time showing again. Rate = K [A2] Thus the rate law will be

Cyclohexene

Methyl

K1 ×100 K1 + K 2

1.26 × 10-4 = 77% 1.26 × 10-4 + 3.8 × 10-5

Percentage equilibrium = 23% 32. At dynamic equilibrium Rate of formation of complex = Rate of its decomposition [Fe(dipy)3 ]2+ 1.45×1013 Ks = = = 1.97×1017 [Fe 2+ ][dipy]3 1.22×10-4 233. At equilibrium d[PtCl4 ] = 0 dt –5 Hence (3.9 × 10 ) [PtCl42–] = (2.1 × 10–3) [Pt(H2O) Cl3–][Cl–]

Or k =

d[PtCl4 2- ] 2.1× 10-3 = = 53.85 [ Pt ( H 2 O)Cl3- ][Cl - ] 3.9 × 10-5

34. We know that T=

2.303(logK1 - logK 2 ) K1 - K 2

8.88

Chemical Kinetics

2.303(log1.78×10-3 - log5.8×10-5 (1.78×10-3 ) - (5.8×10-5 ) = 1990 seconds 35. Let the initial partial pressure of PH3 and the inert gas be P and Pi mm respectively and Pˊ mm of PH3 decomposes at different time intervals Initial partial pressure P 4 PH3  → P4 + 6H2 Partial pressure at different times P-P' P1 4 6P1 4 As given at t = 60 seconds P + Pi = 262.40 (1) P ′ 6P ′ (2) And P-P' + + + Pi = 272 4 4 At t ∝ P + 6P + Pi = 276.40 4

4

(3)

Solving equations (1) (2) and (3) we get P = 18.67 and P1 = 14 Similarly, at t = 120 seconds P ′ 6P ′ P-P' + (4) + + Pi = 275.51 4 4 Solving eqns (1) (4) and (3) we get P = 18.67 and P1 = 17.48 As the given reaction is of the type nA → products where n = 4, we have the following equation for first kinetics 2.303 a 2.303 P t= log = log 4t a-x 4t P - Pi Thus at t = 60s; K1 T = 120s; K2

2.303 18.67 log = 5.8×10-3s -1 4 × 60 18.67-14

2.303 18.67 log = 5.8×10-3 s -1 4×120 18.67 - 17.48

As the values of K are constant the reaction follows the first order. 0.693 0.693 36. K = = = 5.41×10-2 t 0.5 12.8

T =

N 2.303 log 0 = (for radioactive disintegration) K N N a = 0 a-x N

=

2.303 100 log -2 4 5.41× 10

37. t = 0; a ; t = 10 (a-x) = 13.8 K = 20(a-x) = 8.25 2.303 a log K= t a-x

a 2.303 (1) log 10 13.8 a 2.303 (2) When t = 20 K = log 20 8.25 Solving eqns (1) and (2) k = 5.1 × 10–2 min–1, a = 23.09

When t = 10 K -

=

38. Vegetable oil + H2 → Vegetable ghee 2 atm 0.8 atm initial after 50 minute Ni

(i) Rate of reaction =

0.8 Change in pressure = = 50 time

1.6 × 10–2 atm minute (ii) From gas equation PV = nRT; n P 0.8 = = = 0.032v RT 0.0821× 298 Change in molarity 0.0327 Now rate of reaction = = time in second 50× 60 = 1.09 × 10–5 mole litre–1 sec–1

Multiple Choice Questions with only one answer level II -

18 ×103

K1 e RT = 4 ×103 1. K2 e RT =e

-

14 ×103 8.314 × 300

2. A+B→C+D t 1 = 300 Sec 2

∴ .t 3 = 600 Sec 4

2.303 100 log 4 t 2.303 100 log K1 = 50 t K2 = 4.65 K1

3. K 2 =

4. Rate = k [Cl] [CHCl3] Kc =

[Cl]2 Cl2 1

∴ rate = K K c (Cl2 ) 2 [CHCl3 ]

Chemical Kinetics

5. K K c K2 Ea  1 1  =  -  K1 2.303R  T1 T2 

Log

6. Second step is the rate determining step → 3B 7. 2A  640 ─ 3x 640 - x 2

Initial conc Final conc

640 - x = 80 mm X = 560 mm 3x = 840 mm 2 3x Total pressure = 640 - x + = 840+80=920 mm 2 8. Zero order 11. Half-life of reaction is = 30 min  → 2B + C A a — — Initial mole a-x 2x t mole After 30 min 2x1 = a So a ∝ 30 mL 2 7a After 90 min 2x2=   2 8 ∴ 2x2 ∝ 52.5 mL   13. K= K1 +K2+K3= 6.93 × 10–3 sec t 1 = 100 sec

17. Half life ratio of A: B= 2:3 Initial conc of A and B is 3a, 2a After three half lives of A 3a the conc. of A is = 8 2a After two half life of B the conc. of B = 4 3 2 Ratio : = 3 : 4 8 4 18. r = k [SO3]2[O2]

r1 k[a ]2 [a ] = 2 r2 a  a  k    2 2 19. Second order reaction t0.5 ∝ 1 a Log t0.5 = log k- 1 log a So it is second order Log t0.5 = 6.5228 t0.5 = 3.33 × 106 K=

1 = 3 × 10–7 lit/ mol/ sec at 0.5

2

After 100 sec, the pressure of A = 1 atm Pressure of B =

2 × 1= 0.288 atm 6.93

14. Initial conc. = a0 Final conc.= a a -a ∴K = 0 t Kt = a0-a a = a0 -Kt slope = -K 15. K1: K2 = 1:4 K1 Ea1 + K 2 Ea2 Ea = K1 + K 2 16. At 20 min conc.of reactant ∝ 0.8-.02 At 40 min conc.of reactant ∝ 0.65-0.2 ∴ t 3 = 20 min 4

t 1 = 10 min 2

→ X 2 O7 20. 2 XO - 3  a 0...............initial mole a-x x / 2.......... final mole 6a ∝ 30 mol

x ∝ 36 mol 2 ∴ x ∝ 3 mol 2.303 5 ∴K = log 9.212 2 = 0.1 min–1 21. t0.5 = 15 min

6(a-x) + 16

A → 2B + C a 0 0 initial mole a-x 2x mole after time t After 15 min 2x = a ∴ a ∝ 40 m  15  After 60 min 2x =  a 2  16  2 15 × 40 × 2 = 75ml mL 16

8.89

8.90

Chemical Kinetics

22. K΄ = K (OH–)n 61 = K (1)n 120 = K(0.5)n 1 = (2) n 2 n = -1

∴K =

1 =3 1- x 1 = 3 - 3x 2 x= 3 2 1 [B] = × 3 10 1 [A] = 3 B ∴ = 0.2 A

24. Net rate is a reversible the reaction K1-K-1[B]2 dn 1 dB so =dt 2 dt rate in irreversible reaction is = K2B -dB ∴ = 2 K1 - 2 K -1[ B]2 - 3K 2 [ B] dt 25. 2A + B → C0 2C0 C0-2x 2C0–x Here x =

C+D 0 0 …....initial x x …....final

29. P(g )  → 2Q(g ) +

Po P −x

Co 4

2x 2P

−

1

P°+

2

3

( 76 - 57 )10 KP = e 8.314×298 Ka

VP

t Pressure

VP P∝ ressure

3x + 17 = 317 2

∴ x = 40 2.303 240 ∴K= log 23 200 ∴ K = 7.910–3 Min–1

K 49Co 32

( Ea )a - ( Ea ) P 26. K P = e RT Ka

1 S(l ) 2 − Initial Pressure

5P + VP = 617 mm 2 ∴ P° = 240 mm

∴ R = K[A]1[B]2

 C   7C  =K 0   0   2   4 

o

1 R (g ) + 2 − x 2 Po 2

O

C0 7C0 = ,B 4 2

=

−

o

After 30 minutes the concentration of

= A

2.303 1 log 1 1- x

30. CH3OCH3 → CH4 + H2 + CO 0.4 0 0 0……Initial pressure 0.4-x x x x…….pressure after time t After 29 mints 0.4-x = 0.1 X = 0.3 Total pressure = (0.3) 3 + 0.1 = 1 atm - Ea N = e RT NO

27. K5 [NO2NH–] Rate = K5 [NO2NH] Kc = [NO2NH–] [H+]/ [O2N NH2] ∴ Rate = K5 Kc[O2NNH2] [H+]–1

31.

28. K= K1+K2 K = 0.109 + (0.109) 9 = 1.09 hr–1 Assume initial conc. of A is = 1 After I hour the conc. is = 1-x

32. K a = e 8.314 × 500

- Ea

10-5 = e 2×500 −100 × 103

−80 ×103

K p = e 8.314 × T K a = Kp 80 100 80 ∴ = T = 400k 500 tT

Chemical Kinetics

34. 2A X x-2y

47. According to the graph t 3 = 2t 1 4 2 So it is first order

3 2

33. Order of reaction

+ B → C + 2D 2X 0 0 2x-y y 2y

initial conc conc at t

x ∴y = 4 x 7x ∴ [A] = ,[B] = 2 4 ∴ R = K[A]1[B]2  x   7x  R = K     2  4  R = 4.9X3K/32

2

a 2.303 log t a−x 2.303 100 log K= 30 50

35. K =

36. Net activation energy K1 Ea1 + K 2 Ea2 K1 + K 2

42. Net K =K1+K2+K3 = 6.93×10–3 sec–1 t0.5 = 100 sec

K= So:

0.693 = 5.75 × 10−3 min −1 120

dx = k[A] dt

0.7 = 5.75 × 10–3 × 0.4 dt

dt = 173 min 48. C4H8 → 2C2H4 1 0……….initial mol 1-x 2x……...final mol ∴K =

2.303 1 log t 1- x

Hence 1- x = 2x 1 X= 3 2.303 1 log 2.303 × 10–3=

t

2/3

t = 1.76 × 102 sec 49. K p

Ka

( Ea )a − ( Ea ) p

=e

RT

( Ea )a − ( Ea ) p

After 100 sec the pressure of A is 1 atm 2 × 10-3 So partial pressure of B = ×1 6.93 × 10-3 = 0.288 atm 1 1 = aK 4.5 × 10-3 × 0.8 2 = 277 sec. The time required to convert 0.8 M to 0.1 M is t 1 x ∴K = at a - x 1 0.7 × t= 0.8 × 4.5 × 10-3 0.1

44. t 1 =

= 19.44 sec

0.693 -1 sec 40 1.386 K0 = mol / lit / sec 2× 20 K1 = 0.5lit / mol K0

45. K1 =

4000 = e

2 × 300

(Ea)a-(Ea)P= 4.94 Kcal 50. In decomposition process, moles do not change So P1 P2 = T1 T2 2 P = 400 600 P = 3 atm 54. A → nB A0 0 initial conc. A0-x nx final conc. At that point A0-x = nx A0 x= n +1 [B] = nx =

nA 0 n +1

8.91

8.92

57.

Chemical Kinetics H+

a Ratio is constant a-x

64. CH3COOCH3+H2O → CHCOOH+CH3OH Initial titre value ∝ Conc. of HCl Infinite titre value ∝ Conc. of HCl + Conc. of CH3COOH

So it is first order ∴K =

2.303 1 log 5 0.8

K=

K= 4.7 × 10-2 min -1

= 0.0446 min–1 R = KA = 0.0446(1.8) = 0.08 M min–1 58. At low temperature and crystalline form react slowly 59. K = K1+K2 = 3.2 × 10–2 min–1 t 1 = 21.65 min 2

Initial pressure A = 100 +

300 = 1600 0.2

The find pressure of A is = 100 The time required is 4 t 1 = 4 × 21.65 = 86.65 min 2

60. Kp = Ae

−( Ea ) P

K a = A.e

R ( 300 )

- ( Ea ) a RT

= Ae

-9.6

R ( 320 )

∴ ( Ea ) P = 9KJ / mol

61.

K N = A NO 3×10-3 0.001 = A 100 A = 300 Sec

62. Initial molarity H2O2 is = 2 M After 6 hrs, molarity of 10 mL dilute H2O2 is =

20 × 0.02 × 5 = 0.1M 10 × 2

After 6 hrs conc of H2O2 = 1 M t 1 = 6 hrs 2

K = 0.115 hr–1 63.

k=

r −r 2.303 log 0 ∝ t rt − r∝

2.303 45 - ( -15) log 23 15 - ( -15) = 3 × 10-2 =

30 2.303 log 23 10

65. H2O2  → H2O + 1 O2 2 a 0 0 initial mole x a-x — mole after time t 2 a mole at ∝ time — — 2 X ∝ 100 mL 2 a ∝ 400 mL 2 KMnO4 react with H2O2 only Eq H2O2=eq KMnO4 a ∝ 40 mL a-x ∝ 30 mL 66. C12H22O11 +H2O→C6H12O6+C6H12O6 a — — initial mole a-x x x mole after time t 34 = 0.4 a= 85 (a-x) 85 + x (74)+ x (-89) =30 x = 0.04 K=

2.303 0.4 log 10 0.4 - 0.04

67. ri = K [A][B]2 ri = K [1](1)2 rf = K [0.5](0.5)2 rf = 1.2 × 10–3mol/lit/Sec 68. K = 2.303 log a t2 − t1 a−x = 2.303 log 0.04 10 0.03 –2 = 2.87 × 10 min–1 69. It is first order reaction

Chemical Kinetics

Comprehensive Type Questions

Half-life = 10 sec ln 2 K= 10

Passage II

2.

( t 0.5 )1 ( t 0.5 )2

a  = 2   a1 

n −1

K1 =

=

n-1

4280  1.1  =  885  0.5  N=2 3. By increasing temperature rate constant increases, t y 2 decreases

1.

1. If NH3 in excess then -dNH 3 = K1[NH3] dt 2. If NH3 is less then -dNH 3 = k1 k2 dt Passage VI 3. In both cases, K value is same K N = = e − Ea / RT A N0

-dA 1 dB = dt n dt 1 0.1= (0.2) n n=2

2. KC = 3.

Passage V

ln 2 50

Passage VIII

Passage III m 2. r=K(S2 O–2 (I–)n 8 ) –2 3. Here S2O3 is excess then concentration of I– is large So r = K[S2O–2 8 ]

ln 2 1 × 10 5

[ B]2 ( 0.6 ) = = 1.2 A 0.3

-dA 0.1 = = 0.1 mil/lit/sec dt 1

Passage Ix 1. K = K1 (H+)n

1.1×10-4  10-3  =  11×10-4  10-2  n=1

n

2. 1.1×10–4 = k1 (10–3) K1 = 1.1×10–1 lit/mol/sec

Passage VII 1. Initial pressure of A is 1 atm. After infinite time 3x + 2 (1 - x) = 2.2 ∴x = 0.2 So after infinite pressure of C = 2 (1 - x) = 1.6 atm 2. K1 = n = 1 K2 1 - x 4 3. At time t = 10 sec, the pressure equal to 1.6 atm So 1 - x1 + 3x1

K1 K2 +2 (1 - x) =1.6 K1 + K 2 K1 + K 2

3 1 8 x + (1 - x1) = 1.6 5 5 ∴ x1 = 0.5 1 - x1+

Passage x 1. Assume initial mole of A is =a Initial mole of B is =b Half life A  → C is 30 min Half life B  → C is 15 min a After 30 min [A] is = 2 b Conc. B is = 4 a 3b Conc. [C] is = + 2 4 Mole fraction of C is = 0.6 a 3b + 2 4 = 0.6 3: a+b So mole fraction of B at t = 0 min 2 = = 0.4 5

8.93

8.94

Chemical Kinetics

a 2. After 60 min [A]= 4 [B] = b 16

3. t 3/ 4 = 2t y

K=

[C] = 3a + 15b 4 16

0.693 sec–1 20

R = K(A)

Passage xI 1. Rate constant of path I 0.693 = 4 ×10–3 min–1 K1 = 173.25 K2 = 2 × 10–3 min–1 K = 8 × 10–3 min–1 K = K 1 + K 2 + K3 K3 = 2 × 10–3 min–1 At any time B:C:D = 4:4:2 2. Initial rate of consumption of A = -dA = k[ A] = 2 ×10-3 mol/lit/sec dt dB = k1 [ A] = 4 ×10-3 (0.25) dt = l × 10–3 mol/lit/sec dC = 2k2 [ A] = 1 × 10 -3 mol/lit/sec dt dD = K 3 [ A] = 0.5 ×10-3 mol/lit/sec dt 3. Ea =

2

Half-life = 20 sec

k1 Ea1 + k2 Ea2 + k3 Ea3 k1 + k2 + k3

=

0.693 (0.2) = 0.69 × 10 -3 M sec–1 20

4. 1 -dA = 1 dB x dt y dt

x

log -dA = log dB ×log = y dt dt

x =2 y X : Y = 2:1 K K 6. 1 = K 2 = 3 2 3 3K1= 6K2 = 2K3 Passage xIII 1.

Kf Kb

=

2 × 107 = 2 × 109 lit2/mol2 -2 1 × 10

2. It is in equilibrium 4. The reaction cases mean K1[A][B]=K2[C][D] [A][B]=[C][D] Passage xVI

Passage xII -

1.

1 dBro3 - 1 dB = 3 dt 2 dt

1 1 k [ BrO3- ]n = k2 ( BrO3 ) n = k3 [ BrO3- ]n 3 2 ∴ 1 k1 = k2 = 1 k3 3 2 2. dSO3 = 100 mol/lit/min dt 80 -dO 2 1 dSO3 = 2 d2 db 2 -dO 2 100 32 = × = 20 gm min–1 dt 80 2

1. Log A = 14 Ea 2. = 1.25 ×104 2.303R Ea = 57.6 Kcal 4.

10 × 103 Kp = Ka e8.314 × 400

5. r1 = k1[NOBr2] [NO] Kc = [NOBr2]/[NO] [Br2] ∴ r = K1Kc[NO]2 [Br2]

dx =0 dt

Chemical Kinetics

Passage xVII 1.

dC = K [ A]n [ B]n dt 0.0033 = K [0.1]n [0.05]n 25 0.0039 = K [0.1]n [0.1]n 15 0.0077 = K [0.2]m [0.1]n 7.5

Passage xVIII 1. r1 = K11 [A][B] 2. r2 = K1[A]

Ea = 100 KJ Ans: 100 3. r = K A2 B r = K = 0.01 lit2/mol2/sec r = (0.01) (0.5)2 (0.5) 4. Bn +  → B (n + 4) + 4e– a — initial mole a-x x moles at time t 2a ∝ 26 a ∝ 13 2(a-x) + 5x ∝ 35 x∝3 k = 2.303 log 13 = 133 × 10-4 min -1 10 23 5. r = K A2 B0 C rf = K(2A)2 (2B)0(2C)

Passage xIx 1. r = K2[SO3 2H2O] Kc =

SO3 .2 H 2 O [ SO3 ][ H 2 O]2

∴ r = K2Kc [SO3][H2O]2 Passage xx 1. Rate of formation of radical = Rate of removal of radical • So K1[CH3CHO]+K3[CH3CO] • • = K2[CH3CHO][CH3] + 2K4[CH3]2 2 K1[CH3CHO] = 2K4[CH3]

Integer Type Questions 1. They obtain in equi molar ratio means K1 = K2 44 30, 000  1 1  2.303 log = 2x 8.314  300 T  2.303 log

2 38,314  1 1  = 2x 8.314  300 T 

T = 248 K 2. Ka = Ae–Ea/R(500) Kp = Ae–(Ea-20)/R(400) Here Kp = Ka

Ea Ea - 20 = 500 400

6. Volume double mean concentration is halved so order = 1 7. Half life of Ist reaction is 2 hrs Half life of 2nd reaction is 8 hrs 8. K = K1 + K2 1 1 1 = + t 4 12 t = 3 hrs 9. Ea = Ea1 + Ea3 + Ea4 - Ea5 2 10 + 5 + 3 = 9kJ / mol = 2 10. Assume second order reaction K=

1 1 1 -   t a - x a

1 1 [ 2 - 1] = M -1 min -1 5 5 Order = 2 1  11 1  3 − = 11. t75% =   K  25 100  1K (100) K=

1 t50% = 1  1 - 1  = K  50 100  K (100) ∴ t75% = 3 t50%

8.95

8.96

Chemical Kinetics

12. C4H8  → 2 C2H4 a —……initial moles a-x 2x……moles after time t K = 2.303 log a a-x t

a = 1.5 a-x a ∴x= 3 a − x 2a = 3 2 x2

2a =1 3

13. Initial mole of A is 4 after 75% completion. Mole of A is = 1 Mole of B is = 2 Mole of C is = 4 Total mole of A + B + C = 7 14. The time taken for completion of 87.5% reaction is = 300 min so difference is = 20 min X=5 15. So it is zero order 16. r = KAn log r = log k + n log A Tan θ = n = 2 Order = 2 17. Assume half-life of reaction is = T Initial conc. of B = a Initial conc. of A = na r1 = 0.693 [na] T r2 = 1 [a]2 aT r1 = 1.386 r2

0.693 [na] = 1.386 a T T 0.693 n = 1.386 ∴ n = 2.

18. 2AsH3 I Pressure P0 t pressure P0-x

 → 2As + -

-

3H2 3x 2

3P 0 2 Initial pressure P0 + PI = 800 min 3P 0 Infinite time = + PI = 1000 min P0 = 400 min 2 ∝ pressure -

-

PI = 400 min After 10 min 3x + PI = 900 P0-x + 2 X = 200 min So half life = 10 min After x min 3a P0-a + + PI = 950 v 2 a = 300 min The time x = 20 min x ∴ y = = 5 min 4 19. [H + ] = 3×10–6/0.05×10–3 = 6×10–2M 107 mole lit- disappear in/one sec So 6×10–2 mol/lit disappear in 6×10–9 sec

2.303 100 log t 10 2.303 100 K2 = log 2t 1

20. K1 =

K1 =1 K2 22. Concentration of B is excess order with respect to B is zero R = K [A] ∴×=2 x+y=2+0=2 ∴

Chemical Kinetics

Previous Years’ IIT Questions 1. Variation of rate constant k with temperature is given by Arrhenius equation. K = Ae - Ea / RT Equation indicates the variation of k exponentially with temperature (T) - Ea

k ∝ e RT

or

k ∝e

+

Ea ×T R

2. A  → P, a first order reaction log K = log10 AEa (1) 2.303RT Given equation is log K = 6.0 - 2000 (2) T Equating eqs. (1) and (2), log10 A = 6 or A = 106 or 1.0 × 106 s–1 Ea = 2000 2.303R Ea = 2000 × 2.303 × R = 2000 × 2.303 × 8.314 × 10–3 KJ mol–1 = 38.3 KJ mol–1 3. For a first order reaction, K1 = Rate constant K1 = 0.693 (1) 40 For a zero order reaction A  →B 1.386 mol/dm3 1.386 Half conc. = = 0.693 mol dm–3 2

dx 0.693 (2) = dt 20 Divide eq. (1) by (2), K1 = 0.693 × 20 = 0.5 mol–1 dm3 K0 40 0.693

Time = 20 sec, K0 =

4. Let a moles of G combine with b moles of H forming the product aG + bH  → Product Rate of reaction ∝ [G ]a × [ H ]b Rate (r) = K × (G)a × (H)b…….(1) When conc. of both G and H are doubled, the rate becomes 8 times 8r = K × (2G)a × (2 H)b 8r = K × 2a + b [G]a×[H]b……(2) Substitute the value of r in eq. (2) 8 × K × [G]a × [H]b = K × 2a + b× [G]a × [H]b 2a + b = 8 = 23 a+b=3 The order of reaction = 3

8.97

5. The order of reaction can have values 0, 1, 2,3 or even in a fraction. It is always determined experimentally. 6. The concentration reduces from 0.1 M to 0.025 M in 40 minutes = 2t1/2 So half life is 20 min r = k[A] 0.693 [0.01] = 20 = 3.47 × 10–4 M min–1 7. Conc. 800 mol/litre changes to 50 mol/lit in 2×104sec 4 So half life is = 2 ×10 = 5 ×103 sec 4 0.693 = 1.386 ×10-4 sec–1 K= tv 2

8. Rate =

-dN 2 1 -dH 2 1 dNH 3 = = 3 dt 2 dt dt

9. R = K [A] -5 [N2O5] = 2.4 ×10 = 0.8M -5 3 ×10 10. 2N2O5(g) → 4NO2(g) + O2(g) (a) It is a first order reaction and conc. falls exponentially. (b) Rate constant (K) varies with temp. and hence t 1 2

0.693 = also changes with temperature. K 0.693 (d) t 1 = 2 K

(1)

A = 100 a-x = 100-99.6 = 0.1 a 2.303 2.303 100 = log log K a-x K 0.4 2.303 = log 250 K t=

2.303 × 2.398 K Divide eqn. (2) by eqn (1)

T=

11. The relevant expressions are as follows: ∆H 1 +1 Choice (a) log Kp = R T Choice (b) log [X] = log [X]0 + kt Choice (c) P/T = constant (V constant) Choice (d) PV = constant (T constant)

(2)

8.98

Chemical Kinetics

12. In first order reaction, if α is the degree of dissociation therefore 1 Kt = loge = - log e (1 - α ) or e - kt = 1 - α (1 - α ) ∴ α = 1-e–Kt The Arrhenius equation is, K = Ae–Ea/Rt Plot of reciprocal concentration of the reactant vs time is liner. Dimensions of pre = exponential factor ‘A’ are equivalent to dimension of K, which is T–1 for a fist oder reaction

Integer Type Questions 13. If it is a zero order reaction then n K= t K=

0.25 0.6 ,K = 0.05 0.12

So it is zero order 14.

2.303 1 log 1 k 8 2.303 1 log = 1 k 10

t 18 = t 110

t1/ 8 = log 8 = 0.9 t1/10 ∴

t 18 t 110

× 10 = 9

CHAPTER

9 Chemical Equilibrium

S

ome one who had begun to read geometry with Euclid, when he had learned the first proposition, asked Euclid, “But what shall I get by learning these things?” whereupon Euclid called his slave and said “Give him three-pence since he must make gain out of what he learns.” Stobaeus

9.1 IntroductIon In a chemical change, the reactants that are initially present disappear, while at the same time the products, a new set of substances with quiet different properties, appear. The chemical reaction is accompanied by the absorption or release of energy. The speed at which these changes take place varies widely, depending on the nature of the reaction and on the reaction conditions. In some reactions, the change takes place in a fraction of a second, and energy is released with explosive violence. In other reactions, the change take place more gradually and may be observable for many minutes, or hours or days after the reactants are mixed. However, regardless of the speed at which the changes occur initially, sooner or later the changes become slow and eventually they become imperceptible. When the latter state is reached, the properties of the reaction mixture remain constant indefinitely, as long as the temperature, pressure and other external conditions are kept constant. A reaction mixture that had reached a state in which no further change takes place is said to be at equilibrium.

9.1.1 reversibility of reactions An interesting fact about the equilibrium state is that the reaction need not be complete. Consider, for example, the reaction of hydrogen gas with iodine vapour to produce hydrogen iodide gas H2 + I2 → 2HI

(1)

This reaction is very slow at room temperature and progress can be detected only if many days are allowed to elapse between measurements. However, the reaction speeds up considerably as the temperature is raised—the reasons are already considered in the previous chapter (8)

and at temperatures above 300°C the progress to equilibrium can be followed conveniently. It is found that when equilibrium is reached, the reaction mixture contains appreciable quantities of unreacted hydrogen and iodine. Reactions that are incomplete at equilibrium can be reversed easily. Consider for example reaction (2) which is the exact reverse of reaction (1). 2HI → H2 + I2

(2)

When pure hydrogen iodide gas is heated above 300°C, the formation of hydrogen gas and iodine, which is visible as a violet coloured vapour, is soon evident. However, reaction (2) also fails to go to completion and leads to an equilibrium state in which all three substances—H2, I2 and unreacted HI – are present. Moreover, the composition of the equilibrium state resulting from (2) is identical with that resulting from (1) provided that the same numbers of hydrogen and iodine atoms and the same reaction temperature and volume are used. In other words, under comparable conditions, the final equilibrium composition will be the same, regardless of the direction in which the reaction is carried out. Reactions that can be carried out in either direction are said to be reversible. It is convenient to represent reversible reactions, such as (1) and (2) by using double half arrows as shown in (3).   H2 + I2   2HI

(3)

In reversible reactions, the terms reactants and products are ambiguous unless the direction is specified. However, it is customary to use a convention in which the reactants are the substances appearing on the left side of the equation and the products are those appearing on the right side. If the products predominates equilibrium, we say that the equilibrium lies on the right and the reaction has gone

9.2

Chemical Equilibrium

to completion. If the reactants predominate, we say that the equilibrium lies on left. The system at equilibrium is referred as an equilibrium mixture. Any process whether physical or chemical that takes place in the forward direction as well as in the reverse (opposite) direction under the same condition is generally referred to as a reversible process. Equilibrium can be established for both physical process and chemical reactions. The reaction may be fast or slow depending on the experimental conditions and the nature of the reactants. When the reactants in a closed vessel at a particular temperature react to give products, the concentrations of the reactants keep on decreasing, while those of products keep on increasing for some time after there is no change in concentration of either of the reactants or products. This stage of the system is the dynamic equilibrium and the rates of the forward and reverse reactions become equal. It is due to this dynamic equilibrium stage that there is no change in the concentrations of various species in the reaction mixture. In theoretical discussions of chemical equilibrium, one makes no fundamental distinction between reactions that are incomplete at equilibrium and those that appear to have gone to completion. The latter are simply equilibria that lie very far on the right. This theory asserts that some reactant is present at equilibrium in all reactions, even when the amount is too small to be detected. Thus in theory, all chemical reactions are reversible. Based on the extent to which the reactions proceed, the state of chemical equilibrium in a chemical reaction may be classified in to three groups. (i) The reactions that proceed nearly to completion and only negligible concentrations of the reactants are left. In some cases it may not even possible to detect these experimentally. (ii) The reactions in which only small amounts of products are formed and most of the reactants remain unchanged at equilibrium state. (iii) The reactions in which the concentrations of the reactants and products are comparable, when the system is in equilibrium. Thus, in theory all chemical reactions are reversible. The extent of reaction in equilibrium varies with the experimental conditions such as concentrations of reactants, temperature etc. If a reaction is known to be reversible, no matter how slightly there is hope that conditions might be found under which the reaction can be reversed in practice. It is easy to see that this theory greatly enlarges the scope of chemistry because any reaction in which a substance participates as a reactant can if reversed become a method of synthesis for that substance. Optimizations of the operational conditions is very important in industry and laboratory so that the equilibrium is favourable in the direction of the desired product.

9.1.2 Equilibrium in Physical Processes The physical equilibria process can be explained as follows. For example, a liquid taken in a closed vessel incompletely undergoes vaporization and the liquid molecules escape into space above the surface of the liquid in the closed vessel in the form of vapour. The vapour after saturation returns back into the liquid in the form of molecules into the liquid → vapour can be called forward prostate thus liquid  cess and vapour  → liquid as reverse process. If these two processes take place at the same rate, an equilibrium is set up between the two phases (liquid and vapour). This is physical process. Further, it is dynamic process, since under the given set of conditions, both the forward and the reverse process continue to take place but with equal rate. Such a situation can be expected between any two phases of the same substances provided both the phases occur under the given set of conditions and no phase disappears in the process. For example Solid Solid Liquid

     

liquid Vapour Vapour

melting or fusion Sublimation Vapourzation

Similar equilibrium may occur between two different allotropic forms of the substance.  a - sulphur ↽ ⇀  β - sulphur

9.2.1 Solid–Liquid Equilibrium When a pure solid substance is heated, it starts changing into liquid at a certain temperature. At this temperature, the solid and liquid states of the substance coexist under the given conditions of pressure. For any substance at atmospheric pressure, the temperature at which the solid and liquid states can coexist is called the normal melting point or the normal freezing point of the substance. At melting point the solid substance is in equilibrium with liquid state of the substance. If heat energy is added to a mixture of solid and liquid at equilibrium, the solid is gradually converted to liquid while the temperature remains constant. If a solid – liquid system at melting point is taken in a well-insulated container then this constitutes a system in which solid is in dynamic equilibrium with liquid. For example, ice and water kept in a perfectly insulated thermos flask at 273K and the atmosphere pressure are in equilibrium state. In this system, the mass of ice and water do not change with time and the temperature remains constant. However, the equilibrium is not static. The molecules from the liquid water collide against ice and adhere to it and some molecules of ice escape into liquid phase.

Chemical Equilibrium 9.3

There is no change of mass of ice and water; as the rates of transfer of molecules from ice into water, and reverse transfer from water into ice are equal at atmosphere pressure and 273 K. Thus at equilibrium Rate of melting = Rate of freezing

9.2.2 Liquid–Vapour Equilibrium If a sample of water in its solid or liquid phase is placed in an empty container some of it will vapourise to form gaseous water. This change is called evaporation. For the liquid or sublimation for the solid. We can write an equation for evaporation occurring spontaneously as if it were a chemical equation H2O (l) → H2O (g) Since some weak intermolecular bonds in the liquid water break to form the gas, we can consider this a simple kind of a chemical reaction. This reaction can also be reversed. If water vapour is compressed into a smaller volume, spontaneous condensation occurs and droplets of liquid water appear. This reaction is expressed as H2O (g) → H2O (l) Both of these reactions can occur at the same time. If the rate of evaporation equals the rate of condensation, the amount of liquid water and of gaseous water will not change with time. This special circumstance is called phase equilibrium and the liquid plus gaseous water together form an equilibrium system. The equations for the equilibrium system are

of liquid to achieve the same pressure as would be obtained if the volumes were equal. However, it still must be true. If the equilibrium system also includes air or another gas, the vapour pressure represents the partial pressure of the gaseous form that must be represent to give an equilibrium system. Usually, we shall assume that all vapour pressures are independent of the total pressure on a liquid or a solid. However, as we discussed (in chapter 4) osmotic pressure is exactly a result of the effect of total pressure on vapour pressure. Vapour pressure is very strongly dependent on temperature. Solved Problem 1 A stream of nitrogen gas is bubbled through water at 25.0°C. The water is open to the air where the barometric pressure is 759 torr. The N2 bubbles are very small and rise slowly to the surface of the water. What is the composition of these bubbles as they just reach the surface? Solution: Assume that the bubbles reach equilibrium with the liquid water at 25.0°C.  H2O (g) VP = 23.8 torr H2O(l)  At the surface the total pressure on the bubbles equals the barometric pressure. From Dalton’s law of partial pressures, we have Pt = PH2O + PN2 = 759.0 torr Where PH2O = 23.8 torr Therefore PN2 = 759.0 – 23.8 = 735.2 torr The mole fractions of H2O and N2 in the bubbles are

 H2O (g) H2O(l)   H2O (l) Or H2O(g) 

χ N2 =

In order to have an equilibrium system, the pressure of gaseous water must be equal what is called the vapour pressure of the liquid or solid beneath it. At 25°C, H2O (l) will form an equilibrium system with H2O (g) if the pressure of H2O (g) is 23.76 torr. If H2O (g) at -5°C has a vapour pressure of only 3.01 torr, it will be in equilibrium with any H2O(s) in the system. Many liquids have vapour pressures much higher than water has, and they are said to be volatile. Some vapour pressures are very low, and non-volatile solids do not seem to evaporate at all. The pressure necessary to achieve equilibrium does not depend on the amount of solids or liquid that is present or on the volume of the container. The only requirement is that some amount of both phases be present in the final equilibrium system. For this reason, it is difficult to think of this equilibrium in terms of equal rates. It is not easy to imagine how the condensation rate of a large volume of gas could always adjust to the rate of evaporation from a small volume

χH2O =

PN2 Pt

PH

2O

Pt

=

735.2 = 09686 759.0

=

23.8 = 0.0314 759.0

Solved Problem 2 A 20.0 litre bulb is thermostated at 25°C. After being evacuated 1.00 g H2O is injected into the bulb. What will be the pressure in the bulb and how many grams of H2O will remain as liquid? Solution: If some liquid H2O remains in the bulb, the pressure in the bulb would be the vapour pressure of H2O at 25°C or 23.8 torr. If no liquid H2O remains the pressure would be calculated as the ideal gas pressure for 1.00 g of gaseous H2O in 20.0 L bulb at 25°C. Working this kind of problem requires making one of these two assumptions and then proceeding until obtaining a contradiction. If no contradiction is found, the assumption was correct.

9.4

Chemical Equilibrium

Let us assume that some liquid water remains. We now calculate the amount of water in the vapour using the ideal gas law: PV (23.8 / 760) (20.0) = n= RT (0.0821) (298.15) = 2.56 × 10–2 mol Weight of water in the vapour = 18.0 × 2.56 ×10–2 Since this is less than our original 1.00 g our assumption was valid and weight as liquid = 1.00 – 0.46 = 0.54 g If we had made the other assumption, we would have found a pressure for H2O greater than its vapour pressure. This is not possible because the reaction H2O (g) → H2O (l) would then take place.

Energetics of Phase changes Anyone who has stepped from a swimming pool into a brisk freeze knows that when water evaporates, it absorbs heat from its surroundings (the skin in this case). The same effect may be experienced with any other volatile liquid. Some liquids, such as ethyl chloride, can freeze the skin upon evaporation and so are used as local anaesthetics. It is also well known that when a gas condenses to liquid it releases heat to surroundings. The absorption of heat upon evaporation and the release of heat upon condensation are direct demonstrations that the energy of a liquid is lower than that of a gas at the same temperature. In order for a liquid to evaporate, work must be done against the attractive forces between the molecules and this requires that energy be supplied as heat from the surroundings. Conversely, when vapour condenses the system goes to a state of lower energy and this energy is transferred as heat from the system to its surroundings. Now it is possible to draw from our discussion of liquid- vapour equilibrium some very useful generalizations that apply to all situations of physical or chemical equilibrium. The potential energy of molecules in the liquid state is lower than that of molecules in the gas phase. The liquid left undisturbed in a closed container inevitably moves towards a state of equilibrium in which there is a definite concentration of molecules in the vapour phase. Thus it follows that at equilibrium, the system is not in a condition of minimum energy, for the energy of the system can always be lowered by condensing the vapour entirely. This conclusion may seem strong for all our experience with simple mechanical systems suggest that they seek an equilibrium condition in which their energy is as low as possible objects fall, clocks run down and a stirred liquid

stop moving. All these phenomena can be summarized by saying that mechanical system seeks a resting place of minimum energy Minimum Energy: Almost the same thing can be said about molecular systems. It is clear that one of the driving force that determines the molecular system is the tendency to seek a state of lowest possible energy. After all, this reason that a gas condense or that a liquid freezes. It is also certain that the tendency toward minimum energy cannot be the only factor governing the behaviour of molecular systems; if it were no gases would exist at any temperature there is another driving force, just as important as the energy factor. To put it briefly, it is the tendency of the systems to assume a state of maximum molecular chaos or disorder i.e., increases in entropy. The entropy of a liquid is greater than of a solid and that entropy of a gas is greater than that of either a liquid or solid. The natural tendency of systems to reach a state of molecular chaos can be emphasized as that the systems have a tendency to reach a state of maximum entropy. To summarise our discussion, we can list four important features of all equibria that can be illustrated by the liquid–vapour equilibrium. 1. Equilibrium in molecular systems is dynamic and is a consequence of equality of the rates of opposing reactions. 2. A system moves spontaneously toward a state of equilibrium. If a system initially at equilibrium is perturbed by some change in its surroundings, it reacts in a manner that restores it to equilibrium. 3. The nature and properties of an equilibrium state are the same regardless of how it is reached. 4. The condition of a system at equilibrium represents a compromise between two opposing tendencies; the drive for molecules to assume the state of lowest energy and the urge toward molecular chaos or maximum entropy.

9.2.3 Solid–Vapour Equilibrium Like a liquid, a solid can exist in equilibrium with its vapour in a closed container. Thus at any fixed temperature each solid has characteristic fixed vapour pressure. The vapour pressure of a solid increases with increasing temperature. The vapour pressure of the solid increases more rapidly as temperature is raised than does the vapour pressure of the liquid since the solid has the greater heat of vaporization. Consider the apparatus shown in Fig 9.1 one bulb contains a solid the other its liquid and the two bulbs are connected so that vapour can pass freely from one to the other. Now let both bulbs be immersed in a bath at temperature T1. If the vapour pressure of the solid at T1 is

Chemical Equilibrium 9.5

less than that of the liquid, then gas will flow from the bulb containing the liquid to that containing the solid. As this flow persists, the liquid evaporates and the vapour condenses as solid. This continues untill all the liquid is evaporated.

Solid

Liquid

Fig 9.1 Apparatus for the attainment of equilibrium of two phases which are not in contact. Alternatively, if the apparatus is held at a temperature T2 at which the vapour of the solid is greater than that vopour flows from the solid to the liquid. This is accompanied by sublimation of the solids and condensation of the vapour to the liquid and the process continues until all the solid is consumed. Clearly, the solid and liquid are not in equilibrium with each other at temperature T1 or T2 for if they were in equilibrium, the system would not change and both phases would remain indefinitely. In contrast, if the temperature of the apparatus is set at T0 at which the vapour pressure of the liquid and solid are the same, the pressure is uniform and there is no tendency for vapour to flow from one chamber to the other. Thus both the liquid and solid phases remain indefinitely. This persistence of the state of the system indicates that the solid and liquid phases are in equilibrium at a temperature which their vapour pressures are the same. This situation is a specific example of an important general principle; If each of two phases is in simultaneous equilibrium with a third then the two phases are in equilibrium with each other. The temperature, at which liquid, vapour and solid are in simultaneous equilibrium with one another, is called the

triple-point temperature. The triple point is usually very close to what is known as the freezing point which is the temperature at which liquid, vapour and solid are in simultaneous equilibrium in the presence of 1 atm of air pressure. For example, liquid water and ice are simultaneously in equilibrium with water vapour only at a temperature of 0°C, in the presence of air at 1 atm of pressure. If air is completely eliminated from the container then water, ice and water vapour will be simultaneously in equilibrium only at a temperature of 0.01°C. This temperature is the triple point of water and we see that it differs only slightly from normal freezing point. When a vapour is cooled, under certain conditions it will condense directly to a solid. This will happen when the pressure of the vapour is less than the vapour pressure of the solid at the melting point. Thus if the partial pressure of water vapour in the atmosphere is less than 610 Pa (which is vapour pressure of ice at 0°C), then cooling will result in the direct formation of ice. This is the way in which frost formed in a cold night. Conversely, solids which have high vapour pressures will evaporate at a measurable rate, the solid being transformed directly into vapour. This process is termed sublimation. For example, if solid iodine is placed in closed vessel after some time the vessel gets filled up with violet vapour and the intensity of the colour increases with time. After certain time the intensity of the colour becomes constant and at this stage equilibrium is attained. Hence, solid iodine sublimes to give iodine vapour and the iodine vapour condense to give solid iodine. The equilibrium can be represented as  I 2 (vapour ) I 2 ( solid )  Other examples showing this kind of equilibrium are  Camphor ( solid ) ↽ ⇀  Camphor (Vapour )   ⇀ NH 4 Cl ( solid ) ↽ NH 4 Cl (Vapour )

9.2.4 Equilibrium Involving dissolution of Solid or Gases in Liquids Solids in Liquids When a solid is shaken up with a liquid at a constant temperature after a time a point is reached when no more solid will dissolve. The solution is then said to be saturated. We have here an example of a state of equilibrium, where the rate at which solid is dissolving is equal to the rate at which material crystalises out again from the solution. The equilibrium is not affected by the amount of solid present, Since if a small area of surface is considered, the rate at

9.6

Chemical Equilibrium

which molecules escape in that area of surface which remains constant, so that Rate of dissolution = Constant While the rate at which molecules return to this portion of the surface of the solid is proportional to the concentration of those molecules in the solution (C) hence Rate of recrystallisation = Constant × C At equilibrium Rate of dissolution = Rate of crystallization C = Constant The concentration of the solution is therefore constant as long as an excess of solid is present. It follows that a saturated solution is a solution that is in equilibrium with an excess of solid at a given temperature. Equality of the rates of dissolution and crystallization and dynamic nature of equilibrium has been confirmed by adding radioactive sugar, then after some time radioactivity is observed both in the solution and in the solid sugar. Initially, there were no radioactive sugar molecules in the solution but due to dynamic nature of equilibrium, there is exchange between the radioactive and non-radioactive sugar molecules between the two phases. The ratio of the radioactive and non-radioactive molecules in the solution increases till it attains a constant value. Gases in Liquids The rate of dissolution of the gas molecules depends on the rate at which they strike the surface of the liquid, and therefore on the pressure of the gas (P) Rate of dissolution = Constant × P The rate at which dissolved molecules escape from the liquid (i.e., evaporate) will be proportional to the concentration of the solution (c). Rate of evaporation = Constant × C When the gas is in equilibrium with a saturated solution at constant temperature these two rates will be equal, so that C = KP Where K is constant. For example, when a soda water bottle is opened, the carbon dioxide dissolved in it, fizzes out rapidly. This represents an equilibrium situation. At a given pressure, there is an equilibrium between the molecule of the solute in gaseous state and the molecules dissolved in the liquid. This may be expressed as  CO2 (in solution) CO2 (gas) 

Effect of pressure on the solubility of a gas in liquid is given by Henry’s law. This law states that: The mass of a gas dissolved in a given mass of a solvent at a given temperature, is directly proportional to the pressure of the gas above the solvent. Since the soda water bottle is sealed when the gas is at high pressure, there is quite appreciable amount of CO2 dissolved in water and the gas pressure above the solution is high. When the bottle is opened, the pressure of gas above the solution decreases, therefore the dissolved gas escapes (fizzes out) to attain new equilibrium state. Form the discussion made so far involving equilibrium in physical process, the following conclusions can be made:  liquid equilibrium there is only one 1. For solid  temperature (melting point) at 1 atm at which the two phases can coexist. If there is no exchange of heat with the surroundings, the mass of the two phases remain constant.  vapour equilibrium the vapour pres2. For liquid  sure is constant at a given temperature. 3. For dissolution of solids in liquids, the solubility is constant at a given temperature. 4. For dissolution of gases in liquids, the concentration of a gas in liquid is proportional of a gas in liquid is proportional to the pressure of the gas over the liquid.

9.2.5 General characteristics of Equilibrium Involving Physical Process The common characteristics to the system at equilibrium involving physical process discussed so far are summarized as follows: 1. Equilibrium is possible only in a closed system at a given temperature. 2. Both the opposing processes occur at the same rate and there is a dynamic but stable condition. 3. All measurable properties of the system remain constant. 4. When equilibrium is attained for a physical process, it is characterized by constant value of one of its parameters at a given temperature. 5. The magnitude of such quantities at any stage indicates the extent to which the reaction has proceeded before reaching equilibrium.

9.3 chEmIcaL EquILIbrIa If a sample of radioactive matter (not subject to regeneration from another source) is observed for a time sufficiently long, it will be found that the last detectable portion has finally undergone disintegration. The original atoms are all

Chemical Equilibrium 9.7

NaCl + AgNO3  → AgCl + NaNO3 Similarly, when a piece of sodium is added to water, a violent reaction occurs resulting in the formation of sodium hydroxide and hydrogen gas.

C &D

Concentration

destroyed, and it may be said that the reaction has gone to completion. Radioactivity is, in this sense, an irreversible transformation of matter. There are many chemical reactions in which the initial substances are almost completely converted into the products. For example, when equivalent solutions of sodium chloride and silver nitrate are mixed together, the reaction takes place almost instantaneously and equivalent quantity of AgCl is precipitated.

A&B

2Na(s) + 2H2O (l) → 2NaOH (aq) + H2(g) However, it is not possible to carry out the reverse reaction under any known experimental condition, i.e., reduction of aqueous sodium hydroxide by hydrogen to form sodium and water cannot be achieved. All the above process take place unidirectional. Such processes are known as irreversible processes. Those chemical reactions which proceed only in one direction i.e., in which the reaction products do not recombine back to form original reactants are known as irreversible reactions. The reaction between zinc and dilute sulphuric acid to evolve hydrogen and decomposition of potassium chlorate into potassium chloride and oxygen are other well known examples of irreversible reactions. → ZnSO4 + H2 ↑ Zn + H2SO4   → 2KClO3 2KCl + 3O2 ↑ But there are numerous chemical reactions in which the reaction products recombine to form the initial reaction. Earlier in this unit (9.1.1) it was discussed that what is a reversible reaction and now it arrives to equilibrium position by taking the reaction between hydrogen and iodine as example. For a better comprehension, let us consider a hypothetical reversible reaction  C+D A +B  in which all reactants and all products are gases. Quantities of A and B are placed in a container under conditions such that reaction proceeds with measurable velocity. The rate of reaction between A and B will, according to the law of mass action, be proportional to the concentration of A multiplied by the concentration of B. At the start of the experiment, the rate of the reverse reaction will be zero because the concentrations of C and D are zero.

Time

Equilibrium

Fig 9.2 Attainment of chemical equilibrium. As the reaction proceeds the concentrations of both A and B will become progressively less, and the rate of the reaction between the substances will diminish. At the same time, however, the concentration of C and D will increase as these substances are formed through the reaction of A and B. The rate of reaction between C and D increases. Sooner or later the increasing rate of reaction between C and D will equal the diminishing rate of reaction between A and B. After the opposing rates have become equal the concentrations of all four substances will remain constant and equilibrium will have been reached. At equilibrium, not only will the concentration of all reactants and products remain constant, but the pressure will also remain constant, if the forward reaction (the one written toward the right) happens to be exothermic, the reverse reaction will be endothermic. At equilibrium the heat liberated by the forward reaction will be absorbed by the reverse reaction, as a consequence of which the temperature of the reacting system will also remain constant. These outward evidence of inertness must not obscure the fact that both forward and reverse reactions are continuing even though equilibrium has been reached and the concentrations of reactants and products do not change. Proof that the forward and reverse reactions continue at equilibrium may be obtained with the aid of hydrogen isotope. Suppose that the reaction  2NH3 N2 + 3H2  has reached equilibrium and that now a little deuterium gas D2 is added to the system. Almost at once it will be found that the hydrogen in the reaction mixture consists

9.8

Chemical Equilibrium

of H2, D2 and HD molecules and that the ammonia consists of NH3, NH2D, NHD2 and ND3 molecules. These several molecular species could not have been formed if all chemical reactions had stopped. It will be noted that chemical equilibrium bears some resemblance to the processes that established equilibrium in physical processes. It is not be supposed that equilibrium is reached when all substances are present in equal concentration. The relative amounts and concentrations of each substance present at equilibrium may be very different for different reaction, and for the same reactions under different conditions. It may also be mentioned that it is possible to have metastable equilibrium in physical and chemical processes. In certain circumstances, a liquid may be cooled bellow the freezing point without solidification taking place. If water is cooled, without shaking, it may often be taken down to a temperature of 5 or even 10 degree below 0°C, the normal freezing point of water. This effect is known as super cooling. It is shown by many liquids. If now the water is vigorously stirred, or if small piece of ice is added to it, some of water sample will solidify almost instantly. Super cooled liquids are said to be in a metastable state. The reader may wonder if solids may similarly be heated somewhat above the melting point without melting taking place. The reverse of super cooling, does not appear to take place, but sometimes may be heated well above the boiling point without boiling occurring. This is especially true of liquids from which all minute bubbles of air have been removed by centrifusing or by brief application of very high pressure. Such type of equilibrium is called metastable equilibrium and can be represented as  vapour Super cooled water  Thus a system is said to be in metastable equilibrium if on being disturbed by some way, it undergoes a drastic change to some new state. The disturbing factors may be introduction of a catalyst, or heating or cooling. The reaction between hydrogen and oxygen offers an example for metastable equilibrium in chemical reactions. These two gases may be kept mixed together without reactions. For an indefinite time provided, the temperature is not raised and no catalyst is present to help surmount, or lower, the activation energy barrier. Under circumstances, it might erroneously be concluded that equilibrium had been reached. But let spark reach the mixture, or let a few specks of powered platinum be introduced, and the mixture will reach the true equilibrium in a fraction of a second and it will do so with explosive violence? The true equilibrium for this reaction at room temperature lies so far in the direction of the product water that the reaction may be said to go effectively to completion.

One criterion of a true chemical equilibrium is that it may be approached from either side. One may for instance with nitrogen and hydrogen or one may start with ammonia. If a true equilibrium, rather than a false or metastable equilibrium, has been reached it will be found that the relative concentrations of all reactants and all products are the same in either case, provided, of course, that all conditions are maintained the same.

9.3.1 characteristics of chemical Equilibrium 1. Chemical equilibrium is dynamic in nature. In other words, at equilibrium, the forward and backward reaction proceeds simultaneously at equal speeds. 2. Chemical equilibrium can be approached from either direction. 3. The equilibrium can be attained only if the system is closed. In the open system, the products may escape from the container and, therefore backward reaction may not take place. 4. A catalyst does not alter the equilibrium point. In a reversible reaction, catalyst increases the rate of forward as well as backward reaction to the same extent. So equilibrium point is not altered, but the equilibrium is attained fastly. 5. The equilibrium, at constant temperature, is characterized by constancy of certain observable properties, such as pressure, concentration, density or colour. 6. The equilibrium readjusts with the changing conditions and spontaneously goes back to the original state when the disturbing factors are removed.

9.3.2 Limitations of the Equation for chemical Equilibrium 1. The equation for chemical equilibrium gives no idea about how far the reaction proceeds from left to right or from right to left. For example  2SO2 + O2 2SO3  It is evident from the equation that what fraction of SO3 has dissociated into SO2 and O2. The dissociation of SO3 increases with increase in temperature, but still there is no change in equation. It remain the same even if temperature is raised by say 10° or more 2. The equation for the chemical equilibrium also fails to explain how long it will take for a reaction to attain equilibrium. For example, in the reaction  H2(g) + H2O(g) 

1 O2(g) 2

Chemical Equilibrium 9.9

The equilibrium lies very much to the left. Thus hydrogen and oxygen combine readily to establish the above equilibrium. However it has been found that a mixture of hydrogen and oxygen may remain as such for long without reacting to form water. Thus it is not clear from the above equation that how long it would take to establish the equilibrium.

The term active mass used in the above statement implies molecular concentration. In the case of gases and solution it means the number of gm molecules present per litre. The active mass of solids is taken as unity. The active mass is usually expressed by enclosing the symbol or the formula of the substance in square brackets. For example [HCl] =means 36.5 gms of HCl per litre.

9.3.3 types of chemical Equilibria

9.4.1 application of Law of mass action

Chemical equilibria are classified as homogeneous and heterogeneous equilibria depending on the physical states of the reactants and products (a) Homogeneous equilibria: A homogeneous equilibrium is said to be established in a system when only one phase is present in a system i.e., gases or a simple liquid or a solid phase for Example 2SO 2 O2  2SO3 +  gas gas gas

Consider the following reversible reaction taking place at constant temperature,

N 2 + 3H 2  2NH 3  gas gas gas

 CH3COOH + C2 H5 OH ↽ ⇀  CH3COOC 2 H5 + H 2 O soln soln soln liquid Are all homogeneous reversible equilibria (b) Heterogeneous equilibria: A heterogeneous equilibrium is said to be established in a system when different phases are present in the system. For example, reactions 3Fe  Fe3 O4 + 4H 2 ↑ + 4H 2 O  solid gas solid gas CaCO3  CaO + CO2  solid gas solid consist of solid and gas phases and are therefore heterogeneous reversible equilibria.

9.4 Law oF maSS actIon–EquILIbrIum conStant Based on equilibrium observations, the Norwegian chemists’ Cato Maxmillan Guldberg and Peter Waage in 1864 investigated the effect of concentration or mass on the rate of chemical reaction and put forward a law, known as law of mass action. According to this law, The rate at which a substance reacts is directly proportional to its active mass and the rate at which chemical reaction proceeds is proportional to the product of the active masses of the reacting substances.

 C+D A+B 

(1)

The active mass or molecular concentrations of A, B, C, and D are [A], [B], [C] and [D] respectively. According to law of mass action Rate of combination of A and B i.e., rate of forward reaction ∝[A] × [B] = Kf [A] [B] (2) Where Kf is the rate constant for forward reaction Similarly, rate of formation of A and B i.e., rate of backward reaction ∝[C] [D] = Kf [C][D] A dynamic equilibrium is attained when the rate of forward reaction becomes equal to the rate of backward reaction. Thus at equilibrium Kf [A][B] = Kb [C][D] Or

Kf Kb

= Kc =

[C][D] [A][B]

Kc is known as the equilibrium constant. The subscript ‘C’ indicates that Kc is expressed in concentration of mol L–1. At a given temperature, the product of concentrations of the reaction products raised to the respective stoichiometric coefficient in the balanced chemical equation divided by the product of concentrations of reactants raised to their individual stoichiometric coefficients has a constant value. This is known as the equilibrium law or law of chemical equilibrium. The equilibrium constant for a general reaction  cC + dD aA + bB  is expressed as Kc=

[C]c [D]d [A]a [B]b

Where [A], [B],[C] and [D] are the equilibrium concentrations of the reactants and products. For gaseous reactions, it is more convenient to use partial pressures instead of concentrations, since partial pressure of a substance is proportional to its concentration

9.10

Chemical Equilibrium

in the gas phase at a fixed temperature. The equilibrium constant for a general gas reaction

The units of Kc depends upon the molar concentrations represented in equilibrium constant equation and on the difference in the number of moles of the products and reactants

 cC +dD aA+ bB  Kp =

9.4.4 units of Kc and Kp

[ PC ]c [ PD ]d [ PA ]a [ PB ]b

Where Kp is the equilibrium constant in terms of partial pressure.

9.4.2 characteristics of Equilibrium constant 1. Equilibrium constant has a constant value at a given temperature but changes with change in temperature. 2. At a given temperature and pressure, its value remains constant irrespective of concentration or pressure of the reactants and products. 3. If the concentration or pressure change, the equilibrium shifts but equilibrium constant does not change. 4. The value of the equilibrium constant does not change with the presence of catalyst. 5. At a given temperature, the equilibrium constant denotes to which extent the reaction proceeds and in which direction. If Kc>1; forward reaction takes place more than backward reaction; if Kc< 1 backward reaction takes place more than forward reaction. If Kc = 1 both forward and backward reactions proceed equally.

Units of Kc = ( Units of Concentration ) ∆n = Total number moles of products – Total number of moles of reactants The units of Kp depends upon the units of pressure expressed in equilibrium constant equations and difference in the moles of products and reactants Dn

Units of Kp = (Units of Pressure) Dn

9.4.5 relationship between Kc and Kp Consider a reversible gaseous reaction

 m1C + m2D n1A + n2B  Applying law of mass action, we have Kc =

The equilibrium constant Kc is characteristic of a particular reaction is independent of the following factors: 1. The actual quantities of substances involved in the reaction. 2. The presence of inert materials. 3. The presence of a catalyst. 4. The direction from which the equilibrium state is established. Some experimental data showing that the equilibrium constant is independent of the concentrations of the individual reactants are shown in table 9.1.

table 9.1 Some equilibrium concentrations and  CO+H2O (g) constant for the reaction CO2+H2  equilibrium concentrations (mole litre–1) Kc CO2

H2

CO

H2O(g)

0.54 0.12 0.95 0.18

0.34 0.10 0.13 1.05

0.65 0.087 0.51 0.81

0.46 0.22 0.39 0.38

1.6 1.6 1.6 1.6

[C ]m1 [ D]m2

(2)

[ A]n1 [ B ]n2

Using partial pressures in place of molar concentration

Kp =

9.4.3 Factors Influencing Equilibrium Constant

(1)

[ PC ]m1 [ PD ]m2

(3)

[ PA ]n1 [ PB ]n2 Now for an ideals gas PV = nRT or P =

n RT V

or P = CRT where C =

n V

(4)

Substituting the value of P in equation (3) we have Kp =

[CC RT ]m1 [CD RT ]m2 [C A RT ]n1 [CB RT ]n2

(5)

Kp = Kc [ RT ]( m1 + m2 ) - ( n1 + n2 )

(6)

Or Kp = Kc [ RT ]Dn

(7)

Where ∆n = No. of moles of products – No moles of reactants. When a reaction take place with increase in volume i.e., (m1 + m2) > (n1 + n2), Kp is numerically greater than Kc. For example,

 PCl3 + Cl2 PCl5  ∆n = 1 + 1 - 1 = +1, the reaction takes place with the increases in volume thus

Chemical Equilibrium 9.11

Kp = Kc [RT]1 Or Kp = Kc RT

(8)

Similarly, when reaction takes place with a decrease in volume i.e., (m1 + m2) < (n1 + n2), Kp is numerically less than Kc. For example, for the reaction

 2SO3 O2 + 2SO2  ∆n = 2-(2 + 1) = -1 So Kp = Kc [RT]–1

1 1  H 2 + I 2  HI 2 2 The equilibrium constant for the above reaction is given by 1/ 2

  [ HI ]2 [ HI ] =  1/ 2 1/ 22 1/ 2 1/ 2  [H 2 ] [I2 ] [ H 2 ] [ I 2 ] 

KC = 2

(9)

When a reaction takes place with no change in volume i.e., (m1 + m2) = (n1 + n2), Kp is equal to Kc. For example,

 H2 + I2 2HI  ∆n = 2-2 = 0 \Kp = Kc[RT]0 Or Kp = Kc

(10)

1/2

= K C1 or

KC

1

Or K C = [ K C ]2 1

2

Similarly, for other reactions also the Kc and Kp values can be represented as follows: K 1 1 C1  (i) H2+I2  2HI K C = or K C22 = 2 K K C 1

C1

C2 1 1  HI  H2 + I2 2 2

K

9.4.6 change in the Values of Kc and Kp with the change in the Form of chemical Equation The reciprocal of the equilibrium constant value of a chemical reaction in one direction will be the equilibrium constant for the same reaction in the reverse direction  2HI (i) H2+I2  Kc1 =

[ HI ]2 [ H 2 ][ I 2 ]

1

1 KC

K 1 3 C2  N 2 + H 2  NH 3 K P = K P or 2 1 2 2

1

K

C1 (iii) N2+3H2  2NH3 K C2 = 

2

K P 1 = K P2

2

2 or K C2 =

KC

1 KC

1

KP

2 or K P2 =

K

C2 (iv) 2SO2+O2   2 SO3 K C2 =

1 KC

or K C2 = 2

1 KP

1 KC

1

1

C2

1 KP

1

1

SO3   SO2 + 1 O2 K = P2 2

or K P2

2

1 = KP

1

1

2

1 C1 1  N2+3H2  2NH3 K C1 = or K C = 2 KC KC 2 K

1

C2 2NH3   N 2 + 3H 2

1

K C2 1 3  N2 + H 2 KP = NH3  2 2 2

K

(ii) Similarly, for the synthesis of ammonia and for its decomposition the Kc values are related as

K

2

1

1

The equilibrium constant for the reverse reaction  H2+I2 2HI  at the same temperature is [ H ][ I ] K C = 2 22 2 [ HI ] Thus K C =

K

C1  (ii) N2+3H2  2NH3 K C2 = K C1 or K C = K C

KP = 1

1 1 or K P2 = K KP P1 2

If we change the stoichiometric coefficients in a chemical equation by multiplying throughout by a factor then the equilibrium constant also changes. For example, if the reaction  2HI H2+I2  is written as

9.4.7 heterogeneous chemical Equilibria If a system in equilibrium contain more than one phase, it is called heterogeneous equilibrium. Chemical reactions, like phase changes, are reversible. As a consequence, there are conditions of concentration and temperature under which reactants and products exist together at equilibrium. Consider the thermal decomposition of calcium carbonate CaCO3(s)  → CaO(s)+ CO2 (g)

(1)

By carrying out this reaction in an open vessel that allows CO2 to be swept away complete conversion of CaCO3 to CaO can be effected. On the other hand, it is

9.12

Chemical Equilibrium

well known that CaO reacts with CO2 and if pressure of CO2 is high enough the oxide can be completely converted to the carbonate. CaO(s) +CO2(g)  → CaCO3 (s)

(2)

This is of course, just the reverse reaction of (1) thus we must look on reactions (1) and (2) as reversible chemical processes, a fact that we denote by the following notation  CaCO3(s) ↽ ⇀  CaO(s)+ CO2 (g)

[CaO( s )][CO2 ( g )] [CaCO3 ( s )]

The molar concentration of pure solids and liquids are constant i.e., independent of the amount present. So the modified equilibrium constant for thermal decomposition of calcium carbonate will be Kc = [CO2 (g)] Or Kp = PCO2 This shows that at a particular temperature there is a constant concentration or pressure of CO2 in equilibrium with CaO(s) and CaCO3(s).  NH4HS  NH3 + H2S Solid gas gas kp = PNH3 × PH2S The active mass (molar concentration) of NH4HS(s) is taken as constant. In the equilibrium between nickel, carbon monoxide and nickel carbonyl (used in the purification of nickel by Mond’s process)

 Ni (CO)4 (g) Ni(s) + 4CO (g)  The equilibrium constant is Kc =

[ Ni (CO) 4 ] [CO]4

 2AgNO3(aq) + H2O (l) Ag2O(s)+2HNO3(aq)  Kc =

(3)

This chemical system is closely analogous to the ‘physical’ system consisting of a condensed phase and its vapour. Just as a liquid and its vapour come to equilibrium in a closed container there are certain values of the temperature and pressure of CO2 at which CaCO3, CaO and CO2 remain indefinitely. When pure CaCO3 is in closed vessel, it begins to decompose according to reaction (1). As the CO2 accumulates the pressure increases, and eventually reaction (2) begins to occur at noticeable rate, a rate that increases as the pressure of CO2 increases. Finally, the rates of the decomposition reaction and its reverse becomes equal, and the pressure of carbon dioxide remains constant. The system has reached equilibrium on the basis of the stoichiometric equation (3) can derive Kc value. Kc =

It should be noted that in heterogeneous equilibria pure solids or liquids must be present (whatever small it may be) for the equilibrium to exist, but their concentration or partial pressure do not appear in the expression of the equilibrium constant. For example [AgNO3 ]2 [HNO3 ]2

9.5 aPPLIcatIonS oF EquILIbrIum conStantS Equilibrium constant is useful in 1. Predicting the extent of reaction basing on its magnitude 2. Predicting the direction of the reaction 3. Calculating the equilibrium concentrations.

9.5.1 Predicting the Extent of reaction The magnitude of equilibrium constant K indicates the extent to which a reaction can proceed. In other words, it is a measure of the completion of a reversible reaction. Larger the value of Kc, the greater will be the equilibrium concentration proceeds to a greater extent. For example, consider the reaction  N2(g) + 2O2(g) 2NO2(g)  The equilibrium constant for the reaction at 298 K is Kc=

[N 2 ][O 2 ] = 6.7 × 10–16 mol L–1 [NO 2 ]2

The value of Kc is very small which means that the molar concentrations of N2 and O2 in the equilibrium mixture are very small. In other words, the forward reaction has proceeded to small extent only. Thus we can say that NO2 is quite stable and decomposes only slightly. Let us consider another reaction  2CO2(g) 2CO(g)+ O2(g)  The equilibrium constant for the reaction at 1000 k is [CO 2 ]2 = 2.3 × 1022 mol L–1 K= [CO]2 [O 2 ] For this reaction, the value of K is very large which means that the molar concentration of CO2 in the equilibrium mixture is quite large and the molar concentrations of CO and O2 are very small. Thus the reaction goes almost to completion and it may be concluded that carbon dioxide is more stable than carbon monoxide and oxygen under these conditions. Similarly, for the reaction  2HBr(g) H2(g) + Br2(g) 

Chemical Equilibrium 9.13

The value of equilibrium constant at 300 K is [PHBr ]2 Kp = = 5.4 × 1018 [PH ][PBr ] 2

2

The large value of Kp indicates that the concentration of the product, HBr is very high and the reaction goes nearly to completion. Similarly, equilibrium constant for the reaction  2HCl(g) at 300 K is very high H (g) + Cl (g)  2

2

Kc=

[HCl]2 = 4.0 × 1031 [H 2 ][Cl2 ]

This means that the reaction goes almost to completion. Thus large values of Kp or Kc (large than about 103) favour the products more. For the intermediate values of K (approximately in the range of 10–3 to 103 the concentrations of reactants and products are comparable. Smaller values of K (less than 10–3) favours the backward reaction (reactants) more. For example, for the reaction  2NO (g) N2(g) +O2(g)  Kc=

[NO 2 ]2 = 4.8 × 10–31 [N 2 ][O 2 ]

the very small value of Kc indicates that the reactants N2 and O2 will be predominant in the reaction mixture at equilibrium. Similarly, two other similar reactions can be compared  Cu2+(aq) +2Ag(s) (i) Cu (s) +2Ag+ (aq) 

[Cu 2+ (aq)] KC = = 2.0×1015 at.298K [Ag + (aq)]2 ( [Cu(s)]and[Ag(s)] = 1)  Zn (s) + Cu2+ (aq) (ii) Cu(s) + Zn2+(aq) 

9.5.2 Predicting the direction of the reaction The equilibrium constant also helps to find the direction in which an arbitrary reaction mixture of reactants and products will proceed. For this purpose, the reaction quotient Q should be calculated. The reaction quotient is the ratio of product of concentration (or partial pressure) of products to that of the reactants at any stage of the reaction. In other words, the reaction quotient is defined in the same way as the equilibrium constant with molar concentration to give Qc or with partial pressure to give Qp  cC + dD aA+bB  Qc =

[PC ]c [PD ]d [C]c [D]d or.Q = P [A]a [B]b [PA ]a [PB ]b

At equilibrium Qc = Kc Thus If Qc > Kc, the reaction will proceed in the direction of reactants (reverse reaction). If Qc < Kc the reaction will proceed in the direction of products (forward reaction). If Qc = Kc the reaction mixture is already at equilibrium. Thus a reaction has a tendency to form products if QK. For example, in the reaction  2HI(g) H2(g) + I2(g)  The molar concentration of H2, I2 and HI are 0.1, 0.2 and 0.2 mol L–1 respectively at 783 K. The reaction quotient at this stage of the reaction is [HI]2 [0.4]2 = Qc = [H ][I ] [0.1][0.2] = 8 2 2

[Cu 2+ (aq)] = 2.0×10-19 at.298K [Zn 2+ (aq)]2 ( [Cu(s)]and[Zn(s)] = 1) KC =

The comparison of K values of two reactions shows that the second reaction is not possible. In other words, Ag will be precipitated by Cu when copper rod is dipped in a solution containing Ag+ ions, but zinc will not be precipitated by Cu when copper rod is dipped in Zn2+ solution. In fact, the reverse reaction is possible in the second case. The equilibrium constant for the reverse reaction  Cu(s) + Zn2+ (aq) Zn (s) + Cu2+(aq)  1 1 = = 5.0 × 1018 at 298 K. K 2.0 × 10-19 Thus when zinc rod is dipped in a solution containing Cu2+ ions Cu will be precipitated. Will be K1 =

Reactants →

products

Reactants and products at Equilibrium

Reactants ←

Fig 9.3 Predicting the direction of the reaction

products

9.14

Chemical Equilibrium

But the Kc for this reaction at 783 K is 46. Since Qc 1 which indicates that the forward reaction takes place more and products are present predominantly. If ΔG > 0 then –ΔG /RT is negative and e–ΔG /RT A. A  Thus the decreasing order of stability is B>C>A.

(i) When 5 moles/litre of each is taken, calculate the value of ∆G for the reaction at 298 K (ii) Find the direction of reaction and concentration at equilibrium (IIT 2004)  B. the ∆ H for A and B 52. Consider the reaction A  f are 246.4 and 252.3 JK–1 mol–1 respectively. What is the composition of equilibrium mixture at 298 K? O

9.26

Chemical Equilibrium

the formation of products is exothermic (i.e., accompanied by evolution of heat), the reverse reaction i.e., the reaction which favours the formation of reactants must be endothermic (i.e., accompanied by absorption of heat). The effect of change of temperature on chemical equilibria cannot be predicted, therefore, from the law of mass action. It has been established theoretically by Le Chatelier that equilibrium constant and hence the amount of products of a reaction increases with increase in temperature, if the reaction is endothermic. For example, consider the combination of nitrogen and hydrogen to give ammonia

equilibrium at 1000 K for air at atm pressure containing 80% N2 and 20% O2 by volume? 56. Given below are the values of ∆H and ∆S for the reaction at 27°C

It is evident from this equilibrium that the forward reaction is exothermic. Hence backward reaction must be endothermic. Thus according to Le Chatelier’s principle, the increase in temperature will favour the backward reaction i.e., the dissociation of ammonia into nitrogen and hydrogen. Since the production of ammonia i.e., forward reaction is an exothermic reaction, a low temperature will favour its formation. If temperature is increased, the reaction would proceed in the reverse direction in which the heat is absorbed i.e., in that direction in which the reaction is endothermic. Now consider the combination of nitrogen and oxygen to give nitric oxide

φ

φ

φ

φ

53. The density of an equilibrium mixture of N2O4 and NO2 at 1 atm is 3.62 g/litre at 298 K and 1.84 g/litre at  348 K. Calculate the enthalpy of reaction: N2O4  2NO2 also calculate the entropy change during the reaction at 348 K. 54. Calculate the equilibrium concentration ratio C to A, if 2.0 mol each of A and B were allowed to come to equi C+D at 300K ∆G for the change librium A+B  is 460 cal. 1 55. ∆G =77.77 KJ mol–1 at 1000 for the reaction N2 (g) 2 1 + O2 (g). what is the partial pressure of NO under 2

1  SO3(g) SO2(g) + O2 (g)  2 φ

φ

∆H = -98.32 KJ mol–1 and ∆S = -95J K–1 mol–1 find Kp for the reaction.

9.11 LE chatELIEr’S PrIncIPLE The main aim of chemical synthesis in industries is to get maximum yield of the product and to minimize the cost of production. It is possible at low temperatures and pressures. If a reaction cannot take place under such experimental conditions, it is necessary to adjust them There are three important factors which alter the state of equilibrium. These are temperature, pressure and concentration. The addition of a catalyst however has no effect on the state of equilibrium. Le Chatelier, in 1884, put forward a general principle which is very helpful in studying the effect of temperature, pressure and concentration on a system in equilibrium. The principle is known as Le Chatelier principle and states ‘that if a system in equilibrium is subjected to a stress or constrain, the equilibrium gets shifted in such a way as to reduce the effect of the stress or constrain. Let us now consider the effect of temperature, pressure and concentration on some well known chemical equilibrium in the light of the above principle Effect of change of temperature The rate of a reaction increases with the rise in temperature. But increase in the rate of both the forward and backward reactions going on at equilibrium may be quiet different. If the forward reaction i.e., the reaction which favours

 2NH3 (g); ΔH = -22.08 Kcals N2 (g) +3H2 

 2NO(g); ΔH = 43.2 Kcal N2 (g) + O2(g)  Since the forward reaction, i.e., the formation of nitric oxide is endothermic the rise in temperature, therefore favours the formation of nitric oxide. If the temperature is much below 2500°C (at which the above reaction is carried out in an electric furnace) the low temperature will favour the reverse reaction, which is exothermic. Similarly, the reaction  2SO2 (g); ΔH= -23.15 Kcals 2SO2 (g) + O2  is exothermic in the forward direction. Hence the reverse reaction is endothermic and will be favoured by the increase in temperature. A low temperature would favour the forward reaction in order to get better yield, a temperature of 450°C has been suggested.  2NO2(g) N2O4(l)  The above equilibrium is endothermic in the forward direction and hence an increase in temperature favours the formation of product (NO2). The opposite reaction, favouring the reactants (N2O4) must be exothermic. The effect of temperature can also be observed in the following endothermic reaction  [CoCl4]2–(aq) + 6H2O(l) [Co (H2O)6]3+(aq) + 4Cl–(aq)  Pink Colorless Blue

Chemical Equilibrium 9.27

At room temperature, the equilibrium mixture is blue due to [CoCl4]–2. When cooled in a freezing mixture, the colour of the mixture turns pink due to [Co (H2O)6]3+ From the above discussion, it can be concluded that the equilibrium constant for an exothermic reaction (negative ∆H) decreases as the temperature increases. The equilibrium constant for an exothermic reaction (positive ∆H) increases as the temperature increases. Effect of change of Pressure In cases where ∆n = O, i.e., there is no change in the number of moles of reactants and products the law of mass action equation does not have any pressure or volume factor. Such type of equilibria are therefore are not affected by the changes in pressure or volume. For example,  2HI(g) H2(g) + I2(g)  The above equilibrium proceeds in either direction with out any change in volume or in the number of moles. The total number of moles ∆n for the equilibrium is 2-(1 + 1) = 0 According to Lechalelier’s principle the pressure would therefore, have no effect on this equilibrium. Consider the following equilibrium:  2NO2(g) N2O4  ∆n = 2-1 = +1. Thus total number of moles per unit volume increase when the pressure is increased. The volume of the system will, however, decrease. Hence according to Le Chatelier’s principle, the increase in pressure for the above system tends to shift the equilibrium in favour of N2O4 i.e., in that direction in which the degree of dissociation is suppressed (i.e., in which the total number of moles is decreased). Similar is the case with the following equilibrium  PCl3 (g) + Cl2 (g)  PCl5 (g) 1 mole 1 mole 1 mole  N2 (g) + 3H2 (g)  2NH3 (g) 2 moles 1 mole 3 moles  2SO2 (g) + O2 (g) 2SO3 (g)  2 moles 2 moles 1 mole In all the above equilibria, an increase in pressure tends to shift the equilibrium in that direction in which the total number of moles is decreased, i.e., in favour of the reactants. In cases, where ∆n is negative, the pressure terms appear in the denominator of the law of mass action equations. Thus increase of pressure in such cases will result in an increase in the degree of dissociation x, in order to maintain a constant value of Kp. In other words, at higher

pressure more of products will be obtained in such cases. For example, N2(g) + 3H2(g)   2NH3(g) 1 mole 3 moles 2 moles ∆n = 2 - (3 + 1) = -2. The forward reaction is accompanied by a decrease in the number of moles. If the pressure is increased, the volume will decrease and hence the number of moles per unit volume will also increase. The equilibrium will, therefore, shift in the forward direction in which there is a decrease in the number of moles. Hence the formation of ammonia is accompanied by an increase in pressure, and this is in accordance with Le Chatelier’s principle. Thus high pressure is seen to give a high yield of NH3 but the higher the pressure, the greater is the cost of equipment to produce and maintain this pressure. Thus most plants operate at about 200-1000 atmospheres. Similar is the case with the following equilibria  2SO3 (g) 2SO2 (g) + O2 (g)   PCl5 (g) PCl5 (g) + Cl2 (g)  The increase in pressure in the above cases shifts the equilibrium in the direction in which there is a decrease in number of moles. So in order to get good yields of SO3 and PCl5, the above reactions should be carried out at high pressures. If the pressure is low, the equilibrium would be shifted in the reverse direction, i.e., in the direction in which there is an increase in the number of moles. From the above discussion, it can be concluded that in the reactions in which the total number of moles of products is less than the total number of reactants, when volume of the reaction mixture is decreased the partial pressure and the concentration are increased. Then QcKc and the reaction proceeds in the backward direction. Effect of change in concentration According to Le Chaterlier’s principle, the increase of concentration results in shifting equilibrium in such a way as to favour the reaction which proceeds with a decrease in the number of molecules. For example  2SO2 (g) + O2 (g)  2SO3 (g); ΔH= -23.15 Kcals The forward reaction in this case is exothermic and takes place with a decrease in volume. Thus an excess of either oxygen or sulphur dioxide will result in a good yield of sulphur dioxide.

9.28

Chemical Equilibrium

Similarly, the addition of nitrogen or hydrogen will result in a greater yield of ammonia, since the equilibrium  2NH3 (g) N2 (g) + 3H2 (g)  proceeds with a decrease in volume in the forward direction i.e., in favour of the formation of ammonia. The effect of change of concentration can also be explained as below: Consider the following equilibrium: 2+

  Fe (SCN ) ( aq ) Fe3+ ( aq ) + (SCN ) ( aq )    Yellow Colourless Brown -

If either ferric salt (e.g. ferric chloride) or sulphocynide salt (e.g., potassium sulphocynide) is added to the above equilibrium, it has been observed that in either case the concentration of the product [Fe(SCN)]2+ increase in a accordance with the Le Chatelier’s principle. If on the other hand, a small amount of potassium ferrosulphocynide, which furnishes [Fe(SCN)]2+ ion is added to the equilibrium, the solution becomes less dark due to the conversion of dark colored [Fe(SCN)]2+ ion into light colored Fe3+ and SCN– ions. Thus the equilibrium gets shifted in favour of the reactants by increasing the concentration of the products. Hence an increase in the concentration of the reactants will shift the equilibrium in favour of the products, while an increase in the concentration of products will shift the equilibrium in favour of the reactants. Effect of Inert Gas addition If the volume is kept constant and an inert gas such as argon is added which does not take part in the reaction, the equilibrium remains undisturbed. It is because the addition of an inert gas at constant volume does not change the partial pressure or the molar concentrations of the substances involved in the reaction. The reaction quotient changes only if the added gas is a reactant and product. Effect of a catalyst A catalyst increases the rate of the chemical reaction by making available a new low energy pathway for the conversion of reactants to products. It increases the rate of forward and back ward reactions that pass through the same transition state and does not affect equilibrium. Catalysts lower the activation energy for the forward and reverse reactions by exactly the same amount. Catalyst does not affect the equilibrium composition of a reaction mixture. It does not appear in the balanced chemical equation or in the equilibrium constant expression.

If a reaction has an exceedingly small k, a Catalyst would be of little help.

9.11.1 application to physical Equilibrium Le Chatelier’s principle is not restricted only to chemical equilibria but it is also applicable to physical equilibrium. Some of its application to physical equilibria are given below. 1. Effect of pressure on solubility: Consider a solution of a gas in equilibrium with the gas itself. Now suppose the solution is placed in a cylinder and it is compressed with the help of a piston. The pressure on the system, therefore, increases which reduces the volume and as a result some gas dissolves in the solvent without altering the pressure. In other words, solubility of the gas increases when the pressure is increased. Solubility of such solids which dissolve in a solvent with a decrease in volume, also increases with increasing pressure, while the solubility of those solids which dissolve with an increase in volume, decreases with increasing pressure. 2. Effect of temperature on solubility: There are most cases when a solute passes into solution with the absorption of heat i.e., the reaction is endothermic and cooling result. E.g. potassium nitrate in equilibrium with solid potassium nitrate is heated; the change will take place in that direction which will absorb heat in order to balance the effect of the increased temperature. As a result, more of potassium nitrate will dissolve. Hence solubility of a substance will increase with rise in temperature. On the other hand, if the dissolution of a solute is exothermic reaction, i.e., accompanied by evolution of heat e.g. calcium acetate in water, the solubility will decrease with increase in temperature. 3. Vapour pressure of liquid: The conversion of liquid into its vapor is endothermic reaction (heat is absorbed) and conversation of vapour into liquid is exothermic in which heat is evolved. Thus the equilibrium.  Vapour Liquid  Will be shifted towards the right when temperature of the system is raised. In other words, liquid will evaporate as a result of which vapour pressure of the system will also increase. Thus an increase in temperature will increase the vapour pressure of a liquid. 4. Effect of pressure on the melting point of ice: Consider a mixture of ice and water at 0°C and one atmospheric pressure, contained in a cylinder provided with a piston and the pressure on it is increased. The volume of the system decreases without change in pressure. The

Chemical Equilibrium 9.29

volume occupied by water now becomes less than that occupied by ice at the same temperature and so some of the ice melts. Thus the melting point of ice decreases as a result of increase in pressure. In other words, increase of pressure on ice→ water system at a constant temperature causes the equilibrium to shift towards the right. 5. Effect of pressure on the boiling point of a liquid: We know that the vapour pressure of a liquid increases

with increases in temperature, i.e., the conversion of a liquid into vapour is accompanied by increase in vapour pressure of the liquid. If pressure on the system is allowed to increase, as discussed above, some of the vapour will be converted into liquid in order to lower the pressure. On other words, increase of pressure on the system at a constant temperature cause the vapours to condense into liquid state. Thus boiling point of a liquid increases with an increases in pressure.

9.30

Chemical Equilibrium

KEy PoIntS •

• •

• •

•

•

The transformation of one substance into another substance is known as reaction. The chemicals that are used to start the reaction are called reactants and the chemicals that are formed in the reactions are called products. The amount of substance reacting per unit time is called rate of reaction. The concentrations of reactants or products are expressed in moles/litres and is commonly known as activity or effective concentration or effective mass usually expressed in square brackets [x]. Reactions which proceed in one direction till one or both the reactants is completely used up are referred as irreversible or unidirectional reactions. The reactions in which the products can be reconverted into reactants or if the reaction mixture contains both the reactants and products and if the reaction can take place in both directions under a given experimental conditions is called reversible reaction. In a reversible reaction, the reaction proceeding forward in which reactants convert into products is called forward reaction, and the reaction proceeding backward direction in which the products convert into reactants is called backward reaction. Reversible reactions are represented by writing a pair of half headed arrow pointing in opposite directions in between the reactants and products.

• • •

Equilibrium in Physical Processes •

•

Some reversible reactions behave as irreversible or unidirectional reactions and go to completion under some specific conditions. The removal of one of the products of a reversible reaction lead to the completion of the forward reaction. e.g., When lime stone is heated in a lime kiln it decomposes completely since CO2 goes into air but when heated in a closed vessel the decomposition becomes reversible.

•

•

• • •

lime kiln CaCO3 (s)   → CaO(s) + CO2 (g) Closed Vessel   CaCO3(s)   CaO(s) + CO2 (g)

• •

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Reversible reactions must be carried in closed vessel In the beginning in a reversible reaction, the rate of forward reaction is more since the concentrations of reactants are more. As time proceeds the rate of forward reaction decreases as the concentrations of reactants decreases. In the beginning of a reversible reaction the rate of backward reaction is absolutely zero because the concentrations of products are zero.

If an equilibrium exists between two different physical states or between two different allotropic forms of the same substance is called physical equilibria e.g., Solid Solid Liquid a-Sulphur

 Products Reactants  •

As time proceeds the rate of backward reaction increases since the concentrations of products also increases. In a reversible reaction, the stage at which the rate of forward reaction is equal to the rate of backward reaction is called equilibrium state or stage. Reversible reactions does not go to completion in either direction.

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 Liquid   Vapour   Vapour   β-sulphur 

melting or fusion Sublimation Vaporizations Allotropic modification

If a solid–liquid system at melting point is taken in a well insulated container, it constitutes a system in which the solid is in dynamic equilibrium with liquid. It is known as solid – liquid equilibrium. If a liquid or solid is taken in a closed container, the liquid or solid vapourizes called vapourisation or evaporation while the vapour condenses into liquid or solid called condensation. When the rate of vapourisation (sublimation for solid) is equal to rate of condensation a dynamic equilibrium exists between the two physical states constituting the physical equilibria. The pressure necessary to achieve equilibrium in physical equilibria does not depend on the amount of solid or liquid that present in a container. Equilibrium in molecular systems is dynamic and is a consequence of equality of the rates of opposing reactions. A system moves spontaneously toward a state of equilibrium if a system initially at equilibrium is perturbed by some change in its surroundings, it reacts in a manner that restores it to equilibrium. The nature and properties of an equilibrium state are the same regardless of how it is reached. The condition of a system at equilibrium represents a compromise between two opposing tendencies; the drive for molecules to assume the state of lowest energy and the usage toward molecular chaos or maximum entropy. With increase in temperature the vapour pressure of solid increases more rapidly than does the vapour pressure of the liquid since the solid has the greater heat of vaporization.

Chemical Equilibrium 9.31

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The temperature at which, all the three physical states viz solid, liquid and vapour of the same substance exist in equilibrium is called triple point. The triple point is usually very close to the freezing point. In saturated solution, the solution is in equilibrium with an excess of solid present in it at a given temperature. In such cases the rates of dissolution and crystallization are in dynamic equilibrium In a saturated solution of gas in a liquid at constant pressure the rates of dissolution of gas in liquid and escape of gaseous molecules from solution are in dynamic equilibrium. Henry’s law states that the mass of a gas dissolved in a given mass of a solvent at a given temperature, is directly proportional to the pressure of the gas above the solvent.  liquid equilibrium there is only one For solid  temperature (melting point at 1 atm at which the two phases can coexist. If there is no exchange of heat with the surroundings the mass of the two phases remain constant.  Vapour equilibrium the vapour For liquid  pressure is constant at a given temperature. For dissolution of solids in liquids, the solubility is constant at a given temperature. For dissolution of gases in liquids, the mass of a gas dissolved in a given mass of solvent at a given temperature is directly proportional to the pressure of the gas above the solvent.

characteristics of chemical Equilibrium •

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types of chemical Equilibria • •

General characteristics of Equilibria Involving Physical Process • • • • • •

Equilibrium is possible only in a closed system at a given temperature. Both the opposing process occurs at the same rate and there is a dynamic but stable condition. All measurable properties of the system remain constant. When equilibrium is attained for a physical process it is characterized by constant value of one of its parameters at a given temperature. The magnitude of such quantities at any stage indicates extent to which the reaction has proceeded before reaching equilibrium. Super cooling is a process of cooling a liquid below to its freezing point without solidification take place. Super cooled liquids are said to be in metastable state since a slight disturbance such as stirring or the addition of solid piece some of the liquid solidify by instantly. Super cooled liquids in equilibrium with their vapours are called metastable equilibrium.  Vapour Super cooled water 

Though the reversible reaction appears to be stopped at the equilibrium it takes place in both directions with same velocity. Hence it is known as dynamic equilibrium. The concentrations of reactants and products remain constant. Pressure density, colour remain unchanged with time in equilibrium state. Equilibrium can be approached from both sides. The equilibrium readjusts with the changing conditions and spontaneously goes back to the original state when the disturbing factors are removed. By observing pressure, concentration, density or colour the attainment of equilibrium may be identified. The equilibrium equation does not tell us how long it takes for a reaction to attain equilibrium. In the equilibrium state addition of products shifts the equilibrium in the backward direction while the addition of reactants shifts the equilibrium in the forward direction but the equilibrium constant remains constant.

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The equilibrium in which reactants and products are in the same phase is known as homogeneous chemical equilibrium. Homogeneous chemical equilibrium are of two types:  2HI(g) (a) Gas phase e.g., H2(g) + I2(g)  The reactants and products are in the same gaseous phase forming a homogeneous mixture. (b) Liquid phase e.g., CH3COOH (aq) + C2H5OH (aq) → CH3COOC2H5(aq) + H2O The reactants and products are in the same liquid phase form a homogenous mixture. Homogeneous chemical equilibria in gas phase are of two types: (a) Which have total number of moles of reactants are equal be the total number of moles of products  2HI(g) e.g., H2(g) + I2(g)  (b) Which have the total number of moles of reactants not equal to the total number of moles of products e.g.,  2NH3(g) N2(g) + 3H2(g) 

•

The equilibrium in which the reactants and products are in different phases is known as heterogeneous chemical equilibrium e.g.  CaO(s) + CO2(g) CaCO3 (g) 

9.32

Chemical Equilibrium

Law of mass action and its applications

characteristics of Equilibrium constant

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Law of mass action was proposed by C.M. Guldberg and P. Wage in 1813. Law of mass action states that “the rate of a chemical reaction under a given set of conditions is directly proportional to the product of the active masses of the reactants. Since the active masses are directly proportional to their molar concentrations law of mass action can be defined as” the rate of a reaction is proportional to the product of the molar concentrations of the reactants. Law of mass action is applicable to all reactions occurring in gas phase or in the liquid (or solution) phase.  cC + dD For a general reaction aA+bB  The rate of forward reaction ∝[A]a [B]b Or the rate of forward reaction = Kf[A]a[B]b The rate of backward reaction ∝[C]c [D]d Or the rate of backward reaction = Kb [C]c[D]d At the equilibrium, the rate of forward reaction is equal to the rate of backward reaction.

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Kf [A]a[B]b = Kb[C]c[D]d Kf Kb

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=

[C ]c [ D]d = Kc [ A]a [ B]b

Kc is known as equilibrium constant The equilibrium constant is the ratio of the velocity constants of forward and backward reactions Kf Kc = Kb

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units of Kp and Kc •

Equilibrium Constant = Product of the concentrations of the products Product of the concentrations of the reactants

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If the reactants as well as products are gases then partial pressure may be substituted in place of molar concentrations or active masses. Then the equilibrium constant is known as Kp K p=

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PCc × PDd PAa × PBb

The Kc and Kp are related as Kp = Kc [ RT ]Dn ∆n = number of gaseous moles of products-number of gaseous moles of reactants. R is gas constant and T is absolute temperature. If ∆n = 0 then Kp = Kc ∆n>0 then Kp>Kc (∆n is positive) ∆nKc the reaction will proceed in the backward direction. If Qc d2 > d1 (c) d1 > d2 > d3 > d4 (d) (d2 = d1) > (d3 = d4)  SO2(g) + Cl2(g) is 30. The equilibrium, SO2Cl2 (g)  attained at 25°C in a closed container and an inert gas, helium is introduced. Which of the following statements are correct? (a) Concentrations of SO2, Cl2 and SO2Cl2 are changed. (b) More chlorine is formed. (c) Concentration of SO2 is reduced. (d) No change occurs. 31. At a certain temperature, Kc = 1.2 × 103 for the equilibrium  COCl (g). If excess of O were CO (g) +Cl2(g)  2 2 added to that any CO(g) present were removed (by conversion to CO2 and not to COCl2, What would happen to the COCl2, initially present in the equilibrium mixture? (a) It would vanish. (b) More would be formed. (c) Nothing happened. (d) Its amount would be slightly diminished.  CaO (s) + CO (g), 32. For the reaction CaCO3(s)  2 the pressure of CO2 (g) depends on (a) the mass of CaCO3(s) (b) the mass of CaO(s) (c) the masses of both CaCO3(s) and CaO(s) (d) temperature of the system  CaO(s) + CO (g), 33. For the reaction CaCO3(s)  2 the addition of more of CaO(s) causes (a) the decrease in the concentration of CO2(g) (b) the increase in the concentration of CO2(g) φ

 C + Heat, 18. In the gaseous equilibrium, A + 2B  the forward reaction is favoured by (a) low temperature and high pressure. (b) low pressure and low temperature. (c) high pressure and low temperature. (d) high pressure and high temperature. 19. The oxidation of SO2 by O2 to SO3 is exothermic reaction. The yield of SO3 will be maximum if (a) temperature is increased and pressure is kept constant. (b) temperature is reduced and pressure is increased. (c) both temperature and pressure are increased. (d) both temperature and pressure are decreased. 20. For the manufacture of ammonia by the reaction  2NH + 21.9 Kcal, the favourable N2 + 3H2  3 conditions are (a) low temperature, low pressure and catalyst. (b) low temperature, high pressure and catalyst. (c) high temperature, low pressure and catalyst. (d) high temperature, high pressure and catalyst. 21. When pressure is applied to the equilibrium system’  water’ which of the following phenomenon Ice  will happen? (a) More ice will be formed. (b) Water will evaporate. (c) More water will be formed. (d) Ice sublimes. 22. In a vessel containing SO3, SO2 and O2 at equilibrium, some helium gas in introduced so that the total pressure increases while temperature and volume remains constant. According to Le Chatlier principle, the dissociation of SO3 (a) Increases (b) Decreases (c) Remains unaltered (d) Changes unpredictably 23. Which of the following factors will favours the reverse reaction in a chemical equilibrium? (a) Increase in the concentration of one of the reactants. (b) Increase in the concentration of one of the products. (c) Removal of one of the products regularly. (d) None of the above 24. Dissociation of phosphorus pentachloride is favoured by (a) high temperature and high pressure. (b) high temperature and low pressure. (c) low temperature and low pressure. (d) high temperature, high pressure. 25. Pure ammonia is placed in a vessel at a temperature where its dissociation constant is appreciable. At equilibrium (a) Kp does not change significantly with pressure. (b) α does not change with pressure. (c) Concentration of ammonia does not change with pressure. (d) Concentration of hydrogen is less than that of nitrogen.

9.40

34.

35.

36.

37.

38.

Chemical Equilibrium

(c) no change in the concentration of CO2(g) (d) the increase in the concentration of CaCO3(g) Given the following reaction at equilibrium N2(g) +  2NH (g). Some inert gas is added at 3H2(g)  3 constant volume predict which of the following facts will be affected? (a) More of NH3(g) is produced. (b) Less of NH3(g) is produced. (c) No affect on the degree of advancement of the reaction at equilibrium. (d) Kp of the reaction is increased. Given the following reaction at equilibrium  2NH3(g). Some inert gas at conN2(g) + 3H2(g)  stant pressure is added to the system. Predict which of the following facts will be affected? (a) More NH3(g) is produced. (b) Equilibrium shifts to left. (c) No affect on the equilibrium. (d) Kp of the reaction is decreased.  N2(g) + O2(g) For the reaction 2NO(g)  ΔH = –180 KJ mol–1 Which of the following fact does not hold good? (a) The pressure change at constant temperature do not affect the equilibrium constant. (b) The volume change at constant temperature do not affect the equilibrium constant. (c) The dissociation of NO is favoured more at high temperature. (d) The dissociation of NO is favoured less at high temperature.  N2(g) + 3H2(g) For the reaction 2NH3(g)  ΔH = 93.6 KJ mol–1 Which of the following facts does not hold good? (a) The pressure changes at constant temperature does not affect the equilibrium constant. (b) The volume changes at constant temperature does not affect the equilibrium constant. (c) The dissociation of NH3 is more favoured at high temp. (d) The dissociation of NH3 is less favoured at high temp. For an equilibrium, which of the following statement is true? (a) The pressure changes do not affect the equilibrium. (b) More of ice melts if pressure on the system is increased. (c) More of liquid H2O freezes if pressure on the system is increased. (d) The pressure changes may increase or decrease the degree of advancement of the reaction depending upon the temperature of the system.

39. For an equilibrium reaction, which of the following statements is not true? (a) If the reaction quotient of the reaction is greater than Keq, the reaction moves in the backward direction. (b) If the reaction quotient of the reaction is lesser than Keq, the reaction moves in the forward direction. (c) If the reaction quotient of the reaction is equal to Keq, the reaction is at equilibrium. (d) There is no correletation between the reaction quotient and Keq, in predicting the direction in which the reaction proceeds. 40. Solids CaCO3 and CaO and gaseous CO2 are placed in a vessel and allowed to reach equilibrium CaO(s) +  CaCO (s) ΔH = -180 KJ mol-1 CO2(g)  3 The quantity of CaO in the vessel could be increased by (a) adding more of CaCO3 (b) removing some of CO2 (c) Lowering the temperature (d) reducing the volume of vessel 41. The concentration of a pure solid or liquid phase is not included in the expression of equilibrium constant because (a) Solid and liquid concentrations are independent of their quantities. (b) Solids and liquids react slowly. (c) Solids and liquids at equilibrium do not interact with gaseous phase. (d) The molecules of solids and liquids cannot migrate to the gaseous phase.  A(g) + B (g) 42 In the system AB (s)  Doubling the quantity of AB(s) would (a) Increase the amount of A to double its value. (b) Increase the amount of B to double its value. (c) Increase the amount of both A and B to double their values. (d) Cause no change in the amounts of A to B.  2B (g) + 3C (g) if the concentra43. In a system A (s)  tion of C at equilibrium is increased by a factor of 2, it will cause the equilibrium concentration B to change to (a) Two times the original value (b) One half of its original value (c) 2 2 times its original value 1 (d) times the original value 2 2 44. For a given equilibrium reaction, an increase in temperature will (a) Increase the rate of the exothermic reaction more than that of the endothermic reaction. (b) Increase the rate of the endothermic reaction more than that of the exothermic reaction. (c) Increase both rates equally. (d) Decreases both rates equally.

Chemical Equilibrium 9.41

45. The equilibrium constant Kc for the reaction.  SO (g) + NO(g) is 16. If SO2(g) + NO2(g)  3 1 mol of each of all the four gases is taken in dm3 vessel, the equilibrium concentration of NO would be (a) 0.4 M (b) 0.6 M (c) 1.4 M (d) 1.6 M 46. For the decomposition reaction NH2COONH4(s)  2NH3(g) + CO2(g) the Kp = 2.9 × 10–5 atm3. The  total pressure of gases at equilibrium when 1.0 mol of NH2COONH4 (s) was taken to start with would be (a) 0.0194 atm (b) 0.0388 atm (c) 0.0582 atm (d) 0.0776 atm  2HI(g) 47. For the reaction H2(g) +I2(g)  Kc = 66.9 at 350°C Kc = 50.0 at 448°C. The reaction has (a) ΔH = +ve (b) ΔH = –ve (c) ΔH = zero (d) ΔH = whose sign cannot be predicted 48. In a closed vessel 2 moles of SO2 and 1 mole of O2(g) were allowed to react to form SO3(g) under the influence of a catalyst at 1atm. Reaction  2SO (g) occurred. At equilib2SO2(g) +O2(g)  3 rium, it was found that 50% of SO2(g) was converted to SO3(g). Then the pressure of SO3 at equilibrium will be (a) 0.66 atm (b) 0.493 atm (c) 0.33 atm (d) 0.166 atm 49. In a closed container and at constant temperature 0.3 mole of SO2(g) and 0.2 mole of gas at 750 torr, At equilibrium 50% of SO2 (g) is converted to SO3. The partial pressure of SO2(g) will be (a) 375 torr (b) 370 torr (c) 225 torr (d) 150 torr  2NO (g) the numeri50. For the reaction N2O4(g)  2 cal value of Kp when the partial pressures are measured in atmosphere are (i) 1.75 × 103 at 500 K (ii) 1.78 × 104 at 600 K Which of the following statements are correct? (a) The units of Kp are atm–1. (b) The standard enthalpy change of the forward reaction is –ve. (c) The standard enthalpy change of the forward reaction is +ve. (d) The proportion of NO2 in the equilibrium is increased by increasing pressure. 51. A nitrogen–hydrogen mixture initially in the molar ratio of 1:3 reached equilibrium to form ammonia when 25% of the material had reacted. If the total pressure of the system was 21 atm the partial pressure of ammonia at the equilibrium was (a) 4.5 atm (b) 3.0 atm (c) 2.0 atm (d) 1.5 atm

52. A mixture of nitrogen and hydrogen in the ratio of 1:3 reach equilibrium with ammonia when 50% of the mixture has reacted. If the total pressure is P, the partial pressure of ammonia in the equilibrium mixture was: (a) P/2 (b) P/3 (c) P/4 (d) P/6  N O . 53 NO2 associates (or dimerises) as 2NO2  2 4 The apparent molecular weight of a sample of NO2 calculated from the vapour density under certain conditions was 60, the mole fraction of dimers (a) 14/46 (b) 16/46 (c) 28/46 (d) 46/60 (e) 60/92  2CO(g),, the fol54. In the reaction C(s) + CO2(g)  lowing amounts of substances were found in a 2-litre flask at equilibrium C = 0.1 mole CO = 0.05 mole, CO2 = 0.006 mole. The equilibrium constant is (a) 0.208 (b) 4.16 (c) 0.33 (d) 0.416 55. For the reaction 2HI   H2 (g) + I2(g) if we start with pure HI at a concentration of 1.20 moles 1 litre, that of H2 at equilibrium is 0.2 moles per litre, what is the Kc of the reaction? (a) 0.062 (b) 0.031 (c) 0.124 (d) none 56. At 730°C the vapour density of iodine under a pressure of 1 atm is 97.45% of the theoretical, on the assumption that the molecule of gaseous iodine is diatomic. What is the degree of dissociation of I2? (a) 0.0131 (b) 0.0262 (c) 0.082 (d) 0.0324 57. The Keq for the dissociation of atomic iodine I2 (g)  2I(g) is 4 × 10–3  If the equilibrium concentration of atomic iodine is 4 ×10–2 M. What is the concentration of molecular iodine? (a) 0.8 M (b) 0.4 M (c) 0.3 M (d) 0.2 M 58. 0.2 mole of H2 and 0.5 mole of I2 are mixed in a reaction vessel and heated until equilibrium is reached  2HI which of the following stateH2(g) +I2(g)  ment is true? (a) If x mole of HI is present at equilibrium the quantity of I2 present is 0.5 –x mole. (b) If x mole of HI is present at equilibrium the quantity of H2 present is x-0.2 mole. (c) If y mole of I2 is used up in the reaction the quantity of HI present at equilibrium is 2y mole. (d) If y mole of I2 is used up in the reaction the quantity of H2 present at equilibrium is 0.2-(y/2) mole.

9.42

Chemical Equilibrium

59. A mixture of NO2 at initial pressure of 1.0 atm and NO at initial partial pressure of 0.5 atm is brought  to the equilibrium. The reaction is 2NO2(g)  2NO(g)+O2(g) Which of the following is true? (a) If the partial pressure of O2 in the equilibrium mixture is 0.2 atm then Kp = 0.45. (b) If the partial pressure of O2 in the equilibrium mixture is x atm then Kp=

60.

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62.

63.

2 x3 . (1 - 2 x) 2

(c) If at equilibrium, the decrease in the partial pressure of NO2 is y atm, the value of Kp = (0.5+y)2/ (2-2y). (d) None of these Equimolar quantities of HI, H2 and I2 are brought to equilibrium, if the total pressure in the vessel is  1.5 atm and Kp for the reaction. 2HI (g)  H2(g)+I2(g) is 49. Which of the following is the correct for the equilibrium partial pressure of I2? (a) 0.4 (b) 0.9 (c) 0.7 (d) 0.5 Into a one-litre flask at 400°C are placed one mole of N2, 3 moles of H2 and 2 moles of NH3. If Kc for N2(g)  2NH3(g) + 3H2(g)  (a) Left to right (b) right to left (c) system is at equilibrium (d) can not say If we start with one mole of N2, 3 moles of H2 and two moles of NH3 in a one litre container at 500°C (where Kc = 0.08) at equilibrium (a) The number of moles of N2, H2 and NH3 will be in the ratio 1:3:2 (b) The number of moles of N2, H2 will be in the ratio 1:3 (c) The number of moles of N2, will be one (d) The total number of moles will be six For an unknown decompositions chemical

fig. Find the ratio of the stiochiometric coefficient of the reactant and the product. Where D = initial vapour density, d = final vapour density at equilibrium. (a) 1:2 (b) 1:1 (c) 2:1 (d) Data is inadequate to predict 64. A vessel at 1000 K contains CO2 with a pressure of 0.5 atm. Some of the CO2 is converted into CO on the addition of graphite. If the total pressure at equilibrium is 0.8 atm, the value of K is: (a) 0.3 atm (b) 0.18 atm (c) 1.8 amt (d) 3 atm

multiple choice questions with only one answer Level II 1. A certain amount of N2O4(g) is enclosed in a closed container at 127°C when following equilibrium got set up at a total pressure of 10 atm.

 2NO2(g) N2O4(g)  If the concentration (moles) of NO2(g) in the equilibrium mixture be 8 × 105 ppm, the Kc (in mol L–1) for the above reaction at 127°C is equal to (a) 3.189 (b) 2.051 (c) 0.974 (d) 1.842 2. For the following hypothetical chemical equilibrium  2B(aq)+C(aq) A(aq)  It was found that initially only A and B are present 1 . At equilibrium 3 1 mole fraction of A becomes equal to . So Kc for the 2

with mole fraction of B equal to

above equilibrium is. Initial molarity of the ‘B’ is ‘M’. 9 3 M2 (a) M 2 (b) 28 4

α

45° D/d

 nB, operating in gas phase a plot Equilibrium A  between

7 1 (d) M 2 M2 28 7  3. PCl5 (g)  PCl3(g) + Cl2(g). The equilibrium, Kc for the dissociation of PCl5 is 4.0 × 10–2 at 250°C in a 3.0 L flask when equilibrium concentration of Cl2 is 0.15 mol/L. What was the pressure of PCl5 before any dissociation? (R = 0.082 L-atm K–1 mol–1) (a) 37.0 atm (b) 30.59 atm (c) 24.05 atm (d) 6.745 atm

(c)

D and α gives a straight line, as shown in the d

Chemical Equilibrium 9.43

 Na2SO4(s) +10H2O (g) 4. Na2SO4.10H2O(s)  for the equilibrium, Kp = 4.08 × 10–25. If the vapour pressure of water at 0°C is 4.58 torr, at what relative humidity will Na2SO4 be deliquescent (absorb moisture) when exposed to the air at 0°C essentially? (a) 21.25% (b) 35.3% (c) 50% (d) 70% 5. Equilibrium constants are given for the following  AB .2H O(s)+4H O(g); AB2.6H2O(s)  2. 2 2 Kp = 6.89 × 10–12  M N.7H O(s)+5H O(g); M2N.12H2O(s)  2 2 2 –13 Kp = 5.25×10  P2N(s) +10H2O; Kp = 4.08 × 10–25 P2N.10H2O(s)  Which one is the most effective drying agent at 0°C (the vapour pressure of H2O at 0°C is 4.58 torr)? (a) AB2.2H2O (b) M2N.7H2O (c) P2N (d) All are equally effective 6. Assume that the decomposition of HNO3 can be represented by the following equation  4NO (g)+2H O(g)+O (g) 4HNO3 (g)  2 2 2 And at the given temperature 400 K and pressure 30 atm the reaction approaches equilibrium. At equilibrium partial pressure of HNO3 is 2 atm. Find Kc in (mol/lit)3 (a) 32 (b) 24 (c) 18 (d) 16  . LiCl.NH3(s)+2NH3(g) 7. LiCl.3NH3(s)  Kp for the reaction is 9 atm2 at 27°C.A 4.10 lit. flask contains 0.1 mol of LiCl NH3. How many moles of NH3 should be added to the flask at this temperature to drive backward reaction practically to complete? (a) 0.5 (b) 0.6 (c) 0.7 (d) 0.8 8. 0.2 MKCN and 0.06M AgNO3 solutions are mixed in equal volumes. At 25°C Kc for the reaction is  Ag++2CN– is 1.6×10–19. The conc. of Ag(CN)2–  + Ag present in solution is (a) 1.5 × 10–19M (b) 1.5 × 10–18M –19 (c) 3 × 10 M (d) 3 × 10–18M  9. CuSO4.5H2O(s)  CuSO4.3H2O(s) + 2H2O(v) Kp for the reaction is 10–4 atm2. Vapour pressure of water at 25°C is 30.4 mm at 25°C. At what relative humidities will CuSO4.5H2O be efflorescent when exposed to air at 25°C? (a) Below 25% (b) Above 25% (c) At 25% (d) Any 10. When 1 mole of A(g) is introduced in a closed vessel of one litre capacity maintained at constant temperature, the following equilibrium established  B(g)+C(g); Kc =? C(g)   D(g)+B(g); Kc2 =? A(g)  1

If the pressure at equilibrium is twice the initial

pressure calculate

KC1 KC 2

if

[C ]eq [ A]eq

= 2 :1 .

(a) 3:1 (b) 4:1 (c) 2:1 (d) 1:2 11. CoCl42–(aq) is blue in colour while Co(H2O)62+(aq) is pink. The colour of reaction mixture Co (H2O)62+(aq)  CoCl2– +6H O is blue at room tem+4Cl–(aq)  2 (l) 4(aq) perature while it is pink when cooled hence (a) Reaction is exothermic (b) Reaction is endothermic (c) Equilibrium will shift in forward direction on adding water to reaction mixture (d) The reaction is neither endothermic nor exothermic 12. An amount of solid NH4HS is placed in a flask already containing ammonia gas at a certain temperature and 0.50 atm pressure. Ammonium hydrogen sulphide decomposes to yield NH3 and H2S gases in the flask. When the decomposition reaction reaches equilibrium, the total pressure in the flask rises to 0.84 atm. The equilibrium constant for NH4HS decomposition at this temperature is (a) 0.30 (b) 0.18 (c) 0.17 (d) 0.11 13. Haber-Bosch process for the manufacture of NH3 is based on the reaction catalyst ⇀  N 2 ( g ) + 3H 2 ( g ) ↽  2 NH 3 ( g ); ∆H = −46.0 O

K.J/mol and K0P = 14. Which of the following information regarding the above reaction is correct? (a) On adding N2, the equilibrium is shifted to right side with an increase in entropy. (b) The equilibrium constant K0P increases with increase in temperature. (c) At equilibrium 2Gm(NH3)=Gm(N2)+3Gm(H2) where Gm represents the molar Gibbs formation of the species enclosed within the brackets. (d) The use of catalyst helps increasing the rate of forward reaction more than that of backward reactions there by increasing the yield of NH3. 14. Two solid compounds A and C dissociates into gaseous product at the temperature T.   A( s )   B( g ) + D( g )  E ( g ) + D( g ) C ( s )  At this temperature, pressure over excess solid A is 40 atm and that over excess solid C is 80 atm. The total pressure of the gases over the solid mixture is (a) 67.0 atm (b) 89.4 atm (c) 134 atm (d) 178.8 atm

9.44

Chemical Equilibrium

15. The equilibrium constant for the reaction A+B → C is 0.9 How much conc. of A should be added to one litre of 0.1 MB solution so, that 60% B converted into C? (a) 1.67 M (b) 1.73 M (c) 1.82 M (d) 1.91 M 16. 10 g of CaCO3 is taken into a vessel of volume 40 mL. It comes into equilibrium with CaO and CO2 CaCO3⇌CaO+CO2 Kp = 0.324 atm at 27°C Now the vessel is placed in a bigger and evacuated vessel of capacity 800 mL. Now a small hole is made to the smaller vessel. CO2 gas starts effusing into bigger vessel at a rate of 0.044g/sec. How much time (in sec) is required for the effusion of CO2 to stop? (a) 10 sec (b) 15 sec (c) 5 sec (d) 8 sec 17. Consider the equilibrium HgO(s) + 4I–(aq) + H2O(l) – ⇌ HgI2– 4 (aq)+2OH (aq). Which changes will increase the equilibrium concentration of HgI2– 4? (I) Adding 0.2 M HCI (II) Increasing [l–] (III) Increasing no. of moles of HgO(s) (a) I only (b) II only (c) I and II only (d) II and III only 18. For a general gaseous reaction aA(g)+bB(g) ⇌ cC(g) + dD(g) Equilibrium constant Kc,Kp and Kx are represented by the following relations. Kc

Pc Pd X c .X d [ C ]c [ D ]d ; K p  Ca Db and .K x  Ca Db a b [ A] [ B ] PA PB X A .X B

Where [A] represents molar concentration of A PA represents partial pressure of A and P represents total pressure. XA represents mole fraction of A On the basis of above work-up, select the right option. Dng

(a) K p = K c ( RT )

(b) K c = K p ( RT ) (c) K c = K x P

22.

23.

24.

; K p = Kx P

-Dng

1 3P 2 4P 2

(b)

O

Dng

O

O

O

Dng -Dng

; K x = K p ( RT )

19. The activation energies for the forward and reverse element reactions in the system A⇌B are 10.303 and 8.000 Kcal respectively at 500 K. Assuming the preexponential factor to be the same for both the forward and reverse steps and R=2 cal K–1 mol–1 calculate equilibrium constant of the reaction: (a) 1.000 (b) 10.0 (c) 100 (d) 0.1 20. A student determined the value of Ksp for a saturated solution of borax at several different temperatures.

25.

26.

4 3P 2

(d) None 3 The rate of reaction is proportional to its active mass. This statement is termed as (a) Faraday’s Law of Electrolysis (b) Law of Mass Action (c) Le Chatelier Principle (d) Law of Constant Proportions ‘n’ mole of a reactant ‘A’ gives one mole each of B and C in reversible reaction. If degree of dissociation of ‘A’ is independent of initial concentration of ‘A’ then ‘n’ is (a) 1 (b) 2 (c) 3 (d) 4 Consider the following equilibrium for formation of octahedral complex of Co2+ in aqueous medium at 298 K. Reaction (1): Co2+(aq) + 6NH3⇌[Co(NH3)6]+2 K = 2 × 108 Reaction (2): Co2+ + 3en⇌[Co(en)3]+2 K = 4 × 1017 Assuming ∆H 1 = ∆H 2 and ∆S 1 = 0 what is the approximate value of ∆S 2 (a) 95 J/K (b) -90 J/K (c) 180 J/K (d) 150 J/K The degree of dissociation of HI at a particular temperature is 0.8 the volume of 1.5 M sodium thiosulphate solution required to react completely with I2 present at equilibrium in acidic condition. When 0.135 mol each of H2 and I2 are heated at 440 K in a closed vessel of capacity 2.0 L (a) 144 mL (b) 216 mL (c) 108 mL (d) 192 mL For the synthesis of NH3 from the Haber process starting with stoichiometric amount of N2 and H2, the attainment of equilibrium is predicted by which curve? (c)

Dng

; K p = Kx P

(d) K c = K p ( RT )

(a)

; K x = K p ( RT )

-Dng

Dng

The value for ∆S° for the dissolution of borax in water can be determined from: (a) The slope of the line resulting from a plot of ln Ksp versus (I/T) (b) The y-intercept of the line resulting from a plot of ln Ksp versus T (c) The slope of the line resulting from a plot of ln Ksp versus (I/T) (d) The y-intercept of the line resulting from a plot of ln Ksp versus (I/T) 21. N2 and H2 are taken in 1:3 molar ratio in a closed vessel to attain the following equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g) Find Kp for reaction at total pressure of 2P. If PN at 2 equilibrium is P/3.

Chemical Equilibrium 9.45

N2 (g)+3H2(g)→NH3(g) 3

2 H2 NH3 N2

1

amount

amount

2

0

31. The approach to the following equilibrium was observed kinetically from both directions

3 1

0 Reaction

Reaction

(a)

3

(b)

3

NH3

NH3

2 H2

1

N2 Reaction

amount

amount

2

0

N2 NH3 H2

N2

1

0

(c)

H2 Reaction

(d)

27. Rate of disappearance of the reactant A at two different temperature in given set up is -d [ A] = 2 × 10 -2 s -1 [ A] - 4 × 10 -3 s -1 [ B ]at300K dt -d [ A] = 4 × 10 -2 s -1 [ A] - 16 × 10 -4 s -1 [ B ]at400K dt Heat of reaction at the given temperature range, when equilibrium is set up is: 2.303 × 2 × 300 × 400 (a) log 50 cal 100 2.303 × 2 × 300 × 400 (b) log 250 cal 100 (c) 2.303 × 2 × 300 × 400 log 5 cal 100 (d) None 28. At 40°C the vapour pressure in torr, of methyl alcoholethyl alcohol solution is represented by the equation P = 119xA + 135. When XA is mole fraction of methyl alcohol. Then the value of lim

x A →1

PA is XA

(a) 254 torr (b) 135 torr (c) 119 torr (d) 140 torr 29. In an aqueous solution of volume 500 mL, when the reaction of 2Ag++Cu⇌Cu+2+2Ag reached equilibrium. The [Cu+2] was XM. When 500 mL of the water is further added, at the equilibrium [Cu+2] will be (a) 2XM (b) XM (c) Between XM and X/2 M (d) Less than X/2 M 30. For the reaction A+B⇌C+D, equilibrium concentration of [C] = [D] = 0.5 M. If we start with 1 mole each of A and B percentage of A converted to C if we start with 2 mole of A and 1 mole B is: (a) 25% (b) 40% (c) 66.66% (d) 33.33%

PtCl 4-2 +H 2 O ⇌Pt (H2O) Cl3- +Cl -

At 25°C it was found that - d [ PtCl4-2 ] = 3.9 × 10 -5 sec -1 PtCl4-2 - ( 2.1 × 10 -3 sec -1 ) dt [Pt(H2O) Cl3- ] [CI–] What is the value of equilibrium constant when fourth Cl– is complexed? (a) 1.86 × 10–2 (b) 2.1 × 10–3 –5 (c) 3.9 × 10 (d) 0.54 × 102 32. PCl5 is 40% dissociated when pressure is 2 atm. It will be 80% dissociated when pressure is approximately (a) 0.3 atm (b) 0.5 atm (c) 0.6 atm (d) 0.2 atm 33. For N2+3H2 ⇌ 2NH3 one mole N2 and 3 moles H2 are at 4 atm. Equilibrium pressure is found to be 3 atm. Hence Kp is 1 1 (a) (b) 2 ( 0.5 )( 0.15 ) (0.5)(1.5)3 (c)

3×3 ( 0.5 )( 1.5 )3

(d) None

34. 40% of mixture of 0.2 mole of N2 and 0.6 mole of H2 react to give NH3 according to the equation N2+3H2⇌ 2NH3 at constant temperature and pressure. The ratio of final volume to the initial volume of gases are (a) 4:5 (b) 2:4 (c) 7:10 (d) 8:5 35. The equilibrium constant for the reaction A (g) +B(g)⇌2C(g) is 3 × 10–4 at 500 K. In the presence of a catalyst the equilibrium is attained 10 times faster. The equilibrium constant in the presence of a catalyst at 500 K has the value (a) 3 × 10–4 (b) 30 × 10–4 –3 (c) 3 × 10 (d) 30 × 10–3 – – 36. I2 + I ⇌ I 3. This reaction set up in aqueous medium. We start with 1 mole of I2 and 0.5 mole of I– in the litre flask. After equilibrium is reached, excess of AgNO3 gave 0.25 mole of yellow ppt. Equilibrium constant is: (a) 1.33 (b) 2.66 (c) 2.00 4). 3.00 37. If equilibrium constant of CH3COOH+H2O⇌CH3 COO– +H3O+ is 1.8×10–5. If equilibrium constant of CH3COOH+OH–⇌CH3COO–+H2O is (a) 1.8 × 10–9 (b) 5.55 × 1010 –10 (c) 5.55 × 10 (d) 1.8 × 109

9.46

Chemical Equilibrium

φ

38. 2 moles each of A and B were taken in a container and following reaction take place 2A(g)+B(g)⇌2C(g) + D(g). When the system attained equilibrium (a) [A] > [B] (b) [A] < [B] (c) [A] = [B] (d) [A] = [C] 39. Zn/[Zn2+(1M)][Ag+(1M)]/Ag. If equilibrium constant is 6.2 × 1052. Calculate ΔG at 298 K. (a) -301.08 KJ (b) 301.08 KJ (c) -201.08 KJ (d) 201.08 KJ 40. For the following equilibrium N2O4⇌2NO2 in gas phase, NO2 is 50% of the total volume when equilibrium is set up. Hence percentage dissociation of N2O4 is (a) 50% (b) 25% (c) 66.66% (d) 33.33% 41. For a certain equilibrium over the temperature range 500 K to 700 K. The equilibrium constant Kp confirms

φ

6.36 × 10 3 to the equation log Kp = 30.1- 5.7.log .T . T Calculate ΔG at 600 K. (a) -16.4 Kcal/mol (b) 14.60 Kcal/mol (c) -10.064 Kcal/mol (d) -14.00 Kcal/mol Which of the following statement is incorrect for the rate given reaction 4A + B ⇌ 2C + 2D? (a) The rate of disappearance of B is one fourth the rate of disappearance of A. (b) The rate of the appearance of C is half the rate of disappearance of B. (c) The rate of formation of C and D are equal. (d) The rate of formation of D is half the rate of disappearance of A.  CaO(s)+CO2(g) occurThe reaction CaCO3(s)  ring in a closed container is at equilibrium, How is the equilibrium concentration of CO2 affected if mole CaCO3 is added at a fixed temperature? (a) It decreases (b) it remains unaffected (c) It increases (d) it cannot be predicted For the reaction PCl5⇌PCl3+Cl2 in gaseous phase, Kc =4. In a 2 lit flask, there are 2 mole each of PCl3 and Cl2and 0.5 mol of PCl5. Equilibrium concentration of PCl5 is (a) 0.25 mol/lit (b) 0.125 mol/lit (c) 0.75 mol/lit (d) 1 mol/lit  CO(g)+Cl2 Kp for the equilibrium COCl2(g)  –9 is 6.70 × 10 atm at 100°C Calculate the fraction of phosgene dissociated when 1 mole of phosgene is taken in a 100 litre vessel and allowed to reach equilibrium. (a) 6.7 × 10–9 (b) 6.7 × 10–7 –4 (c) 2.48 × 10 (d) 1.48 × 10–4

42.

43.

44.

45.

46. CH3COCH3(g) ⇌CH3-CH3(g) +CO(g) initial pressure of CH3COCH3 is 100 mm. When equilibrium is set up mole fraction of CO(g) is 1/3. Hence Kp is (a) 10 mm (b) 50 mm (c) 25 mm (d) 150 mm 47. For the reaction A⇌B+C the following data were obtained at 30°C.

48.

49.

50.

51.

52.

Experiment

[A] mol/lit

Rate mol lit–1hr–1

1. 2. 3.

0.170 0.340 0.680

0.05 0.10 0.20

The equilibrium constant for the reaction is 0.50 mol lit–1 assuming that the reaction proceeds by a one step mechanism. Calculate the rate constant for the reverse reaction in lit mol–1 hr–1. (a) 0.294 (b) 0.588 (c) 0.50 (d) 0.125 The following equilibrium is set up at 400°C in a 10 lit flask 2NO(g)+O2(g))⇌2NO2(g) K = 10. If no. of mol of NO and NO2 are equal at equilibrium. Determine no. of mole of O2 under equilibrium condition. (a) 10 (b) 1 (c) 100 (d) None For the reaction H2 (g) + l2 (g)⇌2HI(g) the value of the equilibrium constant is 64. If the volume of the container is reduced to one-fourth of its original value, the equilibrium constant will be (a) 64 (b) 32 (c) 276 (d) 16 One mole each of A and B and 3 mole of C and D are placed in 1 lit flask, its equilibrium constant is 2.25 for  C+D, equilibrium concentration of A and A+B  B will be in the ratio (a) 2:3 (b) 3:2 (c) 1:2 (d) 2:1  2C+D initial In the following reaction 3A+B  mole of B is double to that of A. At equilibrium A and C are equal hence% dissociation of B is (a) 105 (b) 20% (c) 40% (d) 5%  PCl3(g)+Cl2(g) and COCl2(g) Two systems PCl5(g)   CO2(g)+Cl2(g) are simultaneously in equilibrium  in a vessel at constant volume. If some CO is introduced into the vessel then the new equilibrium concentration of (a) PCl5 is greater (b) PCl3 remains unchanged (c) PCl5 is less (d) Cl2 is greater

Chemical Equilibrium 9.47

53. The correct relationship between free energy change in a reaction and the corresponding equilibrium constant Kc is (a) -∆G = RT ln Kc (b) ∆G = RT ln Kc (c) -∆G = RT ln Kc (d) ∆G = RT ln Kc 54. Consider the reaction equilibrium:  2SO3(g) ∆H° = –918 KJ. On the 2SO2(g)+O2(g)  basis of the Le Chatelier’s principle, the condition favourable for the forward reaction is (a) Any value of temperature and pressure (b) Lowering of temperature as well as pressure (c) Increasing temperature as well as pressure (d) Lowering the temperature and increasing the pressure 55. A 0.20 M solution of methanoic acid has a degree of ionization of 0.032. Its dissociation constant is (a) 2.1×10–4 (b) 2.1×10–3 –6 (c) 1.1×1010 (d) 1.1×10–3  H2O(g)+CO(g) 56. For the reaction H2(g)+CO2(g)  the equilibrium constant is 1.8 at 1000°C if 1.0 mol of CO2 and 1.0 mol of H2 are taken in a 1-L flask, the final equilibrium concentration of CO at 1000°C will be (a) 0.575 M (b) 0.290 M (c) 0.36 M (d) 0.721 M K1    57. Consider two gaseous equilibria 2NO(g)  k2  4NO(g) where N2(g)+O2(g), 2O2(g)+2N2(g)  K1and K2 are the corresponding equilibrium constant at 298 K The values of the equilibrium constants are related by 1 1 (b) K1 = 2 (a) K 2 = 2 K K1 2 O

O

(c) K2 = K12

(d)

1 K1 = K 2 2

58. A certain quantity of PCl5 is heated in a 10 L vessel K1   at 255°C PCl5 (g)   PCl3(g)+Cl2(g). At equilibrium, the vessel contains 0.10 mol of PCl5 and 0.20 mol each PCl3 and Cl2. The equilibrium constant of the reaction is (a) 0.02 (b) 0.4 (c) 0.04 (d) 0.2 59. The equilibrium partial pressure for the reaction X(g)+Y(g) → Z(g), were recorded as PX = 0.15 atm, Py = 0.10atm. and PZ = 0.30 atm. The volume of the reaction vessel was then reduced such that on re-establishing the equilibrium the partial pressure of X and Y were doubled. The partial pressure of Z would be (a) 1.20 mm (b) 0.60 atm (c) 1.80 atm (d) 0.30 atm

60. When NaNO3 is heated in closed vessel, O2 is liberated and NaNO2 is left behind. At equilibrium (a) The addition of NaNO3 favours the forward reaction. (b) The addition of NaNO2 favours the backward reaction. (c) Increasing pressure favours the forward reaction. (d) Increasing temperature favours the forward reaction. 61. The conditions which favours the manufacture of ammonia by the reaction  2NH3 + 21.9 Kcal, are N2+3H3  (a) Low temperature and high pressure (b) Low temperature and low pressure (c) High temperature and low pressure (d) High temperature and high pressure 62. 1 mol ethyl alcohol was treated with 1 mol of acetic acid at 25°C. At equilibrium, two-thirds of the acid into ester. The equilibrium constant for the reaction is (a) 4 (b) 3 (c) 2 (d) 1 63. The decomposition of N2O4 to NO2 was carried out in chloroform at at 280°C. At equilibrium, 0.2 mol of N2O4 and 2 × 10–3 mol of NO2 ware present in 2 L of solution. The equilibrium constant for the reaction  2 NO2 Is N2O4  (a) 0.01 × 10–4 (b) 2.0 × 10–3 –5 (c) 2.0 × 10 (d) 1.0 × 10–5 64. Solid ammonium carbamate dissociates to give NH3,  2NH3 (g) CO2 according to NH2 COONH4 (s)  + CO2(g). At equilibrium, ammonia is added such that partial pressure of NH3, now equal to the original total pressure. The ratio of total pressure now, to the original total pressure is (a) 27/31 (b) 31/27 (c) 4/9 (d) 9/4 65. The heat of endothermic reaction at constant volume in equilibrium state is 1200 cal more than the heat of reaction at constant pressure at 300 K. The ratio of equilibrium constants Kp and Kc is (a) 3.296 × 10–2 (b) 3.296 × 10–3 –2 (c) 1.648 × 10 (d) 1.648 × 10–3 66. The relation between ∆H , ∆S and equilibrium constant ‘K’ at a given temperature ‘T’ is O

O

O

(a) 2.303 log K =

−∆ H ∆ S + RT RT O

O

(b) 2.303 log K

∆H ∆S − RT RT

(c) 2.30 log K =

−∆H ° ∆S ° + RT R

(d) 2.3032 log K

∆H ° ∆S ° − RT R

O

O

O

O

O

Chemical Equilibrium

67. One mole N2 and 3 mole PCl5 are placed in a 100 litre vessel heated to 227°C. The equilibrium pressure is  PCl3+Cl2 calculate Kp for this 2.05 atm. PCl5  reaction. (a) 0.2 (b) 0.3 (c) 0.4 (d) 0.5 68. For the reaction (1) and (2)  B+C (a) A   2E (b) D  Given Kp1: Kp2 = 9:1 If the degree of dissociation of A and D be same then the total pressure at equilibrium (1) and (2) are in the ratio (a) 3:1 (b) 36: 1 (c) 1:1 (d) 0.5:1 69. Methane on reaction with steam gives CO and H2 CH4(g) + H2O(g)→CO(g)+3H2 (g) The products of the above reaction are used in the preparation of methanol at 500 K.  CH3OH(g)Kp= 10–2 atm–2 CO (g)+2H2(g)  What total pressure at equilibrium is required to convert 50% of CO to CH3 OH at 500 K? (a) 15 atm (b) 10 atm (c) 50 atm (d) 20 atm.  SOCl2 is exothermic and 70. The reaction SO2 + Cl2  reversible. The mixture is at equilibrium in a closed container. Now certain SO2 is added to the container. Which of the following is true? (a) The pressure inside the container will not change (b) The temperature will not change (c) The temperature will increases (d) The temperature will decreases  71. At the equilibrium of the reaction N2O4(g)  2NO2(g) the observed molecular weight of N2O4 is 80 g mol–1 at 350 K. The percentage of dissociation of N2O4(g) at 350 K is (a) 10% (b) 15% (c) 20% (d) 18% d [ A] K1    X +Y,72. For the given reaction A  K2 dt (a) K1[A] (b) K2[X [Y] (c) K1[A]+ k2 [X] [Y] (d) K1[A]- K2 [X] [Y] 73. Which of the following statement concerning the change in ∆G and ∆G during a chemical reaction is most correct? (a) ∆G remains constant while ∆G change and becomes equal to ∆G at equilibrium. (b) Both ∆G and ∆G remain constants during a chemical reaction. (c) ∆G remains constant if the reaction is carried out under non-standard conditions, ∆G remains O

constant if the reaction is carried out under-non standard conditions. (d) ΔG remains constants while ∆G change and becomes equal to zero at equilibrium. The equilibrium concentrations of A and B for the re C(g) are 3 M and 4 M respecaction A(g) + 2B(g)  tively. When the volume of the vessel is doubled and the equilibrium is allowed to reestablish the concentration of B is found to be 3 M. The equilibrium constant Kc for the reaction is (a) 12 (b) 4 (c) 0.25 (d) 0.0833 Consider the following equilibria:  B(s)  B(l) (I) A(s)  (II) 2A(l)   B(g)  B(g) (III) A(s)  (IV) A(g)  Which of the above will not be disturbed by increase in pressure? (a) I and IV (b) I, II and IV (c) II, III and IV (d) IV only 1 mole each of A and D is introduced into one lit container. Simultaneous two equilibriums are established  B+C; Kc = 106 M A   A; Kc = 10–6 M–1 B+D  Equilibrium concentration of ‘A’ will be (a) 10–6 M (b) 10–3 M –12 (c) 10 M (d) 0.5×10–3 M A reaction at 300 K with G = -1743 J/ mol consists of 3 mole of A(g), 6 mole of B(g) and 3 mole of C(g) If A, B and C are in equilibrium in 1 litre container then the reaction may be [Given: 2 = e0.7, R = 8.3 J/ mol]  C  B+2C (a) A+B  (b) A    2C (c) 2A  B+C (d) A+B  φ

9.48

74.

75.

76.

77.

O

multiple choice questions with one or more than one answer  2NO , K = 4 this reversible reaction is 1. N2O4  2 c studied graphically as shown in the figure. Select the correct statements: (a) Reaction quotient has maximum value at point A (b) Reaction proceeds left to right at a point when [N2O4] = [NO2] = 0.1 M (c) Kc = Q when point D or F is reached (d) Reaction quotient has minimum value at a point B A

O

O

O

D C

cmc

F

O

B

time

E G

Chemical Equilibrium 9.49

φ

φ

2. Which of the following is/are the correct statement (s) regarding the catalysed and uncatalysed reaction at constant temperature? (a) Catalyst does not alter ΔH of the reaction (b) ΔG of catalysed and uncataysed reaction is same (c) Catalyst alters equilibrium constant even at constant temperature (d) Catalyst alters Kf and Kb value at constant temperature. 3. Two gases A and B in the molar ratio 1:2 were admitted to an empty vessel and allowed to reach equilibrium at 400°C and 8 atm pressure as A(g) + 2B(g)  2C(g). The mole fraction of ‘C’ at equilibrium  is 0.4. The correct statements are 5 (a) Equilibrium constant Kp = atm -1 8 (b) The pressure at equilibrium when mole fraction of A is 0.1 is 196 atm (c) Partial pressure of ‘C’ is 16/5 atm (d) Partial pressure of A is 4/5 atm 4. Consider the reaction  2SO3 (g) + Heat 2SO2 (g) + O2(g)  What will happen if 0.5 mole of helium gas is introduced into the vessel so that the temperature and pressure remain constant? (a) The equilibrium concentration of SO2 increases (b) The equilibrium will shift to left (c) The equilibrium concentration of O2 increases (d) Equilibrium constant will increase 5. Find out the correct statements: (a) On decreasing pressure on the equilibrium H2O(s)  H2O(l), more ice will be formed  (b) At equilibrium if Kp = 1, then ΔG 0 17. 2CaSO4(s)  Above equilibrium is established by taking sufficient amount of CaSO4(s) in a closed container at 1600 K. Then which of the following may be the correct option? (Assume that solid CaSO4 is present in the container in each case) (a) Moles of CaO(s) will increase with the increase in temperature. (b) If the volume of the container is doubled at equilibrium then partial pressure of SO2 (g) will change at new equilibrium. (c) If the volume of the container is halved partial pressure of O2 (g) at new equilibrium will remain same. (d) If two moles of the He gas is added at constant pressure then the moles of CaO(s) will increase. 18. Hydrazine was taken in a constant volume container at 27°C and 0.3 atm equal moles of oxygen was injected, sealed and finally heated to 1000 K where the following equilibria was established.  2NO +2H O K N2H4+3O2  2 2 p 1  N2H4  N2+2H2 K p2  2NH3+2H2 N2H4+H2  K p3 If the gaseous mixture at equilibrium is passed through moisture absorbent, a decrease of 360 mm in the equilibrium pressure was observed. Now if the dried gaseous mixture is passed through ammonia absorber, a further decrease of 20 mm in the equilibrium pressure was observed. If KP1 = 4, the correct statements from the following are: (a) Equilibrium pressure of H2 is 0.4528 bar (b) Equilibrium pressure of N2H4 is 0.517 bar (c) K P = 9.24 × 10–2 bar2 2

(d) K = 2.95 × 10–3 P 3

 C2H6, 19. For the gas phase reaction, [C2H4+H2  ∆H = –32.7 Kcal] carried out in a vessel, the equilibrium concentration of C2H4 can be increased by (a) Increasing temperature (b) Decreasing pressure (c) Removing some H2 (d) Adding some C2H6  PCl (g)+Cl (g) the for20. For the reaction PCl5(g)  3 2 ward reaction at constant temperature is favoured by (a) Introducing an inert gas at constant volume (b) Introducing chlorine gas at constant volume (c) Introducing an inert gas at constant pressure (d) Increasing the volume of the container

Chemical Equilibrium 9.51

21. Following two equilibria are simultaneously established in a container  PCl3(g) + Cl2(g) PCl5   COCl2(g) CO(g)+ Cl2(g)  If some Ni(s) is introduced in the container establishing another equilibrium  Ni(CO)4(g) 4CO(g)+Ni(g)  When the first two reactions try to regain the equilibrium, which of the following happens? (a) Concentration of COCl2 decreases (b) Concentration of CO increases (c) Concentration PCl3 decreases (d) Concentration PCl5 increases  2NH +CO ; ∆H = +ve 22. NH2COONH4(s)  3(g) 2(g) Correct statement(s) regarding the above equilibrium is/are (a) Increase in temperature drives the reaction in the forward direction (b) Addition of NH2COONH4(s) increases the extent of forward reaction (c) Decrease in temperature decreases the equilibrium constant (d) Introduction of HCl(g) drives the reaction in the forward direction due to the formation of NH4Cl(s)

comprehensive type questions

 2CO(g) is such that 25 mol% of CO2 (g)+C(s)  the gas is CO2 1. What percentage would be of CO2 if the total pressure were 20 atm? (a) Ka2 > Ka3

The very low electrical conductivity of water is due to self ionization of water.  H + + OH H 2 O  [ H + ][OH - ] The dissociation constant of water K = [ H 2 O] Taking the concentration of undissociated water is nearly constant the ionic product of water Kw is given as Kw = [H+][OH–] The ionic product of concentration of H+ and OH+ ions at any temperature of pure water or an aqueous solution is known as ionic product of water and its value is 1 × 10–14 mole litre–2 at 25°C With increase in temperature ionic product of water also increases. In pure water or in neutral solutions the concentrations of H+ and OH+ ions are equal. At 25°C [H + ] = [OH − ] = K W = 1× 10−14 = 1× 10−7

Acidic strength decreases with increases in the strength of bond holding hydrogen atom e.g., H-F > H-Cl bond strength H-Br> H-I Increases in the size of atom with which hydrogen is in bond increases the acidic strength e.g.,

• •

HF < HCl < HBr < HI H 2 O < H 2 S < H 2 Se < H 2Te

•

Increases in the electronegativility of the atom which the hydrogen is in bond, increases the acidic strength CH4 < NH3 < H2O < HF Also the acidic strength increases with increase in the electronegativity of atom to which OH is bound. HOCl > HOBr > HOI For compounds containing OH group attached to the central atom acidic character increases with increase in the number of electronegative atoms attached to the central atom e.g.,

O

Ionization Constant of Water and Its Ionic Product

CCl3COOH > CHCl2COOH > CH2ClCOOH > CH3COOH CFH2COOH > CClH2COOH > CBrH2COOH > CIH2COOH •

O < HO − Cl =O O O

O

Acid catalysed hydrolysis of esters or the inversion of sugar can be used for comparing the strength two acids HA1 and HA2

•

If [ H + ] = [OH - ] in an aqueous solution, it is neutral If [ H + ] > [OH - ] in an aqueous solution, it is called acidic solution If [ H + ] < [OH - ] in an aqueous solution it is called basic solution If we know the concentration of any one of the H + and OH + ion the concentration of the other ion can be calculated. KW 1× 10-14 [H + ] = = [OH ] [OH - ]

[OH - ] =

KW 1× 10-14 = [H + ] [H + ]

ph scale •

pH scale was proposed by P.L Sorensen. To express the low concentration of hydrogen ion is known as pH scale

Ionic Equilibrium 10.45

• •

Whether a solution is neutral or acidic or basic is always expressed in terms of H+ ion concentration or pH. pH is the negative logarithm of H+ ion concentration or logarithm of reciprocal of H+ ion concentration 1 pH = - log[ H + ] or log + [H ]

tion of weak acid will be suppressed due to common ion effect of H+. or anion. This is known as common ion effect. Similarly the ionization of a base give hydroxide and corresponding cation. If any one of the product is added to the base solution the ionization of base will be suppressed e.g.,  CH 3COO - + H + CH 3COOH 

[ H + ] = 10- pH •

Negative logarithms of different terms can be represented as follow .pOH = - log[OH ] pK b = - log K b pM + = - log[M + ] pK e = - log K ( quilibrium coonstant) pKa = l- og Ka pK W = - log K W pH + pOH = 14 pH = 14 - pOH

• • • • •

• • •

•

•

•

•

Since in pure water and in neutral aqueous solutions pH is equal to 7. [ H + ] = [OH - ] = 1× 10-7, pH For strong bases, the concentration of OH– is equal to the normality or molarity × acidity of the base Several industrial processes and biological processes generally occur at specified pH values only. In acid solution, [H+] is greater than 1× 10-7 , so pH is less than 7 and for acidic solutions pH varies from 1 to 7 In alkaline solutions [H+] is less than 1× 10-7 , so pH is greater than 7 and for all basic solutions pH varies from 7 to 14. Since KW change with temperature the pH also change with temperature. With decrease in pH acidic character increases while basic character decreases. If acid is added to pure water the pH decreases and concentration of H+ ion increases. Similarly, if base is added to pure water, the pH increases and the concentration of H+ ion decreases. If the concentration of H+ ion increase by 10 times the pH decreases by one unit or if pH decreases by 1 unit the concentration of H+ ion increases by 10 times. In case of strong acids (which dissociate completely) the concentration of H+ ion is equal to the normality or molarity × basicity of the acid. Whenever equal volumes of two different solutions of pH between 1 and 6 are mixed the pH of the resultant will be 0.3 more than the lower pH. Whenever equal volumes of two different solutions of pH between 8 and 14 are mixed the pH of the resultant solution will be 0.3 less than the higher pH.

 NH 4+ + OH NH 4 OH  If an acid or sodium acetate is added to acetic acid the ionization of acetic acid will be decreased. Similarly, if a base or ammonium chloride is added to ammonium hydroxide, the ionization of ammonium hydroxide will be suppressed.

Types of salts •

• •

•

•

•

•

•

The reaction between an acid and a base to form a salt and water is called neutralization or the neutralization is a process of combining H+ from acid and OH– from base to from neutral water. The salts formed by the loss of all possible protons are called normal salts e.g., NaCl, Na2SO4 Na3PO4 etc. Salts formed by incomplete neutralization of poly basic acid are called acid salts. Such salts still contain one or more replaceable hydrogen atoms. E.g., NaHCO3, NaHSO4, NaH2PO4,Na2HPO4 etc. Salt formed by incomplete neutralization of poly acidic bases are called as basic salt. Such salts still contain one or more hydroxyl group e.g., Mg(OH)Cl, Fe(OH)2Cl, Bi(OH)2Cl etc. Salts formed by the addition of two independent stable salts which exist in solid state but ionize completely in their aqueous solutions and give tests for all the ions present in them are called double salts e.g., KCl. MgCl2 6H2O, alums etc. Salts formed by the addition of two independent stable compounds which exist in both solid and in solutions and do not ionize completely in water are called complex salt e.g., K4[Fe(CN)6]; [Co(NH3)6] Cl3 etc. The salt which furnishes more than one cation or more than one anion when dissolved in water are called mixed salts e.g., Ca (OCl)Cl; Na(NH4)HPO4 etc. The nature of solution formed during the neutralization is not always neutral. It depends on the particular acid and a particular base involved in the reaction and the nature of the salt formed in the neutralization.

Common Ion effect on the Ionization of Acids and Bases

hydrolysis of salt and the ph of Their solutions

•

•

If a strong acid or a salt having common anion to weak acid is added to the solution of weak acid, the ioniza-

The salt hydrolysis is opposite reaction to neutralization reaction of an acid and a base.

10.46

•

•

•

Ionic Equilibrium

Salt-hydrolysis is the reaction in which anion or cation or both of a salt react with solvent water to produce alkalinity or acidity. The alkalinity or acidity produced due to salt hydrolysis is very little. Salts of strong acid and strong bases do not hydrolyze. Thus solutions are neutral, pH of their solution is 7 e.g., KCl, NaCl, KBr K2SO4,KNO3 etc. The solution of salts of strong acid and a weak base hydrolyses in water producing acidic solution due to cationic hydrolysis. The pH of their solutions are less than 7. The hydrolysis constant Kh =

KW Kb

Buffer solution •

•

•

•

Degree of hydrolysis h for such salts is h=

KW Kb × C

pH of such solution can be calculated as

•

1 pH = [ pKW - log C - pK b ] 2 Salts of strong bases and weak acids hydrolyze in water producing basic solution due to anionic-hydrolysis. The pH of their solution are more than 7. The hydrolysis constant.

•

•

 CH 3 COO - + Na + CH 3 COONa 

K Kh = W Ka

•

Degree of hydrolysis h for such salt is

•

h=

•

•

KW Ka × C

pH of such solution can be calculated as 1 pH = [ pKW - log C - pK a ] 2 The nature of the solution of a salt of weak acid and weak base depends on their dissociation constant values. Their hydrolysis constant Kh =

pH of these solutions can be determined by pH =

•

•

KW K a × Kb

Degree of hydrolysis 'h' for such salt is KW h= K a × Kb 1 1 1 pKW + pK a - pK b 2 2 2

If pKa = pKb; pH = 7. If pKa>pKb, pH will be more than 7 and if pKa Ksp the solution is supersaturated solution and (3). If ionic product < Ksp the solution is unsaturated. The suppression of the degree of ionization of a weak electrolyte by the addition of strong electrolyte having an ion common with the weak electrolyte is known as common ion effect. Solubility product is useful in knowing (i) the solubility of an electrolyte (ii) in calculating the solubility of an electrolyte in the presence of a common ion (3). Simultaneous solubility of two electrolytes having common ion (4). Predicting the direction of ionic reactions (5). Precipitation of electrolytes. If a solid having general formula AX BY with molar solubility S, is in equilibrium with its saturated solu-

3

5

5

According to Le Chateliers principle, if the concentration of any one of the ions in equilibrium state of an electrolyte is increased, it should combine with the oppositely charged ion and some electrolyte is precipitated till the Ksp = Qsp where Qsp is the reaction quotient. If the concentration of the one of the ions is decreased more salt will dissolve to increase the concentrations of both the ions till Ksp = Qsp On saponification of oils and fats (hydrolysis with NaOH), soap and glycerol will be formed. To precipitate soap NaCl will be added. Due to common ion effect of Na+, soap will be precipitated and this is known as salting out of soap. oil + NaOH  → Sodium salt + glycerol  → C17 H 35COO - + Na + C17 H 35 COONa ←   → Na + + NaCl Cl Purification of common salt is based on the precipitation of salt by passing HCl gas into saturated solution of common salt (NaCl)

 Na + + Cl NaCl  HCl

 H + + Cl 

Application of solubility Product in Analysis •

By using the concept of solubility product and common ion effect, preferential precipitation of some

Ionic Equilibrium 10.49

•

•

cations from the mixture under controlled concentrations of H+ ion and reagent is carried out. When HCl is added to a solution containg mixture of cations only AgCl, PbCl2 and Hg2Cl2 are precipitated in Ist group. Since their solubility products are less. In the IInd group of qualitative analysis when H2S is passed in the presence of HCl, only second group cations (Pb2+, Hg2+, Cu2+, Cd2+, Bi3+, Sn2+, Sn4+, As3+ and Sb3+) are precipitated. In the presence of HCl, the ionization of H2S is suppressed due to common in effect.  2 H + + S 2 H 2 S 

HCl

•

 → H+

+

•

NH 4 Cl

Cl -

In that low concentration S2– ion, the ionic product of second group metal ion and the S 2– ion exceeds the solubility product of the second group metal sulphides and hence they are precipitated. But the ionic product values of other group metal sulphides are not exceeded their solubility products and thus remain in solution. In the presence of NH4Cl, the ionization of NH4OH is suppressed due to common ion effect of NH+4 ion  NH 4+ NH 4 OH 

•

+

OH -

 NH 4+ + Cl NH 4 Cl  In that low concentrations of OH– the ionic product of OH– ion and third group cation (Fe3+, Al3+ and Cr3+) will exceed the solubility product of their hydroxides and hence are precipitated. But the ionic product of other group metal hydroxides are less than the solubility products of their hydroxides. Thus remain in solution In the presence of NH4OH, the ionization of H2S increases due to the removal of H+ ions from equilibrium by neutralization  2 H + + S 2 H 2 S 

 NH 4+ + OH NH 4 OH 

In that high concentration of S2– ions the ionic product of S2– ion and IV group cation (CO2+, Ni2+, Zn2+ and Mn2+) will exceed the solubility product of their sulphides and hence are precipitated. In the presence of NH4Cl the ionization of (NH4)2 CO3 will be suppressed due to common ion effect  2 NH 4+ + CO32 ( NH 4 ) 2 CO3 

•

 → NH 4+

In that low concentration of CO32– the ionic product of 2– CO 3 ion and Vth group cation (Ba2+, Sr2+ and Ca2+) will exceed the solubility product of their carbonates? So they are precipitated as carbonates. But the ionic 2– products of CO 3 and Mg2+ will not exceed the solubility product of MgCO3, Thus remain in solution. Solubility of salts of weak acids increases at lower pH due to the decrease in the concentration of anion by protonation until Ksp = Qsp 2+  ZnS ↽ ⇀  Zn

S2 − + 2 H +

•

+ Cl -

+ S2 −

 → H 2S

H2S gas being sparingly soluble in water, escape out. So ZnS, MnS are soluble in hot dill HCl. If HCl or HNO3 is added to BaSO4, H2SO4 formed is also strong electrolyte ionises completely and is not removed from the equilibrium 2+ 1−  BaSO 4 ↽ ⇀  Ba + SO 4 SO 24 − + 2H +  → H 2SO 4

•

Hence BaSO4 is insoluble in water. If calcium acetate is added to oxalic acid calcium oxalate will be precipitated completely but if CaCl2 is added to oxalic acid precipitation does not take place completely. This is because when calcium acetate is added to oxalic acid, a weak acid CH3 COOH is formed but when calcium chloride is added a strong acid HCl is formed. So in the presence of HCl, the ionization of oxalic acid will be suppressed resulting in the incomplete precipitation.

10.50

Ionic Equilibrium

PRACTICe exeRCIse Multiple Choice Questions with Only One Answer Level I 1. The pH of 0.001 molar H2SO4 is (a) 3 (b) 11 (c) 2.7 (d) 0 2. NH3 is a base according to (a) Arhenious theory (b) Bronsted theory (c) Lewis theory (d) Both (b) and (c) 3. Which of the following is neither Bronsted acid nor Bronsted base? (a) HSO–4 (b) H3PO4 (c) BF3 (d) OH– 4. The suitable indicator for the titration of weak acid and strong base is (a) Methyl orange (b) Methyl red (c) Phenolphthalein (d) All the above 5. The CuSO4 aqueous solution is ………in nature due to…….. hydrolysis. (a) Basic, anionic (b) Neutral, No (c) Acidic, cationic (d) Amphoteric, both cationic and anionic 6. The aqueous solution of which of the following substances has high pH? (a) Na2CO3 (b) NaCl (c) AlCl3 (d) FeSO4 7. Which of the following maintains constant pH by the addition of small amount of acid or base? (a) NaOH + NaCl (b) H2SO4 + Na2SO4 (c) NaCN + HCN (d) CH3COOH + NH4Cl 8. The pH of a solution having double the alkaline concentration of water is (a) 14 (b) 0 (c) 8.2 (d) 7.3 9. If the ionic product of water at some high temperature is 10–12, then pOH is (a) 6 (b) 8 (c) 7 (d) 14 10. In water the acids HClO4, HCl, H2SO4 and HNO3 exhibit the same strength as they are completely ionsied in water acting as a base. This is called……….. of the solvent water. (a) Strength (b) Capacity (c) Buffer effect (d) Levelling effect 11. Aqueous solution CH3COOH contains (a) CH3COO–, H+ (b) CH3COO–, H3O+, CH3COOH (c) CH 3 COO - , H 3 O + , H + (d) CH 3 COOH , CH 3 COO - , H +

12. Central metal ion in a complex behaves as (a) Lewis base (b) Lewis acid (c) Bronsted acid (d) Arrhenius acid 13. Conjugate base of HPO42- is (a) PO43(b) H 2 PO4(c) H 3 PO4 (d) H 4 PO3 14. Which of the following does not act as Bronsted acid? (a) NH 4+ (b) HSO3(c) HCO3 (d) CH 3 COO 15. Arrange HClO 4 , HClO 3 , HClO 2 , HClO in increasing order of acid strength? (a) HClO < HClO2 < HClO3 < HClO4 (b) HClO4 < HClO3 < HClO2 < HClO (c) HClO4 < HClO2 < HClO < HClO3 (d) None 16. The pKa for acid A is greater than the pKa for acid B. The strong acid is? (a) Acid A (b) Acid B (c) Both a and b (d) Neither a nor b 17. The strength of an acid depends on its (a) Acidity (b) Basicity (c) Degree of dissociation (d) Molecular weight 18. Tribasic acid forms……. anions which exhibit basic and acidic properties. (a) 2 (b) 1 (c) 3 (d) 4 19. Which of the following is not amphoteric? (a) H 2 O (b) HSO43(c) PO4 (d) HPO42 20. A solution of pH 8 is…….. basic than a solution of pH 12. (a) Less (b) More (c) Equally (d) None 21. The buffering action of acidic buffer is maximum when its pH is equal to. (a) 5 (b) 7 (c) pKa/2 (d) pKa 22. pH of K2S solution is (a) >7 (b) CuS > Ag 2 S (c) Hg 2 S > Ag 2 S > CuS (d) Ag 2 S > CuS > HgS 10. pH of 0.1 M BOH (a weak base) is found to be 12. The solution at temperature TK will display an osmatic pressure equal to (a) 0.01 RT (b) 0.10 RT (c) 0.11 RT (d) 1.1 RT 11. The solubility of Mg(OH)2 is increased by adding NH4+ ion Mg(OH)2 +2NH+4  2NH3+2H2O+Mg2+. If Ksp (Mg(OH).) = 1 × 10–11 and Kb for NH4OH = 1.8 × 10–5, then Kc for the reaction is (a) 3.08 × 10–2 (b) 2.71 × 10–2 –3 (c) 5.10 × 10 (d) 5.77 × 10–2 12. In order to prepare a buffer of pH 8.26, the amount of (NH4)2SO4 required to mixed with 1 L of 0.1 MNH3 is [pKb = 4.74] (a) 1.0 mol (b) 10.0 mol (c) 0.50 mol (d) 5 mol 13. Find the ΔpH (initial pH – final pH) when 100 mL 0.01 M HCl is added in a solution containing 0.1 m moles of NaHCO3 solution of negligible volume (Ka1 = 10 –7 Ka2 = 10–11 for H2CO3): (a) 6 + 2 log 3 (b) 6 - log 3 (c) 6 + 2 log 2 (d) 6 - 2 log 3 14. 0.10 M solution of fluoride ions is gradually added to a solution containing Ba2+, Ca2+ and Pb2+ ions, each at a concentration of 1 × 10 –3 M. In what order, from first to last, will the precipitates of BaF2, CaF2, PbF2 form? Solubility Product, Ksp BaF2 1.8 × 10 -7 CaF2 1.5 × 10 -10 PbF2 7.1 × 10 -7 (a) CaF2, PbF2, BaF2 (b) BaF2, CaF2, PbF2 (c) PbF2, BaF2, CaF2 (d) CaF2, BaF2, PbF2 15. Mark the incorrect statements regarding NH4F solution. (a) Solution can act a buffer solution Kw (b) For the solution, K h = K a Kb (c) Solution involves only anionic hydrolysis (d) Degree of hydrolysis is equal to (Kh)1/2

16. Which of the following is correct decreasing order of solubility of AgCl? (I) In water (II) 0.1 M NaCl (III) 0.1 M BaCl2 (IV) 0.1 M NH3

17.

18.

19.

20.

21.

22.

(a) III>II>I>IV (b) IV>I>II>III (c) I>IV>II>III (d) I>II>III>IV Addition of hydrochloric acid to a saturated solution of cadmium hydroxide (Cd(OH)2, Ksp = 2.5 × 10–14) in water would cause: (a) the solubility of cadmium hydroxide to decrease. (b) the OH– concentration decrease and the Cd2+ concentration to increase. (c) The concentrations of both Cd2+ and OH– decrease. (d) The concentration of both Cd2+ and OH– increase. The solubility of CaCO3 = 8 mg/lit. Calculate the solubility product of BaCO3. From this information and from the fact that when Na2CO3 is added slowly to the solution containing equimolar concentration of Ca2+ and Ba2+ no precipitate of CaCO3 is formed until 80% of Ba+2 has been precipitated as BaCO3. (a) 1.28 × 109 (b) 1.28 × 10–10 –11 (c) 1.28 × 10 (d) 1.28 × 10–12 For the reaction [Ag (CN)2]  Ag+ 2CN–, the equilibrium constant at 25°C is 4 × 10–19.If a solution is 0.1 M in KCN and 0.03 M in AgNO3 originally, at equilibrium, the conc. of Ag+ is (a) 7.5 × 10–16 M (b) 7.5 × 1018 M –19 (c) 1.25 × 10 M (d) 1.25 × 10–17 M If 500 mL of 0.4 MAgNO3 is mixed with 500 mL of 2 MNH3 is solution then what is the concentration of [Ag ( NH 3 ) +2 ] in solution? Given K f1 [Ag ( NH 3 ) + = 103 ; K f [Ag( NH 3 ) 2 ]+ = 104 2 (a) 3.33 × 10–7 M (b) 3.33 × 10–5 M (c) 3 × 10–4 M (d) 10–7 M A buffer solution containing 0.04 M Na 2HPO4 and 0.02 M Na3PO4 is prepared. The electrolytic oxidation of 1 millimole of the oraganic compound RNHOH is carried out in 100 mL of the buffer. The reaction is: RNHOH + H 2 O  → RNO2 + 4 H + + 4e Given that for H 3 PO 4 pK a 1 , and pK a3 are 4, 8 and 12 respectively; the dip in pH after electrolytic oxidation will be: (a) 11.7 (b) 8.3 (c) 3.4 (d) 1 Four species are listed below: (I) HCO–3 (II) H3O+ – (III) HSO4 (IV) HSO3F Which one of the following is correct sequence of their acid strength? (a) II MX (c) MX2 > M3X > MX (d) MX > M3X > MX2 26. 2.5 mL of 2 M weak monoacid base (Kb = 1 × 10–12 at 5 2 MHCl in water at 25°C. the 25°C) is titrated with 15

27.

28.

29.

30.

concentration of H+ at equivalence point is (Kw = 1 × 10–14 at 25°C) (a) 3.7 × 10–13 M (b) 3.2 × 10–7 M –2 (c) 3.2 × 10 M (d) 2.7 × 10–2 M During the titration of 100 mL of a weak monobasic acid solution using 0.1 M NaOH, the solution became neutral at 40 mL addition of NaOH and equivalence point was obtained at 50 mL NaOH addition. The Ka of the acid is (log 2 = 0.3) (a) 1 × 10–7 (b) 2 × 10–7 –7 (c) 3 × 10 (d) 4 × 10–7 Equal volumes of the following solutions? are mixed. In which of the following case, the pH of resulting solution will be the average of the two solutions. Given Ka (HCN) = 1010 Ka of acetic acid is equal to Kb of ammonia. (a) HCl (pH = 3) and NaOH (pH = 12) (b) HCl (pH = 2) and HCl (pH = 4) (c) HCN (pH = 2) and NaOH (pH = 12) (d) CH3COOH (pH = 5) and NH3 (pH = 9) Calculate the ratio of pH of a solution containing 1 mole of CH3COONa+1 mole of HCl per litre and of the other solution containing 1 mole of CH3COONa + 1 mole CH3COOH per litre. (a) 1/6 (b) 1/4 (c) 1/8 (d) ½ 20 mL of 0.2 M NaOH are added to 50 mL of 0.2 M acetic acid. pKa = 4.74. Then pH of the solution is (a) 4.74 (b) 4.56 (c) 4.92 (d) 5.10

31. The solubility product of AgCl is 10–10 at 25°C. A solution of Ag+ ion at a concentration of 4 × 10–3 M just fails to yields AgCl with a concentration of 1 × 10–3 M of Cl– ion when the concentration of NH3 in solution is 2 ×10–2 M. The equilibrium constant of Ag+ + 2NH3  [Ag (NH3)2] + is (a) 10–8 (b) 108 –8 (c) 2 × 10 (d) 2 × 108 32. The solubility of silver formate in pure water is 10–2 mol/lit. Ka of formic acid is 1 × 10–5. The solubility of silver formate in buffer solution of pH = 3 is (a) 10–1 mol/lit (b) 10–2 mol/lit –3 (c) 10 mol/lit (d) 10–4 mol/lit 33. 0.2 mole of NH4Cl dissolved in 1000 gm of water lowered the freezing point by 0.71°C. Degree of dissociation of salt is 0.75. Kf for H2O is 2 K kg mol–1. The degree of hydrolysis of salt is (a) 0.033 (b) 0.066 (c) 0.049 (d) 0.099 34. What is the pH of an aqueous solution of 0.1 M ammonium formate assuming complete dissociate? pKa of formic acid = 3.8 and pKb of ammonia = 4.8 (a) 6.5 (b) 7.5 (c) 8.5 (d) 9.5 35. The dissociation constants for aniline, propanoic acid and water are 9 ×10–10, 4 × 10–6 and 10–14 respectively. The degree of hydrolysis of 0.1 N aniline propanoate solution is (a) 16% (b) 32% (c) 47.2% (d) 62.5% 36. Ksp for SrF2 = 5 × 10 -9 at 25°C. How much NaF should be added to 100 mL of solution having 0.015 M Sr+2 ions to reduce its concentration to 0.005 M? (a) 0.1764 gm (b) 0.0882 gm (c) 0.0661 gm (d) 0.0441 gm 37. 25 mL clear saturated solution of PbI 2(aq) requires 12.5 mL of AgNO 3(aq) solution for complete neutralization. Ksp of PbI 2 is 4 × 10 –9. So conc. of AgNO 3 is (a) 2 × 10-3 M (b) 3 ×10 -3 M (c) 4 × 10-3 M (d) 5 ×10-3 M 38. For an indicator, the value of pH = 2 when half of the indicator is present in the unionized form. The pKin of the indicator is (a) 2 (b) 3 (c) 4 (d) 5 39. The dissociation of water at 250°C is 1.9 × 10–7 percent and the density of water is 1.0 g/cm3. The ionization constant of water is: (a) 3.42 × 10–6 (b) 3.42 × 10–8 –14 (c) 1.00 × 10 (d) 2.00 × 10–16

Ionic Equilibrium

40. pKa value of CH3COOH is 4.78. Which of the following statement is correct? (a) pH of 0.1 M CH3 COOH is 4.78 (b) pH = 4.78 then degree of dissociation of CH3COOH is maximum (c) pH = 4.78 then degree of dissociation of CH3COOH is minimum (d) pH = 4.78 then 50% CH3COOH is dissociated 41. One litre of buffer solution of pH = 6.7 can be prepared by 0.005 mole Na H 2PO4 and X mole of Na 2H PO4. K 2 for H 3PO4 = 6 × 10–8. The value of X is (a) 0.1 (b) 0.01 (c) 0.001 (d) 0.0001 42. The solubility of the compound Tl 2S in pure wa ter is 3 × 10 –6 mol/lit. Assume that the dissolved S–2 ion hydrolysis almost completely into HS – and that the further hydrolysis to H 2S can be neglected. K2(H2S) = 10–14. The solubility product of the com pound is (a) 9 × 10–12 (b) 8.1 × 10–22 (c) 8.1 × 10–12 (d) 3.24 × 10–22 43. Approximate pH of 0.10 M aqueous H2S solution having K1 and K2 for H2S at 25°C 10–7 and 10–13 respectively is (a) 4 (b) 5 (c) 9 (d) 8 44. The self-ionization constant for pure formic acid, K = [HCOOH+2] [HCOO–] has been estimated as 10–6 at room temperature. The density of formic acid is 1.22 g/cm3. The percentage of formic acid molecules in pure formic acid converted to formate ion is (a) 0.002% (b) 0.004% (c) 0.006% (d) 0.008% 45. 300 mL of saturated clear solution of CaC2O4 (aq) requires 6 mL of 0.001 M KMnO4 (aq) in acid medium for complete oxidation of C2O4–2 ions. The Ksp of CaC2O4 is (a) 5 × 10–9 (b) 5 × 10–10 –9 (c) 2.5 × 10 (d) 2.5 × 10–10 46. How many number of mole of AgI which may be dissolved in 1 lit. of 1 M CN–solution Ksp for AgI  Ag(CN)2– Kc = 9 = 1 × 10–7 M2? Ag+ + 2 CN–  19 –2 × 10 M (a) 0.245 (b) 0.3675 (c) 0.491 (d) 0.735 47. Both Ag CNS and AgBr is dissolved in a water. Ksp of AgBr = 5 × 10–13 and Ksp of AgCNS = 1 × 10–12. The solubility of AgBr in solution is (a) 4 ×10–7 M (b) 2 × 10–7 M –7 (c) 8.16 × 10 (d) 6.2 × 10–7 M

48. It is found that 0.1 M solution of four sodium salts NaA, NaB, NaC, NaD, have the following pH values. Which one of the corresponding acids is strongest? (a) NaD,11.0 (b) NaC, 10.0 (c) NaB, 9.0 (d) NaA,7.0 49. An Acid –base indicator has Ka = 3.0 × 10–5. The acid form of the indicator is red and the basic form is blue. The change in [H+] required to change the indicator form 75% red to 75% blue is (a) 8 × 10–5 M (b) 9 × 10–5 M –5 (c) 1 × 10 M (d) 3 × 10–4 M 50. At what molar concentrations of HCl will its aqueous solution have an [H+] to which equal contribution come from HCl and H2O at 90°C [Kw of H2O = 10–12 M2 at 90°C] 50 ×10-8

(a)

40 ×10-7 M

(b)

(c) 50 ×10-7 (d) 30 ×10-7 51. 100 mL of 0.5 M hydrazoic acid (N3H Ka = 3.6 × 10–4 and 400 mL of 0.1 M cyanic acid (HOCN, Ka = 8 × 10–4) are mixed. Which of the following is true for final solution? (a) [H+] = 2 × 10–2 M (b) [N3–] = 3.6 × 10–2 M (c) [OCN–] = 4.571 × 10–3 M (d) [OCN–] = 6.4 × 10–3 M 52. Adding a few drops of hydrochloric acid to a beaker containing water results in the formation of hydrated species. Identify which hydrated species can exist in solution. (I) H3O+ (II) H3O–2 + (III) H5O2 (IV) H2CIO– (a) I only (b) I and II (c) I, II, and III (d) I, II, III and IV 53. Consider the titration curve shown below. 12 11 ● ● ● ● ● ●

10

● ●

9

●

8

equialence point ●

7

pH

10.58

●

6

●

5

●

4

● ●

3

● ●

● ●

●

2 1 0 0

1

2

3

4

5

6

7

8

9 10 11 12 13 14 15

volume added (ml)

Ionic Equilibrium 10.59

The titration curve represents the titrations of: (a) a strong acid (breaker) with a strong base (burette). (b) a weak acid (beaker) with a strong base (burette) (c) a strong base (beaker) with a strong acid (burette) (d) a weak base (beaker) with a strong acid (burette) 54. When bromocresol green is dissolved in aqueous solution, an equilibrium is established between bromocresol green (Hln, a weak minoprotic acid), the anion (conjugate base) of bromocresol green (In) and H+ ions: H ln(aq, Yellow) H + (aq, colorless ) + In - (aq, blue - green)

55.

56.

57.

58.

If a small amount of bromocresol green is dissolved in a buffer solution at pH 11, the colour of the resulting solution would be closest to: (a) Colorless (b) Yellow (c) Blue green (d) Red Which of the following combinations of weak-acid dissociation constants and molar concentrations would correspond to the SMALLEST per cent dissociation? 0 (a) K a = 1.7 × 10-4 , CHA = 0.001M -2 0 (b) K a = 1.8 × 10 , CHA = 0.01M 0 (c) K a = 6.5 × 10-5 , CHA = 0.1M -5 0 (d) K a = 6.5 × 10 , CHA = 0.01MS The solubility product of AgCl is 1.8 × 10–10. The minimum volume (in L) of water required to dissolve 1.9 mg of AgCl is approximately (a) 10 (b) 2 (c) 1 (d) 20 What is the aq ammonia concentration of a solution prepared by dissolving 0.15 mole of NH+4 CH3 COO– in -5 ILH2O.[Ka(CH3 COOH) = 1.8 × 10–5 ;; K b ( NH OH ) = 1.8 × 10-5 ] 4 –4 3 (a) 8.3 × 10 (b) 0.15 (c) 6.4 × 10–4 (d) 3.8 × 10–4 A 1 litre solution containing NH 4Cl and NH 4OH has hydroxide ion concentration of 10 –6 mol/lit which of the following hydroxides could be precipitated when the solution is added to 1 litre solution of 0.1 M metal ions. (I) Ag(OH) (K sp = 5 × 10 -3 ) (II) Ca(OH) 2

(K sp = 8 × 10 -6 )

(III) Mg(OH) 2

(K sp = 3 × 10 -3 )

(IV) Fe(OH) 2

(K sp = 8 × 10 -16 )

(a) I,II,IV (b) IV (c) III and IV (d) II, III,IV 59. Liquid ammonia ionizes to a slight extent. At -50°C, its ionic product is 10–30 The number of anions present per mm3 of pure liquid ammonia is (a) 6 × 10–2 (b) 17 × 10–15 23 (c) 6 × 10 (d) 6 × 105

60. The degree of hydrolysis of 102 M NH4 CN in aqueous solution is 0.46 if the concentration of NH4CN increased four folds, the new degree of hydrolysis would be Given Ka (HCN) = 7.2 × 10–20 and Kb (NH4OH) = 1.8 × 105 at 25C (a) 0.23 (b) 0.46 (c) 0.92 (d) 0.100 61. Consider an aqueous solution that contains 0.10 M each of Pb2+, Hg2+ and Ni2+ ions. If a solution containing 2 × 10–20 M S 2– ion is added to the solution containing the metal ions then which sulphides will precipitate from the solution. Given: Ksp(PbS) 8 × 10–28; Ksp(HgS) = 4 × 10–53 and Ksp(NiS) = 3.2 × 10–19 (a) PbS, HgS and NiS (b) PbS and HgS (c) HgS and NiS (d) HgS only 62. A weak monoacidic base is titrated against HCI of 0.3 M and end point has reached on adding 20 mL of the acid. To this solution, 120 mg of NaOH if added. The pH of the resulting solution would be [pKb of the base 4.2] (a) 4.2 (b) 8.4 (c) 9.8 (d) 102 63. Solubility of a sparingly soluble salt Ba3(PO4)2 in mixture of

64.

65.

66.

67.

1 1 M BaCl2 (a = 1) and M Ba(NO3)2 20 20

(a = 1) is [Ksp Ba3 (PO1)2 =10–7 M5. At mass Ba = 137, O = 16] (a) 300.5 mg/lit (b) 601 mg/lit (c) 3005 mg/lit (d) 6010 mg/lit The ionization constants for H2S are K1=1.0 × 10–7, K2 = 1.0 × 10–14. The pH of 0.005 M Na2S solution is (a) 2.3 (b) 11.7 (c) 10.7 (d) 12.7 A buffer solution of pH = 6.7 can be prepared by employing solution of NaH2PO4 and Na2HPO4. If 0.005 mol of NaH2PO4 is weighed out. The no. of moles of Na2HPO4 must be used to make 1 litre of the solution is K2 for H3PO4 = 6 × 10–8 (a) 0.0015 (b) 0.003 (c) 0.0045 (d) 0.006 The self ionization equilibrium of pure formic acid 2HCOOH  HCOOH 2+ + HCOO - + has ionization –6 2 2 product K = 10 mol /lit at 298 K. If percentage ionization is 4 × 10–3. The density of acid in g/mL is (a) 1.15 (b) 1.725 (c) 2.3 (d) 2.875 Solid BaF2 is added to solution containing 0.1 mole of sodium oxalate solution (1 lit) until equilibrium reached if the Ksp of BaF2 and BaC2O4 is 10–6 mol3 lit3. The conc. C2 O −42 in solution is (a) 7.4 × 10–2 M (b) 3.7 × 10–2 M –3 (c) 7.4 × 10 M (d) 3.7 × 10–3 M

10.60

Ionic Equilibrium

68. The solubility product of AB is 4 × 10–10 at 18°C the loss of weight of precipitatation of AB by washing it with 5 litres of water is: (a) 2 × 10–5 mol (b) 10–4 mol (c) 4 × 10–5 mol (d) 2 × 10–4 mol 69. Consider the following reactions of X: (I) [Al (OH)3. (H 2 O)3 ] + OH  [Al(OH) 4 (H 2 O) 2 ]- + H 2 O  (II) [Al(OH)3 (H 2 O)3 ] + H 3 O + [Al(OH) 2 (H 2 O) 4 ]+ + H 2 O 

70.

71.

72.

73.

74.

75.

(a) X is an acid in I and base in II (b) X is a base in I and acid II (c) X is a base in I and II both (d) X is an acid in I and II both pH of a mixture which is 0.1 M in CH3COOH and 0.05 M in (CH3COO)2 Ba is [pKa of CH3COOH = 4.74] (a) 4.74 (b) 2.04 (c) 4.44 (d) 7.00 Solution s a mixture of 0.05 M NaCI and 0.05 M Nal. The concentration of iodide ion in the solution when AgCl just start precipitating is equal to: ( K sp AgCl = 1 × 10 -10 M 2 ; K sp Agl = 4 × 10 -16 M 2 ) (a) 4 × 106 M (b) 2 × 10–8 M 7 (c) 2 × 10 M (d) 8 × 10–15 M Which of the following is most soluble in water? (a) MnS (Ksp = 8 × 10–37) (b) ZnS (Ksp = 7 × 10–16) (c) Bi2S3 (Ksp = 1 × 10–72) (d) Ag3(PO4)(Ksp = 1.8 × 10–18) The pH of 0.2 M NaHCO3 solution at 25°C is 9.2. A 22 mL 2.0 M solution of H2CO3 when treated with 80 mL 0.5 M NaOH results into formation of H2CO3 + NaHCO3 buffer with pH of 8.6 Hence pKa2 of H2CO3 is (a) 10.8 (b) 7.6 (c) 9.2 (d) 8.6 When a salt of weak acid and weak base is dissolved in water, the pH of the resulting sloution will be (a) 7 (b) Depends on the value of Ka and Kb (c) Less than 7 (d) Gretor than 7 The position of equilibrium lies to the right in each of the following reaction: N 2 H 5+ + NH 3  NH +4 + N 2 H 4 NH 3 + HBr

 NH +4 + Br -

N 2 H 4 + HBr  N 2 H 5+ + Br -

Select the correct order of acidic strength: (a) HBr > N 2 H 5+ > NH 4+ (b) HBr > N 2 H 4 > NH 3 (c) HBr > NH 4+ > N 2 H 5+ (d) HBr > NH 3 > N 2 H 4 76. H2CO3 is diprotic acid for which Ka1 = 4.2 × 10–7 and Ka2 = 4.7 × 10–7 and solution will produce a pH close to 9? (a) 0.1 M H2CO3 (b) 0.1 M Na2CO3 (c) 0.1 M NaHCO3 (d) 0.1 M Na2CO3 and 0.1 M NaHCO3

[ Br - ]

does the following cell have its [CO32 - ] reaction at equilibrium? Ag (s) | Ag 2 CO3 (s) | Na 2 CO3 (aq ) || KBr (aq )| Ag (s)

77. At what

K sp = 8 × 10 -12 for Ag 2 CO3 and K sp = 4 × 10 -13 for AgBr (a) 1 × 10 -7 (b) 2 × 10 -7 -7 (c) 3 × 10 (d) 4 × 10 -7 78. Experiment #1: A mixture of water and AgCl(s) is shaken until a saturated solution is obtained. Now the solution is filtered and to 100 mL of the clear filtrate, 100 mL of 0.0003 M NaBr is added. Experiment #2 A mixtures of water and AgBr(s) is shaken unit a saturated solution is obtained. Now the solution is filtered and to 100 mL of the clear filtrate, 100 mL of 0.003 M NaCl is added. Given Ksp (AgCl) = 10–10 and Ksp (AgBr) = 10–14 at 25°C precipitation would happen in: 79. The pH curve of the titration of weak acid with a strong base is given below: S

R pH

Q

P Volume of Titrant (base) Now choose the correct option among the following: (a) pH at point p =

1 1 pKa - log[ A0 ] Where A0 is 3 2

the initial concentration of weak acid 1 ( weak acid ) (b) pH at point Q = pKa. - log 2 ( Salt ) 1 1 1 (c) pH at point R = pKw + pKa + log[ salt ] 2 2 2 1 1 (d) pH at point S = pKw + log[ Base] 2 2

Ionic Equilibrium 10.61

80. If VB litre volume of strong base (MOH) is added to VA litre of a solution of a weak acid (HA) of initial molar concentration A0, then [A] in the titration flask may be given by which of the following expression? (Ka is the dissociation constant of HA) Ao VA K a (a) [ A- ] = (VA + VB ) × (1 + K a ) (b) [ A- ] =

Ao K aVA (VA + VB )([ H 3 O + ] + K a )

(c) [ A- ] =

Ao K aVA (VA + VB )([OH - ] + K a )

(d) [ A- ] =

Ao K aVB (VA + Vb )([ H 3 O + ]) + K a )

81. Depression in freezing point of aqueous solution of 0.1 M NaA and 0.1 M NaCl are in the ratio 21:20. The equilibrium constant of weak acid HA is (a) 2 × 10–9 (b) 2 × 10–10 –12 (c) 9 × 10 (d) 2 × 10–12 82. H2S is bubbled into 0.2 M NaCN solution which is 0.02 M in each [Cd (CN)4]2– and [Ag(CN2]–. The H2S produces 1×10–9 M sulphideion in the solution Given, K sp Ag 2S = 1 × 10 -50 M 3 ; K sp CdS = 7.1 × 10 -28 M 2 K inst [Ag (CN) 2 ]-1 = 1 × 10 -20 M 2 . K inst [Cd (CN) 4 ]-2 = 7.8 × 10 -18 M1 Identify the correct statement. (a) Ag2S precipitates first from the solution (b) CdS precipitates first from the solution (c) None of them precipitates under the given conditions (d) Ag2S precipitates at a sulphide concentration 1 × 1015 M

Multiple Choice Questions with One or More Than One Answer 1. In 0.020 M carbonic acid solution (a) H2CO3 is stronger acid than HCO3(b) H 2 CO3  2H + + CO32 - , K eq = K a .K a 1 2 (c)  HCO3-  ≈ [CO32 - ] (d) it can be said K a  K a 1 2 2. The acid dissociation constant for Al ( H 2 O)36+ is 1.4 × 10–5. It suggests (a) H2O molecules in the hydrated cation are much stronger proton donors than are free solvent water molecules (b) Ionization: Al(H 2 P)36+ (aq)  Al3+ (aq) + 6H 2 O (c) Ionization: Al(H 2 O)36+ (aq) + H 2 O(l)  H 3 O + (aq) + Al(H 2 O)5 (OH) 2 + (aq) (d) Its pH might be more than 7

3. Which of the following statements are correct? (a) Larger the value of dissociation constant, greater is the strength of acid. (b) All Bronsted bases are Lewis bases but all Bronsted acids are not Lewis acids. (c) Degree of dissociation of water increases with the rise of temperature. (d) Degree of hydrolysis of a salt of weak acid and weak base does not change with dilution. 4. Which of the following combination of solutes would result in the formation of a buffer solution? (a) NH3+NH4Cl (b) CH3COOH+NaOH in 2: 1 molar ratio (c) NH3+HCl in 2:1 molar ration (d) CH3COOH+NaOH in 1:2 molar ratio 5. An acid-base indicator Hln exists 50% dissociated at pH = 9. The acid colour of the indicator is yellow while its base colour is red. The red colour predominates when the concentration of conjugate base is at least eight times that of the acid while for yellow colour to predominate, the concentration of the species possessing this colour must be at least twenty times that of the species possessing red colour. Now read the following statements and choose the correct statement (S) from the key given below (a) The pH range of the indicator is 7.7-9.9 (b) The pH range of the indicator is 8.2-10.3 (c) The indicator may serve to indicate the end point of the titration of HCl vs. NH4OH (d) The indicator may serve to indicate the end point of the titration between HCl and NaOH 6. Choose the correct alternatives. (a) Ksp of Fe(OH)3 in aqueous solution is 3.8 × 10-38 at 298 K. The concentration of Fe3+ will increases with decreases in H+ ion concentration (b) In a mixture of NH4Cl and NH4OH in water a further amount of NH4Cl is added. The pH of mixture will decreases. (c) An aqueous solution of each of the following salts NH4I, Cr(NO3)3, KCN will be basic, acidic, and neutral respectively. (d) Degree of hydrolysis of 0.1 M CH3COONH4, 0.2 M CH3COONH4 is same. 7. In photography, equiomole is used as developer according to the following reaction. HO

OH + 2AgBr + 2OH

O=

= O+2Ag+2H2O+2 Br

Which of the following describe (s) the role of equinol in this reaction? (a) It acts as an acid (b) It acts as a weak base (c) It acts as an oxidizing agent (d) It acts as a reducing agent

10.62

Ionic Equilibrium

8. Which of the following is/are true about alkalinity? (a) Alkali metal hydroxides are water-soluble ionic solids, most familiar bases (b) Alkaline earth oxides, such as CaO are weaker bases than corresponding hydroxides (c) 0.28 g of lime (CaO) in 1 lit. water gives a solution of pH =12 (d) 10–7 M NaOH solution has pH value 7 9. An acid-base indicator has Kb equal to 1.0×10–6. The acid form of the indicator is blue and the basic form is red then (a) At pH 8, solution remains complete (99%) red (b) At pH 10, solution remains complete (99%) red (c) At pH 6, solution remains complete (99%) blue (d) At pH 8, solution remains complete (99%) blue 10. On mixing equal volumes of the following solutions, precipitation of AgCl will occur with (Ksp for AgCl is 1.8×10–10 (a) 10–3 M (Ag+) and 10-3 M (Cl - ) (b) 10–4 M(Ag+) and 10-4 M (Cl - ) (c) 10–5 M(Ag+) and 10-5 M (Cl - ) (d) 10–4 M(Ag+) and 10-10 M (Cl - ) 11. For pure water (a) pH increases with increase in temperature (b) pH decreases with increase in temperature (c) pH = 7 at temperature of 25°C (d) pH increases at low temperature but decreases at high temperatures 12. Which is/are the wrong statements? (a) All Arrhenius acids are Bronsted acids but all Arrhenius bases are not Bronsted bases (b) All Bronsted bases are Lewis bases (c) All Bronsted acids are Lewis acids (d) Conjugate acid of strong base is a strong acid 13. In the following reaction [Al (H 2 O)6 ]+3 + H 2 O  H 3 O + + [Al(H 2 O)5 OH]+2 A

B

C

D

(a) A is an acid, B is a base (b) A is a base, B is an acid (c) C is conjucate acid of B and D is conjucate base of A (d) C is conjucate base of B and D is conjucate acid of A 14. Which statement about solubility product is correct? (a) If ionic product < solubility product, solution is unsaturated (b) For AlCl3 Ksp = 27S4, Hence ‘S’ solubility in H2O (c) If ionic product > solubility product, precipitate occurs (d) Solubility AlCl3 in 0.1 M KCl is S. then Ksp = 0.001 S 15. Which of the following statements are correct? (a) NH4OH is a weak base (b) NH4Cl forms an acidic solution in H2O

(c) H3BO3 tripotic acid (d) CH3COONa forms a basic solution in water 16. In mixing equal volumes of the following solutions, precipitation of AgCl will occur (Ksp of AgCl = 1.8 × 10–10) (a) 10–5 M Ag+ and 10–5 M Cl– (b) 10–10 M Ag+ and 10–10 M Cl– (c) 10–3 M Ag+ and 10–6 M Cl– (d) 10–4 M Ag+ and 10–4 M Cl– 17. Which of the following salts will not undergo hydrolysis? (a) NaCl (b) KCl (c) NH4Cl (d) CH3COONa 18. Which of the following is true for alkaline aqueous solution? pk (a) pH >> w (b) pH > pOH 2 pk (c) pOH < w (d) pH < pOH 2 19. 0.56 g CaO lime is dissolved in 100 mL water and made a solution (KW = 10–13). The correct statements are (a) Concentration of OH - = 0.2 M (b) Conc. H+ would be = 5 × 10-13 M (c) Degree of dissociation of water in the solution 9 × 10-15 (d) If 1.12 g of CaO is added to 100 mL at same temperature then conc. OH– is = 0.4 M 20. The variation of PH during the titration of 0.5 N Na2CO3 with 0.5N HCl is shown in the given graph. The following table indicates the colour and pH ranges of different indicators Indicator

Range of Colour in colour change acid

Colour in base

Thymol blue Bromocresol red Bromocresol blue Cresolphethalein

1.2 to 2.8 4.2 to 6.3 6.0 to 7.6 8.2 to 9.8

yelloow Yellow Blue Red

Red Red Yellow Colourless

14 12 10 pH 8 6 4 2 0

10 20 30 40 50 60 vol. of HCl (mL)

Ionic Equilibrium 10.63

21.

22.

23.

24.

25.

Based on the graph and the table, which of the following statements are true? (a) The first equivalence point can be detected by cresolphthalein (b) The complete neutralization can be detected by bromothymol blue (c) The second equivalence point can be detected by bromocresol red (d) The volume of HCl required for the first equivalence point is half the total volume of HCl required for the second equivalence point Let the colour of the indicator Hln (colourless) will be visible only when its ionised form (pink) is 25% or more in a solution. Suppose Hln (pKa = 9.0) is added to a solution of pH = 9.6 predict what will happen? (Take log 2 = 0.3) (a) Pink colour will be visible (b) Pink colour will not be visible (c) % of ionized form will be less than 25% (d) % of ionized form will be more than 25% Titration of 664 mg of pure organic carboxylic acid, H2A, showed a rapid change in pH at 40 mL and at 80 mL of 0.1 M NaOH titrant. When 40 mL of NaOH was added, the pH was 5.85, and 60 mL of NaOH was 8.08. Which of the following statements are correct? (a) The molecular weight of acid is 180 (b) The PKa of acid is 3.62 1 (c) The PKa of acid is 8.08 2 (d) The PKa is 5.85 and PKa is 8.08 1 2 Pick out the correct statement (s). (a) An aqueous solution of a weak base on being diluted its PH decreases (b) pKa of CH 3COOH is 5 and that of HCN is 10. Out of 0.1 M each the aqueous solutions of CH3COONa and NaCN, aq. Solution of NaCN is more basic (c) The molar volume of any real gas at NTP is 22.414 L. (d) Henry’s law is not applicable for the solubility of CO2 gas in water Which of the following mixture can act as a buffer? (a) NaOH + HCOONa (1:1 molar ration) (b) HCOOH + NaOH (2:1 molar ratio) (c) NH4Cl + NaOH (2: 1 molar ration) (d) HCOOH + NH4OH (1:1 molar ratio) Pick out the correct statements among the following (a) pH of NaCN is more than the of CH3COONa in their decinormal solutions [pKa: CH3COOH=5, HCn=10] (b) With the increase of temperature the equilibrium pressure of the reaction,  CaO( s ) + CO2 ( g ) increases CaCO3(s) 

(c) The degree of hydrolysis of NH4CN salt increases with increase of dilution of its solution (d) Addition of water to the equilibrium mixture  2 NH 3( g ) shifts the equilibrium N2(g)+2H2(g)  more towards right hand side. 26. How many of the following reagents dissolve AgCl to a larger extent than water? (a) Con. HCl (b) NH3 (c) Aqueous KCN (d) Dil HCl 27. Which of the following salts tend to dissolve to a higher extent in an acidic buffer compared to pure water? (a) PbI2 (b) Ag3PO4 (c) Mg(OH)2 (d) PbS 28. The degree of ionization of a compound depends on (a) Size of solute (b) Nature of solute (c) Nature of vessel (d) Quantity of electricity passed

Comprehensive Type Questions Passage I Three bottles labelled A, B, C contain 150 mL of 0.1 M weak mono protic acid, 50 mL 0.1 M strong mono acidic base, 150 mL 0.1 M strong monoprotic acid respectively. A student performs some experiments, then the following observations are made: (a) When solution A and B are mixed then it form a buffer of pH = 4.7 (b) When B and C are mixed then solution is acidic (c) When A and C are mixed then solution is acidic and pH is 2 –log 5 1. What would be the ionization constant of weak acid? (a) 2 × 10–4 (b) 10–5 –6 (c) 2 × 10 (d) 5 × 10–6 2. What will be the pH of solution when B and C are mixed completely? (a) 2 – log 5 (b) 5 – log 2 (c) 2 (d) 1 3. What will be degree ionization when weak acid solution is mixed strong acid? (a) 5 × 10–3 (b) 2 × 10–3 –4 (c) 2 × 10 (d) 5 × 10–4 Passage II Aqueous 200 mL 0.1 M H 2A solution when titrated against 0.1 M NaOH shows different results, in presence of different indicators. For phenolphthalein indicator, it is

10.64

Ionic Equilibrium

converted to Na2A and with methyl orange it is converted to NaHA. Ka1 H 2 A = 10-3 , Ka2 H 2 A =10-6  NaHA + H 2 O (methyl Orange) H 2 A + NaOH   Na2 A + H 2 O ( Phenol phtnalein) NaHA + NaOH 

1. When 100 mL of NaOH is added in the presence of methyl orange indicator pH of solution is (a) 3.5 (b) 4.5 (c) 3 (d) 4 2. When 200 mL of NaOH is added the pH of the solution is (a) 3.5 (b) 4.5 (c) 3 (d) 4 3. When 400 mL of NaOH is added in the presence of phenolphthalein indicator. The pH of solution is (a) 8 (b) 4.74 (c) 9.26 (d) 7

strong base is added to it. Total salt is converted to weak base. The total volume was 50 mL. 1. The pKb of base is (a) 4.76 (b) 5.061 (c) 5.23 (d) 5.36 2. The pH of the solution after addition of strong base is (a) 9.22 (b) 10.22 (c) 11.22 (d) 12.22 Passage V AgCNS and AgBr both dissolved in water Ksp AgCNS = 5 × 10–13 Ksp AgBr = 1 × 10–12 1. The solubility of AgCNS in water is (a) 4 × 10–7 M (b) 8.16 × 10–7 M –7 (c) 2 × 10 M (d) 6.08 × 10–7 M 2. The solubility of AgBr in water is (a) 4 × 10–7 M (b) 8.16 × 10–7 M –7 (c) 2 × 10 M (d) 6.08 × 10–7 M

Passage III Arsenic sulphide As2S3, is a sparingly soluble salt and presence of As+3 ion in solution is observed by the formation of its precipitate 1 × 10-9 M The solubility of As2S3 is observed to be 48 in a solution of 10–2 M Na2S solution at 25°C. Assuming no hydrolysis of cationic or anionic part 1. The Ksp value of As2S3 at 25°C

1 ×10-24 576 1 (c) ×10-22 144

(a)

1 ×10-10 36 1 (d) ×10-9 24

(b)

2. The solubility of As2S3 at 25°C in a solution of 3 × 10−3 M AsCl3 solution is, assuming no hydrolysis 1 1 (b) ×10-6 M ×10-7 M 144 180 1 1 (c) (d) ×10-4 M ×10-6 M 96 36 3. If the hydrolysis of sulphide ion is to be considered, then the solubility of As2S3 in pure water is Ka1 H 2S = 10 -7 , Ka 2 H 2S = 10 -14 (a) 1.67 × 10–4 M (b) 5.4 × 10–5 M –3 (c) 7.2 × 10 M (d) 6 × 10–2 M

(a)

Passage VI Ksp AgI = 1.0 × 10–17 Ksp of Hg2I2 = 1 × 10-27 1. The maximum concentration of I – ion at which one of them gets precipitated completely is and first precipitated ion is (a) 10–16 M, Ag+ (b) 10–16 M, Hg2+2 –13 + (c) 10 M, Ag (d) 10–13 M, Hg2+2 2. The percentage of metal ion precipitated is (a) 99% Ag+ (b) 99% Hg2+2 + (c) 99.9% Ag (d) 99.9% Hg2+2 Passage VII The pH value of pure water is 7.0 at 27°C, whereas natural rain water is slightly acidic. This is mainly due to presence of SO3 and NO2 gas obtained by oxidation of SO2 and NO gas. In some causes acid rain due to SO3 and NO2 consist of pH of 4.5 to a value as low as 1.7. At 27°C the acidity constant are SO 2 (aq ) + H 2 O(l)  HSO3− (aq ) + H + (aq ) K a1 =

HSO3− (aq )  SO32 − (aq ) + H + (aq ) Ka2 =

Passage IV A weak base BOH was titrated against a strong acid the pH at ¼, the equalivalent point was 9.24. Now 6 m .eq. of

[H + ][HSO3− ] =10−2 M [SO 2 ] [SO32 − ][H + ] =10−7 M [HSO3 ]

SO 2 (aq ) + H 2 O(aq )  SO32 − (aq ) + 2H + (aq ) K a = K a1K a 2 =10−9 M

Ionic Equilibrium 10.65

H 2 O  H + + OH − K w =10−14 1. The pH of 0.01 M aqueous solution of Na2SO3 will be (Assume hydrolysis proceeds in only one step) (a) 8.5 (b) 9 (c) 9.25 (d) 9.5 2. If rain water consists of 2.463 litre SO3 gas dissolved in 1 litre water at partial pressure of 2 atm, then pH of solution is (R = 0.0821 litre atm/K Mole)(log 2 = 0.3) 1)0.7 (b) 1.4 (c) 0.6 (d) 0.4 3. In acid rain in Assam, it was found that 1 litre rain water consists of 2.463 litre equimolar mixture of SO3 and NO2 gas dissolved at a partial pressure of 0.1 atm. Find out the pH of the solution if HNO2 produced is 60% dissociated under given condition at 27°C. (a) 2 (b) 2.75 (c) 2.25 (d) 1.85

1. The pH at which the MnS will start precipitate (a) 3 (b) 4 (c) 5 (d) 6 2. The conc. of IN 2+ remaining when MnS starts pre cipitate (a) 10–8 mol/lit (b) 10–9 mol/lit –10 (c) 10 mol/lit (d) 10–11 mol/lit Passage xI The acid ionization constant for  Zn(OH ) + + H + is 1.0 ×10-9 Zn +2 + H 2 O  1. The basic dissociation constant of Zn(OH)+ is (a) 10–4 (b) 10–2 (c) 10-6 (d) 10–7 2. The pH of 0.001 M solution of ZnCl2 in above problem is (a) 4 (b) 5 (c) 6 (d) 7 Passage xII

Passage VIII The salt Zn(OH)2(s) is involved in the following two equilibria Zn (OH) 2 (s)  Zn 2 + (aq ) + 2OH - (aq ) K sp = 1 × 10 -17 Zn (OH) 2 (s) + 2OH - (aq )  [ Zn (OH) 4 ]2 - (aq ) K c = 0.1

1. The pH at which the solubility of Zn (OH)2 is minimum (a) 11 (b) 8 (c) 9 (d) 10 2. The solubility of salt is (a) 1 × 10–10 (b) 1 × 10–9 –10 (c) 2 × 10 (d) 2 × 10–9 Passage Ix 0.1 M CH3COOH solution is titrated against 0.05 M NaOH solutions. pH of 0.1 M CH3COOH is 3. 1. The pH of solution at 1/4 th neutralization of acid is (a) 4.21 (b) 4.52 (c) 5.12 (d) 5.47 2. The pH of solution at 3/4 th neutralization of acid is (a) ) 4.21 (b) 4.52 (c) 5.12 (d) 5.47 Passage x A solution containing zinc and manganese ions each at a concentration of 0.01 mol/dm3 is saturated with 0.1 M H2S. Ksp ZnS = 1 × 10–22 mol 2/lit2. Ksp MnS = 1 × 10–16 mol2/lit2. K1, K2, for H2S are 1.0 × 10–7 and 1.0 × 10–14.

