Chemistry : a molecular approach [Third Canadian edition.] 9780134755380, 0134755383, 9780134894829, 0134894820

Citation preview

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Main groups I

1

1

I

DJ 8

2

3

4

5

6

7

Main groups

D

2

D

Metals

D

Metalloids

18 2

He

Nonmetals

13

14 6

15 7

16

17

8

9

10

N

0

F

Ne 20. 18

4.003

3

4

5

Li

Be

B

c

6.941

9.012

10.81

12.0l

14.0l

16.00

19.00

11

12

13

14

15

16

18

Na

Mg

Al

Si

p

s

17

Cl

Ar

22.99

24.31

26.98

28.09

30.97

32.07

35.45

39.95

19

31

32

33

34

35

36

Transition metals I

r---

I

-i

20

3 21

4 22

5 23

6 24

7 25

8 26

9 27

10 28

11 29

12 30

K

Ca

Sc

Ti

v

Cr

Mn

Fe

Co

Ni

Cu

Zn

Ga

Ge

As

Se

Br

Kr

39.10

40.08

44.96

47.87

50.94

52.00

55.85

58.93

58.69

63.55

65.38

69.72

72.64

74.92

78.96

79.90

83.80

37

38

40

41

42

43

44

45

46

47

48

49

50

51

52

53

54

Rb

Sr

39 y

Zr

Nb

Mo

Tc

Ru

Rh

Pd

Ag

Cd

In

Sn

Sb

Te

I

Xe

85.47

87.62

88.91

91.22

92.91

95 .96

[98]

101.07

102.91

106.42

107.87

112.41

114.82

118.71

121.76

127.60

126.90

131.29

55

56

72

73

74

76

77

78

79

80

83

84

85

86

Ba

Hf

Ta

Re

Os

Ir

Pt

Au

Hg

81 TI

82

Cs

w

75

Pb

Bi

Po

At

Rn

132.91

137.33

178.49

180.95

183.84

186.21

190.23

192.22

195.08

196.97

200. 59

204.38

207.2

208.98

[208.98]

[209.99]

[222.02]

87

88

104

105

106

107

108

109

110

111

112

11 3

114

115

116

11 7

118

Fr

Ra

Rf

Db

Sg

Bh

Hs

Mt

Ds

Rg

Cn

Nh

FI

Mc

Lv

Ts

Og

[223.02]

[226.03]

[261.11]

[262.1 1]

[266.12]

[264. 12]

[269.13]

[268.14]

[271 ]

[272]

[277]

[286]

[289]

[289]

[292]

[294]

[294]

~~

Lanthanoid series

Actinoid series

54.94

57

58

59

60

61

62

63

64

65

66

67

68

69

70

71

La

Ce

Pr

Nd

Pm

Sm

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

138.9 1

140. l 2

140.91

144.24

[145]

150.36

151.96

157.25

158.93

162.50

164.93

167.26

168.93

173.05

174.97

92

93

94

95

96

97

98

99

100

10 1

102

103

89

90

91

Ac

Th

Pa

u

Np

Pu

Am

Cm

Bk

Cf

Es

Fm

Md

No

Lr

[227.03]

232 .04

231.04

238.03

[237.05]

[244.06]

[243.06]

[247.07]

[247.07]

[25 1.08]

[252.08]

[257.10]

[258.10]

[259 .10]

[262.11]

Atomic masses in brackets are the masses of the longest-lived or most important isotope of radioactive elements.

List of Elements with Their Symbols and Atomic Masses Element Actinium Aluminum Americium Antimony Argon Arsenic Astatine Barium Berkelium Beryllium Bismuth Bohrium Boron Bromine Cadmium Calcium Californiu m Carbon Cerium Cesium Chlori ne Chromium Cobalt Copernicium Copper Curium Darmstadtium Dubnium Dysprosium Einsteinium Erbium Europium Fermi um Flerovium Fluorine Francium Gadolinium Gallium Germaniu m Gold Hafnium Hassium Helium Holmium Hydrogen Indium Iodine Iridium Iron Krypton Lanthanum Lawrencium Lead Lithium Livermorium Lutetium Magnesium Manganese Meitnerium

Symbol Ac Al Am Sb Ar As At Ba Bk Be Bi Bh B Br Cd Ca Cf

c Ce Cs Cl Cr Co Cn Cu Cm Ds Db Dy Es Er Eu Fm FI F Fr Gd Ga Ge Au Hf Hs He Ho H In I Ir Fe Kr La Lr Pb Li Lv Lu Mg Mn Mt

Atomic Number

Atomic Mass

89 13 95 51 18 33 85 56 97

227.03a 26.98 243.06a 12 1.76 39.95 74.92 209.99a 137.33 247.07a 9.012 208.98 264.12• 10.81 79.90 11 2.41 40.08 25 1.08a 12.01 140.1 2 132.9 1 35.45 52.00 58.93 277• 63.55 247.07a 27 1• 262. 1 l a 162.50 252.08a 167.26 15 1.96 257. l Oa 289a 19.00 223.02• 157.25

4 83 107 5 35 48 20 98

6 58 55 17 24 27 112 29 96 110 105

66 99 68 63 100 114

9 87 64

31 32 79

72 108 2 67 1 49 53 77 26 36 57 103 82 3 116 71 12 25 109

•Mass of longest-lived or most important isotope.

69.72 72.64

196.97 178.49 269.13a 4.003 164.93 1.008 114.82 126.90 192.22 55.85 83.80 138.9 1 262.11 a 207.2 6.941 292• 174.97 24.3 1 54.94 268.14a

Element Mendelevium Mercury Molybdenum Moscovium Neodymium Neon Neptunium Nickel Nihonium Niobium Nitrogen Nobelium Oganesson Osmium Oxygen Palladium Phosphorus Platin um Plutonium Polonium Potassium Praseodymium Promethium Protactinium Radium Radon Rhenium Rhodium Roentgenium Rubidium Ruthenium Rutherfordium Samarium Scandium Seaborgium Selenium Silicon Silver Sodiu m Strontium Sulfur Tantalum Technetium Tellurium Tennessine Terbium Thallium Thorium Thulium Tin Titanium Tungsten Uranium Vanadium Xenon Ytterbium Yttrium Zinc Zirconium

Symbol Md Hg Mo Mc Nd Ne Np Ni Nh Nb N No Og Os

0 Pd p Pt Pu Po

K Pr Pm Pa Ra Rn Re Rh Rg Rb Ru Rf Sm Sc Sg Se Si Ag Na Sr

s Ta Tc Te Ts Tb TI Th Tm Sn Ti

w u v

Xe Yb y Zn Zr

Atomic Number

Atomic Mass

101 80 42 115 60 10 93 28 11 3 41 7 102 118 76 8 46 15 78 94 84 19 59 61 91 88 86 75 45

258. 10• 200.59 95.96 289 144.24 20.18 237.05• 58.69 284 92.9 1 14.01 259.1 0• 294 190.23 16.00 106.42 30.97 195.08 244.06a 208.98" 39.10 140.9 1 145" 23 1.04 226.03a 222.02• 186.2 1 102.9 1

111

272•

37

44 104

62 21 106 34 14 47 11 38 16 73 43 52 117 65 81 90

69 50

22 74 92 23 54 70 39 30 40

85.47 10 1.07 261.11 a 150.36 44.96 266.12• 78.96 28.09 107.87 22.99 87.62 32.07 180.95 93a 127.60 294 158.93 204.38 232.04 168.93 118.7 1 47.87 183 .84 238.03 50.94 13 1.293 173 .05 88.9 1 65.38 9 1.22

CHEMISTRY

A MOLECULAR APPROACH

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NIVALDO J. TRO Westmont Col lege

TRAVIS D. FRIDGEN Memorial University of Newfoundland

LAWTON E. SHAW Athabasca University

TH I RD CANADIAN EDITION

CHEMISTRY A MOLECULAR APPROACH

@ Pearson

Pearson Canada Inc., 26 Prince Andrew Place, North York, Ontario M3C 2H4.

Copyright© 2020, 2017, 2014 Pearson Canada Inc. All rights reserved. Printed in the United States of America. This publication is protected by copyright, and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or otherwise. For information regarding permissions, request forms, and the appropriate contacts, please contact Pearson Canada' s Rights and Permissions Departme nt by visiting www. pearson.com/ca/en/contact-us/permissions.html. Authorized adaptation from Chemistry: A Molecular Approach, Fourth Edition, © 2017, Pearson Education, Inc. Used by permission. All rights reserved . This edition is authori zed for sale only in Canada. Attributions of third-party content appear on the appropriate page within the text. PEARSON, and ALWAYS LEARNING are exclusive trade marks owned by Pearson Education, Inc. or its affiliates in the U.S. and/or other countries. Unless otherwise indicated herein, any third party trademarks that may appear in this work are the property of their respective owners and any refere nces to third party trademarks, logos, or other trade dress are for demonstrative or descriptive purposes only. Such references are not intended to imply a ny sponsorship, endorsement, authorization, or promotion of Pearson Canada products by the owners of such marks, or any relationship between the owner and Pearson Canada or its affiliates, authors, licensees, or distributors.

If you purchased this book outside the United States or Canada, you should be aware that it has been imported without the approval of the publisher or the author. 978-0-13-475538-0 1 20 Library and Archives Canada Cataloguing in Publication Tro, Nivaldo J., author Chemistry : a molecular approach I Nivaldo J. Tro, Westmont College, Travis D. Fridgen, Memorial University, Lawton E. Shaw, Athabasca University. - Third Canadian edition. Issued in print and electronic formats . ISBN 978-0-13-475538-0 (hardcover).- ISBN 978-0-13-489482-9 (looseleaf).-ISBN 978-0- 13-499409-3 (HTML)

1. Chemistry, Physical and theoretical-Textbooks. 2. Textbooks. I. Fridgen, Travis D. (Travis David), 1970-, author II. Shaw, Lawton, 1972-, author Ill. Title. QD453.3.T76 2019

@ Pearson

541

C2018-90595 1-6 C2018-905952-4

To Michael, Ali, Kyle, and Kaden -Nivaldo Tro

To Cailyn, Carter, Colton, and Chloe -Travis Pridgen

To Calvin, Nathan, Alexis, and Andrew -Lawton Shaw

About the Authors ivaldo Tro is a Professor of

N

Chemistry at Westmont College in Santa Barbara, Cali-

fornia, where he has been a faculty member since 1990. He received his Ph.D. in chemistry from Stanford University for work on developing and

using optical techniques to study the adsorption and desorption of molecules to and from surfaces in ultrahigh vacuum. He then went on to the University of California at Berkeley, where he did postdoctoral research on ultrafast reaction dynamics in solution. Since coming to Westmont, Professor Tro has been awarded grants from the American Chemical Society Petroleum Research Fund, from Research Corporation, and from the National Science Foundation to study the dynamics of various processes occurring in thin adlayer films adsorbed on dielectric surfaces. He has been honoured as Westmont's outstanding teacher of the year three times and has also received the college's outstanding researcher of the year award. Professor Tro lives in Santa Barbara with his wife, Ann, and their four children, Michael, Ali, Kyle, and Kaden.

T

ravisFridgen is current!y Associate

Dean of Science and Professor of Chemistry at Memorial Uni-

versity in St. John's, Newfoundland and

Labrador. His research group studies the energetics, reactions, and structures of gaseous self-assembled complexes

Vi

About the Authors

composed of metal ions and biologically relevant molecules such as DNA bases, amino acids, and peptides using a combination of mass spectrometry, tunable infrared lasers, and computational chemistry. Their research is aimed at answering fundamental questions such as why K+ is associated with guanine quadruplexes such as telomeric DNA. He graduated with a B.Sc. (Hons) in chemistry from Trent University and a B.Ed. from Queen's University. His Ph.D. in physical chemistry is from Trent and Queen's Universities, where he studied the spectroscopy of reactive species in a cryogenic matrix environment. During his postdoctoral fellowship at the University of Waterloo, he first began conducting research using mass spectrometric methods. During a brief period as an assistant professor at Wilfrid Laurier University, he initiated a collaboration with a group of researchers from France to spectroscopically determine structures of gas phase proton-bound dimer ions. He has taught physical chemistry courses at all levels, but mostly enjoys teaching first year. He was recently awarded the inaugural Distinguished Teaching Award by the Faculty of Science at Memorial University. He lives in Mount Pearl, Newfoundland and Labrador, with his wife, Lisa; their four children, Cailyn, Carter, Colton, and Chloe; and their three Shih Tzus, Kobe, Jacky, and Joey. They are all avid fans of the Ottawa Senators and enjoy busy, active lives that include outdoor activities such as shovelling snow (good old Newfoundland!).

.

,

I

: .... . k I..; ' f .. " ' . 0(

L

awton Shaw received his Ph.D. in chemistry from the University of Calgary in the area of

photochemical reaction mechanisms of organometallic complexes. Shortly after graduating, he joined the full-time teaching faculty at Mount Royal Col-

lege in Calgary, where he developed one of the first science courses at Mount Royal delivered partially online. This work led to a serious interest in online and distance education. In 2005, he joined the Centre for Science at Athabasca University, where he teaches and coordinates distance-delivered chemistry courses. This experience led to the book Accessible Elements: Teaching Science Online and at a Distance, which he co-edited. His research interests are split between the realms of teaching/education and environmental chemistry. He studies the effects of pharmaceuticals and personal care products on biofilms in freshwater ecosystems. He is a former president of College Chemistry Canada. He has extensive experience with the Athabasca University Faculty Association, serving as President from 2014 to 2017. He lives in St. Albert, Alberta, with his wife, Tanya, and their four children. Their family leisure time is filled with activities such as crosscountry skiing, swimming, and camping.

Brief Contents Units of Measurement for Physical and Chemical Change 2

Atoms and Elements

3

Molecules, Compounds, and Nomenclature

4

Chemical Reactions and Stoichiometry

Gases 6 Thermochemistry 5

7

The Quantum-Mechanical Model of the Atom

8

Periodic Properties of the Elements

9

Chemical Bonding I: Lewis Theory

10

Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

11

Liquids, Solids, and Intermolecular Forces

12

Solutions

13 14

Chemical Kinetics Chemical Equilibrium

15

Acids and Bases

16

Aqueous Ionic Equilibrium

17

Gibbs Energy and Thermodynamics

18

Electrochemistry

19

Radioactivity and Nuclear Chemistry

20

Organic Chemistry I: Structures

21

Organic Chemistry II: Reactions

22

Biochemistry

23

Chemistry of the Non metals

24

Metals and Metallurgy

25

Transition Metals and Coordination Compounds

1 29 57 101 149 197 242 298 334 377 433 491 539 592 636 691 746 795 842 878 920 966 1000 1034 1055

Appendix I: Common Mathematical Operations in Chemistry

A-1

Appendix II: Useful Data

A-7

Appendix Ill: Answers to Selected Exercises

A-17

Appendix IV: Answers to In-Chapter Practice Problems

A-60

Glossary Index

G-1 1-1

vii

Contents Preface

xviii

CHEMISTRY IN YOUR DAY: Where Did Elements 41

Come From?

1

Units of Measurement for Physical and Chemical Change

Physical and Chemical Changes and Physical and Chemical Properties 1.2 Energy: A Fundamental Part of Physical and Chemical Change 1.3 The Units of Measurement

2.5 Atomic Mass: The Average Mass of an Element's Atoms

1

1.1

3

4

CHAPTER IN REVIEW Key Terms 50 Key Concepts 51 Relationships 51 Key Skills 52

11

12

18

19

2

22

23

57 59

Covalent Bonds 60

3.3 Representing Compounds: Chemical Formulas and Molecular Models

60

64

3.4 Formulas and Names

70

3.5 Organic Compounds

2.1 Imaging and Moving Individual Atoms 2.2 Early Ideas About the Building Blocks of Matter 2.3 Modern Atomic Theory and the Laws That Led to It

30 31 32

The Law of Conservation of Mass 32 The Law of Definite Proportions 33 The Law of Multiple Proportions 33 John Dalton and the Atomic Theory 34

viii

57

Ionic Compounds 64 Molecular Compounds 67 Naming Acids 69

29

The Discovery of the Electron 34 The Di scovery of the Nucleus 36 Protons, the Atomic Number, and Neutrons 38 Isotopes: When the Number of Neutrons Varies 39 Ions: Losing and Gaining Electrons 40

3

Molecules, Compounds, and Nomenclature

52

Types of Chemical Formulas 60

Atoms and Elements

2.4 Atomic Structure

EXERCISES Review Questions 52 Problems by Topic 52 Cumulative Problems 55 Challenge Problems 56 Conceptual Problems 56

Ionic Bonds 59

Key Equations and

EXERCISES Review Questions 23 Problems by Topic 23 Cumulative Problems 26 Challenge Problems 27 Conceptual Problems 28

50 Key Equations and

3.1 Hydrogen, Oxygen, and Water 3.2 Chemical Bonds

General Problem-Solving Strategy 19 Order-ofMagnitude Estimations 20 Problems Involving an Equation 20

CHAPTER IN REVIEW Key Terms 22 Key Concepts 22 Relationships 22 Key Skills 23

47

Ions and the Periodic Table 50

Counting Significant Figures 13 Exact Numbers 14 Significant Figures in Calculations 15 Rules for Calculations 15 Rules for Rounding 16 Precision and Accuracy 17

THE NATURE OF SCIENCE: Integrity in Data Gathering 1.5 Solving Chemical Problems

44

The Mole: A Chemist's "Dozen" 44 Converting Between Number of Moles and Number of Atoms 45 Converting Between Mass and Amount (Number of Moles) 45

2.7 The Periodic Table of the Elements

The Standard Units 4 The Metre: A Measure of Length 4 The Kilogram: A Measure of Mass 5 The Second: A Measure of Time 5 The Kelvin : A Measure of Temperature 5 SI Prefi xes 6 Conversions Invol ving the SI Prefi xes 7 Derived Units 8

CHEMISTRY AND MEDICINE: Bone Density 1.4 The Reliability of a Measurement

Mass Spectrometry: Measuring the Mass of Atoms and Molecules 43

2.6 Molar Mass: Counting Atoms by Weighing Them 2

42

Naming Hydrocarbons 72 Cyclic Hydrocarbons 76 Aromatic Hydrocarbons 77 Functionalized Hydrocarbons 78

3.6 Formula Mass and the Mole Concept for Compounds

81

Molar Mass of a Compound 81 Using Molar Mass to Count Molecules by Weighing 81

83

3.7 Composition of Compounds Conversion Factors in Chemical Formulas 84

34

CHEMISTRY IN YOUR DAY: Drug Tablets and Capsules 3.8 Determining a Chemical Formula from Experimental Data

86 87

Calculating Molecular Formulas for Compounds 88 Combustion Analysis 90

CHAPTER IN REVIEW Key Terms 91 Key Concepts 92 Relationships 92 Key Skills 93

91 Key Equations and

CONTENTS

EXERCISES

93

Review Questions 93 Problems by Topic 94 Cumulative Problems 99 Challenge Problems 100 Conceptual Problems 100

4

Chemical Reactions and Stoichiometry

5.4 The Ideal Gas Law 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas

ix 160 162

Molar Volume at Standard Temperature and Pressure 162 Density of a Gas 163 Molar Mass of a Gas 164

101

4.1 Chemistry of Cuisine 4.2 Writing and Balancing Chemical Equations

101 102

How to Write Balanced Chemical Equations 104

4.3 Solutions and Solubility

105

5.6 Mixtures of Gases and Partial Pressures CHEMISTRY IN THE ENVIRONMENT: Stack Sampling

4.4 Precipitation Reactions 4.5 Acid-Base Reactions

109 111

Acid-Base Reactions Evolving a Gas 114

4.6 Oxidation-Reduction Reactions

115

Oxidation States 116 Identifyi ng Redox Reactions 11 8

CHEMISTRY IN YOUR DAY: Bleached Blonde

120

Balancing Oxidation-Reduction Equations 121

4.7 Reaction Stoichiometry: How Much Is Produced?

125

Making Molecules: Mole-to-Mole Conversions 125 Making Molecules : Mass-to-Mass Conversions 125

4.8 Limiting Reactant, Theoretical Yield, and Percent Yield

171

Molar Volume and Stoichiometry 173

174

Kinetic Molecular Theory and the Ideal Gas Law 176 Temperature and Molecular Velocities 177

5.9 Mean Free Path, Diffusion, and Effusion of Gases 5.1O Real Gases: The Effects of Size and Intermolecular Forces

180 181

The Effect of the Finite Volume of Gas Particles 182 The Effect of Intermolecular Forces 183 Van der Waals Equation 184 Real Gases 184

CHEMISTRY IN YOUR DAY: Pressure in Outer Space

185

CHAPTER IN REVIEW 187 Key Terms 187 Key Concepts 187 Key Equations and Relationships 188 Key Skills 188

127

Limiting Reactant, Theoretical Yield, and Percent Yield from Initial Reactant Masses 128

4.9 Solution Concentration and Solution Stoichiometry

167

Collecting Gases over Water 169

5.7 Gases in Chemical Reactions: Stoichiometry Revisited 5.8 Kinetic Molecular Theory: A Model for Gases

Electrolyte and Nonelectrolyte Solutions 106 The Solubility of Ionic Compounds 107

166

132

EXERCISES Review Questions 189 Problems by Topic 190 Cumulative Problems 193 Challenge Problems 195 Conceptual Problems 196

189

Solution Concentration 132

CHEMISTRY IN YOUR DAY: Blended Ethanol Gasoline Using Molarity in Calculations 135 Stoichiometry 137

CHAPTER IN REVIEW Key Terms 139 Key Concepts 139 Relationships 140 Key Skills 140

Solution

139 Key Equations and

EXERCISES Review Questions 141 Problems by Topic 141 Cumulative Problems 145 Challenge Problems 147 Conceptual Problems 148

5

133

6

Thermochemistry

6.1 Chemical Hand Warmers 6.2 The Nature of Energy: Key Definitions

5.1 Breathing: Putting Pressure to Work 5.2 Pressure: The Result of Molecular Collisions

141

6.3 The First Law of Thermodynamics: There Is No Free Lunch CHEMISTRY IN YOUR DAY: Perpetual Motion Machines

Boyle's Law: Volume and Pressure 154 Charles's Law: Volume and Temperature 156 Avogadro's Law: Volume and Amount (in Moles) 159

200 201

Internal Energy 201

149 149 150

Pressure Units 151 The Manometer: A Way to Measure Pressure in the Laboratory 152

CHEMISTRY AND MEDICINE: Blood Pressure 5.3 The Gas Laws: Boyle's Law, Charles's Law, and Avogadro's Law

197 198

Units of Energy 200

6.4 Quantifying Heat and Work

Gases

197

153

154

Heat 205

205

Work: Pressure-Volume Work 209

6.5 Measuring ilrUfor Chemical Reactions: Constant-Volume Calorimetry 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure

212 214

Exothermic and Endothermic Processes: A Molecular View 217 Stoichiometry Involving ilrH: Thermochemical Equations 218

6.7 Constant-Pressure Calorimetry: Measuring ilrH 6.8 Relationships Involving ilrH 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation

220 221 224

X

CONTENTS

Standard States and Standard Enthalpy Changes 224 Calculating the Standard Enthalpy Change for a Reaction 226

8

6.10 Energy Use and the Environment

229

Implications of Dependence on Fossil Fuels 229 CHAPTER IN REVIEW Key Terms 232 Key Concepts 232 Relationships 233 Key Skills 234

232

Key Equations and

EXERCISES Review Questions 234 Problems by Topic 235 Cumulative Problems 239 Challenge Problems 240 Conceptual Problems 241

7

The Quantum-Mechanical Model of the Atom

8.6 Ionic Radii 8.7 Ionization Energy

A Bar Code for Atoms

7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy

248

Electron Affinity 321

320

Metallic Character 321

258

8.9 Some Examples of Periodic Chemical Behaviour: The Alkali Metals, Alkaline Earth Metals, Halogens, and Noble Gases

260

The Alkali Metals (Group 1) 323 The Alkaline Earth Metals (Group 2) 324 The Halogens (Group 17) 325 The Noble Gases (Group 18) 326

322

CHEMISTRY AND MEDICINE: Potassium Iodide

265

Solutions to the Schri:idinger Equation for the Particle in a Box: Quantum Numbers 265 Solutions to the Schri:idinger Equation for the Hydrogen Atom 268

7.6 The Shapes of Atomic Orbitals

306

313 316

8.8 Electron Affinities and Metallic Character 254

The de Broglie Wavelength 261 The Uncertainty Principle 262 Indeterminacy and Probability Distribution Maps 264

7.5 Quantum Mechanics and the Atom

305

Trends in First Ionization Energy 316 Exceptions to Trends in First Ionization Energy 318 Ionization Energies of Transition Metals 319 Trends in Second and Successive Ionization Energies 319

The Wave Nature of Light 244 The Electromagnetic Spectrum 246 Interference and Diffraction 247

7.3 Atomic Spectroscopy and the Bohr Model CHEMISTRY IN YOUR DAY: Atomic Spectroscopy:

301

Effective Nuclear Charge 308 Slater's Rules 310 Atomic Radii of d-Block Elements 312

242

CHEMISTRY AND MEDICINE: Radiation Treatment for Cancer The Particle Nature of Light 250

299 299

Orbital Blocks in the Periodic Table 302 Writing an Electron Configuration for an Element from Its Position in the Periodic Table 303 The d-Block and / -Block Elements 304

8.4 The Explanatory Power of the Quantum-Mechanical Model 8.5 Periodic Trends in the Size of Atoms and Effective Nuclear Charge

243 243

298

8.1 Nerve Signal Transmission 8.2 The Development of the Periodic Table 8.3 Electron Configurations, Valence Electrons, and the Periodic Table

234

7.1 Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small 7.2 The Nature of Light

270

s Orbitals (l = 0) 270 p Orbitals(/= 1) 273 d Orbitals (l = 2) 274 f Orbitals(/ = 3) 274

The Phase of Orbitals 275 Wave Functions 276

Periodic Properties of the Elements

The Hydrogen-Like

in Radiation Emergencies

327

CHAPTER IN REVIEW Key Terms 328 Key Concepts 328 Key Equations and Relationships 328 Key Skills 329

328

EXERCISES Review Questions 329 Problems by Topic 330 Cumulative Problems 331 Challenge Problems 332 Conceptual Problems 333

329

7.7 Electron Configurations: How Electrons Occupy Orbitals 278

g

Electron Spin and the Pauli Exclusion Principle 279 Sublevel Energy Splitting in Multielectron Atoms 280 Electron Configurations for Multielectron Atoms 284 Electron Configurations for Transition Metals 286 Electron Configurations and Magnetic Properties of Ions 288 CHAPTER IN REVIEW Key Terms 290 Key Concepts 291 Relationships 291 Key Skills 292

290

Key Equations and

EXERCISES Review Questions 292 Problems by Topic 293 Cumulative Problems 295 Challenge Problems 296 Conceptual Problems 297

292

9.1 9.2 9.3 9.4

Chemical Bonding I: Lewis Theory

Bonding Models and AIDS Drugs Types of Chemical Bonds Representing Valence Electrons with Dots Lewis Structures: An Introduction to Ionic and Covalent Bonding Drawing Lewis Structures for Molecular Compounds 338 Writing Lewis Structures for Polyatomic Ions 341 Ionic Bonding and Electron Transfer 342

334 335 335 337 338

Xi

CONTENTS

9.5 The Ionic Bonding Model

343

The Born-Haber Cycle 344 Trends in Lattice Energies: Ion Size 345 Trends in Lattice Energies: Ion Charge 346 Ionic Bonding: Models and Reality 347

9.6 Covalent Bond Energies, Lengths, and Vibrations Bond Energy 348 Using Average Bond Energies to Estimate Enthalpy Changes for Reactions 349 Bond Lengths 351 Bond Vibrations 352

348

9.7 Electronegativity and Bond Polarity Electronegativity 355 Bond Polarity, Dipole Moment, and Percent Ionic Character 356 9.8 Resonance and Formal Charge Resonance 359 Formal Charge 361 9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd-Electron Species and Incomplete Octets Odd-Electron Species 364 Incomplete Octets 364 CHEMISTRY IN THE ENVIRONMENT: Free Radicals

354

and the Atmospheric Vacuum Cleaner

9.10 Lewis Structures for Hypercoordinate Compounds CHEMISTRY IN THE ENVIRONMENT: The Lewis Structure of Ozone CHAPTER IN REVIEW

10

10.1 Artificial Sweeteners: Fooled by Molecular Shape 10.2 VSEPR Theory: The Five Basic Shapes Two Electron Groups: Linear Geometry 379

365 366 369 370

372

377 378 378

383

Four Electron Groups with Lone Pairs 383 Five Electron Groups with Lone Pairs 384 Six Electron Groups with Lone Pairs 385

10.4 VSEPR Theory: Predicting Molecular Geometries Predicting the Shapes of Larger Molecules 389 10.5 Molecular Shape and Polarity CHEMISTRY IN YOUR DAY: How Soap Works 10.6 Valence Bond Theory: Orbital Overlap as a Chemical Bond

Key Terms 424 Key Concepts 424 Relationships 425 Key Skills 425

364

Three Electron Groups: Trigonal Planar Geometry 379 Four Electron Groups: Tetrahedral Geometry 380 Five Electron Groups: Trigonal Bipyramidal Geometry 381 Six Electron Groups: Octahedral Geometry 382

10.3 VSEPR Theory: The Effect of Lone Pairs

403

407

424

CHAPTER IN REVIEW

Review Questions 372 Problems by Topic 372 Cumulative Problems 374 Chal lenge Problems 375 Conceptual Problems 376

Chemical Bonding II: Molecular Shapes, Valence Bond Theory, and Molecular Orbital Theory

396

Linear Combination of Atomic Orbitals (LCAO) 408 Period 2 Homonuclear Diatomic Molecules 411 Period 2 Heteronuclear Diatomic Molecules 417 Polyatomic Molecules 418 Larger Conjugated Pi Systems 419 An Extens ion of MO Theory: Band Theory of Solids 421

359

Key Terms 370 Key Concepts 370 Key Equations and Relationships 371 Key Skills 371 EXERCISES

10.7 Valence Bond Theory: Hybridization of Atomic Orbitals sp3 Hybridization 397 sp2 Hybridization and Double Bonds 398 CHEMISTRY IN YOUR DAY: The Chemistry of Vision sp Hybridization and Triple Bonds 403 Writing Hybridization and Bonding Schemes 405 10.8 Molecular Orbital Theory: Electron Delocalization

387 390 393 393

Key Equations and

EXERCISES Review Questions 425 Problems by Topic 426 Cumulative Problems 429 Challenge Problems 431 Conceptual Problems 432

11

Liquids, Solids, and Intermolecular Forces

11.1 Climbing Geckos and Intermolecular Forces 11.2 Solids, Liquids, and Gases: A Molecular Comparison Changes Between States 436 11 .3 Intermolecular Forces: The Forces That Hold Condensed States Together Ion-Induced Dipole Force 438 Dispersion Force 438 Dipole-Dipole Force 440 Hydrogen Bonding 443 Dipole-Induced Dipole Force 445 Ion- Dipole Force 445 CHEMISTRY AND MEDICINE: Hydrogen Bonding in DNA

11.4 Intermolecular Forces in Action: Surface Tension, Viscosity, and Capillary Action Surface Tension 448 Viscosity 449 Capillary Action 450 CHEMISTRY IN YOUR DAY: Viscosity and Motor Oil 11 .5 Vaporization and Vapour Pressure The Process of Vaporization 451 The Energetics of Vaporization 452 Vapour Pressure and Dynamic Equilibrium 454 The Critical Point: The Transition to an Unusual State of Matter 460 11.6 Sublimation and Fusion Sublimation 461 Fusion 462 Energetics of Melting and Freezing 462 11 .7 Heating Curve for Water 11.8 Phase Diagrams The Major Features of a Phase Diagram 465 Navigation Within a Phase Diagram 466 The Phase Diagrams of Other Substances 467

425

433 434 434

437

447 448

450 451

461

463 465

Xii

CONTENTS

11.9 Water: An Extraordinary Substance

468

CHEMISTRY IN THE ENVIRONMENT: Water Pollution 11.10 Crystalline Solids: Determining Their Structure by X-Ray Crystallography 11.11 Crystalline Solids: Unit Cells and Basic Structures Closest-Packed Structures 475 11.12 Crystalline Solids: The Fundamental Types Molecular Solids 478 Ionic Solids 478 Atomic Solids 480

469

CHAPTER IN REVIEW Key Terms 482 Key Concepts 482 Relationships 483 Key Skills 483

470 472

478

482 Key Equations and

EXERCISES Review Questions 483 Problems by Topic 484 Cumulative Problems 488 Challenge Problems 490 Conceptual Problems 490

483

527

12.8 Colloids CHAPTER IN REVIEW Key Terms 530 Key Concepts 530 Relationships 531 Key Skills 532

530 Key Equations and

EXERCISES Review Questions 532 Problems by Topic 533 Cumulative Problems 536 Challenge Problems 537 Conceptual Problems 538

13

Chemical Kinetics

539

13.1 Hibernating Frogs 13.2 The Rate of a Chemical Reaction

540 540

Measuring Reaction Rates 544

13.3 The Rate Law: The Effect of Concentration on Reaction Rate

12

Solutions

12.1 Thirsty Solutions: Why You Shouldn't Drink Seawater 12.2 Types of Solutions and Solubility

491 491 493

CHEMISTRY IN YOUR DAY: Vitamin D in Foods

12.3 Energetics of Solution Formation

497

498

Aqueous Solutions and Heats of Hydration 500

12.4 Solution Equilibrium and Factors Affecting Solubility

501

The Temperature Dependence of the Solubility of Solids 502 Factors Affecting the Solubility of Gases in Water 503

12.5 Expressing Solution Concentration

506 506

and Personal Care Products

51 o

12.6 Colligative Properties: Vapour Pressure Lowering, 513

of Solutions Containing a Volatile (Nonelectrolyte) Solute 516 Freezing Point Depression and Boiling Point Elevation 519

522

12. 7 Colligative Properties of Strong Electrolyte Solutions Strong Electrolytes and Vapour Pressure 525 Colligative Properties and Medical Solutions 526

13.4 The Integrated Rate Law: The Dependence of Concentration on Time

524

549

Half-Life, Lifetime, and Decay Time 554

557

Arrhenius Plots: Experimental Measurements of the Frequency Factor and the Activation Energy 560 The Collision Model: A Closer Look at the Frequency Factor 562

13.6 Reaction Mechanisms

564

Rate Laws for Elementary Steps 565 Rate-Determining Steps and Overall Reaction Rate Laws 565 The Steady-State Approximation 568

571

Homogeneous and Heterogeneous Catalysis 572 Enzymes: Biological Catalysts 574

CHEMISTRY AND MEDICINE: Enzyme Catalysis and the Role of Chymotrypsin in Digestion

577

CHAPTER IN REVIEW Key Terms 578 Key Concepts 578 Key Equations and Relationships 579 Key Skills 579

578

EXERCISES Review Questions 580 Problems by Topic 580 Cumulative Problems 586 Challenge Problems 590 Conceptual Problems 591

580

14

CHEMISTRY IN THE ENVIRONMENT: Antifreeze in Frogs Osmotic Pressure 522

545 Reaction

13.7 Catalysis

CHEMISTRY IN THE ENVIRONMENT: Lake Nyos Molarity 507 Molality 508 Parts by Mass and Parts by Volume 508 Mole Fraction and Mole Percent 510 CHEMISTRY IN THE ENVIRONMENT: Pharmaceuticals

Freezing Point Depression, Boiling Point Elevation, and Osmotic Pressure Vapour Pressure Lowering 513 Vapour Pressures

Determining the Order of a Reaction 546 Order for Multiple Reactants 548

13.5 The Effect of Temperature on Reaction Rate

Nature's Tendency Toward Mixing: Entropy 493 The Effect of Intermolecular Forces 494

and Supplements

532

Chemical Equilibrium

14.1 Fetal Hemoglobin and Equilibrium 14.2 The Concept of Dynamic Equilibrium 14.3 The Expression for the Equilibrium Constant Relating Kp and Kc 596 The Unitless Thermodynamic Equilibrium Constant 598 Heterogeneous Equilibria: Reactions Involving Solids and Liquids 599

592 593 594 596

Xiii

CONTENTS

601

14.4 The Equilibrium Constant (K) Units of Equilibrium Constants 601 The Significance of the Equilibrium Constant 601 Relationships Between the Equilibrium Constant and the Chemical Equation 602

15.8 The Acid-Base Properties of Ions and Salts

CHEMISTRY AND MEDICINE: Life and Equilibrium 14.5 Calculating the Equilibrium Constant from Measured Quantities 14.6 The Reaction Quotient: Predicting the Direction of Change 14.7 Finding Equilibrium Concentrations

603

605 608 610

Finding Partial Pressures or Concentrations at Equilibrium Amounts When We Know the Equilibrium Constant and All But One of the Equilibrium Amounts of the Reactants and Products 610 Finding Equilibrium Concentrations When We Know the Equilibrium Constant and Initial Concentrations or Pressures 611 Simplifying Approximations in Working Equilibrium Problems 615

626

628

636

15.1 Heartburn 15.2 The Nature of Acids and Bases 15.3 Definitions of Acids and Bases

636 637 639

Strong Bases 645

642

CHAPTER IN REVIEW Key Terms 682 Key Concepts 682 Relationships 683 Key Skills 683

680 682

Key Equations and

16

Aqueous Ionic Equilibrium

684

691

16.1 The Danger of Antifreeze 16.2 Buffers: Solutions That Resist pH Change

16.3 Buffer Effectiveness: Buffer Range and Buffer Capacity

692 693

703

Relative Amounts of Acid and Base 703 Absolute Concentrations of the Acid and Conj ugate Base 703 Buffer Range 704

in Human Blood Buffer Capacity 706

16.4 Titrations and pH Curves

The Acid

645 Weak Bases 645

647

The pH Scale: A Way to Quantify Acidity and Basicity 649 pOH and Other p Scales 650

CHEMISTRY AND MEDICINE: Ulcers 15.7 Finding [H 30-t'j, [OW], and pH of Acid or Base Solutions

Amine

Oxyacids 678

CHEMISTRY AND MEDICINE: Buffer Effectiveness

The Brl!)nsted- Lowry

15.6 Autoionization of Water and pH

Cations

Calculating the pH of a Buffer Solution 694 The Henderson- Hasselbalch Equation 695 Calculating pH Changes in a Buffer Solution 698 Buffers Containing a Base and Its Conjugate Acid 701

Acids and Bases

15.5 Base Solutions

674

675

EXERCISES Review Questions 684 Problems by Topic 685 Cumulative Problems 688 Challenge Problems 690 Conceptual Problems 690

Key Equations and

Strong Acids 642 Weak Acids 643 Ionization Constant (Ka) 644

CHEMISTRY IN YOUR DAY: Weak Acids in Wine 15.10 Lewis Acids and Bases

15.12 Ocean Acidification 619

15.4 Acid Strength and the Acid Ionization Constant (Ka)

Finding the pH of Polyprotic Acid Solutions 671 Finding the Concentration of the Anions for a Weak D iprotic Acid Solution 673

Binary Acids 677 Bases 680

EXERCISES Review Questions 628 Problems by Topic 629 Cumulative Problems 633 Challenge Problems 634 Conceptual Problems 635

The Arrhenius Definition 639 Defi nition 640

670

15.9 Polyprotic Acids

15.11 Strengths of Acids and Bases and Molecular Structure 677

The Effect of Changing the Amount of Reactant or Product on Equilibrium 619 The Effect of a Volume Change on Equi librium 622 The Effect of Changing the Pressure by Adding an Inert Gas 623 The Effect of a Temperature Change on Equilibrium 623

15

662

Anions as Weak Bases 663 Cations as Weak Acids 666 Classifying Salt Solutions as Acidic, Basic, or Neutral 667

Molecules That Act as Lewis Acids 676 That Act as Lewis Acids 677

14.8 Le Chatelier's Principle: How a System at Equilibrium Responds to Disturbances

CHAPTER IN REVIEW Key Terms 626 Key Concepts 626 Relationships 627 Key Skills 627

Strong Acids 652 Weak Acids 652 Percent Ionization of a Weak Acid 657 Mixtures of Acids 658 Finding the [OHJ and pH of Basic Solutions 660

652

706

The Titration of a Strong Acid with a Strong Base 706 The Titration of a Weak Acid with a Strong Base 711 The Titration of a Weak Base with a Strong Acid 716 The Titration of a Polyprotic Acid 717 Indicators: pH-Dependent Colours 717

16.5 Solubility Equilibria and the Solubility Product Constant 651

705

K,P and Molar Solubility 720 K,Pand Relative Solubility 723 The Effect of a Common Ion on Solubility 723

720

xiv

CONTENTS

CHEMISTRY AND MEDICINE: Fluoride and Teeth

724

The Effect of an Uncommon Ion on Solubility (Salt Effect) 725 The Effect of pH on Solubility 726

16.6 Precipitation

727

Selective Precipitation 729

16.7 Qualitative Chemical Analysis

730

Group A: Insoluble Chlorides 731 Group B: AcidInsoluble Sulfides 732 Group C: Base-Insoluble Sulfides and Hydroxides 732 Group D: Insoluble Phosphates 732 Group E: Alkali Metals and NH4 + 732

16.8 Complex-Ion Equilibria

733

The Effect of Complex-Ion Equilibria on Solubility 735

CHAPTER IN REVIEW Key Terms 736 Key Concepts 736 Key Equations and Relationships 737 Key Skills 737

736

EXERCISES Review Questions 738 Problems by Topic 738 Cumulative Problems 743 Challenge Problems 744 Conceptual Problems 745

738

17

Gibbs Energy and Thermodynamics

17.1 Spontaneous and Non spontaneous Processes 17.2 Entropy and the Second Law ofThermodynamics

746 747 748

~w~~~~ Calculating Gibbs Energy Changes with Ll,G° = Ll,H° - T Ll,.S" 769 Calculating ll,G0 with Tabulated Values of Gibbs Energies of Formation 771 Calcu lating ll,G0 for a Stepwise Reaction from the Changes in Gibbs Energy for Each of the Steps 772

17.8 Making a Nonspontaneous Process Spontaneous 17.9 What Is Gibbs Energy? 17.10 Gibbs Energy Changes for Nonstandard States: The Relationship Between t1r~ and t1rG

18

Electrochemistry

795

18.1 Pulling the Plug on the Power Grid 18.2 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions

796 796

18.3 Standard Electrode Potentials

799

808

The Relationship Between ll,G0 and E~ell 809 The Relationship Between E~011 and K 811

18.5 Cell Potential and Concentration

812

Concentration Cells 816

CHEMISTRY AND MEDICINE: Concentration Cells in Human Nerve Cells

755

819

819

Dry-Cell Batteries 819 Lead-Acid Storage Batteries 820 Other Rechargeable Batteries 820 Fuel Cells 822

CHEMISTRY IN YOUR DAY: Rechargeable Battery 759 760

Recycling

823

18.7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity

823

Predicting the Products of Electrolysis 826 Stoichiometry of Electrolysis 829

765

18.8 Corrosion: Undesirable Redox Reactions

831

Preventing Corrosion 833

™

CHAPTER IN REVIEW

833

Key Terms 833 Key Concepts 833 Key Equations and Relationships 834 Key Skills 835 EXERCISES Review Questions 835 Problems by Topic 836 Cumulative Problems 839 Challenge Problems 841 Conceptual Problems 841

835

773 776 777

The Gibbs Energy Change of a Reaction Under Nonstandard Conditions 777

17.11 Gibbs Energy and Equilibrium: Relating t1r~ to the Equilibrium Constant (K)

788

18.6 Batteries: Using Chemistry to Generate Electricity

The Effect of Ll,H, Ll,S, and Ton Spontaneity 766

17.7 Gibbs Energy Changes in Chemical Reactions:

EXERCISES Review Questions 788 Problems by Topic 788 Cumulative Problems 791 Challenge Problems 793 Conceptual Problems 794

18.4 Cell Potential, Gibbs Energy, and the Equilibrium Constant

Standard Molar Entropies (S°) and the Third Law of Thermodynamics 760

17.6 Gibbs Energy

785

Predicting the Spontaneous Direction of an OxidationReduction Reaction 805 Predicting Whether a Metal Will Di ssolve in Acid 808

The Temperature Dependence of llSsurr 757 Quantifying Entropy Changes in the Surroundings 757

17.4 Entropy Changes for Phase Transitions 17.5 Entropy Changes in Chemical Reactions: Calculating t1rS'

CHAPTER IN REVIEW Key Terms 785 Key Concepts 786 Key Equations and Relationships 786 Key Skills 787

Electrochemical Cell Notation 799

Entropy 750 The Entropy Change Associated with a Change in State 754

17.3 Heat Transfer and Changes in the Entropy of the Surroundings

The Temperature Dependence of the Equilibrium Constant 783

781

19

Radioactivity and Nuclear Chemistry

19.1 Medical Isotopes 19.2 The Discovery of Radioactivity 19.3 Types of Radioactivity

842 842 843 844

CON TENTS

20.5 Constitutional Isomerism 20.6 Stereoisomerism I: Conformational Isomerism

Alpha (a) Decay 845 Beta (/3) Decay 846 Gamma (y) Ray Emission 847 Positron Emission 847 Electron Capture 847

19.4 The Valley of Stability: Predicting the Type of Radioactivity Magic Numbers 851 Series 851

19.5 Measurements and Units of Radioactivity 19.6 The Kinetics of Radioactive Decay and Radiometric Dating

20.7 Stereoisomerism II: Configurational Isomerism

CHEMISTRY AND MEDICINE: Anesthetics and Alcohol 20.8 Structure Determination

The Integrated Rate Law 853 Radiocarbon Dating: Using Radioactivity to Measure the Age of Fossils and Artifacts 856 Uranium/Lead Dating 857

19.7 The Discovery of Fission: The Atomic Bomb and Nuclear Power

CHAPTER IN REVIEW Key Terms 911 Key Concepts 911 Key Equations and Relationships 912 Key Skills 913

911

913

862

EXERCISES Review Questions 913 Problems by Topic 914 Cumulative Problems 916 Challenge Problems 91 7 Conceptual Problems 918

862

Nuclear Binding Energy 863

19.9 Nuclear Fusion: The Power of the Sun 19.1 O Nuclear Transmutation and Transuranium Elements 19.11 The Effects of Radiation on Life

865 866 868

Acute Radiation Damage 868 Increased Cancer Risk 868 Genetic Defects 868 Measuring Radiation Exposure 868

19.12 Radioactivity in Medicine and Other Applications

870

Diagnosis in Medicine 870 Radiotherapy in Medicine 871 Other Applications 871 CHAPTER IN REVIEW Key Terms 872 Key Concepts 872 Key Equations and Relationships 873 Key Skills 873

872

EXERCISES Review Questions 874 Problems by Topic 874 Cumulative Problems 876 Challenge Problems 877 Conceptual Problems 877

874

20

Perceived Difference Between Organic and Inorganic

20.3 Hydrocarbons

Halides 888 Amines 888 Alcohols 889 Ethers 890 Carbonyls: Aldehydes and Ketones 890 The Carboxylic Acid Family 891

Organic Chemistry II: Reactions 920

21 .1 Discovering New Drugs 21.2 Organic Acids and Bases

921 921

The Range of Organic Acidities 921 Inductive Effects: Withdrawal of Electron Density 923 Resonance Effects: Charge Delocalization in the Conjugate Base 924 Acidic Hydrogen Atoms Bonded to Carbon 925 Mechani sms in Organic Chemistry 925 Acid and Base Reagents 926

928

Redox Reactions 929

CHEMISTRY IN YOUR DAY: Hydrogen and the Oil Sands

931

21.4 Nucleophilic Substitution Reactions at Saturated Carbon

879 879

880

881

Drawing Hydrocarbon Structures 881 Types of Hydrocarbons 884 Alkanes 884 Alkenes 885 Alkynes 885 Conjugated Alkenes and Aromatics 886

20.4 Functional Groups

21

21.3 Oxidation and Reduction

Organic Chemistry I: Structures 878

20.1 Fragrances and Odours 20.2 Carbon: Why It Is Unique THE NATURE OF SCIENCE: Vitalism and the

905

859

CHEMISTRY IN YOUR DAY: Uranium Isotopes and the CANDU Reactor

905

Using the Molecular Formula: The Index of Hydrogen Deficiency 905 Spectroscopic Methods for Structure Determination 908

Nuclear Power: Using Fission to Generate Electricity 860

19.8 Converting Mass to Energy in Nuclear Reactions and Nuclear Binding Energy

897

Cis-Trans Isomerism in Alkenes 898 Enantiomers: Chirality 901 Absolute Configurations 903

851 852

893 894

Conformational Isomerism: Rotation About Single Bonds 894 Ring Conformations of Cycloalkanes 896

849

Radioactive Decay

XV

938

21 .5 Elimination Reactions The E 1 Mechanism 939 The E2 Mechanism 940 Elimination Versus Substitution 940

21.6 Electrophilic Additions to Alkenes Hydrohalogenation 940 Reactions 941

940

Other Addition

21.7 Nucleophilic Additions to Aldehydes and Ketones Addition of Alcohols 942 Reaction 943

887

932

The SN 1 Mechanism 932 The SN2 Mechanism 934 Factors Affecting Nucleophilic Substitution Reactions 935

The Grignard

21.8 Nucleophilic Substitutions of Acyl Compounds 21.9 Electrophilic Aromatic Substitutions 21 .10 Polymerization Step-Growth Polymers 951

942

Addition Polymers 951

945 949 950

xvi

CONTENTS

CHEMISTRY IN YOUR DAY: High-, Medium-, and

23.4 Boron and Its Remarkable Structures

Low-Density Polyethylene CHAPTER IN REVIEW Key Terms 953 Key Concepts 954 Relationships 955 Key Skills 956

952 953 Key Equations and

EXERCISES Review Questions 956 Problems by Topic 957 Cumulative Problems 963 Challenge Problems 964 Conceptual Problems 965

22

956

1009

23.6 Nitrogen and Phosphorus: Essential Elements for Life 1014 Elemental Nitrogen and Phosphorus 1014 Nitrogen Compounds 1016 Phosphorus Compounds 1019

966

Biochemistry Fatty Acids 967 Lipids 970

Fats and Oils 969

966 967

Other

CHEMISTRY AND MEDICINE: Dietary Fat: The Good, 971

22.3 Carbohydrates

972

Simple Carbohydrates: Monosaccharides and Disaccharides 973 Complex Carbohydrates 976

22.4 Proteins and Amino Acids

977

Amino Acids: The Building Blocks of Proteins 978 Peptide Bonding Between Amino Acids 980

22.5 Protein Structure

981

Primary Structure 982

CHEMISTRY AND MEDICINE: The Essential Amino Acids

983

Secondary Structure 984 Tertiary Structure 984 Quaternary Structure 984

22.6 Nucleic Acids: Blueprints for Proteins

985

The Basic Structure of Nucleic Acids 986 Genetic Code 988

The

DNA Replication, the Double Helix, and Protein Synthesis

989

DNA Replication and the Double Helix 989 Protein Synthesis 991

CHAPTER IN REVIEW Key Terms 992 Key Concepts 992

992 Key Skills 993

EXERCISES Review Questions 993 Problems by Topic 994 Cumulative Problems 997 Challenge Problems 998 Conceptual Problems 999

Chemistry of the Nonmetals

23.7 Oxygen

1021

Elemental Oxygen 1021 Uses for Oxygen 1022 Oxides 1022 Ozone 1022

23.8 Sulfur: A Dangerous but Useful Element

1023

Elemental Sulfur 1023 Hydrogen Sulfide and Metal Sulfides 1024 Sulfur Dioxide 1025 Sulfuric Acid 1025

the Bad, and the Ugly

23

23.5 Carbon, Carbides, and Carbonates Carbon 1009 Carbides 1012 Carbon Oxides 1013 Carbonates 1013

22.1 Diabetes and the Synthesis of Human Insulin 22.2 Lipids

22.7

1006

Elemental Boron 1006 Boron- Halogen Compounds: Trihalides 1007 Boron-Oxygen Compounds 1008 Boron- Hydrogen Compounds: Boranes 1008

993

23.9 Halogens: Reactive Elements with High Electronegativity Elemental Fluorine and Hydrofluoric Acid 1027 Elemental Chlorine 1028 Oxides 1028

1026 Halogen

CHAPTER IN REVIEW Key Terms 1029 Key Concepts 1029 Key Skills 1030

1029

EXERCISES Review Questions 1030 Problems by Topic 1030 Cumulative Problems 1032 Challenge Problems 1033 Conceptual Problems 1033

1030

24

Metals and Metallurgy

24.1 Vanadium: A Problem and an Opportunity 24.2 The General Properties and Natural Distribution of Metals 24.3 Metallurgical Processes

1034 1035 1035 1037

Separation 1037 Pyrometallurgy 1038 Hydrometallurgy 1038 Electrometallurgy 1039 Powder Metallurgy 1040

24.4 Metal Structures and Alloys

1040

Alloys 1041

24.5 Sources, Properties, and Products of Some of the 3dTransition Metals

1000

23.1 Insulated Nanowires 23.2 The Main-Group Elements: Bonding and Properties

1001 1001

CHAPTER IN REVIEW Key Terms 1051 Key Concepts 1051 Key Skills 1051

1002

EXERCISES 1052 Review Questions 1052 Problems by Topic 1052 Cumulative Problems 1053 Challenge Problems 1054 Conceptual Problems 1054

Atomic Size and Types of Bonds 1001

23.3 Silicates: The Most Abundant Matter in Earth's Crust Quartz and Glass 1002 Aluminosilicates 1003 Individual Silicate Units, Silicate Chains, and Silicate Sheets 1003

1046

Titanium 1046 Chromium 1047 Manganese 1048 Cobalt 1048 Copper 1049 Nickel 1050 Zinc 1050

1051

CONTENTS

25

Transition Metals and Coordination Compounds

1055

25.1 The Colours of Rubies and Emeralds 25.2 Electron Configurations of Transition Metals

1056 1056

Electron Configurations 1057 States 1057

Oxidation

25.3 Coordination Compounds

1058

Naming Coordination Compounds 1061

25.4 Structure and lsomerization

1063

Structural Isomerism 1063 Stereoisomerism 1063

25.5 Bonding in Coordination Compounds

1067

Ligand Field Theory 1067 Octahedral Complexes 1067 The Colour of Complex Ions and Ligand Field Strength 1068 Magnetic Properties 1070 Tetrahedral and Square Planar Complexes 1071

25.6 Applications of Coordination Compounds

1072

Chelating Agents 1072 Chemical Analysis 1072 Colouring Agents 1072 Biomolecules 1073

CHAPTER IN REVIEW 1075 Key Terms 1075 Key Concepts 1075 Key Equations and Relationships 1076 Key Skills 1076

xvii

EXERCISES 1076 Review Questions 1076 Problems by Topic 1077 Cumulative Problems 1078 Challenge Problems 1079 Conceptual Problems 1079

Appendix I: Common Mathematical Operations in Chemistry A Scientific Notation B Logarithms C Quadratic Equations D Graphs

A-1 A-1 A-3 A-4 A-5

Appendix II : Useful Data A Atomic Colours B Standard Thermodynamic Quantities for Selected Substances at 25°C C Aqueous Equilibrium Constants D Standard Electrode Potentials at 25°C E Vapour Pressure of Water at Various Temperatures

A-7 A-7 A-7 A-12 A-15 A-16

Appendix Ill: Answers to Selected Exercises

A-17

Appendix IV: Answers to In-Chapter Practice Problems

A-60

Glossary Index

G-1

1-1

Preface TO THE STUDENT As you begin this course, think about your reasons for enrolling in it. Why are you taking general chemistry? Why are you pursuing a university or college education at all? If you are like most students taking general chemistry, part of your answer is probably that thi s course is required for your major or you are pursuing your education so that you can get a job some day. Although these are both good reasons, we think there is a better one. The primary reason for an education is to prepare you to live a good life. You should understand chemistry-not for what it can get you-but for what it can do for you. Understanding chemistry is an important source of happiness and fulfillment. Understanding chemistry helps you to live life to its fullest for two basic reasons. The first is intrinsic: Through an understanding of chemistry, you gain a powerful appreciation for just how rich and extraordinary the world really is. For example, one of the most important ideas in science is that the behaviour of matter is determined by the properties of molecules and atoms. With this knowledge, we have been able to study the substances that compose the world around us and explain their behaviour by reference to particles so small that they can hardly be imagined. If you have never realized the remarkable sensitivity of the world we can see to the world we cannot, you have missed out on a fundamental truth about our universe. The second reason is extrinsic: Understanding chemistry makes you a more informed citizen-it allows you to engage with many of the issues of our day. Scientific literacy helps you understand and discuss in a meaningful way important issues from the development of the oil sands in Alberta (Chapter 6) to how the production of pharmaceuticals and personal care products affects our environment and our bodies (Chapter 12). In other words, understanding chemistry makes you a deeper and richer person and makes your country and the world a better place to live. These reasons have been the foundation of education from the very beginnings of civilization. So this is why we think you should take this course and why we wish you the best as you embark on the journey to understand the world around you at the molecular level. The rewards are well worth the effort.

The Strengths of Chemistry: A Molecular Approach Chemistry: A Molecular Approach is first and foremost a student-oriented book. The main goal of the book is to motivate students and get them to achieve at the highest possible level. As we all know, many students take general chemistry because it is a requirement; they do not see the connection between chemistry and their lives or their intended careers. Chemistry: A Molecular Approach strives to make those connections consistently and effectively. Unlike other books, which often teach chemistry as something that happens only in the laboratory or in industry, this book teaches chemistry in the context of relevance. It shows

xvi ii

students why chemistry is important to them, to their future careers, and to their world. Second, Chemistry: A Molecular Approach is a pedagogically driven book. In seeking to develop problem-solving skills, a consistent approach is applied (Sort, Strategize, Solve, and Check), usually in a two- or three-column format. In the twocolumn format, the left column shows the student how to analyze the problem and devise a solution strategy. It also lists the steps of the solution and explains the rationale for each one, while the right column shows the implementation of each step. In the three-column format, the left column outlines the general procedure for solving an important category of problems that is then applied to two side-by-side examples. This strategy allows students to see both the general pattern and the slightly different ways in which the procedure may be applied in differing contexts. The aim is to help students understand both the concept of the problem (through the formulation of an explicit conceptual plan for each problem) and the solution to the problem. Third, Chemistry: A Molecular Approach is a visual book. Wherever possible, images are used to deepen the student's insight into chemistry. In developing chemical principles, multipart images help to show the connection between everyday processes visible to the unaided eye and what atoms and molecules are actually doing. Many of these images have three parts: macroscopic, molecular, and symbolic. This combination helps students to see the relationships between the formulas they write down on paper (symbolic), the world they see around them (macroscopic), and the atoms and molecules that compose that world (molecular). In addition, most figures are designed to teach rather than just to illustrate. They include annotations and labels intended to help the student grasp the most important processes and the principles that underlie them. The resulting images are rich with information but also uncommonly clear and quickly understood. Fourth, Chemistry: A Molecular Approach is a "big picture" book. At the beginning of each chapter, a short paragraph helps students to see the key relationships between the different topics they are learning. A focused and concise narrative helps make the basic ideas of every chapter clear to the student. Interim summaries are provided at selected spots in the narrative, making it easier to grasp (and review) the main points of important discussions. And to make sure that students never lose sight of the forest for the trees, each chapter includes several Conceptual Connections, which ask them to think about concepts and solve problems without doing any math. The idea is for students to learn the concepts, not just plug numbers into equations to churn out the right answer. Finally, Chemistry: A Molecular Approach is a book that delivers the depth of coverage faculty want and students need. We do not have to cut corners and water down the material in order to get our students interested. We simply have to meet them where they are, challenge them to the highest level of achievement, and then support them with enough pedagogy to allow them to succeed.

The Canadian Edition Chemistry: A Molecular Approach, by Nivaldo J. Tro, is widely used in general chemistry courses at colleges and universities

PR EFACE

across North America. So, why do we need a Canadian edition? The short answer is that general chemistry courses in Canada are different from those in the United States. First-year chemistry curricula in Canada are generally at a higher level than what is seen south of the border. There is a need for a strong chemistry textbook that serves Canadian general chemistry courses. The Canadian adaptation of Chemistry: A Molecular Approach drew very heavily on feedback from professors and instructors across Canada. As the Canadian authors, we took the reviews and consultations very seriously and did our best to adapt Tro's textbook accordingly. In general terms, the adaptation involved making the following changes.

International Conventions on Units, Symbols, and Nomenclature The field of chemistry is communicated according to conventions that are determined by the broader international chemistry community, through the International Union of Pure and Applied Chemistry (IUPAC). IUPAC continually releases recommendations on chemical nomenclature, definitions, symbols, and units. IUPAC recommendations are not static; they may evolve over time as new information comes to light. Although many textbooks state that they follow the recommendations of the IUPAC, you will find that the Canadian edition of Chemistry: A Molecular Approach scrupulously follows IUPAC recommendations for chemical names and symbols, nomenclature, and conventions for symbols and units in measurements. In the case of chemical nomenclature, there are a number of non-IUPAC chemical names that are so common that we have to include them along with the IUPAC recommended name. S.I. units of measurement are used exclusively. Imperial units such as the gallon, pound, and the Fahrenheit scale of temperature have not been used in modern science for over a generation. IUPAC recommended defining standard pressure as 1 bar (or 100 kPa) back in 1982. This is the standard that has been adopted by chemists worldwide and is almost exclusive in second-year physical chemistry texts. Only in first-year textbooks does the atmosphere still linger as standard pressure. In this text, standard pressure is the IUPAC-recommended bar. Students will see pressure in various units, but we make little use of the atmosphere. When dealing with ideal gases, the most common value of R is 0.08314 L bar mo1- 1 K- 1• In thermodynamics, we have adopted the recommended notation for enthalpy, entropy, and Gibbs energy changes, placing subscripts for changes after the delta sign rather than after H, S, or G. For example, the standard reaction enthalpy is expressed as !:,,./!" rather than !:,,.H;xn· This is a subtle change that matters. The type of change(!:,,.) is marked on the!:,,. symbol (reaction, !:,,.,; formation, !:,,.c; and so on), rather than the type of thermodynamic quantity. We understand that this notation is not used everywhere. However, we believe that students should use standard notation throughout their education. Students who continue in chemistry or other sciences will eventually come across the standard notation in physical chemistry textbooks and in places like the CRC Handbook of Chemistry and Physics and the NIST Chemistry WebBook (http://webbook. nist.gov/). Furthermore, thermodynamic quantities like !:,,.,If' are always molar quantities and have the units kJ mo1- 1, as recommended by IUPAC. Exclusive use of IUPAC-recommended

xix

units keeps students from getting into unit troubles when doing thermodynamic calculations. Explicitly, we have provided the distinctions and connections between the unitless thermodynamic equilibrium constant, Keq or simply K, and the phenomenological equilibrium constants, Kc and Kp, which can have units in terms of concentration and pressure, respectively, again in accordance with IUPAC recommendations. This is done in the most basic of terms, assuming that gases and solutions are ideal so that their partial pressures and concentrations are assumed to be numerically equivalent to their activities, setting up for a more rigorous treatment in second-year analytical and physical chemistry courses. Following recommendations set out by the IUPAC ensures that we speak a common language-and teach a common language. Otherwise, students who go on in chemistry have to convert from the language learned in first year as soon as the very next year, when they take their first physical chemistry course.

Current Theories We have updated the text so that the most current, consensus scientific view is described. This is most notable in the case of bonding theory and the so-called expanded octet. In this case, evidence shows that the d orbitals have a negligible contribution to bonding, which means that full sp3d and sp3 d2 hybridizations should no longer be included in bonding theories, even though this idea continues to appear in general chemistry textbooks. This Canadian edition reflects the most current understanding of chemical phenomenon, at the first-year level. Organic Chemistry The coverage of organic chemistry has been expanded to two chapters, reflecting the curricula in many Canadian universities, which provide additional organic chemistry coverage in first-year chemistry. The first organic chemistry chapter covers structure and bonding, stereochemistry, and structure determination. The second chapter covers organic reactivity, and it is organized according to reaction mechanisms. Canadian Context Naturally, a Canadian edition will include Canadian examples. In some places, the Canadian content is fun, like the hockey goalie's "Quantum mechanical five hole" in Chapter 7. In other places, Canadian chemistry examples are serious and important, like the chemistry of the oil sands. Wherever Canadian content appears in this edition, it is there to promote student engagement. This book is meant for the Canadian student. End-of-Chapter Problems One of the first things that professors consider when choosing a chemistry textbook is the quality of end-of-chapter problems. This is because, to learn chemistry, students need to work through meaningful exercises and problems. Tro's Chemistry: A Molecular Approach has extensive, high-quality problems. First-year chemistry courses are perhaps the most important courses in chemistry programs, because they lay the foundation for all higher level courses. First-year courses introduce students to the language and discipline of chemistry, and some concepts are not touched on again in the entire undergraduate curriculum. Indeed, many Ph.D. comprehensive questions fall back to ideas learned in first year. This book was prepared with

XX

PREFACE

the fu ll undergraduate curriculum in mind. If you are a student, we hope that the Canadian edition of Chemistry: A Molecular Approach helps you succeed in chemistry. We encourage you to make use of all of the features in this book that are designed to help you learn. If you are a professor, it is our hope that this textbook provides you with the strong content you need to teach first-year chemistry in a way that is true to our discipline.

"Fluoride and Teeth" (Chapter 16), and "Rechargeable Battery Recycling" (Chapter 18). We also added many new end-of chapter problems throughout the book, which gives instructors and students more opportunities to engage with chemistry content and practise problem solving.

Third Canadian Edition

During the development of this book, we obtained many helpful suggestions and comments from colleagues from across the country.

For the third Canadian edition, we had two primary goals. Our first goal was to make focused improvements and write additional content in selected areas. Some of these are described below. In Chapter 7, we have clarified the language and added a brief discussion of what is meant by orbital energies. We improved the discussion of electron configurations of transition metals-a topic that many students find confusing. We also added a whole new section showing the application of the Schrodinger equation to a quantum mechanical system-"the particle in a one-dimensional box." Our aim is to demystify wave functions and quantum numbers. We do this by showing that wave functions are nothing more than mathematical equations representing electrons in an atom. Furthermore, applying the Schrodinger equation to a quantum mechanical system with boundary conditions (i.e., a particle in a box or an electron in an atom) gives rise to quantum numbers. In Chapter 9, we added a brief discussion of homol ytic versus heterolytic bond dissociations. In Chapter 10, we expanded coverage of p-n junctions in diodes and show how these are applied in light emitting diodes (LEDs) and photovoltaic cells. We also address the issue of hybridization of terminal atoms in bonding descriptions. From a shape and structure point of view, when a molecule has a terminal atom with lone pairs of electrons, it is not necessary to assign hybrid orbitals to those lone pairs. However, hybridization of terminal atoms is commonly taught, especially in organic chemistry courses, where reactions result in a bond to the terminal atom. Our continued priority is to show how chemists use different bonding models for different purposes. As well, we added stick-Like drawings to show the shapes of molecules--drawings that students can mimic- along with artistic three-dimensional renderings that students will not be able to reproduce easily. Worked examples are one of the most important and wellused features in this textbook. To continue this strength, we have added some new worked examples, for example on reaction mechanisms in Chapter 13. Finally, we reorganized Chapter 17 slightly by moving the discussion of the third Law of thermodynamics earlier in the chapter with the rest of the quantitative discussion of entropy. We also introduced a new section, including worked examples, on making nonspontaneous processes spontaneous by coupling with exergonic reactions. Our second goal was to update and "evergreen" the book. To do this, we replaced or updated "Chemistry in Your Day" boxes to make them more interesting and relevant to students and thereby enhance learning. New boxes include "Stack Sampling" (Chapter 5), "Weak Acids in Wine" (Chapter 15),

ACKNOWLEDGEMENTS

Editorial Advisory Board We would like to thank the following professors for contributing their valuable time to meet with the author and the publishing team over the course of this project to provide a meaningful perspective on the most important challenges they face in teaching general chemistry and for providing insight and feedback on the third Canadian edition and MasteringChemistry course: Louise Dawe, Wilfrid Laurier University Phil Dutton, University of Windsor Andrew Vreugdenhil, Trent University James Xidos, University of Manitoba

We also thank the following reviewers who offered feedback on the previous edition of this textbook: Francois Caron, Laurentian University Linda Davis, McMaster University Violeta Iosub, University of Calgary Lori Jones, University of Guelph Stephen Kariuki, Nipissing University Jan Matejovic, University of Ontario Institute of Technology Andrew Mosi, Langara College Brad Pavelich, Medicine Hat College Jason Pearson, University of Prince Edward Island

We acknowledge Prof. Dietmar Kennepohl (Athabasca University) and Dr. Nicole Sandblom (University of Calgary), Dr. Neil Anderson (Onyx Pharmaceuticals), Dr. Rebecca Goyan (Simon Fraser University), Drs. Barry Power, Chris Flinn, Karen Hattenhauer, Mike Katz, Ray Poirier, Peter Warburton, and Chris Kozak (Memorial University), Mr. Nicholas Ryan (Memorial University), and Drs. Lucio Gelmini and Robert Hilts (MacEwan University) for helpful discussions and their insightful comments. Dr. Ian Hunt of the University of Calgary worked with us in the early development of the organic chemistry chapters. He provided sage advice on the organization of these chapters and made numerous suggestions on how to present organic chemistry in a way that is both rigorous and accessible to the first-year student. Professor Franitois Caron of Laurentian University provided expert advice on revisions to Chapter 19, improving the presentation of nuclear reaction energetics so that it is consistent with the field of nuclear chemistry.

PREFACE

We would like to thank our wives, Lisa and Tanya, for their encouragement and their continuing patience during all the evenings and weekends we spent working on this book when we could have been with our families. Finally, we would also like to acknowledge the assistance of the many members of the team at Pearson Canada who were involved throughout the writing and production process: Cathleen Sullivan, Executive Acquisitions Editor; Kim Teska,

xxi

Senior Marketing Manager; Kamilah Reid-Burrell, Content Manager; Martina van de Velde, Content Developer; Jessica Misfud, Project Manager; Leanne Rancourt, Production Editor and Proofreader; Susan Broadhurst, Copyeditor; and Anthony Leung, Senior Designer. Travis D. Fridgen Lawton E. Shaw

C

hemistry is relevant to every process occurring around you, at every second. The authors help you understand this connection by weaving specific, vivid examples throughout the text that tell the story of chemistry. Every chapter begins with a brief story that illustrates how chemistry is relevant to all people, at every moment.

you interested in knowing how nerve cells transmit signals?

~ -· Are

See Chapter 8 to learn why periodic properties are essential to understanding this process.

What about the chemistry ···• of everyday life? Chapter 4 illustrates the role chemistry plays in cuisine, from baking a cake to why lemons go well with fish.

_____ __ ___ _ ____ ____ ____ ----·---·----_..,_ __ ________ ___., .. .. 1==~:~~=-

..,..

.. ---.. ----·--..

..,

______ ________ __ ...__ __ -- c :=:=:·~:==~~ __ ... -...__ ___ .. ----··------ii--_...__ _____ ___._,_.,_ .......... .--"'----......._ _____ ........ _,_ ______ _____ _ ... -____ _______ ...._ ----------··__...-_ -________ ______ __ ...... _.._ --___ ___ -- --.--------~----

....,,_,._ _,,

....."_,.....

,..

.......

....,. ...

... -....... _._ M=:= -·--___ .... ..:::=.:::::=:.: __ ,..____ ..... _ ·----------.,-______ ___ .,..__.._ W -I-·__.,___ ______ ._

_.

--·------·-"'

........

...

.,..,... ,...

,, .....

w -----~ ---

1U ~--

""

" ,__..,_ ~-­ M

... '... ' ._

-..

xxii

__

.....

...

-."' __..,. ......,...,,,,, ., _,._.._, ..... ., , __,_...,

_,,, _____ ...,______ Al_., .._.,._..,...... ..._,,, .....

1 :=":=:.::.=~=~ --··---------------i.....--------· .._____ .. ..............--___ ........... ...... ....,-___ ---------......

_ . . . _ .. v..... _,._... _,.-. .. ,,_...,.,,,_

..

,__

........... .................

--.....-.-

,...

- ------·--.. -·-"'-----.... .....--....-.-

These examples make the material more accessible by contextualizing the chemistry and grounding it in the world you live in.

Student Interest

Bone Density Ostcoporosis-tAhich means fHJTUus OOn~-is a rond111on m v.hich bone density becomes too I™. 1be health)' bones of a )OUng adult ha\-c: a density of about 1.0 g cm ,. Pa11enlS suffering from os1eoporosis. ~e\'ct, can ha\e bone dcnsiti~ as low as 0.22 g cm l. These low dcns111c:s mean the bones ha\e dc:1erior.ned and v.eakcncd. rcsuhmg m nicrea..~ SUSttpl1b1hty to fractures, c~pecially hip fracture~. Patients suffering from osteoporosis can also e'perience height loss and d1sfiguration soch as do\\11ger's hump, a cood111on m v.hich the patient bec~s hunched O\erduc: to COllll'f'CSSion of the \·enebrae. Ostcoporlic d11T~ncti in the X-ray imai,.>C:. Treatments for osteoporosis include add1t!Oflal cak1um and 'Ila.mm D. drugs 1h31 pre...c:n1 bone wealenmg, excrci'IC and qrcngth traming. and. in ex1rcme cac;es, hip-replacement ~urgcry.

Question Suppose yoo find a large amnul bone 1n the v.oods. 100 large to fit in a beaker or flasl. How might )OU :ippnmmatc 11s dcnsil)?

Wood frogs (Ram1 syfw11ica) look like most other frogs. They arc a few inches long and have characteristic greenish.brown skin. However. wood frogs survive cold wintc~ in a remarkable way-they partially freeze. In i1s partially

frozen ~ate. the frog has no heanbcat. no blood circu· lt11ion, no brca1hing. :md no brain activity. Within 1- 2 hours of tha.,..ing, howc\·cr, these "ital functions return and the frog hops off 10 find food. How docs the wood frog do this? Most rold·bioodcd ammaJs cannoc SlU'\ ivc frecz. ing iempcraturcs because the waaer within lhcir cells frtt1.cs.Aswc~amcd m Section 11 .9. \\hen water freczes. 11 expands, 1rm'Cl'Sibly damaging cells. When the wood frog hibernates for the winter, how· C\'Cf', it produces large amounts of glucose lhai is sccrcccd into its bloodstream and fills the interior" of its cdls. When 1hc temperature drop5 below freezing. extracellular body fluids. such as mosc in the abdornirul ca''ity. freeze solid. fluids within a:lls. hoY.·c\tt, remain liquid because the high glucose coocentration lowers their frcc7.ing point. In other words. the concentrated glocosc soluA 1be wood fros survh·cs winter by partially frce1ing. It protects tion wi1hin the frog'scells acts as antifrca.e. preventing the water ib cells by flooding them with glucosc. tAhich ac1s as an antifreeze. within the cells from free1.ing and allowing the frog to sur.·i\e.

Question

---

A Magmfied '1ew~ of the bone matrh m a normal femur (left) and one v. c:al.c:ned

..

._ ..,

byosteoporo!>•~(righl).

A Se\'Cl'COSICOpOl'OSl>canl'ICOc:!.·

SIUICSl.-gcry10impianlanlfllficial lnp.1om&.'iCCllmltu1X·ra)1rf1alC

,._ Chemistry and Medicine boxes show applications relevant to biomedical and health-related topics. ,.... Chemistry in the Environment boxes relate chapter topics to current environmental and societal issues.

The wood frog can survhe at body temperatures as low as -8.0 "C. Calculnte the mola1ity int of water to - 8.0 "C. of ll •luoose wlution C H ~o

T Chemistry in Your Day boxes demonstrate the importance of chemistry in everyday situations. CHEMISTRY IN YOUR DAY How Soap Works

I

Imagine eating a greas)' cht...-cscburgcr with both hands and without napkins. By the end of the meal. your hands are coated with grease and oil. If you try to wash them wi1h only water, the)' remain greas)'. However. if you add a little soap. the grease washes awoy. Why? As we just learned. water molecules are polar and the molecu les that compose grease and oil are non polar. As a result. water and grease do no1 mix. 1be molecules lhal compose soap. h™evcr, have a special structure that allows them to interact strongly with both water and gre~. One end of a soap molecule is polar. while the other end is nonpolar. !Son~ 1be nonpolarend is a long h)drocarbon chain. Hydrocarbons are always nonpolar because the electronegati\ily difference bctv.een carbon and hydrogen is small. and because the 1c1rahcdral arrangcmcm about each carbon atom tends to ca~I any small dipole moments of individual bonds. 1be polar head of a soap molecule-usually (1hough not always) ionic-suongly auracts water molecules. v.-h1le the nonpolar tail interacts more strongly v.ith grease and oil molecules (v.-e examine the nature of these interactions in Chapter 11 ). Thus. soap acts as a son of molecular liaison--one end interacting w1th water and the other end imerocting with grease. Soap all™-s water and grease 10 mix. remo,mg the grease from CH3(CHi)110CI 12CH 20H your hands and washing it down the drain.

Question Consider the dc1ergcnt molecule at right. Which end do you think is polar? Wh ich end is nonpolar?

xxiii

Annotated Molecular Art Many illustrations have three parts: • • •

a macroscopic image (what you can see with your eyes) a molecular image (what the molecules are doing) a symbolic representation (how chemists represent the process with symbols and equations)

The goal is for you to connect what you see and experience (the macroscopic world) with the molecules responsible for that world, and with the way chemists represent those molecules. After all, this is what chemistry is all about. 2 Hi(l/)

+ 0 2(l/)

--------+ 2 H20(l/)

Hydrogeoandoxygenreacttoformgaseouswater.

~ •••

• •• • • • • • • • • ••

~ ····· ··· ··· ··

• •• •• • • • •

Symbolic representation

· · · · · · · · ·· Molecular image ~ · · · · · · Macroscopic

image

6. FIGURE 4.10 01ddation-Reduction Reaction The hydrogen in 1he balloon reacts with oxygen upon igni1ion 10 form gaseous water (which is d ispersed in the flame). fbalXln:bnBochsler/Peafso"IEc:klcatietl;~bal:Xln:CharlesO. Wll'll~Scuce]

.. FIGURE4.7 Precipitation of Lead(ll) Iodide When a potassium iodide solution is mixed wi1h a lcad(ll) nitratcsotution,ayellowlead(ll) iodide precipiiatc forms.

2 Kl('

NiC0 3(s)

FOR PRACTICE 4.4 Write a net ionic equation for the precipitation reaction that occurs (if any) when solutions of ammonium chloride and iron(III) nitrate are mixed. FOR MORE PRACTICE 4.4 Write a net ionic equation for the precipitation reaction that occurs (if any) when solutions of sodium hydroxide and copper(II) bromide are mixed.

4.5 Acid-Base Reactions Another important class of reactions that occur in aqueous solution are acid-base reactions. In an acid-base reaction (also called a neutralization reaction), an acid reacts with a base and the two neutralize each other, producing water (or in some cases, a weak electrolyte). As in precipitation reactions, an acid-base reaction occurs when the anion from one reactant combines with the cation of the other. Our stomachs contain hydrochloric acid, which acts in the digestion of food. Certain foods or stress, however, can increase the stomach's acidity to uncomfortable levels, causing what has come to be known as heartburn. Antacids are over-the-counter medicines that work by reacting with and neutralizing stomach acid. Antacids employ different bases-substances that produce hydroxide ions (OH-) in water-as neutralizing agents. Milk of magnesia, for example, contains magnesium hydroxide [Mg(OH) 2], and Mylanta® contains aluminum hydroxide [Al(OH)3]. All antacids, regardless of the base they employ, have the same effect of neutralizing stomach acid and relieving heartburn through acid-base reactions. The Arrhenius definitions (named after Swedish chemist Svante Arrhenius who lived from 1859 until 1927) of acids and bases are as follows: ~ Acid: substance that produces H+ ions in aqueous solution. ~

Base: substance that produces OH- ions in aqueous solution.

In Chapter 15, we will learn more general definitions of acids and bases, but these are sufficient to describe neutralization reactions. According to the Arrhenius definition, HCI is an acid because it produces H + ions in solution: .6. FIGURE 4.8 The Hydronium Ion

An H+ ion is a bare proton. In solution, bare protons associate with water molecules to form hydronium ions, H 30 + (Figure 4.8 ~ ): H +(aq)

+

H 2 0(/)

------'>

H 30 +(aq)

Protons associate with water molecules in solution to form H30+ ions, which in turn interact with other water molecules.

112

Chapter 4

Ch emical Reaction s and Stoichiometry

Chemists often use H +(aq) and H 30 +(aq) interchangeably. The chemical equation for the ionization of HCl is often written to show the association of the proton with a water molecule to form the hydronium ion:

Some acids- called polyprotic acids-contain more than one ionizable proton and release them sequentially. For example, sulfuric acid, H2S0 4, is a diprotic acid. It is strong in its first ionizable proton, but weak in its second: H2S04(aq)

+ H 20 (l )

~ H30 +(aq)

+ HS04- (aq)

HS04- (aq)

+ H20 (/)

~ H30 +(aq)

+

SOl - (aq)

According to the Arrhenius definition , NaOH is a base because it produces OH- in solution: .A. Lemons, limes, and vinegar contain acids. Vitamin C and aspirin are acids. [Eric Schrader/Pearson Education]

Sr(OH)2 , and all group 2 hydroxides, are not very soluble, but all of the Sr(OH)2 that does dissolve dissociates. For this reason they are considered strong bases.

In analogy to diprotic acids, some bases such as Sr(OH)2 produce two moles of OH- per mole of the base: Sr(OHh(aq) ~ sr2+(aq)

In Table 4.3, some common acids and bases are listed. You can find acids and bases in many everyday substances. Foods such as citrus fruits and vinegar contain acids. Soap, baking soda, and milk of magnesia all contain bases. When we mix an acid and a base, the H+ from the acid-whether it is weak or strong-combines with the OH- from the base to form H20 (Figure 4.9 ~ ). Consider the reaction between hydrochloric acid and sodium hydroxide:

H Cl(aq)

+

Acid The word salt in this sense applies to any ionic compound and is therefore more general than the common usage, which refers only to table salt (NaCl).

[Eric Schrader/Pearson Education)

Base

~

H 2 0(~

+

Water

N aCl(aq) Salt

Common Acids and Bases

Name of Acid

are bases.

N aOH(aq)

Acid-base reactions generally form water and an ionic compound-called a salt-that usually remains dissolved in the solution. The net ionic equation for many acid- base reactions is:

TABLE 4.3

.A. Many common household products

+ 2 OH- (aq)

Formula

Name of Base

Formula

Hydrochloric acid

HCI

Sodium hydroxide

NaOH

Hydrobromic acid

HBr

Lithium hydroxide

LiOH

Hydroiodic acid

HI

Potassium hydroxide

KOH

Nitric acid

HN03

Calcium hydroxide

Ca(OH)z

Sulfuric acid

HzS04

Barium hydroxide

Ba(OH)z

Perchloric acid

HCI04

Ammonia*

NH 3 (weak base)

Acetic acid

CH3COOH (weak acid) HF (weak acid)

Hydrofluoric acid

*Ammonia does not contain OW , but it produces OW in a reaction with water that occurs only to a small extent: NH,+(aq) + OW(aq ).

NH3(aq) + H20(/ )

=

4.5 Acid-Base Reactions

HCl(aq)

+

NaOH(aq)

+

H 20(l)

113

... FIGURE 4.9 Acid-Base Reaction The reaction between hydrochloric acid and sodium hydroxide forms water and a salt, sodium chloride, which remains dissolved in the solution.

NaCl(aq)

[© Richard Megna/Fundamental Photographs, NYC]

HCl(aq)

+

NaOH(aq)

j H20(l)

+ NaCl(aq)

EXAMPLE 4.5

WRITING EQUATIONS FOR ACID-BASE REACTIONS

Write a molecular and net ionic equation for the reaction between aqueous HI and aqueous Ba(OH)z. SOLUTION

+ Ba(OHh(aq)

~

You must first identify these substances as an acid and a base. Begin by writing the skeletal reaction in which the acid and the base combine to form water and a salt.

Hl(aq)

Next, balance the equation: this is the molecular equation.

2 Hl(aq)

+ Ba(OHh(aq)

Write the net ionic equation by removing the spectator ions.

2 H +(aq)

+ 2 OH-(aq)

acid

H 20(/)

+ Bal2(aq)

water

base

~

~

2 H20(l)

salt

+ Bal2(aq)

2 H2 0(l)

or simply H +(aq)

+

OH - (aq) ~ H 20(/)

FOR PRACTICE 4.5 Write a molecular and net ionic equation for the reaction that occurs between aqueous H2S04 and aqueous Li OH.

114

Chapter 4

Chem ical Reac t ions and Sto ichiometry

Another example of an acid-base reaction is the reaction between sulfuric acid and potassium hydroxide:

Again, notice the pattern of an acid and base reacting to form water and a salt: Acid

+ Base

------4

Water

+ Salt

When writing equations for acid-base reactions, write the formula of the salt using the procedure for writing formulas of ionic compounds given in Section 4.2.

Acid-Base Reactions Evolving a Gas In some acid-base reactions, two aqueous solutions mix to form a gaseous product that bubbles out of solution. These reactions are sometimes called gas-evolution reactions. Some gas-evolution reactions form a gaseous product directly when the cation of one reactant combines with the anion of the other. For example, when sulfuric acid reacts with lithium sulfide, dihydrogen sulfide gas is formed:

Similarly, when acids composed of ammonium cation react in an acid- base reaction in aqueous solution, ammonia gas is evolved:

Other gas-evolution reactions often form an intermediate product that then decomposes (breaks down into simpler substances) to form a gas. For example, when aqueous hydrochloric acid is mixed with aqueous sodium bicarbonate, the following reactions occur:

intermediate

The intermediate product, H2 C0 3, is not stable and decomposes into H 20 and gaseous C0 2 . The overall reaction can be written as follows: HCl(aq)

+ NaHC03(aq)

------4

H2 0 (/)

+ C02(g) + NaCl(aq)

Like the bicarbonate example above, bases composed of carbonates, sulfites, and bisulfites also form gases in acid- base reactions occurring in aqueous solution, as is summarized in Table 4.4.

