Engineering chemistry-I [7 ed.] 9789339205508, 9339205502

Table of contents :
Title
Contents
Unit 1 Polymer Chemistry
Unit II Chemical Thermodynamic
Unit III Photochemistry and Spectroscopy
Unit IV Phase Rule and Alloys
Unit V Nanochemistry
Question paper

Citation preview

FOHJOFFSJOH!DIFNJTUSZ.J Tfdpoe!Fejujpo

Bcpvu!uif!Bvuips P Krishnamoorthy is presently working as Asst. Professor in postGraduate and Research Department of Chemistry, Dr. Ambedkar Government Arts College, Vyasarpadi, Chennai. He received his M.Sc., M.Phil., M.Ed., in chemistry from Madurai-Kamaraj University and his Ph.D from Anna University. He worked in the Priyadarshini Engineering College, Vaniyambadi, for 10 years and St.Joseph’s College of Engineering for 4 years. He has 20 years of teaching experience. P Krishnamoorthy has published many books and many research papers in various national and international journals. He has also attended and presented papers in several national and international conferences.

FOHJOFFSJOH!DIFNJTUSZ.J Tfdpoe!Fejujpo

P Krishnamoorthy Assistant Professor Post-Graduate & Research Department of Chemistry Dr. Ambedkar Govt. Arts College Vyasarpadi, Chennai Tamil Nadu

McGraw Hill Education (India) Private Limited NEW DELHI McGraw Hill Education Offices New Delhi New York St Louis San Francisco Auckland Bogotá Caracas Kuala Lumpur Lisbon London Madrid Mexico City Milan Montreal San Juan Santiago Singapore Sydney Tokyo Toronto

McGraw Hill Education (India)!Qsjwbuf!Mjnjufe Qvcmjtife!cz!NdHsbx!Ijmm!Fevdbujpo!)Joejb*!Qsjwbuf!Mjnjufe Q.35-!Hsffo!Qbsl!Fyufotjpo-!Ofx!Efmij!221!127 Fohjoffsjoh!Difnjtusz.J Dpqzsjhiu!©!3125-!3123-!cz!NdHsbx!Ijmm!Fevdbujpo!)Joejb*!Qsjwbuf!Mjnjufe/ Op!qbsu!pg!uijt!qvcmjdbujpo!nbz!cf!sfqspevdfe!ps!ejtusjcvufe!jo!boz!gpsn!ps!cz!boz!nfbot-! fmfduspojd-! nfdibojdbm-! qipupdpqzjoh-! sfdpsejoh-! ps! puifsxjtf! ps! tupsfe! jo! b! ebubcbtf! ps! sfusjfwbm!tztufn!xjuipvu!uif!qsjps!xsjuufo!qfsnjttjpo!pg!uif!qvcmjtifst/!Uif!qsphsbn!mjtujoht! )jg!boz*!nbz!cf!foufsfe-!tupsfe!boe!fyfdvufe!jo!b!dpnqvufs!tztufn-!cvu!uifz!nbz!opu!cf! sfqspevdfe!gps!qvcmjdbujpo/ Uijt!fejujpo!dbo!cf!fyqpsufe!gspn!Joejb!pomz!cz!uif!qvcmjtifstNdHsbx!Ijmm!Fevdbujpo!)Joejb*!Qsjwbuf!Mjnjufe/ ISBN (13 digit): 978-93-3920-550-8 ISBN (10 digit): 93-3920-550-2 Nbobhjoh!Ejsfdups;!Lbvtijl!Cfmmboj IfbeÐIjhifs!Fevdbujpo!)Qvcmjtijoh!boe!Nbslfujoh*;!Wjcib!Nbibkbo Tfojps!Qvcmjtijoh!Nbobhfs!)TFN!'!Ufdi/!Fe/*;!Tibmjoj!Kib Fejupsjbm!FyfdvujwfÐBdrvjtjujpot;!T!Wbntj!Effqbl NbobhfsÐQspevdujpo!Tztufnt;!Tbujoefs!T!Cbwfkb Tfojps!Qspevdujpo!Fyfdvujwf;!Tvibjc!Bmj Bttjtubou!Hfofsbm!Nbobhfs!)Nbslfujoh*ÐIjhifs!Fevdbujpo;!Wjkbz!Tbsbuij Bttjtubou!Qspevdu!Nbobhfs;!Ujob!Kbkpsjzb Tfojps!Hsbqijd!EftjhofsÐDpwfs;!Nffov!Sbhibw Hfofsbm!NbobhfsÐQspevdujpo;!Sbkfoefs!Q!Hibotfmb NbobhfsÐQspevdujpo;!Sfkj!Lvnbs

Jogpsnbujpo!dpoubjofe!jo!uijt!xpsl!ibt!cffo!pcubjofe!cz!NdHsbx!Ijmm!Fevdbujpo!)Joejb*-! gspn!tpvsdft!cfmjfwfe!up!cf!sfmjbcmf/!Ipxfwfs-!ofjuifs!NdHsbx!Ijmm!Fevdbujpo!)Joejb*!ops! jut!bvuipst!hvbsbouff!uif!bddvsbdz!ps!dpnqmfufoftt!pg!boz!jogpsnbujpo!qvcmjtife!ifsfjo-!boe! ofjuifs!NdHsbx!Ijmm!Fevdbujpo!)Joejb*!ops!jut!bvuipst!tibmm!cf!sftqpotjcmf!gps!boz!fsspst-! pnjttjpot-!ps!ebnbhft!bsjtjoh!pvu!pg!vtf!pg!uijt!jogpsnbujpo/!Uijt!xpsl!jt!qvcmjtife!xjui!uif! voefstuboejoh!uibu!NdHsbx!Ijmm!Fevdbujpo!)Joejb*!boe!jut!bvuipst!bsf!tvqqmzjoh!jogpsnbujpo! cvu!bsf!opu!buufnqujoh!up!sfoefs!fohjoffsjoh!ps!puifs!qspgfttjpobm!tfswjdft/!Jg!tvdi!tfswjdft!bsf! sfrvjsfe-!uif!bttjtubodf!pg!bo!bqqspqsjbuf!qspgfttjpobm!tipvme!cf!tpvhiu/ Uzqftfu!bu!Ufk!Dpnqptfst-!X[!4:2-!Nbejqvs-!Ofx!Efmij!221!174!boe!qsjoufe!bu Dpwfs!Qsjoufs;!

Dedicated to My Beloved Parents Thiru S Periasamy and Thirumathy P Boomathy

Dpoufout

Preface Acknowledgements

xi xiii

Unit-I

Polymer Chemistry

1. Polymer Chemistry 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9

3

Introduction 3 Functionality and its Importance 4 Classification of Polymers 4 Types of Polymerisation 8 Mechanism of Addition Polymerisation Properties of Polymers 15 Techniques of Polymerisation 19 Nylon 6, 6 Polymer (Polyamides) 22 Epoxide (or) Epoxy Resins 24 Questions 26

Unit-II

11

Chemical Thermodynamics

2. Chemical Thermodynamics Introduction 33 Terminology of Thermodynamics 34 Second Law of Thermodynamics 38 Entropy 39 Entropy Change for an Ideal Gas 40 Entropy Change for Reversible (Nonspontaneous) Process 43 2.7 Entropy Change for Irreversible (Spontaneous) Process 44 2.8 Entropy Change in Physical Transformation

33

2.1 2.2 2.3 2.4 2.5 2.6

45

viii Contents

2.9 2.10 2.11 2.12 2.13 2.14 2.15 2.16 2.17

Free Energy 49 Helmholtz Free-energy Function (A) and Work Gibbs Free-energy Function and Work 51 Criteria of Spontaneity 54 Gibbs–Helmholtz Equation 56 Clausius–Clapeyron Equation 63 Maxwell Relations 65 Van't Hoff Isotherm 69 Van't Hoff Equation (Van’t Hoff Isochore) 71 Exercises 77 Questions 78

Unit-III

50

Photochemistry and Spectroscopy

3. Photochemistry and Spectroscopy 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 3.10 3.11 3.12 3.13 3.14

85

Photochemistry–Introduction 85 Laws of Photochemistry 86 Beer’s Law (or) Beer–Lambert’s Law 90 Quantum Yield or Quantum Efficiency 98 Experimental Determination of Quantum Yield 100 Photophysical Processes (Fluorescence and Phosphorescence) 102 Energy Transfer in Photochemical Reaction (Photosensitisation and Quenching) 106 Chemiluminescence (Cold light) 109 Spectroscopy 114 Electromagnetic Spectrum 116 Absorption and Emission of Radiation 116 Classification of Molecular Spectra 117 Ultraviolet (UV) and Visible Spectroscopy 126 Infrared Spectroscopy 135 Questions 144

Unit-IV Phase Rule and Alloys 4. Phase Rule and Alloys 4.1 Phase Rule: Introduction 151 4.2 Terms Involved in Phase Rule 152

151

Contents ix

4.3 Application of Phase Rule to Various Component Systems 156 4.4 Construction of Phase Diagram by Thermal Analysis 4.5 Phase Diagram of Lead–silver System 163 4.6 Alloys 168 4.7 Importance of Alloys 169 4.8 Applications of Alloy Steels 172 4.9 Heat Treatment of Alloys (Steel) 178 Questions 181

Unit-V

161

Nanochemistry

5. Nanochemistry

191

5.1 Basic Introduction of Nanochemistry 191 5.2 Distinction between Molecules, Nanoparticles and Bulk 191 5.3 Size-dependent Properties of Nanomaterials 192 5.4 Nanoparticles 193 5.5 Nanoclusters 194 5.6 Nanorods 194 5.7 Nanotubes (CNT) 195 5.8 Nanowires 199 5.9 Synthesis of Nanoparticles 200 5.10 Properties of Nanoparticles 205 5.11 Application of Nanoparticles 212 5.12 Applications in Fuel Cells 213 5.13 Catalysis and use of Gold Nanoparticles in Medicine 214 Questions 218 Summary 220 Question Papers

Q.1–Q.24

Qsfgbdf Engineering and technology are inter-disciplinary with chemical, physical and biological sciences contributing immensely to the development of engineering practices. Applied chemistry is a fantastic area with profound implications for engineers as well as biologists. A proper perspective of applied chemistry requires a sound knowledge of the underlying principles. The thoroughly revised and updated form of this book completely covers the syllabus of the Engineering Chemistry course offered to firstyear B.E/B.Tech students of various Indian universities, giving an indepth coverage of a wide range of topics. Hence, this will be useful for all teachers in preparing their lectures. The present book envisages at providing the necessary introduction of various chemical principles involved and applications of various engineering materials and devices in a comprehensive yet understandable manner to the students aspiring to become practicing engineers. The book includes a large number of examples that give students manifold opportunities to understand the topic being presented. Many solved problems are given throughout the book that give students a chance to see how problems can be worked out. The book has been enhanced with clear figures to promote visual and spontaneous learning.

Aims of the Book The aim is to develop the habit of scientific reasoning in students so that they can work with an open and inquiring mind. I also aim to impart intensive and extensive knowledge of the subject so that readers can understand the role of chemistry in the field of engineering. The content endeavours to enhance and develop the analytical and problem-solving capabilities of students.

xii Preface

Salient Features • Offers easy and exhaustive coverage of topics and lucid explanation of concepts as per the present-day needs and practices • Includes compact and accurate study material • Follows an application-based approach—there is a link between theoretical aspects and relevant applications • For more clarity, tables and figures are interspersed throughout the book • All important derivations given stepwise • Fundamental and repeated university questions are arranged chapterwise and extensively dealt in the form of Part-A questions and answers, and given at the end of each chapter • A question bank including Anna University questions and model questions are included at the end of each chapter and also at the end of the book • Important topics such as Phase Rule and Alloys (Unit 2), Photochemistry and Spectroscopy (Unit 3), and Nanochemistry (Unit 5) dealt with appropriately

Unit Organization The content is divided into five units. Unit 1 covers Polymer Chemistry. Chemical Thermodynamics are presented in Unit 2. Photochemistry and Spectroscopy are dealt with in Unit 3. Unit 4 covers Phase Rule and Alloys, while Unit 5 discusses Nanochemistry. Question papers are given at the end of the book.

Feedback Hence, this book will serve its purpose and prove to be a very good book for updating knowledge and creating awareness of recent techniques and processes. I hope this book will fulfill the need of all categories of engineering teachers and students, making them conversant with the various topics in the syllabus. Any corrections and suggestions towards the improvement of the book are most welcome. P Krishnamoorthy

Bdlopxmfehfnfout I am pleased to express my deepest sense of gratitude to our Director, Collegiate Education, for providing me a chance to bring out this book. I also extend my gratitude to the Principal for his encouragement and good wishes. I sincerely thank Dr Y B G Varma, Professor(Retd), Department of Chemical Engineering, IIT Madras, for his continuous guidance and encouragement throughout my life. I am thankful to Mr R Ravichandran, Head, Department of Chemistry, Dr. Ambedkar Govt. Arts College, Vyasarpadi, and all other staff members of the department for their suggestions and support. I am grateful to Dr P Mani, Dean, Faculty of Science and Humanities, Hindustan University, Chennai, for his encouragement and good wishes. I am extremely grateful to Dr Alagar, Dr Sivanesan, Dr Mohan, Professors of Chemistry, A C Tech. Anna University, Chennai, for their kind help, support and encouragement. I would like to acknowledge the editorial and sales staff of McGrawHill Education (India), particularly Vamsi Deepak Sankar, for initiating me into this project. At this stage, I would also like to thank the reviewers who took out time to review the manuscript and provide valuable comments. Their names are given below: • Gandhidasan, KLN College, Madurai • Sampath, Chettinad Enginerring College, Trichy Finally, I thank my friends, family members and beloved wife, Mrs S Nanthini, PG Asst. Govt. Hr. Sec. School, Moovarasampet, Chennai, and daughters, K Santhiya and K Srinisha, my sisters and brothers for their continuous cooperation and support. P Krishnamoorthy

Publisherŭs Note McGraw-Hill Education (India) invites suggestions and comments from you, all of which can be sent to [email protected] (kindly mention the title and author name in the subject line). Piracy-related issues may also be reported.

Qpmznfs!Difnjtusz V O J U J

1.1

Introduction

1.2

Functionality and its Importance

1.3

Classification of Polymers

1.4

Types of Polymerisation

1.5

Mechanism of Addition Polymerisation

1.6

Properties of Polymers

1.7

Techniques of Polymerisation

1.8

Nylon 6, 6 Polymer

1.9

Epoxy Resins

2 D

Qpmznfs!Difnjtusz

I B Q U F S

INTRODUCTION

1.1

Polymers (in Greek poly means many, mers means parts) are macromolecules of high molecular mass formed by the combination of a large number of small molecules of low molecular weight called monomers. Polymers are mainly used in transport, agriculture, construction, etc. Polymeric materials are used as fibres, rubbers, plastics, adhesives, paints, computer components, electrical goods, etc. The important biopolymers are carbohydrates, proteins and nucleic acids. The number of repeating units (n) in a polymer chain is called degree of polymerisation. The molecular mass range of a majority of the polymers lies between 5,000 to 200,000 molecular mass range. For example, polythene is a polymer formed by linking together of a large number of ethene (C2H4) molecules.

where n is the degree of polymerisation, and can be 104 or more. The major properties of polymers depend upon 1. Molecular configuration 2. Degree of polymerisation 3. Branching and cross linking of a polymer The covalent bonds of all the polymers prevent electrical conductivities.

4 Engineering Chemistry-I

FUNCTIONALITY AND ITS IMPORTANCE

1.2

The number of polymerisable groups or reactive sites or bonding sites present in a monomer is known as its functionality. Example

Functionality CH2 = CH2

2

(Bifunctional)

2

(Bifunctional)

3

(Trifunctional)

Ethylene H2N – (CH2)6 – NH2 Hexamethylene diamine CH2OH | CHOH | CH2OH: Glycerol

1. If the functionality of a monomer is two, a straight-chain linear polymer can be produced. Examples

Vinyl monomer, Adipic acid, Ethylene glycol

2. When the functionality of a monomer is three, a three-dimensional network polymer is found. Examples

Glycerol, Melamine

3. When a trifunctional mononer is mixed in small amounts with a bifunctional monomer, a branched-chain polymer is obtained. 4. When a bifunctional monomer with a trifunctional monomer, a 3D network polymer is produced.

CLASSIFICATION OF POLYMERS

1.3

Polymeric materials can be classified into several ways as given below:

1.3.1

Based on Source

Based on origin or source, the polymers are classified into two types.

Polymer Chemistry 5

1.3.2

Based on Application

Based on application, polymers are classified into three types: 1. Plastics 2. Fibres 3. Elastomers

They are high polymeric-type material. Based on response to heat and pressure, these polymers are classified into two types: (a) Thermoplastic resins (b) Thermosets or thermosetting resins (a) Thermoplastic Resins These resins can be softened on heating and harden on cooling reversibly. Repeated heating and cooling do not alter the chemical nature of these materials because the changes are purely of physical nature. These plastics are formed as a result of addition polymerisation and have long-chain molecular structures. Examples

(i) Polyethenic or vinyl resins like polyethylene, polyvinylchoride, polyvinyl acetate, polystyrene etc (ii) Poly carbonates like lexan, merlon (iii) Polyamides like Nylon 6,6 Nylon, 6 (b) Thermosetting Resins These resins can get hardened by heating during moulding process. Once they have solidified, they cannot be softened, since they are permanent setting resins.

6 Engineering Chemistry-I

During moulding, these resins acquire 3D cross-linked structure with strong covalent bonds. They are harder, stronger and more brittle than thermoplastics. Examples

(i) (ii) (iii) (iv) (v) (vi)

Phenolic resins like novolac, bakelite Amino resins like area-formaldelyde Polyester resins Epoxy resins Silicone resins Polyurethanes

1.3.3

Differences between Thermoplastic Resins and Thermosetting Resins

S.No.

Thermoplastic resins

Thermosetting resins

1.

They are formed by addition polymerisation.

They are formed by condensation polymerisation.

2.

They are long linear-chain polymers They have a three-dimensional with less cross-links. network structure.

3.

They soften on heating.

They do not soften on heating.

4.

They are reshaped and reused.

They cannot be reshaped and reused.

5.

They are soft, weak and less brittle. They are hard, strong and more brittle.

6.

They are soluble in some organic They are insoluble in organic solvents. solvents due to strong bond and cross-links

7.

They are reclaimed from waste.

1.3.4

Nomenclature of Polymers

They cannot be reclaimed from waste.

These are polymers containing the same type of monomers. Examples

Polyethylene, Polypropylene

These are polymers containing more than one type of monomer. Examples

Nylon 6, 6, Terylene

Polymer Chemistry 7

Copolymer

Based on the manner in which the monomers are arranged

Random Copolymer

Block Copolymer

Linear sequence of one monomeric unit followed by the sequence of – M1–M2–M2–M1–M2–M1–M1– other unit No definite sequence of monomeric unit

Graft Copolymer Branched structure in which the monomer segments on the branches and the backbone differ

–M1–M1–M1–M2–M2–M2– –M1–M1–M1–M1 | M2 | M2 | M2

The main polymer chain is made up of the same species of atoms called homochain polymer. Example

Polyethylene, – C – C – C – C – chain

The main polymer chain is made up of different species of atoms called heterochain polymer. Example

Terylene, – C – C – O – C – C – O –

8 Engineering Chemistry-I

TYPES OF POLYMERISATION

1.4

The reaction by which monomer units are combined to give a polymer is known as polymerisation.

1.4.1

Addition or Chain-Growth Polymerisation

Monomers containing multiple bonds (double or triple bond) undergo addition polymerisation. In addition polymerisation, no other by-product is obtained. So the molecular weight of the polymer is an integral multiple of the molecular weight of the monomer. This polymerisation reaction takes place by the application of heat, light, pressure or a catalyst for breaking down the double covalent bonds of monomers. ! 1. Ethylene Polymerisation

n(H2C = CH2)

D (or) hn (or) Catalyst

n

H

H

C

C

H

H Polymerisation (Addition)

CH2

CH2

Polyethylene

2. Styrene Polymerisation

n

Polymer Chemistry 9

1.4.2

Condensation (or) Stepwise Polymerisation

Monomers containing same or different types of functional groups undergo condensation polymerisation. In this, polymerisation proceeds by stepwise reaction between reactive functional groups and the elimination of small molecules like H2O, HCl, etc. ! Polyester, Polyamide. 1. Polyester Polymerisation These polymers are obtained by the reaction of dicarboxylic acid with a glycol. n HO – CH2 – CH2 – OH + n HOOC Ethylene glycol

COOH

Terephthalic acid Condensation polymerisation O

––– O – CH2 – CH2 – O – C

C ––– O n

Polyester (or) Terylene (or) Dacron

2. Polyamide (Nylon 6,6) Polymerisation These polymers are obtained by the condensation of diamine with dicarboxylic acid.

10 Engineering Chemistry-I

1.4.3

Copolymerisation

Copolymerisation is the joint polymerisation of two or more monomer species. High-molecular-weight polymers obtained by copolymerisation are called copolymers. In this process, no side products are obtained. ! 1. Styrene-Butadiene Copolymer (SBR)

2. Polyvinyl Chloride-Acrylonitrile-Copolymer These are unsymmetrical and irregular shaped polymers. Copolymers are superior than other polymerisation reactions since this reaction can control various polymer properties like solubility, crystallisation tendency, etc. These polymers are soluble only at high temperature.

11

Polymer Chemistry

1.4.4

Differences between Addition (Chain) Polymerisation and Condensation (Stepwise) Polymerisation

S.No. Addition/chain polymerisation

Condensation/stepwise polymerisation

Monomers having multiple bonds undergo addition polymerisation.

Monomers having same or different types of functional groups undergo condensation polymerisation.

Example Ethylene, Vinyl chloride

Example Glycol, Adipic acid

2.

Number of units decreases steadily throughout the reaction.

Monomer disappears early in the reaction.

3.

High polymer is formed at The molecular weight of the polymer once. increases steadily throughout the reaction.

4.

Mostly thermoplastics are produced. Example Polyethylene, PVC

Mostly thermosetting produced.

Molecular weight of this polymer is an integral multiple of the monomer, longer reaction time gives high yield.

Molecular weight of this polymer need not be an integral multiple of a monomer, longer reaction times are needed to obtain high molecular weight.

1.

5.

plastics

are

Example Phenol-formaldehyde

MECHANISM OF ADDITION POLYMERISATION

1.5

Polymerisation of ethylene and its substituted compounds, CH2 = CHX can be carried out by using any of the following four mechanisms: 1. 2. 3. 4.

1.5.1

Free radical polymerisation mechanism Cationic mechanism of polymerisation Anionic mechanism of polymerisation Coordination polymerisation (or) Ziegler–Natta polymerisation

Free Radical Polymerisation Mechanism

This mechanism involves three major steps. 1. Initiation step 2. Propagation step 3. Termination step

12 Engineering Chemistry-I " This step involves two reactions. The first part is the production of free radicals by the homolytic dissociation of an initiator to yield a pair of radicals R . I 2R •

Initiator

Free radicals

The second part of initiation involves the addition of the free radical (R ) to the first monomer (M) to produce the chain-initiating species (M1) . . R M1 + M Chain-initiating species Free radical Monomer •

•

For the monomer of the type CH2 = CHX where (X = Cl, CN, CH3, COOH, COOCH3, C6H5, etc.)

R – CH2 – C

X

X

–

–

. R + CH2 = CH

–

H

# Monomer radical adds on successively to a large number of monomer molecules and the polymer chain grows in length.

Polymer Chemistry

13

$ The chain length can terminate at a certain point. This occurs by (a) Coupling Two growing radicals may combine and the free valencies are saturated.

–

–

–

–

H H R – CH2 – C + C – CH2 – R X

X

–

–

–

–

H H R – CH2 – C – C – CH2 – R X

X

(Dead polymer)

X

X

Saturated polymer

–

–

–

–

– X

H H H R – CH2 – CH + C = C – R

–

– X

H R – CH2 – C + C – CH2 – R

–

H

–

(b) Disproportionation A hydrogen atom of one radical is transferred to another radical centre. This results in the formation of two polymer molecules, one saturated and the other one unsaturated.

Unsaturated polymer

Widely used initiators are acetyl peroxide (AC2O2) and benzoyl peroxide (Be2O2).

CH3 COO – OOCCH3

70–90°C

2(CH3 COO)

Acetyl peroxide

C6H5 COO – OOCC6H5

1.5.2

80–95°C

2(C6H5 COO)

Cationic Mechanism of Polymerisation

This involves the following three steps. In this mechanism, an acid cation (Y +) influences the mechanism.

14 Engineering Chemistry-I "

+ Y – CH2 – CH

+

–

CH2 = CH

–

Y

+

An acid (Cation)

X New carbocation

X Monomer

#

– X

–

– X

–

+ Y – CH2 – CH – CH2 – CH

+ Y – CH2 – CH + CH2 = CH

X

X

–

Carbocation and + nCH2 = CH X + –– Y –– CH2 – CH CH2 – CH

–

–

n

X

X

Big polymeric chain containing carbocation $

X

Y –– CH2 – CH –– CH2 – CHZ n

X

X

X

Polymer

1.5.3

Anionic Mechanism of Polymerisation

This mechanism is influenced by an anion (base) of the type Zŋ: "

A base (An anion)

CH2 = CH X Monomer

– Z – CH2 – CH

–

+

–

– Z

–

–

–

n

Z– (An anion)

–

+ Y –– CH2 – CH –– CH2 – CH

X New anion (Carbanion)

Polymer Chemistry

15

#

$ In this step, the dead polymer is produced by the addition of a cation (H+).

X

X

(Cation)

Z –– CH2 – CH –– CH2 – CH2 X

n

–

+ H+

–

n

–

–

– Z –– CH2 – CH –– CH2 – CH

X

Polymer

PROPERTIES OF POLYMERS

1.6

The structure of polymers plays an important role in determining the physical, mechanical and chemical properties. The important physical and mechanical properties of a polymer depends on 1. 2. 3. 4.

The size of polymer molecule or chain The shape of the polymer molecule or chain Physical state of the polymer Chemical nature of the monomer unit

The polymeric physical properties can be explained by means of Tg (Glass transition temperature), tacticity, molecular weight of polymer and polydispersity index.

16 Engineering Chemistry-I

1.6.1

Glass Transition Temperature ($#)

All polymers are hard, rigid solids at low temperature and it can be changed from the solid to liquid state either by melting or by softening at higher temperatures. Amorphous polymers with random arrangement of their molecular chains soften and become rubber like solid state at a particular temperature called glass transition temperature, (Tg). The rubber like polymer further softens to a viscous liquid at temperature higher than the Tg. The physical properties like rigidity, toughness and viscosity of a polymer depend on whether the temperature of the polymer is below or above Tg. The Tg for polystyrene and polymethyl methacrylate (PMMA) are higher (Tg ~ 100°C) and remain hard and stiff plastics at room temperature. The Tg of most rubber is well below 0°C (Tg ~ –50°C). The Tg depends on the nature of polymer chain also. The polymer chain undergoes free rotation and exhibits low Tg (Groups like C–C–C or C–O–C bonds). The introduction of aromatic ring stiffens the chain, restricting free rotation and have high Tg value, as more thermal energy is required to bring about thermal motion. Crystalline polymers also exhibit Tg associated with the amorphous regions. Crystalline polymers do not soften just above Tg, since they are still rigid, and then soften only above melting temperature (Tm). %&&

# $#

1. Chain Flexibility

(a) Greater the chain flexibility, smaller will be the Tg. For example,

Tg (polypropylene) > Tg (polyethylene)

Tg is affected by the nature of the substituents. (b) Configuration Tg (polyethylene) > Tg (cis-1, 4-polybutadiene)] since inclusion of double bonds stiffen the chain at the point of inclusion but at the same time increase the flexibility of adjacent bond. (c) Intermolecular Forces Tg (polyethylene) < Tg (PVC) since secondary bonding due to dipole force, H-bonding, etc. decreases the mobility of a chain.

Polymer Chemistry

2. Molecular Weight the polymer.

17

Tg is directly proportional to molecular weight of

Greater the cross-links, the higher the Tg.

3. Cross-links

4. Plasticisers Plasticisers also decrease the Tg, since they cause separation of the chains and increase their mobility. 5. Copolymerisation Random copolymers lower the Tg since they tend to promote disorder, reduce molecular packing and also reduce the interchain forces of attraction.

1.6.2

Tacticity

The orientation of the monomeric units and the spatial arrangement of the functional groups in a polymer molecule give rise to differences in configuration (or) tacticity which has mainly influence on the physical properties of the polymer. Tacticity is explained with polypropylene polymer. Tacticity results in three types of stereo-regular polymers. " The functional groups are arranged on the same side with respect to the main chain.

–

–

–

–

–

– CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – CH3

C H3

CH3

CH3

CH3

Isotactic polypropylene (Isotactic – Greek = Same order)

The functional groups are arranged in alternative sequence with respect to the main chain. –

CH3

–

CH3

–

–

–

– CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – CH3

CH3

CH3

Syndiotactic polypropylene (Syndiotactic – Greek = Contrasting order) % The functional groups are arranged randomly with respect to the main chain.

18 Engineering Chemistry-I

CH3

–

CH3

–

–

CH3

–

–

– CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – CH2 – CH – C H3

CH3

Atactic polypropylene (Atactic – Greek = Without order)

1.6.3

Molecular Weight of a Polymer

Polymers are polydisperse and quite heterogeneous in molecular mass. In other words, polymers are mixtures of molecules of different molecular masses. The molecular weight of a polymer can be determined by different methods like 1. Colligative properties like freezing-point depression (cryoscopy) boiling point elevation (ebullisocopy), osmotic pressure (osmometery) 2. By viscosity measurements 3. Light scattering measurements 4. Ultracentrifuge

1.6.4

Number-Average Molecular Mass (Mn )

This molecular mass is determined by colligative methods. Mn is defined as the total mass (w) of all the molecules in a polymer sample divided by the total number of molecules present. So

Mn =

SNi Mi w = SNi SN

where Ni is the number of molecules of mass Mi . The Mo is a good index of physical properties such as impact and tensile strength but not good as other properties like flow.

1.6.5

Weight-Average Molecular Mass ( Mw )

This molecular mass is obtained from light scattering, ultra centrifugation and sedimentation techniques, which measure molecular size. It is defined as Mw =

SWi Mi SWi

where Wi is the weight fraction of molecules, whose mass is Mi,

Polymer Chemistry

19

Mw can also be defined as 3

Mw = where

SCi Mi SCi Mi SNi M i = = SN i M i C SCi

Ci = Weight concentration of Mi molecules C = Total weight concentration of all polymeric molecules.

The heavier molecules contribute more to the Mw . Hence, the Mw is always greater than the Mn .

1.6.6

Polydispersity Index (PDI)

The ratio of the weight-average molecular weight Mw to that of the number-average molecular weight ( Mn ) is known as the polydispersity index (PDI) and gives a measure of the spread of molecular weight distribution. M PDI = w Mn The higher molecular weight fractions will have a greater effect on Mw and the lower molecular weight fractions will have a greater effect on Mn Hence, Mw ≥ Mn .

TECHNIQUES OF POLYMERISATION

1.7

Polymerisation reactions are exothermic. The main concern during polymerisation is to effectively dissipate this heat from the reaction mixture to avoid explosion. The main polymerisation techniques or processes are given below: Polymerisation Processes

Homogeneous Polymerisation

Heterogeneous Polymerisation

(a) Solution (a) Emulsion (b) Bulk (b) Suspension Polymerisation Polymerisation Polymerisation Polymerisation

1.7.1

Homogeneous Polymerisation

Polymerisation reaction takes place in one homogeneous phase. This occurs in solvent or in a reaction vessel containing only the monomer.

20 Engineering Chemistry-I

There are basically two methods, namely 1. Solution polymerisation 2. Bulk polymerisation

In this polymerisation, the monomer is initially dissolved in a suitable solvent and then the initiator is added to start the polymerisation. Then the polymer is obtained by precipitation. The filtered polymer is dried and used. Example of monomer:

Acrylonitrile and acrylic acid.

Advantages • The solvent reduces the exothermal problem. • The thermal control is easy. Disadvantages • Chain transfer reactions to solvent does not allow the synthesis of high-molecular weight polymers. • There is difficulty in the recovery of the solvent. • The process is costly. • Fire and toxicity hazards are increased by solvent. ' ( Addition and condensation polymerisations can be carried out in a bulk polymerisation reactor. The monomer is taken in the liquid state and the initiator is dissolved in the monomer itself. Polymerisation is then carried out in pre and post-polymerisation stages for controlling the heat of polymerisation. About 10% conversion later the reaction is slow but subsequently it becomes fast. After 20–30% polymerisation, the reaction mixture becomes viscous. In the second stage (post-polymerisation stage), the viscous mixture is transferred to the main reactor maintained at constant temperature. After a known period of time, the contents are poured into methanol, when the polymer gets precipitated out. Monomers Styrene, vinyl chloride, methyl methacrylate can be polymerised by this method.

Polymer Chemistry

21

Advantages

• Low-impurity-level polymer obtained. • Good clarity and good electrical insulation characteristics. Disadvantages

• Polymers have low thermal conductivity and the polymerisation reactions are exothermic, so there is danger of overheating of the reactants. • Auto-acceleration can lead to explosive reactions and hence thermal control is difficult.

1.7.2

Heterogeneous Polymerisation

This polymerisation involves more than one phase. They are of two types. 1. Emulsion polymerisation 2. Suspension polymerisation The monomer is dispersed in an aqueous phase as fine droplets (10–5 to 10– 6 mm in size). They are stabilised by adding surfactants (soaps or detergents) or protective colloids. The surfactants will form micelles when their concentration exceeds critical micelle concentration (CMC) dispersed throughout the solution. The added initiators initiate the reaction in the micelles. Due to formation of polymers, these micelles increase in size. To continue the reaction, the monomer diffuses from its droplets to micelles. Termination reactions are less likely as the polymerisation sites are isolated from each other. Hence, high-moleculer weight polymers can be obtained. The polymers obtained can be used in the emulsion form used in adhesives, surface coatings or textile finishes. Monomers Styrene, vinyl chloride and vinyl acetate can be polymerised by this method.

22 Engineering Chemistry-I Advantages • Temperature can be easily controlled. • Rapid production of polymer of higher molecular weight (directly usable latex.) Disadvantages • Costly and high residual impurity level. • Translucent polymers obtained have inferior electrical insulation characteristic.

In this method, the monomer is dispersed as large droplets (0.1 – 1 mm in size) in water and kept in suspension by mechanical agitation. The stabilisers used are gelatin and cellulose derivatives. The initiators are soluble in monomers and polymerisation continues in each droplet till 100% conversion. The polymer obtained is pearls or spherical beads. Monomers Styrene, vinyl chloride and methyl methocrylate can be polymerised by this method. Advantages • Purity is high. • Viscosity is low throughout. • Temperature can be easily controlled. Disadvantages • Highly sensitive to agitation. • Difficult to control the size of the particles and their surface characteristics.

NYLON 6, 6 POLYMER (POLYAMIDES)

1.8

Polyamides are synthetic polymers. They have amide functional groups. Nylons, which are mostly used for making fibres, belong to this class. Nylon 6, 6 derives its name from its starting materials, adipic acid and hexamethylene diamine, both of which have six carbons.

Polymer Chemistry

23

Preparation of Nylon 6, 6 This can be obtained by the polymerisation of adipic acid with hexamethylene diamine monomers. n H2N – (CH2)6 – NH2 + n HOOC – (CH2)4 – COOH Hexamethylene diamine

Adipic acid Condensation polymerisation

H ––– N – (CH2)6 – N – C —( CH2 — )4 C ––– + 2nH2O n O O H Polyhexamethylene adipate (Nylon 6,6) (Polyamides)

• Translucent, whitish horny and high melting (m.pt. 160°C to 264°C) polymers • High temperature stability • Good abrasion resistance • Insoluble in common organic solvents like methylated spirit, benzene and acetone • Soluble in phenol and formic acid • Good physical strength and self-lubricating properties

• • • • •

&) * They are light, horny and have high melting points. They absorb little moisture and are drip-dry in nature. They are flexible and retain original shape after use. They are resistant to abrasion. On blending with wool, the strength and abrasion resistance of the latter increases.

+ &) * • These fibres are used in making socks, hosiery, undergarments, dresses, carpets, etc.

24 Engineering Chemistry-I

• Nylon bearings and gears work quietly without any lubrication. • They are used for making filaments for ropes, bristles for toothbrushes and films, tyre-cords, etc.

EPOXIDE (OR) EPOXY RESINS

1.9

They are polyethers.

Preparation They are prepared by condensation of epichlorohydrin with bisphenol-A

1.9.1

Preparation of Epichlorohydrin CH2 = CH – CH3 + Cl2 Propylene

400°C

CH2 = CH – CH2Cl

–HCl

Allyl chloride H2O + Cl2 300°C

Cl CH2 – CH(OH) – CH2Cl Glycerol dichlorohydrin 200°C CaO CH2 – CH – CH2Cl + HCl O Epichlorohydrin

1.9.2

Preparation of Bisphenol-A 2 HO

+ CH3COCH3

Phenol

NaOH 50°C

CH3 HO

C CH3 BisPhenol–A

OH

Polymer Chemistry

1.9.3

25

Condensation of Epichlorohydrin with Bisphenol-A CH3 C

n H2C – CH – CH2Cl + n HO O

OH

CH3 Bisphenol–A

Epichlorohydrin

Alkaline catalyst at 60°C (aq. NaOH) CH3 ––– (CH2) – CH – CH2 – O OH

C

O ––– n

CH3 Epoxy resins

The reactive epoxide and hydroxyl groups give a three-dimensional cross-linked structure. The values of n range from 1 to 20. The molecular weight ranges as low as 350 (a mobile liquid) to as high as 8,000 (a solid of m.pt. 145–155°C).

• They are thermosetting resins having good adhesive properties. • Due to the presence of very stable ether linkage, they have high chemical resistance to water, various solvents, acids, alkalies and other chemicals. • They are cured without the application of external heat. • They are sold in the market in liquid or semisolid form with solvents. • They show low shrinkage and excellent adhesive properties. • They contain polar groups like epoxide ring –OH group, so they have good adhesion to polar or metallic surfaces.

26 Engineering Chemistry-I + • It is used to bind a number of substances including metals and glasses • It is sold in the market in the name of araldite. • It is a good surface-coating materials because of its outstanding toughness, flexibility, adhesion and chemical resistance. • These resins are applied over cotton, rayon and bleached fabrics and laminated materials. • Used for skid-resistant surfaces for highways. • Epoxy resin moulds are used for the production of components for aircrafts and automobiles.

QUESTIONS PART-A 1. 2. 3. 4. 5. 6. 7. 8.

9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20.

What is meant by degree of polymerisation? What is functionally of a monomer? Give one example. What are called Copolymer? Give an example. What is glass Transition temperature? What is Tacticity of a Polymer? What is Polydispersity index of a polymer? Write the advantages of emulsion polymeristion? Write the monomers used for the preparation of the following polymers? (a) Nylon 6,6 (b) Epoxy Resins. Write the uses of epoxy resins? Define a polymer. What is meant by polymerization? Define thermoplastic resin? Define thermosetting polymer. What is a monomer? Why cannot thermosetting plastics by reused and reshaped? Write the disadvantages of using solution polymerisation process? What is copolymerisation? What is condensation polymerisation? Write the difference between addition and condensation polymerisation? Write the polymers used by the manufacture of emulsion polymerisation.

