Chemistry: The Central Science

CHEMISTRY THE CENTRAL

SCIENCE

ELEVENTH EDITION

CHEMISTRY THE CENTRAL

SCIENCE

Theodore L. Brown Univenityot Illinois at Urbana-Champaign

H. Eugene LeMay, Jr. UnivenityofNevada, Reno

Bruce E. Bursten UnlvenilyofTennenee

Catherine

J.

Murphy

UnivHsityofSouthCarolina

With Contribution! from

Patrick Woodward The Ohio State University

Upper S.tddle River, NJ 07458

Library of Congress Cataloging-in-Publication Data Chemistry: the central science.-llth ed. I Theodore L. Brown .. [et al.]. p.cm. Includes index. ISBN 978-0-13-600617-6

1. Chemistry-Textbooks. I. Brown, Theodore L.

QD31.3.C43145 2009 540-dc22

2007043227

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ISBN 0-13-600617-5 ISBN 978-0-13-600617-6 Pearson Education Ltd., London

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To our students, whose enthusiasm and curiosity

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Brief Contents Preface xxiv

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25

About the Authors XXXIII

Introduction: Matter and Measurement

36

Atoms, Molecules, and Ions

Stoichiometry: Calculations with Chemical Formulas and Equations 78 Aqueous Reactions and Solution Stoichiometry 118 Thermochemistry 164

210

Electronic Structure of Atoms

Periodic Properties of the Elements

254

Basic Concepts of Chemical Bonding

296

Molecular Geometry and Bonding Theories Gases

340

392

Intermolecular Forces, Liquids, and Solids

436

Modern Materials 480

526

Properties of Solutions Chemical Kinetics

572

Chemical Equilibrium 626

666

Acid-Base Equilibria

Additional Aspects of Aqueous Equilibria 718 Chemistry of the Environment 766 Chemical Thermodynamics

800

Electrochemistry 842 Nuclear Chemistry

892

Chemistry of the Nonmetals Metals and Metallurgy

930

980

Chemistry of Coordination Compounds

1012

The Chemistry of Life: Organic and Biological Chemistry 1050

Appendices A Mathematical Operations 1104

B Properties of Water 1111 C Thermodynamic Quantities for Selected Substances at 298.15 K (25 °C) 1112

D Aqueous Equilibrium Constants 1115

E

Standard Reduction Potentials at 25 oc

Answers to Selected Exercises A-1 Answers to "Give It Some Thought" Glossary

G-1

Photo/ Art Credits Index

vi

l-1

P-1

A-36

1117

Contents Preface xxiv About the Authors

1

xxxm

Introduction: Matter and Measurement 1.1

THE STUDY OF CHEMISTRY 2 Why Study

The Atomic and Molecular Perspective of Chemistry 2 Chemistry?

1.2

1

3

CLASSIFICATIONS OF MATTER 4 States of Matter 4

Pure Substances

5

Elements 6

Compounds 6

Mixtures 8

1.3

PROPERTIES OF MATTER 9 Physical and Chemical Changes 10

1.4

1.5

51 Units 13

Length and Mass 1 4

Volume 1 6

Density 17

Temperature

15

Derived 51 Units 1 6

UNCERTAINTY IN MEASUREMENT 2 0 Precision and Accuracy 20 Calculations

1.6

Separation of Mixtures 11

UNITS OF MEASUREMENT 13

Significant Figures in

Significant Figures 21

23

DIMENSIONAL ANALYSIS 2 4 Using Two or More Conversion Factors 26

SUMMARY AND KEY TERMS 29

Conversions Involving Volume 27

KEY SKILLS 30

KEY EQUATIONS 30 VISUALIZING CONCEPTS 30 EXERCISES 31 ADDITIONAL EXERCISES 33

Chemistry Put to WOrk Chemistry and the Chemical Industry A Closer wok The Scientific Method 13 Chemistry Put ro WOrk Chemistry in the News 18 Strategies in Chemistry Estimating Answers 26 Strategies in Chemistry The !mportance of Practice 28

2

Atoms, Molecules, and Ions 2.1 2.2

4

36

THE ATOMIC THEORY OF MATTER 38 THE DISCOVERY OF ATOMIC STRUCTURE 39 Cathode Rays and Electrons

39

Radioactivity 41

The Nuclear Atom 41

2.3

THE MODERN VIEW OF ATOMIC STRUCTURE 4 3 Atomic Numbers, Mass Numbers, and Isotopes 44

2.4

ATOMIC WEIGHTS 46 The Atomic Mass Scale 46

Average Atomic Masses 47

2.5

THE PERIODIC TABLE 48

2.6

MOLECULES AND MOLECULAR COMPOUNDS 5 1 Molecules and Chemical Formulas Picturing Molecules

2.7

53

Molecular and Empirical Formulas 52

IONS AND IONIC COMPOUNDS 54 Predicting Ionic Charges 55

2.8

52

Ionic Compounds 56

NAMING INORGANIC COMPOUNDS 59 Names and Formulas of Ionic Compounds 59

Names and Formulas of Acids 64

Names and Formulas of Binary Molecular Compounds 65

vii

viii

Contents 2.9

SOME SIMPLE ORGANIC COMPOUNDS 66 Alkanes 66

Some Derivatives of Alkanes 66

SUMMARY AND KEY TERMS 68 VISUALIZING CONCEPTS 69

KEY SKILLS 69

EXERCISFS 70

ADDITIONAL EXERCISFS 75 A Clour Ltlolr Basic Forces 45

Closer Ltlok The Mass Spectrometer 48 Closer Ltlok Glenn Seaborg and Seaborgium 51 Chemistry and Life Elements Required by Living Organisms Strategies in Chemistry Pattern Recognition 58

A A

3

57

Stoichiometry : Calculations with Chemical Formulas an d Equations 78 3.1

CHEMICAL EQUATIONS 80 Balancing Equations 80

Indicating the States of Reactants and Products 83

3.2

SOME SIMPLE PATTERNS OF CHEMICAL REACTIVITY 84

3.3

FORMULA WEIGHTS 87

3.4

AVOGADRO'S NUMBER AND THE MOLE 8 9

Combination and Decomposition Reactions 84

Formula and Molecular Weights 87

Molar Mass 91

Combustion in Air 86

Percentage Composition from Formulas 88

lnterconverting Masses and Moles 9 3

lnterconverting

Masses and Numbers of Particles 94

3.5

EMPIRICAL FORMULAS FROM ANALYSES 9 5 Molecular Formula from Empirical Formula 96

Combustion Analysis 9 7

3.6

QUANTITATIVE INFORMATION FROM BALANCED

3.7

LIMITING REACTANTS 102

EQUATIONS 98 Theoretical Yields 1 05

SUMMARY AND KEY TERMS 107 KEY SKILLS 108 KEY EQUATIONS 108 VISUALIZING CONCEPTS 108 EXERCISFS 109

ADDITIONAL EXERCISFS ll5

INTEGRATIVE EXERCISFS 116

Strategies in Chemistry Problem Solving 89 Chemistry and Life Glucose Monitoring 102 Strategies in Chemistry How to Take a Test 106

4

Aqueous Reactions and Solution Stoichiometry 118 4.1

4.2

GENERAL PROPERTIES OF AQUEOUS SOLUTIONS 120 Electrolytic Properties 1 20

Ionic Compounds in Water 1 20

Compounds in Water 1 22

Strong and Weak Electrolytes 1 22

PRECIPITATION REACTIONS 124 Solubility Guidelines for Ionic Compounds 1 24 Reactions 1 26

4.3

Exchange (Metathesis)

Ionic Equations 1 27

ACID-BASE REACTIONS 1 2 8 Acids 1 29

Bases 1 29

Strong and Weak Acids and Bases 1 30

Identifying Strong and Weak Electrolytes 1 31 Salts 1 32

4.4

Molecular

Neutralization Reactions and

Acid-Base Reactions with Gas Formation 1 34

OXIDATION-REDUCTION REACTIONS 135 Oxidation and Reduction 1 36

Oxidation Numbers 1 3 7

Oxidation of Metals b y Acids and Salts 1 38

T h e Activity Series

1 40

Contents 4.5

CONCENTRATIONS OF SOLUTIONS 1 42 Molarity 146

Expressing the Concentration of an Electrolyte 145

lnterconverting Molarity, Moles, and Volume 145

4.6

Dilution 146

SOLUTION STOICHIOMETRY AND CHEMICAL ANALYSIS 1 4 9 T i trations 150 SUMMARY AND KEY TERMS KEY EQUATIONS EXERCISES

155

156

1 54

KEY SKiLLS 155

VISUALIZiNG CONCEPTS

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

156

161

162

Chemistry Put to \Vc>rk Antacids 135 A CkJSer Look The Aura of Gold 143 Strategies in Chemistry Analyzing Chemical Reactions 143 Chemistry and Life DrinkingToo Much Water Can KillYou 147

5

Thermochemistry 5.1

164

THE NATURE OF ENERGY 166 Kinetic Energy and Potential Energy 166 Surroundings 168

5.2

System and

THE FIRST LAW OF THERMODYNAMICS 170

Relating llE to Heat and Work 171

Internal Energy 170

State Functions 1 72

Exothermic Processes 172

5.3

Units of Energy 167

Transferring Energy: Work and Heat 168

Endothermic and

ENTHALPY 1 74

5.4

ENTHALPIES OF REACTION 177

5.5

CALORIMETRY 179 Heat Capacity and Specific Heat 179

Constant-Pressure Calorimetry 182

Bomb Calorimetry (Constant-Volume Calorimetry) 183

5.6

HESS'S LAW 184

5.7

ENTHALPIES OF FORMATION 1 8 8

5.8

FOODS AND FUELS 1 93

Using Enthalpies o f Formation t o Calculate Enthalpies o f Reaction 190

Foods 193

Fuels 195

Other Energy Sources 197

SUMMARY AND KEY TERMS KEY EQUATIONS EXERCISES

202

201

199

KEY SKiLLS 200

VISUALIZING CONCEPTS

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

201

207

209

CU.ser Look Energy Enthalpy, and P-V W ork 176 Strategies in Chemistry Using Enthalpy as a Guide 180 Chemistry atld Life The Regulation of Human Body Temperature 01emistry Put to \Vc>rk The Hybrid Car 196 ,

A

6

2 10

Electronic Structure of Atoms 6.1

THE WAVE NATURE OF LIGHT 2 1 2

6.2

QUANTIZED ENERGY AND PHOTONS 2 1 5 Hot Objects and the Quantization of Energy 215

185

The Photoelectric Effect

and Photons 216

6.3

LINE SPECTRA AND THE BOHR MODEL 2 1 8 Line Spectra 2 18 Atom 220

6.4

Bohr's Model 2 19

The Energy States o f the Hydrogen

Limitations of the Bohr Model 222

THE WAVE BEHAVIOR OF MATTER 222 The Uncertainty Principle 223

6.5

QUANTUM MECHANICS AND ATOMIC ORBITALS 224 Orbitals and Quantum Numbers 226

ix

x

Contents 6.6

REPRESENTATIONS OF ORBITALS 228 The p Orbitals 231

The s Orbitals 228

6.7

The

d and f Orbitals

232

MANY-ELECTRON ATOMS 232 Orbitals and Their Energies 232

Electron Spin and the Pauli Exclusion

Principle 233

6.8

ELECTRON CONFIGURATIONS 234 Hund's Rule 235

Condensed Electron Configurations 237

Transition Metals 238

6.9

The Lanthanides and Actinides 239

ELECTRON CONFIGURATIONS AND THE PERIODIC TABLE 240 Anomalous Electron Configurations 243

SUMMARY AND KEYTERMS 244

KEY SKILLS 245

KEYEQUATIONS 246 VISUALIZING CONCEPTS 246 EXERCISES 247

ADDITIONAL EXERCISES 251

INTEGRATIVE EXERCISES 253

Clo1er Look The Speed of Light 214 Closer Look Measurement and the Uncertainty Principle 225 A Closer Look Probability Density and Radial Probability Functions 230 A Closer Look Experimental Evidence for Electron Spin 234 Chemistry and Life Nuclear Spin and Magnetic Resonance Imaging 236

A

A

7

Periodic Properties of the Elements 7.1

DEVELOPMENT OF THE PERIODIC TABLE 2 5 6

7.2

EFFECTIVE NUCLEAR CHARGE 257

7.3

SIZES OF ATOMS AND IONS 259

7.4

IONIZATION ENERGY 264

Periodic Trends in Atomic Radii 262

Periodic Trends in Ionic Radii 263

Variations in Successive Ionization Energies 266 Ionization Energies 267

Periodic Trends in First

Electron Configurations of Ions 268

7.5

ELECTRON AFFINITIES 270

7.6

METALS, NONMETALS, AND METALLOIDS 271

7.7

GROUP TRENDS FOR THE ACTIVE METALS 276

Metals 272

Nonmetals 274

Group 1 A: The Alkali Metals 276

7.8

254

Metalloids 276

Group 2A: The Alkaline Earth Metals 279

GROUP TRENDS FOR SELECTED NONMETALS 281 Hydrogen 2 8 1

Group 6 A : T h e Oxygen Group 282

The Halogens 283

SUMMARY AND KEYTERMS 286 KEY EQUATIONS 287 EXERCISES 289

Group 7A:

Group SA: The Noble Gases 284

KEY SKILLS 287

VISUAliZING CONCEPTS 288

ADDITIONAL EXERCISES 292

INTEGRATIVE EXERCISES 294

Closer Look Effective Nuclear Charge 260 Chemistry and Life Ionic Size Makes a Big Difference 265 Chemistry and Life The Improbable Development of Lithium Drugs

A

8

Basic Concepts of Chemical Bonding 8.1

280

296

CHEMICAL BONDS, LEWIS SYMBOLS, AND THE OCTET RULE 298 Lewis Symbols 298

8.2

The Octet Rule 299

IONIC BONDING 299 Energetics of Ionic Bond Formation 301

the S· and p-Biock Elements 302

Electron Configurations of Ions of

Transition-Metal Ions 303

Contents 8.3

COVALENT BONDING 305 Lewis Structures 30S

8.4

Multiple Bonds 307

BOND POLARITY AND ELECTRONEGATIVITY 307 Electronegativity 30B

Electronegativity and Bond Polarity 308

Dipole Moments 310

Differentiating Ionic and Covalent Bonding 3 1 2

8.5

DRAWING LEWIS STRUCTURES 314

8.6

RESONANCE STRUCTURES 319

8.7

EXCEPTIONS TO THE OCTET RULE 322

Formal Charge 316

Resonance in Benzene 321

Odd Number of Electrons 322

Less than an Octet of Valence Electrons 322

More than an Octet of Valence Electrons 323

8.8

STRENGTHS OF COVALENT BONDS 325 Bond Enthalpies and the Enthalpies of Reactions 326 Bond Enthalpy and Bond Length 329 SUMMARY AND KEY TERMS

KEY EQUATIONS 332 EXERCISES

333

331

KEY SKILLS 332

VISUALIZING CONCEPTS

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

332

336

338

ChuFr Loolr Calculation of Lattice Energies: The Bom-Haber Cycle 304 A Goser Look Oxidation Numbers, Formal Charges, and Actual Partial Charges GJtmJistry Put to Work Explosives and Alfred Nobel 328 A

9

Molecular Geometry and Bonding Theories 340 9.1 9.2

MOLECULAR SHAPES 342 THE VSEPR MODEL 344 The Effect of Nonbonding Electrons and Multiple Bonds on Bond Angles 348

Molecules with Expanded Valence Shells 349

Shapes of Larger Molecules 352

9.3

MOLECULAR SHAPE AND MOLECULAR POLARITY 3 5 3

9.4

COVALENT BONDING AND ORBITAL OVERLAP 355

9.5

HYBRID ORBITALS 357

sp Hybrid Orbitals 357

Hybridization Involving

9.6

2 3 sp and sp Hybrid Orbitals 358

d Orbitals

360

Hybrid Orbital Summary 360

MULTIPLE BONDS 362 Resonance Structures, Delocalization, and

7T

Bonding 365

General Conclusions 368

9.7

MOLECULAR ORBITALS 368 The Hydrogen Molecule 368

9.8

Bond Order 370

SECOND-ROW DIATOMIC MOLECULES 371 Molecular Orbitals for Li2 and Be2 371 2p Atomic Orbitals 372

Molecular Orbitals from

Electron Configurations for B2 Through Ne2 375

Electron Configurations and Molecular Properties 376 Heteronuclear Diatomic Molecules 379 SUMMA RY AND KEY TERMS

KEY EQUATIONS 383 EXERCISES

384

381

KEY SKILLS 383

VISUALIZING CONCEPTS

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

383

388

390

Chemistry and Life The Chemistry of Vision 367 A C/osFr Look Phases in Atomic and Molecular Orbitals Chemistry Put to Wrk Orbitals and Energy 380

373

318

xi

xii

Contents

1 Q Gases

392

10.1

CHARACTERISTICS OF GASES 394

10.2

PRESSURE 395 Atmospheric Pressure and the Barometer 395

10.3

THE GAS LAWS 398 The Pressure-Volume Relationship: Boyle's Law 399 Volume Relationship: Charles's Law 400

The Temperature­

The Quantity-Volume Relationship:

Avogadro's Law 400

10.4

THE IDEAL-GAS EQ UATION 402

10.5

FURTHER APPLICATIONS OF THE IDEAL-GAS

Relating the Ideal-Gas Equation and the Gas Laws 404

EQUATION 406 Volumes of Gases in Chemical

Gas Densities and Molar Mass 406 Reactions 408

10.6

GAS MIXTURES AND PARTIAL PRESSURES 4 1 0

10.7

KINETIC-MOLECULAR THEORY 4 1 4

Partial Pressures and Mole Fractions 411

Distributions of Molecular Speed 414

Collecting Gases over Water 412 Application to the Gas Laws 41S

10.8

MOLECULAR EFFUSION AND DIFFUSION 417

10.9

REAL GASES: DEVIATIONS FROM IDEAL BEHAVIOR 420

Graham's Law of Effusion 418

Diffusion and Mean Free Path 419

The van der Waals Equation 422 SUMMARY AND KEYTERMS

KEYEQUATIONS 426 EXERCISES 427

425

KEY SKILLS

426

VISUALIZING CONCEPTS 426

ADDITIONAL EXERCISES 432

INTEGRATIVE EXERCISES 434

Chemistry and Life Blood Pressure 398 Strategies in Chemistry Calculations Involving Many Variables Chemistry Put to Work Gas Pipelines 409 A Closer Look The Ideal-Gas Equation 416 Chemistry Put to Work Gas Separations 420

11

404

Intermolecular Forces, Liquids, and Solids 436 11.1

A MOLECULAR COMPARISON OF GASES, LIQUIDS, AND

11.2

INTERMOLECULAR FORCES 439

SOLIDS 438

ion-Dipole Forces 440 Forces 441 Forces

11.3

London Dispersion

Comparing Intermolecular

446

SOME PROPERTIES OF LIQUIDS 447 Viscosity 447

11.4

Dipole-Dipole Forces 440

Hydrogen Bonding 443

Surface Tension 448

PHASE CHANGES 449 Energy Changes Accompanying Phase Changes 449

Heating Curves 450

Critical Temperature and Pressure 452

11.5

VAPOR PRESSURE 453 Explaining Vapor Pressure on t h e Molecular Level 4 5 4 Pressure, and Temperature 454

11.6

Volatility, Vapor

Vapor Pressure and Boiling Point 455

PHASE DIAGR AMS 4 5 6 T h e Phase Diagrams of H20 and C02 457

Contents 11.7

STRUCTURES OF SOLIDS 458 Unit Cells 460

The Crystal Structure of Sodium Chloride 461

Close Packing of Spheres 463

11.8

BONDING IN SOLIDS 464 Molecular Solids 464

Ionic Solids 467

Covalent-Network Solids 466

Metallic Solids 467 SUMMARY AND KEYTERMS 469

KEYSKILLS 470

VISUALIZING CONCEPTS 471

EXERCISES 472

ADDITIONAL EXERCISES 477

INTEGRATIVE EXERCISES 478

Chemistry Put to WOrk Supercritical Fluid Extraction A Closer Look The Clausius-Clapeyron Equation 456 A Gt>ser Look X-Ray Diffraction by Crystals 465 A Closer Look The Third Form of Carbon 468

12 Modern Materials 12.1

453

480

CLASSES OF MATERIALS 482

12.2

ELECTRONIC STRUCTURE OF MATERIALS 482

12.3

SEMICONDUCTORS 484 Semiconductor Doping 488 Conversion 489

12.4

The Silicon Chip 489

Solar Energy

Semiconductor Light-Emitting Diodes 492

CERAMICS 493 Applications of Ceramics 494

Making Ceramics 494

12.5

SUPERCONDUCTORS 495

12.6

POLYMERS AND PLASTICS 499

12.7

BIOMATERIALS 505

Ceramic Superconductors 497

Making Polymers 499

Structure and Physical Properties of Polymers 502

Characteristics of Biomaterials 507 Heart Repair 508

12.8

Polymeric Biomaterials 507

Vascular Grafts 509

Artificial Tissue 509

LIQUID CRYSTALS 510 Types of Liquid Crystalline Phases 51 1

12.9

NANOMATERIALS 513 Semiconductors on the Nanoscale 513

Metals on the Nanoscale 515

Carbon Nanotubes 515 SUMMARY AND KEYTERMS 517

KEYSKILLS 519

VISUALIZING CONCEPTS 519

EXERCISES 520

ADDITIONAL EXERCISES 524

INTEGRATIVE EXERCISES 525

Clt>ser Look The Transistor 490 Chemistry Put to WOrk Cell Phone Tower Range 498 Chemistry Put to WOrk Recycling Plastics 501 Chemistry Put to Work Toward the Plastic Car 506 Chemistry Put to WOrk Liquid Crystal Displays 513

A

1 3 Properties of Solutions 13.1

526

THE SOLUTION PROCESS 528 The Effect of Intermolecular Forces 528 Formation 529

Energy Changes and Solution

Solution Formation, Spontaneity, and Entropy 531

Solution Formation and Chemical Reactions 533

13.2

SATURATED SOLUTIONS AND SOLUBILITY 534

13.3

FACTORS AFFECTING SOLUBILITY 535 Solute-Solvent Interactions 536 Temperature Effects 541

Pressure Effects 539

xiii

xiv

Contents 1 3 .4

WAYS OF EXPRESSING CONCENTRATION 542 Mass Percentage, ppm, and ppb 542 and Molality 543

13.5

Mole Fraction, Molarity,

Conversion of Concentration Units 544

COLLIGATIVE PROPERTIES 546 Lowering the Vapor Pressure 546 Freezing-Point Depression 550

Boiling-Point Elevation 549 Osmosis 551

Determination of

Molar Mass 555

1 3 .6

COLLOIDS 556 Hydrophilic and Hydrophobic Colloids 557

Removal of Colloidal

Particles 560

SUMMARY AND KEY TERMS 561 KEY EQUATIONS 563 EXERCISES

565

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

A Closer IAolt

KEY SKILLS 562

VISUALIZING CONCEPTS

563

569

570

Hydrates 533

Chemistry and Life Fat- and Water-Soluble Vitamins 538 Chemistry and Life Blood Gases and Deep-Sea Diving 540 A Ooser IAok Ideal Solutions with Two or MoreVolatile Components A Ooser Look Colligative Properties of Electrolyte Solutions 554 Chemistry and Life Sickle-Cell Anemia 559

14 Chemical Kinetics

548

572

14.1

FACTORS THAT AFFECT REACTION RATES 574

14.2

REACTION RATES 575 Change of Rate with lime 577

Instantaneous Rate 577

Reaction Rates

and Stoichiometry 578

14.3

THE RATE LAW: THE EFFECT OF CONCENTRATION ON RATE 580 Reaction Orders: The Exponents in the Rate Law 581 Rate Constants 583

14.4

Units of

Using Initial Rates to Determine Rate Laws 584

THE CHANGE OF CONCENTRATION WITH TIME 585 First-Order Reactions 586

Second-Order Reactions 588

Half-life 589

14.5

TEMPERATURE AND RATE 591 T h e Collision Model 592 Activation Energy 592

T h e Orientation Factor 592 The Arrhenius Equation 594

Determining the

Activation Energy 595

14.6

REACTION MECHANISMS 597 Elementary Reactions 597

Multistep Mechanisms 598

Elementary Reactions 599

The Rate-Determining Step for a Multistep

Mechanism 600

Rate Laws for

Mechanisms with a Slow Initial Step 601

Mechanisms

with a Fast Initial Step 602

14.7

CATALYSIS 605 Homogeneous Catalysis 605

Heterogeneous Catalysis 606

Enzymes 608

SUMMARY AND KEY TERMS 613 KEY EQUATIONS 614 EXERCISES

616

ADDITIONAL EXERCISES

INTEGRATIVE EXERCISES

A Closer Look

KEY SKILLS 614

VISUALIZING CONCEPTS

615

622

624

Using Spectroscopic Methods to Measure Reaction Rates

Chemistry Put to WOrk Methyl Bromide in the Atmosphere G1emistry Put to WOrk Catalytic Converters 608 Chemistry and Life Nitrogen Fixation and Nitrogenase 610

590

580

Contents

15 Chemical Equilibrium

626

15.1

THE CONCEPT OF EQUILIBRIUM 628

15.2

THE EQUILIBRIUM CONSTANT 630 Evaluating Kc 632

Equilibrium Constants in Terms of Pressure, Kp 633

Equilibrium Constants and Units 635

15.3

INTERPRETING AND WORKING WITH EQUILIBRIUM CONSTANTS 635 The Magnitude of Equilibrium Constants 635 Chemical Equation and K 636

The Direction of the

Relating Chemical Equations and

Equilibrium Constants 637

15.4

HETEROGENEOUS EQUILIBRIA 639

15.5

CALCULATING EQUILIBRIUM CONSTANTS 641

15.6

APPLICATIONS OF EQUILIBRIUM CONSTANTS 644 Predicting the Direction of Reaction 644

Calculating Equilibrium

Concentrations 645

15.7

LE CHATELIER'S PRINCIPLE 648 Change in Reactant or Product Concentrations 649 Pressure Changes 649

Effects of Volume and

Effect of Temperature Changes 651

The Effect of

Catalysts 654

SUMMARY AND KEYTERMS 657 KEY EQUATIONS 658 EXERCISES 660

KEYSKILLS 657

VISUALIZING CONCEPTS 658

ADDITIONAL EXERCISES 662

INTEGRAITVE EXERCISES 664

Chemistry Put to "WOrk Chemistry Put to "WOrk

Haber Process 631 Controlling Nitric Oxide Emissions

1 6 Acid-Base Equilibria

666

16.1

ACIDS AND BASES: A BRIEF REVIEW 668

16.2

BR0NSTED-LOWRY ACIDS AND BASES 668

The W lo n in Water 669 Acid-Base Pairs 670

16.3

656

Proton-Transfer Reactions 669

Conjugate

Relative Strengths of Acids and Bases 672

THE AUTOIONIZATION OF WATER 673 T h e l o n Product of Water 674

16.4

THE pH S C ALE 675 pOH and Other "p" Scales 678

16.5

Strong Acids 679

16.6

Strong Bases 680

WEAK ACIDS 681

Calculating K0 from pH 682 pH 684

16.7

Measuring pH 678

STRONG ACIDS AND BASES 679

Percent Ionization 683

Using K0 to Calculate

Polyprotic Acids 688

WEAK BASES 690 Types of Weak Bases 692

16.8 16.9

RELATIONSHIP BETWEEN K. AND Kb 693 ACID-BASE PROPERTIES OF SALT SOLUTIONS 695 An Anion's Ability to React with Water 696 Water 696

16.10

ACID-BASE BEHAVIOR AND CHEMICAL STRUCTURE 699 Factors That Affect Acid Strength 699 Oxyacids 700

16.11

A Cation's Ability to React with

Combined Effect of Cation and Anion in Solution 697

Carboxylic Acids 702

LEWIS ACIDS AND BASES 704 Hydrolysis of Metalions 705

Binary Acids 699

xv

xvi

Contents SUMMARY AND KEYTERMS 707 KEY EQUATIONS 708

EXERCISES 710

KEY SKILlS 708

VISUAliZING CONCEPTS 709

ADDITIONAL EXERCISES 71 5

INTEGRATIVE EXERCISES 716

Cbemistry Put to WOrk Amines and Amine Hydrochlorides 694 Cbemistry and Life The Amphiprotic Behavior of Amino Acids 703

Aspects 1 7 Additional of Aqueous Equilibria

718

17.1

THE COMMON-ION EFFECT 720

17.2

B UFFERED SOLUTIONS 723 Composition and Action of Buffered Solutions 723 Buf fer 724

Calculating the pH of a

Buffer Capacity and pH Range 726

Addition of Strong Acids

or Bases to Buffers 727

17.3

ACID-B ASE TITRATION$ 730 Strong Acid-Strong Base Titrations 730 Base Titrations 733

17.4

Weak Acid-Strong

Titrations of Polyprotic Acids 737

SOLUBILITY EQUILIBRIA 737

The Solubility-Product Constant, Ksp 738

17.5

Solubility and K'P 738

FACTORS THAT AFFECT SOLUBILITY 741 Common-ion Effect 741 Ions 745

Solubility and pH 742

Formation of Complex

Amphoterism 748

17.6

PRECIPITATION A ND SEPA RATION OF IONS 750

17.7

QUALITATNE A NALYSIS FOR METALLIC ELEMENTS 753

Selective Precipitation of Ions 751

SUMMARY AND KEY TERMS 756 KEY EQUATIONS 757 EXERCISES 759

KEY SKILlS 757

VISUAliZING CONCEPTS 757

ADDITIONAL EXERCISES 763

INTEGRATIVE EXERCISES 764

Cbemistry and Life Blood as a Buffered Solution 729 Closer LDDk Limitations ofSolubility Products 741 Chemistry and Life Sinkholes 744 Cbemistry and Life Tooth Decay and Fluoridation 747

A

18 Chemistry of the Environment 18.1

EARTH'S ATMOSPHERE 768

18.2

OUTER REGIONS OF THE ATMOSPHERE 770

18.3

OZONE IN THE UPPER ATMOSPHERE 772

18.4

CHEMISTRY OF THE TROPOSPHERE 775

766

Composition of the Atmosphere 769

Photodissociation 770

Photoionization 772

Depletion of the Ozone Layer 774

Sulfur Compounds and Acid Rain 776

Carbon Monoxide 778

Nitrogen Oxides and Photochemical Smog 779

Water Vapor, Carbon

Dioxide, and Climate 780

18.5

THE WORLD OCEAN 783

18.6

FRESHWATER 785

Seawater 783

Desalination 784

Dissolved Oxygen and Water Quality 786 Supplies 786

Treatment of Municipal Water

Contents 18.7

GREEN CHEMISTRY 788 Solvents and Reagents 789

Other Processes 790

The Challenges of Water

Purification 790 SUMMARY AND KEY T ERMS 792

KEY SKILLS 793

VISUALIZING CONC EPTS 794

EXERCISES 794

ADDI TIONAL EXERCISES 797

INTEGRATIVE EXERCISES 798

A CbJser 1-oolt

Other Greenhouse Gases

A

Water Softening

CbJser Look

782

788

1 9 Chemical Thermodynamics 19.1

800

SPONTANEOUS PROCESSES 802 Seeking a Criterion for Spontaneity 804

Reversible and

Irreversible Processes 804

19.2

ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 806 Entropy Change 806

6.5 f o r Phase Changes 807

T h e Second Law of

Thermodynamics 808

19.3

THE MOLECULAR INTERPRETATION OF ENTROPY 809 Molecular Motions and Energy 809 and Microstates 81 0

Boltzmann's Equation

Making Qualitative Predictions About 6.5 813

The Third Law of Thermodynamics 816

19.4

ENTROPY CHANGES IN CHEMICAL REACTIONS 817 Entropy Changes in the Surroundings 81 8

19.5

GIBBS FREE ENERGY 819 Standard Free-Energy of Formation 822

19.6 19.7

FREE ENERGY AND TEMPERATURE 824 FREE ENERGY AND THE EQUILIBRIUM CONSTANT 826 SUMMARY AND KEY T ERMS 832 KEY EQUATIONS 833 EXERCISES 834

KEY SKILLS 832

VISUALIZING CONC EPTS 833

ADDI TIONAL EXERCISES 839

INTEGRATIVE EXERCISES 840

Closer Look The Entropy Change when a Gas Expands Isothermally Clos,.,-1-oolt Entropy and Probability 812 Chemistry and Life Entropy and Life 815 A Closer 1-ooA What's"Free" about Free Energy? 822 Chemistry and Life Driving Nonspontaneous Reachons 830

A

808

A

2 Q Electrochemistry 20.1

842

OXIDATION STATES AND OXIDATION-REDUCTION REACTIONS 844

20.2

BALANCING OXIDATION-REDUCTION EQUATIONS 846 Half-Reactions 846 of Half-Reactions 846

Balancing Equations by the Method Balancing Equations for Reactions Occurring

in Basic Solution 849

20.3

VOLTAIC CELLS 851

20.4

CELL EMF UNDER STANDARD CONDITIONS 855

A Molecular View o f Electrode Processes 854

Standard Reduction (Half-Cell) Potentials 856

Strengths of Oxidizing and

Reducing Agents 860

20.5

FREE ENERGY AND REDOX RE ACTIONS 862 EMF and 6.G 863

xvii

xviii

Contents 20.6

CELL EMF UNDER NONSTANDARD CONDITIONS 865 The Nernst Equation 865

20.7

Concentration Cells 867

BATTERIES AND FUEL CELLS 870 Lead-Acid Battery 871

Alkaline Battery 872

Metal-Hydride, and Lithium-ion Batteries 872

20.8

CORROSION 874

20.9

ELECTROLYSIS 876

Corrosion of Iron 874

Preventing the Corrosion of Iron 875

Quantitative Aspects of Electrolysis 878

SUMMARY AND KEY TERMS 881 KEY EQUATIONS

EXERCISES 884

Nickel-Cadmium, Nickel· Hydrogen Fuel Cells 873

883

Electrical Work 879

KEY SKILLS 882

VISUAliZING CONCEPTS 883

ADDITIONAL EXERCISES 889

INTEGRATIVE EXERCISES 890

Chemistry and Life Heartbeats and Electrocardiography 868 Cbemistry Put to WOrk Direct Methanol Fuel Cells 874

2 1 Nuclear Chemistry

8 92

21.1

RADIOACTIVITY 894

21.2

PATTERNS OF NUCLEAR STABILITY 898

Nuclear Equations 895

Types of Radioactive Decay 896

Neutron-to-Proton Ratio 898

Radioactive Series 900

Further Observations 901

21.3

NUCLEAR TRANSMUTATIONS 901 Accelerating Charged Particles 902

Using Neutrons 903

Transuranium Elements 903

21.4

RATES OF RAD IOACTIVE DECAY 903

21.5

DETECTION OF RADIOACTIVIT Y 908

21.6

ENERG Y CHANGES IN NUCLEAR REACTIONS 911

21.7

NUCLEAR POWER: FISSION 913

Radiometric Dating 905

Calculations Based on Half-life 905

Radiotracers 91 0

Nuclear Binding Energies 91 2

Nuclear Reactors 91 5

21.8

NUCLEAR POWER: FUSION 918

21.9

RADIATION IN THE ENVIRONMENT AND LIVING SYSTEMS 919 Radiation Doses 920

Radon 921

SUMMARY AND KEY TERMS 924

KEY SKILLS 925

KEY EQUATIONS

925

VISUAliZING CONCEPTS 925

EXERCISES 926

ADDITIONAL EXERCISES 928

INTEGRATIVE EXERCISES 929

Chemistry and Life Medical Appl ications of Radiotracers 910 Closer Look The Dawning of the Nuclear Age 915 Cbemistry Put to WOrk Environmental Applications of Radioisotopes Cbemistry and Life Radiation Therapy 922

A

22 Chemistry of the Nonmetals 22.1

921

930

GENERAL CONCEPTS: PERIODIC TRENDS AND CHEMICAL REACTIONS 932 Chemical Reactions 933

22.2

HYDROGEN 935 Isotopes of Hydrogen 935 Hydrogen 936

Properties of Hydrogen 936

Uses of Hydrogen 937

Preparation of

Binary Hydrogen Compounds 937

Contents 22.3

GROUP SA: THE NOBLE GASES 938

22.4

GROUP ?A: THE HALOGENS 940

Noble-Gas Compounds 939

Properties and Preparation of the Halogens 940 The Hydrogen Halides 942

Uses of the Halogens 941

lnterhalogen Compounds 943

Oxyacids and

Oxyanions 943

22.5

OXYGEN 944 Properties of Oxygen 944 945

22.6

Ozone 945

Preparation of Oxygen 944

Oxides 946

Uses of Oxygen

Peroxides and Superoxides 948

THE OTHER GROUP 6A ELEMENTS: S, Se, Te, AND Po 948 General Characteristics of the Group 6A Elements 949 Preparation of S, Se, and Te 949 Selenium, and Tellurium 949

Occurrence and

Properties and Uses of Sulfur, Su�ides 950

Oxides, Oxyacids, and

Oxyanions of Sulfur 950

22.7

NITROGEN 952 Properties of Nitrogen 952

Preparation and Uses of Nitrogen 952

Hydrogen Compounds of Nitrogen 953

Oxides and Oxyacids

of Nitrogen 954

22.8

THE OTHER GROUP 5A ELEMENTS: P, As, Sb, AND Bi 956 General Characteristics of the Group SA Elements 956 and Properties of Phosphorus 957

Occurrence, Isolation,

Phosphorus Halides 957

Oxy Compounds of Phosphorus 958

22.9

CARBON 960 Elemental Forms of Carbon 961 and Carbonates 963

Oxides of Carbon 962

Carbides 964

Carbonic Acid

Other Inorganic Compounds of

Carbon 965

22.10

THE OTHER GROUP 4A ELEMENTS: Si, Ge, Sn, AND Pb 965 General Characteristics of the Group 4A Elements 965 Preparation of Silicon 966

22.11

Silicates 966

Occurrence and

Glass 968

Silicones 969

BORON 969 SUMMARY AND KEY TERMS 971 KEY SKILLS 972 VISUALIZING CONCEPTS 973 EXERCISES 974 ADDITIONAL EXERCISES 977 INTEGRATIVE EXERCISES 978