1. A one litre buffer solution containing 0.5 M CH3COOH and 0.25 M CH3COONa. The pKa value of acetic acid is 4.74. The solubility product of Ca(OH)2, Al(OH)3, Cr(OH)3 are 5.5 × 10–6, 1.3 × 10–33, 6.3 × 10–31 in pure water respectively. 1. pH of the buffer solution is (a) 5.04 (b) 4.74 (c) 4.4 (d) 4.44 2. If 0.25 mole of HCl is added to the buffer solution then the pH is (a) 3.86 (b) 4.47 (c) 4.86 (d) 2.43 3. If 0.25 mole NaOH is added to the buffer solution then the pH is (a) 4.04 (b) 4.74 (c) 4.44 (d) 5.04 4. AlCl3 is added to the buffer solution then the concentration of Al+3 present in the solution is (a) 1.82 × 10–5 M (b) 6.75 × 10–5 M –4 (c) 6.75 × 10 M (d) 1.82 × 10–4 M 5. One litre solution containing Ca +2 , Al+3 , Cr +3 in 0.1 M concentration each. The solution is mixed with one litre buffer solution. Which ion remains un precipitated? (a) Ca+2 (b) Al+3 +3 (c) Cr (d) All Passage xIII Solid AgNO3 is gradually mixed with equimolar solution of NaBr and NaCl, each having molarity 0.02 M

10.66

Ionic Equilibrium

K sp ( AgBr ) = 5 × 10 -16 K sp ( AgCl ) = 1.0 × 10 -10 1. The minimum [Ag+] required for just the start of precipitation of AgBr (a) 5 × 10–14 M (b) 2 × 10–15 M –14 (c) 2.5 × 10 M (d) 5 × 10–15 M + 2. [Ag ] required to just start the precipitation of AgCl is (a) 5 × 10–8 M (b) 2.5 × 10–8 M –9 (c) 5 × 10 M (d) 2.5 × 10–19 M 3. The concentration of Br– ion left in the solution when precipitation of AgCl just starts. (a) 1.0 × 10–9 M (b) 2 × 10–7 M –9 (c) 5 × 10 M (d) 1.0 × 10–7 M Passage xIV Colligative property measurement is one the techniques used in the measurement of chemical quantities with reasonable accuracy. If a 40.65 g sample of K2SO4 and BaSO4 is dissolved in 900 g of pure water to from a solution ‘A’ is 57°C, its vapour pressure is found to be 39.6 torr while vapour pressure of pure water at 57°C is 40 torr density of solution A is 1.24 g/mL. In a different experiment, when small amount of pure BaSO4 is mixed with water at 57°C it given the osmotic of 4.05 × 10–5 atm. 1. Percentage of K2SO4 in the sample is (a) 65.65% (b) 71.34% (c) 60.35 (d) 78.74% 2. Solubility product of BaSO4 in water at 57°C is (a) 5 × 10–19 (b) 3.125 × 10–13 –13 (c) 5.625 × 10 (d) 5.625 × 10–13 Passage xV The pH of pure water at 25°C and 60°C respectively. HCI gas is passed through water at 25°C till the resulting 1 litre solution acquires a pH of 3. Now 4 × 10–3 mole of NaCN are added to this solution. The fresh 0.1 M HCN solution has pH 5.2. Now half part of solution obtained after addition of NaCN. 0.25 Milli moles of moles of NaOH are added and to other second part add 0.5 Milli mole of HCl is added. 1. The heat of formation of water from H+ and OH– is (a) 13.06 Kcal (b) -16.32 Kcal (c) 16.32 Kcal (d) -13.06 Kcal 2. The volume of HCl passed through the solution at 25°C and 1 atm is (a) 24.46 mL (b) 2.446 mL (c) 0.244 mL (d) 244.6 mL 3. The dissociation constant of HCN is (a) 4.0 × 10–10 (b) 4.0 × 10–6 –4 (c) 4.0 × 10 (d) 4.0 × 10–8

4. The pH of the solution after addition of NaCN is (a) 9.87 (b) 8.87 (c) 6.87 (d) 5.87 5. The pH of solution after addition of 0.5 Milli mole of HCl is (a) 9.68 (b) 9.4 (c) 8.4 (d) 8.7 Passage xVI Acid H2A undergoes the following dissociation reactions: + -7  H2A ↽ ⇀  HA + H K j = 1.0 × 10 + -11 2 HA ↽ ⇀  A + H K 2 = 1.0 × 10 A 20.00 mL aliquot of a solution containing a mixture of Na2A and NaHA is titrated with 0.300 M hydrochloric acid. The progress of the titration is followed with a glass electrode pH meter. Two points on the titration curve are as follows:

mL HCL added pH 1.00 11.00 10.00 9.00 1. What would be the concentration of HA–, on adding 1.00 mL HCl? (a) 0.15 M (b) 0.128 M (c) 0.45 M (d) 0.60 M 2. The amounts (mmol) of Na2A and NaHA initially present is (a) 1.5 m mol and 2.0 m mol respectively (b) 1.5 m mol and 2.4 m mol respectively (c) 3.0 m mol and 1.5 m mol respectively (d) 3.0 m mol and 2.40 m mol respectively 3. Total volume of HCI required to reach the second equivalence point is (a) 30 mL (b) 15 mL (c) 28 mL (d) 47 mL Passage xVII Radio chemical techniques can be used to determine solubility product estimation. The measurement of ra dioactivity can be used to find the concentration in a solubility equilibrium given a fair idea about various equilibrium concentrations. In an experiment, 50.00 mL of 0.010 M AgNO 3 solution containing a silver isotope with a radioactivity of 75,000 counts per min per mL were mixed with 100 mL of a 0.03 M NaIO 3 solution, the mixed solution was diluted to 500 mL and filtered to remove all of the AglO 3 precipitate leaving behind a radioactive solution. Molar mass of AgIO 3 =285 g/mol. There meaning solution was found to have radioactivity of 50 counts per min per mL.

Ionic Equilibrium 10.67

1. Find the mass of the ppt, of AglO3 obtained separated (a) 142 mg (b) 283 mg (c) 383 mg (d) 483 mg 2. The Ksp of AglO3 silver (Radio active) are: (b)b) 3.3 × 10-8 M 2 (a)a ) 3.3 × 10-16 M 2 -10 2 (c)c) 5.3 × 10 M (d)d ) 5.3 × 10-8 M 2 3. The % of unprecipitated silver ions is: (a) 0.15% (b) 0.37% (c) 0.67% (d) 0.97%

Step I II

Volume of NaOH added pH 8.12 mL 16.24 mL

4.57 7.02 (at equivalent point)

1. Which is removed in step I? COOH (a)

COOH

+ H 3O

+ H2O

Passage xVIII Read the following passage and answer the questions at the end of it. Acid rain is an environmental concern all over the world. In assessing the acidity of rainfall, it is important to have an idea of the acidity of natural rainwater. Assuming that natural rain water (that is rain water uncontaminated with nitric acid or sulphuric acid) is in equilibrium with 3.6 ×10-4 atm CO2 (the Henry’s law constant is 1.25 × 106 Torr/mole fraction. Ka1 ( H 2 CO3 ) = 4.3 × 10-7 1. What is the pH of natural rain water? (a) 5.64 (b) 5.70 (c) 7.00 (d) 7.40 2. pH of natural rain water would have been in preindustrial times when the partial pressure of CO 2 was about 2.8 ×10-4 atm (a) 7.0 (b) 5.64 (c) 7.4 (d) 5.70 3. If SO2 content in the atmosphere is 0.12 ppm by weight, pH of rain water is (assume 100% ionization of acid rain as monobasic acid) (a) 5.7 (b) 5.6 (c) 5.4 (d) 2.0

(b)

OH

O(–)

COOH

COO(-)

+ H 3O

+ H2O

OH

OH

(c) Both 50% in each part  H 3 O ⊕ + OH ( - ) (d) H 2 O + H 2 O  (autoprotolysis of H 2 O)

2. pKa1 (= –log Ka1) of p-hydroxybenzoic acids is: (a) 4.57 (b) 9.47 (c) 4.90 (d) 7.00 3. Which is not a correct representation? (a)

COO-

COOH(-)

+ H 3O

+ H2O

Passage xIx Following titration is taken to compute stepwise ionization constant of weak dibasic acid O=COH (b)

OH

O–

COO(-)

COOH

A: + H3O

+ H2O

OH p-hydroxybenzoic acid A has two ionization protons and there can be stepwise neutralization by NaOH 25 mL of a dilute aqueous solution of A in titration with 0.2 M NaOH (aq) pH measured.

OH

(c) Both of the above (d) None of the above

OH

10.68

Ionic Equilibrium

4. pKa2 (= –log Ka2) of p-hydroxy benzoic acids is: (a) 4.57 (b) 7.00 (c) 9.47 (d) 4.90 5. Ion formed in step I is: (a) a base (b) an acid (c) an amphiprotic ion (d) a conjugate base

An acid-base titration is weak acid or weak base. The colour of the ionized form is markedly different from that of unionised form. One form may be colourless, but the other form must be coloured. Assume that the indicator is a weak acid designated HIn +  HIn ↽ ⇀  H + In Colourless

Passage xx Following example specifies use of various indicators in the given pH range. Answer the question given at the end. For a series of indicators, the following colours and pH range over which colour change takes place are as follows: Indicator

Colour change

pH range

U V W X Y

Yellow to blue Red to yellow Red to yellow Yellow to blue Colourless to red

0.0 to 1.6 2.8 to 4.1 4.2 to 5.8 6.0 to 7.7 8.2 to 10.0

1. Consider following statements: (A) Indicator ‘V’ could be used to find the equivalence point for a 0.1 M acetic acid and 0.1 M ammonium hydroxide (ammonia solution) titration. (B) Indicator ‘Y’ could be used to distinguish between 0.1 M and 0.001 M NaOH solutions in water. (C) Indicator ‘X’ could be used to distinguish between solutions of ammonium chloride and sodium acetate. (D) Indicator ‘W’ would be suitable for use in determining the concentration of acetic acid in white vinegar by base titration. (a) All except A (b) All except B and C (c) All except A and D (d) Only C is correct Passage xxI Following passage explains the action of different indicators. Answer the questions given at the end. When there is reaction of an acid with a base, the equivalence point is reached. The point at which the reaction is observed to complete is called the end point. A measurement is chosen such that the end point coincides with or is very close to the equivalence point. End point can be detected by measuring the pH at different point of titration using a pH meter. it is usually more convenient to add indicator to the solution and usually detected a colour change.

+

K In =

Coloured -

[H ][IN ] HIn ] [H

pH = pK In + log

,

[In - ] [HIn ]

The indicator change over a pH range. The transition range depends on the ability to detect small colour changes. With indicators in which both forms are coloured, generally one colour is observed if the ratio of the concentration of the two forms is 10:1 ph transition ranges and colours of some common indicators Colour in Indicator

pH range

Thymol blue (Acid) 1.2 - 2.8 Bromophenol blue

3.1 - 4.6

Acid solution

Alkaline solution

Red

Yellow

Yellow

Purple

Methyl orange 3.1 - 4.5 Red Methyl red 4.2 - 6.3 Red Bromothymol blue 6.0 - 7.5 Orange Phenol red 6.4 - 8.2 Yellow Thymol blue (base) 8.1 - 9.6 Yellow Phenolphthalein 8.0 - 9.8 Colourless Thymophthalein 9.3 - 10.5 Colourless Alizarin yellow R 10.1 - 12.1 Yellow

Yellow Yellow Yellow Red Blue Pink Blue Lilac

1. Midway in the transition of an indicator, behaving as weak acid: (a) pH = pKIn + 1 (b) pH = pKIn – 1 (c) pH = pKIn (d) pH = - pKIn 2. Phenolphthalein is not a suitable indicator for titration of a weak base with acid because: (a) Weak base cannot furnish enough OH– that can ionize phenolphthalein (Weak acid) (b) Salt formed from excess of weak base and indicator is hydrolysed to colourless solution (c) Both the statements are correct (d) None of the above statements is correct

Ionic Equilibrium 10.69

3. Following titration curve is for: (a) NaOH (b) NH4OH (c) Ca(OH)2 (d) Al(OH)3 12 10

Phenolphthalein

8

pH

6

Methyl red

4 2 0

0

20

40

60 80 100 mL 0.1 M HCI

120 140

4. Ka of a certain indicator is 2.0 × 10–6. The colour of HIn is green and that of ln– is red. A few drops of the indicator are added to HCl solution which is then titrated against NaOH solution. Indicator will show colour change at pH: (a) 7.0 (b) 5.7 (c) 3.5 (d) 3.0

Maintenance of pH in blood and in intercellular fluids is absolutely crucial to the processes that occur in living organisms. This is primarily because the functioning of enzymecatalysts for these process-is sharply pH dependent. The normal pH value of blood plasma is 7.4. Severe illness or fear can result from sustained variations of a few tenths of pH unit. Among the factors that lead a condition of acidosis, in which there is decreases in the pH of blood are heart failure, kidney failure, diabetes mellitus, persistent diarrhoea or a long term high protein diet. Temporary condition acidosis may result from prolonged, intensive exercise. Alkalosis, which cause increase in pH of blood, may occur as a result of severe vomiting over breathing or exposure to high altitudes. Several factors are involved in the control pH of blood. A particularly important one is the ratio of dissolved HCO–3 to H2CO3.CO2(g) is moderately soluble in water and in aqueous solution reacts only a limited extent to produce H2CO3.  CO 2 + H 2 O ↽ ⇀  H 2 CO3 (aq ) +  H 2 CO3 + H 2 O ↽ ⇀  HCO3 + H 3 O , pKa1 = 6.11 2+  HCO3- + H 2 O ↽ ⇀  CO3 + H 3 O , pKa 2

Passage xxII

= 10.25

– H2CO3, HCO3

Read the following passage about amino acid. Methionine is an essential amino acid. Its protonated form given below behaves as a dibasic acid. O

H3 N CH C OH CH2CH2SCH3 pK a 1 = 2.28 and pK a 2 = 9.20 1. Zwitterionic form of methionine is: O

(a) H3N CH CO

O –

CH2CH2SCH3 O

(c) H3N C H C OH CH2CH2SCH3

(b) H3 N CH CH CH2CH2SCH3

buffer system we deal only with In the the first ionization step (Ka1): H2CO3 is weak acid and – HCO3 is the conjugate base (salt). CO 2(g) is exchanged for O2(g), which is transported throughout the body by the blood. 1. The pH of blood stream is maintained by a proper balance of H2CO3 and NaHCO3 concentration. The volume of 5 M NaHCO3 solution should be mixed with a 10 mL sample of blood which is 2 M in H2CO3 in order to maintain its pH? (a) 40 mL (b) 38 mL (c) 50 mL (d) 78 mL 2. Important diagnostic analysis in the blood is (a)  H 2 PO4-   HPO4-  (b) [ HCO3- ] / CO2 

O

(d) H3 N C H C O(–) CH2CH2SCH3

2. 30.0 mL of 0.06 M solution of the protonated form of this amino acid is treated with 0.09 M NaOH after addition of 20 mL of NaOH is. Then pH of the solution is: (a) 2.28 (b) 9.20 (c) 5.75 (d) 456

(c) CO32 -  /  HCO32 -  (d)  PO43-  /  HPO42 -  3. Following reaction occurs in the body:  H 2 CO3   H + + HCO3CO2 + H 2 O  If CO2 escapes from the system: (a) pH will decrease (b) pH will increase (c) [H2CO3] remains unchanged (d) Forward reaction is promoted

Passage xxIII

Passage xxIV

Read the following passage giving the role of CO2 buffer in controlling pH of blood.

Following experiment is taken to determine solubility of a sparingly soluble salt. Answer the questions given at the end.

10.70

Ionic Equilibrium

Aqueous calcium chloride solution is mixed with sodium oxalate and precipitate of calcium oxalate formed filtered and dried. Its saturated solution was prepared and 250 mL of this solution was titrated with 0.001 M KMnO4 solution when 6.0 mL of this was required. 1. Which is the indicator in the above titration? (a) Methyl orange (b) Phenolphthalein (c) Sulphuric acid (d) KMnO4 itself 2. Number of mol of KMnO4 required in this titration is: (a) 6 × 10–3 (b) 6 × 10–6 (c) 250 (d) 2.5 × 10–1 3. Number of mol of oxalate present in given saturated solution of calcium oxalate is: (a) 6 × 10-6 (b) 3 × 10-6 -6 (c) 1.5 × 10 (d) 1.5 × 10-5 4. Equivalent of KMnO4 required in the titration is: (a) 3 × 10-5 (b) 6 × 10-6 -5 (c) 1.2 × 10 (d) 1.8 × 10-5 5. Equivalent of oxalate present in calcium oxalate is: (a) 3 × 10-5 (b) 6 × 10-6 -5 (c) 1.2 × 10 (d) 1.8 × 10-5 6. Solubility product of calcium oxalate is: (a) 2.25 × 10-10 (b) 4.0 × 10-9 -12 (c) 2.25 × 10 (d) 3.6 × 10-9 Passage xxV Read the following passage and answer the questions given. If we place a pH meter electrode into a beaker of pure water at 25°C, we find that the water has a pH of 7.00. Then, if we heat the water, the metre shows a de crease in pH. When the water begins to boil at 100°C, the pH meter reads 6.12, if the meter and electrode work as predicated. We find ourselves face to face with a paradox. On one hand, we know that the water sample is neu tral because pure water is neither acidic nor basic. On the other hand, the pH metre reading seems to indicate that the water sample has become acidic because the boiling water has a pH that is less than 7.00. We sometimes hear that all aqueous solutions with pH of less than 7.00 are acidic. 1. pH of boiling water (at 100°C) is (a) 7 (b) Greater than 7 (c) Less than 7 but greater than 6 (d) Between 4 and 6 2. Boiling water thus: (a) Turns blue litmus red (b) Turns red litmus blue (c) Turns turmeric paper brown (d) Neutral to litmus

3. Kw of water at 100°C is: (a) 5.8 × 10-13 (b) 7.6 × 10-7 (c) 1.0 × 10-14 (d) 1.0 × 10-7 4. Increase in temperature: (a) Increases autoprotolysis of water (b) Decreases pH of water (c) Has no effect on neutral nature of water (d) All of the above correct Passage xxVI Read the following passage and answer the questions at the end of it: Phosphoric acid is of great importance in fertilzer production. Besides, phosphoric acid its various salts have a number of applications in metal treatment, food, detergent and toothpaste industries. pKa1 = 2.12, pKa 2 = 7.21, pKa 3 = 12.32 Small quantities of phosphoric acid are exten sively used to impart the sour or tart taste to many soft drinks such as colas and roots beers, in which a density of 1.00 g mL –1 contains 0.05% by weight of phosphoric acid. Phosphoric acid is used as a fertilizer for agriculture, and an aqueous soil digesting 1.00 × 10–3 M phosphoric acid is found to have pH = 7. Zinc is an essential micronutrient for plant growth. Plants can absorb zinc in water soluble form only. In the given soil, zinc phosphate is only the source of zinc and phosphate ions. Ksp (Zinc phosphate) = 9.1 × 10–33 1. Phosphoric acid is a tribasic acid with three-step ionization constants. Thus, its structure is: H

O-H

(a) H P O H

(b) H P O H

O

O O-H

(c) H

O

P O H O

O-H (d) H

O

P O H

O

O

2. What is the pH of the cola assuming that the acidity of the cola arises only from phosphoric acid and second and third ionization constants are of no importance? (a) 2.2 (b) 3.3 (c) 4.4 (d) 1.8

Ionic Equilibrium 10.71

3. Molar concentration of phosphate ion in the soil with pH 7 is (a) 1.2 × 10–4 M (b) 2.2 × 10–4 M –10 (c) 1 × 10 M (d) 1.1 × 10–10 M 2+ 4. Concentration of [Zn ] in the soil is: (a) 9.1 × 10–5 M (b) 2.9 × 10–11 M –10 (c) 4.0 × 10 M (d) 5.65 × 10–9 M Passage xxVII Read the following experiment and answer the questions given at the end of it. An experiment to measure the solubility product constant of CaSO4 involve passing a saturated solution of CaSO4(aq) through an ion exchange column, pictured here Ca2+ ion is retained on the column and two. 25.00 mL saturated CaSO4(aq) + water 8.25 mL 0.0105 M NaOH

Matching Type Questions 1. Match the mixture in Column I with their correspondence characteristic in Column II. Column I (a) CH3COOH (pKa = 4.74; 0.1 M) + CH3COONa (0.1 M) (b) CH3COOH (pKa = 4.74; 0.1 M) + CH3COONa (0.1 M) (c) CH3COOH (pKa = 4.74; 0.1 M) + CH3COONa (0.1 M) (d) CH3COOH (pKa = 4.74; 0.1 M) + CH3COONa (0.1 M)

Column II (p) Acidic buffer at its maximum capacity (q) buffer solution

(r) pH < 7 at 25°C

(s) pH = 7 at 25°C

Note: Equal volume of each component in each mixture is to be taken 2. Match Column I with Column II. Column I

10.00 mL 100.0 mL

H3O+ ions appear in the solution leaving the column for every Ca2+ ion retained. A 25.00 mL. sample of saturated CaSO4(aq) is passed through the column, following by a small volume of water. The solution collected is then diluted to 100.00 mL in a volumetric flask. Finally, 10.00 mL of the solution from the volumetric flask is treated with 8.25 mL of 0.0105 M NaOH. Use this data to obtin a value Ksp for CaSO4. 1. [Ca2+] in the diluted solution is: (a) 2.16 × 10–3 M (b) 1.73 × 10–2 M (c) 4.33 × 10–4 M (d) 1.90 × 10–5 M 2. [SO42–] in the diluted solution is: (a) 8.66 × 10–3 M (b) 7.5 × 10–5 M (c) 4.33 × 10–3 M (d) 1.9 × 10–5 M 3. pH of the solution coming out of volumetric flask is: (a) 2.36 (b) 4.16 (c) 5.00 (d) 2.06 4. Solubility product of CaSO4 is (a) 1.9 × 10–5 M2 (b) 7.5 × 10–5 M2 –4 2 (c) 3.0 × 10 M (d) 4.8 × 10–3 M2

(a) (b) (c) (d)

CuSO4 Na2CO3 NaCl CH3COONH4

Column II (p) (q) (r) (s)

Catonic hydrolysis Anoinc hydrolysis pH < 7 pH = 7

3. H3PO4 is a triprotic acid with K1 = 10–4, K2 = 10–7, and K3 = 10–10 Column I (a) Equimolar mixture of NaH2PO4 (0.1 M) and Na3PO4 (0.1 M) (b) H3PO4 (0.1 mol) + NaOH (0.2 mol) in 1 litre solution (c) Na3PO4 ( 0.1 mol) + NaOH (0.1 mol) in 1 litre solution (d) H3PO4 ( 0.1 mol) + KOH (0.25 mol) in 1 litre

Column II (p) 13

(q) 5.5

(r) 8.5

(s) 10

10.72

Ionic Equilibrium

4. Match the following Column I with Column II. Column I

Column II

(Salt)

(Relation between solubility S mol L–1 and Ksp) 1/5  K sp  (p) S =    108 

(a) Ag2CrO4

7. Match Column I with Column II. Column I (a) Acid indicator

(b) Base indicator (c) Phenolphtalein

(b) AgCNS

(q) Conc. of cation of 2S

(c) Ca3(PO4)2

 K sp   (r) S =   4 

(d) Hg2Cl2

(s) S = ( K sp )1/ 2

1/3

5. Match Column I with Column II. Column I (a) CuSO4 Solution (b) Na2CO3 solution (c) CH3COOH + CH3COONa solution (d) NH4OH + NH4Cl solution

Column II (p) Acidic nature (q) Basic nature (r) Cationic hydrolysis

(d) Methyl orange

Column II (p) Colour of ionized form of indicator appears in acidic medium (q) Colourless in acidic medium (r) Colour of ionized form of indicator appears in basic medium (s) Colour of unionized form of indicator appears in acidic medium

8. For the following matching take the concentrations of the both the solutions being titrated to be equal to 0.1 M, and Ka of CH3COOH = 2 ×10–5, Kal of H2CO3 = 10–7, Ka2 of H2CO3 = 10–11, Kb of NH3 = 2 × 10–5 Column I

Column II

(a) CH3COOH + NaOH

(p) Approximate shape of the titration curve will be

(b) NH4Cl + NaOH

(q) Phephthalein (pKIn = 9) can be used for end point detection (r) At equivalence point

(s) Anionic hydrolysis

6. Match Column I with Column II. Column I

Column II

(a) Equimolar mixture of a strong acid and a strong base (b) Equi-equivalent mixture of a strong acid and a weak base (c) Equi-equivalent mixture of a strong base and a weak acid (d) Equi-equivalent mixture of a weak acid and a weak base

(p) Acidic

(q) Basic

(r) Netural

(s) Can’t be predicted

(c) Na2CO3 + HCl (up to 1st equivalence point) (d) HCl + NaOH (s) At equivalence point, pH > 7

Ionic Equilibrium 10.73

9. Match Column I with Column II. Column I –

(a) pKb for X (Ka of HX = 10–6) (b) pH of 10–8 MHCl (c) pH of 10–2 M acetic acid (Ka = 1.6 x 10–5) (d) pH of solution obtained by mixing equal volumes of solutions with pH = 3 and pH = 5

Column II

Column I

(p) 6.9

(a) NaCl

(q) 8 (r) 3.3

(b) KCN (c) CH3COONa (d) CH3COONH4

(s) 3.4

Column I

Column II

(a) CuSO4 solution (b) Na2CO3 solution (c) CH3COONH4 solution

(p) Cationic hydrolysis (q) Anionic hydrolysis (r) Neutral solution

(d) NaCl solution

(s) Acidic Solution

(a) CH3COOH (b) NH4Cl (c) NH4OH

(d) CH3COONa

11. Match Column I with Column II. Column I (a) CH3COOH CH3COONa mixture in 1:1 molar ration (b) NH4OH, NH4Cl mixture in 1:1 molar ratio (c) CH3COOH solution (d) NH4OH solution

Column II

(a) NH3 (b) O2– (c) HCO–3 (d) KMnO4

(p) pH will not vary with dilution (q) Buffer solution (r) pH is almost 7 (s) Hydrolysis of only anion takes place

Column II (p) Weak electrolyte (q) Strong electrolyte (r) On dilution of aqueous solution, the degree of dissociation increases (s) pH-increases on dilution of aqueous solution. (t) pH-decreases on dilution of aq. Solution

(p) pH increases on addition of HCl

15. Match Column I with Column II.

(q) pH decreases on addition of HCl (r) pH remains same on dilution (s) pH increases on dilution

(a) [H+] in an aqueous solution of NH4CN (b) [H+] in an aqueous solution of NaHCO3

12. Match Column I with Column II. Column I

Column II

14. Match Column I with Column II.

10. Match Column I with Column II. Column I

13. Match Column I with Column II.

Column II (p) Aqueous solution is alkaline in nature (q) Lewis base (r) Does not have a conjugate base (s) Amphoteric in nature

Column I

(c) [H+] in an aqueous solution of NH4HCO3 (d) [CO2–3] in 0.01 M H2CO3

Column II (p) (Ka1 Ka 2 )1/ 2 1/ 2

K K  (q)  w a   Kb 

1

 K 2  (r)  w + Ka1  Ka1    K b  (s) Does not depend on the concentration of salt Ka2.

10.74

Ionic Equilibrium

16. Consider the operations on the left column and match with the solution on the right column: Column I (a) Addition of small amount of HCl decreases the solubility of sparingly soluble salt. (b) Addition of small amount of NaOH decreases the solubility of sparingly soluble salt. (c) Addition of small amount of HCl increases the solubility of sparingly soluble salt. (d) Addition of small amount of AgNO3(s) increases the solubility of sparingly soluble salt.

Column II (p) A saturated solution of AgCl (Ksp = 2 × 10–10)

(a) KMnO4 (b) K2Cr2O7

(c) Na2S2O3 (d) H2O2

Column I (a) 25 mL of NaOH (b) 50 mL of NaOH

(q) A saturated solution of PbCl2. (Ksp = 4 × 10–6) (c) 75 mL of NaOH

(r) AHCl solution saturated with AgCl. (d) 100 mL of NaOH

(s) A NaOH solution saturated with Mg(OH)2. (Ksp of Mg (OH)2 = 4 × 10–7)

17. Match the compounds in Column I with their characteristics in Column II. Column I

18. Match the effect of addition of 0.1 M NaOH to 50 mL of 0.1 M H2C2O4. (Given: Ka1 = 10–4, Ka2 = 10–9). Column I indicates the volume of NaOH added and Column II illustrated the characteristics of the solution after addition of NaOH.

Column II (p) Self indicator in redox titrations (q) One mole transfers 5 moles or more electrons under acidic conditions (r) Reduces I2 to iodide ion in iodimetric titrations (s) Average oxidation state for some atom(s) (t) Can act as oxidant and/ or reductant

Column II (p) pH < 7 (q) pH is independent of the concentrations of the species (r) Addition of small quantities of a strong acid or a strong base does not affect the pH appreciably (s) pH is calculated using the concept of anionic hydrolysis only pH>7

Assertion (A) and Reason (R) Type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): pH of buffer solution changes with change in temperature. Reason (R): K w of water changes with change in temperature. 2. Assertion (A): A solution whose pH is 6.9, must be acidic. Reason (R): In an acidic solution, H+ ion concentration must be greater than K w 3. Assertion (A): Phenolphthalein is used as an indicator during the titration of oxalic acid and sodium hydroxide Reason (R): The pH range of phenolphthalein from 8 to 9.6 4. Assertion (A): Aqueous solution of CH3COONa is alkaline in nature Reason (R): Acetate ion under goes anionic hydrolysis

Ionic Equilibrium 10.75

5. Assertion (A): The addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl Reason (R): The value of Ksp of AgCl MX2 > M3X (b) M3X > MX2 > MX (c) MX2 > M3X > MX (d) MX > M3X > MX2 (2009) Multiple Choice Questions with One or More Than One Answer 9. A buffer solution can be prepared from a mixture of: (a) Sodium acetate and acetic acid in water (b) Sodium acetate and hydrochloric acid in water (c) Ammonia and ammonium chloride in water (d) Ammonia and sodium hydroxide in water (1999)

10.78

Ionic Equilibrium

10. Aqueous solution of HNO3, KOH, CH3COOH, and CH3COOONa of identical concentrations are provided. The pair (s) of solution which from a buffer upon mixing is (are) (a) HNO3 and CH3 COOH (b) KOH and CH3 COOHNa (c) HNO3 and CH3 COONa (d) CH3COOH and CH3COONa (2010) Integer Type Questions 11. Dissociation constant of a substituted benzoic acid at 25°C is 1.0 × 10–4. The pH of 0.01 M solution of its sodium salt is (2009)

12. Amongst the following, the total number of compounds whose aqueous solution turns red litmus paper blue is: KCN

K 2 SO4

( NH 4 ) 2 C2 O4

NaCl

Zn( NO3 ) 2

FeCl3

K 2 CO3

NH 4 NO3

LiCN

(2010) 13. In 1 L saturated solution of AgCl [Ksp (AgCl) = 1.6 × 10–10], 0.1 mol of CuCl  K sp (CuCl ) = 1.0 ×10-6  is added. The resultant concentration of Ag+ in the solution is 1.6 ×10- x . The value of “x” is: (2011)

Ionic Equilibrium 10.79

ANsWeR keys Multiple Choice Questions with Only One Answer

Multiple Choice Questions with One or More Than One Answer

Level I 1. c 2. d 3. c 4. c 5. c 6. a 7. c 8. d 9. a 10. d 11. b 12. b 13. a 14. d 15. a 16. b 17. c 18. a 19. c 20. a 21. d 22. a 23. b 24. a 25. d

1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50.

d d c a b a b d c c a d b d b c b d c c b b a d b

51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75.

b a b b d b a c d b d b c b a c b d c c c a c d c

76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99.

d c d b b a b b b b c c d a b a a b a d c d c b

100. 101. 102. 103. 104. 105. 106. 107. 108. 109. 110. 111. 112. 113. 114. 115. 116. 117. 118. 119. 120. 121. 122. 123.

b d d b a a d a d b d d d a c b d d d d b a d a

11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

b,c c,d a,c a,b,c,d a,b,d c,d a,b a,b,c a,b,c,d a,c,d

Passage I 1. b

2. a

3. c

2. b

3. c

2. c

3. a

Passage II 1. c Passage III 1. a

1. c

2. c

Passage V 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34.

a b b b b c b d d d d d b b a a a

35. 36. 37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51.

d b c a c d c d a d c c a d a c d

52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68.

d c c c a a b a b b c c b a c d b

69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82.

a a c d a b a c b a c b c b

21. 22. 23. 24. 25. 26. 27. 28.

Comprehensive Type Questions

Passage IV

Multiple Choice Questions with Only One Answer Level II 1. b 2. b 3. a 4. c 5. d 6. d 7. c 8. c 9. d 10. c 11. a 12. c 13. a 14. d 15. c 16. b 17. b

a,b,d a,c a,b,c,d a,b,c a,d b,d a,d a,c b,c a,b

1. a

2. b

Passage VI 1. c

2. c

Passage VII 1. d

2. d

Passage VIII 1. d

2. d

Passage Ix 1. b

2. d

3. d

a,d b,c a,b,d b,c,d a,b,d a,b,c b,c,d b,d

10.80

Ionic Equilibrium

Passage x 1. b

Passage xxIII 2. a

1. d

Passage xI 1. b

2. b

3. b

Passage xxIV 2. c

1. d 2. b

3. d 4. a

5. a 6. d

Passage xII 1. d

2. d

3. d

4. b

5. a

Passage xxV 1 c

2. d

3. a

4. d

3. b

4. d

3. d

4. c

Passage xIII 1. c

2. c

Passage xxVI

3. d

1. c

2. a

Passage xIV 1. b

Passage xxVII

2. c

1. a

2. c

Passage xV 1. d

2. a

3. a

4. a

5. b

Passage xVI 1. b

2. d

3. c

Passage xVII 1. a

2. b

3. c

Passage xVIII 1. a

2. d

3. a

Passage xIx 1. b

2. a

3. c

4. c

Passage xx 1. d

2. c

Passage xxII 1. a

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18.

(a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

pqR Pr R qr PR s RS pqrs q PS qr pqs PR PRS qS PqR Pqt PqR

(b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b)

R q R s qs p P pqrs p q qr pqR S qRS PS S qt Pq

(c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

qs S P p p q qrs qrs s pqr qs PqS S PRT RS S Rst qRt

(d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d)

qr pqs S R q s P q R R q PR PqR qrt t qr Rt qSt

Assertion (A) and Reason (R) Type Questions

Passage xxI 1. c

5. c

Matching Type Questions

2. c

3. b

4. a

1. 2. 3. 4.

a d a a

5. 6. 7. 8.

c d c d

9. 10. 11. 12.

a c a b

13. 14. 15. 16.

d a a a

17. c 18. d 19. b

Ionic Equilibrium 10.81

Integer Type Questions 1. 2. 3. 4. 5. 6. 7. 8. 9.

6 1 9 8 9 2 2 5 3

10. 11. 12. 13. 14. 15. 16. 17. 18.

6 8 4 1 2 9 4 5 9

19. 20. 21. 22. 23. 24. 25. 26. 27.

6 6 9 8 7 5 3 6 3

Previous years’ IIT Questions 28. 29. 30. 31. 32. 33. 34. 35. 36.

3 6 9 6 1 2 1 5 7

37. 38. 39. 40. 41. 42. 43.

9 8 8 3 3 9 8

1. 2. 3. 4.

1 1 3 2

5. 6. 7. 8.

2 1 4 4

9. 10. 11. 12.

1,3 3,4 8 3

13. 7

10.82

Ionic Equilibrium

hINTs AND sOLUTIONs hints to Problems for Practice

+ −  [Ag (CN) 2 ]− ↽ ⇀  Ag + 2CN

1. According to Ostwald’s dilution law K=

2

Ka =

α (1 − α ) v 1 = 100 litres. 0.01 0.05 × 0.05 = = 2.63 × 10−5 (1 − 0.05)100

Ka

2. Let the degree of dissociation of CH3COOH be x. Then CH 3 COOH

-

0

0.2 (1 - x )

+

 CH 3COO + H

0.2

Ka =

or [Ag + ] =

V=

α = 0.05, Hence

[Ag + ][CN − ]2 × 4 × 10−19 [Ag (CN) −2 ]

0

0.2 x

initial conc.

0.2 x Equili conc.

=

4.

0.03 × 4 ×10−19 = 7.5 × 10−18 mole / litre. (0.04) 2

  → N 2 H 5+ + OH N2H4 + H2O ←  1

(1 − 2 ) × 9.5 ×10−3

;

K b = Cα 2 Also [ N 2 H 4 ] = C =

V = 900 mL

3. 2 KCN + AgNO3  → [ Ag (CN ) 2 ](0.1 - 0.03 × 2) (0.03 - 0.03)

4 ×10−6 = 4 ×10−4 0.01 α = 2 ×10−2 i.e., α = 0.02 or 2%

C = 0.05

0.2 × 300 = 0.05 × (300 + V )

0.03M

0

initial conc. 0.03

0.16 ×1000 = 0.01 32 × 500

Given K b = 4 × 10−6 M α 2 =

Suppose V mL of water is to be added to dilute 300 mL of CH3 COOH solution to change its concentration from 0.2 M to 0.05 M. As millimoles of CH3COOH before and after dilution will be same.

0.1M

initial conc equili Conc.

Cα 2 (1 − α ) Assu min g 1 − α  1

2

C × ( 2 × 9.5 × 10−3 )

0 a

Kb =

( C × 2x ) ( C × 2x ) C ( 2x ) Ka = = C (1 − 2 x ) (1 − 2x ) 1.8 × 10−5 =

0 a

1- a

0.2 x × 0.2 x = 1.8 × 10-5 ; x = 9.5 × 10-3 0.2 (1 - x )

Now the solution be diluted to C mole per litre in order to double the degree of dissociation, x. Since for the degree of dissociation ‘2x’ the dissociation constant will be the same, therefore, we have

[Ag (CN) −2 ] × 4 × 10−19 [CN − ]2

equili conc.

In this reaction, AgNO3 shall be fully consumed. The concentrations of [Ag (CN)2–] will be 0.03 M. The solution will also contain 0.04 M of free CN– ions as the concentration of unreacted KCN is 0.04 M. The concentration of CN– produced by [Ag(CN)–2] may be neglected because of the very low value of dissociation constant of the equilibrium.

5. Let the degree of dissociation of CH3COOH be x. Thus

 CH 3 COO - + H + CH 3 COOH  0.2 0 0 initial conc. 0.2(1 - x) 0.2x 0.2x equili conc. 0.2x × 0.2x Ka = = 1.8 × 10 -5 ; x = 9.5 × 10 -3 0.2(1 - x) Now, let the solution be diluted to C mole per litre in order to double the degree of dissociation, 2x the dissociation constant will be the same, therefore we have Ka =

( C × 2 x ) ( C × 2 x ) C( 2 x ) 2 = C(1 − 2 x ) (1 − 2 x )

Now suppose V mL of water is to be added to dilute 300 mL of CH3COOH solution to change its concentration from 0.2 M to 0.05 M. As mill moles of CH3COOH before and after dilution will be same 0.2 × 300 = 0.05 × (300 + v) V = 900 mL.

Ionic Equilibrium 10.83

Now per cent dissociation of HCOOH =

6.

HNiC

H+

 ↽ ⇀ 

NiC -

+

1 0 0 initial conc (1 - a ) a a equili conc 0.1 C= = 5 × 10 -2 mol litre -1 ; Ka = 1.4 × 10 -5 2 Ca 2 = Ca 2 ∵ 1 - a ≈ 1 Ka = (1 - a ) 1.4 × 10 -5 5 × 10 -2

Ka = a= C

[HCOO − ] ×100 10−3 = × 100 [HCOOH ] 26.5 Cα α= = 0.004% C(1 − α ) 1.2 × 10 -4 = 548. 4.0 × 10 -10

10. Strength of HCNO = Strength of HCN

+  11. HCOOH + H 2O ↽ ⇀  H 3O + HCOO

C

0

C - 1.6 × 10

= 1.67 × 10 -2 or 1.67%

-3

0

1.6 × 10

-3

initial conc.

1.6 × 10

-3

equili conc.

12. wt of NH 3 10 = wt of solution 100

0.32 × 2 × 1000 = 0.014 M 7. [HC9 H 7 O 4 ] = 180 × 250

∵ 100 g solution contains 10 g NH 3

-9

2.75 × 10 Ka = = 4.43 × 10 -4 0.014 C [H + ] = C. a = 0.014 × 4.43 × 10-4 = 6.21 × 10-6 M. a=

M NH3 = (10 × 1000) / [17 × (100 / 0.99)] = 5.82 (∵V = mass /density) +  → NH 4 OH ↽ NH 3 + H 2 O  ⇀  NH 4 + OH

[C9 H 7 O 4- ] = C. a = 6.21 × 10 -6 M.

1 1- a

8. +  CH 2 FCOOH ↽ ⇀  CH 2 F COO + H 1 (1 - a )

0 a

+

[H ] = C. a = 1.5 × 10

-3

mol litre

0 a

0 a

[OH - ] = C. a = C (Kb / C)

=

0 a

initial conc equili conc

(Kb × C)

[∵C = 5.82 M and Kb = Kw / Ka = 10

-14

/ (5 × 10-10 ) = 2 × 10-5

initial conc

[OH - ] =

equili conc

[H + ] = 10-14 / 1.07 × 10 -2 = 0.9268 × 10 -12 M

[2 × 10 -5 × 5.82] = 1.07 × 10-12 M

-1

Ca 2 (Ca )(Ca ) Ka = = (1 - a ) C(1 - a ) 1.5 × 10 -3 × a 2.6 × 10 = \ a = 0.634 1- a Now Ca = 1.50 × 16 -3 -3

-3

1.50 × 10 = 2.37 × 10 -3 M 0.634 As Ka value is of the order 10–3 M a is not small and cannot be neglected. 9. Density of formic acid = 1.22 g/cm3 Weight of formic acid in 1 litre solution = 1.22 × 103 g

C=

1.22 ×10-3 = 26.5M . Thus [HCOOH] = 46 Since in case of auto ionization [HCOOH+2] [HCOO–] And [HCOO-][HCOOH+2] = 10-6 [HCOO-] = 10–3

13.  NH 4+ NH 3 + H + 

DH 0 = - 52.2 O

 H + + OH 

H2O

 NH 4+ + OH NH 3 + H 2 O 

DH 0 = 56.6 O

DH 0 = 4.4 KJ mol -1 O

Similarly, ΔS for the change = -76.53 Jk–1 mol –1 Or for the change  NH 4+ + OH - DH 0 = 4.4 KJ mol -1 NH 4 OH  O

O

DS 0 = - 76.53 JK mol -1 O

Using the equation DG θ = DH θ - TDSθ O

O

O

= 4.4 - ( -76.53 × 10 -3 ) × 298 = 27.21 KJ mol -1 DG θ = - 2.303R log K b O

27.21 = - 2.303 × 8.314 × 10 -3 × 298 log K b K b 1.7 × 10 -5

10.84

Ionic Equilibrium

14. I

 H + + H 2 PO4H 3 PO4 

K1 = 7.5 × 10 -3

II

 H + + HPO42 H 2 PO4 

K 2 = 6.2 × 10 -8

III

 H + + PO4-3 

HPO4-

H + + H 2 PO40

0

C

C

7.5 × 10-3 =

C2 \ C = 0.024 (0.1 - C )

[ H 2 PO4 ] = 0.024 M [ H 3 PO4 ] = 0.1 - 0.024 = 0.0076 M The value of K1 is much larger than K2 and K3. Also dissociation of II and III steps occurs in presence of H+ ions furnished in Ist step and thus dissociation of II and III steps is further suppressed. Due to common ion effect.  H + + HPO42 For Step II: H 2 PO4-  0.024 (0.024 - Y )

0.024 0 (0.024 + Y ) Y

The dissociation of H2PO–4 occurs in presence of [H+] furnished in step I  H   HPO  (0.024 + Y ) Y Thus K 2 =    -  or 6.2 × 10 -8  H 2 PO4  (0.024 - Y ) \ Y is small 0.024-Y ≈ 0.024 and neglecting y2 0.024Y 6.2 × 10 -8 = \ Y = 6.2 × 10 -8 0.024  HPO 24 -  = K 2 = 6.2 × 10 -8 (insign nificant ) +

24

 HPO42 - 

(6.2 × 10

-8

-x

)

H+ + 0.024 + x

(

conc. before dissociation

10 -7 + x

C-x

x

conc. after dissociation

KW =  H  OH  = (10 + x ) x = 10-14 mol 2 lit -2 -

-7

which reduces to the quadratic equation x 2 + 10-7 x - 10-14 = 0

(10 )

-7 2

-10-7 ±

\x =

+ 4 × 10-14

2 x = 0.62 × 10 mol / litre -7

\  H +  = 10 -7 + x = 10 -7 + 0.62 × 10 -7 = 1.62 × 10 -7 mole litre -1

(

\ pH = - log  H +  = - log 1.62 × 10 -7

)

= 6.7905 pH = 6.7905 (ii) In presence of 10–8 M H+, water dissociates as H+

 H2O ↽ ⇀  C

+ OH -

10 -8

0

conc. before dissociation

10 -8 + x

x

conc. after dissociation

( From HCl )

C-x

\ KW =  H +  OH -  = (10-8 + x ) x = 10-14 mol 2 lit -2 which reduces to the quadratic equation

x 2 + 10-8 x - 10-14 = 0

(10 )

-8 2

-10-8 ±

+ 4 × 10-14

2 x = 0.95 × 10 mol / litre -1

PO43-

-7

\  H +  = (10-8 + x) = (10-8 ) + 0.95 × 10-7  = 1.05 × 10-7

x

 H +   PO43-  (0.024 + x ) x = K3 2 HPO4  6.2 × 10 -8 - x

0

( From HCl )

\x =

For Step III:

+ OH -

10 -7

C

+

[ H + ][ H 2 PO4- ] C.C = (0.1 - C ) [ H 3 PO4 ]

=

H+

 H2O ↽ ⇀ 

For the Ist step  H 3 PO4   0.1 0.1 - C K1

K 3 = 3.6 × 10 -13

(For weak polyprotic acid having no other electrolyte the anion concentration produced in IInd step of dissociation is always equal to K2 if concentration is reasonable) 15. (i) In presence of 10–7 MH+ water dissociates as

)

Again neglecting x2 and assuming 6.2 × 10–8 –x = 6.2 × 10–8 1.024 x \ 3.6 × 10-13 = 6.2 × 10-8 3.6 × 10-13 × 6.2 × 10-8 \x = = 9.3 × 10-19 (insignificant ) 0.024

\ pH = - log  H +  = ( - log1.05 × 10-7 ) pH = 6.9788 -3

16. (i) 10 N KOH KOH 10 -3 N 0

 → K + + OH 0 10

0 -3

10

conc. before ionization -3

conc. after ionization

Ionic Equilibrium 10.85

\ OH -  = 10 -3 moles litre -1 = - log OH -  = - log 10 -3  = 3 \ pH = - pOH = 14 - 3 = 11 \ pOH

(ii) 8 × 10-4 M Ca (OH) 2 Ca ( OH )2  → Ca 2 + 8 × 10-4 M 0

+ 2OH -

0 8 × 10-14

conc. before ionization 2 8 × 10-4 M conc. after ionization 0

(

)

\ OH -  = 2 ( 8 × 10-4 M )

20. We know that for HCl and NaOH Milli equi = milli moles \ m.e of HCl = 0.5 × 25 = 12.5 m.e of NaOH = 0.5 × 10 = 5.0 m.e of HCl in the resultant mixture = 12.5-5.0 = 7.5 Total volume = (25 + 10 + 15) mL = 50 mL m.e 7.50 \normally of HCl = = vol ( mL ) 50

\molarity \ H +  = [ HCl] = \pH = - log

\ pOH = log 16 × 10-4 2.7958 pH = 14 - 2.795 = 11.2042 17. Given pH of solution = 11 pH = - log  H + 

7.5 = 0.8235 50

C (1 - a )

+ OH -

Ca

0 Ca conc. before ionization

4.3 \ H +  = Ca = 3 × 10 -3 × = 1.29 × 10 -4 M 100 \pH = - log  H +  = - log 1.29 × 10 -4 

or  H +  = 10-11 M

pH

KW =  H +  OH -  = 1× 10-14 1× 10-14 \ OH -  = = 1× 10-3 M 10-11 OH -  = 1× 10-3 M

= 3.8894  →

22. CH 3 COOH C C(1 - a )

Strength = morality × molecular weight = 10–3 × 40 = 0.04 g/litre 18. 250 mL of solution contains HCl = 3.65 g 1000 mL of 3.65 × 1000 = 14.6 g 250

14.6 = 0.4 M Molarity (M) of HCl solution = 36.5 0.4 M HCl = 0.4 M H+ pH = – log [H+] = -log 0.4] = 0.3979 19. Given pH = 4.80

pH = - log  H +  = 4.80 or log  H +  = 5.2  H +  anti log of 5.2 = 1.585 × 10 -5 OH -  [OH - ] =

7.5 50

→ NH +4 21. NH 4 OH  C 0

11 = - log  H + 

solution contain contains HCl =

7.5 50

=

10 -14 = 6.3 × 10 -10 M 1.585 × 10 -5

a=

CH 3 COO - + 0

H+ 0

C.a

Ka C

=

1.8 × 10 -5 0.10

Ca

Conc. before ionization Conc. after ionization

= 1.34 × 10 -2

Now  H +  = Ca = 0.10 × 1.34 × 10 -2 = 1.34 × 10 -3 M pH = - log  H +  = - log 1.34 × 10 -3  pH = 2.8728 23. + → (C2 H 5 )2 N H 2 + OH (C2 H5 )2 NH + H 2 O 

C C(1 - a )

[OH ]

0

0

C.a

C.a

= C.a

Given C = 0.05 M = 12 pH pOH = 14 - 12 = 2 pOH

= - log OH -  = 2

or OH -  = 10 -2 M = C.a = 0.05 × a \a

= 0.2

Conc. before ionization Conc. after ionization

10.86

Ionic Equilibrium

+    C H NH ( ) 2 5 2   OH  2  0.05 × (0.2) 2 C.a 2  = = Now K b = (1 - a ) (C 2 H 5 )2 NH  (1 - 0.2)

= 2.5 × 10 -3 [ a = 0.2 so 1 - a ≠ 1]

14.9 M 164.9

+  H2O ↽ ⇀ H

+ OH -

10 -8

0

conc. before dissociation

-8

x

conc. after dissociation

(10 + x )

K w =  H +  OH -  = 10 -14 mol 2 litre -2 = (10 -8 + x) x \ x = 9.5 × 10 -8 M Total  H +  = (10-8 + x) = 10-8 + 9.5 × 10-8 = 1.05 × 10-7 M pH = - log  H +  = - log 1.05 × 10-7 

15 14.9 0.1 = 164.9 164.9 164.9

= 6.06 × 10-4 M pH = - log  H +  = - log 6.06 × 10-4  = 3.2172.

27. Total volume of solution after mixing = 9.9 + 100 = 109.9 mL 9.9 Molarity of HCl after mixing = M 109.9 10 And molarity of NaOH after mixing = M 109.9 OH -  less from NaOH after neutralization = 10 9.9 0.1 = 109.9 109.9 109.9

= 9.099 × 10-4 M pOH = - log OH -  = - log 9.098 × 10-4  = 3.0409

pH = 6.9788.

\ pH = 14 - pOH = 14 - 3.0409 = 10.9591 –12

–4

25. Ka = 2 × 10 , mole of saccharin in 200 mL = 4 × 10 M 4 × 10-4 × 1000 = 2 × 10-3 M [ H sac ] = 200 pH = 3 = 10-3 M  H +  In presence of 10–3 M H+, H sac dissociates as

28. NH +4  NH 3 + H +

K1 = 5.6 × 10−10

H 2 O  H + + OH −

K 2 = 1× 10−14

From the above reactions K

1  ⇀ NH +4 + OH − ↽   NH 3 + H 2 O K = K 2

 H + H sac  2 × 10 -3

Sac -

+

10 -3

-3

(2 × 10 - x) 10 + x 3

+

0

conc.before dissociation

x

conc.after dissociation

-

 H  Sac  \Ka =    [ H Sac] 2 × 10 -12 =

14.9 M 164.9

\ [H+] left from HCl after neutralization =

On dilution 1 mL of this solution to 1 litre the[H+] become 10–8 M and from general formula pH = 8 which is not correct because solution is acidic, its pH can be calculated by considering the H+ from ionization of water as well as from HCl. In presence of 10–8 M H + water dissociates as

C-x

Molarity of HCl after mixing =

And molarity of NaOH after mixing =

24. pH = 5  H +  = 10-5 M

C

26. Total volume of solution = 150 + 14.9 = 164.9 mL

(10

-3

( 2 × 10

-3

5.6 × 10−10 = 5.6 × 104 1× 10−14 Further, at equilibrium K=

K=

)

-x

K1 K2

K2 =

+x x

)

K1 3.4 × 1010 = = 6.07 × 105 K 5.6 × 1010

29. CH 3 COOH  CH 3 COO - + H +

\x = 4 × 10 -12 M –

K1 K2

–12

At equilibrium [Sac ] = 4 × 10

M

1 (1 - x) \Ka =

o x

o conc.before dissociation x conc.after dissociation

x2 = x 2 = 1.8 × 10 -5 ; x = 4.2 × 10 -3 =  H +  1- x

Ionic Equilibrium 10.87

pH = - log  H +  = - log  4.2 × 10-3  = 2.37    

Again for the dissociation of HCO3- we have

HCO3−  CO32 − + H +

Now let 1 litre of 1 M CH3COOH be diluted to V litres so that the pH of the solution doubles. Let the concentration of the diluted solution be C mole/litre. Thus CH 3 COOH  CH 3 COO - + H + C

0

0

C-x

x′

x′

\ 1Ka =

∴  CO32 −  =

(1)

Further, pH = - log x1 [ 2 × 2.37 ] = 4.74 (pH doubles m dilution) Or

log x ′ = -4.74 = 5.26

x ′ = 1.8 × 10 -5 -5 Substituting x ′ in (1) we get, c = 3.6 × 10 M As the number of moles of CH3COOH before and after dilution will be the same. \ moles of CH3COOH = morality × volume in litres \3.6 × 10-5 × V = 1× 1 initial morality = 1 Initial volume = 1 V = 2.78 × 104 litres

31. K h =

4.8 = − log  H +  or log  H +  = −4.18 = 5.82

Taking anti log [H++] = 6.61× 10-5 mole/L Now, for equilibrium H 2 CO3  HCO3− + H +

=

4.69 × 10−11 × 6.73 × 10−5 6.61× 10−5

Kw 1× 10-14 = = 2.2 × 10-5 K a 4.5 × 10-10

 HOCN + NaOH 32. NaOCN + HOH  x=

Ka = C

1 × 10 -14 = 1 × 10 -4 3.33 × 10 -4 × 0.003

Kw = K a .C

% hydrolysis = 1× 10-4 × 100 = 10-2  ⇀  H A sc- + OH 33. A sc- + H 2 O ↽

\ OH -  = C.h = C =

30. pH = − log  H + 

Ka  HCO3− 

 H +  = 4.8 × 10−11 mole / L

conc. before ionization conc. after ionization

x ′. x ′ = .8 × 10 -5 (C - x )

CO32 −   H +  Ka =   HCO3− 

Kh = K h .C = C

KW Ka

10 -14 × 0.02 = 2 × 10 -6 5 × 10 -5

\  H +  = Also h =

1 × 10 -14 = 5 × 10 -9 2 × 10-6 Kh = C

Kw = K a .C

10 -14 = 10 -4 5 × 10 -5 × 0.02

Or 0.01%

 HCO3−   H +  Ka =   H 2 CO3  − 3

−5

 HCO  6.61× 10  or 4.45 × 10−7 =  0.01 −7 × × 4 . 45 10 0.01 ∴ HCO3−  = −5 6.61× 10 = 6.73 × 10

−5

mole / L

(a ) NaBu+H 2 O  NaOH + BuH 34. (a) 1 0 conc. before 0 1-h

h

\ OH -  = C.h = C

h

hydrolysis conc. after hydrolysis

Kh = K h .C = C

10 -14 × 0.2 OH -  = = 10 -10 = 10 -5 2 × 10 -5 pOH = 5 Also pH + pOH = 14

\ pH = 14 - 5 = 9

KW .C Ka

10.88

Ionic Equilibrium

(b) x C6 H 4 COONa + H 2 O  x C6 H 5 COO − + Na + \ OH -  = C.h = C =

KW .C Ka

Kh = K h .C = C

1× 10-14 = 1× 10-8 1× 10-6

\ pH = 8

35. Let V mL of acid and V mL of NaOH be used. Conc. of both acid and NaOH are same. CH 3 COOH + NaOH  → CH 3 COONa + H 2 O

0

1 [log K b - log Ka - log Kw] 2 1 = [ pKa - pKw - pK b ] 2 1 = [3.8 + 14 - 4.8] = 6.5 2

pH =

1× 10-14 × 0.01 = 1× 10-6 1× 10-4

\  H +  =

0.1 × V 2V

37. The pH of salt HCOONH4 (a salt of weak acid + weak base) is given by  HCOOH + NH 4 OH HCOONH 4 + H 2 O 

0.1 × V 2V

0

0

0.1 × V 2V

0

conc. before reaction

0.1V conc. after 2V reaction

0.1 = 0.05M 2 Now calculate pH by hydrolysis of CH 3 COONa \CH 3 COONa  =

CH 3 COONa + H 2 O  CH 3 COOH + NaOH \ OH -  = C.h = C

Kw × C 10 -14 × 0.05 = Ka 1.9 × 10 -5

Kh = C

= 5.12 × 10 -6 \ pOH = 5.29 Hence pH = 14-5.29 = 8.71 36.

Ca (LaC)2 + 2H 2 O ⇌ Ca (OH)2 + 2HLaC 2LaC + 2H 2 O ⇌ 2OH - + 2HLaC Before 1 0 0 1- h

h

h

hydrolysis After hydrolysis

0.13 = 0.26 M 0.5 \ [LaC] = 0.26 × 2 = 0.52 M [∵ 1 mole Ca (LaC )2 gives 2 moles of (LaC)] Now [OH - ] = C.h. K w .C K = C h = K h .C = C Ka Where Cis conc. of anion which undergoes hydrolysis. \ [Ca (LaC)2 ] =

10 -14 × 0.52 \ 10-5.6 = Ka \ Ka = 8.25 × 10-4

+ +  38. Zn 2 + + H 2 O ↽ ⇀  [ Zn (OH)] + H

kh = C

∴ [H + ] = C.h = C

K h .C =

10−9 × 0.001

[Fe(OH) 2 + + H +

Kh = 6.5 × 10−3

∴ pH = 6 39. Fe3+ + H 2 O

 ↽ ⇀ 

1− h

h

h

[H + ] = Ch.

Ch 2 5 and h = 1- h 100 C × 5 × 5 × 100 -3 Thus 6.5 × 10 = 100 × 100 × 95 and K h =

\C = 2.47 \ [ H + ] = 2.47 ×

 K 2 HPO 4 + KOH

40. K 3 PO 4 + H 2 O or PO

3− 4

5 \ pH = 0.9083 100

+ H 2 O  HPO 42 − + OH −

Since hydrolysis proceeds only in Ist step Kw Ka.C

∴[OH − ] = C.h =

=

Kw.C Ka

Ka is III dissociation constant of acid H3PO4 H 3 PO 4  H + + H 2 PO −4

K1

H 2 PO  H

K2

− 4

HPO

2−

+

 H

+ HPO

+

+

+ PO

3− 4

2− 4

K 3 = Ka = 1.3 × 10−12

-

 41. Aniline + Acetate + H 2 O ↽ ⇀  1 1- h

1 1- h

Aniline + Acetic acid 0 h

0 h

Before hydrolysis After hydsolysis

Let concentration of salt be C mol litre-1

Ionic Equilibrium 10.89

[ Aniline][ Acetic acid ] C.h.C.h = C (1 - h).C (1 - h) [ Aniline + ][ Acetate - ]

\Kh = \ Kh

h2 h \ = (1 - h) (1 - h) 2

=

h = 1- h

Kh

Thus the addition of the salt caused the [H+] to

1.008 × 10 -14 1.75 × 10 -5 × 3.83 × 10 -10

Kw = Ka.Kb

h = 54.95%

42. pKa for HCN = 14 - 4.70 = 9.30  NaCN + H 2 O ↽ ⇀  NaOH + HCN 1

0

0

before hydrolysis

1- h

h

h

after hydrolysis

\ [OH - ] = C.h = C pOH =

Kh = kh.C = C

Kw.C Ka

0 h

pH =

10 -4.5 =

10 -14 × C 1.8 × 10 -5

Before hydrolysis After hydsolysis

10 -14 × C 1.8 × 10 -5

Kw .C= Kb

KhC =

\C = 1.8 mol litre -1

\ wt of NH 4 Cl = 1.8 × 53.5 g / litre = 1.8 × 53.5 ×

1 g / 500 mL 2

= 48.15g. 44. First find [H+] Of HCOOH before adding HCOO Na for the equilibrium  H+ + HCOO– HCOOH  +

Ka =

-

 [ H + ] = [ HCOO - ] \ [H ] =

Ka. [ HCOOH ]

=

pKa + log

[ Salt ] [ Acid ]

= - log (1.8 × 10 -5 ) + log

1.8 × 10 -4 × 0.2

= 6 × 10 -3 mole /litre

1.1 1.1 0.9

\ pH = 4.83 46. Let the normality of BOH be n m.e. of BOH = 40 n On the addition of 5 mL of 0.1 N HCl into BOH solution m.e of HCl = 0.1 × 5 = 0.5 m.e of salt formed = 0.5 m.e of BOH used = 0.5 m.e of remaining BOH = (40 n - 0.5) Applying Henderson’s equation [salt ] pOH = pK b + log [base] 0.5 14 − 10.04 = pK b + log .......(1)) 40n − 0.5 Similarly, on the addition of 20 mL of HCl

+ 2

[ H ][ HCOO ] [H ] = [ HCOOH ] [ HCOOH ] +

4 45. Concentration of Na OH = mole per litre = 0.1 M 40 (M.wt of NaOH = 40) Since 0.1 mole of NaOH is added to one litre of the buffer solution containing 1 mole each of CH3COOH and CH3COONa, 0.1 mole of NaOH will neutralize 0.1 mole of CH3COOH, reducing the concentration of CH3COOH from 0.1 M to 0.9 M and increasing the concentration of CH3COONa from 1 M to 1.1 M i.e.,

Now from

43. Let conc. of NH4Cl be C mol litre-1  NH 4 OH + HCl NH 4 Cl + H 2 O 

[H + ] =

3.6 ×10-4 i.e. 1/ 16.6 of its original value. 6 ×10-3

[CH 3 COONa] = 1 + 0.1 = 1.1M

1 [14 + 0.3010 - 9.30] = 2.5 = 2 pH = 14 - 2.5 = 11.5

0 h

diminish to

i.e., [CH 3 COOH ] = 1 - 0.1 = 0.9 M

1 [pK w - log C - pKa ] 2

1 1- h

Now an addition of sodium formate in the acid we have [ Acid ] 0.2 [ H + ] = Ka = 1.8 × 10 -4 × = 3.6 × 10 -4 [ Salt ] 0.1

14 - 9.14 = pK b

+ log

2 .........(2) 40n - 2

Subtracting (1) from (2) we get n = 0.088 Substituting x in (1) we get pKb = 4.7412 -log Kb = 4.7412 Or log Kb = -4.7412 = 5.2588 Taking antilog Kb = 1.82 x 10–5

10.90

Ionic Equilibrium

47. (i) Addition of 0.2 mole of HCl increases the concentration of CH3COOH by 0.2 and reduces the concentration of CH3COO- by 0.2 mole/L \ [CH 3 COOH] = 1 + 0.2 = 1.2 M [CH 3 COO - ] = 1 - 0.2 = 0.8 M = [CH 3 COONa ] [salt ] [acid ] 0.8 = - log (1.8 × 10 -5 ) + log = 4.57 1.2 Using pH = pKa + log

0.2

0

0 Before dissociation

(0.2 - x)

x

x After dissociation

CHCl2 COONa  → CHCl2 COO - + Na + 0.1

x = 0.05 or [ H + ] = 0.05

0.1 0.1 pOH = 4.7447 \ pH = 9.2553

= - log 1.8 × 10 -5 + log

0

0.08

pOH1

[Salt] [Base]

0. 1

0

(0.1 + 0.02)

mole before reaction mole after reaction

0.08 and 1

[ NH 4 Cl] =

0.12 1

0.12 0..08 = 4.9208 \ pH1 = 9.0792 - log 1.8 × 10 -5 + log

=

Change in pH = pH - pH1 = 9.2553 - 9.0792 = + 0.1761 (ii) Now 0.2 mole of NaOH are added NaOH + NH 4 Cl  → NaCl + NH 4 OH 0.02

0.1

0

0

0.08

0.02

pOH 2

=

0 .1

mole before reaction 0.12 mole after reaction

- log 1.8 × 10-5 + log

0.08 0.12

pOH 2 = 4.5686 \ pH 2 = 9.4314 Change in pH = pH - pH 2 = 9.2553 - 9.4314 = - 0.1761 Here pH increases 50. pOH = - log K b + log

5 = 4.7 + log

49. Initial pH of solution when 0.1 0.1 [NH 3 ] = and[NH 4 Cl] = 1 1 pOH = - log 1.8 × 10 -5 + log

0.1

\ [ NH 4 OH] =

 CHCl2 COO - + H + 48. CHCl2 COOH 

For the dissociation of acid [CHCl2 COO - ][ H + ] Ka = 5 × 10 -2 = [CHCl2 COOH ] [0.1 + x][ x] or 0.5 = [0.2 - x]

0.02

 Volume = 1 litre

(ii) In this case 0.1 M of CH3COO- will combine with 0.1 M of HCl forming 0.1 M CH3COOH Total [CH3COOH] = 0.1 + 0.1 = 0.2 M But in the presence of the remaining 0.1 M HCl the dissociation of 0.2 M CH3COOH will be further suppressed by the common ion effect. Hence neglecting [H+] produced by CH3COOH The pH may be calculated for just 0.1 M HCl. pH = -log [H+] = -log (0.1) = 1.0

0.1

(i) Now 0.02 mole of HCl are added, then HCl + NH 4 OH  → NH 4 Cl + H 2 O

[salt] [Base]

a b

a = 2 \ a = 2B b Given a + b = 0.6 2b + b = 0.6 3b = 0.6 or b = 0.2 mole or 0.2 × 17 = 3.4 g / L a = 0.4 mole or 0.4 × 53.5 = 21.4 g / L Thus [Salt] = 0.4 M and [Base] = 0.2 M

Ionic Equilibrium 10.91

51. pH = - log Ka + log

[Salt] [Acid]

4 = - log1.0 × 10-5 + log

Case I

[Salt] (0.5)

[Salt] = - 1 or [Salt] = 0.1 × 0.5 = 0.05 M. (0.5) [Salt] Case - II 6 = - log1.0 × 10 -5 + log (0.5) [salt] \ log =1 (0.5) [Salt] = 10 × 0.5 = 5 M log

Now the two buffers [(I.NaA = 0.05 M and HA = 0.5 M) And (II. NaA = 5 M and HA = 0.5 M)] are mixed in equal proporaion. Thus, new conc. of NaA is mixed buffer 0.05 × V + 5 × V 5.05 = 2V 2 New conc of HA in mixed buffer 0.05 × V + 0.5 × V = = 0.5 M 2V [5.05 / 2] Thus pH = - log1.0 × 10 -5 + log [0.5] pH = 5 + 0.7033 = 5.7033 =

 52. HA + BOH ↽ ⇀  BA weak strong HA + B + OH HA

OH -

+

+

H2O +

 ↽ ⇀  B + A + H2O

 ↽ ⇀ 

A- + H2O

-

K=

[A ] ......(1) [HA][OH - ]

+  Also for weak acid HA : HA ↽ ⇀  H +A

[H + ][A - ] .........(2) [HA] Ka = Kw By Eqs (2) and (1) K Ka 10 -4 \K = = = 1010 Kw 10 -14 Ka =

53. [H2CO3] in blood = 2 M Volume of blood = 10 mL [NaHCO3] = 5 m Let volume of NaHCO3 used = V mL 2 × 10 [H2CO3] in mixture = V + 10 5 ×V [NaHCO3] in mixture = V + 10 [ Salt ] pH = pKa + log [ Acid ] 7.4 = - log 7.8 × 10 -7 + log

(5 × V) (V + 10) (2 × 10) (V + 10)

V = 78.36 mL

→ NH 4 Cl 54. HCl + NH 3  30 40 0 m.m 0 10 30 m.m after reaction pOH = pK b + log

[ NH 4+ ] (pka + [ pK b = 14) [ NH 3 ]

30 = 5.2218 10 pH = 14 - 5.2218 = 8.7782.

= 4.7448 + log

55. pH = l- og K a + log

[Salt ] [Acid ]

Let a mol litre –1 be con concentration of salt, then concentration of acid = (0.29-a) 4.4 = l- og 1.8 × 10 -5 + log

a \ a = 0.9 (0.29 - a )

[Salt ] = 0.09 M [Acid ] = 0.29 - 0.09 = 0.20 M 56. Suppose x moles of sodium propionate is added to one litre of an aqueous solution containg 0.02 mole of propionic acid. [ Salt ] Using pH = pKa + log [ Acid ] x 4.75 = - log (1.34 × 10-5 ) + log 0.02 \ x = 1.5 × 10-2 = 0.015M . Further addition of 0.01 mole of HCl will produce an extra 0.01 mole of propionic acid and remove 0.01 mole of propionate ion in a one – litre solution.

10.92

Ionic Equilibrium +  58. HIn ↽ ⇀  H

[C 2 H 5 COOH ] = 0.02 + 0.01 = 0.03M [C 2 H 5 COONa ] = 0.015 − 0.01 = 0.005M

[H + ][In - ] [HIn ]

Ka =

0.005 = 4.06 0.03 Now since the pH of 0.01 M i.e., 10–2 M HCl is 2, Therofore the pH of the buffer solution is about two times greater than of 0.01 M HCl. pH = − log (1.34 × 10−5 ) + log

When

In -

+

[In - ] = 10 [HIn ]

Ka = [H + ] × 100

[In - ] 1 = [HIn ] 10

[H + ] = 1 × 10 -5 /10 = 10 -6

When

pH = 6

57. (a) As pH = 7, pOH = 14 - 7 = 7 at 25°C

[H + ] = 1 × 10 -5 × 10 = 10 -4

pK b = - log kb = - log (9.8 × 10 -8 ) = 7.0088 Applying pOH = pK b + log 7 = 7.0088 + log

[Salt] [Base]

pH = 4 \ minimum changein pH = 6 - 4 = 2

[Salt] [Base]

 59. HIn ↽ ⇀ 

[Salt] = - 0.0088 log [Base] [Base] log = 0.0088 [Salt] [Base] Taking antilog log = 1.019 [Salt] m.m of base = 1.019 (1) or m m of salt Suppose V1 mL of HCl is mixed V2 mL of imidazazole (base) to make the buffer m.m of HCl = 0.02V1 m.m of imidazole = 0.02 V2 As the buffer is of the base and its salt 0.02 V, m.m. of HCl will combine with 0.02 V1, m.m of base to give 0.02V1 m.m of salt \ m.m of salt = m.m. of HCl = 0.02V1 and m.m of base left = 0.02V2 - 0.02V1 = 0.02(V2 - V1 )

H + + In −

[H + ][In − ] [H + ][Base] = [HIn ] [Acid ] [ ] Acid [H + ] = Ka [Base] K IN = Ka =

For 75%red [H + ] =

(3 ×10−5 )(75) = 9 × 10−5 25

pH = 4.05

( 3 ×10 ) ( 25) = 1×10 -5

For 75% blue  H +  = pH = 5

-5

75

The change in pH = 5-4.05 = 0.95  H + + BPh 60. HBPh  Ka

 H +   BPh -  =   , [ HBPh ]

When BPh - = HBPh, indicator will work. Thus  H +  = 5.84 × 10 -5 \ pH = 4.2336 Also if pH = 4.84 or  H +  = 1.44 × 10 -5 then

\from (1) we get 0.02 (V2 - V1 ) V - V1 = 1.019 or 0 2 = 1.0 V1 V1

(2)

Given that V1 + V2 = 100 mL

(3)

From (2) and (3) we get V1 = 33 mL V2 = 67 mL (b) pH shall remain the same on dilution as both Kb and [Salt]/[Base] will not change.

 H +   BPh -  1.44 × 10 -5 Ca Ka =    or 5.84 × 10 -5 = C(1 - a ) [ HBPh ] a = 0.8 or 8090 61. (a) In pure water Let the solubility of AgCl be S mol/ L  Ag + (aq ) + Cl - (aq ) AgCl ( s )  K sP =  Ag +  Cl -  = S × S S = K sP = 1.5 × 10-10 = 1.224 × 10-5 mol / L

Ionic Equilibrium 10.93

(b) In 0.1 M AgNO3

64. When Mg ( OH )2 starts precipitation

+  AgCl(s) ↽ ⇀  Ag + Cl +

AgNO3  → Ag + NO 0.1

 Mg 2 +  OH -  = K sp of Mg ( OH )2 2

3

[0.1]

0.1

K sP =  Ag +  Cl -  = ( 0.1 + S) S since S is very small, presence of common ion

OH -  2 = 1 × 10 -11 OH -  = 1 × 10 -5 \ pOH = 5

pH = 14 - pOH = 14 - 5 = 9

decreases solubility S ( 0.1) = 1.5 × 10 -10

 65. Na 2 CO3 + 2AgCl ↽ ⇀  2 NaCl + Ag 2 CO3 7.5 7.5 - a

S = 1.5 × 10 -9 mol / L (c) In 0.01 M NaCl +

 AgCl(s) ↽ ⇀  Ag + Cl S S

-

NaCl  → Na + + Cl 0.1 0 .1 K sP =  Ag +  Cl -  = (S) (0.01 + S) S = 1.5 × 10 -8 mol / L 62. For ZnS not to be precipitated from a solution of Zn 2 + and Pb 2 +  Zn 2 +   S 2 -  < K sP of ZnS 10-2   S 2 -  < 1.0 × 10-21

2

\ H +   S 2 -  = 1.1× 10-22 \ H +  10-19  = 1.1× 10-22 \ H +  = 3.3 × 10-2 M

Thus if [H+] = 3.3 × 10–2 or slightly higher the precipitation of ZnS will not take place and only PbS will precipitate. 63. K sp of Pb ( OH )2 = 4S3 = 4 × ( 6.7 × 10 -6 ) = 1.203 × 10 -5 3

The buffer contains pH = 8 \pOH = 6 or OH -  = 10 -6 Now let solubility of Pb(OH) 2 be S mol/L init 2+

-

Thus  Pb  OH  = K sp 2

 Pb 2 +  10 -6  = 1.203 × 10 -5 Buffer pH = 8 \ pOH = 6 and OH -  = 10 -6  Pb 2 +  =

0

0

excess

2a

a

mm added mm left

0.0026 Cl -  = = 7.32 × 10 -5 35.5 millimole 2a Also conc. of Cl - formed = = Vol. in mL 5 2a 0.0026 \ = \a = 1.83 × 10 -4 millimolle 5 35.5 \m. mole of Na 2 CO3 left in 5 mL = 7.5 - 1.83 × 10 -4 = 7.5 7..5 or CO32 -  = 5 K sp of Ag 2 CO3 =  Ag +  2 CO32 -  8.2 × 10 -12 = 5.46 × 10 -12 7.5 5 \  Ag +  = 2.34 × 10 -6 \ Ag +  =

Or the maximum  S 2 -  = 10-19 at which ZnS will begin to precipitate or up to this concentration no precipitation will occur  2 H + + S 2 H 2 S  \ H +  = 11× 10-4

excess

1.203 × 10 -15 = 1.203 × 10 -3 mol / L 10 -12

0.0026 K sp ofAgCl =  Ag +  Cl -  = 2.34 × 10 -6 × 35.5 -10 K sp = 1.71 × 10 66. The K sp values of A g 2 CrO 4 and AgIO3 reveals that CrO 24 - and IO3- will be precipitated on addition of AgNO3 as  Ag +   IO3-  = 10 -13 10 -13  Ag +  = 2 × 10 -11 = required [0.005] 2

 Ag +  CrO 24 -  = 10 -8  Ag +  = required

10 -8 0.1

= 3.16 × 10 -4

10.94

Ionic Equilibrium

Thus AgIO3 will be precipitated first Now in order to precipitate AgIO3 , one can show AgNO3 + NaIO3  → AgIO3 + NaNO3 0.01

0.05

0

0

0.005 0 0.005 0.005 The mole of AgNO3 left are now used to

Ca 2 +  OH -  2 = K sp Ca 2 +  left [ 0.2223] = 4.42 × 10 -5 4.42 × 10-5 Ca 2 +  left = 8.94 × 10-4 mol / L 2 [0.2223] 2

\ mole of Ca ( OH )2 precipitated = Mole of [Ca 2 + ]

preciipitate Ag 2 CrO 4

precitated = 111.5 × 10 -4 - 8.94 × 10 -4 = 102.46 × 10 -4

2AgNO3 + Na 2 CrO 4  → Ag 2 CrO 4 + 2 NaNO3

\ wt of Ca (OH) 2 precipitated

0.005

0.1

0

0

0.0975

0

0.0025

0.005

24

Thus CrO  left in solution = 0.0975 Now solution has AgIO3 (s) + Ag 2 CrO 4 (s) + CrO 24 - ions 0.005

0.0025

\  Ag +  left =

0.0975

K sp Ag 2 CrO 4 24

CrO 

-8

=

10 0.0975

= 3.2 × 10 -4 M K sp AgIO3 10 -13 = \ IO3-  left = 3.2 × 10 -4  Ag +  = 3.1 × 10 -10 M 67. 500 mL of 0.4 M NaOH are mixed with 500 mL of Ca (OH ) 2 a saturated solution having Ca (OH ) 2 solubility as S M 2+  For Ca (OH) 2 ↽ ⇀  Ca + 2OH

K sp = S × (2S) 2 = 4S3 4.42 × 10 -5 = 0.0223 M 4 Now Ca(OH)2 + NaOH are mixed ∴ Solution has Ca2+ and OH– out of which some Ca2+ will precipitate \S = 3

0.0223 × 500 1000 = 0.01115 = 111.5 × 10 -4 M 0.0223 × 2 × 500 500 + 0.4 OH -  = + 1000 1000 [from Ca ( OH )2 ] [ From NaOH ]

On mixing Ca 2 +  =

= 0.2223 M

= 102.46 × 10 -4 × 74 = 7582.04 × 10 -4 g = 758.2 mg 68. For buffer solution NH 4 Cl = 0.25 M and NH 4 OH = 0.05 M pOH

= - log k b + log

[Salt ] [ Base]

\pOH = - log 1.8 × 10 -5 + log

0.25 0.05

\OH -  = 3.6 × 10 -6 M Now both Al(OH)3 and Mg (OH) 2 are mixed thoroughly 3

 Al3+  OH -  = K sp of Al(OH)3 3

 Al3+  3.6 × 10 -6  = 6 × 10 -32  Al3+  = 1.28 × 10 -15 M 2

For  Mg 2 +  OH -  = K sp Mg (OH) 2 2

 Mg 2 +  3.6 × 10 -6  = 8.9 × 10 -12  Mg 2 +  = 0.686 M 69. The minimum [OH–] at which there will be no precipitation of Mg(OH)2 can be obtained by K sp =  Mg 2 +  OH − 

2

9.0 ×10−12 = [ 0.05] OH − 

2

OH −  = 1.34 × 10−5 M

Thus a solution having [OH–]=1.34 × 10–5 M will show precipitation of Mg (OH)2 in 0.05 M Mg2+ solution. These hydroxyl ions are to be derived by a buffer of NH4 Cl and NH4 OH i.e  NH 4+ + OH NH 4 OH   NH 4+ + Cl NH 4 Cl 

Ionic Equilibrium 10.95

From the equations (1) and ( 2)

 NH 4+  OH -  Kb =   NH 4 OH 

For NH 4 OH

In presence of NH4Cl; all the [NH+4] are provided by NH4Cl since common ion effect decreases the dissociation of NH4OH + 4

-5

 NH  1.37 × 10  \ 1.8 × 10-5 =  [0.05]

0.1520

exces

0.0358

0.358 reaction

= 0.1162

At equilibrrium 0.1162 C2 O 24 -  = 0.5 = 0.2324 1M

2

(

Thus  Ag +  C2 O 24 -  > K sp of Ag 2 C 2 O 4 1.29 × 10 -11

)

Ag 2 C 2 O 4 will precipirate out For Ag 2 C2 O 4 precipitation 2

 Ag +  C 2 O 24 -  = K sp Ag 2 C2 O 4 1.29 × 10 -11 = 5.55 × 10 -11 0.2324 Now for Ag 2 CO3 23

K SP =  Ag   CO  = 5.55 × 10

(

+

)

K sp =  Ag +  Cl - 

2

and Cl -

2 0.1 × 0.1 \ NH 3  = = 6.25 \ NH 3  = 2.5 M 16 × 10 -3

Also 0.2 M NH 3 is needed to dissolved 0.1 M Ag + ions. Thus  NH 3  = 2.5 + 0.2 = 2.7 M

 NH 3   Ag +  Kc =  Ag ( NH )+  3 2  

(1)

 Ag + (aq ) + Cl - (aq ) AgCl ( s ) 

(2)

)

In case of simultaneous solubility Ag+ remains same in solution. + Given  NH 3  = 2a = 1M ; also  Ag ( NH 3 )2  = Cl -   

or b 2 =

-

1.8 × 10-10 = 0.29 × 10-2 6.2 × 10-8

b = 0.539 × 10-1 = 0.0539 (1)

 Ag + + 2 NH 3 ↽ ⇀   Ag NH 3 2   

)

+

 NH 3  Kc 1 = = K sp Cl−   Ag ( NH )+  b 2 3 2    

 [ 0.0716]

 71. AgCl + 2 NH 3 ↽ ⇀   Ag NH 3 2  + Cl   +  AgCl (s) ↽ ⇀  Ag + Cl

(

)

2

-11

= 3.97 × 10 -12 mol3 litre -3

 Ag NH +  3 2   Kf =  2 +  Ag   NH 3 

2

 NH 3  Given solubility of AgCl = 0.1M

= b Because Ag+ obtained from AgCl passed in [Ag(NH3) +2 ] state Thus from equation (1) and (2)

2

 Ag +  =

(

a×a

Ksp = [Ag+][Cl–]

Ag 2 CO3 is solid ⋅ Ag 2 C2 O 4 is almost precipitated out.

2

or 1 × 10 -10 × 1.6 × 107 =

2

2

 2 × 0.0358   0.1520  -3  Ag +  C2 O 24 -  =    0.5  = 6.23 × 10 0.5 

+

 NH 3 

 Ag + (aq ) + 2 NH 3 (aq ) 72. Ag ( NH 3 ) 2+ (aq )  a+b 2a

0.0358 CO32 -  = 0.5 = 0.0716 M

2

)

(

 ↽ ⇀  Ag 2 C2 O 4 + K 2 CO3 mole before

( 0.1520 - 0.0358)

(

a = 0.1 M for Ag NH 3

\ NH 4+  = 0.067 or  NH 4 Cl  = 0.067 M

70. Ag 2 CO3 + K 2 C2 O 4

K sp × K f =

 Ag NH +  Cl -  3 2   

+

(2)

 Ag NH )+  = 0.0539 2  (  Ag + (aq) + I - (aq) 73. AgI (s)   

10.96

Ionic Equilibrium

    +   K sp = [Ag ]  I -  = 1.2 × 10 -17

(1)

75.

(i )

K sp =  Zn 2 +  OH - 

-

 Ag + ( aq ) + 2CN - ( aq )    Ag ( CN )2  (aq)    

(ii )



 Ag ( CN )  2 = 7.1 × 1019 Kf =  +  Ag  CN -  Let x mole of AgI be dissolved i -

-

 Ag I (s ) + 2CN - ↽ ⇀   Ag ( CN )2  + I 1 0 0 mole before reaction x x mole after 1 - 2x reaction

from equation (1) and ( 2) K eq = K sp × K b  Ag ( CN )  I    2  K eq =  = 1.2 × 10 -17 × 7.1 × 1019 2 [CN ] K eq = 8.52 × 102 x.x

(1 - 2x )

2

=

x2

(1 - 2x )

2

 Zn ( OH ) Zn ( OH )2 (s ) + 2OH -  4

2-

Let the solubility be S in change (i) and thus solubility will also be S in change (ii) by coupling (i) and (ii) 2+ 2 2 Zn ( OH )2 (s) ↽ ⇀  Zn (aq ) + [ Zn (OH) 4 ] (aq )

K sp × K b = S × S or S = K sp × K b = 1.2 × 10 -17 × 0.12 = 1.2 × 10 -9 M Thus total solubility will be 2S = 2 × 1.2 × 10 -9 = 2.4 × 10 -9 M 2

-

K eq = 8.52 × 102 =

2

 Zn ( OH )2 -  4  Kb =  - 2 OH 

(2)

Let x mole of AgI be dissolved in CN - solution then

-

 Zn 2 + + 2OH Zn ( OH )2 (s ) 

or

x = 29.2 1 - 2x

Thus x = 29.2 - 58.4 x or = 0.49 mole. 2+   74. MBr2 ( g ) ↽ ⇀  MBr2 (aq ) ↽ ⇀  M + 2Br

MBr2 + H 2S → MS + 2HBr K sp of MS =  M 2 +  S2 - 

This solubility is minimum when  Zn 2 +  OH -  = 1.2 × 10 -17 2

1.2 × 10 -9 OH -  = 1.2 × 10 -17 OH -  = 10 -4 M Note : If only I change then solubility of Zn ( OH )2 =

3

K sp

=

4

1.2 × 10 -17 = 1.44 × 10-6 m 4

and at this conc. OH -  = 2 × 1.44 × 10-6 M = 2.88 × 10 -6 M

76. NaCN → Na + + CN 0.2 0 0 0 0.2 0.2

6 × 10 -21 = [ 0.05] S2 - 

+  Ag ( CN )2 ↽ ⇀  Ag + 2CN

\ S2 -  = 1.2 × 10 -19 M

2  Ag +  CN -   Ag +  ( 0.2) = K= = 1.0 × 10 -20 0 02 .  Ag ( CN )  2 

2

Thus MS will be precipitated if H2S provides 1.2 ×10-19 M ions of S2–  2 H + + S 2 Now For H 2 S H 2 S   H +   S 2 -  K1 × K 2 =     H 2 S  2

-7

10 × 1.3 × 10

-13

 H +  1.2 × 10-19  = [0.1]

 H +  = 1.04 × 10-1 and pH = 0.9826

 Ag +  = 5 × 10 -21 M 2+  For Cd (CN) 24 - ↽ ⇀  Cd + 4CN Cd 2 +  CN -  K= Cd ( CN )2 -  4  

4

Cd 2 +  ( 0.2) 7.8 × 10 =  0.02 Cd 2 +  = 9.75 × 10 -17

4

-18

No

2

 Ag +  

2-

 = 1.0 × 10 -50

Ionic Equilibrium 10.97

\S2 -  needed to precipitate Ag 2S =

1.0 × 10 -50

(5 × 10 )

-21 2

= 4 × 10 -10 M

Therefore CdS will be precipitated first

x ′ = 3.2356 or x ′ = 1.80 \ [HCl] = 1.80 M Thus any concentration of HCl greater than 1.80 M will just prevent precipitation of PbS

Sr 2 +  = 2.5 × 10 -3 M left \Sr 2 +  precipitated = (16 - 2.5) × 10 -3 = 13.5 × 10 -3 M -

\ F  needed for this precipitation = 13.5 × 10 2 × 13.5 × 10 -3 M = 27.0 × 10 -3 M + 2F → SrF2

x1 (3.4 × 10 -25 ) x′ = 0.1 2 0.1 × 1.1 × 10-23 x′ 3.4 × 10-25 2

2

77. Sr 2 +  = 16 × 10 -3 Mi Initial

2+

[H + ][S−2 ] [H 2S]

2

7.1 × 10 -28 S  needed to precipitate CdS = = 7.28 × 10 -12 M 9.75 × 10 -17 2-

(\Sr

1.1× 10−23 =

-3

M

)

79. In order to prevent precipitation of ZnS [ Zn 2 + ][S2 − ] < Ksp of ZnS = 1×10−21 Ionic product

Or (0.1) [S2 − ] < 1× 10−21 Or [S2 − ] < 1× 10−20

2

Also Sr 2 +   F-  = K sp SrF . 2 = 2 8 × 10 -9 2 2.8 × 10 -9  F-  = 2.5 × 10 -3

∴[F–] = 1.058 × 10–3M, i.e., The concentration of which will also appear in solution state Thus [F–] needed for 1 litre = [ 27.0 + 1.058] × 10-3 M = 28.058 ×10-3 M \ NaF needed for 1 litre 28.058 ×10-3 × 42 g −3 NaF needed for 100 mL 28.058 ×10 × 42 = 01178 g 10

–

This is the maximum value of S2 before ZnS will precipitate Let [H+] to maintain thus [S2–] be x  2 H + + S 2 Thus for H 2 S  Ka =

[ H + ]2 [ S 2 - ] x 2 (1× 10-20) + + 1.1× 10-21 0.1 [H 2 S ]

Or x = [ H + ] = 0.1 \ ZnS will not be precipitated at a concentration of H+ greater than 0.1 M. 80. First calculate the [I–] to precipitate AgI and Hg2I2 K sp of [AgI] = [Ag + ] [I - ] 8.5 × 10 -17 = [0.1] [I]

78. Ksp of PbS = [Pb2+] [S2–] Since the lead salt is completely dissociated, [Pb2+] is equal to the concentration of lead salt, i.e., [Pb2+] = 0.001 M. If [S2+] is the concentration of S2– required to just start precipitation of PbS -28

[S2–] =

3.4 × 10 0.001

= 3.4 × 10-25

Now the addition of HCl will suppress the dissociation of H2S to the extent that [S2–] = 3.4 × 10-25 M Since HCl is completely ionsied [H + ] = [HCl] Let [HCl] be x ′, Then [H + ] = x ′ 2+  H 2S ↽ ⇀  2H + S At Equi [H 2S] = 0.1 - 3.4 × 10 -25 + x ′ ≈ x ′ [S-2 ] = 3.4 × 10 -25 +

\ Ka =

2-

[H ][S ] [H 2S]

8.5 × 10 -17 = 8.5 ×10 -16 M 0.1 K sp of Hg 2 I 2 = [Hg 22 + ][I - ]2

\ I - to precipitate AgI =

\ [I] to precipitate Hg 2 I 2 = 5.0 × 10 -13 M [I–] to precipitate AgI is smaller, therefore, AgI will start precipitating first. On further addition of I– more AgI will precipitate and when [I–] > 5.0 × 10-13 M MH Hg2I2 will start precipiating. The maximum concentration of Ag+ at this stage will thus be calculated as Ksp of AgI = [Ag+] [I–] 8.5 × 10-17 = [ Ag + ] (5.0 × 10-13 )

8.5 × 10-17  Ag +  = = 1.7 × 10-4 M 5.0 × 10-13 \ Percantage of Ag+ remained unprecipitated = 1.7 × 10-4 × 100 = 0.17% 0.1 Thus percentage of Ag+ precipitated = 99.83 %

10.98

81.

Ionic Equilibrium

K ps of Ag SCN K sp of AgBr

=

[Ag + ][SCN - ] 1 × 10 -12 =2 = [Ag + ][Br - ] 5 × 10 -13

-

Or

[SCN ] =2 [Br - ]

Since the concentration of SCN– ions is twice that of Br– ion concentration in the solution, AgSCN is double that of AgBr. Let the solution of AgBr be S moles per litre \ The solution of AgSCN = 25  Ag + + Br– Now for AgBr   Ag+ + SCN– And AgSCN  2S 2S [Ag+] in solution = S + 2S = 3S [Br–] = S \ Ksp of [Ag Br] = [Ag+] [Br–] 5 × 10–13 = 3S × S S = 4.083 × 10–7 mol/L \ solubility of AgBr = 4.083 × 10–7 mol/L And solubility of AgSCN = 8.166 × 10–7 mol/L

82.

K sp of MgF2 K sp of SrF2

=

 Mg 2 +   F -   Sr 2 +   F - 

Sr 2 + + 2F− S

And for SrSO 4 , K sp = Sr 2 +  SO 24 −  2.8 × 10−7 = Sr 2 +  1.55 × 10−2  Sr 2 +  = 1.8 × 10−5 84. Molar concentration of AgCl =

2S

2+ −  ↽ ⇀  Mg + 2F 2.24S 4.48S Now Ksp of SrF2 = [Sr2+][F–]2 = (S)(2S + 4.48S)2 = 2.9 × 10–9 S = 0.41 ×10–3 Thus  F -  = 2 S + 4.48S = 6.48S = 6.48 × 0.41× 10-3

MgF2

= 2.65 × 10-3

83. From the Ksp data it is clear that SO42 - are mainly produced by Ag 2 SO4 because Ksp of Ag 2 SO4 is much greater than Ksp of SrSO4 and Ksp of Ag 2 SO4 is the product of three ion concentrations Let  SO42 -  = S for Ag 2 SO4 \ K sp of Ag 2 SO4 = (2 S ) 2 S = 1.5 × 10-5

0.01 = 10-4 M 100

As AgCl dissolves completely [Ag+] = 10–4 M. In complexation it is assumed that 10–4Ag+ first combines with Cl– completely to produce 10-4 M AgCl2- and then we consider the equilibrium for the decomposition of AgCl2- in the presence of minimum concentration say x molar NaCl (i.e.,[Cl–] = x) solution. AgCl2-  → Ag + + 2Cl 2

 Ag +  Cl -  K sp AgCl × Cl -  = = 3.3 × 10-6 Kd = -4 10  AgCl2  or

2

Thus the solubility of MgF2 is 2.24 times that SrF2 Let the solubility of SrF2 be S moles/L

 ↽ ⇀ 

\ Ag +  = 2 S = 3.1× 10-2

2

2+ 6.5 × 10-9  Mg  = = 2.24 ; 2.9 × 10-9  Sr 2 + 

SrF2

S = 1.55 × 10-2

10-10 × x = 3.3 × 10-6 10-4

Or x = 3.3 moles/L =3.3×58.5 grams/L =193.05 grams/L \ Amount of NaCl per 100 litre=19305 g=19.3 kg 85.  H +  Formed due to electrolyte oxidation = 4 m moles Na2 HPO4 + Na3 PO4 0.08 × 100 =8

0.02 × 100 2.

initial m.m

 H +  Will be used by PO43- to give H 2 PO4-

PO43- + 2 H +  → H 2 PO42 o

4 o

o 2

Thus a new buffer containing Na2 HPO4 (8 m mole) and H 2 PO4- (2 m mole) will remain in solution.  Na2 HPO4  \ pH = pKa H 2 PO4 + log   H 2 PO4-  8 = 7.2 + log = 7.8% 2

Ionic Equilibrium 10.99

86. From Raoult’s law

2. NH 4 OH + HCl  NH 4 Cl + H 2 O K eq =

p 0 - Ps w × M = Ps m ×W

\K b = 10 -5 NH 4 +

17.540 - 17.536 w = 1.267 × 10-5 17.536 × 18 m ×W m \Molality = × 1000 = 1.267 × 10-5 × 103 m ×W = 1.267 × 10-2 = Molarity = conc of [ BOH ] \

0 a

\ Kb =

0 a

0.05

Initial conc Concc of equili

( NH +4 )(OH - ) ( NH 4 OH)

After dilution pH = 5.046 , H + = 9 × 10 -6 M +  CH 3 COOH ↽ ⇀  CH 3 CO5 + H

Ca 0.01 × 0.267 × 0.267 = = 9.74 × 10 -4 (1 - a ) 1 - 0.267 2

 87. CH 3 COOH + Na + + CN - 

x2 C-x \ C = 1.35 × 10 -5 M M1V1 = M 2 V2 \ Ka =

1 × (0.5) = (1.35 × 10 -5 ) V

CH 3 COO - + Na + + HCN CH 3 COO −  [ HCN ] K eq =  CH 3 COOH  [ HCN ] For CH 3 COOH and HCN in same solution CH 3 COO -   H +  K1 =  CH 3 COOH 

V = 3.36 × 104 lit (1)

4. CN - + HAC  HCN + AC K eq = 1.8 × 10

(2)

-5

4.9 × 10 -10

= 3.674 × 104

5. In the case of PbS ionic product is greater than Ksp 6. Mg (OH) 2  Mg +2 + 2OH -

CN -   H +  K2 =  [ HCN ] From the equations (2) and (3)

(3)

Solubility in buffer = 4 × 10 -4

(

)

\K SP = 4 × 10 -4 (10 -4 ) 2 = 4 × 10 -12

-5

K1 1.8 × 10 = = 3.674 × 104 K 2 4.9 × 10-10

Multiple Choice Questions with Only One Answer Level II

In pure water K sp = 4S3 \ S = 10 -4 M 7. When pH = 2 no. of mole of HCl = 0.01 It pH = 3 the no. of mole HCl = 0.001 No. of mole of HCl removed is = 0.01-0.001 = 0.009  8. NaZn(VO 2 )3 (CH 3 COO)9  Na + + Zn +2 + 3VO +22 + 9CH 3 COO -

1. HCOOH  H + 0.015 0.02 0.015x Ka =

0.05 -

3. pH of 0.5 M CH 3 COOH is 2.523

1.267 × 10 -2 = 1 + a \a = 0.267 \ 1 × 10 -2

K eq =

OH -  NH 4 OH

[OH - ] = 10 -5 pOH = 5, pH = 9

Molarity is also given as 1× 10-2

Now K b =

+

0.1 0.05

+  For BOH ↽ ⇀  B + OH

1 1- a

0.02x

Kb KW

+ HCOO Initial x

(0.02 + x)x (0.015 - x)

[HCOO - ]=1.35×10 -4 M

Conc of equilib

\ K sp = (S)(S) (3S) 2 (9S)9 = 321 S14 9. Solubility of CuS = K spCuS Solubility of Ag 2S

=3

Solubility of HgS = K sp HgS

K spAg S 2

4

10.100 Ionic Equilibrium

10. Π = CST Π = (0.11) RT 11. K C =

20. Ag + + NH 3 0.2 x

K sp

12. pOH = pK b

0.2

+

+ NH 3 →

0.2

[ NH OH ] - log 4

0.8 0.6

y

 NH +4 

+

3

1 0.8

Ag ( NH 3 )

K b2

[ Ag( NH )]

→

initial of conc eq. of conc +

 Ag ( NH 3 )2  initial conc 0.2

final conc

+

 Ag ( NH 3 ) = 3.33 × 10 -5 M

 NH +4  =1mole /litre ( NH 4 )2 SO 4  = 0.5 mole /litre

21. In electrolytes oxidation 4 mili moles of H + is obtained Na 3 PO 4 + H + → Na 2 H PO 4 + Na +

13. pKa1 + pKa 2 7 + 11 = =9 Initial pH 2 = 2 2 NaHCO3 + HCl → NaCl + H 2 O + CO 2 0.1 -

-

1 0.9

-

-

initial mole final mole

2 -

4 2

Na 2 HPO 4 + H 6 4

+

Final mole

→ NaH 2 PO 4 + Na + 2

2 -

pH = pKa 2 - log

0.9 [HCl] = = 9 × 10 -3 100 pH = 3 - log 9 = 3 - 2 log 3 DpH = 6 + 2 log 3

Initial mole

4 6

Initial mole Final mole

2 = 8.3 4

22. In conjugate acid base pair if acid is strong conjugate base is weak 23. PbBr2  Pb +2 + 2Br -

14. 15. 16. 17.

CaF2 having low solubility. So it first ppts Solution involves both cation and anionic hydrolysis AgCl formes complex with NH3 solution Due to common ion effect Cd2+ concentration decreases

(

18. K sp of CaCO3 8 × 10 -5

)

2

= 6.4 × 10 -9

Assume concentration CO3-2 in solution is = x Concentration of Ca +2 = a Concentration of Ba +2 = 0..2a

= 1.28 × 10-9

x 2

x(0.04) 0.03 + [Ag ] = 7.5 × 10 -18 M

\ KC =

0.04

2

\ S = 0.034 M 24. NaCN + HCl → NaCl + HCN 0.01 a - Initial mole 0.01 - a a Final mole

[ HCN ]

CN -  a 8.5 = 9.5 - log 0.61 - a

19. Ag (CN) 2-  Ag + + 2CN - K C = 4 × 10 -19 0.03 0.1 Initial conc 0.03

1.65

K sp = (0.85) (1.65)

pH = PK a - log

\ a.x = 6.4 × 10 -9 (0.2) a.x = K sp BaSO 4 \K sp BaSO 4

0.85

a = 10 0.01 - a a = 9 × 10 -3

eq conc 25. K sp MX

= S1

K sp MX2 = 4S32 K sp

M3 X

= 27S34

So order MX > M 3 X > MX 2

Ionic Equilibrium 10.101

26. At equilibrium point equivalent of base = equivalent of acid 2 2 \ 2.5 × = x 5 15 V = 7.5 mL

32. K sp of silver formed is 10 -4 +  HCOOAg ↽ ⇀  HCOO + Ag x S equi conc.

HCOO -

(

33. NH 4 Cl

\h = 0.27

NH

27. At equivalent point M1V1 = M 2 V2

-

-

4

-

I. mole equi moles

\ Ka = 4 × 10

 A - 

35.

4 × 10

+

10 -7 \ KC =

I m moles F m moles

Acid Salt

h = 1- h

KW K a .K b

K sp

-

5 × 10 -9 = 0.001 0.005

Sr 2 +  Total conc. of NaF required is = 0.021 wt of NaFis = 0.0882 g

37. Conc of PbI 2 in solution is = 10 -3 M  AgI ↓ + Pb (NO3 ) PbI 2 + 2 Ag NO3  M = 4 × 10 -3 M

-

-

I. conc

2 × 10 -2

4 × 10 -3

F. conc

4 × 10 -3 10 -7 2 × 10 -2

I. conc. F. conc.

25 × 2 × 10 -3 = 12.5 M

+  2 NH 3 ↽ ⇀  Ag ( NH 3 )2

(

0.15h equi moolality

1 [ pK W + pK a - pK b ] 2

Conc. of F- in solution =

4

4 -

-3

0.15h

= ( 0.010) 2 = 0.02

30. CH 3 COOH + NaOH → CH 3 COONa + H 2 O

31. Ag

equi molality +

So conc. of F- required to ppt of SrF2

1 \ pH = PKa 2 pH of buffer solution is pKa a

+

0.15

 + H2O ↽ ⇀  NH 4 OH + H

0.015 0.005

29. CH 3 COONa + HCl → CH 3 COOH + NaCl

pH = pKa - log

Cl+

36. Sr 2 + + 2F- → SrF2 ↓

-7

28. Nature of CH3COONH4 solution is neutral

10 6

0.15 (1 - h )

+ 4

34. pH =

[ HA ]

1 4

7 = pKa - log

+

\DTf = K f  m NH4 Cl + m NH+ + m Cl- + m NH4 OH + m H+  4   \ h = 0.033

Molarity of Acid = 0.05 M HA + NaOH  NaA + H 2 O 4

NH +4

→

0.15 (1 - h )

(100) M = 50 (0.1)

log

)

0.05

[H + ] = Ch = 2.7 × 10 -2 M

pH = pKa

equi conc.

K a = 10 -3 x S

C2 h 2 C (1 - h )

5 1

S

\K sp = S.x

K K h = W = 10 -2 Kb Kh =

10 -3

x

[ BCl] = 0.1 M

 H+ ↽ ⇀  HCOOH

+

)

= 10

8

2

38. pH = pKin 39. a = 1.9 × 10 -9

(

)

K W = C2 a 2 = (55.5) 1.9 × 10 -9 = 1 × 10 -14 2

10.102 Ionic Equilibrium

40. If CH 3 COOH is 50% dissociate than pH = pKa 2 41. pH = pKa 2 − log

2

\b = 4 × 10 -7

2-

48. pH = 7 means it is a salt of strong acid and strong base

+ 2 Tl2S ↽ ⇀  2Tl + S x equi conc. 2S

49. pH = pK In - log

 S2 - + H 2 O ↽ ⇀  HS + OH x S S equi conc. -

\x= S

2 2

(

= 4S4 = 4 3 × 10 -6 43. pH =

HI n In -

50. Conc of H + = 2x Conc of OH - = x \( 2x ) ( x ) = 10 -12

( )

\ K sp = (S) 2S 2

\ a ( a + b ) = 1 × 10 -12 b ( a + b ) = 5 × 10 -13

0.005 x

( ) (S )

42. K sp Tl2 S = Tl+

47. Assume solubility of AgCNS = a , AgBr = b

)

4

+  51. N 3 H ↽ ⇀  N3 + H x+y e 0.1 - x x

= 3.24 × 10 -2

+

 HOCN ↽ ⇀  H + OCN x + y ye qui conc 0.08 - y

1 [ pKa1 − log C] 2

1.22 1000 × = 26.52 M 46 1 +  2 HCOOH ↽ ⇀   HCOOH 2   HCOOH 

44. Molarity of HCOOH =

\  HCOO -  = 10 -3 +  2HCOOH ↽ ⇀  HCOOH 2 + HCOO 26.52 - 2 x x x equi conc.

( x )( x + y) (0.1 - x ) y ( x + y) 8 × 10 -4 = 0.08 - y 52. Due to hydrogen bonding all ions exists 53. Titration represents strong base and strong acid 54. Forward reaction favours 3.6 × 10 -4 =

-3

\ decrease off ionization = 2 × 10 26.52 -3 \ Percentage ionization = 8 × 10 H+

+2

45. CaC 2 O 4 + KMnO 4 → Mn + CO 2 + H 2 O 300 × M × 2 = 6 × 0.001 × 5 \ K sp = S2 = 2.5 × 10 -9 +

-

1

-

initial conc.

1 - 2a

a

Final conc.

-17

10 a

\ 9 × 10 =

a

19

10 -17 (1 - 2a )2 a

a = 30 1 - 2a \ a = 0.491 \

Ka C

56. Solubility S = 1.34 × 10-5 mole/lit So 1.34 × 10-4 mol salt dissolved in 10 litres KW K a .K b

h = Kh 1- h [NH 4 OH] = Ch

 2CN - ↽ ⇀  Ag ( CN )2

a

55. Decrease of dissociation a =

57. K h =

Molarity of CaC 2 O 4 = 5 × 10 -5 M

+ 46. Ag

qui conc -

58. Ionic product > Ksp then precipitates is formed  NH +4 + NH 2 59. 2NH 3  \  NH 2-  = 10 -15 mol/lit 60. Degree of hydrolysis is independent of concentration of salt 61. Ionic product > Ksp then precipitate is formed

Ionic Equilibrium 10.103

70. pH = pKa − log

 62. BOH + HCl ↽ ⇀  BCl + H 2 O no. of moles of BClin solution = 6 no. eof m moles of NaOH added = 3 BCl + NaOH ⇌ BOH + NaCl 6 3 Initial mole 3 3 equi mole

0.1 0.1

71. Concentration of Ag + when AgCl starts ppt is 2 × 10 -9 So concentration I - when AgCl start ppt =

\ pOH = pK b then pH = 9.8

72. Solubility is high

63.  Ba +2  = 0.1 Ba 3 ( PO 4 )2 → 3Ba +2 + 2PO 4-2 \ K sp = ( 0.1)

73.

pKa1 + pKa 2 = 9.2 2 H 2 CO3 + NaOH  → NaHCO3 + H 2 O

( 2S)2

3

4 × 10 -16 = 2 × 10 -7 M 2 × 10 -9

4.4 0.4

-4

\ S = 0.5 × 10 \ S = 30.5 mol / lit

pH = pKa1 -log

1 64. pOH = [ pK w − pKa 2 − log C] 2

∴ pKa1 = 7.6 ∴ pKa2 = 10.8

− 4

y

4 -

initial mo ole equi. mole

0.4 4

 H 2 PO  65. pH = pKa 2 − log   HPO −4 

75. When strong acid react with salt weak acid is formed

66.  HCOOH +2   HCOO -  = 10 -6

76. pH =

\ HCOO -  = 10 -6 +  2HCOOH ↽ ⇀  HCOOH 2 + HCO5 C - 2x \a =

x

O

77. Ecell and Ecell are zero

x

2x C

2 × 10-3 C \ [ HCOOH ] = 50 M 4 × 10 -5 =

Conc. of Ag+ is AgBr solution is =

4 × 10 -13 Br -

DTf of NaCl = K f (0.2) -

C 2 O 4-2  = 0.1 - x -

 F  = 2 x 10 -7 0.1 - x \ -6 = 10 ( 2 x )2 68. Solubility = 2 × 10–5 mol/lit 69. Proton donor is acid

8 × 10 -12 CO3-2

79. At ‘R’ point salt undergoes hydrolysis

67. BaF2 + Na 2 C2 O 4 → BaC2 O 4 + 2 NaF 0.1 0.1 - x

Conc. Ag+ in Na2CO3 solution is

78. Ionic product > Ksp then ppt forms

d = 2.3 gm / lit -

pKa1 + pKa 2 2

2x

Initial conc. equi conc.

NaA → Na + + A 81.

A0.1(1 - h )

+ H2O →

HA + OH 0.1h

0.1h

\DTf of NaA = K b (0.1 + 0.1(1 - h ) + 0.1h + 0.1h ) 82. Ionic product > Ksp then ppt forms

10.104 Ionic Equilibrium

Comprehensive Type Questions

Passage III 1. As 2S3  2As3+ + 3S2 -

Passage I

-

1. HA + NaOH → NaA + H 2 O 15 10

5

5 -

pH = pKa - log

-

I. moles F. moles

10 5

-5

2. After mixing [ H+ ] =

15 - 5 1 = = 5 × 102 200 20

→ 3. HA 0.05 0.05 - x

H+ + 0.05 0.05 + x

AI. conc F. conc

x

(0.01)3 =

2As3+

-

-3

1 × 10 -24 576 3S2 -

3 × 10 + 2 x 3x

(

3 × 10 -3

I conc F conc

) (3x ) 2

3

1 × 10 -6 M 36

3. As 2S3  2 As3+ + 3S2 -

\1 = 10 - 5 = 2 × 10 -4 0.05

+

3 × 10 -3

2S

x

S2 + H 2 O  HS- + OH -

-5

\ degree of ionization a =

2

-

\S =

(0.05 + x ) x \ Ka = (0.05 - x ) \x = 10



\ K sp =

pH = 2 - log 5

0.01 I. conc 0.01 + 3S F. conc

  2 \ K sp =  × 10 -9    48 2. As 2S3

\ pKa = 5 \ Ka = 10

2S

eq conc K Kh = W = 1 Ka 2

9S2 x

\ K sp = ( 2S) (9S2 )3 2

\S = 1.67 × 10 -4 M

Passage II 1. H 2 A + NaOH → NaHA + H 2 O 20 10

10 -

10

-

Passage IV I mmoles F mmoles

pH = pKa 2. When 200 mL NaOH is added then solution contains only NaHA \ pH =

pKa1 + pKa 2 = 4.5 2

3. H 2 A + 2 NaOH → Na 2 A + H 2 O 40 20 20 [ Na 2 A] = = 0.0033 600 1 pOH = [pKw - pKa 2 - log 0.033] 2 20 -

I mmoles F mmoles

1. BOH + HCl → BCl + H 2 O 1 3 1 At eq point pOH = pK b - log / 4 4 4 476 = pK b - 0.4771 pK b = 5.23

2. BCl + NaOH → BOH + NaCl a 3a I meq 6 4 4 a F meq a \ =6 4 a = 24 24 = 0.48 [BOH] = 50 1 pOH = [pK b - log 0.48] 2 pH = 11.22

Ionic Equilibrium 10.105

Passage V

Passage VIII

1. Solubility of AgCNS = a Solubility of AgBr = b \ a (a + b) = 5 × 10 -13

1. Solubility S = Zn +2 +  Zn ( OH )4  S=

b ( a + b ) = 1 × 10 -12 , solubility of Ag.CNS = 4 × 10 -7 M 2. ∴ Solubility AgBr = 8.16 × 10–7 M

K sp OH - 

2

+ K C OH - 

-2

2

Solubility minimim means

dS =0 dOH -

\ pH = 10 2. Solubility S = 2 × 10–9 M

Passage VI 1. The conc. of I- required to start ppt of AgI 1 × 10 -17 = = 10 -16 M 0.0 The conc. of I- required to start ppt of Hg 2 I 2

2. Conc. of Ag + in solution when Hg 2 I 2 start ppt is

Passage VII 1 1. pOH = [ pKw − pKa 2 − log C] 2 2. Moles of SO3 dissolved in water is = 0.2 SO3 + H 2 O → H 2SO 4

1 neutralization , solution acts as buffer 4 pH = pKa + log 3 pH = 4.52

Passage x 1. Concentration of S-2 if MnS start ppt is

3. Moles of SO3 dissolved is H 2 O = 0.01 / 2 = 0.005 SO3 + H 2 O → H 2SO 4 \ conc of H + in H 2SO 4 solution is = 0.01 2 NO 2 + H 2 O → HNO3 + HNO 2 \ [ HNO3 ] = 0.0025

[ HNO ] = 0.0025 2

\  H  = 0.01 + 0.0025 + 0.0015 = 0.014 \ pH = 1.85

10 -16 = 10 -14 10 -2

H 2S  2H + + S-2

−1

∴ H  = 0.4 = 4 × 10 M pH = 0.4

+

CH 3 COOH + NaOH → CH 3 COONa + H 2 O

3 2. At neutralization pH = Pka − log 3 4 ∴ pH = 5.47

10 -17 = 10 -4 M 10 -13

+

1. Ka of CH 3 COOH = 10 -5

At

10 -27 = = 10 -13 M 0.1

=

Passage Ix

(H ) (10 ) = + 2

K1K 2 = 10

-21

-14

0.1

\ H + = 10 -4 M pH = 4 2. The conc of Mn 2+ when Mn S start ppt is =

10 -22 = 10 -8 M 10 -14

Passage xI 1. KW = Ka.Kb 2.  H +  = 10−9 ×10−3 = 10−6 M pH = 6

10.106 Ionic Equilibrium

2. K sp BaSO4 = S2

Passage xII 1. pH = pKa − log

0.5 0.25

= (7.47 × 10 -7 ) 2 = 5.625 × 10 -13

2. CH 3 COONa + HCl → CH 3 COOH + NaCl 0.25 -

0.25 -

0.5 0.75

\ pH =

1 [ pKa - log 0.75] 2

I moles F moles

0.25 -

pH = pKa - log 4. Al +3 =

K sp

0.25 0.5

+

I moles F moles

- 3

nRT = 24.16 mL P 1 l[ pKa - og C] 2

3. pH =

5. Ionic product < K sp ppt is not formed

4. NaCN 4 × 10

Passage xIII Br -

3 × 10 = 2.5 × 10 -14 M

HCl

+

−3

10

−3

10 -10 = 5 × 10 -9 M 0.02

3. Conc of Br - remains when AgCl start ppt is =

5 × 10 -16 = 10 -7 M 5 × 10 -9

Passage xIV 1. Π = ICST

4.05 × 10−5 = 2 ( c )( 0.0821) 330 ∴M BaSO 4 = 7.47 × 10−7 M So BaSO 4 is sparingly soluble salt P0 − P n 2i = p n1 3w 40 − 39.6 174 = 900 39.6 18 wt K 2 SO 4 = 29.29 gm % wt of K 2SO 4 = 71.34%

10

-3

1 × 10 -3

0.5 × 10

\ pH = pKa

−3

+

NaCl −

I mole

−

F mole

1 3

+ HCl

5. NaCN

HCN −

−

pH = pKa − log

1.5 × 10

→

−3

2. Conc of Ag + when AgCl start ppt is =

DH = -13.06 Kcal

2.  H +  = 10 -3 m mole of HCl = 10 -3 V=

(OH )

K sp

DH = 13.06 Kcal

-

H + OH  H 2 O

0.25 0.5

1. Conc of Ag + =

Kw 2 DH  1 1 = -   R  T1 T2  Kw1

1. 2.303log

H 2 O  H + + OH -

3. CH 3 COOH + NaOH → CH 3 COONa + H 2 O 0.5 0.25

Passage xV

→ NaCl + HCN -3

-

0.5 × 10 -3 I moles

-

1 × 10 -3

F moles

Passage xVI 1. Na 2 A + HCl  → NaHA + NaCl (1st equivalent point) When 10 mL HCl is added total Na2A converted to NaHA. So initial m moler of Na2A = 3 When 1 mol HCl is added it act as buffer with maximum buffer capacity \ Initial moles of Na HA = 2.4 Conc of NaHA after adding 1 mL of HCl is = 2.7 = 0.128 M 21 2. Initial m molar of NaHA = 2.4, Na2A = 3 3. NaHA + HCl  → NaCl + H 2 A (2nd equivalent point) So total volume of HCl required is = 10 + 10 = 28 mL

Ionic Equilibrium 10.107

Passage xVII

Passage xx

1. Initial count of silver isotope is = 3.75 × 10 Final count of silver in solution = 2.5 × 104

6

m moles of AgIO3 precipitated is = 0.4966 Wt of AgIO3 formed is = 142 mg

Indicator distinguish colour below and above the range Passage xxI pH of the indicator = pK in − log

→ AgIO3 + NaNO3 2. AgNO3 + NaIO3  0.5

3

-

-

-

2.5

-

-

[IO3- ] in solution =

I m moles F m moles

2.5 = 5 × 10 -3 M 500

+

Passage xxII + R

∴K sp =  Ag +   IO3−  = 3.3 × 10−8 M

R

H 3 NC HCOOH + NaOH → H 2 NC HCOOH + H 2O + Na + 1.8

1.8

I. mole

At equivalent point, amphiprotic ion is formed

–6

[Ag ] = in solution = 6.6 × 10 M

HIn In −

Then pH =

pKa1 + pKa 2 2

Passage xxIII 3. Per cent of unprecipitated silver is = 6.67%

10 × 2 5× V V = 78 mL

pH = pKa1 - log Passage xVIII 1. According Henery law P = K X2 XCO2 = 2.18 × 10–7 m = 1.21 × 10 –5 m CO2 + H2O → H2CO3 1 pH = =  pKa1 − log 1.21×10−5  = 5.64 2 2. XCO2 = 1.7 × 10–7 m = 9.45 × 10–6 1 pH = [pKa1 - log C] 2 3. SO 2 + H 2 O  → H 2SO3 0.12 103 × [H 2SO3 ] = =1.875 ×10−6 64 106 [H + ] = 1.875 ×10−6 pH = 5.7 Passage xIx 1. When 16.24 mL NaOH is added then amphiprotic ion is formed So pH =

pKa1 + pKa 2 2

When 8.12 mL of NaOH is added then buffer solution is formed pH = pKa1

Passage xxIV CaCl2 + Na 2 C 2 O 4  → 2 NaCl + CaC 2 O 4 + CaC 2 O 4 + KMnO 4 H → Mn +2 + CO 2

Ca +2 ][C2 O 4-2 ] K sp = [C [1.5 × 10 -6 × 4]2 = 3.6 × 10 -9 Passage xV When temperature increase pH of water decreases Passage xVI Molarity of H 3 PO 4 =

0.05 1000 × = 5.1×10−3 98 100

1 pH = [pKa1 − log 5.1×10−3 ] = 2.2 2 3. H 3 PO 4  3H + + PO 4-3 K = K1 K 2 .K 3 K1 K 2 K 3 =

(10 -7 )3 (PO 4-3 ) 10 -3

PO 4-3 = 2.2 × 10 -4 m 4. K sp = (Zn 2 + )3 (PO 4 -3 ) 2 9.1 × 10 -33 = (Zn 2 + )3 (2.2 × 10 -4 ) 2 Zn +2 = 5.65 × 10 -9 M

10.108 Ionic Equilibrium

5. NH 3 + HCl → NH +4 + Cl -

Passage xxVII

20 5

m moles of NaOH added = 0.0866 m moles of H+ in 100 mL solution = 0.866 m moles of Ca+2 is 25 mL solution = 0.433

I mmoles F m moles

5 15 pOH = 4.57 + 0.477 =5

\ pOH = pK b - log

Integer Type Questions 1. Na 2 CO3 + HCI → NaHCO3 + NaCI 2.5 3.75 - I m moles F m moles 1.25 2.5 NaHCO3 + HCI → NaCI + H 2 CO3

15

15 -

\ pH = 9 6. Ag + 10 -17 a

2.5 1.25 1.25 1.25 Buffer solution. So pH = pKa1

 Ag ( CN )2

-

+ 2CN -

\ 1019 =

1 - 2a

a

K = 10 -19

mole at equil

a2

(1 - 2a )2 × 10-17 a = 10 1 - 2a a = 0.476 moles

2. K sp =  Zn +2  S-2  -21 \ S-2  = 10 10 -19 0.01 = 2

 H +  10 -19  Ka1 : Ka 2 = 0.1 H + = 10 -1 M pH = 1 +2

3. SrCO3 (s)  Sr + CO Conc. at Equlibrium

S

-2 3

10 -8.6

K2 =

( x ) (10

\ K sp =

5

-8.6

)

5 (S) K 2 (S) 10 -8.6

= 9 × 10 -10

0.08

(OH ) (0.08) -

0.02 OH  = 1.25 × 10 -4  H +  = 8 × 10 -11 M -

At equivalent point eq. acid = eq. base HA + NaOH → NaA + H 2 O NaA + HCl → NaCl + HA 1.806 1.806

I m moles 3.612 F m moles 1.806 \pH = pKa = 5 9. K sp = S ( xS)

-

0.08 -

5 × 10 -4 =

a = 0.02 8.

4. CH 3 NH 2 + HCl → CH 3 NH 3+ Cl 0.1 0.02

Kb = C a2

x

CO3-2 + H + → H CO3x

7. N 2 H 4 + H 2 O  N 2 H 5+ OH -

I mole F mole

x

(

27 × 10 -12 = 10 -3 x10 -3

)

x

\x =3 10. K sp = 4S3 = 4 × 10 -6 S = 10 -2 M \ D Ts = 2 × 3 × 10 -2 = 6 × 10 -2 = 0.06 11. 5 × 10 -19 = 4S3 S = 0.5 × 10 -6 M \ OH - = 1 × 10 -6 M pH = 8

Ionic Equilibrium 10.109 -4 12. Ka of HX = 10 pH = pKa = 4

13. 0.2 = 0.79 +

19. At first -half equivalent point pH = pKa1 At Second half equivalent point pH = pKa 2 At equivalent point pH =

0.059 log K sp 1

20. In OH  In + + OH 80 pOH = pK1 - log 20 pH = 8 pH = 14 - 8 = 6 21. pKa of CH 3 COOH = pK b NH 4 OH

\ K sp = 10 -10

( )

K sp = 10 -10 = ( 0.1) Cl -

-9

\ Cl = 10 M

14. Kh =

KW 1 = Ka Kb 9

then pKa = NH +4 = 9

\pK b = 5

2

1  1   =  1- h 9

22. AX 3  A +3 + 3X -

(3a + 3b)

a

1 h = 1- h 3 1 h= 4

BX 3  B

+3

+ 3X

(3a + 3b)

b

(

1.9 × 10 -11 = a 3 × 10 -3

)

3

3a + 3b = 3 × 10 -3

15. H 2 PO 4- + PO 4-3 → HPO 4-2 0.1V 0.1V I mole 0.1V F mole -

a + b = 10 -3 a = 7 × 10 -4 M \b = 3 × 10 -4 M

 HPO 4-2  = 0.05 M pH = pKa 2 + pKa 2 = 9

(

)(

K SP of BX 3 = 3 × 10 -4 3 × 10 -3

Solubillity = 6.4 × 10 -5

24.

K sp = 4 × 10 -9

CH 3 COONa + HCl → CH 3 COOH + NaCl I moles

0.05 0.04 0.01 =5 pH = pK a − log 0.04

17. 2 Na CH 3 ( CH 2 )14 COO  + Ca +2 → 2 Na + 1

0.009 -

0.982

F moes

I moles F moles

Due to palmitate ion solubility is very less \wt of ppt is = 0.009 × 550 = 5 g 18. CH 3 COO Na  = 0.1 M 1 pOH = [ pKw - pKa - log C] 2

0.01 −

Ka = 5.29 × 10−3 C After dilution α = 1.058 × 10−2

25. α =

2

0.009

+

3

−3 − −2 23. H 3 A + A → H 2 A + HA ∴pH = pKa 2

( 250)( M )( 2) = (6.4)(0.001) 5

+ Ca CH 3 ( CH 2 )14 COO 

)

= 8 × 10 -12

16. C2 O 4-2 + MnO 4- + H → Mn +2 + H 2 O + CO 2

-

pKa1 + pKa 2 2

∴1.58 × 10−2 =

2.8 × 10−6 C1

∴Conc. of dilute solution is = 2.5 × 10−2 M ∴M1V1 = H 2 V2

(100 )( 0.1) = 2.5 ×10−2 × V V = 400 mL VH2 O = 300 mL

− 0.01

10.110 Ionic Equilibrium

26. pH =

pKa1 + pKa 2 2

34.

∴10

K 2 HPO 4 + KOH → K 3 PO 4

+ H2O

AgCN  Ag + + CN F conc

s -

+

CN + H  x 10 - y Fconc \K SP = S.x

x HCN s

10 - y.x S Ka.S2 \K SP = 10 - y \H + = 10 -3 , pH = 3 Ka =

29. HCOOH + NaOH → HCOONa + H 2 O

1 At equivalent point pH = pka − log 4 5 4 At equivalent point pH = pKa + log 4 5 Diffence is pH = 2 log 4 \=6

pOH of HCOOK is =

(

)( ) = ( Mg ) (10 )

10

+2

1 14 − 5 + ( log 2 ) + 1 − log 2  2 pOH = 5 pH = 9 38. ( a )( a + b ) = 3.4 × 10 -12

( b)(a + b) = 10-8 \ A +  = a = 3.4 × 10 -8 M 3 -15 39. K SP of Pb ( OH )2 = 4 S = 1.2 × 10

(

In buffer solution 1.2 × 10 -15 = 1.2 × 10 -3 OH -

1 [ pKa + 1] 2 1 After dilution pH = [ pKa + 3] 2 So diffence = 1

32. Initial pH =

33. NaCN, KCN under goes anonic hydrolysis only

)

3

OH - = 10 -6 pH = 8

40. HA  H + + A A-  10 -3.74 = 10 -4.22   [ HA]  A-  \  41. h =

[ HA] = 3

10 -14 × 0.1 = 10 -5 10 -5

42. pOH =

1 [14 - 5 - log 2 + log 2 + 1] = 5 2

pH = 9

-5 2

\Mg +2 = 10 -6 M

−19

0.1

\OH - = 10 -5 M 2

( H ) (10 ) =

37. pOH =

1 [14 − 6 + 2] = 5 2

0.1 =5 0.01

−20

36. Neutral of solution pH = 7

pH = 9

-16

= 10−19

35. OH - = 10 -4  \ Ba 2 ( OH )2  = 5 × 10 -5 M

30. pKa of HCOOH = 10−6

K sp = Mg +2 OH -

0.01

∴pH = 1

KH 2 PO 4 + KOH → K 2 HPO 4 + H 2 O

31. pOH = pK b - log

−21

+ 2

27. H 3 PO 4 + KOH → KH 2 PO 4 + H 2 O

28.

S2 = 10

Ag + +

43. I conc Fconc 108 =

0.02 x 0.02

(0.1)2 x

X = 2 × 10 -8 M

(

2NH 3  Ag NH 3 0.14 0.1

0.02

)

+ 2

K c = 108

Ionic Equilibrium 10.111

Previous years’ IIT Questions  pAq + + qB p 1. Ap Bq  ps

qs

Solubility product Ls = ( pS ) p × (qS ) q = S p+q p p qq 2. An acid is said to be strong if its conjugate base is weak. - H⊕ - H⊕       HClO 4  HClO 2   ClO 4  ClO3 + H⊕ + H⊗

Acid

C.B - H⊕

  HClO 2   ClO + H⊕

Acid  2

C.B - H⊕

   HClO   ClO + H⊕

Oxygen is more electronegegative than chlorine and hence negative charge is dispersed between 4 oxygen atoms in ClO4 . Greater the dispersal of charge, more is the stability and weaker is the conjugate base or stronger is an acid. Charge dispersal decreases as no of oxygenations decreases or acidic strength decreases. The order is HClO < HClO2 < HClO3 < HClO4  2Na ⊕ + SO 4-2 3. Na 2SO 4  0.004 Let x be the degree of dissociation Conc. left

2x 2x

x x

No. of species left in soln. = 0.04-x+x+2x = 0.004+2x The conc. of isotonic, C1 = C2 0.004 + 2x = 0.01 2x = 0.006 X = 0.003 = 3 × 10-3 M % of dissociation = =

Initial conc. 1.0 At neutralisation 0.02 point conc.left

3 × 10-3 × 100 4 × 10-3

4. Since, HX is a weak acid and NaX is its salt with a strong base, degree of hydrolysis of salt of weak acid and strong base h=

= 1× 10-8 = 10-4 % of hydrolysis = 100 × 10-4 = 10-2 % = 0.01%

0.0 0.08

CH3NH2 is a base and forms salt with HCl. HCl is the limiting regent. So, a buffer is formed with lose conc = 0.02 mol. Salt conc. = 0.08 mol [ Salt ] pOH = - pK b + log [ Base] 0.08 0.02 pOH = 3.30 + 0.6021 = 3.9021 \ pH = 14-pOH = 14-3.9021 = 10.098 10.098 = - log  H +   H +  = 8.0 × 10-11 M

6. CO 2 + H 2 O  H 2 CO3 H 2 CO3  H + + HCO3HCO3-  H + + CO3-2 7. During neutralization of mass acidic base (BOH) with HCl, BOH + HCl  → BCl + H 2 O At equivalent point, Meq of BOH = Meq of HCl = Meq of BCl M 1 × V1 = M 2 × V2 2M 2M = V2 × orV2 = 7.5 mL 5 15 One mol of BCl is formed. Total volume of solution after neutralization = 2.5+7.5 = 10 mL 1 = 0.1M Conc. Of BCl= [BCl] = 10 2.5 ×

BCl is a salt of weak base and a strong acid BCl + H 2 O  → BOH + HCl

0.003 × 100 = 75% 0.004

Kw 1× 10-14 = 1× 10-5 × 0.1 Ka × c

0.08 0.00

= - log 5 × 10-4 + log

C.B

x 0.004 - x

+

 CH 3 N H 3 Cl CH 3 NH 2 + HCl 

5.

B⊕ Initial conc h = deg ree of

 BOH + H ⊕ + H 2 O 

0.1 0.1 - 0.1h

hydrolysis + K w [ BOH ] ×  H  KH = = Kb  B+ 

(C = 0.1) 0.h 0.h

10.112 Ionic Equilibrium

10 -14 Ch 2 Ch × Ch = = 10 -12 C(1 - h) (1 - h) 0.1 × h 2 (1 - h)

10 -2 =

8.

(as C = 0.1)

12. The aqueous solution which turns red litmus blue must be basic in nature. Simply the salts of weak acids with strong bases are alkaline and turn red litmus blue. Examples  KOH Turn red KCN + H 2 O  + HCN litmus blue

 2KOH + H 2 CO3 K 2 CO3 + 2H 2 O 

Mx  M + + x -

Strong base weak acid

K sp = S2 S = 2 × 10 -4

 LiCN + H 2 O 

mol / lit Neutral

2S

K sp = 4S

4 2

+

HCN

 2NH 4 OH + C 2 O 4 + 2H 2 O 

H 2 C2 O4

weak base  NaOH NaCl + H 2 O 

\ S = 2 × 10 -5 mol / lit  3M + + x M 3 x  K sp = 27S

( NH )

to litmus

3

3S

LiOH

Strong base weak acid

Mx 2  M +2 + 2x S

Strong base weak acid

weak acid

+ HCl

Strong base Strong acid  2KOH + H 2SO 4 K 2SO 4 + 2H 2 O 

S

Strong base Strong acid

4

S = 1 × 10 -4

 Fe(OH)3 + 3HCl FeCl3 + 3H 2 O 

Acidic

mol / lit

turn blue

Solubility order is M × > M 3 × > M ×2

litmus red

9 (a) A buffer solution is a mixture of weak acid and its conjugate base (for acidic buffer) or it has a weak base and its conjugate acid (for base buffer). (c) NH 3 + NH 4 Cl Weak base Salt of weak base with a strong acid. (c) and (d) 10. Acidic buffer solution is an equimolar mixture of a weak acid and its salt with a strong base. The following pair combine to form acidic buffer as (d) CH 3 COOH + CH 3 COONa

Weak acid

Salt of weak acid with a strong base

→ CH 3 COOH + NaNO3 (c) CH 3 COONa + HNO3  Salt of weak acid Weak acid with a strong base K 10 -14 11. Hydrolysis constant K h = w = - c = 10 -10 K a 10 14

Degree of hydrolysis h = OH - = ch = 10 -6 M pOH = 6, pH = 8

Kh = 10 -4 M C

weak base

Strong acid

 NH 4 OH + HNO3 NH 4 NO3 + H 2 O 

13. 10-7 or x = 7 Conc. Of [AgCl] = 1× 10-10 M To it 0.1 mol of CuCl is added. K sp of CuCl = 1× 10-6 Cu Cl  Cu + + Cl Let the Conc of Cl - = c K sp = Cu +  × Cl -  = x × x x 2 = 1 × 10 -6 x = 1 × 10 -3 Now, [ AgCl] = 1.0 × 10 -10 =  Ag +  × Cl -  10 -10 10 -10  Ag +  = = = 10 -7 Cl -  10 -3 10 - x = 10 -7 or x = 7

CHAPTER

11 Redox Reactions

M

y advice to those who wish to learn the art of scientific prophecy is not to rely on abstract reason, but to decipher the secret language of Nature from nature’s documents: The facts of experience. Max Born

11.1 IntroductIon Nearly all chemical elements may be prepared in two or more different oxidation states. This chapter will be concerned with some general principles involved in oxidation and reduction. The products of a chemical reaction may, in general be found only by experiment. Valuable new reactions are not infrequently discovered by accident. Nevertheless, it is possible to classify the myriad reactions known in inorganic chemistry and in many cases to predict the products. There appear to be only a few main classifications of inorganic reactions. One pattern that was established in the eighteenth century was that many elements combine with oxygen to make oxides. For example, magnesium combines with oxygen when it burns in oxygen.

The nitrogen atom in ammonia loses its hydrogen, while at the same time it has gained oxygen. In fact there are many reactions that show this pattern. The simple definition of oxidation as addition of oxygen or hydrogen and reduction as addition of hydrogen or removal of oxygen. The two processes are complementary. No oxidation process can take place without a corresponding reduction. Consider the following reactions. oxidation

Reduction oxidation

2 Mg ( s ) + O2 ( g )  → 2 MgO( s )

and carbon combines with oxygen

Reduction oxidation

4 NH 3 ( g ) + 5O2 ( g )  → 4 NO( g ) + 6 H 2 O( g )

2HCl + S

H2S+ Cl2 Reduction

CuO( s ) + H 2 ( g )  → Cu ( s ) + H 2 O( g )

In the above equation, hydrogen has gained oxygen: it has been oxidized. Indeed, oxidation and reduction always take place together in a reaction. Many reactions are more complicated than those considered so far. For example, ammonia will burn in oxygen.

2H2O + I2

2HI + H2O2

C ( s ) + O2 ( g )  → CO2 ( g )

Reactions between elements and oxygen were called oxidation reactions. The opposite process, taking oxygen away from an element is known reduction. The simplest reduction reaction is the reduction of an oxide with hydrogen. For example when hydrogen is passed over hot copper (II) oxide. The oxide is reduced

2H2O

2H2 + O2

oxidation Al2O3 + 2Fe

Fe2O3 + 2AI Reduction

The substance that provides the oxygen or removes the hydrogen, e.g., oxygen, hydrogen peroxide and chlorine

Redox Reactions

and so becomes reduced, is the oxidizing agent; similarly, the substance that provides the hydrogen or removes the oxygen, e.g., hydrogen iodide and aluminium and so becomes oxidized is the reducing agent. Consider the reaction between hydrogen sulphide and an aqueous solution of iron (III) chloride. Sulphur is precipitated, so there is no doubt that the hydrogen sulphide is oxidized; but there is only one substance that can play the role of oxidizing agent—the iron (III) chloride, so this must be reduced. Accordingly, the reaction can be written oxidation

In the presence of hydrogen ions, the volumetric reagent potassium permanganate quantitatively oxidizes iron (II) ions to iron (III) ions and is itself reduced to manganese (II) ions. The two partial equations which summarize this change are 5 Fe 2 +  → 5 Fe3+ + 5e − oxidation MnO4− + 8 H + + 5e −  → Mn 2 + + 4 H 2 O reduction

S + 2FeCl3 + 2HCl Reduction

and the iron (III) chloride is reduced to iron II chloride. Since this reaction takes place in aqueous solution where it is known that iron (III) chloride is in the form of ions, it is possible to write a simplified ionic equation to summarize the overall result 3+

2+

H 2 S + 2 Fe  → S + 2 Fe + 2 H

+

The iron (III) ions are reduced to iron (II) ions by gain of electrons; like wise the hydrogen sulphide is oxidized to sulphur and hydrogen ions by the loss of electrons. This can be seen clearly by writing the partial equations: 2 Fe3+ + 2e −  → 2 Fe 2 + reduction H 2 S  → S + 2 H + + 2e − oxidation All reactions involving ions in which electron transfer takes place are classified as oxidation-reduction or redoxreactions. Thus the displacement of copper from a solution of copper(II) ions by zinc can be expressed by the ionic equation. Zn + Cu 2 +  → Zn 2 + + Cu

Or in terms of partial equations Zn  → Zn 2 + + 2e −

oxidation

MnO4− + 8 H + + 5 Fe 2 +  → 5 Fe3+ + Mn 2 + + 4 H 2 O Overall

This reaction involves the breaking of manganese— oxygen bonds and the hydrogen ions are required to combine with the oxygen to give water. The reaction between dichromate ions and iron (II) ions is similar 6 Fe 2 +  → 6 Fe3+ + 6e − 2− 7

Cr2 O

+

3+

+ 14 H + 6e  → 2Cr + 7 H 2 O reduction

2− 7

Cr2 O + 14 H + + 6e −  → 6 Fe3+ + 2Cr 3+ + 7 H 2 O Overall

11.2.1 electron transfer Involving essentially covalent Molecules The reaction between essentially covalent substances to produce a predominantly covalent product can never involve the transfer of electrons; but because the electron pair linking two atoms, together is drawn closer to the more electronegative atom a polarized molecule results, with the partial shift of electrons. Thus consider the reaction between hydrogen and chlorine; the two electrons bonding the hydrogen and chlorine atoms together are drawn closer to the chlorine atom and the chlorine can be considered to be reduced by partial gain of negative charge and the hydrogen oxidized by partial loss of an equal negative charge: δ+

δ−

H : H + Cl×× Cl → 2 H ×× Cl

−

Cu + 2e  → Cu reduction Copper (II) ions (oxidising agent) are reduced to copper by zinc (reducing agent) which itself oxidized to zinc ions. For reactions in which ionic compounds participate, the scope of oxidation – reduction is thus extended and oxidation is defined as loss of electrons and reduction as gain of electrons. Thus we have Oxidation is (i) The gain of oxygen (ii) The loss of hydrogen (iii) The loss of electrons and Reduction is (i) The loss of oxygen (ii) The gain of hydrogen (iii) The gain of electrons

oxidation

−

The arguments developed for ionic redox reactions should not however, be carried over too literally when discussing reactions involving covalent compounds. Thus the formation of carbon monoxide from carbon and oxygen is certainly an oxidation of carbon; yet experiment shows that a very small partial negative charge resides on the carbon atom, despite the fact that the oxygen atom is more electronegative than carbon. Carbon monoxide is considered to have a structure involving the three resonance forms. × ×

+

-

C O: (i)

× ×

: :

H2S + 2FeCl3

2+

11.2 redox reactIons InvolvIng electron transfer and Bond BreakIng

: :

11.2

C O (ii)

× ×

-

C

+

O: (iii)

Redox Reactions

The anamoly is explained as being due to an appreciable contribution from structure (iii) in the resonance hybrid.

11.2.3 competitive electron transfer reactions If a metallic zinc strip is kept in an aqueous solution of copper sulphate, the blue colour of the copper sulphate disappears and the metallic zinc strip will be coated with reddish metallic copper. The Zn2+ ions are formed in the solution while the blue colour of the solution disappears as the Cu2+ ions are disappeared. The formation of zinc ions can be detected by the formation of white precipitate when hydrogen sulphide gas is passed into the solution in alkaline solution with ammonia. The reaction between metallic zinc and aqueous solution of copper nitrate can be represented as

11.3

Experiments shows that at equilibrium, both the reactants and products i.e., Ni2+(aq) and Co2+(aq) are present at moderate concentrations, i.e., neither reactants nor products are greatly favoured. From the discussion made so far it can be understood that there is competition for release of electrons by different elements similar to the release of protons by acids. For example, in the above example, zinc releases electrons easily than copper and hence electrons are transferred from zinc atoms to copper (II) ions and thus reduced. Similarly, copper has more tendency to give electron than silver and therefore the electron releasing tendency of these metals is in the order Zn>Cu> Ag. Thus we can develop a table in which metals and their ions are listed on the basis of their tendency to release electrons. This list which shows the order of the tendency to release electrons is known as electrochemical series or metal activity series.

Zn( s ) + Cu 2 + (aq )  → Zn 2 + (aq ) + Cu ( s )

In the above reaction, zinc has lost electrons to form Zn and therefore, zinc is oxidized. on the other hand Cu2+ ions are reduced by gaining electrons. This redox reaction can be written as. oxidation 2+

2+

Zn (s) + Cu (aq)

Zn(aq) + Cu (s) Reduction

Conversely, when a copper strip is placed in an aqueous solution of zinc sulphate, no reaction will take place. Thus it can be concluded that the state of equilibrium of the above reaction greatly favours the products over the reactants. Suppose if the silver nitrate solution is stirred with a copper rod, the solution develops blue colour due to the formation of Cu2+ ion due to the following reaction. oxidation 2+

+

Cu (s) + 2Ag (aq)

Cu (aq) + 2Ag (s) Reduction

In this reaction, copper solid is oxidized to Cu2+ ions in aqueous solution while the aqueous silver ions are reduced to silver metal. Equilibrium greatly favours the products Cu2+(aq) and Ag(s). In another example if metallic cobalt is placed in nickel sulphate solution, the following reaction occurs. Oxidation 2+

2+

Co(s) + Ni(aq)

Co(aq) + Ni(s) Reduction

11.3 oxIdatIon nuMBers We know that a substance that loses one or more electrons has been oxidized; on the other hand, if it gains one or more electrons it has been reduced. For example, when sodium and chlorine react, the product is the ionic substance sodium chloride. This contains Na+ and Cl– ions. 2 Na + Cl2  → 2 NaCl

When it is unreacted, the sodium is neither oxidized nor reduced; we shall say that its oxidation number is ‘0’ when it is converted into Na+, we shall give an oxidation number of +1. Similarly, chloride atoms start with an oxidation number of 0 when they are present as chlorine molecules Cl2. After they react and turn into chloride ions Cl–, we shall give them an oxidation number of −1. From this discussion, it is easy to assume that the oxidation numbers of atoms when they present as unreacted elements and when they change into ions. The rule is 1. An unreacted element has an oxidation number of 0 and an ion has an oxidation number equal to its charge. Examples illustrating the above rule are given in Table 11.1. Some elements may form more than one type of ion. Generally, these are transition elements such as iron, copper and manganese. However, hydrogen can also give two different ions. The most common one is the H+, which is found in acidic solutions. The hydride ion H– is found in solid hydrides like Na+H– and K+H–. Similarly, oxygen almost always has a charge of −2 when it is an ionic compound. The exceptions are oxygen in peroxides such as barium peroxide BaO2 and super oxides such as KO2. In peroxides O22 − ion and in super oxides O2− ions are present. The average charge on an oxygen atom in peroxide ion is −1 and in super oxide ion is −0.5. So we give the oxidation number a value of −1 and −0.5 as well.

11.4

Redox Reactions

table 11.1 Oxidation numbers of elements in simple ions Unreacted element

Oxidation number

Hydrogen

H2

0

Oxygen

O2

0

Nitrogen Chlorine Bromine Sodium Magnesium Iron

N2 Cl2 Br2 Na Mg Fe

0 0 0 0 0 0

Copper

Cu

0

Ion

Oxidation number

H+ H– O2– O22– O2–

+1 −1 −2 −1 –1/2

N3– Cl– Br– Na+ Mg2+ Fe2+ Fe3+ Cu+ Cu2+

−3 −1 −1 +1 +2 +2 +3 +1 +2

11.3.1 oxidation numbers of elements in covalent compounds Now we will see how to find the oxidation number of an element when it is in a covalent compound. The way to do this is first to pretend that the substance is ionic and then finding what ions would be present. Here, we assume that there is a complete transfer of electrons from a less electronegative atom to a more electronegative atom. For example, in hydrogen chloride, chlorine is more electronegative than hydrogen. Therefore, the shared pair is counted towards chlorine atom as shown below. ⋅⋅

⋅⋅

⋅⋅

⋅⋅

H × + ⋅ Cl : → H | ×. Cl : As a result of this, chlorine gets one extra electron and acquires a unit negative charge. Hence oxidation number of chlorine is −1. On the other hand, hydrogen atom without electron has a unit positive charge. Hence oxidation number of hydrogen in hydrogen chloride is +1. It may be noted that electrons shared between two similar atoms are divided equally between the sharing atoms. For example, in chlorine molecule (Cl2), the electron pair is equally shared between the two chlorine atoms. Therefore, one electron is counted with each chlorine atom as shown below. : :

: :

: Cl : Cl : Now, there is no net charge on each atom of chlorine. In other words, oxidation number of chlorine in Cl2

molecule is zero. Thus the oxidation number (O.N) of the element is defined as the residual charge which its atom has or appears to have when all other atoms from the molecule are assumed to be removed as ions by counting the shared electrons with more electronegative atom. It may be noted that the assumption of electron transfer is made for keeping purpose only. Thus atoms can have positive, zero or negative values of oxidation numbers depending upon their state of combination. Now, since it is not always possible to remember or make out easily in a compound or ion which element is more electronegative than the other, therefore a set of rules have been formulated to determine the oxidation number. If two or more than two atoms of an element are present in the molecule/ion the oxidation number of the atom of the element will then be average of all the atoms of that element.

11.3.2 rules for assigning oxidation number to an atom The following general rules are used for the calculation of oxidation number of an atom in a molecule. 1. The oxidation number of an element in the free state or elementary state is always zero. For example, oxidation numbers of helium in He, hydrogen in H2, oxygen in O2 or O3, iron in Fe, phosphorous in P4, sulphur in S8 are zero. 2. The oxidation number of monoatomic ion is the same as the charge on the ion. For example, oxidation number of K+ is +1, of Ca2+ is +2, of Al3+ is +3. Similarly, the oxidation numbers of Cl–, O2– and N3– are −1, −2 and −3 respectively. 3. In binary compounds of metal and non-metal, the oxidation of metal is always positive while that of the non-metal is negative. For example, in KCl the oxidation number of sodium is +1 and that of chlorine is −1. 4. In compounds formed by the combination of nonmetallic atoms, the atom with higher electronegativity is given negative oxidation number. For example, in HCl, the oxidation number of chlorine is −1 because of its higher electronegativity. 5. Hydrogen is assigned oxidation number +1 in all its compounds except in metal hydrides. In metal hydrides like NaH, MgH2, CaH2 etc. the oxidation number of hydrogen is −1. 6. The oxidation number of fluorine is always -1 in all its compounds. For other halogens, the oxidation number is generally –1 but there are exceptions when these are bonded to a more electronegative halogen atom or oxygen for example, in HI the oxidation number of I is −1 but in IF5 it is +5 and in IF7 it is +7.

Redox Reactions

7. The oxidation number of oxygen is assigned as -2 in most of its compounds however in peroxides (Which contain O−O linkage) like Na2O2, BaO2, H2O2 etc. oxidation number is −1 and in OF2 its oxidation number is +2. In O2F2 which contain O−O (F−O−O−F) bond the oxidation number of oxygen is assigned as +1 8. For neutral molecule, the sum of the oxidation numbers of all atoms is equal to Zero. But in case of polyatomic ion, the sum of oxidation number of all its atoms is equal to the charge on the ion. By applying the above rules, we can find the oxidation number of the desired element in a molecule or in an ion. Generally, the metallic element, have positive oxidation number and non-metallic elements have positive or negative oxidation number. Transition elements usually exhibit several positive oxidation numbers. Representative elements exhibit highest oxidation number equal to its group number, e.g., by the elements of s-block element, But the other representative elements exhibit highest oxidation number equal to group number or group number minus ten. Let us apply the above rules to calculate the oxidation number of some elements. example 1 The oxidation numbers of hydrogen and oxygen in water can be calculated as follows. Hydrogen tends to make H+ ions and oxygen O2– ions. Therefore, if water were ionic, it would contain H+ and O2– ions and we say that the oxidation numbers of hydrogen and oxygen in water are +1 and −2 respectively. Notice that the sum of all the oxidation numbers together is equal to Zero. 2 × (+1) two hydrogen

+ (−2) = 0 one oxygen

example 2 Oxidation numbers of sulphur and oxygen in SO2. In such cases, we have to start with an element whose oxidation number is known certainly. Here it is assumed that the molecule is ionic and oxygen atom would present as oxide ions O2–. Given that sulphur dioxide contains two oxygen atoms, for each sulphur atom, there would be two in the hypothetical ionic compound. In total the two oxide ions carry a charge of −4. Because sulphur dioxide is electrically neutral, the sulphur would have to be present as +4 ions. So the oxidation number of sulphur is +4. example 3 Oxidation number of phosphorous in PO43− ions.

11.5

By using the similar principle as in example 2, except that we have to leave the ion with its charge of −3 we can calculate the oxidation number of phosphorous. Let us assume that oxygen has its normal oxidation number (−2). Now if we balance the charges. Oxidation number (ON) of phosphorous + (4 × Oxidation number of oxygen) = −3 ON(P) +4 (−2) = −3 ON(P) − 8 = −3 ON(P) = +5 example 4 Oxidation number of sulphur in sulphate ion SO42 − ON (S) + 4 × ON (O) = −2 ON(S) + 4 × (−2) = −2 ON (S) −8 = −2 ∴ ON(S) = + 6 example 5 Oxidation number of sulphur in the tetrathionate ion S 4 O62 − We have 4 ON (S) + 6 ON (O) = −2 4 ON (S) + 6(−2) = −2 4 ON (S) = +10 ON (S) = 2.5 To solve the problems, follow the following steps. (i) Write down the formula of a given molecule/ion leaving some space between the atoms. (ii) Write oxidation number on the top of each atom. In case of the atom whose oxidation number has to be calculated, write x. (iii) Beneath the formula, write down the total oxidation numbers of each element. For this purpose, multiply the oxidation numbers of each atom with the number of atoms of that kind in the molecule/ion. Write the product in a bracket. (iv) Equate the sum of the oxidation numbers to zero for neutral molecule and equal to charge on the ion. (v) Solve for the value of x. Problem for Practice 1. Calculate the oxidation number of sulphur in the following molecules/ions. (i) H 2 S (ii) H 2 SO3 (iii) Na2 S 2 O3 (iv) S 2 O72 − (v) H 2 SO4 (vi) S 2 O42 −

11.6

Redox Reactions

2. Calculate the oxidation number of the underlined atoms in each of the following: (i) ClO3– (ii) BrF3 (iii) CH4 (iv) C6H12O6 (v) Na2B4O7 3. Calculate the oxidation number of carbon in the following compounds: (i) C2H2 (ii) CO2 (iii) C2H6 (iv) CH3OH (v) HCOOH (vi) CH2O 4. Calculate the oxidation number of all the atoms in the following compounds/ions: (i) CO2 (ii) SiO2 (iv) ClO −4

(iii) PbSO4

5. Calculate the oxidation number of sulphur in S2 O82 − ion. 6. Calculate the oxidation numbers of the underlined atoms in each of the following: (i) Pb3O4 (ii) Cr2O72– (iii) K2MnO4 (iv) NH4+ (v) NaBH4

11.3.3 average oxidation numbers In example 5, we have seen that the oxidation number of sulphur has a fraction value. The application of general rules give fractional oxidation number to some atoms. Some other examples are C3O2, Br3O8 and Fe3O4 in which the oxidation numbers of carbon, bromine and iron are 4/3, 16/3 and 8/3 respectively. These results are surprising as we know that atoms/ions cannot have fractions of a positive or negative charge. However, oxidation numbers are not properties of atoms in the same way as their charge or mass. We cannot measure oxidation numbers. They are products of our imagination, which as you will see happen to be useful. Also, in the calculation we have just done, we found the average oxidation number of four sulphur atoms. If we look at the arrangement of the atoms in the ion.

[

o

o

o s o

s

s

s o

2-

[

Similarly, the oxidation numbers for carbon in C3O2and bromine in Br3O8can be assigned as follows: -2

+2

0

+2

-2

O=C=C=C=O O o +6 +6 O Br − Br − Br O

O O O

In the case of Fe3O4, the fractional oxidation number can be explained by the fact that Fe3O4 contains Fe atoms of both +2 and +3 oxidation number. It is stoichiometric mixture of ferrous (FeO) and ferric (Fe 2O3) oxides combined as FeO. Fe 2O3. Therefore, the oxidation number found in such cases is average oxidation number. The knowledge of bonding and individual molecular structure is also an important aspect and has to be considered for calculating the oxidation number of the elements. Some interesting examples are as follows. (i) Oxidation numbers of two N atoms in NH4NO3are different NH4NO3 is made of NH +4 and NO3− . In NH +4 ion oxidation number of N is −3 while in NO3− ion oxidation number of N is +5. (ii) In chromium peroxide CrO5 four out of the five O atoms are involved in peroxide linkage (−O−O−). The molecular structure of the compound is O O

O Cr

O

O

Hence oxidation number of Cr can be calculated as x +4 (−1) +1(−2) = 0 or x = +6 (iii) Bleaching powder CaOCl2 is made of Ca2+, ClO– and Cl–. Hence oxidation number of two Cl atoms in CaOCl2 are +1 (in ClO–) and −1 (in Cl– ) respectively. However, average oxidation number of Cl in CaOCl2 is 0. (iv) In HNO4, the oxidation number of N comes out to be +7. This is not possible for N. In fact, HNO4 is peroxy nitric acid which contains one peroxide group (−O−O−)2–. The formula may be written as HNO2(O2). In this case, the oxidation number comes out to be (+1 + x−4−2 = 0 or x = +5).

o

solved Problem Why do the following reactions proceed differently?

Two sulphur atoms in the middle are joined only to other sulphur atoms. This is the situation similar to pure sulphur where the oxidation number of a sulphur atom would be zero. Only two sulphur atoms have oxygen atoms joined to them. If we imagine that these atoms share the charge of +10 we would have two sulphur atoms of oxidation number 0 and two with oxidation number +5.

(a ) Pb3 O4 + 8HCl  → 3PbCl2 + Cl2 + 4 H 2 O (b) Pb3 O4 + 4 HNO3  → 2 Pb( NO3 ) 2 + PbO2 + 2 H 2 O Solution: Pb3O4 is a stoichiometeric mixture of 2 moles of PbO and 1 mole of PbO2. In PbO2, lead is present in +4 oxidation state which can act as an oxidizing agent

Redox Reactions

and oxidizes the Cl − ions in HCl to Cl2 . PbO is basic oxide which can react with acid HCl. Hence the reaction Pb3 O4 + 8 HCl  → 3PbCl2 + Cl2 + 4 H 2 O

can be written in two reactions as follows: 2 PbO + 4 HCl  → 2 PbCl2 + 2 H 2 O

acid − base reaction

0 +4 −1 +2 PbO2 + 4 HCl  → PbCl2 + Cl2 + 2 H 2 O redox reaction

In the second reaction between Pb3 O4 and nitric acid, since HNO3 itself is an oxidizing agent therefore it is unlikely that the reaction may occur between PbO2 and HNO3 . However, the acid-base reaction between PbO and HNO3 occurs as 2 PbO + 4 HNO3  → 2 Pb( NO3 ) 2 + 2 H 2 O

It is the passive nature of PbO2 against HNO3 that makes the reaction different from the one that follows with HCl.

11.3.4 oxidation state Oxidation state is only another way of telling the oxidation number. For example, in CO2, the oxidation state of Carbon is +4 that is also its oxidation number. Similarly, the oxidation state as well as oxidation number of oxygen is −2. Thus the oxidation number denotes the oxidation state of an element in a compound.

11.3.5 distinction Between oxidation number and valency Valency 1. Valency is the combining capacity of the element. It is the number of hydrogen atoms, or chlorine atoms or double the number of oxygen atoms that can combine with one atom of an element. 2. It is the number only and does not carry any sign (plus or minus). For example in CO2 the valency of carbon is 4 and that of oxygen is 2. 3. The valency of an element is always a whole number

Oxidation Number 1. Oxidation number is the residual charge which an atoms has or appears to have when the other atoms from the molecule are removed as ions by counting the shared electrons with more electronegative atom. 2. It refers to the charge, it can be positive, negative or zero. For example, in CO2 the oxidation number of carbon is +4 and that of oxygen is −2 3. Oxidation number may have fractional value. For example, oxidation of sulphur in Na2S4O6 is +2.5.

4. Many elements like C, N and O exhibit fixed valency

11.7

4. Even elements such as C, N, O which exhibit fixed valency can have variable oxidation numbers. For example, valency of carbon is 4 but its oxidation number can vary from −4 to +4 as show below. −4 −2 0 CH4, CH3Cl, CH2Cl2, +4 +2 CHCl3, CCl4

11.3.6 redox reactions in terms of oxidation number Based on the concept of oxidation number, the redox reactions can be defined as Oxidation is a chemical change in which there is an increase in oxidation number. Reduction is a chemical change in which there is a decrease in oxidation number. Similarly, oxidizing and reducing agents can be defined as follows: Oxidizing agent is a substance which increases the oxidation number of other substance in a chemical reaction oxidation number of oxidizing agent decreases. Reducing agent is a substance which decrease the oxidation number of other substances in a chemical reaction The oxidation number of a reducing agent increases. An increase in oxidation number corresponds to oxidation of the element in the given compound that acts as a reductant or reducing agent. A decrease in oxidation number corresponds to reduction of the element in a compound and it behaves as an oxidant or oxidizing agent.

The various terms studied in this chapter can be summarized as Oxidation: Loss of electrons or increase in oxidation number. Reduction: Gain of electrons or decrease in oxidation number. Oxidising Agent: Electron acceptor: Substances whose oxidation number decreases. Reducing Agent: Electron donor: Substances whose oxidation number Increases.

11.3.7 oxidation number and naming of compounds In earlier days, the names of various substances were derived either from some of their common characteristic property. Or

11.8

Redox Reactions

from their source. There was no relationship between the name and the chemical composition of the substances. For example, CuSO4.5H2O was named as blue vitriol and FeSO4. 7H2O was named as green vitriol. Gradually, systematic names which were based on the chemical composition of the compounds replaced these trivial names. For example, the binary compounds were named by writing the names of electropositive element first and the electronegative element later. The suffix ‘ide’ is attached to the name of electronegative elements as described below. NaCl: Sodium chloride Ca3N2: Calcium nitride MgS: Magnesium sulphide If a metal can exhibit variable valency, the compounds in lower oxidation state are called as ‘ous’ compounds while in higher oxidation state are called as ‘ic’ compounds. e.g., Cu2O: Cuprous oxide CuO: Cupric oxide

FeSO4: Ferrous Sulphate Fe2(SO4)3: Ferric Sulphate

German chemist, Alfred Stock suggested that the oxidation number or oxidation state of a metal in a compound can be written as Roman numeral in parenthesis. After the symbol of the metal in the molecular formula. These are popularly known as Stock notation. For example, copper (II) sulphate contains Cu2+ ions and if it were ionic there would be Fe3+ ions in iron (III) chloride. Transition metals in particular can exist in a variety of oxidation states. The oxidation state in the name gives us a guide to the nature of the metal ion in the compound. For example, copper (I) oxide will have Cu+ ions present and the formula of the oxide is Cu2O on the other hand copper (II) oxide contains Cu2+ ions and its formula is CuO. Oxidation states are sometimes written for ions that do not contain transition metals; for example, the bromate (V) ion, BrO3–. If the name of a compound ends with ‘ate’ it always means that the substance contains oxygen; e.g., bromate (V), dichromate (VI) and manganate (VII) ions. By using this clue, we can work out the name of an ion if we are given its formula. The following example illustrates this. Example: The name of ClO– is chlorate (I) ion since it contains chlorine and oxygen it will be chlorate and the oxidation state of chlorine is +1. Previously, it was called hypochlorite ion.

11.4 tyPes of redox reactIons 1. combination reactions In combination reactions, elements or compounds combine to form new substances. These reactions may also be called as addition reactions. For instance, sulphur dioxide may be prepared by the direct union of sulphur and oxygen. S +O2  → SO2

table 11.2 Names and formulae of ions and acids Metal ions

Formula

Non-metal ion

Formula

Aluminium Bismuth (III) Calcium Chromium(III) Chromium (VI) Copper(I) Copper (II) Iron (II) Iron (III) Lead (II) Lead(IV) Magnesium Manganese (II) Potassium Sodium Zinc

Al3+ Bi3+ Ca2+ Cr3+ Cr6+ Cu+ Cu2+ Fe2+ Fe3+ Pb2+ Pb4+ Mg2+ Mn2+ K+ Na+ Zn2+

Bromide Chloride Fluoride Hydride Hydrogen Iodide Oxide Peroxide Sulphide

Br– Cl– F– H– H+ I– O2– O2– S2–

Oxoanions Bromate (I) Bromate (V) Carbonate Chlorate(I) Chlorate(V) Chlorate (VII) Chromate(VI) Dichromate (VI) Iodate(V) Phosphate(V) Nitrate Nitrite Sulphate

Formula BrO– BrO3– CO32– ClO– ClO3– ClO4– CrO42– Cr2O72– IO 3– PO43– NO3– NO2– 2– SO4

Acids Bromic (I) Bromic(v) Carbonic Chloric(I) Chloric(V) Chloric(VII) Chromic (VI)

Formula HBrO HBrO3 H2CO3 HClO HClO3 HClO4 H2CrO4

Iodic Phosphoric (V) Nitric Nitrous Sulphuric

HIO3 H3PO4 HNO3 HNO2 H2SO4

For a combination reaction to be a redox reaction, either one or both the reactants should be in the elemental form. For example, all combustion reactions, which make use of elemental oxygen as well as other reactions involving elements other than oxygen are redox reactions. Some examples of combination reactions which are redox reactions are 0

0

→ C +4 O2−2 ( g ) C ( s ) + O 2 ( g )  0

−3

0

3 M g ( s ) + N 2 ( g )  → Mg 3 N 2 ( s ) −4 +1

0

+4 −2

+1

−2

C H 4 ( g ) + 2 O2 ( g )  → C O 2 ( g ) + 2 H 2 O( l )

2. decomposition reactions The combination process may be considered together with the reverse process namely, decomposition. In the

Redox Reactions

decomposition reaction, a compound breakdown into two or more components. If one of the product in the decomposition reaction is in the elemental state the reaction will be redox reaction. Examples of this type of reactions are: +2 −2

0

0

2 Hg O ( s )  → 2 Hg (l ) + O2 ( g ) +1

−1

0

+1 −1

0

2 K Cl O 3  → 2 K Cl ( s ) + 3 O 2 ( s )

It should be noted that all decomposition reactions are not redox reactions. For example, decomposition of calcium carbonate is not a redox reaction. +2

+2

+4 −2

−2

+4

−2

+1 − 2

0

+1

−2 +1

0

 → 2 N a O H (aq ) + H 2 ( g ) +2

−2 +1

0

Ca ( s ) + 2 H 2 O(l )  → Ca (O H ) 2 (aq ) + H 2 ( g )

Less active metals such as magnesium and iron may not displace hydrogen from cold water but can react with steam to produce H2 gas.

0

2 Na H ( s )  → 2 Na ( s )+ H 2 ( s ) +1 +5 −2

+1

0

2 Na ( s ) + H 2 O (l )

11.9

−2

Ca C O 3 ( s )  → Ca O( s ) + C O 2 ( g )

3. displacement reactions Another class of reaction involves substituting or a change of partners. In displacement reaction, an ion (or an atom) in a compound is replaced by an ion (or an atom) of another element. It may be represented as

0 +1 −2 +2 0  −2 +1  → Mg  O H  ( s ) + H 2 ( g ) Mg ( s ) + H 2 O(l )   2

+1 −2

0

+3

−2

0

→ Fe 2 O 3 ( s ) + 3 H 2 ( g ) 2 Fe( s ) + 3 H 2 O(l )  Several metals can displace dihydrogen from acids. Even those metals which cannot diplace dihydrogen from cold water can displace dihydrogen from acids. For example, cadmium and tin are the examples of such metals. +1 −1

0

+2 −1

0

Zn( s ) + 2 HCl (aq )  → ZnCl2 (aq ) + H 2 ( g ) +1 −1

0

+2

−1

0

X + YZ  → XZ + Y

Mg ( s ) + 2 H Cl (l )  → Mg Cl 2 ( s ) + H 2 ( g )

Displacement reactions are of two types: (i) metal displacement and (ii) non-metal displacement reactions (i) Metal displacement: A metal in a compound can be displaced by another metal in the combined state. Generally, a metal which is a better reducing agent can substitute the metal in the combined state which is a weaker reducing agent. For example, zinc is a better reducing agent than copper can substitute copper from copper sulphate. This type of reactions find many applications in the metallurgy for the extraction of metals.

The rate of liberation of hydrogen from acids by the metals depends on their reactivity. More reactive metals liberate hydrogen more fastly. Very less active metals such as silver gold which occur in native state in nature do not react with hydrochloric acid. From this discussion, we can understand that the metals can be arranged according to their tendency to lose electrons which represent their order of reduction power (Zn> Cu>Ag). Similar to activity series of metals, halogens can also be arranged according to their oxidation power. The oxidation power of halogens decrease from fluorine to iodine while moving down the group. Thus fluorine is more reactive and can displace chloride, bromide and iodide ions in solution. Fluorine is such a powerful oxidising agent that can displace oxygen from water.

+2 +6 −2

o

+2 +6 −2

o

Cu S O 4 (aq ) + Z n( s )  → C u ( s ) + Zn S O4 (aq ) +5 −2

O

+2 −2

O

V2 O5 ( s ) + 5 Ca ( s ) ∆ → 2 V ( s ) + 5 Ca O( s ) +4

−1

O

O

+2

−1

Ti Cl4 (l ) + 2 Mg ( s ) ∆ → Ti ( s ) + 2 Mg Cl2 ( s ) +3

−2

O

+3

−2

O

Cr2 O3 ( s ) + 2 Al ( s )  → Al2 O3 ( s ) + 2 Cr ( s ) In all the above reactions, the metal used as a reducing agent is stronger than the metal reduced and shows more capability to lose electrons as compared to the one that is reduced. (ii) Non-metal displacement: The displacement of a nonmetal in a redox reactions include hydrogen displacement and very rarely oxygen may also be displaced. Highly reactive alkali metals and some alkaline earth metals (Ca, Sr and Ba) displace the hydrogen from cold water by reduction.

+1

−2

0

+1 −1

0

2 H 2 O (l ) + 2 F 2 ( g )  → 4 HF (aq ) + O 2 ( g ) Because of this reason, displacement reactions involving fluorine are not generally carried out in aqueous solution. On the other hand, chlorine can displace bromide and iodide ions, bromine can displace iodide ions in an aqueous solutions as shown below: 0

+1 −1

0

+1 −1

0

−1

+1 −1

0

Cl2 ( g ) + 2 KBr (aq)  → 2 KCl (aq ) + Br2 (l ) +1 −1

0

Cl2 ( g ) + 2 KI (aq)  → 2 KCl (aq) + I 2 (l ) −1

0

Br2 (l ) + 2 I − (aq )  → 2 Br − (aq ) + I 2 ( s )

11.10

Redox Reactions

Generally, halogens are prepared by the oxidation of halides 2 X −  → X 2 + 2e − ( X is halogen)

Cl–, Br– and I– ions can be oxidized chemically but fluorine being strongest oxidizing agent no chemical oxidizing agent can oxidise F– ions to F2. Hence fluorine is prepared only by the electrolytic oxidation of fluoride ions.

4. disproportion reactions These are special types of redox reactions. A disproportion reaction is one by which in an intermediate oxidation state is converted to a mixture of a higher and a lower oxidation state. → Higher oxidation + Lower oxidation Intermediate  oxidation state state state One of the reacting substances in a disproportionation reaction always contains an element that can exist in at least three oxidation states. The element in the form of reacting substance is in the intermediate oxidation state and that both higher and lower oxidation states of that element are formed in the reaction. Some examples of disproportion reactions are as follows: Hydrogen peroxide decomposition is a familiar example for disproportionation. +1

−1

+1 −2

0

2 H 2 O2 (aq )  → 2 H 2 O + O2

Here the oxygen present in peroxide is in-1 oxidation state but converted to zero oxidation state in O2 (an increase) and decreases to –2 oxidation state in H2O. −3

O

+1

→ P H 3 ( g ) + 3H 2 PO2 ( aq ) P4 ( s ) + 3OH − ( aq ) + 3H 2O(l )  −2

O

+2 2 −

→ 4 S 2 − (aq ) + 2 S 2 O3 (aq ) + 6 H 2O(l ) S8 ( s ) + 12OH − (aq )  +1

O

−1

→ XO − (aq ) + X − + H 2O (l ) X 2 ( g ) + 2OH − (aq )  ( X = Cl , Br or I )

cold & dil O

−

−1 −

−5

− 3

3 X 2 ( g ) + 6OH (aq )  → 5 X (aq ) + XO + 3H 2O (l )

+1

0

+2

2 Cu + (aq )  → Cu ( s ) + Cu 2 + (aq) If any lower oxide of manganese such as MnO2 is heated with strong oxidizing agent in the presence of a strong base the manganese is oxidized to the green manganate (VI) oxidation state. 2 MnO2 ( s ) + O2 ( g ) + 4 KOH (aq )  → 2 K 2 MnO4 ( s ) + 2 H 2 O (l )

If the solution in the above reaction is acidified by an acid even as weak as carbonic acid, H2CO3, (passing CO2) disproportionation occur forming yielding the purple permanganate MnO4− ions in which the manganese has a +7 oxidation state plus MnO2. 3MnO42 − + 4 H +  → 2 MnO4− + MnO2 + 2 H 2 O

Certain substances undergo disproportionation as shown below: +4

+3

+5

2 N O2 ( g ) + 2 NaOH (aq )  → Na NO2 + Na NO3 + H 2 O +1

−1

+5

3HO Cl  → 2 HCl + HClO3 +3

−1

+5

3H Cl O2 (eq )  → H Cl (aq ) + 2 H Cl O3 (aq ) +5

4 H Cl O3 (aq )  →

−1

+7

H Cl (aaq ) + H Cl O4 (aq )

Note that in these reactions, a substance (reactant) is not involved in the elemental form

5. comproportionation reactions The reverse of disproportionation is comproportionation. In comproportionation reactions, two species with the same element in two different oxidation states form a single product in which the element is in an intermediate oxidation state. For example, when a solution containing a mixture of potassium bromide and potassium bromate is acidified, bromine is liberated. −1

+5

0

5 K Br + K BrO3 + 3H 2 SO4  → 3K 2 SO4 + 3H 2 O + Br 2

hot & conc.

Among halogens fluorine shows deviation from its reaction with alkali 2 F2 ( g ) + 2OH − (aq )  → 2 F − (aq ) + OF2 ( g ) + H 2 O(l )

Some oxygen will also be formed in the above reaction due to the attack of fluorine on water. This is because fluorine being most electronegative element cannot exhibit any positive oxidation state. So among halogens, fluorine does not exhibit disproportionation Copper (I) ions undergo disproportionation into copper (II) ions and copper metal in aqueous solutions.

6. autoxidation or Induced oxidation Metals such as zinc, palladium, non-metals such as phosphorous turpentine, anthraquinone and some unsaturated compounds absorb oxygen from the air in presence of water converting it into hydrogen peroxide. This oxidation of water to hydrogen peroxide is known as autoxidation. The substances such as phosphorous, zinc etc, are called activators. The activator first combines with oxygen to form addition compound which act as an autoxidator that oxidizes H2O or other substances e.g.,

Redox Reactions

Pb

+ O2  → PbO2

Activator

Problems for Practice

Autooxidator

7. Among the following, which do not show disproportionation? Why?

PbO2 + H 2 O  → PbO + H 2 O2 Acceptor

ClO–

Sodium sulphite solution is oxidized by air but sodium arsenite is not oxidized by air. But when a mixture of sodium sulphite and sodium arsenite is exposed to air, the oxidation reaction of sodium sulphite induces the oxidation of sodium arsenite by air. Na2SO3 + O2 Na2SO5 Na2SO5 + Na2AsO3

Na2SO4 + Na2AsO4

Na2SO3 + Na2AsO3+ O2

Na2SO4 + Na2AsO4

7. Intermolecular redox reactions

SnCl4+ 2FeCl3 oxidation

oxidation -2

+4

0

4MnO2 + 3O2

+5

-2

-1

0

2KCl + 3O2

0

NH4NO2

→ 2 NO( g ) a ) N 2 ( g ) + O2 ( g )  → 2 PbO ( s ) + 4 NO2 ( g ) + O2 ( g ) b) 2 Pb( NO3 ) 2 ( s )  → NaOH (aq ) + H 2 ( g ) c) NaH ( s ) + H 2 O (l )  → NO2− (aq ) + NO3− d ) 2 NO2 ( g ) + 2OH − (aq )  (aq ) + H 2 O (l )

9. Identify the substance oxidized, reduced, oxidizing agent and reducing agent for each of the following reactions:

+2 HBr (aq ) + C6 H 4 O2 (aq ) +

(b) HCHO (l ) + 2  Ag ( NH 3 ) 2  (aq ) + OH − (aq )  → 2 Ag ( s ) + HCOO − (aq ) + 4 NH 3 (aq) + 2 H 2 O (l ) (c) HCHO (l ) + 2Cu 2 + (aq) + 5OH − (aq )  → Cu2 O ( s ) + HCOO − (aq ) + 3H 2 O (l )

 → 2 PbSO4 ( s ) + 2 H 2 O (l ) 10. Identify the oxidizing and reducing agents in each of the following: → 2 HI ( g ) + S ( s ) (a) I 2 ( g ) + H 2 S ( g )  → 4Zn( NO3 ) 2 (aq ) (b) 10 HNO3 (aq ) + 4 Zn ( s ) 

11. Identify oxidants and reductants in each of the following ionic equations: (a) I 2 + 2S 2 O32 −  → 2 I − + S 4 O62 − + 2CO2 + 2 H 2 O

oxidation +3

8. Classify the following redox reactions:

(b) 2 H + + MnO2 + C2 H 2 O4  → Mn 2 +

Reduction

-3

ClO–4

+ NH 4 NO3 (aq ) + 3H 2 O (l )

Reduction oxidation 2KClO3

and

(e) Pb ( s ) + PbO2 ( s ) + 2 H 2 SO4 (aq )

In this reaction, one atom of a molecule is oxidized and another is reduced.

+7

ClO–3

→ N 2 ( g ) + 4 H 2 O (l ) (d) N 2 H 4 (l ) + 2 H 2 O2 (l ) 

8. Intramolecular redox reactions

2Mn2O7

ClO–2

(a) 2 Ag Br ( s ) + C6 H 6 O2 (aq )  → 2 Ag ( s )

If a reaction takes place between two different compounds in which molecules of one compound are oxidized while the molecules of other compound are reduced it is called inter molecular redox reactions e.g., Reduction SnCl2 + 2FeCl3

11.11

(c) Cl2 + 2 Br −  → 2Cl − + Br2

N2 + 2H2O Reduction

11.4.1 Balancing of redox reactions

oxidation -3

+5

+1

N2O + 2H2O

NH4NO3 Reduction

When steam is passed over red hot coke (1000°C) carbon monoxide and hydrogen gas are formed. The reaction is represented by the equation C ( s ) + H 2 O ( g )  → CO ( g ) + H 2 ( g )

11.12

Redox Reactions

On counting the number of atoms of carbon, hydrogen and oxygen on both sides of the equation, it can be seen that they are present in equal numbers before and after the reaction. A chemical equation must have equal number of atoms of each element on either side of the arrow mark in the equation. The equation where the equality of the number of atoms of elements is signified as described above is called a balanced chemical equation. Hydrogen gas burns in oxygen to form water. If we write equation for this reaction as

There are two methods for balancing the redox reactions. One of these methods is based on the change in the oxidation number of reducing agent and the oxidizing agent. The second method is separating the redox reaction into two half reactions—one involving oxidation and the other involving reduction.

H 2 ( g ) + O2 ( g )  → H 2 O (l )

This method is useful for balancing equations of both ionic as well as molecular reactions. This method emphasizes on the atoms of the elements undergoing a change in the oxidation state. In writing equations for oxidation reduction reactions the composition and formulae must be known for the substances that react and for the products that are formed. The steps involved in this method are as follows. (i) Assign oxidation number to each element on the both sides of the given reaction. Identify the element that is oxidized and reduced. (ii) Write the oxidation numbers of the elements participating in the reaction. (iii) Determine the increase and decrease of oxidation number per atom. Multiply the increase or decrease of ON with number of atoms undergoing the change. (iv) Equalize the increase in oxidation number and decrease in oxidation number on the reactant side by multiplying the respective formulae with suitable integers. (v) Balance the equation w. r. t. atoms other than O and H atoms. (vi) Balance ‘O’ by adding equal number of water molecules to the side falling short of O atoms. (vii) H atoms are balanced (in case of ionic equations) depending upon the medium in the following manner. (a) For acid medium, add proper number of H+ ions to the side falling short of H atoms. (b) For basic medium, add proper number of H2O molecules to the side falling short of H atoms and equal number of OH– ions to the other side. Example: Balance the following equation by the oxidation number method.

One can see clearly that this does not obey the law of conservation of atoms. In order to equate the number of atoms of the elements on both sides of the arrow mark, the equation has to be written as 2 H 2 ( g ) + O2 ( g )  → 2 H 2 O (l )

This equation is said to be balanced and is called balanced chemical equation. The numericals placed before the formulae of the reactants and the products in order to balance a chemical equation are known as stoichiometric coefficients. If the value of coefficient is equal to one it need not be indicated in the equation. If we write the equation for the above reaction as H 2 ( g ) + O( g )  → H 2 O (l )

(1)

We may say that this reaction is also balanced. However, this does not represent the reaction under consideration since ‘O’ represents the atomic oxygen which does not exist under normal conditions (only dioxygen exists). Since the elementary gases and compounds exist as molecules, it is only correct to write their molecular formulae while writing chemical equations. However, the above reaction can be represented as 1 H 2 ( g ) + O2 ( g )  → H 2 O(l ) (2) 2 Equations (1) and (2) are known as balanced chemical equations. The coefficients of reactants and products in the equation indicates the ratio of the quantities including the number of molecules) of the respective substance, thus Hydrogen : Oxygen :Water 2 : 1 : 2 or 1 : 1/ 2 : 1 Balanced chemical equations of this type are known as stoichiometric equations and the ratios are called stoichiometric ratios. A balanced chemical equation represents a stoichiometric equation. The exact quantities of the reactants and the products that appear in the balanced chemical equation are known as stoichiometric quantities.

11.4.2 Balancing of redox reactions by oxidation number Method

Cr(s) +Pb(NO3 ) 2 (aq )  → Cr ( NO3 )3 (aq ) + Pb( s )

Step 1: Indicate the ONs of elements participating in the reaction. 0 +2 +3 0 Cr ( s ) + Pb( NO3 ) 2 (aq )  → Cr ( NO3 )3 + Pb( s ) Step 2: Identify the atoms whose oxidation states have changed. (i) The ON of Cr is increased from 0 to +3

Redox Reactions

(ii) The ON of Pb is decreased from +2 to .0 +3

Example: Balance the following equation by the oxidation number method.

+2

∴ Cr → Cr 3+

Pb → Pb

Step 3: Equalize the increase in oxidation number, with the decrease in the ON. Suitable coefficients may be used for this purpose. Thus +3

0

2+

2 Cr  → 2 Cr

MnO42 − + Cl2 → MnO4− + Cl −

Step 1: Indicate the oxidation number of the elements. +6

2Cr ( s ) + 3Pb( NO3 ) 2 (aq ) → 2Cr ( NO3 )3 (aq ) + 3Pb( s )

Example: Balance the following equation by the oxidation number method.

−2

0

+6 − 2

+1− 2

+1−1

0

+1− 2

H 2 SO4 + HI  → H2 S + I2 + H2 O

Step 2: Identify the elements whose oxidation states have changed and write the equation and balance them with respect to the number of atoms. −1 −

I

0

 →I

0

−2 2−

S  →S

→ I2 2 I −  Total increase in ON = +2

Decreasse in ON = −8

Step 3: Equalize the increase in the oxidation number, with the decrease in oxidation number. Increase 4 × ( +2 ) = +8

; Decrease1( −8 ) = −8

Step 4: Write down the equation using these coefficients. H 2 SO4 + 8 HI → H 2 S + 4 I 2 + 4 H 2 O

Step 5: Check up whether the equation is balanced or not. (i) Check whether the charge is balanced or not. All the reactants and products are in molecular state the charge on a molecule is zero therefore the equation is balanced with respect to charge. 0 + 8 × (0) = 0 + 4 × (0) + 0 0=0 (ii) Check whether the numbers of atoms are balanced or not. Reactants Products

10H 10H 1S 1S 4O 4O 8I 8I Now the balanced chemical equation is H 2 SO4 + 8 HI → H 2 S + 4 I 2 + 4 H 2 O

−2

−1

Step 2: Identify the elements whose oxidation states have changed and write the equation and balance them with respect to atoms. +6

+7

−1

0

→ Mn 7 + Mn 6 + 

→ Cl − Cl2 

Increase in ON

→ 2Cl − Cl2 

+7 − 6 = +1

Decreases in ON − 2 − 0 = −2

H 2 SO4 + HI → H 2 S + I 2 + H 2 O

Step 1: Indicate the oxidation numbers of the elements in all the molecules.

+7

Mn O 2 − + Cl 2  → Mn O4− + Cl −

0

3 Pb  → 3Pb

Step 4: Using these coefficients, write the equation to get a balanced redox equation.

+1

11.13

Step 3: Equalize the increase in ON, with the decrease in ON. Increase + 1 Decrease − 2 2 × ( +1)

1 × ( −2)

= +2

= −2

Step 4: Write down the equation using these coefficients. 2 MnO42 − + Cl2 → 2 MnO4− + 2Cl −

Step 5: Check whether the equation is balanced with respect to charge. Reactants Products 2MnO 24 − + Cl2

2MnO 4− + 2Cl −

2 × ( −2) + 0

2 × ( −1) + 2 × ( −1)

= −4+0 = − 4

= −4

Check whether the equation is balanced with respect to number of atoms or not. Reactants Products 2 Mn 2 Mn 8O 8O 2 Cl 2Cl So the balanced equation is 2 MnO42 − + Cl2 → 2 MnO4− + 2Cl −

Problems for Practice 12. Copper react with nitric acid. A brown gas is formed and the solution turns blue. The equation may be written as Cu + NO3− → NO2 + Cu 2 +

Balance the equation by oxidation number method.

11.14

Redox Reactions

13. Balance the following equations.

(i ) Fe2 O3 + C → Fe + CO (ii ) Fe2+ + Cr2 O72 − + H + → Fe3+ + Cr 3+ + H 2 O (iii ) Zn + HNO3 → Zn ( NO3 )2 + NO2 + H 2 O (iv) C6 H 6 + O2 → CO2 + H 2 O 14. Permanganate ion reacts with bromide ion in basic medium to give manganese dioxide and bromate ion. Write the balance chemical equation for the reaction. 15. Balance the following reactions by oxidation number method. (i) FeS 2 + O2 → Fe2 O3 + SO2 (ii) Cr (OH )3 + IO − 3 → I − + CrO4 2 − (in basic medium )

11.4.3 Balancing of redox reactions by Ion – electron Method Ion electron method is also known as half reaction method. The balancing of chemical equation by ion electron method is done by the following steps. Step 1: Find the elements whose oxidation numbers are changed. Choose the substance which act as an oxidizing agent and one that acts as reducing agent. Step 2: Separate the complete equation into two half reactions, one for the change undergo by the oxidizing agent that the other for the change undergo by the reducing agent. Step 3: Balance the half reactions by the following steps. (i) Balance all atoms other than H and O. (ii) Calculate the electrons to whichever side is necessary to make up the difference. (iii) Balance the half equation so that both sides get the same charge. (iv) Add water molecules to complete the balancing of the equation. Step 4: Add two balanced half equations. Multiply one or both half equations by suitable numbers so that on adding the two equations, the electrons are balanced. It may be noted that redox reactions may take place in all the three media viz., acidic or basic or neutral. If H+ ions appear on either side of the equation, the reaction takes place in acidic medium. If OH- ions appear on either side of the equation, the reaction takes place in basic medium. If neither H+ nor OH– ions are present. The reaction occurs in neutral medium. For balancing redox reactions involving

acidic and basic media, the method has to be a modified slightly as illustrated below. (i) For acidic medium add proper number of H+ ions to the side falling short of H atoms. (ii) For basic medium add proper number of H2O molecules to the side falling short of H atoms and equal number of OH– ions to the other side.

example Balance the Reaction

Cr2 O72 − + NO2−  → Cr 3+ + NO3− In acid medium. Solution: Separate the reaction into two half reactions. 1. Cr2 O72 − + NO2−  → Cr 3+ + NO2− 2− 3+ 2. Cr2 O7  → Cr NO2−  → NO3− 3. Balance first the atoms other than O and H in each half reaction.

Cr2 O7 2 −  → Cr 3+

NO 2 −  → NO3−

Cr2 O7 2 −  → 2Cr 3+

NO 2 −  → NO3−

(Cr atoms are balanced) (N Atoms are balanced) 4. Balancing ‘O’ atoms in each half reaction separately by adding required number of ‘H2O’ molecules to the side deficient in oxygen atoms (one H2O for each oxygen atom). Cr2 O72 −  → 2Cr 3+ Cr2 O72 −  → 2Cr 3+ + 7 H 2 O

NO2−  → NO3− NO2− + H 2 O  → NO3−

5. Then balance hydrogen atoms in each half reaction for the reaction in acid medium H+ ions are added in the required number to the side different in H atoms (One H+ for each H). Cr2 O72− → 2Cr 3+ + 7 H 2 O 2− 7

+

3+

Cr2 O + 14 H → 2Cr + 7 H 2 O

NO2− + H 2 O → NO3 NO2− + H 2 O → NO3− + 2 H +

6. The charges are balanced in each half reaction by adding sufficient number of electrons to the side that is deficient in negative charge. Cr2 O72− + 14 H + → 2Cr 3+ + 7 H 2 O

NO2− + H 2 O → NO3− + 2 H +

Cr2 O72− + 14 H + + 6e − → 2Cr 3+ + 7 H 2 O NO2− + H 2 O → NO3− + 2 H + + 2e −

7. Number of electrons is equalized in the two half reactions by multiplying the two half reactions with suitable numbers. Cr2 O72− + 14 H + + 6e − → 2Cr 3+ + 7 H 2 O 3[ NO2− + H 2 O → NO3− + 2 H + 2e] Cr2 O72− + 14 H + + 6e − → 2Cr 3+ + 7 H 2 O 3 NO2− + 3H 2 O → 3 NO3− + 6 H + + 6e −

8. The two half equations are then added and the electrons are cancelled to get the total equation.

Cr2 O72 − + 8 H + + 3 NO2−  → 2Cr 3+ + 3 NO3− + 4 H 2 O

Redox Reactions

9. Add the spectator ions if they are deleted in the beginning. Spectator ions are the ions common on both sides which do not participitate in the reaction.

11.15

The reduction half reaction is: → MnO-4 → MnO2 (On changes from +7 to +4). The two half reactions are as follows Oxidation

Reducion

example

1. I  → I2

1. MnO4−  → MnO2

Balance the equation

2. 2I −  → I2

2. MnO −  → MnO2

(I atoms are balanced)

(Mn atoms are already

−

H2SO4(aq) + HBr(aq) → SO2(g) + Br2(l) Solution: Write skeleton equation in the ionic form for the reaction. Both H2SO4 and HBr are strong. They ionize in solution. Therefore –2

3. MnO4−  → MnO2

3. 2 I  → I2

+ 2H 2 O

–

2H+ (aq) + SO4 (aq) + H+ (aq) + Br (aq) → SO2(g) + Br2(l) The half reactions are Oxidation −1

balanced)) −

−

→ Br2 Br − 

 → MnO2 + 2 H 2 O +4

→ S O2 S O42 − 

→ Br2 1. Br − 

1. SO42 −  → SO2

→ Br2 2. 2 Br − 

2. SO42 −  → SO2

(Balancing Br) 3. 2 Br −  → Br2

+ 4OH − (H atoms are not present)

(H atoms are balanced in basic medium)

5. 2I −  → I 2 +2e −

5. MnO4− + 4 H 2 O + 3e −  → MnO2 + 2 H 2 O

3. SO42 −  → SO2 + 2 H 2 O

+ 4OH −

( Balancing O) → Br2 4. 2 Br − 

4. SO42 − + 4 H +  → SO2 + 2 H 2 O

5. 2 Br −  → Br2 + 2e −

→ Br2 + 2e − 6. 2 Br − 

(Charges are balanced) −

−

6. 3(2I  → I 2 + 2e )

(Balancing charges) 6. SO42 − + 4 H + + 2e −  → SO2 + 2 H 2 O

Electrons are equal in both half reactionns

(Charges are balanced) 6. 2( MnO4− + 4 H 2 O + 3e −  → MnO2 + 2 H 2 O

5. SO42 − + 4 H + + 2e −  → SO2 + 2 H 2 O

(Balaancing charges)

(O atoms balanced) 4. MnO4− + 4 H 2 O

4. 2 I  → I2

Reduction +6

0

(O atoms are not Present)

+ 4OH − (Multiplying the eq.5 with with 3, electrons are balanced)

(Multiplaying the eq.5 with 2, electrons are balanced)

7. 6 I −  → 3I 2 + 6e −

7. 2 MnO4− + 8 H 2 O + 6e −  → 2 MnO2 + 4 H 2 O

7. 2 Br − + SO42 − + 4 H +  → Br2 + SO2 + 2 H 2 O

+ 8OH + On adding the two half equations

example Write the balanced ionic equation which represents the oxidation of iodide (I–) ion with permanganate ion in basic medium to give iodine (I2) and manganese dioxide (MnO2). Solution: The skeleton equation is written as +7

−1

+4

0

Mn O4− + I −  → Mn O2 + I 2 The oxidation half reaction is I −  → I 2 (ON Changes from −1 to 0).

2 MnO4− + 4 H 2 O + 6 I −  → 2 MnO2 + 8OH − + 3I 2

example Write the balanced equation for the oxidation of sulphite ions to sulphate ions in acid medium by permanganate ion. Solution: The ionic skeleton reaction is +7

+4

+2

+6

MnO4− + S O32 −  → Mn 2 + + S O42 − → SO42 − ((+4 to +6) The oxidation half reaction: SO32 − 

11.16

Redox Reactions

→ Mn 2 + ((+7 to +2) The reduction half reaction: MnO4− 

1. C2 O42 − → CO2

1. MnO4− → Mn 2 +

2. C2 O42 − → 2CO2

1. SO32 −  → SO42 −

1. MnO4−  → Mn 2 +

2. SO32 −  → SO42 − (Satoms already balanced)

2. MnO4−  → Mn 2 + (Mn atoms already balanced)

→ SO42 − 3. SO32 − + H 2 O 

3. MnO4−  → Mn 2 + +4 H 2 O

(O atoms balanced) 2− 3

4. SO

(C atoms are balanced )

(Mn atoms are already balanced)

2− 4

3. MnO4− → Mn 2 + + 4 H 2 O

→ 2CO2

3. C2 O

(O atoms are already balanced)

(O atoms are are balanced)

4. C2 O42 − → 2CO2

4. MnO4− + 8 H + → Mn 2 +

(O atoms are balaanced) − 4

+ H2O 2− 4

2. MnO4− → Mn 2 +

4. MnO + 8 H +

+4 H 2 O

+

(H atoms are balanced

2+

 → SO + 2 H (H atoms balanced in acid medium)

 → Mn + 4 H 2 O (H atoms are balanced in acid medium)

→ SO42 − 5. SO O32 − + H 2 O 

5. MnO4− + 8H + + 5e −

+ 2 H + + 2e −

 → Mn 2 + + 4 H 2 O

(Chargees balanced)

(Charges balanced)

→ SO42 − 6. 5( SO32 − + H 2 O 

6. 2( MnO4− + 8 H + + 5e −

+ 2 H + + 2e −

 → Mn 2 + + 4 H 2 O

(Electrons balanced) 7. 5 SO32 − + 5 H 2 O

(Electro ons balanced) 7. 2 MnO4− + 16 H + + 10e −

 → 5SO42 − + 10 H + + 10e −

 → 2 Mn 2 + + 8 H 2 O

in acid medium 2− 4

→ 2CO2 + 2e

5. C2 O

−

5. MnO4− + 8 H + + 5e − → Mn 2 + + 4 H 2 O

(Charges balanced )

(

2− 4

6. 5 C2 O

→ 10CO2 + 2e

−

(Charges balanced )

)

(

6. 2 MnO4− + 8 H + + 5e − → Mn

(electrons balanced ) 2− 4

7. 5 C2 O

2+

)

+ 4H 2 O

(electrons balanced )

→ 10 CO2 + 10e

−

7. 2 MnO4− + 16 H − + 10e − → 2 Mn 2 + + 8 H 2 O

Adding the both half equations 8. 5 C2 O42 − + 2MnO4− + 16 H + → 10CO2 + 2Mn 2 + + 8H 2 O

Adding the both half equations

2 MnO4− + 5C2 O42 − + 16 H + → 10CO2 + 2 Mn 2 + + 8 H 2 O

8. 5SO32 − + 5 H 2 O + 2 MnO4− + 16 H +  → 5SO42 − + 10 H + + 2 Mn 2 + + 8 H 2 O

9.

example

2 MnO4− + 5SO32 − + 6 H +

When white phosphorous is boiled with aqueous sodium hydroxide solution gives phosphine (PH3) and H 2 PO2− write the balanced equation. Solution: Skeleton equation is

 → 2 Mn 2 + + 5SO42 − + 3H 2 O

example Oxalic acid is oxidized by permanganate ion in acid medium to CO2. Balance the reaction by ion-electron method. Solution: Write the skeleton equation. +7

+3

+2

+4

MnO4 + C2 O42 − → Mn 2 + + CO2 Oxidation half reaction: C2 O42 − → CO2 (+3 to +4) Reduction half reaction: MnO–4 → Mn2+ (+7 to +2)

0

−3

−1

P 4  → P H 3 + H 2 P O2− Oxidation half reaction P4 → H2 PO–2 (ON of P changes from 0 to +1) Reduction half reaction: P4 → PH3 (ON of P atom changes from 0 to −3) 1. P4 → H 2 PO2−

1. P4 → PH 3

2. P4 → 4 H 2 PO2

2. P4 → 4 PH 3

( P atoms are balanced )

( P atoms balanced ) − 2

3. P4 + 8 H 2 O → 4 H 2 PO

(O atoms are balanced )

3. P4 → 4 PH 3

( No 'O'atoms)

Redox Reactions

4. P4 + 8OH − →

4. P4 + 12 H 2 O → 4 PH 3

− 2

4 H 2 PO + 8 H 2 O

+ 12OH

(H atoms are balancced in basic medium)

−

(H atoms are balanced in basic medium)

5. P4 + 8OH − + 8 H 2 O →

5. P4 + 12 H 2 O + 12e −

4 H 2 PO2− + 8 H 2 O + 4e −

→ 4 PH 3 + 12OH −

(Charges balanced )

(Charges balanced )

6. 3[ P4 + 8OH − + 8 H 2 O →

6. P4 + 12 H 2 O + 12e −

4 H 2 PO2− + 8 H 2 O + 4e − ]

→ 4 PH 3 + 12OH −

( Electrons balanced )

( Electroons balanced )

5. Cr (OH )3 + H 2 O + 5OH −

(Charges balanced ) (Charges balanced ) − 6. 2[Cr (OH )3 + H 2 O + 5OH → 6. IO3− + 6 H 2 O + 6e − CrO42 − + 5 H 2 O + 3e −

+ 3H 2 O

( Electrons baalanced )

( Electrons balanced )

7. Add the two half equations 2Cr ( OH )3 + 2 H 2 O + 10OH − + IO3− + 6 H 2 O + 6e − →

example

i.e 4 P4 + 12 H 2 O + 12OH −

Balance the following equation.

 → 4 PH 3 + 12 H 2 PO2−

OH → Cr (OH )4  O2 + Cr  −

−

 → PH 3 + 3H 2 PO2−

+3

0

−2 +1

O2 + Cr  →[Cr (OH ) 4 ]−

example

−

Oxidation half reaction: Cr → Cr (OH )4  (ON of   Cr changes from 0 to + 3) − Reduction half reaction: O2 → Cr (OH )4  (ON of   'O' changes from 0 to −2)

Balance the following equation. +5

IO3−

 →

−1

+6

I − + Cr O42 −

Solution: Oxidation half reaction: Cr(OH)3 → CrO42 − (O.N. of Cr changes from +3 to +6) − − Reduction half reaction: IO3 → I (ON of I changes from +5 to −1) 1. Cr (OH )3  → CrO42 −

1. IO3−  →I−

2. Cr (OH )3  → CrO42 − (Cr atoms already balanced )

2. IO3−  →I− (I atoms already balanced

(O atoms balanced ) 4. Cr (OH )3 + H 2 O  → + 5 H 2 O + 5OH

−

4. IO + 6 H 2 O

−

2. O2 + Cr → Cr (OH )4 

2. Cr → Cr (OH )4 

−

(Cr atoms balanced )

(Cr atoms already −

3. O2 + Cr + 2 H 2 O

 H atoms balanced in   basic medium 

→ Cr (OH )4    O atoms balanced ( )

(O atoms balanced ) 4. Cr + 4OH − → Cr (OH )4 

−

4. O2 + Cr + 2 H 2 O → Cr (OH )4 

−

+ 3H 2 O  H atoms balanced in   basic medium 

1. O2 → Cr (OH )4 

−

− 3

 → I + 6OH

−

3. Cr + 4OH − → Cr (OH )4 

(O atoms balanced ) −

1. Cr → Cr (OH )4 

balanced))

3. Cr (OH )3 + H 2 O  → CrO42 − 3. IO3−  → I − + 3H 2 O

CrO

−

Solution: 0

2− 4

→ I − + 6OH −

2Cr ( OH )3 + IO3− + 4OH − → 2CrO42 − + 5 H 2 O + I −

→ 12 H 2 PO2− + 12e − + 4 PH 3 + 12OH −

+3

→ I − + 6OH − + 3H 2 O

2CrO42 − + 13H 2 O + 6e − + I − + 6OH −

3P4 + 24OH − + P4 + 12 H 2 O + 12e − + 24 H 2 O

Cr (OH )3 +

5. IO3− + 6 H 2 O + 6e −

 → CrO42 − + 5 H 2 O + 3e −

7. Adding the two half equations.

or P4 + 3H 2 O + 3OH

11.17

 H atoms balanced  in basic medium 

−

 H atoms alreeady   balanced 

−

11.18

Redox Reactions

5. Cr + 4OH −

5. O2 + Cr + 2 H 2 O + e − −

→ Cr (OH )4  + 3e −

(Chargges balanced ) 6. Cr + 4OH − →

−

→ Cr (OH )4    (Charges balanced ) 6. 3O2 + 3Cr + 6 H 2 O + 3e −

−

Cr (OH )  + 3e − 4 

→ 3 Cr (OH )4   

−

7. Adding the two half reactions Cr + 4 H 2 O + 4OH − + 3O2 + 3e − + 6 H 2 O + 3e − −

→ Cr (OH )4  + 3e − + 3 Cr (OH )4  

−

4 Cr + 4OH − + 6 H 2 O + 3O2 → 4 Cr (OH )4 

−

Problems for Practice 16. Balance the following reaction by ion-electron method. I 2 + S 2 O32 − → I − + S 4 O62 − 17. Balance the following equations by ion electron method. (i) Cr2 O72 − + Fe 2 + + H + → Cr 3+ + Fe3+ + H 2 O (ii) Cr2 O72 − + C2 H 4 O → C2 H 4 O2 + Cr 3+

( acidic medium) (iii) Zn + NO3− → Zn 2 + + NH 4+ (basic medium ) (iv) Br2 + H 2 O2 → BrO3− + H 2 O ( acidic medium ) Significance of Ion–Electron Method 1. Ion electron method applies only to those reactions in which ions participate. It gives importance in expressing the redox reaction in terms of ions, or an ionic equation. 2. In ion electron method, the charge is balanced by putting required number of electrons in both the half reactions, and while adding the two half reactions, the electrons get cancelled. This gives an indication of the number of electrons participating in the redox reaction. This method helps in the study of electrode reactions that take place in an electrochemical cell. 3. This method is not suitable for redox reactions which take place in the molecular state. To balance such reactions, oxidation number method is employed.