Polymer Chemistry

27

PART-B 1. Explain with examples the terms addition Polymerisation, condensation polymerization and copolymeriation. [P.8–10] 2. Write all differences between thermoplasitcs and thermosetting plastics. Give two examples for each type. [P.5–6] 3. Give the manufacture and uses of (i) Nylon 6,6 (ii) Epoxy resin [P.23, 24] 4. Distinguish between addition and condensation polymerisation. [P.11] 5. (a) Define the term functionality of monomers. Explain its significance with a suitable example. [P.4] (b) What are the materials from which an epoxy adhesive is prepared? [P.24] 6. Write on (i) free radical polymerisation, (ii) anionic polymerisation, (iii) Cationic polymerisation. [P.11-15] 7. Describe the Preparation and important Properties of Epoxy resins. [P.24] 8. Define and give examples for (i) monomer, (ii) functionality, (iii) degree of polymerisation (iv) Copolymer. [P.3–4] 9. Explain Copolymerisation with suitable example. [P.10] 10. (a) Explain with examples addition and condensation polymerisation. [P.11] (b) Describe the manufacture and uses of nylon 6 : 6 [P.23] 11. (a) Write the relationship between the properties and structures of thermoplastic and thermosetting plastics. [P.5–6] (b) Explain the engineering properties of thermoplastics. [P.15–17] 12. Write short notes on Preparation reactions, properties and uses of Epoxy resins. [P.24] 13. How molecular weight of a polymer is determined by weight average and number average method. [P.18] 14. Write the mechanism of free radical polymerisation? [P.11–13] 15. Write and explain the cationic mechanism of polymerisation. [P.13–14] 16. Explain all the steps involved in the anionic polymerisation mechanism. [P.14–15] 17. Explain the different techniques used in the polymerisation process. [P.19–22]

28 Engineering Chemistry-I

18. Explain solution polymerisation and Bulk Polymerisation. [P.20] 19. Write and explain the heterogeneous polymerisation process. [P.21–22] 20. What is suspension polymerisation. [P.22]

Two-Mark Questions and Answers 1. What is meant by degree of polymerisation? The number of repeating Units (n) in a polymer chain is called degree of polymerisation. For example. In polythene is a polymer linking to gather of a large number of ethene having n value is of 104 or more. 2. What is functionality of a monomer? Give one example. The number of polymerisable groups or reactive sites or bonding sites present in a monomer is known as its functionality Example: Hexamethylene diamine H2N–(CH2)6 – NH2 is bifunctional. 3. What are the classification of polymers based on source? Based on source they are classified into (i) Natural polymers (e.g) carbolydrate, proteins, etc. (ii) Synthetic polymers. (e.g) PVC, Terylene, etc. 4. What are called copolymers? Give an example. Copolymerisation is the joint polymerisation of two or more monomer species. In this Process, no side products are obtained. Example: Styrene-Butadiene rubber. 5. What is glass transition temperature? Amorphous polymers with random arrangement of their molecular chains soften and become rubber like solid state at a particular temperature called glass transition temperature (Tg). 6. What is tacticity of a polymer? The orientation of the monomeric units and the spatial arrangement of the functional groups in a polymer molecule give rise to differences in configuration (or) tacticity. Tacticity influences the physical properties of the polymer. 7. What is Polydispersity index (PDI) of a polymers? The ratio of the weight-average molecular weight M w to that of the number-average molecular weight is known as the polydispersity index. 8. Write the advantages of emulsion polymerisation. 1. The Temperature can be easily controlled by emulsion polymers. 2. directly usable latex polymer of high molecular weight can be obtained by the polymerisation.

Polymer Chemistry

29

9. Define a polymer. Polymers are macromolecules of high molecular mass formed by the combination of a large number of small molecules of low molecular weight called Monomers. 10. Define thermoplastic resins. These are the polymer resins, can be softened on heating and harden on cooling reversibly. Repeated heating and cooling do not alter the chemical nature of these materials, because the changes are purely of physical nature, e.g. PVC 11. Why thermosetting plastics cannot be reused and reshaped. The molecules of monomers in thermosetting plastics are held together by strong covalent bonds, and these bonds retain their strength, even on heating. So they cannot be softened by heating and cannot be reshaped or reused. 12. What is a monomer? Monomers are small molecules which combine with each other to form polymers. 13. Define thermosetting polymer. These resins can get hardened by heating during moulding process. Once they solidified, they cannot be softened. 14. What is suspension polymerisation? The monomer is dispersed as large droplets (0.1 – 1 mm is size) in water and kept in suspension by mechanical agitation. 15. Write the polymers used by the manufacture of emulsion polymerisation. Styrene, Vinychloride and Vinylacetate.

56 Engineering Chemistry-I

Multiplying by T on both sides, we get TDSTotal = TDSSys – DHSys (or)

DHSys – TDSSys = – TDSTotal

(or)

DGSys = - T DSTotal

…(40)

Equation (40) is the basis for criterion of spontaneity in terms of free energy change of the system only. From Eq. (40), Case I If

DSTotal = +ive,

(DG)T,P = –ive

The process will be spontaneous. Case II

If

DSTotal = 0,

(DG)T,P = 0

The process will be at equilibrium. Case III If

DSTotal = –ive,

(DG)T,P = +ive

The process will be nonspontaneous. Equation (36) is used as a criterion of the feasibility of a process. No process is possible by increase in Gibbs free energy of the system, because according to Eq. (36), the Gibbs free energy always decreases in a spontaneous process. If DH = –ive , DS = +ive then DG = –ive (spontaneous) If DH = –ive , DS = 0, then DG = –ive (spontaneous) If DH = 0, DS = +ive then DG = –ive (spontaneous) If DH = +ive + DS = +ive then DG = –ive (spontaneous) further DH < TDS (v) If DH = –ive + DS = –ive then DG = –ive (spontaneous) further DH > TDS Hence, at irreversible or spontaneous process at constant T and P, DG < 0 and at equilibrium under isothermal and isobaric conditions, DG = 0. (i) (ii) (iii) (iv)

} }

GIBBSŪHELMHOLTZ EQUATION

2.13

The Gibbs–Helmholtz equation helps us determine the value of DG at any given temperature.

Chemical Thermodynamics

57

The simplest form of the equation is DG = DH – TDS G = H – TS

(or)

The differential form of this equation is dG = dH – TdS – SdT

…(41)

We also know that H = E + PV and dH = dE + PdV + VdP So Eq. (41) becomes dG = dE + PdV + VdP – TdS – SdT

…(42)

From the first law of thermodynamics, dqrev = dE + PdV

…(43)

Substituting Eq. (43) in Eq. (42), dG = dqrev + VdP – TdS – SdT

…(44)

We know from the entropy concept DS = \

dq qrev (or) dS = rev T T

dqrev = TdS Substitute dqrev in the place of TdS in Eq. (44), we get dG = dqrev + VdP – dqrev – SdT dG = VdP – SdT

…(45)

At constant pressure, dp = 0. Hence (dG)P = –SdT Ê ∂G ˆ ÁË ˜ = –S ∂T ¯ p

(or)

Considering a change in G from final to initial stage, Ê ∂ (G2 - G1 ) ˆ ˜¯ = – (S2 – S1) ÁË ∂T P

(or)

Ê ∂DG ˆ ÁË ˜ = – DS ∂T ¯ P

Substituting Eq. (46) in DG = DH – TDS

…(46)

58 Engineering Chemistry-I

We get a more useful form of Gibbs–Helmholtz equation Ê ∂DG ˆ DG = DH + T Á Ë ∂T ˜¯ P

…(47)

Similarly, an equation for DA may also be written as Ê ∂DG ˆ DA = DE + T Á Ë ∂T ˜¯ V

…(48)

An alternate form of Gibbs–Helmholtz equation may be obtained by making some changes in Eq. (47). Ê ∂DG ˆ DG = DH + T Á Ë ∂T ˜¯ P

1 Multiplying on both sides by – ÊÁ 2 ˆ˜ , we get ËT ¯ 1 DG ˆ – ÊÁ = – ÊÁ DH ˆ˜ – Ë T 2 ˜¯ Ë T2 ¯ T

Ê ∂( DG) ˆ ÁË ˜ ∂T ¯ P

Rearranging, DG ˆ 1 Ê ∂ ( DG) ˆ Ê DH ˆ + – ÊÁ ˜ =–Á 2 ˜ Á Ë T 2 ˜¯ T Ë ∂T ¯ P ËT ¯ Substituting ∂

∂T

( 1 T ) for – ÊÁË T1 ˆ˜¯ on left hand side 2

( )

DH 1 Ê ∂( DG) ˆ ∂ 1 ˘ DG ¥ ÈÍ ˙ + T ÁË ∂T ˜¯ = – T 2 T T ∂ Î ˚P P

…(49)

Now the left hand side of Eq. (49) appears to be in the form of udv + vdu which is d (uv) ∂ Ê 1ˆ ∂ –1 ÁË ˜¯ = ∂T (T ) ∂T T Mathematically, dxn = nxn–1 ∂ –1 (T ) = –1 ¥ T –2 ∂T 1 = – ÊÁ ˆ˜ Ë T2 ¯

\

\

∂ ∂T

1˘ DH È ÍÎ DG ¥ T ˙˚ = - T 2

…(50)

Chemical Thermodynamics

59

The Gibbs–Helmholtz equations (47), (48) and (50) are applicable to closed systems in thermodynamic equation as DG, DH take up a definite value for a given change. They are state functions.

Applications of GibbsŪHelmholtz Equation 1. Determination of DH in an Electrochemical Cell The heat change (DH) of a redox reaction can be calculated by an equation. DG = –nFE …(51)

The product of n, F and E is the electrical energy output which is equal to the decrease in the Gibbs free energy. ‘n’ refers to the number of electrons. E = emf and F = Faraday = 96,500 columns/g equivalent. The Gibbs–Helmholtz equation is Ê ∂DG ˆ DG = DH + T Á Ë ∂T ˜¯ P

…(52)

Substituting the value of DG from Eq. (51) into Eq. (52), Ê ∂ (- nFE) ˆ –nFE = DH + T Á Ë ∂T ˜¯ P

È ∂E ˘ –nFE = DH – nFT ÍÊÁ ˆ˜ ˙ ÎË ∂T ¯ P ˚ Dividing Eq. (53) by –nF, we get E=–

È ∂E ˘ DH +TÍ ˙ Î ∂T ˚ P nF

…(53)

…(54)

Rearranging Eq. (54),

È Ê ∂E ˆ ˘ DH = - nF ÍE - T Á ˜ ˙ Ë ∂T ¯ P ˚ Î 2. Determination of DS of the Reaction in an Electrochemical Cell In the equation, È Ê ∂E ˆ ˘ DH = –nF ÍE - T Á ˜ ˙ Ë ∂T ¯ P ˚ Î

…(55)

…(56)

Ê ∂E ˆ Equation (55), Á ˜ is the temperature coefficient of emf of the cell. Ë ∂T ¯ P Ê ∂E ˆ The term nF Á ˜ is equivalent to the change in entropy. Ë ∂T ¯ P

60 Engineering Chemistry-I

DS of the cell reaction is arrived as È Ê ∂E ˆ ˘ DH = – nF ÍE - T Á ˜ ˙ Ë ∂T ¯ P ˚ Î Ê ∂E ˆ DH = –nFE + nFT Á ˜ Ë ∂T ¯ P

Ê ∂E ˆ - nFE = DH - nFT Á ˜ Ë ∂T ¯ P The above equation may be compared to the Gibbs free-energy equation. DG = DH – TDS Ê ∂E ˆ DS = nF Á ˜ Ë ∂T ¯ P

If E° is the standard emf then DS° is the standard entropy change. Ê ∂E∞ ˆ DS∞ = nF Á Ë ∂T ˜¯ P Problem 2.16: Calculate the standard entropy change for the reaction Zn(s) + 2H + Æ Zn2++ H2 (g) Ê ∂E∞ ˆ given that the value of Á is +0.85 ¥ 10–4 volt/dg at 25°C. Ë ∂T ˜¯ p Solution

Ê ∂E∞ ˆ DS° = nF ¥ Á Ë ∂T ˜¯ P

= 2 ¥ 23070 ¥ 0.85 ¥ 10–4 DS° = 3.92 entropy units (The value of F = 23,070 in cal/vol◊ g ◊ eqt◊) Problem 2.17: Gibbs free energy of a reaction at 300 K and 310 K are –29 kcal and –29.5 kcal respectively. Determine its DH and DS at 300 K. Solution According to Gibbs–Helmholtz equation, Ê ∂DG ˆ DG = DH + T Á Ë ∂T ˜¯ P

Chemical Thermodynamics

61

Given: T1 = 300 K T2 = 310 K G1 = –29 kcal

G2 = –29.5 kcal

Ê ∂DG ˆ - 29.5 - (- 29) ÁË ˜ = ∂T ¯ p 310 - 300 =–

0.5 10

= – 0.05 kcal \

DH at 300 K,

G1 = –29 kcal

Ê ∂DG ˆ DG = DH + T Á Ë ∂T ˜¯ P –29 = DH + 300 (–0.05) DH = –29 + 15 = –14 kcal And we know,

DG = DH – TDS DS =

(or)

DH - DG - 14 - (-29) = T 300

= + 0.05 kcal K–1 Problem 2.18: The free-energy change DG for process is –138 kJ at 30°C and –135 kJ at 40°C. Calculate the change in enthalpy, DH accompanying the process at 35°C. Solution

From Gibbs–Helmholtz equation, Ê ∂DG ˆ DG = DH +T Á . Ë ∂T ˜¯ P

and

dDG = DG2 – DG1 and dT = T2 – T1

Given: DG1 = –138 kJ DG2 = –135 kJ T1 = 30°C = 30 + 273 = 303 K T2 = 40°C = 40 + 273 = 313 K The value

3 ∂DG - 135 - (- 138) = = 0.3 kJ = 10 ∂T 313 - 303

62 Engineering Chemistry-I

The change in free energy DG at 35°C is taken as the average value as DG at 30°C and 40°C. 303 + 313 = 308 K 2 DG at 308 K =

- 135 + (- 138) 2

= –136.5 kJ Substituting in the equation È ∂ ( DG) ˘ DH = DG – T Í ˙ Î ∂T ˚ P = –137.5 – 308 (0.3) = 228.9 kJ Problem 2.19: Calculate the free energy change (DG) and enthalpy change (DH) for an voltaic cell (Daniel cell) having emf of 1.1 V. The tem∂E perature coefficient È ˘ is found to be –2.6 ¥ 10–4. The cell reaction is Í ∂T ˙ P Î ˚ Zn + CuSO4 � Cu + ZnSO4 Solution Given: Emf (E) = 1.1 V È ∂E ˘ = –2.6 ¥ 10–4 Í ∂T ˙ P Î ˚ F = 96500 coulombs n =2 DG = –nFE = –(2 ¥ 96500 ¥ 1.1) = –212300 joules = –212.3 kJ Calculation of DH Ê ∂ ( DG) ˆ DG = DH + T Á Ë ∂T ˜¯ P

…(i)

∂E –nFE = DH – nFT ÊÁ ˆ˜ Ë ∂T ¯ P

…(ii)

Chemical Thermodynamics

63

On dividing (ii) by –nF, DH +T nF

È ∂E ˘ Í ∂T ˙ P Î ˚

where n = 2 F = 96500 nF = 2 ¥ 96500 DH 1.1 = – + 298 ¥ (–2.6 ¥ 10–4) = 193000 193000 E =–

1.1 = –

DH – 0.0775 1, 93000

–DH = (1.1 ¥ 1,93,000) + 0.0775 = 21230.08 J DH = –21.23 kJ

CLAUSIUSŪCLAPEYRON EQUATION

2.14

A system consisting of one mole of a substance can exist in two phases, A and B. The free energy per mole of the substance in two phases A and B can be GA and GB. At equilibrium, there is no free energy change; so GA = GB

…(57)

The temperature of the system can be raised to T + dT and the pressure becomes P + dP. The new free energies per mole A and B phase are GA + dGA and GB + dGB respectively. At equilibrium, GA + dGA = GB + dGB We know that G = H – TS = E + PV – TS

…(58)

\

dG = dE + PdV + VdP – TdS – SdT

…(59)

But

dq = dE + dw = dE + PdV

…(60)

and

dq = ds or TdS = dq T

…(61)

64 Engineering Chemistry-I

From Eqs (59), (60) and (61), we get dG = VdP – SdT …(62) The work done in various equilibria is due to volume change only. So Eq. (62) may be applied to phase A as well as B. \

dGA = VA dP – SA dT

…(63)

dGB = VB dP – SB dT …(64) where VA and VB are the molar volumes of phases A and B respectively. SA and SB are their molar entropies From Eqs (62), (63) and (64), we get VA dP – SA◊ dT = VB dp – SB ◊ dT (or) (or)

= dp (VA – VB) = – dT (SB – SA) (S - SA ) dP DS = =– B dT (VA - VB ) (VB - VA )

…(65)

…(66)

where DS = Molar entropy change and (VB – VA) is the change in volume when 1 mole of a substance changes from A to B. If molar heat of transition from phase A to B is q then DS = q/T From Eq. (66) and Eq. (67), we get q dP = dT T (VB - VA )

…(67)

…(68)

Equation (68) is the Clausius–Clapeyron equation.

Applications 1. To Determine the Latent Heat from Vapour Pressure By knowing the vapour pressure of two phases at two different temperatures in equilibrium, the latent heat can be calculated by using the Clausius–Clapeyron equation. In vapourisation, Vvapour >> Vliquid

(or)

Vvapour – Vliquid ª Vvapour

(The molar volume Vliquid is neglected in comparison with Vvapour). \ the Clausius–Clapeyron equation takes the form dP LV = dT TVVapour where LV is the molar heat of vaporisation.

65

Chemical Thermodynamics

RT

Assuming the gas law (PV = RT) in vapour, so we substitute V = , P in the above equation. L ◊P dP = V2 dT RT L ◊dT dP = V 2 P RT Integrating, by assuming LV is a constant over a small temperature change, we get 1˘ P L È1 ln 2 = V Í - ˙ P1 R Î T1 T2 ˚ (or)

(or)

ln

P2 LV È T2 - T1 ˘ = Í ˙ P1 R Î T1 T2 ˚

…(69)

where P1 and P2 are the vapour pressures at temperatures T1 and T2 respectively. By substituting these values in Eq. (69), the molar heat of vaporization (LV) can be calculated. 2. To Study the Effect of Pressure on Boiling Point If the boiling point of a liquid at one pressure and its latent heat of vaporization are known, then boiling point of liquid at another pressure is calculated from Eq. (69). 3. The Effect of Temperature on Vapour Pressure of Liquid If the latent heat (LV) and vapour pressure of a liquid at one temperature are known, the vapour pressure of liquid at another temperature can be calculated from Eq. (69).

MAXWELL RELATIONS

2.15

The change in pressure (P), volume (V) and temperature (T) in relation to change in entropy (S) has been given by Maxwell relations. This relation finds applications in thermodynamic considerations of change in equilibrium systems. 1. The mathematical form of the Ɲrst law is dE = dq – w

dE = dq – PdV dq = dE + PdV

…(70)

66 Engineering Chemistry-I

The mathematical expression for the second law is dS = dqrev T dqrev = TdS

…(71)

From Eqs (70) and (71), we can write TdS = dE + PdV dE = TdS – PdV

…(72)

Now E is a function of entropy and volume. On differentiating Eq. (72) with respect to entropy S at constant V, we get Ê ∂E ˆ = T ÁË ˜¯ ∂S V

…(73)

On differentiating Eq. (72) with respect to volume at constant S, we get Ê ∂E ˆ = – P …(74) ˜ ÁË ∂V ¯ S Upon differentiating Eq. (73) with respect to V at constant entropy S, we get ∂ ÈÊ ∂E ˆ ˘ Ê ∂T ˆ …(75) Í ˙ =Á Ë ∂V ˜¯ S ∂V ÎÁË ∂S ˜¯ V ˚ S

Parallely, on differentiating Eq. (74) with respect to S at constant volume (V), ∂ ÈÊ ∂E ˆ ˘ Ê ∂P ˆ …(76) ÍÁ ˜ ˙ = – ÁË ˜¯ ∂S ÎË ∂V ¯ S ˚ ∂S V

where E is the state function and dE is exact differential function. ∂2 E ∂2 E = ∂V ∂S ∂S ◊ ∂V On comparing Eqs (75) and (76), we get Ê ∂P ˆ Ê ∂T ˆ ÁË ˜¯ = - ÁË ˜¯ ∂S V ∂V S

…(77)

Equation (77) is Maxwell’s relation based on the first law of thermodynamics. 2. Based on Enthalpy Equation

H = E + PV

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67

On differentiating, dH = dE + PdV + VdP

…(78)

dE + PdV = TdS

…(79)

dH = TdS + VdP

…(80)

Since Putting (79) in (78), we get Now H is a function of entropy and pressure. On differentiating Eq. (80) with respect to S at constant P Ê ∂H ˆ = T ÁË ˜ ∂S ¯ P

…(81)

Upon differentiating Eq. (80) with respect to P at constant S, Ê ∂H ˆ = V ÁË ˜ ∂P ¯ S

…(82)

Differentiating Eq. (81) with respect to P at constant S, we get ∂ ∂P

ÈÊ ∂H ˆ ˘ Ê ∂T ˆ ÍÁË ˜¯ ˙ = ÁË ˜¯ ∂ S ∂P S P ˚S Î

…(83)

Differentiating Eq. (82) with respect to S at constant P, ∂ ∂S

ÈÊ ∂H ˆ ˘ Ê ∂V ˆ ÍÁË ˜¯ ˙ = ÁË ˜ ∂S ¯ P Î ∂P S ˚ P

…(84)

H is a state function and dH is the exact differential. ∂2 H ∂2 H = ∂P ◊ ∂S ∂S ◊ ∂P On comparing the Eqs (83) and (84), we get the following Maxwell relation. Ê ∂T ˆ = Ê ∂V ˆ ÁË ˜¯ Á ˜ ∂P S Ë ∂S ¯ P

…(85)

3. Helmholtz Energy or Work Function (A) A = E – TS On differentiating, we get

dA = dE – TdS – SdT

…(86)

TdS = dE + PdV

…(87)

We know that

68 Engineering Chemistry-I

Substituting TdS from Eq. (87) into Eq. (86), we get dA = dE - dE – PdV – SdT dA = – SdT – PdV

…(88)

Now A is a function of temperature and volume on differentiating Eq. (88) with respect to T at constant V. Ê ∂A ˆ = – S ÁË ˜ ∂T ¯ V

…(89)

Upon differentiating Eq. (88) with respect to V at constant T, Ê ∂A ˆ = – P ˜ ÁË ∂V ¯ T

…(90)

On differentiating Eq. (89) with respect to V at constant T, we get ∂ ∂V

ÈÊ ∂A ˆ ˘ Ê ∂S ˆ ÍÁË ˜¯ ˙ = – ÁË ˜ ∂V ¯ T Î ∂T V ˚

…(91)

T

and On differentiating Eq. (90) with respect to T at constant V, ∂ ∂T

ÈÊ ∂A ˆ ˘ Ê ∂P ˆ ÍÁË ˜¯ ˙ = – ÁË ˜¯ ∂T V Î ∂V T ˚

…(92)

V

A is a state function and dA is exact differential. ∂2 A ∂2 A = ∂V ∂T ∂T ◊ ∂V On comparing Eqs (91) and (92), Ê ∂S ˆ Ê ∂P ˆ ÁË ˜ =Á ˜ ∂V ¯ T Ë ∂T ¯ V

…(93)

Equation (93) is the Maxwell relation. 4. The Relationship for Gibbs Free Energy

G = H – TS On differentiating dG = dH – TdS – SdT We know that H = E + PV dH = dE + PdV + VdP dH = TdS + VdP (Since TdS = dE + PdV)

…(94)

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69

Substituting the value of dH in Eq. (94), we get dG = TdS + VdP – TdS – SdT dG = –SdT + VdP

…(95)

Now G is a function of T and P. On differentiating Eq. (95) with respect to T at constant P Ê ∂G ˆ = – S ˜ ÁË ∂T ¯ P

…(96)

Upon differentiating Eq. (95) with respect to P at constant T, Ê ∂G ˆ = V ˜ ÁË ∂P ¯ T

…(97)

On differentiating Eq. (97) with respect to P at constant T, ∂ ÈÊ ∂G ˆ ˘ Ê ∂S ˆ ÍÁË ˜¯ ˙ = – ÁË ˜¯ ∂P Î ∂T P ˚T ∂P T

…(98)

On differentiating Eq. (97) with respect to T at constant P, ∂ ÈÊ ∂G ˆ ˘ Ê ∂V ˆ ÍÁË ˜¯ ˙ = ÁË ˜ ∂T Î ∂P T ˚P ∂T ¯ T

…(99)

G is a state function and dG is exact differential. ∂2G ∂2G = ∂P ◊ ∂T ∂T ◊ ∂P On comparing Eqs (98) and (99), we get Ê ∂V ˆ Ê ∂S ˆ ÁË ˜¯ = - ÁË ˜¯ ∂T P ∂P T

…(100)

Equation (100) is the Maxwell relation.

VAN'T HOFF ISOTHERM

2.16

The difference between standard Gibbs energies of products and reactants is their standard states at the given temperature is standard Gibbs energy of a reaction (DG°). Van't Hoff has derived a simple relation between the activities of the species concerned in the reaction and the standard Gibbs energy of a reaction (DG°). Let us consider a reversible general reaction A+B�C+D

70 Engineering Chemistry-I

Gibbs free energy of reactant A is given by GA = G°A + R + ln [A]

[A] = Activity of A G°A = Standard Gibbs energy of A

Similarly, we can write GB = G°B + RT ln [B] GC = G°C + RT ln [C] GD = G°D + RT + ln [D] DG of the reaction = GProducts – GReactants …(101) DG = [GC + GD] – [GA + GB] DG = (G°C + G°D + RT ln [C] [D]) – (G°A + G°B + RT ln [A] [B]) = [(G°C + G°D) – (G° A + G° B)] + RT ln = (GºP – G°R) + RT ln

[C] [D] [ A] [B]

[C] [D] [ A] [B]

(where P = Product, R = Reactants)

DG = DG∞ + RT ln

[C] [D] [ A] [B]

…(102)

(where DG° = standard Gibbs free energy of the reaction. [A], [B], [C] and [D] are activities of A, B, C and D at the given instant). When the reaction is at equilibrium, DG = 0, so, we have

(or) where

È [C] [D] ˘ 0 = DG° + RT ln Í ˙ Î [ A] [B] ˚ eq

…(103)

DG∞ = - RT ln Keq

…(104)

Keq = Equilibrium constant È [C] [D] ˘ =Í ˙ Î [ A] [B] ˚ eq

Substituting the value of DG° from Eq. (104) to Eq. (102),

DG = - RT ln k + RT ln

[C] [D] [ A] [B]

Equation (105) is known as Van't Hoff isotherm. The equation DG° = – RT ln k gives (Since Keq = k)

…(105)

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71

the relation between standard free energy and equilibrium constant. DG° is the standard free energy change when [A] = [B] = [C] = [D] = 1 Applications of Van't Hoff Isotherm 1. If DG° and activities of substances involved in the reaction are known, DG can be calculated. 2. DG° = – RT ln k is used to calculate equilibrium constant if DG° is known and vice versa.

VAN'T HOFF EQUATION (VAN'T HOFF ISOCHORE)

2.17

The effect of temperature on equilibrium constant is quantitatively expressed by Van't Hoff equation. This is obtained by combining Van't Hoff isotherm and Gibbs– Helmholtz equation. DG° = – RT ln k …(106) DG∞ RT Differentiating Eq. (107) with respect to T at constant P, ln k = –

…(107)

∂ ln kP 1 È ∂ ( DG∞ / T ) ˘ =– Í ˙ ∂T RÎ ∂T ˚ d ln kP 1 d ( DG∞ / T ) ˘ = – ÈÍ …(108) ˙ dT RÎ dT ˚ The effect of temperature on free-energy change is predicted by Gibbs–Helmholtz equation. (i.e.)

d ( DG∞ / T ) DH ∞ =– 2 dT T

…(109)

Comparing Eqs (108) and (109), we have d ln kP DH ∞ = dT RT 2

…(110)

Equation (110) is known as Van't Hoff equation which predicts the effect of T on the equilibrium constant.

72 Engineering Chemistry-I

If the reaction is exothermic (DH°= –ive), the equilibrium constant decreases with increase in temperature. If the reaction is endothermic (DH° = +ive), the equilibrium constant increases with decrease in temperature. On integrating Eq. (110) between the limits K 1 at T 1 , K 2 at T2, K2

Ú

DH ∞ d ln k = R

T2

1

ÚT

2

dT

T1

K1

Assuming that DH° is independent of T, T

ln

k2 DH ∞ È 1 ˘ 2 =– k1 R ÍÎ T ˙˚T1

(or)

ln

1˘ DH ∞ È 1 k2 =– ÍT - T ˙ R Î 2 k1 1˚

(or)

ln

DH ∞ È 1 k2 1˘ = - ˙ Í R Î T1 T2 ˚ k1

(or)

ln

k2 DH ∞ È T2 - T1 ˘ = Í ˙ k1 R Î T1 T2 ˚

…(111)

Equation (111) is another form of Van't Hoff isochore. If equilibrium constants at two different temperatures are known, the heat of reaction can be calculated. Similarly, the variation of k (equilibrium constant) with temperature of a reversible reaction at constant volume is

∂ ln k DE∞ = RT 2 ∂T

…(112)

where DE° = Standard internal energy change of a reaction. On integrating Eq. (112), we get ln

k2 DE∞ È T2 - T1 ˘ = Í ˙ k1 R Î T1 T2 ˚

Equations (113) also another form of Van't Hoff isochore. For reactions consisting of only solids and liquids, DH° = DE° \ Eqs (111) and (113) are the same.

…(113)

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73

Integrating Eq. (110) without using limit, we get,

Ú d ln k (or)

=

log k =

DH ∞ R

dT

ÚT

2

- DH ∞ + Constant 2.303 RT

…(114)

The value of constant can be calculated for any reaction by substituting a known value of K at a given T. Equation (114) shows that, a plot of log k vs 1/T should give a straight - DH ∞ . From the slope, DH° can be calculated. line with slope = 2.303 R DH° = – 2.303 R ¥ slope Applications 1. Knowing k1 at one T, we will be able to calculate k2 at another T if DH° of the reaction is known. 2. Easy to predict whether a reaction is favourable at a particular temperature. If Keq value is high, forward reaction is favourable. 3. If forward reaction is exothermic, k will decrease on increasing the T, i.e. low T will favour the forward reaction. Similarly, increase of T will favour endothermic reaction. Problems based on Van't Hoff isotherm and isochore

Problem 2.20: The standard molar reaction enthalpy for N2(g) + 3H2(g) � 2NH3(g) DH°r = – 92.2 kJ at 298 K and equilibrium constant at this temperature is 6 ¥ 105. Calculate k at 400 K assuming DH° is constant for the given temperature range. Solution Given: DH°r = – 92.2 kJ DH°r = – 92200 J K298 = 6 ¥ 105 K400 = ? k DH ∞ È T2 - T1 ˘ ln 2 = Í ˙ k1 R Î T1 T2 ˚ ln k2 – ln k1 =

- 92200 È 400 - 298 ˘ Í ˙ 8.314 Î 400 ¥ 298 ˚

74 Engineering Chemistry-I

ln k2 – ln (6 ¥ 105) = – 11089.7 ¥

102 400 ¥ 298

ln k2 – 13.304 = – 9.4895 ln k2 = – 9.4895 + 13.304 = 3.815 k2 = 45.38 Problem 2.21: At 400°C, the value of equilibrium constant of the reaction N2 + 3H2 � 2NH3 is 1.64 ¥ 10–4 atm. At 500°C, the value of the equilibrium constant is 0.145 ¥ 10–4 atm. Calculate the heat of reaction. Solution Given: T1 = 400 + 273 = 673 K T2 = 500 + 273 = 773 K K1 at 400°C = 1.64 ¥ 10–4 atm K2 at 500°C = 0.145 ¥ 10–4 atm R = 8.314 JK–1 mol–1 DH È T2 - T1 ˘ k2 = Í ˙ 2.303 R Î T1 T2 ˚ k1

log log

0.145 ¥ 10-4 1.64 ¥ 10

-4

=

È 773 - 673 ˘ DH Í ˙ 2.303 ¥ 8.314 Î 673 ¥ 773 ˚

DH = –104.9 kJ Problem 2.22: Calculate the equilibrium constant at 298 K of the reaction N2O4 (g) Æ 2NO2 (g) given that the standard free energies of formation at 298 K are 97540 J mol–1 for N2O4 and 51310 J mol–1 for NO2. Solution

The standard free energy change for the reaction is DG° = DG°Product – DG°Reactant = 2 ¥ 51310 – 97540 = 5080 Jmol –1

Now we know DG° = – RT ln k - DG∞ ln k = RT - 5080 ln k = 8.314 ¥ 298

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75

ln k = –2.0504 k = 0.1286 Problem 2.23: The equilibrium constant K P for a reaction at 327°C and 347°C are 1 ¥ 10 –12 and 5 ¥ 10 –12 respectively. Considering DH° to be constant in the above temperature range, calculate DH° and DS° for the reaction. Solution

According to Van't Hoff equation, log

DH ∞ È T2 - T1 ˘ k2 = Í ˙ 2.303 R Î T1 T2 ˚ k1

Given: k 1 = 1 ¥ 10 –12, k 2 = 5 ¥ 10 –12 R = 1.987 cal k –1 mol –1 T 1 = 600 K, T 2 = 620 K Ê 5 ¥ 10-12 ˆ DH∞ log Á = -12 ˜ 2.303 ¥ 1.98 Ë 1 ¥ 10 ¯

È 620 - 600 ˘ Í ˙ Î 600 ¥ 620 ˚

Ê 5 ¥ 10-12 ˆ Ê 600 ¥ 620 ˆ DH° = 2.303 ¥ 1.987 Á log Á 1 ¥ 10-12 ˜ Ë 620 - 600 ˜¯ Ë ¯

We know At book, \

= = = DG° = k = DG° = = =

2.303 ¥ 1.987 ¥ 18600 ¥ 0.6990 59285.6 cal 59.28 kcal –2.303 RT log k 1 ¥ 10 –12 –2.303 ¥ 1.987 ¥ 600 ¥ log (1 ¥ 10 –12 ) –2.303 ¥ 1.987 ¥ 600 ¥ (–11) +30095.6 kcal (or) 30.09 kcal

DG° = DH° –TDS°

Now,

(o r) DS° =

59.28 - 30.09 DH ∞ - DG∞ = 600 T

= 5 9 . 2 2 k c a l K –1 Problem 2.24: The value of equilibrium constant for a reaction is found to be 10,000 at 25°C. Calculate DG° for the reaction. Solution: DG° = –2.303 RT log k

76 Engineering Chemistry-I

Given: R T K DG°

\

= 1.987 cal k–1 mol–1 = 298 K = 10,000 = –2.303 ¥ 1.987 ¥ 298 ¥ log 10,000 = –2.303 ¥ 1.987 ¥ 298 ¥ 4 = –5454.66 cal

Problem 2.25: Calculate DG° at 300 K for a reaction when KP = 100. Solution According to Van't Hoff isotherm DG° = –2.303 RT log k Given: \

R = 1.987 cal k–1 mol–1, T = 300 K, K = 100 DG° = –2.303 ¥ 1.987 ¥ 300 ¥ log 100 = –2.303 ¥ 1.987 ¥ 300 ¥ 2 = –2745.63 cal

Problem 2.26: The equilibrium constant of a reaction doubles on raising the temperature from 25°C to 35°C. Calculate DH° for the reaction. Solution Given:

T1 = 275 + 25 = 298 K T2 = 273 + 35 = 318 K R = 8.314 k–1 mol–1 ln

DH ∞ È T2 - T1 ˘ k2 = Í ˙ R Î T1 T2 ˚ k1

Ê k2 ˆ ÁË k ˜¯ = 2 1 È TT ˘ DH° = R Í 1 2 ˙ ln2 = 2.303 R Î T2 - T1 ˚ = 53137 J = 58.137 kJ mol–1

È T1 T2 ˘ Í ˙ log2 Î T2 - T1 ˚

Chemical Thermodynamics

77

EXERCISES 1. The equilibrium constant KP for the following reaction is 20.2 atm–1 at 945°C and 9.21 atm–1 at 1065°C. H2(g) + S(g) � H2S(g) Calculate the DH° value for the above reaction. [Ans: – 88.634 kJ mol–1] 2. The equilibrium constant Keq for a reaction is 2 ¥ 10–4 at 473 K and at 483 K is 4 ¥ 10–4. Calculate the value of DH for the reaction (R = 8.314 J). [Ans: 131.668 kJ mol–1] 3. Calculate the entropy change when two moles of an ideal gas expand isothermally and reversibly from an initial volume of 1 litre to a final volume of 10 litres at 300 K. [If R = 1.987 cal K–1 then DS = 9.15 eu, R = 8.314 JK–1, then DS = 38.3 eu] 4. Calculate the entropy change involved in the expansion of 5 moles of an ideal gas at 25°C from a pressure of 10 atm to a pressure of 4 atm. (Given: R = 8.314 Jk–1 mol–1) [Ans: 38.09 JK–1] 5. Calculate the entropy change of n-hexane when one mole of it evaporates at 341.7 K. Enthalpy of vaporization of n-hexane is [Ans: 84.87 JK–1 mol–1] 29 kJ mol–1. 6. Calculate the standard Gibbs free energy change at 298 K for the reaction C(s) + O2(g) Æ CO2(g). Given that DH° = –393.4 kJ mol–1 and DS° = 2.9 JK–1 mol–1 predict whether the reaction is spontaneous or not. [Ans: –394.2 kJ mol–1 , Spontaneous] 7. For a chemical reaction, the value of DH and DS are – 22.6 kcal and 45.2 cal K–1 respectively at 27°C. What is the value of DG for the reaction at 27°C. [Ans: –9040 cal] 8. The free-energy change (DG) accompanying a given process is – 85.77 at 25°C and – 83.68 kJ at 35°C. Calculate the change in enthalpy (DH) for the process at 30°C. [Ans: –148.25 kJ] 9. The free-energy change (DG) for a reaction is found to be – 3.138 ∂DG ˆ kcal at 300 K and ÊÁ for it is –14.39 cal k–1. Find DH for the Ë ∂T ˜¯ reaction at 300 K. [Ans: +1179 cal k–1] 10. The equilibrium constant for a reaction CO2(g) + H2(g) � CO(g) + H2O(g) is 73. Calculate the value of standard free energy change. [Ans: –10.632 kJ mol–1]

78 Engineering Chemistry-I

11. The equilibrium constant kP for the reaction A + B Æ C + D is 1.5 ¥ 10–10 at 300°C and 1.0 ¥ 10–8 at 450°C. Calculate the enthalpy of the reaction (R = 8.314 Jk–1 mol–1)

[Ans: 96.45 kJ]

QUESTIONS Short Questions 1. What do you understand by the term internal energy? [P.38] 2. What are spontaneous processes? Give examples. [P.54] 3. Define entropy. What is its unit? [P.39] 4. State which of the following will have more entropy and why. [P.45–49] (a) Ice and water (b) Water at 100°C (or) steam at 100°C (c) A gas placed in a container of 1 litre or of 2 litre capacity. 5. State the second law of thermodynamics in two forms. [P.39] 6. Write briefly on the physical significance of entropy. [P.45–47] 7. What is meant by entropy? Mention some of the important formulas relating the entropy calculations. [P.42–47] 8. Define the terms (i) system, and (ii) surroundings. [P.34] 9. Define the terms closed system and isolated systems. [P.35] 10. Mention the applications of Gibbs–Helmholtz equation. [P.59] 11. Give the following equations: (i) Gibbs–Helmholtz equation [P.56–59] (ii) Van't Hoff isotherm. [P.69–71] 12. What are path functions? Give examples. [P.36] 13. What are state functions? Give examples. [P.36] Descriptive Questions 1. Derive the Gibbs–Helmholtz equation and discuss its application. [P.56–59] ∂ T P ∂ ˆ =–Ê ˆ [P.65–66] 2. Prove that ÊÁ ÁË ˜¯ Ë ∂V ˜¯ S ∂S V 3. Derive the Van't Hoff equation. [P.71–73] 4. Explain the term free energy and maximum work function. Derive the relationship between two. [P.50–51] 5. Derive an expression for entropy change in an isothermal expansion of an ideal gas. [P.42–43] ∂S ∂V ˆ 6. Prove ÊÁ ˆ˜ = – ÊÁ Ë ∂P ¯ T Ë ∂T ˜¯ P

[P.68–69]

Chemical Thermodynamics

79

7. Derive an expression for the variation of equilibrium constant of a reaction with temperature. [P.71–72] 8. Explain the thermodynamic criteria of spontaneity. [P.55–56] 9. Prove that (i) dE = TdS – SdT

[P.65]

(ii) dH = TdS + VdP

[P.68]

10. Prove that (i) – dA = PdV + SdT

[P.67]

(ii) dG = VdP – SdT

[P.63]

11. Derive an expression for Vant Hoff's isochore. [P.71] 12. What is the influence of DH and DS value in the spontaneity of the reaction? [P.55–56]

Two-Mark Questions and Answers 1. What are homogeneous and heterogeneous systems? The system is completely uniform throughout. (i.e) a single phase is called homogeneous system. e.g pure single solid (or) liquid or mixture of gases. The system is not uniform throughout (i.e) consists of two or more phases. e.g: Ice in contact with water. 2. What are state and path variables? Give an example. The state of a system changes with change in any of the thermodynamic variables like pressure, temperature, volume and composition. e.g: Ice melts into liquid water. Path variables which depend on the path through which a system attains the final state. e.g: Heat, Work, etc. 3. Why is entropy a state function? The entropy change (DS) in a process depends only on the initial and final state. So entropy is a state function. 4. Define intensive properties. It is defined as the properties which are independent of the amount of the substance present in the system. e.g: Temperature, Pressure, density, etc. 5. Define extensive properties. It is defined as the properties which depends on the amount of the substance present in the system. e.g: Mass, Volume, Internal energy (E), etc.