TABLE 4.4

Types of Compounds That Undergo Gas-Evolution Reactions

Reactant Type

Intermediate Product

Gas Evolved

Example

Sulfides

None

H2S

2 HCl(aq)

+ K2S(aq) -

Carbonates and bicarbonates

H2C03

C02

2 HCl (aq)

+ K2C03(aq) -

H20(/)

Sulfites and bisulfites

H2S03

S02

2 HCl(aq)

+ K2S03(aq) -

H20(/)

Ammonium

NH40H

NH3

NH 4Cl (aq)

+ KOH(aq) -

+ 2 KCl(aq)

H2S (g)

H20(/)

+ C0 2 (g ) + 2KCl(aq) + S02 ( g) + 2KCl (aq)

+ NH3(g) + KCl(aq)

4.6

EXAMPLE 4.6

Oxidation-Reduction Reactions

115

WRITING EQUATIONS FOR GAS-EVOLUTION REACTIONS

Write a molecular equation for the gas-evolution reaction that occurs when you mix aqueous nitric acid and aqueous sodium carbonate. Begin by writing a skeletal equation in which the cation of each reactant combines with the anion of the other. You must then recognize that H2 C03 (aq) decomposes into H 20(l) and C0 2(g), and write these products into the equation. Finally, balance the equation.

~ + Na2C03(aq) ~ HN0 3(aq) + Na2 C03(aq) HN03(aq)

2 HN0 3(aq)

~

H2C03(aq)

~

H 20 (l)

+ Na2 C03(aq)

~

+ NaN03(aq)

+ C02 (g) +

H 20 (l )

NaN0 3(aq)

+ C02(g) + 2 NaN03(aq)

FOR PRACTICE 4.6 Write a molecular equation for the gas-evolution reaction that occurs when you mix aqueous hydrobromic acid and aqueous potassium sulfite. FOR MORE PRACTICE 4.6 Write a net ionic equation for the reaction that occurs when you mix hydroiodic acid with calcium sulfide.

4.6

Oxidation-Reduction Reactions

Oxidation-reduction reactions or redox reactions are reactions in which electrons transfer from one reactant to the other. The rusting of iron , the bleaching of hair, and the production of electricity in batteries involve redox reactions. Many redox reactions (for example, combustion reactions) involve the reaction of a substance with oxygen (Figure 4.10 T):

+ 3 0 2(g) C 8H 18 (l) + f,f 0 2 (g)

~

+ 0 2 (g)

~

4 Fe(s)

2 H 2(g)

Applications of oxidation-reduction reactions

I are covered in Chapter 18.

2 Fe20 3(s)

~ 8 C02 (g)

+ 9 H20 (g )

2 H20(l)

Hydrogen and oxygen react to fo rm gaseous water.

+

A FIGURE 4.10 Oxidation-Reduction Reaction (which is dispersed in the flame).

The hydrogen in the balloon reacts with oxygen upon ignition to form gaseous water

[balloon: Tom Bochsler/Pearson Education; exploding balloon: Charles D. Winters/Science Source]

116

Chapter 4

Ch emical Reaction s and Sto ichiometry

2 N aCl(s)

Electrons are transferred from sodium to chlorine, forming sodium chloride. Sodium is oxidized and chlorine is reduced.

2 N a(s)

..&. FIGURE 4.11 Oxidation-Reduction Without Oxygen When sodium reacts with chlorine, electrons transfer from the sodium to the chlorine, resulting in the formation of sodium chloride. In this redox reaction, sodium is oxidized and chlorine is reduced. [© Richard Megna/Fundamental Photographs, NYC]

The ability of an element to attract electrons in a chemical bond is called electronegativity. We cover electronegativity in more detail in Section 9.7.

However, redox reactions need not involve oxygen. Consider, for example, the reaction between sodium and chlorine to form sodium chloride (NaCl), depicted in Figure 4.11 .A : 2 Na(s)

+

Cl2(g)

~

2 NaCl(s)

This reaction is similar to the reaction between sodium and oxygen, which forms sodium oxide: 4 Na(s)

+

0 2(g)

~

2 Na20 (s)

In both cases, a metal (which has a tendency to lose electrons) reacts with a nonmetal (which has a tendency to gain electrons). In both cases, metal atoms lose electrons to nonmetal atoms. A fundamental defin ition of oxidation is the loss of electrons, and a fundamental definition of reduction is the gain of electrons. The transfer of electrons need not be a complete transfer (as occurs in the formation of an ionic compound) for the reaction to qualify as oxidation-reduction. For example, consider the reaction between hydrogen gas and chlorine gas: H2(g)

+

Cl2(g)

~

2 HCl(g)

Hydrogen loses electron density (oxidation) and chlorine gains electron density (reduction).

Even though hydrogen chloride is a molecular compound with a covalent bond, and even though the hydrogen has not completely transferred its electron to chlorine during the reaction, you can see from the electron density diagrams (Figure 4. 12 ..,.) that hydrogen has lost some of its electron density-it has partially transferred its electron to chlorine. In the reaction, hydrogen is oxidized and chlorine is reduced and, therefore, this is a redox reaction.

..&. FIGURE 4.12 Redox with Partial Electron Transfer When hydrogen bonds to chlorine, the electrons are unevenly shared, resulting in an increase of electron density (reduction) for chlorine and a decrease in electron density (oxidation) for hydrogen.

Identifying whether or not a reaction between a metal and a nonmetal is a redox reaction is fairly straightforward because of ion formation. But how do we identify redox reactions that occur between nonmetals? Chemists have devised a scheme to track electrons before and after a chemical reaction. In this scheme- which is like bookkeeping for electronseach shared electron is assigned to the atom that attracts the electrons most strongly.

Oxidation States

4.6

Then a number, called the oxidation state or oxidation number, is given to each atom based on the electron assignments. In other words, the oxidation number of an atom in a compound is the "charge" it would have if all shared electrons were assigned to the atom with the greatest attraction for those electrons. For example, consider HCI. Since chlorine attracts electrons more strongly than hydrogen, we assign the two shared electrons in the bond to chlorine; then H (which has lost an electron in our assignment) has an oxidation state of+ 1, and Cl (which has gained one electron in our assignment) has an oxidation state of -1 . You can use the following rules to assign oxidation states to atoms in elements and compounds.

Oxidation-Reduction Reactions

117

Do not confuse oxidation state with ionic charge. Unlike ionic charge- which is a real property of an ion- the oxidation state of an atom is merely a theoretical (but useful) construct.

Rules for Assigning Oxidation States These rules are hierarchical. If any two rules conflict, follow the rule with the smaller number (i.e. , rule 1 takes precedence over rule 2): 1. The oxidation state of each atom in an element is 0.

2. The oxidation state of the atom in a monoatomic ion is equal to the ion's charge. 3. The sum of the oxidation states of all atoms in: Ill> a neutral molecule is always 0. Ill> a polyatomic ion is always equal to the charge of the ion.

4. In their compounds, metals have positive oxidation states: Ill> Group 1 metals always have an oxidation state of + 1. Ill> Group 2 metals always have an oxidation state of +2. 5. The oxidation state of hydrogen in a compound is usually + l. 6. In their compounds, the nonmetals typically have negative oxidation states: Ill> Fluorine always has an oxidation state of -1. Ill> Oxygen usually has an oxidation state of -2. Ill> The other group 17 elements usually have an oxidation state of -1. Ill> The other group 16 elements usually have an oxidation state of -2. Ill> Group 15 elements usually have an oxidation number of -3. Another point to keep in mind is that when assigning oxidation states to elements that are not covered by the rules , such as carbon, use rule 3 to deduce their oxidation state once all other oxidation states have been assigned. For example, when assigning oxidation states to C and H in methane, CH4 , rule 5 says that the oxidation state of hydrogen is + 1. According to rule 3, the sum of all oxidation states for methane must be 0, so the oxidation state of carbon must be -4.

EXAMPLE 4.7

ASSIGNING OXIDATION STATES

Assign an oxidation state to each atom in each element, ion, or compound: (b) Na+ (a) Cl2 (c) KF (d) C02 (e) so/

-

(t) K20 2

SOLUTION (a) Since Cl 2 is an element, the oxidation state of both Cl atoms is 0 (rule 1).

Cl2 CIC! 00

(b) Since Na+ is a monoatomic ion, its oxidation state is equal to its charge, 1 + (rule 2).

Na+ Na+ +l

(c) The oxidation state of K is + 1 (rule 4). Since there are only two

atoms in this compound, we can use rule 3 to assign an oxidation number to F. The sum of the oxidation numbers must be 0, so the oxidation number of F is - 1, which agrees with rule 6.

KF KF +I - I

sum: + l- 1 = 0

(continued )

118

Chapter 4

EXAMPLE 4.7

Chemical Reactions and Stoichiometry

(CONTINUED}

(d) The oxidation state of each oxygen is -2 (rule 6). The sum of the oxidation states must be 0 (rule 3), so the oxidation state of carbon is +4.

C02 (Cox state) + 2(0 ox state) (Cox state) + 2(-2) = 0 C ox state

=0

= +4

C02 +4 -2

=

sum: + 4 + 2(- 2)

0

so/-

(e) The oxidation state of each oxygen is -2 (rule 6). We might expect the oxidation state of sulfur to be -2. However, if that were the case, the sum of the oxidation states would not equal the charge on the ion. Since 0 is mentioned higher on the list of rules than S, it takes priority and we compute the oxidation state of sulfur by setting the sum of all the oxidation states equal to -2 (the charge of the ion, rule 3).

(S ox state) + 4(0 ox state) (S ox state) + 4(-2) = -2 S ox state = +6

= -2

so/+6-2

sum: + 6 + 4(-2) = - 2

(t) The oxidation state of potassium is + 1 (rule 4 ). We might ordinar-

K202 2(K ox state) + 2(0 ox state) 2( + 1) + 2(0 ox state) = 0 0 ox state = -1

ily expect the oxidation state of 0 to be -2 (rule 6), but group 1 metals are mentioned higher on the list of rules (rule 4), so we deduce the oxidation state of 0 by setting the sum of all the oxidation states equal to 0 (rule 3).

=0

K20 2 +I - I

sum:2(+ 1) + 2(- 1)

=

0

FOR PRACTICE 4. 7 Assign an oxidation state to each atom in each element, ion, or compound: (a) Cr

(b) Cr3+

(c) CC14

(d) SrBr2

In most cases, oxidation states are positive or negative integers; on occasion, an atom within a compound can have a fractional oxidation state. Consider K02. The oxidation states are assigned as follows: + I -!

sum: +I +2(- !) = 0

-!

In K0 2, oxygen has a oxidation state. Although this seems unusual, it is acceptable because oxidation states are merely an imposed electron bookkeeping scheme, not an actual physical quantity.

Identifying Redox Reactions We can use oxidation states to identify redox reactions, even between nonmetals. For example, is the following reaction between carbon and sulfur a redox reaction? C + 2S

~

CS 2

If so, what element is oxidized? What element is reduced? We can use the oxidation state rules to assign oxidation states to all elements on both sides of the equation.

Oxidation states:

C o

+

2S o

~

CS 2 +4 - 2

L LJ LOxidation ~ Reduction

4.6

Oxidation-Reduction Reactions

119

Carbon changes from an oxidation state of 0 to an oxidation state of +4. In terms of our electron bookkeeping scheme (the assigned oxidation state), carbon loses electrons during the conversion of reactants to products and is oxidized. During the conversion of reactants to products, sulfur changes from an oxidation state of 0 to an oxidation state of -2. In terms of our electron bookkeeping scheme, sulfur gains electrons and is reduced. In terms of oxidation states, oxidation and reduction are defined as follows: ~

Oxidation: an increase in oxidation state

~

Reduction: a decrease in oxidation state

Notice that oxidation and reduction must occur together. If one substance loses electrons (oxidation), then another substance must gain electrons (reduction). A substance that causes the oxidation of another substance is called an oxidizing agent. Oxygen, for example, is an excellent oxidizing agent because it causes the oxidation of many substances. In a redox reaction, the oxidizing agent is always reduced. A substance that causes the reduction of another substance is called a reducing agent. Hydrogen, for example, as well as the group 1 and group 2 metals (because of their tendency to lose electrons) are excellent reducing agents. In a redox reaction, the reducing agent is always oxidized. You will learn more about redox reactions below, including how to balance them. For now, you need to be able to identify redox reactions, as well as oxidizing and reducing agents, according to the following guidelines:

Remember that a reduction is a reduction in

I oxidation state.

Redox reactions: ~

Any reaction in which there is a change in the oxidation states of atoms in going from reactants to products.

In a redox reaction: ~

The oxidizing agent oxidizes another substance (and is itself reduced).

~

The reducing agent reduces another substance (and is itself oxidized).

EXAMPLE 4.8

IDENTIFYING REDOX REACTIONS, OXIDIZING AGENTS, AND REDUCING AGENTS

Determine whether each reaction is an oxidation- reduction reaction. For each oxidation-reduction reaction, identify the oxidizing agent and the reducing agent. (a) 2 Mg(s)

+

+

(d) C2H6(g)

~

2 MgO(s)

~ 2 H20(l) + CaBri(aq) 2 2 Fe +(aq) ~ Zn 2+(aq) + 2 Fe(s)

(b) 2 HBr(aq)

(c) Zn(s)

0 2(g)

+

+ Ca(OH)i(aq)

~Oz(g) ~ 2 C02(g)

+

3 H20(/)

SOLUTION (a) This is a redox reaction because magnesium increases in oxidation number (oxidation) and oxygen decreases in oxidation number (reduction).

2 Mg(s)

+

0 2 (g)

~

L . _L 0

0

2 MgO(s) +2 -2

-+-1'

R eduction Oxidation _ _ _ _ __, _ Oxidizing agent: 0 2 Reducing agent: Mg

(b) This is not a redox reaction because none of the atoms undergoes a change in oxidation number. This is an acid- base or neutralization reaction. (c) This is a redox reaction because zinc increases in oxidation number (oxidation) and iron decreases in oxidation number (reduction).

2 HBr(aq)

+

Ca(OH)z(aq)

+ I - I

~

2 H 20(l)

+2- 2 + I

Zn(s)

+

Fe 2 +(aq) +2 I

~

CaBr2(aq ) +2 - 1

Zn2 +(aq) +2

L . -------+t

0

+

+I - 2

+ Fe(s) 0

Reduction .!

Oxidation _ _ __._

Oxidizing agent: Fez+ R educing agent: Zn

(continued)

120

Chapter 4

EXAMPLE 4.8

Chemical Reactions and Stoichiometry

(CONTINUED}

(d) This is a redox reaction because carbon increases in oxidation number (oxidation) and oxygen decreases in oxidation number (reduction). Combustion reactions are redox reactions.

+

C 2H 6(g)

t 0 2(g)

L . .L

-3 +I

~ 2 C0 2(g)

0

+

3 H 20(0

+4 -2

+I - 2

-if-*-------*

Reduction Oxidation _ _ __._

Oxidizing agent : 0 2 Reducing agent: C 2H 6

FOR PRACTICE 4.8 Determine whether each reaction is an oxidation-reduction reaction. For all redox reactions, identify the oxidizing agent and the reducing agent. (a) 2 Li(s) (b) 2Al(s)

+ +

CJ2(g) ----" 2 LiCJ(s) 3Sn2 +(aq)----" 2Al3+ (aq)

(c) Pb(N0 3h(aq) (d) C(s)

+

+ 2 LiCJ(aq)

+

3Sn(s)

----" PbCl 2(s)

+ 2 LiN0 3(aq)

0 2(g) ----" C02(g)

Which statement is true? (a) A redox reaction involves eitherthe transfer of an electron ora change in the oxidation state of an element. (b) If any of the reactants or products in a reaction contain oxygen, the reaction is a redox

reaction. (c) In a reaction, oxidation can occur independently of reduction. (d) In a redox reaction, any increase in the oxidation state of a reactant must be accompanied

by a decrease in the oxidation state of a reactant.

CHEMISTRY IN YOUR DAY

Bleached Blonde

·~

Have you ever bleached your hair? Most home kits for hair bleaching contain hydrogen peroxide (H 20 2) , an excellent oxidizing agent. When applied to hair, hydrogen peroxide oxidizes melanin, the dark pigment that gives hair its colour. Melanin is an oligomer of 5,6-dihydroxyindole and 5,6-dihydroxyindole-2-carboxylic acid, which means that these two molecules are Linked together to make chains of varying lengths. The structure is highly conjugated, meaning that it has many alternating single and double bonds. These kinds of organic structures often absorb visible Light and give rise to colours. We will cover this in Chapter 10.

HO~

HO~~) H

5,6-dihydroxyindole

HOXX>-

I

HO

\

~

~

~

O

HO

"-oH

5,6-dihydroxyindole-2-carboxy lic acid

HO HO /)-0

When melanin is oxidized with hydrogen peroxide, the ring structure can be broken. This reduces the number of alternating single and double bonds and removes the colour, leaving the hair with the fam iliar bleached look.

c,y HO HO

Oxidation

HOOC~

HOOCv1---~:( '

'

H

HO

Melanin Sample Structure

'DH

4.6

Oxidation-Reduction Reactions

121

Hydrogen peroxide also oxidizes other components of hair. For example, protein molecules in hair contain -SH groups called thiols. Hydrogen peroxide oxidizes these thiol groups to sulfonic acid groups, -S03H. The oxidation of thiol groups to sulfonic acid groups causes changes in the proteins that compose hair, making the hair more brittle and more likely to tangle. Consequently, people with heavily bleached hair generally use conditioners containing compounds that form thin, lubricating coatings on individual hair shafts. These coatings prevent tangling and make hair softer and more manageable.

Question The following is a reaction of hydrogen peroxide with an alkene:

Can you see why this reaction is a redox reaction? Can you identify the oxidizing and reducing agents? ~ The bleaching of hair involves a redox reaction in which melanin-the main pigment in hair-is oxidized.

[Olga Sapegina/Fotolia]

Balancing Oxidation-Reduction Equations Consider the following reaction between calcium and water: Ca(s) 0

L

+

2 H20(/) ~ Ca(OH) 2(aq) + H2(g) +1-2 +2-2 +1 0 ._!- - - - - -t + -- Reduction J Oxidation _ _ __._

Since calcium increases in oxidation state from 0 to +2, it is oxidized. Since hydrogen decreases in oxidation state from + 1 to 0, it is reduced. Balancing redox reactions can be more complicated than balancing other types of reactions because both the mass (or number of each type of atom) and the charge must be balanced. Redox reactions occurring in aqueous solutions can be balanced by using a special procedure called the half-reaction method of balancing. In this procedure, the overall equation is broken down into two half-reactions: one for oxidation and one for reduction. The half-reactions are balanced individually and then added together. The steps differ slightly for reactions occurring in acidic and in basic solution. The following example demonstrates the method used for an acidic solution, and Example 4.10 demonstrates the method used for a basic solution.

EXAMPLE 4.9

HALF-REACTION METHOD OF BALANCING AQUEOUS REDOX REACTIONS IN ACIDIC SOLUTION

Balance the following redox reaction in acidic solution: Fe2+(aq)

+

1. Assign oxidation states to all atoms and identify the substances being oxidized and reduced. This step simply allows you to categorize, with confidence, the half-reactions as oxidation or reduction in the next step.

Mn04-(aq) ~ Fe3+(aq)

+2

I

+ Mn2+(aq)

+7 -2

l.__ Reduction

+3

+2

- -tt----_.t

" " · - - - - Oxidation _ _ _ __._

Fe2+ is being oxidized by Mn04 - to Fe3+. Mn04- is the oxidizing agent. The reducing agent, Fe2+, reduces Mn04- to Mn2+.

(continued)

122

Chapter 4

EXAMPLE 4.9

Ch emi cal Re action s and Stoichiometry

(CONTINUED}

2. Separate the overall reaction into two halfreactions: one for oxidation and one for reduction.

Oxidation: Fe 2 +(aq) -

3. Balance each half-reaction with respect to mass in the following order: • Balance all elements other than H and 0 . • Balance 0 by adding H20. • Balance H by adding W .

All elements other than 0 and H are balanced, so proceed to balance 0 and then H. The oxidation half-reaction does not contain 0 or H, so it remains the same for this step.

Fe3+(aq) Mn 2 +(aq)

Reduction: Mn04 - (aq) -

Fe2 +(aq) -

Fe3+(aq)

Since the left side of the reaction contains 4 0, we add 4 H 2 0 to the right side. This results in 8 H on the right side, so we add 8 H+ to the left side. All elements should now be balanced. 8 H+(aq)

4. Balance each half-reaction with respect to charge by adding electrons. (Make the sum of the charges on both sides of the equation equal by adding as many electrons as necessary.)

+

Mn2 +(aq)

Mn0 4 - (aq) -

Fe2 +(aq) -

Fe3+(aq)

Mn2 +(aq) + 4 H 2 0 (1)

Multiply the oxidation reaction by 5 so that there are an equal number of electrons on the left side of the oxidation half-reaction as there are electrons on the right side of the reduction half-reaction.

6. Add the two half-reactions together, cancelling electrons and other species as necessary.

5 Fe3+ (aq) + SeMn2 +(aq) + 4 H 20 (l)

5 Fe2 +(aq) -

x

+ 8 H+(aq) + Mn0 4-(aq) -

5 Fe 2 +(aq)

+ 8 H+(aq) +

7. Verify that the reaction is balanced both with respect to mass and with respect to charge.

Reactants

Products

5 Fe

5 Fe

8H 1 Mn

8H 1 Mn

40 +17 charge

40 +17 charge

FOR PRACTICE 4 .9 Balance the following redox reaction in acidic solution:

+

Cr(s) -

H2(g)

+ cr2+(aq)

FOR MORE PRACTICE 4.9 Balance the following redox reaction in acidic solutio n:

+ N0 3- (aq) -

Cu2 +(aq)

+

5 Fe3+(aq)

+

x

Mn2 +(aq) + 4 H20(l)

Mn04- (aq)

~ 5 Fe3+(aq)

Cu(s)

+ e-

The reduction reaction has a net 7+ on the left side and 2+ on the right side. To balance the charge, we add five electrons to the left side:

5 Fe2 +(aq) 5 e- + 8 H+(aq) + Mn04- (aq) -

H+(aq)

4 H20(l)

For the oxidation reaction, there is a net positive charge on the right side, so we add one electron to the right to balance this charge:

Se - + 8 H+(aq) + Mn0 4 - (aq) -

5. Make the number of electrons in both halfreactions equal by multiplying one or both half-reactions by a small whole number.

+

N02(g)

+

Mn2 +(aq)

+ 4 H20

(l)

4.6

Oxidation-Reduction Reactions

123

When a redox reaction occurs in basic solution, you can balance the reaction in exactly the same way, except that you must add an additional step to neutralize any H+ with OH- . The H+ and the OH- combine to form H20 , as shown in the following example.

BALANCING REDOX REACTIONS OCCURRING IN BASIC SOLUTION

EXAMPLE 4.10

In Example 4.9, we balanced the reaction between iron(II) and the permanganate ion in acidic solution: 5 Fe2 +(aq)

+

8 H +(aq)

+

Mn04(aq) ~ 5 Fe3 +(aq)

+

Mn 2 +(aq)

+ 4 H 2 0(l)

Balance this reaction in basic solution.

SOLUTION To balance redox reactions occurring in basic solution, follow the half-reaction method outlined above for acidic solutions, but add an extra step to neutralize the W with 011 as shown in step I below. 1. Starting with the balanced redox reac-

5 Fe 2 +(aq)

tion in acidic solution, neutralize H+ by adding the same number of 011 ions to both sides of the equation.

~

+ 8 OH-(aq) + Mn04(aq) 5 Fe3+ (aq) + Mn 2 +(aq) + 4 H2 0(l) + 8 OH-(aq) 8 H+(aq)

+ .8" H 2 0(/) + Mn04 (aq) ~ 5 Fe3+ (aq) + Mn 2 +(aq) +

~

+ 4 H 20 (l) + Mn04(aq) ~ 5 Fe3+ (aq) + Mn 2 +(aq) +

8 OH-(aq)

2. The side with both H+ and 011 will combine to form water.

5 Fe 2 +(aq)

Then cancel as much water as possible from both sides of the equation.

5 Fe 2 +(aq)

3. Verify that the reaction is balanced.

+

Reactants

Products

5 Fe

5 Fe

8H

8H

50

50

1 Mn

I Mn

- 9 charge

- 9 charge

FOR PRACTICE 4.10 Balance the following redox reaction occurring in basic solution. CIO-(aq)

+ Cr(OH)4-(aq)

~ Crol -(aq)

+ Ci-(aq)

Disproportionation Reactions A disproportionation reaction is one in which a reactant is both oxidized and reduced. For example, hydrogen peroxide (H20 2) decomposes to form H 20 and 0 2. Some H20 2 is reduced, which forms H20 , and some H20 2 is oxidized, which forms 0 2 : 2H202(aq) -1

~

2H20(0 -2

+

02(g) 0

~'~:~_J To balance disproportionation reactions we follow the same method as laid out in Examples 4.9 and 4.10.

+

8 OH- (aq)

124

Chapter 4

EXAMPLE 4.11

Chem ical React ions and Sto ichiometry

BALANCING DISPROPORTIONATION REACTIONS

Balance the disproportionation reaction of chlorine to form chloride and chlorate ions in basic solution: ~

Cl2(aq)

1. Assign oxidation states to help categorize the half-reactions.

Cl-(aq) + ClO)(aq)

Cl2(aq)

~

Ci-(aq)

+

Cio;(aq)

+5

-1

0

C:'.='~::::!_J

2. Separate the overall reaction into half-reactions.

Reduction: Cl2 (aq) Oxidation: Cl2 (aq)

~

Cl- (aq)

~

Cl03 (aq)

3. Balance each half-reaction with respect to mass: • Balance all elements other than H andO.

Cl2 (aq)

~

2 Cl- (aq)

Cl2 (aq)

~

2 ClO) (aq)

• Balance 0 by adding H20.

Cl2 (aq)

~

2 Cl-(aq)

Cl2 (aq) + 6 H20(l) • Balance H by adding H+.

~

2 Cl0 3 (aq)

Cl2 (aq) ~ 2 Cl- (aq) Cl2 (aq) + 6 H 20(l) ~ 2 ClO)(aq) + 12 H +(aq)

4. Balance each half-reaction with respect to charge.

Cl2 (aq) + 2e-

5. Make the number of electrons in both equations the same by multiplying each by a small whole number.

5 Cl 2 (aq) + lOe- ~ 10 Cl-(aq) Cl2 (aq) + 6 H 20(l) ~ 2 ClO)(aq) + 12 H +(aq) + lOe-

6. Add the half-reactions together and make necessary cancellations.

5 Cl2 (aq) +

7. Neutralize H+ by adding the same number of 0!1 ions to both sides of the equation.

6 Cl2 (aq) + 6 H20(l) + 12 OH- (aq)

The side with both Wand 0!1 will cornbine to form water.

6 Cl2 (aq) + fi-H20(l) + 12 OH- (aq)

Then cancel as much water as possible from both sides of the equation. In this case, we can divide each stoichiometric coefficient by a factor of two.

6 Cl2 (aq) + 12 OH-(aq) ~ 2 CIO)(aq) + 10 Cl-(aq) + 6 H20 (/) 3 Cl2 (aq) + 6 OH-(aq) ~ CIO)(aq) + 5 Cl - (aq) + 3 H20 (l)

8. Verify that the reaction is balanced with respect to mass and charge.

~

2 Cl- (aq)

Cl2 (aq) + 6 H20(l) ~ 2 ClO)(aq) + 12 H +(aq) + lOe-

we

~

10 Cl- (aq)

Cl2 (aq) + 6 H 20(l) ~ 2 CIO) (aq) + 12 H +(aq) + we 6 Cl2 (aq) + 6 H20(l) ~ 2 Cl03 (aq) + 10 Cl-(aq) + 12 H+ (aq) ~ 2 ClO) (aq) + 10 Cl- (aq) + 12 H+ (aq) + 12 OH- (aq)

~

6

2 CIO)(aq) + 10 Cl-(aq) + .ll H20 (/)

Reactants

Products

6Cl

6Cl

6H

6H

60

60

-6 charge

-6 charge

FOR PRACTICE 4.11 Balance the following disproportionation reaction of chlorine dioxide in acidic medium. Cl02 ~ Cl02 + Cl03

4.7

4.7

Reaction Stoichiometry: How Much Is Produced?

Reaction Stoichiometry: How Much Is Produced?

A balanced chemical equation provides the exact relationship between the amount of reactant and the amount of product. For example, in a combustion reaction, the chemical equation provides a relationship between the amount of fuel burned and the amount of carbon dioxide emitted. In the following discussion, we use octane (a component of gasoline) as our fuel. The balanced equation for the combustion of octane is:

The balanced equation shows that 16 C0 2 molecules are produced for every 2 molecules of octane burned, or that 8 C02 molecules are produced for every mole of octane burned. It is equally correct to write a combustion reaction such as the preceding one to show only one molecule of the fuel by dividing all the coefficients by a factor of 2: CsH1 s(g)

+

6f-0 2(g) ~ 8 C02(g)

+

9 H20(l)

Nothing has changed; there are still 8 C02 molecules produced for every mole of octane burned. We can extend this numerical relationship between molecules to the amounts in moles as follows: The coefficients in a chemical reaction specify the relative amounts in moles of each of the substances involved in the reaction. In other words, from the equation, we know that eight moles of C02 are produced for every mole of octane burned. The use of the numerical relationships between chemical amounts in a balanced chemical equation are called reaction stoichiometries. Stoichiometry allows us to predict the amounts of products that will form in a chemical reaction based on the amounts of reactants that react. Stoichiometry also allows us to determine the amount of reactants necessary to form a given amount of product. These calculations are central to chemistry, allowing chemists to plan and carry out chemical reactions in order to obtain products in the desired quantities.

Making Molecules: Mole-to-Mole Conversions A balanced chemical equation is simply a "recipe" for how reactants combine to form products. From our balanced equation for the combustion of octane, for example, we can write the following stoichiometric ratio:

1 mol C 8H 18 : 8mol C02 We can use this ratio to determine how many moles of C0 2 form when a given number of moles of C8H 18 burns. A typical car or small truck requires 50.0 L, which is about 300 mol of octane, to fill the tank. Suppose we burn this entire tank of gas, 300 mol of C8H 18; how many moles ofC02 form? We use the ratio from the balanced chemical equation. The ratio acts as a conversion factor between the amount in moles of the reactant C8H 18 and the amount in moles of the product:

300 ~

X

8 mol C02 1~

= 2.40

X

103 mol C02

The combustion of 300 mol of C8H 18 (1 tank of gas) adds 2400 mol of C0 2to the atmosphere!

Making Molecules: Mass-to-Mass Conversions Respiration is a chemical reaction where glucose reacts with oxygen to form carbon dioxide and water:

Let's estimate the mass of C02 that is produced by respiration of the glucose equivalent to the sugar in one chocolate bar, approximately 36.0 g. The calculation is similar to the one done in the previous section, but this time we are given the mass of fuel (C6H 120 6 ) instead of the number of moles. Consequently, we must first convert the mass (in grams) to the amount (in moles). The general conceptual plan for calculations where you are

Stoichiometry is pronounced

I stoy-kee-om-e-tree.

125

126

Chapter 4

Chemical Reactions and Stoichiometry

given the mass of a reactant or product in a chemical reaction and asked to find the mass of a different reactant or product takes the form: ( AmountA (in moles)

AmountB (in moles)

MassB

where A and B are two different substances involved in the reaction. We use the molar mass of A to convert from the mass of A to the amount of A (in moles). Then, we use the appropriate ratio from the balanced chemical equation to convert from the amount of A (in moles) to the amount of B (in moles). Finally, we use the molar mass of B to convert from the amount of B (in moles) to the mass of B. To calculate the mass of C0 2 produced from the respiration of 36.0 g of glucose, we use the following conceptual plan:

Conceptual Plan

44.01 g co,

6mol C02

1 molC 6H 120 6 180.16 g c , H 12 0,

1 mol C 6 H 12 0

6

1 mol C02

Relationships Used I mol C 6H 120 6 : 6 mol C02 (from the chemical equation) molar mass ofC6H 120 6 = 180 .16 g mol - 1 molarmass ofC02 = 44.0lgmol - 1

Solution We follow the conceptual plan to solve the problem, beginning with the mass of C 6H 120 6 and cancelling units to arrive at the mass of C02: _ ~ 36 0

1 rnol--GoHI206 180 .16 ~

x

=

x

rool-e02 x 1 rnol--GoHI206 6

44.01 g C02 1 rool-e02

52.8gC02

The following are additional examples of stoichiometric calculations. EXAMPLE 4.12

STOICHIOMETRY

In photosynthesis, plants convert carbon dioxide and water into glucose (C6H 120 6) according to the following reaction: 6C02(g)

+

6H20(/) sunlight6 0 2(g)

+

C6H1206(aq)

Suppose you determine that a particular plant consumes 37 .8 g of C02 in one week. Assuming that there is more than enough water present to react with all of the C02, what mass of glucose (in grams) can the plant synthesize from the C02?

SORT The problem gives the mass of carbon dioxide and asks you to find the mass of glucose that can be produced. STRATEGIZE The conceptual plan follows the general pattern of mass A ~ amount A (in moles) ~ amount B (in moles) ~ mass B. From the chemical equation, d educe the relationship between moles of carbon dioxide and moles of glucose. Use the molar masses to convert between grams and moles.

GIVEN: 37 .8gC02 FIND: g C6H 1206 CONCEPTUAL PLAN

1 mol C0 2

1 molC 6 H 120 6

180.1 6 g c,H 120,

44.0 1 gC02

6mol C02

1 molC, H 1206

RELATIONSHIPS USED molarmass C02 = 44.0lgmol- 1 6 mol C02 : 1 mol C 6H 120 6 molar mass C 6H 120 6

=

180.16 g mol- 1

4.8

SOLVE Follow the conceptual plan to solve the problem. Begin with g C0 2 and use the conversion factors to arrive at g C6 H 120 6 .

127

Limiting Reactant, Theoretical Yield, and Percent Yield

SOLUTION 37 .8 n ori--x 2

~'-'

1 mel-e(J:;2 44.01 g.-(202

1 mr. 1 c.._u..-rr 180 .16 g C6H 120 6 X ~ 6 " 1 2 '"' 6 X 6..n:ml-e02 1~

= 25.8

g

C6H 120 6

CHECK The units of the answer are correct. The magnitude of the answer (25.8 g) is less than the initial mass of C02 (37 .8 g). This is reasonable because each carbon in C02 has two oxygen atoms associated with it, while in C6H 120 6 , each carbon has only one oxygen atom and two hydrogen atoms (which are much lighter than oxygen) associated with it. Therefore, the mass of glucose produced should be less than the mass of carbon dioxide for this reaction. FOR PRACTICE 4.12 Magnesium hydroxide, the active ingredient in milk of magnesia, neutralizes stomach acid, primarily HCl, according to the following reaction: Mg(OH)z(aq)

+ 2 HCl(aq)

~

2 H20(l)

+ MgCl2(aq)

What mass of HCl, in grams, can be neutralized by a dose of milk of magnesia containing 3.26 g Mg(OH)2 ? FOR MORE PRACTICE 4.12 One component of acid rain is nitric acid, which forms when N02 , also a pollutant, reacts with oxygen and water according to the following simplified equation: 4 N02 (g)

+ 0 2(g) + 2 H 20(/)

~

4 HN0 3 (aq)

The generation of the electricity used by a medium-sized home produces about 16 kg of N02 per year. Assuming that there is adequate 0 2 and H2 0 , what mass of HN03 , in kg, can form from this amount of N0 2 pollutant?

Under certain conditions, sodium can react with oxygen to form sodium oxide according to the following reaction:

A flask contains the amount of oxygen represented by the diagram on the right ~ . Which diagram best represents the amount of sodium required to completely react with all of the oxygen in the flask, according to the above equation?

(b)

(a)

4.8

(d)

(c)

Limiting Reactant, Theoretical Yield, and Percent Yield

It is intuitive to begin with an example to understand three more important concepts in

reaction stoichiometry: limiting reactant, theoretical yield, and percent yield. We will use the combustion of propane (C3H 8), the most common barbeque fuel, as our example. The balanced equation for the combustion of propane is: C3Hg(g)

+

5 02(g)

~

-.

3 C02(g)

+

4 H 20 (/)

•• -

128

Chapter 4

Chemical Reactions and Stoichiometry

Supposing we start with 2 molecules of C 3H8 and 15 molecules of Oz, what is the reactant that limits the amount of the products that can be formed (the limiting reactant)? What is the maximum amount of C02 that can be produced from the limiting reactant (the theoretical yield)? First we must calculate the number of C02 molecules that can be made from 2 molecules of C3H 8 :

-.d'sf 2 C,"3"-•

x

3 C02 = 1 Mt>M2 "

V2

M 1V1 = M 2 V2

RELATIONSHIP USED

SOLVE Begin with the solution dilution equation and solve it for V2.

M 1V1 V2=-M2

Substitute in the required quantities and compute V2 .

15.0 ~ X 0.200 L

Make the solution by diluting 0.200 L of the stock solution to a total volume of 1.00 L (V2). The resulting solution will have a concentration of 3.00 mol L- 1•

3.00 ~

= 1.00 L

CHECK The final unit (L) is correct. The magnitude of the answer is reasonable because the solution is diluted from 15.0 mol L- 1 to 3.00 mol L- 1, a factor of five. Therefore, the volume should increase by a factor of five. FOR PRACTICE 4.17 To what volume (in mL) should you dilute 100.0 mL of a 5.00 mol L- 1 CaCl 2 solution to obtain a 0.750 mol L - I CaC12 solution? FOR MORE PRACTICE 4.17 What volume of a 6.00 mol L- 1 NaN03 solution should you use to make 0.525 L of a 1.20 mol L- 1 NaN0 3 solution?

Solution Stoichiometry In Section 4.7, we learned how the coefficients in chemical equations are used as conversion factors between the amounts of reactants (in moles) and the amounts of products (in moles). In aqueous reactions, quantities of reactants and products are often specified in terms of volumes and concentrations. We can use the volume and concentration of a reactant or product to calculate its amount in moles. We can then use the stoichiometric coefficients in the chemical equation to convert to the amount of another reactant or product in moles. The general conceptual plan for these kinds of calculations begins with the volume of a reactant or product:

( Volume A .. - -..

Amount A (in moles)

AmountB (in moles)

VolumeB

We make the conversions between solution volumes and amounts of solute in moles using the molarities of the solutions. We make the conversions between amounts in moles of A and B using the stoichiometric coefficients from the balanced chemical equation. The following example demonstrates solution stoichiometry.

138

Chapter 4

EXAMPLE 4.18

Chemical Reactions and Stoichiometry

SOLUTION STOICHIOMETRY

What volume of a 0.150 mol L- 1 KC! solution will completely react with 0 .150 L of a 0.175 mol L- 1 Pb(N03) 2 solution according to the following balanced chemical equation?

SORT You are given the volume and concentration of a Pb(N03)2 solution. You are asked to find the volume of KC! solution (of a given concentration) required to react with it. STRATEGIZE The conceptual plan has the following form: volume A ~ amount A (in moles) ~ amount B (in moles) ~ volume B. The molar concentrations of the KC! and Pb(N03h solutions can be used as conversion factors between the number of moles of reactants in these solutions and their volumes. The stoichiometric coefficients from the balanced equation are used to convert between number of moles of Pb(N03h and number of moles of KC!.

GIVEN: 0.150 L of Pb(N0 3h solution, 0.175 mol L- 1 solution, 0.150 mol L- 1 KC! solution FIND: volume KC! solution (in L) CONCEPTUAL PLAN r ; ;b(N03 )i

l

solution 0.175 mol Pb(N03},

2mol KC I

1 L Pb(N03), solution

1 mol Pb(N03),

l

VKCI solution

molKCI '

'

1 L KC I solution 0.150 mol KCI

RELATIONSHIPS USED [ Pb(N03h ] =

0.175 mol Pb(N03h 1 L Pb(N03) 2 solution

2 mol KC! : 1 mol Pb(N03h [KC!]= 0.150mol KCI 1 L KC! solution Don't forget that [X] is a shorthand way of writing the concentration of X in units of mol L- 1•

SOLVE Begin with L Pb(N03h solution and follow the conceptual plan to arrive at the volume of KC! solution.

SOLUTION 0.150 L Pb(W0 3h soluuon x X

2 rool-f{CI 1 m.ol-PbfN03J2

X

0 .175 m.ol-PbfN03J2 1 L Pb("N03h so!Unon

1 L KCI solution 0 .150 rool-f{CI

= 0.350 L KC! solution

CHECK The final units (L KCI solution) are correct. The magnitude (0.350 L) seems reasonable because the reaction stoichiometry requires 2 mol of KCI per mole of Pb(N0 3)2 . Since the concentrations of the two solutions are not very different (0. 150 mol L- 1 compared to 0.175 mol L- 1), the volume of KCJ required should be roughly two times the 0.150 L of Pb(N03h given in the problem. FOR PRACTICE 4 .18 What volume (in mL) of a 0.150 mol L- 1 HN0 3 solution will completely react with 35.7 mL of a 0.108 mol L- 1Na2C03 solution according to the following balanced chemical equation?

FOR MORE PRACTICE 4.18 In the reaction above, what mass (in grams) of carbon dioxide forms?

Chapter in Review

139

CHAPTER IN REVIEW Key Terms Section 4.2 chemical reaction (102) combustion reaction (I 02) chemical equation ( 102) reactants ( 103) products ( 103) balanced chemical equation (103)

Section 4.3 solution ( 105) solvent ( 105) solute ( l 05) aqueous solution (I 05) electrolyte ( 106) strong electrolyte ( 106) nonelectrolyte (106)

strong acid ( l 06) weak acid ( 107) weak electrolyte (107) soluble (l 07) insoluble (I 07)

Section 4.4 precipitation reaction ( I 09) precipitate ( l 09) molecular equation (109) complete ionic equation ( I09) spectator ion (110) net ionic equation ( 110)

Section 4.5 acid-base reaction ( 11 1) neutralization reaction ( 111)

Arrhenius definitions ( 111) hydronium ion ( 111 ) polyprotic acid ( 112) diprotic acid ( 112) salt ( 112) gas-evolution reaction (114)

Section 4.6 oxidation-reduction reaction (1 15) redox reaction (115) oxidation ( 116) reduction (116) oxidation state ( 117) oxidizing agent (119) reducing agent ( 119)

disproportionation reaction ( 123)

Section 4.7 stoichiometry (125)

Section 4.8 limiting reactant ( 128) theoretical yield ( 128) actual yield ( 130) percent yield ( 130)

Section 4.9 molarity (M) (132) stock solution ( 136) concentrated solution (136) diluted solution (136)

Key Concepts Writing and Balancing Chemical Equations (4.2)

Oxidation-Reduction Reactions (4.6)

In chemistry, we represent chemical reactions with chemical equations. The substances on the left-hand side of a chemical equation are called the reactants, and the substances on the right-hand side are called the products. Chemical equations are balanced when the number of each type of atom on the left side of the equation is equal to the number on the right side.

In oxidation-reduction reactions, one substance transfers electrons to another substance. The substance that loses electrons is oxidized, and the substance that gains them is reduced. An oxidation state is a fictitious charge given to each atom in a redox reaction by assigning all shared electrons to the atom with the greater attraction for those electrons. Oxidation states are an imposed electronic bookkeeping scheme, not an actual physical state. The oxidation state of an atom increases upon oxidation and decreases upon reduction. A combustion reaction is a specific type of oxidation- reduction reaction in which a substance reacts with oxygen-emitting heat and forming one or more oxygen-containing products.

Aqueous Solutions and Solubility (4.3) An aq ueous solution is a homogeneous mixture of water (the solvent) with another substance (the solute). Solutes that completely dissociate (or completely ionize in the case of acids) to ions in solution are strong electrolytes and are good conductors of electricity. Solutes that only partially dissociate (or partially ionize) are weak electrolytes, and solutes that do not dissociate (or ionize) at all are nonelectrolytes. A substance that dissolves in water to form a solution is soluble. The solubility rules are an empirical set of guidelines that help predict the solubilities of ionic compounds; these rules are especially useful in Section 4.4 when determining whether or not a precipitate will form.

Reaction Stoichiometry (4.7) Reaction stoichiometry refers to the numerical relationships between the reactants and products in a balanced chemical equation. Reaction stoichiometry allows us to predict, for example, the amount of product that can be formed for a given amount of reactant, or how much of one reactant is required to react with a given amount of another.

Limiting Reactant, Theoretical Yield, and Percent Yield (4.8) Precipitation Reactions (4.4) In a precipitation reaction, we mix two aqueous solutions and a solid-or precipitate-forms. Aqueous reactions, such as precipitation reactions, can be represented with a molecular equation, which shows the complete neutral formula for each compound in the reaction. Alternatively, a reaction can be represented with a complete ionic equation, which shows the di ssociated nature of the aqueous ionic compounds. Finally, a third representation is the net ionic equation, in which the spectator ions-those that do not change in the course of the reaction-are left out of the equation.

When a chemical reaction actually occurs, the reactants are usually not present in the exact stoichiometric ratios specified by the balanced chemical equation. The limiting reactant is the one that is available in the smallest stoichiometric quantity-it will be completely consumed in the reaction and it limits the amount of product that can be made. Any reactant that does not limit the amount of product is said to be in excess. The amount of product that can be made from the limiting reactant is the theoretical yield. The actual yield-always equal to or less than the theoretical yield-is the amount of product that is actually made when the reaction is carried out. The percentage of the theoretical yield that is actually produced is the percent yield.

Acid-Base Reactions (4.5) An acid is a substance which produces H+ in solution and a base is a substance which produces Orr in solution. In an acid-base reaction, the acid and base neutralize each other, producing water (or in some cases, a weak electrolyte). In gas-evolution reactions, the acid and base combine in solution and a gas is produced.

Solution Concentration and Solution Stoichiometry (4.9) We often express the concentration of a solution in molarity, the number of moles of solute per litre of solution. We can use the molarities and volumes of reactant solutions to predict the amount of product that will form in an aqueous reaction.

140

Chapter 4

Ch emi cal Re action s and Stoichiometry

Key Equations and Relationships Mass-to-Mass Conversion: Stoichiometry (4.7) mass A -

amount A (in moles) -

amount B (in moles) -

mass B

Percent Yield (4.8) actual yield % yield = h . I . Id x 100% t eoret1ca y1e

Molarity (4.9) . amount of solute (in mol) M = !!._ molarity = volume of solution (in L) or V

Solution Dilution (4.9) Solution Stoichiometry (4.9) volume A ---+ amount A (in moles) --+amount B (in moles) -

volume B

Key Skills Balancing Chemical Equations (4.2) • Examples 4.1 , 4.2 • For Practice 4. 1, 4.2

• Exercises 25-34, 59, 60

Predicting Whether a Compound Is Soluble (4.3) • Example 4.3

• For Practice 4.3

• Exercises 35-38

Writing Equations for Precipitation Reactions (4.4) • Example 4.4

• For Practice 4.4 •For More Practice 4.4 • Exercises 39-42

Writing Equations for Acid-Base Reactions (4.5) • Example 4.5

• For Practice 4.5

•Exercises 49-50

Writing Equations for Gas-Evolution Reactions (4.5) • Example 4.6

•For Practice 4.6

•For More Practice 4.6

•Exercises 5 1-52

Assigning Oxidation States (4.6) • Example 4.7

•For Practice 4.7

•Exercises 53- 56

Identifying Redox Reactions, Oxidizing Agents, and Reducing Agents, and Using Oxidation States (4.6) • Example 4.8

• For Practice 4.8

•Exercises 57- 58

Balancing Redox Reactions in Acidic or Basic Solution; Balancing Disproportionation Reactions (4.6) • Examples 4.9, 4. 10, 4. 11

• For Practice 4.9, 4. 10, 4. 11

• For More Practice 4.9

• Exercises 6 1-68

Calculations Involving the Stoichiometry of a Reaction (4.7) • Example 4.12 • For Practice 4. 12

• For More Practice 4.12 • Exercises 69-80

Determining the Limiting Reactant and Calculating Theoretical and Percent Yields (4.8) • Examples 4.1 3, 4.14

• For Practice 4.13, 4.14

•Exercises 81-96

Calculating Solution Concentration and Using Molarity in Calculations (4.9) • Examples 4.15, 4.1 6

• For Practice 4.15, 4.1 6

• For More Practice 4.15, 4. 16

• Exercises 97-102

Solution Dilutions (4.9) • Example 4. 17

• For Practice 4.17

• For More Practice 4.1 7

• Exercises 103-106

• For More Practice 4.1 8

• Exercises 107-1 10

Solution Stoichiometry (4.9) • Example 4. 18

• For Practice 4.1 8

Exercises

141

EXERCISES Review Questions 1. What is reaction stoichiometry? What is the significance of the

coefficients in a balanced chemical equation? 2. In a chemical reaction, what is the limiting reactant? The theoretical yield? The percent yield? What do we mean when we say a reactant is in excess?

3. The percent yield is normally calculated using the actual yield and theoretical yield in units of mass (g or kg). Would the percent yield be different if the actual yield and theoretical yield were in units of amount (moles)? 4. What is an aqueous solution? What is the difference between the solute and the solvent? 5. What is molarity? How is it useful? 6. Explain how a strong electrolyte, a weak electrolyte, and a nonelectrol yte differ. 7. Explain the difference between a strong acid and a weak acid. 8. What does it mean for a compound to be soluble? Insoluble? 9. What are the solubility rules? How are they useful? 10. What cations and anions have compounds that are usually soluble? What are the exceptions? What anions have compounds that are mostly insoluble? What are the exceptions? 11. What is a precipitation reaction? Give an example.

12. How can you predict whether a precipitation reaction will occur upon mixing two aqueous solutions? 13. Explain how a molecular equation, a complete ionic equation, and a net ionic equation differ. 14. What are the Arrhenius definitions of an acid and a base?

15. What is an acid-base reaction? Give an example. 16. Why can you not change the subscripts on the chemical formulas in order to balance a chemical reaction? 17. What is a gas-evolution reaction? Give an example. 18. What reactant types give rise to gas-evolution reactions? 19. What is an oxidation-reduction reaction? Give an example. 20. What are oxidation states? 21. How can oxidation states be used to identify redox reactions? 22. What happens to a substance when it becomes oxidized? Reduced? 23. In a redox reaction, which reactant is the oxidizing agent? The reducing agent? 24. What is a combustion reaction? Why are they important? Give an example.

Problems by Topic Writing and Balancing Chemical Equations

Q

Sulfuric acid is a component of acid rain formed when gaseous sulfur dioxide pollutant reacts with gaseous oxygen and Liquid water to form aqueous sulfuric acid. Write a balanced chemical equation for this reaction. 26. Nitric acid is a component of acid rain that forms when gaseous nitrogen dioxide pollutant reacts with gaseous oxygen and liquid water to form aqueous nitric acid. Write a balanced chemical equation for this reaction.

'l) In a popular classroom demonstration, solid sodium is added to liquid water and reacts to produce hydrogen gas and aqueous sodium hydroxide. Write a balanced chemical equation for this reaction.

28. When iron rusts, solid iron reacts with gaseous oxygen to form solid iron(III) oxide. Write a balanced chemical equation for this reaction. G) Write a balanced chemical equation for the fermentation of sucrose (C 12H 22 0 11) by yeasts in which the aqueous sugar reacts with water to form aqueous ethyl alcohol (C2 H5 0H) and carbon dioxide gas. 30. Write a balanced equation for the photosynthesis reaction in which gaseous carbon dioxide and liquid water react in the presence of chlorophyll to produce aqueous glucose (C6 H 120 6) and oxygen gas. Write a balanced chemical equation for each reaction: a. Solid lead(Il) sulfide reacts with aqueous hydrobromic acid to form solid lead(Il) bromide and dihydrogen monosulfide gas.

e

b. Gaseous carbon monoxide reacts with hydrogen gas to form

gaseous methane (CH 4) and Liquid water. c. Aqueous hydrochloric acid reacts with solid manganese(IV) oxide to form aqueous manganese(Il) chloride, liquid water, and chlorine gas. d. Liquid pentane (C5H 12) reacts with gaseous oxygen to form carbon dioxide and liquid water. 32. Write a balanced chemical equation for each reaction: a. Solid copper reacts with solid sulfur (S 8) to form solid copper(!) sulfide. b. Solid iron(Ill) oxide reacts with hydrogen gas to form solid iron and liquid water. c. Sulfur dioxide gas reacts with oxygen gas to form sulfur trioxide gas. d. Gaseous ammonia (NH 3) reacts with gaseous oxygen to form gaseous nitrogen monoxide and gaseous water. ~ Balance each chemical equation: a. C0 2(g) + CaSi03(s) + H20(/) Si02(s) + Ca(HC03h(aq) b. Co(N0 3h(aq) + (NH4)iS(aq) Co 2S3(s) + NH 4 N03(aq) Cu(s) + CO(g) c. Cu20(s) + C(s) d. H2(g) + Cl2(g) HCl(g) 34. Balance each chemical equation: a. Na2 S(aq) + Cu(N03h(aq) NaN03(aq) + CuS(s) b. N 2H4 (1) NH 3(g) + N2(g) c. HCl(aq) + 02(g) H10(l ) + Cl2(g) d. FeS(s) + HCl(aq) FeCl 2(aq) + H 2S(g)

142

Chapter 4

Chemical Reac t ions and Stoichiometry

Types of Aqueous Solutions and Solubility

0

For each compound (all water soluble), would you expect the resulting aqueous solution to conduct electrical current? a. CsCl b. CH 30H c. Ca(N02)z d. C6H 12 06 36. Classify each compound as a strong electrolyte or nonelectrolyte: a. MgBr2 b. C12H220 11 c. Na2C03 d. KOH G) Determine whether each compound is soluble or insoluble. If the compound is soluble, list the ions present in solution . a . AgN03 b. Pb(C2H30 2h c. KN03 d. (NH4)2S

38. Determine whether each compound is soluble or insoluble. For the soluble compounds, list the ions present in solution. a. AgI b. Cu3(P04)z c. CoC0 3 d. K3P04

show the reaction of aqueous Hg2(N0 3)z with aqueous sodium chloride to form solid Hg2Cl2 and aqueous sodium nitrate. 46. Lead ions can be removed from solution by precipitation with sulfate ions. Suppose that a solution contains lead(Il) nitrate. Write complete ionic and net ionic equations to show the reaction of aqueous lead(Il) nitrate with aqueous potassium sulfate to form solid lead(Il) sulfate and aqueous potassium nitrate.

$

48. Write balanced molecular and net ionic equations for the reaction between nitric acid and calcium hydroxide.

G)

Precipitation Reactions

G) Complete

and balance each equation. If no reaction occurs, write NO REACTION.

a. b. c. d.

LiI(aq) + BaS(aq) ------> KC1(aq) + CaS(aq) ------> CrBri(aq) + Na2C03(aq) ------> NaOH(aq) + FeC13(aq) ------>

40. Complete and balance each equation. If no reaction occurs,

write NO REACTION. a. NaN0 3(aq) + KCl(aq) ------> b. NaCl(aq) + Hg2(C2H30 2h(aq) ------> c. (NH4 )zS0 4 (aq) + SrClz(aq) ------> d. NH4Cl(aq) + AgN0 3(aq) ------>

Q

Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write NO REACTION. a. potassium carbonate and lead(Il) nitrate b. lithium sulfate and lead(Il) acetate c. copper(Il) nitrate and magnesium sulfide d. strontium nitrate and potassium iodide

42. Write a molecular equation for the precipitation reaction that occurs (if any) when each pair of aqueous solutions is mixed. If no reaction occurs, write NO REACTION. a. sodium chloride and lead(Il) acetate b. potassium sulfate and strontium iodide c. cesium chloride and calcium sulfide d. chromium(III) nitrate and sodium phosphate

(D Write balanced complete ionic and net ionic equations for each reaction:

a. b. c. d.

HCl(aq) + LiOH(aq) ------> H 20(1) + LiCl(aq) MgS(aq) + CuCl2(aq) ------> CuS(s) + MgC12(aq) NaOH(aq) + HN0 3(aq) ------> H 20(1) + NaN0 3 (aq)

Na3P04(aq) + NiClz(aq) ------> Ni 3(P04h(s) + NaCl(aq) 44. Write balanced complete ionic and net ionic equations for each reaction: a. K 2S04(aq) + CaI2(aq) ------> CaS04(s) + KI(aq) b. NH4Cl(aq) + NaOH(aq) ------> H20(/) + NH 3(g) + N aCl(aq) c. AgN0 3(aq) + NaCl(aq) ------> AgCl(s) + NaN0 3(aq) d. HC2Hp2(aq) + K2C03(aq) ------> H 20 (1) + C02(g) + KC2H30 2(aq) Dimercury (I+) ions (Hg22+) can be removed from solution by precipitation with Cl - . Suppose that a solution contains aqueous Hg2(N03)z. Write complete ionic and net ionic equations to

CD

Complete and balance each acid-base equation : a. H 2 S04 (aq) + Ca(OH)z(aq) ------> b. HCI04(aq) + KOH(aq) ------> c. H 2S04 (aq) + NaOH(aq) ------>

50. Complete and balance each acid-base equation: a. HJ(aq) + LiOH(aq) ------> b. HC 2H30 z(aq) + Ca(OH) 2(aq) ------> c. HC1(aq) + Ba(OH)z(aq) ------>

0

Complete and balance each gas-evolution equation: a. HBr(aq) + NiS(s) ------>

b. NH4I(aq) + NaOH(aq) ------> c. HBr(aq) + Na2S(aq) ------> d. HC104(aq) + Li2C03(aq) ------> 52. Complete and balance each gas-evolution equation: a. HN03(aq) + Na2S0 3(aq) ------> b. HC1(aq) + KHC0 3(aq) ------> c. HC 2H30 2(aq) + NaHS03(aq) ------> d. (NH4 )iSOiaq) + Ca(OH)z(aq) ------>

Oxidation-Reduction and Combustion

9

Assign oxidation states to each atom in each element, ion, or compound: a. Ag d. H 2S

54. Assign oxidation states to each atom in each element, ion, or compound: a. Cl 2 b. Fe3+ c. CuC1 2 d. CH4 f. HS04e. Cr20 iWhat is the oxidation state of Cr in each compound? a. CrO b. Cr03 c. Cr20 3 56. What is the oxidation state of Cl in each ion? a. CJOb. Cl02c. Cl03f) Determine whether each reaction is a redox reaction. For each redox reaction, identify the oxidizing agent and the reducing agent. a. 4 Li(s) + 0 2(g) ------> 2 Li20(s) b. Mg(s) + Fe2+(aq) ------> Mg2+(aq) + Fe(s) c. Pb(N0 3h(aq) + Na2S04(aq) ------> PbS04(s) + 2 NaN03(aq) d. HBr(aq) + KOH(aq) ------> H 20 (/) + KBr(aq) 58. Determine whether each reaction is a redox reaction. For each redox reaction, identify the oxidizing agent and the reducing agent. a. Al(s) + 3 Ag+(aq) ------> A13+(aq) + 3 Ag(s) b. S03(g) + H20 (I) ------> H2S04(aq) c. Ba(s) + Cl2(g) ------> BaCiz(s) d. Mg(s) + Br2(1) ------> MgBr2(s)

•

Ionic and Net Ionic Equations

Write balanced molecular and net ionic equations for the reaction between hydrobromic acid and potassium hydroxide.

Exercises

0

Complete and balance each combustion reaction equation: a. S(s) + Oz(g) b. C3H6(g) + Oz(g) c. Ca(s) + 0 2(g) d. CsH12S(/) + 0 2(g) 60. Complete and balance each combustion reaction equation: a. C4H6(g) + 0 2(g) b. C(s) + 02(g) c. CS2(s) + Oz(g) d. C3HsO(/) + 02(g) -

fD

143

Balance the equation and determine how many moles of Ba(OH)z are required to completely neutralize 0.461 mol CH 3COOH. Calculate how many moles of N02 form when each quantity of reactant completely reacts:

a. 2.S mol N20 5 c. lS .2 g NzOs

b. 6.8 mol NzOs d. 2.87 kg Nz0 5 72. Calculate how many moles of NH3 form when each quantity of reactant completely reacts:

Balancing Redox Reactions ~ Balance each redox reaction occurring in acidic aqueous solution. Cr(s) + K +(aq) a. K(s) + cr3+(aq) A l3+(aq) + Fe(s) b. Al(s) + Fe2+(aq) Br- (aq) + Nz(g) c. Br03- (aq) + N2Hig) -

62. Balance each redox reaction occurring in acidic aqueous solution. Zn2+(aq) + Sn(s) a. Zn(s) + Snz+ (aq) Mgz+ (aq) + Cr(s) b. Mg(s) + Cr3+(aq) Mn2+(aq) + Al3+ (aq) c. Mn04- (aq) + Al(s) BaCl2(s) b. CaO(s) + C02(g) ---> CaC0 3(s) c. 2 Mg(s) + 0 2(g) ---> 2 MgO(s) d. 4 Al(s) + 3 0 2(g) ---> 2 AI20 3(s) 78. For each of the reactions, calculate the mass (in grams) of the product formed when 15.39 g of the underlined reactant completely reacts. Assume that there is more than enough of the other reactant. a. 2 K(s) + Cl 2(g) ---> 2 KCJ(s) b. 2 K(s) + Br2(l) ---> 2 KBr(s) c. 4 Cr(s) + 3 Oz(g) ---> 2 Cr20 3(s) d. 2 Sr(s) + 0 2(g) ---> 2 SrO(s) G) For each of the acid-base reactions, calculate the mass (in grams) of each acid necessary to completely react with and neutralize 4.85 g of the base. a. HCl(aq) + NaOH(aq) ---> H20 (l) + NaCl(aq) b. 2 HN03(aq) + Ca(OH)i(aq) --->

~ 84. Consider the reaction:

fl

c. H2S04 (aq) + 2 KOH(aq)

--->

2 H2 0 (l ) 2 H20 (l)

+ Ca(N03h(aq) + K 2S04 (aq)

80. For each precipitation reaction, calculate how many grams of the first reactant are necessary to completely react with 55.8 g of the second reactant. a. 2 Kl(aq) + Pb(N0 3h(aq) ---> Pbl2(s) + 2 KN0 3(aq) b. Na2C03(aq) + CuCl2(aq) ---> CuC03(s) + 2 NaCl(aq) c. K 2S04(aq) + Sr(N0 3h(aq) ---> SrS04(s) + 2 KN03(aq)

Limiting Reactant, Theoretical Yield, and Percent Yield

fJD

For the reaction, fi nd the limiting reactant for each of the initial amounts of reactants. 2Na(s)

+ Br2(g)

--->

2 NaBr(s)

2 CH30H(g)

4 Al(s)

e

a. b. c. d.

+ 3 0 2(g)

--->

2 Al 20 3(s)

I mol Al, I mol 0 2

2 COz(g)

+ 4 H20 (g)

(b)

(a)

(c)

~ For the reaction, compute the theoretical yield of the product (in moles) for each initial amount of reactants. Ti(s)

+ 2 CJ 2(g)

--->

TiC14(s)

a. 4 mol Ti, 4 mol Cl 2 b. 7 mol Ti, 17 mol Cl2 c. 12.4 mol Ti, 18.8 mol Cl2 86. For the reaction, compute the theoretical yield of product (in moles) for each initial amount of reactants. 2 Mn(s)

+ 2 0 2(g)

--->

2 Mn02(s)

a. 3 mol Mn, 3 mol 0 2 b. 4 mol Mn, 7 mol 0 2 c. 27 .5 mol Mn, 43 .8 mol 0 2

Q)

Zinc sulfide reacts with oxygen according to the reaction : 2 ZnS(s)

+ 3 0 2(g)

--->

2 ZnO(s)

+ 2 S02(g)

A reaction mixture initially contains 4.2 mol ZnS and 6.8 mol 0 2. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left? 88. Iron(II) sulfide reacts with hydrochloric acid according to the reaction: FeS(s)

E:D

+ 2 HCl(aq)

--->

FeC12 (s)

+ H2 S(g)

A reaction mixture initially contains 0.223 mol FeS and 0.652 mol HCI. Once the reaction has occurred as completely as possible, what amount (in moles) of the excess reactant is left? For the reaction, compute the theoretical yield of product (in grams) for each initial amount of reactants. 2 Al(s)

+ 3 CJ2(g)

--->

2 AICl 3(s)

90. For the reaction, compute the theoretical yield of the product (in grams) for each initial amount of reactants.

Consider the reaction:

+ 0 2(g)

--->

a. 2 .0 g Al, 2 .0 g Cl2 b. 7 .5 g Al, 24 .8 g Cl2 c. 0.235 g Al, 1.15 g Cl2

4 mol Al, 2.6 mol 0 2 16 mol Al, 13 mol 0 2 7 .4 mol Al, 6 .5 mol 0 2 4 HCl(g)

+ 3 0 2(g)

Each of the molecular diagrams represents an initial mixture of the reactants. How many C02 molecules would be formed from the reaction mixture that produces the greatest amount of products?

a. b. c. d.

2 mol Na, 2 mol Br2 I .8 mol Na, I .4 mol Br2 2 .5 mol Na, 1 mol Br2 12 .6 mol Na, 6.9 mol Br2 82. For the reaction, find the limiting reactant for each initial amount of reactants.

(c)

(a)

--->

2 H 2 0 (g)

+ 2 Cl2(g)

Each molecular diagram represents an initial mixture of the reactants. How many molecules of Cl2 would be formed from the reaction mixture that produces the greatest amount of products?

Ti(s)

+ 2 F2 (g)

a. 5 .0 g Ti, 5 .0 g Fz b. 2 .4 g Ti, I .6 g Fz c. 0.233 g Ti, 0.288 g Fz

--->

TiF4(s)

145

Exercises

~ Iron(III) oxide reacts with carbon monoxide according to the equation:

A reaction mixture initially contains 22.55 g Fei0 3 and 14.78 g CO. Once the reaction has occurred as completely as possible, what mass (in grams) of the excess reactant is left?

98. Calculate the molarity of each solution: a. 0.38 mol LiN0 3 in 6.14 L of solution b. 72.8 g C 2H 6 0 in 2.34 L of solution c. 12.87 mg Kl in 112.4 mL of solution

G)

92. Elemental phosphorus reacts with chlorine gas according to the equation:

A reaction mixture initially contains 45.69 g P 4 and 13 l.3 g CI2 . Once the reaction has occurred as completely as possible, what mass (in grams) of the excess reactant is left?

100. What volume of a 0.200 mol L - I ethanol solution contains each amount in moles of ethanol? a. 0.45 mol ethanol b. 1.22 mol ethanol c. 1.2 X 10- 2 mol ethanol

49 A

laboratory procedure calls for making 400.0 mL of a l. l mol L - I NaN0 3 solution. What mass of NaN0 3 (in grams) is needed?

~ Lead ions can be precipitated from solution with KC! according to the reaction: Pb 2+(aq)

+ 2 K CI(aq)

-->

PbCI2(s)

+ 2 K +(aq)

When 28.5 g KC! is added to a solution containing 25.7 g Pb2 + , a PbCI2 precipitate forms. The precipitate is filtered and dried and found to have a mass of 29 .4 g. Determine the limiting reactant, theoretical yield of PbCl2 , and percent yield for the reaction.

94. Magnesium oxide can be made by heating magnesium metal in the presence of oxygen. The balanced equation for the reaction is: 2 Mg(s)

+

0 2 (g)

-->

2 MgO(s)

When l O. l g Mg is allowed to react with 10.5 g 0 2, l l.9 g MgO is collected. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

How many moles of KCl are contained in each solution?

a. 0.556 L of a 2.3 mol L - I KC! solution b. l .8 L of a 0.85 mol L - I KCl solution c. 114 mL of a l.85 mol L - I KC I solution

102. A chemist wants to make 5.5 L of a 0.300 mol L- 1 CaCl 2 solution. What mass of CaCl2 (in grams) should the chemist use?

G If 123 mL of a

1.1 mol L- 1 glucose solution is diluted to 500.0 mL, what is the molarity of the diluted solution? 104. If 3.5 L of a 4 .8 mol L- 1 SrC12 solution is diluted to 45 L, what is the molari ty of the diluted solution?

GI) To what volume should you dilute 50.0 mL of a

12 mol L - I stock HN0 3 solution to obtain a 0.100 mol L- 1 HN03 solution?

106. To what volume should you dilute 25 mL of a 10.0 mol L- 1 H2S0 4 solution to obtain a 0. 150 mol L- 1 H 2S0 4 solution?

G Consider the precipitation reaction:

~ Urea (CH4 N20 ) is a common fertilizer that can be synthesized

2 Na 3P04(aq)

+ 3 CuCl2(aq)

-->

Cu3(P0 4 h(s)

by the reaction of ammonia (NH3) with carbon dioxide:

In an industrial synthesis of urea, a chemist combines 136.4 kg of ammonia with 2 l l .4 kg of carbon dioxide and obtains l 68.4 kg of urea. Determine the limiting reactant, theoretical yield of urea, and percent yield for the reaction. 96. Many compu ter chips are manufactured from silicon, which occurs in nature as Si02 . When Si02 is heated to melting, it reacts with solid carbon to form liquid silicon and carbon monoxide gas. In an industrial preparation of silicon, 155.8 kg of Si0 2 reacts with 78.3 kg of carbon to produce 66. l kg of silicon. Determine the limiting reactant, theoretical yield, and percent yield for the reaction.

Solution Concentration and Solution Stoichiometry

G Calculate the molarity of each solution:

+ 6 NaCI(aq)

What volume of a 0. 175 mol L- t Na3P04 solution is necessary to completely react with 95.4 mL of 0. 102 mol L- 1 CuCI2 ? 108. Co nsider the reaction: Li2S(aq)

+

Co(N0 3 h(aq)

-->

2 LiN0 3(aq)

+

CoS(s)

What volume of a 0.150 mol L- t Li2 S solution is required to completely react with 125 mL of 0. l 50 mol L- I Co(N0 3)z?

G) What is the minimum amount of 6.0 mol L-

1 H 2S0 4 necessary to produce 25.0 g of H2 (g) according to the reaction between aluminum and sulfuric acid?

110. What is the molarity of ZnCl 2 that forms when 25.0 g of zinc completely reacts with CuCl 2 according to the following reaction? Assume a final volume of 275 mL.

a. 3.25 mol LiCI in 2.78 L of solution b. 28 .33 g C 6 H 120 6 in I .28 L of solution c. 32.4 mg NaCl in 122.4 mL of solution

Zn(s)

+ CuC12(aq)

-->

ZnCl2 (aq)

+

Cu(s)

Cumulative Problems

GD The density of a 20.0%-by-mass ethylene glycol (C H 0 tion in water is l.03 g mL-

I.

2 6 2 ) soluFind the molarity of the solution.

112. Find the percent by mass of sodium chloride in a l.35 mol L- 1 NaCl solution. The density of the solution is l.05 g mL- 1.

@

People often use sodium bicarbonate as an antacid to neutralize excess hydrochloric acid in an upset stomach. What mass of hydrochloric acid (in grams) can 2.5 g of sodium bicarbonate neutralize? (Hint: Begin by writing a balanced equation for

146

Chapter 4

Ch emical Reac t ion s and Sto ichiometry

the reaction between aqueous sodium bicarbonate and aq ueous hydrochloric acid.) 114. Toilet bowl cleaners often contain hydrochloric acid, which dissolves the calcium carbonate deposits that accumulate within a toilet bowl. What mass of calcium carbonate (in grams) can 3.8 g of HCI dissolve? (Hint: Begin by writing a balanced equation for the reaction between hydrochloric acid and calcium carbonate.)

120. A hydrochloric acid solution will neutralize a sodium hydroxide solution. Look at the molecular views showing one beaker of HCI and four beakers of NaOH. Which NaOH beaker will j ust neutralize the HC I beaker? Begin by writing a balanced chemical eq uation for the neutralization reaction.

G The combustion of gasoline produces carbon dioxide and water.

Assume gasoline to be pure octane (C 8H 18) and calculate the mass (in kilograms) of carbo n dioxide that is added to the atmosphere per 1.0 kg of octane burned. (Hint: Begin by writing a balanced equation for the combustio n reaction.) 116. Many home barbeques are fuelled with propane gas (C 3H8) . What mass of carbon dioxide (in kilograms) is produced upon the complete combustion of 18.9 L of propane (approximate contents of one 5-gallon tank)? Assume that the density of the liquid propane in the tank is 0.621 g mL- 1. (Hint: Begin by writing a balanced equation for the combustion reaction.)

G Aspirin can be made in the laboratory by reacting acetic anhy-

dride (C4H60 3) with salicylic acid (C7H 60 3) to form aspirin (C9Hg0 4) and acetic acid (C2H40 2). The balanced equation is: C4H60 3

+ C7 H60 3 ---->

C9Hs0 4 + C2H40 2

In a laboratory synthesis, a student begins with 3.00 mL of acetic anhydride (density = 1 .08 g mL- 1) and 1.25 g of salicylic acid. Once the reaction is complete, the student collects 1.22 g of aspirin. Determine the limiting reactant, theoretical yield of aspirin, and percent yield for the reaction. 118. The combustion of liquid ethanol (C2H 50H) produces carbon dioxide and water. After 4.62 mL of ethanol (density = 0 .789 g mL- 1) was allowed to burn in the presence of 15.55 g of oxygen gas, 3.72 mL of water (density = I .00 g mL- 1) was collected. Determine the limiting reactant, theoretical yield of H 20 , and percent yield for the reaction. (Hint: W rite a balanced equation for the combustio n of ethanol.)

fl) A loud classroom demonstration involves igniti ng a hydrogenfilled balloon. The hydrogen within the balloon reacts explosively with oxygen in the air to form water. If the balloon is filled with a mixture of hydrogen and oxygen, the explosion is even lo uder than if the balloon is filled only with hydrogen-the intensity of the explosion depends on the relative amounts of oxygen and hydrogen within the balloon. Look at the molecular views representing different amounts of hyd rogen and oxygen in four different ball oons. Based on the balanced chemical equation, which balloon will make the lo udest explosio n?

(a)

(b)

(c)

(d)

9

Predict the products and write a balanced molecular equation for each reaction. If no reaction occurs, write NO REACTION. a. H Cl(aq) + Hgz(N0 3)2(aq) ----> b. KHS03(aq) + HN0 3(aq) ----> c. aqueous ammonium chloride and aqueous lead(II) nitrate d . aqueo us amm oni um chloride and aq ueous ca lc ium hydroxide 122. Predict the products and write a balanced molecular equation for each reaction. If no reaction occurs, write NO REACTION. a. H 2S0 4 (aq) + HN0 3(aq) ----> b. Cr(N03)J(aq) + LiOH(aq) ----> c. liquid pentanol (C5 H 120 ) and gaseous oxygen d . aqueous strontium sulfide and aqueous copper(II) sulfate Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms insoluble precipitates with calcium and magnesium ions, removing them from solution. Suppose that a solution is 0.050 mol L- 1 in calcium chloride and 0.085 mol L- 1 in magnesium nitrate. What mass of sodium phosphate would have to be added to 1.5 L of this solution to completely eliminate the hard water ions? Assume complete reaction. 124. An acid solution is 0. 100 mol L- 1 in HCI and 0.200 mol L- 1 in H2S0 4. What volume of a 0. 150 mol L- 1 KOH solution would completely neutralize all the acid in 500.0 mL of this solution?

f»

G Find the mass of barium metal (in grams) that must react with

0 2 to produce enough barium oxide to prepare 1.0 L of a 0.1 0 mol L - l solution of OH - . 126. A solution contains cr3+ ions and Mg 2+ ions. The addition of 1.00 L of 1.5 1 mol L- 1 NaF solution is requi red to cause the complete precipitation of these ions as CrF3(s) and MgF 2(s). The total mass of the precipitate is 49.6 g. Find the mass of Cr3+ in the original solution.

G The nitrogen in sodium nitrate and in ammonium sulfate is

available to plants as fertilizer. Which is the more economical source of nitrogen, a fertilizer containing 30.0% sodium nitrate by weight and costing $ 18 per I 00 kg or one containing 20.0% ammonium sulfate by weight and costing $ 16.20 per 100 kg? 128. Find the volume of 0.1 10 mol L- l of hydrochloric acid necessary to react completely with 1.52 g Al(OH)J. (a)

(b)

(c)

(d)

fJ) Treatment of gold metal with BrF3 and KF produces Br2 and KAuF4, a salt of gold. Identify the oxidizing agent and the reducing agent in this reaction. Find the mass of the gold salt that forms when a 73.5 g mixture of equal masses of all three reactants is prepared. 130. We prepare a solution by mixing 0. 10 L of0.12 mol L- l sodium chloride with 0.23 L of a 0. 18 mol L - l MgCl2 solution. What volume of a 0 .20 mol L- 1 silver nitrate solution do we need to precipitate all the Cl- ions in the solution as AgCl?

Exercises

GD A solution contains one or more of the following ions: Ag+, Caz+, and Cuz+. When you add sodium chloride to the solution, no precipitate forms. When you add sodium sulfate to the solution, a white precipitate forms. You filter off the precipitate and add sodium carbonate to the remaining solution, producing another precipitate. Which ions were present in the original solution? Write net ionic equations for the formation of each of the precipitates observed. 132. A solution contains one or more of the following ions: Hg/+, Baz+, and Fez+. When potassium chloride is added to the solution, a precipitate forms. The precipitate is fi ltered off, and potassium sulfate is added to the remaining solution, producing no precipitate. When potassium carbonate is added to the remaining solution, a precipitate forms. Which ions were present in the original solution? Write net ionic equations for the formation of each of the precipitates observed.

@

147

134. An important reaction that takes place in a blast furnace during the production of iron is the formation of iron metal and COz from Fez03 and CO. Find the mass of Fez03 required to form 910 kg of iron. Find the amount of COz that forms in this process.

0

A liquid fuel mixture contains 30.35% hexane (C6 H 14), 15.85% heptane (C7H 16), and the rest octane (C8H 18) . What maximum mass of carbon dioxide is produced by the complete combustion of 10.0 kg of this fuel mixture? 136. Titanium occurs in the magnetic mineral ilmenite (FeTi03), which is often fou nd mixed with sand. The ilmenite can be separated from the sand with magnets. The titanium can then be extracted from the ilmenite by the following set of reactions: FeTi03(s)

+ 3 Clz(g) + 3 C(s) - -

TiC14 (/)

The reaction of NH3 and Oz forms NO and water. The NO can be used to convert P4 to P40 6 , forming Nz in the process. The P40 6 can be treated with water to form H3P0 3, which forms PH3 and H3P04 when heated. Find the mass of PH3 that forms from the reaction of 1.00 g NH 3.

3 CO(g) + FeClz(s) + TiC14 (g) + 2 Mg(s) --- 2 MgClz(/) + Ti(s)

Suppose that an ilmenite-sand mixture contains 22.8% ilmenite by mass and that the first reaction is carried out with a 90.8% yield. If the second reaction is carried out with an 85.9% yield, what mass of titanium can be obtained from 1.00 kg of the ilmenite-sand mixture?

Challenge Problems

G A mixture of C H

3 8 and CzHz has a mass of 2.0 g. It is burned in excess Oz to form a mixture of water and carbon dioxide that contains 1.5 times as many moles of COz as of water. Find the mass of CzHz in the original mixture. 138. A mixture of 20.6 g P and 79.4 g Clz reacts completely to form PCl3 and PC15 as the only products. Find the mass of PC1 3 formed.

0

A solution contains Ag+ and Hgz+ ions. The addition of0.100 L of 1.22 mol L- 1 Nal solution is j ust enough to precipitate all the ions as Agl and Hglz. The total mass of the precipitate is 28. 1 g. Find the mass of Agl in the precipitate.

140. Lakes that have been acidified by acid rain (HN0 3 and HzS04 ) can be neutralized by a process called liming, in which limestone (CaC0 3) is added to the acidified water. What mass of limestone (in kg) would completely neutralize a 15.2 billion-litre lake that is 1.8 X l0- 5 mol L- 1 in HzS04 and 8.7 X 10- 6 mol L- 1 in HN03?

0

0

determine the appropriate dose for succimer treatment of lead poisoning. What minimum mass of succimer (in milligrams) is needed to bind all of the lead in a patient's bloodstream? Assume that patient blood lead levels are 45 µg dL- 1, that total blood volume is 5.0 L, and that one mole of succimer binds one mole of lead. A particular kind of emergency breathing apparatus-often placed in mines, caves, or other places where oxygen might become depleted or where the air might become poisonedworks via the following chemical reaction: 4 KOz(s)

+ 2 COz(g) --- 2 KzC03(s) + 3 Oz(g)

Notice that the reaction produces Oz, which can be breathed, and absorbs COz, a product of respiration. Suppose you work for a company interested in producing a self-rescue breathing apparatus (based on the above reaction) that would allow the user to survive for 10 minutes in an emergency situation. What are the important chemical considerations in designing such a unit? Estimate how much KOz would be required for the apparatus. (Find any necessary additional information-such as human breathing rates-from appropriate sources. Assume that normal air is 20% oxygen.)

We learned in Section 4.4 that sodium carbonate is often added to laundry detergents to soften hard water and make the detergent more effective. Suppose that a particular detergent mixture is designed to soften hard water that is 3.5 X 10- 3 mol L- 1 in Caz+ and 1.1 X 10- 3mol L - I in Mgz+ and that the average capacity of a washing machine is 75 L of water. If the detergent requires using 0.65 kg detergent per load of laundry, determine what percentage (by mass) of the detergent should be sodium carbonate in order to completely precipitate all of the calcium and magnesium ions in an average load of laundry water.

144. Metallic aluminum reacts with MnOz at elevated temperatures to form manganese metal and aluminum oxide. A mixture of the two reactants is 67.2% mole percent Al. Find the theoretical yield (in grams) of manganese from the reaction of 250 g of this mixture.

142. Lead poisoning is a serious condition resulting from the ingestion of lead in food, water, or other environmental sources. It affects the central nervous system, leading to a variety of symptoms such as distractibility, lethargy, and loss of motor coordination. Lead poisoning is treated with chelating agents, substances that bind to metal ions, allowing it to be eliminated in the urine. A modem chelating agent used for this purpose is succimer (C4H6 0 4 Sz). Suppose you are trying to

5 9 forms boric acid, H 3B03. Fusion of boric acid with sodium oxide forms a borate salt, NazB4 0 7 . Without writing complete equations, find the mass (in grams) of B5H9 required to form 151 g of the borate salt by this reaction sequence. 146. A mixture of carbon and sulfur has a mass of 9.0 g. Complete combustion with excess Oz gives 23.3 g of a mixture of COz and SOz. Find the mass of sulfur in the original mixture.

G Hydrolysis of the compound B H

148

Cha pter 4

Chem ical React ions and Stoichiometry

Conceptual Problems

G Consider the reaction: 4 K(s)

+

150. Consider the reaction: 0 2 (g) -

2 K 20(s)

The molar mass of K is 39. 10 g mol- 1 and that of 0 2 is 32.00 g mol- 1• Without doing any extensive calculations, pick the conditions under which potassium is the limiting reactant and explain your reasoning. a. 170 g K , 3 1 g 0 2 b. 16 g K, 2.5 g 0 2 c. 165 kg K, 28 kg 02 d. 1.5 g K, 0.38 g 02

Consider also this representation of an initial mixture of N 2H4 and N 20 4 :

•

•

148. Consider the reaction: 2 NO(g)

+

5 H 2(g) -

2 NH3(g)

+

2 H20(g)

A reaction mixture initially contains 5 moles of NO and 10 moles of H 2. Without doing any calculations, determine which set of amounts best represents the mixture after the reactants have reacted as completely as possible. Explain your reasoning. a. l mol NO, 0 mol H 2 , 4 mol NH3, 4 mol H 20 b. 0 mol NO, I mol H2 , 5 mol NH3, 5 mol H 20 c. 3 mol NO, 5 mol H 2 , 2 mol NH3, 2 mol H 20 d. 0 mol NO, 0 mol H 2 , 4 mol NH3, 4 mol H 20

Which diagram best represents the reaction mixture after the reactants have reacted as completely as possible?

G) The circle below represents 1.0 L of a solution with a solute concentration of 1 mol L- 1:

(a)

., Explain what you would add (the amount of solute or volume of solvent) to the solution to obtain a solution represented by each diagram:

(a)

(b)

Q•• (c)

. ..

, .,

9

•

~

9

.,

.

••

~(c) ·

•

•

• e The buildup of pressure, which results from the constant collisions of gas molecules with the suifaces around them, expels the cork in a bottle of champagne. (Quade Paul/Pearson Education]

W

E CAN SURVIVE FOR WEEKS without food , days without water, but only minutes without air. Fortunately, we live at the bottom of a vast ocean of air, held to Earth by gravity. We inhale a lungful of this air

every few seconds, keep some of the molecules for our own needs, add some molecules our bodies no longer have a use for, and exhale the mixture back into the surrounding air. The air around us is matter in the gaseous state. What are the fundamental properties of these gases? What laws describe their behaviour? What theory explains these properties and laws? The scientific method proceeds in this way-from observations to laws to theories-exactly the way we will proceed in this chapter. The gaseous state is the simplest and best-understood state of matter. In this chapter, we examine that state.

5.1

Breathing: Putting Pressure to Work

Every day, without even thinking about it, you move approximately 8500 litres of air into and out of your lungs. The total weight of this air is about 25 kilograms. How do you do it? The simple answer is pressure. You rely on your body's ability to create pressure differences to move air into and out of your lungs. Pressure is the force exerted per unit area by gas molecules as they strike the surfaces around

5.1 Breathing: Putting Pressure to Work 149 5.2 Pressure: The Result of Molecular Collisions 150 5.3 The Gas Laws: Boyle's Law, Charles's Law, and Avogadro's Law 154 5.4 The Ideal Gas Law 160 5.5 Applications of the Ideal Gas Law: Molar Volume, Density, and Molar Mass of a Gas 162 5.6 Mixtures of Gases and Partial Pressures 166 5.7 Gases in Chemical Reactions: Stoichiometry Revisited 111 5.8 Kinetic Molecular Theory: A Model for Gases 174 5.9 Mean Free Path, Diffusion, and Effusion of Gases 180 5.10 Real Gases: The Effects of Size and Intermolecular Forces 181

149

150

Chapter 5

Gases

Gas molecules

• Collision s with surfaces create pressure.