11.5 redox reactIons as the BasIs for tItratIons In Chapter 10, we have seen that a titration method for finding out the strength of one solution against the other using a pH sensitive indicator. Similarly, in redox systems the titration method can be adopted to determine the strength of reductant/oxidant using a redox sensitive

indicator. The usage of indicators in redox titration is illustrated below. (a) Titration of iron (II) with potassium dichromate (VI). Iron (II) ammonium sulphate, Fe(NH4)2(SO4)2, (Mohr’s Salt) solution is a suitable source of iron (II) ions, Fe2+. This substance should normally be made up by dissolving it in water mixed with dilute sulphuric acid (The acid helps to prevent the iron (II) ions being hydrolysed. A 20 mL aliquot is placed in a conical flask in the normal way, and usually about 10 mL of dilute sulphuric acid is added. A few drops of diphenylamine solution are added followed by 5 mL of phosphoric acid. Diphenylamine is a redox indicator. It is colourless unless it is oxidized, in which case it is converted into a violet – blue dye. When dichromate (VI) ions run into the solution, they oxidize iron (II) ions to iron (III). When all the iron (II) ions have been oxidized, the next one or two drops of the dichromate (VI) solution attack the indicator molecules. This is when the violet – blue colour appears and the end point has been reached. The phosphoric acid combines with the iron (III) ions made during the reaction to give complex ions. This stops the iron (III) ions themselves taking part in a redox reaction with the indicator, which might mask the end point. The equation for this reaction is 6 Fe3+ (aq ) + Cr2 O72 − (aq ) + 14 H + (aq )  → 6 Fe3+ (aq ) + 2Cr 3+ (aq ) + 7 H 2 O (l )

This shows 6 mol Fe2+ = 1 mol Cr2O72– (b) Titration of oxalate (ethanedioate) ions with potassium manganate (VII). A solution of sodium oxalate (also known as sodium ethanoate) Na2C2O4 can be used as a source of oxalate ions C2O42–. A 20 mL aliquot should be placed in a titration flask and 10 mL of dilute sulphuric acid added. To ensure that the reaction takes place at suitable rate, the solution must be heated to nearly 60°C before potassium manganate (VII) solution is run in from a burette. The nice thing about manganate (VII) ions in water is that they give an intense purple colur. When they oxidize another chemical in acidic solution, the product is a solution of manganese (II) ion, which is colourless. This means that potassium manganate (VII) solution can act as its own indicator (self indicator). While oxalate ions are still present, the manganate (VII) solution run into the flask loses its colur. Once the oxalate ions have reacted, the next drop of manganate (VII) solution gives a permanent purple colour, which makes the end point. The equation for the reaction is 5C2 O42 − (aq ) + 2 MnO4− (aq ) + 8H + (aq )  →10CO2 ( g ) + 2 Mn 2 + + 4 H 2 O (l ) Therefore 5mol C2 O42 − = 2mol MnO4−

Redox Reactions

(c) Titrations involving iodine Compounds that contain iodine are widely used in titrations. The titrations make use of one (Or more) of the following changes. (i) Iodide ions can be oxidized to iodine 2 I − (aq ) − 2 e −  → I 2 (s) Colourless black

(ii) Iodate (v) ions IO–3, will oxidize iodide ions to iodine IO3− (aq ) + 5 I − (aq ) + 6 H + (aq )  → 3I 2 ( s ) + 3H 2 O Colourless Colourless black 2– (iii) Thiosulphate ions, S2O3, can reduce iodine to iodide ions. 2 S 2 O32 − (aq ) + I 2 ( s )  → S 4 O62 − (aq ) + 2 I − (aq ) Colourless black colourless Colourless Although solid iodine is black and insoluble in water, it is often the case that reactions are done in which many iodide ions are about. If iodine is produced in the presence of iodide ions, it is converted into soluble triodide ions I3–. I 2 ( s ) + I − (aq )  I 3− (aq ) black dark brown Small amounts of iodine molecules, I2 can be detected by using starch as an indicator. Depending on the amount of iodine present, starch will give a blue to almost black colour.

an Iodate (v)/Iodide/thiosulphate titration Suppose we have a solution of a potassium iodate (v) whose concentration is to be determined. The method is to react the iodate (V) ions with an excess of iodide ions so that all the iodate (V) ions are converted into iodine. Having done this, we can titrate the solution with thiosulphate ions in order to find the amount of iodine released. In practice, we place 20 mL of the iodate (V) solution in a conical flask together with at least 10 mL of dilute sulphuric acid. Then an excess of potassium iodide solution is added. This liberates iodine; but because iodide ions are left over, the triodide ions are produced rather than solid iodine. Now sodium thiosulphate of known concentration is run in from a burette. Gradually, the colur of the solution fades owing to iodine being converted back into iodide ions. When the solution is pale straw colour, starch solution is added. This gives a strong black colour. When the solution goes clear we know that all the iodine has been used up by the thiosulphate ions. This is the end point of the titration. Therefore 1 mol I2 = 6 mol S2O3–2 and overall 1 mol – IO 3 = 6 mol S2O3–2

11.19

11.5.1 limitations of concepts of oxidation number The concept of oxidation number though is wider and simple in its application yet puts us in some perplexing situations. Two such situations are (i) Many reactions are known as redox reaction according to classical approach where as it is not possible to do same by applying the idea of oxidation number concept. For example, consider the reaction

SiCl4 (l ) + LiAlH 4 ( s )  → SiH 4 ( g ) + LiCl ( s ) + AlCl3 ( s ) According to classical approach silicon is reduced here and lithium and aluminium both are oxidized, but according to oxidation number change approach, none of these species finds a change in its oxidation number. (ii) Many times it may be difficult to interpret the reaction as redox reaction according to all the three approaches namely; classical, electronic, and oxidation number approach. For example CH 3 NCO( g ) + H 2 O(l )  → CH 3 NH 2 ( g ) + CO2 ( g )

In view of the above quoted reactions and many other similar type of reactions, a fresh relook into the discussion for redox processes is necessary. Very recently in 1997, the concept of redox processes has been re-examined in the light of changes in electron density. A broader and more useful definition has been suggested which classifies oxidation as a decrease of electron density and reduction as an increase of electron density around the atom(s) involved in the reaction. A careful application of the above concept reveals that in first reaction silicon is reduced and lithium and aluminium are oxidized. It is because replacing more electronegative chlorine atoms in SiCl4 by less electro negative hydrogen atoms increases the charge density about silicon. Further in the second reaction nitrogen of methyl isocyanate (MIC) is reduced as the nitrogen is converted to amino group and carbon of MIC is oxidized as it is converted to carbon dioxide. Another advantage with this concept seems to be the easy comparativeness among oxidants and reductants. Let us illustrate this by taking the example of halogenation of methane. CH 4 ( g ) + 4 X 2 ( g )  → CX 4 ( g ) + HX ( g )

For simplicity the reaction is being considered as gaseous phase only. Now according to the classical concept as well as the oxidation number concept, methane is oxidized whereas halogen is reduced. The electron density concept however still goes one step forward and conveys that the oxidative trend among the halogens should follow the sequence F2 > Cl2 >Br2> I2. It is because for considering the change in electron density, the bonds under consideration C−F,

11.20

Redox Reactions

C−Cl, C−Br and C−I bond. The decrease in electron density around carbon of methane follows the same sequence in which these bonds are written. This is a recent concept and it has yet to be accepted by the wider chemists.

mechanism as that occurring when the two solutions are mixed in a beaker, the overall chemical change is the same; namely

11.6 redox reactIons and electrode Processes

11.6.1 electrode Potential

All reactions which involve loss and gain of electrons can be performed under conditions in which an electric current generated. Such an arrangement is called electrochemical cell. Two beakers, one containing potassium iodide solution and the other a solution of iron(III) chloride are connected by a salt bridge containing potassium chloride (a device that prevents excessive mixing of the two solutions). A platinum electrode is dipped into each solution and joined together through a sensitive galvanometer. The galvanometer shows a deflection indicating the flow of current; at the same time, the yellow colour of liberated iodine will appear around the electrode dipping into the potassium iodide solution and iron (II) ions (detected by adding potassium hexacyanoferrate (III) solution) will appear at the electrode. The reaction is thus 2 I −  → I 2 + 2e − oxidation 2 Fe3+ + 2e −  → 2 Fe 2 + reduction

Electrons flow from the electrode dipping into the potassium iodide solution to the other through the external circuit (Opposite to conventional current flow) current being conducted through the solution itself by the slow migration of ions. Although we cannot be sure that the reaction carried out under these conditions involves the same

2 Fe3+ + 2 I −  → 2 Fe 2 + + I 2

Any metal placed in an ionizing solvent such as water has a tendency to dissolve with the formation of ions. This reaction is reversible.  M n + + ne − M 

For active metals such as sodium, it is easy to cause the reaction to go to completion to the right, as indicated above, but for inactive metals such as platinum, the tendency is so slight that perceptible formation of ions may be brought about only with difficulty. Thus whenever a solid substance is kept in water, it tends to pass into the solution at constant temperature. This tendency of a solid is known as solution pressure. If a slightly, larger amount of solid passes into the solution and when the dissolution stops, the solution becomes super saturated. On addition of further amount of solid in the solution, it will separate as a solid rather than going into the solution. This tendency of solid separation is greater than dissolution when the solution is super saturated. The pressure exerted by dissolved solid substance is its osmotic pressure. Whether a solid passed into solution or separated as a solid depends upon the balance of solution and osmotic pressure. Nernst assumed that all metals possess solution pressure and tend to pass into solution. When a metal is dipped in pure water, a number of positive ions will pass into the solution and the metal acquires negative charge. Since the positive ions remain

G Salt bridge

Potassium iodide Solution

Iron (III) chloride solution

fig 11.1 Reactions between ionic substances in solution can be made to generate an electric current

Redox Reactions

11.21

Metal Rod Metal Rod

+ + + + + + + + +

− − − − − − − −

+ + + +

+ + + + + + + + +

Solution (a)

Metal Rod

+ + + + + + + +

+ + + + + + + +

Solution

Solution

(b)

(c)

− − − − − − − −

fig 11.2 Nernst theory of solution pressure and the origin of helmholtz double layers (a) P>π (b) P=π (c) P π, metal will continue to dissolve into solution until the accumulated charge opposes the further action. With the result, metal becomes negatively charged while the solution is positive (Fig 11.2 a) such metals are zinc cadmium, alkali metals etc. 2. When P = π, no potential will be set up between metal and the solution. Such systems are known as null electrodes. (Fig 11.2 b). 3. When P < π, the positive ions of the dissolved salt separate on the metal surface until the accumulated charge opposes the further action. At equilibrium, metal acquires positive charge while the solution is negative one (Fig 11.2C). Such metals are copper, mercury, gold. Nernst further observed that the theory of solution pressure is also applicable to non metals i.e., the substances which produce negative ions in the solution. For example, oxygen and chlorine electrodes give O2– and Cl– ion in the solution in such cases. 1. If P > π the electrode becomes positively charged while the solution contains negative charges. 2. If P = π no potential will be set up

3. If P< π the electrode will be negatively charged while the solution acquires positive charges. Now if we consider a zinc rod was placed into a solution containing zinc ions, an equilibrium would be set up between them. There is a tendency for zinc atoms on the surface of the rod to be attracted into the solution. However, they do not enter the solution as atoms, but as zinc ions, Zn2+. In this guise, they can be solvated by water molecules. The electrons left behind when a zinc atom is transformed into a positive ion remain on the rod. As a result, the region of solution very close to the rod suffers an increase in positive charge (owing to the extra Zn2+ ions) while the rod carries a layer of negative charge (the electrons left behind). Other metals dipping into solutions of their ions undego a similar but opposite change. For them, some of the ions in the solution cling on to the metal and attract electrons. This leaves the rod with a positive charge. Because the solution near to the rod loses positive ions, it is slightly negatively charged. An electric double layer is set up. This layer is known as the Helmholtz double layer. If we call the metal M and we assume that it makes a positively charged ion Mn+ the same equilibrium is involved in both cases.  M ( s ) M n + (aq ) + ne −  The difference is that for some metals the equilibrium lies to the left (these give extra positive ions into the solution), while for others it lies to the right (these take positive ions out of the solution).

Redox Reactions

11.6.2 the standard hydrogen electrode The S.H.E consists of hydrogen gas bubbling over a platinum electrode immersed in a solution of hydrochloric acid.

At standard conditions, the hydrogen must be at a pressure of 1 atm and the acid must be 1 mol dm–3 in concentration (strictly the acid should be at a unit activity which is approximately 1.18 mol dm–3 but we can ignore the difference). The temperature should be 25°C. The platinum electrode is usually coated with finely divided platinum called platinum black. This acts as a catalyst to allow equilibrium between the gas and the solution to be established quickly The reaction that takes place in the S.H.E is   2 H + (aq ) + 2e −   H2 (g)

φ

Under standard conditions the electromotive force emf of the S.H.E is defined to be exactly zero volts. A standard emf. is given the symbol E , so we have E S.H.E = 0.000V The more systematic way of writing this is to put E H +/ H2= 0.000V. φ

From now on, we shall call the metal rods as electrodes. Whenever there is a separation of positive and negative charges, we should be able to measure a voltage. In this case, there should be a voltage between the electrode and the surrounding solution. At the first sight, it might seem an easy thing to measure this voltage. All we need is a voltmeter and two pieces of wire. We connect one piece to the electrode, the other we dip into the solution. Unfortunately, this idea has a flaw too. As soon as we dip the metal wire into the solution, another equilibrium is set up. This time it is between the metal from which the wire is made and ions it gives in solution. We have introduced another Helmholtz double layer. The best we can do now is to measure the difference in voltage between the two double layers. This is an unavoidable state of affairs; we cannot directly measure the voltage between the Helmholtz double layers. This is unfortunate because if we could measure the voltage it would tell us something about how good a metal is at releasing electrons. Metals that release electrons easily are good reducing agents so by comparing the voltage for each metal, we would be able to put them in order of their reducing power. Given that we are found to measure a voltage between two double layers the most convenient thing to do is to agree to keep one of them constant and always measure the difference between this one and the others. The system that has been chosen as a standard is the standard hydrogen electrode. We shall refer to this as the S.H.E.

φ

11.22

11.6.3 standard electrode Potentials As a standard electrode is already established the other metal electrodes are connected to the S.H.E and measure the voltage between the two. Fig 11.4. Shows an arrangement using a zinc electrode. At standard condition, the zinc rod should be dipping into 1 mol dm–3 solution of the zinc ions (strictly the solution should have an activity of 1. Using activities makes an allowance for the way ions influence each other solution). The entire arrangement makes up an electrochemical cell with the S.H.E making one half cell and the zinc in a solution of zinc ions the other half cell. The two half-cells are connected by Hydrogen gas at 1 atm pressure

Temperature 25°

Platinum electrode coated with platinum black

Hydrochloric acid of unit activity about 1.18 mol dm-3 fig 11.3 A standard hydrogen electrode (S.H.E)

Redox Reactions

11.23

-0.76 Digital voltmeter Hydrogen gas 1 atm.

zinc electrode

Saltbridge

porous plug

Platinum electrode zinc sulphate solution 1 mol dm-3

Hydrochloric acid 1 mol dm-3

fig 11.4 Measurement of standard electrode potential using standard hydrogen electrode

φ

φ

φ

a salt bridge. A typical salt bridge is made by dissolving an ionic substance such as potassium chloride in agar. The warm agar is used to fill a U− tube and when it cools, it sets into a jelly. The charge on the potassium and chloride ions provides electrical contact between the two half-cells. Now we have another convention that we must know. The emf of a combination of two half-cells to be the difference between the emf of the half-cell on the right-hand side minus the emf of the half-cell on the left hand side. Ecell = Eright – E left

φ

φ

φ

A S.H.E. is always used as the left-hand half-cell. In this case we have

φ

φ

E cell = E Zn2+/ Zn – E H +/ H2 φ

so E cell = E Zn2+/ Zn because E H +/ H2 = 0 V by definition.

φ

While performing this experiment, the voltmeter would only give a reading if its negative terminal was connected to the zinc. This means that the zinc is negative compared to the S.H.E. indeed E Zn2+/ Zn = − 0.76v. This value is known as the standard electrode potential. The electrode potential is a measure of tendency of an electrode potentials and may be of two types (i) Oxidation Potential: The tendency of an electrode to lose electrons or to get oxidized is called oxidation potential thus oxidation potentials give the tendency of

an electrode to lose electrons i.e.,  M n + (aq ) + ne − M ( s )  e.g .,

 Zn 2 + (aq ) + 2e − Zn( s )  H2

 2 H + (aq ) + 2e − 

(ii) Reduction Potential: The tendency of an electrode to gain electrons or to get reduced is called its reduction potential. Therefore, it measures the tendency of an electrode to gain electrons i.e.,  M ( s ) M n + (aq ) + ne −  2+  Cu ( s ) e.g ., Cu (aq ) + 2e −   H 2 ( g ) 2 H + (aq ) + 2e −  It is evident that the oxidation potential is the reverse of reduction potential. For example, if reduction potential of Zn is −0.76 volts, its oxidation potential is +0.76 volts. According to the present convention, the half cell reactions are always written as reduction half reaction and their potentials are represented as reduction potentials. In the reaction taking place at an electrode i.e., at the interface of the metal and its salt solution at each electrode both reduced and oxidized forms of the same species are present. These represent the species in the reduction and oxidation half reactions. The oxidized and reduced forms of a substance taking part is an oxidation or reduction halfreaction is called redox couple.

11.24

Redox Reactions

From the discussion made so far, it can be concluded that in case the electron accepting tendency of the metal electrode is more than that of a S.H.E., its standard reduction potential gets a positive sign and in case the electron accepting tendency of the metal electrode is lesser than that of S.H.E its standard reduction potential gets a negative sign. It must be remembered that according to the latest convention, all standard potentials are taken as reduction potentials. The electrode at which reduction occurs with respect to S.H.E. has +ve reduction potential. The electrode at which oxidation occurs with respect to S.H.E. has –ve reduction potential.

11.6.4 reference electrodes The standard cell must be capable of giving constant and reproducible e.m.f and its variation with temperature should be negligibly small. Since a standard hydrogen electrode is difficult to prepare and maintain, the Weston cell is considered as widely accepted standard cell. The anode consists of a layer of solid cadmium amalgam containing 12.5 % cadmium and the cathode consists of a thick paste of Hg2SO4. The solution is saturated with 8 CdSO4 ⋅ H 2 O with some excess crystals present on both 3

sides to maintain saturation schematically, it can be written The standard electrode potentials are very important and give a lot of useful information. If the standard electrode potential is more than zero then its reduced form will be more stable compared to hydrogen gas. In other words it will be readily reduced in comparison to hydrogen and therefore acts as strong oxidizing agent. On the other hand; if the standard electrode potential is negative than hydrogen then hydrogen gas is more stable than the reduced form of the series. In other words, it will be readily oxidized in comparison to hydrogen and therefore acts as strong reducing agent. Similarly, we can compare two species. The species having higher electrode potential will be readily reduced (acts as oxidizing agent) in comparison to the other.

8 as Pt| Cd(Hg)| CdSO4 ⋅ H 2 O( s ) | Hg 2 SO4 ( s ) | Hg| pt 3 8 The cell reaction is Cd ( s ) + HgSO4 ( s ) + H 2 O(l ) 3 8  → CdSO4 . H 2 O( s ) + 2 Hg (l ) 3

cork sealed with parafin or wax saturated solution on of CdSO4

difference Between emf and Potential difference The potential difference is the difference in the electrode potentials of the two electrode present in the cell under any conditions while emf is the potential generated by a cell when it draws no current or it is under conditions of zero electron flow. The distinction between emf and potential difference may be given as below emf 1. It is the potential difference between the two electrodes when no current is flowing in the circuit, i.e., in an open circuit. 2. It is maximum voltage obtainable from the cell. 3. It is responsible for the flow of steady current in the cell.

Potential differnce 1. It is the difference of the electrode potentials of the two electrodes when the cell is sending current through the circuit. 2. It is less than the maximum voltage obtainable from the cell (i.e., emf of the cell) 3. It is not resposible for the flow of steady current in the cell.

CdSO4·8/3H2O crystals Hg +

Cd+Hg

−

fig 11.5 The weston cell The emf of the cell at 298 K is 1.01463V. The temperature dependence is very small i.e., 4.05 × 10−5 VK −1 . The Weston cell is reversible. It is not subjected to permanent damage due to the passage of current through it. Reference electrode: An electrochemical cell consists essentially of two parts defined by two electrodes. In a cell, one electrode is under investigation and the other is a conventional electrode of known property. The last one is called a reference electrode. A very common and easily

(1atm )

11.25

φ

constructed one is the calomel electrode. It consists of a tube at the bottom of which there is a layer of mercury and mercurous chloride. The whole tube is then filled with a solution of known strength of KCl (generally saturated 1N or 0.1 N). A platinum wire is fixed in the bottom of the tube. Electrolytic connection must be made to the rest of the cell and this is done through a side arm in which KCl solution has usually been stiffened with agar- agar or gelatin. This type of electrolytic connection is known as a salt bridge. A complete cell is shown Fig.11.6. The cell diagram is Pt| Hg| Hg2Cl2(s)|1 N KCl or saturated KCl|HCl(m) | H 2 | Pt

φ

Redox Reactions

E 298 = 0.280 V (1N KCl) and E 298= 0.242 V (saturated KCl) The emf of the cell depends at each temperature on the strength of KCl solution. Three different types of electrodes are used i.e., 0.1 N, 1 N and saturated. The potentials of these electrodes as a function of temperature, the emf at 298 K and the corresponding electrode reaction are given in Table 11.3. Suppose it is required to measure the electrode potential of zinc immersed in 0.1 N solution ZnSO4. The Zn half cell is connected with the calomel half cell through a salt bridge of KCl. The complete cell is

Zn | ZnSO4 ( 0.1N ) || KCl ( saturated ) | Hg 2 Cl2 ( s ) | Hg

If the HCl is at unit activity then

Zinc rod

salt bridge Saturated solution of KCl

ZnSO4 solution KCl Calomel Hg2Cl2 Mercury

fig 11.6 Calomel electrode combined with electrode whose potential is to measured

Electrode

Symbols

0.1 N Calomel

Hg |Hg2Cl2 (s) KCl (0.1N) Hg |Hg2Cl2 (s) KCl (1 N) Hg |Hg2Cl2 (s) KCl (saturated)

1 N Calomel Saturated calomel

φ

table 11.3 Potentials of calomel electrodes emf

E 25°C

Reaction

E = 0.338−7 × 10-5(t−25)

0.338

Hg2Cl2(s)+2e– → 2Hg(l)+2Cl– (0.1N)

E = 0.2800–2.4 × 10-4(t−25)

0.2800

Hg2Cl2(s)+2e– → 2Hg(l)+2Cl– (IN)

E = 0.2451–7.6 × 10-4(t−25)

0.2415

Hg2Cl2(s)+2e– → 2Hg(l)+2Cl– (saturated)

11.26

Redox Reactions

φ

φ

φ

The calomel electrode is written on the right as reduction takes place at this electrode. The emf of the cell is determined by the usual potentiometric method. At 298 K it is found to be 1.0022 volt. Now according to convention

φ

E cell = E cell (right) – E cell (left) 1.0022 = 0.2422 – E Zn \E

φ

2+

Zn2+/ Zn =

/ Zn

0.2422 – 1.002 = – 0.76 volt

11.6.5 factors affecting the values of standard electrode Potentials The electrode potentials of the group 1A metals decrease in the order lithium > potassium> sodium; thus lithium appears to occupy an anomalous position since an increase metallic character – taken as an increasing tendency for the metal atom to lose electrons and measured quantitatively as ionization energy increases in the order lithium> sodium > potassium. In order to see why lithium has an unexpectedly high electrode potential, it is necessary to consider the individual energy changes that take place in the process +

Li ( s )  → Li (aq ) + e − Where Li+(aq) denotes hydrated lithium ions The energy change involved when one mole of lithium metal passes into solution as hydrated lithium ions can be determined by considering the process to take place in a series of hypothetical stages each one of which involves an energy change. (a) Conversion of solid lithium into gaseous lithium atoms. This is an endothermic process and the sublimation energy, the energy required to convert one mole of solid lithium into gaseous atoms, is absorbed Li ( s )  → Li ( g )

S = +159.0 KJ / mol

(b) Removal of the outer electron of the lithium atom to give a lithium ion. The energy required for this is the ionization energy I Li ( g )  → Li + ( g ) + e −

I = +520.0 KJ / mol

(c) Hydration of the lithium ion This is an exothermic process and the heat liberated is known as the heat of hydration ∆Hh Li + ( g ) + water  → Li + (aq ) ∆H h = −507.1KJ / mol The overall change in the complete reaction is the sum of the separate energy changes (law of conservation energy). Thus: ∆H = S + I + ∆H h = (+159.0 + 520.0 − 507.1) KJ / mol = +171.9 KJ / mol

The change Li ( s )  → Li + (aq ) + e − is endothermic to the extent of 171.9 KJ/mol. A similar series of calculations could be done for sodium and potassium: the appropriate energy changes together with those of lithium are set out in Table 11.4. table 11.4 Energy change for the process M(s) → Mn+ (aq) + ne– Enthalpy Changes (KJ/mol) Heat of sublimation (S) Ionisation energy(I) Heat of hydration (∆Hh) Overall energy change

Lithium Sodium Potassium +159.0 +520.0 −507.1 +171.9

+108.7 +493.8 −395.9 +206.6

+90.00 +418.4 −317.2 +191.2

The process of an alkali metal passing into solution as its hydrated ions thus becomes less endothermic and neglecting entropy effects more likely along the series sodium, potassium, lithium in accordance with their standard electrode potentials. The anomalous behavior of lithium is therefore due to the very high heat of hydration of its ions, a consequence of the fact that smaller ions can bind water molecules more tightly than larger ones.

11.6.6 salt Bridge and its function A salt bridge has an important role to play in an electrochemical cell. It is usually an inverted U tube filled with concentrated solution (generally saturated solution) of inert electrolytes. An inert electrolyte is the one whose ions are neither involved in any electrochemical change nor do they react chemically with the electrolytes electrically connected by them. Generally salts like KCl, KNO 3, NH4NO3 etc. are used. For preparing a salt bridge gelatin or agar is dissolved in a hot aqueous solution of KCl, KNO3 or NH4NO3 and filled in a U-tube. On cooling the agar solution becomes solid and used as salt bridge. The important functions of the salt bridge are: 1. Salt bridge connects the solution in the two half-cells and completes the cell circuit: As already explained the solutions of the half cells are connected by a salt bridge and their electrodes by means of a wire. Therefore the salt bridge completes the circuit. 2. Salt bridge keeps the solutions in the two half-cells electrically neutral: To understand consider the ZnCuSO4 cell. In the oxidation half-cell, Zn atoms lose electrons and change into Zn2+ ions. As more and more Zn2+ ions are formed extra positive charge accumulates in the solution which prevents the further formation of Zn2+ ions and flow of electrons from the

Redox Reactions

11.6.7 electrochemical series

φ

φ

In the previous section, it was already discussed about the measurement of standard reduction potential (E ) of zinc electrode. In a similar manner the E values of other electrodes can also be measured. The standard reduction potentials can be arranged in the increasing order and this order is called as electrochemical series. The arrangement of elements in order of decreasing reduction potential values is called electrochemical series. The electrochemical series, also called activity series, of some typical electrodes is given in Table 11.5. It may be noted that the standard oxidation potential will have the same magnitude but opposite sign. For example, the standard oxidation potential of first electrode Li, Li+ is +3.05V and standard oxidation potential of the electrode F–, F2 is −2.87V.

11.6.8 application of electrochemical series 1. relative strengths of oxidizing and reducing agents The electrochemical series helps us to pick out substances that are good reducing agents. In the electrochemical series the substances are arranged in the decreasing order of reduction potential i.e., decreasing tendency for reduction to occur or power as oxdising agent. Therefore the elements at the top of the table have maximum tendency to get reduced and subsequently they will act as good oxidizing agents. On the other hand, the substances at the bottom of the table have lower reduction potential therefore they have least tendency to get reduced. Consequently, they may be oxidized and act as good reducing agents. Thus, the substances which have lower reduction potentials are stronger reducing agents while those which have higher reduction potentials are stronger oxidizing agents. 2. Comparison of reaction of metals: A metal having smaller reduction potential can displace metals having larger reduction potentials from the solutions of their salts. For example, copper having lower reduction potential than silver, displace silver from silver nitrate solution. In general, a metal occupying lower position in the series can displace the metals lying above it from the solutions of their salts. In other words, the metals occupying lower positions in the electrochemical series are more reactive in displacing the other metals from the solutions of their salts. 3. Predicting feasibility of a redox reaction: The electrochemical series helps in finding out whether a given reaction is feasible or not from the E values of the two electrodes. In general, a redox reaction is feasible only if the species which has higher reduction potential is reduced i.e., accepts the electrons and the species which has lower reduction potential is oxidized i.e., loses the electrons. Otherwise, a redox reaction is not feasible. In other words, the species to release electrons must have lower reduction potential as compared to the species which is to accept electrons. The electrochemical series (Table 11.5) gives the decreasing order of electrode potentials (reduction of different electrodes on moving down the table). This means that the species to accept the electrons (getting reduces) must be higher in the electrochemical series as compared to the other which is to lose electrons (getting oxidized). This can be explained with the following examples: φ

zinc rod. Similarly, in the reduction half cell Cu2+ ions get reduced to Cu resulting the accumulation of negative charge around cathode due to excessive SO42 − ions. This prevents the flow of electrons to the copper ions. Thus in the absence of salt bridge, the flow of electrons occur only for a short while and the cell will stop working. When salt bridge is used sufficient number of Cl– ions migrate from salt bridge to the anode half-cell and neutralize the excessive positive charge. Similarly to neutralize the excessive negative charge at cathode due to the additional SO42 − ions in cathode half cell sufficient number of K+ ions migrate from the salt bridge to this half cell. Thus salt bridge provides cations and anions to replace the ions produced in the two half-cells to maintain electrical neutrality. 3. Prevention of Liquid-Junction Potential: When two salt solutions of different concentrations are in contact with one another, the junction between the two solutions contribute a potential to the cell. This potential is called liquid junction potential. This potential is caused by the difference in the rates of diffusion of the two ions the more dilute solution acquiring a charge corresponding to that of faster moving ion. If for example, a concentrated solution of HCl forms a junction with dilute solution both H+ ion and Cl– ions diffuse from the concentrated solution into the dilute solution. The H+ ion moves faster and thus the dilute solution becomes positively charged because of an excess of H+ ions. The more concentrated solution is left with an excess of Cl– ions and thus acquires negative charge. The actual separation of charge is very small but shows an appreciable potential difference. Salt bridge eliminates this liquid junction potential by neutralizing the charges.

11.27

Cu 2 + (aq ) + 2 Ag ( s )  → Cu ( s ) + 2 Ag + (aq )

11.28

Redox Reactions

table 11.5 Standard Electrode Potentials at 298 K

Ions are present as aqueous species and H2O as liquid; gases and solids are shown by g and s.

Increasing strength of oxidizing agent

3+

→ Reduced form  → 2F

−

E /V

−

2.87

2+

Co + e H 2 O2 + 2 H + + 2e − MnO4− + 8 H + + 5e − Au 3+ + 3e − Cl2 ( g ) + 2e − Cr2 O72 − + 14 H + + 6e − O2 ( g ) + 4 H + + 4e − MnO2 ( s ) + 4 H + + 2e − Br2 + 2e − NO3− + 4 H + + 3e − 2 Hg 2 + + 2e − Ag + + e − Fe3+ + e − O2 ( g ) + 2 H + + 2e − I 2 + 2e − Cu + + e − Cu 2 + + 2e − AgCl ( s ) + e − AgBr ( s ) + e − 2H + + 2e − Pb 2 + + 2e − Sn 2 + + 2e − Ni 2 + + 2e − Fe 2 + + 2e − Cr 3+ + 3e − Zn 2 + + 2e − 2 H 2 O + 2e − Al 3+ + 3e − Mg 2 + + 2e − Na + + e − Ca 2 + + 2e − K + + e− Li + + e −

 → Co  → 2H 2 O  → Mn 2 + + 4 H 2 O  → Au ( s )  → 2Cl −  → 2Cr 3+ + 7 H 2 O  → 2H 2 O → Mn 2 + + 2 H 2 O  → 2Br −  → NO( g ) + 2 H 2 O  → Hg 22 +  → Ag ( s )  → Fe 2 +  → H 2 O2  → 2I −  → Cu ( s )  → Cu ( s )  → Ag ( s ) + Cl −  → Ag ( S ) + Br −  → H 2 (g)  → Pb( s )  → Sn( s )  → Ni ( s )  → Fe( s )  → Cr ( s )  → Zn( s )  → H 2 ( s ) + 2OH − (aq )  → Al ( s )  → Mg ( S )  → Na ( s )  → Ca ( s )  → K (s)  → Li ( s )

Increasing strength of reducing agent

F2 ( g ) + 2e

−

φ

Reaction) Oxidised form + ne–)

1.81 1.78 1.51 1.40 1.36 1.33 1.23 1.23 1.09 0.97 0.92 0.80 0.77 0.68 0.54 0.52 0.34 0.22 0.10 0.00 −0.13 −0.14 −0.25 −0.44 −0.74 −0.76 −0.83 −1.66 −2.36 −2.71 −2.87 −2.93 −3.05

φ

φ

φ

a. A negative E means that the redox couple is a stronger reducing agent than the H + / H 2 couple. b. A positive E means that the Redox couple is a weaker reducing agent than the H + / H 2 couple.

φ

From the electrochemical series (Table 11.5) E value of Cu = +0.34 V and the E value of Ag = +0.80V. Since the reduction potential of Ag is more than that of Cu, this means that silver has greater tendency to get reduced as compared to copper. Thus the reaction Ag + (aq ) + e −  → Ag ( s )

occurs more readily than the reaction Cu 2 + (aq ) + 2e −  → Cu ( s ) Similarly, since reduction potential of copper is less than that of Ag, this means that copper will be oxidized in comparison to Ag thus the reaction Cu ( s )  → Cu 2 + (aq ) + 2e −

Redox Reactions

occurs more readily than +

Ag ( s )  → Ag (aq ) + e

−

Therefore silver will be reduced and copper will be oxidized and the above reaction is not feasible. Rather, the reverse reaction +

2+

Cu ( s ) + 2 Ag (aq )  → Cu (aq ) + 2 Ag ( s ) ccan occur.

11.7 IMPortance of the redox reactIons In huMan actIvIty Redox reaction plays an important role in the human activity, some of them are as follows: 1. Many metal oxides are reduced to metals by using suitable reducing agents. For example, Al2O3 is reduced to aluminium by cathodic reduction in electrolytic cell. Fe2O3 is reduced to iron in blast furnace using coke.

11.29

2. Photosynthesis is used to convert carbon dioxide and water by chlorophyll of green plants in the presence of sunlight to carbohydrate. Chlorophyll 6CO2 ( g ) + 6 H 2 O(l )  → C6 H12 O6 (aq ) + 6O2 ( g ) sunlight

In this case, CO2 is reduced to carbohydrates and water is oxidized to oxygen. The light provides the energy required for the reaction 3. Oxidation of fuels is an important source of energy which satisfies our daily need of life Fuels + O2  → CO2 + H 2 O + energy

In living cells glucose (C6H12O6) is oxidized to CO2 and H2O in the presence of oxygen and energy is released. C6 H12 O6 (aq ) + 6O2 ( g )  → 6CO2 ( g ) + 6 H 2 O(l ) + Energy

4. In fuel cells involving reaction between a fuel and oxidizing agent meet our demand of electrical energy in space capsule.

11.30

Redox Reactions

key PoInts •

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The reaction that involves gain of oxygen or loss of hydrogen or loss of electrons is called oxidation reaction. The reactions that involve loss of oxygen or gain hydrogen or gain of electrons is called reduction reaction Reactions involving both oxidations–reduction processes are called redox reactions and always oxidation and reduction take place together in a redox reaction. In covalent molecules, the more electronegative atom which gets partial negative charge is considered as reduced while the atom which acquires partial positive charge is considered as oxidized. There is competition for release of electrons by different elements. For example, zinc can release electrons easily than copper and hence electrons can transfer from zinc atoms to Cu2+ ions. The list which shows the order of the tendency to release electrons is known as electrochemical series or metal active series. Oxidation number of an element is the residual charge which its atom has or appear to have when all the atoms from the molecule are assumed to be removed as ions by counting the shared electrons with more electronegative atom. The oxidation number of an element in the free state or elementary state is always zero. The oxidation number of monatomic ion is the same as the charge on the ion. In binary compounds of metals and non-metals, the oxidation number of metals is always positive while that of the non-metal is negative. In compounds formed by the combination of non- metallic atoms, the atom with higher electronegative is given negative oxidation number. Hydrogen is assigned oxidation number +1 in all its compounds except in metal hydrides The oxidation number of fluorine is always −1 in all its compounds. For other halogens, the oxidation number is generally −1, but there are exceptions when these are bonded to a more electronegative halogen atom or oxygen. The oxidation number of oxygen is assigned as −2 in most of its compounds however in peroxides superoxides 1 and in fluorides O2F2 and OF2 it exhibit −1, − , +1 and 2 +2 oxidation number respectively. For neutral molecules, the sum of the oxidation number of all atoms is equal to zero.

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In the case of polyatomic ion, the sum of oxidation numbers of all atoms is equal to the charge on the ion. If a molecule or ion contain more than one atoms of the same element with different oxidation numbers the oxidation number of the element is taken as average oxidation number. The oxidation number of an element cannot exceed its group number However, if the calculated oxidation number of an element is exceeding its group number it may contain peroxybonds. Oxidation state is another way of expressing oxidation number. Unlike oxidation number, valency of an element is only a number and do not carry any sign and it is always a whole number. Oxidizing agent is a substance which increases the oxidation number of other substance in a chemical reaction. It accepts electrons. Its oxidation number decreases during redox reaction. Reducing agent is a substance which decreases the oxidation number of other substance in a chemical reaction. It donates electrons. Its oxidation number increases during redox reaction.

naming of Inorganic compounds •

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Binary compounds were named by writing the name of electropositive element first and the electronegative element later with a suffix added to the name of electronegative element e.g., NaCl sodium chloride; Ca3N2, calcium nitride etc. If a metal exhibitt variable valency, the compounds in the lower oxidation state are called as “ous” compounds while in higher oxidation state are called as “ic” compounds e.g., FeSO4: Ferrous sulphate Cu2O: Cuprous oxide Fe2(SO4)3: Ferric sulphate CuO: Cupric oxide

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According to Stock notation, the oxidation number or oxidation state of the metal should be written as Roman numeral in parenthesis after the name of the metal in the name of the compound e.g., copper (II) sulphate; iron (III) chloride. Oxidation states are sometimes written for ions that − do not contain transition metals e.g., BrO3 : bromate (v) ion. If the name of a compound ends “ate”, it always means that the substance contains oxygen e.g., bromate (V), dichromate (VI); mangnate (VII) ion.

Redox Reactions

types of redox reactions •

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Combination reactions are those in which elements or compounds combine to form new substances. For a combination reaction to be a redox reaction, either one or both the reactants should be in the elemental from. If a compound breakdowns into two or more components, it is called decomposition reaction. If one of the product in decomposition reaction is an element, it is redox reaction and it is an opposite process of combination. If an ion/atom in a compound is replaced by an ion/ atom of another element it is called displacement reaction. If a metal in a compounds is displaced by another metal, a better reducing metal can displace a weaker reducing metal. Non-metal displacement reactions involve displacement of hydrogen by metal like Na, Ca from cold water; Mg, Fe from steam and by metals from acids. More electronegative non-metal displaces less electronegative non-metal from their compounds e.g., F2 can displace O 2 from water, F2 can displace Cl2 Br2, or I2 ; Cl2 can displace Br2 or I2 and Br2 can displace I2 from the respective chlorides,bromides or iodides. A disproportion reaction is the one in which an intermediate oxidation state is converted to a mixture of a higher and a lower oxidation states. Comproportion reaction is reverse of disproportion reaction. Two species with the same element in two different oxidation states a single product in which the element is in an intermediate oxidation state. If a compound (water) is oxidized by air in the presence of certain substances (phosphorous, metals such as Zn, Pd and certain unsaturated compounds), it is called autoxidation reaction. The substances in the presence of which water is oxidized are called activators. The intermediate combining with oxygen are called autoxidator. If the oxidation of one compound induces the oxidation of another compound it is called induced oxidation reaction. If the molecules of substance is oxidized while the molecules of another substance are reduced in a redox reaction, it is called intermolecular redox reaction. If one atom of the same molecule is reduced and another atom of the same molecule is oxidized in a redox reaction, it is called intramolecular redox reaction.

11.31

Balancing of redox reactions •

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The equation where the equality of the number of atoms is signified is called balanced chemical equation. The numerical placed before the formulae of the reactants and the products in order to balance a chemical equation are known as stoichiometric coefficients. The steps involved in balancing of redox-reactions by oxidation number method are as follows: (i) Assign oxidation number to each element on both sides of the given reaction. Identify the element that is oxidized and reduced. (ii) Write the oxidation number of the elements participating in the reaction. (iii) Determine the increase and decrease of oxidation number per atom. Multiply the increase and decrease of ON with number of atoms undergoing change. (iv) Equalize the increase in ON and decrease in ON on the reactant side by multiplying the respective formulae with suitable integers. (v) Balance equation w.r.t atoms other than ‘O’ and ‘H’ atoms (vi) Balance ‘O’ atoms by adding equal number of water molecules to the side falling short of ‘O’ atoms (vii) H atoms are balanced (in case of ionic equations) depending upon the medium in the following manner. (a) For acid medium add proper number of H+ ions to the side falling short of H atoms (b) For basic medium add proper number of H2O molecules to the side falling short of H atoms and equal number of OH– ions to the other side. Various steps involved in the balancing of redox reactions by ion-electron method are as follows The ion electron method is also known as half-reaction method (i) Find the elements whose oxidation number are changed. Choose the substance which acts as an oxidizing agent and one that acts as reducing agent. (ii) Separate the complete equation into two half reactions one for the change undergone by oxidizing agent and one that acts as reducing agent (iii) Balance the half reactions by following steps (a) Balance all atoms other than H and O. (b) Calculate the oxidation number on both sides of the equation. Add electrons to whichever side is necessary to make up the difference. (c) Balance the half equation so that both sides get the same charge. (d) Add water molecules to complete the balancing of the equation (iv) Add the two balanced half equations. Multiply one or both half equations by suitable number so that on adding the two equations the electrons are balanced.

11.32

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Redox Reactions

If H+ ions appear on either side of a redox reaction it takes place in acid medium. If OH– ions appear on either side of the equation then the reaction is taking place in basic medium. If neither H+ nor OH– ions are present the reaction occurs in neutral medium. For balancing Redox reactions involving acidic and basic media the method has to be modified as (i) For acidic medium add proper number of H+ ions to the side falling short of H atoms. (ii) For basic medium add proper number of H2O molecules to the side falling short of H atoms and equal number of OH– to the other side. Ion-electron method applies only to those reactions in which ions participate. It gives importance in expressing the redox reaction in terms of ions or an ionic equation. The half equation method gives the information about the number of electrons participating in the redox reaction. Ion-electron method is not useful for redox reactions which take place in the molecular state. For such reactions oxidation number method can be used. The strength of the solution of reductant/oxidant can be determined by the titration method using redox sensitive indicator.

limitations of concept of oxidation number •

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Certain Reactions which are considered as redox reactions according to classical approach cannot be explained by applying the oxidation number approach e.g. SiCl4 (l ) + LiAlH 4 ( s ) → SiH 4 ( g ) + LiCl ( s ) + AlCl3 ( s )

•

According to classical approach silicon is reduced (since it gained hydrogen) while lithium and aluminium are oxidized (since they lost hydrogen) but their oxidation numbers are not changed. Several reactions cannot be interpreted as redox reactions according to any of the three classical electronic or oxidation number approach. CH 3 CNO( g ) + H 2 O(l ) → CH 3 NH 2 ( g ) + CO2 ( g )

redox reactions and electrode Potentials • •

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Electric current is considered as flow of electrons or flow of electric charge through a conducting medium. If electron transfer in a redox reaction takes place through a copper wire instead directly from reductant to oxidant electric current is generated. The solutions of oxidant and reductant are taken separately in two beakers in which platinum electrodes are

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dipped and connected by a copper wire through a galvanometer. Deflection in galvanometer indicates the flow of current. Electrons flow from the electrode dipped in reductant solution to the electrode dipped in oxidant solution. When a solid substance is kept in water, it tends to pass into solution at a constant temperature and this tendency is called solution pressure. The pressure exerted by dissolved solid substance is called its osmotic pressure. When an electrode is dipped into a solution of its own the metal tends to pass into solution (solution pressure P) while the ions in solution tend to deposit on the electrode (osmotic pressure π) the p is opposed by π When P > π metal continue to dissolve until the accumulated charge opposes further dissolution. So the metal becomes negatively charged while the solution is positive e.g., Zn, Cd alkali metals. When P = π no potential will be set up between metal and the solution. Such systems are known as null electrodes. When P < π the metal ions from solution deposit on the metal electrode until the accumulated charge opposes the further deposition. At equilibrium metal acquires positive charge while the solution acquires negative charge e.g., Cu, Hg, Au, etc. Non-metals also follow the solution pressure theory. In such cases. (i) If P > π the electrode becomes positively charged while the solution contain negative charges. (ii) If P = π no potential will be set up. (iii) If P < π the electrode will be negatively charged while the solution acquire positive charge. An electrical double layer known as Helmholtz double layer is set up at the interface of metal and solution. The potential developed at the interface of an element in contact with its own ions is called single electrode potential. Single electrode potential cannot be measured or determined experimentally. The potential difference between two single electrodes is measured using potentiometer. The electrode of known potential used to determine the potential of another electrode in the potentiometric method is called reference electrode. The potential of hydrogen electrode when hydrogen gas at one atmosphere pressure in contact with H+ ions of unit activity in solution is assumed as zero. This is known as normal hydrogen electrode (NHE) or standard hydrogen electrode (SHE). Hydrogen electrode is a reversible electrode.

Redox Reactions

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Any element in contact with its own ions as called single electrode. If two single electrodes are connected a galvanic cell will be formed in which electrons flow from oxidation electrode to reduction electrode due to potential difference. In this galvanic cell, each single electrode is called halfcell. The single electrode potential measured at 25°C when an electrode is placed in a solution of its own ions of unit activity (or concentration) is called standard electrode potential. If the element is a gas it should be at 1 atm pressure. On combining an electrode with SHE, if reduction take place at the electrode the potential of the electrode is called as reduction potential and it is given +ve sign as reduction potential or –ve sign as oxidation potential. On combining an electrode with SHE, if oxidation take place at the electrode the potential of the electrode is called as oxidation potential and it is given +ve sign as oxidation potential or –ve sign as reduction potential. SRP (Standard reduction potential) = −SOP (standard oxidation potential) SOP (standard oxidation potential) = −SRP (Standard reduction potential) According to the present convention the half cell reactions are always written as reduction half reactions and their potentials are represented as reduction potentials The oxidized and reduced form of a substance taking part in an oxidation or reduction half – reaction is called redox couple and its standard electrode potential is written as Eox/red The difference in the electrode potentials of the electrodes present in the cell under any conditions is known as the potential difference. The electromotive force (emf) is the potential generated by a cell when it draws no current and it is under conditions of zero electron flow. Potential difference is responsible for the flow of steady current in the cell but not related to e m f SHE is called primary standard reference electrode but since its construction is difficult some other reference electrodes called secondary reference electrodes are used. A saturated calomel electrode is a secondary reference electrode. It is Hg/Hg2Cl2(s):KCl (saturated) and its potential is −0.2422V. The standard electrode potentials are influenced by the sublimation, ionization energies of the element and the hydration energies of their ions.

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11.33

Though the sublimation and ionization enthalpies of lithium are high, very high heat of hydration energy of its ions leads to easy formation of ions from the metal. So it will act as a strongest reducing agent not only among alkali metals but among all reducing agents. While measuring the single electrode potentials the two half cells are connected by using a salt bridge. A typical salt bridge is made by dissolving potassium chloride in agar. The charge on the potassium and chloride ions provides electrical contact between the two half-cells. When salt bridge is used sufficient number of Cl– ions migrates from salt bridge to anode half cell and neutralize the excessive positive charge. Similarly, to neutralize the excessive negative charge in cathode half-cell sufficient number of K+ ions migrates from the salt bridge. When two salt solutions of different concentrations are in contact with one another due to the difference in the rates of diffusion of the two ions, the more dilute solution acquires a charge corresponding to that of the faster moving ion. Thus, the junction between the two solution contribute a potential called liquid junction potential to the cell. Salt bridge eliminates the liquid junction potential by neutralizing the charges. Generally, potassium chloride is used in salt-bridge because both K+ and Cl– ions have same transport number. Transport number is the fraction of current carried by the ion As per the IUPAC nomenclature the single electrodes are represented as follows: (i) Metal ion symbol is first written followed by metal with a vertical line between the two (a) Zinc electrode Zn2+ | Zn (b) Copper electrode Cu2+ | Cu (ii) For non-metal electrodes the non-metal symbol (X2) is first written followed by non-metal ion symbol (Xn–) with concentration and a vertical line between two phases. Platinum symbol Pt is written before non-metal symbol. (a) Hydrogen electrode Pt|H2|H+ (b) Chlorine electrode Pt|Cl2|Cl– (c) Oxygen electrode Pt|O2|OH– In the IUPAC system, the standard potential given are standard reduction potentials. The arrangement of various electrodes in the order of increasing reduction potential values is called electrochemical series or electromotive series. The elements which are above hydrogen in emf series cannot reduce H+ ions and liberate H2 gas from acids.

11.34

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Redox Reactions

The elements which are below hydrogen in emf series can reduce H+ ions and they cannot displace hydrogen from acids. The metals at the top are strongly electropositive and act as strong reducing agents. An element above in the e.m.f series can displace an element below to it from its salt solution but not vice versa.

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Hydroxides of the metal in the upper part of the series are strongly basic and the salts of these metals do not undergo hydrolysis. Hydroxides of metals at the lower part of the series are weakly basic and the salts of these metals undergo hydrolysis.

Redox Reactions

11.35

PractIce exercIse Multiple choice Questions with only one answer level I 1. In the unbalanced reaction: Al + KMnO4 + H2SO4 → KHSO4 + Al2(SO4)3 + MnSO4 + H2O If the stoichiometric coefficients of Al, H2SO4, MnSO4 and H2O are w, x, y and z respectively, the numerical x z value of  −  will be  y w (a) 2.0 (b) 2.1 (c) 1.2 (d) 2.4 2. The oxidation number of carbon is zero in (I) HCHO (II) CH2Cl2 (III) C6H12O6 (IV) C12H22O11 (a) I, II Only (b) I, II, III Only (c) All (d) None 3. Which of the following have been arranged in order of decreasing oxidation number of sulphur? (i) H2S2O7>Na2S4O6>Na2S2O3>S8 (ii) SO2+ > SO42– > SO32– >HSO4 (iii) H2SO5> H2SO3> SCl2> H2S (iv) H2SO4 > SO2>H2S>H2S2O8 The correct set is (a) I, IV (b) I, III (c) II, IV (d) II, III 4. For the redox reaction, Zn + NO3− → Zn 2 + + NH 4+ in basic medium, the coefficients of Zn, NO3− and OH– in the balanced equation are respectively (a) 4,1,7 (b) 7,4,1 (c) 4,1,10 (d) 1,4,10 5. Oxidation states of all atoms present in potassium tri-iodide are 1 1 1 1 (a) + , , , 3 3 3 3

1 1 1 (b) +1, + , − , − 3 3 3

(c) +1,−1,0,0 (d) +1,−1,−1,−1 6. A solution containing Cu +2 , C2 O4 −2 ions. The solution requires 22.6 mL of 0.02 M KMnO4 in the presence of H2SO4 for oxidation. The resulting solution is neutralized and add excess KI. The iodine is liberated according to this equation Cu2+ + I– → Cu2I2 +I2

The iodine liberated requires 11.3 mL of 0.05 M Na2S2O3 solutions. The molar ratio of Cu +2 , C2 O4 −2 in solution is (a) 1:1 (b) 1:2 (c) 2:1 (d) 1:3 7. In a reaction FeS2 is oxidised by O2 to Fe2O3 and SO2. If the equivalents of O2 consumed are ‘y’ then the equivalents of FeS2 consumed and moles of Fe2O3 and SO2 produced respectively (a) y,

y 2y , 22 11

y y (b) y, , 2 5

(c) y, y, y (d) None of these 8. A solution of 0.1 M KMnO4 is used for the reaction S 2 O3 −2 + 2 MnO4− + H 2 O → MnO2 + SO4 −2 + OH − . The volume of KMnO4 required to react 0.158 g of Na2S2O3 is (MW = 158) (a) 13.33 mL (b) 6.66 mL (c) 3.33 mL (d) 26.67 mL 9.

10.

11.

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H+ C2 H 2 + MnO − 4 → Mn 2 + + CO2

The number of H+ ions in the balanced equation of the above redox reaction is (a) 8 (b) 6 (c) 4 (d) 2 What are the oxidation numbers of nitrogen in NH4NO3? (a) +3, −5 (b) −3, +5 (c) +3, −6 (d) +2, +2 Which one of the following elements show different oxidation state? (a) sodium (b) fluorine (c) chlorine (d) potassium How many moles of acidified FeSO4 can be completely oxidized by one mole of KMnO4? (a) 10 (b) 5 (c) 6 (d) 2 MnO2+4HCl → MnCl2+H2O+Cl2. The equivalent weight of HCl is (a) 73 (b) 36.5 (c) 18.25 (d) None In a reaction, 4 moles of electrons are transferred to one mole of HNO3. The possible product obtained due to reaction (a) 20.5 mole of N2 (b) 1 moles of NO2 (c) 0.5 mole of N2O (d) 1 mole of NH3

11.36

Redox Reactions

15. When ammonium nitrate is gently heated, an oxide of nitrogen formed. What is the oxidation state of nitrogen in this oxide? (a) +4 (b) +5 (c) +2 (d) +1 16. ClO3− + 6 H + + X → Cl − + 3H 2 O . The value of X is (a) 4e– (b) 5e– – (c) 6e (d) 7e– 17. When thiosulphate ion is oxidized by iodine, which one of the following ions is produced? (b) SO4−2 (a) SO3−2 (c) S 4 O6−2

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(d) S 4 O6−2

18. The oxidation number of carbon in CH2O is (a) −2 (b) +2 (c) 0 (d) +4 19. If three electrons are lost by an ion M+3. Its final oxidation number would be (a) 0 (b) +2 (c) +5 (d) +6 20. Oxidation number of phosphorous in Mg2P2O7 is (a) +1 (b) +3 (c) +5 (d) +7 21. Which one show +6 oxidation state (a) MnO–4 (b) Cr (CN)6–2 –2 (c) NiF6 (d) CrO2CI2 22. In acid solutions, the reaction MnO4– → Mn+2, MnO4– in this is (a) oxidized by 3 electrons. (b) reduced by 3 electrons. (c) oxidized by 5 electrons. (d) reduced by 5 electrons. 23. In which of the following compounds, transition metal is in oxidation state of zero? (a) [Co ( NH 3 )6 ]CI 2 (b) [ Fe ( H 2 O )5 SO4 ] 24.

28.

(c) [ Ni ( CO )4 ] (d) [ Fe ( H 2 O )6 ](OH ) 2 The oxidation number of sulphur in S8, S2 F2 and H2S respectively are (a) 0, +1 and −2 (b) +2, +1 and −2 (c) 0, +1 and +2 (d) −2, +1 and −2 The number of electrons to balance the following equation NO3− + 4 H + + e − → 2 H 2 + NO is (a) 5 (b) 4 (c) 3 (d) 2 C2 H 6 ( g ) + NO2 → CO2 ( g ) + H 2 O ( I ) . In this equation the ratio of the coefficients of CO2 and H2O is (a) 1:1 (b) 2:3 (c) 3;2 (d) 1:3 The reaction, 3ClO − ( aq ) → ClO3− ( aq ) + 2Cl–(aq) is an example of (a) oxidation reaction (b) auto oxidation

31.

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(c) disproportionation reaction (d) decomposition reaction In a reaction between zinc and iodine, zinc iodide is formed. What is being oxidized? (a) Zinc ions (b) Iodide ions (c) Zinc atom (d) Iodine A reducing agent is a substance which can (a) accept electrons (b) donate electrons (c) accept protons (d) donate protons In the reaction H2O2 + Na2CO3 → Na2O2 + CO2 + H2O the substance undergoing oxidation is (a) H2O2 (b) Na2O2 (c) Na2CO3 (d) None of these Carbon is in the lowest oxidation state in (a) CH4 (b) CCl4 (c) CF4 (d) CO2 The oxidation state of nitrogen in N3H is 1 (b) +3 (a) + 2 1 (c) −1 (d) − 3 Oxidation number of iron in potassium ferrocyanide K 4 [ Fe ( CN ) 6 ] is (a) +2 (b) +3 (c) +4 (d) +1 Oxidation state of oxygen atom in potassium peroxide is 1 (a) Zero (b) − 2 (c) −1 (d) −2

35. Chlorine is in +1 oxidation state in (a) HCl (b) HClO4 (c) ICl (d) Cl2O 36. In the balanced equation 5H2O2 + XClO2 + 2OH– → XCl– + YO2 + 6H2O. The reaction is balanced if (a) X = 5, Y = 2 (b) X = 2, Y = 5 (c) X = 4, Y = 10 (d) X = 5, Y = 5 37. In the chemical reaction Ag 2 O + H 2 O + 2e − → 2OH − + 2Ag (a) Water is oxidized (b) Silver is oxidized (c) Silver is reduced (d) Hydrogen is reduced 38. The compound which could act both as oxidizing as well reducing agent is (a) SO2 (b) MnO2 (c) Al2O3 (d) CrO3 39. The coefficients of I–, IO3– and H+ in the redox reaction I − + IO3− H + → I 2 + H 2 O in the balanced form respectively are (a) 5, 1, 6 (b) 1, 5, 6 (c) 6, 1, 5 (d) 5, 6, 1

Redox Reactions

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φ

φ

51. The position of some metals in the electrochemical series in decreasing electropositive character is given as Mg>Al>Zn>Cu>Ag. What will happen if a copper spoon is used to stir a solution of aluminium nitrate? (a) The spoon will get coated with aluminium. (b) The alloy of aluminium and copper is formed. (c) The solution becomes blue. (d) There is no reaction. 52. Which of the following statement is wrong? (a) F2 is the strongest oxidizing agent as its reduction potential is high. (b) Li is the weakest reducing agent as its reduction potential is low. (c) Li is the strongest reducing agent as its oxidation potential is high. (d) F– ion does not show reducing property. 53. Out of Cu, Ag, Fe and Zn the metal which can displace all others from their salt solution is: (a) Ag (b) Cu (c) Fe (d) Zn 54. The standard reduction potential at 25°C of Li+/ Li, Ba2+/Ba, Na+/Na and Mg2+/Mg are −3.05, −2.73, −2.71, and −2.37 volt respectively. Which one of the following is the strongest oxidizing agent? (a) Na+ (b) Li+ +2 (c) Ba (d) Mg2+ 55. Which of the following metal does not react with CuSO4 solution? (a) Ag (b) Mg (c) Zn (d) Sn 56. The following facts are available: 2 A− + B2 → 2 B − + A2 2C − + B2 → No reaction 2 D − + A2 → 2 A + D2 Which of the following statement is correct? (a) E C–/C2>E B–/B2>E A–/A2>E D–/D2 (b) E C–/C2Y>X, Then: (a) Y will oxidize X and not Z (b) Y will oxidize Z and not X (c) Y will oxidize both Z and X (d) Y will reduce both X and Z

Multiple choice Questions with one or More than one answer 1. 15 g of KMnO4 in acidic medium is equal to (a) 0.09493 moles (b) 0.4446 g equivalents (c) 9.492 L of 0.05 N KMnO4 (d) 10 mL of 0.05 M KMnO4 2. Choose the correct option(s). (a) KClO4 and KMnO4 are isomorphous in nature. (b) Equivalent weight of Fe0.9 O when it is converted to FeO is 332. (At wt. of Fe = 56) (c) Molar volume of water at STP is 22.4 litre. (d) When ‘w’ g Na2SO4.10H2O is added in X g of water, molality of SO2–4 is

1000 . 332X

3. Which of the following reactions represent disproportionation? (a) F2 + NaOH → NaF + H 2 O + O2 (b) 2Cu + → Cu +2 + Cu ( s ) (c) 2 H 2 O2 ( l ) → 2 H 2 O ( l ) + O2 ( g ) (d) Cl2 + H 2 O → HCI + HOCI

4. In which of the molecules, oxygen shows integer oxidation number? (a) Fe0.93 O1.00 (b) F2O2 (c) Fe3O4 (d) P4O10 5. Which shows oxidation? (a) A+ n1 → A+ n2 + (n2 − n1 )e − ( n2 > n1 ) (b) A + ne − → A− n (c) A2 + 2e − → 2 A− (d) A− n1 + ( n2 − n1 ) e − → A− n2 ( n1 > n2 ) 6. For the balanced redox reaction aNO3− + bAs2 S3 + 4 H 2 O → xAsO43− + yNO + zSO4 2 − + 8 H + Which of the following statements are correct? (a) Equivalent weight of As2S3 is M/24 where ‘M’ is molecular weight of As2S3 (b) The value of a:b = 8:1 a + 2b is 1 (c) The value of x+ y (d) The value of

z−x is 1 3

7. An oxidising agent in a chemical reaction under goes. (a) Gain of electrons (b) Loss of electrons (c) A decrease in oxidation number. (d) An increase in oxidation number. 8. The oxidation number of Mn is +2 in (a) Manganese oxide (b) Manganese chloride (c) Manganese sulphate (d) Potassium permanganate 9. Oxygen has an oxidation state of −1 in (a) H2O2 (b) BaO2 (c) OF2 (d) H2S2O8 10. Consider the redox reaction, S 22O32 − + I 2 → S 4 O6 2 − + 2 I − 2S (a) 2S2O32– gets reduced to S4O62– (b) 2S2O32– get oxidized to S4O62– (c) I2 gets reduced to I– (d) I2 gets oxidized to I– 11. Which of the following reactions do not involve oxidation reductions? (a) 2 Rb + 2 H 2 O → 2 RbOH + H 2 (b) 2CuI 2 → 2CuI + I 2 (c) NH 4 Cl + NaOH → NaCl + NH 3 + H 2 O (d) 4 KCN + Fe (CN 2 ) → K 4 [ Fe (CN )6 ] 12. Oxidation numbers of carbon is correctly given for Compound O. No (a) HN = C +2 (b) H−C ≡ N +4 (c) CCl4 +4 (d) C6H12O6 0

Redox Reactions

13. When Cl2 is passed through NaOH in cold, the oxidation number of Cl changes from (a) 0 to −1 (b) 0 to +1 (c) 0 to −2 (d) 0 to +2

comprehensive type Questions Passage I Redox reactions are those in which oxidation and reduction take place simultaneously. Oxidizing agent can gain electron where as reducing agent can lose electron easily. The oxidation state of any element can never be in fraction. If oxidation number of any element comes out be in fraction, it is an average oxidation number of that element which is present in different oxidation states. 1. The oxidation state of Fe3O4 is (a) 2 and 3 (b) 8/3 (c) 2 (d) 3 2. 1 N

3

N-H N 2

In this compound HN3 (hydrazoic acid), oxidation state of N 1 , N 2 and N 3 are (a) 0, 0, 3 (b) 0, 0, −1 (c) 1, 1, −3 (d) −3, −3, −3 3. Equivalent weight of chlorine molecule in the equation 3Cl2 + 6NaOH → 5 NaCl + NaClO3 + 3H 2 O (a) 42.6 (b) 35.5 (c) 59.1 (d) 71 Passage II In molecules containing covalent and coordinate bonds, the rules assigned to calculate oxidation number of atoms are• For each covalent bond between dissimilar atoms, the less electronegative element is assigned the oxidation number of +1 whereas the more electronegative atom is assigned the oxidation number of -1. • For co-ordinate bond between similar or dissimilar atoms if the donor atom, is less electronegative than the acceptor atom, the donor atom is assigned the oxidation number of +2 while the acceptor atom assigned the oxidation number of −2. • In contrast to the above statement, if the donor atom is more electronegative than the acceptor atom, the contribution of the co-ordinate bond is neglected. 1. On the basis of structure, the oxidation of two Cl atoms in CaOCl2 respectively are (a) –1 and +1 (b) +2 and −2 (c) −2 and +2 (d) −1 and +3

11.39

2. The oxidation number of sulphur in Na2S4O6 is (a) 1.5 (b) 2.5 (c) 3 (d) 2 3. The oxidation number of S in H2S2O8 is (a) +2 (b) +4 (c) +6 (d)+7 4. The oxidation number of iron in Fe3O4 is (a) +2 (b) +3 (c) 0 (d) −8/3 Passage III Valency is the combining capacity of an element. It is defined as the number of hydrogen atoms or double the number of oxygen atoms with which an atom of the element combines, whereas the oxidation number is the charge which an atom has or appears to have when present in the combined, valency of an element can neither be zero nor fractional whereas oxidation number of an element can be zero or fractional. 1. A metal ion M3+ loses 3 electrons, its oxidation number will be (a) +3 (b) +6 (c) 0 (d) −3 2. In which of the following pairs, there is greatest difference in the oxidation number of the underlined elements? (a) NO2 and N2O4 (b) P2O5 and P4O10 (c) N2O and NO (d) SO2 and SO3 3. The oxidation number and covalency of sulphur in sulphur molecule (S8) is (a) 0 and 2 (b) +6 and 8 (c) 0 and 8 (d) +6 and 2 4. The oxidation number and valency of C in sucrose (C12H22O11) is (a) +4 and 4 (b) +3 and 0 (c) +2 and 4 (d) 0 and 4

Match the columns type Questions 1. Match the following: Column I (Reaction) (a) (b) (c) (d)

NH 3 → NO3− Fe2 S3 → 2 FeSO4 + SO2 KMnO4 in acidic medium CuS → CuSO4

Column II (Equivalent wt.) (p) (q) (r) (s)

M/20 M/5 M/8 M

11.40

Redox Reactions

2. Match the reactions in Column I with nature of the reactions/type of the products in Column II. Column I (a) (b) (c) (d)

+3

Column II +

Sb → SbO CrO4 −2 → Cr2 O7 −2 MnO4 −2 → MnO4 − B4 O7 −2 → B (OH )3

(p) (q) (r) (s)

Acidification Heating Hydrolysis Oxidation

assertion (a) and reason (r) type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): In a redox reaction, the oxidation number of oxidant increases, while that of reductant decreases. Reason (R): The reactant which undergoes oxidation is called reductant and the reactant which undergoes reduction is called oxidant. 2. Assertion (A): Cu+ ion does not exist in solution. Reason (R): Cu+ ion undergoes disproportionation in aqueous solution. 3. Assertion (A): The oxidation number of phosphorous in P4 is zero. Reason (R): The oxidation number of an element in the free state is zero. 4. Assertion (A): Amount KMnO4 used against a particular titrate is more with H2SO4 than with HNO3. Reason (R): HNO3 is a good oxidizing agent like KMnO4. Hence it will oxidize the substance to be oxidized by KMnO4. 5. Assertion (A): Oxidation number of carbon in CH2O is zero. Reason (R): CH2O is formaldehyde. It is a covalent compound. 6. Assertion (A): White P reacts with NaOH, the products are PH3. NaH2PO2. The reaction is an example of disproportination reaction. Reason (R): In disproportionation reaction one substance acts as oxidising and reducing agent.

7. Assertion (A): HNO2 acts as an oxidising as well as reducing agent. Reason (R): The oxidation number of nitrogen remains same. 8. Assertion (A): Oxidation is the process in which electrons are lost. Reason (R): Reduction is the process in which electrons are gained. 9. Assertion (A): Spectator ions are the species that are present in the solution but do not take part in the reaction. Reason (R): The phenomenon of formation of H2O2 by the oxidation of H2O is known as autoxidation. 10. Assertion (A): White P reacts with caustic soda; the products are PH3 and NaH2PO2. This reaction is an example of disproportionation reaction. Reason: In disproportionation reaction the same substance may be simultaneously act as an oxidising agent and as a reducing agent. 11. Assertion (A): Fluorine acts as a strongest oxidising agent. Reason (R): Fluorine is most electronegative. 12. Assertion (A): In NH4NO3, the oxidation number of the two N atoms is −3 and +5 respectively. Reason (R): One N atom is present in the cation while the other is present in the anion. 13. Assertion (A): Bromide ion is serving as a reducing agent in a reaction. 2MnO–4 (aq) + Br– (aq) + H2O(l) → 2MnO2 (aq) + BrO–3 (aq) + 2HO–(aq) Reason (R): Oxidation number of Br increases from −1 to 5. 14. Assertion (A): Oxidation number of Ni in [ Ni ( CO )4 ] is Zero. Reason (R): Nickel is bonded to neutral ligand, carbonyl.

Integer type Questions 1. What will be the n-factor of the reactant in the following reaction? ∆ → N 2 + Cr2 O3 + 4 H 2 O ( NH 4 )2 Cr2O7  2. Valency factor of HNO3 in the following reaction is X/5. Then the value of x is Zn + HNO3 → Zn ( NO3 )2 + N 2 O + H 2 O

Redox Reactions

Previous years’ IIt Questions 1. Amongst the following, identify the species with atom in +6 oxidation state: 3− (a) MnO4− (b) Cr ( CN ) 6 2− (c) NiF6 (d) CrO2 CI 2 (2000) 2. The reaction 3ClO– (aq)  → ClO3– (aq) + 2Cl– (aq) is an example of: (a) oxidation reaction (b) reduction reaction (c) disproportionation reaction (d) decomposition reaction (2001) 3. The pair of compounds in which both the metals are in the highest possible oxidation state is: (a) [ Fe ( CN )6 ]3− ,[Co ( CN )6 ]3− (b) CrO2 Cl2 , MnO4 − (c) TiO3 , MnO2 (d) [Co ( CN )6 ]3− , MnO3 (2004)

11.41

4. Match the following Column I − 2

Column II 2− 2

(a) O → O2 + O (b) CrO42 − + H + → (c) MnO4 − + NO2− + H + → (d) NO3− + H 2 SO4 + Fe 2 + →

(p) redox reaction (q) one of the products has trigonal planar structure (r) dimeric bridged tetrahedral metal ion (s) disproportionation

(2007) 5. Integer Type Questions 1. The difference in the oxidation numbers of the two types of sulphur atoms in Na2S4O6 is: (2011)

11.42

Redox Reactions

answer keys Multiple choice Questions with only one answer level I 1. b 2. c 3. b 4. c 5. c 6. b 7. a 8. d 9. b 10. b 11. c 12. b 13. a

14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26.

c d c c c d c d d c a d b

27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39.

c c b d a d a c d b c a a

40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52.

b a c c b b c c c b d d b

53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63.

d d a a a a a b c a a

Multiple choice Questions with one or More than one answer 1. 2. 3. 4. 5. 6. 7.

a, c a, b b, c, d a, b, c, d a, d c, d a, c

8. 9. 10. 11. 12. 12. 12.

a, b, c a, b, d b, c c, d c, d c, d c, d

13. a, b

comprehensive type Questions Passage I 1. a

2. b

3. a

2. b

3. c

4. d

2. d

3. a

4. d

Passage II 1. a Passage III 1. b

Matching type Questions 1. (a) r 2. (a) r

(b) p (b) p

(c) q (c) s

(d) r (d) r

assertion (a) and reason (r) type Questions 1. d 2. a 3. a

4. a 5. b 6. a

7. c 8. b 9. b

10. a 11. a 12. a

13. a 14. b

Integer type Questions 1. 6

2. 4

Previous years’ IIt Questions 1. 4 2. 3 3. 2

4. a-ps, b–r c-p q, d-p 5. 5

Redox Reactions

11.43

hInts and solutIons hints to Problems for Practice 1. (i) Oxidation number of hydrogen is +1. Let the oxidation number of sulphur be x. +1 x H2 S ; 1× 2 + x = 0 ∴ x = −2 1× 2 x Oxidation number of S in H2S is −2. (ii) Oxidation number of hydrogen is +1 and that of oxygen is −2. Let the oxidation number of S be x x −2 +1 H2 S O3 (+1 × 2) + x + ( −2 × 3) = 0 ( −1 × 2) x ( −2 × 3) ∴ x = +4 Oxidation number of S in H2SO3 = + 4 (iii) Oxidation number of Na is +1 and that of oxygen is −2. Let the oxidation of S be x +1 x −2 Na2 S2 O3 ; (+1 × 2) + ( x × 2) + ( −2 × 3) = 0 ( +1 × 2) ( x × 2) ( −2 × 3) ∴ x = 2 Oxidation number of S in Na2S2O3 is +2 (iv) The oxidation number of oxygen is −2 and let the oxidation number of sulphur be x  x −2   S 2 O7 

( x × 2)

2−

;

( x × 2) + ( −2 × 7) = −2 ∴x = 6

( −2 × 7)

Oxidation number of S in S2O72 − is + 6 (v) Oxidation number of hydrogen is +1 and that of oxygen is −2 let the oxidation number of S be x x −2 +1 H 2 S O4 ; (+1 × 2) + x + ( −2 × 4) = 0 ( +1 × 2) x ( −2 × 4) ∴ x == +6 Oxidation number of S in H2SO4 is + 6 (vi) S 2 O42 − The oxidation number of oxygen is −2 and let the oxidation number of sulphur be x

(ii) BrF3

x = +3 (iii) CH 4

−2

2−

  O4   S2   x×2 −2×4

( x × 2) + ( −2 × 4) = −2 ∴ x = +3

∴ oxidation number of s in S 2 O42 − is + 3

2. (i)

ClO3−

x + ( −2 × 3) = −1 x − 6 = −1 or x = +5

x + (1 × 4) = 0 x = −4

(iv) C6 H12 O6 (6 × x) + (1 × 12) + ( −2 × 6) = 0 x=0 (v) Na2 B4 O7 (1 × 2) + ( x × 4) + ( −2 × 7) = 0 x = +3

3. In all these compounds, the ON (oxidation number) of oxygen is −2 and that of hydrogen is +1. So oxidation number of carbon assuming x can be calculated as (i ) C2 H 2 : 2 x + (2 × 1) = 0 ∴ x = 1 (ii ) CO2 : x − (2 × 2) = 0 ∴ x + 4 (iii ) C2 H 6 :2 x + (1 × 6) = 0 ∴ x = −3 (iv) CH 3 OH : x + (1 × 4) − 2 ∴ x = −2 (v) HCOOH : x + (2 × 1) + ( −2 × 2) = 0 ∴ x = +2 (vi ) CH 2 O : x + (1 × 2) + ( −2) = 0 ∴ x = 0 x −2

4. (i ) C O2 x( −2 × 2) x

x + ( −2 × 2) = 0 or x = +4 oxidation number of carbon is + 4

−2

(ii ) Si O2 x( −2 × 2)

x + ( −2 × 2) = 0 or x = +4 oxidation number of silicon is + 4

+2 x −2

(iii ) Pb S O 4 (+2) + x + ( −2 × 4) = 0 (+2 × 1) x ( −2 × 4) x = +6 Oxidation number of sulphur = +6 x

x

x + ( −1 × 3) = 0

−2

(iv) Cl O4− x( −2 × 4)

x + ( −2 × 4) = −1 or x = +7 oxidation number of Cl = +7

5. If the oxidation number of sulphur is calculated considering the oxidation number of oxygen is −2 we get +8 for sulphur. The oxidation number of an element cannot exceed its group number. So S2O82– contain a peroxide (−O−O−) bond in which the oxidation number of oxygen is −1. For the other oxygen atoms, the oxidation number is −2.

11.44

Redox Reactions

S2 O8 2 x + ( −2 × 6) + ( −1 × 2) = −2 2 x = −2 + 14 = +12 ∴ x = +6 So oxidation number of S in S 2 O82 − is + 6 x

−2

; 3 x + ( −2 × 4) = 0 8 ∴x = 3 2 Oxidation number of Pb = 2 3

6. (i ) Pb3 O 4 3× x

−2×4

−2

x

(ii)

Cr2 O72 −

; 2 x + ( −2 × 7) = −2 ∴ x = +6 Oxidation number of Cr = +6

2×x

(iii)

−2×7

+1

x

−2

(+1 × 2) + x + ( −2 × 4) = 0 ∴ x = +6 oxidation Number of Mn = +6

K 2 Mn O4 ;

+1 × 2

x −2×4

x

+1

+1

x

(iv) N H 4+

; x + (+1 × 4) = +1 x + (1 × 4) ∴ x = −3 Oxidation number of N = −3

(v)

i

(+1× 1) + x + (−1× 4) = 0 x = +3

−1

+1

+5

→ 2 Cl − + ClO3− 3Cl O −  −1

+5

→ Cl − +2 Cl O3− 3Cl O 2  +5

+5 − 2

+4 − 2

−1

+7

→ Cl − +3Cl O 4− 4 Cl O3− 

8. (a) Nitrogen combine with oxygen to form NO. So it is a combination redox reaction. (b) Lead nitrate breaks up into there components. So it is a redox decomposition reaction. (c) Hydrogen of water has been displaced by hydride ion into dihydrogen gas. So it is displacement redox reaction. (d) NO2 (+4 state) disproportionate in alkaline medium into NO –2 (+3 state) and NO –3 (+5 state). So it is an example of disproportionation redox reaction.

+2

Step 3. Identify the atoms whose oxidation states have changed. (i) The O.N of cooper increase from 0 to + 2 (ii) The ON of nitrogen decreases from + 5 to + 4 +2

0

7. ClO4– do not show disportionation because the ‘Cl’ in ClO4– is in its highest oxidation state i.e., +7. The remaining three disproportionate as follows.

+3

0

Cu + NO3−  → NO2 + Cu 2 +

Cu  → Cu 2 +

Na B H 4 +1× 1 x − 1× 4

9. (a) Silver is reduced while C of compound C6H6O2 is oxidized Therefore oxidizing agent is Ag and reducing agent is C6H6O2. (b) Oxidising agent is Tollen’s reagent; reducing agent is HCHO. (c) Copper undergoes reduction (oxidising agent); HCHO undergoes oxidation (reducing agent) (d) Nitrogen in N2H4 undergoes oxidation (reducing agent) H2O2 undergoes reduction (oxidising agent) (e) Pb is oxidized (reducing agent); PbO2 is reduced. (Oxidising agent). 10. (a) I2 is oxidising agent; H2S is reducing agent. (b) HNO3 is oxidising agent; Zn is reducing agent. 2– 11. (a) Oxidant is I2, reductant is S2O3 (b) Oxidant is MnO2 reductant is C2H2O4 (c) Oxidant is Cl2 reductant is Br– 12. Step 1. Skelton equation Cu + NO–3 → NO2 +Cu2+ Step 2. Write the oxidation number of each atom

increase in ON + 2 − 0 = +2

+5

+4

NO3−  → NO2 Decrease in ON = +5 − (+4) = +1

Step 5. Equalize the increase in the oxidation number with the decrease in oxidation number. Increase 1 × (+2) = +2 Decrease 2 × (+1) = +2 Step 6. Write down the equation using these coefficient Cu+ 2NO–3 → Cu2+ + 2NO2 Step 7. Balance the other atoms except H and O as Cu + 2NO–3 → Cu2+ + 2NO2 Step 8. Reaction takes place in acid medium. So add H+ ions to the side deficient in H+ and balance H atoms and O atoms. Cu + 2 NO3− + 4 H +  → 2 NO2 + Cu 2 + + 2 H 2 O 13. (i) Step 1. Write the skeleton equation with oxidation number +3

−2

0

0

+2 −2

Fe2 O3 + C  → Fe+ C O

Step 2. The oxidation number of carbon increases from 0 to +4 while that of iron from +3 to +2

Redox Reactions 3+

0

Increase in ON = 1 × (+2)=2 Decrease in ON = 2 × (+1) = 2 Step 4. Write down the equation using these coefficients Zn + 2 HNO3  → Zn( NO3 ) 2 + 2 NO2 + H 2 O Step 5. To balance the two NO3− ion add two HNO3 more on the left hand side.

+2

0

Fe  → Fe

C  → CO

Decreases in ON + 3 − 0 = +3

Increases in ON + 2 − 0 = +2

(+3 × 2) − (0 × 2) + 6( For 2atoms ) + 2 − 0 = +2

Step 3. Equalize the increase in ON with the decrease in ON.

Decrease + 6

3 × (+2) = +6

Step 4. Write down the equation using these coefficients Fe2 O3 + 3C  → 2 Fe + 3CO Step5. Check all the atoms are balanced. Write the final equation. Fe2 O3 + 3C  → 2 Fe + 3CO (ii) Step 1. Write the skeleton equation along with ON of each atom. +2 2+

Fe

+6

+

−2 Cr2 O72 −

+1 +

+3 3+

+3 3+

Step 2. Oxidation number of iron increases from +2 to +3 while that of chromium decrease from +6 to +3 +1

+3

+6

+3

Fe 2 +  → Fe3+

Cr 6 +  → Cr 3+

Increase in ON=1

Decrease in ON=3 For 2 Cr atoms in Cr2O2-7. decrease in ON=6

Step 3. Equalise the increase in ON with the decrease ON. Increase in ON = 6 × (+1) = +6 Decrease in ON = 1 × (+6) = +6 Step 4. Write down the equation using these coefficients 6 Fe2+ + 2Cr2O72– + H + → 6Fe3+ + 2Cr3+ + H2O Step 5. Balance the number of hydrogen ions and O atoms 6 Fe2+ + Cr2O72– + 14H + → 6Fe3+ + 2Cr3+ + 7H2O (iii) Step 1. Write the skeleton equation with ON of each atom O

+1 + 5 − 2

+2

+5 − 2

+4 − 2

+1 − 2

Zn + HNO3  → Zn ( NO3 ) 2 + NO2 + H 2 O Step 2. Oxidation number of Zn increases from O to +2 and Oxidation number of nitrogen decreases from +5 to +4, while a part of nitrogen undergoes no change in ON. O

2+

Zn  → Zn 2 +

+5 − 2

+4

Zn + 4 HNO3  → Zn( NO3 ) 2 + 2 NO2 + H 2 O Step 6. Balance the number of H + ions and O atoms. Zn + 4 HNO3  → Zn ( NO3 )2 + 2 NO2 + 2 H 2 O (iv) Step 1. Write the skeleton equation with ON of each atom −1

−2

NO3−  → NO2

increase in ON = +2 Decrease in ON = +1 Step 3. Equalize the increase in ON with decrease in ON

+1

O

+4 − 2

+1

−2

C6 H 6 + O2  → CO2 + H 2 O Step 2. Oxidation number of carbon increases from −1 to +4 while the oxidation number oxygen decrease from O to −2.

+1 −2

+ H  → Fe + Cr + H 2 O

11.45

−1

+4

−2

O

−2

C  → C O2

O2  → O+ O

increase in ON = 5 Total increase in ON = 6 × 5 = 30

Decrease in ON = −2 Total decrease = 6(4 in CO2 and 2 in H 2 O)

Step 3. Equalize the increase in ON with decrease in ON by multiplying C6 H 6 by 2 and O2 by 15

2C6 H 6 + 15O2  → CO2 + H 2 O Step 4. Balance all atoms 2C6 H 6 + 15O2  →12CO2 + 6 H 2 O 14. Step 1. Write the skeleton equation with ON of each element. +7

−2

−1

+4

−2

+5

−2

MnO4− + Br −  → MnO2 + BrO3−

Step 2. Oxidation number of Mn change +7 to + 4 (a decrease of +3 and the ON of Br increase from −1 to +5) (an increase of +6) +7

+4

−1

+5

Mn 7 +  → Mn 4 + Br −  → Br O3− Decrease ON==6 6 DerceaseininON ON==33 Increase IncreaseinON Step 3. Equalize the increase in ON with decrease in ON Decrease in ON = 2 × 3 = 6 Increase in ON = 1 × 6 = 6

Step 4. Write down the equation using these coefficients. 2MnO4− + Br −  → MnO2 + BrO3− Step 5. Balance all the other atoms except H and O 2 MnO4− + Br −  → 2 MnO2 + BrO3− Step 6. Since the reaction occurs in basic medium add OH– ions on the side deficit of O atoms and H2O on the side deficit of H atoms.

11.46

Redox Reactions

2 MnO4− + Br − + H 2 O  → 2 MnO2 + BrO3− + OH − Step 7. Write the balanced equation. 2 MnO4− + Br − + H 2 O  → 2 MnO2 + BrO3− + 2OH − 15. (i) Step 1. Write the skeleton with ON of each atom +2 −1

+3

O

−2

+4 − 2

2Cr (OH )3 + IO3− + 4OH −  → I − + 2CrO42 − + 5 H 2 O 16. Separate the reaction into half reaction → S 2 O62 − The oxidation half reaction: S 2 O32 −  →I− The reduction half reaction: I 2  Oxidation Reduction

FeS 2 + O2  → Fe2 O3 + SO2

1. S2 O32-  → S4 O62-

(ON of S in FeS2 −1 and not −2) Step 2. The oxidation number of iron as well as sulphur increase from +2 to +3 and −1 to +4 respectively. Since Fe and S must maintain their atomic ratio, the change in ON will be considered together. On the other hand the ON of Oxygen decrease from O to −2 thus Increase in ON 1 for Fe Decrease in ON 2 for ‘O’ atom. 5 For S atom i. e., or 4 per O2 1+ (2 × 5) = 11 per FeS2 Step 3. To balance increase and decrease of ON multiply FeS2 by 4 and O2 by 11 4 FeS 2 + 11O2  → Fe2 O3 + SO2 Step 4. Balance all atoms 4 FeS 2 + 11O2  → 2 Fe2 O3 + 8SO2

2.Balance the number of atoms

(ii) Step 1. Write the skeleton equation with ON of each atom +3 − 2+1

+5 − 2 − 3

Cr(OH)3 + IO

−1 −

−2 2− 4

+6

 → I + CrO

Step 2. ON of Cr from +3 to +6 (an increase of 3) and the ON of I decreases from +5 to −1 (a decrease of 6) +3 3+

Cr

+6 6+

 → Cr

increase in ON = 3

+5

− 3

−1 −

→I IO 

1.I 2  → I-

2S2 O32-  → S4 O62-

2.I 2  → 2I-

3.Balance The charges 2S2 O32-  → S4 O62- +2e-

3.I 2 +2e-  → 2I-

4.Add the two half reactions I 2 + 2S2 O32-  → S4 O62- +2I17. (i) Separate the equation into two half reactions → Fe3+ The oxidation half reaction: Fe 2 + 

→ Cr 3+ The reduction half reaction: Cr2 O72 −  Oxidation Reduction 1. CrO72 −  → Cr 3+

→ Fe3+ 1. Fe 2+  2. Balance the atoms

2. CrO72 −  → Cr 3+

Fe 2+  → Fe3+ 3. Balance the charges Fe 2+  → Fe3+ +e −

3. CrO72 − +6e −  → 2Cr 3+

4. Since the reaction takes place in acid medium add 14 H + on the left to equate the net charge in reduction reaction Fe 2+  → Fe3+ +e- Cr2 O72- +6e- +14H +  → 2Cr 3+

Decrease in ON = 6

Step 3. Equalize the increase in ON with decrease in ON. Increase in ON 2 × 3 = 6 Decrease in ON = 1 × 6 Step 4. Write the equation with these coefficients 2Cr(OH)3 + IO3-  → I- +2CrO 2-4 Step 5. Balance all other atoms except H and O. 2Cr(OH)3+ IO3–  → I– + 2CrO2– 4 Step 6. Balance O atoms by adding OH − ions on the side deficient in O− atoms and one molecule of water for each H− atom needed on the side deficient in H atom. At the same time, add an equal number of OH − ions to the opposite side. 2Cr(OH)3 + IO3- +5OH -  → I- +2CrO 2-4 + 5H 2 O+OH Step 7. Balance the number of oxygen and hydrogen atoms and write the balanced equation.

5. Balance H atoms add 7H2O molecules in reduction half reaction Fe 2 +  → Fe3+ + e − CrO72 − + 6e − + 14H +

 → 2Cr 3+ + 7 H 2 O 6. To balance the number of electrons in both oxidation reduction reactions multiply the oxidation half reactions. 6Fe 2 +  → 6 Fe3+ + 6e − CrO72 − + 6e − + 14H +  → Cr 3+ 7. Add the two half equations 6Fe 2 + + CrO72 − + 14H +  → 6 Fe3+ + 2CrO 3+ + 7H 2 O (ii) Separate the equation into two half reactions −1

O

Oxidation half reaction: C2 H 4 O  → C2 H 4 O2 +6

+3

Reduction half reaction: Cr2 O72 −  → Cr 3+

Redox Reactions

Oxidation half reaction 1. C2 H 4 O → C2 H 4 O2

Reduction Half reaction 1. Cr2 O7 2 − → Cr 3+

5. Balance the H and O atoms (no need)

5. Add water molecules to complete balancing NO − 3 + 8e − + 7 H 2 O

2. Balance the carbon atoms 2. Balance the Cr atom Cr2 O7 2 − → 2Cr 3+ C2 H 4 O → C2 H 4 O2 C atoms already balanced 3. The ON of C on the left 3. The ON of Cr atom is +6 is −1 while on the right is and on the right is three. 0. To account for the difThus each Cr atom has ference oxidation number, gained 3 electrons i.e, add two electrons on the total number of electrons right side. gained is 6 on the left side. C2 H 4 O → C2 H 4 O2 + 2e − Cr2 O7 2 − + 6e − → 2Cr 3+ 4. Since the reaction is taking 4. Since the reaction is takplace in acid medium add ing acid medium add 2H+ on the right and H2O 14H+ on the left and 7H2O molecule on the left side. on the right side. C2 H 4 O → C2 H 4 O2 Cr2 O7 2 − + 6e − + 14 H + +2e − + 2 H + → 2Cr 3+ + 7 H 2 O Adding the reactions after multipying the oxidation equation with 3 to equate the number of electrons 3C2H4O + Cr2O72– + H+  → 3C2H4O2 + 2Cr3+ (iii) Separate the equation into two half reactions Oxidation half reaction: Zn → Zn 2 + Reduction half reaction: NO3− → NH 4+ Oxidation half reaction 1. Zn → Zn

2+

Reduction half reaction − + 1. NO 3 → NH 4

2. Balance all atoms other 2. Balance all atoms oththan H and O (already er than H and O atoms done) (already balanced)

→ NH 4 + + 10 OH − 6. Balance the number of electrons by multiplying with 4 to equate the electrons on both sides 4 Zn → 4 Zn 2 + + 8e − Add the two reactions to the balanced equations

4 Zn + NO3 − + 7 H 2 O → 4Zn 2 + + NH + 4 + 10 OH − Adding the reactions after multiplying the oxidation equation with 3 to equate the number of electrons 3C2 H 4 O+Cr2 O72- +8H +  → 3C2 H 2 O 2 +2Cr 3+ (iv) Separate the equation into two half reactions Oxidation half reaction: Cr (OH )3  → CrO42 − − − →I Reduction half reaction: IO3  Oxidation half reaction

Reduction half reaction

1. Cr ( OH )3 → CrO4

1. IO3− → I −

2−

2. The ON of Cr on the left 2. The ON of iodine on the is +3 and on the right it left is +5 and on the right is +6 to account for the is −1. To account for the difference add 3 elec- difference add 6 electrons trons on the right on the left 2− − IO − 3 + 6e − → I − Cr ( OH )3 → CrO4 + 3e 3. Since the reaction is tak- 3. Since the reaction is taking ing place in basic medium place in basic medium 2-

Cr (OH)3 + OH- → CrO 4 + 3e4. Balance the H and O atoms

3. Add electrons to make up 3. Nitrogen changes its ON the difference on ON from +5 to −3 and the difference is 8 electrons. To balance − − + 2+ − NO Zn → Zn + 2e 3 + 8e → NH 4

Cr (OH)3 + OH- + 4OH- → CrO2-4 + 4H2O + 3e–

4. Balance the charges (already balanced)

5. To equate the number of electrons with reduction half reaction multiply with 2

4. Balance the charges. Since the reaction has taken place in basic medium, add 10 OH– ions on the right side. NO − 3 + 8e −

→ NH 4 + + 10 OH −

11.47

or Cr ( OH )3 + 5OH − → CrO4 2 − + 4 H 2 O + 3e −

IO-3 + 6e- → I- + 3OH4. Balance the H and O atoms IO3− + 3H 2 O + 6e − → I − + 3OH − + 3OH − or IO3− + 3H 2 O + 6e − → I − + 6OH −

2Cr(OH)3 +10OH  → 2CrO 2-4 +8H 2 O+6eAdd the two half reactions 2Cr(OH)3 +IO3- +4OH -  → 2CrO 2-4 +5H 2 O+I-

11.48

Redox Reactions

Multiple choice Questions with only one answer level I 1. 10 Al + 6KMnO4 + 30H2SO4 − 6KHSO4+5Al2(SO4)3 + 6MnSO4 + 24 H2O 2. In all compounds oxidation number of carbon is zero. 3. In H2S2O7,H2SO5 sulphur oxidation numbers are +6 4. Zn → Zn +2 + 2e − NO − 3 + H 2 O + 8e − → NH + 4 + 10 OH − 4 Zn + NO − 3 + 7 H 2 O → 4 Zn +2 + NH 4 + + 10 OH −

(s) It is a disproportionation reaction involving change in oxidation state as shown in the reaction 2Oθ 2 → Oθ 2 + O22 −

( B) → r 2CrO42 − + 2 H + → Cr2 O72 − + H 2 O Structure of is Cr2 O72 − a dimeric bridge tetrahedral metal ion as o o o

5. In KI3 it contain K+, I–,I2 6. mole of Ca+2 = 11.3 × 0.05 1000 Mole of C2O–2 4 =

22.6 × 0.02 × 5 2000

11 O2 → Fe2 O3 + 4SO2 2 8. Volume of KMnO4 is V 0.3V 0.158 ∴ = ×8 1000 158 V = 26.67 mL

7. 2 FeS 2 +

9. MnO − 4 + 8 H + + 5e − + → Mn +2 + 4 H 2 O C2 H 2 + 4 H 2 O → 2CO2 + 10 H + + 10e − 2 MnO4 − + C2 H 2 → 2 M +2 + 2CO2 + 4 H 2 O 13. n factor of HCl is 0.5 EW=36.5/0.5=73

Matching type Questions 4. (A) → p,s tion state. (p)

involving increase and decrease of OxidaO2 − → O2 + e − (Loss of electron=Oxidation)

O2 − + e → O2 2 − (Gain of electron=reduction)

o

Cr

Cr

o

o

o

(C) → pq

Integer type Questions 1. Oxidation number of nitrogen changes from −3 to 0 2. Oxidation number of nitrogen changes from +5 to +1

Previous years’ IIt Questions 1. (d) In CrO2Cl2, Cr oxidation number is +6 2. (c) 3ClO − → ClO3− + Cl − It is a disproportionation reaction involving both increase and decrease of oxidation number. 3. (b) In CrO2Cl2 the Cr Oxidation number is +6. In MnO–4 the Mn oxidation number is +7.

CHAPTER

12 Electrochemistry

T

he Philosopher should be a man willing to listen to every suggestion but determined to judge for himself. He should not be biased by appearances, have no favorite hypothesis, be of no school, and to doctrine have no master. He should not be a respect of persons, but things. Truth should be his primary object. If to these qualities be added industry, he may indeed hope to walk with the veil of the temple of nature. Michael Faraday

12.1 IntroDuCtIon Electrical energy plays an important role in many chemical reactions. The branch of science which deals with the relationship between electrical energy and chemical energy and inter-conversation of one form into another is called electrochemistry. These may include chemical reactions which occur due to passage of electrical energy as well as the generation of electrical energy due to chemical reactions. The basis of these types of process are redox reactions which you may have learnt in the previous chapter. These are very important class of reactions. Large number of chemical and biological reactions are redox reactions. These include the burning of fuels for obtaining energy for domestic, transport of industrial purposes, digestion of food in animals; photosynthesis to capture energy from the sun, many industrial processes for extracting metals from their ores and manufacture of important chemicals, function of dry and wet batteries. Electrochemistry is very vast and interdisciplinary branch of chemistry which finds tremendous applications. Its study is very important as it helps in development of new technologies which are economical and environment-friendly.

12.2 ElECtroChEmICal CElls It is a well known fact that energy manifest itself in different forms which are interconvertible into one another. Among different forms of energy, the electrical energy plays a very significant role in our daily life. Many chemical transformations and industrial processes are based on electrical energy and its relationship with chemical energy. There are large numbers of spontaneous redox reactions

which form the basis of production of electrical energy. The device in which such chemical process are carried out is called electrochemical cell or galvanic cell. For example, Daniel cell is based on the following redox reaction.

Cu(2aq+ ) + Zn( s )  → Cu ( s ) + Zn 2 + + Electrical energy Galvanic Cells The device in which chemical energy is converted into electrical energy is called galvanic cell or electrochemical cell or voltaic cell in honour of Galvani and Volta respectively. When plates of two dissimilar metals are placed in a conducting liquid such as an aqueous solution of a salt or an acid, the resulting system becomes a source of electricity. If the plates are connected by a wire, the wire becomes warm. Further, a magnetic needle brought near to the wire is deflected. If this wire is cut and two ends are inserted in a conducting solution of chemical substance, chemical action will be observed in the solution where the wire comes into contact with the liquid. This action may be a gas evolution or liberation of a metal whose salt is present in the conducting medium. These effects of wire deflection of magnetic needle and chemical decompositions are due to flow of current in the wire connecting the plates. Daniel Cell is one example for an electrochemical cell. The spontaneous oxidation-reduction reaction, made use of in the working of Daniel cell is Zn( s ) + CuSO4 (aq )  → ZnSO4 (aq ) + Cu ( s )

In the electrochemical cell, the redox reaction is carried in such a way that oxidation and reduction processes are carried out in separate vessels. So every electrochemical

12.2

Electrochemistry

cell consists of two half cells and these are connected to each other by a salt bridge. Oxidation reaction occurs in one half cell and reduction occurs in the other half cell. The arrangement of Daniel and Voltaic cells are shown in Fig 12.1. For example in the Daniel cell, the two half cells are Zn/ZnSO4 (aq) (Zinc rod dipped in zinc sulphate solution) and Cu/CuSO4 (aq) (copper rod dipped in CuSO4 solution) In Zn/ZnSO4 (aq) half-cell, oxidation reaction occurs  Zn 2 + + 2e Zn  In Cu/CuSO4 half-cell reduction reaction occurs  Cu Cu 2 + + 2e -  The total oxidation –reduction reaction is written as  Zn 2 + (aq ) + Cu Zn + Cu 2 + (aq )  (Cu2+ and Zn2+ are from CuSO4 and ZnSO4 respectively)  or Zn (s) + CuSO 4 (aq ) ↽ ⇀  ZnSO 4 (aq ) + Cu (s) The two half cells are also referred to as single electrodes. The electrode at which oxidation occurs is anode and that at which reduction occurs is cathode. In the Daniel cell shown in Fig 12.1 zinc rod is anode and the copper rod is cathode. Due to the oxidation process occurring at the anode it becomes a source of electrons and

acquires a negative charge in the cell. Similarly, due to reduction process occurring at the cathode it acquires positive charge and becomes a receiver of the electrons. Thus in the electrochemical cell anode electrode acts as negative terminal and the cathode electrode acts as positive terminal. When the two electrodes are connected, the electrons given by zinc atoms at anode (zinc rod) flow to the cathode (copper rod). This can be detected by keeping an ammeter in the circuit. Ammeter is used to know the passage of current which moves in opposite direction to the flow of electrons. The two half cells are connected with a salt bridge to complete the circuit. When the circuit is completed by inserting the key in the circuit, it is observed that the electric current flows through external circuit as indicated by the ammeter. The zinc rod dipping into a ZnSO4 solution is oxidation half cell and the copper electrode dipping into CuSO4 solution is reduction half cell. The working of the electrochemical cell and the functions of salt bridge are represented in Fig 12.2.

12.2.1 representation of an Electrochemical Cell An electrochemical cell or galvanic cell consists of two electrodes: anode and cathode. The electrolyte solution containing these electrodes are called half-cells. When these

e-

−

electrons flow Current flow

Ammeter

Cathode (Cu)

Salt bridge

Anode (Zn)

+

Cu

Zn2+ 1 M ZnSO4 (aq)

1 M CuSO4 (aq) Fig 12.1 A simple Galvanic cell

Electrochemistry

Ammeter

electron flow key

Current flow

e-

_

+

Anode (Zn)

1 M ZnSO4 (aq)

12.3

Cathode (Cu)

Salt bridge

Zn2+

Porous Plugs

Zn2+ Cl-

Zn2+

Oxidation Zn → Zn2+ + 2 e-

SO42– K

+

1 M CuSO4 (aq)

SO42–

Cu

Cu2+

Reduction Cu2++ 2 e– → Cu

Fig 12.2 Working of an electrochemical cell half cells are combined a cell is formed. The following conventions are used in representing an electrochemical cell: 1. A galvanic cell is represented by writing the anode (where oxidation occurs) on the left hand side and cathode (where reduction occurs) on the right hand side. 2. The anode of the cell is represented by writing metal or solid phase first and then the electrolyte (or cation of the electrolyte) while the cathode is represented by writing the electrolyte first (or cation) and then metal or solid phase. The metal and the cation are separated either by a semicolon (;) or by a vertical line. The concentration of the electrolyte is also mentioned within the bracket after the cation. For example, Zn ; Zn 2 + or Zn Zn 2 + or Zn Zn 2 + (1 M ) (anode) 2+

Cu ; Cu or Cu

2+

2+

Cu or Cu (1 M ) Cu (Cathode)

12.2.2 Electrochemical Changes: Electrolytic Cells and Galvanic Cells The chemical changes that take place due to electric current flow are called electrochemical changes these are mainly of two types. 1. Electrolytic cells and 2. Galvanic cells

1. Electrolytic cells: In these cells, electrical energy is converted into chemical energy. Here a chemical reaction takes place at the expense of electrical energy. The electrical energy drives the redox reactions which have positive reaction free energy (ΔG) and are non-spontaneous. The decomposition of an electrolyte by means of the electric current is known as electrolysis. The assembly in which the electrolysis is carried are called electrolytic cells. For example, when electric current is passed through molten sodium chloride, sodium is liberated at the cathode while chlorine is liberated at anode. 2 NaCl Electric  → 2 Na + Cl2 current cathode

anode

2. Electrochemical cells or Galvanic cells: In these cells, chemical energy is converted into electrical energy. Here the spontaneous redox reactions are used to generate electric current. This is reverse of electrolysis. The cells in which chemical energy is converted into electrical energy are called electrochemical cells or galvanic cells. The early example of a galvanic cells is Daniel cell which was devised by the British chemist John Daniel in 1836.

12.4

Electrochemistry

3. The salt bridge which separated the two half cells is indicated by two vertical lines. For examples, Zn Zn 2 + (1M) | Cu2 + (1M) Cu –

Zn | Zn2+(1M) Anode oxidation occurs

Hydrogen electrode Pt, H2 (g)|H–1(aq) Chlorine electrode Pt, Cl2 (g)|Cl–1(aq) +

Electrons flow

||

Cu2+ (1M) | Cu

Salt bridge

Cathode Reduction occurs

Sometimes negative and positive signs are also put on the electrodes to show the release and loss of electrons taking place on them, Anode is a negative pole while cathode acts as positive pole. The electrons flow from the negative pole (anode) to the positive pole (cathode) in the external circuit on the other hand, conveniently, the current is said to flow in the opposite direction i.e., from cathode to anode in the external circuit (outside the cell) while from anode to cathode inside the cell (i.e., in the electrolyte). The cell representation can be easily written by noting the following points: Left side Anode Oxidation Negative

Salt Bridge

in such a manner that gas and its ions are brought in contact at the surface of the inert metal e.g.,

Right side Cathode Reduction Positive

The initial letters of the above mentioned terms on left side LAON comes before the initial letters of the terms on right side RCRP in alphabetical order. So left–right, anodecathode, oxidation–reduction, negative–positive may be written in alphabetical order and thus cell can be written.

4. Metal-metal insoluble salt-salt anion: In this type of electrodes, a metal and its sparingly soluble salt is dipped in an aqueous salt of some other soluble salt containing the common anion as that of sparingly soluble salt of the metal e.g., Ag-AgCl(s) | KCl (aq). In this electrode, silver and sparingly soluble salt of silver i.e., AgCl is put in aqueous solution of KCl. The concentration of Ag+ ions in KCl solution is controlled by concentration of KCl and it can be calculated from the knowledge of Ksp of AgCl as follows:  Ag +  =  

K sp of AgCl -

[Cl ]

=

K sp of AgCl [KCl]

The electrode reactions are Ag ( s ) + Cl - (aq )  → AgCl + e - oxidation AgCl ( s ) + e -  → Ag ( s ) + Cl - (aq ) Reduction Another important example of the type of electrode is calomel electrode 2Hg (l) + 2Cl - (aq )  → Hg 2 Cl2 (s) + 2e - oxidation Hg 2 Cl2 (s) + 2e -  → 2Hg (l) + 2Cl -

reduction

5. Redox electrodes: These electrodes include a platinum wire dipped in a solution of a mixture of two salts of the same metal having different oxidation states some common examples are as follows: Pt Fe 2+ |Fe3+ Fe 2+  → Fe3+ + e - or Fe3+ + e -  → Fe 2+ Pt Sn 2 + |Sn 4 + Sn 2 + → Sn 4 + + 2e - or Sn 4 + 2e - → Sn 2 +

12.2.3 types of Electrodes The frequently used eletrodes in electrochemical cells are as follows: 1. Metal-metal ion electrode: Such type of electrodes include a metal rod dipped in the solution of its own ions some examples are Zn/Zn2+, Cu/Cu2+; Ag/Ag+ etc. 2. Amalgam electrodes: These are similar to metal-metal ion electrodes, but here the metal is used in the form of its amalgam with Hg. Amalgam is formed to modify the activity of metal Zn-Hg/Zn2+ is one of the common example of this type. 3. Gas electrodes: Such electrodes involve inert metal such as platinum dipped in the solution containing ions of the gaseous element the arrangement is made

If the metallic electrodes dissolve or form, as the cell reaction proceeds, they are called active electrodes. As an example, we already have zinc and copper rod electrodes which are respectively consumed and formed as the reaction  → Zn 2 + + Cu Zn + Cu 2 + ← 

proceed from left to right. The electrodes which are left unchanged by the net cell reaction are known as inert or sensing electrodes. For example, in the redox electrodes the platinum strips are immersed in the ferrous and ferric and stannous and stannic solutions. As the cell operates, ferric ions are reduced while stannous ions are oxidized 2 Fe3+ + Sn 2 +  → 2 Fe 2 + + Sn 4 +

Electrochemistry

Thus at one platinum electrode, ferric ions acquire electrons and becomes ferrous ions and stannous ions become stamic ions at other platinum electrode, while the platinum electrodes remain unchanged. Such electrodes which remain unchanged during cell reaction are called inert or sensing electrodes. E.g., platinum and graphitic carbon are the two substances most commonly used.

12.2.4 Electromotive Force of the Cell In the previous chapter (chapter 11) we have already learnt about the origin of potential, electrode potential, standard electrode potentials and the arrangement of different electrodes as emf series. Every electrode in the emf series has its own potential. By combining any two electrodes (half cells) we can get an electrochemical cell. When the circuit is completed, the loss of electrons occurs at the electrode having lower reduction potential where as the gain of electrons occur at the electrode with higher reduction potential. The difference in the electrode potentials of the two electrode of the cell is termed as electromotive force (abbreviated as emf) or cell voltage (E cell). Mathematically, it can be expressed as EMF = E red (cathode) - E red (anode) or simply EMF = E cathode - E anode Since in the representation of the cell, the anode is written on the left hand side and cathode is written on

the right hand side, the emf of the cell is also sometimes written as EMF = E right - E left or E R - E L EMF of the cell may be defined as the potential difference between the two terminals of the electrochemical cell when either no or very little current is drawn from. It is measured with the help of potentiometer or vacuum tube voltmeter. If the galvanometer shows zero deflection when the sliding contact is at position B i.e., no current is drawn from the cell. The emf of the cell is given by the expression (for Daniel cell) O

E Cell = 2XB/XY = 1.10v (The cell being at 25°C) If now the contact is made at position A then current is drawn from the cell: similarly, with the contact at position C the accumulator drives current through the cell in the opposite direction, i.e., the normal cell reaction is reversed. Slight variations in position of the sliding contact thus allows the cell to operate or to be driven backwards, i.e., a reversible change can be made to occur. With sliding contact at position B, the cell is theoretically capable of delivering the maximum amount of electrical work. In practice, however, no current is drawn from the cell under these conditions and hence no chemical reaction can occur; but if the sliding contact is only fractionally off balance an extremely small current can be drawn from the cell and the amount of electrical work obtained can be made to approach the theoretical maximum.

2V +

–

Potentiometer wire A

X

B

C

Y Sliding Contact

G

12.5

Galvanometer

+ cell – Cu / Cu2+ || Zn2+ | Zn

Fig 12.3 Determination of the emf of an electrochemical cell

12.6

Electrochemistry

12.2.5 reversible and Irreversible Cells In dealing with energy relations of cells, the thermodynamic principles find very extensive application. In thermodynamic sense, a cell is said to be reversible if it fulfills the following conditions: (i) If the emf (exactly equal to its own) is applied by some external source, the cell reaction stops. (ii) If the external emf applied to it is slightly greater to its own emf, the current begins to flow in opposite direction and the cell reaction is reversed. (iii) If the external emf is slightly less than that of the cell itself, a very small amount of the current will flow and a corresponding small amount of chemical reaction takes place in the cell. When these requirements are satisfied by a cell, it is a reversible cell and if these conditions are not satisfied the cell is said to be irreversible. The difference between reversible and irreversible cells can be illustrated by taking two examples. First consider the Daniel cell Zn / ZnSO 4 / /CuSO 4 / Cu The net cell reaction is Zn + Cu

2+

 → Zn

2+

+ Cu

The emf is 1.09 volts. If the emf is increased infinitesimally beyond 1.09 volts the current flows in opposite direction and cell reaction is reversed i.e., Cu + Zn 2 +  → Cu 2 + + Zn If the emf is less than 1.09 volts, the cell reaction continues Zn + Cu 2 +  → Zn 2 + + Cu And a smaller amount of current is given out by the cell. Another cell which does not satisfy the above condition is represented below Zn + / H 2SO 4 solution / Ag + When the two electrodes are connected zinc dissolves with evolution of hydrogen to form zinc sulphate i.e., Zn(s) + H 2SO 4  → ZnSO 4 + H 2 (g) However, when the cell is connected with an external source of potential slightly greater than its own, silver dissolves at one electrode, hydrogen is evolved at the other and the cell reaction becomes 2Ag(s) + H 2SO 4  → Ag 2SO 4 + H 2 (g) The cell however fulfills the first condition of reversibility but does not satisfy the second condition and hence it is an irreversible.

table 12.1 Difference between EMF and potential difference EMF

Potential Difference

1. It is the potential difference between the two electrodes when no current is flowing in the circuit i.e., in an open circuit 2. It is maximum voltage obtainable from the cell

1. It is the difference of the electrodes potentials of the two electrodes when the cell is sending current through the circuit 2. It is less than the maximum voltage obtainable from the cell (i.e., emf of the cell) 3. The work calculated from potential difference is less than maximum work obtainable from the cell 4. It is not responsible for the flow of steady current in the cell

3. The work calculated from emf is the maximum work obtainable from the cell 4. It is responsible for the flow of steady current in the cell

12.3 ElECtrICal EnErGy The passage of electricity through a conductor is accompanied by the evolution of heat. According to the first law of thermodynamics, the heat liberated must be exactly equal to the electrical energy. The heat is found to be proportional to the quantity of electricity passed and the emf of the cell. This is the maximum amount of electrical work obtained from the cell. The maximum amount of electrical energy or electrical work is the product of potential difference (in volts) and charge flowing (in coulombs) Welec = nFE

Where Welec is the work obtained n is the number of moles of electrons that flow during the chemical reaction and F is the Faraday (96500 coulombs/mol of electrons). For example, in the Daniel cell, the emf is 1.10 volts. The cell reaction involves liberation as well as taking up of 2 electrons. Therefore the quantity of electricity produced according to Faraday’s second law is 2 Faradays i.e., 2 × 96500 coulombs. Hence the maximum amount of electrical work that can be obtained Wuseful is Wuseful = 2 × 96500 × 1.10 = 212300 joules = 212.3 KJ/mol Zn A decrease in free energy, -ΔG was defined to be equal to the maximum useful work that a system could perform. Thus -ΔG = Wuseful = 212.3 KJ/mol Zn or ΔG = –212.3 KJ/mol Zn

Electrochemistry

Since the electrochemical cell contains the substances in their standard states i.e., the zinc and copper (II) ions are at unit concentration and the temperature is 25°C the free energy change is the standard free energy change ∆G O

O

∆G = –212.3 KJ/mol Zn

12.3.1 standard Free Energies of half-Cell reactions

The value -0.76 Faraday-volts is the standard free energy for Fe3+/Fe2+ electrode Fe3+ + e– → Fe2+ Since only one electron is involved in this change, the standard redox potential of this system is + 0.76 V. The use of standard free energies of half-cell reactions explains why electrode potential cannot always be combined by simple addition and subtraction. Thus Fe3+ + 3e– → Fe Fe2+ + 2e– → Fe

If we define the standard free energy of a half-cell reaction to be: O

G = –nFE

12.7

O

E = – 0.04V E = – 0.44V O

By subtraction, we have

O

Where ‘n’ is the number of moles of electrons involved F is the Faraday and E is the standard electrode potential then the free energy change for any cell reaction can be easily calculated. Since the standard electrode potential of hydrogen is arbitrarily fixed as zero, the standard free energy of the hydrogen electrode is also zero. Consider the following two half-cells reactions. O

2Ag+ + 2e– → 2Ag G Ag = –2F(+0.08) = –1.60 Faraday – volts

Fe3+ + e– → Fe2+ But the standard redox potential of this system cannot be obtained by subtraction of the standard redox potential, i.e., it is not +0.40 V but 0.76 V as we have seen above. In order to obtain the correct value, the number of electrons involved in each half-cell reaction must also be considered as explained above.

O

Cu2+ + 2e– → Cu G Cu = –2F (0.34) = 0.68 Faraday – volts O

By subtraction and rearrangement, we have: Cu + 2Ag+ → Cu2+ + 2Ag ∆G = ∆G Ag – ∆G Cu = –1.60 – (–0.68) = –0.92 Faraday – volts O

O

O

The negative value for the change in free energy means that the reaction as written will take place from left to right. By using the appropriate conversion factor the free energy change can be converted into KJ. Now consider the two separate half-cell reactions: Fe3+ + 3e– → Fe ∆G Fe3+/Fe2+ = –3F (–0.04) = +0.12 Faraday – volts O

Fe2+ + 2e– → Fe ∆G Fe2+/Fe = –2F (–0.44) = +0.88 Faraday – volts By subtraction, we have O

Fe3+ + e– → Fe2+ ∆G = ∆G Fe3+/Fe – G Fe2+/Fe = +0.12–(+0.88) = 0.76 Faraday–volts O

O

O

12.3.2 Dependence of redox Potential on Ionic Concentration and on temperature When a metal is placed in a solution of its ions there is a tendency for metal ions to leave the metal lattice and pass into solution, thus leaving an excess of electrons and hence a negative charge on the metal. There is also a reserve tendency for metal ions from solution to deposit on the metal, leading to a positive charge on the metal. In practice one effect is greater than the other, so a potential difference is set up between the metal and a solution of its ions. If the concentration of metal ions in contact with the metal is decreased there will be a smaller tendency for metal ions to deposit on the metal, i.e., the electrode potential of the metal will become less positive. Similarly, if the metal ion concentration is increased, the electrode potential will become more positive. The way in which the electrode potential of a metal is related to the metal ion concentration and to the temperature is given by Nernst equation; which can be derived as follows. We have already seen that standard free energy change ∆G and the emf the cell are related as O

O

O

–∆G = nFE cell The free energy of the chemical reaction is related to the equilibrium constant K of the redox reaction as per thermodynamics. ΔG = ∆G + RT ln K O

12.8

Electrochemistry

Let the equilibrium reaction of the cell be

(ii) Anion Electrode Example Pt , A2 ( g )| A(naq- )

 cC + dD aA + bB 

The electrode reaction A + ne -  → An -

[C ]c [ D]d Then K eq = [ A]a [ B]b

O

ΔG = ∆G + RT ln Keq O

O

= ∆G + RT ln

O

[C]c [D]d [A]a [B]b

But – ΔG = nFEcell O

or

O

and – ∆G = nFE cell ∴ –Ecell = –E cell + O

E = E –O

RT log K eq nF

[C ]c [ D]d 2.303RT log nF [ A]a [ B]b

This equation is applicable to even single electrode reactions substituting the value of R = 8.314 JK–1 mol–1 T = 298 K and F = 96500 coulombs we get = E –E= O

[A n - ] 2.303RT log nF [A] 2.303RT E = E° log[A n - ] nF 2.303RT E = E° log C nF 0.059 = E° log C n E = E° -

[C ]c [ D]d 0.059 log n [ A]a [ B]b

O

O

12.3.4 Equilibrium Constant from nernst Equation Consider the reaction between copper and a molar solution of silver ions: Cu + 2 Ag +  → Cu 2 + + 2 Ag

This reaction can be carried out in an electrochemical cell by using Cu/Cu2+ (molar) as one half-cell and Ag/Ag+ (molar) as the other half-cell. The standard free energy change for this reaction at 25°C is -0.92 Faraday –Volts -0.92 × 96500 KJ 1000

O

12.3.3 nernst Equation for single Electrode reaction Single electrodes can be classified into two categories. These are (i) The electrodes whose potential is dependent on the concentration of cations or metal ions (H+ or Mn+) (reversible with cations) (ii) The electrodes whose potential is dependent on the concentration of anions (An–) or non-metal ions (Reversible with anions) (i) Cation or Metal ion Electrodes n+ Example M (aq) | M (s)

The electrode reaction n+ M (aq) + ne -  →M

∆G = –0.92 Faraday–Volts =

= –88.74 KJ The electrode potentials of the silver and copper halfcells are given by the Nernst equation i.e., RT log e [Ag + ] (n = 1) F RT or EEAg ==EEAg log e [Ag + ]2 Ag + 2F RT log e [Cu 2 + ] (n = 2) E=Cu = EθuCu + 2F EEAg ==EEAg Ag + O

O

O

If this cell is allowed to deliver current, silver is deposited and copper (II) ions are formed; the emf of the cell under these conditions is given by [ Ag + ]2 RT log e 2F [Cu 2 + ]

(EAg – ECu) = (EθAg –- EECuθ )) + O

2.303RT [M] log n + nF [M ] 2.303RT θ log[M n + ]  [M] = 1 or E == EE ++ nF 2.303RT log C ==EE ++ nF 0.059 E° + log C  R = 8.314 JK -1 mol n F = 96500 Coulomb

 [A] = 1

O

E == EE –O

Multiplying each side of the equation by –2F we have

O

O

–2F(EAg – ECu) = –2F(E θAg – E θCuu ) - RT Log e O

O

[Ag + ]2 [Cu 2 + ]

But –2F(EAg–ECu) is the free energy change ΔG, and –2F(E Ag – E Cu) is the standard free energy change ∆G , thus O

O

O

G=∆ G θ − RT log e ΔG ∆G O

[Ag + ]2 [Cu 2 + ]

Electrochemistry

At equilibrium ΔG = 0, i.e., the cell is fully discharged and its emf is zero, therefore G θ = RT log e ∆G O

+

But when we write the reaction as 2Zn(s) + 2Cu 2 + (aq) → 2 Zn 2 + (aq) + 2Cu(s)

2+

2

[Ag equili] [Cu equili] = - RT log e 2+ [Cu equili] [Ag + equili]

or

12.9

∆G = - 4FE If the activity of all the reacting species is unity then E = E and we have O

∆G G θ = -2.303 RT log10 O

[Cu 2 + equili] = -2.303 RT log10 K [Ag + equili]

Since R is 8.314 J/mol –deg (or 0.008314 KJ/mol-deg) and the temperature is taken to be 25°C (298 K) We have -88.74 = - 2.303 × 0.008314 × 298log10 K 88.74 = 15.45 2.303 × 0.008314 × 298

or log10 K =

K = 1015.45 This exceedingly large value of K means that the reaction between copper and silver ions can be considered to go to completion. In general, we can write EEcellll = O

2.303RT log K c nF

This equation gives the relationship between equilibrium constant of the reaction and standard potential of the cell in which that reaction takes place. Thus, equilibrium constants of the reaction, difficult to measure otherwise, can be calculated from the corresponding E value of the cell. O

12.3.5 Electrochemical Cell and Gibbs Energy of the reaction In electrochemical cells, the chemical energy is converted into electrical energy. The cell potential is related to Gibbs energy change. In an electrochemical cell, the system does work by transferring electrical energy through an electrical circuit. Electrical work done in one second is equal to the electrical potential multiplied by the total charge passed. We have learnt that ΔG for a reaction is measure of the maximum useful work that can be obtained from a chemical reaction ΔG = maximum work It was already shown that the reaction between free energy change and maximum work obtainable from an electrochemical cell as O

O

∆G = Wmax = –nFE cell It may be noted that E is an intensive property but ΔG is an extensive property and the value depend on n. For example, for the reaction

Zn( s ) + Cu 2 + (aq )  → Zn 2 + (aq ) + Cu ( s ) ∆G = -2 FE

∆G = –nFE

O

O

Thus by measuring E value we can calculate an important thermodynamic property. From the temperature dependence of E , we can also calculate ΔH and ΔS . From the standard Gibbs energy, we can also calculate equilibrium constant by the equation O

O

O

O

∆G = –RT ln K The above equation helps us to predict the feasibility of the cell reaction. For a cell reaction to be spontaneous ΔG must be negative. This means that E must be positive for a spontaneous cell reaction

12.3.6 relationship Between Electrical Energy and Enthalpy Change of Cell reaction According to the Gibbs–Helmholtz equation which is applied to reversible changes  ∂∆G  ∆G = ∆H + T   ∂T  P  ∂ - ( nFE )  - nFE = ∆H + T    ∂T P ∂ E   or - nFE = ∆H - TnF    ∂T  P  ∂E  or nFE = - ∆H + nFT    ∂T  P -∆H  ∂E  ∴E = +T   nF  ∂T  P The above equation permits the calculation of the heat of reaction from the measured emf and temperature coefficient of emf of the reaction. Evidently, whether the electrical energy Viz, nFE, is equal to or greater or less than the heat of reaction (i.e., –∆H) depends upon the sign of the

∂E value   P , the temperature coefficient. ∂  T

∂E (i) If   is positive i.e., if the emf of the cell increases ∂  T P with rise in temperature the electrical energy will be greater than the heat of reaction i.e., nFE > ΔH

Electrochemistry

activity is a1 to the amalgam of the electrode on left side while a2 is the activity of Zn to the amalgam of the electrode on right side. Further, this equation also shows that the emf depends on the ratio of the zinc activities in the two amalgams and not at all on the activity of the zinc ions in the solution. In the final equation E does not appear. Thus it is concluded that E is zero and the simplified equation is φ

∂E (ii) If   is negative i.e., the emf of the cell decreases ∂  T P with increase in temperature the electrical energy will be smaller than the heat of the reaction i.e., nFE < ΔH

φ

12.10

 ∂E  (iii) If   is equal to zero the electrical energy will be  ∂T  P equal to heat of reaction i.e., nFE = ΔH

12.4 ConCEntratIon CElls 1. Electrode concentration cells The concentration cells in which the electrode material having different concentration are dipped in the solution of the metal ions are called electrode concentration cells. These may be of two types: (a) Amalgam concentration cell: In these cells, concentration of the electrode is changed by taking the amalgams. Two amalgams of the same metal at two different concentrations are immersed in the same electrolytic solution. Let us take the following cell having the base metal Zn where the activity of Zn in each amalgam is a1 and a2 dipped in a solution of Zn2+ of activity aZn2+

(

)

Zn(Hg)(a Zn = a1 )| Zn 2 + a Zn 2+ | Zn ( Hg ) (a Zn = a 2 )

Cell reactions

Right: Zn

(

(

)

)

-

a Zn 2+ + 2e  → Zn(Hg)(a 2 ) Reduction

Net reaction Zn(Hg)(a1 )  → Zn(Hg)(a 2 ) The emf for the electrode at the left is E1 RT a Zn 2+ 0 E1 = EEZn ln zn 2F Zn and for the electrode at the right is E2 a RT ln 2 E2 = E 0ZZn' n' 2F a Zn 2+ Here E Zn' = –E Zn The cell emf, E is given by O

O

O

O

a RT a Zn 2+ RT ln + (– E Zn) ln 2 2F a1 2F a Zn 2+ RT a 2 =ln 2F a1

E = E Zn O

or

0.0591 E 2

a 0.0591 log 1 n a2

(b) Gas concentration cell: In these type of cells, the electrode used is a gas material at different activities and dipped in the solution of gas ions. The potential of these cells depend on the pressure of the gases and concentration of its ions in solution. Consider the following cell of this type:

(

)

(

(

)

)

H 2 PH = P1 | H + aH + | H 2 PH = P2 ; P1 > P2 2

2

Cell reactions 1 Left H 2 ( P2 )  → H + aH + + e - Oxidation 2 1 → H 2 ( P2 ) Reduction Right H + a H+ + e -  2 1 1 → H 2 ( P2 ) Net reaction H 2 ( P1 )  2 2 RT a H+ E1 = ln 1 ( E H2 = 0) Oxidation F P1 2

(

)

( )

O

Left : Zn ( Hg ) (a1 )  → Zn 2 + a Zn 2+ + 2e - Oxidation 2+

Ecell =

O

a1 = a2

° log

The above equation indicates that the emf of this cell is due to the transfer of Zn from the amalgam where its

1

E2 = −

RT P2 2 ln F a H+

E Cell = E1 + E 2 1

RT a H+ RT P2 2 ln 1 − ln E Cell = − F F a H+ P 2 1

RT P2 E Cell = − ln 2F P1 P 0.0591 log 2 at 25°C n P1 The cell reaction for the spontaneous process is the expression of hydrogen gas from a pressure P1 to P2. The emf from this expression depends only on the two pressures and is independent of the activity of the H+ ion in which the electrodes are immersed 2. Electrolytic concentration cell In these cells, electrodes are identical but these are immersed in solutions of the same electrolyte of different concentration. The source of electrical energy in the cell is the tendency of the electrolyte to diffuse from a solution of or E Cell =

Electrochemistry

2 H 2 O  → H 3 O + + OH - at 25°C

The emf of the cell is given by the following equation C ( R.H .S ) 0.0591 log 2 at 25°C n C1 ( L.H .S )

The concentration cells are used to determine the solubility of sparingly soluble salts, valency of the cation of the electrolyte and transition point of two allotropic forms of a metal used as electrodes etc. solved Problem 1

7.

 2.303RT  CuSO4 in the other cell  = 0.06  . F   (IIT 2003) Calculate the equilibrium constant for the reaction Fe2+ + Ce4+ → Fe3+ + Ce3+ Given E Ce4+/Ce = 1.44V and E Fe3+/Fe2+ = 0.68V (IIT 1997) Calculate the equilibrium constant for the reaction  2 Fe 2 + + I 3- The standard reduction 2 Fe3+ + 3I -  potentials in acidic conditions are 0.77 and 0.54 V respectively for Fe3+/Fe2+ and I3- / I - couples. (IIT 1998) Find the equilibrium constant for the reaction In 2 + + Cu 2 +  → In3+ + Cu + at 298 K E In3+/In+ = –0.42V Given E Cu2+/Cu+ = 0.15V E In2+/In+ = –0.40V (IIT 2004) Find the solubility product of standard solution of Ag2CrO4 in water at 298 K if the emf of the cell Ag|Ag+ (stand Ag2CrO4 sol) || Ag+ (0.1 M) | Ag is 0.164 V at 298 K. (IIT 1998) A silver electrode is immersed in saturated Ag2SO4(aq). The potential difference between the silver and the standard hydrogen electrode is found to be 0.711 V. Determine Ksp (Ag2SO4). Given E Ag+/Ag = 0.799V. (Roorke 2000) An excess of liquid Hg was added to 10–3 M acidified solution of Fe3+ ions it was found that only 5% of the ions remained as Fe3+ at equilibrium at 25°C calculate E for 2Hg/Hg22 + at 25°C 2Hg + 2Fe3+ → Hg22 + + 2Fe2+ and E Fe2+/Fe3+ = –0.77V (IIT 1995) φ

The standard electrode potential of Cu, Cu2+ is -0.34 Volt. At what concentration of Cu2+ ions will this electrode potential be zero? Solution: The electrode reaction (reduction)

(IIT 1989) 5. The standard reduction potential for Cu2+/Cu is +0.34V calculate the reduction potential at pH = 14 for the above couple Ksp [Cu (OH)2] = 1 × 10–19 (IIT 1996) 6. Two students use same stock solution of ZnSO4 and a solution of CuSO4. The emf of one is 0.03 V higher than the other. The conc of CuSO4 in the cell with higher emf value is 0.5 M. Find out the conc of

0.0591 1 log 2 [Cu 2 + ]

0.0591 log[Cu 2 + ] 2 0.0591 log[Cu 2 + ] 0 = 0.34 + 2 — or log [Cu2+] = –11.5059 = 12 .4941

8.

φ

=E +

φ

φ

9.

Taking anti-log [Cu2+] = 3.12 × 10–12 M 10. Problems for Practice

11.

φ

φ

φ

1. A galvanic cell is constructed with Ag+/Ag and Fe3+/ Fe2+ electrode. Find the concentration of Ag+ at which the emf of the cell is zero at equimolar concentrations of Fe2+ and Fe3+ (EAg/Ag = 0.80V and E Fe3+/Fe2+ = 0.77V). 2. The standard reduction potential of Cu2+/Cu and Ag+/ Ag electrodes are 0.337 and 0.799 volt respectively. Construct a galvanic cell using these electrodes so that its standard emf is positive. For what concentration of Ag+ will the emf of the cell at 25°C be zero if the concentration of Cu2+ is 0.01 M? (IIT 1990) 3. Zinc granules are added in excess to 500 mL of 1 M nickel nitrate solution at 25°C until equilibrium

12.

φ

φ

ECu2+/Cu = E Cu2+/Cu –

E = +0.34 Volt φ

 Cu2+ + 2e– ↽ ⇀  Cu

φ

ECell =

φ

or simply Cu | Cu 2 + (C1 )|| Cu 2 + (C2 )| Cu

is reached. If the standard reduction potentials of Zn2+/Zn and Ni2+/Ni are -0.75 V and -0.24 V respectively find out the concentration of Ni2+ in solution at equilibrium. (IIT 1991) 4. The standard reduction potential at 25°C of the reaction  H 2 + 2OH - is - 0.8277 VVolt. Cal2 H 2 O + 2e -  culate the equilibrium constant for the reaction

φ

higher concentration to that of lower concentration. With the expiry of time the two concentrations tend to become equal. Thus at the start, the emf of the cell is maximum and is gradually fall to zero. Such cell is represented in the following manner by taking the copper electrode as an example Cu | CuSO 4 (C1 ) || CuSO 4 ( C 2 ) | Cu C2 > C1

12.11

12.12

Electrochemistry

If G f For OH− and H2O are -157 KJ mol−1 and –237.2 KJ mol−1, determine G f for [Al(OH)4]– 20. Calculate the emf of the cell φ

φ

13. For the galvanic cell

O

O

φ

[Ag(NH3)2]+ + e → Ag + 2NH3 Given Ag+ + e → Ag has E = 0.799V 23. The overall formation constant for the reaction of 6 moles of CN– with cobalt (II) is 1 × 1019. Calculate the formation constant for the reaction of 6 moles of CN– with cobalt (II). Given that φ

[Co(CN)6]3– + e → [Co(CN)6]4–; E [Co(CN)6]4– = –0.83 Co3+ + e– → Co2+; E Co3+/Co2+ = 1.82 V 24. The emf of a cell corresponding to the reaction Zn + 2H+ (aq) → Zn2+ (0.1 M) + H2(g) 1 atm is 0.28 volt at 25°C. Write the half-cell reactions and calculate the pH of the solution at the hydrogen electrode. E Zn2+/Zn = –0.76 volt and E H+/H2 = 0 (IIT 1986) 25. The following electrochemical cell has been set up

Pt(1)|Fe3+ , Fe 2 + (a = 1) ||Ce 4 + , Ce3+ (a = 1) |Pt(2)

φ

saturated solution of AgCl find the value of log10  Zn 2 +  How many moles of Ag will be precipated 2  Ag +  in this reaction? Given E Zn2+/Zn = –0.76V (IIT 2005) 17. The standard potential of the following cell is 0.23 v at 15°C and 0.21 V at 35°C. Pt, H2(g) | HCl(aq) || AgCl(s) | Ag(s) (i) Write the cell reaction. (ii) Calculate ΔH and ΔS for the cell reaction by assuming that these quantities remain unchanged in the range is to 35°C. (iii) Calculate the solubility of AgCl in water at 25°C. Given the standard reduction potential of the Ag+(aq)/Ag(s) couple is 0.80 V at 25°C. (IIT 2001) 18. The emf of cell Zn|ZnSO4 || CuSO4| Cu at 25°C is 0.03 V and the temperature coefficient of emf is -1.4 × 10– per degree. Calculate the heat of reaction for the change taking place inside the cell. 19. E Cell for reaction 4Al(s) + 3O2(g) + 6H2O + 4OH– → 4 [Al(OH)4]– is 2.73V.

Ka for CH3 COOH = 1.8 × 10–5, Kb for NH4OH = 1.8 × 10–5 21. Two weak acid solutions HA1 and HA2 each with the same concentration and having pKa values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25°C) and are interconnected through a salt bridge. Find emf of cell. 22. Dissociation constant for [Ag(NH3)2]+ into Ag+ and NH3 is 6 × 10–14. Calculate the E for half reaction.

O

O

O

φ

O

Pt1H 2 | CH 3 COOH || NH 4 OH | Pt , H 2 1 atm 0.1M 0.01M 1 atm

φ

E (Fe3+/Fe2+) = 0.77V E (Ce4+/Ce3+) = 1.61V If an ammeter is connected between the two platinum electrodes predict the direction of flow of current. Will the current increase or decrease with time. 26. Determine the degree of hydrolysis and hydrolysis constant of aniline hydrochloride if Pt ( H 2 ) H + C6 H 5 NH 3Cl ( H 2 ) Pt 1 1atm 1M M 1atm 32

φ

Calculate the emf generated and assign correct polarity to each electrode for a spontaneous process after taking an account of cell reaction at 25°C. Given KspAgCl = 2.8 × 10–10, KspAgBr = 3.3 × 10–13 (IIT 1992) 14. The standard reduction potential for the half cell NO3 (aq) + 2H+(aq) + e– → NO2(g) + H2O is 0.78V (i) Calculate the reaction potential in 8M H+ (ii) What will be the reduction potential of the halfcell in a neutral solution? Assume all the other species to be at unit concentration. (IIT 1993) 15. The standard reduction potential of the Ag+/Ag electrode at 298 K is 0.799V Given that for AgI, Ksp = 8.7 × 10–17, evaluate the potential of the Ag+|Ag electrode in a saturated solution of AgI. Also calculate the standard reduction potential of the I–|AgI|Ag electrode (IIT 1994) 16. For the reaction Ag + (aq ) + Cl - (aq ) → AgCl ( s ); the ∆G Values for Ag+ (aq), Cl– (aq) and AgCl(s) are + 77, -129 and -109 KJ mol–1. Write the cell representation of above reaction and calculate E at 298 K. Also calculate Ksp of AgCl at 298 K. If 6.539 × 10–2 g of metallic zinc is added to 100 mL

φ

0.001M

0.2 M

φ

Ag | AgCl(s) KCl || KBr , AgBr (s) | Ag

E cell = -0.188V at 300 K

27. If it is desired to construct the following voltaic cell to have Ecell = 0.0862 V, what concentration of chloride ion must be present in the cathodic half-cell to achieve the desired emf given Ksp of AgCl and AgI are 1.8 × 10 -10 and 8.5 × 10 -17 respectively? Ag(s)| Ag+ [saturated AgI (aq)] || Ag+ (saturated AgCl x MCl–) | Ag(s)

Electrochemistry

28. Construct a cell in which the disproportionation reaction 2CuCl → CuCl2 + Cu takes place. Also calculate the equilibrium constant for the reaction if Cu2+/Cu+ and Cu+/Cu are 0.153 v and 0.518 V respectively. 29. The standard oxidation potential of Ni/Ni2+ electrode is 0.236 V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution will the measured emf be zero at 25°C ? (Assume [Ni2+] = 1 M and PH2 = 1 atm)

be found that the solution was no longer of the same concentration throughout. The mechanism by which electricity is conducted in the copper sulphate solution is obviously is the same as in the case of a metal. This is because copper sulphate is an ionic compound in solution the copper and the sulphate ions are free to move individually so that when electric field applied the copper ions move towards the negative electrode (or cathode)and the sulphate ions move towards the positive electrode (or anode). The current is therefore carried across the solution by ions and not by electrons. Liquids in which this type of conduction takes place are known as electrolytes.

φ

φ

30. If NO3- → NO 2 (acid medium); E = 0.790V and NO3- → NH 2 OH (acid medium); E = 0.731V

12.13

At what pH the above two half reactions will have the same E values? Assume the concentrations of all the species to the unity.

Anode

+

–

Cathode

12.5 ConDuCtanCE A substance which allows an electric current to flow through it is called conductor while the one which does not allow any electric current to flow through it is called an insulator. Conductors may be divided into two categories: (i) Metallic or electronic conductor: Metals are good conductors of electricity because they contain a certain number of electrons which are free to move in an electric field applied. If the two ends in a piece of metal wire are connected to the terminals of a battery, the difference in potential causes electrons to flow towards the positive electrode. This movement of electrons causes certain quantity of electricity (equal to the number of electrons traveling multiplied by the charge on an electron) to be transferred from one end of the wire to the other. This does not cause any change in the composition of the metal because the electrons that flow into the battery from one end of the wire are replaced by an equal number received from the battery at the other end. (ii) Electrolytic conductors: Certain liquids can also conduct electricity for example, if a battery is connected to two platinum electrodes immersed in a solution of copper sulphate a current will flow. At the same time, however, it will be noticed that a deposit of metallic copper forms on the electrode connected to the negative pole of the battery, while the bubbles of gas (which could be shown to be oxygen) form at the other electrode Furthermore, if samples of solution taken from near either electrode were analysed, it would

2+ Cu 2SO4

Fig 12.4 Conduction in electrolyte

In an electrolyte anions (i.e., atoms or groups which have received one or more electrons during their formation) migrate towards the anode where electrons are given up. These flow toward the battery (or some other source of direct current) and thence to the cathode to which the positively charged cations are attracted. Both kinds of ions play a part in the conductance. Anions carry negative electricity in one direction and cations carry positive electricity in the opposite direction (which amounts to the same process) the total transfer of electricity is not necessarily shared equally between the two kinds of ions for it is known that different ions move at different speeds. The total conductivity of an electrolyte is this sum of the conductance of the anions and cations. Conduction of current through a metallic conductor and an electrolytic conductor may briefly be distinguished as follows:

Electrochemistry

table 12.2 Difference between metallic and electrolytic conductor Metallic conduction 1. It occurs due to flow of electrons 2. No changes in the chemical properties of the conductor occurs 3. It does not involve the transfer of any matter 4. It show an increase in resistance as temperature is increased

Electrolytic conduction 1. it occurs due to movement of ions in solution or a fused electrolyte 2. It involves chemical reaction which take place at electrodes 3. It involves transfer of matter in the form of ions 4. It show a decrease in resistance as the temperature is increased.

12.5.1 Conductance of Electrolytes When cation (metal ions) reach the cathode they are discharged i.e., they absorb electrons and become metal atoms which adhere to the electrode. At the anode, negative ions discharge and liberate electrons. Thus the flow of electrons in the external circuit is balanced by the removal of electrons from the cathode and the liberation of electrons at the anode. What is actually occurring at the electrodes is the decomposition of the electrolyte by means of the electric current, a process known as electrolysis. Therefore cations and anions get discharged at respective electrodes and are collected to neutral particles. This is essentially a redox reaction and it is known as primary change. The primary products may be collected as such or they undergo further changes to from molecules or compounds. These are called secondary products and the change is known as secondary change. mechanism of Electrolysis The process of electrolysis was explained by the theory of ionization. According to ionic theory, the electrolytes are present as ions in solution and the function of electricity is only to direct these ions to their respective electrodes. The electrolytes can be electrolyzed in the dissolved or molten state. the nature of Electrolytes From the definition of electrolytic conductance given above, it is evident that an electrolyte must contain ions. It is known that electrovalent compounds consist of ions even in the solid state if an electrovalent substance (e.g., sodium chloride) is melted, the ions will be free to move and the liquid will therefore conduct electricity. The same will also be true for a solution of the substance. It is in fact found

that many salts are electrolytes both in the fused state and in solution. There are however other types of compounds which give conducting solutions i.e., acids, bases and covalent salt (e.g., Al2Cl6). These are all covalent substances so that in the pure state they consist of molecules and not ions. The fact that their solutions will conduct electricity shows that the process of solution has caused ionization of the molecules. Hydrogen chloride for example is a non-conducting liquid (when below its boiling point) and if dissolved in dry benzene forms a non-conducting solution. When it is dissolved in water however the solution (hydrochloric acid) is strongly conducting and shows the usual acidic properties that we associate with the presence of hydrogen ions. This is due to ionization as a result of a reaction between the HCl molecules and water HCl + H 2 O  → H 3 O + + Cl This reaction takes place because of the strong affinity between a hydrogen ion and water molecule and produces the hydronium ion H3O+. The latter is a chemical compound with the following electronic structure.

[

H

: :

12.14

: O: H

H

]

+

A base (e.g., ammonia) will also react with water forming ammonium and hydroxyl ions NH 3 + H 2 O  → NH 4+ + OH Covalent salts also tend to ionize in solution because of affinity between the resulting metal ion and water e.g., 3+ Al2 Cl6 + 12 H 2 O  → 2  Al ( H 2 O)6  + 6Cl Where the water is held by dative bonds. Ions in aqueous solution always have a certain number of water molecules attached to them i.e., they are hydrated to a greater or lesser extent. Some of the water may be attached by chemical bonds as in the two examples above; in addition some will be held by electrostatic attraction between the charged ions and the polar water molecules (ion-dipole attraction). When the chemical formula of the ion is written the attached water is sometimes shown but more often omitted: thus the hydrogen ion in aqueous solution is alternatively written as H+ or H3O+. It must be remembered that the latter is correct, the former being an abbreviated way of writing it. In the same way, the aluminum ion is usually shown as Al3+ whereas it actually exists in the hydrated form shown above.

Electrochemistry

The above reactions where by a solvent cause’s ionization of the dissolved material are all reversible and the extent to which ionization occurs varies very widely according to the nature of both the dissolved material and the solvent. If a substance is completely (or almost completely) ionized in solution it is known as a strong electrolyte. Hydrochloric acid is one example and solution of all electrovalent salts are strong electrolytes since the latter are completely ionized even before they are dissolved. On the other hand, many acids are classified as weak electrolytes. Since ionization involves a reaction between solute and solvent the extent to which any particular compound is ionized will depend on the solvent used: a substance which is a strong electrolyte in one solvent may become a weak electrolyte in a different solvent. In this book, only solutions in water will be considered.

12.5.2 Factors affecting Electrical Conductivity of Electrolytic solutions The electrical conductivity of the solution of electrolytes depends upon the factors: (i) Interionic attractions: These depend upon the interaction between the ions of the solute molecules, i.e., solutesolute interactions. If the solute-solute interactions are large the extent of ionization will be less. These interactions are also responsible for the classification of electrolytes as strong electrolytes and weak electrolytes. (ii) Salvation of ions: These depend upon the interactions between the ions of the solute and the molecules of the solvent and are called solute-solvent interaction as described already. If the solute-solvent interactions are strong, the ions of the solute will be highly solvated and their electrical conductivity will be low. (iii) Viscosity of the solvent: The viscosity of the solvent depends upon the solvent-solvent interactions. Larger the solvent-solvent interactions, larger will be the viscosity of the solvent. All the above factors decrease with increase in temperature. Therefore the average kinetic energy of the ions of the electrolyte increases with increase in temperature consequently conductance of electrolytic solution increases with rise in electronic conductors (metallic conductors). The conductance of electronic conductors decrease with increase in temperature. some basic terms commonly used are as follows When voltage is applied to the electrodes dipped into an electrolytic solution ions of the electrolyte move, therefore, electric current flows through the electrolytic solution. The power of electrolytes to conduct electric current is termed

12.15

conductance or conductivity. Like metallic conductors electrolytic solution also obey ohms law. Ohm Law: The strength of current (I) passing through a conductor varies (i) directly to the potential difference (E) applied across the conductor and (ii) inversely proportional to the resistance (R) of the conductor E Thus I = .......................................................(i ) R Where I is in amperes E in volts and R is measured in ohms. Ampere: It is defined as the current strength which when passed through an aqueous solution of silver nitrate deposits 0.001118 gm of silver in one second. Ohm: An ohm is the resistance (at 273 K) of a column of mercury of uniform cross-section 106.3 cm long and weighing 14.4521 cm. Volt: A volt is the difference of potential that is necessary to make a current of one ampere flow through a resistance of one ohm. Resistance (R): The resistance of any conductor varies (i) directly as the length of the conductor (l) and (ii) inversely as its area of cross-section (a) Thus R ∝ l......................................(ii) l R ∝ .....................................(iii) a Combining (ii) and (iii) we get l a l R = ρ. a R ∝

Where ρ is constant and depends upon the nature of the material. It is called specific resistance of the material If l = 1 cm a = 1 sq. cm, then Resistance (R) = specific resistance ρ Thus the specific resistance is defined as equal to the resistance in ohms of a material which is 1 cm in length and 1 sq cm in area of cross-section. In other words, specific resistance of one cm cube of the material. IUPAC recommends the use of term resistivity over specific resistance. It can be seen that 1 Ωm = 100 Ωcm or 1 Ωcm = 0.01 Ωm Units: the units of resistivity are cm 2 a ρ = R. = ohm = ohm. cm l cm Its SI units are ohm meter (Ω m). But quite often ohm centimeter (Ω cm) is also used.

12.16

Electrochemistry

Conductivity and conductance (C): This term is more frequently used in electrochemistry. The term ‘conductance ‘means the ease with which current flows through the 1 conductor. It is usually written as C = . Thus conductR ance is the reciprocal of resistance. Units: The units of conductance are reciprocal ohm (ohm–1) of mho. ohm is also abbreviated as Ω so that ohm–1 may be written as Ω–1 According to SI system, the units of electrical conductance are Siemens S (i.e., 1 S = 1 Ω–1) The inverse of resistivity is called conductivity (or specific conductance). It is represented by the symbol κ (Greek Kappa). The IUPAC has recommended the use of term conductivity over specific conductance. It may be defined as the conductance of a solution of 1 cm length and having 1 cm2 as the area of cross-section. In other words conductivity is the conductance of one centimeter cube of a solution of an electrolyte. Thus κ = 1/ρ +

–

1 cm

Solution of an electrolyte.

1 cm 1 cm

Fig. 12.5 Illustration of conductivity From (iii), we know R = ρ. ∴

l a

or

ρ=

a l

I 1 I =k= . a R ρ

Thus, Specific Conductance (κ) = cell constant × conductance ∴ Cell Constant =

Specific Conductance Conductance

table 12.3 Conductivities of some substances at 298 K

Material Sodium Copper Silver Iron Gold Graphite

Glass Teflon

Conductivity (Sm–1)

Conductors 2.1 × 103 5.9 × 103 6.2 × 103 1.0 × 103 4.5 × 103 12.0

Insulators 1 × 10–16 1 × 10–18

Material

Conductivity (Sm–1)

Aqueous Solutions Pure Water 3.5 × 10–5 0.1 M HCl 3.91 0.1 M NaCl 0.20 0.01 M NaCl 0.12 0.01 M KCl 0.14 0.1 M 0.047 CH3COOH 0.016 0.01 M CH3OOH Semi Conductors Si 1.0 × 10–2 Ge 2.0 CuO 1 × 10–7

12.5.3 molar Conductivity or molar Conductance Molar conductivity is defined as the conducting power of all the ions produced by dissolving one mole of an electrolyte in solution. It is denoted by Λ (lambda). Molar conductance is related to specific conductance (κ) as k Λ= M Where M is the molar concentration. If M is in the units of molarity i.e., moles per litre (mol L–1), the Λ may be expressed as k × 1000 Λ= M Relation between conductivity and molar conductivity: The relation between conductivity and molar conductivity can be easily obtained from the definitions of the terms. Suppose 1 cm3 of a solution of an electrolyte is placed between two large electrodes of 1 cm2 area of cross-section lying 1 cm apart. The measured conductance of the solution will be its conductivity (by definition because it gives conductance of 1 cm cube of solution). Further, suppose that this solution contains one gram mole of the electrolyte, then the measured conductance of the solution will be equal to the molar conductance (Λ). Thus, for this solution containing 1 gm mole of electrolyte placed between two parallel electrodes of 1 cm2 area of cross – section and one cm apart. Conductance = Conductivity (κ) = Molar conductivity (Λ) Now suppose that solution is diluted to 100 cc, there are now 100 cm cubes of the solution. The conductance of

Electrochemistry

each one cm cube will be conductivity so that the conductance of the solution will be 100 times of its conductivity. But even now, the solution contains 1 gram mole of the electrolyte therefore the measured conductance will be the molar conductivity. Thus Molar conductivity (Λ) = 100 × conductivity In other words Λ = κ × V Where V is the volume of the solution in cm3 containing one gram mole of the electrolyte. 1 mole of electrolyte is present in 1000 cm3 of solution M Thus Λ = κ × V in cm3 containing 1 mole of electrolyte. =

units of molar Conductance The units of molar conductance can be derived from the formula Λ=

k × 1000 M

The units of κ are s cm–1 and units of Λ are cm3 Λ = S cm -1 × mol = S cm 2 mol -1

12.17

Equivalent conductance is defined as the conducting power of all the ions produced by one gram equivalent of an electrolyte in a given solution. It is denoted by Λe. Equivalent Conductance = Specific conductance x V where V is the volume in CC containing one gram equivalent of the electrolyte. Thus Λe = κ × V The units of equivalent conductance are ohm–1 cm2 (gram equiv)–1 which can be obtained as follows. Λe = κ × V cm3 –1 –1 = ohm cm × g. eq = ohm–1 cm–1 (g. eq)–1

12.5.5 measurement of the Conductance of solutions Conductivity Water: Water used for the preparation of solutions should be absolutely pure. It should not give conductance due to dissolved impurities. Such water is called conductivity water. It is prepared by distilling a number of times to which a little KMnO4 and KOH has been added in a hard glass distillation assembly. The conductance of this water is as low as 4.3 × 10–8 ohm–1. Conductivity Cells: The solution whose conductivity is to be determined is placed in a special type of cell known a conductivity cells. These are various forms as shown in Fig 12.6.

According to SI system, molar conductance is expressed as s m2 mol–1, if concentration is expressed as mol m–3. Both types of units are used in literature and are related to each other as 1 S m 2 mol -1 = 10 4 S cm 2 mol -1 or 1 S cm 2 mol -1 = 10 -4 S cm 2 mol -1

12.5.4 Equivalent Conductance It is a property which is suitable for characterizing electrolytic conductors. In such cases, the value of the conductance also depends upon the concentration of the electrolyte or the number of ions present in solution. One mole of KCl ionizes to give one mole of K+ ions and one mole of Cl– ions. The total charge carried by these ions is equal to two Faradays. On the other hand, one mole of potassium sulphate ionizes to give two moles of K+ ions and one mole of SO42– ions. The total charge carried by these ions is equal to 4 faradays. Clearly, two solutions (i) one containing one mole of KCl and other containing half mole of potassium sulphate will give comparable results. Hence for measuring conductance of electrolytes a more significant term is commonly used, which is equivalent conductivity.

Fig 12.6 Conducting cells These are made of pyrex glass and fitted with glass tubes which in some cases are finely fixed in an abonite cover so that the distance between electrodes is not altered during the experiment. The electrodes are connected to the circuit by means of mercury placed in the tubes. These electrodes are platinized by putting a solution of chloroplatinic acid in the cell and passing the alternating current for some time. The electrodes become black due to

12.18

Electrochemistry

a coating of finely divided platinum. This fine deposit gives a larger surface to the electrodes. Cell constant: We know that 1 1 . R A 1 = L× A

k =

equation the same cell constant applies to a measurement with any other solution. table 12.4 Specific conductance κ of KCl solution

l k = A L Specific conductance ,k Or cell constant = Observed conduction L Cell constant =

Specific conductance = κ = L × cell constant Units of cell constant =

k

L Ohm -1m -1 = = m -1 ( S .I ) Ohm -1 The resistance of the cell is measured when filled with a standard solution (Table 12.4) (Say

N KCl solution) at 10

a given temperature and cell constant is calculated from the

Resistance Box

k (ohm–1 cm–1)

Concentration (N)

273 K

291 K

293 K

298 K

1.0

0.06543

0.09820

0.10202

0.11173

0.1

0.007154

0.011192 0.011667 0.012886

0.01

0.0007751 0.001227 0.001275 0.0014111

Measurement of conductance: The wheatstone bridge is generally employed for this purpose. It measures the resistance of solution the reciprocal of which is the conductance. A schematic diagram of the apparatus is shown in Fig. 12.7. The unknown solution of electrolyte taken in a suitable conductivity cell placed in a thermostat.The resistance of an electrolyte cannot be measured by applying direct current because of following reasons to polarization effects. (i) Electrode reaction change the concentration of electrolyte arround the electrode.

R

B

Z

A R

Telephone head

Induction coil Thermostat Test Solution

Fig 12.7 Conductivity determination

Electrochemistry

Since R is known and lengths ZB and ZA can be read from the scale fixed below the wire AB, the resistance of the solution can be calculated. The various defining equations are summarized in table . table 12.5 Definitions of electrical quantities units Quantity

Symbol

c.g.s

SI

Resistance

R

ohm

ohm

Specific resistance

ρ

ohm-cm

ohmcm

Conductance

L

ohm–1 (or) mho

ohm–1

Specific conductance

κ

ohm–1 cm–1

ohm–1 m–1

Cell constant Equivalent conductance

κ´

cm–1 cm2 equiv–1 ohm–1

m–1 m2 equiv–1 ohm–1

E=1R

R.A l 1 L= R

ρ=

1 1 = p AR 1 κ´ = A 1000K Λ= C

κ=

table 12.6 Molar conductivity (s cm2 mol–1) of a few electrolytes in water at 298 K C

HCl KCl KNO3 CH3 COOH NH4OH

0.1 0.05 0.01

391.3 129.0 120.4 399.1 133.4 126.3 412.0 141.3 132.8

0.005 0.001 0.0005 Infinite dilution

415.8 421.4 422.7 426.2

143.5 146.9 147.8 149.9

131.5 141.8 142.8 146.0

5.2 16.3

3.6 11.3

49.2 67.7 390.7

34.0 46.9 271.0

The values in the table indicate that molar conductance of strong (HCl, KCl, KNO3) as well as weak electrolytes (CH3COOH, NH4OH) increases with decrease in concentration or increase in dilution. The variation is different for strong and weak electrolyte. (i) Variation of conductivity with concentration of strong electrolytes: A strong electrolyte (e.g., KCl) shows a small drop in Λ as the concentration increase. Since the KCl is completely ionized and Λ measures the conductance due to a constant number of ions it might be expected that there would be no change with change in concentration. 150 KCl

100

4 10

Λ

Defining relationship

solutions have high conductance. Weak electrolytes ionize to lesser extent so their solutions have low conductance. 2. Concentration of the solution: Generally, the molar conductance of an electrolyte increases with decrease in concentration or increase in dilution. The molar conductance of few electrolytes in water at different concentrations are given in table 12.6.

S m2 mol-1

(ii) The products of electrolysis produce an opposing potential. In order to avoid these difficulties, the resistance of electrolytes is measured by applying an alternating current with the frequency of the order of kilo hertz from an induction coil. R is the resistance box and a telephone head is used to direct the current. AB is a uniform wire and a sliding contact Z moves over it. When current flows, a known resistance R is intro duced through the resistance box. The resistance should be of about same order as that of the solution. The sliding contact Z is moved along the wire AB until no sound (or minimum sound) is heard in the detector at this null point. Resistance of the solution Length ZB = Resistance R Length ZA

12.19

In general, conductance of an electrolyte depends upon the following factors. 1. Nature of electrolyte: The conductance of an electrolyte depends upon the number of ions present in the solution. Therefore, the greater the number of ions in the solution, the greater is the conductance. Strong electrolytes ionize completely and hence their

50

Λ

12.5.6 Factors affecting Variation of molar Conductance

Acetic acid

0

0.05 Concentration (mol dm–3)

0

Fig 12.8 Variation of molar conductance with concentration for a strong electrolyte (KCl) and a weak electrolyte (acetic acid)

12.20

Electrochemistry

Debye and Huckel explained the behaviours of strong electrolytes. In concentrated solution of strong electrolytes, there are vast number of ions but they are very close together indeed owing to the attraction between the opposite charges a positive ion will on average find itself surrounded by a sphere of negative ions. Similarly, a negative ion will be surrounded by positive. Each ion interferes with its neighbors. Therefore if a positive ion attempts to move in one direction under the influence of the electric field between the plates in a conductivity cell the surrounding negative ions will hold it back. Likewise the negative ions are restrained by the positive ions. Now if the solution is diluted (increase in dilution) the ions are on average further apart and the amount of interference between them decrease (Fig 12.9) this is why the molar conductivity of strong electrolyte increases with dilution. To set against this, is the fact that, as the dilution increase, there comes a time when the amount of interference is so small that further dilution has little or no effect. When this happens, the molar conductivity remains constant. It can be seen as the curve for potassium chloride level off. The limit that the line approaches is the molar conductivity at infinite dilution. If the graph is extrapolated to zero concentration the limiting molar conductivity (Λ 0) is obtained. The extrapolation is more accurate if Λ is plotted against C since this gives a straight line. Λ0 represents the molar conductivity of the electrolyte in the absence of any ionic atmosphere effect. It has been observed that the variation of molar conductivity with concentration may be given by the expression. Λ = Λ0 – AC1/2

Where A is a constant and Λ0 is the molar conductivity at infinite dilution. (ii) Variation of molar conductivity with concentration for weak electrolytes: The variation of Λ with dilution can be explained on the basis of number of ions in solution. The number of ions furnished by an electrolyte in solution depends upon the degree of dissociation with dilution. In weak electrolytes, a minority of the molecules are dissociated into ions. This is why the molar conductivity is very small. As more water is added, more of the molecules dissociate into ions. So the molar conductivity increases. The graph for ethanoic acid show that the increase in conductivity is not as rapid as for strong electrolytes. The line approaches a limiting value very slowly and attempts to predict the limiting value that it might reach can give only very approximate answers. The degree of dissociation can be calculated at any concentration as a ∝=

Λc Λ0

Where a the degree of ionization, ΛC is the molar conductance at concentration C and Λ0 is the molar conductance at infinite dilution. We shall call Ka the equilibrium constant for the dissociation of acetic acid in water. +  CH 3 COOH (aq ) ↽ ⇀  CH 3 COO (aq ) + H (aq )

CH 3 COO - (aq )   H + (aq )  Ka =  CH 3 COOH ( aq ) Let us suppose that 1 mol of acetic acid is in a volume V of solution. The acid is only partially dissociated into ions and we shall say that at equilibrium a fraction i.e.,

— —

—

—

—

— —

+

—

+

—

— —

—

— —

— —

Fig 12.9 Ionic atmosphere effect

Electrochemistry

α (alpha) is converted into acetate and hydrogen ions. Then we have the following pattern of concentrations: +  CH 3 COOH (aq ) ↽ ⇀  CH 3 COO (aq ) + H (aq )

1- a a V V Conc. at equilibrium mol dm -3

a V

(a v)(a V ) ∴Ka = (1 - a ) V

or Ka =

a2 (1 - a )V

Now , if we put a = Λ Λ 0 Ka =

(Λ Λ )

2

0

(1 - Λ Λ 0 )V

By using this equation we can calculate the equilibrium constant from conductivity measurements.

12.6 KohlrausCh’s law

table 12.7 Kohlrausch‘s law of independent migration of ions Λe (298 K)

Difference

KOH LiOH

271.5 236.7

34.8

KCl LiCl

149.9 115

34.9

KNO3 LiNO3

145 110.1

34.9

KCl NaCl

149.9 126.5

23.4

KBr NaBr

156.6 128.2

23.4

KNO3 NaNO3

145 121.6

23.4

HCl HNO3

426.2 421.3

4.9

Electrolyte

Λe (298 K)

Difference

LiCl LiNO3

115 110.1

4.9

KCl KNO3

149.9 145

4.9

NaCl NaNO3

126.5 121.6

4.9

Electrolyte

From these observations, Kohlrausch concluded that at infinite dilution where dissociation for all electrolytes is complete and where all inter ionic effects disappear, each ion migrates independently and contributes to total equivalent conductance of an electrolyte a definite share which depends only on its own nature and not at all and that of the ion with which it is associated. Kohlrausch noticed this fact and put forward the law of independent migration of ions as. The equivalent conductance at infinite dilution is composed of two independent quantities one contributed by anion and the other by cation. Consequently o

It has already been seen that equivalent conductivity of solution increase with dilution until it becomes constant. This limiting value Λe is known as the equivalent conductivity at infinite dilution. Kohlrausch (1875) observed that the difference between the equivalent conductivity at infinite dilution of pairs of salts having an ion in common is constant at a constant temperature (Table 12.7)

12.21

o

Λ0 = Λa + Λc o

o

Where Λ a and Λ c are the ionic equivalent conductivities at infinite dilution of anion and cation respectively. The ionic conductivity can also be related to the speed at which the ions travel. The current carried by an ion will depend on the rate at which it conveys electrical charge across the solution i.e., Total current ∝ Number of ions × Charge × Velocity. The ionic conductivity represents the current carried by one mole of ion, so that the number of ions is equal to the Avogadro constant (L). Supposing the electrolyte to contain only singly charged ions, i.e., each ion carries a charge equal in magnitude to that of a portion (e) then the total charge will be Le. This is a fixed quantity of electricity known as the Faraday constant. It follows that Λ + ∝ Velocity of cation

Λ - ∝ Velocity of anion

The velocities given in this relationship are actually equal to the speed at which the ions move when there is a potential gradient of 1 V m–1 ; this is known as the ionic mobility. Some ionic conductivities are given in Table 12.8

12.22

Electrochemistry

table 12.8 Ionic conductivities at 298 K (Sm2 mol–1) Cations +

H Na+ K+ Ag+ Li Ca2+ Mg2+ Cu2+

Anions −

0.0350 0.0050 0.0074 0.0062 0.00387 0.0119 0.01062 0.01072

OH Cl− NO−3 HCOO− F− Br− I− SO2− 3 CO2− 3

0.0199 0.0076 0.0072 0.0045 0.00554 0.00781 0.00768 0.01596 0.01386

12.6.1 applications of Kohlrausch's law

Comparison of ionic conductivity figures can throw some light on the extent to which ions in solution are hydrated. The force acting on each ion carrying a single charge due to a unit potential gradient will be the same. Assuming that the ions are all spherical in shape, the speed of an ion will depend on size: a small ion will travel faster than a large one. Now Na+ is smaller than K+ as far as the actual ions themselves are concerned but the figures show that Na+ in solution has lower conductivity than that of K+ so that the sodium ion when in solution must be larger than the potassium ion. This is because the former attract more water molecules i.e., it is hydrated to greater extent than the latter. The conductivity figures also show that hydrogen and hydroxyl ions conduct electricity at a much higher rate than other ions-too high in fact to be accounted for by the movement of the ions themselves. Hydrogen and hydroxyl ions appear to conduct so much better than sodium ions because of some subterfuge. They employ a different method of getting about. They indulge in molecule hopping. Hydrogen ions which are really protons attach themselves to lone pairs on the water molecules. A proton on one water molecule can very easily find itself on a neighboring molecule simply by swapping from one lone pair to another. The result is that conduction takes place by protons being passed from one

+ H + :O

H

H

H

+ O

H

H

H

– O + H

O

O

– H + O

H

H

H

H

molecule to another. The conduction process is something like the knocking down of giant layouts of dominoes. The wave of movements just as the conduction by protons in water is the result of many small changes in the arrangements of the protons. Same is the case with hydroxyl ions. This process can be continued through the electrolyte and causes the charge to travel faster than the solution than the speed at which ions can move.

1. Determination of limiting equivalent conductance of weak electrolytes: Equivalent conductance at infinite dilution can be obtained graphically by extrapolation for strong electrolytes. This procedure cannot be applied for weak electrolytes. Such solutions have lower conductances at higher concentrations but the values increase greatly with increasing dilution. Because of the steep rise in Λ at high dilution the extrapolation to zero concentration is uncertain and may result in large errors. Kohlrausch’s law gives a different procedure for determining Λ0 of weakly dissociated electrolytes. Suppose we want to determine Λ0 values of acetic acid. From Kohlrausch’s law Λ 0 CH 3 COOH = Λ 0 CH 3 COONa + Λ 0 HCl - Λ 0 NaCl 0 0 0 = Λ 0 Na + + Λ 0 CH 3 COO - + Λ 0H + + Λ Cl - Λ Cl - - Λ Na +

= Λ 0 CH 3 COO - + Λ 0H + Thus from the conductance of completely dissociated sodium acetate, hydrochloric acid and sodium chloride in water solution the equivalent conductance of weakly dissociated acetic acid in water solution at infinite dilution can be calculated. 2. Calculation of solubility of sparingly soluble salts: Conductance offers a very simple and convenient method for determining the solubility of sparingly

+ O+ H

O

H

H

+ H

O

H

H

H

+ O

H

H

O

H

H

O + H

– H + O

H

H

+

Electrochemistry

soluble salts such as barium sulphate, lead, sulphate, silver chloride etc. in water. The procedure involved is as follows: A saturated solution of the salt in water of known specific conductance kH2O is prepared and its specific conductance ksolution is measured. The specific conductance is due to both the salt and the water. The specific conductance of the salt alone = ksalt = ksolution –kwater. The equivalent conductance of the salt is 1000 k salt Λ0 = C Where C is the concentration in equivalent dm–3 and hence the solubility. Since the solubility is very low the salt is supposed to be completely ionized and the equivalent conductivity may be taken as the equivalent conductivity at infinite 0 dilution i.e., Λ e . Therefore 1000 k salt Λ = C 0 e

Now

Λ 0e salt = Λ + + Λ -

From the value of Λ obtained from the table of ionic conductance’s C can be calculated from the above equation. 3. Calculation of degree of dissociation of weak electrolytes: The degree of dissociation of weak electrolytes such as NH 4 OH , CH 3 COOH etc. at any dilution can be calculated by measuring the equivalent conductivity of solution Λ C and Λ 0 of the electrolyte by the relationship

12.23

Since water dissociates very slightly the concentration of undissociated water may be taken as constant. ∴  H +  OH -  = Constant. This constant is represented by Kw which is called ionic product of water. Thus KwK=w  H +  OH -  The product of concentrations of H + and OH - ions in water is constant at constant temperature. K w can be determined by a number of methods. Conductivity measurement is one of them. Ionic product of water depends on temperature. Table 12.9 consists of the best values of Kw at different temperature table 12.9 Ionic product of water Kw at various temperatures Temperature (K)

Kw × 1014

273 283 298 310 333

0.114 0.292 1.008 2.919 9.614

0

a ∝=

ΛC Λ0

4. Calculation of ionic conductance’s: By combining Kohlrausch’s law and the ionic mobilities absolute ionic mobility of ions can be calculated. Absolute ionic mobility =

Ionic Conductance 96500

5. Ionic product of water: Conductance measurements and other evidences support that water ionises according to the equation  H + + OH H 2 O 

For the ionization the equilibrium constant  H +  OH -  K =    H 2 O 

12.6.2 Conductometric titrations Titrations followed conductometrically are often used in chemistry. The principle involved in such titrations is that electrical conductance depends upon the number and mobility of ions. For practical purposes, it is not necessary to know the actual specific conductances of the solution. The conductance readings correspond to the various added amounts of titrants are plotted against the latter. The intersection of the lines gives the required end point. The method conductometric titration is capable of considerable accuracy provided that there is good temperature control and a correction is applied for the volume changes during the titration. The most common conductometric titrations are given here. 1. Titration of strong acid against strong base: When a strong alkali i.e., NaOH added to a solution of strong acid i.e., HCl the reaction is (H + + Cl - ) + (Na + + OH - )  → Na + + Cl - + H 2 O Here the fast moving hydrogen ions are replaced by slower moving sodium ions; hence the conductance decreases with the addition of NaOH Solution. After reaching the equivalence point, the fast moving hydroxyl ions will be present unneutralised in excess which will increase the conductance. The equivalence point is determined from the break in the curve.

Electrochemistry

A

Conductivity

D

C

B x

End point

Volume in mL

Fig 12.10 Titration of strong acid against strong base 2. Titration of weak acid against strong base: If a moderately weak acid is titrated against strong base such as sodium hydroxide, the form of conductance titration curve is as shown in fig 12.11. The neutralization reaction may be represented as (CH 3 COO - + H + ) + ( Na + + OH - )  → CH 3 COO - + Na + + H 2 O

replaces, and so the specific conductance of the solution increases. When the acid is completely neutralised, further addition of NaOH introduces of fast moving OH–. The conductance therefore increase more rapidly after the equivalence point. The break in the conductance titration curve gives the equivalence point. 3. Titration of strong acid against weak base: If a strong acid is titrated with a weak base i.e., aqueous solution of HCl against NH4OH the reaction occurs as

( H + + Cl - ) + ( NH 4+ + OH - )  → NH 4+ + Cl - + H 2 O The conductance will fall at first due to replacement of fast moving H+ by slow moving NH4+ ions. When the equilibrium point is passed, however, the conductance will remain almost constant, since the free base is a weak electrolyte and does not dissociate much in the absence of acid (due to common ion NH4+ from the salt formed during neutralization). The curves obtained are shown in Fig 12.12

End point Conductanctivity

12.24

Conductivity

End point x Volume of weak base added Fig 12.12 Titration of strong acid against weak base

Volume of base added

4. Titration of a weak acid versus weak base: The reaction between a weak acid say, CH3COOH and a weak base NH4OH is  NH 4+ + CH 3 COO - + H 2 O CH 3 COOH + NH 4 OH 

The initial solution of weak acid has a low conductance and the adding of alkali may first result in further decrease inspite of the formation of a salt i.e., sodium acetate with a high conducting power. The reason for this is that the common anion i.e., CH3COO– Suppresses the dissociation of the acetic acid with further addition of NaOH, however, the conductance of the highly ionized salt soon exceeds that of the weak acid which it

Conductance

Fig 12.11 Titration of weak acid against strong base

Volume of weak base added

Fig 12.13 Titration of weak acid against weak base

Electrochemistry

NaO

HC l

H vs N

aO

Conductivity

H

CH 3

H COO

aOH vs N

volume of NaOH added

Fig 12.14 Titration of a mixture of HCI and CH3 COOH against NaOH

6. Titrations involving precipitation reactions: In reactions of the type

( K + + Cl ) + ( Ag + + NO3- )  → AgCl + K + + NO3When a precipitate is formed, one salt is replaced by an equivalent of another e.g. KCl by KNO 3 and show that the conductance remains almost constant in the initial stages. After the end point is passed however, the excess of added salt causes a sharp rise in the conductance. The end point of the reaction can be determined from the plots. (Fig 12.15)

End Point Conductivity

The conductivity in the beginning decreases due to formation of a salt, CH3COONH4 which decreases the dissociation of weak acid further. With more addition of NH4OH the conductance increases due to increased concentration of the salt, which is highly ionized. This continues until at end point, considerable hydrolysis of the salt occurs. However, beyond this, the salt hydrolysis is suppressed by the excessive base added. Thus conductance remains nearly constant Fig 12.13 5. Titration of a mixture of strong and weak acids: One of the valuable features of the conductance method of analysis is that it permits the analysis of a mixture of HCl and CH3COOH is to be titrated against an alkali, the curves are shown in Fig 12.14. The initial decrease is due to the neutralization of strong acid first. The titration of acetic acid will start only after HCl has been completely neutralised. When the neutralization is complete there is further increase in conductance due to the excess of strong base.

12.25

vol of KCl added

Fig 12.15 chloride

Titration of silver nitrate against potassium

Advantages of conductometric titrations: conduct metric titrations have several advantages: 1. Coloured solutions which cannot be titrated with the help of indicators can be successfully titrated conductometrically. 2. These titrations can be employed successfully for the titration of weak acid against weak base which do not give the sharp colour in ordinary volumetric titrations. 3. It can be also employed in case of very dilute solutions. 4. No special precaution is necessary near the end point as it is determined graphically. 5. These titrations can be used to study the precipitation reactions. solved Problem 2 The equivalent conductance of sodium propionate, hydrochloric acid and sodium chloride at infinite dilution are 85.9, 426.1 and 126.4 ohm–1 cm2 eq–1 respectively at 298 K. Calculate the equivalent conductance at infinite dilution for propionic acid. Solution: According to Kohlrausch’s law (i) (Λ∝) C2H5COONa = (Λ∝) C2H5COO + (Λ∝) H+ (ii) (Λ∝) HCl = (Λ∝) H+ + (Λ∝)Cl– (iii) (Λ∝) NaCl = (Λ∝)Na+ + (Λ∝)Cl–

12.26

Electrochemistry

To obtain (Λ∝) C2H5COOH = Λ∝ C2H5COONa + (Λ∝) HCl– (ΛNaCl) = 85.9 + 426.1–126.4 = 385.60 hm–1 cm2 eq–1 solved Problem 3 The molar conductivities at infinite dilution of AgNO3, NaCl and NaNO3 are 116.5, 110.3 and 105.2 ohm –1 cm2 mol–1 respectively. The conductivity of AgCl in water is 2.4 × 10–6 ohm–1 cm–1 and of water used in experiment is 1.16 × 10 –6 ohm–1 cm–1. Find out the solubility of AgCl in gm dm–3 Solution: 1000 k salt C Now (Λ ∝ ) AgCl = (Λ ∝ ) AgNO3 + (Λ ∝ ) NaCl - (Λ ∝ ) NaNO3

Λ∝ =

= 116.5 + 110.3 - 105.2 = 121.6 k salt = k solution - k H O

solved Problem 5 At 291 K the ionic velocities of Ag+ is 0.00057 cm s–1 and that NO 3 is 0.00063 cm s–1. What is the value of Λ∝ for AgNO3 at 291 K? If the specific conductivity of 0.1 N AgNO3 solution is 0.00947 ohm–1 cm–1 find the degree of dissociation at this dilution. Solution: Ionic conductance Absolute mobility = 96500 Λ Ag+ = 0.00057 × 96500 Λ NO- = 0.00063 × 96500 3

∴Λ Ag+ + Λ NO- = 96500 (0.00057 + 0.00063) 3

= 96500 × 0.0012 = 115.8 1000 × k 1000 × 0.00947 Λc = = = 94.7 c 0.1 Λ 94.7 ∴a = c = = 0.817 Λ ∝ 115.8

2

= 2.4 × 10-6 - 1.16 × 10-6 ohm -1 cm -1

solved Problem 6

Substituting the values in the above equation.

At 298 K the specific conductivity of water is 5.51 × 10–8 ohm–1 cm–1 the ionic conductance’s of H+ and OH– ions at the same temperature are 349.8 and 198.5 respectively. Calculate the value of ionization constant of water. Solution: 1000 × k 1000 × 5.51 × 10 -8 = Λc c c Λ ∝ = Λ H+ + Λ OH - = 369.8 + 198.5 = 548.3

1000 × 1.24 × 10 -6 121.6 = C 1000 × 1.24 × 10 -6 C= 121.6 = 1.019 × 10 -5 equivalent per dm3

= (1.019 × 10 -5 × 143.5) g per dm3 = 1.4622 × 10 -3 g dm -3

Water may be considered to be the dilute solution of H+ and OH– ions in dissociated molecule of H2O. Therefore Λc will be taken as Λ∝

solved Problem 4 Calculate the degree of dissociation of sodium chloride in 0.1 N Solution. The equivalent conductivity at this dilution is 98.4 ohm–1 cm2 equivalent –1 and ionic conductances of sodium and chloride ions are 43.4 and 65.5 ohm–1 cm2 equiv–1 respectively Solution: −1

2

Λ c for NaCl = 98.4 ohm cm equivalent

1000 × 5.51 × 10 -8 c 1000 × 5.51 × 10 -8 = 1.005 × 10 -7 c= 548.3 K w = [H + ][OH - ] = 1.005 × 10 -7 × 1.005 × 10 -7 ∴548 =

= 1.01 × 10 -14

−1

Λ ∝ for NaCl = Λ Na + + Λ Cl− = 43.4 + 65.5 =108.9 Λ 98.4 α= c = = 0.903. Λ ∝ 108.9

Problems for Practice 31. The resistance of a solution ‘A’ is 50 ohms and that of solution ‘B’ is 100 ohms, both solutions being taken in the same conductivity cell. If equal volumes of

Electrochemistry

solution A and B are mixed, what will be the resistance of the mixture using the same cell? (Assume that there is no increase in the degree of dissociation of A and B on mixing) 32. The equivalent conductivity of 0.05 N Solution of a monobasic acid is 15.8 s cm2 eq–1. If equivalent conductivity of the acid at infinite dilution is 350 s cm2 eq–1. Calculate the 1. Degree of dissociation of acid 2. Dissociation constant of acid. 33. At a certain temperature the saturated solution of silver chloride has conductivity 1.12 × 10–6 s cm–1. The ionic conductances of Ag+ and Cl– ions at infinite dilution are 54.3 and 65.5 s cm2 eq–1 at the same temperature. Find the solubility of AgCl at this temperature. 34. Calculate the conductivity (in 1 × 10 –7 s m–1 unit) of a solution prepared by dissolving 10 –7 mole of AgNO3 in 1 litre of saturated aqueous solution of AgBr. K sp of AgBr = 12 × 10 -14 and Λ oAg+ , Λ oBr - and Λ oNO3

-3

-3

-3

are 6 × 10 , 8 × 10 and 7 × 10 sm mol

35.

36.

37.

38.

39.

2

-1

respectively. Neglect the contribution of solvent (IIT 2006) A water sample of swimming pool shows the resistance of 9.2 × 103 ohm at 25°C using a conductivity cell. The same cell dipped in 0.02 M KCl has the resistance of 85 ohm. Now 0.5 kg sodium chloride is dissolved in pool and thoroughly stirred. The solution sample now shows the resistance of 7.6 × 103 ohm. If Λ0M for NaCl and ΛV for 0.2 M KCl are 126.5 s cm2 mol–1 and 138.0 s cm2 mol–1, calculate the volume of water in pool. The molar conductivities at infinite dilution of AgNO3, NaCl and NaNO3 are 116.5, 110.3 and 105.25 cm2 mol–1 respectively. The conductivity of AgCl in water is 2.40 × 10–6 s cm–1 and of water used is 1.16 × 10–6 s cm–1 Find the solubility of AgCl in g litre –1 Conductivity of a saturated aqueous solution of Co2[Fe(CN)6] is 2.06 × 10–6 s cm–1. If ionic molar conductivities of Co2+ and [Fe (CN)6]4– are 86 s cm2 mol–1 and 444.0 s cm2 mol–1 respectively, calculate the solubility product of Co2 [Fe (CN)6]. A big irregular shaped vessel contained water, the conductivity of which was 2.56 × 10 –5 s cm–1. 500 g of NaCl was then added to the water and the conductivity after the addition of NaCl was found to be 3.10 × 10 –5 s cm–1. Find the capacity of the vessel if it is fully filled with water ( Λ oM of NaCl = 149.9) The equivalent conductivity of 0.10 N solution of MgCl2 is 97.1 s cm2 eq–1 at 25°C. A cell with electrodes that are 1.50 cm2 in surface area and 0.50 cm apart is filled with 0.1 N MgCl2 solutions. How much

40.

41.

42.

43.

12.27

current will flow when the potential difference between the electrodes is 5 volts? At 298 K the conductivity of pure water is 5.51 × 10–8 s cm–1. The ionic conductances of H+ and OH– at this temperature are 349.8 and 198.5 s cm2 eq–1 respectively. Calculate the ionic product of water. At 18°C the mobilities of NH 4+ and ClO4– ions are 6.6 × 10 –4 and 5.7 × 10 –4 cm2 volt–1 s–1 at infinite dilution. Calculate conductance of ammonium perchlorate solution. For H+ and Na+ the values of Λ° are 349.8 and 50.11 respectively. Calculate the mobilities of these ions and their velocities if they are in a cell in which the electrodes are 5 cm apart and to which a potential of 2 volts is applied. The equivalent conductance of an infinitely dilute solution of NH4Cl is 150 and the ionic conductance’s of OH– and Cl– ions are 198 and 76 respectively. What will be the equivalent conductance of the solution of NH4OH at infinite dilution. If the equivalent conductance’s of a 0.01 N solution of NH4OH is 9.6, what will be its degree of dissociation.

12.7 ElECtrolysIs Before we explain electrolysis we must understand that if an electric current is to pass between two points there must be a movement of charged particles between the points. In a metal the particles are free electrons. If a voltage is applied across the ends of a metal wire the free electrons will being to move along the wire and the current can be measured. If however a current is to pass through a liquid there must be ions that are free to move. Thus the substances which allow electric current to flow through them are called electric conductors. The substances whose solution in water or their fused liquids allow electric current to flow through them and undergo a chemical change are called electrolytes e.g fused salts such as NaCl, KCl, aqueous solutions of salts such as NaCl, KCl, K 2 SO4 , CuSO4 , acids such as HCl, H 2 SO4 , HNO3 and bases such as NaOH, KOH etc. There are many chemical substances which do not conduct electric current through them either in the solid or in the molten state or in aqueous solutions. These are called non- electrolytes e.g. glucose, sugar, urea etc. Thus electrolysis (Olysis = decomposition, electro = by emf) means the decomposition of a chemical substance by applied emf we have already seen that no chemical reaction occurs when electric current is passed through metallic conductors, but electrolytes undergo chemical reactions at the electrode by the passage of electric current.

12.28

Electrochemistry

Phenomenon of Electrolysis

12.8 oVEr VoltaGE

Two platinum wires or rods are placed in the melt or solution of an electrolyte taken in a container. These rods are called electrodes. The two rods are then connected with the help of the copper wires to the two poles of a battery (+ve and –ve poles)

During electrolysis cations migrate to the oppositely charged cathode where electrons are received and the atoms discharged. At the anode one of two things may happen viz anions are discharged or if the electrode be attackable e.g. copper atoms may be converted into ions Both actions are accompanied by the release of electrons. If the electrolyte be a salt in water solution then hydrogen and hydroxyl ions are present also, and thus discharge of either hydrogen or the metal or both at the cathode is possible. At the anode there is a third possibility of discharge of oxygen (from the hydroxyl ions). The factor which determines which kind of ion is discharged is its discharge potential. Suppose two zinc rods are placed in a solution of a zinc salt. Then both rods will have the same potential difference with the solution and when connected no current will flow. Suppose an exceedingly small emf be applied then at one electrode A (say) the cathode zinc atoms are deposited and at the anode B, zinc atoms go into solution as ions to replace them. If the emf is reversed then A becomes the anode and B the cathode. Here the discharge potential is almost the same as the electrode potential for this concentration. The result of electrolysis in the above example taken to explain the phenomenon of electrolysis with the cathode is copper electrode while anode is chlorine electrode. The electrode potentials of copper and chlorine are 0.34 V and 1.36 V respectively so that the cell will acquire an emf of 1.36-0.34= 1.02 V, with the chlorine the positive pole i.e., the cell emf will be in opposition to the applied emf. If at the start of the experiment, the applied emf is small, only a very small current will pass (limited by the rate at which the products of electrolysis can diffuse away from the electrodes). As the applied voltage is increased there will be increasing concentration of copper and chlorine at the electrodes until a voltage is reached which can overcome the emf of the copper – chlorine cell after which the current through the cell will increase rapidly (see Fig 12.17) and copper will be continuously deposited on the cathode and chlorine gas liberated at the anode. The applied emf. required to cause continuous electrolysis is known as decomposition potential. We see that in this example taking place at the electrodes are Cathode Cu 2 + + 2e -  → Cu (reduction )

+

– Ammeter Cathode (Pt)

Volt meter

Cu2+

Cl-

Anode (Pt)

Copper Choride Solution

Fig. 12.16 Electrolysis of copper (II) Chloride solution between platinum electrodes As soon as these connections are made electrolysis starts. Chemical reactions occur at the electrodes with the formation of the decomposition products and the current flows through the solution. For example if an aqueous solution of copper chloride is taken in the container and the electrolysis is carried out copper metal is formed at the cathode and chlorine gas is liberated at anode. In aqueous copper chloride solution there are free Cu2+ and Cl– ions. Cu2+ ions move to this cathode (–ve electrode) and undergo reduction (electronation to the metallic copper) At Cathode : Cu2+ + 2e– → Cu Cl– ions travel towards the anode (+ ve electrode) and undergo oxidation (de electronation) to give Cl2 gas → Cl2 ( g ) + 2e - (deelectronation) At anode 2Cl -  Reduction always takes place at the negative electrode Oxidation always takes place at the positive electrode Thus in the electrolysis of aqueous copper chloride between platinum electrodes the following reaction occur CuCl2  → Cu 2 + + 2Cl At cathode Cu 2+ + 2e-  → Cu (electronation or reducion) → Cl2 + 2e - (deelectronation or At cathode 2Cl -  oxidation ) → Cu + Cl2 Over all CuCl2 

Anode 2Cl -  → Cl2 + 2e - (oxidation) So that in electrolysis, reduction occurs at the cathode and oxidation at the anode. The overall reaction in the cell illustrated therefore

CuCl2  → Cu + Cl2

Electrochemistry

12.29

Current density

followed by the combination of hydrogen atoms to form molecules 2H  → H2

Decomposition Potential

Applied Potential

Fig 12.17 Current voltage curve for the decomposition of copper (II) chloride The external battery supplies the electrical energy necessary to maintain the flow of electrons from anode to cathode at a potential difference equal to back emf of the cell. The current and therefore the rate at which electrolysis occurs, depends on the applied emf in excess of the decomposition potential (see fig 12.17 the steep portion of the graph) and obey ohms law current

Excess potential = Internal resistance of cell

It might be expected that the decomposition potential would be equal to the normal emf of the cell set up by the products of electrolysis which has already been shown to be 1.02 V for the copper-chlorine cell. In fact the decomposition potential is 1.34 V which is higher than the theoritical value by 0.32 V. This additional potential is called the over potential or over voltage and arises because the standard electrode potentials are measured under conditions of equilibrium where as when ions are being continuously discharged, the electrode may no longer be in equilibrium with the ions in solution. It is found that over potentials are particularly high when gases (especially hydrogen and oxygen) are liberated. This is thought to be due to one stage in the process of discharge of the ions being relatively slow. The liberation of hydrogen at an electrode for example involves first the removal of a proton from the hydroxonium ion and its neutralization by an electron: H 3O +  → H + + H 2O H + + e -  →H

It is probable that the particular stage that is mainly responsible for the over potential is not always the same, but depends on the current density. The discharge of a metal ion is a simpler process, and the deposition of metal on an electrode does not usually give rise to an appreciable over potential. Table 12.10 shows some figures for hydrogen and oxygen over potentials at electrodes of different metals. Over potentials increases if the current is increased, and the values given in the table all refer to the same current density. It will be seen that the figures vary with the nature of electrode, and that the physical state of the metal is also important, as shown by the difference between smooth platinum and an electrode that has been ‘Platinised” (i.e., covered with a layer of platinum black) which increases the effective surface area of the electrode. Over potentials are of considerable practical factor as to the nature of the products that will be obtained at the electrodes. table 12.10 Hydrogen and oxygen overpotentials (current density = 10 A m–2) Electrode Platinum (blacked) Platinum (smooth) Nickel Iron Silver Chromium Lead Mercury

Over Potential Hydrogen Oxygen 0.01 0.09 0.33 0.4 0.44 0.5 0.67 1.04

0.46 1.11 0.6 0.41 0.6 0.58 0.8 –

(a) The hydrogen over-potential. The variation of over potential of hydrogen from metal to metal is perhaps by the following step in the discharge of hydrogen ions (i.e., hydroxonium ions)

H 3 O + + e -  → H 2 O + M  H Here M…….H represents a hydrogen atom adsorbed on the metal M, and the strength of the M…..H bond will depend on M. The next stage is: M……H + M…….H→ H2 + 2M i.e., the formation of hydrogen molecule on the metal surface. It is therefore the formation of M….H which influences the over potential.

12.30

Electrochemistry

We can now consider the importance of over-potential in electrolysis. If in an electrolytic cell we have to apply a voltage of Ed to start the process, then Ed = ( E A + eA ) - ( Ec + ec ) + IR Where EA + eA is the anode discharge potential, EA being the reversible potential, and eA the over-potential. The corresponding terms for the cathode are Ec + ec. I is the current and R the resistance (Usually IR is quite small) Ed is then called the decomposition voltage. The products to be obtained at cathode and anode can now be considered. Consider the electrolysis of a neutral aqueous solution of a salt MaAm using unattackable electrodes, e.g. platinum. The cations will be Mm+ and H+, the anions Aa– and OH–. It is thus a question of whether one or both cations and one or both anions will be discharged and at cathode and anode respectively. Since the solution is neutral [H+] = [OH–] = 10–7 The discharge potential of hydrogen, by Nernst equation will be 0.058 log 10 [10-7 ] 1 = - 0.41Volt EH = 0 +

This is the least possible discharge potential of hydrogen from such a solution, and platinised platinum, with hydrogen over potential of nil, would need to be the cathode. If the cathode were shiny platinum, with a hydrogen over-potential of 0.25, then the discharge potential would be (–0.41 – 0.25) = –0.66. At the anode, hydroxyl ions are discharged, the following reactions may occur OH -  → OH + e 2OH  → H2O + O 2O  → O2

The overall process is

4OH -  → 2 H 2 O + O2 + 4e We must therefore consider the factors affecting the discharge of hydroxyl ions. These are (a) the pH:- The value of the reversible potential O2/OH will be given by 0.058 log 10 [OH - ] 1 Thus, in neutral solution where [OH–] = 10–7 In acid solution for example where [OH–] < 10–7 E = +0.81 volts In acid solution of (for example) pH = 1, pOH = 13 E = +0.4 -

∴ - log10 [OH - ] = 13 Hence E = 0.4 + (0.058 × 13) = 1.2Volts. Thus the potential increases as the pH falls. (b) The oxygen over-potential: Again, as with hydrogen, this varies with conditions: Some values for different metals have been given Table 12.10. Thus the overpotential for platinum is about + 0.5 Volt; hence in acid solution, the discharge potential for oxygen is (1.2 + 0.5), i.e., about 1.7 Volts. This is the operative potential for discharge of anions in the electrolysis of sulphates (and nitrates) in acid solution, for sulphate and nitrate ions discharge at a far more positive potential than does the hydroxyl ion.

12.8.1 Cathode Products Suppose we electrolyse with platinum electrodes, a molar solution of copper (II) sulphate acidified*1 to a pH of say, 1. Then the anode potential EA + eA will have to be +1.7 volts. The values of the cathode potential E c + ec required can be obtained from the Fig 12.18. which shows the cathode potential (in acid solution) for various metals plotted against – log Cion i.e., the lines are plots of the Nernst equation hence in molar solution C = 1 and – log C = 0. The cathode potentials of the metals are simply the Ec (reversible potential) values at C = 1, as given in Table 10 (emf series) and these decrease i.e., become more negative, with decreasing values of C. (i.e., as – log C increase) For copper at C = 1, Ec is 0.34 volt, so that when the applied voltage reaches (1.70 – 0.34) i.e., 1.36 volts, deposition of copper will begin. As copper deposits, the concentration of Cu2+ ions in solution (C) must fall and –log10 Cion increase, and so the cathode potential moves along the ‘Cu’ line in the diagram. At – log Cion = 7 the concentration of Cu+2 ions remaining in solution is negligibly small and at this value (the end of the line in Fig. 12.18) the cathode potential has become about 0.15 volt, and so the applied voltage rises too (1.70 – 0.15) i.e., 1.55 volts, when virtually all the copper has been deposited. But the applied voltage must increase well beyond this before hydrogen appears at the cathode, for the diagram (H2 on Cu in acid line) shows that the cathode potential for hydrogen discharge from the now copper coated cathode is –0.6 volt. Hence the applied voltage must be at least 1.70 – (0.6) = 2.3 volts for hydrogen to be liberated. 1

* In a neutral solution of Copper (II) Sulphate, there is always the chance that some quite small effect of electrolysis may alter the pH of the solution with a consequent marked change in electrode potentials. Such small effects produce little change in acid solution. It is for this reason that such a solution is considered – the value of pH = 1 is purely arbitrary, being sufficiently acid to counter act any slight change.

Electrochemistry

0

2

pH 6

4

8

,

- 1.2

H2

- 1.0

on

Cu

ry va Zn

10

ing

12

pH

- 0.8

Cathode potential

- 0.6

g

n ryi

- 0.4 - 0.2

H2

on

Fe H p

a t. v

H2 on Cu in acid

H2 on Pt in acid

0 Cu Ag

+ 0.4 + 0.6 + 0.8 0

2

4

6 8 -log Cion

concentrations of silver, applied voltage will discharge silver ions in preference to other metal ions present. Hence silver can be plated from a bath of potassium silver cyanide K[Ag(CN)2] the article to be plated being the cathode and a piece of silver the anode. The complex ion [Ag (CN)2]– is only slightly ionized -

p

+ 0.2

12.31

10

12

Fig 12.18 Electrode potentials (The dotted lines refer to the pH axis)

If we wish to electrolyze a solution containing 1 gram –ion per litre of zinc, as (say) zinc sulphate, using platinum electrodes, the diagram shows that, in acid solution, as soon as the cathode potential reaches 0 (H2 on Pt in acid’ line) i.e., as soon as the applied voltage reaches 1.70–0 = 1.7 volts, hydrogen is liberated, before the zinc is deposited; hence hydrogen is the only cathode product in acid solution. The deposition of zinc can also be achieved by using a copper cathode (or, of course, copper-plated platinum) to make the cathode potential for hydrogen more negative (–0.6 volt in acid medium, see ‘H2 on Cu in acid’ in acid line in fig 12.18) and further, by using a solution which is slightly alkaline so that the cathode potential for hydrogen is moved up the dotted ‘H2 on Cu, varying pH’ line in the figure, to a value above (i.e., more negative than)that for the deposition of zinc (‘Zn’ Line) i.e., above – 1.0 volt. The decomposition voltage required for deposition of zinc is then initially, 1.70 – (–0.75) = 2.45 Volts, for virtually complete deposition of zinc (end of the Zn line) Hence it is possible, from a solution containing both cupric and zinc ions, to deposit first copper in acid solution, and then zinc in alkaline solution. The quantitative application of this fact is used in electroplating, analysis of alloys containing these metals. Similar considerations explain electroplating in silver plating, the Fig 12.18, indicates that even with very low

 Ag + + 2CN  Ag (CN ) 2   The low discharge potential of silver explains its deposition in spite of its low concentration so low in fact that its presence is not detectable by a chloride ion since the solubility product of silver chloride cannot be reached. Low metal ion concentrations are preferred in plating for they give rise to a very adherent film of metal. Although the concentration of silver ions is low, plating proceeds uninterrupted since the equilibrium concentration of silver ions is maintained from a large reserve concentration of cyanide ions [Ag(CN)2]–. From a mixed solution of potassium tetracyanocuprate (I), K3[Cu(CN)4]. and zinc sulphate with careful control of concentrations it is possible to deposit both metals i.e., to give a plating of brass the composition of which can be controlled. Since metals such as sodium, calcium and aluminum lie above zinc in the electrochemical series they are never deposited at the cathode in acid or alkaline solutionhydrogen only being liberated.

12.8.2 anode Products It has been mentioned that in acid solution with the platinum anode the discharge potential of oxygen is 1.70 volts and that the discharge potential of sulphate and nitrate ions are much higher; hence in such sulphate and nitrate solutions when anions are discharged oxygen is the anode product. Bromide and iodide ions are discharged at +0.54 and 1.17 volts respectively hence in acid solution electrolysis of a bromide or iodide will yield bromine or iodine at the anode rather than oxygen. For chloride ions an anode potential of about 1.3 volts is required for discharge from a molar solution. Hence in neutral solution where oxygen is also formed at an anode potential of (+0.8) + (+0.5) = 1.3 volts, oxygen or chorine are the anode products; but if the solution is more concentrated in chloride ions then chlorine is the major anode product. Thus electrolysis of concentrated sodium chloride solution yields chlorine not oxygen at the anode. Similarly electrolysis of dilute hydrochloric acid yields cathodic hydrogen and anodic oxygen-but from concentrated acid the products are hydrogen at the cathode and chlorine at the anode. The high discharge potential of the acetate ion is lower than that for the hydroxyl ion and hence the acetate ion is

12.32

Electrochemistry

discharged at an anode: the overall anode reaction can be written:

2CH 3 COO - - 2e -  → C2 H 6 ↑ +2CO ↑ This is the Kolbe synthesis of hydrocarbons containing •

an even number of carbon atoms: the first-formed CH 3 CO O

hydrogen is present as an anion consisting of a proton and two electrons, i.e., H– In the extraction of aluminum by the electrolysis of alumina dissolved in cryolite (Na3AlF6) the aluminum with a lower discharge potential than sodium is liberated at the cathode and oxygen with a lower discharge potential than fluorine is evolved at the anode.