80 Engineering Chemistry-I

6. Why is internal energy of a system a state function? Internal energy of a system depends only on the state of the system and not upon how the system attains the state. 7. Under what condition does the entropy of a substance become zero? The substance should be perfectly crystalline and the temperature is absolute zero. 8. Why is work not a state function? Work depends upon the path followed. So work is not a state function. 9. Define entropy. Entropy is a measure of degree of disorder (or) randomness in a system. It is also considered a measure of unavailable form of energy. Entropy is a measure of degree of disorder (or) randomness in a system. It is also considered a measure of unavailable form of energy. Unavailable energy Entropy = Temperature 10. Explain spontaneous process. A process which occurs of its own, without any outside help, is called spontaneous reaction. e.g: evaporation of water. 11. What is entropy of vaporisation? It is the entropy change when one mole of a liquid is transformed to the vapour state at its boiling point. D Svapour = Svap – Sliquid =

DH vapour Tb

12. Define entropy of fusion. It is defined as the change in entropy when one mole of a solid substance is transformed into the liquid state at its melting point. D Sfusion = DS liquid – DS Solid =

DHfusion Tm

13. Explain nonspontaneous process? It is a process which cannot proceed by itself and is known as non spontaneous process. e.g: Formation of ice in a refrigerator, Electrolysis of water.

Chemical Thermodynamics

81

14. Write the applications of Van’t hoff isotherm. 1. If DG° and activities of substances involved in the reaction are known DG can be calculated. 2. DG° = – RT ln k is used to calculate equilibrium constant if DG° is known and vice verga. 15. How is internal energy change related to heat and work? DE = q – w. 16. How does the entropy changes when (a) a solid is melted? (b) a gas is liquefied? (c) H2O is frozen? (a) A solid is melted entropy increases. (b) A gas is liquefied, the entropy decreases. (c) H2O is frozen entropy decreases.

Difnjdbm Uifsnpezobnjdt V O J

2.1

Introduction

2.2

Terminology of Thermodynamics

2.3

Second Law of Thermodynamics

2.4

Entropy

2.5

Entropy Change for an Ideal Gas

2.6

Entropy Change for Reversible Process

2.7

Entropy Change for Irreversible Process

2.8

Entropy Change in Physical Transformation

2.9

Free Energy

2.10 Helmholtz Free Energy Function and Work

U

2.11 Gibbs Free Energy Function and Work 2.12 Criteria of Spontaneity 2.13 Gibbs-Helmholtz Equation 2.14 Clausius-Clapeyron Equation

JJ

2.15 Maxwell Relations 2.16 Vant Hoff Isotherm 2.17 Vant Hoff Isochore

3 D

Difnjdbm!Uifsnpezobnjdt

I B Q U F S

INTRODUCTION

2.1

Thermo means heat and dynamics means flow. Thermodynamics means the study of flow of heat. Thermodynamics is a branch of science which mainly gives a quantitative treatment of energy changes from one form to another during physical and chemical transformation. It tells us the direction in which changes take place in nature but does not say anything about the rate of change of a process. The thermodynamic principles are explained in four laws. 1. The first law of thermodynamics tells about the conservation of energy and the concept of internal energy. 2. The second law of thermodynamics explains the limits of converting internal energy into work and explains the concept of entropy and reaction feasibility. 3. The third law of thermodynamics provides data for the measurement of entropy. 4. The zeroth law of thermodynamics governs thermal equilibrium among systems in physical contact. Thermodynamic study can predict the feasibility of a chemical reaction. It provides important information regarding the stability, yield or efficiency, optimum range of temperature and pressure. Thermodynamics helps determine the extent a process can proceed before attainment of equilibrium. The major applications of thermodynamics are in all fields of energy technology as in steam engines, internal combustion engines, air conditioning, refrigeration, chemical industrial plants, etc.

34 Engineering Chemistry-I

Limitations of Thermodynamics 1. Thermodynamics is mainly applicable to macroscopic systems. The microscopic internal structure of atoms, ions and molecules cannot be explained. 2. Thermodynamics does not say anything about the rate of physical or chemical change of reaction. It concerns only the initial and final states of the system.

TERMINOLOGY OF THERMODYNAMICS

2.2

The major terms used frequently in thermodynamics are the following: 1. System A system is any part of the universe which is isolated from its surrounding by means of a real or imaginary boundary. A system contains a definite quantity of substance or matter. 2. Surroundings The remaining part of the universe other than the portion of the system is called surroundings. The surrounding is separated from the system by a definite boundary. 3. Boundary It is the region or interface separating the system from the surrounding. 4. Types of Systems (a) Open System An open system can exchange heat as well as matter with the surroundings.

Example Open beaker containing hot water. The heat and water (matter) can be transferred either from the system to the surroundings or from the surroundings to the system. There is an open boundary between the system and surroundings. Vapour Heat

Heat Water

Heat

Fig. 2.1 Open system

Chemical Thermodynamics

35

(b) Closed System A closed system which can exchange heat energy but not matter with its surroundings.

Example Water is taken in a closed non-insulated vessel, the heat energy can be exchanged but not matter with the surroundings. Heat

Vapour Heat

Heat Water

Heat

Fig. 2.2

Closed system

(c) Isolated System An isolated system is totally separated from the surroundings by an insulated boundary.

This system cannot exchange both heat energy and matter with its surroundings. It has no mechanical and thermal contact with the surroundings. Example Water in contact with its vapour in a closed insulated vessel.

Fig. 2.3

Isolated system

36 Engineering Chemistry-I

Depending on the physical state of matter, systems can further classified as follows: (a) Homogeneous System The system is completely uniform throughout, i.e. a single phase is called homogeneous system.

Example (i) Pure single solid (or) liquid (or) mixture of gases (ii) True solution (solid solute miscible in liquid solvent) (b) Heterogeneous System The system is not uniform throughout, i.e. consists of two or more phases.

Example (i) Ice in contact with water (liquid). (ii) Ice in contact with water (vapour). 5. State Variables (or) Functions The state of a system changes with change in any of the thermodynamic variables like pressure, temperature, volume and composition. Ice melts into liquid water by applying heat. The reverse reaction takes place when liquid water is cooled. This change depends on temperature and so temperature is a state variable. The other state variables are pressure, volume, energy, enthalpy, entropy and free energy. 6. Path Variables (or) Functions Path variables which depend on the path through which a system attains the final state.

Example Heat (q), work (w), etc. 7. Properties of a System The properties of a system can be divided into two types: (a) Extensive Properties Such properties depend on the amount of the substance present in the system.

Example Mass, volume, internal energy (E), enthalpy (H), entropy (S), free energy (G)

Chemical Thermodynamics

37

(b) Intensive Properties Such properties are independent of the amount of the substance present in the system.

Example Temperature, pressure, density, concentration, surface tension, boiling point, freezing point, refractive index, etc. 8. Types of Processes of a System A system may change from one state to another by path. The path by which this change of state occurs is called a process.

When a system changes from one state to another, it is accompanied by change in energy. In an open system, there may be change of matter also. (a) Isothermal Process The temperature of the system remains constant during each stage of the process.

This can achieved by placing the system in a thermostat (dT = 0). (b) Adiabatic Process It is a process in which no heat can flow into or out of the system.

This can achieved by carrying out the process in an insulated container (dq = 0). (c) Isobaric Process The process which occurs at constant pressure is known as isobaric process.

DE = q – PDV (d) Isochoric Process The process which occurs at constant volume is known as isochoric process (dV = 0). (e) Cyclic Process If a system in a given state goes through a number of different process and returns to its final state, the overall process is known as cyclic process: dE = 0, dH = 0

Example Carnot cycle (f) Reversible Process It is a process in which the state of the system is changed infinitesimally slowly so that its direction at any point can be reversed by a small change in any of the variables.

Ice(solid)

Heat Cool

Water(liquid)

38 Engineering Chemistry-I (g) Irreversible Process It is a process in which the state of the system is changed rapidly so that its direction at any point cannot be reversed by a small change in any of the variables.

Example Naturally occurring processes like rain, waterfalls from hills, etc. (h) Internal Energy (E) It is the total energy content of the system. The total energy possessed by the molecule is the sum of electronic energy, translational kinetic energy, rotational kinetic energy, vibrational and nuclear energy.

E = Ee + Et + Er + Ev + En

First Law of Thermodynamics It is the law of conservation of energy. This law states that “energy can neither be created nor destroyed, but can be converted from one form to another form.” The total energy of the system and its surroundings must remain constant. Mathematically, where

DE DE q w

=q–w = Change in internal energy = Quantity of heat absorbed = Work done

1. Limitations of the First Law (or) Need for the Second Law of Thermodynamics.

(a) The first law is concerned only with energy changes but not the direction in which the energy change takes place. (b) It does not say whether a reaction is feasible or not. 2. Advantages of the Second Law of Thermodynamics

(a) It gives the direction of flow of heat. (b) It predicts whether a reaction is spontaneous or not. (c) It helps us to calculate the maximum fraction of heat that can be converted into work.

SECOND LAW OF THERMODYNAMICS

2.3

Work can always be converted into heat but heat cannot be completely converted into work without leaving a permanent change in the system or surroundings.

Chemical Thermodynamics

39

The second law talks about not only conversion of energy but also the extent of conversion with direction. In the second law, only a fraction of heat can be converted into work and the rest remains unavailable and unconverted. This law is based mainly on the observations and experiences and is applicable to cyclic process only. Statements of Second Law (a) In Terms of Flow of Heat Heat never flows from a cooler to a hotter body on its own. (b) In Terms of Entropy The entropy of the universe is increasing continuously and attains maximum randomness. (c) Clausius Statement It is impossible to construct a heat engine which can transfer heat in a cyclic process from a cold reservoir to a hot reservoir, unless some external work is done on the engine. (c) KelvinŪPlanck Statement It is impossible to take heat from a hot reservoir and convert it into work by a cyclic process without transferring a part of the heat to a cold reservoir.

ENTROPY

2.4

Entropy is a measure of degree of disorder or randomness in a system. It is also considered a measure of unavailable form of energy. The second law says that conversion of the entire amount of energy is not possible, but only partial conversion is practically achieved. The remaining unavailable energy is used to create randomness in a system. This unavailable energy is entropy. The unit for entropy is calories deg–1 mol–1 (or) entropy unit. In SI units, it is joules K–1 mol–1. Entropy is a state function, it is an extensive property, and its value depends upon amount of the substance involved. Entropy =

Unavailable energy Temperature

40 Engineering Chemistry-I Mathematical Expression of Entropy From the Carnot cycle, the efficiency of an engine is

h= (or)

1-

q2 - q1 T2 - T1 = T2 q2

q1 T = 1- 1 T2 q2 q q1 = 2 T2 T1

and

q2 = heat absorbed by the system at temperature T2 q1 = heat rejected to the sink at temperature T1 Heat adsorbed ¸ -Heat rejected ˝ = to the sink by the system ˛ q q2 =- 1 T1 T2

\ \

q1 q + 2 =0 T1 T2 When the system exchanges heat reversibly and infinitesimally, S

dq =0 T

S is entropy and change in entropy ds, for infinitesinally small change ds =

dq (or) T

DS =

qrev T

…(1)

This is the mathematical statement of entropy. According to the Carnot cycle, the entropy of the universe is ever increasing.

ENTROPY CHANGE FOR AN IDEAL GAS

2.5

Entropy is a state function and its value depends on variables such as temperature (T), pressure (P) and volume (V).

2.5.1

When Temperature (T) and Volume (V) are the Variables

When n moles of an ideal gas occupy a volume (V) at temperature (T), it absorbs an infinitesimally small amount of heat dqrev at temperature T.

Chemical Thermodynamics

41

The increase in entropy is dqrev …(1) T = dE – dw (according to first law of thermodynamics)

ds = dqrev

The work involved is expansion of gas dw = –PdV Thermodynamically, we know, dE Cv = ÊÁ ˆ˜ Ë dT ¯ v \

…(2)

dE = n Cv dT

where Cv is the molar heat capacity at constant volume of the gas. \

dqrev = n Cv dT + PdV

…(3)

According to ideal gas law, PV = nRT

…(4)

dqrev = n Cv dT + \

ds =

nRT dV V

dqrev T

…(5) …(6)

Substituting Eq. (5) in Eq. (6), nCV dT + nRT . ds =

dV V

T

ds = n CV

dT dV + nR . T V

…(7)

Integrating Eq. (7) with respect to limits, S2

Ú ds = nCV

S1

T2

Ú

T1

DS = n Cv ln or

V

2 dT dV + nR Ú T V V

…(8)

1

T2 V + nR ln 2 T1 V1

DS = 2.303 n Cv log

T2 V + 2.303 n R log 2 T1 V1

…(9) …(10)

42 Engineering Chemistry-I

2.5.2

When Temperature (T) and Pressure (P) are the Variables P1V1 = n R T1 and P2 V2 = nRT2 PV P1V1 = 2 2 T2 T1 PT V2 = 1 2 P2T1 V1

Substituting the value of DS = nCv ln = nCv ln

…(11)

V2 from Eq. (11) in Eq. (9), V1 PT T2 + nR ln 1 2 P2T1 T1 T2 P T + nR ln 2 + nR ln 1 T1 P2 T1

…(12)

We know Cp – Cv = R Cv = Cp – R

…(13)

Substituting the value of CV from Eq. (13) to Eq. (12), P T T2 + nR ln 2 + nR ln 1 P2 T1 T1 T P DS = n Cp ln 2 + nR ln 1 T1 P2 DS = n (Cp – R) ln

DS = nCp ln

T2 P – nR ln 2 T1 P1

DS = 2.303 nCp log

2.5.3

…(14) …(15) …(16)

P T2 – 2.303 nR log 2 P1 T1

…(17)

Entropy Changes of an Ideal Gas in Various Processes

1. Isothermal Process This process is carried out at constant temperature.

DT = T2 – T1 = 0 T1 = T2 Hence, Eqs (10) and (17) become DS T = 2.303 nR log

V2 V1

…(18)

Chemical Thermodynamics

DS T = – 2.303 nR log DS T = 2.303 nR log

43

P2 P1

P1 P2

…(19)

2. Isobaric Process The pressure remains constant in this process (DP = 0)

\

P1 = P2 Hence, Eq. (17) becomes

\

DSp = 2.303 nCp log 10

T2 T1

…(20)

3. Isochoric Process The volume remains constant in this process (DV = 0) \ V 1 = V2 Hence, Eq. (10) becomes

DSv = 2.303 nCv log

T2 T1

…(21)

In open or closed systems, the total entropy (Eq. 17) change (DStotal) will be equal to the sum of the change in entropy of the system (DSSystem) and the change in entropy of the surroundings (DSSurroudings). DStotal = DSSystem + DSSurroudings For a spontaneous process, DStotal should be greater than zero. The entropy of the universe always increases due to spontaneous changes. DStotal > 0

(Spontaneous)

For a system in equilibrium, the entropy change should be equal to zero. DS = 0 (Equilibrium) For a nonspontaneous process, the entropy change DStotal will always be negative DStotal < 0

ENTROPY CHANGE FOR REVERSIBLE (NONSPONTANEOUS) PROCESS

2.6

In an isothermal and reversible process, the system absorbs q heat from the surroundings at temperature T and the entropy change is

44 Engineering Chemistry-I

q T At the same time, the entropy change in the surroundings decreases due to the loss of the same quantity of heat ‘q’. q DSSurroudings = T Hence, the net change in the entropy in the process is given as DSSystem =

DS(rev) = DSSystem + DSSurroundings q Ê - qˆ + Á ˜ = 0 eu T Ë T¯ Therefore, in a reversible isothermal process, there is no net change in entropy. DS(rev) =

ENTROPY CHANGE FOR IRREVERSIBLE (SPONTANEOUS) PROCESS

2.7

When a system at higher temperature T2 transfers q amount of heat irreversibly to its surroundings maintained at a temperature T1. -q Then the decrease in entropy DSSystem = T2 The increase in entropy \ change in entropy

DSSurroudings =

+q T1

DSirreversible = DSSystent + DSSurroudings =

-q q + T2 T1

1. Entropy of Vaporisation It is the entropy change when one mole of a liquid is transformed to the vapour state at its boiling point.

DSvapour = Svap – Sliquid =

DH vapour Tb

where DHvap is the enthalpy of vaporisation per mole. Tb is the boiling-point temperature of the liquid in kelvin.

Chemical Thermodynamics

45

2. Entropy of Sublimation It is the change in entropy involved in the transformation of a solid directly into the vapour state at its sublimation temperature.

DSSubli = Svap – SSolid =

DHSubli TS

where DHSubli is the enthalpy of sublimation per mole, TS is the sublimation temperature in kelvin. È T - T1 ˘ = qÍ 2 ˙ Î T1T2 ˚ Since T2 > T1, (T2 – T1) is a positive value. \

DSirrever > 0

All spontaneous processes are irreversible and hence the entropy increase will accompany every spontaneous process.

ENTROPY CHANGE IN PHYSICAL TRANSFORMATION

2.8

Entropy change takes place even when the system undergoes physical transformations such as fusion, vaporization and sublimation from one crystalline form into another.

2.8.1

Entropy of Fusion

It is defined as the change in entropy when one mole of a solid substance is transformed into the liquid state at its melting point. DSfusion = DS liquid – DS Soild =

DHfusion Tm

where DHfusion is the enthalpy of fusion per mole and Tm is the melting temperature at kelvin. Problems

Problem 2.1: Helium, weighing 4 g, is expanded from 1 atm to one tenth to the original pressure at 30°C. Calculate the change in its entropy assuming it to be an ideal gas.

46 Engineering Chemistry-I

Solution:

Entropy change of varying pressure of an ideal gas is DS = 2.303 nR log

Number of moles

n=

P1 P2

Weight in gram 4 = Molecular weight 4

= 1 mole R = 1.987 cal k–1 mol–1

P1 = 1 atm P2 =

DS = 2.303 ¥ 1 ¥ 1.987 ¥ log

1 atm 10

1 1 10

= 4.57 eu Problem 2.2: vided

Calculate the entropy change for the reaction at 25°C pro2H2 + O2 Æ 2 H2O (g)

(g)

(l)

The standard entropies of gaseous hydrogen, oxygen and water liquid are 130.6 J k–1 mol –1, 205.1 J k–1 mol–1 and 70 J k–1 mol–1 respectively. Is the reaction spontaneous of the temperature? Solution:

The entropy change of the reaction is DS = S S Product – S Sreactant = S 2S°H2O – S 2S°H2 + S°O 2 = (2 ¥ 70) – [(2 ¥ 130.6) + 205.1] = –326.3 J k–1 mol–1

DS is – ive which indicates that the reaction is not spontaneous. Problem 2.3: One mole of an ideal gas expands isothermally to twice its original volume at 27°C. Calculate the entropy change during the process. Solution:

Entropy change in an isothermal expansion of an ideal gas DS = 2.303 nR log

V2 V1

Chemical Thermodynamics

47

Data: n = 1 mole, V1 = 1 and V2 = 2 R = 1.987 cal K–1 mol–1 DS = 2.303 ¥ 1 ¥ 1.987 log 2/1 = 1.377 cal K–1 mol–1 or eu Problem 2.4: What are the entropy changes for the reversible vaporization and fusion of one mole of water at 100°C and 0°C respectively? DH(vaporization) = 9.7171 kcal mol–1

Given that

DH(fusion) = 1.4363 kcal mol–1 Solution DS(System) = – DSSurroundings + DS(Universe) = 0 For the vaporisation, DS(System) =

n.DH( v ) TK

=

1.0 ¥ 9717.1 cal mol -1 = 26.04 eu 373.15 K

For the fusion, DS(System) =

1.0 ¥ 1436.3 cal mol -1 273 K

= 5.258 eu Problem 2.5: 36 g of water is heated from 300 K to 310 K at constant pressure. Calculate the increase in entropy for this process. The molar heat capacity at constant pressure for water is 75.3 JK–1 mol–1. Solution DS = n Cp ln

T2 T1

36 = 2 mol 18 Cp = 75.3 J k–1 mol–1 n=

T1 = 300 K, T2 = 310 K DS = 2 mol ¥ 75.3 J K–1 mol–1 ¥ 2.303 log = 4.94 J K–1 mol–1

310 300

Problem 2.6: Calculate the entropy changes for the fusion and vaporisation of 1 mol of water at 273 K and boiling point at 373 K. The molar enthalpy of fusion is 6.01 kJ mol–1 and the molar entropy vaporisation is 40.79 kJ mol–1.

48 Engineering Chemistry-I

Solution (i) Fusion entropy change (DSfusion) DS(fusion) =

DHfusion Tf

DHfusion = 6.01 k J mol–1 = 6010 J mol–1, Tf = 273 K \

DHfusion =

6010 J mol -1 = 2201 J k–1 mol–1 273 K

(ii) Vaporisation entropy change DSvap = =

DH vap Tb 40700 J mol -1 373 k

DHvap = 40.7 k J/mol–1 = 40700 J mol–1 Tb = 373 K

= 109.1 J k–1 mol–1 Problem 2.7: 2 moles of an ideal gas at 27°C expand isothermally from a volume of 20 litres to 60 litres. Calculate the change in entropy. Solution DST = 2.303 nR log

V2 V1

Ê 60 ˆ DS300 K = 2.303 ¥ 2 ¥ 8.314 log Á ˜ Ë 20 ¯ = 18.3 J K–1 Problem 2.8: Calculate the change in entropy accompanying the isothermal expansion of 4 moles of an ideal gas at 300 K until its volume has increased three times. Solution DS = 2.303 nR log

V2 V1

= 2.303 ¥ 4 ¥ 1.987 ¥ log = 8.733 cal

3 1

Chemical Thermodynamics

49

Problem 2.9: Calculate the entropy change when one mole of water is converted into steam, reversibly at its boiling point. Latent heat of vaporisation is 540 cal g–1. Solution DS =

DH vap Tb

= =

Lvap ¥ M T 540 ¥ 18 373

DS = 20.06 cal K–1 mol–1 Problem 2.10: Calculate the entropy change accompanying the transformation of 1 mole of rhombic sulphur to monoclinic sulphur at 93°C, the heat of transformation being 412 J/mol. Solution DStrans =

=

DH trans Ttrans 412 J mol -1 (273 + 93) K

DStrans = 1.125 J/mol/k Problem 2.11: Calculate the entropy change accompanied by melting of 5 kg of ice at 0°C, given that the heat of fusion is 120 calories per gram. Solution DSfusion =

-1 -1 DHfusion 120 cal g ¥ 4.18 J cal ¥ 1000 g/kg = (273 + 0) K Tfusion

DSfusion = 1837.36 J kg–1 K–1

FREE ENERGY

2.9

To express the spontaneity of a process more directly, the another thermodynamic function, free energy has been introduced by Jossiah Willard Gibbs in 1876. The free-energy concept comes from the second law of thermodynamics. According to this law, the total energy is not completely converted into work and a part of it is used to increase the randomness, which is the unavailable part of energy.

50 Engineering Chemistry-I

Total energy ¸ Ê Energy convertedˆ Ê Energy lost byˆ ˝=Á ˜¯ - ÁË giving to sink ˜¯ supplied ˛ Ë into work The energy converted into work is otherwise termed as available energy to do useful work. This is known as free energy. The two free-energy functions are (1) Helmholtz free-energy function (A), and (2) Gibbs free-energy function (G) The Helmholtz free-energy function is known as work function which reveals the maximum work. The Gibbs free-energy function is known as net work function. The importance of both free energy functions are given below.

HELMHOLTZ FREE-ENERGY FUNCTION (A) AND WORK

2.10

The Helmholtz free-energy function A is given by A = E – TS

…(22)

where E is the internal energy and S is the entropy. The change in free energy A is given by DA = DE – TDS We know that change in entropy q DS = rev T

…(23)

… (24)

Substituting Eq. (24) in (23), \

DA = DE – qrev

…(25)

But from the mathematical statement of the first law of thermodynamics, q = DE + PDV we get (or)

DE = q rev – PDV

…(26)

Chemical Thermodynamics

51

Substituting the value of DE from Eq. (26) in Eq. (25), we get

DA = qrev – PDV – qrev DA = – PDV fi – Wrev

(or)

–DA = Wrev

Wrev is the maximum work and the negative sign for DA tells us that decrease in the Helmholtz free-energy function is the maximum work obtainable from the system under isothermal conditions.

GIBBS FREE-ENERGY FUNCTION AND WORK

2.11

The Gibbs free-energy function (G) is otherwise simply called the free energy and is given by G = H – TDS …(27) where H is the enthalpy, and S is the entropy function. The change in free energy with respect to the initial and final stages of the systems is DG = DH – TDS …(28) We know that entropy is related to internal energy and work function by the relation \

DH = DE + PDV

…(29)

DG = DE + PDV – TDS

…(30)

But from Helmholtz free energy, DA = DE – TDS

…(31)

Substituting DA form Eq. (31) in Eq. (30), (or)

DG = DA + PDV DG = –Wrev + PDV

(or)

DG = –Wrev + PDV

(at constant pressure) (Since DA = –Wrev)

The two work functions in the RHS cannot get cancelled because both are different. The term Wrev represents electrical work, surface work other than the work of expansion. The term PDV represents only the work of expansion. The term Wmax is greater than PDV.

52 Engineering Chemistry-I

Therefore, (Wrev – PDV) may be called net work represented by Wnet . Now, – DG = Wrev – DG fi Wnet This means Decrease in Gibbs ¸ ˝ = [ Net work ] ˛

free energy

Problem 2.12: DH and DS for a reaction at 27°C are –10 kcal and 20 cal K–1 respectively. What is DG for the reaction? Predict whether the reaction will be feasible or not. Solution DG = DH – TDS DH = –10 kcal = –10,000 cal

Given:

DS = 20 cal K–1 T = 27 + 273 = 300 K \

DG = –10,000 – 300 (20) = –16000 cal DG = –16 kcal (The reaction is feasible)

Problem 2.13: 25°C.

Predict whether the following reaction is spontaneous at C (s) + H2O (l) Æ CO (g) + H 2 (g)

Given:

DH = 31.4 kcal mol–1 and DS = 32 cal deg–1 at 25°C

Solution We know that DG = DH – T D S Given: DH = 31.4 kcal

or

31.4 ¥ 103 cal mol–1

DS = 32 cal deg–1 T = 25 + 273 = 298 K \

DG = DH – TDS = 31.4 ¥ 10–3 – 298 (32)

Chemical Thermodynamics

53

= + 21864 DG = + 2.186 kcal (The reaction is non-spontaneous) Problem 2.14: DG for a reaction at 300 K is –16 kcal. DH for the reaction is –10 kcal. What is the entropy of the reaction and what will be DG at 330 K? Solution DG = DH – TDS DH - DG T -10 - (-16) = 300

DS =

DG = – 16 kcal, DH = –10 kcal T = 300 K

DS = 0.02 kcal K–1 At 330 K,

DG = DH – TDS = –10 – 330 (0.02) = –10 – 6.6 DG = –16.6 kcal

Problem 2.15: Calculate the standard free energy change for the reaction H2(g) + I2(g) Æ 2HI (g). Given that the standard enthalpy change and standard entropy change for the reaction are 52 k J and 165.2 J k–1 respectively. Is this reaction spontaneous at 298 K? Solution DG° = DH° – TDS° Given: DH° = 52 kJ or 52 ¥ 103 J DS° = 165.2 J K–1 T = 298 K \

DG° = 52 ¥ 103 – 298 (165.2) = + 2770.4 k J DG° = + 2.77 k J (Non-spontaneous reaction)

54 Engineering Chemistry-I

CRITERIA OF SPONTANEITY

2.12

1. Spontaneous Process A process which proceeds of its own under a given set of conditions is called as spontaneous process. It occurs in one direction only and is called a natural process. The tendency of a process to occur spontaneously is called spontaneity. Examples • Evaporation of water • Ice melting • Heat flows from hot body into a cold body • Water falls from hills 2. Nonspontaneous Process It is a process which cannot proceed by itself and is known as nonspontaneous process. Examples

• Formation of ice in a refrigerator takes place as long as the electric current is supplied to it. • Electrolysis of water takes place as long as the electric current is passed to produce H2 and O2. The feasibility (spontaneity) of physical and chemical processes can be explained by two new thermodynamic functions: entropy and free energy. Increase in entropy means change from an ordered to less ordered or disordered state. For spontaneous process, the entropy increases. Spontaneous Change

‘A’ State Highly ordered Low entropy (less probable)

Fig. 2.4

‘B’ State Highly disordered High entropy [more probable]

Spontaneous and nonspontaneous processes

Chemical Thermodynamics

55

The change from A state to B is spontaneous. So we can say that a change in a system which is accompanied by an increase in entropy tends to be spontaneous.

Thermodynamic Criteria for the Spontaneous Process at Constant Temperature and Pressure From thermodynamics, H = E + PV

…(32)

The isothermally available energy present in a system is called free energy (G) G = H – TS

…(33)

Substitute Eq. (32) in Eq. (33), we get G = E + PV – TS The change in free energy (DG) is DG = DE + D(PV) – D(TS)

…(34)

For constant T and P, D(PV) becomes P and V, and D (TS) becomes TDS. Now, Eq. (34) becomes (DG)T,P = DE + PDV – TDS

…(35)

At constant T and P, DH = DE + PDV So Eq. (35) becomes (DG)T,P = DH – TDS

…(36)

To develop the criteria of spontaneity in terms of free energy change, DSTotal = DS Sys + DS Surr ¸ -q Ô change of the surroundings ˝ DS Surr = T Ô is given by ˛

…(37)

The decrease in entropy

…(38)

(Negative sign indicates that heat has been lost by the surroundings) Heat gained by the system q = (DH)sys. (or) DSSur =

- q -( DH )Sys = T T

…(39)

Qipupdifnjtusz boe!!Tqfdusptdpqz V O

3.1

Photochemistry-Introudction

3.2

Laws of Photochemistry

3.3

Beer-Lambert's Law

3.4

Quantum Yield

3.5

Experimental Determination of Quantum Yield

J U

3.6

Photophysical Process

3.7

Energy Transfer in Photochemical Reaction

3.8

Chemiluminescence

3.9

Spectroscopy

3.10 Electromagnetic Spectrum 3.11 Adsorption and Emission of Radiation 3.12 Classification of Molecular Spectra 3.13 Ultraviolet and Visible Spectroscopy

JJJ

3.14 Infrared Spectroscopy

4 D

I B Q U F S

3.1

Qipupdifnjtusz!boe! Tqfdusptdpqz

PHOTOCHEMISTRYŪINTRODUCTION

3.1

In general, chemical reactions are categorized into two major categories depending on the source of energy used. They are 1. Photochemistry

2. Radiation chemistry

1. Photochemistry Photochemistry mainly deals with the study of chemical reactions produced directly or indirectly by photons or light absorption. Many photochemical reactions of different kinds are

• • • • •

Synthesis Decomposition Polymerisation Isomeric change Reduction and oxidation

These reactions can be initiated and carried out by the absorption of radiation lying in the wavelengths approximately between 200 nm to 800 nm regions. The study of chemical reactions which occur after electronic excitation of atoms or molecules with electromagnetic radiation, having wavelengths approximatly 200 nm to 800 nm (UV and visible regions), are called photochemical reactions. For example, (a) Reaction between H2 and Cl2 : dark æææ Æ No reaction (i) H2 + Cl2 æwithout light

hg

(ii) H2 + Cl2 ææÆ 2 HCl

86 Engineering Chemistry-I

(b) Simple molecules like CH4, NH3, H2O and CO2 exposed with light hg undergo complex photochemical reactions, for example proteins and nucleic acid. (c) Photosynthesis in the leaves of plants by means of sunlight. Sunlight (hg )

Æ C6H12O6 + 6 CO2 6 CO2 + H2O æææææ Chlorophyll glucose Thousands of these glucose molecules are then combined to form one or other of two giant carbohydrate molecules, for example cellulose and starch. 2. Radiation Chemistry The chemical reaction effects are produced by high-energy radiations like x-rays and g-rays. These reactions are studied under radiation chemistry. (a) Thermal Reactions These chemical reactions takes place by the absorption of thermal (heat) energy. (b) Dark Reaction These chemical reactions takes place in the absence of light radiation. Table 3.1 Differences between photochemical and thermal reactions S. No.

Photochemical reactions

Thermal reactions

1.

These reactions involve absorption These reactions involve the of light radiation. absorption or evaluation of heat.

2.

These occur only in the presence of These occur in dark as well as in light. light.

3.

Change in free energy (DG) for DG for these thermochemical these reactions may be positive or reactions is always negative. negative.

4.

The effect of temperature has very Temperature has a significant effect little influence. on the rate of such reactions.

LAWS OF PHOTOCHEMISTRY

3.2

We can expect some relationship between the light absorbed and the chemical change because all photochemical reactions take place as a result of the absorption of radiation.

Photochemistry and Spectroscopy

87

Photochemical reactions are governed by three basic laws. 1. The Grotthuss–Draper law 2. The Stark–Einstein law of photochemical equivalence 3. The Lambert–Beer law

3.2.1

The GrotthussŪDraper Law

Any molecule which absorbs light will undergo photochemical reaction. (or) Only the light radiation absorbed by the reacting system (molecules) are effective in producing chemical change. It is not necessary to mean that the absorbed light will always bring about the chemical reaction. Some molecules that absorb light are converted to kinetic energy and in a majority of the cases, the absorbed light may get re-emitted in the form of light of lower or the same frequency, known as fluorescence or phosphorescence.

3.2.2

The StarkŪEinsteinŭs Law of Photochemical Equivalence

Based on quantum theory of radiation, Einstein proposed the law of photochemical equivalence. The quantitative aspect of photochemical reactions was studied by Stark and Einstein as the quantum theory of light. Each molecule which takes part in a photochemical primary process absorbs one photon or quantum of light causing the reaction.” The general primary photochemical process is R + hg (Reactants) Light molecule

R* Excited reactant molecule

P Product molecule

According to the equation, one mole of the reacting species absorbs one mole of photons or one Einstein of energy (E). E = Nhg The energy absorbed per mole is known as one einstein. This is given by c E = Nhg = Nh l

88 Engineering Chemistry-I

where N is Avogadro number (6.023 ¥ 1023 mol–1) h is Planck’s constant (6.626 ¥ 10–34 J.S) C is the velocity of light (3 ¥ 108 ms–1) l is the wavelength of light To calculate the value of einstein for a radiation of 250 nm wavelength, E=

3.2.3

Nhc 6.023 ¥ 1023 ¥ 6.627 ¥ 10-34 ¥ 3 ¥ 108 = = 478.97 kJ l 250 ¥ 10-9

The LambertŪBeer Law (Absorption Law)

The absorption of light by a molecule is governed by two laws. 1. Lambert’s law

3.2.4

2. Beer’s law

Photophysical Law of Photochemistry

The absorption of light in the visible and near UV region by a sample in solution is clearly explained by a photophysical law known as Beer–Lambert’s law. Interaction of electromagnetic radiation with matter may lead to absorption, refraction, scattering and transmission. If Io be the intensity of the incident radiation, Io = IAbs + ITrans + IScat + IRefr By suitable designing of the optics, the scattering and refraction can be eliminated. Io = IAbs + ITrans. Lambertŭs Law The relationship between the extent of light absorbed and the thickness of the absorbing material, in the case of a pure substance, is given by Lambert’s law.

When a beam of monochromatic radiation passes through a homogeneous absorbing medium, the rate of decrease of intensity of the radiation with thickness of absorbing medium is proportional to the intensity of the incident radiation. This law may be mathematically expressed as –

dI μI dx

…(1)

Photochemistry and Spectroscopy

–

dI = KI dx

89

…(2)

(or) – where

dI = k.dx I

…(3)

I = the intensity of radiation x = thickness of the medium k = absorption coefficient

The value of k depends upon the nature of the absorbing medium. If Io is the intensity of radiation before entering the absorbing medium (when x = 0) then the intensity of radiation I after passing through any finite thickness, x of the medium can be obtained by integrating Eq. (3) between limits. I

Ú

x

dI = - k dx I

Ú

ln

…(4)

0

Io

I = –kx Io I = e–kx Io I = Io e–kx

…(5)

where Io is the intensity of incident radiation and I is the intensity of transmitted radiation. Equation (5) is known as Lambert’s law. Further, Eq. (5) becomes I = e–kx Io \

2.303 log

I = –kx Io

log

Io k = x 2.303 I

log

Io = Œx I

90 Engineering Chemistry-I

where Œ = extinction coefficient of the absorbing medium. It is related to absorption coefficient k by the expression Œ=

k = 0.4342 k and k = 2.303 Π2.303

When the absorbing substance is in solution, the relationship between the intensities of the incident and the transmitted radiation is given by Beer’s law (or) Lambert–Beer’s law.

BEERŭS LAW (OR) BEERŪLAMBERTŭS LAW

3.3

When a beam of monochromatic light is passed through a solution of an absorbing substance, the rate of decrease of intensity of radiation with thickness of the absorbing solution is proportional to the intensity of incident radiation (I) as well as to the concentration (C) of the solution. This law is mathematically written as –

dI μ IC dx

…(6)

–

dI = k¢ IC dx

…(7)

On rearranging, Eq. (7) becomes – where

dI = k¢C dx I

…(8)

K = molar absorption coefficient C = concentration of the solution I = intensity of radiation

Let Io be the intensity of radiation before entering the absorbing solution (i.e., when x = 0) Equation (8) is integrated with suitable limits. I

Ú

Io

(or)

x

dI = - k ¢C dx I

Ú 0

ln

I = –k¢ Cx Io

2.303 log

I = –k¢Cx Io

…(9)

Photochemistry and Spectroscopy

log

k¢ Io = Cx 2 . 303 I

A = ŒCx

(or) Œ=

where

91

…(10)

k¢ = molar absorptive coefficient. 2.303

Io = A = absorbance (or) optical density I Equation (10) is called Beer–Lambert’s law. From Eq. (10), the absorbance (A) is directly proportional to molar concentration (C) and thickness (x) or path length. log

3.3.1

Explanation of Terms used in Absorption Spectra

1. Absorbance, A The negative logarithm to the base 10 of the transmittance of the solution is called absorbance.

Io P = log o = –log10 T = Œbc …(11) I P where Io and I are the intensities of incident radiation and transmitted radiation respectively. Io is called opacity. Po and P are the power of incident radiation and I transmitted radiation. A is also known as optical density or absorbance, Œ is extinction coefficient. Absorbance or optical density A = log 1/T (or) – log T. A = log

2. Transmittance, T The fraction of the incident light that is transmitted by a given species is called transmittance.

T=

I P = Io Po

…(12)

3. Absorptivity (or) Molar Absorptivity CoefƝcient When C = 1 m and b = 1 cm then A = Œ (i.e.) molar absorptivity is defined as the absorbance of a one molar solution placed in a cell of one cm path length. log Io I A Œ= = log dm3 mol–1 cm–1 …(13) xC xC It is expressed in units of litre mol–1 cm–1.

92 Engineering Chemistry-I

3.3.2

Applications of BeerŪLambertŭs Law

It is used to determine unknown concentrations. The absorbance ‘Asoln’ of a standard solution of known concentration ‘Csoln’ is measured. According to Beer–Lambert’s law, Asoln = Œ Csoln x Asoln = Œx …(14) Csoln Then the absorbance ‘Aunknown’ of a solution of unknown concentration ‘Cunknown’ is measured. Aunknown = ŒCunknown x Aunknown = Œx Cunknown

…(15)

From equations (14) and (15), we get Asoln A = unknown Csoln Cunknown A Cunknown = unknown ¥ Csoln Asoln

\

…(16)

The values Aunknown and Asoln. are experimentally known and Csoln. is also known. The value of unknown concentration Cunknown can be calculated from Eq. (16).

Limitations of BeerŪLambertŭs Law 1. It is applicable only for dilute solutions because in a concentrated solution, the refractive index of the solution alters. 2. It is not valid if temperature changes during measurements. 3. This law is valid only when monochromatic radiation is used. Problems

Problem 3.1 A solution shows a transmittance of 20% when taken in a cell of 2.5 cm thickness. Calculate its concentration if the molar absorption coefficient is 12000 dm3 mol–1 cm–1. Formula: A = ŒCx Given Data: x = 2.5 cm, Œ = 12000 dm3 mol–1 cm–1. %T = 20% \ T =

20 = 0.2 100

A = –log T = –log 0.2 = 0.6990

Photochemistry and Spectroscopy

93

A = ŒCx

Solution

C=

A 0.6990 = 12000 ¥ 2.5 Œx

= 2.33 ¥ 10–5 mol dm–3. Problem 3.2 Compound X exhibits a molar extinction coefficient of 245 m2 mol–1 at 450 nm. What concentration of X in a solution will cause a 25% decrease in the intensity of 450 nm radiation, when the solution is placed in a 0.01 m absorption cell? Io = ŒCl I Io = 100% then I = 75% because there is 25%

Formula: Data:

log

reduction in the intensity. (i.e.,)

Io 100 = 75 I

Solution

log

\

Io = ŒCl I log Io I log(100 /75) = C= Œl 245 m2 mol-1 ¥ 0.01 m = 0.0510 mol m–3.