• FIGURE 5.1 Gas Pressure Pressure is the force per unit area exerted by gas molecules colliding with the surfaces around them.

them (Figure 5.1 ... ). Just as a ball exerts a force when it bounces against a wall, a gaseous atom or molecule exerts a force when it collides with a surface. The sum of all these molecular collisions is pressure-a constant force on the surfaces exposed to any gas. The total pressure exerted by a gas depends on several factors, including the concentration of gas molecules in the sample-the higher the concentration, the greater the pressure. When you inhale, the muscles that surround your chest cavity expand the volume of your lungs. The expanded volume results in a lower concentration of gas molecules (the number of molecules does not change, but since the volume increases, the concentration goes down). This in tum results in fewer molecular collisions, which results in lower pressure. The external pressure (the pressure outside of your lungs) remains relatively constant and is now higher than the pressure within your lungs. As a result, gaseous molecules flow into your lungs, from the region of higher pressure to the region of lower pressure. When you exhale, the process is reversed. The chest cavity muscles relax, which decreases the lung volume, increasing the pressure within the lungs and forcing air back out. In this way, within the course of a normal human lifetime, you will take about half a billion breaths and move about 250 million litres of air through your lungs. With each breath, you create pressure differences that allow you to obtain the oxygen that you need to live.

5.2

Pressure: The Result of Molecular Collisions

Air can hold up a jumbo jet or knock down a building. How? As we just discussed, air contains gaseous atoms and molecules in constant motion. The particles collide with each other and with the surfaces around them. Each collision exerts only a small force, but when the forces of the many particles are summed, they quickly add up. The result of the constant collisions between the atoms or molecules in a gas and the surfaces around them is pressure. Because of pressure, we can drink from straws, inflate basketballs, and move air into and out of our lungs. Variation in pressure in Earth's atmosphere creates wind, and changes in pressure help us to predict weather. Pressure is all around us and even inside of us . The pressure that a gas sample exerts is the force per unit area that results from the collisions of gas particles with the surrounding surfaces: Pressure

Lower pressure

force area

F

= -- = -

A

[5.1)

Higher pressure

• FIGURE 5.2 Pressure and Particle Density A low density of gas particles results in low pressure. A high density of gas particles results in high pressure.

• Pressure variations in Earth's atmosphere create wind and weather. The H's in this map indicate regions of high pressure, usually associated with clear weather. The L's indicate regions of low pressure, usually associated with unstable weather. The map shows a typhoon off the northeast coast of Japan. The isobars, or lines of constant pressure, are labelled in hectopascals ( 100 Pa).

The pressure of a gas sample depends on several factors, including the number of gas particles in a given volume-the fewer the gas particles, the lower the pressure (Figure 5 .2 ... ). Pressure decreases with increasing altitude because there are fewer molecules per unit

5.2

Pressure: The Result of Molecular Collisions

151

CF3 + Cl Cl+ 0 3 ----> CIO + 0 2 0

fD The zinc within a copper-plated penny will dissolve in hydrochloric acid if the copper coating is filed down in several spots (so that the hydrochloric acid can get to the zinc). The reaction between the acid and the zinc is 2 H +(aq) + Zn(s) ----> H 2(g) + Zn2 +(aq). When the zinc in a certain penny dissolves, the total volume of gas collected over water at 25 °C was 0.95 1 L at a total pressure of 0.990 bar. What mass of hydrogen gas was collected? 72. A heliox deep-sea diving mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.5 bar?

What mass of H2 0 is required to form 1.4 L of 0 2 at a temperature of 3 15 Kand a pressure of 0.957 bar?

i)

CH 30 H can be synthesized by the reaction: CO(g) + 2 H2 (g) ----> CH3 0 H(g)

What volume of H 2 gas (in L), at 748 Torr and 86 °C, is required to synthesize 25.8 g CH30 H? How many litres of CO gas, measured under the same conditions, are required? 76. Oxygen gas reacts with powdered aluminum according to the reaction:

What volume of 0 2 gas (in L), measured at 782 Torr and 25 °C, completely reacts with 53.2 g Al?

G)

Automobile air bags inflate following a serious impact. The impact triggers the chemical reaction:

If an automobile air bag has a volume of 11.8 L, what mass of NaN 3 (in g) is required to fully inflate the air bag upon impact? Assume STP conditions. 78. Lithium reacts with nitrogen gas accordi ng to the reaction: 6 Li(s) + N2 (g) ----> 2 Li 3N(s) What mass of lithium (in g) reacts completely with 58.5 mL of N 2 gas at STP?

+ UV light ----> 0 2 + 0

What total volume of ozone at a pressure of 25.0 Torr and a temperature of 225 K is destroyed when all of the chlorine from 15.0 g of CF3CI goes through ten cycles of the above reactions?

Kinetic Molecular Theory •

G) Consider the chemical reaction:

74. Consider the chemical reaction:

3

C IO + 0 ----> Cl + 0 2

Reaction Stoichiometry Involving Gases

How many litres of hydrogen gas are formed from the complete reaction of 15.7 g C? Assume that the hydrogen gas is collected at a pressure of 1.0 bar and a temperature of 355 K.

Hydrogen gas (a potential future fuel) can be formed by the reaction of methane with water according to the equation:

Consider a 1.0 L sample of helium gas and a 1.0 L sample of argon gas, both at roo m temperature and atmospheric pressure. a. Do the atoms in the helium sample have the same average kinetic energy as the atoms in the argon sample? b. Do the atoms in the helium sample have the same average velocity as the atoms in the argon sample? c. Do the argon atoms, because they are more massive, exert a greater pressure on the walls of the container? Explain. d . W hich gas sample would have the faster rate of effusion?

82. A flask at room temperature contains exactly equal amounts (in moles) of nitrogen and xenon. a. Which of the two gases exerts the greater partial pressure? b. The molecules or atoms of which gas have the greater average velocity? c. The molecules of which gas have the greater average kinetic energy? d. If a small hole were opened in the flask, which gas would effuse more quickly?

G)

Calculate the root mean square velocity and kinetic energy of F 2 , Cl 2, and Br2 at 298 K. Rank these three halogens with respect to their rates of effusion.

84. Calculate the root mean square velocity and kinetic energy of CO, C02 , and S0 3 at 298 K. Which gas has the greatest velocity? The greatest ki netic energy? The greatest effusion rate? ~ We obtain urani um-235 from U-238 by fluorinating the uranium to form UF6 (which is a gas) and then taking advantage of the different rates of effusion and diffusion for compounds containing the two isotopes. Calculate the ratio of effusion rates for 238UF6 and 235UF6 . The atomic mass of U-235 is 235.054 amu and that of U-238 is 238.05 1 amu.

86. Calculate the ratio of effusion rates for Ar and Kr.

Q)

A sample of neon effuses from a container in 76 seconds. The same amount of an unknown noble gas requires 155 seconds. Identify the gas.

88. A sample of N20 effuses from a container in 42 seconds. How long would it take the same amount of gaseous I2 to effuse from the same container under identical conditions?

Exercises

CD The graph shows the distribution of molecular velocities for two different molecules (A and B) at the same temperature. Which molecule has the higher molar mass? Which molecule would have the higher rate of effusion?

500

0

1000 1500 Molecular velocity (m s- 1)

2000

193

92. The following is a plot of the distribution of molecular speeds for N2 at 298 K. Sketch the distribution of molecular speeds for: a. N2 at 150 K b. N2 at 500 K c. He at 298 K

2500

90. The graph shows the distribution of molecular velocities for the same molecule at two different temperatures (T1 and T2 ). Which temperature is greater? Explain.

0

1000 1500 500 Molecular velocity (m s- 1)

2000

Real Gases

G)

Which postulate of the kinetic molecular theory breaks down under conditions of high pressure? Explain. 94. Which postulate of the kinetic molecular theory breaks down under conditions of low temperature? Explain.

G Use the van der Waals equation and the ideal gas equation to 0

G

1000 2000 Molecular velocity (m s- 1)

3000

Oxygen (02 ) and chlorine (Cl2) have molar masses of 32.00 and 70.90 g mo1- 1, respectively. In one hour, 0.315 mol of chlorine effuses through a tiny hole. How much oxygen effuses through the same hole in the same amount of time?

calculate the volume of 1.000 mol of neon at a pressure of 500.0 bar and a temperature of 355.0 K. Explain why the two values are different. (Hint: One way to solve the van der Waals equation for Vis to use successive approximations. Use the ideal gas law to get a preliminary estimate for V.) 96. Use the van der Waals equation and the ideal gas equation to calculate the pressure exerted by I.OOO mol of Cl2 in a volume of 5.000 L at a temperature of 273.0 K. Explain why the two values are different.

Cumulative Problems G Modem

American pennies are composed of zinc coated with copper. A student determines the mass of a penny to be 2.482 g and then makes several scratches in the copper coating (to expose the underlying zinc). The student puts the scratched penny in hydrochloric acid, where the following reaction occurs between the zinc and the HCI (the copper remains undissolved): Zn(s)

+ 2 HCl(aq)

----->

H2 (g)

+ ZnCl 2(aq)

The student collects the hydrogen produced over water at 25 °C. The collected gas occupies a volume of0.899 Lat a total pressure of 791 mmHg. Assuming that all the Zn in the penny dissolves, calculate the percent zinc by mass in the penny. 98. A 2.85 g sample of an unknown chlorofluorocarbon decomposes and produces 564 mL of chlorine gas at a pressure of 752 Torr and a temperature of 298 K. What is the percent chlorine (by mass) in the unknown chlorofluorocarbon? Q) The mass of an evacuated 255 mL flask is 143.187 g. The mass of the flask filled with 356 mbar of an unknown gas at 25 °C is 143.289 g. Calculate the molar mass of the unknown gas. 100. A 118 mL flask is evacuated and found to have a mass of97 .129 g. When the flask is filled with 1024 mbar of helium gas at 35 °C, it has a mass of 97. 17 1 g. Was the helium gas pure?

•

A gaseous hydrogen and carbon-containing compound is decomposed and found to contain 82.66% carbon and 17.34% hydrogen by mass. The mass of 158 mL of the gas, measured at 556 Torr and 25 °C, was 0.275 g. What is the molecular formula of the compound?

102. A gaseous hydrogen and carbon-containing compound is decomposed and found to contain 85.63% C and 14.37% H by mass. The mass of 258 mL of the gas, measured at STP, was 0.646 g. What is the molecular formula of the compound?

G) Consider the reaction: 2 NiO(s)

----->

2 Ni(s)

+ 0 2 (g)

If 0 2 is collected over water at 40.0 °C and a total pressure of

745 mmHg, what volume of gas will be collected for the complete reaction of 24.78 g of NiO? 104. Consider the reaction:

If this reaction produces 15.8 g of Ag(s), what total volume of

gas can be collected over water at a temperature of 25 °C and a total pressure of 752 Torr?

194

0

Chapter 5

Gases

When hydrochloric acid is poured over potassium sulfide, 42.9 mL of hydrogen suLfide gas is produced at a pressure of 752 Torr and 25.8 °C. Write an equation for the gas-evolution reaction and determine how much potassium sulfide (in grams) reacted.

112. The emission of N02 by fossil fuel combustion can be prevented by injecting gaseous urea into the combustion mixture. The urea reduces NO (which oxidizes in air to form N0 2) according to the reaction: 2 CO(NH2h(g)

106. Consider the reaction: 2 SOi(g)

+

+

4 NO(g)

+

0 2 (g) ----> 4 Ni(g)

+

2 COi(g)

+

4 H20(g)

0 2 (g) ----> 2 S03(g)

Suppose that the exhaust stream of an automobile has a flow rate of 2.55 L s- 1 at 655 K and contains a partial press ure of NO of 16.5 mbar. W hat total mass of urea is necessary to react completely with the NO formed during 8.0 hours of driving?

a. If 285.5 mL of S02 reacts with 158.9 mL of 0 2 (both measured at 3 15 K and 67 mbar), what is the limiting reactant and the theoretical yield of S03? b. If 187.2 mL of S03 is collected (measured at 3 15 K and 67 mbar), what is the percent yield for the reaction?

G A mmonium carbonate decomposes upon heating according to

@ Trinitrotoluene (TNT, C7H5N30 6) undergoes complete combustion according to the following balanced chemical equation:

the balanced equation:

Calculate the total volume of gas produced at 22 °C and 1.02 bar by the complete decomposition of 11 .83 g of ammonium carbonate.

108. Ammonium nitrate decomposes explosively upon heating according to the balanced equation:

If 25.0 g of TNT is combusted in a 0.450 L container filled with

0 2 at a pressure of 7 .00 bar and a temperature of 298 K, calculate the maximum mass of C02 that could be produced.

114. Buzz Lightyear has found his spaceship! However, it was out of fuel. By searching on Google, he found a way to make hydrazine (N2H4 ) from ammonia (NH 3) and hydrogen peroxide (H20 2 ):

Calculate the total volume of gas (at 125 °C and 748 Torr) produced by the complete decomposition of 1.55 kg of ammonium nitrate.

Buzz has 27 .2 L of ammonia at a temperature of 295 K and a pressure of 2.75 bar, as well as 115 g of peroxide. What is the maximum mass of hydrazine Buzz can make?

G) Olympic cyclists fill their tires with helium to make them lighter. Calculate the mass of air in an air-filled tire and the mass of helium in a helium-filled tire. What is the mass difference between the two? Assume that the volume of the tire is 855 mL, that it is filled to a total pressure of 125 psi, and that the temperature is 25 °C. Also assume an average molar mass for air of 28.8 g mol- 1•

110. In a common classroom demonstration, a balloon is filled with air and submerged in liquid nitrogen. The balloon contracts as the gases within the balloon cool. Suppose the balloon initially contains 2.95 L of air at a temperature of 25.0 °C and a pressure of 0.998 bar. Calculate the expected volume of the balloon upon cooling to -1 95.8 °C (the boiling point of liquid nitrogen). When the demonstration is carried out, the actual volume of the balloon decreases to 0.61 L. How does the observed volume of the balloon compare to your calculated value? Can you explain the difference?

GD An ordinary gasoline can measuring 30.0 cm by 20.0 cm by 15.0 cm is evacuated with a vacuum pump. Assuming that virtually all of the air can be removed from inside the can, and that atmospheric pressure is 1.0 l bar, what is the total force (in Newtons) on the surface of the can? Do you think that the can could withstand the force? 116. Twenty-five millilitres of liquid nitrogen (density = 0.807 g mL - I) is poured into a cylindrical container with a radius of 10.0 cm and a length of 20.0 cm. The container initially contains only air at a pressure of 1.01 bar and a temperature of 298 K. If the liquid nitrogen completely vaporizes, what is the total force (in N) on the interior of the container at 298 K?

QD A 160.0 L helium tank contains pure helium at a pressure of 1855 psi and a temperature of 298 K. How many 3.5 L helium balloons will the helium in the tank fill? (Assume an atmospheric pressure of 1.0 I bar and a temperature of 298 K.)

[Tom Bochsler/Pearson Education]

0

Gaseous ammonia can be injected into the exhaust stream of a coal-burning power plant to reduce the pollutant NO to N 2 according to the reaction:

Suppose that the exhaust stream of a power plant has a flow rate of 335 L s- 1 at a temperature of 955 K, and that the exhaust contains a partial pressure of NO of 22.4 Torr. What should be the flow rate of am monia delivered at 1.007 bar and 298 K into the stream to react completely with the NO if the ammonia is 65.2% pure (by volume)?

118. An 11.5 mL sample of liquid butane (density = 0.573 g mL- I) is evaporated in an otherwise empty container at a temperature of 28.5 °C. The pressure in the container following evaporation is 892 Torr. What is the volume of the container? A scuba diver's expiration of air creates a spherical bubble with a radius of 2.5 cm at a depth of 30.0 m where the total pressure (including atmospheric pressure) is 4.00 atm. What is the radius of the bubble when it reaches the surface of the water? (Assume that the atmospheric pressure is 1.0 l bar and the temperature is 298 K.)

GI)

120. A particular balloon can be stretched to a maximum surface area of 1257 cm 2. The balloon is fi lled with 3.0 L of helium gas at a pressure of 1.007 bar and a temperature of 298 K. The balloon is then allowed to rise in the atmosphere. If the atmospheric temperature is 273 K, at what pressure will the balloon burst? (Assume the balloon to be in the shape of a sphere.)

Exercises

e

A catalytic converter in an automobile uses a palladium or platinum catalyst (a substance that increases the rate of a reaction without being consumed by the reaction) to convert carbon monoxide gas to carbon dioxide according to the reaction: 2 CO(g)

+ Oi(g)

---->

2 C02 (g)

A chemist researching the effectiveness of a new catalyst combines a 2.0 : l.O mole ratio mixture of carbon monoxide and oxygen gas (respectively) over the catalyst in a 2.45 L flask at a total pressure of 993 mbar and a temperature of 552 °C. When the reaction is complete, the pressure in the flask has dropped to 552 Torr. What percentage of the carbon monoxide was converted to carbon dioxide?

122. A quantity of N2 occupies a volume of 1.0 L at 300 K and l.O bar. The gas expands to a volume of 3.0 L as the result of a change in both temperature and pressure. Find the density of the gas at these new conditions. AmixtureofCO(g)and0 2(g)in a l.OLcontainerat 1.0 X 103 K has a total pressure of 2.2 bar. After some time, the total pressure falls to l.9 bar as the result of the formation of C02 . Find the mass (in grams) of C02 that forms. 124. The radius of a xenon atom is 1.3 x 1o- s cm. A 100 mL flask is filled with Xe at a pressure of 1.0 bar and a temperature of 273 K. Calculate the fraction of the volume that is occupied by Xe atoms. (Hint: The atoms are spheres.) A natural gas storage tank is a cylinder with a movable top whose volume can change only as its height changes. Its radius remains fixed. The height of the cylinder is 22.6 m on a day when the temperature is 22 °C. The next day, the height of the cylinder increases to 23.8 m when the gas expands because of a heat wave. Find the temperature on the second day, assuming that the pressure and amount of gas in the storage tank have not changed. 126. A mixture of 8.0 g CH4 and 8.0 g Xe is placed in a container and the total pressure is found to be 0.44 bar. Find the partial pressure of CH4 . A steel container of volume 0.35 L can withstand pressures up to 88 bar before exploding. What mass of helium can be stored in this container at 299 K? 128. Binary compounds of alkali metals and hydrogen react with water to liberate H2 (g). The H 2 from the reaction of a sample of

f»

e

9

C)

195

NaH with an excess of water fills a volume of 0.490 Labove the water. The temperature of the gas is 35 °C and the total pressure is 758 Torr. Determine the mass of H 2 liberated and the mass of NaH that reacted. In a given diffusion apparatus, 15.0 mL of HBr gas diffused in 1.0 min. In the same apparatus and under the same conditions, 20.3 mL of an unknown gas diffused in 1.0 min. The unknown gas is a hydrocarbon. Find its molecular formula.

130. A sample of N20 3(g) has a pressure of 0.017 bar. The temperature (in K) is then doubled and the N20 3 undergoes complete decomposition to N02 (g) and NO(g). Find the total pressure of the mixture of gases assuming constant volume and no additional temperature change. When 0.583 g of neon is added to an 800 cm3 bulb containing a sample of argon, the total pressure of the gases is found to be 1.17 bar at a temperature of 295 K. Find the mass of the argon in the bulb. 132. A gas mixture composed of helium and argon has a density of 0.670 g L- 1 at l.01 bar and 298 K. What is the composition of the mixture by volume? A gas mixture contains 75.2% nitrogen and 24.8% krypton by mass. What is the partial pressure of krypton in the mixture if the total pressure is 993 mbar? 134. Lightn ing McQueen's tires have a maximum rating of 270 kPa. When a tire is cold ( 12.0 °C), it is inflated to a volume of 11.8 L and a pressure of 250 kPa. Whi le racing on a hot day, the tire warms to 65.0 °C and its volume expands slightly to 12. l L. Does the pressure in the tire exceed the maximum rating? a. A liquid hydrocarbon (containing only C and H) is analyzed by combustion analysis. 1.75 g of the substance was found to produce 3 .17 L of C02 at l.O bar, 298 K, and l.92 g H2 0 (molar mass= 18.015 g mol- 1). What is the empirical formula of the compound? b. To obtain the actual molecular formula, the liquid was analyzed by the Dumas method to determine the molar mass. The following data were obtained:

0

0

0

temperature = 100 °C volume of flask= 247 mL mass of empty flask= 7 1.814 g pressure= 813 Torr mass of flask and compound vapour= 72.523 g What is the molar mass and the molecular formula of the compound?

Challenge Problems 136. A I 0 L container is filled with 0.10 mol of H2 (g) and heated to 3000 K, causing some of the H 2(g) to decompose into H(g). The pressure is found to be 3.0 bar. Find the partial pressure of the H(g) that forms from H2 at this temperature. (Assume two significant figures for the temperature.) A mixture of NH 3(g) and N2 H 4(g) is placed in a sealed container at 300 K. The total pressure is 0.50 bar. The container is heated to 1200 K at which time both substances decompose completely according to the equations 2 NH3(g) ----> N2 (g) + 3 H2 (g) and N 2 H4 (g) ----> N2 (g) + 2 H 2 (g). After decomposition is complete, the total pressure at 1200 K is found to be 4.5 bar. Find the percent of N2H 4(g) in the original mixture. (Assume two significant figures for the temperature.) 138. A quantity of CO gas occupies a volume of 0.48 L at l.O atm and 275 K. The pressure of the gas is lowered and its temperature is raised until its volume is l.3 L. Find the density of the CO under the new conditions.

G

@

When C02(g) is put in a sealed container at 701 Kand a pressure of 10.0 bar and is heated to 1401 K, the pressure rises to 22.5 bar. Some of the C02 decomposes to CO and 0 2 . Calculate the mole percent of C02 that decomposes. 140. The world burns approximately 9.0 X 10 12 kg of fossil fuel per year. Use the combustion of octane as the representative reaction and determine the mass of carbon dioxide (the most significant greenhouse gas) formed per year. The current concentration of carbon dioxide in the atmosphere is approximately 387 ppm (by volume). By what percentage does the concentration increase each year due to fossi I fuel combustion? Approximate the average properties of the entire atmosphere by assuming that the atmosphere extends from sea level to 15 km and that it has an average pressure of 381 Torr and average temperature of 275 K. Assume Earth is a perfect sphere with a radius of 637 1 km.

196

Chapter 5

Gases

G The atmosphere slowly oxidizes hydrocarbons in a number of

air-filled balloon has decreased by 5.0%. By what percent has the volume of the helium-filled balloon decreased? (Assume that the air is four-fifths nitrogen and one-fifth oxygen, and that the temperature did not change.)

steps that eventually convert the hydrocarbon into carbon dioxide and water. The overall reaction of a number of such steps for methane gas is: CH4(g)

+

5 0 2(g)

+

5 NO(g) ---> C02(g)

+

+

H20 (g)

5 N02(g)

+

@A

mixture of CH4(g) and C 2H 6(g) has a total pressure of 0.53 bar. Just enough 0 2(g) is added to the mixture to bring about its complete combustion to C02(g) and H20 (g). The total pressure of the two product gases is fo und to be 2.2 bar. Assuming constant volume and temperature, find the mole fraction of CH4 in the mixture .

2 OH(g)

Suppose that an atmospheric chemist combines 155 mL of methane at STP, 885 mL of oxygen at STP, and 55.5 mL of NO at STP in a 2.0 L flask. The flask is allowed to stand for several weeks at 275 K. If the reaction reaches 90.0% of completion (90.0% of the limiting reactant is consumed), what is the partial pressure of each of the reactants and products in the flask at 275 K? What is the total pressure in the flask?

142. Two identical balloons are filled to the same volume, one with air and one with helium. The next day, the volume of the

144. A sample of C 2H2(g) has a pressure of 7 .8 kPa. After some time, a portion of it reacts to form C 6H 6(g). The total pressure of the mixture of gases is then 3.9 kPa. Assume that the volume and temperature do not change. What fraction of C2 H2(g) has undergone reaction?

Conceptual Problems

G When the driver of an automobile applies the brakes, the passengers are pushed toward the front of the car, but a helium balloon inside the car is pushed toward the back of the car. Upon forward acceleration, the passengers are pushed toward the back of the car, but the helium balloon is pushed toward the front of the car. W hy?

146. Suppose that a liquid is 10 times more dense than water. If you were to sip this liquid at sea level using a straw, what could be the maximum length of the straw? This reaction occurs in a closed container:

G

A(g)

+

2 B(g) ---> 2 C(g)

A reaction mixture initially contains 1.5 L of A and 2.0 L of B. Assuming that the volume and temperature of the reaction mixture remain constant, what is the percent change in pressure if the reaction goes to completion?

148. One mole of nitrogen and one mole of neon are combined in a closed container at STP. How big is the container? Exactly equal amounts (in moles) of gas A and gas B are combined in a 1 L container at room temperature. Gas 8 has a molar mass that is twice that of gas A . Which statement is true for the mixture of gases and why? a . The molecules of gas B have greater kinetic energy than those of gas A. b. Gas B has a greater partial pressure than gas A. c. The molecules of gas 8 have a greater average velocity than those of gas A. d. Gas B makes a greater contribution to the average density of the mixture than gas A. 150. Which gas would you expect to deviate most from ideal behaviour under conditions of low temperature: F2, Cl 2, or Br2? Explain.

@

W

E HAVE SPENT THE FIRST FEW CHAPTERS of this book examining one of

the two major components of our universe-matter. We now turn our

attention to the other major component-energy. As far as we know,

matter and energy-which can be interchanged but not destroyed-make up the physical universe. Unlike matter, energy is not something we can touch or hold in our hand, but we experience it in many ways. The warmth of sunlight, the feel of wind on our faces , and the force that presses us back when a car accelerates are all manifestations of energy and its interconversions. And, of course, energy is critical to society and to the world. The standard of living around the globe is strongly correlated with the access to and use of energy resources. Most of those resources, as we shall see, are chemical ones, and we can understand their advantages as well as their drawbacks in terms of chemistry.

6.1

Chemical Hand Warmers

Winters can be long in Canada, and many people enjoy participating in outdoor winter activities such as hockey, skating, skiing, snowboarding, ice fishing, winter hiking, and just playing in the snow. Most of us, however, don' t like being cold, especially in our hands and feet. One solution is the chemical hand warmer, a small

6.1 Chemical Hand Warmers

197

6.2 The Nature of Energy: Key

Definitions 198 6.3 The First Law of Thermodynamics: There Is No Free Lunch 200 6.4 Quantifying Heat and Work 205 6.5 Measuring ~ ,U for Chemical Reactions: Constant-Volume Calorimetry 212 6.6 Enthalpy: The Heat Evolved in a Chemical Reaction at Constant Pressure 214 6.7 Constant-Pressure Calorimetry: Measuring ~ ,H 220 6.8 Relationships Involving ~ , H 221 6.9 Determining Enthalpies of Reaction from Standard Enthalpies of Formation 224 6.10 Energy Use and the Environment 229

197

198

Chapter 6

Thermochemistry

pouch that comes sealed in a plastic package. The package, which is opened and placed in gloves or boots, slowly warms up and keeps your hands or feet warm all day long. Warming your hands with chemical hand warmers involves many of the principles of thermochemistry, the study of the relationships between chemistry and energy. When you open the package that contains the hand warmer, the contents are exposed to air, and an exothermic reaction occurs. The most commonly available hand warmers use the oxidation of iron as the exothermic reaction:

(a)

(b)

The most important product of this reaction is not a substance-it is heat. We' ll define heat more carefully later, but in general, heat is what you feel when you touch something that is warmer than your hand (in this case, the hot hand warmer). Although some of the heat is lost through the minute openings in your gloves, most of it is transferred to your hands and to the pocket of air surrounding your hands, resulting in a temperature increase. The magnitude of the temperature increase depends on the size of the hand warmer and the size of your glove (as well as some other details). But in general, the size of the temperature increase is proportional to the amount of heat released by the reaction. In this chapter, we examine the relationship between chemical reactions and energy. Specifically, we look at how chemical reactions can exchange energy with their surroundings and how to quantify the magnitude of those exchanges. These kinds of calculations are important, not only for chemical hand warmers, but also to many other important processes such as the heating of homes and the production of energy for our society.

6.2 The Nature of Energy: Key Definitions

(c)

A FIGURE 6.1 (a) A curling stone (also known as a "rock") sliding down the ice has kinetic energy due to its motion. (b) When this stone collides with another, it does work and transfers energy to the second stone. (c) The second stone now has kinetic energy as it slides away from the collision. [Canadian Curling Association/Courtesy of lSN]

Energy is the capacity to do work, and work (w) is the result of a force acting through a distance. When you push a box across the floor, you have done work. Consider another example of work: a curling stone sliding down the ice and colliding with a second stationary curling stone (Figure 6.1 ~). The curling stone sliding down the ice has energy due to its motion. When it collides with another stone, it does work to get the second stone moving, resulting in the transfer of energy from one stone to the other. The second curling stone absorbs the energy and begins to move. As we just saw with chemical hand warmers, energy can also be transferred through heat (q) , the flow of energy caused by a temperature difference. For example, if you hold a cup of coffee in your hand, energy is transferred in the form of heat from the hot coffee to your cooler hand. Think of energy as something that an object or set of objects possesses. Think of heat and work as ways that objects or sets of objects exchange energy. The energy contained in the curling stone moving down the ice is an example of kinetic ener gy, the energy associated with the motion of an object. The energy contained in a hot cup of coffee is thermal en ergy, the energy associated with the temperature of an object. Thermal energy is actually a type of kinetic energy because it arises from the motions of atoms or molecules within a substance. The water at the top of a waterfall contains potential energy, which is the energy associated with the position or composition of an object. The potential energy of the water at the top of the waterfall is a result of its position in Earth's gravitational field, as shown in Figure 6.2 ~ . Gravity pulls the water down from the top of the falls, releasing the potential energy. The amount of potential energy is related to how high the falls are (or the object is). The greater the height, the more energy is stored as potential energy. Another example of potential energy is the energy contained in a compressed spring. When you compress a spring, you push against the forces that tend to maintain the spring's uncompressed shape,

6.2 The Nature of Energy: Key Definitions

199

storing energy as potential energy. Chemical energy, the energy associated with the relative positions of electrons and nuclei in atoms and molecules, is also a form of potential energy. Some chemical compounds, such as the methane in natural gas or the iron in a chemical hand warmer, are like a compressed springthey contain potential energy, and a chemical reaction can release that potential energy. The law of conservation of energy states that energy can be neither created nor destroyed. However, energy can be transferred from one object to another, and it can assume different forms. For example, as the water from the top of the waterfall drops toward the river below, as shown in Figure 6.2, some of its potential energy is converted to kinetic energy due to its velocity. This kinetic energy can be used to tum a turbine in a power plant to transform it to electricity, another form of energy. If you release a compressed spring, ..6. FIGURE 6.2 Energy Transformation: Potential and Kinetic Energy I The water at the top of Niagara Falls has gravitathe potential energy becomes kinetic energy as the spring expands tional potential energy. When the water drops over the fall s, the outward, as shown in Figure 6.3 T . When iron reacts with oxygen potential energy is transformed into kinetic energy-the energy within a chemical hand warmer, the chemical energy of the iron and of motion. The kinetic energy from water falling can be used to oxygen becomes thermal energy that increases the temperature of generate hydroelectricity. your hand and glove. [Jodie Johnson/Fotolia] A good way to understand and track energy changes is to define the system under investigation. For example, the system may be a beaker of chemEinstein showed that it is mass- energy that is conserved; one can be converted icals in the lab, or it may be the iron reacting in a hand warmer. The system's surroundinto the other. This equivalence becomes ings are everything with which the system can exchange energy. If we define the important in nuclear reactions, discussed in chemicals in a beaker as the system, the surroundings may include the water that the Chapter 19. In ordinary chemical reactions, chemicals are dissolved in (for aqueous solutions), the beaker itself, the lab bench on however, the interconversion of mass and energy is not a significant factor, and we can which the beaker sits, the air in the room, and so on. For the iron in the hand warmer, the regard mass and energy as independently surroundings include your hand, your glove, the air in the glove, and even the air outside conserved. of the glove. In an energy exchange, energy is transferred between the system and the surroundings, as shown in Figure 6.4 T . If the system loses energy, the surroundings gain the same exact amount of energy, and vice versa. When the iron within the chemical hand warmer reacts, the system loses energy to the surroundings, producing the desired temperature increase within your gloves.

System Energy Gauge

Kinetic energy

Before Energy Transfer

After Energy Transfer

Surroundings Energy Gauge

~ Empty

Full

Empty

Full

Full

Empty

Full

~I Empty

The surroundings gain the exact amount of energy lost by the system.

(a)

(b)

..6. FIGURE 6.3 Energy Transformation: Potential and Kinetic Energy II (a) A compressed spring has potential energy. (b) When the spring is released, the potential energy is transformed into kinetic energy.

..6. FIGURE 6.4 Energy Transfer If a system and surroundings had energy gauges (which would measure energy content in the way a fuel gauge measures fuel content), an energy transfer in which the system transfers energy to the surroundings would result in a decrease in the energy content of the system and an increase in the energy content of the surroundings . The total amount of energy, however, must be conserved.

200

Chapter 6

Thermochemistry

Units of Energy We can deduce the units of energy from the definition of kinetic energy. An object of mass m, moving at velocity v, has a kinetic energy KE given by: 1

- mv2

KE =

[6.1]

2t \ kg m s- 1

The SI unit of mass is kg, and the unit of velocity is m s- 1• The SI unit of energy is therefore kg m2 s-2 , defined as the joule (J), named after the English scientist James Joule (1818- 1889). lkgm2 s- 2

3.6 X 105 J or 0.10 kWh used in 1 hour

Ji.. A watt (W) is l J s - I , so a lOOW light bulb uses 100 J every second, or 3.6 X 105 J every ho ur. (Quade Paul/Pearson Education]

The "calorie" referred to on all nutritional labels (regardless of the capitalization) is I always the capital CCalorie.

= lJ

One joule is a relatively small amount of energy- for example, a I 00-watt light bulb uses 3.6 X 105 Jin I hour. Therefore, we often use the kilojoule (kJ) in our energy discussions and calculations (1 kJ = 10001). A second commonly used unit of energy is the calorie (cal), originally defi ned as the amount of energy required to raise the temperature of 1 g of water by 1 °C. The current definition is 1 ea! = 4.184 J (exact); a calorie is a larger unit than a joule. A related energy unit is the nutritional, or uppercase "C" Calorie (Cal), equivalent to 1000 lowercase "c" calories. The Calorie is the same as a kilocalorie (kcal): 1 Cal = 1 kcal = 1000 cal. Electricity bills typically are based on another, even larger, energy unit, the kilowatt-hour (kWh): 1 kWh = 3.60 X 106 J. Table 6.1 shows various energy units and their conversion factors. Table 6.2 shows the amount of energy required for various processes. TABLE 6.1 Energy Conversion Factors* 1 calorie (cal)

=

4.184 joules (J)

1 Calorie (Cal) or kilocalorie (kcal)

=

1000 cal = 4184 J

1 kilowatt-hour (kWh)

= 3.60 x 106 J

*All conversion factors in this table are exact.

TABLE&.2

Energy Uses in Various Units Amount Required to Raise Temperature of 1 g of Water by 1 °c

Amount Required to Light 100W Bulb for 1 Hour

Amount Used by Human Body in Running 1 km (Approximate)

Average Amount of Household Electricity Used in 1 Day per Canadian

joule (J)

4.184

3.60 x 105

2.6 x 105

4.3 x 107

calorie (cal)

1.00

8.60 x 104

6.2 x 104

1.0 x 107

Calorie (Cal)

0.00100

86.0

62

1.0 x 104

1.16x10- 5

0.100

0.072

12

Unit

kilowatt-hour (kWh)

[Based on Natural Resources Canada: http://oee.nrcan.gc.ca/corporate/statistics/neud/dpa/data_e/sheu07/sheu_048_1.cfm)

6.3 The First Law of Thermodynamics: There Is No Free Lunch We call the general study of energy and its interconversions thermodynamics. The laws of thermodynamics are among the most fundamental in all of science, governing virtually every process that involves change. The first law of thermodynamics is the Jaw of energy conservation, which we can state as follows: The total energy of the universe is constant. In other words, since energy is neither created nor destroyed, and since the universe does not exchange energy with anything else, its energy content does not change. The first Jaw

6.3 The First Law of Thermodynamics: There Is No Free Lunch

CHEMISTRY IN YOUR DAY

Perpetual Motion Machines

~·

In 1812, a man named Charles Redheffer appeared in Philadelphia with a machine that he claimed could run forever without any energy input-a perpetual motion machine. He set up the machine on the edge of town and charged admission to view it. He also appealed to the city for money to build a larger version of the machine. When city commissioners came out to inspect the machine, Redheffer did his best to keep them from viewing it too closely. Nonetheless, one of the commissioners noticed something suspicious: the gears that supposedly ran to an external driveshaft were cut in the wrong direction. The driveshaft that the machine was allegedly powering was instead powering the machine. Redheffer left Philadelphia exposed as a fraud. In 2008, a man by the name of Richard Willis, from Magnacoaster Motor Company Inc. in Kitchener, Ontario, appeared on Dragon 's Den, a popular CBC show where inventors and budding entrepreneurs pitch their ideas for a company to a panel of "dragons" (venture capitalists) in hopes that they will invest in their company. Magnacoaster claims to have invented a device in which power is injected between a set of magnets and emerges at a "higher voltage, a higher amperage, and a higher [The Franklin Institute] frequency of the power." In other words, it claims to have built a perpetual motion machine of the first kind that produces more energy than is put into the machine. If it works, this would provide unlimited free energy; however, the technology violates the law of conservation of energy. It is not surprising that there have been no independent verifications of the technology, and according to reports, no units have been delivered to customers as of 2015.

has many implications: the most important one being that, with energy, you do not get something for nothing. The best you can do with energy is break even-there is no free lunch. According to the first law, a device that would continually produce energy with no energy input, sometimes known as a perpetual motion machine, cannot exist. Occasionally, the media speculate or report on the discovery of a machine that can produce energy without the need for energy input. For example, you may have heard someone propose an electric car that recharges itself while driving, or a new motor that can create additional usable electricity as well as the electricity to power itself. Although some hybrid (electric and gasoline-powered) vehicles can capture energy from braking and use that energy to recharge their batteries, they could never run indefinitely without additional fuel. As for the motor that powers an external load as well as itself- no such thing exists. Our society has a continual need for energy, and as our current energy resources dwindle, new energy sources will be required. And those sources, whatever they may be, will follow the first law of thermodynamics-energy must be conserved.

Internal Energy The internal ener gy (U) of a system is the sum of the kinetic and potential energies of all of the particles that compose the system. According to the first law of thermodynamics, the change in the internal energy of the system (11U) is the sum of the heat transferred (q) and the work done (w):

11U

=

q

+

[6.2]

w

In the above equation, and from this point forward , we follow the standard convention that 11U (with no subscript) refers to the internal energy change of the system. As shown in Table 6.3, energy entering the system through heat or work carries a positive sign and

TABLE 6.3 Sign Conventions for q, w, and 4.U

+ system gains thermal energy + work done on the system 11 U (change in internal energy) + energy flows into the system

q (heat)

w(work)

- system loses thermal energy - work done by the system - energy flows out of the system

201

202

Chapter 6

Thermochemi stry

A FIGURE 6.5 Combustion of Hydrogen The internal energy change when burning hydrogen depends only on the states of the initial and final products, not on how the hydrogen is burned. The product of the reaction, H 20, initially is formed as a hot gas in (a) and (b), but will eventually cool and condense to the liquid state. In (c), the car is powered by a hydrogen fuel cell in which H 2 and 0 2 are separate from one another. [(a) Photo Researchers/Science History lmages/Alamy Stock Photo; (b) NASA; (c) Stephen Barnes/TransporVAlamy Stock Photo]

energy leaving the system through heat or work carries a negative sign. The system is like a chequing account-withdrawals are negative and deposits are positive. Internal energy is a state function, which means that its value depends only on the state of the system, not on how the system arrived at that state. The state of a chemical system is specified by parameters such as temperature, pressure, concentration, and physical state (solid, liquid, or gas). Let's consider three different ways of reacting hydrogen and oxygen, depicted in Figure 6.5 A . The first way is exemplified in the historical disaster where the Hindenburg burst into flames. The H 2 from the blimp reacted with the 0 2 in the atmosphere to produce water. Almost all of the internal energy for this reaction was released as heat. In engines of the space shuttle, H2 is also reacted with 0 2' but in a significantly more controlled manner than the explosion of the Hindenburg. In the space shuttle engine some of the energy is released as heat but a significant amount is used to do work, propelling the space shuttle. The third example is what happens in a fuel cell, for example, in a toy car. The H 2 and 0 2 gases do not actually mix in a fuel cell. Rather electrons from H 2 in one side of the fuel cell are transferred to the 0 2 side of the fuel cell where H+, 0 2' and electrons combine to form water. In the fuel cell, very little heat is produced and almost all of the energy from the H 2 and 0 2 reaction is used to do work, moving the vehicle. In all three of these examples, the chemical reaction is the same: H2(g)

1

+ 2 0 2(g)

~

H 20 (l)

As long as the states (temperature and pressure) of the reactants and products are the same before and after the reaction, respectively, the change in the internal energy is the same. The change in internal energy is the difference in internal energy between the final and initial states:

D.rU

=

Uproducts -

U reactants

[6.3]

For the reaction of gaseous H 2 and 0 2 to produce liquid water, the change in internal energy is - 293 kJ mo1- 1, no matter which path is taken. The three different paths taken can have very different amounts of heat and work associated with them, but the sum of the heat and the work (q+w), the internal energy change, is the same for all three paths. We can portray the energy changes that occur during a reaction with an energy diagram, which compares the internal energy of the reactants and products: For a chemical reaction, the difference in internal energy is usually written as !l,U and has units of kJ mo1- 1 , indicating the change in internal energy per mol of reaction as written. !l u is an internal energy change for a process in units of kJ. In this text we use !l,U to indicate a molar internal energy change. Note that for any process the sign of !l U and !l,U will be the same.

1

H2(g);z Oz(g) (reactants) Internal energy

!::.,U < 0 (negative) (product)

The vertical axis of the diagram is internal energy, which increases as you move up on the diagram . For this reaction, the reactants are higher on the diagram than the product because they have higher internal energy. As the reaction occurs, the reactants become products, which have lower internal energy. Therefore, energy is given off by the reaction and D. rU (that is, Uproctucts - Ureactan1s) is negative.

6.3 The First Law of Thermodynamics: There Is No Free Lunch

Where does the energy lost by the reactants (as they transform to products) go? If we define the thermodynamic system as the reactants and products of the reaction, then energy flows out of the system and into the surroundings. System

Surroundings

ll U,y, < 0 (negative)

ll U,urr > 0 (positive)

According to the first law, energy must be conserved. Therefore, the amount of energy lost by the system must exactly equal the amount gained by the surroundings: [6.4] In the engine of the space shuttle, where H2 and 0 2 are reacted together to form water, both quantities q and w are negative as heat is released from the engine to air around the engine (part of the surroundings) and work is done on the space shuttle, moving it through the sky. Now, suppose the reaction is reversed:

The energy level diagram is nearly identical with one important difference: H20(/) is now the reactant and H 2(g) and Oi(g) are the products. Instead of decreasing in energy as the reaction occurs, the system increases in energy. H2~), Internal energy

1

2 Oz(g)

(products)

11,U> 0 (positive) (reactant)

In this reversed reaction, D.,U is positive and energy flows into the system and out of the surroundings. System

Surroundings Energy flow .__ _ _ _,,

ll U,y, > 0 (positive)

fl U,urr < 0 (negative)

Summarizing Energy Flow: ~

If the reactants have a higher internal energy than the products, D. U sys is negative and energy flows out of the system into the surroundings.

~

If the reactants have a lower internal energy than the products, D. U sys is positive and energy flows into the system from the surroundings.

Again, recall the chequing account analogy. Energy flowing out of the system is like a withdrawal and therefore carries a negative sign. Energy flowing into the system is like a deposit and carries a positive sign.

Consider the fictitious internal energy gauges for a chemical system and its surroundings: Empey Full I I'--·-----'------'-----'-------' Chemical system Empey Full I I'--·--~-----'---~------'Surroundings

203

204

Chapter 6

Thermochem istry

Which of the following best represents the energy gauges for the same system and surroundings following an energy exchange in which ~ Usys is negative?

IEmpty

Full

I: Chemical system

IEmpty

Full

I: Surroundings

IEmpty

Full

I: Chemical system

IEmpty

Full

I: Surrou ndings

IEmpty

Full

I: Chemical system

IEmpty

Full

I: Surroundings

(a)

(b)

(c)

EXAMPLE 6.1

INTERNAL ENERGY, HEAT, AND WORK

The firing of a potato cannon provides a good example of the heat and work associated with a chemical reaction. In a potato cannon, a potato is stuffed into a long cylinder that is capped on one end and open at the other. Some kind of fuel is introduced under the potato at the capped end-usually through a small hole-and ignited. The potato then shoots out of the cannon, sometimes flying hundreds of metres, and the cannon emits heat to the surroundings. If the burning of the fuel performs 855 J of work on the potato and produces 1422 J of heat, what is 6.U for the burning of the fuel? (Note: A potato cannon can be dangerous and should not be constructed without proper training and experience.) SOLUTION To solve the problem, substitute the values of q and w into the equation for 6.U. Since work is done by the system on the surroundings, w is negative. Similarly, since heat is released by the system to the surroundings, q is also negative.

6.U

=q+w = -1422 J - 855 J = -2277 J

FOR PRACTICE 6.1 A cylinder and piston assembly (defined as the system) is warmed by an external flame. The contents of the cylinder expand, doing work on the surroundings by pushing the piston outward against the external pressure. If the system absorbs 559 J of heat and does 488 J of work during the expansion, what is the value of 6. U?

Identify each of the following energy exchanges as heat or work and determine whether the sign of heat or work (relative to the system) is positive or negative. (a) An ice cube melts and cools the surrounding beverage. (The ice cube is the system.) (b) A metal cylinder is rolled up a ramp. (The metal cylinder is the system.) (c) Steam condenses on skin, causing a burn. (The condensing steam is the system.)

6.4 Quantifying Heat and Work

6.4

205

Quantifying Heat and Work

In the previous section, we calculated Li U based on given values of q and w . We now tum to calculating q (heat) and w (work) based on changes in temperature and volume.

Heat As we saw in Section 6.2, heat is the exchange of thermal energy between a system and its surroundings caused by a temperature difference. Notice the distinction between heat and temperature. Temperature is a measure of the thermal energy within a sample of matter. Heat is the transfer of thermal energy. Thermal energy always flows from matter at higher temperatures to matter at lower temperatures. For example, a hot cup of coffee transfers thermal energy-as heat-to the lower-temperature surroundings as it cools down. Imagine a world where the cooler surroundings actually got colder as they transferred thermal energy to the hot coffee, which got hotter. Such a world exists only in our imagination (or in the minds of science fiction writers), because the spontaneous transfer of heat from a hotter object to a colder one is a fundamental principle of our universe-no exception has ever been observed. The thermal energy in the molecules that compose the hot coffee distributes itself to the molecules in the surroundings. The heat transfer from the coffee to the surroundings stops when the two reach the same temperature, a condition called thermal equilibrium. At thermal equilibrium, there is no additional net transfer of heat.

The reason for this one-way transfer is related to the second law of thermodynamI ics, which we discuss in Chapter 17.

Temperature Changes and Heat Capacity When a system absorbs heat (q) , its temperature changes by LiT: Heat (q)

System

!::.T

Experiments show that the heat absorbed by a system and its corresponding temperature change are directly proportional: q rx: Li T. The constant of proportionality between q and LiT is the system's heat capacity (C), a measure of the system's ability to absorb thermal energy without undergoing a large change in temperature:

q

= C

X

LiT

[6.5]

) Heat capacity Notice that the higher the heat capacity of a system, the smaller the change in temperature for a given amount of absorbed heat. We define the heat capacity (C) of a system as the quantity of heat required to change its temperature by 1 °C. As we can see by solving Equation 6.5 for heat capacity, the units of heat capacity are those of heat (typically J) divided by those of temperature (typically C): 0

c = _!j__ = _!__ (or J c- 1)

TABLE 6.4 Specific Heat Capacities of Some Common Substances

Substance

Specific Heat Capacity, C.(J 9 -1 oc-1)*

Elements Lead

0.128

0

LiT °C In order to understand two important concepts related to heat capacity, consider putting a steel saucepan on a kitchen flame. The saucepan's temperature rises rapidly as it absorbs heat from the flame. However, if you add some water to the saucepan, the temperature rises more slowly. Why? The first reason is that when you add the water, the same amount of heat must now warm more matter, so the temperature rises more slowly. In other words, heat capacity is an extensive property-it depends on the amount of matter being heated (see Section 1.3). The second (and more fundamental) reason is that water is more resistant to temperature change than steel- water has an intrinsically higher capacity to absorb heat without undergoing a large temperature change. The measure of the intrinsic capacity of a substance to absorb heat is its specific heat capacity (C.), the amount of heat required to raise the temperature of 1 gram of the substance by 1 °C. The units of specific heat capacity (also called specific heat) are J g- 1 0 c- 1. Table 6.4 shows the values of the specific heat capacity for several substances. Heat capacity is sometimes reported as molar heat capacity, the amount

Gold

0.128

Silver

0.235

Copper

0.385

Iron

0.449

Aluminum

0.903

Compounds Ethanol

2.440

Water

4.184

Materials Glass (Pyrex)

0.75

Granite

0.79

Sand

0.84

•At 298 K

206

Chapter 6

Th ermochem istry

of heat required to raise the temperature of 1 mol of a substance by 1 °C. The units of molar heat capacity are J mol- 1 0 c- 1. You can see from these definitions that specific heat capacity and molar heat capacity are intensive properties-they depend on the kind of substance being heated, not on the amount. Notice that water has the highest specific heat capacity of all the substances in Table 6.4--changing the temperature of water requires that a lot of heat be exchanged. If you have ever experienced the drop in temperature that occurs when travelling from an inland region to the coast during the summer, you have experienced the effects of water's high specific heat capacity. For example, the average daily high temperature for July in Kamloops, BC (an inland city), is 28.3 °C, whereas the average .A The high heat capacity of the water surrounding Victoria, daily high temperature for July in Victoria, BC (a coastal city), is BC, results in relatively cooler summer and warmer winter only 21.9 °C, even though the amount of sunlight falling on these temperatures. two cities is similar. In January, the average daily temperature in [B Calkins/Shutterstock] Kamloops is -4.2 °C and that in Victoria is 3.8 °C! In fact, the record low in Kamloops is -37.2 °C, but in Victoria the record low is -15.6 °C. Why the large temperature difference? Victoria is at the south end of Vancouver Island, surrounded by water of the Strait of Juan de Fuca connected to the Pacific Ocean. Water, with its high heat capacity, absorbs much of the sun's heat without undergoing a large increase in temperature, which keeps Victoria cooler in the summer. Similarly, in the winter, when there is not much sunlight, the water releases heat without decreasing its temperature much and keeps Victoria warm. Kamloops is about 400 km inland. The land surrounding Kamloops, with its low heat capacity, will undergo larger fluctuations in temperature and cannot moderate the climate as the water does for Victoria. It is quite typical of coastal regions that their climates are moderate compared to inland regions. The specific heat capacity of a substance can be used to quantify the relationship between the amount of heat added to a given amount of the substance and the corresponding temperature increase. The equation that relates these quantities is:

Heat (J) --+ q = m X C, X Li T

t

M ass (g)

11 T in °Cis equal to 11 T in K, but not equal

I to 11 Tin °F (Section 1.3).

EXAMPLE 6.2

~ Temperature

"

change (oC)

[6.6)

Specific heat capacity (J g- 1 0 c-1)

where q is the amount of heat in J, m is the mass of the substance in g, Cs is the specific heat capacity in J g- 1 °C- 1, and LiT is the temperature change in °C. The following example demonstrates the use of this equation.

TEMPERATURE CHANGES AND HEAT CAPACITY

Suppose you find a 2006 commemorative pure silver (99.99%) dollar in the snow. How much heat is absorbed by the silver dollar as it warms from the temperature of the snow, which is - I 0.0 °C, to the temperature of your body, which is 37 .0 °C? The silver dollar has a mass of 25.175 g.

SORT You are given the mass of silver as well as its initial and final temperatures. You are asked to find the heat required for the given temperature change.

= 25. 175gsilver = - 10.0 °C Tr = 37.0 °C

GIVEN: m T; FIND: q

6.4

STRATEGIZE The equation q = mCs!:l.T gives the relationship between the amount of heat (q) and the temperature change (/:l.T).

Quantifying Heat and Work

207

CONCEPTUAL PLAN

c,, m, .iT

q

l

RELATIONSHIPS USED q = mC5 !:1T (Equation 6.6) Cs= 0.235 Jg- 1 c- 1 (Table6.4) 0

SOLVE Gather the necessary quantities for the equation in the correct units and substitute them into the equation to compute q.

SOLUTION

= TF - Ti= 37.0 °C - (- 10.0 °C) = 47.0 °C q = mC /:l.T = 25.175 g X 0.235 J ~OC?'" X 47.0 X' = 278 J

!:l.T

5

CHECK The unit J is correct for heat. The sign of q is positive, as it should be since the silver dollar absorbed heat from the surroundings. FOR PRACTICE 6.2 To determine whether a shiny gold-coloured rock is actually gold, a chemistry student decides to measure its heat capacity. She first weighs the rock and finds it has a mass of 4.7 g. She then finds that upon absorption of 57.2 J of heat, the temperature of the rock rises from 25 °C to 57 °C. Find the specific heat capacity of the substance composing the rock and determine whether the value is consistent with the rock being pure gold. FOR MORE PRACTICE 6.2 A 55.0 g aluminum block initially at 27.5 °C absorbs 725 J of heat. What is the final temperature of the aluminum?

Suppose you are cold-weather camping and decide to heat some objects to bring into your sleeping bag for added warmth. You place a large water jug and a rock of equal mass near the fire. Over time, both the rock and the water jug warm to about 50 °C. If you could bring only one into your sleeping bag, which one should you choose to keep you the warmest? Why?

Thermal Energy Transfer As we noted earlier, when two substances of different temperatures are combined, thermal energy flows as heat from the hotter substance to the cooler one. If we assume that the two substances are thermally isolated from everything else, then the heat Jost by one substance exactly equals the heat gained by the other (according to the Jaw of energy conservation). If we define one substance as the system and the other as the surroundings, we can quantify the heat exchange as follows:

Suppose a block of metal initially at 55 °C is submerged into water initially at 25 °C. Thermal energy transfers as heat from the metal to the water:

-qmeul

=

qwater

208

Chapter 6

Thermochem istry

The metal will get colder and the water will get warmer until the two substances reach the same temperature (thermal equilibrium). The exact temperature change that occurs depends on the masses of the metal and the water and on their specific heat capacities. Since q = m X Cs X 11T, we can arrive at the following relationship: -qmetal

=

qwater

-mmetal X Cs, metal X /1 T metal

=

mwater X Cs, water X /1 T water

The following example shows how to work with thermal energy transfer.

EXAMPLE 6.3

THERMAL ENERGY TRANSFER

A 32.5 g cube of aluminum initially at 45 .8 °C is submerged into I 05.3 g of water at 15.4 °C. What is the final temperature of both substances at thermal equilibrium? (Assume that the aluminum and the water are thermally isolated from everything else.)

SORT You are given the masses of aluminum and water and their initial temperatures. You are asked to find the final temperature.

GIVEN: mAI

=

32.5 g, mH,O = 105.3 g

Ti, Al

=

45.8 °C, Ti, H,O = 15.4 °C

FIND: Tr STRATEGIZE The heat lost by the aluminum (qA 1) equals the heat gained by the water (qH 20 ) . Use the relationship beween q and 11T and the given variables to find a relationship between /1 T Al and /1 TH,O·

CONCEPTUAL PLAN -qAI

=

( mAh

C,, Ah mH2o C,, H,o'

qH,O

-mAI X

Use the relationship between 11TA1 and l1T8 20 (that you just found in the previous step) along with the initial temperatures of the aluminum and the water to determine the final temperature. Note that at thermal equilibrium, the final temperature of the aluminum and the water is the same; that is, Tr. Al = Tr. H,o = Tr.

l

Cs, Al

X

~TAI= constant x ~TH,o-

~TAI =

Ti,Ah Ti,HzO "" ~ T Al =

=

SOLUTION -qAI = qH20

constant x

Substitute the values of m (given) and Cs (from Table 6.4) for each substance, and solve the equation for 11TAI· (Alternatively, you can solve the equation for 11TH,o·)

- ( 32.5 g X

-(mAI X Cs, AI X l1TA1)

11TAI

=

~ TH, o

=

0.903 J g- 1 oc- 1 (Table 6.4)

=

105.3 g X

4.184 J

g· '.X'.:'. · 11TH,o

= 44Q.57' 11TH,O

-15.Ql2(Tr - Ti, H,o)

=

+

-

'

2

16.012 17.3 °C

15.Ql2 ·Ti, H,o

+

Ti, Al

15.Ql2' Ti, H,O + Ti, Al

15.012' Ti Ho

f -

=

)

g· '.X'.:'. · l1TA1 =

-15.Ql2 · Tr

16.Ql2 · Tr T. -

mH,O X Cs, H2 0 X 11TH,O

- 15.Ql2' 11TH,O

Tr - Ti, Al

=

=

0.903 J

-29.~47 · 11TAI

Tr

~TH,O

m X Cs X 11T (Equation 6.6)

SOLVE Write the equation for the relationship between the heat lost by the aluminum (qA1) and the heat gained by the water (q8 20 ) and substitute q = m X Cs X 11T for each substance.

Substitute the initial temperatures of aluminum and water into the relationship from the previous step and solve the expression for the final temperature (Tr). Remember that the final temperature for both substances will be the same.

X

Tr

RELATIONSHIPS USED Cs, H,o = 4. 184 J g- 1 oc- 1, Cs, Al q

C,, H20

mH,O X

+

Ti Al '

15.012 · 15.4 °C + 45.8 °C = ---------16.012

6.4 Quantifying Heat and Work

209

CHECK The unit °C is correct. The final temperature of the mixture is closer to the initial temperature of the water than the aluminum. This makes sense for two reasons: (1) water has a higher specific heat capacity than aluminum , and (2) there is more water than aluminum. Since the aluminum loses the same amount of heat that is gained by the water, the greater mass and specific heat capacity of the water make the temperature change in the water less than the temperature change in the alurninum.

FOR PRACTICE 6.3 A block of copper of unknown mass has an initial temperature of 65.4 °C. The copper is immersed in a beaker containing 95.7 g of water at 22.7 °C. When the two substances reach thermal equilibrium, the final temperature is 24.2 °C. What is the mass of the copper block?

Substances A and B, initially at different temperatures, come in contact with each other and reach thermal equilibrium. The mass of substance A is twice the mass of substance B. The specific heat capacity of substance B is twice the specific heat capacity of substance A. Which statement is true about the final temperature of the two substances once thermal equilibrium is reached? (a) The final temperature will be closer to the initial temperature of substance A than substance B. (b) The final temperature will be closer to the initial temperature of substance Bthan substance A.

(c) The final temperature will be exactly midway between the initial temperatures of substances A and B.

Work: Pressure-Volume Work Energy transfer between a system and its surroundings can occur via heat (q) or work (w). We just saw how to calculate the heat associated with an observed temperature change. We now turn to calculating the work associated with an observed volume change. Although a chemical reaction can do several different types of work, for now we will limit our discussion to pressure-volume work. We have already defined work as a force acting through a distance. Pressure-volume work occurs when the force is caused by a volume change against an external pressure. For example, pressure-volume work occurs in the cylinder of an automobile engine. The combustion of gasoline causes gases within the cylinders to expand, pushing the piston outward and ultimately moving the wheels of the car. We derive an equation for the value of the pressure-volume work from the definition of work as a force (F) acting through a distance (d): w

=

-F X d

[6.7]

The negative sign appears because the force of the expanding gases is acting in the opposite direction of the piston's motion. When the volume of a cylinder increases (Figure 6.6 T), it pushes against an external force. That external force is pressure (P), which is defined as force (F) divided by area (A):

F

P= -

A

..A. The combustion of gasoline within an engine's cylinders does pressurevolume work that ultimately results in the motion of the car.

The force in this equation must be a

I constant force. or

F=PXA

If we substitute thi s expression for force into the definition of work given by Equation 6. 7, we arrive at the following: w =-PX A X d

The distance through which the force acts is the change in the height of the piston as it moves during the expansion (6.h). Substituting 6.h ford, we get: w =-PX AX 6.h

210

Chapter 6

Thermochem istry

Since the volume of a cylinder is the area of its base times its height, then A X 11h is actually the change in the volume (11 V) that occurs during the expansion. Thus, the expression for work becomes:

_______ ..,,.

w=-Pl1V

[6.8)

A V = A X Ah

,.,..-- -- --[ Volume change ) Cross-sectional area = A

11 V is the final volume minus the initial volume, Vf - Vi or V2 - V1, so during an expansion /1 V is positive and the work is negative, in agreement with the convention that if the system (the expanding gases) does work on the surroundings, its sign is negative .

.&. FIGURE 6.6 Piston Moving Within a Cylinder Against an External Pressure

EXAMPLE 6.4

PRESSURE-VOLUME WORK

To inflate a balloon, you must do pressure- volume work on the surroundings. If you inflate a balloon from a volume of 0.100 L to 1.85 L against an external pressure of 1.00 bar, how much work is done (in joules)? SORT You know the initial and final volumes of the balloon and the pressure against which it expands. The balloon and its contents are the system. STRATEGIZE The equation w = -P 11 V specifies the amount of work done during a volume change against an external pressure.

GIVEN: V 1 = 0.100 L, V2 = 1.85 L, P = 1.00 bar FIND: w CONCEPTUAL PLAN

r;,Av ---{ iv=

SOLVE To solve the problem, compute the value of /1 V and substitute it, together with P, into the equation.

~I

-Pil. V

SOLUTION

11 V

w

Convert the units of the answer (bar L) to J using 100 J = 1 bar L.

w

= V2 - Vi = 1.85 L - 0.100 L = 1.75 L = -Pl1V = -1.00 bar x 1.75 L 100 J

= - 1.7 5 J:IBrr: x 1..barL = - 175 J

CHECK The unit J is correct for work. The sign of the work is negative, as it should be for an expansion: Work is done on the surroundings by the expanding balloon.

FOR PRACTICE 6.4 When fuel is burned in a cylinder equipped with a piston, the volume expands from 0.255 L to 1.45 L against an external pressure of 1.02 bar. In addition, 875 J is emitted as heat. What is 11U for the burning of the fuel?

6.4

Quantifying Heat and Work

211

We can use Equation 6.8 to determine the work for a volume change under constant pressure. What about the work done when a chemical reaction occurs? Consider the combustion of propane at constant pressure and at 298 K:

When this reaction occurs, six equivalents of gas are converted to three equivalents of gas and four equivalents of liquid water. If we assume the volume of the liquid water is negligible compared to the volume of the gases, then the system contracts by three gas equivalents (6 products - 3 reactants =contraction of 3 equivalents). If we assume that the gases behave ideally, the volume change for this chemical system is due only to the change in the amounts of gases: w

=

-Pt:i.V

=

-t:i.nRT

The work done during the chemical reaction can be computed using the equation: w

=

[6.9]

-t:i.nRT 1

mol- 1),

T is the absolute temperature, and where R is the ideal gas constant (8.314 J Kt:i.n is the difference between the stoichiometric coefficients (v, which are unitless) of gaseous products and gaseous reactants. For the combustion of propane:

tJ.n =

Vproduct gases -vreactant gases

=

Vco2 -vc, H8 -vo,

= 3-1-5= -3

Using Equation 6.9: w = -t:i.nRT= -(-3)(8.314JVmol- 1)(298K) = 7.43 X 103 Jmol-

1

We see that 7.43 X 103 J of work is done on the system per mol of reaction that occurs for the combustion of propane. EXAMPLE 6.5

PRESSURE-VOLUME WORK IN A CHEMICAL REACTION

Methanoic acid (CH20 2 ) is a liquid that bums in oxygen to produce gaseous carbon dioxide and liquid water. Compute the P-V work associated with the combustion of methanoic acid at 298 K.

SORT You are given the reactants and products for a chemical reaction as well as the temperature at which you are asked to compute the P-V work associated with the chemical reaction.

GIVEN: CH2 0 2(l) and 0 2(g) produces C02 (g) and H2 0 (l), T=298 K FIND: w

STRATEGIZE The equation w = - t:i.nRT specifies the amount of work done during a chemical equation involving gases. To compute the change in the stoichiometric coefficient of gaseous products and reactants you also need a balanced equation.

CONCEPTUAL PLAN . H20 (g) t::.,I-f' = 44.0 1 kJ mol- 1

.A. When carbon dioxide sublimes, the gaseous C0 2 is cold enough to cause water vapour in the air to condense, forming fog.

Estimate the mass of water that must evaporate from the skin to cool the body by 0.50 °C. Assume a body mass of 95 kg and assume that the specific heat capacity of the body is 4 .0 (J g- 1oC- 1). 104. Propane gas burns according to the exothermic reaction:

C3Hg(g)

+ 5 0 2(g)

----->

3 C0 2(g)

+ 4 H10 (g) t::. ,I-f' = - 2044 kJ mol- 1

What mass of propane gas is necessary to heat 1.5 L of water from room temperature (25.0 °C) to boiling (100.0 °C)? Assume that during heating, 15% of the heat emitted by the propane combustion reaction goes to heat the water. The rest is lost as heat to the surroundings.

G Use standard enthalpies of formation to calculate the standard

change in enthalpy for the melting of ice. (The !::.rl-f' for H20 (s) is - 291.8 kJ mol- 1.) Use this value to calculate the mass of ice required to cool 355 mL of a beverage from room temperature (25.0 °C) to 0.0 °C. Assume that the specific heat capacity and density of the beverage are the same as those of water. 106. Dry ice is solid carbon dioxide. Instead of melting, solid carbon dioxide sublimes under normal atmospheric pressure according to the equation: C0 2(s)

G

----->

C02(g)

When dry ice is added to warm water, heat from the water causes the dry ice to sublime more quickly. The evaporating carbon dioxide produces a dense fog often used to create special effects. In a simple dry ice fog machine, dry ice is added to warm water in a Styrofoam cooler. The dry ice produces fog until it evaporates away, or until the water gets too cold to sublime the dry ice quickly enough. Suppose that a small Styrofoam cooler holds 15.0 L of water heated to 85 °C. Use standard enthalpies of formation to calculate the change in enthalpy for dry ice sublimation, and calculate the mass of dry ice that should be added to the water so that the dry ice completely sublimes away when the water reaches 25 °C. Assume no heat loss to the surroundings. (The !::.rl-f' for C02(s) is -427.4 kJ mol- 1.) A 25.5 g aluminum block is warmed to 65.4 °C and plunged into an insulated beaker containing 55.2 g water initially at 22.2 °C. The aluminum and the water are allowed to come to

[Md8speed/Shutterstock]

thermal equilibrium. Assuming that no heat is lost, what is the fi nal temperature of the water and aluminum? 108. If 50.0 mL of ethanol (density= 0.789 g mL- 1) initially at 7.0 °C is mixed with 50.0 mL of water (density= 1.0 g mL- 1) initially at 28.4 °C in an insulated beaker, and assuming that no heat is lost, what is the fi nal temperature of the mixture?

f) Palmitic acid (C 16H320 2) is a dietary fat fo und in beef and butter. The caloric content of palmitic acid is typical of fats in general. Write a balanced equation for the complete combustion of pal mitic acid and calculate the standard enthalpy of combustion. What is the caloric content of palmitic acid in Cal g- 1? Do the same calculation for table sugar (sucrose, C 12H220 11). Which dietary substance (sugar or fat) contains more Calories per gram? The standard enthalpy of formation of palmitic acid is -208 kJ mol- 1 and that of sucrose is -2226.1 kJ mol- 1. (Use H 20 (l) in the balanced chemical equations because the metabolism of these compounds produces liquid water.) 110. Hydrogen and methanol have both been proposed as alternatives to hydrocarbon fuels. Write balanced reactions for the complete combustion of hydrogen and methanol and use standard enthalp ies of formation to calculate the amount of heat released per kilogram of the fuel. Which fuel contains the most energy in the least mass? How does the energy of these fuels compare to that of octane (CsH1s)?

GD Derive a relationship between t::.H and !::. U for a process in

which the temperature of a fixed amount of an ideal gas changes. 112. Under certain nonstandard conditions, oxidation by 0 2(g) of 1 mol of S02(g) to S03(g) absorbs 89.5 kJ. The enthalpy of formation of S0 3(g) is -204.2 kJ mol- 1under these conditions. Find the enthalpy of formation of S0 2(g). One tablespoon of peanut butter has a mass of 16 g. It is combusted in a calorimeter whose heat capacity is 120.0 kJ c- 1• The temperature of the calorimeter rises from 22.2 °C to 25.4 °C. Find the food caloric content of peanut butter. 114. A mixture of 2.0 mol of H 2(g) and 1.0 mol of 0 2(g) is placed in a sealed evacuated container made of a perfect insulating

0

0

240

Chapter 6

Thermochem istry

material at 25 °C. The mixture is ignited with a spark and it reacts to form liquid water. Find the temperature of the water.

G A 20.0 L volume of an ideal gas in a cylinder with a piston is

at a pressure of 3.0 bar. Enough weight is suddenly removed from the piston to lower the external pressure to 1.5 bar. The gas then expands at constant temperature until its pressure is 1.5 bar. Find !::.. U, !::..H, q, and w for this change in state. 116. When 10.00 g of phosphorus is burned in 0 2 (g) to form P4 0 10(s), enough heat is generated to raise the temperature of 2950 g of water from 18.0 °C to 38.0 °C. Calculate the enthalpy of formation of P4 0 10(s) under these conditions.

G The t::..rH for the oxidation of S in the gas phase to S0

away. How much water boils away? Assume that no heat is lost to the surroundings. The heat of vaporization of water is 40.7 kJ mo1- 1 and the heat capacity of water is 4.184 J g- 1 0 c-1•

G Instant cold packs use ammonium nitrate as the active ingredi-

ent. When the device separating the ammonium nitrate from the water is broken, the ammonium nitrate dissolves in the water, which is an endothermic process. The enthalpy of dissolution of ammonium nitrate is 28.1 kJ mol- 1,

is -204 kJ mol- 1 and for the oxidation of S0 2 to S0 3 is 89.5 kJ mol- 1. Find the enthalpy of formation of S02 under these conditions. 118. The !::.. rH° of Til 3(s) is -328 kJ mol- 1 and the !::..rH 0 for the reaction 2 Ti(s) + 3 12 (g) ---> 2 Til3(s) is -839 kJ. Calculate the !::..rH of sublimation of I2 (s), which is a solid at 25 °C. 3

@ A gaseous fuel mixture contains 25.3% methane (CH4), 38.2%

ethane (C2 H6), and the rest propane (C3H8) by volume. When the fuel mixture contained in a l.55 L tank, stored at 755 Torr and 298 K, undergoes complete combustion, how much heat is emitted? (Assume that the water produced by the combustion is in the gaseous state.) 120. A gaseous fuel mixture stored at 745 Torr and 298 K contains only methane (CH4) and propane (C3H 8) . When l l.7 L of this fuel mixture is burned, it produces 769 kJ of heat. What is the mole fraction of methane in the mixture? (Assume that the water produced by the combustion is in the gaseous state.)

G A copper cube measuring 1.55 cm on edge and an aluminum

cube measuring 1.62 cm on edge are both heated to 55.0 °C and submerged in 100.0 mL of water at 22.2 °C. What is the final temperature of the water when equilibrium is reached? (Assume a density of 0.998 g mL- 1 for water.) 122. A gaseous solution of methane and propane having a mass of 4.00 g was burned in excess oxygen. Twenty-five percent of the heat produced was used to raise the temperature of 125 g of liquid water from 0.0 °C to its boiling point at 100.0 °C. What is the mass of methane and propane in the mixture?

@

What volume of methane at STP would be required to convert a 2 .0 kg block of ice at - 15 °C to water at 15 °C, ass uming that no heat is lost? The heat of fusion for ice is 6.0 I kJ mol- 1• The heat capacity of ice is 2.108 J g- I °C-I and the heat of combustion for methane is 89 1 kJ mol- 1•

124. A 1.0 kg block of aluminum (specific heat capacity = 0.897 J g- 1 0 c- 1 at 925 °C) is dropped into 2.0 kg of water at 25 °C. The water is heated to its boiling point, and some of the water boils

[Shutterstock]

Smarty Pants decides to design an instant cold pack to try to quickly cool his beer. Assume that a beer (including bottle) has a heat capacity of 1.5 kJ 0 c- 1 and the cold pack contains 250 g of water (Cs = 4 .1 84 J g- 1 0 c- 1) , which also needs to be cooled. What mass of NH4N0 3 is required in each cold pack to cool the beer and cold pack from 20 °C to a drinkable 9 °C? Assume also that d eeer t (.O\ the pack is only \nsta~ orM 35% efficient (i.e., Sin9\e us only 35 % of the "cooling power" goes into cooling the beer and cold pack).

126. Magnesium metal reacts with acid in solution according to the following reaction:

Mg(s)

+ 2 tt+(aq)

---+

M g2+(aq)

+ H2 (g)

When 0. 1021 g of Mg(s) is combined with enough acid solution to make 150.0 g of solution in a coffee-cup calorimeter, all of the magnesium reacts, raising the temperature from 21.80 °C to 24.90 °C. Determine !::..Ji and t::.. rU for the reaction (at 298 K). Assume that the solution is mostly water, so that the heat capacity is 4 .184 J g- 1 0 C- 1•

G During a fire rescue, a helicopter got far too low and far too hot;

the temperature reading was 75.5 °C! Spotting a backyard pool in the distance, the pilot flew there and landed in it. After cooling down (and while still sitting in the pool), the helicopter's temperature gauge read 25. 1 °C. If the helicopter's heat capacity was 6637 kJ 0 c- 1 and there was 7.01 X 104 kg of water in the pool, what was the initial temperature of the pool?

Challenge Problems 128. A typical frostless refrigerator uses 655 kWh of energy per year in the form of electricity. Suppose that all of this electricity is generated at a power plant that burns coal containing 3.2% sulfu r by mass and that all of the sulfur is emitted as S02 when the coal is burned. If all of the S02 goes on to react with rainwater to form H 2 S04 , what mass of H 2S04 does the annual operation of the refrigerator produce? (Hint: Assume that the remaining percentage of the coal is carbon, and begin by calculating !::..rH 0 for the combustion of carbon.)

@A

large sport utility vehicle has a mass of 2.5 x 103 kg. Calculate the mass of C0 2 emitted into the atmosphere upon accelerating the SUV from 0.0 mph to 65.0 mph. Assume that

the required energy comes from the combustion of octane with 30% efficiency. (Hint: Use KE = 1/i mv2 to calculate the kinetic energy required for the acceleration.) 130. Combustion of natural gas (primarily methane) occurs in most household heaters. The heat given off in this reaction is used to raise the temperature of the air in the house. Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of methane requi red to heat the air in a house by 10.0 °C. Assume the following: house dimensions are 30.0 m x 30.0 m X 3.0 m; molar heat capacity of air is 30 J K- 1 mol- 1; and l.00 mol of air occupies 22.7 L for all temperatures concerned.

Exercises

0

When backpacking in the wilderness, hikers often boil water to sterilize it for drinking. Suppose that you are planmng a backpacking trip and will need to boil 35 L of water for your group. What volume of fuel should you bring? Assume the following: the fuel has an average formula of C7 H 16; 15% of the heat generated from combustion goes to heating the water (the rest is lost to the surroundings); the density of the fuel is 0.78 g mL- 1; the initial temperature of the water is 25.0 °C; and the standard enthalpy of formation of C7H 16 is -224.4 kl mol- 1•

132. An ice cube of mass 9.0 g is added to a cup of coffee. The coffee's initial temperature is 90.0 °C and the cup contains 120.0 g of liquid. Assume that the specific heat capacity of the coffee is the same as that of water. The heat of fusion of ice (the heat associated with ice melting) is 6.0 kl mol- 1• Find the temperature of the coffee after the ice melts.

@

The optimum drinking temperature for a Shiraz is 15.0 °C. A certain bottle of Shiraz having a heat capacity of 3 .40 kl 0 c- 1 is 23.1 °C at room temperature. The heat of fusion of ice is 6.0 I kJ mol- 1 and the heat capacity of ice is 2. 108 1 g- 1 0 c- 1• Assume that no heat is lost to the rest of the surroundings. What minimum mass of ice, originally at - 5.0 °C, is required to bring the final temperature to 15.0 °C?

134. Find 11,H, 11,U, q, and w for the freezing of water at - 10.0 °C. The specific heat capacity of ice is 2.04 1 g- 1 0 c- 1 and its heat of fusion (the quantity of heat associated with melting) is -332 1 g- 1•

0

241

Starting from the relationship between temperature and kinetic energy for an ideal gas, find the value of the molar heat capacity of an ideal gas when its temperature is changed at constant volume. Find its molar heat capacity when its temperature is changed at constant pressure.

136. An amount of an ideal gas expands from 12.0 L to 24.0 L at a constant pressure of 1.0 atm. Then the gas is cooled at a constant volume of 24.0 L back to its original temperature. Then it contracts back to its original volume. Find the total heat flow for the entire process. The heat of vaporization of water at 373 K is 40.7 kl mol- 1• Find q, w, 11U, and 11H for the evaporation of 454 g of water at this temperature. 138. Find /1 U, 11H, q, and w for the change in state of 1.0 mol H 2 0(/) at 80 °C to H 2 0 (g) at 110 °C. The molar heat capacity of H 20 (/) = 75.3 1 mol- 1 K - 1, molar heat capacity of H 20 (g) = 25.01 mol- 1 K - 1, and the heat of vaporization of H 20 is 40.7 X 103 1 mol- 1 at 100 °C.

G

@ The heat of combustion of liquid octane (C8 H 18) to carbon dioxide and liquid water at 298 K is -1303 kJ mol- 1• Find 11,U for this reaction. 140. Find 11, H for the combustion of ethanol (C2H 60) to carbon dioxide and liquid water from the following data: the heat capacity of the bomb calorimeter is 34.65 kJ K - 1 and the combustion of 1.765 g of ethanol raises the temperature of the calorimeter from 294.33 K to 295.84 K.

Conceptual Problems

0

Which statement is true of the internal energy of the system and its surroundings following a process in which /1 Usys = +65 kl? Explain. a. The system and the surroundings both lose 65 kJ of energy. b. The system and the surroundings both gain 65 kl of energy. c. The system loses 65 kl of energy and the surroundings gain 65 kl of energy. d. The system gains 65 kl of energy and the surroundings lose 65 kl of energy.

142. The internal energy of an ideal gas depends only on its temperature. Which statement is true of an isothermal (constanttemperature) expansion of an ideal gas against a constant external pressure? Explain. a. /1 U is positive b. w is positive d. 11U is negative c. q is positive

0

146. Two substances, A and B, initially at different temperatures, are thermally isolated from their surroundings and allowed to come into thermal contact. The mass of substance A is twice the mass of substance B, but the specific heat capacity of substance B is four times the specific heat capacity of substance A. Which substance will undergo a larger change in temperature?

G When I mol of a gas bums at constant pressure, it produces 241 8 1 of heat and does 5 1 of work. Identify 11,U, 11,H, q, and w for the process.

@ Which

expression describes the heat evolved in a chemical reaction when the reaction is carried out at constant pressure? Explain. a. 11 U - w b. 11 U c. 11U - q

144. Two identical refrigerators are plugged in for the first time. Refrigerator A is empty (except for air) and refrigerator B is filled with jugs of water. The compressors of both refrigerators immediately turn on and begin cooling the interiors of the refrigerators. After two hours, the compressor of refrigerator A turns off, while the compressor of refrigerator B continues to run. The next day, the compressor of refri gerator A can be heard

turning on and off every few minutes, while the compressor of refrigerator B turns off and on every hour or so (and stays on longer each time). Explain these observations. A I kg cylinderofaluminum and I kg jug of water, both at room temperature, are put into a refrigerator. After one hour, the temperature of each object is measured. One of the objects is much cooler than the other. Which one is cooler and why?

148. In an exothermic reaction, the reactants lose energy and the reaction feels hot to the touch. Explain why the reaction feels hot even though the reactants are losing energy. Where does the energy come from?

@

Which statement is true of a reaction in which /1 V is positive? Explain.

a. 11H = 11 U c. 11H < 11 U

b. 11H > 11U

7.1 Quantum Mechanics: The Theory

That Explains the Behaviour of the Absolutely Small 243 7.2 The Nature of Light 243

7.3 Atomic Spectroscopy and the Bohr Model

254

7.4 The Wave Nature of Matter: The de Broglie Wavelength, the Uncertainty Principle, and Indeterminacy 260 7.5 Quantum Mechanics and the Atom 265

7.6 The Shapes of Atomic Orbitals 210 7.7 Electron Configurations: How

Electrons Occupy Orbitals 278

Our universe contains objects that span an almost unimaginable range of sizes. This chapter is focused on the behaviour of electrons, one of the smallest particles in existence. [right to left: NASA; NASA; NASA; Alan Gignoux/LatitudeStock/Alamy Stock Photo; Multichill; Travis D. Fridgen]

HE EARLY PART OF THE 20TH century brought changes that revolutionized

T

how we think about physical reality, especially in the atomic realm.

Before that time, all descriptions of the behaviour of matter had been

deterministic- the present set of conditions completely determining the future.

Quantum mechanics changed that. This new theory suggested that for subatomic particles-electrons, neutrons, and protons- the present does NOT completely determine the future. For example, if you shoot one electron down a path and measure where it lands, a second electron shot down the same path under the same conditions will most likely land in a different place! Quantum theory was developed by several unusually gifted scientists, including Albert Einstein, Niels Bohr, Louis de Broglie, Max Planck, Werner Heisenberg, P. A. M. Dirac, and Erwin Schri:idinger. These scientists did not necessarily feel comfortable with their own theory. Bohr said, "Anyone who is not shocked by quantum mechanics has not understood it." Schrodinger wrote, "I don't like it, and I'm sorry I ever had anything to do with it." Albert Einstein disbelieved the very theory he helped create, stating, "God does not play dice with the universe." In fact, Einstein attempted to disprove quantum mechanics-without success- until he died. But quantum mechanics

242

7.2

was able to account for fundamental observations, including the very stability of atoms, which could not be understood within the framework of classical physics. Today, quantum mechanics forms the foundation of chemistry-explaining the periodic table and the behaviour of the elements in chemical bonding-as well as providing the practical basis for lasers, computers, and countless other applications.

7.1

Quantum Mechanics: The Theory That Explains the Behaviour of the Absolutely Small

In everyday language, small is a relative term: Something is small relative to something else. A car is smaller than a house, and a person is smaller than a car. But smallness has limits. For example, a house cannot be smaller than the bricks from which it is made. Atoms and the particles that compose them are unimaginably small. As we have learned, electrons have a mass of less than a trillionth of a trillionth of a gram, and a size so small that it is immeasurable. A single speck of dust contains more electrons than the number of people that have existed on Earth over all the centuries of time. Electrons are small in the absolute sense of the word-they are among the smallest particles that make up matter. And yet, an atom's electrons determine its chemical and physical properties. If we are to understand these properties, we must try to understand electrons. The absolute smallness of electrons makes it a challenge to understand them through observation. Consider the difference between observing a baseball, for example, and observing an electron. You can measure the position of a baseball by observing the light that strikes the ball, bounces off it, and enters your eye. The baseball is large in comparison to the disturbance caused by the light, so the baseball is virtually unaffected by your observation. By contrast, imagine observing the position of an electron. If you attempt to measure its position using light, the light itself disturbs the electron. The interaction of the light with the electron actually changes its position, the very thing you are trying to measure. The inability to observe electrons without disturbing them has significant implications. It means that when you observe an electron, it behaves differently than when you do not observe it- the act of observation changes what the electron does. It means that our knowledge of electron behaviour has limits. It means that the absolutely small world of the electron is different from the large world that we normally experience. Therefore, we need to think about subatomic particles in a different way than we think about the macroscopic world. In this chapter, we examine the quantum-mechanical model of the atom, a model that explains how electrons exist in atoms and how those electrons determine the chemical and physical properties of elements. We have already learned much about those properties. We know, for example, that some elements are metals and that others are nonmetals. We know that the noble gases are chemically inert and that the alkali metals are chemically reactive. We know that sodium tends to form 1 + ions and that fluorine tends to form I - ions. But we do not know why. The quantum-mechanical model explains why. In doing so, it explains the modern periodic table and provides the basis for our understanding of chemical bonding.

7.2 The Nature of Light Before we explore electrons and their behaviour within the atom, we must understand some of the properties of light. As quantum mechanics developed, light was (surprisingly) found to have many characteristics in common with electrons. Chief among these is the wave-particle duality of light. Certain properties of light are best described by thinking of it as a wave, while other properties are best described by thinking of it as a particle. In this section, we will first explore the wave behaviour of light, and then its particle behaviour. We will then turn to electrons to see how they also display the same wave-particle duality.

The Nature of Light

243

244

Chapter 7

The Quantum-Mechan i ca l Model of the Atom

Electric field component

Magnetic field component

-cl:>-27---- ~,~:~:i" .A. FIGURE 7.1 Electromagnetic Radiation Electromagnetic radiation can be described as a wave composed of oscillating electric and magnetic fields. The fields oscillate in perpendicular planes.