•

radicals lose carbon dioxide to give C H 3 radicals which •

•

→ C2 H 6 dimerise to give ethane, i.e., CH 3 + CH 3  We have so far considered anodic processes with an unattackable anode. Because of the high over potential of oxygen at many metal surfaces, the anodic process may be attack on the electrode rather than evolution of oxygen; thus in the electrolysis of a solution using a copper anode e.g. a solution of copper (II) sulphate the anode process is

Cu - 2e -  → Cu 2 + At an anode potential of only +0.34 volt; hence copper atoms form ions in this way oxygen is not evolved. Controlled attack on a metal anode can be used to produce a smooth surface on it. This is known as electro-polishing. In all the above cases, it has been assumed that anode and cathode products are kept separate. If diffusion of these should occur chemical interaction may take place. Consider the electrolysis of brine with an unattackable anode, e.g. carbon. Removal of hydrogen ions at the cathode results in the increase in OH– ion concentration to give an alkaline solution. If these ions diffuse to the anode they react with the chlorine liberated to give a mixture of chloride and hypochlorite:

Cl2 + 2OH -  → Cl - + ClO - + H 2 O

(1)

or in hot solution

3Cl2 + 6OH -  → 5Cl - + ClO3- + 3H 2 O

(2)

Electrolysis under both conditions are carried out on the industrial scale using a stirrer between the electrodes to mix the products. (1) is used in the manufacture of bleaching fluid, and (2) to make sodium chlorate from which the potassium compound is obtained. Fused Electrolytes Here is a general rule only one kind of cation and one kind of anion are present and the question of selective discharge does not arise. It may be mentioned that in the electrolysis of the fused lithium and sodium hydrides hydrogen is evolved at the anode for in these ionic compounds the

12.8.3 mercury Cathodes In castner - Kellner or Kellner – Solvay process for the manufacture of caustic soda, chlorine and hydrogen, saturated brine is electrolyzed using a carbon anode and mercury cathode. At the cathode sodium is liberated and dissolves in the mercury to form sodium amalgam which is withdrawn from the cell and treated with water* 2 to give caustic soda and hydrogen. Two factors determine the liberation of sodium instead of as might be expected hydrogen at the cathode. The very high over-potential of hydrogen at a mercury electrode raises its discharge potential very considerably but this in itself will not bring the figure to –2.72 volts which is that for the discharge of pure solid sodium. Here however the sodium dissolves in the mercury to give a dilute solution and the electrode potential of such a dilute amalgam is about a volt less than that of solid Na/Na+ The ability of mercury cathode to discharge most metal ions in preference to hydrogen is made use of in polarography in which a cathode of mercury dropping from a jet is used so that the metal ions discharge on fresh mercury surface. The potential at which each metal discharges is then characteristic of that metal and polarography is thus a valuable method of electrolytic analysis of a mixture of metal in the same solution. Zinc has a fairly high hydrogen over-potential and thus pure zinc has little action with a dilute acid. There is even less action if the zinc be amalgamated for not only is the hydrogen over-potential greater, but since the zinc is in solution as an amalgam its tendency to form ions is decreased. If some pure zinc is coated with copper (by dipping into copper (II) sulphate solution) then evolution of hydrogen from a dilute acid proceeds readily for the hydrogen over-potential of copper is much less than that of zinc. If we are to predict which ions will be discharged during electrolysis, we should know their over voltages. But it is not necessary at this level of chemistry. However, the more easily a metal ion is reduced i.e., the more easily it 2

*A carbon or metal strip touches the amalgam in the water; this strip then becomes the cathode and the amalgam the anode of a ‘cell’ which is short-circuited since the electrodes are in contact. Hydrogen is then rapidly liberated at anode. This device is called “denuder”

Electrochemistry

12.33

gains electrons; the more easily it should be discharged at the cathode. For example Cu2+ ions should be more easily discharged than Na+ ions. On the other hand, the negative ions that are oxidized the most easily i.e.give up their electrons easily; will be most easily discharged at the anode. For example I– ions should be more easily discharged than Cl– ions. The order of discharge applies in the particular cases are summarized in Table 12.11.

The order of discharge is only approximate and changes with concentration. In aqueous solutions, water moles are always available for discharge. But in acidic solutions, where the concentration of hydrogen ion is appreciable, H+ ions are discharged in preference to water molecules, likewise in alkaline solutions, where the concentration of hydroxide ions is appreciable the reactions are

table 12.11 Selective discharge of ions during an electrolysis

Thus if more than one type of ions are attracted towards a particular electrode, then the one discharged is the ion which requires least energy. This is called the preferential discharge. Products of electrolysis and the electrode reactions of fused salts and aqueous solution of some salts are given in table 12.12. It can be seen that if more than one ion is present in the electrolyte, the reaction with higher value of E is prefered and the product in that reaction is liberated at that electrode. It can also be seen that when an electrolyte is electrolyzed between the electrodes made with the metal common in electrolyte metal is deposit at the cathode and an equal amount of metal acting as anode dissolves into solution.

At the Cathode Na

+

cathode 2H + (aq) + 2e -

At the Anode F–

Most difficult

Al3+

SO2–4

Zn2+

NO–3

Pb2+

Cl–

H+

Br–

Cu2+

I–

Ag+

anode 4OH - (aq)  → 2H 2 O(l) + O 2 (g) + 4e -

O

OH–

Least difficult

 → H 2 (g)

E

Product Anode Anode Reaction

1

Fused NaCl

Pt

Na + e → Na

-2.71

Na

Pt

2

Aqueous NaCl

Pt

-2.71 Na+(aq) + e– → Na H+(aq) + e– → 1/2 H2 0.00

H2

Pt

3

Fused lead bromide Water

Pt

Pb2+ +2e–→Pb

-0.13

Pb

Pt

Pt

H+ +e–→1/2 H2

0.0

H2

Pt

4

+

–

E

φ

S. No Electrolyte Cathode Cathode reaction

φ

table 12.12 Electrode reactions and products of electrolysis Product

Cl → Cl + e ; Cl + Cl→Cl2 Cl– → 1/2 Cl2+ e– 2H2O(l) →O2(g) + 4H+(aq) + 4e– Br– →1/2Br2 + e–

+1.36 Cl2(g)

4OH– → 2H2O + O2 + 4e– 4OH– → 2H2O + O2 + 4e–

+1.23 O2 (g)

+0.36 Cu2+(aq)

–

–

+1.36 Cl2(g) +1.23 +1.09 Br2(g)

5

Aqueous CuSO4

Pt

Cu2+(aq) +2e–→Cu(s) 0.36 H+(aq)+ e–→1/2H2 0.0

Cu

Pt

6

Cu

Cu2+ +2e–→ Cu

Cu

Cu

Cu → Cu2+ + 2e–

7

Aqueous CuSO4 Sulphuric acid

Pt

H+ + e–→1/2 H2

0.00

H2

Pt

8

Fused NaOH

Pt

Na+ + e– → Na

-2.71

Na

Pt

4OH– → 2H2O + O2 +1.23v + 4e–; 2SO42–(aq) → S2O82– 1.96v Dil H2SO4 gives O2 (aq) + 2e– but in higher concentration S2O82– – 4OH → 2H2O + O2 +1.23 O2(g) + 4e–

+1.23 O2 (g)

12.34

Electrochemistry

12.9 FaraDay’s laws oF ElECtrolysIs From the examples discussed so far, it can be understood that fused electrolytes or aqueous solutions of electrolytes between platinum electrodes or other electrodes (generally iron and graphite) undergo electrolysis. Metals are deposited or H2 gas is liberated at the cathode and O2 gas or non-metals are liberated at the anode. This means that during the electrolysis of fused salts or aqueous solution of electrolytes, the electrolytes undergo decomposition as a result of the electrode reactions (electronation or reduction and de-electronation or oxidation). Michael Faraday, in 1833 carried a number of electrolysis experiments and on the basis of the results he formulated two laws. These are known as Faraday’s laws of electrolysis. These laws deal with the mathematical relationship between the quantity of electricity (time × current = coulombs) passing through the solution and the quantity of the metals deposited or gases liberated at the electrodes during electrolysis.

12.9.1 Faraday's First law of Electrolysis The mass of the substance liberated or deposited or dissolved or underwent electrode reaction at an electrode during the electrolysis of an electrolyte is directly proportional to the quantity of electricity passing through the electrolyte. Faraday’s first law can be expressed mathematically as m ∝ Q or m = eQ Q = Quantity of electricity in coulombs Q=C×t C is current in amperes and t is time in seconds

∴m = ect ‘e’ is known as electrochemical equivalent of the metal deposited or of the gas liberated at the electrode e=

m gm / coulomb Q

If Q = 1; then e = m Therefore ‘e’ is the mass in grams of the element deposited or liberated at the electrode for the passage of one coulomb of electricity through the electrolyte solution. Electrochemical equivalent (e) may be defined as The mass of a substance deposited or liberated or dissolved or underwent electrode reaction at an electrode during the passage of one coulomb of electricity during the electrolysis of electrolytic solutions or melts is known as electrochemical equivalent. Faraday (F): The quantity of electric charge carried by one mole of electrons (i.e., 6.022 × 10 23 electrons) is

called Faraday constant and is denoted by E in the honour of Michael Faraday who discovered laws of electrolysis. Mathematically F = N × e = 6.022 × 1023 × 1.602 × –19 10 coulomb (The charge on one electron is 1.602 × 10–19 Coulombs) 1 F = 96488 C mol–1 or ≅ 96500 C mol–1 Faradays first law therefore, is written mathematically m=

ect Formula mass of species or 96500 Valency of species × 96500 Formula mass E = Valency

E is chemical equivalent of the species; C is the number of coulombs of electricity passed; t is the time in seconds Thus the chemical equivalent can be defined as The mass of the substance deposited or liberated or dissolved or underwent electrode reaction at an electrode during the passage of one Faraday of electricity in the electrolysis is called chemical equivalent

12.9.2 Faraday’s second law of Electrolysis If the same quantity of electricity is passed through different electrolytic cells connected in series containing different electrolytic solutions or melts the masses of the different species deposited or liberated or dissolved or underwent electrode reactions at the electrodes are directly proportional to the chemical equivalent of the substances Faraday’s second law can be Explained as follows. If Q coulombs of electricity are passed through three electrolyte cells connected in series containing respectively aqueous solutions of CuSO4 , AgNO3 and H 2 SO4 solutions as the electrolytes, the masses of Cu (deposited) or O2 gas liberated in the first cell (mCu, mO2) and the mass of Ag (deposited) or O2 gas (liberated) in the cell (m Ag , m 'O2 ) and the masses of H2 and O2 gases (liberated) in the third cell

(m

H2

, mO"

2

) are

proportional respectively to their

chemical equivalents (ECu, EO2, EAg, EH2). This can be expressed mathematically m Cu E Cu m Ag E Ag = = ; m O2 E O2 m O2 E O2 m H2 m

" O2

=

E H2 E O2

or m Cu : m Ag : m H2 = E Cu : E Ag : E H2

Or m1 : m 2 : m3 = E1 : E 2 : E 3

Electrochemistry

+

–

+

CuSO4 (aq)

–

+

AgNO3 (aq)

12.35

–

H2SO4 (aq)

Fig 12.19 An electrolysis experiment to illustrate the Faraday’s second law table 12.13 Difference between galvanic cell and electrolytic cell Galvanic Cell

Electrolytic Cell

1. In galvanic cell, the redox reaction takes place indirectly and electrical energy is produced 2. In galvanic cell, reaction is spontaneous

1. In electrolytic cells, the electrical energy is used and electrochemical change is carried but 2. In electrolytic cell, reaction is not spontaneous

3. In galvanic cell, anode is negative and cathode is positive electrodes 4. Two electrodes are set up in two different vessels

3. In electrolytic cell, anode is positive and cathode is negative electrodes 4. Both the electrodes are suspended in the same vessel

5. The electrolytes taken in two half-cells are different

5. Only one electrolyte is taken

6. The electrodes taken in two half-cells are of different materials

6. Electrodes taken may be of same or different material

7. Salt bridge/porous pot is used

7. Salt bridge is not required

12.9.3 application of Electrolysis The phenomenon of electrolysis has wide applications. Some important application are given here. 1. Determination of equivalent masses of elements: By using the Faraday's second law the equivalent weight of an element can be determined WA EquivalentWeight of A = WB EquivalentWeight of B Knowing the weights of the elements A and B liberated at the cathodes in a series and the equivalent weight of one element, the equivalent weight of the other element can be calculated. By using this relationship, the

2.

3. 4. 5.

equivalent weights of gaseous non metals liberated at the anode can be determined Electrometallurgy: Metals which are highly eletropositive such as sodium, potassium, magnesium aluminum etc which cannot be extracted can prepared by electrochemical reduction of their fused salts. Preparation of non-metals: Non-metals such as fluorine, hydrogen can be prepared by the electrolysis method Refining of metals: The metals like copper, silver, gold, aluminum, tin etc are purified by electrolysis. Manufacture of compounds: Compounds such as NaOH, KOH, Na2CO3, KClO3, white lead, KMnO4etc are manufactured by electrolysis

12.36

Electrochemistry

6. Electroplating: The process of coating an inferior metal with a superior metal by electrolysis is known as electroplating. Electroplating is done to prevent the inferior metal from corrosion and to make it attractive in appearance. The metal to be electroplated is taken as cathode and the metal to be deposited is made the anode in an electrolyte both containing a solution of a salt of the anode metal. On passing electric current in the cell the metal of the anode dissolves out and is deposited on the cathode article in the form of a thin film. The following conditions are maintained for fine coating 1. The surface of the article should be clean. It should be free from grease or oil and its oxide layer. The surface is cleaned with chromic acid or detergents 2. The surface of the article should be rough so that the metal deposited sticks permanently 3. The concentration of the electrolyte should be so adjusted as to get smooth coating 4. Current density must be the same throughout. table 12.14 Electroplating of certain objects with metals Metal to be electroplated Anode Cathode

Electrolyte

Copper

Cu

Object

CuSO4 + dil H2SO4

Silver

Ag

Object

K  Ag CN 2  

Nickel

Ni

Object

Nickel ammonium sulphate

Gold

Au

Object

K  Au (CN ) 2 

Zinc

Zn

Iron objects ZnSO4

Tin

Sn

Iron objects SnSO4

Thickness of coated layer Let the dimensions of metal sheet to be coated be (a cm × b cm) Thickness of coated layer = c cm Volume of coated layer = (a × b × c) cm3 Mass of the deposited substance = volume × density = (a × b × c) × d g I ×t × E ∴ (a × b × c) × d = 96500 Using the above relation the thickness of the coated layer can be calculated. solved Problem 7 An electric current is passed through three cells in series containing respectively solutions of CuSO4, AgNO3 and KI. What weight of Ag and I2 be liberated while 1.25 g of Cu are being deposited

Solution: From Faraday's second law Weight of Cu Weight of Ag = Equilivalent weight of Cu Equivalent of weight of Ag =

Weight of I 2 Equivalent weight of I 2

1.25 g weight of Ag Weight of I 2 = = 63.5 / 2 108 127 1.25 × 2 × 108 = 4.28 g 63.5 1.25 × 2 × 127 = 5.00 g Weight of Ag = 63.5

∴Weight of Ag =

solved Problem 8 What current strength in amperes will be required to liberate 10 g iodine from KI solution in one hour? Solution: Equivalent weight of KI = 127 g ∴ 127 g of I2 will be liberated by 96500 coulombs

∴ 10 g of I2 will be liberated by

96500 × 10 Coulombs 127

∴Q = It Q 96500 × 100 1 ∴I = = × = 2.11amp t 127 1× 60 × 60

Problems for Practice 45. 30 mL of 0.13 M NiSO4 is electrolyzed using a current of 360 milliamperes for 35.3 minutes. How much of the metal would have been plated out if current efficiency was only 60%? (Ni = 58.7) 46. Calculate the quantity of electricity that would be required to reduce 12.3g of nitrobenzene to aniline, if the current efficiency for the process is 50%. If the potential drop across the cell is 3volts how much energy will be consumed (IIT 1990) 47. A 100-watt, 110 volt incandescent lamp is con nected in series with an electrolytic cell containing cadmium sulphate solution. What weight of cadmium will be deposited by the current flowing for 10 hours? (IIT 1987) 48. A current of 1.70A is passed through 300mL of a 0.16M solution of ZnSO4 for 230s with a current efficiency of 90%. Find out molarity of Zn2+ after the deposition of Zn. Assume the volume of the solution to remain constant during electrolysis. (IIT 1991)

Electrochemistry

49. Copper sulphate (250 mL) was electrolyzed using a platinum anode and a copper cathode for 16 minute. It was found that after electrolysis the absorbance of the solution was reduced to 50% of it original value. Calculate the concentration of copper sulphate in the solution to begin with. (IIT 2000) 50. Electrolysis of a solution of MnSO4 in aqueous sulphuric acid is a method for the preparation of MnO2 as per reaction Mn2+(aq)+2H2O→MnO2(s)+2H+(aq)+H2(g) Passing a current of 27A for 24hrs gives one Kg of Mn. What is the value of current efficiency? Write the reaction taking place at the cathode and at the anode. (IIT 1997) 51. Chromium metal can be plated out from an acidic solution containing CrO3 according to following equation.

56.

57.

58.

CrO3 (aq ) + 6H + + 6e −  → Cr (s) + 3H 2 O Calculate (a) How many grams of chromium will be plated out by 24000 coulomb? (b) How long will it take to plated out 1.5 g of Cr by using 12.5 ampere current? (IIT 1993) 52. An aqueous solution of NaCl on electrolysis gives H2(g), Cl2 (g) and NaOH according to reaction 2Cl - (aq) + 2H 2 O  → 2OH - (aq) + H 2 (g) + Cl2 (g) A direct current of 25 ampere with a current efficiency of 62% is passed through 20 litre of NaCl solution 20% by weight (a) Write down the reaction taking place at the electrodes. (b) How long will it take to produce 1 Kg of Cl2? (c) What will be the molarity of solution with respect to OH– Assume no loss in volume due to evaporation (IIT 1992) 53. How many grams of silver could be plated out on a serving tray by electrolysis of solution containing silver in +1 oxidation state for a period of 8.0 hr at a current of 8.46 ampere? What is the area of the tray if the thickness of the silver plating is 0.00254 cm? (Density of silver is 10.5 g/cm3) (IIT 1997) 54. A cell Ag| Ag+ || Cu2+| Cu initially contains 1 M Ag+ and 1 M Cu2+ ions calculate in the cell potential after the passage of 9.65 A of current for 1 hour. (IIT 1999) 55. How long has a current of 3 ampere to be applied through a solution of silver nitrate to coat a metal

59.

12.37

surface of 80 cm2 with 0.005mm thick layer? Density of silver is 10.5/cm3. (IIT 1985) An acidic solution of Cu2+ salt containing 0.4g Cu2+ is electrolysed until all the copper is deposited The electrolysis is continued for seven more minutes with volume of solution kept at 100 mL and the current at 1.2 amp. Calculate the gases evolved at NTP during the entire electrolysis. (IIT 1989) Assume that impure copper contains iron, gold and silver as impurities. After passing a current of 140 ampere for 482.5 seconds the mass of anode decreased by 22.260 g and the cathode increased in mass by 22.011 g. Estimate the percentage of iron and copper originally present. 19 g of molten SnCl2 is electrolyzed for some time using inert electrodes until 1/0.119 g of Sn is deposited at the cathode. No substance is lost during electrolysis. Find the ratio of the masses of SnCl2 : SnCl4 after electrolysis. Potassium chlorate is prepared by electrolysis of KCl in basic solution. 6OH - + Cl -  → ClO3- + 3H 2 O + 6e -

If only 60% of the current is utilized in the reaction what time will be required to produce 10 g KClO3 using a current of 2 amp? 60. The density of Cu is 8.94 g mL Find out the charge in coulombs to plate an area 10 cm×10 cm to a thickness of 10–2 cm using CuSO4 solution as electrolyte

12.10 BattErIEs One of the milestone is the development of batteries by using galvanic cells, which are portable and used for the generation of electrical energy. The term battery is generally used for two or more galvanic cells connected in series. A useful battery should have the following characteristics: (i) It should be light and compact so that it can be easily transported (ii) It should survive for longer periods whether it is in use or not (iii) The voltage of the battery should not change much during in use. Types of commercial cells: The commercial cells are mainly of two types (i) primary cells (ii) secondary cells. (i) Primary cell: Primary cell is an electrochemical cell which act as a source of electrical energy without previously charged up by an electric current from an external source of current.

12.38

Electrochemistry

In these cells, the chemical reaction take place only once and cannot be reversed by an external electric energy source therefore they are not rechargeable and they become dead over a period of time and the chemical reaction stops Examples of primary cell are: dry cell, mercury cell. (ii) Secondary cells: Secondary cell is one in which electric energy from an external source is first converted into chemical energy (electrolysis). Then the external source of current is removed and the cell is made to operate in the reverse direction. In the secondary cells, the reactions can be reversed by an external electric energy source. Therefore these cells can be recharged by passing electric current and used again and again. These are also called storage cells. Examples of secondary cell are: lead storage battery and nickel cadmium storage cell.

12.10.1 Primary Cells 1. Dry cells: This is a modification of Le Clanche cell. The Le Clanche cell was invented by G Le Clanche in 1868. A dry cell is shown in the Fig 12.20 +

Seal

The zinc ions (Zn2+) so produced combine with ammonia librated in cathodic reaction to from diammine zinc (II) cation 2+

→  Zn ( NH 3 )2  Zn 2 + + 2 NH 3  It gives a voltage of approximately 1.2 to 1.5 V. There are many varieties of dry cells such as silver cell or lithium cell. Greater voltage can be achieved by using multiple dry cells arranged in a series (plus-to-minus-to-plus). When cells are connected in series (with cathode of one attached to the anode of other) the battery produces voltage that is the sum of the emfs of the individual cells. This dry cell does not have an indefinite life because NH4Cl being acidic corrodes the zinc container even when not in use. These are used in torches, toys, flash lights, calculators etc. 2. Mercury cell: this cell contain zinc mercury amalgam as anode and a paste of mercury (II) oxide and carbon as cathode. The electrolyte is a paste of KOH and ZnO. The reaction occurrs in the cell as follows

Anode

Cathode HgO(s) + H 2 O + 2e -  → Hg (l) + 2OH Overall

Graphite (cathode) MnO2+ C

Paste of NH4Cl+ZnCl2

Zinc anode

–

Zn + 2OH -  → ZnO(s) + H 2 O + 2e -

amalgam

→ ZnO(s) + Hg (l) Zn + HgO(s) 

amalgam

In this case, the overall cell reaction does not involve any ion in solution whose concentration can change. Therefore it has the advantage that its potential remains almost constant throughout its life. This is its special feature. By contrast, the potential of the ordinary dry cell decrease slowly but continuously as it is used. The voltage of mercury cell is approximately 1.35 V. These miniature cells find a frequent use for watch, video cameras, hearing aids and other compact devices.

Fig 12.20 A dry cell The liquid state electrolytes of Le Clanche cell are replaced by semi-solid (paste) electrolytes. A cylindrical Zn vessel is covered with a cardboard. This is sealed with pitch. Zn vessel act as negative electrode. A carbon rod is introduced at the centre of the Zn vessel. The carbon rod acts as positive electrode. This is surrounded by a paste of (C+MnO2). The remaining space is filled with ( NH 4 Cl + ZnCl2 ) paste. The two pastes are separated by a porous sheet. The reaction taking place at the electrodes are: Cathode : MnO2 + NH 4+ + e -  → MO(OH ) + NH 3 (oxidation state of Mn changes from + 4 to + 3)

Anode : Zn  → Zn 2 + + 2e -

12.10.2 secondary Cells 1. Lead storage cell: When an electric current is passed through an electrolytic cell, chemical changes occur and electric energy is converted into chemical energy. The phenomenon is electrolysis and the cell is called electrolytic cell. If this cell is reversible, then removing the source of current and connecting the electrodes of the electrolysis cell by means of a conductor electrical energy will flow through the conductor. This form of cell is called a battery or a storage battery or a secondary cell. Chemical changes occur during the charging of the cell with current. These changes are reversed during “discharging”.

Electrochemistry

Theoretically, any reversible galvanic cell should be able to store electrical energy. But most of them are practically unsuitable. This is because of low electrical energy storing capacity and incomplete reversibility. One such reversible storage cell of practical importance is acid storage cell or lead accumulator. This is the most commonly used battery in automobiles. Each battery consists of a number of voltaic cells connected in series. Three to six such cells are generally combined to get 6 to 12 volt battery. In each cell, the anode is a grid of lead packed with finely divided spongy lead and the cathode is a grid of lead packed with PbO2. The electrolyte is aqueous solution of sulphuric acid (38% by mass) having a density 1.30 g/mL sulphuric acid. When the lead plates are kept for some times a deposit of lead sulphate is formed on them. At the anode, lead is oxidized to Pb2+ ions insoluble PbSO4 is formed. At the cathode PbO2 is reduced to Pb2+ ion and PbSO4 is formed. The following reactions takes place in the lead storage cell. Anode : Pb + SO 24 -  → PbSO4 + 2e -

It is clear from the above reaction that during the working of the cell PbSO4 is formed at each electrode and sulphuric acid is used up. As a result, the concentration of the density of H2SO4 falls below 1.2 g/ mL battery needs recharging. recharging the Battery The battery can be recharged by connecting it to an external source of direct current with voltage greater than 12V. It forces the electrons to flow in opposite direction resulting in the deposition of Pb on the anode and PbO2 on the cathode. During recharging operation, the cell behaves as electrolytic cell. The recharging reactions are → Cathode( -Ve) : PbSO4 ( s )  Pb( s ) + SO42 - (aq ) Anode(+Ve) :

E red = 1.69V → 2PbSO4 + 2H 2 O Overall : Pb + PbO 2 + 4H + + 2SO 4-2  E Cell = 2.05V

Anode

→ PbSO4 ( s ) + 2 H 2 O(l )  PbO2 ( s ) + SO42 - (aq ) + 4 H + (aq ) + 2e -

Overall :

2 PbSO4 ( s ) + 2 H 2 O(l )  → Pb( s ) + PbO2 ( s ) + 4 H + (aq) + 2 SO4-2

E Oxi = 0.36V Cathode : PbO 2 + SO 24 - + 4H + + 2e -  → PbSO 4 + 2H 2 O

12.39

It may be noted that storage battery acts as voltaic cell as well as electrolytic cell. For example, when it is used to start the engine of the automobile, it acts as a voltaic cell and produces electric energy. During recharging it acts as an electrolytic cell.

+ cathode

Lead plates PbO2 paste

H2SO4 (Aqueous)

Fig 12.21 Lead storage battery

12.40

Electrochemistry

The complete cell reaction in both the directions of charging and “discharging” can be written as seen of the individual electrode process. disch arg e   2 PbSO4 + 2 H 2 O Pb + PbO2 + H 2 SO4  Ch arg e

for 2 Faradays Hence this theory was earlier known as “double sulphation theory. This was proposed by Glasstone and Trube in 1883. The voltage of the accumulator increases with the increasing concentration. Thus during the “discharge” process of the lead accumulator sulphuric acid is replaced by an equivalent quantity of water. Therefore as the accumulator, produce current (during usage) H2SO4 concentration decreases. On recharging reverse reaction takes place and H2SO4 is generated and water is consumed original acid strength is restored. Thus the changes discharge or charge are associated with change in specific gravity of H2SO4. Hence the cell is tested for the extent of charge or discharge by determining the specific gravity of H2SO4 of the battery. The voltage varies from 1.88V (5% H2SO4) to 2.15 V (40% H2SO4) 2. Nickel cadmium storage cell It is another rechargeable cell. It consists of cadmium anode and cathode is made of a metal grid containing nickel (IV) oxide. These are immersed in KOH solution. The reaction taking place during discharge and charge are

Fuel cells are electrochemical devices which convert the energy of fuel oxidation reactions into electrical energy. In the fuel cells the chemical energy (combustion of fuel) is converted directly into electrical energy. In the fuel cells, the fuel is oxidized at the anode like any other electrochemical cell; the fuel cell has two electrodes and an electrolyte. But the fuel and the oxidant are continuously and separately supplied to the two electrodes of the cell at which they undergo reaction. These primary cells are capable of supplying current for as long as they are provided with a supply of the reactants namely the fuel and the oxidant. The very first fuel cell is known to be developed by Sir William grove in 1839. He used platinum electrodes and H2 and O2 as reactants. Fuel cells based on combustion of hydrocarbons are also operated. These are discussed below. (i) Hydrogen – Oxygen fuel cell: The cell is shown in the Fig 12.22. It consists of two electrodes made of porous graphite. Platinum is coated on the surface of the electrodes. The electrodes are placed in aqueous solution of the electrodes. The electrodes are placed in aqueous solution of KOH or NaOH. Hydrogen and oxygen are bubbled into the cell under a pressure of 50 atm. When the electrodes are connected, a flow of electric current takes place. H2O anode

+

−

cathode

Disch arg e

 ⇀ Anode : Cd (s) + 2OH - ↽   Ch arg e Cd (OH) 2 (s) + 2e Disch arg e

 ⇀ Cathode : NiO 2 (s) + 2H 2 O + 2e - ↽   Ch arg e Ni(OH) 2 + 2OH Disch arg e  ⇀ Overall Cd (s) + NiO 2 (s) + 2H 2 O ↽   Ch arg e

Cd(OH) 2 + Ni(OH) 2 (s) In these reactions, there is no formation of gaseous products. The reaction products generally remain sticking to the electrodes and can be reconverted by recharging the cell. The charging process is similar to lead storage battery. The cell is is called nicad cell and has voltage of 1.4V.

12.10.3 Fuel Cells Recent advances in electrochemistry and technology have led to the introduction of compact source of power in which fuels are used to produce electricity without the intervention of thermal devices such as generators, turbines, boilers etc. These fuel cells are much more efficient than thermal sources. These are used in space-craft and may become the basis of pollution free transport.

H2

OH

-

porous carbon electrodes

Fig 12.22 Hydrogen – oxygen fuel cell The electrode reactions are: Anodic reaction H 2  → 2H + + 2e 2H + + 2OH -  → 2H 2 O

The net half cell reaction is H 2 + 2OH -  → 2H 2 O

O2

Electrochemistry

The overall reaction

Cathodic reaction 1 O2 + H 2 O + 2e -  → 2OH 2

C3 H 8 + 5O2  → 3CO2 + 4 H 2 O

Now the overall cell reaction is

2 H 2 ( g ) + O2 ( g )

 → 2H 2 O

The emf of the cell is found to be 1.23 volt. The cell reaction is the same as combustion of H2 in air or oxygen. The energy of the fuel oxidation reaction has not been liberated as heat but it has directly converted into electrical energy. Efficiency of the cell: When the oxidation of fuel occurs at a constant temperature and pressure the maximum available heat is ΔH. Only a part of this energy i.e., ΔG is converted into electrical energy. The ability of a fuel to convert the energy of the fuel oxidation reaction into electrical energy is expressed in terms of the efficiency of the cell. It is defined as ε=

Hydrocarbon fuel – air cells normally use platinum electrodes, concentrated phosphoric acid electrolyte, and operate at a temperature in the region of 373K. These cells can produce about 0.1W cm–2 of electrode surface. Thermodynamic characteristics of some possible fuel cells are given in table. advantages of Fuel Cells 1. High efficiency of energy conversion (nearly 100%). 2. The absence of moving parts in the cell and so elimination of wear and tear problems. 3. Silent operation 4. Absence of harmful waste products 5. The exhaust from the H2 – O2 fuel cell is water which is not a pollutant.

∆G × 100 ∆H

Saturn V, the rocket that took Neil Armstrong to the moon was powered by liquid hydrogen fuel. Oxygen (needed for combustion) as well as hydrogen were carried in the rocket in separate tanks. In spite of its high heat of combustion (242 J mol–1), its use as fuel is limited due to lack of availability of free hydrogen in nature and the difficult of its storage and distribution. (ii) Hydrocarbon – Oxygen fuel cell: Hydrocarbon and their oxygenated derivatives such as CH4, C2H2, C2H8, alcohols etc., are often used as fuels in fuel cells. The half cell reactions with propane are given below. Anodic reaction C3 H8 + 6H 2 O  → 3CO 2 + 2OH - + 20 e [H + + OH -  → H 2 O] × 20

Cathodic reaction [O2 + 2 H 2 O + 4e -  → 4OH - ] × 5

12.11 CorrosIon Metals occur in nature as their compounds (oxides, carbonates, sulphides etc). These are called minerals or ores. But man for his own purposes, extracting the metal from these compounds. Now nature again try to get back the material taken by man. The metals try to convert back into the compound from which it is prepared. This natural tendency of a metal for conversion into its mineral compound form is called corrosion. This may take place in many ways but in most cases the metals interacts with its environment (in the presence of polluted air, water, associated other metal etc). Therefore the natural tendency of conversion of a metal into its mineral compound by interacting with the environment is known as corrosion. Or The process of deterioration of a metal as a result of its reaction with air or water (environment) surrounding it.

table 12.15 Thermodynamic characteristics of some fuel cells Cell

Cell Reaction

∆G

C - O2 CH4 - O2 CH3OH - O2

∆H

E

∈

( KJ mol )

( KJ mol -1 )

(V )

1 H 2 + O2  → H 2O 2 C + O2  → CO2

-237.2

-258.9

1.229

0.83

-137.3

–110.5

0.712

1.24

CH 4 + 2O2  → CO2 + 2 H 2 O 3 CH 3 OH + O2  → 2 CO2 + 2 H 2 O

-818.0

-890.4

1.060

0.92

-706.9

-764.0

1.224

0.93

-1

H2 - O2

12.41

12.42

Electrochemistry

For example, iron converts itself into its oxide (Fe2O3, hematite), Copper converts itself into its carbonate (malachite) Silver converts itself into its sulphide (Ag2S, argentite). Some special names are given for some types of corrosion. For example, the corrosion of iron by conversion into iron oxide known as rusting. Corrosion of silver by conversion into its sulphide is known as tarnishing.

mechanism of Corrosion The process of corrosion may be chemical or electrochemical in nature. The electrochemical corrosion considered as the anodic dissolution of the metal undergoing corrosion. The anodic dissolution is an oxidation process of the metal and the metal getting corroded undergoes oxidation to the metal ion under the conditions of corrosion M  → M n + + ne Hence, corrosion can be defined in electrochemical terms as anodic dissolution of the metal. The anodic dissolution of a metal under the conditions of corrosion is electrochemical corrosion. The corrosion, therefore, occurs if the environmental conditions of the metal favor the formation of a voltaic cell with the metal acting as anode. The dissolution of the metal i.e., the metal corroded shall form a voltaic cell with another metal or a part of its own under the given conditions is basically of two types: (1) hydrogen evolution type (2) Differential oxygenation type.

12.11.1 hydrogen Evolution type This type of corrosion is exhibited by metals which can displace hydrogen gas from aqueous solutions. This is possible when the electrode potential of the metal under the conditions of corrosion is more negative than that of the hydrogen electrode under the given conditions. The hydrogen electrode potential depends on the nature of metal used as the cathode and the pH of the solution. It is almost zero in 1 M acid solutions if the cathode metal is paltinised platinum (Black platinum). But if the cathode metal is mercury it is about 0.8 V for other metals the value lies in between these values i.e., O to 0.8 V. Therefore, in aqueous solutions the pH of the medium, the chemical natures of the metal undergoing corrosion and the impurity in metal serves as cathode are important to know whether the anodic metal corrodes or not under these conditions. For example, pure Zn does not corrode in salt solution but in presence of Cu as impurity Zn corrodes. Zn corrodes in 2 M acid but not in neutral salt solutions.

12.11.2 Differential oxygenation Corrosion If O2 concentration is not uniformly distributed on the surface of the metal, corrosion of the metal may take place at the point where O2 concentration is less. This is because the position of the metal with access to high concentration of O2 functions as anode. Hence the metal with differential oxygenation, functions as galvanic cell. The portion for less access of O2 functions as anode. Hence metal undergoes anodic dissolution and is said to be corroded. Rusting of iron: Chemically, rust is hydrated form of ferric oxide Fe2O3.xH2O. Rusting of iron is generally cause by moisture, carbon dioxide and oxygen present in air. Iron rusts only in the presence of moist air but does not rust in dry air and in vacuum. It is believed that a non-uniform surface of metal or impurities present in iron behave like small electric cells (called corrosion couples) in the presence of water containing dissolved oxygen or carbon dioxide. A film of moisture with dissolved carbon dioxide constitutes electrolytic solution covering the metal surface at various places. The schematic representation of mechanism of rusting of iron is shown in Fig 12.23 istur

f mo

o Drop

t

Rus

Fe Anode

2+ Fe + 2e -

e

4H++O2+4e-

2H2O

Flow of electrons Iron

Fig 12.23 Mechanism of rusting of iron At the cathodes of each cell, the electrons are taken up by hydrogen ions (reduction take place) the H+ ions are obtained either from water or from acidic substance (e.g., CO2) in water. H2O → H++ OH– Or CO2 + H2O → H++HCO3– At cathode H++e– → H The hydrogen atoms on the iron surface reduce dissolved oxygen 4H + O2 → 2H2O Therefore the overall reaction at cathode of different electrochemical cell may be written as 4H++ O2 + 4e– → 2H2O

Electrochemistry

The overall redox reaction may be written by multiplying reaction at anode by 2 and adding reaction at cathode to equalize number of electrons lost and gained i.e., Oxidation half-reaction: Fe(s) → Fe2+(aq) + 2e–] × 2 (E = –0.44V) Reduction half-reaction: 4H+ + O2 + 4e– → 2H2O (E = 1.23V) Overall reaction: 2Fe(s) + 4H+ + O2 → 2Fe2+(aq) + 2H2O (E cell = –1.67V) The ferrous ions are oxidized further by atmospheric oxygen to form rust. O

O

O

4Fe2+ (aq) + O2 (g) + 4H2O → 2Fe2O3 +8H+ And Fe2O3 + xH2O → Fe2O3.xH2O Rust The salt water accelerates corrosion. This is mainly due the fact that salt water increases the electrical conduction of electrolyte solution formed on the metal surface. Therefore rusting becomes more serious problem where salt water is present. Factors affecting corrosion (i) Position of metals in emf series: The reactivity of metal depends upon its position in the emf series. More the reactivity of the metal more will be the possibility of the metal getting corroded. (ii) Presence of impurities in metals: The impurities help in setting up voltaic cells which increase the speed of corrosion. (iii) Presence of electrolytes: Presence of electrolytes in water also increase the rate of corrosion e.g., corrosion of iron in salt water takes place rapidly and to larger extent than in pure water. (iv) Presence of CO2 in water: Presence of CO2 in natural water increases rusting of iron. Water containing CO2 acts as an electrolyte and increases the flow of electrons from one place to another. (v) The nature of the impurity: Cu favours corrosion of Zn because zinc is more anodic than copper, zinc disfavors corrosion of iron (galvanization). (vi) Concentration of O2 in contact with the surface of the metal: A metal rod half immersed in aqueous salt solution gets corroded at the surface not exposed to the O2 of the air, but the immersed part of the metal gets easily corroded. (vii) Highly conducting solution favor rapid corrosion: “Atmospheric corrosion” resulting from exposure of metal to air generally containing H2S, SO2, CO2, H2O vapour is explained in terms of chemical reactions. But this type can also be explained electrochemically if carefully examine.

12.43

Prevention of corrosion There are several methods for protecting metals from corrosion (iron from rusting). Some of these methods are as follows. 1. Barrier protection: In this method, a barrier film is introduced between iron and atmosphere oxygen and moisture. Barrier protection can be achieved by any of the following methods: (i) The surface is coated with paint (ii) The surface is protected by applying a thin film of oil or grease (iii) The metal is electroplated with metals like tin, nickel, zinc, chromium, aluminum etc. In this type of protection, if scratches or cracks appear in the protective layer then surface of iron may get exposed and get rusted. The rusting beneath the protective layer and eventually peals off the protective layer. 2. Sacrificial protection: In this method, the surface of iron is covered with a layer of more active metal like zinc. The active metals lose electrons in preference to iron and goes into ionic state thus preventing the rusting. However, the covering metal gets consumed in due course of time but so long as it is present iron is not rusted. This type of process in which rusting of iron is protected is called sacrificial protection. Zinc metal is commonly used for covering iron surfaces and the process is called galvanization. The galvanized iron materials maintain their lustre due to the coating of invisible layer of basic zinc carbonate, ZnCO3. Zn (OH)2 on the zinc film due to reaction between zinc, oxygen, CO2 and moisture in air. If some scratches occur on the protection zinc film on coated iron, even then iron will not be rusted. This is due to the fact that because of scratches, zinc oxidises in preference to iron. This is so because the reduction potential of zinc is less than the reduction potential of iron. Zn2+(aq) + 2e– → Zn(s) Fe2+(aq) + 2e– → Fe(s)

O

E = –0.76V E = –0.44V O

Hence, zinc undergoes oxidation in preference to iron. However, if tin is coated on iron, the film will be effective as long as it is intact. When scratches occur at the coating surface both metals are exposed to oxygen and iron is preferably oxidized and is rusted. This is due to the fact the reduction potential of tin is more than that of iron. Sn2+(aq) + 2e– → Sn(s) Fe2+(aq) + 2e– → Fe(s)

O

E = –0.14V E = –0.44V O

Thus iron will be oxidized in preference to tin. 3. Electrical protection: This method is used for protecting iron articles which are in contact with water such

12.44

Electrochemistry

as underground water pipes. The article of iron is connected with more active metals like magnesium or zinc. The active metal has lower reduction potential than iron and will lose electrons in preference to iron. For example, magnesium has lower reduction potential than iron O

E = –2.37V E = –0.44V O

Drop of moisture

B

Current density

Mg2+(aq) + 2e– → Mg(s) Fe2+(aq) + 2e– → Fe(s)

Zn → Zn2+ (Anode)

Flow of electrons

C

Iron (Cathode)

A

Fig 12.24 Rusting of iron is prevented by galvanization (Coating with zinc) sacrificial protection Ground level

Magnesium (anode)

Iron pipe (cathode)

Fig 12.25 Sacrificial protection with magnesium Therefore magnesium will be oxidized in preference to iron and therefore it will protect iron from being rusted. 4. Using Anti-rust solutions: To retard the corrosion, certain anti-rust solutions are used. For example, solutions of alkaline phosphates and alkaline chromates are generally used as anti-rust solutions. Due to the alkaline nature of this solution, the H+ ions are removed from the solutions and rusting is prevented. For example, iron articles are dipped in boiling alkaline sodium phosphate solutions when a protective insoluble sticking film of iron phosphate is formed.

12.11.3 Passivity of metals Most metals dissolve as anodes when the potential is just slightly more positive than the reversible value in particular solution. If the current density is increased, however, a point is reached at which the anode potential rises suddenly and there is corresponding decrease of current, at the same time, anode practically ceases to dissolve although its appearance is unchanged. The metal is then said to be in passive state.

Anode Potential

Fig 12.26 Anodic passivity A state of non-reactivity reached with time of action after an initial state of reactivity is passivity. The general nature of the variation is shown in the Fig 12.26 AB represents the change of anode potential with current density when the electrode is active and dissolves quantitatively. At B, the electrode becomes passive and the potential rises while the current decrease to C.

Passivity of metal is of three types 1. Electrochemical passivity: As discussed above, metals dissolve at anodes only when the potential at the anode exceeds the reversible electrode potential of the metal. The passivity of iron, cobalt and nickel is favored by alkalinity and also by the presence of anions like NO3- , ClO3- etc. Molybdenum and tungsten are rendered passive more easily in acid than in alkaline solutions and oxidizing agents inhibit the passivity of tin. 2. Chemical passivity: Passivity can be produced without the action of an electric current. If for example, iron is dipped into concentrated nitric acid, there may be instantaneous reaction but the metal does not continue to dissolve; the iron has thus become passive. The passivity can be removed by scratching the surface or by touching it under the surface of an electrolyte. This type of passivity can be induced by nitric acid and other oxidizing agents not only in iron but also in nickel, cobalt and chromium. 3. Mechanical passivity: In certain instances the dissolution of a metal is prevented by a visible film of metal oxide. For example PbO2 on Pb. This is called

Electrochemistry

mechanical passivity. This type of passivity is shown by Fe, Co, Ni, Mn, Pb etc. Several theories were put forward to explain passivity but no single theory could explain the facts satisfactorily. One of such theory proposes the formation of invisible metal oxide layer on the metal. The oxide layer is so thin that it is not at all visible to the naked eye. Its presence can be proved only through chemical reactions. A thick oxide film and underneath a thin invisible oxide film are formed. solved Problem 9

Cathode :2 H 2 O + O2 + 4e -  → 4OH : H 2 + 2OH -  → 2 H 2 O + 2e -

Solution: 67.2 No. of moles of H2 reacting – =3 22.4 Eq of H2 used = 3 × 2 = 6 W I.t I × 15 × 60 Now = ;6 = E 96500 96500 ∴ I = 643.33 ampere

Eq of H2 = Eq of Cu formed Eq of Cu deposited = 6 WCu = 6 ×

Solution: Adding the charging and discharging reactions Pb + PbO2 + 4 H + + 2SO42 -  → PbSO4 + 2 H 2 O

NH

2 SO4

= MH

2 SO4

( Since 2SO42 - requires 2 electrons )

i.e., Normality = Molarity Before discharge 39 × 1.294 × 1000 M H2SO4 ( I ) = = 5.15 98 × 100 Mole of H 2SO 4 = 5.15 × 3.5 = 18.025 After discharge

In a fuel cell H2 and O2 react produce electricity. In the process, H2 gas is oxidized at the anode and O2 at cathode. If 67.2 litre H2 at STP reacts in 15 minutes, what is average current produced? If the entire current is used for electro deposition of Cu from Cu2+ how many gram of Cu are deposited?

Anode

12.45

63.5 = 190.5 g 2

Wt of Cu deposited = 190.5 g

20 × 1.139 × 1000 = 8.1375 98 × 100 Mole of H 2SO 4 = 2.325 × 3.5 = 8.1375 M H2SO4 ( II ) =

∴ Mole or equivalent of H2SO4 used = 18.025−8.1375 = 9.8875 W I.t ∴ = E 96500 I.t = 9.8875 × 96500

= 954143.75 ampere sec = 265.04 ampere hr Problems for Practice 61. A lead storage cell is discharged which causes the H2SO4 electrolyte to change from a concentration of 34.6% by weight (density 1.261g mL–1 at 25°C) to one of 27% by weight. The original volume of electrolytes is one litre. How many Faradays have left the anode of battery? Note the water is produced by the cell reaction as H2SO4 is used up. Overall reaction. Pb(s) + PbO 2 (s) + 2H 2SO 4 (l)  → 2PbSO 4 (s) + 2H 2 O 62. The electrode reaction for charging of a lead storage battery is PbSO4 + 2e - → Pb + SO42 -

solved Problem 10 During the discharge of a lead storage battery, the density of sulphuric acid fell from 1.294 mL–1 to 1.139 g mL–1 Sulphuric acid of density 1.294 g mL–1 is 39% by weight and that of density 1.139 g mL–1 is 20% by weight. The battery holds 3.5 litre acid and the volume practically remained constant during the discharge. Calculate the number of ampere hour for which the battery must have been used. The charging and discharging reactions are (IIT 1986) Pb + SO42 -  → PbSO4 + 2e - Charging PbO2 + 4 H + + 2 SO42 - + 2e -  → PbSO4 + 2 H 2 O Discharging

PbSO4 + 2 H 2 O → PbO2 + SO42 - + 4 H + + 2e -

The electrolyte in the battery is an aqueous solution of sulphuric acid before charging the specific gravity of the liquid was found to be 1.28 (36.9% H2SO4 by wt). If the battery contained two liters of the liquid; calculate the average current used for charging the battery.Assume that the volume of the battery liquid remained constant during charging. 63. A lead storage battery has initially 200g of lead and 200 g of PbO2, plust excess H2SO4. Theoretically, how long could this cell deliver a current of 10 amp, without recharging, if it were possible to operate it so that the reaction to completion.

12.46

Electrochemistry

KEy PoInts •

• •

•

•

•

•

•

•

• •

•

In an electrolytic cell electrical energy is converted into chemical energy while in galvanic cell or voltaic cell chemical energy is converted into electrical energy. Daniel cell is a typical example for galvanic or voltaic cell. In a galvanic or voltaic cell a spontaneous chemical oxidation reduction reaction occurs and generates electrical energy. Voltaic cell contain two half cells that are electrically connected to each other. Each half-cell in a voltaic cell is called single electrode. A simple half-cell is made by dipping a metal rod into its own ions M/Mn+(aq) In Daniel cell one half-cell is made by dipping zinc rod in aqueous solution of zinc ions [Zn/Zn2+(aq)] and known as zinc electrode. Another half-cell is made by dipping copper rod in aqueous solution of copper ions [Cu/Cu2+(aq)] and known as copper electrode. In the Daniel cell electrons flow from zinc metal electrode to copper electrode through external circuit. Metal ions flow from one half-cell to the other through salt bridge. In voltaic cell at one electrode the element goes into solution as ions by losing electrons which is an oxidation process or deelectronation. The electrode at which oxidation takes place is known as oxidation electrode or anode. At the second electrode of voltaic cell the ions gain electrons and discharged at the electrode which is reduction process or electronation. The electrode at which reduction take place is called reduction electrode or cathode. In voltaic cell, the anode will have negative charge because electrons will flow from it while the cathode will have positive charge because the electrons are drawn from it. In contrast in electrolytic cell the anode will have positve charge while cathode will have negative charge. In both galvanic and electrolytic cells oxidation take place at anode and reduction take place at cathode. In Daniel cell zinc electrode is the anode and copper electrode is the cathode. Electrons flow from zinc electrode to copper electrode i.e., anode to cathode in the external circuit while current flows from copper electrode to zinc electrode i.e., cathode to anode in the external circuit. In the Daniel cell at zinc electrode (or anode), zinc undergoes oxidation or deelectronation take place and Zn2+ ions goes into solution Zn( s )  → Zn 2 + (aq ) + 2e While at copper electrode (or cathode), Cu2+ ions gains electrons and copper atoms deposit on the cathode. 2+

-

Cu (aq ) + 2e  → Cu ( s )

•

The two half cells in a voltaic cell are connected to each other through a salt bridge for completion of electric circuit. • A voltaic cell may be reversible or irreversible. • In a reversible voltaic cell the cathodic and anodic reactions are always in equilibrium state. • A reversible cell should satisfy the following conditions i. If a voltaic cell is connected to an external source of emf equal to that of voltaic cell no current flows in the voltaic cell and the cell reaction stops. ii. If the emf of the external source is more than the emf of voltaic cell, current flows from the external source into the voltaic cell and the reaction in the cell will be reversed. iii. if the emf of the external source is less than the emf of the voltaic cell, current flows from the voltaic cell into the external source. If a cell do not satisfy the above conditions it is said to be irreversible. Daniel cell is a reversible cell.

representation of an Electrochemical Cell •

A galvanic cell is represented by writing the anode on the left hand side and cathode on the right hand side. Anode of the cell is represented by writing the metal or solid phase first and then electrolyte while the cathode is represented by writing the electrolyte first and then metal or solid phase. The metal and cation are separated either by a semicolon (;) or by a vertical line. The concentration of the electrolyte is also mentioned within the bracket after the cation e.g., Zn; Zn 2 + or Zn | Zn 2 + or Zn | Zn 2 + (1 M) (anode)

Cu 2 + ;Cu or Cu 2 + | Cu

or Cu 2 + (1 M) | Cu (cathode)

The salt bridge which separates the half cells is indicated by two vertical limes e.g. Zn | Zn 2 + (1 M ) Anode : Oxidation Occurs

|| Salt Bridge

Cu 2 + (1 M ) | Cu Cathode : Reduction Occurs

types of Electrodes • •

Metal-metal ion electrode constitute a metal rod dipped in a solution of its own ions e.g. Zn/Zn2+; Cu/Cu2+. If a metal is used as its amalgam in a metal-metal ion electrode, it is called amalgam electrode. Amalgam is prepared to modify the activity of the metal e.g., Zn-Hg/Zn2+

Electrochemistry

•

•

•

Gas electrode involve an inert metal such as platinum dipped in the solution containing ions of the gaseous element. E.g., Hydrogen electrode. If a metal and its sparingly soluble salt is dipped in an aqueous salt of some other soluble salt containing the common anion as that of sparingly soluble salt of the metal, it is called metal-metal insoluble salt-salt anion electrode. If a platinum wire is dipped in a solution of a mixture of two salts of the same metal having different oxidation states, it is called redox electrode e.g., Pt, Fe2+/Fe3+.

•

•

nernst Equation •

•

O

The difference in the electrode potentials of the two electrodes of the cell is termed as electromotive force (emf) or cell voltage (Ecell). Mathematically, emf = Ered(cathode) – Ered(anode) Simply emf = Ecathode–Eanode = Eright–Eleft or ER–EL The emf of a cell can be measured by using potentiometer when the emf applied from outside is equal to the emf of the cell is equal, the galvanometer shows no deflection.

Nernst equation gives the dependence of the electrode potential on the concentration of ions with which the electrode is reversible.  M For a metal electrode the reduction M n + + ne -  The Nernst equation to calculate the electrode potential at different concentration is

E=E -

•

Electromotive Force of the Cell

12.47

2.303RT [ M ] or E = E + 2.303RT log C log nF nF  M n +  O

Since M is solid, its activity will be unity. For a non-metal electrode the reduction reaction is  An - . The Nernst equation to calculate A + ne -  electrode potential at different concentration is n-

E=E O

 A  2.303RT log  nF [A]

2.303RT log  A n -  . nF Since the activity of A is unity General equation for any electrode is E = or E = E O

•

E

O

–

-

2.303RT [ Products] . Where E is the log nF [ Reactants] O

Electrical Energy •

•

•

•

The passage of electricity through a conductor is accompanied by the evolution of heat. According to the first law of thermodynamics, the heat librated must be exactly equal to the electrical energy. This heat evolved is proportional to the quantity of electricity passed and the emf of the cell. The maximum amount of electrical work obtained from the cell is the product of potential difference (in Volts) and charge flowing. Wele = nFE where n is the number of moles of electrons that flow during the chemical reaction and F the Faraday (96500 coulombs/ mol of electrons). The Wuseful of the Daniel cell can be calculated as Wuseful = 2 × 96500 × 1.10 = 212300 Joules or 212.3 KJ/mol Zn. The maximum useful work obtained from a system is equal to the decrease in Gibbs energy –ΔG. So for Daniel cell. –ΔG = Wuseful = 212.3 KJ/ molZn Or ΔG = –212.3 KJ/ mol Zn. If the emf of the cell is standard potential, the Gibbs energy change ΔG also standard Gibbs energy change ΔG .. The standard Gibbs energy for single electrode also can be calculated in a similar way using the standard electrode potential G = –nFE .

•

electrode potential at required concentration E is the standard electrode potential, R is the gas constant, T is the absolute temperature n is the number of electrons involved in the reaction, F is the faraday i.e., 96500 coulombs [M] or [A] are activities of metal or non- metal respectively [Mn+] is activity of metal ion and [An–] is the activity of non-metal ion taken as molor concentrations At 25°C by substituting the value of R, T and F value the Nernst equation is E = E O

0.059 [ Products] log n [ Reactants]

Cell reaction •

At anode always oxidation take place while at cathode reduction take place combining the two half reactions we get the cell reaction e.g., for Daniel cell Zn ( s )  → Zn 2 + (aq ) + 2e - at anode Cu 2 + (aq ) + 2e -  → Cu ( s ) 2+

at cathode

2+

Zn( s ) + Cu (aq )  → Zn (aq ) + Cu ( s ) cell reaction

O

•

O

O

Calculation of Cell Potential •

O

The standard emf (E ) of the galvanic cell is calculated using the equation

12.48

Electrochemistry O

O

O

E = E cathode – E anode or = E reduction electrode – E oxidation electrode or = E RHE – E LHE O

relationship Between Electrical Energy and Enthalpy Change of Cell reaction

O

O

O

• •

O

E Cell is always positive so that the cell reaction is spontaneous. If the emf of the cell is more, the reaction is faster. The emf of the cell (E ) depends on the intensity of the reaction in the cell. The emf of a cell at different concentrations can be calculated using the equation

The enthalpy change in a cell reaction can be obtained by the equation E=

O

•

)

θ θ –-E anode E = (E cathode node ) + O

O

0.059 [oxidant ] . log 2 [ Reductant ]

O

•

•

The Nernst equation to calculate the cell potential is ECell = E Cell -

 ∂E  If   is positive i. e if the emf of the cell increases  ∂T  P

with rise in temperature the electrical energy will be greater than the heat of reaction i.e., nFE > ΔH

Equilibrium Constant from nernst Equation •

•

-∆H  ∂E  +T   . nF  ∂T  P

0.0591 [ Products] (at 298 K) . log n [ Reactants ]

 ∂E  If   is negative i.e., if the emf of the cell  ∂T  P

decreases with rise in temperature the electrical energy will be smaller than the heat of the reaction i.e., nFE < ΔH. •

For a cell reaction of the type

 ∂E  If   is equal to zero, the electrical energy will be  ∂T  P

equal to heat of reaction i.e., nFE = ΔH.

 cC + dD aA + bB  0.0591 [ C] [ D] . log Ther nernst equation E = E n [ A ]a [ B]b c

d

Electrode Concentration Cells

O

When the cell reaction in an electrochemical cell is at equilibrium the cell emf (E) is zero since the system does no network then Nernst equation E=

•

•

2.303RT 0.0591 log K or E = log K at 298 K . nF n O

• •

The electrical work done in one second is equal to electrical potential multiplied by total charge passed. The reversible work done by a Galvanic cell is equal to decrease in its Gibbs energy ΔG = –nFE. If the activity of all reacting species is unity then O

E=E ∴ ΔG = –nFE. O

•

From the standard Gibbs energy the equilibrium constant by the equation ∴ ΔG = –RT ln K. O

a 0.0591 log 1 . 2 a2 The emf of this cell is due to transfer of Zn from the amalgam of activity a1 to the amalgam of the activity a2 and the emf depends on the ratio of the zinc activities in the this amalgams and not on the activity of Zn2+ ions. E of the electrode is zero. If the electrode used is a gas material at different activities and dipped in the solution of gas ions it is known as gas concentration cell. The potential of these cells depends on the pressure of the gas, concentration of its ions in solution e.g., E=

Electrochemical Cell and Gibbs Energy •

The cells in which the electrode material having different concentration are dipped in the solution of the metal ions are called electrode concentration cells. If two amalgams of the same metal at two different concentrations are immersed in the same electrolytic solution it is called amalgam concentration cells e.g., Zn (Hg )(a zn = a1 ) | zn 2 + (a zn 2+ ) | zn (Hg ) (a zn = a 2 )

O

•

(

)

H 2 PH2 = P1 | H + (a H+ )| H 2 (PH2 = P2 ); P1 > P2 .

The cell reaction for the spontaneous process is the expansion of hydrogen gas from P1 to P2. The emf of this cell depends only on the two pressures and is independent of the activity of the H+ ion in which the electrodes are immersed.

Electrochemistry

•

•

•

•

•

•

•

•

•

•

If identical electrodes are immersed in solutions of the same electrolyte of different concentrations it is known as electrolytic concentration cell. The electrolyte diffuse from a solution of higher concentration to that of lower concentration due to which electrical energy is produced. At the beginning, the emf of the cell is more and gradually fall to zero when the concentrations of the solutions become equal. Using concentration cells, the solubility of sparingly soluble salts, valency of the cation of the electrolyte and transition point of two allotropic forms of a metal can be determined. The substances which allows the electric current to pass through it called electrical conductor. Electrical conductors are two types (i) metallic conductors or electrical conductors (ii) electrolytic conductors or electrolytes. Electronic conductors or metallic conductors are those in which flow of electric current is due to the movement of free electrons from a higher negative potential region to a lower potential region without producing any chemical change in the conductor e.g., metals, alloys, graphite, solid salts like CdS, CuS. Electrolytic conductors are in which flow of electric current is due to the movement or migration of ions towards oppositely charged electrodes and is accompanied by chemical changes at the electrodes e.g., fused salts, aqueous solutions of salts, acids and bases In electronic conduction, electrons flow from negative end to positive end while in electrolytic conduction, ions move towards the oppositely charged electrodes Increase in temperature increases the resistance of electronic conductors while the resistance of electrolytic conduction decreases since the viscosity of solution decreases, mobility of ions increase and degree of ionization of electrolyte increases. Substances whose melts or aqueous solutions conduct electric current are called electrolytes. All salts, acids and bases are referred to as electrolytes. Substances whose melts or aqueous solutions do not conduct electric current are called non-electrolytes e.g. non-polar covalent substances like urea, glucose, sugar etc., are non electrolytes.

• •

•

•

positive pole of the battery is called anode or positive electrode. The current enters the electrolyte through the anode and leaves through cathode. At cathode, electronation of cation or reduction take place and at anode deelectronation of anions or oxidation take place The flow of electrons across the boundary is accompanied by chemical reaction i.e., oxidation–reduction. Such a reaction is called electrolysis. Electrolysis take place only at electrodes. The electrolyte as a whole remains neutral during the process of electrolysis as equal number of charges are neutralized at the electrodes.

nature of Electrolytes •

•

• •

•

•

•

•

If substances ionize completely, it is known as strong electrolyte while the substances which show very little ionization in solution are called weak electrolytes. The ionization of substance depends on the nature of solvent since ionization involves a reaction between solute and solvent. A substance which is a strong electrolyte in one solvent may become a weak electrolyte in a different solvent. If the solute – solute interactions are strong the extent of ionization will be less. If the solute – solvent interaction are strong ionization increases but their electrical conductivity decreases as they are highly solvated. If the solvent – solvent interactions are stronger, the viscosity of the solvent will be more, thus the conductivity of ions decreases. The conductivity of electrolytic solution depend on (i) the nature of the electrolyte (ii) size of the ions produced and their solvation (iii) concentration of the electrolyte. (iv) Temperature. Conductivity of electrolytes is measured by using conductivity cell which contain two electrodes separated by a fixed distance ‘l’ and have area of cross-section A The resistance R of conductivity cell is given by the equation R=

Electrolysis •

•

The rods, plates or metal foils through which electric current enter or leave the electrolyte are called as electrodes. The electrode which is connected to negative pole of the battery is called as cathode or negative electrode and the one which is connected to the

12.49

•

1 p. l = . A kA

The quantity l/A for a particular conductivity cell is constant denoted by the G* and is called cell constant. The cell constant can be determined by using a KCl solution whose conductivity is known accurately at various concentrations The cell constant G* = l/A = Rκ

12.50

• •

Electrochemistry

The conductances of different solutions can be determined by using wheatstone bridge principle. The specific conductance of a solution κ is given by κ =

•

•

G the total conductance of the solution is the R

product of specific conductance and volume of the solution κ × V. If the amount of electrolyte dissolved in solution is equal to the gram equivalent weight of the electrolyte, then the total conductance is known as equivalent conductance. Λeq = 1000 κ/C where C is the concentration of the solution in gram equivalent per litre. The unit of equivalent conductivity is ohm–1 cm2 eq–1 If the amount of electrolyte dissolved in solution is equal to the gram molecular weight of electrolyte then the total conductance is known as molar conductivity (ΛM)

a∝ =

•

•

•

•

•

•

•

Where Λc is the molar conductivity at concentration C and Λ0 is the molar conductance at infinite dilution. Using the value of a (the degree of dissociation) the equilibrium constant for a weak electrolyte can be calculated k=

α2 (1 − α) v

Kohlrausch's law •

Kohlrauschs law states that the equivalent conductivity of an electrolyte at infinite dilution (Λ0) is the sum of the equivalent conductivities of the cations and anions Λ 0 = λ0+ + λ0-

Mol wt ΛM = Λe × . eq.wt. The unit of molar conductivity is ohm–1 cm2 eq–1. Specific conductance always decreases with the decrease in concentration both for strong and weak electrolytes due to the decrease in the number of ions per unit volume that carry the current in a solution. The resistance of an electrolyte cannot be measured by applying direct current because of the following reasons due to polarization effect such as the change in concentration of electrolyte due to electrolysis and the products formed during electrolysis produce an opposing potential. Hence alternating current is used. Ions in a solution are surrounded by oppositely charged ions and hence their movement towards electrodes in an applied electric field will be restrained. With dilution the number of oppositely charged ions surrounding a particular ion decreases and hence the conductivity increases for strong electrolytes which already ionized completely. The molar conductivity will be maximum at infinite dilution which can be obtained by extrapolating the line showing relation between conductivity and concentration. This is known as limiting molar conductivity. The variation of molar conductivity with concentration may be given by the expression Λ = Λ0 – AC1/2 where A is a constant and Λ0 is the molar conductivity at infinite dilution. The molar conductivity of weak electrolytes increases with increase in ionization. The degree of dissociation (a) of weak electrolytes can be calculated at any concentration by

Λc Λ0

+

•

•

Where λ0 is the ionic conductance of the cation and λ0- is the ionic conductance of the anion. The degree of dissociation can also be represented in terms of molar conductivity a∝= = Λ c / Λ 0 where C refers to the concentration of the electrolyte in gram equivalent per litre. The degree of dissociation can also be represented in terms of molar conductivity a∝= = Λ cM / Λ 0M where Λ cM is the molar conductivity at concentration C and Λ 0c is the molar conductivity at infinite dilution. From a the dissociation constant can be calculated k=

•

•

Kohlrauch law is also useful to calculate Λ0 for any electrolyte from the λ0 of individual ions. The molar conductivities of H+ and OH– ions are very high because these ions are passed from one molecule to another and released at the electrodes without travelling. Equivalent conductance of weak electrolytes can be calculated from the conductances of completely dissociated strong electrolytes e.g. 0 Λ CH

•

C α2 (1 − α)

3 COOH

0 = Λ CH

3 COONa

+ Λ 0HCl - Λ 0NaCl

Solubility of sparingly soluble salts can be calculated from the specific conductance of its saturated solution and from the equivalent conductivity at infinite dilution obtained from Λ 0+ + Λ 0Λ 0e =

Absolute ionic mobility =

1000 K salt C

Λ 0± Ionic Conductance = 96500 96500

Electrochemistry

• • •

•

•

•

•

•

Using ionic conductance measurements the ionic product of water can be determined as 1 × 10–14 at 25°C Titrations followed by conductance are called conductomtric titrations In the titration of strong acid vs strong base, conductance decreases till the equivalence point since more conducting H+ ions are replaced by less conducting cations from the base. But after the end point, conductance increases rapidly due to the more conducting OH– ions from strong base. In the titration of weak acid vs strong base, the conductance increases slowly since the less ionizing weak base is replaced by its more ionizing salt formed by neutralization with strong base. After the equivalence point the conductance increase rapidly due to the more conducting OH– ions form the strong base In the titration of strong acid vs weak base, conductance decreases till the equivalence point since more conducting H+ ions are replaced by less conducting cations of the base. But after equivalence point conductance does not change much because the weak base do not ionize due to the common ion effect of its salt formed during the titration. In the titration of weak acid vs weak base conductance decrease in the beginning due to the formation of salt which further decreases the ionization of weak acid due to common ion effect. With increase in the addition base, conductivity increases due to the formation of ionizing salt. Also some hydrolysis of the salt occurs. Beyond the end point, conductance remains constant. In the titration of a mixture of strong acid and weak acid with a strong base the conductance decreases until the strong acid is neutralized due to replacement of more conducting H+ ions with less conducting cations of base. After the neutralization of strong acid and during the neutralization of weak acid, the conductance remains nearly constant but after complete neutralization conductivity increases due to excess of strong base. In precipitation titrations, conductance remain almost constants since one salt is replaced by another but after the end point, conductance rapidly increases due to the addition of excess of salt.

Elecrolysis •

•

Electrodes used in the electrolysis of different electrolytes are of two types: (i) inert electrodes (ii) active electrodes. The potential at which an ion loses its charge and liberated at the electrode is called discharge potential.

•

•

• •

• •

•

• •

12.51

The applied emf required to cause continuous electrolysis of an electrolyte such as CuCl2 into elements is known as decomposition potential. The potential required for decomposition is always greater than the decomposition potential and is called over potential or over voltage. This is because the standard electrode potentials are measured at equilibrium conditions but in electrolysis ions are continuously discharged. Over voltages are particularly more for gases and particularly for H2 and O2. The high hydrogen over voltage is due to the more energy required to remove H+ ion from H3O+ ion followed by the addition of electron and formation of H2 molecule. Over voltages depend on the applied emf and also on the nature of electrode. When inert electrodes such as platinum are used in the electrolysis of solution of electrolytes the products formed at the cathode and anode depend on the nature of electrolyte. If more than one type of ions are attracted to the given electrode then the ion discharged is the ion which requires less energy or the ion which is least active. This is known as preferential discharge. If the electrode is active, metal deposits at cathode while at anode metal is dissolved. When fused salts are electrolyzed using inert electrode they dissociate into their constituent elements.

Faraday's laws of Electrolysis • •

•

•

•

Faraday explained that the decomposition of electrolytes by an electric current is governed by two laws. First law: The amount of the substance liberated or deposited or dissolved at an electrode during electrolysis of an electrolyte is directly proportional to the quantity of electricity passing through the solution of electrolyte or the melt. Mathematically, Faraday’s first law is m ∝ q or m = eq or m = ect where m is the mass of the substance liberated or deposited or dissolved, q is the quantity of electricity in coulombs, t is the time in seconds, c is the strength of current in amperes and e is the electrochemical equivalent of the ion or metal or molecule deposited or liberated or dissolved at the electrode. Electrochemical equivalent of a substance is the amount deposited or liberated or dissolved or underwent electrode reaction at an electrode by passing one ampere current for one second i.e., one coulomb. Chemical equivalent of a substance is the amount of substance deposited or liberated or dissolved or had undergone electrode reaction at an electrode during

12.52

Electrochemistry

the passage of one faraday of electricity during the electrolysis of electrolyte solution or melt chemical equivalent of an element or ion

=

•

Atomic Weight Valency or charge of the ion

In the dry cell a zinc container act as the anode and the container is filled with moist paste of ZnCl2 , NH 4 Cl and MnO2 . In the middle of the container a graphite electrode is placed which acts as anode. The electrode reactions in dry cell are. Anode Zn  → Zn 2 + + 2e -

The electrochemical equivalent of an element is directly portional to its chemical equivalent E.

Cathode 2 MnO2 + 2 NH 4+ + 2e  → Mn2 O3 + 2 NH 3 + 2 H 2 O

e ∝ E or E = F × e

•

Units of electrochemical equivalent is gram coulomb–1 One faraday i.e., 96,500 coulombs is equal to the charge present on one mole (6.023×1023) of electrons or protons. m=

•

•

•

Zn 2 + + 4 NH 3 + 2Cl -  →  Zn( NH 3 ) 4  Cl2 ( s ) •

E×c×t E e = 96500 96500

Second law of faraday states that if the same quantity electricity is passed through different electrolyte solutions or melts the amount of the different substances liberated or deposited or dissolved or had undergone reaction at the electrode are directly proportional to their chemical equivalents. W1 W2 W3 = = E1 E2 E3

•

The secondary reaction which is not involved in the electrode reaction occurs.

Anode

•

Faraday’s laws are useful for (i) preparing several pure metals (ii) to separate metals from metals (iii) to prepare several compounds (iv) to electroplating of one metal on the other.

•

•

 → ZnO(s) + H 2 O + 2e  → Hg(l)

+ 2OH -

 → ZnO(s) + Hg(l)

A secondary cell is a cell in which the electrode reactions are reversed by the application of an external current. Hence such a cell once used can be recharged. The lead storage cell which generates a voltage of 6V or 12V is an example of the secondary cell. The lead storage battery contain a lead electrode which acts as anode and a grid of lead covered with PbO2 acting as cathode. The electrolyte is 40% H2SO4 solution. The electrode reactions are: Anode

Pb(s) + SO 24 -  → PbSO 4 (s) + 2e -

Cathode Pb(s) + SO 24 - + 4H + + 2e -  → PbSO 4 (s) + 2H 2O

Batteries

•

+ H 2 O + 2e -

Overall Zn(Hg) + HgO(s) Reaction

The chemical equivalents depend on the number of electrons participated at the electrode reaction. The chemical equivalents or equivalent weights of NaCl, KCl, KBr, NaOH etc are equal to their molecular weights since only one electron take part in electrode reaction. The equivalent weights of other electrolytes depend on the number of electrons. Equivalent Weight = Molecular Weight No of Electrons involved in electrode reaction

The commercial cells are of two types (i) primary cells and (ii) secondary cells. A primary cell is a cell in which the electrical energy is obtained at the expense of chemical reaction. A primary cell works as long as the active chemicals reacting. The dry cell which generates a voltage of ~1.25-1.50V is an example of the primary cell.

Zn(Hg) + 2OH -

Cathode HgO

•

•

Mercury cell is another primary cell used for the low current devices like hearing aids, calculators, quartz watches etc. the mercury cell contains a zinc container which serves as the anode, a carbon rod the cathode and a moist paste of HgO and KOH electrolyte. A porous paper lining keeps the electrolyte separated from the anode. The electrode reactions are

In the lead storage battery a number of anodes are joined together and are arranged alternatively with a number of cathode plates. The cathode plates are joined together. The chemical reactions for discharging and charging of the cell are as follows Charge   Pb + PbO 2 + 2H 2SO 4   2Pb SO 4 + 2H 2 O Discharge

• •

•

During the discharge of the cell sulphuric acid is consumed and hence the density of sulphuric is decreased. During the charge of the used battery sulphuric acid is regenerated Pb is deposited at the anode and PbO2 is formed at the cathode. A lead storage battery is known as the lead accumulator.

Electrochemistry

12.53

The overall reaction is

Corrosion

Fe + O2 + H2O → Fe2+ + 2OH–; E = 1.67V O

•

•

•

•

•

• •

Corrosion is a process of deterioration and consequent loss of solid metallic material through an unwanted chemical or electrochemical attack by its environment starting at the surface. The chemical corrosion occurs due to the direct chemical action of the environment (ex inorganic liquid) or atmospheric gases as O2, H2S, SO2 halogens and ammonia. The extent of chemical corrosion depends on the chemical affinity of solid metal with the corrosive environment and ability of the reaction product to form protective film on the metal surface. The electrochemical corrosion occurs when a metal is in contact with the conducting liquid or when two dissimilar metals or alloys are dipped partially or completely in a solution. The electrochemical corrosion occurs due to the existence of separate anodic or cathodic areas between which there occurs a flow of current through the conducting solution. The corrosion always occurs at the anodic areas. The rusting of iron occurs due to corrosion. The electrode reactions in the rusting process are as follows. Anode Fe → Fe2+ + 2e–; E = –0.44V Cathode O2 + 2H2O + 4e– → 4OH–; E = 1.23V O

O

•

The Fe2+ and OH– ions combine to form Fe(OH)2 which oxidizes to Fe(OH)3 in excess of oxygen. The product formed correspond to Fe2O3.xH2O. If the supply of oxygen is limited, the corrosion product is black magnetite (Fe3O4).

Protection of Corrosion •

• •

•

•

Corrosion is prevented by coating the metal surface with a thin film of paint grease metal (eg. Zn, Sn, Ni, Cu, Cr) metal oxides (eg., Fe3O4). Alloying the base metals like Fe, Cu, Al etc., with Cr, Ni, V, W, Au, Pt etc., leads to the protection of metals. The metal to be protected is made to behave as a cathode by the application of an external current. Since corrosion does not occur at cathode but occurs only at anode the metal will be protected. When a metal surface is connected to a more anodic metal by a wire so that the corrosion occur at the more active metal preventing the less active metal from corrosion. Metals which are used as sacrificial anode are Zn, Mg, Al and their alloys. Mg is attached to steel frames of ships for preventing corrosion of Fe.

12.54

Electrochemistry

PraCtICE ExErCIsE multiple Choice Questions with only one answer level I 1. Passage of one ampere current through 0.1 M Ni(NO3)4 solution using Ni electrodes brings in the concentration of solution to ________ in 60 seconds. (a) 0.2 M (b) 0.5 M (c) 0.1 M (d) 0.025 M 2. Strong electrolytes are those which (a) Dissolve readily in water (b) Conduct electricity (c) Dissociate into ions at high dilution (d) Completely dissociate into ions at all dilutions 3. The one that is a good conductor of electricity in the following list of solids is (a) Sodium chloride (b) Graphite (c) Diamond (d) Silica 4. The charge required for the reduction of 1 mole of Cr2O72– ions to Cr3+ is (a) 96500 C (b) 2×965000 (c) 3×96500 C (d) 6×96500 C 5. 10800 C of electricity through the electrolyte deposited 2.977 g of metal with atomic mass 106.4 g mol–1. The charge on the metal cation is (a) +4 (b) +3 (c) +2 (d) +1 6. One coulomb of charge passes through solution of AgNO3 and CuSO4 connected is series and the conc. of two solutions being in the ratio 1:2. The ratio of amount of Ag and Cu deposited on Pt electrode is (a) 107:63.54 (b) 54:31.77 (c) 107.9:31.77 (d) 54:63.54 7. Electrolysis of dil. H2SO4 liberates gases at anode and cathode. (a) O2 and SO2 respectively (b) SO2 and O2 respectively (c) O2 and H2 respectively (d) H2 and O2 respectively 8. Charge of ion N–3 is: (a) F (b) 2F (c) 3F (d) 4F 9. During electrolysis of H2O the molar ratio of H2 and O2 formed is (a) 2:1 (b) 1:2 (c) 1:3 (d) 4:1 10. 3 faradays of electricity is passed through the three electrolytic cells connected in series containing Ag+,

11.

12.

13.

14.

15.

16.

17.

18.

Ca2+ and Al+3 ions respectively. The molar ratio in which the three metal ions are liberated at the electrodes is (a) 1:2:3: (b) 3:2:1 (c) 6:3:2 (d) 3:4:2 Conducting power of a solution is directly proportional to (a) Dilution (b) Number of ions (c) Current density (d) Volume of solution The conductivity of a strong electrolyte (a) Increase on dilution (b) Does not change considerably on dilution (c) Decreases on dilution (d) Depends upon density Faraday's Laws of electrolysis are related to the (a) Atomic number of cation (b) Atomic number of anion (c) Speed of ion (d) Equivalent weight of electrolyte A dilute aqueous solution of Na2SO4 is electrolyzed using platinum electrodes. The products at the anode and cathode are (a) O2/H2 (b) S2O82–, Na (c) O2, Na (d) S2O82–, H2 In the electrolysis of an aqueous solution of sodium sulphate, 2.4 lit of oxygen at STP was liberated at anode. The volume of hydrogen at STP liberated at cathode would be (a) 1.2 L (b) 2.4 L (c) 2.6 L (d) 4.8 L In the electrolysis of CuCl2 solution using copper electrode if 2.5 gm Cu deposited at cathode, then at anode (a) 890 mL of Cl2 at STP is liberated (b) 445 mL of O2 at STP is liberated (c) 2.5 g of copper is deposited (d) A decrease of 2.5 g of mass taken place The reaction occurring at anode when the electrolysis of an aqueous solution containing Na2SO4 and CuSO4 is done using Pt electrodes is → Cu 2 + + 2e (a) Cu  2− (b) 2SO 4 + 2H 2 O  → 2H 2SO 4 + O 2 + 4e − (c) 2H 2 O  → O 2 + 4 H + + 4e − (a) Cl  → Cl2 + 2e In the electrolysis of alkaline water a total of 1 mol of gas is evolved. The amount of water decomposed would be (a) 1 mol (b) 2 mol 2 1 (c) mol (d) mol 3 3

Electrochemistry

19. In an electrolysis of metallic chloride 3.283 g of the metal (molar mass 197 g mol–1) was deposited on the cathode by the passage of 4825 C of electric charge. The charge number of metal ion is (a) 0.5 (b) 1 (c) 2 (d) 3 20. In acidic medium, MnO4- is converted to Mn2+ while acting as an oxidizing agent. The quantity of electricity required to reduce 0.05 mol of MnO4- would be (a) 0.01 F (b) 0.05 F (c) 0.25 F (d) 0.15 F 21. When CuCl2 was electrolyzed using Cu electrodes the mass of cathode is increased by 3.18 g. What happened at the anode? (a) 0.56 lit of O2 was liberated (b) 0.56 lit of Cl2 was liberated (c) 0.1 mole of dissolves Cu+2 (d) 0.05 mol dissolves as Cu+2 22. On electrolysis of H2SO4 solution, the weight of product obtained at cathode and anode are in the ratio (a) 1:8 (b) 8:1 (c) 1:16 (d) 16:1 23. 10 amp. of current passed for 10 minutes deposited 6 g of metal from its salt solution. The ECE of that metal is: (a) 10–3 (b) 6 × 10–2 –6 (c) 2 × 10 (d) 6.02 × 10–3 24. Certain quantity of current has liberated 11.2 lit O2 same current can deposit how many moles of Mg? (a) 1 (b) 2 (c) 0.5 (d) 4 25. How long will it take a current of 2 amp to deposit all the silver from 500 mL of normal AgNO3 solution? (a) 24125 sec (b) 48250 sec (c) 12000 sec (d) 96600 sec 26. For the electrolytic product of NaClO4 according to the reaction NaClO3 + H 2 O → NaClO 4 + H 2 how many faradays of electricity would be required to produce 0.5 mole of NaClO4? (a) 1 (b) 2 (c) 3 (d) 1.5 27. When an aqueous solution of lithium chloride is electrolyzed using graphite electrodes, (a) pH of the resulting solution increases. (b) pH of resulting solution decrease. (c) As the current flows, pH of the solution around the cathode increases. (d) As the current flow, pH of the solution around the anode decreases. 28. Electrochemical equivalent of a divalent metal is 3.0 × 10–4 what is the approximate atomic weight of the metal: (a) 107.8 (b) 63.6 (c) 58 (d) 55.8

12.55

29. At the time of charging, lead accumulator act as a (a) An electrolytic cell (b) A Galvanic cell (c) A Daniel cell (d) None of these 30. The standard oxidation potentials E for the half reactions are as Zn → Zn 2 + + 2e, E θ = +0.76 V O

O

Fe → Fe 2 + + 2e, E θ = +0.41 V O

The emf of the cell Fe 2 + + Zn → Zn 2 + + Fe is

(a) 0.35 V (b) -0.35 V (c) +1.17 (d) -1.17 V 31. Strongest reducing agent is (a) K (b) Mg (c) Al (d) I 32. The reaction

1 H 2 (g) + AgCl(s) → H + (aq) + Cl- (aq) + Ag(s) 2 occurs in the galvanic cell (a) Ag / AgCl(s) / KCl(soln) // AgNO3 (soln) / Ag (b) Pt / H2(g) / HCl(soln) // AgNO3 (soln) / Ag (c) Pt / H2(g) / HCl(soln) // AgCl(s)KCl / Ag (d) Pt / H2(g) / KCl(soln) // AgCl(s) / Ag 33. The emf of the cell (Zn | ZnSO4 (0.1M) || CdSO4 (0.01M) | Cd) is

(E 34.

35.

36.

O

Zn 2+ Zn

O

= −0.76V, E Cd2+

Cd

= 0.40 V at 298 K

(a) +0.33V (b) +0.36V (c) +1.13V (d) -0.36V A solution containing one mole per litre of each Cu (NO3)2; AgNO3; Hg2(NO3)2; Mg(NO3)2 is being electrolyzed by using inert electrodes. By increasing voltage, the sequence of deposition of metals on the cathode will be (a) Ag, Hg, Cu (b) Cu, Hg, Ag (c) Ag, Hg, Cu, Mg (d) Mg, Cu, Hg, Ag Which one will liberate Br2 from KBr? (a) HI (b) I2 (c) Cl2 (d) SO2 From the following E values of half cells (I) A → A + + e; E = +1.2V (II) B− → B + e; E = −2.1V (III) C → C2 + + 2e; E = −0.38V (IV)D −2 → D − + e; E = −0.59V What combination of two half cell would result in a cell with the largest potential? (a) I and IV (b) II and III (c) III and IV (d) I and II E for F2 + 2e − → 2F− is 2.8V, E for 1 / 2 F2 + e − = F− is (a) 2.8V (b) 1.4V (c) -2.8V (d) -1.4V O

O

O

O

O

37.

38.

)

O

O

12.56

Electrochemistry

39. The value of emf for a feasible cell reaction is (a) < 1 (b) 0 (c) =1 (d) > 0 40. The half cell reduction potential of a hydrogen electrode at pH = 10 will be (a) 0.59V (b) -0.59V (c) 0.059V (d) -0.059V 41. How much will the reduction potential of a hydrogen electrode change when its solution initially at pH = 0 is neutralized to pH = 7? (a) Increase by 0.059V (b) Decrease by 0.059V (c) Increase by 0.41V (d) Decrease by 0.41V 42. Out of Cu, Ag, Fe and Zn the metal which can displace all others from their salt solution is: (a) Ag (b) Cu (c) Zn (d) Fe 43. Dilute H2SO4 on electrolysis shows the anodic reaction with Pt electrodes (a) SO −42 → SO 4 + 2e − (b) 2 SO4-2 + 2 H 2 O → 2 H 2 SO4 + O2-2 (c) 4 H + + 4e - → 2 H 2 (d) 2 H 2 O - 4e - → 4 H + + O2 44. The electrolyte used in gold plating of copper articles containing (a) Molten gold (b) CuSO4 (c) AuCl3 (d) AuCl3 + NaCN 45. A cell is spontaneous when (a) ΔG is +Ve (b) E is –Ve (c) E is +Ve (d) ΔG is -Ve 46. The most powerful reducing agent is (a) HCl (b) HI (c) HF (d) HBr 47. The most powerful oxidizing agent is (a) F2 (b) Cl2 (c) Br2 (d) I2 48. A galvanic cell not necessarily contains (a) A cathode (b) An anode (c) Ions (d) Porous plate 49. When a lead storage battery is discharged (a) SO2 is evolved (b) Lead sulphate is consumed (c) Lead is formed (d) H2SO4 is consumed 50. If a salt bridge is removed between the two half cells, the voltage (a) Drops to zero (b) Does not change (c) Increases gradually (d) Increase rapidly 51. The reference electrode is made from which of the following? (a) HgCl2 (b) Hg2Cl2 (c) ZnCl2 (d) CuSO4 O

O

52. Stronger is oxidizing agent, greater is the (a) Standard reduction potential (b) Ionic nature (c) Standard oxidation potential (d) None 53. CuSO4 is not stored in aluminum bottles because (a) Cu gets oxidized (b) Cu gets reduced (c) Cu+2 gets reduced (d) CuSO4 decomposes 54. When a copper wire is placed in a solution of AgNO3, the solution acquires blue colour. This is due to the formation of (a) Cu+2 ions (b) Cu+ ions (c) Soluble complex of Cu with a AgNO3 (d) none 55. Which of the following will increase the voltage of the cell? Sn(s) +2Ag+ (aq)→Sn+2 (aq) +2Ag(s) (a) Increase in the size of the silver (b) Increase in the concentration of Sn 2 + (c) Increase in the concentration of Ag+ (d) None of these 56. E RP for Fe+2/Fe and Sn+2/Sn is –0.44 and –0.14 volts respectively. The standard emf of cell Fe2++ Sn → Sn+2 + Fe is (a) +0.30 V (b) -0.58 V (c) +0.58 V (d) -0.30 V 57. In electroplating, the article to be electroplated serves as (a) Cathode (b) Anode (c) Electrolyte (d) None of these 58. The standard potential of Zn/Zn+2 and Ag+/Ag are +0.763 and 0.799V. The standard emf of cell is (a) 1.562V (b) 0.036V (c) -1.562V (d) 0.799 59. The hydrogen electrode is dipped in a solution of pH = 3 at 25°C. The reduction potential of half cell would be (a) 0.177V (b) -0.177V (c) 0.0887V (d) 0.159V 60. The standard reduction potentials of Cu2+/Cu and Cu2+/ Cu+ are 0.337 and 0.153 V respectively. The standard electrode potential of Cu+/Cu half cell is (a) 0.184V (b) 0.826V (c) 0.521V (d) 0.4790V 61. A standard hydrogen electrode has zero electrode potential because (a) Hydrogen is easier to oxidize. (b) This electrode potential is assumed to be zero (c) Hydrogen atom has only one electron. (d) Hydrogen is the lightest element. O

Electrochemistry

62. On the basis of position in the electrochemical series, the metal which does not displace hydrogen from water and acids is (a) Hg (b) Al (c) Pb (d) Ba 63. The correct electrochemical series can be obtained from K, Na, Ca, Al, Mg, Zn, Fe, Pb, H, Cu, Hg, Ag, and Au by interchanging (a) Only Al and Mg (b) Only Na and Ca (c) Both 1 and 2 (d) None 64. The correct order of chemical reactivity with water according to electrochemical series is (a) K>Mg>Zn>Cu (b) Mg>Zn>Cu>K (c) K>Zn>Mg>Cu (d) Cu>Zn>Mg>K 65. Molar aluminum electrode coupled with normal hydrogen electrode gives an emf of 1.66 volts, so the standard electrode potential of aluminum is (a) -1.66V (b) +1.66V (c) -0.83V (d) +0.83V 66. Which of the following statement is correct? (a) Cathode is positive terminal in an electrolytic cell. (b) Cathode is negative terminal in a galvanic cell. (c) Reduction occurs at cathode in either of cells. (d) Oxidation occurs at cathode in either of cells. 67. Which of the following statement is correct? (a) Cathode is negative terminal both in galvanic and electrolytic cells. (b) Anode is positive terminal both in galvanic and electrolytic cells. (c) Cathode is negative terminal in an electrolytic cell whereas anode is negative terminal in a galvanic cell. (d) Anode is negative terminal in an electrolytic cell where as cathode is positive terminal in a galvanic cell. 68. In electrochemical corrosion, the metal undergoing corrosion (a) Acts as anode (b) Acts as cathode (c) Undergoes reduction (d) Liquefies 69. Given E Cu2+/Cu = 0.337 V and E Sn2+/Sn = 0.136 V Which of the following statements is correct? (a) Cu2+ ions can be reduced by H2(g) (b) Cu can reduce Sn2+ (c) Cu can be oxidized by H+ (d) Sn2+ ions can be reduced by H2 (g) 70. The reduction potential of the hydrogen half-cell will be negative if (a) P(H2) = 1 atm and [H+] = 1 M (b) P(H2) = 1 atm and [H+] = 2 M (c) P(H2) = 2 atm and [H+] = 1 M (d) P(H2) = 2 atm and [H+] = 2 M O

O

12.57

71. By how much will be potential of half-cell Cu2+/ Cu change if the solution is diluted to 100 times at 298K? (a) Increase by 59 mV (b) Decrease by 59 mV (c) Increase by 29.5 mv (d) Decrease by 29.5 mv 72. Given that EH 2O , H 2, Pt = 0 at 298 K. the pressure of H2 gas (a) 10–7 atm (b) 10–10 atm –12 (c) 10 atm (d) 10–14 atm 73. The value of EH 2O , H 2 (1atm ) Pt at 298 K would be (a) -0.2.7V (b) -0.414V (c) 0.207V 4)0.414V 74. Given that E Fe3+/Fe and E Fe2+/Fe are -0.036 V and -0.439 V respectively. The value of E Fe3+/Fe2+Pt would be (a) (-0.036-0.439) V (b) (-0.036+0.439) V (c) [3(-0.0-36)+2(-0.439)] V (d) [3(-0.036-2(-0.439)] V 75. A current of 9.65 A is drawn from a Daniell cell for exactly 1 hr. If molar masses of Cu and Zn are 63.5 g mol–1 and 65.4g mol –1, respectively, the loss in mass at anode and gain in mass at cathode respectively are (a) 11.43 g, 11.7 g (b) 11.77 g,11.43 g (c) 22.86 g, 23.54 g (d) 23.54 g, 22.86 g 76. A cell contains two hydrogen electrodes. The negative electrode is in contact with a solution of 10–6 M hydrogenous solution. The EMF of the cell 0.118 V at 298 K. The concentration of H+ ions at the positive electrode is (a) 10–5 M (b) 10–6 M –4 (c) 10 M (d) 10–3 M 77. Electrode potential of copper electrode where the conc. of Cu2+ is 5 M is …………………. V if E of Cu2+/Cu is 0.34 V (a) 0.36 V (b) -0.36V (c) 0.45 V (d) -0.45 V 78. Standard EMF of the cell Fe/Fe2+ is 0.44V where as standard EMF of the cell Cu/Cu2+ is –0.32V then (a) Cu oxidizes Fe2+ ion (b) Cu2+ oxidizes iron (c) Cu reduce Fe2+ ion (d) Cu2+ ion reduces iron 79. Given the E values of Ag+/ Ag, K+/ K, Mg2+/Mg, and Cr+3/Cr are 0.80V, -2.93 V, -2.38V and -0.74 V respectively. Which of the following order regarding the reducing powers of metal is correct? (a) Cr>Ag>Mg>K (b) AgMg (d) Cr>Ag>Mg>K O

O

O

O

O

Electrochemistry

O

88. Zinc is coated over iron to prevent rusting of iron because (a) It is cheaper than iron (b) E Zn2+/Zn = E Fe2+/Fe (c) E Zn2+/Zn < E Fe2+/Fe (d) E Zn2+/Zn > E Fe2+/Fe O

O

O

O

O

O

Conductance

89. In the conductometric titration of silver nitrate against KCl, the graph obtained is Conductance

80. Two electrolytic cells, one containing acidified ferrous chloride and another acidified ferric chloride are in series. The ratio of iron deposited at cathode in the two cells will be (a) 3:1 (b) 2:1 (c) 1:1 (d) 3:2 81. During the charging of a lead storage battery, the reaction occurring at the cathode is represented by (a) Pb2+ + 2e → Pb (b) Pb2+ + SO–2 4 → PbSO4 (c) Pb → Pb2+ + 2e + (d) PbSO4+2H2O→2PbO2+ 2H 2 +SO4 + 2e 82. The values of the standard oxidation electrode potentials of the following reactions are given: (a) Cu → Cu 2 + + 2e − E = − 0.345V E = + 0.136V (b) Sn +2 + 2e − → Sn (c) Fe → Fe 2 + + 2e − E = + 0.440V (d) Zn → Zn 2 + + 2e − E = + 0.762V Which one of the following is most easily reduced? (a) Cu+2 (b) Sn+2 +2 (c) Fe (d) Zn+2 83. A galvanic cell is made up with S.H.E. and another hydrogen electrode. The other hydrogen electrode should be immersed in which of the following solutions to get maximum values of emf? (a) 0.1 M HCl (b) 0.1M CH4COOH (c) 0.1M H2SO4 (d) 0.1M H3PO4 84. By which of the following change it is possible to increase the voltage of a cell in which the reaction is Zn( s ) + Cu 2 + ( aq ) → Zn 2 + ( aq ) + Cu( s ) (a) By increasing the concentration of Zn2+. (b) By increasing the concentration of Cu2+. (c) By changing the size of Zn electrode. (d) By increasing the size of Cu electrode. 85. The emf of a concentration cell consisting of two zinc electrodes one dipping into M/4 solution of zinc sulphate and the other M/16 solution of the same salt at 25°C is (a) 0.0124 V (b) 0.0250 V (c) 0.0178 V (d) 0.0356 V 86. Given that E = -0.035V, the free energy change ΔG for the reaction Fe3+ + 3e - → Fe( s ) is : (a) 10.12 KJ (b) 5.12 KJ (c) 20.84 KJ (d) 3.47 KJ 87. The measured voltage of cell Cd/Cd+2 (1.0M/H+(?)/ H2(1.0 atm) Pt is 0.220 V. What is the pH of this solution?  → Cd E0 = –0.400V I. Cd +2 (aq) + 2e - ←  + -  → H 2 (g) E = –0.00V II. 2H (aq) + 2e ←  (a) 3.04 (b) 3.72 (c) 1.52 (d) 2.23

(a)

(b) volume of KCl

O

volume of KCl

O

O

O

O O

Conductance

O

Conductance

12.58

(c)

(d) volume of KCl

volume of KCl

90. What is the overall cell reaction for the cell? Cr(s)/Cr+3(1M)// Br– (1M)/Ag Br(s) Ag (a) Cr + 3AgBr → Cr+3+3 Ag + 3Br– (b) 3Ag+Cr+3 → 3Ag+ + Cr (c) Cr+3Ag+ → Cr+3 + 3Ag (d) 2Cr+3Br2 → 2Cr+3 + 6Br– 91. Hydrogen electrodes are placed in two solutions of pH = 3 and pH = 8 at 25°C. A salt bridge links the solutions. What is the emf of the cell thus constructed? (a) 0.327 V (b) 0.295 V (c) -0.277 V (d) 0.177 V 92. A hydrogen electrode is dipping in a 0.01 N solution of H2SO4. It is coupled through a salt bridge with a normal calomel electrode (SRP = 0.28 V). What is the emf of the cell at 25°C (a) 0.3982 V (b) 0.2803 V (c) 0.3038 V (d) 0.2083 V 93. E Ni/Ni+2 = 0.236 V if this electrode is clubbed with hydrogen electrode. What should be the PH of the acid in H2 electrode to give zero cell potential at 25°C (a) 2 (b) 1 (c) 8 (d) 4 94. A metal has ERP = -0.76V and metal B has ERP = 0.8V which is correct? (a) A can liberate H2 from acid (b) B can liberate H2 from acid (c) Both A and B can liberate H2 from acid (d) None can liberate H2 from acid O

Electrochemistry

95. The molar conductance of HCl, NaCl and CH3COONa at infinite dilution are 426, 126 and 91 s cm2 mole–1 respectively. The molar conductance of CH 3 COOH at infinite dilution is …….. s cm2 mol–1 (a) 561 (b) 391 (c) 261 (d) 612 96. The value of molar conductance of HCl is greater than that of NaCl at a particular temperature and dilution because (a) MW of HCl < MW of NaCl (b) UH+> UNa+ (c) HCl is acid (d) HCl ionizes more than NaCl 97. The formula a = λ∨ / λ∞ is valid for (a) weak electrolytes. (b) strong electrolytes. (c) salts. (d) all 98. The increase in the molar conductivity of HCl with dilution is due to (a) increase in the self ionization of water. (b) hydrolysis of HCl. (c) decrease in the self ionization of water. (d) decrease the inter ionic forces. 99. The unit of cell constant is (a) ohm–1 (b) ohm–1 cm–1 –1 (c) cm (d) ohm–1cm2 eq–1 100. Unit of specific conductance is (a) ohm cm (b) ohm–1 cm –1 –1 (c) ohm cm (d) ohm cm–1 101. Which has maximum conductivity? (a) [Cr(NH3)3 Cl3] (b) [Cr(NH3)4Cl2]Cl (c) [Cr(NH3)5Cl]Cl2 (d) [Cr(NH3)6]Cl3 102. The specific conductance of 0.01 N KCl is X ohm–1 cm–1 having conductance Y ohm–1. The specific conductance of 0.01 N NaCl having conductance Z ohm–1 is (a) YX/Z (b) ZX/Y (c) Z/YX (d) None 103. Equivalent conductance of 1M CH3COOH is 10 ohm1 cm–2eq–1 and that at infinite dilution is 200 ohm–1cm2 eq–1. Hence % of ionization of CH3COOH is (a) 5% (b) 2% (c) 1% (d) 4% 104. Equivalent conductance of saturated BaSO4 is 400 ohm–1 cm–2 eq–1 and specific conductance is 8×10–5 ohm–1 cm–1. Hence Ksp of BaSO4 is (a) 4×10–8 M2 (b) 1×10–8 M2 –4 2 (c) 2×10 M (d) 1×10–4 M2

12.59

multiple Choice Questions with only one answer level II 1. The standard reduction potentials for two reactions are given below: AgCl( s ) + e -  → Ag ( s ) + Cl - ( aq )

E 0 = 0.22V O

Ag (+aq ) + e -  → Ag ( s ) E 0 = 0.80V The solubility product of AgCl under standard conditions of temperature (298 K) is given by (a) 1.6×10–5 (b) 1.5×10–8 –10 (c) 3.2×10 (d) 1.5×10–10 2. At 298 K the standard free energy of formation of H2O(l) is -237.20 KJ/mole while that of its ionization into H+ ion and hydroxyl ions is 80 KJ/mole. The emf of the following cell at 298 K will be (take F= 96500 C) OH - /H 2 (g,1bar)|H + (1M)||OH - (1M)|O 2 (g,1bar) (a) 0.40 V (b) 0.81 V (c) 1.23 V (d) -0.40 V 3. The oxide of a metal (R) can be reduced by the metal (P) and metal (R) can reduce the oxide of metal (Q). Then the decreasing order of the reactivity of metal (P), (Q) and (R) with oxygen is (a) P>Q>R (b) P>R>Q (c) R>P>Q (d) Q>P>R 4. There are two reactions O

 A(s) - E10 A + + e - 

O

Then log

 A(s) - E 02 A 2 + + 2e -  O

k2 is equal to (where k1 and k2 are equilibrium k1

constants of 1st and 2nd reactions)

(

0 0 (a) 2 E 2 - E1 0.059

O

O

0 0 (c) 2E 2 - E1 0.059 O

O

)

0 0 (b) E1 - E 2 0.059 O

O

0 0 (d) 2E1 - E 2 0.059 O

O

5. For the fuel cell reaction 2H2(g) + O2(g) → 2H2O(l); ∆H 298 (H2O,l) = –285 KJ what is ∆S 298 for the given fuel cell reaction? Given: O2(g) + 4H+(aq) + 4e– → 2H2O(l); E = 1.23 V (a) 0.635 KJ/K (b) –0.635 KJ/K (c) 3.51 KJ/K (d) -0.322 KJ/K 6. The dissociation constant of n-butyric acid is 1.6 × 10–5 and the molar conductivity at infinite dilution is 380×10–4 sm2 mol–1. The specific conductance of the 0.01 M acid solution is (a) 1.52 × 10–5 sm–1 (b) 1.52 × 10–2 sm–1 (c) 1.52 × 10–3 sm–1 (d) 1.52 × 10–8 sm–1 O

O

O

12.60

Electrochemistry

7. In a hydrogen electrode the pressure of H2 gas has been increased from 1 to 10 atm keeping the conc. of H+ as one molar. Hence the potential of the electrode at 298 K. (a) Remains zero (b) Decreases 59.1 mV (c) Decreases by 29.55 mV (d) Increased by 59.1 mV 8. We have taken a saturated solution of AgBr.Ksp of AgBr is 12 × 10–14. If 10–7 mole of AgNO3 are added to 1 litre of this solution, then the conductivity of this solution in terms of 10–7 Sm–1 units will be. [Given: λ 0 (Ag + ) = 4 × 10 -3 sm 2 mol -1 , λ 0 (Br - )

6 × 10 -3 sm 2 mol -1 , λ°(NO- ) = 5 × 10-3 sm 2 mol -1 3

9.

10.

11.

12.

13.

(a) 39 (b) 55 (c) 15 (d) 41 The solution copper sulphate, in which copper rod is dipped, is diluted 10 times. The reduction potential of copper. (a) Increases by about 30 mV. (b) Decreases by about 30 mV. (c) Increases by about 60 mV. (d) Decreases by about 60 mV. The standard redox potentials (reduction reaction) of Pt / Cr2 O72 - , Cr 3 ; Pt / MnO 4- , Mn 2 + ; Pt / Ce +4 , Ce3+ in the presence of acid are 1.33 V, 1.51 V and 1.61 V respectively at 25°C. The oxidizing power of these systems decrease in the order. (a) Cr2O72 - > MnO 4- > Ce 4 + (b) MnO 4- > Cr2O72 - > Ce 4 + (c) Ce 4 + > MnO 4- > Cr2O72 (d) MnO 4- > Ce 4 + > Cr2O72 For completely reducing 20 mL of 0.5 M MnO–4 to Mn2+(aq), the charge required is. (a) 4825 C (b) 48250 C (c) 48.25 C (d) 5 F Pure water is saturated with pure solid AgCl, a silver electrode is placed in the solution and the potential is measured against normal calomel electrode at 25°C. This experiment is then repeated with a saturated solution of AgI. If the difference in potential in the two cases is 0.177 V, what is the ratio of solubilities of AgCl and AgI at the temperature of experiment? (a) 103 (b) 106 4 (c) 10 (d) 102 Reduction potentials of some half-reactions are given as Fe2+ + 2e– → Fe; E = –0.47 volts Fe3+ + 3e– → Fe; E = –0.57 volts Fe2+ + e– → Fe2+; E = –0.77 volts O O O

2–

FeO4 + 3e– + 8H+ → Fe3+ + 4H2O; E = +2.20 volts Which of the following statements is correct? (a) Neither Fe3+ nor Fe2– have any tendency to reduce to Fe. 2– (b) FeO4 is the weakest oxidizing agent. (c) ΔG value for FeO42 -  → Fe3+ will be large and positive (d) None of these 14. Ksp of BaSO4 is 1 × 10–10. If the ionic conductance of Ba++and SO4- - ions are 64 and 80 ohm– cm2 mol–1 respectively then its specific conducatance is. (a) 1.44 × 10–8 ohm–1 cm–1 (b) 144 × 10–8 ohm–1 cm–1 (c) 1.44 × 108 ohm–1 (d) 144 × 108 ohm–1 cm–1 15. For the reaction: O

4 2 Al + O2  → Al2 O3 , ∆G = −827 ∆G = –827 KJ 3 3 Minimum emf required to carry out electrolysis of Al2O3 (F = 96500 C mol–1) (a) 8.56 V (b) 2.14 V (c) 4.28 V (d) 6.41 V 16. A gas X at 1 atm pressure is bubbled through a solution containing a mixture of MY– ions and 1 M Z– ions at 25°C. If reduction potentials of Z>Y; X>Z then (a) Y will oxidize Z and not X (b) Y will oxidize X and not Z (c) Y will oxidize X and Z (d) Y will reduce both X and Z 17. Standard emf of the galvanic cell constructed from zinc and chlorine electrodes 2.12 V, then the emf of the galvanic cell Zn/Zn2– (0.01 M)// Cl–(0.01M)/Cl2 Pt is (a) 2.12 V (b) 2.24 V (c) 2.30 V (d) 1.94 V 18. Given E = 0.08 V for the Fe3+ (cyt b)/Fe2+ (cyt b) couple, and E = 0.22 V for the Fe3+ (cyt c1)/ Fe2+ (cyt c1) couple, at pH = 7.0 at 25°C and cyt is an abbreviation for cytchromes. The equilibrium constant for the following equation is O

O

 Fe 2 + (cyt c1 ) + Fe3+ (cyt b) Fe 2 + (cyt c1 ) + Fe 2 + (cyt b)  (a) 264 (b) 254 (c) 244 (d) 234 19. The potential for the reaction O2(g) + 4H+ + 4e– → 2H2O is 1.23 V in 0.1 N strong acid solution. The potential of this couple in a solution of pH = 10 is (a) 0.699 (b) 0.599 (c) 0.463 (d) 0.363

Electrochemistry

20. At 25°C the free energy of formation of H2O(l) is -56700 cal/mol. While that of its ionisation to H+, OH– is 19050 cal/mol. H2(1 atm) H+//OH– / O2 (1 atm). The EMF of the above cell is (a) 0.70 (b) 0.50 (c) 0.40 (d) 0.30 21. E Sn/Sn2+ = 0.14 V, E Sn/Sn4+ = 0.15 V The E value of Sn 4 + + 2e  → Sn +2 is (a) 0.16 V (b) -0.16 V (c) 0.32 V (d) -0.32 V 22. Zn + 2 AgCl  → ZnCl2 + 2 Ag , ∆H of this reaction at 27°C is -222 KJ. If the EMF of the cell is 1.015 V. the temperature coefficient of this cell at 27°C is (a) 4.5 × 10-4 vol / deg (b) –-4.5 × 10-4 vol / deg (c) 6.5 × 10-4 vol / deg (d) – 6.5 × 10-4 vol / deg 23. In a zinc manganese dioxide cell, the anode is made up of Zn and cathode of carbon rod surrounded by a mixture of MnO2, carbon, NH4Cl,ZnCl2. The cathodic reaction is O

O

O

2MnO 2 + Zn 2 + + 2e  → ZnMn 2 O 4 8.7 g MnO2 present is in cathodic compantment. How many seconds the dry cell will continue to give 4 × 102 amp (a) 241 sec (b) 482 sec (c) 24.1 sec (d) 48.2 sec 24. At a certain temperature, the saturated solution of silver chloride has electrolytic conductivity 1.2 × 10–6 mho cm–1. The ionic conductance of Ag+ and Cl– ions at infinite dilution are 55, 65 mho cm2 eq–1. The solubility product of AgCl is (a) 10–5 (b) 10–8 –9 (c) 10 (d) 10–10 25. The equivalent conductance of 0.1N solution of MgCl2 is 100 mho cm2 eq–1 at 25°C. A cell with electrodes that are 1.5 cm2 in surface area and 0.5 cm a part is filled with 0.1 N MgCl2 solution. How much current will flow then the potential difference between the electrodes is 5 volt? (a) 0.075 amp (b) 0.10 amp (c) 0.15 amp (d) 0.20 amp 26. An oxide of metal (AW = 112) contains 12.5% oxygen by weight. 0.965 amp current was passed for a period of 1 minute through the molten liquid. The weight of metal deposit is (a) 0.0336 gm (b) 0.0504 gm (c) 0.0672 gm (d) 0.0840 gm

12.61

27. A 200 watt, 100 volt incandescent lamp is connected in series with a ZnCl2 electrolytic cell. The weight of zinc deposited on passing current for 965 sec. is (a) 0.3275 gm (b) 0.655 gm (c) 0.9825 gm (d) 1.31 gm 28. Electrolysis of dilute aqueous NaCl solution was carried our by passing 10 milli ampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is (1 Faraday = 96500 C mol–1) (a) 9.65 × 104 sec (b) 19.3 × 104 sec 4 (c) 28.95 × 10 sec (d) 38.6 × 104 sec 29. For the cell (at 1 bar H2 pressure) Pt/H2(g)HX(m1), NaX(m 2 ) / / NaCl(m3 ) / AgCl / Ag it is found that  mHX .mCl -   approaches the value of E - E 0 + RTF -1 ln   mx-  0.2814 in the limit of zero concentration. Calculate O

ln

1 for the acid HX at 27°C. (R=8.3 J Mole–1 K–1, Ka

F = 96500 C) (a) 10.9 (b) 19.3 (c) 4.825 (d) 96.5 30. EMF of the following cell is 0.67 V at 298 K Pt H 2 (1 atm) / H + ( pH = 6.5) / / KCl (1N ) / Hg 2 Cl2 / Hg calculate E Cl– / Hg2Cl2 / Hg (a) 0.28 V (b) 0.3 V (c) 0.26 V (d) 0.4 V 31. The calomel cell Hg, Hg 2 Cl2 / Cl - (aq ) values of electrode oxidation potentials are plotted at different log [Cl–] variation is represented by O

(a)

(b)

E

E

Log Cl–

Log Cl– (d)

(c)

E

E

Log Cl–

Log Cl–

32. The hydrogen electrode when placed in a buffer solution of CH3COONa and CH3COOH in the ratio x:y and y:x has oxidation electrode potential E1 and E2 volts respectively at 25°C (PH2 = 1 atm). pKa For CH3COOH will be (b) E1 - E2 (a) E1 + E2 (c)

E1 + E2 0.0591× 2

(d)

E1 - E2 0.0591× 2

12.62

Electrochemistry

33. The standard reduction potentials are given +3 +2  Fe ( H O )  + e -   Fe ( H O )  E 0 = 0.77V → 2 2   6 6  

OH

O

O

-3

-4

 Fe(CN )6  + e -  →  Fe(CN ) 4  E 0 = 0.33V O

S1 : Fe(CN )6-4 can easily oxidize in comparison with Fe( H 2 O)6+2

(

S 2 : Fe H 2 O

)

+3

6

+ Fe (CN )6

-4

(

 → Fe H 2 O

35.

36.

37.

38.

O

39.

+2

6

+ Fe (CN )6 is not feasible reaction -4 S3 : Fe–C Bond length in Fe ( CN )6 is shorter than -3 that in Fe ( CN )6 (a) F F T (b) T F F (c) T F T (d) F T F The cell Pt / H 2 (1 atm )  H + , pH = X  normal calomel electrode, has EMF of 0.67 V at 25°C. The standard oxidation potential of calomel electrode is –0.28V, then pH of solution will be (a) 6.6 (b) 3.3 (c) 13.2 (d) 1.1 Given the following molar conductivities at 25°C; HCl, 426Ω -1 cm 2 mol -1 ; NaCl ,126Ω -1 cm 2 mol -1 ; NaC (sodium crotonate), 83Ω -1 cm 2 mol -1 . The conductivity of 0.001 mol / dm3 acid solution is 3.83 × 10-5 Ω -1cm -1 . The dissociation constant of crotonic acid is (a) 4.11 × 10–5 M (b) 3.11 × 10–5 M –5 (c) 2.11 × 10 M (d) 1.11 × 10–5 M Following cell has EMF 0.7995 V. Pt / H 2 (1 atm) / HNO3 (1M ) / / AgNO3 (1M ) / Ag If we add enough KCl to the Ag cell so that the final Cl– is 1 M. Now the measured emf of the cell is 0.222 V. The Ksp of AgCl would be: (a) 1 × 10–9.8 (b) 1 × 10–19.6 –10 (c) 2 × 10 (d) 2.64 × 10–14 Calculate the equivalent conductivity ( Ω -1cm 2 eQ -1 ) at infinite dilution of the salt KOOC.COONa. Given the ionic conductivities at infinite dilution of C2 O4 -2 , K + and Na+ as 148.2, 50.1 and 73.5Ω -1 cm 2 mol -1 , respectively. (a) 135.9 (b) 271.8 (c) 182.3 (d) 212.7 The standard EMF of Mn+2/Mn electrode is. +1.18 V at 25°C. The E OH–|Mn(OH)2(s)|Mn(s) is (Given Ksp Mn(OH)2 = 2 × 10–13 and log 2 = 0.3) (a) 1.56 V (b) 1.18 V (c) 0.805 (d) None of these Standard free energies of formation (in KJ/mol) at 298 K are -237.2, -394.4 and -8.2 for H 2 O(l ) ' CO2( g ) and pentane (g) respecpectively. The value of E cell for the pentane-oxygen fuel cell is (a) 1.0968 V (b) 0.0968 V (c) 1.968 V (d) 2.0968 V -3

34.

)

O

+2H++2e, E =1.30 V O

40. For the half cell OH

O

At pH = 2, electrode potential is (a) 1.36 V (b) 1.30 V (c) 1.42 V (d) 1.20 V 41. At which of the following concentration of Cl–, the potential of Ag / AgCl / Cl– is maximum? (a) 1 N (b) 0.1 N (c) 2 N (d) 0.01 N 42. In which of the following case, the increase in concentration of ion cause increase in E value? (a) Pt(H2)/H+(aq) (b) Ag / AgCl / Cl–(aq) (c) Pt / Quinhydrone / H+(aq) (d) Ag / Ag+(aq) 43. Standard electrode potential (E ) for OCl - / Cl - and O

1 Cl - / Cl2 are respectively 0.94 V and -1.36 V. 2 The E OCl - / Cl2 will be (a) -0.42V (b) -2.20 V (c) 0.52 V (d) 1.04 V 44. Aqueous solution of Na2SO4 containing a small amount of phenolphthalein is electrolysed using Pt electrodes. The colour of solution after some time will (a) remains colourless. (b) changes from pink to colourless. (c) changes colourless to pink. (d) remains pink. 45. Consider the following standard reduction potentials, Half-reaction E ,V O

O

 Ni(s) Ni 2 + (aq) + 2e - 

-0.23

 Co(s) Co 2 + (aq) + 2e - 

-0.28

 Fr(s) Fe (aq) + 2e 

-0.41

 Cr(s) Cr 3+ (aq) + 3e - 

-0.74

 Mn(s) Mn (aq) + 2e 

-1.03

2+

2+

-

-

Which of the following metals could be used successfully to galvanize iron? (a) Ni only (b) Ni and Co (c) Mn only (d) Mn and Cr

Electrochemistry

46. Consider the following standard reduction potentials, Half-reaction V Half Reaction EE ,,V O

 Cu(s) Cu 2 + (aq) + 2e - 

0.34

MnO 4 (aq) + 2H 2 O(l) + 3e  MnO 2 (s) + 4OH - (aq) 

0.59

+

-

O 2 (g) + 2H 3 O (aq) + 2e  H 2 O 2 (aq) + 2H 2 O(l) 

0.68

And the following redox reactions, (i) H 2 O 2 (aq ) + 2H 2 O(1) + Cu 2 + (aq ) +  ↽ ⇀  O 2 (g ) + 2H 3 O (aq ) + Cu (s)

(ii) 2MnO −4 (aq ) + 4H 2 O(1) + 3Cu (s) 2+ −  ↽ ⇀  3Cu (aq ) + 2MnO 2 (s) + 8OH (aq )

(iii) 2MnO 4− (aq ) + 10H 2 O(1) + 3H 2 O 2 (aq ) +  ↽ ⇀  3O 2 (g ) + 6H 3 O (aq ) + 2MnO 2 (s)

+8OH − (aq ) Which of the redox reactions above will be spontaneous under conditions? (a) i only (b) ii only (c) i and iii (d) ii and iii 47. In the standard hydrogen electrode, a platinum wire is used as the electrode. Consider the following standard reduction potentials, Half-reaction E ,V

12.63

operation of this cell? (a) The anode compartment contains a 0.10 Fe2+ solution which decreases in concentration as the cell operates and the cathode compartment contains a 1.0 M Fe2+ solution which also decreases in concentration as the cell operates. (b) The anode compartment contains a 0.10 M Fe2+ solution which increases in concentration as the cell operates, and the cathode compartment contains a 1.0 M Fe2+ solution which decreases in concentration as the cell operates. (c) The anode compartment contains a 0.10 M Fe2+solution which decreases in concentration as the cell operates, and the cathode compartment contains a 1.0 M Fe2+ solution which increases in increases in concentration and the as the cell operates. (d) The anode compartment contains a 1.0 M Fe2+ solution which decrease in concentration as the cell operates and the cathode compartment contains a 0.10 M Fe2+solution which increase in concentration as the operates. 49. Molar conduction at infinite dilution of Al+3 and SO4–2 ions are 180 and 160 ohm–1 cm2 mol–1. The equivalent conductance of Al2(SO4)3 at infinite dilution is (a) 340 (b) 170 (c) 240 (d) 140 1 50. On plotting Λ c2 / Λ ∝ (Λ∝ - Λ c ) versus of chloro c

aceticacid a straight line with slope equal to 0.0014 is obtained. [ Λ c , Λ ∝ are molar conductivity of the acid at concentration. C and at infinite dilution]. Then the Zn 2 + (aq) + 2e -  → Zn(s) -0.76 value of its dissociation constant is 2H 3 O + (aq) + 2e -  → (a) 0.005 (b) 0.0014 (c) 10–5 (d) 0.0007 2H 2 O(l) + H 2 (g) 0.00 51. The ECell = 1.18 V for Zn (s)| Zn +2 | (1M) || Cu +2 (1M) | C Pt 2 + (aq) + 2e -  → Pt(s) 1.20+2 +2 (1M) | Cu(s) Which of the following statements best describes what The value of x if when excess granulated zinc is added would happen if the platinum wire in the standard to 1 M Cu2+ solution the [Cu2+]eq becomes 10–x M, is hydrogen electrode were replaced with a zinc wire? 2.303T   Note: assume that this is the only change made to the = 0.059   T = 298K, F   half-cell. (a) Zinc ion in the solution would be reduced; the (a) 40 (b) 30 mass of the zinc electrode would increase. (c) 20 (d) 10 (b) The pH of the solution would decrease. 52. True or false. (c) The zinc electrode would be oxidized; the mass of (a) KMnO4 liberates O2. the zinc electrode would decrease. Given E 0 MnO- ,Mn +2 ,H+ / Pt = 1.51V E 0 H+ |O |Pt = 1.223V 2 4 (d) This change would have no effect-the standard hydrogen electrode would continue to function (b) O2 Will reduce [Ag (CN)2]– in presence of CN– in properly. alkaline solution. 48. A concentration cell is made up of the Fe/Fe2+ couple Given E 0[ Ag(CN) ]- ,CN- |Ag = -0.7V, E 0Pt|O |OH- = 0.4V 2+ 2 where the Fe concentrations are 1.0 M and 0.10 M. 2 which of the following statements best describes the O

O

O

O

O

O

12.64

Electrochemistry

59. The electrode MnO4- (1M ) / Mn +2 (1 M) is present in a solution of H+ concentration of 1 M. The concentration decreases to 10–4 M at 25°C. The cell potential Ag - AgCl E 0red = 0.222 (a) Decreased by 0.38 V (b) Decreased by 0.19 V Ag - AgBr = 0.03 (c) Increased by 0.38 V Ag - AgI = -0.151 (d) Increased by 0.19 V Ag - Ag 2S = -0.69 60. EBi+3/Bi = 0.226 V, E Cu+2/Cu = 0.344 V. A mixture conThen the Ksp values must follow: tains 1 M Bi+3, 1 M Cu+2 is electrolyzed at 25°C. At what concentration of Cu+2, bismuth starts deposit K sp (AgCl) > K sp (AgBr) > K sp (AgI) > K sp (Ag 2S) (b) 10-2 M (a) 10-1 M -3 (a) TFT (b) TTF (c) 10 M (d) 10-4 M (c) TFF (d) TTT 61. Consider the cell Ag | AgBr | Br– || Cl– | AgCl | Ag The equivalent conductance of CH 3 COONa, HCI and Na at 25°C the solubility product of AgCl and AgBr are 1× 10-10 and 5 × 10-13 respectively. At which ratio of NaCl at infinite dilution are 90, 425, 125 mho cm2 eq–1 respectively. The degree of dissociation of 0.1 M aceconcentration of Br– and Cl– ions would the emf of tic is 0.001. The equivalent conductance of acetic acid cell be zero? at this concentration is in mho cm2 eq–1 (a) 100: 1 (b) 1:100 (a) 0.039 (b) 0.0039 (c) 200:1 (d) 1:200 (c) 0.39 (d) 3.9 62. After electrolysis of sodium solution with inert elecThe dissociation constant for Ag ( NH 3 ) 2+ into Ag+ trodes for a certain period of time, 700 mL of 1 N and NH3 is 6 × 10–14. E0Ag+ | Ag = 0.8 volt. The value NaOH left. During the same time 31.80 gm Cu was ° of E Ag (log 6 = 0.7782) deposited in copper voltmer. The percentage of NaOH + ( NH 3 )2 | Ag obtained is (a) −0.019 V (b) 0.019 V (a) 50% (b) 60% (c) -0.039V (d) 0.039V (c) 70% (d) 80% 63. During an electrolysis of conc. H2SO4 perdisulphuric +0.15v + +0.50v +2 Cu Cu Cu acid (H2S2O8) and O2 form in equimolar amount. The The value of amount of H2 that will form simultaneously will be E = X volt ( 2H 2 SO4 → H 2 S2O8 + 2H + + 2e- ) (a) Thrice that of O2 in moles. ‘X’ is (b) Twice that of O2 in moles. (a) -0.65v (b) 0.65V (c) Equal to that O2 in moles. (c) -0.325V (d) 0.325V (d) Half of that of O2 in moles. A cell Ag | Ag + || Cu +2 | Cu initially contains 1 M 64. A constant current was flown for 965 sec. through a Cu+2, 1 M Ag+ 9.65 amp current is passed for 4000 sec. KI solution oxidizing iodide ion to iodine. At the end into the cell. The change in cell potential is (log 2.45 = of experiment 20 mL of 0.1 Na2S2O3 solutions re0.3892) quired for reduction change. The average rate of cur(a) Increased by 0.0229 rent flown is (b) Decreased by 0.0229 (a) 0.4 amp (b) 0.3 amp (c) Increased by 0.0114 (c) 0.2 amp (d) 0.1 amp (d) Decreased by 0.0119 65. The resistance of solution A is 50 ohm that of B is 100 E Value of half cell NO3- + 2 H + + e -  ohm. Both are taken in same cell. If equal volume of → NO2 + H2O is 0.8. The reduction potential half cell in neutral A and B are mixed and the mixture is taken in same solution is cell then the resistance is (a) -0.026V (b) 0.026V (a) 150 ohm (b) 75 ohm (c) -0.387 V (d) 0.387V (c) 66.67 ohm (d) 33.3 ohm The EMF of the cell Zn | ZnSO4 || CuSO4 | Cu at 25°C 66. The density of copper is 6.35 gm/mL. The number of couis 0.03 V and temperature coefficient of emf is –1.2 × lomb needed to plate an area of 10 × 10 cm2 to a thickness –4 10 V/degree the heat of reaction is of 10–2 cm using CuSO4 solution as electrolytes is (a) -12.7 KJ/mol (b) 12.7 KJ/mol (a) 19300 coulomb (b) 14475 coulomb (c) -13.8 KJ/mol (d) 13.8 KJ/mol (c) 12062 coulomb (d) 9650 coulomb (c) The standard potentials of some metal – insoluble salt-anion electrode are as given below O

O

53.

54.

55.

O

56.

57.

58.

O

O

Electrochemistry

67. During the discharge of lead storage battery the density of H2SO4 fall 1.3 gm/mL to 1.1 gm/mL. Sulphuric acid of density 1.3 gm/mL is 40% by weight and that of density 1.1 gm/mL is 20% by weight. The battery contains 2.0 litres of acid. The volume remains unchanged during discharge. How many faraday of current is liberated? (a) 6.12 (b) 4.92 (c) 5.34 (d) 3.06 68. Calculate the quantity of electricity that would be required to reduce 12.3 gm of nitrobenzene to aniline if current efficiency is 50% (a) 347400 coulomb (b) 115800 coulomb (c) 23160 coulomb (d) 694800 coulomb 69. Perdisulphuric acid can be prepared by the electrolytic → H 2 S 2 O8 + 2 H + oxidation of H 2 SO4 as 2 H 2 SO4  – +2e . Oxygen and hydrogen are byproducts. 8.96 litres of H2 and 2.24 litres of O2 were generated at STP. The weight of H2S2O8 formed is (a) 19.8 g (b) 79.2 g (c) 59.4 g (d) 39.6 g 70. Two weak acids HA1 and HA2 having same concentrations with pKa values 3 and 5 are placed in contact with hydrogen electrode (1 atm, 25°C). They are inter connected through salt bridge. The emf of the cell is (a) 0.059 V (b) –0.059 V (c) 0.11 V (d) –0.11 V 71. Pt Cl2 / HCl / Cl2 , Pt cell reaction will be spontaneous

12.65

KJ for (H + + CI - )(aq) mol (b) 654 V (d) None of these

-130.79 (a) 456 V (c) 564 V

75. Given Ag X (s ) + e -  → Ag ( s ) + X - (aq ) E 0 E = 0.062V Ag + (aq ) + e -  → Ag ( s ) E 0 = 0.80V O

Hence the solubility product of AgX would be : [Given: Anti log 0.7 = 5.0 and (a) 4×10–12 (c) 4×10–17

2.303 × RT = 0.06] F (b) 5×10–13 (d) 4×10–14

76. Given: Zn 2 + (aq ) + 4CN - (aq )  →[ Zn(CN ) 4 ]2 - K 20 Kf = 1 × 10 Zn 2 + (aq ) + 2e -  → Zn( s ) E 0 = -0.76V Find the E value of the reaction: [ Zn(CN ) 4 ]2 - + 2e -  → Zn( s ) + 4CN - (aq ) [Given: 2.303 × 8.3 × 298 = 5700] (a) -1.46V (b) 2.92V (c) -1.35V (d) +1.46V 77. Calculate the EMF of the cell at 298 K. O

O

Pt | H 2 (1atm) | NaOH(xM) // NaCl (xM) | AgCI(s) | Ag If E 0Cl- |AgCl|Ag = +0.222V O

P1 0.1 M P2 if (a) P1 = P2 (b) P1 < P2 (c) P1 > P2 (d) P1 = P2 = 1 atm 72. The galvanic cell is Ag/AgBr, KBr //KCl/ AgCl/ Ag 0.001 M

0.2 M

The solubility product of AgCl, AgBr are 2 × 10–10, 1.0×10–13. The EMF of the cell is (a) -0.0295 V (b) 0.0295 V (c) -0.059 V (d) 0.059 V 73. The standard reduction potential at 25°C for the reaction 2H 2 O + 2e −  → H 2 + 2OH − is − 0.8277 V calculate ionic product of water from this data is (a) 1.0×10–14 (b) 9.35×10–15 –14 (c) 1.1×10 (d) 9.9×10–15 74. Calculate the cell EMF in mV for Pt | H 2 (1 atm) | HCl(0.01 M) | AgCl(s) | Ag(s) at 298 K If ∆G 0f values are at 25°C O

-109.56

KJ for AgCl (s) and mol

(a) (b) (c) (d)

1.048V -0.04V -0.604V emf depends on ‘x’ and cannot be determined unless value of ‘x’ is given 78. Electrolysis of a solution HSO4- ions produces S 2 O82 . Assuming 75% current efficiency what current should be employed to achieve a production rate of 1 mole of S 2 O82 per hour? (a) +71.5 amp (b) 35.7 amp (c) 142.96 amp (d) 285.93 amp 79. An electrochemical cell is prepared by coupling half cell X (0.1gm AgCl in 10 l H2O) to half cell Y (10 gm AgBr in 0.1 l H2O) at 298 K

K sp ( AgCl) = 10 -10 K sp ( AgBr ) = 10 -14 In the external circuit between the silver electrodes (a) No current will flow. (b) Current will flow from X to Y. (c) Current will flow Y to X. (d) Current flow cannot be determined.

12.66

Electrochemistry

80. In a Cu voltameter, mass deposited in 30 seconds is 100 grams. Carefully analyse the current – time graph shown below and identify the incorrect statement.

(d) During electrolysis cations and anions move towards anode and cathode respectively. 5. Zinc is used to protect iron from corrosion because (a) E of zinc is less than that of iron (b) E of zinc is more than that of iron (c) Both zinc and iron have same E value (d) Zinc under goes oxidation in preference to iron and hence prevents rusting of iron. 6. During the purification of copper by electrolysis: (a) The anode used is made of cooper ore. (b) Pure copper is deposited on the cathode. (c) The impurities such as Ag, Au are present in solutions as ions. (d) Concentrations of CuSO4, solution remains constant during dissolution of Cu. 7. For the cell at 298 K O O

O

100 mA

10s

20s

30s

(a) Electrochemical equivalent for Cu is 50 (b) a constant current of 66.66 mA would also discharge the same amount in the same time (c) 33.33 grams got discharge in 10 seconds (d) 50 grams got discharge in 15 seconds

multiple Choice Questions with one or more than one answer 1. Which one is/are correct among the following given, 0 0 E Cu = 0.521, E Cu = 0.153 +1 2+ / Cu / Cu + O

O

(a) Cu+ disproportionate (b) Cu+2 disproportionate (c) E 0 + +2 + E 0 +1 is positive O

O

Cu /Cu

Cu /Cu

(d) All of above 2. Two half cells have potential – 0.76 V and -0.13V respectively. A galvanic cell is made from these two halfcells. Which of the following statements is correct? (a) Electrode of half-cell potential -0.76V serves as cathode. (b) Electrode of half-cell potential -0.76V serves as anode. (c) Electrode of half-cell potential -0.13V serves as cathode. (d) Electrode of half-cell potential -0.76V serves as positive electrode and 0.13V as negative electrode. 3. In which case Ecell– E cell is zero? (a) Cu | Cu 2 + ( 0.01M ) || Ag + ( 0.1M ) | Ag (b) Pt ( H 2 ) | pH = 1|| Zn 2 + ( 0.01M ) | Zn (c) Pt ( H 2 ) | pH = 1|| Zn 2 + (1M ) | Zn (d) Pt ( H 2 ) | H + ( 0.01M ) || Zn +2 ( 0.01M ) | Zn 4. Which one of the following is/are correct? (a) Ions from the electrolyte in the salt bridge flow into each half cell to maintain electrical neutrality. (b) In saturated solution of KCl, ionic mobility of cation is equal to ionic mobility of anion. (c) ΔG is positive for chemical process during electrolysis. O

Ag(s) AgCl(s) KCl aq AgNO3 Ag(s) Which of the following wrong? (a) The EMF of the cell is zero when [Ag+]anodic = [Ag+]cathodic (b) The amount of AgCl(s) precipitate in anodic compartment will decrease with the working of the cell (c) The concentration of [Ag+] is constant in anodic compartment with the working of cell (d) E cell = E 0 Ag + O

-

– E 0Cl- /AgCl/Ag O

/Ag

0.059 1 log anodic 1 [Cl - ]

8. When a galvanic cell starts, working, with the passage of time (a) Spontaneity of the cell reaction decreases, Ecell decreases. (b) Reaction quotient Q decreases. Ecell increases. (c) Reaction quotient Q increases Ecell increases. (d) At equilibrium Q = Kc and Ecell = 0. 9. Some statements are given below. The correct statements are (a) The electrolysis of aqueous NaCl produces hydrogen gas at the cathode and chlorine gas at the anode. (b) The electrolysis of a dilute solution of sodium fluoride produces oxygen gas at the anode and hydrogen gas at the cathode. (c) The electrolysis of concentrated sulphuric acid produces SO2 gas at the anode and O2 gas at the cathode. (d) After the electrolysis of aqueous sodium sulphate, the solution becomes acidic.

Electrochemistry O

10. The standard redox potentials E of the following systems are: i.

MnO 4-

ii. Sn

Pb - Hg (1.0 M ) Pb 2 + (aq )(1M ) Pb 2 + (aq )(1M ) Pb - Hg (0.5M )

E (Volts) +

-

+ 8H + 5e  →

Mn 2 + + 4H 2 O 2+

14. During the working of the cell

0O

System

 → Sn

4+

+ 2e

-0.15

→ iii. Cr2 O72 - + 14H + + 6e - 

15.

2Cr 3+ + 7H 2 O

1.33

→ Ce4 + + e iv. Ce3+ 

-1.61

The oxidizing power of the various species are related as (a) Cr2 O72 - > Ce 4 + (b) Ce 4 + > Sn 4 + (c) Ce 4 + > MnO 4- (d) MnO 4- > Sn 4 + 11. Consider the cell: Ag(s) | AgCl (standard solution) || AgNO3 (aq) (1.0M)|Ag(s) EMF of the above cell is given by (Ksp of AgCl = 1.0 × 10–10 M2) 1 (a) E cell = E θAs+ / Ag + 0.0592 log[K sp (AgCl)] 2 O

(b) Ecell = 0.0592 log

O

2

Br2 (aq) + 2e

-

-

I 2 (s) + 2e 28

-

16.

17.

1 [ K sp ( AgCl )]1/ 2

(c) Ecell = 0.0592 × 5V (d) Ecell = 296mV 12. The emf of the following cell is 0.22 V. Ag(s)|AgCl(s)|KCl(1M)| H+ (1M) | H2(g)(1 atm); Pt(s). Which of the following will decrease the EMF of cell? (a) Increasing pressure of H2(g) from 1 atm to 2 atm. (b) Increasing Cl– concentration in anodic compartment. (c) Increasing H+ concentration in cathodic compartment. (d) Decreasing KCl concentration in anodic compartment. 13. Pick out the correct statements among the following from inspection of standard reduction potentials (Assume standard state conditions) Cl2 (aq) + 2e -  → 2Cl - (aq) E 0Cl / Cl= +1.36 V  → 2Br (aq) E

0O Br2 / Br -

 → 2I - (aq)

O

24

E 0I / I-

S2 O (aq) + 2e  → 2SO (aq) E

2

0O S2 O82- /SO24-

= +1.09 V = +0.54 V = +2.00 V

(a) Cl2 can oxidize SO42 - from solution. (b) Cl2 can oxidize Br– and I– from aqueous solution. (c) S2O82– can oxidize Cl–, Br– and I– from aqueous solution. (d) S2O82– is added slowly, Br– can be reduced in presence of Cl–.

[Pb2+] in right half cell decreases. [Pb2+] in left half cell decreases. [Pb2+] does not change in either of the half cell. Morality of lead amalgam in right half cell increases while that of left half cell decrease. Which of the following cells give the cell potentials to be their standard values? (a) Zn(s)|Zn2+(aq)(0.01M)||H3O+(aq)(0.1M)| |H2g(1 atm), Pt| (b) Cu(s)|Cu2+(aq)(0.25M)||Ag+(aq)(0.5M)|Ag (c) Cd(s)|Cd2+(aq)(0.01M)||pH = 1|H2(g)(1 atm), Pt (d) Zn(s)|Zn2+(aq)(0.1M)||pH = 1|H2(g)(1 atm), Pt Which of the following statement (s) is /are correct? (a) F2 is the strongest oxidizing agent. (b) Li is the strongest reducing agent. (c) Li+ is the weakest oxidizing agent. (d) F2 is highest reduction potential. For the redox reaction A+ +B→B+ + A (a) The reaction would be spontaneous if oxidation potential of B is greater than oxidation potential of A. (b) The reaction will be spontaneous if E A0+ / A - EB0+ / B is + ve . (c) The reaction will be spontaneous E A0+ / A - EB+ / B is –-ve. (d) The reaction will be spontaneous if A is a gas. The values of E of some reactions are given (a) (b) (c) (d)

1.51 -

12.67

18.

O

O

O

O

O

→ 2I I 2 + 2e - 

E 0 = 0.54 volts

→ Sn 2 + Sn +4 + 2e - 

E 0 = 0.152 volts

→ 2Cl Cl2 + 2e - 

E 0 = 1.36 volts

→ Fe 2 + Fe3+ + e - 

E 0 = 0.76 volts

→ Ce +3 Ce +4 + e - 

E 0 = 1.6 volts

O

O

O

O

O

Hence (a) Fe3+ oxidizes Ce+3 (b) Ce4+ can oxidize Fe2+ (c) Sn2+ will reduce Fe3+ to Fe2+ (d) Cl2 will be liberated from KCl by passing I2. 19. Identify the correct statements. (a) Λm increases with increase in temperature (b) Λm decreases with increase in concentration (c) Specific conductance increases with increase in concentration (d) Specific conductance decrease with increasing temperature

12.68

Electrochemistry

20. The function of salt bridge in a galvanic cell is (a) To keep the electrolytic solution neutral. (b) To allow the ions into solution. (c) To complete the circuit by allowing electrons to flow through it. (d) To increase a potential difference.

2. Concentration of H+ after electrolysis is (a) 0.094 M (b) 0.012 M (c) 0.1 M (d) 0.06 M 3. Concentration of SO4–2 after electrolysis is (a) 0.094 M (b) 0.012 M (c) 0.1 M (d) 0.06 M

Comprehensive type Questions

Passage III

Passage I

The following diagram shows an electrochemical cell in which the respective half cells contain aqueous solution of the salts XCl2 and YCl3. Given that

O

The standard cell potential (E cell) of a reaction is related as ΔG = nFE cell, where –ΔG equals maximum electrical work. n = number of moles of electrons exchanged as per the balanced chemical equation. O

O

O

ΔG r = ΔH r – TΔS O

O

O

3X(s) + 2Y3+(aq) → 3X2+(aq) + 2Y(s) Based on this, answer the given questions

(1)

r

Volt meter V

And also O

O

(2) Aqueous XCl2 0O

δE 1. The temperature coefficient of the emf of cell is δT given by.

a.

b.

∆S0r nF

d.

- nEF0

O

nF

∆S0r ∆S0 c. nFT O

Metal Y, T = 298 K

Metal X

∂( ∆G 0r ) = -∆S0r ∂T

O

O

2. At 300 K, for the reaction. Zn(s)+2AgCl(aq)→ZnCl2(aq)+2Ag(s) is –218 KJ  δE 0  while the E of the was 1.015 V   of the  δT  P cell is. O

O

(a) –4.2×10–4 VK– (b) –3.81×10–4 VK–1 (c) 0.11 VK–1 (d) +7.62×10–4 VK–1 3. Calculate ΔS for the given cell reaction in above problem. (a) –73.53 JK–1 (b) 83.53 JK–1 –1 (c) 100 JK (d) None of these. O

Salt Bridge

Aqueous YCl3 (0.1 M)

(0.01 M)

1. Which of the following statements is correct? (i) The electrode made from metal X has positive polarity. (ii) Electrode Y is the anode. (iii) The flow of electrons is form Y to X. (iv) The reactions at electrode X is an oxidation. (v) The salt bridge would most likely contain silver nitrate (a) Only I (b) Only IV (c) Only V (d) All except V 0 2. If E 0 X / X 2+ = 0.20V , EY03+ /Y = 0.40V then Ecell of the above designed cell is (a) 0.60 V (b) –0.60 V (c) 0.20 V (d) –0.20 V 3. Ecell of the above designed cell is: 0.059 0.059 (a) 0.6+ V (b) 0.6+ ×2V 6 6 O

(c) –0.2+

O

O

0.059 0.059 ×V (d) –0.2+ ×2V 6 6

Passage II

Passage IV

50 mL of 0.1 M CuSO4 solution is electrolysed by using Pt. electrodes with a current of 0.965 ampere for a period of 1 minute. Assume volume of solution does not change during electrolysis. 1. Concentration of Cu2+ after electrolysis is (a) 0.094 M (b) 0.012 M (c) 0.1 M (d) 0.06 M

The potential of an electrode when each species involved in it exists in the standard state is called its standard potential. The standard reduction potential of a couple is the measure of its tendency to get reduced. A series obtained by arranging the various couples in order of their decreasing standard potential is called electrochemical series. Any of the two couples of this series joined together gives an

Electrochemistry

electrochemical cell in which reduction occurs at the electrode which occupies the higher position. The standard potential of hydrogen electrode is taken on to be zero by convention. Given below are the sequence of half-reactions (acidic media) with relevant E values in volt at 298 K.

12.69

2. Standard emf experienced by the person with dental filling is (a) +2.89 V (b) –2.89V (c) –0.93 V (d) +0.43 V 3. The standard reduction potential of the reaction

O

MnO4-  → MnO42 - → MnO2 → 0.560V

2.26V

0.95V

-1.18V

Mn 2 +  → Mn Also given some more E298 data are: O

-0.44V +0.036V Fe 2 +  → Fe( S )  → Fe3+

1. The E for MnO4- → Mn 2 + is equal to (a) 1.054 V (b) 1.506 V (c) 5.27 V (d) 7.53 V O

2+ + → Mn 2 + + 5 Fe3+ + 4 H 2 O, 2. MnO 4 + 5 Fe + 8H  is equal to (a) 0.734 V (b) 0.282 V (c) 4.498 V (d) Can’t be calculated as concentrations of species are not given 3. For the cell Fe(s) |Fe2+ (0.1M) ||100 mL 0.3 M HA (pKa = 5.2) Mixed with 50 mL 0.4M NaOH| H2 (1 atm), Pt

RT The Ecell at 298 [2.303 = 0.059V and log10 20 = F 1.3] is equal to (a) –0.15 V (b) +0.185 V (c) –0.44 V (d) +0.145 V

H 2O + e- →

1 H 2 + OH - at 298 K is 2

F ln K w RT RT ln[PH 2 ]1/2 [OH - ] (b) E 0 = F (a) E 0 = O

O

1/2

RT  PH 2  (c) E = ln F H+    0O

(d) E 0 = O

RT ln K w F

Passage VI Corrosion of metals is associated with electrochemical reactions. This also applies for the formation of rust on iron surface, where the initial electrode reactions usually are: (i) Fe(s) → Fe2+(aq) + 2e– (ii) O2(g) + 2H2O(l) + 4e–→ 4OH− (aq) An electrochemical cell in which these electrode reactions take place is constructed. The temperature is 25°C. The cell is represented by the following cell diagram Fe(s)|Fe2+ (aq) || OH–(aq), O2 (g) | Pt(s) Standard electrode potentials (at 25°C):

Passage V

Fe 2 + (aq ) + 2e −  → Fe(s)E θ = −0.44V O

Accidentally chewing on a stray fragment of aluminum foil can cause a sharp tooth pain if the aluminium comes in contact with an amalgam filling. The filling, an alloy of silver, tin and mercury, acts as the cathode of a tiny galvanic cell, the aluminium behaves as the anode, and saliva serves as the electrolyte. When the aluminium and the filling come in contact, an electric current passage from the aluminium to the filling which is sensed by a nerve in the tooth. Aluminium is oxidized at the anode, and O2 gas is reduced to water at the cathode. 0 E Al = -1.66V , EO0 3+ / Al O

O

2, H

+

/ H 2O

= 1.23V

1. Net reaction taking place when amalgam is in contact with aluminium foil (a) 4Al + 3O2 + 6H2O → 4Al(OH)3 (b) 4Al + 3O2 + 12H+ → 4Al3+ + 6H2O (c) 4Al + 3O2 → 4Al2O3 (d) 2H2 + O2 → 2H2O

O 2 (g ) + 2H 2 O (l) + 4e −  → 4OH − (aq )E θ = 0.40V O

Nernst factor: RT ln 10/F = 0.05916 Volt (at 25°C) Faraday constant: F = 96485 C mol–1 1. The overall reaction (indicated above) is allowed to proceed for 24 h under standard conditions and at constant current of 0.12 A. What mass of Fe converted to Fe2+ after 24 h? (a) 1.0 g (b) 1.5 g (c) 2.0 g (d) 3.0 g 2. The equilibrium constant at 25°C for the overall cell reaction is 2Fe + O 2 + 2H 2 O  → 2Fe+2 + 4OH (a) (b) (c) (d)

1.75 × 1056 (M6 bar–1) 6.2 × 1056 (M6 bar–1) 3.25 × 1055 (M6 bar–1) 3.25 × 1056 (M6 bar–1)

12.70

Electrochemistry

3. For the following conditions: [Fe2+] 0.015 M, pH (right hand half-cell) = 9.0 P(O2) = 0.700 bar The Ecell at 25°C is (a) 1.90 V (b) 1.19 V (c) 2.10 V (d) 2.70 V

2. Calculate the equivalent conductivity (Λ) for HCl (a) 450 s cm2 mol–1 (b) 398 s cm2 mol–1 2 –1 (c) 550 s cm mol (d) 112 s cm2 mol–1 3. Calculate the equivalent conductivity (Λ) for H+ and OH– together (a) 450 s cm2 mol–1 (b) 307 s cm2 mol–1 2 –1 (c) 507 s cm mol (d) 112 s cm2 mol–1

Passage VII Redox reactions play a vital role in chemistry and biology. The values of standard redox potential (E ) of two half – cell reactions decide which way the reaction is expected to proceed. A simple example is a Daniel cell in which zinc goes into solution and copper gets deposited. Given below are a set of half-cell reactions (acidic medium) along with their E (V) with respect to normal hydrogen electrode values. Using this data I 2 + 2e -  → 2I E 0 = 0.54;

Passage Ix

O

O

KCl(aq) Zn

Ni

O

Cl2 + 2e -  → 2Cl -

E 0 = 1.36;

Mn3+ + e -  → Mn 2 +

E 0 = 1.50;

Fe3+ + e -  → Fe 2 +

E 0 = 0.77;

O

O

Zn2+, Cl−

Ni2+, Cl−

ANODE

CATHODE

O

O2 + 4 H + + 4e -  → 2 H 2 O E 0 = 1.23; O

1. Among the following, identify the correct statement: (a) Chloride ion is oxidized by O2 (b) Fe2+ is oxidized by iodine (c) Iodide ion is oxidized by chlorine (d) Mn2+ is oxidized by chlorine 2. While Fe3+ is stable, Mn3+ is not stable in acid solution because (a) O2 oxidises Mn2+and Mn3+ (b) O2 oxidises both Mn2+ and Fe2+ to Fe3+ (c) Fe3+ oxidizes H2O to O2 (d) Mn3+ oxidizes H2O to O2 3. Sodium fusion extract, obtained from aniline, on treatment with iron (II) sulphate and H2SO4 in presence of air gives a prussian blue precipitate. The blue colour is due to the formation of (a) Fe4[Fe(CN)6]3 (b) Fe3[Fe(CN)6]2 (c) Fe4[Fe(CN)6]2 (d) Fe3[Fe(CN)6]3 Passage VIII The specific conductivity (κ) of a 0.1 M NaOH solution is 0.0221 s cm–1. On addition of an equal volume (V) of 0.1 M HCl solution, the value of ‘κ’ falls to 0.0056 s cm–1. On further addition of the same volume ‘V’ of 0.1 M HCl, the value of ‘κ’ rises to 0.0170 s cm–1. 1. Calculate the equivalent conductivity (Λ) for NaOH. (a) 112 s cm2 mol–1 (b) 221 s cm2 mol–1 2 –1 (c) 432 s cm mol (d) 295 s cm2 mol–1

1. The spontaneous reaction that takes place in this cell is (a) Zn + Ni → Zn 2 + + Ni 2 + (b) Zn + Ni 2 + → Zn 2 + + Ni (c) Ni + Zn 2 + → Ni 2 + + Zn (d) Ni 2 + + 2 K → 2 K + + Ni 2. When the cell is operating (a) e– flow through the external circuit to the Zn electrode while K+ diffuses into the Zn/Zn2+ half cell and Cl– diffuses into the Ni/Ni2+ half cell. (b) e– flow through the external circuit to the Zn electrode while K+ diffuses into the Ni/Ni2+ half cell and Cl– diffuses into the Zn/Zn2+ half cell. (c) e– flow through the external circuit to the Ni electrode while K+ diffuses into the Zn/Zn2+ half cell and Cl– diffuses into the Ni/Ni2+ half cell. (d) e– flow through the external circuit to the Ni electrode while K+ diffuses into the Ni/Ni2+ half cell and Cl– diffuses into the Zn/Zn2+ half cell. 3. Pt electrodes are often used in galvanic cells. The above cell could continue to operate if: (a) The Zn electrode were replaced by Pt; however, the Ni electrode is needed. (b) The Ni electrode were replaced by Pt; however, the Zn electrode is needed. (c) Either electrode is replaced by Pt and the other electrode remains as before. (d) The cell cannot operate if there is a change in the electrodes

Electrochemistry

12.71

Passage x

Passage xIII

The standard potential of the following cell is 0.23 V at 15°C and 0.21 V at 35°C

The standard reduction potential of the Ag+/Ag electrode at 298 K is 0.8 V. The solubility product of AgI is 10–18. Saturated solution of AgI is taken in the cell 1. The electrode potential of the half cell is (a) 0.269 V (b) –0.269 V (c) –0.534 V (d) 0.534 V 2. The standard electrode potential of I–/AgI/Ag is (a) 0.262 V (b) –1.862 V (c) 1.862 V (d) –0.262 V 3. Instead saturated solution, AgI dissolved in 0.1 MKI solution then electrode potential is (a) 1.803 V (b) –0.203 V (c) –1.803 V (d) 0.203 V 4. The above half cell is coupled with standard hydrogen electrode. Then the reaction take place at anode is (a) 2H+ + 2e → H2 (b) Ag+ + e → Ag (c) H2 → 2H+ + 2e (d) Ag → Ag+ + e

Pt H 2( g ) HCl( aq ) AgCl(s ) Ag (s ) 1. The cell reaction is (a) H 2 + AgCl  → Ag + + Cl (b)

1 H 2 + AgCl  → H + + Ag + Cl 2

(c) H 2 + 2AgCl  → Ag + + 2HCl (d) H 2 + 2Cl -  → 2H + + Cl2 2. In the above question, the value of ΔS is (a) –80.50 J (b) –90.50 J (c) –96.50 J (d) –100 J 3. The value of ΔH in the above question is (a) 49.987 KJ (b) 50.987 KJ (c) 48.987 KJ (d) 40.987 KJ O

O

Passage xIV The galvanic cell is

Passage xI The EMF of the cell Zn|Zn+2|(0.1M)||HCl|(1 lit) H2 (1 atm), Pt is 0.701 V at 25°. O E0 Zn / Zn +2

= 0.760V .

(

0.001 = 0.0316

)

1. The weight of NaOH required to neutralize H+ presents in RHS of the cell is (a) 12.64 (b) 1.264 gm (c) 0.1264 (d) 0.01264 gm 2. The EMF of the cell after neutralization of H+ in the cell is (a) 0.3765 V (b) –0.3765 V (c) 0.1882 V (d) –0.1882 V Passage xII The cell is Mg Mg +2 Ag + Ag E 0 O

Ag + /Ag

Ag E 0 O

Mg/Mg 2 +

= 0.80 V

= 2.37 V

1. The value of log Kc is (a) 107.45 (b) 80.5 (c) 67.15 (d) 53.725 2. The maximum work obtained by the cell is (a) 2 × 102 KJ (b) 3 × 102 KJ 2 (c) 5 × 10 KJ (d) 6 × 102 KJ

Zn / Zn ( NO3 )2 ( aq ) || Cu ( NO3 )2 ( aq ) | Cu 100 ml

1M

100 ml

1M

The cell is charged by 0.48 amp for 10 hours then the cell allowed to function as a galvanic cell at 25°C

E0

O

Cu +2 /Cu

= 0.34V

E0 O

Zn +2 / Zn

= -0.76V

1. Which reaction takes place at anode during cell is charged (a) Zn → Zn 2 + (b) Zn +2 + 2e - → Zn +2 (c) Cu + 2e → Cu (d) Cu → Cu 2 + + 2e 2. Before charging the free energy change of the cell is (a) 212.3 KJ (b) –212.3 KJ (c) 106.15 KJ (d) –106.15 KJ 3. After charging the EMF of the cell is (log 1.9 = 0.2788) (a) 1.137 V (b) 1.062 V (c) 1.175 V (d) 1.024 V 4. The weight of zinc deposited by 0.48 amp in 10 hours is (a) 11.79 gm (b) 5.895 gm (c) 2.9475 gm (d) 17.68 gm 5. Before charging the temperature coefficient of the cell is –4.311 × 10–4 Volt/degree. The heat of reaction is (a) 118 KJ (b) –237 KJ (c) –118 KJ (d) 237 KJ

12.72

Electrochemistry

Passage xV Tollen’s reagent is used for the detection of aldehyde when a solution of AgNO3 is added to glucose with NH4OH then gluconic acid formed 0 Ag + + e -  → Ag ; Ered = 0.8V O

C6 H12O6 + H 2 O  → Gluconic acid (C6 H12O7 ) + 2 H + + 2e - ; 0 Eoxd = -0.05V O

Ag ( NH 3 ) 2+ + e -  → Ag ( s ) + 2 NH 3 ; 0 Ered = 0.337V O

RT F   Use 2.303 × F = 0.0592 and RT = 38.92 at 298K   

→ 2 Ag ( s ) + C6 H12O7 + 2 H + 1. 2 Ag + + C6 H12O6 + H 2 O  Find lnK of this reaction. (a) 66.13 (b) 58.38 (c) 28.30 (d) 46.29 2. The pH of the solution is 11. Then EMF of cell is (a) Increased by 0.65 (b) Decreased by 0.65 (c) Increased by 0.325 (d) Decreased by 0.325 3. Ammonia is always is added in this reaction. Which of the following must be incorrect? (a) NH3 combines with Ag+ to form a complex (b) Ag ( NH 3 ) 2+ is a weak oxidizing reagent than Ag+ (c) In absence of NH3 silver salt of gluconic acid is formed (d) NH3 has affected the standard reduction potential of glucose/gluconic acid electrode

Passage xVI Two liter solution of a buffer mixture containg 1 M NaH2PO4 and 1 M Na2HPO4 is placed in two compartments one liter each of an electrolytic cell. Two platinum electrodes are inserted in compartments. 1.6 ampere current is passed for 200 minutes. Assume electrolysis of water takes place at each compartment pKa for H 2 PO4- = 2.2 1. The reaction takes place at anode is (a) 2 H + + 2e  → H2 (b) H 2  → 2 H + + 2e 1 (c) 2OH -  → H 2 O + O2 + 2e 2 1 (d) O2 + H 2 O + 2e -  → 2OH 2 2. The number of equivalents of H+ (or) OH– will be discharged at electrodes are (a) 0.05 (b) 0.1 (c) 0.15 (d) 0.2

3. The pH of the solution in anode compartment is (a) 1.89 (b) 2.02 (c) 2.2 (d) 2.37 4. The pH of the solution in cathode compartment is (a) 1.89 (b) 2.02 (c) 2.2 (d) 2.37 Passage xVII Electrochemical series in a very condensed form is given below. Moving down the series the strength of the reducing agent increases. A negative E means that the redox couple is a stronger reducing agent than H+/H2 couple. On the other hand, a positive E means that the redox couple is a weaker reducing agent than the H+ H2 couple. All E values are measured w.r.t E 0 + 1 H , H2 2 which has been arbitrarily fixed to be zero. O

O

O

O

Half − cell reaction (redn.)

E θ / V at 25°C

MnO −4 + 8H + + 5e −  → Mn 2 + + 4H 2 O

1.51

Cl2 (g ) + 2e  → 2Cl 2− 7

+

O

−

1.36 3+

−

Cr7 O + 14H + 6e  → 2Cr + 7 H 2 O 1.33 O 2 (g ) + 4H + + 2e −  → 2H 2 O +

−

MnO 2 (s) + 4H + 2e  → Mn +

−

3+

−

+

−

1.23 2+

+ 2H 2 O 1.23

Ag + e  → Ag (s) Fe + e  → Fe

0.80

2+

0.77

Cu + e  → Cu (s) 2+

−

2+

−

0.52

Fe + 2e  → Fe(s) Zn

Mg

2+

− 0.44

+ 2e  → Zn (s)

− 0.76

−

+ 2e  → Mg (s)

− 2.71

1. If Kw at 25°C is 10–14. Calculate E for the half reaction O

1 H 2 O + e −  → H 2 (g ) + OH − , E at 25°C is nearly 2 equal to (a) –0.64 V (b) +0.23 V (c) –0.83 V (d) –0.38 V 2. The EMF of the cell: Fe | Fe3+ (0.8M ) || Ag + (0.2M ) | Ag ( s ), at 25°C is equal to (a) 0.799 V (b) 0.728 V (c) 0.536 V (d) 0.289 V

Electrochemistry

3. Which one of the following reactions is not feasible? The concentration of ion wherever involved is 1.0 M and pressure of gas is 1 atm 25°C (a) Mg ( s ) + 2Cu +  → Mg 2 + + 2Cu ( s ) 2+ (b) Cr2 O7 + 14 H + 6Cl -  → 2Cr 3+ + 7 H 2 O + 3Cl2 + (c) Zn + MnO2 + 4 H  → Zn 2 + + Mn 2 + + 2 H 2 O (d) 2Cl2 + 2 H 2 O  → 4 HCl + O2 Passage xVIII

  Au ( CN )  Au + + 2CN -  2 

-

After equilibrium has been reached the aqueous phase is pumped off and metallic gold is recovered from it by reacting the gold complex with zinc which is converted to [Zn(CN)4]2– + Au -

Zn + 2  Au ( CN )2   →  Zn ( CN )4     

2-

+ 2 Au

Gold in nature is frequently alloyed with silver which is also oxidized by aerated sodium cyanide solution    Zn (CN )4 

2-

2. [Au(CN)2]– is a very stable complex under certain conditions. The concentration of cyanide ion which is required to keep 99% of the gold in the form of the cyanide complex is Given K  Au CN  - = 4 × 1028  ( )2  f    -28 -14 (a) 2 × 10 M (b) 3 × 10 M (c) 5 × 10-28 M

Metallic gold is frequently found in aluminosilicate rocks and it is finely dispersed among other minerals. It may be extracted by treating the crushed rock with aerated sodium cyanide solution during this process metallic gold is slowly converted to [Au(CN)2]– which is soluble in water

+ 2e -  → Zn + 4CN - ; E 0 = -1.26V O

-

 Au (CN )  + e 2  -

 Ag (CN )  + e 2 

 → Au + 2CN - ; E 0 = -0.60V O

 → Ag + 2CN - ; E 0 = -0.31V O

1. Five hundred litres (500 L) of a solution 0.01 M in [Au(CN)2]– and 0.003 M in [Ag(CN)2]– was evaporated to one third of the original volume and was treated with zinc (40 g). Assuming that deviation from standard conditions is unimportant in this case and all the concerned redox reaction go essentially to completion. Hence the concentrations of [Au(CN)2]– and of [Ag(CN)2]– after reaction are (a)

[ Au (CN )2 ]- = 0.03; [ Ag (CN )2 ]-

(b)

[ Au (CN )2 ]- = 0.01M ; [ Ag (CN )2 ]- = 0.009M

(c)

[ Au (CN )2 ]

(d)

[ Au (CN )2 ]- = [ Ag (CN )2 ]-

-

= 0.003M

= 0.03M ; [ Ag (CN ) 2 ] = 0.002 M -

= 0.03M

12.73

(d) 5 × 10-14 M

3. There have been several efforts to develop alternative gold extraction process which could replace this one because (a) Sodium cyanide solution corrode mining machinery (b) Sodium cyanide escapes into ground water and produces cyanide which is toxic to many animals (c) Gold obtained by this process is not pure (d) All the gold present in the ore cannot be extracted completely Passage xIx Water can be thought of as hydrogen oxidized by oxygen. Thus hydrogen can be recovered by reduction of water, using an aqueous solution of sodium sulphate at a platinum electrode connected to the negative terminal of a battery. The solution near the electrode becomes basic. Water can also be thought of as oxygen reduced by hydrogen. Thus oxygen can be recovered by oxidation of water at the ‘Pt’ electrode connected to the positive terminal. When copper is used at both electrodes, gas is generated only at one electrode during the initial stage of electrolysis. Another species in solution that can be reduced is sodium ion. The reduction of sodium ion to metallic sodium does not occur in aqueous solution, because water is reduced first. The electrode potential is affected by other reactions taking place around the electrode. The potential of the Cu2+/CU electrode in a 0.100 M Cu2+ solution changes as Cu(OH)2 precipitates. Precipitation of Cu(OH)2 begins at pH = 5 [temperature = 25°C, K w = 1.00 × 10-14 at 25°C] 1. Based on the above observations connect the following half reactions with the standard reduction potentials (in volts) (1) reduction of copper ion Cu 2 + i) + 0.34 (2) reduction of oxygen ii) - 2.710 (3) reduction of water iii) - 0.83 + (4) reduction of sodium ion Na iv) 0.000 (5) reduction of hydrogen ion v) + 1.230 (a) a - i, b - ii, c - iii, d - v, e - iv (b) a - i, b - v, c - ii, d - iii, e - iv (c) a - i, b - iii, c - ii, d - v, e - iv (d) a - i, b - v, c - iii, d - ii, e - iv

(

)

(

)

12.74

Electrochemistry

2. Standard reduction potential for Cu ( OH )2( s ) + 2e -  → Cu ( s ) + 2OH (a) –0.431 V (b) –0.331 V (c) –0.22 V (d) –0.131 V 3. Reduction potential of copper electrode at pH = 1 in the above solution is (a) + 0.51 V (b) + 0.41 V (c) + 0.21 V (d) + 0.31 V

3. The pH in cathodic compartment of cell after passing 1.25 A0 current for 643.3 minute (a) 4.2 (b) 5.22 (c) 4.74 (d) 5.04 Passage xxII Given the reaction 2M + 6H+ → 2M3+ + 3H2, ΔH 298 = –3.0 Kcal. The entropies are 6.5 cal/K for M , -22.2 cal / K for Mn3+, 31.2 Cal/K for H2 and –10.0 cal/K for H+, ∆G 0f for H+ in 0.00. 1. The standard free energy of formation of M3+ is (a) −31.7 Kcal (b) 96.2 Cal/K (c) –15.8 Kcal (d) –63.4 Kcal 2. E For the half-reaction, M 3+ + 3e  → M is (a) 0.229 V (b) –0.229 V (c) –0.114 V (d) +0.114 V 3. The reaction 2 M + 6 H +  → 2 M 3+ + 3H 2 is (a) Spontaneous at 25°C (b) Non-spontaneous at low temperatures (c) Non-spontaneous at all temperatures (d) Non-spontaneous at 25°C O

O

Passage xx A half cell is prepared by K2Cr2O7 in a buffer solution of pH = 1. Concentration of K2Cr2O7 is 1 M. To 3 liter of this solution 570 g of SnCl2 is added which is oxidised completely 0 to SnCl4 Given : ECr = 1.33V ; O -2 / Cr +3 . H + O

2

7

2.303 RT = 0.06; F

Atomic of mass of Sn = 119; Esn0 +4 / Sn+2 = 0.15 O

1. Number of moles of Cr+3 formed are (a) 2 (b) 6 (c) 4 (d) 3 0 after the reaction of 2. Half cell potential ECr -2 +3 + 2 O7 / Cr . H SnCl2 is (a) 1.187 (b) 1.191 (c) 1.285 (d) None of these 3. Emf of the cell O

2+

4+

27

+

3+

+

Pt | Sn , Sn , H ||Cr2 O , Cr , H |Pt 0.1 M 0.2 M 1 M

(a) 1.194 V (c) 1.18 V

0.2 M

1M

O

matching type Questions 1. Match the following

1M

(b) 1.176 V (d) 1.164 V

Column I (a) Na+(aq)+e– → Na(s)

Passage xxI A cell designed below contains one liter of mixture of HCl and NaCN 1 M and 2 M respectively. The two components using Pt electrodes. pKa for HCN is 4.74 Pt[H 2 (1atm) ] HCl + NaCN HCl + NaCN Pt H 2 (1 atm) 1M

2M

1M

2M

If temperature is constant at the given condition then answer the following questions: 1. pH in right side compartment is (a) 5.74 (b) 4.74 (c) 3.74 (d) 5.04 2. EMF of the cell is: (a) 0 (b) 0.0591 pH (c) –0.0591 pH (d) Cannot be calculated

(b) The concentration

of CuSO4 solution does not change on electrolysis (c) Cu(s) → Cu2+(aq) + 2e– (d) 4OH– → 2H2O + O2 + 4e–

Column II (p) Reaction taking place

at anode during electrolysis of very dilute solution of NaCl using electrodes. (q) Reaction taking place at anode during electrolysis of CuSO4 using Cu electrode. (r) Electrolysis of aq CuSO4 using Cu electrodes. (s) Reaction taking place at cathode when electrolysis of aq NaCl is done using Hg electrode.

Electrochemistry

5. Match the following

2. Match the following Column I

Column II

(a) κ specific conductance

(p) Increase with

(b) Λm (molar conductance)

(q) Decreases with

(c) α (degree of

dissociation) (d) Resistance

dilution. dilution. (r) Decrease with increase in concentration of electrolyte. Λm Λ ∝m

(s)

Column I (Electrolytes which are electrolysed) (a) NaCl(aq)

(p) H2 Liberates

(b) AgNO3(aq)

(q) O2 Liberates

(c) CuSO4(aq) (Pt electrode)

(r) pH Increases

(d) H2SO4(aq)

(s) pH Decreases

Column I (a) Boric acid +

Borax Column II

(a) Electrolysis of NaCl

(p) Chlorine evolves at

solution. (b) Electrolysis of Na2SO4 solution. (c) Electrolysis of CuSO4 solution using Pt electrodes.

anode. (q) Oxygen evolves at anode. (r) Hydrogen evolves at cathode.

(d) Electrolysis of H2SO4

(s) Nature of solution is

solution.

Column II (Characteristics)

6. Match Column I with Column II

3. Match the following Column I

Column II Kw (p) h= KaC (q) pH =

formate (c) Sodium acetate

(r) Λ ∞ = λ +∞ + λ -∞

(d) Meta boric acid

(s) pH = pKa + log log

basic after electrolysis.

Column I (a) Zn | Zn2+(C) || Zn2+ (b) H2(P = 1 atm) | HCl (1

O

M)|| H+ (1 M) | H2 (0.01 Bar) pt (b) Ag | AgCl (KCl 0.1 M) || Ag+ (0.01 M) Ag (c) Cu | Cu2+ (0.1 M) || Cu2+

(0.01 M) | Cu (d) Pt | Cl2 (1 bar) | HCl (0.1

M) || NaCl (0.1 M) | Cl2 (1 bar) pt

[ salt ] [ acid ]

7. Match column I with column II.

4. E 0Ag+ / Ag = 0.8 V K sp AgCl = 10 -10

(a) Pt H2 (0.1 Bar) | H+ (0.1

1 (pK w +pK a - pK b ) 2

(b) Ammonium

Column II (p) Spontaneous cell

(2C) | Zn

Column I

12.75

Column II

N) || H2SO4(1 N) | H2 (P = 1 atm) (c) Cu | Cu2+ (0.01 M) || Ag+ (0.1 M) | Ag

reaction. (q) Working cell

representation. (r) By increaseing

concentration of cation in cathodic compartment then cell works.

(p) Concentration cell

(q) Ecell>0 (r) E 0cell = 0 but cell is

(d) Ag | AgCl | (KCl, 0.1

(s) Concentration cell.

M) || Ag+(0.01 M) | Ag

O

working (s) Non working condi-

tion

8. Match column I with column II. Column I (a) Electrode reversible with

respect to cation. (b) Electrode reversible with respect to anion. (c) Redox electrode. (d) Reference electrode.

Column II (p) Pt | Fe2+, Fe3+ (aq) (q) Pt | H2 | H+ (a = 1) (r) Pt|Hg|Hg2Cl2|Cl–(aq) (s) Zn | ZnSO4(aq)

12.76

Electrochemistry

9. Match column I with column II. Column I (Quantities)

assertion (a) and reason (r) type Questions

Column II (Factors on which dependency exist)

(a) Molar conductance

(p) Temperature

(b) emf of a cell in

(q) Concentration of species

operation (c) Electrode potential

(r) Nature of substance

(d) Standard reduction

(s) No. of electrodes lost or

involved involved

potential

gained in the reaction

10. Match column I with column II. Column I (Reaction) (a) AgCl(s)+e– →

O

Column II (E values for the corresponding reaction/s) O

O

O

O

(p) E = E

Ag+/Ag

Ag(s) + Cl(aq) (b) AgNO3(aq)+e– →

Ag(s) + NO3– (aq)

(c) AgCl(s) → Ag+(aq)

(q) E = E

Ag+/Ag

+ 0.059

Ksp(AgCl) O

(r) E = 0.059 log Ksp (AgCl)

+ Cl(aq) (d) H2O → H+(aq) + –

OH (aq)

O

(s) E = 0 [If it is taken as

concentration cell] (t) E = 0.059 log Kw O

11. Match column I with column II.

“In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): The conductivity of metals decreases with increases of temperature where as that of electrolytic solution increases. Reason (R): Electrons in metals are very tightly held by the nucleus and are not free to move. 2. Assertion (A): Molar conductivity of a weak electrolyte at infinite dilution cannot be determined experimentally. Reason (R): Kohlrausch law helps to find the molar conductivity of a weak electrolyte at infinite dilution. 3. Assertion (A): Electrolysis of an aqueous solution of KI gives I2 at the anode but that of KF gives O2 at the anode not F2. Reason (R): I– ions have much higher oxidation potential than water while F– ions have much lower oxidation potential than water. 4. Assertion (A): Zinc displaces copper from copper sulphate solution. Reason (R): The E of zinc is –0.76 V and that of copper is +0.34 V. 5. Assertion (A): Zn cannot be obtained by the electrolysis of mixture of aqueous ZnCl2, AgNO3 and CuSO4. Reason (R): Standard oxidation potential of Zn is higher than that of Ag and Cu. 6. Assertion (A): Copper metal does not displace hydrogen from dilute H2SO4 solution. Reason (R): Standard reduction potential of copper electrode is negative. 7. Assertion (A): Specific conductance decreases with dilution whereas equivalent conductance increases. Reason (R): On dilution, number of ions per milli liter decrease but total number of ions increases considerably. 8. Assertion (A): An electrochemical cell can be set up only if the redox reaction is spontaneous. Reason (R): A reaction is spontaneous if free energy change is negative. 9. Assertion(A): A non-spontaneous reaction takes place in an electrolytic cell. Reason (R): In an electrolytic cell, external source of voltage is used to bring about chemical reaction. O

Column I (Electrolytes which are electrolysed in presence of inert electrodes)

Column II (Characteristic)

(a) NaCl(aq)

(p) H2 liberates

(b) AgNO3(aq)

(q) O2 liberates

(c) CuSO4(aq)

(r) pH increases

(d) CH3COONa(aq)

(s) pH decreases

12. Match column I with column II. Column I

Column II

(a) H+

(p) 350

(b) Na+

(q) 50

+

(c) Li

(d) Cs

+

(r) 39 (s) 77

Electrochemistry

10. The potential of the cell

Integer type Questions 1. The pKsp of AgI is 10. If the E0 of Ag+ / Ag is 0.8 V. Calculate the E0 of AgI(s) + e → Ag + I–. Answer  2.303RT  = 0.06  . Multiply  F  

expressed in volts

the answer with 10 and fill in the answer key. 2. Calculate the useful work done during the reaction

1 Ag ( s ) + Cl2( g ) → AgCl( s ) . 2

12.77

Given

that

ECl0 O

2

/ Cl -

=

+1.36V and ECl0 - / Ag + Cl / Ag = +0.330V If PCl2 = 1 and T = O

298 K. Answer x (102) KJ/mol. The value of x is____. 3. The chlorate ion can disproportionate in basic solution according to reaction: 2ClO3-  ClO2- + ClO4. What is the equilibrium concentration of perchlorate ions from a solution initially at 0.1 M in chlorite ions at 298 K?. The equilibrium constant is 10–x Than x is 0 Given : EClO = 0.33 V at 298 K / ClO -

Pt, H 2 (1 atm)

0.1M BCl Cu 2 + Cu is 0.88 V at 25°C 0.2 M BOH (1M)

E 0 Cu +2 / Cu = 0.34 V, O

2.303 RT = 0.06 F

If 20 mL of 0.1 M BOH is titrated with 0.1 M HCl, the pH at equivalent point is ______. 11. Calculate the cell potential of half cell having the reaction: M 2 S + 2e - → 2M + S 2 - at 27°C in a solution of pH = 3 and saturated with 0.1M H2S. For H2S, K1 =10–8 K 2 =10–13, Ksp (M2S) = 1.0×10–50 E0M+/M = 1 V assume R = 10 J/K/ mol and

2.303RT 0.06 . Express = nF n

the magnitude of your answer after multiplication with 100. 12. The voltage of the cell given is –0.46 V Pt (s)H 2 (g ) | NaHSO3 (aq ), Na 2SO3 || Zn 2 + (aq ) | Zn (s) 0.4 M

Also Zn

2+

-

6.44 ×10-3 M

0.3 M

θO

+ 2e → Zn (s), E = -0.736 V

O

4

3

4. 3 amphere current was passed through an aqueous solution of an unknown salt of Pd for 1 hour. 2.977 gm of Pdn+ was deposited at cathode. The value ‘n’ is (AW of Pd is 106)_______. 5. The K = 4.95 × 10–5 s cm–1 for a 0.00099 M solution. Calculate the reciprocal of the degree of dissociation ∞ of acetic acid, if λM for acetic acid is 400 s cm 2 –1 mol . 6. How many moles of electrons are gained by one mol of nitrobenzene to convert into aniline under acidic condition? 7. It is observed that the voltage of a galvanic cell using the reaction M ( s ) + xH + → M x + +

x H 2 varies linearly 2

with the log of the square root of the hydrogen pressure and the cube root of the Mx+ concentration. The value of ‘x’ is_______. 8. The electrolysis of cold sodium chloride solution produces NaOH and Cl2. The Cl2 produced, disproportionate in NaOH solution to give sodium hypochlorite (NaClO) and sodium chloride. How long will a cell operate to produce 1.00 × 103 L of 7.45% (w/w) solution of NaClO if the cell current is 9.65 ampere? Assume that the density of solution is 1.00 g/mL. The required time is 1×10x sec. The value of x is______. 9. The emf of the cell: M (s) | MnXn (0.1M/M)/(s) is found to be 0.0295 V at 25°C. If MnXn is soluble salt, Find the value of n.

+

 H  SO32 -  K =    -  = 6.44 × 10 - x . What will be 'x' ?  HSO3  13. EMF of the following electrochemical cell at 25°C is –0.021 V Pt

H 2 (g) H + (aq) H 2 (g) Pt 1atm pH = 1 x atm

RT   Value of x is  given 2.303 = 0.06  . F  

14. The potential of the cell containing two hydrogen electrodes as represented below Pt, H2(g) / H+(10–6M) // H+(10–4 M) / H2(g), Pt at 298 K is a × 0.059 V then (a) is ____. 15. A metal ‘S’ is added in excess to 500 mL of 1 molar solution of another metal ion Bn+ at 25°C until the equilibrium, S+Bn+  Sn++ B is reached. If ES0n+ / S and EB0n+ / B are –0.75V and –0.24 V respectively. Given at equilibrium, the concentration of Bn+ is 5.15×10–18 M. The value of n is ____. Given, log 5 = 0.699 and log 2.575 =0.4108 16. Pt : H 2 HCOOK NH 4Cl H 2 : Pt O

O

(1 atm, 25°C) (0.01 M )

(0.01 M ) (1 atm, 25°C)

K b Values of

HCOO– and NH3 are 10–6 and 10–8 Respectively. The potential of the given cell is “0.0591 X volts”. What is the value of “X”?

12.78

Electrochemistry

17. The osmatic pressure of a 0.1 M metal suplhate (M2SO4) solution is 4.926 atmospheres at 27°C. A rod of that metal ‘M’ is dipped in this solution to obtain an electrode of reduction potential given by (E 0M+ / M - 0.0591 x) volts what is x? O

(0.0821 × 300 = 24.63) 18. Number of moles of electrons needed to pass through 2 liters of AlCl3 solution containing 267 g of salt to discharge 2/3rd of the chloride ions present in the solution______. 19. Reduction potential of hydrogen electrode is –0.355V. The pH of the HCl solution used in the hydrogen electrode is_______. 20. Given the cell: Cd(s)|Cd(OH)2(s)|NaOH (aq, 0.01 M) | H2(g, 1 bar)|Pt(s) with Ecell.= 0.0 V. If E 0Cd2+ |Cd = -0.39 V then K sp of Cd(OH 2 ) is10 -3n . Where 'n' is_____. 21. An inaccurate ammeter and copper voltmeter are connected in series and the ammeter shows 0.525 amperes. If 0.635 gm of Cu is deposited in 1hr, the percentage error in ammeter is______. 22. After electrolysis of NaCl solution with inert electrodes for a certain period of time, 600 mL of the 1 N solutions was left which was found to be NaOH. During the same time 31.8 g of Cu was deposited in copper voltmeter in series with the electrolytic cell. The percentage of NaOH obtained is 10 Xx. Then ‘X’ value is_______. 23. Ag(s)|Ag 2 CO 3 (s)|Na 2 CO 3 (aq)||KBr(aq)|AgBr(s) Ag(s) Ksp = 8 × 10 –12 for Ag2CO3 and Ksp = 4 × 10–13 for for Ag 2 CO3 andK sp = 4 ×10-13 for AgBr O

[ Br - ] 23

=

x × 10-7 when the reaction in the cell is

CO at equilibrium. What is ‘x’? 24. The equilibrium constant for the reaction In 2 + + Cu 2 + → In 3+ + Cu +

O

Pt | H 2 (1 atm) | CH 3 COOH(1 M) + CH3 COONa(1 M) / / CH 3 COOH(1M) + CH 3 COONa(1M) / Pt / H 2 (1 atm) 28. The specific conductance of saturated solution of AgCl is found to be 1.86×10–6 ohm cm–1 and that of water is 6 × 10–8 ohm–1 cm–1. The solubility of AgCl in mol/L is 1.31 × 10–x, x is equal to (Given Λ0 AgCl = 137.2 ohm1 cm2 eq–1)________. 29. The equilibrium constant of the reaction Cu (s ) + 2Ag + ( aq )  Cu +2 ( aq ) + 2Ag (s ) ; O

'x' is (E = 0.46 V at 298 K)_______. 30. For a saturated solution of AgCl at 25°C, specific conductance is 3.41 10–6 ohm–1 cm–1 and that of water used for preparing the solution was 1.6 10–6 ohm–1 cm–1. Then solubility product of AgCl is __ × 10–10 moles2 lt–2. λ~AgCl = 90.5 ohm–1 cm2 eq–1. 31. “ E MnO- / Mn +2 , H+ ” differs by 0.48 volts if the pH is 4

changed from “x” to “y”, keeping the concentrations of all other species equal. What is the numerical difference between “x” and “y”? 32. The voltage of the cell Pb(s)|PbSO4(s)|NaHSO4 (0.600 M)|Pb2+(2.50×10–5 M)Pb(s)| is E = +0.061 V and K2 = [H + ][SO 24 - ][HSO 4- ] . The dissociation constant for HSO4- is 1.0 × 10- M . Find the value of M. – Given, Pb(s) + SO2– 4 (aq) → PbSO4(S) + 2e (E = 0.356) 2+ E (Pb / Pb) = –0.126V. 33. At equimolar concentration of Fe2+ and Fe3+, what must [Ag+] be so that the voltage of the galvanic cell made from Ag+/Ag and Fe3+ / Fe2+ electrodes equals to zero? The reaction is  Fe2++A+ ↽ ⇀  Fe3+ + Ag. Determine the equilibrium constant at 25°C for the reaction. Given, O

O

E 0 Ag+ / Ag = 0.799 V and E 0 Fe3+ / Fe2+ = 0.771 V . O

At 25°C, is given as K = a × 1010. Find the value of ‘a’. Given E (Cu2+|Cu+) = 0.15, E (In3+|In+) = –0.42, E (In2+|In+) = –0.4 V 25. How many grams of water will be electrolysed by 96500 columb charge? 26. 241.25 ampere current was passed through an aqueous solution of an unknown salt of manganese for 1 minute 40 seconds. 6.875g Mn+ x was deposited at Cathode. The value of “x” is (At mass of Mn = 55)_______. 27. A cell designed below contains one liter of buffer mixture of acetic acid and sodium acetate each 1 M two compartments using platinum electrodes. pKa of acid O

is 4.74. The emf of the cell at 298 K is; neglecting the liquid – liquid junction potential.

O

O

34. The standard oxidation potential of Ni/Ni2+ electrode is 0.236V. If this is combined with a hydrogen electrode in acid solution, at what pH of the solution will be measured emf be zero at 25°C? Assume (Ni 2+ = 1) M. 35. The pOH of 0.5 L of 1.0 M NaCl after the electroly sis for 965 s using 5.0 A current (100% efficiency), is 36. The value of the reaction quotient Q, for the cell is M × 10–2 Cr ( S ) | Cr 3+ ( aq )( 0.1 M ) Cu 2 + aq (0.5 M ) | Cu ( s ) Find the value of M.

Electrochemistry

Previous years’ IIt Questions multiple Choice Questions with one Correct answer 1. For the electrochemical cell, M | M+ || X– | X, E M+ | M = 0.44V and E (X/X–) = 0.33V M From this data one can deduce that (2000S) (a) M+X  → M + + X - is the spontaneous reaction → M + X is the spontaneous reac(b) M + + X -  tion O

O

5. In the electrolytic cell, flow of electrons is from (2003S) (a) Cathode to anode in solution (b) Cathode to anode through internal supply (c) Cathode to anode through internal supply (d) Anode to cathode through internal supply 6. The emf of the cell Zn | Zn 2 + ( 0.01M ) || Fe 2 + (0.001M ) | Fe (2004S) A 298 K is 0.2905 then the value of equilibrium constant for the cell reaction is 0.32

(c) Ecell = 0.77V

0.32

(a) e 0.0295

(d) Ecell = -0.77V 2. Saturated solution of KNO3 is used to make ‘saltbridge’ because (2001S) (a) velocity of K+ is greater than that of NO3(b) velocity of NO3- is greater than that of K+ (c) velocities of both K + and NO3- are nearly the same (d) KNO3 is highly soluble in water 3. The correct order of equivalent conductance at infinite dilution of LiCl, NaCl and KCl is (2001S) (a) LiCl > NaCl >KCl (b) KCl>NaCl>LiCl (c) NaCl>KCl>LiCl (d) LiCl>KCl>NaCl 4. Standard electrode potential data are useful for understanding the suitability of an oxidant in a redox titration. Some half cell reactions and their standard potentials are given below: (2002S) MnO 4- ( aq.) + 8H + ( aq.) + 5e -  → Mn 2 + ( aq.) +4H 2 O ( I ) E  = 1.51 V O

Cr2 O72 - ( aq.) + 14H + ( aq.) + 6e -  → 2Cr 3+ ( aq.) +7H 2 O ( l) = E 0 = 1.38 V O

Fe3+ ( aq.) + e -  → Fe 2 + ( aq.) E 0 = 0.77 V O

Cl2 ( g ) + 2e -  → 2Cl - ( aq.) E 0 = 1.40 V O

Identify the only incorrect statement regarding the quantitative estimation of aqueous Fe(NO3)2. (a) MnO4- can be used in aqueous HCl (b) Cr2 O72 - can be used in aqueous HCl (c) MnO4- can be used in aqueous H 2 SO4 (d) Cr2 O72 - can be used in aqueous H 2 SO4

12.79

(b) 10 0.0295

0.26

0.32

(c) 10 0.0295 (d) 10 0.0591 7. The rusting of iron takes place as follows: 1 2H - + 2e - + O 2  E 0 = +1.23V → H 2 O(l) : 2 Fe 2 - + 2e -  E 0 = -0.44V → Fe(s) : O

O

Calculate ΔG = (in KJ) for the overall reaction. (a) –322 KJ mol–1 (b) –161 KJ mol–1 (c) –152 KJ mol–1 (d) –76 KJ mol–1 (2005) 8. Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 milliampere current. The time required to liberate 0.01 mol of H2 gas at the cathode is: (1 Faraday =96500 C mol–1) (a) 9.65 × 104 sec (b) 19.3 × 104 sec 4 (c) 28.95 × 10 sec (d) 38.6 × 104 sec (2008) 9. Consider the following cell reaction: O

+ 2+ 2Fe(s) + O 2(g) + 4H (aq)  → 2Fe(aq) + 2H 2 O(l)

E 0 = 1.67 V O

At  Fe 2 +  = 10-3 M , P (O2 ) = 0.1 atm and pH = 3, the cell potential at 25°C is: (a) 1.47 V (b) 1.77 V (c) 1.87 V (d) 1.57 V (2011) 10. AgNO3(aq) was added to an aqueous KCl solution gradually and the conductivity of the solution was measured. The plot of conductance (Λ) versus the volume of AgNO3 is: (a) Λ

(b) Λ

Volume

(P)

(a) P (c) R

(d) Λ

(b) Λ

Volume

Volume

(Q)

(R)

Volume

(S)

(b) Q (d) S (2011)

12.80

Electrochemistry

Comprehensive type Questions Read the passage given below and answer the questions that follows. Passage I

The concentration of potassium ions inside a biological cell is at least twenty times higher than the outside. The resulting potential difference across the cell is important in several processes such as transmission of nerve impulses and maintaining the ion balance. A simple model for such a concentration cell involving a metal M is: M ( s ) | M + (aq; 0.05 molar ) || M + (aq ;1molar ) | M ( S )

For the above electrolytic cell, the magnitude of the cell potential | Ecell | = 70 mV. (2010) 11. For the above cell: (a) Ecell < 0; ΔG > 0 (b) Ecell > 0; ΔG < 0 (c) Ecell < 0; ΔG > 0 (d) Ecell > 0; ΔG < 0 12. If the 0.05 molar solution of M+ is replaced by a 0.0025 molar M+ solution, then the magnitude of the cell potential would be: (a) 35 mV (b) 70 mV (c) 140 mV (d) 700 mV O O

Electrochemistry

12.81

answEr KEys multiple Choice Questions with one or more than one answer

multiple Choice Questions with only one answer level I 1. c 2. d 3. b 4. d 5. a 6. c 7. c 8. c 9. a 10. c 11. b 12. c 13. d 14. a 15. d 16. d 17. c 18. d 19. d 20. c 21. d

22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

a a a a a a c a a a c c a c d b a d d d c

43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60. 61. 62. 63.

d d d b a d d a b a c a c d a a b c b a a

64. 65. 66. 67. 68. 69. 70. 71. 72. 73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84.

a a c c a a c d d b d b c a b b d a a b b

85. 86. 87. 88. 89. 90. 91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. 103. 104.

d a a c b a b a d a b b a d c c d b a b

1. 2. 3. 4. 5. 6. 7.

a,c b,c a,b a,b,c a,d a,b,d b,c,d

8. 9. 10. 11. 12. 13. 14.

a,d a,b b,c,d b,c,d a,d b,c a,d

15. 16. 17. 18. 19. 20.

a,b,c a,b,c,d a,b b,c a,b,c a,b

Comprehensive type Questions Passage I 1. b

2. b

3. a

2. b

3. c

2. a

3. b

2. a

3. d

2. a

3. d

2. b

3. b

2. d

3. a

2. b

3. c

2. d

3. b

Passage II 1. a Passage III 1. b Passage IV

multiple Choice Questions with only one answer level II 1. d 2. a 3. b 4. a 5. a 6. b 7. c 8. a 9. b 10. c 11. a 12. a 13. a 14. b

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

b d c d a c b b c d c a b b

29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

a a d c c a d a a c a c c b

43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.

c a d b c b d b a a c b d c

57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.

a a a d d c a c c a a b d a

71. 72. 75. 74. 75. 76. 77. 78. 79. 80.

1. b

b d b a b c a a b c

Passage V 1. b Passage VI 1. d Passage VII 1. c Passage VIII 1. b Passage Ix 1. b

12.82

Electrochemistry

Passage x 1. b

Passage xxI 2. c

3. a

Passage xI 1. b

2. a

1. c

2. d

3. b 4. c

Passage xIV 1. d 2. b

3. a

2. b

3. a

matching type Questions

Passage xIII 1. a 2. d

2. a

Passage xxII

Passage xII 1. a

1. b

3. a 4. b

5. b

2. a

3. d

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

(a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a) (a)

s q prs pqr pr s pqs, qs pqr q pr p

(b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b) (b)

r pr qr pqr qs qr rs r pqrs p qs q

(c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c) (c)

qr prs q ps qs pr pq p pqrs rs qs r

(d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d) (d)

p q qr ps pqs r pqs qr rs t pr s

Passage xV 1. b

1. c 2. b 3. a

Passage xVI 1. c 2. d

3. b 4. d

2. a

3. b

2. d

3. b

Passage xVIII 1. c

2. c

3. a

2. a

3. d

Passage xx 1. a

7. c 8. b 9. a

1. 2. 3. 4. 5. 6. 7.

2 1 1 4 8 6 3

8. 9. 10. 11. 12. 13. 14.

7 2 5 2 8 5 2

15. 16. 17. 18. 19. 20. 21.

2 6 1 4 6 5 2

22. 23. 24. 25. 26. 27. 28.

6 2 1 9 2 0 5

29. 30. 31. 32. 34. 35. 36.

4 4 5 3 8 3 8

Previous years' IIt Questions

Passage xIx 1. d

4. a 5. b 6. c

Integer type Questions

Passage xVII 1. c

assertion (a) and reason (r) type Questions

1. 1 2. 3

3. 2 4. 1

5. 3 6. 2

7. 1 8. 2

9. 4 10. 4

11. 2 12. 3

Electrochemistry

12.83

hInts anD solutIons hints to Problems for Practice

3. The cell reaction for the cell

O

1. As E for silver electrode is more positive the cell may be represented as Fe 2 + / Fe3+ / Ag + / Ag E cell = E right - E left 0.0591   E =  E θAg+ / Ag + log  Ag +   n   3+   Fe   0.0591 log - E θFe3 + / Fe2 + +  n  Fe 2 +    O

Zn | Zn 2 + || Ni 2 + | Ni Zn + Ni

0 = 0.03 + 0.0591 log  Ag +  0.03 = -0.5085 = 1.4915 0.0591 ∴  Ag +  = 0.3121 M Or log  Ag +  = -

O

O

2. As E Ag+/Ag is more positive than E Cu2+/Cu the former will act as cathode and the latter anode. Thus the cell may be represented as Cu | Cu 2 + || Ag + | Ag [Cu 2 + ] = 0.01 M and E cell = 0 ∴ E cell = E Ag+ / Ag - E Cu 2+ / Cu = 0 0.0591   Or = E Ag+ / Ag + log  Ag +   n   0.0591   - E Cu 2+ / Cu + log Cu 2 +   = 0 2   0.0591   Or = 0.799 + log  Ag +   1   O

O

0.0591   - 0.0337 + log ( 0.01) = 0 1   +   Or log  Ag  = -8.8172 - 9.1828 Taking antilog  Ag +  = 1.52 × 10 -9 M

 Zn

E cell = Ecell O

2+

is

+ Ni For which

 Zn 2 +  0.0591 log  2 +  2  Ni 

At equilibrium E cell = 0

O

0.0591   0 = 0.80 + log  Ag +   1   . 0 0591   - 0.77 + log 1   Fe 2 +  = Fe3+ 1  

2+

∴ E cell = O

 Zn 2 +  0.0591 log  2 +  2  Ni 

E Ni2+ / Ni - EZn 2+ / Zn = O

O

 Zn 2 +  0.0591 log  2 +  2  Ni 

 Zn 2 +  log  2 +  = 17.259  Ni  Taking anti log  Zn 2 +  = 1.816 × 1017  Ni 2 + 

This concentration ratio shows that almost whole of the Ni2+ ions are reduced to Ni and therefore the concentration of Zn2+ produced from Zn would be nearly 1M  Ni ( NO3 ) = 1M  . Thus   2 1

= 1.816 × 1017  Ni 2 +  or  Ni 2 +  = 5.5 × 10-18 M

4. Let the equilibrium constants for the following equations be as follows: H 2 O + H 2 O  H 3 O + + OH - ......... K1 (1)  1 H 3 O + + e -  H 2 + H 2 O............ K 2 ( 2) 2  Adding them we get 1 H 2 O + e -  H 2 + OH - ............. K 3 (3) 2  2 To calculate K1 For equation ( 2) E H O+ / H = 0 O

3

2

12.84

Electrochemistry

0.0591   log K  ∴ K 2 = 1 E =   1

Cell II

O

Zn ZnSO 4 CuSO 4 Cu C1′

For equation (3) E = -0.8277 volt ( given ) O

0 E cell + ′ = E cell O

C′2

C′ 0.06 log 2 ...............(2) 2 C1

0.0591   log K  ∴ K 2 = 1  E =   1

If E cell > E cell ′ , then E cell - E cell ′ = 0.03 V

For equation (3) E = -0.8277 volt ( given )

and C 2 = 0.5 M

0.0591 log K 3 = -0.8277 1 08277 log K 3 = ; K 3 = 1 × 10 -14 0.0591

∴ By Eqs (1) and (2) 0.03 =

O

O

∴E = O

Further as equation (3) is the sum of equation (1) and (2) we have

Or C2′ = 0.05 M 7.

O

O

5. Given that pH = 14, i.e.,  H +  = 10-14 M and OH -  = 1M

Cu 2 +  =

10-19 OH - 

2

=

10-19 = 10-19 1

For the half-cell reaction Cu 2 + + 2e -  Cu E = Eθ O

0.0591 1 log 2 Cu 2 + 

= 0.34 -

0.0591 1 log -19 2 10

= -0.22V .

6. Given Cell I : Zn | ZnSO 4 || CuSO 4 | Cu C1

C2

Cu 2 +  0.06 E cell + log 2  Zn 2 +  E cell = E E cell

θO cell

Cu 2 +  0.06 log + 2  Zn 2 + 

C 0.06 log 2 .............(1) = E θcell + 2 C1 O

O

O

= 1.44 - 0.68 = 0.76V 0.76 ∴ Log10 K c = = 12.8814 0.059 ∴ K c = 7.6 × 1012

∴ K1 = 1× 10-14

2

0.059 log 10 K c 1 θ E θcell = E ce - E θFe3+ / Fe2+ 4+ / Ce3+

E θcell =

K 3 = K1 .K 2 = K1

K sp = Cu 2 +  OH -  = 10-19

0.06 0.5 log 2 C 2′

8. For the equilibrium  2 Fe 2 + + I 3- , E = 0 2 Fe3+ + 3I -  E = E0 O

0 Also Ecell O

0.059 0.059 log K c or E 0 = log10 K c 2 2 0 = EFe - EI0/ I 3+ / Fe2+ O

O

O

3

= 0.77 - 0.54 = 0.23V 0.059 log10 K c Thus 0.23 = 2 ∴K c = 6.26 × 107 9. Given In 3+ + 2e - → In +

E10 = - 0.42V.......(1)

In 2 + +

E 02 = - 0.40V.......(2)

e - → In +

O

O

By subtracting eq (2) from eq (1) a third half cell reaction can be obtained as E 30 × 1 × F = E10 × 2 × F - E 0 × 1 × F O

O

O

or E 30 = 2 × ( -0.42) - 1 × ( -0.40) O

= - 0.44V ∴ In

3+

+ e - → In 2 +

E 30 = - 0.44V O

For the reactions Cu 2 + + e - → Cu + E 04 = 0.15 O

In 2 + → In 3+ + e - E 50 = + 0.44 O

Electrochemistry

  K sp = [Ag + ]2 [SO 42 - ]

The net redox change Cu 2 + + In 2 + → Cu + + In 3+ E

0O cell

 3.2 × 10 -2  = (3.2 × 10-2 ) 2  = 1.6 × 10 -5  2  

= E + E = 0.15 + 0.44 = 0.59V 0O 4

0O 3

Also E 0cell = O

0.059 log K c 1

12. For  Hg 22 + + 2Hg + 2Fe3+ 

0.059 0.59 = log K c 1 ∴ K c = 1010

excess excess

10. For the cell Ag/Ag+(Ag2CrO4 sol. saturated) || Ag+ 0.11n | Ag; Ecell = 0.164 V at 298 K 0.059 We have E cell = E 0Ag+ / Ag - E 0Ag / Ag+ + log10 1 [Ag + ] R.H.S [Ag + ]L.H.S O

O

or

0.164 = 0 +

 2Ag + CrO Now K sp for Ag 2 CrO 4  K sp = [Ag]2 [CrO 42 - ] Since [Ag ]L.H.S = 1.66 × 10

×

5

0 95

100

2 × 100

×10

-3

0 95 100

Before reaction ×10

-3

After reaction

For the cell at equilibrium E cell = 0 = E Fe3+ / Fe2+ - E Hg2+ / Hg 0 = E θHg2+ / Hg O

0.059 log10 [Hg 22 + ] + E θFe3+ / Fe2+ 2 O

2

 Fe3+  0.059 + log10  2 2  Fe 2 +  0=E

24

θO Hg 22+ / Hg

 Fe3+  0.059 + 0.77 + log10 2 [Fe 2 + ]2 [Hg 22 + ]

(  E θFe3+ / Fe2+ = 0.77 V) 2

E θHg2+ / Hg O

-4

2

M.