Oosu Look The Hydrogen Economy 937 Chemistry and Life How Much Perchlorate Is Too Much? 944 Chemistry and Life Nitroglycerin and Heart Disease 956 Chemistry and Life Arsenic in Drinking Water 960 Chemistry Put to Work Carbon Fibers and Composites 962

A

2 3 Metals and Metallurgy

980

23.1

OCCURRENCE AND DISTRIBUTION OF METALS 982

23.2

PYROMETALLURGY 984

23.3

HYDROMETALLURGY 987

23.4

ELECTROMETALLURGY 988

Minerals 982

Metallurgy 983

The Pyrometallurgy of Iron 985

Hydrometallurgy of Gold 987

Formation of Steel 986

Hydrometallurgy of Aluminum 988

Electrometallurgy of Sodium 988

Electrometallurgy of Aluminum 989

Electrorefining of Copper 990

23.5

METALLIC BONDING 991 Physical Properties of Metals 992 for Metallic Bonding 992

23.6

Electron-Sea Model

Molecular-Orbital Model for Metals 994

ALLOYS 995 lnterrnetallic Compounds 997

xix

xx

Contents 23.7

TRANSITION METALS 998 Electron Configurations and Oxidation States 1 000

Physical Properties 999 Magnetism 1001

23.8

CHEMISTRY OF SELECTED TRANSITION METALS 1002 Chromium 1 002

Iron 1 003

Copper 1 004

SUMMARY ANDKEYTERMS 1006 KEYSKILLS 1007 VISUALIZING CONCEPTS 1007 ADDITIONAL EXERCISES 1010

A Ooser l.Dok A Closer Look

EXERCISES 1008 INTEGRATIVE EXERCISES 1011

Charles M. Hall 990 Shape-Memory Alloys 998

of Coordination 24 Chemistry Compounds 1012 24.1

METAL COMPLEXES 1 0 1 4 The Development o f Coordination Chemistry: Werner's Theory 1014 The Metal-Ligand Bond 1 01 6 Geometries

24.2

Charges, Coordination Numbers, and

1 01 6

LIGANDS WITH MORE THAN ONE DONOR ATOM 1 01 9 Metals and Chelates i n Living Systems 1 021

24.3

NOMENCLATURE OF COORDINATION CHEMISTRY 1025

24.4

ISOMERISM 1 0 26

24.5

COLOR AND MAGNETISM 1 031

Structural Isomerism 1 027

Color 1 031

24.6

Stereoisomerism 1 027

Magnetism 1 033

CRYSTAL-FIELD THEORY 1 033 Electron Configurations in Octahedral Complexes 1 037

Tetrahedral and

Square-Planar Complexes 1 038

SUMMARY AND KEYTERMS 1042 KEYSKILLS 1043 VISUALIZING CONCEPTS 1043 ADDITIONAL EXERCISES 1046

A Closer l.Dolr

EXERCISES 1044 INTEGRATIVE EXERCISES 1048

Entropy and the Chelate Effect 1021

Chemistry and Life The Battle for Iron in Living Systems A Closer Loolt Charge-Transfer Color 1040

25

1024

The Chemist�y of Life: Organic and Biological Chemistry 25.1

1050

SOME GENERAL CHARACTERISTICS OF ORGANIC MOLECULES 1052 T h e Structures o f Organic Molecules 1052 Substances 1 053

T h e Stabilities o f Organic

Solubility and Acid-Base Properties of Organic

Substances 1 054

25.2

INTRODUCTION TO HYDROCARBONS 1054

25.3

ALKANES, ALKENES, AND ALKYNES 1 055 Structures of Alkanes 1 056 Alkanes 1 058

1062

Structural Isomers 1 057

Cycloalkanes 1 060

Alkynes 1 064

Alkenes

Addition Reactions of Alkenes and Alkynes 1 065

Mechanism of Addition Reactions 1 066

25.4

Nomenclature of

Reactions of Alkanes 1 061

Aromatic Hydrocarbons 1 068

ORGANIC FUNCTIONAL GROUPS 1070 Alcohols 1 072

Ethers 1 073

Acids and Esters 1 074

Aldehydes and Ketones 1 073

Amines and Amides 1 077

Carboxylic

Contents 25.5

CHIRALITY IN ORGANIC CHEMISTRY 1078

25.6

INTRODUCTION TO BIOCHEMISTRY 1080

25.7

PROTEINS 1080 Amino Acids 1 080

Polypeptides and Proteins 1 082

Protein Structure 1 084

25.8

CARBOHYDR ATES 1086 Disaccharides 1 087

25.9 25.10

Polysaccharides 1 088

LIPIDS 1 089 NUCLEIC ACIDS 1090 SUMMARY ANDKEYTERMS 1095 KEYSKILLS 1096 VISUALIZING CONCEPTS 1096 EXERCISES 1097 ADDITIONAL EXERCISES 1102 INTEGRATIVE EXERCISES 1103

Oumtistry Put to �rk Gasoline 1061 Chemistry and Life Polycyclic Aromatic Hydrocarbons 1070 Chemistry Put to �rk Portrait of an Organic Chemical 1076 Chemistry and Life The Origins of Chirality in Living Systems 1085 Strategies in Chemistry What Now? 1094

Appendices II 04

A

Mathematical Operations

B

Properties of Water

C

Thermodynamic Quantities for Selected Substances

D

Aqueous Equilibrium Constants 1115

E

IIII

at2981 . 5 K (25°C) 1112 Standard Reduction Potentials at 25 oc 1117 Answers to Selected Exercises A-1 Answers to "Give It Some Thought" A-36 Glossary G-1 Photo/Art Credits P-1 Index

1-1

xxi

Chemical A

lications and Essa s

Strate ies in Chemis Estimating Answers

26

The Importance of Practice Pattern Recognition Problem Solving

58

89 106

180

143

Calculations Involving ManyVariables

1094

Chemistry and the Chemical Industry

404

Antacids

135

The Hybrid Car

196

Explosives and Alfred Nobel

Chemistr and Li e Elements Required by Living Organisms

102

DrinkingToo Much Water Can Kill You

Gas Pipelines

409 420

380

Gas Separations

57

147

185 Nuclear Spin and Magnetic Resonance Imaging 236 Ionic Size Makes a Big Difference 265 The Improbable Development of Lithium Drugs 280 The Chemistry ofVision 367 Blood Pressure 398 Fat- and Water-Soluble Vitamins 538 Blood Gases and Deep-Sea Diving 540 Sickle-Cell Anemia 559 Nitrogen Fixation and Nitrogenase 610 The Amphiprotic Behavior of Amino Acids 703 Blood as a Buffered Solution 729 Sinkholes 744 Tooth Decay and Fluoridation 747 Entropy and Life 815 Driving Nonspontaneous Reactions 830 Heartbeats and Electrocardiography 868 Medical Applications of Radiotracers 910 Radiation Therapy 922 How Much Perchlorate Is Too Much? 944 Nitroglycerin and Heart Disease 956 Arsenic in Drinking Water 960 The Regulation of Human Body Temperature

xxii

18

Chemistry in the News

Orbitals and Energy

Glucose Monitoring

1085

28

Analyzing Chemical Reactions

What Now?

Polycyclic Aromatic Hydrocarbons

The Origins of Chirality in Living Systems

How to Take a Test

Using Enthalpy as a Guide

1024 1070

The Battle for Iron in Living Systems

Supercritical Fluid Extraction

498

Cell Phone Tower Range

501

Recycling Plastics

Toward the Plastic Car Liquid Crystal Displays

4

328

453

506 513

Methyl Bromide in the Atmosphere

608 631

590

Catalytic Converters The Haber Process

Controlling Nitric Oxide Emissions Amines and Amine Hydrochlorides Direct Methanol Fuel Cells

874

656 694

Environmental Applications of Radioisotopes Carbon Fibers and Composites Gasoline

1061

962

Portrait of an Organic Chemical

The Scientific Method Basic Forces

45

13

The Mass Spectrometer

48

Glenn Seaborg and Seaborgium The Aura of Gold

1076

143

Energy, Enthalpy, and P- V Work

51 176

921

Chemical Applications and Essays The Speed of Light

214

Colligative Properties of Electrolyte Solutions

Measurement and the Uncertainty Principle

225

Probability Density and Radial Probability Functions Experimental Evidence for Electron Spin

230

234

Calculation of Lattice Energies: The Born-Haber Cycle

304

Oxidation Numbers, Formal Charges, and Actual Partial Charges 318 Phases

416

The Clausius-Oapeyron Equation

The Transistor Hydrates

782

788

The Entropy Change when a Gas Expands Isothermally 808

812

What's "Free" About Free Energy?

456

The Dawning of the Nuclear Age

X-Ray Diffraction by Crystals 465 The Third Form of Carbon

Water Softening

Entropy and Probability

in Atomic and Molecular Orbitals 373

The Ideal-Gas Equation

741

Limitations of Solubility Products Other Greenhouse Gases

260

Effective Nuclear Charge

The Hydrogen Economy

468

Charles M. Hall

490

Ideal Solutions with Two or More Volatile Components

937

998

Entropy and the Chelate Effect

548

822

915

990

Shape-Memory Alloys

533

554

Using Spectroscopic Methods to Measure Reaction Rates 580

Charge-Transfer Color

1040

1021

xxiii

Preface TO T H E I N S T RU C T O R Philosophy This is the eleventh edition of a text that has enjoyed unprecedented success over its many editions. It is fair to ask why there needs to be yet another edi­ tion. The answer in part lies in the nature of chemistry itseU, a dynamic science in a process of continual discovery. New research leads to new applications of chemistry in other fields of science and in technology. In addition, environmen­ tal and economic concerns bring about changes in the place of chemistry in so­ ciety. We want our textbook to reflect that dynamic, changing character. We also want it to convey the excitement that scientists experience in making new dis­ coveries and contributing to our understanding of the physical world. In addition, new ideas about how to present chemistry are being offered by teachers of chemistry, and many of these new ideas are reflected in how the textbook is organized and the ways in which individual topics are presented. New technologies and new devices to assist students in learning lead to new ways of presenting learning materials: the Internet, computer-based classroom projection tools, and more effective means of testing, to name just a few. All of these factors impact on how the text and the accompanying supplementary ma­ terials are modified from one edition to the next. Our aim in revising the text has been to ensure that it remains a central, in­ dispensable learning tool for the student. It is the one device that can be carried everywhere and used at any time, and as such, it is a one-stop source of all the information that the student is likely to need for learning, skill development, reference, and test preparation. We believe that students are more enthusiastic about learning chemistry when they see its importance to their own goals and interests. With this in mind, we have highlighted many important applications of chemistry in every­ day life. At the same time, the text provides the background in modern chem­ istry that students need to serve their professional interests and, as appropriate, to prepare for more advanced chemistry courses.

If the text is to support your role as teacher effectively, it must be addressed

to the students. We have done our best to keep our writing clear and interesting

and the book attractive and well-illustrated. Furthermore, we have provided nu­ merous in-text study aids for students, including carefully placed descriptions of problem-solving strategies. Together, we have logged many years of teaching experience. We hope this is evident in our pacing and choice of examples. A textbook is only as useful to students as the instructor permits it to be. This book is loaded with many features that can help students learn and that can guide them as they acquire both conceptual understanding and problem­ solving skills. But the text and all the supplementary materials provided to sup­ port its use must work in concert with the instructor. There is a great deal for the students to use here, too much for all of it to be absorbed by any one student. You, the instructor, are the guide to a proper use of the book. Only with your ac­ tive help will the students be able to fully utilize all that the text and its supple­ ments offer. Students care about grades, of course, but with encouragement, they can also care about learning just because the subject matter is interesting. Please consider emphasizing features of the book that can materially enhance student appreciation of chemistry, such as the Chemistry Put to Work and Chemistry and Life boxes that show how chemistry impacts modem life and its relationship to health and life processes. Learn to use, and urge students to use, xxiv

Preface the rich Internet resources available. Emphasize conceptual understanding, and place less emphasis on simple manipulative, algorithmic problem solving. Spending less time on solving a variety of gas law problems, for example, can open up opportunities to talk about chemistry and the environment.

Organization and Contents The first five chapters give a largely macroscopic, phenomenological view of chemistry. The basic concepts introduced-such as nomenclature, stoichiome­ try, and thermochemistry-provide necessary background for many of the lab­ oratory experiments usually performed in general chemistry. We believe that an early introduction to thermochemistry is desirable because so much of our understanding of chemical processes is based on considerations of energy change. Thermochemistry is also important when we come to a discussion of bond enthalpies. We believe we have produced an effective, balanced approach to teaching thermodynamics in general chemistry.

It is no

easy matter to walk

the narrow pathway between-on the one hand-trying to teach too much at too high a level and-on the other-resorting to oversimplifications. As with the book as a whole, the emphasis has been on imparting conceptual under­ standing, as opposed to presenting equations into which students are supposed to plug numbers. The next four chapters (Chapters

6-9)

deal with electronic structure and

bonding. We have largely retained our presentation of atomic orbitals. For more advanced students, Closer Look boxes deal with radial probability functions and the nature of antibonding orbitals. In Chapter

7 we have

improved our discus­

sion of atomic and ionic radii. The focus of the text then changes to the next level of the organization of matter: the states of matter (Chapters solutions (Chapter

13).

10 and 11)

and

Also included in this section is an applications chapter

on the chemistry of modern materials (Chapter

12),

which builds on the stu­

dent's understanding of chemical bonding and intermolecular interactions. This chapter has again received substantial revision, in keeping with the rapid pace of change in technology. It has been reorganized to emphasize a classifica­ tion of materials based on their electronic bonding characteristics. This chapter provides an opportunity to show how the sometimes abstract concept of chem­ ical bonding impacts real world applications. The modular organization of the chapter allows you to tailor your coverage to focus on those materials (semi­ conductors, polymers, biomaterials, nanotechnology, etc.) that are most rele­ vant to your students. The next several chapters examine the factors that determine the speed and extent of chemical reactions: kinetics (Chapter thermodynamics (Chapter

14),

equilibria (Chapters

15-17),

19), and electrochemistry (Chapter 20). Also in this

section is a chapter on environmental chemistry (Chapter

18), in which the con­

cepts developed in preceding chapters are applied to a discussion of the atmos­ phere and hydrosphere. After a discussion of nuclear chemistry (Chapter 21), the final chapters sur­ vey the chemistry of nonmetals, metals, organic chemistry, and biochemistry (Chapters

22-25). Chapter 22 has been shortened slightly. Chapter 23 now con­

tains a modern treatment of band structure and bonding in metals. A brief dis­ cussion of lipids has been added to Chapter

25.

These final chapters are

developed in a parallel fashion and can be treated in any order. Our chapter sequence provides a fairly standard organization, but we recog­ nize that not everyone teaches all the topics in exactly the order we have chosen. We have therefore made sure that instructors can make common changes in teaching sequence with no loss in student comprehension. In particular, many instructors prefer to introduce gases (Chapter

10)

after stoichiometry or after

thermochemistry rather than with states of matter. The chapter on gases has been written to permit this change with no disruption in the flow of material. It is also possible to treat the balancing of redox equations (Sections

20.1

and

20.2)

xxv

xxvi

Preface earlier, after the introduction of redox reactions in Section instructors like to cover organic chemistry (Chapter (Chapter

9). This, too, is a largely seamless move.

25)

4.4. Finally, some

right after bonding

We have introduced students to descriptive organic and inorganic chem­ istry by integrating examples throughout the text. You will find pertinent and relevant examples of "real" chemistry woven into all the chapters as a means to illustrate principles and applications. Some chapters, of course, more directly address the properties of elements and their compounds, especially Chapters 4,

7, 12, 18, and 22-25. We also incorporate descriptive organic and inorganic chemistry in the end-of-chapter exercises.

Changes in This Edition Some of the changes in the eleventh edition made in individual chapters have already been mentioned. More broadly, we have introduced a number of new features that are general throughout the text. Chemistry: The Central Science has traditionally been valued for its clarity of writing, its scientific accuracy and

currency, its strong end-of-chapter exercises, and its consistency in level of cov­

erage. In making changes, we have made sure not to compromise these charac­

teristics. At the same time, we have responded to feedback received from the faculty and students who used previous editions. To make the text easier for students to use, we have continued to employ an open, clean design in the lay­ out of the book. Illustrations that lend themselves to a more schematic, bolder presentation of the underlying principles have been introduced or revised from earlier versions. The art program in general has been strengthened, to better convey the beauty, excitement, and concepts of chemistry to students. The chapter-opening photos have been integrated into the introduction to each chapter, and thus made more relevant to the chapter's contents. We have continued to use the What's Ahead overview at the opening of each chapter, introduced in the ninth edition, but we have changed the format to make the materials more useful to students. Concept links (=) continue to provide easy-to-see cross-references to pertinent material covered earlier in the text. The essays titled Strategies in Chemistry, which provide advice to students on problem solving and "thinking like a chemist," continue to be an important feature. The Give It Some Thought exercises that we introduced in the tenth edi­ tion have proven to be very popular, and we have increased their number. These are informal, rather sharply focused questions that give students oppor­ tunities to test whether they are actually "getting it" as they read along. We have continued to emphasize conceptual exercises in the end-of-chapter exer­ cise materials. The Visualizing Concepts category of exercise has been continued in this edition. These exercises are designed to facilitate concept understanding through use of models, graphs, and other visual materials. They precede the regular end-of-chapter exercises and are identified in each case with the rele­ vant chapter section number. We continue to use multi-focus graphics to depict topics in macroscopic, microscopic, symbolic, and conceptual representation so students learn to see chemistry the way scientists do, from a variety of perspec­ tives. The Integrative Exercises, which give students the opportunity to solve more challenging problems that integrate concepts from the present chapter with those of previous chapters, have also been increased in number.

New essays in our well-received Chemistry Put to Work and Chemistry and Life series emphasize world events, scientific discoveries, and medical break­ throughs that have occurred since publication of the tenth edition. We maintain our focus on the positive aspects of chemistry, without neglecting the problems that can arise in an increasingly technological world. Our goal is to help stu­ dents appreciate the real-world perspective of chemistry and the ways in which chemistry affects their lives. A minor change that you will see throughout the text is the use of con­ densed structural formulas for simple carboxylic acids. For example, we now write CH3COOH for acetic acid instead of HC2H302.

Preface

You'll also find that we've •

Revised or replaced some of the end-of-chapter Exercises, with particular focus on the black-numbered exercises (those not answered in the Appendix).

•

Integrated more conceptual questions into the end-of-chapter material. For the convenience of instructors, these are identified by the CQ anno­ tation in the Annotated Instructor's Edition, but not in the student edition of the text.

•

Carried the stepwise Analyze, Plan, Solve, Check problem-solving strategy into nearly all of the Sample Exercises of the book to provide additional guidance in problem solving.

•

Expanded the use of dual-column problem-solving strategies in many Sample Exercises to more clearly outline the process underlying mathemat­ ical calculations, thereby helping students to better perform mathematical calculations.

•

Added both Key Skills and Key Equations sections to the end of chapter material to help students focus their study.

TO THE S T U D ENT Chemistry : The Central Science, Eleventh Edition, has been written to introduce you to modem chemistry. As authors, we have, in effect, been engaged by your instructor to help you learn chemistry. Based on the comments of stu­ dents and instructors who have used this book in its previous editions, we be­ lieve that we have done that job well. Of course, we expect the text to continue to evolve through future editions. We invite you to write to us to tell us what you like about the book so that we will know where we have helped you most. Also, we would like to learn of any shortcomings, so that we might further improve the book in subsequent editions. Our addresses are given at the end of the Preface.

Advice for Learning and Studying Chem istry Learning chemistry requires both the assimilation of many new concepts and the development of analytical skills. In this text we have provided you with nu­ merous tools to help you succeed in both. If you are going to succeed in your course in chemistry, you will have to develop good study habits. Science cours­ es, and chemistry in particular, make different demands on your learning skills than other types of courses. We offer the following tips for success in your study of chemistry: Don't fall behind! As your chemistry course moves along, new topics will build on material already presented . If you don't keep up in your reading and problem solving, you will find it much harder to follow the lectures and discus­ sions on current topics. "Cramming" just before an exam has been shown to be an ineffective way to study any subject, chemistry included. Focus your study. The amount of information you will be expected to learn can sometimes seem overwhelming. It is essential to recognize those concepts and skills that are particularly important. Pay attention to what your instructor is emphasizing. As you work through the Sample Exercises and homework as­ signments, try to see what general principles and skills they deal with . Use the What's Ahead feature at the beginning of each chapter to help orient you to what is important in each chapter. A single reading of a chapter will simply not be enough for successful learning of chapter concepts and problem-solving skills. You will need to go over assigned materials more than once. Don't skip the Give It Some Thought features, Sample Exercises, and Practice Exercises.

xxvii

xxviii

Preface They are your guides t o whether you are actually learning the material. The

Key Skills and Key Equations at the end of each chapter should help you focus your study.

Keep good lecture notes.

Your lecture notes will provide you with a clear and

concise record of what your instructor regards as the most important material to learn. Use your lecture notes in conjunction with this text; that's your best way to determine which material to study.

Skim topics in the text before they are covered in lecture.

Reviewing a topic be­

fore lecture will make it easier for you to take good notes. First read the intro­ duction and Summary, then quickly read through the chapter, skipping Sample Exercises and supplemental sections. Pay attention to the titles of sections and subsections, which give you a feeling for the scope of topics. Try to avoid think­ ing that you must learn and understand everything right away.

After lecture, carefully read the topics covered in class. As you read, pay atten­ tion to the concepts presented and to the application of these concepts in the Sample Exercises. Once you think you understand a Sample Exercise, test your understanding by working the accompanying Practice Exercise.

Learn the language of chemistry.

As you study chemistry, you will encounter

many new words. It is important to pay attention to these words and to know their meanings or the entities to which they refer. Knowing how to identify chemical substances from their names is an important skill; it can help you avoid painful mistakes on examinations. For example, "chlorine" and "chlo­ ride" refer to very different things.

Attempt the assigned end-of-chapter exercises. Working the exercises that have been selected by your instructor provides necessary practice in recalling and using the essential ideas of the chapter. You cannot learn merely by observing;

you must be a participant. In particular, try to resist checking the

Student Solutions Manual (if you have one) until you have made a sincere effort to solve

the exercise yourself. If you really get stuck on an exercise, however, get help

from your instructor, your teaching assistant, or another student. Spending more than

20 minutes on a single exercise is rarely effective unless you know

that it is particularly challenging.

Make use of the online resources. Some things are more easily learned by dis­ covery, and others are best shown in three dimensions. If your instructor has included Mastering Chemistry with your book, take advantage of the unique tools it provides to get the most out of your time in chemistry. The bottom line is to work hard, study effectively, and use the tools that are available to you, including this textbook. We want to help you learn more about the world of chemistry and why chemistry is the central science. If you learn chemistry well, you can be the life of the party, impress your friends and parents, and . . . well, also pass the course with a good grade.

ACKN OWLE D G M EN T S The production of a textbook is a team effort requiring the involvement of many people besides the authors. Many peo­ ple contributed hard work and talent to bring this edition to life. Although their names don't appear on the cover of the book, their creativity, time, and support has been instrumental in all stages of its development and production. Each of us has benefited greatly from discussions with colleagues and from correspondence with both instructors and students both here and abroad. Colleagues have also helped immensely by reviewing our materials, sharing their insights, and providing suggestions for improvements. On this edition we were particularly blessed with an exception­ al group of accuracy checkers who read through our materials looking for both technical inaccuracies and typographi­ cal errors.

Preface

xxix

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xxx

Preface

Ann Cartwright Dana Chatellier Stanton Ching Paul Chirik William Cleaver Beverly Clement Robert D. Cloney john Collins Edward Werner Cook Elzbieta Cool Thomas Edgar Crumm Dwaine Davis Ramon Lopez de Ia Vega Nancy De Luca Angel de Dios john M. DeKorte Daniel Domin James Donaldson Bill Donovan Ronald Duchovic David Easter joseph Ellison George 0. Evans II Clark L. Fields Jennifer Firestine jan M. Fleischner Michelle Fossum Roger Frampton joe Franek David Frank Ewa Fredette Karen Frindell john I. Gelder Paul Gilletti Peter Gold james Gordon Thomas J. Greenbowe Michael Greenlief Eric P. Grimsrud john Hagadorn Randy Hall jolm M. Halpin Marie Hankins Robert M. Hanson Daniel Haworth lnna Hefley David Henderson Paul Higgs Gary G. Hoffman Deborah Hokien Robin Homer Roger K. House William jensen Janet Johannessen Andrew Jones Booker juma Ismail Kady

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Siam Kahmis Steven Keller john W. Kenney Neil Kestner Leslie Kinsland Donald Kleinfelter David Kort George P. Kreishman Paul Kreiss Manickham Krishnamurthy Brian D. Kybett William R. Lammela john T. Landrum Richard Langley N. Dale Ledford Ernestine Lee David Lehmpuhl Donald E. Linn, Jr.

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We would also like to express our gratitude to our many team members at Prentice Hall whose hard work, imagi­ nation, and commitment have contributed so greatly to the final form of this edition: Nicole Folchetti, our Editor in Chief, who has brought energy and imagination not only to this edition but to earlier ones as well; Andrew Gilfillan, our Chemistry Editor, for his enthusiasm and support; Jonathan Colon, Editorial Assistant, who coordinated all reviews and accuracy checking; Ray Mullaney, our Developmental Editor in Chief, who has been with us since the very first edition of this textbook, helping guide its evolution, infusing it with a sense of style and quality; Karen Nein, our Development Editor, whose diligence and careful attention to detail were invaluable to this revision; Marcia Youngman, our copy editor, for her keen eye; Maureen Eide, our Art Director, who managed the complex task of bringing our sketches into final form; Jennifer Hart, who coordinated all the supplements that accompany the text; Donna Young, our Production Editor, who always seemed to be available for questions any time of the day or night, and who man­ aged the complex responsibilities of bringing the design, photos, artwork, and writing together with efficiency and good cheer. The Prentice Hall team is a first-class operation. There are many others who also deserve special recognition including the following: Jerry Marshall (Truitt and Marshall), our photo researcher, who was so effective in finding photos to bring chemistry to life for students; Linda Brunauer (Santa Clara University) for her marvelous efforts in preparing the Annotated Instructor's Edition of the text; Roxy Wilson (University of illinois) for performing the difficult job of working out solutions to the end-of-chapter exercises. Finally, we wish to thank our families and friends for their love, support, encouragement, and patience as we brought this eleventh edition to completion. Theodore L. Brown

H. Eugene LeMay, Jr.

Bruce E. Bursten

Catherine ). Murphy

Patrick M. Woodward

School of Chemical Sciences

Department of Chemistry

College ofArts and Sciwces

Department of Cizemistry

Department of Chemstry i

University of Illinois,

University ofNevada

University of Tennessee

Reno, NV 89557

Knoxville, TN 37996

University of South Carolina

Columbus, OH 43210

[email protected]

[email protected]

Columbia, SC 29208

woodward@chemistry.

[email protected]

ohio-state.edu

Urbana-Champaign Urbana, IL 61801 [email protected]

and Biochemistry

T7te Ohio State University

List of Resources For Students MasteringChem istry: MasteringChemistry is the first adaptive-learning online homework system . It provides selected end-of-chapter problems from the text, as well as hundreds of tutorials with automatic grading, immediate answer-specific feedback, and simpler questions on request. Based on extensive research of precise concepts students struggle with, MasteringChernistry uniquely responds to your immediate learning needs, thereby optimizing your study time. Solutions to Red Exercises (0-13-600287 -0) Prepared by Roxy Wilson of the University of Illinois-Urbana­ Champaign. Full solutions to all of the red-numbered exercises in the text are provided. (Short answers to red exercises are found in the appendix of the text) . Solutions to Black Exercises (0-13-600324-9) Prepared by Roxy Wilson of the University of illinois-Urbana­ Champaign. Full solutions to all of the black-numbered exercises in the text are provided.

Student's G u ide (0-13-600264-1) Prepared by James C. Hill of California State University. This book assists

students through the text material with chapter overviews, learning objectives, a review of key terms, as well as self­ tests with answers and explanations. This edition also features the addition of MCAT practice questions.

Laboratory Experiments (0-13-600285-4) Prepared by John H. Nelson and Kenneth C. Kemp, both of the University of Nevada. This manual contains 43 finely tuned experiments chosen to introduce students to basic lab techniques and to illustrate core chemical principles. This new edition has been revised to correlate more tightly with the text and now features a guide on how to keep a lab report notebook. You can also customize these labs through Catalyst, our custom database program. For more information, visit www.prenhall.com/catalyst.

Virtual Chemlab: An easy to use simulation of five different general chemistry laboratories that can be used to supplement a wet lab, for pre-laboratory and post­ laboratory activities, for homework or quiz assignments, or for classroom demonstrations.

For Instructors Annotated Instructor's Edition (0-13-601 250-7) Prepared by Linda Brunauer of Santa Clara University. Provides marginal notes and information for instructors and TAs, including transparency icons, suggested lecture demonstrations, teaching tips, and background references from the chemical education literature.

Full Solutions Manual (0-13-600325-7) Prepared by Roxy Wilson of the University of Illinois-Urbana­ Champaign. This manual contains all end-of-chapter exercises in the text. With an instructor's permission, this manual may be made available to students. I nstructor's Resou rce Center on CD-DVD (0-13600281-1) This resource provides an integrated collection of resources to help you make efficient and effective use of your time. This CD/DVD features most art from the text, including figures and tables in PDF format for high-resolution prin­ ting, as well as four pre-built PowerPointTM presentations. The first presentation contains the images embedded within PowerPoint slides. The second includes a complete lecture outline that is modifiable by the user. The final two presentations contain worked "in chapter" sample exercises and questions to be used with Classroom Response Systems. This CD/DVD also contains movies, animations, and electronic files of the Instructor's Resource Manual.

Test Item File (0-13-601251-5) Prepared by Joseph P.

Laurino and Donald Cannon, both of the University of Tampa. The Test Item File now provides a selection of more than 4000 test questions.

Instructor's Resou rce Manual (0-13-600237-4) Prepared by Linda Brunauer of Santa Clara University and Elizabeth Cook of the University of Louisiana. Organized by chapter, this manual offers detailed lecture outlines and complete descriptions of all available lecture demonstra­ tions, the interactive media assets, common student misconceptions, and more.

Transparencies (0-13-601256-6) Approximately 300 full­ color transparencies put principles into visual perspective and save you time when preparing lectures. Annotated Instructor's Edition to Laboratory

Experiments (0-13-6002862) Prepared by John H. Nelson and Kenneth C. Kemp, both of the University of Nevada. This AlE combines the full student lab manual with appendices covering the proper disposal of chemical waste, safety instructions for the lab, descriptions of standard lab equipment, answers to questions, and more. BlackBoard and WebCT: Practice and assessment materials are available upon request in these course management platforms. xxxii

About the Authors THEODORE L. BROWN received his Ph.D. from Michigan State University in 1956. Since then, he has been a member of the faculty of the University of Jllinois, Urbana-Champaign, where he is now Professor of Chemistry, Emeritus. He served as Vice Chancellor for Research, and Dean, The Graduate College, from 1980 to 1986, and as Founding Director of the Arnold and Mabel Beckman Institute for Advanced Science and Technology from 1987 to 1993. Professor Brown has been an Alfred P. Sloan Foundation Research Fellow and has been awarded a Guggen­ heim Fellowship. In 1972 he was awarded the American Chemical Society Award for Research in Inorganic Chem­ istry, and received the American Chemical Society Award for Distinguished Service in the Advancement of Inorganic Chemistry in 1993. He has been elected a Fellow of both the American Association for the Advancement of Science and the American Academy of Arts and Sciences. H. EUGENE LEMAY, JR., received his B.S. degree in Chemistry from Pacific Lutheran University (Washington) and his Ph.D. in Chemistry in 1966 from the University of Illinois (Urbana). He then joined the faculty of the Univer­ sity of Nevada, Reno, where he is currently Professor of Chemistry, Emeritus. He has enjoyed Visiting Professor­ ships at the University of North Carolina at Chapel Hill, at the University College of Wales in Great Britain, and at the University of California, Los Angeles. Professor LeMay is a popular and effective teacher, who has taught thou­ sands of students during more than 35 years of university teaching. Known for the clarity of his lectures and his sense of humor, he has received several teaching awards, including the University Distinguished Teacher of the Year Award (1991) and the first Regents' Teaching Award given by the State of Nevada Board of Regents (1997). BRUCE E. BU RSTEN received his Ph.D. in Chemistry from the University of Wisconsin in 1978. After two years as a National Science Foundation Postdoctoral Fellow at Texas A&M University, he joined the faculty of The Ohio State University, where he rose to the rank of Distinguished University Professor. In 2005, he moved to his present position at the University of Tennessee, Knoxville as Distinguished Professor of Chemistry and Dean of the College of Arts and Sciences. Professor Bursten has been a Camille and Henry Dreyfus Foundation Teacher-Scholar and an Alfred P. Sloan Foundation Research Fellow, and he has been elected a Fellow of the American Association for the Advancement of Science. At Ohio State he has received the University Distinguished Teaching Award in 1982 and 1996, the Arts and Sciences Student Council Outstanding Teaching Award in 1984, and the University Distinguished Scholar Award in 1990. He received the Spiers Memorial Prize and Medal of the Royal Society of Chemistry in 2003, and the Morley Medal of the Cleveland Section of the American Chemical Society n i 2005. He was elected President of the American Chemical Society for 2008. In addition to his teaching and service activities, Professor Bursten's re­ search program focuses on compounds of the transition-metal and actinide elements. CATHERI N E J. M U RPHY received two B.S. degrees, one in Chemistry and one in Biochemistry, from the Uni­ versity of Illinois, Urbana-Champaign, in 1986. She received her Ph.D. in Chemistry from the University of Wiscon­ sin in 1990. She was a National Science Foundation and National Institutes of Health Postdoctoral Fellow at the California Institute of Technology from 1990 to 1993. In 1993, she joined the faculty of the University of South Car­ olina, Columbia, where she is currently the Guy F. Lipscomb Professor of Chemistry. Professor Murphy has been honored for both research and teaching as a Camille Dreyfus Teacher-Scholar, an Alfred P. Sloan Foundation Re­ search Fellow, a Cottrell Scholar of the Research Corporation, a National Science Foundation CAREER Award win­ ner, and a subsequent NSF Award for Special Creativity. She has also received a USC Mortar Board Excellence in Teaching Award, the USC Golden Key Faculty Award for Creative Integration of Research and Undergraduate Teaching, the USC Michael J. Mungo Undergraduate Teaching Award, and the USC Outstanding Undergraduate Research Mentor Award. Since 2006, Professor Murphy has served as a Senior Editor to the journal of Plzysical Chem­ istry. Professor Murphy's research program focuses on the synthesis and optical properties of inorganic nanomate­ rials, and on the local structure and dynamics of the DNA double helix.

Contributing Author

PATRICK M . WOODWARD received B.S. degrees in both Chemistry and Engineering from Idaho State Uni­ versity in 1991. He received a M.S. degree in Materials Science and a Ph.D. in Chemistry from Oregon State Univer­ sity in 1996. He spent two years as a postdoctoral researcher in the Department of Physics at Brookhaven National Laboratory. In 1998, he joined the faculty of the Chemistry Department at The Ohio State University where he cur­ rently holds the rank of Associate Professor. He has enjoyed visiting professorships at the University of Bordeaux, in France, and the University of Sydney, in Australia. Professor Woodward has been an Alfred P. Sloan Foundation Re­ search Fellow and a National Science Foundation CAREER Award winner. He currently serves as an Associate Edi­ tor to the journal of Solid State Chemistry and as the director of the Ohio REEL program, an NSF funded center that works to bring authentic research experiments into the laboratories of first and second year chemistry classes in 15 colleges and universities across the state of Ohio. Professor Woodward's research program focuses on understand­ ing the links between bonding, structure, and properties of solid state inorganic functional materials.

xxxili

INTRODUCTION : MATTER AND MEASUREMENT

HUBBLE SPACE TELESCOPE IMAGE o f the Crab Nebula, a 6-light-year-wide expanding remnant o f a star's supernova explosion. The orange filaments are the tattered remains of the star and consist mostly of hydrogen, the sim plest and most plentiful element in the un iverse. Hydrogen occurs as molecules in cool regions, as atoms in hotter regions, and as ions in the hottest regions. The processes that occur within stars are responsible for creating other chemical elements from hydrogen.

W H AT ' S 1.1

1.2

1.3

1.4

A H EA D numbers and units. The units of measurement used throughout science are those of the metric system, a decimal system of measurement.

The Study of Chemistry

We begin by providing a brief perspective of what chemistry is about and why it is useful to learn chemistry.

1.5

Classifications of Matter

Next, we examine some fundamental ways to classify materials, distinguishing between pure substances and mixtures and noting that there are two fundamentally different kinds of pure substances: elements and compounds. Properties of Matter

We then consider some of the different kinds of characteristics, or properties, that we use to characterize, identify, and separate substances. U nits of Measurement

1.6

Uncertainty in Measurement

We also observe that the uncertainties inherent in all measured quantities are expressed by the number of significant figures used to report the number. Significant figures are also used to express the uncertainties associated with calculations involving measured quantities. Dimensional Analysis

We recognize that units as well as numbers are carried through calculations and that obtaining correct units for the result of a calculation is an important way to check whether the calculation is correct.

We observe that many properties rely on quantitative measurements, involving both

HAVE YOU EVER WON D ERED why ice melts and water

evaporates? Why do leaves turn colors in the fall, and how does a battery generate electricity? Why does keeping foods cold slow their spoilage, and how do our bodies use food to maintain life? Chemistry answers these questions and countless others like them. Chemistry is the study of materials and the changes that materials undergo. One of the joys of learning chemistry is seeing how chemical principles operate in all aspects of our lives, from everyday activities like lighting a match to more far-reaching matters like the development of drugs to cure cancer. Chemical principles also operate in the far reaches of our galaxy (chapter-opening photo) as well as within and around us. This first chapter lays a foundation for our study of chemistry by provid­ ing an overview of what chemistry is about and dealing with some fundamen­ tal concepts of matter and scientific measurements. The list above, entitled "What's Ahead," gives a brief overview of the organization of this chapter and some of the ideas that we will consider. As you study, keep in mind that the chemical facts and concepts you are asked to learn are not ends in themselves; they are tools to help you better understand the world around you.