Result: Concentration of the solution = 0.0510 mol m–3. Problem 3.3 A solution of 2 cm thickness transmits 40% incident light. Calculate the concentration of the solution given that e = 6000 dm3 mol–1 cm–1. Formula: A = ŒCx Given data: x = 2 cm Œ = 6000 dm3 mol–1 cm–1 %T = 40%

\T=

40 = 0.40 100

We know A = –log T = –log 0.40 = 0.3979 Solution Concentration, C =

A 0.3979 = Œ x 2 ¥ 6000

94 Engineering Chemistry-I

Result:

Concentration of the solution = 3.316 ¥ 10–5 mol dm–3.

Problem 3.4 A monochromatic radiation is incident on a solution of 0.05 M of an absorbing substance. The intensity of the radiation is reduced to one-fourth of the initial value after passing through 10 cm length of the solution. Calculate the value of molar extinction coefficient of the substance. A = ŒCx

Formula:

Œ=

A log Io I = Cx Cx

Given data: I = 0.25; C = 0.05 m Io x = 10 cm. Solution Œ=

log 0.25 0.05 ¥ 10

0.5 Π= 0.6021 \

Œ=

0.6021 0.5

= 1.204 dm3 mol–1 cm–1. Result: The value of molar extinction coefficient Œ = 1.204 dm3 mol–1 cm–1. Problem 3.5 Calculate the molar absorption of a 1 ¥ 10–4 M solution, the absorbance of which is 0.20 when the path length is 2.5 cm. Formula:

A = ŒCx

Given data: Solution

A = 0.20; C = 1 ¥ 10–4; x = 2.5 cm

Molar absorptivity,

Œ= =

A Cx

0.20 1 ¥ 10-4 ¥ 2.5

= 800 dm3 mol–1 cm–1.

Photochemistry and Spectroscopy

95

Result: The molar absorption of a solution = 800 dm3 mol–1 cm–1. Problem 3.6 When a monochromatic light is passed through a cell of 1 cm length, the intensity of the radiation is reduced to 10%. If the same radiation is passed through a cell of 8 cm length, what is the transmittance radiation? Calculate the length of a cell in order to have 25% absorbance. Formula:

According to Lambert’s law, 2.303 log

I = –kx Io

Io = kx I Given data: Io = 100%, I = 90%, x = 1 cm, k = ? 2.303 log

2.303 log

Solution

100 =K¥1 90

2.303 ¥ 0.0458 = K Absorption coefficient (K) = 0.1055 Second Given data: Io = 100%, I = x, x = 8 cm, K = 0.1055 log

\

100 0.1055 ¥ 8 = = 0.3664 2.303 I 100 = Anti log [0.3664] = 2.363 I 100 = 43% I= 2.363

(or)

I=

43 = 0.43 100

From the above data, we know Io = 100%, I = 25%, K = 0.1055x = ? 2.303 log

100 = 0.1055x 25

2.303 ¥ 0.6021 = 0.1055x x=

2.303 ¥ 0.6021 = 13.1 cm 0.1055

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Result: (i) The transmission radiation (I) = 0.43 (ii) The length of a cell for 25% absorbance = 13.1 cm Problem 3.7 Find out the absorbance (or) optical density of a solution, if the transmittance of a solution is 18.5%. Formula: Data:

A = log 1/T or –log T

Transmittance, T = 18.5% =

18.5 = 0.185 100

Solution Absorbance or optical density

(A) = log 1/T (or) – log T = –log 0.185 = 0.733 Result: Absorbance = 0.733 Problems based on LambertŪBeerŭs Law

Problem 3.8 The percentage transmittance of an aqueous solution of unknown compound is 20% at 25°C and 300 nm for a 4 ¥ 10–5 M solution in a 2 cm cell. Calculate the absorbance (A), the molar extinction coefficient and the percentage transmittancy of a 2 ¥ 10–5 M solution in a 4 cm cell. Solution I 100 ˆ (i) Absorbance (A) = log ÊÁ o ˆ˜ = log ÊÁ = 0.699. Ë 20 ˜¯ Ë I¯ (ii) Molar extinction coefficient Given x C A A

= 2 cm = 4 ¥ 10–5 M = 0.699 = ŒCx

Œ= A = Cx

0.699 4 ¥ 10-5 M ¥ 2 cm

= 8.7 ¥ 103 M–1 cm–1 (iii) Percentage transmittancy = Given x = 4 cm C = 2 ¥ 10–5 M

I ¥ 100. Io

Photochemistry and Spectroscopy

97

Œ = 8.7 ¥ 10–5 m–1 cm–1 Io = ŒCx I = 8.7 ¥ 103 M–1 cm–1 ¥ 2 ¥ 10–5 M ¥ 4 cm

A = log

Io = 0.696 I Io = Anti log (0.696) I Io = 4.966 I 1 = 0.2 I/Io = 4.966 I Percentage transmittancy = 100 ¥ Io = 100 ¥ 0.2 = 20% log

Result: (i) Absorbance (A) = 0.699 (ii) The molar extinction coefficient = 8.7 ¥ 103 m–1 cm–1 (iii) Percentage transmittance of a 2 ¥ 10–5 M solution in a 4 cm cell = 20% Problem 3.9 A solution of 3 cm thickness transmits 30% incident light. Calculate the concentration of solution. (Given: Œ = 4,000 dm3 mol–1 cm–1) Solution

Œ = 4,000 dm3 mol–1 cm–1 30 = 0.3 100 x = 3 cm

I/Io =

Absorbance

1 A, = log Io/I = log ÊÁ ˆ˜ Ë 0.3 ¯

A = ŒCx C=

= 0.523

A 0.523 = 3 Œ x 4, 000 dm mol-1 cm-1 ¥ 3 cm

= 4.35 ¥ 10–5 mol dm–3

98 Engineering Chemistry-I

Result: Concentration of solution = 4.35 ¥ 10–5 mol dm–3

QUANTUM YIELD OR QUANTUM EFFICIENCY

3.4

The relationship between the number of molecules reacting with the number of photons absorbed can be expressed by means of quantum yield. Quantum efficiency is defined as the ratio of number of molecules of reactants consumed or the number of molecules of product formed to the quanta of light absorbed. f=

Number of molecules consumed or formed in a given time Numbeer of photon (quanta) absorbed in the same time

Stark–Einstein photochemical equivalence law is applied to only primary photochemical reactions. For primary absorption of one photon, one molecule undergoes dissociation, and such cases f will be unity. Mostly, f ranges from 0.04 to 106. When the quantum yield is less than unity, it is called low quantum yield, and when it is greater than one, it is called high quantum yield. In some cases, the activated photochemically molecule initiates a series of thermal reactions, called secondary processes. As a result of secondary processes, the reactant molecules undergo chemical changes by absorbing only one quanta. For such cases, f will be greater than unity. In certain cases, the molecules may be photochemically deactivated. For such cases, less than one molecule may react per quanta. So f will be less than one.

3.4.1

Classification of Photochemical Reactions Based on Quantum Yield

Based on quantum yield, the photochemical reactions are classified into the following: 1. Reaction with Very Low f (Less Than One)

(a) Formation of HBr H2 + Br2 Æ 2HBr

f = 0.01

(b) Dissociation of NH3 2 NH3 Æ N2 + 3 H2

f = 0.2

Photochemistry and Spectroscopy

99

(c) Dissociation of nitrogen dioxide 2 NO2 Æ 2 NO + O2 2. Reaction with Moderate f (Simple Integers)

(a) Dissociation of H2S H2S Æ H2 + S

f=1

(b) Dissociation of hydrogen bromide 2 HBr Æ H2 + Br2

f=2

(c) Organisation of oxygen 3O2 Æ 2O3

f = 3.

3. Reaction with Very High f

(a) Formation of phosgene f = 103

Co = Cl2 Æ CoCl2 (b) Formation of HCl H2 + Cl2 Æ 2 HCl

3.4.2

f = 10 4 – 106.

Reason for High Quantum Yield

1. Primary process might generate free radical which generates chain reaction. . A + hg Æ A free radical This free radical again produces free radicals in secondary reaction process. Hence, f will be greater than one. 2. The intermediate formed acts as a catalyst. 3. The exothermic reaction may activate other molecules and these molecules react without absorption of additional photon of radiation.

3.4.3

Reason for Low Quantum Yield

1. The excited molecules can get deactivated before forming products. 2. Primary photochemical reactions may be reversed in a dark reactions. 3. Molecules are not receiving sufficient energy to enable them to react. 4. The dissociated fragment may recombine to give low f.

100 Engineering Chemistry-I

EXPERIMENTAL DETERMINATION OF QUANTUM YIELD

3.5

We know that f=

Number of molecules decomposed or formed Number of quanta ab bsorbed per unit time

=–

dP dA (or) + dt dt Io (fraction of light absorbed)

In order to calculate quantum yield, we can 1. Determine the number of molecules reacted in a given time. 2. Determine the amount of photons absorbed by the same substance in the same time.

3.5.1

Determination of Number of Moles Reacting

This can be determined by the usual analytical technique used in kinetics. The extent of photochemical reaction depends upon the intensity of light used irrespective of its wavelength. A small quantity of the sample is pipetted out from time to time and its concentration is measured time to time by the changes in some physical property like absorption, optical rotation, etc. It may be remembered that energy of light radiation falling per square centimetre of surface is given by the product of intensity of light and the time of exposure of the substance.

3.5.2

Determination of Number of Einsteins of Light Absorbed

To measure the quantum efficiency, we know the intensity of incident and the emitted light. They can be measured by actinometry. Actinometry is used to determine the intensity of light comes from the reaction vessels. The intensity of light measured with the empty cell and with a reaction mixture. The difference between those two readings will give the amount of energy absorbed by the reaction system. The intensity of light can be measured by using thermopile or by a chemical actinometer. 1. Thermopile This thermopile can be calibrated with standard light sources. It consists of a series of unlike metals connected in a galvanometer. One end of it is

Photochemistry and Spectroscopy 101

coated with Pt black and the other end is kept as such. Light coming from the reaction cell is incident on the black surface. It gets heated up resulting in a difference in temperature of the black end and the other end. Thus, the current flows in the circuit. Current produced μ Intensity of light 2. Chemical Actinometers Chemical actinometers consist of a gas mixture or solutions which are more sensitive to light. The substance is exposed with radiation, a chemical reaction takes place and the extent of the reaction is a direct measure of energy absorbed. The important two types of actinometers are

(a) Uranyl oxalate actinometer (b) Ferrioxalate actinometer (a) Uranyl Oxalate Actinometer uranyl sulphate in water.

UO22+ +

hg

0.05 m oxalic acid is mixed with 0.01 m

(UO22+ )*

excited uranyl ion

(UO22+ )*

+ COOH

UO22+ + CO2 + CO + H2O

COOH

The residual unreactive concentration of oxalic acid can be found out by titrating the mixture with std Km nO4. Here, the amount of oxalic acid consumed is the measure of the intensity of radiation. (b) Ferrioxalate Actinometer When potassium ferrioxalate solution is exposed with light, the Fe3+ ion is reduced to Fe2+. The iron can be estimated calorimetrically using O- phenanthroline as a complexing agent. The wavelength is up to 4000 A and we can use the solution of 0.006 M k3Fe(ox)3 used. For longer wavelength, the solution of 0.15 m is used. Here, f varies from 1.2 to 1.1.

102 Engineering Chemistry-I

Fig. 3.1

Apparatus for study of photochemical reaction.

PHOTOPHYSICAL PROCESSES (FLUORESCENCE AND PHOSPHORESCENCE) 3.6.1

3.6

Fluorescence

When a beam of light is incident on certain substances, they emit visible light or radiation and they stop emitting light or radiations as soon as the incident light is cut off. This is known as fluorescence. Fluorescence is the action of stimulating light by fluorescent substances. Examples

1. Minerals—fluorite, petroleum, vapours of Na 2. Organic dyes—Fluorescein, Eosin 3. Compounds—Chlorophyll, ultramarine, Riboflavin Applications

1. Fluorescent tubes 2. The difference in fluorescence caused by UV rays in different types of ink enables the detection forged documents. 3. When UV light is incident on newly laid eggs, they exhibit fluorescence with rosy color, while bad eggs appear blue. Types of Fluorescence

1. Resonance fluorescence 2. Sensitised fluorescence 1. Resonance Fluorescence When an incident light falls on a substance and it emits light of same frequency as that of incident light, this phenomenon is called resonance fluorescence.

Photochemistry and Spectroscopy 103

Example

= 2537 A Hg + hg ælææææ Æ Hg* Hg* Æ Hg + hg

when the excited atom returns to ground state, it emits radiations of the same wavelength of 2537 Å. 2. Sensitised Fluorescence A substance which is normally non-fluorescent may be made fluorescent in the presence of other fluorescent substances. This is called sensitised fluorescence. Example

Hg (normal) + hg (absorbed radiation) Æ Hg* (excited) Hg* + Tl (normal) Æ Hg (normal) + Tl* (excited) Tl* Æ Tl + hg (emitted radiation)

3.6.2

Phosphorescence

When light radiation is incident on certain substances, they emit light continuously even after the incident light is cut off. This type of delayed fluorescence is called phosphorescence and the substances are called phosphorescent substances. Examples

1. Cadmium sulphide, barium sulphide 2. Minerals—ruby, emarald 3. Dyes—when dissolved in boric acid or glycerol and cooled to form a glassy solids Applications

1. It is used widely for dials on clocks and watches because they glow in the dark. 2. The luminous paints contain 2.5% alkali chlorides and traces of heavy metal sulphides. This mixture is used for painting watch dials, electric switches, etc. The other photophysical process is photoelectric effect used in wireless communication and as detectors in electronic equipment. The alkali metals like Na, K, Li, etc. exhibit photoelectric effect.

104 Engineering Chemistry-I

1. Non-radiative Transition

Singlet higher to lower (S3 Æ S2 Æ S1) or Triplet higher to lower T3 Æ T2 Æ T1 (a) Vibrational Cascade (VC) Transition from upper lower vibrational state to the lower upper vibrational state. (b) Internal Conversion (IC) Energy of the activated molecules is dissipated in the form of heat through molecular collisions. This occurs less than 10–11 seconds. (c) Intersystem Crossing (ISC)

S2 to T2 ¸ ˝ forbidden transition takes place at relatively slow rate. S1 Æ T1 ˛ 2. Radiative Transition

1. Fluorescence lifetime 10–8 – 10–6 second and it is allowed transition (hgf ). 2. Phosphorescence 10–6 to few seconds This is forbidden transition (hg p).

3.6.3

Mechanism of Fluorescence and Phosphorescence

Jablonski Diagram

Fig. 3.2

Jablonski diagram for photophysical processes

Photochemistry and Spectroscopy 105

1. Spin Multiplicity:

2S + 1 S is total electron spin.

2. Singlet Ground State Even number of e–s in the G state (2, 4, 6, …) all e–s are spin paired. Upward orientation of the e– spin gets cancelled by the downward orientation S = 0. 1 1 S1 = + ; S2 = – 2 2 1 1 S = S 1 + S2 = + - = 0 2 2 2S + 1 = 2 ¥ 0 + 1 = 1 Spin multiplicity = 1

The molecule is in singlet ground state. higher

lower 3. Triplet Excited State

Spin orientation of the two e–s maybe parallel (≠≠)

Parallel ≠Ø 1 1 S = S1 + S2 = ÊÁ + ˆ˜ + ÊÁ + ˆ˜ = 1 Ë 2¯ Ë 2¯ 2S + 1 = 2 ¥ 1 + 1 = 3 Triplet 4. Singlet Excited State

≠Ø

1 -1 S = S1 + S2 = + + =0 2 2 2S + 1 = 2 ¥ (0) + 1 = 1 Singlet excited states

106 Engineering Chemistry-I

(a) A molecular electronic state in which all of the electron spins are paired is called a singlet state. (b) An atom or molecule may contain two unpaired electrons both having the same spin, it is then a triplet state.

ENERGY TRANSFER IN PHOTOCHEMICAL REACTION (PHOTOSENSITISATION AND QUENCHING) 3.7.1

3.7

Photosensitisation

A substance which when added to the reaction mixture helps to start the photochemical reaction but itself does not undergo any chemical change is called a photosensitiser and the process is called photosensitisation. Photosensitisation acts as a carrier of energy. Examples (a) Atomic photosensitisers Mercury, Cadmium, Zinc (b) Molecular photosensitisers Benzophenone, SO2 (c) Photosynthesis of Carbohydrates in Plants Chlorophyll is a sensitising agent. Its conjugation leads to absorption in the visible region. ( 600 nm -700 nm ) x CO2 + xH2O ænhg æææææææ Æ [CH2O] x + x O2 D G° = –ive Chlorophyll

Chl + hg Æ Chl* Absence of chlorophyll, D G° = + 2875 kJ so the reaction is not produced due to high +ve value. (d) Combination of CO and H2 sensitised by Hg Vapour formaldehyde and some glyoxal

Products are

traces of hg (UV = 253.7 nm)

CO and H2 + Hg ææææææææææ Æ Hg atoms are excited Hg* transfer energy by collisions to hydrogen molecules and dissociates them into H atoms. Hg + hg Æ Hg* Hg* + H2 Æ Hg + 2H Primary step energy transfer H + CO Æ HCO ˘ HCO + H2 Æ HCHO + H ˙ ˙ reactions 2HCO Æ HCHO + CO ˙ 2HCO Æ CHO – CHO ˙ (glyoxal) ˚

Photochemistry and Spectroscopy 107

Other Examples Hg with decomposition of H2, NH3, H2O, PH3, various hydrocarbons, alcohols, ethers, acids and amines.

3.7.2

Dissociation of Hydrogen Hg + hg Æ Hg* Hg* + H2 Æ H2* + Hg . H2* Æ 2H

1. Quenching (Stopped) When a photochemically excited atom has a chance to undergo collision with another atom or a molecule before its fluorescence, the intensity of the fluorescence radiation may be diminished or stopped, a phenomenon called the quenching of fluorescence. (a) Internal Quenching Quenching may also occur when the molecule changes from the singlet excited state to the triplet excited state—internal quenching. (b) External Quenching Quenching occurs from the addition of an external substance, which absorbs energy from the excited molecule—external quenching. Transfer of energy from the molecule to atom with which it undergoes collision.

(i) The photochemically excited atom may activate the atom with which it undergoes collision. Hg* + Tl Æ Hg + Tl* Thallium (ii) The excited atom may undergo collision with another molecule to form the products. Hg* + H2 Æ Hg + 2H + hg (iii) This reaction is a photosensitised reaction with Hg acting as a sensitising agent. The photochemically excited atom may activate a molecule by collision. Cd* + H2 Æ Cd + H2* 2. Uses Quenching of fluorescence is that excited molecules or atoms obtained in the quenching may undergo further reaction, so as to continue the photochemical change, such as

108 Engineering Chemistry-I

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

3.7.3

Photosensitisation Sensitised fluorescence Photochemical reactions Dissociation of molecules to form atoms

Mechanism of Photosensitisation S1

D¢ Isc D3

S1

A¢ T1

Higher energy

A3 Transfer of energy from the sensitiser to reactant

hg So Sensitiser (Donor)

Low energy

T1

So Reactant (Acceptor)

Fig. 3.3 Donor-accept system

D + hg Æ D¢ ISC Æ D3 D¢ ææ

3

D + A Æ D + A3

A3 Æ Products (photosensitisation) D3 Æ Products (quenching)

Photo Inhibitors

1. Certain substances which are able to retard the rate of photochemical reactions when present in small amounts are called photo inhibitors. This phenomenon is known as photochemical inhibition. 2. Photo inhibitors interrupt the chain reaction by removing chain carrier atoms or radicals.

Photochemistry and Spectroscopy 109

Examples

(a) Traces of NO and propylene lower the f of photochemical combination of H2 and Cl2. (b) Traces of NH3 which when present in H2 and Cl2 reactions lowers it f. (c) SO2 and O2 act as inhibitors in various photochemical reactions.

CHEMILUMINESCENCE (COLD LIGHT)

3.8

This chemiluminescence reaction is the reverse of an ordinary photochemical reaction. In a photoreaction, light is absorbed, while in chemiluminescence, light is emitted at ordinary temperature as a result of a chemical reaction. The emission occurs at ordinary T, the emitted radiation is known as cold light. The energy released during the chemical reaction makes the product molecule electronically excited. The excited molecule then emits radiation, as it returns to the ground state. Examples 1. Oxidation of phosphorous vapours by atmospheric oxygen produces a greenish yellow glow.

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

4P + 5O2 Æ 2 P2O5 + hg Oxide of 5-aminophthalic hydroxides or cyclic hydroxides by H2O2 emits bright green light. Bioluminescence—Emission of cold light by fireflies (glow-worm) due to the aerial oxides of luciferon (a protein) in the presence of enzyme. (luciferase) Grignard reagent (Rmgx) produces greenish blue light, on oxidation by air. Oxidation of decaying wood containing certain bacteria. Alkali metal vapours react with halogen and with organic halides at low temperature. SrCl2 + H2SO4 (dilute) Æ SrSO4 + 2HCl + hg (feeble glow)

Mechanism of Chemiluminescence Chemiluminescence can be explained by considering anion-cation reactions.

110 Engineering Chemistry-I Example –

+

Ar + Ar Aromatic Aromatic anion cation 1

Ar*

1

Ar* + Ar

Ar + hg

Problem 3.10 When propionaldehyde is irradiated at 200 torr and 30°C with light of wavelength 3020 Å, the f for CO production is found to be 0.54. If the incident light intensity is 15, 000 ergs/second. Calculate the rate of CO formation what is the light intensity in einsteins per second. Solution \ E=

1 einstein light of l = 3020 Å consists of Nhc 6.023 ¥ 10 23 ¥ 6.625 ¥ 10 -27 ¥ 3 ¥ 1010 = ergs of energy l 3020 ¥ 10 -8 = 3.964 ¥ 1012 ergs

So

15,000 ergs/second =

15, 000 3.964 ¥ 1012

einstein

= 3.784 ¥ 10–9 einsteins/second Given f = 0.54 =

No. of moles of CO 3.784 ¥ 10 -9

produced

(a) Rate of form of CO = 0.54 ¥ 3.784 ¥ 10–9 = 2.04 ¥ 10–9 moles/second (b) Intensity of hg absorbed = 3.784 ¥ 10–9 einsteins/second Problem 3.11 For the photochemical formation of ethylene from di-npropylketone using a radiation of 313 nm wavelength, the quantum yield is 0.21. Calculate the number of moles of ethylene formed when the sample is irradiated with 50 watts of this radiation assuming all the radiation is absorbed by the sample.

Photochemistry and Spectroscopy 111

Solution (i) f = 0.21 (ii) Energy of one quantum corresponding to 313 nm, E=

hc 6.626 ¥ 10 -34 Js ¥ 3 ¥ 108 ms -1 = = 6.35 ¥ 10 -19 J l 313 ¥ 10 -9 m

50 watts = 50 J s–1

(1 watt = 1 J s–1)

No. of

¸ 50 Js -1 Ô Total energy absorbed = quanta ˝ = Energy of one quantum 6.35 ¥ 10 -19 J absorbed Ô˛ = 7.87 ¥ 1019 s–1 f= 0.21 =

No. of ethylene molecules formed No. of quanta absorbed No. of ethylene molecules formed 7.87 ¥ 1019 s -1

No. of ethylene molecules formed = 0.21 ¥ 7.87 ¥ 1019 s–1 = 1.65 ¥ 1019 molecules \ Amount of ethylene formed =

1.65 ¥ 1019 molecules 6.023 ¥ 10 23 molecules/mol

= 274 ¥ 10–5 mol Problem 3.12 of 85 nm.

Calculate the energy per mole of light having a wavelength E=

Solution Given: l h C NA

N A hc l

= 85 nm = 85 ¥ 10–9 m = 6.62 ¥ 10–34 J.s = 3 ¥ 108 ms–1 = 6.023 ¥ 1023 mol–1

E=

6.023 ¥ 10 23 mol -1 ¥ 6.62 ¥ 10 -34 J.s ¥ 3 ¥ 108 ms -1 85 ¥ 10 -9 m

= 1.407 ¥ 106 J mol–1 = 1.407 ¥ 103 kJ mol–1

112 Engineering Chemistry-I

Problem 3.13 When a substance x was exposed to light. 0.002 moles of it reacted in 20 minutes and 4 seconds. At the same time, x absorbed 2.0 ¥ 106 photons of light per second. Calculate the quantum yield of the reaction (Avogadro’s Number NA = 6.023 ¥ 1023) Solution (i) Number of molecules x reacted = 0.002 ¥ NA = 0.002 ¥ 6.023 ¥ 1023 (ii) Number of photons absorbed per second = 2.0 ¥ 106 \ Number of photons absorbed in 20 min and 4 s (1204 seconds) = 2.0 ¥ 106 ¥ 1204 f=

No. of molecules reacted 0.002 ¥ 6.023 ¥ 10 23 = No. of photons absorbed 2.0 ¥ 106 ¥ 1204

= 5.0 ¥ 1011 Problem 3.14 In a photochemical reaction,

CH2Cl COOH + H2O

OH | H2C – COOH + HCl

hg

It was found that after irradiating the solution at 253.7 nm for 837 minutes, 34.36 J of energy was absorbed and 2.296 ¥ 10–5 mol of HCl were formed. Calculate the quantum yield of the reaction. Solution The quantum corresponds to Wavelength =

hc 6.626 ¥ 10-34 JS ¥ 3 ¥ 108 MS-1 = l 253.7 ¥ 10-9 M = 7.84 ¥ 10–19 J

\ Number of quanta absorbed =

34.36 J 7.84 ¥ 10-19 J

= 4.39 ¥ 1019 Number of moles of HCl ¸ –5 ˝ = 2.296 ¥ 10 mol formed ˛ = 2.296 ¥ 10–5 ¥ 6.023 ¥ 1023 molecules Number of molecules reacting = 1.38 ¥ 1019 molecules

Photochemistry and Spectroscopy 113

\

Quantum yield (f) =

1.38 ¥ 1019

= 0.314 4.39 ¥ 1019 Problem 3.15 The f of the photochemical reaction

6 H2(g) + Cl2(g) æhg æÆ 2HCl (g) is 1 ¥ 10 with a wavelength of 480 nm. Calculate the number of moles of HCl(g) produced per joule of radiation energy absorbed.

Solution E = hg =

-34 8 -1 hc 6.626 ¥ 10 JS ¥ 3 ¥ 10 MS = l 480 ¥ 10-9 M

= 4.14 ¥ 10–19 J \ Energy per mole = ENA = 4.14 ¥ 10–19 J ¥ (6.023 ¥ 1023 mol–1) = 2.5 ¥ 105 J mol–1 In the above reaction, one photon (hg ) produces 2 moles of HCl and f is 106 \ Number of moles of HCl produced = 2 ¥ 106 moles Number of moles of HCl produced per joule of radiation energy =

Result:

2 ¥ 106

= 8 mol 2.5 ¥ 105 The number of moles of HCl = 8 mol.

Problem 3.16 If a photon of wavelength 2450 A° dissociates a bond having a bond energy of 95 k/cal/mol, what is the energy required for dissociation? Solution Energy of a photon

E= =

hc l 6.626 ¥ 10-34 Js ¥ 3 ¥ 108 Ms-1

2450 ¥ 10-10 M

= 8.11 ¥ 10–19 J Ê 1 cal ˆ = 1.94 ¥ 10–19 cal E = 8.11 ¥ 10–19 J ¥ Á ˜ Ë 4.184 J ¯

114 Engineering Chemistry-I

Now the energy required for the dissociation of one bond (E¢) E¢ =

95 ¥ 103 cal mol 6.023 ¥ 1023 mol

= 1.58 ¥ 10–19 cal

Energy E > Energy E¢ So the bond will be dissociated.

SPECTROSCOPY

3.9

Spectroscopy is scientific discipline that develops and applies methods, instruments and strategies to obtain information on the composition and nature of matter in space and time. The types of problems encountered in analytical techniques are: 1. Qualitative analysis of elements, ions, atomic groups and functional groups in mixtures of substances 2. Qualitative analysis of a single or mixtures of chemical species 3. Quantitative analysis for elements, ions, atomic groups and functional groups in mixtures of substances 4. Quantitative analysis of a single chemical species or mixtures of species throughout the sample or at its surface The field of spectroscopy is divided into emission and absorption spectroscopy. (Fig. 3.4). An emission spectrum is obtained by spectroscopic analysis of some light source, such as a flame or an electric arc. Fluoroescent lights and colors obtained by heating salts of certain elements in a flame are common examples of emission spectra. An absorption spectrum is obtained by placing the substance between the spectrometer and some source of energy that provides electromagnetic radiation in the frequency range being studied. The spectrometer analyzes the transmitted energy relative to the incident energy for a given frequency. Again, the high-energy states are usually short-lived. The major fact of absorbed energy in the infrared region is heat. Thus, the temperature of the substance (or solution) increases while the spectrum is being determined. The major fact of absorbed energy in the ultraviolet region is re-emission of light.

Photochemistry and Spectroscopy 115 E2

E2 hv

hv

E E1

E1

Fig. 3.4

(a) Absorption spectrum

(b) Emission spectrum

Spectroscopy deals with the study of interaction of electromagnetic radiation with the matter (atoms and molecules). The interaction of electromagnetic radiation with matter gives rise to two types. 1. Atomic spectroscopy 2. Molecular spectroscopy 1. Atomic Spectroscopy It arises from the transition of electrons from one energy level to another due to absorption or emission of energy in the atom. 2. Molecular Spectroscopy It involves the transition of electrons between rotational and vibrational energy levels in addition to electronic transition. So molecular spectra is much more complicated than atomic spectra. Molecular spectra is useful in determining the shape and size of molecule, bond length, strength of bond, bond dissociation energy, bond angle and to determine the structure and constitution of compounds. The summary of the various types of molecular spectra is tabulated in Table 3.2. Table 3.2 Molecular spectral transitions S. No.

Spectra

Transitions

Region of electromagnetic spectrum

Criteria

1.

Rotational Between (or microwave) rotational energy levels

Microwave (1–100 cm–1)

Molecule must possess permanent dipole moment, e.g. HCI, H2O, etc.

2.

Between Vibtational and vibrational vibrational energy levels rotational (or infrared)

Infrared (500–4000 cm–1)

Dipole moment of molecule must change during vibrations.

(Contd.)

116 Engineering Chemistry-I Table 3.2 (Contd.) 3.

Electronic between (or UV–Visible) electronic energy levels

Presence of chromophore in a molecule.

Visible (12500–25000 cm–1) and UV (25000–70000 cm–1)

ELECTROMAGNETIC SPECTRUM

3.10

The electromagnetic spectrum consists of electromagnetic radiations of different wavelengths, frequencies and energies. The spectrum consists of different types of radiations arranged in the order of increasing wavelength (l) (or) decreasing the frequencies. X -ray Cosmic rays < g - ray < rays < visible light < Ultraviolet 1444424444 3 144444 42444444 3 100 pm 10 nm l = 1 pm - 100 pm 10 nm - 1000 nm < infrared rays l = 1000 nm – 100 mm

< microwaves 100 mm – 1 cm

< radiowaves ESR

NMR

1 cm – 100 cm

100 cm – 10 m

Cosmic rays have the highest frequencies and radiowaves have the lowest frequencies. All electromagnetic radiations travel with the same speed of light but they differ in their wavelengths from each other.

ABSORPTION AND EMISSION OF RADIATION

3.11

E2 (Ex. st) 1. Absorption Spectrum Beam of electromagnetic radiation is alhg lowed to fall on the molecule in the ground state. The molecule absorbs hg energy and undergoes a transition from the E1 (G. S) lower energy level to the higher energy Fig. 3.5 Absorption spectrum level. The measure of this decrease in the intensity of radiation is the absorbtion spectroscopy.

2. Emission Spectrum Molecules coming down from the excited state to the ground state with the emission of photons of energy hg is known as emission spectrum. Fig. 3.6 Emission spectrum

Photochemistry and Spectroscopy 117

Regions of UV and Visible Spectrum

Region Vacuum UV Ultraviolet Visible

Wavelength range 2–180 nm 180–400 nm 400–780 ææææææÆ Ï 420 nm — Violet Ô 470 — Blue Ô ÔÔ530 — Green Ì Ô580 — Yellow Ô620 — Orange Ô ÔÓ700 — Red

Regions of IR Spectrum

1 lm = 10–3 nm Region

Wavelength range

Wave number range

For IR Middle IR Near IR

14.3–50 mm 2.5–15 mm 0.7–2.5 mm

700–200 cm–1 4000–666 cm–1 14290–4000 cm–1

CLASSIFICATION OF MOLECULAR SPECTRA

3.12

These are of three types—corresponding to three types of energy changes in a molecule. 1. Electronic Spectra

• Caused by the absorption of high-energy photons. • Send electrons to high energy level. • Within the visible and UV regions of the electromagnetic spectrum. 2. Vibrational Spectra

• Caused by still lower energy photons which cause changes in vibrational energy levels. • They are in the near IR region.

118 Engineering Chemistry-I 3. Rotational Spectra

• Caused by still lower energy photons. • Cause changes in the rotational level in the molecules. • They are in the microwave region. Energy of the incident photon is large enough to produce changes in electronic-vibrational–rotational spectra. Photons capable of bringing vibrational level can also cause rotational energy changes Æ vibrational–rotational spectra Pure rotational spectra involves only small energy changes which cannot bring about vibrational or electric changes. Different types of electromotive radiations interact with matter and give rise to various types of spectroscopy. Absorption of electromagnetic radiation by matter in radiofrequency region

NMR spectroscopy (or) ESR spectroscopy

Absorption of radiation by molecules in microwave region. (It causes transition between the rotational level of the molecules)

Rotational spectroscopy

Absorption of radiation by atoms or molecules in visible and UV region

Electronic transition

Electronic spectroscopy

3.12.1

The Width and Intensity of Spectral Transitions

1. Width of Spectral Lines

Two main factors Collision broadening Ø

Doppler broadening Ø

Takes place more in liquid Rapid motion of molecules towards or phase than in gaseous phase away from the observer or detector Ø Deform the charge clouds of the outer electrons Ø

Photochemistry and Spectroscopy 119

Spectral transition broadened Liquid, gases, molecules moving liquid-interaction is more so in high speed, hence, spectral line broadened broadened Ø Gas spectra sharper than liquid spectra; Solid more sharp perfect crystal, no interaction because the motion of particles is more restricted 2. Intensity of Spectral Lines

(a) Boltzman population of the energy levels (b) Transition probability between the energy level (c) Path length and concentration of the sample (a) Boltzman Population of the Energy Levels

N

No N DE K

–DE/KT

No = e = Number of molecules in the ground state = Number of molecules in the excited state. = Energy difference between ground state of excited state = Boltzman constant

DE is large and N N is small. o Number of molecules in the excited state is less than the ground state. (b) Transition Probability Selection rules determine the allowed transitions and forbidden transitions (intense spectral lines). (c) Path Length and Concentration of the Sample Found experimentally that the intensity of spectral lines decreases exponentially as the concentration and path length of the sample.

3.12.2

Spectral Selection Rules

1. Represent the changes in quantum number during a particular transition. (a) Pure rotational transition is D J = ±1 where J is the rotational quantum number. (b) Pure vibrational transition is DV = ±1. where V is the vibrational quantum number. (c) Combined vibrational rotational transitions are DV = ±1 and DJ = ±1

120 Engineering Chemistry-I

2. Gross selection rules represent the general features of molecules like dipole moment, polar, polarisability, etc. (a) Molecules give pure rotational spectrum—it must be polar (permanent electrical dipole moment). (b) Vibrational selection rule represents that the electric dipole moment of the molecule must change when the atoms of a molecule are displaced relative to one another. (c) Vibration Raman transition; the polarisability must change as the molecule vibrates. Spectral transitions obey a given selection rule, called allowed transitions (more intense and stronger). If it violates selection rule, forbidden transitions.

3.12.3

Dipole Moment (l)

• Homonuclear diatomic molecules (H2, Cl2, N2, O2, etc.) nonpolar (H–H, H:H) • Heteronuclear molecules (H2O, HCl, SO2, etc.) These molecules are different in electronegativity of the two atoms. They have electric dipole (H+ Cl–) +

–

H Æ Cl ; Hd , Cl d The degree of polarity or strength of which is expressed as dipole moment (vector quantity).

3.12.4

Rotational Transition (Rotational Spectra) (Microwave Spectroscopy)

Microwave spectra occur in a for IR region or microwave region of the electromagnetic spectrum. The molecules only possess a permanent dipole moment to give rotational spectra. Example HCl, HBr, CO, NO, H2O possess a permanent dipole moment and interact with microwave radiation and are microwave active. Homonuclear diatomic molecules (H2, Cl2, O2 and N2) are microwave inactive since they do not possess permanent dipole moment. They are used to determine bond length in heteronuclear molecules. Consider the angular momentum, rotational quantum. h2 Number J, rotational energy, E J = J( J + 1) J = 0, 1, 2, 3… 8p 2 I and so on.

Photochemistry and Spectroscopy 121

Limitations

1. Only molecules which possess permanent dipole moment give rotational spectra. 2. Heteronuclear molecules exhibit spectrum only when they are in gaseous state. Gaseous state internuclear attractive forces are negligible and molecules can rotate around the centre of gravity. In solids and liquids, internal attraction is strong and hence they do not exhibit rotational spectra.

3.12.5 VibrationalŪRotational Transition (Infrared Spectroscopy) This spectra is observed in the near IR region. Energy changes in vibrational transitions are always larger than those in pure rotational transitions. Molecules which possess either a permanent dipole moment or the dipole moment arises due to vibration of atoms in molecules. These molecules are infrared active e.g. (CO, NO, CN, HCl). The homonuclear diatomic molecules like O2, N2 and Cl2 have no permanent dipole moment of their own. However, when they are subjected to electromagnetic radiation it induces an oscillating dipole moment and excite them to undergo vibrational–rotational transitions. • Pure vibrational spectra is observed only in liquids where interactions between the molecules inhibit rotation. IR is useful in the determination of molecular structure (stiffness or elasticity of the molecules—force constant, bond length and bond angles). It is used for identifying functional groups in organic compounds. IR frequencies are expressed in terms of wave number, cm–1.

3.12.6

Vibrational Motion of Diatomic Molecules (Simple Harmonic Oscillator)

In diatomic molecules, atoms are linked by a covelent bond. The force constant is a measure of the bond strength between two atoms. According to Hooke’s law, the restoring force, f pulling or pushing the spring to its equilibrium position is directly proportional to displacement, x. fμx K = force constant f = –kx x = displacement The negative sign indicates that the force is the restoring force and has a direction opposite to that of the displacement.

122 Engineering Chemistry-I Frequency of Oscillations (f ) g depends on the reduced mass m of the system and force constant, k

1 K –1 S 2p m In vibrational spectrum, the oscillation frequency expressed in terms of wave number, w Frequency of oscillations g =

(i.e) w = g cm –1

c w=

1 2p c

K cm-1 m

In lower state of vibration, the diatomic molecules behave as a simple harmonic oscillator, but, in higher quantum levels, there is a deviation from a SHO and the vibration becomes anharmonic, (DV = ±1, ±2, ±3, etc.) The portability is very low. Theoretically, vibrational transition for harmonic motion are restricted to change of unity DV = ±1. For vibrational-rotational spectra, DETotal = DEvib ± DErot Vibration of Polyatomic Molecules Polyatomic molecules show more than one fundamental vibrational absorption bands.

Sum of coordinates ¸ Ô Number of degree of freedom = to locate all the atoms ˝ of a molecule in space Ô˛ Each atom has three degrees of freedom corresponding to three cartesion coordinates (x, y, z) Total number of degrees of freedom in a molecule containing N–atoms is equal to 3 N (includes rotational, vibrational and translational) 3N = Trans + Vib + Rot. The centre of gravity of the molecule has 3 independent translational degrees of freedom. If the molecule is linear, two independent modes of rotation about x-axis and y-axis are possible. \

(3N) = 3 + 2 + vibrational

Photochemistry and Spectroscopy 123

(or)

Vibrational degrees of freedom = 3N – 5 (Linear molecules) 3N – 6 (for non-linear molecules).

3N – 5 fundamental modes of vibration (corresponding to fundamental band) are possible for linear molecules. 3N – 6 vibrational degrees of freedom are available for a non-linear molecule containing N-atoms. Each vibrational degree of freedom corresponds to the fundamental mode of vibration and each fundamental mode corresponds to a bond.