1.66 × 10 -4 M 2 1.66 × 10 -4 K sp = [1.66 × 10 -4 ]2 [ ] 2 K sp = 2.287 × 10 -12 mol3 litre -3 [CrO 24 - ]L.H.S =

 5 -3  100 × 10  0.059 = 0.77 + log10 2  95 × 10 -3   95 × 10 -3   100   2 × 100    

= -0.792 V

13.

E cell = E Ag+ /Ag - E Ag+ /Ag R.H.S L.H.S 0.059  O  log10 [Ag + ]R.H.S  = E 0Ag+ /Ag + 1    0O  0.059 - E Ag+ /Ag + log[Ag + ]L.H.S  1   ++ [Ag ] 0.059 [Ag ]R.H.S 0.059 E cell = log10 ................(i) 1 [Ag + ]L.H.S

11. The given cell is Pt H 2 H + Ag 2SO 4 (aq) Ag 1 M Saturated → 2H + + 2e The reactions are, H 2 

Now for L.H.S K sp AgCl = 2.8 × 10 -10

2Ag + + 2e -  → 2Ag E Ag+ / Ag - E H+ / H

10

-3

2Fe3+

O

+

+

10

-3

2

0.059 0.1 log10 1 [Ag + ]L.H.S

[Ag + ]L.H.S = 1.66 × 10 -4 M.

Thus

12.85

∴ [Ag + ] = 2

2.8 × 10 -10 2.8 × 10 -10 = × 1.4 × 10 -9 M [Cl - ]] 0.2 [Cl 0.2

For R.H.S K sp AgBr = 3.3 × 10 -13

0.0591 log [Ag + ]2 2 1 [0.799 - 0.711] × 2 = =3 log + 2 0.059 [Ag ]

[Ag + ][Br - ] = 3.3 × 10 -13

0.711 = 0.799 +

+ 2

[Ag ] = 10

-3

+

∴ [Ag ] = 3.2 × 10

Now the solubility equilibrium is   2Ag + + SO 24 Ag 2SO 4   K

= [Ag + ]2 [SO 2 - ]

∴ [Ag + ] =

-2

3.3 × 10 -13 3.3 × 10 -13 = = 3.3 × 10 -10 M. [Br - ] 0.001

∴By eq.1 E cell =

0.059 3.3 × 10 -10 log10 = - 0.037 V 1 1.4 × 10 -9

12.86

Electrochemistry

Thus, to get Ecell positive, polarity of cells should be reversed i.e., cell is Ag | AgBr(s) KBr || AgCl, KCl | Ag and

 E cell = 0 at equilibrium , thus from eq. 3

E = +0.037 V

0.059 log K sp AgI 1 0.059 E 0I- / AgI/ Ag - 0.799 = log 8.7 × 10 -17 1 or E 0I- / AgI/ Ag = -0.948 + 0.799 = 0.149 V

0.001 M

0.2 M

14. (i) In 8M H+ solution, conc. of all other species is unity 0.0591 E = E0 + log10 [ H + ]2 1 = 0.78 + 0.059 Log10 (8) 2 = 0.78 + 0.1062 = 0.8862V (ii) In case of neutral solution; concentration of [H+] = 10–7 M and conc. of all other species are unity, then 0.059 log10 [ H + ]2 E = E0 + 1 0.059 log10 [10-7 ]2 = 0.78 + (-0.826) = 0.78 + 1 = - 0.046V O

O

E 0I- / AgI/ Ag - E 0Ag+ / Ag

0.059 log10 [Ag + ]...................(1) 1 AgI = [Ag + ][I - ]

E Ag+ / Ag = E θAg+ / Ag + O

Also K sp

O

O

1 16. Ag (s ) + Cl2 (g) → AgCl (s) ∆G1θ = -109KJ  (1) 2 Ag (s) → Ag + (aq) + e ∆G θ2 = +77KJ  (2) O

O

1 Cl2 (g) + e - → Cl - (aq) 2 By eqs (1) - (2) - (3)

= -109 - 77 + 129 = -57KJ - ∆G θ = nF E θ

O

57 × 103 = 1 × 96500 × E θ ∴ E θCell = 0.59V O

(anode)

E θAg+ / Ag = 0.799 V O

→ AgI(s) + e Ag + I -  -

Ag + I  → AgI

∴ E cell

(cathode)

 1  Also E Cell = E θCl- / AgCl/ Ag - 0.059 log Cl -    O

0.059 ∴ By eq (1) E Ag+ / Ag = 0.799 + log10 (9.32 × 10 -9 ) 1 = 0.799 - 0.474 = 0.32 V

+

O

The cell is Ag | AgCl(s) | Cl - (aq) || Ag + (aq) | Ag

..................(2)

Also Ag + e -  → Ag

O

O

∴ [Ag + ] = K sp (AgI) = 8.7 × 10 -17

+

∆G 3θ = -129KJ ..(3)

Ag + (aq) + Cl - (aq) → AgCl(s); ∆G θ4

For saturated solution [Ag + ] = [I - ] = 9.32 × 10 -9

0.059 log  Ag +   I -  1

=

O

15.

=

O

O

0.059   log[Ag + ] + =  E θAg+ / Ag   1

{E

+ 0.059 log  Ag + 

}

At equilibrium E Cell = 0 E θAg+ / Ag - E θCl- / AgCl/ Ag = -0.059 log  Ag +  Cl -  O

O

E θCell = -0.059 log K sp AgCl O

0.59 = -0.059 log K sp AgCl K sp AgCl = 1 × 10-10 M 2 Let solubility of AgCl be S then S = K Sp = 10-10 = 10-5 M

O

 θ 1  0.059  E I- / Ag I / Ag + 1 log [I - ]  ..................(3)

θO Ag + / Ag

Mole of AgCl in its 100 mL saturated solution

O

= 10-5 ×

100 = 10-6 1000

Mole of Zn added in it =

6.539 × 10-2 = 10-3 65.39

Electrochemistry

E 0Cl- / AgCl/ Ag = 0.22 at 25°C

For Ag → Ag + + e - ; ∆G 0 = 77 KJ

O

O

- ∆G 0 = nFE 0

O

O

Also E 0Cl- / AgCl/ Ag = E 0Ag+ / Ag + 0.059 log K sp AgCl O

77 × 10 = 0.080 1× 96500 For the redox reaction on addition of Zn to AgCl saturated solution 3

0 = E Ag + / Ag O

0 Zn 2 + + 2e - → Zn EZn = -0.76V 2+ / Zn O

0 2 Ag → 2 Ag + + 2e - E Ag = +0.80V + / Ag O

0 Zn 2 + + 2 Ag → Zn + 2 Ag + ECell = -1.56V O

+

0 Also Ecell = ECell O

2

 Ag  0.0591 log  2 + 2  Zn 

at equilibriumEcell = 0  Zn 2 +  0.059 log  2 2  Ag +   Zn 2 +  1.56 × 2 ∴log  = = 52.88 2 0.059  Ag +  0 ∴ ECell =O

 Zn 2 +  = 7.61×1052 and K C =  2  Ag + 

O

0.22 = 0.80 + 0.059 log K sp AgCl K sp AgCl = 1.47 × 10 -10 solubility of AgCl = K sp = 1.47 × 10 -10 = 1.21 × 10 -5 mol litre -1 18. According to Gibbs – Helmholtz equation heat of reaction ΔH given as   ∂E   ∆H = nF T   - E T ∂     P T = 273 + 25 = 298 K , n = 2, F = 96500C , E = +0.03V  ∂E  -4 -1 and   = -1.4 × 10 VK  ∂T  P ∴ ∆H = 2 × 96500  298 × ( -1.4 × 10-4 ) - 0.03 = -13842 Joule = -13.842 KJ mol -1 19. For the given cell reaction ∆G° = - nFE ° O

O

∆G° = -12 × 96500 × 2.73 = -3.1613 × 103 KJ O

Since Kc is appreciably high thus nearly whole of Ag+ is converted to Ag thus mole of Ag formed = mole of Ag+ in 100 mL solution = 106. Note that Zn is in excess. 17. Pt, H 2 (g) | HCl(aq) || AgCl(s) | Ag(s) 1 (i) H 2 → H + + e - anode 2 AgCl + e - → Ag + Cl - cathode 1 H 2 + AgCl → H + + Ag + Cl 2 -∆G° = nFE ° = 1 × 96500 × 0.23 = 22195 J (at 15°C) -∆G° = nFE ° = 1 × 96500 × 0.21 = 20265 J (at 35°C) O

O

O

O

Also ∆G° = ∆H ° - T ∆S ° ∴ -22195 = ∆H ° - 288 × ∆S ° -20265 = ∆H ° - 308 × ∆S ° O

O

O

O

O

∴ ∆S ° = -96.50 J Also - 22195 = ∆H ° - 288 × ( -96.5) = -49987 J ∴ ∆H ° = -49.987 O

O

(iii) Consider the following reaction at AgCl (s)/ Cl– /Ag electrodes E Cell = 0 at equilibrium Also E E0

0O Cl - / AgCl/ Ag

n = 12  4 Al → 4 Al 3+ + 12e 3O2 + 12e - → 6O 2 Now for given reaction ∆G° = 4 × ∆G °f  Al (OH )4  O

O

-

-6 × G °f [ H 2 O ] - 4 × G °f OH -  O

O

(G

°O f

for elements is zero

-3.1613 × 103 = 4 × G °f  Al (OH )4  -6 × ( -237.2) - 4 × ( -15) O

)

-

-

∴ G°  Al (OH )4  = 1303KJ mol -1 O

O

O

-E

0O H+ / H2

= 0.22 at 25°C

= 0.22 at 25°C

12.87

20. At L.H.S From CH 3 COOH  CH 3 COO - + H +  H +  = C × a = C

Ka = Ka.C C

= 1.8 × 10 -5 × 0.1 = 1.342 × 10 -3 mol litre -1 At R.H. S; From NH 4 OH  NH +4 + OH [OH - ] = C × a = C

Kb = K b .C C

 12.88

Electrochemistry

1.8 × 10 -5 × 0.01 = 0.424 × 10 -3 mol litre -1 ∴  H +  =

10 -14 = 2.359 × 10 -11 mol litre -1 0.424 × 10 -3

1 Now for cell H 2 → H + + e - at anode i.e., L.H.S 2 1 H + + e - → H 2 at cathode i.e., R.H.S 2 E Cell = E R.H.S - E L.H.S   + 0.059  0  H  R.H.S  = E H+ / H + log  1 1   PH  2    2   2   PH   0.059  0  2  log + - E H+ / H +  1  H   L.H.S    +  H  0.059 R.H.S log10 = 1  H +  R.H.S  PH2 = 1 atm on both sides O

O

=

0.059 2.359 × 10 -11 log10 = -0.4575 volts 1 1.342 × 10-3

2

2

( C are same) E Cell = E right - E left 1 1  = -0.059  pKa1 - pKa 2  2 2  0.059 = [5 - 3] = +0.059 volts. 2 22. Ag  → Ag + + e E θAg+ / Ag = 0.799 V O

 Ag ( NH 3 )+  → Ag + 2 NH 3    Eθ =? +  Ag ( NH )+  / Ag 3 2    Ag ( NH 3 )  Ag + 2 NH 3 O

E cell E cell

 Ag ( NH 3 )+  0.059 2 =E + log10  + 2 1  Ag  [ NH 3 ] = 0 and E θcell = E θ - E θAg+ / Ag + θO cell

O

E θcell = O

E θ

O

O

O

 Ag ( NH3 )  / Ag  2

0.059 0.059 log10 K c = log10 6 × 10-14 = -0.780 V 1 1 = 0.780 - 0.799 = 0.019 V +

Ag ( NH3 )2  / Ag  

323. Co ( CN )4 -  → Co ( CN )6 + e 6

E θRP = -0.83 V O

Co3+ + e -  → Co 2 + Co ( CN )6 + Co 4-

21. The cell can be written as Pt , H 2 (1 atm ) | HA 2 || HA1 | H 2 (1 atm ) Pt At L.H.S E H+ / H = E θH+ / H + O

0.059 log 10  H +  2 1

∴ E H+ / H = E θH+ / H − 0.059 ( pH )2 O

0.059 log  H +  1 1 − 0.059 (pH)1

At R.H.S E H+ / H = E θH+ / H + O

O

For acid HA1 HA1  H + + A1−  H +  = C.α = Ka.c 1 1 ( pH )1 = pKa1 = log10 C 2 2 Similarly, (pH) 2 =

 Co

E θRP = 1.82 V O

+ Co ( CN )6

3-

2+

4Co3+  Co ( CN )6  0.059   log10 + 32+  1  Co  Co ( CN )6  

Co3+  Co ( CN )6  CN -  0.059   + log10 6 32+  1  Co  Co ( CN )6 CN -    4-

− log  H +  = pH

E H+ / H = E θH+ / H

E cell = E

θO cell

3+

1 1 pKa 2 - log10 C 22 22

( C are same) E Cell = E right - E left 1 1  = -0.059  pKa1 - pKa 2  2 2  0.059

or E cell = E

θO cell

Also Co 2 + + 6CN -  Co ( CN )6  Co(CN)64 -  And K f1 =  6 Co 2 +  CN - 

4-

→ Co (CN )6 and Co3+ + 6CN -  Kf 0.059 0 + E cell = E cell log10 1 1 Kf

3-

O

2

At equilibrium E = 0 0 = 0.83 + 1.82 +

1019 0.059 log10 1 Kf 2

K f2 1019

= 8.23 × 1044 ∴ K f = 8.23 × 1063 2

6

Electrochemistry 0 24. Ecell = 0.76

12.89

O

[Ag + ] = K sp of AgI = 8.5 × 10 -17 = 9.22 × 10 -9 M

Applying Nernst equation

 Ag +  0.059 R.H.S log ∴ 0.0860 = 1 9.22 × 10-9

2+

 Zn   H 2  0.0591 log  2 2  H +  ( 0.1) ×1 0.0591 0.28 = 0.76 log 2 2  H +  0 Ecell = Ecell O

 Ag +  R.H.S or = 28.68 9.22 × 10 -9

(

2

log 0.1 - log  H +  = 16.2436  - log  H +  = pH

)

2 pH = 16.2436 - log 0.1 17.2436 pH = = 8.6218 2 25. Current will flow from higher reduction potential electrode to lower reduction potential electrode i.e., from Pt (2) electrode to Pt (1) electrode 0 Ecell = 1.61 - 0.77 = 0.84 volt O

∴ Ag +  R.H.S = 28.68 × 9.22 × 10 -9 M Also for R.H.S  Ag +   Cl -  = K sp of AgCl ∴ Cl -  =

K sp AgCl  Ag + 

=

1.8 × 10 -10 28.68 × 9.22 × 10 -9

or  MCl -  = 6.8 × 10-4 M

28. Pt | CuCl || CuCl2 | Pt Anode Cu +  → Cu 2 + + e -

26. 1 H 2  → H+ + e2 1 H + + e -  → H2 2 0.059   log H +  ∴ E cell = E θH+ / H + 2 1  

Cathode

0 ECu = 0.153V 2+ / Cu + O

Cu + + e -  → Cu 0 ECu = 0.518V + / Cu → Cu 2 + + Cu 2Cu +  O

E 0cell = E cathode - E anode = 0.518 - 0.153 = 0.365 V O

O

0.059 log K c 1 0.059 0.365 = log K1 1 ∴ K c = 1.50 × 106 Also E 0 = O

0.059   log  H +   - 0.188 - E θH+ / H + 2 1   0.0591 log  H +  = 0+0+ 1 ∴  H +  = 6.51 × 10 -4 M O

+ 3

Now C6 H 4 NH + H 2 O  C6 H 5 NH 2 + H 3 O 1 ∴  H +  = C.h or 6.51 × 10 -4 = ×h 32 ∴ h = 2.08 × 10 -2

+

1 × 2.08 × 10 -2 32

(

)

2

= 1.325 × 10 -5

E θNi2+ / Ni = −0.236V O

E θH+ / H = 0

Cathode 2H + + 2e −  → H2 E

θO cell

O

2

= E cathode − E enode = 0 − ( −0.236 ) = 0.236V 2

E cell = E

Also K H = Ch 2 KH =

29. Anode Ni  → Ni 2 + + 2e −

θO cell

0.059 log10 + 2

0 = 0.236 +

 H +   Ni 2 + 

2 0.059 log10  H +  2 2

27. E cell = E 0Ag+ / Ag + E 0Ag+ / Ag O

or 0.0860 =

O

 Ag +  0.059 R.H.S log + 1  Ag +  L.H.S

+ 0.059 0.059  Ag  R.H.S log 1 1  Ag +  L.H.S

Also  Ag +  L.H.S can be derived as

Or − log  H +  = 4 ∴ pH = 4

→ NO2 + H 2 O 30. NO3- + 2 H + + e - 

E 0 = 0.790V O

NO3- + 7 H + + 6e -  → NH 2 OH + 2 H 2 O E 0 = 0.731V O

12.90

Electrochemistry

Since the two half cell reactions have same E values E NO- / NO 2 = E NO- / NH2 OH 3

3

2

E θNO- / NO 2 + O

3

 H +   NO3-  0.0591 log 1 [ NO2 ]

=E

∴ Λ 0eq Agcl = Λ 0Ag+ + Λ 0Cl- = 54.3 + 65.5 = 119.8 s cm 2 eq -1

( Λ

 H +   NO3-  0.0591 log + 6 [ NH 2 OH ]

2 7 0.0591 0.0591 log  H +  = 0.731 + log  H +  1 6 0.790 + 0.118log  H +  = 0.731 + 0.0688log  H + 

0.0591 = 1.1992 0.0492 pH = 1.1992

31. Let κ1 and κ2 be the specific conductance of the solutions A and B respectively and the constant of the cell be x ∴ For solution A: sp. Conductance = Conductance × Cell constant 1 × x.......................................(1) 50 For solution B: sp conductance 1 k2 = × x............................(2) 100 When equal volumes of A and B are mixed both the solutions get doubly diluted; hence their individual contribution towards the sp. Conductance of the

k1 =

k1 k and 2 respectively and the sp. 2 2 1 conductance of the mixture will be k1 + k 2 2 1 1 ∴ For the mixture k1 + k 2 = × x (R is resistances 2 R mixture will be

)

)

of the mixture)……………….(3) From equations (1) (2) and (3) we get R = 66.67 ohms Λ eq

15.8 32. 1. Degree of dissociation, ∝ a= 0 = = 0.04514 Λ eq 350 2. For monobasic acid

HA  H + + A − K=

Cα 2 2 = Cα 2 = 0.05 × ( 0.04514 ) (1 − α ) = 1.019 × 10−4

(Since a = 0.04514 and 1–a =1 neglecting a concentration c in mol litre–1. Since HA is monobasic acid its normality = molarity).

1000 N

0 for sparingly soluble salt = Λ eq

)

1000 N

1.12 × 10-3 eq liter -1 119.8 1.12 × 10-3 = × 143.5g litre -1 = 1.34 × 10 -3 g litre -1 119.8

∴N =

- log  H +  =

(

eq

∴119.8 = 1.12 × 10-6 ×

Or 0.790 +

(

0 Λ Cl - = 65.5

We know that Λ eq = k × V = k ×

7

θO NO3- / NH 2 OH

33. Λ 0Ag+ = 54.3

34. Solubility of AgBr in presence of 10–7 mole of AgNO3 solution. K sp of AgBr =  Ag +   Br −  12 × 10−14

= S + 10−7  × S

S = 3 × 10−7 M ∴ Ag +  = 3 × 10−7 + 10−7  = 4 × 10−7 M  Br −  = 3 × 10−7 M  NO3−  = 1× 10−7 M Λ × M 1000 2 Λ in sm mol−1 = Λ × 104 s cm 2 mol−1 ) (

∵K =

∴κ Ag+ κ Br − κ NO− 3

∴ κ total

6 × 10−3 × 10−4 × 4 × 10−7 = 24 × 10−9 Scm −1 1000 8 × 10−3 × 104 × 3 × 10−7 = 24 × 10−9 Scm −1 = 1000 7 × 10−3 × 104 × 10−7 = = 7 × 10−9 Scm −1 1000 = κ Ag+ + κ Br − + κ NO− =

3

−9

= 24 × 10 + 24 × 10−9 + 7 × 10−9 = 55 × 10−9 s cm −1 = 55 × 10−7 Sm −1 In the unit of 1× 10−7 S m −1 , the conductivity is 55

Electrochemistry

1000 M 1000 138 = κ × ∴ κ = 2.76 × 10−3 0.2 l Also κ KCl = c × a l  1 1  1 2.76 × 10−3 = ×  C = =  R 85  85 a  l c onstant = 0.2346 cm −1 ∴ Cell a I = 1.087 × 10−4 s C Pool water = 9.2 × 103 I C NaCl ( aq ) = = 1.316 × 10−4 s 7.6 × 103 ∴ C NaCl = C NaCl ( aq ) − C pool water

1000 N 1000 -6 121.6 = 1.24 × 10 × M 1.24 × 10-3 ∴ M = N = 121.6 = mol litre -1 121.6 1.24 × 10-3 = × 143.5 gm litre -1 121.6 Solubility of AgCl = 1.463 × 10 -3 g litre -1

35. Λ KCl = κ ×

= [1.316 − 1.087 ] × 10 = 2.29 × 10 −4

Therefore Λ M = k ×

37. In the saturated solution the complex is assumed to be 100% ionized For Co 2  Fe ( CN )6  ; Λ 0M = 2 × Λ 0Co2+ + Λ Fe (CN)46

= 2 × 86 + 444 = 616s cm mol 2

2

= 1.65 × 10 s cm

l = 2.29 × 10−5 × 0.2346 a = 5.37 × 10−6 s cm −1

∴ κ NaCl = c ×

Λ 0M = k ×

Cond.

=

116.5 = Λ Ag+ + Λ

Λ 0M NaCl

=

110.3 = Λ 0Na + + Λ 0Cl- ....... ( 2)

Λ 0M NaNO3

=

105.2 = Λ 0Na + + Λ 0NO- ...... (3)

∴Λ

616 =

1000 CM

1.65 × 10 -6 × 1000 CM

∴ CM = 2.68 × 10 -6 mol litre -1 2+  ∴ Co 2  Fe ( CN )6    2Co +  Fe ( CN )6    2s s

(

K sp = 4s3 = 4 × 2.68 × 10 -6

0 NO3-

N=

3

= 116.5 + 110.3 + 105.2 = 121.65 cm mol

3

= 7.699 × 10 -17

wt in grams 500 = = 8.547 Eq.wt 58.5

If 8.547 Equivalents of NaCl are present in v litres

.......(1)

2

)

4-

38. The volume of the vessel is assumed to be V litres Eq. of NaCl =

36. For monovalent electrolytes Equivalent Cond. = Molar

0 M Agcl

-1

if molarity is C M , Then

500 8.55    M = 58.5 × V = V  (l) (l)   1000  V is volume of water ≈ volume  ∴Λ 0NaCl = κ ×  M  of solution       V 1000 × (l) 126.5 = 5.37 × 10−6 × 8.55 ∴ V( l ) = 2.01× 105 litre

Λ

-1

Also k Complex = k Sol - k H O = 2.06 × 10 -6 - 0.41 × 10 -6

−5

-6

0 M AgNO3

12.91

-1

For sparingly soluble salt Λ M = Λ 0M ∴ Λ M AgCl = 121.6 S cm 2 mol -1 Now k AgCl + water = 2.40 × 10 -6 S cm -1 k water = 1.16 × 10 -6 S mol -1 k AgCl = ( 2.40 - 1.16) × 10 -6 S cm -1 = 1.24 × 10 -6 S cm -1

8.547 V

The conductivity of the NaCl solution (only due to presence of Na+ and Cl– ions) k V = 3.10 × 10 -5 - 2.56 × 10 -5 k V = 0.54 × 10 -5 s cm -1 1000 × V 8.547 As the vessel is large in size, the resulting solution may be supposed to be diluted. Λ M NaCl = 0.54 × 10 -5 ×

12.92

Electrochemistry 0 42. U H+ =

1000 × V = 149.9 8.547 V = 2372.5 × 102 litre = 2372.5 × 102 litre

0.54 × 10 -5 ×

50.11 F 96500 = 5.20 × 10−4 cm 2 volt −1 s −1 Ionic velocity c( m/s ) U0 = pot..diff v( olt ) /distance between the electrodes ( cm )

0.50 1 = 1.50 3 Equivalent conductivity ∴ Conductivity = Volume ( cc ) containing 1 eq. 97.1 × 0.1 = 0.00971s cm -1 1000 Conductvity ∴ Conductance = cell constant 0.00971 s = 0.02913 s = 1 3 1 ohm ∴ Resistance = 0.02913 ∴ Current in ampere

40. Λ

1

Velocity of Na +

0.02913 =Λ

+Λ

Now Λ 0eq = Λ eq for water Λ eq = 548.3 s cm 2 eq -1 1000 N 5.51 × 10 -8 × 1000 = 548.3 N ∴ N = 1.005 × 10 -7 eq litre -1 also Λ eq = k V ×

or  H +  = OH -  = 1.005 × 10 -7 for water K w =  H +  OH -  1.005 × 10 -7  K w = 1.01 × 10 -14 41. From Kohlrausch's law we have Λ

=Λ

0 NH 4+

+Λ

0 ClO4−

(

Λ 0NH

∴ Λ 0NH

=

0 Λ 0NH + + Λ OH - = 74 + 198 = 272

4 OH

4 Cl

2

0 - Λ Cl - = 150 - 76 = 74

Λc Λ0

9.6 = 0.0353 272

45. Charge in coulombs = current in amperes × time in seconds

= 0.36 × 35.3 × 60 0.36 × 35.3 × 60 = F = 0.0079 F 96500 ∴ 1 mole of electric charge (1 F) produces 1 equivalent of the substance ∴ eq. of Ni which should be deposited for 100% current efficiency = 0.079 0.0079 × 58.7 = g 2 = 0.2318g ( eq. wt of Ni = 58.7 / 2) But the current efficiency is 60%

= 0.1391g

4

0 = F U 0NH+ + U ClO − 4

=

4

Actual amount of Ni deposited = 0.2318 ×

0 = FU 0NH+ + FU ClO − 4

∴ Λ 0NH +

=

0 OH -

= 349.8 + 198.5 = 548.3s cm 2 eq -1

0 NH 4 ClO4

0 Λ 0NH + + Λ Cl -

Degree of dissociation =

= 0.1456 ampere. 0 H+

2 5 = 1.45 × 10−3 cm s −1 2 = 5.20 × 10−4 × 5 = 2.08 × 10−4 cm s −1

=

4 Cl

4

Resistance ( ohm )

0 eq H 2 O

Λ 0NH

44.

Potential difference ( by ohms low ) 5

=

∴ Velocityy of H + = 3.62 × 10−3 ×

=

=

Λ 0Na +

U 0Na + =

39. Cell constant =

=

Λ 0H+

349.8 = F 96500 = 3.62 × 10−3 cm 2 volt −1 s −1

∴Λ M NaCl = Λ 0M NaCl

4

)

= 96500 ( 6.6 × 10−4 + 5.7 × 10−4 ) = 118.67 mho cm 2

60 100

Electrochemistry

46. No. of eq. of nitrobenzene converted to aniline =

Wt in g 12.3 = = 0.6 123 / 6 eq. wt of C6 H 2 NO 2

C6 H 5 NO 2 → C6 H 5 NH 2 ; Change in O.N = 6 M.wt 123 = Change in ON 6 ∴ mole of electricity for 100% Current efficiency = 0.6F

∴ eq wt. of C6 H 5 NO 2 =

But the current efficiency = 50% ∴ mole of electricity used = 0.6 × 2 = 1.2 F = 1.2 × 965600 Coulombs = 115800 Coulombs The energy consumed = electricity in coulomb × Pot. drop in volt (1 joule = 1 volt × 1 coulombs) = 115800 × 3 J = 347400 J = 347.40 KJ wattage 100 = 47. Current in amperes = voltage 100 Charge in coulombs = Current in amperes × time in seconds 100 × 10 × 60 × 60 = 32727.27 110 32727.27 ∴ Charge in faraday = F = 0.34 F 96500 ∴ mole of electricity = 0.34 F =

Initial mole of Zn (Or ZnSO4) = 0.16 × 0.3 = 0.048 Mole of Zn remained undeposited = 0.048 – 0.0018234 = 0.0461766 0.0461766 = 0.154 M . Molarity after electrolysis = 0.3 49. Equivalent of Cu2+ lost during electrolysis 2 ×10-3 ×16 × 60 I× t = = =1.989 ×10-5 96500 96500 1.989 ×10-5 Or mole of Cu2+ lost during electrolysis = 2 This value is 50% of the initial concentration of solution Thus initial mole of CuSO4 =

2 × 1.989 ×10-5 = 1.989 × 10-5 2

Thus initial concentration of CuSO4 =

1.989 ×10-5 ×1000 250

[CuSO4] = 7.95 × 10–5 M E.I.T 96500 87 × I × 24 × 60 × 60 1000 = 2 × 96500 I = 25.6 ampere. 25.6 Current efficiency = × 100 = 94.8% 27 Reactions Anode : Mn 2 + → Mn 4 + + 2e -

50. W =

Cathode : 2H + + 2e - → H 2

Amount of Cd deposited = 0.34 eq = (0.34 × eq.wt) g = 0.34 ×

112.4 = 19.11 g 2

51. Eq. wt of Cr At. wt. 52 = 6 no. of e lost or gained by 1 molecule or Cr -

(eq wt of cadmium = 112.4/2) (a)  96500 coulombs deposit 1.70 × 230 = 0.004052 F 96500 Eq. of Zn to be deposited for 100% current efficiency = 0.004052. Or mole of Zn to be deposited for 100% current efficiency= 0.004052/2 = 0.002026 Or mole of Zn to be deposited for 90% current efficiency = 0.9 ×0.002026 = 0.0018234

48. Mole of electric charge =

12.93

=

52 g of Cr 6

52 24000 × g Cr 6 96500 = 2.1554 g of Cr

24000 coulomb deposit =

(b) Also given WCr = 1.5 g, I = 12.5 ampere, t = ?, ECr = 52/6 E.I .T 96500 52 × 12.5 × t 1.5 = 6 × 96500 t = 1336.15 second w=

12.94

Electrochemistry

Cu 2 + + 2e - → Cu

2Cl -  → Cl2 + 2e -

52. (a) Anode

2Ag → 2Ag + + 2e -

Cathode 2H 2 O + 2e -  → 2OH - + H 2 (b) w =

9.65 × 60 × 60 eq 96500 = 0.36eq = 0.36 mole Ag + ions formed =

E.I.T 96500

 WCl = 103 g, E Cl = 35.5 2

Thus [Ag + ]left = 1 + 0.36 = 1.36 M

2

35.5 × 25 × 62 × t 100 × 96500  Current efficiency = 62%

[Cu 2 + ]right = 1 - 0.18 = 0.82 M

103 =

Thus new cell is Cu / Cu 2 + || Ag + / Ag 0.82 M

t = 175374.83sec 25 × 62 ∴i= ampere 100 t = 48.71hr

0 Thus E cell = E cell + O

103 formed = = 28.17 35.5

Mole of OH– formed = 28.17 (∴ monovalent) mole 28.17 = = 1.408 mol litre vol in litre 20

107.8 × 8.46 × 8 × 60 × 60 96500 = 272.18 g . 272.18 = 25.92 mL Volume of Ag = 10.5 25.92 ∴ Surface area = = 1.02 × 104 cm 2 0.00254

53. WAg =

EIT 96500

O

108 × 3× t 96500 0.42 × 96500 or t = = 125.09 second 108 × 3

So 0.42 =

O

56. 0.4 g of Cu2+ =

0.4 = 0.0126 gm equivalent 31.75

At the same time, the oxygen deposited at anode

Thus emf of cell Cu/Cu2+ || Ag+ |Ag will be [ Ag + ]2 0.059 0 log10 - ECu 2+ / cu 2 [Cu 2 + ] O

∴ [ Ag + ] = 1M and [Cu 2 + ] = 1M O

108 g. 96500

If the current passed for t seconds We know that W = Z × I × t

2 Ag + + 2e -  → 2 Ag

∴ Ecell = E 0 +

55. Mass of silver to be deposited = Volume × density = Area × thickness × density Area = 80 cm2; thickness = 0.0005 cm and density = 10.5 g/cm3 Mass of silver to be deposited = 80 × 0.0005 × 10.5 = 0.42 g Applying to silver E = 2 × 96500 Z=

Cu  → Cu 2 + + 2e -

O

Thus E cell increase by 0.01 V

=

0 0 54. This cell will not work because since ECu < E Ag 2+ + / Cu / Ag The equation for electrochemical cell will be

0 Ecell = E Ag + + / Ag

(1.36) 2 0.059 log10 2 0.82

= E θcell + 0.01 Volt

(c) Eq. of OH– formed = eq. of Cl2 formed = eq. of Cl2

∴ [OH - ] =

1.36 M

0.059 1 log10 2 1

0 Ecell = Ecell O

After the passage of 9.65 ampere for 1 hr, i.e., 9.65 × 60 × 60 coulomb charge, during which the cell reaction is reversed thus, Cu2+ are discharged from solution and Ag metal passes to ionic state. The reaction during passage of current are

= 0.0126 g - equivalent 8 = × 0.0126 = 0.00315 g - mol 32 After the complete deposition of copper , the electrolysis will discharge hydrogen at cathode and oxygen at anode. The amount of charge passed = 1.2 × 7× 60 = 504 coulomb

Electrochemistry

So, oxygen liberated 1 = × 504 = 0.00523 g equivalent 96500 8 × 0.00523 = 0.001307 g mole 32 Hydrogen liberated = 0.00523 gm equivvalent 1 = × 0.00523 = 0.00261 g mole 2 Total gases evolved = (0.00315 + 0.001307 + 0.00261) g mole = 0.007067 mole volume of gases at NTP = 22400 × 0.007067 mL =

57. The increase in weight of cathode is due to the deposition of copper only. So the mass of impurities will the difference in the weight of anode dissolved and the increase in weight at cathode. Mass of impurities = 22.260- 22.011= 0.249 g At anode copper and iron are oxidized; the gold and silver collect below anode in the form of anode mud M → M 2 + + 2e - ( M = Cu or Fe) 140 × 482.5 = 0.35 No. of moles of metal oxidized = 2 × 96500 No. of moles of copper =

22.011 = 0.3466 63.5

No. of moles of iron = 0.35-0.3466 = 0.0034 58. The reaction that taking place during electrolysis is 2 SnCl2  → SnCl4 + Sn 2 × 190 g

59. According to the given equation Eq. Wt of KClO3 =

261g 119 g

119 g of Sn is deposited by the decomposition of 380 g of SnCl2. So 0.119 g of Sn is deposited by the 380 × 0.119 = 0.380 g of SnCl2 decomposition of 119 Remaining amount of SnCl2 = (19–0.380) = 18.62 g 380 g of SnCl2 produce = 261g of SnCl4 So 0.380 g of SnCl2 produce = 261 × 0.380 = 0.261 g of SnCl4 380 ∴ The ratio of SnCl 2 SnCl 4 =

18.62 i.e., 71.34 :1 0.261

M .wt 122.5 = = 20.4 6 6

∴ equivalent of KClO3 to be produced =

10 20.4

Since current efficiency is 60%; hence 1 faraday (i.e., 96500 coulombs) shall produce 0.6 equivalent instead of 1 equivalent 10 ∴ Production of eq. of KClO3 shall require 20.4 96500 10 × i.e., 78839.86 coulombs 0.6 20.4 ∴ time required =

= 158.3 mL

12.95

no of coulombs Current in amp

78839.86 = 39419.93 seconds 2 ≈ 39420 second 39420 hours = 60 × 60 = 10.95 hours =

60. Wt of Cu to be coated = Volume of Cu deposited × density = 10 × 10 × 10–2 × 8.94 = 8.94 g 8.94 63.5 (eq. Wt of Cu = ∴ eq of Cu = = 31.8) 31.8 2 = 0.281 ∴ mole of electric charge = 0.281 faraday = 0.281 × 96500 Coulombs = 27116.5 coulombs 61. Before electrolysis Volume of solution = 1 litre = 1000 mL Wt of solution = 1000 × 1.261= 1261g (∴ w = v × d) ∴ Wt of H 2 SO4 =

34.6 × 1261 = 436.306 g 100

After electrolysis Now during reaction wt of water formed = xg x 18 x Mole of H2SO4 used = 18 (∴ mole ratios of H2SO4 : H2O = 1:1) Mole of water formed =

Wt of H2SO4 left = (436.306-5.44x)g 436.306 - 5.44 x 27 = New wt of new solution = 1261 + x - 5.44 x 100 x = 22.59 g

12.96

∴

Electrochemistry

22.59 mole of H2O are formed 18

∴ Mole of H2O = Eq of H2O (∴ 2H2O consume 2 electrons) Now 1 mole of H2O formed by passage of 1 Faraday ∴ 22.59 mole of H2O formed by the passage of 18 22.59 faraday = 1.255 faraday 18 62. The overall battery reaction is 2 PbSO4 + 2 H 2 O → Pb + PbO2 + SO2H 4 2SO4 ∴ Two moles of electrons are involved in the production two moles of H2SO4 ∴ equivalent wt of H2SO4 = M. Wt of H2SO4 = 98 Before electrolysis  100 g of H2SO4 contains 15.7g of H2SO4 100 15.7 mL of H2SO4 contains eq of H2SO4 1.11 98 ∴ 2000 mL H2SO4 contains 15.7 × 1.11 × 2000 = 98 100 3.556 eq) Similarly, we get Number of equivalent of H2SO4 after electrolysis 36.9 1.28 × × 2000 = 9.6387 = 98 100 ∴ No. of eq of H2SO4 produced = (9.6387–3.5560) = 6.0827 ∴ moles of electrical charge used = 6.0827 faradays Average current used = charge in coulombs time in seconds 6.0827 × 96500 = amp = 1.629 amp 100 × 60 × 60 Or

multiple Choice Questions with only one answer level I 33. E = 0.76 + 0.4 = 1.16 O

∴ Zn + Cd +2  → Zn +2 + Cd ∴E = 1.16 -

E = 1.13 1 → H2 40. H + + e  2 0.059 1 E = E0 log +1 1 H ∴ E = -0.059V O

55. Sn + 2 Ag +  → Sn +2 + 2 Ag E = E0 + O

Pb 200 207

+ mole

PbO 2 + 200

mole

2H 2SO 4  → 2PbSO 4 + 2H 2 O excess

239

As mole of PbO2 is less than that of Pb, PbO2 is the limiting reactant which shall be totally consumed. Number of faraday delivered by the battery = no. of eq of PbO2 lost 200 = ×2 239 400 ∴ charge = ×96500 coulom 239 400×96500 1 = × hr 239×10 60×60 = 4.486 hr

0.059 ( Ag + ) 2 log 2 Sn 2 +

→ Cu 60. Cu +2 + 2e  Cu

+2

∆G1 = -2(0.337)F

+ e  → Cu

+

Cu + + e  → Cu

∆G 2 = -1(0.153)F ∆G = ∆G1 - ∆G 2 = -0.521F

∴ ∆G = - nFE

O

63. Discharging of battery takes place through the reaction

0.059 0.1 log 2 0.01

∴ E = 0.521V

70. H + + e − →

1 H2 2 1/ 2

E=0−

PH 0.059 log +2 1 H

71. Cu +2 + 2e - → Cu 0.059 E = Eθ + log Cu 2 + 2 O

72. H + + e − →

1 H2 2 1/ 2

PH 0.059 log +2 1 H θ E =E =0 E = Eθ − O

O

∴ PH1/22 = (H + )

Electrochemistry

1 H2 2 0.059 1 E = 0− log 1 10 ( −7 )

73. H + + e − →

E = − 0.413V 65.5 9.65 × 60 × 60 × of = 11.70 2 96500 63.5 9.65 × 60 × 60 × = 11.43 g Gain in mass at Cu = 2 96500

75. Loss

+

76. H 2 | H || H | H 2

0.05a log K sp 1 ∴ K sp = 1.5 × 10 -10 M 2

1. E 0Cl|AgCl|Ag = E 0Ag+ / Ag + O

O

Agcl

Zn= g → H 2 O; ∆G1 = 237.2 KJ 2. H 2 + 1 2 O 2  H 2 O  → H + + OH - ; ∆G 2 = 80 KJ Cell reaction is

∴ ∆G = - nFE ∴ E = 0.4 V

0.059 10 -6 log 1 x + -4 ∴ [ H ] = 10 E= 0-

+   →A 4. A + e ← 

77. Cu 2 + + 2e - → Cu 0.059 1 E = 0.34 log = 0.36V 2 5

0.22 - 0.4 = - 0.059 log

1 H+

92. H2/H+|| calomel electrode 0.059 1 log + 1 H E = 0.3982 V

E = 0.25 +

93. Ni +2H + → Ni 2+ + H 2 0.059 1 0 = 0.236 log + 2 2 (H ) pH=4

κ1000 = 2 × 10−4 eq λ S = 10−4 mol lit −1 S=

∴ K sp =1× 10−8 M 2

∆G1 = −1FE10

  → A ∆G 2 = −2FE 02 A +2 + 2e − ←    → 2A + ∆G = ∆G 2 − 2∆G1 A +2 + A ←  = −F  2E 02 − 2E10 

+2

87. Cd + 2 H → Cd + H 2 0.059 1 E = 0.4 log 2 2 ( x)

104.

level II

∴ ∆G = ∆G1 + 2 ∆G 2 = -77 KJ

x

+

multiple Choice Questions with only one answer

H 2 + 1 2 O 2 + H 2 O  → 2H + + 2OH -

+

10 -6

12.97

∴ E 0 = 2 ( E 02 − E10 )

E0 =

K 0.059 log 22 2 K1

∴log

K 2 2(E 02 − E10 ) = 0.059 K12

5. O 2 + 4H + + 4e − → 2H 2 O E 0 = 1.23V 2H 2  → 4H + + 4e −

E0 = 0

2H 2 + O 2  → 2H 2 O

E 0 = 1.223V

∆G 0 = −474.7 KJ 6. Dissociation constant Ka = Ca2 1.6 × 10–5 = 0.01a2 a = 0.04 λ Degree of d issociation ∝ = V λ∝ ∴λ V = 0.04×380×10-4 =15.2 × 10 -4 sm 2 mol -1 Specific conductance K=1.52 × 10 -2 sm 2 mol -1 7. 2H + + 2e −  → H2 E= 0−

0.059 log PH2 2

12.98

Electrochemistry 0 18. E cell = 0.22 - 0.08 = 0.14 0.059 log K eq E0 = 1 log K eq = 2.37 O

8. After adding AgNO3 to saturated AgCl the conc Ag+ =4 × 10-7 m -

O

-7

Br = 3 × 10 m NO3- = 1 × 10 -7 M k 1000 Equivalent conductance = in λ M ∴K Ag+ = 16 × 10 -7 sm -1 K Br - = 18 × 10 -7 sm -1

cgs

19. E = E 0 O

K No- = 5 × 10 -7 sm -1

O

-7

∴ Specific conductance =39 × 10 sm

-1

Cu +2 + 2e −  → Cu 0.05a log 0.1 E = Eθ + 2

21. ΔG Values are additive

O

22.

→ Mn +2 + 4H 2 O 11. MnO −4 + 8H + + 5e −  so chargee required is = 5 × 96500 × 0.01 = 4825coulombs 12. Assume solubility of AgCl in pure water is S1 solubility of AgBr in pure water is S2

24. λv =

-

∂E − E] ∂t

k 1000 S

1.2 × 10 -6 × 1000 5 -5 ∴ S = 10 , K sp = 10 -10

The cellis Ag |Ag || Cl | Hg 2 Cl2 | Hg 0.059 log S1 1 0.059 E2 = E0 + log S2 1

∆H = nF[T

∂E   − 222 × 103 = 2 × 96500 300 − 1.015 ∂T   ∂E = −4.5 × 10−4 Vol / deg ∂T

No. of moles of MnO 4− is = 0.01

+

0.059 1 log + 4 4 (H )

E 0 = 123 + 0.059 = 1.289 The RPValue at pH = 10 is E = 1.289 - 0.059 ( pH ) = 0.699 V

3

9.

K eq = 234 units

E1 = E 0 + O

O

S E1 - E 2 = 0.177 = 0.059 log 1 S2 ∴

25.

S1 = 103 S2

l a 0.5 0.01 = c 1.5 c = 0.03 mho ∴Current = 0.15 amp

14. Λ BaSO4 = 64 + 80 = 144 ohm -1 cm 2 mol -1

∴k = c

solubility of BaSO 4 = K sp = 10 -5 M k1000 M ∴k = 144 × 10 -8 ohm -1cm -1 Λ=

15. ΔG = –4FE

26.

→ Zn +2 + 2Cl 17. Zn + Cl2  0.059 log ( Zn 2 + )(Cl - ) 2 2 0.059 E = 2.12 log (0.01)(0.01) 2 = 2.3V 2 E = E0 O

k 1000 N k 1000 100 = 0.1 ∴ k = 0.01 mho cm -1

λv =

87.5 12.5 = 112 / n 8 So valency of metal = 2 112 0.965 × 60 Weight of metal W = × = 0.0336 gm 2 96500

Electrochemistry

27. 200 = 100 Q Q = 2 Coulombs EQ ∴W = = 0.655 gm F

36. 0.222 = 0.7995 +

Λ = 148.2 + 50.1 + 73.5 = 271.8ohm -1cm 2 mol -1 Equivalent Conductance 271.8 = = 135.9 ohm -1cm 2 eq -1 2

O

38. E 0OH - |Mn(OH)

= E 0 Mn 2+ |Mn + O

O

O

2 |Mn

0.059 log K sp 2

→ 5CO2 + 6 H 2 O 39. C5 H12 + 8O2 

RT m H + × mCl− E−E + ln F mx− θO

∆G = 6(-237.2) + 5(-394.4) - (-8.2) ∴∆G = -3387 KJ

1 RT m Hx × mCl− RT = ln In F mx− F Ka

O

0.059 log K sp 1

37. Molar Conductance

+ − 29. H 2 | H || Cl || AgCl | Ag RT E = Eθ − ln m H + .mCl− F RT K a .m H + E = Eθ − ln m Cl − F mx−

E − Eθ +

12.99

1 1 RT = 0.2814 ln = 10.9 ln Ka Ka F

∴∆G = -nFE = -3387 × 103 = -32 × 96500 × E ∴E = 1.0968V 41. The half cell reaction is Ag + Cl → AgCl + e 0.059 log Cl ∴ E = Eθ + 1 O

0.059 1 log + 1 H

30. E = E 0 + O

51.

0.67 = E 0 + 0.059(6.5) O

E 0 = 0.28 V O

0.28 = E 0Cl- Hg O

∴

+ E0H O

2 Cl2

O E 0Cl- Hg Cl Hg 2 2

Hg

2 /H

+

= 0.28 V

31.

2 Hg + 2Cl -  → Hg 2 Cl2 + 2e

32.

0.059 log Cl 1 CH 3 COOH = CH 3COO - + H + E = E0 + O

∴ Ka = ∴ H+ =

(CH 3COO - )( H + ) CH 3COOH K a [CH 3COOH ]

Zn + Cu +2 → Zn 2 + + Cu Initial 1 Conc Final a 1 Conc 0.059 1 E0 = log 2 a 0.059 1 1.18 = log 2 a 1 log +2 = 40 Cu O

54. Ag + + e → Ag E 0 Ag + / Ag O

Ag ( NH 3 ) 2 + + e  → Ag + 2 NH 3

[CH 3COO - ]

Ag ( NH 3 ) 2+  Ag + + 2 NH 3

1 H 2  → H + + e2 y 0.059 E1 = 0 log K a x 1 Ka x 0.059 E2 = 0 log y 1 1 E1 + E2 = 0.059 × 2 log Ka

0.059 1 0 ∴ E = -078V

Half Cell Reaction is

∴ pK a =

E1 + E2 (0.059)2

E0 = O

O

° -0.78V = -0.8 + ENH O

° ENH O

+ 3 / Ag /( NH 3 ) 2 / Ag

+ 3 / Ag /( NH 3 ) 2 / Ag

= 0.019 V

° E Ag ( NH O

+ 3 )2

/ Ag

12.100 Electrochemistry

56. The Galvanic cell is Cu|Cu+2|Ag+|Ag| When current passed through the cell it acts as electrolytic cell ∴ Silver electrode dissolved in electrolyte. Cu+2 deposit on copper electrode. 9.65 × 4000 eq Ag + = eq Cu +2 = = 0.4 96500 ∴Cu Cu +2 Ag + Ag

68. C6 H 5 NO2 + 6 H + + 6e -  → C6 H 5 NH 2 + 2 H 2 O moles of C6H5NO2= 0.1 Eq of C6H5NO2 = 0.6 no. of Faradays = 0.6 Current efficiency = 0.5 So actual no. of Faradays = 1.2

0.8 1.4 0.059 0.8 log E = E0 2 (1.4) 2 O

58. ∆H = nF[T

δE - E] δT

69. Eq H2 = eq O2 + eq H2S2O8 → Cl2 + 2e.........At LHS 71. 2Cl - 

→ Mn +2 + 4H 2 O 59. MnO 4- + 8H + + 5e -  E = E0 O

67. Initial wt of H2SO4 solution = (1.3)(2000) = 2600 g Initial wt of H2SO4 = 2600 × 0.4 = 1040 g Final wt of H2SO4 = 440 g Wt of H2SO4 consumed = 600 g moles of H2SO4 consumed = 6.1224 moles of H2SO4 consumed = eq of H2SO4 = no. of Faradays

Cl2 + 2e  → 2Cl - .........At RHS

0.059 1 log + 8 5 (H )

So P1 < P2 +2

60. When bismuth starts deposit then E value of Cu /Cu is 0.226x 0.059 1 log +2 ∴ 0.226 = 0.344 2 Cu Cu +2 = 10-4 M

→ H 2 + 2OH - E θ = -0.8277 V 73. 2H 2 O + 2e -  O

H 2  → 2H + + 2e -

Eθ = 0

2H 2 O  2H + + 2OH -

E θ = -0.8277 V

O

O

0.059 log K w 1 ∴ K w = 9.35 × 10 -15 Eθ = O

Ag

Ag

61. Conc at equilibrium

+

Ag

K sp Ag Br Br

-

+

Ag

K sp AgCl Cl

-

Ag + LHS 0.059 log E = Eθ 1 Ag + RHS O

∴

Ag + LHS Ag + RHS 5 × 10 -13 -

O

∆G 0 = - nFE 0 O

=1

∴ ∆G 0 = -130.79 - ( -109.56) = -21.23 O

=

1 × 10 -10

E 0 = -21.23 × 103 / 1 × 96500 = 0.22 V 0.059 E = 0.22 log (0.01) (0.01) = 0.456 V 1 O

-

Cl Br Br - 5 × 10 -13 1 = = -10 200 Cl 1 × 10

1 H 2 + AgCl  → Ag + H + + Cl 2 0.059 10 -14 log x = 1.048V ∴ E = 0.222 1 x

77. The cell reaction is

63. Eq H2 = eq O2 + eq H2S2O8 tant = 65. Cell constant

1 H 2 + AgCl → H + + Ag + Cl 2 ∴ ∆G 0 = -130.79 - ( -109.56) = -21.23

74. Cell reaction is

l =x a

Specific conductance of A =

x 50

Specific conductance of B =

x 100

x   x When equal volumes are mixed net K =  + /2 50 100   So net resistance R = 66.67 ohm

Electrochemistry 12.101

Passage IV

78. 2HSO −4  → S2 O8−2 + 2H + + 2e −

1. ΔG values are additive

Ct F C × 60 × 60 2= 96500 C = 53.611 amp Curreent efficiency = 0.75 So requried current = 71.5 amp eq =

0 0 2. E 0 = EMnO + EFe +2 / M +2 / Fe+3 O

O

4

HA +

3.

NaOH → NaA + H 2 O

In moles 30 F moles 10 +  HA  H + A

20

20 -

-

 20  Ka = (H + )    10 

79. Ksp of AgCl is greater than Ksp of AgBr So Cell is

K a 6.3 × 10 -6 = = 3.1 × 10-6 M 2 2 Fe | Fe +2 || H + | H 2

Ag AgBr AgCl Ag satd

O

H+ =

satd

Comprehensive type Questions

0.059 0.1 log 2 (3.1 × 10-6 ) 2 E = 0.145V E = 0.44 -

Passage I ∂∆G 0 ∂E 0 = nF ∂T ∂T O

1. ∆S0 = O

O

Passage V 1. 4 Al + 3O2 + 12 H + → 4 Al +3 + 6 H 2 O

 ∂E 0  2. ∆H 0 = nF  t - E0  t ∂   O

O

O

0 2. Ecell = E Al0 / Al +3 + EO0 , H + / H O = 2.89 xV O

O

O

2

∂E ∂T

0O

3. ∆S0 = nF O

3.

H 2O + e- →

0.965 × 60 = 6 × 10 -4 96500 Mole of Cu +2 deposit = 3 × 10 -4

1. Eq. of Cu +2 deposit =

∴ E0 = O

Initial moles of Cu +2 = 5 × 10 -3 Final moles of Cu +2 = 4.7 × 10 -3 conc of Cu

2. Equivalent of Cu ∴  H +  =

E0 = x O

E0 = 0 O

E0 = x O

RT RT ln  H +   OH -  = ln K w F F

Passage VI

= 0.094 M +2

1 H 2 + OH 2

1 H2 → H + + e2  H + + OH H 2 O 

Passage II

+2

2

deposit = Equivalent of H formed +

1. W =

56 0.12 × 24 × 60 × 60 × = 3 gm 2 96500

2. E 0 =

0.059 log K eq 4

-4

6 × 10 × 1000 = 0.012 M 50

3. Conc of SO4-2 remain unchanged in electrolysis

O

3.

2Fe + O 2 + 2H 2 O → 2Fe +2 + 2OH 2

Passage III 1. Oxidation reaction taken place anode ( - ) 0 2. Ecell = 0.2 + 0.4 = 0.6V O

2

Y +3  0.059 3. E = E + log 3 6  X +2  0O

(

(0.015) 10-5 0.059 E = 0.84 log 4 0.7 E = 1.19 V

)

2

12.102 Electrochemistry

Passage VII

Passage xI

1. Reduction potential value E 0Cl O

1. Zn | Zn +2 |H + | H 2

> E 0I / IO

2

/ Cl-

2

2. Reduction potential value of E 0Mn +3 / Mn +2 > E O0 O

3. Blue colour is Fe4  Fe (CN )6   

0.701 = 0.76 -

O

2

/ H+ / H2 O

0.059 0.1 log 2 1 + H+

(

)

2

H + = 0.0316M

3

mole of NaOH required is = 0.0316 wt NaOH = 1.264 gm

Passage VIII k 1000 = 221 s cm 2 mol -1 1. Molar conductance Λ = M 2. At 0.1M concentration k Na + + k OH - = 0.0221 s cm -1 When 0.1 M HCl is added NaOH converted to NaCl The concentration NaCl = 0.05 At 0.05 M concentration k Na + + k Cl- = 0.0056s cm -1 At 0.1 M concentration k Na + + k Cl- = ( 0.0056)2

2. After neutralization H+ concentration is 10–7 M Passage xII 1. E 0 = 2.37 + 0.8 = 3.17 V O

0.059 log K 2 log K = 107.45 E0 = O

2. Maximum work = nFE = 6 × 102 KJ

= 0.0112 s cm -1 On further addition 0.1 M HCl solution contains 0.1 M + 0.2 M H , Cl 3 3 0.1 M Na + specific conduction is 0.017 s cm -1 3 At 0.1 M concentration k Na + + 2k Cl- + k H+ = ( 0.017 ) 3 = 0.051s cm -

1. Ag + + e → Ag 0.059 log K sp = 0.269 V 1 0.059 log K sp = E θAg+ / Ag + 1

E = 0.8 + 2. E θI- |Ag I/ Ag O

3. E = 0.08 +

At 0.1 M HCl k HCl = 0.0398 molar conduction =

Passage xIII

0.0398 × 1000 = 398 s cm 2 mol -1 0.1

O

10-18 0.059 log = -0.203V 1 0.1

4. Oxidation of H2 at anode takes place Passage xIV

+

3. Molar conduction of H + OH

-

0.057 × 1000 = = 570Scm 2 mol -1 0.1

Passage x 1 1. H 2 + AgCl → H + + Cl - + Ag 2  ∂E  2. ∆S0 = nF    ∂T  O

 ∂E  3. ∆H 0 = nF T - E T ∂   O

1. When cell is charged Cu electrode dissolved in electrolyte 0 = 1.1V , ∆G = -nFE = -212.3KJ 2. Ecell O

3. eq of Cu +2 formed = eq of Zn +2 consumed = 0.18 Zn | Zn +2 ||Cu +2 | Cu 0.01 0.19 E = 1.1 4. wt of Zn =

0.059 0.01 log = 1.137V 2 0.19

65.5 0.48 × 10 × 60 × 60 × = 5.89 gm 2 96500

 ∂E  5. ∆H = nF T - E  ∂T 

Electrochemistry 12.103

Passage xV

Passage xVIII → Zn (CN )4 + 2 Au E 0 = 0.66V 1. 2 Au (CN )2 + Zn  -

1. E = 0.8 - 0.05 = 0.75V 0O

-

O

O

O

2 Ag (CN )2 + Zn  → Zn (CN )4 + 2 Ag E 0 = 0.95V

0.059 log K 2 ∴ln K = 58.38

E0 =

2. E = E 0 -

-2

-2

O

so zinc react with silver complex form 2 Ag (CN )2 +

Zn  → Zn (CN )4 + 2 Ag

-

2 0.059 log ( H + ) 2

-2

0.5 0.30

I moles F mole

-

0.66 -

-

-

∴  Au (CN )2  = ( 0.01) 3 = 0.03M  

3. When NH3 is added H+ concentration changes

-

 Ag (CN )  = 0.02 M 2 

Passage xVI 1. The reaction takes place at anode is oxidation 2. No of equivalents =

ct 1.6 × 200 × 60 = = 0.2 F 96500

2.

(

Na 2 HPO 4 + H +  → NaH 2 PO 4 + Na +

3. I moles F moles

1 0.8

0.2 -

  Au ( CN )  Au + + 2CN -  2  99 Kf = 2 1 CN -

1.2 0.8 = 2.2 - log 1.5 = 2.02 V

Passage xIx

(

2. K sp of Cu ( OH )2 = ( 0.1) 10 -9 E 0Cu (OH ) O

4

-

-2 4

4. H 2 PO + OH  → HPO + H 2 O ∴ pH = pKa 2 + log

1.2 = 2.37 V 0.8

2

/ Cu

Eθ = 0

0.059 log10 -19 = -0.22 V 2

= 0.34 +

2 -0.059 log (10-13 ) = 0.51V 2 0.059 -0.22 2

Passage xx

O

Eθ = X O

Cr2 O7-2 + 14 H + + 3Sn +2  → 3Sn +4 + 2 Cr +3 + 7 H 2 O

1. I moles F moles

3

10 -1

3

2

-1

-

2. E = 1.33 -

2. EFe+3 / Fe = -0.036V

3.

0 Fe | Fe +3 || Ag + | Ag Ecell = 0.036 + 0.8 = 0.836V

10

0.059 2 log   6 3

0.059 0.8 log = 0.799 3 (0.2) 3

→ 2Cr +3 + 7 H 2 O + 3Cl2 3. Cr2 O7-2 + 14 H + + 6Cl - 

2

-

-

-

3

2

-

14 2 10-1 ) = 1.187V ( 3

E 0 = 1.33 - 0.15 = 1.18 O

O

O

= 10 -19

O

0.059 log K W 1 X = -0.83 V

0 Ecell = 1.33 - 1.36 = -0.03V

2

Eθ = X

X=

E = 0.836 -

)

3. E = -0.22

Passage xVII 1 → H 2 + OH 1. H 2 O + e -  2 1 H 2  → H+ + e 2 +  H2O ↽ ⇀  H + OH

)

∴ CN - = 5 × 10 -14 M

1 1.2

pH = pKa 2 - log =

-

E = 1.18 -

 ( 0.2 )3 (1)2  0.059  = 1.16V log  3 6  ( 0.1) ( 0.2 ) 

12.104 Electrochemistry

1.

106 3 × 60 × 60 × 96500 n n=4

4. 2.977 =

Passage xxI NaCN + HCl  → HCN + NaCl ∴ pH = pka

2. pH in two compartment is same. So E = 0 1.25 × 643.3 × 60 = 0.5 96500 → HCN CN - + H + 

5.

K 1000 = 50 M 1 λ ∝ 400 = = =8 50 α λV

λ V=

+ 3. eq H formed =

I mole F mole

1 0.5

0.5 -

∴pH = pK a + log

1 1.5

0.5 = 4.2 1.5

Passage xxII O

O

O

= 96.2 cal / K

2. ∆G = - nF E O

→ NaOCl + NaCl + H 2O 8. Cl2 + 2 NaOH  eq Cl2 = mole of Cl2 gas = eq NaOCl = Mol of NaOCl

0.01 - 0059 log 0.1 n ∴n= 2 10. H 2 H + Cu +2 Cu

3. It is spontaneous

0.059 log ( H + ) 2 2 ∴ [ H + ] = 1 × 10 -2 M  B + + OH BOH  0.88 = 0.34 -

Integer type Questions 1. EI0- / AgI / Ag = 0.8 + O

9. n M + nX -  → M n X n + ne - ......LHS

∴E=

0O

= +0.229 V

0.06 log K sp 1

EI0- / AgI / Ag = 0.2

10 -5 × 0.1 = 5 × 10 -6 0.2 As equivalent point conc. of BCl = 0.05

O

Kb =

2. Ag + Cl - → AgCl + e 1 Cl2 → Cl - - e 2 1 Ag + Cl2 → AgCl E 0 = 1.36 - 0.33 = 1.03 V 2 ∴ ∆G = - nFE.

1 pOH = [ pK w - pK b - log C ] 2

O

work = nFE = 1 × 96500 × 1.03 = 105 J  ClO2- + ClO4- E 0 = 0.33 - 0.36 = -0.03V 3. 2ClO3-  O

0.06 log K 2 0.06 -0.03 = log K 2 ∴ Log K = -1 E0 = O

∴ K = 10 -1

0.059 log[ M x + ] PH 2 x / 2 x

M n X n + ne -  → nM + nX -

∆G = -31.7Kcal 0O

E0

O

Time = 1 × 107 sec

∆S0 = ( -22.2 ) 2 + ( 31.2 ) 3 - ( -10 ) 6 - ( 6.5 ) 2

0O

7. E = E 0 -

Wt of NaOCl = 7.45 × 104 g

1. ∆G 0 = ∆H 0 - T∆S0 O

→ C6 H 5 NH 2 + 2H 2 O 6. C6 H 5 NO 2 + 6H + + 6e − 

11. K1 K 2 =

( H + ) 2 (52 ) [H 2 S ]

(10 -3 ) 2 (5-2 ) 0.1 2 -16 ∴ 5 = 10

10 -21 =

E ° S 2- | H O

= EM° + / M + O

2 |S |M

0.06 log K sp 2

0.06 × ( -50) = -1.5 2 0.06 E = -1.5 log10 -16 = 2 x 2 = 1+

Electrochemistry 12.105

12. H 2 | H + || Zn +2 | Zn

( )

H+ 0.059 - 0.46 = -0.763 log 2 0.3

( ) H+

2

2

19. E = 0 20.

Cd |Cd +2 || H + |H 2

= 5.37 × 10 -11 0.3  H +  = 4 × 10-6 M

log

0 = 0.39 -

Cd +2 0.059 log 2 2 (10-12 )

Cd +2 = 10-11

∴ K 2 = 4 × 10-6 × 6.44 × 10-3 0.4 = 6.44 × 10-8 0.06 x log 2 1 ∴ x = 5 atm.

0.059 1 log + 1 H

K sp = (10-11 ) ( 0.01) = 10-15 2

13. - 0.021 = 0 -

14. E = 0 -

10 -6 0.059 log -4 1 10

Cl F 0.635 C × 60 × 60 = 63.5 / 2 96500 C = 0.536 amp Percentage error = 2%

21. eq =

15. E 0cell = 0.75 - 0.24 = 0.51 O

0.51 =

0.059 1 log n 5.15 × 10 -18

∴n = 2

16. pOH of HCOOK is 1 pOH = [14 - 8 - log 0.01] = 4 2 pH = 10

22. Equivalent of NaOH = 0.6 Equivalent of Cu deposit = 1 23. E = E 0 O

∴Ag +LHS = Ag +RHS

pH of NH 4 Cl solution is 1 pH = [14 - 8 - log 0.01] = 4 2 10 -10 0.059 log -4 ∴E = 01 10 ∴ E = 0.059 × 6 17.

π, = iCST 4.926 = 1( 0.1)( 0.0821)(300) i=2 i = 1+ 2a ∴a = 0.5 +

 M  = 0.1M 0.0591 1 E = E0 log 0.1 1 E = E 0 - 0.059 O

O

8 × 10-12 4 × 10-13 = CO3-2 Br 24. E 0 = O

0.059 log K 1

25. Equivalent weight of H2O = 9 26. w =

M Cl− 2F

27. pH in two compartments is same. So E value is Zero 28. K AgCl =1.86×10 -6 - 6×10 -8 =1.8×10 -6 ohm -1 cm -1 K1000 5 1.8×10-6 ×1000 137.2= S SolubilityS=1.3×10 -5 mol/lit Λ=

18. mole of salt = 2 dissolved Cl - ions = 4

Ag + 0.059 log +LHS 1 Ag RHS

29. E 0 = O

0.059 log K 2

12.106 Electrochemistry

30. K = K AgCl + K H

Previous years’ IIt Questions

2O

1. For M+X– → M+X, Ecell = 0.44–0.33 = 0.11 V is positive, hence reaction is spontaneous. 2. The salt used to make ‘salt-bridge’ must be such that the ionic mobility of cation and anion are of comparable order so that they can keep the anode and cathode half cells neutral at all times. 3. As we go down the group 1 (i.e., from Li+ to K+), the ionic radius increases, degree of salvation decreases and hence effective size decreases resulting in increase in ionic mobility. Hence equivalent conductance at infinite dilution increases in the same order. 4. MnO4- will oxidise Cl– ion according to the following equation 2 MnO4- + 16 H + + 10Cl -  → 2 Mn 2 + + 8 H 2 O + 5Cl2 ↑ O

5Cl K 1000 Λ= 5 31. MnO 4- + 8H + + 5e -  → Mn +2 + 4H 2 O E1 = E 0 O

0.059 1 log 5 H+

( )

8

0.059 (8x ) 5 0.059 E2 = E0 (8y) 5 Difference between x, y = 5 E1 = E 0 O

O

32. Pb + SO 4-2  → PbSO 4 + 2e -

E 0 = 0.356 V O

Pb +2 + 2e  → Pb

0.059 log Pb 2 + SO 4-2 2 = (2.5 × 10 -5 ) (SO 4-2 )

0.061 = 0.23 + 1.87 × 10 -6

E 0 = -0126 V O

(

)(

)

O

O

(SO ) = 0.0748 M -2 4

K2 =

The cell corresponding to this reaction is as follows: E cell = Pt , Cl2 (1 atm ) | Cl - || MnO4- , Mn 2 + , H + | Pt 1.51–1.40 = 0.11V E cell being +ve, ΔG will be –ve and hence the above reaction is feasible. MnO4- will not only oxidise Fe2+ ion but also Cl– ion simultaneously. So the quantitative estimation of aq Fe(NO3)2 cannot be done by this. 5. The electrons flow from cathode to anode through internal supply in an electrochemical cell.

(0.0748)(0.0748) = 10-2 0.6

O

6. Zn | Zn +2 ( 0.01M ) || Fe +2 ( 0.001 M ) | Fe

0.059 log K 33. E = 1 0O

Eθ

= E θcell

O

0.059 1 0 = 0.236 log 2 (1 + x )2

35. equivalent of OH - formed is

O

5 × 965 = 5 × 10 -2 96500

OH -  = 10 -3 ∴ pOH = 3

(Cr ) =quotient=8 (0.1) = × 10 (Cu ) (0.5) 3+ 2

2

2+ 3

3

-2

O

O

0.0591 10 -3 log -2 2 10 0.0591 = 0.2905 log 10-1 2 0.0591 = 0.2905 + 2 = 0.2905 + 0.02950 = 0.32 V RT Eθ = ln K 2F 0.0591 Eθ = log K 2 0.32 ⇒ K = 100.32 / 0.0295 log K = 0.0295 Eθ

∴ pH = 8

36. Reaction

= E θcell -

O

34. The cellis Ni | Ni +2 || H + | H 2

RT ln K nF RT ln K nF

E θcell

O

O

= 0.2905 -

Electrochemistry 12.107

7. ∆G θ = - nFE θ

O

O

( n = 2)

 Fe +2 + 2e - ↽ ⇀  Fe E

θO cell

=E

θO H+ / H2 O

-E

θO Fe+2 / Fe

23 - ( -0.44) = 1.67 V = 1.2

10. When AgNO3 is added slowly to KCl soln., part of KCl changes to AgCl (ppt). No change in electrical conductivity take place. After the equivalence point, when all the KCl is ppted the electrical conductivity increases. Passage I

96500 × 1.67 = -322.3 KJ 1000 ∆G θ = -322 KJ / mol ∆G θ = -2 × O

O

8. No. of Faraday's of electricity needed to liberate 0.01 Mole = 0.01 × 2F = 0.02F = 2 × 10 -2 × 96500 = 1930 coulomb A = 10 milliampere = 0.01 ampere Q = I × t Q 1930 t= = = 1.93 × 105 = 19.3 × 104 sec I 0.01

11. The electrochemical concentration cell given is + M (s ) | M + ( 0.05M ) || M aq (1M ) | M (s)

cell voltage of concentration cell given = 70 mV +  M At cathode M aq + e -  n =1 Apply Nernst eqn. RT ln 1.0 E 0.0591 log 1.0 = E 0M+ / M + 1 +  M + e -  n =1 At anode M aq Apply Nernst eqn. E anode E 0 M+ / M + 0.0591 log ( 0.05) E cathode = E 0M+ M + O

O

O

2+  9. 2Fe(s) + O 2 (g ) + 4H + (aq ) ↽ ⇀  2Fe (aq ) + 2H 2 O(l)

n=4  Fe 2 +  [ H 2 O ] Q = 4 [ Fe]2 [O2 ]  H +  2

2

2

10 -3  × 12 10 -6 = -13 = 107 = 4 10 12 × 0.1 × 10 -3  0.059 log10 Q n 0.059 log 107 = 1.67 4 0.059 × 7 = 1.67 4

E = Eθ O

= 1.57 V

Cell voltage =

(1)

(2)

Ecathode - Eanode 1.0 0.05 = 0.0591 log 20 = 0.0591 × 1.3010 = + ve

70 mV = 0.0591 log

(3v Since the cell voltage is + ve ∆Ecell > 0. It is a spontaneous process (ΔG < 0). 12. On replacing anodic concentration from 0.05 M to 0.0025 M, cell voltage is 1.0 E cell = 0.0591 log = 0.0591 log 400 0.0025 = 0.0591 log ( 20)

2

= 2 × 0.0591 log 20 = 2 × 70 = 140 mV

(4)

This page is intentionally left blank.

CHAPTER

13 Stoichiometry

H

appy the man, studying Nature’s laws, though known effects can trace the secret cause.” Virgil

13.1 Atomic weights And equivAlent weights It is a logical sequel of to the publication of the Atomic Theory that attempts should be made to obtain a list of the relative weights of the atoms of the different elements. Much data was already available, for there were gravimetric results by which the laws of conservation of mass and constant composition had been discovered. Further, much work on reacting quantities which ultimately led to the law of reciprocal proportions, had already been done. It was known within a reasonable degree of accuracy that 1 unit of weight of hydrogen combined with 8 units of oxygen to form water (the only compound of these two elements then known), but of the relative numbers of atoms in the water molecule, nothing was known. From the simple combining ratios suggested in his theory, Dalton assumed that one atom of hydrogen combined with one oxygen, to give the formula HO*. He suggested that the lightest element, hydrogen should be taken as the standard and that its atomic weight should be regarded as 1. From the formula HO for water, the atomic weight of oxygen would thus be 8. Similarly from the formula NH for ammonia, that nitrogen would be 4.7. In other cases, assumptions of simple composition did actually give correct values of atomic weights. But as we now know, all these values were in fact equivalent or combining weights. More than any other scientists, the Swedish chemist Berzelius supported Dalton in his attempt to build up a table of atomic weights. He analyzed numerous inorganic compounds and from the results obtained, tables of atomic *At first Dalton represented the atoms by diagrams. It was Berzelius who suggested a letter or letters to replace these, obtaining them from the Latin names for the elements-Latin being the one language common to all the European nations. It is for this reason that the ending-um is so common e.g., aluminium, magnesium.

weights were built up and revised from time to time. Like those of Dalton, however, the weights of Berzelius were equivalents, for at this time there were no means of knowing whether, for example, the formula for the oxide of an element was E2O or EO or EO2 etc. There was much disagreement on the subject among the chemists of the time. The fact that different standards of measurements were used added to the confusion. Dalton took H = 1 as the standard and thus the atomic weight of an element was the number of times one atom of that element was as heavy as one atom of hydrogen. Since oxygen combined more frequently with other elements than did hydrogen, some maintained that oxygen as standard was preferable. Berzelius used O = 100 as standard; the English men Thomson and Wollaston took O = 1 and O = 10 respectively. It was the German Stas who adopted the standard O=16 which was used in the International Table of Atomic Weights until recently. The new standard is the common isotope of carbon 12C = 12; on this basis the atomic weight 'n' of an element means that its atom is n/12 times as heavy as that of carbon 12. Certain discoveries were soon made which helped to establish the relationship between equivalent weights and the true atomic weights. The first of these came in 1819, the French chemists Dulong and Petit, who found that there was an inverse numerical relationship between the specific heats of solid elements and the so-called atomic weights of the day or some simple multiple of them. Secondly, from similarity in crystalline form of related compounds, Mitcherlish was able to make certain deductions concerning atomic weights and in 1818 he formulated these as law. At the beginning of the nineteenth century, measurements were being made on the specific gravities (or vapour densities) of gases and the vapours of volatile liquids.

13.2

Stoichiometry

Notable among the workers of the period on this subject were Berzelius and Dumas. Distinct relationship between the results obtained and the atomic weights of the time were seen. It was not until half way through the century when the atomicities of the gases became known that this work on vapour densities, in the hands of the Italian chemist Cannizzaro provided a method for determination of atomic weights. A fuller account of each of these methods is given later in this chapter. The publication in 1869 by Mendeleef’ of his classification of elements made the necessary corrections of atomic weights in those few cases of elements which for one reason or another were exceptional in some way. The discovery in 1827 by Faraday, that if the same current be passed successfully through a number of different electrolytes for the same time, the weights of products in each cell were in direct proportion to their equivalents, merely provided another method of obtaining these values, but gave no direct help in finding atomic weights. The equivalent of an element may be defined as the number of units of weight of the element which can combine with or displace unit weight of hydrogen or the equivalent of any other element or group of elements. Since unit weight of hydrogen combines with 8 units of oxygen to form water, the equivalent of oxygen is 8, and the weight of any other element combining with 8 units of oxygen is thus, by definition, the equivalent of the element. In the two oxides of sulphur of formula SO2 and SO3 the weights of sulphur combining with 8 units of oxygen are 8 and 5.33 respectively. Experimentally, therefore sulphur has at least two equivalents. But sulphur has only one atomic weight; hence its combining power differs in the two oxides, i.e., it has more than one valency. Most elements are found to have more than one equivalent and therefore have more than one valency. The general relationship between equivalent, valency and atomic weight is as follows.

The product of these gives the atomic weight. Since an equivalent weight is a reacting weight, and since elements can react with compounds, such reactions may also provide methods of determining equivalents. The equivalent of a compound is that weight which reacts with one equivalent of an element but since different classes of compounds react in different ways, some specific definitions are sometimes given. The equivalent of an acid is the formula weight containing unit weight of hydrogen replaceable by a metal to form a salt, i.e., the number of units of weight capable of giving rise to unit weight of ionic hydrogen. Since the number of atoms of hydrogen replaceable by a metal to form a salt in each molecule of acid is its basicity, then:

Equivalent weight × Valency = Atomic weight

2 + 32 + 64 = 49 g per litre 2 46 +112 + 48 = 53 g per litre 1 N Na 2 CO3 Contains 12

Since the valency of an element is the number of atoms of hydrogen that one atom of the element combine with or replace, it must be a whole number. Thus the equivalent of an element is either equal to or some sub-multiple of, its atomic weight, according to whether its valency is one or some whole number greater than one. Since there is no known chemical method of obtaining directly an accurate value for the atomic weight of an element the procedure followed in atomic weight determinations by chemical methods has often been: 1. Determination of equivalent 2. Determination of valency which has to be a whole number

Molecular weight of an acid = Equivalent of acid Basicity The equivalent of sulphuric acid is half its molecular weight and that of tribasic orthophosphoric acid H3PO4 is one third of its molecular weight. In those salts which undergo metathesis with acids, for example, the equivalent is the weight reacting with one equivalent of the acid. The equivalent of sodium carbonate (Na2CO3) is thus half its molecular weight, for: 2H + 2 equiv.

CO32 −  → H 2 O + CO 2

+

2 equiv.

Standard solution (i.e., solution of which the concentration is known) of some of the above compounds are used in volumetric analysis, some of the excercises of which provide methods of determining the equivalents of elements. A frequently used concentration is the normal one where the equivalent weight in grams is dissolved to make one litre of solution. Thus 1 N H 2SO 4

Contains

13.1.1 determination of equivalents 1a. Preparation of the oxide: A known weight of the element is burned in a stream of oxygen and the weight of oxide is found. The weight of element combining with 8 units of oxygen is the equivalent. The method is more suitable for finding the equivalents of combustible non-metals, e.g., carbon, phosphorous, sulphur, the acidic anhydrides formed being absorbed

Stoichiometry 13.3

in alkali, e.g., caustic soda which is weighed before and after the combustion. Dumas and Stas (in 1839) determined the equivalent of carbon in carbon dioxide by burning known weights of graphite (natural and artificial) and diamond in oxygen.

0.324 0.406 = Eq. wt of copper oxide E Cu 0.324 0.406 0.406 = = Eq. wt of Cu + Eq. wt of O E Cu + 8 E Cu E Cu = 31.60

solved Problem 1 0.5 g of metal upon combustion (direct oxidation) yields a single oxide weighing 0.666 g calculate the equivalent weight of the metal. Solution: M + O 2  → MO Equivalent of metal Wt of oxygen = Eq.wt.of metal Eq.wt of oxygen 0.5 0.166 = (∴wt 060 = 0.666 − 0.5 = 0.1669 ) 8 EM ∴ EM = 24 Thus equivalent wt of metal = 24 1b. By an indirect method: A known weight of the metal is dissolved in nitric acid to form the nitrate. The solution is evaporated to dryness to obtain the solid nitrate, which on subsequent heating, decomposes to the oxide which is weighed. To ensure complete decomposition, heating is continued until two consecutive weightings are the same. This method is suitable for all those metals which dissolve in nitric acid to from a nitrate which decomposes to the oxide. It cannot be used therefore, for the metals iron, cobalt, nickel and chromium (which are rendered ‘passive’ by concentrated nitric acid) not for potassium and sodium (the nitrates of which decompose to the nitrite only) nor for silver and mercury (whose metals decompose to the metal). If a metal forms two oxides in general, the higher one is obtained, e.g., tin is converted to stannic oxide, SnO2. solved Problem 2 0.324 g of copper was dissolved in nitric acid and the copper nitrate so produced was burnt till all copper nitrate converted to 0.406 g of copper oxide. Calculate the equivalent weight of copper. Solution: Cu 0.324 g

+ HNO3  → Copper nitrate  → Copper oxide 0.406 g

Let the equivalent weight of copper (Cu) is ECu Equivalent of Cu = Equivalent of copper oxide.

Hence the equivalent weight of Cu=31.60. 2. Reduction of the oxide A Known weight of the oxide is heated in a stream of hydrogen to find the loss in weight, which is the oxygen content. Completion of the reduction is tested, as is usual in gravimetric exercise by continuing unitil there is no further change in weight. The method is suitable for those metals low down in the electrochemical series, e.g., iron, lead, copper. No reducing agent is needed for the metal mercury and silver, lower down the series. The equivalent of silver, but not that of mercury, can be found by heating a weighed amount of the oxide in a crucible and weighing the silver left. solved Problem 3 Dry hydrogen was passed over 1.58 g of red hot copper oxide till all of it completely reduced to 1.26 g of copper (Cu). If in this process 0.36 g of H2O is formed, what will be the equivalent wt of copper and oxygen (H = 1). Solution: Copper oxide H +

2

→ Cu + H 2 O

∴Eq. of copper oxide = Eq. of copper = Eq. of H 2 O ∴

Wt. of H 2 O Wt. of Cu Wt. of copper oxide = = u Eq. wt. of H 2 O Eq. wt. of copper oxide Eq. wt. of Cu

Suppose eq. wt. of Cu and O are a and b repectively and eq. wt. of H is 1. 1.58 1.26 0.36 ∴ = = a+b a 1+ b ∴ a = 31.5, b=8

3. Displacement of hydrogen from dilute acids (Other than nitric) or an alkali A known weight of metal is allowed to react with an excess amount of acid or alkali and the volume of hydrogen liberated is measured at the barometric pressure and room temperature. A wide variety of apparatus (described in most elementary text books of chemistry) can be used. The method is suitable for the metals zinc, magnesium and iron with dilute acid, or aluminum and zinc with an alkali (Heat being required in the latter actions).

13.4

Stoichiometry

26.5 mL of 0.1 N NaOH = 2.65 mL N H2SO4

solved Problem 4 A piece of metal weighting 0.620 g yielded 622 mL of hydrogen, collected over water at 15°C and under a barometric pressure of 758 mm. The pressure of water vapour at 15°C is 12.7 mm. Solution: Since the pressure of hydrogen and water vapour together equal to barometric pressure. Pressure of hydrogen = (758–12.7) mm = 745.3 mm 745.3 273 × mL Volume of hydrogen at S.T.P = 622 × 760 285

Then, total excess 1 N H2SO4 = 26.5 mL Volume reacting with zinc = (50–26.5) mL = 23.5 mL Since 1 litre of 1 N H2SO4 contains 1 g equivalent, the amount of zinc reacting with this volume of acid will be the gram-equivalent. 23.5 mL of 1 N H2SO4 react with 0.764 g of zinc. ∴ 1000 mL of 1 N H 2SO 4 react with

= 578.4 mL Now 11.2 litres of hydrogen at S.T.P weigh 1 g. Hence the weight of metal displacing 11.2 litres S.T.P. of hydrogen is the gram–equivalent. 578.4 mL hydrogen is displaced by = 0.620 g metal. ∴ 11, 200 mL hydrogen is displaced by

11200 × 0.620 g 578.4

= 12.01 g ∴ Equivalent of metal = 12.01 An easier method than the above method is as follows. A suitable known weight of the metal is added to a known volume say 50 mL, of normal sulphuric acid (an amount in excess of that required to dissolve the metal). When the reaction between the metal and acid is finished, the solution is made up to a volume of 250 mL in a graduated flask and volumes of 25 mL withdrawn for titration with decinormal alkali, using methyl orange (screened or unscreened) as indicator. solved Problem 5 A piece of zinc weighing 0.764 g was added to 50 mL of 1 N sulphuric acid. The resulting solution was made up to 250 mL with distilled water. 25 mL of this solution required on the average 26.5 mL of 0.1 N NaOH for neutralization of the acid in excess. Calculate the equivalent weight of zinc. Solution: Since 10 mL of a decinormal solutions react with mL of a normal solution.

1000 × 0.764 g 23.5

= 32.5 g \ Equivalent of zinc = 32.5 4. Synthesis of the chloride This method is more suitable for metals than for nonmetals. It consists of heating a known amount of the metal in a stream of dry chorine. If a metal should form two different chlorides, the equivalent of the metal in the higher chloride is determined. Stas, and later Richards used the method to synthesise silver chloride in the atomic weight determination. 5. Analysis of a soluble chloride of a metal A known weight of the anhydrous chloride is dissolved in distilled water and made up to a known volume. Portions are then withdrawn and titrated with decinormal silver nitrate, potassium chromate being used as the indicator Ag +

+

Cl −  → AgCl

1 equiv 1 equiv 10 litres 0.1 N 35.5 g

From the above equation, the weight of chlorine, and hence by subtraction the weight of metal in the known weight of chloride taken can be calculated. The equivalent of the metal is that weight combining with 35.5 units (the equivalent) of chlorine. This method is particularly suitable for the determination of the equivalents of sodium, potassium and ammonium, since their chlorides are readily obtainable in the only anhydrous condition. Bromides and iodides can be used in this method and if the equivalent of the metal in the salt used is known, those of bromine or iodine can be obtained.

Stoichiometry 13.5

solved Problem 6 When excess of silver nitrate is added to 1.5276 g of cadmium chloride, 2.3885 g of silver chloride was precipitated. What is the equivalent weight of cadmium? Solution: 2.3885×35.5 Wt of chlorine present in 2.3885 g of AgCl = 143.55 = 0.5909 g ∴ Wt of chlorine present in cadmium chloride = 0.5909 g Wt of cadmium = 1.5276 – 0.5909 = 0.9367 g Eq of Cd = Eq of Cl 0.9367 0.590 = E 35.5 ∴ Equivalent weight of cadmium = 56.275 6. Estimation of the CO 2− 3 in a normal carbonate (a) For Soluble carbonates: e.g., those of sodium and potassium A known weight of anhydrous carbonate is dissolved in the distilled water of a known volume and portions of known volume are withdrawn from solution for titration with decinormal acid, using methyl orange as indicator. 2H + 2 equiv 20 litres 0.1 N

+

CO32 −  → H 2 O+CO 2 60 g

From the above equation, the weight of the CO32– ion in a known weight of carbonate taken, can be calculated, as can also, by difference, the weight of the metal. Hence the weight of metal combining with 30 g of the carbonate ion can be found. This is the equivalent. (b) For insoluble carbonates A weighed amount of the carbonate is added to a known volume of normal acid which is in excess of that required to react with all the carbonate. The resulting solution is diluted is a known volume and portion withdrawn for titration with decinormal alkali, methyl orange being the indicator. solved Problem 7 A piece of calcite (Calcium carbonate) weighting 3.82 g is added to 100 mL 1 N HCl and the resulting solution diluted to a volume of 250 mL. 250 mL of this solution required, on the average, 23.7 mL of 0.1 N NaOH for neutralization of the acid in excess.

Solution: 23.7 mL 0.1 N NaOH = 2.37 mL 1 N HCl Total excess 1 N HCl = 23.7 mL 1 N HCl Volume of 1 N HCl used for carbonate = (100 – 23.7) mL = 76.3 mL 2H+ + CaCO3  → Ca2++CO2 + H2O 2 Litres 1 N 40 + 60 g 2 litres in 1 N HCl react with 60 g CO32– ion 76.3×60 g = 2.29 g ∴ 76.3 mL in HCl react with 2000 ∴ Weight of calcium in 3.82 g of calcium carbonate = (3.82–2.29) g = 1.53 g Since the equivalent of the CO32– ion is 30 Equivalent of calcium =

30 × 1.53 = 20.05 2.29

7. Electrolysis The same current may be passed successively for the same time through the following three cells. A — Containing dilute sulphuric acid with platinum electrodes. B — Containing aqueous copper sulphate with copper electrodes. C — Containing aqueous silver nitrate with silver electrodes. In A, the hydrogen evolved at the cathode and the oxygen at the anode may be collected and measured. In B, the weight of copper deposited on the cathode may be found by weighing this electrode before and after electrolysis. In C, the weight of silver deposited at the cathode is similarly found. By Faraday’s Law these quantities are in direct proportion to the equivalents and the equivalents of oxygen, silver and copper can be found. If the volume of hydrogen liberated is 11.2 L S.T.P. (1 g), the volume of oxygen set free is 5.6 L S.T.P. (8 g), The weight of copper deposited is 31.75 g and of silver 107.9 g.

13.1.2 determination of Atomic weights (chemical methods) There is no absolute method of determining valency, but the value for a particular element in a particular compound has been inferred by one or more of the following methods: 1. Dulong and Petit 2. Mitscherilich 3. Cannizzaro – now the only important method 4. Periodic classification

13.6

Stoichiometry

Of these methods, the first depends upon obtaining an approximate atomic weight, and methods 2 and 4 are comparative, i.e., the valency is inferred by reference to other substances. Cannizzaro’s method (3) if carried out accurately, is the most direct method of obtaining atomic weight without using equivalent or valency. 1. The method of Dulong and Petit In 1819, these two French chemists discovered an inverse relation between the specific heat of a solid element and its atomic weight, which may be expressed mathematically as Atomic weight × Specific heat = (approximately) constant = 6.4 The value 6.4 is known as the atomic heat. Alternatively the rule may be stated; Atoms of all solid elements have the same heat capacity, for the amount of heat required to raise the temperature of the gram atomic weight (i.e., the atomic weight expressed in grams) of any solid element 1°C is 6.4 calories. While the rule holds good for the heavier elements, particularly the metals, there are certain notable exceptions. e.g., carbon, silicon boron. These are, incidentally, elements which have high boiling points. The specific heat of an element varies with the temperature increasing to a constant value as the temperature rises. The atomic heat therefore rises with temperature to the constant 6.4 the temperature at which this value is reached varying for different elements. In the exceptions above, the temperature is high. The rule of Dulong and Petit is, in fact, the limining case of law which has now been explained on the quantum theory. solved Problem 8 A piece of tin weighing 2.510 g was dissolved in concentrated nitric acid after ignition, the residue weighed 3.186 g. The specific heat of tin is 0.054. Solution: The first results enable the equivalent of tin to be calculated Weight of oxygen in residue = (3.186–2.510) g = 0.676 g 0.676 g of oxygen combines with 2.510 g tin ∴ 8 g of oxygen combines with ∴ Equivalent of tin = 29.71

8×2.51 =29.71g 0.676

From Dulong and Petit’s rule Atomic weight of tin 6.4 (approx ) 0.054 = 119 (approx ) ∴ Valency of tin in oxide =

119 =4 29.71

Because valency must by definition, be a whole number, an approximate value of the atomic weight is to find it ∴ Accurate value of the atomic weight = 29.71 × 4 (exactly) = 118.8 2. Mitcherlich’s Method: Law of Isomorphism Towards the end of the eighteenth century, Abbé Haüy first put forward the view that each chemical compound had its own particular crystalline character. The belief was growing that there was some connection between crystalline structure and chemical similarity. At the beginning of the nineteenth century, the young chemist Mitscherlich was working in Berlin on the analysis of the acid potassium and ammonium phosphates and arsenates. He noticed a similarity in crystalline shape among these compounds and his interest in crystalline structure developed, and continued when he went to work under Berzelius. Certain naturally occurring minerals were found to possess similar crystalline shape. Berzelius suggested the term isomorphous (Greek iso = same, morphe = shape) to describe such substances. At first Mitcherlich thought the phenomenon of isomorphism depended on the numbers and not the kinds of atoms in the molecules, for he had noticed, for example, the one of the crystalline forms of calcium carbonate had the same shape as the crystals of sodium nitrate. Later he realized that the nature rather than the number of atoms was the determining factor and in 1818, he formulated the Law of Isomorphism which may be stated as chemically similar substances crystalise in the same form. An exchange of arsenic and phosphorous caused no alteration to crystalline form, because the two elements were chemically similar and hence possessed the same valency. Both the element potassium and the group ammonium had the same valency. The Law was tremendous help to Berzelius who, by means of it was able to publish in 1826 a revised table of atomic weights. An example may serve to show the application of the law. The sulphate, selenate and chromate of potassium are isomorphous. If it is accepted that the formula for sulphur trioxide SO3, then the other acid anhydrides are SeO3 and CrO3. But the other oxide of chromium is found on analysis

Stoichiometry 13.7

to contain half the amount of oxygen per gram of chromium and thus, its formula is Cr2O3. This chromium oxide gives rise to chromium (III) salts, the sulphate of which is isomorphous with ferric sulphate, the basic anhydride of which therefore be Fe2O3. The lower oxide of iron which gives rise to ferrous salts contains two–thirds the amount of oxygen per gram iron, and its formula must therefore be FeO. Ferric oxide is similar to aluminum oxide whose formula will thus be Al2O3, and so on. Here is a method of establishing molecular formula and getting from the equivalent to the atomic weight. In order that substances may be considered to be isomorphous they must possess the following three characteristics: 1. Have similar crystalline shape: This need not mean identical angles between crystal faces – a difference of 2° can be accepted. 2. Form ‘overgrowths’: If a crystal of one isomorphous substance is suspended in a saturated solution of another with which it is isomorphous, this second substance will deposit round the crystal in a layer of uniform thickness and thus cause no alteration in shape. 3. Form mixed crystals: If a solution of two isomorphous substances be evaporated to crystallizing point, then the crystal which separate have a composition similar to that of the solute. Such crystals are solid solutions. The best examples of isomorphism are provided by the double sulphates known as alums. The general formula of which is XY (SO4)2. 12H2O where X is univalent and is commonly Na, K, NH4 and Y is trivalent and commonly Al, Fe, Cr. A crystal overgrowth is readily made by suspending a crystal of (purple) chrome alum KCr (SO4)2. 12H2O in a solution of (colourless) potash alum KAl(SO4)2.12H2O of suitable concentration. Another example is the (red) crystal of the double salt cobalt magnesium sulphate in a solution of the double magnesium potassium sulphate which develops as a colourless growth. The following conclusions can be deduced from the phenomenon of isomorphism. 1. Masses of two elements that combine with same mass of other elements in their respective compounds are in the ratio of their atomic masses. * Now a days crystals are said to be isomorphous if they possess of similar arrangement of geometrically similar structural units. Thus CaCO3 and NaNO3 are not chemically similar but are isomorphous because the crystals are made up of the ions Ca2+ and Na+ which are spheres of about the same size, and NO–3 and CO32–which are the same shape and size. Another pair of isomorphous substances with similar ions is MgO and NaF (Mg+2 and Na+ and O2– and F–)

Mass of one element (A) that combines with certain mass of other element Atomic mass of A = Mass of other element (B) Atomics mass of B that combines with certain mass of other elements 2. The valencies of the elements forming isomorphous compounds are the same. solved Problem 9 Potassium chromate is isomorphous to potassium sulphate (K2SO4) and is found to contain 26.78 % chromium. Calculate the atomic mass of chromium (K = 39.10) Solution: Since the formula of potassium sulphate is K2SO4. So the formula potassium chromate should be K2CrO4 as it is isomorphous to K2SO4. If the atomic mass of chromium is A, then formula mass of potassium chromate should be = 2 × 39.1 + A + 64 = (142.2 + A) % of Chromium = So

A × 100 (142.2 + A )

100 = 26.78 (142.2 + A )

100 A = 26..78 (142.2 + A) 26.78 × 142.2 or A = = 52.00 73.22 3. Cannizzaro’s method Half way through the nineteenth century, Cannizzaro drew attention to Avogadro’s hypothesis and used it to determine atomic weights. The method consists in determinng the vapour density of as many gaseous or readily volatile compounds of the element in question is possible. Until recently, this method was generally applicable only to non-metals, for few volatile compounds of the metals were known, now however, many of these have been discovered. From the relationship Vapour density (H = 1) × 2 = Molecular weight. The molecular weight of each compound is found. Analyses of these compounds were then made to determine the weight of the element in question, in molecular weight of each compound. Table13.1 for the element carbon, may be taken as illustration. If a sufficiently large number of compounds is taken, it is extremely likely that in the molecule of at least one such compounds there is only one atom of carbon. The

13.8

Stoichiometry

table 13.1 Analysis of carbon compounds showing weight of carbon in molecular weight of each compound Vapour density (H = 1) Carbon monoxide Carbon dioxide Methane Ethylene Carbon disulphide Chloroform Acetylene

Molecular weight

Weight of carbon in molecular weight of each compound

14 22 8 14

28

12

44

12

16

12

28

24

38 59.75 13

76

12

119.5

12

26

24

smallest weight therefore in the final column is the atomic weight. If the molecular weight is approximate, it serves to find the valency, like Dulong and Petit’s method. But measurements of vapour density can now be made accurately, and Cannizzaro’s method is then of fundamental importance, giving atomic weight directly from molecular weight. 4. The periodic classification When Mendeliev drew up his table, the equivalent of beryllium was 4.55, and since its oxide was isomorphous with aluminum oxide, its valence was throught to be 3 and its atomic weight, therefore, is 13.65. This atomic weight would place the element in series between carbon and nitrogen, a position so unlikely that Mendeliev maintained that its valency must be 2 with an atomic weight of 9.10 which placed the element between lithium and boron, a position in keeping with it’s chemical behavior. This was later confirmed, When the vapour density of its chloride was found to be 40 giving a molecular weight of 80 which agrees with 9 + 71 (At wt of Cl × 2 = 35.5 × 2 = 71) (The specific heat of beryllium rises rapidly with temperature, and at ordinary temperature the element does not follow the Dulong and Petit rule). The atomic weight of uranium was also corrected from 120 to 240. Other examples of similar corrections were indium, cerium and yttrium.

13.1.3 Physical methods of determining Atomic masses Two methods are available: 1. The method of limiting densities In the case of gases only, the molecular weight is determined by the method of limiting densities (described later).

The molecular weight is divided by the atomicity gives the atomic mass. Since some noble gases, e.g., helium, argon, etc. form no chemical compounds they have no equivalents. Chemical methods of obtaining atomic weights, such as those described above are not possible and the method employed is that of limiting densities. 2. The mass spectrograph This is the only direct method of obtaining an accurate value for the atomic weight of an element. The discussion of the details of this method is beyond the scope of this book. Problems for Practice 1. A sample of pure calcium metal weighing 1.35 g was quantitatively converted to 1.88 g of pure CaO. What is the atomic weight? 2. 1.00 g of EuCl2 is treated with excess aqueous AgNO3 and all the chlorine is recovered as 1.29 AgCl. Calculate the atomic weight of Eu (Cl = 35.5, Ag = 108). 3. 7.38 g of sample of metal oxide is quantitatively reduced to 6.84 g of pure metal. If the specific heat of the metal is 0.0332 Cal/ g. Calculate the valency and the accurate atomic weight of the metal. 4. The following data were obtained from the analysis of five volatile phosphorous compounds. The percentages of phosphorous in these compounds having the molecular weights 222, 154, 140, 300 and 126 are respectively 27.9%, 20.2% 22.5% 43.7% and 24.6%. Calculate the atomic weight of phosphorous. 5. A compound contains 28% of nitrogen and 72% of a metal by weight. 3 atoms of the metal combine with 2 atoms of nitrogen. Find the atomic of the metal (N = 14).

Stoichiometry 13.9

6. Potassium selenate is isomorphous with potassium sulphate and contains 45.42% selenium by weight. Calculate the atomic weight of selenium. 7. 1.5276 g of CdCl2 was found to contain 0.9367 g of Cd. Calculate at. wt of Cd. 8. On dissolving 2.0 g a metal in H2SO4, 4.51 g of metal sulphate was formed. The specific heat of metal is 0.057 Cal/ g. Find the exact atomic wt of the metal. 9. A metal M has density of 7.42 g/mL. Calculate the atomic weight of the metal, if the radius of the atom of metal = 1.432×10–8 cm. Assume the atom of this metal is a sphere. 10. The element ytterbium forms a compound YbBr3 To a solution of 1.3209 g of YbBr3 in water, excess silver nitrate is added precipitating all the bromides as AgBr, an insoluble pale yellow solid. The AgBr is collected, dried and found to weight 1.8027 g. Calculate the atomic weight of Yb(Br = 80.0; Ag = 108). 11. The mass spectrum of carbon shows that 98.892% of carbon atoms are C-12 with a mass of 12 amu and 1.108% are C-13 with a mass of 13.00335 amu. Calculate the atomic weight of naturally occurring carbon. 12. Bromine has two naturally occurring isotopes, one of them Br-79, has a mass of 78.9183 amu and an abundance of 50.54%. Calculate atomic mass of other isotope Br-81 (Average atomic mass = 79.904 g mol–1). 13. The percentage of chlorine in the chloride of an element is 44.71% 158.5 g of this chloride on vaporization occupies a volume of 22.4 litre at NTP. Calculate the atomic weight and valency of the element. 14. The equivalent weight of an element is 13.16. It forms a salt, isomorphous with K2SO4. Deduce the atomic weight of the element. 15. Cu2S and Ag2S are isomorphous in which percentage of suphur are 20.14% and 12.94% respectively. Calculate the atomic weight of silver (Cu = 63.5). 16. Two elements P and Q form the compounds P2Q3 and PQ2. If 0.15 mole of P 2Q3 weighs 15.9 g and 0.15 mole of PQ2 weighs 9.3 g, find the atomic weights of P and Q. 17. A compound which contains one atom X and two atoms of Y for each three atoms Z is made by mixing 5.00 g of X, 1.15 × 1023 atoms of Y and 0.03 mole of Z atoms. Given that only 4.40 g of the compound is formed calculate the atomic weight of Y, if the atomic weights of X and Z are 60 and 80 amu respectively.

13.2 molecular weights and Formulae of gases The density of a gas may be given in two ways. Firstly it may be given as weight per unit volume at S.T.P. This is usually referred to as normal density and the units

commonly used are g per cm3 (some times written as g per cc or g per mL, g cm–3, g mL–1) or g per litre (g L–1). Secondly, the density may be described by the vapour density. This a density relative to that of some gas, usually hydrogen, taken as standard. The vapour density may therefore be defined as The number of times a given volume of a gas is as heavy as an equal volume of hydrogen, both volumes being measured at the same temperature and pressure. From the density of a gas, no matter how expressed, the molecular weight can be obtained. 1. From the normal density The gas equation PV = nRT The number of moles(n) =

weight W = molecular weight M

Substituting for n, we get PV =

WRT M

But the density d =

and therefore P =

W V

dRT dRT and M = M P

Hence if the density is known, the molecular weight can be calculated. If the density is expressed as grams litre–1, the value of R to be taken must be 0.0821 litre atmospheres mole–1 degree–1. Then substituting d × 0.0821× 273 M= 1 = d ×22.41 The figure 22.41 (usually taken as 22.4 ) is the grammolecular volume. This is the volume occupied by the molecular weight in grams of all gases at S.T.P. If this figure is used, then the molecular weight can be calculated from the normal density as follows. Taking the normal density of hydrogen chloride, say as 1.639 g litre–1 The molecular weight of hydrogen chloride = 22.4 × 1.639 = 36.7 2. Vapur density method Suppose two vessels of equal volume be filled at the same temperature and pressure with a gas X and hydrogen, respectively. Then since the volumes are equal, the number of moles of gas in each volume is, by Avogadros law, the same. Let the number be n, and

13.10

Stoichiometry

the molecular weights of the two gases be MX and MH respectively Then, the weight of gas X = MX ×n Weight of hydrogen = MH × n By definition Vapour density of X =

Mx × n Mx = MH × n MH

Since the hydrogen molecule is diatomic MH = 2 (or more correctly 2 ×1.008). Hence, since X is any gas, we can write the general relation from which the molecular weight can be obtained as Vapour density × 2 = molecular weight Both these methods of determining molecular weight depend on one or other of the gas laws, all of which are ideal or limiting. The values obtained are thus approximate only, but they are nevertheless adequate for the purposes for which they are required. If however, instead of using the ideal gas equation, that of Van der Waal’s is used, and if the constants a and b are known, then a molecular weight nearer the correct value is obtained.

13.3 methods for determination of density 1. Regnault’s method (1842) A large glass globe is evacuated and weighed. It is then filled at a known temperature and pressure with the gas of which the density is required and reweighed. The globe is now filled with some liquid (e.g., water) of which the density is known, and again weighed. From this weight and the density of the liquid, the volume V of the globe is calculated. Hence, the weight of a volume V of the gas at a known temperature and pressure is obtained and thus the density is calculated. Extreme care is required for an accurate result. A globe of volume about 2 litres must be used in order to get a measurable weight of gas, and a similar globe of known weight should be placed on the other end of the balance beam as part counter poise. Weights should be corrected for buoyancy before being used to calculate the density. The vapour density can be determined by repeating the experiment with the globe filled with hydrogen under the same temperature and pressure and comparing the two weights. Towards the end of the last century, Lord Rayleigh used this method to determine the density of nitrogen, with such a degree of accuracy that he was able to detect the slight discrepancy between the density of nitrogen

obtained from air and that of nitrogen prepared by chemical methods. This work led to the discovery of argon, the first of the inert gases. The chief difficulty is weighing the gas, owing to the relatively large volume of a gas in comparison to its weight. Morley avoided this difficulty in his experiment on the composition of water by absorbing the hydrogen in palladium and weighing the palladium. A comparable laboratory method, giving satisfactory result for the density of oxygen, consists in heating a known weight of potassium chlorate until a convenient volume of oxygen has been collected and measured on reweighing the loss in weight of the chlorate, which is equal to the weight of oxygen can be found. 2. Diffusion method Soret (1866) determined the vapour density and hence the molecular weight of ozone, comparing its rate of diffusion with that of a gas of known density-making use of Graham’s law. Two similar vessels were filled respectively with ozonized oxygen and a mixture of chlorine and oxygen (the percentage of oxygen in both cases being the same). The two vessels were separated by means of glass plates, each containing a small hole. The plates were moved so that the holes coincided and gas then diffused in both directions. After a given time, the amounts of ozone and chlorine which had escaped from their containers were measured and these amounts taken as the rates of diffusion. Like the other gas laws, Grahams’ law is a limiting law and molecular weight by means of it is approximate only. solved Problem 10 100 mL of oxygen diffused through a membrane in 5 minutes. 120 mL of an unknown gas under the same conditions, diffused through the membrane in 10 minutes. Find (a) the molecular weight of the unknown gas and (b) the volume of chlorine which, under the same conditions, would diffuse through the membrane in 15 minutes. Solution: (a) The volume of the unknown gas which would diffuse in 5 minutes is 60 mL. Substituting in Graham’s law and taking the vapour density of the unknown gas is D, we get: 100 D = 60 16 100 × 4 i.e. D = 60 D = 44.4 ∴ molecular weight = 88.8

Stoichiometry 13.11

(b) Let the volume of chlorine which would diffuse in 5 minutes be ‘a’ mL 100 35.5 Then = a 16 100 × 4 ∴a = = 67.1 35.5 \ Volume of chlorine diffusing in 15 minutes = 201.3 mL

The value P1V1/P0V0 is obtained by measuring the volumes of a given amount of gas, under a constant temperature of 0°C, at pressures of one atmosphere and below. The product of pressure and volume is then plotted against pressure. For most gases, the graph is straight line and takes the form shown in Fig 13.1. By extrapolation, the value of PV at zero pressure i.e., P0V0 is obtained.

solved Problem 11 A certain volume of a gas X diffused in 20.5 minutes. Under the same conditions, an equal volume of hydrogen diffused through the membrane in 3 minutes. What deduction can be made? Solution: Let the volume of hydrogen diffusing in three minutes be V mL 3V Then the volume of x diffusing in 3 minutes = mL 20.5 Now substituting in Graham’s equation 3V = 20.5 V

1 Dx

20.5 3 ∴ Dx = 46.7 Molecular weight = 93.4 3. The method of limiting densities For a given weight of gas w, g occupying V litre at 0°C and a pressure of P atmospheres: Density per unit pressure in g per litre=W/PV If the gas is ideal, the product PV would be constant, but since gases are not ideal, the value W/PV varies with pressure. The lower the pressure of gas, the more nearly does it behave as an ideal one, and at zero pressure (P0) the value W/P0V0 would of an ideal gas. The expression W/P0V0 is called the limiting density of the gas in contrast to W/P1V1-obtained under a pressure of one atmosphere which is numerically equal to normal density of the gas PV W W = × 1 1 P0 V0 P0 V0 P1 V1 =

PV W × 1 1 P1 V1 P0 V0

i.e., limiting densityy = normal density ×

0

Pressure (atmospheres)

1

Fig 13.1 Graph of PV against P From the above relation between normal and limiting densities the latter can be calculated. The limiting density of oxygen is 1.42762 g L–1 atmosphere–1 against the normal density of 1.42900 g L–1 atmosphere–1. Burt and Edgar, at the beginning of the century, determined the atomic weight of chlorine. They determined the normal density of hydrogen chloride (1.63915 g L–1 atmosphere–1) the value of P1V1 (54, 803) and P0V0 (55,213).

i.e. Dx =

Now

PV

P1 V1 P0 V0

Problems for Practice 18. A hydrated sulphate of metal contained 8.1% metal and 43.2% SO2− 4 by weight. The specific heat of metal is 0.24 Cal/g. What is the molecular formula of hydrated sulphate. 19. The weight of 1 litre sample of ozonised oxygen at N.T.P was found to be 1.5 g. When 100 mL of this mixture at N.T.P. were treated with trepentine oil, the volume was reduced to 90 mL. Calculate the molecular weight of ozone. (MLNR 1996) 20. 1.878 x, when heated in a stream of HCl gas was completely converted to chloride MClx which weighed 1.0 g. The specific heat of metal is 0.14 cal/g. What is the molecular weight of metal bromide. 21. Cupric ammonium sulphate was found to contain 27.03% water of crystallization. Upon strongly heating it gave cupric oxide corresponding to 19.89% of starting mass. Find the empirical formula of cupric ammonium sulphate. (Cu = 63.5, N = 14, S = 32, H = 1, O = 16)

13.12

Stoichiometry

22. The molecular mass of an organic acid was determined by the study of its barium 4.290 g of salt quantitatively converted to free acid by the reaction with 21.64 mL of 0.477 M H2SO4, the barium salt was found to have 2 moles of water of hydration per Ba2+ ion and the acid is monobasic. Calculate the molecular weight of anhydrous acids? 23. When 2.96 g of mercuric chloride is vapourised in a 1 litre bulb at 680 K, the pressure is 458 mm. What is the molecular weight and molecular formula of mercuric chloride vapour (Hg = 200.6). 24. A mixture contains NaCl and an unknown chloride MCl (i) 1 g of this is dissolved in water. Excess of acidified AgNO3 solution is added to it. 2.567 g of a white precipitate is formed. (ii) 1 g of the original mixture is heated to 300°C. Some vapours come out which are absorbed in acidified AgNO3 solution. 1.341 g of a white precipitate is obtained. Find the molecular weight of the unknown chloride. 25. In a Victor Meyer apparatus 0.168 g of a volatile compound displaced 49.4 mL of air measured over water at 20°C and 740 mm of pressure. Calculate the molecular weight of the compound (aqueous tension at 20°C = 18 mm). 26. When 2 g of a gas ‘A’ is introduced into an evacuated flask kept at 25°C, the pressure is found to be one atmosphere. If 3 g of another gas ‘B’ is then added to the same flask, the total pressure becomes 1.5 atm. Assuming ideal gas behavior, calculate the ratio of the molecular weights, MA : MB. 27. A Dumas bulb was filled with 0.4524 g of vapour at 136°C and 758 mm of pressure. Calculate molecular weight of the vapour, if the capacity of the bulb is 127 mL. 28. 0.3151 g of a substance when introduced into a Hofmann’s tube generated 128.5 mL of vapour at 30°C, the level of Hg inside being 430 mm higher than outside the tube, and the barometer reading was 758 mm. Calculate the molecular weight of the substance.

13.4 eudiometry or gas Analysis By exploding one volume of hydrogen with one volume of chlorine in a strong glass bulb (by exposure to sunlight or the light of burning magnesium), two volumes, at the same temperature and pressure, of hydrogen chloride are obtained. The experiment provides a convenient method of illustrating Gay-Lussac’s Law of volumes.

1 vol. of hydrogen + 1 vol. of chlorine → 2 vol. of hydrogen chloride. Writing the unknown formula of hydrogen chloride as HxCly then by Avogadro’s Law, H2 + Cl2 → 2HCl And x = y = 1 to give the molecular formula HCl The molecular formula of carbon monoxide and any of the gaseous hydrocarbons may be found by exploding known volumes of these with known volumes (in excess) of oxygen in a eudiometer over mercury. In each case, the residue in the eudiometer consist of carbon dioxide and unused oxygen, and by introducing a little caustic potash to dissolve the carbon dioxide, the volume of excess oxygen can be found (since the experiment is carried out at laboratory temperature, the water formed, in the case of hydrocarbons condenses as a liquid of negligible volume). After correcting all volumes to the same pressure, the temperature will be the same throughout if time is left for cooling after the explosion-it will be found in the case of the carbon monoxide that: 2 vol carbon monoxide (CxOy)+1 vol oxygen → 2 vol carbon dioxide. By Avogadro’s Law: 2CxOy + O2 → 2CO2 and x = y = 1 to give molecular formula CO By a similar explosion of known volumes of hydrogen and oxygen, and measurement and determination of the nature of the gas in excess, it will be found that two volumes of the hydrogen gase combine with one volume of oxygen. If a mixture of the two gases of this composition (known as electrolytic gas) be exploded in a eudiometer maintained at a temperature above 100°C (by surrounding it with the vapour of, for example amyl alcohol of boiling point 138°C), then the volume of the resulting steam can be found, to give: 2 vol H2 + 1 vol O2 → 2 vol steam And as above 2H2 + O2 → 2HxOy And x = 2, y = 1 showing the molecular formula of steam to be H2O. The molecular formula of ammonia may be obtained by explosion with oxygen, but the gas must be decomposed into nitrogen and hydrogen, otherwise oxides of nitrogen as well as water will be formed. A volume of ammonia is collected over mercury in a eudiometer and sparks are passed until there is no further increase in volume. The volume is almost doubled (since 98 percent of the gas is decomposed). Oxygen is then

Stoichiometry 13.13

introduced and a spark passed to convert the hydrogen into water. The remaining volume of nitrogen and excess oxygen is treated with alkaline pyrogallol, to dissolve the excess oxygen and enable the volume of nitrogen to be read. It will be found that: 2 vol of ammonia → 4 vol of a mixture of nitrogen and hydrogen and that: 1 1 vol of oxygen were used to combine with the 2 hydrogen. Since hydrogen and oxygen combine in the proportion of 2:1 by volume, 3 vol. of hydrogen were produced by the decomposition of 2 vol of ammonia, and therefore also 1 vol of nitrogen, i.e., 2 vol. ammonia → 1 vol. nitrogen + 3 vol. hydrogen and by Avogadro’s law, 2NxHy → N2 + 3H2 and x = 1, y = 3, and the molecular formula of ammonia is NH3. Analytical methods are used to determine the molecular formulae of the oxides of nitrogen and hydrogen sulphide since quantitative synthesis of these gases are not possible. Known volumes of the gases, enclosed in a tube over mercury, are reduced by heating with metals. The heating may be either direct or electrical. Any metal which does not react with hydrogen can be used for hydrogen sulphide, and similarly any metal not combining directly with nitrogen can be used for the nitrogen oxides. Davy used sodium to determine the composition of the oxides of nitrogen, at the beginning of last century, but less active metals such as iron and copper are more suitable. In the case of hydrogen sulphide, for example, it will be found that a volume of hydrogen equal to that of the hydrogen sulphide is obtained applying Avogadro’s law. HxSy + [2YCu] → [YCu2S] + H2 Equating for x, we get x=2 and a molecular formula of H2 Sy. No further deduction can be made from this experiment since nothing is known of the number of atoms in the solids (in square brackets). The experiment must be followed by a vapour density determination to obtain the molecular weight, which enables the value of y to be found as 1. By similar argument from experimental results, the molecular formulae of the oxides of nitrogen can be found.

13.4.1 Abnormal vapour densities In some cases, particularly at high temperatures, the vapour density obtained gives a molecular weight different from the expected value and often bearing little numerical relation to it. When less than expected the abnormality is due to dissociation. A higher value than expected, points to association. Numerous cases are known and a few examples are given: 1. Ammonium chloride: This dissociates into ammonia and hydrogen chloride  NH 4 Cl ↽ ⇀  NH 3 + HCl Such a dissociation is common to ammonium salts of volatile acids. On cooling, the reverse process of association occurs and this is the cause of the property of sublimation common to such salts. 2. Phosphorous pentachloride: This dissociates into the trichloride and chlorine  PCl5 ↽ ⇀  PCl3 + Cl2 3. Dinitrogen tetroxide: (N2O4) The dissociations take place   N 2 O4 ↽ ⇀  2 NO 2 ↽ ⇀  2 NO + O 2 4. Aluminium chloride: At 400°C, this exists as Al2Cl6 molecules; at a temperature of 1,100°C thermal dissociation to the AlCl3 molecule is complete.  Al2 Cl6 ↽ ⇀  2AlCl3 At higher temperatures still, it dissociates into aluminum and chlorine. In some cases, it is possible to identify the dissociation products or to show that some new substances with different properties are formed. Problems for Practice 29. 20 mL of CO was mixed with 50 mL of oxygen and the mixture was exploded. On cooling, the resulting was shaken with KOH. Find the volume. 30. One litre of a mixture of CO and CO2 is passed through red hot charcoal in tube. The new volume becomes 1.4 litres. Find out % composition of mixture by volume. All measurements are made at same P and T. 31. A mixture of 20 mL of CO, CH4 and N2 was burnt in excess of O2 resulting in reduction of 13 mL of volume. The residual gas was then treated with KOH solution to show a contraction of 14 mL in volume. Calculate the volume of CO, CH4 and N2 in mixture.

13.14

32.

33.

34.

35.

36.

37.

38.

39.

40.

Stoichiometry

All measurements are made at constant pressure and temperature. (IIT 1995) 16 mL of a gaseous aliphatic compound CnH3nOm was mixed with 60 mL O2 and sparked. The gas mixture on cooling occupied 44 mL. After treatment with KOH solution. The volume of gas remaining was 12 mL. Deduce the formula of compound. All measurements are made at constant pressure and room temperature. 50 mL of pure and dry oxygen was subjected to a silent electric discharge and on cooling to the original temperature, the volume of ozonized oxygen was found to be 47 mL. The gas was then brought into contact with terpentine oil, when after the absorption of ozone, the remaining gas occupied a volume of 41 mL. Find molecular formula of ozone; all measurements are made at constant P and T. 60 mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 mL of N2 was formed, calculate the volume of each gas in mixture. All measurements are made at constant P and T. Assume H2O in liquid phase. 40 mL ammonia gas taken in eudiometer tube was subjected to sparks till the volume did not further change. The volume was found to increase by 40 mL of oxygen was then mixed and the mixture was further exploded. The gases remained were 30 mL. Deduce formula of ammonia. All measurements are made at constant P and T. Assume H2O in liquid phase. A gaseous alkane is exploded with oxygen. The volume of O2 for complete combustion to CO2 formed is in the ratio of 7:4. Deduce molecular formula. When a mixture of 10 mole of SO2, 15 mole of O2 was passed over catalyst, 8 mole of SO3 was formed. How many mole of SO2 and O2 did not enter into combination? 100 mL of a gas at NTP was heated with tin. Tin converted into stannous sulphide and hydrogen was left. This hydrogen when passed over hot CuO produced 0.081 g of water. If the vapour density of the gas is 17, find its formula. The percentage by volume of C3H8, CH4 and CO is 36.5. Calculate the volume of CO2 produced when 100 mL of the mixture is burnt in excess of O2. 9 volumes of a gaseous mixture consisting of a gaseous organic compound A and just sufficient amount of oxygen required for complete combustion, yielded on burning, 4 volumes of CO2, 6 volumes of water vapour and 2 volumes of N2, all volumes measured at the same temperature and pressure. If the compound A contained only C, H and N, (i) how many volumes of oxygen are required for complete combustion (ii) what is the molecular formula of the compound.

13.5 BAlAnced chemicAl equAtion The quantitative work which may be said to have been started by Black, Cavendish and Lavoisier was now bearing much fruit. That branch of chemistry dealing with composition of substances, by weight or volume is called stiochiometry. This term was first used by Richter. The laws of chemical combination discussed earlier (chapter 1 basic concepts of chemistry) are Laws of Stiochiometry. A chemical reaction may be represented through symbols and formulae of the substances involved as reactants and products. Then it is called chemical equation. For example, when zinc reacts with hydrochloric acid, zinc chloride and hydrogen are produced. This reaction may be represented using symbols and formulae as follows. Zn + HCl  → ZnCl2 + H 2 In the above equation the number of atoms of various elements on both sides are not equalized. Such equations are called skeleton equations. The chemical equation is always balanced as per the law of conservation of mass. The number of atoms of each element on both sides must be same for which the equation is balanced. In order to equalize the number of atoms of various elements, the various species are multiplied with appropriate numbers. The above equation can be balanced by multiplying the HCl with 2. Zn+2HCl→ZnCl2+H2 A chemical equation in which the number of atoms of each element is equal on the reactant side and the product side is called a balanced equation. Here, we consider the relation between the quantities of the substances consumed and the quantities of the substances produced in a chemical reaction. This area of study is known as stiochiometry (pronounced stoy key-omuh-tree) a name derived from Greek word stanchions (element) and matron (measures). “The quantitative relationship existing between the quantities of the reactants and the products in a chemical reaction is known as stoichiometry”. In order to carry out stoichiometric calculations, one should be able to write a balanced chemical reaction. The balanced chemical equation must fulfill the following conditions. 1. It should represent a true chemical change, i.e., if a reaction is not possible between certain substances, it cannot be represented by a chemical equation. 2. It should be balanced. 3. It should be molecular, i.e., all the species be represented in their molecular form. For example, elementary gases like hydrogen, oxygen etc should be represented as H2 and O2.

Stoichiometry 13.15

13.5.1 information conveyed by chemical equation Consider the reaction between nitrogen and hydrogen to form ammonia as N 2 (g ) + 3H 2 (g )  → 2 NH 3 (g ) The stoichiometric coefficients in the equation indicate that 1 molecule of nitrogen combines with 3 molecules of hydrogen to form 2 molecules of ammonia. N 2 (g )

+ 3H 2 (g )  → 2 NH 3 (g )

1 molecule

3 molecules

2 molecules

Multiplying by 6.022 × 1023 the entire equation we get 1× 6.022 × 1023

3 × 6.022 × 1023

2 × 6.022 × 1023

molecules

molecules

molecules

This means 1mol of nitrogen reacts with 3 mols of hydrogen to form 2 mols of ammonia. By taking molar masses into consideration it can be said that 28 g of nitrogen reacts with 6 g of hydrogen to form 34 g of ammonia. Thus a balanced chemical equation represents a stiochiometric equation. The exact quantities of the reactants and the products that appear in the balanced chemical equation are known as stoichiometric quantities. A chemical equation conveys both qualitative and quantititative information. Qualitatively a chemical equation tells the names of the various reactants and the products. Quantitatively a chemical equation represents (i) The relative number of reactant and product species (atoms or molecules) taking part in the reaction. (ii) The relative number of moles of the reactants and products. (iii) The relative masses of the reactants and products. (iv) The relative volumes of gaseous reactants and products. The chemical equation can be made more informative by the following modifications. (i) The physical states of reactants and products can be indicated by using the abbreviations; s for (solid), (l) for liquid, (g) for gas and (aq) for aqueous solution. For example, Zn (s) + 2HCl(aq )  → ZnCl2 (aq ) + H 2 (g ) (ii) To indicate the strength of acid or base dil for dilute or conc for concentrated is written before the formula of acid or base. Zn (s) + dil 2HCl  → ZnCl2 (aq ) + H 2 (g)

(iii) The reaction conditions such as presence of catalyst temperature, pressure etc, may be written above the arrow between the reactants and products N 2 (g ) + 3H 2 (g )

Fe / M 0 725K 2 NH 3 600 atm

(iv) Heat change taking place during the reaction may be expressed in any of the following two ways N 2 (g) + 3H 2 (g) → 2 NH 3 + 93.6 KJ N 2 (g) + 3H 2 (g) → 2 NH 3 ; ∆H = − 93.6 KJ (v) The formation of a precipitate, if any, in a reaction can be expressed by writing the word ppt. Or by an arrow pointing downward. AgNO3 (aq ) + NaCl(aq ) → Ag Cl(s) ↓ + NaNO3 (aq ) ppt

Similarly, the evolution of a gas in a chemical reaction can be indicated by an arrow pointing upward. Zn (s) + 2HCl(aq ) → ZnCl2 (aq ) + H 2 (g ) ↑ (vi) The distinction between slow and fast reactions can be made by writing word slow and fast on the arrow head. (vii)The reversible nature of the reaction may be indicated by two opposite half arrows which indicates that the reaction is occurring in the forward as well as in the backward direction.  ⇀ N 2 (g ) + 3H 2 (g ) ↽   2 NH 3 (g ) Backward Forward

13.5.2 Balancing of chemical equation There are many methods to balance a chemical equation. Some of these methods are (i) Hit and trial method or trial and error method (ii) Partial equation method (iii) Oxidation number method (iv) Ion-electron method The first two methods are discussed below while the other two methods are discussed in chapter 10 Redox reactions. hit and trial method

This method involves the following steps. 1. Write the symbols and formulae of the reactants and the products in the form of a skeleton equation. 2. If elementary gases such as hydrogen, oxygen etc. are involved in the equation, these are represented in their atomic form in the beginning and after balancing the equation is changed to molecular form. 3. The formula which contains maximum number of elements is selected first and the atoms present in it are balanced.

13.16

Stoichiometry

4. In case, the above method fails, then start balancing the atoms which appear minimum number of times. 5. Verify that the number of atoms of each element is balanced in the final equation. 6. While balancing the chemical equation, the chemical formula of any compound should not be changed for the sake of convenience because each compound has a fixed chemical formula. To understand the application of above points let us consider the following equation. Fe + H 2 O → Fe3 O 4 + H 2 1. Changing the elementary substance hydrogen to atomic form Fe + H 2 O  → Fe3 O 4 + H 2. Selecting Fe3O4 which contain largest number of atoms first multiply H2O with 4 to balance oxygen atoms. In 4 molecules of H2O there are 8 H atoms. So multiply H with 8 on right hand side. There are 3 Fe atoms on right hand side. So Fe on left hand side is multiplied by 3. Thus the balanced equation in atomic form will be 3Fe + 4H 2 O  → Fe3 O 4 + 8H Convert to molecular form 3Fe + 4H 2 O  → Fe3 O 4 + 4H 2 It is the balanced equation. Partial equation method

This method is better method than hit and trial method. The following steps are involved in this method. 1. The given reaction is supposed to take place in different steps. These different steps are called partial equations. 2. Each partial equation is balanced separately by hit and trial method as described in the earlier method. 3. The partial equations are multiplied by suitable numbers if necessary, so as to cancel out the species which are not involved in the original reactions and the final products. 4. Partial equations are then added to get the final balanced chemical equation. To apply these points let us consider the reaction between copper and concentrated nitric acid. Cu + HNO 3 → Cu ( NO3 ) 2 + NO 2 + H 2 O 1. The probable partial equations for the above reactions are Cu + HNO 3 + O → Cu ( NO3 ) 2 + H 2 O HNO 3 →

H 2 O + NO 2 + O

2. Balance the partial chemical equations separately by hit and trial method as Cu + 2HNO 3 + (O) → Cu ( NO3 ) 2 + H 2 O 2HNO 3 →

H 2 O + 2 NO 2 + (O)

3. These are balanced chemical equations. 4. Add the above two equations. After cancelling the ‘O’ atoms in the two equations as they do not appear in the final equation Cu + 2HNO 3 + (O) → Cu ( NO3 ) 2 + H 2 O 2HNO 3 →

H 2 O + 2 NO 2 + (O)

Cu + 4HNO3  → Cu ( NO3 ) 2 + 2H 2 O + 2 NO 2

13.6 numericAl cAlculAtion BAsed on chemicAl equAtions It is possible to calculate the amounts of products formed in a chemical reaction, if the amounts of the reactants participating in the reaction and the stoichiometric equation are known. Such calculations help us to know beforehand whether a particular chemical reaction under consideration is economically viable for the production of a required product or not. Stoichiometric calculations are of several types like the following: 1. Mole to mole relationships: In these problems, the moles of one of the reactants/products is to be calculated if that of other reactants/products are given. 2. Mass-mass relationships: In these problems, the mass of one of the reactants/products is to be calculated if that of the other reactants/products are given. 3. Mass-volume relationship: In these problems, mass, or volume of one of the reactants or products is calculated from the mass or volume of other substances. 4. Volume-volume relationship: In these problems, the volume of one of the reactants/products is given and that of the other is to be determined. 5. Mass-volume-energy relationship: In these problems, mass or volume of one of the reactants or products and the energy liberated is calculated from the mass or volume of other substances. The main steps for solving such problems are (i) Write down the balanced chemical equation. (ii) Write down the moles or gram atomic or gram molecular masses of the substances whose quantities are given or have to be calculated. In case, there are two or more atoms or molecules of a substance, multiply the mole or gram atomic mass

Stoichiometry 13.17

or molecular mass by the number of atoms or molecules. (iii) Write down the actual quantities of the substances given. For the substances whose weights/ volume have to be calculated, write the sign interrogation (?) (iv) Calculate the result by an unitary method. In many cases during the chemical reaction, one of the reactants is taken in excess. This is to ensure the completion of reaction. However, on completion of reaction, some of the reactant, taken in excess, is left over. For example, consider the combustion hydrogen 2 H 2 (g ) + O 2 → 2 H 2 O(g ) Suppose that 2 moles of H2 and 2 moles of O2 are available for reaction. It follows from the equation that only 1 mole of O2 is required for complete combustion of 2 moles of H2, 1 mole of O2 will, therefore, be left over on completion of the reaction. The amount of the product obtained is determined by the amount of the reactant that is completely consumed in the reaction. This reactant is called the limiting reagent Thus, limiting reagent may be defined as the reactant which is completely consumed during the reaction. The reactant which is not completely consumed is referred to as excess reactant. In the above example, H2 is the limiting reagent and oxygen is excess reactant. The amount of water formed will therefore be determined by the amount of H2. Since 2 moles of H2 are taken, it will form 2 moles of H2O on combustion. 1. Calculation based on mole to mole relationship

Solution: In Haber’s process, ammonia is manufactured as per the reaction N 2 + 3H 2 → 2 NH 3 8.2 mol

?

2 mol of NH3 are produced from 1 mol of N2 1 8.2 mol of NH3 are produced from × 8.2 = 4.1 mol of N2 2 2. Mass to mass relationship solved Problem 14 Chlorine is prepared in the laboratory by treating manganese dioxide (MnO2)with aqueous hydrochloride acid (HCl) according to the reaction MnO 2 (s) + 4HCl(aq ) → MnCl2 (aq ) + 2H 2 O(s) + 5Cl2 (g ) How many grams of HCl react with 5.0 g of manganese dioxide? Solution: The balanced equation is MnO 2 (s) + 4HCl(aq) → MnCl2 (aq) + 2H 2 O(s) + 5Cl2 (g ) 87 g

146 g

87 g of MnO2 react with 146.0 g of HCl 5 g of MnO2 react with =

146 × 5 = 8.39 g 87

3. Mass-volume relationship solved Problem 15

solved Problem 12 How many moles of iron can be made from Fe2O3 by the use of 16 mol of carbon monoxide in the following reaction Fe 2 O3 + 3CO → 2Fe + 3CO 2 Solution: The reaction Fe 2 O3 + 3CO  → 2Fe+ 3CO 2 16 mol

?

3 mol of CO are used to make 2 mol Fe 2 16 mol of CO can produce × 16 = 10.67 mol Fe 3

When 50 g of a sample of sulphur was burnt in air 4% of the sample was left over. Calculate the volume of air required at STP containing 21% oxygen by volume. Solution: Wt of sample = 50 g Amount left over after combustion = 4% For 100 g of sample 4 g are left over 4 50 g of sample × 50 g = 2 g 100 \ 2 g impurity is present in 50 g of sulphur sample \ The weight of sulphur in the sample = 50−2 = 48 g The stoichiometric equation S + O 2  → SO 2

solved Problem 13 How many moles of nitrogen are required to produce 8.2 moles of ammonia in Haber’s process by reaction with hydrogen.

32g 22.4 L STP

32 g of sulphur will completely burn in 22.4 L O2 at STP

13.18

Stoichiometry

22.4 L × 48 g = 33.62 of 48 g of sulphur requires 32 g O2 at STP Volume of O2 in air = 21% 21 lit of O2 is available in 100 L air (STP) 33.6 lit of O2 is available in

100 L × 33.6 L = 160 L air 21 L

46. A polystyrene, having formula Br3C6H2(C8H8)n was prepared by heating styrene with tribromo benzyl peroxide in the absence of air. If it was proved to contain 10.46% bromine by weight, find the value of n. 47. The chief component of glass is silica (SiO2) It can be dissolved by hydrofluoric acid HF, according to the following reaction that produce silicon tetrafluoride SiF4, a gas at room temperature.

4. Volume-volume relationship

SiO2+4HF→SiF4+2H2O

solved Problem 16 What volume of ammonia is formed when 2.24 lit of N2 combine with 4.48 at STP? Solution: The balance equation N2

48. 49.

+ 3H 2  → 2 NH 3

1 mol 3 mol 22.4 L 3 × 22.4 = 67.2 L

2 mol 2 × 22.4 = 44.8 L

Hence the 2.24 L of N 2 require 6.72 L of hydrogen, but the available H2 is 4.48 L. So it is the limiting reagent. \ 4.48 L of H2 forms

50.

51.

2 × 22.4 × 4.48 = 2.987 L 3 × 22.4

\ The volume of ammonia formed is 2.987 L. 52. Problems for Practice 41. How many grams of oxygen are required to burn completely 570 g of octane? 42. Calculate the weight of iron which will be converted into its oxide by the action of 18 g of steam. (MLNR 1996) 43. A 1.0 g sample of KClO3 was heated under such conditions that a part of it decomposed according to the equation (i) 2KClO 3 → 2KCl + 3O 2 and the remaining underwent a change according to the equation (ii) 4KClO 3 → 3KClO 4 + KCl If the amount of oxygen evolved was 146.8 mL at STP, what is the percentage by weight of KClO4 in the residue? 44. Calculate the residue obtained on strongly heating 2.67g of Ag2CO3. 45. Calculate the weight of lime (CaO) obtained by heating 200 Kg of 90% pure lime stone(CaCO3).

53.

How many grams and how many moles of SiF4 can be produced from 63.4 g of HF? (H = 1, O = 16, F = 19, Si = 28) How much KClO3 is needed to produce enough oxygen for burning 10g of coke? Determine the amount of litharge (PbO) produced by heating 132.4 g lead nitrate of 65% purity (Pb = 207.2, O = 16). An hourly requirement of an astronaut can be satisfied by the energy released when 34.2 g of sucrose (Cl2H22O11) are burnt in his body. How many grams of oxygen would be needed in a space capsule to meet his requirement for one day? A compound of iron and chlorine is soluble in water. An excess of silver nitrate was added to precipitate the chloride ion as silver chloride. If 134.8 mg of the compound gave 304.8 mg of AgCl, what is the formula of the compound? (Fe = 56, Ag = 108, Cl = 35.5) Sphalerite is a zinc sulphide (ZnS) mineral and an important commercial source of zinc metal. The first step in the processing of the ore consists of heating zinc sulphide with oxygen to give zinc oxide, ZnO and SO2. How many Kg of oxygen gas combine with 5.00 × 103 g of zinc sulphide in this reaction? Aspirin (an analgesic drug) is prepared by heating salicylic acid C7H6O3, with acetic anhydride C4H6O3, the other product is acetic acid C2H4O2. C7H6O3+ C4H6O3→C9H8O4+C2H4O2

(i) What is the theoretical yield (in g) of aspirin, C9H8O4, when 2.00 g of salicylic acid is heated with 4.00g of acetic anhydrid? (ii) If the actual yield of aspirin is 2.10g what is the percentage yield? 54. Calcium carbide, CaC2 used to produce acetylene, C2H2 is prepared by heating calcium oxide, CaO and carbon to high temperature CaO(s) + 3C(s) → CaC 2 (s) + 2CO(g ) If a mixture contains 1.15 kg of each reactant, how many grams of calcium carbide can be prepared?

Stoichiometry 13.19

55. Titanium, which is used to make aeroplane engines and frames, can be obtained from titanium tetrachloride, which in turn is obtained from titanium oxide by the fallowing process. 3TiO2(s) + 4C(s) + 6Cl2(g) → 3TiCl4(g) + 2CO2(g) + 2CO(g)

56.

57.

58.

59.

60

A vessel contains 4.15 g TiO2, 5.67 g C and 6.78 g Cl2, suppose the reaction goes to completion as written. How many grams of TiCl4 can be produced (Ti = 48)? What volume of hydrogen at NTP would be liberated by the action of zinc on 50 mL of dilute sulphuric acid of 40% purity and having specific gravity of 1.3 (Zn = 65)? A mixture of aluminum and zinc weighing 1.67 g was completely dissolved in acid. 1.69 litre of hydrogen measured at 0°C and 1atm pressure was evolved. What was the weight of aluminum in the mixture? Gastric juice contains 3.0 g HCl per litre. If a person produce 2.5 litres of gastric juice per day, How many antacid tablets containing 0.4 g of Al(OH)3 are needed to neutralize all the acid produced per day? A mixture of formic acid and oxalic acid is heated with conc. H2SO4. The gas produced is collected and on its treatment with KOH solution the volume of the gas decreases by one sixth. Calculate the molar ratio of the two acids in the original mixture. In a rocket motor fueled with butane C4H10, how many kg of liquid oxygen should be provided with each kg of butane to provide complete combustion?

64. A mixture of NaCl and KCl weighted 5.4892 g. The sample was dissolved in water and treated with an excess of silver nitrate in solution. The resulting AgCl weighed 12.7052 g. Calculate the percentage of NaCl in the mixture? 65. A mixture of NaCl and NaBr weighing 3.5084 g was dissolved and treated with enough AgNO3 to precipitate all the chloride and bromide as AgCl and AgBr. The washed precipitate was treated with KCN to solubilize the silver, and the resulting solution was electrolyzed. NaCl + AgNO3→AgCl + NaNO3 NaBr + AgNO3→AgBr+ NaNO3 AgCl + 2KCN→K[Ag(CN2)] + KCl AgBr + 2KCN→K[Ag(CN2)]+ KBr 4K[Ag(CN)2] + 4 KOH→4Ag + 8KCN2 + 2H2O + O2 After the final step was complete, the deposit, of metallic silver weighed 5.5028 g, determine the composition in the initial mixture. 66. The insecticide DDT is made by the reaction Cl CCl3CHO + 2C6H5Cl

2C4H10 + 13O2→8CO2+ 10H2 O 61. How much 83.4 % pure salt cake (Na2SO4) could be produced from 250 kg of 94.5% pure salt in the reaction 2NaCl+ H2SO4 →Na2SO4+ 2HCl? 62. A 1.2048 g sample of impure Na 2CO3 is dissolved and allowed to react with a solution of CaCl2. The resulting CaCO3 after precipitation, titration and drying was found to weigh 1.0362 g. Assuming that the impurities do not contribute to the weight of the precipitate, calculate the percent purity of the Na2CO3. 63. A particular 100 – octane aviation gasoline used 1.00 cm3 of tetra ethyl lead (C2H5)4Pb of density 1.66 g/ cm3, per litre of product. This compound is made as follows 4C2 H5 Cl + 4 NaPb  → (C2 H5 ) 4 Pb + 4 NaCl + Pb How many grams of ethyl chloride C2H5Cl is needed to make enough tetraethyl lead for 1.00 L of gasoline? (Pb = 207, Cl = 35.5, C = 12, H = 1)

CHCl3 + H2O Cl

67.

68.

69.

70.

71.

If 10 g of chloral were treated with 10 g of chorobenzene, how much DDT would, be formed? Assume the reaction goes to completion without side reaction or losses. Find the ratio of Hg 2I2 : HgI2 formed Equal masses of mercury and iodine are allowed to react completely to from a mixture of mercurous and mercuric iodides formed (I = 127; Hg = 201). 60 mL of a mixture of nitrous oxide and nitric oxide was exploded with excess of hydrogen. If 38 mL of N2 was formed, calculate the volume of each gas in mixture. All measurement are made at constant P and T, assume H2O in liquid phase. When a mixture of 10 moles of SO2 and 15 moles of O2 was passed over catalyst, 8 moles of SO3 was formed. How many moles of SO2 and O2 left uncombined during the reaction? A mixture of FeO and Fe3O4 when heated in air to a constant weight gains 5% in weight. Find the composition of the initial mixture. What weight of CO is required to form Re2 (CO)10 from 2.50 g of Re2O7 according to the unbalanced reaction Re2O7 + CO→Re2 (CO)10 + CO2 (Re = 186.2, C = 12 and O = 16)

13.20

Stoichiometry

72. Three different brands of liquid chlorine are available in the market for use in purifying water in swimming pools. All are sold at the same rate of Rs 10 per litre and are water solutions. Brand A contains 10% hypochlorite (ClO; Wt/Vol), brand B contains 7% available chlorine (Cl) and brand C contains 14% sodium hypochlorite (NaOCl). Which of the three would you buy? 73. In the analysis of a 0.50 g sample of feldspar, a mixture of the chlorides of Na and K is obtained, which weights 0.1180 g. Subsequent treatment of the mixed chlorides with silver nitrate gives 0.2451 g of AgCl. What is the percentage of sodium oxide and potassium oxide in feldspar? 74. A solid mixture (5 g) consisting of lead nitrate and sodium nitrate was heated below 600°C until the weight of the residue is constant. If the loss in weight is 28%, find the amount of the lead nitrate and sodium nitrate in the mixture. (IIT 1990) 75. From the following reaction sequence CaC2 + H 2 O  → CaO + C2 H 2 C2 H 2 + H 2  → C2 H 4 n C2 H 4  → (C 2 H 4 ) n Calculate the mass of polyethylene which can be produced from 10 kg of CaC2. 76. From the following series of reactions Cl2 + 2KOH → KCl + KClO + H 2 O 3KClO → 2KCl + KClO3 4KClO3 → 3KClO 4 + KCl Calculate the mass of chlorine needed to produce 100 g of KClO4.

13.7 titrimetric method oF AnAlysis

to know when to stop the addition of titrant, some substances known as indicator which change colour by the excess titrant are used. The point where the indicator change colour by the excess titrant is called the end point. It is desirable that the end point must be as close as possible to the equivalency point, choosing the indicator to make the two points coincide. The two points is one of the important aspects of titrimetric analysis and is known as selection of indicator. The term titration refers to the process for measuring the volume of titrant required to reach the equivalency point. For longer time, the term volumetric analysis was used rather than titrimetric. The word titrimetric is preferable because volume measurements need not be confirmed to titrations. In some analyses, for example, one might measure the volume of a gas.

reactions used for titrations The chemical reactions which serve as the basis for titrimetric determinations are conveniently grouped into four types. 1. Acid-base: There are large number of acids and bases which can be determined by titrimetry. If HA represents the acid to be determined and BOH the base the reactions are HA+OH– → A– +H2O And BOH+H3O+ → B+ +2H2O The titrants are generally standard solution of strong electrolytes such as sodium hydroxide and hydrochloric acid. 2. Oxidation-reduction (redox): Chemical reactions involving oxidation reduction are widely used in titrimetric analyses. For example, iron in the +2 oxidation state can be titrated with a standard solution of cerium (IV) sulphate. Fe2++Ce4+→Fe3++Ce3+ Another oxidizing agent which is widely used as titrant is potassium permanganate KMnO4. Its reaction with iron (II) in acid solution is.

A titrimetric method of analysis is based on a chemical reaction like

5Fe 2 + + MnO 4− + 8H +  → 5Fe3+ + Mn 2 + + 4H 2 O

a A + t T → Products

3. Precipitation: The precipitation of silver cation with the halogen anions is a widely used titrimetric procedure. The reaction is

Where ‘a’ molecules of analyte ‘A’ react with t molecules of the reagent ‘T’ called as titrant. It is added incrementally normally from a burett in the form of a solution of known concentration. The solution whose concentration is exactly known is called as standard solution and its concentration is determined directly from the weight of the substance dissolved or by a process called standardization. The addition of the titrant is continued until the amount of T chemically equivalent to that of A has been added. It is then said that the equivalency point of the titration has been reached. In order

Ag++X–→AgX(s) Where X– can be chloride, bromide, iodide or thiocyanate (SCN–) ion. 4. Complex formation: The complex formation reactions can also be used in titrimetry. An example of reaction in which a stable complex is formed is that between silver and cyanide ions. Ag++2CN– → [Ag(CN)2]–

Stoichiometry 13.21

This reaction is the basis of the so called Liebig method for the determination of cyanide. Certain organic reagents like ethylene damine tetra acetate (EDTA) form stable complexes with a number of metal ions, and are widely used for the titrimetric determination of those metals. requirement of a reaction to use in titration All the known reactions cannot be used as the basis for titration. A reaction must satisfy the following requirements to use for a titration. 1. The reaction must proceed to virtual completion at the equivalency point i.e., the equilibrium constant must be very high. 2. The reaction must proceed according to a definite chemical equation. There should be no side react ions 3. Some method like indicator or instrumental method must be available to know when the equivalency point is reached. 4. The reaction must be rapid so that the titration can be completed in a few minutes. The reaction between HCl and NaOH has equilibrium constant 1014 and goes to virtual completion very fast. At the equivalency point the pH changes by several units and a number of indicators are available. So this reaction is well suited for titration. On the other hand the reaction between boric acid and sodium hydroxide has an equilibrium constant 6 × 104 which is less and hence the reaction is not complete. Similarly the reaction between acetic acid and ethyl alcohol also does not go to completion and takes place slowly. Such reactions are not suitable for titrations.

equivalent weights

Molecular weight Basicity of the acid

1 1 mole of a divalent cation, 2 3

mole of a trivalent cation and so on. The equivalent weight of a substance is called an equivalent, just as the molecular weight is called mole. The equivalent and molecular weights are related as E.W. =

MW n

Where n is number of moles of hydrogen ions, electrons or univalent cations furnished or combined with the reacting substance. One equivalent of any acid react with one equivalent of any base; one equivalent of any oxidizing agent reacts with one equivalent of any reducing agent and so on. Some compounds may undergo more than a single reaction and hence can have more than one equivalent weight. For example, the permanganate ion can undergo the following reactions: MnO 4− + e −  → MnO 42 − MnO 4− + 4H + + 3e −  → MnO 2 + 2H 2 O MnO 4− + 8H + + 4e −  → Mn 3 + + 4 H 2 O MnO 4− + 8H + + 5e −  → Mn 2 + + 4H 2 O The equivalent weight of KMnO4 in such reactions is its molecular weight divided by 1, 3, 4 or 5 depending on the reactions occurred. The reaction of phosphoric acid with a base can be stopped when the following reaction has occurred.

Here the equivalent weight of the acid is the same as molecular weight. But if the reaction is carried further. H 3 PO 4 + 2OH −  → 2H 2 O + HPO 42 − Here the equivalent weight is half the molecular weight.

Basicity is the number of replaceable hydrogen’s in an acid. For bases, Equivalent weight =

of a univalent cation,

H 3 PO 4 + OH −  → H 2 O + H 2 PO −4

The equivalent weight of a substance involved in a reaction used as the basis for a titration is defined as follows 1. Acid-base: For acids, equivalent weight =

substance required to furnish or react with one mole

Mole cular weight Acidity of the base

Acidity of a base is the number of replaceable OH– ions in a base. 2. Redox: The gram-equivalent weight is the weight in grams of the substance required to furnish or react with one mole of electrons. 3. Precipitation or complex formation: The gram – equivalent weight is the weight in grams of the

concentration terms In titrimetric analysis, the concentration terms molarity and normality are most frequently used. There are other systems like formality, mole fraction, percent by weight, percent by volume, parts per million (ppm) are also employed. They are explained in solutions (Chapter 4). Molarity (M): The concentration system is based on the volume of solution and hence is convenient in laboratory procedures where the volume of solution is the quantity measured. It is defined as the the number of moles of solute per litre of solution represented by M.

13.22

Stoichiometry

M=

W W 1000 or × MW × V lit MW Vcc

Normality (N): Normality is the number of gram equivalent of solute per litre of solution and is represented by N. W N= EW × V lit

or

Equivalent mass of H3PO4

W × 1000 EW × Vcc

Where W is the weight of solute. MW is the molecular weight, EW is the equivalent weight of solute and V is the volume of solution. As mentioned earlier that the process by which the concentration of a solution is accurately ascertained is known as standardization. A standard solution can be sometimes be prepared by dissolving an accurately weighed sample of the desired solute in an accurately measured volume of solution. This method is not generally applicable since very few substances are available in pure form. Such substances. which can be used to prepare standard solution are called primary standards. 1. Equivalent mass of an acid Eq. wt of an acid =

For H3PO4 depending upon the no of hydrogen’s participated in neutralization reaction, equivalent mass changes (i) For the reaction H3PO4 + NaOH → NaH2PO4+H2O

=

98 = 98 (∴ one H + is neutralized ) 1

(ii) For the reaction H3PO4+2NaOH→Na2HPO4+2H2O Equivalent mass of H3PO4 =

98 = 49 (∴two H + are neutralized ) 2

(iii) For the reaction H3PO4+3NaOH→Na3PO4+3H2O Equivalent mass of H3PO4 =

98 = 32.66 3

2. Equivalent mass of a base

Molecular mass of acid Basicity of acid

Eq. mass of a base =

Molecular mass of base Acidity of base

Basicity of acid = No. of replaceable hydrogen atoms present in one molecule of acid

Acidity of base = No. of replaceable OH– ions present in one molecule of the base

table 13.2 Equivalent weight of certain acids

table 13.3 Equivalent weights of some common bases

Equivalent mass

Base

Acidity

Molecular mass

Equivalent mass

63

63 = 63 1

NaOH

1

40

40 = 40 1

36.5 = 36.5 1

KOH

1

56

36.5

56 = 56 1

Ca(OH)2

2

74

74 = 37 2

NH4OH

1

35

35 = 35 1

Acid

Basicity

Molecular mass

HNO3

1

HCl

1

H3PO2

1

66

66 = 66 1

CH3COOH

1

60

60 = 60 1

H2SO4

2

98

98 = 49 2

H2C2O4.2H2O

2

126

126 = 63 2

H3PO3

2

82

82 = 41 2

3. Equivalent mass of a salt Equivalent mass of salt =

Mol. wt of salt Total no. of valeencies of acidic/ basic radicals

Stoichiometry 13.23

table 13.4 Equivalent weights of some common salts Total no. of valencies of acidic/ Molecular Equivalent basic radicals mass mass

Salt NaCl

1

58.5

CaCl2

2

111

AlCl3

3

133.5

Ca3(PO4)2

6

310.0

Al2(SO4)3

6

342

58.5 = 58.5 1 111 = 55.5 2 133.5 = 45.17 3 310 = 51.67 6 342 = 57 6

4. Equivalent mass of an ion

formula wt of ion Equivalent mass of ion = charge of ion

(i) For example for KMnO4 (a) in acid medium

No. of charges Equivalent on the ion Mass of ion mass

Cl–

1

35.5

NO3−

1

62

SO 2− 4

2

96

3− 4

3

95

CH3COO–

1

59

Na+

1

23

Mg2+

2

24

Al3+

3

27

PO

35.5 = 35.5 1 62 = 62 1 96 = 48 2 95 = 31.67 3 59 = 59 1 23 = 23 1 24 = 12 2 27 = 9 3

+2

+7

2 KMnO 4 +3H 2SO 4  → K 2SO 4 + 2 Mn O 4 +3H 2 O +5(O) Eq. wt of KMnO 4 =

+7

Eq. wt of KMnO 4 =

Mol of compound 2 × No. of 'O'atoms given by one molecule or

Formula wt of oxidising Total changein oxidation number of an elemeent that undergoes reduction

158 = 52.66 3

(c) Alkaline medium (strong alkaline) +6

+7

2 KMnO 4 +2KOH  → K 2 Mn O+H 2 O +(O) Eq.wt of KMnO 4 =

158 =158 1

In dilute alkaline medium MnO 4− +2H 2 O +3e − → MnO 2 +OH − Eq.wt of KMnO 4 =

158 =52.66 3

(ii) For K2Cr2O7 K2Cr2O7+4H2SO4→K2SO4+Cr2(SO4)3+4H2O+3(O) Eq.wt of K 2 Cr2 O7 =

294 = 49 6

6. Equivalent weight of reducing agent Eq. wt of reducing agent = =

5. Equivalent weight of an oxidizing agent

=

+2

2 KMnO 4 +H 2 O  → 2 MnO 2 +2KOH +3(O)

mol.wt of the compound 2 × no of 'O' atoms used by one molecuule formula wt. of reducing agent mber of an Total change in oxidation num element that undergoes oxidation =

Eq. wt of oxidizing agent

158 =31.6 g 5

(b) Neutral medium

table 13.5 Equivalent weight of some common ions Ion

Molecular wt of the the compound No. of 'e − 'gained by1 mole

or

Molecular weight of compound No of 'e' lost by 1 mole

(i) For oxalic acid H2C2O4·2H2O+(O)→2CO2 + 3H2O 126 g 16 g Equivalent mass of oxalic acid =

6 Mol. wt of compound 126 = 63 = 2× No. of 'O' atoms used per 1 mole 2

13.24

Stoichiometry

(ii) Ferrous suphate 2FeSO4+H2SO4+(O)→Fe2(SO4)3+H2O 2×152 Eq wt. of FeSO 4 = =152 2

13.7.1 Acid-Base titrations Acid-base equilibrium is an extremely important topic througout chemistry and in other fields such as agriculature, biology and medicine. Titrations involving acids and bases are widely employed in the analytical control of many products. In an acid-base titration when an acid is added to a base or vice-versa the pH of the solution changes. Now let us see how the pH of solution changes when a base is added to an acid or vice-versa in an acid-base titration. For example 50 mL of 0.1 M HCl is titrated with 0.1 M NaOH. The initial pH will be [H3O+] = 0.1 pH =1.0 When 10 mL of base is added the following reaction takes place −

H 3 O +OH  2H 2 O +

To calculate the pH of solution first the concentration of the solution must be known as follows V1 N1 − V2 N 2 Total volume Where V1 is the volume of acid, N1 is the normality of the acid while V2 and N2 are the volume and normality of base respectively  H 3 O +  =

(50×0.1) − (10×0.1) 60

=6.67×10−2 m mol/mL

pH = 2−log 6.67 = 1.18 The pH values for other volumes of titrant can be calculated in a similar manner. The pH values when different volumes of 0.1 M NaOH is added to 0.1 M HCl are given in Table 13.6.

table 13.6 pH values of solution in the titration of 0.1 M HCl with 0. 1 M NaOH S. No.

NaOH mL Total volume added of solution

pH

1

0.00

50.00

1.00

2

10.00

60.00

1.18

3

20.00

70.00

1.37

4

30.00

80.00

1.60

5

40.00

90.00

1.95

6

49.00

99.00

3.00

7

49.90

99.90

4.00

8

49.95

99.95

4.30

9

50.00

100.00

7.00

10

50.05

100.05

9.70

11

50.10

100.10

10.00

12

51.00

101.00

11.00

13

60.00

110.00

11.96

14

70.00

120.00

12.23

From the table, it can be seen that initially the pH rises gradually as the titrant is added, rises more rapidly as the equivalent point is approached and then increases by about 5.2 units for the addition of only 0.1 mL of bases at the equivalency point. Beyond the equivalency point, the pH again increases only slowly as titrant is added. In this titration both the acid and base are strong. Depending on the strength of the acid and base involved in the titration, the acid-base titrations are of four types. 1. Titration of strong acid with strong base e.g., HCl vs NaOH 2. Titration of strong acid with weak base e.g., HCI vs NH4OH 3. Titration of weak acid with strong base e.g., CH3COOH vs NaOH 4. Titration of weak acid with weak base e.g., CH3COOH vs NH4OH The change in the pH of solution at the equivalence point depends on the strength of the acid and base involved in the titration. Depending on the change in pH of the solution at the equivalence point, the indicator should be selected as explained in Chapter 10 (Ionic equilibria).

Stoichiometry 13.25

Analysis of a mixture of carbonates and hydroxide When a strong acid reacts with a mixture of two or more basic substances, the strongest base is neutralized first and during this neutralization, other bases remain inert. Once the reaction with strong base is completed, the next strongest base starts neutralizing. Thus neutralization of a mixture of acids or bases takes place sequentially starting from the strongest and ending with the weakest. When a solution containing both NaOH and Na2CO3 is titrated with HCl using phenolphthalein indicator, the amount of acid used corresponds to complete the following neutralization reactions. NaOH+HCl→NaCl+H2O

(1)

Na2CO3+HCl→NaHCO3+NaCl

(2)

When these two reactions (1 and 2) are complete phenolphthalein changes in colour from pink to colourless. If methy orange is added to this solution it will be yellow in colour due to NaHCO3. Then by adding HCl further neutralization takes place as follows. NaHCO3+HCl→NaCl+H2O+CO2

(3)

When the neutralization is complete the solution acquires red colour. Now the amounts of NaOH and Na2CO3 can be calculated as follows. Let the volume of HCl consumed by NaOH in the reaction (1) is X mL and the volume of HCl consumed by Na2CO3 in the reaction 2 is Y mL, the total volume consumed in the two reactions is x+y mL. The volume of NaOH consumed after phenolphthalein end point and to the methyl orange end point will be again Y mL as per the reaction 3. So from theses values the volume of NaOH consumed by NaOH is X mL and the volume of NaOH consumed by Na2CO3 will be Y mL. Analysis of a mixture of carbonate and Bicarbonate Here sodium carbonate is more basic than sodium bicarbonate. When a mixture of sodium carbonate and sodium bicarbonate is titrated with a strong acid solution, stronger base Na2CO3 is neutralized first. +

+

Na 2 CO3 + H  → 2 Na + HCO

− 3

When all carbonates are converted into bicarbonates, the solution is still basic but pH falls from 13 to 10 (approximately) since carbonate can turn the phenolphthalein to pink, the volume of acid consumed in the phenolphthalein end point corresponds to the conversion of carbonate to bicarbonate. Then the total bicarbonate present in the solution is estimated by using methyl orange as indicator.

The volume of acid consumed in the phenolphthalein end point corresponds to the amount of carbonate while the volume of acid consumed in the methyl orange end point corresponds to both carbonate and bicarbonate amounts. From these values the amounts of carbonate and bicarbonate can be calculated. Analysis of Phosphoric Acid and its salt The neutralization of phosphoric acid with sodium hydroxide take place in the following steps. H3PO4→NaH2PO4+H2O Phenolphthalein can change its colour at this point. In the second step NaH2PO4+NaOH→Na2HPO4+H2O In this case methyl orange can change its colour at this point. The ionization of HPO2− 4 is very small and it cannot be titrated by a strong base. But in reverse if salt of phosphoric acid is titrated against HCl or H2SO4 following three step neutralizations occur Na3PO4+HCl→Na2HPO4+NaCl Na2HPO4+HCl→NaH2PO4+NaCl NaH2PO4+HCl→H3PO4+NaCl

(1) (2) (3)

At the end of the reaction 1, the pH of the solution drops to approximately 9-10 and phenophthanleim can be used as suitable indicator. After the second reaction is complete, pH drops to approximately 5 and methyl orange can be used as indicator to detect the end point. When the reaction-3 is complete, the pH drops to very low and thymol blue can be used to detect the end point. solved Problem 17 What is the strength in g per litre of solution of sulphuric acid 12 mL of which neutralize 15 mL of N/10 sodium hydroxide solution. (Roorke 1987) Solution: V1 N NaOH

= V2 N 2 H 2SO 4

1 × 15 = N 2 × 12 10 15 N2 = = 0.125 10 × 12 Normality × Eq.m mass = Strength ( g / L ) Strength = 0.125 × 49 = 6.125 g / L

13.26

Stoichiometry

solved Problem 18 25 mL of mixture containing Na2CO3 and NaOH requires 19.5 mL of 1/200 N HCl using phenolphthalein as indicator. If methyl orange is indicator then 25 mL of solution requires 25.9 mL of the same HCl for end point. Calculate the strength of each substance in mixture. Solution: For phenolphthalein as indicator Meq. of HCl used for 25 mL of mixture 1 = 19.5 × = 0.0975 200 1 Meq of Na2CO3 = 0.0975 2 For methyl orange as indicator 1 = 0.1295 Meq of HCl used = 25.9 × 200 ∴ Meq. of NaOH +

∴ Meq of NaOH + Meq of Na2CO3 = 0.1295 By Eqns (1) and (2) Meq of NaOH = 0.0655 WNaOH × 1000 = 0.0655 40

Meq. of Na2CO3 = 0.064 WNa CO × 1000 × 2 2

3

106

= 0.064

WNa2CO3 = 3.392×10–3 g per 25 mL

∴ WNaOH = 2.62×10–3 g per 25 mL

13.7.2 redox titrations It is interesting to compare redox with acid-base reactionselectron transfer on one hand and proton transfer on the other. We can think Fe2+and Fe3+, for example, as conjugate pair by analogy to conjugate Bronsted acids and bases on the other hand, there are important differences. For instance, electrons can travel through wires while protons cannot. Thus to effect a proton transfer the donor and acceptor must encounter each other directly, where as in redox reactions the electron donor and acceptor can be in separate solutions if we wish. The direct reaction is also possible. There is a marked difference in the reaction rates. Acid-base reactions are very fast and are known as instantaneous. Redox reactions are some times slow and titrimetric methods based on these reactions require elevated temperatures, addition of catalyst or perhaps an excess reagent followed by back titration. Slowness shows that redox reactions are complex and the electron transfer is only one part of multistep sequence which involve forming or breaking covalent bonds, protonation and various sorts of rearrangements.

Another marked difference between acid-base and redox reaction of water, the common solvent. Protons are rapidly transferred to H2O to form H3O+ and these protons can be easily passed to base to regenerate H2O. Similarly removal of proton from water to form OH– is readily reversed if protons are supplied by another acid. Further all species H3O+, H2O and OH–are highly soluble. By contrast, the addition of electrons to water results in the formation of hydrogen, a slightly soluble gas. 2H 2 O + 2e −  → 2OH − + H 2 The products tend to escape from the solution and moreover even if remained, many of its reactions are so slow under reasonable conditions that it would seldom be suitable titrant. Similar considerations arise if water is oxidized. 2H 2 O  → O 2 + 4 H + + 4e − Thus while strong acids and bases that undergo proton exchange with water are good titrants, reagents that oxidise or reduce water are usually avoided; i.e., the strongest oxidants and reductants are impractical titrants. We do not encounter the leveling effect such as that occur in acid-base chemistry. Redox reactions are widely used for titrimetric determinations of both inorganic and organic substances. In addition, prelimanary redox steps to establish desired oxidation states precede the measurements in much analysis. Redox potential: Metal ions act as oxidizing or reducing agents as a result of their ability to gain or lose electrons forming ions of lower or higher valency which represent different stages of oxidation. An atom of iron, for example, may lose two electrons to form the ferrous ion or three electrons to form the ferric ion. Fe  → Fe 2 + + 2e − → Fe3+ + 3e − Fe  A ferric ion acts as an oxidizing agent because it can gain one electron and be converted into the lower state of oxidation, i.e., ferrous ion. Fe3++e– →Fe2+ Similarly, the ferrous ion acts as a reducing agent because it can lose one electron to be converted to the ferric ion. Fe 2 +  → Fe3+ + e − However, if we take a mixed solution of ferric and ferrous ions the solution will possess reducing or oxidizing properties according as whether it can lose or gain electrons. If a platinum plate be immersed in such a solution and the

Stoichiometry 13.27

solution tends to oxidise the plate. Electrons will be removed from the plate which acquire a positive charge. If on the other hand, the solution tends to reduce the plate it will give up electrons to the plate and the plate acquires negative charge. The magnitude and sign of the charge given to the plate, therefore measure the oxidizing or reducing power of the solutions. If we set up a cell of the type Pt|Fe 2+||Fe3+ | Pt and join the two electrodes by a wire, the ferrous ions will give electrons to the platinum electrode and thus pass into ferric state, whereas these electrons will travel by the connecting wire and be taken up by the ferric ions to be reduced to the ferrous state. The potential difference so produced will go on developing until it puts a stop to this process. The potential E of the platinum electrode dipping in solution containing ferric and ferrous ions when measured against a standard electrode will be given by nernst equation.  Fe3+  2.303RT log10  2 +  E=E + nF  Fe  Where E is the potential which is set up when the platinum electrode is dipped in 50% mixture of ferrous and ferric ions and is known as the redox potential or formal potential. R is gas constant, F is faraday (96,500 coulombs), T is absolute temperature, n is the difference in the valencys of the ion. On Substituting the values of R, F = 96500 coulombs and T = 298 K we have. O

O

0.0591 log10 E=E − n O

 Fe 2 +   Fe3+ 

A higher positive redox potential indicates that the ion is in the higher oxidation sate and is acting as a strong oxidizing agent while high negative redox potential indicates the higher reducing power of the ion in the lower state of oxidation. So by using an oxidizing agent, we can titrate a reducing agent. In the titration of reducing agent with an oxidizing agent the redox potential changes just as the pH changes in the titration of an acid with a base. Here how the redox potential changes is explained by taking the example of the titration of 50 mL of 0.05 M Fe2+ solution with 0.1 M Ce4+ solution. In the initial stages there is no cerium. So it is impossible to use the cerium couple. The system contains Fe2+ the formal potential of iron couple is 0.771 V (vs. S.H.E.) but as per the Nernst equation. E = 0.771 −

0.059 0.05 =−∝ log 1 0

But a little Fe3+ will present due to the oxidation by air. Let us assume 0.1% of the iron is present as Fe3+ which

gives Fe2+ ratio 1000:1. In such condition the concentration of Fe3+ is equal to 0.05/1000, i.e., 5×10–5 M. E = 0.771 − 0.059 log

0.05 = +0.594 ν 5 × 10 −5

When 10 mL of 0.01 M Ce4+ solution is added, the system contains both unreacted Fe2+ and Fe3+ formed by the oxidation of Fe2+ by Ce4+ as per reaction. 3+ 3+  Fe 2 + + Ce 4 + ↽ ⇀  Fe + Ce

Since the equilibrium constant is so high 5.3×1015, the reaction virtually goes to completion. Then the concentration of Fe2+ and Fe3+ can be calculated as described in the acid base titrations by using the formula. V M − V2 M 2 ( 0.05 × 50 ) − ( 0.1× 10 )  Fe 2 +  = 1 1 = = 2.55 × 10−2 Total volume ( 50 + 10 ) Ce 4 +  0.1× 10  Fe3+  = = = 1.7 × 10−2 Total volume 50 + 10 Since x moles of Ce4+ will oxidize x moles of Fe2+ to produce x moles of Fe3+ from the Nernst equation E = 0.771 − 0.0591 log

2.5 × 10−2 = 0.761V 1.7 × 10−2

At the equivalency point, the reaction truly goes to completion then [Fe 2+] and [Ce4+] are equal to zero. So at the equivalency point [Fe 2+] = [Ce4+] and [Fe 3+] = [Ce3+].  Fe 2 +  Ce3+  Therefore log  3+   4 +  = 0  Fe   Ce  0 And 2×Eeq pt = E 0Fe + E Ce O

0 E 0Fe + E Ce +0.771 + 1.70 = = 1.236V 2 2 O

Eeq.pt =

O

O

O

So the average of the two E values gives the potential at the equivalency point E 0eq pt . For any reaction in which the number of electrons lost by the reductant is the same as the number gained by the oxidant. The potential at the equivalency point is simply the arithmetic mean of the two standard potentials. O

E10 + E 02 2 O

E eq.pt =

O

After the equivalency point, all the iron exists as Fe3+, if Ce4+ solution is added there will be cerium couple [Ce3+]/[Ce4+] and its potential can be calculated in a similar way. The changes in the redox potentials in the titration of 50 mL of 0.05 M Fe2+ solution with 0.1 M Ce4+ solution are shown in the Table 13.7.

13.28

Stoichiometry

and possess property of being coloured in their oxidation form and colourless in the reduced state as “leuco” compounds. If the oxidation potential of the solution is more positive than the redox potential of the indicators, the indicator will present fully in the oxidized form and thus the solution will be coloured. If the oxidation potential of the solution is less than the redox potential of the indicator then the indicator will present in colourless reduced form and the solution will be colourless. A redox indicator must be selected suitable for a redox titration depending upon the redox potential of the indicator and the potential change at the end point.

table 13.7 Changes in potentials in the titrations of 50 mL of 0.05M Fe2+ solution with 0.1 M Ce4+ solution Volume of Titrant

E.V.

0.00 5.00 10.00 12.50 20.00 23.00 24.00 25.00 26.00 30.00 50.00

– +0.735 +0.761 +0.771 +0.807 +0.834 +0.853 +1.126 +1.62 +1.66 +1.70

Selection of the indicator

From the table, it can be seen that initially the potential rises gradually as the titrant is added, rises more rapidly as the equivalency point is approached and then increases by about 0.56 V for the addition of 0.1 mL of Ce4+ at the equivalency point. Beyond the equivalency point again increase slowly as the titrant is added.

For simplicity, the redox couple of a redox indicator is designated as follows  In + + e− ↽ ⇀  In Colour A

Colour B

The equation for the potential of this system is E = E 0In − 0.0591 log10 O

redox indicator 1. Self indictor: A coloured substance may act as its own indicator. For example, potassium permanganate solutions are so deeply coloured that a slight excess of this reagent in a titration can be easily detected. 2. Specific indicator: A specific indicator as a substance, which reacts in a specific manner with one of the reagents in titration to produce colour. Examples are starch, which forms a deep blue colour with iodine and thiocyanate ion which forms a red colour with iron (III) ion. 3. External or spot test indicators: These are once employed when no internal indicator is available. For example ferricyanide ion was used to detect iron (II) ion by formation of iron (II) ferricyanide (Turn bull's blue) on a spot plate outside the titration vessel. 4. Redox indicators: These undergo oxidation-reduction reactions in the titration. These are organic substances

[ In ]  In + 

Where E 0In is the standard potential of the indicator couple. If the ratio of  In +  [ In ] is 10:1 or greater only colour A can be seen. Also if the ratio is 1:10 or smaller only coloured B can be observed. That is O

10 = E 0In = − 0.0591 1 1 Colour B E = E 0In − 0.0591 Log10 = E 0In = + 0.0591 10 ______________________________________________ Subtracting ∆E = ± 2 × 0.0591 = ± 0.12 V Colour A

E = E 0In − 0.0591 log10 O

O

O

O

Thus, a change in potential about 0.12 V is required to bring about a change in colour of the indictor. The potential where a redox indicator changes colour is known as transition potential. Some redox indicators, their colour, transition potentials and conditions are given in the Table 13.8

table 13.8 Transition potentials of some redox indicators Indicator

Colour reductant

Colour oxidant

Transition potential V

Conditions

Diphenyl amine Diphenyl benzidine Diphenyl amine sulphonic acid Ferroin Phenosafrine Methylene blue

Colourless Colourless Colourless

Violet Violet Red Violet

0.76 0.76 0.85

1 M H2SO4 1 M H2SO4 Dilute acid

Red Colourless Colourless

Faint Blue Red Blue

1.11 0.28 0.36

1 M H2SO4 1 M acid 1 M acid

Stoichiometry 13.29

Obviously an indicator should change colour at or nearest equivalency potential. If the titration is feasible, there will be a large change in potential at the equivalency point. So an indicator whose transition potential lies in this range of potential change at the equilibrium must be selected. For example in the titration of iron (II) with cerium the potential changes from about 0.86 V to 1.26 V at the end point so the indicator ferroin whose transition potential falls in this ranges can be employed. solved Problem 19 How many mL of a 0.05 M KMnO4 solution are required to oxidize 2.0 g of FeSO4 in dilute solution (acidic). (Roorke 1986) Solution: 2KMnO4 + 10FeSO4 + 8H2SO4 → K2SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O 10 × 151.8 g of FeSO4 require KMnO4 = 2 × 158 g Amount of KMnO4 required by 2 g of FeSO4 =

2 × 158 × 2 10 × 151.8

Let V mL of KMnO4 solution (0.05 M) is required Amount of KMnO4 in this solution =

Thus

158 × 0.05 ×V 1000

158 × 0.05 × V 2 × 158 × 2 = 1000 10 × 151.8 ∴ V = 52.7 mL

solved Problem 20 0.5 g mixture of K2Cr2O7 and KMnO4 was treated with excess of KI in acidic medium. Iodine liberated required 100cm3 of 0.15N sodium thiosulphate solution for titration. Find the per cent amount of each in the mixture. (Roorke 1995) Solution: Let ‘a’ g of K2Cr2O7 be present in the mixture Mass of KMnO4 = (0.5 − a)g Eq. wt of K 2 Cr2 O7 = Eq. wt of KMnO 4 =

Mol. mass 294 = = 49 6 6

Mol. mass 158 = = 31.6 5 5

No. of equivalents of KMnO 4 =

0.5 − a 31.6

No. of equivalents of Na2S2O3 in 100 cm3 of 0.15 N solution 100 × 0.15 = = 0.015 1000 Equivalents of K2Cr2O7 + Equivalents of KMnO4 ≡ Equivalents of iodine ≡ Equivalents of Na2S2O3 ( 05 − a ) a = 0.015 + 49.0 31.6 17.4a = 1.274 or a = 0.0732 0.0732 × 100 = 14.64 % of K 2 Cr2 O 7 = 0.5 %KMnO 4 = 85.36

13.7.3 Precipitation titrations Precipitation reactions have been widely used in analytical chemistry in titrations. But titrations involving precipitation reactions are not so numerous as acid-base or redox titrations. In the beginning, they are limited to those involving precipitation of silver ion with anions such as the halogens and thiocyanate. The reason for such limited reactions are lack of suitable indicators and in some cases, particularly in the titration of dilute solutions the rate of reaction is too slow and is not suitable to carry a titration. Sometimes, the composition of the precipitate is frequently not known because of coprecipitation. We shall limit our discussion here to precipitation titrations involving silver salts with particular emphasis on the indicators which have been successfully employed in such titrations. Equilibrium calculations are based on the solubility product constant and this was discussed in detail in chapter 10 (ionic equilibrium). Here the titration of 50 mL of 0.1 M sodium chloride with 0.1 M silver nitrate was described. As the silver nitrate is added to sodium chloride solution, the silver chloride will be precipitated according to the reaction. NaCl + AgNO3  → AgCl ↓ + NaNO3 If the addition of silver nitrate is added gradually to sodium chloride the precipitate gradually increases and completes at the equivalency point where the concentration of both sodium chloride and silver nitrate are equal. The results are listed in Table.13.9. It can be seen from the table 13.9 that precipitation ceases completely after the end point. It is necessary to detect the equivalency point where the precipitation is complete for which a suitable indicator is required. In titrations involving silver salts, there are three indicators which have been successfully employed for several years. The Mohr’s

13.30

Stoichiometry

method uses chromate ion CrO 2− to precipitate brick red 4 Ag2CrO4. The volhard method uses Fe3+ ion to form a coloured complex with thiocyante ion SCN–. Fajan’s method utilizes adsorption indicators.

by EDTA with several metal ions having coordination number six. The structure of an EDTA metal complex is shown below.

table 13.9 Titration of 50 mL of 0.1 M NaCl with 0.1 M AgNO3 AgNO3mL

[Cl–]

%Cl– ppt.

0.00 10.00 20.00 30.00 40.00 49.00 49.9 50.0 50.1 51.0 60.0

0.1 0.067 0.043 0.025 0.011 0.001 1×10–4 1×10–5 1×10–6 1×10–7 1×10–8

0.00 20.0 40.0 60.0 80.0 98.0 99.8 100.0 100.0 100.0 100.0

13.7.4 complexometric titrations One of the type of chemical reactions which may serve as the basis of a titrimetric determination involves the formation of a soluble but slightly dissociated complex or complex ion. An example is the reaction of silver ion with cyanide ion to form the very stable [Ag(CN) 2]− complex ion. Ag + + 2CN −  →  Ag ( CN )2   

−

The metal ion in the complex is called as the central atom and the group attached to the central atom is called ligand. The number of bonds formed by the central metal atom is called the coordination number of the metal. Analytical applications based on the use of chealating agents (which can form heterocyclic rings with metal ions) as titrants for metal ions have shown remarkable growth in recent years. The widely used complexing agent for complexometric titrations is ethylenediaminetetracetic acid also known as dinitrilotetra acetate and given the abbreviation EDTA. It has the structure HOOC – CH2

CH2 – COOH N

HOOC – CH2

CH2

CH2

N CH2 – COOH

It is a hexadentate ligand since it can donate six pairs of electron pairs from two nitrogen atoms and four carboxylic groups. So 1:1 complexes are usually formed

2–

O C

CH2

C=O

CH2 CH2

O

N

O

M2+

O

N

CH2

O CH2

CH2

C=O C=O

The complexation reactions are really a competition between H2O and the ligand, or between the ligands present in solution. The ligand which can form stable complex will be formed. The reaction between bivalent cation M2+ And EDTA can be written as M 2+ +H 2 Y 2 −  MY 2 − +2H + For simplicity ethylenediaminetetraacetic acid is given the formula H4Y; the sodium salt Na2 H2Y and yields the complex forming H2Y2– in aqueous solution. The general equation may be written as n − 4+

M n+ + H 2 Y 2 −  [ MY ]

+ 2 H+

So as the EDTA is gradually added to the metal ion solution metal–EDTA colourless complex will be formed as above. Now the detection of equivalency point where the complexation is completed has to be done. For this purpose metallochromic indicators are used. Basically, the metallochromic indicators are coloured organic compounds that can form chealates with metal ions. The metal-metallochromic indicator chealate will have different colour from the colour of the free indicator. The metal-metallochromic indicator chelate is less stable than metal EDTA complex and hence the indicator will release the metal ion to EDTA titrant. The reactions which result in the colour change can be written as M 2+ + HIn 2 − colour A

 → MIn − +H + colour B

MIn +H 2 Y  → MY 2 − + H 2 In − −

colour B

2

colour A

Stoichiometry 13.31

So, in the beginning when a metallochromic indicator is added to metal ion solution, it will give colour B due to the formation of metal ion metallochromic indicator chelate. At the end point due to its release by the formation of stable metal ion-EDTA complex, it will have colour A. The most widely used indicator in complexometric titrations is Eriochrome black-T and its structure is HO

OH –O S 3

N= N CH3

It will form blue coloured solution. When added to a metal ion solution it will form a wine red coloured solution due to the formation chelate. At the end point due to the release from chealate by formation of stable metal ion-EDTA chealate the solution will have blue colour. Several other metallochromic indicators that can be used in the complexometric titrations of different metal ions with EDTA are known. They are murexide, xylenol orange, solochrome black etc. Application of complexometric titrations Complexometric titrations have been carried out successfully on nearly all common cations. These titrations have virtually replaced the former tedious gravimetric analyses for many metals in a variety of samples. There are several procedures employed 1. Direct titrations: These are carried out with at least 25 cations using metallochromic indicators. In order to prevent the precipitation of metal ions as metal hydroxides, complexing agents like citrate and tartrate are often added. The ammonia – ammonium chloride buffer is used for metal which form complexes with ammonia. The total hardness of water, calcium and magnesium can be determined by direct titration with EDTA using eriochrome black-T or calmagnite indicator. 2. Back titrations: These are used when the reaction between the cation and EDTA is slow or when a suitable indicator is not available. Excess EDTA is added and the excess is titrated with standard solution of magnesium using eriochrome black-T or calagmite indicator. This method can also be used to determine metals in precipitates such as lead sulphate and calcium in calcium oxalate.

3. Replacement titrations: These are used when a suitable indicator is not available for the metal ion being determined. An excess of the solution containing magnesium – EDTA complex is added to the metal ion solution to be determined. The metal ion displaces the magnesium from relatively weak EDTA complex. M 2+ + MgY 2 −  MY 2 − + Mg 2+ This displaced Mg2+ is then titrated with standard EDTA solution using eriochrome black-T or calgamite as indicator. Indirect determinations of several types are reported. Sulphate has been determined by adding excess barium ion to precipitate BaSO4. The excess Ba2+ is then titrated with EDTA. Phosphate has been determined by titration of the Mg2+ in equilibrium with the moderately soluble Mg (NH4) PO4. Since the metal ions differ in the stabilities of their EDTA complexes, it is occasionally possible to obtain consecutive end points for mixture of ions in a single titration. Hardness of water: The hardness of water is due to the presence of soluble chlorides, sulphates and bicarbonates of calcium and magnesium. The permanent hardness is due to the presence of soluble chloride and sulphates of calcium and magnesium. The sum of the temporary and permanent hardness is called the total hardness and it is represented in ppm. 100 mL of hard water is buffered with ammonia – ammonium chloride buffer of pH 10 and then titrated with 0.01 M EDTA, using eriochrome black-T. For example, 100 mL of a sample of water required 15.0 mL of 0.01 M EDTA then calculations can be made as follows M 2+ + [ H 2 EDTA ] → [ MEDTA ] + 2H + 2−

2−

From this equation, 1000 mL of 1 M EDTA 1 mL of 1 M EDTA 15 mL of 0.01 M EDTA

= 100 g of CaCO3 = 100 mg of CaCO3 = 0.15 mL of 1 M EDTA = 15 mg of CaCO3

This is present in 100 mL sample. Therefore, total hardness of the sample = 150 mg L–1 as CaCO3 1 litre water = 1000 g = 1000 × 1000 mg = 1×106 mg 1 ppm = 1 part per million, i.e., 1 × 106 \ 150 mg in 1 litre is 150 ppm

13.32

Stoichiometry

solved Problem 21

Problems for Practice

How much AgCl will be formed by adding 200 mL of 5N HCl to a solution Containing 1.7 g AgNO3? Solution: AgNO3 + HCl  → AgCl + HNO3 M eq. mixed

1.7 ×1000 200×5 170 =10 =1000

M eq. after 0 990 reaction ∴ M eq.of AgCl formed =10 W ×1000 =10 143.5 ∴ WAgCl =1.435

0

0

10

10

solved Problem 22 10 mL of tap water containing Ca2+and Mg2+ in the presence of HCO3− was properly buffered and the indicator murexidé is added. The sample was diluted and heated to 60°C Titration 0.01 M EDTA solution changed the indicator colour at 7.50 mL. This complexed Ca2+ only. A second 10 mL sample was made basic and eriochrome black-T indicator is added. Titration with 0.01 M EDTA solution changed the indicator colour at 13.02 mL. Under these conditions both Ca2+ and Mg2+ are complexed. If the 10 mL of water sample were to be evaporated to dryness, what weight of CaCO3 + MgCO3 would be formed? Solution: All EDTA complexes are formed on a one to one basis with dispositive ions. Mole of Ca 2+ + Mg 2+ = mole of CaCO3 + mole of MgCO3 0.01×13.02 =13×10−5 1000 = moleof CaCO3

= Mole of Ca 2+

0.01×7.50 = 7.5×10−5 1000 = 13×10−5 − 7.50×10−5 = 5.5×10−5 =

∴ Mole of MgCO3

∴ Wt. of CaCO3 +MgCO3 = 7.50×10−5×100+5.5×10−5×84 = 1.21×10−2g (Mol Wt of CaCO3 =100, MgCO3 = 84)

77. The two acids H2SO4 and H3PO4 are neutralized separately by the same amount of an alkali when sulphate and dihydrogen orthophosphate are formed respectively. Find the masses of H2SO4 and H3PO4. 78. 0.376 g of aluminum reacted with acid to displace 0.468 litre of hydrogen at NTP. Find the equivalent volume of hydrogen if the equivalent weight of Al is 9. 79. Dry hydrogen was passed over 1.58 g of red hot copper oxide till all of it completely reduced to 1.26g copper (Cu). If in this process 0.36 g of H2O is formed, what will be the equivalent weight of Cu and O? (H = 1) 80. 0.13 g of Cu, when treated with AgNO3 solution, displaced 0.433 g of Ag.0. 13 g of Al, when treated with CuSO4 solution displaced 0.47 g of Cu. 1.17 g of Al displaces 0.13 g of hydrogen from an acid. Find the equivalent weight of Ag if equivalent weights of Cu and Al are not known. 81. The equivalent weight of MnSO4 is half its molecular weight when it is converted to (a) Mn2O3 (b) MnO 2 (c) MnO 4− (d) MnO24 − . Indicate the correct answer. (IIT 1988) 82. The equivalent weight of a metal is double that of oxygen. How many times is the weight of its oxide greater than the weight of the metal? 83. One gram of an alloy of Al and Mg when treated with excess of dil HCl forms MgCl2, AlCl3 and hydrogen. The liberated hydrogen collected over Hg at 0°C has a volume of 1.20 litres at 0.92 atm pressure. Calculate the composition of the alloy. (Al = 27, Mg = 24) 84. 0.324 g of copper was dissolved in nitric acid and the copper nitrate so produced was burnt till all copper nitrate converted to 0.406 g of copper oxide. Calculate the equivalent weight of copper. 85. Exactly 50 mL of Na2CO3 solution is equivalent to 56.3 mL of 0.102 N HCl in acid-base neutralization. How many g CaCO3 would be precipitated if an excess of CaCl2 solution were added to 100 mL of this Na2CO3 solution? 86. Mn2+(aq) can be determined by titration with MnO −4 (aq) 3Mn 2+ +2MnO 4− → 6MnO 2 +2H 2 O A 25 mL sample of Mn2+(aq) requires 34.77 mL of 0.05876 M KMnO4(aq) for its titration. What is the molarity of Mn2+(aq)?

Stoichiometry 13.33

87. A particular acid – rain water has SO32− . If a 25.0 mL sample of this water requires 34.08 mL of 0.01964 M KMnO4 for its titration. What is the molarity of SO32− in acid – rain? 2MnO 4− +5SO32 − +6H + → 5SO 24 − +2Mn 2+ +3H 2 O

88. A 1.1 g sample of copper ore is dissolved and the Cu2+(aq) is treated with excess KI. The liberated I2 requires 12.12 mL of 0.10 M Na2S2O3 solution for titration. Calculate % of copper by weight in the ore. 89. The electrolyte in a lead storage battery is dil sulphuric acid H2SO4(aq). This acid must have concentration between 4.8 M and 5.3 M if the battery is to be most effective. A 5.0 mL sample of a particular battery acid requires 46.40 mL of 0.875 M NaOH for its complete reaction. Does the concentration of the battery acid will fall with in range for the most effective operation of the battery? 90. Calculate the weight of K2Cr2O7 required to produce from excess oxalic acid 5.0 L CO2 at 750°C and 1.07 atm. Pressure (Cr = 52, K = 39) Cr2 O72- +3C2 O 2-4 +14H + → 6CO 2 +2Cr 3+ +7H 2 O 91. A 0.608g sample of fertilizer contained nitrogen as ammonium sulphate (NH4)2SO4. It was analyzed for nitrogen by heating sulphate with sodium hydroxide.

( NH )

4 2

SO 4 (s) + 3NaOH(aq )  → Na 2SO 4 (aq ) + 2H 2 O(l) + 2 NH 3 (g )

The ammonia was collected in 46.3 mL of 0.213 M HCl NH3(g) + HCl(aq) → NH4Cl(aq) This solution was titrated for excess HCl with 44.3 mL of 0.1284 NaOH. Calculate the % of nitrogen in the fertilizer. 92. On ignition, Rochelle salt, KNa C4H4O6·4H2O (mol. wt 282) is converted into KNaCO3 (mol. wt 122). A sample of the original salt weighs 0.9546 g and the ignition product is titrated with H2SO4. Calculate the percentage purity of sample from the following data. H2SO4 used = 41.72 mL 10.27 mL of H2SO4 ≡ 10.35 mL of 0.129 N NaOH NaOH used for back titration = 1.91 mL 93. Chile salt petre is a natural source of NaNO3 which also contain NaIO3. The NaIO3 can be used as a source of iodine produced in the following reactions: IO3− + 3HSO3−  → I − + 3H + + 3SO 24 −

(1)

5I − + IO3− + 6H +  → 3I 2 (s) + 3H 2 O

(2)

1.0 L of the starting solution which contains 5.80 g NaIO3/L is treated with stoichiometric quantity of NaHSO3. Then a further quantity of the starting solution is added to the reaction mixture to bring about the second reaction. How many 'g' of NaHSO3 are required in step (1) and what addition volume of the starting solution in unit be added in step (2)? 94. What is the percent of free SO3 in oleum (H2S2O7) that is labelled 109% H2SO4? 95. Copper sulphate reacts with KI in acidic medium to liberate I2. 2CuSO4+4Kl → Cu2I2+K2SO4+I2 Mercuric periodate Hg5(IO6)2 reacts with a mixture of KI and HCl according to the following equation: Hg5(IO6)2 + 34 KI + 24 HCl → 5K2 Hg I4 + 8I2 + 24KCl + 12H2O The liberated iodine is titrated against Na2S2O3 solution, 1 mL of which is equivalent to 0.0499 g of CuSO4·5H2O. What volume in mL of the Na2S2O3 solution will be required to react with the I2 liberated from 0.7245 g of Hg5 (IO6)2? Mol wt of Hg5 (IO6)2 = 1448.5; CuSO4.5H2O = 249.5 96. 1.44 g pure FeC2O4 was dissolved in dil HCl and solution diluted to 100 mL. Calculate the volume of 0.01 M. KMnO4 required to oxidize FeC2O4 solution completely. 97. For titrating a certain volume of a reducing substance by 1 M KMnO4, it is found 20 mL was used in acidic medium, 33.4 mL ussed in alkaline medium and 100 mL was used in neutral medium. If MnO −4 is reduced to Mn2+ in acidic medium, to what state MnO −4 is reduced in alkaline medium and neutral medium. 98. A small amount of CaCO3 completely dissolves in 525 mL of (N/10) HCl and no acid is left at the end. After converting all calcium chloride to CaSO4, how much plaster of paris can be obtained? (Dhanbad 1991) 99. A certain solution consists of Na2CO3 and NaHCO3.30 mL of this required 12 mL of 0.1 N H2SO4 using phenolphthalein as indicator. In pesence of methyl orange, 30 mL of same solution needed 40 mL of 0.1 N H2SO4.What is the amount of Na2CO3 and NaHCO3 per litre in mixture? 100. 200 mL of a solution of mixture of NaOH and Na2CO3 was first titrated with phenolphalein and 0.1 N HCl. 17.5 mL of HCl was required for the end point. After this methyl orange was added and 2.5 mL of same HCl was again required for another end point. Calculate the amounts of NaOH and Na2CO3 per litre in mixture.

13.34

Stoichiometry

101. Zinc can be determined volumetrically by the precipitation reactions: 3Zn2++2K4[Fe(CN)6] → K2Zn3[Fe(CN)6]2+ 6 K+

102.

103.

104.

105.

106.

A sample of zinc ore weighing 1.5432 g was prepared for reaction and needed 34.68 mL of 0.1043 M K4[Fe(CN)6] for titration. Calculate the % of zinc in the ore sample. 1.25 g of a bleaching powder sample is dissolved in 100 mL of water and 25 mL of which are treated with KI solution. The iodine so liberated required 12.5 mL of 0.04 Na2S2O3 solution in titration. Calculate the percentage of chlorine available from the sample of bleaching powder. To a 25 mL H2O2 solution, excess of acidified solution of KI was added. The iodine liberated required 20 mL of 0.3 N Na2S2O3 solution. Calculate the volume strength of H2O2 solution. (IIT 1997) One litre of a mixture of O2 and O3 at NTP was allowed to react with an excess of acidified solution of KI. The iodine liberated required 40 mL of M/10 sodium thosulphate solution for titration. What is the weight percent of ozone in the mixture. Ultraviolet radiation of wavelength 300 nm can decompose ozone? Assuming that one photon can decompose one ozone molecule, how many photons would have been required for the complete decomposition of ozone in the original mixture? (IIT 1997) A sample of hard water contains 96 ppm of SO 2− 4 and 183.ppm of HCO3− with Ca2+ as the only cation. How many moles of CaO will be required to remove HCO3− from 1000 Kg of this water? If 1000 Kg of this water is treated with the amount of CaO calculated above what will be the concentration (in ppm) of residual Ca2+ ions (Assume CaCO3 to be completely insoluble in water). If Ca2+ ions in one litre of the treated water are completely exchanged with hydrogen ions, what will be its pH (One ppm means one part of the substance in one million parts of water weight/weight)? (IIT 1997) A 3 g sample containing Fe3O4, Fe2O3 and inert impure substance is treated with excess of KI solution in presence of dilute H2SO4. The entire iron is converted into Fe2+ along with the liberated of iodine. The resulting solution is diluted to 100 mL. A 20 mL of the dilute solution requires 1 mL of 0.5 M Na2S2O3 solution to reduce the iodine present. A 50 mL of the dilute solution after complete extraction of the iodine requires 12.8 mL of 0.25 M KMnO4 solution in dilute

H2SO4 medium for the oxidation of Fe3+ calculate the percentage of Fe2O3and Fe3O4 in the original sample. (IIT 1996) 107. A mixture in which the mol ratio of H2 and O2 is 2:1 is used to prepare water by the reaction 2H2(g)+O2(g)→2H2O(g) The total pressure in the container is 0.8 atm at 20°C before the reaction. Determine the final pressure at 120°C after reaction assuming 80% yield of water. (Roorke 1999) 108. A sample of Mg was burnt in air to give a mixture a MgO and Mg3N2. The ash was dissolved in 66 M eq of HCl and the resulting solution was back titrated with NaOH. 12 M eq of NaOH were required to reach the end point. An excess of NaOH was then added and the solution distilled. The ammonia released was then trapped in 10 M eq of IInd acid solution. Back titration of this solution required 6 M eq of the base. Calculate the percentage of Mg burnt to the nitride. (Roorke 1998) 109. 1.20 g sample of Na2CO3 and K2CO3 was dissolved in water to from 100 mL of a solution. 20 mL of this solution required 40 mL of 0.1 N HCl for complete neutralization. Calculate the weight of Na2CO3 in mixture. If another 20 mL of the solution is treated with excess of BaCl2, what will be the weight of precipitate ? (Roorke 1997) 110. If 100 g of V2O5 is dissolved in acid and is reduced to V2+ by zinc metal, how many moles of I2 could be reduced by the resulting solution if it is further oxidized to VO2+ ions. (V = 51, O = 16, I = 127) 111. 12 g of an impure sample of arsenious oxide was dissolved in water containing 7.5 g of NaHCO3 and the resulting solution was diluted to 250 mL. 25 mL of this solution was completely oxidized by 22.4 mL of a solution of iodine. 25 mL of this iodine solution reacted with same volume of a solution contained 24.8 g of sodium thiosulphate (Na2S2O3.5H2O) in one litre. Calculate the percentage of arsenious oxide in the sample. (Atomic mass of As = 75) (Roorke 1999) 112. 1.6 g of pyrolusite ore was treated with 50 cm3 of 1.0 N oxalic acid and some sulphuric acid. The oxalic acid left undecomposed was raised to 250 cm3 in a flask. 25 cm3 of this solution when titrated with 0.1 N KMnO4 required 32 cm3 of the solution. Find out the percentage of pure MnO2 in the sample and also the percentage of available oxygen. (Roorke 1996)

Stoichiometry 13.35

113. 5 mL of 8 N nitric acid, 4.8 mL of 5 N hydrorochloric acid and a certain volume of 17 M sulphuric acid are mixed together and made up to 2 litres. 30 mL of this acid mixture exactly neutralize 42.9 mL of sodium carbonate solution containing one gram of Na2CO3. 10H2O in 100 mL of water. Calculate the amount in grams of the sulphate ions in solution. (IIT 1985) 114. 0.50 g of a mixture of K 2CO3 and Li2CO3 requires 30 mL of a 0.25 N HCl solution for neutraliza tion. What is the percentage composition of the mixture. (Roorke 1989) 115. (i) What is the mass of sodium bromate and molarity of the solution necessary to prepare 85.4 mL of 0.672 N solution when the half reaction is BrO3− + 6H + + 6e −  → Br − + 3H 2 O (ii) What would be the mass as well as molarity if the half cell reaction is 2BrO3− + 12H + + 10e −  → Br2 + 6H 2 O (IIT 1987) 116. 50 mL of an aqueous solution of H2O2 was treated with an excess of KI solution and dilute H2SO4. The liberated iodine required 20 mL 0.1 N Na2S2O3 solution for complete interaction. Calculate the concentration of H2O2 g/L. (Roorke 1981) 117. A mixture of H2C2O4 (oxalic acid) and NaHC 2O4 weighing 2.02 g was dissolved in water and the solution made up to one litre. 10 mL of the solution required 3 mL of 0.1 N NaOH solution for com plete neutralization. In another experiment 10 mL of the same solution in hot dilute H2SO4 medium, required 4 mL of 0.1 N KMnO 4 solution for complete reaction. Calculate the amount of H2C2O4 and NaHC2O4 in the mixture. (IIT 1990) 118. A solution of 0.2 g of a compound containing Cu2+ and C2 O 2− 4 ions on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na2CO3, acidified with dilute acetic acid and treated with excess of KI. The liberated I2 required 11.3 mL of 0.05 M Na2S2O3 solution for complete reduction. Find out the mole ratio of Cu2+ and C2 O 2− 4 in the compound. Write down the balanced redox reactions involved in the above titrations. (IIT 1991) 119. A 1.0 g sample of Fe2O3 solid of 55.2 per cent purity is dissolved in acid and reduced by heating the

solution with zinc dust. The resultant solution is cooled and made up to 100 mL. An aliquot of 25 mL of this solution requires 17 mL of 0.0167 M solution of an oxidant for titration. Calculate the number of electrons taken up by the oxidant in the reaction of the above titration. (IIT 1991) 120. Hydrogen peroxide solution (20 mL) reacts quantitatively with a solution of KMnO4 (20 mL) acidified with dilute H2SO4. The same volume of the KMnO4 solution is just decolourised by 10 mL of MnSO4 in neutral medium forming a dark brown precipitate of hydrated MnO2. The brown precipitate is dissolved in 10 mL of 0.2 M sodium oxalate under boiling condition in the presence of dilute H2SO4. Write the balanced equations involved in the reaction and calculate the motality of H2O2. (IIT 2001) 121. The Mn3O4 formed on strong heating of a sample of MnSO4·4H2O was dissolved in 100 cm3 of 0.1 N FeSO4 containing dilute H2SO4. The resulting solution reacted completely with 50 cm 3 of KMnO4 solution. 25 cm 3 of this KMnO4 solution requires 30 cm3 of 0.1 N FeSO 4 solution for complete reaction. Calculate the amount of MnSO4·4H2O in the sample. (Roorke 2001) 122. The CrO 2− ion may be present in waste water from 4 a chrome plating plant. It is reduced to insoluble Cr(OH)3 by S2 O 2− 4 in basic medium. 3 S2 O 24 − + 2CrO 24 − + 2H 2 O + 2OH − → 6SO32 − + 2Cr ( OH )3 50 L of water require 193.5 g of Na2S2O4. Calculate in waste water molarity and normality of CrO 2− 4 (Cr = 52, S = 32, O = 16). 123. Borax in water gives B4 O72 − + 7 H 2 O  → 4H 3 BO3 + 2OH − How many g of borax (Na2B4O7·10H2O) are required to (i) Prepare 100 mL of 0.2 M solution? (ii) Neutralise 25 mL of 0.1934 M of HCl and H2SO4 separately? 124. A mineral CuFeS2 was analyzed for Cu and Fe percentage. 20 g of it was boiled with dil. sulphuric acid and diluted to 1 litre. 10 mL of this solution required 2 mL of 0.01 M MnO −4 in acidic medium. In second titration 25 mL of the same solution required 5 mL of 0.01 M S2 O32− solution idometrically. What is the % of Cu and Fe in the mineral?