The Wave Nature of Light Light is electromagnetic radiation, a type of energy embodied in oscillating electric and magnetic fields. An electric field is a region of space where an electrically charged particle experiences a force. A proton, for example, has an electric field around it. If you bring another charged particle into that field, that particle will experience a force . A magnetic field is a region of space where a magnetic particle experiences a force. A magnet, for example, has a magnetic field around it. If you bring another magnet into that field, that magnet will experience a force. Electromagnetic radiation can be described as a wave composed of oscillating, mutually perpendicular electric and magnetic fields propagating through space, as shown in Figure 7.1 .... In a vacuum, these waves move at a constant speed of 3.00 X 108 m s- 1fast enough to circle the Earth in one-seventh of a second. This great speed is the reason for the delay between the moment when you see a flash of lightning and the moment you hear thunder. The light from the lightning flash reaches your eye almost instantaneously. The sound, travelling much more slowly (340 m s- 1), takes longer.

.A. Because light travels nearly a million times faster than sound, the flash of lightning reaches your eyes before the roll of thunder reaches your ears. [ExcellentPhoto/iStock/Getty Images]

The symbol A is the Greek letter lambda,

I pronounced "lamb-duh."

We can characterize a wave by its amplitude and its wavelength. In the graphical representation shown at the top of the next page, the amplitude of the wave is the vertical height of a crest (or depth of a trough). The amplitude of the electric and magnetic field waves in light determines the light's intensity or brightness- the greater the amplitude, the greater the intensity. The wavelength (A) of the wave is the distance between adjacent crests (or any two analogous points) and is measured in units such as metres, micrometres, or nanometres.

7.2 The Nature of Light

Wavelength and amplitude are both related to the quantity of energy carried by a wave. Imagine trying to swim out from a shore pounded by waves. Waves of greater amplitude (higher waves) or shorter wavelength (more closely spaced, and thus steeper, waves) will make the swim more difficult. Notice also that amplitude and wavelength can vary independently of one another, as shown in Figure 7.2 T . A wave can have a large amplitude or a small amplitude and a short wavelength or a long wavelength. The most energetic waves have large amplitudes and short wavelengths. Like all waves, light is also characterized by its frequency (v), the number of cycles (or wave crests) that pass through a stationary point in a given period of time. The units of frequency are cycles per second (cycle s- 1) or simply s- 1• An equivalent unit of frequency is the hertz (Hz), defined as 1 cycle s- 1• The frequency of a wave is directly proportional to the speed at which the wave is travelling- the faster the wave, the more crests will pass a fixed location per unit time. Frequency is also inversely proportional to the wavelength (A)-the farther apart the crests, the fewer will pass a fixed location per unit time. For light, therefore, we can write: c

v= -

[7.1]

A

where the speed of light, c, and the wavelength, A, are both expressed in the same unit of distance. Therefore, wavelength and frequency represent different ways of specifying the same information-if we know one, we can readily calculate the other. For visible light-light that can be seen by the human eye-wavelength (or, alternatively, frequency) determines colour. White light, produced by the sun or by a light bulb, contains a spectrum of wavelengths and therefore a spectrum of colours. We can see these colours-red, orange, yellow, green, blue, indigo, and violet-in a rainbow or when white

Different wavelengths, different colours.

Different amplitudes, different brightness.

dim ~As--'>

\

(\ (\ (\

/

vvvv

bright

\

/\

V\J

f

.&. FIGURE 7.2 Wavelength and Amplitude Wavelength and amplitude are independent properties. The wavelength of light determines its colour. The amplitude, or intensity, determines its brightness.

The symbol v is the Greek letter nu,

I pronounced "noo."

245

246

I nano

Chapter 7

=

The Quantum-Mechan i ca l Model of the Atom

10-9

.&. FIGURE 7.3 Components of White Light We can pass white light through a prism and separate it into its constituent colours, each with a different wavelength. The array of colours makes up the spectrum of visible light.

.&. FIGURE 7.4 The Colour of an Object A red shirt is red because it reflects predominantly red light while absorbing most other colours.

[Richard Griffin/Fotolia)

[Uluc Ceylani/Shutterstock)

light is passed through a prism (Figure 7.3 ...). Red light, with a range of wavelengths centred at about 750 nanometres (nm), has the longest wavelength of visible light; violet light, with a range of wavelengths centred at about 400 nm, has the shortest. The presence of a variety of wavelengths in white light is responsible for the way we perceive colours in objects. When a substance absorbs some colours while reflecting others, it appears coloured. For example, a red shirt appears red because it reflects predominantly red light while absorbing most other colours (Figure 7.4 ...). Our eyes see only the reflected light, making the shirt appear red.

The Electromagnetic Spectrum

Gamma rays are discussed in more detail in

I Chapter 19.

Frequency, v(Hz)

104

106

Radio

Low energy

1010

10s

1012

1014

1016

10 1s

Visible light~ Microwave Infrared Ultraviolet

1022

1020

X-ray

1024

Gamma ray

AM r1,

Wavelength,

A (m)

Visible light makes up only a tiny portion of the entire electromagnetic spectrum, which includes all wavelengths of electromagnetic radiation. Figure 7 .5 T shows the main regions of the electromagnetic spectrum, ranging in wavelength from 10- 15 m (gamma rays) to 105 m (radio waves). In Figure 7.5, short-wavelength, high-frequency radiation is on the right and long-wavelength, low-frequency radiation is on the left. As you can see, visible light constitutes only a small region in the middle. As we saw previously, short-wavelength light inherently has greater energy than long-wavelength light. The most energetic forms of electromagnetic radiation have the shortest wavelengths. The form of electromagnetic radiation with the shortest wavelength is the gamma (y) ray. Gamma rays are produced by the sun, other stars, and certain unstable atomic nuclei on Earth. Excessive exposure to gamma rays is dangerous to humans because the high energy of gamma rays can damage biological molecules.

10- 9

10s

750 Red

700

650

.&. FIGURE 7.5 The Electromagnetic Spectrum

550 600 Wavelength, A (nm)

500

1 o ~u

10- 13

450

High energy

10- 1s

400

Violet

The right side of the spectrum consists of high-energy, high-frequency, shortwavelength radiation. The left side consists of low-energy, low-frequency, long-wavelength radiation. Visible light constitutes a small segment in the middle.

7.2

EXAMPLE 7.1

The Nature of Light

247

WAVELENGTH AND FREQUENCY

Calculate the wavelength (in nm) of the red light emitted by a barcode scanner that has a frequency of 4.62 X 10 14 s- 1• SOLUTION You are given the frequency of the light and asked to find its wavelength. Use Equation 7.1, which relates frequency to wavelength. You can convert the wavelength from metres to nanometres by using the conversion factor between the two (1 nm = 10-9 m).

c v= -

A c v

- - A-

3.00 X 108 m

v-1'

4.62 x 10 14 v-1'

= 6.49 x 10- 7 m = 6.49 x 10-7 m x

1 nm = 649 nm 10- m

-9-

FOR PRACTICE 7 .1 A laser dazzles the audience in a rock concert by emitting green light with a wavelength of 515 nm. Calculate the frequency of the light.

Next on the electromagnetic spectrum, with longer wavelengths than gamma rays, are X-rays, familiar to us from their medical use. X-rays pass through many substances that block visible light and are therefore used to image bones and internal organs. Like gamma rays, X-rays are sufficiently energetic to damage biological molecules. While several yearly exposures to X-rays are relatively harmless, too much exposure to X-rays increases cancer risk. Sandwiched between X-rays and visible light in the electromagnetic spectrum is ultraviolet (UV) radiation, most familiar to us as the component of sunlight that produces a sunburn or suntan. While not as energetic as gamma rays or X-rays, ultraviolet light still carries enough energy to damage biological molecules. Excessive exposure to ultraviolet light increases the risk of skin cancer A U ltraviolet light from the sun and cataracts and causes premature wrinkling of the skin. A To produce a medical X-ray, shortthe patient is exposed to produces suntans and sunburns. Next on the spectrum is visible light, ranging from wavelength electromagnetic radia[Juefraphoto/Fotolia] violet (shorter wavelength, higher energy) to red (longer tion that can pass through the skin wavelength, lower energy). Visible light- at low to modto create an image of bones and erate intensity-does not carry enough energy to damage internal organs. High-intensity visible light, such as that biological molecules. It does, however, cause certain mol(Poznyakov/Shutterstock] emitted by a laser, can damage biological ecules in our eyes to change their shape, sending a signal I tissue by burning it. to our brains that results in our ability to see. Beyond visible light lies infrared (IR) radiation. The heat you feel when you place your hand near a hot object is infrared radiation. All warm objects, including human bodies, emit infrared light. Although infrared light is invisible to our eyes, infrared sensors can detect it and are often employed in night vision technology to "see" in the dark. Beyond infrared light, at longer wavelengths still, are microwaves, used for radar and in microwave ovens. Although microwave radiation has longer wavelengths and therefore lower energies than visible or infrared light, it is efficiently absorbed by water and can therefore heat substances that contain water. The longest wavelengths are those of radio waves, which are used to transmit the signals responsible for AM and FM radio, cellular telephone, television, and other forms of communication. A Warm objects emit infrared Light,

Interference and Diffraction Waves, including electromagnetic waves, interact with each other in a characteristic way called interference: They can cancel each other out or build each other up, depending on

which is invisible to the eye but can be captured on film or by detectors to produce an infrared photograph. [Compliments of SPI CORP: www.x20.org]

248

Chapter 7

The Quantum-Mechan ica l Model of the Atom

CHEMISTRY AND MEDICINE

Radiation Treatment for Cancer

X-rays and gamma rays are sometimes called ionizing radiation because their short wavelengths correspond to high energies that can ionize atoms and molecules. When ionizing radiation interacts with biological molecules, it can permanently change or even destroy them. Consequently, we normally try to limit our exposure to ionizing radiation. However, doctors use ionizing radiation to destroy molecules within unwanted cells, such as cancer cells. In radiation therapy (or radiotherapy), doctors aim X-ray or gamma-ray beams at cancerous tumours (groups of cells that divide uncontrollably and invade surrounding healthy tissue). The ionizing radiation damages the molecules within the tumour's cells that carry genetic information-information necessary for the cell to grow and divide. Consequently, the cell dies or stops dividing. Ionizing radiation also damages molecules in healthy cells, but cancerous cells divide more quickly than normal cells, making them more susceptible to genetic damage. Nonetheless, harm to healthy tissues during treatments can result in side effects such as fatigue, skin lesions, hair loss, and organ damage. Medical workers try to reduce such effects by appropriate shielding (of healthy tissue) and by targeting the tumour from multiple directions, minimizing the exposure of healthy cells while maximizing the exposure of cancerous cells. Another side effect of exposing healthy cells to radiation is that they too may become cancerous. If a treatment for cancer may cause cancer, why do we continue to use it? In radiation therapy, as in most other disease therapies, there is an associated risk. We take risks all the time, many of them for lesser reasons. For example, every time we fly in an airplane or drive in a car, we risk injury or even death. Why? Because we perceive the benefit-the convenience of being able to travel a significant distance in a short time-to be worth the relatively small risk. The situation is similar in cancer therapy, or any other therapy for that matter. The benefit of cancer therapy (possibly curing a cancer that might otherwise kill you) is worth the risk (a slight increase in the chance of developing a future cancer).

Healthy tissue

Tumour

.A. During radiation therapy, a tumour is targeted from multiple directions in order to minimize the exposure of healthy cells while maximizing the exposure of cancerous cells. [Pearson Education)

Question

.A. In radiation therapy, highly energetic gamma rays are aimed at cancerous tumours.

Why is visible light (by itself) not used to destroy cancerous tumours?

[Fermilab/Reidar Hahn]

Understanding interference in waves is critical to understanding the wave nature of the I electron , as we will soon see.

their alignment upon interaction. For example, if two waves of equal amplitude are in phase when they interact- that is, they align with overlapping crests-a wave with twice the amplitude results. This is called constructive interference.

Constructive interference

On the other hand, if two waves are completely out of phase when they interact-that is, they align so that the crest from one source overlaps with the trough from the other source-the waves cancel by destructive interference .

.A. When a reflected wave meets an incoming wave near the shore, the two waves interfere constructively for an instant, producing a large amplitude spike. [Bonita R. Cheshier/Shutterstock]

Waves out of phase

Destructive interference

7.2

Diffracted wave

Particle trajectory

Particle Behaviour

I

w ith slit

:--

·~~~~~~~~~~~~~---­ ·~~~~~~~~~~~~~----

I

=

Waves also exhibit a characteristic behaviour called diffraction (Figure 7.6 .&.). When a wave encounters an obstacle or a slit that is comparable in size to its wavelength, it bends (or diffracts) around it. The diffraction of light through two slits separated by a distance comparable to the wavelength of the light, coupled with interference, results in an interference pattern, as shown in Figure 7.7 T . Each slit acts as a new wave source, and the two new waves interfere with each other. The resulting pattern consists of a series of bright and dark lines that can be viewed on a screen (or recorded on a film) placed at a short distance behind the slits. At the centre of the screen, the two waves travel equal distances and interfere constructively to produce a bright line. A small distance away from the centre in either direction, the two waves travel slightly different distances, so that they are out of phase. At the point where the difference in distance is one-half of one

Slits

Film (top view)

249

Na+(g) + o -(g) b. Li(g) + Cl(g) ----> Li+(g) + Ci-(g)

~ You believe you have cracked a secret code that uses elemental symbols to spell words. T he code uses numbers to designate the elemental symbols. Each number is the sum of the atomic number and the highest principal quantum number of the highest occupied orbital of the element whose symbol is to be used. The message may be written forward or backward. Decode the following messages: a. 10, 12, 58, 11, 7, 44, 63, 66 b. 9, 99, 30, 95, 19, 47, 79 98. T he electron affinity of sodium is lower than that of lithium, while the electron affinity of chlorine is higher than that of fl uorine. Suggest an explanation for this observation.

Challenge Problems Q) Consider

He

32

0.1 6

d . Use the densities and molar masses of krypton and neon to calculate the nu mber of atoms of each fo und in a volume of 1.0 L. Use these values to estimate the number of atoms that occur in 1.0 L of Ar. Now use the molar mass of argon to estimate the density of Ar. How does this estimate compare to that in part (a)?

Ne

70

0 .8 1

100. As we have seen, the periodic table is a result of empirical obser-

Ar

98

Kr

112

Xe

130

the densities and atomic radii of the noble gases at 1 bar and 25 °C:

Element

Rn

Atomic Radius (pm)

Density (g L- 1)

3.38

8.96

a. Estimate the densities of argon and xenon by interpolation fro m the data. b. Estimate the density of the element with atomic number 118, Og, by extrapolation of the data. c. Use the molar mass of neon to estimate the mass of a neon atom. Then use the atomic radius of neon to calculate the average density of a neon atom. How does this density compare to the density of neon gas? What does this comparison suggest about the nature of neon gas?

vation (i.e., the periodic law), but quantum theory explains why the table is so arranged. Suppose that, in another universe, quantum theory was such that there were one s orbital but only two p orbitals (instead of three) and only three d orbitals (instead of five). D raw out the first four periods of the periodic table in this alternative universe. W hic h elements would be the equivalent of the noble gases? Halogens? Alkali metals?

41» Consider the metals in the first transition series. Use periodic

trends to predict a trend in density as you move to the right across the series.

102. Only trace amounts of the synthetic ele ment darmstadtium, atomic number 110, have been obtained . The element is so highly unstable that no observations of its properties have been possible. Based on its position in the periodic table, propose three different plausible valence electron configurations for this element.

Exercises

41» What is the atomic number of the as yet undiscovered element

333

magnesium that are diagonal to each other have comparable metallic character. Suggest an explanation for this observation.

in which the 8s and 8p electron energy levels fill ? Predict the chemical behaviour of this element.

G) The heaviest known alkaline earth metal is radium, atomic num-

104. Consider the following isoelectronic series of ions and the ionic radii:

ber 88. Find the atomic numbers of the as yet undiscovered next two members of the series.

S 2- ( 184 pm)

0

Ci-( 18 lpm)

K+( 133pm)

Ca2+(99pm)

For each ion. estimate Zeff using Slater's rules. Does Zeff explain the variation in ionic radius? Explain why the ionic radii of the two cations are much smaller than the ionic radii of the anions.

Unlike the elements in groups 1 and 2, those in group 13 do not show a regular decrease in first ionization energy in going down the column. Explain the irregularities. 106. Using the data in Figures 8.12 and 8.13, calculate t:.E for the reaction: Na(g)

+ Cl(g)

~

Na+(g)

+ Ci-(g)

G Despite the fact that adding two electrons to 0 or S forms an ion with a noble gas electron configuration, the second electron affinity of both of these elements is negative. Explain. 108. In Section 2.7, we discussed the metalloids, which form a diagonal band separating the metals from the nonmetals. There are other instances in which elements such as lithium and

110. Predict the electronic configurations of the first two excited states (next higher energy states above the ground state) of Pd.

QD Table 8.2 does not include francium because none of its isotopes are stable. Predict the values of the entries for Fr in Table 8.2. Predict the nature of the products of the reaction of Fr with (a) water, (b) oxygen, and (c) chlorine. 112. From its electronic configuration, predict which of the fi rst 10 elements would be most similar in chemical behaviour to the as yet undiscovered element 165. Using Slater's rules, estimate Zeff for the group I metals (lithium through to rubidium). Compare these values with the atomic radii of group I metals shown in Figure 8.10. What consideration, besides Zeff• is required to explain periodic trends down a group?

@

114. The trend in second ionization energy for the elements from lithium to fluorine is not a regular one. Of the elements N, 0 , and F, 0 has the highest and N the lowest second ionization energy. Explain.

Conceptual Problems @

Imagine that in another universe, atoms and elements are identical to ours, except that atoms with six valence electrons have particular stability (in contrast to our universe, where atoms with eight valence electrons have particular stability). Give an example of an element in the alternative universe that corresponds to: a. a noble gas. b. a reactive nonrnetal. c. a reactive metal.

116. According to Coulomb's law, rank the interactions between the charged particles from lowest potential energy to highest potential energy: a. A I + charge and a I - charge separated by JOO pm. b. A 2 + charge and a 1- charge separated by 100 pm. c. A I + charge and a I + charge separated by JOO pm. d. A I + charge and a I - charge separated by 200 pm.

G Use the trends in ionization energy and electron affinity to explain why calcium fluoride has the formula CaF2 and not Ca 2F or CaF.

The AIDS drug lndinavir-shown here as the missing piece in a puzzle depicting the protein HIV-protease-was developed with the help of chemical bonding theories.

9.1 Bonding Models and AIDS Drugs 335 9.2 Types of Chemical Bonds

[Quade Paul/Pearson Education]

335

9.3 Representing Valence Electrons with Dots 337 9.4 Lewis Structures: An Introduction to Ionic and Covalent Bonding 338 9.5 The Ionic Bonding Model

343

HEMICAL BONDING IS AT THE HEART of chemistry. The bonding theories

C

that we are about to examine are-as Karl Popper eloquently states in the above quote- nets cast to understand the world. In this chapter and

the next one, we will examine three theories with successively finer "meshes." The first is Lewis theory, a simple model of chemical bonding, which can be

9.6 Covalent Bond Energies, Lengths, and Vibrations 348

carried out on the back of an envelope. With just a few dots, dashes, and chemical

9.7 Electronegativity and Bond Polarity 354

symbols, Lewis theory can help us to understand and predict a myriad of chemical

9.8 Resonance and Formal Charge 359 9.9 Exceptions to the Octet Rule: Drawing Lewis Structures for Odd-Electron Species and Incomplete Octets 364 9.10 Lewis Structures for Hypercoordinate Compounds 366

observations. The second is valence bond theory, which treats electrons in a more quantum-mechanical manner, but stops short of viewing them as belonging to the entire molecule. The third is molecular orbital theory, essentially a full quantummechanical treatment of the molecule and its electrons as a whole. Molecular orbital theory has great predictive power, but at the expense of great complexity and intensive computational requirements. Which theory is "correct"? Remember that theories are models that help us understand and predict behaviour. All three of these theories are extremely useful, depending on exactly what aspect of chemical bonding we want to predict or understand.

334

9.2

9.1

Types of Chemical Bonds

335

Bonding Models and AIDS Drugs

In 1989, researchers used X-ray crystallography-a technique in which X-rays are scattered from crystals of the molecule of interest-to determine the structure of a molecule called HIV-protease. HIV-protease is a protein (a class of large biological molecules) synthesized by the human immunodeficiency virus (HIV). This particular protein is crucial to the virus's ability to multiply and cause acquired immune deficiency syndrome, or AIDS. Without HIV-protease, HIV cannot spread in the human body because the virus cannot replicate. In other words, without HIV-protease, AIDS can't develop. With knowledge of the HIV-protease structure, pharmaceutical companies set out to create a molecule that would disable HIV-protease by sticking to the working part of the molecule, called the active site. To design such a molecule, researchers used bonding theories-models that predict how atoms bond together to form molecules-to simulate the shape of potential drug molecules and how they would interact with the protease molecule. By the early 1990s, these companies had developed several drug molecules that seemed to work. Since these molecules inhibit the action of HIV-protease, they were named protease inhibitors. In human trials, protease inhibitors, when given in combination with other drugs, have decreased the viral count in HIV-infected individuals to undetectable levels. Although protease inhibitors do not cure AIDS, many AIDS patients are still alive today because of these drugs. Bonding theories are central to chemistry because they explain how atoms bond together to form molecules. They explain why some combinations of atoms are stable and others are not. For example, bonding theories explain why table salt is NaCl and not NaCl2 and why water is H20 and not H3 0. Bonding theories also predict the shapes of molecules-a topic in our next chapter-which, in turn, determine many of the physical and chemical properties of compounds. The bonding model we examine in this chapter is called Lewis theory, named after the American chemist G. N. Lewis (1875-1946). In Lewis theory, valence electrons are represented as dots, and we can draw Lewis electron-dot structures (or simply, Lewis structures) to depict molecules. These structures, which are fairly simple to draw, have tremendous predictive power. With minimal computation, Lewis theory can be used to predict whether a particular set of atoms will form a stable molecule and what that molecule might look like. Although we will also examine more advanced theories in the following chapter, Lewis theory remains the simplest model for making quick, everyday predictions about most molecules.

9.2 Types of Chemical Bonds We begin our discussion of chemical bonding by asking the question, "why do bonds form in the first place?" This seemingly simple question is vitally important. Imagine our universe without chemical bonding. There would be just 9 1 different kinds of substances (the 91 naturally occurring elements). With such a poor diversity of substances, life would be impossible, and we would not be around to wonder why. The answer to this question, however, is not simple and involves not only quantum mechanics but also some thermodynamics that we do not introduce until Chapter 17. Nonetheless, we can address an important aspect of the answer now: Chemical bonds form because they lower the potential energy between the charged particles that compose atoms. As we already know, atoms are composed of particles with positive charges (the protons in the nucleus) and negative charges (the electrons). When two atoms approach each other, the electrons of one atom are attracted to the nucleus of the other according to Coulomb's law (see Section 7.7) and vice versa. However, at the same time, the electrons of each atom repel the electrons of the other, and the nucleus of each atom repels the nucleus of the other. The result is a complex set of interactions among a potentially large number of charged particles. If these interactions lead to an overall net reduction of energy between the charged particles, a chemical bond forms. Bonding theories help us to predict the circumstances under which bonds form and also the properties of the resultant molecules. We can broadly classify chemical bonds into three types, depending on the kind of atoms involved in the bonding (Figure 9.1 T).

X-ray crystallography is discussed in more

I detail in Section 11.10.

Proteins are discussed in more detail in

I Chapter 22.

.A. G. N. Lewis [Lawrence Berkeley National Laboratory/Science Source]

336

Chapter 9

Chem ical Bonding I: Lew is Theory

Ionic bonding

Covalent bonding

Ice, H 20(s) Table salt, NaCl(s}

Metallic bonding

e sea

Sodium metal, Na(s}

.&. FIGURE 9.1 Ionic, Covalent, and Metallic Bonding [top left: Madlen/Shutterstock: top right: Valentyn Volkov/123RF: bottom: Tim Ridley/DK Images]

Types of Atoms

Type of Bond

Characteristic of Bond

Metal and nonmetal

Ionic

Electrons transferred

Nonmetal and nonmetal

Covalent

Electrons shared

Metal and metal

Metallic

Electrons pooled

We learned in Chapter 8 that metals tend to have low ionization energies (their electrons are relatively easy to remove) and that nonmetals tend to have high electron affinities (they readily gain electrons). When a metal bonds with a nonmetal, it transfers one or more electrons to the nonmetal. The metal atom becomes a cation and the nonmetal atom an anion. These oppositely charged ions then attract one another, lowering their overall potential energy as described by Coulomb's law. The resulting bond is an ionic bond. We also learned in Chapter 8 that nonmetals tend to have high ionization energies (their electrons are relatively difficult to remove). Therefore, when a nonmetal bonds with another nonmetal, neither atom transfers electrons to the other. Instead, the two atoms share some electrons. The shared electrons interact with the nuclei of both of the bonding atoms, lowering their potential energy in accordance with Coulomb's law. The resulting bond is a covalent bond. Recall from Section 3.2 that we can understand the stability of a covalent bond by considering the most stable arrangement (the one with the lowest potential energy) of two positively charged particles separated by a small distance and a negatively charged particle. As you can see from Figure 9.2 .,,., the arrangement in which the negatively charged particle lies between the two positively charged ones has the lowest potential energy because, in this arrangement, the negatively charged particle interacts most strongly with both of the positively charged ones. In a sense, the negatively charged

9.3

~

/ / / /

+

,/ .

Lowest potential energy (most stable)

""

""

Representing Valence Electrons with Dots

I

I -

+

-

337

N(g) + N(g) It is a very strong and stable bond, which explains nitrogen's relative inertness.

The bond energy of a particular bond in a polyatomic molecule is a little more difficult to determine because a particular type of bond can have different bond energies in different molecules. For example, consider the C-H bond. In CH4, the energy required to break one C-H bond is 438 kJ mo1- 1:

!J.,H = 438 kJ mo1- 1 However, the energy required to break a C- H bond in other molecules varies slightly, as shown here: F3 C- H(g)

------'>

Br3C - H(g)

------'>

CI 3C- H(g)

------'>

+ H(g) Br3C(g) + H(g) Cl 3C(g) + H(g) F3C(g)

!J.,H = 446 kJ mol- 1 !J.rH = 402 kJ mol- 1 !J.,H = 401 kJ mo1- 1

We can calculate an average bond energy for a chemical bond, which is an average of the bond energies for that bond in a large number of compounds. For example, for the limited number of compounds listed above, we calculate an average C- H bond energy of 422 kJ mo1- 1• Table 9. 1 lists average bond energies for a number of common chemical bonds averaged over a large number of compounds. Notice that the C- H bond energy is listed as 414 kJ mo1- 1, which is not too different from the value we calculated from our limited number of compounds. Notice also that bond energies depend not only on the kind

9.6

TABLE9.1

Average Bond Energies

Bond

Bond Energy (kJ mo1- 1)

H- H H-C H-N H-0 H-S H- F H-CI H-Br H- 1

c-c C=C C=C C-N C=N C=N C-0 C=O C=O C-CI

436 414 389 464 368 565 431 364 297 347 61 1 837 305 615 891 360 736* 1072 339

Covalent Bond Energies, Lengths, and Vibrations

349

I Bond Energy (kJ mo1- 1)

Bond

Bond

163 418 946 222 590 272 200 243 159 142 498 190 203 234 159 253 243

N-N N=N N=N N-0 N=O N-F N-CI N-Br N-1 0-0 O=O 0-F 0 -CI 0-1 F- F Cl-F Cl-Cl

Br-F Br-CI Br-Br I-Cl 1-Br 1- 1 Si- H Si-Si Si-C Si-0 Si-Si Si=O Si-Cl S-F S-0 S=O S-CI S-Br

s-s S=S

Bond Energy (kJ mo1- 1)

237 218 193 208 175 151 323 226 301 450 226 523 391 285 265 515 253 218 266 418

'799 in C02

of atoms involved in the bond, but also on the type of bond: single, double, or triple. In general, for a given pair of atoms, triple bonds are stronger than double bonds, which are, in turn, stronger than single bonds. For example, consider the bond energies of carboncarbon triple, double, and single bonds listed at left.

Using Average Bond Energies to Estimate Enthalpy Changes for Reactions We can use average bond energies to estimate the enthalpy change of a reaction. For example, consider the following reaction: H 3C-H(g)

+ Cl-Cl(g)

~

H 3C-Cl(g)

+ H-Cl(g)

We can imagine this reaction occurring by the breaking of a C - H bond and a Cl - Cl bond and the forming of a C- Cl bond and an H - Cl bond. We know that when bonds break, the process is endothermic (positive bond energy), and when bonds form, the process is exothermic (negative bond energy). So, we can calculate the overall enthalpy change as a sum of the enthalpy changes associated with breaking the required bonds in the reactants and forming the required bonds in the products, as shown in Figure 9.6 T . H 3C-H(g)

+ Cl-Cl(g)

~

H 3C-Cl(g)

Bonds Broken +414 kJ mo1- 1 Cl - Cl break +243 kJ mo1- 1

+ H-Cl(g)

Bonds Formed -339 kJ mo1- 1 H - Cl form -43 1 kJ mo1- 1

C - H break

C- Cl form

Sum (l) !::./f' s bonds broken: +657 kJ mol- 1 Sum (l) !::./f' s bonds formed: -770 kJ mol- 1 !::./f

=

l(!::./f's bonds broken)

+

l(!::.,H's bonds formed)

1

= +657 kJ mol- - 770 kJ mol- 1 = -113 kJ mo1- 1 We find that !::./f = -113 kJ mol- 1• Calculating !::./f from tabulated enthalpies of formation- as we learned in Chapter 6-gives !::.,H0 = -101 kJ mol- 1- fairly close to 0

Bond

Bond Energy (kJ mol- 1)

C=C

837 611 347

C=C

c-c

350

Chapter 9

Chemi cal Bonding I: Lew is Th eory

.... FIGURE 9.6 Estimating a 1Hfrom Bond Energies We can approximate the enthalpy change of a reaction by summing up the enthalpy changes involved in breaking old bonds and forming new ones.

+

H

f) Make C-CI and H-CI bonds

0 Break C-H and Cl -Cl bonds

the value we obtained from average bond energies. In general, you can calculate !:i./f from average bond energies by summing the changes in enthalpy for all of the bonds that are broken and adding the sum of the enthalpy changes for all of the bonds that are formed. Remember that !:i./f is positive for breaking bonds and negative for forming them:

f:i. 1 H = 2.(!:i.1H 's bonds broken)

t

Positive

+ 2.(!:i. H 's bonds formed) 1

t

Negative

As you can see from the above equation: ~

A reaction is exothermic when weak bonds break and strong bonds form.

~

A reaction is endothermic when strong bonds break and weak bonds form.

Scientists often say that "energy is stored in chemical bonds or in a chemical compound," which may make it sound as if breaking the bonds in the compound releases energy. For example, in biology, we often hear that energy is stored in glucose or in ATP. However, breaking a chemical bond always requires energy. When scientists say that energy is stored in a compound, or that a compound is energy rich, it means that the compound can undergo a reaction in which weak bonds break and strong bonds form, releasing energy. It is always the forming of chemical bonds that releases energy.

The reaction between hydrogen and oxygen to form water is highly exothermic. Which of the following is true of the energies of the bonds that break and form during the reaction? (a) The energy needed to break the required bonds is greater than the energy released when

the new bonds form. (b) The energy needed to break the required bonds is less than the energy released when the

new bonds form. (c) The energy needed to break the required bonds is about the same as the energy released when the new bonds form.

9.6

EXAMPLE 9.5

Covalent Bond Energies, Lengths, and Vibrations

351

CALCULATING J),,rff FROM BOND ENERGIES

Hydrogen gas, a potential fuel , can be made by the reaction of methane gas and steam:

Use bond energies to calculate 11,H for this reaction.

SOLUTION Begin by rewriting the reaction using the Lewis structures of the molecules involved.

H

H-?-H+2H-Q-H

~

4H-H+Q=C=Q

H

Determine which bonds are broken in the reaction and sum the bond energies of these.

H

I

H- C- H + 2 H- 0 - H

I

H

'l(11,Hs bond broken) = 4(C-H) + 4(0-H) = 4(414 kJ mol- 1) + 4(464 kJ mol- 1) = 3512 kJ mol- 1

Determine which bonds are formed in the reaction and sum the negatives of their bond energies.

..

..

4H - H + O .. = C = O..

'l(l1,Hs bonds formed) = -4(H-H)-2(C=O) = -4(436 kJ mol- 1) - 2(799 kJ mol- 1) = -3342 kJ mol- 1

Find 11.H by summing the results of the previous two steps.

11,H = l(11,H's bonds broken) + l(11,H 's bonds formed)

= 3512 =

- 3342

170 kJ mol- 1

FOR PRACTICE 9.5 Another potential future fuel is methanol (CH30H). Write a balanced equation for the combustion of gaseous methanol and use bond energies to calculate the enthalpy of combustion of methanol in kJ mol- 1• FOR MORE PRACTICE 9.5 Use bond energies to calculate 11,H for this reaction: N2(g)

+ 3 H2(g)

~

2 NH3 (g).

Bond Lengths Just as we can tabulate average bond energies, which represent the average energy of a bond between two particular atoms in a large number of compounds, we can tabulate average bond lengths (Table 9.2). The average bond length represents the average length of a bond between two particular atoms in a large number of compounds. Like bond energies, bond lengths depend not only on the kind of atoms involved in the bond, but also on the type of bond: single, double, or triple. In general, for a particular pair of atoms, triple bonds are shorter than double bonds, which are, in turn, shorter than single bonds. For example, consider the bond lengths (along with bond strengths, repeated from earlier in this section) of carbon-carbon triple, double, and single bonds: Bond

Bond Length (pm)

Bond Strength (kJ mol- 1)

C=C C=C

120 134 154

837 611 347

c-c

352

Chapter 9

Chem ica l Bond ing I: Le wis Th eory

f'TABLE9.2 Bond

I

Average Bond Lengths Bond Length (pm)

H- H H- C H- N H- 0 H- S H-F H- CI H- Br H- 1 Br- Br

74 110 100 97 132 92 127 141 161 228

Bond

c- c C=C C= C C- N C=N C=N C- 0 C= O C- CI 1- 1

Bond Length (pm)

154 134 120 147 128 11 6 143 120 178 266

Bond

N- N N=N N= N N- 0 N=O 0-0 O= O F- F Cl- Cl

Bond Length (pm)

145 123 110 136 120 145 121 143 199

Notice that as the bond gets longer, it also becomes weaker. This relationship between the length of a bond and the strength of a bond does not necessarily hold for all bonds. Consider the following series of nitrogen- halogen single bonds: Bond

( Bond Lengths )

Fz

NNNN-

F CI Br 1

Bond Length (pm)

Bond Strength (kJ mol- 1)

139 191 214 222

272 200 243 159

Although the bonds generally get weaker as they get longer, the trend is not a smooth one. 143 pm

C l2

199 pm

Br 2

12

266 pm

.A. Bond lengths in the diatomic halogen molecules.

Bond Vibrations Bond lengths are not static. That is, when two atoms are bonded together, the distance between the atoms is not the same all the time. In fact, the two atoms move toward and away from each other continuously, like two balls attached to each other by a spring. This type of motion is called a bond stretching vibration, or a stretching vibration. Literally, the bond stretches and then contracts. For a bond vibration to take place, energy must be supplied to the molecule. Energy can originate from molecular collisions or from the absorption of electromagnetic radiation in the infrared region because the infrared frequency range of electromagnetic radiation corresponds to the energies (E = h v) of molecular vibrations. The absorption of infrared radiation by a compound can be measured easily by irradiating the sample with an infrared source (called a glow bar) and measuring the intensity of radiation transmitted through the sample at each wavelength or frequency. This is called infrared (IR) spectroscopy. In infrared absorption spectroscopy, rather than reporting the wavelength of absorptions, chemists use the reciprocal of wavelength, called the wavenumber, which is usually given in units of reciprocal centimetres (cm- 1) . This unit is proportional to energy and frequency. Thus, the higher the wavenumber, the greater the energy of light that was absorbed. In Figure 9.7 .,., the wavenumber values of IR absorptions of HX (X=F, Cl, Br, I) molecules are plotted against the H - X bond energies. As the bond strength increases, so too does the wavenumber value of the absorption. In other words, it takes higher energy photons to excite the stretching vibration of HF, with a bond energy of 565 kJ mol- 1, than it does to excite the stretching vibration of HI, with a bond energy of 297 kJ mol- 1•

9.6

Covalent Bond Energies, Lengths, and Vibrations

4200

Lone pair-bonding pair > Bonding pair-bonding pair Most re pulsive

Least repulsive

Molecular geometry: bent

384

Chapter 1O

Chem ical Bond i ng II: Molecular Shapes, Valence Bond Theory , and Molecular Orbital Theory

We see the effects of this ordering in the progressively smaller bond angles of CH4, NH3, and H20, as shown in Figure 10.4 ..-. The relative ordering of repulsions also helps to determine the geometry of molecules with five and six electron groups when one or more of those groups are lone pairs, as we shall now see. ( No lone pairs )

Ideal tetrahedral geometry

( O ne lone pair )

( Two lone pairs )

Actual molecular geometry

(a) ... FIGURE 10.4 The Effect of Lone Pairs on Molecular Geometry (a) The bond angles get progressively smaller as the number of lone pairs on the central atom increases from zero in CH4 to one in NH3 to two in H 20. (b) Sketches of the tetrahedral, trigonal pyramidal, and bent molecular geometries.

,,.N .. H,,. \'''1 H

(b)

H

Five Electron Groups with Lone Pairs Consider the Lewis structure of SF4 : :f:

.I .. I ..

:f--'S- F:

:f:

The seesaw molecular geometry is some-

Itimes called an irregular tetrahedron.

The central sulfur atom has five electron groups (one lone pair and four bonding pairs). The electron geometry, due to the five electron groups, is trigonal bipyramidal. In determining the molecular geometry, notice that the lone pair could occupy either an equatorial position or an axial position within the trigonal bipyrarnidal electron geometry. Which position is most favourable? To answer this question, we must consider that, as we have just seen, lone pair-bonding pair repulsions are greater than bonding pair-bonding pair repulsions. Consequently, the lone pair should occupy the position that minimizes its interaction with the bonding pairs. If the lone pair were in an axial position, it would have three 90° interactions with bonding pairs. In an equatorial position, however, it has only two 90° interactions. Consequently, the lone pair occupies an equatorial position. The resulting molecular geometry is called seesaw, because it resembles a seesaw (or teeter-totter): Three go0 lone pair-bonding pair repulsions

Axial lone pair does not occur

Two go0 lone pair-bonding pair repulsions

Equatorial lone pair

Molecular geometry : seesaw

10.3 VSEPR Theory: The Effect of Lone Pairs

When two of the five electron groups around the central atom are lone pairs, as in BrF3, the lone pairs occupy two of the three equatorial positions-again, minimizing 90° interactions with bonding pairs and also avoiding a lone pair-lone pair 90° repulsion. The resulting molecular geometry is T-shaped:

:f:

F

I

I ..Br-F:

·.

Br-F

..

I

F

Electron geometry: trigonal bipyramidal

Molecular geometry: T-shaped

When three of the five electron groups around the central atom are lone pairs, as in XeF2, the lone pairs occupy all three of the equatorial positions, and the resulting molecular geometry is linear:

:f:

F

I. :xe:

I I

Xe

1•

:f:

F

Electron geometry: trigonal bipyramidal

Molecular geometry: linear

Six Electron Groups with Lone Pairs The Lewis structure of BrF5 is shown below. The central bromine atom has six electron groups (one lone pair and five bonding pairs). The electron geometry, due to the six electron groups , is octahedral. Since all six positions in the octahedral geometry are equivalent, the lone pair can be situated in any one of these positions. The resulting molecular geometry is square pyramidal:

:f:

.. .I

..

"'....

:f----=Br-F:

..

/ :f:

..

:f:

Electron geometry: octah edral

Molecular geometry: square pyramidal

385

386

Chapter 1O

TABLE 10.1 Electron Groups*

Chem ical Bonding II: Molecular Shapes, Valence Bond Theory , and Molecular Orbital Theory

Electron and Molecular Geometries Bonding Groups

Lone Pairs

Electron Geometry

Molecular Geometry

Approximate Bond Angles

2

2

0

Linear

Linear

180°

3

3

0

Trigonal planar

Trigonal planar

120°

Example

.. .. :o=c=o:

•• •

:p:

..

..

I ..

..

:F-B-F:

3

Trigonal planar

2

Bent

A 1.00 mol L-1 NaCl solution

-

.,....OJ:J i::

~

H3C-N==C

~ Reactant H3C-C==N

Reaction progress

(b)

Product

.6. FIGURE 13.13 The Activated Complex (a) The reaction pathway includes a Reaction progress

transitional state-the activated complex-that has a higher energy than either the reactant or the product. (b) Each wag is an approach to the activation barrier.

(a)

The energy required to reach the activated complex is the activation energy. The higher the activation energy, the slower the reaction rate (at a given temperature).

The Frequency Factor We just saw that the frequency factor represents the number of approaches to the activation barrier per unit time. Any time that the NC group begins to rotate, it approaches the activation barrier. For this reaction, the frequency factor represents the rate at which the NC part of the molecule wags (vibrates side-to-side). With each wag, the reactant approaches the activation barrier (Figure 13. l 3(b) ). However, approaching the activation barrier is not equivalent to surmounting it. Most of the approaches do not have enough total energy to make it over the activation barrier. The Exponential Factor The exponential factor is a number between 0 and 1 that represents the fraction of molecules that have enough energy to make it over the activation barrier on a given approach. Another way of saying this is that the exponential factor is the fraction of approaches that are energetic enough to be successful and result in the product. For example, if the frequency factor is 109 s- 1 and the exponential factor is 10-7 at a certain temperature, then the overall rate constant at that temperature is 109 s- 1 x 10- 7 = 102 s- 1. In this case, the NC group is wagging at a rate of 109 s- 1. With each wag, the activation barrier is approached. However, only 1 in 107 molecules has sufficient energy to actually make it over the activation barrier. The exponential factor depends on the temperature (7), which you can control, and the activation energy (Ea) of the reaction is a property of the reacting species, which you cannot control. Exponential factor

=

e-E,/ RT

A low activation energy and a high temperature make the negative exponent small, so that the exponential factor approaches one. For example, if the activation energy is zero, then the exponent is zero, and the exponential factor is exactly one (e-0 = I)-every approach to the activation barrier is successful. By contrast, a large activation energy and a low temperature make the exponent a very large negative number, so that the exponential factor becomes very small. For example, as the temperature approaches 0 K, the exponent approaches an infinitely large number, and the exponential factor approaches zero (e- = 0) and the reactants can never get over the energy barrier to form products. As the temperature increases, the number of molecules having enough thermal energy to surmount the activation barrier increases. At any given temperature, a sample of molecules will have a distribution of energies, as shown in Figure 13.14 T . Under common circumstances, only a small number of the molecules have enough energy to make it over the activation barrier. Because of the shape of the energy distribution curve, however, a small change in temperature results in a large difference in the number of molecules having enough energy to surmount the activation barrier. This explains the sensitivity of reaction rates to temperature. 00

Summarizing Temperature and Reaction Rate: ~

The frequency factor is the number of times that the reactants approach the activation barrier per unit time.

560

Chapter 13

Chemical Ki netics

.... FIGURE 13.14 Thermal Energy Distribution At any given temperature, the atoms or molecules in a gas sample will have a range of energies. The higher the temperature, the wider the energy distribution and the greater the average energy. The fraction of molecules with enough energy to surmount the activation energy barrier and react increases sharply as the temperature rises.

As temperature increases, the fraction of m olecules with enough energy to surmount the activation energy barrier also increases.

E.

~

"u "5

1

0"

i

\+.;

I

a

Activation energy

0

c 0

·;:: u

"' ~ Energy ~

The exponential factor is the fraction of approaches that are successful in surmounting the activation barrier and forming products.

~

The exponential factor increases with increasing temperature, but a large activation energy results in a small exponential factor.

Arrhenius Plots: Experimental Measurements of the Frequency Factor and the Activation Energy The frequency factor and activation energy are important quantities in understanding the kinetics of any reaction. To see how we measure these factors in the laboratory, consider again Equation 13.24: k = Ae- EJRT. Taking the natural logarithm of both sides of this equation, we get the following:

= In(Ae- EJ RT) Ink = In A + In e- E,/RT

I Remember that ln(A x B) = In A + In B.

Ink

I Remember that In e' = x.

Ea Ink = InA - -

RT

In k In an Arrhenius analysis, the pre-exponential factor (A) is assumed to be independent of temperature. Although the pre-exponential factor does depend on temperature to some degree, its temperature dependence is much less than that of the exponential factor and is often ignored.

EXAMPLE

13.7

[1 3.25)

Ea = - R

(1) T +

In A

y=mx+ b

Equation 13.26 is in the form of a straight line. A plot of the natural logarithm of the rate constant (ln k) versus the inverse of the temperature in kelvin ( l/T) yields a straight line with a slope of-Ea / R and a y-intercept of ln A. Such a plot is called an Arrhenius plot and is commonly used in the analysis of kinetic data, as shown in the following example.

USING AN ARRHENIUS PLOT TO DETERMINE KINETIC PARAMETERS

The decomposition of ozone shown here is important to many atmospheric reactions: 0 3(g) Temperature (K)

[13.26)

Rate Constant (L mo1-

~ 1

s- 1)

Oz(g)

+

O(g)

Temperature (K)

Rate Constant (L mo1- 1 s- 1)

600

3.37 x 103

1300

7.83 x 107

700

1400

l.45 x 108

800

4.85 x 104 3.58 x 105

1500

2.46 x 108

900

l.70 x 106

1600

3.93 x 108

1000

106

1700

5.93 x 108

1100

7

l.63 x 10

1800

8.55 x 108

1200

3.81 x 107

1900

l.1 9 x 109

5.90 x

A study of the kinetics of the reaction results in the data table. Determine the value of the frequency factor and activation energy for the reaction.

13.5 The Effect of Temperature on Reaction Rate

561

SOLUTION To find the frequency factor and activation energy, prepare a graph of the natural logarithm of the rate constant (In k) versus the inverse of the temperature (1 / T).

'-...,._

15 --

PS Cs

Cl2 + HN0 3

When the sun rises in the Antarctic spring, the sunlight dissociates the chlorine molecules into chlorine atoms:

CJ2

-------'>

light

2 CJ

The chlorine atoms then catalyze the destruction of ozone by the mechanism discussed previously. This continues until the sun melts the stratospheric clouds, allowing chlorine atoms to be reincorporated into their reservoirs. The result is an ozone hole that forms every spring and lasts about 6- 8 weeks (Figure 13.19 T). A second example of heterogeneous catalysis involves the hydrogenation of double bonds within alkenes. Consider the reaction between ethene and hydrogen, which is relatively slow at normal temperatures:

However, in the presence of finely divided platinum, palladium, or nickel, the reaction happens rapidly. The catalysis occurs by the four-step process depicted in Figure 13.20 T . 1. Adsorption: the reactants are adsorbed onto the metal surface. 2. Diffusion: the reactants diffuse on the surface until they approach each other. 3. Reaction: the reactants react to form the products. 4. Desorption: the products desorb from the surface into the gas phase.

The large activation energy of the hydrogenation reaction-due primarily to the strength of the hydrogen-hydrogen bond in H2- is greatly lowered when the reactants adsorb onto the surface. May

September

products What is the rate law if the reaction is zero order with respect to A? First order? Second order? For each case, explain how a doubling of the concentration of A would affect the rate of reaction. 7. How is the order of a reaction generally determined? 8. For a reaction with multiple reactants, how is the overall order of the reaction defined? 9. Explain the difference between the rate law for a reaction and the integrated rate law for a reaction. What relationship does each ki nd of rate law express? 10. Write integrated rate laws for zero-order, first-order, and second-order reactions of the form A ----> products. 11. What do the terms half-life, lifetime, and decay time mean? Write the general expressions for the decay time and half-life for zero-, first-, and second-order reactions. Write expressions for the amount of time it takes for zero-, first-, and second-order reactions to deplete to 1/5 the original reactant concentration.

12. How do reaction rates typically depend on temperature? What part of the rate law is temperature dependent? 13. Explain the meaning of each term within the Arrhenius equation: activation energy, frequency factor, and exponential factor. Use these terms and the Arrhenius equation to explain why small changes in temperature can result in large changes in the reaction rate.

14. What is an Arrhenius plot? Explain the significance of the slope and intercept of an Arrhenius plot. 15. Explain how a chemical reaction occurs according to the collision model. Explain the meaning of the orientation factor within this model. 16. Explain the difference between a normal chemical equation for a chemical reaction and the mechanism of that reaction. 17. In a reaction mechanism, what is an elementary step? Write down the three most common elementary steps and the corresponding rate law for each one. 18. What are the two requirements for a proposed mechanism to be valid for a given reaction? 19. What is an intermediate within a reaction mechanism? 20. What is a catalyst? How does a catalyst increase the rate of a chemical reaction? 21. Explain the difference between homogeneous catalysis and heterogeneous catalysis. 22. What are the four basic steps involved in heterogeneous catalysis? 23. What are enzymes? What is the active site of an enzyme? What does the term substrate mean? 24. What is the general two-step mechanism by which most enzymes work?

Problems by Topic Reaction Rates

fD Consider the reaction: a. Express the rate of the reaction in terms of the change in concentration of each of the reactants and products.

b. In the first 25.0 s of this reaction, the concentration of HBr dropped from 0.600 mol L_, to 0.512 mol L- 1• Calculate the average rate of the reaction in this time interval. c. If the volume of the reaction vessel in part b was 1.50 L, what amount of Br2 (in moles) was formed during the first 15.0 s of the reaction?

581

Exercises 26. Consider the reaction: 2 N 20(g)

------->

2 N2(g)

+ 0 2 (g)

a. Express the rate of the reaction in terms of the change in concentration of each of the reactants and products. b. In the first 15.0 s of the reaction, 0.015 mol of 0 2 is produced in a reaction vessel with a volume of 0.500 L. What is the average rate of the reaction over this ti me interval? c. Predict the rate of change in the concentration of N20 over this time interval. In other words, what is !:l.[N 20] / flt?

Q

For the reaction 2 A(g) + B(g) -------> 3 C(g), a. Determine the expression for the rate of the reaction in terms of the change in concentration of each of the reactants and products. b. When A is decreasing at a rate of 0.100 mol L - I s- 1, how fast is B decreasing? How fast is C increasing?

!

28. For the reaction A(g) + B(g) -------> 2 C(g), a. Determine the expression for the rate of the reaction in terms of the change in concentration of each of the reactants and products. b. When C is increasing at a rate of 0.025 mol L- 1 s- 1, how fast is B decreasing? How fast is A decreasing?

G)

+

20

0.904

30

0.860

40

0.8 18

50

0.778

60

0.740

70

0.704

80

0.670

90

0.637

JOO

0.606

Between 50 and 60 s? b. What is the rate of formation of 0 2 between 50 and 60 s?

G)

Consider the reaction:

3 F2(g)

------->

+ Br2(g)

-------> 2

HBr(g)

The graph below shows the concentration of Br2 as a function of time.

2 CIF3(g)

1.2

!:l.[Cl2]/ M

Rate

MF2]/ M

-0.012 mol L - s-

1.000 0.95 1

H 2(g)

Complete the table.

1

0 10

a. What is the average rate of the reaction between I0 and 20 s?

Consider the reaction: Clz(g)

[NOz] (mol L- 1)

Time (s)

'S

1.0

0

1

._§, 0.8

""- "-..... ""-..

c:

'-,...

.g 0.6 rl c 0.4

30. Consider the reaction:

Br

l

.......... ""-r--.

OJ

u

c: 0

Complete the table.

u 0.2 0

0

1SO

100 Time(s)

a. Use the graph to calculate each quantity:

Consider the reaction:

The following data were collected for the concentration of C4H8 as a function of time: Time (s)

[C4H8] (mol L- 1)

0

I.OOO

10

0 .9 13

20

0.835

30

0.763 0.697

50

0.637

34. Consider the reaction:

1.2

a. What is the average rate of the reaction between 0 and I0 s? Between 40 and 50 s? b. What is the rate of formation of C2H4 between 20 and 30 s? 32. Consider the reaction: -------> NO(g)

(i) The average rate of the reaction between 0 and 25 s. (ii) The instantaneous rate of the reaction at 25 s. (iii) The instantaneous rate of formation of HBr at 50 s. b. Make a rough sketch of a curve representing the concentration of HBr as a function of time. Assume that the initial concentration of HBr is zero.

The graph below shows the concentration of H 20 2 as a function of time.

40

N02(g)

so

0

-0.080 mol L - 1 s- 1

'S

1.0

0 ._§, 0.8 c:

.g

"'

0.6

"

' i-.

""- ' r--.

rl

""'

~ 0.4 u

c:

I

+ 2 02(g)

fH O ?l

u 0.2

............

----

0

The tabulated data were collected for the concentration of N02 as a function of time:

0

10

20

30

40 T ime(s)

so

60

70

80

582

Chapter 13

Chem ical Kine ti cs

Use the graph to calculate each quantity: a . The average rate of the reaction between I 0 and 20 s. b. T he instantaneous rate of the reaction at 30 s. c. The instantaneous rate of formation of 0 2 at 50 s. d. If the initial volume of the H20 2 is 1.5 L, what total amount of 0 2 (in moles) is formed in the first 50 s of reaction?

order? (Assume the same numerical value for the rate constant with the appropriate units.)

G)

The Rate Law and Reaction Orders

e The graph below shows a plot of the rate of a reaction versus the concentration of the reactant A for the reaction A

~

products.

0.012 0.010

'~ 'i

0.008

/

....l

v

/

0 0.006

40. A reaction in which A, B, and C react to form products is zero

/

_§_

order in A, one-half order in B, and second order in C. a . Write a rate law for the reaction. b. What is the overall order of the reaction? c. By what factor does the reaction rate change if [AJ is doubled (and the other reactant concentrations are held constant)? d. By what factor does the reaction rate change if [BJ is doubled (and the other reactant concentrations are held constant)? e. By what factor does the reaction rate change if [CJ is doubled (and the other reactant concentrations are held constant)? f. By what factor does the reaction rate change if the concentrations of all three reactants are doubled?

I/

!l 0.004 "' i:i:;

I/ I/

0.002 0

v

/ /

0

0.2

0.4 0.6 [A](mol L- 1)

0.8

a . What is the order of the reaction with respect to A? b. Make a rough sketch of how a plot of [AJ versus time would appear. c. Write a rate law for the reaction including an estimate for the value of k.

Q

36. The graph below shows a plot of the rate of a reaction versus the concentration of the reactant.

'i

~

'i

A reaction in which A, B, and C react to form products is first order in A, second order in B, and zero order in C. a. Write a rate law for the reaction. b. W hat is the overall order of the reaction? c. By what factor does the reaction rate change if [AJ is doubled (and the other reactant concentrations are held constant)? d. By what factor does the reaction rate change if [BJ is doubled (and the other reactant concentrations are held constant)? e. By what factor does the reaction rate change if [CJ is doubled (and the other reactant concentrations are held constant)? f. By what factor does the reaction rate change if the concentrations of all three reactants are doubled?

Consider the data below showing the initial rate of a reaction (A ~ products) at several different concentrations of A. What is the order of the reaction? Write a rate law for the reaction including the value of the rate constant, k.

0.012

[A] (mol L- 1)

Initial Rate (mol L- 1 s- 1)

0.010

0.100

0.053

0.008

0.200

0.210

0.300

0.473

....l

0 0.006

_§_

42. Consider the data below showing the initial rate of a reaction (A ~ products) at several different concentrations of A. What is the order of the reaction? Write a rate law for the reaction including the value of the rate constant, k.

!l 0.004 "' i:i:;

0.002 0 0

0.2

0.4 0.6 [A](mol L- 1)

0.8

[A] (mol L- 1)

Initial Rate (mol L- 1 s- 1)

0. 15

0.008

a. What is the order of the reaction with respect to A? b. Make a rough sketch of how a plot of [AJ versus time would appear. c. Write a rate law for the reaction including the value of k.

Q) What are the units of k for each type of reaction?

$

0.30

0.016

0.60

0.032

The data below were collected for this reaction:

a . first-order reaction b. second-order reaction c. zero-order reaction 38. This reaction is first order in N2 0 5:

The rate constant for the reaction at a certain temperature is 0.053 s- 1• a . Calculate the rate of the reaction when [N 20 5J = 0.055 molL- 1• b. What would the rate of the reaction be at the same concentration as in part a if the reaction were second order? Zero

[N0 2] (mol L- 1)

[F21 (mol L- 1)

Initial Rate (mol L- 1 s- 1)

0.1 00

0.100

0.026

0.200

0.100

0.05 1

0.400

0.400

0.411

Write an expression for the reaction rate law and calculate the value of the rate constant, k. What is the overall order of the reaction?

Exercises 44. The data below were collected for this reaction:

n me(s)

[N20sl (mol L- 1)

0

I.OOO

25

0.822

50

0.677

[CH3CI] (mol L- 1)

[Clzl (mol L- 1)

Initial Rate (mol L- 1 s- 1)

0.050

0.050

0.014

75

0.557

0. 100

0.050

0.029

100

0.458

0.200

0.200

0. 115

Write an expression for the reaction rate law and calculate the val ue of the rate constant, k. What is the overall order of the reaction?

The Integrated Rate Law and Half-Life

G Indicate the order of reaction consistent with each observation. a. A plot of the concentration of the reactant versus time yields a straight line. b. T he reaction has a half-life that is independent of initial concentration. c. A plot of the inverse of the concentration versus time yields a straight line. 46. Indicate the order of reaction consistent with each observation. a. The half-life of the reaction gets shorter as the initial concentration is increased. b. A plot of the natural log of the concentration of the reactant versus time yields a straight line. c. The half-life of the reaction gets longer as the initial concentration is increased. G) The data below show the concentration of AB versus time for this reaction: AB(g)

-----> A(g)

Time (s)

+

B(g)

[AB] (mol

0

0.950

50

0.459

100

0.302

150

0.225

200

0.180

250

0. 149

300

0.128

L- 1)

350

0. 11 2

400

0.0994

450

0.0894

500

0.08 12

Determine the order of the reaction and the value of the rate constant. Predict the concentration of AB at 25 s. 48. The data below show the concentration of N20 5 versus time for this reaction:

G)

583

125

0.377

150

0.3 10

175

0.255

200

0.210

Determine the order of the reaction and the value of the rate constant. Predict the concentration of N2 0 5 at 250 s. The data below show the concentration of cyclobutane (C4 H8) versus time for this reaction:

Time (s)

[C4 H8] (mol L- 1)

0

I.OOO

10

0.894

20

0.799

30

0.714

40

0.638

50

0.571

60

0.510

70

0.456

80

0.408

90

0.364

100

0.326

Determine the order of the reaction and the value of the rate constant. What is the rate of reaction when [C4H8] = 0.25 mol L - I? 50. A reaction in which A -----> products was monitored as a function of time. The results are shown below.

Time (s)

[A) (mol L- 1)

0

I.OOO

25

0.914

50

0.829

75

0.744

100

0.659

125

0.573

150

0.488

175

0.403

200

0.318

Determine the order of the reaction and the value of the rate constant. What is the rate of reaction when [A] = 0. 10 mol L- I?

584

8

Chapte r 13

Chemical Kine t ics

take for 25% of the C-14 atoms in a sample of C-14 to decay? If a sample of C-14 initially contains l .5 mmol of C-14, how many millimoles are left after 2255 years?

This reaction was monitored as a function of time: A----+B+C A plot of ln[A] versus time yields a straight line with slope - 0.0045 s- 1• a. What is the value of the rate constant (k) for this reaction at this temperature? b. Write the rate law for the reaction. c. What is the half-life? d. What is the lifeti me for this reaction? e. If the initial concentration of A is 0.250 mol L- 1, what is the concentration after 225 s?

The Effect of Temperature and the Collision Model

e

The diagram shows the energy of a reaction as the reaction progresses. Label each of the following in the diagram:

52. This reaction was monitored as a function of time: AB -----+ A+ B

e

A plot of l/[AB] versus time yields a straight line with slope 0.055 L mol- 1 s- 1• a. What is the value of the rate constant (k) for this reaction at this temperature? b. Write the rate law for the reaction. c. What is the half-life when the initial concentration is 0.55 mol L- 1? d. How long will it take for the concentration of AB to decrease from 0.55 mol L- 1 to 0.15 mol L- I? e. If the initial concentration of AB is 0.250 mol L- 1, and the reaction mixture initially contains no products, what are the concentrations of A and B after 75 s?

The decomposition of S02Cl2 is first order in S02Cl2 and has a 4 s- 1 at a certain temperature. rate constant of 1.42 X a. What is the half- life for this reaction? b. How long will it take for the concentration of S02Cl2 to decrease to 30% of its initial concentration? c. If the in itia l concentration of S02Cl2 is 1.00 mol L- 1, how long will it take for the concentration to decrease to 0.7 8 mol L- 1? d. If the initial concentration of S02Cl2 is 0. 150 mol L- 1, what is the concentration of S02 Cl2 after 2.00 X 102 s? After 5.00 x 102 s? 54. The decomposition of XY is second order in XY and has a rate 3 L mol- 1 s- 1 at a certain temperature. constant of 7.02 X a. What is the half-life for this reaction at an initial concentration of 0.100 mol L- 1? b. How long will it take for the concentration of XY to decrease to 10% of its initial concentration when the initial concentration is O. IOO mol L- 1? When the initial concentration is 0.200 mol L - I? c. If the initial concentration of XY is 0.150 mol L- 1, how long will it take for the concentration to decrease to 0.062 mol L - i? d. lfthe initial concentration ofXY is 0.050 mol L - 1, what is the concentration of XY after 5.0 x I0 1 s? After 5.50 x I02 s?

w-

w-

8

The half-life for the radioactive decay of U-238 is 4.5 billion years and is independent of initial concentration. How long will it take for I0% of the U-238 atoms in a sample of U-238 to decay? If a sample of U-238 initially contained 1.5 X I0 18 atoms and was formed 3.8 billion years ago, how many U-238 atoms does it contain today? 56. The half-life for the radioactive decay of C- 14 is 5730 years and is independent of the initial concentration. How long does it

Reaction progress

reactants products activation energy (Ea) enthalpy of reaction (ll, H) e. reverse activation energy Ea,r 58. A chemical reaction is endothermic and has an activation energy that is twice the value of the enthalpy of the reaction. Draw a diagram depicting the energy of the reaction as it progresses. Label the position of the reactants and products and indicate the activation energy and enthalpy of reaction. a. b. c. d.

G)

The activation energy of a reaction is 56.8 kJ mol- 1 and the frequency factor is 1.5 X 10 11 s- 1• Calculate the rate constant of the reaction at 25 °C. 60. The rate constant of a reaction at 32 °C is 0.055 s- 1. If the frequency factor is 1.2 X I0 13 s- 1, what is the activation barrier?

(9

The rate constant (k) for a reaction was meas ured as a function of temperature. A plot of In k versus l/T (in K) is linear and has a slope of - 7445 K. Calculate the activation energy for the reaction.

62. The rate constant (k) for a reaction was measured as a function of temperature. A plot of In k versus l/T (in K ) is linear and has a slope of - 1.0 I X I 04 K. Calculate the activation energy for the reaction.

~ The data shown below were collected for the first-order reaction:

Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction.

Temperature (K)

Rate Constant (s- 1)

800

3.24

x 10- 5

900

0.00214

1000

0.0614

1100

0.955

Exercises

64. The data below show the rate constant of a reaction measured at several different temperatures. Use an Arrhenius plot to determine the activation barrier and frequency factor for the reaction.

Temperature (K)

Rate Constant (s- 1)

300

0.0134

310

0.0407

320

0. 114

330

0.303

340

0.757

72. Which of these two reactions would you expect to have the smaller orientation factor? Explain. a. O(g) + N 2 (g) NO(g) + N(g) b. NO(g) + C l2(g) NOCl(g) + Cl(g)

Reaction Mechanisms

f» Consider this overall reaction which is experimentally observed to be second order in AB and zero order in C: AB + C -

data shown below were collected for the second-order reaction: HCl(g) + H(g)

Rate Constant (L mo1- 1 s- 1)

90

0.00357

100

0.0773

110

0.956

120

7.781

AB + AB ---;;;-"" AB2 + A

Slow

AB2 + C ---;:;-" AB + BC

Fast

74. Consider this overall reaction which is experimentally observed to be second order in X and first order in Y:

Use an Arrhenius p lot to determine the activation barrier and frequency factor for the reaction.

Temperature (K)

A + BC

Show that the following proposed mechanism fits the observations.

(D The

Cl(g) + Hi(g) -

585

X+Y -

XY

a. Does the reaction occur in a single step in which X and Y collide? b. Use the steady-state approximation to determine the rate law predicted by the following mechanism. Is this mechanism valid? Under what conditions?

k

X2 + Y --2...+ XY + X

66. The data below show the rate constant of a reaction measured at several different temperatures. U se an Arrhenius plot to determine the activation barrier and freq uency factor for the reaction.

G)

Nitrosyl chloride, N02Cl, is a powerful nitrating agent. It is thought to decompose via the following mechanism.

Temperature (K)

Rate Constant (s- 1)

3 10

0.00434

320

0.0140

a. What is the overall reaction that occurs?

330

0.042 1

b. Use the steady-state approximation to determine the rate law for this reaction.

340

0. 118

350

0.3 16

76. Consider this two-step mechanism for a reaction: N02(g) + Cl2(g) -

A reaction has a rate constant of 0.0 117 s- 1 at 400.0 K and 0.689 s- 1 at 450.0 K. a. Determine the activation barrier for the reaction. b. What is the value of the rate constant at 425 K?

68. A reaction has a rate constant of 0.000 122 s- 1 at 27 °C and 0.228 s- 1 at 77 °C. a. Determine the activation barrier for the reaction. b. What is the value of the rate constant at 17 °C? (D If a temperature increase from l 0.0 °C to 20.0 °C doubles the rate constant for a reaction, what is the value of the activation barrier for the reaction? 70. If a temperature increase from 20.0 °C to 35.0 °C triples the rate constant for a reaction, what is the value of the activation barrier for the reaction?

0

fj

Consider these two gas-phase reactions: a. AA(g) + BB(g) 2 AB(g) b. AB(g) + CD(g) AC(g) + BD(g) If the reactions have identical activation barriers and are carried out under the same conditions, which one would you expect to have the faster rate?

N02(g) + CI(g) -

k,

k,

ClN02(g) + Cl(g) ClN02(g)

Slow Fast

a. What is the overall reaction? b. Identify the intermediates in the mechanism. c. What is the predicted rate law?

fj

In the gas phase, the reaction between 0 3 and N02 produces 0 2 and N 20 5 :

The reaction is proposed to occur via the following mechanism:

a. Define the reaction rate in terms of the concentration of N 20 5 , and use the steady-state approximation to determine the rate law predicted by this mechanism. b. What would the rate law be if the concentration of N02 was very high?

586

Chapter 13

Chem ical Kine t ics

a. Show that the mechanism sums to the overall reaction.

78. In the gas phase, the reaction between 0 2 and NO produces N02 :

b. Use the mechanism and the information about the rates of the first and second elementary steps to predict the rate law for the reaction. c. Define the reaction rate in terms of the appearance of TI+, and use the steady-state approximation to determine the rate law predicted by this mechanism. d. The answer in (b) assumes that the second step is slow. Show that the rate law from the steady-state approximation in (c) predicts the same answer as in (b) if ki is very small (resulting in a very slow second step). e. Use the rate law determined by the steady-state approximation to determine the rate law if the concentration ofTI 3+ was so large that the second step of the reaction became very fast.

0 2 + 2 NO ------> 2 N02 . The reaction is proposed to occur via the following mechanism: k,

2NO ~ N102 L,

a. Define the reaction rate in terms of the disappearance of 0 2 , and use the steady-state approximation to determine the rate law predicted by this mechanism. b. What would the rate law be if the concentration of 0 2 was very high?

fJ)

In the gas phase, the reaction between IC! and H 2 produces HCI and I2 :

The reaction is proposed to occur via the following mechanism: HI + HCI HI+ IC!

Catalysis

e

2 IC! + H2 ------> 2 HCI + I2 .

Fast Slow

a. Show that the mechanism sums to the overall reaction. b. Use the mechanism and the information about the rates of the first and second elementary steps to predict the rate law for the reaction. c. Define the reaction rate in terms of the appearance of I2 , and use the steady-state approximation to determine the rate law predicted by this mechanism. d. The answer in (b) assumes that the second step is slow. Show that the rate law from the steady-state approximation in (c) predicts the same answer as in (b) if k2 is very small (resul ting in a very slow second step) .

80. In solution, the mechanism for the reaction : Hg~+ + Tl3+ ------> 2 Hg2+ + TI+

Many heterogeneous catalysts are deposited on high surfacearea supports. Why? 82. Suppose that the reaction A ------> products is exothermic and has an activation barrier of 75 kJ mol- 1• Sketch an energy diagram showing the energy of the reaction as a function of the progress of the reaction. Draw a second energy curve showing the effect of a catalyst.

tD Suppose that a catalyst lowers the activation barrier of a reaction from 125 kJ mo1- 1 to 55 kJ mol- 1• By what factor would you expect the reaction rate to increase at 25 °C? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical.)

84. The activation barrier for the hydrolysis of sucrose into glucose and fructose is l 08 kJ mol- 1• If an enzyme increases the rate of the hydrolysis reaction by a factor of l million, how much lower must the activation barrier be when sucrose is in the active site of the enzyme? (Assume that the frequency factors for the catalyzed and uncatalyzed reactions are identical and a temperature of 25 °C.)

is proposed to be: Hg+ Hg2+ Hg + Tl3+ ~ Hg 2 + + TI+

Fast Slow

Cumulative Problems (9 The data below were collected for this reaction at 500 °C:

Time (h)

[CH3 CN] (mol L- 1)

0.0

l.000

5.0

0.794

10.0

0.63 1

15.0

0.501

20.0

0.398

25.0

0.3 16

a. Determine the order of the reaction and the value of the rate constant at this temperature. b. What is the half-life for this reaction (at the initial concentration)? c. How long will it take for 90% of the CH3NC to convert to CH3CN? 86. The data below were collected for this reaction at a certain temperature:

Exercises

Time (h)

[X2Y] (mol L- 1)

0.0

0.100

1.0

0 .0856

2.0

0.0748

3.0

0 .0664

4.0

0 .0598

5.0

0 .0543

0

a. Determine the order of the reaction and the value of the rate constant at this temperature. b. W hat is the half-life fo r this reac tio n (at the initial concentration)? c. What is the concentration of X after 10.0 hours?

f)

Consider the reaction : A+ B +C___,.D The rate law for this reaction is: [A)[C]2 Rate = k [B] 112 Suppose the rate of the reaction at certain initial concentrations of A, B, and C is 0.0115 mol L- 1 s- 1. What is the rate of the reaction if the concentrations of A and C are doubled and the concentration of B is tripled?

88. Consider the reaction :

The rate law for this reactio n is:

Suppose that a 1.0 L reaction vessel initially contains 1.0 mol of 0 3 and 1.0 mol of 0 2. What fraction of the 0 3 will have reacted when the rate falls to one-half of its initial value? ~ At 700 K acetaldehyde decomposes in the gas phase to methane and carbon monoxide. The reaction is:

A sample of CH 3CHO is heated to 700 K and the pressure is measured as 220 mbar before any reaction takes place. The kinetics of the reaction are then followed by measurements of total pressure and these data are obtained: t (s) ?Total

(mbar)

0

1000

3000

7000

220

240

270

310

Find the rate law, the specific rate constant, and the total pressure after 2.00 x I 04 s. 90. At 400 K oxalic acid decomposes according to the reaction:

In three separate experi ments, the intial pressure of oxalic acid and final total pressure after 20 OOO s are measured.

Experiment PH,c,o, at t = 0 (kPa) ? Total

at t = 20 OOO s (kPa)

2

3

65. 8

92.1

111

94.6

132

160

Find the rate law of the reaction and its specific rate constant.

587

Dinitrogen pentoxide decomposes in the gas phase to form nitrogen dioxide and oxygen gas. The reaction is first order in dinitrogen pentoxide and has a half-life of 2.81 h at 25 °C. If a 1.5 L reaction vessel initially contains 745 Torr of N 20 5 at 25 °C, what partial pressure of 0 2 will be present in the vessel after 215 minutes?

92. Cyclopropane (C3H 6 ) reacts to form propene (C 3H6) in the gas phase. The reaction is first order in cyclopropane and has a rate constant of 5.87 X 10- 4 s- 1 at 485 °C. If a 2.5 L reaction vessel initially contains 722 Torr of cyclopropane at 485 °C, how long will it take for the partial pressure of cyclopropane to drop to below 1.00 x 102 Torr? ~ Iodine atoms combine to form I2 in liquid hexane solvent with a rate constant of 1.5 X 10 10 L mol- 1 s- 1. The reaction is second order in I. Since the reaction occurs so quickly, the only way to study the reaction is to create iodine atoms almost instantaneously, usually by photochemical decomposition of I2. Suppose a flash of light creates an initial [I] concentration of 0.0100 mol L- 1. How long will it take for 95% of the newly created iodine atoms to recombine to form I2?

94. The hydrolysis of sucrose (C 12H 220 11) into glucose and fructose in acidic water has a rate constant of 1.8 X I 0-4 s- 1 at 25 °C. Assuming the reaction is first order in sucrose, determine the mass of sucrose that is hydrolyzed when 2.55 L of a 0.150 mol L- 1 sucrose solution is allowed to react for 195 minutes.

C) The

reaction AB(aq) _____,. A(g) + B(g) is second order in AB and has a rate constant of 0.0118 L mol- 1 s- 1 at 25.0 °C. A reaction vessel initially contains 250.0 mL of 0.100 mol L- 1 AB which is allowed to react to form the gaseous product. The product is collected over water at 25.0 °C. How much time is required to produce 200.0 mL of the products at a barometric pressure of 755. l mmHg? (The vapour pressure of water at this temperature is 23.8 mmHg.)

96. The reaction 2 H 20 2(aq) _____,. 2 H 20(l) + 0 2(g) is first order in H20 2 and under certain conditions has a rate constant of 0.00752 s- 1at20.0 °C.A reaction vessel initially contains 150.0 mL of 30.0% H20 2 by mass solution (the density of the solution is 1.11 g mL). The gaseous oxygen is collected over water at 20.0 °C as it forms. What volume of 0 2 will form in 85 .0 seconds at a barometric pressure of 742.5 mmHg. (The vapour pressure of water at this temperature is 17 .5 mmHg.)

G Consider this energy diagram:

- - Reaction progress -

a. How many elementary steps are involved in this reaction? b. Label the reactants, products, and intermediates. c. Which step is rate limiting? d. Is the overall reaction endothermic or exothermic?

98. Consider the reaction in which HCl adds across the double bond of ethene:

588

Chapter 13

Chemical Kine t ics

The following mechanism, with the following energy diagram, has been suggested for this reaction:

a. Is the overall reaction exothermic or endothermic? b. There are six activation energies, one associated with each of the rate constants in the mechanism. Two of the elementary steps are barrierless, in that they have activation energies equalling zero. Which of the six activation energies, E a. I• E a,- 1' E a,2• Ea,- 2• E a,3• E a,- 3• are equal to zero? c. Place the other four activation energies in the potential energy diagram. d. On the potential energy diagram, draw an arrow indicating the enthalpy of reaction, ti ,H.

100. Below is an Arrhenius plot for the isomerization of cyclopropane to propene, showing the equation for the linear best fit to the data. Arrhenius P lo t for the Isomerization of C yclopropane o .--~~.--~~.--~---.~~---.~~---.~~---.~~-. - 1 1--~~b ~- ~--jf-~---f~~---t~~---t~~-t~~-t -21--~~1----~1--~---f~~---t~~---t~~-t~~-t

\ C= I

f.:'

I C + HCl \

I I H-C - C I I

Cl

~

-7>--~~l--~--il--~--l~~--1~~---.+~~-+~~-t

-9 >--~~1--~~1--~--1~~--1~~-+~~- ~-+-~--l - 1 0 1--~~1--~~l--~--!~~--1~~--1~~-+~~-t

0.0015

G

following reaction studied in the gas phase (Fridgen, T. D., et al. J. Phys. Chem. A 2005, 109, 75 19): CICH3

0.0016

0.0017

0.0018

0.0019

0.0020

0.0021

0.0022

T - 1/K- 1

G) For the

----->

-61--~~1--~--11--~---f~~---t-,.~---t~~-t~~-t

-8 >--~~1--~~1--~--1~~--1~~-+~"-.,--+~~-t

a. Based on the energy diagram, determine which step is rate limiting. b. What is the rate law and order of the reaction based on your answer to part (a)? c. Use the steady-state approximation to determine the rate Jaw and the order of the reaction. How does it differ from the answer in part (b )? d. Is the overall reaction exothermic or endothermic?

+ CH3CN

-41--~~1--~~1----=-"-

N2(g) + 3 H2(g)

2 NH3(g)

N2(g) + 3 H2(g)

0

.... FIGURE 14.9 Le Chatelier's Principle: The Effect of a Pressure Change (a) Decreasing the volume increases the pressure, causing the reaction to shift to the right (fewer moles of gas, lower pressure). (b) Increasing the volume reduces the pressure, causing the reaction to shift to the left (more moles of gas, higher pressure).

e

4 mol of gas

2 mol of gas

2 NH3(g)

0

e

4 mol of gas

2 mol of gas

R eaction shifts right (toward side w ith fewer moles of gas particles).

Reaction shifts left (toward side with more moles of gas particles).

(a)

(b)

14.8

Le Chatelier's Principle: How a System at Equilibrium Responds to Disturbances

~

Increasing the volume causes the reaction to shift in the direction that has the greater number of moles of gas particles.

~

If a reaction has an equal number of moles of gas on both sides of the chemical

623

equation, then a change in volume produces no effect on the equilibrium.

The Effect of Changing the Pressure by Adding an Inert Gas Consider again a gaseous reaction mixture at equilibrium. What happens if this time we keep the volume the same, but increase the pressure by adding an inert gas to the mixture? Although the overall pressure of the mixture increases, the partial pressures of the reactants and products do not change. Since the partial pressure of the inert gas is not in the equilibrium-constant expression Q = K, there is no effect and the reaction does not shift in either direction.

EXAMPLE 14.14

THE EFFECT OF A VOLUME CHANGE ON EQUILIBRIUM

Consider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture? Adding an inert gas at constant volume? SOLUTION The chemical equation has 3 mol of gas on the right and zero moles of gas on the left. Decreasing the volume of the reaction mixture increases the pressure and causes the reaction to shift to the left (toward the side with fewer moles of gas particles). Increasing the volume of the reaction mixture decreases the pressure and causes the reaction to shift to the right (toward the side with more moles of gas particles.) Adding an inert gas has no effect. FOR PRACTICE 14.14 Consider the following reaction at chemical equilibrium:

What is the effect of decreasing the volume of the reaction mixture? Increasing the volume of the reaction mixture?

The Effect of a Temperature Change on Equilibrium According to Le Chatelier's principle, if the temperature of a system at equilibrium is changed, the system will shift in a direction to counter that change. So, if the temperature is increased, the reaction will shift in the direction that tends to decrease the temperature and vice versa. Recall from Chapter 6 that an exothermic reaction (negative t.iH) emits heat: Exothermic reaction:

A

+

B

~

C

+

D

+

heat

We can think of heat as a product in an exothermic reaction. In an endothermic reaction (positive !'.iH), the reaction absorbs heat. Endothermic reaction:

A

+

B

+

heat

~

C

+

D

We can think of heat as a reactant in an endothermic reaction. At constant pressure, raising the temperature of an exothermic reaction-think of this as adding heat-is similar to adding more product, causing the reaction to shift left. For example, the reaction of nitrogen with hydrogen to form ammonia is exothermic.

In considering the ettect of a change in temperature, we are assuming that the heat I is added (or removed) at constant pressure.

624

Chapter 14

Chem ica l Equil i br i um

Reaction shifts left. Smaller K

Raising the temperature of an equilibrium mixture of these three gases causes the reaction to shift left, absorbing some of the added heat and forming less products and more reactants. Note that, unlike adding additional NH 3 to the reaction mixture (which does not change the value of the equilibrium constant), changing the temperature does change the value of the equilibrium constant. The new equilibrium mixture will have more reactants and fewer products and therefore a smaller value of K. Conversely, lowering the temperature causes the reaction to shift right, releasing heat and producing more products because the value of K has increased. Remove heat

Reaction shifts right. Larger K

In contrast, for an endothermic reaction, raising the temperature (adding heat) causes the reaction to shift right to absorb the added heat. For example, the following reaction is endothermic. Add heat

!

N 204(g) + heat colourless

Reaction shifts right. Larger K

Raising the temperature of an equilibrium mixture of these two gases causes the reaction to shift right, absorbing some of the added heat and producing more products because the value of K has increased. Since N20 4 is colourless and N02 is brown, the effects of changing the temperature of this reaction are easily seen (Figure 14.10 ~ ). On the other hand, lowering the temperature (removing heat) of a reaction mixture of these two gases causes the reaction to shift left, releasing heat, forming less products, and lowering the value of K. Remove heat

t

N204(g) + heat colourless

R eaction shifts left. Smaller K

14.8

N z0 4 (g) colourless

+

Le Chatelier's Principle: How a System at Equilibrium Responds to Disturbances

625

h eat

• • ••• '•.' Lower temperature: N20 4 favoured

Higher temperature: N0 2 favoured

.&. FIGURE 14.10 Le Chatelier's Principle: The Effect of a Temperature Change Because the reaction is endothermic, raising the temperature causes a shift to the right, toward the formation of brown N02 . [© Richard Megna/Fundamental Photographs, NYC]

Summarizing the Effect of a Temperature Change on Equilibrium: In an exothermic chemical reaction, heat is a product. ~

Increasing the temperature causes an exothermic reaction to shift left (in the direction of the reactants); the value of the equilibrium constant decreases.

~

Decreasing the temperature causes an exothermic reaction to shift right (in the direction of the products); the value of the equilibrium constant increases.

In an endothermic chemical reaction, heat is a reactant. ~

Increasing the temperature causes an endothermic reaction to shift right (in the direction of the products); the equilibrium constant increases.

~

Decreasing the temperature causes an endothermic reaction to shift left (in the direction of the reactants); the equilibrium constant decreases.

EXAMPLE 14.15

Adding heat favours the endothermic direction. Removing heat favours the I exothermic direction.

THE EFFECT OF A TEMPERATURE CHANGE ON EQUILIBRIUM

The following reaction is endothermic:

What is the effect of increasing the temperature of the reaction mixture? Of decreasing the temperature?

SOLUTION Since the reaction is endothermic, we can think of heat as a reactant: Heat

+ CaC03(s) ;;::::::==: CaO(s) + COig)

Raising the temperature is equivalent to adding a reactant, causing the reaction to shift to the right. Lowering the temperature is equivalent to removing a reactant, causing the reaction to shift to the left.

FOR PRACTICE 14.15 The following reaction is exothermic:

What is the effect of increasing the temperature of the reaction mixture? Of decreasing the temperature?

626

Chapter 14

Chem ica l Equ i lib r i um

CHAPTER IN REVIEW Key Terms Section 14.2

Section 14.3

Section 14.6

Section 14.8

reversible (594) dynamic equilibrium (594)

equilibrium constant (Kc) (596) law of mass action (596) activity (598)

reaction quotient (Qc) (608)

Le Chatelier's principle (619)

Key Concepts The Equilibrium Constant (14.1) The relative concentrations of the reactants and the products at equilibrium are expressed by the equilibrium constant, K. The equilibrium constant measures how far a reaction proceeds toward products: A large K (much greater than I ) indicates a high concentration of products at equilibrium, and a small K (less than l ) indicates a low concentration of products at equilibrium.

components, we can drop the units from Kc and Kp to obtain an estimate of K from meas ured quantities. And from the unitless thermodynamic equilibrium constant, we can obtain quantitative values in bar for gas-phase reactions and mol L- 1 for solutionphase reactions. When the equations for a chemical reaction are reversed, multiplied, or added to another equation, K must be modified accordingly.

Dynamic Equilibrium (14.2)

Calculating K (14.5)

Most chemical reactions are reversible; they can proceed in either the forward or the reverse direction. When a chemical reaction is in dynamic equilibrium, the rate of the forward reaction equals the rate of the reverse reaction, so the net concentrations of reactants and products do not change. However, this does not imply that the concentrations of the reactants and the products are equal at equilibrium.

The equilibrium constant can be calculated from equilibrium concentrations or partial pressures by substituting measured values into the expression for the equilibrium constant (as obtained from the law of mass action). In most cases, the equilibrium concentrations of the reactants and products-and therefore the value of the equilibrium constant-can be calculated from the initial concentrations of the reactants and products and the equilibrium concentration of j ust one reactant or product.

The Equilibrium-Constant Expression (14.3) The equilibrium-constant expression is given by the law of mass action. For reactions in solution, the equilibrium constant, Kc, is equal to the concentrations of products, raised to their stoichiometric coeffic ients, divided by the concentrations of reactants, raised to their stoichiometric coefficients. For reactions in the gas phase, it is more convenient (and correct) to write the equilibrium constant, Kp, which is equal to the partial pressures of the products, raised to their stoichiometric coefficients, divided by the partial pressure of the reactants, raised to their stoichiometric coefficients. These two equilibrium constants can be related by Equation 14.5. The thermodynamic equilibrium constant, K, is written in terms of activities of the components, and for our purposes it is equal to the numerical value of Kc for solution-phase reactions or to the numerical value of Kp for gas-phase reactions. The equilibrium constant contains only partial pressures or concentrations of reactants and products that exist as gases or solutes, respectively. Pure liquids are not included in the expression for the equilibrium constant.

The Reaction Quotient, Q (14.6) The ratio of the concentrations (or partial pressures) of products raised to their stoichiometric coefficients to the concentrations of reactants raised to their stoichiometric coefficients at any point in the reaction is called the reaction quotient, Q. Like K, Q can be expressed in terms of concentrations CQc) or partial pressures ( Q p). At equilibrium, Q is equal to K; therefore, the direction in which a reaction will proceed can be determined by comparing Q to K. If Q < K , the reaction moves in the direction of the products; if Q > K, the reaction moves in the reverse direction.