2

C HA P T E R

1

Introduction: Matter and Measurement

1 . 1 THE STUDY OF CHEMISTRY Before traveling to an unfamiliar city, you might look at a map to get some sense of where you are heading. Because chemistry may be unfamiliar to you, it's useful to get a general idea of what lies ahead before you embark on your

journey. In fact, you might even ask why you are taking the trip.

The Atomic and Molecular Perspective of Chem istry Chemistry is the study of the properties and behavior of matter. Matter is the physical material of the universe; it is anything that has mass and occupies space. A property is any characteristic that allows us to recognize a particular type of matter and to distinguish it from other types. This book, your body, the clothes you are wearing, and the air you are breathing are all samples of matter. Not all forms of matter are so common or so familiar. Countless experiments have shown that the tremendous variety of matter in our world is due to com­

100 very basic, or elementary, substances called elements. As we proceed through this text, we will seek to relate the properties of matter to its composition, that is, to the particular elements it contains. Chemistry also provides a background to understanding the properties of matter in terms of atoms, the almost infinitesimally small building blocks of matter. Each element is composed of a unique kind of atom. We will see that the properties of matter relate to both the kinds of atoms the matter contains (composition) and to the arrangements of these atoms (structure). Atoms can combine to form molecules in which two or more atoms are joined together in specific shapes. Throughout this text you will see molecules represented using colored spheres to show how their component atoms con­ nect to each other (Figure 1.1 T). The color provides a convenient and easy way to distinguish between the atoms of different elements. For examples, compare the molecules of ethanol and ethylene glycol in Figure 1.1. Notice that these molecules have different compositions and structures. Ethanol contains only one oxygen atom, which is depicted by one red sphere. In contrast, ethylene glycol has two atoms of oxygen. Even apparently minor differences in the composition or structure of mole­ cules can cause profound differences in their properties. Ethanol, also called grain alcohol, is the alcohol in beverages such as beer and wine. Ethylene gly­ col, on the other hand, is a viscous liquid used as automobile antifreeze. The ,. Figure 1 .1 Molecular models. properties of these two substances differ in many ways, including the tempera­ The white, dark gray, and red spheres tures at which they freeze and boil. The biological activities of the two mole­ represent atoms of hydrogen, carbon, cules are also quite different. Ethanol is consumed throughout the world, but and oxygen, respectively. you should never consume ethylene glycol be­ cause it is highly toxic. One of the chal­ lenges that chemists undertake is to alter the composition or structure Ill of molecules in a controlled way, (a) Oxygen creating new substances with ., • different properties. , • Every change in the world­ observable (d) Ethanol • binations of only about

•

••

..

(b) Water

•

.,,

•

,

from

boiling

water to

the changes that occur as

our bodies

combat

invading viruses-has its basis in the world of atoms and molecules. Thus, as we pro­ ceed with our study of chemistry,

(c) Carbon dioxide

(e) Ethylene glycol

(f) Aspirin

we will find ourselves thinking in two realms: the

macroscopic

realm

1.1

The Study of Chemistry

3

=

of ordinary-sized objects (macro large) and the submicroscopic realm of atoms and molecules. We make our observations in the macroscopic world-in the laboratory and in our everyday surroundings. To understand that world, however, we must visualize how atoms and molecules behave at the sub­ microscopic level. Chemistry is the science that seeks to understand the proper­ ties and behavior of matter by studying the properties and behavior of atoms and molecules. G IVE

IT

S O M E

THOUGHT

(a) In round numbers, about how many elements are there? (b) What submicroscopic particles are the building blocks of matter?

Why Study Chem istry? Chemistry provides important understanding of our world and how it works. It is an extremely practical science that greatly impacts our daily lives. Indeed, chemistry lies near the heart of many matters of public concern: improvement of health care; conservation of natural resources; protection of the environment; and provision of our everyday needs for food, clothing, and shelter. Using chemistry, we have discovered pharmaceutical chemicals that enhance our health and prolong our lives. We have increased food production through the use of fertilizers and pesticides, and we have developed plastics and other ma­ terials that are used in almost every facet of our lives. Unfortunately, some chemicals also have the potential to harm our health or the environment. As ed­ ucated citizens and consumers, it is in our best interest to understand the pro­ found effects, both positive and negative, that chemicals have on our lives and to strike an informed balance about their uses. Most of you are studying chemistry, however, not merely to satisfy your cu­ riosity or to become more informed consumers or citizens, but because it is an essential part of your curriculum. Your major might be biology, engineering, pharmacy, agriculture, geology, or some other field. Why do so many diverse subjects share an essential tie to chemistry? The answer is that chemistry, by its very nature, is the central science, central to a fundamental understanding of other sciences and technologies. For example, our interactions with the materi­ al world raise basic questions about the materials around us. What are their compositions and properties? How do they interact with us and our environ­ ment? How, why, and when do they undergo change? These questions are im­ portant whether the material is part of high-tech computer chips, a pigment used by a Renaissance painter, or the DNA that transmits genetic information in our bodies (Figure 1.2 T). By studying chemistry, you will learn to use the powerful language and ideas that have evolved to describe and enhance our understanding of matter. The language of chemistry is a universal scientific language that is widely used

(a)

(b)

"f'

Figure 1 .2 Chemistry helps us better understand materials. (a) A

microscopic view of an EPROM (Erasable Programmable Read-Only Memory) silicon microchip. (b) A Renaissance painting, Young Girl Reading, by Vittore Carpaccio (1 472- 1 S26). (c) A long strand of DNA that has spilled out of the damaged cell wall of a bacterium.

(c)

4

CHAPTER 1

Chemis

Introduction: Matter and Measurement

Rut to Work

C H E M I ST RY A N D T H E C H E M I CAL I N D USTRY

M icals such as those shown in Figure 1.3

any people are familiar with common household chem­ II>, but few realize the size and importance of the chemical industry. Worldwide sales of chemicals and related products manufactured in the United States total approximately $550 billion annually. The chemical industry employs more than 10% of all scientists and engineers and is a major contributor to the US economy. Vast amounts of chemicals are produced each year and serve as raw materials for a variety of uses, including the manufacture of metals, plastics, fertilizers, pharmaceuticals, fuels, paints, ad­ hesives, pesticides, synthetic fibers, microprocessor chips, and numerous other products. Table 1.1 T lists the top eight chemicals produced in the United States. We will discuss many of these substances and their uses as the course progresses. People who have degrees in chemistry hold a variety of po­ sitions in industry, government, and academia. Those who work in the chemical industry find positions as laboratory chemists, carrying out experiments to develop new products (research and development), analyzing materials (quality control), or as­ sisting customers in using products (sales and service). Those with more experience or training may work as managers or company directors. A chemistry degree also can prepare you for alternate careers in teaching, medicine, biomedical research, in­ formation science, environmental work, technical sales, work with government regulatory agencies, and patent law.

TABLE

1 .1 • The Top Eight Chemicals Produced by the Chemical Industry in 2006'

Rank 1

2 3

4

5 6 7

8 aMost

""' Figure 1 .3 Household chemkals. Many common supermarket products have very simple chemical compositions.

data from

Chemical

Formula

Sulfuric acid Ethylene Lime Propylene Phosphoric acid Ammonia Chlorine Sodium hydroxide

H2S04 C2H4 CaO C3H6 H3P04 NH3 Cl2

2006 Production (billions of pounds)

NaOH

79

55 45

35

24

23 23 18

Principal End Uses

Fertilizers, chemical manufacturing Plastics, antifreeze Paper, cement, steel Plastics Fertilizers Fertilizers Bleaches, plastics, water purification Aluminum production, soap

Chemical and Engineering News, July 2, 2007, pp. 57, 60.

in other disciplines. Furthermore, an understanding of the behavior of atoms and molecules provides powerful insights into other areas of modern science, technology, and engineering.

1 .2 CLASS IFICATI O N S OF MATTER Let's begin our study of chemistry by examining some fundamental ways in which matter is classified and described. Two principal ways of classifying matter are according to its physical state (as a gas, liquid, or solid) and accord­ ing to its composition (as an element, compound, or mixture). States of Matter A sample of matter can be a gas, a liquid, or a solid. These three forms of matter are called the states of matter. The states of matter differ in some of their simple

1.2

Classifications of Matter

5

observable properties. A gas (also known as vapor) has no fixed volume or shape; rather, it conforms to the volume and shape of its container. A gas can be compressed to occupy a smaller volume, or it can expand to occupy a larger one. A liquid has a distinct volume independent of its container but has no specific shape. A liquid assumes the shape of the portion of the container that it occupies. A solid has both a definite shape and a definite vol­ ume. Neither liquids nor solids can be com­ pressed to any appreciable extent. The properties of the states of matter can be understood on the molecular level (Figure 1.4 .,. ). In a gas the molecules are far apart and are moving at high speeds, colliding re­ peatedly with each other and with the walls of the container. Compressing a gas decreases the amount of space between molecules, increases the frequency of collisions between molecules, but does not alter the size or shape of the molecules. In a liquid the molecules are packed close­ ly together but still move rapidly. The rapid movement allows the molecules to Gas Liquid Solid slide over each other; thus, a liquid pours eas.6. Figure 1 .4 The three physical ily. In a solid the molecules are held tightly together, usually in definite states of water-water vapor, liquid arrangements in which the molecules can wiggle only slightly in their other­ water, and Ice. In this photo we see wise fixed positions. both the liquid and solid states of water. Pure S u bstances

Most forms of matter that we encounter-for example, the air we breathe (a gas), gasoline for cars (a liquid), and the sidewalk on which we walk (a solid)-are not chemically pure. We can, however, resolve, or separate, these forms of matter into different pure substances. A pure substance (usually referred to simply as a substance) is matter that has distinct properties and a composition that does not vary from sample to sample. Water and ordinary table salt (sodium chloride), the primary components of seawater, are examples of pure substances. All substances are either elements or compounds. Elements cannot be de­ composed into simpler substances. On the molecular level, each element is composed of only one kind of atom [Figure l.S(a and b) T )]. Compounds are

(a) Atoms of an element

(b) Molecules

of an element

(c) Molecules of a compound

We cannot see water vapor. What we see when we look at steam or clouds is tiny droplets of liquid water dispersed in the atmosphere. The molecular views show that the molecules in the gas are much further apart than those in the liquid or solid. The molecules in the liquid do not have the orderly arrangement seen in the solid.

(d) Mixture of elements and a compound

.6. Figure 1 .5 Molecular comparison of element, compounds, and mixtures. Each element contains a unique kind of atom. Elements might consist of individual atoms, as in (a), or molecules, as in (b). Compounds contain two or more different atoms chemically joined together, as in (c). A mixture contains the individual units of its components, shown in (d) as both atoms and molecules.

6

CHAPTER 1

Introduction: Matter and Measurement

.,. Figure 1 .6 Relative abundances of elements. Elements in percent by mass

Aluminum

Other

7.5%

7%

in (a) Earth's crust (including oceans and atmosphere) and (b) the human body.

I

Earth's crust

Human body

(a)

(b)

Hydrogen

/ 10%

substances composed of two or more elements; they contain two or more kinds of atoms [Figure l.S(c)]. Water, for example, is a compound composed of two el­ ements: hydrogen and oxygen. Figure l.S(d) shows a mixture of substances. Mixtures are combinations of two or more substances in which each substance retains its own chemical identity. Elements

Currently, 117 elements are known. These elements vary widely in their abun­ dance, as shown in Figure 1.6 .a.. For example, only five elements-oxygen, silicon, aluminum, iron, and calcium-account for over 90% of Earth's crust (including oceans and atmosphere). Similarly, just three elements-oxygen, carbon, and hydrogen-account for over 90% of the mass of the human body. Some of the more common elements are listed in Table 1 .2 .., along with the chemical abbreviations, or chemical symbols, used to denote them. The symbol for each element consists of one or two letters, with the first letter capitalized. These symbols are mostly derived from the English name for the element, but sometimes they are derived from a foreign name instead (last column in Table 1.2). You will need to know these symbols and learn others as we encounter them in the text. All the known elements and their symbols are listed on the front inside cover of this text. The table in which the symbol for each element is enclosed in a box is called the periodic table. In the periodic table the elements are arranged in vertical columns so that closely related elements are grouped together. We describe the periodic table in more detail in Section 2.5. ,

GIVE

IT

S O M E

THOUGHT

Which element is most abundant in both Earth's crust and in the human body? What is the symbol for this element?

Compounds

Most elements can interact with other elements to form compounds. For exam­ ple, consider the fact that when hydrogen gas burns in oxygen gas, the

TABLE 1 .2

• Some Common Elements and Their Symbols

Carbon Fluorine Hydrogen Iodine Nitrogen Oxygen Phosphorus Sulfur

c F H I N 0

p s

Aluminum Bromine Calcium Chlorine Helium Lithium Magnesium Silicon

AI Br Ca Cl He Li Mg Si

Copper Iron Lead Mercury Potassium Silver Sodium Tin

Cu (from cuprum) Fe (from ferrum) Pb (from plumbum) Hg (from hydrargyrum) K (from knlium) Ag (from argentum) Na (from natrium) Sn (from stannum)

1.2

Classifications of Matter

Hydrogen gas, H2 .A

Figure 1 .7 Electrolysis of water. Water decomposes into its component elements, hydrogen and oxygen, when a direct electrical current is passed through it. The volume of hydrogen, which is collected in the right tube of the apparatus, is twice the volume of oxygen, which is collected in the left tube.

elements hydrogen and oxygen combine to form the compound water. Con­ versely, water can be decomposed into its component elements by passing an electrical current through it, as shown in Figure 1.7 .A. Pure water, regardless of its source, consists of 11% hydrogen and 89% oxygen by mass. This macroscop­ ic composition corresponds to the molecular composition, which consists of two hydrogen atoms combined with one oxygen atom: Hydrogen atom

Oxygen atom

Water molecule

(written 0)

(written H)

The elements hydrogen and oxygen themselves exist naturally as diatomic (two-atom) molecules: Oxygen molecule

Hydrogen molecule

As seen in Table 1.3 �. the properties of water bear no resemblance to the proper­ ties of its component elements. Hydrogen, oxygen, and water are each a unique substance, a consequence of the uniqueness of their respective molecules.

TABLE 1 .3

• Comparison of Water, Hydrogen, and Oxygen

State' Normal boiling point Density' Flammable

Water

Hydrogen

Oxygen

Liquid

Gas

Gas

100 'C 1000 g/L

- 253 'C 0.084 g/L

No

Yes

11At room temperature and atmospheric pressure. (See Section

10.2.)

-183 'C

1.33

No

g/L

7

8

CHAPTER 1

Introduction: Matter and Measurement The observation that the elemental composition of a pure compound is always the same is known as the law of constant composition (or the Jaw of definite proportions). French chemist Joseph Louis Proust (1754-1826) first put forth the law in about 1800. Although this law has been known for 200 years, the general belief persists among some people that a fundamental difference exists between compounds prepared in the laboratory and the corresponding compounds found in nature. However, a pure compound has the same compo­ sition and properties regardless of its source. Both chemists and nature must use the same elements and operate under the same natural laws. When two ma­ terials differ in composition and properties, we know that they are composed of different compounds or that they differ in purity. GIVE

IT

S O M E

T H O U G HT

Hydrogen, oxygen, and water are all composed of molecules. What is it about a molecule of water that makes it a compound, whereas hydrogen and oxygen are elements?

Mixtures

Most of the matter we encounter consists of mixtures of different substances. Each substance in a mixture retains its own chemical identity and its own prop­ erties. In contrast to a pure substance that has a fixed composition, the compo­ sition of a mixture can vary. A cup of sweetened coffee, for example, can contain either a little sugar or a lot. The substances making up a mixture (such as sugar and water) are called components of the mixture. Some mixtures do not have the same composition, properties, and appear­ ance throughout. Both rocks and wood, for example, vary in texture and ap­ pearance throughout any typical sample. Such mixtures are heterogeneous [Figure 1.8(a) T]. Mixtures that are uniform throughout are homogeneous. Air is a homogeneous mixture of the gaseous substances nitrogen, oxygen, and smaller amounts of other substances. The nitrogen in air has all the properties that pure nitrogen does because both the pure substance and the mixture con­ tain the same nitrogen molecules. Salt, sugar, and many other substances dis­ solve in water to form homogeneous mixtures [Figure 1.8(b)]. Homogeneous mixtures are also called solutions. Although the term solution conjures an image of a liquid in a beaker or flask, solutions can be solids, liquids, or gases. Figure 1.9 IJJo summarizes the classification of matter into elements, compounds, and mixtures.

IJJo

Figure 1 .8 Mixtures. (a) Many common materials, including rocks, are heterogeneous. This close-up photo is of malachite, a copper mineral. (b) Homogeneous mixtures are called solutions. Many substances, including the blue solid shown in this photo (copper sulfate), dissolve in water to form solutions.

(a)

(b)

1.3

Properties of Matter



0

:::;

- O 'C

Kelvin scale

l

- Water boils

- 98 .6 'F ., �0

- Normal body temperature

�

--

Celsius scale

32 'F

l

- Water freezes

Fahrenheit scale

*Mass and weight are not interchangeable terms but are often incorrectly thought to be the same. The weight

ofat1 object is the force that its mass exerts due to gravity. In space, where gravitational forces are very weak, an astronaut can be weightless, but he or she cannot be massless. In fact, the astronaut's mass in space is the

same as it is on Earth.

1 .1 7 Australian stamp.

i': 2 .5 dJ OJ

�

.5 dJ OJ

- 37.0 'C ., �-

f

Figure

Many countries employ the Celsius temperature scale in everyday use, as illustrated by this stamp.

.!!) "'

-;;;

� 2 .5 dJ OJ 0

-

"'

.!!) "'

310 K - ., �-

f

.li.

7c

) are used to indicate the number of atoms of each element. The prefix mono- is never used with the first element. When the prefix ends in a or o and the name of the second element begins with a vowel (such as oxide), the a or o of the prefix is often dropped. The following examples illustrate these rules:

Cl20 N204

dichlorine monoxide dinitrogen tetroxide

NF3 P4510

nitrogen trifluoride tetraphosphorus decasulfide

It is important to realize that you cannot predict the formulas of most mol­ ecular substances in the same way that you predict the formulas of ionic com­ pounds. For this reason, we name molecular compounds using prefixes that explicitly indicate their composition. Molecular compounds that contain hy­ drogen and one other element are an important exception, however. These compounds can be treated as if they were neutral substances containing H + ions and anions. Thus, you can predict that the substance named hydrogen chloride has the formula HCl, containing one H + to balance the charge of one Cl-. (The name hydrogen chloride is used only for the pure compound; water solutions of HCl are called hydrochloric acid.) Similarly, the formula for hydro­ gen sulfide is H2S because two H + are needed to balance the charge on 52-.

Relating the Names and Formulas of Binary Molecular Compounds I Name the following compounds: S02, PCls, N203.

• SAMPLE EXERCISE 2.15

(a)

(b)

(c)

SOLUTION

The compounds consist entirely of nonmetals, so they are molecular rather than ionic. Using the prefixes in Table 2.6, we have (a) sulfur dioxide, (b) phosphorus pentachlo­ ride, and (c) dinitrogen trioxide. - PRACTICE EXERCISE

Give the chemical formula for (a) silicon tetrabromide, (b) disulfur dichloride. SiBr4, (b) S2Cl2

Answers: (a)

TABLE 2.6 • Prefixes Used in Naming Binary Compounds Formed between Nonmetals Prefix

MonoDiTriTetraPentaHexaHeptaOctaNonaDeca-

Meaning

1 2 3

4 5

6

7

8 9

10

65

66

C HAPTER 2

Atoms, Molecules, and Ions

2.9 S O M E S I M P LE O RGANI C C O M P O UN D S The study of compounds o f carbon is called organic chemistry, and as noted earlier in the chapter, compounds that contain carbon and hydrogen, often in combination with oxygen, nitrogen, or other elements, are called organic com­ pounds. We will examine organic compounds and organic chemistry in some detail in Chapter 25. You will see a number of organic compounds throughout this text; many of them have practical applications or are relevant to the chem­ istry of biological systems. Here we present a very brief introduction to some of the simplest organic compounds to provide you with a sense of what these molecules look like and how they are named. Alkanes Compounds that contain only carbon and hydrogen are called hydrocarbons. In the most basic class of hydrocarbons, each carbon atom is bonded to four other atoms. These compounds are called alkanes. The three simplest alkanes, which contain one, two, and three carbon atoms, respectively, are methane (CH4), ethane (C2H6), and propane (C3H8). The structural formulas of these three alkanes are as follows: H

H

I

H

I

H

I

H

I

H-C-C-H

H-C-C-C-H

H Methane

H H Ethane

H H H Propane

I

I

H

H-C-H

I

I

I

I

I I

We can make longer alkanes by adding additional carbon atoms to the "skele­ ton" of the molecule. Although the hydrocarbons are binary molecular compounds, they are not named like the binary inorganic compounds discussed in Section 2.8. Instead, each alkane has a name that ends in -ane. The alkane with four carbon atoms is called butane. For alkanes with five or more carbon atoms, the names are de­ rived from prefixes such as those in Table 2.6. An alkane with eight carbon atoms, for example, is called octane (C8H18), where the acta- prefix for eight is combined with the -ane ending for an alkane. Gasoline consists primarily of octanes, as will be discussed in Chapter 25. Some Derivatives of Alkanes Other classes of organic compounds are obtained when hydrogen atoms of alkanes are replaced with functional groups, which are specific groups of atoms. An alcohol, for example, is obtained by replacing an H atom of an alkane with an -OH group. The name of the alcohol is derived from that of the alkane by adding an -ol ending: H

I

H

I

H

I

H

I

H

I

H

I

H-C-OH

H-C-C-OH

H-C-C-C-OH

H Methanol

H H Ethanol

H H H 1-Propanol

I

I

I

I

I

I

Alcohols have properties that are very different from the properties of the al­ kanes from which the alcohols are obtained. For example, methane, ethane, and propane are all colorless gases under normal conditions, whereas methanol, ethanol, and propanol are colorless liquids. We will discuss the reasons for these differences in properties in Chapter 11.

2.9

Some Simple Organic Compounds

67

The prefix "1" in the name 1-propanol indicates that the replacement of H with OH has occurred at one of the "outer" carbon atoms rather than the "mid­ dle" carbon atom. A different compound called 2-propanol (also known as iso­ propyl alcohol) is obtained if the OH functional group is attached to the middle carbon atom: H

I

H

I

H

H

I

I

H-C-C-C-OH

I

H

I

H

H

H

I

(a)

I

H-C-C-C-H

I

I

H

H

I 0

I

H

H

2-Propanol

1-Propanol

Ball-and-stick models of these two molecules are presented in Figure 2.29 ... . Much of the richness of organi c chemistry is possible because organic com­ pounds can form long chains of carbon-- N02

(unbalanced)

This equation has three 0 atoms on the left side of the arrow and two 0 atoms on the right side. We can increase the number of 0 atoms by placing a coefficient 2 on the product side: 02

+ NO --->

2 N02

(unbalanced)

Now there are two N atoms and four 0 atoms on the right. Placing the coefficient 2 in front of NO balances both the number of N atoms and 0 atoms: 02 + 2 NO ---> 2 N02

(balanced)

(c) The left box (reactants) contains four 02 molecules and eight NO molecules.

Thus, the molecular ratio is one 02 for each two NO as required by the balanced equation. The right box (products) contains eight N02 molecules. The number of N02 molecules on the right equals the number of NO molecules on the left as the bal­ anced equation requires. Counting the atoms, we find eight N atoms in the eight NO molecules in the box on the left. There are also 4 X 2 = 8 0 atoms in the 02 molecules and eight 0 atoms in the NO molecules, giving a total of 16 0 atoms. In the box on the right, we find eight N atoms and 8 x 2 = 16 0 atoms in the eight N02 molecules. Because there are equal numbers of both N and 0 atoms in the two boxes, the draw­ ing is consistent with the law of conservation of mass.

3.1

Chemical Equations

83

- PRACTICE EXERCISE

In the following diagram, the white spheres represent hydrogen atoms, and the blue spheres represent nitrogen atoms. To be consistent with the law of conservation of mass, how many NH 3 molecules should be shown in the right box?

?

Answer:

Six N H3 molecules

Indicating the States of Reactants and Products Additional information is often added to the formulas in balanced equations to indicate the physical state of each reactant and product. We use the symbols (g), (1), (s), and (aq) for gas, liquid, solid, and aqueous (water) solution, respectively. Thus, Equation 3.4 can be written CH4(g)

+ 2

Oz(g)

�

COz(g)

+ 2

HzO(g)

[3.5]

Sometimes the conditions (such as temperature or pressure) under which the reaction proceeds appear above or below the reaction arrow. The symbol !J. (the Greek uppercase letter delta) is often placed above the arrow to indicate the ad­ dition of heat. - SAMPLE EXERCISE 3.2 J Balancing Chemical Equations Balance this equation:

Na(s) + H20(1) --> NaOH(aq) + H2(g) SOLUTION Begin by counting each kind of atom on both sides of the arrow. The Na and 0 atoms are balanced-one Na and one 0 on each side-but there are two H atoms on the left and three H atoms on the right. Thus, we need to increase the number of H atoms on the left. To begin balancing H, let's try placing the coefficient 2 in front of H20: Na(s) + 2 H20(/) Beginning this way does not balance H but does increase the number of H atoms among the reactants, which we need to do. Adding the coeffi­ cient 2 causes 0 to be unbalanced; we will take care of that after we bal­ ance H. Now that we have 2 H20 on the left, we can balance H by putting the coefficient 2 in front of NaOH on the right:

Balancing H in this way fortuitously brings 0 into balance. But notice that Na is now unbalanced, with one Na on the left and two on the right. To rebalance Na, we put the coefficient 2 in front of the reactant:

Na(s) + 2 H20(/)

-->

NaOH(aq) + H2(g)

--> 2

NaOH(aq)

+ H2(g)

2 Na(s) + 2 H20(1) --> 2 NaOH(aq) + H2(g)

Finally, we check the number of atoms of each element and find that we have two Na atoms, four H atoms, and two 0 atoms on each side of the equation. The equation is balanced.

Comment Notice that in balancing this equation, we moved back and forth placing a coefficient in front of H20, then NaOH, and finally Na. In balancing equations, we often find ourselves following this pattern of moving back and forth from one side of the arrow to the other, placing coefficients first in front of a formula on one side and then in front of a formula on the other side until the equation is balanced. You can always tell if you have balanced your equation correctly, no matter how you did it, by checking that the number of atoms of each element is the same on both sides of the arrow. - PRACTICE EXERCISE Balance the following equations by providing the missing coefficients: (a) _ Fe(s) + _Oz(g) --> _Fez0 3(s) (b) _CzH 4 (g) + _02(g) --> _ C02(g) + _ H zO(g) (c) _Al(s) + _ HCI(aq) --> _A1Cl 3(aq) + _ H2(g) Answers: (a) 4, 3, 2; (b) 1, 3, 2, 2; (c) 2, 6, 2, 3

84

C HA P T E R 3

Stoichiometry: Calculations with Chemical Formulas and Equations

3.2 S O M E S I M P LE PATTERNS O F CHEMI CAL REACTIVITY In this section we examine three simple kinds of reactions that we will see fre­ quently throughout this chapter. Our first reason for examining these reactions is merely to become better acquainted with chemical reactions and their bal­ anced equations. Our second reason is to consider how we might predict the products of some of these reactions knowing only their reactants. The key to predicting the products formed by a given combination of reactants is recogniz­ ing general patterns of chemical reactivity. Recognizing a pattern of reactivity for a class of substances gives you a broader understanding than merely mem­ orizing a large number of unrelated reactions.

Combination and Decomposition Reactions

Table 3.1 ., summarizes two simple types of reactions: combination and decom­ position reactions. In combination reactions two or more substances react to form one product. There are many examples of combination reactions, especial­ ly those in which elements combine to form compounds. For example, magne­ sium metal burns in air with a dazzling brilliance to produce magnesium oxide, as shown in Figure 3.5 � :

2 Mg(s)

+

02 (g)

---->

2 MgO(s)

[3.6]

This reaction is used to produce the bright flame generated by flares and some fireworks. When a combination reaction occurs between a metal and a nonmetal, as in Equation 3.6, the product is an ionic solid. Recall that the formula of an ionic compound can be determined from the charges of the ions involved. coo (Section 2.7) When magnesium reacts with oxygen, for example, the mag­ nesium loses electrons and forms the magnesium ion, Mg2+. The oxygen gains 2 electrons and forms the oxide ion, 0 -. Thus, the reaction product is MgO. You should be able to recognize when a reaction is a combination reaction and to predict the products of a combination reaction in which the reactants are a metal and a nonmetal. GIVE IT SOME THOUGHT When Na and S undergo a combination reaction, what is the chemical formula of the product?

In a decomposition reaction one substance undergoes a reaction to pro­ duce two or more other substances. Many compounds undergo decomposition reactions when heated. For example, many metal carbonates decompose to form metal oxides and carbon dioxide when heated: CaC03(s)

�

CaO(s)

+

C02{g)

[3.7]

TABLE 3.1 • Combination and Decomposition Reactions Combination Reactions A + B ----> C C(s) + Oz(g) ----> COz(g) N2(g) + 3 H2(g) ----> 2 NH3(g) CaO(s) + H20(1) ----> Ca(OH)z(s)

Two reactants combine to form a single product. Many elements react with one another in this fashion to form compounds.

Decomposition Reactions C ----> A + B 2 KC103(s) ----> 2 KCl(s) + 3 02(g) PbC03(s) --+ PbO(s) + C02(g) Cu(OH)z(s) ----> CuO(s) + H20(1)

A single reactant breaks apart to form two or more substances. Many compounds react this way when heated.

3.2

Some Simple Patterns of Chemical Reactivity

85

COMBI NATION REACTION

In combination reactions, two or more substances react to form one product.

+

2 Mg(s)

The ribbon of magnesium metal is surrounded by oxygen in the air, and as it bums, an intense flame is produced. A

2 Mg0(s)

When magnesium metal bums, the Mg atoms react with 02 molecules from the air to form magnesium oxide, MgO, an ionic solid.

At the end of the reaction, a rather fragile ribbon of white solid, MgO, remains.

Figure 3.5 Combustion of magnesium metal In air.

The decomposition of CaC03 is an important commercial process. Limestone or seashells, which are both primarily CaC03, are heated to prepare CaO, which is known as lime or quicklime. About 2 X 10 10 kg (20 million tons) of CaO is used in the United States each year, principally in making glass, in ob­ taining iron from its ores, and in making mortar to bind bricks. The decomposition of sodium azide (NaN3) rapidly releases N2(g), so this reaction is used to inflate safety air bags in automobiles (Figure 3.6 ... ): [3.8] The system is designed so that an impact ignites a detonator cap, which in turn causes NaN3 to decompose explosively. A small quantity of NaN3 (about 100 g) forms a large quantity of gas (about 50 L). We will consider the volumes of gases produced in chemical reactions in Section 10.5. • SAMPLE EXERCISE 3.3

Balanced Equations for Combination and Decomposition Reactions J Writing

Write balanced equations for the following reactions: (a) The combination reaction that occurs when lithium metal and fluorine gas react. (b) The decomposition reac­ tion that occurs when solid barium carbonate is heated. (Two products form: a solid and a gas.)

SOLUTION (a) The symbol for lithium is Li. With the exception of mercury, all metals are solids at room temperature. Fluorine occurs as a diatomic molecule (see Figure 2.19). Thus, the reactants are Li(s) and F2{g). The product will be composed of a metal and a nonmetal, so we expect it to be an ionic solid. Lithium ions have a 1 + charge, u+,

A Figure 3.6 An automobile air bag. The decomposition of sodium azide, NaN3(s), is used to inflate automobile air bags. When properly ignited, the NaN3 decomposes rapidly, forming nitrogen gas, N2(9), which expands the air bag.

86

C HA P T E R 3

Stoichiometry: Calculations with Chemical Formulas and Equations whereas fluoride ions have a 1- charge, r. Thus, the chemical formula for the prod­ uct is LiF. The balanced chemical equation is 2 Li(s)

+ F2(g)

----->

2 LiF(s)

(b) The chemical formula for barium carbonate is BaC03. As noted in the text, many metal carbonates decompose to form metal oxides and carbon dioxide when heated. ln Equation 3.7, for example, CaC0 decomposes to form CaO and C02. Thus, we 3 would expect that BaC03 decomposes to form BaO and C02. Barium and calcium are both in group 2A in the periodic table, which further suggests they would react in the same way: BaC03(s) -----> BaO(s)

+ C02(g)

- PRACTICE EXERCISE Write balanced chemical equations for the following reactions: (a) Solid mercury(II) sulfide decomposes into its component elements when heated. (b) The surface of alu­ minum metal undergoes a combination reaction with oxygen in the air. Answers: (a) HgS(s) -----> Hg(l) + S(s); (b) 4 Al(s) + 3 02(g) -----> 2 Al20 3(s)

Combustion i n Air Combustion reactions are rapid reactions that produce a flame. Most of the combustion reactions we observe involve 02 from air as a reactant. Equation 3.5 illustrates a general class of reactions that involve the burning, or combustion, of hydrocarbon compounds (compounds that contain only carbon and hydro­ gen, such as CH4 and C2H4). cx:o (Section 2.9) When hydrocarbons are cornbusted in air, they react with 02 to form C02 and H20.* The number of molecules of 02 required in the reaction and the number of molecules of C02 and H20 formed depend on the composition of the hydrocarbon, which acts as the fuel in the reaction. For example, the com­ bustion of propane (C3H8), a gas used for cooking and horne heating, is de­ scribed by the following equation: [3.9]

& Figure 3.7 Propane burning In air. The liquid propane, C3H8, vaporizes and mixes with air as it escapes through the nozzle. The combustion reaction of C3Hs and Oz produces a b l ue flame.

The state of the water, H20(g) or H20(I), depends on the conditions of the reaction. Water vapor, H20(g), is formed at high temperature in an open con­ tainer. The blue flame produced when propane burns is shown in Figure 3.7 ... . Combustion of oxygen-containing derivatives of hydrocarbons, such as CH30H, also produces C02 and H20. The simple rule that hydrocarbons and related oxygen-containing derivatives of hydrocarbons form C02 and H20 when they bum in air summarizes the behavior of about 3 million compounds. Many substances that our bodies use as energy sources, such as the sugar glu­ cose (C6H1206), similarly react in our bodies with 02 to form C02 and H20. In our bodies, however, the reactions take place in a series of intermediate steps that occur at body temperature. These reactions that involve intermediate steps are described as oxidation reactions instead of combustion reactions. - SAMPLE EXERCISE 3.4

Writing Balanced Equations Combustion Reactions I for

Write the balanced equation for the reaction that occurs when methanol, CH30H(I), is burned in air.

SOLUTION When any compound containing C, H, and 0 is combusted, it reacts with the 02(g) in air to produce C02(g) and H20(g). Thus, the unbalanced equation is CH 30H(J)

+ 02(g)

-----> C02(g) + H20(g)

*When there is an insufficient quantity of02 present, carbon monoxide (CO) will be produced along with the C02; this is called incomplete combustion. If the amount of 02 is severely restricted, fine particles of carbon that we call soot will be produced. Complete combustion produces only C02 and H20. Unless specifically stated to the contrary, we will always take combustion to mean complete combustion.

3.3 In this equation the C atoms are balanced with one carbon on each side of the arrow. Because CH30H has four H atoms, we place the coefficient 2 in front of HzO to bal­ ance the H atoms: CH30H{l) + 02(g) -----> C02(g) + 2 H20(g) Adding the coefficient balances H but gives four 0 atoms in the products. Because there are only three 0 atoms in the reactants (one in CH30H and two in Oz), we are not finished yet. We can place the fractional coefficient in front of 02 to give a total of four 0 atoms in the reactants (there are X 2 = 3 0 atoms in 02): CH30H(I) +

�

�

� � 02(g) -----> C02(g) + 2 H20(g)

Although the equation is now balanced, it is not in its most conventional form be­ cause it contains a fractional coefficient. If we multiply each side of the equation by 2, we will remove the fraction and achieve the following balanced equation:

2 CHpH(I) + 3 02(g) -----> 2 C02(g) + 4 H20(g) - PRACTICE EXERCISE Write the balanced equation for the reaction that occurs when ethanol, C2H50H(I), is burned in air. Answer: C2H50H(I) + 3 02(g) -----> 2 C02(g) + 3 H20(g)

3.3 FORM ULA WEI GHTS Chemical formulas and chemical equations both have a quantitative signifi­ cance; the subscripts in formulas and the coefficients in equations represent precise quantities. The formula H20 indicates that a molecule of this substance (water) contains exactly two atoms of hydrogen and one atom of oxygen. Simi­ larly, the coefficients in a balanced chemical equation indicate the relative quan­ tities of reactants and products. But how do we relate the numbers of atoms or molecules to the amounts we measure in the laboratory? Although we cannot directly count atoms or molecules, we can indirectly determine their numbers if we know their masses. Therefore, before we can pursue the quantitative aspects of chemical formulas or equations, we must examine the masses of atoms and molecules, which we do in this section and the next.

Formula and Molecular Weights The formula weight of a substance is the sum of the atomic weights of each atom in its chemical formula. Using atomic masses from a periodic table, we find, for example, that the formula weight of sulfuric acid (H2S04) is 98.1 amu:* FW of H2S04

=

2(AW ofH) + (AW of S)

+

4(AW of O)

= 2(1.0 amu) + 32.1 amu + 4(16.0 amu) = 98.1 amu For convenience, we have rounded off all the atomic weights to one place be­ yond the decimal point. We will round off the atomic weights in this way for most problems. If the chemical formula is merely the chemical symbol of an element, such as Na, then the formula weight equals the atomic weight of the element, in this case 23.0 amu. If the chemical formula is that of a molecule, then the formula weight is also called the molecular weight. The molecular weight of glucose (C6H1 206), for example, is MW of C6H 1 206

=

6(12.0 amu)

+

12(1.0 amu) + 6(16.0 amu)

=

180.0 amu

Because ionic substances, such as NaCI, exist as three-dimensional arrays of ions (Figure 2.23), it is inappropriate to speak of molecules of NaCI. Instead, "The abbrevation i AW is usedfor atomic weight, FWfor formula weight, and MWfor molecular weight.