3.12.7

Types of Molecular Vibrations

1. Stretching vibration

2. Bending vibration

1. Stretching Vibration Two bonded atoms continuously oscillate between the bond by changing the distance without altering the bond axis or bond angles. Diatomic: H – H H – Cl – Vibrate only one way Triatomic:

CO2

O=C=O Two different stretching modes (Symmetry or unsymmetry)

O=C=O

O=C=O

Symmetry

Unsymmetry

2. Bending Two bonded atoms continuously oscillate between the bond with a common atom by changing the bond angles in and out of the bond axis plane. These are of four types: Scissoring

Rocking

In-plane bending vibrations

Twisting

Wagging

Out-plane bending vibrations

124 Engineering Chemistry-I

In some cases, because of symmetry factors, two or more vibrational modes may be identical with same energy. Such vibrations are generally referred to as degenerate modes.

3.12.8

Modes of Vibration for some Simple Molecules

The dipole changes during vibration, the molecule is IR active. 1. Water, H2O O

(non-linear) triatomic molecule H

H

3N – 6 = 3 ¥ 3 – 6 = 9 – 6 = 3 fundemental modes of vibration: bending, symmetry stretching or unsymmetry stretching mode. All the three modes of vibration involve change in dipole moment and hence are IR active.

2. Carbon Dioxide, CO2: Linear Triatomic Molecules 3N – 5 = 3 ¥ 3 – 5 = 4 fundamental modes of vibration: Symmetry stretching (IR inactive), antisymmetry stretching (inactive) and two bending modes During the stretching vibration, there is no change in the dipole moment. Hence symmetry stretching will not absorb radiation; hence, it is IR inactive. During asymmetrical stretching, there is a change in the dipole moment; hence this mode is IR active. Two bending modes have the same energy and are said to be degenerate (i.e.) only one absorption bond is expected due to the bending modes of vibration.

Photochemistry and Spectroscopy 125

3. Sulphur Dioxide, SO2: Non-linear Triatomic Molecules Dipole moment of about 1.6. 3N – 6 = 3 ¥ 3 – 6 = 3 fundamental modes of vibration: S O

S O

Symmetry stretching

O

S O

Asymmetry stretching

O

O

Bending mode

NO2 bent (non-linear) triatomic molecule 3N – 6 = 3 ¥ 3 – 6 = 3 All three vibrational modes involve a change in dipole moment and hence are IR active.

Hydrogen, H2 : Linear Diatomic 3N – 5 = 3 ¥ 2 – 5 = 1

H–H stretching is not accompanied by a change in dipole moment and is IR inactive.

3.12.9

Electronic Transition (or) Electronic Spectroscopy (Ultraviolet and Visible Spectroscopy)

It is studied both in emission and absorption and involves transition of an electron from one electronic state (energy level) to another. It is observed in UV and visible region of spectrum. Absorption spectra electrons in the molecule are promoted from the lower electronic state to high electronic state by the absorption of radiation.

126 Engineering Chemistry-I

Emission spectra involves transition of electrons from higher excited state to lower electronic state. Electronic transitions in molecules are accompanied by vibration transitions and the spectra consists of a series of bands. Each band has a number of lines due to rotational transitions. Hence, rotational–vibrational spectrum is superimposed on the electronic spectrum and this process produces a very complex band system.

3.12.10

Born-Oppenheimer Approximation

Total energy of the molecules can be expressed by Born-Oppenheimer approximation. ETotal = Eelectronic + Evibrational + Erotational (Each energy's are independent of each other) Change in energy, DETotal = DEelect + DEvibn + DErotational The frequency of radiation (in terms of wave number, cm–1) emitted or absorbed during electronic transition is DETotal g = hc Wave number associated with the transition can be written as a sum of wave number for each type of transition. g = g elect + g vib + g rot

ULTRAVIOLET (UV) AND VISIBLE SPECTROSCOPY

3.13

Absorption of radiation in the ultraviolet (200 to 4000 nm) and visible (400–750 nm) regions of the electromagnetic spectrum results in transitions between electronic levels, because the energy changes are large. Every molecule can undergo electronic transitions. The large energy changes involved in electronic transitions also cause simultaneous change in vibrational and rotational energies, Eelec > Evib > Erot. The permitted energy levels are shown in Fig. 3.7.

Energy change in ultraviolet and visible region

Photochemistry and Spectroscopy 127

E2

E1

V=3 V=2 V=1 V=0

V=3 V=2 V=1 V=0

J = 0,1, etc.

Fig. 3.7 Permitted energy levels in ultraviolet and visible regions

The rotational–vibrational spectrum is superimposed on the electronic spectrum and this process produces a very complex band system. So this is otherwise known as electronic spectra. The actual amount of energy required depends on the difference in energy between the ground state and the excited state of the electrons. E1 – Eo = hn. 3.13.1

Energy Level Diagram

Energy absorbed in the visible and UV region by a molecule undergoes transitions of valence electrons in the molecules. These transition are, s–s*; n–s*; nÆp* and p–p* Based on the spectral selection rule, n–p* transition is forbidden and all other transitions are allowed. The energy-level diagram for a molecule is shown in Fig. 3.8. The energy values for different transitions are in the following order: n–p* < pÆp* < nÆs* < sÆs*

128 Engineering Chemistry-I s*(antibonding)

Energy

p*(antibonding)

n

s*

n

p

p*

n (antibonding)

p*

p

p (bonding)

s*

s (bonding) s

s*

Fig. 3.8

3.13.2

s

p*

Relative energies of various electronic transitions

Various Types of Electronic Transitions in Organic Molecules

1. r-r * Transition Transition of an electron from bonding sigma orbital to the higher energy antibonding sigma orbital. Since sigma bonds are very strong, the energy required for s Æ s* transition is very high and the absorption occurs only at very short wavelength (around 150 nm). Example

Alkanes

(a) CH4 : sÆs* Æ lmax = 121.9 nm (b) C2H6 : sÆs* Æ lmax = 135 nm. 2. nãr * Transition This type of transitions occurs in the saturated compounds having one hetero atom with lone pair of electrons (n-electrons). The energy required for these transitions are lesser than that required for sÆs* transitions and therefore absorption takes place at a longer wavelength (around 185 nm). Example

N(CH3)3

sÆs* transition occurs at l = 99 nm while nÆs* transition occurs at l = 227 nm

Photochemistry and Spectroscopy 129

3. o ão* Transition This type of transition can be seen in compounds with unsaturated centres (i.e., compounds with p-electron system). Examples

Alkenes, aromatics, carbonyl compounds

In ethylene pÆp* transition occurs at l = 171 nm In acetone p Æ p* transition occurs at l = 189 nm 4. não* Transition This type of transitions are shown by unsaturated molecules containing hetero atoms like N, O and S. This transition involved least amount of energy than all the transitions and therefore absorption take place at longer wavelengths. Example (a) Saturated aldehydes and ketones

For CH3–CO–CH3, nÆp*

transition occurs at l = 279 nm

(b) Unsaturated aldehydes and ketones For CH2=CH–CO–CH3, nÆp* transition occurs at l = 324 nm

3.13.3

Chromophores

Chromophore is a functional group which has a characteristic absorption in the visible or UV region. Such a group contains double or triple bonds. Colour usually appears in an organic compound when it contains unsaturated groups which should more appropriately be called groups with multiple bonds. Example Benzene, nitro, nitroso, azo, carbonyl, etc. Chromophores undergo pÆp* transitions in the short wavelength regions of UV radiations. 1. Chromogen The compound containing the chromophoric group is called chromogen. Depth of colour increases with the number of chromophores. A single C=C group, as in ethylene, does not produce any colour. The colour develops as if a number of these groups are present in conjugation. For example,

CH3(CH=CH)6 – CH3 is yellow in colour. 2. Auxochrome They are functional groups that increase the colour-imparting properties of chromophores without themselves acting as chromophores.

130 Engineering Chemistry-I

For example, –OH

–OR

hydroxyl alkoxy

3.13.4

–NH2

–NHR

–NR2

amino

alkylated

amino

Absorption and Intensity Shifts

The lmax of an absorption may be shifted to higher or lower wavelength for various reasons. Similarly, the value of emax becomes higher or lower (Fig. 3.9). 1. Bathochromic Shift or Red Shift The shift of absorption maximum (lmax) towards longer wavelength due to the presence of auxochrome is called bathochromic shift, (e.g.) n Æ p* transition for carbonyl compounds experience bathochromic shifts. 2. Hypsochromic Shift or Blue Shift It leads to the shift of absorption maximum towards shorter wavelength, (e.g.) absorption maximum of aniline shifts from 280 mm to 200 mm. 3. Hyperchromic Effect (or) Shift It leads to the increased intensity of absorption. 4. Hypochromic Effect or Shift It leads to the decreased intensity of absorption. Hyperchromic shift Hypsochromic shift

Bathochromic shift

emax Hypochromic shift

Wave lenght

Fig. 3.9 The various shifts in emax and lmax

3.13.5

Instrumentation of UV Spectrometer

The important components of UV-visible spectrophotometer are as follows. 1. Radiation Source UV region—Deuterium discharge lamp or xenon arc lamp Visible region—Tungsten filament or tungsten iodide lamp

Photochemistry and Spectroscopy 131

Criteria of Radiation Source

(a) Intensity of radiation must be sufficient (b) Stable and continuous radiation 2. Monochromators Monochromators are used to disperse the radiation according to the wavelength. The essential components of a monochromator are

(a) Entrance slit (b) Dispersing element (a prism or a grating) (c) Exit slit 3. Beam Splitter It is a rotating disc (chopper). Half of the chopper is mirrored. This helps in getting the splitted beams of equal intensity by means of reflection and transmission. 4. Cells (Sample and Reference Cells) Visible region—Sample container must be of matched pair of glass cuvets

UV region—Quartz cuvets Specific care must be taken that reflection/refraction interferences are negligible. 5. Detectors The following three type of detectors are commonly used:

(a) Photovoltaic or barrier-layer cell (b) Photocell or photoemissive cell (c) Photomultiplier tube The detector converts the radiation into electric current. The magnitude of the current is directly proportional to the concentration of the solution. 6. Recorder The signal from the detector is received by the recording system provided with a recorder pen.

3.13.6

Functioning of UV-visible Spectrophotometer

A beam of light from a source is allowed to pass through a monochromator unit in a mirror system. The radiation of narrow range of wavelengths coming out of the monochromator (exit slit) is received by the rotator system which divides the beam into two identical beams.

132 Engineering Chemistry-I

The two beams, one emerging from the sample cell and the other from the reference cell, are focussed on the detector. The instrument detector is so designed that it can compare the intensities of the two beams at each wavelength of the region. The output from the detector is connected to a phase-sensitive amplifier. The signals transmitted by the amplifier are transmitted to a recorder. If the compound absorbs light at a particular wavelength then the intensity of the sample beam (I) will be less than that of the reference beam (I0). A plot of wavelength of the entire region versus the absorbance (A) of the light at each wavelength gives the graph, where absorbance (A) = log I0/I. Such a graph is known as absorption spectrum. The chart drive is coupled to the rotation of the prism. Thus, the absorbance and transmittance of the sample is recorded as a function of wavelength. The schematic block diagram of a double-beam visible UV spectrophotometer is given in Fig. 3.10. Sample

Source

Detection unit

Monochromator

Recorder

Reference Beam splitter

Fig. 3.10

3.13.7

Visible UV spectrophotometer

Applications of UV Spectroscopy

1. Structural Elucidation The presence or absence of a particular functional group can be detected by observing UV spectrum. From the l max and e max values, an idea about the nature of the molecule is obtained (Fig. 3.11). Examples

(a) Conjugation (b) Carbonyl group (aldehyde/ketones) (c) Benzene or aromatic compounds (d) Tautomerism (keto and enol form)

Photochemistry and Spectroscopy 133

Absorption

Absorption

–C– ||

200 250 300 350 Wavelength (nm)

O

200 250 300 350 Wavelength (nm)

Fig. 3.11 Benzene (lmax = 255 nm) and carbonyl absorption (lmax = 300-350 nm)

2. Detection of Impurities This is one of the best methods to detect the impurities present in organic compounds because of the following reasons.

(a) Presence of UV active impurities related to the sample shows strong absorption band and have characteristic emax values. (b) Extra absorption peaks other than original compound indicates the presence of impurities. Examples

(a) The most common impurity in cyclohexane, i.e., benzene, can be easily detected by the UV absorption band at 255 nm (Fig. 3.11). (b) Aromatic and unsaturated impurities present in the raw materials of adipic acid and hexamethylene diamine, used for the manufacture of Nylon 6,6 can be easily detected by UV spectra. Only the pure raw materials give high quality product. 3. Reaction Progress Progress of the reaction can be monitored by recording the spectrum at different time intervals and observing for the absence of absorption band corresponding to starting material at a particular wavelength, e.g. Reduction of aldehydes/ketones to corresponding alcohols H] C6H5CHO æ[æ Æ C6H5CH2OH

If the reaction is complete then the absorption peak due to nÆp* of carbonyl group will be absent. 4. Geometrical isomerism Cis-trans isomerism can be followed by UV spectroscopy because trans isomers normally absorb at a longer wavelength (with larger emax value) than the cis isomer.

134 Engineering Chemistry-I

For example, stilbene H5C6

C6H5

H5C6 C=C

C=C H

H

H

H

C6H5 trans-isomer lmax = 295 nm

cis-isomer lmax = 283 nm

Absorbance

5. Quantitative Analysis The concentration of a substance in solution can be determined by recording the absorbance of solutions prepared at different concentrations and also solution of the substance of unknown concentration. A graph of concentration vs absorbance is drawn (Fig. 3.12).

Unknown concentration Concentration

Fig. 3.12

Calibration graph

From Beer’s law, I log ÊÁ o ˆ˜ = A ŒCl Ë I¯ where,

Π= molar extinction coefficient C = concentration l = length of the cell

A straight line passing through the origin is obtained. From the graph, the unknown concentration is determined. Determination of concentration of transition metals Co, Ni, Al, etc., can be done by this method. 6. Determination of Molecular Weight Molecular weight determination is possible provided suitable derivatives could be prepared which absorb in the UV-visible range. For example, molecular weight of amine can be determined by first converting it into amine picrate. A known weight of it is dissolved per litre of solution and its absorbance is determined at lmax.

Photochemistry and Spectroscopy 135

7. Determination of calcium in blood serum can be determined by this spectroscopy. 8. Fluorescence and phosphorescence can be identified in substance by UV spectroscopy. 9. Charge transfer complexes can be studied by UV-visible spectroscopy.

INFRARED SPECTROSCOPY

3.14

IR Spectroscopy deals with the study of impact of infrared radiation on molecules. When a beam of IR radiation is passed through a molecule, some of the frequencies are absorbed and transitions take place between vibrational levels. Hence, IR spectroscopy is also called vibrational spectroscopy.

3.14.1

Vibrational Spectroscopy

It involves the transitions between the vibrational energy levels of a molecule on the absorption of radiations falling in the spectral range of 10 cm–1 – 12800 cm–1 (infrared region). A single vibrational energy change is accompained by a large number of rotational energy changes. Thus, the vibrational spectra appear as vibrational–rotational bands. This region is further divided into three subregions: 1. The region from 12500 to 4000 cm–1 is near infrared. 2. The region from 4000 to 667 cm–1 is the infrared or ordinary IR. 3. The region from 667 to 50 cm–1 is the far IR. The sources of IR are the electrically heated rod of rare-earth oxides.

3.14.2

Molecular Vibrations of IR Spectrum

The following are some of the regions of stretching frequencies of the bonds: C–C, C–O, C–N C=C, C=O, C=N, N=O C∫C, C=N

1300–800 cm–1 (7.7–12.5 mm) 1900–1500 cm–1 (5.3–6.7 mm) 2300–2000 cm–1 (2.6–3.7 mm)

1. Stretching vibrations There are mainly two types of vibrations, namely stretching and bending, or deforming, vibrations. In stretching vibrations, the atoms

136 Engineering Chemistry-I

move essentially along the bond axis so that the bond length increases or decreases at regular intervals but the atoms remain in the same bond axis. Stretching vibrations are of two types (Fig. 3.13).

Asymmetrical Stretching

Symmetrical Stretching

Fig. 3.13

In-plane Out-of-plane Bending Bending or or Scissoring Wagging

In-plane Bending or Twisting

In-plane Bending or Rocking

Stretching and bending vibrations

(a) Symmetrical Stretching In symmetrical stretching, with respect to a particular atom the other two atoms in a molecule move in the same direction, i.e. either away (or) towards with respect to a particular atom. For example in methylene group, the two hydrogen atoms move away from the central carbon atom without change in the bond angle. (b) Asymmetric Stretching In asymmetric stretching, one atom moves away from the central atom, while the other atom moves towards the central atom. For example in methylene group, one hydrogen atom approaches the carbon and the other hydrogen atom moves away from the carbon atom. Methylene group has both symmetrical and asymmetrical stretching frequencies. 2. Bending or Deforming Vibrations Bending or deforming vibrations may produce change in bond angle between bonds with a common atom. There are mainly four types of bending vibrations and they are given above (Fig. 3.13).

(a) Scissoring (b) Rocking (c) Wagging (d) Twisting (a) Scissoring In the scissoring kind of bending vibrations, the two atoms connected to the central atom move toward and away from each other with the change in valency angle along the same plane. It is a in-plane bending. (b) Rocking In the rocking type of bending motion, the atoms attached to the central atom together swing back and forth in the plane of the molecule. This is also an in-plane bending.

Photochemistry and Spectroscopy 137

(c) Wagging In wagging type of bending motion. The atoms attached to the central carbon atom swings back and forth, out of plane of the molecule. This is an out-of-plane bending. (d) Twisting In twisting, the atoms attached to the central carbon atom move front and back, in and out of plane. Examples of Stretching and Bending Vibrations

The number of fundamental (or) normal vibrational modes of a molecule can be calculated as follows: (i) For Non-linear Molecule A non-linear molecule containing ‘n’ atoms has (3n – 6) fundamental vibrational modes. Examples

CH4 Æ (3 ¥ 5 – 6) = 9 Fundamental vibrational modes

Ŵ Water Water is a bend (non-linear) triatomic molecule and has 3n – 6 = (3 ¥ 3 – 6) = 3 fundamental vibrational modes. These modes and their corresponding frequencies are shown below (Fig. 3.14).

(a) g sym = 3652 cm–1

(b) g asym = 3756 cm–1

(c) Bending (g = 1596) cm –1

Fig. 3.14 Fundamental vibrational modes in water molecules

The IR spectrum of water exhibits 3 absorption bands at 1596 cm–1, 3652 cm–1 and 3756 cm–1 corresponding to the bending, symmetric stretching and the asymmetric stretching vibrations respectively. Thus, for a vibration to be IR active, the dipole moment of the molecule must change.

138 Engineering Chemistry-I

(ii) For linear molecules A linear molecule containing ‘n’ atoms has (3n – 5) fundamental vibrational modes. Example Carbon dioxide CO2 is a linear triatomic molecule and has 3n – 5 (3 ¥ 3 – 5) = 4 fundamental vibrational modes. They are • Symmetric stretching (g = 1340 cm–1) • Asymmetric stretching (g = 2350 cm–1) • In-plane bending (g = 666 cm–1) and • Out-of-plane bending (g = 666 cm–1).

O¨CÆO Symmetric stretching OÆCÆO Asymmetric stretching

≠ O—C—O Ø Ø and – O — C+ — O– Bending

From the four absorption fundamental vibrations, both in-plane and out-of-plane bending and asymmetric stretching vibrations involve change in dipolemoment (all are IR-active). But the symmetric stretching vibration does not involve any change in the dipole moment (IR inactive). In addition to the fundamental vibrations such as stretching and bending vibrations, overtones and beats are also observed in IR. 3. Overtones These are weak absorptions with frequencies corresponding to the integral multiple of fundamental vibrations. 4. Beats There may be the absorptions corresponding to the combination of two frequencies (g1 + g2) or difference of two frequencies (g1 – g2).

3.14.3

Analysis of IR Spectrum

1. An IR spectrum gives the following informations. Frequency range corresponding to 900–1400 cm–1 contains a large number of unassigned vibrations. This is called fingerprint region and is useful for the identity of a compound.

Photochemistry and Spectroscopy 139

2. Frequency range of 1400–2300 cm–1 furnishes evidence for the presence of different functional groups such as C=C, –C∫C–, benzene, –CHO, C=O, –COOH, –COOR, –CONH2, –C∫N, etc. This region is called the group frequency region. 3. Frequency range of 2800–3200 cm–1 provides information on the stretching vibrations of aliphatic and aromatic –CH bonds. 4. Study of intermolecular and intramolecular hydrogen bonding due to –OH, –NH and –SH groups in molecules is made in the frequency range of 3300–3700 cm–1. Thus, most of the organic applications of infrared spectroscopy is concentrated in the frequency range of 1400–4000 cm–1.

3.14.4

Principle of IR Spectra

1. IR spectra is produced due to the energy released by different modes of vibration and rotation of molecule in a compound. 2. Infrared spectra are usually plotted between percentage transmittance as ordinate (y-axis) and wave numbers as absica (x-axis). This makes absorption bands appear as dips in the curve form instead of peaks as seen the UV spectra. 3. Each dip in the spectrum is called as a band or peak and represents absorption of IR radiation at that frequency by the sample. 4. The absorption of IR radiation is based on the increasing the energy of vibration and rotation associated with a covalent bond in a molecule provided there is a change in the dipole moment of the molecule. 5. Most of the organic molecules undergo such change in dipole moment except a few. For example diatomic molecules such as O2, N2, H2, Cl2, etc., have no absorption in the IR region since there is no net change in dipole moment occurs during the vibration of these molecules.

3.14.5

Instrumentation

1. Components Infrared spectrophotometers have the components similar to that of ultraviolet spectrophometers. They are (a) light source, (b) monochromators, (c) sample holder or cell, (d) detector, and (e) recorder. (a) Light Source For infrared radiation, an incandescent material is used as a source of IR radiation. Generally, the solid material is heated electrically to 1200–1500°C. Some of the important sources are (i) Nernst glower, (ii) Globar, (iii) carbon arc, and (iv) heated nichrome or rhodium wire.

140 Engineering Chemistry-I

The Nernst glower consists of a hollow tube about 2 cm long and 1 mm in diameter made up of sintered cerium oxide, zirconium oxide, thorium oxide and yttrium oxide mixtures. Globar consists of a silicon carbide rod which when heated to approximately 1200°C, emits IR radiation (1–40 m region). Globar is more stable than the Nernst glower. (b) Monochromator Quartz can be used as a prism material to get IR radiation from 0.76–8 mm. But generally, sodium chloride is used in the IR spectrophotometer as the prism material. It can be used in the range of 2.5 mm to 20 mm. For the far IR region, potassium bromide or cerium bromide prism materials can also be used. (c) Sample Holder or Cell The sample holder is generally made up of a glass or metal cylinder (10 cm long). The sample holders contain sodium chloride windows. Sample holders are also purely made up of alkali metal halides such as sodium chloride (NaCl) or potassium bromide (KBr). These sample holders being hygroscopic, should be protected against moisture at a suitable temperature. Polishing with buffer powders maintains them in their original condition. (d) Detector IR detectors convert thermal radiant energy into electrical energy. The important detectors are

(i) Photoconductivity cell (or) photodetectors (ii) Thermocouple or Thermal detectors (iii) Pyroelectric detectors (e) Recorder detector.

The recorder records the signal coming out from the

2. Working of IR Spectrophotometer In a single-beam spectrophotometer, the radiation emitted from the source passes through the sample, collimating mirror and then through a prism and a littrow mirror. The prism and the littrow mirror selects the desired wavelength of radiation and allows it to fall on the detector and finally the recording system (Fig. 3.15).

Photochemistry and Spectroscopy 141

Sampling area Condensing optics and beam chopper

Radiation source

Detector

Monochromator

Amplifier and recorder

Fig. 3.15 Block diagram of IR spectrophotometer

In the double-beam spectrophotometer, the source is split into two identical beams having equal intensity; one of the beams passes through the sample and the other beam passes through the reference for compensation. Sample

Recorder Detector

Source Monochromator Reference Beam splitter

Fig. 3.16

Block diagram of double-beam IR spectrometer

The two beams are recombined onto a common axis and are alternately focussed by sector mirror on to the entrance slit of the monochromator. Finally, the beam reaches the detector. The signal from the detector then passes through a servometer to the recorder (Fig. 3.16).

3.14.6

Application of Infrared Spectroscopy

1. IdentiƝcation of a Substance The fingerprint region of IR (900–1400 cm–1) helps in identifying a substance. This is illustrated in Table 3.3. Table 3.3 Group frequencies of vibrations Bond

Vibration

Frequency (cm–1)

Si–H

Stretching bending

2200 800–950

Si–OH

Si–O stretching

830–110 (strong)

Si–Cl

Stretching

666 cm–1

Si–F

Stretching

800–1000

P=O

Stretching

1150–1250

P–OH

P–O stretching

910–1040 (strong)

142 Engineering Chemistry-I 2. Study of Geometrical Isomerism Presence of cis and trans isomers of a compound can be identified by IR spectroscopy, e.g., Disubstituted alkenes of the form CHR1 = CHR2 with different –C=C– stretching frequencies for the cis and trans isomer, are given below. H

H

R1

C=C R1

H C=C

R2

cis-alkene 1662–1626 cm–1 730–665 cm–1

H

R2 trans-alkene 1678–1668 cm–1 980–960 cm–1

3. Checking the Presence of Moisture in the Given Sample The IR spectrum of a substance shows a broad band around 3300–3600 cm–1 if the sample contains moisture. Eliminating the moisture by drying and recording the IR spectrum again reveals the absence of the peak at 3300–3600 cm–1. 4. Study of Hydrogen Bonding in Molecules IR spectroscopy is a powerful tool for the study of intermolecular and intramolecular hydrogen bonding. 5. Assessment of Purity of a Sample The IR spectroscopy helps in detecting the presence of trace impurities in substances. Bands appearing at frequencies other than that of the sample, reveals the absorptions due to impurities. From the position of the absorption bands, the type of impurity can be identified. For example, commercial acetone might contain traces of acetic acid as impurity. The IR spectrum of such a sample shows bands in the region of 2500–3300 cm–1 for –OH vibration superimposed with –CH stretching and a band in the region of 1710 cm–1 corresponding to the carbonyl group of the acid. Acetone shows very intense bands at 1720 cm–1 for the carbonyl and sharp –CH stretching vibrations (2900–3050 cm–1). 6. Characterisation of Organic Substances Two samples having the same molecular formula and molecular weight can be differentiated by IR spectroscopy through functional-group analysis.

Photochemistry and Spectroscopy 143

For example, a molecule C3H6O exhibiting functional isomerism. The two substances having the formula C3H6O are analysed by IR spectroscopy. One of the sample shows a strong absorption band at 1720 cm–1 revealing the presence of carbonyl group. In the other sample, band at 3300– 3500 cm–1 is seen which confirms the presence of a –OH group. Hence, it is inferred that Sample 1 might be a ketone and Sample 2 might be an alcohol. 7. Differentiation of Aliphatic and Aromatic Compounds For example, CH3COCH3 Aliphatic–Absorption band at 1720 cm–1

C6H5COC6H5 Aromatic–Absorption band at 1665 cm–1 Nowadays, FT (Fourier-Transform) IR spectrometers are used in most laboratories. The major advantages of FT-IR instruments over dispersive spectrometers include better speed and sensitivity, better light gathering power, more accurate wavelength calibration, simple mechanical design, and the virtual elimination of the problems of stray light and IR emission. Because of these advantages, nearly all the new IR instruments are FT-IR systems. FT–IR spectrometers detect all the wavelengths at all the times. The comparison between a spectrophotometer and colorimeter is given in Table 3.4. Table 3.4 S. No

Components of spectrophotometer and colorimeter

Components

Spectrophotometer

Colorimeter

1.

Radiation source

Hydrogen or deuterium Tungsten filament lamp lamp

2.

Optical system

Prism or diffraction grating or monochromator

3.

Materials used in the Quartz or fused silica optical system

4.

Sample holders

Quartz or fused silica, Round glass cells rectangular cells

5.

Detector

Photomultiplier

Tinted glass filters

Round glass cells

Photovoltaic cell

144 Engineering Chemistry-I

QUESTIONS PART-A 1. What are called photochemical reactions? Give one example. 2. Write any two differences between photochemical and thermal reactions. 3. State Lamberty Beer Law? 4. What is stark-Einstein law of photochemical equivalence? 5. What is Grotthuss-Draper law? 6. Define quantum efficiency or quantum yield? 7. Define fluorescence? 8. Define the term Phosphorescences. 9. What are called chemiluminescence? Give one example. 10. Define Photosensitization. 11. Write the relationship between Absorbance (A) and transmittance (T). 12. Calculate the molar absosption of a 1 ¥ 10–4 M solution, the absorbance of which is 0.20 when the path length is 2.5 cm. 13. What are chromophores? Give some examples. 14. What are auxochrome? Give some examples. 15. What is finger print region? Mention its important uses. 16. Define the term Bathochromic shift? 17. Write any two applications of IR Spectroscopy. 18. Write the frequency regions of near IR, IR and far IR. 19. Define the term Hypsochromic shift or Blue Shift. 20. List out the various types of electronic transitions in organic molecules. 21. What is Born-oppenheimer Approximation? 22. What are Absorption spectrum? 23. What are the sources of uv light in uv visible spectrometers. 24. What is Electromagnetic radiation? 25. Explain how uv-visible spectrometric methods are useful for quantitative analysis. 26. What is finger print region? Mention its important uses.

PART-B 1. List out and explain the fundamental photochemical laws. [P.86–90] 2. Derive and explain Beer-lambert‘s laws. [P.90–92] 3. When a monochromatic light is passed through a cell of 1 cm length, the intensity is reduced to 10%. If the same radiation is passed

Photochemistry and Spectroscopy 145

4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17.

18.

through a cell of 8 cm length, what is the transmittance radiation? Calculate the length of a cell in order to have 25% absorbance. [Ans: (l) = 0.43, 13.1 cm] How photochemical reactions are classified based on quantum yield? [P.98–99] Write the possible reasons for high and low quantum yield. [P.99] What are called chemical Actinometers. List out and explain two types of chemical Actinometers. [P.101–102] Write the mechanism of fluorescence and Phosphorescence by using Jablonski diagram. [P.104] What is photo sensitisation? Write its mechanism. [P.106] What is photo chemical inhibitions. Explain with examples. [P.108–109] Calculate the energy per mole of light having wavelngth of 85 nm. [Œ = 1.407 ¥ 103 kJ/mol] What are the various types of electronic transitions? [P.128–129] Explain the various components and working of uv-visible spectrometers. [P.130–132] Write the principles and working of a uv-visible spectrometer giving the block diagram. [P.131] Outline the applications of IR spectroscopy. [P.141–143] Derive Beer-Lambert’s law and write all the limitations observed in the quantitative analysis. [P.90–92] Discuss the phenomena of fluosescence and phosphosescence. [P.102–103] Explain [P.129] (i) Chromophores (ii) chromogen (iii) Auxochromes. Write the principles of IR spectra. [P.139]

Two-Mark Questions and Answers 1. What are called photochemical reactions? Give one example. The study of chemical reactions which occur after electronic excitation of atoms or molecules with electromagnetic radiation, having wavelengths approximately 200 nm to 800 nm (uv and visible regions) are called photochemical reactions. hg (e.g) H2 + Cl2 ææÆ 2 Hcl.

146 Engineering Chemistry-I

2. Write two differences between thermal and photochemical reactions. S.No.

Thermal reactions

Photochemical reactions

1.

This reaction involve the absosption or evaluation of heat.

These reactions involve absorption of light radiation.

2.

DG is always Negative

DG may be +ive or –ive.

3.

Temperature has a major role on Temperature has very little the rate of such reactions influence.

3. What a Grotthuss-Draper law? Any molecule which absosbs light will undergo photochemical reaction. 4. What is Stark-Einstein‘s law of photochemical equivalence? Each molecule which takes part in a photochemical primary process absorbs one photon or quantum of light causing the reaction. 5. Define Beer-Lambert’s law. When a beam of monochromatic light is passed through a solution of an absorbing substance, the rate of decrease of intensity of radiation with thickness of the absorbing solution is proportional to the intensity of incident radiation (I) as well as to the concentration (c) of the solution. A = Œcx where Œ is the molar absorptivity coefficient 6. What is absorbance (A)? The negative logarithm to the bage 10 of the transmittance of the solution is called absorbance (A). I A = log o = – log10 T = Œbc I Where Io and I are the intensity of incident and transmitted radiation respectively. Œ is extinction coefficient. 7. A solution of 2 cm thickness transmits 40% incident light. Calculate to concentration of the solution given that Œ = 6000 dm3 mol cm–3. Formula: A = Œcx Given data: x = 2 cm Œ = 6000 dm3 mol–1 cm–1 % T = 40% \ T = We know

40 = 0.40 100 A = – log T = – log 0.40 = 0.3979

Photochemistry and Spectroscopy 147

Solution C=

A 0.3979 = Œ x 2 ¥ 6000

= 3.316 ¥ 10–5 Result: Concentration of the solution (c) = 3.316 ¥ 10–5 mol dm–3. 8. Define quantum yield. It is defined as the ratio of number of molecules of reactants consumed or the number of molecules of product formed to the quanta of light absosbed. f=

Number of molecules consumed or formed in a given time Number of photon n(quanta) absosbed in the same time

9. What is fluorescence? When a beam of light is incident on certain substances, they emit visible light or radiation and they stop emitting light or radiations as soon as the incident light is cut off. This is known as fluorescence. e.g: organic dyes like Flurescein Eosin 10. What are chromophores? Give some examples. [A.u.May-2003] Chromophores is a functional group which has a characteristic absorption in the visible or UV region. Such group contains double or triple bonds. Example: Benzene. nitro, nitoso, azo, carbonyl etc ., 11. What are auxochrome? Give some examples. They are functional groups that increase the colour imparting properties of chromophores without being itself acting as a chromophore. e.g: –OH –OR –NH2 –NHR –NR2 hydroxyl

alkoxy

amino

alkylated

amino

12. Define absorbance (A) and Transmittance (T). Absorbance, A: The negative logarithm to the base 10 of the transmittance of the solution is called Absorbance. Transmittance, T: The fraction of the incident light that is transmitted by a given species is called transmittance. 13. What is fingerprint region? Mention its important uses. IR spectrum gives the following informations Frequency range corresponding to 900-1400 cm-1 contains a large number of unassigned vibrations. This is called fingerprint region and is useful for the identity of a compound.

148 Engineering Chemistry-I

14. Define the term Bathochromic shift. It leads to the shift of absorption maximum (l max) towards longer wave length due to the presence of auxochrome is called bathochromic shift. (e.g.,) n Æ p* transition for carbonyl compounds experience bathochromic shifts. 15. Write any two applications of IR spectroscopy. (i) Checking the presence of moisture in the given sample: The IR spectrum of a substance shows a broad band around 3300 - 3600 cm–1 if the sample contains moisture. Eliminating the moisture by drying and recording the IR spectrum again reveals the absence of the peak at 3300 – 3600 cm–1. (ii) Study of hydrogen bonding in molecules: IR spectroscopy is a powerful tool for the study of intermolecular and intramolecular hydrogen bonding.

Qibtf!Svmf boe!Bmmpzt V

4.1

Phase Rule—Introduction

4.2

Terms Involved in Phase Rule

4.3 Application of Phase Rule to One Component

O

Systems 4.4

Construction of Phase Diagram by Thermal Analysis

4.5

J

Phase Diagram of Lead-Silver System

4.6 Alloys 4.7

Importance of Alloys

4.8 Applications of Alloy Steels

U JW

4.9

Heat Treatment of Alloys (Steel)

5 D

Qibtf!Svmf!boe!Bmmpzt

I B Q U F S

PHASE RULE: INTRODUCTION

4.1

Physiochemical systems can be identified as homogeneous and heterogeneous systems. • A solution of sodium chloride in water, for example, is uniform throughout in physical and chemical properties. • The system is CH3COOH (l) + C2H5OH(l)

CH3COOC2H5 (l) + H2O (l)

Such systems which contain single phase are called homogeneous systems. A heterogeneous system is one which consists of parts which have different physical properties. For example, in ice-water-vapour system, there are three mechanically separable and physically distinct portions called phases. So, heterogeneous systems contain more than single phase, each phase being homogeneous in itself. The phase rule, enunciated by Willard Gibbs (1876), is an important generalisation dealing with the behaviour of heterogeneous systems at equilibrium. The relationship between phases in equilibrium in a system as a function of pressure, temperature and composition can be represented graphically called phase diagrams. The study of phases relationships is useful in understanding the properties of materials in general and has several practical applications in metallurgical processes, separation of components by solvent extraction, steam distillation and zone refining.

152 Engineering Chemistry-I

Phase Rule Statement "The equilibrium between any number of phases is not influenced by gravity or electrical or magnetic forces, or by surface action. It is influenced only by temperature, pressure and concentration then the number of degrees of freedom (F) of the system is related to the number of components (C) and of phases (P) by the phase rule equation. F=C–P+2

This rule is properly applied for any system at equilibrium at a definite temperature and pressure.

TERMS INVOLVED IN PHASE RULE 4.2.1

4.2

Phase

A homogeneous physically distinct parts and mechanically separable portion of a system which is separated from other parts of the system by definite boundary surfaces is called a phase. Examples 1. Gaseous Phase A mixture of two or more gases is homogeneous. So, in this system, the phase is one, e.g. air (a mixture of O2, H2, N2, CO2 and water vapour, etc., constitutes a single phases). 2. Liquid Phase (a) Two immiscible liquids in contact form two different phases, e.g. carbon tetrachloride and water system. When carbon tetrachloride and water are mixed, they form two distinct layers which are physically distinct and mechanically separable. (b) If two liquids are miscible (alcohol and water) then they will form only one liquid phase. 3. A solution (solute dissolved in a solvent) consists of one phase only, e.g. glucose solution in water. 4. Each solid makes up a separate phase, except in the case of solid solutions, e.g. many forms of sulphur can exist together, but these are all separate phases.

S(rhombic)

S(monoclinic)

S(liquid)

S(vapour)

5. A heterogeneous mixture like CaO(s) + CO2(g) CaCO3(s) consists of three phases, i.e. two solids CaCO3 and CaO, and one gaseous, CO2.

Phase Rule and Alloys 153

6. In the equilibrium reactions, Fe(s) + H2O(g) FeO(s) + H2(g) Here, there are two solid phases (Fe, FeO) and one gaseous phase consisting of H2O(g) and H2(g). So, three phases exist in equilibrium. 7. The freezing point of water consists of three phases. Ice(s) Water(1) Water vapour(g)

4.2.2

Component

The smallest number of independent variable constituents, taking part in the state of equilibrium, by means of which the composition of each phase can be expressed in the form of chemical equation is called component. Examples

1. In the water system, Ice(s) Water(l) Vapour(g) the chemical composition present in the equilibrium is H2O/ (water). So, it is a one-component system. 2. The sulphur system consists of four phases: rhombic, monoclinic, liquid and vapour. Here, the chemical composition of the phases is S (sulphur). So, it is a one-component system. SRh

Smono

Sliq

Svap

3. In a dissociation of NH4Cl in a closed vessel, NH4Cl(s)

NH4Cl(g)

NH3(g) + HCl(g)

the proportions of NH3 and HCl are equivalent and the composition of both phases (solid and gaseous) can be expressed in terms of NH4Cl alone. So, the number of components is one. Suppose if NH3 or HCl is in excess, the system becomes a two-component system. 4. Thermal Decomposition of CaCO3 CaCO3(s)

CaO(s) + CO2(g)

The composition of each of the three phases can be expressed in terms of at least any two of the independently variable constituents such as CaCO3, CaO and CO2. Suppose CaCO3 and CaO are chosen as the two components then the composition of different phases is CaCO3 = CaCO3 + 0CaO CaO = 0CaCO3 + CaO CO2 = CaCO3 – CaO Thus, it is a two-component system.

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5. Saturated Solution of NaCl It consists of solid salt, salt solution and water vapour. Here, chemical composition of all the three phases can be expressed in terms of NaCl and H2O. Hence, it is two-component system. 6. Dissociation Reaction CuSO4.3H2O(s) + 2H2O(g) CuSO4.5H2O(s) Here, the components are CuSO4 and H2O as the chemical constituents. So, it is two-component system.

4.2.3

Degree of Freedom (F) or Variance

Pressure, temperature and composition of the phases, etc., are the factors which can affect the equilibrium of a system. The minimum number of independently variable factors, such as temperature, pressure and composition of the phases, which must be arbitrarily specified in order to represent perfectly all the conditions of a system is known as degree of freedom. Consider the following examples to make the definition clear. 1. Ice(s)

Water(1)

Vapour(g)

Here, the three phases will be in equilibrium only at a particular temperature (T) and pressure (P). Hence, the system does not have any degree of freedom. So it is nonvariant or invariant. 2. Consider the equilibrium Water(1)

Water vapour(g)

Here, we must state either T or P to define the system completely. Hence, the degree of freedom is one (or) univariant system. 3. NaCl(s)

NaCl-water(aq)

water vapour(g)

Here, we must state that either temperature or pressure to define the system completely because the saturation solubility is fixed at a particular temperature and pressure. Hence, the degree of freedom is univariant.