Finding Equilibrium Concentrations (14.7) There are two general types of problems in which K is given and one (or more) equilibrium amounts can be found: (1) K, initial amounts, and (at least) one equilibrium amount is given; and (2) K and only initial amounts is given. The first type is solved by rearranging the law of mass action and substituting given values. The second type is solved by using a variable, x , to represent the change in amounts.

The Thermodynamic Equilibrium Constant (14.4) While formally the equilibriu m constants Kc and Kp have units, the thermodynamic equilibrium constant, K , written in terms of unitless activities, is unitless. As long as the equilibrium-constant expressions for K are written in terms of partial pressures in bar for gases and concentrations in mol L- 1 for solution-phase

Le Chatelier's Principle (14.8) When a system at equilibrium is disturbed-by a change in the amount of a reactant or product, a change in volume, or a change in temperature-the system shifts in the direction that minimizes the disturbance.

Chapter In Review

627

Key Equations and Relationships Expression for the Equilibrium Constant, Kc (14.3)

+ bB(aq)

aA(aq)

Kc =

[CJc[W

~

cC(aq)

+ dD(aq)

(equilibrium concentrations only)

-b

[A] 8 [B]

Expression for the Equilibrium Constant, Kp (14.3) aA(g) KP

+

bB(g)

~

cC(g)

+

dD(g)

- (Pc) c(Po)d (Pp.J8(Ps)b (equilibrium partial pressures only)

-

Relationship Between the Equilibrium Constants, Kc and Kp(14.3)

Relationship Between the Equilibrium Constant and the Chemical Equation (14.4) 1. If you reverse the equation, invert the equilibrium constant. 2. If you multiply the coefficients in the equation by a factor, raise the equilibrium constant to the same factor. 3. If you add two or more individual chemical equations to obtain an overall equation, multiply the corresponding equilibrium constants by each other to obtain the overall equilibrium constant. The Reaction Quotient, De (14.6)

+ bB(aq)

aA(aq)

~

cC(aq)

+ dD(aq)

[C]c[D]d

De = [A]a[B]b (concentrations at any point in the reaction) The Reaction Quotient, Op (14.6) aA(g) _ (PcnPo)d QP - (PtJa(Ps)b

+ bB(g)

~

cC(g)

+ dD(g)

(partial pressures at any point in the reaction)

Relationship of Q to the Direction of the Reaction (14.6) Q Q

< K Reaction goes to the right. > K Reaction goes to the left.

Q=

K Reaction is at equilibrium.

Key Skills Expressing Equilibrium Constants for Chemical Equations (14.3) • Example 14. l

• For Practice 14. l

• Exercises 21, 22

Relating Kp and Kc (14.3) • Example 14. l

• For Practice 14. l

• Exercises 31, 32

Writing Equilibrium Expressions for Reactions Involving a Solid or a Liquid (14.3) • Example 14.2

• For Practice 14.2 • For More Practice 14.2

• Exercises 33, 34

628

Ch a pter 14

Chem ica l Equi li br i um

Manipulating the Equilibrium Constant to Reflect Changes in the Chemical Equation (14.4) •Example 14.3

•For Practice 14.3

•For More Practice 14.3

•Exercises 27-30

Finding Equilibrium Constants from Experimental Partial Pressure and Concentration Measurements (14.5) •Examples 14.4, 14.5

•For Practice 14.4, 14.5

•Exercises 35, 36, 41 , 42

Predicting the Direction of a Reaction by Comparing Q and K (14.6) • Example 14.6

• For Practice 14.6

• Exercises 47-50

Calculating Equilibrium Concentrations from the Equilibrium Constant and One or More Equilibrium Concentrations (14.7) • Example 14.7

• For Practice 14.7

• Exercises 37-46

Finding Equilibrium Partial Pressures from Initial Partial Pressures and the Equilibrium Constant (14.7) • Examples 14.8, 14.9

• For Practice 14.8, 14.9

• Exercises 53, 56, 58-60

Calculating Equilibrium Concentrations from the Equilibrium Constant and Initial Concentrations (14.7) •Example 14.10

•For Practice 14.10

•Exercise 57

Finding Equilibrium Amounts in Cases with a Small Equilibrium Constant (14.7) • Examples 14.11 , 14.1 2

•For Practice 14.11, 14 .12

•Exercises 6 1, 62

Determining the Effect of Changing the Amount of a Reactant or Product on Equilibrium (14.8) • Example 14. 13

• For Practice 14 .1 3

• Exercises 63-66

Determining the Effect of a Volume Change on Equilibrium (14.8) • Example 14.14

• For Practice 14. 14

• Exercises 67, 68

Determining the Effect of a Temperature Change on Equilibrium (14.8) • Example 14.15

• For Practice 14. 15

• Exercises 69, 70

EXERCISES Review Questions 1. How does a developing fetus in the womb get oxygen? 2. W hat is dynamic equilibrium? Why is it called dynamic? 3. G ive the general expression for the equilibrium constant for the following reactions:

a. aA(aq) + bB(aq) b. aA(g)

+

~

cC(aq)

bB(g) ~ cC(g)

+

+

dD(aq)

dD(g)

4. W hat is the significance of the equilibrium constant? What does a large equilibrium constant tell us abo ut a reaction? A small one?

8. W hat units should be used w hen expressing concentrations or partial pressures in the equilibrium constant? What are the units of K, Kp, and Kc? Explain. 9. W hy are the concentrations of pure solids and pure liquids omitted from equilibrium constant expressions? 10. Does the value of the equilibrium constant depend on the initial amounts of the reactants and products? Do concentrations or partial pressures of the reactants and products at eq uilibrium depend on their initial amounts? Explain.

5. What happens to the value of the equilibrium constant for a reaction if the reaction equation is reversed? Multip lied by a constant?

11. Explain how you might deduce the equilibrium constant for a reaction in which you know the initial amounts of the reactants and products and the equilibri um amount of only one reactant or product.

6. If two reactions sum to an overall reaction, and the equilibrium constants for the two reactions are K 1 and K 2, w hat is the equilibrium constant for the overall reaction?

12. W hat is the definition of the reaction quotient (Q) for a reaction? W hat does Q measure?

7. Explain the difference between K, K0 a nd Kp. For a given reaction, how are the two constants related ?

13. What is the value of Q when each reactant and product is in its standard state?

Exercises

629

14. In what direction will a reaction proceed for each condition: (a) Q < K; (b) Q > K; and (c) Q = K?

but 2.5x """ 2.5. Explain why a small x can be ignored in the first case, but not in the second.

15. Many equilibrium calculations involve finding the equilibrium am ounts of reactants and products given their initial concentrations or partial pressures and the equilibrium constant. Outline the general procedure used in solving these kinds of problems.

17. What happens to a chemical system at equilibrium when that equilibrium is disturbed?

16. In equilibrium problems involving equilibrium constants that are small relative to the initial concentrations or partial pressures of reactants, we can often assume that the quantity x (which represents how far the reaction proceeds toward products) is small. When this assumption is made, the quantity x can be ignored when it is subtracted from a large number, but not when it is multiplied by a large number. In other words, 2.5 - x = 2.5,

19. What is the effect of a change in volume on a chemical reaction (that includes gaseous reactants or products) initially at equilibrium?

18. What is the effect of a change in the amount of a reactant or product on a chemical reaction initially at equilibrium?

20. What is the effect of a temperature change on a chemical reaction initially at equilibrium? How does the effect differ for an exothermic reaction compared to an endothermic one?

Problems by Topic Equilibrium and the Equilibrium-Constant Expression

Q

Find and fix each mistake in the equilibrium-constant expressions.

reaction at the same temperature for the three different halogens. Rank the equilibrium constants for the three reactions from largest to smallest.

PH,Ps2 Kp = - -PH,S

a. 2 H2S(g) ~ 2 H2 (g) + S2(g)

b. Cu(s) + 2 Ag+(aq) ~ Cu2+(aq) + 2 Ag(s) [Cu2+][Ag]2 K = ----c [Cu][Ag+]2 c. Li(s)

K

=

+

2 H2 0(1) ~ Li +(aq)

+ 2 OH- (aq) + 2 H2 (g) (b)

(a)

1 [Li +][OH - JPH,

22. Write an expression for the equilibrium constant of each chemical equation.

a. CO(g) + Cl 2(g) ~ COCl 2(g) b. SbCl5 (g) ~ SbCl3(g) + Cl2(g) c. CHig) + 2 H2S(g) ~ CS2(g) + 4 Hz(g) d. HS04-(aq) + H20(1) ~ SOi - (aq) + H30 +(aq) e. Au3+ (aq) + 3 I - (aq) ~ Au(s) + 3/2 I2(s) f. 2 Mn04- (aq) + 6 H +(aq) + 5 H 2C 20 4(aq) ~ 2 Mnz+(aq) + 8 H 20(1) + 10 C0 2(g)

f» When the reaction below comes to equilibrium, will the concentrations of the reactants or products be greater? Does the answer to this question depend on the initial concentrations of the reactants and products? A(aq)

+

B(aq) ~ 2 C(aq) K = 1.4 X 10- 5

24. Ethene (C2H 4) can be halogenated by this reaction :

where X 2 can be Cl 2 (green), Br2 (brown), or I2 (purple). Examine the three figures representing equilibrium concentrations in this

(c)

fD H

2 and I 2 are combined in a flask and allowed to react according to the reaction:

Hz(g)

+

I2(g)

~

2 Hl(g)

Examine the figures on the next page (sequential in time) and answer the questions: a. Which fig ure represents the point at which equilibrium is reached? b. How would the series of figures change in the presence of a catalyst? c. Would the final figure (vi) have different amounts of reactants and products in the presence of a catalyst?

630

Chapter 14

. Q.

Chem ica l Equ i li br i um

.,

• • • • •• • • • • • • 0

0

0

•

w

•• ., • • • • 0

0

0

w

(}

(}

w

(i)

0

•

~

e

Use these reactions and their equilibrium constants to predict the equilibrium constant for the following reaction:

0

0

0

0

~

.

•

•

30. Use the reactions below and their equilibrium constants to predict the equilibrium constant for the reaction, 2 A(s) ~ 3 D (g) . A (s) ~ ! B(g)

w

3 D(g)

•

8

(}

4'

0

e

• • • • • • • • • • • • • • • • w

'9

'9

CD Calculate Kc for each reaction:

Kp = 6.26 x 10- 22 (at 298 K) b. CH4(g) + H20 (g) ~ CO(g) + 3 Hz(g) Kp = 7.7 X 1024 (at 298 K ) c. 12(g) + Cl2(g) ~ 2 ICl(g) Kp = 81.9 (at 298 K )

a. I2(g)

w.

0

w

.,

8

(iv)

.,

, •• , • 9

0

8

e

0

9

• •• e

e

• •• • • •e •• • • ~

~

0

.

~

Kz = 2.35

Kp, Kc, and Heterogeneous Equilibria

(}

(iii)

B(g)

K 1 = 0.0334

(ii)

61

0

~

+ C(g) + 2 C(g)

~

~

0

(v)

•

26. A chemist trying to synthesize a particular compound attempts two different synthesis reactions. The equilibrium constants for the two reactions are 23.3 and 2.2 X 104 at room temperature. However, upon carrying out both reactions for 15 minutes, the chemist finds that the reaction with the smaller equilibrium constant produces more of the desired product. Explain how this might be possible.

fD The reaction below has an equilibrium constant of K

32. Calculate Kp for each reaction : a. Nz0 4(g) ~ 2 N02(g) Kc = 5.9 X 10- 3 (at 298 K ) b. N 2(g) + 3 H 2(g) ~ 2 NH 3(g) Kc = 3.7 X 108 (at 298 K) c. N 2 (g) + 0 2 (g) ~ 2 NO(g) Kc = 4.10 X 10- 3! (at 298 K)

G Write an equilibrium-constant expression for each chemical equation involving one or more solid or liquid reactants or products.

~

(vi)

= 9. 1 X 106

~ 2 I (g)

a. COl - (aq) + H 20(1) ~ HC03- (aq) + OW(aq)

b. 2 KC10 3(s) ~ 2 KCl (s) + 3 0 2(g) c. HF(aq) + H20(l) ~ H30 +(aq) + F(aq) d. NH3(aq) + H20(1) ~ NH/(aq) + OW(aq) 34. Find the mistake in the equilibrium-constant expression and fix it.

Relating the Equilibrium Constant to Equilibrium Concentrations and Equilibrium Partial Pressures

@

Consider the reaction:

at 300 K.

Calculate K for each of the reactions and predict whether reactants or products will be favoured at equilibrium: ~ CO(g) + 2 H 2(g) b. !CO(g) + Hz(g) ~ !CH30H(g) c. 2 CH30H(g) ~ 2 CO(g) + 4 H2(g) 28. The reaction below has an equilibrium constant of K = 2.2 X 106 at 298 K.

a. CH30H(g)

An equilibrium mixture of this reaction at a certain temperature was found to have Pco = 0.105 bar,PH, = 0.114 bar, and P CH,OH = 0. 185 bar. What is the value of the equilibrium constant at this temperature?

36. Consider the reaction:

An equilibrium mixture of this reaction at a certain temperature was found to have PNH, = 0.105 bar and PH,s = 0.114 bar. What is the value of the equilibrium constant at this temperature?

Ci) Consider the reaction: Calculate K for each of the reactions and predict whether reactants or products will be favoured at equilibrium:

a. COF2(g) ~ !C0 2(g) + ! CF4(g)

Complete the table. Assume that all partial pressures are equilibrium values and in bar.

b. 6 COF2(g) ~ 3 C02(g) + 3 CF4(g) c. 2 C02(g) + 2 CF4 (g ) ~ 4 COF2(g)

Q

Consider the reactions and their respective equilibrium constants: NO(g)

+ ! Br2(g) ~ NOBr(g) 2 NO(g) ~ N2(g) + Oz(g)

K = 5.3 K = 2 .1

X 1030

T(K)

PN,

350

0.121

4 15

0. 11 0

5 15

0.120

PNH3

0.105

0.565 0.128

0.140

K 13.2 0.05 12

Exercises

38. Consider the following reaction: H2(g)

+ I1(g)

~

2 HI(g)

Complete the table. Assume that all partial pressures are equilibrium values and in bar.

G)

T(K)

PH,

Pi,

PHI

298

0.0302

0.0388

0 .922

600

0.0421

5 15

0.0525

0.387

Ptt,s = 0.266 bar . Will more of the solid form or will some of the existing solid decompose as equilibrium is reached? 48. Consider the reaction: 2 H2S(g)

+ Br2(g)

2 BrNO(g)

78.2

G)

A solution is made containing an initial [Fe3+] of 2.4 X 10- 4 mol L- 1 and an initial [SCN-] of 8.0 X 10- 4 mol L - 1• At equilibrium, [FeSCN2+] = I. 7 X I 0- 4 mol L - I. Calculate the value of the equilibrium constant (Kc). S02(g)

+

+ I1(g)

~

A 2.25 L container contains 0.055 mol of N02 and 0.082 mol of N20 4 at 298 K. Is the reaction at equilibrium? If not, in what direction will the reaction proceed?

Finding Equilibrium Concentrations from Initial Concentrations and the Equilibrium Constant •

Consider the reaction and associated equilibrium constant: aA(aq)

Cl2(g)

G

+ 2 H2(g)

~

reaction.

a. a= I ; b = I b. a= 2; b = 2 c. a = I; b = 2 52. Consider the reaction and associated equilibrium constant: aA (aq)

A reaction mixture in a 5. 19 L flask at 500 K contains 9.02 g of CO and 0.55 g of H2. At equilibrium, the flask contains 2.3 1 g of CH30H. Calculate the equilibrium constant at this temperature.

2 NH3(g)

~

3 H2(g)

+

Consider the reaction: NH4HS(s)

~

NH3(g)

bB (aq)

~

cC(aq) K = 5.0

i)

For the following reaction, K = 0.538 at 313 K. N10 4(g)

~

2 N02(g)

If a reaction vessel initially contains an N20 4 concentration of

0.0500 mol L- 1 at 313 K, what are the equilibrium concentrations of N20 4 and N02 at 313 K? (Hint: Ensure use of proper units for the equilibrium constant.)

54. For the following reaction, K = 24.6 at 668 K.

N1(g)

The Reaction Quotient and Reaction Direction

G)

+

Find the equilibrium concentrations of A, B, and C for each value of a, b, and c. Assume that the initial concentrations of A and B are each 1.0 mol L- 1 and that no C is present at the beginning of the reaction. a. a = I; b = I; c = 2 b. a = I; b = l; c = I c. a = 2; b = I; c = I (set up equation for x; don ' t solve)

CH30H(g)

46. Ammonia, NH3(g), at a pressure of 1.35 bar, is placed in a container at a certain temperature. When equilibrium is established at that temperature, the pressure of Hi{g) is 0.630 bar. Determine the value of Kp for the decomposition of NH3 at that temperature:

bB(aq) K = 4 .0

Find the equilibrium concentrations of A and B for each value of

2 HI(g)

HCI, at a pressure of 2.300 bar, and 0 2, at a pressure of I .OOO bar, are placed in a container, allowed to react, and come to equilibrium at 750 K. When equilibrium is established, the total pressure of the gases present is 2.835 bar. Determine the value of Kp at 750 K for the reaction: 4 HCI(g) + 0 2(g) ~ 2 Cl2(g) + 2 H20(g)

~

a and b. Assume that the initial concentration of A in each case is 1.0 mol L- 1 and that no B is present at the beginning of the

44. Consider the reaction:

CO(g)

x 10- 5 at 298 K

SO. Nitrogen dioxide dimerizes according to the reaction:

6) Consider the reaction: H2(g)

SO/ - (aq)

A 1.5 L solution contains 6.55 g of dissolved silver sulfate. If additional solid silver sulfate is added to the solution, will it dissolve?

A reaction mixture is made containing an initial Pso,ci, = 2.6 bar. At equilibrium, PCI, = 0.21 bar. Calculate the value of the equilibrium constant.

A reaction mixture in a 3.67 L flask at 500 K initially contains 0.37 1 g H2 and 17.93 g I2. At equilibrium, the flask contains 17 .72 g HI. Calculate the equilibrium constant at this temperature.

+

K = I. I

42. Consider the reaction: ~

x 10- 4 at 1073 K

Silver sulfate dissolves in water according to the reaction:

K = 32 .9 at 315 K

In a reaction mixture at equilibrium, the partial pressure of NO is 108 mbar and that ofBr2 is 126 mbar. What is the partial pressure of BrNO in the equilibrium mixture? 40. Consider the reaction: S02Cl2(g) ~ S02(g) + Cli(g) K = 2.97 x 103 at 625 K In a reaction mixture at equilibrium, the partial pressure of S02 is 137 mbar and that of CI 2 is 285 mbar. What is the partial pressure of S02Cl2? Q Consider the reaction: Fe3+ (aq) + SCN- (aq) ~ FeSCN2+(aq)

+ S2(g)

A reaction mixture contains 0.112 bar of H 2, 0.055 bar of S2, and 0.445 bar of H2S. Is the reaction mixture at equilibrium? If not, in what direction will the reaction proceed? Ag2S04Cs) ~ 2 Ag+(aq)

~

S02Cl2(g)

2 H2(g)

K = 4.4

Consider the reaction: 2 NO(g)

~

K

33.0

0.386

631

+ H2S(g)

At a certain temperature, K = 8.5 X I 0- 3. A reaction mixture at this temperature containing solid NH4HS has PNH, = 0.266 bar,

If a reaction mixture initially contains a CO concentration of

0. 1500 mol L- 1 and a Cl2 concentration of 0.175 mol L- 1 at 668 K, what are the equilibrium concentrations of CO, Cl 2, and COCI2 at 668 K? (Hint: Ensure use of proper units for the equilibrium constant.)

632

Chapter 14

Chem ica l Equ i lib r i um

'1) Consider the reaction: NiO(s)

+ CO(g)

~

64. Consider this reaction at equilibrium: Ni(s)

+

C02(g) K = 80 .3 at 1500 K

2 BrNO(g)

If a mixture of solid nickel(II) oxide and 2.28 bar carbon mon-

G)

+ H20 (g)

~

C0 2(g)

+

K = 130 at 500 K

Hi(g)

If a reaction mixture initially contains 0.550 bar each of CO and H 20 , what will be the partial pressures of each of the reactants and products at equilibrium? Consider the reaction:

+

CH 3COOH(aq)

H 20(l) ~ H 30 +(aq)

+

(I

K = 1.8 X 10- 5 at 25 °C

If a solution initially contains 0.2 10 mol L -

i CH3COOH, what is the equilibrium concentration of H30 + at 25 °C?

S02(g)

+ Cl2(g)

+ Br2(g)

Consider this reaction at equilibrium:

Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance. a. 0 2 is removed from the reaction mixture. b. KCI is added to the reaction mixture. c. KC10 3 is added to the reaction mixture. d. 0 2 is added to the reaction mixture.

CH 3COO- (aq)

58. Consider the reaction:

2 NO(g)

Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance. a. NO is added to the reaction mixture. b. BrNO is added to the reaction mixture. c. Br2 is removed from the reaction mixture.

oxide is allowed to come to equilibrium at 1500 K, what will be the partial pressure of CO and C02 at equil.ibrium? 56. Consider the reaction: CO(g)

~

66. Consider this reaction at equilibrium: K = 6 .9 X 10- 6 at 1000 K

~ S02Cl2(g)

If a reaction mixture initially conta.ins 0.710 bar each of S02

@

and Cl2, what is the partial pressure of S02Cl2 at equilibrium? Consider the decomposition reaction of chlorine monofluoride at 1000 K: 2 ClF(g) ~ Cl2(g)

+

Fi(g) K = 1.75 X 10- 6 at 1000 K

If a reaction mixture initially contained 0.500 bar of CIF, what are

the partial pressures of all reactants and products at equilibrium? 60. Consider the reaction: CO(g)

+

H10 (g) ~ C02(g)

+

H1(g)

K = 0.l l8at4000 K A reaction mixture initially contains a CO partial pressure of 1344 mbar and a H20 partial pressure of 1766 mbar at 4000 K. Calculate the equilibrium partial pressures of each of the products. ~ Consider the reaction: A(aq)

~

B (aq)

+

C(aq)

Find the equilibrium concentrations of A, B, and C for each value of K. Assume that the initial concentration of A in each case is 1.0 mol L - I and that the reaction mixture initially contains no B or C. Make any appropriate simplifying assumptions. a. K = 1.0 b. K = 0.010 c. K = 1.0 X 10- 5

(i

Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance. a. C is added to the reaction mixture. b. H2 0 is condensed and removed from the reaction mixture. c. CO is added to the reaction mixture. d. H2 is removed from the reaction mixture. Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change. a. Ii(g) ~ 2 l(g) (volume is increased) b. 2 H2S(g) ~ 2 H2(g) + S2(g) (volume is decreased) c. Ii(g) + Cl2(g) ~ 2 ICl(g) (volume is decreased)

68. Each of the following reactions is allowed to come to equilibrium and then the volume is changed as indicated. Predict the effect (shift right, shift left, or no effect) of the indicated volume change. a. CO(g) + H10(g) ~ C02(g) + H1(g) (volume is decreased) b. PC13(g) + Cl2(g) ~ PC15 (g) (volume is increased) c. CaC03 (s) ~ CaO(s) + C02(g) (volume is increased) @ This reaction is endothermic. C(s)

+

C02 (g) ~ 2 CO(g)

62. Consider the reaction: A(g)

~

2 B(g)

Find the equilibrium partial pressures of A and B for each value of K. Assume that the initial partial pressure of B in each case is 1.0 bar and that the initial partial pressure of A is 0.0 bar. Make any appropriate simplifying assumptions. a . K = 1.0 b. K = 1.0 X 10- 4 c. K = 1.0 X 105

Le Chatelier's Principle

(D Consider the reaction at equilibrium: CO(g)

+ Cl2(g)

~

Predict the effect (shift right, shift left, or no effect) of increasing and decreasing the reaction temperature. How does the value of the equilibrium constant depend on temperature?

70. This reaction is exothermic.

Predict the effect (shift right, shift left, or no effect) of increasing and decreasing the reaction temperature. How does the value of the equilibrium constant depend on temperature?

fD Coal, which is primarily carbon, can be converted to natural gas, primarily CH4 , by the exothermic reaction:

COCli(g)

Predict whether the reaction will shift left, shift right, or remain unchanged after each disturbance: a. COC12 is added to the reaction mixture. b. Cl2 is added to the reaction mixture. c. COC12 is removed from the reaction mixture.

Which disturbance(s) will favour CH4 at equilibrium? a. adding more C to the reaction mixture b. adding more H2 to the reaction mixture c. raising the temperature of the reaction mixture

Exercises

d. lowering the volume of the reaction mixture e. adding a catalyst to the reaction mixture f. adding neon gas to the reaction mixture 72. Coal can be used to generate hydrogen gas (a potential fuel) by the endothermic reaction:

633

gas, the formation of less hydrogen gas, or have no effect on the quantity of hydrogen gas. a. adding more C to the reaction mixture b. adding more H 20 to the reaction mixture c. raising the temperature of the reaction mixture d. increasing the volume of the reaction mixture e. adding a catalyst to the reaction mixture f. adding an inert gas to the reaction mixture

If this reaction mixture is at equilibrium, predict whether each

disturbance will result in the formation of additional hydrogen

Cumulative Problems fi)

Carbon monoxide replaces oxygen in oxygenated hemoglobin according to the reaction: Hb0 2(aq)

+

CO(aq) ~ H bCO(aq)

+

+

0 2(aq)

H b(aq)

+

CO(aq) ~ HbCO(aq)

~

Hb0 2(aq)

3

K = 0.042. A 10.0 L container at 650 K has 1.0 g of MgO(s) and C0 2 at P = 0.0260 bar. The container is then compressed to a volume of 0.100 L. Find the mass of MgC03 that

0 2 (aq)

a. Use the reactions and associated equilibrium constants at body temperature to find the equilibrium constant for the above reaction. H b(aq)

fD At 650 K, the reaction MgC0 (s) ~ MgO(s) + C02(g) has

K = 1.8

is formed. 78. A system at equilibrium contains 12(g) at a pressure of 0.21 bar and l(g) at a pressure of0.23 bar. The system is then compressed to half its volume. Find the pressure of each gas when the system returns to equilibrium.

G)

K = 306

Consider the exothermic reaction:

b. Suppose that an air mixtu re becomes polluted with carbon

monoxide at a level of 0.10%. Assuming the air contains 20.0% oxygen, and that the oxygen and carbon monoxide ratios that dissolve in the blood are identical to the ratios in the air, what would be the ratio of HbCO to Hb02 in the bloodstream? Comme nt on the toxicity of carbon monoxide. 74. Nitrogen monoxide is a pollutant in the lower atmosphere that irritates the eyes and lungs and leads to the formation of acid rain. Nitrogen monoxide forms naturally in atmosphere according to the endothermic reaction: N2(g)

+

0 2(g) ~ 2 NO(g)

K = 4.1 X 10- 31 at 298 K.

Use the ideal gas law to calculate the concentrations of nitrogen and oxygen present in air at a pressure of 1.0 bar and a temperature of 298 K. Assume that nitrogen composes 78% of air by volume and that oxygen composes 21% of air. Find the "natural" equilibrium concentration of NO in air in units of molecules/ cm3 . How would you expect this concentration to change in an automobile engine in which combustion is occurri ng?

fD The

reaction C0 2(g)

+

C(s) ~ 2 CO(g) has K

= 48.3 at

1200 K. a. Calculate the total pressure at equilibrium when 4.45 g of C02 is introduced into a 10.0 L container and heated to 1200 K in the presence of 2.00 g of graphite. b. Repeat the calculation of part (a) in the presence of 0.50 g of graphite. 76. A mixture of water and graphite is heated to 890 K in a 10.0 L container. When the system comes to equilibrium it contains 0. 875 mol each of CO gas and H 2 gas and 0.289 mol of H 20 gas and some graphite. Some 0 2 gas is added to the system and a spark is applied so that the H2 reacts completely with the 0 2, forming H 20 . Find the amount of CO in the flask when the system returns to equilibrium.

If you were a chemist trying to maximize the amount of C2H 4 Cl2

produced, which tactic(s) might you try? Assume that the reaction mixture reaches equilibrium. a. increasing the reaction volume b. removing C2H 4 Cl2 from the reaction mixture as it forms c. lowering the reaction temperature d . addi ng Cl2 80. Consider the endothermic reaction:

If you were a chemist trying to maximize the amount of C2H412

produced, which tactic(s) might you try? Assume that the reaction mixture reaches equilibrium. a. decreasing the reaction volume b. removing 12 from the reaction mixture c. raising the reaction temperature d . adding C2H 4 to the reaction mixture ~ Consider the reaction: H2(g)

+ l2(g)

~

2 Hl(g)

A reaction mixture at equilibrium at 600 K contains 0.1 70 bar H2, 0.1 20 bar 12 , and 1.26 bar HI. A second reaction mixture, also at 600 K, contains 0.135 bar each of H2 and 12 and 1.10 bar of HI. Is the second mixture at equilibrium? If not, what will be the partial pressure of HI when the reaction reaches equili brium at 600 K? 82. Consider the reaction: 2 H 2S(g)

+ S02(g)

~

3 S(s)

+

2 H p(g)

A reaction mixture initially containing 0.600 bar each of H2 S and S02 was found to contain 0.0054 bar of H 20 at equilibrium at 1300 K. A second reaction mixture at the same temperatu re initially contains 0.500 bar of H 2S and 0.350 bar of S02. Calculate the equilibrium concentrations of all reactants and products at equilibrium.

634

Q)

Chapter 14

Che m ical Equi li br i um

Ammonia can be synthesized according to the reaction: N1(g)

+ 3 H1(g)

~

2 NH3(g) K = 5.3 X 10- 5 at 725 K

A 200.0 L reaction container initially contains 1.27 kg of N 2 and 0.310 kg of H2 at 725 K. Assuming ideal gas behaviour, calculate the mass of NH 3 (in g) present in the reaction mixture at equilibrium. What is the percent yield of the reaction under these conditions? 84. Hydrogen can be extracted from natural gas according to the reaction: CHig)

+ C02(g)

~

G)

2 CO(g)

Find the starting pressure of CC14 at this temperature that will produce a total pressure of 1.0 bar at equilibrium. 88. The equilibrium constant for the reaction S02(g) + N02(g) ~ S0 3(g) + NO(g) is 15.6 at 1000 K. Find the amount of N0 2 that must be added to 2.4 mol of S02 in order to form l .2 mol of S0 3 at equilibrium.

0

+ 2 H1(g)

K = 4.5 X 102 at 1115 K

An 85.0 L reaction container initially contains 23.3 kg of CH4 and 55.4 kg of C02 at l l 15 K. Assuming ideal gas behaviour, calculate the mass of H 2 (in g) present in the reaction mixture at equilibrium. What is the percent yield of the reaction under these conditions?

G) The system described by the reaction

At 693 K, CC14 decomposes to carbon and chlorine. The K for the decomposition is 0.76.

A sample of CaC03(s) is introduced into a sealed container of volume 0.654 L and heated to 1000 K until equilibrium is reached. The K for the reaction

is 3.9 X 10- 2 at this temperature. Calculate the mass of CaO(s) that is present at equilibrium. 90. An equilibrium mixture contains N20 4 (P = 0.28 bar) and N02 (P = l. l bar) at 350 K. The volume of the container is doubled at constant temperature. Calculate the equilibrium pressures of the two gases when the system reaches a new equilibrium.

G Carbon monoxide and chlorine gas react to form phosgene: is at equilibrium at a given temperature when Pco = 0.30 bar, PCI, = 0.10 bar, and PcoCI, = 0.60 bar. An additional pressure of Cl2(g) = 0.40 bar is added. Find the pressure of CO when the system returns to equilibrium. 86. A reaction vessel at 1215 K contains a mixture of S02 (P = 3.00 bar) and 0 2 (P = 1.00 bar). When a catalyst is added this reaction takes place:

At equilibrium the total pressure is 3.75 bar. Find the value of K.

CO(g)

+ Cli(g)

~

COCl i(g) K = 3.10 at 746 K

If a reaction mixture initially contains 215 mbar of CO and 245 mbar of Cl2, what is the mole fraction of COC12 when equilibrium is reached?

92. Graphite can react with gaseous water to form carbon monoxide gas and hydrogen gas. The equilibrium constant for the reaction at 700.0 K is Kp = 1.60 X 10- 3. If a 1.55 L reaction vessel initially contains 145 mbar of water at 700.0 Kin contact with excess graphite, find the percent by mass of hydrogen gas of the gaseous reaction mixture at equilibrium.

Challenge Problems G) Consider the reaction:

96. At a given temperature a system containing 0 2(g) and some oxides of nitrogen can be described by these reactions:

a. A reaction mixture at 652 K initially contains 522 mbar of NO and 421 mbar of 0 2. At equilibrium, the total pressure in the reaction mixture is 748 mbar. Calculate K at this temperature. b. A second reaction mixture at 652 K initially contains 255 mbar of NO and 185 mbar of 0 2. What is the equilibrium partial pressure of N02 in this mixture? 94. Consider the reaction: 2 S02(g)

G

+

+ 02(g) 2 N02(g)

G

~ 2 N02(g) ~

N20 4 (g)

104 K = 0.10

K

=

A pressure of l bar of N20 4(g) is placed in a container at this temperature. Predict which, if any, component (other than N20 4 ) will be present at a pressure greater than 0.2 bar at equilibrium. A sample of pure N02 is heated to 337 °C at which temperature it partially dissociates according to the equation

02(g) ~ 2 S03(g) K = l 1.4 at 950 K

A 2.75 L reaction vessel at 950 K initially contains 0.100 mol of S02 and O. l 00 mol of 0 2. Calculate the total pressure in the reaction vessel when equilibrium is reached. Nitric oxide reacts with chlorine gas according to the reaction: 2 NO(g)

2 NO(g)

+ C12(g)

~

At equ ilibrium, the density of the gas mixture is 0.520 g L _ , at 0.750 atm. Calculate Kc for the reaction. 98. When N20 5(g) is heated, it dissociates into N20 3(g) and 0 2 (g) according to the reaction:

2 NOCl(g) K = 0.27 at 700 K

A reaction mixture initially contains equal partial pressures of NO and Cl2. At equilibrium, the partial pressure of NOCl was measured to be 115 mbar. What were the initial partial pressures of NO and Cl2?

The N20 3(g) dissociates to give N20(g) and 0 2(g) according to the reaction: N103(g) ~ N10(g)

+ 0 2(g)

K = 1.69 X 106 at 273 K.

A 1.25 L reaction vessel initially contains 0.500 mol of N20 5 and 273 K.

Exercises

a. Determine the partial pressures of N20 5 , N20, and 0 2 at equilibrium. b. Determine the partial pressure of N2 0 3 at equilibrium.

G) a. Silver chloride is only slightly soluble in pure water at 25 °C: AgCl(s) ~ Ag+(aq)

+

Ci-(aq) K = LS X 10-

10

Calculate the concentration of Ag+ and Cl- in a solution that is saturated with AgCl (i.e., the system is at equilibrium and there is still solid AgCI visible). b. The addition of ammonia to the solution results in an increased solubility of AgCI. Ammonia complexes with Ag+:

635

Determine the concentration of Cl- in a solution containing silver chloride and 0.100 mol L- 1 NH 3 at equilibrium.

100. A sample of S03 is introduced into an evacuated sealed container and heated to 1233 K. The following equilibrium is established:

The total pressure in the system is found to be 3.0 bar and the mole fraction of 0 2 is 0.12. Find K.

Conceptual Problems

0

A reaction A(aq) ~ B (aq) has an equilibrium constant of 1.0 x 10- 4 . For which of the initial reaction mixtures is the x is small approximation most likely to apply? a. [AJ = 0.0010 mol L- 1; [BJ = 0 mol L - 1 b. [AJ = 0 mol L- 1; [BJ = 0.10 mol L - 1 c. [AJ = 0.10 mol L - I; [BJ = 0.10 mol L - I d. [AJ = 0. 10 mol L- 1; [BJ = 0 mol L- 1

104. Consider the reaction: aA(g)

~

bB(g)

Each of the entries in the table below represents equilibrium partial pressures of A and B under different initial conditions. What are the values of a and bin the reaction?

102. The reaction A (g) ~ 2 B(g) has an equilibrium constant of Kc = 1.0 at a given temperature. If a reaction vessel contains equal initial amounts (in moles) of A and B, will the direction in which the reaction proceeds depend on the volume of the reaction vessel? Explain .

PA(bar)

P8 (bar)

4.0

2.0

A particular reaction has an equilibrium constant of Kp = 0.50. A reaction mixture is prepared in which all the reactants and products are in their standard states. In which direction will the reaction proceed?

•

2.0

1.4

LO

LO

0.50

0.71

0.25

0.50

15.1 Heartburn

636

Milk of magnesia contains a base that can neutralize stomach acid and relieve heartburn.

15.2 The Nature of Acids and Bases 637

[Quade Paul/Pearson Education]

15.3 Definitions of Acids and Bases 639 15.4 Acid Strength and the Acid Ionization Constant (Ka) 642 15.5 Base Solutions

645

I

N THIS CHAPTER, we apply the equilibrium concepts learned in the previous chapter to acid-base phenomena. Acids are common in many foods , such as limes, lemons, and vinegar, and in a number of consumer products, such as

toilet cleaners and batteries. Bases are less common in foods but are key ingredients

15.6 Autoionization of Water and pH 647

in consumer products such as drain openers and antacids. We will examine three

15.7 Finding [H30+i, [OHl. and pH of Acid or Base Solutions 652

different models for acid- base behaviour, all of which define that behaviour dif-

15.8 The Acid-Base Properties of Ions and Salts 662

ferently. In spite of their differences, the three models coexist, each being useful at

15.9 Polyprotic Acids

explaining a particular range of acid-base phenomena. We will also examine how

670

15.10 Lewis Acids and Bases

675

to calculate the acidity or basicity of solutions and define a useful scale, called the

15.11 Strengths of Acids and Bases and Molecular Structure 677

pH scale, to quantify acidity and basicity. These calculations are very similar to the

15.12 Ocean Acidification

kind of equilibrium problems that we explored in Chapter 14.

680

15.1

Heartburn

Heartburn is a painful burning sensation in the esophagus (the tube that joins the throat to the stomach), just below the chest. The pain is caused by hydrochloric acid (HCI), which the stomach excretes to kill microorganisms and to activate enzymes that break down food. Hydrochloric acid sometimes backs up out of the stomach and into the esophagus, a phenomenon known as acid reflux. Recall from Section 4.5 that acids are substances that-by one definition that we will elaborate on shortlyproduce H+ ions in solution. When hydrochloric acid from the stomach comes in

636

15.2 The Nature of Acids and Bases

contact with the lining of the esophagus, the H+ ions irritate the esophageal tissues, resulting in the burning sensation. Some of the acid can work its way into the lower throat and even the mouth, producing pain in the throat and a sour taste (characteristic of acids) in the mouth. Almost everyone experiences heartburn at some time, most commonly after a large meal when the stomach is full and the chances for reflux are greatest. Strenuous activity or lying in a horizontal position after a large meal increases the likelihood of stomach acid reflux and the resulting heartburn. The simplest way to relieve mild heartburn is to swallow repeatedly. Saliva contains the bicarbonate ion (HC03- ), which acts as a base and, when swallowed, neutralizes some of the acid in the esophagus. Later in this chapter, we will see how bicarbonate acts as a base. You can also treat heartburn with antacids such as Tums, milk of magnesia, or Mylanta. These over-the-counter medications contain more base than our saliva does and therefore are more effective at neutralizing esophageal acid. For some people, heartburn becomes a chronic problem. Gastroesophageal reflux disease (GERD) is the medical condition associated with chronic heartburn. In patients with GERD, the band of muscles (called the esophageal sphincter) at the bottom of the esophagus just above the stomach does not close tightly enough, allowing the stomach contents to leak back into the esophagus on a regular basis. Medical researchers have developed a wireless sensor to help diagnose and evaluate treatment of GERD. Using a tube that goes down through the throat, a physician attaches the sensor to tissues in the patient's esophagus. The sensor reads pH-a measure of acidity that we discuss in Section 15.6-and transmits the readings to a recorder worn on the patient's body. The patient goes about his or her normal business for the next few days while the recorder monitors esophageal pH. The physician then reads the record of esophageal pH to make a diagnosis or evaluate treatment. In this chapter, we examine acid and base behaviour. Acids and bases are not only important to our health (as we have just seen), but are also found in many household products, foods , medicines, and of course in nearly every chemistry laboratory. Acid-base chemistry is central to much of biochemistry and molecular biology. The building blocks of proteins, for example, have acidic and basic properties (called amino acids) and the molecules that carry the genetic code in DNA are bases.

637

I Bases were first defined in Section 4.5.

The concentration of stomach acid varies

I from about 0.01 to 0. 1 mol L- 1 .

15.2 The Nature of Acids and Bases Acids have the following general properties: a sour taste, the ability to dissolve many metals, the ability to tum blue litmus paper red, and the ability to neutralize bases. Some common acids are listed in Table 15 .1.

TABLE 15.1

Litmus paper contains certain dyes that change colour in the presence of acids I and bases.

Some Common Acids

Name

Occurrence/Uses

Hydrochloric acid (HCI)

Metal cleaning; food preparation; ore refining; primary component of stomach acid Fertilizer and explosives manufacturing; dye and glue production; automobile batteries; electroplating of copper

Nitric acid (HN0 3)

Fertilizer and explosives manufacturing; dye and glue production

Ethanoic acid (acetic acid) (HC2H302)

Plastic and rubber manufacturing; food preservative; active component of vinegar

Citric acid (H3C5H507)

Present in citrus fruits such as lemons and limes; used to adjust pH in foods and beverages Found in carbonated beverages due to the reaction of carbon dioxide with water

Hydrofluoric acid (HF)

Metal cleaning ; glass frosting and etching

Phosphoric acid (H3P04)

Fertilizer manufacture; biological buffering; preservative in beverages

For a review of acid naming, see

I Section 3.4.

The formula for acetic acid is commonly

I written as CH3COOH.

638

Chapter 15

Acids and Bases

You can find hydrochloric acid in most chemistry laboratories. In industry, it is used to clean metals, to prepare and process some foods, and to refine metal ores. As we have just seen, hydrochloric acid is also the main component of stomach acid. HCl ( H ydrochloric acid )

Sulfuric acid and nitric acid are also common in the laboratory. They play major roles in the manufacture of fertilizers, explosives, dyes, and glues. Sulfuric acid, produced in larger quantities than any other chemical, is contained in most automobile batteries. 0

II I

H-0-S=O 0

0

II

I

H-O-N-0

H

HN0 3 ( Sulfuric acid

( Nitric acid

You can probably find acetic acid in your home-it is the active component of vinegar. It is also produced in improperly stored wines. The word vinegar originates from the French words vin aigre, which means sour wine. Wine experts consider the presence of vinegar in wines a serious fault, since it makes the wine taste like salad dressing. 0

II

H

I I

H- 0 - C- C- H H

CH3COOH ( Acetic acid )

Acetic acid is a carboxylic acid, an acid that contains the following grouping of atoms: 0

II H-o-c_.. Acetic acid makes vinegar taste sour. [GOIMAGES/Alamy Stock Photo]

Coffee is acidic overall , but bases present in coffee-such as caffeine-impart a bitter I flavour.

Carboxylic acid group

Carboxylic acids are often found in substances derived from living organisms. Other examples of carboxylic acids are citric acid, the main acid in lemons and limes, and malic acid, found in apples, grapes, and wine. Bases have the following general properties: a bitter taste, a slippery feel, the ability to turn red litmus paper blue, and the ability to neutralize acids. Because of their bitterness, bases are less common in foods than are acids. Our aversion to the taste of bases is probably an evolutionary adaptation to warn us against alkaloids, organic bases found in plants that are often poisonous. (For example, the active component of hemlock- the poisonous plant that caused the death of the Greek philosopher Socrates- is the alkaloid coniine.) Nonetheless, some foods, such as coffee and chocolate (especially dark chocolate), contain small amounts of base. Many people enjoy the bitterness, but only after acquiring the taste over time.

15.3

Definitions of Acids and Bases

639

[H30+]. Notice that changing [H3 0 +] in an aqueous solution produces an inverse change in [OH- ] and vice versa.

Summarizing Kw: ~ A neutral solution contains [H 30 +]

EXAMPLE 15.2

=

= 1.0

[OH- ]

10- 7 mol L - l (at 25 °C).

X

~

An acidic solution contains [H30 +] > [OH- ].

~

A basic solution contains [OH- ] > [H30 +].

~

In all aqueous solutions both H 30 + and OH- are present, with [H 30 +][0H- ] Kw = 1.0 x 10- 14 (at 25 °C).

=

USING Kw IN CALCULATIONS

Calculate [OH-] at 25 °C for each solution, and determine whether it is acidic, basic, or neutral. (a) [H30 +] = 7.5 X 10-5 mol L- 1 (c) [H30 +] = 1.0 X 10-7 mol L- 1 (b) [H30 +]

= 1.5 X 10-9 mol L -l

SOLUTION (a) To find [OH- ], use the ion product constant. Substitute the given value for [H30+] and solve the equation for [OH-].

Since [H30 +] > [OH- ], the solution is acidic.

=

1.0 x 10- 14

(1.5 X 10-9)[0H-] [OH-]

=

1.0 x 10- 14

X 10- 7, the solu-

(1.0 X 10-7)[0H- ] [OW]

=

(b) [OH-]

= 1.5 X 10-2 mol L -l = 1.0 X 10-7 mol L - l

(c) [OH-]

1.0 x 10- 7 Neutral solution

= 8.2 X 10- 10 mol L -l

=

= 1.0

1.0 x 10- 14

FOR PRACTICE 15.2 Calculate [H30 +] at 25 °C for each solution, and determine whether it is acidic, basic, or neutral.

= 1.3 x 10- 10 mol L - 1

= 1.0

1.5 x 10-9 Basic solution

(c) Substitute the given value for [H30 +] and solve the acid ionization equation for [OH- ].

(a) [OW]

= 1.0 X 10- 14 = 1.0 x 10- 14

Kw

7.5 x 10-5 Acidic solution

Since [H30+] < [OH-], the solution is basic.

= 1.0

=

(7.5 x 10- 5 )[0H- ] [OH- ]

(b) Substitute the given value for [H3 0 +] and solve the acid ionization equation for [OH-] .

Since [H30 +] = 1.0 X 10-7 and [OH- ] tion is neutral.

[H30 +][0W]

X 10- 14

6.7 X 10-6 mol L- 1

X 10- 14

= 1.0

X

10- 7 mol L-

1

649

15.6 Autoionization of Water and pH

The pH Scale: AWay to Quantify Acidity and Basicity The pH scale is a compact way to specify the acidity of a solution. We define pH as follows:

A solution with [H30+]

=

The log of a number is the exponent to which 10 must be raised to obtain that number. Thus, log 101 = 1; log 102 = 2; log 10- 1 = - 1; log 10- 2 = - 2, etc. (see Appendix IB).

1.0 x 10-3 mol L - l (acidic) has a pH of: pH

=

- log[H 30 +] - log(l.0 X 10-3) -(-3.00)

=

3.00

Notice that the pH is reported to two decimal places here. This is because only the numbers to the right of the decimal point are significant in a logarithm. Since our original value for the concentration had two significant figures, the log of that number has two decimal places. 2 significant digits

log@ \

When you take the log of a quantity, the result should have the same number of decimal places as the number of significant figures in the original quantity (see Section 1.4).

2 decimal places

'

X 10- 3 = 3.@

If the original number had three significant digits, the log would be reported to three decimal places: 3 significant digits

3 decimal places

log@>< 10- 3 = \ A solution with [H 3 0 +]

=

TABLE 15.8 The pH of Some

'

Common Substances

3@

Substance

1.0 X 10- 7 mol L - l (neutral) has a pH of pH

=

- log[H 30 +] - log(l.0

X 10- 7)

-(-7.00)

=

7.00

In general, at 25 °C:

pH

Gastric juice (human stomach)

1.0-3.0

Limes

1.8-2.0

Lemons

2.2-2.4

Soft drinks

2.0-4.0

Plums

2.8-3.0

Wines

2.8-3.8

Apples

2.9-3.3

Peaches

3.4-3.6

~

pH < 7

The solution is acidic.

~

pH > 7

The solution is basic.

Cherries

3.2-4.0

~

pH = 7

The solution is neutral.

Beers

4.0-5.0

Table 15.8 lists the pH of some common substances. As we discussed in Section 15.2, many foods, especially fruits, are acidic and have low pH values. Relatively few foods are basic. The foods with the lowest pH values are limes and lemons, and they are among the sourest. Since the pH scale is a logarithmic scale, a change of 1 pH unit corresponds to a 10-fold change in H 30 + concentration (Figure 15.10 T). For example, a lime with a pH of 2.0 is 10 times more acidic than a plum with a pH of 3.0 and I 00 times more acidic than a cherry with a pH of 4.0.

2

Acidic

.&. FIGURE 15.10 The pH Scale

3

4

5

6

7

8

9

10

11

Rainwater (unpol luted)

5.6

Human blood

7.3-7.4

Egg whites

7.6-8.0

Milk of magnesia

10.5 10.5- 11 .5

Household ammon ia

4% NaOH solution

12

14

13

pH

An increase of l on the pH scale corresponds to a decrease in [H+] by a factor of 10.

Basic

650

Chapter 15

EXAMPLE 15.3

Acids and Bases

CALCULATING pH FROM [Hl OR [OHl

Calculate the pH of each solution at 25 °C and indicate whether it is acidic or basic. (a) [H30 +) = 1.8 X 10-4 mol L -I (b) [OH-] = 1.3 X 10-2 mol L -

I

SOLUTION (a) To calculate pH, substitute the given [H30 +] into the pH equation.

Since pH < 7, this solution is acidic.

pH = = = =

- log[H30 +] - log(l .8 x 10- 4) -(-3.74) 3.74 (acidic)

[H30 +][0H- ] = Kw = 1.0 X 10- 14 [H30 +](l.3 X 10- 2) = 1.0 X 10- 14 1.0 x 10- 14 [HO+] = = 7.7 X 10- 13 mol L- 1 3 1.3 x 10-2 Then substitute [H30 +] into the pH equation to find pH.

pH = - log[H30 +] = -log(7.7 x 10- 13)

= -(-12.11) Since pH > 7, this solution is basic.

= 12.1 1 (basic)

FOR PRACTICE 15.3 Calculate the pH of each solution and indicate whether it is acidic or basic. (b) [OH- ]= 7.1 X 10- 3 molL- 1 (a) [H30 +] = 9.5 X 10- 9 mol L- 1

EXAMPLE 15.4

CALCULATING [H 301 FROM pH

Calculate the H30+ concentration for a solution with a pH of 4.80. SOLUTION To find the [H 30 +] from pH, start with the equation that defines pH. Substitute the given value of pH and then solve for [H 30 +]. Since the given pH value was reported to two decimal places, the [H30 +] is written to two significant figures. (Remember that 1010gx = x (see Appendix IB). Some calculators use an inv log key to represent this function.)

pH = - log[H30 +] 4.80 = - log[H30 +] -4.80 = log[H 30 +] 10-4.80 = 101og[H 30 +] 10- 480 = [H30 +] [H30 +) = 1.6 X 1o-smol L -

I

FOR PRACTICE 15.4 Calculate the H 30+ concentration for a solution with a pH of 8.37.

pOH and Other p Scales The pOH scale is analogous to the pH scale but is defined with respect to [OH-] instead of [H30 +]. pOH = - log[OW] Notice that p is the mathematical operator I - log; thus, pX = - log X.

A solution having an [OH- ] of 1.0 X 10- 3 mol L - I (basic) has a pOH of 3.00. On the pOH scale, a pOH less than 7 is basic and a pOH greater than 7 is acidic. A pOH of 7 is

15.6 Autoionization of Water and pH

0 .0

1.0

2.0

3.0

4.0

5.0

6.0

7.0

9.0

10.0

11.0

12.0

13.0

pH

Acidic

14.0

8.0

13.0

12.0

11.0

10.0

9.0

8.0

7.0

651

14.0 Basic

6.0

5.0

4.0

3.0

2.0

1.0

0.0

pOH .&. FIGURE 15.11 pH and pOH

neutral (Figure 15.11 • ).We can derive a relationship between pH and pOH at 25 °C from the expression for Kw:

Taking the log of both sides, we get log([H30+][0W]) log[H 30+]

+

=

log[OH- ]

-14.00

-log[H 30+] - log[OH-] pH

+

pOH

log(l.0 x 10- 14)

14.00

=

14.00

The sum of pH and pOH is always equal to 14.00 at 25 °C. Therefore, a solution with a pH of 3 has a pOH of 11. Another common p scale is the pKa scale defined as follows :

pKa = - log Ka The pKa of a weak acid is another way to quantify its strength. The smaller the pKa, the stronger the acid. For example, chlorous acid, with a Ka of 1.1 X 10- 2, has a pKa of 1.96, and methanoic acid, with a Ka of 1.8 X 10- 4 , has a pKa of 3.74. This tells us that chlorous acid is stronger than methanoic acid.

CHEMISTRY AND MEDICINE Ulcers An ulcer is a lesion on the wall of the stomach or small intestine. Under normal circumstances, a thick layer of mucous lines the stomach wall and protects it from the hydrochloric acid and other gastric juices in the stomach. If that mucous layer is damaged, however, stomach juices come in direct contact with the stomach wall and begin to digest it, resulting in an ulcer. The main symptom of an ulcer is a burning or gnawing pain in the stomach. Acidic drugs, such as aspirin, and acidic foods, such as citrus fruits and pickling fluids, irritate ulcers. When consumed, these substances increase the acidity of the stomach juices, exacerbating the irritation to the stomach wall. On the other hand, antacidswhich contain bases-relieve ulcers. Common antacids include Tums (active ingredient: calcium carbonate) and milk of magnesia (active ingredient: magnesium hydroxide). The causes of ulcers are manifold. For many years, a stressful lifestyle and a rich diet were blamed. More recent research, however, has shown that a bacterial infection of the stomach lining is responsible for many ulcers . (The 2005 Nobel Prize in Physiology or Medicine was awarded to Australians Barry J. Marshall and J. Robin Warren for their discovery of the bacterial cause of ulcers.) Long-term use of some over-the-counter pain relievers, such as aspirin, is also believed to produce ulcers.

Question Which dessert would be less likely to irritate an ulcer, key lime pie or meringue (made of egg whites)? Use the K,Pvalues in Table 16.2 to calculate the molar solubility of each compound in pure water. a. AgBr b. Mg(OH)z c. CaF2

94. Use the K,P values in Table 16.2 to calculate the molar solubility of each compound in pure water. a. CuS b. Ag2Cr04 c. Ca(OH) 2 Use the given molar solubilities in pure water to calculate Ksp for each compound. a. NiS; molar solubility = 3.27 X 10- 11 mol L- 1 b. PbF2 ; molar solubility = 5.63 X 10- 3 mol L- 1 c. MgF2 ; molar solubility = 2.65 X 10-4 mol L- 1

G

96. Use the given molar solubilities in pure water to calculate Ksp for each compound. a. BaCr04 ; molar solubility = 1.08 X 10- 5 mol L- 1 b. Ag2S0 3; molar solubi lity = 1.55 x 1o-s mol L- 1 c. Pd(SCN)z; molar solubi lity = 2.22 X 10- 8 mol L- I

G Two

compounds with general formulas AX and AX2 have K,P = 1.5 X l o-s. Which of the two compounds has the higher molar solubility? 98. Consider the compounds with the generic formulas listed and their corresponding molar solubilities in pure water. Which compound will have the smallest value of K,P? a. AX ; molar solubility = 1.35 X 10- 4 mol L- I b. AX2 ; molar solubility = 2.25 X 10- 4 mol L- 1 c. A2X; molar solubility = 1.75 X 10- 4 mol L- 1

Q)

Use the K,P value from Table 16.2 to calculate the solubility of iron(II) hydroxide in pure water in grams per 100.0 mL of solution. 100. Based on the solubility product constants given below, which salt is the least soluble? (i) Fe(OH)J Ksp = 6.3 x 10- 38 (ii) Pb3(P04)i Ksp = 3.0 X 10-44 (iii) Ca3(P04}z Ksp = 1.0 x 10- 26 Consider fi ve different saturated solutions containing one of the following salts. Without using a calculator, determine which would have the lowest pH. (You may use a calculator if you wish, but a "back of the envelope" calculation-an estimateshould suffice.) (i) KOH (ii) Pb(OH)i Ksp = 1.4 x 10- 20 (iii) Be(OH)i Ksp = 6.9 x 10- 22 (iv) Sc(OH)J Ksp = 4.2 x 10- 18 (v) Y(OH)J Ksp = 1.0 X 10- 22 102. The solubility of copper(!) chloride is 3.9 1 mg per 100.0 mL of solution. Calculate K,P for CuCI. Calculate the molar solubility of barium fluoride in each liquid or solution. c. 0.15 mol L- 1 NaF a. pure water b. 0.10 mol L- I Ba(N0 3)z

9

@

104. Calculate the molar solubility of copper(II) sulfide in each liquid or solution. a. pure water b. 0.25 mol L- 1 CuC1 2 c. 0.20 mol L- I K2S

G Calculate the molar solubility of calcium hydroxide in a solution buffered at each pH. a. pH = 4 b. pH = 7 c. pH = 9 106. Calculate the solubility (in grams per 1.00 X 102 mL of solution) of magnesium hydroxide in a solution buffered at pH = 10. How does this compare to the solubility of Mg(OH)z in pure water? Determine whether or not each compound will be more soluble in acidic solution than in pure water. Explain. a. BaC03 b. CuS c. AgCl d . Pbl 2

9

Exercises

108. Determine whether or not each compound will be more soluble

9

in acidic solution than in pure water. Explain. a. Hg2 Br2 b. Mg(OH)z c. CaC0 3

d. Agl

For each of the following compounds, use chemical equations to show how aqueous solubility is increased when dilute HzS04(aq) is added. a. CaC03(s) b. BaF2 (s) c. Ca3(P04h(s)

110. For each of the following compounds, use chemical equa-

tions to show how aqueous solubility is increased when dilute HzS04(aq) is added. a. Be(OHh(s) b. MgC03(s) c. PbS(s)

Precipitation and Qualitative Analysis

GD A solution containing sodium fluoride is mixed with one con-

taining calcium nitrate to form a solution that is 0.015 mol L- 1 in NaF and 0.010 mol L- 1 in Ca(N03)z. Will a precipitate form in the mixed solution? If so, identify the precipitate.

112. A solution containing potassium bromide is mixed with one

containing lead acetate to form a solution that is 0.013 mol L- 1 in KBr and 0.0035 mol L-I in Pb(CH3C00)2. Will a precipitate form in the mixed solution? If so, identify the precipitate.

cation will precipitate first? What minimum concentration of Na2S04 will trigger the precipitation of the cation that precipitates first? b. What is the remaining concentration of the cation that precipitates first, when the other cation begins to precipitate? 118. Consider a solution that is 0.022 mol L- 1 in Fe 2 + and 0.014 mol L- 1 in Mgz+. a. If potassium carbonate is used to selectively precipitate one of the cations while leaving the other cation in solution, which cation will precipitate first? What minimum concentration of K2C03 will trigger the precipitation of the cation that precipitates first? b. What is the remaining concentration of the cation that precipitates first, when the other cation begins to precipitate?

Complex-Ion Equilbria

0

A solution is made that is 1.1 X 10- 3 mol L- 1 in Zn(N0 3)2 and 0.150 mol L- 1 in NH 3. After the solution reaches equilibrium, what concentration of Zn2 +(aq) remains? 120. A 120.0 mL sample of a solution that is 2.8 X 10- 3 mol L- 1 in AgN03 is mixed with a 225.0 mL sample of a solution that is 0.10 mol L- 1 in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

@ Predict whether or not a precipitate will form if you mix 75.0 mL of a NaOH solution with pOH = 2.58 with 125.0 mL of a 0.018 mol L- 1 MgC12 solution. Identify the precipitate, if any. 114. Predict whether or not a precipitate will form if you mix 175.0 mL of a 0.0055 mol L- 1 KCl solution with 145.0 mL of a 0.0015 mol L- 1 AgN0 3 solution. Identify the precipitate, if any.

G Potassium hydroxide is used to precipitate each of the cations

from their respective solution. Determine the minimum concentration of KOH required for precipitation to begin in each case. a. 0.015 mol L- 1 CaC12 b. 0.0025 mol L- 1 Fe(N03)z c. 0.00 18 mol L- I MgBr2 116. Determine the minimum concentration of the precipitating

agent on the right to cause precipitation of the cation from the solution on the left. a. 0.035 mol L- I BaN03; NaF b. 0.085 mol L- I Cal2; K1S04 c. 0.00 18 mol L- 1 AgN0 3; RbCl

G Consider

743

G Use the appropriate values of K,Pand Kr to find the equilibrium constant for the reaction: FeS(s)

+ 6 CW(aq)

~ Fe(CN) 64- (aq)

+

S2- (aq)

122. Use the appropriate values of K,P and Kr to fi nd the equilibrium

constant for the reaction: PbCI2(s)

@

+ 3 OW(aq)

~

Pb(OH)3- (aq)

+ 2 Ci-(aq)

Calculate the molar solubility of Ag2S (Ksp = 6.2 X 10- 5 1) solution that is also 5.0 mol L- 1 in NH3 . (Kr Ag(NH3) / = 1.6 x 107 .)

124. Silver iodide has a Ksp of 8.51 X 10- 17 and therefore is quite

insoluble. Ag(CN)z- has a formation constant of 1.0 X 1021 . What is the solubility of silver iodide in a 0. 100 mol L -I solution of NaCN?

a solution that is 0.010 mol L- 1 in Ba2+ and 0.020 mol L- 1 in Ca2 +. a. If sodi um sulfate is used to selectively precipitate one of the cations while leaving the other cation in solution, which

Cumulative Problems

9

A 150.0 mL solution contains 2.05 g of sodium benzoate and 2.47 g of benzoic acid. Calculate the pH of the solution.

126. A solution is made by combining 10.0 mL of 17.5 mol L- 1 ace-

tic acid with 5.54 g of sodium acetate and diluting to a total volume of 1.50 L. Calculate the pH of the solution.

G Abuffer is created by combining 150.0 mL of 0.25 mol L-

1

HCOOH with 75.0 mL of 0.20 mol L- 1 NaOH. Determine the pH of the buffer. 128. A buffer is created by combining 3.55 g of NH 3 with 4.78 g of HCl and diluting to a total volume of 750.0 mL. Determine the pH of the buffer. A 1.0 L buffer solution initially contains 0.25 mol of NH 3 and 0.25 mol of NH4CI. In order to adj ust the buffer pH to 8.75, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?

e

130. A 250.0 mL buffer solution initially contains 0.025 mol of

HCOOH and 0.025 mol of NaCH02. In order to adjust the buffer pH to 4.10, should you add NaOH or HCl to the buffer mixture? What mass of the correct reagent should you add?

GD In analytical chemistry, bases used for titrations must often be standardized; that is, their concentration must be precisely determined. Standardization of sodium hydroxide solutions can be accomplished by titrating potassium hydrogen phthalate (KHC8 H4 0 4 ) , also known as KHP, with the NaOH solution to be standardized. a. Write an equation for the reaction between NaOH and KHP. b. The titration of 0.5527 g of KHP required 25.87 mL of an NaOH solution to reach the equivalence point. What is the concentration of the NaOH solution? 132. A 0.5224 g sample of an unknown monoprotic acid was titrated with 0.0998 mol L- I NaOH. The equivalence point of the

744

Chapter 16

Aqueous Ionic Equ ili brium

titration occurs at 23.82 mL. Determine the molar mass of the unknown acid .

G A 0.25 mol sample of a weak acid with an unknown pK was 3

combined with lO.O mL of 3.00 mol L- 1 KOH and the resulting solution was diluted to 1.500 L. The measured pH of the solution was 3.85. What is the pKa of the weak acid?

134. A 5.55 g sample of a weak acid with Ka = 1.3 X 10- 4 was combined with 5.00 mL of 6.00 mol L- 1 NaOH and the resulting solution was diluted to 750 mL. The measured pH of the solution was 4.25. What is the molar mass of the weak acid?

0

A 0.552 g sample of ascorbic acid (vitamin C) was dissolved in water to a total volume of 20.0 mL and titrated with 0.1103 mol L- 1 KOH. The equivalence point occurred at 28.42 mL. The pH of the solution at 10.0 mL of added base was 3.72. From this data, determine the molar mass and Ka for vitamin C.

136. Sketch the titration curve from Problem 135 by calculating the pH at the beginning of the titration, at one-half of the eq uivalence point, at the equivalence point, and at 5.0 mL beyond the equivalence point. Pick a suitable indicator for this titration from Table 16.1.

G One of the main components of hard water is CaC03. When

hard water evaporates, some of the CaC0 3 is left behind as a white mineral deposit. If a hard water solution is saturated with calcium carbonate, what volume of the solution has to evaporate to deposit 1.00 X 10 2 mg of CaC03?

138. Gout-a condition that results in j oint swelling and pain- is caused by the formation of sodium urate (NaC5H 3N4 ) crystals within tendons, cartilage, and ligaments. Sodium urate precipitates out of blood plasma when uric acid levels become abnormally high. This could happen as a result of eating too many rich foods and consuming too much alcohol, w hich is w hy gout is sometimes referred to as the " disease of kings." If the sodium concentration in blood plasma is 0. 140 mol L- 1, and Ksp for sodium urate is 5.76 X 10- 8, what minimum concentration of urate would result in precipitation?

Q) Pseudogout, a condition with symptoms similar to those of gout (see Problem 138), is caused by the formation of calcium diphosphate (Ca2 P2 0 7 ) crystals within tendons, cartilage, and ligaments. Calcium diphosphate will precipitate out of blood plasma when diphosphate levels become abnormally high. If the calcium concentration in blood plasma is 9 .2 mg dL- 1, and K sp for calcium diphosphate is 8.64 X 10- 13 , what minimum concentration of diphosphate results in precipitation?

140. Calculate the solubility of silver chloride in a solution that is O. lOO mol L- 1 in NH3.

0

Calculate the solubility of copper(II) sulfide in a solution that is 0. 150 mol L- 1 in NaCN.

142. Aniline, abbreviated cpNH 2 , where cp is C 6H 5 , is an important organic base used in the manufacture of dyes. It has

0

Kb = 7.5 X 10- w In a certain manufacturing process it is necessary to keep the concentration of cpNH3+ (aniline 's conjugate acid, the anilinium ion) below 1.0 X 10- 9 mol L- 1 in a solution that is 0.10 mol L- 1 in aniline. Find the conce ntration of NaOH necessary for this process. The Kb of hydroxylamine, NH20H, is 1.10 X 10- 8. A buffer solution is prepared by mixing lOO.O mL of a 0.36 mol L- 1 hydroxylamine solution with 50.0 mL of a 0.26 mol L -I HCI solution. De termine the pH of the resulting solution.

144. A 0.867 g sample of an unknow n acid requires 32.2 mL of a 0.1 82 mol L- 1 barium hydroxide solution for neutralization. Assuming the acid is diprotic, calculate the molar mass of the acid.

0

A 25.0 mL volume of a sodium hydroxide solution req uires 19.6 mL of a 0.189 mol L- 1 hydrochloric acid for neutralization. A lO.O mL volume of a phosphoric acid solution requires 34.9 mL of the sodium hydroxide solution for complete neutralization. Calculate the concentration of the phosphoric acid solution.

146. Find the mass of sodium formate that must be dissolved in 250.0 cm 3 of a 1.4 mol L-I solution of formic acid to prepare a buffer solution with pH = 3.36.

@ What relative masses of dimethyl amine and dimethyl ammonium chloride do you need to prepare a buffer solution of pH = 10.43?

148. You are asked to prepare 2.0 L of a HCN/NaCN buffer that has a pH of 9.8 and an osmotic pressure of 1.35 bar at 298 K. What masses of HCN and NaCN should you use to prepare the buffer? (Assume complete dissociation of NaCN.) What should the molar concentrations of benzoic acid and sodium benzoate be in a solution that is buffered at a pH of 4.55 and has a freezing point of -2.0 °C? (Assume complete dissociation of sodium benzoate and a density of 1.0 1 g mL- 1 for the solution.)

@

150. If two solutions of equal vo lume, one saturated with CrP04 (Ksp = 2.4 X l0- 23) and the other saturated with Ca(I03)z (Ksp = 1.0 X 10-7 ), are mi xed (no solids from either solution are mixed), will Ca3(P04)z (Ksp = 1.3 X 10- 32 ) precipitate?

G A student measures the pH of a saturated solution of iron(II) hydroxide (Fe(OH)z) to be 9.41. Based on this measure ment, what is the K sp of Fe(OH)z?

152. Dolomite, CaMg(C03)z, is a mineral that is similar to limestone, CaC03. The K sp of dolomite is 2.0 X 10- 17 . What is the molar solubility of dolomite in pH-neutral water? How does this compare to the molar solubility of CaC03?

9

Aluminum hydroxide, Al(OH)J(s), has a K sp of approximately I x I 33 . Calculate the molar solubility of aluminum hydroxide in solutions that are buffered at pH 9, 8, 7, 6, 5, and 4. By what factor does the molar solubility change with every unit of decreasing pH?

o-

Challenge Problems 154. Derive an equation similar to the Henderson-Hasselbalch equation for a buffer composed of a weak base and its conjugate acid. Instead of relating pH to pKa and the relative concentrations of an acid and its conjugate base (as the Henderson- Hasselbalch equation does), the equation should relate pOH to pKb and the relative concentrations of a base and its conjugate acid.

0

Since soap and detergent action is hindered by hard water, laundry fo rmulations usually include water softeners-called builders--designed to remove hard water ions (especially Ca2 +

and M g 2 +) from the water. A common builder used in North America is sodium carbonate. Suppose that the hard water used to do laundry contains 75 ppm CaC03 and 55 ppm MgC0 3 (by mass). What mass of Na2C03 is required to remove 90.0% of these ions from I 0.0 L of laundry water?

156. A 0.558 g sample of a diprotic acid with a molar mass of 255.8 g mol- 1 is dissolved in water to a total volume of 25.0 mL. The solution is then titrated with a saturated calcium hydroxide solution.

Exercises

0

a. Assuming that the pKa values for each ionization step are sufficiently different to see two equivalence points, determine the volume of added base for the first and second equivalence points . b. The pH after adding 25.0 mL of the base was 3.82. Find the value of Ka,· c. The pH after adding 20.0 mL past the first equivalence point was 8.25. Find the value of

K.,.

When excess solid Mg(OHh is shaken with 1.00 L of 1.0 mol L- 1 NH4 CI solution, the resulting saturated solution has pH = 9.00. Calculate the K,P of Mg(OH)2 . 158. What amount of solid NaOH must be added to 1.0 L of a 0.10 mol L- 1 H 2 C03 solution to produce a solution with [H+] = 3.2 X 10- 11 mol L- 1? (There is no significant volume change as the result of the addition of the solid.) Calculate the solubility of Au(OH)J in (a) water and (b) l.O mol L- 1 nitric acid solution. (K,P = 5.5 X l0- 46)

0

745

160. Calculate the concentration of 1- in a solution obtained by shaking 0. 10 mol L- 1 Kl with an excess of AgCl(s).

e

What volume of 0. 100 mol L- 1 sodium carbonate solution is required to precipitate 99% of the Mg from 1.00 L of0.100 mol L- 1 magnesium nitrate solution?

162. Find the solubility of Cul in 0.40 mol L- 1 HCN solution. The K,Pof Cul is l. I X 10- 12 and the Kr for the Cu(CNh- complex ion is l X 1024 . • Find the pH of a solution prepared from 1.0 L of a 0.10 mol L- I solution of Ba(OHh and excess Zn(OHh(s). The K,P of Zn(OHh is 3 X 10- 15 and the Kr ofZn(OH)/- is 2 X 10 15 . 164. What amount of HCI gas must be added to l .00 L of a buffer solution that contains [acetic acid] = 2.0 mol L-I and [acetate] = 1.0 mol L- 1 in order to produce a solution with pH= 4.00?

Conceptual Problems •

Without doing any calculations, determine whether pH = pKa, pH > pK., or pH < pKa. Assume that HA is a weak monoprotic acid. a. 0.10 mol HA and 0.050 mol of A- in 1.0 L of solution b. 0. l 0 mol HA and 0. l 50 mol of A- in l .O L of solution c. O. lO mol HA and 0.050 mol of OH- in 1.0 L of solution d. O. lO mol HA and 0.075 mol of OH- in 1.0 L of solution

166. A buffer contains O. lO mol of a weak acid and 0.20 mol of its conjugate base in 1.0 L of solution. Determine whether or not each addition exceeds the capacity of the buffer. a. adding 0.020 mol of NaOH b. adding 0.020 mol of HCI c. adding O.lO mol ofNaOH d. adding 0.0 I0 mol of HCI

G Consider three solutions:

(i) 0. 10 mol L- 1 solution of a weak monoprotic acid (ii) O. lO mol L- I solution of strong monoprotic acid (iii) 0. 10 mol L- 1 solution of a weak diprotic acid

Each solution is titrated with 0 .15 mol L- 1 NaOH. Which quantity will be the same for all three solutions? a. the volume required to reach the final equivalence point b. the volume required to reach the first equivalence point

c. the pH at the first equivalence point d . the pH at one-half the first equivalence point 168. Two monoprotic acid solutions (A and B) are titrated with identical NaOH solutions. The volume to reach the equivalence point for solution A is twice the volume requi red to reach the equivalence point for solution B, and the pH at the equivalence point of solution A is higher than the pH at the equivalence point for solution B. Which statement is true? a. The acid in solution A is more concentrated than in solution B and is also a stronger acid than that in solution B. b. The acid in solution A is less concentrated than in solution B and is also a weaker acid than that in solution B. c. The acid in solution A is more concentrated than in solution B and is also a weaker acid than that in solution B. d . The acid in solution A is less concentrated than in solution B and is also a stronger acid than that in solution B.

G) Describe the solubility of CaF

2 in each solution compared to its solubility in water. a. in a O.lO mol L- 1 NaCl solution b. in a 0. 10 mol L- 1 NaF solution c. in a 0. 10 mol L- 1 HCl solution

In this clever illusion, it seems that the water can perpetually flow through the canal. However, perpetual motion is f orbidden under the laws of thermodynamics. [Quade Paul/Pearson Education]

17.1 Spontaneous and Nonspontaneous Processes 74 7 17.2 Entropy and the Second Law of Thermodynamics 748

HROUGHOUT THIS TEXT, we have examined and learned much about

17.3 Heat Transfer and Changes in the Entropy of the Surroundings 755

T

17.4 Entropy Changes for Phase Transitions 759

expand to fill their containers. We now turn to the following question: Why do these

17.5 Entropy Changes in Chemical Reactions: Calculating ~ rS

0

760

17.6 Gibbs Energy

~rG

occur (kinetics) and how to predict how far they will go (through the use of

equilibrium constants). We have learned that acids neutralize bases and that gases

changes occur in the first place? What ultimately drives physical and chemical changes in matter? The answer may surprise you. The driving force behind chemical and physical change in the universe is a quantity called entropy, which

765

17.7 Gibbs Energy Changes in Chemical Reactions: Calculating 0

chemical and physical changes. We have studied how fast chemical changes

is related to the dispersion (or spreading out) of energy. Nature tends toward that state in which energy is spread out to the greatest extent possible. Although it does

769

17.8 Making a Nonspontaneous Process Spontaneous 773

not seem obvious at first glance, the freezing of water below 0 °C, the dissolving

17.9 What Is Gibbs Energy?

of a solid into a solution, the neutralization of an acid by a base, and even the

776

17.10 Gibbs Energy Changes for Nonstandard States: The Relationship Between ~rG and 0

~rG

111

(they all result in greater energy dispersion). In our universe, entropy always increases.

17.11 Gibbs Energy and Equilibrium: Relating ~rG to the Equilibrium Constant (K) 781 0

746

development of a person from an embryo all increase the entropy in the universe

747

17.1 Spontaneous and Nonspontaneous Processes

17.1

Spontaneous and Nonspontaneous Processes

A fundamental goal of thermodynamics is to predict spontaneity. For example, will rust spontaneously form when iron comes into contact with oxygen? Will water spontaneously decompose into hydrogen and oxygen? A spontaneous process is one that occurs without ongoing outside intervention (such as the performance of work by some external force). For example, when you drop a book in a gravitational field, it spontaneously drops to the floor. When you place a ball on a slope, it spontaneously rolls down the slope. For simple mechanical systems, such as the dropping of a book or the rolling of a ball, predicting spontaneity is fairly intuitive. A mechanical system tends toward lowest potential energy, which is usually easy to see (at least in simple mechanical systems). However, the prediction of spontaneity for chemical systems is not so intuitively obvious. To do so, we need to develop a criterion for the spontaneity of chemical systems. In other words, we want to develop a chemical potential that predicts the direction of a chemical system, much as mechanical potential energy predicts the direction of a mechanical system (Figure 17.1 T).

Solid NaCl

Potential energy

Direction of spontaneous change

Direction of spontaneous change

C hemical potential

:::> "' 0

:::> "' 0

OJ

c:

OJ

ro

c: .(!l c:

c: 0

0..

0

c: "' 0

0..

V'l

z D issolved ions

(a)

(b)

.&. FIGURE 17.1 Mechanical Potential Energy and Chemical Potential (a) Mechanical potential energy predicts the direction in which a mechanical system will spontaneously move. (b) We seek a chemical potential that predicts the direction in which a chemical system will spontaneously move.

-

(

.~ ~,~~\ I. ,~,

·" "":' .

••

j/1-

I -.-

.,

-· We must not confuse the spontaneity of a chemical reaction with the speed of a chemical reaction. In thermodynamics, we study the spontaneity of a reactionthe direction in which and extent to which a chemical reaction proceeds. In kinetics, we study the speed of the reaction-how fast a reaction takes place (Figure 17 .2 T).

KINETICS Intermediate states, speed Products

Reactants

~ Initial and final -

states, spontaneity Reaction progress

-

.&. Iron spontaneously rusts when it comes in contact with oxygen. [top: Darling Kindersley, Ltd.; bottom: Henrik Winther Ander/Fotolia]

""' FIGURE 17.2 Thermodynamics and Kinetics Thermodynamics deals with the relative chemical potentials of the reactants and products. It enables us to predict whether a reaction will be spontaneous, and to calculate how much work it can do. Kinetics deals with the chemical potential of intermediate states, and enables us to determine why a reaction is slow or fast.

748

Chapter 17

Gibbs Energy and Therm odynam ics

A reaction may be thermodynamically spontaneous but kinetically slow at a given temperature. For example, the conversion of diamond to graphite is thermodynamically spontaneous. But your diamonds will not become worthless anytime soon because the process is extremely slow kinetically. Although the rate of a spontaneous process can be increased by the use of a catalyst (Section 13.7), a nonspontaneous process cannot be made spontaneous by the use of a catalyst. Catalysts affect only the rate of a reaction, not the spontaneity. One last word about nonspontaneity- a nonspontaneous process is not impossible. The extraction of iron metal from iron ore is a nonspontaneous process; it does not happen if the iron ore is left to itself, but that does not mean it is impossible. As we will see later in thi s chapter, a nonspontaneous process can be made spontaneous by coupling it to another process that is spontaneous, or by supplying energy from an external source. Iron can be separated from its ore if external energy is supplied, usually by means of another reaction (that is itself highly spontaneous) .

.... Even though graphite is thermodynamically more stable than diamond, the conversion of diamond to graphite is kinetically so slow that it does not occur at any measurable rate. [diamond: Igor Kali/Fotolia; graphite: Daniel Schwen]

Diamond

Graphite

17.2 Entropy and the Second Law of Thermodynamics

I See Section 6.6 for the definition of enthalpy.

The first candidate in our search for a chemical potential might be enthalpy, which we defined in Chapter 6. Perhaps, just as a mechanical system proceeds in the direction of lowest potential energy, so a chemical system might proceed in the direction of lowest enthalpy. If this were the case, all exothermic reactions would be spontaneous and all endothermic reactions would not. However, although most spontaneous reactions are exothermic, some spontaneous reactions are endothermic. For example, above 0 °C, ice spontaneously melts, but the process is endothermic. So enthalpy must not be the sole criterion for spontaneity. We can learn more about the driving force behind chemical reactions by considering several processes (like ice melting) that involve an increase in enthalpy. These processes are enthalpically uphill-meaning that they are endothermic- yet they occur spontaneously. What drives them? .... the melting of ice above 0 °C .... the evaporation of liquid water to gaseous water

The use of the word disorder here is analogous to our macroscopic notions of disorder. The definition of molecular disorder, which is covered shortly, is very specific.

.... the dissolution of sodium chloride in water Each of these processes is endothermic and spontaneous. Do they have anything in common? Notice that, in each process, disorder or randomness increases. In the melting of

17.2

Entropy and the Second Law ofThermodynamics

749

ice, the arrangement of the water molecules changes from a highly ordered one in ice to a somewhat disorderly one in liquid water.

Increasing

entropy

0) for a process to be spontaneous. The entropy of the system can therefore decrease ( dSsys < 0) as long as the entropy of the surroundings increases by a greater amount ( dSsurr > - dSsys), so that the overall entropy of the universe undergoes a net increase. For liquid water freezing or water vapour condensing, we know that the change in entropy for the system (dSsys) is negative because the water loses ways of storing energy (rotations and translations), resulting in fewer microstates for the particular macrostate. For dSuniv to be positive, therefore, dSsurr must be positive and greater in absolute value (or magnitude) than dSsys as shown graphically here:

Even though (as we saw earlier) enthalpy by itself cannot determine spontaneity, the increase in the entropy of the surroundings caused by the release of heat explains why exothermic processes are so often spontaneous.

Low temperature

More available energy levels at higher temperatures

Kinetic energy

.&. FIGURE 17.5 Maxwell-Boltzmann Distribution of Kinetic Energies for Two Different Temperatures

But why does the freezing of ice or the condensation of water increase the entropy of the surroundings? The first clue in answering this question is that both processes, freezing and condensation, are exothermic: they give off heat to the surroundings which increases the temperature of the surroundings. If we look at a Maxwell-Boltzmann distribution of kinetic energies for two different temperatures, as in Figure 17 .5 ~ , it is apparent that higher energy levels are accessible at higher temperatures. With more available energy levels, there are more possible microstates available that can describe the macrostate. Another way of saying this is that the energy can be di spersed over more possible energy levels. The freezing of water below 0 °C and the condensation of water vapour on a cold night both increase the entropy of the universe because the heat given off to the surroundings increases the entropy of the surroundings to a sufficient degree to overcome the entropy decrease in the water.

Summarizing Entropy Changes in the Surroundings: ~

An exothermic process increases the entropy of the surroundings.

~

An endothermic process decreases the entropy of the surroundings.

17.3 HeatTransfer and Changes in the Entropy of the Surroundings

757

The Temperature Dependence of LlSsurr We have just seen how the freezing of water increases the entropy of the surroundings by dispersing heat energy into the surroundings. Yet we know that the freezing of water is not spontaneous at all temperatures. The freezing of water becomes nonspontaneous above 0 °C. Why? Because the magnitude of the increase in the entropy of the surroundings due to the dispersal of energy into the surroundings is temperature dependent. The greater the temperature, the smaller the increase in entropy for a given amount of energy dispersed into the surroundings. Recall that the units of entropy are joules per kelvin: energy units divided by temperature units. Entropy is a measure of energy dispersal (joules) per unit temperature (kelvins). We can understand the temperature dependence of entropy changes due to heat flow with a simple analogy. Imagine that you have $1000 to give away. If you gave the $1000 to a rich person the impact on his net worth would be negligible (because they already have so much money). If you gave the same $1000 to a poor person, however, their net worth would change substantially (because they have so little money). Similarly, if you disperse 1OOO J of energy into surroundings that are hot, the entropy increase is small (because the impact of the I OOO J is small on surroundings that already contain a lot of energy). If you disperse the same I OOO J of energy into surroundings that are cold, however, the entropy increase is large (because the impact of the 1OOO J is great on surroundings that contain little energy). For this same reason, the impact of the heat released to the surroundings by the freezing of water depends on the temperature of the surroundings-the higher the temperature, the smaller the impact. We can now understand why water spontaneously freezes at low temperature but not at high temperature. For the freezing of liquid water into ice, the change in entropy of the system is negative at all temperatures. ASuniv

=

LlSsys

t

+ ASsurr

Negative

\

Positive and large at low temperature Positive and small at high temperature

At low temperatures, the decrease in entropy of the system is overcome by the large increase in the entropy of the surroundings (a positive quantity), resulting in a positive ASuniv and a spontaneous process. At high temperatures, on the other hand, the decrease in entropy of the system is not overcome by the increase in entropy of the surroundings (because the magnitude of the positive ASsurr is smaller at higher temperatures), resulting in a negative ASuniv; therefore, the freezing of water is not spontaneous at high temperature as shown graphically here. ASuniv

=

ASsys

Low Temperature: Spontaneous

+ ASsurr

(for water freezing) ( High Temperature: Nonspontaneous

ASuniv

ASuniv

Quantifying Entropy Changes in the Surroundings We have seen that when a system exchanges heat with the surroundings, it changes the entropy of the surroundings. At constant pressure, we can use q sys to quantify the change in entropy for the surroundings (ASsurr). In general, ~

A process that emits heat into the surroundings (qsys negative) increases the entropy of the surroundings (positive ASsurr).

Condensation is also spontaneous at temperatures below 100 °C, but nonspontaneous at higher temperatures. The positive entropy change in the surroundings is greater at lower temperature than at higher temperature.

758

Chapter 17

Gibbs Energy and Th ermod ynam ics ~

A process that absorbs heat from the surroundings (qsys positive) decreases the entropy of the surroundings (negative ilSsurr) .