Formula Weights

87

88

C HA P T E R 3

Stoichiometry: Calculations with Chemical Formulas and Equations we speak of formula units, represented by the chemical formula of the sub­ stance. The formula unit of NaCI consists of one Na+ ion and one Cl- ion. Thus, the formula weight of NaCI is the mass of one formula unit: FW of NaCI = 23. 0 amu + 35.5 amu = 58.5 amu

- SAMPLE EXERCISE 3.5

I Calculating Formula Weights

Calculate the formula weight of (a) sucrose, C12H22011 (table sugar), and (b) calcium nitrate, Ca(N03)z. SOLUTION

(a) By adding the atomic weights of the atoms in sucrose, we find the formula weight to be 342.0 amu:

(b) If a chemical formula has parentheses, the subscript out­ side the parentheses is a multiplier for all atoms inside. Thus, for Ca(N03)z, we have

12 C atoms = 12(12.0 amu) = 144.0 amu 22 H atoms = 22(1.0 amu) = 22.0 amu 11 0 atoms = 11(16.0 amu) = 176.0 amu 342.0 amu 1 Ca atom = 1(40.1 amu) = 40.1 amu 2 N atoms = 2(14.0 amu) = 28.0 amu 6 0 atoms = 6(16.0 amu) = 96.0 amu 164. 1 amu

- PRACTICE EXERCISE Calculate the formula weight of (a) Al(OHh and (b) CHpH. Answers: (a) 78.0 amu, (b) 32.0 amu

Percentage Composition from Formulas Occasionally we must calculate the percentage composition of a compound-that is, the percentage by mass contributed by each element in the substance. For ex­ ample, to verify the purity of a compound, we can compare the calculated per­ centage composition of the substance with that found experimentally. Forensic chemists, for example, will measure the percentage composition of an un­ known white powder and compare it to the percentage compositions for sugar, salt, or cocaine to identify the powder. Calculating percentage composition is a straightforward matter if the chemical formula is known. The calculation de­ pends on the formula weight of the substance, the atomic weight of the element of interest, and the number of atoms of that element in the chemical formula:

(

number of atoms

)(

)

atomic weight of that element of element % element = --'---,-----,-------,--------, -'formula weight of compound

- SAMPLE EXERCISE 3.6

x

1 00%

[3.1 0]

I Calculating Percentage Composition

Calculate the percentage of carbon, hydrogen, and oxygen (by mass) in C12H220n. SOLUTION Let's examine this question using the problem-solving steps in the "Strategies in Chemistry: Problem Solving" essay that appears on the next page. Analyze We are given a chemical formula, C1 2H22011, and asked to calculate the per­ centage by mass of its component elements (C, H, and 0). Plan We can use Equation 3.10, relying on a periodic table to obtain the atomic weight of each component element. The atomic weights are first used to determine the formula weight of the compound. (The formula weight of C12H22011, 342.0 amu, was calculated in Sample Exercise 3.5.) We must then do three calculations, one for each element. Solve Using Equation 3.10, we have %C =

(12)(12.0 amu) X 100% = 42.1% 342.0 amu

%H =

(22)(1.0 amu) X 100% = 6.4% 342.0 amu

3.4

%0

=

(11)(16.0 amu) 342.0 amu

X

100%

=

Avogadro's Number and the Mole

89

51.5%

Check The percentages of the individual elements must add up to 100%, which they do in this case. We could have used more significant figures for our atomic weights, giving more significant figures for our percentage composition, but we have adhered to our suggested guideline of rounding atomic weights to one digit beyond the deci­ mal point. - PRACTICE EXERCISE Calculate the percentage of nitrogen, by mass, in Ca(N03h. 17.1%

Answer:

3.4 AVO GADRO 'S NUMBER AND THE M OLE Even the smallest samples that we deal with in the laboratory contain enor­ mous numbers of atoms, ions, or molecules. For example, a teaspoon of water 23 (about 5 mL) contains 2 X 10 water molecules, a number so large that it al­ most defies comprehension. Chemists, therefore, have devised a special count­ ing unit for describing such large numbers of atoms or molecules. In everyday life we use counting units such as a dozen (12 objects) and a gross (144 objects) to deal with modestly large quantities. In chemistry the unit for dealing with the number of atoms, ions, or molecules in a common-sized sample is the mole, abbreviated mol.* A mole is the amount of matter that con­ tains as many objects (atoms, molecules, or whatever objects we are consider­ 2 ing) as the number of atoms in exactly 12 g of isotopically pure 1 C. From 23 experiments, scientists have determined this number to be 6.0221421 X 10 Scientists call this number Avogadro's number, in honor of the Italian scientist Amedeo Avogadro (1776- 1856). Avogadro's number has the symbol NA, 23 which we will usually round to 6.02 X 10 mol- t. The unit mol-l ("inverse 23 mole" or "per mole") reminds us that there are 6.02 X 10 objects per one mole. A mole of atoms, a mole of molecules, or a mole of anything else all contain Avogadro's number of these objects: 23 1 2 1 mol 12C atoms = 6.02 x 10 C atoms 23 1 mol H20 molecules = 6.02 X 10 H20 molecules 23 1 mol N03 - ions = 6.02 x 10 N03- ions

P RO B L E M S O LVI N G

P

ractice is the key to success in problem solving. As you prac­ tice, you can improve your skills by following these steps:

Step 1: Analyze the problem.

Read the problem carefully for under­

standing. What does it say? Draw any picture or diagram that will

help you to visualize the problem. Write down both the data you are given and the quantity that you need to obtain (the unknown).

Step 2: Dwelop a plan for solving the problem.

Consider the possi­

ble paths between the given information and the unknown. What principles or equations relate the known data to the un­

known? Recognize that some data may not be given explicitly in

the problem; you may be expected to know certain quantities

(such as Avogadro's number) or look them up in tables (such as

atomic weights). Recognize also that your plan may involve ei­ ther a single step or a series of steps with intermediate answers.

Step 3: Solve the problem.

Use the known information and suit­

able equations or relationships to solve for the unknown. Di­

mensional analysis (Section 1.6) is a very useful tool for solving a

great number of problems. signs, and units.

Step 4: Check tlze solution.

Be

careful with significant figures,

Read the problem again to make sure

you have found all the solutions asked for in the problem. Does

your answer make sense? That is, is the answer outrageously

large or small, or is it in the ballpark? Finally, are the units and

significant figures correct?

*The term mole comes from the Latin word moles, meaning "a mass." The term molecule is the diminutive fonn of this word and means "a small mass."

90

C HA P T E R 3

Stoichiometry: Calculations with Chemical Formulas and Equations Avogadro's number is so large that it is difficult to imagine. Spreading

6.02 X 10 23 marbles over the entire surface of Earth would produce a continu­

ous layer about 3 miles thick. If Avogadro's number of pennies were placed side by side in a straight line, they would encircle Earth 300 trillion (3 X 1014) times. - SAMPLE EXERCISE 3.7

I Estimating Numbers of Atoms

Without using a calculator, arrange the following samples in order of increasing 2 numbers of carbon atoms: 12 g 1 C, 1 mol C2H:u 9 X 1023 molecules of C02. SOLUTION Analyze We are given amounts of different substances expressed in grams, moles, and number of molecules and asked to arrange the samples in order of increasing numbers of C atoms. 12 Plan To determine the number of C atoms in each sample, we must convert g C, 2 1 mol C2H2, and 9 X 10 3 molecules C02 all to numbers of C atoms. To make these conversions, we use the definition of mole and Avogadro's number. Solve A mole is defined as the amount of matter that contains as many units of the 2 12 matter as there are C atoms in exactly 12 g of C. Thus, 12 g of 1 C contains 1 mol of 2 2 C atoms (that is, 6 . 02 x 10 3 C atoms). One mol of C2H2 contains 6 x 10 3 C2H2 mole­ cules. Because there are two C atoms in each C2H2 molecule, this sample contains 2 12 X 10 3 C atoms. Because each C02 molecule contains one C atom, the sample of 2 2 C02 contains 9 X 10 3 C atoms. Hence, the order is 12 g 12C (6 X 10 3 C atoms) < 9 2 2 2 X 10 3 C02 molecules (9 X 10 3 C atoms) < 1 mol C2H2 (12 x 10 3 C atoms). Check We can check our results by comparing the number of moles of C atoms in each sample because the number of moles is proportional to the number of atoms. Thus, 12 g of 1 ZC is 1 mol C; 1 mol of C2H2 contains 2 mol C, and 9 X 1023 molecules of C02 contain 1.5 mol C, giving the same order as above: 12 g 1ZC (1 mol C) < 9 X 2 10 3 C02 molecules (1.5 mol C) < 1 mol C2H2 (2 mol C). - PRACTICE EXERCISE Without using a calculator, arrange the following samples in order of increasing num­ 2 ber of 0 atoms: 1 mol H20, 1 mol C02, 3 X 10 3 molecules 03. Answer: 1 mol H20 (62 x 10 23 0 atoms) < 3 X 10 23 molecules 03 (9 x 10 23 0 atoms) < 1 mol C02 (12 X 10 3 0 atoms)

- SAMPLE EXERCISE 3.8

I Converting Moles to Number of Atoms

Calculate the number of H atoms in 0.350 mol of C6H 1 206. SOLUTION

Analyze We are given both the amount of a substance (0.350 mol) and its chemical formula (C6H 1206). The unknown is the number of H atoms in the sample. Plan Avogadro's number provides the conversion factor between the number of moles of C6H 1206 and the number of molecules of C6H 1 206. Once we know the num­ ber of molecules of C6H 1206, we can use the chemical formula, which tells us that each molecule of C6H 1206 contains 12 H atoms. Thus, we convert moles of C6H 1206 to molecules of C6H 1206 and then determine the number of atoms of H from the number of molecules of c� 1206: Moles C� 1206 Solve H atoms

---->

= atoms H 2

6.02 X 10 3 moleeales- Et;HU()6 1�

)(

12 H atoms 1 moleeule€6HU()6

)

Check The magnitude of our answer is reasonable. It is a large number about the magnitude of Avogadro's number. We can also make the following ballpark calcula­ 2 2 tion: Multiplying 0.35 x 6 x 10 3 gives about 2 X 10 3 molecules. Multiplying this re­ 2 2 sult by 12 gives 24 X 10 3 = 2.4 X 10 4 H atoms, which agrees with the previous, more detailed calculation. Because we were asked for the number of H atoms, the units of our answer are correct. The given data had three significant figures, so our answer has three significant figures.

3.4

Avogadro's Number and the Mole

91

- PRACTICE EXERCISE How many oxygen atoms are in (a) 0.25 mol Ca(N03h and (b) 1.50 mol of sodium carbonate? x 1023, (b) 2.71 X 1024

Answers: (a) 9.0

Molar Mass A dozen (12) is the same number whether we have a dozen eggs or a dozen ele­ phants. Clearly, however, a dozen eggs does not have the same mass as a dozen 23 elephants. Similarly, a mole is always the same number (6.02 X 10 ), but 1-mole samples of different substances will have different masses. Compare, for exam­ 2 24 ple, 1 mol of 12C and 1 mol of Mg. A single 1 C atom has a mass of 12 amu, whereas a single 24Mg atom is twice as massive, 24 amu (to two significant fig­ ures). Because a mole always has the same number of particles, a mole of 24Mg 2 2 must be twice as massive as a mole of 1 C. Because a mole of 1 C has a mass of 24 12 g (by definition), then a mole of Mg must have a mass of 24 g. This exam­ ple illustrates a general rule relating the mass of an atom to the mass of Avogadro's number (1 mol) of these atoms: The mass of a single atom ofan element

(in amu) is numerically equal to the mass (in grams) of 1 mol of that element.

This statement is true regardless of the element: 2 12 1 atom of 1 C has a mass of 12 amu � 1 mol C has a mass of 12 g 1 atom of Cl has an atomic weight of 35.5 amu 1 atom of Au has an atomic weight of 197 amu

�

�

1 mol Cl has a mass of 35.5 g 1 mol Au has a mass of 197 g

Notice that when we are dealing with a particular isotope of an element, we use the mass of that isotope; otherwise we use the atomic weight (the average atomic mass) of the element. (Section 2.4) For other kinds of substances, the same numerical relationship exists be­ tween the formula weight (in amu) and the mass (in grams) of one mole of that substance: ceo

1 H 20 molecule has a mass of 18.0 amu 1 No3- ion has a mass of 62.0 amu 1 NaCl unit has a mass of 58.5 amu

�

1 mol H 20 has a mass of 18.0 g

�

1 mol NaCI has a mass of 58.5 g

�

1 mol N03- has a mass of 62.0 g

Figure 3.8 T illustrates the relationship between the mass of a single molecule of H20 and that of a mole of H 20. GIVE

IT SOME

THOUGHT

(a) Which has more mass, a mole of water (H20) or a mole of glucose (C6H1206)? (b) Which contains more molecules, a mole of water or a mole of glucose?

The mass in grams of one mole of a substance (that is, the mass in grams per mol) is called the molar mass of the substance. The molar mass (in g/mol) of any substance is always numerically equal to its formula weight (in amu). The sub­ stance NaCI, for example, has a formula weight of 58.5 amu and a molar mass

Laboratory-size sample

Single molecule

... Figure 3.8 Comparing the mass of 1 molecule H20 and 1 mol H20. Notice that both masses have the same number but have different units (1 8.0 amu compared to 1 8.0 g) representing the huge difference in mass.

1 molecule H20 (18.0 amu)

1 mol H20 (18.0 g)

92

C HA P T E R 3

Stoichiometry: Calculations with Chemical Formulas and Equations

TABLE 3.2 • Mole Relationships Name of Substance

Formula

Formula Weight (amu)

Molar Mass (g/mol)

Atomic nitrogen

N

14.0

14.0

Molecular nitrogen

Nz

28.0

28.0

Silver

Ag Ag+

107.9

107.9

Silver ions

107.9"

107.9

Barium chloride

BaCiz

208.2

208.2

Number and Kind of Particles in One Mole

{

\

6.02 X 1023 N atoms 6.02 x 1023 N2 molecules 2(6.02 X 1023 ) N atoms 23 6.02 X 10 Ag atoms 6.02 x 1023 Ag+ ions 6.02 x 1023 BaCI2 units 6.02 X 1023 Ba2+ ions

2(6.02 x 1023) Cl- ions

"Recall that tile electron has negligible massi thus, ions and atoms have essentially the same mass.

of 58.5 g/mol. Further examples of mole relationships are shown in Table 3.2 .;.. Figure 3.9 ..,. shows 1 -mole quantities of several common substances. The entries in Table 3.2 for N and N2 point out the importance of stating the exact chemical form of a substance when we use the mole concept. Suppose you read that 1 mol of nitrogen is produced in a particular reaction. You might interpret this statement to mean 1 mol of nitrogen atoms (14.0 g). Unless other­ wise stated, however, what is probably meant is 1 mol of nitrogen molecules, N 2 (28.0 g), because N2 is the most common chemical form of the element. To avoid ambiguity, it is important to state explicitly the chemical form being dis­ cussed. Using the chemical formula N 2 avoids ambiguity. - SAMPLE EXERCISE 3.9 I Calculating Molar Mass What is the mass in grams of 1.000 mol of glucose, C6H1206?

.;. Figure 3.9 One mole each of a solid, a liquid, and a gas. One mole of NaCI, the solid, has a mass of 58.45 g. One mole of H20, the liquid, has a mass of 1 8.0 g and occupies a volume of 1 8.0 ml. One mole of 02, the gas, has a mass of 32.0 g and occupies a balloon whose diameter is 35 em.

SOLUTION Analyze We are given a chemical formula and asked to determine its molar mass. Plan The molar mass of a substance is found by adding the atomic weights of its component atoms. Solve 6 C atoms 12 H atoms

6 0 atoms

6(12.0 amu) 12(1.0 amu) 6(16.0 amu)

72.0 amu 12.0 amu 96.0 amu 180.0 amu

Because glucose has a formula weight of 180.0 amu, one mole of this substance has a mass of 180.0 g. ln other words, C6H1 206 has a molar mass of 180.0 g/mol. Check The magnitude of our answer seems reasonable, and g/ mol is the appropri­ ate unit for the molar mass. Comment Glucose is sometimes called dextrose. Also known as blood sugar, glu­ cose is found widely in nature, occurring in honey and fruits. Other types of sugars used as food are converted into glucose in the stomach or liver before the body uses them as energy sources. Because glucose requires no conversion, it is often given in­ travenous] y to patients who need immediate nourishment. People who have diabetes must carefully monitor the amount of glucose in their blood (See "Chemistry and Life" box in Section 3.6). - PRACTICE EXERCISE Calculate the molar mass of Ca(N03)z.

Answer: 164.1 g/mol

3.4

Avogadro's Number and the Mole

lnterconverting Masses and Moles Conversions of mass to moles and of moles to mass are frequently encountered in calculations using the mole concept. These calculations are simplified using dimensional analysis, as shown in Sample Exercises 3.10 and 3.11.

- SAMPLE EXERCISE 3.10

I Converting Grams to Moles

Calculate the number of moles of glucose (C6H1206) in 5.380 g of C6H1206. SOLUTION Analyze We are given the number of grams of a substance and its chemical formula and asked to calculate the number of moles. Plan The molar mass of a substance provides the factor for converting grams to moles. The molar mass of C6H1206 is 180.0 g/mol (Sample Exercise 3.9) .

Solve Using 1 mol C�1206 = 180.0 g C6H1206 to write the appropriate conversion factor, we have

Check Because 5.380 g is less than the molar mass, a reasonable answer is less than one mole. The units of our answer (mol) are appropriate. The original data had four significant figures, so our answer has four significant figures. - PRACTICE EXERCISE How many moles of sodium bicarbonate (NaHC03) are in 508 g of NaHC03? Answer: 6.05 mol NaHC03

- SAMPLE EXERCISE 3.1 1

I Converting Moles to Grams

Calculate the mass, in grams, of 0.433 mol of calcium nitrate. SOLUTION

Analyze We are given the number of moles and the name of a substance and asked to calculate the number of grams in the sample. Plan To convert moles to grams, we need the molar mass, which we can calculate using the chemical formula and atomic weights. 2 Solve Because the calcium ion is Ca + and the nitrate ion is N03-, calcium nitrate is Ca(N03)z. Adding the atomic weights of the elements in the compound gives a for­ mula weight of 164.1 amu. Using 1 mol Ca(N03)z = 164.1 g Ca(N03)z to write the ap­ propriate conversion factor, we have

Check The number of moles is less than 1, so the number of grams must be less than the molar mass, 164.1 g. Using rounded numbers to estimate, we have 0.5 X 150 = 75 g. The magnitude of our answer is reasonable. Both the units (g) and the number of significant figures (3) are correct. - PRACTICE EXERCISE What is the mass, in grams, of (a) 6.33 mol of NaHC03 and (b) 3.0 furic acid? 3 Answers: (a) 532 g, (b) 2. 9 x 10- g

x

10-s mol of sul­

93

94

C HAPTER 3

II> Figure 3.1 0 Procedure for lnterconvertlng the mass and the number of formula units of a substance. The number of moles of the substance is central to the calculation; thus, the mole concept can be thought of as the bridge between the mass of a substance in grams and the number of formula units.

Stoichiometry: Calculations with Chemical Formulas and Equations Use molar mass

Use Avogadro's number

lnterconverting Masses and Numbers of Particles The mole concept provides the bridge between mass and the number of parti­ cles. To illustrate how we can interconvert mass and numbers of particles, let's calculate the number of copper atoms in an old copper penny. Such a penny weighs about 3 g, and we will assume that it is 100% copper: ( �-""" ( 1 rne+eu ) ( 6.02 X 1023 Cu atoms ) Cu atoms = 3 5��) g-Etr 1 rnel-etr 63.5 = 3 X 1022 Cu atoms

We have rounded our answer to one significant figure, since we used only one significant figure for the mass of the penny. Notice how dimensional analy­ sis (Section 1.6) provides a straightforward route from grams to numbers of atoms. The molar mass and Avogadro's number are used as conversion factors to convert grams --> moles atoms. Notice also that our answer is a very large number. Any time you calculate the number of atoms, molecules, or ions in an or­ dinary sample of matter, you can expect the answer to be very large. In contrast, the number of moles in a sample will usually be much smaller, often less than 1. The general procedure for interconverting mass and number of formula units (atoms, molecules, ions, or whatever is represented by the chemical formula) of a substance is summarized in Figure 3.10 "'· C1CD

-->

- SAMPLE EXERCISE 3.1 2

[ Calculating the Number of Molecules and Number of Atoms from Mass

(a) How many glucose molecules are in 5.23 g of C6H1206? (b) How many oxygen atoms are in this sample? SOLUTION Analyze We are given the number of grams and the chemical formula and asked to calculate (a) the number of molecules and (b) the number of 0 atoms in the sample.

(a) Plan The strategy for determining the number of molecules in a given quantity of a substance is summarized in Figure 3.10. We must convert 5.23 g C6H1206 to moles c�1 20& which can then be converted to molecules c�1206· The first conver­ sion uses the molar mass of C6H1 206: 1 mol C6H1 206 = 180.0 g C6H1 206. The second conversion uses Avogadro's number. Solve Molecules C�1 206 = (5.23 �

= 1.75

X

(

1 m�

180.0 � 22 10 molecules C6H1206

)(

6.02 X 1o23 molecules C6H1 206 1�

)

Check The magnitude of the answer is reasonable. Because the mass we began with is less than a mole, there should be fewer than 6.02 x 1023 molecules. We can make a ballpark estimate of the answer: 5/200 = 2.5 X 10-2 mol; 2.5 X 10-2 X 6 X 1023 = 1 15 x 102 = 1.5 x 1022 molecules. The units (molecules) and significant figures (three) are appropriate.

(b) Pl an To determine the number of 0 atoms, we use the fact that there are six 0 atoms in each molecule of C�1 206. Thus, multiplying the number of molecules C6H1 206 by the factor (6 atoms 0/1 molecule C6H1206) gives the number of 0 atoms. Solve

Atoms 0 = (1.75 X 1022 mQiee>IIesC6II120(;)

= 1.05 x 1023 atoms 0

( � : ��! 0

2 H2C6Hs04(/) + 2 H20(g)

(a) Assume that you carry out this reaction starting with 25.0 g of cyclohexane and that cyclohexane is the limiting reactant. What is the theoretical yield of adipic acid? (b) If you obtain 33.5 g of adipic acid from your reaction, what is the percent yield of adipic acid? SOLUTION Analyze We are given a chemical equation and the quantity of the limiting reactant ( 5 . g of CJ-In). We are asked first to calculate the theoretical yield of a product (H2C6H804) and then to calculate its percent yield if only 33.5 g of the substance is actually obtained.

20

Plan (a) The theoretical yield, which is the calculated quantity of adipic acid formed in the reaction, can be calculated using the following sequence of conversions: g C6H 12

---->

mol C6H 12

---->

mol H2C6Hs04

---->

g H2C6Hs04

(b) The percent yield is calculated by comparing the actual yield (33.5 g) to the theo­ retical yield using Equation 3.14. Solve

=

(25 .

II2C6II804 )( 146.0 H2C6Hs04 ) 0�) (184.mel-EgH1z)(2=>1 0 � 2 meJ.Cof'ln l m<JIIIzC61Is04 g

= 3 .5 g H2C6Hs04

4

·

(b) Percent y1eld

=

actual yield X . . theoretlca1 y1eld

100°'o = "

33 5 g

43.5 g

X

100°'o = 77.0°'o "

"

Check Our answer in (a) has the appropriate magnitude, units, and significant figures. In (b) the answer is less than as necessary.

100%

Summary and Key Terms

2.

Calculations in which you must show your work

107

Your in­

come later in the course, but it is useful to talk about them

structor may present you with a numerical problem in

here. (You should review this box before each exam you

which you are to show your work in arriving at a solu­

take, to remind yourself of good exam-taking practices.)

tion. In questions of this kind, you may receive partial credit even if you do not arrive at the correct answer, de­ pending on whether the instructor can follow your line of

Be sure to label your drawing as completely as possible.

4.

Other types of questions

Other exam questions you might

encounter include true-false questions and ones in which

reasoning. It is important, therefore, to be as neat and or­

you are given a list and asked to indicate which members

ganized as you can be, given the pressures of exam tak­

of the list match some criterion given in the question.

ing. It is helpful in approaching such questions to take a

Often students answer such questions incorrectly be­

few moments to think about the direction you are going

cause, in their haste, they misunderstand the nature of the

to take in solving the problem. You may even want to

question. Whatever the form of the question, ask yourself

write a few words or a diagram on the test paper to indi­

this: What is the instructor testing here? What material

cate your approach. Then write out your calculations as

am I supposed to know that this question covers?

neatly as you can. Show the units for every number you write down, and use dimensional analysis as much as

3.

Finally, if you find that you simply do not understand how

you can, showing how units cancel.

to arrive at a reasoned response to a question, do not linger over

Questions requiring drawings

the question. Put a check next to it and go on to the next one. If

Sometimes a test question

will require you to draw a chemical structure, a diagram

time permits, you can come back to the unanswered questions,

related to chemical bonding, or a figure showing some

but lingering over a question when nothing is coming to mind

kind of chemical process. Questions of this kind will

is wasting time you may need to finish the exam.

- PRACTICE EXERCISE

Imagine that you are working on ways to improve the process by which iron ore con­ taining Fe203 is converted into iron. your tests you carry out the following reaction on a small scale:

In

Fep3(s)

+ 3 CO(g) -----+ 2 Fe(s) + 3 C02(g)

(a) If you start with 150 g of Fep3 as the limiting reagent, what is the theoretical yield of Fe? (b) If the actual yield of Fe in your test was 87.9 g, what was the percent yield? Answers: (a) 105 g Fe, (b) 83.7%

C H APTER REV I E W SUMMARY AND KEY TERMS Introduction and Section 3.1 The study of the quan­ titative relationships between chemical formulas and chemical equations is known as stoichiometry. One of the important concepts of stoichiometry is the law of conser­ vation of mass, which states that the total mass of the prod­ ucts of a chemical reaction is the same as the total mass of the reactants. The same numbers of atoms of each type are present before and after a chemical reaction. A balanced chemical equation shows equal numbers of atoms of each element on each side of the equation. Equations are bal­ anced by placing coefficients in front of the chemical for­ mulas for the reactants and products of a reaction, not by changing the subscripts in chemical formulas.

Section 3.2 Among the reaction types described in this chapter are (1) combination reactions, in which two reac­ tants combine to form one product; (2) decomposition reactions, in which a single reactant forms two or more products; and (3) combustion reactions in oxygen, in

which a hydrocarbon or related compound reacts with 02 to form COz and HzO. Section 3.3 Much quantitative information can be determined from chemical formulas and balanced chemi­ cal equations by using atomic weights. The formula weight of a compound equals the sum of the atomic weights of the atoms in its formula. If the formula is a molecular formula, the formula weight is also called the molecular weight. Atomic weights and formula weights can be used to determine the elemental composition of a compound.

Section 3.4 A mole of any substance is Avogadro's 2 number (6.02 X 10 3) of formula units of that substance. The mass of a mole of atoms, molecules, or ions (the molar mass) equals the formula weight of that material expressed in grams. The mass of one molecule of H20, for example, is 18 amu, so the mass of 1 mol of H20 is 18 g. That is, the molar mass of H20 is 18 g/ mol.

108

C HA PTER 3

Stoichiometry: Calculations with Chemical Formulas and Equations

Section 3.5 The empirical formula of any substance can be determined from its percent composition by calculat­ ing the relative number of moles of each atom in 100 g of the substance. 1f the substance is molecular in nature, its molecular formula can be determined from the empirical formula if the molecular weight is also known. Sections 3.6 and 3.7 The mole concept can be used to calculate the relative quantities of reactants and products in chemical reactions. The coefficients in a balanced equa­ tion give the relative numbers of moles of the reactants and products. To calculate the number of grams of a

product from the number of grams of a reactant, first convert grams of reactant to moles of reactant. Then use the coefficients in the balanced equation to convert the number of moles of reactant to moles of product. Finally, convert moles of product to grams of product. A limiting reactant is completely consumed in a reac­ tion. When it is used up, the reaction stops, thus limiting the quantities of products formed. The theoretical yield of a reaction is the quantity of product calculated to form when all of the limiting reagent reacts. The actual yield of a reaction is always less than the theoretical yield. The percent yield compares the actual and theoretical yields.

KEY SKILLS • Balance chemical equations. • Calculate molecular weights. Convert grams to moles and moles to grams using molar masses. • Convert number of molecules to moles and moles to number of molecules using Avogadro's number. Calculate the empirical and molecular formula of a compound from percentage composition and molecular weight. Calculate amounts, in grams or moles, of reactants and products for a reaction. • Calculate the percent yield of a reaction. •

• •

KEY EQUATIONS

• % element =

• % yield

(

number of atoms of that element

)(

atomic weight of element

formula weight of compound

(actual yield) .

.

(theoretical y:teld)

X 100%

)

[3.14]

x 100%

[3.10]

This is the formula to calculate the mass percentage of each element in a compound. The sum of all the percentages of all the elements in a compound should add up to 100%. This is the formula to calculate the percent yield of a reaction. The percent yield can never be more than 100%.

VISUALIZING CONCEPTS 3.1 The reaction between reactant A (blue spheres) and reac­ tant B (red spheres) is shown in the following diagram:

drawing below represents a sample of H2. Make a corre­ sponding drawing of the CO needed to react completely with the H2. How did you arrive at the number of CO molecules in your drawing? [Section 3.2]

Based on this diagram, which equation best describes the reaction? [Section 3.1]

(a) A2 + B ---> A2B (b) A2 + 4 B ---> 2 AB2 (c) 2 A + B4 ---> 2 AB2 (d) A + B2 ---> AB2 3.2 Under appropriate experimental conditions, H2 and CO undergo a combination reaction to form CH30H. The

3.3 The following diagram represents the collection of ele­ ments formed by a decomposition reaction. (a) 1f the blue spheres represent N atoms and the red ones repre­ sent 0 atoms, what was the empirical formula of the original compound? (b) Could you draw a diagram rep-

Exercises resenting the molecules of the compound that had been decomposed? Why or why not? [Section 3.2]

3.4 The following diagram represents the collection of C02 and H20 molecules formed by complete combustion of a hydrocarbon. What is the empirical formula of the hy­ drocarbon? [Section 3.2]

3.5 Glycine, an amino acid used by organisms to make pro­ teins, is represented by the molecular model below. (a) Write its molecular formula. (b) Determine its molar mass. (c) Calculate the mass of 3 moles of glycine. (d) Calculate the percent nitrogen by mass in glycine. [Sections 3.3 and 3.5]

109

how many moles of each product can be obtained start­ ing with 4.0 mol CH4? [Section 3.6]

3.7 Nitrogen (N2) and hydrogen (H2) react to form ammo­ nia (NH3). Consider the mixture of N2 and H2 shown in the accompanying diagram. The blue spheres represent N, and the white ones represent H. Draw a representa­ tion of the product mixture, assuming that the reaction goes to completion. How did you arrive at your repre­ sentation? What is the limiting reactant in this case? [Section 3.7]

3.8 Nitrogen monoxide and oxygen react to form nitrogen dioxide. Consider the mixture of NO and 02 shown in the accompanying diagram. The blue spheres represent N, and the red ones represent 0. (a) Draw a representation of the product mixture, assuming that the reaction goes to completion. What is the limiting reactant in this case? (b) How many N02 molecules would you draw as products if the reaction had a percent yield of 75%? [Section 3.7]

3.6 The following diagram represents a high-temperature reaction between CH4 and H20. Based on this reaction,

EXERCISES B alancing Chemical Equations 3.9 (a) What scientific principle or law is used in the process of balancing chemical equations? (b) In balancing equa­ tions, why should you not change subscripts in chemi­ cal formulas? (c) How would one write out liquid water, water vapor, aqueous sodium chloride, and solid sodi­ um chloride in chemical equations? 3.10 (a) What is the difference between adding a subscript 2 to the end of the formula for CO to give C02 and adding a coefficient in front of the formula to give 2 CO?

(b) Is the following chemical equation, as written, con­ sistent with the law of conservation of mass? 3 Mg(OHh (s) + 2 H3P04(aq) -----+ Mg3(P04h(s) + 6 Hp( l) Why or why not?

3.11 Balance the following equations: (a) CO(g) + 02(g) -----> C02(g) (b) N205(g) + H20(1) -----> HN03(aq) (c) CH4(g) + Cl2(g) -----> CC14( 1) + HCJ(g)

110

C HA PTER 3

Stoichiometry: Calculations with Chemical Formulas and Equations

+

+

poses to form solid potassium chloride and oxygen gas.

CH 4(g) (d) AI4C3(s) + H20(/) ---> AI(OH)J(s) (e) C5H1002(1) + 02(g) ---> C02(g) H20(g)

(f)

Fe(OH)J(s) + H2S0 4(aq)

(g) Mg3N2(s) + H zS04(aq)

3.12

+

--->

Fe2(S04)J(aq)

---> MgS0 4(aq)

Balance the following equations:

+

+

+

(c) Solid zinc metal reacts with sulfuric acid to form hy­

drogen gas and an aqueous solution of zinc sulfate.

+

(f)

--->

AgN03(aq) + Na2S04(aq)

--->

(g) CH3NH2(g)

3.13

+

02(g)

Ca3(P04)z(s)

AgzS04(s)

--->

C02(g)

+

+

+

3.14

+

each of the following descriptions: forms.

(a) When sulfur tri­

(b) Boron sulfide, B2S3(s), reacts violently with

water to form dissolved boric acid, H3B03, and hydro­

H20(1)

gen sulfide gas. (c) When an aqueous solution of Jead(II) nitrate

NaN03(aq)

H20(g)

Write balanced chemical equations to correspond to oxide gas reacts with water, a solution of sulfuric acid

is mixed with an aqueous solution of sodium io­

dide, an aqueous solution of sodium nitrate and a yel­ low solid, lead iodide, are formed.

Nz(g)

(d) When solid

mercury(II) nitrate is heated, it decomposes to form solid mercury(II) oxide, gaseous nitrogen dioxide, and

Write balanced chemical equations to correspond to each of the following descriptions:

(e) When

hydrogen sulfide gas is passed over solid hot iron(Ill) sulfide and gaseous water.

+

H3P04(aq)

H3P03(aq), and aqueous hydrochloric acid.

hydroxide, the resultant reaction produces solid iron(Ill)

02(g) H20(g) (c) NH4N03(s) ---> N2(g) (d) Ca3P2(s) + HzO(I) ---> Ca(OH)z(aq) + PH3(g)

(e) Ca(OH)z(aq)

water, it reacts to form aqueous phosphorous acid,

(NH4)zS04(aq)

(a) Li(s) Nz(g) ---> Li3N(s) H20(1) ---> La(OH)J(aq) (b) Laz03(s)

+

(d) When liquid phosphorus trichloride is added to

H20(1)

oxygen.

(a) Solid calcium car­

(e) Copper metal reacts with hot concentrated

sulfuric acid solution to form aqueous copper(II) sul­

bide, CaC2, reacts with water to form an aqueous solu­

fate, sulfur dioxide gas, and water.

tion of calcium hydroxide and acetylene gas, C2H2.

(b) When solid potassium chlorate is heated, it decom-

Patterns of Chemical Reactivity 3.15 (a)

When the metallic element sodium combines with

the nonmetallic element bromine, Br2(1), how can you

3.18

Write a balanced chemical equation for the reaction that occurs when

(a) aluminum metal undergoes a combina­ (b) copper(II) hydroxide de­

determine the chemical formula of the product? How do

tion reaction with 02(g);

you know whether the product is a solid, liquid, or gas

composes into copper(II) oxide and water when heated;

at room temperature? Write the balanced chemical

(c) heptane, C7H16(I), burns in air;

equation for the reaction.

tive MTBE (methyl tert-butyl ether), CsH120(1), burns

(b) When a hydrocarbon

(d) the gasoline addi­

in air.

bums in air, what reactant besides the hydrocarbon is involved in the reaction? What products are formed?

Write a balanced chemical equation for the combustion of benzene, C6H6(1), in air.

3.16 (a)

3.19

Determine the chemical formula of the product

(a) Al(s) Cl2(g) ---> AICJ3(s) (b) C2H4(g) + Oz(g) ---> COz(g)

with the nonmetallic element oxygen, 02. Write the bal­

(b) What

(c) Li(s)

0 is completely combusted in air? Write a balanced acetone,

C3H60(l), in air.

3.17

3.20

Write a balanced chemical equation for the reaction that occurs when

(a) Mg(s) reacts with Cl2(g); (b) barium

carbonate decomposes into barium oxide and carbon

dioxide gas when heated; (c) the hydrocarbon styrene, C8H8(1),

is

combusted

in

air;

(d) dirnethylether,

CH30CH3(g), is combusted in air.

N2(g)

+ HzO(g)

---> Li3N(s) (d) PbC03(s) ---> PbO(s) + C02(g) (e) C7H802(1) + 02(g) ---> C02(g)

products form when a compound containing C, H, and chemical equation for the combustion of

+ +

reactions:

formed when the metallic element calcium combines anced chemical equation for the reaction.

Balance the following equations, and indicate whether

they are combination, decomposition, or combustion

+

H20(g)

Balance the following equations, and indicate whether they

+

are combination, decomposition, or combustion reactions:

+

(a) C3H6(g) Oz(g) ---> COz(g) + HzO(g) H20(g) (b) NH�03(s) ---> N20(g) (c) CsH60(1) + Oz(g) ---> COz(g) HzO(g) (d) N2(g) H2(g) ---> NH3(g) (e) KzO(s) HzO(I) ---> KOH(aq)

+ +

+

Formula Weights 3.21

Determine the formula weights of each of the following

gas and used as an anesthetic in dentistry;

compounds:

acid, HC7H 0z, a substance used as a food preservative; 5 (c) Mg(OH)z, the active ingreclient in milk of magnesia;

(c)

3.22

(f)

(a) nitric acid, HN03; (b) KMn04; Ca3(P04)z; (d) quartz, SiOz; (e) gallium sulfide,

chromium(Ill) sulfate, (g) phosphorus trichloride.

Determine the formula weights of each of the following compounds:

(a )

nitrous oxide, N20, known as laughing

(b) benzoic

(d) urea, (NH2)zCO, a compound used as a nitrogen fer­ (e) isopentyl acetate, CH3C02C5H11, responsible

tilizer;

for the odor of bananas.