13.36

Stoichiometry

Key Points Atomic weights and equivalent weights • •

•

•

•

•

Atomic weight scale is a table that lists the weights of all the elements relative to some common standard. The mass of C-12 is taken as standard and its mass is assigned exactly 12.0 atomic mass units (amu) and masses of all other atoms are given relative to this standard. One amu is defined as a mass exactly equal to one twelfth the mass of one carbon-12 atom 1 amu = 1.66056 × 10–24 g. Many naturally occurring elements consists of more than one isotope. For such elements the atomic weights determined are the average atomic weights of isotopes. Equivalent weight of an element is the number of units of weight of the element which can combine with or displace unit weight of hydrogen or the equivalent of any other element or group of elements. Valency of an element is the number of atoms of hydrogen that one atom of the element combine with or replace. It must be a whole number.

•

•

•

•

•

Equivalent weight × Valency = Atomic weight. •

The basicity of an acid is the number of atoms of hydrogen replaceable by a metal to form a salt in each molecule of acid. Molecular weight of an acid = Equivalent of acid Basicity

•

•

•

The acidity of a base is the number of replaceable OH− ions in a base. Molecular weight of a base = Equivalent of base acidity

•

•

The equivalent weight of a salt that undergo metathesis with acids is the weight reacting with one equivalent of the acid. Solutions whose concentrations are known exactly are called standard solutions.

•

determination of Atomic weights •

determination of equivalents •

•

In the oxide method the weight of an element that combines with 8 units of oxygen is taken as the equivalent and it is determined by combining a known weight of the element in a stream of oxygen. Metal nitrates are prepared by dissolving known weight of metal in nitric acid which on heating decomposes to metal oxide (except 1A group metals, silver and mercury nitrates). From this metal oxide the weight of metal that combines with 8 units of oxygen will be determined.

In the reduction of oxide method a known weight of oxide is reduced by hydrogen to find the loss in weight from which the equivalent weight of element that combines with 8 units of oxygen will be determined. In hydrogen displacement method the equivalent weight of metal can be determined by measuring the volume of hydrogen displaced by a metal with an acid or base, e.g., Mg, Zn, Fe with dil acid or Al and Zn with an alkali. In the chloride method a known weight of the metal is heated in a current of chlorine from which the amount of metal that can react with one equivalent of chlorine will be determined. The metals which can form two different oxides or chlorides, the equivalent of the metal in the higher oxide or chloride is determined. In soluble chloride method a known weight of metal chloride is dissolved in water and the weight of metal that combines with equivalent of chlorine (35.5) is determined by estimating the chloride ion by titrating with standard AgNO3 solution, e.g., equivalents of Na, K, ammonium salts can be determined. In soluble carbonate method the equivalent of metal is determined by estimating the amount of the metal that combines with one equivalent of CO32− (30 g) by titrating with dilute acid. For insoluble carbonates a known weight of metal carbonate is dissolved in known volume of normal acid from which the weight of metal that combines with one equivalent of carbonate (30 g) is determined. In electrolysis method the weight of an element liberated at the electrode by passing electric current which liberates 11.2 L of hydrogen at STP is taken as equivalent of the element.

•

•

According to Dulong and Petit’s law atoms of all elements have same heat capacity and from this the atomic weight can be determined. Atomic weight × Specific heat = Constant 6.4 (approximately) According to Mitcherlich’s law of isomorphism. chemically similar substances crystalise in the same form, i.e., isomorphism. According to law of isomorphism masses of two elements that combine with same mass of other elements in their respective compounds are in the ratio of their atomic masses.

Stoichiometry 13.37

Mass of one element A ( ) that combines with certain mass of other element Atomic mass of A = Mass of one element B ( ) that Atomic mass of B combines with the same mass of other elements

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The valencys of the elements forming isomorphic compounds are the same. According to Cannizzaro’s method, if an element forms more than one compound, in the molecule of at least one such compound there is only one atom of the element and the smallest weight therefore is the atomic weight. Preparation of periodic table by Mendeliev helped the correction of atomic weights of Be, U, In, Ce and Yt. In the limiting densities method the volumes of given amount of gas under constant temperature of 0°C at pressures of one atmosphere and below. The graph showing the reaction between the product of pressure and volume vs pressure is a straight line. By extrapolation of the graph the value of PV at zero pressure i.e., P0V0 is obtained from which molecular weight can be obtained. Molecular weight divided by the atomicity gives atomic weight. Atomic weights can be determined directly by mass spectrograph method.

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Balancing of chemical equation •

methods for the determination of density •

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In Regnaults method by knowing the weights of a gas and the weight of a liquid whose density is known, the density of the gas can be calculated. Gas diffusion method (making use of Graham’s law) can be used for the determination of density of an unknown gas by measuring the rates of diffusion of a known and unknown gases and knowing the density of the known gas. Eudiometry or Gas Analysis method is used to determine the molecular formula of gases. At high temperatures the molecular weights determined using vapour density method may be low or high which are known as abnormal molecular weights and the densities as abnormal vapour densities. This is due to either dissociation or association of gas molecules. Vapour densities of NH4Cl, PCl5, N2O4 obtained are less due to dissociation while the vapour density of aluminium chloride is more due to association to Al2Cl6.

stoichiometry •

The quantitative relationship existing between the quantities of the reactants and the products in a chemical reaction is known as stoichiometry.

A chemical equation in which the number of atoms of each element is equal on the reactant side and the product side is called a balanced equation. Balanced chemical equation represents a stoichiometry equation. The exact quantities of the reactants and the products that appear in the balanced chemical equation are known as stoichiometric quantities. A chemical equation conveys both qualitative and quantitative information. Quantitatively a chemical equation tells the names of the various reactants and products. Quantitatively a chemical equation represent i) the relative number of reactant and product species) atoms or molecules) taking part in reaction; ii) the relative number of moles of the reactants and products; iii) the relative masses of the reactants and products and iv) the relative volumes of gaseous reactants and products.

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Hit and trial or Trial and Error Method involves the following steps: (i) Write the symbols and formula of the reactants and the products in the form of skeleton equation. (ii) If elementary gases such as hydrogen, oxygen etc are involved in the equation, these are represented in their atomic form in the beginning and after balancing the equation is changed to molecular form. (iii) The formula which contain maximum number of elements is selected first and the atoms present in it are balanced. (iv) In case, the above method fails, then start balancing the atoms which appear minimum number of times. (v) Verify that the number of atoms of each element is balanced in the final equation. (vi) While balancing the chemical equation the chemical formula of any compound should not be changed for the sake of convenience because each compound has a fixed chemical formula. Partial equation method involves the following steps. (i) The given reaction is supposed to take place in different steps. These different steps are called partial equations. (ii) Each partial equation is balanced separately by hit and trial method as described in the earlier method. (iii) The partial equations are multiplied by suitable numbers if necessary, so as to cancel out the species which are not involved in the original reactants and the final products. (iv) Partial equations are then added to get the final balanced chemical equation.

13.38

Stoichiometry

chemical reaction and numerical calculations •

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Stoichiometry calculations help in calculating whether the production of a particular substance is economically cheap or not. Stoichiometry calculations are of the following types: (i) Mole to mole relationships (ii) Mass-mass relationships (iii) Mass-volume relationships (iv) Volume-volume relationships (v) Mass-volume-energy relationship If the amount of the reactant in a particular reaction is known, then the amount of the other substance required in the reaction or the amount of the product formed in the reaction can be calculated. For stoichiometry calculations the following steps should be considered. (i) A balanced chemical equation using chemical formula of reactants and products must be written. (ii) The coefficients of balanced chemical equation give the mole ratio of the reactants and products. (iii) The mole ratio can be converted into weight – weight, weight-volume, volume-volume ratios. Limiting reactant or limiting reagent is the reactant that is entirely consumed when a reaction goes to completion. A reactant that is not completely consumed in a reaction is called as excess reactant.

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Acid-Base titrations • •

titrimetric method of Analysis •

A titrimetric method of analysis is based on a chemical reaction like aA + tT → Products

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where a moles of analyte A react with t molecules of the reagent T called titrant. Titrant is added incrementally from a burette in the form of a solution and the process is called titration. The point at which the addition of the titrant T is exactly chemically equivalent to that of A is called equivalence point. The substance that changes colour by the addition of excess titrant is called indicator. The point where the indicator changes its colur is called end point. The term titration refers to the process for measuring the volume of titrants required to reach the equivalence point.

The chemical reactions which serve as basis for titrimetric determination are of four types: (i) Acid-base (ii) Oxidation-reduction (redox) (iii) Precipitation and (iv) Complex formation All known reactions cannot be used as the basis for titration. For using in titration a reaction must satisfy the following conditions: (i) The reaction must proceed to virtual completion at the equivalency point, i.e., the equilibrium constant must be very high. (ii) The reaction must proceed according to a definite chemical equation. There should be no side reactions. (iii) Some method like indicator or instrumental method must be available to know when the equivalency point is reached. (iv) The reaction must be rapid so that the titration can be completed in a few minutes. Substances which can be used to prepare standard solutions directly by dissolving an accurately weighed sample of the desired substance in an accurately measured volume of solution are called primary standard.

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In an acid-base titration when an acid is added to a base or vice-versa the pH of the solution changes. Depending on the strength of acid and base involved in the titration, the acid base titrations are four types. (i) Titration of strong acid with strong base, e.g., HCl vs NaOH. (ii) Titration of strong acid with weak base e.g., HCl vs NH4OH. (iii) Titration of weak acid with strong base, e.g., CH3COOH vs NaOH. (iv) Titration of weak acid with weak base, e.g., CH3COOH vs NH4OH. The change in pH of the solution at equivalence point depends on the strength of the acid and base involved in the titration. Depending on the change in pH of the solution at the equivalency point, the indicator should be selected. When a strong acid reacts with a mixture of two or more basic substances, the strongest base is neutralized first while the other bases remain inert during neutralization of strong base. Thus neutralization of a mixture of acids or bases takes place sequentially starting from the strongest and ending with weakest. When a solution containing both NaOH and Na2CO3 is titrated with HCl using phenolphthalein indicator the amount of acid used corresponds to the conversion of NaOH to NaCl and Na2CO3 to NaHCO3. If the

Stoichiometry 13.39

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titration is continued using methyl orange indicator, the amount of acid used corresponds to the NaHCO3. From these values the amounts of NaOH and Na2CO3 can be calculated. When a solution containing a mixture of sodium carbonate and sodium bicarbonate is titrated with HCl using phenolphthalein, the amount of acid used corresponds to the conversion of Na2CO3 to NaHCO3. If the titration is continued using methyl orange indicator the amount of acid used corresponds to the total NaHCO3, i.e., NaHCO3 present initially and NaHCO3 formed from Na2CO3. From these values the amounts of Na2CO3 and NaHCO3 present in the mixture can be calculated. Neutralization of phosphoric acid with NaOH takes place in three steps forming NaH2PO4, Na2HPO4 and Na3PO4. Phenolphthalein can change its colour in the step of conversion of H3PO4 to NaH2 PO4 while methyl orange can change its colour in the conversion of NaH2PO4 to Na2HPO4. The ionization of HPO 2− is 4 very small and cannot be titrated by a strong base. If Na3PO4 is titrated with HCl or H2SO4 phenolphthalein changes its colour when the formation of NaH2PO4 is complete, methyl orange changes its colour when the formation of Na2HPO4 is complete and thymol blue changes its colour when Na2HPO4 is completely converted into H3PO4.

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redox titrations • •

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Acid-base reactions involve proton transfer while the redox reaction involves electron transfer. Acid-base reactions are instantaneous while some oxidation-reduction reactions are slow and titrimetric methods based on these reactions require elevated temperatures, addition of catalyst or an excess of reagent followed by back titration. Redox reactions are sequence of multistep reactions and transfer of electron in only one part of multistep sequence which involves forming or breaking covalent bonds, protonation and various sorts of rearrangements. In acid-base reactions and in redox reactions the common solvent is water. In acid-base reactions protons can be transfered from acid to water to form H3O+ and these protons can be easily passed to base to regenerate H2O. Similarly removal of proton from water to from OH– which readily converts to H2O by the protons supplied by acid. Further all the species H3O+, OH– and H2O are highly soluble. The addition of electron to water results in the formation of H2 and oxidation of water results in the formation of oxygen which tend to escape from solution. Hence the titrants that oxidises or reduces water are

usually avoided. Strongest oxidants and reductant are impractical titrants. Levelling effect such as that occurs in acid-base chemistry is not encountered in redox reactions. Metal ion act as oxidizing or reducing agents as a result of their ability to gain or lose electrons forming ions of lower or higher valency which represent different stages of oxidation, e.g., Fe2+ can act as reducing agent while Fe3+ can act as oxidizing agent. The power of oxidation or reduction of a metal ion depends on its redox potential, i.e., the potential developed when a platinum rod is immersed in a solution containing equal amount of both oxidation states (Reduced form and oxidized form). A higher positive redox potential indicates that the ion is in the higher oxidation state and can act as strong oxidizing agent while high negative redox potential indicates the higher reducing power of the ion in the lower oxidation state. In the titration of reducing agent with an oxidizing agent the redox potential changes just as pH changes in the acid base titration. The changes in redox potential in a redox titration can be calculated using Nernst equation. At the equilibrium point the potential in a redox reaction is equal to the average E values of the two redox couples formed in the redox titration formed from oxidizing and reducing agents. For any reaction in which the number of electrons lost by a reductant is the same as the number gained by the oxidant. Hence, the potential at the equivalency point is simply the arithmetic mean of the two standard potentials E10 + E 02 2 O

E eq , pt = •

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After the equivalency point only either oxidant or reductant will present in excess depending on which is added to what. Indicators used for the detection of end point in a redox titrations are of the following types. A coloured substance may act as its own indicator and is called self indicator, e.g., potassium permanganate acts as self indicator in its reactions. Specific indicators are the substances which react in a specific manner with one of the reagents in titration to produce colour, e.g., starch gives blue colour with iodine, thiocyanate ion gives red colour with Fe3+ ions. Redox indicators undergo oxidation-reduction reactions in titration and have different colours in oxidized and reduced forms. A redox indicator must be selected suitable for a redox titration depending upon the redox potential of the indicator and the potential change at the end points.

13.40

•

Stoichiometry

The potential where a redox indicator changes colour is known as transition potential. An indicator whose transition potential lies in the range of potential change at the equilibrium must be selected. •

Precipitation titrations • •

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Equilibrium calculations in precipitation titrations are based on the solubility product constant. Precipitation titrations are used to determine silver ion with sodium chloride or chloride, bromide or iodide ions with silver nitrate. At the equivalency point precipitation is complete. The equivalency point in the titrations involving silver salts, in Mohr’s method potassium chromate is used as adsorption indicator. In Volhard method, Fe3+ ion which form coloured complex with thiocyanate ion is used. Fajan’s method also utilizes adsorption indicators.

complexometric titrations •

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Titrations involving complex formation by the metal ions with ligands are called Complexometric titrations. Chelating ligands which can form heterocyclic rings with metal ions are the better titrants for the titration

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of metal ions. The widely used complexing agent is ethylenediaminetetracetic acid (EDTA) which is a hexa-dentate ligand and forms 1:1 complex with several metal ions. The complexation reactions in complexometric titrations are really a competition between H2O and ligand or between the ligands present in solution. The ligand which can form stable complex will be formed. The equivalency point in complexometric titrations can be detected by using metallochromic indicators. The metallochrmic indicators are coloured organic compounds that can form chelates with metal ions forming metal–metallochromic indicator chelate with different colour from the colour of the free molecule. The metal-metallochromic indicator chelate is less stable than metal-EDTA complex and hence the indicator will release the metal ion to the EDTA titrant. Which results in the colour change. Most widely used metallochromic indicators are eriochrome black-T, murexidé, xylenol orange, solochrome black etc. Complexometric titrations are much useful in the estimation of several metal ions in particular for determination of hardness (Mg 2+ and Ca2+ ions) of water.

Stoichiometry 13.41

PrActice exercise multiple choice questions with only one Answer level i 1. 25 mL of H2O2 solution when added to excess acidified KI solution liberates iodine, which require 20 mL of 0.3125 N sodium thiosulphate solution. The volume strength of H2O2 is (a) 1.4 (b) 1.9 (c) 2.3 (d) 5.6 2. What volume of 0.2 M Ba(MnO4)2 solution is required for complete oxidation of 25 g of 89.6% pure FeCr2O4 in acidic medium according to the reaction H+

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MnO4− +FeCr2O4 → Fe+3 +Cr2O7–2+Mn+2 (a) 700 mL (b) 175 mL (c) 350 mL (d) 200 mL POCl3 hydrolyses in water to give H3PO4 and HCl only. What volume of 2 M Ca(OH) 2 is needed to completely neutralize the 50 mL of 0.1 M POCl 3 solution. (a) 5 mL (b) 7.5 mL (c) 15 mL (d) 10 mL 18.2 g of V2O5 (MW=182) was dissolved in acid and reduced to V2+ by treatment with zinc metal. The number of mole of I2 which could be reduced by the resulting V2+ solution obtained above as it is reoxidized to V+4 state is (a) 0.05 (b) 0.1 (c) 0.15 (d) 0.2 100 mL of 0.1 M aqueous ferric alum solution is completely reduced with iron to give ferrous ions. The volume of 0.1 M acidified KMnO4 solution required for complete reaction with ferrous ions in the solution will be (a) 60 mL (b) 30 mL (c) 40 mL (d) 100 mL 100 mL of oxygen is subjected to silent eclectic discharge to get 80 mL oxygen and ozone mixture. The volume of ozone and oxygen in the mixture in mL (a) 40, 40 (b) 20, 60 (c) 60, 20 (d) 50, 30 10 grams of impure sample of arsenious oxide dissolved in water and diluted to 250 mL. 25 mL of this solution was completely oxidized by 20 mL of iodine solution reacted with same volume of a solution containing 24.8 grams of Na 2S2O3. 5H 2O in one litre. The percentage of arsenious oxide in the sample is (a) 0.99% (b) 9.9% (c) 99% (d) none

8. MnO2 was boiled with 65 mL of 1 N oxalic acid in acid medium. The solution is filtered. The filtrate is diluted to 500 mL. 100 mL of this solution required 50 mL of N/10 KMnO4. The weight of MnO2 is (a) 0.87 g (b) 1.5 g (c) 1.74 g (d) 1.95 g 9. 0.75 mole of solid A4 and 2 moles of O2 are heated in a sealed vessal to produce only one gaseous compound B without leaving A4 and O2. After the compound in the vessel is brought to initial temperature the pressure is found to be half the initial pressure. The formula of the compound is (a) A2O5 (b) A3O4 (c) AO2 (d) AO3 10. Zinc can be determined by the precipitation reaction 3Zn+2+2K4[Fe(CN)6]→K2Zn3[Fe(CN)6]2+6K+. A sample of zinc weighting 2.45 g was prepared for this reaction and required 25 mL of 0.1 M K4[Fe(CN)6] for titration. The % of zinc in the ore is (a) 10% (b) 20% (c) 15% (d) 13% 11. 1.149 grams of a sample containing KCl and NH4Cl is heated until constant weight. The residue is dissolved in 20 mL N/10 AgNO3 solution. It is required to cause precipitation of chloride ions. The weight of NH4Cl in the sample is (a) 0.25 g (b) 0.5 g (c) 0.075 g (d) 1.0 g 12. In a gravimetric determination of ‘P’ an aqueous solution of H2PO4− is treated with a mixture of ammonium and magnesium ions to precipitate as Mg(NH4) PO4·6H2O. This is heated and decomposed to Mg2P2O7. A solution of H2PO4– yielded 1.11 g Mg2P2O7. The weight of NaH2PO4 was present in solution is (a) 0.6 g (b) 1.2 g (c) 1.8 g (d) 2.4 g 13. A sample containing 0.496 g of (NH4)2C2O4 (MW = 124) and inert material was dissolved in water and made strongly alkaline with KOH which conserved NH4+ into NH3. The liberated NH3 was distilled into exactly 50 mL of 0.05 M H2SO4 The excess H2SO4 was back titrated with 10 mL of 0.1 M NaOH. The percentage of (NH4)2C2O4 with sample is (a) 40% (b) 50% (c) 60% (d) 75% 14. 2 grams of a mixture of C2H2 and C3H8 is subjected to combustion. If the products CO2 and H2O are in 3:2 mole ratio. The weight of C2H2 in the initial mixture is (a) 1 g (b) 1.2 g (c) 1.56 g (d) 1.82 g

13.42

Stoichiometry

15. 5 g of a sample of Na2CO3 and Na2SO4 is dissolved in 250 mL solution. 25 mL of this solution neutralizes 20 mL of 0.2 M H2SO4. The % Na2CO3 in this sample is (a) 84.8% (b) 8.48% (c) 15.2 (d) 42.4% 16. Number of mole of K2Cr2O7 reduced by one mole of Sn2+ ions is (a) 1/3 (b) 3 (c) 1/6 (d) 6 17. For a given mixture of NaHCO3 and Na2CO3, volume of given HCl required is x mL with phenolphthalein indicator and further y mL required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO3 is (a) 2x (b) x/2 (c) y (d) (y–x) 18. A sample of ferrous sulphate and ferrous oxalate was dissolved in dil. H2SO4. The complete oxidation of reaction mixture required 40 mL of N/15KMnO4. After oxidation, the reaction mixture was reduced by zinc and H2SO4. On again oxidation by same KMnO4. 25 mL ware required. The mole ratio of FeSO4: FeC2O4 in the sample is (a) 3:7 (b) 2:3 (c) 7:3 (d) 3:5 19. The reaction Cl2(g)+S2O3–2 → SO4–2 + Cl– is to be carried out in basic medium. Starting with 0.15 mole of Cl2 0.01 mole of S2O3–2 and 0.3 mole of OH– how many moles of OH– will be left in solution after the reaction is complete? (a) 0.1 (b) 0.15 (c) 0.2 (d) 0.25 20. 0.2828 g of iron wire was dissolved in excess dilute H2SO4 and the solution was made up to 100 mL. 20 mL of this solution required 30 mL of N/30 K2Cr2O7 for exact oxidation. The percentage purity of Fe in wire is (a) 99% (b) 90% (c) 100% (d) 92% 21. H2O2 oxidises Sn+2 into Sn+4. H2O2 slowly decompose at STP to give O2. 100 g of 17% by mass of H2O2 in water is treated with 100 mL of 2M Sn+2 and then the mixture is allowed to stand until no further reaction takes place. Then the volume of O2 liberted at STP is (a) 4.67 lit (b) 3.36 lit (c) 2.24 lit (d) 1.12 lit 22. 20 mL of a solution containing FeSO4 and Fe2(SO4)3 is acidified with H2SO4 and reduced with zinc. The solution requires 30 mL of N/10 K2Cr2O7 for oxidation. Before reduction with zinc 20 mL of same solution

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requires 20 mL of N/10 K2Cr2O7 for oxidation. The molarities of FeSO4, Fe2(SO4)3 are (a) 0.1, 0.05 (b) 0.05, 0.025 (c) 0.1, 0.1 (d) 0.1, 0.025 20 mL of H2O2 solution react with 20 mL of KMnO4 in acid medium. The same volume of KMnO4 just depolarised by 10 mL of MnSO4 in neutral medium to form MnO2. This MnO2 is dissolved in 10 mL of 0.2 M Na2C2O4 in acid medium. So molarity of H2O2 is (a) 0.05 (b) 0.025 (c) 0.075 (d) 0.1 The loss in weight when 10.5 g of MgCO3 is heated in an open crucible is (a) 4.4 g (b) 5.5 g (c) 2.2 g (d) 8.9 g Consider the reaction 2X(g)+3Y(g)→Z(g), where gases X and Y are insoluble and inert to water and Z form a basic solution. In an experiment 3 mole each of X and Y are allowed to react in 15 lit flask at 500 K. When the reaction is complete, 5 L of water is added to the flask and temperature is reduced to 300 K. The pressure in the flask is (neglect aqueous tension) (a) 1.64 atm (b) 2.46 atm (c) 4.92 atm (d) 3.28 atm The percentage of oxalate ion in given sample of oxalate salt of which 0.3 g dissolved in 100 mL of water required 90 mL of N/20 KMnO4 for complete oxidation is (a) 33% (b) 66% (c) 90% (d) 100% 5 g of mixture of NaCl and Na2CO3 was dissolved in water and the volume made up to 250 mL. 25 mL of this solution required 50 mL of N/10 HCl for completed reaction. The percentage of NaCl in the mixture is (a) 53% (b) 47% (c) 63% (d) 37% A mixture of methane and ethylene in the volume ratio X:Y has a total volume 30 mL. On completed combustion it gave 40 mL of carbon dioxide. If the ratio in the original mixture had been Y:X instead of X:Y, the volume of CO2 would have been produced on combustion is (a) 40 mL (b) 60 mL (c) 50 mL (d) 30 mL 2×10–3 g of green algae absorbs 7×10–4 moles of CO2 per hour by photosynthesis as per the following reaction 6nCO2 + 5nH2O → (C6H10O5)n + 6nO2. If all the carbon in CO2 is converted into starch, how long will it take for the algae to increase its mass by 100% (a) 6.34 hr (b) 6.34 min (c) 63.4 min (d) 3.33 hr

Stoichiometry 13.43

30. In an experiment to determine the formula of a nonmetallic bromide of known relative molecular mass, 0.1 mole of the bromide was dissolved in 500 mL of water, 50 mL of this solution reacted exactly with 300 mL of 0.1 MAgNO3 (aq). If the other element present is denoted by the letter Z, the most probable formula for the bromide is (a) Z3Br (b) Z2Br6 (c) ZBr6 (d) ZBr3 31. A certain mass of pure KIO3 was dissolved in water and volume of solution was made to 250 mL. To 25 mL of this solution excess KI(s) was added and the solution was acidified with dil. HCl. The liberated iodine requires 120 mL 0.5 M Na2S2O3 solution for complete reduction in acetic acid medium. Hence the number of molecules of KIO3 dissolved in 250 mL solution is equal to (a) 3.0115×1021 (b) 1.2046×1022 21 (c) 2.4049×10 (d) 6.023×1022 32. How many mL of 0.2 M aluminium permanganate solution will be needed to oxidize completely a 30 mL 0.5 M solution of ferrous oxalate in dil. H2SO4 medium? (a) 20 (b) 30 (c) 40 (d) 15 33 A solution contains mixture of H2SO4, H2C2O4. 20 mL of this solution requires 40 mL of neutralization and 20 mL of

M NaOH for 10

N KMnO4 for oxidation. 10

The molarity of H2C2O4, H2SO4 are (a) 0.1, 0.1 (b) 0.1, 0.05 (c) 0.05, 0.1 (d) 0.05, 0.05 34. The chromium in 2.17 g chromate [FeO.Cr2O3] was oxidized to a (+6) state by fusion with sodium peroxide. The fused mass was treated with water and boiled to destroy excess of peroxide. After acidification the sample was treated with 70 mL of 0.17 M Fe2+. A back titration of 3 mL of 0.05 N K2Cr2O7 was required to oxidize the excess iron (II). What is the % of chromate in the mixture? [Cr = 52, Fe = 56, K =39] (a) 10% (b) 20% (c) 30% (d) 40% 35. A solution contains Na2CO3 and NaOH. By using phenolphthalein indicator 25 mL of mixture requires 20 mL of 0.005 N HCl for the end point. If methyl orange indicator is used 25 mL of mixture requires 25 mL of 0.005 N HCl. The molarity of NaOH, Na2CO3 are (a) 0.003 M, 0.001 M (b) 0.001 M, 0.003 M (c) 0.003 M, 0.002 M (d) 0.002 M, 0.003 M

36. The mass of oxalic acid H2C2O4 which can be oxidized to CO2 by 100 mL of an MnO4– solution 10 mL of which is capable of oxidizing 50 mL of 1 N I – to I2 is (a) 22.5 g (b) 25 g (c) 20 g (d) 45 g 37. An equimolar mixture of ferrous oxalate & stannous chloride is treated with decinormal acidic KMnO4 giving ferric, stannic and chlorate ions along with CO2 gas. If the titre value is 225 mL, number of millmoles of chlorate ions produced are (a) 1.5 m moles (b) 7.5 m moles (c) 3 m moles (d) 2.64 m moles 38. A saturated solution of CuSO4.5H2O were diluted to 100 mL and treated with excess of KI. The liberated iodine was sufficient to react with 56 mL of 0.092 N sodium thiosulphate. What is the concentration of the copper sulphate solution in moles of CuSO4.5H2O per litre, in the diluted solution: (a) 0.0338 (b) 0.0515 (c) 0.103 (d) 0.02575 39. A mixture of NaHC2O4 and KHC2O4.H2C2O4 required equal volumes of 0.2 N KMnO4 and 0.12 N NaOH separately. What is the molar ratio of NaHC2O4 and KHC2O4.H2C2O4 in the mixture? (a) 6:1 (b) 1:6 (c) 1:3 (d) 3:1 40. The NH3 evolved from 1.40 g sample of protein was absorbed in 45 mL of 0.4 N HNO3. The excess acid required 20 mL of 0.1 M NaOH. The % N in the sample is (a) 8 (b) 16 (c) 19.42 (d) none 41. How many litres of methane would be produced at STP when 0.6 g of CH3MgBr is treated with excess of C4H9NH2. (a) 0.8 litre (b) 0.08 litre (c) 0.112 litre (d) 1.12 litre 42. 0.214 g of KIO3 (MW=214) was treated with an excess of KI solution. The solution was acidified. Iodine was liberated. The liberated iodine was decolourised by 45 mL of thiosulphate solution. The molarity of sodium thiosulphate solution is (a) 0.133 M (b) 0.066 M (c) 0.111 M (d) 0.055 M 43. 20 mL of M/60 solution of KBrO3 was added to a fixed volume of SeO3–2 solution. Bromine evolved according to this reaction. SeO3–2 +BrO3– +H+→SeO4–2+ Br2+H2O. The bromine is removed by boiling and excess KBrO3 was back titrated with 5.1 mL of M/25 solution of NaAsO2 according to this reaction BrO3– +

13.44

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Stoichiometry

AsO2– + H2O→ Br–+AsO4–3 + H+. Atomic Weight of Se = 79. Calculate amount of SeO3–2 in solution. (a) 0.080 g (b) 0.040 g (c) 0.084 g (d) 0.042 g V1 mL NaOH of molarity X and V2 mL Ba(OH)2 of molarity Y are mixed together. The mixture is completely neutralized by 100 mL of 0.1 N HCl. If V1/V2 = 1/4 and X/Y = 4, then, what fraction of acid is neutralized by Ba(OH)2? (a) 0.5 (b) 0.25 (c) 0.33 (d) 0.67 20 g of impure Mg(NH4)PO4 on heating produces a gas. The gas is dissolved in H2O and on titration consumes 10 cc 2 M pyrophosphoric acid for complete reaction. The percentage purity of Mg(NH4) PO4 is (Mg = 24, P = 31) (a) 54.8% (b) 33.8% (c) 67.6% (d) 49% A 0.518 g sample of lime stone is dissolved in HCl and then the calcium is precipitated as CaC2O4. After filtering and washing the precipitate, it requires 40.0 mL of 0.250 N KMnO4, solution acidified with H2SO4 to titrate it as, MnO4–+H++C2O42–→Mn2++CO2+2H2O The percentage of CaO in the sample is: (a) 54.0% (b) 27.1% (c) 42% (d) 84% I M NaOH solution was slowly added to 1000 mL of 196 g impure H2SO4 solution and the following plot was obtained. The percentage purity of H2SO4 sample and slope of the curve respectively

H+ mol/lit

3 2 1 1

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Vol of NaOH added in lit

(a) 75% −

1 3

(b) 80% −

1 2

(c) 80% –1 (d) 75%–1 48. 0.759 g of a silver salt of dibasic organic acid on ignition left 0.463 g metallic silver. The equivalent mass of the acid is (a) 70 (b) 108 (c) 60 (d) 50 49. 0.7 g of sample Na2CO3.XH2O ware dissolved in water and the volume was made to 100 mL. In this

20 mL of this solution requires 19.8 mL of N/10 HCl for neutralization. The value of X is (a) 7 (b) 3 (c) 2 (d) 5 50. 20 mL of XM HCl neutralises completely 10 mL of 0.1 M NaHCO3 solution and a further 5 mL of 0.2 M Na2CO3 solution to methyl orange end point. The value of X is (a) 0.167 M (b) 0.133 M (c) 0.15 M (d) 0.2 M 51. RH2 can replace Ca+2 in hard water RH2+Ca+2 →RCa+2H+ 1 lit. of hard water after passing through RH2 has H+ = 0.01 M. Hence hardness of water in ppm Ca+2 is (a) 100 (b) 200 (c) 50 (d) 125 52. N2+3H2→2NH3. Molecular weight of NH3 and N2 are x1, x2 and their equivalent weights are y1 and y2. Then y1–y2 is (b) 2 x 2 − x1 (a) 2 x1 − x1 6 6 (c) 3x1-x2 (d) 3x2-x1 53. One gram of a mixture Na2CO3, NaHCO3 in 2:1, molar ratio requires x equivalents HCl for neutralization. The mixture is strongly heated and react with HCl. How many equivalents of HCl is required? (a) 2x (b) x (c) 3x/4 (d) 3x/2 54. A solution contains Na2CO3, and NaHCO3 10 mL of the solution required 2.5 mL of 0.1 M H2SO4 for neutralization using phenolphthalein as indicator. Then methyl orange is added 2.5 mL of 0.2 M H2SO4 is required. The amounts of Na2CO3, NaHCO3 in 1 lit solution is (a) 5.3 g, 4.2 g (b) 5.3 g, 6.2 g (c) 4.2 g, 5.3 g (d) 6.3 g, 5.3 g 55. 100 mL of 1 M KMnO4 oxidized 100 mL H2O2 in acidic medium. Volume of same KMnO4 required (MnO4–→MnO2) in alkaline medium to oxidize 100 mL of H2O2 will be (a)

100 mL 3

(b)

(c)

300 mL 3

(d) none

500 mL 3

56. Atomic weight of barium is 137.34. the equivalent weight of barium in BaCrO4 used as oxidizing agent in acidic solution is (a) 137.34 (b) 45.78 (c) 114.45 (d) 68.67

Stoichiometry 13.45

57. A mixture of carbonates of two bivalent metals contains 13.6% of one metal. If 2.5 g of the mixture liberates 1.33 g of CO2. The % by weight of the other metal is (a) 13.6 (b) 12.6 (c) 13.84 (d) 12.84 58. One gram of Na3AsO4 is boiled with excess of solid KI in presence of strong HCl. The iodine evolved is absorbed in KI solution and titrated against 0.2 N hypo solution AsO4–3+2H++2I– → AsO3–3+H2O+I2 Calculate the volume of thiosulphate consumed. AW of As = 75. (a) 48.1 mL (b) 38.4 mL (c) 24.7 mL (d) 30.3 mL 59. Sulphuryl chloride SO2Cl2 reacts with water to give a mixture of H2SO4 and HCl. No of moles of Al(OH)3 required to neutralize the solution formed by adding 3 mol SO2Cl2 to excess water (a) 2 (b) 4 (c) 6 (d) 3 60 10 mL of H2O2 solution (volume strength=x) required to 10 mL of

61.

62.

63.

64.

N KMnO4 solution in acidic medium. 0.56

Hence ‘x’ is (a) 0.1 (b) 10 (c) 100 (d) 0.01 Mass of KHC2O4 required to reduces 100 mL of 0.2 N KMnO4 in acidic medium is X g and to neutralise 50 mL of 0.04 M Ca (OH)2 is Y g then (a) 5x = 2y (b) 5x = y (c) 2x = 5y (d) 5y = x A mixture of (NaHCO3+Na2CO3) in 1:1 molar ratio requires a vol of HCl x mL, with methyl orange and ‘y’ mL with phenolphthalein indicator in same titration. The relation between x and y is (a) x = 3y (b) y = 3x (c) x = 6y (d) y = 6x A mixture of NaHCO3 and Na2CO3 requires a volume of HCl x mL with phenolphthalein indicator and further ‘y’ mL required with methyl orange indicator. Hence volume of HCl for complete reaction of NaHCO3 is (a) y-x (b) x-y (c) 2y-x (d) 2x-y When one gram mol of KMnO4 reacts with HCl, the volume of chlorine liberated at NTP will be (a) 11.2 lit (b) 22.4 lit (c) 44.8 lit (d) 56 lit

65. 3.18 gr. of pure Cu heated in a stream of oxygen for some time to gain weight with the formation of CuO. The final weight is 3.92. What percent of copper remains unoxidized (a) 6.5 (b) 6.9 (c) 7.6 (d) 7.9 66. 0.5 g of fuming H2SO4 (Oleum) is diluted with water. This solution is completely neutralised by 26.7 mL of 0.4 N NaOH. The percentage of free SO3 in the sample is (a) 30.6% (b) 40.6% (c) 20.6% (d) 10.6% 67. In an experiment 50 mL of 0.1 M solution of a metallic salt reacted exactly with 20 mL of 0.1 M solution of Na2SO3. In reaction SO3–2 is oxidized to SO4–2. The original oxidation number of metal in the salt is +2. What would be the new oxidation number of the metal? (a) 0 (b) 2 (c) 4 (d) 1 68. On heating H2O2 decomposes to give oxygen. Under these conditions 1 mol gas occupies 24 lit. 100 mL of XM H2O2 produces 3 lit. of O2. The value of X is (a) 0.125 (b) 1.25 (c) 2.5 (d) 0.25 69. The no. of moles of KMnO4 that will be needed to react completely with 1 mole ferrous oxalate in acidic solution is (a) 2/5 (b) 3/5 (c) 4/5 (d) 1/5 70. Oxalic acid reacts with concentrated H2SO4 to give a mixture of two gases. When this mixture is passed through caustic potash, one of the gases is absorbed. What is the product formed by absorbed gas with caustic potash? (a) K2SO4 (b) K2HCO3 (c) K2CO3 (d) KOH

multiple choice questions with one or more than one Answer 1. H2C2O4. 2H2O (Mol. = wt = 126 ) can be oxidized into CO2 by acidified KMnO4. 6.3 g of oxalic acid can be oxidized by (MW of KMnO4 = 158) (a) 3.16 g of KMnO4 (b) 200 mL of 0.1 M KMnO4 (c) 0.1 mole of KMnO4 (d) 0.02 moles of KMnO4 2. One mole of FeC2O4 is separately oxidized by KMnO4 and K2Cr2O7 in acid medium into Fe+3 ions and CO2 molecules. Which statements are true? (a) The mole ratio of KMnO4 and K2Cr2O7 required is 6:5. (b) The ratio of number of equivalents of KMnO4 and K2Cr2O7 required is 1:1.

13.46

3.

4.

5.

6.

7.

Stoichiometry

1 (c) Equivalent weight of FeC2O4 is rd of its formu3 la weight. (d) FeC2O4 undergoes disproportionation. Which of the following samples of reducing agent (s) are chemically equivalent to 25 mL of 0.2 N KMnO4 to be reduced to Mn+2 and H2O? (a) 25 mL of 0.2 M FeSO4 to be oxidized to Fe+3. (b) 50 mL of 0.1 M H 3 AsO 3 to be oxidized to H 3 AsO 4 . (c) 30 mL of 0.1 M H2O2 to be oxidized to O2 and H+. (d) 25 mL of 0.1 SnCl2 to be oxidized to Sn+4. An oleum, labeled ‘109% H2SO4” suggests (a) 109 g of H2SO4 is produced on reacting 100 g of sample with H2O. (b) 100g of H2SO4 is produced on reacting 109 g of sample with H2O. (c) 40 g of free SO3 in 100 g oleum. (d) 9 g free SO3 in 100 g oleum. A solution containing Cu2+ and C2O42– ions is titrated with 20 mL of M/4 KMnO4 solution in acidic medium. The resulting solution is treated with excess of KI after neutralization. The evolved I2 is then reacts with 25 mL of M/10 hypo solution. (a) The difference of the number of m mol Cu2+ and C2O42–ions in the solution is 10 m mol. (b) The difference of the number of m mole Cu2+ and C2O42– ions the solution is 22.5 m mole. (c) The equivalent weight of Cu2+ ions in the titration with KI is equal to atomic weight of Cu2+. (d) The equivalent weight of KI in the titration is M/2 (M = Molecular weight of KI). 20 mL of H2O is reacted completely with acidified K2Cr2O7 solution. 40 mL of K2Cr2O7 solution required to oxidize the H2O2 completely. Also, 2.0 mL of the same K2Cr2O7 solution required 5.0 mL of a 1.0 MH2C2O4 solution to reach equivalence point (a) The H2O2 solution is 5 M. (b) The volume strength of H2O2 is 56 V. (c) The volume of strength oh H2O2 is 112 V. (d) If 40 mL of M/8 H 2O2 is further added to the 10 mL of above H2O2 solution, the volume strength of the resulting solution is changed to 16.8 V. Which of the following statements is/are correct for given reaction? 3Cu+8HNO3→3 Cu (NO3)2 +2NO+4H2O is (a) Equivalent mass of HNO3 is 84. (b) n factor of HNO3 is I. (c) n factor of HNO3 is

3 . 4

(d) Cu is oxidized to Cu2+.

8. Which of the following has/ have same percentage of carbon? (a) C2H5OH (b) C6H12O6 (c) CH3COOH (d) C2H5NH2 9. 100 mL

M 10

Ba(MnO4)2 in acidic medium can

oxidize completely (a) 100 mL of 1 M FeSO4 (b)

10.

11.

12.

13.

14.

100 mL of M FeC2O4 3

(c) 25 mL of 1 M K2 Cr2O7 solution (d) 75 mL of 1 M C2O42– solution In the reaction, A+2B→2C, 5 moles of A and 8 moles of B are reacted, then (a) Whole A is consumed. (b) Whole B is consumed. (c) 8 moles of C are formed. (d) 8 equivalents of C are formed. Correct statements (a) The oxidation states of Na in NaHg, Na2Hg Na3Hg are +1, +2, +3 respectively. (b) At STP, the volume occupied by 4.5 g water is 5.6 L. (c) 0.1 mole P4 contains 3.6×1023 P-P bonds. (d) 3.33×1025 water molecules are present in one litre of water (d=1g/mL). 4Mg+ 10HNO3 → 4Mg (NO3)2 + NH4NO33H2O Incorrect statements regarding this reaction is/are (a) 4 g atoms of Mg is oxidized by 10 mole of HNO 3. (b) 4 g at atom of Mg is oxidized by 1 mole HNO3. (c) 1 mole of HNO3 is reduced by 0.4 atoms of Mg. (d) 1 mole of HNO3 is reduced by 4 atoms of Mg. 100 mL of 0.15 N H2O2 is completely oxidized by (a) 150 mL of 0.1 N KMnO4 solution. (b) 2.5×10–3 moles of K2Cr2O7 in acidic medium. (c) 15×10–3 moles of KMnO4 in basic medium. (d) 15 moles of O3 in acidic medium. Which of the following statement(s) is/are correct? (a) 0.2 moles of KMnO4 will oxidize one mole of ferrous ions to ferric ions in acidic medium. (b) 21.5 moles of KMnO4 will oxidize 1 moles of ferrous oxalate to ferric oxalate in acidic medium. (c) 0.6 moles of KMnO4 will oxidize 1 mole of ferrous oxalate to one mole of ferric ion and carbon dioxide in acidic medium. (d) 1 mole of K2Cr2O7 will oxidize 2 moles of ferrous oxalate to ferric ions and carbon dioxide in acidic medium.

Stoichiometry 13.47

15. 11.2 g of mixture of MCl (volatile) and NaCl gave 28.7 g of white ppt. with excess of AgNO3 solution. 11.2 g of same mixture on heating gave a gas that on passing into AgNO3 solution gave 14.35 g of white ppt. Hence, (a) Atomic mass of M is 18. (b) Mixture has equal mole fraction of MCl and NaCl. (c) MCl and NaCl are in 1:2 molar ratio. (d) Ionic mass of M is 10. 16. HIO4 absorbs water to form H5IO6. If 96 g of pure anhydrous HIO4 absorbs water, its mass increases to 105 g. Which of the following is (are) true for final substance? (a) It contains both periodic acid (HIO4) and paraperiodic acid, H5IO6. (b) Half of original HIO4 is converted to H5IO6. (c) HIO4 has been completely converted to H5IO6. (d) Only 25% of original HIO4 has been converted to H5IO6. 17. 0.1 mol of MnO4– (in acidic medium) can (a) Oxidise 0.5 mol of Fe2+ (b) Oxidise

1 ×mol of FeC2O4 6

(c) Oxidise 0.25 mol of C2O42– (d) Oxidise 0.6 mol of Cr2O72– 18. H2C2O4 and NaHC2O4 behave as acids as well as reducing agents. Which is/are correct statement(s)? (a) Equivalent wt of H2C2O4 and NaHC2O4 are equal to their molecular weights when behaving as reducing agent. (b) 100 mL of 1 N solution of each is neutralized by equal volumes of 1 M Ca(OH)2. (c) 100 mL of 1 N solution of each is neutralized by equal volumes of 1 N Ca(OH)2. (d) 100 mL of 1 M solution of each is oxidized by equal volumes of 1 M KMnO4. 19. 0.2 mole of K3PO4 and 0.3 mole of BaCl2 are mixed in 1 L of solution. Which of these is/are correct? (a) 0.2 mole of Ba3 (PO4)2 will be formed. (b) 0.1 mole of Ba3 (PO4)2 will be formed. (c) 0.6 mole of KCl will be formed. (d) 0.3 mole of KCl will be formed. 20. For unbalanced reactions S–2 + (O.A)1 → SO2 + other products and all SO2 produced is reacting with (O.A)2 SO2 + (O.A)2 → SO42– + other product. What is the correct option. Given (O.A)1, (O.A)2 & other products do not contains ‘S’. O.A1 & O.A2 are different oxidizing agent (a) Equivalent of SO2 in reaction I = equivalents of O.A1

(b) Equivalents of SO2 in reaction II = equivalents of O.A1 (c) Moles of sulphide = Moles of sulphate (d) Moles of (O.A)1 ÷ Moles of (O.A)2 =

2 6

21. If 4 moles of A2(g) and 2 moles of B4(g) at 300 K are reacted in a closed container so that none of the reactants are remained, and only one product and that too in gaseous phase, is formed and temperature reaches 450 K, then which option would be correct? (a) Product is A4B4 if ratio of initial and final pressure is 2:1. (b) Product is A2B2 if ratio of initial and final pressure is 1:1. (c) Product is A2B4 if ratio of initial and final pressure is 2:1. (d) Product is AB if ratio of initial and final pressure is 1:2. 22. A mixture of 20 mL of CO, CH4 and N2 was burnt in excess of O2 resulting in reduction of 13 mL of volume. The residual gas then treated with KOH solution to show a reduction in volume of 14 mL. Then the correct statement among the following is (a) Volume of CO in mixture is 10 mL. (b) Volume of CH4 in mixture is 4 mL. (c) Volume of CH4 in mixture is 6 mL. (d) Volume of N2 in mixture is 6 mL. 23. 0.1 M solution of KI reacts with excess of H2SO4 and KIO3 solutions, according to equation 5I– + IO3– + 6H+ → 3I2 + 3H2O; which of the following statement is correct (a) 200 mL of the KI solution reacts with 0.004 mole KIO3. (b) 100 mL of the KI solution reacts with 0.006 mole of H2SO4. (c) 0.5 litre of the KI solution produced 0.005 mole of I2. (d) Equivalent weight of KIO3 is equal to  Molecular weight  .   5  

comprehensions type questions Passage i 50 mL of a solution containing 1 g each of Na2CO3, NaHCO3, NaOH was titrated with HCl having 1 N. 1. The titre value when phenolphthalein is used (a) 43.86 mL (b) 34.4 mL (c) 17.2 mL (d) 21.9 mL

13.48

Stoichiometry

2. The titre value when methyl orange indicator used form beginning (a) 55.8 mL (b) 43.86 mL (c) 34.4 mL (d) 25.0 mL 3. The titre value when methyl organce indicator is used after first end point (a) 34.4 mL (b) 21.3 mL (c) 30.7 mL (d) 25.0 mL Passage ii 3.6 g of magnesium is burnt in limited supply of oxygen. The residue was treated with 100 mL of H2SO4 (35% by mass, 1.26 g/mL. density) and 2.463 litre of H2 at 1 atm and 27°C was evolved. After the reaction H2SO4 was found to have a density of 1.05 g/mL. Assuming no volume change in H2SO4. 1. The final % by weight of H2SO4 was (a) 25% (b) 28% (c) 36.78% (d) 30% 2. The % by mass of magnesium converted to oxide (a) 25% (b) 30% (c) 33.33% (d) 50% 3. Volume of the used O2 at STP was (a) 1.12 L (b) 0.56 L (c) 2.24 L (d) 4.48 L Passage iii 20 mL of a solution containing 0.2 g of impure sample of H2O2 reacts with 0.316 g of KMnO4 in acid medium 1. The purity of H2O2 is (a) 70% (b) 80% (c) 85% (d) 91% 2. The volume of dry O2 evaluated at STP is (a) 112 mL (b) 56 mL (c) 168 mL (d) 224 mL Passage iv 13.4 g dry analytical grade sodium oxalate (mol wt = 134) was dissolved in 10 mL H2O and to that 100 mL 2 M H2SO4 was added. The solution was cooled to 25-30°C. In the resulting solution H2SO4 can be considered to be completely dissociated. Now to this solution 0.1 M KMnO4 solutions was added till a very faint pink colour persisted. 1. How many mL of KMnO4 solution must have been used in order to get faint pink colour at the end point? (a) 400 mL (b) 200 mL (c) 100 mL (d) 300 mL

2. The solution was cooled to 25-30°C, therefore temperature must have gone up. It was due to (a) Chemical reaction between sodium oxalate and H2SO4 (b) High heat of hydration of H+ ions (c) Breaking of H-bonds in H2SO4 (d) Decrease in the viscocity of H2SO4 3. It was observed that in the initial additions of KMnO4 pink colour disappeared slowly but later on it disappeared quickly. The possible reason can be (a) The reaction between MnO4– and H2SO4 is slow at low concentration of MnO4–. (b) Oxidizing power of H2SO4 is much lower than that of KMnO4. (c) The reaction between MnO4– and C2O4– is slow at room temperature but it gets catalysed by the production of Mn2+ ions. (d) None of these Passage v A definite amount of BaCl2 was dissolved in HCl solution of unknown normality. 20 mL of this solution was heated with 21.4 mL of N/10 NaOH for complete neutralization. Another 20 mL of this solution was added to 50 mL of N/10 Na2CO3 and the precipitate was filtered off. The filtrate react with 10.5 mL of 0.8/10 (NH4)2SO4 using phenolphthalein indicator. M.W of BaCl2 is 208 1. The strength of HCl is (a) 1.95 g/lit (b) 2.925 g/lit (c) 5.85 g/lit (d) 3.9 g/lit 2. The strength of BaCl2 in solution is (a) 3.067 g/lit (b) 6.13 g/lit (c) 4.601 g/lit (d) 9.134 g/lit Passage vi 0.36 g sample of FeS2 results Fe3+ and SO2 upon oxidation. Released SO2 is estimated using 40 mL of barium permanganate in acid medium resulting SO42– and Mn2+. 10 mL of filtate of Ba(MnO4)2 in acid medium was treated with excess KI and I2 produced required 5 mL of 0.05 M hypo solution. In another titration 25 mL of same solution of Ba(MnO4)2 treated with KI in alkaline medium. Produced I2 required 20 mL of same hypo solution resulting S4O62– and Mn6+. 1. The number of equivalents of Ba (MnO4)2 in the original 40 mL solution are (a) 8 × 10–3 (b) 4 × 10–4 –4 (c) 8 × 10 (d) 6 × 10–4 2. Moles of SO2 are (a) 5.5 × 10–3 (b) 8 × 10–4 –3 (c) 3.5 × 10 (d) 2.75 × 10–3

Stoichiometry 13.49

3. % of FeS2 in the sample will be (a) 100% (b) 58.3% (c) 25% (d) 5% Passage vii In presence of chloride ion, Mn+2 can be titrated with MnO4−, both reactants being converted to a complex of Mn (III). A 0.545 g sample containing Mn3O4 was dissolved and all manganese was converted to Mn+2. Titration in presence of chloride ion consumed 31.1 mL of KMnO4 that was 0.117 N against oxalate. 1. No. of equivalents of Mn+2 reacted is (a) 3.63 × 10–3 (b) 0.72 × 10–3 –3 (c) 14.5 × 10 (d) 2.9 × 10–3 +2 2. No. of moles of Mn produced form the sample is (a) 3.63 × 10–3 (b) 2.9 × 10–3 –3 (c) 14.5 × 10 (d) 0.72 × 10–3 3. % of Mn3O4 in this sample is (a) 40.75% (b) 20.38% (c) 122.25% (d) 12.225% Passage viii MnO2 react with HCl according to this equation MnO2 + 4HCl → MnCl2 + 2H2O + Cl2 Certain volume of 2 M HCl produce 2.24 lit of Cl2 at STP 1. The volume of HCl used in this reaction is (a) 100 mL (b) 150 mL (c) 200 mL (d) 250 mL 2. The weight of MnO2 used in above process is (a) 4.35 g (b) 6.52 g (c) 8.7 g (d) 13.0 g 3. 2 g of bleaching powder was suspended in 500 mL of water. 50 mL of this solution on treatment with KI and HCl. Liberated Iodine react with 20 mL of

N Na2S2O3. 10

Percentage of available chlorine in bleaching powder is (a) 7.1% (b) 3.55% (c) 71% (d) 35.5% Passage ix A 5 g sample containing Pb3O4, PbO2 and some inert inpurity is dissolved in 250 mL dilute HNO3 solution and 2.68 g of Na2C2O4 was added so that all lead converted into Pb2+. The resultant solution needs two experiments to find out composition of initial sample

(i) A 10 mL of this solution required 8.0 mL, 0.1 N oxidizing agent for titration of excess of oxalate and this solution requires 4 mL of permanganate solution for oxidation of Pb2+ to Pb4+. (ii) 10 mL of this permanganate solution is equivalent to 4.48 mL of 5 Vol of H2O2 solution [Atomic weight of Pb = 207 Na = 23, C = 12, O = 16]. Answer the following questions: 1. Mass percentage of PbO2 in the original sample was (a) 68.5 (b) 62 (c) 38 (d) 23.9 2. Normality of the KMnO4 solution used from oxidation of Pb2+ to Pb4+ was (a) 0.6 (b) 0.4 (c) 0.2 (d) 0.8 3. Weight of inert impurity in the original sample was (a) 0.42 g (b) 0.32 g (c) 0.38 g (d) 0.52 g Passage x Iodine index is defined as the amount mg of commercial acid consumed per g of I2. A 200 mL sample of a citrus fruit drinks containing ascorbic acid (vitamin C molecular weight 176, C6H8O6) was acidified with H2SO4 and 10 mL of 0.025 M I2 was added. Some of the I2 was reduced by the ascorbic acid to HI. The excess of I2 required 4.6 mL of 0.01 M S2O3–2 for reduction. The reactions are C6H8O6+I2 → C6H6O6+2HI 5H2O+S2O3–2 +4I2 → 2SO4–2 +8I– +10H+ 1. Find the iodine index (a) 693 (b) 346.5 (d) 1039.5 (d)` 1386 2. What was the vitamin C content of the drink in mg vitamin per lit drink (a) 11.616 (b) 58 (c) 29 (d) 14.5 Passage xi KMnO4 oxidizes the Xn+ ion into XO3− , itself changing to Mn+2 in acid medium. 2.68×10−3 mole of Xn+ requires 1.61×10−3 moles of KMnO4. The equivalent weight of X Cln is 56. 1. The value of n is (a) 1 (b) 2 (c) 3 (d) 4 2. Atomic weight of metal X in above data (a) 41 (b) 97 (c) 153 (d) 69

13.50

Stoichiometry

Passage xii

Passage xiv

1.5 g of brass containing Cu and Zn reacts with 3M HNO3 solution. The following reactions take place.

Pyrolusite, MnO2 is the main ore from which manganese is produced. The manganese content of the ore may be determined by reducing the MnO2 under acidic conditions to Mn2+ with the oxalate ion, C2O42–, the oxalate ion being oxidized to carbon dioxide during the reaction. The analytical determination is carried out by adding a known excess volume of oxalate solution to a suspension of the pyrolusite and digesting the mixture on hot water bath until all the MnO2 has been reduced. The excess, unreacted, oxalate solution is then titrated with standardised potassium permanganate, KMnO4 solution, after which the manganese content of the ore can be calculated. Potassium permanganate solution is standardised, under acid conditions, against oxalate ion where the MnO4– ion is reduced to Mn2+ and the oxalate ion is oxidized to carbon dioxide. 1. A student prepared a standard solution of sodium oxalate by weighing 3.2000 g of the dry anhydrous salt, dissolving it in distilled water and making the solution up to 500.0 mL. Two successive titrations, under acidic conditions, of 25.0 mL aliquots of the oxalate solution required 24.80 mL, 24.73 mL respectively of potassium permanganate solution. What is the molarity of the permanganate solution? (a) 4.776 × 10–3 (b) 1.929 × 10–2 (d) 7.1924 × 10–2 (d) 4.776 × 10–2 2. In determining the manganese in pyrolusite a student weighed 0.2123 g of the ore into a 250 mL conical flask added 50 mL of 2M H2 SO4 and a 50.0 mL aliquots of the standard sodium oxalate by pipette to the flask. The flask was carefully heated on a water bath and when all the MnO2 had been reduced the excess oxalate ion was titrated with the standardised potassium permanganate solution where a litre of 12.87 mL was required to with the excess oxalate. How many moles of MnO2 were contained in the sample of pyrolusite? (a) 1.767 × 10–3 (b) 6.207 × 10–4 (c) 1.767 × 10–2 (d) 1.99 × 10–2 3. What is the percent (%) purity (by mass) of pyrolusite? (a) 23.20 (b) 46.11 (c) 61.25 (d) 72.36

Cu+HNO3 → Cu+2 + NO2(g)+H2O Zn+HNO3 → Zn+2 +NH4+ + H2O 0.0425 moles of NO2 gas evolved in this reaction. 1. The percentage of zinc present in the brass is (a) 90% (b) 80% (c) 20% (d) 10% 2. The volume of HNO3 required to react with zinc in above reaction is (a) 0.232 mL (b) 0.064 mL (c) 0.128 mL (d) 0.192 mL 3. The volume of HNO3 required to react brass is (a) 9.55 mL (b) 7.12 mL (c) 4.82 mL (c) 14.33 mL Passage xiii H2O2 acts both as oxidising and reducing agent. H2O and O2 are products when H2O2 acts as oxidant and reducing agent respectively. The strength of H2O2 is expressed in terms of molarity, normality, % strength and volume strength. H2O2 decomposes as H2O2 → H2O + 1/2 O2(g) i.e., one mole O2 is released form 2 mole H2O2.X ‘volume’ strength of H2O2 means 1 volume (mL or litre) of H 2O2 sample releases X volume (mL or litre) O 2 gas at NTP on its decomposition. Hence molarity = X/11.2 moles per litre, i.e., normality of H2O2 = X/5.6. Thus volume strength, i.e., X = 5.6 X normality. Weight of H2O2 (in g) present in 100 mL H2O2 solution is called percentage strength of H2O2. 1. The percentage strength of 20 ‘vol’ H2O2 is which of the following: (a) 10% (b) 6.07% (c) 22% (d) 15% 2. What volume of H2O2 solution of 22.4 ‘vol’ strength is required to oxidise 6.3 g oxalic acid: (a) 10 mL (b) 11.2 mL (c) 25 mL (d) 30 mL 3. 50 mL of H2O2 solution were diluted to 200 mL and 10 mL of this diluted H2O2 solution reduced 10 mL of 0.1 M KMnO4 acidic solution. The volume strength of H2O2 is: (a) 2.8’vol’ (b) 5.6’vol’ (c) 11.2’vol’ (d) 22.4’vol’

Stoichiometry 13.51

matching type questions 1. Matching the following: Column I

Column II

(a) 16g of O2 (b) Gram Equivalent Volume of H2 (c) 18 g of H2O (d) ½ mole of O2+1 g atom of H2

(p) 1 g atom of O (q) 22.4 L at STP (r) 18 mL (s) 11.2 L at STP

5. One gram of mixture of Na2CO3 and K2CO3 was made up to 250 mL in aqueous solution. 25mL of this solution was neutralized by 20 mL of HCl of unknown concentration. The neutralized solution required 16.24 mL of 0.1 N AgNO3 for precipitation. (a) (b) (c) (d)

Column I

Column II

Molarity of HCl is Wt of Na2CO3 in grams Wt of K2CO3 in grams Conc. of HCl in g/250 mL

(p) (q) (r) (s)

0.74 0.6 0.4 0.0812

2. Matching the following: Column I

Column II

(a) 4.1g H2SO3

(p) 200 mL of 0.5 N NaOH used for complete neutralization (q) 200 milli moles of oxygen atoms (r) Central atom has its highest oxidation number (s) May react with an oxidizing agent

(b) 4.9 g H3PO4 (c) 4.5g H2C2O4

(d) 5.3g Na2CO3

3. Matching the following: Column I (a) EW of Fe3O4 in the reaction Fe3O4+KMnO4 → Fe2O3+MnO2 (b) EW of FeC2O4 in FeC2O4 → Fe+3+CO2 (c) EW of NaNO3 in NO3− → N2 (d) EW of NaBrO3 in BrO3− + H+ → Br−+ H2O

Column II (p) 48

(q) 25.16 (r) 232 (s) 17

4. 0.4N K2Cr2O7 required to liberate all Cl2 from 1.17 g of NaCl in a solution acidified with H2SO4 Cr2O7–2+Cl–+H+ → Cr+3Cl2+H2O. Column I (a) The volume (in mL) of K2Cr2O7 required is (b) The no of milligrams of K2Cr2O7 Present in 1 mL solution is (c) The no of milli moles of Cl2 liberated is (d) The no of milli moles of H2 used in the reaction is

Column II (p) 10 (q) 19.6

(r) 46.66 (s) 50

6. Given two mixtures: (i) NaOH and Na2CO3 (ii) NaHCO3 and Na2CO3 100 mL of mixture (i) required w and x mL of 1 M HCl in separate titrations using phenolphthalein and methyl orange indicators while 100 mL of mixture (ii) required y and z mL of same HCl solution in separate titrations using the same indicators. Match the following: Column I (Substance) (a) Na2CO3 in (1) (b) Na2CO3 in (2) (c) NaOH in (1) (d) NaHCO3 in (2)

Column II (Molarity in solution)

Mixture

(p) (2w–x)×10–2

Mixture

(q) (z–2Y)×10–2

Mixture

(r) y×10–2

Mixture

(s) (x–w)×10–2

7. To 1 Mol FeO.Fe2O3 excess KI(aq) is added in acidic medium and the resulting mixture is titrated with hypo using starch as indicator. Match the following: Column I

Column II

(a) Change in oxidation number of Fe atoms per mol of FeO. Fe2O3 (b) Change in oxidation number of S-atom in hypo (c) Oxidation number of S-atom in the final product (d) Moles of iodine liberated that is titrated with hypo

(p) 1

(q) 2

(r) +0.5

(s) +2.5

13.52

Stoichiometry

Assertion (A) and reason (r) type questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): Equivalent weight of NH 3 in the reaction N2 + H2 → NH3 is 17/3 which that of N 2 is 28/6 Reason (R): Equivalent Weight = Molecular weight no.of electrons lost or gained per molecule of reacttant 2. Assertion (A): 5 molal aqueous solution of NaOH equals to 5 molar at the same temperature if density is 1.2 g/cc Reason (R): At same temperature molarity equal to molality. 3. Assertion (A): Consider the following reaction Cu2S + 2O2 → 2CuO + SO2 ‘a’ g equivalents of O2 are used to produce ‘a’ g equivalents each of CuO and SO2 Reason (R): In any chemical reaction, number of g equivalents of each reactant consumed and those of each product produced are same. 4. KIO3 reacts with KI to liberate iodine and liberated iodine is titrated with standard hypo solution. The reactions are: (i) IO3− + I− → I2 (valency factor = 5/3) (ii) I2 + S2O32− → S4O62− + I− (valency factor = 2) meq of hypo = meq of I2, meq of IO3− = meq of I– ∴ IO3− react with I− ⇒ meq of IO3− = meq of I– Assertion (A): Meq of hypo = 2×meq of IO3– Reason (R): Valency factor of I2 in both the equation are different therefore we can not equate mill equivalents in sequence.

integer type questions 1. A polyvalent metal weighting 0.1 g and having atomic weight 51 reacted with dil H2SO4 to give 43.9 mL of H2 gas at STP The solution containing the metal in the lower oxidation state was found to require 58.8 mL of 0.1 N KMnO4 for complete oxidation. The higher valency of metal is __________.

2. A 5 mL solution of H2O2 liberates 0.566 g of iodine from acidified KI Solution. The volume strength of H2O2 is __________. 3. KMnO4 oxidizes Xn+ to XO3– and it self changing to Mn+2 is acid medium. 2.68×10–3 moles of Xn+ requires 1.61×10–3 mole of KMnO4. The value of ‘N’ is _____. 4. A solution of 0.2 g of a compound containing Cu+2 and C2O4–2 ions on titration with 0.02 M KMnO4 in presence of H2SO4 consumes 22.6 mL oxidant. The resulting solution is neutralized by Na2CO3 acidified with CH3COOH and titrated with excess of KI. The liberated I2 required 11.3 mL of 0.05M Na2S2O3 solution. The molar ratio of C2O4–2to Cu+2 is __________. 5. 1 g sample of Fe2O3 solid of 55.2% purity is dissolved in acid reduced by heating the solution with zinc dust. The resultant solution is cooled and made up to 100 mL. An aliquot of 25mL of this solution required 17 mL of 0.0167 M solution of an oxidant for titration. The no of electrons taken up by oxidant in the above titration is __________. 6. 150 g of Ba(MnO4)2 sample containing inert impurity is completely reacting with 100 mL of 11.2 vol of H2O2 in the presence of H2SO4. What will be the percentage purity of Ba(MnO4)2 In the sample is __________. 7. 0.4 g of a polybasic acid HnA (all the hydrogens are acidic) requires 0.5 g NaOH for complete neutralization. The number of replaceable hydrogen atoms in the acid would be. (Molecular weight of the acid is 96 g) __________. 8. 5 mL of a gaseous hydrocarbon was exposed to 30 mL of O2. The resultant gas, on cooling is found to measure 25 mL, of which 10mL was absorbed by NaOH and the remainder by pyrogallol. All measurements are made at constant pressure and temperature. The sum of number of carbon and hydrogen atoms in the hydrocarbon is __________. 9. 5×10–5 moles KMnO4 in acidic medium is completely consumed to oxidize 5×10–5 moles of Xn+ to XO4–, then what is the value of n? 10. A 50 mL 1.92% (W/N) solution of a metal ion Mn+ (atomic weight = 60) was treated with 5.332g hydrazine hydrate (90%pure) and mixture was saturated with CO2 gas when entire metal gets precipitated as a complex [M(H2N-NHCOO)n]. The complex was filtered off and filtrate was titrated with

M KIO3 in 10

the presence of conc. HCl according to the following equation: N2H4+IO3− +2H++CI−→ Cl+3H2O+N2 ↑ The volume of

M KIO3 solution needed for the end 10

point to arrive was 480 mL. Find the value of 'n'.

Stoichiometry 13.53

11. When ammonium vanadate is heated with oxalic acid solution, a compound (Z) is formed. A sample of (Z) was titrated with KMnO4 solution in hot acidic solution. The resulting liquid was reduced with SO2, the excess SO2 boiled off and the liquid again titrated with the same KMnO4 solution. Assume that KMnO4 oxidizes all oxidation states of vanadium to vanadium (+V) and SO2 reduces vanadium (+V) to (+IV). If the ratio of volumes of KMnO4 used in two titrations was 5:1, then the oxidation state of vanadium in compound (Z) is __________. 12. 0.36 g of an element ‘X’ is heated with NaOH and NaNO3 to produce Na2XO2 and NH3. Ammonia produced is absorbed in 100 mL of 0.11 M H2SO4 and excess of acid is back titrated with 48mL of 0.25 M NaOH. The atomic weight of element X is __________. 13. A mixture of NaOH and Na2CO3 required 25 mL 0.1 M HCI using phenolphthalein as the indicator. However, the same amount of mixture required 30 mL of 0.1 M HCl when methyl orange was used as the indicator. The molar ratio of NaOH and Na2CO3 in the mixture is x:y then the x+y is __________. 14. An+ is maximum oxidized by acidified KMnO4 solution into AO3–. If 2.68 m moles of A+(n+1) requires 32.16mL of 0.05 M acidified KMnO4 solution for complete oxidation, value of ‘n’ is__________. 15. A 6.772 g sample of a pesticide was decomposed by fusion with Na2CO3 and leaching the residue with hot water. The fluoride ion in the sample was precipitated as PbClF by HCI and Pb(NO3)2. The precipitate was filtered, washed and dissolved in 5% HNO3 the chloride ion was precipitated as AgCl by 50mL 0.2 M AgNO3. The excess Ag+ was completely neutralized with 10mL 0.15 N NH4SCN. What is the percentage of Na2SeF6 in the sample? (atomic weight of Se = 79 g mol–1) 16. A 150 mL of solution of I2 is divided into two unequal parts. Ist part reacts with hypo solution in acidic medium. 15 mL of 0.4 M hypo was consumed. IInd part was added with 100mL of 0.3 M NaOH solution. Residual base required 10mL of 0.3 M H2SO4 solution for complete neutralization. The initial concentration of I2 was x×10–1 M, then the value of x is __________. 17. 12.72 g Na2CO3 was treated by a series of reagents so as to convert all of its carbon to KxZny[Fe(CN)6]2 (M. Wt=700) then the weight KxZny [Fe(CN)6]2 formed in ‘g’ is (there is no loss(or) gain in mass of carbon in above changes). 18. 60 mL of a mixture of equal volumes of Cl2 and oxide of chloride was heated to react completely and then brought to original temperature and pressure. The resulting gas mixture was found to have volume of 75 mL. On treatment with alkaline pyrogallol solution, the volume contracted by 15 mL. Assume that all measurement are

19.

20.

21. 22.

23.

24.

25.

made at the same ‘T’ and pressure. If the formula of this oxide is Cl2On What will be the value of ‘n’? 100 mL of 5.6 V H2O2 is decolourise 60 mL of KMnO4 in acidic medium. How many mL of the same KMnO4 will be decolourised by 50 mL of 0.56 V H2O2 in dilute alkaline medium. A 0.495 g sample of As2O3 requires 5mL of ‘x’ normal solution of KMnO4 for complete titration, whereas as a 0.83 g sample of KI requires 10 mL of ‘y’ normal solution of KMnO4 for complete conversion of I– to I2 The ratio x/y is __________. A mole of KMnO4 oxidizes how many moles of Fe(HC2O4)2 in acidic solution? 0.01 moles of FeSn [iron (II) sulphide] requires 0.06 moles AO4–3 for complete oxidation. The species formed are FeO,SO2 and A+2 Calculate the value of n. Na2CO3+HCI→NaHCO3+NaCl (mol Wt of Na2CO3 = 106) when 0.53 g of Na2CO3 dissolved in 100 mL. the normality of the solution is x×10–2 then ‘x’ is __________. An unknown cupric salt with formula Cux(CO3)y (OH)z where x:y:z is a sample ratio is analyzed to determine the exact formula. A 1.7225 g sample of salt was dissolved in 100 mL of pure water. A 50 mL portion of this solution required 10 mL 1 N H2SO4 solution to reach the equivalence point if phenolphthalein was used as indicator. Another 50 mL portion was titrated using methyl orange as indicator and 15 mL acid of same strength was required. Find the value of x+y+z. 2 mole of calcium is burnt in excess air to give to give an equimolar of CaO and Ca3N2. The volume in L of 1 M HCl required to react completely react with the product mixture is__________.

Previous years’ iit questions 1. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N NaOH required to completely neutralize 10 mL of this solution is: (2001) (a) 40 mL (b) 20 mL (c) 10 mL (d) 4 mL 2. In the standardization of Na2S2O3 using K2Cr2O7 by iodometry, the equivalent weight of K2Cr2O7 is: (2001) (a) (molecular weight)/2 (b) (molecular weight)/6 (c) (molecular weight)/3 (d) same as molecular weight 3. Mixture X = 0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5Br]SO4 was prepared in 2 litre of solution. 1 litre of mixture X +excess AgNO3 →Y 1 litre of mixture X +excess BaCl2 →Z

13.54

Stoichiometry

No. of moles of Y and Z are: (2003) (a) 0.01, 0.01 (b) 0.02, 0.01 (c) 0.01, 0.02 (d) 0.02, 0.02 4. Consider a titration of potassium dichromate solution with acidified Mohr’s salt solution using diphenylamine as indicator. The number of moles of Mohr’s salt required per mole of dichromate is: (2007) (a) 3 (b) 4 (c) 5 (d) 6 Passage Bleaching powder and bleach solution are produced on a large scale and used in several household products.

The effectiveness of bleach solution is often measured by iodometry. 5. Bleaching powder contains a salt of an oxoacid as one of its components. The anhydride of that oxoacid is (2012) (a) Cl2O (b) Cl2O7 (c) ClO2 (d) Cl2O6 6. 25 mL of household bleach solution was mixed with 30 mL of 0.50 M KI and 10 mL of 4 N acetic acid In the titration of the liberated iodine, 48 mL 0.25 N Na2S2O3 was used to reach the end point. The molarity of the household bleach solution is (a) 0.48 M (b) 0.96 M (c) 0.24 M (d) 0.024 M

Stoichiometry 13.55

Answer Keys multiple choice questions with only one Answer level i 1. a 2. c 3. b 4. d 5. a 6. a 7. b 8. c 9. b 10. a 11. d 12. b 13. b 14. c

15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25. 26. 27. 28.

a a d c c a b d d b b b b c

29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. 41. 42.

b d d d d b a a d b d b c a

c d a a d a c b b a b a b b

43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54. 55. 56.

57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70.

c a b b c a a d c c d c b c

Passage v 1. d

2. b

Passage vi 1. a

1. 2. 3. 4. 5. 6. 7. 8.

a, b, d a, b, c a, d a, c a, c a, b, d a, c, d b, c

9. 10. 11. 12. 13. 14. 15. 16.

a, b b, c c, d b, d a, b, c a, c, d a, b a, b

17. 18. 19. 20. 21. 22. 23.

a, b, c b, c, d b, c a, c a, b, d a, b, d a, b, d

3. b

2. b

3. a

Passage vii 1. d

Passage viii 1. c

2. c

3. d

2. b

3. c

Passage ix 1. d

multiple choice questions with one or more than one Answer

2. c

Passage x 1. a

2. b

Passage xi 1. b

2. b

Passage xii 1. d

2. d

3. d

comprehensive type questions Passage xiii

Passage i

1. b 1. b

2. a

2. c

3. d

3. b Passage xiv

Passage ii 1. b

1. b 2. c

2. a

3. d

3. b

matching type questions Passage iii 1. c

2. a

Passage iv 1. a

2. b

3. c

1. 2. 3. 4. 5. 6. 7.

(a) (a) (a) (a) (a) (a) (a)

ps ps r s s s p

(b) (b) (b) (b) (b) (b) (b)

s qr p q r r r

(c) (c) (c) (c) (c) (c) (c)

pr pqs s p q p s

(d) (s) (d) (d) (d) (d) (d)

pq r q r p q r

13.56

Stoichiometry

Assertion (A) and reason (r) type questions 1. a

2. c

3. a

4. d

1. a 2. b

integer type questions 1. 2. 3. 4. 5.

5 5 2 2 3

6. 7. 8. 9. 10.

5 3 6 2 3

11. 12. 13. 14. 15.

0 9 5 1 5

16. 17. 18. 19. 20.

1 7 1 5 4

Previous years' iit questions

21. 22. 23. 24. 25.

1 4 5 7 5

3. a 4. d

5. a 6. c

Stoichiometry 13.57

hints And solutions hints to Problems for Practice 1. The formula of calcium oxide is CaO. From the formula moles of Ca = moles of O Weight of Ca Weight of O = At. Wt of Ca At. Wt of O 1.35 1.88 − 1.35 = At. Wt of Ca 16 ∴ At. wt of Ca = 40.75 2. EuCl2 + 2AgNO3 → 2AgCl + Eu(NO3)2 1.0 g 1.29 g Moles of Cl in EuCl2 = moles of Cl in AgCl 2 × moles of EuCl2 = 1× mole of AgCl 2 × 1 1× 1.29 = (x = at. wt of Eu ) x + 71 143.5 ∴ x = 152.48 3. Let the formula of the oxide is M2Ox where x is the valency of the metal M. x × moles of M = 2 × moles of O Atomic weight × Specific heat = 6.4 (Dulong and Pett’s law) 6.4 ∴ Approximate atomic wt = = 193 0.0332 x × 6.84 2 × (7.38 − 6.84) = 193 16 ∴ x = 1.9 So valency = 2 ( Valency is always a whole number) 2 × 6.4 2 × (7.38 − 6.84) Now the accurate at wt = At. wt = 16 = 202.07 4. According to Cannizarro’s principle the smallest weight of an element present in one mole of its various compounds gives either atomic weight of the element or a simple multiple of the atomic weight. (i) Wt of P in 1 mole (222 g) =

27.9 × 222 = 61.94 100

202 × 154 = 31.08 (ii) Wt of P in 1 mole (154 g) = 100 (iii) Wt of P in 1 mole (140 g) =

22.5 × 140 = 31.50 100

(iv) Wt of P in 1 mole (300 g) =

43.7 × 300 = 131.10 100

(v) Wt of P in 1 mole (126 g) =

24.6 × 126 = 30.99 100

∴ The at. wt of P = 31 (the nearest whole number) 5. The formula of the compound is M3N2 (M = metal) 2 × moles of M=3 × moles of N If the wt of the compound is 1g. then weight of M = 0.729 and weight of N = 0.28 g ∴2×

0.72 0.28 = 3× At.wt. of M 14

At. Wt of the metal = 24 6. Potassium selenate is isomorphus with potassium sulphate (K2SO4). Thus the molecular formula is K2SeO4 Let x is the at. wt of selenium. Mol. wt of K2SeO4 = (39 × 2 + x + 16 × 4) = (142 + x) (142 + x)g K2SeO4 has = xg of Se 100 g. K2SeO4 has =

x × 100 (142 + x )

Also x × 100 = 45.42 142 + x x = 118.2 Thus at. wt of Se = 118.2 7. Wt of cadmium = 0.93679 Wt of chlorime = 1.5276–0.9367 = 0.5909 g For CdCl2; eq of Cd = Eq. of Cl 0.9367 0.5909 = E 35.5 ∴ Eq wt = 56.275 ∴ At. wt = Eq. wt × Valency = 56.275 × 2 = 112.55 8. Eq. of metal = Eq of metal sulphate. 2 4.51 ( EW of SO 24 − = 48) = EW EW + 48 EW = 38.24 Now At. Wt =

6.4 6.4 = = 112.28 Sp. heat 0.057

At.Wt 112.28 = = 2.94 ≈ 3 Eq.Wt 38.24 ∴ Exacct at. wt. of metal = Eq.Wt×Valency = 38.24×3=114.72 ∴Valency of the metal =

13.58

9.

Stoichiometry

At. Wt of metal = Volume of atom × density No At. Wt 4 = πr 3 × d 23 3 6.022 × 10 ∴

At Wt 4 22 = × ×(1.432×10−8 )3 × 7.42 = 54.96 6.022 × 1023 3 7

∴ At Wt = 54.96 10. YbBr3 + AgNO3  → AgBr + Yb( NO3 )3 1.3209 g

1.80287 g

Here moles of Br in YbBr3 = moles of Br in AgBr 3 × moles of YbBr3 = 1× moles of AgBr

1.3209 1× 1.8207 3× = ( x + 80 × 3) (108 + 80) Where x = at. wt of Yb ∴ x = 173.26 Thus at. wt of Yb = 173.26 g mol–1 11. Average at. wt =

A1 x1 + A 2 x 2 x1 + x 2

Where A1 and A2 are atomic masses of C-12 and C-13. X1 and X2 are the % of C-12 and C-13 =

(12.0 × 98.892) + (13.00335 × 1.108) = 12.011 amu 100

A1X1 +A 2 X 2 X1 +X 2 78.9183 × 50.54 + A 2 × 49.46 79.904 = 100 A 2 = 80.89 amu

12. Average atomic mass =

Thus, the atomic mass of Br-81 = 80.89 amu 13.  1 mol of a gas S.T.P occupies 22.4 litres, the wt of 1 mol is the mol. wt in grams. ∴ M. wt of chloride = 158.5 Let the formula of the chloride be MClx (x is valency of M) x × moles of M = 1 × moles of Cl Or x ×

wt of M wt of Cl = 1× At. Wt. of M At. Wt. of Cl

Now, mol. wt of MClx = at.wt of M + 35.5x = 158.5 ∴ At. Wt of M = 158.5–35.5x Wt of Cl = 0.4471 g Wt of M = 1–0.4471 = 0.5529 g

0.5529 0.4471 = 158.5 − 35.5x 35.5 x = 1.996 ≈ 2 ∴Valency = 2 x×

14. Isomorphous substances have same molecular formula. So the salt of the element x will have the formula X2SO4 of the type of K2SO4. So the valency of x is equal to K, i.e., At. Wt = eq. Wt × Valency = 13.16 × 1 = 13.16 15. Let the At. Wt of S and Ag are x and y respectively. For Cu2S 1 × moles of Cu = 2 × moles of S or 1 ×

0.7986 0.2014 = 2× 63.5 x

(1)

( 1g of Cu2S contains 0.7986 g and 0.2014 g of Cu and S respectively) For Ag2S 1 × moles of Ag = 2 × moles of S 1×

0.8706 0.1294 = 2× Y x

(2)

(1 g of Ag2S contains 0.8706 g and 0.1294 g of Ag and S respectively) From eqns (1) and (2) we get Y = 107.7 ∴ atomic weight of silver is 107.70 16. Mol. wt of P2Q3 = Wt of 1 mole =

15.9 × 1 = 106 0.15

9.3 × 1 = 62 0.15 Suppose the at.wt of P and Q are respectively x and y Thus 2x + 3y = 106 (P2Q3 = 106) x + 2y = 62 (PQ2 = 62) On solving x = 26 Y = 18 5 = 0.08 mole 17. Moles of x = 60 1.15 × 1023 = 0.19 mole Moles of Y = 6.022 × 1023 Mol. Wt of PQ2 = Wt of 1 mole =

Moles of Z = 0.03 mole Since the formula of the compound is XY2Z3 Moles of X : moles of Y : moles of Z = 1:2:3 = 0.01:0.02:0.03 Comparing the ratio with the moles of X, Y and Z, the moles X and Y are in excess and therefore, moles of

Stoichiometry 13.59

x and y associated with 0.03 mole of z are 0.01 and 0.02 mole respectively. Now Wt of x + wt of y + wt of z = wt of x y2 z3 or 0.01 × 60 + 0.02 × at. wt of y + 0.03 × 30 = 4.4 ∴ At wt of y = 70 amu 18. Let hydrated sulphate be M2(SO4)x · yH2O Where x is the valency of metal 6.4 At. wt of metal = = 26.67 0.24 Also equivalent of metal = Equivalent of SO 2− 4 8.1 43.2 = = 48 ( Eq wt of SO 2− 4 26.67 / x 48 43.2 × 26.67 ∴x = = 2.96 8.1 × 48 x = 2.96 ≈ 3 ∴ Exact At. wt of metal = 9 × 3 = 27 ∴ Hydrated sulphate M2(SO4)3·y H2O Mol wt of hydrated metal sulphate = (2 × 27) + (96 × 3) + 18 y  (342 + 18 y) g M2(SO4)3 · y H2O has 18y g H2O ∴ 100 g. M2(SO4)3 · y H2O has = ∴

18y × 100 g H2O 342 + 18y

18y × 100 = % of H2O = 100–(8.1 +43.2) = 48.7 342 + 18y

y = 18 ∴ The hydrated sulphate is M2(SO4)3 · 18H2O 19. Volume of ozone absorbed by turpentine oil = 10 mL ∴ Volume of O2 = 100–10 = 90 mL WRT 1.5×0.0821×273 Molecular weight of ozone = = PV 1×1 = 33.62 Total volume of ozonised oxygen sample = 1000 mL ∴ Mole of volume ratio of O2 and O3 = 900:100 ∴ Mol. wt of ozonized oxygen =

or

(900 × 32) + 100 × x 1000

= 33.62

x = 48.2 Mol wt of ozone = 48.2

( 900 × 32 ) + 100 × x 1000

6.4 =45.71 0.14 MBrx + x HCl → MCl x + xHBr

20. At.wt of metal =

1.878 g 1.0 g Equivalent of MBrx = Equivalent of MCl x 1.878 1.0 = E+80 E+35.5 E = 15.18 At. wt 45.71 = = 3.01 ≈ 3 Eq. wt 15.18 ∴ Exact atomic weight of metal = 15.18 × 3 = 45.54 Valency of the metal =

Molecular weight MBrx = 45.54 + (80 × 3) = 285.54 21. CuSO 4 ⋅ ( NH 4 )2 SO 4 ⋅ xH 2 O ∆ → CuO mol of CuSO 4 = mol of CuO 19.89 ×100 = 19.89 g CuO 100 19.89 = = 0.25 mol CuO = 0.25 mol of CuSO 4 =39.875 g 79.5 H 2 O = 27.030 g =

Wt.of ( NH 4 )2 SO 4 =100 − ( 39.875 + 27.030 ) g ∴

=33.095 CuSO 4 : (NH 4 ) 2 SO 4

:

H2O

39.875g 33.095g 27.03g 0.25 0.25 1.5 mol ratio or 1 1 6 Thus, the empirical formula is CuSO 4 ⋅ (NH 4 ) 2SO 4 ⋅ 6H 2 O 22. Meq. of Ba salt = M eq of organic Acid W Ba salt ×1000 = V×N M 2 4.290 ×1000 = 21.64 × 0.4771 × 2. M 2 ∴ Mol.wt of salt =415.61 415.61-137-36 =121.31 2 ∴ Molecular weight of organic acid =121.31+1=122.31 Mol.wt of anion=

13.60

Stoichiometry

23. Let M be the molecular weight of mercuric chloride 2.96 No. of mole of the mercuric chloride vapour = M Again from the gas equation PV = nRT  458  × 1  PV 760   = No. of moles n = RT 0.0821× 680

2 3 Mole of B = MA MB

Total moles =

2 3 + MA MB 2 2

MA

MA +

3

×1.5 MB 3

And partial pressure of B =

atm K–1 mol–1 and T = 680 K. 458 ×1 2.96 760 = Hence M 0.0821×680

2

MB

MB +

3

×1.5 MB

As given partial pressure of A = 1atm partial pressure of B = 1.5–1 = 0.5 atm 2

M = 274 Since HgCl2 has a molecular weight (200.6 + 2 × 35.5) = 271.6, which is nearly equal to the calculated value of 274, HgCl2 is the molecular formula for mercuric chloride 24. Suppose the molecular weight of MCl is M NaCl + MCl → AgCl (1−x)g xg 2.567 g Moles of Cl in NaCl + moles of Cl in MCl = moles of Cl in AgCl; 1 × moles of NaCl + 1 × moles of AgCl (1)

Further at 300°C MCl is supposed to undergo sublimation while NaCl does not then 3 MCl AgNO  → AgCl 300° C x

Moles of A =

Partial pressure of A =

458 atm, V = 1 litre, R = 0.0821 lit. Where P = 760

1 − x x 2.567 + = 58.5 M 143.5

26. Total pressure = 1.5 atm

1.341 g

Applying principle of atom conservation for Cl atoms moles of Cl in MCl = 1 × moles of Cl in AgCl 1 × moles of MCl = 1 × moles of AgCl x 1.341 = (2) M 143.5 From equations 1 and 2 M = 53.5 ∴ Mol wt of MCl = 53.5 25. Pressure due to dry air only = (740-18) mm Volume of vapor of 0.168 g of compound at NTP = Volume of air displaced by 0.168 g at NTP (740 − 18) × 49.4 × 273 = 43.72 mL = 293 × 760 43.72 ∴ mole of Vapour = 22400 Wt. in grams 22400 Mol wt. = = 0.168 × No. of moles 43.72 = 866.06

2

MA

MA

3 2

3

+

×1.5 MB

MB

MA +

3

= ×1.5

1 =2 0.5

MB

M 2 M 1 or × B =2 or A = 3 MA MB 3 27. Volume of the vapour at NTP = 84.57 mL 84.57 Mole of vapour = 22400 22400 ∴ Mol wt=0.4524× =119.8 84.57

758 × 127 × 273 = 409 × 760

28. Pressure of vapour in Hofmann's tube = 758–430 = 328 mm 328 × 128.5 × 273 = 50 mL Volume of vapour at NTP = 303 × 760 Mole of vapour =

50 22400

Mol wt = 0.3151 ×

22400 = 141.16 50

29. 2CO+ O2→2CO2 According to principle of atom conservation moles of CO = moles of CO2 (1) And moles of CO + 2x moles of O2 = 2x moles of CO2 (2) From eqns (1) and (2) Moles of CO = 2× moles of O2 Moles of CO =2. or Moles of O 2 or

Volume of CO =2. (Avogadro's law) Volume of O 2

Stoichiometry 13.61

Since the volume of CO is twice that of O2, for 20 mL of CO the volume of O2 will be 10 mL, which will react with 20 mL of CO. Thus the volume of O2 remaining unreacted is 40 mL. Further, as KOH absorbs CO2 produced in the reaction, the only gas left is O2, the volume of which is 40 mL. 30. On passing over charcoal only CO2 is reduced to CO CO + C → no reaction CO2 + C → 2CO b 0 vol. before reaction 0 2b vol. after reaction a + b = 1 and a + 2b = 1.4 0.4 ×100=40% ∴ b = 0.4 litre ∴ % of b = 1 0.6 ∴ a = 0.6 litre ∴ % of a = ×100=60% 1

 Volume of CO = a; ∴ volume of CO2 = a CH4+ 2O2→ CO2 + 2H2O(l) Volume of CH4 = b Volume of CO2 = b N2 + O2 → NO reaction Volume of CO2 formed = Volume absorbed by KOH a + b = 14 mL (2) Now Initial volume of CO + CH4 + N 2 + vol. of O2 taken – Vol. of CO2 formed – volume of N2 – volume of O2 left = 13 (The contraction) ∴ a + b + c + vol O2 taken – vol of O2 left – (a + b) – c = 13 Vol of O2 used = 13 (3)

From eqns (1), (2) and (3) a = 10 mL; b = 4 mL; c = 6 mL 32. 3n  3n m  Cn H 3n O m +  n+ −  O 2 → nCO 2 + H 2 O(l) 4 2 2  16 60 0 0 vol taken 3n m − ] 16n vol left 4 4 ∴ Vol of CO2= 16 n = vol absorbed by KOH 16 n = 44 – 12 = 32. n = 2. Vol of O2 left = 12 [60 − 16 n +

 3n m  16  n + −  = 48 ∴ m = 1 4 2  Formula of compound = C2H6O 33. nO2 50 mL (50–a) mL

31. Let a mL CO, b mL CH4 and c mL N2 be present in mixture Then a+b+c = 20 (1) 1 CO + O 2  → CO 2 2

a a +2 b = 13 (  vol of O 2 used= + 2 b) 2 2

∴ vol of O2 used = 60 – 12= 48

 → 2On

0 2a mL n

vol. of before reaction vol. after reaction (Let a mL of O2 converts to On)

Vol. of O2 left = 50 – a 41 = 50 – a a=9 Also volume of O3 formed = 47 – 41 = 6 mL 2a = 6 or n = 3 n 34. Let the vol of NO and N2O be a and b mL respectively then a + b = 60 (1) 3 NO + N 2 O + H 2 → N 2 + H 2 O (l) 2 a mL b mL excess 0 0 vol. before reaction 0 0 38 mL vol. after reaction 1 mole or 1 vol. NO gives

1 vol N2 2

and 1 mole or 1 Vol. N2O gives 1 vol N2 a + b = 38 2

(2)

From eqns (1) and (2) a = 44 mL and b = 16 mL 35. Assume the formula of ammonia NaHb a b N2 + H2 2 2 40 0 0 vol. before reaction 0 20a 20b vol. after reaction 20a + 20b = 40+40 = 80 (∴ an increase in volume occurs by 40 mL) Or a + b = 4 (1) Now 40 mL = O2 is added 1 H2 +  → H 2 O(l) O2 2 20b 40 vol. before reaction 20b 0 40 − vol. after reaction 2 NaHb →

13.62

Stoichiometry

Gases at the end of the reaction = 30 mL Vol of O2 left + vol of N2 left = 30 mL 40 – 10b + 20a = 30 b–2a = 1 From eqans (1) and (2) a = 1 and b = 3 ∴ Molecular formula of ammonia = NH3

Vol of CO2 formed = 3a +b + c = 3 × 36.5 + (100–36.5) = 173 mL (2)

36. Consider formula of alkane be Cn H2n+2  n+1 O 2  → nCO 2 +(n+1) H 2 O(l) Cn H 2 n + 2 +  n+ 2   Vol. of O 2 used 7 = Vol. of CO 2 formed 4 n+(n+1)/2 7 = n=2 n 4 Alkane is C2H6 37. 2SO2 + O2 → 2SO3 10 15 0 10–2x 15–x 2x Given 2x = 8 or x = 4 Mole of SO2 left = 10–(2 × 4) = 2 Mole of O2 left = 15–4 = 11 38. Tin is converted into sulphide and hydrogen is left, this gas contains H and S say HaSb a a H a Sb + bSn  → bSnS + H 2 CuO  → H2O 2 2 This reaction suggests that Mole of H2: mole of H2O formed::1:1 and mole of HaSb: Mole of H2::1:a/2 100 0.081 2 = × 22400 81 a a=2 Mol wt of HaSb = V.D × 2 = 17 × 2 = 34 (1 × a) + (32 × b) = 34 (1 × 2) + (32 × b) = 34 b=1 ∴ gas is H2S 39. C3H8+5O2→3CO2+4H2O(l) CH4+2O2→ CO2+2H2O(l) CO+1/2O2→ CO2 Let a mL, b mL and c mL be volumes of C3H8, CH4 and CO respectively in 100 mL given sample, then a + b + c = 100 a = 36.5 Now CO2 is formed as a result of combustion of volume  1 Vol C3H8 gives 3 vol. CO2: 1 vol CH4 gives 1 vol CO2 and 1 vol CO gives 1 vol CO2

40. Let the compound A is CxHyN2 (i) C x H y N 2 + O 2 → CO 2 + H 2 O( vapor ) + N 2 (9 − v) vol or (9 − v)mol

v vol 4 vol v mol 4 mol

6 vol 6 mol

2 vol 2 mol

2 × mol of O2 = 2 × moles of CO2 + 1 × moles of H2O 2V = (2 × 4) + (1 × 6) = 14; v = 7 vol (ii) C x H y N z + O 2 → CO 2 + H 2 O( vapor ) + N 2 or

2 vol 2 mo l l

7 vol 7 mol

4 vol 6 vol 4 mol 6 mol

2 vol 2 mol

x × mol of CxHyNz=1 × mol of CO2 x ×2 = 2 × 4; x = 2 y×moles of C x H y N z = 2 × mol of H 2 O(vapour) y×2=2×6;Y = 6 z×mol of C x H y N z = 2 × mol of N 2 z×2 = 2×2; Z = 2 Hence, the compound is C2 H 6 N 2 41. 2C8H18 + 25O2 → 16CO2 + 18H2O 2 mole 25 mole 2 × 114 25 × 32 = 228 g = 800 g For burning 228 g of C8H18, O2 required = 800 g For burning 570 g of C8H18, O2 required = = 2000 g

800 × 570 228

42. The reaction 3Fe + 4H2O → Fe3O4 + 4H2 3×56 4 × 18  4 × 18 g of steam reacts with 3 ×56 g of Fe 3 × 56 × 18 = 42 g ∴ 18 g of steam reacts = 4 × 18 Wt of iron = 42 g 43. From equation (i) 2KClO3 → 2KCl + 3O2 2 × 225 2 × 74.5 3 × 22400 mL = 245.0 = 149.0 = 67,200 mL  67,200 mL O2 produced from 245.0 g KClO3 ∴ 146.8 mL O2 produced = 245 × 146.8 = 0.5352 g KClO3 67.200

Stoichiometry 13.63

Thus wt of KClO3 decomposed according to eq (ii) = (1.0 – 0.5352) = 0.4648 g ∴ 245.0 g KClO3 gives =149.0 g KCl ∴ 0.5352 g KClO3 gives

48.

12 g

149.0 × 0.5352 = 0.3255 g KCl 245.0 From eq (ii) 74.5 × 0.4648 = 0.0704 g KCl Wt of KCl formed = 490.0 Total residue = (0.394) + 0.3255 + (0.0707) = 0.7903 g Thus % of KClO4 in the residue 0.3941× 100 = = 49.87% 0.7903 1 44. Ag 2 CO3 → 2Ag + CO 2 + O 2 2 2 × 108  276 g of Ag 2 CO3 gives 216 of Ag 216 × 2.67 ∴ 2.67 g of Ag 2 CO3 gives = 2.09 g 276 ∆ → CaO+CO 2 45. CaCO3  100 g 56 g 44 g

100 Kg of CaCO3 gives 56 Kg of CaO 200×90 56×200×90 Kg of CaCO3 gives =100.8 Kg 100 100×100 46. Let the weight of polystyrene prepared be 100 g 10.46 ∴ No of mole of Br in 100g of polystyrene = 79.9 = 0.1309 mole From the given formula of polystyrene, we have No of mole of Br = 3 × mole of Br3C6H2 (C8H8) 0.1390 = 3 ×

47.

3 × 100 Wt = mol. mass 314 + 104n

n = 19 SiO 2 +4HF  → SiF4 +2H 2 O 4×20 =80 g

104 g

 80 g of HF gives 104.0 g of SiF4 104.0×63.4 = 82.42 g SiF4 80 82.42 moles of SiF4 = =0.793moleSiF4 104.0

63.4 g of HF gives

32 g

12 g of coke requires 32 g of O2 32×10 =26.66 g O 2 12 Also 2KClO3  → 2KCl+O 2

=

heat

+ O 2  → CO 2

C

10 g of coke requires

2×122.5 =145.0 g

3×32 = 96 g

 96 g of O 2 is produced from 245 g of KClO3 26.66 g of O 2 is produced from 49.

Pb(NO3 ) 2 ∆ → PbO + 331.2 g

245×26.66 = 68.04 g 96

NO 2

223.2 g

 331.2 g Pb(NO3 ) 2 produces 223.2 g of PbO 132.4 g Pb(NO3 ) 2 of 65% purity produces 223.2×132.4×65 =57.99 g 331.2×100 50. C12H22O11 + 12O2 → 12CO2 + 11H2O 342 g 12×32 = 384 g 342 g of sucrose require 384 g O2 gas 34.2 g of sucrose/hr require

384 × 34.2 = 38.4 g 342

Thus O2 required/day =38.4 × 24 = 921.6 g 51. Let formula of iron chloride be FeClx FeCl x

+x AgNO3  → xAgCl+Fe (NO3 )x

(56+35.5x)g 143.5xg Mass of FeClx mol mass of FeClx = Mass of AgClx mol mass of AgCl 134.8 56+35.5xx = 304.8 143.5x x=2 Thus iron chloride is FeCl2. 52. 2ZnS + 3O2→2ZnO+2SO2 2 × 97 3 × 32 = 194 g = 96 g 194 g of ZnS required 96 g of O2 96 × 5.00 × 103 5.00 × 103 g of ZnS requires = 194 = 2.424 Kg of O 2

13.64

Stoichiometry

53. (i) C7H6O3 + C4H6O3 → C9H8O4 + C2H4O2 138 g 102 g 108 g 138 g of salicylic acid requires 102 g of acetic anhydride (Ac2O) 2 g of salicylic acid requires 102 × 2 = 1.48 g of acetic anhydride 138 Since 1.48 g of Ac2O is sufficient for 2 g of salicylic acid we have more than enough AC2O. Hence salicylic acid is the limiting reagent. Now, 138 g of salicylic acid gives

180 × 2 = 2.61 g 138

Thus, theoretical yield of aspirin = 2.61 g (ii) % yield of aspirin =

54. CaO(s)

2.10 × 100 = 80.46% 2.61

+ 3C(s)  → CaC2 (s) + CO(g )

56 g 36 g 64 g 56 g of CaO reacts with 36 g of carbon 36 × 1.15 1.15 Kg of CaO reacts with = 0.74 Kg of carbon 56 Thus, CaO is the limiting reagent here Now, 56 g CaO produces 64 g of CaC2 1.15 Kg Cao produces

64 × 1.15 = 1.31 Kg of CaC 2 56

55. 3TiO2(s)+4C(s)+6Cl2(g)→3TiCl4(g)+2CO2(g)+2CO(g) From the reaction, 4.15 4.15 g TiO 2 = = 0.052 mol of TiO 2 . 80 = 0.052 mol of TiCl4 5.67 g

C=

6.78 g Cl2 =

5.67 0.4725 × 3 = 0.4725 mol of C = 12 4 = 0.3543 mol of TICl4 6.78 0.0955 × 3 = 0.0955 mol of Cl2 = 71 6 = 0.0477 mol of TiCl4

Hence Cl2 is limiting reagent ∴TiCl4 produced = 0.0477 mol = 0.0477×190 = 9.063 g 56. Weight of 50 mL dilute H2SO4 = 50×1.3 = 65.0 g It is of 40% purity. Then wt of dil H2SO4 = 65 × 40 = 26 g 100

Zn+H 2SO 4  → ZnSO 4 +H 2 1 mole 98

1 mole 22.4 lit (NTP)

98 g of H 2SO 4 liberates 22.4 litres(NTP) of H 2 26 g of H 2SO 4 liberates =

22.4×26 =5.94 litre at NTP 98

57. Let the mixture contain ‘a’ g of aluminium Zinc = (1.67-a) g Suppose the mixture is dissolved in H2SO4 2Al + 3H 2SO 4  → Al2 (SO 4 )3 +3H 2 ↑ 2×27 = 54 g

3×22.4 = 67.2 litre (NTP)

At NTP 54 g of Al liberates 67.2 litre of H2 ‘a’g of Al liberates =

67.2 × a litre 54

Also, Zn + H2SO4 → ZnSO4 + H2 65 g 22.4 litre (NTP) 65 g of Zn liberates 22.4 litre of H2 at NTP (1.67−a)g of Zn liberates ? litre of H2 at NTP 22.4 × (1.67 − a ) = litre 65 67.2 22.4 × (1.67 − a ) ∴ ×a + = 1.69 54 65 a = 1.24 Thus wt of Al in the mixture = 1.24 g. 58. Wt of HCl in 2.5 litres of gastric juice = 2.5 × 3 = 7.5 g Al(OH) 3 + 3HCl  → AlCl3 +3H 2 O 1 mole

3 mole

78 g

3 × 36.5 = 109.5 g

109.5 g of HCl completely neutralized by 78 g of Al(OH)3 7.5 g of HCl would be completely neutralized by 78×7.5 = 5.342 g 109

∴ No of tablets of antacid =

5.342 =13.36 ≈14 4

→ CO + H 2SO 4 .H 2 O 59. HCOOH + H 2SO 4  1 mol

1 mol

H 2 C2 O 4 +H 2SO 4  → CO + CO 2 +H 2SO 4 .H 2 O 1 mol

1 mol 1 mol

Stoichiometry 13.65

Let the total vol. of the gas collected be V litre. Volume of CO2 in the mixture = v ( CO2 is absorbed in KOH ) 6 V 5V Volume of total CO = V − = 6 6 V Volume of CO produced from H2C2O4 = 6 V 1 = × mol 6 22.4 5V V 4V − = Vol. of CO produced from HCOOH = 6 6 6 4V 4 = 6×22.4 = V 1 1 × 6 22.4 Thus HCOOH:H2C2O4 is 4:1

63. Mol. mass of (C2H5)4Pb = 12×8+20×1+207 = 323 g Mass of 1.00 cm3 of (C2H5)4Pb = volume × density = 1.00 × 1.66 = 1.66 g 1.66 = = 0.00514 mol. 323 1 mol of (C2H5)4 Pb requires 4 mol of C2H5Cl ∴ 0.00514 mol of (C2H5)4Pb requires = 4×0.00514 = 0.2056 mol = 0.02056×64.5 = 1.33 g C2H5Cl 64. NaCl+AgNO3  → AgCl+NaNO3 KCl+AgNO3  → KCl+NaNO3 moles of AgCl formed =

In this case No.of moles of NaCl+KCl=N No moles of AgCl formed =0.08854 mol.

60. 2C4H10 + 13O2 → 8CO2+10H2O 2×58 13×32 = 116 g = 416 g For burning of 116 g of butane, O2 required = 416 g For burning 1000 g of butane O2 required 416×1000 = = 3586.2 g = 3.59 Kg O 2 116

or

61. 2NaCl+H 2SO 4  → NaSO 4 +2HCl

WNaCl =40.062 g

2×58.5 =117.0 g

WNaCl WKCl =0.0885 mol + 58.5 74.5 WNaCl +WKCl =5.4892 g

Thus % of NaCl=

142×250×103 94.5 × = 287×103 117 100 = 287 Kg of Na 2SO 4 Na 2SO 4 formed is only83.4% pure. 287×100 83.4

x 3.5084 − x + 58.5 103 x = 2.2885 g

0.05095 =

= 344.1Kg

62. Na 2 CO3 + CaCl2 → CaCO3 + 2NaCl

(

Composition of mixture = 2.2885 % of NaCl = × 100 = 65.23% 3.5084

1 mol = 100.09 g

)

n CaCO3 =

1.0362 = 0.10353mol 100.09

From the equation n(NaCO3) = n(CaCO3) = 0.010353 mol Na2CO3 Mass of pure Na2CO3 in sample = 0.010353 × 105.99 = 1.0973 g. Na2CO3 1.0973 ∴ % purity of Na 2 CO3 = × 100 = 91.07% 1.2048

4.062 ×100=73.99% 5.4892

x 58.5 Then mass of NaBr = (3.5084-x) g; mol of NaBr 3.5084 − x = 103 5.5028 = 0.05095 mol of Ag Mol of Ag Formed= 108 = mol of NaCl + Mol of NaBr

=

1 mol = 105.99 g

( 2)

65. Let the mass of NaCl = xg; Mol of NaCl =

94.5% pure 250×103 of NaCl produce Na 2SO 4

∴ wt of 83.4% pure Na 2SO 4 produced =

(1)

from eqns (1) and ( 2)

142 g 142 g

94.5% pure117.0 g of NaCl produce Na 2SO 4 =142 g

12.7052 =0.08854 mol 1435

% of NaBr = 100–65.23 = 34.77% 66. Cl CHCl3 + H2O

CCl3·CHO + 2C6H5Cl Cl 1 mol = 147.5 g

1 mol = 2 × 112.5 = 225 g

1 mol = 354.5 g

13.66

Stoichiometry

147.5 g of chloral required 225 g of chlorobenzene 10 g of chloral required 225 × 10 = 15.25 g chlorobenzene 147.5 Since in this case only 10 g chlorobenzene is available. Hence, chlorobenzene is the limiting reagent. Now, 225 g of C6H5Cl gives = 354.5 g of DDT 354.5 × 10 ∴10 g of C6 H 5 Cl gives = = 15.75 g DDT 225 Thus, wt of DDT produced = 15.75 g 67. Hg

+

201 g 2Hg

I2

 →

254 g +

402 g

I2

HgI 2 455 g

 →

254 g

Hg 2 I 2 656 g

Let the masses of Hg 2 I 2 formed be 'a'g and HgI 2 formed be 'b'g In the formation of Hg 2 I 2 , Hg 402 254 ×a,I 2 required = ×a 656 656 Similarly,in the formation of HgI 2 , Hg required = required =

201 254 ×b; I 2 required = ×b 455 455 402 201 254 24 So ×a+ ×b= a+ b 656 455 656 455 148 53 or a= b 656 453 a 53 656 or = × =0.5163 b 455 148 Thuss ratio of Hg 2 I 2 :HgI 2 formed =0.5163:1 68. Let the vol of NO and N2O be x and y mL respectively, then x + y = 60 mL 3 NO + N 2 O + H 2  → N 2 +H 2 O(l) 2 0 before reaction x mL y mL excess 0 0 0 38 mL after reaction Since 1 mol/1 vol. of NO gives 0.5 vol. of N2 and 1 mol/1 vol. of N2O gives 1 vol. of N2 0.5x + y = 38 From eq. (1) and (2) x = 44 mL and y = 16 mL

69. 2SO 2

+

O 2  → 2SO3

10

15

0

(10-2x )

(15-x )

2x Final moles

Here

Initial moles

2x = 8 ∴x = 4

∴moles of SO 2 left =10 − 2×4 = 2 and moles of O 2 left =15 − 4 = 11 70. Where FeO and Fe3O4 are heated, both change to Fe2O3. Let the weights of FeO and Fe3O4 be x g and y g ∴ Total weight of the reactant = (x+y)g due to the weight increase 5% on heating change of both FeO and Fe3O4 to Fe2O3. 105 The wt of Fe 2 O3 = × ( x+y ) =1.05 ( x+y ) g 100 FeO + Fe3 O 4  → Fe 2 O3 xg

yg

1.05 ( x+y ) g

Moles of Fe in FeO + moles of Fe in Fe3O4 = moles of Fe in Fe2O3 1 × mole of FeO + 3 × moles of Fe3O4 = 2 × moles of Fe2O3 x 3y 2×1.05 ( x+y )  FeO=72, Fe3 O 4 =232,  + =   72 232 160   Fe 2 O3 =160 Dividing by y 1 x 3 2 × 1.05 x 2 × 1.05 × + = × + y 72 y 232 160 160 x  1 2.1  2.1 3 − =  − y  72 160  160 232 x 81 = y 319 81 ∴ % of FeO = × 100 = 20.02% 81 + 319 319 and % of Fe3 O 4 = × 100 = 79.98% 81 + 319 71. Consider that the relative moles of each reactant and product are as Re2O7 + CO → Re2(CO)10 + CO2 a b c d moles moles moles moles Moles of Re in Re2O7= moles of Re in Re2(CO)10 2×moles of Re2O7=2 × moles of Re2(CO)10 2a = 2c a=c (1) Similarly, moles of C atoms in CO = moles of C in Re2(CO)10 + moles of C in CO2

Stoichiometry 13.67

Or b = 10 c + d (2) Also, Moles of O in Re2O7 + moles of O in CO = moles of O in Re2(CO)10 + moles of O in CO2 7 × moles of Re2O7 + 1 × moles of CO = 10 × moles of Re2(CO)10 + 2 × moles of CO2 or 7a + b = 10c + 2d (3) From eqns 1, 2 and 3 17a = b i.e., 17 × moles of Re2O7 = moles of CO 2.5 17 × 484.4 Wt of CO in gm  M.Wt of Re 2 O7 = 484.4 and  =   28  M.Wt of CO = 28  Wt of CO = 2.46 g 72. Let each of the solutions be 100 mL Then for brand A Wt of ClO = 10 g For brand B Wt of Cl = 7 g For brand C Wt of NaOCl = 14 g (i) Moles of Cl in ClO in brand A = 1 × moles of ClO = 1×

10 ( ClO = 51.5) 51.5

Brand B contains more weight of Cl. So it should be bought 73. Let the weight of NaCl is xg I Step I Na 2 O + K 2 O Step  → NaCl + KCl AgNO  → AgCl 3 xg ( 0.118 − x ) g

1 0.0343 × × 62 = 0.018 g 2 58.5 1 0.0837 × 94 = 0.053 g and wt of K2O = × 2 74.5 ∴% of Na2O =

0.018 × 100 = 3.6% 0.5

and % of K2O =

0.053 × 100 = 10.6% 0.50

74. The loss in weight (28% of 5 g), i.e., 1.4 g is due to the formation of the gases NO 2 and O2 which escape out

)

2

 → PbO + NO 2 + O 2

( x − y) g

xg

14 × 35.5 = 6.67 g Wt of Cl in brand C= 74.5

0.2451 g

Moles of Cl in NaCl + moles of Cl in KCl = moles of Cl in AgCl. 1 × moles of NaCl + 1 × moles of KCl = 1 × moles of AgCl x 0.118 − x 0.2451 + = ; x = 0.0343 g 74.5 143.5 58.5 (M. wts of NaCl = 58.5; KCl = 74.5 and AgCl = 143.5) ∴ Wt of NaCl = 0.0343 g Wt of KCl = 0.0837 g

1 Wt of KCl × × M.W of K2O 2 M.W of KCl

∴Wt of Na 2 O =

(

(ii) Wt of Cl in brand B = 7 g (iii) Moles of Cl in NaOCl in brand C = 1 × moles of NaOCl 14 = 1× (M. wt of NaOCl = 74.5) 74.5

in feld spar

And Wt of K2O =

Pb NO3

10 × 35.5 = 6.89 g 51.5

1 Wt of NaCl × × M.W of Na 2 O 2 M.W of NaCl

∴Wt of Na2O =

NaNO3  → NaNO 2

}

Wt of Cl in brand A =

Now calculate the weight of Na2O and K2O for getting Na and K atoms. 2 × moles of Na2O = moles of NaCl 2 × moles of K2O = moles KCl

yg

O2

+

(3.6 − x + y) (1.4 − y)

5−x

(

Moles of Pb NO3

)

2

= moles of PbO

x x−y = M.Wts of Pb NO3 = 331 and PbO = 223 2 331 223 and moles of NaNO3 = moles of NaNO2[NaNO3= 85, NaNO2 = 69] from which we get X = 3.3246 g 5 − x 3.6 − x + y = ( NaNO3 = 85, NaNO2 = 69 ) 85 69

(

(

)

Thus Wt of Pb (NO3)2=3.3246 g Wt of NaNO3=5–3.3246 = 1.6754 g

)

13.68

Stoichiometry

75. By using factor-label method (Dimensional Analysis) (1mol CaC2 ) Wt of C ( 2 H 4 )n = (10000 g CaC2 ) × ( 64 g CaC2 )

×

1 mole C 2 H 2 1 mol C2 H 4 × 1 mole CaC 2 1 mol C2 H 2

×

1 mole (C2 H 4 ) n 28n g ( C2 H 4 )n × 1 mole C2 H 4 1 mole ( C2 H 4 )

n

= 4375 g 76. Wt of Cl2 = 100 g KClO 4 ×

1 mol KClO 4 139 g KClO 4

2

)

1.58 1.26 0.36 = = 1+ y x+y x ∴ x = 31.5 and y = 8 ∴

→ Ag 80. Cu + AgNO3  0.13 g 0.433 g

(1)

Al + CuSO 4  → Cu

(2)

0.13 g

0.47 g

Al + acid  → H2 1.17 g

(3)

0.13 g

4 mol KClO3 3 mol KClO 4 × × 3 mol KClO 4 1 mol KClO3

Eq. of Al = Eq. of H 2

1 mol Cl2 71g Cl2 × × 1 mol KClO 1 mol Cl2

∴

= 204.5 g

0.13 0.47 = ; Eq. Wt of Cu = 32.5 Eq. Wt of Cu 9 Eq. of Cu = Eq. of Ag 0.13 0.433 ∴ = ; Eq. Wt of Ag 32.5 Eq. Wt of Ag

→ 2H + + SO 24 − 77. H 2SO 4  H 3 PO 4  → H + + H 2 PO 4− Basicity of H2SO4 is 2 while that of H3PO4 is 1 Eq. of alkali = Eq of H2SO4 = Eq of H3PO4 Wt of H 3 PO 4 Wt of H 2SO 4 ∴ = Eq.wt. of H 2SO 4 Eq.wt. of H 3 PO 4 ∴

(

Eq. Wt of H2O = 1+y E H O = E H + E O

Wt of H 2SO 4 49 1 = = Wt of H 3 PO 4 98 2

78. Eq. wt of Al = Eq. wt of hydrogen Wt of Al Vol of H 2 atNTP ∴ = E.wt Al Eq.vol of H 2 0.376 0.468 = 9 V ( litres ) ∴ V = 11.2 litres 79. Copper oxide + H2 → Cu + H2O ∴ Eq of copper oxide = Eq of Cu = Eq. of H 2 O Wt. of H 2 O Wt. of Cu Wt of copper oxide = = ∴ Eq. wt of copper oxide Eq. wt of Cu Eq. wt of H 2 O If Eq. wt of Cu and O are x and y respectively and since Eq. wt of H is 1, we have Eq. wt of copper oxide = x+y Eq. wt of copper = x

0.17 0.13 = ; Eq.Wt of Al = 9 Eq.Wt of Al 1 Eq. of Al = Eq of Cu

= 108.25 81. In the case of MnO2 MnSO4 → MnO2 +2 +4 Change in oxidation number per mole of MnSO4 = 2 ∴Eq. Wt of MnSO4 =

Mol. Wt 2

82. Eq. of metal = Eq of oxygen Wt. of metal Wt. of oxygen ∴ = Eq. Wt. of metal Eq. Wt. of oxygen Wt. of oxygen Eq. Wt. of oxygen 1 = = ( given ) Wt.of metal Eq. Wt. of metal 2 1 wt of oxygen 3 + 1 = + 1 = = 1.5 2 wt of metal 2 wt of oxygen + wt of metal = 1.5 wt of metal wt of oxide ∴ = 1.5 wt of metal

Stoichiometry 13.69

83. Vol of H2 produced at S.T.P. 1.2 × 0.92 273 = × = 1.104 litres 273 1 Both Al and Mg combinedly produ uce 1.104 litres

86. 3Mn 2 + + 2MnO 4−  → 5MnO 2 + 2H 2 O Mol wt 3 Molarity × 3 = normality Mol wt and Eq. wt of Mn 2 + = 2 2+ ∴Molarity of Mn × 2 = normality of Mn 2 + Eq. wt of MnO 4− =

of H 2 ∴Eq of Al +Eq. of Mg = Eq of H 2 wt of Al Wt of Mg = Eqq wt. of Al Eq. wt of Mg = Eq.of H 2 =

. of Mn 2 + = Meq of MnO 4− M eq

vol. of H 2 at S.T.P. Eq. vol of H 2

N × 25.0 = 0.05876 × 3 × 34.77 N = 0.24517 0.24517 ∴ Molarity of Mn 2 + = = 0.1225 M 2

Let the weight of Al is xg ∴

x 1 − x 1.104  Eq. wt of Al = 27 / 3 and  + =   27 / 3 24 / 2 11.2  Eq. wt of Mg = 24 / 2 

87. 2MnO 4− + 5SO32 − + 6H +  → 5SO 24 − + 2Mn 2 + + 3H 2 O

x = 0.55 and 1−x = 0.45 Thus the wt of Al = 0.55g and wt of Mg = 0.45 g

Mol wt 5 Molarity × 5 = Normality Mol wt and Eq. wt of SO32 − = 2 Molarity × 2 = Normality Eq. wt of MnO 4− =

84. Cu + HNO3  → Copper nitrate  → Copper oxide 0.324 g

0.406 g

Let the Eq. wt of Cu is E Cu Eq. of Cu = Eq. of copper oxide 0.324 0.406 = Eq. wt of copper oxide E Cu

Now Meq. of SO32 − = Meq of MnO 4− N × 25.0 = 0.01964 × 5 × 34.08 N = 0.1339 0.1339 ∴Molarity = = 0.06685 M 2

0.324 0.406 0.406 = = E Cu Eq. wt of Cu + Eq. of O E Cu + 8

ECu = 31.60 85. M eq. of Na 2CO3 = M eq of HCl

→ Cu 2 I 2 + I 2 88. 2Cu 2 + + 4H − 

N × 50.0 = 0.102 × 56.3 ∴ N = 0.11485

2S2 O32 − + I 2  → S4 O62 − + 2I − 2S2 O32 − ≡ I 2 ≡ 2Cu 2 +

Na 2CO3 + CaCl2  → CaCO3 + 2 NaCl 1 mol = 106 g

or S2 O32 − = I = Cu 2 +

1mol = 100 g

Eq. wt. of S2 O32 − = Mol wt

Now M.E . of CaCO3 Eq of Na 2CO3 = M eq

(

0.11485 × 100 = 11.485 ∴Eq wt of CaCO = 100 / 2 = 50 3 11.485 × 50 = = 0.574 g CaCO3 1000

Eq. wt. of Cu 2 + = Mol wt

)

Eq. of Na 2S2 O3 = Eq. of I 2 = Eq. of Cu 2 + NEV 0.1 × 63.5 × 12.12 = = 0.077 g 1000 1000 0.077 × 100 = 70% % of Cu in 1.1 g sample (ore) = 1000 W(Pure Cu ) =

13.70

Stoichiometry

→ Na 2SO 4 + H 2 O 89. 2 NaOH + H 2SO 4  m mol of NaOH = meq of NaOH 46.40 × 0.875 4M = 46.40 × 0.875 N

92.

Given 10.27 mL of N H 2SO 4 ≡ 10.35 mL of 0.1297 N NaOH

0.875 × 46.40 = N × 5.0

10.35 × 0.1297 = 0.1307 N 10.27 Let unreated H 2SO 4 be V mL

N H SO = 8.12

∴V mL of 0.1307 N H 2SO 4 ≡ 1.91 mL of 0.1297 N NaOH

Meq. of NaOH = Meq. of H 2SO 4

2

4

∴ M H SO = 2

4

8.12 = 4.06 M 2

N H SO = 2

4

VmL used = 41.72 − 1.89 = 39.83 mL 39.83mL of . 0 1307 N H 2SO 4 = eq of K NaCO3 39.83 × 0.1307 × 61 g 1000 = 0.31759 =

The concentration of acid falls below the range for the most effective operation of battery. PV 1.07 × 5 = 0.1873 mol CO 2 = RT 0.0821 × 348 → 6CO 2 + 2Cr 3+ + 7 H 2 O Cr2 O72 − + 3C2 O 24 − + 14H + 

90. n CO2 = 1 mol

6 mol

6 mol of CO 2 formed from 1mole K 2 Cr2 O7 1 × 0.1873 6 = 0.0312 mol

0.1873 mol of CO 2 formed from =

= 0.0312 × 294 = 9.18 g K 2 Cr2 O7 91. Volume of unreacted HCl = Vol. (V mL) of 0.213 M HCl ≡ 44.3 mL of 0.128 M NaOH 44.3 × 0.128 = 26.62 mL 0.213 HCl reacted = 46.3−26.62 = 19.68 mL Eq. of acid utilized for NH3 ≡ Eq. of NH3 produced from fertiliser 19.68 mL of 0.213 M HCl =19.68 mL of 0.213 M nitrogen 19.68 × 0.213 × 14 = 0.0587g 1000 0.0587 × 100 = 9.65% ∴% nitrogen in fertiliser = 0.608 V=

K Na C4 H 4 O6 4H 2 O  → KNaCO3 = 282 g

= 122 g

∴ wt of Rochelle salt in 0.3175 g KNaCO3 282 × 0.3175 = 0.7340g 122 % of pure Rochelle salt in sample =

=

0.7340 × 100 = 76.89% 0.9546

93. IO3− + 3HSO3−  → I − + 3H + + 3SO 24 − ...........(1) 1 mol

3 mol

1 mol

1molof NaIO3 reacts with 3molof NaHSO3 3 × 5.80 = 0.0879 mol 1 × 198 mLof NaIO3 reacted = molof NaIformed = 0.0293 mol

∴5.80g / L NaIO3 reacts with 5I − + 5 mol

IO3− + 6H +  → 3I 2 (s) + 3H 2O....( 2) 1 mol

5 molof NaI reacts with 1molof NaIO3 ∴ 0.0293 of NaI reacts =

1 × 0.0293 = 5.86 × 10−3 mol 5 = 1.16g NaIO3

Given , 5.8 g NaIO3 is present in 1 litreof solution. 1.16 g of NaIO3 is present in 1 litreof solution =

1 × 1.16 = 0.2 litre solution 5.8

Hence 0.2 litre solution NaIO3 is additionally needed in step (2)

Stoichiometry 13.71

97. In acid medium 94. Labelled 109% H2SO4 refers to the total wt of pure H2SO4, 109g that would present after dilution of 100g of the oleum when all free SO3 would combine with water to form H2SO4. Thus 9 g of H2O is added to 100 g of oleum +

H2O

+7 +2 MnO 4− + 8H + + 5e −  → Mn 2 + + 4H 2 O Eq. wt of MnO −4 = M / 5 ∴1 M MnO 4− = 5 N MnO 4−

SO3  → H 2SO 4

1 mol = 18 g

Let normality of MnO −4 in alkaline medium be N1 and neutral medium N2. 20 mL of 5N KMnO −4 ≡ 334 mL of N1 MnO −4 ≡ 100 mL of N2 MnO −4

1 mol = 80 g

18 g of H 2 O combine with 80 g of SO3 80 × 9 = 40 g SO3 18 Thus100g of oleum contains 40g SO3 or 40% free SO3 present in oleum 95. Given 1 mL of Na 2S2 O3 ≡ 0.0499 g of CuSO 4 . 5H 2 O

∴ N1 ( MnO −4 ) = 2.99 ≅ 3

∴9 g of H 2 O combines with

N2 ( MnO −4 ) =1 Hence change in oxidation in alkaline medium = 3 +7 +4 MnO 4− + 2H 2 O + 3e −  → MnO 2 + 4OH −

0.0499 = 2 × 10 −4 mol of CuSO 4 .5H 2 O 249.5 2 Na 2S2 O3 + I 2  → 2 NaI + Na 2S4 O6 =

2 mol

In neutral medium change in oxidation state is 1 +7 +6 MnO 4− + e −  → MnO 42 −

1 mol

mol of Hg 5 (IO6 ) 2 =

0.7245 = 5 × 10 −4 1448.5

From the reaction 8 mol of I2formed from 1mol of Hg5 (IO6)2 ∴5×10–4 mol of Hg5(IO6)2 gives =8×5×10–4 =40×10–4 mol 1 mol of I2 contains with 2mol of Na2S2O3 ∴40 ×10−4 mol contains with 2×40×10–4=80×10–4mol Na2S2O3 80 × 10 −4 = 40 mL ∴ mL of Na2S2O3 needed = 2 × 10 −4 96. 5FeC 2 O 4 + 3MnO −4 + 24H + 5 mol

98. Meq of CaCO3 = Meq of HCl = Meq of CaCl2 formed. = 525 mL × 0.1 N = 52.5 1 CaCl2  → CaSO 4  → CaSO 4 . H 2 O 2 ∴ Meq of CaCl2 = Meq of CaSO4 formed from CaCl2 1 = Meq of CaSO 4 . H 2 O 2 W×2 = × 1000 = 52.5 145 ∴ Wt of plaster of pairs = 3.81 g 99. For phenolphthalein as indicator Meq. of H2SO4 used for 30 mL mixture = 12 × 0.1 = 1.2 ∴ 1 Meq of Na2CO3 = 1.2 (1) 2 For methyl orange as indicator Meq of H2SO4 used for 30 mL mixture = 40 × 0.1 = 4 ∴Meq of Na2CO3+Meq of NaHCO3 = 4 (2) From eq (1) and (2)

3 mol 3+

 → 5Fe + 10CO 2 + 3Mn

2+

+ 12H 2 O

5 mol of Fe2C2O4 required 3 mol of MnO −4 for oxidation 1.44 mL of FeC2O4 required= ∴ 144 3 × 1.44 = 0.006 mol MnO −4 5 × 144 0.01 mol of KMnO4 present in 1000 mL solution 1000 × 0.006 ∴ 0.006 mol KMnO4 present in 0.01 − = 600 mL MnO 4 solution

Meq of Na 2CO3 = 2.4 ∴

WNa CO × 2 2

3

106

× 1000 = 2.4

Meq. of NaHCO3 = 1.6 WNaHCO

3

84

× 1000 = 1.6

or WNa CO = 0.1272 in 30 mL

WNaHCO = 0.1344 g in 30 mL

∴ Strength of Na 2CO3

Strength of NaHCO3

2

3

0.1272 × 1000 30 = 4.24 g / litre =

3

0.11344 × 1000 30 = 4.48 g / litre =

13.72

Stoichiometry

100. For phenolphalein as indicator Meq of HCl used for 200 mL mixture=17.5×0.1=1.75 1 ∴Meq of NaOH + Meq of Na2CO3=1.75 (1) 2 After 1st end point methyl orange is added to the solution of mixture. Now Meq of HCl used for the solution after 1st endpoint on addition of methyl orange as indicator. = 2.5 × 0.1 = 0.25 1 ∴ Meq of Na 2 CO3 = 0.25 2 By Eqns (1) and (2) Meq of Na 2 CO3 = 0.5 2

WNaOH = 0.06g / 200 mL WNa CO 2

∴ Strength of NaOH 0.06 × 1000 200 = 0.30g litre −1 =

NH O = 2

(2)

= 0.50 106 = 0.0265g / 200 mL

2

6 = 0.24 eq / litre 25

Let the volume strength of H2O2 be xV Thus 1 litre of H2O2 produces x litre of O2 at NTP ∴Eq. of H 2 O 2 litre −1 = Eq of O 2 =

WNa CO × 1000 × 2

Meq of NaOH = 1.5

103. Meq of 25 mL of H2O2 solution = Meq of I2 = Meq of Na2S2O3 = 20×0.3=6.0

VNTP x = Eq. Vol of O 2 5.6

x = 0.24 5.6 x = 1.3444

3

3

Strength of Na 2 CO3 0.0265 × 1000 200 = 0.1325 litre −1

=

101. 3Zn + 2K 4  Fe ( CN )6    3 mol 2 mol 2+

 → K 2 Zn 3  Fe ( CN )6  + 6K +  2

Hence volume strength of H2O2 is 1.344 V 104. The reactions are 2KI + H2O + O3 → 2KOH + I2 + O2 1 mol 1 mol I2 + 2Na2S2O3 → Na2S4O6 + 2NaI mol of O3 = mol of I2 =

1 mol of Na2S2O3 2

1 40 1 × × = 0.002 2 1000 10 = 0.002 × 48 = 0.096 g O3 =

The given reaction is not a redox one

= 0.002 × 22.4 = 0.0448 lit O3 at NTP

3 m M of Zn used= mM of K 4  Fe ( CN )6  used 2 Wzn × 1000 3 = × 34.68 × 0.1043 65 2 ∴Wzn = 0.3527g

∴Volume of O 2 = 1 − 0.0448 = 0.9552 litre

0.33527 × 1000 = 22.85% 1.5432 102. In 25 mL of bleaching powder Meq of Cl2 = Meq of iodine liberated =Meq. Of Na2S2O3 solution =12.5×0.04=0.5 ∴Meq of available chlorine in 100 mL = 0.5 × 100 = 2.0 25 ∴ Eq. of available chlorine =2×10–3 ∴ Wt of available chlorine in 1.25 g sample of bleaching powder 2 × 10–3 × 35.5 [∴ Eq Wt of chlorine = 35.5] = 0.071 g 0.071× 1000 = 5.68% ∴ % of available chlorine = 1.25 ∴ % of Zn in ore sample =

0.9552 × 32 = 1.3654 g O 2 22.4 Wt of O3 ∴ Wt of O3in the mixture = Wt of O3 + Wt of O 2 =

0.096 × 100 0.096 + 1.3654 = 6.57%

=

Also number of photons N=

umber of O3 molecules

= Av. No × moles of O3 = 6.022 × 1023 × 0.02 = 1.20 × 1021 105. Ca(HCO3)2+CaO→2CaCO3+H2O 1 mol 1 mol Mol of CaO = Mol of Ca(HCO3)2 in 106 g of solution 1 1 183 = × Mol of HCO3− = × = 1.5 2 2 61

Stoichiometry 13.73

Now Ca2+ ion left in the solution are from CaSO4 only. Mol of 96 Ca 2 + / 106 g = mol of SO 24 − / 106 g = = 1 mol / 106 g 96 = 40g / 106 g or 40 ppm. After removal of Ca(HCO3)2 1 litre of water contains Ca2+ = mol of SO 2− 4 =

1× 103 = 10−3 M 106

If Ca2+ are completely exchanged with H+ [H+] = 2 × 10–3 pH = –log[H+] = 2.6989 106. Fe3 O 4 + 8H + + 2I −  → 3Fe 2 + + I 2 + 4H 2 O Fe 2 O3 + 6H + + 2I −  → 2Fe 2 + + I 2 + 3H 2 O I 2 + 2S2 O − 4

2− 3

2− 6

 → S4 O + 2I

−

+

MnO + 8H + 5Fe 2 +  → Mn 2 + + 5Fe3+ + 4H 2 O From the reaction I2 formed from Fe3O4 and Fe2O3 will be determined by Na2S2O3 and Fe2+ by MnO −4 Fe3O4 is an equimolar mixture of FeO and Fe2O3 Meq of I2 = Meq of Na2S2O3 used = Meq of Fe3+ in Fe2O3 and Fe3O4 11× 0.5 × 100 = 27.5 Meq of Fe2+ in 100 mL = 20 Fe2+ is oxidized into Fe3+ with acidified KMnO4 ∴ M eq of Fe2+ = M eq of KMnO4 used 12.80 × 0.25 × 5 × 100 = 32 50 ∴ Meq of FeO in Fe3 O 4 = 32 − 27.5 = 4.5

=

→ 2H2O(g) 107. 2H2(g)+O2(g)  2a mol a mol 0 initial 2a−2x a−x 2x final

80 Yeild of water is 80% = 2a × = 1.6a 100 (2x) ∴x = 0.8a H2 remain after the reaction = 2a- 1.6a = 0.4a mol O2 remain after the reaction = 0.2a mol And H2O formed after the reaction = 1.6a mol At 120°C total no mol in gaseous phase = 0.4a+0.2 a+1.6 a+2.2 a Given p = 0.8 atm T = 293 K nRT 3a × 0.0821× 293 ∴ V= = 90.21a = P 0.8 The volume of the container remains same After the reaction P = nRT 2.2a × 0.0821× 393 = = 0.7868 V 90.21a 108. Let mol of Mg used for the formation of MgO and Mg3N2 be x and Y respectively. The reactions are 2Mg+O 2  → 2MgO x o

0 x

inintial final

3Mg+N 2  → Mg 3 N 2 Y o

0 y

inintial final

3

Mol of MgOandMg 3 N 2 present in mixture = x+y/3 MgO + 2HCl  → 2MgCl2 +H 2 O

4.5 × 72 = 0.324 g FeO 1000 → Fe3O4 ∴ FeO + Fe2O3  72 160 232 72g of FeO is present in 232g of Fe3O4 232×0.324 0.324g FeO present in = = 1.044g Fe3O4 72 160×0.324 wt of Fe2O3 = = 0.72g Fe2O3 72

x 2x x → 2MgCl2 +2NH 4 Cl Mg 3 N 2 +8HCl 

Total Meq of Fe3+ as Fe2O3 in (Fe2O3 and Fe3O4) = 27.5 Thus Meq. of Fe2O3 + Meq of Fe2O3 (in Fe3O4) = 27.5

3 Meq of HCl = 60 − 12 = 48

=

WFe2 O3 0.72 ×1000+ ×100 = 27.5 160 160 2 2 WFe2 O3 =1.49g 1.044×100 = 34.8% 3 1.49×100 %of Fe 2 O3 = = 49.67% 3 Thus %of Fe3O 4 =

y

8y Y 3 3 Mol of MgCl2 = mol of MgO = x Mol of MgCl2 = mol of Mg 3 N 2 =Y Mol of NH 4 Cl =

2y 3

Mol of HCl used = 2x+ 8y

or 2x+ 8y = 48 3 Also mol of NH4Cl formed = mol of ammonia liberated = mol of HCl consumed for absorbing NH3

13.74

Stoichiometry

2y =4orY=6 3

24.8 × 25 248 × 1 1 N(I 2 ) = 10 N × 25 =

8× 6 = 48 or x = 16 3 6 Hence %Mg burnt to Mg 3 N 2 = ×100=27.27% 6+16 Now from (1) 2 x +

109. Let wt of Na2CO3 = xg and K2CO3 = Yg ∴ x+y = 1.20 g For 100 mL solution mixture Meq of Na2S2O3 + Meq of K2CO3 = Meq of HCl used x×2 y×2 40 × 0.1× 100 × 1000 + × 1000 = 106 138 20 69x + 53y = 73.14 From eqns (1) and (2) x = 0.5962Na2CO3 y = 0.6038 g K2CO3

(1)

(2)

Na 2 CO3 +K 2 CO3 +2BaCl2 → 2BaCO3 +2KCl + 2NaCl M eq of BaCO3 =M eq of Na 2 CO3 + Meq of K 2 CO3 (present in 20 mL solution) = M eq of HCl (for 20 mL solution) = 40 × 0.1 = 4 WBaCO3 ×2 Now ×1000 = 4 197 WBaCO3 = 0.394 g

24.4 × 1 = 2.24 10 2.24 × 250 ∴ M eq of As 2 O3 in 250 mL = = 22.4 25 22.4 × 198 WAs2 O3 = = 1.1088 4 × 1000 1.1088 % of As 2 O3 = × 100 = 9.24% 12 M eq of As 2 O3 in 25 mL =

112. Meq of oxalic acid added = 1.0 × 50 = 50 Meq of oxalic acid left in 25 mL = Meq of KMnO4 = 0.1×32 = 3.2 ∴Meq of oxalic acid used = 50–32 = 18 Now M eq MnO2= M eq of oxalic acid used WMnO2 × 1000 =18 Eq.Wt WMnO2 ×1000×2 86.9 WMnO2 =0.782g

=18

0.782 ×48.875% 1.6 M eq of MnO 2 =M eq of O 2 =18

∴ %of MnO 2 = WO2 ×1000

110. V2O5 + 10H+ + 3Zn → 3Zn2+ + 2V2+ + 5H2O  2O+ I 2 + V 2 + → 2I − + VO 2 + + 2H +  × 2 H V2O5 + 6H + + 3Zn + 2I 2 → 3Zn 2 + + 4I − + 3H 2O + 2VO 2 +

8

=18

18×8 =0.144g 1000 ∴ % of available O 2 in pyrolusite sample WO2 =

2 mol

1 mol = 182 g

=

g 2O5 = 100 V

100 = 0549molV2O5 182

113. Mol wt Na2CO3 10H2O=286

From the reaction 1 mol of V2O5 reduces = 2 mol of I2 0.549 mol of V2O5 reduces = 2 × 0.549 = 0.1098 mol of I2 111. As2O3 + 6NaHCO3 → 2Na3AsO3 + 3H2O + 6CO2 +3

0.144 ×100=9 1.6

+5

Na3AsO3 + I2 + H2O → Na3 AsO4 + 2HI Meq of As2O3 = Meq of I2 = 22.4 × N I2 + Na2S2O3 → Nal + Na2S4O6 Meq of I2 = Meq of hypo

286 = 143 2 100 mL solution of sodium carbonate contains 1 g 1000 mL solution of sodium carbonate contains = 10 g Eq. wt of Na2CO3 10H2O =

10 Normality of the solution = 143 Applying V1N1=V2V2 10 × 42.9 = 0.1 Normality of the acid solution = 143 × 30 Let V mL be volume of H2SO4 Taken (8×5) + (4.8×5) + (34×V) = 0.1×2000 ∴ V = 4 mL

Stoichiometry 13.75

Amount of SO 2-4 = =

Normality×Eq.mass × volume 1000

34×48×4 = 6.528 g 1000

114. The reactions are K 2 CO3 + 2HCl → 2KCl+H 2 O +CO 2 138

2×36.5

69 36.5 Let x g of K 2 CO3 be present in the mixture Wt of Li 2 CO3 =(0.50-x) x 69 (0.50 − x ) No. of gram equivalents of Li 2 CO3 = 37 No.of gram equivalentsin 30 mL of 0.25N HCl ×30 3 Normality×volume 0.25× = = = 400 1000 100 x (0.50-x) 3 At the equivalent point + = 69 37 400 ∴x = 0.4 48 K 2 CO3 = 0.48 g; or 96% No. of gram equivalents of K 2 CO3 =

Li 2 CO3 = 0.02 g; or 4%

115 (i) Mol mass of NaBrO3= 151 Since 6 electrons are involved in the reaction Eq mass of NaBrO3 =

151 6

Amount of NaBrO3 in 85.5 mL of 0.672 N solution 0.672 151 × ×85.5=1.446g 1000 6 Normality 0.672 molarity = = =0.112 2M n 6 =

(ii) Each bromate ion take up 5 electrons ∴ Eq mass of NaBrO3=

mol mass 151 = 5 5

Amount of NaBrO3in 85.5 mL of 0.672 N solution 151 0.672 × ×85.5 = 1.7352 g 5 1000 Normality 0.672 = =0.1344 M Molarity = n 5 =

116. H2O2 + 2KI + H2SO4 → K2SO4 + 2H2O + I2 2Na2S2O3 + I2 → Na2S4O6 + 2NaI 34 = 17 Eq. mass of H2O2 = 2 20 mL 0.1 N Na2S2O3 = 20 mL 0.1 N I2 solution = 20 mL 0.1 N I2 solution Amount of H 2 O 2 in 50 mL eq. solution 0.1×17 × 20 = 0.034 g 1000 0.034 Concentration in g/L = ×1000 = 0.68 50 =

117. Let the wt of H2C2O4 in 10 mL of the solution be x g. The wt of NaHC2O4 in 10 mL will be (0.0202-x) g. In the 1st expt. H2C2O4 and NaHC2O4 are neutralized by NaOH changing to Na2C2O4. The eq. wt of H2C2O4and NaHC2O4 will be 90 and 112 respec2 tively. M eq of H 2 C2 O 4 +M eq of NaHC2 O 4 =M eq of NaOH x (0.0202-x) ×1000+ (1) ×1000=0.1×3 90 112 2 In the IInd expt. both H2C2O4 and NaHC2O4 are oxidized to CO2 by KMnO4 The eq wt of H2C2O4 and NaHC2O4 will therefore be 90/2 and 112/2 respectively. Thus, M eq of H 2 C2 O 4 + M eq of NaOH C2 O 4

=M eq KMnO 4 x (0.0202 − x ) × 1000 + × 1000 × 0.1× 4 90 112 / 2 2 subtacting (1) from (2) 0.1 0.0202 − x = 1000 112 ∴x = 0.009g/10 mL of solution The 1000 mL of solution contains 0.9 g of H 2 C2 O 4

(2)

and NaHC2 O 4 =2.02-0.9=1.12 g 118. At first stage C2 O 2-4 is oxidized to CO2 by KMnO4 according to the following reaction 5C2 O-4 +2MnO-4 +16H + → 10CO 2 +2Mn 2+ +8H 2 O Normality of KMnO 4 =0.02×5=0.1N In the next stage Cu2+ librates I2 from KI and this librated I2 requires 11.3 mL of 0.05 M Na2S2O3 solution complete reaction.

13.76

Stoichiometry

2Cu 2+ +2I- → 2Cu + +I 2 2S2 O +I 2 → S4 O + 2I 23

26

M eq of H 2 O 2 in 20 mL solution = M eq of 20 mL of KMnO 4

-

Normality of Na2S2O3 solution = 0.05 × 1= 0.05 N M eq of Cu 2+ M eq of Na 2 C2 O 4 0.05×11.3 1 = = = M eq of KMnO 4 0.1×22.3 4 M eq of C2 O 2-4 As Cu 2+ → Cu + and C2 O 2-4 → 2CO 2 m.mol of Cu 2+ ×1 1 = m.mol of C2 O 4 ×2 4 or

=M eq of M nO 2 = M eq of Na 2 C 2 O 4 5 4 5 ∴ mM of H 2 O 2 = m M of KMnO 4 × = × =2 2 5 2 ∴ M×20=2 or M H2 O2 =

mole of Cu 2+ 1 = 2 mole of C2 O 2-4

119. Weight of Fe2O3 = 0.552 0.552 80 160 Fe2O3→2 FeO (In this change eq wt of Fe2O3= 2 = 80) Suppose the number of electrons taken up by the oxidant be ‘n’ Normality of the oxidant = 0.0167 nN or 0.0167n ×17 Meq No of eq. Fe2O3=

∴ M eq of 25 mL of Fe 2+ solution = 0.0167 n×17 ∴ M eq of 100 mL of Fe 2+ solution = 4×0.0167n×17 68×0.0167n ∴ Eq of 100 mL of Fe 2+ solution = 1000 Eq of Fe 2 O3 =Eq of FeO 0.552 68×0.0167n = ;∴ n=6 80 1000 120. (i) MnO 2 +Na 2 C2 O 4 +2H 2SO 4 ppt 1 mole 1 mole → MnSO4 +Na 2SO 4 +2CO 2 +2H 2 O (ii) 2KMnO 4 + 3MnSO 4 +2H 2 O

2 = 0.1M 20

→ Mn 3 O 4 121. MnSO 4 ⋅ 4H 2 O heat When Mn3O4 reacts with FeSO4 it is reduced to Mn2+ during which the oxidation state changes from 8/3 to 2. The excess of FeSO4 is determined by using standard KMnO4 solution standardised by another FeSO4 solution. Normality of KMnO 4 = 25 × N = 30×0.1 30×0.1 3 = 25 25 Suppose the value of unreacted FeSO4 = V mL N=

V mL of 0.1 N FeSO4 = 50 mL of Or V =

50 × 3 = 60 mL 0.1× 25

∴ Vol of FeSO4 used for Mn3O4 = (100–60) = 40 mL 40 mL of 0.1 N FeSO4 = 40 mL of 0.1 N Mn3O4= 40 mL of 0.1 N MnSO4.4H2O Eq. mass of MnSO4.4H2O =

Eq. mass of MnSO4.4H2O = ∴ Mass of MnSO 4 .4H 2 O =

2mole

→ 5MnO 2 +K 2SO 4 +2H 2 O ppt 5 mole 2 mole

5 mole

→ K 2SO 4 +2MnSO 4 +8H 2 O+5O 2

E×0.1×40 E = g 1000 250 M

( 83 ) − 2

=

3M 3 × 223 = 2 2

3×223 =1.338 g. 2×250

122. 3S2 O 24 − + 3CrO 24 − +H 2 O+2OH − → 6SO32 − +2Cr(OH)3 3 mol

(iii) 2KMnO 4 + 3H 2SO 4 + 5H 2 O 2

3 N KMnO4 25

2 mol

3mol of S2 O 24 − require =2 mol of CrO 24 − 193.5 2 193.5 mol of S2 O 24 − requires = × 174 3 174 = 0.74 mol CrO 24 − 0.74 mol CrO 24 − arein 50 L waste water 0.74 =0.0148 M 50 N of CrO 24 − = 0.0148×3 = 0.0444 N

∴M of CrO 24 − =

Stoichiometry 13.77 2− 123. (i) B4 O7 +7H 2 O → 4H 3 BO3 +2OH −

N= M×2 ( M= N/no.of replaceable OH ) M eq of borax insolution =100 × 0.2 × 2 = 40 Wborax ×1000×2 =40 M Wboorax ×1000×2 =40 382 Wborax =7.64 g (ii) For neutralisation of HCl M eq of HCl = M eq of borax Wborax ×1000×x 382 ∴Wborax =0.9235g

25×0.1934=

For neutralisation of H 2SO 4 Wborax ×1000×2 =25×0.1934×2 382 Wborax =1.847 g

124. 1st expt MnO 4− +8H + +5Fe 2+ → Mn 2+ +5Fe3+ +4H 2 O M eq of Fe 2+ = M eq of MnO 2-4  N = M × 5 for MnO4−  10 mL ×N =2×0.05    in acidic medium N=0.01 ∴Fe 2+ in mineral = 0.01×56 = 0.56 g/litre in 20g /L solution ∴ % of Fe in mineral =

0.56 × 100 = 2.8 % 20

2nd Expt 2Cu 2+ +4I − → Cu 2 I 2 +I 2 I 2 +2S2 O3 → S4 O62 − +2I − M eq of Cu 2+ = M eq of S2 O32 − 25×N = 5×0.01 0.05 = 0.002 25 ∴ Cu 2+ in mineral = 0.002×635 N=

multiple choice questions with only one Answer level i 1. Eq H2O2 = Eq Na2S2O3 25 × N = 20 × 0.3125 NH2O2= 0.25 N 2NH2O2 equal to 11.2 Vol ∴ 0.25 N H2O2 equal to 1.4 Vol 2. Wt of FeCr2O4 = 25 × 0.89 = 22.40 g mole of FeCr2O4 = 0.1 Eq Ba(MnO4)2= eq FeCr2O4 (0.2)(10)V = (0.1)7 1000 V = 350 mL 3. POCl3 +3H2O → H3PO4 + 3HCl m moles of POCl3 = 5 m moles of H3PO4 formed is = 5 m moles of HCl formed is = 15 m moles of H+ = 30 m moles of OH− required is = 30 30 = 2 × 2 × V V = 7.5 mL 4. moles of V2O5 = 0.1 V+2 + I2 →V+4 + I– Eq V+2 = eq I2 0.2×2 = eq I2 mole of I2 = 0.2 5. Fe2(SO4)3 + Fe → 3FeSO4 m mole of Fe2(SO4)3 is = 10 m moles of FeSO4 Formed is =30 Eq FeSO4 = eq KMnO4 30=0.1×5×V 6. 3O2 → 2O3 100 – I. Vol 100–3x 2x F. Vol 100–× = 80 × =20 7. Eq I2= eq Na2S2O3.5H2O 0.1× 25 = 2.0 × 10–3 1000 Eq I2 in 25 mL solution =2×10–3 As2O3 +I2 → I– + As+5 Eq As2O3 in 25 mL is = 2 × 10–3 Eq As2O3 in 250 mL = 2 × 10–2

∴Eq I2 in 20 mL =

= 0.127g/litre present in 20 g/L solution 0.127 % of Cu = ×1 100 = 0.635% 20

0.02 =

W 198

= 0.99 4 So Purity = 9.9%

13.78

Stoichiometry

8. m eq of H2C2O4 added is = 65 Unreacted m eq of H2C2O4 = 50 × 1 × 5 = 25 10 reacted m eq of H2C2O4 = 65–25 = 40 40 W = 1000 87 2 Wt of MnO2 = 1.74 g 9. Pressure becomes half mean 1 mole compound obtained according to this data ∴ A4 + 2O2 → A3O4 10. m mole of K4[Fe(CN)6] = 2.5 m mole of Zn+2 = 3.75 3.75 W ∴ = 1000 65.5 Wt of Zn = 0.245 g ∴ 50% of zinc in the ore is 10% 11. On heating NH4Cl decomposes. ∴ KCl + AgNO3 → AgCl+KNO3 W 1 1 × = 20 × 74.5 10 1000 Wt of KCl =0.149 g Wt of NH4Cl= 1 g 1 12. NaH2PO4 → Mg (NH4)PO4.6H2O → Mg2P2O7 2 mole of Mg2P2O7=5×10–3 –2 mole of NaH2PO4=10 Wt of NaH2PO4=1.2 g 13. m moles of H2SO4 added = 50×0.05=2.5 Unreacted m moles of H2SO4 =0.5 Reacted m moles of H2SO4 = 2.0 m moles of NH3 liberated = 4 m moles of (NH4)2C2O4 =2.0 2.0 W = 1000 124 Wt of compound = 0.248 % Purity =50% 14. Assume Wt of C2H2 = x Wt of C3H8 = 2 –x C 2H 2 +

2 x 6 − 3x + 26 44 = 3 x 8 − 4x 2 + 26 44 ∴ x = 1.56 g 15. Na2CO3 + H2SO4 →Na2SO4 + H2O + CO2 m moles of Na2CO3 in 25 mL solution = (20) (0.2) = 4 m moles in 250 mL solution is = 40 40 w = 1000 106 WNa2CO3 = 4.24 g % weight Na2CO3 in sample is = 84.8% 16. Cr2O7–2 + 3Sn+2 + 14H+ →3Sn+4 + 2Cr+3 + 7H2O 17. Assume m moles of NaHCO3 = a m moles of Na2CO3 = b When phenolphthalein indicator is used then b = XN When methyl orange indicator is used then a+b = YN Then = a = (y-x)N So vol of HCl is = y-x 18. Assume m moles of FeSO4 = a, FeC2O4 = b 40 a+3b = 15 25 a+b = 15 ∴ a:b = 7:3 19. 4Cl2 + S2O3–2 + 10OH–→2SO4–2 + 8Cl– + 5H2O 20. Fe + H2SO4 → FeSO4 + H2 + FeSO4 + K2Cr2O7 H→ Fe+3+Cr3+ 30 +2 m moles of Fe in 20 mL Solution is =1 30 m moles of Fe+2 in 100 mL Solution = 5 5 w = 1000 56 Wt Fe = 0.28 % Purity of Fe = 99%

5 O → 2CO2 + H2O 2 2

C3H8 + 5O2 → 3CO2 + 4H2O Moles of CO2 formed is

2 x 3(2 − x ) + 26 44

Mole of H2O formed is

x 4(2 − x ) + 44 26

21. I moles of H2O2 =

17 = 0.5 34

Reacted mole of H2O2 = 0.2t Unreacted mole of H2O2 = 0.3 H2O2→ H2O +

1 O2 2

Vol. of O2 obtained at STP = 3.36 lit

Stoichiometry 13.79

22. Assume m moles of Fe SO4 = a, m moles of Fe2(SO4)3 =b a +2b = 30 × a = 20 ×

1 =3 10

1 =2 10

2 = 0.1 M 20 0.5 = 0.025 M Molarity of Fe2 (SO4)3 = 20 Molarity of FeSO4 =

23. 2MnO4– + 3Mn+2 → 5MnO2 + MnO2 + C2O4–2 H→ CO2+ Mn+2 m mole of MnO2 = 10×0.2 = 2 2× 2 = 0.8 5 0.8 Molarity of KMnO = = 0.04 4 20 m mole of MnO–4 =

H2O2+KMnO4+H2SO4 → K2SO4 + MnSO4 + H2O+O2 N1V1 = N2V2 2M1 (20) = (0.04)5(20) M = 0.1 M 24. MgCO3 →MgO+CO2 25. 2x + 3y → z 3 3 – I. mole 1 – – F. mole 5 litre water is added then Z (g) dissolve in it and 1 mole x occupies 10 lit volume 1× 0.0821× 300 = 2.46 atm 10 1 1 26. m moles of C2O4–2 = 90 × × = 2.25 20 2 Wt of C O –2 = 0.198 P=

2

4

% of C2O4–2 = 66% 27. Na2CO3+ 2HCl →NaCl + H2O+CO2 1 1 m moles of Na2CO3 is 25 mL solution=50 × × =2.5 10 2 m moles of Na2CO3 in 250 mL solution = 25 25 w × 1000 106 Wt of Na2CO3 = 2.65 g Wt of NaCl = 2.35 g 28. CH4+2O2 →CO2+2H2O. C2H4+3O2→ 2CO2+2H2O so volume ratio of CH4:C2H4 = 20:10 so = x:y = 2:1

29. 162 n g of starch formed by 6 n moles of CO2 2×10−3 g starch formed by = mole of CO2

2 × 10 −3 × 6n = 0.074×10−3 162n

60 × 0.074 × 10−3 = 6.34 min 7 × 10−4 300 × 0.1 30. No. of moles of Br−= × 10 = 0.3 1000 The required time =

0.1 moles Compound given 0.3 moles of Br− ∴ Compound is ZBr3 31. KIO3+5KI+6HCl → 6KCl +3H2O+3I2 60 = 30 m moles of I2 formed is 120 ×0.5 = 2 30 m moles of KIO3 is 25 mL solution = =10 3 m moles of KIO3 is 250 mL solution = 200 no of molecules of KIO3 = 6.023×1022 +

H 32. Al (MnO4)3 + FeC2O4 → Fe+3+CO2+Mn+2 n factor of Al (MnO4)3=15 n factor of FeC2O4 =3 (0.2)(15)V = (30)(0.5)3. V = 15 mL 33. Assume m moles of H2SO4 = a, m moles of H2C2O4 =b 1 (a+b)2 = 40 × 10 1 2b = 20 × 10 1 = 0.05 M Molarity of H2C2O4 = 20 1 = 0.05 M Molarity of H2SO4 = 20

34. m eq Cr+3 = 70×0.17–3×0.05 = 11.75 m mole of FeO.Cr2O3 =1.958 Wt of FeO.Cr2O3 =0.436 g % wt of FeO.Cr2O3 = 20% 35. m mole of Na2CO3 = a, m mole is NaOH =b a+b=20×0.005=0.1 2a+b=25×0.005=0.125 0.025 Molarity of Na2CO3 = = 0.001 M 25 Molarity of NaOH =

0.075 = 0.003 M 25

13.80

Stoichiometry

36. MnO4– + I– → I2 + Mn+2 10N =50×1 N MnO4– = 5N 100 × 5 =0.5 ∴eq H2C2O4 = 1000 Wt of H2C2O4=22.5 g 37. m moles of FeC2O4 = a, m moles of SnCl2 = a 3a+14a = 225×0.1 a=1.32 m moles ClO3― formed is =(1.32)2=2.64 38. n factor of CuSO4.5H2O= 1 m mole of CuSO4.5H2O=a a=56×0.092= 5.152 m moles of CuSO4.5H2O in 100 mL=51.52 morality of =0.0512 m 39. m mole of NaHC2O4= a m mole of KHC2O4.H2C2O4= b 2a+4b = 0.2 V a+3b = 0.12 V a:b = 3:1 40. m moles of NH3=45×0.4―20×0.1 = 16 So wt of nitrogen = 0.224 g 0.224 % wt of nitrogen = × 100 = 16% 1.4 41. CH3MgBr + C4H9NH2 →CH4+ C4H9NHMgBr moles of CH3MgBr = 5×10–3 Vol of CH4 at STP = 0.112 lit 42. KIO3 +5KI+6HCl →6KCl +3H2O+3I2 I2+2Na2S2O3 → Na2S4O6+2NaI mole of KIO3 = 1×10–3 mole of I2 liberated = 3×10–3 eq I2 = eq Na2S2O3 45M 6×10–3 = 1000 M = 0.133 M 1 43. m mole of KBrO3 added is = 20 × = 0.333 60 5.1 2 × = 0.068 Unreacted m mole of KBrO3 = 25 6 Reacted m moles of KBrO3 =0.265 5 m moles of SeO3–2= 0.265= × 0.0625 2 Wt of SeO3–2 = 0.084 g 44. m eq NaOH =X V1 m eq of Ba(OH)2= 2yV2 x V1 + 2y V2 = 100×0.1=10 V1 1 x = , =4 V2 4 y

45. On heating NH3 gas is liberated m mole of NH3 = 10×2×4 = 80 m mole of Mg (NH4)PO4 = 80 Wt of Mg (NH4) PO4 = 10.96 g 46. m moles of CaO = m moles of CaC2O4 = a 2a = 40×0.25 = 10 Wt of CaO = 0.28 g 47. Molarity of H2SO4 = 1.5 M Wt of H2SO4 = (1.5)98 = 147 g 48. Equivalent of silver =

0.463 = 4.287×10–3 108

Equivalent of dibasic acid = 4.287×10–3 E wt of anion =

0.296 = 69 4.287 × 10−3

E wt of anion acid = 69+1 = 70 49.

50. 51.

52.

53.

54.

0.7 1 1 ×5 = 19:8 × × MW 10 100 2 Mw =142 X=2 20x = 10×0.1+5×0.2×2 X = 0.15 M mole of H+ = 0.01×1=0.01 Mole of Ca+2 = 0.005 Wt of Ca+2 = (0.005)40 = 0.2 g Hardness = 200 ppm n factor N2 is = 6 n factor of H2 is = 2 n factor of NH3 = 3 x x y1 = 1 , y2 = 2 6 3 Mole of Na2CO3 = 2a NaHCO3 = a Na2CO3+2HCl →NaCl + H2O+CO2 Na2CO3+HCl →NaCl+H2O+CO2 Eq of HCl = 4a+a When mixture is strongly heated NaHCO3 converted Na2CO3 2NaHCO3 →Na2CO3 + H2O+CO2 m mole of Na2CO in 10 mL Solution = a m mole of NaHCO3 in 10 mL Solution = b a = 2.5×0.1×2 = 0.5 a+b = 2.5×0.1×2 = 1 m mole present in 100 mL solution =100 a, 100 b 100 × 0.5 WNa 2 CO3 = 1000 106 WNa2CO3 = 5.3 g Wt of NaHCO3 = 4.2 g

Stoichiometry 13.81

55. n factor of KMnO4 in acidic medium = 5 n factor of KMnO4 in alkaline medium = 3 56. n factor of BaCrO4 = 3 137.34 = 45.78 E wt of Ba = 3 −2 57. 60 g of CO3 ion gives 44 g of CO2 So wt of CO3−2 present in sample = 1.33×60/44 = 1.81 g % wt of CO3−2 is sample = 72.4% % wt of metal ion = 100 − 72.4 −13.6 = 13.84% 1 58. moles of Na3 AsO4 = = 4.8×10–3 208 Eq Na3 AsO4 = eq hypo 4.8×10–3×2 = 0.2 V 59. SO2Cl2 + 2H2O → H2SO4 + 2HCl 3moles SO2Cl2 hydrolysis to give 12 moles H+ No of moles of Al(OH)3 required is = 4 1 60. 10 N = 10× 0.56 1 NH2O2 = N 0.56 Vol of H2O2 = 10  X  100 × 0.2 ×2 = 61.   Mw  1000 50 × 0.04 × 2 Y ×1= 1000 Mw 62.

63.

64. 65.

66.

∴ 2x = 5y m moles of NaHCO3 = a, m moles of Na2CO3 = a 3a = XN a = YN ∴ x = 3y m moles of NaHCO3 = a, m moles of Na2CO3 = b b = XN a+b = YN ∴ a = (Y–X) N 5 KMnO4 + 8HCl → KCl + MnCl2 + 4H2O + Cl2 2 Vol of Cl2 gas = (2.5)(22.4) = 56lit Weight of oxygen consumed = 3.92 – 3.18 = 0.74 Weight of Cu converted to CuO = 2.93 g % Cu oxidized = 92.3% % of unreacted Cu = 7.6% Oleum is mixture of H2SO4 + SO3 2 x (0.5 − x )2 26.7 × 0.4 + = 98 80 1000

% Weight of free SO3 = 20.6% 67. (50) (0.1)(2–x) = (25)(0.1)2 X=1

68. H2O2 → H2O +

1 O2 2

No. of moles of oxygen = No. of moles of H2O2 = M=

3 1 = 24 8

1 4

1 1000 × =2.5 M 4 100

69. n factor of FeC2O4 = 3 70. CO2 + 2KOH → K2CO3 + H2O

multiple choice questions with one or more than one Answer 1. Mole of H2C2O4.2H2O = 0.05 Eq of H2C2O4. 2H2O = 0.1 So eq of KMnO4 = 0.1 2. N-factor of FeC2O4 = 3 N-factor of KMnO4 = 5 N-factor of K2Cr2O4 = 6 3. m eq of KMnO4 = (25) (0.2) = 5 4. 109% H2SO4 means SO3+H2O→H2SO4 40 g SO3 react with 9 g of H2O to form H2 SO4 5. m moles of Cu+2 = a, m mole of C2O4–2=b 1 ∴ 2b = 20 × × 5 = 25 4 1 a = 25 × = 2.5 10 6. Eq of K2Cr2O4 = eq H2C2O4 2N = 5×1×2 NK2Cr2O4 = 5 Eq K2Cr2O4 = eq H2O2 40×5 = 20 N NH2O2 = 10 (d) Final Molarity of solution M= 40 × 5 + 10 × 5 8 = 1.5M 50 7. n factor HNO3 =

3 4

8. (a) In C2H5OH, 46 g of alcohol contain 24 g carbon % wt of carbon = 52.17% (b) C6H12 O6, CH3 COOH having some empirical formula 1 9. m eq Ba(MnO4)2 = 100 × × 10 = 100 10

13.82

Stoichiometry

10. B is limiting Value 11. One white P4 molecule contains 6 P – H bonds moles of H2O =

1000 = 55.5 18

In this reaction 9 moles of HNO3 is a spectator m eq of H2O2 = 100 × 0.15 = 15 n – factor of FeC2O4 = 3 On heating MCl converted into gas HIO4 + 2H2O→H5IO6 eq of KMnO4 = (0.1) 5 = 0.5 n factor of H2C2O4 = 2, NaHC2O4 = 1 in acid medium n factor of H2C2O4 = 2, NaHC2O4 = 2 when acid a reducing agent 19. 2K3PO4 + 3BaCl2→Ba3(PO4)2 +6KCl 20. In a chemical reaction all compounds having same equivalents

12. 13. 14. 15. 16. 17. 18.

comprehensive type questions Passage i 1 = 9.43 × 10–3 106 1 = 0.0119 Moles of NaHCO3 = 84 1 Moles of NaOH = = 0.025 40

1. Moles of Na2CO3 =

When phenolphthalein indicator is used 1V 0.025 + 0.00943 = 1000 V = 34.4 mL 2. When methyl orange indicator is used then 1V (0.00943) 2 + 0.0119 + 0.025 = 1000 V = 55 mL 3. After first end point 1V 0.00943 + 0.0119 = 1000 V = 21.3 mL

Passage ii 1. 2Mg + O2 → 2MgO Mg + H2SO4 → MgSO4 + H2 MgO + H2SO4 → MgSO4 + H2O Moles of Hydrogen = 0.1

Weight of unreacted Mg = 2.4 Moles of H2SO4 consumed = 0.1 + 0.05 = 0.15 Weight of H2SO4 consumed = 14.7g Weight of H2SO4 unreacted = 44.1 – 14.7 = 29.4 g 105 g solution contains 29.4 g H2SO4 % weight H2SO4 = 28% 2. 1.2 g of Mg is converted to MgO 3. Moles of O2 required is = 0.025 Vol of O2 of STP = 0.56 L

Passage iii 1. Moles of KMnO4 =

0.316 = 2 × 10–3 158

Eq KMnO4 = eq H2O2 = 0.01 Weight of H2O2 = 0.17 g % purity of H2O2 = 85% 2. 5H2O2 + 2KMnO4 + 3H2SO4 → K2SO4 + 2Mn SO4 + 8H2O + 5O2 Moles of H2O2 = 0.005 Moles of O2 = 0.005 Vol of O2 = 112 mL Passage iv 1. Moles of Na2C2O4 = 0.1 Eq Na2C2O4 = eq KMnO4 (0.1) 2 =

(0.1)5v 1000

V = 400 mL Passage v 1. M moles of HCl in 20 mL solution = 21.4 × Weight of HCl in 20 mL = ∴ Strength = 3.9 g/lit

1 = 2.14 10

2.14 × 36.5 = 0.07811 g 1000

2. M eq of Na2CO3 added = 5 Un reacted m eq of Na2CO3 = 10.5 × 0.08 × 2 = 1.68 M eq of BaCl2 = 5 – 1.68 – 2.14 = 1.18 M moles of BaCl2 in 20 mL = 0.59 Wt of BaCl2 in 20 mL = 0.122 g Wt of BaCl2 in lit = 6.13 g

Stoichiometry 13.83

Passage vi 1. FeS2 → Fe+3 + 2SO2 + SO2 + Ba(MnO4)2 H→ SO4–2 + Mn+2 Ba (MnO4)2 + I– + OH– →I2 + Mn+6 (25)(2M) = 20 × 0.05 MBa(MnO4)2 = 0.02 M M eq of Ba (MnO4)2 added = 40 × 0.02 × 10 = 8 2. Unreacted m eq of Ba(MnO4)2 = 5×0.05×4 = 1 Reacted m eq of Ba (MnO4)2 = 7 M eq of SO2 = 7 M moles of SO2 = 3.5 3. Moles of FeS2 = 1.75×10–3 Wt of FeS2 = 0.21 g % purity of FeS2 = 58.3%

Reacted moles of Na2C2O4 = 0.01 Moles of Pb3O4 = a, PbO2 = b 2a + 2b = (0.01) 2 + H KMnO4 + H2O2 → Mn+2 + H2o + O2 5M (10) = 4.48×0.89 M KMnO4 = 0.08 6a + 2b = 4×0.08×5/1000×25 = 0.0016×25 = 0.04 a = b = 0.005 Wt PbO2 = 0.005×239 = 1.195 g Mass percentage of PbO2 = 23.9% 2. Molarity of KMnO4 = 0.08 Normality of KMnO4 = 0.4 3. Weight of Pb3O4 = (0.005) 685 = 3.425 g Weight of impurity = 5 – 1.195 – 3.425 = 0.38 g Passage x 1. Moles of I2 added = 2.5×10–4

Passage vii +

–

1. MnO4 + C2O4

–2

H→ Mn+2 + CO2

Molarity of KMnO4 =

0.117 = 0.0234M 5

Mn+2 + MnO4– → Mn+3 Eq Mn+2 = 0.0234×4×31.1/1000 = 2.9×10–3 2. Moles of Mn+2 = eq Mn+2 3. Moles of Mn3O4 = 9.666×10–4 Wt of Mn3O4 = (9.66×10–4) 229 = 0.222 g % wt of Mn3O4 = 40.75%

4.6 × 0.01× 4 = 1.84×10–4 1000

Reacted moles of I2 = 0.66×10–4 Weight of I2 = 0.01674 g Weight of commercial salt = 0.0116 Iodine index = 693 2. Moles of vitamin C = 0.66×10–4 Weight of vitamin C is 200 mL = 0.0116 g Vitamin C = 58 mg/mL Passage xi 1. Eq KMnO4 = eq xn+ (1.61×10–3) 5 = (2.68×10–3)(5-n) n=2 Formula = XCl2

Passage viii 1. Moles of Cl2 = 0.1 Moles of HCl = 0.4 Volume of HCl = 200 mL

2. Ew =

2. Moles of MnO2 = 0.1 Wt of MnO2 = 8.7 g

MW 3

MW = 3(56) = 168 Aw of X = 168 – 71 = 97

3. Eq of Cl2 = eq of I2 = eq of Na2S2O3 = 2×10–3 in 50 mL Eq of Cl2 in 500 mL = 0.02 Weight of Cl2 = 0.71 g % weight of available Cl2 = 35.5% Passage ix 1. Moles of Na2C2O4 added is = 0.02 Unreacted moles Na2C2O4 =

Unreacted moles of I2 =

8 × 0.1 25 × = 0.01 1000 2

Passage xii 1. Weight of Cu = X g Weight of Zn = 1.5 – X g Eq Cu = eq NO2 2x = 0.0425 63.5 X = 1.35 g Weight of Zn = 0.15 g So % of zinc in brass = 10%

13.84

Stoichiometry

integer type questions 2. Eq Zn = eq HNO3 0.15 3× 8× v = 65.5 / 2 1000 V = 0.192 mL

1. mole of H2 =

mole of metal M =

1.35 1× 3 × v = 63.5 / 2 1000

(n-2)

V = 14.17 mL Total volume = 14.17 + 0.192 = 14.33 mL 2. Passage xiii 1. 3.4% H2O2 solution equaled to 11.2 vol So 20 Vol equal = 6.07% 6.3 = 0.1 63

3.

Eq H2O2 = 0.1 4v 0.1= 1000 V = 25 mL

4.

3. m eq KMnO4 = (10)(0.1) 5 = 5 m eq H2O2 in 200 mL = 100 m eq in 50 mL solution = 100 Normality of H2O2 = 2 N Vol strength of H2O2 = 22.4 vol Passage xiv 1. Eq Na2C2O4 = eq KMnO4 3.2 1 24.73 × 5 × m × ×2 = 134 20 1000 M = 1.929×10–2 2. Moles of Na2C2O4 in 50 mL =

0.1 = 1.96×10–3 51

∴ so M+H2SO4 → MSO4 +H2 + M+2 + MnO4– H→ Mn2+ + Mn+

3. Eq Cu = eq HNO3

2. Eq oxalic acid =

43.9 × 10−3 = 1.959 × 10−3 22.4

3.2 1 × = 0.00238 134 10

1.929 × 10−2 × 5 × 12.87 Unreacted moles of Na2C2O4 = 1000 × 2 =0.00062 Reacted moles of Na2C2O4 = 0.00176 3. Weight of MnO2 = (0.00176)87 = 0.153 g % purity = 72.36%

5.

0.1 0.1× 58.8 = 51 1000

∴ n-2 = 3 n=5 Mole of I2 = 2.22×10–3 Eq H2O2 = eq I2 5× N = 2.22×10–3×2 1000 N = 0.891 Vol of H2O2 = 5 Eq KMnO4 = eq × n+ 1.61×10–3×5 = 2.68×10–3(5-n) 5–n = 3 m mole of C2O4–2 = a, m moles of Cu+2 = b 2a = 0.02×22.6×5 b = 0.05×11.3 a:b = 2:1 mole of Fe2O3 = 3.45×10–3 3.45 × 10−3 17 × 0.0167(n ) = 4 1000

n=3 100 × 2 = 0.2 6. Eq H2O2 = eq Ba (MnO4)2 = 1000 0.2 Wt of Ba (MnO4)2 = × 375 = 7.5 gm 10 % purity = 5% 7. Eq NaOH = eq HnA 0.5 0.4 = 40 EW EW = 32 MW= 96 96 n= =3 32 y y  8. CxHy  x +  O2 →xCO2 + H2O(l) 4 2   5x = 10 ∴ x = 2 y   x +  5 = 15 ∴y = 4 4  9. (5×10–5)5 = 5×10–5(7–n)

Stoichiometry 13.85

10. mole of Mn+ = 0.016 5.322 × 0.9 = 0.0957 mole of N2H4.H2O added = 50 1 1 Unreacted mole of N2H4.H2O = 480 × × = 0.048 10 100 reacted mole of N2H4.H2O = 0.0477 ∴ Mn+:N2H4.H2O = 1:3

17. Moles of Na2Co3 = 0.12 M moles of KxZny[Fe(CN)6]2 = 0.01 Weight of Kx Zny[Fe(CN)6]2 = 0.01×700 = 7 18. Vol of Cl2 = 30 mL Vol of Cl2Ox = 30 mL Cl2Ox → Cl2 +

11. Assume oxidation number of vanadium in Z is x 5− x 5 = 1 1 ∴x=0 12. Mole of NH3 =

100 × 0.11× 2 48 × 0.25 − =0.01 1000 1000

X + NaOH + NaNO3 →Na2xO2 + NH3 Eq x = eq NH3

Vol of O2 formed = 15 ∴X=1 19. 100×1 = 60×M×5 M KMnO4 = 0.333 In dilute alkaline medium KMnO4 converts into MnO2 0.333×3(v) = 50×0.1 V = 5mL 20. Eq As2O3 = eq KMnO4 (2.5×10–3)4 =

8.36 = (0.01)8 AW / 2 AW = 9 13. m mole of NaOH = a m mole of Na2CO3 = b a+b = 2.5 a+2b = 3 ∴=5 14. eq A+(n+1) = eq KMnO4

mole of Cl– =

21. N factor of Fe (HC2O4)2 = 5 ∴=1 22. Eq FeSn = eq AO4–3 (6.01)(X) = (0.06)3 X = 18  2  + 4  n = 18  n N=4

I2 + 2NaOH → NaI + NaIO3 + H2O m moles of NaOH used = 100×0.3-10×0.6 = 24 m moles of I2 in part II = 12 Total m moles of I2 = 15 M=

0.53 1000 × = 5 × 10−2 100 100 ∴=5

23. N =

50 × 0.2 10 × 0.15 − = 8.5 ×10–3 1000 1000

mole of Na2SeF6 = 1.41×10–3 Wt of Na2SeF6 = (1.41×10–3)239 = 0.336 g %wt = 5% 15 × 0.4 =3 16. M moles of I2 in part I = 2

15 = 10–1m 150

5x 1000

10 y 5×10–3 = 1000 ∴=4

2.68(5 − n − 1) = 32.16×0.05×5/1000 1000 n=1 15. M mole of Cl– = mole of F–

x O2 2

24.

1.7225 1 10 × 1 × ( y + Z) = MW 2 1000 1.7225 1 15 × 1 × (2 y + Z) = MW 2 1000

∴=7 25. Moles of CaO, Ca3N2 = 0.5 CaO+2HCl → CaCl2 + H2O Ca3N2+8HCl → 3CaCl2 + 2NH4Cl

13.86

Stoichiometry

Previous years’ iit questions 1. Normality of oxalic acid is 0.400 N1V1 = N2V2 10×0.4 = 0.1V V = 40 mL 2. Cr2O7–2 → Cr+3 EW=

MW 6

3. Mixture X=0.02 mol of [Co(NH3)5SO4]Br and 0.02 mol of [Co(NH3)5 Br]SO4 in 2 litre. No. of moles of [Co(NH3)5 Br] SO4 per litre = 0.02/2 = 0.01 litre No. of moles of [Co(NH3)5SO4]Br per litre = 0.02/2 = 0.01 litre

(

)

(

)

[Co(NH3 )5SO 4 ]Br + AgNO3  → AgBr ↓ +[Co NH 3 5 SO 4 ]NO3

(Y )

0.01 mole/L

0.01 mole

(

)

[Co NH 3 5 Br ]SO 4 + BaCl2  → BaSO 4 ↓ +[Co NH 3 5 Br ]Cl2 0.01mol / L

( Z)

0.01mole

4. Mohr’s salt FeSO4.(NH4)2SO4. 6H2O has Fe2+ as active reducing part. 6Fe 2 + + Cr2 O72 − + 14H + → 6Fe3+ + 2Cr 3+ + 7 H 2 O Hence, 1 mole dichromate requires 6 moles of Mohr’s salt. 5. Oxy acid is HOCl 6. m eq of I2 = 48×0.25 = 12 m mole of I2 = 6 m mole of Cl2 = 6 6 Molarity of CaOCl2 = = 0.24 M 25

CHAPTER

14 Surface Chemistry

T

he chemists are a strange class of mortals, impelled by almost insane impulse to seek their pleasure among smoke, vapour, soot and flame poisons and poverty; yet among all these evils, I seem to live so sweetly that I may die if I would change place with the Persian king. Johann Joachim Belcher

14 A ADSORPTION 14.1 INTRODucTION Several properties of solids and liquids depend on the nature of their surface. The surface is a boundary that represents the separation of two bulk phases. The surface is also called as interface of the two bulk phases. The chemistry that deals with the phenomenon that occurs at the surface and the surface is represented by separating the bulk phases by a hyphen or a slash. e.g., a solid and a gas interfaces may be represented as solid-gas or solid/gas. Since the gases are completely miscible there is no interface between the gases. Some of the important phenomena that occur at the surface include corrosion electrode processes, heterogeneous catalysis, dissolution, crystallization etc. Thus the study of the surface chemistry is of great importance as it has many applications in industry and as well as in our daily life. The surface of a solid and a liquid is always in a state of strain or tension due to the presence of unbalanced forces on the surface. The numbers of such forces increase with the increase in surface area. To satisfy these unbalanced forces, molecules of the other substances have a tendency to accumulate on the surface. We know that surface is an important property of liquids. According to the property, the molecules on the surface of the liquid experience a strong inward pull and there exists some residual unbalanced forces acting along the surface of the liquid. This property is also noticed in case of solids. The surface of the finely divided solid may have residual forces because inter atomic bonds get cleared when the solid is broken into smaller pieces. In order to satisfy these residual forces surface of the solid attracts the

molecules of the gases or other substances which come into its contact. After considering gaseous reactions in the presence of solids, Faraday in 1833 suggested that the gases collected on the surface of the solid to form a film in which, owing to the high concentration of reactants, the chemical change occurred at a faster rate. It is now known that such films are formed on all surfaces in contact with gases The phenomenon of higher concentration of molecular species (gases or liquids) at the surface of a solid in the bulk is called adsorption The solid substances on the surface of which gets adsorbed due to molecular attractions is called adsorbent while the substances which gets adsorbed on the solid surface due to molecular attractions is called adsorbate. The adsorbent may be solid or a liquid and the adsorbate, may be a gas or a solute in solution. For example activated charcoal (finely divided charcoal) adsorbs a number of gases like ammonia phosgene etc. here charcoal is adsorbent and the gas is adsorbate. Water vapours are adsorbed by silica gel. In this case, silica gel is adsorbent and water vapours are adsorbate: Similarly animal charcoal acts as an adsorbent when it is used to decolourise a number of organic compounds. Desorption is the reverse of adsorption i.e., it is the process which involves the removal of adsorbate from the surface of the adsorbent. The phenomenon of desorption is very common in gases which are absorbed on the surface of finely divided metals. It can be brought about

14.2

Surface Chemistry

by heating. The heat energy absorbed weakens the forces binding the molecules of the adsorbent and adsorbate. When the molecules of a substance are uniformly distributed throughout the body of another substance the phenomenon is called absorption. For example when ammonia gas is passed through water ammonia gets uniformly distributed in water to give a solution and we say that ammonia has been absorbed. Therefore one can conclude that absorption is bulk phenomenon while adsorption is surface phenomenon. In practice it will be expected that in the case of adsorption equilibrium will be attained rapidly whereas in the case of absorption it will be reached more or less slowly. Some more examples of absorption phenomenon are (i) If chalk piece is dipped into a solution of coloured ink and kept for some time the chalk piece absorbs the coloured substance. (ii) A sponge placed in water absorbs water into it. Some times adsorption may be followed by dissolution of the adsorbate in adsorbent i.e., passing from the surface into the bulk of the adsorbent (absorption). Mc Bain has suggested the term sorption to describe a process in which both absorption and adsorption take place simultaneously. In such cases the substance gets uniformly distributed into the bulk of the solid but at the same time its concentration is higher at the surface than in bulk Table 14.1 Difference between adsorption and absorption Adsorption

Absorption

1. Adsorption is a surface 1. In absorption the subphenomenon. The adstance penetrates into sorbing substance is the bulk of the other subcalled adsorbate and is stances only concentrated on the surface of adsorbent 2. The rate of adsorption is 2. The rate of absorption rapid to start with and its is slow and it occur at a rate decreases slowly uniform rate.

14.2 ENThAlPy Of ADSORPTION The amount of heat evolved when one mole of an adsorbate (gas or liquid) is adsorbed on the surface of an adsorbent is called enthalpy of adsorption. During adsorption, there is always a decrease in residual forces of the surface, i.e., there is decrease in surface energy which appears in heat. Adsorption therefore is invariably an exothermic process and therefore ΔH is negative for adsorption and favours the process. On the

other hand, the molecules of the adsorbate (gas) are held on the surface of the adsorbent and therefore, they have lesser tendency to move about freely. In other words, entropy decrease i.e., ΔS is negative and the entropy factor opposes the process. According to Gibbs-Helmholtz equation. ΔG = ΔH − TΔS Thus for the adsorption to occur, ΔG must be negative, which is possible only when ΔH>TΔS in magnitude. This is true in the beginning. However as the adsorption continues ΔH keeps on decreasing and TΔS keeps on increasing and ultimately ΔH becomes equal to TΔS so that ΔG becomes zero. This state is called adsorption equilibrium

Positive and Negative Adsorption It was shown theoretically by Gibbs that those substances which lower the surface tension of a solvent in which they are dissolved become concentrated in the surface layer, whilst the concentration of substances which raise the surface tension is less in the surface layer than in bulk of the solution Thus when many inorganic salts are dissolved in water they increase the surface tension slightly, so that the concentration of the salt in the top layer of the solution will be less than in the bulk of the liquid on the other hand many organic substances, such as carboxylic acids, phenols and amines lower the surface tension and are therefore more concentrated in the top layer than in the bulk of the solution. When the concentration of adsorbate is more on the surface of adsorbent relative to its concentration in the bulk it is called positive adsorption on the other hand if the concentration of the adsorbate is less on the surface relative to its concentration in the bulk is known as negative adsorption

14.3 TyPES Of ADSORPTION There are two types of adsorption. Each type differs in the extent how tightly the adsorbate molecules are held on the surface of the adsorbent (a) Physical adsorption (or Van der Waals adsorption) (b) Chemisorption When the gas molecules (adsorbate) adsorbed on the surface of adsorbent heat is liberated. It is called the heat of adsorption. Higher the value of heat of adsorption stronger will be the forces of attraction between adsorbent molecules and adsorbate (a) Physical adsorption: In this type of adsorption molecules of adsorbate are held by the weak forces of

Surface Chemistry 14.3

attraction on the surface of adsorbent. These weak forces are Van der Waals forces. The heat of adsorption in this type is of the order of 20 to 40 KJ mol–1. If the temperature is raised the kinetic energies of the gas molecules increase and they leave the surface of the adsorbent. Thus rise in temperature lowers the extent of adsorption. Moreover, in physical adsorption the equilibrium is reversible and is established rapidly. Physical adsorption does not depend upon the chemical nature of substance which is adsorbed. Physical adsorption increases with the increase in pressure. (b) Chemisorption or chemical adsorption or Longmuir adsorption: Unlike physical adsorption chemisorption involves the formation of a chemical linkage between the adsorbed molecule and the surface of adsorbent. Thus it is highly selective. It is found that only particular type of molecules are adsorbed by a solid in chemisorptions. In other words, this type of adsorption depends upon the chemical properties of gas and the adsorbent. Moreover chemisorption is accompanied by much higher heat changes (40-400 KJ mol–1). Unlike physical adsorption it is not reversible. In many cases it is found that physical adsorption takes place at low temperature, but as the temperature is raised it changes into chemisorptions. Some times chemisorptions is also called activated adsorption Some important points of differences between physical adsorption and chemical adsorption are given Table 14.2.

14.4 fAcTORS AffEcTINg ThE ADSORPTION Of gAS by SOlID Following are the factors which determine the extent to which a gaseous substance is adsorbed by a solid adsorbent: 1. Nature of the gas (adsorbate): Since physical adsorption is non-specific in nature, every gas will get adsorbed on the surface of any metal to some extent. However, the gases like CO2, SO2, NH3, HCl etc., which get liquefied easily are adsorbed to a greater extent than the gases like H2, O2, N2, etc., which are liquefied with difficulty. This is because the easily liquefiable gases have stronger inter-molecular forces which is an important factor for adsorption to occur. Since chemisorption is specific in nature, it occurs only if the gas can form a chemical bond with the solid. 2. Nature of the solid (adsorbent): It is observed that the same gas is adsorbed to different extents by different solids at the same temperature. Further, the volume of gas adsorbed increases with increase in surface area of the adsorbent. Consequently, the substances like silica gel and charcoal are excellent adsorbents because they have large surface area due to their porous structures. Finely divided substances have larger adsorption power than when they are present in the compact form. This is again due to the larger surface area of the finely divided substances.

Table 14.2 Difference between the physical and the chemical adsorptions Property

Physical adsorption

Chemical adsorption

1

Nature of adsorption

Weak

Strong

2

Enthalpy of adsorption ∆H; KJ mol–1 Reversibility of adsorption

Low 20-40

High 40-400

Reversible and occurs rapidly

Irreversible and occurs slowly

5

Temp. at which adsorption occurs Effect of temp

Low temperature (below the b.pt. of the adsorption gas) Decreases with rise in temperature

High temperature (generally above the b. pt of the adsorption gas) increase with rise in temperature

6

Specificity

Not specific generally. Takes place on all surface

7

Nature of adsorbate layers

Forms multimolecular layers

Highly specific. Take place on specific surface Forms monomolecular layers

8

Effect of pressure

Increases with increase in pressure

Pressure of adsorbate has a negligible effect

9

Activation energy

Activation energies are small

Significant i.e., relatively high

10

Nature of adsorbate

11

Ease of desorption

Depends on the nature of adsorbate only. Easily liquefiable gases adsorb more readily Easily desorbs since Van der Waals forces are weak

Depends on the adsorbate as well as adsorbent. No correlation can be given Not easy since chemical forces are involved

12

Compound formation

No surface compound is formed

Surface compounds are formed

3 4

14.4

Surface Chemistry

3. Pressure: At constant temperature, the adsorption of a gas increases with increase of pressure. However, it has been found that the increase in adsorption is not directly proportional to the applied pressure and is found to be somewhat less. In order to study the effect of pressure on the adsorption of a gas on the surface of a solid, we must remember that adsorption is a reversible process. When the adsorbent and the adsorbate are enclosed in the same vessel, a stage is reached, when the amount of the gas adsorbed becomes equal to the amount of gas desorbed. At this stage, equilibrium is said to be established i.e., the surface of the adsorbent get saturated with the adsorbed gas and no more adsorption occurs. At the equilibrium state, pressure of the gas becomes constant and is called saturation pressure (P0). The extent of adsorption is generally expressed as x/m where x is the mass of the adsorbate and ‘m’ that of adsorbent. The curve obtained by plotting the amount of gas adsorbed x/m against gas pressure is called adsorption isotherm. From the graph we find that at the lower value of pressure the graph is nearly a straight line. At the equilibrium pressure or the saturation pressure (Ps), x/m reaches its maximum value i.e., no more adsorption takes place even if the pressure is further increased—such isotherms are obtained when the adsorbate simply forms a unimolecular layer on the surface of the adsorbent (physical adsorption).

4. Temperature: Physical adsorption is stronger at lower temperature. The reason is that Van der Waals forces are stronger at low temperatures, on the other hand chemisorption is strongly exothermic and thus should be favoured by decrease in temperature. But as its energy of activation is very high, we say that the rate of reaction is so slow at lower temperature that it cannot be observed. Due to this we say that a gas is physically adsorbed at lower temperature and chemisorbed at high temperature. If a plot is drawn between the amount of a gas adsorbed (x/m) and the temperature (T) at constant equilibrium pressure then the curve obtained is called isobar.

x/m

T (a)

x m =K

x/m x m

T

(b)

fig 14.2 (a) Physical adsorption (b) Chemical adsorption

x m = KP P fig 14.1 Adsorption isotherm

Ps

Adsorption isobar for physical adsorption shows that there is regular decrease in adsorption with the rise in temperature. Adsorption isobar for chemisorptions shows first an increase and then a regular decrease.

Surface Chemistry 14.5

This is an equation of the straight line in which log K

14.5 ADSORPTION ISOThERmS The variation of adsorption with pressure at a constant temperature is represented by a curve known as adsorption isotherm. The curve showing the effect of temperature on the extent of adsorption at a given pressure is called an adsorption isobar and the curve showing the variation of pressure with temperature for a given amount of adsorption is called isostere.

1 is the slope n of this straight line. Now if a plot is drawn between x m against log P a straight line is obtained, provided the results at low and high pressure are neglected. Thus, Freundlich isotherm holds only at intermediate pressures.

14.5.1 freundlich Isotherm

limitation

Freundlich gave an empirical relationship between the quantity of gas adsorbed by a given amount of solid adsorbent surface and pressure at a particular temperature. Freundich isotherm can be expressed mathematically by an empirical equation. 1 x = a = KP n m The above relationship is commonly known as Freundlich adsorption isotherm. In this relation x m is the amount of gas adsorbed per gram of the adsorbent at the pressure P and K and n are constants depending upon the nature of the gas and adsorbent. The relation can be derived as follows: At low temperature the graph is linear.

This isotherm fails in case the concentration of the adsorbate

∴

x ∝P m

or

x = KP ′ m

(1)

log x/m

At very high pressure, the adsorption becomes almost independent of pressure i.e., x ∝ P0 (2) m x In an intermediate range of pressure, m is proportional to pressure raised to some fraction power between 1 and 0 i.e., 1 1 x x ∝ P n or = KP n (3) m m

}

Intercept = log K

log P fig 14.3 A graph between log x m and log P

is the intercept of the line on the Y-axis and

1 can have values between 0 and 1 n 1 x (i) When = 0= , constant which shows that the n m

is very high. Constant

adsorption is independent of pressure. (ii) When

1 x x = 1, = KP i.e., ∝ P. the adsorption n m m

varies directly with pressure. Both the above conditions are supported by experimental results.

14.5.2 Adsorption from Solution Solids can adsorb solutes from solution also. It is observed that solid adsorbents adsorb certain solutes from solution in preference to other solutes and solvents. For example, animal charcoal decolourises impure sugar solution by adsorbing colouring dye in preference to sugar molecules. Similarly, when a dilute litmus solution is shaken with charcoal the solution becomes colourless. This is due to adsorption. This type of adsorption is also affected appreciably by increasing the temperature. The increase in concentration of solution also cause an increase in the magnitude of adsorption at a given temperature. Regarding the adsorption from solution, the following conclusions can be made: (i) The magnitude of adsorption decreases with increase in temperature. (ii) The magnitude of adsorption increase with increase in concentration of the solute. (iii) The magnitude of adsorption increases with increase in the surface area. (iv) The magnitude of adsorption depend on the nature of the adsorbent and adsorbate. The effect of concentration on the adsorption magnitude is given by the Freundlich isotherm. 1 x = KC n m

(4)

14.6

Surface Chemistry

Where C is the concentration of the solution when the adsorption is complete. On taking logarithm of equation (4) x 1 log =log K+ log C (5) m n Plotting log x m against log C a straight line is obtained (similar to Fig 14.3) which shows the validity of Freundlich isotherm. This can be experimentally tested by shaking equal weight (= m) of activated charcoal with 50 cm3 of acetic acid solution of different concentrations in different flasks. After shaking for some time, it is allowed to stand for equilibrium to be established. The difference between the initial and final concentrations give the value of x using the equa tion (5), the validity of Freundlich isotherm can be demonstrated.

14.5.3 langmuir Adsorption Isothermlangmuir Adsorption Equation Langmuir (1916) was the first to give a quantitative theory of adsorption. The main points of this theory are: 1. The solid surface consists of a fixed number of adsorption sites where only adsorption of gaseous molecules can take place. 2. Each site can hold only one gaseous molecule. Thus a gas on being absorbed by solid surface cannot form more than one layer i.e., adsorption is unimolecular. 3. Adsorption is localized and takes place only when gas molecules collide with vacant solid surfaces. Magnitude of adsorption of a gas on a solid surface goes on increasing with increase in pressure until the entire surface of the solid gets completely covered by a unimolecular layer of the gas. 4. Thickness of the adsorbed layer is the same as the thickness of the molecule of the adsorbate. 5. The gaseous molecules adsorbed at different sites do not interact with each other. 6. Heat of adsorption is the same for all sites and is independent of pressure and fraction of the adsorbent surface covered. 7. The phenomenon of adsorption involves a dynamic equilibrium and can be represented as: G + S  GS Where G, S and GS represent the unadsorbed gaseous molecules, the vacant site on the surface of adsorbent and the adsorbed gaseous molecules respectively. Thus adsorption consists of two opposing processes, namely condensation of the gas molecules on the solid surface and evaporation of the gas molecules from the surface, back into the

gaseous phase. At equilibrium, the rates of these two opposing processes become identical, i.e., Rate of condensation or adsorption = rate of evaporation or desorption. The concentration of adsorbed molecules GS depends only on one factor namely the number of occupied adsorption sites, which in turn is proportional to fraction θ of the surface covered with gaseous molecules. The rate of desorption or evaporation is proportional to the fraction of the surface that is covered with gaseous molecules. The concentration of vacant sites on the surface of an adsorbent will be directly proportional to the fraction of the surface that remains uncovered, i.e., (1-θ): Therefore, the rate of adsorption or condensation will be proportional to (1-θ) i.e., the fraction of surface uncovered. It also depends upon the pressure of the gas as according to the kinetic theory, the number of molecules striking per unit area is proportional to the gas pressure P. Thus we can write that the rate of adsorption or condensation ∝ (1 − θ ) P = K1 (1 − θ ) P Where K1 is a constant of proportionality. Rate of desorption or evaporation ∝ θ = K 2 θ Where K2 is another constant of proportionality. Since at equilibrium, the rate of condensation must be equal to the rate of evaporation. Hence K1 (1 − θ P) = K 2 θ or K1 P − K1 θP = K 2 θ K1 P = K 2 θ + K1 θP = θ ( K1 P + K 2 ) θ=

K1 P

(K P + K ) 1

2

(6)

Dividing the numerator and denominator both by K2 K1 P K2 K′ P θ= = 1 1+K1′ P K1 P +1 K2 Where

K1 K2

(7)

is another constant K´

Since, according to Langmuir’s theory, only a single layer of molecules can be adsorbed, the fraction θ is proportional to the amount of gas adsorbed per unit mass of the adsorbent, representing this quantity by a, a∝θ

Surface Chemistry 14.7

a=

x = K ′2 θ m

(8)

Where x is the amount of gas adsorbed on m gram of adsorbent and K ′2 is another proportionality constant. K ′2 K1′ θ K ′ K ′ P αP P i.e., 2 1 = (9) ′ 1+K1 1+K1′ 1+βP The above equation (9) is known as Langmuir’s adsorption isotherm. The values of constants α and β depend upon the nature of the solid adsorbent, gas adsorbate and temperature. Their values can be obtained experimentally. Verification: To verify the validity of Langmuir adsorption equation, divide both sides of the equation (8) by P so that we get x m= α P (1+ β ) P a = K ′2 θ=

Now taking the reciprocal, we get P 1 β = + P (10) x α α m P As α and β are constants, therefore, on plotting x m versus P, We should get a straight line with slope equal to β/α and intercept on the y-axis equal to 1/ α. This has been actually found to be so (Fig 14.4) for the adsorption of N2 on mica at 90 K.

Explanation of adsorption isotherm: Equation (9) can be used to correlate the following experimental facts regarding the adsorption of gases. (i) At low pressure, the amount of gas adsorbed is proportional to gas pressure. (ii) At high pressure, the extent of adsorption is independent of pressure. (iii) In the intermediate range of pressure, the extent of adsorption is not directly proportional to pressure but is somewhat less. (i) At low pressure: The factor K1′ P is much smaller than 1 and hence can be ignored. Thus we have. x ≈ K1′ K ′2 P = K ′ P m (11) x or ∝ P. m That is the extent of adsorption and is directly proportional to the pressure of the gas. (ii) At higher pressure: In this case K1′ P will be much greater than 1 and therefore the factor 1 can be neglected in comparison to K1P. Thus we have. x K1′ K ′2 P = = K ′2 = K ′2 P 0 m K′ P

(12)

Hence the extent of adsorption is independent of pressure. (iii) In the intermediate range of pressure: From equation (11) and (12) we have. x At very low pressure ∝ P1 m x At very high pressure ∝ P0 m Hence for intermediate pressure we must have 1 x ∝P n m Where 1 is a value lying between 0 and 1 i.e., n 1 x = KP n m

6

12

18

24

30

36

P fig 14.4 Langmuir adsorption isotherm for adsorption of N2 on mica at 90 K

which is a Freundlich isotherm. Thus Freundlich adsorption isotherm is a special case of Langmuir isotherm. However, Langmuir adsorption isotherm is superior to Freundlich adsorption in the following respects: (i) Results obtained from Langmuir adsorption isotherm are in better agreement with experimental values. (ii) It is applicable over a wide range of pressure. Limitation of Langmuir Theory: The theory is valid only at comparatively low pressure and high temperature. With the increase in pressure or decrease in temperature, additional layers are formed and the adsorption is no longer unimolecular. This has led to modern concept of multi molecular adsorption.