~

The magnitude of the change in entropy of the surroundings is proportional to the magnitude of qsys·

We can summarize these three points with the proportionality: [17.2) We have also seen that, for a given amount of heat exchanged with the surroundings, the magnitude of ilSsurr is inversely proportional to the temperature. In general, the higher the temperature, the lower the magnitude of ilSsurr for a given amount of heat exchanged: [17.3) Combining the proportionalities in Equations l 7 .2 and l 7 .3, we get the foll owing general expression at constant temperature: [17.4]

Remember from Chapter 6 that the units for enthalpy of reaction are kJ mo1- 1, where "mol- 1" means per mole of reaction. Similarly, the units for entropy of reaction are J 1\1 mo1- 1.

For any chemical or physical process occurring at constant temperature and pressure, the entropy change of the surroundings is equal to the energy dispersed into the surroundings (- ilHsys) divided by the temperature of the surroundings in kelvin. This equation provides insight into why exothermic processes have a tendency to be spontaneous at low temperatures-they increase the entropy of the surroundings. As temperature increases, however, a given negative ilH produces a smaller positive ilSsurr; thus, exothermicity becomes less of a determining factor for spontaneity as temperature increases. Under conditions of constant pressure q,ys = ilHsys; therefore,

LlSsurr

EXAMPLE 17 .2

- ilHsys

= ---

(constant P, T)

T

[17.5]

CALCULATING ENTROPY CHANGES IN THE SURROUNDINGS

Consider the combustion of propane gas:

(a) Calculate the entropy change in the surroundings associated with this reaction occurring at 100 °C. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?

SOLUTION (a) The entropy change of the surroundings is given by Equation 17.5. Substitute the value of il, H and the temperature in kelvins and calculate LlSsurr·

T

= 273 + 100 = 373 K

LlSsurr

=

- Li H

----:j:-

- -(-2044 kJ mol- 1) 373 K = + 5.48 kJ K- 1mol- 1 = 5.48 X 103 J K- 1mol- 1 (b) Determine the number of moles of gas on each side of the reaction. An increase in the number of moles of gas implies a positive ilSsys·

6 mol gas

ilSsys is positive.

7 mol gas

17.4

(c) The change in entropy of the universe is the sum of the entropy changes of the system and the surroundings. If the entropy changes of the system and surroundings are both the same sign, the entropy change for the universe will also have the same sign.

Entropy Changes for Phase Transitions

b.Suniv = b.Ssys

I Positive

759

+ b.Ssurr

\ Positive

Therefore, b.Suniv is positive and the reaction is spontaneous.

FOR PRACTICE 17.2 Consider the reaction between nitrogen and oxygen to form dinitrogen monoxide:

(a) Calculate the entropy change in the surroundings associated with this reaction occurring at 25 °C. (b) Determine the sign of the entropy change for the system. (c) Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?

FOR MORE PRACTICE 17.2 A reaction has b.,H = -107 kJ mol- 1 and b.,S = 285 J K- 1mol- 1• At what temperature is the change in entropy for the reaction equal to the change in entropy for the surroundings?

Do biological systems contradict the second law of thermodynamics? By taking energy from their surroundings and synthesizing large, complex biological molecules, plants and animals tend to concentrate energy, not disperse it. How can this be so?

17.4 Entropy Changes for Phase Transitions Let's return to the example of water freezing to ice. We said that at low temperatures, below 273.15 K (0 °C), Suniv is positive, and the process is spontaneous. At higher temperatures, Suniv is negative, and the freezing of water is nonspontaneous. What happens, though, at exactly 273.15 K? The process is neither spontaneous nor nonspontaneous. In fact, at 0 °C, if you have a mixture of water and ice, the system is at equilibrium. As long as no heat is added or removed, no more water freezes, and no more ice melts. When the system is at equilibrium, Suniv = 0 J K- 1 mol- 1• Equation 17 .1 can be written as follows: [17.6] While this system is at equilibrium, if we add a little bit of heat, a small amount of ice melts. On the other hand, if we remove a little bit of heat, some of the water will freeze. In both cases, the temperature remains at 0 °C, and the system is still at equilibrium. This process of melting ice by adding a little heat and freezing water by removing a little heat in an ice-water mixture at 0 °C is called a reversible process (we will return to the subject of a reversible process in Section 17.9). The small amount of energy added or removed from the system in this reversible process is called qrev· We can write a similar equation to Equation 17.4 for the entropy change in the surroundings:

b.S surr

=

- qrev

T

[17.7]

By combining Equations 17.6 and 17.7, we obtain the following expression for the entropy change of the system:

b.S

sys

=

qrev

T

[17.8]

760

Chapter 17

Gibbs Energy and Therm odynamics

Again, at constant pressure, q = t::.H, so for any process occurring reversibly at constant temperature, we can determine the entropy change for the system by measuring the change in enthalpy for the system using the relationship in Equation 17.9:

D.rH t::.,S= T

[ 17 .9]

Types of processes that can be considered to occur at constant temperature and reversibly are phase changes. For example, water freezes at 0 °C (273 .1 5 K) and the enthalpy change associated with the freezing of water (enthalpy of fusion) is -6.01 kJ mol- 1• Using Equation 17 .9, we calculate the entropy changes associated with freezing to be the following:

t::. S = r

-6.01 kJ mol- 1 = -0 .0220 kJ K- 1 mo1- 1 273.15 K

= -22.0 J K- 1 mo1- 1

Note that this entropy change corresponds to a loss of entropy, which is sensible for freezing.

17.5 Entropy Changes in Chemical Reactions: Calculating a rS° In Chapter 6, we learned how to calculate standard changes in enthalpy (D.rH 0 ) for chemical reactions. We now tum to calculating standard changes in entropy for chemical reactions. Recall from Section 6.9 that the standard enthalpy change for a reaction (f::.,H 0 ) is the change in enthalpy for a process in which all reactants and products are in their standard states. Recall also that the standard states of substances are defined as follows: ~

For a Gas: The standard state for a gas is the pure gas at a pressure of exactly 1 bar.

~

For a Liquid or Solid: The standard state for a liquid or solid is the pure substance at a pressure of 1 bar and at the temperature of interest (often taken to be 25 °C).

~

For a Substance in Solution: The standard state for a substance in solution is a concentration of 1 mol L- 1•

We now define the standard entropy change for a reaction (ArS0 ) as the change in entropy for a process in which all reactants and products are in their standard states. Since entropy is a function of state, the standard change in entropy is therefore the standard entropy of the products minus the standard entropy of the reactants:

t::.,S° = S°(products) - S°(reactants) But how do we find the standard entropies of the reactants and products? Recall from Chapter 6 that we defined standard molar enthalpies offormation (D.rH°) to use in calculating t::.,H We now need to define standard molar entropies (S to use in calculating t::.,S°. 0

0

•

)

Standard Molar Entropies (S0 ) and the Third Law of Thermodynamics In Chapter 6, we defined a relative zero for enthalpy. Recall that we assigned a value of zero to the standard enthalpy of formation for an element in its standard state. This was necessary because absolute values of enthalpy cannot be determined. In other words, for enthalpy, there is no absolute zero against which to measure all other values; therefore, we always have to rely on enthalpy changes from an arbitrarily assigned standard. For entropy, however, there is an absolute zero. The absolute zero of entropy is established by the third law of ther modynamics, which states that: T he entropy of a perfect cr ystal at absolute zero (0 K) is zero. Earlier we learned that in any distribution of energy among molecules, no molecule can have 0 J of energy. Even at absolute zero, no particles can have 0 J of kinetic energy.

761

17.5 Entropy Changes in Chemical Reactions: Calculating t;./?

If the particles did stop moving completely, this would disobey Heisenberg 's uncertainty

principle, which is stated in Equation 7.10:

Lix X miiv

~

h

47T

If any molecule stopped moving completely, the uncertainty in its velocity, Liv, would

Perfect

be 0 and the left-hand side would equate to 0, which is not greater than h/4n as required by Heisenberg's uncertainty principle. So, even at 0 K, all substances have what we call zero-point energy, tiny vibrations with respect to one another (Figure 17 .6 ~ ). However, the zero-point energy is distributed evenly throughout the crystal and there is only one way to distribute the energy-the crystal's lowest energy state. The number of microstates is one (W = 1), and therefore S = 0. We can measure all entropy values against the absolute zero of entropy as defined by the third law. Table 17 .1 shows values of standard entropies at 25 °C for selected substances. A more complete list can be found in Appendix IIB. Standard entropy values are listed in units of joules per kelvin per mol (J K- 1 mol- 1) . The units of mole in the denominator is required because entropy is an extensive property-it depends on the amount of the substance. At 25 °C, the standard entropy of any substance is the amount of energy dispersed into one mole of that substance at 25 °C. This amount of energy depends on the number of places or "energy levels" to disperse energy within the substance. The factors that affect the number of energy levels to disperse energy-and therefore the standard entropy-include the state of the substance, the molar mass of the substance, the particular allotrope, its molecular complexity, and its extent of dissolution. We examine each of these separately.

crystal at O K W=1 5=0

Relative Standard Entropies: Gases, Liquids, and Solids As we saw in Section 17 .2, the entropy of a gas is generally greater than the entropy of a liquid, which is in turn greater than the entropy of a solid. We can see these trends in the tabulated values of standard entropies. For example, consider the relative standard entropies of liquid water and gaseous water at 25 °C. S0 (J K- 1 mo1 - 1 ) H20(/)

70.0

H20(g)

188.8

Gaseous water has a much greater standard entropy because, as we discussed in Section 17 .2, it has more energy levels available to disperse energy at 25 °C.

TABLE 17.1

Standard Molar Entropy Values (S for Selected Substances at 298 K

Substance

S° (J K- 1 mol- 1)

0

)

Substance

S° (J K- 1 mol- 1)

Liquids

Gases H2(g)

130.7

Ar(g)

154.8

CH 30H(/)

CH4(g)

186.3

Br2(/)

H20(g)

188.8

CsHs(/)

N2(g)

191.6

H20(/)

Substance

S° (J K- 1 mol- 1)

Solids MgO(s)

27.0

126.8

Fe(s)

27.3

152.2

Li(s)

29.1

173.4

Cu(s)

33.2

Na(s)

51 .3

70.0

NH3(g)

192.8

K(s)

64.7

F2(g)

202.8

NaCl(s)

72.1

02(g)

205.2

CaC03(S)

91.7

C2H 4(g)

219.3

FeCl3(S)

142.3

Cl 2(g)

223.1

zero-point energy

.A. FIGURE 17.6 A perfect crystal at 0 K has only one way to distribute energy to the particles in the system. All particles have only zero-point energy.

Some elements exist in two or more forms,

I called al/otropes, within the same state.

762

Chapter 17

Gibbs Energy and Th erm od ynamics

Relative Standard Entropies: Molar Mass Consider the standard entropies of the noble gases at 25 °C: S0 (J K- 1 mo1- 1 ) He(g)

126.2

Ne(g)

146.1

Ar(g)

154.8

Kr(g)

163.8

Xe(g)

169.4

•

The more massive the noble gas, the greater its entropy at 25 °C. A complete explanation of why entropy increases with increasing molar mass is beyond the scope of this text. Briefly, the energy states associated with the motion of heavy atoms are more closely spaced than those of lighter atoms. The more closely spaced energy states allow for greater dispersal of energy at a given temperature and therefore greater entropy. This trend holds only for elements in the same state. The effect of a state change-from a liquid to a gas, for example- is far greater than the effect of molar mass.

Relative Standard Entropies: Allotropes As mentioned previously, some elements can exist in two or more forms-called allotropes-in the same state of matter. For example, the allotropes of carbon include diamond and graphite-both solid forms of carbon. Since the arrangement of atoms within these forms is different, their standard molar entropies are different:

C(s, diamond)

24.

C(s, graphite)

In diamond the atoms are constrained by chemical bonds in a highly restricted threedimensional crystal structure. In graphite the atoms bond together in sheets, but the sheets have freedom to slide past each other. The less constrained structure of graphite results in more energy levels into which energy can be dispersed and therefore greater entropy compared to diamond.

Relative Standard Entropies: Molecular Complexity For a given state of matter, entropy generally increases with increasing molecular complexity. Consider the standard entropies of the following two gases: Molar Mass (g mol- 1)

S 0 (J K- 1 mol- 1)

Ar(g)

39.948

154.8

NO(g)

30.006

210.8

17.5 Entropy Changes in Chemical Reactions: Calculating t;./?

-

--

Translational motion

R o tational motion

763

0), then the change in Gibbs

17.6 Gibbs Energy

767

energy will be negative at all temperatures and the reaction will be spontaneous at all temperatures.

'

lire = lirH - T lirS

/

\

Negative

Negative at all temperatures

Positive

As an example, consider the dissociation of N2 0:

lirH' = -163.2 kJ mol- 1 2 mol gas

3 mol gas

The change in enthalpy is negative- heat is emitted, increasing the entropy of the surroundings. In Section 17.5, we learned how to determine the value of li,s", but for the present discussion we just need to look at the reaction and determine whether lirs> is positive or negative. For the above example, the number of moles of gas increases, therefore there are more ways to disperse the given energy of the system, more translations, vibrations, and rotations. Therefore, li,s" is positive. Since the entropy of both the system and the surroundings increases, the entropy of the universe must also increase, making the reaction spontaneous at all temperatures.

Case 2: il.r H Positive, il.r S Negative If a reaction is endothermic lir H, and if the change in entropy for the reaction is negative lirS, then the change in Gibbs energy will be positive at all temperatures and the reaction will be nonspontaneous at all temperatures.

'

lire= lirH- T lirS

/

\

Positive

Positive at all temperatures

Negative

As an example, consider the formation of ozone from oxygen: 3 0 2(g) ~ 2 0 3(g) 3 mol gas

lirF = 285.4 kJ mol- 1

2 mol gas

The change in enthalpy is positive-heat is therefore absorbed, decreasing the entropy of the surroundings. The change in entropy is negative, which means that the entropy of the system decreases. (We can see that the change in entropy is negative from the balanced equation-the number of moles of gas decreases.) Since the entropy of both the system and the surroundings decreases, the entropy of the universe must also decrease, making the reaction nonspontaneous at all temperatures.

Case 3: il.rH Negative, il.rS Negative If a reaction is exothermic lirH, and if the change in entropy for the reaction is negative lirS, then the change in Gibbs energy will depend on temperature. The reaction will be spontaneous at low temperature, but nonspontaneous at high temperature.

'

lire = lirH- T lirs

/

Negative at low temperatures Positive at high temperatures

\

Negative

Negative

As an example, consider the freezing of liquid water to form ice:

lirH' = -6.01 kJ mol - 1 The change in enthalpy is negative-heat is emitted, increasing the entropy of the surroundings. The change in entropy is negative, which means that the entropy of the system decreases. (We can see that the change in entropy is negative from the balanced equation-a liquid turns into a solid.) Unlike the two previous cases, where the changes in entropy of the system and of the surroundings had the same sign, the changes here are opposite in sign. Therefore, the overall change in Gibbs energy depends on the relative magnitudes of the two changes. At

Recall from Chapter 6 that !!i.,H° represents the standard enthalpy change. The definition of the standard state was first given in Section 6.9 and is summarized in Section 17.5.

768

Chapter 17

Gibbs Energy and Thermodynamics

a low enough temperature, the heat emitted into the surroundings causes a large entropy change in the surroundings, making the process spontaneous. At high temperature, the same amount of heat is dispersed into warmer surroundings, so the positive entropy change in the surroundings is smaller, resulting in a nonspontaneous process.

Case 4: tl. rH Positive, tl.rS Positive If a reaction is endothermic t:..rH, and if the change in entropy for the reaction is positive li.rS, then the change in Gibbs energy will again depend on temperature. The reaction will be nonspontaneous at low temperature but spontaneous at high temperature.

'

Ii.re = li.rH - T li.rS

/

Positive at low temperatures Negative at high temperatures In this example t:.,H0 is the same as

\

Positive

Positive

As an example, consider the vaporizing of liquid water to gaseous water:

I tlfusHo.

H 20 (l ) ~ H 20 (g)

11rH = 40.7 kJ mol - 1 (at 100 °C) 0

The change in enthalpy is positive- heat is absorbed from the surroundings, so the entropy of the surroundings decreases. The change in entropy is positive, which means that the entropy of the system increases. (We can see that the change in entropy is positive from the balanced equation- a liquid turns into a gas.) The changes in entropy of the system and the surroundings are again of opposite sign, only this time the entropy of the surroundings decreases while the entropy of the system increases. In cases such as this, high temperature favours spontaneity because the absorption of heat from the surroundings has less effect on the entropy of the surroundings as temperature increases. The results of this section are summarized in Table 17.2. Notice that when li. rH and t:..r Shave opposite signs, the temperature will not affect whether or not the reaction is spontaneous. When li.rH and li.rS have the same sign, however, the spontaneity does depend on temperature. The temperature at which the reaction changes from being spontaneous to being nonspontaneous (or vice versa) is the temperature at which t:..rG changes sign, which can be found by setting li.rG = 0 and solving for T, as shown in part (b) of Example 17.4. TABLE 17.2 The Effect of tl.rH, tl.r S, Ton Spontaneity

A,H

A,S

Low Temperature

+ + +

+

EXAMPLE 17 .4

High Temperature

Example

Spontaneous (!l,G negative)

Spontaneous (ti ,G negative)

2 N20(g) -

Nonspontaneous (ti,G positive)

Nonspontaneous (ti,G positive)

3 02(g) -

Spontaneous (ti,G negative)

Nonspontaneous (ti,G positive)

H20(/) -

H20(s)

Nonspontaneous (ti, G positive)

Spontaneous (ti, G negative)

H20(/) -

H20(g)

2 N2(g) + 02(g) 2 03(g)

COMPUTING GIBBS ENERGY CHANGES AND PREDICTING SPONTANEITY FROM il.rH AND il.rS

Consider the reaction for the decomposition of carbon tetrachloride gas: CC14(g) ~ C(s, graphite)+ 2 Cl2(g)

t:..rH = 95.7 kJ mol- 1;

t:..rS = 142.2 J K- 1mo1-

1

(a) Calculate t:.. rG at 25 °C and determine whether the reaction is spontaneous. (b) If the reaction is not spontaneous at 25 °C, determine at what temperature (if any) the reaction becomes spontaneous.

SOLUTION

(a) Use Equation 17 .15 to calculate 11,G from the given values of li.rH and li.rS. The temperature must be in kelvins . Be sure to express both li.rH and li.rS in the same units (usually joules) .

T = 273 + 25 = 298 K

li.rG = li.rH - Tli.rS = 95.7 X 103 J mol- 1 - (298K)142.2 J K- 1 mo1- 1 = 95.7 X 103 Jmo1- 1 - 42.4 X 103 Jmo1- 1 = +53.3 X 103 J mol- 1 The reaction is not spontaneous.

17.7

(b) Since t:i.rS is positive, t:i.rG will become more negative with increasing temperature. To determine the temperature at which the reaction becomes spontaneous, use Equation 17.15 to find the temperature at which t:i.rG changes from positive to negative (set t:i.rG = 0 and solve for T) . The reaction is spontaneous above this temperature.

769

Gibbs Energy Changes in Chemical Reactions: Calculating !:i../I '

f:i.rG 0

= =

f:i.rH- Tf:i.rS 95.7 X 103 J mol-

1

-

(T)l42.2 J K- 1 mol-

1

3 1 T= 95.7 X 10 1J mol- 1 ~~~~~~~-

142.2 J K- mol-

= 673 K

FOR PRACTICE 17.4 Consider the reaction: f:i.r H

=

- 137.5 kJ mol-

1 ;

Calculate t:i.rG at 25 °C, and determine whether the reaction is spontaneous. Does tive as the temperature increases?

)1 n

111

=

t:i.rS

,

I

Which statement is true regarding the sublimation of dry ice (solid C02 )? (a) li,H is positive, li ,S is positive, and li ,G is positive at low temperature and negative at high temperature. (b) li,H is negative, li,S is negative, and li,G is negative at low temperature and positive at high temperature. (c) li,H is negative, li,S is positive, and li ,G is negative at all temperatures. (d) li,H is positive, li ,S is negative, and li ,G is positive at all temperatures.

17.7 Gibbs Energy Changes in Chemical Reactions: Calculating ar G0 Section 17 .5 discussed how to calculate the standard change in entropy for a chemical reaction Cf:i.rS However, a more useful quantity that relates to spontaneity of a chemical reaction is the standard chan ge in Gibbs energy (f:i.rG0 ) . This section examines three methods to calculate the standard change in Gibbs energy for a reaction (f:i.rG0 ). In the first method, we calculate t:i.rH0 and t:i.rS0 from tabulated values of t:i.rH 0 and S°, then use the relationship t:i.rG0 = t:i.rH° - Tt:i.rS° to calculate t:i. rG0 • In the second method, we use tabulated values of Gibbs energies of formation to calculate t:i.rG0 directly. In the third method, we determine the Gibbs energy change for a stepwise reaction from the Gibbs energies of each of the steps. Finally, we discuss what exactly the Gibbs energy is. Remember that t:i.rG is extremely useful because it tells us about the spontaneity of a process under standard conditions. The more negative t:i.rG is, the more spontaneous the process is and the further it will go toward products to reach equilibrium. 0

).

0

0

Calculating Gibbs Energy Changes with

~rG

0

=

~rH

0

-

1

tip become more negative or more posi-

CONCEPTUAL CONNECTION 17.2

r"""'""'r.,..'

- 120.5 J K- 1 mol-

T ~r~

In Chapter 6 (Section 6.9), we learned how to use tabulated values of standard enthalpies of formation to calculate t:i.rH 0 • In Section 17.5, we learned how to use tabulated values of standard entropies to calculate t:i.rS0 • We can use these calculated values of t:i.rH 0 and t:i.rS0 to determine the standard Gibbs energy change for a reaction by using the equation: [17.17] 0

Since tabulated values of standard enthalpies of formation and standard entropies (S are usually applicable at 25 °C, strictly speaking the equation should be valid only when T = 298 K (25 °C). However, the changes f:i.rH 0 in t:i.rS0 over a limited temperature range are small when compared to the changes in the value of the temperature itself. For this reason, Equation 17.17 can be used to estimate changes in Gibbs energy at temperatures other than 25 °C. )

770

Chapter 17

EXAMPLE 1 7 .5

Gibbs Ene rgy and Thermodynamics

CALCULATING THE STANDARD CHANGE IN GIBBS ENERGY FOR A REACTION USING arG° = arlf' - T arS°

One of the possible initial steps in the formation of acid rain is the oxidation of the pollutant S0 2 to S03 by the reaction: S02(g)

+

~02 (g) ~ S03(g)

0

Calculate /:,.rG at 25 °C and determine whether the reaction is spontaneous. SOLUTION Begin by looking up (in Appendix IIB) the standard enthalpy of formation and the standard entropy for each reactant and product.

Calculate /:,.rlf' using Equation 6.17.

Reactant or product

t:,.1H 0 (kJ mol - 1) -296.8 0 -395.7

S02 (g) Oz(g) S03(g) /:,.rH

0

S 0 (J mo1 - 1 K- 1 ) 248.2 205.2 256.8

= ~ Vp /:,.rH;roducts - ~ Vr /:,.rH;'.,actants = [t:,.rHso,(g)] - [ t:,.rHso,(g) + ~t:,.rHo,(g)] = -395.7 kJ mol- 1 -(-296.8 kJ mol- 1 +

~

X

0 kJ mol- 1)

= -98.9 kJ mol- 1 Calculate /:,.rS° using Equation 17 .10.

= [ t:,.S'So,(g)J - [ t:,.S'S o,(g) + ~t:,.so,(g) ] = 256.8 J K- 1 mol - 1 - [248 .2 J K- 1 mol- 1 +

~(205.2 J K- 1 mol-

1) ]

= -94.0J K- 1 mo1- 1 Calculate tJ.rG0 using the calculated values of /:,.rH° and /:,.rS° and Equation 17.17. The temperature must be converted to kelvins.

T /:,.rGo

= = = = =

25

+ 273 = 298 K

/:,. rH o - T /:,. rS o

-98.9 X 103 J mol- 1 - 298 K (-94.0 J K- 1 mol- 1) -70.9 X 103 J mo1- 1 -70.9 kJ mol- 1 The reaction is spontaneous at this temperature, because t:,.p0 is negative.

FOR PRACTICE 17.5 Consider the oxidation of NO to N02 :

Calculate /:,.rG 0 at 25 °C and determine whether the reaction is spontaneous.

EXAMPLE 17.6

ESTIMATING THE STANDARD CHANGE IN GIBBS ENERGY FOR A REACTION AT A TEMPERATURE OTHER THAN 25 °C USING arG° = arlf' - TarS°

For the reaction in Example 17 .5, estimate the value of /:,.rG0 at 125 °C. Does the reaction become more or less spontaneous at this elevated temperature; that is, does the value of /:,.rG0 become more negative (more spontaneous) or more positive (less spontaneous)? SOLUTION Estimate /:,.rG0 at the new temperature using the calculated values of /:,.rlf' and /:,. rS° from Example 17.5. For T , convert the given temperature to kelvins. Make sure to use the same units for /:,.rH° and /:,.rS° (usually joules).

= /:,.rGo = = = = T

125 K

+ 273 K = 398 K

/:,.rH o - T /:,.rSo

-98.9 X 103 J mol- 1- 398 K (-94.0 kJ K- 1 mol- 1) -61.5 X 103 J mol- 1 -61.5 kJ mol- 1

17.7

Gibbs Energy Changes in Chemical Reactions: Calculating !:i../I'

771

Since the value of !J.,G0 at this elevated temperature is less negative (or more positive) than the value of !J.,G 0 at 25 °C (which is - 70.9 kJ mol- 1), the reaction is less spontaneous. FOR PRACTICE 17 .6 For the reaction in For Practice 17.5, calculate the value of /J.,G0 at -55 °C. Does the reaction become more spontaneous (more negative !J.,G0 ) or less spontaneous (more positive !J.,G0 ) at the lower temperature?

Calculating

~r G

0

with Tabulated Values of Gibbs Energies of Formation

0

Because /J.,G is the change in Gibbs energy for a chemical reaction-the difference in Gibbs energy between the products and the reactants-and because Gibbs energy is a state function, we can calculate /J.,G0 by subtracting the Gibbs energies of the reactants of the reaction from the Gibbs energies of the products of the reaction. Also, since we are interested only in changes in Gibbs energy (and not in absolute values of Gibbs energy itself), we are free to define the zero of Gibbs energy as conveniently as possible. By analogy with our definition of enthalpies of formation, we define the Gibbs energy of formation (.lirG0 ) as follows: 0

The Gibbs energy of formation (.lir G is the change in Gibbs energy when 1 mol of a compound forms from its constituent elements in their standard states. The Gibbs energy of formation of pure elements in their standard states is zero. )

We can measure all changes in Gibbs energy relative to pure elements in their standard states. To calculate /J.,G0 , we subtract the Gibbs energies of formation of the reactants multiplied by their stoichiometric coefficients from the Gibbs energies of formation of the products multiplied by their stoichiometric coefficients. In the form of an equation: [17.18] In Equation 17 .18, vP are the unitless stoichiometric coefficients of the products, v, are the unitless stoichiometric coefficients of the reactants, and /J.rG represents the standard Gibbs energies of formation. Table 17.3 shows /J. rG0 values for selected substances. You can find a more complete list in Appendix IIB. Notice that, by definition, elements have standard Gibbs energies of formation of zero. Notice also that most compounds have negative standard Gibbs energies of formation. This means that those compounds spontaneously form from their elements in their standard states. Compounds with positive Gibbs energies of formation do not spontaneously form from their elements and are therefore less common. Example 17.7 demonstrates the calculation of /J.,G0 from /J.rG0 values. This method of calculating /J. ,G0 works only at the temperature for which the Gibbs energies of formation are tabulated, namely, 25 °C. Estimating !J.,G at other temperatures requires the use of /J. ,G 0 = !J. ,H 0 - T !J. ,S0 , as demonstrated previously. 0

0

TABLE 17.3 Standard Molar Gibbs Energies of Formation (a1 G0 )

for Selected Substances at 298 K Substance

a1 G (kJ mol- 1) 0

Substance

a 1 G (kJ mol - 1) 0

H2(g)

0

CH4(g)

-50.5

02(9)

0

H20(g)

-228.6

N2(g )

0

H20(/)

-237.1

C(s, graphite)

0

NH3(g)

- 16.4

C(s, diamond)

NO(g)

87.6

CO(g)

- 137.2

2.900

N02(g)

51.3

C02(g)

-394.4

NaCl(s)

384.6

772

Chapter 17

EXAMPLE 17. 7

Gibbs Energy and Th ermod ynam ics

CALCULATING

~rG°

FROM STANDARD GIBBS ENERGIES OF FORMATION

Ozone in the lower atmosphere is a pollutant that can be formed by the following reaction involving the oxidation of unburned hydrocarbons:

0

Use the standard Gibbs energies of formation to determine ti ,G for this reaction at 25 °C.

SOLUTION Begin by looking up (in Appendix IIB ) the standard Gi bbs energies of formation for each reactant and product. Remember that the standard Gibbs energy of formation of a pure element in its standard state is zero.

Reactant or product

ti1G0 (kJ mol- 1) -50.5

CH4(g)

0 - 394.4 - 228.6 163.2

Oz(g) COz(g) H 20 (g) 0 3(g)

Calculate ti ,G0 by substituting into Equation 17.1 8.

= [tirGC:o,(g) + 2( tirG'H,o(g)) + 4(tirGo,(g))] - [!irGc tt (g) + 8(!irGo,(gl)] = [-394.4 kJ mo1- 1 + 2(- 228.6 kJ mol- 1) + 4(163 .2 kJ mol- 1) ) 4

- [-50.5 kJ mol- 1 =

- 198.8 kl mol- 1

+ 8(0.0 kJ mol- 1) ]

+ 50.5 kJ mol- 1

= - 148.3 kl mol- 1 (at 25 °C) FOR PRACTICE 17.7 One of the reactions occurring within a catalytic converter in the exhaust pipe of a car is the simultaneous oxidation of carbon monoxide and reduction of NO (both of which are harmful pollutants).

Use standard Gibbs energies of formation to calculate ti,G0 for this reaction at 25 °C. Is the reaction spontaneous?

FOR MORE PRACTICE 17.7 In For Practice 17. 7, you calculated ti,G for the simultaneous oxidation of carbon monoxide and reduction of NO using standard Gibbs energies of formation. Calculate ti,G0 for that reaction again at 25 °C, only this time use ti,G0 = ti ,H° - Tti,S°. How do the two values compare? Use your results to calculate ti,G0 at 500.0 K, and explain why you could not calculate ti,G at 500.0 K using tabulated standard Gibbs energies of formation. 0

0

0

Calculating firG for a Stepwise Reaction from the Changes in Gibbs Energy for Each of the Steps Recall from Section 6.8 that, since enthalpy is a state function, we can calculate ti,H° for a stepwise reaction from the sum of the changes in enthalpy for each step (according to Hess 's law). Since Gibbs energy is also a state function, the same relationships that we covered in Chapter 6 for enthalpy also apply to Gibbs energy: 0

1.

If a chemical equation is multiplied by some factor, then ti p the same factor.

2. 3.

If a chemical equation is reversed, then ti p changes sign. If a chemical equation can be expressed as the sum of a series of steps, then ti,G for the overall equation is the sum of the Gibbs energies of reactions for each step.

is also multiplied by

0

The following example illustrates the use of these relationships to calculate ti p stepwise reaction.

0

0

for a

17.8 Making a Nonspontaneous Process Spontaneous

773

CALCULATING '1rG° FOR A STEPWISE REACTION

EXAMPLE 17 .8

Find /J,.rG0 for the following reaction:

3 C(s) Use the following reactions with known

D,.p

+ 5 02(g) C(s) + Oi(g)

C3Hg(g)

0

+4

H 2(g)

-------'>

C 3Hg(g)

' s:

-------'>

3 C02(g)

-------'>

C02(g)

+4

H20(g)

= -2074 kJ mol- 1 D,.p = -394.4 kJ mol - 1

/J,.rG 0 0

D,.p = 0

1

-457.1 kJ mol -

SOLUTION To work this problem, manipulate the reactions with known /J,.rG0 's in such a way as to get the reactants of interest on the left, the products of interest on the right, and other species to cancel.

+ 4 H 20 (g)

Since the first reaction has C 3H 8 as a reactant, and the reaction of interest has C 3H 8 as a product, reverse the first reaction and change the sign of /J,.rG0 •

3 C0 2(g)

ThesecondreactionhasCasareactantand C0 2 as a product, just as required in the reaction of interest. However, the coefficient for C is 1, and in the reaction of interest, the coefficient for C is 3. Therefore, multiply this equation and its !J,.rG0 by 3.

3 x [C(s)

The third reaction has H 2(g) as a reactant, as required. However, the coefficient for H2 is 2, and in the reaction of interest, the coefficient for H 2 is 4 . Multiply this reaction and its D,.p0 by 2 .

2 x [2 H 2(g)

+

-------'>

c3H 8(g) + 5 Q2(g)

0 2(g) -------'> C02(g)]

/J,.,G

+

0 2(g)

2 H20(g)]

-------'>

3 C(s)

+

3--0Z(g)

-------'>

3-e0Z(g)

FOR PRACTICE 17.8 0

Ni(g)

+

0 2(g)

2 N20(g)

17.8

0 2(g)

-------'>

-------'>

0

values:

2 NO(g)

= -71.2 kJ mol- 1 D,.p = 175.2 kJ mol - 1

+

/J,.rG = -207.4 kJ mol- 1

-------'>

2 N2(g)

kJ mol-

=2

1)

1

X (-457.1 kJ mol -

!J,.rG0 0

Find !J,.rG for the reaction :

+

X (-394.4 kJmol -

= - 1183

D,.p =

0

2 NO(g)

/J,.,G0

= 2074 kJ mo1- 1

1 )

= -914.2 kJ mol- 1

0

D,.p

= 3

= - 1183

Lastly, rewrite the three reactions after multiplying through by the indicated factors and show how they sum to the reaction of interest. /J,.rG for the reaction of interest is then the sum of the D,.p 's for the steps.

Use the following reactions with known

0

/J,.,G0

2 N02(g)

0 2(g)

/J,.rG0 0

0

Making a Nonspontaneous Process Spontaneous

A nonspontaneous process can be made spontaneous by coupling it with a process that is highly spontaneous. For example, hydrogen gas is a potential future fuel to generate electricity because it can be used in a fuel cell, a type of battery in which the reactants are constantly supplied. (Fuel cells are discussed in Chapter 18.) The main problem with using hydrogen as a fuel is securing a source of it. Where can we get the very large amounts of hydrogen gas that would be needed to meet the world's energy demands? Earth's oceans and lakes

kJmol- 1

-914.2 kJ mol-

1

774

Chapter 17

Gibbs Energy and Thermodynamics

A process with a positive t:i. ,r;> is often called endergonic, whereas one with a I negative t:i.,rJ> is called exergonic.

contain vast amounts of hydrogen, but that hydrogen is locked up in water molecules, and its decomposition to hydrogen and oxygen gases has a positive Li,G0 and is nonspontaneous. Li,G0 (298 K)

= 228.6 kJ mol- 1

To obtain hydrogen from water, we need to find another reaction with a highly negative Lip0 that could couple with the decomposition reaction to give an overall reaction with a negative Lip0 • For example, the oxidation of carbon monoxide to carbon dioxide has a large negative Li,G0 and is highly spontaneous. Li,G0

= -257.2 kJ mol- 1

If we add the two reactions together, we get a negative Li,G0 : H20(g) CO(g)

+

H2(g)

1

+

1

Lip0

~(g)

Lip0 = -257.2 kJ mol- 1

C02(g)

~(g)

= 228.6 kJ mol- 1

Lip0 = -28.6 kJ mol- 1

Processes that convert chemical fuels, such as glucose, to molecules that the cell can use for energy, such as ATP (see Example 17.9), are called catabolic reactions. Anabolic reactions are those that require energy to occur.

The reaction between water and carbon monoxide is thus a spontaneous way to generate hydrogen gas. The coupling of nonspontaneous reactions with highly spontaneous ones is also important in biological systems. The synthesis reactions that create the complex biological molecules, such as proteins and DNA needed by living organisms, for example, are themselves nonspontaneous. Living systems grow and reproduce by coupling these nonspontaneous reactions with highly spontaneous ones. The main spontaneous reaction that ultimately drives the nonspontaneous ones is the metabolism of food . The oxidation of glucose is highly spontaneous:

Lip = -2880 kJ mol- 1 0

Spontaneous reactions such as these ultimately drive the nonspontaneous reactions necessary to sustain life. The oxidation of glucose drives the overall production of a molecule called adenosine triphosphate (ATP), which in turn is used to drive specific nonspontaneous reactions. For example, converting glucose to glucose-6-phosphate- a key step in the beginning of glycolysis- is an endergonic process and, therefore, nonspontaneous. glucose + Pi ~ glucose-Pi

Li,G0

= 13.8 kJ mol- 1

However, coupling the reaction with the hydrolysis of ATP to adenosine diphosphate (ADP) and inorganic phosphate (Pi): ATP ~ ADP + Pi

Li,G0

= -30.5 kJ mol- 1

makes the process spontaneous: ~

glucose-Pi

Lip0 = 13.8 kJ mol- 1

ATP

~

ADP +Pi

Lip0 = - 30.5 kJ mol- 1

glucose+ ATP

~

glucose-Pi + ADP

Lip0 = -16.7 kJ mol- 1

glucose + Pi

EXAMPLE 17.9

COUPLING NONSPONTANEOUS AND SPONTANEOUS REACTIONS TO FORM OVERALL SPONTANEOUS REACTIONS

One of the steps in glycolysis, a process that produces stored chemical energy in cells, is the conversion of phosphoenolpyruvate (EPy-Pi) to pyruvate (Py), which converts ADP to ATP. The symbolic thermochemical reactions at physiological temperatures (3 10 K) are: ADP +Pi ~ ATP

Li,G0

EPy-Pi ~ Py + Pi

Li,G0

= 30.5 kJ mol- 1 = -61.9 kJ mol- 1

where Pi is inorganic phosphate. Determine the standard change in Gibbs energy for the coupled reaction producing ATP.

17.8 Making a Nonspontaneous Process Spontaneous

775

SOLUTION To determine the overall standard Gibbs energy change for the coupled reaction, add the two reactions together, including the individual Gibbs energies of reaction. When adding the reactions together, Pi is cancelled out to provide the overall reaction. Addition of the D.,0° values provides the overall !i,0° for the conversion of ADP to ATP coupled with the conversion of EPy-Pi to Py. Although the conversion of ADP to ATP is nonspontaneous, when coupled with the conversion of EPy-Pi to Py, the overall process is spontaneous.

ADP +Pi

~

ATP

EPy-Pi

~

Py + Pi

tip = tip =

EPy-Pi + ADP

~

Py + ATP

0

30.5 kJ mol- 1

0

-61.9 kJ mol- 1

D.rG0

= -31.4 kJ mol- 1

FOR PRACTICE 17.9 The overall oxidation of glucose to C0 2 and H 20 provides 2880 kJ mol- 1 of Gibbs energy, and the conversion of ADP to ATP requires 30.5 kJ mol- 1 of Gibbs energy. Given that oxidation of 1 glucose can produce as many as 38 ATP molecules, which can then be used to drive other nonspontaneous cellular processes, how much Gibbs energy is extracted from the oxidation of glucose to be used later, and how much Gibbs energy is left over?

Although it is common to write the reaction of ATP to ADP in the following manner:

tip =

ATP ~ ADP + Pi

0

-30.5 kJ mol- 1

it is important to note that the reaction is a hydrolysis reaction. ATP reacts with water to form ADP and Pi. The actual reaction is shown below. The hydrolysis transforms the relatively high energy triphosphate group into a lower energy diphosphate and inorganic phosphate. The large exergonicity of the reaction is due to stronger bonds and higher entropy in the products. H

H "'--..N/

I

c

N....._C/ ~N

0

0

H-C

0

...

I I I N---0-i-o-i-o-i-o-~;_, 0 " I o

-

o

-

o

-

c

c H;I c-cl/ .H I I OH OH

I \H H I

+

0

+

II

-o-P-OH

I o-

+

30.5 KJ mol- 1

II

I

c

c "N-?

"H

776

Cha pter 17

Gibbs Energy and Th e rmod ynamics

17.9 What Is Gibbs Energy?

Maximum work = 5296.2 kJ

We often want to use the energy released by a chemical reaction to do work. For example, in an automobile, we want to use the energy released by the combustion of gasoline to move the car. The change in Gibbs energy of a chemical reaction represents the maximum amount of energy available to do work, if D..rG0 is negative. For many reactions, the amount of Gibbs energy is less than the change in enthalpy for the reaction. Consider the combustion of octane at 25 °C: CsH1 s(/)

25

+2

/j. rH

Oig) ~ 8 C0 2(g) 0

D..rS

0

l

+ 9 H20 (l)

= - 5470.3 kJ mol-I = - 584.3 JK- 1 mol- 1

D..,0° = - 5296.2k:Jmol- 1

8 C0 2(g) + 9 H20 (Q

Minimum heat lost to surroundings= 174.1 kJ (at 298 K)

.A. FIGURE 17.10 Gibbs Energy Although the reaction produces 5470.3 kJ of ene rgy, only 5296.2 kJ is available to do work. The rest of the energy is lost to the surroundings, increasing the entropy of the surroundings.

More formally, a reversible reaction is one that will change direction upon an infinitesimally small change in a variable (such as temperature or pressure) related to the reaction.

The reaction is exothermic and gives off 5470.3 kJ of heat energy. However, the maximum amount of energy available for useful work is 5296.2 kJ (Figure 17 .10 ~) . Why? We can see that the change in entropy of the system is negative. Nevertheless, the reaction is spontaneous. This is possible only if some of the emitted heat goes to increase the entropy of the surroundings by an amount sufficient to make the change in the entropy of the universe positive. The amount of energy available to do work (the Gibbs energy) is what is left after accounting for the heat that must go into the surroundings. The change in Gibbs energy for a chemical reaction represents a theoretical limit and is the maximum work that can be done by the reaction. In thermodynamics, a reaction that achieves the theoretical limit with respect to Gibbs energy is called a reversible reaction. A reversible reaction occurs infinitesimally slowly, and the Gibbs energy is drawn out in infinitesimally small increments that exactly match the amount of work that the reaction is being used to do during that increment (Figure 17. 11 T). Theoretically, burning octane in an internal combustion engine (like those in an automobile) would provide 5296.2 kJ of useful work. The combustion of octane in a real internal combustion engine is not done reversibly; it is done irreversibly. An amount of fuel is drawn into the cylinder and mixed with oxygen, and a spark initiates the combustion that is over extremely quickly. This results in far less work being done than the value of D..rG because additional energy produced by the reaction is lost to the surroundings as heat. Furthermore, the pistons on a real engine and the motion of the car parts are hindered by friction, and overcoming this friction produces heat that is energy lost to the surroundings. The effect of the reaction being done irreversibly and losses due to friction is to further reduce the efficiency of the reaction to do useful work. 0

Weight of sand exactly matches p ressure at each increment.

Incrementally remove sand

Sand

bn

t

Incrementally rem ove sand

Gas

.A. FIGURE 17.11 A Reversible Process In a reversible process, the Gibbs energy is drawn out in infini tesimally small increments that exactly match the amount of energy that the process is producing in that increment. In this case, grains of sand are re moved one at a time, resul ting in a series of small expa nsions in whic h the weight of sand almost exactly matches the pressure of the expandi ng gas. This process is close to reversibleeach sand grain would need to have an infi nitesimall y small mass for the process to be fully reversible.

17.10 Gibbs Energy Changes for Nonstandard States: The Relationship Between t:i. ,G" and t:i.,G

777

Another way to convert the energy from a chemical reaction into useful work is to use a fuel cell, which can be much more efficient than an internal combustion engine, mainly because the Gibbs energy from the reaction is done so that it produces electricity directly, and this electricity is used to do work. There is no heat lost to friction in the engine, and the reaction, being done more slowly and in a more controlled manner, does not lose as much of the energy due to the reaction being done irreversibly. What is the same, whether a mole of octane is combusted in an internal combustion engine or in a fuel cell, is that the absolute maximum amount of work that can be done is 5296.2 kJ. The theoretical limit to the amount of useful work that can be done by any reaction is equal to the Gibbs energy. If the change in Gibbs energy of a chemical reaction is positive, 11,G0 represents the minimum amount of energy required to make the reaction occur. Again, 11p0 is a theoretical limit, and making a real nonspontaneous reaction occur always requires more energy than the theoretical limit.

17.10

Gibbs Energy Changes for Nonstandard States: The Relationship Between iirG° and iirG

We have learned how to calculate the standard Gibbs energy change for a reaction (l1p0 ) . However, the standard Gibbs energy change applies only to a very narrow set of conditions, namely, those conditions in which the reactants and products are in their standard states. Consider the standard Gibbs energy change for the evaporation of liquid water to gaseous water:

The standard Gibbs energy change for this process is positive, so the process is nonspontaneous. But you know that if you spill water onto the floor under ordinary conditions, it spontaneously evaporates. Why? Because ordinary conditions are not standard conditions, and l1p0 applies only to standard conditions. For a gas, such as the water vapour in the above reaction, standard conditions are those in which the pure gas is present at a partial pressure of 1 bar. In a flask containing liquid water and water vapour under standard conditions (PH,o = 1 bar) at 25 °C the water would not vaporize. In fact, since l1p0 is negative for the reverse reaction, the reaction would spontaneously occur in reversewater vapour would condense. In open air under ordinary circumstances, the partial pressure of water vapour is much less than atmospheric pressure. The conditions are not standard and therefore the value of 11p0 does not apply. For nonstandard conditions, we must calculate 11P (as opposed to l1,G to predict spontaneity. 0

)

The Gibbs Energy Change of a Reaction Under Nonstandard Conditions

..., PLOOR

&

PLANCHER MOUILL~

We can calculate the Gibbs energy change of a reaction under nonstandard conditions (11p) from 11p0 using the following relationship, where Q is the reaction quotient (defined in Section 14.6), T is the temperature in kelvins, and R is the gas constant in the appropriate units (8.314 J K- 1 mol- 1) : [17.19] In all thermodynamic equations, remember that Q must be expressed in terms of pressures in bar for gases and concentrations in mol L- I for solutes. We drop the units, as we did in Chapter 14, because we assume that the unitless activities are equal to the magnitudes of the pressures and concentrations. For liquids and solids, we assume that they are relatively pure and their amounts are not expressed in Q (their activities are equal to 1). We can demonstrate the use of this equation by applying it to the liquid-vapour water equilibrium under several different conditions, as shown in Figure 17.12 "'·Note that by the law of mass action, for this equilibrium, Q = Ptt,O· H20(/) ~ H 20(g)

Q

= Ptt,o

A Spilled water spontaneously evaporates even though llrG0 for the vaporization of water is positive. Why? [Stephen Finn/Fotolia]

778

Chapter 17

Gibbs Energy and Thermodynamics

.,... FIGURE 17.12 Gibbs Energy Versus Pressure for Water The Gibbs energy change for the vaporization of water is a function of pressure.

Q=O

Q=1

Water condenses b.,G positive

Water evaporates b., G negative

Q=K Equilibrium b.,G = 0

Ptt2 o

Ptt2o = 0 bar

= 0.0317 bar

PH

2

o

= 1 bar

Ptt 2o (Not to scale)

Standard Conditions Under standard conditions, PH2o = 1 bar, and therefore Q = 1. Substituting, we get this expression: 11,G

=

l1,G 0 +RT In Q

= 8.56 kJ mol- 1 + RTln(l) = 8.56 kJ mol- 1 Under standard conditions, Q will always be equal to 1, and since ln( l ) = 0, the value of 11,G will be equal to l1,G0 • For the liquid-vapour water equilibrium at 298 K, since 11rG > 0, the reaction is not spontaneous in the forward direction but is spontaneous in the reverse direction. As stated previously, under standard conditions at 298 K, water vapour condenses into liquid water. 0

Equilibrium Conditions At 25 .00 °C, liquid water is in equilibrium with water vapour at a pressure of 0.0317 bar; therefore, Q = PH2o = 0.0317. Substituting, we get this expression:

11rG

11rG + RT In Q 0

8 .56 kJ mo1- 1

+ 8.314 J V

8 .56 kJ mol- 1

+ (-8.56

0 kJ mol-

X

mo1- 1 (298 K) ln(0 .03 17) 103 J mol- 1)

(~) 1000 J

1

Under equilibrium conditions, the value of RT in Q will always be equal in magnitude but opposite in sign to the value of l1,G0 • Therefore, the value of 11,G will always be zero at equilibrium at any temperature. Since 11,G = 0, the reaction is not spontaneous in either direction, as expected for a reaction at equilibrium. A water partial pressure of 5 .DO x 10-3 bar corresponds to a relative humidity of I 16% at 25 °c.

Other Nonstandard Conditions To calculate the value of l1,G0 under any other set of nonstandard conditions, calculate Q and substitute the value into the equation. For example, the partial pressure of water vapour in the air on a dry (nonhumid) day might be 5.00 X 10- 3 bar, so Q = 5.00 X 10- 3 . Substituting, we get this expression: 11,G

+ RTln Q 8.56kJmol- 1 + 8.314JVmol- 1 (298K)ln(5.00 l1,G 0

8 .56 kJ mol-

1

-4.6 kJ mol-

+ (-1.3!.2 x 104 J mol- 1) C~o~ ) 1

1

X

10- 3)

17.1 O Gibbs Energy Changes for Nonstandard States: The Relationship Between t:i..,G° and t:i..,G

779

Under these conditions, the value of 11rG < 0, so the reaction is spontaneous in the forward direction, consistent with our experience of water evaporating when spilled on the floor.

EXAMPLE 17.10

CALCULATING Ll rG UNDER NONSTANDARD CONDITIONS

Consider the following reaction at 298 K:

+

2 NO(g)

0 2(g)

~

11rG = - 71.2 kJ mol 0

2 N0 2(g)

1

Calculate 11,G under the following conditions: PNo

=

0.100 bar;

=

P0 ,

0.100 bar;

PNo,

=

2.00 bar

Is the reaction more or less spontaneous under these conditions than under standard conditions?

SOLUTION Use the law of mass action to calculate Q. Q

P~o = _2_,_ =

(2 00)2

.2 (0.100) (0.100)

PNoPo,

11rG = l1 p

0

Substitute Q, T, and l1,G into Equation 17.16 to calculate 11,G. (Since the units of R include joules, write 11rG0 in joules.)

0

+ RT In

=

4.00 x 103

Q

=

- 7 1.2

X

103 J mol- 1

+

8.314 -

= = =

- 7 1.2

X

103 J mol- 1

+

20.5

-50.7

X

103 J mo1- 1

1 -

mol·K

X

(298 .K) ln(4.00

X

103)

103 J mol- 1

- 50.7 kJ mol- 1

The reaction is spontaneous under these conditions, but less spontaneous than it was under standard conditions (because 11,G is less negative than l1,G0 ) .

CHECK The calculated result is consistent with what we would expect based on Le Chatelier's principle; increasing the concentration of the products and decreasing the concentration of the reactants relative to standard conditions should make the reaction less spontaneous than it was under standard conditions. FOR PRACTICE 17.10 Consider the following reaction at 298 K: 2 H 2S(g)

+

S02(g)

~

3 S(s, rhombic)

+

11rG = - 102 kJmol- 1 0

2 H 20 (g)

Compute 11rG under the following conditions: P H,s

=

2.00 bar;

Pso,

=

1.50 bar;

P H,o

=

0.0100 bar

Is the reaction more or less spontaneous under these conditions than under standard conditions?

FOR MORE PRACTICE 17.10 As we saw in Example 17.9, the standard Gibbs energy for the hydrolysis of ATP to ADP can provide 30.5 kJ mol- 1 of energy at 37 °C (3 10 K) and at standard conditions-that is, when ATP, ADP, and P; are all at 1 mol L- I . More realistic concentrations of ATP, ADP, and P; are around 5 mmol L- I . Determine 11,G for the conversion of ATP to ADP at 3 10 K when [ATP] = [ADP] = [P;] = 5 mmol L- 1.

fl

jfT

n CONCEPTUAL CONNECTION 17.3

E [[ II 1mniat@MIH1fli[.[ff!.!.1AH@Gtmmm1According to Le Chatelier's principle and the dependence of Gibbs energy on reactant and product concentrations, which statement is true? Assume that both the reactants and the products are gaseous. {a) A high concentration of reactants relative to products results in a more spontaneous reac-

tion than one in which the reactants and products are in their standard states.

(continued )

780

Chapter 17

Gibbs Energy and Th erm od ynamics

(b) A high concentration of products relative to reactants results in a more spontaneous reac-

tion than one in which the reactants and products are in their standard states. (c) A reaction in which the reactants are in standard states, but in which no products have formed, will have a t..,Gthat is more positive than t.. ,G0 •

Equation 17 .17 can be substituted into Equation 17 .19 to compute the Gibbs energy change under nonstandard conditions and at temperatures other than 298 K:

t:,,.p = /:,,.J-f' - T/:,,.,S 0 + RT In Q

[17.20)

This equation is valid under the assumption that the standard entropy and enthalpy changes for a reaction are independent of temperature, which is a satisfactory assumption for small temperature changes and that we use in this chapter.

EXAMPLE 17.11

CALCULATING ilrG UNDER NONSTANDARD CONDITIONS AND AT TEMPERATURES OTHER THAN 298 K

Consider the following reaction: Ca(OH)i (s) ~ Ca2+(aq)

+ 2 OH-(aq)

/:,,.,H0

= -17.6kJmol- 1

/:,,.,s' = -158.3 J K- 1 mol- 1 A saturated solution of Ca(OHh contains 1.2 X 10- 2 mol L- 1 at 298 Kand is at equilibrium with the solid in the container. The solution is quickly heated to 350 K. Calculate t:,,.p at 350 K with the 298 K concentrations and state whether Ca(OHh

will precipitate or more can be dissolved at the higher temperature.

SOLUTION We can use Equation 17 .20 to compute t:,,.p at nonstandard conditions and at temperatures other than 298 K. We are given the enthalpy and entropy changes for the reaction and the new temperature. We first need to use the concentration of the solution to compute Q. Due to stoichiometry, the concentration of OH- is twice that of Ca2- . Now we can substitute Q, /:,,.J-f', /:,,.,s', T, and R into Equation 17 .20. Ensure the units are the same before adding terms; since /:,,.,s' and R are in J K- 1 mol- 1, it is best to write !:,,.,H° in J mol- 1•

[Ca2 +] Q

= 1.2 X 10- 2 mol L - I [OH-] = 2.4 X 10- 2

= [Ca2 -] [OH-) 2 = (1.2 x 10- 2) (2.4 x 10- 2) 2 = 6.91 x 10- 6

t:,,.,G

=

0

t:,,.p

-

T/:,,.,S° + RTin Q

= - 17.6 X 103 Jmol - 1

+ (8.314J V

-(350K)(-158.3J ~ mol - 1 )

mol- 1)(350K)(ln6.91X10- 6)

= -17.6 X 103 Jmol- 1 + 55.4 X 103 Jmo1- 1 -34.6 X 103 J mol- 1

= 3.2 X 103 Jmo1- 1 The reaction is not spontaneous under these conditions; some Ca(OHh will precipitate upon heating.

CHECK The reaction was saturated at 298 K and we increased the temperature. The reaction is exothermic (heat is a product of the reaction as written), so by Le Chatelier's principle, adding heat by increasing the temperature would cause the reaction to shift toward reactants, precipitating Ca(OH)i. Similarly, we learned above that since /:,,.,S 0 is negative, an increase in temperature favours the reverse reaction. The positive t:,,.p we computed, therefore, seems reasonable. FOR PRACTICE 17.11 Consider the dissolution of lead(II) chloride: PbCI2(s) ~ Pb2 +(aq)

+ 2 Cl-(aq)

/:,,.J-f'

= 26.12 kJ mol- 1

/:,,.,s' = -4.3 J K- 1 mol- 1

17.11

Gibbs Energy and Equilibrium: Relating li,G· to the Equilibrium Constant (K)

781

Compute 11p at 360 K when the concentration of Pb 2+ is 0.0030 mol L- 1 and the concentration of Cl- is 0.0060 mol L- 1• Could more PbCl2 be dissolved if some solid was present, or will it precipitate under these conditions?

FOR MORE PRACTICE 17.11 Consider the dissolution of lead(II) chloride, the reaction and standard enthalpy and entropy changes for the reaction are provided above in For Practice 17 .11. At what temperature would the concentrations provided above represent an equilibrium mixture if some solid were present?

17.11

0

Gibbs Energy and Equilibrium: Relating '1rG to the Equilibrium Constant (K)

We have learned throughout this chapter that 11,G0 determines the spontaneity of a reaction when the reactants and products are in their standard states. In Chapter 14, we learned that the equilibrium constant (K) determines how far a reaction goes toward products, a measure of spontaneity. Therefore, as you might expect, the standard Gibbs energy change of a reaction and the equilibrium constant are related-the equilibrium constant becomes larger as the Gibbs energy change becomes more negative. In other words, if the reactants in a particular reaction undergo a large negative Gibbs energy change as they become products, then the reaction will have a large equilibrium constant, with products strongly favoured at equilibrium. If, on the other hand, the reactants in a particular reaction undergo a large positive Gibbs energy change as they become products, then the reaction will have a small equilibrium constant, with reactants strongly favoured at equilibrium. We can obtain a relationship between 11p0 and K from Equation 17.19. We know that at equilibrium Q = Kand 11P = 0. Making these substitutions,

11p = l1,G + RT In Q 0

0 = l1p +RT In K 0

[17.21] 0

We can better understand the relationship between l1,G and K by considering the following ranges of values for K, as summarized in Figure 17 .13 T . ~

When K < 1, In K is negative and 11p is positive. Under standard conditions (when Q = 1) the reaction is spontaneous in the reverse direction.

~

When K (when Q

~

0

> I, In K is positive and 11p is negative. Under standard conditions = 1) the reaction is spontaneous in the forward direction. When K = I, In K is zero and 11,G is zero. The reaction happens to be at equilib0

0

The relationship between li,G and Kis logarithmic-small changes in li,G· have I a large effect on K.

0

rium under standard conditions.

EXAMPLE 17.12

THE EQUILIBRIUM CONSTANT AND iirG0

Use tabulated Gibbs energies of formation to calculate the equilibrium constant for the following reaction at 298 K:

SOLUTION Begin by looking up (in Appendix IIB) the standard Gibbs energies of formation for each reactant and product.

Reactant or product

11 16° (kJ mol - 1)

N10 4(g)

99.8

N02(g)

51.3 (continued )

782

Chapter 17

Gibbs Energy and Th erm od ynam ics

EXAMPLE 17.12

(CONTINUED)

Calculate tip0 by substituting into Equation 17 .18.

= 2[!irG°No2(g)] - l:irGN,04 (g) = 2(51.3 kJ mol- 1) - 99 .8 kJ mol - 1 = 2.8 kJ mol- 1 Calculate K from tip0 by solving Equation 17.21 for Kand substituting the values of ti,G0 and temperature.

-ti G In K= - -'-

0

RT

-2.8 x 103 J/ mol 1 8.314- - (298K) mol·K = -1.13 K

=

e- 1.1 3

= 0.32 FOR PRACTICE 17.12 Calculate tip0 at 298 K for the following reaction:

K

[ A(g)

~ B(g)

K< 1J

= 6.0

[ A(g)

X

105

~ B(g)

Standard conditions PA = P8 = 1 bar

K> 1

J

Standard conditions PA= Ps = 1 bar

Q=l

R everse reaction spontaneous

~! Pure A

Forward reaction spontaneous PureB

Q=l

!~ PureB

Pure A

Extent of reaction

Extent of reaction

(b)

(a)

[ A(g)

~

B(g) K = 1 J

Standard conditions PA = P8 = 1 bar

Q=l At equilibrium

Pure A

PureB Extent of reaction

(c)

-

+

Cl- (aq)

-4 11

- 240

- 167

72

59

57

QD Below are some thermochemical data for the dissolution equilibrium of solid Al(OH)J in water: Al(OH)J(s)

->

- 1284 85.4

A13+(aq)

+ 3 OW(aq)

- 538.4

-230.02

-325

- 10.9

0

Calculate the 298 K standard Gibbs energy change for this reaction, the equilibrium constant, and the solubility in mol L - 1 of solid NaCl.

110. Consider the following equilibrium: Ni(g)

+ 3 Hi(g)

->

2 NH3(g)

a. Using the thermodynamic data from Appendix IIB, compute 1:!..,H°, l:!..,s°, and l:!..,G0 at 420 K. b. Compute the equilibrium constant for this reaction at 420 K. c. Compute l:!..,G at 420 K when PN, = 4.00 bar, PH, = 2.50 bar, and PNH, = 5.00 bar. d. At what temperature would the reaction mixture in (c) be at

a. Compute 1:!..,H°, l:!..,s°, and l:!..,G at 298 K, and state whether the reaction is spontaneous under standard conditions. b. Determine the equilibrium constant for the dissolution of Al(OH)J at 298 K. c. Determine the Gibbs energy change for the dissolution of Al(OHh when the concentrations of Al 3+ and Ol1 are both 1.00 X 10- 7 mol L - 1, and state whether the reaction is spontaneous under these conditions. d. At what temperature will Al(OH)J(s) be at equilibrium with Al3+ and OH- , each at 1.00 X 10- 7 mol L - 1? e. Explain why the entropy of dissolution for this reaction, solid Al(OH)J dissolving to form Al 3+ and 3 OH- , is so negative.

equilibrium?

Conceptual Problems 112. Which is more efficient, a butane lighter or an electric lighter (such as can be fou nd in most automobiles)? Explain. Which statement is true? a. A spontaneous reaction is always a fast reaction. b. A spontaneous reaction is always a slow reaction. c. The spontaneity of a reaction is not necessarily related to the speed of a reaction. 114. Which process is necessarily driven by an increase in the entropy of the surroundings? a . the condensation of water b. the sublimation of dry ice c. the freezing of water Which statement is true? a. A reaction in which the entropy of the system increases can be spontaneous only if it is exothermic. b. A reaction in which the entropy of the system increases can be spontaneous only if it is endothermic.

0

0

c. A reaction in which the entropy of the system decreases can be spontaneous only if it is exothermic. 116. Which process is spontaneous at 298 K? a. H 2 0(l) ------> H 20 (g, l bar) b. H2 0(l) ------> H 20 (g, 0.10 bar) c. H 2 0(l) ------> Hp(g, 0.010 bar)

G The Gibbs energy change of the reaction

A(g) ------> B(g) is zero under certain conditions. The standard Gibbs energy change of the reaction is - 42.5 kJ mol- 1• Which statement must be true about the reaction? a. The concentration of the product is greater than the concentration of the reactant. b. The reaction is at equilibrium. c. The concentration of the reactant is greater than the concentration of the product.

The MP3 player shown here is powered by a hydrogen-oxygen fuel cell, a device that generates electricity from the reaction between hydrogen and oxygen to form water. (Quade Paul/Pearson Education]

18.1 Pulling the Plug on the Power Grid 796 18.2 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions 796 18.3 Standard Electrode Potentials 799

T

minister responsible for all financial matters) wanted to know how Michael Fara-

18.4 Cell Potential, Gibbs Energy, and the Equilibrium Constant 808 18.5 Cell Potential and Concentration 812

day's apparently esoteric investigations of electricity would ever be useful to the

18.6 Batteries: Using Chemistry to Generate Electricity 819

HIS CHAPTER'S OPENING QUOTE FROM MICHAEL FARADAY illustrates an important aspect of basic research (research for the sake of understanding how nature works). The Chancellor of the Exchequer (the British cabinet

empire. Faraday responded in a way that the Chancellor would understand- he pointed out the eventual financial payoff. Today, of course, electricity is a fundamental form of energy, powering our entire economy. Although basic research does not always lead to useful applications, much of the technology our society relies

18.7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity 823 18.8 Corrosion: Undesirable Redox Reactions 831

on has grown out of basic research. The history of modern science shows that you must first understand nature (the goal of basic research) before you can harness its power. In this chapter, we seek to understand oxidation-reduction reactions (first introduced in Chapter 4) and how we can exploit them to generate electricity. The applications range from the batteries that power flashlights to the fuel cells that may one day power our homes and automobiles.

795

796

Chapter 18

Elect ro chemistry

18.1

The kilowatt-hour is a unit of energy first

I introduced in Section 6.2.

Pulling the Plug on the Power Grid

The power grid distributes centrally generated electricity throughout the country to homes and businesses. When you turn on a light or electrical appliance, electricity flows from the grid, through the wires in your home, to the light or appliance. The electrical energy is converted into light within the lightbulb or into work within the appliance. The average Canadian household consumes about 1000 kilowatt-hours (kWh) of electricity per month. The local electrical utility, of course, monitors your electricity use and bills you for it. In the future, you may have the option of disconnecting from the power grid. Several innovative companies are developing small, fuel-cell power plants-each no bigger than a refrigerator- to sit next to homes and quietly generate enough electricity to meet each household's power needs. The heat produced by a fuel cell's operation can be recaptured and used to heat water or the space within the home, eliminating the need for a hot-water heater and a furnace. Similar fuel cells could also be used to power cars. Fuel cells are highly efficient and, although some obstacles to their development and use must yet be overcome, one day they may supply a majority of our power while producing less pollution than fossil fuel combustion. Fuel cells are based on oxidation-reduction reactions (see Section 4.6). The most common type of fuel cell--called the hydrogen- oxygen fuel cell- is based on the reaction between hydrogen and oxygen.

In this reaction, hydrogen and oxygen form covalent bonds with one another. Recall that, according to Lewis theory, a single covalent bond is a shared electron pair. However, since oxygen is more electronegative than hydrogen, the electron pair in a hydrogenoxygen bond is not equally shared: Oxygen gets the larger portion (see Section 9.7). In effect, oxygen has more electrons in H2 0 than it does in elemental 0 2- it has gained electrons in the reaction and has therefore been reduced. In a direct reaction between hydrogen and oxygen, oxygen atoms gain electrons directly from hydrogen atoms. In a hydrogen-oxygen fuel cell, the same redox reaction occurs, but the hydrogen and oxygen are separated, forcing electrons to travel through an external wire to get from hydrogen to oxygen. These moving electrons constitute an electrical current. Fuel cells employ the electron-gaining tendency of oxygen and the electron-losing tendency of hydrogen to force electrons to move through a wire to create the electricity that can provide power for a home or an electric automobile. Smaller fuel cells can also replace batteries and be used to power consumer electronics such as laptop computers, cell phones, and MP3 players. The generation of electricity from spontaneous redox reactions (such as a fuel cell) and the use of electricity to drive nonspontaneous redox reactions (such as those that occur in gold or silver plating) are examples of electrochemistry, the subject of this chapter. ..&. BC Transit had 20 hydrogen fuelcell buses in service in Whistler, BC, for the 2010 Winter Olympic games. These buses, which were manufactured in Winnipeg, MB, have top speeds of 90 km/h, a life expectancy of 20 years, and produce zero emissions. The only exhaust from running the bus is water vapour. While the buses themselves do not emit C02, since there are no terrestrial reservoirs of H2 , there may have been significant emissions in the production and transportation of the of H2 fuel. Cited as using triple the operating costs of diesel buses, the hydrogen fuel cell buses have been put in storage and are being sold off. [Marcel Sloover]

18.2 Voltaic {or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions Electrical current is the flow of electric charge. Electrons flowing through a wire or ions flowing through a solution both constitute electrical current. Since redox reactions involve the transfer of electrons from one substance to another, they have the potential to generate electrical current. For example, consider the spontaneous redox reaction: Zn(s)

+

Cu2 +(aq) ~ Zn2 +(aq) 2

+

Cu(s)

When Zn metal is placed in a Cu + solution, the greater tendency of zinc to lose electrons results in Zn being oxidized and Cu 2 + being reduced-electrons are transferred directly from the Zn to the Cu2 + (Figure 18.1 T). Although the actual process is more complicated, we can imagine that--on the atomic scale-a zinc atom within the zinc metal transfers two electrons to a copper ion in solution. The zinc atom then becomes a zinc ion dissolved in the solution. The copper ion accepts the two electrons and is deposited on the zinc as solid copper. Suppose we could separate the zinc atoms and copper ions and force the electron transfer to occur another way- not directly from the zinc atom to the copper ion, but through a wire connecting the two half-reactions. The flowing electrons would constitute an electrical current and could be used to do electrical work.

18.2 Voltaic (or Galvanic) Cells: Generating Electricity from Spontaneous Chemical Reactions

797

Cu(s)

798

Chapter 18 e

Electrochemistry

e

.._ FIGURE 18.3 An Analogy for Electrical Current Just as water flows downhill in response to a difference in gravitational potential energy, electrons flow through a conductor in response to an electrical potential difference, creating an electrical current. [Alejandro Diaz Diez/AGE Fotostock]

I The ampere is often abbreviated as amp. The continual flow of electrical current in a voltaic cell requires a pathway by which counterions can flow to neutralize charge buildup; this is discussed above.

If the two half-cells are connected by a wire running from the zinc, through a lightbulb or other electrical device, to the copper, electrons spontaneously flow from the zinc electrode (which is more negatively charged and therefore repels electrons) to the copper electrode. As the electrons flow away from the zinc electrode, the Zn/ Zn2 + equilibrium shifts to the right (according to Le Chatelier's principle) and oxidation occurs. As electrons flow to the copper electrode, the Cu/ Cu2 + equilibrium shifts to the left, and reduction occurs. The flowing electrons constitute an electrical current that lights the bulb. We can understand electrical current and why it flows by analogy with water current in a stream (Figure 18.3 ... ). The rate of electrons flowing through a wire is analogous to the rate of water moving through a stream. Electrical current is measured in units of amperes (A). One ampere represents the flow of one coulomb (a measure of electrical charge) per second:

lA = lCs- 1 Since an electron has a charge of 1.602 X 10- 19 C, 1 A corresponds to the flow of 6.242 X 10 18 electrons per second. The driving force for electrical current is analogous to the driving force for water current. Water current is driven by a difference in gravitational potential energy (caused by a gravitational field). Streams flow downhill, from higher to lower potential energy. Electrical current is also driven by a difference in potential energy (caused by an electric field resulting from the charge difference on the two electrodes) called potential difference. Potential difference is a measure of the difference in potential energy (usually in joules) per unit of charge (coulombs) . The SI unit of potential difference is the volt (V), which is equal to one joule per coulomb: lV=lJC- 1 A large potential difference corresponds to a large difference in charge between the two electrodes and therefore a strong tendency for electron flow (analogous to a steeply descending streambed). Potential difference, since it gives rise to the force that results in the motion of electrons, is also referred to as electromotive force (emf). In a voltaic cell, the potential difference between the two electrodes is the cell potential (Ecen) or cell emf. The cell potential depends on the relative tendencies of the reactants to undergo oxidation and reduction. Combining the oxidation of a substance with a strong tendency to undergo oxidation and the reduction of a substance with a strong tendency to undergo reduction produces a large difference in charge between the two electrodes and therefore a high positive cell potential. The cell potential also depends on the concentrations of the reactants and products in the cell and the temperature (which we will assume to be 25 °C unless otherwise noted). Under standard conditions (1 mol L- 1 concentration for reactants in solution and 1 bar pressure for gaseous reactants), the cell potential is called the standard cell potential (E~.11 ) or standard emf. For example, the standard cell potential in the zinc and copper cell described previously is 1.10 volts: Zn(s) + Cu 2 +(aq) ~ Zn 2 +(aq) + Cu(s)

E~ell =

1.10 V

If the zinc is replaced with nickel (which has a lower tendency to be oxidized) the cell

potential is lower: Ni(s) + Cu 2 +(aq) ~ Ni2+(aq) + Cu(s)

E~. 11 = 0.62 V

The cell potential is a measure of the overall tendency of the redox reaction to occur spontaneously-the lower the cell potential, the lower the tendency to occur. A negative cell potential indicates that the forward reaction is not spontaneous. In all electrochemical cells, we call the electrode where oxidation occurs the anode and the electrode where reduction occurs the cathode. In a voltaic cell, the anode is the source of electrons or negative charge and it is convention to assign the anode as the negative electrode and label it with a negative sign (-).The cathode receives the electrons, is the positive electrode, and is labelled with a positive sign(+). In the voltaic cell, electrons flow spontaneously from the anode to the cathode-from the negative electrode to the positive electrode-through the wire connecting the electrodes. As electrons flow out of the anode, positive ions (Zn 2 + in the preceding example) form in the oxidation half-cell, resulting in a buildup of positive charge in the solution. As electrons flow into the cathode, positive ions (Cu2 + in the preceding example) are reduced at the reduction half-cell, resulting in a buildup of negative charge in the solution.

18.3

If the movement of electrons from anode to cathode were the only flow of charge, the build up of the opposite charge in the solution would stop electron flow almost immediately. The cell needs a pathway by which counterions can flow between the half-cells without the solutions in the half-cells totally mixing. One such pathway is a salt bridge, an inverted, U-shaped tube that contains a strong electrolyte such as KN0 3 and connects the two half-cells (see Figure 18.2). The electrolyte is usually suspended in a gel and held within the tube by permeable stoppers. The salt bridge allows a flow of ions that neutralizes the charge buildup in the solution. The negative ions within the salt bridge flow to neutralize the accumulation of positive charge at the anode, and the positive ions flow to neutralize the accumulation of negative charge at the cathode. In other words, the salt bridge completes the circuit, allowing electrical current to flow.

Electrochemical Cell Notation Electrochemical cells are often represented with a compact notation called a cell diagram or line notation. For example, the electrochemical cell discussed on the previous page in which Zn is oxidized to Zn2 + and Cu 2 + is reduced to Cu is represented as follows: Zn(s) IZn 2 +(aq) 11 Cu 2 +(aq) ICu(s) In this representation: ~

We write the oxidation half-reaction on the left and the reduction on the right. A double vertical line, indicating the salt bridge, separates the two half-reactions.

~

Substances in different phases are separated by a single vertical line, which represents the boundary between the phases.

~

For some redox reactions, the reactants and products of one or both of the half-reactions may be in the same phase. In these cases (which are explained further below), the reactants and products are simply separated from each other with a comma in the line diagram. Such cells use an inert electrode, such as platinum (Pt) or graphite, as the anode or cathode (or both).

Consider the redox reaction in which Fe(s) is oxidized and Mn04 - (aq) is reduced: 5 Fe(s)

+ 2 Mn04-(aq) + 16 H+ (aq) -

5 Fe 2 +(aq)

+ 2 Mn2 +(aq) + 8H20(l)

The half-reactions for this overall reaction are as follows:

Oxidation: Fe(s) Fe2 +(aq) + 2 eReduction: Mn04-(aq) + 8 H+ (aq) + 5 e- -

Mn2 + (aq)

+ 4 H 20(l)

Notice that, in the reduction half-reaction, the principal species are all in the aqueous phase. In this case, the electron transfer needs an electrode on which to occur. An inert platinum electrode is employed, and the electron transfer takes place at its surface. Using line notation, we represent the electrochemical cell corresponding to the above reaction as follows: Fe(s) 1Fe2 +(aq) I 1Mn04- (aq), H+(aq), Mn 2 +(aq), H 2 0(l) IPt(s) The Pt(s) on the far right indicates that an inert platinum electrode acts as the cathode in this reaction, as depicted in Figure 18.4 T .

18.3

Standard Electrode Potentials

As we have just seen, the standard cell potential (E~etD for an electrochemical cell depends on the specific half-reactions occurring in the half-cells, and is a measure of the potential energy difference (per unit charge) between the two electrodes. We can think of the electrode in each half-cell as having its own individual potential, called the standard electrode potential. The overall standard cell potential (E~en) is the difference between the two standard electrode potentials. We can better understand this idea with an analogy. Consider two water tanks with different water levels connected by a common pipe, as shown in Figure 18.5 ..-. The water in each tank has its own level and corresponding potential energy. When the tanks are connected, water flows spontaneously from the tank with the higher level (higher potential energy) to the tank with a lower water level (lower potential energy). Similarly, each

Standard Electrode Potentials

799

800

Chapter 18

Electrochemistry

.... FIGURE 18.4 Inert Platinum Electrode When the chemical species in a half-reaction are all in the aqueous phase, a conductive surface is needed for electron transfer to take place. In such cases, an inert electrode of graphite or platinum is often used . In this electrochemical cell, an iron strip acts as the anode and a platinum strip acts as the cathode. Iron is oxidized at the anode and Mn04- is reduced at the cathode.

Anode Fe(s)

Cathode Pt(s)

Oxidation

Fe(s)

High potential energy

Low potential energy

Direction of spontaneous

----->

Reduction 5 e- -----> Mn2+(aq)

Fe2+(aq) +

+

for Electrode Potential

4 H20(I)

half-cell in an electrochemical cell has its own electrode potential. When the cells are connected, electrons flow spontaneously from the electrode with the higher potential for oxidation to the electrode with a higher potential for reduction. The half-cell electrode that is arbitrarily chosen to have a potential of zero is the standard hydrogen electrode (SHE). This half-cell consists of an inert platinum electrode immersed in 1 mol L- 1 HCl with hydrogen gas at 1 bar bubbling through the solution over the platinum electrode, as shown in Figure 18.6 ~ -When the SHE acts as the cathode, the following half-reaction occurs:

2H+(aq)

flow A FIGURE 18.5 An Analogy

+

+ 2e-

---4

H 2(g)

If we connect the standard hydrogen electrode to an electrode in another half-cell of interest, we can measure the potential difference (or voltage) between the two electrodes. Since the SHE is assigned a standard electrode potential of exactly zero volts, we can determine the standard electrode potential of the other half-cell. For example, consider the electrochemical cell shown in Figure 18.7 ~ -In this electrochemical cell, Zn is oxidized to Zn2+, so the Zn2+rzn electrode is the anode. At the SHE, W is being reduced to produce gaseous H 2 , as expected, since it must be the cathode. We have set up this electrochemical cell so that all solutes are present at a concentration of 1 mol L-I and all gases are at 1 bar pressure (everything in their standard states). The measured cell potential for this cell is 0.76 V. The positive value for the standard cell potential means that the electrochemical cell runs spontaneously with the Zn2+rzn as the anode and the SHE as the cathode. By convention, the standard electrode potentials are always written for reduction half-reactions. We write the standard electrode potentials for the two half-reactions in Figure 18.7 as follows:

+ 2 eZn +(aq) + 2 e2H+(aq) 2

---4 ---4

H2 (g) Zn(s)

E

0

E

0

= O.OOV = ?V,

even though the reverse reaction is occurring at the Zn2+/zn electrode and where we are trying to determine the standard electrode potential for the Zn2+/zn electrode. This convention allows us to write the standard cell potential, E~ell• as the difference between the standard electrode potentials of the cathode and the anode: Fce11

=

E~athode

-

E : node

[18.1)

18.3

Standard Electrode Potentials

801

In Equation 18.1 , remember that both E~athode and £,;node are the standard electrode potentials for the half-reactions written as reductions (reduction potential). The negative sign in Equation 18.1 takes into account that at the anode the oxidation half-reaction is occurring and the oxidation potential is the negative of the reduction potential. Since the SHE cathode has a potential of zero volts, we can determine the standard electrode potential for the Zn2+/Zn half-cell (the anode) from the measured standard cell potential: Foell

0.76 V Ez 0 2+;z0

= = =

Fcathode -

£,;node

O.OOV - Ez0 2+;z 0 -0.76V

The negative standard electrode potential means that the Zn2+/zn electrode will be the anode when it is paired with a SHE in a voltaic cell. In fact, we will see shortly that the Zn 2+/zn electrode will be the anode in any voltaic cell when it is paired with an electrode with a more positive electrode potential. Now let's consider an electrochemical cell composed of a SHE and a Cu2+/Cu electrode with a Cu2+ concentration of 1 mol L-1• When these two electrodes are paired as in Figure 18.8 T, the SHE is the anode, and the Cu2+/Cu electrode is the cathode with electrons spontaneously flowing from the SHE anode to the Cu 2+/Cu cathode. The standard cell potential is measured to be 0.34 V. In the same way as before, we can determine the standard cell potential for the Cu2+/Cu electrode: E~ell

= = E'Cu2+ ;cu = 0.34 V

E~athode - E ; node E~u' +;cu V - 0.00 V

0.34 V

The positive standard electrode potential means that reduction is favoured at the Cu2+/Cu electrode-it will be the cathode-when paired with an electrode with a more negative standard cell potential, such as the SHE.

Summarizing Standard Electrode Potentials: ~

The electrode potential of the standard hydrogen electrode (SHE) is exactly zero.

~

The electrode in any half-cell with a greater tendency to undergo reduction is positively charged relative to the SHE and therefore has a positive E 0

•

Anode Zn(s)

Zn(s) -

Oxidation Zn2 +(aq)

Reduction

+

2 e-

2 H+(aq)

+

2 e- -

H 2(g)

A FIGURE 18.7 Measuring Standard Electrode Potential Since the electrode potential of the SHE is known (0.00 V), the standard electrode potential for the Zn2+/zn electrode can be determined.

A FIGURE 18.6 The Standard Hydrogen Electrode The standard hydrogen electrode (SHE) is arbitrarily assigned an electrode potential of zero. All other electrode potentials are then measured relative to the SHE.

802

Chapter 18

Elect ro chemistry

Eo - Eo - E"anode cell cathode 0.34 V = E~u2·;cu V - 0.00 V E~u2·;cu = 0.34 V

( 0.34 V) -

e

Oxidation

H 2 ~ 2H++ 2 e-

Cu2+

+

R edu cti on 2 e- ~

Cu(s)

.A FIGURE 18.8 Electrochemical Cell with SHE as the Anode In this pairing of electrodes, the potential for Cu2+ to be reduced is higher than for H+ (as in Figure 18.7), so the SHE is the anode. ~

The electrode in any half-cell with a lesser tendency to undergo reduction (or greater tendency to undergo oxidation) is negatively charged relative to the SHE and therefore has a negative£°.

~

The cell potential of any electrochemical cell (E~ell) is the difference between the electrode potentials of the cathode and the anode (E~ell = E~at - E;n).

~

Foell is positive for spontaneous reactions and negative for nonspontaneous reactions.

We write the standard electrode potentials for the three half-reactions just discussed as follows: Cu 2 +(aq) + 2 e- - - - Cu(s) E° = 0.34V

2H+(aq) 2

Zn +(aq)

+

2 e- - - - H 2 (g)

+ 2 e- ---

Zn(s)

E° =

o.oov

E° = -0.76V

We can see that the Cu2+/Cu electrode has the most positive standard electrode potential, followed by the SHE, and the Zn2+/zn electrode is the most negative. This means that the Cu2+/Cu electrode has the highest potential of the three to be the cathode, whereas the Zn2+/ Zn electrode has the lowest potential of the three to be the cathode and the highest of the three to be the anode. If we were to pair the Cu2+/Cu and Zn2+/zn electrodes, electrons would spontaneously flow from the Zn2+/zn anode to the Cu 2+/Cu cathode. We can calculate the overall standard cell potential of this electrochemical cell using Equation 18.1: Foell = E~athode - F.node Foell = Ecu2+ /Cu - Ezn2+ /Zn Foell

= 0.34 V - (-0.76 V) = 1.lOV

The standard electrode potentials for a number of common half-reactions are shown in Table 18.1.

18.3

TABLE 18.1

803

Standard Electrode Potentials at 25 °C

Reduction Half-Reaction Stronger oxidizing agent

Standard Electrode Potentials

E0 (V)

+ 2 eH202(aq) + 2 H+(aq) + 2 eF2(g)

Pb0 2(s)

+ 4 W(aq) +

soi - (aq)

+ 4 W(aq) + 3 eMn04- (aq) + 8 W(aq) + 5 eMn04- (aq)

+ 3 e-

Au3+(aq)

+ 4 H+ (aq) + 2 e-

Pb02(S) Cl2(g)

+ 2 e-

Cr20/- (aq)

+ 14 W (aq) + 6 e-

+ 4 W (aq) + 4 eMn02(s) + 4 W (aq) + 2 e02(g)

+ 6 H+ (aq) + 5 e-

103- (aq) Br2(/)

+ 2 e+ 2 W (aq) +

N0 3- (aq)

+ 4 W (aq) + 3 e-

+

e-

e-

+ 4 W(aq) + 4 eAg+ (aq) + e02 (g)

+

Fe3 +(aq)

e-

+ 2 W(aq) + 2 eMn04- (aq) + e02(g)

+ 2 ecu+(aq) + e02(g) + 2 H20(/) + 4 eCu2+(aq) + 2 e-

l2(S)

S04 2- (aq)

2 F- (aq)

2.87

---7

2 H20(/)

1.78

---7

PbS04 (s)

+ 2 H20(/)

1.69

---7

Mn02(s)

+ 2 H20(/)

1.68

---7

Mn2+(aq)

---7

Au(s)

---7

Pb2+(aq)

---7

2 Ci-(aq)

---7

2 cr3+(aq)

---7

2 H20(/)

---7

---7

+ 4 H+ (aq) + 2 e-

+ 4 H20(/)

1.50

Mn 2+(aq) ~1 2 (aq)

+ 2 H20(/)

1.46 1.36

+ 7 H20(1)

1.33 1.23

+ 2 H20(/)

+ 3 H20(/)

1.21 1.20 1.09

---7 ---7

NO(g)

---7

Cl02- (aq)

0.95

---7

2 H20(1)

E = 0.82V(pH

---7

Ag(s)

0.80

---7

Fe 2+(aq)

0.77

---7

H202(aq)

0.70

---7

Mnoi- (aq)

0.56

---7

2 i-(aq)

0.54

---7

Cu(s)

0.52

---7

4 oH- (aq)

0.40

---7

Cu(s)

0.34

---7

H2S03(aq)

---7

Cu + (aq)

0.16

Sn2+(aq)

0.15

+ H20(I)

+ 2 H20(/)

+ H20(/)

1.00 0.96

+

Sn4+(aq)

+ 2 e-

---7

2 W(aq)

+ 2 e-

---7

H2(g)

Fe3+(aq)

+ 3 e-

---7

Fe(s)

-0.036

Pb2+(aq)

+ 2 e-

---7

Pb(s)

-0.13

Sn2+(aq)

+ 2 e-

---7

Sn(s)

- 0.14

Ni 2+(aq)

+ 2 e-

---7

Ni(s)

-0.23

---7

Cd(s)

- 0.40

---7

H2(9)

---7

Fe(s)

- 0.45

---7

Cr2+(aq)

-0.50

---7

Cr(s)

- 0.73

---7

Zn(s)

-0.76

---7

H2(g)

---7

Mn(s)

Cd2+(aq)

+ 2 e2 H20(/) + 2 eFe2+(aq) + 2 ecr3+(aq)

+

cr3+(aq)

+ 3 e-

e-

Zn 2+(aq)

+ 2 e2 H20(/) + 2 e-

Mn2+(aq)

+ 2 e-

= 7)

0.20

Cu2+(aq)

e-

Weaker reducing agent

1.51

2 snaq) vo 2+(aq)

---7

V02+ (aq) Cl0 2(g)

+ 2 e-

---7

0

+ 2 OW(aq)

+ 2 OW(aq)

E

= - 0.41 V(pH = 7)

-0.83 - 1.18

(continued)

804

Chapter 18

TABLE 18.1

El ect ro chemistry

(Continued)

Reduction Half-Reaction

f 0 (V)

+ 3 e-

-----+

Al(s)

- 1.66

Mg2+(aq)

-----+

Mg(s)

-2.37

Na+ (aq)

+ 2 e+ e-

---+Na(s)

- 2.71

---+Ca(s)

- 2.76

---+ Ba(s)

- 2.90

---+K(s)

- 2.92

---+ Li(s)

- 3.04

Al3 +(aq)

+ 2 eBa2+(aq) + 2 eCa2+(aq)

+ eu+(aq) + e-

K+ (aq) Weaker oxidizing agent

Stronger reducing agent

The highlighted reactions are for nonstandard state, but are more common. The electrode potentials are for the half-reactions at pH 7 ([H"'] = 1 x 10- 7 or [OHl = 1 x 1o-7). See Section 18. 7 on electrolysis of aqueous solutions for the significance of these concentrations.