111

Exercises H

3.23 Calculate the percentage by mass of oxygen in the following compounds: (a) morphine, C17H19N03; (b) codeine, C18H21N03; (c) cocaine, C17H21N04; (d) tetracycline, C22H2�Ps; (e) digitoxin, C41HM013; (f) vancomycin, C66H7sCl2N9024· 3.24 Calculate the percentage by mass of the indicated ele­ ment in the following compounds: (a) carbon in acety­ lene, C2Hz, a gas used in welding; (b) hydrogen in ascorbic acid, HC6H�6, also known as vitamin C; (c) hydrogen in ammonium sulfate, (NH4hS04, a sub­ stance used as a nitrogen fertilizer; (d) platinum in PtCl2(NH3h, a chemotherapy agent called cisplatin; (e) oxygen in the female sex hormone estradiol, C1sH240� (f) carbon in capsaicin, C1sH27N03, the com­ pound that gives the hot taste to chili peppers.

I

H

I

H

I

0

II

(c) H3C-C-C-C-0-C-CH3

I

I

I

H3C H H Isopentyl acetate (banana flavor)

3.26 Calculate the percentage of carbon by mass in each of the compounds represented by the following models:

3.25 Based on the following structural formulas, calculate

(a)

(b)

(c)

(d)

the percentage of carbon by mass present in each compound: H H \ I o C=C I \ II (a) H-C C-C-H

I \ c-c

I H

Benzaldehyde (almond fragrance)

\ H

H H3CO I \ c=C 1 \ (b) HO-C C-C-H

I \ c-c

I H

?J

Vanillin (vanilla flavor)

\ H

Avogadro's Number and the Mole 3.27 (a) What is Avogadro's number, and how is it related to

the mole? (b) What is the relationship between the for­ mula weight of a substance and its molar mass? 12 3.28 (a) What is the mass, in grams, of a mole of C? (b) How 12 many carbon atoms are present in a mole of C?

3.29 Without doing any detailed calculations (but using a pe­

riodic table to give atomic weights), rank the following samples in order of increasing number of atoms: 23 0.50 mol H20, 23 g Na, 6.0 X 10 N2 molecules. 3.30 Without doing any detailed calculations (but using a pe­ riodic table to give atomic weights), rank the following samples in order of increasing number of atoms: 23 3.0 x 10 molecules of H202, 2.0 mol CH4, 32 g 02.

3.31 What is the mass, in kilograms, of an Avogadro's number

of people, if the average mass of a person is 160 lb? How 24 does this compare with the mass of Earth, 5.98 x 10 kg? 3.32 If Avogadro's number of pennies is divided equally among the 300 million men, women, and children in the United States, how many dollars would each receive? How does this compare with the gross domestic prod­ uct of the United States, which was $13.5 trillion in 2006? (The GOP is the total market value of the nation's goods and services.)

3.33 Calculate the following quantities: (a) (b) (c) (d)

mass, in grams, of 0.105 moles sucrose (C12H220n) moles of Zn(N03h in 143.50 g of this substance number of molecules in 1 .0 X 10� mol CH3CH20H number of N atoms in 0.410 mol NH3

3.34 Calculate the following quantities

3 (a) mass, in grams, of 5.76 x w- mol of CdS (b) number of moles of NH4Cl in 112.6 g of this substance 2 (c) number of molecules in 1 .305 x 10- mol C6H6 3 (d) number of 0 atoms in 4.88 x 1 0- mol Al(N03)3 3

3.35 (a) What is the mass, in grams, of 2.50 x 10- mol of (b) (c) (d)

3.36 (a) (b)

ammonium phosphate? How many moles of chloride ions are in 0.2550 g of aluminum chloride? 2 What is the mass, in grams, of 7.70 X 10 0 molecules of caffeine, CsH10N402? What is the molar mass of cholesterol if 0.00105 mol weighs 0.406 g? What is the mass, in grams, of 0.0714 mol of iron(II1) sulfate? How many moles of ammonium ions are in 8.776 g of ammonium carbonate?

112

C HA PTER 3

Stoichiometry: Calculations with Chemical Formulas and Equations

21 (c) What is the mass, in grams, of 6.52 X 10 molecules of aspirin, C9H804? (d) What is the molar mass of diazepam (Valium� if 0.05570 mol weighs 15.86 g? 3.37 The molecular formula of allicin, the compound respon­ sible for the characteristic smell of garlic, is C6H100S2. (a) What is the molar mass of allicin? (b) How many moles of allicin are present in 5.00 mg of this substance? (c) How many molecules of allicin are in 5.00 mg of this substance? (d) How many S atoms are present in 5.00 mg of allicin? 3.38 The molecular formula of aspartame, the artificial sweetener marketed as NutraSweet®, is C 14HtsN205. (a) What is the molar mass of aspartame? (b) How many moles of aspartame are present in 1 .00 mg of aspartame? (c) How many molecules of aspartame are present in 1.00 mg of aspartame? (d) How many hydrogen atoms are present in 1.00 mg of aspartame? 21 3.39 A sample of glucose, C6H 1206, contains 1.250 x 10 car­ bon atoms. (a) How many atoms of hydrogen does it

contain? (b) How many molecules of glucose does it contain? (c) How many moles of glucose does it con­ tain? (d) What is the mass of this sample in grams? 3.40 A sample of the male sex hormone testosterone, 2 C19H280:u contains 7.08 X 10 0 hydrogen atoms. (a) How many atoms of carbon does it contain? (b) How many molecules of testosterone does it contain? (c) How many moles of testosterone does it contain? (d) What is the mass of this sample in grams? 3.41 The allowable concentration level of vinyl chloride, C2H3Cl, in the atmosphere in a chemical plant is 2.0 X 10-6 g/L. How many moles of vinyl chloride in each liter does this represent? How many molecules per liter? 3.42 At least 25 J.lg of tetrahydrocannabinol (THC), the active ingredient in marijuana, is required to produce intoxica­ tion. The molecular formula of THC is C21H3o02. How many moles of THC does this 25 J.lg represent? How many molecules?

Empirical Formulas 3.43 Give the empirical formula of each of the following compounds if a sample contains (a) 0.0130 mol C, 0.0390 mol H, and 0.0065 mol 0; (b) 11 .66 g iron and 5.01 g oxy­ gen; (c) 40.0% C, 6.7% H, and 53.3% 0 by mass. 3.44 Determine the empirical formula of each of the follow­ ing compounds if a sample contains (a) 0.104 mol K, 0.052 mol C, and 0.156 mol (b) 5.28 g Sn and 3.37 g F; (c) 87.5% N and 12.5% H by mass.

0;

3.45 Determine the empirical formulas of the compounds with the following compositions by mass: (a) 10.4% C, 27.8% S, and 61.7% Cl (b) 21.7% C, 9.6% 0, and 68.7% F (c) 32.79% Na, 13.02% AI, and 54.19% F

3.46 Determine the empirical formulas of the compounds with the following compositions by mass: (a) 55.3% K, 14.6% P, and 30.1% 0 (b) 24.5% Na, 14.9% Si, and 60.6% F (c) 62.1% C, 5.21% H, 12.1% N, and 20.7% 0

3.47 What is the molecular formula of each of the following compounds? (a) empirical formula CH2, molar mass = 84 g/ mol (b) empirical formula NH2Cl, molar mass = 51.5 g/mol 3.48 What is the molecular formula of each of the following compounds? (a) empirical formula HC02, molar mass = 90.0 g/mol (b) empirical formula C2H40, molar mass = 88 g/mol 3.49 Determine the empirical and molecular formulas of each of the following substances: (a) Styrene, a compound substance used to make Styro­ foam® cups and insulation, contains 92.3% C and 7.7% H by mass and has a molar mass of 104 g/mol. (b) Caffeine, a stimulant found in coffee, contains 49.5% C, 5.15% H, 28.9% N, and 16.5% 0 by mass and has a molar mass of 195 g/mol. (c) Monosodium glutamate (MSG), a flavor enhancer in certain foods, contains 35.51% C, 4.77% H, 37.85%

0, 8.29% N, and 13.60% Na, and has a molar mass of 169 g/mol. 3.50 Determine the empirical and molecular formulas of each of the following substances: (a) Ibuprofen, a headache remedy, contains 75.69% C, 8.80% H, and 15.51% 0 by mass, and has a molar mass of 206 g/mol. (b) Cadaverine, a foul smelling substance produced by the action of bacteria on meat, contains 58.55% C, 13.81% H, and 27.40% N by mass; its molar mass is 102.2 g/mol. (c) Epinephrine (adrenaline), a hormone secreted into the bloodstream in times of danger or stress, con­ tains 59.0% C, 7.1% H, 26.2% 0, and 7.7% N by mass; its MW is about 180 amu. 3.51 (a) Combustion analysis of toluene, a common organic solvent, gives 5.86 mg of C02 and 1.37 mg of H20. lf the compound contams only carbon and hydrogen, what is its empirical formula? (b) Menthol, the substance we can smell in mentholated cough drops, is composed of C, H, and 0. A 0.1005-g sample of menthol is combust­ ed, producing 0.2829 g of C02 and 0.1159 g of H20. What is the empirical formula for menthol? lf menthol has a molar mass of 156 g/mol, what is its molecular formula? 3.52 (a) The characteristic odor of pineapple is due to ethyl butyrate, a compound containffig carbon, hydrogen, and oxygen. Combustion of 2.78 mg of ethyl butyrate produces 6.32 mg of C02 and 2.58 mg of Hp. What is the empirical formula of the compound? (b) Nicotme, a component of tobacco, is composed of C, H, and N. A 5.250-mg sample of nicotine was combusted, producing 14.242 mg of C02 and 4.083 mg of H20. What is the em­ pirical formula for nicotine? lf nicotine has a molar mass of 160 ± 5 g/mol, what is its molecular formula? 3.53 Washing soda, a compound used to prepare hard water for washing laundry, is a hydrate, which means that a

Exercises certain number of water molecules are included in the solid structure. Its formula can be written as Na2C03·xH20, where x is the number of moles of H20 per mole of Na2C03. When a 2.558-g sample of washing soda is heated at 25 °C, all the water of hydration is lost, leaving 0.948 g of Na2C03. What is the value of x?

113

3.54 Epsom salts, a strong laxative used in veterinary medi­ cine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as MgS04·xH20, where x indicates the number of moles of H20 per mole of MgS04. When 5.061 g of this hydrate is heated to 250 oc, all the water of hydration is lost, leaving 2.472 g of MgS04• What is the value of x?

Calculations Based on Chemical Equations 3.55 Why is it essential to use balanced chemical equations when determining the quantity of a product formed from a given quantity of a reactant? 3.56 What parts of balanced chemical equations give infor­ mation about the relative numbers of moles of reactants and products involved in a reaction? 3.57 Hydrofluoric acid, HF(aq), cannot be stored in glass bot­ tles because compounds called silicates in the glass are attacked by the HF(aq). Sodium silicate (Na2Si03), for example, reacts as follows: Na2Si03(s) + 8 HF(aq) ----> H2SiF6(aq)

+ 2 NaF(aq) + 3 H20(1)

(a) How many moles of HF are needed to react with 0.300 mol of NazSi03? (b) How many grams of NaF form when 0.500 mol of HF reacts with excess N a2Si03? (c) How many grams of Na2Si03 can react with 0.800 g of HF? 3.58 The fermentation of glucose (C6H1206) produces ethyl alcohol (C2H50H) and C02: C6H1206(aq)

---->

2 CzHsOH(aq) + 2 COz(g)

(a) How many moles of C02 are produced when 0.400 mol of C6H 1206 reacts in this fashion? (b) How many grams of C�1206 are needed to form 7.50 g of C2H50H? (c) How many grams of C02 form when 7.50 g of C2H50H are produced? 3.59 Several brands of antacids use Al(OHh to react with stomach acid, which contains primarily HCI: AI(OH)J(s) + HCI(aq) ----> AIC13(aq) + H20(1)

(a) Balance this equation. (b) Calculate the number of grams of HCl that can react with 0.500 g of AI(OHlJ. (c) Calculate the number of grams of A1Cl3 and the number of grams of H20 formed when 0.500 g of AI(OHh reacts. (d) Show that your calculations in parts {b) and (c) are consistent with the law of conservation of mass. 3.60 An iron ore sample contains Fe203 together with other substances. Reaction of the ore with CO produces iron metal: Fe203(s)

+ CO(g)

---->

Fe(s)

+ C02(g)

(a) Balance this equation. (b) Calculate the number of grams of CO that can react with 0.150 kg of Fe203. (c) Calculate the number of grams of Fe and the num­ ber of grams of C02 formed when 0.150 kg of Fe203 reacts.

(d) Show that your calculations in parts {b) and (c) are consistent with the law of conservation of mass. 3.61 Aluminum sulfide reacts with water to form aluminum hydroxide and hydrogen sulfide. (a) Write the balanced chemical equation for this reaction. (b) How many grams of aluminum hydroxide are obtained from 14.2 g of aluminum sulfide? 3.62 Calcium hydride reacts with water to form calcium hy­ droxide and hydrogen gas. (a) Write a balanced chemi­ cal equation for the reaction. (b) How many grams of calcium hydride are needed to form 8.500 g of hydrogen? 3.63 Automotive air bags inflate when sodium azide, NaN3, rapidly decomposes to its component elements: 2 NaN3(s) ----> 2 Na(s) + 3 Nz(g)

(a) How many moles of N2 are produced by the decom­ position of 1.50 mol of NaN3? (b) How many grams of NaN3 are required to form 10.0 g of nitrogen gas? (c) How many grams of NaN3 are required to produce 10.0 tt3 of nitrogen gas, about the size of an automo­ tive air bag, if the gas has a density of 1.25 g/L? 3.64 The complete combustion of octane, C8H18, the main component of gasoline, proceeds as follows: 2 C8H18(1)

+

25 02(g) ----> 1 6 C02(g) + 1 8 H20(g)

(a) How many moles of 02 are needed to burn 1.25 mol of C8H18? (b) How many grams of 02 are needed to burn 10.0 g of CaH ta? (c) Octane has a density of 0.692 g/mL at 20 oc. How many grams of 02 are required to burn 1 .00 gal of C aH1a?

3.65 A piece of aluminum foil l.OO em square and 0.550 mm thick is allowed to react with bromine to form alu­ minum bromide as shown in the accompanying photo.

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Stoichiometry: Calculations with Chemical Formulas and Equations

(a) How many moles of aluminum were used? (The 3 density of aluminum is 2.699 g/ cm .) (b) How many grams of aluminum bromide form, assuming the alu­ minum reacts completely? 3.66 Detonation of nitroglycerin proceeds as follows: 4 C3H sNpg(l) -----> 12 C02(g) + 6 N2(g) + 02(g) + 10 HzO(g)

(a) If a sample containing 2.00 mL of nitroglycerin (den­ sity = 1.592 g/mL) is detonated, how many total moles of gas are produced? (b) If each mole of gas occupies 55 L under the conditions of the explosion, how many liters of gas are produced? (c) How many grams of N2 are produced in the detonation?

Limiting Reactants; Theo retical Yields 3.67 (a) Define the terms limiting reactant and excess reactant. (b) Why are the amounts of products formed in a reac­ tion determined only by the amount of the limiting reac­ tant? (c) Why should you base your choice of what compound is the limiting reactant on its number of ini­ tial moles, not on its initial mass in grams?

tant remain after the limiting reactant is completely consumed?

3.68 (a) Define the terms theoretical yield, actual yield, and percent yield. (b) Why is the actual yield in a reaction al­ most always less than the theoretical yield? (c) Can a re­ action ever have 110% actual yield? 3.69 A manufacturer of bicycles has 4815 wheels, 2305 frames, and 2255 handlebars. (a) How many bicycles can be manufactured using these parts? (b) How many parts of each kind are left over? (c) Which part limits the production of bicycles? 3.70 A bottling plant has 121,515 bottles with a capacity of 355 mL, 122,500 caps, and 40,875 L of beverage. (a) How many bottles can be filled and capped? (b) How much of each item is left over? (c) Which component limits the production? 3.71 Sodium hydroxide reacts with carbon dioxide as follows: 2 NaOH(s) + C02(g) -----> Na2C03(s) + H20(/) Which reagent is the limiting reactant when 1.85 mol NaOH and 1.00 mol C02 are allowed to react? How many moles of Na2C03 can be produced? How many moles of the excess reactant remain after the completion of the reaction? 3.72 Aluminum hydroxide reacts with sulfuric acid as follows: 2 Al(OHh(s) + 3 H2S04(aq) -----> Alz(S04)3(aq) + 6 HzO(I) Which reagent is the limiting reactant when 0.500 mol Al(OHh and 0.500 mol H2S04 are allowed to react? How many moles of Al2(S04h can form under these conditions? How many moles of the excess reactant re­ main after the completion of the reaction? 3.73 The fizz produced when an Atka-Seltzer® tablet is dis­ solved in water is due to the reaction between sodium bicarbonate (NaHC03) and citric acid (H3C6H507): 3 NaHC03(aq) + H3C6Hg07(aq) -----> 3 COz(g) + 3 HzO(I) + Na3C6Hs07(aq)

In

a certain experiment 1.00 g of sodium bicarbonate and 1.00 g of citric acid are allowed to react. (a) Which is the limiting reactant? (b) How many grams of carbon dioxide form? (c) How many grams of the excess reac-

3.74 One of the steps in the commercial process for convert­ ing ammonia to nitric acid is the conversion of NH3 to NO: 4 NH3(g) + 5 02(g) -----> 4 NO(g) + 6 H20(g)

In a certain experiment, 1.50 g of NH3 reacts with 2.75 g

of 02 (a) Which is the limiting reactant? (b) How many grams of NO and of H20 form? (c) How many grams of the excess reactant remain after the limiting reactant is completely consumed? (d) Show that your calculations in parts (b) and (c) are consistent with the law of conser­ vation of mass. 3.75 Solutions of sodium carbonate and silver nitrate react to form solid silver carbonate and a solution of sodium ni­ trate. A solution containing 3.50 g of sodium carbonate is mixed with one containing 5.00 g of silver nitrate. How many grams of sodium carbonate, silver nitrate, silver carbonate, and sodium nitrate are present after the reaction is complete? 3.76 Solutions of sulfuric acid and lead(ll) acetate react to form solid lead(ll) sulfate and a solution of acetic acid. If 7.50 g of sulluric acid and 7.50 g of lead(ll) acetate are mixed, calculate the number of grams of sulfuric acid, lead(ll) acetate, lead(ll) sulfate, and acetic acid present in the mixture after the reaction is complete. 3.77 When benzene (C6H6) reacts with bromine (Br2), bro­ mobenzene (C6H5Br) is obtained: C6H6 + Brz -----> C6HsBr + HBr (a) What is the theoretical yield of bromobenzene in this reaction when 30.0 g of benzene reacts with 65.0 g of bromine? (b) If the actual yield of bromobenzene was 42.3 g, what was the percentage yield? 3.78 When ethane (C2H6) reacts with chlorine (Cl2), the main product is C2H5Cl; but other products containing Cl, such as C2H4Cl:u are also obtained in small quantities. The formation of these other products reduces the yield

Additional Exercises of C2H5Cl. (a) Calculate the theoretical yield of C2H5Cl when 125 g of C2H6 reacts with 255 g of Cl2, assuming that C2H6 and Cl2 react only to form C2H5Cl and HCI. (b) Calculate the percent yield of C2H5Cl if the reaction produces 206 g of C2HsCI. 3.79 Hydrogen sulfide is an impurity in natural gas that must be removed. One common removal method is called the Claus process, which relies on the reaction:

115

Under optimal conditions the Claus process gives 98% yield of 58 from H2S. If you started with 30.0 grams of H2S and 50.0 grams of Ov how many grams of Ss would be produced, assuming 98% yield? 3.80 When hydrogen sulfide gas is bubbled into a solution of sodium hydrmdde, the reaction forms sodium sulfide and water. How many grams of sodium sulfide are formed if 1.50 g of hydrogen sulfide is bubbled into a so­ lution containing 2.00 g of sodium hydroxide, assuming that the sodium sulfide is made in 92.0% yield?

ADDITI ONAL EXERC ISES 3.81 Write the balanced chemical equations for (a) the com­ plete combustion of acetic acid (CH3COOH), the main active ingredient in vinegar; (b) the decomposition of solid calcium hydroxide into solid calcium(ll) oxide (lime) and water vapor; (c) the combination reaction be­ tween nickel metal and chlorine gas. 3.82 The effectiveness of nitrogen fertili2ers depends on both their ability to deliver nitrogen to plants and the amount of nitrogen they can deliver. Four common nitrogen-contain­ ing fertili2ers are ammonia, ammonium nitrate, ammoni­ um sulfate, and urea [(NH2hCO]. Rank these fertilizers in terms of the mass percentage nitrogen they contain. 3.83 (a) Diamond is a natural form of pure carbon. How many moles of carbon are in a 1.25-carat diamond (1 carat = 0.200 g)? How many atoms are in this dia­ mond? (b) The molecular formula of acetylsalicylic acid (aspirin), one of the most common pain relievers, is C9H804. How many moles of C9H804 are in a 0.500-g tablet of aspirin? How many molecules of C9Hs04 are in this tablet? 3.84 (a) One molecule of the antibiotic known as penicillin G 21 has a mass of 5.342 X 10- g. What is the molar mass of penicillin G? (b) Hemoglobin, the oxygen-carrying pro­ tein in red blood cells, has four iron atoms per molecule and contains 0.340% iron by mass. Calculate the molar mass of hemoglobin. 3.85 Very small crystals composed of 1000 to 100,000 atoms, called quantum dots, are being investigated for use in electronic devices. (a) A quantum dot was made of solid silicon in the shape of a sphere, with a diameter of 4 nm. Calcu­ late the mass of the quantum dot, using the density 3 of silicon (2.3 g / cm ). (b) How many silicon atoms are in the quantum dot? 3 (c) The density of germanium is 5.325 g/ cm If you made a 4 nm quantum dot of germanium, how many Ge atoms would it contain? Assume the dot is spherical. 3.86 Serotonin is a compound that conducts nerve impulses in the brain. It contains 68.2 mass percent C, 6.86 mass percent H, 15.9 mass percent N, and 9.08 mass percent O. lts molar mass is 176 g/mol. Determine its molecular formula. 3.87 The koala dines exclusively on eucalyptus leaves. Its di­ gestive system detoxifies the eucalyptus oil, a poison to other animals. The chief constituent in eucalyptus oil is a substance called eucalyptol, which contains 77.87% C,

11 .76% H, and the remainder 0. (a) What is the empiri­ cal formula for this substance? (b) A mass spectrum of eucalyptol shows a peak at about 154 amu. What is the molecular formula of the substance? 3.88 Vanillin, the dominant flavoring in vanilla, contains C, H, and 0. When 1.05 g of this substance is completely combusted, 2.43 g of C02 and 0.50 g of H20 are pro­ duced. What is the empirical formula of vanillin? [3.89] An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of C02 was formed. In a separate experiment the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCI. De­ termine the empirical formula of the compound. [3.90] An oxybromate compound, KBrO., where x is un­ known, is analyzed and found to contain 52.92% Br. What is the value of x? [3.91] An element X forms an iodide (XI3) and a chloride (XC13). The iodide is quantitatively converted to the chloride when it is heated in a stream of chlorine: 2 Xl3 + 3 Cl2

-->

2 XCl3 + 3 !2

If 0.5000 g of XI3 is treated, 0.2360 g of XC13 is obtained. (a) Calculate the atomic weight of the element X. (b) Identify the element X.

3.92 If 1.5 mol of each of the following compounds is com­ pletely combusted in oxygen, which one will produce the largest number of moles of H20? Which will produce the least? Explain. C2H50H, C3H8, CH3CH2COCH3. 3.93 A method used by the U.S. Environmental Protection Agency (EPA) for determining the concentration of ozone in air is to pass the air sample through a "bub­ bler" containing sodium iodide, which removes the ozone according to the following equation: 03(g) + 2 Nai(aq) + HP(l) --> 02(gl + l2(s) + 2 NaOH(aq) (a) How many moles of sodium iodide are needed to 6 remove 5.95 x 10- mol of 03? (b) How many grams of sodium iodide are needed to remove 1 .3 mg of 03? 3.94 A chemical plant uses electrical energy to decompose aqueous solutions of NaCl to give Clv Hv and NaOH: 2 NaCl(aq) + 2 HP(I)

-->

2 NaOH(aq) + H2(g) + Cl2(g) If the plant produces 1.5 X 106 kg (1500 metric tons) of Cl2 daily, estimate the quantities of H2 and NaOH produced.

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Stoichiometry: Calculations with Chemical Formulas and Equations

3.95 The fat stored in the hump of a camel is a source of both energy and water. Calculate the mass of HzO produced by metabolism of 1.0 kg of fat, assuming the fat consists entirely of tristearin (C57H11006), a typical animal fat, and assuming that during metabolism, tristearin reacts with Oz to form only COz and HzO.

[3.96] When hydrocarbons are burned in a limited amount of air, both CO and COz form. When 0.450 g of a particular hydrocarbon was burned in air, 0.467 g of CO, 0.733 g of CO:u and 0.450 g of HzO were formed. (a) What is the em­ pirical formula of the compound? (b) How many grams of Oz were used in the reaction? (c) How many grams would have been required for complete combustion?

3.97 A mixture of Nz(g) and Hz(g) reacts in a closed container to form ammonia, NH3(g). The reaction ceases before ei­ ther reactant has been totally consumed. At this stage 3.0 mol Nz, 3.0 mol Hz, and 3.0 mol NH3 are present. How many moles of Nz and Hz were present originally?

[3.98] A mixture containing KC103, KzC03, KHC03, and KCl was heated, producing CO:u Oz, and HzO gases accord­ ing to the following equations:

2 KC103(s)

->

2 KHC03(s)

->

2 KCl(s) + 3 Oz(g) KzO(s) + HzO(g)

KzC03(s)

->

KzO(s) + COz(g)

+

2 COz(g)

The KCl does not react under the conditions of the reac­ tion. If 100.0 g of the mixture produces 1.80 g of HzO, 13.20 g of COz, and 4.00 g of Oz, what was the composi­ tion of the original mixture? (Assume complete decom­ position of the mixture.)

3.99 When a mixture of 10.0 g of acetylene (CzHz) and 10.0 g of oxygen (Oz) is ignited, the resultant combustion reac­ tion produces COz and HzO. (a) Write the balanced chemical equation for this reaction. (b) Which is the lim­ iting reactant? (c) How many grams of CzHz, 02, CO:u and HzO are present after the reaction is complete?

3.100 Aspirin (C9H804) is produced from salicylic acid (C7H603) and acetic anhydride (C4H603): C7H603 + C4H603

->

CgHs04 + HCzH Pz

(a) How much salicylic acid is required to produce 1.5 x 10z kg of aspirin, assuming that all of the salicylic acid is converted to aspirin? (b) How much salicylic acid would be required if only 80% of the salicylic acid is converted to aspirin? (c) What is the theoretical yield of aspirin if 185 kg of salicylic acid is allowed to react with 125 kg of acetic anhydride? (d) If the situation described in part (c) produces 182 kg of aspirin, what is the per­ centage yield?

INTEGRATIVE EXERCI SES (These exercises require skills from earlier chapters as well as skills from the present chapter.)

3.101 Consider a sample of calcium carbonate in the form of a cube measuring 2.005 in. on each edge. If the sample has a density of 2.71 g/ cm3, how many oxygen atoms does it contain?

3.102 (a) You are given a cube of silver metal that measures 1 . 000 em on each edge. The density of silver is 10.5 3 g/ cm . How many atoms are in this cube? (b) Because atoms are spherical, they cannot occupy all of the space of the cube. The silver atoms pack in the solid in such a way that 74% of the volume of the solid is actually filled with the silver atoms. Calculate the volume of a single silver atom. (c) Using the volume of a silver atom and the formula for the volume of a sphere, calculate the ra­ dius in angstroms of a silver atom.

3.103 (a) If an automobile travels 225 mi with a gas mileage of 20.5 mi/ gal, how many kilograms of COz are produced? Assume that the gasoline is composed of octane, C8H18(1), whose density is 0.69 g/ mL. (b) Repeat the cal­ culation for a truck that has a gas mileage of 5 mil gal.

3.104

1n

1865 a chemist reported that he had reacted a

verted into sulfur dioxide gas, which is a pollutant. To reduce sulfur dioxide emissions, calcium oxide (lime) is used. The sulfur dioxide reacts with calcium oxide to form solid calcium sulfite. (a) Write the balanced chemi­ cal equation for the reaction. (b) If the coal is burned in a power plant that uses 2000 tons of coal per day, what mass of calcium oxide is required daily to eliminate the sulfur dioxide? (c) How many grams of calcium sulfite are produced daily by this power plant?

3.106 Copper is an excellent electrical conductor widely used

in making electric circuits. 1n producing a printed circuit board for the electronics industry, a layer of copper is laminated on a plastic board. A circuit pattern is then printed on the board using a chemically resistant poly­ mer. The board is then exposed to a chemical bath that reacts with the exposed copper, leaving the desired cop­ per circuit, which has been protected by the overlaying polymer. Finally, a solvent removes the polymer. One re­ action used to remove the exposed copper from the cir­ cuit board is Cu(s) + Cu(NH3)4Clz(aq) + 4 NH3(aq)

->

2 Cu(NH3)4Cl(aq)

weighed amount of pure silver with nitric acid and had recovered all the silver as pure silver nitrate. The mass ratio of silver to silver nitrate was found to be 0.634985. Using only this ratio and the presently accepted values for the atomic weights of silver and oxygen, calculate the atomic weight of nitrogen. Compare this calculated atomic weight with the currently accepted value.

A plant needs to produce 5000 circuit boards, each with a surface area measuring 2.0 in. X 3.0 in. The boards are covered with a 0.65-mm layer of copper. 1n subsequent processing, 85% of the copper is removed. Copper has a 3 density of 8.96 g/ cm Calculate the masses of Cu(NH3)4 Clz and NH3 needed to produce the circuit boards, as­ suming that the reaction used gives a 97% yield.

3.105 A particular coal contains 2.5% sulfur by mass. When

3.107 Hydrogen cyanide, HCN, is a poisonous gas. The lethal dose is approximately 300 mg HCN per kilogram of air

this coal is burned at a power plant, the sulfur is con-

Integrative Exercises when inhaled. (a) Calculate the amount of HCN that gives the lethal dose in a small laboratory room measur­ ing 12 x 15 x 8.0 ft. The density of air at 26 oc is 0.00118 g/cm 3 (b) If the HCN is formed by reaction of NaCN with an acid such as H2S04, what mass of NaCN gives the lethal dose in the room?

2 NaCN(s) + H2S04(aq) -----> Na2S04(aq) + 2 HCN(g) (c) HCN forms when synthetic fibers containing Orion®

® ® or Acrilan burn. Acrilan has an empirical formula of CH2CHCN, so HCN is 50.9% of the formula by mass. A rug measures 12 X 15 ft and contains 30 oz of Acrilan® fibers per square yard of carpet. If the rug burns, will a lethal dose of HCN be generated in the room? Assume that the yield of HCN from the fibers is 20% and that the carpet is 50% consumed.

3.108 The source of oxygen that drives the internal combus­ tion engine in an automobile is air. Air is a mixture of gases, which are principally N2 (-79%) and 02 (-20%).

117

In the cylinder of an automobile engine, nitrogen can react with oxygen to produce nitric oxide gas, NO. As NO is emitted from the tailpipe of the car, it can react with more oxygen to produce nitrogen dioxide gas. (a) Write balanced chemical equations for both reac­ tions. (b) Both nitric oxide and nitrogen dioxide are pol­ lutants that can lead to acid rain and global warming; collectively, they are called "NOx'' gases. In 2004, the United States emHted an estimated 19 million tons of ni­ trogen dioxide into the atmosphere. How many grams of nitrogen dioxide is this? (c) The production of NOx gases is an unwanted side reaction of the main engine combustion process that turns octane, C8H18, into C02 and water. If 85% of the oxygen in an engine is used to combust octane, and the remainder used to produce ni­ trogen dioxide, calculate how many grams of nitrogen dioxide would be produced during the combustion of 500 grams of octane.

AQUEOUS REACTIONS AND SOLUTION STOICHIOMETRY

A VIEW OF THE PACIFIC OCEAN along the Cal ifornia coastline.

1 18

W H AT ' S 4.1

4.2 4.3 4.4

A H E A D

General Properties of Aqueous Solutions

We begin by examining the nature of the substances dissolved in water, whether they exist in water as ions, molecules, or some mixture of the two. This information is necessary to understand the nature of reactants in aqueous solutions. Precipitation Reactions

We will identify reactions in which soluble reactants yield an insoluble product. Acid-Base Reactions

We will explore reactions in which protons, ions, are transferred between reactants.

4.5

4.6

Concentrations of Solutions

We will recognize that the amount of a compound in a given volume of a solution is called its concentration. Concentration can be expressed in a number of ways, the most usual of which is moles of compound per liter of solution (molarity). Solution Stoichiometry and Chemical Analysis

We will use what we have learned about stoichiometry and concentration to calculate the amounts or concentrations of substances in solution.

H+

Oxidation-Reduction Reactions

We will examine reactions in which electrons are transferred between reactants.

TH E WATERS OF T H E PAC I F I C OCEAN,

seen in this chapter-opening

photograph of the California coast, are part of the World Ocean that covers almost two-thirds of our planet. Water has been the key to much of Earth's evolutionary history. Life itself almost certainly originated in water, and the need for water by all forms of life has helped determine diverse biological structures. Your own body is about 60% water by mass. We will see repeatedly throughout this text that water possesses many unusual properties essential to supporting life on Earth. The waters of the World Ocean may not appear to be any different from those of Lake Tahoe or the water that flows from your kitchen faucet, but a taste of seawater is all it takes to demonstrate that there is an important differ­ ence. Water has an exceptional ability to dissolve a wide variety of substances. Water on Earth-whether it is drinking water from the tap, water from a clear mountain stream, or seawater-invariably contains a variety of dissolved sub­ stances. A solution in which water is the dissolving medium is called an aqueous solution. Seawater is different from what we call "freshwater" be­ cause it has a much higher total concentration of dissolved ionic substances. Water is the medium for most of the chemical reactions that take place within us and around us. Nutrients dissolved in blood are carried to our cells, where they enter into reactions that help keep us alive. Automobile parts rust

1 19

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C HA PTER 4

Aqueous Reactions and Solution Stoichiometry when they come into frequent contact with aqueous solutions that contain var­ ious dissolved substances. Spectacular limestone caves (Figure 4.1 �) are formed by the dissolving action of underground water containing carbon dioxide, C02(aq): CaC03(s) + HzO(l) + COz (aq)

.A. Figure 4.1 Limestone cave. When C02 dissolves in water, the resulting solution is slightly acidic. Limestone caves are formed by the dissolving action of this acidic solution acting on CaC03 in the limestone.

------+

Ca(HC03)z(aq)

[4.1]

In Chapter 3 we saw a few simple types of chemical reactions and their descriptions. In this chapter we continue to examine chemical reactions by fo­ cusing on aqueous solutions. A great deal of important chemistry occurs in aqueous solutions, so we need to learn the vocabulary and concepts used to describe and understand this chemistry. In addition, we will extend the con­ cepts of stoichiometry that we learned in Chapter 3 by considering how solu­ tion concentrations are expressed and used.

4. 1 GENERAL P ROPERTIES OF AQUEOUS SOLUTIONS Recall that a solution i s a homogeneous mixture o f two o r more substances. cx:o (Section 1.2) The substance present in the greatest quantity is usually called the solvent. The other substances in the solution are called the solutes; they are said to be dissolved in the solvent. When a small amount of sodium chloride (NaCI) is dissolved in a large quantity of water, for example, the water is the solvent and the sodium chloride is the solute.

Electrolytic Properties Imagine preparing two aqueous solutions-one by dissolving a teaspoon of table salt (sodium chloride) in a cup of water and the other by dissolving a tea­ spoon of table sugar (sucrose) in a cup of water. Both solutions are clear and colorless. How do they differ? One way, which might not be immediately obvi­ ous, is in their electrical conductivity: The salt solution is a good conductor of electricity; the sugar solution is not. Whether a solution conducts electricity can be determined by using a de­ vice such as that shown in Figure 4.2 � . To light the bulb, an electric current must flow between the two electrodes that are immersed in the solution. Al­ though water itself is a poor conductor of electricity, the presence of ions caus­ es aqueous solutions to become good conductors. Ions carry electrical charge from one electrode to the other, completing the electrical circuit. Thus, the con­ ductivity of NaCI solutions indicates the presence of ions in the solution. The lack of conductivity of sucrose solutions indicates the absence of ions. When NaCI dissolves in water, the solution contains Na+ and cr ions, each sur­ rounded by water molecules. When sucrose (C12H22011) dissolves in water, the solution contains only neutral sucrose molecules surrounded by water molecules. A substance (such as NaCI) whose aqueous solutions contain ions is called an electrolyte. A substance (such as C12H22011) that does not form ions in solu­ tion is called a nonelectrolyte. The difference between NaCI and C1 2H220n arises largely because NaCl is ionic, whereas C12H22011 is molecular.

Ionic Compounds in Water Recall from Section 2.7 and especially Figure 2.23 that solid NaCl consists of an orderly arrangement of Na+ and Cl- ions. When NaCI dissolves in water, each ion separates from the solid structure and disperses throughout the solution, as shown in Figure 4.3(a) �. The ionic solid dissociates into its component ions as it dissolves.

4.1

General Properties of Aqueous Solutions

ELECTROLYTIC PROPERTIES

One way to differentiate two aqueous solutions is to employ a device that measures their electrical conductivities. The ability of a solution to conduct electricity depends on the number of ions it contains. An electrolyte solution contains ions that serve as charge carriers, causing the bulb to light.

No ions A nonelectrolyte solution does not contain ions, and the bulb does not light.

Few ions

Many ions

If the solution contains a small number of ions, the bulb will be only dimly lit.

If the solution contains a large number of ions, the bulb will be brightly lit.

.A. Figure 4.2 Measuring lon concentrations using conductivity.

(a)

(b)

.A. Figure 4.3 Dluolutlon In water. (a) When an ionic compound dissolves in water, H20 molecules separate, surround, and disperse the ions into the liquid. (b) Methanol, CH30H, a molecular compound, dissolves without forming ions. The methanol molecules contain black spheres, which represent carbon atoms. In both parts (a) and (b) the water molecules have been moved apart so that the solute particles can be seen more clearly.