4.2.4

Merits of Phase Rule

1. Phase rule is applicable to both physical and chemical equilibria and it requires no information regarding molecular structure. 2. In phase rule, there is no information regarding microstructure. So, it is applicable to macroscopic systems. 3. Phase rule is a specific method of classifying equilibrium states in terms of phases, components and degrees of freedom.

Phase Rule and Alloys 155

4. Phase rule helps us predict the behaviour of a system under different sets of variables. 5. Phase rule is useful to analyse the properties of materials in the heterogeneous equilibrium system. 6. The study of low-melting eutectic alloys used in soldering is based on the principle of phase rule. 7. It indicates that different systems with same degrees of freedom behave similary.

4.2.5

Limitations of Phase Rule

1. It can be applied only for heterogeneous equilibrium systems. Thus, it is not applicable to the systems which are slow in reaching the equilibrium state. 2. It considers only three variables, namely, temperature, pressure and composition. The other variables such as gravity, surface forces, electric and magnetic influence have not been considered. 3. It applies only to a single-equilibrium system and provides no information regarding any other possible equilibria in the system. 4. It has restriction that all phases of the system must be present simultaneously under the identical conditions of temperature and pressure. 5. It has conditions that solid and liquid phases must not be in finely divided state, otherwise deviations occur from the phase rule.

4.2.6

Phase Diagram

A phase diagram is a graphical representation obtained by plotting one variable against another. Thus, it gives the conditions of equilibrium between various phases of a substance. Shape of phase diagrams, depend on the slope of the boundary line between various phases. The effect of changes in variables like temperature, pressure and composition or changes of phase in a system can be understood from the diagram. It gives the properties like melting point, boiling point, transition temperature, triple point, eutectic point, etc. A line in the diagram indicates the equilibrium between two phases. A point denotes that three phases are in equilibrium. In the case of a one-component system, temperature is plotted against pressure taking temperature in the X-coordinate and pressure in the Ycoordinate and the diagram is called P-T phase diagram. In a two-component system, a plot of temperature against composition is known as T-C diagram.

156 Engineering Chemistry-I

The alloy composition is expressed mostly in atomic percentage. These phase diagrams help us in studying and controlling the various processes such as phase separation, solidification of metals, change of structure during heat treatment process like annealing quenching, tempering, etc.

APPLICATION OF PHASE RULE TO VARIOUS COMPONENT SYSTEMS 4.3.1

4.3

One-Component System

The water system and sulphur system are the best examples for onecomponent systems. According to phase rule, F = C – P + 2, it is clear that when there is one component in the system existing in one phase rule, i.e. C = 1 and P = 1 then the degree of freedom (F) is calculated from phase rule F = 1 – 1 + 2 = 2. Hence, all the systems of one component can be completely described graphically on paper, by taking two variable factors, viz. pressure and temperature, on appropriate axes. From the mathematical statement of phase rule in systems of one component, the following is observed. 1. When the number of phases is reduced to one, the systems will have two degrees of freedom and are known as bivariant systems. 2. When the number of phases is reduced to two, the systems will have one degree of freedom and are known as univariant systems. 3. When the three phases co-exist, systems will possess no degree of freedom and are known as nonvariant systems. In the phase diagram of a one-component system, univariant, bivariant and nonvariant systems are represented by line, areas and points respectively. Water System The example of the one-component system, as we know, is H2O which is the only chemical individual involved in it. The three phases in the system are Ice Water Vapours (solid) (liquid) (gas)

The above three single phases may occur in four possible combinations in equilibrium as follows. (a) Liquid–Vapour (c) Solid–Vapour

(b) Liquid–Solid (d) Solid–Liquid–Vapour

Phase Rule and Alloys 157

Fig. 4.1

Phase diagram of water system

The existence of phases in equilibrium at any time depends upon the condition of pressure and temperature. These conditions have been determined experimentally. After plotting these on graph paper by taking pressure and temperature on appropriate axis, we get a phase diagram as in Fig. 4.1. This phase diagram has the following parts: • • • •

The stable curves OA, OB and OC (shown by thick lines) One metastable curve OB' (shown by a dotted line) Three areas AOC, BOC and AOB. Define point O known as triple point.

(a) Explanation of Curves

(i) OA (Vaporization Curve) This is known as vaporization curve. It represents the equilibrium between liquid water and vapour at different temperatures. At every point of the curve, the two phases are present. The starting point of the curve is O, which is the freezing point of water (0.0075°C at 4.58 mm). This curve ends at A, the critical temperature (374°C at 218 atm) beyond which the two phases merge into each other. Water liquid

Water vapour

The water-vapour system is univariant. It may also be shown by phase-rule equation. F = C – P + 2 or F = 1 – 2 + 2 = 1 (since C = 1, P = 2)

158 Engineering Chemistry-I

(ii) OB' Curve The dotted curve OB' is a continuation of OA curve and represents the vapour pressure of supercooled water. This curve is known as metastable curve. When there is slight disturbance, the supercooled phase at once changes to solid ice and the curve merges into OB. Supercool water

Water vapour

(iii) OB (Sublimation Curve) This curve represents the equilibrium between ice and vapour. It shows the effect of pressure on the freezing point of water. The curve starts from the point O and ends at B. This system (solid ice water vapour) has one degree of freedom as shown in the curve (Fig. 4.1). It can also be shown by phase rule equation. F=C–P+ 2=1–2+ 2=1

(univariant)

(iv) OC (Fusion Curve or Melting Point Curve) This curve shows the equilibrium between ice and liquid water at various pressures. It shows the effect of pressure on the melting point of ice. It is to be noted that the line inclines towards the pressure axis which shows that expansion takes place on freezing of water and melting point of ice is decreased by increase of pressure. The upper point of this curve OC goes up to a point (2000 atm, 20°C), where one type of ice changes into another solid–but the solidliquid equilibrium still remains there. This system is a univariant system, as at any point of the curve there are two phases (ice and water). The phase rule equation is Solid ice Water liquid F = C – P + 2 or F = 1 – 2 + 2 or F = 1 [Since C = 1, P = 2] (b) Explanation of Areas Pressure and temperature are the two quantities to define the system completely at any point in an area. Thus, the system which is represented by area has two degrees of variance. In the water system, there are three areas.

(i) AOB—vapours of water only (ii) BOC—ice only (iii) COA—water in liquid from only From the above, it is clear that for a system represented by area Number of phases = 1 (vapour or water or ice) Number of components = 1 (as all the phases can be represented by the chemical formula H2O)

Phase Rule and Alloys 159

According to phase rule equation, F=C–P+2=1–1+2=2 So, systems are bivariant. (c) Triple Point At this point, the three curves OA, OB and OC meet together. At this, all the three phases are in equilibrium. At the point O, temperature and pressure are fixed at 0.0075°C and 4.58 mm. One of the three phases will disappear, when either pressure or temperature is changed. The system had zero degree of freedom at the point O. This can be shown by phase rule equation.

F=C–P+2=1–3+ 2=0 At O, P=3 C=1

(C = 1, P = 3) (Ice, liquid and water vapour) (only H2O can represent all phases)

Concluding the above discussion, the main features of the phase diagram are as in Table 4.1 given below. Table 4.1 Main features of phase diagram System

Phase

Degree of freedom

Curves (OA, OA', OB, OC)

2

1

Areas (AOB, BOC, COA)

1

2

Triple point (O)

3

0

4.3.2

Two-component Systems (Condensed Phase Rule)

For a two-component system, C = 2 and the phase rule equation is written as F=C–P+2 F=2–P+2=4–P ...(1) Since the minimum number of phases for any system is one, therefore from Eq. (1), it follows that the maximum number of degrees of freedom in a two-component system is F=4–1=3 This means that in order to define a two-component system completely, all the three variables—pressure, temperature and concentration— must be specified.

160 Engineering Chemistry-I

In order to represent these relations graphically, we require a threedimensional diagram in which three co-ordinate axes representing pressure, temperature and concentration are at right angles to each other. This will lead to a space model which is difficult to construct on paper. In order to remove this difficulty, the common practice is to fix one of the three independent variables, i.e. any two of three variables which will graphically represent the conditions of equilibrium for a two-component system. The various equilibria which can be possible for a two-component system are (a) solid–liquid equilibria, (b) solid–gas equilibria, (c) liquid– liquid equilibria, and (d) liquid–gas equilibria. We shall be limiting ourselves to the study of only solid–liquid equilibria having no gas phase. Such systems having no gas phase are known as condensed systems. Since the effect of pressure is small on this type of equilibria, therefore experiments are usually conducted under atmospheric pressure, thus keeping the pressure constant. This will reduce the degree of freedom of the system by one and for such system, the phase rule becomes F =C–P+1

(2)

This is known as reduced phase rule having two variables, namely, temperature and concentration of one of the constituents. Therefore, solid–liquid equilibria are represented on temperature composition diagrams. Types of Two-component systems (Solid-Liquid Equilibria)

The two-component solid-liquid systems are divided into several types based on their miscibility and on the nature of the solids separating from the solution. The different types are • Simple eutectic formation • Compound formation Types of Compound Formation

(a) Formation of Compound with Congruent Melting Point Here, the formed compound melts exactly at a constant temperature into liquid, having the same composition as that of the solid. (b) Formation of Compound with Incongruent Melting Point Here, the formed compound decomposes completely below its melting point giving out a new solid with a composition different from that of the original.

Phase Rule and Alloys 161

(iii) Solid Solution Formation Here, two metals miscible in both the solid and liquid states form a solid solution. However, here only one type is discussed in which the two components do not react with each other but simply mix into each other in the molten state or in the solution. Such system is known as eutectic system, e.g. silver–lead system.

CONSTRUCTION OF PHASE DIAGRAM BY THERMAL ANALYSIS

4.4

Construction of phase diagram from cooling curves is called thermal analysis. It involves the study of cooling curves for pure metals and for the various compositions of two components of a system during solidification. The form of cooling curve indicates the composition of the solid. When a pure substance in fused or liquid state is allowed to cool slowly and the temperature is noted at definite time intervals, the graphic representation of the rate of cooling will be a continuos curve. The graph represented by plotting fall of temperature with time is called cooling curve.

4.4.1

One Substance

When a pure substance in liquid state is allowed to cool slowly and the temperature is noted at definite times, a continuous curve is obtained (Fig. 4.2a).

Fig. 4.2 Cooling curves

When freezing point is reached and the solid makes its appearance, it is indicated by a break in the continuity of the cooling curve.

162 Engineering Chemistry-I

At this point, the temperature will remain constant until the liquid is completely solidified. Thereafter, the fall in temperature will again become continuous due to heat dissipation.

4.4.2

Two Substances

1. When a mixture of two solids in fused state is allowed to cool slowly, the cooling curve is obtained in a similar manner as described earlier. A continuous line is obtained as long as the mixture is in liquid state. When a solid phase begins to form, the rate of cooling suddenly alters and the cooling curve exhibits a break. But, the temperature does not remain constant as in the previous case. It decreases continuously, but at a different rate. When the mixture forms an eutectic, the fall of temperature continues till the eutectic point is reached (Fig. 4.2b). The temperature remains constant, until the solidification is complete. Thereafter, fall of temperature become uniform, but is different from the previous one. 2. In actual practice, the pure metals A and B and number of mixtures of metal A and B of various compositions between 100% of A and 100% of B are prepared. These are heated above their respective melting point to give homogeneous liquid. Each liquid is then allowed to cool slowly and the temperature is recorded after small intervals of time. In this way, several cooling curves are obtained. B

B A

Temperature

A

O

Time

100% A

100% B Composition

Fig. 4.3

Construction of phase diagram of two solids by using thermal analysis

Phase Rule and Alloys 163

A break in the curve indicates the appearance of a new phase. The first break in each curve occurs at the freezing point. The pure metal will freeze at one constant temperature and the cooling curve will show a horizontal line. For any mixture, it is possible to obtain its freezing point and eutectic point. As the mixtures differ in their compositions, their freezing point will commence at different temperature. The second break occurs at which the temperature remains constant. This gives eutectic temperature, which will be same in all mixtures irrespective of their initial composition. The freezing point of pure A and B (100%) are also determined in a similar manner. Now, the cooling curve of an alloy system of same metals, but of unknown composition is determined and its freezing point located in the T– C diagram. The composition corresponding to this freezing point yields the composition of that alloy (Fig. 4.3).

4.4.3

Eutectic Mixture

Eutectic mixture is a solid solution of two or more substances having the lowest freezing point of all the possible mixtures of the components. This is taken advantage of in alloys of low melting point which are generally eutectic mixtures.

4.4.4

Eutectic Point

Two or more solid substances capable of forming solid solutions with each other have the property of lowering each other's freezing point, and the minimum freezing point attainable corresponding to the eutectic mixture is termed the eutectic point (means lowest melting point). 1. All the eutectic points are melting points, but all the melting points need not be the eutectic points. 2. All the eutectic points are triple points, but all the triple points need not be eutectic points.

PHASE DIAGRAM OF LEADŪSILVER SYSTEM

4.5

It is a two-component system with four possible phases namely solid Ag, solid Pb, solution of Ag + Pb and vapour. As the gaseous phase is practically absent and one variable (pressure) is neglected, the condensed phase rule (reduced phase rule equation) will be applicable. Thus, when F = 0, P is 3, the maximum number of phases (other than vapour) that can co-exist would be three.

164 Engineering Chemistry-I

The phase diagram of the system is drawn taking temperature and composition along the axes (Fig. 4.4). The equilibrium diagram consists of two curves OA and OB, intersecting at the point O.

Temperature °C

961° A

F¢ = C – P + 1 X Liquid (Pb + Ag)

B 327°

Solid Ag + Liquid Y O

Solid Ag + Liquid

303° Solid Ag + Pb 0 Pb 100 Ag

Composition

Eutectic Point 97.4 Pb 100% Pb 2.6 Ag 0 Ag

Fig. 4.4 Phase diagram of lead–silver system

1. Curve AO It represents the freezing point curve of Ag on the successive addition of small quantities of lead. The starting point A represents the melting point (961°C) of pure silver; thus, at this point, pure silver coexists as solid and liquid (vapour being neglected). It is noted that the melting point of silver is lowered gradually by the addition of lead into it. All along this curve, the added lead goes into the solution and the separation of silver occurs till the point O is reached, where no more lead goes into solution. Hence, the point O represents the lowest possible temperature (303°C) in the system which corresponds to fixed composition, viz. 2.6% Ag and 97.4% Pb. This point O is called the eutectic point. Thus, the curve AO represents the effect of addition of lead on the melting point of pure silver. All along the curve AO, two phases (solid silver and liquid) are present in equilibrium neglecting the vapour phase. Thus, the system is univariant. F=C–P+ 1=2–2+1=1 2. Curve BO It represents the freezing point curve of lead to which successive amounts of silver are added. The point B represents the melting point of pure lead (327°C). It is noted that the addition of silver to lead lowers the melting point of lead. All along this curve, the added silver goes into solution and separation of lead occurs till the point O is reached. At this point, no more silver goes into the solution, i.e. the solution is saturated with respect to silver. Hence, the melting point of lead does not fall any more.

Phase Rule and Alloys 165

All along this curve, two phases (solid lead and liquid melt) are present in equilibrium. Therefore, as in the curve AO, the system is univariant here also. F=C–P+1=2–2+1=1 3. Eutectic Point O This is the point where the two freezing point curves AO and BO intersect each other. The point O represents a fixed composition of the two constituents (2.6% Ag and 97.4% lead) and is called eutectic point; the composition at this point is called eutectic composition and the temperature at this point is called eutectic temperature (303°C). No mixture of lead and silver has a melting point lower than the eutectic temperature. At this point, the temperature remains constant, until the whole of the melt solidifies to become solid of eutectic composition. Since a eutectic point has three phases (solid lead, solid silver and their solution) in equilibrium, the system will be invariant, i.e. it has no degree of freedom as evident from the reduced phase rule equation.

F=C–P+1=2–3+1=0 4. Application of the Lead-Silver System to Pattinson's Process The above principle is applied in the Pattinson's process for the desilverisation of lead. A sample of argentiferrous lead, containing less than 2.6% Ag is heated to a temperature well above is melting point so that the system consists only of liquid phase. Let the composition of the alloy (liquid melt) corresponds to point X in Fig. 4.4. The melt is now allowed to cool gradually along the line XY till the point Y is reached where lead starts separating out as solid and thus the residual solution (melt) will comparatively be richer in silver. Further cooling will shift the system along the line YO with the result lead continues to separate out and the melt continues to be richer and richer in silver, till the point O is reached, where an alloy containing about 2.6% Ag is obtained. Areas The area above the line AOB containing both lead and Ag is in molten state (single phase). So the phase rule becomes

F=C–P+1=2–1+1=2

(bivariant)

So, both the temperature and composition is used to specify the system completely. The area below the line AO (solid Ag + liquid melt) The area below the line BO (solid Pb + liquid melt) The area below the point O (eutectic + solid Ag (or) solid Pb)

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From these three areas, we have two phases. Hence, the system is univariant. F = C – P + 1 = 2 – 2 + 1 = 1. 5.

Uses of Eutectic System • Eutectic systems are used in solders for joining two metal pieces together, e.g. Pb–Sn solders. • Suitable alloy compositions are predicted by using eutectic systems, e.g. woods metal (alloy containing 50% Bi, 25% Pb, 12.5% Cd and 12.5% Cu) melts at 65°C only. • Very low melting alloys can be obtained and are used in safety devices, e.g. as plugs in automobiles. • They are used in safety fuses used in buildings to protect them against fire hazards. • They are used in fire sprinklers and as fail safe devices in boilers.

4.5.1

Two Component System with Congruent Melting Point

A compound melts at a constant temperature to give a liquid having the same composition as that of the solid compound. Zinc-magnesium system shows the formation of a congruent melting compound (MgZn2). 1. ZincŪMagnesium Phase Diagram This phase diagram may be the combination of two simple eutectic diagrams joined to yield a maximum at a point B in the figure 4.5. The two eutectic points are E1 and E2 .

Fig. 4.5 Phase diagram of Zn–Mg system

Phase Rule and Alloys 167

Eutectic point E1 is for Zn–MgZn2 system. Molar composition at E1 is 79% Zn and 21% MgZn2 At eutectic point E1, MgZn2(s)

Zn(s)

solution (melt)

There are three phases, namely Zn(s), MgZn2(s) liquid melt and there are two components, namely Zn and MgZn2. At E1,

phase = 3,

component = 2

\ F=C–P+1 =2–3+1=0 So, E1 is invariant point. 2. Eutectic Point, E2 It is for MgZn2–Mg system. The molar composition at E2 is 61% MgZn2 and 39% Mg. The equilibrium phases are

Mg(s)

MgZn2(s)

Liquid melt.

Component = 2 (MgZn2 and Mg) Phase = 3 (Mg(s), MgZn2(s)), Liquid melt.) \

F = C – P + 1 = 2 – 3 + 1 = 0. So the eutectic point E2 also an invariant point.

3. Curves Curve AE1 is the melting point curve of Zn in the presence of Mg, along AE1 with two phases,

Zn(s)

Liquid melt, are in equilibrium.

Similarly, the curve CE2 is the melting point curve (or frequency point curve) of Mg in the presence of zinc along CE2; two phases, Mg(s) liquid melt are in equilibrium. Curve E1BE2 has maximum at the point B (590°C) where the liquid has the same composition as the solid in equilibrium. Thus, temperature at the point B represents the melting point of the compound MgZn2, which is the congruent melting point. The sharpness of maxima gives the stability of the compound formed, flat maxima indicates an appreciable dissociation. At the point B, two phases (solid and liquid) have the same composition. At this temperature, the two-component system becomes one component since the compound is MgZn2. (Component = 1 at point B) So,

F =C+1–P =1+1–2=0

Hence, the congruent melting point is also a nonvariant point.

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Curve BE1 represents depression in freezing point of compound MgZn2 in the presence of Zn. Curve BE2 represents depression in freeing point of the compound MgZn2 in the presence of Mg.

ALLOYS

4.6

Alloys are the homogeneous substances formed by mixing two or more elements, at least one of them being a metal. Types of Alloys Based on the constituents, alloys are of three types. (a) Alloys Formed by Two or More Metals.

Example Brass (Cu–Zn) (b) Alloys Formed Between a Metal and a Nonmetal.

Example WC alloy (Wolframite–Carbon) (c) Alloy Amalgam Formed Between Mercury and Another Metal

Examples Zinc amalgam The characteristics of a metal can be modified by alloying the metal with another metal or nonmetal. Metals and alloys are used in various fields like aircrafts, automobiles, building and bridge construction, railway, light and heavy mechinery and equipment, shipping and transportations, etc. Metals and alloys used in industry possess some definite physical characteristics like specific gravity (mass per unit volume), specific heat (water has the highest specific heat), weldability, elasticity, ductility, malleability, toughness, hardness, brittleness, machinability, refractoriness, tensile strength, thermal conductivity, electrical conductivity, magnetic property, etc. Note: Alloys are homogeneous in molten state but they may not be homogeneous upon solidification. So, an alloy in solid state may be either homogeneous or heterogeneous. Metals Employed for Making Alloys The important metals used for making alloys are copper, lead, zinc, tin nickel, aluminium, chromium and tungsten. Sometimes antimony and bismuth are also used.

Phase Rule and Alloys 169

IMPORTANCE OF ALLOYS

4.7

The important characteristics of alloys are the following: 1. Alloys are metallic in nature. So, alloying improves the flexibility of the base metal. 2. Alloys possess good hardness than the respective metals. Example 1. Gold alloyed with copper is much harder than soft gold. 2. Pure iron is soft in nature and when it alloys with carbon, nickel and chromium, its hardness increases. Alloying is also done for the following reasons. 1. To Lower the Melting Point of a Metal The melting point of an alloy is lower than those of its constituent elements. For example, Wood's metal is an alloy of Bi, Pb, Cd and Sn. It melts at only 71°C which is much lower than those of its components. 2. To Enhance Tensile Strength Alloy formation increases the tensile strength of the parent or base metal. Example Addition of 1% carbon increases the tensile strength of pure iron by about ten times. 3. To Increase Corrosion Resistance Metals in pure forms are reactive and easily affected by surrounding corrosive gases, moisture, etc. This reduces the life of the metal. But, when the metal is alloyed with other metals, it resists corrosion and increases its life. Generally, alloys are more resistant to corrosion than pure metals. Examples (a) Bronze (an alloy of copper with tin) is more corrosion-resistant than copper. (b) Pure iron gets corroded easily but stainless steel (an alloy of Fe with Cr and Ni) is resistant to corrosion from acids. 4. To Modify Colour of the Metal The colour of a metal can be modified by alloying it with another suitable element. Thus, the prepared alloy has a colour quite different from the colour of the base metals.

170 Engineering Chemistry-I Example Brass (an alloy of copper) is red in colour and zinc (silver white) is yellow in colour. 5. To Provide the Casting Nature of Metal Pure molten metals undergo contraction on solidification. In order to get good castings, metals have to be alloyed with other elements. Then, the alloys expand on solidification. Alloys are also easily fusible. So, they give good castings. Example Pure Pb, on alloying with 5% tin and 2% antimony, yields type metal which is used for casting printing type, due to its extraordinary casting properties.

4.7.1

Steel

Steel is an alloy of iron (as ferrite) containing 0.25–2.5% of carbon (as cementite, Fe3C) and traces of S, P, Si and Mn. The percentage of carbon in steel is intermediate between that in wrought iron and in cast iron. S and P are objectionable impurities. Silicon (0.15 to 0.35%) is in the form of iron silicide. Varieties of Steel Varieties mainly depending on the amount of carbon present in steel are as follows. 1. Mild and Quenched Steels This contains lower percentage of carbon prepared by quenching or hardening process. 2. Hard Steel This contains higher percentage of carbon. It is hard and brittle like glass. 3. Alloy Steels or Special Steels Addition of small amount of metals like Cr, Mn, Ni, Al, Mo, Si and V gives special properties (hardness, tensile strength, resistance to corrosion and coefficient of expansion) to the steel. Such steels are called alloy steels or special steels. They are plaincarbon steels which find only limited use in all fields. The main purposes of alloying steel are to improve tensile strength, improve ductility, improve toughness, improve elasticity, improve heat resistance, improve corrosion resistance, improve hardenability, improve resistance to abrasion, improve acid and chemical-resisting property, improve hardness, minimise the coefficient of expansion, improve cutting ability, etc.

Phase Rule and Alloys 171

4.7.2

Functions and Specific Effects of Alloying Elements

The properties of steels are greatly affected by the presence of alloying elements. Examples

• Nickel It improves tensile strength, ductility, toughness, elasticity, heat and corrosion resistances. • Chromium It is added in proportions up to 18%. Below 1.5%, added chromium increases tensile strength, while 12% addition gives high corrosion resistance. • Manganese Addition of 1–1.5% manganese increases strength, toughness and brittleness. Addition of 11–14% provides a high degree of hardness. • Molybdenum Addition of Mo increases corrosion resistance and abrasion resistance. • Tungsten Addition of tungsten increases red hardness, toughness, abrasion resistance and shock resistance. • Vanadium Addition of vanadium increases the tensile strength, ductility and shock resistance. Table 4.2 S. No.

Name of the alloy steel

Some important alloy steels

Metal added

Properties

Uses

(i)

Manganese steel

1.2–1.5% Mn

Hard, tough, resistant to wear

Rock crushing machinery, helmets, etc.

(ii)

Nickel steel

2.5–5% Ni

Hard, tough, rustless

Armour plate, cables, propeller shafts, automobile parts, heavy guns

(iii)

Invar

3.6 % Ni

Same coefficient of expansion as glass

Pendulum, measuring tapes, instruments

(iv)

Chrome steel

1.5–2% Cr

Extremely hard

Cutting tools, crushing machinery

(v)

Stainless steel

13–14% Cr 1% Ni

Resists corrosion

Food-processing machinery

(vi)

Tungsten

14–20% W

Hard and strong

High-speed tools

(vii)

Chrome vanadium steel

1% Cr 0.15% V High, tensile strength

For making spring, shafts, axles, etc.

172 Engineering Chemistry-I Types of Alloys Alloys are classified in a number of ways: 1. Based on the Principal Metal in the Alloy For example, copper alloys, lead alloys, iron alloys, etc. A copper alloy, has copper as the major constituent (e.g. bronze). 2. Based on Phases Single-phase alloys consist of a uniform phase, e.g. brass (60% Cu + 40% Zn). 3. Based on Iron Content Ferrous alloys contain Fe as a constituent; nonferrous alloys do not contain iron. Stainless steel (Fe + Cr) is a ferrous alloy; Solder (Pd + Sn) is a nonferrous alloy.

APPLICATIONS OF ALLOY STEELS 4.8.1

4.8

Ferrous Alloys or Alloy Steels

Ferrous alloys contain iron as the base metal. Other than iron, elements like A1, B, Cr, Co, Cu, Mn are present in this type of alloy to improve the properties of steels. The important examples of ferrous alloys are • Nichrome alloy • Stainless steel, etc. 1. Nichrome Alloy It is an alloy of iron, nickel and chromium. Its percentage composition is

Nickel Chromium Iron and Manganese

-

(50–70%) (11–25%) (5–39%) (0–2%)

Many types of nichrome alloys are possible based on the percentages of these metals. A steel alloying with 10% nickel and 20% chromium can be used at 1100°C. Steel containing 16–20% chromium with low carbon possesses oxidation resistance up to 900°C.

Phase Rule and Alloys 173

Properties They possess • Good temperature resistance • High electrical and corrosion resistance Uses • They are used in making furnace heating coils, heating elements in electric irons, toasters, etc. • They are used in making gas turbines, aero engine valves, retorts, annealing boxes, etc. • They are used in the construction of equipment exposed to high temperatures. • They are used for winding in electric heaters. 2. Stainless Steel It is a corrosion resistant alloy steel, immune to attack by chemicals and the atmosphere. It is a silvery white metal possessing high tensile strength. It contains mainly chromium (16% or more) together with other elements like nickel, molybdenum, etc. The protection against corrosion is due to the formation of a dense tough strong film of chromium oxide at the surface of the metal. There are two important types of stainless steel.

(a) Heat-treatable stainless steels (b) Non-heat treatable stainless steels (a) Heat-Treatable Stainless Steels These steels contains 1.2% carbon and 12–16% chromium. They are magnetic in nature and have tough material and they can be worked in cold state. They can be used up to 800°C. They have good weather and waterresisting property.

Uses

They find use in making surgical instruments scissors, blades, etc

(b) Non-heat-treatable Stainless Steels They show less strength at high temperature and are more resistant to corrosion. Based on the composition, they are classified into two types.

(i) Magnetic stainless steels (ii) Nonmagnetic stainless steels

174 Engineering Chemistry-I Table 4.3 Magnetic and nonmagnetic stainless steels Magnetic stainless steels

Nonmagnetic stainless steels

1.

They contain 12–22% chromium They contain 18–26% chromium, 8–21% and 0.35% C. Nickel 0.15% Carbon. (Total % of Cr and Ni is more than 23%)

2.

Corrosion resistance is better than Here, corrosion resistance is increased by heat-treatable stainless steel. adding little molybdenum.

3.

Used for making chemical equip- Used for making household utensils, ment and automobile parts. dental and surgical instruments.

Advantages of Stainless Steels • They are nonstaining, rustproof and resistant to wear and tear. The most widely used common stainless steel, containing 18% chromiun and 8% nickel (called 18/8 stainless steel), has maximum corrosion resistance. • They are tough, hard, strong and very durable. • They have an attractive appearance and are rustproof. • The surface can be restored to its original finish, brightness and appearance. • If the protective film of chromium oxide is removed somehow, by scratching, a new film is developed on the surface instantaneously and the steel again changes to stainless. • They are used in making several articles such as blades, valves, steam-turbine parts, surgical instruments, cutlery, etc. Disadvantage The important disadvantage is it has no scrap value as that of other alloys like brass and bronze. 3. Low Melting Alloys These alloys melt or fuse at lower temperature (melting point is less).

Wood's metal and rose metal are two examples (a) Wood's Metal

It contains Bi = 50%, Pb = 25%, Sn = 12.5% and Cd = 12.5%

Properties It is readily fusible (m.pt: 70°C).

Phase Rule and Alloys 175

Uses • • • •

It is used for making fire alarms and automatic sprinkles. It is used for making safety plugs for cookers, milkpots. It is used as a soft solder for joining two metallic parts. It is used as castings for dental works.

(b) Rose Metal

Properties

It contains Bi = 50%, Pb = 28% and Sn = 22%

It is readily fusible (melting point = 89°C).

Uses It is used for making fuse wires, fire alarms, castings for dental works, etc.

4.8.2

Nonferrous Alloys

The main constituents of nonferrous alloys are copper, lead, tin, aluminium, etc. They do not contain iron as one of the main components. The melting points of nonferrous alloys are lower than those of ferrous alloys. These alloys find wide applications due to their special properties like good formability, softness, attractive colour, electrical and magnetic properties, low coefficient of friction and low density, corrosion resistance, etc. The important nonferrous copper alloys are • Brass • Bronze 1. Copper Alloys Copper and its alloys rank next to steel as engineering materials. The important commercial copper alloys are listed as follows: Table 4.4 Alloys (a) Brasses

Important commercial copper alloys

Composition

Characteristics

Cu = 60–90% Zn = 40–10%

They have lower melting point than Cu and Zn. They have good corrosion resistance against water.

Uses

(Contd.)

176 Engineering Chemistry-I

Table 4.4 (Contd.) The main forms of the brasses are the following: (i) Commercial brass or guilding metal

Cu = 90% Zn = 10%

Stronger and harder Rivets, hardwares, than pure copper. screws. costumes, Golden in colour. jewellery, etc.

(ii) Dutch-metal or low brass

Cu = 80% Zn = 20%

Golden colour. Suit- Cheap jewellery, able for all drawing m u s i c a l i n s t r u operations. ments, battery caps, flexible hoses, tubes, nameplates, etc.

(iii) Cartridge brass or spinning brass

Cu = 70% Zn = 30%

Soft, ductile, harder General-purpose and stronger than brass, cartridge cascopper. es, condenser tubes, sheet fabrication, household articles, etc.

(iv) Muntz-brass

Cu = 60% Zn = 40%

Stronger than 70/30% Marine fittings, brass. condenser tubes, radiator cores, screws, springs, chains, etc.

(b) Special Brasses Contain metal(s) other than Cu and Zn (i) High-tensile brass

Very strong, hard and Switchgears, autoCu = 60% tough. claves. Zn = 40% With small additions of Fe, Al, Sn, Mn and Ni.

(ii) Admiralty brass

Cu = 70% Zn = 29% Sn = 1%

Possesses high corrosion and abrasion resistances.

Propellers and marine works.

(iii) Naval brass

Cu = 62% Zn = 37% Sn = 1%

Possesses high abrasion and corrosion resistances.

Condenser tubes and marine works.

(Contd.)

Phase Rule and Alloys 177

Table 4.4 (Contd.) (iv) German silver or cupro-nickel

(c) Bronzes

Cu = 50% Zn = 20% Sn = 30%

It possesses good strength and corrosion resistance to salt water. It is extremely ductile, malleable, and looks like silver.

Cu = 80–95% Sn = 20–5% Besides these, other metals are also added

They are tough, strong, corrosion resistant, can be readily casted and machined.

Utensils, tablewares, ornaments, corrosion-resistant implements, coinage, decorative articles, etc.

The main bronzes are the following: (i) Coinage-bronze or common bronze (ii) Gunmetal

Cu = 89–92% Sn = 11–8% Cu = 85% Zn = 5% Sn = 5% Pb = 5%

It is soft, ductile and durable.

Pumps, valves, wires, flanges, utensils, coins, etc.

It is hard, tough, strong to resist the force of explosion.

Hydraulic fittings, heavy-load bearings, parts of high-pressure steam plants, water fittings, marine pumps, etc.

(iii) Phosphorus bronze (a) Low-phosphorus bronze

Cu=rest Sn=0.7% P=up to 0.4%

Spindles for valves They possess great strength, resistance to and pumps, boiler fittings, sheets, etc. corrosion. They give good castings.

(b) High-phosphorus bronze

Cu = rest Sn = 10–13% P = 0.4 to 1%

They are hard, brittle and abrasion resistant. They possess low coefficient of friction.

For making bearings and gears, taps turbine blades, fibres for moving coil galvanometers, fuses, etc.

(iv) Aluminium bronze

Cu = 90–93% Al = 10–7%

It is quite strong. readily fusible, gives good castings, resistant to corrosion, even at high temperatures. It possesses good abrasion resistance.

For all casting operations, bushes and bearings, which retain strength up to 400∞C, jewellery, utensils, coins, photoframes, etc.

(Contd.)

178 Engineering Chemistry-I

Table 4.4 (Contd.) (v) Nickel bronzes

Cu = 90% Ni = 9% Fe = 1%

They are hard, higher in tensile strength and better corrosion resistant than copper.

For rolling purposes, unhardened shafts, valves, and general-purpose semi-hard bearings.

(vi) High-nickel bronze

Cu = rest Ni = 15–60% Zn = 12% Sn = up to 1% Mg = 0.1%

They are quite hard, very good in tensile strength. They possess superior corrosion-resisting properties.

For hydraulic components (like bearings, used in contact with sea water.

(vii) Grid-metal

Cu = 90% Sb = 9% Sn = 1%

Hard and possess good casting properties.

Used for casting perforated plates and storage battery plates.

HEAT TREATMENT OF ALLOYS (STEEL)

4.9

Heat treatment is defined as, the process of heating and cooling of solid steel article under carefully controlled conditions, thereby developing in it certain physical properties, without altering its chemical composition. Heat treatment causes • Refinement of grain structure • Removal of the imprisoned gases • Removal of internal stresses The steps in the heat-treatment process are • • • • • • • •

4.9.1

Annealing Hardening or quenching Tempering Normalising Carburising Flame hardening Nitriding Cyaniding

Annealing (Softening)

This is done by heating the metal to a certain high temperature, followed by very slow cooling in a planned manner in a furnace.

Phase Rule and Alloys 179

The purpose of annealing is to increase the mechinability and also remove the imprisoned gases and internal stresses. Types of Annealing 1. Low-temperature Annealing or Process Annealing It involves heating steel to a temperature below the lower critical point, followed by slow cooling.

Main Purpose (a) It improves machinability by relieving the internal stresses or internal strains. (b) It increases ductility and shock resistance. (c) But it reduces the hardness. 2. High-temperature Annealing or Full Annealing It involves heating steel to a temperature (about 30°C–50°C) above the higher critical temperature; then holding it at that temperature, for sufficient time to allow the internal changes and then cooled gradually to room temperature. The approximate annealing temperatures of various grades of carbon steel are • Mild steel—840–870°C • Medium-carbon steel— 780–840°C • High-carbon steel—760–780°C

Purpose Full annealing increases the ductility and machinability. This process makes the steel softer, together with an appreciable increase in its toughness.

4.9.2

Hardening (or) Quenching

Hardening is the process of heating steel beyond critical temperature and then suddenly cooling it either in oil or brine water or some other fluids. Hardening increases the hardness of the steel. The faster the rate of cooling, harder will be the steel produced. Medium-and high-carbon steels can be hardened, but low-carbon steels cannot be hardened. Main Purpose • Hardening the steel increases its resistance to wear, ability to cut other metals and strength, but steel becomes extra brittle. • The main aim of hardening is to increase abrasion resistance, so that it can be used for making cutting tools.

180 Engineering Chemistry-I

4.9.3

Tempering

It consists in heating the already hardened steel to a temperature lower than its own hardening temperature and then allowing it to cool slowly. In tempering, the temperature to which hardened steel is reheated controls the development of the final properties. Thus, 1. For retaining strength and hardness, the reheating temperature should not exceed 400°C. 2. For developing better ductility and toughness, the reheating temperature should be within 400–600°C. Main Purpose • Tempering removes any stress and strains that might have developed during quenching. • It reduces the brittleness and also some hardness but toughness and ductility are simultaneously increased. • Cutting tools like blades, cutters, chisels, and tool-bits always require tempering.

4.9.4

Normalising

Normalising involves the process of heating steel to a definite temperature (above its higher critical temperature) and allowing it to cool gradually in still air. Main Purpose • The homogeneity of the steel structure is recovered. • Grains are refined. • Internal stresses are removed. • Toughness is increased. Normalising takes much lesser time than annealing. Normalised steel is used in engineering works.

4.9.5

Carburising

The mild steel article is taken in a cast iron box containing small pieces of charcoal. It is heated up to 900–950°C and is allowed to keep it as such for sufficient time, so that the carbon (charcoal) is absorbed to the required depth. The article is then allowed to cool slowly within the iron box itself. The outer skin of the article is converted into high-carbon steel containing about 0.8–1.2% carbon.

Phase Rule and Alloys 181

Main Purpose The purpose for carburisation is to increase hard-wearing surface on steel articles.

4.9.6

Flame Hardening

It is essentially a localised hardening. It consists in heating an area to be surface-hardened by means of an oxy-acetylene (or oxy-hydrogen) flame, followed by abrupt cooling by spraying water on it.

4.9.7

Nitriding (Process of Getting Super-hard Surface)

It is the process of heating the metal alloy in presence of ammonia at a temperature to about 550°C. The nitrogen (obtained by the dissociation of ammonia) combines with surface of the alloy to form hard nitride. The purpose of nitriding is to get super-hard surface.

4.9.8

Cyaniding

It is also a type of case-hardening or carburising process. It consists in producing a layer of hard surface on low or medium-carbon steels by immersing the metal in a molten salt, containing cyanide (like KCN or NaCN) at a temperature of about 870°C and then quenching in oil or water. The hard surface is produced due to the absorption of carbon and nitrogen by the metal surface.

QUESTIONS PART-A 1. Calculate the number of phases present in the following systems.

2. 3. 4. 5.

(a) MgCO3(s) MgO(s) + CO2(g) (b) Rhombic sulphur(s) Monoclinic sulphur(s) (c) Ice(s) Water(l) Water vapour(g) (d) An emulsion of oil in water State phase rule and explain the terms involved. Define eutectic point. State reduced phase rule. How many phases and components are present in the following system? CaCO3(s) CaO(s) + CO2(g)

182 Engineering Chemistry-I

6. 7. 8. 9. 10. 11. 12. 13. 14. 15.

16. 17. 18. 19. 20. 21. 22. 23.

Define degree of freedom. What is an invariant system? Give an example. What is triple point? Mention its significance. Define eutectic point. State the conditions under which two substances can form a simple eutectic. Differentiate between melting point, triple point and eutectic point. Mention the merits of phase rule. Mention the uses of cooling curves. What are the limitations of phase rule? NH4Cl(s) NH3(g) + HCl(g). Write the values of phases, degree of freedom and component for this system. What is the effect of pressure on the melting point of ice? Eutectic is a mixture and not a compound explain. What is an eutectic alloy? Mention any two advantages of alloy making. What is meant by quenching in heat treatment of metals? Name some important heat treatment processes. What is tempering? What are stainless steels?