Example 18.1 shows how to calculate the potential of an electrochemical cell from the standard electrode potentials of the half-reactions. EXAMPLE 18.1

CALCULATING STANDARD POTENTIALS FOR ELECTROCHEMICAL CELLS FROM STANDARD ELECTRODE POTENTIALS-I

Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurri ng in an electrochemical cell at 25 °C (the equation is balanced): Al(s)

+ N0 3-(aq) + 4 H +(aq)

~ A13+(aq)

+

NO(g)

+ 2 H 20(/)

SOLUTION

Begin by separating the reaction into oxidation and reduction half-reacti ons. In this case, it is apparent thatAl(s) is oxidized. In cases in which it is not as clear, you may want to assign oxidation states to help determine the correct half-reactions.

Oxidation:

Next, look up the standard electrode potential for each half-reactio n. Remember, these reactions are written as reduction half-reactions. Use Equation 18.1 to calculate the standard cell potential.

(Cathode):

Al(s) ~ Al3+ (aq)

Reduction:

N0 3- (aq)

+ 4 H +(aq) + 3--e

~ NO (g )

+ 3-e +

2 H 20(/)

Overall:

£

(Anode):

E~ell

=

E~ell

= £No3/ NO - £~13+ / Al = 0.96 V - (- 1.66 V) = 2.62 v

£~ell

Fcathode -

0

= -1.66 V

£ ; node

FOR PRACTICE 18.1 Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C (the equation is balanced): Mn04- (aq)

+ ! I2 (aq) + 2 H +(aq)

~ Mn2+(aq)

+ 103- (aq) + H20(/)

In electrochemical cells where you must multiply one or both of the half-reactions by a constant in order to cancel out the electrons, you do not multiply the standard electrode potential. That is because E° is an intensive quantity-it doesn' t depend on the amount

18.3 Standard Electrode Potentials

805

of substance. Potential is a ratio of energy to charge; increasing the amount of substance increases both the energy content and the amount of charge by the same amount, so the ratio remains unchanged. Density (mass/volume) is also an intensive quantity. The following example is slightly more complicated than the previous one in that one or both of the half-reactions need to be multiplied by a factor in order to balance the equation.

EXAMPLE 18.2

CALCULATING STANDARD POTENTIALS FOR ELECTROCHEMICAL CELLS FROM STANDARD ELECTRODE POTENTIALS-II

Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C (the equation is balanced):

SOLUTION Again, begin by separating the reaction into oxidation and reduction half-reactions. The oxygen in H 20 2 goes from an oxidation state of -1 to 0 and therefore is oxidized while Cr goes from +6 to + 3 and is therefore reduced. Next, look up the standard electrode potential for each half-reaction. These reactions are written as reduction half-reactions . Note that although the 0 2/ H 20 2 cell is one-third of that used to balance the redox reaction , when using Equation 18.1 to calculate the standard cell potential, you do not multiply the standard cell potential by any factor.

Oxidation:

3 H2 0 2 (aq) Cr2 0/-(aq)

Reduction:

+ 14 H +(aq) +

+ 6 H+(aq) + (j..e"' 2 Cr3+(aq) + 7 H20(/)

3 0 2 (g)

(j..e"' -

1.33 v Eo

(Anode):

E~ell

=

E~e11

= Ecr20?- ;cf'+ - Eo,JH20 2 = 1.33 V - 0.70V = 0.63V

E~e11

E~athode

-

= 0.70V

E : node

FOR PRACTICE 18.2 Use tabulated standard electrode potentials to calculate the standard cell potential for the following reaction occurring in an electrochemical cell at 25 °C (the equation is balanced): Sn2 +(aq)

+ 2 cr2+(aq)

-

Sn(s)

+ 2 Cr3+(aq)

Predicting the Spontaneous Direction of an Oxidation-Reduction Reaction An oxidation-reduction reaction will proceed spontaneously in the direction that provides a positive cell potential. We can use Equation 18.1 and the standard electrode potentials of the two relevant half-reactions in Table 18.1. For example, consider an

806

Chapter 18

El ect ro chemistry

electrochemical cell composed of electrodes represented by the following two reduction half-reactions: Ni 2 +(aq)

Mn 2 +(aq)

+ 2 e+ 2 e-

E° = -0.23 v E° = -1.18 v

~ Ni(s) ~ Mn(s)

Equation 18.1 says: ~ell

=

~athode

-

E : node

In order to have an overall positive cell potential, the electrode with the more negative standard electrode potential must be the anode, and the one with the more positive standard electrode potential must be the cathode. We can confirm that this provides a positive cell potential: E~ell

= £Ni2+/Ni - E Mn2+/Mn = -0.23 V - (-1.18 V) = 0.95V

We can write the overall reaction, remembering that the Mn2+/Mn electrode is the anode and must be written in the opposite direction-as an oxidation: Ni 2 +(aq)

Cathode (reduction):

+ 2---e'"'

~ Ni(s)

Mn(s) ~ Mn2 +(aq)

Anode (oxidation): 2

Ni +(aq)

+ Mn(s)

~ Ni(s)

+ 2---e'"'

+ Mn +(aq) 2

The overall cell potential is positive, indicating a spontaneous reaction. The electrochemical cell corresponding to this spontaneous redox reaction is shown in Figure 18.9 T. We draw the manganese half-cell on the left as the anode and the nickel half-cell on the right as the cathode. Electrons flow from the anode to the cathode. Another way to predict the spontaneity of a redox reaction is to note the relative positions of the two half-reactions in Table 18.1. Since the table lists reduction half-reactions in order of decreasing electrode potential, the half-reactions near the top of the tablethose having large positive electrode potentials-tend to occur in the forward direction. The half-reactions near the bottom of the table-those having large negative electrode potentials-tend to occur in the reverse direction. In other words, as you move down Table 18.1 , the half-reactions become less likely to occur in the forward direction and more likely to occur in the reverse direction. As a result, any reduction half-reaction listed will be spontaneous when paired with the reverse of a half-reaction that appears below it in Table 18.1.

Summarizing the Prediction of Spontaneous Direction for Redox Reactions: ~

The half-reaction with the more positive electrode potential will undergo reduction (so substances listed at the top of Table 18.1 tend to undergo reduction; they are good oxidizing agents) .

.... FIGURE 18.9 Mn/Ni2 + Electrochemical Cell The reduction

-

e

of Mn2 + is listed below the reduction of Ni2 + in Table 18.l; the reduction of N iz+ is spontaneous when paired w ith the oxidation of Mn.

1

Salt bridge

Anode

Cathode

18.3 ~

The half-reaction with the more negative electrode potential will undergo oxidation (so substances listed near the bottom of Table 18. l tend to undergo oxidation; they are good reducing agents).

~

Any reduction reaction in Table 18.1 is spontaneous when paired with the reverse of the reaction listed below it.

EXAMPLE 18.3

Standard Electrode Potentials

807

PREDICTING SPONTANEOUS REDOX REACTIONS AND SKETCHING ELECTROCHEMICAL CELLS

Without calculating E~e!I> predict whether each of the following redox reactions is spontaneous. If the reaction is spontaneous as written, make a sketch of the electrochemical cell in which the reaction could occur. If the reaction is not spontaneous as written, write an equation for the spontaneous direction in which the reaction would occur and sketch the electrochemical cell in which the spontaneous reaction would occur. In your sketches, make sure to label the anode (which should be drawn on the left), the cathode, and the direction of electron flow.

+ Mg2 +(aq) ~ Fe2 +(aq) + Mg(s) 3 Mg(s) + 2 Cr3+(aq) ~ 3 Mg2 +(aq ) + 2 Cr(s)

(a) Fe(s) (b)

SOLUTION

+ Mg2 +(aq)

(a) Fe(s)

~ Fe2 +(aq)

+ Mg(s)

This reaction involves the reduction of Mg 2+ and the oxidation of Fe. The reduction half-reactions along with the standard electrode potentials are:

+ 2 eMg +(aq) + 2 eFe2 +(aq) 2

~

Fe(s)

~ Mg(s)

E' = - 0.45 v £

0

= -2.37 V

The magnesium half-reaction has the more negative standard electrode potential and is the reaction that will occur as an oxidation (in the reverse). The reaction will not be spontaneous as written. However, the reverse reaction would be spontaneous: 2

Fe +(aq)

+ Mg(s)

~ Fe(s)

+ Mg +(aq) 2

The electrochemical cell is shown in Figure 18. 10 ~ ­ (b) 3 Mg(s)

+ 2 Cr3+(aq)

~ 3 Mg2 +(aq )

Anode

Cathode

.& FIGURE 18.10 Electrochemical Cell Composed of Mg 2+/Mg and Fe2+/Fe Electrodes

+ 2 Cr(s)

This reaction involves the oxidation of Mg and the reduction of c r3+. From Table 18.1 , the reduction half-reactions and standard electrode potentials are:

Cr3+ (aq) + 3 e- ~ Cr(s) Mg 2 +(aq) + 2 e- ~ Mg(s)

E' = -0.73 v E' = -2.37 v

With this pairing of half-reactions, the Mg2+/Mg half-reaction has the more negative standard electrode potential and will occur as an oxidation as written in the overall reaction. The reaction is spontaneous as written. Another way of saying this is that the reaction pairs the Cr3+/Cr reduction with the reverse of a half-reaction below it in Table 18.1-such pairings are always spontaneous. The corresponding electrochemical cell is shown in Figure 18.1 1 ~ .

Anode

.& FIGURE 18.11 Electrochemical Cell Composed of Mg2+/Mg and cr3+1cr Electrodes

FOR PRACTICE 18.3 Will the following redox reactions be spontaneous under standard conditions?

+ Ni2 +(aq) ~ Zn2 +(aq) + Ni(s) 3 Zn(s) + 2 A13+ (aq) ~ 3 Zn2 +(aq) + 2 Al(s)

(a) Zn(s) (b)

Cathode

808

Chapter 18

Electrochemistry

A solution contains both Nal and NaBr. Which oxidizing agent could you add to the solution to selectively oxidize i-(aq) but not sn aq )? Cl2

(a)

(b)

H202

(c)

CuCl2

(d)

HN03

Predicting Whether a Metal Will Dissolve in Acid In Chapter 15, we learned that acids dissolve metals. Most acids dissolve metals by the reduction of H + ions to hydrogen gas and the corresponding oxidation of the metal to its ion. For example, if solid Zn is dropped into hydrochloric acid, the following reaction occurs:

+

2 H+ (aq)

Zn(s)

Zn(s)

+

2H+(aq)

~

Zn2+(aq)

+ Hz(g)

.A When zinc is immersed in hydrochloric acid, the zinc is oxidized, forming ions that become solvated in the solution. Hydrogen ions are reduced, forming bubbles of hydrogen gas. [© Richard Megna/Fundamental Photographs, NYC]

2--e ~ H2(g) Zn(s) ~ Zn2 +(aq)

+ 2 H+ (aq)

~ Zn +(aq) 2

+ 2--e + H2(g)

We observe the reaction as the dissolving of the zinc and the bubbling of hydrogen gas. The zinc is oxidized and the H+ ions are reduced. Notice that this reaction involves the pairing of a reduction half-reaction (the reduction of H+) with the reverse of a half-reaction that falls below it in Table 18. l. Therefore, this reaction is spontaneous. What would happen, however, if we paired the reduction of H+ with the oxidation of Cu? The reaction would not be spontaneous because it involves pairing the reduction of H+ with the reverse of a half-reaction that is listed above it in the table. Consequently, copper does not react with H+ and will not dissolve in acids such as HCI. In general, metals whose reduction half-reactions are listed below the reduction of F to H2 in Table 18.1 will dissolve in acids, while metals listed above it will not. An important exception to this rule is nitric acid (HN03), which can oxidize metals and produces NO gas through the reduction half-reaction: N0 3-(aq)

+ 4 H+ (aq) +

3 e- ~ NO(g)

+

2 H20(l)

E'

= 0.96 V

Since this half-reaction is above the reduction of H+ in Table 18.l, HN03 can oxidize metals (such as copper) that can 't be oxidized by HCI.

Which metal dissolves in HN03 but not in HCI? (a)

18.4

Fe

(b)

Au

(c)

Ag

Cell Potential, Gibbs Energy, and the Equilibrium Constant

We have seen that a positive standard cell potential (£~ 011 ) corresponds to a spontaneous oxidation-reduction reaction. And we know (from Chapter 17) that the spontaneity of a reaction is determined by the sign of t:,.rG, or under standard state condtions by the sign of t:,.rG'. Therefore, £~ 011 and t:,.rG' must be related. We also know from Section 17. I I that t:,.rG' for a reaction is related to the equilibrium constant (K) for the reaction. Since £~011 and t:,.rG' are related, then £~ 011 and K must also be related.

18.4

Cell Potential, Gibbs Energy, and the Equilibrium Constant

Before we look at the nature of each of these relationships in detail, let's consider the following generalizations. For a spontaneous redox reaction (one that will proceed in the forward direction when all reactants and products are in their standard states): ~

!::.,G is negative ( < 0)

~

Fcen is positive ( > 0)

~

K

0

> 1

For a nonspontaneous reaction (one that will proceed in the reverse direction when all reactants and products are in their standard states): ~

!::.p is positive ( > 0)

~

E~ell

~

K < 1

0

is negative ( < 0)

The Relationship Between ~rG and f~e 11 0

We can derive a relationship between !::.,G and Fcenby briefly returning to the definition of potential difference from Section 18.2-a potential difference is a measure of the difference of potential energy per unit charge (q): 0

Maximum work (Wmaxl

l potential energy difference (in J)

E

charge (in C)

t

Charge (q)

Since the potential energy difference represents the maximum amount of work that can be done by the system on the surroundings, we can write: Wmax

=

-qE~ell

[18.2]

The negative sign follows the convention used throughout this text that work done by the system on the surroundings is negative. We can quantify the charge (q) that flows in an electrochemical reaction by using Faraday's constant (F), which represents the charge in coulombs of 1 mol of electrons. F

= 96485 C mol- 1

The total charge is q = nF, where n is the number of electrons transferred in the balanced chemical equation (a unitless value) and Fis Faraday's constant. Substituting q = nF into Equation 18.2, Wmax

=

-qE~ell -nFE~ell

[18.3]

Finally, recall from Chapter 17 that the standard change in Gibbs energy for a chemical reaction (11,G°) represents the maximum amount of work that can be done by the reaction. Therefore, Wmax = !::.,G Making this substitution into Equation 18.3, we get the following important result: 0

•

f::.,G = -nFE~ell 0

[18.4]

where 11,G° is the standard change in Gibbs energy for an electrochemical reaction, n is the number of electrons transferred in the balanced equation, Fis Faraday's constant, and E~ell is the standard cell potential. The following example shows how to apply this equation to calculate the standard Gibbs energy change for an electrochemical cell.

809

810

Chapter 18

EXAMPLE 18.4

Electrochemistry

RELATING 4rG 0 AND £ 0 cell 0

Use the tabulated electrode potentials to calculate l1,G for the reaction between sodium metal and neutral water (pH= 7.0): 2Na(s)

+ 2 H2 0(l)

~ H 2 (g)

+ 20W(aq) + 2Na+(aq )

Is the reaction spontaneous?

+ 2 H20(l)

+ 20W(aq) + 2Na+(aq)

SORT You are given a redox reaction and asked to find l1,G

GIVEN: 2Na(s)

STRATEGIZE Use the tabulated values of standard electrode potentials to calculate Fcen using Equation 18.1. Then use Equation 18.4 to calculate l1,G from E~ell ·

CONCEPTUAL PLAN

SOLVE Break the reaction up into oxidation and reduction half-reactions and find the standard electrode potentials for each from Table 18.1. Find Fcen using Equation 18.1.

SOLUTION The way the reaction is written, Na is being oxidized, so the Na+/Na reaction is the anodic reaction and water reduction is the cathodic reaction. The half-reactions and standard electrode potentials are as follows:

~ H 2 (g)

0

•

FIND: l1,G

0

E~.n

0

2 H20(/) + 2 e- ~ H2 (g) Na+(aq) + e- ~ Na(s)

Cathode: Anode: E~ell

+ 2 OH- (aq)

E° = -0.41 v E°

= -2.71 v

= Fcathode - E;node

Fceu = -0.41 V - (-2.71 V)

= 2.30 v Calculate l1,G from E~ell using Equation 18.4. Two electrons would be cancelled in the half-reaction to balance the overall reaction (the Na+/Na half-reaction would be multiplied by two); therefore, n = 2. In order for units to cancel, 2.30 V should be expressed as 2.30 J c- 1• 0

= -2(96 485 IZ mol- 1)(2.30 J ~) = -4.44 X 105 Jmol- 1( X l kJ/ 10001) = -444 kJ mol- 1 0

Since l1,G is negative, the reaction is spontaneous in the direction written.

CHECK The answer is in the correct units (kJ mol- 1) and seems reasonable in magnitude (sodium ignites when put in water and must be stored under oil). The sign is negative, which means the reaction is spontaneous and is consistent with a positive Foell· FOR PRACTICE 18.4 Use tabulated electrode potentials to calculate l1,G for the reaction: 0

12 (s)

+ 2 Br- (aq)

~

2 i-(aq)

+ Br2 (/)

Is the reaction spontaneous?

ci ~

fi\lil

CONCEPTUAL CONNECTION 18.3

ii IMl@HJl.! i!itl•lii@ii J.l,ll1'!.!11fl.@ijtitl;§!tifi;M@ii·i&W 0

0

When you do the For Practice Problem 18.4, you will see that the reaction of 12 with Br- to form 1and Br2 is not spontaneous. Based on conceptual reasoning, which of the following best explains why 12 does not oxidize Br-? (a)

Br has a larger electon affinity than I; therefore, we do not expect Br- to give up an electron to 12.

(b) I has a larger electron affinity than Br; therefore, we do not expect 12 to give up an electronto Br-. (c) Br- is in solution, and 12 is a solid. Solids do not gain electrons from substances in solution.

18.4

The Relationship Between

Cell Potential, Gibbs Energy, and the Equilibrium Constant

811

£;

811 and K We can derive a relationship between the standard cell potential (Ecell) and the equilibrium constant for the redox reaction occurring in the cell (K) by returning to the relationship between /:,.rG and K that we learned in Chapter 17. Recall from Section 17 .11 that 0

= -RT In K

0

/:,.rG

[18.5]

By setting Equations 18.4 and 18.5 equal to each other, we get:

-

= -RT In K

nFE~ell

RT =-In K nF

E~ell

[18.6]

Equation 18.6 is usually simplified for use at 25 °C by making the following substitutions: R = 8.314 J mol- 1 K- 1; T = 298.15 K; F = 96485 C mol- 1

Substituting into Equation 18.6, we get the following important result: Foell

=

0.0257 v In K n

[18.7]

where Foell is the standard cell potential, n is the number of electrons transferred in the redox reaction, and K is the equilibrium constant for the balanced redox reaction at 25 °C. The following example demonstrates how to use Equation 18.7.

EXAMPLE 18.5

RELATING E;en AND K

Use the tabulated electrode potentials to calculate K for the oxidation of copper by H+ (at 25 °C). Cu(s)

SORT You are given a redox reaction and asked to find K. STRATEGIZE Use the tabulated values of electrode potentials to calculate E~ell· Then use Equation 18.7 to calculate K from Fce11·

+

2 H+(aq)

GIVEN: Cu(s)

+

------+

Cu 2 +(aq)

2 H+(aq)

------+

+

H2(g)

Cu2 +(aq)

+

H2 (g)

FIND: K CONCEPTUAL PLAN E~.11

(

E~.u 'ii--.ir-;;-' o 0.0257 V Jn K E cell = n

SOLVE Break the reaction up into oxidation and reduction half-reactions and find the standard electrode potentials for each from Table 18.1. Find Fce11 using Equation 18.1.

SOLUTION The way the reaction is written, Cu is being oxidized, so the Cu 2+/Cu reaction is the anodic reaction and the W reduction is the cathodic reaction. The half-reactions and standard electrode potentials are as follows :

Cathode: Anode: E~ell E~ell

= =

E~e11 =

=

2 H+(s) Cu 2 +(aq )

E~athode -

+ 2 e-------+ H 2 (g) + 2 e- ------+ Cu(s)

E = O.OOV E = 0.34 V

E~node

E8+/H, - EC.u2 +/cu

0.00 V - 0.34 V -0.34 v

The reaction has a negative E~ell and is not spontaneous under standard conditions. (continued)

812

Chapter 18

EXAMPLE 18.5

Electrochemistry

{CONTINUED)

Calculate K from Fce11 using Equation 18.7. Two electrons would be cancelled in the half-reaction to balance the overall reaction; therefore, n = 2.

E~ell

=

lnK =

=

0.0257 v n

lnK

nE~ell

0.0257 v 2 X (-0.34 V) =-26.46 0.0257 v

K = e- 2!?..46 = 3.2 X 10- 12

CHECK The answer has no units, as expected for an equilibrium constant. The K value is much less than one, indicating that the reaction lies far to the left at equilibrium, as expected for a reaction in which Fce 11 is negative. FOR PRACTICE 18.5 Use the tabulated electrode potentials to calculate K for the oxidation of iron by H+ (at 25 °C):

2 Fe(s) + 6 H+(aq)

-----'>

2 Fe3+ (aq) + 3 H 2(g)

Notice that the fundamental quantity in the above relationships is the standard change in Gibbs energy for a chemical reaction (LirG From that quantity, we can calculate both E~ell and K. The relationships between these three quantities is summarized with the following diagram: 0

).

~,G

0

=

-nFE°cell

~ell

K

Ff.

_ 0.0257 V In K

cell -

18.5

n

Cell Potential and Concentration

We have learned how to find Foell under standard conditions. For example, we know that when [Cu2 +] = I mol L- 1 and [Zn2 +] = I mol L- 1, the following reaction produces a potential of 1.10 V. Zn(s)

+ Cu2 +(aq, 1 mol L- 1)

-----'>

Zn 2 +(aq, 1 mol L- 1)

+ Cu(s)

E~ell

= 1.10 V

However, what if [Cu 2 +] > 1 mol L- I and [Zn2 +] < 1 mol L- I? For example, how would the cell potential for the following conditions be different from the potential under standard conditions? Zn(s)

+

Cu 2 +(aq, 2 mol L- 1)

-----'>

Zn2 +(aq, 0.01 mol L- 1)

+ Cu(s)

E cell

= ?

Since the concentration of a reactant is greater than standard conditions, and since the concentration of product is less than standard conditions, we can use Le Chatelier's principle to predict that the reaction has an even stronger tendency to occur in the forward direction and that Ecell is therefore greater than 1. 10 V (Figure 18.12 ~ ).

813

18.5 Cell Potential and Concentration

Standard conditions

Zn(s)

-

[zn2+]

= 1 mol L- 1

(Cu 2 +]

= 1 molL-1

[ N onstandard conditions

C u(s)

Oxidation

Zn(s)

Zn2 +(aq)

Reduction + 2 e-

Cuz+(aq)

+

2 e-

Zn(s)

-

(Cu 2 +]

-

Oxidation

Zn(s)

Zn2 +(aq)

Cu(s)

+

C u(s)

= 2 mol L- 1

Reduction + 2 e-

Cu2 +(aq) 2 e-

-

C u(s)

A FIGURE 18.12 Cell Potential and Concentration This figure compares the Zn/ Cuz+ electrochemical cell under standard and nonstandard conditions. In this case, the nonstandard conditions consist of a higher Cuz+ concentration ([Cu 2 +] > I mol L- 1) at the cathode and a lower Zn2 + concentration at the anode ([Znz+] < 1 mol L- 1). According to Le Chatelier's principle, the forward reaction has a greater tendency to occur, resulting in a greater overall cell potential than the potential under standard conditions.

We can derive an exact relationship between E cell (under nonstandard conditions) and by considering the relationship between the change in Gibbs energy (d ,G) and the standard change in Gibbs energy (il,G that we learned in Section 17.10: E~ell

0

)

d,G = d,G

0

+

[18.8]

RT In Q

where R is the gas constant (8.314 J/ mol · K), T is the temperature in kelvins, and Q is the reaction quotient corresponding to the nonstandard conditions. Since we know the relationship between d,G and Ecell (Equation 18.4), we can substitute into Equation 18.8:

d ,G = il,G -nFEcell

=

+

0

RT In Q

-nFE~ell + RT In Q

We can then divide each side by -nF to arrive at: E cell

=

E~ell -

As we have seen, R and Fare constants; at T

RT

[18.9]

-In Q nF

RT

= 25 °C, -

nF

0.0257 v n

Substituting into Equation 18.9, we arrive at the Nernst equation: E cell

=

0

E cell -

0.0257 v In Q n

[18.10]

where E cell is the cell potential in volts, Ecen is the standard cell potential in volts, n is the number of electrons transferred in the redox reaction, and Q is the reaction quotient. Notice that, under standard conditions, Q = 1, and (since log 1 = 0) E cell = E~eII> as expected. Example 18.6 shows how to calculate the cell potential under nonstandard conditions.

The Nernst equation is sometimes written in terms of the base 10 logarithm . Since In Q = 2.303 log Q, the Nernst equation can be expressed as: 0.0592 v fce11 = EZ.11 - - - - log Q

n

Both forms of the Nernst equation will give the same answer.

814

Chapter 18

EXAMPLE 18.6

Electrochemistry

CALCULATING fcen UNDER NONSTANDARD CONDITIONS

Determine the cell potential for an electrochemical cell based on the following two half-reactions:

Cathode: Anode:

MnOi(aq)

+ 4 H+(aq) + 3 eCu2+(aq) + 2 e-

SORT You are given the halfreactions occurring at the cathode and anode as well as the concentrations of the aqueous reactants and products. STRATEGIZE Use the tabulated values of electrode potentials to calculate Fce 11 and then calculate Ecell using Equation 18.10.

~ Mn02 (s) ~

GIVEN: [Mn0 4-J FIND:

+ 2 H20(l)

Cu(s)

= 2.0mo!L- 1, [H+J = l.Omo!L- 1,and [Cu2 +] = O.Olmo!L- 1•

Ecell

CONCEPTUAL PLAN E~ell

(

E~0u, [Mn04-],[H+],[cu2+] ' ___.{ 0.0257

Ecell '

v

Eccll = E;cll - - - In Q

"

SOLVE Using the tabulated values for the reduction reactions and Equation 18.1, compute the standard cell potential.

SOLUTION

Cathode: Mn04-(aq) + 4H+(aq) + 3 e- ~ Mn02 (s) + 2 H2 0(l) E = 1.68 V Anode: Cu 2 +(aq) + 2 e- ~ Cu(s) E° = 0.34 V 0

E~ell

=

E~alhode - E;node

E~ell = E~n04 (Mn02 E~ell

-

E~u2 + / Cu

= 1.68 V - 0.34 V = 1.34 v

Equation 18.10 contains the reaction quotient, Q, which requires the balanced equation to write. Balance the equation by writing the anodic reaction in reverse and multiplying each reaction by a factor so that the electrons cancel. Calculate Ecell from E~ell· The value of n is the number of electrons transferred in the balanced equation, which is 6. The expression for Q can be determined from the balanced equation.

Mn04- (aq)

2 Mn04-(aq)

~ Mn02 (s) + 2 H 20(l) Cu(s) ~ Cu 2 +(aq) + 2 e-

+ 4 H +(aq ) + 3 e-

+ 8 H+(aq) + 6-£

3 Cu(s) ~ 3 Cu

Ecell

0

=

Ecell -

=

E

0

cell

-

0.0257 v In Q n 0.0257 V [Cu2 +]3 In - - - - - n [Mn04- J2[H+]8

= 1.34 V -

0.0257 v 6

In

= 1.34 V-(-0.065 V) = 1.41 v

(0.010) 3 (2.0) 2 (1.0) 8

X

+ 4 H20(l) +(aq) + 6-ef

~ 2 Mn02 (s) 2

X ) 3

18.5 Cell Potential and Concentration

815

CHECK The answer has the correct unit (V). The value of Ecell is larger than E~eII> as expected based on Le Chatelier's principle because one of the aqueous reactants has a concentration greater than standard conditions, and the one aqueous product has a concentration less than standard conditions. Therefore, under the conditions given, the reaction has a greater tendency to proceed toward products than under standard conditions and has a greater cell potential. FOR PRACTICE 18.6 Determine the cell potential for an electrochemical cell based on the following two half-reactions,

+ 2 e-

Ni 2 +(aq)

Cathode: Anode:

~ Ni(s)

+ 2 H+(aq) + e- ~ V0 2 +(aq) + H 20(/) 0.010molL- 1, [ V02 +] = 2.0molL- 1 [H+J = 1.0molL- 1,and [Ni 2+] = 2.0molL- 1• VO/(aq)

when [VO/]

From the above examples, and from Equation 18.10, we can conclude the following: ~

When a redox reaction within a voltaic cell occurs under standard conditions, Q = 1; therefore E cell = Foell· Ecell

=

0.0257 v ln Q n

0

Ecell -

0.0257 v

~ ln(l)

=O

tl

- - - ln(l) n

~

When a redox reaction within a voltaic cell occurs under conditions in which Q < 1, the greater concentration of reactants relative to products drives the reaction to the right, resulting in E cell > E~ell ·

~

When a redox reaction within an electrochemical cell occurs under conditions in which Q > 1, the greater concentration of products relative to reactants drives the reaction to the left, resulting in Ecell < £~011 .

~

When a redox reaction reaches equilibrium, Q = K. The redox reaction has no tendency to occur in either direction and E cell = 0. Ecell

= E~ell -

E~ell -

0.0257 v In Q n ~ Ecell 0

0.0257 V In K n

(see Equation 18.7)

ov

=

This last point explains why batteries do not last forever- as the reactants are depleted, the reaction proceeds toward equilibrium and the potential tends toward zero.

In an electrochemical cell, Q = 0.0010 and K = 0.10. Which statement is true? (a) f cenis positive and

(b) f cenis negative and

fi:; Both f cenand fi:;

fi:; is negative. fi:; is positive. 11

11

(c) Both f cenand

11

are positive.

(d}

11

are negative.

816

Chapter 18

Electrochemistry

Concentration Cells Since cell potential depends not only on the half-reactions occurring in the cell, but also on the concentrations of the reactants and products in those half-reactions, we can construct a voltaic cell in which both half-reactions are the same, but in which a difference in concentration drives the current flow. For example, consider the electrochemical cell shown in Figure 18.13 T , in which copper is oxidized at the anode and copper ions are reduced at the cathode. The second part of Figure 18.13 depicts this cell under nonstandard conditions, with [Cu 2 +] = 2.0 mol L- 1 in one half-cell and [Cu 2 +] = 0.010 mol L- 1 in the other: Cu(s)

+

Cu 2 +(aq, 2.0 mol L- 1) ~ Cu 2 +(aq, 0.010 mol L- 1)

+ Cu(s)

The half-reactions are identical and the standard cell potential is therefore zero.

Cathode:

Cu 2 +(aq)

+ 2 e-

~ Cu(s)

Anode:

Cu 2 +(aq)

+ 2 e-

~ Cu(s)

Foell Foell

E

0

= 0.34 V

E

0

= 0.34 V

= Fcathode - E ; node = 0.34 V - 0.34 V = o.oov

The overall equation can be determined from the cathode and anode reactions, where c and a denote whether the species is in the anode or cathode:

Cathode:

Cu 2 +(aq, c)

[ Standard conditions

~

Cu(s, c)

Cu(s, a) ~ C u2+(aq, a)

Anode: Overall

+ 2--e'"'

Cu2 +(aq, c)

+ Cu(s, a)

~ Cu(s, c)

+

+ 2--e'"'

Cu 2 +(aq, a)

[ N onstandard conditio ns

C u(s) ------> C u 2 +(aq) + 2 eSolution becomes more concentrated with flow of curre nt .

C u 2 + + 2 e- ------> Cu(s) Solution becomes less concentrated with flow of current.

.A. FIGURE 18.13 Cu/Cu2 + Concentration Cell If two half-cells have the same Cu2 + concentration, the cell potential is zero. If one half-cell has a greater Cu2 + concentration than the other, a spontaneous reaction occurs. In the reaction, Cu2 + ions in the more concentrated cell are reduced (to solid copper), while Cu2 + ions in the more dilute cell are formed (from solid copper). The concentration of copper ions in the two half-cells tends toward equality.

18.5 Cell Potential and Concentration

817

Because of the different concentrations in the two half-cells, the cell potential must be calculated using the Nemst equation:

Ecell

0

=

Ecell -

=

Ecell -

0

0.0257 v In Q n 0.0257 V

= O.OOOV =

n

[Cu (aq, a)] In - - - - [Cu (aq, c)]

0.0257 2

v

(0.010) ln - (2.0)

0.068 v

The cell produces a potential of 0.068 V. Electrons spontaneously flow from the ha/feel! with the lower copper ion concentration to the half-cell with the higher copper ion concentration. You can imagine a concentration cell in the same way you think about any concentration gradient. If you mix a concentrated solution of Cu2+ with a dilute solution, the Cu 2 + ions flow from the concentrated solution to the dilute one. Similarly, in a concentration cell, the transfer of electrons from the dilute half-cell results in the forming of Cu2 + ions in the dilute half-cell. The electrons flow to the concentrated cell, where they react with Cu2 + ions and reduce them to Cu(s). Therefore, the flow of electrons has the effect of increasing the concentration of Cu 2 + in the dilute cell and decreasing the concentration of Cu 2 + in the concentrated half-cell.

EXAMPLE 18. 7

DETERMINING THE Ksp OF A SLIGHTLY SOLUBLE SALT USING A CONCENTRATION CELL

A concentration cell is set up as in the diagram to the right. The cathode is a standard Zn2+ (i.e., Zn(N0 3 )i)/Zn electrode), and the anode is composed of a Zn electrode immersed in a saturated solution of Zn(OH) 2 . The cell potential is measured to be 0.169 V. Determine the Ksp for Zn(OH)i.

SORT You are given the cell potential and a concentration cell with an unknown Zn2+ concentration.

GIVEN:

E cell

=

0.169 V

FIND: Ksp (continued)

818

Chapter 18

EXAMPLE 18. 7

Electrochemistry

(CONTINUED)

STRATEGIZE Since it is a concentration cell £~ 011 = 0 V, determine the expression for Q from the balanced equation and calculate the [Zn2+] in the saturated Zn(OH)2 solution using Equation 18.10. Then determine the expression for Ksp in terms of the solubility, S, and calculate K sp·

CONCEPTUAL PLAN

SOLVE Determine the expression for the reaction quotient, Q, in terms of the concentrations of the solutes.

SOLUTION

0.0257 v

E cell =

E~ell- - n - - In Q

Zn 2 +(aq, c)

Cathode:

Zn (s, c) Zn(s, a) ~ Zn2+(aq, a)

Anode: Zn 2 +(aq, c)

Overall

Q= Use the Nernst equation (Equation 18.10) to determine [Zn2+(aq, a)].

E cell

=

E~ell

= E

0.169 V

-13.15

Determine the expression for terms of the solubility, S.

• cell

+ Zn(s, a)

~ Zn(s, c)

Zn2+(aq, a)

[Zn2+ (aq, c)]

v

0.0257 n

-

0.0257 V [Zn2+(aq, a)] In n [Zn2+(aq, c)] -

+

+ 2--.e

[Zn2+(aq, a)]

-

= 0.000 V =

+ 2--.e ~

In Q

0.0257 V 2

In

[Zn2+(aq, a)] [1.0 mol L- 1]

[Zn2+(aq, a)] ln 1.0molL- 1

[Zn2+(aq, a)] 1.0 mol L- 1

=e

[Zn2+(aq, a)]

= 1.94

-1 3 15 ·X 10-6 mol L- 1

Zn(OH)z(s) ~ Zn 2+(aq)

K sp Ill

+ 2 OH- (aq)

All of the OH- comes from the dissolution of Zn(OH)2. Since [Zn2+] = S, from the stoichiometry of the equilibrium [Olr] = 2S. K sp

= [Zn2 +][oH-]2 = (S)(2S)2

K sp

= 4S3 = 4(1 .94 x 10-6 ) 3 = 3 X 10- 17

CHECK Zn(OH)2 is slightly soluble and the value of the

K sp

is small, so the answer seems reasonable.

FOR PRACTICE 18.7 A concentration cell similar to the one in the diagram in Example 18.7 is constructed to measure the concentration of Fe3+. The cathode is a standard Fe3+/Fe electrode and the anode is composed of Fe electrode immersed in a saturated solution of Fe(OHh- The cell potential is measured to be 0.7887 V. Determine the K sp for Fe(OHh-

819

18.6 Batteries: Using Chemistry to Generate Electricity

CHEMISTRY AND MEDICINE Concentration Cells in Human Nerve Cells Recall from Section 8.1 that tiny pumps in the membranes of human nerve cells pump ionsespecially sodium (Na+ ) and potassium (K+)through those membranes establishing a concentration gradient for each ion: The concentration of sodium ions is higher outside the cell than within, while exactly the opposite is true for potassium ions. These concentration gradients result in an electrical potential across the cell membrane, called the resting potential, of about - 70 mV. (The interior of the cell is negative with respect to the exterior.) When the nerve cell is stimulated, certain channels in the membrane open, allowing Na+ ions to rush into the cell and causing the potential to temporarily rise to about +30 mV (Figure 18.14 T ). This is followed by the opening of other channels that allow K+ ions to rush out of the cell, bringing the potential back down to near its resting potential. The

result is a spike in the electrochemical potential across the membrane, which provides the stimulus for a similar spike in the neighbouring segment of the membrane (Figure 18.15 T ). In this way, an electrical signal moves down the length of a nerve cell. When the electrical signal reaches the end of the nerve cell, it triggers the release of a chemical neurotransmitter, which travels to the neighbouring nerve cell and stimulates the same kind of electrochemical spike. In this way, neural signals travel throughout the brain and nervous system of a human being. Sodium channels --- Potassium channels close open Sodium channels open Potassium channels close

Stimulation

}

Extracellular fluid

+30

~ 3 Nerve cell membrane

c

0

0"

0..

"i::

i"'

}

Cell

:'"E - 70 -+---

--'

interior

A FIGURE 18.14 Concentration Changes in Nerve Cells In a nerve cell at rest, the concentration of sodium ions is higher outside the cell than inside. The reverse is true for potassium ions. When a nerve cell is stimulated, sodium channels open and Na+ ions flood into the cell. A fraction of a second later, the sodium channels close and potassium channels open, allowing K+ ions to leave the cell.

0

2

3

4

Time(ms)

A FIGURE 18.15 Potential Changes Across the Nerve Cell Membrane The changes in ion concentrations that take place when a nerve cell is stimulated result in a spike in the electrochemical potential across the membrane.

18.6 Batteries: Using Chemistry to Generate Electricity We have seen that we can combine the electron-losing tendency of one substance with the electron-gaining tendency of another to create electrical current in a voltaic cell. Batteries are simply voltaic cells conveniently packaged to act as portable sources of electricity. The actual oxidation and reduction reactions depend on the particular type of battery. In this section, we examine several different types.

Dry-Cell Batteries Common batteries, such as the type you find in a flashlight, are called dry-cell batteries because they do not contain large amounts of liquid water. The most common dry cells are alkaline batteries (Figure 18.16 T). Alkaline batteries employ a Zn anode in a basic

820

Chapter 18

Electrochemistry

medium (therefore the name alkaline), where it is oxidized according to the following oxidation reaction: ~

Oxidation (Anode):

Zn(s) + 2..-0tr(aq)

Reduction (Cathode):

2 Mn02(s) + 2 H20 (l) +

Zn(OHh(s) +

z.-e

z.-e ~ 2 MnO(OH)(s) + 2..-0tr(aq)

Overall reaction:

Zn(s) + 2 Mn02(s) + 2 H20 (l ) ~ Zn(OHh(s) + 2 MnO(OH)(s)

The two half-reactions produce a voltage of about 1.5 V. Alkaline batteries have a longer working life and a longer shelf life than their nonalkaline counterparts.

Lead-Acid Storage Batteries The batteries in most automobiles are lead-acid storage batteries. These batteries consist of six electrochemical cells wired in series (Figure 18. 17 T). Each cell produces 2 V for a total of 12 V. Each cell contains a porous lead anode where oxidation occurs and a lead(IV) oxide cathode where reduction occurs according to the reactions:

Oxidation (Anode): .&. FIGURE 18.16 Alkaline Batteries In the common dry-cell battery, zinc acts as the anode and a graphite rod immersed in a paste of Mn02 and a base.

Reduction (Cathode):

Pb02(s) + HS04- (aq) + 2

m + (aq) +

2-e

~

PbSOis) + 2 H 20 (l)

Overall reaction:

[Dorling Kindersley, Ltd.)

Pb(s) + Pb0 2(s) + 2 HS04- (aq) + 2 H+(aq) ~ 2 PbS0 4(s) + 2 H20 (l)

Both the anode and the cathode are immersed in aqueous sulfuric acid (H2S04 (aq)). As electrical current is drawn from the battery, both electrodes become coated with PbSOis). If the battery is run for a long time without recharging, too much PbS04(s) develops on the surface of the electrodes and the battery goes dead. The lead-acid storage battery can be recharged by an electrical current (which must come from an external source such as an alternator in a car). The current causes the preceding reaction to occur in reverse, converting the PbSOis) back to Pb(s) and Pb02(s).

Other Rechargeable Batteries The ubiquity of power electronic products such as laptops, cell phones, and digital cameras, as well as the growth in popularity of hybrid electric vehicles, has driven the need for efficient, long-lasting, rechargeable batteries. The most common types include the nickel-cadmium (NiCad) battery, the nickel-metal hydride (NiMH) battery, and the lithium-ion battery. The Nickel-Cadmium (NiCad) Battery Nickel-cadmium batteries consist of an anode

composed of solid cadmium and a cathode composed of NiO(OH)(s). The electrolyte

..... FIGURE 18.17 Lead-Acid Storage Battery A lead-acid storage battery consists of six cells wired in series. Each cell contains a porous lead anode and a lead oxide cathode, both immersed in sulfuric acid.

Anode(-) : Lead grid packed with finely divided spongy lead

Electrolyte: 30% solution ofH2S04

Cathode(+): Lead grid packed with Pb0 2

18.6 Batteries: Using Chemistry to Generate Electricity

821

is usually KOH(aq). During operation, the cadmium is oxidized and the NiO(OH) is reduced according to the equations: Oxidation (Anode):

Cd(s)

Reduction (Cathode):

+

2 OH- (aq)

2 NiO(OH)(s)

+

~

Cd(OH)i(s)

2 H20(l)

+ 2 e-

+

2 e-

~

2 Ni(OH)i(s)

+

2 OH- (aq)

The overall reaction produces about 1.30 V. As current is drawn from the NiCad battery, solid cadmium hydroxide accumulates on the anode and solid nickel(II) hydroxide accumulates on the cathode. But by running current in the opposite direction, the reactants can be regenerated from the products. A common problem in recharging NiCad and other rechargeable batteries is knowing when to stop. Once all of the products of the reaction are converted back to reactants, the charging process should ideally terminate-otherwise the electrical current will drive other, usually unwanted, reactions such as the electrolysis of water to form hydrogen and oxygen gas. These reactions typically damage the battery and may sometimes even cause an explosion. Consequently, most commercial battery chargers have sensors designed to measure when the charging is complete. These sensors rely on the small changes in voltage or increases in temperature that occur once the products have all been converted back to reactants.

The Nickel-Metal Hydride (NiMH) Battery Although NiCad batteries were the standard rechargeable battery for many years, they are being replaced by others, in part because of the toxicity of cadmium and the resulting disposal problems. One of these replacements is the nickel-metal hydride or NiMH battery. The NiMH battery employs the same cathode reaction as the NiCad battery but a different anode reaction. In the anode of a NiMH battery, hydrogen atoms held in a metal alloy are oxidized. If we let M represent the metal alloy, we can write the half-reactions as follows: Oxidation (Anode):

MH(s)

+

OH-(aq)

Reduction (Cathode): NiO(OH)(s)

+

~

H 20(l)

M(s)

+

e-

+

H2 0(l)

~

+

e-

Ni(OHh(s)

+

OH- (aq)

In addition to being more environmentally friendly than NiCad batteries, NiMH batteries also have a greater energy density (energy content per unit battery mass), as we can see in Table 18.2. In some cases, a NiMH battery can carry twice the energy of a NiCad battery of the same mass, making NiMH batteries the most common choice for hybrid electric vehicles.

The Lithium-Ion Battery The most common type of rechargeable battery is the lithiumion battery. Since lithium is the least dense metal (0.53 g cm- 3) , lithium batteries have high energy densities (see Table 18.2). The lithium battery works differently than the other batteries we have examined so far, and the details of its operation are beyond the scope of our current discussion. Briefly, you can think of the operation of the lithium battery as being due primarily to the motion of lithium ions from the anode to the cathode. The anode is composed of graphite into which lithium ions are incorporated between layers of carbon atoms. Upon discharge, the lithium ions spontaneously migrate to the cathode, which consists of a lithium transition-metal oxide such as LiCo0 2 or LiMn2 0 4 . The transition metal is reduced during this process. Upon recharging, the transition metal is oxidized, forcing the lithium to migrate back into the graphite (Figure 18.18 T). The flow of lithium ions from the anode to the cathode causes a corresponding flow of electrons in the external circuit. Lithium-ion batteries are commonly used in applications where light TABLE 18.2 Energy Density and Overcharge Tolerance of Several

Rechargeable Batteries Battery Type

Energy Density (W h kg - 1)

Overcharge Tolerance

NiCad

45-80

Moderate

NiMH

60-120

Low

Li ion

110-160

Low

30-50

High

Pb storage

.A. Several types of batteries, including NiCad, NiMH, and lithium-ion batteries, are recharged by chargers that use household current. [COV, LLC/ Pearson Science]

822

Chapter 18

Elect ro chemistry

weight and high energy density are important. These include cell phones, laptop computers, and digital cameras.

Fuel Cells

tUiflllh •

~Q

~

•

W1i11i1!Q~

We discussed the potential for fuel cells in the opening section of this chapter. Fuel cells may one day replace-or at least work in combination with-centralized power grid electricity. In addition, electric vehicles powered by fuel cells may one day usurp vehicles powered by internal combustion engines. Fuel cells are like batteries; the key difference is that a battery is self-contained, while in a fuel cell the reactants need to be constantly replenished from an external source. With use, normal batteries lose their ability to generate voltage because the reactants become depleted as electrical current is drawn from the battery. In a fuel cell, the reactants-the fuel provided from an external source-constantly flow through the battery, generating electrical current as they undergo a redox reaction . The most common fuel cell is the hydrogen-oxygen fuel cell (Figure 18.19 T). In this cell, hydrogen gas flows past the anode (a screen coated with platinum catalyst) and undergoes oxidation:

QQ

Oxidation (Anode) : 2 H2 (g)

Q

Graphite

+ 4 OW(aq)

~

4 H2 0(/)

+ 4 e-

Oxygen gas flows past the cathode (a similar screen) and undergoes reduction:

Lithium transitionmetal oxide

Lithium ions Charge

.&. FIGURE 18.18 Lithium-Ion Battery In the lithium-ion battery, the spontaneous flow of lithium ions from the graphite anode to the lithium-transition metal oxide cathode causes a corresponding flow of electrons in the external circuit.

The half-reactions sum to the following overall reaction:

Discharge

Notice that the only product is water. In the space shuttle program, hydrogen-oxygen fuel cells consume hydrogen to provide electricity and astronauts drink the water that is produced by the reaction. In order for hydrogen-powered fuel cells to become more widely used, a more readily available source of hydrogen must be developed. •

Electron

Electric circuit

Proton •

Oxygen atom Anode

Cathode

Electrolyte

Oxidation

.,... FIGURE 18.19 Hydrogen-Oxygen Fuel Cell In this fuel cell, hydrogen and oxygen combine to form water.

Reduction

2 H 2(g) + 4 Off(aq) -->

4 H 2 0(1)

+

4 e-

Oz(g) + 2 H20(I) + 4 e->

40ff(aq)

18. 7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity

CHEMISTRY IN YOUR DAY

823

Rechargeable Battery Recycling

~·

Have you ever dropped off old rechargeable batteries or an old car battery at a recycling depot? Have you wondered what happens to them after you do? Automotive lead-acid storage batteries are perhaps the oldest and most widely used type of rechargeable battery. They are relatively easy to recycle: The plastic battery case is broken open and the sulfuric acid electrolyte is drained. The lead plates and connectors can be removed whole. The lead is melted and purified, and then it is used to make new batteries. More than 99% of the lead metal in lead-acid storages batteries is recycled in this way. This is important because lead is a toxic heavy metal that should be kept out of landfills. Compared to lead-acid storage batteries, NiCad, NiMH, and lithium-ion batteries are more difficult to dismantle. Often, these Pressed powdered batteries consist of very thin sheets of anode, cathode, and separator negative electrode materials that are rolled tightly together and immersed in an electrolyte. This is known as a "jelly roll" construction. Furthermore, these batteries have a large number of distinct materials, which makes Sintered positive recycling more challenging. For example, a typical lithium-ion bat- electrode tery has LiCo02 cathode material on aluminum foil, graphite anode material on copper foil, a LiPF6 electrolyte in an organic solvent, polyethylene plastic to separate the electrodes, a steel case, and "' Jelly roll construction of a rechargeable battery. other smaller parts such as an electronic switch that prevents over[Based on 2007 Encyclopedia Britannica. https://www.britannica.com/technology/ charging. Dismantling the batteries is neither practical nor economi- battery-electronics/images-videos] cal, so other approaches are used to recycle the battery materials. Some commercial recyclers recover the metals in rechargeable batteries through metallurgical processes (see Chapter 24): Organic components such as plastics can be removed through high-temperature combustion, which also converts metals into metal oxides. After this process, metals can be extracted and separated as individual metal salts through hydrometallurgical techniques, which involve leaching the metals with acid and then precipitating the metal salts. Valuable cobalt, copper, and nickel can be recovered in this way. Lithium, which has less value, is not recovered in this particular process. Retriev Technologies in Trail, British Columbia, is an example of a commercial recycler that recovers some of the lithium from batteries. Its process includes mechanically shredding the batteries into small pieces. The shredded battery materials are then "' Technician sorting rechargeable batteries for recycling. immersed in water, which dissolves Li+ ions. The water also reacts [Wolfgang Rattay/REUTERS] with lithium metal to produce LiOH(aq) and H 2 (g), which is burned. Shredded plastic and metal pieces are placed on a shaker table, which separates materials on the basis of density (lighter plastic pieces go to the top and heavier metal pieces end up on the bottom). Valuable metals are separated. To recover some of the lithium that was dissolved in water, sodium carbonate is added. This precipitates Li 2C0 3(s), which is filtered off. Lithium carbonate has a number of industrial uses, including the manufacture of new lithium-ion batteries.

Question Write a chemical equation for the precipitation of Li2 C03 from a Li+(aq) solution to which sodium carbonate is added. What drives the precipitation of Li 2 C03 ?

18.7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity In a voltaic cell, a spontaneous redox reaction produces electrical current. In an electrolytic cell, electrical current drives an otherwise nonspontaneous redox reaction through a process called electrolysis. We have seen that the reaction of hydrogen with oxygen to form water is spontaneous and can be used to produce an electrical current in a fuel cell.

824

Chapter 18

Electrochemistry

.,... FIGURE 18.20 Electrolysis of Water Electrical current can decompose water into hydrogen and oxygen gas. Oxygen gas

Hydrogen gas

Water with soluble salt

Oxygen bubbles

~

%

Hydrogen bubbles

Anode

Cathode

2H 20(/) ----> Oz(g) + 4H+ + 4 e-

2H2 0(/) + 2e- ----> Hz(g) + 2 Oir(aq}

External source

By supplying electrical current, we can cause the reverse reaction to occur, separating water into hydrogen and oxygen (Figure 18.20 A).

(spontaneous-produces electrical current; occurs in a voltaic cell) (nonspontaneous-consumes electrical current; occurs in an electrolytic cell)

+

Silver electrode

Anode Ag(s} ----> Ag +(aq) + e-

Object ro be plated

Cathode Ag +(aq} + e----> Ag(s)

.&. FIGURE 18.21 Silver Plating Silver can be plated from a solution of silver ions onto metallic objects in an electrolytic cell.

Recall from the last section that one of the problems associated with the widespread adoption of hydrogen fuel cells is the scarcity of hydrogen. Where will the hydrogen to power these fuel cells come from? One possible answer is to obtain hydrogen from water through solar-powered electrolysis. A solarpowered electrolytic cell can produce hydrogen from water when the sun is shining. The hydrogen made in this way can be converted back to water to generate electricity and could also be used to power fuel-cell vehicles. Electrolysis also has numerous other applications. For example, most metals are found in Earth's crust as metal oxides. Converting an oxide to a pure metal requires that the cationic form of the metal be reduced, a nonspontaneous process. Electrolysis can be used to produce these metals. Thus, sodium is produced by the electrolysis of molten sodium chloride (discussed in the following subsection). Electrolysis is also used to plate metals onto other metals. For example, silver can be plated onto a less expensive metal using the electrolytic cell shown in Figure 18.21 .... In this cell, a silver electrode is placed in a solution containing silver ions. An electrical current causes the oxidation of silver at the anode (replenishing the silver ions in solution) and the reduction of silver ions at the cathode (coating the less expensive metal with solid silver) . Oxidation (Anode): Ag(s) ~ Ag + (aq) Reduction (Cathode): Ag + (aq)

+

+ e-

e- ~ Ag(s)

Since the standard cell potential of the reaction is zero, the reaction is not spontaneous under standard conditions. An external power source can be used to drive current flow and cause the reaction to occur.

18. 7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity

Voltaic Cell

Electrolytic Cell

Zn 2+(aq} Zn(s}

Zn(s)

Cu(s}

Cu2+(aq)

-

2

zn2+(aq) + 2e-

Cu +(aq)

+

2 e-

zn2+(aq)

Cu(s}

+

2 e-

Zn(s}

Cu(s} Cu2+(aq)

+

2 e-

.&. FIGURE 18.22 Voltaic Versus Electrolytic Cells In a Zn/ Cu2+ voltaic cell, the reaction proceeds in the spontaneous direction. In a zn2+ / Cu electrolytic cell, electrical current drives the reaction in the nonspontaneous direction. The voltage required to cause electrolysis depends on the specific half-reactions. For example, we have seen that the oxidation of zinc and the reduction of Cu2 + produces a voltage of 1.10 V under standard conditions.

(Cathode): Cu2 +(aq) (Anode) : Zn2 +(aq)

+ 2 e- ~ Cu(s) + 2 e- ~ Zn(s)

E

0

E

0

= 0.34 V = -0.76 V

E~ell = E~athode - E ; node E~eu

= 0.34 V - ( -0.76 V) = 1.10 v

If a power source producing more than I.JO Vis inserted into the Zn/ Cu2 + voltaic cell, electrons can be forced to flow in the opposite direction, causing the reduction of zn2+ and the oxidation of Cu, as shown in Figure 18.22 .... Notice in Figure 18.22 that Zn is the anode in the voltaic cell and the cathode in the electrolytic cell. Recall that the anode and cathode are defined in terms of the reaction that occurs at their surface (oxidation always occurs at the anode). In a voltaic cell, the anode is the source of electrons and is therefore labelled with a negative charge. The cathode draws electrons and is therefore labelled with a positive charge. In an electrolytic cell, however, the source of the electrons is the external power source. The external power source must draw electrons away from the anode; thus, the anode must be connected to the positive terminal of the battery (as shown in Figure 18.22). Similarly, the power source drives electrons toward the cathode (where they will be used in reduction), so the cathode must be connected to the negative terminal of the battery. The charge labels (+ and-) on an electrolytic cell are therefore opposite of what they are in a voltaic cell.

Summarizing Characteristics of Electrochemical Cell Types: In all electrochemical cells: ~

Oxidation occurs at the anode.

~

Reduction occurs at the cathode.

In voltaic cells: ~

The anode is the source of electrons and has a negative charge (anode -).

~

The cathode draws electrons and has a positive charge (cathode

+ ).

825

826

Chapter 18

Electrochemistry

In electrolytic cells: .,.. Electrons are drawn away from the anode, which must be connected to the positive terminal of the external power source (anode +). .,.. Electrons are forced to the cathode, which must be connected to the negative terminal of the power source (cathode - ). e

~

Predicting the Products of Electrolysis

e

=n

+

J

Inert electrode

Ine rt electrode

Molten

NaCl

Predicting the products of an electrolysis reaction is in some cases relatively straightforward and in other cases more complex. We cover the simpler cases first and follow with the more complex ones. Pure Molten Salts Consider the electrolysis of a molten salt such as sodium chloride, shown in Figure 18.23

..&. FIGURE 18.23 Electrolysis of Molten NaCl In the electrolysis of a pure molten salt, the anion (in this case Cl-) is oxidized and the cation (in this case Na+) is reduced.

Throughout this discussion, more positive

I means the same thing as less negative.

+

2 e2 Na(s)

Although the reaction as written is not spontaneous, it can be driven to occur in an electrolytic cell by an external power source. We can generalize as follows: .,.. In the electrolysis of a pure molten salt, the anion is oxidized and the cation is reduced.

Mixtures of Cations or Anions What if a molten salt contains more than one anion or cation? For example, suppose our electrolysis cell contained both NaCl and KCL Which of the two cations would be reduced at the cathode? In order to answer this question, we must ask which of the two cations is more easily reduced. Although the values of electrode potentials for aqueous solutions given in Table 18.1 do not apply to molten salts , the relative ordering of the electrode potentials does reflect the relative ease with which the metal cations are reduced . We can see from the table that the reduction of Na+ is listed above the reduction of K+; that is, Na+ has a more positive electrode potential. Na+(aq) K+(aq)

+ e+ e-

~ Na(s)

E

0

= -2.71 V (for aqueous solution)

~

E'

= -2.92 V (for aqueous solution)

K(s)

Therefore, Na+ is easier to reduce than K+. Consequently, in a mixture of NaCl and KC!, Na+ will have a greater tendency to be reduced at the cathode. Similarly, what if a mixture of molten salts contained more than one anion? For example, in a mixture of NaBr and NaCl, which of the two anions would be oxidized at the cathode? The answer is similar: the anion that is more easily oxidized (the one with the more negative electrode potential). Throughout this discussion, more negative

I means the same thing as less positive.

Cl 2(g) Br2(l)

+ 2 e+ 2 e-

~

2 Ci-(aq)

E

= 1.36 V

~

2 Br- (aq)

E'

= 1.09 v

Since the standard electrode potential for the bromine half-reaction is more negative, the reverse reaction will proceed more easily. Therefore, electrons are more easily extracted from the bromide rather than the chloride. We can generalize as follows: .,.. The cation that is most easily reduced (the one with the more positive electrode potential) is reduced first. .,.. The anion that is most easily oxidized (the one with the more negative electrode potential) is oxidized first.

18. 7 Electrolysis: Driving Nonspontaneous Chemical Reactions with Electricity

827

Aqueous Solutions Electrolysis in an aqueous solution is complicated by the possibility of the electrolysis of water itself. Recall that water can be either oxidized or reduced according to the following half-reactions:

E° = 1.23 v E = 0.82 V (pH = 7) Cathode: 2 H 20 (l)

+ 2 e-

-----+

H 2 (g)

E° = -0.83 v = -0.41 V (pH = 7)

+ 2 OW(aq)

E

The electrode potentials under standard conditions are shown to the right of each half-reaction. However, in pure water at room temperature, the concentrations of H+ and OH- are not standard. The electrode potentials for [H+] = 10-7 mol L- 1 and [OH- ] = 10- 7 mol L- 1 (at pH = 7) are shown in blue. Using those electrode potentials, we can calculate Ecell for the electrolysis of water as follows: Ecell

= Ecat - Ean = - 0.41 V - 0.82 V = - 1.23 V

When a battery with a potential of several volts is connected to an electrolysis cell containing pure water, no reaction occurs because the concentration of ions in pure water is too low to conduct any significant electrical current. When an electrolyte such as Na2 S0 4 is added to the water, however, electrolysis occurs readily. In any aqueous solution in which electrolysis is to take place, the electrolysis of water at either electrode or at both electrodes is also possible. For example, consider the electrolysis of a sodium iodide solution, as shown in Figure 18.24 T . For the electrolysis of molten Nal, we can readily predict that i- is oxidized at the anode and that Na+ is reduced at the cathode. In an aqueous solution, however, two different reactions are possible at each electrode. At the cathode, it is possible to reduce either Na+ or H20, according to the following:

2 H20(/) + 2 e- -----+ Hz(g) Na+(aq) + e------+ Na(s)

+ 2 OH- (aq)

E E

0

= -0.41 V (pH = = -2.71 V

7)

The reaction that occurs at the cathode is the one that has the most positive electrode potential. The reaction with the most positive electrode potential is the reduction of water; therefore, water will be reduced at the cathode, not sodium. At the anode, it is possible to oxidize either H20 or r, according to the reverse of the following: 0 2(g)

+ 4 H +(aq) + 4 eI 2(aq) + 2 e-

-----+ -----+

= 0.82 V (pH = E° = 0.54 v

2 H20 (/) 2 r-(aq)

E

.A. Pure water is a poor conductor of electrical current, but the addition of an electrolyte allows electrolysis to take place, producing hydrogen and oxygen gas in a stoichiometric ratio. [Charles D. Winters/Science Source]

7)

The reaction that will occur at the anode is the one with the most negative standard electrode potential (because that will be the easiest to run in reverse). In this case, 1 will be oxidized to 12 at the anode. The convention is to write the half-reactions as reductions and e

e

=n

---+ + J

Inert electrode

Inert electrode

"t Hz(g) Aqueous :

Nal

-

I .,_ I (aq) I H20 -+ I Anode I Cathode 2

q aq) I 2(s)

-----'>

+

2 e-

2 H 2 0(/) H 2 (g)

+ +

2 e-

-----'>

2 Oir(aq)

Radium-226 (atomic mass

= 226.025402 u) decays to radon3

222 (a radioactive gas) with a half-life of 1.6 X 10 years. Assuming that radon gas does not decay, what volume of radon gas (at 25.0 °C and 1.0 bar) does 25.0 g of radium produce in 5.0 days? (Report your answer to two significant digits.)

84. In one of the neutron-induced fission reactions of U-235 (atomic mass = 235.043922 u), the products are Ba-140 and Kr-93 (a radioactive gas). What volume of Kr-93 (at 25.0 °C and 1.0 bar) is produced when 1.00 g of U-235 undergoes this fission reaction?

~ When a positron and an electron annihilate one another, the resulting mass is completely converted to energy. Calculate the energy associated with this process in MeV.

86. A typical nuclear reactor produces about 1.0 MW of power per day. What is the minimum rate of mass loss required to produce this much energy?

f)

Find the binding energy in an atom of 3He, which has a mass of 3.016030 u.

88. The overall hydrogen burning reaction in stars can be represented as the conversion of four protons to one a particle. Use the data for the mass of H- 1 and He-4 to calculate the energy released by this process.

~ The nuclide 247Es can be made by bombardment of 238 U in a reaction that emits five neutrons. Identify the bombarding particle. 90. The nuclide 6Li reacts with 2H to form two identical particles. Identify the particles.

Q

The half-life of 238U is 4.5 X 109 y. A sample of rock of mass 1.6 g produces 29 Bq. Assuming all the radioactivity is due to 238 U, fi nd the percent by mass of 238U in the rock.

92. The half-life of 232Th is 1.4 X 10 10 y. Find the number of disintegrations emitted by 1.0 mol of 23 2Th in l minute.

e

A 1.50 L gas sample at 0.980 bar and 25.0 °C contains 3.55 % radon-220 by volume. Radon-220 is an alpha emitter with a half-life of 55.6 s. How many alpha particles are emitted by the gas sample in 5.00 minutes?

94. A 228 mL sample of an aqueous solution contains 2.35% MgCI 2 by mass. Exactly one-half of the magnesium ions are Mg-28, a beta emitter with a half-life of 21 hours. What is the decay rate of Mg-28 in the solution after 4.00 days? (Assume a density of 1.02 g mL- 1 for the solution.)

Exercises

e

When a positron and an electron collide and annihilate each other, two photons of equal energy are produced. Find the wavele ngth of these photons. 96. The half-life of 235U, an alpha emitter, is 7 .1 X 108 y. Calculate the number of alpha particles e mitted by 1.0 mg of this nuclide in 1.0 minute.

877

process occurs at muc h lower temperatures than e ithe r of the first two. 98. The nuclide 18F decays by both electron capture and 13+ decay. Find the differe nce in the e nergy released by these two processes. The atomic masses are 18F = 18.000950 u and 180 = 17 .999 1598 u.

G Given that the e nergy released in the fu sion of two deuterons to 3

a He and a neutron is 3.3 MeV and in the fusion to tritium and a proton it is 4.0 MeV, calculate the energy cha nge fo r the process 3He + 1n ------> 3H + 1p. Suggest an explanation for why this

Challenge Problems G)

Before the retirement of the space shuttle in 2011, it carried about 72 500 kg of solid aluminum fuel, which is oxidized with ammonium perchlorate according to the reaction: 10 A l(s)

+

6 NH 4Cl04 (s) ------> 4 A l20 3(s) + 2 AlCl3(s)

+

12 H 20(g)

+

8

3 N 2(g)

The space shuttle also carrie d about 608 000 kg of oxygen (which reacts with hydrogen to form gaseous wate r). a. Assum.ing that aluminum and oxygen are the limiting reactants, determine the total ene rgy produced by these fue ls. (l!..rH for solid ammonium perchlorate is -295 kJ mol- 1.) b. Suppose that a future s pace shuttle is powered by matterantimatter annihilation. The matter could be normal hydrogen (containing a proton and an electron) and the antimatter could be antihydrogen (containing an antiproton and a positron). What mass of antimatter is required to produce the energy equivalent of the aluminum and oxygen fuel previously carried on the space shuttle? 0

BF3 contained in a 3.0 L container at 298 K is bombarded with neutrons until half of the BF3 has reacted. What is the pressure in the container at 298 K ? In addition to the natu ral radioactive decay series that begins w ith U-238 and e nds with Pb-206, there are natural radioactive decay series that begin with U -235 and Th-232. Both of these series end with nuclides of Pb. Predict the likely end product of each series and the number of a decay steps that occur.

102. The hydride of an unstable nuclide of a group 2 metal, MH 2(s), decays by a-emission. A 0.025 mol sample of the hydride is placed in an evacuated 2.0 L container a t 298 K. After 82 minutes, the pressure in the container is 0.55 atm. Find the half-life of the nuclide. The nuclide 38Cl decays by beta emission with a half -life of 40 .0 m.inutes. A sample of 0.40 mol of H 38Cl is placed in a

@

6.24 L container. After 80.0 minutes, the pressure is 1650 mmHg. What is the temperature of the container?

100. When BF3 is bombarde d with neutrons, the boron undergoes an alpha decay, but the F is unaffected. A 0.20 mol sample of

Conceptual Problems 104. Approximately how many half-lives must pass for the amount of radioactivity in a substance to decrease to below 1% of its initial level?

fJD Closely examine the diagram representing the beta decay of fluorine-2 1 and draw in the missing nuc leus.

2JF

106. Identical amounts of two different nuclides, an alpha emitter and a gamma e mitter, w ith roughly equal half-lives are spilled in

a building adjacent to your bedroom. Which of the two nuclides poses the greater health threat to you while you sleep in your bed? If you accidentally wander into the building and ingest equal amounts of the two nuc lides, which one poses the greater health threat?

G A person is exposed for three days to identical am ounts of two different nuclides that em.it positrons of roughl y equal energy. The half-life of nuclide A is 18.5 days and the half-life of nuclide B is 255 days. Which of the two nuclides poses the greater health risk?

20.1 20.2 20.3 20.4 20.5

Fragrances and Odours 879 Carbon: Why It Is Unique 879 Hydrocarbons 881 Functional Groups 887 Constitutional Isomerism 893

20.6 Stereoisomerism I: Conformational Isomerism 894 20.7 Stereoisomerism II: Configurational Isomerism 897 20.8 Structure Determination 905

About half of all men 's colognes contain at least some patchouli alcohol (C 15 H 260), an organic compound (pictured here) derived from the patchouli plant. Patchouli alcohol has a pungent, musty, earthy fragrance. (Quade Paul/Pearson Education]

0

RGANIC CHEMISTRY IS THE STUDY of carbon-containing compounds. Since life has organized itself around organic compounds, organic chemistry is critical to the study of life. Carbon is unique in the sheer

number of compounds that it forms. Millions of organic compounds are known, and researchers discover new ones every day. Carbon is also unique in the diversity of compounds that it forms. In most cases, a fixed number of carbon atoms can combine with a fixed number of atoms of another element to form many different compounds. For example, 10 carbon atoms and 22 hydrogen atoms can form 75 distinctly different compounds. With carbon as the backbone, nature can take the same combination of atoms and bond them together in slightly different ways to produce a huge diversity of substances. It is not surprising that life is based on the chemistry of carbon because life needs diversity to exist, and organic chemistry is nothing if not diverse. In thi s chapter, we peer into Friedrich Wohler's "primeval tropical forest" (see chapter-opening quotation) and discover the most remarkable things .

878

20.2

20.1

Carbon: Why It Is Unique

879

Fragrances and Odours

Have you ever ridden an elevator with someone wearing too much perfume? Or found yourself too close to a skunk? Or been overwhelmed by a whiff of rotting fish? What causes these fragrances and odours? When we inhale certain molecules called odourants, they bind with olfactory receptors in our noses. This interaction sends a nerve signal to the brain that we experience as a smell. Some smells, such as that of perfume, are pleasant (when not overdone). Other smells, such as that of the skunk or rotting fish , are unpleasant. Our sense of smell helps us identify food, people, and other organisms, and alerts us to dangers such as polluted air or spoiled food. Smell (olfaction) is one way we probe the environment around us. Odourants, if they are to reach our noses, must be volatile. However, many volatile substances have no scent at all. Nitrogen, oxygen, water, and carbon dioxide molecules, for example, are constantly passing through our noses, yet they produce no smell because they do not bind to olfactory receptors. Most common smells are caused by organic molecules, molecules containing carbon combined with several other elements, such as hydrogen, nitrogen, oxygen, and sulfur. Organic molecules are responsible for the smells of roses, vanilla, cinnamon, almond, jasmine, body odour, and rotting fish. When you wander into a rose garden, you experience the sweet smell caused in part by geraniol, an organic compound emitted by roses. Men's colognes often contain patchouli alcohol, an earthy-smelling organic compound extracted from the patchouli plant. If you have been in the vicinity of skunk spray (or have been unfortunate enough to be sprayed yourself), you are familiar with the unpleasant smell of but-2-en-l -thiol and 3-methylbutan-1 -thiol, two particularly odouriferous compounds present in the secretion that skunks use to defend themselves. The study of compounds containing carbon combined with one or more of the elements mentioned previously (hydrogen, nitrogen , oxygen, and sulfur), including their properties and their reactions, is known as organic chemistry. Besides composing much of what we smell, organic compounds are prevalent in foods, drugs, petroleum products, and pesticides. Organic chemistry is also the basis for living organisms. Life has evolved based on carbon-containing compounds, making organic chemistry of utmost importance to any person interested in understanding living organisms.

20.2 Carbon: Why It Is Unique Why did life evolve based on the chemistry of carbon? Why is life not based on some other element? The answer may not be simple, but we know that life-in order to existmust entail complexity, and carbon chemistry is certainly complex. The number of compounds containing carbon is greater than the number of compounds containing all of the other elements combined. The reasons for carbon's unique and versatile behaviour include its ability to form four covalent bonds, its ability to form double and triple bonds, and its tendency to catenate (that is, to form chains).

Carbon's Tendency to Form Four Covalent Bonds Carbon-with its four valence electrons- forms four covalent bonds. Consider the Lewis structure and space-filling models of two simple carbon compounds, methane and ethane. H

H

H

H

H

I H-C-H I

I I H-C-C-H I I

Methane

Ethane

H

The geometry about a carbon atom forming four single bonds is tetrahedral, as shown above for methane. Carbon's ability to form four bonds, and to form those bonds with a number of different elements, results in the potential to form many different compounds. As you learn to draw structures for organic compounds, always remember to draw carbon with four bonds.

Carbon's Ability to Form Double and Triple Bonds Carbon atoms also form double bonds (trigonal planar geometry) and triple bonds (linear geometry), adding even more diversity to the number of compounds that carbon forms.

CH3CH= CHCH2SH But-2-en-1-thiol

CH3

I

CH3CHCH2CH2SH 3-Methylbutan-1-thiol

..i. The smell of skunk is caused primarily by the molecules shown here. [CDV, LLC/Pearson Education]

880

Chapter 20

Organic Chemistry I: Structures

H

"C=C/

H

"

H/

H-c==c-H

H

Ethyne

Ethene

In contrast, silicon (the element in the periodic table with properties closest to that of carbon) does not readily form double or triple bonds because the greater size of silicon atoms results in a Si - Si bond that is too long for much overlap between nonhybridized p orbitals.

Carbon's Tendency to Catenate Carbon, more than any other element, can bond to itself to form chain, branched, and ring structures.

H

H

H

I I I H-C-C-C-H I I I H

H

H

H

I

H

I

H

I

H-C-C-C-H

~

I

~

H-C-H

I

H Propane

Cyclohexane

2-Methylpropane

Vitalism and the Perceived Difference Between Organic and Inorganic By the end of the l8th century, chemists had learned that compounds could be broadly categorized as either organic or inorganic. It was believed that organic compounds came from living things while inorganic compounds came from the nonliving things on Earth. Sugar---obtained from sugarcane or the sugar beet-is a common example of an organic compound. Salt-mined from the ground or extracted from ocean wateris a common example of an inorganic compound. Organic and inorganic compounds are different, not only in their origin, but also in their properties. Organic compounds are easily decomposed. Sugar, for example, readily decomposes into carbon and water when heated. (Think of the last time while cooking you burned something sugary in the pan or in the oven.) Inorganic compounds are more difficult to decompose. Salt decomposes only when heated to very high temperatures. Even more curious to these early chemists was their inability to synthesize a single organic compound in the laboratory. Although they were able to synthesize many inorganic compounds, despite concerted efforts, they were not able to synthesize any organic compounds. The origin and properties of organic compounds led early chemists to postulate that organic compounds were unique to living organisms. They hypothesized that living organisms contained a vital force-a mystical or supernatural power- that allowed them to produce organic compounds. They thought that producing an organic compound outside of a living organism was impossible because the vital force was not present. This belief- which became known as vitalismexplained why no chemist had succeeded in synthesizing an organic compound in the laboratory.

An experiment performed in 1828 by German chemist Friedrich Wohler ( 1800-1882) marked the beginning of the end of vitalism. Wohler heated ammonium cyanate (an inorganic compound) and formed urea (an organic compound). NH4 0CN Ammonium cyanate

heat

~

H 2NCONH2 Urea

Urea was a known organic compound that had previously been isolated only from urine. Although it was not realized at the time, Wohler's simple experiment was a key step in opening all of life to scientific investigation. He showed that the compounds composing living organisms-like all compoundsfollow scientific laws and can be studied and understood. Today, known organic compounds number in the millions, and modern organic chemistry is a vast field that produces substances as diverse as drugs, petroleum products, and plastics and many other compounds that are not Friedrich Wohler found in nature. Ill> The synthesis of urea in 1828 by German chemist Friedrich Wohler marked the beginning of the end for vitalism.

Urea

20.3

Hydrocarbons

Although other elements can form chains, none beats carbon at this ability. Silicon, for example, can form chains with itself. However, silicon's affinity for oxygen (the Si-0 bond is 142 kJ mol- 1 stronger than the Si - Si bond) coupled with the prevalence of oxygen in our atmosphere means that silicon-silicon chains are readily oxidized to form silicates (the silicon-oxygen compounds that compose a significant proportion of minerals). By contrast, the C-C bond (347 kJ mol - 1) and the C-0 bond (359 kJ mol - 1) are nearly the same strength, allowing carbon chains to exist relatively peacefully in an oxygen-rich environment. Silicon's affinity for oxygen robs it of the rich diversity that catenation provides to carbon.

20.3 Hydrocarbons Hydrocarbons are compounds that contain only carbon and hydrogen. They are the simplest type of organic compounds and are commonly used as fuel s. Candle wax, oil, gasoline, and natural gas are all composed of hydrocarbons. Hydrocarbons provide the structural backbone for all organic compounds, which makes them good starting materials in the synthesis of many different products such as dyes, pharmaceuticals, plastics, and rubber.

Drawing Hydrocarbon Structures Before we examine the different types of hydrocarbons, it is useful to discuss how we usually draw their structures. We will use these representations throughout the rest of this chapter and the next. H

I

H

H

H

H

H H- C -H H

I I

I I

I I

I I

H-C--C--C-H

H-C-C-C-C-H H

H

H

I I

H

H

I I

H

I I

H

2-Methylpropane

Butane

Butane and isobutane are structural isomers, molecules with the same molecular formula but different structures. Because of their different structures, they have different properties-they are indeed different compounds. Isomerism is ubiquitous in organic chemistry. Butane has two structural isomers. Pentane (C5H 12) has three, hexane (C6 H 14) has five, and decane (C 10H 22) has 75! The structure of a particular hydrocarbon is represented with a structural formula, a formula that shows not only the numbers of each kind of atoms, but also how the atoms are bonded together. Organic chemists use several different kinds of structural formulas. For example, we can represent butane and 2-methylpropane in each of the following ways:

Structural formula H

Butane

H

H

Condensed structural formula H

I I I I H-C-C-C-C-H I I I I H

H

H

H

H

I

H H-C-H H 2-Methylpropane

I I

I I

I I

H-C--C--C-H H

H

Carbon skeleton formula

H

CH3 -

CH2 -CH 2 -CH3

/'-.../

Ball-andstick model

Space-filling model

881

882

Chapter 20

Organic Chemistry I: Structures

The structural formula shows all of the carbon and hydrogen atoms in the molecule and how they are bonded together. The condensed structural formula groups the hydrogen atoms with the carbon atom to which they are bonded. Condensed structural formulas may show some of the bonds (as on the previous page) or none at all. The condensed structural formula for butane can also be written as CH3CH2CH2CH3. The carbon skeleton formula (also called a line formula) shows the carbon-carbon bonds only as lines. Each end or bend of a line represents a carbon atom bonded to as many hydrogen atoms as necessary to form a total of four bonds. Carbon skeleton formulas allow you to draw complex structures quickly. Note that structural formulas are generally not three-dimensional representations of the molecule-as space-filling or ball-and-stick models are-but rather two-dimensional representations that show how atoms are bonded together. As such, the most important feature of a structural formula is the connectivity of the atoms, not the exact way the formula is drawn. For example, consider the following two condensed structural formulas for butane and the corresponding space-filling models below them:

A carbon skeleton formula is called a line formula because it uses lines to represent a I molecule.

Same molecule

Since rotation about single bonds is relatively unhindered at room temperature, the two structural formulas are identical, even though they are drawn differently. We represent double and triple bonds in structural formulas with double or triple lines. For example, we draw the structural formulas for C 3H6 (propene) and C 3H4 (propyne) as follows:

Structural formula

H

I

Propene

H

I

Condensed structural formula

Carbon skeleton formula

Ball-andstick model

Space-filling model

H

I

H-C=C-C-H

I

H

H Propyne

I I

H-C-C-C-H H

The kind of structural formu la we use depends on how much information we want to portray. The following example illustrates how to write structural formulas for a compound.

20.3

EXAMPLE 20.1

883

Hydrocarbons

WRITING STRUCTURAL FORMULAS FOR HYDROCARBONS

Write the structural formulas and carbon skeleton formulas for the five isomers of C 6H 14 (hexane). SOLUTION

To start, draw the carbon backbone of the straightchain isomer.

C-C-C-C-C-C

Next, determine the carbon backbone structure of the other isomers by arranging the carbon atoms in four other unique ways.

c- c- c - c - c I c c I c- c- c - c I c

Fill in all the hydrogen atoms so that each carbon forms four bonds.

c - c - c - c- c I c c I c-c-c-c I c H

H

H

H

H

H

H

H

H

H

H

H

I I I I I I H-C-C-C-C-C-C-H I I I I I I H

H

I

H

I

H

I

H

I

I

H - C- C- C- C- C- H

I

I

H

I

I

H

I

H H H- C- H

H

H

I

I

I

H

I

I I

I

I

I

I

I

H- C- C- C- C- H

I

I

I

H

H H H-C-H

I

H-C-H

I

I

I

H

I I

H- C- C- -C- C- H H

I

I

Write the carbon skeleton formulas by using lines to represent each carbon-carbon bond. Remember that each end or bend represents a carbon atom.

FOR PRACTICE 20.1 Write the structural and condensed structural formulas for each of the following carbon skeleton formulas:

o

I

I

H H H-C-H

H

(c)

H

H

H-C-H

(b) ~

I

I

I

(a)~

H

H

H

I

I

H- C- C- C- C- C- H

H

H

H

I

I

H H-C-H

I

H

884

Chapter 20

TABLE 20.1 Type of Hydrocarbon

Organic Chemistry I: Structures

Alkanes, Alkenes, Alkynes Type of Bonds

Generic Formula*

Example

Average Carbon-Carbon Bond Energy (kJ mol- 1)

Typical Bond Length (pm)

347

154

611

134

837

120

H H Alkane

I I

I I

H-C-C-H

All single

H H Ethane

H Alkene

H

\=c/ / "- H H

One double

Ethene

Alkyne

H-C==C-H

One triple

Ethyne

*n is the number of carbon atoms. These formulas apply only to noncyclic structures containing no more than one multiple bond.

Section 20 .3 makes use of valence bond theory and molecular orbital theory, which were covered in Chapter 10. We recommend that you review this material.

Types of Hydrocarbons Hydrocarbons can be classified as open chain hydrocarbons or cyclic hydrocarbons. Open chain hydrocarbons consist of open chains of carbon atoms. For example, butane and 2-methylpropane are open chain hydrocarbons. Cyclic hydrocarbons involve carbon atoms that are arranged to form one or more ring structures (e.g., cyclohexane).

0 Cyclohexane (carbon skeleton formula)

Hydrocarbons are also classified according to the type of bonds between carbon atoms. Alkanes have only single bonds between carbon atoms, alkenes have a double bond, and alkynes have a triple bond. These categories are outlined in Table 20. 1. Saturated hydrocarbons are hydrocarbons with the maximum number of hydrogen atoms for the number of carbon atoms present. Unsaturated hydrocarbons have fewer hydrogen atoms than the maximum for that number of carbon atoms. Unsaturation occurs whenever rings or TT bonds are present in the structure.

Alkanes The simplest alkane is methane, CH4 . In Chapter 10, we saw that the carbon atom in methane is sp3 hybridized and the four C - H bonds are ) . The two are mirror images of one another, but you cannot superimpose one on the other. For this reason, a right-handed glove does not fit on your left hand and vice versa. This type of isomerism is known as chirality. The word derives from the Greek word cheir, which means "hand." Any carbon atom with four different substituents in a tetrahedral arrangement is a chirality centre. Consider 3-methylhexane below. The molecules on the left and right are nonsuperimposable mirror images and are enantiomers of one another. ..6. FIGURE 20.7 Mirror Images The left and right hand are nonsuperimposable mirror images, just as are optical isomers.

Optical isomers of 3-methylhexane

Chirality is important, not only to organic chemistry, but also to biology and biochemistry. Most biological molecules are chiral and usually only one or the other enantiomer is active in biological systems. For example, glucose, the primary fuel of cells, is chiral. Only one of the enantiomers of glucose has that familiar sweet taste and only that enantiomer can fuel our cellular functioning; the other enantiomer is not even metabolized by the body. Some of the physical and chemical properties of enantiomers are indistinguishable from one another. For example, both of the optical isomers of 3-methylhexane have identical melting points, boiling points, and densities. However, the properties of enantiomers

902

Chapter 20

Organic Chemistry I: Structures Directio n of light propagation ,'1

,,1

I I I I

I

I I I I

Polarizer

I ; ·I · I I I I

,

I I I

Normal light

I

',

I I I I

I I

Light source

I

I I I

I

I

I

I

1,'

I

O p tically active sample

Planepolarized ligh t

Planepolarized light

Analyzer

Viewer

.&. FIGURE 20.8 Rotation of Plane-Polarized Light Plane-polarized light rotates as it passes through a sample containing only one of two enantiomers.

differ from one another in two important ways: (1) the direction in which they rotate plane-polarized light and (2) in their chemical behaviour in a chiral environment.