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Aqueous Reactions and Solution Stoichiometry

8+ 88+

Water is a very effective solvent for ionic compounds. Although water is an electrically neutral molecule, one end of the molecule (the 0 atom) is rich in elec­ trons and has a partial negative charge, denoted by 8-. The other end (the H atoms) has a partial positive charge, denoted by 8+ as shown in the margin. Positive ions (cations) are attracted by the negative end of H20, and negative ions (anions) are attracted by the positive end. As an ionic compound dissolves, the ions become surrounded by H20 mol­ ecules, as shown in Figure 4.3(a). The ions are said to be solvated. We denote these ions in chemical equations by writing them as Na+(aq) and Cl-(aq), where "aq" is an abbreviation for "aqueous." (Section 3.1) The solvation process helps stabilize the ions in solution and prevents cations and anions from recom­ bining. Furthermore, because the ions and their shells of surrounding water molecules are free to move about, the ions become dispersed uniformly throughout the solution. We can usually predict the nature of the ions present in a solution of an ionic compound from the chemical name of the substance. Sodium sulfate (Na2S04), for example, dissociates into sodium ions (Na+) and sulfate ions 2 (S04 -). You must remember the formulas and charges of common ions (fables 2.4 and 2.5) to understand the forms in which ionic compounds exist in aque­ ous solution. 000

GIVE IT SOME THOUGHT What dissolved species are present in a solution of (a) KCN, (b) NaCI04?

Molecular Compounds in Water When a molecular compound dissolves in water, the solution usually consists of intact molecules dispersed throughout the solution. Consequently, most molecular compounds are nonelectrolytes. As we have seen, table sugar (su­ crose) is an example of a nonelectrolyte. As another example, a solution of methanol (CH30H) in water consists entirely of CH30H molecules dispersed throughout the water [Figure 4.3(b)]. A few molecular substances, however, have aqueous solutions that contain ions. Acids are the most important of these solutions . For example, when HCI(g) dissolves in water to form hydrochloric acid, HCl(aq), it ionizes; that is, it dissociates into H+(aq) and Cqaq) ions.

Strong and Weak Electrolytes Two categories of electrolytes-strong and weak-differ in the extent to which they conduct electricity. Strong electrolytes are those solutes that exist in solu­ tion completely or nearly completely as ions. Essentially all soluble ionic com­ pounds (such as NaCI) and a few molecular compounds (such as HCl) are strong electrolytes. Weak electrolytes are those solutes that exist in solution mostly in the form of molecules with only a small fraction in the form of ions. For example, in a solution of acetic acid (CH3COOH) most of the solute is pre­ sent as CH3COOH(aq) molecules. Only a small fraction (about 1%) of the CH3COOH is present as H+(aq) and CH3Coo-(aq) ions.* We must be careful not to confuse the extent to which an electrolyte dis­ solves with whether it is strong or weak. For example, CH3COOH is extremely soluble in water but is a weak electrolyte. Ba(OHh, on the other hand, is not very soluble, but the amount of the substance that does dissolve dissociates almost completely. Thus, Ba(OHh is a strong electrolyte. "The chemical fonnula of acetic acid is sometimes written as HC2H302 with the acidic H written in front of the chemical fomwla, so the formula looks like that of other common acids such as HCl. The formula CH,COOH conforms to the molecular structure ofacetic acid, with the acidic H on the 0 atom at the end of theformula.

4.1

General Properties of Aqueous Solutions

When a weak electrolyte such as acetic acid ionizes in solution, we write the reaction in the following manner:

(4.2 ] The half-arrows in both directions mean that the reaction is significant in both directions. At any given moment some CH3COOH molecules are ionizing to form H+ and CH3Coo- ions. At the same time, H+ and CH3Coo- ions are recombining to form CH3COOH. The balance between these opposing processes determines the relative numbers of ions and neutral molecules. This balance produces a state of chemical equilibrium in which the relative numbers of each type of ion or molecule in the reaction are constant over time. This equilibrium condition varies from one weak electrolyte to another. Chemical equilibria are extremely important. We will devote Chapters 15-17 to examining them in detail. Chemists use half-arrows in both directions to represent the ionization of weak electrolytes and a single arrow to represent the ionization of strong elec­ trolytes. Because HCl is a strong electrolyte, we write the equation for the ion­ ization of HCl as follows: HC1(aq)

-->

H+(aq) + Cqaq)

(4.3 ]

The absence of a reverse arrow indicates that the H+ and Cl- ions have no ten­ dency to recombine in water to form HCl molecules. In the following sections we will look more closely at how we can use the composition of a compound to predict whether it is a strong electrolyte, weak electrolyte, or nonelectrolyte. For the moment, you need only to remember that soluble ionic compounds are strong electrolytes. We identify ionic compounds as those composed of metals and nonmetals-such as NaCl, FeS04, and Al(N03)3-or compounds containing the ammonium ion, NH4+-such as NH4Br and (NH4)zC03. GIVE

IT SOME THOUGHT

Which solute will cause the lightbulb in the experiment shown in Figure 4.2 to glow more brightly, CH30H or MgBrz?

- SAMPLE EXERCISE 4.1

Relative Numbers of Anions and Cations Chemical Formulas I toRelating

The diagram on the right represents an aqueous solution of one of the following com­ pounds: MgC12, KCI, or K2S04. Which solution does the drawing best represent?

SOLUTION Analyze: Plan:

We are asked to associate the charged spheres in the diagram with ions present in a solution of an ionic substance. We examine the ionic substances given in the problem to determine the rela­ tive numbers and charges of the ions that each contains. We then correlate these charged ionic species with the ones shown in the diagram. The diagram shows twice as many cations as anions, consistent with the for­ mulation KzS04.

Solve: Check:

Notice that the total net charge in the diagram is zero, as it must be if it is to represent an ionic substance.

- PRACTICE EXERCISE

If you were to draw diagrams (such as that shown on the right) representing aqueous solutions of each of the following ionic compounds, how many anions would you show if the diagram contained six cations? (a) NiS04, (b) Ca(N03)z, (c) Na3P04, (d) Alz(S04h

Answers:

(a) 6, (b) 12, (c) 2, (d) 9

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Aqueous Reactions and Solution Stoichiometry

4.2 PRECIPITATION REACTIONS Figure 4.4 T shows two clear solutions being mixed. One solution contains lead nitrate, Pb(N03)z, and the other contains potassium iodide (KI). The reaction between these two solutes produces an insoluble yellow product. Reactions that result in the formation of an insoluble product are called precipitation reactions. A precipitate is an insoluble solid formed by a reaction in solution. In Figure 4.4 the precipitate is lead iodide (Pbl2), a compound that has a very low solubility in water: Pb(N03h(aq) + 2 KI(aq)

--+

Pbl 2(s) + 2 KN03(a q)

[4.4]

The other product of this reaction, potassium nitrate (KN03), remains in solution. Precipitation reactions occur when certain pairs of oppositely charged ions attract each other so strongly that they form an insoluble ionic solid. To predict whether certain combinations of ions form insoluble compounds, we must con­ sider some guidelines concerning the solubilities of common ionic compounds.

Solubility Guidelines for Ionic Compounds The solubility of a substance at a given temperature is the amount of the substance that can be dissolved in a given quantity of solvent at the given temperature.

T Figure 4.4 A precipitation reaction.

PREC I P ITATION REACTION

Reactions that result in the formation of an insoluble product are known as precipitation reactions.

2 KI(aq) The addition of a colorless solution of potassium iodide (KI)

+

to a colorless solution of lead nitrate

Pbi2(s) + 2 KN03(aq) produces a yeUow precipitate of lead iodide (Pbl2) that slowly settles to the bottom of the beaker.

4.2

TABLE 4.1 • Solubility Guidelines for Common Ionic Compounds in Water Soluble Ionic Compounds

Compounds containing

Important Exceptions

N03CH3COO­ CIBrr-

so/-

None None Compounds of Ag+, Hgl+, and Pb 2+ Compounds of Ag+, Hgl+, and Pb 2+ Compounds of Ag+, Hgl+, and Pb 2+ Compounds of Sr2+, Ba2+, Hg22+, and Pb2+

Insoluble Ionic Compounds

Important Exceptions

Compounds containing

Compounds of NH4 +, the alkali metal cations, and Ca 2+, Sr2+, and Ba 2+ Compounds of NH4 + and the alkali metal cations Compounds of NH4 + and the alkali metal cations Compounds of the alkali metal cations, and NH4+, Ca 2+, Sr2+, and Ba2+

For instance, only 1 .2 X 10-3 mol of Pbi 2 dissolves in a liter of water at 25 oc. In our discussions, any substance with a solubility less than 0.01 moljL will be referred to as insoluble. In those cases the attraction between the oppositely

charged ions in the solid is too great for the water molecules to separate the ions to any significant extent; the substance remains largely undissolved. Unfortunately, there are no rules based on simple physical properties such as ionic charge to guide us in predicting whether a particular ionic compound will be soluble. Experimental observations, however, have led to guidelines for predicting solubility for ionic compounds. For example, experiments show that all common ionic compounds that contain the nitrate anion, N0 3-, are soluble in water. Table 4.1 • summarizes the solubility guidelines for common ionic compounds. The table is organized according to the anion in the compound, but it also reveals many important facts about cations. Note that all common ionic compounds of the alkali metal ions (group lA of the periodic table) and of the

ammonium ion (NH4+) are soluble in water.

- SAMPLE EXERCISE 4.2 J Using Solubility Rules

Classify the following ionic compounds as soluble or insoluble in water: (a) sodium carbonate (Na2C03), (b) lead sulfate (PbS04).

SOLUTION Analyze: We are given the names and formulas of two ionic compounds and asked to predict whether they are soluble or insoluble in water. Plan: We can use Table 4.1 to answer the question. Thus, we need to focus on the anion in each compound because the table is organized by anions. Solve:

(a) According to Table 4.1, most carbonates are insoluble. But carbonates of the alka­ li metal cations (such as sodium ion) are an exception to this rule and are soluble. Thus, Na2C03 is soluble in water. (b) Table 4.1 indicates that although most sulfates are water soluble, the sulfate of Pb2+ is an exception. Thus, Pb504 is insoluble in water.

- PRACTICE EXERCISE

Classify the following compounds as soluble or insoluble in water: hydroxide, (b) barium nitrate, (c) ammonium phosphate. (a) insoluble, (b) soluble, (c) soluble

Answers:

(a)

cobalt(II)

Precipitation Reactions

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Aqueous Reactions and Solution Stoichiometry To predict whether a precipitate forms when we mix aqueous solutions of two strong electrolytes, we must (1) note the ions present in the reactants, (2) consider the possible combinations of the cations and anions, and (3) use Table 4.1 to determine if any of these combinations is insoluble. For example, will a precipitate form when solutions of Mg(N03h and NaOH are mixed? Both Mg(N03h and NaOH are soluble ionic compounds and strong elec­ trolytes. Mixing Mg(N03)z(aq) and NaOH(aq) first produces a solution contain­ ing Mg 2+, N03-, Na+, and OH- ions. Will either of the cations interact with either of the anions to form an insoluble compound? In addition to the reac­ tants, the other possible interactions are Mg2+ with OH- and Na+ with N03 -. 2 From Table 4.1 we see that hydroxides are generally insoluble. Because Mg + is not an exception, Mg(OHh is insoluble and will thus form a precipitate. NaN03, however, is soluble, so Na+ and N03- will remain in solution. The bal­ anced equation for the precipitation reaction is Mg(N03h(aq) +

2 NaOH(aq)

-----4

Mg(OHh(s)

+ 2 NaN03(aq)

[4.5]

Exchange (Metathesis) Reactions Notice in Equation 4.5 that the cations in the two reactants exchange anions­ Mg2+ ends up with OH-, and Na+ ends up with N03 -. The chemical formulas of the products are based on the charges of the ions-two OH- ions are needed to give a neutral compound with Mg2 +, and one N03- ion is needed to give a neutral compound with Na+. cx:c (Section 2.7) The equation can be balanced only after the chemical formulas of the products have been determined. Reactions in which positive ions and negative ions appear to exchange partners conform to the following general equation: AX + BY

Example:

AgN0 3(aq) + KCl(aq)

-----4 -----4

AY + BX

[4.6]

AgCl(s) + KN03 (aq)

Such reactions are called exchange reactions, or metathesis reactions (meh­ TATH-eh-sis, which is the Greek word for "to transpose"). Precipitation reactions conform to this pattern, as do many acid-base reactions, as we will see in Section 4.3. To complete and balance a metathesis equation, we follow these steps:

1. Use the chemical formulas of the reactants to determine the ions that are present. 2. Write the chemical formulas of the products by combining the cation from one reactant with the anion of the other. (Use the charges of the ions to de­ termine the subscripts in the chemical formulas.) 3. Finally, balance the equation.

- SAMPLE EXERCISE 4.3 I Predicting a Metathesis Reaction

(a) Predict the identity of the precipitate that forms when solutions of BaC12 and K2504 are mixed. (b) Write the balanced chemical equation for the reaction.

SOLUTION Analyze: Plan:

We are given two ionic reactants and asked to predict the insoluble prod­ uct that they form.

We need to write down the ions present in the reactants and to exchange the anions between the two cations. Once we have written the chemical formulas for these products, we can use Table 4.1 to determine which is insoluble in water. Know­ ing the products also allows us to write the equation for the reaction.

Solve:

(a) The reactants contain Ba2+, o-, K+, and 5042- ions. 1f we exchange the anions, we will have Ba504 and KCI. According to Table 4.1, most compounds of sol- are 2 soluble but those of Ba + are not. Thus, Ba504 is insoluble and will precipitate from solution. KCl, on the other hand, is soluble.

4.2 (b) From part (a) we know the chemical formulas of the products, BaS04 and KCI. The balanced equation with phase labels shown is BaCI2(aq)

+

K2504(aq) ---> Ba504(s)

+

2 KCI(aq)

- PRACTICE EXERCISE (a)

What compound precipitates when solutions of Fe2(S04h and LiOH are mixed?

(b) Write a balanced equation for the reaction. (c) Will a precipitate form when solu­

tions of Ba(N03)z and KOH are mixed? (a) Fe(OH)J; (b) Fe2(S04h(aq) + 6 LiOH(aq) ---> 2 Fe(OH)J(s) + 3 Li2S04(aq); (c) no (both possible products are water soluble)

Answers:

Ionic Equations In writing chemical equations for reactions in aqueous solution, it is often use­ ful to indicate explicitly whether the dissolved substances are present predom­ inantly as ions or as molecules. Let's reconsider the precipitation reaction between Pb(N03h and 2 KI, shown previously in Figure 4.4: Pb(N03h(aq) + 2 KI(aq)

�

Pb1 2(s) + 2 KN03(aq)

An equation written in this fashion, showing the complete chemical formulas of the reactants and products, is called a molecular equation because it shows the chemical formulas of the reactants and products without indicating their ionic character. Because Pb(N03h, KI, and KN03 are all soluble ionic compounds and therefore strong electrolytes, we can write the chemical equation to indicate explicitly the ions that are in the solution: 2 Pb +(aq) + 2 N03-(aq) + 2 K+ (aq ) + 2 qaq) � Pbl 2 (s) + 2 K+(aq ) + 2 N03-(aq) (4.7] An equation written in this form, with all soluble strong electrolytes shown as ions, is called a complete ionic equation. Notice that K+(aq) and N03-(aq) appear on both sides of Equation 4.7. Ions that appear in identical forms among both the reactants and products of a com­ plete ionic equation are called spectator ions. These ions are present but play no direct role in the reaction. When spectator ions are omitted from the equation (they cancel out like algebraic quantities), we are left with the net ionic equation: 2 Pb + (aq) + 2 qaq) � Pblz(s) (4.8] A net ionic equation includes only the ions and molecules directly involved in the reaction. Charge is conserved in reactions, so the sum of the charges of the ions must be the same on both sides of a balanced net ionic equation. In this case the 2 + charge of the cation and the two 1 - charges of the anions add to give zero, the charge of the electrically neutral product. If every ion in a complete ionic

equation is a spectator, then no reaction occurs. GIVE

IT SOME THOUGHT

Are any spectator ions shown in the following chemical equation? Ag +(aq) + Na+(aq) + Cqaq) ---> AgCl(s) + Na +(aq)

Net ionic equations are widely used to illustrate the similarities between large numbers of reactions involving electrolytes. For example, Equation 4.8 expresses the essential feature of the precipitation reaction between any strong 2 2 electrolyte containing Pb + and any strong electrolyte containing r-: The Pb + (aq ) and r-(aq) ions combine to form a precipitate of Pbl2. Thus, a net ionic equation demonstrates that more than one set of reactants can lead to the same net reaction. For example, aqueous solutions of KI and Mgl2 share many chemical similarities because both contain r- ions. The complete equation, on the other hand, identifies the actual reactants that participate in a reaction.

Precipitation Reactions

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Aqueous Reactions and Solution Stoichiometry The following steps summarize the procedure for writing net ionic equations:

1. Write a balanced molecular equation for the reaction.

2. Rewrite the equation to show the ions that form in solution when each soluble strong electrolyte dissociates into its component ions. Only strong electrolytes

dissolved in aqueous solution are written in ionic form.

3. Identify and cancel spectator ions.

- SAMPLE EXERCISE 4.4 I Writing a Net Ionic Equation

Write the net ionic equation for the precipitation reaction that occurs when solutions of calcium chloride and sodium carbonate are mixed.

SOLUTION Analyze: Our task is to write a net ionic equation for a precipitation reaction, given the names of the reactants present solution. Plan: We first need to write the chemical formulas of the reactants and products and in

then determine which product is insoluble. We then write and balance the molecular equation. Next, we write each soluble strong electrolyte as separated ions to obtain the complete ionic equation. Finally, we eliminate the spectator ions to obtain the net ionic equation. Calcium chloride is composed of calcium ions, Ca2+, and chloride ions, Cl-; hence an aqueous solution of the substance is CaC12(aq). Sodium carbonate is com­ posed of Na + ions and C032- ions; hence an aqueous solution of the compound is Na2C03(aq). In the molecular equations for precipitation reactions, the anions and cations appear to exchange partners. Thus, we put Ca2+ and col- together to give CaC03 and Na+ and o- together to give NaCI. According to the solubility guide­ lines in Table 4.1, CaC03 is insoluble and NaCI is soluble. The balanced molecular equation is CaCI2(aq) + Na2C03(aq) -----> CaC03(s) + 2 NaCI(aq) In a complete ionic equation, only dissolved strong electrolytes (such as soluble ionic compounds) are written as separate ions. As the (aq) designations remind us, CaClz, Na2C03, and NaCI are all dissolved in the solution. Furthermore, they are all strong electrolytes. CaC03 is an ionic compound, but it is not soluble. We do not write the formula of any insoluble compound as its component ions. Thus, the complete ionic equation is Ca 2+(aq) + 2 Cqaq) + 2 Na+(aq) + C032-(aq) -----> CaC03(s) + 2 Na+(aq) + 2 Cqaq) CC and Na+ are spectator ions. Canceling them gives the following net ionic equation: Ca2+(aq) + C032-(aq) -----> CaC03(s)

Solve:

We can check our result by confirming that both the elements and the electric charge are balanced. Each side has one Ca, one C, and three 0, and the net charge on each side equals 0. If none of the ions in an ionic equation is removed from solution or changed in some way, then they all are spectator ions and a reaction does not occur.

Check:

Comment:

- PRACTICE EXERCISE

Write the net ionic equation for the precipitation reaction that occurs when aqueous solutions of silver nitrate and potassium phosphate are mixed. Ag3P04(s) Answers: 3 Ag+(aq) + Pol-(aq) ----->

4.3 ACID-BASE REACTIONS .6. Figure 4.5 Some common household acids (left) and bases (right).

Many acids and bases are industrial and household substances (Figure 4 .5 ._), and some are important components of biological fluids. Hydrochloric acid, for example, is an important industrial chemical and the main constituent of gastric juice in your stomach. Acids and bases are also common electrolytes.

4.3

Acid-Base Reactions

129

Acids Acids are substances that ionize in aqueous solutions to form hydrogen ions, thereby increasing the concentration of H+(aq) ions. Because a hydrogen atom consists of a proton and an electron, H+ is simply a proton. Thus, acids are often called proton donors. Molecular models of three common acids, HCl, HN03 and CH3COOH, are shown in the margin. Just as cations are surrounded and bound by water molecules (see Figure 4.3[a]), the proton is also solvated by water molecules. The nature of the proton in water is discussed in detail in Section 16.2. In writing chemical equations in­ volving the proton in water, we represent it simply as H+(aq). Molecules of different acids can ionize to form different numbers of H+ ions. Both HCI and HN03 are monoprotic acids, which yield one H+ per mole­ cule of acid. Sulfuric acid, H2S04, is a diprotic acid, one that yields two H+ per molecule of acid. The ionization of H2S04 and other diprotic acids occurs in two steps: [4.9] H2S04(aq) --> H+(aq) + HS04-(aq) HS04 -(aq)

�

H+(aq) + SO/-(aq)

HCl

[4.10]

Although H2S04 is a strong electrolyte, only the first ionization is complete. Thus, aqueous solutions of sulfuric acid contain a mixture of H+(aq), HS04 -(aq), 2 and S04 -(aq). The molecule CH3COOH (acetic acid) that we have mentioned frequently is the primary component in vinegar. Acetic acid has four hydrogens, but only one of them is capable of being ionized in water. Only the hydrogen that is bound to oxygen in the COOH group will ionize in water; the other hydrogens are bound to carbon and do not break their C - H bonds in water. We will dis­ cuss acids much more in Chapter 16. GIVE

IT SOME THOUGHT

The structural formula of citric acid, a main component of citrus fruits, is shown here:

H

I

H-C-COOH

I

HO-C-COOH

I

H-C-COOH

I

H How many H+(aq) can be generated by each citric acid molecule when citric acid is dissolved in water?

Bases Bases are substances that accept (react with) H+ ions. Bases produce hydroxide ions (OH-) when they dissolve in water. Ionic hydroxide compounds such as NaOH, KOH, and Ca(OH}z are among the most common bases. When dis­ solved in water, they dissociate into their component ions, introducing OH­ ions into the solution. Compounds that do not contain OH- ions can also be bases. For example, ammonia (NH3) is a common base. When added to water, it accepts an H+ ion from the water molecule and thereby produces an OH- ion (Figure 4.6 � ): [4.11 ] Ammonia is a weak electrolyte because only a small fraction of the NH3 (about 1%) forms NH4+ and OH- ions.

"" Figure 4.6 Hydrogen lon transfer. An H20 molecule acts as a proton donor (acid), and N H 3 as a proton acceptor (base). Only a fraction of the NH3 reacts with H20; NH3 is a weak electrolyte.

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Aqueous Reactions and Solution Stoichiometry

TABLE 4.2 • Common Strong Acids and Bases Strong Acids

Hydrochloric, HCl Hydrobromic, HBr Hydroiodic, HI Chloric, HC103 Perchloric, HCI04 Nitric, HN03 Sulfuric, HzS04

Strong Bases

Group 1A metal hydroxides (LiOH, NaOH, KOH, RbOH, CsOH) Heavy group 2A metal hydroxides [Ca(OH)z, Sr(OH)z, Ba(OH)z]

Strong and Weak Acids and Bases Acids and bases that are strong electrolytes (completely ionized in solution) are called strong acids and strong bases. Those that are weak electrolytes (partly ion­ ized) are called weak acids and weak bases. Strong acids are more reactive than weak acids when the reactivity depends only on the concentration of H+(aq). The reactivity of an acid, however, can depend on the anion as well as on H+(aq). For example, hydrofluoric acid (HF) is a weak acid (only partly ionized in aque­ ous solution), but it is very reactive and vigorously attacks many substances, in­ cluding glass. This reactivity is due to the combined action of H+(aq) and F-(aq). Table 4.2 .a. lists the common strong acids and bases. You should commit these to memory. As you examine this table, notice that some of the most com­ mon acids, such as HCI, HN03, and H2S04, are strong. (For H2S04, as we noted earlier, only the first proton completely ionizes.) Three of the strong acids are the hydrogen compounds of the halogen family. (HF, however, is a weak acid.) The list of strong acids is very short. Most acids are weak. The only common strong bases are the hydroxides of Li+, Na+, K+, Rb+, and Cs+ (the alkali metals, group lA) and the hydroxides of Ca2+, Sr 2+, and Ba 2+ (the heavy alkaline earths, group 2A). These are the common soluble metal hydroxides. Most other metal hydroxides are insoluble in water. The most common weak base is NH3, which reacts with water to form OH- ions (Equation 4.11). GIVE IT SOME THOUGHT

Which of the following is a strong acid: H2S03, HBr, CH3COOH?

- SAMPLE EXERCISE 4.5 I Comparing Acid Strengths

The following diagrams represent aqueous solutions of three acids (HX, HY, and HZ) with water molecules omitted for clarity. Rank them from strongest to weakest.

4.3

SOLUTION Analyze: We are asked to rank three acids from strongest to weakest, based on schematic drawings of their solutions. Plan: We can examine the drawings to determine the relative numbers of uncharged

molecular species present. The strongest acid is the one with the most H+ ions and fewest undissociated acid molecules in solution. The weakest acid is the one with the largest number of undissociated molecules. The order is HY > HZ > HX. HY is a strong acid because it is totally ionized (no HY molecules in solution), whereas both HX and HZ are weak acids, whose solu­ tions consist of a mixture of molecules and ions. Because HZ contains more H+ ions and fewer molecules than HX, it is a stronger acid.

Solve:

- PRACTICE EXERCISE

Imagine a diagram showing 10 Na+ ions and 10 OH� ions. If this solution were mixed with the one pictured on the previous page for HY, what would the diagram look like that represents the solution after any possible reaction? (H+ ions will react with OH� ions to form Hp.) Answer: The final diagram would show 10 Na+ ions, 2 OH� ions, 8 y� ions, and 8 H20 molecules.

Identifying Strong a nd Weak Electrolytes If we remember the common strong acids and bases (Table 4.2) and also remember that NH3 is a weak base, we can make reasonable predictions about the electrolytic strength of a great number of water-soluble substances. Table 4.3 T summarizes our observations about electrolytes. To classify a solu­ ble substance as a strong electrolyte, weak electrolyte, or nonelectrolyte, we simply work our way down and across this table. We first ask ourselves whether the substance is ionic or molecular. If it is ionic, it is a strong elec­ trolyte. The second column of Table 4.3 tells us that all ionic compounds are strong electrolytes. If the substance we want to classify is molecular, we ask whether it is an acid or a base. (Does it have H first in the chemical formula or contain a COOH group?) If it is an acid, we rely on the memorized list from Table 4.2 to determine whether it is a strong or weak electrolyte: All strong acids are strong electrolytes, and all weak acids are weak electrolytes. If an acid is not listed in Table 4.2, it is probably a weak acid and therefore a weak elec­ trolyte. For example, H3P04, H2S03, and HC7H502 are not listed in Table 4.2 and are weak acids. If the substance we want to classify is a base, we again turn to Table 4.2 to determine whether it is one of the listed strong bases. NH3 is the only molecular base that we consider in this chapter, and Table 4.3 tells us it is a weak electrolyte. (There are compounds called amines that are related to NH3 and are also molecular bases, but we will not consider them until Chapter 16.) Finally, any molecular substance that we encounter in this chapter that is not an acid or NH3 is probably a nonelectrolyte.

TABLE 4.3 • Summary of the Electrolytic Behavior of Common Soluble Ionic and Molecular Compounds

Strong Electrolyte

Weak Electrolyte

Nonelectrolyte

Ionic

All

None

None

Molecular

Strong acids (see Table 4.2)

Weak acids Weak bases

All other compounds

Acid-Base Reactions

131

132

C HA PTER 4

Aqueous Reactions and Solution Stoichiometry

- SAMPLE EXERCISE 4.6 I Identifying Strong, Weak, and Nonelectrolytes

Classify each of the following dissolved substances as a strong electrolyte, weak elec­ trolyte, or nonelectrolyte: CaCl:u HN03, C2H50H (ethanol), HCOOH (formic acid), KOH.

SOLUTION Analyze: Plan:

We are given several chemical formulas and asked to classify each sub­ stance as a strong electrolyte, weak electrolyte, or nonelectrolyte. The approach we take is outlined in Table 4.3. We can predict whether a sub­ stance is ionic or molecular based on its composition. As we saw in Section 2.7, most ionic compounds we encounter in this text are composed of a metal and a nonmetal, whereas most molecular compounds are composed only of nonmetals.

Solve: Two compounds fit the criteria for ionic compounds: CaC12 and KOH. Be­ cause Table 4.3 tells us that all ionic compounds are strong electrolytes, that is how we classify these two substances. The three remaining compounds are molecular. Two, HN03 and HCOOH, are acids. Nitric acid, HN03, is a common strong acid, as shown in Table 4.2, and therefore is a strong electrolyte. Because most acids are weak acids, our best guess would be that HCOOH is a weak acid (weak electrolyte). This is correct. The remaining molecular compound, C2H50H, is neither an acid nor a base, so it is a nonelectrolyte. Although C2H50H has an OH group, it is not a metal hydroxide; thus, it is not a base. Rather, it is a member of a class of organic compounds that have C -OH bonds, which are known as alcohols. = (Section 2.9) . The COOH group is called the "carboxylic acid group" (Chapter 16). Molecules that have this group are all weak acids.

Comment:

- PRACTICE EXERCISE

Consider solutions in which 0.1 mol of each of the following compounds is dissoived in 1 L of water: Ca(N03)z (calcium nitrate), C6H1206 (glucose), CH3COONa (sodium acetate), and CH3COOH (acetic acid). Rank the solutions in order of increasing elec­ trical conductivity, based on the fact that the greater the number of ions in solution, the greater the conductivity. C�1206 (nonelectrolyte) < CH3COOH (weak electrolyte, existing main­ ly in the form of molecules with few ions) < CH3C00Na (strong electrolyte that provides two ions, Na+ and CH3COO-) < Ca(N03)z (strong electrolyte that provides 2 three ions, Ca + and 2 N03-)

Answers:

Neutralization Reactions and Salts The properties of acidic solutions are quite different from those of basic solu­ tions. Acids have a sour taste, whereas bases have a bitter taste. Acids can change the colors of certain dyes in a specific way that differs from the effect of a base (Figure 4.7 ... ) . The dye known as litmus, for example, is changed from blue to red by an acid and from red to blue by a base. In addition, acidic and basic solutions differ in chemical properties in several important ways that we will explore in this chapter and in later chapters. When a solution of an acid and a solution of a base are mixed, a neutralization reaction occurs. The products of the reaction have none of the characteristic properties of either the acidic solution or the basic solution. For example, when hydrochloric acid is mixed with a solution of sodium hydroxide, the following reaction occurs: •

A Figure 4.7 The acld-ba•e Indicator bromthymol blue. The indicator is blue in basic solution and yellow in acidic solution. The left flask shows the indicator in the presence of a base, aqueous ammonia (labeled as ammonium hydroxide). The right flask shows the indicator in the presence of hydrochloric acid, HCI.

HCl(aq) + NaOH(aq) (acid)

(base)

---4

H 20(l) (water)

+

NaCl(aq) (salt)

[4.12]

Water and table salt, NaCl, are the products of the reaction. By analogy to this reaction, the term salt has come to mean any ionic compound whose cation comes from a base (for example, Na+ from NaOH) and whose anion comes from an acid (for example, Cl- from HCl). In general, a neutralization reaction

between an acid and a metal hydroxide produces water and a salt.

*Tasting chemical solutions is not a good practice. However, we have all had acids such as ascorbic acid (vita­ min C), acetylsnlicylic acid (aspirin), and citric acid (in citrus fruits) in our mouths, and we arefamiliar with their characteristic sour taste. Soaps, which are basic, have the characteristic bitter taste of bases.

4.3

(b)

(a)

+ Cqaq) + Na+(aq) +

OH - (aq ) -->

H 20(!)

Therefore, the net ionic equation is

H +(aq)

+ OH (aq) -

-->

+ Na+ (aq)

133

(c)

Because HCl, NaOH, and NaCI are all soluble strong electrolytes, the com­ plete ionic equation associated with Equation 4.12 is

H + (aq)

Acid-Base Reactions

+ Cqaq)

H 20(!)

[4.13] [4.14]

"" Figure 4.8 Reaction of Mg(OH)z(s) with acid. (a) Milk of magnesia is a suspension of magnesium hydroxide, Mg(OHh(s), in water. (b) The magnesium hydroxide dissolves upon the addition of hydrochloric acid, HCI(aq). (c) The final clear solution contains soluble MgCI2(aq), as shown in Equation 4 . 1 5 .

Equation 4.14 summarizes the essential feature of the neutralization reaction between any strong acid and any strong base: H +(aq ) and OH -(aq) ions com­ bine to form H20. Figure 4.8 • shows the reaction between hydrochloric acid and the base Mg(OHh, which is insoluble in water. A milky white suspension of Mg(OHh called milk of magnesia is seen dissolving as the neutralization reaction occurs:

+ 2 HCl(aq) Mg(OHh(s) + 2 H +(aq)

Molecular equation:

Mg(OHh(s)

Net ionic equation:

--> -->

+ 2 H20(!) 2 Mg +(aq) + 2 H20(!) MgC1 2(aq)

[4.15] [4.16]

Notice that the OH- ions (this time in a solid reactant) and H + ions combine to form H 20. Because the ions exchange partners, neutralization reactions be­ tween acids and metal hydroxides are also metathesis reactions. •

SAMPLE EXERCISE 4.7 I Writing Chemical Equations for a Neutralization Reaction

(a) Write a balanced molecular equation for the reaction between aqueous solutions of acetic acid (CH3COOH) and barium hydroxide, Ba(OHh. (b) Write the net ionic equation for this reaction.

SOLUTION Analyze: Plan: Solve:

We are given the chemical formulas for an acid and a base and asked to write a balanced molecular equation and then a net ionic equation for their neutralization reaction.

As Equation 4.12 and the italicized statement that follows it indicate, neutralization reactions form two products, H20 and a salt. We examine the cation of the base and the anion of the acid to determine the composition of the salt.

Z+ (a) The salt will contain the cation of the base (Ba ) and the anion of the acid (CH3Coo-). Thus, the formula of the salt is Ba(CH3C00h. According to the solubility guidelines in Table 4.1, this compound is soluble. The unbalanced molecular equa­ tion for the neutralization reaction is

CHAPTER 4

134

Aqueous Reactions and Solution Stoichiometry

To balance this molecular equation, we must pro­ vide two molecules of CH3COOH to furnish the two CH3Coo- ions and to supply the two H+ ions needed to combine with the two OH- ions of the base. The balanced molecular equation is

(b) To write the net ionic equation, we must deter­ mine whether each compound in aqueous solution is a strong electrolyte. CH3COOH is a weak elec­ trolyte (weak acid), Ba(OHh is a strong electrolyte, and Ba(CH3C00h is also a strong electrolyte (ionic compound). Thus, the complete ionic equation is

2

2 CH3COOH(aq) + Ba +(aq) + 2 OH-(aq) ---> 2 2 H20(1) + Ba +(aq) + 2 CH3coo-(aq) 2 CH3COOH(aq) + 2 OH-(aq)

Eliminating the spectator ions gives

--->

2 Hp(l) + 2 CH3COO-(aq)

Simplifying the coefficients gives the net ionic equation: We can determine whether the molecular equation is correctly balanced by counting the number of atoms of each kind on both sides of the arrow. (There are 10 H, 6 0, 4 C, and 1 Ba on each side.) However, it is often easier to check equations by count­ ing groups: There are 2 CH3COO groups, as well as 1 Ba, and 4 additional H atoms and 2 additional 0 atoms on each side of the equation. The net ionic equation checks out because the numbers of each kind of element and the net charge are the same on both sides of the equation.

Check:

- PRACTICE EXERCISE (a) Answers: (a) H2C03(aq) + 2 KOH(aq)

Write a balanced molecular equation for the reaction of carbonic acid (H2C03) and potassium hydroxide (KOH). net ionic equation for this reaction.

2 Hp(l) +

(b) Write the

K2C03(aq); (b) H2C03(aq) + 2 OH-(aq) 2 H20(/) + COl-(aq). (H2C03 is a weak acid and therefore a weak electrolyte, whereas KOH, a strong base, and K2C03, an ionic compound, are strong --->

--->

electrolytes.)

Acid-Base Reactions with Gas Formation Many bases besides OH- react with H+ to form molecular compounds. Two of these that you might encounter in the laboratory are the sulfide ion and the car­ bonate ion. Both of these anions react with acids to form gases that have low solubilities in water. Hydrogen sulfide (H2S), the substance that gives rotten eggs their foul odor, forms when an acid such as HCl(aq) reacts with a metal sulfide such as Na2S:

Molecular equation: Net ionic equation:

2 HCl(aq) + Na 2S(aq)

�

2 H+(aq) + S 2-(aq)

H2S(g) + 2 NaCl(aq) �

H2S(g)

[4.17] [4.18]

Carbonates and bicarbonates react with acids to form C02 gas. Reaction of

A-:11 ·/""'

--:... -� .. i.

,, -

col- or HC03- with an acid first gives carbonic acid (H2C03). For example,

when hydrochloric acid is added to sodium bicarbonate, the following reaction occurs:

HCl(aq) + NaHC03(aq)

�

NaC!(aq) + H2C0 3 (aq)

[4.19]

Carbonic acid is unstable. If carbonic acid is present in solution in sufficient concentrations, it decomposes to form H20 and C02, which escapes from the solution as a gas.

[4.20] -

A Figure 4.9 Carbonates react with adds to form carbon dioxide gas. Here NaHC03 (white solid) reacts with hydrochloric acid. The bubbles contain C02.