14.8

Surface Chemistry

14.6 APPlIcATIONS Of ADSORPTION Some important applications of adsorption are: 1. In gas masks: Activated charcoal is used in gas masks. The poisonous gas vapours are adsorbed by charcoal and pure air passes through its pores. 2. In dehumidizers: Silica and alumina gels are used as adsorbents, they remove moisture and control the humidity of rooms. 3. Removal of colouring matter: The colouring matter in sugar juice and vegetable oils is removed by substances like activated charcoal. During recrystallization, coloured impurity is generally removed with animal charcoal. 4. Creating high vacuum: Adsorption of gases on solids is employed for creating high vacuum between the walls of the Dewar containers designed for storing liquid air. 5. In chromatography: Chromatography is based on selective adsorption of different substances by an adsorbent. 6. Adsorption Indicators: Certain dyes such as fluorescein are used as indicators in titrations involving precipitation reactions. Such an indicator is adsorbed more by the precipitate than by the solution. 7. In qualitative analysis: In the qualitative analysis, the confirmation of Al3+ ions by Lake Test upon adsorption. Al (OH) 3 adsorbs blue colour from the solution. 8. Dyeing of cloth: Cloth is immersed in a mordant such as alums which adsorb the dye particle. Thus cloth gets dyed. 9. Separation of inert gases: Due to the difference in degree of adsorption of gases by charcoal, a mixture of inert gases can be separated by adsorption on coconut charcoal at different low temperatures.

10. Heterogeneous Catalysts: Adsorption of reactants on the solid surface of the catalysts affects the rate of reaction between the reactants. The reaction proceeds more rapidly after adsorption. There are many gaseous reactions of industrial importance involving solid catalysts. Manufacture of ammonia using iron as catalyst, manufacture of H 2SO4 by contact process and use of finely divided nickel in the hydrogenation of oils are excellent examples of heterogeneous catalysis. 11. In curing diseases: A number of drugs are adsorbed on the germs and kill them or these are adsorbed on the tissues and heat them. 12. Froth floatation process: A low grade sulphide is concentrated by separating it from silica and other earthen impurities. The collectors added to the slurry of the sulphide ore adsorbs on ore particles and prevents them from wetting. So they will be collected in the froth. The impurities are wetted by water and sink down 13. Cleaning agents: Soaps and detergents get adsorbed on the interface and thus reduce the surface tension between dirt and cloth, subsequently the dirt is removed from the cloth. 14. Surfactants: Surfactants work as emulsifier in the manufacture of emulsion. The emulsifiers work on the principle of adsorption. 15. Softening of hard water: In permutit process hard water is passed through sodium alumino silicate (Zeolite). Ca2+ and Mg2+ ions are adsorbed by zeolites which gives out Na+ to water. Similarly, when hard water is passed through ion exchange resins, the different types of ions present in water are removed by the ion exchange resins by exchanging with H+ or OH–ions.

Surface Chemistry 14.9

KEy POINTS Adsorption and Adsorption Isotherms

•

•

•

• •

•

• •

• •

• • •

• •

•

•

• • •

•

The surface area of a solid is referred to as the atoms in the topmost layer of the solid but not in the interior or bulk of the solid. The solid surface area includes the atoms present up to a depth of 100 nm only. The phenomenon of attraction and consequent accumulation or adherence of molecules of a substance on the surface of a liquid or solid is called adsorption. The phenomenon of concentration of molecules of a gas or liquid on a surface of solid or liquid is called adsorption. The substance adsorbed on the surface of a liquid or solid is called adsorbate. The substance on whose surface the adsorption occurs is called adsorbent e.g., in the adsorption of acetic acid on charcoal, adsorbate is acetic and charcoal is adsorbent. For the effective adsorption the surface area of adsorbent (solid or liquid) should be clean and free from impurities. The removal of impurities and the unwanted pre adsorbent materials from the surface of adsorbent is called activation of the adsorbent. Activated charcoal is prepared by heating charcoal at 623 K-1273 K in vacuum or in presence of inert gas. Activation of adsorbent increases the surface area due to which adsorption increase. Adsorption is surface phenomenon while absorption is bulk phenomenon in which a gas or liquid is uniformly distributed in the bulk of the solid. Adsorption is of two types, physical adsorption and chemical adsorption. Physical adsorption mainly depends on the nature of adsorbate but chemical adsorption depends on both adsorbate and adsorbent. Physical adsorption depends on critical temperature of gas while chemical adsorption does not depend on critical temperature. The amount of a gas adsorbed on metal surface or a solid depends on (i) surface area of the adsorbent (ii) nature of gas (iii) pressure of gas (iv) temperature. Greater the surface area, greater is the adsorption. Finely divided metals or porous substances adsorb more substance due to large surface area. The gases which can be liquefied easily will be adsorbed more, thus the gases having high critical temperature will be adsorbed more. The gases having high critical temperatures such as SO2, NH3, HCl, CO2 will be adsorbed more. Gases like H2, O2, N2 etc., which have low critical temperatures are adsorbed less.

• •

•

•

• •

•

•

At low pressure mono layered and at high pressure multi layered adsorption takes place in physical adsorption. At low temperature (at 453K) the adsorption of nitrogen on iron is physical adsorption but at high temperatures (at 773 K) it is chemical adsorption due to the formation of metal nitride. Sometimes with increase in temperature physical adsorption converts into chemical adsorption. The graph showing relation to pressure (p) and to the ratio of masses of adsorbate (x) and adsorbent (m) at a constant temperature is called adsorption isotherm. Freundlich adsorption isotherm is

1 x = KP n which m

indicates that at any given temperature the amount of gas (x) adsorbed by unit mass of the adsorbent (m) is related to the adsorption equilibrium pressure (P). For Freundlich adsorption isotherm, if a graph is drawn showing the relation between log x/m and log P, a straight line with slope equal to 1/n is obtained. The graph showing the relation between x/m and temperature at constant pressure is called as adsorption isobar. With increase in temperature the physical adsorption decreases while the chemical adsorption increase to a maximum and then decreases. Longmuir adsorption isotherm equation is mathematically x = αP where α and β are constants x is m m 1+βP the magnitude of adsorption and P is the equilibrium pressure. Freundlich adsorption isotherm is a special case of Langmuir adsorption isotherm.

Adsorption from Solutions • •

• • •

•

Porous and finely divided substances adsorb dissolved substances from their solutions. Activated charcoal is used to remove coloured impurities and dye stuffs from solutions and removes the colour of the raw cane sugar juice. Adsorption from solution also follows the same principles as in the case of gases by metals. Some adsorbents selectively adsorb some solutes more effectively than the others. Adsorption equilibrium exists between the amount adsorbed (x/m) and concentration of the solute at equilibrium. 1 x P is = K×C n or m 1 x log = log k+ log C m n

14.10

Surface Chemistry

by converting some inactive molecules into active ones. From chemical kinetics, the activation energy can be calculated as follows:

14 b cATAlySIS 14.7 INTRODucTION At the beginning of the last century when Thenard first prepared hydrogen peroxide, he found that it is decomposed more rapidly in the presence of certain metals e.g., platinum which were apparently unchanged. Platinum, in one form or another, was also known to bring about the spontaneous combustion of certain gases. It was known, too, that acids were capable of speeding up certain hydrolytic actions, remaining unchanged at the end of the process. The term catalysis (meaning loosening down ) was introduced by Berzelius in 1835 to describe this phenomenon. A catalyst may be defined as a substance which alter the rate of a chemical change without itself undergoing any permanent chemical change. Catalysis is therefore concerned primarily with chemical kinetics.

Positive and Negative catalyst Usually a catalyst accelerates the rate of a reaction and is therefore known as positive catalyst.

In the absence of catalyst

Potential energy

∆E

In the presence of positive catalyst

P

R Collision number

fig 14.5 Potential energy diagram for the reaction Positive catalyst increases the rate of reaction by lowering activation energy of reaction. Catalyst changes the mechanism by changing the intermediate i.e., intermediate of low energy is formed. It increases the rate

K1 ∆E = e RT K2 Where K1 is rate constant in presence of catalyst and K2 is the rate constant in the absence of a catalyst. log e

K1 ∆E = K 2 RT

 ∆E  K1 = antilog   K2  2.303 RT  In some relatively few known cases, a catalyst may slow down a reaction e.g., 1. Small amount of glycerol, sulphuric acid and acetanilide retard the decomposition of hydrogen peroxide. 2. The oxidation of chloroform is retarded by adding a small quantity of alcohol. 3. Pre-ignition of gases in the cylinder of an internal combustion engine (i.e., knocking) is prevented by the presence of substances added to petrol e.g., lead tetraethyl, Pb(C2H5)4 and nickel carbonyl Ni (CO)4. 4. Oxidation of sodium sulphite by atmospheric oxygen is retarded by adding a small amount of ethyl alcohol. Such substances which decrease the rate of reaction or retard the reaction are called negative catalysts or inhibitors. The negative catalyst may act in one of the following ways. 1. In many cases, it acts by poisoning or merely destroying the catalyst which already happens to be present in the reaction mixture. For example, the decomposition of hydrogen peroxide is catalysed by traces of alkali dissolved from the glass container. But the addition of an acid would destroy the alkali and thus the decomposition of hydrogen peroxide would also be prevented. 2. It dislocates the mechanism by which the reaction would ordinarily proceed. For example, lead tetra ethyl or nickel carbonyl is added to petrol to retard the ignition of petrol vapour in compression in an internal combustion engine thus lead tetraethyl or nickel carbonyl prevents the combination of petrol and oxygen on compression. Negative catalysts decrease the rate by increasing the activation energy of reaction. Mechanism is altered by altering the intermediate: the new intermediate lies at high energy state.

Surface Chemistry 14.11

14.7.2 catalytic Poisons or Anticatalysts In the presence of negative catalyst

Potential energy

∆E

In the absence of catalyst

R

P collision number

fig 14.6 Potential energy diagram for the reaction ΔE = Increased activation energy.

14.7.1 Promoters Many cases are known where the addition to a positive catalyst of a second substance which by itself has no catalytic activity will increase the activity of the catalyst. Such a substance is called promoter or activator. A promoter may thus be regarded as a catalyst for catalyst. Examples are 1. Aluminum oxide in Haber process helps the iron. Other promoters to the catalyst in this particular reaction are molybdenum, the oxides of potassium and uranium (the last named being the first one used by Haber). 2. In the manufacture of hydrogen from water gas in Bosch’s process, finely divided iron acts more efficiently as a catalyst in presence of metallic copper as promoter. 3. Tellurium acts as promoter to nickel (catalyst) in the manufacture of vegetable ghee from vegetable oils. Although the action of a promoter is not clearly understands, the increased catalytic activity may probably be due to the increased adsorption of the reactant at the interfaces formed between the promoter and the catalyst. In some cases, a catalyst has hardly any effect unless a second substance known as co-catalyst is present. For example, the presence of a trace of water a cocatalyst is essential, if boron trifluoride is to act as a catalyst in the polymerisation of many unsaturated organic compounds.

Those substances which reduce or even completely destroy the activity of a catalyst are known as catalytic poisons or anti catalysts. In their presence, the efficiency or activity of the catalyst is greatly reduced and the catalyst becomes sluggish. The poisoning of a catalyst may either be temporary or permanent. When the poison is added, it will be adsorbed on the surface of the catalyst in preference to the reactants, the poisoning is called temporary poisoning. On the other hand, if poison reacts chemically with the catalyst and forms a new catalytically inert surface, the poisoning is called permanent poisoning. In such cases, the catalyst is regenerated in its original form by some suitable means and in this manner the catalyst regains its activity. In temporary poisoning, the surface decrease its activity or loses it entirely, only for the period that the poison, is in contact with it. The activity is restored as the poison is removed from the catalyst. The diminition in activity is probably due to a strong preferential adsorption of the poision on the catalyst surface. With sufficiently strong adsorption the reactant may be displaced completely and the whole surface becomes covered with inactive coating of poison. Retarding effect of CO2 on the hydrogenation of ethylene in presence of copper catalyst, is an example of catalytic temporary poisoning. Permanent poisoning involves a chemical interaction between the surface and the poison to form a new surface which is catalytically inert. Activity may be re-established by chemical rejuvenation only. Volatile silicon and sulphur cause permanent poisoning in case of many catalysts. Arsenic compounds are also fatal especially to platinum, causing permanent poisoning. Examples, 1. In the manufacture of ammonia by Haber process carbon monoxide acts as anticatalyst. 2. In the manufacture of H2SO4 by contact process traces of arsenic oxide poison the platinum acting as catalyst in the oxidation of SO2 to SO3. 3. Decomposition of H2O2 catalysed by platinum is poisoned by HCN. 4. Traces of bromine vapours act as anticatalyst to finely divided nickel in the hydrogenation of oils. 5. Small amount of carbon monoxide renders the copper catalyst inert in the action of hydrogen on ethylene.

14.7.3 Autocatalyst There are certain reactions, where one of the products of the reaction acts as catalyst itself. Such reactions are called autocatalysis reactions and the type of reaction in which one of the products acts as a catalyst is known as autocatalysis.

14.12

Surface Chemistry

14.8 TyPES Of cATAlySIS

Examples 1. In the hydrolysis of an ester, the reaction is catalysed by the acetic acid which is one of the products of the hydrolysis.  CH 3 COOC 2 H 5 + H 2 O ↽ ⇀  CH 3 COOH + C 2 H 5 OH autocatalyst 2. The reaction between copper and nitric acid is slow in the beginning, but the reaction becomes faster gradually due to the formation of nitrous acid as a side product during the reaction which acts as catalyst. 3 Cu + 8HNO3  → 3Cu ( NO3 ) + 4H 2 O + 2 NO 2

2 NO + H 2 O + O  → 2HNO 2

14.8.1 homogeneous catalytic Reactions When the catalyst and reactants are in the same phase and the reacting system is homogeneous the catalyst is said to be homogeneous catalyst and the reaction is homogeneous catalytic reaction. Examples 1. Combination of sulphur dioxide and oxygen in the presence of nitric oxide as catalyst (lead chamber’s process for sulphuric acid).

3. The reaction 2AsH 3  + → 2As autocatalyst

Catalytic reaction may be divided into two kinds: (I) Homogeneous catalysis and (II) Heterogeneous catalysis.

3H 2

is catalysed by arsenic. 4. Oxidation of oxalic acid by acidified potassium permanganate is catalysed by MnSO4 which is one of the products of the reaction. 2KMnO 4 + 5H 2 C2 O 4 + 3H 2SO 4  → 2MnSO 4 + K 2SO 4 + 8H 2 O + 10CO 2 autocatalyst It is readily noticeable that in the titration involving the above reaction the time taken to decolourize the first few drops of permanganate is longer than that for the next volume added when an appreciable concentration of manganese (II) ions is present.

14.7.4 Induced catalysis When one reaction influences the rate of other reaction which does not occur under ordinary conditions, the phenomenon is known as induced catalysis. Example: 1. Sodium arsenite solution is not oxidised by air. If however air is passed through a mixture of the solution of sodium arsenite and sodium sulphite both of them undergo simultaneous oxidation. The oxidation of sodium sulphite thus induces the oxidation of sodium arsenite. 2. The reaction of mercuric chloride (HgCl2) with oxalic acid is very slow but potassium permanganate is reduced readily with oxalic acids. If however oxalic acid is added to a mixture of potassium permanganate and mercuric chloride both are reduced simultaneously. The reduction of potassium permanganate thus induces the reduction of mercuric chloride.

2SO 2 (g)+O 2 (g) + [ NO ] (g) → 2SO3 (g) + [ NO ] Catalyst Catalyst 2. Combination of carbon monoxide and oxygen to form carbon dioxide in presence of nitric oxide as catalyst. 2CO(g)+O 2 (g) + [ NO ] (g) → 2CO 2 + [ NO ] Catalyst Catalyst 3. Combination of hydrogen and chlorine to form hydrogen chloride in presence of water vapours as catalyst. H 2 (g) +Cl2 (g) + [ H 2 O ] (g) → 2HCl ( g ) + [ H 2 O ] ( g ) Catalyst Catalyst 4. Hydrolysis of cane sugar in aqueous solution in the presence of H+ ions from a mineral acid (liquid phase). + H + (fromH 2SO 4 ) → aqueous catalyst

C12 H 22 O11 aqueous

+

H2O liqiud

C6 H12 O 6 aqueous

+

C6 H12 O 6 + aqueous

H + (From H 2SO 4 ) aqueous catalyst

14.8.2 heterogeneous catalytic Reactions When the catalyst and reactants are present in different phases the catalyst is said to be heterogeneous catalyst and the resulting reactions are heterogeneous catalytic reactions. Examples 1. Decomposition of potassium chlorate into oxygen in presence of manganese dioxide. 2KClO3 + [ MnO 2 ] → Solid Solid catalyst

2KCl +O 2 + [ MnO 2 ] solid gas solid catalyst

Surface Chemistry 14.13

2. Decomposition of hydrogen peroxide in aqueous solution in presence of colloidal platinum as catalyst. 2H 2 O 2 + [ Pt ] → 2H 2 O +O 2 + [ Pt ] aqueous solid catalyst liqiud gas solid catalyst 3. Manufacture of ammonia by Haber process. N 2 +3H 2 + [ Fe+Mo ] gas gas solid catalyst

→ 2NH 3 + [ Fe+Mo ] gas solid catalyst

4. Combination of SO2 and O2 in presence of finely divided Pt (Contact process for H2SO4) 2SO 2 +O 2 + Pt  2SO3  gas gas solid gas Catalyst

Pt Solid Catalyst

14.9 chARAcTERISTIcS Of cATAlyST 1. The catalyst remains unchanged in amount and chemical composition at the end of the reaction: This does not mean that it remains so throughout the course of the reaction for in some derivatives of a catalyst have been identified. There may however be a permanent physical change For example, granular manganese dioxide is left as a powder after its catalytic decomposition of potassium chlorate. Smooth platinum becomes roughened and pitted after catalysing the combination of hydrogen and oxygen. 2. A trace of catalyst usually produces a large amount of reaction: In many reactions only a small amount of catalyst is sufficient to bring about a large change in the quantities of reactions, provided the activity of the catalyst also remains unchanged. (i) A concentration of copper (II) ion (Cu2+) as low as 1g in 109 litres will appreciably accelerate the oxidation of sodium sulphate by atmospheric oxygen. (ii) Merely a trace of the Cu2+ ion catalyses the nitrogen forming reaction in the preparation of hydrazine. (iii) As little as 1 gram – atom of colloidal platinum in 108 litres will bring about the decomposition of hydrogen peroxide. (iv) One milligram of fine platinum powder is sufficient to cause the combination of 2.5 litres of a mixture of H2 and O2 to form water. In some cases and within limits, the rate of reaction is directly proportional to the concentration of the catalyst e.g., hydrolysis in presence of hydrogen and hydroxyl ions. Measurement of the rates of some particular hydrolysis reactions e.g., sucrose with equivalent concentration of different

acids provides a means of measuring the hydrogen ion concentrations and hence the relative strength of the acids. In some gaseous reactions catalysed by contact with a solid surface the rate varies distinctly with the effective area of the catalytic surface. This is why higher rates are obtained if the catalyst is in a finely divided condition and deposited on some inactive material. 3. A catalyst generally cannot initiate a reaction: A catalyst generally cannot initiate a reaction; it merely alters the rate of a reaction which is already in progress. Ostwald gave this as an essential characteristic but is difficult to maintain. In theory, a reaction can proceed if the free energy change ΔG is favorable. In practice such a reaction may proceed at an immeasurably slow rate yielding such a small amounts of products that identification by the ordinary methods of analysis is not possible. Thus even if a catalyst increases the rate (say) tenfold, the reaction may still not appear to ‘go’. It will be realized that in terms of energy the function of a catalyst is to alter the activation energy and ΔG is not affected. However, platinum black used as a catalyst in the combination of hydrogen and oxygen has been found to initiate the reaction but the rate is so small, as to be negligible. 4. A catalyst does not affect the position of equilibrium in a reversible reaction: A catalyst affects the rate at which equilibrium is established but not the position of equilibrium in a reversible reaction. It also alters the rate of forward and reverse reactions equally. For example, Sabatier and Sendersen found that nickel will catalyze the addition of hydrogen to certain organic compounds. Thus the double bonds in oils are saturated and the oils there by converted into solid fats by bubbling hydrogen through the oil containing finely divided nickel suspension. But nickel also a dehydrogenating catalyst converting alcohols to aldehydes and ketones – the reverse process. C2 H 5 OH Ni → CH 3 CHO+H 2 Other hydrogenation catalysts cobalt, platinum, iron copper are capable of catalysing dehydrogenations also. There are odd apparent exceptions to the rate that a catalyst does not alter the equilibrium position. In the hydrolysis of an ester, more ester is present at equilibrium if only water is added than if concentrated hydrochloric acid is used. Here however with the concentrated hydrochloric acid, water molecules are removed from the sphere of action. HCl + H 2 O → H 3 O + + Cl −

And thus by the equilibrium law, the removal of the product causes the hydrolysis to proceed further.

14.14

Surface Chemistry

5. Catalytic activity is more or less specific: Catalytic activity is more or less specific i.e., a catalyst for one reaction will not necessarily catalyse another reaction. This is certainly true of enzyme catalysts, for cases are known where if the reactant shows optical isomerism, the catalyst will bring about the reaction of one isomer but not the other. Manganese dioxide catalyses the evolution of oxygen from potassium nitrate on the other hand the weakly electropositive metals iron cobalt nickel platinum palladium will catalyse many varied reactions. 6. Catalyst does not alter the nature of products: It is well known fact that some reactions proceed whether a catalyst is added or not. For example, it is true that potassium chloride and oxygen are products of the decomposition of potassium chlorate whether manganese dioxide is present or not, that sulphur trioxide is formed from sulphur dioxide and oxygen in the absence of platinum or vanadic oxide, that a mixture of nitrogen and hydrogen yield a small amount of ammonia with or without the catalyst. But some reactions have actually been observed in which a catalyst has been found to change the nature of the product. For example, AlCl3 HCOOH   H 2 O +CO Cu HCOOH   H 2 +CO 2

7. Change of temperature alters the rate of catalytic reaction: By increasing the temperature, there is an increase in the catalytic power of a catalyst but after a certain temperature its power begins to decrease. A catalyst has thus a particular temperature at which its catalytic activity is maximum. The temperature is termed as optimum temperature. However, in the case of colloidal solution acting as catalysts, the catalytic activity decreases by the rise of temperature as it may cause coagulation of the colloidal solution.

14.10 ThEORIES Of cATAlySIS When ever a reaction occurs, reactant molecules acquire extra energy to reach activated state before forming the end products. An alternative path is provided for the reaction so that it requires a lower activation energy, the reaction will be faster. How can the path of reaction be changed? It is not possible to give uniform explanation of the mechanism of the phenomenon of catalysis as catalytic reactions are of varied nature. However, a catalyst is considered to bring about the change according to one of the following theories. 1. Intermediate compound formation theory 2. Adsorption theory

14.10.1 Intermediate compound formation Theory

Similarly, CO + 3H 2 Ni → CH 4 + H 2 O + Cr2 O3 CO + 2H 2 ZnO  → CH 3 OH

This theory was advanced by Clement and Desormes in 1806 in case of homogeneous catalysts. According to this theory

u → HCHO CO + H 2 C

It is evident in the above reactions that a catalyst in each case alters the nature of the product, i.e., the nature of the product formed depends upon the nature of catalyst used. In sunlight, chlorine substitutes in the methyl group of toluene molecule. CH3Cl2

CH2Cl Cl2

CH2 Cl2

hv

hv

hv

“the catalyst first forms an intermediate compound with one of or more of the reactants which then either by interaction with other reactants or by decomposition reproduces the catalyst in its original chemical composition and liberates the products of the reaction.”

CCl3

In the presence of a catalyst e.g., iodine, iron (a’ halogen carrier’) substitution occurs in the benzene ring.

Let two reactants A and B combine in the presence of a catalyst x forming a product AB. A+B  → AB

Cl CH3 Cl2

CH3 and

Fe o-chloro toluene

Cl

CH3 p-chloro toluene

Where there are two possible reaction paths and velocity of one is increased so much relative to the other, then other products are obtained.

or

( uncatalysed reaction )

intermediate   A+X  → AX   compound  catalysed reaction A+B+X  → ABX  formation  AX+B  → AB+X   reproduction of catalyst ABX  → AB+X 

Surface Chemistry 14.15

This theory affords good explanation for the following: (i) The catalyst remains unchanged in the reaction. (ii) The rate of reaction varies with the increase in concentration of the catalyst. (iii) Since the catalyst undergoes chemical reaction, specificity in the reaction of a catalyst is expected. Intermediate compounds are unstable and very reactive. A very fast technique such as mass spectrograph has detected intermediates in some cases. The slight change in the overall composition of the recovered catalyst and a possible decrease in the amount is a proof of the formation of intermediate compounds. Many catalytic processes can be explained on the basis of this theory. 1. Friedel-Crafts reaction: Consider the interaction of benzene and ethyl chloride. The reaction is believed to occur as follows. (only relevant electrons are shown). −

Cl ..

..

Cl : Al + Cl C2 H 5 ..

Cl

Cl   ..    → Cl : Al : Cl + C2 H5+ ..     Cl

+

Then C 2 H 5 + C6 H 6  → C6 H 5 C 2 H 5 + H and

+

– Cl   ..   : Cl  + H +  → AlCl3 + HCl Cl : Al ..   Cl  

Here a series of reactions occurs in which a reactive intermediate compound of one reactant and catalyst is first formed. This reacts with the other reactant and finally the catalyst is reformed. The velocity of the whole reaction is determined by that of the slowest intermediate step which is thus referred to as the rate determining step. Support for this intermediate compound theory of catalysis is given by the fact that in some cases, intermediate compounds have been isolated. Further evidence also comes from the tracing of these reactions e.g., Friedel-Crafts with aluminum chloride containing radioactive chlorine. 2. In the absence of a catalyst, nitric oxide and chlorine combine to form nitrosyl chloride. 2 NO + Cl2  → 2 NOCl In the presence of bromine, the reaction goes much faster, proceeding by the following steps: → 2 NOBr 2 NO + Br2  → 2 NOCl + Br2 2 NOBr + Cl2 

3. The hydrolysis of esters under the influence of the hydrogen ion probably proceeds by the formation of intermediate compounds by virtue of the fact that the proton readily adds on to neutral molecules, as it does to form the hydroxonium ion: H + + H 2 O  → H3O+ The hydrolysis is believed to occur by the following steps. +

  O ||   + CH 3 − C − O − R+H 3O  → CH3 − C − O − O − R  + H 2O |   H   O ||

+

+

   O O || ||     → CH 3 − C  + ROH CH 3 − C − O| − R       H     +

   O O || ||     → CH 3 − C − O − H  CH 3 − C  + : O| − H  |     H H    

+

+

  O O || ||    → CH − C − O − H CH − C − ΟΗ + Η +  3  3 |   H  

4. Preparation of ethyl ether from ethanol using concentrated sulphuric acid involves the formation of ethyl bisulphate as intermediate compound. → C2 H 5 HSO 4 + H 2 O C2 H 5 OH +  H 2SO 4   → C2 H 5 OC2 H 5 +  H 2SO 4  C2 H 5 HSO 4 + C2 H 5 OH  The ether formation takes place first by protonation and then by removal of water molecule forming an intermediate compound ethyl bisulphate as shown below. → HSO 4− + H 3 O + H 2SO 4 + H 2 O  →[C2 H 5 − O − H]+ + H 2 O C 2 H 5 − O − H + H 3 O +  |

H +

[C2 H 5 − O − H]+  → C 2 H 5 − O  + H 2 O |

H +

C2 H 5 − O  + HSO 4−  → C2 H 5 HSO 4 C 2 H 5 HSO 4 + C2 H 5 OH  → C2 H 5 O C2 H 5 + H 2SO 4

5. Iodination of propanone (acetone) takes place in the presence of acid. In the first step, protonation of acetone takes place forming an intermediate. O ||

+

OH ||

CH 3 − C − CH 3 + H 3 O  → CH 3 − C − CH 3 + H 2 O +

14.16

Surface Chemistry

Now the protonated intermediate donate the proton from α carbon. +

OH

OH

|

|

CH 3 − C − CH 3 + H 2 O  → CH 2 = C − CH 3 + H 3 O +

6. In the manufacture of sulphuric acid by Lead Chamber’s process, nitroso sulphuric acid is formed as intermediate compound. N 2 O3 + 2SO 2 + H 2 O + O 2  → 2 NO HSO 4

}

2 NO HSO 4 + H 2 O  → H 2SO 4 + NO + NO 2

It can be seen that the hydrogen has been removed from a different position in the molecule so that the later is now in what is called the enol form and can readily react with iodine. OH

7. Thermal decomposition of CH3CHO in presence of iodine vapour

O

|

||

CH 2 = C − CH3 + I 2  → ICH 2 − C − CH3 + HI Two points should be especially noted in these examples. 1. First, the reaction has been made easier. (and therefore quicker) as a result of the addition of a proton to the reactant molecule and its subsequent removal. 2. It is necessary for both an acid and a base to be present for this type of catalysis to take place. Water can act as either as an acid, in conjunction with a strong base or as a base in conjunction with a strong acid. This type of catalysis sometimes is called acid-base catalysis. The keto-enol tautomerism also occurs by the acid-base catalysis as shown below. +

O ||

OH

OH

||

+

|

H CH 3 − C − CH 3 → CH 3 − C − CH 3  → CH 2 = C − CH3 + H +

The following example illustrates the catalytic reactions in the presence of OH– ions. −

− 2

NH 2 NO 2 + OH  → NHNO + H 2 O Nitroamine

↓ N 2 O3

↓ N 2 O + OH −

Or in presence of CH3COO– ions.

NH 2 NO 2 + CH 3 COO −  → NHNO 2 + CH 3 COOH ↓ N 2 O + OH − → CH 3 COO − + H 2 O. CH 3 COOH + OH −  It can be noted all the acid-base catalysed reactions are taking place through the formation of intermediate either by protonation or deprotonation. Other example of intermediate compounds formation catalysed reaction are as follows.

→ CH 3 I + HI + CO CH 3 CHO +  I 2   → CH 4 +  I 2  CH 3 I + HI  8. Decon’s process for the manufacture of Cl2  2 Cu 2 Cl2  + O 2  → 2Cu 2 OCl2 → 2CuCl2 + H 2 O Cu 2 OCl2 + 2HCl  2CuCl2  → Cu 2 Cl2 + Cl2 9. Laboratory method for preparation of O2 → 2KMnO 4 + Cl2 + O 2 2KClO 3 +  2MnO 2   2KMnO 4  → K 2 MnO 4 + MnO 2 + O 2 K 2 MnO 4 + Cl2  → 2KCl +  MnO 2  + O 2

14.10.2 Adsorption Theory So far some of the examples of homogeneous catalysis are discussed and shown that how the catalyst combines with the reactant molecule to from an intermediate, thus offering an alternative path for the reaction to take. Heterogeneous catalysis works in the same way, but in this case it is the surface of the solid catalyst with which the reacting molecule must combine. A heterogeneous catalytic reaction may involve the following stages: 1. Diffusion of reactants to surface. 2. Adsorption of reactants. 3. Reaction of the adsorbed layer. 4. Desorption of products. 5. Diffusion of products from surface. Steps 2 and 3 are usually the important ones. The slowest of the five is the rate determining step. Where there are two reactants, reaction may occur by adsorption of two molecules adjacent to one another followed by the formation of an intermediate complex and finally the products.

Surface Chemistry 14.17 B

A

=

CH

H C 2

CH

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

A

B

A

B

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

Ni

Ni

H

2

2

B

M

The atoms at the surface of the solid have unsatisfied valencies since there are no atoms above them while the atoms at the centre of the solid are surrounded by atoms in all directions thus all their valencies are completely satisfied. Hence the atoms at the surface can form chemical bonds with the adsorbed molecules, but this necessitates a change in the bond structure of the molecules. With an ethene molecule, for example, the double bond can open to provide two bonds with the surface. Hydrogen and oxygen on the other hand, dissociate, so that they are adsorbed as single atoms. H C 2

A

2

H

H

Ni

Ni

The result of adsorption is to make the molecules more chemically reactive than they are in the normal state. In other words the reaction in which they take part will have a lower activation energy than the same reactions between normal molecules. Hydrogenation catalysts adsorb hydrogen very strongly; dehydrating catalysts (e.g., alumina on alcohol giving ethylene) are strong adsorbents of water. In certain cases where there are two reactants, it does not necessarily mean that both are adsorbed to the same extent, nor at the same rate. In the Haber process, the rate of adsorption of nitrogen is slow and probably the rate determining step. In hydrogenation reactions, there is evidence that nickel splits up the hydrogen molecule in to atoms, as we have seen above, and in reactions where hydrogen and carbon monoxide react at a metal surface, hydrogen atoms can

occupy position between the chemisorbed CO groups as shown below. O O O O || || || || C H C H C H C || || || || M M M M Molecular distortions may occur and in those reactions where different products are formed from the same reactants as already seen in the reaction between CO and H2, it may be that distortion at the catalyst surface is different. The formation of defect solid structures can be important. When carbon monoxide is oxidized by oxygen using a metal oxide (e.g., NiO), it is believed that oxygen is taken up by the oxide to form a semiconductor and that this then reacts readily with the carbon monoxide molecules attached to the oxide surface. It is of course an essential condition that the surface complexes or distorted molecules react more readily than the normal reactants. It is desirable that the product of the reaction should be able to leave the surface quickly, otherwise the amount of surface, free for further reaction becomes restricted and the reaction rate falls. This can be regarded as the poisoning of the catalyst by a product of the reaction. It is probable that most cases of poisoning are due to preferential adsorption of the poison which may also form a surface compound, thereby preventing reactants from reaching the surface. Volatile compounds of sulphur (Hydrogen sulphide and carbon disulphide) hydrogen cyanide, as well as arsenic compounds are poisonous to many catalysts. The prevention of decomposition of hydrogen peroxide by acetanilide may be due to the preferential adsorption of the latter on the glass walls of the containing vessel. Inhibition may also occur by interaction of inhibitor with one of the intermediate products to hold up the

14.18

Surface Chemistry

reaction chain. Lead tetraethyl probably prevents knocking’ by breaking up the long reaction chains which tend to from during compression. It has been found that the amount of a poison required to stop the reaction completely is less than the theoretical quantity needed to cover the whole of the surface assuming a unimolecular layer to be formed. This indicates that catalytic activity is confined to a comparatively small area of the surface, and it is now recognized that reaction takes place at certain point on the surface known as active centers. Since the adsorption of reacting molecules depends upon the free valencies the whole surface is not capable of activating the molecules but only the active centers can bring about this change. The number of free valencies can be easily increased in the following ways. 1. By making the surface of catalyst rough: Rough surface of a catalyst is more effective than a smooth one. This is due to the fact that the rough surface has large number of free valencies crowded at the edges, cracks, corners and peaks. These free valences are the active centers and the activity of the catalyst is mainly due to these active centers. M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

M

Rough surface 2. By subdividing the catalyst: The catalyst in finely divided state will be more effective due to very rich in free valencies and hence it’s large surface area. Thus finely divided nickel and colloidal platinum are known to be more efficient catalysts.

The adsorption theory can be summarized as follows: 1. The surface of the catalyst is used again and again due to alternate adsorption and desorption. Thus a small quantity of the catalyst can catalyse large amounts of reactants. 2. Chemical adsorption depends on the nature of the adsorbent and adsorbate. Hence catalysts are specific in action. 3. At the end of reaction since the products desorb from the surface of catalyst it remains unchanged. 4. The energy of adsorption compensate the activation energy of the reacting molecules to some extent. Thus the reactions occur at faster rate. 5. Greater efficiency of the catalyst in finely divided state and rough surface is due to more number of active centers. 6. The adsorption theory clearly explains the poisoning of catalysts. The poisons are preferentially adsorbed at the active centers of the catalyst thus decreases the active centers for the reactant molecules. Hence the catalytic activity decreases. 7. The action of promoters which alone have no catalytic power is not understood. Alumina prevents the melting and sintering of the iron surface in the Haber process but its effect is greater than can be explained solely by the maintenance of the larger area of the rough ended surface.

14.10.3 Activity and Selectivity of heterogeneous catalysis Activity is the ability of a catalyst to accelerate chemical reaction. In certain cases the activity can be as high as 1010 i.e., the catalysed reaction is 1010 times faster than the uncatalysed reaction. H 2 (g) +O 2 (g)  No reaction Pt 2H 2 (g) +O 2 (g)   2H 2 O (g) 

Selectivity of a catalyst is its ability to direct the reaction in such away as to yield particular product excluding others. e.g, CH3 CH 3 ( CH 2 )5 CH 3 Pt(s)  →

M

M

M

M

+H 2 (g)

Propylene and oxygen selectively give acrolein over bismuth molybdate as catalyst. Bismuth

M

M

M

M

molybdate CH 3 − CH = CH 2 + O 2   CH 2 = CH − CHO + H 2 O BiMO 4 acrolein

Acetylene on hydrogenation in presence of Pt or Ni or Pd catalyst gives ethane. Subdivision of the surface.

Pt / Pd / Ni HC  CH+2H 2 (g)   CH 3 − CH 3

Surface Chemistry 14.19

Hydrogenation of acetylene in pressure of Lindlar’s catalyst gives ethylene. Pd+BaSO4 HC  CH+H 2   H 2 C = CH 2 Sulphur or Quinoline

14.10.4 Shape Selective catalysis by zeolites Certain catalytic reaction depends upon the structure of the pores and the size of the reactant and product molecules. Such catalytic reactions are called shape – selective catalysts. Zeolites which are three dimensional silicates made with the replacement of certain silicon atoms by aluminum atoms. These zeolites have honey-comb structures containing micro pores. These zeolites can act as molecular sieves and allow certain molecules to pass through the pores and certain others are not. The molecules which are smaller than the pore size can pass through them while the molecules bigger than the pore size cannot pass through the pores. While certain reactants are passing through the pores of zeolites, reaction may takes place catalysed by the zeolite. The reaction taking place in zeolites depends upon the size and shape of reactant and product molecules as well as the cavities and pores of the zeolites. Zeolites occur naturally and can be synthesized for catalytic selectivity. Zeolites are being very widely used as catalysts in petrochemical industries for cracking of hydrocarbons and isomerisation. An important catalyst used in the petroleum industry is ZSM – 5 which converts alcohols directly into gasoline (petrol) by dehydrating them to give a mixture of hydrocarbons. Dehydration of methanol forms hydrocarbons such as gasoline and other fuels

1, 3 – dimethyl benzene (m-xylene) isomerizes to 1, 4 = dimethyl benzene (p-Xylene) in presence of zeolite. Zeolites stabilize only transition states that fit properly in the pores and thus the zeolites derive their selectivity. Zeolites are micro porous aluminosilicates with general form n+

 x [ H 2 O ]m n

} {[AlO ] [SiO ] } 2 x

Product

Starting with CH3OH (%)

Starting with n-C7H15 OH (%)

Methane

1

0

Ethane

0.6

0.3

Isobutene

18.7

19.3

n-butane

5.6

Iso-pentane

7.8

8.7

Benzene

1.7

3.4

Toluene

10.5

14.3

Xylene

17.2

11.6

11

14.11 ENzymE cATAlySIS Almost all enzymes are protein molecules with molecular weight ranging to over million amu. These are complex nitrogenous organic compounds which are produced by living plants and animals. An enzyme has enormous catalytic activity converting a thousand or so reactant molecules to products in a second that occur in living organisms. Hence the enzymes are termed as biochemical catalysts and the phenomenon is known as biochemical catalysis. Enzymes are also highly specific each enzyme acting only on a specific substance or a specific type of substance catalysing it to undergo a particular reaction. Examples (i) The enzyme ‘alcohol dehydrogenate’ reduces acetaldehyde to ethanol in the yeast cell.

nCH 3 OH Zeolite → ( CH 2 )n +nH 2 O

{M

Table 14.3 Percentage of different hydrocarbons in the mixture obtained from two alcohols catalysed by zeolites

x−

2 y

where Mn+ cations and H2O molecules bind inside cavities or pores of the Al-O-Si frame work. Small neutral molecules such as CO2, NH3 and hydrocarbons (including aromatic compounds) can also adsorb to the internal surface which makes the utility of zeolites as catalysts. The pore size in a zeolite varies between 260 pm to 740 pm. The percentage of different hydrocarbons in the mixture of hydrocarbons obtained from two different alcohols i.e., CH3OH and n C7H15OH are given in the Table 14.3.

dehydrogenate CH 3 CHO alcohol  → CH 3 CH 2 OH

(ii) The enzyme urease catalyses the hydrolysis of urea to CO2 and water. NH 2 CONH 2 Urease → 2NH 3 +CO 2 (iii) Inversion of cane sugar into glucose and fructose is catalysed by the enzyme invertase. C12 H 22 O11 +H 2 O Invertase  → C6 H12 O6 +C6 H12 O6 Sucrose glucose fructose (iv) Maltose is converted into glucose is catalysed by maltase. C12 H 22 O11 +H 2 O Maltase → 2C6 H12 O6 Maltose

14.20

Surface Chemistry

(v) Diastage catalyses the conversion of starch into maltose. Diastase

2 C6 H10 O5 n + nH 2 O  nC12 H 22 O11 starch Maltose (vi) In the fermentation of sugar, zymase converts glucose and fructose to ethyl alcohol. C6 H12 O6 Zymase → 2C 2 H 5 OH + 2CO 2 (vii) The enzyme lipase present in caster seed oil catalyses the fermentation of sugar to produce glycerol instead of alcohol. (vii) Lactic bacilli present in curd catalyse the fermentation of milk. (ix) In stomach, the pepsin enzyme converts proteins into peptides while in intestine the pancreatic trypsin converts proteins into amino acids by hydrolysis. (x) Catalase (origin: Plant juice, blood) brings about rapid decomposition of hydrogen peroxide. Digestion of food is a very complicated process depending on the consecutive influences of a whole series of enzymes. The enzymes pepsin and ptylin (origin: saliva) in gastric juice work together to increase the rate of breaking down of larger molecules of proteins and starch into small ones. The latter are easily utilized by the cells of the body. Enzymes are thus all important in the control of biological processes.

14.11.1 characteristics of Enzyme catalysis Though very similar to heterogeneous catalysts, enzyme catalysts have some peculiar characteristics also. They are as follows: 1. They do not take part in the reaction, but act catalytically i.e., they are organized catalysts of nature. 2. Enzymes like inorganic catalysts do not disturb the final state of equilibrium in a reversible reaction. 3. The rate of an enzyme reaction is proportional to the concentration of the reactant (substrate) provided the concentration is small. At high concentrations, the rate of reaction is independent of concentration i.e., the kinetics are of zeroth order. 4. A small quantity of enzyme can bring about the decomposition of large amount of the substrate. For example, urease extracted from soya bean, will catalyse the hydrolysis of urea when present only to the extent of 1 part in 19 millions. 5. Their action is highly specific i.e., a particular enzyme can bring about a particular reaction. For example, urease catalyses the hydrolysis of urea but it has no effect on the hydrolysis of methyl urea. 6. An enzyme is most reactive at a particular temperature called optimum temperature. In many cases, this temperature is 35-40 °C.

7. Enzymes like other catalysts are influenced by the presence of other substances. Some enzymes called co-enzymes act as promoters while HCN, CS2 etc, poison them. 8. They are destroyed by ultraviolet rays and by heat. At temperature too low (0°C) or too high (70-76°C), the enzymes are most inactive. 9. They bring about many complex reactions e.g., oxidation reduction hydrolysis. 10. The rate of an enzyme-catalysed reaction is maximum at a particular pH called optimum pH, which is between pH values 5-7. 11. Metal ions such as Na+, Mn2+, Co2+, Cu2+ etc activate the enzyme catalysis and are called activators. These metal ions when weakly bonded to enzyme molecules increase their catalytic activity. Amylase in the presence of sodium chloride i.e., Na+ ions are catalytically very active.

14.11.2 mechanism of Enzyme catalysis The substance whose reaction, the enzyme catalyzes is called the substrate. For example, the enzyme sucrase acts on the substrate sucrose. Fig 14.7 shows schematically how an enzyme acts. The enzyme molecule is a protein chain that tends to fold into a roughly spherical form with an active site at which the substrate binds and the catalysis takes place. The substrate molecule, S, fits into the active site on the enzyme molecule, E, somewhat in the way a key fits into a lock forming an enzyme substrate complex, ES (The lock and key model only a rough approximation because the active site on an enzyme deforms somewhat to fit the substrate molecule). In effect, the active site “recognizes” the substrate and gives the enzyme its specificity. On binding to the enzyme, the substrate may have bonds that weaken or new bonds form that helps yield the products P. Thus the enzyme-catalysed reactions may be considered to proceed in two steps: Step1: Binding of enzyme to substrate to form an activated complex E  S  ES* Step 2: Decomposition of the activated complex to form product ES*  E  P The formation of the enzyme substrate complex provides a new path way to products with a lower activation energy. Fig 14.8. compares the potential energy curves of the uncatalysed and catalysed reactions. Note the lower activation energy of the catalysed reaction.

Surface Chemistry 14.21

P

}

Products

P

Active site

S

S

E

E

Enzyme-substrate complex

Enzyme

Enzyme

fig 14.7 Mechanisms of enzyme catalysed reaction

Activation Energy Energy

Energy

Activation Energy

S

ES E+S

P

E+P

Progress of reaction

Progress of reaction B

A

fig 14.8 Potential-energy diagrams for the reaction of substrate S and product P (A) the uncatalysed reaction (B) the enzyme catalysed reaction. Table 14.4 Some enzymatic reactions

1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12.

Enzyme

Source

Enzymatic reaction

Invertase Zymase Diastase Maltase Bacillus Lacti Micoderm aceti Urease Lipase Pepsin Trypsin Amylase Ptylin

Yeast Yeast Malt Yeast Curd Vinegar Soyabean Castor seed Stomach Intestine Saliva Saliva

Sucrose → Glucose and fructose Glucose → Ethyl alcohol and CO2 Starch → Maltose Maltose → Glucose Fermentation of milk Ethyl alcohol → Acetic acid Urea → Ammonia and CO2 Fat → Glycerol Proteins → Amino acids Proteins →Amino acids Starch → Glucose Starch → Sugar

14.22

Surface Chemistry

Table 14.5 Some industrial catalytic processes Process 1. Haber’s process for the manufacture of ammonia N 2 + 3H 2  → 2 NH 3 2. Ostwald’s process for the manufacture of nitric acid. 4 NH 3 + 5O 2  → 4 NO + 6H 2 O

Catalyst Finely divided iron mixed with molybdenum as promoter; conditions 200 bar pressure; 723-773 K temperature. Platinized asbestos Temperature 573 K.

2 NO + O 2  → 2 NO 2 4 NO 2 + 2H 2 O + O 2  → 4HNO3 3. Contact process for the manufacture of sulphuric acid. → 2SO3 2SO 2 + O 2 

Platinized asbestos or vanadium pentoxide (V2O5) Temperature 673–723 K

→ H 2S2 O7 (oleum) SO3 + H 2SO 4  → 2H 2SO 4 H 2S2 O7 + H 2 O  4. Lead chamber’s process for the manufacture of sulphuric acid 2SO 2 + O 2  → 2SO3

Oxides of nitrogen

SO3 + H 2 O  → H 2SO 4 5. Synthesis of methanol CO + 2H 2  → CH 3 OH 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

Preparation of acetone from acetylene. Preparation of vinyl chloride Preparation of HCHO from CH4 Oxidation of methyl pyridines to nicotinic acids Straight chain alcohol (Seynol Process) Higher alcohol synthesis Preparation of branched hydrocarbon Manufacture of vanaspathi Preparation of ethyl benzene Manufacture of ethyl alcohol by manufacture of sugar Decon’s process for manufacture of chlorine High quality petrol fuel by hydrogenation of coal, coaltar etc.

ZnO, Cr2O3 FeCl3 Cu2Cl2 Fe2O3, Cr2O3,Cu-As alloy V2O5–MoO3 Fe ZnO, Cr2O3 alkali ThO2, SnO, H2O3 Ni SiO2, Al2O3, AlCl3 Invertase, Zymase CuCl2 Sulphides and oxalates of Sn, Mo, W etc.

Surface Chemistry 14.23

Table 14.6 Examples of homogeneous catalysis Process

Catalyst

1. Decomposition of ethanoic acid to ketene at 650°C CH 3 COOH  → CH 2 CO + H 2 O

Triethyl phosphate

2. Combination of sulphur dioxide and oxygen 2SO 2 + O 2  → 2SO3

Nitrogen oxides (NO & NO2)

3. Hydrolysis of esters CH 3 COOCH 3 + H 2 O  → CH 3 COOH + CH 3 OH

Acid (H+ ions) or Base (OH–ions)

4. Iodination of acetone CH 3 COCH3 + I 2  → ICH 2 COCH3 + HI

Acid (H+ ions)

5. Friedel Crafts alkylation or acylation CH 3 Cl + C6 H 6  → C6 H 5 CH 3 + HCl

AlCl3

CH 3 COCl + C6 H 6  → C6 H 5 COCl + HCl 6. Hydrolysis of sugar C12 H 22 O11 + H 2 O  → C6 H12 O6 + C6 H12 O6

Acid (H+ ions)

Table 14.7 Examples of heterogeneous catalysis Process

Catalyst

1. Haber process for manufacture of ammonia N 2 ( g ) + 3H 2 ( g )  → 2 NH 3 ( g )

Iron powder as catalyst Mo or oxides of Al and K as promoters

2. Contact process for manufacture of sulphuric acid 2SO 2 ( g ) + O 2 ( g )  → 2SO3 ( g )

Platinum or vanadium pentoxide

SO3 + H 2SO 4  → H 2 S2 O 7 H 2S2 O7 + H 2 O  → 2 H 2 SO 4 3. Hydrogenation and dehydrogenation i. Hydrogenation of oils. ii. Styrene (to prepare polystyrene) is obtained by the dehydrogenation of ethyl benzene. C6 H 5 CH 2 CH 3  → C6 H 5 CH = CH 2 + H 2

Nickel Metal oxides at 600°C

4. Cracking of petroleum. 5. Manufacture of methanol CO + 2H 2  → CH 3 OH

Mixture of alumina and silica ZnO+Cr2O3

6. Manufacture of ethanol CH 2 = CH 2 ( g ) + H 2 O ( g )  → C 2 H 5 OH

Phosphoric acid on a solid support

7. Manufacture of nitric acid by Ostwald’s process by oxidation of ammonia 4 NH 3 + 5O 2  → 4 NO + 6H 2 O

Platinum

→ 2 NO 2 2 NO + O 2  → 4 HNO3 4 NO 2 + H 2 O + O 2 

14.24

Surface Chemistry

Table 14.8 Example of positive catalysis 1. Decomposition of H2O2 in the presence of colloidal sulphur. 2H 2 O 2 ( l ) Pt→ 2H 2 O ( l ) + O 2 ( g ) 2. Decomposition of KClO3 in the presence of manganese dioxide. 2 2KClO 3 MnO  → 2KCl + 3O 2

3. Oxidation of ammonia in the presence of platinum gauze. 4 NH 3 + 5O 2 Pt→ 4 NO + 6H 2 O 4. Oxidation of SO2 to SO3 in the presence of nitrogen oxides. + NO 2 2SO 2 + O 2 NO  → 2SO3

5. Oxidation of SO2 to SO3 in the presence of platinized asbestos or vanadium pentoxide. Pt (or ) 2SO 2 + O 2  → 2SO3 V O 2

5

6. Oxidation of hydrochloric acid to chlorine by Deacon’s process in the presence of cupric chloride. 2 4HCl + O 2 CuCl  → 2H 2 O + 2Cl2 450° C

7. Hydrogenation of oils in the presence of Nickel. Vegetable oil + H 2 Ni → Ghee 8. Synthesis of ammonia by Haber process in the presence of a mixture of iron and molybdenum. Fe N 2 + 3H 2 450  → 2 NH 3 °C

9. Manufacture of methyl alcohol from water gas in the presence of zinc oxide and chromium oxide. + Cr2 O3 CO + 2H 2 ZnO  → CH 3 OH

10. Formation of methane in the presence of Nickel. CO + 3H 2 Ni → CH 4 +H 2 O

Surface Chemistry 14.25

KEy POINTS catalysis and Types of catalysis •

•

•

•

•

•

•

•

•

•

•

•

For the first time, Berzelius noticed the catalytic activity of substances. He gives the name catalysis for the phenonmetion of increasing the rate of reaction. Catalysis means loosening the bonds which hold the atoms in the reacting molecules. The substances which accelerate the rate of a reaction is known as positive catalyst. Positive catalysts increase the rate of reaction by lowering the activation energy. Positive catalyst changes the mechanism by changing the intermediate i.e., intermediate of lower energy. It increases the rate by converting some inactive molecules into active ones. Substances which decrease the rate of a reaction or retard the reaction are called negative catalysts or inhibitors. A negative catalyst decreases the rate of reaction by increasing the activation energy of the reaction or by poisoning or destroying the catalyst. Sometimes, it dislocates the mechanism. Substance which itself has no catalytic activity but will increase the activity of catalyst is called promoter or activator. A promoter thus be regarded as a catalyst for catalyst e.g., Al2O3 K2O, Mo powder act as promoter to iron catalyst in Haber process for the manufacture of ammonia. The substances which reduce or even completely destroy the activity of a catalyst are known as catalytic poisions or anti-catalysts. Temporary poisioning is due to adsorption of catalytic poision on the surface of catalyst while permanent poisioning is due to chemical reaction with catalyst. Carbon monoxide in the manufacture of ammonia by Haber’s process, arsenic oxide in the manufacture of H2SO4 by contact process, HCN in the decomposi tion of H 2O2 catalysed by Pt, CO in the action H 2 on ethylene in presence of Ni act as catalytic poisons. If one of the products or intermediate formed in a reaction acts as a catalyst for the same reaction it is known as autocatalyst and the phenomenon is known as autocatalysis. Mn2+ ion formed in the reduction of permanganate by oxalic acid, arsenic formed in the decomposition of arsine, acetic acid produced in the ester hydrolysis will act as catalysts in those reactions. When one reaction influences the rate of other reaction which does not occur under ordinary conditions, it is known as induced catalysis e.g., oxidation of sodium sulphite by air induces the oxidation

•

•

•

•

of sodium arsenite by air. Reduction of KMnO4 with oxalic acid induces the reduction of HgCl 2 with oxalic acid. If the catalyst and the reactants are in the same phase it is called homogeneous catalysis. They occur either in the gas phase or in the liquid state e.g., oxidation of SO2 to SO3 using oxides of nitrogen in lead chambers process conversion of CO to CO2 in the presence of NO vapour phase decomposition of CH 3CHO into CH4 and CO in the presence of I2 vapours are examples for gas phase homogeneous catalysis. Decomposition of H2O2 in the presence of aqueous iodide ions, hydrolysis of ester in the presence of acid or alkali and hydrolysis of sugar in the presence of mineral acids are examples for liquid phase homogeneous catalysis. If the catalyst and reactants are not in the same phase, then it is known as heterogeneous catalysis. Generally, the catalyst is solid while the reactants are liquids or gases. Most of the solid catalysts are either transition metals or their compounds. Oxidation of SO2 to SO3 in the presence of V2O5 in contact process, synthesis of ammonia by Haber’s process using iron catalyst, decomposition of H 2O2 in the presence of MnO2 or Pt are some examples for heterogeneous catalysis.

characteristics of catalysts • • • • • •

A catalyst cannot initiate a reaction but can only alter the velocity of a chemical reaction. A catalyst remains chemically unaffected at the end of a chemical reaction but may change physically. A small amount of the catalyst is sufficient to catalyze the reaction. A catalyst generally functions under the optimum conditions only like temperature, pressure, pH etc. Catalyst has no influence on the position of equilibrium but helps to attain the equilibrium fastly. Catalytic activity is more specific. Change of catalyst in a particular reaction may give different products. When C2H5OH is heated in the presence of Al2O3, dehydration takes place, but in the presence of copper dehydrogenation takes place.

Theories of catalysis •

The action of catalyst can be explained by (i) intermediate compound theory (ii) adsorption theory.

14.26

•

• •

• •

•

•

•

•

• •

Surface Chemistry

In the intermediate compound theory, the catalyst combines with one or more reactants forming intermediate compound. The intermediate compound, directly or by reacting with other reactant molecules form products liberating catalyst. Friedel Craft’s reaction, acid-base catalysed reaction, decomposition of acetaldehyde in the presence of iodine vapours, manufacture of chlorine by Deacon’s process, decomposition of KClO3 into KCl and O2 are some examples for catalysis through the formation of intermediate compound. Adsorption theory is applicable to heterogeneous catalytic reactions occurring on the surface of solids. The steps involved in the heterogeneous catalysis are i. diffusion of reactants to surface ii. adsorption of reactants on the solid surface of the catalyst at the appropriate centers called active centers iii. reaction between the adsorbed reactants iv. desorption of products v. diffusion of products from surface. Where there are two reactants, reaction may occur by adsorption of the molecules adjacent to one another followed by the formation of an intermediate complex and finally the products. The atoms at the surface of solid have unsatisfied valencies and can form bonds with reactant molecules and thus decrease the activation energy of the reaction. In the reactions involving only two reactants, the reactant molecules of both reactants may not adsorb at the same rate or to the same extent, e.g., in Haber’s process the rate of adsorption of nitrogen is slow and it is the rate determining step. In the reaction where different products are formed from the same reactants in the presence of different catalysts molecular distortions may occur i.e., the distortions at the surface of different catalysts are different. The energy of adsorption compensate the activation energy of the reacting molecules to some extent. Greater efficiency of the catalyst in finely divided state and with rough surface is due to more number of active centers.

Shape-Selective catalysis by zeolites •

•

•

•

Enzyme catalysis •

•

•

• •

• •

•

Activity and Selectivity of heterogeneous catalysis • •

Activity is the ability of the catalyst to accelerate chemical reactions. Selectivity of a catalyst is its ability to direct the reaction in such a way as to yield a particular product excluding others.

The catalytic reactions that depend upon the structure of the pores and the size of the reactant and product molecules are called shape-selective catalysts. Zeolites which have honey comb structures containing micropores are examples for the shape selective catalysts. Zeolites can act as molecular sieves which allow certain molecules which are smaller than the pore size to pass through them while the molecules bigger than the pore size cannot pass through them. When certain reactant molecules are passing through the pores of zeolites reaction may take place catalysed by the Zeolite. ZSM-5 a zeolite converts alcohols directly into gasoline by dehydrating them to a give a mixture of hydrocarbons.

•

Enzymes are protein molecules which are complex nitrogenous organic compounds which are present in the living plants and animals. Enzymes have enormous catalytic activity and are termed as biochemical catalysts and their catalytic activity is similar to heterogeneous catalyst. The rate of an enzyme catalysed reaction is proportional to the concentration of the reactant provided the concentration is small, but at high concentrations the rate is independent of concentration i.e., the kinetics are of zeroth order. Enzyme catalysis is highly specific i.e., a particular enzyme can bring about a particular reaction only. An enzyme is most reactive at a particular tem perature called optimum temperature generally at 35-40°C. Metal ions such as Na+, Mn2+, Co2+,Cu2+ etc which activate the enzyme catalysis are called activators. An enzyme first bind to substrate to form an activated complex (intermediate) which then decomposes to form product. The enzyme and reactant molecule have suitable shapes like lock and key which can fit in one another to form an activated complex. Hence the enzyme catalysis is highly specific because different enzymes and reactant molecules have different shapes. Only enzyme and reactant molecule having suitable shape, participate in reaction. Formation of the enzyme-substate complex provides a new path way to products with a lower activation energy.

Surface Chemistry 14.27

14.13 cOllOIDAl STATE–AN INTERmEDIATE STATE

14 c cOllOIDS 14.12 INTRODucTION During his work on diffusion (about 1861) Graham found that the solute of an aqueous solution of common salt diffused through a parchment membrane about four hundred times as fast as did: Gum Arabic from its aqueous solution. Both solutions appeared homogeneous and both passed through the filter paper without any separation of solute. Those substances showing rapid diffusion possessed crystalline structure and Graham called them as crystalloids, in contrast to those such as gum arabic having no such structure, which he called colloids (from Greek, meaning glue like). The classification was unsound for it has since been shown (by x-ray work) that certain colloids e.g., gold and barium sulphate posses a crystalline structure and further, that many of the substances which Graham considered as crystalloids are obtainable in the colloidal condition. Neither is it true to consider, as was thought later, that colloidal condition was a fourth state of matter. The difference in the diffusion through a parchment membrane can only be explained by assuming that ‘colloids’ in the dissolved state form bigger particles which cannot pass through small pores of the parchment membrane. On the other hand, crystalloids are broken down in solution to form ions which can diffuse quickly. It is interesting to note that the particle size of the same substance may be different in different solvents. For example, NaCl is a crystalloid when dissolved in water but forms colloidal solution in benzene. Similarly, soap forms a colloidal solution in water but acts as a crystalloid in alcohol. In general, a colloid can be converted into crystalloid and vice versa.

A colloidal state of a substance can be regarded as an intermediate state between a true solution and a suspension. 1. True solution: When a soluble substance like NaCl sugar etc is dissolved in water it forms a homogeneous solution in which molecules ions are uniformly dispersed in water. Such solutions are called true solution. These are the solutions in which the molecules or ions of solute are uniformly distributed in the entire mass of the solvent. The molecules or ions of the true solutions cannot be seen with a naked eye or with a powerful microscope. The particle size of the true solution is less than 1 mµ. Such solutions can pass through the filter paper as well as through a parchment membrane. 2. Suspensions: When an insoluble substance such as calcium carbonate is shaken with enough water, a heterogeneous mixture results. The particles of calcium carbonate are seen in the solvent. The dispersion of an insoluble substance in a solvent is suspension. The particles of diametre greater than 10–4 cm (10 µ) would give a suspension. These particles can be seen with a naked eye and cannot pass through an ordinary filter paper. 3. Colloidal solution: When a soluble substance such as gum is dissolved in water, then a heterogeneous solution results. The particle size in this case is found to be greater than that of solute molecules in true solutions and smaller than that of suspension. Such a solution is colloidal solution and its particle size is in between 1 mµ–0.1 µ. The size of the colloidal particles is less than the wave length of visible light (380-760 mµ). Thus these particles cannot be seen with a naked eye or through an ordinary microscope. In contrast to a

Table 14.9 Characteristics of solutions, colloids and suspensions Property

True solution

Colloidal solution

Suspension

1

Particle size

Less than 1 mµ

1 mµ -0.1 µ

more than 10 µ

2

Separating by Not possible Not possible Particles do not settle

Possible Possible Settles on centrifusing

Possible

3

(a) Ordinary filter paper (b) ultra filtration Setting

4 5 6

Diffusion

Diffuse rapidly

Diffuse slowly

Does not diffuse

Appearance

Transparent

Almost transparent

Opaque

Tyndall effect

–

Shows

Shows

7

Brownian movement

Negligible

Shows

May or may not show

8

Nature

Hemogeneous

Heterogeneous

Heterogeneous

Possible Settle under the effect of gravity

14.28

Surface Chemistry

suspension from which the solid particles separate out on standing, under the influence of gravity, a colloidal solution is a stable system i.e., does not undergo sedimentation, and owes its stability, in part, to the size of particles in it. In a true solution, particles of solute are either molecules or ions. In a colloidal solution, they are aggregates of these (sometimes called micelles).

14.13.2 Classification Based on Physical State of Dispersed Phase and Dispersion medium Depending upon whether the dispersed phase and the dispersion medium are solids, liquids or gases eight types of colloidal systems are possible. The different types of colloids along with examples are given in the Table 14.10.

Phases of colloids

14.13.3 Classification Based Upon Appearance

A colloidal solution is heterogeneous in nature. It consists of two phases (a) Dispersed phase: The extremely small particles which are dispersed in the solvent, form a dispersed phase. Consider a colloidal solution of silver in water. The particles of silver form a dispersed phase. (b) Dispersion medium: The solvent in which the particles are dispersed form a dispersion medium. It is generally the component which is present in excess. Eight types of colloidal solutions are formed. Colloidal solutions of different types are listed in the Table 14.10.

(a) If the appearance of colloidal solution is like a fluid, it is called a sol. Such sols in which water is present as the dispersion medium are called hydrosols. Those which contain benzene as the dispersion medium are called benzosols. If alcohol, is a dispersion medium, it is called alcosol. (b) The colloidal solutions which have more rigid structures are called gels. Gelatin behaves both as a sol and gel. At high temperature and low concentration of gelatin, the colloid is a hydrosol (water is dispersion medium). At low temperature and high concentration of gelatin, the hydrosol changes into gel.

Table 14.10 Some colloidal systems Dispersed Dispersion Colloidal phase medium system

Examples

Solid

Solid

Solid sol

Minerals, gem stones

Solid

Liquid

Sols

Solid

Gas

Liquid

Solid

Aerosols of solids Gels

Paints, gold sol, muddy water Smoke, dust, storm

Liquid

Liquid

Emulsion

Milk, medicines

Liquid

Gas

Gas

Solid

Gas

Liquid

Aerosols of liquids Solid foam Foam, Froth

Fog, clouds, insecticide sprays Pumice stone, foam, rubber Soda water, Whipped cream froth etc.

Butter, jellies, boot polish

Note: The gas in gas mixtures are always true solutions

14.13.1 Classification of Colloids Colloids are classified on the basis of the following criteria: i. Physical state of dispersed phase and dispersed medium ii. Nature of appearance iii. Nature of interaction between dispersed phase and dispersion medium iv. Type of particles of the dispersed phase v. Based upon charge

14.13.4 Classification Based on Interaction of Phases (a) Lyophilic colloids (solvent loving): These are defined as the colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium. These are generally stable solutions. These are self stabilized due to strong attractive forces operating between the two phases. These are reversing on evoporating the solution, the solid obtained can be reconverted into the solution by simply agitating it with the dispersion medium e.g., gums, gelatin starch form lyophylic sols. In lyophilic colloids, particles of the disperse phase being surrounded by molecules of the dispersion medium (or sometimes dispersion medium molecules are enmeshed in a net work of particles). The micelles of hydrated silica proteins (e.g., albumen, gelatin) higher carbohydrates and high polymers are heavily solvated and thus examples of lyophilic sols. (b) Lyophobic sols: These are the colloidal solutions in which particles of the dispersed phase have no affinity for the dispersion medium. Lyophobic meaning solvent hating. Such solutions are relatively less stable and are not easily prepared. They can be easily precipitated by heating the sol or on adding small amount of electrolyte to it. These are irreversible. The solid obtained by precipitation cannot be reconverted into colloidal solution by simply shaking it with the

Surface Chemistry 14.29

Table 14.11 Distinction between lyophilic and lyophobic sols

1 2 3 4 5 6

Property

Lyophilic sols

Lyophobic sols

Preparation Nature Size of particles Stabilization Visibility Viscosity

Can be easily prepared by direct mixing. Reversible Comparable size as true molecules Self stabilized Particles are not visible under ultra microscope Viscosity is much higher than of dispersion medium. It is less than that of dispersion medium Particles may move towards anode or cathode or may not move at all. For coagulation of sol large amounts of electrolytes are required. It does not exhibit Tyndall effect. Have high values of colligative properties such as osmotic pressure elevation in b. pt depression in f. pt. etc.

Prepared only by special methods Irreversible Particles constitute aggregates of molecules. Less stable and require traces of stabilizers Particles are visible under ultra microscope. Same viscosity as of dispersion medium.

7 Surface tension 8 Action of electric field 9 Action of electrolyte 10 Tyndall effect 11 Colligative properties

dispersion medium, e.g., Metals, metallic oxides (hydrated) and sulphides are common example of lyophobic sols. If the dispersion medium is water, the lyophilic sols may be described as hydrophilic sols while the lyophobic sols are described as hydrophobic sols.

14.13.5 Classification Based on Type of Particles of Dispersed Phase Depending upon the molecular size, colloids are of three types (a) Multimolecular colloids: These are such colloids in which the individual particles consists of aggregate of atoms or small molecules having molecular size less than one mµ (= 10−7 cm). In these colloids, the particles are held together by weak Van der Waal’s forces. Some examples of such colloids are (i) Sols of gold atoms (ii) Sulphur molecules (S8) (b) Macro molecular colloids: These are such colloids, in which the size of the particles of the dispersed phase is of the order of colloidal dimensions. Some examples are (i) Sols of starch, cellulose etc. (ii) Sols of proteins. (c) Associated colloids: These are such substances which behave as normal electrolytes at low concentration but undergo association at higher concentration and behave as colloidal solution. Soap sol is an example

Same surface tension as that of dispersion medium Particles may move towards cathode or anode depending upon the charge. For coagulation small amounts of electrolytes are required. Exhibit Tyndall effect They show high osmotic pressure less elevation in b pt and low depression in f.pt.

of associated colloids. When soap (sodium stearate) is dissolved in water it ionises as follows: − +  C17 H 35 COONa ↽ ⇀  C17 H 35 COO + Na Sodium stearate Stearate ion (soap)

In the concentrated solution of soap, the stearate ions get associated to form particle of colloidal size. Table14.12 Difference between multi-molecular colloids and macro-molecular colloids Multi-molecular colloids

Macro-molecular colloids

1. The colloidal particles consist of aggregates of atoms or molecules with diametre less than 1 nanometre. 2. In a multi-molecular colloid, the particles are held by Van der Waal’s forces. Examples are sulphur sol, gold sol etc.

In these, the molecules of the dispersed phase are of colloidal size They have very high molecular masses which are of the order of millions. Examples are proteins, starch, cellulose etc.

14.13.6 Classification Based on Charge The colloidal solutions are called positive or negative sols depending upon the nature of charge on the dispersed phase particles. For example, sols of Fe(OH)3, Al(OH)3 basic dyes like methylene blue are positive sols. Sols of Cu, Ag metal sulphides such as As2S3, CdS etc., are negative sols.

14.30

Surface Chemistry

14.14 PREPARATION Of cOllOIDS

+

–

Metal rod

3

Colloidal state (10 A° to 10 A°) is an intermediate one between suspension (>103 A°) and true solution (1−10 A°). To get a substance from either substances in bulk are broken into fine particles or brought about the molecular particles to form the larger aggregates. The methods which involve the breakage of coarse particles into colloidal dimensions are called dispersion methods. The methods which involve the union of a large number of molecules to form bigger particles of colloidal range are called condensation methods. Some substances which have particles of colloidal size but could not get suspended in dispersion medium for lack of affinity between the two are stabilsed by some other substances called stabilizers. The following methods are generally used for the preparation of sols. A. Dispersion methods 1. Mechanical dispersion: Substance to be dispersed is finely ground by usual methods. It is then shaken with dispersion medium which gives a coarse suspension. This suspension is now fed in a colloidal mill (Fig 14.9) which consists of two steel discs having very small distance between them. These discs are rotated at very high speed in opposite directions. The suspension passing through the discs are further broken apart giving colloidal size particles due to a large shearing effect. Driving belt

Metal disc Discharge

Metal Disc

fig 14.9 Colloidal mill 2. Electro-dispersion or Bredig’s arc method: This method is particularly suitable for preparing the colloidal solutions of metals like silver, gold, platinum etc.

Electric arc Ice Bath Water + KOH

fig 14.10 Bredig's arc method for preparing metal sols Metal electrodes are immersed in the dispersion medium. The dispersion medium is further placed in the freezing mixture when an electric arc is struck; an intense heat of the arc turns the metal into vapours. The vapours get condensed by the ice cold water to give particles of colloidal size as desired. A little KOH may be added to water (dispersion medium) in order to stabilize the sol. The solution may be filtered to remove still bigger particles. This method is not suitable when the dispersion medium is an organic liquid in which considerable charring occurs. A modified method known as Svedberg’s method is used for such medium. In this method, the direct current is replaced by alternating current from an induction coil provided with a condenser. 3. Peptisation: It is the process by which a stable colloidal solution can be prepared from substances originally present in massive form when the colloidal particles pre-exist in the substance to be dispersed. In this process, the precipitate passes into colloidal solution by treating it again and again with the solvent alone or in presence of a foreign substance called peptizing agents. Peptisation can be brought about by the following ways: i. By adding electrolytes: Certain substances are converted into the colloidal solution when shaken with water containing a small trace of electrolyte which acts as the peptizing agent. The electrolyte supplies suitable ions to be adsorbed on the fine particles of precipitate. It increases the boundary potential of the precipitate and they form a colloidal solution. For example, freshly prepared ferric hydroxide precipitate on treatment with a small quantity of ferric chloride solution (peptizing agent) produces a dark reddish brown colloidal solution of Fe(OH)3. Similarly, a colloidal solution of Al(OH)3 is obtained when fresh;y precipitated Al(OH)3 is treated with a small quantity of dil HCl. The amount of acid added being sufficient to convert the hydroxide completely into chloride AgCl can be peptized by AgNO3.

Surface Chemistry 14.31

The process of transferring back a precipitate into colloidal form is called peptisation. It is the reverse of coagulation. In other words, it is the redispersion of a coagulated solution. FeCl3, AgNO3 etc., are known as peptizing agents. The Peptizing action is due to preferential adsorption of one of the ions of the electrolyte which then gives to the colloidal particle a positive or negative charge according to the charge of the adsorbed ion. For example, Fe(OH)3 adsorbs Fe3+ ions from FeCl3 (peptizing agent) and thereby gets a positive charge the surface. The charged particles get separated, yielding smaller sized colloidal particles of the type [Fe(OH)3] Fe3+ ii. By dispersion medium: Substances like gelatin and gum, give sols when they are shaken in water. They are said to be peptized by the medium. iii. By another colloid: Lamp black is peptized by gums. iv. By washing the precipitate: If a precipitate of BaSO4 or CuS is washed continuously a stage is reached in each case when the filtrate coming out from the filter paper begins to carry some of the particles of the substance in the colloidal form. 4. By application of heat: Lyophilic but not lyophobic sols may be made by merely warming together disperse phase and medium, e.g., gelatin in water. The gelatin can be removed from the sol by evaporation and on addition of water to the residue, the sol is obtained again. b. condensation methods Condensation methods consist of those chemical reactions in which normally the disperse phase is one of the insoluble products, the reactions being carried out under conditions such that building up of the particle is controlled to within the limits given above. Conditions governing the formation of aggregates were set forth by Von Weimarn. The rate of aggregation must be very rapid in order to obtain a large number nuclei that exhaust the remaining solute before too much particulate growth occurs. This can be achieved by working at a high degree of super saturation defined as (C-S)/S where C is the concentration of the solute to be aggregated and S is the solubility. If super saturation is greatly increased by increasing C, the nuclei formed may be close enough together to form gel, instead of sol. It is preferred to work under conditions in which S is as small as possible so that nuclei will remain highly dispersed. The following methods are used to prepare sol by condensation process. 1. Change of dispersion medium: In this method, a substance is dissolved in a solvent and then the solution is added to another solvent in which it is less soluble.

For example, if an alcoholic solution (True solution) of sulphur is added in excess of water, a colloidal solution of sulphur results since sulphur is insoluble in water. Similarly, by adding water to a solution of phosphorous in alcohol a sol is produced; or by adding about ten parts by volume of alcohol to one part of concentrated solution of calcium acetate, a solid gel (Solid alcohol) is formed. Silver iodide dissolves in aqueous potassium iodide owing to the formation of the complex ion [AgI2]–. If the solution is poured into water, when the complex ion breaks up, a colloidal solution of silver iodide is obtained. 2. Change of physical state: Sols of certain elements i.e., sulphur and mercury are obtained by passing their vapour from boiling elements into cold water containing suitable stabilizers such as ammonium salt for mercury sols. 3. Chemical methods i. Metathesis or double decomposition: A silver chloride sol is obtained by adding silver nitrate solution to that of a chloride Ag + + Cl−  → AgCl Von Weimarn prepared a barium sulphate sol by adding manganese sulphate to barium thiocyanate Ba 2 + + SO 24 −  → BaSO 4 One of the essential conditions is high concentration of reactant solution (Von Weimarn chose barium thiocyanate because it is one of the more soluble barium salt) for the particles of insoluble product from such solution are small. Hot and dilute solution cause aggregation to the point of precipitation. In the above cases, the resulting systems must be dialyzed to remove free ions, otherwise in time; the colloidal particles would be precipitated. A (yellow) hydrosol of arsenic (III) sulphide is made by passing hydrogen sulphide through a cold aqueous solution of arsenic (III) oxide. At the completion of the reaction, a little arsenic (III) oxide solution is added until there is no smell of hydrogen sulphide. By removing excess hydrogen sulphide (an electrolyte) the sol is stabilized and dialysis is unnecessary. It is to be noted that the solution of arsenic (III) oxide is cold, since a hot solution would give a yellow precipitate of arsenic (III) sulphide. This is why in group II of the classical analysis table hydrogen sulphide is passed through a hot (and acid) solution to get complete precipitation. Regarding salt hydrolysis as a type of metathesis, a (red) sol of hydrated iron (III) oxide is obtained by adding drop by drop a concentrated solution of iron (III) chloride to a large volume of boiling water. +  Fe ( OH ) ( H O )   → Fe ( OH )3 + 3H 2 O + H + 2 2  4 

14.32

Surface Chemistry

Here again, the resulting ions are removed by dialysis to give the stable sol commonly called dialyzed iron. Hydrated oxides of other weakly electropositive metals can be prepared by this method. ii. Reduction: Hydrosols of many metals e.g., silver, gold, platinum may be prepared by reducing aqueous solution of their salts. Reducing agents which are not electrolytes, e.g., carbon monoxide, phosphine, phosphorous, formaldehyde, phenyl hydrazine, tannic acid and the like preferred. Faraday (1857) prepared red and blue gold sols by reducing gold (III) chloride by a solution of white phosphorous in ether. The different colours of the sols depend on the size and shape of the gold micelles (Blue micelles are large and assymmetrical and red ones are small and spherical) and were determined by the amounts of reducing agent used. Using phosphine as reducing agent it is possible to produce particles of colloidal gold containing only fourteen atoms each. iii. Oxidation: Sulphur sols may be obtained by the oxidation of hydrogen sulphide by merely allowing its aqueous solution to stand in air or by careful addition of hydrochloric acid to sodium thiosulphate. Wackenroder’s solution obtained by passing hydrogen sulphide through an aqueous solution of sulphurous acid contains colloidal sulphur as well as some of the thionic acid. Presence of oxidizing agents e.g., Cr O 24 − may cause the formation of colloidal sulphur in group II of the analysis table, with the consequent difficulty of removing it. (This is why if an oxidizing agent is present it may be preferable it reduce it with sulphur dioxide and then boil to remove excess sulphur dioxide before passing hydrogen sulphide.  → As 2S3 (sol) + 3H 2 O As 2 O3 + 3H 2S Metathesis SO 2 + 2H 2S oxidation  → 3S (sol) + 2H 2 O duction  → 2AuCl3 + 3HCHO + 3H 2 O Re

2Au (sol) + 3HCOOH + 6HCl FeCl3 + 3H 2 O Hydrolysis  → Fe ( OH )3 (sol) + 3HCl

described above, hydrochloric acid is also present. It is often desirable to remove these impurities particularly when they are electrolytes, because they reduce the stability of the sol. Separation can be achieved by taking advantage of the difference in size between dissolved molecules or ions and the colloidal particles. The following methods are used for the purification of colloids. 1. Dialysis: We know that the particles of the colloid solution cannot pass through the parchment membrane while those of the true solution easily pass through it. The process of separation of a crystalloid from a colloid by diffusion through a parchment membrane is called dialysis. The apparatus which is employed to effect such separation is called dialyzer. The process of dialysis can be carried but by using two types of dialyzers. (a) Graham’s dialyzer Colloidal solution

Water + crystalloid

Parchment membrance Water

fig 14.11 Purification of colloidal solution by dialysis

It consists of shallow cylinder open at both the ends. A membrane is fitted over one of the ends. An impure colloidal solution is taken in this dialyzer and it is then suspended in a large vessel containing distilled water. Fresh water is continuously added from one end which flows out from the other and the rate of flow is maintained so as to keep its level constant. The electrolytes or impurities are thus continuously removed from the shallow cylinder as the process of dialysis proceeds. The progress of dialysis can be noted after intervals by testing and estimating the impurity in the cylinder. A colloidal solution of silicic acid can be prepared by treating sodium silicate with dilute hydrochloric acid. Na2SiO3 + 2HCl → H2SiO3 + 2NaCl

14.15 PuRIfIcATION Of SOlS When a sol has been prepared there are often other materials left in solution besides the colloidal particles; for example in iron (III) oxide sol, made by the method

The whole solution is put to dialysis when sodium chloride and unused HCl diffuse away. Thus pure colloidal solution of silicic acid is left behind. It is a long process and takes many days to complete.

Surface Chemistry 14.33

(b) Electro dialyzer

Anode

+

(a) (b) (c) (d) (e) Water + crystalloid – Cathode Dailysing parchment membrance

Water

Sol

fig 14.12 Purification of colloid by electrodialysis The process of dialysis can be hastened when carried under the influence of electric field. In this method, the outer vessel is fitted with two electrodes. The impure colloidal solution is taken in the shallow cylinder. The application of electric field increases the movement of the ions to the oppositely charged electrodes. Limitation: This method is not-suitable for the fast removal of molecular particles of sugar and other non-electrolytes from the colloidal solution. The process of dialysis is useful (i) in distinguishing a colloid from a crystalloid (ii) for the purification of the colloidal solution containing the dissolved substance in the molecular and the ionic from. 2. Ultra-filtration: The separation of the crystalloid from colloidal system can also be carried out by ultrafiltration. Ordinarily the filter papers have pores larger than one micron (10–6m). Thus the colloidal particles can pass through these filter papers along with the ions or molecules. The pores of the filter paper can be made smaller by soaking it in a solution of gelatin or of collodion and then in turn hardening them by soaking in formaldehyde. Now the pores of the filter paper become smaller and the colloidal particles cannot pass through them. The filter papers thus prepared are known as ultra-filters. By soaking filter paper in solutions of collodion of different concentrations a series or graded filter papers can be prepared. Even the finest filter paper will allow the passage of crystalloids through them. Graded ultra-filters permit the separation of colloidal particles of different sizes from one another.

14.16 PROPERTIES Of cOllOIDAl SOluTIONS The properties of colloidal solutions are best considered under a number of specific heads, namely

Physical properties Colligative properties Optical properties Kinetic properties Electrical properties.

14.16.1 Physical Properties (i) Physical properties: The physical properties of sol depend on whether the sols are lyophobic or lyophilic. For dilute lyophobic sols such properties as density, surface tension, and viscosity are not very different from those of medium. On the other hand, lyophilic sols show a high degree of solvation and as a result the properties of medium are modified. For example, the viscosity is much higher for the sols than for the medium. (ii) Heterogeneous nature: Colloidal particles are larger than molecules. They form heterogeneous mixture consisting of dispersed phase and dispersed medium. (iii) Non-setting: Sols are stable systems. Suspended particles remain suspended indefinitely. In London Museum, a gold sol made in 1857 by Michael Faraday has not settled down upto this date. (iv) Filterability: Colloidal particles can easily pass through the ordinary filter papers but it can be filtered by means of ultra filter paper. (v) Visibility: It is not possible to see colloidal particles with naked eye or with the help of microscope. The size and shape of the colloidal particles can be determined (i) by scanning electron microscope (SEM) (ii) transmission electron microscope (TEM) and (iii) scanning transmission electron microscope (STEM).

14.16.2 colligative Properties Sols exhibit colligative properties, but the effects observed are very much smaller than ordinary solution. The effects are practically negligible with the exception of osmotic pressure. The reason for this lies in the difference in particle size of two types of dispersions. For example, if 0.01 mol of solute is dissolved in 1000 g of the solvent to form true solution the total number of particles will be 0.01 N where N is the Avogadro’s number. But if the same substance is colloidally dispersed in the same amount of solvent to form aggregates containing 100 molecules per particle the number of particles present will be 1/100 of that present in the true solution. Therefore the colligative effect is reduced by the same amount. Osmotic pressure is larger in magnitude than other colligative properties therefore it shows some appreciable value.

14.34

Surface Chemistry

14.16.3 Optical Properties (i) Tyndall effect: It has been noted that when a beam of light enters a dark room it lights up the dust particles floating in the air. In the same way when a strong beam of light is made to fall on a colloidal solution the path of the beam is illuminated by a bluish light. In 1869 Tyndall found that the path of a beam of light is seen when passed through a colloidal solution but not through a true solution. The luminosity by the path of a beam is known as Tyndall effect. The visibility results from the scattering of light from the surface of the colloidal particles. The scattering of light cannot be due to simple reflection as the size of the particles is smaller than the wavelength of visible light. This is why the colloidal particles cannot be seen under the microscope. In this phenomenon, the particles become self luminous due to the absorption of light energy and then scatter light of shorter wave length. In case a convergent beam of light is made to fall upon a solution by means of a lens a bright blue cone is observed. It is called Tyndall cone. The Tyndall effect is used in establishing the heterogeneity of colloidal solution and it is good example of the optical properties of colloids. eye Microscope Tyndal cone

Light source

Tyndal effect can be observed during the projection of picture in the cinema hall due to scattering of light by dust and smoke particles present there. This effect can be used to distinguish between a true solution and a colloidal solution. (ii) Colour: The colour of the sol depends on (a) Wavelength of the light scattered by the sol. (b) Size and shape of the sol particle. (c) Specific selective absorption power of the dispersed phase and dispersed medium. (d) The way an observer receives the light. In the case of red or blue gold sols, some of the light is absorbed as well as scattered and more than fifty years ago it was shown that the colour is related to the size and shape of the particles. In recent years, light scattering (the variation of the intensity of the light with angle of scatter is measured) has become an important method for determining the size and shape of particles in colourless sols e.g., sulphur and more particularly the high polymer molecules form colloidal solutions. The Tyndall effect has been used by Zsigmondy in the ultra microscope. A powerful beam of light concentrated by a system of lenses is passed through the solution and the beam viewed through a microscope the axis of which is at right angles to beam. Spots of light (not the actual colloidal particles for they are too small) are seen in continual movement. From the number of light spots in the field of vision depth of liquid concentration and density of solution Zsigmondy was able to calculate the diametre of the colloidal particles. The size and shape of colloidal particles have more recently been investigated by means of the electron microscope. The ultra microscope has also been used to study aerosols, i.e., smokes. The effect of particle size on colour of the sol can be observed from the experimental data of silver sol. Table 14.13 Colour changes in silver sol

Colloidal solution

fig 14.13 Tyndall cone

Diametre of particles (metre) –8

Tyndall effect is observed when the following two conditions are satisfied: 1. The refractive indices of the dispersed phase and dispersion medium must differ greatly in magnitude. In dilute solution, scattering depends on the size of the colloidal particles present and on the refractive index ratio; m = n/n0 where n is the refractive index of the particles dissolved and n0 that of medium. The scattering is more pronounced when larger size of the particles are present and higher the value of m. 2. The diametre of the dispersed particles is not much smaller than the wave length of the light used.

6 ×10 9 ×10–8 13 ×10–8 15 × 10–8

Colour of silver sol Orange-yellow Orange-red Purple Violet

When the light is passed through milk and water mixture, it appears blue when it is viewed in the same direction as that of light passing but looks red when viewed at right angles. This is due to the way an observer receives the light. Some examples of Tyndall effect are (i) Blue colour of sky and sea water. (ii) Visibility of tails of comets. (iii) Twinkling of stars.

Surface Chemistry 14.35

14.16.4 Kinetic Properties

14.16.5 Electrical Properties

Brownian movement: In 1827, the botanist RobertBrown noticed that pollen grains in aqueous suspension were in continual motion. After the discovery of ultra microscope, it was observed that suspended particles in colloidal solutions are also in constant rapid zig-zag motion which is known as Brownian movement named after its discoverer.

Electrophoresis: When a colloidal solution is placed in an electric field the particles being electrically charged move towards one or other electrode. The migration of electrically charged sol particles under an applied electric field is called electrophoresis or strictly cataphoresis or anaphoresis according to the electrode to which the particles move. Reservoir Cathode

fig 14.14 Brownian movement of colloidal particle Cause of Brownian movement: The kinetic theory of molecular motion furnishes a nice explanation of this phenomenon. The Brownian motion of sol particles is due to the bombardment by the molecules of the medium surrounding the particle. This colloidal particle acquires the

(

Anode

Water despersion medium

initial level

)

same kinetic energy ≈ 3 KT as all other molecules but 2 due to its heavy mass its velocity will be much less as compared to that of the molecule of the medium. The motion can be seen under suitable microscope. The movement slows down with the increase in the size of the particles. With increase of particle size the probability of unequal bombardment diminishes. The average number of collisions per unit area from all directions is the same with no net transfer of the momentum and hence they do not exhibit Brownian movement. Brownian movement of colloidal particles is useful in following respects. (i) Confirmation of kinetic theory: It is a direct demonstration of ceaseless motion of molecules as pictured by kinetic theory. (ii) Determination of Avogadro’s number: Einstein developed an equation based on the Brownian movement measurement. This equation helps us to calculate the number of particles in a given mass of colloidal particles from which the Avogadro’s number can be calculated. (iii) In stabilizing colloidal solution: The rapid and random movement prevents the setting of colloidal particles by gravity and thus helps in stabilizing colloidal solution to some extent.

Stop cock

fig 14.15 Electrophoresis of colloidal solution The apparatus consists of a U-tube provided with two platinum electrodes and a stop-cock through which it is connected to a funnel shaped reservoir. A small amount of water is first placed in the U-tube and a reasonable quantity of the sol is taken in the reservoir. The stop-cock is then slightly opened and the reservoir is gradually raised so as to introduce the sol in the U-tube gently. The water is displaced upwards producing a sharp boundary. When an electric current of 50-200 volts is passed by connecting the electrodes to the terminals of a battery the colloidal particles move towards the electrodes. The movement of the colloidal particles can be followed by observing the position of the boundaries by means of a naked eye or a lens. When the particles are negatively charged (as in the case of arsenic sulphide sol), the boundary on the cathode side is seen to move down and that on the anode side to move upward showing that the particles move towards the anode. Thus by noting the direction

14.36

Surface Chemistry

of motion of the particles in an electric field it is possible to determine the sign to the charge on the colloidal particles. With an iron (III) oxide sol, such an experiment shows that all the particles move towards the negative electrode showing that all the particles carry a positive charge. A particular type of particle does not necessarily always carry a charge of the same sign as it often depends on the way in which the sol has been prepared. For example, a silver chloride sol made by adding silver nitrate solution to a solution of chloride there being a slight excess of chloride ions, is negatively charged; but if a slight excess of silver nitrate is now added the sol becomes positive. Electrophoresis experiments have established the fact that the particles in a stable sol always carry a charge in fact without this charge the stability is lost. The question now arises how the charge is acquired.

(iii) Due to preferential adsorption of ions: The particles adsorb preferentially positive or negative ions present in the medium. When two or more ions are present in the dispersion medium, preferential adsorption of the ions common to the colloidal particle usually takes place. This can be explained by following examples as. If a dilute solution of AgNO3 is added to a slight excess of KI solution a negatively charged sol of AgI i.e., (Ag I) I−: K+ is obtained. But when a dilute solution of KI is added to a slight excess of AgNO3 solution a positively charged sol of AgI is formed. In this sol, Ag+ is preferentially adsorbed forming a sol as (AgI) Ag+ : NO3− . However, no sol is produced by adding equivalent amounts of KI and Ag NO3 solutions.

14.16.6 Origin of charge

+ +

As seen above the colloidal particles carry either positive or negative charges and all particles of the same colloid carry the same charge. The charge on the particle is due to one or more of the following reasons. (i) Due to self ionization: When colloidal particles such as soaps and detergents are dissolved in water ionized molecules associate to form a micelle. The outer surface will thus be charged depending upon the charge of the ions from which it is formed. Thus sodium palmitate solution will have negative charge on sol particles. (ii) Due to direct ionization of surface groups: Proteins contain many ionisable acidic and basic groups. Thus the particles of protein sol can either have positive charge or negative charge depending upon the pH of the medium. A protein molecule has a carboxylic group and a basic NH2 group. A general formula of protein can be written as COOH − X − NH2. How the charge on the protein particles vary with the change in pH of medium can be shown as follows. COOH ×

+ HCl

COOH

NH2

×

+

Cl–

(Acidic medium)

+ NH3 Positively charged

NH2

×

COOH

+ NaOH

COO– ×

+

NH2 Negatively charged

Na

+

( Alkaline medium )

+

+

-

-

-

+

-

+

+ +

-

-

-

+

+

-

Ag I

-

+

-

-

+

-

+ -

(b)

-

+

+

-

-

+

+

-

- = I-

-

Ag I

+ = K-

(a)

+

+ -

+

-

-

-

+

+

-

-

+ = Ag + - = NO 3

fig 14.16 Silver iodide sol particle (a) In the presence of excess of potassium iodide solution (b) in the presence of excess of silver nitrate solution If FeCl3 is added to excess of hot water and dialyzed a positively charged sol of ferric hydroxide is formed. Fe3+ is preferentially adsorbed on sol particles as (Fe2 O3· XH2O) Fe3+ : 3Cl− When FeCl 3is added to NaOH with constant stir ring a negatively charged sol as (Fe 2O 3 xH 2O)OH −: Na + is obtained where x H 2O indicates that Fe 2O 3 is hydrated. In the case of silver halides, it is believed that the charge on the particle originates inside the crystal surface. When for example, silver bromide particles are in an electrolyte for containing a high concentration of bromide ions the lattice has certain silver ions missing (it is said to be a ‘defect’ structure) and so has a net negative charge. In the presence of excess of silver ions, however, the lattice contains an excess of these ion which are present interstitially between the existing ions as shown in the Fig 14.17.

Surface Chemistry 14.37

Br Br -

Br Br -

Br -

- + - + - × - + - + - + - + - × x = Missing Ag+ ions

Ag +

Ag + Ag +

-

-

Ag + Ag +

+ - + -+ + -+ + + - + - + - + + = Interstitial Ag+ ions

fig 14.17 Silver bromide: electrolyte and crystal (iv) Due to the nature of colloid formation: If the particle is made up of neutral molecules e.g., a particle of a solid hydrocarbon or a particle of sulphur then it acquires its charge by adsorbing ions from the dispersion medium on its surface. Thus the hydrocarbon if dispersed in pure water might adsorb hydroxyl ions and become negatively charged, or hydroxonium ions H3O+ and become positively charged. A sulphur sol if prepared from sulphide might adsorb sulphide ions or H3O+ if obtained by addition of acid to a thiosulphate, sulphite or thiosulphite ions. In the case of colloidal metals such as gold and platinum the adsorption is more like chemisorptions; thus chloride ions attach themselves to a gold particle to form negatively charged complex gold ions on the surface, such as AuCl2− ; similarly hydroxyl ions (which are necessary) to obtain a platinum sol by Bredig’s method from hydroxylcomplexes with the platinum. Colloidal particles also acquire charge due to electrification of particles by friction due to electron capture by the sol particles e.g., during electro dispersion of metals. Based on the nature of the charge the colloidal solutions have been classified into positively charged and negatively charged colloids. Some common sols with the type of charge on their particles are given in table 14.14 Table 14.14 Some common sols with their charges Positively charged i. Oxides, e.g., TiO2 sol ii. Metallic hydroxides e.g., Fe(OH) 3, Cr(OH) 3 and Al(OH) 3 sols iii. Basic dyestuffs e.g., Methylene blue sol iv. Hemoglobin

Negatively charged i. Metals e.g., Copper, silver and gold sols ii. Metallic sulphides e.g., As2S3, CdS and Sb2S3 iii. Acid dye stuffs e.g., eosin, congo red iv. Sols obtained from gums, starch kaolin, charcoal etc.

14.16.7 Electrical Double layer Whatever the nature of the charge on the particle, it must be balanced by ions of opposite sign in the electrolyte. These ions form a diffuse layer round the particle. Helmholtz (1879) suggested that an electrical double layer of positive and negative charge is generally formed at the surface of separation between two phases. Consider solid-liquid surface where a solid is in contact with a liquid, a double layer of ions appears at the surface of separation. One part of this double layer is fixed on the surface of the solid. This is called the fixed part of the double layer. The fixed layer may consist of positive or negative ions. The second part of the double layer consist of a mobile layer of ions which extends into the liquid phase. The mobile layer consists of ions of both the signs but the net charge on it is equal and opposite to that on the fixed part of the double layer. Since separation of charge is a seat of potential the charges of opposite signs on the fixed and the diffused parts of the double layer results a difference in potential between these layers. This potential difference between the fixed layer and the diffused layer having opposite charge is called the Electro kinetic potential or Zeta potential and the double layer is called as Helmholz electrical double layer. The presence of equal and similar charges on colloidal particles is largely responsible in providing stability to the colloidal solution because the repulsive forces between charged particles having same charge prevent them from coalescing or aggregating when they come closer to one another. As the solid particles and the liquid carry opposite charges it is clear that on the application of electric field, the colloidal particles and the liquid (dispersion medium) will move in opposite directions. The electrophoretic mobilities of different colloidal substances are different. This properties help to separate different colloidal particles from their mixture. Proteins, polysaccharides etc, can be separated from their mixture on the basis of their different electrophoretic mobilities.

14.38

Surface Chemistry

Fixed layer Fixed layer -

+ + +

+

-

+

+

+ +

-

-

-

+

+ +

-

-

-

+ -

+ Diffused layer

{

+

+

-

+ +

-

-

+ +

+ -

+

+

+ -

-

-

-

+ +

-

+ +

-

-

-

+

+

-

-

-

Colloidal Particle Solid

+ -

-

+

+

-

+

- +

+

-

+

Colloidal Particle Solid

-

+

-

+

+ -

-

-

-

{

+ +

Diffused layer

(a)

(b) fig 14.18 The electrical double layer

If experiment is arranged in such a way that particles of solid (colloidal) can move but not the medium then the phenomonon is called electrophoresis. But if the experiment is arranged in such a way that the particles of the medium may move but not the colloidal particles, then the phenomenon is called electrosmosis. Isoelectric point: Hardy noticed that the speed of negatively charged sol of albumin decreased on the addition of an acid. On gradually adding this acid, it was observed that the direction of cataphoretic motion got reversed. It indicates reversal of charge on the sol particles from negative to positive, it is reasonable to believe that there would be a concentration of acid at which the sol particle will have no charge. In general, proteins exhibit cataphoresis but the direction of movement would depend on pH of the dispersion medium.At a particular pH, the sol particles become neutral and exhibit no movement in an electric field. This is called the isoelectric point of the sol. Albumin from horse serum has isoelectric point of pH 4.9. At the isoeletric point, the electrical conductivity, solubility and viscosity of protein sols become minimum.

14.16.8 Electrosmosis When electrophoreses of dispersed particles is prevented by suitable means, the medium can be made to move under the influence of an applied potential. This phenomenon is referred to as electro osmosis. A simple apparatus is shown in Fig 14.19.The colloid is placed in the compartment separated by membranes.

Outside the membranes, water is filled. When potential is applied across the two electrodes placed close to the membranes, the liquid level is observed to fall on one side and rise on the other owing to passage of water through membranes. The direction of flow of water depends on the change of the colloid.

Water

–

+ Electro osmosis

fig 14.19 Electro osmosis For positively charged sols, the medium is negatively charged, and hence the flow will take place from negative compartment to positive compartment. For negatively charged colloids, the reverse will be true and the level on the capillary of negative electrode compartment will rise. By measuring the difference in water level (h) the pressure (p) due to electro osmosis can be calculated, i.e., ρ = hdg where d is the density of the particles and g is the acceleration due to gravity. Application of electro osmosis: This phenomenon has got a number of technical applications such as (i) in

Surface Chemistry 14.39

drying pastes and slurries (ii) in tanning of hides and similar products (iii) in the removal of water from moist clay etc. (IV), in the preparation of pure sols, i.e., colloidal sol of silicic acid of low molecular weight may be prepared.

14.17 STAbIlITy Of cOllOIDS A question of considerable importance and interest arises that why are lyophobic and lyophilic sols are stable? The answer is that lyophobic colloids are stable due to their electric charge and for lyophilic sols the stability is due to their charge and extensive solvation.

Colloidal Particle

-

Lyophilic sol particle

-

-

-

+

-

+

-

+

+

or

Colloidal Particle -

Solvent layer

-

+ +

Colloidal Particle +

-

+

+

+ +

Lyophobic sol particles

fig 14.20 Stability of Sols It is clear that the similar charges on colloidal particles of lyophobic colloids will show electrostatic repulsion, prevent them from coalescing or aggregating when they come closer to one another thus the lyophobic colloids are stable. Lyophilic colloids are very much hydrated or solvated by the solvent molecules. It is assumed that the colloidal particles are covered by sheath of the liquid in which they are dispersed. The sheath acts as barrier which prevents the colloidal particles to come together and form bigger aggregates. This fact is supported by the high viscosity of the lyophilic colloids. Sometimes stabilizers are used to stabilize the sols. Then what are these stabilizing agents? Colloidal particles almost invariably hold on their surfaces these stabilizing substances attachable on one hand and attachable to molecules of the medium on the other. Such intermediates also serve to prevent the particles from coming nearer enough to each other. These substances are called stabilizers. The stabilizing action is due to one of the following reasons. (i) It may decrease the surface tension between the immiscible liquids and bring them into colloidal form. (ii) It may produce charge on the neutral atoms or molecules as a lyophobic sols and cause repulsion for example. Cl–in gold sols. (iii) It may convert immiscible liquids into emulsion by getting preferentially wetted in one. In the preparation of water-kerosene emulsion, soot acts as a stabilizer.

14.17.1 coagulation Coagulation or flocculation is the process of breaking up of collidal solution by which the collidal particles come close and result in the precipitation of the dispersed phase. It may be noted here that a colloid can precipitate by growth as well as by coagulation. In a silver halide sol, there are both small and very small particles; if the sol is left for some time, the smaller particles dissolve and crystallize out on to the larger, thus the sol coarsens and may in time precipitate. This is known as “ageing” process. To bring about precipitation of a dispersed phase, conditions must be created which are opposite to those which lead to stability for sols. This means that the charge on the particles of lyophobic sols must be removed while with lyophilic sols removal of charge (if present) and the adhering layer of solvent is necessary. Coagulation of lyophobic sols: The stability of lyophobic sol is due to the electrical double layer. When the charges on the particles are neutralized the zeta potential around the particle drops. This reduces the electrical repulsion between the particle and thus the forces of attraction between the particles (Which are Van der Waals forces like those between noble atoms) bring them together thus causing coagulation or flocculation of the sol. The removal of charges can be done by different ways: (a) By electrolysis: In cataphoresis, the particles of the dispersed phase move toward the oppositely charged electrodes. When it comes in contact with such electrode for long, it is discharged and precipitated. (b) By mutual precipitation: When oppositely charged sols, e.g., positively charged Fe(OH)3 and negatively charged, As2S3 are mixed in almost equal proportions, their charges are neutralized with each other. Both sols may be partially or completely precipitated. (c) On boiling the sol: Precipitation of a colloid can occasionally be effected by boiling the sol. In boiling the sol the amount of electrolyte adsorbed by the sol is reduced and if this reduction is sufficient the sol precipitates. (d) On freezing the sol: Freezing the sol results in removal of medium. If it is carried for enough time there may not be enough medium left to keep the sol stable. (e) By persist dialysis: On prolonged dialysis, the traces of electrolytes present in the sol are removed and the colloid becomes unstable. The particles of red colloidal gold sol turn blue if electrolyte is added and finally the gold is precipitated as a dark-blue solid. Here the blue colour is due to the larger particles formed on coagulation. If a red sol is dialyzed it also becomes blue,

14.40

Surface Chemistry

does not precipitate. In this case, the blue colour is produced by the process.

Table 14.15 Flocculation values of common electrolytes.

O + O → OO red

red

Blue

which produces a non-spherical particle, coagulation does not proceed further. (f) By addition of other electrolytes: Addition of electrolytes readily brings about the coagulation of a colloidal solution. A few drops of CaCl2 solution can coagulate As2S3 sol. The explanation is very simple. The particles of As2S3 sols carry negative charges and are kept apart due to mutual electrical repulsion. Calcium chloride in solution produces Ca2+ ions and Cl− ions. As2S3 Particles being negatively charged will attract and adsorb some of the Ca2+ ions and with the result their negative charges get neutralized. The sol coagulates. Hence positive ions precipitate negative charged sols and negative ions precipitate positively charged sols. Coagulation will be most rapid at the isoelectric point where there is no charge left and it is interesting that same rate law applies to coagulation of a sol at its isoelectric point and to the coagulation of a smoke, where the particles are uncharged (In both cases the coagulation is followed by counting the particles with the ultra microscope).

14.17.2 hardy–Schulz law Since it is the charge on the ions added is responsible for the precipitation of colloidal solution the power of an ion to coagulate a sol depends on its velocity. The greater the charge of the ion added, the greater its power to cause precipitation. The quantity of the electrolyte required to coagulate a fixed amount of sol depends upon the valency of ion bearing a charge opposite to that of colloidal particle. This is known as Hardy-Schulz law. It has been observed that in the coagulation of As2S3 sol the flocculating power of cations is in the order. Al3+ > Ba 2 + > Na + Similarly, in the coagulation of Fe (OH)3 sol the coagulating power of anions is in the order: 4-

 Fe ( CN )6  >PO3-4 >SO 2-4 >Clflocculation Value

The minimum concentration of an electrolyte which is able to cause coagulation or flocculation of sol is termed as flocculation value. It is usually expressed in terms of milli moles of electrolyte layer per litre of the sol. The flocculation value of some common electrolytes for As2S3, Fe(OH)3 Fe2O3 sols is given in Table 14.15.

Colloidal solution As2S3

Charge on Milli moles of colloidal electrolytes for particles Electrolyte precipitation Negative

Fe(OH)3 or Fe2O3. x H2O Positive

NaCl KCl MgSO4 CaCl2 AlCl3 Al(NO3)3 KCl KBr KNO3 KBrO3 K2SO4 K2CrO4 K3[Fe(CN)6]

58 50 0.81 0.65 0.093 0.095 102 134 132 133 0.21 0.325 0.096

14.17.3 coagulation of lyophilic Sols Lyophilic sol is stabilized by two distinct factors – its charge and its hydration. Lyophilic colloids may or may not have electrical character and it would seem that particle charge is not an important factor contributing to the stability of such systems. Small amounts of electrolytes e.g., a little salt added to an albumen solution have no effect. High concentrations do however cause coagulation. This process is known as 'salting out' and is possibly quite different from the coagulation effect of an ion on a lyophobic sol. Here, the ions added in large concentration may remove the layers of solvent molecules surrounding the colloidal particle (dehydration by hydration of ions of salt) layers which have contributed to the original stability. Removal of these layers of water molecules (in hydrophilic sol) may be achieved by adding a dehydrating agent e.g., alcohol; by adding about ten parts by volume of alcohol to one part of a concentrated solution of calcium acetate a solid gel (solid alcohol) is formed. The salting out effect depends upon the nature of the ions and the salts of a given metal. A series is set up in the order of their effectiveness. It is called Holfmeister series or Lyotropic series. The order of cation is Mg 2+ >Ca 2 + >Sr 2+ >Ba 2 + >Na + >K + >Rb + >Cs + And the order of anion is (Citrate)3− > ( tartrate)2 − >SO24 − > CH3COO − >Cl − >NO3− >ClO 3− >I − >CNS−

Surface Chemistry 14.41

Protection of colloid

14.17.5 gold Number

It has been observed that lyophobic sols are easily flocculated in comparison to lyophobic sols on addition of small amounts of electrolytes. When the larger amount of lyophilic sol is added to lyophobic sol, it becomes less sensitive to coagulation. It is said that lyophobic sol is protected by lyophilic sol and it has becomes more stable. For example, the coagulation of gold or silver sol can be prevented by addition of lyophilic sols like gelation or albumin. Thus the phenomenon of protection can be defined as The process by which the sol particles are prevented from coagulation by electrolyte due to the previous addition of some lyophilic sol. Then how a sol is protected? The answer is that the sol particles adsorb the protective colloid particles and form a protective layer above them. This layer prevents the precipitating ions from reacting the sol particles which in this way are preserved from precipitation Lyophobic sol particle (Protected particle)

The protective action is compared in terms of Zsigmondy’s Gold number. It is defined as The number of mili grams of protective colloid which just prevents the coagulation (accompanied by a change in colour from red to blue) of 10 mL of given gold sol when 1 mL of 10% solution of NaCl is added to it. Therefore, the smaller the gold number of lyophilic colloid, the greater is its protective power.

Lyophilic sol (Protective colloid)

Table 14.16 Gold numbers and their reciprocals Lyophilic sols Gelatin

Gold Number

Reciprocal

0.005-0.01

200-100

Caseinate

0.01

100

Hemoglobin

0.03

33

Gum arabic

0.15

6.6

Sodium oleate

0.4

2.5

Gum tragacanth

2

0.5

Potato starch

25

0.04

Egg albumin

0.08-0.10

12.5-10

Gelatin and starch have maximum and minimum protective power

According to a recent view, the sol becomes stable merely due to mutual adsorption of their particles no matter whether the sol particle is adsorbed on the protecting colloid or protecting colloid is adsorbed on the sol particle. But it is assumed that if the protected particles are smaller than the protecting particle, the former will be adsorbed by the latter (Fig 14.21 a) and if the protected particles are bigger than the protecting particles the later will be adsorbed by the former (Fig 14.21 b).

congo Rubin Number Ostwald proposed that it is the amount of protective colloid in mg which prevents the colour change in 100 mL of 0.01% congo rubin dye solution to which 0.16 g equivalent of KCl added when observed after 10-15 min.

Protecting particle

Protected particle Protected particle Protecting particle (a)

(b) fig 14.21 Protection of a sol

14.42

Surface Chemistry

14.18 EmulSIONS

Emulsifying Agents

A colloidal system involving one liquid dispersed in another is known as emulsion. Any two immiscible liquids can form an emulsion. These emulsions are generally unstable and form two separate layers on standing for some time. They can be stabilized by a third substance known as emulcifier or emulsifying agent.

The emulsifying agents are classified as follows: (i) Agents for oil/water (O/W) emulsions: The principle O/W promoting agents are the hydrophilic colloids such as the proteins, gums, carbohydrates, natural and synthetic soaps etc. The natural and synthetic soaps which are soluble in water form the most widely used O/W emulsifying agent. The natural soaps are chiefly the alkali metal salts of the fatty acids but also includes the bile salts (e.g., sodium oleate and glycooleate). (ii) Agents for water/oil (W/O) emulsions: The principal W/O promoting agents are all members of the hydrophobic class of colloids as for example, the heavy metal salts of the fatty acids, long chain alcohols, soot or lamp black etc. Of the heavy metal salts of fatty acids, those in common use are the oleates and stearates of magnesium, aluminum and calcium etc.

Types of Emulsions (a) Oil in water type: In this type, oil droplets are dispersed in water which form a continuous phase, for example, phenol and cresol used as disinfectants are emulsions of this type. More examples are milk and vanishing creams. Milk is an emulsion of fat in water. The emulsifying agent is protein called casein. (b) Water in oil type: In this type, water droplets are dispersed in oil which form a continuous phase. Cod liver oil is an emulsion of water in oil type in which caseins, gums and egg yolk have been added as emulsifying agent.

oil oil

H2 O

H2 O

oil

H2 O

H2 O

oil oil

H2 O

H2 O

oil oil

oil in water

H2 O

water in oil

fig 14.22 Emulsion Table 14.17 Comparison in the properties of two types of emulsions. Oil in water type

Water in oil type

1. Dilution test: This type of emulsions can be diluted with water. No separate layer results. 2. Dye test: When an oil soluble dye is added to the emulsion no coloured droplets are seen with a microscope. 3. This type of emulsion is less viscous and its electrical conductivity is more.

It cannot be diluted with water. On dilution a separate layer results.

4. This type of emulsion can be spread on the surface of water.

On adding an oil soluble dye coloured droplets are seen with a microscope. This type of emulsion is more viscous and its electrical conductivity is less. It cannot be spread on the surface of water.

Preparation of Emulsions The preparation of emulsions is carried out by the following methods: (i) Agents in oil methods: The emulsifying agent is dispersed in oil and water is added drop by drop which yields emulsion of W/O type. (ii) Agent in water method: The emulsifying agent is dissolved in water and oil is added drop by drop which results the O/W type emulsion. (iii) Nascent soap method: The emulsions which are stabilized by soap are prepared by this method. The fatty acid part is dissolved in oil and the alkaline part in water. When these two are mixed the formation of soap occurs which stabilizes the emulsion.

14.18.1 Theories of Emulsification One type of emulsifier decreases the interfacial tension between the two liquids. For example, soap decreases the interfacial tension and gets concentrates at the surface. Hence the liquid whose surface tension with soap film is greater, rolls in water type emulsion. Calcium oleate and magnesium oleate cause lowering of interfacial tension at oil soap film interface and thus form oil water type emulsion. (b) Certain finely divided sols which are unequally wetted by two liquids also act as emulsifying agents. The surface of the solid in contact with the other liquid forms the internal phase. Soot particles are wetted more by kerosene oil than with water. Thus it produces an emulsion of water in kerosene.

Surface Chemistry 14.43

Adsorbed film theory: Emulsifiers of detergents type, for example, soaps usually consist of a hydrocarbon chain and a polar group when they are added to an oil-water system the hydrocarbon chain will attach itself to oil while the polar group attracts the water. This means that the soap tends to concentrate at the interface between the oil and water and form a coherent film. This film will have two interfaces one between the oil and soap, and other between soap and water. Soaps of monovalent cations are readily dispersed in water but not in oil, form a film. Film [Emulsifying agent]

oil

Water

fig 14.23 Consequently, the surface tension is lower in water side than the oil side. Since the area of inside phase of the film surrounding a sphere is smaller than that of outside phase, the film tends to curve so that it encloses globules of oil in water. On the other hand a film composed of soaps of di and trivalent cations is freely dispersed in oil but not in water side and the film tends to curve in a such a manner as Paraffin chain

to enclose the globule of water in an outer and continuous oil phase.

Oil

Water

Oil

Water

Oil

Water

14.24 Formation of emulsion The oriented wedge theory: The orientation concept leads to a simple explanation of the effect of emulsifying agent on emulsion type. In this theory the geometrical consideration plays a part. Reference to the Fig 14.25, portraying situation in an emulsion stabilized by a bivalent soap gives a clear idea of how geometry virtually requires the formation of water-oil type of emulsion. An examination of the figure shows that if the cross sectional area of the terminal atom of the polar group (in this case the metal atom) is greater than the hydrocarbon chain the water surface of the absorbed film necessarily be greater than the oil surface, i.e., the water will surround the oil and an oil-water emulsion is obtained. Conversely, if the cross-sectional area of the metal atom is less than that of hydrocarbon chain a water-oil emulsion is formed. Further, the type of emulsion formed will depend on the size of metal atom and also upon its valency since an atom with valency greater than one will have more hydrocarbon chain attached to it. Properties of Emulsions Emulsions are usually opaque owing to the difference in refractive index between the two liquids. The particles tend COOH group

COOH group

Paraffin chain

oil

Water

W/O type

O/W type

fig 14.25 Emulsification

14.44

Surface Chemistry

to be larger than those in sols and are at the upper end of the colloidal size range. A certain number of large particles are frequently present in an emulsion and if it is of the oil in water type they will tend to rise to the top since they are lighter than water. This is termed creaming and occurs in milk. The large particles can be broken up by the process of homogenisation in which the emulsion can be made with as much as 90% of the volume occupied by the disperse phase; by comparison a lyophobic sol is very dilute (usually less than 1%). Emulsions show all the properties of colloidal solution. uses of Emulsion 1. Concentration of ores: In the froth flotation process, the finely powdered ore is treated with an oil emulsion. On passing air, it foams in such a way that the particles of the ore are carried to the surface and then collected. 2. In medicines and cosmetics: Various pharmaceuticals, lotions, creams and ointments are nothing but emulsions. Only drugs are mostly available in the form of emulsions. 3. Digestion of fats: The digestion of fats in the intestines is helped by emulsification. Demulsification It is defined as a process which involve the breaking down of an emulsion to yield constituent liquids. The process of demulsification can be brought about by the following physical methods 1. By boiling or freezing or centrifuging the emulsion, the demulsification results. Cream is separated from the milk by the centrifugation method. 2. By adding a substance which destroys the emulsifier 3. Water in oil type emulsion can be demulsified by adding a dehydrating agent in it.

14.19 ASSOcIATED cOllOIDS When molecules or ions that have both a hydrophobic and a hydrophilic end are dispersed in water they associate or aggregate to form colloidal-sized particles or micelles. A micelle is a colloidal–sized particle formed in water by the association of molecules or ions that each have a hydrophobic end and a hydrophilic end. The hydrophobic ends point inward toward one another and the hydrophilic ends are the outside of the micelle facing the water. A colloid in which the dispersed phase consists of micelles is called an associated colloid.

Ordinary soap in water provides an example of an association colloid. Soap consists of compounds such as sodium stearate C17H35 COONa. The stearate ion has a long hydrocarbon end that is hydrophobic (because it is non polar ) and a carboxyl group COO− at the other end that is hydrophilic (because it is ionic). CH 3CH 2CH 2CH 2CH 2CH 2CH 2CH 2CH 2CH 2 Hydrophobic end

CH 2CH 2CH 2CH 2CH 2CH 2CH 2 C

O O–

Hydrophilic end

Stearate ion

In water solution, the stearate ions associates into micelles in which the hydrocarbon end point inward toward one another and away from water and ionic carboxyl group are on the outside of the micelle facing the water (Fig 14.26). The cleaning action of soap occurs because oil and grease can be absorbed into the hydrophobic centres of soap micelles and washed away. Synthetic detergents also form associated colloids. Sodium lauryl sulphate is a synthetic detergent present in toothpastes and shampoos. CH 3CH 2CH 3CH 2CH 2CH 2CH 2CH 2CH 2CH 2CH 2CH 2SO3− Na +

Sodium lauryl sulphate The micelle particles are pegged into the grease or dirt and converts the grease or dirt in colloidal form (emulsion in the case grease or oil) and removes from the cloth or body Some other examples of micelle system are Cetyl trimethyl ammonium bromide CH 3 ( CH 2 )15 N + (CH 3 )3 Br −

p-Dodecyl benzene C12 H 25 sulphonate

O

SO3− Na +

Surfactants: Surfactants are those substances which are preferentially adsorbed at the interface like air – water, oil –water and solid –water interface. Thus it is the surfactant which is responsible for micellisation and emulsification. Surfactants are divided into three categories: (i) Cationic surfactants. These on ionization give a cation having hydrophobic and hydrophilic group e.g., CH3

CH2(CH2)13CH3 N

CH3

CH2(CH2)13CH3

Hydrophobic end C16 H13 Hydrophobic end

Hydrophilic end N + Cl– Hydrophilic end

Surface Chemistry 14.45

Na+

Na+ Na+

C

C

O

–

O

O–

O

C

O

Na+

Na+

O–

Na+

Na+

Na+

O

O–

C

–

O

C

O

Na+

Na+

Na+ C

O

C

O–

Na+

O–

C

O

O

O–

Na+

Na+

Na+

Water Fig 14.26 A stearate micelle in a water solution

fig 14.27 Cleaning action of soap

Na+

14.46

Surface Chemistry

(ii) Anionic surfactants: These are already described such as sodium palmitate, sodium stearate, sodium oleate and salts of sulphonic acid have the molecular formula. Cn H 2n+1SO3 M where M = Na + , K + , NH +4 etc (iii) Non ionogenic surfactants: These do not ionize or dissociate in aqueous medium but these molecules also have hydrophobic and hydrophilic end. Alcohols having high molecular weights add to several molecules of ethylene oxide to form hydroxy surfactants. Cn H 2n+1OH+x CH 2 − CH 2 → Cn H 2n+1 (OCH 2 CH 2 ) x OH Hydrophobic end

Hydrophilic end

The formation of micelles takes place only above a particular temperature called Kraft temperature (TK) and above a particular concentration called critical micelle concentration (CMC). On dilution these colloids revert back to individual ions. For soaps, the CMC is 10–4 to 10–3 mol L–1. Many cationic detergents also have germicidal properties and are used in hospital disinfectants and in mouth washes.

14.20 gElS The colloidal system constituting the liquid as dispersed phase and the solid as dispersion medium is known as Gel. There are some sols which have high concentration of dispersed solid and change spontaneously into semisolid form on cooling. Such sols are called gels. Gelatin dissolves in water to form colloidal solutions and this solution on cooling sets into jelly. In fact, gels are formed by the interlocking of the particles of solid dispersion medium in the form of loose frame work inside which liquid is present. In certain conditions, the Lyophilic sols may be co-agulated to give a semi-solid jelly which encloses all the liquid present in the sol. The process of gel formation is called gelation. some common examples of gels are: gelatin, gum arabic, silicic acid, ferric hydroxide. Gels can be classified into two types: (a) Elastic gels: These are the gels which posses the property of elasticity. Such gels can change their shape on applying force and return to the original shape when the applied force is removed. Some examples of this type are gelatin, starch, agar-agar etc. (b) Non-elastic gels: Non-elastic gels are rigid and do not posses elasticity. An example is silica gel. It is important to note when the gel is allowed to stand for some time, it shrinks and loses the liquid present in it. This shrinking of gel is termed as synerisis or weeping of

the gel. Elastic gel on losing water become elastic solid. Such solids on mixing with water again get converted into gel. This is known as imbibation. Non-elastic gels on dehydration become glassy powder and this glassy powder converted back into gel by adding water. Some gels (elastic which are partially dehydrated) swell when dipped in water. It is due to the increase in volume of the gel and this process called swelling. Some gels get converted into liquid on shaking and reset when allowed to stand. This transformation (sol – gel) is known as thixotropy. The softening behaviour of marshy land under trading is due to thixotropic behaviour of bentonite clay present in such soils.

14.21 APPlIcATIONS Of cOllOIDS There are many applications of colloids in daily life research and industry. These are mainly two types 1. Natural applications 2. technical applications 1. Natural applications (i) Blue colour of the sky: Colloidal particles scatter only blue light. The dust particles suspended in air scatter blue light which reaches our eyes and thus the sky looks blue to us. (ii) Food: Many of our food are colloidal in nature Milk, starch, proteins, fruit jellies etc., are colloids. (iii) Blood: Blood is a colloid solution of an albuminoidal substance. The styptic action of alum and ferric chloride solution is due to coagulation of blood forming a clot which stops further bleeding. (iv) Agriculture: Soil is a mixture of clay organic material and salts. The study of colloids is very important in understanding the nature of soil, exchange of ions. Fertile soils are colloidal in nature in which humus acts as a protective colloid. On account of colloidal nature, soils adsorb moisture and nourishing materials. (v) Fog mist and rain: When a large mass of air containing dust particles is cooled below its dew point the moisture from the air condenses on the surfaces of the particles forming droplets. These droplet being colloidal in nature continue to float in the air in the form of mist or fog. Clouds are aerosols having small droplets of water suspended in air. On account of condensation in the upper atmosphere the colloidal droplets of water grow bigger and bigger in size till they come down in the form of rain. Some times, the rainfall occurs when two oppositely charged clouds meet. (vi) Formation of delta: River water is a colloidal solution of clay. Sea water contains a number of electrolytes. When river water meets the sea water,

Surface Chemistry 14.47

the electrolytes present in sea water coagulate the colloidal solutions of clay which gets deposited with formation of delta. 2. Technical applications (i) Medicines: A number of medicinal and pharmaceutical preparations are emulsions, i.e., colloidal in nature. Such medicines are easily absorbed by the body tissue and are therefore more effective. (ii) Smoke precipitation: Smoke in big cities create problem in industrial areas. These are dispersion of electrically charged colloidal particles in air and are removed from the air by electrophoresis principal. To achieve this, Cottrell devised an electrical precipitator. The smoke along with gases and dust are led through a chamber provided with a high voltage wire. The charged particles are discharged and fall down and gases free from smoke and dust pass up the chimney.

High voltage Flue or Stock

Smoke

Carbon particles

× × × × × × fig 14.28 Cottrell precipitator (iii) Sewage disposal: Water from sewage contains dust particles and many other suspended impurities. These impurities can be coagulated when water is allowed to flow between oppositely charged plates.

(iv) Water treatment: Water resources (rivers, lakes or canals) for city water supply contain a large amount of impurities. When alum is dissolved in such water, the coagulation of colloidal impurities take place since alum supplies Al3+ ions which discharge the negatively charged sol particles. The clear water is then decanted. (v) Tanning: Skin and hides are gel structures containing proteins in the colloidal state. Colloidal particles in hides are positively charged. On soaking the hides in tannin, which is negatively charged sol, mutual coagulation takes place. The process is called tanning. It imparts hardness to leather. (vi) Soaps and detergents: Soaps and detergents when dissolved in water form associated colloids. Their colloidal chemistry is very important for understanding detergent action. (vii) Dyeing: It is essentially an adsorption of dye by the colloidal particles of the mordant frequently used. (viii) Rubber industry: Rubber is obtained from trees as an emulsion of negatively charged rubber particles in water. On boiling, the protective layer of the protein is broken and rubber particles are coagulated. The coagulated mass is vulcanized by treating it with sulphur and sold as rubber. Various goods are rubber plated by making them anode in a plating bath when negatively charged rubber particles are deposited on them. (ix) Photographic plates: Sensitive emulsion deposited on photographic plates, films and paper contains fine silver bromide particles suspended in gelatin emulsion. (x) Artificial rain: Clouds consist of charged particles of water dispersed in air. Rain is caused by the aggregation of these minute particles. Bancroft succeeded in causing artificial rain by throwing electrified sand from an aeroplane and thus coagulating the mist hanging in the air. (xi) Chemical war fare: Smoke screens are used in war fare. It generally consists of very fine particles of titanium oxide dispersed in air and is ejected from aeroplanes. As titanium oxide is very heavy, the smoke screen is observed as a curtain of dazzing whiteness.

14.48

Surface Chemistry

KEy POINTS colloids and Their Types •

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According to Graham substances like sugar, urea, acids, bases etc are crystalloids (crystal like) and gelatin, albumin, glue, gums etc are colloids (glue like). Graham’s classification is not appropriate because crystalloids also form colloids. Any substances can be brought into the colloidal state by adopting suitable experimental conditions. If the particle size of the solute in the binary system is less than 1 mμ it is a true solution. If the particle size of the solute in the binary system is greater than 1 μ a suspension is formed. If the particle size of the solute in the binary system is in the range of 1 μ to 1 mμ a colloidal solution is formed. In true solution, the particles are either single molecules or ions, cannot be separated from solution by filtration and do not settle due to gravitational forces or by centrifugation. In colloidal solution, the particles are aggregates of the molecules or ions. Colloidal solutions are opaque. The particles from a suspension can be separated by filtration. The particle in a suspension settles down due to gravitational forces on standing. The properties of true solutions, colloidal solutions and suspensions are different due to the difference in their particle sizes. Ultra filters are the filter papers soaked in a solution of gelatin or collodion and subsequently hardened by soaking in formaldehyde. The filtration using ultra filters is called ultra filtration. Separation of particles from colloidal solution is not possible by ordinary filtration but can be separated by ultra filtration. Particles in colloidal solution settle on centrifugation. The particles in a true solution diffuse rapidly, in a colloid diffuse slowly but in suspension do not diffuse. The suspension and colloidal solutions are heterogeneous in nature while the true solutions are homogeneous in nature. Colloids and suspensions exhibit Tyndall effect but true solutions do not exhibit Tyndall effect. In the colloids, the phase constituting the colloidal particles is called dispersion phase while the medium in which the colloidal particles are distributed is called dispersion medium. The dispersion medium or dispersion phase may be solid or liquid. Depending upon whether the dispersed phase and the dispersion medium are solids, liquids or gases, eight types of colloidal systems are possible. The gas in a gas mixture is always a true solution.

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If the appearance of a colloidal solution is like a fluid, it is called a sol. Depending on the dispersion medium a sol may be hydrosol, benzosol, alcosol, etc. The colloids having more rigid structure are called gels. The dispersion medium in a colloid is named as lyo. Colloidal solutions in which the particles of the dispersed phase have great affinity for the dispersion medium are called lyophillic colloids. The colloidal solutions in which particles of the dispersed phase have no affinity for the dispersion medium are called lyophobic colloids. If water is dispersion medium lyophilic colloids are called hydrophilic colloids while lyophobic colloids are called hydrophobic colloids. Lyophilic colloids can be prepared by directly mixing the dispersion phase but lyophobic colloids cannot be prepared. The colloidal particles in lyophillic colloids do not carry charge but in lyophobic colloids carry charge. The Tyndall effect in lyophilic colloids is weak while in lyophobic colloid is strong. Lyophilic colloid particles are solvated while lyophobic colloid particles are not solvated. Lyophilic colloids are stable but the lyophobic colloids are stable only in the presence of protective colloids. Viscosity of lyophilic colloids is more when compared to lyophobic colloids. Lyophilic colloids are reversible while lyophobic colloids are irreversible. Lyophilic colloids cannot be precipitated or coagulated by the addition of electrolytes but lyophobic colloids are easily precipitated by adding small amounts of electrolytes. Depending upon the molecular size, colloids are three types. 1. Multimolecular colloids consist of aggregate of atoms or small molecules held together by weak Van der Waals forces e.g., sols of gold atoms, sulphur molecules (S8) 2. Macromolecular colloids consist particles of macromolecules having colloidal particle size such as starch, cellulose, proteins etc 3. Associated colloids consist electrolytes which undergo association at higher concentration and behave as colloidal solution e.g., soap (sodium stearate) solution. The colloidal solutions are either positive or negative sols depending upon the nature of charge on the colloidal particles. e.g., Al(OH)3, Fe(OH)3, basic dyes such as methylene blue are positive sols while sols of Cu, Ag, metal sulphides such as As2S3, CdS etc are negative sols.

Surface Chemistry 14.49

Preparation of colloids • •

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Colloids may be prepared either by dispersed methods or by condensation methods. In mechanical dispersion method, colloidal solutions are prepared by grinding the coarse suspension between oppositely rotating discs in the presence of a stabilizing agent. In Bredig’s arc method colloidal solutions of metals such as gold, silver are prepared by producing electric arcs between metal electrodes immersed in the dispersion medium. Due to intense heat of arc, metals vaporize and condense in dispersion medium as colloidal particles in the presence of KOH as stabilizing agent. The process in which a stable colloid is prepared by treating the substance in massive form repeatedly with solvent or with a foreign substance called peptizing agent, is called peptisation. The process of transferring back a precipitate into colloidal form is called peptisation. It is the reverse of coagulation. The peptizing action is due to preferential adsorption of one of the ions of the electrolyte which then gives to the colloidal particle a positive or negative charge depending on the charge of adsorbed ion. E.g., Fe(OH)3 adsorbs Fe3+ ions from FeCl3 (peptizing agent) and thereby gets positive charge. In condensation methods smaller particles are made to convert into bigger particles size involving chemical reactions. In the change of dispersion medium method, a substance is dissolved in a solvent and then added to another solvent in which it is less soluble. e.g., colloids of sulphur, phosphorous can be prepared by dissolving them in organic liquids and then by adding to water. In metathesis or double decomposition method colloidal solutions are prepared by the reaction between two different solutions which can form insoluble compounds in colloidal state. E.g., by mixing AgNO3 & NaCl or BaCl2 and H2SO4 the sols of AgCl and BaSO4 can be prepared. Colloids of certain metal hydroxides can be pre pared by the hydrolysis of their salts e.g., Fe (OH) 3 colloid can be prepared by the hydrolysis of FeCl3 in water. Hydrosols of metals like Ag, Au, Pt etc can be prepared by the reduction of their soluble salts with suitable reducing agent in aqueous solutions. The different colours of the gold sols depend on the size and shape of the gold micelles. Blue micelles are large or asymmetrical and red ones are small and spherical.

•

In oxidation methods certain sols are prepared by oxidizing certain compounds e.g., Wackenroder’s solution is prepared by the oxidation of aqueous solution of H2S by air or by careful addition of HCl to hypo solution.

Purification of Colloids •

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During the preparation of colloids some electrolytes may left in the colloidal solution which decrease the stability of colloids. The removal of these impurities is known as purification of colloids Removal of the electrolytes by taking the advantage that colloidal particles cannot pass through parchment membrane while electrolytes can pass through it is called dialysis In the electro dialysis method electrolytes are removed by taking colloidal solution in a parchment membrane between two electrodes. The ions in colloidal solution are attracted by the oppositely charged electrodes in an electric field. Colloids can also be purified by ultra filtration.

Properties of colloidal Solutions •

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When a beam of light is made to fall on a colloidal solution the path of the bean of light is illuminated by a bluish light. The luminosity by the path of beam is known Tyndall effect. It is due to the scattering of light from the surface of the colloidal particles. The scattering of light is due to the absorption of light energy and then scatter the light of shorter wavelength. In case, a convergent beam of light is made to fall upon a solution by means of lens a bright blue cone is called Tyndall cone is observed. Tyndall effect is observed when the following two conditions are satisfied. (i) The refractive indices of the dispersed phase and dispersion medium must differ greatly in magnitude (ii) The diametre of the dispersed particles is not much smaller than the wave length of the light used. The colour of the sol depends on (a) wave length of the scattered light by the sol (b) size and shape of the sol particle (c) specific selective absorption power of the dispersed phase and (d) the way an observer receives the light.

Kinetic Properties •

The constant rapid zig–zag motion of colloidal particles in a colloidal solution is known as Brownian movement named after its discoverer.

14.50

•

•

Surface Chemistry

The Brownian movement is due to the continuous bombardment of the molecules of dispersion medium on colloidal particles The Brownian movement decreases with increase in the size of the colloidal particle due to the decrease in possibility of unequal bombardment diminish.

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When electrophoresis of dispersed particles is prevented by suitable means, the medium can be made to move under the influence of an applied potential. This phenomenon is referred to as electro osmosis. Electro osmosis is used in drying pastes, in tanning of hides and similar products in the removal of water from moist clay etc.

Electrical Properties •

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The migration of electrically charged sol particles under an applied electric field is called electrophoresis. Strictly cataphoresis or anaphoresis according to the electrode to which the particles move. The charge on colloidal particles may be due to the following reasons. 1. When colloidal particles such as soaps and detergents are dissolved in water, self ionization takes place leading to the association of ionized particles. Depending on the charge of ion the colloidal particles get charge. 2. Proteins contain ionisable acidic and basic group. Thus the particles of protein sol can either have positive charge or negative charge depending on the pH. In solutions having pH less than 7 the protein sols may get positive charge. 3. The colloidal particles may get charge due to preferential adsorption of ions present in solution. When two or more ions are present in the dispersion medium preferential adsorption of the ion common to the colloidal particle usually take place. When a dilute solution of KI is added to a slight excess of AgNO3 solution a positively charged sol of AgI is formed due to preferential adsorption of Ag+ ions. If a dilute solution of AgNO3 is added to a slight excess of KI solution a negatively charged sol AgI is formed due to preferential adsorption of I− ions. No sol is formed if equivalent amounts of KI and AgNO3 solutions are mixed. Whatever the nature of the charge on the particle it must be balanced by ions of opposite sign in the electrolyte which forms a diffuse layer round the particle thus forming an electrical double layer of positive and negative charges called Helmholtz double layer is formed. The potential difference between the fixed layer and the diffused layer having opposite charge is called the electro kinetic potential or zeta potential and the double layer is called electrical double layer. The electrophoresis mobilities of different colloidal substances are different which helps the separation of different colloidal particles from mixtures. At a particular pH, the sol particles become neutral and exhibit no movement in an electric field and this pH is called isoelectric point.

Stability of colloids •

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Lyophobic colloids are stable since the similar charges on colloidal particles will show electrostatic repulsion and prevent the colloidal particles from coalescing or aggregating Lyophilic colloidal particles are much solvated or hydrated which act as a barrier which prevents the colloidal particles to come together to form bigger aggregates. Stabilizers stabilize the colloids by decreasing the surface tension between the immiscible liquids and bring them into colloidal form, or may produce charge on the neutral atoms or molecules which cause repulsion between colloidal particles.

coagulation •

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Coagulation or flocculation is the process of breaking up of a colloidal solution by which colloidal particle come close and result in the precipitation of the dispersed phase. If a colloidal solution is stored for longer periods smaller particles dissolves and crystallize out on the larger particles causing the coagulation of colloid. This is known as ageing. Coagulation of lyophobic colloids may be caused by adding an electrolyte having opposite charge of the colloidal particles. The electrolyte having opposite charge to the colloidal particle neutralizes the charge on colloidal particles, thus causing the aggregation of particles and coagulation. Mutual coagulation is a process of coagulation by mixing two different colloids with opposite charges in equal proportions. Coagulation can also be carried by boiling, freezing or by persist dialysis. Greater the charge on the ion added, greater is its power to cause precipitation. This is known Hardy-Schulz law Order of coagulation power Al3+ > Ba 2 + > Na +  Fe ( CN )  6 

4−

> PO34− > SO 24 − > Cl−

Surface Chemistry 14.51

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The minimum concentration of an electrolyte which is able to cause coagulation or flocculation of sol is termed as flocculation value. The coagulation of lyophilic colloid by adding salt solutions of high concentrations is called ‘salting out’. This may be due to the removal of solvent layer around the lyophilic colloid by the salt. The salting out effect depends upon the nature of ion and the salts of a given metal and the series is called Holfmeister series or Lyotropic series. The order of cation is Mg

2+

> Ca

2+

> Sr

2+

> Ba

2+

+

+

+

> Na > K > Rb > Cs

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+

The order of anion is (Citrate)3− > ( tartrate)2 − > SO24 − > CH3COO − > Cl − > NO3− > ClO3− > I − > CNS−

Associated colloids •

Protection of colloids •

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The process by which the sol particles are prevented from coagulation by electrolyte due to the previous addition of some lyophilic sol is called protection of colloid. The lyophilic colloids form as a layer above the lyophobic colloid particle and prevent the coagulation. The number of milligram’s of protective colloid which just prevents the coagulation of 10 mL of given gold sol when 1 mL of 10% solution of NaCl is added to it is called gold number. Smaller the gold number of lyophilic colloid, the greater is its protective power. Gelatin and starch have maximum and minimum protecting power.

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Emulsions • • • •

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A colloidal system involving one liquid dispersed in another is known as emulsion. Any two immiscible liquids can form an emulsion. The substance that stabilises an emulsion is called emulcifier or emulsifying agent. Emulsions are two types i) oil in water (O/W) (ii) water in oil (W/O) depending on the dispersion medium and dispersion phase. Emulsifier such as soap decreases the interfacial tension between the liquids causing emulsion particles roll easily. A certain number of large particles are frequently present in an emulsion and if it is of the oil in water type they will tend to rise to the top since they are lighter than water. This is termed as creaming and occurs in milk. Emulsions show all the properties of colloidal solution.

In the froth flotation process the ore particles adsorb collector and prevent from wetting and collected in the foam formed by passing air into an oil emulsion. The digestion of fat in intestine is aided by emulsification. It converts into soap with alkaline solution of intestine and emulsifies the fat. Demulsification is a process of breaking down the emulsion to yield constituent liquids. Demulsification can be carried by freezing or by boiling or by centrifuging the emulsion. Conversion of cream into butter by churning is a process of demulsification of emulsion of fat in water. In natural oil wells oil and water form emulsions which should be demulsified.

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The aggregates formed in water by association of normal simple molecules, each having a hydrophobic end and a hydrophilic end is called associated colloid or micelle. A colloid in which the dispersed phase consists of micelles or aggregated colloidal particles is an associated colloid. If the molecule of a substance aggregates spontaneously in a given solvent to form colloid, it is called associated colloid. The stearate ions associate themselves in high concentrations to form ionic micelle. In the stearate micelle the hydrophobic group is alkyl group end and the hydrophilic group is carboxylate ion end. Ordinary soap, synthetic detergents belong to micelle. The cleaning action of soap is based on solubilisation of grease or dirt on the cloth into micelle, which is known as emulsification of grease. The hydrophilic group in stearate ion is the COO− group which is known as head of stearate ion. The hydrocarbon chain of the stearate anion has affinity for grease, oil or dirt which is hydrophobic and is called tail of the anion. The tail of the stearate anion dissolves in the grease or dirt and encapsulates to from micelles which are removed by washing process. Soap is an emulsifying agent in the cleaning process. All synthetic detergents function as emulsifying agents. Surfactants are the substances which preferentially adsorbed at the interfaces like air-water, oil-water and solid-water interfaces and are responsible for micellisation and emulsification. Surfactants may be cationic, or anionic or non ionogenic. The formation of micelles takes place only above a particular temperature called Kraft temperature (Tk) and above a particular concentration called critical micelle concentration (CMC).

14.52

Surface Chemistry

gels •

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The colloidal system constituting the liquid as dispersed phase and the solid as dispersed medium is known as gel. Gels are two types i) elastic gels and ii) non-elastic gels. Elastic gels can change their shape on applying force and return to the original shape when the applied force is removed. e.g., gelatin, starch, agar-agar. Non – elastic gels are rigid and do not possess elasticity, e.g., silica gel. Synerisis or weeping of gel is a process of shrinking of gel by losing liquid on standing for some time. Elastic gels on losing water become elastic solid. These on mixing with water again get converted into gel. This is known as imbibition. Non-elastic gels on dehydration become glassy powder and this glassy powder cannot be converted back into gel by adding water. Some gels (elastic which are partially dehydrated) swell when dipped in water and is known as swelling of gel.

Application of colloids • • •

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The colloidal dust particles suspended in air scatter only blue light and hence the sky looks blue. Many of our food materials such as milk, starch, proteins, fruit jellies etc are colloids. Blood is a colloid of albuminoid. The styptic action of alum and ferric chloride solution is due to coagulation of blood forming a clot which stops further bleeding. When a large mass of air, containing dust particles, is cooled below its dew point, the moisture from air condenses on the surfaces of the particles forming

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droplets. These droplets being colloidal in natural continue to float in the air form mist or fog. Clouds are aerosols having small droplets of water suspended in air. The colloidal droplets of water grow bigger due to which rain fall occurs. Sometimes rain fall occurs when two oppositely charged clouds meet. River water containing colloidal clay when meets the sea water containing electrolytes coagulates and the clay deposits forming delta. The colloidal particles present in the smoke going out of chimney can be removed by passing the smoke through electrodes with opposite charges using Cottrell precipitator, where carbon particles get discharged and precipitated. Water from sewage is colliod of many impurities and can be coagulated when it is allowed to flow between oppositely charged plates. Muddy water can be purified by adding alum which coagulates the colloidal impurities. Negatively charged latex particles of rubber under suitable conditions can be electroplated. The latex is coagulated by adding electrolytes such as sodium acetate and acetic acid. Smoke screens used in warfare consists of fine particles of titanium oxide being heavier form smoke screen. Artificial rains can be produced by throwing electrified sand from an aeroplane which coagulates the mist hanging in the air. Dyeing is essentially an adsorption of dye by the colloidal particles of the mordant. Colloidal particles in hides are positively charged sol and can be mutually coagulated by soaking the hides in tannin which is negatively charged. Photographic plates are coated with sensitive emulsion containing fine silver bromide particles suspended in gelatin emulsion.

Surface Chemistry 14.53

PRAcTIcE ExERcISE

multiple choice Questions with Only One Answer level I 1. Sorption is the term used when (a) Adsorption takes place. (b) Absorption takes place. (c) Both takes place. (d) Desorption takes place. 2. Which of the following is not a characteristic of chemisorption? (a) Adsorption is irreversible. (b) ΔH is of the order of 40 KJ. (c) Adsorption is specific. (d) Adsorption increases with increase of surface area. 3. Which is correct regarding the physical adsorption of a gas on surface of solid? (a) On increasing temperature adsorption increases continuously. (b) Enthalpy and entropy change is negative. (c) Adsorption is more for some specific substance. (d) Irreversible. 4. Which one the following characteristic is not correct for physical adsorption? (a) Adsorption on solid is reversible. (b) Adsorption increase with increase in temperature. (c) Adsorption is spontaneous. (d) Both enthalpy and entropy of adsorption are negative. 5. The heats of adsorption in physisorption lie in the range (KJ/mol) (a) 40-400 (b) 40-100 (c) 20-40 (d) 1-10 6. In physical adsorption, the forces associated are (a) Ionic (b) Covalent (c) Van der Waals (d) H-bonding 7. How many layers are adsorbed in chemical adsorption? (a) 1 (b) 2 (c) 3 (d) 4 8. Chromatography is a technique based on (a) Solubilities of solute (b) Adsorption of solute (c) Chemical adsorption followed by dispersion (d) Differential adsorption of different constituents of a mixture

9. Which characteristic of adsorption is wrong? (a) Physical adsorption in general decreases with temperature. (b) Physical adsorption in general increases with temperature. (c) Physical adsorption is a reversible process. (d) Adsorption is limited mainly to the surface only. 10. The volumes of gases H2, CH4, CO2 and NH3 adsorbed by 1 g of charcoal at 288 K are in the order (a) H2 > CH4 > CO2 >NH3 (b) CH4 > CO2 > NH3 >H2 (c) CO2 >NH3 >H2 > CH4 (d) NH3 > CO2 > CH4 >H2 11. The slope of the straight line graph between log x/m and log P for the adsorption of a gas on solid is (a) K (b) log k 1 (c) n (d) n 12. According to Langmuir adsorption isotherm, the amount of gas adsorbed at very high pressure. (a) Reaches a constant limiting value. (b) Goes on increasing with pressure. (c) Goes on decreasing with pressure. (d) Increases first and decreases later with pressure 13. Which among the following statement is false? (a) Increase of pressure may increase the amount of physical adsorption. (b) Increase of temperature may decrease the amount of adsorption. (c) The adsorption may be monolayered or multilayered. (d) Particle size of the adsorbent will not affect the amount of adsorption. 14. The graph relates the adsorption with pressure at constant temperature is called (a) an isobar (b) an isotherm (c) an isostere (d) none of these. 15. Which plot is the adsorption isobar for chemisorptions?

x m

x m

(a)

T

(b)

(c)

T

x m

Adsorption

ADSORPTION

T

(d)

T

Surface Chemistry

24. Following is the variation of physical adsorption with temperature.

T

(b)

(c)

T

(d)

T

Adsorption

(a)

Adsorption

Adsorption

16. There are certain properties related to adsorption (a) Reversible (b) Formation of unimolecular layer (c) Low heat of adsorption (d) Occurs at low temperature and decreases with increasing temperature. Which of the above properties are for physical adsorption? (a) I, II, III (b) I, III, IV (c) II, III, IV (d) I, III 17. Rate of physisorpion increases with (a) Decrease in temperature (b) Increase in temperature (c) Decrease in pressure (d) Decrease in surface area 18. Which of the following is less than zero during adsorption? (a) ΔG (b) ΔS (c) ΔH (d) All of the above 19. When a graph is plotted between log x/m and log P, it is straight line with an angle 45° and intercept 0.3010 on y-axis. If initial pressure is 0.4 atm, what will be the amount of gas adsorbed per gm of adsorbent? (a) 0.4 (b) 0.8 (c) 0.1 (d) 0.6 20. 1 gm of charcoal adsorbs 100 mL of 0.5 M CH 3COOH to form a mono layer and there by the molarity of CH3COOH reduces to 0.49 M. Surface area of char coal is 3.015 × 102 m2/gm.The surface area of the charcoal adsorbed by each molecule of acetic acid is (a) 3×10–19 m2 (b) 2×10–19 m2 –19 2 (c) 4×10 m (d) 5×10–19 m2 21. The Langmuir adsorption isotherm is deduced using the assumption (a) The adsorption sites are equivalent in their ability to adsorb the particles. (b) The heat of adsorption varies with coverage. (c) The adsorbed molecules interact with each other. (d) The adsorption takes place in multilayers. 22. Adsorption is multilayer in case of (a) physical adsorption (b) chemisorption (c) both (a) and (b) (d) none of these 23. Which of the following statements about physical adsorption is not correct? (a) It is usually monolayer. (b) It is reversible in nature. (c) It involves Van der Waals interactions between adsorbent and adsorbate. (d) It involves small value of adsorption.

Adsorption

14.54

T

multiple choice Questions with One or more Than One Answer 1. Select correct statements about physical adsorption (a) Both enthalpy and entropy of adsorption are negative. (b) Adsorption of gas on solids is reversible. (c) Adsorption increases with increase in temperature. (d) Adsorption is spontaneous. 2. Which of the following statement(s) is/are true (a) The volume of an ideal gas measured at the experimental pressure, that is dissolved in a given volume of the solvent has a constant value. (b) The solubility of a gas, in a liquid increases with the pressure of the gas. 1 (c) Frenudlich equation is log x/m = log K+ log P n (here n>1) and it considers multilayer adsorption. (d) According to Langmuir, the ability of particle to bind on the surface is indepenendent of whether or not nearby sites are occupied. 3. Which among the following statements are correct with respect to adsorption of gases on a solid? (a) The extent of adsorption is equal to KPn according to Freundilich isotherm. (b) The extent of adsorption is equal to KP1/n according to Freundilich isotherm. (c) The extent of adsorption is equal to (1+bP)/aP according to Langmuir isotherm. (d) The extent of adsorption is equal to aP/(1+bP) according to Langmuir isotherm 4. Which of the following statements are true for physisorption? (a) Extent of adsorption increases with increase in pressure. (b) It needs activation energy. (c) It can be reversed easily. (d) It occurs at high temperature.

Surface Chemistry 14.55

5. Which one of the following is a correct statement? (a) Physical adsorption is reversible in nature. (b) Physical adsorption involves Van der Waals forces. (c) Rate of physical adsorption increases with increase of pressure on the adsorbate. (d) High activation energy is involved in physical adsorption. 6. Adsorption is accompanied by (a) Decrease in entropy of the system (b) Decrease in enthalpy of the system (c) Decrease in free energy of system (d) No change in free energy of system

Assertion (A) and Reason (R) Type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): Gases are easily adsorbed on the surface of metals especially transition metals. Reason (R): Transition metals have free valencies. 2. Assertion (A): NH3 is more adsorbed on activated coconut charcoal than H2. Reason (R): Critical temperature for NH3 is more than that of H2. 3. Assertion (A): When a finely divided active carbon or clay is stirred into a dilute solution of a dye, the intensity of colour in the solution is decreased. Reason (R): The dye is adsorbed on the solid surface. 4. Assertion (A): According to Freundlich; x/m = k.p1/n Reason (R): The isotherm shows variation of the amount of gas adsorbed by the adsorbent with temperature.

5. Assertion (A): Adsorption of a gas over solid surface is generally accompanied by decrease in enthalpy and decrease in entropy both. Reason (R): Higher the critical temperature, TC of a gas the more easily it will be liquefied and also more readily it will be adsorbed on the solid. 6. Assertion (A): The enthalpy of physisorption is greater than chemisorption. Reason (R): Molecules of adsorbent and adsorbate are held by Van der Waals forces in physical adsorption and by chemical bonds in chemisorptions. 7. Assertion (A): For adsorption ∆G, ∆S and ∆H all have negative values. Reason (R): Adsorption is spontaneous process accompanied by decreases in randomness. 8. Assertion (A): Physical adsorption of molecules on the surface requires activation energy. Reason (R): Because the bonds of adsorbed molecules are broken.

Integer Type Questions 1. Two grams of charcoal is stirred in 100 mL of 0.5 M CH3COOH Solution. Acetic acid adsorbed on charcoal and forms monolayer. As a result, concentration of acetic acid decreases to 0.48 M. The surface area of charcoal is 3.01×102 m2/g. The surface area of charcoal adsorbed by each molecule of acetic acid is n × 10–19 m2.What is the value of n? 2. The extent of adsorption (x/m) at a pressure of 1 atm from the following graph is  x log    m

45°

}

0.6020

log P →

3. 47.7 mL of 1 M oxalic acid is shaken with 0.5 g wood charcoal. The final conc. of the solution after adsorption is 0.5 M. What is the amount of oxalic acid adsorbed per gram of carbon?

14.56

Surface Chemistry

cATAlySIS multiple choice Questions with Only One Answer level I 1. Which of the following statement is more correct? (a) Catalyst only accelerates the rate of a chemical reaction. (b) A catalyst can retard the rate of a chemical reaction. (c) A catalyst can control the speed of a reaction. (d) A catalyst alters the speed of a reaction. 2. Which one of the following is not the example of homogeneous catalysis? (a) Formation of SO3 in the chamber process. (b) Formation of SO3 in the contact process. (c) Hydrolysis of an ester in presence of acid. (d) Decomposition of KCIO3 in presence of MnO2. 3. The decomposition of hydrogen peroxide can be slowed down by addition of a small amount of acetamide. The latter act as (a) Inhibitor (b) Promoter (c) Moderator (d) Poison 4. In the reaction: KMnO4+H2SO4+H2C2O4 → products, Mn++ ions act as (a) Positive catalyst (b) Negative catalyst (c) Auto catalyst (d) Enzyme catalyst 5. In the Haber process of synthesis on NH3 (a) Mo acts as a catalyst and Fe as a promoter. (b) Fe acts as a catalyst and Mo as a promoter. (c) Fe acts as inhibitor and Mo as a catalyst. (d) Fe acts as promoter and Mo as auto-catalyst. 6. TEL minimizes the knocking effect when mixed with petrol. It acts as (a) Positive catalyst (b) Negative catalyst (c) Auto-catalyst (d) Induced catalyst 7. Platinised asbestos is used as a catalyst in the manufacture of H2SO4. It is an example of (a) Homogeneous catalyst (b) Heterogeneous catalyst (c) Auto-catalyst (d) Induced catalyst 8. The catalyst used in the manufacture of hydrogen by Bosch’s process is (a) Fe2O3 (b) Cr2O3 (c) Fe2O3+ Cr2O3 (d) Cu 9. In the Ostwald’s process for the manufacture of HNO3, the catalyst used is (a) Fe (b) Pt (c) V2O5 (d) Cu

10. Efficiency of the catalyst depends on its (a) Molecular weight (b) Number of free valencies (c) Physical state (d) Amount used 11. Which of the following types of metals make the most efficient catalysts? (a) Transition metals (b) Alkali metals (c) Alkaline earth metals (d) Radioactive metals 12. Which is false for catalyst? (a) A catalyst can initiate a reaction. (b) It does not alter the position of equilibrium in a reverible reaction. (c) A catalyst remains unchanged in quality and composition at the end of reaction. (d) Catalysts are sometimes very specific in respect of a reaction. 13. A catalyst (a) Increases the free energy change in the reaction. (b) Decrease the free energy change in the reaction. (c) Does not increases and decreases the free energy change in the reaction. (d) Can either decreases or increases the free energy change depending on what catalyst we use. 14. A catalytic poison renders the catalyst ineffective because (a) It is preferentially adsorbed on the catalyst. (b) It adsorbs the molecules of the reactants. (c) It combines chemically with the catalyst. (d) It combines with one of the reactant. 15. According to adsorption theory of catalysis, the reaction rate increase, because (a) Adsorption produces heat which increases the rate of reaction. (b) In the process of adsorption, the kinetic energy of the molecules increases. (c) The concentration of reactants at the active centers becomes high due to adsorption. (d) The activation energy of the reaction becomes high due to adsorption. 16. The modern theory of catalysis is based on (a) Active masses (b) Atomic or molecular weights (c) Size of the particle (d) Number of free valencies 17. In the manufacture of H2SO4 by contact process the presence of As2O3 acts as (a) Catalytic promoter (b) Catalytic poison (c) Induced catalyst (d) Auto catalyst

Surface Chemistry 14.57

18. In oxidation of oxalic acid by KMnO4, the colour of KMnO4 disappears slowly in the start of reaction but disappears very fast afterwards. This is an example of (a) Auto catalysis (b) Negative catalysis (c) Induced catalysis (d) Positive catalysis 19. Which statement is wrong? (a) Haber’s process of NH3 requires iron as catalyst. (b) Friedel-Craft’s reaction requires anhydrous AlCl3. (c) Hydrogenation of oils requires iron as catalyst. (d) Oxidation of SO2 to SO3 requires V2O5. 20. Regarding criteria of catalyst, which one of the following statements is not true? (a) The catalyst is unchanged chemically after the reaction. (b) A small quantity of catalyst is often sufficient to bring about a considerable amount of the reaction. (c) In reversible reaction, the catalyst alters the equilibrium position. (d) The catalyst accelerates the rate of reaction. 21. Which is not correct for heterogeneous catalysis? (a) The catalyst decrease the energy of activation. (b) The surface of catalyst plays an important role. (c) The catalyst actually forms a compound with reactants. (d) There is no change in the energy of activation. 22. Which of the following statements about a catalyst is/ are true? (a) A catalyst accelerates the reaction by bringing down the energy of activation. (b) A catalyst does not take part in the reaction mechanism. (c) A catalyst makes the reaction more feasible by making the ∆G0 more negative. (d) A catalyst makes the equilibrium constant of the reaction more favorable for the forward reaction. 23. Which of the following kind of catalysis can be explained by the adsorption theory? (a) Homogeneous catalysis (b) Acid-base catalysis (c) Heterogeneous catalysis (d) Enzyme catalysis 24. Catalytic poison (a) Combines with catalyst (b) Adsorbed on the active centers of the surface of the catalyst (c) Coagulate the catalyst (d) It combines with the any one of the reactants

25. A catalyst in the finely divided form is most effective because (a) Less surface area is available (b) More active centers are formed (c) More energy gets stored in the catalyst (d) None of these 26. Which is wrong in case of enzyme catalysis? (a) enzymes work best at an optimum temperature (b) enzymes work at an optimum pH (c) enzymes are highly specific for substrates (d) an enzyme raises activation energy

multiple choice Questions with One or more Than One Answer 1. The following are some statements about a catalyst. The correct statements are: (a) Small amount of catalyst is sufficient to catalyse the reaction. (b) A catalyst initiates a chemical reaction. (c) A catalyst functions under the optimum conditions like pH, temperature. (d) A catalyst has no effect on the equilibrium state. 2. The activity and selectivity of zeolites as catalyst is based on (a) Their pore size (b) Their surface area (c) Size of their cavities on the surface (d) Their amount 3. In chemical reaction, the catalyst (a) Alters the amount of products (b) Lowers the activation energy (c) Decreases Ea for forward reaction (d) Increases ∆H of forward reaction

Assertion (A) and Reason (R) Type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): NO acts as homogenous catalyst in the oxidation of CO by O2. Reason (R): NO increases the rate of oxidation of CO into CO2.

14.58

Surface Chemistry

2. Assertion (A): Enzyme catalysed reactions are of first order. Reason (R): Enzymes never undergoes any change 3. Assertion (A): The presence of catalyst increases the speed of the forward and backward reactions to the same extent. Reason (R): Activation energy for both the forward and backward reactions is lowered to the same extent. 4. Assertion (A): A catalysts speed up a reaction but doesn’t participate in its mechanism. Reason (R): A catalyst provides an alternative path of lower activation energy of the reactants. 5. Assertion (A): Catalysts are always transition metals. Reason (R): Transition metals have variable oxidation state.

cOllOIDS multiple choice Questions with Only One Answer level I 1. Which among the following will bring about coagulation of TiO2 sol quickest and in the least of molar concentration? (a) NaCl (b) Mg SO4 (c) Al2(SO4)3 (d) K4[Fe(CN)6 2. On adding few drops of dil. HCI to freshly precipitated ferric hydroxide, a red coloured colloidal solution is obtained. This phenomenon is known as (a) Protective action (b) Coagulation (c) Peptisation (d) Dialysis 3. Butter is a colloid in which (a) Fat is dispresed in fat (b) Fat is dispresed in water (c) Water is dispresed in fat (d) Suspension of casein in water 4. Lyophobic colloids are (a) Reversible (b) Irreversible (c) Water loving (d) Solvent loving 5. Sulphur sol contains (a) Discrete sulphur atoms (b) Discrete sulphur molecules (c) Water dispersed in solid sulphur (d) Large aggregates of sulphur molecules 6. Greater the valency, the higher is the coagulating power of ion. This rule was introduced by (a) Hardy-Schulze (b) Graham (c) Kossel and Lewis (d) Faraday

7. All colloidal solution show (a) Very high osmotic pressure (b) High osmotic pressure (c) Low osmotic pressure (d) No osmotic pressure 8. An examples of micelle is (a) As2O3 solution (b) Ruby glass (c) Na2CO3 solution (d) Sodium stearate concentrated solution 9. Which is not shown by sols? (a) Adsorption (b) Tyndall effect (c) Paramagnetism (d) Flocculation 10. Which of the following is an emulsifier? (a) Soap solution (b) Water (c) Oil (d) NaCl 11. Micelles are (a) Adsorbent solutes (b) associated colloids (c) adsorbed surfaces (d) Ideal solution 12. Substances whose solutions can readily diffuse through animal membranes are called (a) Colloids (b) Crystalloids (c) Electrolytes (d) Non-electrolytes 13. The size of the colloidal particles is in between (a) 10–7 –10–9 cm (b) 10–9 –10–11 cm –5 –7 (c) 10 –10 cm (d) 10–2 – 10–3 cm 14. Peptization is a process of (a) Precipitating the colloidal particles (b) Purifying the colloidal sol (c) Dispersing the precipitate in to colloidal sol (d) Movement of colloidal particles towards the opposite charged electrodes 15. Colloids are purified by (a) Brownian motion (b) Precipitation (c) Dialysis (d) Filtration 16. When excess of electrolyte is added to a colloid it? (a) Coagulates (b) Gets diluted (c) Dissolved (d) Does not change 17. Bleeding is stopped by application of ferric chloride. This is because (a) The blood starts flowing in opposite direction (b) The ferric chloride seals the blood vessel (c) The blood reacts and forms a solid which seals the blood vessel (d) The blood is coagulated and thus, the blood vessel is sealed

Surface Chemistry 14.59

18. Gold number is a measure of (a) The amount of gold present in the colloidal solution (b) The amount of gold required to break the colloid (c) The amount of gold required to protect the colloid (d) The protective power of the lyophilic colloid 19. On addition of one mL solution of 10% NaCl to 10 mL. gold sol in presence of 0.0025 g of starch, the coagulation is just prevented. The gold number of starch is (a) 25 (b) 2.5 (c) 0.25 (d) 0.025 20. Which of the following has minimum flocculation value? (a) Pb2+ (b) Pb4+ 2+ (c) Sr (d) Na+ 21. Brownian motion shown by colloidal particle is its ………property. (a) Optical (b) Electrical (c) Kinetic (d) Chemical 22. A freshly prepared Fe (OH)3 precipitate is peptized by adding FeCl3 solution. The charge on the colloidal particle is due to preferential adsorption of (a) CI- ions (b) Fe+++ ions (c) OH ions (d) None 23. Hardy-Schulze rule state that (a) Non-electrolytes have better coagulating action on colloids than electrolytes (b) Sols are coagulated by effective ions whose charge is opposite to that of sol and the ions of higher charge are much more effective than ions of lower charge. (c) Charge of the ions has no effect on the coagulation of a sol. (d) Sols are coagulated only by those ions whose charge is similar to that of the sol. 24. Which is correct statement in case of milk? (a) Milk is an emulsion of fat in water. (b) Milk in emulsion of protein in water. (c) Milk is stabilized by protein. (d) Milk is stabilised by fat 25. The gold numbers of A, B, C and D are 0.04, 0.002, 10 and 25 respectively. The protective power of A, B, C and D are in the order (a) A>B>C>D (b) B>A>C>D (c) D>C>B>A (d) C>A>B>D 26. The movement of dispersion medium in an electric field when the dispersed particles are prevented from moving is called (a) Cataphoresis (b) Electrophoresis (c) Electro-osmosis (d) Brownian movement

27. The potential difference between the fixed charged layer and the diffused layer having opposite charge is called (a) Colloidal potential (b) Zeta potential (c) Electrostatic potential (d) None of these 28. Colloidal particles carry charge. This can be show by (a) Tyndall effect (b) Cataphoresis (c) Brownian movement (d) Dialysis 29. Which metal sol cannot be prepared by Bredig’s arc method? (a) K (b) Cu (c) Au (d) Pt 30. On adding AgNO3 solution into KI solution, a negatively charged colloidal sol is formed when they are in (a) 100 mL of 0.1 M AgNO3 + 100 mL of 0.1 M KI (b) 100 mL of 0.1 M AgNO3 + 50 mL of 0.2 M KI (c) 100 mL of 0.1 M AgNO3 + 50 mL of 0.1 M KI (d) 100 mL of 0.1 M AgNO3 + 100 mL of 0.15 M KI 31. Micelles have (a) Higher colligative properties compared colloidal sols (b) Lower colligate properties (c) Same colligative properties (d) Low molar mass 32. Which of the following ion has minimum flocculation value? (a) CI(b) SO4-2 3(d) [Fe (CN)6]4(c) PO4 33. If 1000 mg of lyophilic sol prevent the coagulation of 1 L lyophobic sol, then protection number is (a) 10 mg (b) 1 mg (c) 25 mg (d) None 34. A liquid is found to scatter a beam of light but leaves no residue when passed through the filter paper. The liquid can be described as (a) A suspension (b) Oil (c) A colloidal sol (d) True solution 35. Gelatin is mostly used in making ice creams in order to (a) Prevent making of colloid (b) Stabilise the colloid and prevent crystallization (c) Stabilise the mixture (d) Enrich the aroma 36. At critical micelle concentration (CMC) the surfactant molecules (a) Decomposes (b) Becomes completely soluble (c) Associates (d) Dissociates

14.60

Surface Chemistry

37. According to Hardy-Schulze law the order of coagulation power of cations will be (a) Na+>Ba2+>Al+3 (b) Al+3>Ba+2>Na+ (c) Ba+2>Al+3>Na+ (d) Al+3>Na+>Ba+2 38. Which one of the following method is commonly used for destruction of colloid? (a) Dialysis (b) Condensation (c) Filtration by animal membrane (d) By adding electrolyte 39. Position of non polar and polar part in micelles (a) Polar at outer surface but non polar at inner surface (b) Polar at inner surface but non polar at outer surface (c) Distributed over all the surface (d) Are present in the surface only 40. Milk is a Colloid in which (a) Liquid is dispersed in a liquid (b) Solid is dispersed in a liquid (c) Gas is dispersed in a liquid (d) Sugar is dispersed in a liquid 41. Surface tension of lyophilic sol is (a) Lower than that of H2O (b) More than that of H2O (c) Equal than that of H2O (d) None of these 42. Flocculation value is expressed in terms of (a) Milimole per litre (b) Mol per litre (c) Gram per litre (d) Mol per millilitre 43. A soap solution (C17H35COO Na) becomes a colloidal sol at a concentration of 1.2×10–3 M. On an average 2.4×1013 colloidal particles are present in 1 mm3. The average number of stearate ions in one colloidal particle (micelle) is (a) 30 (b) 60 (c) 90 (d) 120 44. Fixed parts of a colloidal sol. of AgI are respectively [AgI]Ag+ and [AgI]I- In presence of (a) KI and AgNO3 (b) KI and KI (c) AgI and KI (d) AgNO3 and KI 45. The colligate property of a colloidal sol. compared to solution of non-electrolyte of same concentration will be: (a) Same (b) Higher (c) Lower (d) Higher or Lower 46. Coagulation or demulsification can be done by some of the methods given below I. By addition of a substance which would destroy the emulsifier II. By addition of an electrolyte which would destroy the charge III. By heating, freezing and centrifuging

47.

48.

49.

50.

51.

52.

53.

Select correct methods (a) I, II only (b) I, II, III (c) II only (d) III only Electro-osmosis is observed when: (a) Dispersion medium begins to move in an electric field (b) Dispersed phase begins to move in an electric field (c) In both 1 and 2 (c) None of these Colloid solutions of gold prepared by different methods are of (a) Different diametres of colloidal gold particle (b) Variable valency of gold (c) Different concentration of gold particles (d) Impurities produced by different methods In electrophoresis: (a) Sol. Particles move towards opposite electrodes (b) Medium moves towards opposite electrodes (c) Neither 1 nor 2 (d) Both 1 and 2 Lyophilic sols are more stable than lyophobic sols because (a) The colloidal particles have positive charge. (b) The colloidal particles have no charge. (c) The colloidal particles are solvated. (d) There are strong electrostatic repulsions between the negatively charged colloidal particles. The formation of micelles, which occur only beyond a certain temperature, is called (a) Critical temperature (b) Critical sol temperature (c) Consulate temperature (d) Kraft temperature Fog is an example of colloidal system of (a) liquid in a gas (b) gas in a liquid (c) gas in s soild (d) soild in a liquid The dispersed phase in colloidal iron (III) hydroxide and colloidal gold is positively and negatively charged respectively. Which of the following is NOT correct? (a) MgCl2 solution coagulates the gold sol more readily than iron (III) Hydroxide sol (b) Na2SO4 solution causes coagulation in both sol (c) Mixing of the sol has no effect (d) Coagulation in both sol can be brought by electrophoresis

Surface Chemistry 14.61

54. The physical states of dispersing phase and dispersion medium in colloid like pesticide spray respectively are (a) Solid, gas (b) Gas, liquid (c) Liquid, gas (d) Liquid, solid 55. Electro-osmosis is observed when: (a) Dispersion medium begin to move in an electric field (b) Dispersed phase begins to move in an electric field (c) In both 1 and 2 (d) None of these 56. Gold number of a lyophilic sol is such property that: (a) The larger its value, the greater is the peptizing power. (b) The lower its value, the greater is the peptizing power. (c) The lower its value, the greater is the protecting power. (d) The larger its value, the greater is the protecting power. 57. A colloidal solution can be purified by which of following the method of (a) Dialysis (b) Peptization (c) Filtration (d) Oxidation 58. Select correct statement(s): (a) Hydrophilic colloid is a colloid in which there is a strong attraction between the dispersed phase and water. (b) Hydrophobic colloid is a colloid in which there is a lack of attraction between the dispersed phase and water. (c) Hydrophobic sols are often formed when a solid crystallized rapidly from a chemical reaction or a supersaturated solution. (d) All of the above. 59. Which one of the following statements is false for hydrophilic sols? (a) They do not require electrolytes for stability. (b) Their coagulation is reversible. (c) Their viscosity is of the order of that of water. (d) Their surface tension is usually lower than that of dispersion medium. 60. Which one of the following statements is/are correct? (a) Brownian movement is more pronounced for smaller particles than for bigger ones. (b) Sols of metal sulphides are lyophilic. (c) Schulze-Hardy law states that the bigger the size of the ion the greater is its coagulating power. (d) One would expect charcoal to adsorb chlorine more strongly than hydrogen sulphide.

multiple choice Questions with One or more Than One Answer 1. Some following statements are about hydrophilic sols: The correct statements are: (a) They donot require electrolytes for stability. (b) Their coagulation is reversible. (c) Their viscosity is in the order of that of water. (d) Their surface tension is usually lower that of dispersion medium. 2. Pick out the correct statements (a) Gold sol does not act as a protective colloid. (b) In a reaction that follows first order kinetics the fraction reacted is given by (1-e–kt) where ‘k’ is the rate constant. (c) The temperature above which micellization takes place is known as reduced temperature. (d) Hydrolysis of an ester by an alkali follows second order kinetics. 3. Which of the following statement is correct? (a) In the coagulation of positively charged solution the flocculating power is in the order [Fe(CN)6 ]4 − > PO34−SO 24 − (b) Physiosorption decreases with increase of temperature. (c) The formation of micelles takes place only below a particular temperature called Kraft temperature. (d) The minium concentration at which micelles forms is called CMC. 4. Which of the following are correct statements? (a) Spontaneous adsorption of gases on solid surface is an exothermic process as entropy decreases during adsorption. (b) Formation of micelles takes place when temperature is below Kraft temperature (Tk) and concentration is above critical micelle concentration (CMC). (c) A colloid of Fe (OH)3 is prepared by adding NaOH in FeCl3 solution, the particles of this sol will move towards cathode during electrophoresis. (d) According to Hardy-Schulze rule the coagulation (flocculating) value of Fe3+ ion will be more than Ba2+ or Na+. 5. Which of the following are correct statements? (a) Hardy schulz rule is related to coagulation. (b) Brownian moment and Tyndall effect are the characteristic of colloids. (c) In gel, the liquid is dispersed in liquid. (d) Lower the gold number, more is the protective power of lyophillic sols.

14.62

Surface Chemistry

6. Select the correct statements: (a) In multimolecular colloidal solution atoms or molecules are held together by Van der Waals forces. (b) Cleansing action of soap is due to adsorption of oily and greasy material at hydrophilic centres. (c) Hydrophobic part of soap penetrates into oil and hydrophilic part in water to impart stability of emulsion. (d) The aggregation of ions i.e. micelle formation is dependent of temperature. 7. Which of the following statements(s) is true? (a) Lyophobic sol shows greater light scattering effect than lyophilic sol. (b) The stability of lyophobic sol is due to presence of charge on sol particles. (c) A catalyst increases the value of rate constant. (d) The absolute entropy of an element is zero at 25°C and 1 atm pressure. 8. Identify correct statement among the following: (a) Gold and As2S3 sols are multimolecular colloids 273 k 244 k X M

(b)

195 k

P

This graph is for Van der Waal’s adsorption as explained by freundlich (c) Gas Van der Waals constants a b X 140.842 0.03913 Y 137.802 0.03183 Under given conditions, more volume of gas ‘X’ is adsorbed than gas ‘Y’ per given unit mass of adsorbent (d) Coagulation of gold sol is effective with Al3+ than with Mg2+ 9. Which is/are correct for metal sulphide sols? (a) The solution particles are positively charged due to preferential adsorption of metal ions. (b) The solution particles are negatively charged due to preferential adsorption of sulphide ions. (c) The cations of added electrolytes are effective in causing the coagulation of the solution. (d) The solution is unstablilsed due to both electric charges and hydration of the particles.

10. Which of the following is/are a property of hydrophilic sols? (a) High concentration of dispersed of phase can be easily attained. (b) Coagulation is reversible. (c) Viscosity and surface tension are nearly as that of water. (d) The charge on the particles depends on the pH value of the medium; it may be positive, negative or even zero. 11. The capacity of an ion to coagulate a colloidal solution depends on (a) its shape. (b) the amount of its charge. (c) the sign of the charge. (d) its mass. 12. During electro osmosis of Fe(OH)3 sol (a) Sol particles move towards anode. (b) Sol particles move towards cathode. (c) the dispersion medium moves towards anode. (d) the sol particles do not move in either direction. 13. Consider the following statements. Which are correct? (a) The colour of hydrophobic sol depends on the wavelength of light scattered by the dispered particle. (b) The smaller the gold number value of a hydrophilic colloid, the greater is its protective power. (c) Equivalent conductance increases with dilution for an electrolyte solution while its specific conductance decreases. (d) Aluminium is obtained from Al2O3 by reduction with carbon monoxide. 14. Which of the following statements is are/correct? (a) SO 2− ion possesses greater flocculation value 4 than [Fe(CN)6]3– ions, for ferric hydroxide sol. O ||

(b) CH 3 − (CH 2 )14 − C − O − Na + has higher critical micellar concentration (CMC) than O ||

CH 3 − (CH 2 )10 − C − O − Na + . (c) Peptisation works on the principle of relative adsorption of common ions from electrolyte. (d) Above Kraft temperature with increase in temperature the micellar concentration decreases.

Surface Chemistry 14.63

comprehensive Type Questions Passage I

3. The charge on particles of aerosols is due to (a) formation and subsequent interruption of particle-particle contact. (b) by spraying conc. solution of non hygroscopic substances on clouds. (c) by spraying dry ice on super cooled aerosols. (d) throwing electrified sand or fumes of Agl.

Lyophilic colloidal sols are much more stable than lyophobic colloidal sols. This is due to the extensively solvation of lyophilic colloidal sols, which forms a protective layer outside it and thus prevents it from forming associated colloids. Lyophillic colloidal sols protect lyophobic colloidal sols from precipitation by the action of electrolytes. This is due to formation of a protective layer by lyophilic sols outside lyophobic sols. Lyophiclic colloidal sols are called protective sols. Gelatin (lyophilic) protects gold sol (lyophobic) from coagulation on the addition of sodium chloride solution. Protective powers of different colloidal sols are measured in terms of ‘gold number’ (Zsigmondy) It is defined as, the amount of protective sol in milligrams that prevents the coagulation of 10 mL. of a given gold sol, on adding 1 mL of 10 per cent solution of sodium chloride. Thus smaller the gold number of a lyophilic sol, the greater is the protective power. 1. 0.025 g of starch sol is required to prevent coagulation of 10 mL gold sol when 1 mL of 10% NaCl solution is added. What is gold number of startch sol? (a) 0.025 (b) 30 mg (c) 0.25 (d) 25 2. A compound has 0.03 as its gold number. Hence, 200 mL of gold sol will require that compound so that gold is not coagulated by 20 mL of 10 % NaCl solution. (a) 0.03 mg (b) 30 mg (c) 0.60 mg (d) 6 mg 3. A suitalble cation can get [AgI]I– coagulated from its colloidal sol. The ratio of moles of AgNO3, Pb(NO3)2, and Fe(NO3)3 required to coagulate 1 mole of [AgI] I– respectively is: (a) 1:1:1 (b) 1:2:3 (c) 6:3:2 (d) 4:2:1

Lyophilic solutions protect lyophobic solutions from precipitation by the action of electrolyte. The lyophilic solution forms a protective layer outside lyophobic solution. The protective power measures in terms of gold number. It is defined as the amount of protective solution in milligrams that prevents the coagulation of 10 mL of gold solution on adding 1 mL of 10% solution of NaCl. 1. The gold numbers for gelatin, haemoglobin, sodium acetate are 0.005, 0.03, 0.7 respetively. Which has greater protective action. (a) Gelatin (b) Haemoglobin (c) Sodium acetate (d) All are same protective power 2. Gold number of haemoglobin is 0.03. Hence 100 mL of gold solution will require haemoglobin so that gold is not coagulated by 10 mL of 10% NaCl selection. (a) 0.03 mg (b) 30 mg (c) 0.3 mg (d) 3 mg 3. What happens when gelatin is added to gold solution. (a) Gold solution is destabilized and peptisation occurs. (b) Gold solution is destabilized and cougulation occurs. (c) Gold solution is destabilized and no coagulation takes place. (d) Gold solution is stabilized and no coagulation takes place.

Passage II

Passage IV

The clouds consists of charged particles of water disperssed in air. Due to the aggregation of these particles heavy rain occurs by aggregation of minute particles. Artificial rain is possible by throwing electrified sand or silver iodide from the aeroplane which coagulate the mist hanging in air. 1. Clouds are colloidal dispersion of (a) water in air. (b) air in solid. (c) solid in air. (d) air in water. 2. The dispersion of liquid or solid in air is (a) gels (b) aerosols (c) foam (d) solution

Colloidal particles being bigger, aggragates the number of particles in colloidal solution is comparatively small as compared to a true solution. Hence, the values of colligative properties (osmotic pressure, lowering in vapour pressure, depression in freezing point and elevation in boiling point) are of small order as compared to values shown by true solutions at same concentrations. If a homogenous solution placed in dark is observed in the direction of light, it appears clear and, if it is observed form a direction at right angles to the direction of light beam, it appears perfectly dark. The Tyndall effect is due to the fact that colloidal

Passage III

14.64

Surface Chemistry

particles scatter light in all diections in space, This scattering of light illuminates the path of beam in the colloidal dispersion. Tyndall effect is used to distinguish between a colloidal and true solution. When colloidal solutions are viewed under a powerfull ultra microscope, the colloidal particles appear to be in a state of continuous zig-zag motion all over the field of view. This motion is known as Brownian movement. This motion is independent of the nature of the colloid but depends on the size of the particles and viscosity of the solution. Smaller the size and lesser the viscosity, faster is the motion. Colloidal particles always carry an electric charge. The nature of this charge is the same on all the particles in a given colloidal solution and may be eighter positive or negative. Positively charged solutions

matching Type Questions 1. Match the following Column I with Column II Column I

Column II

(a) Coagulation

(p) Due to presence of charge (q) Due to neutralization of charge (r) Due to scattering of light (s) Due to impact of the molecules of the dispersion medium with sol particles

(b) Electrophoresis (c) Tyndall effect (d) Brownian movement

Negatively charged solutions

(a) Hydrated metallic (a) Metals, e.g., copper, siloxides, eg. Al2O3.xh2O, ver, gold solutions, etc. CrO3.xH2O,Fe2O3. x H2O, etc. (b) Basic dye stuffs, e.g., (b) Metallic sulphides, methylene blue e.g., As2S3,Sb2S3,CdS sol solution (c) Hemoglobin (blood) (c) Acid dye stuffs, e.g., eosin, congo red sols (d) Oxides, e.g., TiO2 (d) Sols of starch, gum, solution gelatin, clay, charcoal, etc.

2. Match the following Column I with Column II Column I (Colloid)

Column II (Type)

(a) As2S3 (b) Sulphur

(p) Lyophobic sol (q) Multi molecular colloid (r) Macro molecular colloid (s) Associate colloid

(c) Starch (d) Soap

3. Match the following Column I with Column II Column I

1. The zig–zag movement of the colloidal particles in a colloidal sol is due to (a) air currents striking the sol (b) lightness of their mass. (c) net force of surface tension acting downwards. (d) hitting of the colloidal particles by molecules of the dispersion medium. 2. Arsenic sulphide sol is prepared by passing H2S through arsenic oxide solution. The charge developed on the particles is due to adsorption of (a) H+ (b) S2– (c) OH– (d) O2– 3. A negatively charged suspension of clay in water will need for precipitation the minimum amount of (a) Aluminium chloride (b) Potassium sulphate (c) Sodium hydroxide (d) Hydrochloric acid

Column II (p) Scattering of light (q) Purification of colloidal solution (r) Addition of electrolyte (s) Precipitation of colloidal Solution

(a) Coagulation (b) Peptization (c) Tyndall effect (d) Dialysis

4. Match the following Column I with Column II Column I (Colloidal solution) (a) (b) (c) (d)

Collodion Fog Butter Milk

Column II (Dispersed phase) (p) (q) (r) (s)

Water Cellulose Fat Oil

Column III (dispersed medium) (u) Water (v) Oil (w) Air (x) Fat

Surface Chemistry 14.65

Assertion (A) and Reason (R) Type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): A colloidal sols of Fe (OH)3 formed by peptization carries positive charge by using FeCl3. Reason (R): During formation of Fe(OH)3 sols electrons are lost by the particles. 2. Assertion (A): The formation of micelles takes place only above a particular temperature called Kraft temperature (Tk). Reason (R): Associated colloids revert back to individual ions on dilutions. 3. Assertion (A): Micelles are formed by surfactant molecules above the critical micellar concentration (CMC). Reason (R): The conductivity of solution having surfactant molecules decreases sharply at the CMC. 4. Assertion (A):Addition of AgNO3(aq) to KI(aq) gives negative sol whereas addition of KI(aq) to AgNO3 gives positive sol of AgI. Reason (R): The sol particles adsorb the common ions present in solution and acquire their charge. 5. Assertion (A): Sol particles show Tyndall effect. Reason (R): The scattering is directly proportional to size of sol particle.

Integer Type Questions 1. Gold number of starch sol is 25. If 10–x kg of starch is added to 40 mL of standard gold sol to avoid coagulation when 4 mL of 10% NaCl is added. Find out of ‘X’ value. 2. When 10–4 g of a protective colloid was added to 20 mL of a standard gold sol, the precipitation of latter was just prevented on addition of 2 mL of 10% NaCl solution. The gold number of protective colloid is ……x 10–2. 3. Among the following statements how many are correct (a) Colloidal silica is a protective colloid. (b) Lyophilic colloids are protective colloids. (c) Greater the coagulation power lesser the coagulating value. (d) Lyophobic colloids are highly solvated. (e) Metal sols are negatively charged. (f) Milk is an emulsion.

4. A detergent (C2H25SO4-Na+) solution becomes a colloidal solution at a concentration of 10-3 M. On an average of 1013 colloidal particles are present in 1 mm3. What is the average number of ions are contain particle (micelle). Express your answer by dividing with 10. 5. How may of the following emulsion have dispersed phase as an oil (a) Milk (b) Cold cream (c) Vanishing cream

Previous years’ IIT Questions 1. Among the electrolytes, coagulating agent for Sb2S3 sol is: (a) Na2SO4 (b) CaCl2 (c) Al2(SO4)3 (d) NH4Cl 2. Among the following, the surfactant that will form micelles in aqueous solution at the lowest molar concentration at ambient condition is: (2008) (a) CH3–(CH2)15N+(CH3)3Br(b) CH3–(CH2)11OSO3–Na+ (c) CH3–(CH2)6COO3–Na+ (d) CH3–(CH2)11 N+(CH3)3Br– 3. Lyophillic sols: (2005) (a) Irreversible sols (b) They are prepared from inorganic compound (c) Coagulated by adding electrolytes (d) Self stabilizing 4. Adsorption of gasses on soiled surface is generally exothermic because (a) Enthalpy is positive (b) Entropy decreases (c) Entropy increases (d) Free energy decreases (2004) 5. Rate of physiosorption increases with: (a) Decrease in temperature (b) Increase in temperature (c) Decrease in pressure (d) Decrease in surface area (2003) 6. The correct statement (s) pertaining to the adsorption of a gas on a solid surface is (are) (a) Adsorption is always exothermic. (b) Physisorption may transform into chemisorption at hight temperature. (c) Physisorption increases with increasing temperature but chemisorptions decreases with increasing temperature. (d) Chemisorption is more exothermic than physiosorption, however it is very slow due to higher energy of activation.

14.66

Surface Chemistry

7. The given graphs/data I, II, III and IV represent general trends observed for different physisorption and chemisorption processes under mild conditions of temperature and pressure. Which of the following choice (s) about I, II, III and IV is (are) correct? (2012) II

P constant

Amount of gas adsorbed

P constant

Amount of gas adsorbed

I

III

IV

200 K 250 K

Potential Energy

T

Amount of gas adsorbed

T

0

Distance of molecule from the surface Δ Hads = 150 KJ mol-1

P

(a) (b) (c) (d)

I is physisorption and II is chemisorption I is physisorption and III is chemisorption IV is chemisorption and II is phisisorption IV is chemisorption and III is phisisorption

8. Choose the correct reason(s) for the stability of the lyophobic colloidal particles. (2012) (a) Preferential adsorption of ions on their surface from the solution. (b) Preferential adsorption of solvent on their surface from the solution. (c) Attraction between different particles having opposite charges on their surface. (d) Potential difference between the fixed layer and the diffused layer of opposite charges around the colloidal particles. 9. Statement-1: Micelles are formed by surfactant molecules above the critical micelle concentration (CMC). Statement-2: The conductivity of a solution having surfactant molecules decreases sharply at CMC 10. 20% of surface sites are occupied by N2 molecules. The density of surface site is 6.023×1015 cm–2 and total surface area is 1000 cm2. The catalyst is heated to 300 K while N2 is completely desorbed into a pressure of 0.01 atm and volume of 2.46 cm3. Find the number of active sites occupied by each N2 Molescule. (IIT 2005) 11. 1 gm of charcoal adsorbs 100 mL 0.5 M CH3COOH to form a monolayer, and thereby the molarity of CH3COOH reduces to 0.49. Calculate the surface area of the charcoal adsorbed by each molecule of acetic acid. Surface area of charcoal = 3.01×102 m2/gm. (IIT 2003)

Surface Chemistry 14.67

ANSwER KEyS Assertion (A) and Reason (R) Type Questions

ADSORPTION multiple choice Questions with Only One Answer level I 1. c 2. b 3. b 4. b 5. c

6. 7. 8. 9. 10.

c a d b d

11. 12. 13. 14. 15.

d a d b c

16. 17. 18. 19. 20.

b a d b d

21. 22. 23. 24.

a a a b

multiple choice Questions with One or more Than One Answer 1. a, b, d 2. a, b, c, d

3. b, d 4. a, b, c

5. a, b, c 6. a, b, c

Assertion (A) and Reason (R) Type Questions 1. a 2. a

3. a 4. c

5. b 6. d

7. b 8. c

1. b 2. d

3. a 4. d

5. b

cOllOIDS multiple choice Questions with Only One Answer level I 1. d 2. c 3. c 4. b 5. d 6. a 7. c 8. d 9. c 10. a 11. b 12. c

13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24.

a c c a d d b b c b b c

25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36.

b c b b a d b d a c b c

b d a a c a a d c b a a

37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48.

49. 50. 51. 52. 53. 54. 55. 56. 57. 58. 59. 60.

a c d a c c a c a d c a

Integer Type Questions 1. 5

2. 4

3. 6

cATAlySIS multiple choice Questions with Only One Answer level I 1. d 2. b 3. a 4. c 5. b 6. b 7. b 8. c 9. c

10. 11. 12. 13. 14. 15. 16. 17. 18.

b a a c a c d b a

19. 20. 21. 22. 23. 24. 25. 26.

c c d a c b b d

2. a, b, c

1. 2. 3. 4. 5.

a, b, d a, b, d a, b, c, d a, c a, b, d

3. b, c

6. 7. 8. 9. 10.

a, b, d a, b, c a, d b, c a, b, c

11. 12. 13. 14.

b, c c, d a, b, c, d a, c, d

comprehensive Type Questions Passage I 1. d

2. c

3. c

2. b

3. a

2. c

3. d

2. b

3. a

Passage II 1. a

multiple choice Questions with One or more Than One Answer 1. a, c, d

multiple choice Questions with One or more Than One Answer

Passage III 1. a Passage IV 1. d

14.68

Surface Chemistry

matching Type Questions 1. 2. 3. 4.

(a) (a) (a) (a)

q p,q r,s q,v

(b) (b) (b) (b)

p p,q r p,w

(c) (c) (c) (c)

Integer Type Questions r r p s

(d) (d) (d) (d)

s s q s,v

1. 4 2. 5

2. b

3. b

4. a

5. 2

Previous years’ IIT Questions

Assertion (A) and Reason (R) Type Questions 1. c

3. 4 4. 6

5. a

1. 2. 3. 4.

c a d b

5. 6. 7. 8.

a a,b,d a,c a,d

9. b 10. 2 11. 5 × 10–19 m2

Surface Chemistry 14.69

hINTS AND SOluTIONS

ADSORPTION

3. Initial m mole of H2C2O4.2H2O = 47.7 Final m moles of H2C2O4.2H2O = (4.77)(0.5) = 23.85 m moles of H2C2O4.2H2O adsorbed = 23.85 The weight of H2C2O4.2H2O adsorbed is

multiple choice Questions with Only One Answer level I

=

x = KP1/ n 19. m 1 x Log = log K + log P m n Log K = 0.3010 ∴K=2 1 Tan 45° = =1 n n=1 x = KP1/ n ∴ m ∴ = x = 2(0.4)1 = 0.8

(23.85)126 = 3.0 gm 1000

The weight of oxalic acid adsorbed per gm char coal is = 6 gm

cOllOIDS multiple choice Questions with Only One Answer level I 19. Gold number expressed in terms of milligrams. So the amount of starch required to prevent the coagulation of sol is 2.5 mg ∴ Gold number = 2.5

20. Intial m moles of CH3COOH is = 100 × 0.5 = 50 Find m moles of CH3COOH is = 100 × 0.49 = 49 m moles of CH3COOH adsorbed is = 1 no. of molecules of CH3COOH adsorbed is = 6.023 × 1020 Surface area of char coal is = 3.015×102 m2 The adsorbed all of each molecule is =

3.015 × 102 6.023 × 1020

= 5×10

–19

2

m

Integer Type Questions 1. No. of molecules of CH3COOH adsorbed is = 2(6.023 × 1020) Surface area of char coal is = 2×3.01×102 m2 The adsorbed area of each molecule is = 5×10–19 m2

25. If gold number is less protecting power is mole 30. m moles of AgNO3 = 10 m moles of KI = (100) (15) = 15 KI is excess So I– is adsorbed on AgI Sol 43. 106 mm3 solution contains 1.2×10–3×6×1023 molecules ∴ 1 mm3 solution contain 7.2×1014 molecules No. of stearate Ions is one colloidal particle is =

7.2 × 1014 = 30 24 × 1013

Integer Type Questions 1. The amount of starch required is (25)4 = 100 mg.

2. Tan 45° = 1 ∴

1 =1 n

n=1 Log K = 0.6020 K=4 x = KP1/ n m x = 4(1)1 = 4

2. The weight of protective colloid required for 10 mL 1 solution is = × 10−4 gm = 5 × 10−5 gm = 0.05 mg 2

14.70

Surface Chemistry

Previous years’ IIT Questions 10. no. of surface sites = 6.023×1015×103 = 6.023×1019 No. of N2 molecules =

(0.01)2.46 × 6.023 × 1023 = 6.023×1017 82.1× 300

No. of sites occupied by N2 molecules = (6.023×1018) (0.2) No. of sites occupied by each N2 molecules = 2

11. The no. of m moles of CH3COOH adsorbed = 1 The no. of molecules of CH3COOH adsorbed = 6.023×1020 Surface area of char coal = 3.01×102 The Each molecule adsorbed area =

3.01× 102 = 5 × 10−19 m 2 6.02 × 1020

CHAPTER

15 Nuclear Chemistry

T

o him who is discoverer in the field (of science) the products of his imagination appear so necessary and natural that he regards them, and would like to have them regarded by others, not as creations of thought but as given realities. Albert Einstein

Chemical reactions involve mainly the rearrangement of the outer most electrons, i.e., atoms combine by sharing electrons (covalency) and by transferring electrons from one atom to another atom (electrovalency). In these processes atomic nuclei remain unchanged. Nuclear chemistry, however, investigates changes that occur in the atomic nucleus itself.

15.1 DiscoveRy of NatuRal RaDioactivity Around the turn of century, several discoveries were made that completely reshaped the fields of chemistry and physics. One of these was the discovery of radioactivity by Henri Becquerel in 1896. The discovery was quite accidental. Becquerel was studying phosphorescence in various compounds. He carried an experiment by keeping a photographic film between black papers above which potassium uranium sulphate was kept and left in sun light for a few hours. When the film was developed, Becquerel found that the film was similar to that exposed to direct sunlight. He wanted to conduct some more similar experiments, but due to bad weather, cloudy for several days, he could not continue. In the mean time, he stored his photographic plates and uranium compound in a drawer. Later, probably due to impatience, he developed one of the photographic plate and surprisingly, he found that it was blackened, an unexpected result. Becquerel and others then thought, that it was a new type of fluorescence. But later it was found that it was due to the radiations coming from uranium compound. Thus Becquerel had discovered radioactivity accidentally, but in fact the name radioactivity was given later by Marie Curie. Marie Curie won the Nobel Prize twice, once for physics shared along with her husband Pierre Curie and Henri Becquerel in 1903, and for chemistry on her own in 1911. Bequerel’s discovery attracted the attention of Marie Curie working on pitch blend, a black mineral from Bohemia. She found it be far more radioactive than it should be on the

basis of uranium content. She suspected the presence of some other more radioactive element in the mineral and decided to separate this hypothetical element. For three long hectic years she and her husband laboured feverishly, hardly stopping to eat or sleep. From over a tone of the mineral of pitch blende, which an ore containing primarily U3O8, Madam Curie isolated one new source of radioactivity by precipitating bismuth as sulphide from a solution of the ore. Since bismuth itself is not radioactive, this activity was due to new element whose sulphide was similar to that of bismuth sulphide. This new element was named polonium. In another separation, barium chloride was fractionally precipitated from the pitch blende residue and found to show high activity. This new element which behaves like barium was called radium. Some more radioactive elements were discovered are actinium (Debierne 1900), Protactinium (Hahn and Mietner 1918) and Samarium (Heversy and Paul 1933).

15.1.1 Radioactive Radiations Bequerel, while studying the properties of the radiations which he had discovered, observed in 1889 that a part of these radiations was deflected in a magnetic field in the same direction as cathode rays. Thus suggested that the radiations contained some negatively charged particles like cathode rays. Shortly afterwards, Rutherford while studying penetrating power of these radiations, arrived at the conclusion that the emanations were of two types, which he designated alpha (α) rays and beta (β) rays. In 1900, P. Willard of France discovered a third type of radiation having greater penetrating power. These radiations have been named as gamma (γ) rays. Rutherford kept a sample of radioactive substance (uranium in a cavity of a lead block (Fig 15.1). The lead block stops all the radiations except those which can pass through the slit and constituted a beam. These radiations coming from the narrow slit were allowed to pass through strong electric and magnetic fields. The deflections

15.2

Nuclear Chemistry

of the rays were recorded on a photographic plate. The radiations were found to split into the three types of rays as mentioned above. (i) The rays deflected towards the negative electrode indicates that they are positively charged particles. These are α rays. (ii) The rays deflected towards the positive electrode indicates that they are negatively charged particles. These are β rays. (iii) The rays which are not deflected at all are therefore neutral. These were called γ-rays. γ -rays

γ -rays

– – – – – – – – – – – – – – – – – – – – – –

α -rays β -rays

(a)

+ + + + + + + + + + + + + + + + + +

β -rays

α -rays

(b)

fig 15.1 (a) Deflection of radioactive rays in electric filed (b) Deflection of radioactive rays in magnetic field characteristic properties of these three types of radiations 1. Alpha (α) rays: These consist of a stream of small positively charged particles which are four times as heavy as hydrogen atom and carry two units of positive charge each. The alpha particles may be considered as helium nuclei (He2+) because the atomic mass of helium is four and its atom has only two electrons. Thus alpha particles may be regarded as fast moving helium nuclei with energy about (6-16) × 10–13 J. The characteristics of α-particles–1 are: (i) The α-particles are emitted by radioactive substances with high speed. The average velocity of an α-particle is about 1/10th of the velocity of light. (ii) Each α-particle has a mass of 4 a.m.u. and carries two units of positive charge (+2). Thus, α-particles may be considered as helium nuclei (He2+). (iii) These rays can be deflected by electric field and magnetic field. (iv) Since the mass of α-particle is sufficiently large and their velocity is very high, they posses very large kinetic energy.

(v) These rays cause ionization of the gas through which they pass. (vi) Being relatively large in size, they cannot easily penetrate through the solid matter. They can penetrate only through the thin layers of solids such as mica, aluminum etc. (vii) The α-rays produce luminosity when they fall on photographic plate. 2. Beta (β) rays: These consist of negatively charged particles which have the same e/m value as the cathode rays. Therefore, β-rays were considered to be merely the streams of electrons. Their energies are about (0.03-5.0)×10–13 J. The characteristic properties of these rays are: (i) These rays are found to consist of negatively charged particles each having mass and charge equal to that of an electron. (ii) They are emitted with high speed and their velocity is about 10 times the velocity of α-rays or the same as that of light. (iii) Although the velocity of β-rays is very high, yet because of their small size, the kinetic energy is much less than that of α-rays. (iv) These rays are deflected by electric and magnetic fields but in opposite direction to that of α-rays. (v) The β-rays have more penetrating power than α-rays. The penetration power is about 100 times more than that of α-rays. (vi) Due to their small kinetic energy, they produce less ionization of gas through which they pass. (vii) These rays affect the photographic plate but they produce less luminosity when allowed to fall on zinc sulphide. 3. Gamma (γ) rays: These are neutral particles and consist of radiation of very short wave length (and therefore very high energy) that often accompanies α-and β-emission. The energies of γ-radiation are in the same range as those of β-particles. The main characteristics are given below: (i) γ-rays travel with very high speed, almost equal to that of light. (ii) They are neutral particles and are not deflected by electric and magnetic fields. (iii) γ-rays are weak ionizers of gases. (iv) They are the most penetrating of three types of radiations. The penetration power of γ-rays is 100 times more than that of β-rays and 10,000 times more than that of α-rays. (v) Their power to affect photographic plate and cause luminosity on zinc sulphide are also poor. The main characteristics of α, β and γ-rays are summarized below.