PART-B 1. Explain the terms component and degree of freedom. [P.153–154] 2. What is thermal analysis? Draw the cooling curves of a pure substance and a mixture; discuss. [P.161] 3. State phase rule and explain the terms involved. [P.152–153] 4. Why is condensed phase rule used for a two-component alloy system? [P.159–160] 5. Discuss in detail the application of phase rule to lead-silver system. [P.163–166] 6. Discuss the formation of simple eutectic system with applications. [P.165–166] 7. State the phase rule and discuss its application to Ag-Pb system. [P.152, 165–166] 8. Discuss the application of phase rule to water system. [P. 156–158] 9. Draw a neat phase diagram and explain the lead-silver system. Briefly write about Pattinson's process. [P. 163–166] 10. What is condensed phase rule? What is the number of degrees of freedom at the eutectic point for a two component system? [P.165] 11. Draw the phase diagram of simple eutectic A-B system, with the following data and discuss. [P.163–166]

Phase Rule and Alloys 183

12. 13.

14. 15.

16. 17. 18.

19. 20. 21.

A (m.pt) 961° C Eutectic temperature 305°C B (m.pt) 327°C Eutectic composition 2.6% A State phase rule. Explain the following. [P.152, 158–165] (i) Metastable equilibrium (ii) Eutectic point What is meant by phase diagram? With the help of phase diagrams, explain the following terms. [P.159–165] (i) Triple point, (ii) Eutectic point Apply phase rule to a one-component system. [P.156–157] Draw and explain the phase diagram of ice-water vapour system. How does the melting point of ice change with variation of pressure? [P.158] Apply phase rule to binary alloys forming eutectics. [P.162–163] Explain the differences among melting point, eutectic point and triple point with suitable examples for each. [P.158, 165, 159] What is the number of phases in the following systems? (i) Saturated solution of NaCl (ii) Mixture of rhombic and monoclinic sulphur (iii) Decomposition equilibrium of CaCO3 (iv) Mixture of O2 and N2 (v) Mixture of benzene and water Explain desilverisation of lead. [P.165] The eutectic is a mixture and not a compound. Justify. [P.163] Discuss different heat treatment methods and their effects on metals and alloys. [P.178–181]

22. What is meant by annealing of steel? 23. Explain any two non-ferrous alloys. 24. Write a short note on nickel alloys.

[P.178] [P.175–177] [P.178]

25. Mention the composition and uses of nichrome and stainless steel. [P.172–173] 26. What are the main purposes of alloying steel?

[P.169–170]

27. What are the nonferrous alloys? Give their properties? [P.175–178] 28. What happens during carburising and nitriding of steel? [P.180–181] 29. Give a brief account of [P.175–177] (i) Brass (ii) Bronze 30. Explain [P.178–180] (i) Hardening (ii) Tempering (iii) Annealing 31. Describe types of annealing hardening in detail. [P.179] 32. Discuss the thermal analysis of alloys. [P.178] 33. Write a brief note on classification of alloys. [P.172–177]

184 Engineering Chemistry-I

Two-Mark Questions and Answers 1. State phase rule. [A.U. Jan-2006] The equilibrium between any number of phases is not influenced by gravity or electrical or magnetic forces, or by surface action and is influenced only by temperature, pressure and concentration, then the number of degrees of freedom (F) of the system is related to the number of components (C) and of phases (P) by the phase rule equation. F=C–P+2 2. Mention any two merits of phase rule.

[A.UJan-2004]

• Phase rule is applicable to both physical and chemical equilibriums and it requires no information regarding molecular structure. • In phase rule, there is no information regarding micro structure. So, it is applicable to macroscopic systems. 3. Calculate the number of phases present in the following systems. (a) (b) (c) (d)

MgCo3(s) MgO(s) + CO2(g) Rhombic sulphur(s) monoclinic sulphur(s) Ice(s) water(I) water vapour(g) An emulsion of oil in water. [A.U. May-2007]

Ans. (a) 3, (b) 2, (c) 3, (d) 2 4. Write the limitations of phase rule. • It can be applied only for heterogenous equilibrium systems. • It considers only three variables, namely, temperature, pressure and composition. • It applies only to a single-equilibrium system and provides no information regarding any other possible equilibria in the system. 5. State reduced phase rule. [A.U.May-2006] The solid-liquid equilibria having no gas phase. Such systems having no gas phase are known as condensed systems. Since the effect of pressure is small on the type of equilibria, therefore experiments are usually conducted under atmosperic pressure, thus keeping the pressure as constant. This will reduce the degree of freedom of the system by one and for such system, the phase-rule becomes F=C–P+1

Phase Rule and Alloys 185

6.

7.

8. 9.

This is known as reduced phase rule having two variables, namely, temperature and concentration of one of the constituents. Define eutectic point. [A.U.June-2006] Two or more solid substances capable of forming solid solutions with each other have the property of lowering each other's freezing point and the minimum freezing point attainable corresponding to the eutectic mixture, is termed the eutectic point (means lowest melting point). Define degree of freedom. [A.U.Dec-2003] The term degree of freedom is meant by the minimum number of independently variable factors, such as temperature, pressure and composition of the phases, which must be arbitrarily specified in order to represent perfectly all the conditions of a system. What is the effect of pressure on the melting point of ice? The melting point of ice decreases with increase of pressure. State the conditions under which two substances can form a simple eutectic. (i) The two substances must be completely miscible in the liquid state but completely immiscible in the solid state. (ii) They are not chemically react with each other.

10. What is triple point? [A.U.Dec-2005] The point at which three phases (of any types) like solid, liquid and vapour phases are in equilibrium. The system has zero degree of freedom at the triple point. 11. NH4Cl(s) Æ NH3(g) + HCl(g). Write the values of P, F and C for this system. P = 2 (one solid, two gaseous phage) C = 1 (one component system) F=C–P+2=1–2+2=1 F = 1. 12. What is the difference among melting point, triple point and eutectic point? [A.U.May-2005] Melting point - solid and a liquid having the same composition are in equilibrium. Triple point - Three phases (of any types) are in equilibrium. Eutectic point - Two solids and their solution are in equilibrium. 13. What are called eutectic mixture? [A.U.Jan-2004] Eutectic mixture is a solid solution of two or more substances having the lowest freezing point of all the possible mixture of the

186 Engineering Chemistry-I

components. This is taken advantage of in "alloys of low melting point" which are generally eutectic mixtures. 14. What is the effect of pressure on the melting point of ice? [A.U.May-2007] The effect of pressure on the melting point of ice is to be noted that the line inclines towards the pressure axis which shows that expansion takes place on freezing of water and melting point of ice is decreased by increase of pressure. 15. What are the uses of eutectic system? • Eutectic systems are used in solders for joining two metal pieces together. (e.g.,) Pb-Sn solders. • Suitable alloy compositions are predicted by using eutectic systems (e.g.,) woods metral (alloy containing 50% Bi, 25% Pb, 12.5% Cd and 12.5% Cu) melts at 65°C only. 16. What are called alloys? Mention its types. [A.U.Jan-2005] Alloys are the homogeneous substances formed by mixing two or more elements, atleast one of them being a metral. Types of alloys: Based on the constituents, alloys are of three types. (i) Alloys formed by two or more metals. Examples: brass (Cu-Zn) (ii) Alloys formed between a metal and a non-metal. Examples: WC alloy (Wolframite–Carbon) (iii) Alloy amalgam formed between mercury and another metal. Examples: Zinc amalgam 17. What is meant by ferrous alloys? [A.U.Jan-2006] Ferrous alloys contain iron as the base metal. Other than iron, elements like Al, B, Cr, Co, Cu, Mn are present in this type of alloy to improve the properties of steels. 18. What is meant by non-ferrous alloys? [A.U.April-2004] The main constituents of non-ferrous alloys are copper, lead, tin, aluminium, etc., They do not contain iron as one of the main components. The melting points of the non-ferrous alloys are lower than those of ferrous alloys. 19. Give the name and composition of an alloy possessing zero co-efficient of expression. Invar. The composition is 0.3 – 0.5% C 30 – 60% Ni Rest Fe

Phase Rule and Alloys 187

20. What is the basic difference between brass and bronze? [A.U.May-2004] Brass is an alloy of copper and zinc. Bronze is an alloy of copper and tin. 21. What are fuses? Give example. [A.U.May-2003] Fuses are alloys melts or fusible at lower temperature (melting point is less). For example, Wood's metal: They contain Bi = 50%, Pb = 25%, Sn = 12.5% and Cd = 12.5% 22. Define heat treatment of alloys. [A.U.May-2005] Heat treatment is defined as, "the process of heating and cooling of solid steel article under carefully controlled conditions, thereby developing in it certain physical properties, without altering its chemical composition. 23. What is meant by carburising? [A.U.Jan-2007] The mild steel article is taken in a cast iron box containing small pieces of charcoal. It is heated upto 900-950°C and is allowed to keep it as such for sufficient time, so that the carbon (charcoal) is absorbed to the required depth. The article is then allowed to cool slowly within the iron box itself. 24. What are the advantages of alloy making? [A.U.Jan-2006] (i) It enhances the hardness of the metal. (ii) It lowers the melting point. (iii) It enhances corrosion-resistance.

Obopdifnjtusz V

5.1

Basic introduction of Nanochemistry

5.2

Distinction between Molecules, Nanoparticles and Bulk

O J U

5.3

Size-dependent Properties of Nanomaterials

5.4

Nanoparticles

5.5

Nanoclusters

5.6

Nanorods

5.7

Nanotubes (CNT)

5.8

Nanowires

5.9

Synthesis of Nanoparticules

5.10 Properties of Nanoparticules 5.11 Application of Nanoparticules 5.12 Application in Fuel Cells 5.13 Catalysis and Use of Gold

W

Nanoparticules in Medicine

6 D

Obopdifnjtusz

I B Q U F S

BASIC INTRODUCTION OF NANOCHEMISTRY

5.1

The term nanotechnology is defined as development and utilisation of structures and devices with organisational features at the intermediate scale between individual molecules and about 100 nanometres, where novel properties occur as compared to the bulk materials. The terms nanomaterial and nanotechnology mean that the design, characterisation, production, and application of structures, devices and systems by controlled manipulation of size and shape at the nanometre scale (atomic, molecular and macromolecular scale) that produces, structures, devices and systems with at least one novel characteristic or property. Nanotechnology is a fundamental understanding of how nature works at the atomic scale. New industries will be generated as a result of this understanding. Further, we understand how electrons are moved in a conductor by applying a potential difference led to electric lighting, the telephone, computing, the Internet and many other industries, all of which would not be possible without it. The real value of the nanotube would be in their application, whether within the existing industry, or to enable the creation of a whole new one.

DISTINCTION BETWEEN MOLECULES, NANOPARTICLES AND BULK

5.2

Nanoparticles are generally considered to be a number of atoms or molecules bonded together with a radius of < 100 nm.

192 Engineering Chemistry-I

A nanometre is 10–9 m or 10 Å, and so particles having a radius of about £ 1000 Å can be considered nanoparticles. Figure 5.1 gives an arbitrary classification of atomic clusters according to their size showing the relationship between the number of atoms in the cluster and its radius. For example, a cluster of one nanometre radius has approximately 25 atoms, but most of the atoms are on the surface of the cluster. Number of Atoms 1

Radius (nm) 1 molecules

10 102

Nanoparticles

103 104

10

105 106

Bulk

Fig. 5.1 Distinction between molecules, nanoparticles and bulk according to the number of atoms in the cluster

The field of nanochemistry is very interesting and their unique properties in their size is smaller than critical length, thermal diffusion length or a scattering length, that characterise many physical phenomena. The electrical conductivity of a metal is strongly determined by the distance that the electrons travel between collisions with the vibrating atoms or impurities of the solid. This distance is called the mean free path or the scattering length. If the sizes of particles are less than these characteristic lengths, it is possible that new nanochemistry may occur. The working definition of a nanoparticle is an aggregate of atoms between 1 and 100 nm viewed as a subdivision of a bulk material, and of dimensionless than the characteristic length of some phenomena.

SIZE-DEPENDENT PROPERTIES OF NANOMATERIALS

5.3

Nanomaterials and nanotechnology mainly depend on size-dependent properties:

Nanochemistry 193

They are recognized as 1. 2. 3. 4. 5. 6.

Chemical Properties — reactivity, catalysis Thermal Properties — melting temperature Mechanical Properties — adhesion, capillary forces Optical Properties — absorption and scattering of light Electrical Properties — tunneling of current Magnetic Properties — super paramagnetic effect

Other than this, many other sensing and biochemical properties and functions are also favourable. Normally, the size of a nanometre is compared to that of the human hair which is 80,000 nm wide. The chemist synthesises the engineering aspects of preparing minute pieces of matter with nanometre sizes in one or two or three dimensions. Chemists strive towards this objective from the atom up and nanophysicists operate from the bulk down. Building and organising nano-objects under mild and controlled conditions one atom at a time instead of manipulating the bulk in principle provide the chemist with simple, reproducible and cost-effective synthetic approaches to materials of perfect atom sizes and shapes rather than having to use the complex and sophisticated instrumental techniques of the engineering physicist.

NANOPARTICLES

5.4

The particle size of a material has significant influence on its physical and chemical properties. When the particle size is reduced to nanometre scale, it behaves completely different from its bulk properties. The physico-chemical properties of nanoparticles are due to three main reasons. 1. The size of nanoparticles is comparable to the Bohr radius of the excitation. This significantly alters the optical, luminescent and redox properties of nanoparticles when compared to the bulk material. 2. The surface atoms constitute a fraction of the total number of the atoms of the nanoparticle. As the particle size decreases, the net internal cohesive force increases. The surface energy depends on the internal cohesive force that should increase with decreasing particle size.

194 Engineering Chemistry-I

3. The natural size of the nanoparticle is comparable with the size of molecules. In general, the reduction in the particle size results in 1. Increase in surface-to-bulk-atom ratio 2. Increase in the surface area

NANOCLUSTERS

5.5

Nanoclusters constitute an intermediate state of matter between molecules and solids. Nanocluster sizes range from sub-nanometre to about 10 nanometres in diameter and are of technological interest in numerous areas of applied science such as material sciences, catalysis, optoelectronics. Nanocluster consisting up to a couple of hundred atoms and large aggregates containing 103 or more atoms are more often called nanoparticles. Nanoclusters have properties and structures which are very sensitive to their composition and size. The important nanocluster inorganic bulk materials are ZnO, SiO 2, CdS. Non-bulklike properties of nanoclusters may provide the inspiration for novel low-density materials for many applications.

NANORODS

5.6

They are nanostructures shaped like long sticks with a diameter in the nanoscale but having a length that is very much longer. Each nanorod dimension ranges from 1–100 nm. They may be synthesised from metals or semiconducting materials. Their standard aspect ratios (length divided by width) are 3 to 5. Nanorods are produced by direct chemical synthesis. A combination of ligands act as shape-control agents and bond to different faces of the nanorod with different strength. This allows different faces of the nanorod to grow at different rates producing an elongated object. The nanorod's present diameters are as small as 12 nanometres. The important commercial available nanorods are aluminium, bismuth, cadmium, gold, nickel, platinum, palladium, silicon, TiO2 and zinc oxide nanorods.

Nanochemistry 195

Applications They are used mainly in display technologies to microelectro-mechanical systems. The reflectivity of the rods can be changed by changing their orientation with an applied electric field. The other applications are in optical, sensing, solar cells, magnetic and electronic devices.

NANOTUBES (CNT)

5.7

The more interesting nanostructures with large application potential are carbon nanotubes. The carbon nanotubes are a sheet of graphite rolled into a tube with bonds at the end of the sheet forming the bonds that close the tube. Figure 5.2 shows the structure of a tube formed by rolling the graphite sheet about an axis parallel to C–C bonds.

Fig. 5.2

Structure of carbon nanotubes, depending on how graphite sheets are rolled. (a) Armchair structure (b) Zigzag structure (c) Chiral structure

A single-walled nanotube (SWNT) can have a diameter of 2 nm and a length of 100 mm, making it effectively a one-dimensional structure called a nanowire.

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5.7.1

Fabrication of Carbon Nanotubes

Carbon nanotubes are fabricated by the following methods. 1. Laser evaporation 2. Carbon arc method 3. Chemical vapour deposition 1. Laser Evaporation Figure 5.3 illustrates the apparatus for making carbon nanotubes laser evaporation. Furnace Water cooled copper collector

Laser

Argon gas Graphite target Quartz tube

Fig. 5.3

Experimental set-up for synthesising carbon nanotubes by laser evaporation

A quartz tube containing argon gas and a graphite target are heated to 1200°C contained in the tube, but outside the furnace is a water-cooled copper collector. The graphite target contains little amounts of cobalt and nickel acting as catalytic nucleation sites for the formation of the tubes. An intense pulsed laser beam is incident on the target, evaporating carbon from the graphite. The argon then sweeps the carbon atoms from the high-temperature zone to the colder copper collector on which they condense into nanotubes. Nanotubes of 10–20 nm in diameter and 100 mm long can be fabricated by using this method.

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2. Carbon Arc Method A potential of 20–25 V is applied across carbon electrodes of 5–20 mm diameter and separated by 1 mm at 500 torr pressure of flowing helium. The carbon atoms are ejected from the positive electrode and nanotubes formed on the negative electrode. As the nanotubes forms, the length of the positive electrode decreases, and a carbon deposit on the negative electrode increases. In order to produce single-walled nanotubes, a small amount of cobalt, nickel or iron is incorporated as a catalyst in the central region of the positive electrode. If no catalysts are used, the tubes are nested or multiwalled type (MWNT), which are nanotubes within nanotubes as shown in Fig. 5.4. Multiwalled Carbon Nanotubes

Fig. 5.4

Nested nanotube in which one tube is inside the another

The carbon-arc method can produce single-walled nanotubes of 1–5 nm diameters with a length of 1 mm. 3. Chemical-Vapour Deposition Method This method involves decomposing a hydrocarbon gas such as methane (CH4) at 1100°C. During the gas decomposition, the carbon atoms are produced and they condense on a cooler substrate that may contain various catalysts such as iron. This method produces tubes with open ends, which does not occur when other methods are used. This method may be the most favourable method and allows continuous fabrication. Nanotubes are synthesised, containing a mix of metallic and semiconducting materials. A group of IBM has developed a method to separate the semiconducting from the metallic nanotubes.

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The separation was accomplished by depositing bundles of nanotube containing metallic and semiconducting material on a silicon wafer. Metal electrodes were then deposited over the bundle. Using the silicon wafer as an electrode, a small bias voltage was applied that prevents the semiconducting tubes from conducting, effectively making them insulators. A high voltage is then applied across the metal electrodes, thereby sending a high current through the metallic tube but not the insulating tubes. This causes the metallic tubes to vaporise leaving behind only the semiconducting tubes.

5.7.2

Structure of Carbon Nanotubes

The various structures of carbon nanotubes have different properties. A nanotube can be formed when a graphite sheet is rolled up about the axis T as shown Fig. 5.5. The Ch vector is called the circumferential vector, and it is at right angles to T. Three examples of nanotube structures constructed by rolling the graphite sheet about the T vector having different orientations in the graphite sheet are as shown in Fig. 5.2. When T is parallel to the C–C bonds of the carbon hexagons, it is referred to as the armchair structure. [Fig. 5.2(a)] The tubes mentioned in Fig. 5.2(b) and 5.2(c) are zigzag and the chiral structures, formed by rolling about a T vector having different orientations in the graphite plane, but not parallel to C–C bonds.

Fig. 5.5

Graphite sheet showing the basic vectors a1 and a2 of the two-dimensional unit cell, the axis vector T about which the sheet is rolled to get the armchair structure nanotube [Fig. 5.2(a)] and the circumferential vector CR at right angles to T. Other orientations of T on the sheet generate the zigzag and chiral structures of Fig. [5.2(b) and (c)].

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Generally, nanotubes are closed at both ends, which involves the introduction of a pentagonal topological arrangement on each end of the cylinder. The tubes are essentially cylinders with each end attached to half of a large fullerene-like structure. In the case of single-walled nanotubes, metal particles are found at the ends of the tubes, which is evidence for the catalytic role of the metal particles in their formation.

NANOWIRES

5.8

A nanowire is a nanostructure with a diameter of the order of a nanometre (10–9 metres). The different types of nanowires exist including 1. Metallic nanowires, e.g. Ni, Pt, Au 2. Semiconducting nanowires, e.g. Si, InP, GaN, etc. 3. Insulating nanowires, e.g. SiO2, TiO2 The nanowires can be used in the near future to link tiny components into very small circuits. Using nanotechnology, such components could be created out of chemical compounds.

Synthesis of Nanowires There are two methods available: 1. Top-down, and 2. Bottom-up approach Most synthesis uses a bottom-up approach. Nanowire production uses several techniques like suspension, electrochemical deposition, vapour deposition, etc. Common technique for creating a nanowire is vapour-liquid-solid synthesis. This process can produce crystalline nanowires of some semiconductor materials. It is used as a source material either as a laser-ablated particles or as a feedgas such as silane.

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SYNTHESIS OF NANOPARTICLES

5.9

The preparation of nanoparticles with well-defined size and morphology is more important for various industrial applications like fine ceramics, abrasive powders, high-reactivity, pigments, catalysts, etc. A variety of techniques have been applied for the synthesis of nanoparticles. Several points have to be considered by the synthetic process. 1. 2. 3. 4.

Range of applicability for different class of materials Reproducibility of average size and shape of particles Control over the average particle size and range of size Homogeneity of product phases

Among chemical methods, several low-temperature techniques like polymeric citrate precursor method, precipitation, sol-gel synthesis and solvothermal methods have been used to avoid the agglomeration and grain growth to produce nanoparticles. Some of the popular methods of synthesis of nanoparticles are summarised below.

5.9.1

Precipitation Nano Synthesis

It is a technique useful for the preparation of ceramic oxide powders. In multicomponent system, this technique is usually limited to the cations of chemically similar properties. In the precipitation method, the required amount of the aqueous solution of desired metal ions were mixed together. The pH is adjusted in the region where the metal ion gets precipitated by adding aqueous solution of ammonia, NH 4OH or ammonium carbonate. The precipitate was washed with distilled water, dried and ground to obtain particles of sizes smaller than 5 mm. The main disadvantages for this method is 1. The two reactants have very different solubilities in water 2. The reactants do not precipitate at the same time

5.9.2

Hydrothermal Synthesis

Thermolysis is a method of synthesising nanoparticles by applying heat. It is a method of synthesis of single crystals which depends on the solubility of minerals in hot water under high pressure. The crystal growth is performed in an apparatus consisting of a steel pressure vessel called autoclave, in which a nutrient is supplied along with water.

Nanochemistry 201

A gradient of temperature is maintained at the opposite ends of the growth chamber so that the hotter end dissolves the nutrient and the cooler end causes seeds to take additional growth. The advantage of this method is to include the ability to create crystalline phases which are not stable at the melting point. The materials which have a high vapour pressure near their melting points can also be grown by this method. This method is suitable for the growth of large good-quality crystals while maintaining good control over their composition. Process The supersaturation state is achieved by reducing the temperature in the crystal growth zone. The nutrient is placed at the lower part of the autoclave with specific amount of solvent. The autoclave is heated in order to create two temperature zones. The nutrient in the lower part dissolves in the hotter zone and is transported to the upper part by convective motion of the solution. The cooler and dense solution in the upper part of the autoclave descends and the counterflow of solution ascends. The solution becomes supersaturated in the upper part of low temperature and crystallises out. The compounds synthesised under hydrothermal conditions are simple and complex oxides, tungstates, molybdates, carbonates, silicates, emeralds, etc. This synthesis is commonly used to grow synthetic quartz, gems and other single crystals with commercial value.

5.9.3

Solvothermal Synthesis

In this method, a solvent is mixed with certain metal precursors and the solution mixture is placed in an autoclave kept at relatively high temperature and pressure in an oven to carry out the crystal growth and assembly process. The developed pressure in the vessel due to the solvent vapour, elevates the boiling point of the solvent. The solvents used are ethanol, methanol, toluene and t-butanol. This is a versatile method for the synthesis of variety of nanoparticles with narrow size distribution and dispersity. This technique is further used for the synthesis of zeolites, inorganic structures and solid materials.

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Nanophase materials are produced by chemical reactions in an aqueous or organic medium under the simultaneous application of heat and pressure in the presence of an alkali or acid. The properties like viscosity or dielectric constant are considerably reduced by pressure and temperature. Solvothermal conditions are advantageous over other chemical methods, namely 1. It is easy to control particle size and morphology by varying the synthesis conditions like temperature, pressure, time, concentration, pH, nature of additives. 2. Many nanomaterials can be directly synthesised in the desired crystalline phase at low temperature in one step. This method is one of the simple techniques for the preparation of nanomaterials with high yield and the number of materials have been grown which are well documented.

5.9.4

Electrodeposition

Template-assisted electrodeposition is the best technique for synthesising nanomaterials with controlled shape and size. Arrays of nanostructured materials with specific arrangements can be prepared by this method, employing active or restrictive template as a cathode in an electrochemical cell. The formation of nuclei on the electrode substrate was studied on the basis of macroscopic thermodynamic considerations. The nucleation of nanostructures on the electrode substrate during electrodeposition is influenced by the crystal structure of the substrate, specific free surface energy, adhesion energy, lattice orientation of the electrode surface, and crystallographic lattice mismatch at the nucleus-substrate interface boundary. The size distribution of the electrodeposits strongly depends on the kinetics of the nucleation and growth. The electrodeposition process involves the formation of either an instantaneous or a progressive nucleation. In instantaneous nucleation, all the nuclei form instantaneously on the electrode substrate, and subsequently grow with the time of electrodeposition. These nuclei gradually grow and overlap, and therefore, the progressive nucleation process exhibits zones of reduced nucleation rate around the growing stable nuclei. The electrodeposition method consists of an electrochemical cell and accessories for applying controlled current at a certain voltage. The cell

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contains a reference electrode, a specially designed cathode, and an anode or counter electrode. The cathode substrate on which electrodeposion of the nanostructure takes place can be made of either nonmetallic or metallic materials. Using the cathode surface as a template, various nanostructures or morphologies can be synthesised for specific applications.

5.9.5

Chemical Vapour Deposition Method (CVD)

This method is the process of chemically reacting a volatile compound of a material to be deposited, with other gases, to produce a nonvolatile solid that deposits atomistically on a suitably placed substrate. Gas-phase (homogeneous) reactions and surface (heterogeneous) reactions are intricately mixed. Gas-phase reactions become progressively important with increasing temperature and partial pressure of the reactants. An extremely high concentration of the reactants will make gasphase reactions predominant leading to homogeneous nucleation. A solid is deposited on a heated surface via a chemical reaction from the vapour or gas phase requiring activation energy. This energy can be provided by a high temperature above 900°C in thermal CVD. In plasma CVD, the reaction is activated by plasma at temperature ranges from 300 to 700°C. In laser CVD, pyrolysis occurs when laser thermal energy heats an absorbing substrate. In photolaser CVD, the chemical reaction is induced by UV radiation. In this process, the reaction is photon-activated and deposition occurs at room temperature. Carbon fibres and filaments are prepared by hydrocarbons over a metal catalyst by chemical-vapour deposition method.

Fig. 5.6

Chemical-vapour deposition

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Figure 5.6 shows a chemical-vapour deposition furnace where the reacting gases are introduced into quartz tube, water on which the particle has to be deposited are kept in a zone furnace. Aligned BN nanotubes are obtained when aligned multiwall nanotube are used as templates. The examples of nanoparticles developed by chemical vapour depositions are 1. Formation of CNTs from ethylene supported catalysts such as Fe, Co and Ni 2. The production of single-walled nanotubes as well as doublewalled CNTs on Mo and Mo–Fe alloy catalysts 3. High yields of single-walled nanotubes have been obtained by catalytic decomposition of an H2|CH4 mixture over well-dispersed metal particles such as Co, Ni and Fe on magnesium oxide at 1000°C 4. Particles of ZrO2, Y2O3 and nanowhiskers have been produced by CVC method

5.9.6

Laser Ablation

In this method, high-power laser pulses are used to evaporate matter from a target surface such that the stoichiometry of the material is preserved in the interaction. As a result, a jet of particles is ejected normal to the target surface. The ablated species condense on the substrate placed opposite to the target as shown in Fig. 5.7. La

se

rb ea m

Heater Substrate

Vaccum chamber

Fig. 5.7

Target

Q2

Principle of pulsed laser deposition

The ablation process takes place in a vacuum chamber either in vacuum or in the presence of some background gases. In oxide films, oxygen is the background gas. For example, the oxygen pressure for YBCO thin film is approximately 0.2 m bar. The substrate temperature is sufficiently large (700–800°C) and

Nanochemistry 205

uniform across the whole substrate area to obtain epitaxial films. Laserpulse energy density on the target should also be larger than a certain threshold value (for YBCO, 1 J/cm2). This method is applicable to high-melting-point elements and transition metals. This method is suitable to carry out in situ studies on highly toxic and radioactive materials, often in trace amounts. Elements like iron, gold, palladium and compounds of sulphides have been prepared by this method. This method is capable of high rate of production of 2–3 g/min.

PROPERTIES OF NANOPARTICLES

5.10

The important and most interesting properties of carbon nanotubes are 1. Electrical properties 2. Vibrational properties 3. Mechanical properties

5.10.1

Electrical Properties

Carbon nanotubes have the important property that they are metallic or semiconducting, depending on the diameter and chirality of the tube. Chirality refers to how the tubes are rolled with respect to the direction of T vector in the graphite plane. Figure 5.8 is a plot of the energy gap of semiconducting chiral carbon nanotubes versus the reciprocal of the diameter, showing that as the diameter of the tube increases, the band decreases. 1.4

Energy Gap (eV)

1.2 1 0.8 0.6 0.4 0.2 0 0

1

2

3

4

5

6

100/D [1/A]

Fig. 5.8

Plot of the magnitude of the energy gap of a semiconducting nanotube vs the reciprocal of the diameter of the tube

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3

Differential Conductance (normalised)

Differential Conductance (normalised)

Scanning Tunneling Microscopy (STM) has been used to investigate the electronic structure of carbon nanotubes. In this measurement, the position of the STM tip is fixed above the nanotube, and the voltage 'V' between the tip and the sample is swept while the tunneling current 'I" is monitored. The measured conductance G = I/V is a direct measure of the local electronic density of states. The density of states is a measure of how close together the energy levels are to each other. Figure 5.9 (a) shows the STM data as the differential conductance, (dI/dv)/(I/V) versus the applied voltage between the tip and carbon nanotube. The data shows clearly the energy gap in metallic materials at voltages where very little current is observed. The voltage width of this region measures the gap, which for the semiconducting material shown in Fig. 5.9 (b).

2

1

2

1

0

0 –1

0 Voltage (v) (a)

Fig. 5.9

3

1

–1

0

1

Voltage (v) (b)

Plot of differential conductance vs tunneling current of (a) metallic nanotubes, and (b) semiconducting nanotubes

The electronic states of the tube do not form a single wide electronic energy band, but instead split into one-dimensional sub-bands that are evident in the data in Fig. 5.9. Electron transport has been measured on individual single-walled carbon nanotubes. The measurements at a millikelvin (T = 0.001 K) on a single metallic nanotube lying across two metal electrodes show steplike features in the current-voltage (I-V) measurements. The steplike features in the I-V curve are due to single-electron tunneling and resonant tunneling through single molecular orbitals. Single-electron tunneling occurs when the capacitance of the nanotubes is so small that adding a single electron requires an electrostatic charging energy greater than the thermal energy.

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Electron transport is blocked at low voltages, which is called coulomb blockade. By gradually increasing the gate voltage, electrons can be added to the tube one by one. Generally, in one-dimensional systems, the presence of a defect will cause a localisation of the electrons. However, a defect in a nanotube will not cause localisation because the effect will be averaged over the entire tube circumference because of the doughnut shape of the electron-wave function. In the metallic state, the conductivity of the nanotubes is very high, estimated that they can carry a billion amperes per square centimetre. Copper wire fails at one million amperes per square centimetre because resistive heating melts the wire. The main reason for the high conductivity of the carbon tubes is that they have very few defects to scatter electrons, and thus a very low resistance. High currents do not heat the tubes in the same way they heat copper wires. Nanotubes also have a very high thermal conductivity, more than that of diamond. This means that they are also very good conductors of heat. Magnetoresistance is a phenomenon whereby the resistance of a material is changed by the applications of a dc magnetic field. Carbon nanotubes display magnetoresistive effects at low temperature. Figure 5.10 shows a plot of the magnetic field dependence of the change in resistance DR of nanotubes at 2.3 K and 0.35 K, compared to their resistance R in zero magnetic field.

Fig. 5.10

Effect of a dc magnetic field Vs the resistance of nanotubes at the temperatures of 0.35 and 2.3 K

This is a negative magnetoresistance effect because the resistance decreases with increasing dc magnetic field, so its conductance increases.

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This occurs because when a dc magnetic field is applied to the nanotubes, the conduction electrons acquire new energy levels associated with their spiraling motion about the field, called Landan levels, lying very close to the topmost filled energy levels (the Fermi level). Thus, there are more available states for the electrons to increase their energy, and the material is more conducting.

5.10.2

Vibrational Properties

The atoms in a nanoparticle continually vibrate back and forth. Each nanoparticle has a specific set of vibrational motions, called normal fundamental modes of vibration. These are determined by the symmetry of the molecule. For example CO2 has a linear structure: O = C = 0 The total fundamental normal modes of vibrations are calculated as 3N – 5 where N is the number of atoms in the molecule.

= 3 ¥ 3 – 5 = 9 – 5= 4 Normal modes of vibrations. They are (a) Symmetric stretch, consists of an in-phase elongation of the two C = O bonds (b) Asymmetric stretch, consists of out-of-phase stretches of the C = O bond length, where one bond length increases while the other decreases (c) Two bending modes of vibration (in-plane and out of plane bending) Similarly, carbon nanotubes also have normal modes of vibrations. Figure 5.11 shows two normal modes of nanotubes.

E2g 17 cm–1

A1g 165 cm–1

Fig. 5.11 Examples of two normal modes of vibration of carbon nanotubes

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(a) One mode, labelled A1g, involves an "in and out" oscillation of the diameter of the tube. (b) Another mode, the E2g mode, involves a squashing of the tube where it squeezes down in one direction and expands in the perpendicular direction oscillating between a sphere and an ellipse. The frequencies of these two modes are Raman active and depend on the radius of the tube. Figure 5.12 is a plot of the frequency of the A1g mode as a function of this radius. The dependence of this frequency on the radius is now routinely used to measure the radius of nanotubes. 400 350

Frequency (cm–1)

300

250

200

150

100 2

2.5

3

3.5

4

4.5

5

5.5

6

6.5

7

. Radius (A)

Fig. 5.12 Plot of the frequency of the Raman A1g vibrational normal mode versus the radius of the nanotube

5.10.3

Mechanical Properties

Carbon nanotubes are very strong. When weight 'W' is attached to the end of a thin wire nailed to the roof of a room, the wire will stretch. The stress S occuring on the wire is defined as the load, (or) the weight per unit cross-sectional area A of the wire. S = W/A The strain e is defined as the amount of stretch DL of the wire per unit length L e = DL

L

where L is the length of the wire before the weight is attached.

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Generally, we say that stress S is proportional to strain e. S = Ee E is the proportionality constant = LW/ADL = Young's modulus. Young's modulus is a property of a given material. It characterises the elastic flexibility of a material. The larger the value of Young's modulus, the less flexible the material. Carbon nanotubes have Young's moduli ranging from 1.28 to 1.8 TPa. One terapascal (TPa) is a pressure very close to 107 times atmospheric pressure. Young's modulus of steel is 0.12 TPa, which means that Young's modulus of carbon nanotubes is almost 10 times that of steel. This implies that carbon nanotubes are very stiff and hard to bend. But actually carbon nanotubes are very thin. The deflection D of a cylindrical hollow beam of length L with a force F on the end and the inner and outer radii of ri and ra has been shown to be FL3 D= 3EI p (ro4 - ri4 ) where I is the areal moment of inertia given by 4 The wall thickness of carbon nanotubes is about 0.34 nm, r 4a – r 4i is very small, somewhat compensating for the large value of E. While bending, the carbon nanotubes are very resilient and buckle like straws but they do not break and can be straightened back without any damage. Carbon nonotubes have little defects like dislocations or grain boundaries in the structure of their walls, so fracture on bending does not occur. The other reason is that they are bent severely, the almost hexagonal carbon rings in the walls change in structure but do not break. The carbon-carbon bonds in the nanotubes are sp2 hybrids. These sp2 bonds can rehybridise as they are bent. Strength is not the same as stiffness. The tensile strength of carbon nanotubes is about 45 billion pascals. High-strength steel alloys break at about 2 billion pascals. Thus, carbon nanotubes are about 20 times stronger than steel. Nested nanotubes have better mechanical properties. For example, multi-walled nanotubes of 200 nm diameter have a tensile strength of 0.007 TPa (i.e., 7 GPa) and a modulus of 0.6 Tpa.

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The other properties of nanoparticles are the following: 1. Elastic Properties Hardness values for nanocrystalline pure metals are 7 to 10 times higher than those of large grains (> 11 mm) metals. The intrinsic elastic moduli of nanostructured materials are essentially the same as those for conventional grain size materials until the grain size becomes very small (< 5 nm). Thus, for most nanostructured materials (grain size > 10 nm), the elastic modules are not unique properties and not a negative value. 2. Hardness and Strength To enhance this property of polymeric materials, nanofillers are also very active applications of nanomaterials. The improvements are in yield stress (30%) and Young's modulus (170%) in polypropylene filled with ultrafine SiO2. Nylon 6 is filled with 50 nm silica particle increase in tensile strength (15%), strain-to-failure (150%), Young's modulus (23%) and impact strength (78%). 3. Ductility and Toughness Nanocrystalline ceramics can be pressed and sintered into various shapes at significantly lower temperatures. Example A hard brittle ceramic, zirconia, can be rendered superplastic in nanoscale. The electrodeposited copper consists of large grains or domains of a few nanometres in extent that are subdivided into grains ranging from a few nanometres in size (100 nm). It was observed that during rolling, the low misorientation angles increase with deformation and the dislocation density builds up at the grain boundaries. Transparent plasticlike polymethyl methacrylate is used for optical components. Some disadvantages of plastic components like low hardness and abrasion resistance may be optimised through the modification with nanomaterials. 4. Magnetic Properties The magnetic strength is measured in terms of coercivity and saturation magnetisation values. This value increases with a decrease in the grain size and an increase in the specific surface area (per unit volume) of the grains. The magnet made by nanocrystalline yttrium-samarium-cobalt grains possess very unusual magnetic properties due to the large surface area. They are used in submarines, automobile alternators, motors for ships and Magnetic Resonance Imaging (MRI) in medical diagnostics.

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APPLICATION OF NANOPARTICLES 5.11.1

5.11

Applications of Carbon Nanotubes

The single and multiwalled carbon nanotubes are used in many fields like battery electrodes, electronic devices, reinforcing fibres, development of flat panel display in television and computer monitors and also in military, for developing a highly digitised battlefield for command, control and communication. Chemical gas sensors can also be the basis for employing nanotubes. For the application potential to be realised, methods for large-scale production of single-walled carbon nanotubes will have to be developed. The methods used to scale up the multiwalled tubes should provide the basis for scaling up synthesis of single-walled nanotypes. Because of the enormous application potential, it might be reasonable to hope that large-scale synthesis methods will be developed, resulting in a decrease in the cost of the order of $10 per pound.

5.11.2

Field Emission and Shielding

When a small electric field is applied parallel to the axis of a nanotube, electrons are emitted at a very high rate from the ends of the tube. This is called field emission. The application of this effect is the development of flat-panel displays. Television and computer monitors use a controlled electron gun to impinge electrons on the phosphors of the screen, which then emit light of the appropriate colours. Samsung, a Korean company, is developing a flat-panel display using the electron emission of carbon nanotubes. A Japanese company is using this electron emission effect to make vacuum-tube lamps that are as bright as conventional light bulbs, longer-lived and more efficient. Other researches are using the effect to develop a way to generate microwaves. The high electrical conductivity of carbon nanotube means that they will be poor transmitters of electromagnetic energy. A plastic composite of carbon nanotubes could provide lightweight shielding material for electromagnetic radiation. This is a matter of much concern to the military. Computers and electronic devices that are a part of this system need to be protected from weapons that emit electromagnetic pulses.

5.11.3

Nanotubes in Computers

The feasibility of designing field-effect transistors (FET's), the switching components of computers, is based on semiconducting nanotubes connecting two gold electrodes.

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A major objective of computers developers is to increase the number of switches on a chip. Carbon nanotubes with diameters of 2 nm have extremely low resistance, and thus can carry large currents without heating, so they could be used as interconnects. Their very high thermal conductivity means that they can also serve as heat sinks, allowing heat to be rapidly transferred away from the chip.