Optical Activity Plane-polarized light is light in which electric field waves oscillate in only one plane, as shown in Figure 20.8 A . When a beam of plane-polarized light is directed through a sample containing only one of two enantiomers, the plane of polarization of the light is rotated, as shown in Figure 20.8. This property is known as optical activity. One of the two enantiomers rotates the plane of polarization of the light clockwise and is called the dextrorotatory isomer (or the d isomer). The other isomer rotates the plane of polarization of the light counterclockwise and is called the levorotatory isomer (or the l isomer). An equimolar mixture of both enantiomers does not rotate the polarization of light at all and is called a ra cemic mixture.

Dextrorotatory means turning clockwise, or to the right. Levorotatory means turning I counterclockwise, or to the left.

Chemical Behaviour in a Chiral Environment Chiral compounds also exhibit different chemical behaviour when they are in a chiral environment (a chiral environment is simply one that is not superimposable on its mirror image). Enzymes are large biological molecules that catalyze reactions in living organisms and provide chiral environments. Consider the following simplified picture of two enantiomers in a chiral environment:

Enantiomer does not fit.

Enantiomer fits.

One of the enantiomers fits the template, but the other does not, no matter how it is rotated. In this way, an enzyme is able to catalyze the reaction of one enantiomer because that particular enantiomer fits the "template."

Which structure is chiral?

(a}

H

I

H- C- CI

I

Br

(b}

H

H

I

I

Br - C- C- H

I

Cl

I

H

(c}

H

H

I

I

I

I

Br - C- C- CI

H H

(d}

H

Cl

I

I

I

I

Br - C- C- CI

H H

903

20.7 Stereoisomerism II: Configurational Isomerism

Absolute Configurations

2-

Enantiomers have identical physical properties except in the way that they interact with planepolarized light. Thus, it is difficult to determine experimentally the actual structure of an enantiomer. How does a chemist determine whether an enantiomer is one configuration or its mirror image? The actual configuration of atoms at a chirality centre is known as the absolute configuration. There was no way to know the absolute configuration of stereoisomers until 1951, when the Dutch chemist Johannes Martin Bijvoet and coworkers used an X-ray crystallography technique to determine the absolute configuration of sodium rubidium d-tartrate. Once the absolute configuration is known, it is labelled according to the R,S system. In this system, each group around the chirality centre is assigned a priority according to the Cahn-Ingold-Prelog rules that are used in the E,Z system for alkenes. Priorities are assigned as the numbers 1, 2, 3, and 4, with 1 being the highest-priority group and 4 the lowest. The molecule is "viewed" from the opposite side as the lowest-priority group. (This is similar to a Newman projection, where we look along the bond.) If the other three groups are arranged in a clockwise sequence from 1-3, then the configuration is R (from the Latin rectus, meaning right). If the sequence is counterclockwise, then the configuration is S (from the Latin sinister, meaning left). The Rand S configurations ofbutan-2-ol are shown in Figure 20.9 T .

:J

(1) OH

(4)H11,...

L_. J2)

/ CH3

CH2CH3

(4) Hil 1'" 1,2,3 C lockwise

c

(1) OH

~'-... (3)

/ CH3 CH2CH 3

1,2,3 Counterclockwise

(2)

(3)

(R)-Butan-2-ol

EXAMPLE 20.3

(S)-Butan-2-ol

Na+Rb+

H-c-oH

I

Ho-c-H CO ff Sodium rubidium d-tartrate

R2 > R 3 > L, and the substrate is the S enantiomer. After dissociation of the leaving group, the stereochemical configuration of the substrate is lost because the intermediate now has a planar geometry. Nucleophilic attack can occur from either side of the carbocation, giving a mixture of R and S enantiomers as products. The process of converting a single enantiomer into an equal mixture of enantiomers is called racemization.

Ri

Ri

I~

cI+ l \

c ...,,,,L

Step 1

R: \

+

L-

.::;

R3

R1

S e nantiomer

R3

Trigonal planar

R1

I

Step 2

R:t''"C'-/

R3

Nu

S enantiomer

or

R1

,,.-----...,._ I

Nu: -

c+

~·

R~

~

\

R3 R enantiomer R acemic mixture

The SN2 Mechanism The SN2 mechanism involves a single step. The nucleophile "attacks" the substrate from the side opposite the leaving group. The reaction proceeds through a transition state with simultaneous bond formation and bond breaking.

,,.-----...,._ I ,...

Nu: - + -

I

C-L

Nu - C -

I Nucleophile

Substrate

+

:L-

I Transition state

Product

Leaving group

The substitution reaction of bromomethane with hydroxide follows an SN2 mechanism. Figure 21.3 .,,. shows the energy profile for this reaction mechanism. :Br:Hydroxide (nucleophile)

Bromomethane (substrate)

Methanol (product)

Bromide ion (leaving group)

The " 2" in the SN2 designation means bimolecular. The single step in the mechanism is bimolecular and therefore the rate law for a SN2 substitution reaction is second order. In this case, rate= k[CH3Br)[OH- ].

21.4

Nucleophilic Substitution Reactions at Saturated Carbon

935

Transition state

Ea

I

Nu: - + - C - L

I I

Nu-C-

+

:L-

I R eaction coordinate

.&. FIGURE 21.3 Energy Profile for SN2 Mechanism The rate of reaction depends on the activation energy for the formation of the unstable transition state.

If the carbon atom at the site of substitution is a chirality centre, the SN2 mechanism results in stereochemical inversion. This means that the substitution product has the opposite absolute configuration to the initial substrate. Inversion occurs because the nucleophile attacks from the opposite side from the leaving group. As the molecule passes through the transition state, the other three groups must "flip" away from the direction of nucleophilic attack. This is a lot like an umbrella being inverted in high winds. In the figure below, you can see how the absolute configuration of the product must be opposite from that of the substrate:

.&. The inversion of an umbrella in high winds is similar to the stereochemical inversion that occurs in SN2 substitution reactions. S e nantiome r

Transition state

R enantiomer

Factors Affecting Nucleophilic Substitution Reactions The Nucleophile In SNl reactions, the rate-limiting step is the formation of the carbocation intermediate. The nucleophile is only involved in the second step. Therefore, the nucleophile has no effect on rates of SN1 reactions. On the other hand, in SN2 reactions, the nature of the nucleophile is very important because it "displaces" the leaving group by forming a new bond at the site of substitution. The effectiveness of a nucleophile, or nucleophilicity, in SN2 reactions varies depending on the substrate and reaction conditions. The nucleophile appears in the rate law, so changes in the choice of nucleophile affect the rate of an SN2 reaction. An SN2 substitution with a good nucleophile will take place much faster than with a poor nucleophile. In fact, a poor nucleophile will often not react with a substrate via an SN2 pathway to any appreciable extent. Table 21.3 lists some common nucleophiles and their relative strengths. The Leaving Group The leaving group is important in both SN1 and SN2 mechanisms. In both mechanisms, as the bond with the carbon atom is broken, the leaving group must develop and accept a negative charge. Therefore, the stability of the corresponding anion is the most important criterion for the effectiveness of the leaving group. A convenient way to assess the relative stabilities of anions is to compare the pK3 values of the corresponding acids. The best leaving groups are the conjugate bases of strong acids.

[Robert Hoetink/Shutterstockl

936

Chapter 21

Organic Chemistry II: Reactions

TABLE 21.3 Selected Nucleophiles and Their Relative Effectiveness 1- ,sr- ,cN-

Good

Rs - (e.g., CH3S- ) OW, Ro- (e.g., CH30- ) Moderate

Rcoo- (e.g., CH3COO- ) RSH (e.g., CH 3SH) NH 3 RNH 2, R2NH, R3N H2 0

Poor

ROH (e.g., CH30H) RCOOH (e.g., CH 3COOH)

Common leaving groups:

r

> Br-

> c1-

>> F-

> CH3COO- > HO- > CH30- > NH2-

The leaving group is always part of the substrate, which is part of the rate law for both SNI and SN2 substitutions. The leaving group directly affects the rate of substitution.

Look up the pKavalues of the conjugate acid of each of the common leaving groups listed above. Confirm that the best leaving groups are conjugate bases of strong acids. Is a good leaving group a weak base or a strong base?

Substrate Structure In SN1 reactions, the rate of reaction is strongly affected by the stability of the carbocation intermediate. The stability of a carbocation depends on the number of alkyl groups it is bonded to. H

H-

I C+ \

H

Methyl carbocation

CH3

H

I H c- c+

I Hc- c +

H

H

3

\

Primary (1°) carbocation

3

\

Secondary (2°) carbocation

CH 3

I Hc- c + 3

\

CH3

Tertiary (3°) carbocation

Increasing carbocation stability

The order of carbocation stabilities is due to inductive effects. However, it is not as straightforward as the inductive effects discussed in Section 21.2. The carbocation species shown above do not contain any highly electronegative atoms bonded to carbon atoms, which would result in bond dipoles. In these carbocations, the positive charge on the carbon atom induces dipoles in adjacent sigma bonds and draws electron density toward it. This somewhat delocalizes the charge over nearby atoms. With more adjacent sigma bonds, the charge can be better delocalized. This explains the order of carbocation stability- the number of adjacent sigma bonds increases from methyl to primary, secondary, and tertiary carbocations. SN 1 reactions only occur with secondary and teritiary substrates because of the stability of the carbocations. Methyl and primary substrates never undergo SN 1 substitutions because of the very low stability of methyl and primary carbocations.

21.4

937

Nucleophilic Substitution Reactions at Saturated Carbon

In SN2 reactions, the rate is affected by the amount of steric hindrance at the substitution site. Consider two substrates: bromomethane and 2-bromo-2-methylpropane. When a nucleophile attacks the carbon atom in bromomethane, it must come in close proximity with the three hydrogen atoms, which is not energetically favourable. However, if the same nucleophile attacks the carbon in 2-bromo-2-methylpropane, it must experience even closer contact with all of the atoms on the three methyl groups. The degree of steric hindrance is so unfavourable at tertiary carbons that SN2 substitution does not occur. Instead, SN 1 substitution occurs because of the stability of the tertiary carbocation. SN2 substitution only occurs in methyl, primary, and secondary substrates. H

H 3C

H 3C

H 111° .. ·C -Br

H111 .... C -Br

H111 .... C -Br

H Bromomethane

H Bromoethane (1 ° halide) SN2 only

\

'

SN2 only

\

'

H 3C

\

\

H c111 .... C-Br

'

3H 3C' 2-Bromo-2-methy!propane (3° halide) SNl only

H 3C 2-Bromopropane (2° halide) SNl or SN2

In the substitution of methyl and primary substrates, the mechanism is always SN2. In tertiary substrates, the mechanism is always SN I. With secondary substrates, substitution can be either SN I or SN2, and the mechanism depends on the other factors that affect nucleophilic substitution: the nucleophile and the leaving group. The prediction of which substitution mechanism occurs in secondary substrates is often difficult. In some cases, both substitution mechanisms take place simultaneously.

EXAMPLE 21.5

PREDICTING NUCLEOPHILIC SUBSTITUTIONS

Predict the substitution product for the following reaction, and predict whether the mechanism is SN 1 or SN2. Draw the mechanism.

SOLUTION SORT You are given a substitution reaction. You are asked to predict the mechanism and then draw the mechanism.

STRATEGIZE To predict the mechanism, identify the substrate and nucleophile, and determine whether the substrate is primary, secondary, or tertiary. Determine whether it is a strong or weak nucleophile. Once all of this is known, the mechanism can be predicted and drawn.

SOLVE The substrate, CH3CH2Br, is a 1° substrate. The substitution will occur by an SN2 mechanism.

GIVEN: The reactants CH3 CH 2Br and CNFIND: Predict the mechanism of substitution and draw the mechanism. CONCEPTUAL PLAN

Determine Nature of Substrate and Nucleophile Predict Mechanism ~ Draw Mechanism

~

RELATIONSHIPS USED ~

I 0 substrate gives SN2

~

2° substrate gives SNl or SN2, depending on Nu and L

~

3° substrate gives SN 1

The reaction will proceed by an SN2 mechanism:

The mechanism can be drawn by comparing with the general SN2 mechanism in this section.

l (continued)

938

Chapter 21

EXAMPLE 21.5

Organic Chemistry II: Reactions

(CONTINUED)

CHECK The mechanism gives a balanced chemical reaction. Charges are balanced. FOR PRACTICE 21.5 Predict the substitution product for the following reaction, and predict whether the mechanism is SN1 or SN2.

Elimination Reactions

21.5

Elimination reactions occur when atoms are removed from a molecule. In organic chemistry, an important type of elimination is the loss of two atoms or groups from adjacent carbon atoms in a molecule. These reactions are therefore often referred to as 1,2-eliminations. The removal of atoms leads to the formation of multiple bonds in the molecule: L

H

I I - c-c 1

Base

\

~

I

I

C= C

I \

+

HL

2

In this section, we will focus on two types of elimination reaction-dehydration and dehydrohalogenation. Dehydration is the loss of water from an alcohol. This reaction occurs in the presence of a strong acid and heat. H 3P04

Heat

Butan- 1-ol

But- 1-ene

Dehydrohalogenation is the loss of HX (X = Cl, Br, I ) from an alkyl halide. This reaction often requires the presence of a strong base and heat.

1- Bromopropane

Sodium ethoxide

Heat

Propene

Ethanol

Sodium bromide

Many elimination reactions give mixtures of products because there are often two or more elimination sites around the - X or - OH groups. For example, 2-bromo3-methylbutane yields two alkene products. The more substituted alkene, 2-methylbut2-ene, is the major product. CH3

CH3

/CH........._ /CH3 CH 3 CH

I

Br 2- Bromo-3-methylbutane An exception to Zaitsev's rule occurs when large, bulky bases, such as ( CH3 )3CO - , are used. The bulky base extracts an Hatom from a less-substituted carbon atom , which leads to a less-substituted alkene.

CH3

I

I

/C~ /CH 3 CH 3

C

I

H

2-Methylbut-2-ene

I

+

/

CH 3

CH

"'CHz

.......... ~

C

I

H 2-Methylbut-1-ene

Major product

Whenever isomeric alkenes are produced in an elimination reaction, the major product is usually the most substituted alkene. This is known as Zaitsev's rule.

21.5

The E1 Mechanism The El mechanism for elimination is similar to the SNl mechanism for substitution in that they share the same initial, rate-determining step, the dissociation of an atom or group to give a carbocation intermediate. Similarly, the designation "El" means elimination, unimolecular. For example, the elimination of HBr from 2-bromo-2-methylpropane in methanol follows an El mechanism. The initial step is: CH 3

Step 1

I CH-C-CH 3 I 3 ( sr:

CH 3

Rate de termining

I+

CH3-C -CH3

+

.... ·~[ ·

In the second step, a hydrogen atom adjacent to the carbocation is removed by a base, leaving a C-C double bond. In this particular reaction, the solvent methanol can act as the base:

Step 2

In the previous section, we saw that - OH is a poor leaving group. Thus, in dehydration reactions, the leaving group is not -OH. A much better leaving group is formed in the presence of a strong acid like H2 S04 . The -OH becomes protonated to give an oxonium ion, which is an ion in which oxygen has a positive charge because it is bonded to three other atoms. Recall from Table 21. 1 that H2 S04 is a stronger acid than CH3CH2 0H2+. So, in concentrated sulfuric acid, virtually all of the -OH groups are protonated. This creates a very good leaving group, namely H 20: CH3

CH3

Step 1

H

+

+

I

CH-C-CH3

C:~-H

~

I I o~ /••'-.,

CH-C-CH 3

H

3

~

H

Oxoniumion The conjugate base of H2 S04 , HS04- , removes a hydrogen atom from the carbocation and thus generates the alkene: CH3 Step 2

I + ~,,..-..... ;:

CH3- C -CH2 - H

0

II II

+ :o-S-OH ••

~

0

In Step 1 of the dehydration mechanism above, H20 is produced. In concentrated H2S04 , what species is H20 converted to? Write a chemical equation. Use Table 21.1 to confirm that HS04 - is the strongest base in the reaction mixture. What does this tell you about the pKa of a carbocation?

Just like the SNl mechanism, the El mechanism generally does not occur in primary alcohols or primary alkyl halides, due to the instability of primary carbocations. The strength of the base is not a concern in El reactions because it is only involved in the second step, which does not affect the rate. If a weak base is used, the elimination will probably proceed via an El mechanism. This is different from the E2 mechanism on the next page, which generally requires a strong base.

Elimination Reactions

939

940

Chapter 21

Organic Chemistry II: Reactions

The E2 Mechanism The E2 mechanism involves a single step and is bimolecular. A base removes a hydrogen ato m, the C-C double bond is formed, and the leaving group dissociates. All of this occurs in a single step. The removal of the hydrogen atom is integral to the E2 mechanism. Therefore, this pathway is strongly favoured when a strong base, B- , such as an alkoxide (e.g., CH 3CH2 0 - Na +) is used. E2 eliminations can occur at primary, secondary, and tertiary carbon centres.

B:

_(+ - I-, CI

10

C- L

\ I C=C + / \

---'>

I

BH

+

L :-

Elimination Versus Substitution We already know that elimination reactions can produce mixtures of isomeric alkene products. Another complication is that bases, which promote elimination, are also nucleophiles. This opens up the possibility of competing elimination and substitution reactions. For example, when 2-bromopropane reacts with a weak base (ethanol) versus a strong base (sodium ethoxide), the yields of elimination and substitution products are very different: CH3CH 20H

I

CH 3CH CH 3

The reagent CH3CH20- Na +, sodium ethoxide, is a very strong base (see Table 21.1 ). As a reagent, it is often listed with CH3CH20H because CH3CH20-Na+ is prepared by reacting sodium metal, Na, with an excess of CH3CH20H: 2CH3CH20H(/) + 2 Na(s) ~ 2CH3CH20- Na+ + H2(g)

97%

3%

Br

CH3CH=CH 2

OCH2CH3

+

I

CH3C H CH3

2-Bromopropane 80%

20%

Elimination product

Substitution product

C H 3CH 20 - Na + CH 3CH 20H

The general rule is that strong bases favour the elimination pathway over substitution.

21.6

Electrophilic Additions to Alkenes

In Section 21.4, we defined a nucleophile as an atom or group that could form new bonds by donating a lone pair of electrons to an atom with a partial positive charge. Nucleophiles are Lewis bases in that they are electron donors. The opposite of a nucleophile is an electrophile. An electrophile forms a new bond by accepting a lone pair of electrons. Electrophiles are Lewis acids, or electron acceptors. In the previous two sections, we saw how carbocations are intermediates in SN1 and El pathways. In each of these pathways, the carbocations were electrophiles in that they formed new bonds with nucleophiles by accepting a lone pair of electrons. This is really a sort of acid-base reaction. Electrophiles are not limited to reactions with typical nucleophiles. In this section, we will describe the reactions of electrophiles with alkenes. Carbon-carbon double bonds are high in electron density and can donate electrons to an electrophile, which leads to an electrophilic addition reaction.

Hydrohalogenation Hydrohalogenation is the addition of HX to an alkene to give an alkyl halide. It is the reverse reaction of the elimination of HX, covered in the preceding section. The addition of HX is a two-step process. The first step is the reaction of the electrophile (HX) and the alkene to form a carbocation intermediate:

Step 1

\

I

(/\ C= C + H\

Alkene

v

H

X

---'>

Electrophile

• I - c - c - + :x: I I .. Carbocation intermediate

The second step is the same as what we saw in the SN1 mechanism- the reaction of the carbocation with x-:

21.6

H

Step 2

·~ :x: + - c+I - c..

I

I

~

X

H

1

I

I I - c- c-

Many alkenes are unsymmetrical in that there are different groups on each end of the double bond. In such cases, we would normally expect to see two different addition products, where the halogen atom is bonded to either of the two carbon atoms. In reality, there is usually only one addition product, the one in which the halogen atom is bonded to the most substituted carbon. For example, when propene reacts with HCl, the only product observed is 2-chloropropane:

CH3CH=CH 2

+

Cl HCl

I

~

Propene

H

I

Cl

H

I

+

I

CH3CH-CH2

CH3CH-CH2

2-Chloropropane

1-Chloropropane not observed

This product selectivity arises from the stabilities of carbocation intermediates. As we have seen, carbocation stabilities are in the order 3° > 2° > 1°. In the first step of the addition mechanism, the hydrogen atom adds to the double bond in such a way as to generate the most stable carbocation, which subsequently reacts to form a C- X bond. This pattern of reactivity is known as Markovnikov's rule, after the Russian chemist who first proposed it.

Draw the bond dipole of HX (e.g., HCI) to show why it is the electrophile in the first step of the hydrohalogenation mechanism.

Other Addition Reactions Several other addition reactions involve alkenes. A selection is listed in Table 21.4. It is worth noting that many addition reactions can be formally considered to be oxidation

TABLE 21.4 Selected Additions to Alkenes for the Reaction: \

I

/ C=C

R..eagent ~

\

Reaction

Reagent

Product(s)

Hydrogenation

H

I

H

I

-c-c1

Halogenation

Cl

I

I

Br

I

I

- c- c1 Epoxidation

RC03H (peroxy acid)

0

II

R- C- 0- 0-H Dihydroxylation

KMn04, NaOH

I

Cl

or

1

lo\

-c-c/

\

OH OH

I

I

1

I

- c- c-

I

- c- c-

, an epoxide

I

Br

Electrophilic Additions to Alkenes

941

942

Chapter 21

Organic Chemistry II: Reactions

or reduction reactions. For example, we have previously shown that hydrogenation is a reduction reaction. In halogenation, epoxidation, and hydroxylation reactions, the formal charges of the carbon atoms become more positive (i.e., oxidation). In contrast, hydration (the addition of water) is not formally a reduction or oxidation reaction.

Follow the mechanistic steps for hydrohalogenation to write a similar mechanism for the hydration of alkenes with H20, H+.

Nucleophilic Additions to Aldehydes and Ketones

21.7

In Chapter 20, we saw that the carbonyl group is highly polarized. The carbon atom in a carbonyl group has a large partial positive charge due to the double bond to the more electronegative oxygen. Carbonyl groups are subject to nucleophilic attack at the electrophilic carbon atom. Unlike at a saturated carbon, where nucleophilic attack leads to substitution, nucleophilic attack on a carbonyl group gives an addition product. As we will see, this is due to the carbon-oxygen double bond. The general mechanism is as follows. The first step in the reaction is the formation of a bond between the nucleophile and the carbon atom of the aldehyde or ketone. The intermediate formed contains a negatively charged oxygen, which reacts with a proton, H +, to give the - OH group.

OH

(~~ /

C

'-....

+:Nu-

I

~

-C -

Nu

I

Addition of Alcohols OH

OH

R'-C-OR"

R'-C-OR"

I I

H

Hemiacetal

I I

R

An example of nucleophilic addition to aldehydes and ketones is the formation of hemiacetals and hemiketals, which are compounds with an -OH and an -OR group on the same carbon atom. Hemiacetals are produced from the reaction between an aldehyde and an alcohol. Hemiketals are produced from the reaction between a ketone and an alcohol. These reactions are reversible and are usually catalyzed by acid.

Hemiketal

OH

I

H+ CH3CH2 CHO

+

Propanal

CH 30H

CH3CH 2 CH - OCH3

Methanol

1-Methoxypropan-1-ol (a hemiacetal)

OH

0

II

CH3CCH3 In IUPAC nomenclature, all compounds with the general structure RR'C(OH)(OR") are known as hemiacetals. The class term "hemiketal" is considered a subclass of hemiacetals.

Propanone

H+

+

CH30H

I I

H 3C-C-OCH 3 CH3 2-Methoxypropan-2-ol (a hemiketal)

21.7

Nucleophilic Additions to Aldehydes and Ketones

The mechanisms for formation of hemiacetals and hemiketals are identical, so we will only present the mechanism for hemiacetals. In acidic conditions, the oxygen atom in the carbonyl group can be protonated to give a type of oxonium ion (step 1). The protonated carbonyl has a resonance structure in which the carbon atom is positively charged. This carbon atom is very electrophilic and is attacked by the nucleophilic oxygen atom of the alcohol (step 2). The final step is the release of a proton, regenerating the acid catalyst (step 3).

Resonance structures for protonated carbonyl oxonium ion

rCH,CH}c

CH,CH,Cc 1

Step 1 Aldehyde

Oxoniumion (protonated carbonyl)

Step 2

Oxoniumion OH

OH

I

Step 3

CH3CH2C-H

I I

CH3CH2C-H

I+

vo"

H

+

H+

O-CH3 Hemiacetal

CH3

Hemiacetals are important in biological systems because many monosaccharides (e.g., glucose) primarily exist in the hemiacetal form:

H

I

H

I

OH H

I

I

HO

HOCH 2-C-C-C-C-CHO

I

I

I

OH OH H

I

OH

Glucose (C6H1206)

Glucose herniacetal

The Grignard Reaction A very useful reaction involving nucleophilic addition to aldehydes and ketones is the Grignard reaction. This reaction involves a carbon-centred nucleophile, called a Grignard reagent, having the general formula RMgX (X = I, Br, Cl). The Grignard reagent is typically prepared by reacting a halogenated compound with magnesium metal in anhydrous ether. Anhydrous conditions are important because water reacts with the Grignard reagent.

CH3CH2Br

+ Mg

Diethyl ether

CH3CH2MgBr Ethylmagnesium bromide (a Grignard reagent)

943

I Anhydrous =

"no water"

944

Chapter 21

Organic Chemistry II: Reactions

Grignard reagents are "organometallic compounds," meaning that they contain a metal-carbon bond. The Mg-C bond has significant ionic character. Carbon is more electronegative than magnesium, so the carbon atom has a large partial negative charge and can act as a nucleophile. For example,

o- o+ CH 3CH2 -MgBr Grignard reagents (RMgX) react with aldehydes and ketones to give alcohols. The mechanism involves transfer of the R group from the Mg atom to the carbonyl carbon atom. Once all of the Grignard reagent is consumed, the reaction mixture is "worked up" (i.e., neutralized) with dilute acid, which converts the alkoxide salt to the alcohol.

\C~ = O + R-MgX

I v

R

R

I

I

-c-o-

~

-C-OH + Mg 2 + +

x-

1

1

Alkoxide salt

The Grignard reaction is useful in organic synthesis because it results in new C - C bonds. Thus, it provides a way to join smaller organic units together into larger molecules.

EXAMPLE 21.6

GRIGNARD REACTIONS

Predict the organic products for the following Grignard reactions: 0 CH3CH2MgBr +

(1) Diethyl ether

II

(2) H 3 o +

/C......__

H

H (a)

+

0-

/

0

c

\

(1) Diethyl ether

CH3

(2) H30 +

(b)

SOLUTION

(a) The R group in the Grignard reagent is an ethyl group. This transfers to the carbonyl carbon to give the alcohol. Since the two other groups on the carbonyl carbon are hydrogen atoms, the alcohol must be primary. (b) The R group in the Grignard reagent is a methyl group, which transfers to the carbonyl carbon to give the alcohol. Since the carbonyl is a ketone, the final product must be a tertiary alcohol.

H

CH CH 3

2

I I

CH

OH

21.8

Nucleophilic Substitutions of Acyl Compounds

FOR PRACTICE 21.6 Predict the organic products for the following Grignard reactions: C 6H 5MgBr

+

(1) Diethyl ether CH3 CHO

(2) HJO+ (a)

(1) Diethy1ether

(b)

21.8 Nucleophilic Substitutions of Acyl Compounds The acyl compounds described in Chapter 20 are very similar to aldehydes and ketones, in that they all contain a C = 0 bond. However, unlike carbonyl compounds, acyl compounds react with nucleophiles to give substitution reactions. Why the difference? In aldehydes and ketones, the carbon atom of the carbonyl unit is bonded to hydrogen and/or carbon atoms, which are not good leaving groups. In contrast, in acyl compounds, the carbon atom is bonded to a potentially good leaving group such as an alkoxy group and this changes the reactivity. Substitution is the predominant reaction observed in acyl compounds. Consider the general case of a nucleophile reacting with an acyl group. The first step of the reaction is the same as the addition step that occurs with carbonyls, forming an alkoxide intermediate in which the central carbon atom is sp3 hybridized. Subsequent to this addition step, the leaving group departs, giving the substitution product. 0

// -C \

+ L:-

Nu

An example of acyl substitution is Fischer esterification, the reaction of carboxylic acids with alcohols to give esters. The reaction is reversible, which means that esters can be hydrolyzed by water. Both reactions are catalyzed by acid. Esterification

0

//

R-C

+

R'OH

\OH Carboxylic acid

Ester Hydrolysis

The mechanism of esterification has similarities to the mechanism for the formation of hemiacetals and hemiketals, discussed in the previous section. For example, the first step involves the double-bonded oxygen atom reacting with H+ to form an oxonium ion. Also similarly, the next step is nucleophilic attack by the alcohol. After this, the mechanisms diverge. The next steps in esterification involve movements of H+ to ultimately generate an H20 leaving group. The loss of H2 0 completes the substitution. Each step in the esterification mechanism can proceed in either direction.

945

946

Chapter 21

Organic Chemistry II: Reactions

The Mechanism of Fischer Esterification

Carboxylic acid

Oxoniumion

Ester

Esterification is a reversible equilibrium process, and the equilibrium constants are usually small. K is typically between 1 and 10 when the alcohol is primary or secondary, and K < 1 when the alcohol is tertiary. In order to obtain practical yields of ester, the equilibrium must be shifted toward products. This can be done by using an excess of the alcohol or by removing the ester and/or water through distillation. This is a practical application of Le Chatelier's principle from Chapter 14. The esterificati on of carboxylic acids is an example of the interconversion between acyl groups. Direct interconversion between groups is not always possible because an order of reactivity exists among acyl groups, with acyl chlorides being the most reactive and amides being the least reactive. For the most part, this can be explained by the ability of various groups to function as leaving groups, which was discussed in Section 21.4. 0 0

//

R-C

\Cl Acy! choride

0

0

0

//

II II /c...____ /c...____

R

0 R Acid anhydride

II

0

//

R-C

R-C

\

\

0-R 0 - H Ester Carboxylic acid

/

R

C

H

'---··/ N

I

H Amide

Increasing reactivity

Generally speaking, it is possible to convert a more reactive acyl group into a less reactive acyl group. Interconversions of acyl groups are summarized in Table 21.5. It is possible to convert acyl chlorides (most reactive) to any of the other groups. This is important because conversion to the acyl chloride allows for an indirect way to convert carboxylic acids into any of the other groups. That is, once the acyl chloride is made, it is possible to generate any of the other acyl groups in high yields. Acy! chlorides can be prepared by reacting the corresponding carboxylic acid with either SOCl2 (thionyl chloride) or (COClh (oxalyl chloride). A disadvantage of using SOCl2 is that one of

21.8

Nucleophilic Substitutions of Acyl Compounds

the by-products is S0 2, a very toxic gas. ( COCIh is considered a milder reagent, and the by-products include CO and C02 , which are not as toxic as S02 . 0

0

//

R-C

\

+

SOC12

0 -H

\

0-H

Carboxylic acid

+

~

o, I

R- C

Acy1 chloride

0

l l

c-c

\

Cl

+ HCl + S02

\1

Thionyl chloride

Carboxylic acid

//0 R-C

//

~

//

R-C

+ HCl + C0 2 +CO

\1

0

Acy 1 chloride

Oxalyl chloride

TABLE 21.5 Reagents for Common lnterconversions of Acyl Groups The reagent for an interconversion is found where the reactant and product acyl groups meet in the table. Reactant Acyl Group

Product Acyl Group

0

R-C

\1

0

0

c

c

II

//

R/

II

"'-o/ "'- R

0

R-C

//

0

R-C

\ 0-R

0

II

//

\0- H

R/

C

H "'-N/

I

H

0

0

R-C

//

-

R'-C

\1 R/

0

c

c

II

"'-o/

R' OH

H20

2 NH 3

R'OH

H20, W

2 NH 3

R'OH, H+

H20, H+

NH 3

\ o_

0

II

//

-

*

"'- R

0

R-C

//

-

-

\ 0-R 0

//

R- C

SOCl 2or (COCl)2

*

R'OH, H+

-

t

\ 0-H 0

II

/

R

C'-.,, .. / H N

-

-

-

H20, H+, heat

-

I

H •Anhydride exchange is a means of generating symmetrical acid anhydrides. Two moles of carboxylic acid react with acetic anhydride to give the new anhydride and 2 moles of acetic acid: 2 RCOOH + (CH3CO )20 --> ( RCO )20 + 2 CH3COOH. t The reaction between a carboxylic acid and an ammonia is an acid-base reaction, and the product is an ammonium carboxylate (NH4+Rcoo-), which does not have a good nucleophile or electrophile. Al high temperatures, the ammonium carboxylate can be converted to the amide. This is an industrially important reaction, but is not used in the laboratory.

947

948

Chapter 21

Organic Chemistry II: Reactions

The conversion of one ester into another is called transesterification. This equilibrium reaction is entirely analogous to Fischer esterification, with the first ester reacting with an alcohol to give an ester with a different alkoxy group. The reaction is catalyzed by acid, and follows the same basic mechanism as for esterification: 0 //

+

R-C

\

EXAMPLE 21.7

H+ ~

R ' OH

0

//

R-C

\

0-R

+

ROH

0-R'

ACYL GROUP INTERCONVERSIONS

Suggest appropriate reagents for the following interconversion . (Hint: This interconversion requires more than one step.)

0-

Cu z+(aq)

+

2 e-

+

2 e-

------>

Cu(s)

95. 1.8 x 102 s 97. 1.2 X 103 A 99. 2Mn04- (aq) + 5 Zn(s) + 16 H+(aq) ~ 2 Mn2 +(aq) + 5 Zn2 +(aq) + 8 HzO(l) 34.9 mL 101. The drawing should show that several Al atoms dissolve into solution as AI 3 + ions and that several Cu2 + ions are deposited on the Al surface as solid Cu. b. cannot be dissolved 103. a. 68.3 mL c. cannot be dissolved 105. 0.3 107. There are no paired reactions that produce more than about 5 or 6 V. 109. a. 2.83 V b. 2.71 V c. 16 h 111. 176 h 113. 0.71 v 115. a. t:./i ' = 461 kJ mol- 1, K = 1.4 x 10-81 b. t:.p' = 2.7 X 102 kJ mo1 - 1, K = 2.0 X 10- 43 117. MCl4 119. 51.3% 121. pH = 0.85 123. 0.86 mol L - I 125. 4.1 X 105 L 127. 435 s 129. 8.46% u 131. The overall cell reaction for both cells is 2 Cu+(aq) ~ Cu2 +(aq) + Cu(s). The difference in E 0 is because n = 1 for the first cell and n = 2 for the second cell. For both cells, t:. ,d = -35.1 kJ mol - 1• 133. a.

Chapter 19 31. a. 2§iu ~ iHe + 2§8 Th b. 2§8Th ~ iHe + 2§~Ra

A-47

2

~iPb ~

2

33. 2§6Th ~ iHe + 2 §~Ra 2 2 ~~Ra ~ -~e + ~~Ac 2 2 §~Ac ~ -?e + §8Th 2 §gTh ~ iHe + 2~:Ra 35. a. 2 §~Fr b. _9e c. +9e d. _9e 37. a. stable, N / Z ratio is close to 1, acceptable for low Z atoms b. not stable, N / Z ratio much too high for low Z atom c. not stable, N /Zratio is less than 1, much too low d. stable, N / Z ratio is acceptable for this Z 39. Sc, V, and Mn each have odd numbers of protons. Atoms with an odd number of protons typically have less stable isotopes than those with an even number of protons. b. positron emission 41. a. beta decay d. positron emission c. positron emission b. Fe-62 43. a. Cs- 125 45. 2.34 X 109 years 47. 0.57 MBq 49. 10.8 h 51. 2.66 x 103 y 53. 2.40 x 104 y 55. 2.7 x 109 y 57. 2§iu + bn ~ + j~Sr + 2 bn 59. fH + fH ~ ~He + bn 61. 2§~u + bn ~ 2§iu 2 §~u ~ 2§jNp + _~e 2 §jNp ~ 2§2Pu + -~ e 63. 2~§cf + '~c ~ fblRf + 4 bn 65. 9.0 x 10 13 J 67. a. mass defect = 0.13701 u binding energy = 7.976 MeV/nucleon b. mass defect = 0.54369 u binding energy = 8.732 MeV/nucleon c. mass defect = 1.16754 u binding energy = 8.431 MeV/nucleon 69. 7.228 x 10 10 J g- 1 U-235 71. 7.84 X 10 10 J g- 1 H-2 73. radiation: 25.5 J; fall: 370 J 75. 68 km 77. a. lP +~Be ~ ~Li + iHe 1.615 MeV b. 2 ~Bi + ~~Ni ~ m Rg + bn 236.4 MeV c. '?£w + -?e ~ '~jTa 1.573 MeV 79. a. '!:Ru ~ _9e + '!1Rh 2 2 b. ~~Ra ~ +?e + J~Fr c. jgzn ~ + ?e + ~~Cu d. TbNe ~ -?e + 1lNa

'ttxe

A-48 81. 83. 85. 87. 89. 91. 93. 95. 97. 99. 101.

APPENDIX Ill: Answers to Selected Exercises

2.9 X 10 18 beta emissions, 37 Gy 1.6 X 10- 5 L 1.022 MeV 7.72MeV 14N 0. 15% 1.24 x 1021particles 2.42 X 10- 12 m -0.7 MeV, there is no coulombic barrier for collision with a neutron. a. 1.164 x 10 10 kJ b. 0.1299 g U-235 forms Pb-207 in seven a-decays and four {3-decays, and Th-232 forms Pb-208 in six a-decays and four f3 -decays. 3.0 X 102 K 21p ~ 21 Ne + oe 9 10 - I

43. a.

b.

c.

d.

40 CH 3

Chapter 20 b. alkene

c. alkyne

sp

sp

s/

s/

CH3

CH3-

CH -

CH = CH2

sp3

sp2

c.

sp3

H2

c

s/

s/ HC/ '-...CH s/ \\ I/ s/ HC-CH s/

CH 3-c==c-CH3

s/

d. alkene

b.

d.

I

CH3-

CH=CH2

sp3

sp2

sp2

sp2

b.

CH3

41. a.

b. NH23

sp

0 sp

//

CH - C 33 sp2 \

CH3 3

sp

sp

I I

H 3C-C-CH-CH3

2

c.

0 sp2

II

CH3- C sp3 sp2

CH 3 sp3

d.

I

CH 3 CH3

O-CH2-CH3 3 3 3 sp

sp

sp

0 s/

// -CH2-C

CH3 3

sp

sp

3

s/ \

Cl sp3

2

\

\ c1

45. CH3CH2CH3 < CH 3COCH3 < CH3CHOHCH3 47. CH2=CH-CH2-CH2-CH2-CH3 CH3-CH=CH -CH2-CH2-CH3 CH3-CH2-CH =CH -CH2-CH3

103. 105• 107. Nuclide A is more dangerous because the half-life is shorter (18.5 days) and so it decays faster.

35. a. alkane 37. a.

CH - C

51. a. not constitutional isomers b. constitutional isomers c. constitutional isomers

A-49

APPEN DIX Ill : Answers to Selected Exercises

77. a. 3 79. a.

53. For example,

0 0-- c(

H 2C = CH 2 2

sp

sp

t~

sp

3

c = c -NH2 3 2 2

sp

sp

sp

sp

01

H 2C = CH 2 2

sp

55. For example,

sp

+

C = NH2 2 2

sp

sp

sp2

b. 0

-1->-

H 2c 2

NH2

sp

Ht,-

+

-l->-

C2

'bx(

D

/ N~

:;:0 :;:

0

~OH ~OH

HOLJOH

b. 6

sp2

57.

H

Cl~H H

H H

c.

H@-:

H

~H

H sp sp

81.

H

Staggered

/ ,Cl

a. Examples:

Eclipsed

HN~OH 2

59.

Cl

HAH

b. Examples: 0

HY-H Antistaggered

Eclipsed

Gauche-staggered

H~C l CH3 ~H Cl

c. Examples:

0

H H

Eclipsed

Gauche-staggered

H

OH

~

CH3 H

HO~Cl

Cl~

CH3

H H Eclipsed

61. a. Z b. E c. Z d. E 63. a. enantiomers b. same c. enantiomers b. enantiomers 65. a. same c. same b. S;R 67. a. achiral c. s 69. a. S b. R c. s d. R 71. a. IHD = 3 b. IHD = 3 c. IHD = 4 d. IHD = 6 73. a. IHD = 1; a carboxylic acid or ester could be present, or a combination of a ketone or aldehyde group along with an alcohol or ether group. b. IHD = 0; an alcohol or ether group must be present. 75. a. N-H; 3100-3500cm- 1 (medium) b. C=O, 1630-1800 cm- 1; C-0, 1050-1250 cm- 1; 0-H, 3200-3500 cm- 1

83. a.

CH3-

CH2-

CH -

I

CH = CH 2

CH3 Can exist as a stereoisomer.

b.

CH3

CH3

I

I

CH3-CH=C-CH2-CH2-CH3 Can exist as a stereoisomer.

c. H 3C-CH=C-CH2-CH2-CH3

I

CH2CH2CH3 Cannot exist as a stereoisomer.

85. l >

d.

/CH3 CH3CH= C-CH

I

\

CH3 CH3 minor

c. (CH3CH2lzNH + ClCH 2CH20H (CH3CH2}zN -

49. a. SN 1

NaOH

/CH3

~

CH3CH2C = C

I

CH2CH20H + H20 + NaCl

CH3

b. SN2

+ H20

\ CH3

major

51. a. (CH3hC _0Br

(CH3hC+ + Br-

..==="

57. a.

~. (CH3hC• + CH3CH2RH ----+

major

minor

H~

( I.

..

(CH3hC-O-CH2CH3 + HzO :

----+

(CH3hCOCH2CH3 + H30+

minor

major

major

H

I ,.,...c ...,,,,CH CH I

\

2

+ Br- inversion 3

CH3

c~

c. H3C,,,,,.l/Br

+

H20

----+

H3C",,.. U O H • H3C11,... racemization

_ . 0

,,,0H • HBr

minor

+

A-52

APPENDIX Ill : Answers to Selected Exercises

a.

H

Cl

I

I

CH3-CH -

0 - M g B r + CH3CHO

CH -

CH3

OH

~ ~-rn-rn-rn - ~+~ - rn-rn-rn-~

I

CH3

CH3-CH2-

CH3

H

o-~H-CH3

I

I

Br

H

Br

Br

c.

I

I

I

Br

b.

I

I

CH3-CH-MgBr

I

CH-CH3

CH -

1) Diethyl ether 2} H3 o•

1) Diethyl ether +

2)H30 +

CH3

65. a. CH3-CH-CH = CH2

I

c.

Pd/C

+ H2 ~

1) Diethyl ether 2}H3 0 +

CH3 OH CH3

CH3- , H - CH2-

I

CH3CH2CCH2CH3

CH3

I

CH3

75. a.

67. a.

I

CH -CH-CH-CH 3

I

77. a.

CH 3 0H

b.

0-!

I

II

CH2 -C-OCH2 CH3 + H 2 0

0

"'

-

OH

I

I I

0 CH3-CH2 - CH 2 -

(Pa

b.

0-MgBr

0-MgBr 3

CH3

c.

b.

OH

+HO

OCH3

2

79.

CH -C-CH-CH 2 3

0

CH3

69. a. H+, H 20

b. HCI

71. a.

I

d. Br2

c. H 2, Pd/C

// c- c

3

2

//

CH3-CH2-CH2-CH 2-CH2- C \

H + CH 3CH20H

C-

Cl

CH 2

I I

C0 2 + CO + HCI

C-H

b.

OCH 2CH3

CH3-

0 // C\

CH2 -

H• + CH 3CH20H

------->

O-CH3

b.

0

//

+ CHpH

CH 3-CH2- C \ OH

CH 3

CH 2

I I

Cl

0

II

OH CH -CH -

\

Cl

0 CH3-CH2-CH2-

0

~

C-CH

O-CH2CH3

c.

3

0

CH 3-

O-CH(CH 3)z

//

+ NH3 ------->

C\

O-CH3

O-t-H d.0- 0 OH

// c

OCH3

\

O-CH3

H 20.H•

+

A-53

APPENDIX Il l : Answers to Selected Exercises

81. a.

~ ~:

101. a.

b. 1) (COCl)2 ; 2) two equivalents NH(CH3 h

~

v

b.

('YI

87. a.

c.

O

II

Cl

~

v

( I 'Y

( J C ' C H,

C(CH3)J

O

AICl~Cl

d.

c.

,...,..c......__ ,...,..cH2CH3 :CH2 0

103.

·· oz b.

0

0

//

3

II

F, / FF, /FF, /F 89. ... "'- /c"-. /c"-. /c"-. c c c F/ ' FF/ 'FF/ "-F

ONa

CH 3

H3C-1 ) CH=CH2

II

-->

I+

H 2C-C-CH=CH2

Cl

t t

Carbocation resonance structures:

CH3

I +~

H3C - C - CH = CH2

CH3 t - CH2

CH3

..

11

dispersion forces dispersion, dipole-dipole forces dispersion, dipole-dipole forces dispersion, dipole-dipole forces, and hydrogen bonding acid-base nucleophilic substitution nucleophilic addition nucleophilic substitution OH

CH3

I+

H 20 : + H 3C-C-CH=CH2 ~

-->

CH2CH3

I

f

c ....,,,CH CH \

2

CH3

3 +

( J c "" CH3 OH #

I

racemic mixture

b.

111. a. K2Cr20 7 , H2 S04 b. CH3 -CH-CH2 -

I

CH3 R enantiomer

II

CH 3-CH2 -C -O-C-CH3

I

91. HO-C

(J

"

CH3

0

95. a. b. c. d. 97. a. b. c. d. 99. a.

2

0

105. 558 g 107. a. 0.92 kg b. SOCl2 reaction by-products: 0.56 kg HCI, 0.98 kg S02 (COClh reaction by-products: 0.56 kg HCI, 0.67 kg C02, 0.43 kg CO 109.

~Cl

c. Cl2, FeCl3

93. a.

0

II

0-0H

CH -CH -C

b. AlC13 ,

0

CH -CH -CH-C 3 2 I \ CH3 Cl

0

85. a.

b.

0

//

c. CH3 CH20H, H +

COOH

A-54

APPENDIX Ill: Answers to Selected Exercises

113. a.

c.

0 0 II II n H O-C-R-C-OH

(CH2 CO)i0

HO-CH2-

H

d. H

0 II C

1 / 0 ~1 ,,,,,, s

C OH C 1\1 H C - C OH

11;1

H+OH

~\ 6~R

HO-CH2

s

43. b.

0 0 0 0 II II + II II n HO-C-R-C-OH n Cl - C - R " - C -

0 II CH I H-C-OH

CI

H zC-OH

0

I HO-C-H

I I

H-C-OH

115. Excess NH 3 reacts with the HCI by-product. 117. Reduction

H-C-OH

Chapter 22 a. c. 33. a. c. 35. 31.

45.

b. d. b. d.

not a fatty acid saturated fatty acid saturated fatty acid not a fatty acid

OH

0

H zC -OH

0

OH H 2C (CH 2)4 - (CH =CHCH 2)z I I I HO-CH + H 3C (CH2 )6 -

OH

H zC-OH

not a fatty acid steroid not a fatty acid monounsaturated fatty acid

I

H

I HzC-OH

47.

0

OH

H

OH

H 2 C-OH

II C -OH

r----o

I

H 2C

I

0

H

OH

H

0

II

H

HzC-O-C-(CH2)6 - (CH2CH =CH)z - (CH2)4 -

I

~ ~

OH

(CH2)6 -

(CH2CH = CH )z - (CH 2)4 -

HN -

II

b.

CH - C - 0

_

0 +

HN -

I

3

HO -CH (CH2)6 - (CHzCH = CH)z - (CH2)4 -

HO"-

R

CH2 HO I /

c-o

R -....... 1/H H\I HC OH I c

\1

,,,,,- s

11

s - 1- 1~H H

c. HN 3

HC H

ROH

H

ROH

H

ROH

d.

0 +

~OH

b.

/

H

_

I

CH3

b. not a carbohydrate d. not a carbohydrate b. aldose, pentose d. aldose, tetrose

monosaccharide disaccharide aldose, hexose ketose, tetrose

II

CH- C - 0 CH3

I

CH3

Triglyceride is expected to be an oil.

41. a.

H

Fructose

0 +

CH3

3

H zC - O - C -

37. a. c. 39. a. c.

OH

Glucose

49. a.

H -C - O - C -

I

CH3

II

CH - C - 0

I

CH2

I H 3C -CH I

CH3

_

0

II

+

HN -CH-C-0 3

I

CH2

I I CH2 I CH2 I CH2

NH2

OH " 'R H 2C"-

OH

_

A-55

APPEN DIX Il l: Answers to Selected Exercises

H

51.

0

I

HzNij,,,

·c -

H

II

C - OH

65.

0

I

HJClj,

II

'··c - C -

~ H3C

NH2

I

OH

N

~ HzN

,f HC \

53. 6: Ser-Gly-Cys, Ser-Cys-Gly, Gly-Ser-Cys, Gly-Cys-Ser, Cys-Ser-Gly, Cys-Gly-Ser 55.

H

0

H

I II HzN -C-C-OH I CH2 I

0

+

Q

OH

0

I I

75.

H

I I

~CH N

H

H-N-

(

0

I

....-C""'-

I

H

II

N

~N

ACATGC G 154 codons, 462 nucleotides a. protein b. carbohydrate c. lipid A codon is composed of three nucleotides. A codon codes for a specific amino acid while a gene codes for an entire protein.

OH H

II H

67. 69. 7 1. 73.

I II HN-C-C-OH z I

---c,....-

C~

0

C-C -

d~ 3

Trigonal pyramidal H

II

Q

"

Bent

Tetrahedral

77. valine, leucine, isoleucine, phenylalanine 79. Gly-Arg-Ala-Leu-Phe-Gly-Asn-Lys-Trp-Glu-Cys

CH2

bH

OH

\

H 2N-C-C-NH-C-C-OH CH2

Trigonal planar

II ~

I

81.•.H~~OH

f3

·nz~If:

OH

OH

H

57. a.

HN2

NH2

H

b.

H

0

I II C- CI CH 2 I CH 2 I C= O I 0

NH -

H

0

I II C- CI CH2 I CH2 I s I

NH-

OH

0

H

OH

0

I II I II I II HN-C - C-NH-C- C-NH-C- C-OH 2 I I I CH 2 CH2 CH2 I I I H3C-CH OH SH I

83. As the temperature increases, the favourable entropy for uncoiling a chain becomes dominant. On cooling, the favourable enthalpy of forming hydrogen bonds between paired bases is dominant. 85. a. C02H H 3C -

H

0

I - CII HN-C 2 I CH 2 I SH

H

0

I II NH-C - CI CH2 I H3C-CH I

H

2

L-alanine

CH3

59. tertiary 61. primary 63. a. A c. T

b. not a nucleotide d. not a nucleotide

L-cysteine

0

I II NH-C - CI CH2 I OH

I

C-·.,,,,H

\ NH

CH3

c.

a

'TI:rz~~H

SH

CH3 H

b H~~H

0

I II C- CI CH2 I

c.

I

OH HO -

C02H

CH2- C \ ''H

d.

~

HO -

C-

F02H CH2-

C \ 'H

NH2 L-serine

NH2 L- aspartic acid

87. When the fake thymine nucleotide is added to the replicating DNA, the chain cannot continue to form because the -N=N+=NH group on the sugar prevents future phosphate linkages.

A-56

APPENDIX Ill: Answers to Selected Exercises

89. V max = 47.6, K, = 1.68 91. H3N+ -CH2Coo- + H + ~ (H 3N+ -CH2COOH [HA]/ [A - ] = 2, H3N+ -CH2Coo- ~ H2NCH2COO- + H + [HAJ/[A- ] = 0.4, pH = 6.0 93. A three-base codon codes for a single amino acid. If there are only three bases, there could be 27 different three-base codon arrangements. Therefore, you could theoretically code for the 20 different amino acids needed.

Chapter 23 17. 19. 21. 23. 2S. 27. 29.

31. 33.

3S.

37.

39.

41.

43. 4S. 47. 49.

Sl.

a. +4

b. +4 c. +4 Ca3Al2(Si04)J 4 tetrahedrons stand alone, orthosilicates amphibole or double-chain structure; Ca2+, Mg2+, Fe2+, Al3+ 950 g NC13 has a lone pair that BC1 3 lacks, giving it a trigonal pyramidal shape, as opposed to BC13' s trigonal planar shape. a. 6 vertices, 8 faces b. 12 vertices, 20 faces closo-Boranes have the formula B,,H,,2- and form fully closed polyhedra, nido-boranes have the formula B,,H,,+4 and consist of a cage missing a corner, and arachno-boranes have the formula B,,H,, +6 and consist of a cage missing two or three corners. Graphite consists of covalently bonded sheets that are held to each other by weak interactions, allowing them to slip past each other. Diamond is not a good lubricant because it is an extremely strong network covalent solid, where all of the carbon atoms are covalently bonded. Activated charcoal consists of fine particles, rather than a lump of charcoal, and subsequently has a much higher surface area. Ionic carbides are composed of carbon, generally in the form of the carbide ion, cl-, and low-electronegativity metals, such as the alkali and alkaline earth metals. Covalent carbides are composed of carbon and lowelectronegativity nonmetals or metalloids, such as silicon. a. solid ~ gas b. gas ~ liquid ~ solid c. solid ~ gas a. CO(g) + CuO(s) ~ C02(g) + Cu(s) b. Si0 2(s) + 3 C(s) ~ SiC(s) + 2 CO(g) a. +2 b. +4 c. +4/ 3 Fixing nitrogen refers to converting N2 to a nitrogencontaining compound. White phosphorus consists of P4 molecules in a tetrahedral shape with the atoms at the corners of the tetrahedron. This allotrope is unstable because of the strain from the bond angles. Red phosphorus is much more stable because one bond of the tetrahedron is broken, allowing the phosphorus atoms to make chains with bond angles that are less strained. saltpeter: 13.86% N by mass Chile saltpeter: 16.48 % N by mass

S3. HN3 has a positive~rG", meaning that it spontaneously decomposes into H 2 and N2 at room temperature. There are no temperatures at which HN 3 will be stable. ~rH is positive and ~rS is negative, so ~rG will always be positive. SS. a. NH4N0 3(aq) + heat ~ N 20(g) + 2 H 20(/) b. 3 N02(g) + H 20(l) ~ 2 HN0 3(l) + NO(g) c. 2 PC13(l) + 0 2(g) ~ 2 POC13(l) S7. N03- , N02- , N3- , N2H/, NH/

S9.

..

:c1:

I

..

..

:cl:

..

:cl .. -P.. - cl: ..

:cl:

.. \. I ..

:c1-P-c1:

..

I

..

:c1: Trigonal pyramidal

Trigonal bipyramidal

61. CO(NH2h + 2 H20 ~ (NH4hC03 14 g 63. P40 6 forms if there is only a limited amount of oxygen available, while P40 10 will form with greater amounts of oxygen. 6S. The major source of oxygen is the fractionation of air by which air is cooled and liquefied and oxygen is separated from the other components. 67. a. superoxide b. oxide c. peroxide 69. Initially, liquid sulfur becomes less viscous when heated because the S 8 rings have greater thermal energy, which overcomes intermolecular forces. Above 150 °C the rings break and the broken rings entangle one another, causing greater viscosity. 71. a. 4.3 x 10-22 g b. 4 x 10- 19 g heat

73. 2 FeS 2(s) ~ 2 FeS(s) + S2(g) 5.2 X 102 L 7S. CS 2 + 3 Cl2 ~ CC14 + S2Cl2; Cl reduced, S oxidized 77. 8 kg from lignite, 39 kg from bituminous coal 79. Chlorine is much more electronegative than iodine, allowing it to withdraw electron density from oxygen more than iodine can, and ultimately making the H very positive. The H-0 bond is more like H + - o- , which ionizes in solution more easily. 81. a. rateHcifratec 1, = 1.394 b. rateHc 1/ rateHF = 0.7408 c. rateHc 1/ rateH1 = 1.873 83. 4 Na20 2 + 3 Fe ~ 4 Na20 + Fe30 4 8S. The bond length of the 0 2 species increases as electrons are added because they are added to the 1T* antibonding orbital. o}- is diamagnetic. 87. 2.0 mol of C-C bonds, 714.8 kJ mol- 1, 6.9 X 102 kJ mol- 1• This value, calculated from the bond energy, is too low because it doesn't include van der Waals attractions between C atoms not directly bonded to each other. 89. -50 kJ mol - 1 b. -11.0kJmol- 1 91. a. -l3.6kJmo1- 1 1 c. -24.8 kJ mol Fe20 3 is the most exothermic because it has the highest oxidation state and is therefore able to oxidize the most CO per mol Fe.

APP ENDI X Il l : Answers to Selected Exercises

93. a. c. 95. a. b.

O= C = C = C = Q b. sp ~ 92 kJmol - 1 7.6 x 10- 22 1.2 x 10- s c. [N2H4] = 0.009 mol L - i, [N2H5+] = 0.0025 mol L [N 2Hr ] = 7 .0 X 10- 13 mol L - 1

97. The acid is HO

/N~ / OH

N"

i,

. and the base 1s

0

II

N+

H2N /

" o-

The acid is weaker than nitrous acid because of electron donation by resonance in contributing structures such as

!'?

EE>

/ N°"- ~RH HO N°

99.

101.

103.

105.

The base is weaker than ammonia because of electron withdrawal by the electronegative nitrogen group. The triple bond in nitrogen is much stronger than the double bond in oxygen, so it is much harder to break. This makes it less likely that the bond in nitrogen will be broken. Sodium dihydrogen phosphate (NaH2P04) can act as a weak base or a weak acid. A buffer can be made by mixing it with either Na2HP04 or with Na3P04, depending on the desired pH of the buffer solution. F is extremely small, and so there is a huge driving force to fill the octet by adding an electron, giving a - 1 oxidation state. Other halogens have access to d orbitals, which allows for more hybridization and oxidation state options. S03 cannot be a reducing agent because the oxidation state of S is + 6, the highest possible oxidation state for S. Reducing agents need to be able to be oxidized. S02 can be a reducing agent or an oxidizing agent because the oxidation state of S is + 4.

Chapter 24 15. Typically, metals are opaque, are good conductors of heat and electricity, and are ductile and malleable, meaning they can be drawn into wires and flattened into sheets. 17. aluminum, iron, calcium, magnesium, sodium, potassium 19. Fe: hematite (Fe20 3), magnetite (Fe30 4) Hg: cinnabar (HgS) V: vanadinite [Pb 5(V04)Cl], carnotite [K2(U0 2h(V04)2 · 3 H10 J Nb: columbite [Fe(Nb03)2] 21. MgC0 3(s) + heat ~ MgO(s) + C0 2(g) Mg(OHh(s) + heat ~ MgO(s) + H 20 (g) 23. The flux is a material that will react with the gangue to form a substance with a low melting point. MgO is the flux. 25. Hydrometallurgy is used to separate metals from ores by selectively dissolving the metal in a solution, filtering out impurities, and then reducing the metal to its elemental form.

A-57

27. The Bayer process is a hydrometallurgical process by which Al 20 3 is selectively dissolved, leaving other oxides as solids. The soluble form of aluminum is Al(OH)4- . 29. Sponge powdered iron contains many small holes in the iron particles due to the escaping of the oxygen when the iron is reduced. Water-atomized powdered iron has much more smooth and dense particles as the powder is formed from molten iron. 31. a. 50% Cr, 50% V by moles; 50.5% Cr, 49.5 % V by mass b. 25 % Fe, 75% V by moles; 26.8% Fe, 73.2% V by mass c. 25% Cr, 25% Fe, 50% V by moles; 24.8 % Cr, 26.6% Fe, 48.6% V by mass 33. Cr and Fe are very close to each other in mass, so their respective atomic radii are probably close enough to form an alloy. Also, they both form body-centred cubic structures. 35. A: solid, 20 % Cr, 80% Fe B: liquid, 50% Cr, 50% Fe 37. A: solid, 20% Co and 80% Cu overall. Two phases; one is the Cu structure with 4 % Co, and the other is the Co structure with 7% Cu. There will be more of the Cu structure. B: solid Co structure, 90 % Co, 10% Cu 39. C would fill interstitial holes; Mn and Si would substitute for Fe. 41. a. Mo 2N b. CrH2 43. a. zinc c. manganese b. copper 45. - 19.4 kJ mol- 1 47. When Cr is added to steel it reacts with oxygen to form a thin and highly adherent oxide layer on the surface, which acts as an effective barrier against corrosion. A Cr-steel alloy would be used in any situation where the steel might be easily oxidized, such as when it comes in contact with water. 49. rutile: 33.3% Ti by moles, 59.9 % Ti by mass ilmenite: 20.0 % Ti by moles, 31.6% Ti by mass 51. Titanium must be arc-melted in an inert atmosphere because the high temperature and flow of electrons would cause the metal to oxidize in a normal atmosphere. 53. Ti02 is the most important industrial product of titanium and it is often used as a pigment in white paint. 55. The Bayer process is a hydrometallurgical process used to separate Al20 3 from other oxides. The Al 20 3 is selectively dissolved by hot, concentrated NaOH. The other oxides are removed as solids and the Al20 3 precipitates out of solution when the solution is neutralized. 57. cobalt and tungsten 59. 3.3 kg Fe, 2.0 kg Ti 61. Four atoms surround a tetrahedral hole and six atoms surround an octahedral hole. The octahedral hole is larger because it is surrounded by a greater number of atoms. 63. Mn has one more valence electron than does Cr, which allows it to attain higher positive oxidation states. 65. Ferromagnetic atoms, like paramagnetic ones, have unpaired electrons. However, in ferromagnetic atoms, these electrons align with their spin oriented in the same direction, resulting in a permanent magnetic field. 67. The nuclear charge of the last three is relatively high because of the lanthanide series in which the 4f subshell fall s between them and the other six metals of the group.

A-58 69. 71. 73. 7S.

APPENDIX Il l : Answers to Selected Exercises

a. 16.0 cm

b. 4.95 cm

c. 14%

92%

5.4 x 107 First, roast to form the oxide. 4 CoAsS(s) + 9 02(g) 4 CoO(s) + 4 S02(g) + As406(s) Then reduce the oxide with coke. Co(s) + CO(g) CoO(s) + C(s) The oxides of arsenic are relatively volatile and can be separated, but they are poisonous. 77. Au and Ag are found in elemental form because of their low reactivity. Na and Ca are group 1 and group 2 metals, respectively, and are highly reactive as they readily lose their valence electrons to obtain noble gas configurations.

Chapter 25 lS. a. b. c. d. 17. a. 19. a. c. 21. a. b. c. d. 23. a. c. 2S. a. b. 27.

[Ar] 4s23d8, [Ar] 3d8 [Ar] 4s23d5, [Ar] 3d3 [Kr] 5s24d 1, [Kr] 5s 14d 1 [Xe] 6s24f 4 5d3, [Xe] 4f 4 5d3 +5 b. +7 c. +4 +3, 6 b. +2, 6 +2, 4 d. + 1, 2 hexaaquachromium(III) tetracyanocuprate(II) pentaaminebromoiron(III) sulfate aminetetraaquahydroxycobalt(III) chloride [Cr(NH3 ) 6]3+ b. K 3 [Fe(CN)6] d. [Pt(H20)4][PtCl6] [Cu(en)(SCNh] [Co(NH3 MCN)3 ] , triaminetricyanocobalt(III) [Cr(en)3] 3 + , tris(ethylenediamine)chromium(III) O'-._

/0

I _......,NH3 Mn H3N/ I "NH3 NH3

0

I

H3N"

_......,NH3

Mn

I "NH3 NH3

H 3N/

29. [Fe(H 20)5 Cl]Cl • H 2 0, pentaaquachloroiron(II) chloride monohydrate [Fe(H 20)4Cl2] • 2 H2 0 , tetraaquadichloroiron(II) dihydrate 31. b, c, e. b. 2 (enantiomers) 33. a. 3 3S. a. NH3 NH3 oc0,,

I

H3N

CO NH3

,,,"co

;O>H 3.17. C2 H 5 3.18. C2H40

Chapter 4 4.1. 4.2. 4.1.

4.3.

4.4. 4.4.

4.5. 4.6. 4.6. 4.7.

Si02(s) + 3 C(s) SiC(s) + 2 CO(g) 2 C2H6 (g) + 7 Oz(g) 4 C02(g) + 6 H20 (g) Conceptual Connection Both (a) and (d) are correct. When the number of atoms of each type is balanced, the sum of the masses of the substances involved will be the same on both sides of the equation. Since molecules change during a chemical reaction, their number is not the same on both sides, nor is the number of moles necessarily the same. a. Insoluble. b. Insoluble. c. Soluble. d. Soluble. For this reaction, according to the solubility rules, no precipitation should occur. Therefore, there is no net reaction since all of the ions remain soluble. For More Practice For this reaction, the copper(II) hydroxide product will be insoluble, but the sodium bromide will be completely soluble. Therefore, the net ionic equation is: Cu(OH)z(s) Cu2+(aq) + 2 OH- (aq) H2S04Caq) + 2 LiOH(aq) 2 H20(l) + Li2S04(aq) H+(aq) + OH-(aq) H20(aq) 2 HBr(aq) + K2S03(aq) H20(l) + S02(g) + 2 KBr(aq) For More Practice 2 H+(aq) + S2 - (aq) H 2S(g) a. Cr = 0 b. Cr3+ = +3 c. c1- = -1, c = +4 d. Br = -1, Sr = +2

e. 0 = -2, S = +6 4.8.

4.2.

4.9.

f. 0 = -2, N = +5 a. This is a redox reaction in which Li is the reducing agent (it is oxidized) and Cl 2 is the oxidizing agent (it is reduced). b. This is a redox reaction in which Al is the reducing agent and Snz+ is the oxidizing agent. c. This is not a redox reaction because no oxidation states change. d. This is a redox reaction in which carbon is the reducing agent and oxygen is the oxidizing agent. Conceptual Connection (d) Since oxidation and reduction must occur together, an increase in the oxidation state of a reactant will always be accompanied by a decrease in the oxidation state of a reactant. Cr(s) + 2 H +(aq) 2 Cr2+(aq) + H2 (g)

APP EN DIX IV: Answers to In-Chapter Practice Problems

4.9.

For More Practice Cu(s) + 4 H +(aq) + 2 N0 3- (aq) ~ Cu2 +(aq) 4.10. 3 CIO- (aq) + 2 Cr(OH)4- (aq) + 2 OH- (aq) ~

+

3 Ci-(aq)

2 N02 (g)

+ 2 Croi

+

2 H 20(l)

- (aq)

+

5 H 2 0 (l)

4.11. 2 Cl0 2 + H 2 0 - Cl02 + Cl0 3 + 2 H + 4.12. 4.08 g HCl 4.12. For More Practice 22 kg HN03 4.3. Conceptual Connection (c) Since each 0 2 molecule reacts with 4 Na atoms, 12 Na atoms are required to react with 3 0 2 molecules. 4.4. Conceptual Connection (c) Nitrogen is the limiting reactant, and there is enough nitrogen to make 4 NH3 molecules. Hydrogen is in excess, and two hydrogen molecules remain after the reactants have reacted as completely as possible. 4.13. H2 is the limiting reactant, since it produces the least amount of NH3 . Therefore, 29.4 kg NH3 is the theoretical yield. 4.14. CO is the limiting reactant, since it only produces 114 g Fe. Therefore, 114 g Fe is the theoretical yield ; percentage yield = 63.4 % yield 4.5. Conceptual Connection The limiting reactant is the 1 mol H2 0, which is completely consumed. The 1 mol of H 20 requires 3 mol of N02 to completely react; therefore, 2 mol N02 remain after the reaction is complete. 4.15. 0.214 mol L- 1 NaN03 4.15. For More Practice 44.6 g KBr 4.16. 402 g C 12 H22011 4.16. For More Practice 221 mL of KC! solution 4.6. Conceptual Connection (b) The mass of a solution is equal to the mass of the solute plus the mass of the solvent. Although the solute seems to disappear, it really does not, and its mass becomes part of the mass of the solution, in accordance with the law of mass conservation. 4.17. 667 mL 4.17. For More Practice 0.105 L 4.18. 51.4 mL HN03 solution 4.18. For More Practice 0.170 g C0 2

Chapter 5 5.1. 5.1. 5.2. 5.3. 5.4. 5.5. 5.6. 5.6. 5.1.

5.7. 5.7.

1.005 bar, 753.8 Torr For More Practice 1.3 X 10-2 mbar 2.1 bar at a depth of approximately 11 m. 123 mL 11.3 L 1.65 bar, 23.9 psi 16.0 L For More Practice 1.30 X 103 mbar Conceptual Connection (a) Since 1 g of H2 contains the greatest number of moles (due to H 2 having the lowest molar mass of the listed gases), and since one mole of any ideal gas occupies the same volume, the H 2 will occupy the greatest volume. d = 4.91 g L- 1 For More Practice 44.6 g mol- 1

A-63

A-64

APPEN DIX IV: Answers to In-Chapter Practice Problems

5.8. 5.9. 5.10. 5.11. 5.12. 5.12.

70.8 g mol- 1

5.15.

2 --

5.4.

Conceptual Connection The temperature of the samples must follow the order of A < B K, therefore the reaction will proceed left toward reactants. 14.3. Conceptual Connection (c) Since N20 4 and N0 2 are both in their standard states, they each have a partial pressure of 1.0 bar. Consequently, Qp = I . Since Kp = 0.15, Qp > Kp, and the reaction proceeds to the left. 14.7. P 1 = 0.077 bar 14.8. PN, = 0.092 bar, P0 , = 0.092 bar, PNo = 0.066 bar 14.9. PNo, = 0.150 bar, PN,o. = 0.125 bar 14.10. [Fe 3 +] = 0.106mol L- 1, [SCN-] =0.006 mol L- 1, and [Fe(SCN)2+] = 0.569 mol L- 1

14.11. PH,s = 1.00 bar, PH, = 1.00 bar, Ps, = 5.13 X 10-6 bar 14.12. Using the x is small approximation, PH,s = 1.00 X 10-3 bar, PH, = 7.58 X 10-6 bar, P 82 = O.lOObar 14.4. Conceptual Connection (a) The x is small approximation is most likely to apply to a reaction with a small equilibrium constant and an initial concentration of reactant that is not too

A-73

A-74

APPEND IX IV: Answers to In-Chapter Practice Problems

small. The bigger the equilibrium constant and the smaller the initial concentration of reactant, the less likely that the x is small approximation will apply. 14.13. Adding Br2 increases the concentration of Br2, causing a shift to the left (away from the Br2) . Adding BrNO increases the concentration of BrNO, causing a shift to the right. 14.14. Decreasing the volume causes the reaction to shift right. Increasing the volume causes the reaction to shift left. 14.15. If we increase the temperature, the reaction shifts to the left. If we decrease the temperature, the reaction shifts to the right.

Chapter 15 15.1. a. H 2 0 donates a proton to C5 H5 N, making it the acid. The conjugate base is therefore OH-. Since C 5H5N accepts the proton, it is the base and becomes the conjugate acid C 5H5 NH+. b. Since HN03 donates a proton to H20 , it is the acid, making NO) the conjugate base. Since H 20 is the proton acceptor, it is the base and becomes the conjugate acid, H 30 + 15.1. Conceptual Connection (b) H 2 S04 and H2 S03 are two different acids, not a conjugate acid- base pair. 15.2. Conceptual Connect ion c10- , because the weaker the acid, the stronger the conjugate base. 15.2. a. [H30 +] = 6.7 X 10- 13 mol L - 1 Since [H30 +] < [OH-], the solution is basic. b. [H30 +] = 1.0 X 10- 7 mol L- 1 Neutral solution c. [H 0 +] = 1.2 X 10- 5 mol L - 1 3

Since [H30 +] > [OH- ], the solution is acidic. 15.3. a. 8.02 (basic) b. 11. 85 (basic) 15.4. 4.3 X 10-9 mol L- 1 15.5. 1.8 X 10-2 mol L- 1 15.6. 3.28 15.7. 2.72 15.8. 1.8 x 10- 6 15.3. Conceptual Connection (c) The validity of the x is small approximation depends on both the value of the equilibrium constant and the initial concentration- the closer that these are to one another, the less likely the approximation will be valid. 15.9. 0.85% 15.4. Conceptual Connection Solution (c) has the greatest percent ionization because percent ionization increases with decreasing weak acid concentration. Solution (b) has the lowest pH because the equilibrium H 30 + concentration increases with increasing weak acid concentration. 15.10. 2.7 X 10-7 mol L- 1 15.5. Conceptual Connection (a) A weak acid solution will usually be less than 5% dissociated. Therefore, since HCl is the only strong acid, the 1.0 mol L- 1 solution is much more acidic than either a weak acid that is twice as concentrated or a combination of two weak acids with the same concentrations. 15.11. [OW]= 0.020molL- 1 pH= 12.30 15.12. [OW] = 1.2 x 10- 2 mol L- 1 pH = 12.08 15.13. a. weak base b. pH-neutral 15.14. 9.07 15.15. a. pH-neutral b. weak acid c. weak acid

APP EN DIX IV: Answers to In-Chapter Practice Problems

15.16. a. basic b. acidic c. pH-neutral d. acidic 15.17. 3.83 15.18. [SO/ - ] = 0.00386 mol L- 1 pH= 1.945 15.19. 5.6 X 10- 11 mol L- 1 15.6. Conceptual Connection (a) Since the carbon atom in (a) is bonded to another oxygen atom, which draws electron density away from the C-H bond (weakening and polarizing it), and the carbon atom in (b) is bonded only to other hydrogen atoms, the proton in structure (a) is more acidic. 15.7. Conceptual Connection The most basic nitrogen is (d), because it is attached to an electron-donating methyl group. The least basic nitrogen is (b), because it is attached to a highly electronegative fluorine atom. Since nitrogen (c) is in close proximity to the electrondonating methyl group, it is more basic than (a), which is close to the electron withdrawing fluorine. d > c > a > b

Chapter 16 16.1. pH = 4.44 16.1. For More Practice pH= 3.44 16.2. 9.04 16.1. Conceptual Connection (a) Since the pH of the buffer is less than the pKa of the acid, the buffer must contain more acid than base ([HA] > [A- ]). In order to raise the pH of the buffer from 4.25 to 4 .72, you must add more of the weak base. (Adding a base will make the buffer solution more basic.) 16.3. 4.87 16.3. For More Practice 4.65 16.2. Conceptual Connection (b) Since acid is added to the buffer, the pH will become slightly lower (slightly more acidic). Answer (a) reflects too large a change in pH for a buffer, and answers (c) and (d) are in the wrong direction. 16.4. 9.68 16.4. For More Practice 9.56 16.5. hypochlorous acid (HCIO); 2.4 g NaClO 16.3. Conceptual Connection (a) Adding 0.050 mol of HCl will destroy the buffer because it will react with all of the NaF, leaving no conjugate base in the buffer mixture. 16.6. 1.74 16.7. 8.04 16.4. Conceptual Connection (c) Since the volumes and concentrations of all three acids are the same, the volume of NaOH required to reach the first equivalence point (and the only equivalence point for titrations i and iii) is the same for all three titrations. 16.8. 2.30 X 10-6 mol L- 1 16.9. 5.3 x 10- 13 16.10. 1.21 X 10-5 mol L- 1 16.11. FeC0 3 will be more soluble in an acidic solution than PbBr2 because the co}ion is a basic anion, whereas Br- is the conjugate base of a strong acid (HBr) and is therefore pH-neutral. 16.12. Q > Ksp; therefore, a precipitate forms . 16.13. 2.9 X 10- 6 mol L- 1

A-75

A-76

APPEN DIX IV: Answers to In-Chapter Practice Problems

16.14. a. AgCl precipitates first; [NaCl] = 7. 1 X 10-9 mol L- 1 b. [Ag+] is 1.5 X 10-s mol L - I when PbC1 2 begins to precipitate, and [Pb2l is 0.085 mol L- 1• 16.15. 9.6 X 10- 6 mol L- 1 16.5. Conceptual Connection (c) Only NaCN contains an anion (CN- ) that forms a complex ion with Cu2+. [From Table 16.3 we can see that Kr = 1.0 X 1025 for [Cu(CN)i- .J Therefore, the presence of CN- will drive the dissolution reaction of CuS.

Chapter 17 17.1.

17.2.

17.2. 17.1.

17.3. 17.4.

17.2.

17.5. 17.6.

17.7. 17.7.

a. positive b. negative c. positive a. -548 J K- 1 mo1- 1 b. !::,.Ssys is negative. c. !::,.Suniv is negative, and the reaction is not spontaneous. For More Practice 375 K Conceptual Connection Biological systems do not violate the second law of thermodynamics. The key to understanding this concept is realizing that entropy changes in the system can be negative as long as the entropy change of the universe is positive. Biological systems can decrease their own entropy, but only at the expense of creating more entropy in the surroundings (which they do primarily by emitting the heat they generate by their metabolic processes). Thus, for any biological process, !::,.Suniv is positive. -153.2 J K- 1 mol- 1 !::,.G = -101.6 X 103 J mo1- 1 Therefore, the reaction is spontaneous. Since both !::,./f and t::,.,S are negative, as the temperature increases, t::,.,G will become more positive. Conceptual Connection (a) Sublimation is endothermic (it requires energy to overcome the intermolecular forces that hold solid carbon dioxide together), so t::,.,H is positive. The number of moles of gas increases when the solid turns into a gas, so the entropy of the carbon dioxide increases and t::,.,S is positive. Since t::,.p = !::,.,H - T !::,.,S, !::,.,G is positive at low temperature and negative at high temperature. t::,. ,G' = -36.3 kJ mol- 1 Since t::,.,G' is negative, the reaction is spontaneous at this temperature. t::,.,G' = -42.1 kJ mo1- 1 Since the value of t::,.,G' at the lowered temperature is more negative (or less positive) (which is -36.3 kJ mol- 1), the reaction is more spontaneous. !::,.,G' = -689.6 kJ mol- 1 Since t::,.,G' is negative, the reaction is spontaneous at this temperature. For More Practice !::,.,G' = -689.7 kJ mol- 1 (at 25 ' C) The value calculated for t::,.p' from the tabulated values (-689.6 kJ mol- 1) is the same, to within 1 in the least significant digit, as the value calculated using the equation for t::,.p' . t::,.p = -649.7 kJ mol- 1 (at 500.0 K) You could not calculate t::,.p' at 500.0 K using tabulated !::,.rG values because the tabulated values of free energy are calculated at a standard temperature of 298 K, much lower than 500 K. + 107. 1 kJ mo1- 1 30.5 kJ mo1- 1 X 38 = 1160 kJ mo1- 1 of Gibbs energy is stored, and 2880 kJ mo1- 1 - 1160 kJ mol- 1 = 1720 kJ mol- 1 is left over. t::,.p = -129 kJ mol- 1 The reaction is more spontaneous under these conditions than under standard conditions because t::,.,G is more negative than t::,.,G 0

0

17.8. 17.9. 17.10.

0

•

APPEN DIX IV: Answers to In-Chapter Practice Problems

17.10. For More Practice 8.314 J K- 1 mol X 310 K X In (

0.005 x 0.005) 0.005

tJ.p = -30.5 kJ mol- 1 + - - - - - - - - - - - - - - - - - - 1000 J/k:J

= -44.2 kJ mol- 1 17.3.

17.11. 17.11. 17.12. 17.13. 17.14.

Conceptual Connection (a) A high concentration of reactants relative to products will lead to making the term RT In Q in Equation 17 .14 negative. tJ.rG will be more negative than t:,.p•, and the reaction will be more spontaneous. t:,.p = -20.3 kJ mol- 1, more PbC12 could dissolve For More Practice T=331 K -33.0 kJ mo1- 1 tJ.rH' = -60.4 kJ mol- 1; tJ.rS' = -68.4 J K- 1 mol- 1 Pco2 > 0.011 bar

Chapter 18 18.1. 0.31 v 18.2. E~en = 0.36 V 18.3. (a) The reaction will be spontaneous under standard conditions. (b) The reaction will not be spontaneous under standard conditions. 18.1. Conceptual Connection (d) The reduction of HN03 is listed below the reduction of Br2 and above the reduction of 12 in Table 18.1. Since any reduction half-reaction is spontaneous when paired with the reverse of a half-reaction below it in the table, the reduction ofHN03 is spontaneous when paired with the oxidation ofi-, but is not spontaneous when paired with the oxidation of Br- . 18.2. Conceptual Connection (c) Ag falls above the half-reaction for the reduction of H+ but below the halfreaction for the reduction of N0 3- in Table 18.1. 18.4. tJ.rG' = 1.1 X 102 kJ mol- 1 Since tJ.rG' is positive, the reaction is not spontaneous. 18.3. Conceptual Connection (a) Br is more electronegative than I. If the two atoms were in competition for the electron, the electron would go to the more electronegative atom (Br). Therefore, 12 does not spontaneously gain electrons from Br-. 18.5. 4.5 x 10 3 18.6. K = 4.2 x 1045 18.4. Conceptual Connection (a) Since K < 1 , £;;en is negative. (Under standard conditions, the reaction is not spontaneous.) Since Q < K,Ecell is positive. (The reaction is spontaneous under the nonstandard conditions of the cell.) 18.7. Ksp = 2.8 x 10-39 18.8. Anode: 2 H20(l) ~ 0 2 (g) + 4 H +(aq) + 4 eCathode: 2 H20(/) + 2 e- ~ H 2(g) + 2 OH-(aq) 18.9. 6.0 X 10 1 min 18.5. Conceptual Connection Cu is the only metal in the list that does not act as a sacrificial anode. This is because Cu is a weaker reducing agent than Fe. Mn, Mg, and Zn are all stronger reducing agents than Fe and thus can serve as sacrificial anodes.

Chapter 19 19.1. 2JgPo ~ 2 g Pb + iHe 19.2. a. 2§iu ~ 2§JTh + iHe 2 §6Th ~ 2§lPa + -~e 2 §lPa ~ 2~JAc + iHe

A-77

A-78

APPENDIX IV: Answers to In-Chapter Practice Problems

b. fTNa ~ f6Ne

+_pe

c. ~~Kr + _~e ~ ~~Br 19.2. For More Practice Positron emission (n K ~ n Ar + +~e) or electron capture

(igK + +~e ~ igAr) 19.1. Conceptual Connection (c) The arrow labelled x represents a decrease of 2 neutrons and 2 protons, indicative of alpha decay. The arrow labelled y represents a decrease of 1 neutron and an increase of I proton, indicative of beta decay. 19.3. a. positron emission b. beta decay c. positron emission 19.2. Conceptual Connection (b) The half-life is the time it takes for the number of nuclei to decay to one-half of their original number. 19.4. 10.7 y 19.5. t = 964 y No, the C-14 content suggests that the scroll is from about C.E. 1000, not 500 s. c .E. 19.6. 1.0 x 109 y 19.3. Conceptual Connection x _ 11 . 2 7593 10

J x 6.0221X1023 U-235ate1Tls U-235 atom 1 mol U-235

= 1.6617 X 10 13 J/mol U-235 The fission of 1 mol of U-235 produces about 17 billion kJ. This is 17 million times more energetic than a chemical reaction that produces 1000 kJ mo1- 1.

19.7.

Mass defect= 1.934 u Nuclear binding energy= 7 .569 MeV/nucleon

19.4. Conceptual Connection Lawrencium-256 19.5. Conceptual Connection Nuclide A. Because nuclide A has a shorter half-life, more of the nuclides will decay, and therefore produce radiation, before they exit the body.

Chapter 20 20.1.

a.

H

I

H

H

I

H

I

I

H

I

H

I

H - C- C- C- C- C- C- H

I

H b.

I

I

H

I

H

H

I

H

I

H

H

I

H- C- H

I

I I

I

H - C- C- C- C- H

I

H c.

H H

I

I

H

I

H

H

H \/

H2

H

'-- /c". /

H-c

\

c-H

I

H-c-c_H

I

H

\

H

/c"-

CH2

\

CH2

I

H2C - CH2

APP EN DIX IV: Answers to In-Chapter Practice Problems

20.1. Conceptual Connection (c) and (d), which also have the molecular formula C6H 12 20.2. Conceptual Connection gauche-staggered 20.2. a. Z b. E c. E 20.3. Conceptual Connection (b) This structure is the only one that contains a carbon atom (the one on the left) with four different substituent groups attached (a Br atom, a Cl atom, an H atom, and a CH3 group). 20.3. a. S

b. s c. R

20.4. a. IHD = 2; possible combinations are two rings, one ring and a double bond, two double bonds, or one triple bond. b. IHD = 2; possible combinations are two rings, one ring and a double bond, two double bonds, or one triple bond. 20.4. For More Practice a. IHD = 3 b. IHD = 4 20.5. propanone (or acetone)

Chapter 21 21.1. Ho:

..

f

~ ' H ...C'o ·- H

I

H

21.2. 21.3.

HO-H

+

O-

H

I

H

ow -3

+3

a. CH 3 - C ==N c.

0

-3 /

.Jc

CH3

-....... - 3

CH3

21 .4. LiAIH4 or H2 , Pd/C 21.1. Conceptual Connection This can be confirmed by determining the oxidation states for carbon or nitrogen atoms in the functional group. All reactions involve a change in oxidation state. 21 .2. Conceptual Connection From Table 21.1 , the pK3 values of the conjugate acids are in the increasing order: HI < HBr < HCl < HF < CH3COOH < H 2 0 < CH30H < NH 3 . This means that good leaving groups are the conjugate bases of strong acids. That is, good leaving groups are weak bases. 21.5. SNl (weak nucleophile, secondary alkyl halide). Product is a racemic mixture: +

21.3. Conceptual Connection H2 0 + H 2S04 ~ H3 0 + + HS04- ; in fact, HS04- is the only conjugate base in this reaction mixture. Given that HS04- (which is also an acid) is reacting as a base towards a carbocation, a carbocation must be extremely acidic (pKa