Nuclear Chemistry 15.3

table 15.1 Characteristics of radioactive rays α-rays

Property 1 2 3 4 5 6 7 8

9

β-rays –24

Mass Charge Identity Velocity Energy Effect of electric and magnetic fields Power to ionize gases Penetrating power

6.64 × 10 g or 4 amu +2 units Helium nuclei (He2+) Nearly 1/10 of light About (6-16) x 10–13 J Deflected towards negative pole

Effect on photographic plate and zinc sulphide

Both are affected strongly more than β- and α-rays

Very large Small. Can be stopped by an aluminum foil of 0.1 mm thickness

15.1.2 Nuclear stability The reason why some atomic nuclei undergo spontaneous disintegration while others do not is still far from being completely understood. Some clues to nuclear stability may be found from the following facts. For instance, of the 265 stable nuclides known, 156 have an even number of protons plus an even number of neutrons, but only 5 have an odd number of each. Some of this information is summarized in Fig 15.2 where the number of protons present in the nuclei of the atoms is plotted against the number of neutrons also present. The points shown represent only those atomic nuclei that are stable.

140

stable nuclei

unstable region

120

1)

unstable region

100

Number of neuton

)= /p

80

f oo

60 40

No

20 0

.o

f

ns to u ne

=

20 40 No. of protons

s(n

pr

on ot

N

60

80

fig 15.2 Variation of neutrons with number of protons. Stable nuclei lie in the zone of stability in which n/p ratio raises with increasing atomic number

γ-rays –28

9.11 × 10 g –1 Unit Electrons same as that of light (0.03–5.0) × 10–13 J Deflected towards positive pole Small Large, 100 times that of α-rays. Can be stopped by an aluminum sheet of 1 cm thickness Effect less than α-rays

Negligible 0 Same as that of light Same as that of light Same range as that of β-particles Not deflected Very small Very large, 10,000 times of rays Least effect on both

It will be noted that the elements of lower atomic number, the stable isotopes are, for the most part, those in which the number of protons is approximately equal to the number of neutrons. But as the atomic number becomes larger the proportion of neutrons to protons begins to rise until in, for instance, in the element lead there are about three neutrons for every two protons in all the stable nuclei. Fig 15.2 represents what may be called the zone of stability or the ratio of neutrons to protons that tends to yield stable nuclei, and this ratio rises with rising atomic number. It should also be noted that there are no stable isotopes of elements with more than 83 protons that is, with atomic number 84 or higher. The several kinds of radioactive emission described may be thought of as changes that have as their purpose the shift of the nucleus in question into the zone of stability. The kind of ray emitted will then depend on whether the nucleus has too high or too low a neutron to proton ratio, or whether it lies in the region where all nuclei are radioactive. Thus two cases arise. (i) n/p ratio is higher than required for stability: If the neutron to proton ratio is too high for stability, then the obvious solution is for the nucleus to emit a neutron, but such simple processes are fairly rare. Lowering of the neutron to proton ratio occurs more frequently, by emission of a beta ray. Krpton-87 decays by this mechanism, which may be represented as 87 36

− Kr → 87 37 Rb + e

in which the electron (or negatron) has, of course, a charge of –1 and an almost negligible mass. It may be wondered how an electron can come from a nucleus that is supposed to contain only neutrons

15.4

Nuclear Chemistry

table 15.2 Neutron-proton ratio in some stable nuclei Isotope

12 6

C

14 7

N

16 8

O

20 10

Ne

40 20

64 30

Ca

90 40

Zn

120 50

Zr

Sn

150 60

Nd

202 80

Hg

n

6

7

8

10

20

34

50

70

90

122

p

6

7

8

10

20

30

40

50

60

80

n/p

1

1

1

1

1

1.13

1.25

1.40

1.50

1.53

energy released during the change from the excited to the normal state of a nucleus is considerably greater than that involving electronic states. This accounts for the very high frequency short wavelength and dangerously high penetrating power of gamma rays. The following chart summarizes the several changes described. The diagram shows the effect of each particle on both atomic number and the neutron number (that is the number of neutrons present) as it leaves the nucleus.

1+

Change in atomic number

and protons. The probability is that a neutron in the nucleus is converted to a proton plus an electron. The net increase of nuclear charge by one unit changes the krypton into the element of next higher atomic number namely rubidium. (ii) n/p ratio is lower than required for stability: If the nucleus of an atom has too low a neutron to proton ratio for stability, it may conveniently lower the number of protons by the process of K capture. If the nucleus “swallows” so to speak, one of the electrons that normally surround the nucleus. As this electron generally comes from the K level, the process is called K capture, but L capture also occurs. An example of K capture is found in argon −37, as follows. 37 37 18 Ar → 17 Cl n/p ratio increases from 2.06 to 2.18, argon being transmuted to chlorine. Another way in which the neutron to proton ratio may be raised to bring the nucleus nearer the zone of stability is by emission of a positive electron or positron. This occurs for nitrogen-13, which decays as follows. 13 13 7 N → 6 C n/p ratio increase from 1.86 to 2.17 becoming carbon-13. Alpha ray emission occurs frequently for elements of atomic number higher than 83. It represents a more complicated mechanism for diminishing the number of protons, even though the alpha particle itself contains two neutrons in addition to two protons. An example of alpha emission is found in polonium – 212, which decays as follows. 212 208 4 n/p ratio increase from 84 PO → 82 Pb + ;2 He 2.52 to 2.54 The product, a stable nuclide that also occurs naturally, is lead-208. Alpha emission is generally followed by several additional steps, some of which will be described later. The emission of gamma rays is different from those processes described above, since no change occurs in mass number or atomic number. A gamma ray is a form of energy, and its emission represents the discharge of excess energy from the nucleus. Just as the electron system surrounding the nucleus can have excited states from which electrons fall to normal states with emission of electromagnetic energy, so the nuclei may have excited states. But the

0

e-

1 0

n

original nucleus

e+

112-

4 2

0

1+

He

2Change in number of neutrons fig 15.3 Transmutations caused by nuclear decay

15.1.3 Measurement of Radioactivity A device of major importance in the study of radioactivity is the Geiger Müller counter. A small metal cylinder is arranged inside a glass tube, and thin metal wire passes through the center of the cylinder. The wire insulated from the cylinder and a high voltage is applied across the cylinder and wire. The glass tube contains a small amount of vapour such as a mixture of alcohol and argon. If now

Nuclear Chemistry 15.5

any ionizing radiation, as from radioactive substance enters the counter tube. It sets up a slight electric discharge from cylinder to wire. To amplify and recording circuitss

Geiger tube

However, the Curie is such a large unit becquerel Bq is selected as SI unit. One Bq is defined as one disintegration per second. The radioactivity is much harmful to living things. Depending on the effect on living organisms another two types of units are proposed. They are gray, Gy, and sievert, Sv. One gray is equivalent to 1 Kg of tissue receiving a dose whose energy equivalent is one joule. The sievert is related to gray as

Cylinder

Sievert = gray × quality number. Alpha particles have a quality number of 20 which reflects their danger compared to beta particles, which have a quality number of 1

High voltage source

Incoming radiation

Ground

fig 15.4 Gieger-Muller counter tube and part of the electric circuit for counting ionizing radiations from radioactive substances This discharge may be amplified and recorded in a number of ways and it becomes possible to measure the intensity of the radiation from any radioactive source by placing it near the Geiger tube. The intensity of the radiation may be recorded as a dial reading or as number of clicks or may be recorded graphically on a sheet of paper. Such instruments of many varieties are available commercially and find application in all types of nuclear research. γ - radiations are detected by scintillation counter. A phosphor is used in this counter which produces flash of light when it is struck by electromagnetic radiation like γ - rays for detection of γ - rays. Sodium iodide (NaI) and thallium iodide (TlI) are used as phosphor. Rutherford first used zinc sulphide (ZnS) as phosphor in detection of α-particles.

15.1.4 units of Radioactivity Radium is one of the most radioactive substances known. One gram (1 g) of pure radium emits about 3.7 × 1010 alpha particles each second. This is also the number of disintegrations of radium atoms each second. It was taken as one Curie Ci, i.e., one Ci is defined as 3.7×1010 disintegrations per second.

Name

Symbol

Definition

Curie

Ci

3.7 × 1010 disintegration per second

Becquerel Gray Sievert Rutherford

Bq Gy Sv Rd

One disintegration per second. 1 Kg of tissue receiving 1 J of energy gray × quality number of radiation 106 disintegration per sec

= 3.7×1010 disintegration per sec = 3.7×107 disintegration per sec = 3.7×104 disintegration per sec = 106 disintegration per sec = 103 disintegration per sec = 1 disintegration per sec = 3.7×104 Rutherford = 3.7×1010 bequerel Specific activity of a radioactive nuclide is its activity per kilogram (or dm3) of radioactive material.

Curie 1 milli Curie 1 micro Curie 1 Rutherford 1 milli Rutherford 1 micro Rutherford 1 Curie

15.2 NucliDes Experimental studies show that an atom possesses a small, dense, positively charged nucleus with a radius of about 10–13 to 10–12 cm, surrounded by electrons. The resulting atomic radius is about 10–8 cm. Since the mass of the electrons is very small, the nucleus possesses practically all the atomic mass. The nucleus may be represented as composed of two types of particles, neutrons and protons. The neutron, n, is an uncharged elementary particle, having a free rest mass of 1.67482×10–24 g or 1.0086654 atomic mass unit. An atomic mass unit (amu) is 1.66043×10–24 g or one-twelfth the mass of the predominant type of carbon atom. The Proton, p, is an elementary particle having a positive charge of 4.80298 × 10–10 electrostatic unit (esu) or 1.60210×10–19 columb, and free rest mass of 1.0072766 amu. The electron e has a negative charge of magnitude e = 4.80298×10–10 esu or 1.60210×10–19 coulomb and a free rest mass of 0.000548597 amu.

15.6

Nuclear Chemistry

It should be emphasized that the masses given above are for the individual particles at rest. If they are combined in an atom, or if they are moving rapidly, their mass differs from the free rest mass. Neutral atoms always contain as many electrons as protons. Atomic masses obtained with the mass spectrograph include the masses of the electrons. For many calculations, it is convenient to combine the mass of a proton with that of an electron to obtain 1.0078252 amu, the free rest atomic mass of a proton. Protons and neutrons, since they are constituents of nuclei, are called nucleons. Any individual atomic species is called nuclide. A nuclide is identified by its atomic number and its mass number. The atomic number (Z) is the number of protons in the nucleus-that is, the nuclear charge. The mass number (A) is defined as the total number of nucleons in the nucleus. Since the masses of individual combined nucleons are approximately 1 amu, every atomic mass, measured in atomic mass units, is approximately an integer. The mass number, therefore, is also defined as the integer lying closest to the atomic mass. The neutron number (N) is the number of neutrons present in the nucleus. Therefore, N = A–Z A nuclide is represented by the chemical symbol preceded by the atomic number as a subscript and followed by the mass number as superscript. Thus 16 8 O represents the predominant oxygen nuclide with atomic number 8 and mass number 16. The subscript is frequently omitted, since a given element always has the same atomic number. The complete notation for proton is 11 p ; for a neutron 10 n . In present notation the decay of radium is written 226 88

222

Ra  →86 Rn + 42 He

This may be considered the reaction of the nuclei or the net reaction of neutral atoms. The decaying nuclide 22 − 226 88 Ra − is called parent nuclide and the product −86 Rn − is called the daughter nuclide. The daughter nuclide in one reaction can be the parent in a later reaction. Nuclides having the same atomic number (Z) but different mass numbers are called isotopes. Therefore they have different neutron number, e.g., 11 H, 12 H and 13 H . Nuclides having the same mass number (A) but different atomic numbers are called isobars. These also have dif17 17 ferent neutron number, e.g., 17 7 N, 8 O and 9 F . Although isobars have the same mass number, their actual masses differ 17 17 slightly. Thus the actual masses of isobars 17 7 N, 8 O and 9 F are 17.008580, 16.999133, and 17.002093 amu respectively. Nuclides having the same neutron number (N) but different mass numbers are called isotones. These have different atomic numbers.

table 15.3 Characteristics of nuclides Types of nuclides

Characteristics

Examples

Isotopes

Z = same; A = different

Isobars

Z = different, A = same

1 2 1 1 228 88

238 H, H, 31 H; 235 92 U 92 U 228 Ra 228 89 Ac 90 Th

Isotones

N = same, Z = different, 39 40 18 Ar 19 K A = different U−X2, U−Z Isomers N = same, P = same, Z = same, A = same. Shows different types of radioactivity CO2, N2O Isosters No. of electrons same No. of atoms same. Physical properties same 235 231 Isodiaphers Isotopic excess mass 92 U 90Th (N-P) same

15.2.1 types of Radioactive Decay 1. Alpha decay: With rare exception α-activity has been observed only with nuclei (Z>60). When a nucleus emits an α-particle, the atomic number (Z) and the neutron number (N) of the nucleus are each reduced by 2. Since Z/N ratio is less than unity in the region of α-active nuclei, this removal of an equal number of protons and neutrons reduces Z/N ratio. In the α-decay isodiapheres are formed.

Atomic mass number Atomic number

Parent nuclide α → Daughter nuclide A A−4 Z

Z-2

Examples 226 88

4 Ra  → 222 86 Rn+ 2 He

238 92

U  → 23490Th+ 42 He

2. Beta-decay: The β-particle emitted by a nuclide undergoing β-decay results from the conversion of a nuclear neutron into a proton. 1 → 11 p + −10 e 0 n  The electron thus produced escapes as a beta particle leaving the proton in the nucleus, e.g., 214 82

0 → 214 Pb  83 Bi + −1 e 0 Th  → 234 91 Pa + −1 e

234 90

In the β-decay the daughter nuclide formed is an isobar of parent nuclide. If a nuclide decays by emitting an alpha particle and 2 beta particles successively, the resulting nuclide will be

Nuclear Chemistry 15.7

the isotope of parent nuclide since they have same atomic number but their masses differ 4 units. 3. Wolfgang Pauli in 1931 suggested that an unobserved particle is emitted simultaneously with the β-particle. This particle was named the neutrino from the Italian for “small neutral one”. In 1956, Frederick Reines and Clyde Cowan Jr., obtained experimental evidence of the conversion of protons into neutrons and positrons by neutrinos. 4. A small part of the energy loss which was attributed previously entirely to neutrinos has been found to result from emission of a low – intensity continuous spectrum of electromagnetic radiation called inner bremstrahlung which apparently accompanies all β-decay and covers the same energy range as the neutrinos. Because of its low intensity it is observed with difficulty. 5. Internal–conversion electrons: In 1914, Rutherford suggested that a γ-photon from nucleus of an atom might transfer all its energy to an electron of the same atom. Such an internal photoelectric effect is frequently observed. The energy of each ejected electron equals that of the γ-photon minus the binding energy of the electron in the atom. These electrons, called internal–conversion electrons. They can be easily distinguished from those emitted during β-decay because they are each emitted with one of several definite energies, giving a line spectrum rather than the continuous spectrum characteristic of β-emission. Measurement of the energies of these electrons and knowledge of the binding energies of electrons in the various electron shells of the atom permit determination of the energy of the photon. The probability of internal conversion of γ - photon is a function of the electron density at and near the nucleus. Therefore the fraction of photons internally converted during decay of a given nuclide is influenced by the chemical state of the nuclide. Internal conversion removes electrons from their positions near the nucleus. When an electronic position near the nucleus is vacant, another planetary electrons falls into it and x-ray photon is emitted. Therefore x-radiation accompanies internal conversion of γ-photons. 6. Positron decay: Though no natural nuclides decay by positron emission, a great many artificial nuclides do. In several respects positron decay is similar to β-decay, the significant difference being that annihilation radiation always accompanies positron decay. The positron emitted by a radioactive nuclide from conversion of a nuclear proton into neutron. 1 1

→ 01 n+ 01β p 

Example 30 15

0 P  → 30 14 Si + 1 β

7. Electron capture: In 1938, Luis Alvarez observed evidence of a second mechanism converting a nuclear proton into a neutron – the capture of planetary electron by the nucleus. Removal of electron from near the nucleus results in emission of x radiation as an internal conversion of γ-photons. The wavelengths of the observed x-rays are characteristic of the x-ray spectrum of the daughter nuclide. In electron capture the change in the nucleus results from 0 1 → 01 n −1 e+ 1 p  Example → + 73 Li Be + 01 e  Neutrinos are emitted during electron capture and are probably more energetic. In the decay of 7Be and 37Ar by electron capture, the neutrino emission has been found to give the respective daughter nuclei 7Li and 37Cl. Gamma radiation accompanies electron capture by some nuclides—for example 74 Be _ but others only evidence of such nuclear change is a spontaneous emission of xradiation and inner bremsstrahlung. Nuclides having identical atomic numbers and identical mass numbers but different radioactive properties are called radioactive isomers – first observed by Otto Hahn in 1921. By careful measurements, Hahn learned that the β-decay curve of the daughter of UX, was apparently the sum of two components, the UX2 β– emission with a half- life of 1.18 min and a small amount of β– emission decaying with a half life of 6.7 hours. He attributed the latter to a nuclide, UZ, having the same atomic and mass number as UX2. Later investigations showed that UZ is formed from about 0.15 percent of the UX2 by the emission of 0.394 Mev. γ- photons which changes neither atomic number nor mass number, but does produce a somewhat more stable radioactive nuclide. Conversion of one isomeric form into another by emission of a γ- photon is called isomeric transition and like other radioactive decay, always occurs with a characteristic half-life for the parent isomer. Numerous isomeric transitions have been observed artificially produced nuclides with half-lives ranging from 2.8 ×10–10 sec to 3.65 years. 7 4

15.2.2 Disintegration theory Rutherford and Soddy in 1903, as a result of their extensive studies on radioactivity of substances, particularly of uranium and actinium, proposed the disintegration theory. It was suggested that on emission of α or β-rays which are usually but not always accompanied by emission of γ-rays the parent nuclides are get transformed into those of new elements, the daughter nuclides. These daughter nuclides

15.8

Nuclear Chemistry

themselves are radioactive and hence disintegrate. In this way the process of radioactive disintegration continues, until the original parent nuclide gets transformed into a stable one. A partial chain of such transformation is indicated in the following example. →

226

Ra α → 222 Rn α → 218 Po α → 214 Pb β → 214 Bi →

This example clearly indicates that after a period of time the decay series, which involve large number of atoms, would result in a mixture of atomic species such as radium, radon, polonium, lead and bismuth. Some of these atoms would be emitting alpha rays and others β- and γ-rays. Hence as a result, any radioactive substance will emit a mixture of α- β- and γ-rays.

211

The radioactive bismuth 83 Bi emits a beta particle 211 resulting in the formation of polonium 84 Po , which belongs to group VIA, i.e., one place to the right of the parent element bismuth These examples are summarized into the following way.

IIIA

IVA

(

)

215 84

211 Po  → P b 82

Which belongs to group IVA, i.e., two places to the left of the parent element polonium. (ii) When a β-particle is emitted, the daughter element will have atomic number one unit more than that of the parent element and will thus belong to the next group to the right in the periodic table. Example 210 82

Pb β →

210 83

Bi

Radioactive lead ( Pb ) emits beta particle resulting 211 Bi ) which bein the formation of radioactive bismuth ( 83 longs to group VA, i.e., one place to the right of the parent 211 element lead ( 82 Pb ) . 210 82

VIA

–α

211 Pb 82

–β

15.2.3 soddy–fajan–Russel’s Group Displacement law Since radioactive disintegration is a process in certain elements by which one chemical element is transformed into another, hence there must be a direct relation between the particle given off and the type of the element that is produced. Soddy, Fajan and Russell discovered in 1913 a simple law known as the group displacement law. Which gives the relationship between the particle emitted and the element produced. It can be stated as follows. (i) When an α-particle is emitted, a new atom is formed, which has the mass number less by four units and atomic number less by two units than those of the parent atom. This is due to the fact that α-particle has an atomic mass of four units and nuclear charge of two. Therefore the decay product (daughter) will be present two groups left to the original (parent) atom in the Mendeleev periodic table, e.g., Polonium 215 84 Po is in the VIA of the periodic table. On losing alpha particle it is transformed into radioactive 211 lead ( 82 Pb ) .

VA

VIIA

215 Po 84

211 Po 84

211 84 Bi

211 211 211 Here 215 84 Po and 84 Po are isotopes while 82 Pb, 83 Bi 211 and 84 Po are isobars. From this, we conclude that a beta particle emission results in the formation of an isobar while an isotope is produced as a result of the combined emission of one alpha particle and two beta particles. 234 90 Th is a member of actinide series. All the fourteen members of actinides have been placed in the III B group along with protactinium. It emits a beta particle and 234 is transformed to protactinium 91 Pa which also belongs actinide series, i.e., group III of the periodic table. 234 90

234 Th  → e91 Pa +

0 −1

Here group displacement law is not followed hence, group displacement law should be applied with great care, especially in the case of lanthanides (57 to 71) actinides (89 to 103, VIII group 26 to 28; 44 to 46; 76 to 78), IA and IIA groups. solved Problem i Calculate no. of α- and β-particles emitted when 238 92 U Pb changes into radioactive 206 . 82 (IIT 2000) Solution: Suppose the number of α-particles emitted be a and β-particles be b 238 92

U  →

206 82

Pb + a 42 He + b

0 −1

e

Balancing the mass no. on both sides of the above equation.

Nuclear Chemistry 15.9

238 = 206 + 4a + 0 × b ∴a=8 Hence number of α-Particles emitted = 8 and on balancing the above equation with respect to atomic number on both sides 92 = 82+2a+b (−1) = 82+2×8−b = 98−b ∴b=6 So the no. of β-particles emitted = 6

−

dN = λdt N

This on integration gives −ln N = λt + C (3) Where C is the integration constant. To start with, i.e., when t = 0, let the number of disintegrating atoms present be N0. Substituting the values in the above equation, we get − ln N = λt − ln N 0

Problems for Practice

or ln N 0 − ln N = λt

1. Calculate the mass no. and group in the periodic table for RaC in the following change.

N0 = λt N 1 N ∴ λ = ln 0 t N or ln

α α β α Ra − → Rn − → RaA − → RaB − → RaC

2. Calculate the group of element formed in the final stages of radioactive changes given below. 235 92

α

U  →

231 90

−β

Th →

231 91

−α

X →

227 89

Ac

3. How many α and β particles are emitted in passing 208 down from 230 90 Th to 82 Pb?

15.3 Rate of DisiNteGRatioN The activity of a radioactive element is measured by the rate at which it changes into its daughter element. Rate of disintegration depends on the nature of the radioactive element and is independent of external factors such as temperature pressure or the state of chemical combination. According to the law, the radioactive decay at an instant is directly proportional to the number of disintegrating nuclides existing in the sample under study (or), the quantity of a radioactive element which disappears in unit time (rate of disintegration) is directly proportional to the amount present. Let us consider a radioactive element A which is changing into another element B. Let N atoms of A be present initially. →B A 

If in a small time dt, number of atoms which decay is dN then –dN/dt gives the rate of disintegration which must be proportional to the number of atoms present (=N). Mathematically, dN ∝N − dt dN or − = λN dt (1) Where λ is the proportionality constant called the radioactive constant or decay constant or disintegration constant. Rewriting the equation (1)

(4)

Changing the base logarithm from e to 10, we get N 2.303 (5) log 0 t N From this equation the dimensions of λ are time–1 (s–1, min–1 etc) From equation (4) N0 N = e λt or = e − λt (6) N N0 λ=

∴ N = N 0 e − λt The equation (6) shows that the number of active nuclides decreases exponentially with time.

Radio activity

226 88

(2)

Time fig 15.5 Exponential manner of radioactive decay This fundamental law is statistical one which shows that the probability of decay of a given nuclide in a short interval of time dt is dNt/Nt= -λdt which is independent of

15.10

Nuclear Chemistry

the age of the nuclide. It is not possible to tell that which of the Nt nuclide will disintegrate at the instant t but only infer that the rate of disintegration will be proportional to Nt. The activity or the strength of radioactive substance is defined as  d Nt  (7) A=  = λNt  dt  The number of atoms disintegrating in unit time can be obtained by finding the number of particles, such as α, emitted in this time by the sample under study. The activity per unit mass of the sample is called its specific activity Combining the equations (5) and (7),  d Nt  − λt A= (8)  = λN 0 e  dt  The activity of a sample decreases exponentially with time, so that a plot of the logrithan of activity, log A against time t gives a straight line. The slope of this is the value of λ.

15.3.1 Half-life Period The nuclei of some atoms have a certain inherent instability. Out of every large group of atoms—say, of radium— some will disintegrate soon; others, later. For a single atom of radium it would impossible to predict the moment of disintegration; but for the very large numbers of atoms present in weighable amounts of radium, the number of atoms that will have disintegrated in any given time interval is predictable. The mass of radioactive element that will disintegrate in a given time depends on the inherent instability of the nuclei and on the number of atoms taken at the start of the measurements. The time necessary for half the mass of any sample to decay is called the half-life of the nuclide under consideration. If one were to start with 1 gram of radium, then at the end of one half-life, there would be 0.5 gram of radium left. At the end of the second half-life, there would be 0.25 gram left, and soon. The concept of half-life period was introduced by Rutherford in 1904. It is a measure of the radioactivity of the element since shorter the half-life period of the radioactive element larger the number of its atoms that are disintegrating in a unit time. Let t0.5 be the half-life period (the time taken by half of the atoms to disintegrate), i.e., after time t0.5=N=No/2

∴

N0 N0 N0 = = =2 N N0 − N0 / 2 N0 / 2

Substituting λ

N0 =2 in the equation (5), we get N 2.303 log 2 = t 0.5 =

∴ t 0.5 =

2.303 0.693 × 0.3010 = t 0.5 t 0.5 0.693 λ

(9)

Since λ is a constant it is evident from the above that the rate of decay of a nuclide, and hence its half-life period, is virtually independent of temperature, pressure or any other ascertainable condition. Thus the half-life period t0.5 of a particular radioactive element is independent of the amount of the radioactive element.

15.3.2 average life or Mean life Period Rate of disintegration of a radioactive element is proportional to the number of atoms of this element present. As disintegration proceeds in a given sample, the number of its atoms goes on decreasing and with this the rate of disintegration also goes on diminishing. The process of disintegration will therefore, continue indefinitely. Total decay period of any element is therefore infinite use of total decay periods for radioactive elements would thus be meaningless. Instead of total decay period or total life period another term generally used is average lifer period (tav) which is the reciprocal of the disintegration constant λ 0.693 From equation (5) λ= t 0.5 t 1 = 0.5 = 1.44 t 0.5 (10) λ 0.693 Thus average life period = 1.44 × Half-life period Average life period tav =

15.3.3 Radioactive equilibrium Let us suppose that there is a large quantity of a nuclide, that because of its long half-life, decays at a rate that remains almost constant over many years. Let this nuclide A decay into B, which has a short half-life and which in turn decays to a stable nuclide C thus 0.5 long 0.5 short A t → B t  → C ( stable )

Because A is present in large amount and has a long half-life, the number of atoms of B being formed per unit time is practically constant. But B, with its short half-life, is decaying at a rate that depends on the number of atoms of B present. It is clear that a steady state will be reached in which the number of atoms of B disintegrating per unit

Nuclear Chemistry 15.11

time is just equal to the number being generated from A. Whether the number of atoms of B present at the steady state is large or small depends on the quantity and half-life of A and on the half-life of B. This phenomenon of the steady state is frequently encountered. It makes possible the natural occurrence of many radioactive elements, which, if deprived of the parent element (A as above), would quickly disappear entirely. The steady state is sometimes refered to as a radioactive equilibrium, but the equilibrium that occurs is quite different from the reversible equilibrium as described in chemical equilibrium. Radioactive equilibrium is set up as a result of processes that operate in one direction only-namely, that of decay. This situation is not unlike the steady state of mountain lake where water flows at one end out at the other. The attainment of radioactive equilibrium can be shown as follows A  → B  → C  → D Long lived

( Radioactive descendants)

Parent After a sufficient length of time, the rate at which a daughter element is formed in the decay series becomes equal to the rate at which it decays. In other words, a state of equilibrium, termed radioactive equilibrium is reached. For example, let us say ‘C’ is being formed at the same rate as the rate of decay. If λ1 and λ2 represent the disintegration constants of B and C respectively and N1 and N2 represent the number of atoms of these elements present at any time, then during a certain period. No. of atoms of B which disintegrate = No. of atoms of C produced = λ1 N1 and No. of atoms of C which disintegrate = λ2 N2 In the state of radioactive equilibrium λ 1 N1 = λ2 N2 N1 λ 2 ( t av )1 = = N 2 λ1 ( t ) av

(t

i.e., λ1>λ2. As t becomes very large e–λt will approach zero faster than the e –λ2t Accordingly the equation (11) simplifies to λ1 λ 2 − λ1

N2 =

(N o, e

− λ1 t

1

N1o, 1 e − λ1 t = N1

(12)

)

λ1 N2 = N1 λ 2 − λ1 i.e., when equilibrium is reached, both parent and daughter activities decrease at equal rates, the rate of decrease. Being dependent of the half-life period of the parent. This type of equilibrium is called transient equilibrium and there is appreciable decrease in the total activity. Case II If λ1>λ2 i.e., the average life of the daughter is greater than that of parent: If one starts with pure radioactive parent, the activity of daughter will increase up to tmax and then decreases with its own rate. Here e–λt approaches zero after certain time, i.e., the parent substance vanishes practically and no equilibrium is established.

15.3.4 Parallel Path Decay Sometimes certain radioactive elements (say A) may decay in two different parallel paths (say into B and C).

A

B C

Decay constant of ‘A’ = decay constant of ‘B’ + Decay constant of ‘C’ λA = λB + λC Here λB = [fractional yield of B]X λA λC = [fractional yield of C]X λA

15.3.5 Maximum yield of Daughter Nuclide av

= 1/ λ)

(11)

2

Therefore, the elements in radioactive equilibrium are present in the ratio of their average life-periods. The element is present in larger amount and the relative number of atoms of parent and daughter inversely proportional to their decay constants. Under these conditions the daughter breaks up as fast as it is formed. Hence daughter is said to be in a secular or permanent equilibrium with parent. Case-I λ183 even extra neutrons are not able to maintain stability. This is the reason why all elements with a nuclide of Z>83 are unstable.

16O

9

12 C

8

4 He

7 6 5 4 3 2 1 0

20 40 60 80 100 120 140 160 180 200 220 240 260 Mass Number

fig 15.6 Nuclear binding energy curve A graph of binding energy per nucleon v/s the mass number is given in Fig 15.6. It shows that the atom with mass number in between 60-65 have the highest value of binding energy for their nuclei. Hence the nuclei of such atoms (medium sized nuclei) are more stable. It is found that to become much more stable, the lighter elements with low binding energy tend to convert into comparatively heavier elements (medium sized nuclei) while the heavy elements also with low binding energy tend to convert into comparatively lighter elements (medium sized nuclei). Thus, they attain high binding energy. These types of conversions lead to the phenomenon of nuclear fission and nuclear fusion

Nuclear Chemistry 15.17

table 15.4 Binding energies of typical elements Binding Binding energy energy particle mev per mev

Hydrogen

2 1

H

0.00235

2.20

1.09

Helium

4 2

He

0.0302

28.12

7.03

Lithium

7 3

Li

0.0420

39.10

5.59

Beryllium

9 4

Be

0.0623

58.00

6.44

Boron

11 5

B

0.0814

75.80

6.89

Carbon

12 6

C

0.0986

91.80

7.66

Nitrogen

14 7

N

0.112

104.00

7.46

Oxygen

16 8

O

0.137

127.00

7.94

Argon

40 18

Ar

0.368

342.00

8.57

Iron

56 26

Fe

0.523

487.00

8.70

Krypton

84 36

K

0.893

729.00

8.66

Tin

118 50

Sn

1.080

1,011.00

8.52

Xenon

132 54

Xe

1.19

1,110

8.38

Lead

204 82 238 92

Pb U

1.700

1,580

7.77

1.900

1780

7.45

Uranium

15.5.3 Mass Defect and Packing fraction The mass defect of the atom is the difference of its atomic mass and its mass number. For carbon

12 6

C , the mass defect

is zero because its mass M = 12 amu and A = 12. ∴ Mass defect ∆ = M–A A mass defect gives the deviation of the atomic mass from the whole number A. The packing fraction P is the mass defect per elementary particle in the nucleus. ∆ M−A = A A A graph between packing fraction p and mass number A is shown in Fig 15.7. The graph shows the interesting results. (i) The packing fraction is positive for elements having mass number below 20. (ii) The packing fraction is negative for elements having mass number between 20 and 200. (iii) The packing fraction is positive for elements having mass number beyond 200. p=

+ Packing Fraction

Element

Mass Nucleus defect

20

40

0 Carbon –

Mass number A fig 15.7 Packing fraction curve

The packing fraction is zero for carbon A = 12 and Z = 6. This does not mean that its binding energy is zero. The mass defect and packing fraction only show their relations with respect to carbon. A nucleus having positive packing fraction has its mass M greater than its mass number A. It means that the loss of mass is due to binding energy arrangement in this nucleus is less than carbon (It is common to represent packing fraction in terms of P×104. Packing fraction P is defined as 104 times the average mass excess per nucleon.

15.6 NucleaR ReactioNs Like chemical reactions, nuclear reactions are also accompanied by release or absorption of energy. This is represented by addition on the right hand side. Thus the nuclear reaction between aluminum and α-particle may be represented as 27 13

30

Al + 42 He  →15 P +10 n + Q

Similarly, interaction between nitrogen and α-particle is written as 14 7

N + 42 He  →

17 8

O + 11H + Q

If the energy is released in a nuclear reaction, it is called exoergic and if the energy is adsorbed in a nuclear reaction, it is called endoergic reaction.

15.18

Nuclear Chemistry

15.6.1 Nuclear Reactions versus chemical Reactions Nuclear reactions differ from chemical reactions in the following respects: 1. Chemical reactions involve some loss, gain or overlap of outer orbital electrons of the reactant atoms. On the contrary, nuclear reactions involve emission of alpha particles, beta particles or positrons from inside the nucleus. 2. The nuclear reactivity of radioelement is independent of its state of chemical combination. The radium atom in elementary radium and radium ion in RaCl2 are similar in their radioactive behavior. 3. A chemical reaction is balanced in terms of mass only, but any nuclear reaction must be balanced in terms of both mass and energy. 4. Nuclear reactions are accompanied by energy changes which far exceed the energy changes in chemical reactions. 5. The chemical reactions are dependent upon the number of extra nuclear electrons whereas nuclear reactions are independent of the electrons but depend on the nature of the nucleus. 6. The chemical reactions are dependent on external conditions such as temperature, pressure whereas nuclear reactions are not. 7. The chemical reactivity of an element depends upon the nature of the bonds present in the concerned compound on the other hand, the nuclear reactivity of the element is independent of its state of chemical combination, e.g., radium whether present as such or in the form of its compound exhibit similar radioactivity.

Problems for Practice 27. With what velocity should an α particle travel towards the nucleus of a copper atom so as to arrive at a distance 10–13 meter from the nucleus of the copper atom? (IIT 1997) and 28. The radioactive disintegration of 239 94 Pu α-emission process is accompanied by loss of 5.24 M 4 ev/dis. If t0.5 of 239 94 Pu is 2.44 × 10 year, calculate the energy released per year from 1.0 g sample of 239 94 Pu in KJ. 29. Calculate the energy released in M ev and joules in the following nuclear reaction: 2 1

30.

31.

32.

33.

34.

solved Problem 5 35 The mass defect for 17 Cl is found to be 0.32 amu. Calculate the binding energy per nucleon. Solution:

Binding Enegy (BE) = ∆m × 931.48 Mev = 0.32 amu × 931.48 = 298.07 Mev B.E / nucleon = =

Total BE No.Of nucleons

298.07 = 8.516 Mev 35

35.

36.

H + 21 H  → 23 He + 01 N

Assume that the masses of 21 H , 23 He and neutron are 2.0141, 3.0160 and 1.0087 amu respectively. (MLNR 1997) Consider an α-particle just in contact with a 238 92 U nucleus. Calculate the columbic repulsion energy (i.e., the height of columbic barrier between 238U and α-particle) assuming that the distance between them is equal to be sum of their radii. One of the most stable nuclei is 55Mn. Its nucleide is 55Mn. Its nucleide mass is 54.938 amu. Determine binding energy per nucleon. When a positron and electron collide, they are annihilated and two gamma photons of equal energy are emitted. Calculate the wave length corresponding to this gamma emission. 0 0 → 2 gamma photons of equal energy. +1 e + −1 e  ΔE for the combustion of 1 mol ethylene in oxygen is –1.4 × 103 K.J. What would be the loss in mass (expressed in amu) accompanying the oxidation of one molecule of ethylene? it is proposed to use the nuclear fusion reaction 2 21 H  → 31 He + 11 H + energy to produce industrial electricpower. If the output is to be 80 Mw and the energy of the reaction is used with 40% efficiency, how many g deuterium fuel be needed per day? Natural nitrogen is 99.63% 14N with mass 14.00307 amu and 0.37% 15N with mass number 15.00011 amu. Which is more stable 14N or 15N? Calculate the volume of H2 gas at N.T.P. which would be produced per minute from radiolysis of water in a 1000 Kw reactor assuming that 100 ev of energy produces 1 molecule of water 2H 2 O → H 2 + H 2 O 2

Nuclear Chemistry 15.19

37. Calculate the energy released in the reaction 35 17

1 0

35 16

1 1

Cl + n  → S+ H

[(Cl-35) = 34.9688 amu, (S-35) = 34.9690 amu; (H-1) = 1.0078 amu, 01 n = 1.0087 amu] 38. Calculate the maximum energy of a β-particle emitted from a 14C nucleus which does not emit any γ-radiation. The masses of C-14 and N-14 isotopes are 14.003242 amu and 14.003074 amu respectively. 14 6

C  →

14 7

N + −10 e

39. An isotopic species of lithium hydride 6Li2H is a potential nuclear fuel on the basis of the reaction 6 3

Li + 21 H  → 2 42 He

Calculate the expected power production in Mw associated with 1.00 g of 6Li2H per day (process is 90% efficient). = [36 Li 6= .0152 U , 2 H 2.0140 U, 42 He = 4.00260 U ] 40. Calculate the packing fraction, mass defect and energy 40 released in the formation of argon atom 18 Ar . Isotopic mass of Ar = 39.96 2384 amu. 41. The sun radiates energy at the rate of 4 ×1026 J sec–1. If the energy of fusion process 4 11H  → 42 He + 2

0 −1

e

atomic particle. In all cases, except those involving neutrons, electrons, or gamma rays, the bombarding particle has a positive charge. Since the nuclei of atoms are also positively charged, it is obvious that there will be electrostatic repulsion to be overcome before the bombarding particle (projectile) can enter the target nucleus. This repulsion is overcome if the projectile is given a high velocity and, therefore, considerable kinetic energy. In the earliest experiments involving nuclear transformation, the only projectiles available were those ejected during the course of the disintegration of naturally radioactive atoms. The first artificial transmutation to be achieved was announced by Lord Rutherford in 1919. The alpha particles emitted by some of the decay products from radon are given off at high velocities. He was able to show that in a small fraction of the atomic collisions between the alpha particles and the nitrogen atoms, hydrogen atoms were produced. The nuclear equation for the reaction is 14 7

N + 42 He  → 11H +

17 6

O

Such an equation is often written in an abbreviated form 14

N(α p) 17O

And is referred to as an α, p reaction. Another example of an α, p reaction is the conversion of aluminium-27 to silicon-30 27

is 27 Mev. Calculate the amount of hydrogen that would be consumed per day for the given process? 42. One mole of fuel on combustion liberates 1800 KJ of energy. Calculate the loss in mass during the combustion.

15.6.2 Artificially Induced Nuclear Reactions So far we have discussed just one type of nuclear reaction: that which occurs during the disintegration of a naturally radioactive substance. In addition to this type, there are a wide variety of other nuclear reactions that can be brought about the bombardment of nuclei with such subatomic particles as protons, alpha particles, neutrons and electrons or with gamma rays. These include (i) Transmutation to a new stable isotope (ii) Transmutation to a new radioactive isotope (iii) Nuclear fission, and (iv) Nuclear fusion Which of these several reactions occurs depends on (i) the kind of nucleus being bombarded, (ii) the nature of the bombarding particle and (iii) the energy of the bombarding particle. In order for an artificially induced nuclear reaction to occur, the target nucleus must be bombarded by some sub

Al(α p) 30Si

Bombardment by α-particles does not necessarily yield a proton as one product. A famous reaction, which led J. Chadwick to the discovery of the neutron in 1932, was the α, n reaction of beryllium-9. If beryllium-9 is bombarded with high energy alpha particles, the product is carbon-14 plus a neutron. 9 4

Be + 42 He  → 126 C + 01 n

High-velocity protons may cause the conversion of oxygen-18 to fluorine-18 18 8

O + 11H  →

18 9

F + 10 n

and deuterons (deuterium nuclei) may cause changes such as the conversion of sodium-23 into another isotope of sodium. 23 11

24

Na +12 H  →11 Na +11 H

The neutron has unique advantage as projectile for nuclear bombardment. Neutrons, having no charge, are repelled neither by the electron cloud surrounding the nucleus nor by the nucleus itself. Furthermore, neutrons are readily produced by the comparatively simple method of mixing beryllium with a little radium, which yields neutrons by the 9Be(α, n) 12C reaction mentioned above.

15.20

Nuclear Chemistry

Atomic number change

Occasionally, it is possible to use particles considerably heavier than those mentioned. For instance, the nucleus of the carbon atom has been used to produce elements of very high atomic number. It should also be noted that the process of nuclear fission produces elements of medium atomic number. It will be convenient to summaries and amplify the preceding discussion by a chart showing the changes produced in any nucleus by various kinds of bombardment. The changes are indicated by the change (gain or loss) of neutrons and the change (gain or loss) of atomic number. It will be recalled that the mass number is the sum of the atomic number plus the number of neutrons these changes are shown Fig 15.8. 2+

α, 2n

1+

p, n

0

γ, n n, 2n

p, γ d, n Original nucleus

1–

d, α

γ, P

2–

n, α

0

α, n

15.6.3 types of Nuclear Reactions The nuclear reactions may be classified either in terms of the overall energy transformation that occurs or in terms of the types of projectiles that are used in the bombardment of target nuclei. 1. Nuclear reactions induced by charged projectile: The penetration of a nucleus by a charged particle will not only increase the energy of the nucleus but also tend to increase the proton neutron ratio within the nucleus. So a compound nucleus thus formed may undergo positron emission or orbital-electron capture as well the ordinary modes of decay. 2. (i) Nuclear reactions induced by protons: Fast moving protons have been found to bring about at least four types of transmutations giving (a) neutrons (b) deuterons (c) α-particle and (d) γ-rays. Examples of these types of transmutations are as follows.

n, γ d, P n, γ d, P 1+

(a ) ( p, n ) type ( b) ( p, d ) type (c) ( p, α ) type (d ) ( p, γ ) type

1– Neutron change

fig 15.8 Change of atomic number and number of neutrons produced by various bombardments For example, deuteron bombardment of nucleus may liberate a neutron, giving a, d, n reaction. This raises the atomic number by one unit, but makes no net change in the number of neutrons. The mass number is raised by one unit. The limitation inherent in the use of naturally available particles is that their energy, is determined by the nature of the radioactive decay process and cannot be varied at will. To overcome this limitation and to have available particles of very high kinetic energy various devices such as the cyclotron, the Van De Graff accelerators and the betatron have been invented. The operating principle of all accelerators is the same, although the details of construction and operation vary widely, depending on the energy range and the type of particle employed. In every case, because the particle is charged, it can be accelerated if one take advantage of the attractive and repulsive forces that exist between unlike and like charges respectively These attractive and repulsive forces serve to push the particles to high velocities so that on encountering the target nucleus, they are able to produce the desired nuclear reaction.

23 11 9 4

23 Na +11 H  → 12 Mg + 10 n

→ 84 Be + 12 H Be +11 H 

14 7

4 → 11 N +11 H  6 C + 2 He

27 13

28 Al +11 H  → 14 Si + γ

(ii) Nuclear reactions induced by deuterons: Deuterons are the most effective charged projectiles. It gives 16 Mev additional energy to the structure. The binding energy within the deuteron is only 2 Mev, hence the fusion of the deuteron with target nucleus involves a net gain of energy of 14 Mev. The transmutation carried by accelerated deuterons gives (a) proton (b) neutron and (c) α-particle. Examples are as follows. (a) (d, p) type: This type of transmutation is most common and occurs in light and heavy elements. 2 1

H + 12 H  → 13 H + 11H

209 83

→ Bi + H  2 1

210 83

Bi + H 1 1

(light element ) ( heavy element )

(b) (d, n) type: This type of transmutation is also important and occurs most easily with heavier elements. 27 13

28 Si +1 n Al +12 H  →14 0

209 83

210

Bi +12 H  →84 Po +10 n

(c) (d, α) (type): This type of transmutation occurs in light elements using high speed deuteron 12 6

10

C +12 H  →5 B + 42 He

Nuclear Chemistry 15.21

(iii) Nuclear reactions induced by α-particle: α-particles are able to bring about two types of transmutations leading to the production of (a) proton (α, p) and (b) neutron (α, n) emission. The emission of neutron is particularly marked with beryllium although it has been observed with other elements even of high atomic number of α-particles of very high velocities are used. Examples of two types of transmutation shown by α-particles are 19 9

( a ) ( α, p ) type ( b ) ( α, n ) type

9 4

F + 42 He  →

22 10

Ne + 11H

B + 10 n  → 37 Li + 42 He

(b) (n, p) type: These reaction are produced by only fast neutrons 14 7

N + 10 n  →

24 12

Mg + 10 n  →

14 6

C + 11H 24 11

Na + 11H

(c) (n, 2n) type: 63 29

Cu + 10 n  →

62 29

Cu + 2 10 n

(d) (n, γ) type: These are most common type 40 18

Ar + 10 n  →

107 47

1 0

Ag + n  →

41 18

Ar + γ

108 47

Ag + γ

3. Nuclear reactions in terms of energy transformations: The various transformations which may take place in nuclear reactions can be classified into five types. (i) Projectile capture reactions: Almost all nuclides will absorb the slow neutrons. Frequently the neutron capture is followed by gamma emission. It is also possible to have proton capture with gamma emission but such reactions occur only with lighter

85 37

86 Rb + 01 n  → 37 Rb+γ

12 6

1 C+ H  → 137 N+γ 1

(ii) Projectile capture and emission of other particles: Most of the nuclear reactions are of this type. After the projectile entered the nucleus, one or more particles will be emitted from the compound nucleus to attain stability. The nature of emitted particle or particles depends on the energy of the projectile.

1 Be + 42 He  → 12 6 C + 0n

2. Nuclear reactions induced by neutrons: Neutrons are considered to be very effective particles for transmutation because they carry no repulsion in penetrating atomic nuclei. All elements except helium 42 He have been transmuted by neutron bombardment. Neutrons differ from other projectiles because they can be produced only by nuclear reactions and once produced cannot be accelerated. Their kinetic energies generally ranges between 1 to 13 Mev. It is also observed that slow neutrons are more effective transmuting agents than the more energetic neutron. In transmutation, it gives as (a) α-particles (b) a proton (c) more than one neutron and (d) γ rays. (a) (n, α) type: It is generally takes place with light elements 10 5

elements.

14 7

N+ n01  → 146 C+ 11 H

19 9

F+ n01  → 167 N+ 42 He

11 5

1 B+ H  → 116 C + 01 n 1

27 13

4 → 24 Al+ 11 H  12 Mg+ 2 He

16 8

2 O+ H  → 147 N+ 42 He 1

10 5

→ 137 N+ 01 n B+ 42 He 

(iii) Spallation reactions: High-speed projectiles with energies approximately 400 Mev may chip fragments from a heavy nucleus leaving smaller nucleus. Usually the nucleus so produced has an atomic number between 10 and 20 units and some times even 30 units lower than the original nucleus. In spallation reactions, the nucleons will leave either singly or in a bunch. Spallation differs from ordinary nuclear reactions to the extent that in the ordinary nuclear reactions the nucleus is rarely lowered in atomic number by more than 2 spallation and also the reaction is not self sustaining. Further, the change in mass for a spallation reaction is not so great as that in fission. A hypothetical spallation reaction may be indicated by the following reaction 63 29

Cu + 42 He (400 Mev) →

37 17

Cl+14 11 H +16 01 n

R. Serber at Berkely suggested a mechanism for spallation. According to him the high energy projectile enters a target and knocks on a single nucleon. The later may rush out of the nucleus or may knock on another nucleon with in the nucleus sharing its energy with it. This knock on process may be repeated number of times. This process may proceed in a cascade until several knock on nucleons are expelled from the target nucleus. 4. Fission reaction: In a fission reaction, a heavy nucleus will be broken into two or more medium – heavy fragment. Fission reactions are generally accompanied by the emission of neutrons. Rarely other particles will be emitted. Most of the nucleide with mass numbers more than 200 will undergo fission when subjected to the bombardment with neutrons. A few nuclei undergo fission when bombarded with slow or thermal

15.22

Nuclear Chemistry

neutrons of little energy. A typical reaction is indicated by the equation 235 92

U+ 01 n →

141 56

92 1 Ba+ K 36 r + 2-3 0 n+200 Mev

5. Fusion reactions: Isotopes of the very high elements may react with one another to form heavier and more stable nuclei. The potential barrier degree reactions of this type are extremely high. Millions of degree of temperature is required to overcome the potential barrier. It is believed that the energies of the stars may be derived from reactions of this kind. A typical fusion reaction takes place between the heavy isotopes of hydrogen, deuterium and tritium, in which helium nucleus is the product. 2 1

H+ H  → He+ n+17.6 Mev 2 1

4 2

1 0

15.6.4 Artificial or Induced Radioactivity In the artificial transmutation described in the previous section, in many cases the product nucleus is unstable and decays further with the emission of electrons until a stable element is formed in a manner exactly analogous to the heavy radioactive elements. In other words, a stable nucleus due to phenomenon of artificial transmutation is transformed to a radioactive nuclei. Since this property of radioactivity is produced by artificial means, it is called artificial or induced radioactivity. Artificial radioactivity is defined as the process by which a new radioactive isotope of a known element can be prepared. The artificial radioactivity is a part of transmutation, it was first discovered by Madam Irene Curie and her husband M. Frederick Joliot in 1934 in the course of their studies of transmutation by α- particles on boron, magnesium and aluminum. They found that the bombarded substance continued to emit radiations even after the source of α-particles had been withdrawn. These radiations from the targets were identified as positron (+ve electrons) and their intensity falling of exponentially with time which is a characteristic of radioactive disintegration. For example, when 24 is bombarded with 12 Mg α-particle, it will be absorbed by the nucleus and form a very unstable nucleus and hence disintegrates instantaneously with the emission of neutrons. 24 12

Mg

1 + 42 He  → 27 14 Si + 0 n

The product 27 14 Si called silicon, is unstable and disintegrates spontaneously and behaves as ordinary radioactive element and emits positrons instead of α- or β- particles. Thus 27 0 → 27 14 Si  13 Al + +1 e

As soon as Curie and Joliot announced the phenomenon of artificial radioactivity, the study was taken up in many laboratories and some five hundred radioactive isotopes of all elements from helium to uranium with few exceptions have been discovered. It was confirmed that α-particle is not only the projectile but other projectiles such as protons, deuterons and neutrons are able to produce radio elements which emit not only positrons but also several particles.

15.6.5 Cause of Artificial Radioactivity – Bohr theory of compound Nucleons In 1936, Bohr developed a theory to explain what happens when nuclei are bombarded by particles of moderate energies. The projectile particles will have kinetic energies upto 40 Mev and when these particles enter the target nucleus, the energy of the projectile particle is distributed throughout the nucleus called compound nucleus which will have a life time of 10 –14 to 10 –12 sec. This compound nucleus to become stable, it emits the extra energy associated with a particular nucleon or combination of nucleons. If the energy of compound nucleus is not sufficient to produce the ejection of a particle the excess energy may be emitted in the form of gamma radiation. The compound nucleus may be formed in a variety of ways associated with different energies. This can be explained by the formation of compound nucleus in different ways as shown below. The disintegration of the compound nucleus may also take place in different ways depending on the energy associated with it but not on the mode of its formation. 24 11

Na 42 He

27 13

Al  10 n

26 12

Mg 12 H

27 12

Mg 11 H

28 13

Al

26 13

Al  210 n

28 13

Al  

27 12

Mg 11 H

24 11

Na 42 He

If the projectile is a slow neutron the compound nucleus will be in a low degree of excitation. Such a compound nucleus may not eject a nuclear particle in its decay. But emit a gamma radiation to restore stability by loosing the extra energy. If the projectile is a high speed alpha particle, the compound nucleus will be in high degree of excitation and the resulting product would be expected to emit one or more particles.

Nuclear Chemistry 15.23

by propagating and repeating fission reactions as shown below.

15.6.6 Nuclear fission Very heavy nuclei have a lower binding energy per nucleon than nuclei with an intermediate mass. The heavy nuclei are thus less stable than the nuclei of intermediate mass. If a heavy nucleus is bombarded with a slow neutron, the extra energy may cause the nucleus to split into two fragments and spontaneously emit two or more neutrons. This phenomenon is called nuclear fission. This was first discovered by Hahn and Strassmann. They found that 235 92 U atom was bombarded with neutrons, the nucleus of uranium split up into two nuclei. The splitting of 235 92 U could be in different ways and depends on exactly how the nucleus splits up. Three of the more common reactions are 144 56 Ba

235 92 U

+ 10 n

138 53 I

140 55

+

+

90 36 Kr

95 39 Y

+2

1 0n

+ 3 10 n

1 Cs 92 37 Rb +40 n

In all the above it can be seen that the daughter nuclei formed are of two types. The first type will have masses of about 130 to 160 while the second type will have masses of about 86 to 110. It is rare for the two daughter nuclei of about the same mass to be formed. The total mass of the fission products is nearly about 0.22 mass units less than the mass of uranium and neutron. This mass will be converted into energy and hence in the fission of uranium nucleus by a neutron about 200 Mev (according to E = mc2) will be released. This energy is about twelve times more than that liberated in normal nuclear reactions. If 1 Kg of uranium was completely fissioned, the energy liberated will be almost equal to the energy liberated in the explosion of 17500 tonnes of TNT. The primary nuclei formed in the fission of uranium will have high n/p ratio and decay by β-emission causing the decrease in n/p ratio. After undergoing several such steps, stable nuclei will be formed. Thus each of the decay products is associated with a decay chain. For example, 138 53 95 39

β β I β → 138 → 138 → 138 54 Xe  55 Cs  56 Ba (Stable )

Y β → 9405 Zr β → 95 → 95 41 Nb  42 Mo (Stable )

Chain reactions: The neutrons produced in the fission of uranium atom by bombarding with a neutron are called as secondary neutrons. These neutrons may again cause the fission of more uranium nuclei, there

n

3n U n

U

U

U

U

U n n n

n

n

n n n UU

U

n n n UU

U

U

Such self sustaining reactions are called chain reaction or autocatalytic reaction. If E be the energy released per fission, the total energy liberated from the chain reactions goes on multiplying in each step and ultimately becomes enormous. Thus large amount of energy can be exploited for useful purpose. Critical mass: There must be a minimum amount of fissionable material that can produce a self sustaining chain reaction. This is called critical mass. The critical mass depends on several factors. (i) The purity of the sample (ii) The density of the material (iii) Its geometric shape and its surroundings These all effect the neutron propagation ratio. If the sample is impure, some neutrons are lost by colliding with non-fissionable atoms. If the material is more denser there, will be more chance for neutrons to collide with another nucleus. The shape of sample is also important since neutrons are more likely to escape from a long thin strip of material than from a sphere. In its safety position, an atomic bomb consists of a number of pieces of pure 235U, all of sub-critical size. At the time of explosion, all these pieces are made to combine by high explosive so that they will form a single piece above critical size. Now neutrons from air (Produced by cosmic rays) trigger the bomb. When fission occurs tremendous energy will be released in the form of heat. Air in contact with the bomb expands suddenly and a shock wave of great destructive impulse travel across. The highly penetrating γ-rays produced here are exceedingly dangerous. The radioactive fragments scatter over a wide area causing radioactive dangers for longer periods. Atom bombs can be made with U-235 and Pu-239. The atom bombs dropped on the cities Hiroshima and Nagasaki of Japan during world war II were made one with U-235 called Little Boy and the other with Pu-239 called Fat Man.

15.24

Nuclear Chemistry

Uranium bomb was dropped on Hiroshima on 6th August 1945 and plutonium bomb was dropped on Nagasaki on 9th August 1945. Four square miles were devastated by uranium bomb and cause 70,000 immediate deaths. Two square miles were devastated by plutonium bomb causing 45,000 immediate deaths. The fission chain reaction can be controlled and its explosive effect can be considerably damped by absorbing the slow neutrons. Certain substances like highly purified graphite and heavy water (D2O) is employed for slowing down the secondary neutrons. These substances are called moderators. The high speed neutrons collide with and rebound from the nuclei of carbon atoms. As a result, neutrons lose their kinetic energy and their velocity is reduced. The whole device in which this process is carried is called as nuclear reactor, where the nuclear reactions are carried under controlled conditions and the energy liberated is used for several peaceful purposes. Uranium contains 235U, 238 U of which 235U is only fissile. Natural uranium contains of 99.28% of uranium -238 and only 0.72% of uranium -235. As most of the mass of natural uranium consists U-238, the neutrons released during nuclear fission will try to bombard the nuclei of U-238 mostly and very few neutrons can cause fission of U-238 but neutrons of all possible energies can cause fission of U-235. So it has to be separated. Seaborg discovered that plutonium is also fissile. Plutonium is produced by irradiating the relatively plentiful 238U with neutrons in a nuclear reactor. 238 92

U + 10 n  →

237 92

U α →

239 93

N P  →

239 94

Pu

All isotopes of plutonium are fissile and it became the important nuclear fuel. The materials which can produce chain reaction directly by bombarding with slow neutrons are called fissile materials. The materials which by themselves are nonfissile in nature but can be converted into a fissile material by reactions with neutrons are called fertile materials. Boron or cadmium steel rods are used as control rods. These rods absorb neutrons and thereby control the rate of fission. Liquid alloy of sodium and potassium is used as coolant. This transfers the heat to the exchanger.

15.6.7 Nuclear fusion The atoms with lighter nuclei are comparatively less stable because of their low binding energies. Fusion of two such nuclei results in the production of a nucleus of lighter mass with high binding energies which is more stable. This phenomenon is known as nuclear fusion and

may be defined as the process of combining of two lighter nuclei to give a heavy nucleus. For example, when the deuterium nuclei are allowed to combine (by raising the temperature to several million degrees) the fusion takes place as follows. 2 1

H + 12 H  → 13 H + 11H + energy

Like fission, the fusion is also accompanied by the release of a tremendous amount of energy and it is also a chain reaction. In fact the above fusion continues in the following two manners: 3 1

H + 12 H  → 42 He + 10 n + energy

and may be written to be overall 2 1

H + 12 H  → 42 He + γ

It should be remembered that a large amount of energy is required to initiate the fusion. This energy is obtained by nuclear fission and then used for fusion. Fusion reactions involve the release of extremely high amount of energy and produce much greater amount of energy per particle in comparison to fission reactions. The energy of sun and stars is produced by fusion. In sun and stars, hydrogen atoms continuously fuse to give helium atoms. The large energy liberated is responsible for extremely high temperature of sun and stars. Fusion reactions taking place in sun and stars are supposed to be following. 12 6

C + 11H  →

13 7

N  → 136 C + 10 e

13 6

C + n11H  →

14 7

N + 11H  → 158 O + energy

15 8

O  →

15 7

N + 11H  → 126 C + 42 He

15 7

13 7

14 7

N + energy

N + energy

N + 10 e

Hydrogen bomb is based on fusion reactions. Energy released is so enormous that it is about 1000 times that of an atom bomb. In hydrogen bomb, a mixture of deuterium oxide (D2O) and tritium oxide (T2O) is enclosed in a space surrounding an ordinary bomb. The temperature produced by the explosion of the atomic bomb initiates the fusion reaction between 13 H and 12 H releasing huge amount of energy. The first hydrogen bomb was exploded in 1952.

Nuclear Chemistry 15.25

So far, it has not been possible to bring about fusion under controlled conditions. The main drawbacks in using the energy released in nuclear fusion reactions are (i) They are uncontrollable. (ii) They have to be carried only at very high temperatures. Since these reactions have to be carried at very high temperatures, they are called thermonuclear reactions. The advantages of the energy released in the nuclear fusion reactions as a potential source of commercial electrical power are. (i) The quantity of energy released in the fusion reactions is much greater than in fission reaction. (ii) The products of fusion are non-radioactive and are gases thus pollution can be minimized.

15.6.8 Plutonium and the actinides The development of nuclear reactors has made possible the synthesis of many new elements and of new isotopes of elements that occur in nature. This discussion will be restricted to those elements called actinides. For all practical purposes the natural element of highest atomic number is uranium, the transuranium elements (of atomic number 93 or higher) must be made synthetically. The synthesis of transuranium elements is one of the most spectacular achievements in all science. The first synthetic element prepared was technicium but the first transuranium element to be prepared was synthesized by E. M. Mc Millan and P. H. Abelson in 1940. They bombarded uranium with neutrons to obtain uranium – 239 by a neutron capture (n, γ) reaction. 238 92

U + 01 n  → 239 92 U + γ

Uranium–239 then disintegrated by beta emission, with a half-life of 23.5 min, to produce an element of atomic number 93. It was named neptunium. 239 92

239 U → 93 Np +

0 −1

e

Elements of still higher atomic number have been synthesized chiefly by G. T Seaborg and his associates. By far the greatest interest has centered on element 94 named plutonium. This element is produced in substantial quantities as a nuclear explosive for atomic bombs and as a power source for nuclear reactors. The production of plutonium in the uranium nuclear reactor will be in some detail. The sequence of reactions involved are those already given for neptunium–239 plus a further beta disintegration of the neptunium with a half-life of 2.3 days to yield plutonium –239.

238 93 Np

 → 238 94 Pu +

0 −1 e

Although the plutonium–239 may be made undergo the nuclear fission reaction, yet its normal spontaneous alpha-ray activity has a half-life of 24, 000 yr When natural uranium containing only 0.7 percent, uranium-235 is placed in a graphite–moderated reactor, the spontaneous fission of the 235 92 U liberates enough neutrons to sustain a chain reaction, provided the size and arrangement of the reactor and its fuel elements are appropriate. The neutrons given off during fission have such a high energy that they cannot be captured by more abundant uranium–238 to produce 239 92 U as required first for the production of plutonium. Another way of starting this is: while the fission cross-section of 235 92 U may be low for fast neutrons, the capture cross-section of 239 92 U for fast neutrons is very low (The fission capture cross-section of 238 92 U is very low for all neutrons, but the capture capture cross-section of 238 92 U for slow neutrons is high). For this reason a nuclear reactor must contain a moderator such as graphite if any plutonium is to be obtained. After natural uranium has been kept in the reactor for some time, at an appropriate neutrons flux level, it will be found that the sequences of reactions given 239 239 above, leading from 238 92 U through 92 U to 93 Np to 239 94 Pu has taken place and that the uranium metal sample, if analysed, contains a small proposition of plutonium. The uranium plus plutonium accumulated fission-product impurities may then be removed from the reactor and treated chemically to concentrate and purify the plutonium. Plutonium which is itself may serve as a target for the production of elements of still high atomic number. Thus intense slow-neutron bombardment of plutonium-239 yields successively plutonium isotopes of higher mass number. Some of these isotopes decay by beta emission to form elements of higher atomic number. Plutonium–239 bombarded with slow neutrons may, for instance, yield plutonium–241, which decays to form americium–244, which decays to curium–244. In this manner it has been possible to extend the periodic table. Furthermore, neutrons are not only possible bombardment particles that may be used for this purpose. Plutonium itself was, first made by bombarding uranium-238 with high-energy deuterons, the isotope obtained being 238 94 Pu . High-energy alpha particles (helium ions) have been used to produce elements of atomic number over 100. Element 103 has been made by bombarding the artificial element 98 with high-energy boron nuclei.

15.26

Nuclear Chemistry

table 15.5 Transuranium Elements Atomic number 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111

Elements Neptunium Plutonium Americium Curium Berkelium Californium Einsteinium Fermium Mendelevium Nobelium Lawrencium Rutherfordium Dubnium Seaborgium Bohrium Hassnium Meitnerium Darmstadium Rontgenium

Symbol Np Pu Am Cm Bk Cf Es Fm Md No Lw Rf Dg Sg Bh Hs Mt Ds Rg

15.7 RaDioactive isotoPes as tRaceRs If an isotope has characteristic property which permits it to be identified among other isotopes of the same elements, then that isotope may act as a tag or label to follow its sequences of physical and chemical changes. Two properties of isotopes are used to detect them, they are (1) mass and (2) radioactivity An isotope added for this purpose is known as an isotopic tracer and the element which is labeled by the particular isotope is called a tracer element. The possibility for the isotopic method of solving problems in science appear almost infinite, only a few of them we explained. Radioactive isotopes are favoured over stable isotope in tracer studies because the radioisotopes can be easily detected and followed through a sequence of changes. Further, radioisotopes are readily available for most of the elements. They can be obtained from four principle sources (1) members of one of the natural radioactive decay series (2) irradiation of stable elements with high speed projectiles from accelerators such as the cyclotron (3) irradiation of stable elements with neutrons from an atomic reactor and (4) the product of fission. It is important to note that practically all artificial radio isotopes used in tracer investigation are beta emitters and the continuous decay of these nuclides results in the

concentration of each tracer at a constant rate. So the time factor may be quite important in the practical use of such tracer. Radio isotopes of either very short lives or very long lives are not usually suited to tracer work. Short lived radio isotopes limit the time for observation while long lived radioisotopes may emit such feeble radiation that it is difficult to detect. Two types of instruments are used to determine the amounts of isotopes. In experiments involving radioactive isotopes. The measuring instrument is the electroscope, the electrometer or the Geiger-Müller counter. In experiments using stable isotopes, the measuring instrument is mass spectrometer. In this instrument, atoms under analysis are ionized and passed through a magnetic field. Isotopes are separated by using difference in mass deflections. Methods for detecting radioisotopes are more sensitive (1000) times than those for isotopes which differ in mass.

15.8 aPPlicatioNs of RaDioactive isotoPes 15.8.1 carbon Dating The technique in which radio carbon is used for estimation of ages of archeological specimens is known as radio carbon dating. The neutrons in cosmic rays when slowed down are captured by the nitrogen in the air by the following reaction 14 1 → 146 C + 11 H 7 N+ 0 n  Radioactive carbon

The radioactive carbon 146 C decays by emitting an electron with a half- life period 5600 years 14 → 147 N+ -10 e 6 C 

The radio carbon formed from the atmospheric 147 N soon gets converted to CO2. This CO2 containing radioactive carbon will be taken up by plants in photosynthesis where it converts into carbohydrates. These carbohydrates are taken by all living animals in their food. The radioactive carbon present in these carbohydrates will return to the atmosphere through respiration and decays back to form 14 7 N . Depending on the rate of production of the radioactive carbon from the atmosphere by the cosmic ray neutrons and decay back into 147 N through the plant animal carbon cycle, an equilibrium will be established with all living matter containing a certain percentage of radioactive C-14. This conclusion was verified by the analysis of carbon contents of fresh samples of wood. Libby suggested that once a living organism is dead, it ceases the consumption of radiocarbon and the radio carbon present in the dead body goes on decaying which will have a half-life of

Nuclear Chemistry 15.27

5600 years. So its activity decreases with time. From this suggestion, it had become possible to estimate the age of carbonaceous matter which has been dead for period varying 1000 years to probably 1,000,000 years, if it is assumed that the relative proportion of 14 6 C in it at the time of its death is the same as that which exists at present time in fresh samples. The fresh sample gives 16.1 disintegrations per minute per gram of carbon. If the dead samples give Nt disintegrations per minute per gram carbon. Then λ=

N 2.303 log10 0 t Nt

In this equation, λ the disintegration constant of 14C can be determined from its known half life N0=16.1 disintegrations per minute, time t of the old age sample is calculated. The 14C dating method can be applied to the samples which are dead long before.

45. A charcoal sample taken from a pit in an archeologist’s excavation of a rock shelter was believed to have been when early occupants of the shelter burned wood for cooking. A 200 mg sample of pure carbon from the charcoal was found in 1987 to have a disintegration rate of 0.25 count/min. The activity of a contemporary sample taken from a freshly cut tree was found to be 15.3 counts per min per g carbon. How long ago did the tree grow from which the archeological sample was taken? t0.5 of 14 C is 5730 year. Assume that no new 14C was formed after the tree that supplied the wood was felled and the 14 C level in living tree has been the same overtime. 46. The activity of hair of an Egyptian mummy is 10 dis/ min–1 g–1 of 14C. Calculate the age of the mummy. t0.5 of 14C is 5730 years and the disintegration rate of fresh sample of 14C is 15 dis-min–1 g–1.

solved Problem 6

15.8.2.the age of the earth

A piece of wood, reportedly from a Mughal emperor’s tomb, was burned and 7.32 g CO2 was collected. The total radioactivity in the CO2 was 10.8 disintegration/minute. How old was the wood sample? The initial radioactivity was 15.3 dis/min (r0.5 = 5730Y). Solution: The mass of carbon in the sample is 44 g of CO2 contains = 12 g of C 12 × 7.32 7.32 g of CO2 contains= = 1.966 g C 44

When uranium decays, helium gas is given off, and finally lead is formed. For every uranium-238 atom that decays, one atom of lead-206 is produced at the end of the decay chain. Along the way eight alpha-particles, i.e., helium atoms, are released. Also, we know the half-lives of all the isotopes in the decay series of uranium-238. This means that, if we are able to count the number of lead or helium atoms in a sample of uranium, we should be able to calculate how long the uranium has been decaying. Needless to say, there are many difficulties in performing the calculation. For one thing we must be sure that the lead or helium has not arrived by some other means, for example as the product of other decay schemes. Rocks can also be dated by comparing the amounts of various other isotopes such as lead-206 and lead-204, potassium-40 and argon-40, rubidium-87 and strontium-87. Using isotope dating methods, the oldest rocks on earth appear to be about 3.7×109 years old. Therefore we can say that the earth is at least this old, because the isotope method can only work from the time that the rocks had solidified. For example, any helium formed while the rocks were still liquid would have boiled off. By comparison, rocks from the moon and meteorites appear to be around 4.6×109 years old. The absence of ages greater than this suggests that the universe it self was formed no more than 5×109 years ago.

Ao = 15.3 dis/min g 0.693 0.693 year −1 λ= = 5730 t 0.5 A=

As we know, t =

10.8 × 5.4 dis / min g 2.00 A 2.303 2.303 × 5730 15.3 log 0 = log λ A 0.693 5.4 = 8.6127 years.

Problem for Practice 43. The bones of prehistoric bison was found to have a 14C activity of 2.8 dis/min g carbon. Approximately how long ago did the animal live. Given A0 =15.3 dis/min g t0.5 = 5730 years. 44. A freshly cut piece of wood gives 16100 counts of β-ray emission per minute per Kg and an old wooden bowl gives 13200 counts per minute per Kg. Calculate the age of wooden bowl. The half-life period of 14C is 5568 years.

solved Problem 7 (a) On analysis a sample of uranium ore was found to contain 206 238 Pb and 1.667 g of 92 U . The half-life period 0.277 g of 82 238 9 of U is 4.51×10 year. If all lead were assumed to have 238 come from decay of 92 U , what is the age of the earth?

15.28

Nuclear Chemistry

238 236 206 (b) An ore of 92 U is found to contain 92 U and 82 Pb 238 in the weight ratio of 1:0.1. The half-life of 92 U is 4.51×109 year. Calculate the age of ore. (IIT 2000) Solution: 1.667 238 (a ) GIven 92 U = 1.667 g= mole = 7.004 × 10 −3 mole 238 0.277 206 = 1.345 × 10 −3 mole 82 Pb = 0.277 g = 206

In this case all the lead comes from the decay of uranium. ∴ No. of mole of Pb formed =

0.277 = 1.345 × 10−3 mole 206

No. of mole of U decayed= 1.345×10–3 mole Initial no of mole of U before decay = 7.004×10–3 +1.345×10–3 N0 = 8.349×10–10 mole Moles of U after decay N = 7.004×10–3 year–1 λ=

0.693 0.693 = = 1.536 × 10 −10 year −1 t 0.5 4.5 × 109

As we know N 8.349 × 10 −3 2.303 2.303 log 0 = log t= λ N 1.536 × 10 −10 7.004 × 10 −3 ∴ eAge of earth ( t ) = 1.143 × 109 year.

(b)

1 0.1 + = 46.85 × 10 −4 mole 238 206 1 = 42.0 × 10 −4 mole N= 238 0.693 λ= = 1.54 × 10 −9 year −1 4.5 × 109 N0 =

t=

46.85 × 10 −4 2.303 log −10 1.54 × 10 42.0 × 10 −4

t = 7.097 × 108 year. Problems for Practice 47. A sample of 238U (half-life= 4.5 × 109 year) ore is found to contain 23.8 g of 238U and 20.6 g of 206Pb. Calculate the age of the ore. (Roorkee 1996) 48. A sample of pitch blende is found to contain 50% uranium and 2.425% lead. Of this only 93% was 206Pb isotope. If the disintegration constant is 1.52 × 10–10 year, how old could be the pitch blend deposits.

232 49. In nature decay chain series starts with 90 Th and 208 finally terminates at 82 Pb . A thorium ore sample was found to contain 8 × 10–5 mL of STP and 5 × 10–7 g of 232 Th. Find the age of ore sample assuming that source of He to be only due to decay 232Th. Also assume complete retention of He with in the ore. t0.5 for 232Th = 1.39 × 10–10 year.

15.8.3. applications in industry 1. Detection of impurity: Addition of trace quantity of 32 P to the iron ore has been used in steel industry for detecting the presence of phosphorous impurity. The disappearance of radioactivity from the molten steel means the removal of all the phosphorous. 2. Detection of wear in an engine: Radio isotopes are also used in studying the wear and lubrication of motor car engines. In practice, piston rings are made radioactive by irradiating them in a nuclear reactor. Then they are assembled into an engine and it is allowed to run in the presence of a lubricating oil. The oil is tested regularly for the appearance of radioactivity due to the presence of tiny particles (even less than millionth) of metal from the wearing of the piston rings. Such tests are successfully used in developing more efficient oils. 3. Detection of the level of a liquid in a storage tank: A gamma emitting isotope is placed on a float (like cork) with in the tank. The float will move up and down inside the tank as the level of the liquid increases and decreases respectively. The position of the float can be located easily from outside with the help of GeigerMüller counter at the same height so the level of the liquid inside the tank can be detected. Radioisotopes are used to know the level of change in blast furnace. 4 Detection of leakage of liquid in transportation: The moving marker technique is used in the transportation of oil in pipe lines and detecting leakage in pipelines. The same idea can be used in tracing movement of oils through the complex pipe lines in an oil refinery and in the transportation of oil in cross-country pipe lines to identify a change – over in shipments. 5. Controlling of thickness of sheets: The β-ray thickness gauges can be used for continuous measurements of sheet metals, paper, plastic and rubbery type fabrics. As a film emerges from rollers, it passes between a beta source and a detector. Any change in thickness will produce corresponding change in radioactivity transmitted between the beta source and its detector. The beta gauge activates an automatic mechanism for resetting the rollers to a predetermined level for uniform film thickness.

Nuclear Chemistry 15.29

6. In metallurgy: In metallurgy, tracer atoms are used to control the steel smelting process. They permit in making an exceedingly rapid check of the chemical composition of the steel. It is also possible to determine the origin of harmful impurities in the metal. In the production of metal and alloys, it is necessary to subject them to various different treatments like annealing, quenching, cold-rolling etc. By incorporating radio isotopes it is possible to know that what is taking place when a metal is subjected to a particular treatment. 7. In laundry: Another important use of radioisotopes is in laundering industry. A good detergent after cleaning the fabric should get rid of it. Retention of the detergent to the fabric will lead to deterioration of the fabric. Labelled detergents have been used in study of retention of detergents by fabrics. Other industrial applications of radioisotope are (a) radio autographs are replacing the use of X-ray method in detecting internal flaws in large piece of metal (b) with the help of radioisotopes valuable information regarding corrosion and its inhibition can be known (c) optimum mixing time for materials such as paint, ink, plastic products, powder etc may be determined by using radioisotopes (d) Radio carbon -14 is used as a tracer in studying the mechanism involved in alkylation, polymerization, catalytic cracking, catalytic synthesis as well as many other reaction of industrial importance.

15.8.4 application in agriculture 1. Manner of absorption of fertilizer: Radioactive phosphorous when present in the fertilizer is being used to investigatie the rate and manner of the absorption of fertilizer. Moreover, by such studies it is possible to know separately about the proportion of the plants phosphorous coming from the soil and from the fertilizer and thus about the kind of the fertilizer which is best for a given soil and crop. The whole process is studied by radio autography. Radio autography is based on the fact that β-particles and γ-rays emitted in the decay of radio isotopes produce images on a special photographic plate. For example, a tomato leaf fed with radio active phosphorous is placed on a photographic plate. The picture obtained clearly exhibit the area of concentration of radioactive phosphorous. By using radio autograph technique it is possible to follow the 35S supplied as sulphate to a plant. The movement of liquids in certain plants has been traced by using radio active phosphates.

2. Detection of necessary elements for plant: The use of radio isotopes gave the information concerning the use and need of elements like iron, cobalt, copper, manganese, zinc and molybdenum. 3. Effectiveness and harmfulness of fungicides: Many fungicides contain sulphur. Using 35S, the effectiveness of fungicide and in some cases the damage to the fruit resulting from its application can be known. For example, tracer technique shows that sulphur dusted on lemon fruit will penetrate the peal to a sufficient depth to cause injury to the fruit. 4. Mechanism of photosynthesis: The mechanism of photosynthesis occurring in green plants under the influence of sunlight has been elucidated with 18O and 14 C isotopes. Using 18O it had been established by Ruben and coworkers (1943) that the oxygen released in photosynthesis is from water (mechanism I) and not from CO2 (mechanism II). 6 CO 2 + 12H 218 O  → C6 H12 O6 + 6H 2 O + 6 18 O 2 .......I 6 C18 O 2 + 12H 2 O  → C6 H12 O6 + 6H 218 O + 6O 2 ........ II

15.8.5 application in Medicine 1. In studying the circulation of blood: Radio sodium 24 Na is being used in studying the circulation of blood. Normally, when radioactive sodium 24Na in the form of sodium chloride is injected in the blood stream, it should pass from the fore arm to the foot in its maximum concentration within half an hour. The course of the tracer can be followed by means of Gieger – Muller counter. If there is any obstruction the time period will be longer. Also it is possible to detect the area of obstruction and the degree of obstruction, from which it can be known whether medical treatment is sufficient or amputation is required. In a severe obstruction, the counter may serve to indicate at what level amputation should be performed. 2. Detection of tumors: Certain labeled atoms, for example, radioactive iodine are preferentially used in tumorous growths. This property has been utilized in locating brain tumors and sometimes their limit of growth can also be determined. Other applications of radio isotopes in medicine may be listed as (i) By means of tracer iron, it has been possible to improve methods for storing blood. (ii) Tracer iron has been used in studying disturbances associated with pregnancy. (iii) From tracer studies it was found that phospholipids, a fatty phosphorous compound found in blood plasma is formed in liver.

15.30

Nuclear Chemistry

(iv) Tracer studies shows that thyroxine, the iodine containing amino acid which is important in the functioning of thyroid is produced in muscular and intestinal tissues. (v) Tracer studies show that reserve iron is stored in the mucous membrane of the intestine with in the body.

15.8.6 therapeutic uses In cancer: Many radioisotopes have been successfully used in checking growth of or removing the cancer cells. The most important active isotopes have been cobalt-60, radium, iodine (for thyroid cancer), phosphorous -32 and gold-198. Radioactive phosphorous in the form of phosphate is widely used in the treatment of certain blood disorders. The oral ingestion of 32P isotope in the form of a phosphate is used in the control of polycythemia in which the bone marrow produces an excess of red blood cell. The β-activity of the phosphorouss prevents the formation of excess redblood corpuscles. The same treatment is used for another abnormal blood condition known as leukemia which is caused by the over production of white corpuscles. Radioactive cobalt 60Co is used now-a-days as a source of radiation for the radiotherapy in the treatment of cancer. Radioactive iodine 131I has been for a long time in the treatment of thyroid complications such as hyperthyroid and cancer of thyroid. The malignancy is cured by the radioactive radiation of 131I.

15.8.7 applications in biochemistrys The application of tracer studies in biological fields can be classified into four areas. 1. Study of the dynamic steady state: At one time it was believed that the food taken by living organism is to supply energy to the body and to repair or replace the damaged parts of the body. Also it was thought that the excess food taken will be stored in the body and left undisturbed until there is need. By the tracer study, it was found that these ideas are erroneous. From the tracer study, it was found that there is constant exchange between the elements, supplied in the diet and those which make up the body tissues. The atoms with in a living organism are undergoing a continuous replacement and these reactions are dynamic in nature in constant state of flux. 2. Study of transport of ions across cell membranes: Tracer studies helped to know the movement of ions across cell boundaries. The transport of ions accross

the cell boundaries is governed by some active mechanism which is different from ordinary physicochemical equilibrium process the principles which apply to ordinary osmotic process do not necessarily apply to the passage of ions through cellular membranes. For example, the cells maintain high concentration of potassium ions inside in the presence of low concentration outside. Also the cell excludes sodium ions even when the extra cellular concentrations are high. 3. Metabolic interconversions: The tracing of a metabolic process involves two basic steps. (i) The labeling of a food or food like material. (ii) Analyzing the intermediate or ultimate products that may be formed to determine the amount of tracer isotopes if present. The mechanism of metabolism is so complex that the appearance of a labeled isotope in the final product may be misleading. For example, under certain conditions carbon dioxide may be either a waste product or may enter into the synthesis of a more complex substance. If a labeled carbon atom is in the carbonylgroup of the tracer compound, it may appear as labeled carbon dioxide among the waste products. On the other hand, it is possible that labeled carbon dioxide may enter into a number of metabolic processes, in such a manner that the labeled carbon atom appear in another compound. From tacer studies, it was found that carbon dioxide may enter into a number of metabolic inter conversions.. From the tracer studies the protein metabolism was well understood. In the living organisms, the synthesis and breakdown of proteins are controlled by enzymes. These reactions are followed by the addition of labeled glycine to the protein extract. Any synthesis which may take place will include the tracer compound and after a period of time the protein may be removed and analyzed for tracer content; from this a protein system can be known before its breakdown. 4. Mineral metabolism: The application of tracer isotopes to study the mineral metabolism is of five types. (i) The movement of ions across cell membranes (ii) The metabolism of bone tissues (iii) The metabolism of trace elements. (iv) The metabolism of hemoglobin (v) The toxicology of heavy elements. These investigations are usually followed by absorption, distribution and excretion patterns by using tracer isotopes in small quantities so as not to upset normal metabolic equilibria. From these studies, it was found that calcium metabolism will take place in the intestine. By tracer studies the role of trace elements in the body has also be established. For example copper is utilized in the formation

Nuclear Chemistry 15.31

of hemoglobin, cobalt is present in vitamin B12 and manganese and zinc seem to play important roles in the enzyme chemistry of the body.

15.8.8 applications of Radioisotopes in chemistry Applications of Radioisotopes in chemistry are numerous. A few of them are described. 1. Solubility’s of sparingly soluble salt: The solubility of sparingly soluble salts like lead sulphate, silver chloride etc. can be determined. For example in the determination of the solubility of lead sulphate, a known quantity of radio isotope of 212Pb as nitrate is dissolved in a solution of a known quantity of ordinary lead nitrate. The activity of the resulting mixture was measured from which the activity per milligram lead is calculated. From this solution, lead is precipitated as lead sulphate and the solution is kept at constant temperature until the saturated solution of lead sulphate is formed. A sample of definite volume of the clear supernatant liquid is withdrown at constant temperature and then evaporated to dryness. The activity of the residue is determined in exactly the same manner as for the original sample before the precipitation of the salt. From the activity of the residue and from the exact volume of the supernatant liquid which was evaporated, the amount of lead in the solution could be ascertained easily and hence the solubility of lead sulphate is obtained accurately. The solubility of water in benzene or other hydrocarbons is too small to be measured by ordinary physicochemimical methods. If however a definite quantity of radioactive tritium 31 H in the form of tritium oxide is added to a given amount of water, the resulting mixture has an activity that be measured in the minute amounts that may dissolve in organic solvents. Isotopic dilution method: It is used to determine the quantity of constituent (radioactive or non radioactive) in a mixture of closely related compounds which are difficult to separate and to estimate quantitatively, by the usual conventional methods. For example, the blood volume in an animal or human body of donor can be determined. Sufficient time is allowed to incorporation (some–time weeks) of the tracer into the circulatory cells. A sample is withdrawn and its activity is determined per milliliter of blood. Then an aliquot of the sample is introduced into the recipient. After a lapse of proper time, a sample from the recipient is analyzed for radioactivity. From the dilution of activity it is possible to calculate the volume of blood in the recipient. Isotopic dilution analysis is becoming increasingly important in analytical biochemistry. For example, the glycine

percentage in protein hydrolysis can be determined. This method is much time saving than ordinary methods. In this method, synthetic glycine containing radio carbon 14C is added to a definite weight of protein material. After allowing sufficient time for mixing a protein is removed from the mixture and purified as to glycine content. This will contain labeled and unlabelled molecules. By measuring the activity the amount of labeled glycine and the extent of dilution can be calculated. From these values, it is possible to calculate the total amount of unlabelled glycine in the original protein sample Activation analysis: It is one of the sensitive and specific methods available for the determination of trace quantities of wide range of elements. This method was first proposed by G. Hevery and H. Lewis Denmark in 1936 and improved by Clark and Overman in 1947. It should be possible to irradiate a mixture of elements so that their radio activities are different. Therefore, it is some time possible to detect very small amounts of an impurity in a sample by subjecting it to irradiation in such a manner that the impurity is selectively converted into a radioactive isotope. This method has been used for the analysis of mixtures of the rare earths metals and for the detection of gallium in iron, copper in nickel and hafnium in zirconium. Studies on reaction mechanisms. Mechanisms of many reactions have been elucidated with the application of isotopes. The hydrolysis of an ester may occur in either of the following two ways leading to the same products. CH3 CO OC2H5+H 18OH → CH3CO18OH + C2H5OH

(1)

CH3COO C2H5+H18O H→CH3COOH + C2H518OH Using water labeled with 18O, it has been established that in majority of the cases, the 18O becomes associated with the acid and not with the alcohol formed in the reaction. This confirms the mechanism. Structure of compounds: One of the earliest application to structural problems was the establishment of the structure of thiosulphate ion. If sulphur is heated with a solution of sulphite ion labeled with radioisotope of sulphur the labeled thiosulphate ions are formed. S+SO32-  S2 O32When the labeled thiosulphate ion is decomposed by acidifying the products are sulphur and sulphite and all activity remains associated with the sulphite ion. No trace of activity is found in the precipitated sulphur. This clearly indicates that S-O bonds remain unaffected in the synthesis and decomposition of thiosulphate and that of two sulphur atoms in thiosulphate are not equivalent which is according to the following structure.

15.32

Nuclear Chemistry

O 1 S =S = O 1 O

solved Problem 8 In order to determine the volume of blood in an animal without killing it, a 1.0 mL sample of an aqueous solution containing tritium is injected into the animals blood stream. The sample injected has an activity 1.8 ×106 CPs. After sufficient time for the sample to be completely mixed with the animals blood due to normal blood circulation, 2.0 mL of blood are withdrawn from animal, and the activity of the blood sample withdrawn is found to be 1.2 ×104 CPs. Calculate volume of blood Solution: Let the volume of blood in animals body= V mL Before mixing activity = 1.8 ×106 CPs. After mixing in mL blood activity

=

1.2×104 VmL = 6×103 V mL 2

Assuming activity remains constant ∴6×103 V = 1.8×106 V=

1.8×106 = 300 mL 6×103

Problems for Practice 50. A small amount of solution containing 24Na radionucleide with activity A=2×103 dPs was administered into blood of a patient in a hospital body. After 5 hours a sample of the blood drawn out from the patient body showed an activity of 16 dpm per cc. Half-life period of 24Na is 15 hr find : (i) volume of the blood of patient (ii) activity of blood sample drawn after a further time of 5h 51. Fallout from nuclear explosion contains 131I and 90Sr calculate the time required for the activity of each of these isotopes to fall to 1.0% of its initial value. Radio iodine and radio strontium tend to concentrate in the thyroid and the bones respectively of mammals which ingest them out of these two isotope (t0.5 for 131I = 8 days; t0.5for90Sr = 20 days). 52. 40K consists 1.2×10–2 % of the naturally occurring potassium. The human body contains 3.5×10–10 % potassium by weight. Calculate the total radioactivity resulting from 40K decay in a 80 Kg human. t0.5 for40K is 1.3×109year. 53. Tritium or 31 H is formed in the upper atmosphere by cosmic rays, similar to the formation of radioactive carbon. Tritium has been used to determine the age of wines. A particular wine that has been aged in a bottle has a tritium content only 87% of that in a similar wine of the same mass that has just been bottled. How long has the aged wine been in the bottle? t0.5 for 31 H is 12.2 years.

Nuclear Chemistry 15.33

aNsweR Keys Natural Radioactivity

units of Radioactivity

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•

The spontaneous, natural disintegration of un stable atomic nuclei is known as natural radioactivity. Nuclei of some heavy atoms are unstable and disintegrate spontaneously giving radiations known as α, β and γ radiations. α-rays contain α-particles with four units mass and two units positive charge (helium nuclei) and are deflected towards negative electrode. β-rays emitted in natural radioactivity contain negatively charged particles (electrons. These are deflected towards positive electrodes. γ-rays are high energy electromagnetic radiations which hive high penetration power.

Nuclear stability • •

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• • •

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The plot showing the relation to protons and neutrons gives the zone of stability. The zone or belt in which stable nuclides fall in the plot between mass number and N/P ratio is known as zone of stability. In the nucleides of lighter elements up to z = 20 the n/P ratio is 1 and are more stable. The lower portion of the stability belt coincides with the line of n/p = 1. In the heavier nucleides, i.e., Z>20, the number of neutrons are more than the number of protons and n/p>1. The excess neutrons in nucleides Z>20, increases the attractive forces and also reduce the repulsion between the protons. The heavier elements above atomic number 84 fall outside the zone of stability and are radioactive. Isotopes with high n/p ratio disintegrate by β-decay thus decreasing n/p ratio. Isotopes with low n/p ratio disintegrate by α-decay or 0 by losing a positron ( +1 e ) or by capturing K electron thus increase n/P ratio. Nuclei with 2,8,20,28,50,82, or 126 number of protons or neutrons are founds stable. These numbers are called magic number. Nucleides with even number of protons and even number of neutrons are stable. When the number of protons or neutrons are even number their spins are paired otherwise some resultant spin remain.

• • • • • • • • •

•

Activity of radioactive nucleide is the number of disintegration emitted by it in a given time. Radioactivity is measured in curie. Curie is equivalent to 3.7×1010 disintegration per second or 3.7×1010 atoms disintegratied in one second. 1 milli curie (m Ci) is 3.7×107 disintegrations per second. 1 micro curie (µ Ci) is 3.7×104 disintegrations per second. Rutherford (Rd) is the quantity of a radioactive material that undergoes 106 disintegrations per second. In SI system the unit of radioactivity is Becquerel (Bq) 1 Bq = 1 disintegration per second. One gray is equivalent to 1 Kg of tissue receiving a dose whose energy equivalent is one joule. The sievert = gray × quality number. Alpha particles have a quality number of 20 which reflects their danger compared to beta particles which have a quality number of 1. Specific activity of a radioactive nucleide is its activity per kilogram (or dm3) of the radioactive material.

Nuclides • • •

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• • • •

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•

The relation between volume of nucleus and the number of nucleons was proposed by Rutherford. The small central part of the atom is called nucleus. The study and the use of reactions occurring be tween the nuclei of atoms is the study of nuclear chemistry. The fundamental particles present in the nucleus are proton and neutron. The particles present in the nucleus are known as nucleons. The nuclear particles other than proton and neutron are called subatomic particles. Volume of the nucleus is proportional to mass or the number of nucleons present in it. Units of nuclear radius is Fermi 1 Fermi = 1 × 10–13cm. Radius of the nucleus can be calculated from. r = R0. A1/3 cm where r is the radius of the nucleus, R0 is constant 1.4 × 10–13cm and A is mass number. Atomic radii are of the order 10–10m or 10–8cm while the radius of the nuclei is of the order 2 × 10 –15m or 2 × 10 –13cm. Radius of the nucleus is nearly 105 times less or 1/100000 times of the radius of the atom. Volume of the nucleus is approximately 1 × 10–36 cc.

15.34

•

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• •

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•

•

Nuclear Chemistry

Volume of the atom is about 1 × 10–24 cc. Density of the nucleus is about 1014gm/cc or 1011 Kg/cc or 108 tons/cc and is about 1013 times greater than the density of atom. The term isotopes was introduced by Soddy. Isotopes of many elements were identified and also separated by Aston. Isotopes are the atoms of the same element having the different mass numbers (A) and the same atomic numbers (Z). Isotopes are supposed to be formed when a radioactive element emits one α- and two β-particles in a succession. Isotopes have same number of electrons, protons, same atomic number and same chemical properties but show differences in the atomic weights, number of neutrons, mass number, physical properties and radioactive properties. Heavier isotopes participates in chemical reactions slowly compared to lighter isotopes. The atomic weights of the elements are fraction al but not whole numbers due to the presence of isotopes. The atomic weights of the elements are the average of the mass numbers of different isotopes. 3 The lightest radioactive isotope is tritium ( 1 H ) . Isobars are the atoms of different elements with same mass numbers (A) but with different atomic numbers (Z). Isobars have different number of protons, neutrons, electrons and have different atomic numbers physical, chemical and radioactive properties but have same mass number and atomic weights. Isotones are atoms of different elements having the same number of neutrons (n) but different mass numbers (A) and different atomic numbers (Z). Isotones have same number of neutrons but have different number of electrons, protons and differences in the mass number, atomic weight, physical, chemical and radioactive properties. The difference in the number of neutrons and protons (A-2Z) is called isotopic number. The atoms having same isotopic number are called isodiaphers. Isodiaphers have difference in number of electrons, protons neutrons, mass number, atomic number, atomic weight, physical and chemical properties. Nuclides having identical atomic numbers and identical mass numbers, but different radioactive properties are called radioactive isomers.

Radioactive Disintegration and its Rate •

Radioactive disintegration theory was proposed by Rutherford and Soddy.

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•

Radioactive elements have unstable nuclei and undergoes spontaneous disintegration from one chemical atom to another. The disintegration of radioactive elements accompanies the emission of α, β or γ-rays. The first or starting radioactive element emits only one type of particles α, or β or γ-rays. The starting radioactive element is called parent nuclide. The new element formed in the radioactive disintegration is called product nuclide. If the product element also radioactive and again disintegrates to another new element, the product element is called daughter element. The number of atoms that disintegrate per second is directly proportional to the number of remaining unchanged radioactive atoms present at any time. The number of atoms of a radioactive element that disintegrates (or disappears) in unit time is known rate of disintegration (–dN/dt). −dN/dt ∝N or − dN/dt = λN Where λ is called decay constant. N/No = e−λt or N = No e−λt where No is the number of atoms of radioactive element initially present (at time t = 0) and N is the number of radioactive atoms present at the time t. Decay constant is the average fraction of the number of atoms disintegrating per second. All radioactive disintegration reactions are Ist order. No 2.303 this is Known as log t N

•

Rate equation λ =

•

radioactive disintegration equation or radioactive decay equation. Units of decay constant λ is t–1 (i.e., sec–1 or min–1, or hr–1 yr–1). Log No/N = 0.4343 λt.

•

Half–life • •

The term half – life was introduced by Rutherford. The time required to decay exactly one half of the initial amount of a radioactive element is known as half – life of that radioactive element.

•

Half life equation λ =

•

Half life of a radioactive element is independent of initial concentration. Radioactive substances left behind m=m0/2n where n is the number of half lives. If n is the number of half-lives then the amount of substance reduces to (1/2)n.

• •

0.693 0.693 or t 0.5 = . λ t 0.5

Nuclear Chemistry 15.35

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The amount after a given number of half-lives is given n N 1 by N = n0 or N = N 0 ×   . 2 2 The time required for complete disintegration is always infinity. Half – life indicates the radioactivity and stability of an element. Shorter the half life greater is the radioactivity of the element or vice-versa. Average or mean life of a radioactive element is the ratio of the total life time of all the radioactive atoms to the total number of atoms present in it. Total life time Average life (τ ) = Total number of atoms Average life is the reciprocal of the disintegration constant λ 1 τ= . λ Relation between average life, half-life and decay constant is t 1 τ = = 0.5 = 1.44 × t 0.5 . λ 0.693 Average life τ =1.44 × Half life or Half life = 0.693 × average life. The activity per unit mass of the sample is called its specific activity. The rate at which the number of atoms of any member in a radioactive chain become constant, it is said to be in radioactive equilibrium with its parent element. A  → B  →C N A λ B τA = = Or λ A N A = λ B N B N B λ A τB

•

•

NA and NB are the number of atoms of A and B,λAand λB are decay constants τA and τB are average lives of A and B. At the radioactive equilibrium the number of atoms of different radioactive elements present in the reaction series are (i) Inversely proportional to their radioactive contents (or) (ii) Directly proportional to their half–lives or average lives. If the radioactive decay takes place in different parallel paths the decay constant of parent nucleide is the sum of the decay constants of the different parallel paths.

Disintegration series •

The series of successive stages in the disintegration of radio active element till a final stable nuclide is formed is known as radioactive disintegration series.

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• • • • •

There are four disintegration series of which three are natural and one is artificial. The radioactive disintegration series are named after the longest lived member or prominent member of the series. The three natural radioactive disintegration series are uranium series, thorium series and actinium series In the actinium series the longest lived parent element is 235U. The radioactive disintegration series are also named as per the mass number values of their member nucleids If the mass number is divisible by 4 without any reminder then the radioactive disintegration series is called as 4 n series. If the reminder is 1,2 or 3 on division of mass number by the series are named as 4n+1, 4n+2 and 4n+3 respectively. The actinium (Z = 231) series is 4n+3 series. The uranium (Z = 238) series is 4n+2 series. The neptunium (Z = 241) series is 4n+1 series. The thorium (Z = 232) series is 4n series. Important aspects of radioactive series. Series

• •

Initial nuclide

End No. of particles product ejected α

β

4n

Th

Pb

6

4

4n+1

Pu

Bi

8

7

4n+2

U

Pb

8

6

4n+3

U

Pb

6

5

The end product of all the three natural series is stable isotope of lead. The 4n+1 or neptunium series is artificial and the end product in this series is the stable isotope of Bi. In this series gases are not formed.

soddy-fajan’s Group Displacement law • •

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Soddy, Fajan and Russell proposed the group displacement law. When an α-Particle is emitted the position of the product nuclide element will fall two groups to the left of the parent element in the periodic table. When a β-particle is emitted the position of the product nuclide element will fall one group to the right of the parent element in the periodic table. Some elements deviate from group displacement law. Actinides are the elements with Z=89, 90 to 103 belong to III group. Elements 89 to 102 when they emit β-particle, present in the same group. The group displacement law is not observed in the case of same III B elements.

15.36

Nuclear Chemistry

theories of Nuclear stability

Packing fraction

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Packing fraction =

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Packing fraction is equivalent to the difference between the actual mass of the isotope and the nearest whole number divided by this whole number. The values of packing fraction may be positive or negative or zero. The packing fraction of C−12 is zero because it is taken as the reference for atomic mass scale. Nucleides having −ve packing fraction are more stable while those having +ve packing fraction are unstable and are radioactive. Though the packing fraction of He, C, O are positive they are exceptionally stable.

The stability of the nucleus depends on its com position and it can be explained by the following factors. (i) Mass defect (∆m), binding energy and average binding energy (ii) n/P ratio (iii) Magic numbers of neutrons and protons (iv) Packing fraction.

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Mass defect (∆m) Binding Energy and average binding energy

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Mass defect is the difference in the sum of the masses of nucleons and the actual mass of nucleus. ∆m = Calculated mass-actual mass. Or ∆m = (mass of neutrons + mass of protons + mass of electrons )−(Actual mass). Binding energy is the energy released during the formation of a nucleus of the atom from the constituent protons and neutrons. (or) the energy required to disrupt the nucleus into constituent protons and neutrons. The mass that converted into binding energy is called as the mass defect. Binding energy of a nucules can be calculated from the mass defect by using Einstein’s mass-energy equation E=mc2. 1 amu mass defect of the nucleus releases an energy of 931 Mev. Greater the mass defect greater is the binding energy and also greater is the stability. Binding energy per nucleon is called average binding energy. Binding energy per nucleon is obtained by dividing the binding energy with mass number and it is the measure of the stability of nucleus. The binding energy per nucleon is less for lighter nuclides and increase with mass number. The binding energy per nucleon decreases with increase in the mass number after reaching a maximum. The nuclides of mass number 56 have maximum binding energy per nucleon. Since the binding energy per nucleon is less in heavier nuclides their fission into two or more nuclides of low mass numbers release energy and become stable. Nuclides of low mass number having low binding energy per nucleon, when fused to form heavier nuclides having higher binding energy per nucleon release energy.

Isotopic mass − Mass number × 104 Mass number

Nuclear Reactions •

The reactions taking place between atoms involving nuclide changes are known as nuclear reactions. The first nuclear reaction first reported by Rutherford was 14 7

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N + 42 He  →

17 8

O + 11H

The nucleus which was bombarded by a particle is called target nucleus. The particle which is used to bombard the target nucleus is called projectile. The nuclide which was formed in the nuclear reaction is called product nuclide. The particle produced in the nuclear reaction is called ejected particle. The projectiles and ejectiles in a nuclear reaction are always represented by their symbols instead of nu14 17 clides e.g. 7 N ( α, β ) 8 O . A nuclear reaction must be balanced on both sides of the equation with respect to the mass numbers and atomic numbers. Energy change of nuclear reaction is represented by +Q if it is exoergic (exothermic) or –Q if it is endoergic (endothermic) on products side. In a nuclear reaction the number of the nucleons of the target nucleide changes but in a chemical reaction nuclei are unaffected. Only outer electrons participate Nuclear reactions brings about the atomic transformation or transmutation but in a chemical reaction no change occurs in atoms. The energy released in a nuclear reaction is very high when compared to the energy released in a chemical reaction. In a nuclear reaction there will be loss in small amount of mass but a chemical reaction occurs without loss of mass

Nuclear Chemistry 15.37

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Nuclear reactions does not depend on physical state, combination but chemical reaction depends on nature of bonds in the compound physical state etc. Rate of nuclear reactions is independent of external factors such as temperature, pressure and catalyst but chemical reaction depend on them. Nuclear reactions involve the emission of particles like α, β–, β+ etc but in a chemical reaction no particles are ejected but only exchange of electrons takes place. Nuclear reactions are classified in two ways (i) depending on the overall change in atoms and energy (ii) depending on the nature of projectile particle. If the projectile particle is absorbed by the target nucleus with emission of γ-rays it is called projectile capture reaction e.g., 12 1 13  → N +γ 6 C+ H 1 7 If the projectile particle is absorbed by the target nucleus and a new nucleus is formed with the ejection of a new particle it is known as particle-particle reaction. e.g., 24 11

•

Na + 42 He  →

26 13

Al + 2 10 n

Depending on the nature of projectile particles used, the nuclear reactions are of different types. (α, p); (p, α); (n, p); (d, n); (γ, n)

Nuclear fission •

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144 56 Ba

235 92 U

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Artificial radioactivity was discovered by E. Joliot and I Curie. If a stable nucleus is converted into unstable nucleus by bombarding with fast moving particles like proton, deuteron,α-particle it is called artificial radioactivity or induced radioactivity. Natural radioactive elements occur in nature while artificial radioactive elements are prepared by bombardment of stable nuclides with α, β, γ, n-Particles. Natural radioactivity is generally exhibited by heavier nuclides (Z>83), but artificial radioactivity can be exhibited even by lighter elements. Natural radioactive elements emit α, β and γ radiations while artificial radioactive elements emit β– or β+ particles during their decay. Disintegration of natural radioactive elements oc cur in several steps but the disintegration of artificial radioactive elements generally occur in a single step. Half-life of natural radioactive elements is generally high while the half-life of artificial radioactive element is generally low. Both natural and artificial radioactive elements disintegrate exponentially.

+ 10 n

138 53 I

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90 + 36 Kr + 2 10 n

95 + 39 Y + 3 10 n

140 55 Cs

Artificial Radioactivity •

Nuclear fission reactions are those in which heavier nucleus breaks down into two or more lighter nuclei of almost equal size with release of large amount of energy Nuclear fission reactions were first observed by Hahn, Strassman in the nuclear fission of U-235 by bombarding with slow neutrons. Nuclear fission may takes in several ways of which the three common ways are as follows

1 + 92 37 Rb + 4 0 n

Among the three isotopes of uranium 232U, 235U and 238 U, only 235U can undergo nuclear fission reaction with slow neutron. When 238U is bombarded with fast moving neutrons it converts into 239Pu. 239Pu can undergo nuclear fission reaction. The materials which can produce chain reactions directly by bombarding with slow neutrons are called fissile materials. The materials which by themselves are non-fissile in nature but can be converted into a fissile material by reactions with neutrons are called fertile materials. During the nuclear fission reaction 235U about 0.215 amu, i.e., about 1% mass of uranium converts into energy. The total mass of products in the nuclear fission reaction of 235U when bombarded with a neutron is about 0.215 a mu less than the sum of the masses of 235U and neutron. 1 gm of U-235 releases about 7.94 × 107 KJ energy in the fission reaction. The energy released in a nuclear fission reaction is several million times greater than ordinary combustion reactions. In the fission of each U-235 nucleus on an average 3 neutrons are released. The neutrons released in nuclear fission reactions are called secondary neutrons. The successive nuclear fission reactions produced by the secondary neutrons is called chain reactions which can be used in making atomic bomb. The controlled nuclear reactions can be used in the generation of nuclear powered electricity. Control rods such as boron or cadmium or graphite rods absorb neutrons and there by control the rate of fission.These reactions are called self perpetuating chain reactions.

15.38

• • • •

Nuclear Chemistry

Liquid alloy of sodium and potassium is used as coolant which transfers the heat to the exchanger. The minimum amount of fissionable material that can produce self sustaining reaction is called critical mass. Critical mass depends on purity, density and geometric shape of fission material. High density and high purity of fissionable material is favorable for propagation of chain reations.

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Nuclear fusion reaction involve the process of fusing two lighter nuclei into a stable heavier nuclei with release of energy. In the fusion reaction about 0.7% of mass converts into energy where as in fission reaction about 0.1% of mass only converts into energy. The energy released in nuclear fusion is greater than that in nuclear fission reactions. In the fusion reaction since there is repulsion between the two nuclei due to the presence of positive charge they should be carried at very high temperature of that order of million degrees. Since the fusion reactions are carried at high temperature, they are called as thermonuclear reactions. The solar energy i.e., the energy liberating from sun and stars is due to nuclear fusion. In the hydrogen bomb the materials used are a mixture of tritium, deuterium and tritium oxides. Nuclear fission reactions take place in heavier elements while fusion reactions occur in lighter elements. Nuclear fission reactions can takes place even at room temperature while nuclear fusion reactions takes place only at high temperatures. Chain reactions of nuclear fission reactions can be controlled while that of nuclear fusion reactions cannot be controlled. Nuclear fusion reactions are more promising as future source of energy because it is 1000 times more powerful than atom bomb, it has no limit of critical mass and cannot explode unless ignited.

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Neptunium the first transuranium element was synthesized by E. McMilan and P.H. Abelson in 1940 by bombarding uranium with neutrons. 238 1 239 92 U + 0 n → 92 U + γ 239 92

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U→

239 93

Np + e −

Elements of higher atomic number are synthesized mainly by G.T. Seaborg and his associates.

U through

239 92

U to

239 93

U to

239 94

Pu

Plutonium, which itself synthetic, may serve as a target for the production of elements of still higher atomic number. The intense slow neutron bombardment of plutonium –239 yields successively plutonium isotopes of high mass number, some of which decay by beta emission to form elements of higher atomic number. Plutonium–239 bombarded with slow neutrons may for instance yield plutonium–241 which decays to form americium 241 which in turn may be bombarded to form americium 244 that decays to curium–244. In this manner it is possible to synthesize new elements and extend periodic table. High energy deuterons, alpha particles can also be used to produce elements of atomic number over 100.

Radioactive isotopes as tracers • •

• •

The isotopes of elements which exhibit radioactivity are called radioactive isotopes. The radioactive isotopes can be easily traced in the experiments involving them, by their radioactivity. So they are called radioactive tracers. The heavier isotopes can be traced in the experiments involving them by following their mass. An isotope added for this purpose is known as isotopic tracer and the element which is labeled by the particular isotope is called a tracer element.

applications of Radio isotopes •

•

Plutonium and the actinides •

If natural uranium is kept in nuclear reactor for some time at an appropriate neutron flux level, due to the sequence of reactions leads to the formation from 238 92

•

Nuclear fusion •

•

•

• •

The technique in which radio-carbon is used for the estimation of ages of archeological and biological specimen is known as radio carbon dating. The nitrogen in air capture the neutrons in cosmic rays when slowed down and disintegrate into radioactive carbon 14 6 C by emitting an electron with a half life period 5600 years. The radioactive carbon gets converted to CO2 and taken up by plants during photosynthesis where it converts into carbohydrates. These carbohydrates are taken by all living animals in their food. Once the living organism is dead, it ceases the consumption of radiocarbon and it goes on decaying in the dead body. From the relative proportion of 14C in it at the time of its death is the same as that which exists in fresh samples.

Nuclear Chemistry 15.39

•

14

C dating method can be applied to calculate the age of samples which are dead before by using the equation.

•

N 2.303 log10 0 λ Nt

•

t=

• •

Where λ is the disintegration constant of 14C (can be determined from its known half life) N0 = 16.1 disintegrations per minute (activity of fresh sample) t is the age of sample. Calculation of age of rocks is called uranium dating Since 238U decays to the final product 206Pb, by knowing their ratio in the rock its age can be calculated. Number of atoms 206 Pb = e −λt Number of atoms 238 U, left

•

•

In the chemical analysis radioactive isotopes are extensively used. (i) The solubility of sparingly soluble salts like PbSO4 and AgCl can be determined by using radioisotopes. (ii) Trace amounts (impurities) of elements in industrial raw materials and products can be determined. By using tracer technique it was found that the oxygen released by plants during photosynthesis is from water but not from CO2.

•

• • • • •

• •

The mechanism of acid catalysed ester hydrolysis is studied by using H218O. Radioactive iodine is used in the detection of location of brain tumor and for its diagnosis. Radioactive sodium 24Na is used for the study of the restricted circulation of blood and the volume of blood in a living organism. Pumping condition of heart are studied using radio isotope of iodine or sodium. 131 I is used to detect and curing the cancer. Blood disorders are treated with radio phosphorous 32P. 131 I is used in treating the thyroid. In the agriculture radioactive isotopes are used (i) To study how phosphorous is absorbed and distributed to various parts of plants by using 32 P (ii) To study how minerals are transported in plants using radio sulphur 34S When seeds are irradiated with γ-radiations, yield of the crop is improved. In industry radioactive isotopes are used (i) in detecting the level of liquid in big storage tanks. (ii) in selecting the suitable lubricating oils for machines. (iii) in detecting the flaws in the huge structures.

15.40

Nuclear Chemistry

PRactice exeRcise Multiple choice Questions with only one answer level i 1. The activity of radioactive isotopes change with (a) Temperature (b) Pressure (c) Chemical environment (d) None of the above 2. Radioactivity was discovered by (a) M curie (b) Rutherford (c) Villard (d) H. Becquerel 3. Decrease in atomic number is not observed during (a) α-emission (b) β-emission (c) Positron emission (d) electron capture 4. The largest stable nucleus is (a) U-238 (b) U-236 (c) U-235 (d) Pb-206 5. One Curie of activity is equivalent to (a) 3.7×107 dps (b) 3.7×1010 dps 4 (c) 3.7×10 dps (d) 3.7×10 dps 6. The group displacement law is given by (a) Rutherford (b) Soddy and Fajan (c) Becquerel (d) Iron Curie 7. Emission of a β-particle by an atom of an element results in the formation of (a) Isotone (b) Isobar (c) Isomer (d) Isomorph 8. Loss of β-particle is equivalent to (a) Increase of one proton only (b) Decrease of one neutron only (c) Both (a) and (b) (d) None of these 9. The 4n+2 disintegration series is also known as (a) Uranium series (b) Actino-Uranium series (c) Neptunium series (d) Thorium series 227 10. The isotope of 231 90 Th can be converted to 90 Th by (a) One alpha emission (b) Four beta emission (c) Two alpha and beta emissions (d) One alpha and two beta emissions 11. The radioactive series to which 224 88 Ra belongs is (a) Actinium series (b) Thorium series (c) Uranium series (d) Neptumium series 12. Actinium series starts with A and ends at Z. A and Z are 206 207 (a) 232 (b) 235 90 Th , 82 Pb 92 U, 82 Pb (c)

235 90

Th , 208 82 Pb

(d)

227 89

Ac, 209 83 Bi

13. β-emission is caused by the transformation of one neutron into a proton. This results in the formation of a new element having (a) Nuclear charge higher by 1 unit (b) Nuclear charge lower by 1 unit (c) Same nuclear charge (d) None of these 14. The average life of a radionuclide is (a) Always less than unity (b) Reciprocal of the decay constant (c) Constant for all radioelements (d) All the above 15. A radioactive element has atomic mass 90 amu and half-life of 28 years. The number of disintegrations per second per gram of element is (a) 5.24×1010 (b) 5.24×108 –10 (c) 5.24×10 (d) 5.24×1012 16. The radioactivity of old wood [C-14, t0.5 = 6000 years] is half the radioactivity of the new wood. What is the age of the wooden house (made up old wood) (a) 3000 year old (b) 6000 year old (c) 9000 year old (d) 12000 year old 17. Half-life period is equal to (a) 0.693/K (b) 2.303/K (c) 0.639 K (d) log a/(a-x) 18. A sample of a radioactive isotope with a half life of 20 days weighs 1 g. After 40 days, the weight of the remaining element is (a) 0.25 g (b) 0.50 g (c) 0.00 g (d) 0.16 g 19. Half-life period of a radioelement is independent of (a) Temperature and pressure (b) State of chemical combination (c) Amount of radioelement (d) All of the above 20. The disintegration constant of a certain radioactive element is 1.54 × 10–10 years–1. Its half-life period would be (a) 9×1018 yrs (b) 1.54×109 yrs 9 (c) 4.5×10 yrs (d) 15.4×109 yrs 21. If 8.0 g of radioactive isotope has a half life of 10 h, the half-life of 2.2g of the same substance is (a) 2.6 h (b) 5 h (c) 10 h (d) 40 h 22. The half-life period of a radioactive element is 140 days. After 560 days, one gram of the element will reduce to (a) 1/2 g (b) 1/4 g (c) 1/8 g (d) 1/16 g

Nuclear Chemistry 15.41

(

)

23. 1 gm atom of an α-emitting nuclide AZ X t1 2 =10 hours was placed in a sealed container. The time required for the accumulation of 4.52×1023 helium atoms in the container is (a) 4.52 hrs (b) 9.40 hrs (c) 10.0 hrs (d) 20.0 hrs 24. If the amount of radioactive substance is increased three times, the number of disintegrating atoms per unit time will be (a) Doubled (b) One-third (c) Triple (d) Unchanged 25. Consider the following statements (A) All isotopes are radioactive (B) Emission of β-rays is accompanied by the emission of anti neutrinos (C) Alpha particles have a mass number of 4 and carry a charge of +2 (D) γ-rays have a higher penetrating power than X-rays. Of these statements (a) A, B & C are correct (b) B, C & D are correct (c) C and D are correct (d) B & D are correct 26. What are the α and β-particle are emitted in the nuclear reaction 22890Th  → 212 83 Bi (a) Four alpha and one beta (b) Three alpha and seven beta (c) Eight alpha and one beta (d) One alpha and four beta 27. In the change 238 → 202 92 U  82 Pb α and β-particles have been emitted. How many α-particles were emitted in this change (a) 10 (b) 5 (c) 9 (d) 7 28. 238 U emits an α-particle; the product has mass num92 ber and atomic number respectively (a) 236,92 (b) 234,90 (c) 238,90 (d) 236,90 29. In hydrogen bomb, hydrogen is converted into (a) Barium (b) U-235 (c) U-238 (d) He 30. The end product of 4n series is(a)

208 82

Pb

(b)

206 82

Pb

(c)

207 82

Pb

(d)

204 83

Bi

31. Which of the following is used in dating archeological findings (a)

238 92

(c)

1 1

U

H

(b)

14 6

C

(d)

56 26

Fe

32. Transmutation of C-11 into the isotope B-11 is an example of (a) α-decay (b) β-decay (c) γ-decay (d) positron decay 33. 14C has a half-life of 5760 years. 100 mg of its sample will reduce to 25 mg in (a) 1440 years (b) 2880 years (c) 11520 years (d) 17280 years 34. Nuclear energy is the result of following change (a) Neutron → Proton (b) Proton → Neutron (c) Mass → Energy (d) None of these 35. Which of the following is an advantage of nuclear power plants over coal-burning plants? Nuclear plants (a) form numerous radioactive fission products (b) Do not pollute the air with SO2, soot, and fly-ash (c) Produce more thermal pollution than coal plants (d) Use more fuel 36. Nuclear fusion occurs in (a) Atomic bomb (b) Hydrogen bomb (c) Neutron bomb (d) None of theses 37. In the nuclear reactor to slow down neutrons the use is made of (a) Heavy Water (b) Saltish water (c) Fused caustic soda (d) Ordinary water 38. In carbon dating (a) The decay rate of 146 C is studied. (b) The rate of accumulated of 146 C is studied. (c) The rate of of 126 C is studied. (d) The rate of formation of 136 C is studied. 39. Radioactive I-131 is not used (a) In the treatment of thyroid diseases (b) Pumping conditions of heart (c) Increasing absorption of Ca in the body (d) For locating tumors in brain 40. What will be the total number of electrons, protons and neutrons in the product formed by loss of one α-ray from 238 92 U (a) 326 (b) 333 (c) 324 (d) 332 41. If 235U is bombarded with neutron, atom will split into (a) Sr+Pb (b) Cs+Rb (c) Kr+Cd (d) Ba+Kr 42. The radioactive decay follow (a) Zero order (b) First order (c) Second order (d) Third order

15.42

Nuclear Chemistry

43. The triad of nuclei is isotonic in(a)

14 6

(c)

14 6

C,

15 7

C,

14 7

N,

17 9

N,

17 9

F F

(b)

12 6

(d)

14 6

C, C,

14 7 14 7

N,

N,

18 9 19 9

F

F

α β 44. In the reaction Po − → Pb − → Bi , if Bi belongs to group 15, to which group P0, belongs (a) 14 (b) 15 (c) 13 (d) 16 234 45. 234 Pa   → U occurs with the emission of 91 92 (a) α-particle (b) β-particle (c) γ-rays (d) positron 46. Bismuth is the end product in radioactive disintegration series known as (a) 4n (b) 4n+1 (c) 4n+2 (d) 4n+3 47. The energy released in nuclear reactions corresponding to one amu is (a) 280 MeV (b) 931.61 MeV (c) 8.314 J (d) 4.183 MeV 206 48. The number of α-particles emitted 218 88 Ra → 82 Pb is (a) 3 (b) 4 (c) 6 (d) 2 49. For treatment of cancerous tumors, the ____________ isotope used was (a) Co-60 (b) U-235 (c) Pu-239 (d) Th-231 4 30 1 50. In the nuclear reaction 27 13 Al + 2 He → 15 X + 0 n , the element X is (a) sulphur (b) carbon (c) phosphorus (d) silicon 51. The half life period of four isotopes (i) 6.7 Yr (ii) 8 × 103 Yr (iii) 5760 Yr (iv) 2.35 × 105 Yr the most stable isotope is (a) (i) (b) (ii) (c) (iii) (d) (iv) 52. The 4n series stars from Th-232 and ends at(a) Pb-208 (b) Bi-209 (c) Pb-206 (d) Pb 207 53. How many α and β-particles are emitted in the trans234 formation 238 92 U → 92 U (a) 1, 1 (b) 1, 0 (c) 1, 2 (d) 2, 1 54. In the reaction 94 Be + X → 105 B + γ , X is (a) proton (b) deutron (c) α-paricle (d) neutron 55. The reaction 21 D + 31T → 42 He + 01 n , is an example of (a) Nuclear fission (b) Nuclear fusion (c) Artificial radioactivity (d) Radioactive disintegration

α β β 56. In the decay series A − → B − → C − →D (a) A and B are isobars (b) A and C are isobars (c) A and D are isotopes (d) B and C are isotopes 57. In the radioactive change AZ P →AZ +1 Q →AZ −−14 R →AZ −−14 S the radiations emitted in sequence are (a) α, β, γ (b) β, α, γ (c) γ, α, β (d) β, γ, α 58. With passing of time, the radioactive disintegration (a) Increase (b) decreases (c) remains same (d) may increases or decreases 59. Atom bomb is based on (a) Nuclear fusion (b) Nuclear fission (c) Induced radioactivity (d) Disintegration 60. A radioisotope has half life of 10 years. What percentage of the original amount of it would you expect to remain after 20 years (a) 0 (b) 12.5 (c) 25 (d) 8 61. Which of the following has the maximum penetrating power (a) α-particle (b) proton (c) γ-radiation (d) positron 62. Which of the following nuclear changes is incorrect

(a)

40 20

Ca + 10 n  →

(b)

24 12

Mg + α  →

(c)

113 20

(d)

43 20

40 19 27 14

Cd + 10 n  →

Ca + α  →

46 21

K + 11H

Si + 10 n

112 20

Cd +

0 −1

e

Sc + 11H

63. If 235 92 U is assumed to decay only by emitting two α-and one β-particles. The possible product of decay is (a) 211 (b) 235 89 Ac 89 Ac (c) 236 (d) 227 89 Ac 89 Ac 64. In a certain radioactive decay, an electron is emitted. It comes out from (a) Outermost orbit of the atom (b) Inner shells of the atom (c) Nucleus of the atom (d) None of these 65. When passing through a magnetic field the largest deflection is experienced by (a) α-rays (b) β-rays (c) γ-rays (d) all equal

Nuclear Chemistry 15.43

66. The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnetic field in one direction are (a) definitely α-rays (b) definitely β-rays (c) both α and β-rays (d) either α and β-rays 67. Which one of the following particles is used to bom28 31 bared 13 Al to give 15 P and neutron? (a) 12 H (b) γ (c) α (d) β 68. Which of the following nuclei is unstable?

69.

(a)

10 5

B

(c)

14 7

N

(b)

10 4

Be

(d)

16 8

O

219 86

Rn is member of actinium series. Another member of the same series is (a)

235 92 225 89

U

(b)

232 90

Th

35 15

(c) Ac (d) P 70. Helium may be produced by the fusion of isotopes of (a) radium (b) uranium (c) plutonium (d) hydrogen 71. The number of neutrons accompanying the formation 94 of 139 54 Xe and 38 Sr from the absorption of slow neutron

72.

73.

74.

75.

235 by 92 U by nuclear fission is (a) 0 (b) 2 (c) 1 (d) 3 Nuclear reactors are based on (a) Natural activity (b) Nuclear fission (c) Nuclear fusion (d) Spontaneous chemical reaction Hydrogen bombs are based in(a) Natural activity (b) Nuclear fission (c) Nuclear fusion (d) Spontaneous chemical reaction Which is the missing particle in the following unclear 2 reaction? 37 Li + H  → 83 Li + _______ 1 (a) proton (b) deutron (c) positron (d) α-particle Rutherford’s experiment which established the nuclear model of the atom used a beam of (a) β-Particles, which impinged on a metal foil and got absorbed. (b) γ-rays, Which impinged on a metal foil and ejected electrons. (c) Helium atoms which impinged on a metal foil and got scattered. (d) Helium nuclei which impinged on a metal foil and got scattered.

235 U nuclide is bombarded with slow neutrons, 76. The 92 the uranium-undergoes (a) Fusion (b) Fission (c) Radioactivity decay (d) No change 77. What is X in the nuclear reaction14 7

15

N +11 H  →8 O + X

(a)

1 1

H

(c) γ

(b)

1 0

n

(d)

0 1

e 206

238 78. In the nuclear reaction 92 U  →82 Pb . The number of α and β-particles emitted are (a) 7α, 5β (b) 6α, 4β (c) 4α, 3β (d) 8α, 6β 238 79. 92 U emits two β particles and an α-particle the product has the atomic and mass number as (a) 92,236 (b) 90,234 (c) 90,238 (d) 96,236 80. C-14 has half life of 5760 year, 100 mg of sample containing C-14 is reduced to 25 mg in (a) 11,520 yrs (b) 2880 yrs (c) 1440 yrs (d) 17280 yrs 81. A radioactive isotope decays at such a rate that after 96 min only 1/8th of the original amount remains. The value t1/2 of this nuclide is (a) 12.0 min (b) 32.0 min (c) 24.0 min (d) 48.0 min 82. Which of the following is used in neutron absorber in the nuclear reactor (a) Water (b) Deuterium (c) Some compound of Uranium (d) Cadmium 83. Which of the following elements belongs to 4n-series (a) Pb-206 (b) Bi-209 (c) Pb-208 (d) Pb-207 7 84. In a reaction 37 Li+Z  →4 Be+10 n . The bombarding projectile Z is (a) α-particle (b) Deutron (c) Neutron (d) Proton 85. An element X looses one α and two β-particles in three successive stages. The resulting element will be (a) an isobar of X (b) isotope of X (c) X itself (d) an isotone of X 86. Insert the missing figure in the following 55 25 Mn ( n, γ ) → ____________ (a) 56 (b) 55 25 Mn 24 Cr

(c)

55 25

Mn

(d)

56 24

Cr

15.44

Nuclear Chemistry

87. A radioactive sample has half life of 1500 years. A sealed tube containing 1 g of a sample will contain…….g of the sample after 3000 years. The missing figure is (a) 1 g (b) 0.5 g (c) 0.25 g (d) 0 g 88. According to nuclear reaction 4 Be + 24 He  → 12 1 C + , n the mass number of Be atom is 6 0 (a) 4 (b) 8 (c) 6 (d) 9 89. Half life of radioactive element is 100 yrs.. the time in which it disintegrate 50% of its mass will be (a) 50 yrs (b) 200 yrs (c) 100 yrs (d) 25 yrs 90. Energy released in the nuclear fusion reaction is 2 1

H + 13 H  → 42 He + 10 n. Atomic mass of

2 1

H = 2.014; 13 H = 3.016, 42 He = 4.003;

1 0

n = 1.009 ( all in amu )

(a) 16.60 MeV (b) 500 J (c) 4×106 KCal (d) 8.30 eV 91. Positron has mass equal to(a) Electron (b) α-particle (c) proton (d) deutron 92. Which of the following is (n, p) type reaction 13 1 → 14 (a) 5 C + 1H  6 C (b)

14 7

N + 11H  → 15 8 O

(c)

27 13

27 Al + 10 n  → 12 Mg + 11H

235 94 (d) 92 U + 10 n  → 140 54 Xe + 88 Sr 93. Half-life of radioactive substance is 60 min. During 3 hrs, the fraction of total no of atoms that have decayed would be (a) 12.5% (b) 87.5% (c) 8.5% (d) 25% 94. The composition of tritium is (a) 1 electron, 1 proton, 1 neutron (b) 1 electron, 2 protons, 1 neutron (c) 1 electron, 1 proton, 2 neutrons (d) 1 electron, 1 proton, 3 neutrons 95. The radioactive decay of 88 35 X by a β-emission produces an unstable nucleus which spontaneously emits a neutron. The final product is

(a)

88 37

(c)

88 34

X Z

(b)

89 35

Y

(d)

87 36

W

96. On emission of one α-particle followed by one 238 β-particle from 92 X . Then number of neutrons in the atom will be (a) 142 (b) 146 (c) 144 (d) 143

97. 1 g of radioactive element reduces to 125 mg after 24 hours. Half life of isotope is (a) 8 hours (b) 24 hours (c) 6 hours (d) 4 hours 98. The radioactive element has t0.5 of 60 minutes. The amount remaining after 3 hours is (a) 17.5% (b) 12.5% (c) 25% (d) 50% 99. Which does not take place by α-particle disintegration (a) (c)

238 92

234

U  →90 Th

226 88

222

Ra  →86 Rn

228

(b)

232 90

Th  →88 Ra

(d)

213 83

213 Po Bi  →84

100. A radioactive element X emits 3α, 1β and 1γ-particles 225 and forms 76 Y element X is (a)

237 81

(c)

236 81

X X

(b)

237 80

X

(d)

236 80

X

101. Positron is (a) electron with +ve charge (b) a helium nucleus (c) a nucleus with two protons (d) a nucleus with one neutron and one proton 102. Which element is the end product of every natural radioactive series (a) Pb (b) Sn (c) C (d) Bi 103. 226Ra disintegrates at such a rate that after 3160 years only one-fourth of its original amount remains. The half-life of 226Ra will be (a) 790 years (b) 3160 years (c) 1580 years (d) 6230 years 104. The end product of (4n+2) disintegration series is (a)

204 82

(c)

209 82 228 90

Pb Pb

(b)

208 82

Pb

(d)

206 82

Pb

212 83

105. If Th disintegration to Bi then the number of and β-particles emitted is (a) 4α and 7β (b) 4α and 1β (c) 4α only (d) 7β only 106. The t1/2 of radioactive element is 140 days. In 560 days a sample of this element would be reduced to… of its initial mass (a) ½ (b) ¼ (c) 1/8 (d) 1/16 107. Number of neutrons present in a parent nucleus X, which gives 14 7 N after successive β-emissions would be (a) 6 (b) 7 (c) 8 (d) 9 108. A 300 gram radioactive sample has initial half life is 3 hours. The remaining quantity after 18 hour’s will be (a) 4.68 g (b) 2.34 g (c) 3.34 g (d) 9.37 g

Nuclear Chemistry 15.45

Multiple choice Questions with only one answer level ii 1. Which of the following statement (s) are correct about half-life period? (i) It is proportional to initial concentration for zeroth order. (ii) Average life = 1.44 half-life for first order reaction. (iii) Time of 75% reaction is thrice of half-life period in second order reaction. (iv) 99.9% reaction takes place in 100 minutes for the first order reaction then rate constant is 0.0693 min–1. (a) (i) only (b) (ii), (iii) only (c) (i), (ii), (iii) only (d) (i), (ii), (iii), (iv) 2. 227Ac has a half life 20 years. It decays in two parallel paths leading to 227Th and 223Fr. The percentage yields of these two nucleids are 2%, 98% respectively. The rate constant of the formation of Th is (a) 6.93×10–3 Yr–1 (b) 6.93×10–4 Yr–1 (c) 3.46×10–3 Yr–1 (d) 3.46×10–4 Yr–1 3. 27 mg of pure PuO2 has an activity of 6.02×107 dps atomic weight of Pu is 238. The decay constant of Pu is (a) 10–11 sec–1 (b) 10–12 sec–1 –13 –1 (c) 10 sec (d) 10–14 sec–1 4. An archaeological specimen containg 14C gives 40 counts in 5 minutes per gm of carbon. Freshly cut wood gives 20 counts per gm of carbon per minute. The counter gives 5 counts per minute in the absence of any sample. Half life of carbon is 5000 years. The age of specimen is (a) 1.1×102 years (b) 1.1×103 years 4 (c) 1.1×10 years (d) 1.1×105 years 40 5. 19 K consists of 0.01% of the potassium in nature. The human body contains 0.5% potassium by weight. Half 40 life of 40K is 1.38×109 years. The radioactivity of 19 K in 80 Kg of human body is (a) 3×109 dpy (b) 3×1010 dpy 11 (c) 3×10 dpy (d) 3×1012 dpy 238 6. The isotopes U and 235U occur in nature in the ratio 100:1 assuming that at the time of earth formation, they are present is equal ratio. Half life of 238U and 235 U are 5×109 and 5×108 years respectively. The age of earth of (a) 3.7×107 years (b) 3.7×108 years 9 (c) 3.7×10 years (d) 3.7×1010 years

7. Radium lose an α-particle and leaves a gaseous element radon. How many mL of radon at 25°C and 1 atm are in equilibrium with 1 g of radium? (t1/2 of Ra = 1590 year, t1/2 of Rn = 3.82 days, R = 0.0821 litre atm/mole K; At. wt. of Ra = 226) (a) 710×10–5 mL (b) 710×10–4 mL (c) 650×10–6 mL (d) 710×10–6 mL 1 8. The 146 C and 126 C ratio in a piece of wood is part 16 14 that of atmosphere. Half life at 6 C is 5500 years. The age of wood is (a) 5500 year (b) 11000 year (c) 1650 years (d) 22000 years 9. In the chain reaction 238 92

U +10 n  → Nuclide A + Nuclide B + 310 n + Energy E

10.

11.

12.

13.

Neutrons and energy produced at the (n-1)th step will be (a) 3n, nE (b) 3n–1, 3n–2 E n n–1 (c) 3 , 3 E (d) None of these For a certain reaction 10% of the reactant decomposes in one hour 20% in two hrs. and 30% in 3 hrs. The unit of the rate constant will be (a) hours–1 (b) moles litre–1 sec–1 –1 –1 (c) litre mole sec (d) mole sec–1 Lead is the final product formed by a series of changes in which the rate determining stage is the radioactive decay of uranium-238. This radioactive decay is first order reaction with a half-life of 4.5×109 years. What would be the age of a rock sample originally lead free, in which the molar proportion of uranium to lead is now 1: 3? (a) 1.5×109 years (b) 2.25×109 years 9 (c) 4.5×10 years (d) 9.0×109 years Two radioactive nuclides A and B have half lives of 50 min and 10 min respectively. A fresh sample contains the nuclides of B to be eight time that of A. How much time should elapse so that the number of nuclides of becomes double of B. (a) 30 (b) 40 (c) 50 (d) None The average (mean) life at a radio nuclides which decays by parallel path is λ

λ

1 → B; λ =1.8×10−2 sec −1 , 2A  2 → C; = λ 2 10−3sec −1 A  1

(a) 52.63 sec (b) 500 sec (c) 50 sec (d) None 14. The ratio of activities of two radionuclides X and Y in a mixture at time t= 0 was found to be 8:1 After two hour’s the ratio of activities become 1:1. If the t1/2 of radio nuclides X is 20 min. find t1/2 (in minutes) of radionuclide Y. (a) 20 min (b) 30 min (c) 40 min (d) 50 min

15.46

15.

16.

17.

18.

19.

20.

21.

Nuclear Chemistry

I is a β-emitter (t1/2 = 7 days) employed to prepare KI. A sample of activity 10 µ Ci is injected in a patient diagnosed with thyroid disorder. If 90% of the injected KI is transferred to the thyroid, what would be the activity of thyroid gland after two weeks? (a) 2.5 µ Ci (b) 10 µ Ci (c) 4.5 µ Ci (d) 2.25 µ Ci After the emission of one alpha and one beta particles from the nucleus 238 92 X , the number of neutrons in the product nucleus would be (a) 138 (b) 140 (c) 143 (d) 146 To obtain an isotope of a given radioactive element, the element must emit (a) One alpha and one beta particles (b) One alpha and two beta particles (c) Two alpha and two beta particles (d) One alpha and four beta particles Consider the decay series: A→B→C→D, where A, B and C are radioactive elements with half- life of 4.5 seconds, 15.0 days 1.00 second respectively and D is non- radioactive element. Starting with 1.00 mol of A, the number of moles A, B, C and D left 30 days are (a) One more of D none of A, B, or C (b) 3/4mol of B, 1/4mol of D and none of A or C (c) 1/4 mol B, 3/4mol of D and none of A or C (d) 1/2 mol of B, 1/4mol of C, 1/4mol of D and none of A A proportional counter is filled in succession with the same amount of 3 samples of CO2 at a fixed pressure and the following counts were noted (i) Dead CO2from coal: 9000 counts in 900 min (ii) CO2from contemporary wood: 22000 counts in 200 min (iii) CO2from test sample: 24, 000 counts in 400 min Then the age of the sample will be (of 14C = 5570 years) (a) 5570 years (b) 4997 years (c) 11140 years (d) 9994 years In a particle scattering experiment, Rutherford took a radioisotope 293X and observed that the minimum activity is 30 Rutherford. Find out the minimum weight of radioisotope required to carry out the experiment up to 10 hours. (Given t1/2 of 293 X = 23.5 hours) (a) 2.36×10–8 g (b) 8.58×10–8 g –8 (c) 5.72×10 g (d) 11.44×10–8 g The counting rate observed from a radioactive source at t = 0 second was 1600 counts/ sec and at t = 8 sec it was 100 counts /sec. The counting rate as count per sec at t = 6 sec will be (a) 400 (b) 300 (c) 200 (d) 150 131

22. Half-life period of 10 days. Amount of 1 mol of A is (a) 1/2 (c) (1/2)1/11 23.

the radioactive element A is A left on 11th day starting with (b) (1/2)12/11 (d) (1/2)11/10

Column 1 (I)

24

(A) Studying transport of minerals in plants (B) Treatment of cancer

Na

(II)

131

(III)

60

(IV)

Column 2

I

Co

35

(C) Treatment of thyroid complaints (D) Study of restricted circulation of blood

S

(a) I-D, II-C, III-A, IV-B (b) I-A, II-C, III-D, IV-B (c) I-D,II-C, III-B,IV- A (d) I-C, II-D, III- B, IV-A 24. Shorter the radioactive half-life (a) greater is the number of atoms disintegrating per second. (b) smaller is the decay constant. (c) less reactive is the parent nucleus. (d) greater is the mass-energy change. 25. In a radio active equilibrium, the ratio between two atoms of radio active elements A and B are 2×109:1. If half-life period of A is 2×1010 years. The half life ratio of B is (a) 5 years (b) 7.5 years (c) 10 years (d) 12.5 years U 26. 238 (III B) undergoes following emissions: 92 238 92 U

−α

−α

−β

→ A → B → C

Which is/ are correct statements? (a) A will be IB group. (b) A will be IIIB group. (c) B will be IA group. (d) C will be III A group. 27. The mean lives of a radio active substance are 1620 years and 405 years for α emission and β emission respectively. If it is decaying α, β emission simultaneously, the time required to decay

3 th of the sample is 4

(a) 22.5years (b) 225 years (c) 45.0 years (d) 450 years 28. 64Cu (half life = 13.86 hours) decay by β– emission (40%) β+ emission (20%) and electron capture (40%) half life value of β– emission (a) 69.3 hr (b) 17.3 hr (c) 51.9 hr (d) 34.65 hr

Nuclear Chemistry 15.47

29. Maximum number of stable nuclides are when (a) p is even, n is odd. (b) p is even and n is even. (c) p is odd and n is even. (d) p is odd and n is odd. 30. A radioactive isotope 94X (t1/2= 15 day ) gives 78Y, a stable atom along with α-particles. If 94 g of X are taken and kept in a sealed tube, how much He will accumulate in 30 days at S.T.P. (a) 67.2 litre (b) 56 litre (c) 22.4 litre (d) 44.8 litre 31. The half life period of a radioactive element ‘P’ is same as the mean life time of another radioactive element ‘Q’. Initially both of them have the same number of atoms. (a) P and Q decay at same rate always. (b) P will decay at a faster rate than Q. (c) Q will decay at a faster rate than P. (d) P and Q have the same decay rate initially. 32. Fluorine – 18 is a radio active isotope that decays by positron emission to form oxygen – 18 with a half life of 109.7 min. Physician’s use 18F for the study of brain by injecting a quantity of fluoro-substituted glucose in to the blood of patient. The glucose accumulates in the region where the brain is active and needs nourishment. If a sample of glucose that contains 18F is injected into the blood, how long it take for 99% decay of 18F? (a) 730 minutes (b) 326 minutes (c) 219.4 minutes (d) 10 hours 33. A 0.500 mL sample of an aqueous solution containing 0.48 µ Ci per mL of 24Na is injected in to the blood stream of an animal. After allowing sufficient time for complete circulatory mixing, a 0.100 mL aliquot of blood is removed and found to have a net activity of 466 dpm of 24Na. The volume of the blood of the animal is (a) 63 mL (b) 681 mL (c) 1247 mL (d) 114 mL

Multiple choice Questions with one or More than one answer 1. Identify the correct statements: (a) On bombarding 14 7 N nuclei with α-particle, the nuclei of the product formed after release of proton would be 17 8 O. 228 89 Ac

and 229 90 Th belongs respectively to Actinium and neptunium series. (c) Nuclide and its decay product after α-emission are called isodiaphers. (d) Half life of radium is 1580 years. Its average life will be 1097.22 years. (b)

2. Identify the false statements: (a) If the equilibrium constants for the reaction Br2→2Br at 500 K and 700 K are 10–10 and 10–5 respectively. Then reaction is exothermic. (b) When a solution prepared by mixing 90 mL of HCl solution (pH = 3), 10 mL solution of NaOH (pH=11) and 60 mL water. The pH will be 3.1. 235 207 U → 82 Pb , the total (c) In the nuclear reaction 92 number of α and β particles lost would be 11. (d) An aqueous solution of an alcohol (B.P. 56°C) in water (B.P 100°C) has vapour pressure more that of water. 3. Select correct statements: (a) Smaller the packing fraction of an element, the greater will be the stability of its nucleus. (b) Packing fraction is negative when the isotopic mass is less than the mass number. (c) Packing fraction is positive when the isotopic mass is less than the mass number. (d) Packing fraction is highest for hydrogen and least for iron. 4. A radioactive element decays as follows 2β α X 30 → Y 2 →Z min days

Which of the following statements are correct set? (a) After two hours, less than 10% of the initial ‘X’ is left. (b) Atomic number of X and Z are same. (c) Maximum amount of Y present at any time before 30 minutes is less than 50% of the initial amount of X. (d) The mass number of Y is greater than X. 5. In a zero-order reaction (a) The rate constant has the unit mol L–1 s–1. (b) The rate is independent of the concentrations of the reactants. (c) The half-life depends on the concentrations of the reactants. (d) The rate is independent of the temperature of the reaction. 6. Which of the following statements are correct? (a) 235 92 U undergoes β-decay preferentially. (b) The nuclear fusion 411 H → 42 He + Q occuring at sun and stars is catalyzed by 14C. (c) At steady state, the element with the highest half life is present in largest amount. (d) Helium, carbon and oxygen do not lie on the curve of the plot of mass number and binding energy of nucleon (MeV) indicating that they are exceptionally stable.

15.48

Nuclear Chemistry

7. The order of reaction A → product can be given by the expression [where r = rate of reaction; [A]1 = concentration at time t1: [A]2 = concentration at time t2] (a)

ln r2 − ln r1 ln[A]2 − ln[A]1

(b)

 −d[A]  (b) ln   / ln[A]  k.dt 

ln [A 0 ]2 − ln[A 0 ]1 ln[ t1/ 2 ]2 − ln[ t1/ 2 ]1 (d)

ln (r / k ) ln[A]

8. Pick out the correct statement(s) from among the following: (a) One gram each of radium elemental and RaSO4 will have the same activity. (b) The beta particle emitted by a radioactive element is from valence shell of the atom. (c) Nuclear isomers will have the same mass numbers as well as atomic number. (d) The fraction decayed during ‘n’ half lives is 2n − 1 2n . 9. The unit of radioactivity is (a) Curie (b) Rutherford (c) Becquerel (d) None of the above 10. Which one of the following notations shows the product incorrectly 242 243 10 13 (a) 96 Cm (α, 2n) 97 Bk (b) 5 B, (α, 2n ) 7 N 28 29 (c) 147 N, (n , p)146 C (d) 14 Si, (d, n )15 P 11. It is observed that only 0.39% of the original radioactive sample remains undecayed after eight hours. Hence (a) The half-life of that substance is 1 hour. 1 (b) The mean life of the substance is hour. log e 2 (c) Decay constant of the substance is (loge 2) hour–1. (d) if the number of radioactive nuclei of this substance at a given instant is 108 then the number left after 30 min would be 2 × 109 .

comprehensive type Questions Passage i A small amount of solution containing a radioactive nucleide Ax was administered into blood of a patient. The activity of the nuclide is 2 × 103 dps. Its half life is 15 hours. After 5 hours a sample of the blood drawn out from the patient. It’s activity was 16 dpm per mL. 1. What is the volume of the ‘blood’ in the patient? (Log (1.26) = 0.1) (a) 2.92 lit (b) 3.95 lit (c) 4.92 lit (d) 5.95 lit

2. What is activity of the sample after another 5 hours time? (Log 1.59 = 0.2) (a) 11.18 dpm per mL (b) 1.118 dpm per mL (c) 12.71 dpm per mL (d) 1.271 dpm per mL 3. Radioactive nuclide Ax decays by β– emission (42%), β+ emission (58%) in the patient body. Then half life value of β– decary path is (a) 27.2 hours (b) 10.2 hours (c) 3.57 hours (d) 35.72 hours 4. At Ax emits α ray. How many helium atoms present in 30 hours? (a) 1.558 × 108atoms (b) 1.168 × 108 atoms (c) 7.79 × 107atoms (d) 3.895 × 10 atoms 5. If Ax decays and finally By unradioactive element obtained. After some time the sample contains 20% Ax and 10% By gm atoms present in it. The time taken for the attainment of conversion is (a) 8.78 hours (b) 15.0 hours (c) 30.1 hours (d) 40.2 hours Passage ii A follows parallel path first order reaction to give B and C K1 = 1.5 × 10–5 sec–1

B

5A K2 = 5 × 10–6 sec–1

2C

If initial conc of A is 0.236 M (anti log of 0.072 = 1.180) 1. The rate constant of the reaction is (a) 2 × 10–5 sec–1 (b) 2 × 10–4 sec–1 –6 –1 (c) 2× 10 sec (d) 2 × 10–7 sec–1 2. The concentration of ‘C’ after 5 hours is (a) 3.6 × 10–2 M (b) 3.6 × 10–3 M –3 (c) 7.1 × 10 M (d) 7.1 × 10–2 M 3. The concentration of ‘B’ after 5 hours is (a) 5.4× 10–2 M (b) 1.08 × 10–2 M –4 (c) 7.1 × 10 M (d) 7.1 × 10–3 M Passage iii A rock type mineral of a radioactive element

210 84 A

(t1/2 =

6930 years) contains another element 206 82 B supposed to be formed by the radioactive disintegration of the former by α-decay mode. The mineral contains another radioactive isotope of A-210 where mass number is 211 and t1/2 = 693 years. This isotope is also α-emitter decaying to

Nuclear Chemistry 15.49

non-radioactive isotope of element B. The mineral is also containing helium gas trapped into its cavities formed during the radioactive disintegration of the two isotopes of A. It is believed that at the time of the formation of the mineral the number of atoms of the two radioactive isotopes of A were equal. A chemical analysis of the mineral done today reveals that number of atoms of A-210: Number of atoms of A-211 is 100:1. 1. How much old is the mineral? (a) 2.2 × 103 years (b) 5.1 × 103 years (c) 3.7 × 105 years (d) 4.5 × 109 years 2. The ratio of the number of atoms of A-210 to A-211 after a period of 6930 years from its very inception is approximately equal to (a) 607.5:1 (b) 458.4:1 (c) 512.3:1 (d) 700.0:1 3. The ratio of the mass of A-210 to the mass of its product of disintegration after the mineral will grow to the age of 2.079 × 104 years will be approximately equal to (a) 1:0.99 (b) 1:8.3 (c) 1:2.98 (d) 1:6.866 Passage iv Tritium T (an isotope of H) combines with fluorine to form weak acid TF, which ionizes to give T+. Tritium is radioactive and is β-emilter. A fresh sample prepared in aqueous solution of TF has PT (same of pH) of 2 and fraezer at freezes at – 0.42°C Kf for water is 2 K Kg mol–1. 500 mL of freshly prepared solution were allowed to stand for 24.8 years. Half life of tritium is 12.4 years. 1. The ionization constant of T F is. (a) 5.2 × 103 (b) 5 × 10–3 –4 (c) 5 × 10 (d) 5.2 × 10–4 2. No. of β-particles emitted are. (a) 4.5 × 1021 (b) 3.0 × 1021 22 (c) 4.5 × 10 (d) 3.0 × 1022 Passage v A radioactive substance ‘A’ converts to stable nuclei D by following series of reaction. A→B→C→D Given: t1/2 for ‘A’ = 0.0693 days t1/2 for ‘B’ = 6930 days t1/2 for ‘C’ = 6.93 days 1. Number of nuclei of ‘C’ formed (approximately) in the first 10 days are, if initially 10 20 nuclei of A is taken (a) 1018 (b) 1016 17 (c) 10 (d) 1019

2. Number of nuclei of ‘D’ present after 6930 days are, if initially 1020 nuclei of A is taken 1 (a) 1010 (b) × 1020 2 1 17 (c) × 10 (d) 109 2 3. To determine the age of a stone, it was analysed for the nuclei of ‘B’ and ‘D’ and were found to be 1018 and 3 × 1018 respectively. Assuming that all the ‘D’ was produced by the disintegration of ‘A’ only, the age of the stone is [Given: log2 = 0.3, in 3 = 0.45] (a) 6930 days (b) 10395 days (c) 13860 days (d) 3465 days Passage vi The variety of radioactive isotopes arises from the fission ‘daughters’ reaction of neutrons with building material and formation of transuranium elements from bombardment of 238 U by neutron and α-particles. The longer half lives present a greater problem as they remain radioactive for even hundreds of years. The total amount of radioactive waste in United States exhibits activity about 1.5 billion Curies consider the β-decay of 90Sr (89.907738 amu) 90 38

0 Sr →90 39 Y + −1 β t 0.5 = 28.1 yrs

The 90Y (89.907152 amu) further decay’s as follows: 90 39

0 Y →90 40 Zr + −1 β t 0.5 = 64 yrs

90

Zr (89.904703 amu) is a stable isotope. 1. Starting with one mole of 90Sr, calculate the number of moles of 90Sr that will decay in a year? (a) 0.24 moles (b) 0.48 moles (c) 0.976 moles (d) 0.024 moles 2. If one microgram of 90Sr was absorbed by a newborn child, how much 90Sr will remain in bones after 20 years? (a) 6.1 × 10–7 g (b) 2.8 × 10–7 g –5 (c) 2.1 × 10 g (d) 3.2 × 10–4 g 3.

Sr & 90 39 Y both undergoes β-decay due to n (a) Low ratio p (b) To increase nuclear binding energy per nucleon n (c) To make ratio equal to 1 p (d) To decrease nuclear binding energy per nucleon 90 38

Passage vii Natural radioactivity is defined as the spontaneous disintegration of naturally occurring, heavy elements (with

15.50

Nuclear Chemistry

atomic number more than 82) with the emission of invisible radiations (α, β-particles and γ-rays). The radiations emitted by radioactive elements were classified into three different types, depending upon the effect of the applied magnetic or electric field on the radio-active elements. These different types of radiations are α (alpha) rays (α-particles), β (beta) rays, (β-particles) and γ (gamma) rays. When an α-particle is emitted from a nucleus, the mass number is decreased by four units and the atomic number is decreased by two units. When a β-particle is emitted form a nucleus, the atomic number of the new element is increased by one unit but, the mass number does not change. When a γ-ray is emitted from a nucleus, the mass number and the atomic number do not change. Gamma rays are electromagnetic radiations. 1. The radiations from a naturally occurring radioactive substance, as seen after deflection by a magnetic field in one direction are (a) either α or β-rays (b) definitely β-rays (c) both α and β-rays (d) definite α-rays 2. On bombarding 147 N with α-particles, the nuclei of the product formed after the release of a proton will be (a) 189 F (b) 178 O (c) 17 (d) 188 O 9 F 218 206 3. The number of α-particles emitted by 84 Ra → 82 Pb is (a) 3 (b) 4 (c) 6 (d) 2 4. Of the following radiations, the one most easily stopped by air is (a) β-rays (b) γ-rays (c) α-rays (d) X-rays Passage viii The binding energy per nucleon is smaller for light nucleides and increases steadily in kcal/mole with nuclear mass, because more the particles there are in the nucleus, the greater should be total energy required to break it apart. Instability of nucleus is due to high n/p ratio. The range of n/p value for stability is 1 to 1.56. If a nucleide has n/p ratio greater than 1.5, it emits at β-particle. n/p ratio is zero for H. n/p ratio for 209 83 Bi is 1.52. If n/p ratio is greater than 1.56, that nuclide becomes radioactive. There is no naturally occurring nuclide having n/p ratio less than 1. If a nucleide has n/p ratio outside the belt of stability it shows radioactivity. Nuclide with highest n/p ratio of 2.0 is 13 H . Lighter nucleides, upto 40Ca have n/p ratio equal to 1. For heavier nuclides n/p ratio increases and becomes greater than 1, in order to overcome the

proton-proton repulsion. Positron decay increases the n/p ratio, while beta-particle decay decreases the n/p ratio. 60 1. 27 Co is radioactive because (a) It has high p/n ratio. (b) It has high n/p ratio. (c) Its atomic number is high. (d) None of these. 2. Maximum number of stable nucleides are when (a) p is even and n is even. (b) p is even and n is odd. (c) p is odd and n is odd. (d) p is odd and n is even. 3. An unstable nuclide (Z>70) with n/p ratio less than that required for stability can attain stability by (a) K-electron capture (b) β+-decay (c) α-decay (d) All of these Passage ix For a radioactive decay A→B,

−dN = λN dt

[λ = decay constant, N = Number of atoms at any time t] or N 2.303 2.303 a log10 0 or λ = log10 λ= t N t (a − x ) Half-life period (t1/2 or t0.5) is the time required for disintegration of one half of the original amount of the radioactive substance. If t = 1/2, then λ =

2.303 0.693 log10 2 or t1/ 2 = = λ t 1/ 2

constant. Half-life t0.5 of a radioactive substance is constant and is independent of the initial amount taken. Amount of radioactive substance left after n half-life periods is N = No (1/2)n. Total time t = n × t0.5. where n is a whole number. Half-life of a radioactive substance is useful in identifying a nuclide and also predicting its stability, higher the halflife, more stable is the substance. According to GeigerNuttal rule, activity of a nucleus is inversely proportional to its half-life/average life. Shorter the half-life, greater is the radioactivity of the element. 1. With the passage of time, the rate of radioactive disintegration (a) may increase or decrease (b) remains same (c) decreases (d) increases 2. The t1/2 of radioactive K-40 is 5.27 years (λ=2.5×10–7 min–1). The decay activity of 2.0 g of the sample is about (a) 7.5 × 1015 dpm (b) 5 × 105 dpm (c) 5 × 1010 dpm (d) 7.5 × 1020 dpm 3. Mark the incorrect relation (a) N0 = Neλt (b) τ = 1.44 t1/2 (c) N = N0 (1/2)n (d) t0.5 = λ/ln2

Nuclear Chemistry 15.51

Matching type Questions 1. Match the following Column I

Column II

(a)

14 6C

(b)

214 88 Ra

(c)

7 5B

(r) β-emitter

(d)

18 8O

(s) posiron emission

(p) Number of neutrons is equal to magic number (q) α-emitter

2. A part of the radioactive disintegration series is given in Column I. Match Column I with Column II and select the correct the correct match. Column I

3. Assertion (A): The nuclear isomers are the atoms with the same atomic number and same mass number, but with different radioactive properties. Reason (R): The nucleus in the excited state will evidently have a different half-life as compared to that in ground state. 4. Assertion (A): Radioactivity is a nuclear phenomenon. Reason (R): Radioactivity is the phenomenon of spontaneous emission of certain invisible radiations from the nuclei of the atoms of some elements. 5. Assertion (A): Rate of disintegration of uranium increases with the increase in amount of uranium. Reason (R): Rate of disintegration nuclide does not depends upon temperature pressure state of combination etc. 6. Assertion (A): 146 C is s −10β emitter. Reason (R): Unstable nucleus having n/p >1 are −10 α emitter.

Column II

(a)

221 237 → 87 Fr 93 N P 

(b)

234 91 Pa

218  → 85

(c)

230 90 Th

 → 84 Po

(d)

229 90 Th

213  → 84

At

214

Po

(p) 4α, 2β (q) 4n+2 (r) Pb206 (end product of the series) (s) 4n+1

assertion (a) and Reason (R) type Questions “In each of the following questions a statement of Assertion (A) is given followed by a corresponding statement of Reason (R) just below it. Mark the correct answer from the following statements. a. If both assertion and reason are true and reason is the correct explanation of assertion b. If both assertion and reason are true but reason is not the correct explanation of assertion c. If assertion is true but reason is false d. If assertion is false but reason is true 1. Assertion (A): In a nuclear fission process, the total mass of fragments is always greater than the mass of the original nuclei. Reason (R): Difference in the mass due to fission of a heavy nucleus is converted into energy according to mass-energy conversion. 2. Assertion (A): The activity of 1 g pure uranium 235 will be same than the same amount present as U3O8. Reason (R): In the combined state, the activity of the radioactive element decreases.

integer type Questions 1. In nature a decay chain series starts with 232 90 Th and finally terminate at 208 Pb . A thorium sample was 82 found to contain 23.2 gm of 232 Th and 13.44 litres of 90 He gas at NTP. Half life of 232 90 Th is 2 years. The age of thorium sample in years is. 2. The number of neutrons accompanying the formation 90 of 144 54 Xe and 38 Sr by the absorption of a slow neutron by 235 92 U followed by nuclear fission is. 3. A human body excrets [removes by discharges, sweating etc.] certain material by a law similar to radio activity. If technetium is injected in some form in the human body, the body excrets half of the amount in 24 hours. A patient is given an in jection containing 99Tc. This isotope is radioactive with a half life 6 hrs. The activity from the body just after injection is 6 mCi. How many hours will elapse before the activity falls to 3 m Ci? Express your answer after dividing with 2.4 4. One mole of A undergoes decay as, m m-8 4 A   → B+2 He . If the volume of helium z z-4 2 collected in 18 days is 33.6 L at STP, the half life period of A will be in days. 5. How many of the following nuclides are expected to be β-emitters

(

)

3 14 24 30 39 65 1 H; 6 C; 11 Na ; 15 P; 20 Ca ; 29 Cu

6. In the sequence of following nuclear reaction, 238 98

−α

−β

−β

nα

→ Y → Z → L → 218 90 M

What is the value of ‘n’?

15.52

Nuclear Chemistry

7. The ratio of active nuclides ‘X’ and ‘Y’ in a mixture at time t = 0 was found to be 4:1. After two hours, the ratio of activities becomes 1:1. If the t1/2 of radio nuclide ‘x’ is 20 min then t1/2 (in minutes of radio nuclide ‘Y’ is represented as

Previous years’ iit Questions 1.

n × t 1 of ‘x’ then ‘n’ is. 2 2

8. To determine the age of a stone, it was analyzed for 1020 nuclei of A and 7 ×1020 nuclei of B. If nuclei of B are assumed to be obtained only due to radioactive decay of A and no B was present initially, find the age of stone in years (Half life of A = 1095 days). 9. At 27°C 100 mL of 0.1 M radium chloride is 10–3Curies active. The activity of 500 mL of 0.2 M radium phosphate solution at 77°C is x × 10–2 Curie then ‘x’ is__________. 10. The counting rate observed from radioactive source at t = 0 sec was 1600 counts/sec and t = 8 sec it was 100 counts/sec. The counting rate observed as count per sec at t = X sec is 200 count/sec. The value of ‘X’ is? 11. A fresh radioactive mixture containing short lives species A and B, both emitting α-particles initially of 8000 α-particles per minute, 20 minute later they emitts at the rate of 3500 α-particles per minute. If the half- lives of the species. A and B are 10 minutes and 500 hours respectively, and the ratio of activities of A and B in the initial mixture was a:b; then a+b is __________. 12. A sample of 238U (t1/2=4.5×109) are found to contain 23.8 g of 238U and 20.6 g of 206Pb. The age of the ore is

2.

3.

4.

5.

27 13

29 Al is a stable isotope, 13 Al is expected to disintegrate by: (a) α-emission (b) β-emission (c) positron emission (d) proton emission (1996) The number of neutrons accompanying the formation 94 of 139 54 Xe and 38 Sr due to the absorption of a slow neu235 tron by 92 U , followed by nuclear fission is: (a) 0 (b) 2 (c) 1 (d) 3 (1999) 23 Na is more stable isotope of Na. Find out the process 23 Na can undergo radioactive decay. by which 11 1 (a) 0 n Emission (b) α-emission (c) β+ emission (d) K-electron capture (2003) 23 A positron is emitted from 11 Na . The ratio of atomic mass and number of the resulting nuclide is: (a) 22/10 (b) 22/11 (c) 23/10 (d) 23/12 (2007) Bombardment of aluminum by α-particle leads to its artificial disintegration in two ways, (i) and (ii) as shown. Products X, Y and Z respectively are, (ii) 27 30 13 Al 15 P + Y

(i)

x × 10 x years. The value of ‘x’ is__________. 2 13. At a given instant there are 25% undecayed radioactive nuclei in a sample. After 10 second, the number of undecayed nuclei reduces to 12.5% the time in which the number of undecayed nuclei further reduce to 6.25 % of the reduced number is 10x seconds. The x is__________. 14. A radio active element had an initial activity of 56Ci. After 69.3 sec it was found to have an activity of 28Ci. The number of atoms in the element having an activity of 100 dps is 10x. What is the value of x? 15. 100 mL of a saturated solution of M2SO4 is giving 24 disintegration per hour due to radio active metal “M” (λ = 2×10–17 hour–1, N0 = 6×1023 a). If the solubility product of the salt is x×10–3y What is (x+y)?

30 14

(a) (b) (c) (d)

Si + X

30 14

Si + Z

proton, neutron, positron neutron, positron, proton proton, positron, neutron positron, proton, neutron

(2011) 30 6. Assertion (A): Nuclide 13 Al is less stable than 40 20 Ca , Reason (R): Nuclide having odd number of protons and neutrons are generally unstable. 7. Assertion (A): The plot of atomic number (y-axis) versus number of neutron (x-axis) for stable nuclei shows a curvature toward x-axis from the line of 45° slope, as the atomic number is increased. Reason (R): Proton-proton electrostatic repulsion being to overcome attractive forces involving protons and neutrons in heavier nuclide. (2008)

Nuclear Chemistry 15.53

Passage One of the most interesting substances is 14 6 C (half-life 5760 years) which is used in determining the age of carbonbearing material (e.g., wood animal fossils, etc). Carbon-14 is produced by the bombardment of nitrogen atoms present. In the upper atmosphere with neutrons (from cosmic rays) 14 1 7 N +0

1 n →14 6 C + 1H

Thus, carbon-14 is oxidized to CO2 and eventually ingested by plants and animals. The death of plants or animals put an end to the amount of 14C in the dead tissues starts decreasing due to its disintegration as per the following reaction: 14 6 C

→14 7 N+

0 −1β

The 14C isotope enters the biosphere when carbon dioxide is taken up in plant photosynthesis. Plants are eaten by animals, which exhale 14C as CO2. Eventually, 14C participates in many aspects of the carbon cycle. The 14C lost by radioactive decay is constantly replenished by the production of new isotopes in the atmosphere. In this decay- replenishment process, a dynamic equilibrium is established whereby the ratio of 14C remains constant in living matter. But when an individual plant or an animal dies the 14C isotope in it is no longer replenished, so the ratio decreases as 14 C (after the death of living matter) would be less than in a living matter. The decay constant can be calculated using the following formula. 0.693 t1/ 2 = λ The intensity of the cosmic rays have remained the same for 30,000 years. But since some years the changes in this are observed due to excessive burning of fossil fuel and nuclear tests. 8. Why do we use the carbon dating to calculate the age of the fossil? (a) Rate of exchange of carbon between atmosphere and living is slower than decay of 14C. (b) It is not appropriate to use 14C dating to determine age.

(c) Rate of exchange of 14C between atmosphere and living organism is so fast that an equilibrium is set up between the intake of 14C by organism and its exponential decay. (d) None of the above. 9. What should be the age of the fossil for meaningful determination of its age? (a) 6 years (b) 6000 years (c) 60,000 years (d) Cannot be used to calculate any age 10. A nuclear explosion has taken place leading to increase in concentration of 14C in near by areas. 14C concentration if C1 in nearby areas and C2 in areas far away. If the age of the fossil is determined to be T1 and T2 at the respective plane then: (a) The age of the fossil will increase as the place where explosion has taken and T1 − T2 =

1 C1 ln λ C2

(b) The age of the fossil will decrease as the place where explosion has taken and T1 − T2 =

1 C1 ln λ C2

(c) The age of fossil will be determinted to be same. (d) T1/T2 = C1/C2 11. The total number of α and β-particles emitted in the nuclear reaction 238 → 214 92 U  82 Pb is (2009) 12. The number of neutrons emitted when 235 U undergoes 92 Xe and 90 38 Sr is ______. (2010) 13. The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element X as shown below. To which group, element X belongs in the periodic table? 63 1 1 1 29 Cu +1 H → 6 0 n + α + 2 1H + X controlled nuclear fission to

142 54

(2012)

15.54

Nuclear Chemistry

aNsweR Keys Multiple choice Questions with only one answer level i 1. d 2. d 3. c 4. d 5. b 6. b 7. b 8. c 9. a 10. d 11. b 12. b 13. a 14. b 15. d 16. b 17. a 18. a

19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 30. 31. 32. 33. 34. 35. 36.

d c c c d c b a c b d a a d c c b b

37. 38. 39. 40. 41. 42. 43. 44. 45. 46. 47. 48. 49. 50. 51. 52. 53. 54.

a a c c d b a d b b b a a c d a c a

55. 56. 57. 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 68. 69. 70. 71. 72.

b c b b b c c c d c b d c b a d d c

73. 74. 75. 76. 77. 78. 79. 80. 81. 82. 83. 84. 85. 86. 87. 88. 89. 90.

c a d b c d b a b d c d d a c d c a

91. 92. 93. 94. 95. 96. 97. 98. 99. 100. 101. 102. 103. 104. 105. 106. 107. 108.

Passage iii 1. b a c b c d d a b d a a a c d b c d a

2. c

Passage iv 1. d

2. c

Passage v 1. c

level ii 1. d 7. d 2. b 8. d 3. b 9. b 4. c 10. b 5. c 11. d 6. c 12. c

1. d

a c d c b c

19. 20. 21. 22. 23. 24.

a a c d c a

25. 26. 27. 28. 29. 30.

c b d d b a

31. c 32. a 33. d

3. c

2. a

3. b

2. b

3. a

Passage vii 1. a

4. c

Passage viii 2. a

3. d

2. a

3. a

Passage ix 1. c

13. 14. 15. 16. 17. 18.

2. b

Passage vi

1. b

Multiple choice Questions with only one answer

3. d

Matching type Questions 1. (a) 2. (a)

p,q p,s

(b) p,q,r (b) p,q,r

(c) p,s (c) p,q,r

(d) r (d) p,s

integer type Questions More than one answers 1. 2. 3. 4.

a, c a, b a, b, d a, b, c

5. 6. 7. 8.

a, b, c c, d a, b, d c, d

9. a, b, c 10. a, b 11. a, b, c

1. 2 2. 2 3. 2

4. 9 5. 4 6. 4

7. 3 8. 9 9. 3

10. 6 11. 4 12. 9

13. 4 14. 4 15. 9

assertion and Reason type Questions 1. 4 2. 3

comprehension type

3. 1 4. 1

5. 2 6. 1

Passage i 1. d

2. c

3. d

2. c

3. b

Passage ii 1. a

4. b

5. a

Previous years' iit Questions 1. 2 2. 4 3. 1

4. 3 5. 1 6. 2

7. 1 8. 3 9. 2

10. 1 11. 8 12. 4

13. 8

Nuclear Chemistry 15.55

HiNts aND solutioNs Hints to Problems for Practice 1. α-particle emission shows a loss 4 units in mass and 2 units in atomic number, β-particle emission shows a gain of 1 unit in atomic number and no change in mass. 226 88

Ra  → −α

II gp

222 86

Rn  → −α

Zero gp

218 84

RaA  → −β

VI gp

218 85

RaB  → −α

VIIB

214 83

a 2.303 log t a−x a 2.303 0.0104 = log a−x 6.9

We know that λ =

RaC

VB

a = 1.07438 a−x a = 14.4438 × 143244 = 2.07 × 106 β-Particles

Thus atomic number of RaC = 83 Mass no of RaC = 214 Group of RaC = V 2.

235 92

α U − →

III gp

231 90

α Th − →

III gp

231 88

β X − →

III gp

227 89

III gp

N 2.303 log 0 t N A 2.303 2.303 100 log 0 = log Here λ = 366 10 t A −3 −1 = 6.29 × 10 minute

6. λ =

3. Let the no of α-particles emitted be a and β-particles be b Th  →

208 82

Pb + a 42 He + b −10 e

Balancing the mass No. on both sides 232 = 208 + 4a + 0 × b = 208 + 4a 232 − 208 A= =6 4 ∴ No of α-particles emitted = 6 Balance the above equation on both side for atomic number 90 = 82 + 2a + b(−1) = 82 + 2 × 6−b = 82+12–b B = 82 + 12−90 = 4 Thus the no of β-particles emitted = 4 4. Given t0.5 for 90Sr = 28.1 years; N0 = 10–6 gm t = 20 years, N = ? 0.693 0.693 λ= = = 2.466 × 10 −2 years 28.1 t 0.5 2.303 No log Also t = λ N

20 =

2.07 × 106 × 90 = 3.402 × 10 −16 g 6.02 × 1023

∴wt of 90 Mo =

Ac

Elements with atomic number 89 and 90 to 103 are in III gp of periodic table and known as actinides.

23 90

0.693 0.693 = = 0.0104 year −1 66.6 t 0.5

λ=

10−6 2.303 log N 2.466 × 10−2

N = 6.1 × 10−7 g 5. β-Particle emission rate = 346 per minute β-Particles required in 6.9 hours = 346 × 6.9 × 60 = 143244 β−particles  t0.5 = 66.6 hrs

∴ t 0.5 = 7.

0.693 0.693 = = 110 minute. λ 6.29 × 10−3

( i ) Fraction of

3 1

H left after 49 years

T 49 = =4 t 0.5 12.25 Y

4

N 1 1 1 = =  = N 0  2  2 16   1 = × 8 × 10−18 = 5 × 10−19 16

But protium atoms

( H ) remains unchanged = 1 1 1

ratio of 13 H : 11H = 5 × 10 −19 : 1

10 mole H 2 O 18 10 = ×6.02×1023 atoms of 11 H 9 10 = × 6.02 × 1023 × 8 × 10−18 atoms of 13 H 9 = 5.35 × 106 atoms of 13 H

( ii ) 10 g of H 2 O =

Y

4

N 1 1 1 =  = = No  2   2  16 N × 1 5.35 × 106 × 1 ∴ N= 0 = 16 16 = 3.34 × 105 atoms of 13 H

15.56

Nuclear Chemistry

8. A = Ae−λt A 2.303 log = −λt A0

11. Activity of 1 g a substance is called specific activity 1g of naturally occurring K contains 0.0118 atom % 40 K(1.18×10–4g 40K) and non-radioactive isotopes. 0.693 0.693 λ= = = 5.33 × 1010 year −1 1.3 × 109 t 0.5

A = Ae − λt 2.303 log 2.303 log ∴

A = − λt A0

5.33 × 10−10 × 1.18 × 10−4 × 6.02 × 1023 40.0 = 9.465 × 108 year −1

A = λN =

10 0.693 0.693 = × 3.0 = × 3.0 A0 0.5 t 0.5

10 = 0.0156 A0

A 0 = 640 min −1 A 640 × 30 = 2.8 × 104 N0 = 0 = 0.693 λ 9. A − λN = For 239 Pu A =

0.693 N t 0.5

= 2.27 × 10 s For

t=

g

=

107 = 2.7 × 10 −2 = 0.027 m curie 3.7 × 1010

6.02 × 1023 0.693 × × 365 × 24 × 3600 Pu : A = 240 6.58 × 104

Let the fraction of 240Pu = x, then the fraction of 239Pu = 1–x ∴ (8.37 × 109)x + (1+x)(2.27 × 109) = 6.0 × 109 (6.1 × 109)x + (2.27 × 109) = 6.0 × 109 x = 0.71 = 71% 240Pu and 39% 239Pu 0.693 = 1.145 × 10−2 min −1 10. λ for daughter 212 Bi = 60.5 0.693 λ for parent 212 Pb = = 1.089 × 10−3 min −1 10.6 × 60 λdd 2.303 log As we known t max = λd − λp λp =

2.303 1.145 × 10−2 − 1.089 × 10−4

log

1.145 × 10−2 1.089 × 10−4

= 226.8 min = 3.78 hr.

104 × 600 × 1000 = 107 dps 60 × 10

We know that 3.7×1010 dps is shown by 2 Curie of 14C 107 dps is shown by

−1

= 8.37 × 109 s −1 g −1

100 2.303 log = 6.1 year. −1 20 2.645 × 10

104 dps 13. Rate of decay of 10 mL gas = 104 dpm = 60 Thus rate of decay of 600 litre gas

6.02 × 1023 0.693 × × 365 × 24 × 3600 4 240 2.44 × 10 9 −1

240

0.693 0.693 = = 2.645 × 10−1 year −1 2.62 t 0.5

12. λ =

14. Given No=4750 dpm, N=2760dpm, t=10min, t0.5=? N 2.303 2.303 4750 log 0 = log λ= = 5.4 × 10−2 min −1 10 2760 t N 0.693 0.693 ∴t 0.5 = = = 12.8 min λ 5.4 × 10−2 15. λ = t= 16.

0.693 0.693 = = 4.68 × 10 −2 hr −1 14.8 t 0.5 2.303 2.303 No 100 log = log = 25.73 hrs −2 λ N 4.68 × 10 30

λ=

0.693 0.693 = = 0.144 min −1 4.8 t 0.5 N 2.303 log 0 t N 1 2.303 log 0.144 = 19.2 N 1 0.144 × 19.2 log = = 1.20 2.303 N 1 = 15.86 N N = 0.063 mg 1 0.144 × 48 = 3.00 ( ii ) log = N 2.303 N = 9.9 × 10−4 mg

(i ) λ =

Nuclear Chemistry 15.57

17. 0.693 0.693 λ= = = 1.3734 × 10−11 s −1 1600 × 635 × 24 × 3600 t 0.5 No.of atomss in1g of radium = Also

1× 6.02 × 1023 = 2.66 × 1021 226

dN = λN = 1.3734 × 10−11 × 2.66 × 1021 dt = 3.65 × 1010 dps g −1

t = 120 days 18. t 0.5 = 60 days T 120 ∴n = = =2 60 t 0.5

23.

A z X

 →

Let X moles of X give x moles of He after 20 days ∴ moles of x after 20 days = 1-x From the eqn λ =

or

0.693 = 1.23 × 10−4 year −1 5760 No 2.303 log Also, t = λ N 2.303 1 log = = 11525 years. −4 0.25 1.203 × 10

20. Given N0 = 3150, N = 3055, t =1 hour λ= t 0.5

N 2.303 2.303 3150 log 0 = log = 3.06 × 10−2 year −1 1 3055 t N 0.693 = 22.64 year = 3.06 × 10−2

21.

N 0 100 = N 1 N 2.303 λ= log 0 t N N 2.303 log 0 or t = N λ 2.303 1000 log = = 141.5 days. 0.693 14.2 1

x = 0.75 mole ∴Vol. of He at NTP = 0.75 × 22.4 = 16.8 litres. 24. The amount of X after 10 days be x moles

0.693 0.693 = = 5.84 × 10−4 sec −1 1187 t 0.5

Also t =

N 2.30 2.303 2.303 = 600 seconds log 0 = log −4 N 5.84 × 10 1.62 λ

λ=

N 2.303 0.693 log 0 = t N t 0.5

or

2.303 0.1 0.693 log = 10 x 5

x = 0.025 mole. No. of atoms = 0.025 × 6.02 × 1023 = 1.505 × 1022 Thus the number of atoms on the eleventh day = 1.505 × 1022 Now number of atoms decaying on the eleventh day = No. of disintegration per day d( N) 0.693 Or − ×N = λ ( N) = dt t 0.5 d ( N) 0.693 or − = ×1.505 × 1022 = 2.086 × 1021 dt 5 25. We know, N = no. of atoms = moles × Av const =

1 × 6.022 × 1023 200

d( N) = λ( N) dt 0.693 1 0.002 = × × 6.02 × 1023 200 t 0.5 ∴t 0.5 = 1.0436 × 10244 Seconds = 3.334 ×1016 years −

26. λ Cu =

22. λ=

N 2.303 0.693 log 0 = λ N t 0.5

2.303 1 0.693 log = 20 1− x 10

% of radioactivity remain after 2 half-lives No 100 = 2 = = 4% 25 2 19. Given t 0.5 of 14 C = 5760 year λ=

A−4 4 Z − 2 Y + 2 He

64 29

Cu

0.693 = 0.054 12.8 64 30

Zn + –10 e (38%)

64 28

Ni + 01e (19%)

64 28

Ni (43%)

15.58

Nuclear Chemistry

∴ λ1 = Fractional yield of Zn × λcu 38 = × 0.054 = 0.0205 100 ∴ t0.5 for β– emission =

32.

=

=2×9.1×10-31×(3×108 ) 2 =16.38×10−14 J

0.693 = 33.8 h 0.0205

6

× (4 × 1.672 × 10

−27

)

= 6.3×10 ms 28. Rate = λN 0.693 6.02 × 1023 = × = 7.153 × 1016 dis / year 239 2.44 × 104 loss in energy per year in emission of α-particles = 5.24 × 7.153 × 1016 Mev = 37.482 × 1016 × 1.602 × 10−16 KJ = 60.05 KJ 29. ∆m = 2 × m 12 H − m 32 He + m n = (2×2.041)–(3.016+1.0087) = 0.0035 amu ∆E = ∆m × 931.48 Mev = 0.0035 × 931.48 = 3260 Mev = 5.223 × 10−13 J ( 1 Mev = 1.60 × 10−13) 30. rnucleus = 1.3 ×10−13 × (A)1/3 Where A = mass No. r 238 U = (1.3 × 10−13) × (238)1/3 = 8.06 × 10−13 cm r 4He = (1.3 × 10−13) × (4)1/3 = 2.06 × 10−13 cm Total distance in between uranium and α-nuclei = 8.06 × 10−13 + 2.06 × 10−13 = 10.12 × 10−13 cm θ θ coulombic repulsion energy = 1 2 r 92×4.8×10-10 ×2×4.8×10-10 = =418.9×10-7 erg 10.12×10-13 =418.9 × 10−7 × 6.242 × 1011 ev = 26.14 Mev 31. Mass of 55Mn = (25 × mp) + (30 × mn) = (25 × 1.00783) + 30 × 1.00867) = 25.19575 + 30.26010 = 55.45585 amu Mass defect ∆m = 55.938−55.45585 = 0.48215 amu

= 8.16 Mev/nucleon

∆E C

2

=

-1.4 × 103 × 103 × 103 (3 × 108 ) 2

=1.6 × 10−8 amu/molecule.

–1

B.E. per nucleon =

= 2.42×10 −12 m = 2.42 Pm 33. ∆m =

29 × (1.6 × 10 −19 ) 2 3.14 × 8.85 × 10

16.38×10 −14 =8.19×10-14 J 2 hc 6.62×10−34 ×3×108 We know that λ = = E 8.19×10-14

∴Energy of one photon =

Ze 2 4π ∈0 mr

−12

→ 2 gamma photons of equal energy

E= 2 MeC2

In a similar way t 0.5 for β+ emission = 67.6 h And t 0.5 for electron capture = 29.85h 27. At nearest distance K.E. should be equal to repulsion energy 2 Ze 2 1 1 mv 2 = × 2 4π ∈0 r V2 =

0 0 +1 e+ -1 e

0.48215 × 931.48 55

34. 2 21 H → 31 H+ 11 H+energy Mass defect (∆m) = (2 × md)−mt + mp = (2 × 2.01410)−3.01605 + 1.00783 = 4.02820 − 4.02368 = 0.00432 amu B.E = 0.00432 × 931.48 Mev = 4.0239 Mev = 4.0239 × 106 × 1.602 × 10−9 J = 6.45 × 10−13J Energy produced per mol of 21 H =

6.45×10-13 ×6.02×1023 =19.4×1010 J/mol of 21 H 2

40% efficiency means that =

19.4×10-10 ×40 =7.76×1010 J/mol of 21 H is useful. 100

Energy required per day = 80 × 106 × 24 × 3600 = 6.91 × 102 J 7.76 × 1010 J of energy produced from 1 mol of 21 H 6.91 × 102 J of energy produced from 1×6.91×1012 7.76×1010

=89.07 mol /day

Deuterium fuel required per day in g = 89.07 × 2 = 178.14 21 H /day 35. 14N contains 7n, 7P and 7e Mass = 7mn + 7mp + 7me = 7 × 1.0086650 + 7 × 1.0072765 + 7 × 0.0005486 = 14.1154 amu ∆m = 14.11543 − 14.00307 = 0.11236 amu B.E. per nucleon =

0.11236 × 931.48 = 7.48M ev 14

15 N contains 8n+7p+7e− Mass of 15N = 8mn + 7mp + 7me = 8 × 1.0086650 + 7 × 1.0072765 + 7 × 0.0005486 = 15.12409 amu Δm = 15.12409–15.00011 = 0.12398 amu

Nuclear Chemistry 15.59

B.E. per nucleon =

0.12398×931.48 =7.698 Mev 15

Thus 15N is more stable due to its large B.E. per nucleon. 36. 100 ev = 100 ×1.602×10–19= 16.02×10–18J Also 1000 Kw= 106Js–1 = 6×107 J min–1 16.02×10–18J of energy produces 1 molecule of H2 ∴ 6 × 107 energy produces 1×6×107 = =3.745×1024 16.02×10-18 3.745×1024 = =6.22 mol of H 2 6.02×1023 One mol of H2 occupies 22400 mL volume at N.T.P ∴6.22 mol of H2 occupies 22400 × 6.22 = 139328 mL = 139.328 litre 37. Mass defect ∆m=[m(Cl)+m(n)]-[m(S)+m(H)] = [34.9688+1.0087]-34.9690+1.0078] = 0.0007 amu Energy released = 0.0007 × 931.48 = 0.6520 M ev 38. Mass defect ∆m = m14C − m14N = 14.003242 − 14.003014 = 0.000168 amu ∴ Energy released = 0.000168 × 931.48 = 1565 Mev 6  39. Mass defect ∆m = m  Li  + ( m12 H ) − 2m ( 42 He ) 3  = (6.01512 + 2.014101) − 2 × 4.0026 =0.02402 amu The energy released during the mass defect or mass decay =∆m × 931.48 Mev = 0.02402 × 931.48 = 22.374 Mev = 22.374 × 106 ev = 22.374 × 106 × 1.602 × 10−19 = 35.84 × 10−13 J Energy relased per mol LiH = 35.84 × 10−13 × 6.02 × 1023 = 21.58 × 1011 j mol−1 Energy released per g of LiH 21.58 × 1011 = = 2.6975 × 1011 J g −1 day −1 8 Energy released per g of LiH per second 2.6975 × 1011 = 3.122 × 106 J g −1 s −1 24 × 3000 = 3.122 × 106 Wg −1 ( J s −1 = 1W )

=

3.122 × 106 × 90 = = 2.8098 × 106 Wg −1 100

m−A ×104 A 39.962384 − 40 = × 104 = −9.404 40

40. Packing fraction ( p ) =

Mass defect = (m−A) m = 22 × mn + 18 × mp + 18me − A = 40.3314 − 39.962384 = 0.369016 amu B.E. = ∆m × 931.48 Mev = 0.369016 × 931.48 = 343.73 Mev 41. 27 Mev = 27 × 106 × 1.602 × 10−19 J = 43.25 × 10−13J And 4 × 1026 J sec−1 = 4 × 1026 × 24 × 3600 = 3.456 × 1031 J day−1 ∴ 43.25 × 10−13 J of energy is obtained from 4 amu of H ∴ 3.456 × 1031 J day−1 of energy is obtained from 3.456 × 1031 4 = × 6.02 × 1023 43.25 × 10−13 = 5.31× 1019 g of H. 42. ∆E = ∆m × c2 ∆E 1800 × 103 ∴∆m = 2 = = 2 × 10−11 kg 8 2 c ( 3 ×10 ) 0.693 0.693 = = 1.209 × 10 −4 year −1 5730 t 0.5

43. λ = t=

2.303 15.3 log −4 2.80 1.209 × 10

= 14049 years.

44. Here λ = t=

0.693 0.693 = = 1.244 × 10−4 year −1 5568 t 0.5

N 2.303 16100 2.303 log 0 = log = 1596 years −4 λ N 1.244 × 10 13200

45. λ =

0.693 0.693 = = 1.209 × 10−4 year −1 5730 t 0.5

No = 15.3 counts per min/g carbon For 200 mg sample N0 = 3.06 counts/200 mg ∴t = 46. λ =

2.303 3.06 log = 20720 year. 0.25 1.209 × 10−4

0.693 0.693 = = 1.209 × 10−4 year −1 5730 t 0.5

age of mummy t =

15 2.303 log = 3354.3 years. −4 10 1.209 × 10

15.60

Nuclear Chemistry

23.8 = 0.1 atm 238 N 0 = U present + U decayed

47. N = U present =

23.8 20.6 = 0.1 + 0.1 = 0.2g atoms + 238 206 0.693 λ= = 1.54 × 10−10 year −1 4.5 × 109 =

0.2 2.303 t= log = 4.5 × 109 year. 0.1 1.54 × 10−10 48. Uranium present in pitch blend sample 50 = = 2.10 × 10 −3 g atoms 100 × 238 2.425 Pb present = = 1.17 × 10 −4 g atom 100 × 206 Pb formed due to decay of uranium 1.17 × 10−4 × 93 = = 1.09 × 10−4 g atom 100 Hence N = 2.10 × 10−3g atom No = (wt of uranium + wt of lead formed) = (2.10 + 0.109)×10−3g atom = 2.209 × 10−3g atom Now t =

N 2.209 × 10−3 2.303 2.303 log O = log N 1.52 × 10−10 λ 2.10 × 10−3

= 3.3 × 108 year 49.

232 90

4 0 Th  → 208 82 Pb + 6 2 He + 4 −1 e

∴ 6 mol of He is formed by 1 mol of Th decay Or 6 × 22400 mL He is formed by 232g Th decay ∴ 8 × 10−5 mL He is formed by =

A0 N0 = A N 0.693 0.693 = = 4.62 × 10−2 hour −1 λ= t 0.5 t 0.5 As we know

232 × 8 × 10−5 6 × 22400

= 1.38 × 10−7 g = Th decay At t = t, sample has Th = 5 × 10−7g = N At t = 0, sample Th = (5× 10−7 + 1.38 × 10−7) = N0 = 6.38 × 10−7g 0.693 0.693 Also λ = = = 4.98 × 10−11 year −1 1.39 × 1010 t 0.5 6.38 × 10−7 2.303 t= log 4.98 × 10−11 5 × 10−7 = 4.89 × 109 year 50. Let V mL blood be present in patient (i) A0 for 24Na = 2 × 103 dps = 2 × 103 × 60 = 120 × 103 dpm for 24 A for Na = 16 dpm/mL at t = 5 hours

∴t =

N 120 × 103 2.303 2.303 log = log 0 = −2 λ N 4.62 × 10 16 v

V = 5.95 × 103 mL (ii) Activity of blood sample after 5 hours duration t = 5 + 5 = 10 hours A 2.303 t= = log 0 A λ 120 × 103 2.303 log 10 = A 4.62 × 10−2 A=

75.6 × 103 dpm / mL 5.95 × 103

= 12.705 dpm / mL =

12.705 = 0.2118 dps / mL 60

0.693 = 8.66 × 10−2 day −1 8 2.303 100 log t= = 53.17 days 1 8.66 × 10−2

51. For

131

Iλ=

0.693 = 3.46 × 10−2 day 20 2.303 100 log = 132.9 days t= 1 3.46 × 10−2

For 90Sr λ =

Radioactive 90Sr is to be more serious because it will take longer time to fall its activity to 1%. 52. Total No. of 40K atoms in a 80 kg human body 1.2 × 10−2 3.5 × 10−1 6.02 × 1023 × × 80 × 103 × 100 100 40 20 = 5.06 × 10 atoms =

and λ =

0.693 0.693 = 1.3 × 10−9 × 365 × 24 × 60 t 0.5

= 1.014 × 10−15 min−1 A = λN = 1.014 × 10−15 min−1 A = λN = 1.014 × 10−15 × 5.06 × 1020 = 5.13 × 105 dpm 53. λ =

0.693 0.693 = = 5.68 × 10−2 year −1 12.2 t 0.5

N 100 2.303 2.303 log log 0 = λ N 5.68 × 10−2 87 = 2.4522 year t=

Nuclear Chemistry 15.61

Multiple choice Questions with only one answer

13. λ = λ1+ λ2 = 1.9×10–2 sec–1 Average life = 1/ λ=52.63 sec

level ii

14. N=N0 e– λt

0.693 = 0.03465 yr−1 20 K Percentage yeild of Th = 1 × 100 K K1 2= ×100 0.03465

 0.693 0.693    120 − 

2. Rate Constant K =

K1 = 6.93 × 10−4yr−1 3.

−dN =λN dt 6.02×107 = λ

27×10 270

−3

15. After two weaks, the activity is 2.5 µCi 90% transferred into thyroid gland ∴ so activity in thyroid gland is 2.25 µCi. 16.

( 6.023×10 ) 23

∴λ = 10−12 sec −1 4. No. of counts of fresh wood is = 20–5 = 15 per minute No. of counts of archeological specimen is = 8–5 = 3 per minute 15 0.693 2.303 log = ∴ 3 t 5000 ∴ t = 1.1 × 104 years 5. Wt of

1 8  t 1 2 20 = e 1 1 ∴ t1/2 = 40 min

40

K present in human body is 80 × 103 ×

0.5 0.01 = 0.04 g × 100 100 −dN 0.693 0.04 = × × 6.023×1023 = 3 × 1011 dpy 9 dt 1.38×10 40 −λt

α

238 92 X

234

 0.693

0.693 

7. λ1N1 = λ2N2 of equilibrium 1 8. part remain means the time taken is four half-lives. 16 9. Neutron and energy are in 3:1 ratio. 10. The given data is zero order. 11. Molar ratio of U: Pb is 1:3 requires two half-lives.

18. Half lifes of A and C are very less. 19. Activity of fresh wood is 110–10 = 100 counts Activity of test sample is 60–10 = 50 counts So age of sample is 5570 years. 20. After 10 hours the acitivity of sample is 30 Rutherford A 0.693 2.303 = log 0 23.5 10 30 A0 = 40.29 Rutherford, ∴

dN = λN dr

40.29 × 106 =

0.693  W 23 ×  × 6.023 × 10 23.5 × 60 × 60  293 

W = 2.39 × 10–8 g 21. t = 8 sec is four half-lifes So counting rate after 6 hrs is 200.  1 22. a n =    2

11/10

27. λ = λ1 +λ 2 1 1 λ= + = 3.08×10−3 years 1620 405 ∴ t 1 = 225 years 2

12. Initial moles of A:B = a:8a a a After 50 minute A:B = : 2 4

234

17. Isotopes are formed by emission of 1α, 2β.

6. Nt = N0 e

100  5×108 − 5×109  t =e 1 ∴ t = 3.7 × 109 years

β

 → 90 Y  → 91 Z So no. of neutrons is 234–91=143

∴ t 3 = 450 years 4

15.62

Nuclear Chemistry

28. Decay constant = 0.05 hours–1 Percentage yield of β– emission is = 40% λ 40 = 1 ×100 0.05 λ1 = 0.02 hours −1

5.

∴ t = 8.78 hours

Half-life = 34.65 hours 30.

94

X → 4 42 He7 +Y 78

Passage ii

3 th X disintegrate 4 3 So vol. of He gas at STP = 4 × × 22.4 = 67.2 litre. 4 0.693 2.303 100 32. = log 109.7 t 1 t = 730 minutes After 30 days

33.

0.48×3.7×104 466 v = × 0.5 60 0.1 v =114 mL

comprehensive type Questions Passage i

1. Rate constant K = K1+K2 K=1.5×10–5 +5×10–6 = 2×10–5 sec–1 2. 2 × 10 −5 =

1. Initial activity = 2 × 10 dps after 5 hours the activity is = x 2 × 103 0.693 2.303 log ∴ = x 15 5 ∴ x = 1.588 × 103 dps So 1.588 × 103 × 60 = 16V V = 5.95 lit 2. Activity of sample after another 5 hours is y 0.693 2.303 16 = log y 15 5 y = 12.71 dpm mL λ1 × 100 0.693 15

λ1 = 0.0194 hr–1 t1/2 of β− emission process = 35.72 hrs 4.

−dN = λN dt 0.693 N 15 × 60 × 60 Initial no. of Ax atom = 1.5584×108 atom After 30 hours Ax atom = 3.897×108 atom No. of He atom formed in 30 hours = 1.168×108 atom 2 × 103 =

2.303 0.236 log 5 × 60 × 60 At

At = 0.1646 M The amount of A consumed is = 0.0714 2 1 [c] = 0.0714 × × = 7.1×10−3 M 5 4 1 3 −2 3. Conc. of B = 0.0714 × × = 1.08 × 10 M 5 4 Passage iii 1. N = N 0 e − λt 100 (λ − λ )t =e 2 1 1 t = 5.1 × 103 years

3

3. 42 =

Final Ax → By Composition 0.2 0.1 Initial composition = 0.3 0.693 2.303 3 = log 15 1− 2

2. After a period of 6930 years 210A present is

1 fraction 2

of initial amount. After a period of 6930 year 211A present approximates 0.001 fraction of initial amount. 1 3. In three half lives period 210A present fraction and 8 7 206 B present fraction of initial amount. 8 Passage iv Conc at TF  T + + F− equilibrium c −x x x ∴ ∆Tf = Kf (c+x) 0.42 = 2(C+0.01) 2 C = 0.2 M ( 0.01) Ionization constant = 0.19 = 5.2 × 10−4 2. No. at moles of TF and T+ total present in solution is = 0.1 In two half-lives period no. of β-particles emitted on = 0.075 moles.

1.

Nuclear Chemistry 15.63

Passage v

4.

1. Half life of ‘A’ is very less when compared to 10 days. In 10 days total A converted to B In 10 days no of B atom converted to ‘C’ is 1020 0.693 2.303 = log 20 6930 10 10 − x The formed ‘C’ slowly disintegrates. The no. of nuclei of ‘C’ formed approximately is = 1017 2. Half life of A, and C are very less. So B→D is 6930 1030 year No of nuclei ‘D’ formed is 2 3. The given data exist in two half lives.

0.693 2.303 1 = log 28.1 1 1− x X = 0.024 moles

2.

10−6 0.693 2.303 = log 28.1 1 an

∴ an = 6.1 × 10−6 gm 3. In β decay binding energy per nucleon increases

integer type Questions 1.

232 90 Th

6 4  → 208 82 Pb + 2 He

Mixture contain 0.1 mole 232 90 Th and 0.6 mole of He gas ∴ Initial mole of 232 90 Th = 0.2 ∴ The time required for this change equal to half – life 2.

235 1 92 U+ 0 n

 → m-8 z-4 B +

2

( He) 4 2

a – – initial mole a-x – 2x moles after time t. No. of moles of He gas = 1.5 moles Initial moles of A = 1 moles After 18 days moles of A remains is = 0.25 So half life = 9 days 5. β-emitters 13 H;

14 24 65 6 C; 11 Na; 29 Cu

6. Total No. of α-particles emitted are = 5 So the value of n = 4 7. Initially x, y are in 4a, a ratio a After two hours the amount of x is 16 a So the amount of y in two hours is 16 So half life of y = 30 minutes

Passage vi 1.

m zA

90 1  → 144 54 Xe + 38Sn+ 2 0 n

3. It is parallel path mechanism λ =λ1 +λ 2 0.693 0.693 0.693 = + t 0.5 24 6 ∴ t 0.5 = 4.8 hours

8. Final composition A → B 1020 7×1020 In this conversion three half lifes time is required 9. m moles of radium in RaCl2 is 10 and its activity is 10–3 Curie. m moles of radium in Ra3(PO4)2 is 500 × 0.2×3 = 300 So its activity is 3 × 10–2 Curie. 10. In four half lifes 1600 counts converted 100 counts So in three half lifes it converted to 200 counts. 11. Half life of B is very high when compared to A. Initial activity a + b = 8000 a After 20 minutes the + b = 3500 A = 6000, b = 2000 4 ∴ a:b = 3:1; a+b = 4 238 206 12. U → Pb I moles a – t moles a–x x a–x = x = 0.1 a = 0.2 moles The time taken in this process is half life 13. Half life is 10 sec The time required to convert 6.25 % is 40 sec.

15.64

Nuclear Chemistry

14. Half life is 69.3 sec. −dN = λN dT 0.693 N 100 = 69.3 N = 104 atom

Previous years’ iit Questions 1. β-emission in

the ratio of

N 16 is high, only = P 13

those reactions are favoured which help in lowering N ratio P 29 β − decay 29 →14 Si + −01e 13 Al 

−dN 15. = λN dt 24 = 2×10–17 × (moles) 6×1023 moles of M+ = 2×10–6 molarity of M+ = 2×10–5 M M2SO4  2M+ + SO–24 ∴ 2S = 10–5 M Ksp = 4S3 = 4 × 10–15

29 13 Al

12. 13.

92 U

235

+ 0 n1 →

63 1 29 Cu + 1H

142 54 x

+

-(β emission ) . 38Sr

90

→ 610 n + 42 He + 211 H +

+ 40 n1 52 26 Fe

So 26Fe belongs to 8th group in Periodic table.