APPLICATIONS IN FUEL CELLS

5.12

Carbon nanotubes have wide applications in battery technology. Lithium is the important charge carrier in many batteries. This can be stored inside nanotubes. In general, it is estimated that one lithium atom can be stored for every six carbons of the tube. The other application is storing hydrogen in nanotubes, that is related to the development of fuel cells as sources of electrical energy for future automobiles. A fuel cell consists of two electrodes separated by a special electrolyte that allows hydrogen ions only and not electrons to pass through it. Hydrogen is sent to the anode, and gets ionised. The free electrons go to the cathode through an external circuit. The hydrogen ions diffuse through the electrolyte to the cathode and take electrons and oxygen, combining to form water. The possible hydrogen needed for the fuel cell is to store inside carbon nanotubes. In general, it is estimated that the carbon nanotubes need to hold 6.5% hydrogen by weight. At present, only about 4% hydrogen by weight has been put inside the tubes. An elegant method to put hydrogen into carbon nanotubes employs the electrochemical cell as shown in Fig. 5.13.

Fig. 5.13

An electrochemical cell used to inject hydrogen into carbon nanotubes

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The negative electrode is single-walled nanotubes in the form of paper electrolyte in KOH solution. A counter electrode consists of Ni(OH)2. The water of the electrolyte is decomposed into positive hydrogen ions (H+) that are attracted to the negative SWNT electrode. The presence of hydrogen bonded to the tubes is indicated by a decrease in the intensity of a Raman-active vibration.

CATALYSIS AND USE OF GOLD NANOPARTICLES IN MEDICINE

5.13

A catalytic agent is a material (metal or alloy) that enhances the rate of a reaction between chemicals. Nanotubes serve as catalysts for some chemical reactions. For example, nested nanotubes with ruthenium metal bonded to the outside have been demonstrated to have a strong catalytic effect in the hydrogenation reaction of cinnamaldehyde (C6H5CH=CHCHO) in the liquid phase, compared with the effect when the same metal RU is attached to other carbon substances. Chemical reactions have also been carried out inside nanotubes, such as the reduction of nickel oxide (NiO) to the base metal Ni, and the reduction of AlCl3 to its base metal Al. A stream of hydrogen gas at 475°C partially reduces MoO3 to MoO2, with the accompanying formation of steam H2O, inside multiwalled nanotubes. Cadmium sulphide (CdS) crystals have been formed inside nanotubes by reacting cadmium oxide (CdO) crystals with hydrogen sulphide gas (H2S) at 400°C. Catalysis by core-shell nanoparticles has a tremendous impact in the chemical industry, pharmaceutical products and the fuel sector. Most industrial processes are catalytically controlled. The catalytic activity and selectivity of coreshell particles can either be due to the core or due to the shell. The specific features of the nanoparticles in the catalytic industry is due to their large surface-to-volume ratio and the specific binding sites on them. Nanosized gold has high catalytic activity in the oxidation and reduction of hydrocarbons. Gold and alloy core-shell nanoparticles are model building blocks for devising nanostructred catalysts.

Nanochemistry 215

Enormous variation can be achieved in the catalytic activity by changing the composition of the core, size, shape and surface properties. Catalysis by shell includes asymmetric catalysis and mediated electron transfer. The types of catalysis by nanoparticle cores are 1. Nanoparticles supported on oxides including coreshell nanoparticles or polymers, 2. Nanoparticles encapsulated in dendrimers, and 3. Nanoparticles encapsulated in alkane thiolate monolayers. Nanosized gold supported on oxide surfaces showed a high degree of catalytic activity as compared to bulk gold, which is a poor catalyst, e.g. oxidation of CO in presence of nanosized Au. Further, the nanoparticles incorporated in the dendrimer cavities exhibit remarkable catalytic activity and selectivity towards hydrogenation and C–C coupling reactions. In core-shell nanoparticle-based catalysis, the most important issue that must be addressed is the surface passivation of nanoparticles. This can be enhanced by place exchange, interparticle linking, size processing, electrochemical, thermal and photochemical annealing.

Use of Gold Nanoparticles in Medicine Gold nanoparticles characterisation can be done by UV-visible spectroscopy in which a surface plasomon band appears at 500–700 nm. This happens because of electronic oscillations in the conduction band of metals, on exposure to electromagnetic waves. The surface plasmon resonance phenomenon occurs due to a matching of the frequency of the oscillation of the electron cloud and that of the incident light. For gold nanoparticles, the frequency is in the visible region which imparts intense colour to the nanoparticle solution.

Fig.5.14

Gold nanoshell modified using an antibody with an analyte

216 Engineering Chemistry-I

1. Gold Nanoshells for Blood Immunoassay, and Cancer Detection and Therapy Nanoshells are conjugated with antibodies that act as recognition sites for a specific analyte. The analyte causes the formation of dimers, which modify the plasmon-related absorption feature in a known way (Fig. 5.14). These nanoshells have a large optical scattering cross section, they can be used as potential contrasting agents for photonics-based imaging modalities. Among the methods, Reflectance Confocal Microscopy (RCM) and Optical Coherence Tomography (OCT) is used to facilitate early cancer detection. Optical properties of nanoshells can be tuned in such a way that they can be used for both imaging and therapy. Selective accumulation of the nanoshell can be used to image the tumour by using the high permeability and retention properties of the cancer cells. SKBr3 breast cancer cells imaged by using targeted HER2 nanoshells. HER2 is an acronym and refers to human epidermal growth factor receptor (a protein). The gold nanoshells are found to have applications in many biological fields. These are nonbiodegradable and also catalytically active which may create problems for the biological systems. 2. Gold Nanoshells for Enhancing the Raman Scattering Raman scattering is used to study different surfaces. This technique has a very weak sensitivity. However, roughened surfaces can be used for enhancing the Raman signal. Interactions of light with individual nanoshells allowed a 10-billion-fold increase in the Raman effect. 3. Gold nanoparticles are used as labels for imaging cell lines and tissues. Immunoglobulin G-capped gold nanoparticles are used to image pathogenic organisms like staphylococcus pyrogenes, etc. 4. Antibody-capped nanoparticles can be used for the selective destruction of tumours. 5. The mechanism of drug resistance, drug action and drug uptake by microorganisms can be probed by nanoparticle-based spectroscopic techniques. 6. Gold nanoparticles are extraordinarily efficient for clinical diagnostic purposes as they give strong signatures in optical absorption, fluorescence spectroscopy, X-ray diffraction and electrical conductivity.

Nanochemistry 217

7. Gold nanoparticles interact strongly with biomolecules containing thiol or amine groups and can be suitable modified with a number of small molecules, proteins, DNA and polymers. Various biomolecules bound to the gold nanoparticle surface can be detected by using various analytical measurement tools such as MALDI-TOF MS and Confocal Raman Spectroscopy. 8. Other Major Applications (a) Medical implants, such as orthopaedic implants and heart valves, are made of titanium end stainless steel nanoalloys. (b) The nanocrystalline tungsten heavy alloys lend themselves to such a self-sharpening mechanism because of their unique deformation characteristics, such as grain-boundary sliding. Hence, nanocrystalline tungsten heavy alloys and composites are being evaluated as potential candidates to replace Depleted Uranium (DU) penetrators to protect from residual radioactivity and carcinogenics. (c) The microelectronics industry emphasising miniaturisation in circuits, such as transistors, resistors and capacitors, are reduced in size in order to enable computations at great speed. (d) In televisions, the resolution improves with a reduction in the size of the pixel, or the phosphors. Nanocrystalline zinc selenide, zinc sulphide, cadmium sulphide and lead telluride synthesised by the sol-gel technique are candidates for improving the resolution of monitors. Nanophosphors reduce the cost of these displays to render High-Definition Televisions (HDTV's) and personal computers affordable to be purchased by an average household. (e) The flat-panel displays constructed out of nanomaterials possess much higher brightness and contrast than the conventional ones owing to their enhanced electrical and magnetic properties. (f) The rate and the extent of reactions are greatly increased by a decrease in the grain singe. Hence, high-sensitivity sensors made nanocrystalline materials are extremely sensitive to the change in their environment. The typical sensors are smoke detectors, like detectors on aircraft wings, automobile engine performance sensor, etc. In general, nanomaterials are used in longer-lasting satellites, food industry, fabric industry, pollutants controller, aerospace components with enhanced performance characteristics, cutting tools, high-power magnets, weapon manufacture, ete.

218 Engineering Chemistry-I

QUESTIONS Part-A 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What are nanoparticles? List out the methods used for the fabrication of carbon nanotubes. What are the properties of nanotubes? What are the other materials which can form nanotubes? What are the industrical applications of carbon nanotubes? What are carbon nanotubes? What is field emission in nanotubes and what are its applications? Write any two applications of gold nanoparticles in medicine. List out the structures of carbon nanotubes. What are nanocatalysts?

Part-B 1. What are the properties of nanotubes and how would one study those? [P.205–211] 2. What are carbon nanotubes? Write the fabrication and structure of carbon nanotubes. [P.195–199] 3. Write the possible applications of carbon nanotubes in fuel cells. [P.213] 4. What are the uses of gold nanoparticles in medicine? [P.215–216] 5. Explain in detail the fabrication of carbon nanotubes. [P.196–198]

Two-Mark Questions and Answers 1. What are called nanoparticles? Nano particles are generally considered to be a number of atoms or molecules bonded together with a radius of < 100 nm. A nanometer is 10-9 m or 10 B¡, so particles having a radius of about £ 1000 Å can be considered to be nanoparticles. 2. List out the methods used for the fabrication of carbon nanotubes. (i) Laser evaporation (ii) Carbon are method (iii) Chemical vapour deposition 3. List out the important properties of carbon nanotubes. (i) Electrical properties (ii) Vibrational properties (iii) Mechanical properties

Nanochemistry 219

4. What are the industrial application of carbon nanotubes? The single and multiwalled carbon nanotubes are used in many fields like battery electrodes, electronic devices, reinforcing fibers, development of flat panel display in Television and Computer monitors and also in military, for developing a highly digitized battle field for command, control and communication. Chemical gas sensor also be the basis for bases for employing nanotubes. 5. What are carbon nanotubes? The more interesting nanostructure with large application potential are carbon nanotubes. The carbon nanotubes are a sheet of graphite rolled into a tube with bonds at the end of the sheet forming the bonds that close the tube. 6. What are called field emission in nanotubes? What are its application? When a small electric field is applied parallel to the axis of a nanotube, are emitted at a very high rate from the ends of the tube. This is called field emission. The application if thus effect us the development of flat panel displays. 7. What is the application of nanotubes in fuel cells? Carbon nanotubes have wide applications in battery technology. Lithium is the important change carrier in many batteries. This can be stored inside nanotubes. The other application is storing hydrogen in nanotubes, that is related to the development of fuel cells as sources of electrical energy for future automobiles. 8. What are called nanocatalyst? Nanotubes serve as catalysts for some chemical reactions. For example, nested nanotubes with ruthenium metal bonded to the outside have been demonstrated to have a strong catalytic effect in the hydrogenation reaction of cinnamaldehyde (C6H5CH=CHCHO) in the liquid phase compared with the effect with the same metal Ru is attached to other carbon substances. 9. What is the application of gold nanoparticles in medicine? ! –! Gold nanoshells for blood immunoassay and cancer detection and therapy. ! –! Gold nanoparicles and used as labels for imaging cell lines and tissues. Immunoglobulin G-capped gold nanoparticles are used to image pathogenic organisms like staphylococcus pyrogenes etc. ! –! Antibody-capped nanoparticles can be used for the selective destruction of tumors.

220 Engineering Chemistry-I

10. What are called normal fundamental modes of vibration. Give an example. The atoms in a nanoparticle continually vibrate back and forth. Each nanoparticle has a specific set of vibrational motions, called normal fundamental modes of vibration. These are determined by the symmetry of the molecule. For example CO2 is linear structure: O = C = 0 The total fundamental normal modes of vibrations are calculated as 3N – 5 where N is number of atoms in the molecule. = 3 ¥ 3 – 5 = 9 – 5 = 4 Normal modes of vibrations.

SUMMARY * Polymers are the materials used in transport, agriculture, constructions etc. Thermoplastics and thermosetting polymers are used in daily life. Nylon 6, 6 and epoxy resin type polymers are explained elaborately. * The detailed terminology of second law of thermodynamics and entropy for an ideal gas, phase transitions are discussed in detail. The thermodynamical equations like Gibbs Helmholtz, clausiusclapeyron, maxwell and vant’ Hoff equations are explained elaborately. * Photo chemical fundamental laws and physical process of light reactions. Photosensitisation and chemiluminescance reactions are explained well. Spectroscopic principles and instrumental set up are explained elaborately. * Physical Phase transitions of one component systems, two component systems and the terms involved are analysed well. Alloys are the materials used to improve the physical properties of the individual metals. Ferrous and non ferrous alloys, stainless steel alloys and the various heat treatment techniques adopted to improve the physical properties can be explained elaborately. * Nanotechnology is approximately 100 nanometers or smaller and involves developing materials or devices within that size. * Introduction of nonmaterial, carbon nanotables and its types, fabrications, properties includes electrical, vibrational and mechanical are discussed in detail. * Carbon nanotubes are used in various fields like fuel cells, catalysis and medicine.

B.E/B.TECH DEGREE EXAMINATION, JANUARY 2014 FIRST SEMESTER (Common to all branches-Regulation 2013) CY6151-ENGINEERING CHEMISTRY-I Time:

Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. What is meant by degree of polymerization? 2. Mention any two uses of epoxy resin. 3. Calculate the entropy change for the reversible isothermal expansion of 10 moles of an ideal gas to 50 times its original volume at 298 K. 4. Define entropy for an ideal gas. 5. What is Chemiluminescence? 6. What is meant by adsorption of radiation? 7. What are alloys? 8. What is condensed phase rule? 9. What are carbon nanotubes? 10. What is laser ablation? PART B (5 ë 16 = 80 Marks) 11. (a) (i) Discuss cationic polymerization mechanism in detail. (8) (ii) Distinguish thermoplastics and thermosetting plastics. (8) (OR) 11. (b) (i) Explain any four properties of polymers in detail. (8) (ii) Discuss the preparation, properties and uses of Nylon 6:6. (8) 12. (a) (i) Derive Gibbs – Helmholtz equation and Explain. (8) (ii) Compute free energy change when 5 moles of an ideal gas expands reversibly and isothermally at 300 K from an initial volume of 50 L to 1000 L. (8) (OR) 12. (b) (i) What is meant by Vant Hoff’s reaction isotherm? Derive the expression for a reaction isotherm of the general reaction: aA + bB Æ cC + dD. (8)

Q.2 Engineering Chemistry-I

(ii) Discuss the criteria for chemical reaction to be spontaneous. (8) 13. (a) (i) State and explain the laws of photochemistry in detail. (8) (ii) Explain the principle and instrumentation of UV-visible spectroscopy with Neat block diagram. (8) (OR) 13. (b) (i) Explain the principle and instrumentation of IR spectroscopy with Neat block diagram. (8) (ii) Discuss 1. Fluorescence and 2. Phosphorescence in detail. (4+4) 14. (a) (i) Draw a neat one component water system and explain in detail. (8) (ii) Discuss the heat treatment of steel in detail. (8) (OR) 14. (b) (i) Draw a neat zinc–magnesium system and explain in detail. (8) (ii) Discuss composition, properties and uses of any two nonferrous alloys. (8) 15. (a) (i) How are carbon nanotubes are synthesized? Explain in detail. (8) (ii) Distinguish molecules, nanoparticles and bulk materials. (8) (OR) 15. (b) (i) Discuss various types of synthesis involved in preparation of nanomaterials. (8) (ii) Explain 1. Nanocluster 2. Nanowire with examples (4+4)

B.E/B.TECH DEGREE EXAMINATION, DEC2013/JAN2014 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111/183101 -ENGINEERING CHEMISTRY-I Time:

Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. What is break point chlorination? 2. Mention the salts responsible for temporary and permanent hardness of water. 3. What is Teflon? How is it formed? 4. What is co-polymerisation? Give one example. 5. What is adsorption? 6. What is heterogeneous catalysis? 7. Write an equation of a nuclear fission reaction. 8. What are fuel cells? 9. What are refractories? How are they classified? 10. What are solid lubricants? Give one example. PART B (5 ë 16 = 80 Marks) 11. (a) (i) How will you determine the hardness of water by EDTA method? Explain. (8) (ii) Describe the process of demineralization of water. (8) (OR) (b) (i) What do you understand by internal conditioning? Explain phosphate and calgon conditioning. (8) (ii) With a neat diagram explain reverse osmosis method of desalination. (8) 12. (a) (i) Explain the mechanism of Free Radical polymerization. (8) (ii) How will you obtain 1. Nylon 6,6 2. Polyurethane (4+4)

Q.4 Engineering Chemistry-I

(OR) (b) (i) Explain addition and condensation polymerization. Give at least 2 examples each. (8) (ii) What is vulcanization? How does vulcanization improve properties of rubber? Discuss. (8) 13. (a) (i) Derive Langmuir adsorption isotherm. (8) (ii) What are the factors affecting the rate of adsorption? (8) (OR) (b) (i) Write the differences between physisorption and chemisorption. (8) (ii) Derive Gibbs adsorption equation. (8) 14. (a) (i) Explain the construction and working of Hydrogen oxygen fuel cell. (8) (ii) Write a brief account on solar cell. (8) (OR) (b) (i) What are the functions of the following in a nuclear reactor. 1. D2O 2. Cd steel rods 3. Molten alloy of Na-K (8) (ii) Constitute a lead acid battery. Discuss its functioning. (8) 15. (a) (i) Explain the significance of the following properties exhibited by refractory materials. 1. Porosity 2. Dimensional stability 3. Thermal spalling. (8) (ii) Discuss the following characters of lubricating oils. 1. Flash and fire point 2. Cloud and pour point. (8) (OR) (b) (i) Write a brief note on the abrasive property of 1. Diamond 2. SiC 3. Quartz (8) (ii) What are nano materials? What are their advantages? Mention their applications. (8)

B.E/B.TECH DEGREE EXAMINATION, JUNE 2013 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111/183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What is alkalinity? What are its types? What is reverse osmosis? What is vulcanization? What is its use? Give the structures of (i) Nylon 6,6 (ii) Butyl rubber What is an isotherm? What are its types? What is an adsorbent? What is a fuel cell? What are its advantages? What is breeder reactor? Define refractoriness. Differentiate SWNT and MWNT. PART B (5 ë 16 = 80 Marks)

11. (a) (i) Define hardness. How is it determined? (10) (ii) Give a brief note on disadvantages of using hard water in boiler. (6) (OR) (b) (i) What is internal conditioning? What are the various methods of internal conditioning? Explain. (10) (ii) Write a note on demineralization process. (6) 12. (a) (i) What is polymerization? Distinguish between addition & condensation polymerization. (10) (ii) What are composites? Explain its various types. (6) (OR) (b) (i) What is free radical polymerization? Explain its mechanism in detail. (12) (ii) Write a note on polyurethanes. (4)

Q.6 Engineering Chemistry-I

13. (a) (i) Explain the Langmuir and Hinshelwood mechanism and explain the isotherm and cases in detail. (12) (ii) Write a note on Freundlich isotherm. (4) (OR) (b) (i) Discuss the role of adsorbents in catalysis. (10) (ii) Write about ion exchange adsorption process. (6) 14. (a) (i) What is nuclear reactor? Explain the process of power generation using a neat sketch. (12) (ii) Write a note on Lithium batteries. (4) (OR) (b) (i) What are solar cells? What are the challenges involved in the conversion of solar energy into useful energy? (10) (ii) Explain the mechanism of hydrogen oxygen fuel cell. (6) 15. (a) Explain the following: (i) Natural and synthetic abrasives (ii) Refractories and their properties ( 8 + 8) (OR) (b) Write a note on the following: (i) Mechanism of lubrication (ii) Applications of nano materials

(8 + 8)

B.E/B.TECH DEGREE EXAMINATION, JANUARY 2013 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111/183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Define alkalinity in water. How is alkalinity classified? Distinguish between soft water and demineralised water. What is co-polymerisation? Give one example. How is Teflon prepared? Mention its uses. Write the differences between physisorption and chemisorption. Give the principle of ion exchange adsorption. What are fissile nucleides and fertile nucleides? What are secondary cells? Give an example. How are refractories classified? Give one example each. Name some important methods of manufacturing CNT. PART B (5 ë 16 = 80 Marks)

11. (a) (i) Describe the methods of internal treatment of water. (ii) Draw and explain break point chlorination curve.

(8) (8)

(OR) (b) (i) Explain the boiler troubles. 1. Scale and sludge 2. Caustic embrittlement (8) (ii) What is desalination? Explain one method of desalination in detail. (8) 12. (a) (i) Define the term monomer and functionality. Explain condensation polymerization with a suitable example. How does it differ from chain polymerization? (8) (ii) Discuss the preparation, properties and uses of polyurethane and poly ethylene terephtahlate. (8) (OR) (b) (i) What are thermoplastics and thermosetting plastics? Distinguish between the two. (8)

Q.8 Engineering Chemistry-I

(ii) What are the drawbacks of raw rubber? Describe the process to improve the properties of raw rubber in detail. (8) 13. (a) (i) Enumerate the factors influencing adsorption of gases on solids. (8) (ii) Derive an expression for Langmuir adsorption isotherm. What are its limitations? (8) (OR) (b) (i) Explain the role of Ni catalyst in the hydrogenation of ethylene. What are the role of promoters in catalysis? (8) (ii) Describe the process of treatment of effluent by activated sludge process. Give any four applications of activated carbon. (8) 14. (a) (i) What are nuclear chain reactions? Explain how the amount of nuclear energy can be improved. (8) (ii) Explain the construction and working of lead acid battery. (8) (OR) (b) (i) What are fuel cells? Explain the construction and working of a fuel cell. (8) (ii) State the principle and applications of solar batteries. (8) 15. (a) (i) Explain the following properties of a lubricant. 1. Viscosity and viscosity index 2. Flash and fire point (8) (ii) What are abrasives? Explain the Moh’s scale of hardness. How is silicon carbide prepared? (8) (OR) (b) (i) Explain the manufacture of alumina and magnesite bricks. (8) (ii) Write a note on solid lubricants. (8)

B.E/B.TECH DEGREE EXAMINATION, MAY/JUNE 2012 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111/183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What do you mean by hardness of water? How is it classified? What is caustic embrittlement? How can it be avoided? How is polyurethane prepared? What is the role of sulphur in the vulcanisation of rubber? What is an adsorption isotherm? Mention any two uses of activated carbon. What are non-conventional energy sources? Give two examples. What are the applications of lithium batteries? What are abrasives? Give two examples for natural abrasives. Under what situations solid lubricants are used? PART B (5 ë 16 = 80 Marks)

11. (a) (i) 50 ml of sample of hard water required 35 ml of 0.01 M EDTA in the titration. 50 ml of the same sample of water after boiling required 12 ml of 0.01M EDTA. Calculate total and temporary hardness of water. (6) (ii) What is disinfection? Explain the mechanism of disinfection by chlorination. What are the disadvantages of bleaching powder over other disinfectants? (10) (OR) (b) (i) Explain the problems associated with the use of hard water in boilers. (8) (ii) What is demineralisation? Explain the methods of demineralisation. (8) 12. (a) (i) Distinguish between addition and condensation polymerizations with one example each. (8)

Q.10 Engineering Chemistry-I

(ii) Give the preparation and uses of Lexan. (iii) Mention the properties of engineering plastics.

(4) (4)

(OR) (b) (i) What is vulcanisation of rubber? List out the changes caused due to the vulcanisation of rubber. (8) (ii) What are Synthetic rubbers? How is butyl rubber prepared? (8) 13. (a) (i) Compare Physisorption and Chemisorption. (8) (ii) Adsorption of gases on solids is greatly influenced by temperature, pressure and nature of the adsorbent and adsorbate. Justify. (8) (b) (i) Describe the role of adsorbents in catalysis with examples. (8) (ii) How is the ion exchange adsorption useful in demineralisation of water? Explain. (8) 14. (a) (i) Give an account on the different methods by which solar energy can be harnessed? (8) (ii) What are the components of a nuclear reactor? Write briefly about each component. (8) (OR) (b) (i) What is reversible battery? Describe the construction and working of Lead acid. Storage battery with reactions occurring during chargomg and discharge cycles. (8) (ii) What are fuel cells? Describe the construction and working of H2 – O2 fuel cell. (8) 15. (a) (i) What is meant by refractoriness? How is it measured? (6) (ii) Write a note on the preparation, properties and uses of fire clay bricks. (6) (iii) How is norbide synthesised? Mention its properties and uses. (4) (OR) (b) (i) Explain the mechanism of lubrication. (8) (ii) What are nanomaterials? Discuss the types of carbon nanotubes and their Applications. (8)

B.E/B.TECH DEGREE EXAMINATION, NOVEMBER 2011/JAN 2012 FIRST SEMESTER (Common to all branches-Except Marine engineering) 183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Explain the role of Eriochrome Black – T in the EDTA titration. Name an internal method used for high pressure boiler and explain. How is AIBN initiating polymerization? Mention the monomers used for the preparation of Nylon-6,6 and PET. Write any two limitations of Freundlich isotherm. Draw an adsorption isotherm and explain. What are breeder reactors? Write any two advantages of fuel cells. Classify refractories with examples. Define flash and fire points of lubricating oils. PART B (5 ë 16 = 80 Marks)

11. (a) (i) What is alkalinity? Describe the principle and determination of alkalinity by mixed indicator titration method. (8) (ii) Write the role of coagulants in the domestic water treatment and explain the disinfection method using chlorine. (8) (OR) (b) (i) Discuss the demineralization method for the external conditioning of boiler feed water. (8) (ii) Define desalination and describe reverse osmosis. (8) 12. (a) (i) Differentiate between addition and condensation polymerization. (8) (ii) Write the preparation , properties and uses of PVC and Teflon. (8) (OR)

Q.12 Engineering Chemistry-I

(b) (i) Identify each one rubber used for tyre and tube manufacturing and discuss about them. (8) (ii) What are composites? Describe polymer matrix composites. (8) 13. (a) (i) Differentiate between Physisorption and Chemisorption with examples. (8) (ii) Illustrate effects of pressure and temperature on adsorption of gases on solids. (8) (OR) (b) (i) Derive and discuss Langmuir adsorption isotherm. (8) (ii) Discuss the zeolite process for water pollution abatement. (8) 14. (a) (i) Distinguish between nuclar fission and nuclar fusion. (8) (ii) Draw a neat diagram of photovoltaic cell and explain its working principle. (8) (OR) (b) (i) Construct a H2 – O2 fuel cell and explain. (8) (ii) Describe in detail on Nickel – Cadmium rechargeable cell. (8) 15. (a) (i) Discuss any four properties of refractories. (8) (ii) Write the preparation, properties and uses of silicon carbide and boron carbide. (8) (OR) (b) (i) Illustrate the mechanism of thick film and thin film lubrications. (8) (ii) Write an account of carbon nano tubes and its applications. (8)

B.E/B.TECH DEGREE EXAMINATION, JUNE 2011 FIRST SEMESTER (Common to all branches-Except Marine engineering) 183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What is Calgon conditioning? Distinguish between soft water and demineralized water. What are homochain and heterochain polymers? What are composites? Define the terms adsorbent and adsorbate. Differentiate between adsorption and absorption. What is nuclear fission? What is fuel cell? What are lubricants? Define the term nanochemistry. PART B (5 ë 16 = 80 Marks)

11. (a) (i) Describe demineralization process with its advantages. (8) (ii) Briefly explain the four disinfection methods. (8) (OR) (b) (i) How is scale formed in boilers? What are its drawbacks? (8) (ii) Explain reverse osmosis with its advantages (8) 12. (a) (i) Explain the mechanism of free radical polymerization reaction. (8) (ii) Discuss the preparation , properties and uses of PVC. (8) (OR) (b) (i) Differentiate between thermoplastics and thermosetting plastics with examples. (8) (ii) What is Butyl rubber? Give the properties and uses of Butyl rubber. (8)

Q.14 Engineering Chemistry-I

13. (a) (i) Derive the Freundlich’s adsorption isotherm. (8) (ii) Write in brief about the role of activated carbon in pollution control. (8) (OR) (b) (i) Discuss the factors that influence adsorption of gases on solids. (8) (ii) Distinguish between Physisorption and Chemisorption. (8) 14. (a) (i) Explain the functioning of nuclear reactor. (8) (ii) Discuss briefly about the principle and functions of lithium battery. (8) (OR) (b) (i) Explain the principle and functions of lead- acid battery. (8) (ii) Discuss briefly about the principle and functions of a fuel cell. (8) 15. (a) (i) What are refractories? How are they classified? Give examples. (8) (ii) Write explanatory notes on silicon carbide. (8) (OR) (b) (i) Discuss four important properties of lubricants. (8) (ii) Explain the fabrication and applications of carbon nano tubes. (8)

B.E/B.TECH DEGREE EXAMINATION, JANUARY 2011 FIRST SEMESTER (Common to all branches-Except Marine engineering) 183101-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

What are the constituents responsible for alkalinity of water? What is hardness of water? What do you mean by vulcanization of rubber? Write down the monomeric units that present in polyurethane. Define physical adsorption. What is an adsorption isotherm? What are nuclear moderators? Give two examples for moderators. What are the advantages of storage batteries? How is hardness of abrasives expressed? What are the carbon nanotubes? PART B (5 ë 16 = 80 Marks)

11. (a) (i) What is caustic embrittlement? How does it arise? What are the methods of prevention of caustic embrittlement? (8) (ii) Explain the determination of total hardness of water by EDTA method. (8) (OR) (b) (i) Explain the desalination of water by electro dialysis method. (8) (ii) What is break – point chlorination? Bring out its significance. (8) 12. (a) (i) Discuss briefly on the methods of preparation of Nylon 6, 6 and mention its uses. (8) (ii) Explain vulcanization of rubber and its advantages. (8) (OR) (b) (i) Explain the different types of classification of plastics with suitable examples. (8)

Q.16 Engineering Chemistry-I

(ii) What is FRP? Discuss the properties and uses of FRP. (8) 13. (a) (i) Derive the Langmuir adsorption isotherm. (8) (ii) Write in brief about the role of ion exchangers in pollution control. (8) (OR) (b) (i) Discuss the various types of adsorption isotherms. (8) (ii) Write a note on adsorption of solute from solutions. (8) 14. (a) (i) Explain the nuclear chain reaction with a suitable example. (8) (ii) How is Nicad battery constructed? Explain the cell reactions involved. (8) (OR) (b) (i) Write briefly about the advantages and the limitations of the Wind energy. (8) (ii) Discuss the principle and functions of an alkaline battery. (8) 15. (a) (i) Discuss any FOUR important properties of refractories. (8) (ii) Write explanatory notes on synthetic abrasives. (8) (OR) (b) (i) Give a brief account on solid lubricants. (8) (ii) Explain the properties and advantages of carbon nano tubes. (8)

B.E/B.TECH DEGREE EXAMINATION, MAY / JUNE 2010 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

How is water sterilized by ozone? What is calgon conditioning? Why thermosetting plastics cannot be remoulded? Mention two advantages of polymer matrix composites. What is Freundlich’s adsorption isotherm? Mention any four applications of adsorption. Distinguish between nuclear fission and fusion reactions. What are the applications of lithium batteries? What is meant by thermal spalling with respect to a refractory? Under what situations solid lubricants are used? PART B (5 ë 16 = 80 Marks)

11. (a) (i) 0.28 g of CaCO3 was dissolved in HCl and the solution was made upto one litre with distilled water. 100 ml of the above solution required 28 ml of EDTA solution on titration. 100 ml of a hard water sample required 33 ml of same EDTA solution on titration. Another 100 ml of the same water, after boiling, cooling and filtering required 10 ml of EDTA solution on titration. Calculate the temporary and permanent hardness of water. (8) (ii) What is break point chlorination? Indicate its significance. (4) (iii) 100 ml of raw water sample on titration with N/50 H2SO4 required 12 ml of the acid to phenolphthalein endpoint and 15.5 ml of the acid to methyl orange indicator. Determine the type and amount of alkalinity present in the water sample. (4) (OR)

Q.18 Engineering Chemistry-I

11. (b) (i) Explain the demineralization of water by ion-exchange process. How are exhausted cation and anion exchange resins regenerated? (8) (ii) What is reverse osmosis? Explain desalination of water by reverse osmosis method. What are the advantages of RO method? (8) 12. (a) (i) Distinguish between addition and condensation polymerizations with one example each. (8) (ii) Give the preparation and uses of Lexan. (4) (iii) Mention the properties of engineering plastics. (4) (OR) 12. (b) (i) What is vulcanization? Explain why natural rubber needs vulcanization. How is it carried out? (8) (ii) What are composites? Write a detailed note on fibrereinforced composites. (8) 13. (a) (i) Differentiate between physisorption and chemisorption. (8) (ii) What are the various factors influencing the adsorption of a gas on a solid? (8) (OR) 13. (b) (i) What is the role of adsorbents in catalysis? (8) (ii) What i s the role of activated carbon in air and water pollution control? (8) 14. (a) (i) With a neat sketch explain the working of light water nuclear power plant. (8) (ii) Write a note on solar cells. (8) (OR) 14. (b) (i) What is reversible battery? Describe the construction and working of Lead acid storage battery with reactions occurring during charging and discharging cycles. (8) (ii) What are fuel cells? Describe the construction and working of H2-O2 fuel cell. (8) 15. (a) (i) Explain any five properties of refractories? (10) (ii) What are abrasives? How are they classified? Give one example for each. (6) (OR) 15. (b) (i) What are fluid and boundary lubrications? Explain. (8) (ii) What are carbon nano tubes? Explain any three of their important applications. (8)

B.E/B.TECH DEGREE EXAMINATION, JANUARY 2010 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. Calculate the hardness of a water sample containing 2.4 mg of calcium chloride in 500 ml of water? 2. What is Calgon? How does it function in water treatment? 3. Why thermosetting plastics can not be remoulded?. 4. What is the role of Sulphur in the vulcanization of rubber? 5. Compare absorption and adsorption. 6. Mention any four applications of adsorption. 7. Furnish the sequence of reactions in proton cycle nuclear fusion. 8. Give any two advantages of alkaline battery over dry cell. 9. Define refractoriness of a refractory. 10. What are nano materials? Mention any two of their characteristic properties. PART B (5 ë 16 = 80) 11. (a) (i) How is temporary hardness of water estimated by EDTA method? (ii) What are the requirements of potable water? How will you purify water for drinking purpose?. (OR) 11. (b) (i) What are ion exchange resins? How are they useful in removing hardness of water? (ii) What is desalination? With a neat diagram, describe the ‘reverse osmosis’ method for the desalination of brackish water. 12. (a) (i) How are the following polymers prepared? 1. Teflon 2. Polystyrene 3. PET 4. Nylon 6, 6 (ii) Explain the mechanism of free radical addition polymerization.

Q.20 Engineering Chemistry-I

(OR) 12. (b) (i) What is natural rubber? Explain why natural rubber needs vulcanization. How is it carried out? (ii) What are composites? Give the preparation and uses of glass fiber reinforced composites and carbon fiber reinforced composites. 13. (a) (i) Compare physisorption and chemisorption. (ii) Adsorption of gases on solids is greatly influenced by temperature, pressure and nature of the adsorbent and adsorbate justify. (OR) (b) (i) Describe the role of adsorbents in catalysis with examples. (ii) How is ion exchange adsorption useful in demineralization of water? Explain. 14. (a) (i) Explain with a neat diagram the parts and functions of a nuclear reactor. (ii) Write a note on photovoltaic cell. (OR) 14. (b) (i) (ii) 15. (a) (i) (ii)

Explain the working of hydrogen oxygen fuel cell. Write a short note on lithium batteries. How are alumina and carborundum manufactured? What are refractories? How are they classified? (OR)

15. (b) (i) What are fluid and boundary lubrication? Explain. (ii) What are carbon nano tubes? Explain any three of their important applications.

B.E/B.TECH DEGREE EXAMINATION, MAY/JUNE 2009 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111-ENGINEERING CHEMISTRY-I Time: Three Hours

Maximum:

100 Marks

Answer ALL Questions PART A (10 ë 2 = 20 MARKS) 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

How is the hardness of water expressed? What are the disadvantages of scale formation? How is polymerization classified? Give one example for each class. Write any two uses of PVC. Define the terms adsorbent and adsorbate giving suitable examples. What is an adsorption isotherm? What is nuclear energy? Explain using a suitable example. List any two advantages of lithium batteries. Mention any four natural abrasives. How are lubricants classified? Give one example for each class. PART B (5 × 16 = 80 marks)

11. (a) (i) Describe briefly the various methods of internal conditioning of boiler feed water (10) (ii) 100 ml of a water sample required 20 ml of 0.01 M EDTA for the titration with Eriochrome Black-T indicator. 100 ml of the same water sample after boiling and filtering required 10 ml of 0.01 M EDTA. Calculate the total, carbonate and non carbonate hardness of the sample. (6) (OR) (b) (i) Explain the necessity for sterilization of domestic water and discuss the various methods of sterilization. Discuss in detail about the break point chlorination. (10) (ii) Describe the reverse osmosis method for desalination of sea water with a neat sketch. (6) 12. (a) (i) Write the preparation properties and uses of PET and polyurethane. (8)

Q.22 Engineering Chemistry-I

(ii) Write a note on FRP.

(8)

(OR) (b) (i) Distinguish between addition and condensation polymerization. (8) (ii) Write the mechanism of free radical polymerization. (8) 13. (a) (i) What are the factors that influence the adsorption of gases on solids? Discuss in detail. (8) (ii) Explain the role of adsorption in catalysis using an appropriate example. (8) (OR) (b) (i) Derive the Langmuir’s adsorption isotherm and discuss the effect of pressure on the isotherm. (12) (ii) Illustrate how the ion exchange adsorption is useful in the demineralization of water (only the process without diagram). (4) 14. (a) (i) What are the components of a nuclear reactor? Write briefly about each component. (10) (ii) Discuss the characteristics of the reaction when uranium undergoes nuclear fission. (6) (OR) (b) (i) Describe the principle and functioning (charging and discharging) of lead acid battery. (10) (ii) Write a note on wind energy. (6) 15. (a) (i) Give a brief account of neutral refractories. (6) (ii) What are grinding wheels? How are they manufactured? (10) (OR) (b) (i) Discuss the following properties of lubricants. (1) Flash point. (5) (2) Pour point. (5) (ii) Explain the properties and applications of carbon nanotubes. (6)

B.E/B.TECH DEGREE EXAMINATION, JANUARY 2009 FIRST SEMESTER (Common to all branches-Except Marine engineering) CY2111-ENGINEERING CHEMISTRY-I Time : Three Hours

Maximum : 100 Marks Answer ALL Questions PART A (10 ë 2 = 20 MARKS)

1. What is hardness of a solution containing 0.585 grams of NaC1 and 0.6 grams of MgSO4 per litre? 2. How is the exhausted zeolite softener bed regenerated? 3. Explain functionality of a monomer with a suitable example. 4. Explain condensation polymerization with a suitable example. 5. What is the role of adsorbent in catalysis? 6. What is the effect of increase in temperature and increase in pressure on the absorption of a gas on a solid? 7. What is breeder reactor? 8. State the reaction when a lead storage battery is recharged? 9. Give two examples for neutral refractory. 10. Explain the lubrication action of graphite. PART B (5 x 16 = 80 MARKS) 11. (a) (i) Explain the softening of water by deionization process. (ii) How is the temporary and permanent hardness of water determined? (OR) (i) What is meant by reverse osmosis? Explain the purification of water by reverse Osmosis. (ii) Explain the softening of water by zeolite process. 12. (a) (i) Write the mechanism of free radical polymerization. (ii) Discuss the synthesis and uses of SBR and Butyl rubbers. (OR) (i) Give an account on fibre reinforced platics. (ii) Describe the synthesis of polyurethane and state its uses.

Q.24 Engineering Chemistry-I

13. (a) (i) Derive the Langmuir Adsorption Isothern and interpret the results 1. at Low pressure and 2. at High pressure. (ii) Distinguish between Physisorption and Chemisorption (OR) (b) (a) (i) State Friendlich adsorption isotherm and explain the terms in it. Explain the different types of adsorption isotherms of gases on solid. (ii) Discuss the role of adsorbands in pollution abatement. 14. (a) (i) Give an account of H2 – O2 fuel cell. (ii) Explain the construction and working of Cd-Ni cell. (OR) (b) (i) Describe using a block diagram the light water nuclear reactor for power generation. (ii) Give an account of solar cells. 15. (a) (i) What are the characteristics of a good refractory? Write a note on carborundum. (ii) What are lubricants? Discuss the different types of lubrication. (OR) (b) Write short notes on any TWO of the following: (i) Carbon nanotubes. (ii) Refractoriness. (iii) Synthetic abrasives.