The decomposition of H2C03 produces bubbles of C02 gas, as shown in Figure 4.9 -

+ 2 Hp(l)

+ NO(g)

All the gold ever mined would easily fit in a cube 19 m on a side and weighing about

1.1

x

108 kg (125,000

tons). More

than 90% of this amount has been produced since the beginning

ANALY Z I N G C H E M I CA L REAC T I O N S

I

n this chapter you have been introduced to a great number of

chemical reactions. A major difficulty that students face in try­

ing to master material of this sort is gaining a "feel" for what

•

If metathesis cannot occur, can the reactants possibly en­

gage in an oxidation-reduction reaction? This requires

that there be both a reactant that can be oxidized and

happens when chemicals are allowed to react. In fact, you might

one that can be reduced.

tant can figure out the results of a chemical reaction. One of our

By asking questions such as these, you should be able to

marvel at the ease with which your professor or teaching assis­

goals in this textbook is to help you become more adept at pre­

predict what might happen during the reaction. You might not

ical intuition" is understanding how to categorize reactions.

you, you will not be far off. As you gain experience with

dicting the outcomes of reactions. The key to gaining this "chem­ Attempting to memorize the many individual reactions in

chemistry would be a futile task.

It is far

more fruitful to try to

always be entirely correct, but if you keep your wits about

chemical reactions, you will begin to look for reactants that might not be immediately obvious, such as water from the so­

recognize patterns to determine the general category of a reac­

lution or oxygen from the atmosphere.

are faced with the challenge of predicting the outcome of a chem­

experimentation. If you perform an experiment in which two

tion, such as metathesis or oxidation-reduction. Thus, when you ical reaction, ask yourself the following pertinent questions: • • • •

What are the reactants in the reaction?

One of the greatest tools available to us in chemistry is

solutions are mixed, you can make observations that help

you understand what is happening. For example, using the

information in Table

4.1 to predict whether a precipitate will

Are they electrolytes or nonelectrolytes?

form is not nearly as exciting as actually seeing the precipi­

If the

tory portion of the course will make your lecture material

Are they acids and bases?

reactants are electrolytes, will metathesis produce

a precipitate? Water? A gas?

tate form, as in Figure

easier to master.

4.4. Careful observations in the labora­

144

C HAPTER 4

Aqueous Reactions and Solution Stoichiometry

(a)

& Figure 4.1 6 Procedure for

(b)

preparation of 0.250 L of 1 .00 M solution of Cu504. (a) Weigh out 0.250 mol (39.9 g) of CuS04 (formula weight = 1 59.6 amu). (b) Put the CuS04 (solute) into a 250-ml volumetric flask, and add a small quantity of water. (c) Dissolve the solute by swirling the flask. (d) Add more water until the solution just reaches the calibration mark etched on the neck of the flask. Shake the stoppered flask to ensure complete mixing.

(d)

(c)

Molarity Molarity (symbol M) expresses the concentration of a solution as the number of moles of solute in a liter of solution (soln): Molarity =

moles so�ute . volume of solution m liters

[4.33]

A 1.00 molar solution (written 1.00 M) contains 1.00 mol of solute in every liter of solution. Figure 4.16 & shows the preparation of 250.0 mL of a 1.00 M solution of CuS04 by using a volumetric flask that is calibrated to hold exact­ ly 250.0 mL. First, 0.250 mol of CuS04 (39.9 g) is weighed out and placed in the volumetric flask. Water is added to dissolve the salt, and the resultant solution is diluted to a total volume of 250.0 mL. The molarity of the solution is (0.250 mol CuS04)/(0.250 L soln) = 1.00 M. GIVE

IT SOME THOUGHT

2 4 Which is more concentrated, a 1.00 x 10- M solution of sucrose or a 1.00 x 10- M solution of sucrose?

- SAMPLE EXERCISE 4.1 1

I Calculating Molarity

Calculate the molarity of a solution made by dissolving 23.4 g of sodium sulfate (Na2S04) in enough water to form 125 mL of solution.

SOLUTION Analyze: We are given the number of grams of solute (23.4 g), its chemical formula (Na2S04), and the volume of the solution (125 ml). We are asked to calculate the molarity of the solution. Plan: We can calculate molarity using Equation 4.33. To do so, we must convert the number of grams of solute to moles and the volume of the solution from milliliters to liters. Solve: The number of moles of Na2S04 is obtained by using its molar mass:

( ( 10�0�) =

Moles NazS04 = (23.4 �)

Converting the volume of the solution to liters:

Liters soln

=

Thus, the molarity is

. Molanty

0.165 mol Na 2S04 0.125 L soln

=

(125 mt)

=

1 mol NazS04 � 142

1.32

)

= 0.165 mol NazS04

0.125 1. mo1 Na2504

L soln

=

132 M

Because the numerator is only slightly larger than the denominator, it is reasonable for the answer to be a little over 1 M. The units (moljL) are appropriate for molarity, and three significant figures are appropriate for the answer because each of the initial pieces of data had three significant figures.

Check:

4.5

- PRACTICE EXERCISE

Concentrations of Solutions

145

Calculate the molarity of a solution made by dissolving 5.00 g of glucose (C6H1 206) in sufficient water to form exactly 100 mL of solution. Answer: 0.278 M

Expressing the Concentration of an Electrolyte When an ionic compound dissolves, the relative concentrations of the ions in­ troduced into the solution depend on the chemical formula of the compound. For example, a 1.0 M solution of NaCl is 1.0 M in Na + ions and 1.0 M in Cl ­ ions. Similarly, a 1.0 M solution of Na2S04 is 2.0 M in Na + ions and 1.0 M in 2 S04 - ions. Thus, the concentration of an electrolyte solution can be specified either in terms of the compound used to make the solution (1.0 M Na2S04) or in terms of the ions that the solution contains (2.0 M Na + and 1.0 M S042-).

I Calculating Molar Concentrations of Ions • What are the molar concentrations of each of the ions present in a 0.025 M aqueous solution of calcium nitrate?

SAMPLE EXERCISE 4.12

SOLUTION Analyze: We are given the concentration of the ionic compound used to make the solution and asked to determine the concentrations of the ions in the solution. Plan: We can use the subscripts in the chemical formula of the compound to deter­ mine the relative concentrations of the ions. Solve: Calcium nitrate is composed of calcium ions (Ca2+) and nitrate ions (N03 -), so 2+

its chemical formula is Ca(N03)z. Because there are two N03- ions for each Ca ion in 2 the compound, each mole of Ca(N03)z that dissolves dissociates into 1 mol of Ca + 2 and 2 mol of N03 -. Thus, a solution that is O .Q25 M in Ca(N03)z is 0.025 M in Ca + and 2 X 0.025 M = 0.050 M in N03 -,

2 The concentration of N03- ions is twice that of Ca + ions, as the subscript 2 after the N03- in the chemical formula Ca(N03)z suggests it should be.

Check:

- PRACTICE EXERCISE

What is the molar concentration of K+ ions in a 0.015 M solution of potassium carbonate? Answer: 0.030 M K +

lnterconverting Molarity, Moles, and Volume The definition of molarity (Equation 4.33) contains three quantities-molarity, moles solute, and liters of solution. If we know any two of these, we can calcu­ late the third . For example, if we know the molarity of a solution, we can calcu­ late the number of moles of solute in a given volume. Molarity, therefore, is a conversion factor between volume of solution and moles of solute. Calculation of the number of moles of HN03 in 2.0 L of 0.200 M HN03 solution illustrates the conversion of volume to moles: moles HN03 = (2.0 k Mn2+(aq) + 5 Fe3+(aq) + 4 H 20(!) (a) How many moles of Mn04- were added to the solution? (b) How many moles of Fe2+ were in the sample? (c) How many grams of iron were in the sample? (d) If the sample had a mass of 0.8890 g, what is the percentage of iron in the sample? Answers: (a) 1.057 X 10-3 mol Mn04- (b) 5.286 x 10-3 mol Fe2+, (c) 0.2952 g, (d) 33.21%

•

SAMPLE EXERCISE 4.17 [ Determining Solution Concentration Via an Acid-Base Titration

One commercial method used to peel potatoes is to soak them in a solution of NaOH for a short time, remove them from the NaOH, and spray off the peel. The concentra­ tion of NaOH is normally in the range of 3 to 6 M. The NaOH is analyzed periodically. In one such analysis, 45.7 mL of 0.500 M H S04 is required to neutralize a 20.0-mL 2 sample of NaOH solution. What is the concentration of the NaOH solution?

4.6

Solution Stoichiometry and Chemical Analysis

SOLUTION Analyze:

We are given the volume (45.7 mL) and molarity (0.500 M) of an H2S04 solution that reacts completely with a 20.0-mL sample of NaOH. We are asked to calculate the molarity of the NaOH solution. We can use the volume and molarity of the H2S04 to calculate the number of moles of this substance . Then, we can use this quantity and the balanced equation for the reaction to calculate the number of moles of NaOH. Finally, we can use the moles of NaOH and the volume of this solution to calculate molarity.

Plan:

The number of moles of H2S04 is given by the product of the volume and mo­ larity of this solution:

Solve:

moles H2S04

=

(45 . 7 mb-sulfi)

=

2.28

X

2

(

1 I. mol Ag3P04 => g Ag3P04. We use the coefficients in the balanced equation to convert moles of AgN03 to moles Ag3P04, and we use the molar mass of Ag3P04 to convert the number of moles of this substance to grams.

(7.5

X

4

10- mei-AgN03)

( 1 mel-Amf'04 ) ( 418.7 g Ag3P04 3 mel-AgN<Jj

l

melc-A:gJP04

)

= 0.10 g Ag3P04

The answer has only two significant figures because the quantity of AgN03 is given to only two significant figures.

CHAP T E R R E V I E W SUMMARY AND KEY TERMS Introduction and Sectio n 4.1 Solutions in which water is the dissolving medium are called aqueous solu­ tions. The component of the solution that is in the greater quantity is the solvent. The other components are solutes. Any substance whose aqueous solution contains ions is called an electrolyte. Any substance that forms a solu­ tion containing no ions is a nonelectrolyte. Electrolytes that are present in solution entirely as ions are strong electrolytes, whereas those that are present partly as ions and partly as molecules are weak electrolytes. Ionic com­ pounds dissociate into ions when they dissolve, and they are strong electrolytes. The solubility of ionic substances is made possible by solvation, the interaction of ions with polar solvent molecules. Most molecular compounds are nonelectrolytes, although some are weak electrolytes, and a few are strong electrolytes. When representing the ionization of a weak electrolyte in solution, half-arrows in both directions are used, indicating that the forward and reverse reactions can achieve a chemical balance called a chemical equilibrium.

Section 4.2 Precipitation reactions are those in which an insoluble product, called a precipitate, forms. Solubility guidelines help determine whether or not an ionic com­ pound will be soluble in water. (The solubility of a sub­ stance is the amount that dissolves in a given quantity of solvent.) Reactions such as precipitation reactions, in which cations and anions appear to exchange partners, are called exchange reactions, or metathesis reactions. Chemical equations can be written to show whether dissolved substances are present in solution predominant­ ly as ions or molecules. When the complete chemical for­ mulas of all reactants and products are used, the equation is called a molecular equation. A complete ionic equation shows all dissolved strong electrolytes as their component ions. In a net ionic equation, those ions that go through the reaction unchanged (spectator ions) are omitted. Section 4.3 Acids and bases are important electrolytes. Acids are proton donors; they increase the concentration of H +(aq) in aqueous solutions to which they are added.

Key Equations Bases are proton acceptors; they increase the concentra­ tion of OH-(aq) in aqueous solutions. Those acids and bases that are strong electrolytes are called strong acids and strong bases, respectively. Those that are weak elec­ trolytes are weak acids and weak bases. When solutions of acids and bases are mixed, a neutralization reaction results. The neutralization reaction between an acid and a metal hydroxide produces water and a salt. Gases can also be formed as a result of acid-base reactions. The re­ action of a sulfide with an acid forms H 2S{g); the reaction between a carbonate and an acid forms C02{g) . Section 4.4 Oxidation is the loss of electrons by a sub­ stance, whereas reduction is the gain of electrons by a substance. Oxidation numbers keep track of electrons during chemical reactions and are assigned to atoms using specific rules. The oxidation of an element results in an increase in its oxidation number, whereas reduction is accompanied by a decrease in oxidation number. Oxi­ dation is always accompanied by reduction, giving oxidation-reduction, or redox, reactions. Many metals are oxidized by 0 2, acids, and salts. The redox reactions between metals and acids and between metals and salts are called displacement reactions. The products of these displacement reactions are always an element (H2 or a metal) and a salt. Comparing such reac­ tions allows us to rank metals according to their ease of oxidation. A list of metals arranged in order of decreasing

155

ease of oxidation is called an activity series. Any metal on the list can be oxidized by ions of metals {or H+) below it in the series. Section 4.5 The composition of a solution expresses the relative quantities of solvent and solutes that it contains. One of the common ways to express the concentration of a solute in a solution is in terms of molarity. The molarity of a solution is the number of moles of solute per liter of solution. Molarity makes it possible to interconvert solu­ tion volume and number of moles of solute. Solutions of known molarity can be formed either by weighing out the solute and diluting it to a known volume or by the dilution of a more concentrated solution of known con­ centration (a stock solution). Adding solvent to the solu­ tion (the process of dilution) decreases the concentration of the solute without changing the number of moles of solute in the solution (Mconc X Vconc = Mctil X VctiJ) . Section 4.6 In the process called titration, we combine a solution of known concentration (a standard solution) with a solution of unknown concentration to determine the unknown concentration or the quantity of solute in the unknown. The point in the titration at which stoichio­ metrically equivalent quantities of reactants are brought together is called the equivalence point. An indicator can be used to show the end point of the titration, which coincides closely with the equivalence point.

KEY SKILLS • Recognize compounds as acids or bases, and as strong, weak, or nonelectrolytes. • Recognize reactions as acid-base, precipitation, metathesis, or redox. • Be able to calculate moles or grams of substances in solution using molarity. • Understand how to carry out a dilution to achieve a desired solution concentration. • Understand how to perform and interpret the results of a titration.

KEY EQUATIONS • Molarity =

moles solute volume of solution in liters

• Mconc X Vconc = Mctil

X

Vctu

[4.35 ]

[4.33]

Molarity is the most commonly used unit of concentra­ tion in chemistry. When adding solvent to a concentrated solution to make a dilute solution, molarities and volumes of both concen­ trated and dilute solutions can be calculated if three of the quantities are known.

156

C HAPTER 4

Aqueous Reactions and Solution Stoichiometry

VISUALIZING CONCEPTS 4.1 Which of the following schematic drawings best de­

scribes a solution of Li2S04 in water (water molecules

4.4 A 0.1 M solution of acetic acid, CH3COOH, causes the

4.2 to glow about as 0.001 M solution of HBr. How do you ac­ count for this fact? [Section 4.1]

Ughtbulb in the apparatus of Figure

not shown for simplicity)? [Section 4.1]

brightly as a

4.5

You are presented with three white solids, A, B, and C,

which are glucose (a sugar substance), NaOH, and

AgBr. Solid A dissolves in water to form a conducting

+

solution. B is not soluble in water. C dissolves in water to form a nonconducting solution. Identify A, B, and C.

[Section 4.2] (a)

(b)

(c)

4.6 We have seen that ions in aqueous solution are stabi­ lized by the attractions between the ions and the water

molecules. Why then do some pairs of ions in solution

4.2 Methanol, CH30H, and hydrogen chloride, HCl, are

both molecular substances, yet an aqueous solution of methanol does not conduct an electrical current, where­

as a solution of HCl does conduct. Account for this dif­

form precipitates? [Section 4.2]

4.7

sol-.

ference . [Section 4.1]

4.3 Aqueous solutions of three different substances, AX, AY, and AZ, are represented by the three diagrams below.

Which of the following ions will

always be a spectator

ion in a precipitation reaction? (a) CC (b) N03-, 2 (c) NH4+, (d) S -, (e) Explain briefly. [Section 4.2]

4.8 The labels have fallen off two bottles, one containing Mg(N03)z and the other containing Pb(N03)z. You have

Identify each substance as a strong electrolyte, weak

a bottle of dilute H2S04. How could you use it to test a

electrolyte, or nonelectrolyte. [Section 4.1]

portion of each solution to identify which solution is

which? [Section 4.2]

4.9

Explain how a redox reaction involves electrons in the same way that an acid-base reaction involves protons.

[Sections

4.3 and 4.4]

4.10 If you want to double the concentration of a solution, how could you do it? [Section 4.5]

(a)

(b )

(c)

EXERCISES Electrolytes 4.11 When asked what causes electrolyte solutions to con­ duct electricity, a student responds that it is due to the

movement of electrons through the solution. Is the stu­

dent correct? If not, what is the correct response? 4.12 When methanol, CH30H, is dissolved in water, a noncon­ ducting solution results. When acetic acid, CH3COOH,

dissolves in water, the solution is weakly conducting and

acidic in nature. Describe what happens upon dissolution

in the two cases, and account for the different results.

4.13 We have learned in this chapter that many ionic solids

4.16 Specify what ions are present upon dissolving each of the following substances in water: (a) Mgiz, (b) Al(N03)J, (c) HC104, (d) NaCH3COO. 4.17 Formic acid, HCOOH, is a weak electrolyte. What solute particles are present in an aqueous solution of this com­

pound? Write the chemical equation for the ionization

of HCOOH.

4.18 Acetone, CH3COCH3, is a nonelectrolyte; hypochlorous acid, HClO, is a weak electrolyte; and ammonium

chloride, NH4Cl, is a strong electrolyte. (a) What are the

dissolve in water as strong electrolytes, that is, as sepa­

solute particles present in aqueous solutions of each

tate this process?

solved in solution, which one contains

rated ions in solution. What properties of water facili­

4.14 What does it mean to say that ions are hydrated when an ionic substance dissolves in water?

4.15 Specify what ions are present in solution upon dissolv­ ing each of the following substances in water: (a) ZnCI2,

(b) HN03, (c) (NH4)zS04, (d) Ca(OH)z.

compound?

(b) If 0.1 mol of each compound is dis­

0.2 mol of solute

particles, which contains 0.1 mol of solute particles, and

which contains somewhere between solute particles?

0.1

and

0.2 mol of

Exercises

157

Precipitation Reactions and Net Ionic Equations 4.19

4.20

4.21

4.22

4.23

4.24

Using solubility guidelines, predict whether each of the following compounds is soluble or insoluble in water: (a) NiC12, (b) Ag2S, (c) Cs3P04, (d) SrC03, (e) PbS04. Predict whether each of the following compounds is sol­ uble in water: (a) Ni(OHh, (b) PbBr:z, (c) Ba(N03h, (d) AlP04, (e) AgCH3COO. Will precipitation occur when the following solutions are mixed? If so, write a balanced chemical equation for the reaction. (a) Na2C03 and AgN03, (b) NaN03 and NiS04, (c) FeS04 and Pb(N03)z. Identify the precipitate (if any) that forms when the following solutions are mixed, and write a balanced equation for each reaction. (a) Ni(N03h and NaOH, (b) NaOH and K2S04, (c) Na2S and Cu(CH3COO)z. Name the spectator ions in any reactions that may be involved when each of the following pairs of solutions are mixed. (a) Na2C03(aq) and MgS04(aq) (b) Pb(N03)z(aq) and NazS(aq) (c) (NH4)JP04(aq) and CaC12(aq ) Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the specta­ tor ion or ions in each reaction.

+ (NH4}zC03(aq) ---> Ba(N03h(aq) + K2S04(aq) ---> Fe(N03h(aq) + KOH(aq) --->

(a) Crz(S04)J(aq) (b) (c) 4.25

4.26

4.27

4.28

Separate samples of a solution of an unknown salt are treated with dilute solutions of HBr, H2SO"' and NaOH. A precipitate forms in all three cases. Which of the follow­ ing cations could the solution contain: K+; Pb 2+; Ba2+? Separate samples of a solution of an unknown ionic compound are treated with dilute AgN03, Pb(N03):z, and BaC12. Precipitates form in all three cases. Which of the following could be the anion of the unknown salt: Br-; col-; N03-? You know that an unlabeled bottle contains a solution of one of the following: AgN03, CaCl:z, or Alz(S04)J. A friend suggests that you test a portion of the solution with Ba(N03h and then with NaCI solutions. Explain how these two tests together would be sufficient to de­ termine which salt is present in the solution. Three solutions are mixed together to form a single solu­ tion. One contains 0.2 mol Pb(CH3COOh, the second contains 0.1 mol Na2S, and the third contains 0.1 mol CaC12• (a) Write the net ionic equations for the precipita­ tion reaction or reactions that occur. (b) What are the spectator ions in the solution?

Acid-Base Reactions 4.29

4.30

4.31

4.32

4.33

4.34

4.35

aqueous solution of an unknown solute is tested with litmus paper and found to be acidic. The solution is weakly conducting compared with a solution of NaCI of the same concentration. Which of the following sub­ stances could the unknown be: KOH, NH3, HN03, KClO:z, H3P03, CH3COCH3 (acetone)?

Which of the following solutions has the largest concen­ tration of solvated protons: (a) 0.2 M LiOH, (b) 0.2 M HI, (c) 1.0 M methyl alcohol (CH30H)? Explain. Which of the following solutions is the most basic? (a) 0.6 M NH3, (b) 0.150 M KOH, (c) 0.100 M Ba(OH)z. Explain.

4.36 An

What is the difference between (a) a monoprotic acid and a diprotic acid, (b) a weak acid and a strong acid, (c) an acid and a base? Explain the following observations: (a) NH3 contains no OH- ions, and yet its aqueous solutions are basic; (b) HF is called a weak acid, and yet it is very reactive; (c) although sulfuric acid is a strong electrolyte, an aque­ ous 2solution of H2S04 contains more HS04- ions than 504 - ions.

4.37

HO, HBr, and HI are strong acids, yet HF is a weak acid. What does this mean in terms of the extent to which these substances are ionized in solution? What is the relationship between the solubility rules in Table 4.1 and the list of strong bases in Table 4.2? Another way of asking this question is, why is Cd(OHh, for example, not listed as a strong base in Table 4.2? Label each of the following substances as an acid, base, salt, or none of the above. Indicate whether the sub­ stance exists in aqueous solution entirely in molecular form, entirely as ions, or as a mixture of molecules and ions. (a) HF; (b) acetonitrile, CH3CN; (c ) NaCI04; (d) Ba(OH)z.

4.38

4.39

Classify each of the following substances as a nonelec­ trolyte, weak electrolyte, or strong electrolyte in water: (a) H2S03, (b) C2H50H (ethanol), (c ) NH3, (d) KCI03, (e) Cu(N03)z. Classify each of the following aqueous solutions as a nonelectrolyte, weak electrolyte, or strong electrolyte: (a) HC104, (b) HN03, (c) NH4Cl, (d) CH3COCH3 (acetone), (e) CoS04, (f) CnH220u (sucrose). Complete and balance the following molecular equa­ tions, and then write the net ionic equation for each: (a) HBr(aq) + Ca(OH)z(aq) ---> (b) Cu(OH)z(s) + HCI04(aq) ---> (c) Al(OH)J(s) + HN03(aq) Write the balanced molecular and net ionic equations for each of the following neutralization reactions: (a) Aqueous acetic acid is neutralized by aqueous potassium hydroxide. (b) Solid chromium(Ill) hydroxide reacts with nitric acid. (c) Aqueous hypochlorous acid and aqueous calcium hydroxide react. --->

4.40

158

C HAPTER 4

Aqueous Reactions and Solution Stoichiometry

4.41 Write balanced molecular and net ionic equations for the following reactions, and identify the gas formed in each:

(a) solid cadmium sulfide reacts with an aqueous

solution of sulfuric acid;

(b) solid magnesium carbonate

reacts with an aqueous solution of perchloric acid.

4.42 Because the oxide ion is basic, metal oxides react readily with acids. (a) Write the net ionic equation for the fol­ lowing reaction:

FeO(s) + 2 HC104(aq) -----> Fe(ClO.J2(aq) + H20(1)

(b) Based on the equation in part (a), write the net ionic

equation for the reaction that occurs between NiO(s)

4.43 Write a balanced molecular equation and a net ionic equation for the reaction that occurs when (a) solid

CaC03 reacts with an aqueous solution of nitric acid;

(b) solid iron(II) sulfide reacts with an aqueous solution

of hydrobromic acid.

4.44 As K20 dissolves in water, the oxide ion reacts with water molecules to form hydroxide ions. Write the mol­

ecular and net ionic equations for this reaction. Based

on the definitions of acid and base, what ion is the base in this reaction? What is the acid? What is the spectator

ion in the reaction?

and an aqueous solution of nitric acid.

Oxidation-Reduction Reactions 4.45 Define oxidation and reduction in terms of (a) electron

4.52 Which of the following are redox reactions? For those

4.46 Can oxidation occur without accompanying reduction?

is reduced. For those that are not, indicate whether they

transfer and

(b)

oxidation numbers.

Explain.

4.47 Which circled region of the periodic table shown here

contains the most readily oxidized elements? Which contains the least readily oxidized?

that are, indicate which element is oxidized and which

are precipitation or acid-base reactions.

(a) (b)

Cu(OH)z(s)

+ 2 HN03(aq)

----->

Cu(N03)z(aq) + 2 H20(1)

Fe203(s) + 3 CO(g) -----> 2 Fe(s) + 3 C02(g)

(c) Sr(N03)z(aq) + H2S04(aq) -----> SrS04(s) + 2 HN03(aq)

(d) 4 Zn(s) + 10 H+(aq) + 2 N03 -(aq) -----> 2

4 Zn +(aq) + NzO(g) + 5 HzO(I)

4.53 Write balanced molecular and net ionic equations for the reactions of (a) manganese with dilute sulfuric acid;

(b) cluomium with hydrobromic acid; (c) tin with

hydrochloric acid; HCOOH.

(d) aluminum with formic acid,

4.54 Write balanced molecular and net ionic equations for 4.48 From the elements listed in Table 4.5, select an element that lies in region A of the periodic table shown above and an element that lies in region C. Write a balanced

oxidation-reduction equation that shows the oxidation

of one metal and reduction of an ion of the other.

You will need to decide which element is oxidized and which is reduced.

4.49 Determine the oxidation number for the indicated ele­ ment in each of the following substances: (a) S in 502, (b) C in COC12, (c) Mn in Mn04-, (d) Br in HBrO, (e) As in As4, (f) 0 in K202. 4.50 Determine the oxidation number for the indicated ele­ ment in each of the following compounds: (a) Ti in Ti02, (b) Sn in SnC13 -, (c) C in C20l-, (d) N in N2H4, (e) N in HNOz, (f) Cr in Cr20l-. 4.51 Which element is oxidized and which is reduced in the following reactions?

(a) N2(g) + 3 H2(g) -----> 2 NH3(g) (b) 3 Fe(N03)z(aq) + 2 Al(s) -----> 3 Fe(s) + 2 Al(N03)J(aq) (c) Cl2(aq) + 2 Nal(aq) -----> I2(aq) + 2 NaCl(aq) (d) PbS(s) + 4 H202(aq) -----> PbS04(s) + 4 HzO(l)

the reactions of (a) hydrochloric acid with nickel;

lute sulfuric acid with iron;

magnesium;

(b) di­

(c) hydrobromic acid with

(d) acetic acid, CH3COOH, with zinc.

4.55 Using the activity series (Table 4.5), write balanced chem­ ical equations for the following reactions. If no reaction occurs, simply write NR.

(a)

solution of copper(II) nitrate;

Iron metal is added to a

(b) zinc metal is added to

a solution of magnesium sulfate; (c) hydrobromic acid is added to

tin metal; (d) hydrogen gas is bubbled through (e) aluminum

an aqueous solution of nickel(II) chloride;

metal is added to a solution of cobalt(ll) sulfate.

4.56 Based on the activity series (Table 4.5), what is the out­ come

(a) (b)

(if any) of each of the following reactions? + NiC12(aq)

Mn(s)

----->

Cu(s) + Cr(CH3COO)J(aq) ----->

(c) Cr(s) + NiS04(aq) ----->

(d) Pt(s) + HBr(aq) -----> (e) Hz(g) + CuClz(aq) -----> 2

4.57 The metal cadmium tends to form Cd + ions. The follow­ ing observations are made: (i) When a strip of zinc metal

is placed in CdCI2(aq), cadmium metal is deposited on

Exercises the strip. (ii) When a strip of cadmium metal is placed in Ni(N03h(aq), nickel metal is deposited on the strip. (a) Write net ionic equations to explain each of the ob­ servations made above. (b) What can you conclude about the position of cadmium in the activity series? (c) What experiments would you need to perform to locate more precisely the position of cadmium in the activity series?

159

4.58 (a) Use the following reactions to prepare an activity series for the halogens: Br2(aq)

+

Cl2(aq) +

2 Nal(aq)

2 NaBr(aq)

-->

-->

2 NaBr(aq) + I2(aq)

2 NaCl(aq) +

Br2(aq)

(b) Relate the positions of the halogens in the periodic table with their locations in this activity series. (c) Predict whether a reaction occurs when the following reagents are mixed: Cl2(aq) and KI(aq); Br2(aq) and LiCl(aq).

Solution Composition; Molarity 4.59 (a) Is the concentration of a solution an intensive or an extensive property? (b) What is the difference between 0.50 mol HCl and 0.50 M HCI? 4.60 (a) Suppose you prepare 500 mL of a 0.10 M solution of some salt and then spill some of it. What happens to the concentration of the solution left in the container? (b) Suppose you prepare 500 mL of a 0.10M aqueous solu­ tion of some salt and let it sit out, uncovered, for a long time, and some water evaporates. What happens to the concentration of the solution left in the container? (c) A certain volume of a 0.50 M solution contains 4.5 g of a salt. What mass of the salt is present in the same volume of a 2.50 M solution? 4.61 (a) Calculate the molarity of a solution that contains 0.0250 mol NH4Cl in exactly 500 mL of solution. (b) How many moles of HN03 are present in 50.0 mL of a 2.50 M solution of nitric acid? (c) How many milliliters of 1.50 M KOH solution are needed to provide 0.275 mol of KOH? 4.62 (a) Calculate the molarity of a solution made by dissolv­ ing 0.750 grams of Na2S04 in enough water to form exactly 850 mL of solution. (b) How many moles of KMn04 are present in 250 mL of a 0.0475 M solution? (c) How many milliliters of 11.6 M HCl solution are needed to obtain 0.250 mol of HCl? 4.63 The average adult human male has a total blood volume of 5.0 L. If the concentration of sodium ion in this aver­ age individual is 0.135 M, what is the mass of sodium ion circulating in the blood? 4.64 A person suffering from hyponatremia has a sodium ion concentration in the blood of 0.118 M and a total blood volume of 4.6 L. What mass of sodium chloride would need to be added to the blood to bring the sodium ion concentration up to 0.138 M, assuming no change in blood volume? 4.65 The concentration of alcohol (CH3CHzOH) in blood, called the "blood alcohol concentration" or BAC, is given in units of grams of alcohol per 100 mL of blood. The legal definition of intoxication, in many states of the United States, is that the BAC is 0.08 or higher. What is the concentration of alcohol, in terms of molarity, in blood if the BAC is 0.08? 4.66 The average adult male has a total blood volume of 5.0 L. After drinking a few beers, he has a BAC of 0.10 (see Exercise 4.65). What mass of alcohol is circulating in his blood?

4.67 Calculate (a) the number of grams of solute in 0.250 L of 0.175 M KBr, (b) the molar concentration of a solution containing 14.75 g of Ca(N03)z in 1.375 L, (c) the volume of 1.50 M Na3P04 in milliliters that contains 2.50 g of solute. 4.68 (a) How many grams of solute are present in 50.0 mL of 0.488 M K2Cr207? (b) If 4.00 g of (NH4)zS04 is dissolved in enough water to form 400 mL of solution, what is the molarity of the solution? (c) How many milliliters of 0.0250 M CuS04 contain 1.75 g of solute? 4.69 (a) Which will have the highest concentration of potassi­ um ion: 0.20 M KCI, 0.15 M K2Cr04, or 0.080 M K3P04? (b) Which will contain the greater number of moles of potassium ion: 30.0 mL of 0.15 M K2Cr04 or 25.0 mL of 0.080 M K3P04? 4.70 1n each of the following pairs, indicate which has the higher concentration of Cl- ion: (a) 0.10 M CaC12 or 0.15 M KCI solution, (b) 100 rnL of 0.10 M KCI solution or 400 rnL of 0.080 M LiCI solution, (c) 0.050 M HCl solution or 0.020 M CdC12 solution. 4.71 Indicate the concentration of each ion or molecule pre­ sent in the following solutions: (a) 0.25 M NaN03, 2 (b) 1.3 X 10- M MgS04, (c) 0.0150 M Ct;H120IV (d) a mix­ ture of 45.0 mL of 0.272 M NaCI and 65.0 mL of 0.0247 M (�)zC03. Assume that the volumes are additive. 4.72 Indicate the concentration of each ion present in the so­ lution formed by mixing (a) 42.0 rnL of 0.170 M NaOH and 37.6 rnL of 0.400 M NaOH, (b) 44.0 rnL of 0.100 M and Na2S04 and 25.0 mL of 0.150 M KCI, (c) 3.60 g KCl in 75.0 rnL of 0.250 M CaC12 solution. Assume that the volumes are additive. 4.73 (a) You have a stock solution of 14.8 M NH3. How many milliliters of this solution should you dilute to make 1000.0 mL of 0.250 M NH3? (b) If you take a 10.0-mL portion of the stock solution and dilute it to a total vol­ ume of 0.500 L, what will be the concentration of the final solution? 4.74 (a) How many milliliters of a stock solution of 10.0 M HN03 would you have to use to prepare 0.450 L of 0.500 M HN03 ? (b) If you dilute 25.0 mL of the stock so­ lution to a final volume of 0.500 L, what will be the concentration of the diluted solution? 4.75 (a) Starting with solid sucrose, C12H22011, describe how you would prepare 250 mL of a 0.250 M sucrose solu­ tion. (b) Describe how you would prepare 350.0 mL of 0.100 M C12H22011 starting with 3.00 L of 1.50 M C12HzzOu .

160

C HAPTER 4

Aqueous Reactions and Solution Stoichiometry

4.76 (a) How would you prepare 175.0 mL of 0.150 M AgN03 solution starting with pure solute? (b) An experiment calls for you to use 100 mL of 0.50 M HN03 solution. All you have available is a bottle of 3.6 M HN03. How would you prepare the desired solution?

[4.77] Pure acetic acid, known as glacial acetic acid, is a liquid with a density of 1 .049 g/mL at 25 °C. Calculate the mo­ larity of a solution of acetic acid made by dissolving

20.00 mL of glacial acetic acid at 25 oc in enough water to make 250.0 mL of solution. [4.78] Glycerol, C3H803, is a substance used extensively in the manufacture of cosmetics, foodstuffs, antifreeze, and plastics. Glycerol is a water-soluble liquid with a densi­ ty of 1.2656 g/L at 15 oc. Calculate the molarity of a so­ lution of glycerol made by dissolving 50.000 mL glycerol at 15 oc in enough water to make 250.00 mL of solution.

Solution Stoichiometry; Titrations 4.79 What mass of KCl is needed to precipitate the silver ions from 15.0 mL of 0.200 M AgN03 solution? 2 4.80 What mass of NaOH is needed to precipitate the Cd + ions from 35.0 mL of 0.500 M Cd(N03)z solution?

4.81 (a) What volume of 0.115 M HCI04 solution is needed to neutralize 50.00 mL of 0.0875 M NaOH? (b) What vol­ ume of 0.128 M HCI is needed to neutralize 2.87 g of Mg(OH)z? (c) If 25.8 mL of AgN03 is needed to precipi­ tate all the o- ions in a 785-mg sample of KCl (forming AgCI), what is the molarity of the AgN03 solution? (d) If 45.3 mL of 0.108 M HCl solution is needed to neu­ tralize a solution of KOH, how many grams of KOH must be present in the solution?

4.82 (a) How many milliliters of 0.120 M HCl are needed to completely neutralize 50.0 mL of 0.101 M Ba(OH)z solu­ tion? (b) How many milliliters of 0.125 M H2S04 are needed to neutralize 0.200 g of NaOH? (c) If 55.8 mL of BaCI2 solution is needed to precipitate all the sulfate ion in a 752-mg sample of Na2S04, what is the molarity of the solution? (d) If 42.7 mL of 0.208 M HCl solution is needed to neutralize a solution of Ca(OH)z, how many grams of Ca(OHh must be in the solution? 4.83 Some sulfuric acid is spilled on a lab bench. You can neutralize the acid by sprinkling sodium bicarbonate on it and then mopping up the resultant solution. The sodi­ um bicarbonate reacts with sulfuric acid as follows: 2 NaHC03(s) + HzS04(aq) ->

NazS04(aq)

+

2 HzO(I) + 2 COz(g)

Sodium bicarbonate is added until the fizzing due to the formation of C02(g) stops. If 27 mL of 6.0 M HzS04 was spilled, what is the minimum mass of NaHC03 that must be added to the spill to neutralize the acid?

4.84 The distinctive odor of vinegar is due to acetic acid, CH3COOH, which reacts with sodium hydroxide in the following fashion:

CH3COOH(aq)

+

NaOH(aq)

->

HzO(l)

+

NaCzHPz(aq)

If 3.45 mL of vinegar needs 42.5 mL of 0.115 M NaOH to reach the equivalence point in a titration, how many grams of acetic acid are in a 1.00-qt sample of this vinegar?

4.85 A sample of solid Ca(OH)z is stirred in water at 30 oc until the solution contains as much dissolved Ca(OH)z as it can hold. A 100-mL sample of this solution is with­ drawn and titrated with 5.00 x 10-2 M HBr. It requires 48.8 mL of the acid solution for neutralization. What is the molarity of the Ca(OH)z solution? What is the solu­ bility of Ca(OH)z in water, at 30 oc, in grams of Ca(OH)z per 100 mL of solution?

4.86

ln a laboratory, 6.82 g of Sr(N03)z is dissolved in enough water to form 0.500 L of solution. A 0.100-L sample is withdrawn from this stock solution and titrated with a 0.0245 M solution of Na2Cr04. What volume of Na2Cr04 solution is needed to precipitate all the Sr2+(aq) as SrCr04?

4.87 A solution of 100.0 mL of 0.200 M KOH is mixed with a solution of 200.0 mL of 0.150 M NiS04. (a) Write the balanced chemical equation for the reaction that occurs. (b) What precipitate forms? (c) What is the limiting reac­ tant? (d) How many grams of this precipitate form? (e) What is the concentration of each ion that remains in solution?

4.88 A solution is made by mixing 12.0 g of NaOH and 75.0 mL of 0.200 M HN03. (a) Write a balanced equation for the reaction that occurs between the solutes. (b) Calculate the concentration of each ion remaining in solution . (c) Is the resultant solution acidic or basic? [4.89] A 0.5895-g sample of impure magnesium hydroxide is dissolved in 100.0 mL of 0.2050 M HCI solution. The excess acid then needs 19.85 mL of 0.1020 M NaOH for neutralization. Calculate the percent by mass of magnesium hydroxide in the sample, assum­ ing that it is the only substance reacting with the HCI solution.

[4.90] A 1.248-g sample of limestone rock is pulverized and then treated with 30.00 mL of 1 .035 M HCl solution. The excess acid then requires 11.56 mL of 1.010 M NaOH for neutralization. Calculate the percent by mass of cal­ cium carbonate in the rock, assuming that it is the only substance reacting with the HCl solution.

Additional Exercises

161

ADDITIONAL EXERCISES 4.91 Explam why a titration experiment is a good way to measure the unknown concentration of a compound in solution. 4.92 The accompanying photo shows the reaction between a solution of Cd(N03)z and one of Na2S. What is the identi­ ty of the precipitate? What ions remam in solution? Write the net ionic equation for the reaction.

4.93 Suppose you have a solution that might contain any or 2 2 2 all of the following cations: Ni +, Ag+, Sr +, and Mn + Addition of HCl solution causes a precipitate to form. After filtering off the precipitate, H2S04 solution is added to the resultant solution and another precipitate forms. This is filtered off, and a solution of NaOH is added to the resulting solution. No precipitate is ob­ served. Which ions are present in each of the precipi­ tates? Which of the four ions listed above must be absent from the original solution?

4.94 You choose to investigate some of the solubility guide­ lines for two ions not listed in Table 4.1, the chromate ion (Crol-) and the oxalate ion (C2ol-). You are given O.Ql M solutions (A, B, C, D) of four water-soluble salts: Solution

Solute

Color of Solution

A

Na2Cr04

Yellow

c

AgN03

Colorless

B

D

(NH4)zC204

CaC12

Colorless Colorless

When these solutions are mixed, the following observa­ tions are made: Expt Number

2

Solutions Mixed

Result

A +B

No precipitate, yellow solution

A+C

3

A + D

5

B+D

4

6

Red precipitate forms

No precipitate, yellow solution

B + C

White precipitate forms

C + D

White precipitate forms

White precipitate forms

(a) Write a net ionic equation for the reaction that occurs in each of the experiments. (b) Identify the precipitate formed, if any, in each of the experiments. (c) Based on these limited observations, which ion tends to form the more soluble salts, chromate or oxalate? 4.95 Antacids are often used to relieve pain and promote healing in the treatment of mild ulcers. Write balanced net ionic equations for the reactions between the HCl(aq) in the stomach and each of the following substances used in various antacids: (a) Al(OHh(s), (b) Mg(OH)z(s), (c) MgC03(s), (d) NaAl(C03)(0H)z(s), (e) CaC03(s). [4.96] Salts of the sulfite ion, sol-, react with acids in a way similar to that of carbonates. (a) Predict the chemical formula, and name the weak acid that forms when the sulfite ion reacts with acids. (b) The acid formed in part (a) decomposes to form water and a gas. Predict the molecular formula, and name the gas formed. (c) Use a source book such as the CRC Handbook of Chemistry and Physics to confirm that the substance in part (b) is a gas under normal room-temperature conditions. (d) Write balanced net ionic equations of the reaction of HC!(aq) with (i) Na2S03(aq), (ii) Ag2S03(s), (iii) KHS03(s), and (iv) ZnS03(aq). [4.97] The commercial production of nitric acid involves the following chemical reactions: 4 NH3(g) + 5 02(g) -----> 4 NO(g) + 6 H20(g) 2 NO(g) + 02(g) -----> 2 N02(g) 3 N02(g) + H20(1) -----> 2 HN03(aq) + NO(g)

(a) Which of these reactions are redox reactions? (b) In each redox reaction identify the element undergoing oxidation and the element undergoing reduction. 4.98 Use Table 4.5 to predict which of the following ions can be reduced to their metal forms by reacting with +zinc: + 2 2+ 2+ (a) Na2 (aq), (b) Pb (aq), (c) Mg (aq), (d) Fe (aq), 3 (e) Cu +(aq), (f) Al +(aq). Write the balanced net ionic equation for each reaction that occurs. [4.99] Lanthanum metal forms cations with a charge of 3+. Consider the following observations about the chemistry of lanthanum: When lanthanum metal is exposed to air, a white solid (compound A) is formed that contains lan­ thanum and one other element. When lanthanum metal is added to water, gas bubbles are observed and a differ­ ent white solid (compound B) is formed. Both A and B dissolve in hydrochloric acid to give a clear solution. When either of these solutions is evaporated, a soluble white solid (compound C) remains. If compound C is dissolved in water and sulfuric acid is added, a white precipitate (compound D) forms. (a) Propose identities for the substances A, B, C, and D. (b) Write net ionic equations for all the reactions described. (c) Based on the preceding observations, what can be said about the posi­ tion of lanthanum in the activity series (Table 4.5)? 4.100 A 35.0-mL sample of 1.00 M KBr and a 60.0-mL sample of 0.600 M KBr are mixed. The solution is then heated to evaporate water until the total volume is 50.0 mL. What is the molarity of the KBr in the final solution?

162

C HAPTER 4

Aqueous Reactions and Solution Stoichiometry

4.101 Using modern analytical techniques, it is possible to de­ tect sodium ions in concentrations as low as 50 pgjmL. What is this detection limit expressed in (a) molarity of Na+, (b) Na+ ions per cubic centimeter? 2 2 2 4.102 Hard water contains Ca +, Mg +, and Fe +, which inter­ fere with the action of soap and leave an insoluble coating on the insides of containers and pipes when heated. Water softeners replace these ions with Na+ If 1500 L of 2 2 hard water contains 0.020 M Ca + and 0.0040 M Mg +, how many moles of Na+ are needed to replace these ions?

4.103 Tartaric acid, H2C4H406, has two acidic hydrogens. The acid is often present in wines and precipitates from solu­ tion as the wine ages. A solution containing an un­ known concentration of the acid is titrated with NaOH. It requires 24.65 mL of 0.2500 M NaOH solution to titrate both acidic protons in 50.00 mL of the tartaric acid solution. Write a balanced net ionic equation for the

neutralization reaction, and calculate the molarity of the tartaric acid solution.

4.104 The concentration of hydrogen peroxide in a solution is determined by titrating a 10.0-mL sample of the solution with permanganate ion. 2 Mn04-(aq) + 5 H202(aq) + 6 H+(aq) --->

2 MnZ+(aq)

+

5 02(g) + 8 H20(!)

If it takes 14.8 mL of 0.134 M Mn04- solution to reach the equivalence point, what is the molarity of the hydro­ gen peroxide solution?

[4.105] A solid sample of Zn(OHh is added to 0.350 L of 0.500 M aqueous HBr. The solution that remains is still acidic. It is then titrated with 0.500 M NaOH solution, and it takes 88.5 mL of the NaOH solution to reach the equiva­ lence point. What mass of Zn(OHh was added to the HBr solution?

INTEGRATIVE EXERCISES 4.106 (a) By titration, 15.0 mL of 0.1008 M sodium hydroxide is needed to neutralize a 0.2053-g sample of an organic acid. What is the molar mass of the acid if it is monopro­ tic? (b) An elemental analysis of the acid indicates that it is composed of 5.89% H, 70.6% C, and 23.5% 0 by mass. What is its molecular formula?

4.107 A 3.455-g sample of a mixture was analyzed for barium ion by adding a small excess of sulfuric acid to an aque­ ous solution of the sample. The resultant reaction pro­ duced a precipitate of barium sulfate, which was collected by filtration, washed, dried, and weighed. If 0.2815 g of barium sulfate was obtained, what was the mass percentage of barium in the sample? 3 [4.108] A tanker truck carrying 5.0 X 10 kg of concentrated sulfuric acid solution tips over and spills its load. If the sulfuric acid is 95.0% H2S04 by mass and has a density of 1.84 g/mL, how many kilograms of sodium carbonate must be added to neutralize the acid?

4.109 A sample of 5.53 g of Mg(OHh is added to 25.0 mL of 0.200 M HN03. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) How many moles of Mg(OHh, HN03, and Mg(N03h are present after the reaction is complete? 4.110 A sample of 1.50 g of lead(ll) nitrate is mixed with 125 mL of 0.100 M sodium sulfate solution. (a) Write the chemical equation for the reaction that occurs. (b) Which is the limiting reactant in the reaction? (c) What are the concentrations of all ions that remain in solution after the reaction is complete?

4.111 A mixture contains 76.5% NaCI, 6.5% MgC12, and 17.0% Na2S04 by mass. What is the molarity of Cl- ions in a solution formed by dissolving 7.50 g of the mixture in enough water to form 500.0 mL of solution?

[4.112] The average concentration of bromide ion in seawater is 65 mg of bromide ion per kg of seawater. What is the molarity of the bromide ion if the density of the seawa­ ter is 1.025 g/mL?

[4.113] The mass percentage of chloride ion in a 25.00-mL sam­ ple of seawater was determined by titrating the sample with silver nitrate, precipitating silver chloride. It took 42.58 mL of 0.2997 M silver nitrate solution to reach the equivalence point in the titration. What is the mass per­ centage of chloride ion in the seawater if its density is 1 .025 g/mL? 4.114 The arsenic in a 1 .22-g sample of a pesticide was con­ 3 verted to As04 - by suitable chemical treatment. It was then titrated using Ag + to form Ag3As04 as a preci­ 3 pitate. (a) What is the oxidation state of As in As04 -? (b) Name Ag3As04 by analogy to the corresponding compound containing phosphorus in place of arsenic. (c) If it took 25.0 mL of 0.102 M Ag+ to reach the equiva­ lence point in this titration, what is the mass percentage of arsenic in the pesticide?

[4.115] The newest U.S. standard for arsenate in drinking water, mandated by the Safe Drinking Water Act, required that by January 2006, public water supplies must contain no greater than 10 parts per billion (ppb) arsenic. If this ar­ 3 senic is present as arsenate, As04 -, what mass of sodi­ um arsenate would be present in a 1 .00-L sample of drinking water that just meets the standard?

[4.116] The safe drinking water standard for arsenic (which is usually found as arsenate, see 4.115) is 50 parts per bil­ lion (ppb) in most developing countries. (a) How many grams of sodium arsenate are in 55 gallons of water, if the concentration of arsenate is 50 ppb? (b) In 1993,

Integrative Exercises naturally occurring arsenic was discovered as a major contaminant in the drinking water across the country of Bangladesh. Approximately 12 million people in Bangladesh still drink water from wells that have higher concentrations of arsenic than the standard. Recently, a chemistry professor from George Mason University was awarded a $1 million Grainger Challenge Prize for Sus­ tainability for his development of a simple, inexpensive system for filtering naturally occuring arsenic from drinking water. The system uses buckets of sand, cast iron, activated carbon, and wood chips for trapping ar­ senic-containing minerals. Assuming the efficiency of such a bucket system is 90% (meaning, 90% of the ar­ senic that comes in is retained in the bucket and 10% passes out of the bucket), how many times should water that is 500 ppb in arsenic be passed through to meet the 50 ppb standard?

163

[4.117] Federal regulations set an upper limit of 50 parts per mil­ lion (ppm) of NH3 in the air in a work environment [that is, 50 molecules of NH3(g) for every million molecules in the air]. Air from a manufacturing operation was drawn 2 through a solution containing 1.00 x 10 mL of O.D105 M HCI. The NH3 reacts with HCI as follows:

NH3(aq) + HCl(aq)

�

NH4Cl(aq)

After drawing air through the acid solution for 10.0 min at a rate of 10.0 L/min, the acid was titrated. The re­ maining acid needed 13.1 mL of 0.0588 M NaOH to reach the equivalence point. (a) How many grams of NH3 were drawn into the acid solution? (b) How many ppm of NH3 were in the air? (Air has a density of 1.20 g/L and an average molar mass of 29.0 g/mol under the conditions of the experiment.) (c) Is this man­ ufacturer in compliance with regulations?

THERMOCHEMISTRY

WILDFLOWERS . Plants use solar energy to carry out photosynthesis of carbohydrates, which provide the energy needed for plant growth. Among the molecules produced by plants are 02 and ethylene, C2H4.

1 64

W H AT ' S 5.1

5.2

5.3

A H E A D

The Nature of Energy

We begin by considering the nature of energy and the forms it takes, notably kinetic energy and potential mergy. We will also discuss the units used in measuring energy and the fact that energy can be used to accomplish work or to transfer heat. To study energy changes, we focus on a particular part of the universe, which we call the system. Everything else is called the surroundings.

5.4

5.5

The First Law of Thermodynamics

We will then explore the first law of thermodynamics: Energy cannot be created or destroyed, but it can be transformed from one form to another or transferred between systems and surroundings. The energy possessed by a system is called its internal energ�;. Internal energy, £, is a statefunction, a quantity whose value depends only on the state, or condition, of a system now, not on how the system came to be in that state.

5.6

5.7

Enthalpy

Next, we will encounter a state function called enthalpy, H, that relates to energy. This function is useful because the change in enthalpy, !J.H, measures the quantity of heat energy gained or lost by a system in a process occurring under a constant pressure.

5.8

Enthalpies of Reaction

We will see that the enthalpy change associated with a chemical reaction (!J.H,xnl is given by the enthalpies of the products minus the enthalpies of the reactants. This quantity is directly proportional to the amount of reactant consumed in the process. Calorimetry

We will next examine calorimetry, which is an experimental technique that we use to measure heat changes in chemical processes. Hess's Law

We will observe that the enthalpy change for a reaction can be calculated using appropriate enthalpy changes for other reactions. To do so, we apply Hess's law, which uses the fact that enthalpy, like energy, is a state function. Enthalpies of Formation

Then we will discuss how to establish standard values for enthalpy changes in chemical reactions and how to use them to calculate enthalpy changes for reactions (!J.Hcxnl · Foods and Fuels

Finally, we will examine foods and fuels as sources of energy and discuss some related health and social issues.

MODERN SOCI ETY D E P E N D S ON EN ERGY

for its existence.

Energy is used to drive our machinery and appliances, to power our transportation vehicles, and to keep us warm in the winter and cool in the summer. It is not just modern society, however, that depends on energy. Energy is necessary for all life. Plants, such as those in the chapter-opening photograph, use solar energy to carry out photosynthesis, allowing the plants to grow. The plants, in turn, provide food, from which we humans derive the energy we need to move, to maintain body temperature, and to carry out bodily functions. What exactly is energy, though, and what principles are involved in its many transactions and transformations, such as those from the sun to plants to animals? In this chapter we begin to explore energy and its changes. We are moti­ vated partly by the fact that energy changes invariably accompany chemical reactions. Indeed, sometimes we use a chemical reaction specifically to obtain energy, as when we burn fuels. Thus, energy is very much a chemical topic. Nearly all of the energy on which we depend is derived from chemical reac­ tions, whether those reactions are associated with the combustion of fuels, the discharge of a battery, or the metabolism of our foods. If we are to properly 1 65

166

C HA PTER 5

Thermochemistry understand chemistry, we must also understand the energy changes that accompany chemical reactions. The study of energy and its transformations is known as thermodynamics (Greek: therme-, "heat"; dy'namis, "power"). In this chapter we will examine an aspect of thermodynamics that involves the relationships between chemical reactions and energy changes involving heat. This portion of thermodynamics is called thermochemistry. We will discuss additional aspects of thermody­ namics in Chapter 19.

5 . 1 THE NATURE O F ENERGY Although the idea of energy is a familiar one, it is a bit challenging to deal with the concept in a precise way. Energy is commonly defined as the capacity to do work or to transfer heat. This definition requires us to understand the concepts of work and heat. We can think of work as the energy used to cause an object with mass to move against a force and heat as the energy used to cause the temperature of an object to increase (Figure 5.1 �tial• indicat­ ing the system has gained energy from its surroundings. A negative value of !!J.E is obtained when Etinal < £initial• indicating the system has lost energy to its surroundings. Notice that we are taking the point of view of the system rather than that of the surroundings in discussing the energy changes. We need to remember, however, that any change in the energy of the system is accompa­ nied by an opposite change in the energy of the surroundings. These features of energy changes are summarized in Figure 5.4 T. -

£initial

£final < Einitial t.E < 0 (negative)

Energy gained from surroundings

Final state

------- Efmal

------- Emitial

E of system increases

E of system decreases (a)

(b) *The symbol by t.h.

6.

is commonly used to denote change. For example, a change in height, h, can be represented

5.2

The First Law of Thermodynamics

In a chemical reaction, the initial state of the system refers to the reactants, and the final state refers to the products. When hydrogen and oxygen form water at a given temperature, the system loses energy to the surroundings. Because energy is lost from the system, the internal energy of the products (final state) is less than that of the reactants (initial state), and t.E for the process is negative. Thus, the energy diagram in Figure 5.5 � shows that the internal en­ ergy of the mixture of H2 and 02 is greater than that of H20. GIVE

IT SOME

Relating

11 £ to

H2(g), 02(g)

� � �

Heat and Work

As we noted in Section 5.1, a system may exchange energy with its surround­ ings as heat or as work. The internal energy of a system changes in magnitude as heat is added to or removed from the system or as work is done on it or by it. If we think of internal energy as the system's bank account of energy, we see that deposits or withdrawals can be made either in terms of heat or in terms of work. Deposits increase the energy of the system (positive t.E), whereas with­ drawals decrease the energy of the system (negative t.E). We can use these ideas to write a very useful algebraic expression of the first law of thermodynamics. When a system undergoes any chemical or physi­ cal change, the magnitude and sign of the accompanying change in internal energy, t.E, is given by the heat added to or liberated from the system, q, plus the work done on or by the system, w: t.E = q + w

AE < 0

(negative)

AE > 0

(positive)

�

THOUGHT

The internal energy for Mg(s) and Clz{g) is greater than that of MgClz(s). Sketch an energy diagram that represents the reaction MgCl2(s) -> Mg(s) + Cl 2(g).

171

.s A

Figure 5.5 Energy diagram for the lnterconverslon of H2(g), 02(g), and H20(I). A system composed of H 2(9) and 02(9) has a greater internal energy than one composed of H20(/). The system loses energy (Af < 0) when H2 and 02 are converted to H20. It gains energy (Af > 0) when H20 is decomposed into H2 and 02.

(5.5]

When heat is added to a system or work is done on a system, its internal energy in­ creases. Therefore, when heat is transferred to the system from the surroundings,

q has a positive value. Adding heat to the system is like making a deposit to the energy account-the total amount of energy goes up. Likewise, when work is done on the system by the surroundings, w has a positive value (Figure 5.6 � ). Work also is a deposit, increasing the internal energy of the system. Conversely, both the heat lost by the system to the surroundings and the work done by the system on the surroundings have negative values; that is, they lower the inter­ nal energy of the system. They are energy withdrawals and as a result, lower the total amount of energy in the energy account. The sign conventions for q, w, and t.E are summarized in Table 5.1 1'. Notice that any energy entering the sys­ tem as either heat or work carries a positive sign.

A

Figure 5.6 Sign conventions for heat and work. Heat, q, gained by a system and work, w, done on a system are both positive q uantities. Both increase the internal energy, E, of the system, causing Af to be a positive quantity.

TABLE 5. 1 • Sign Conventions for q, w, and AE For q For w For AE

•

+ means system gai11s heat

+ means work done 011 system + means 11et gai11 of energy by system

- means system loses heat - means work done by system - means 11et loss of energy by system

SAMPLE EXERCISE 5.2 1 Relating Heat and Work to Changes of Internal Energy

Two gases, A{g) and B{g), are confined in a cylinder-and-piston arrangement like that in Figure 5.3. Substances A and B react to form a solid product: A{g) + B(g) ---> C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decrea­ ses under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system?

172

C HA PTER 5

Thermochemistry

SOLUTION

Analyze: The question asks us to determine !1E, given information about q and w.

Plan: We first determine the signs of q and w (Table 5.1) and then use Equation 5.5, !1E = q + w, to calculate !1E.

Solve: Heat is transferred from the system to the surroundings, and work is done on the system by the surroundings, so q is negative and w is positive: q = -1150 J and w = 480 kJ. Thus, !1E = q

+

w = (-1150 J) + (480 J) = -670 J

The negative value of !1E tells us that a net quantity of 670 J of energy has been trans­ ferred from the system to the surroundings. Comment: You can think of this change as a decrease of 670 J in the net value of the system's energy bank account (hence the negative sign); 1150 J is withdrawn in the form of heat, while 480 J is deposited in the form of work. Notice that as the volume of the gases decreases, work is being done on the system by the surroundings, result­ ing in a deposit of energy.

- PRACTICE EXERCISE

Calculate the change in the internal energy of the system for a process in which the system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. Answer: +55 J

Endothermic and Exothermic Processes (a)

When a process occurs in which the system absorbs heat, the process is called (Endo- is a prefix meaning "into.") During an endothermic process, such as the melting of ice, heat flows into the system from its surround­ ings. If we, as part of the surroundings, touch a container in which ice is melt­ ing, it feels cold to us because heat has passed from our hands to the container. A process in which the system loses heat is called (Exo- is a pre­ fix meaning " out of.") During an exothermic process, such as the combustion of gasoline, heat exits or flows out of the system and into the surroundings. Figure 5.7 � shows two examples of chemical reactions: one endothermic and the other highly exothermic. In the endothermic process shown in Figure 5.7(a), the tem­ perature in the beaker decreases. In this example the system consists of the chemical reactants and products. The solvent in which they are dissolved is part of the surroundings. Heat flows from the solvent, as part of the surround­ ings, into the system as reactants are converted to products. Thus, the tempera­ ture of the solution drops.

endothermic.

exothermic.

GIVE IT SOME THOUGHT 4

(b)

Figure 5.7 Examples of

endothermic and exothermic reactions. (a) When ammonium thiocyanate and barium hydroxide octahydrate are mixed at room temperature, an endothermic reaction occurs: 2 NH4SCN(s) + Ba(OHh · 8 H20(s) -----> Ba(SCNh(aq) + 2 NH3(aq) + 1 0 H20(/). As a result, the temperature of the system drops from about 20 'C to -9 'C. (b) The reaction of powdered aluminum with Fe203 (the thermite reaction) is highly exothermic. The reaction proceeds vig o ro usly to form Al203 and molten iron: 2 Al(s) + Fe203(s) -----> AI203(s) + 2 Fe(/).

Using Figure 5.5 as a reference, indicate whether the reaction 2 H20(1) ----> 2 + is exothermic or endothermic. What feature(s) of the figure indicate whether the reaction is exothermic or endothermic?

H2(g) 02(g)

State Fu nctions Although we usually have no way of knowing the precise value of the internal energy of a system, E, it does have a fixed value for a given set of conditions. The conditions that influence internal energy include the temperature and pres­ sure. Furthermore, the total internal energy of a system is proportional to the total quantity of matter in the system because energy is an extensive property. (Section 1.3) Suppose we define our system as 50 g of water at 25 °C, as in Figure 5.8 10> . The system could have arrived at this state by cooling 50 g of water from 100 ac or by melting 50 g of ice and subsequently warming the water to 25 °C. cx:c

5.2

50 g

50 g

50 g

1oo •c

25 ·c

o ·c

Hp (l)

H20 (l)

173

The First Law of Thermodynamics

..,. Figure 5.8 Internal energy,

f,

a state function. E depends only on the present state of the system and not on the path by which it arrived at that state. The internal energy of 50 g of water at 25 •c is the same whether the water is cooled from a higher temperature to 25 •c or warmed from a lower temperature to 25 •c.

Hp (s)

The internal energy of the water at 25 •c is the same in either case. Internal en­ ergy is an example of a state function, a property of a system that is determined by specifying the system's condition, or state (in terms of temperature, pres­ sure, and so forth). The value of a state function depends only on the present state of the system, not on the path the system took to reach that state. Because E is a state function, !1E depends only on the initial and final states of the sys­ tem, not on how the change occurs. An analogy may help to explain the difference between quanti­ ties that are state functions and those that are not. Suppose you are traveling between Chicago and Denver. Chicago is 596 ft above sea level; Denver is 5280 ft above sea level. No matter what route you take, the altitude change will be 4684 ft. The distance you travel, however, will depend on your route. Altitude is analogous to a state function because the change in altitude is independent of the path taken. Distance traveled is not a state function. (a) Some thermodynamic quantities, such as E, are state functions. Other quantities, such as q and w, are not. Although !1E = q + w does not depend on how the change occurs, the specific amounts of heat and work produced depend on the way in which the change is carried out, analogous to the choice of travel route between Chicago and Denver. Nevertheless, if changing the path by which a system goes from an initial state to a final state increases the value of q, that path change will also decrease the value of w by exactly the same amount. The result is that the value for !1E for the two paths will be the same. We can illustrate this principle with the example of a flashlight battery as our system. In Figure 5.9 ... , we consider two possible ways of discharging the battery at constant temperature. If a coil of wire shorts out the battery, no work is accomplished because nothing is moved against a force. All the energy is lost from the battery in the form of heat. (The wire coil will get warmer and release heat to the surrounding air.) On the other hand, if the battery is used to make a small motor tum, the discharge of the battery produces work. Some heat will be released as well, although not as much as when the battery is shorted out. The magnitudes of q and w are different for these two cases. If the initial and final states of the battery are identical in both cases, however, then Ll.E = q + w must be the same in both cases because !1E is a state function. Thus, !1E depends only on the initial and final states of the system, regardless of how the transfers of energy occur in terms of heat and work.

\ I . '"�:�.,t )T

GIVE

IT SOME

THOUGHT

In what ways is the balance in your checkbook a state function?

Ch.., .

!

Heat

T wr

Discharged battery

(b)

_! !1£

Energy lost by battery

.A. Figure 5.9 Internal energy Is a state function, but heat and work are not. The amounts of heat and work transferred between the system and the surroundings depend on the way in which the system goes from one state to another. (a) A battery shorted out by a wire loses energy to the surroundings only as heat; no work is performed by the system. (b) A battery discharged through a motor loses energy as work (to make the fan turn) and also loses energy as heat. Now, however, the amount of heat lost is much less than in (a). The value of t.E is the same for both processes even though the values of q and w in (a) are different from the values of q and w in (b).

174

C HA PTER 5

Thermochemistry

5.3 ENTHALPY The chemical and physical changes that occur around us, such as photosynthe­ sis in the leaves of a plant, the evaporation of water from a lake, or a reaction in an open beaker in a laboratory, occur at essentially constant atmospheric pres­ sure. The changes can result in the release or absorption of heat or can be ac­ companied by work that is done by or on the system. The heat flow is the easiest change to measure, so we will begin to focus on that aspect of reactions. Nevertheless, we still need to account for any work that accompanies the process. Most commonly, the only kind of work produced by chemical or physical changes open to the atmosphere is the mechanical work associated with a change in the volume of the system. Consider, for example, the reaction of zinc metal with hydrochloric acid solution: 2 [5.6] Zn(s) + 2 H +(aq) --> Zn +(aq ) + H 2(g) If we carry out this reaction in the laboratory hood in an open beaker, we can see the evolution of hydrogen gas, but it may not be so obvious that work is being done. Still, the hydrogen gas that is being produced must expand against the existing atmosphere, which requires the system to do work. We can see this better by conducting the reaction in a closed vessel at constant pressure, as illustrated in Figure 5.10 T. In this apparatus the piston moves up or down to maintain a constant pressure in the reaction vessel. If we assume for simplic­ ity that the piston has no mass, the pressure in the apparatus is the same as the atmospheric pressure outside the apparatus. As the reaction proceeds, H 2 gas forms, and the piston rises. The gas within the flask is thus doing work on the surroundings by lifting the piston against the force of atmospheric pressure that presses down on it. The work involved in the expansion or compression of gases is called pressure-volume work (or P-V work). When the pressure is constant, as in our example, the sign and magnitude of the pressure-volume work is given by

w = -P ilV

[5.7] (llV =

where P is pressure and LlV is the change in volume of the system "'final - Vinitia l)· The negative sign in Equation 5.7 is necessary to conform to the sign conventions given in Table 5.1. Thus, when the volume expands, Ll V is a positive quantity and w is a negative quantity. That is, energy leaves the system as work, indicating that work is done by the system on the surround­ ings. On the other hand, when a gas is compressed, Ll V is a negative quantity

II> Figure 5.1 0 A system that does work on Its surroundings. (a) An apparatus for studying the reaction of zinc metal with hydrochloric acid at constant pressure. The piston is free to move up and down in its cylinder to maintain a constant pressure equal to atmospheric pressure inside the apparatus. Notice the pellets of zinc in the L-shaped arm on the left. When this arm is rotated, the pellets will fall into the main container and the reaction will begin. (b) When zinc is added to the acid solution, hydrogen gas is evolved. The hydrogen gas does work on the surroundings, raising the piston against atmospheric pressure to maintain constant pressure inside the reaction vessel.

(a)

(b)

5.3

Enthalpy

175

(the volume decreases), which makes w a positive quantity. That is, energy en­ ters the system as work, indicating that work is done on the system by the sur­ roundings. The "A Closer Look" box discusses pressure-volume work in more detail, but all you really need to keep in mind for now is Equation 5.7, which applies to processes occurring at constant pressure. We will take up the prop­ erties of gases in more detail in Chapter 10.

GIVE IT SOME THOUGHT If a system does not change its volume during the course of a process, does it do

pressure-volume work?

A thermodynamic function called enthalpy (from the Greek word

enthalpein, meaning "to warm") accounts for heat flow in processes occurring at constant pressure when no forms of work are performed other than P-V work. Enthalpy, which we denote by the symbol H, equals the internal energy plus the product of the pressure and volume of the system:

[5.8]

H = E + PV

Enthalpy is a state function because internal energy, pressure, and volume are all state functions. When a change occurs at constant pressure, the change in enthalpy, !:J.H, is given by the following relationship: !:J.H = !:J.(E + PV)

[5.9]

= !:J.E + P !:J.V

That is, the change in enthalpy equals the change in internal energy plus the product of the constant pressure times the change in volume. We can gain further insight into enthalpy change by recalling that !:J.E = q + w (Equation 5.5) and that the work involved in the expansion or compression of gases is w = -P !:J. V. If we substitute -w for P !:J. V and q + w for !:J.E into Equation 5.9, we have !:J.H = !:J.E + P !:J. V = (qp +

w) - w =

qp

[5.10]

where the subscript P on the heat, q, emphasizes changes at constant pressure. Thus, the change in enthalpy equals the heat gained or lost at constant pressure. Because qp is something we can either measure or readily calculate and because so many physical and chemical changes of interest to us occur at constant pressure, enthalpy is a more useful function than internal energy. For most reac­ tions the difference in !:J.H and !:J.E is small because P !:J. V is small. When !:J.H is positive (that is, when qp is positive), the system has gained heat from the surroundings (Table 5.1), which is an endothermic process. When !:J.H is negative, the system has released heat to the surroundings, which is an exothermic process. These cases are diagrammed in Figure 5.11 �. Because H is a state function, !:J.H (which equals qp) depends only on the initial and final states of the system, not on how the change occurs. At first glance this statement might seem to contradict our earlier discussion in Section 5.2, in which we said that q is not a state function. There is no contradic­ tion, however, because the relationship between !:J.H and heat (qp) has the spe­ cial limitations that only P-V work is involved and the pressure is constant.

GIVE IT SOME THOUGHT What is the advantage of using enthalpy rather than internal energy to describe energy changes in reactions?

Surroundings

!J.H > O (End oth ermic) Surroundings System

!J.H < O (Exothermic)

.;. Figure 5. 1 1 Endothermic and exothermic processes. (a) If the system absorbs heat (endothermic process), !J.H will be positive (!J.H > 0). (b) If the system loses heat (exothermic process), !J.H will be negative (!J.H < 0).

C HA PTER 5

176

Thermochemistry E N E RGY, E N T H A L P Y, AND P- V WORK

n chemistry we are interested mainly in two types of work:

I electrical work and mechanical work done by expanding

gases. We will focus here only on the latter, called pressure­ volume, or P-V, work. Expanding gases in the cylinder of an automobile engine do P-V work on the piston; this work even­ tually turns the wheels. Expanding gases from an open reac­ tion vessel do P-V work on the atmosphere. This work accomplishes nothing in a practical sense, but we must keep track of all work, useful or not, when monitoring the energy changes of a system. Consider a gas conlined to a cylinder with a movable pis­ ton of cross-sectional area A (Figure 5.12T). A downward force, F, acts on the piston. The pressure, P, on the gas is the force per area: P = FIA. We will assume that the piston is weightless and that the only pressure acting on it is the atmospheric pressure that is due to the weight of Earth's atmos­ phere, which we will assume to be constant. Suppose the gas in the cylinder expands and the piston moves a distance, IJ.h. From Equation 5.3, the magnitude of the

P

=

Initial state

F/A

P

r l

Cross-sectional area = A

=

IJ.h

[5.11]

Magnitude of work = =

F X IJ.h = P X A X IJ.h P X IJ.V

Because the system (the confined gas) is doing work on the surroundings, the work is a negative quantity: w=

[5 . 12]

-P IJ.V

Now, if P-V work is the only work that can be done, we can substitute Equation 5.12 into Equation 5.5 to give

IJ.E

=

q+

w=

q

-

P fJ. V

[5.13]

The subscript V indicates that the volume is constant. Most reactions are run trnder constant-presstue condi­ tions. In this case Equation 5.13 becomes

T

qp =

t.h

Final state

Magnitude of work = force x distance = F x

We can rearrange the definition of pressure, P = FIA, to F = P X A. In addition, the volume change, fl.V, resulting from the movement of the piston, is the product of the cross­ sectional area of the piston and the distance it moves: fl. V = A x IJ.h. Substituting into Equation 5.11,

When a reaction is carried out in a constant-volume container (fl. V = 0), the heat transferred equals the change in internal energy: [5.14] (constant volume)

F/A

_l

work done by the system equals the distance moved times the force acting on the piston:

Volume change

.A. Figure 5.12 Pressure-volume work. A piston moving upward, expanding the volume of the system against an external pressure, P, does work on the surroundings. The amount of work done by the system on the surroundings is w = - P IJ. V.

IJ.E = qp - P fJ. V or IJ.E + P fl.V (constant pressure)

[5.15]

But we see from Equation 5.9 that the right-hand side of Equa­ tion 5.15 is just the enthalpy change under constant-pressure conditions. Thus, IJ.H = qp, as we saw earlier in Equation 5.10. In summary, the change in internal energy is equal to the heat gained or lost at constant volume; the change in enthalpy is equal to the heat gained or lost at constant pressure. The dif­ ference between IJ.E and IJ.H is the amount of P- V work done by the system when the process occurs at constant pressure, -P fl. V. The volume change accompanying many reactions is close to zero, which makes P IJ.V, and therefore the difference between IJ.E and IJ.H, small. It is generally satisfactory to use IJ.H as the measure of energy changes during most chemical processes.

Related Exercises: 5.33, 5.34, 5.35, 5.36

- SAMPLE EXERCISE 5.3 I Determining the Sign of IJ.H

Indicate the sign of the enthalpy change, IJ.H, in each o f the following processes car­ ried out under atmospheric pressure, and indicate whether the process is endothermic or exothermic: (a) An ice cube melts; (b) 1 g of butane (C�1 0) is combusted in suffi­ cient oxygen to give complete combustion to C02 and H20.

SOLUTION Analyze: Our goal is to determine whether IJ.H is positive or negative for each

process. Because each process appears to occur at constant pressure, the enthalpy change of each one equals the amount of heat absorbed or released, IJ.H = qp.

5.4

Enthalpies of Reaction

Plan: We must predict whether heat is absorbed or released by the system in each process. Processes in which heat is absorbed are endothermic and have a positive sign for those in which heat is released are exothermic and have a negative sign for

!l.H. !l.H; Solve: In (a) the water that makes up the ice cube is the system. The ice cube absorbs heat from the surroundings as it melts, so !l.H is positive and the process is endother­ rllic. In (b) the system is the 1 g of butane and the oxygen required to combust it. The combustion of butane in oxygen gives off heat, so !l.H is negative and the process is exotherrllic.

- PRACTICE EXERCISE

Suppose we confine 1 g of butane and sufficient oxygen to completely combust it in a cylinder like that in Figure 5.12. The cylinder is perfectly insulating, so no heat can escape to the surroundings. A spark initiates combustion of the butane, which forms carbon dioxide and water vapor. If we used this apparatus to measure the enthalpy change in the reaction, would the piston rise, fall, or stay the same? Answer: The piston must move to maintain a constant pressure in the cylinder. The products contain more molecules of gas than the reactants, as shown by the balanced equation

2 C4H w(g) + 13 02(g) -----> 8 C02(g) +

10 H20(g)

As a result, the piston would rise to make room for the additional molecules of gas. Heat is given off, so the piston would also rise an additional amount to accommodate the expansion of the gases because of the temperature increase.

5.4 ENTHALPIES OF REACTION Because !lH = Hfinal - HinitiaJ, the enthalpy change for a chemical reaction is given by the enthalpy of the products minus the enthalpy of the reactants:

!lH

=

(5.16]

Hprod ucts - Hreactants

The enthalpy change that accompanies a reaction is called the enthalpy of reaction, or merely the heat of reaction, and is sometimes written !lHrxn' where "rxn" is a commonly used abbreviation for "reaction." The combustion of hydrogen is shown in Figure 5.13 T. When the reaction is controlled so that 2 mol H2{g) burn to form 2 mol H20{g) at a constant pres­ sure, the system releases 483.6 kJ of heat. We can summarize this information as 2 H2{g) + 02{g) ---> 2 H20{g)

!lH

= -483.6 kJ

[5.17]

2 H2(g) + Oz(g)

!l.H < O

(exothermic)

2 H20(g) (a)

(b)

(c)

.A. Figure 5.1 3 Exothermic reaction of hydrogen with oxygen. (a) A candle is held near a balloon filled with hydrogen gas and oxygen gas. (b) The H2(g) ignites, reacting with 02(g) to form H20(g). The resultant explosion produces a ball of flame. The system gives oft heat to its surroundings. (c) The enthalpy diagram for this reaction, showing its exothermic character.

177

178

C HAPTER 5

Thermochemistry

tJ.H is negative, so this reaction is exothermic. Notice that tJ. H is reported at the end of the balanced equation, without explicitly mentioning the amounts of chemicals involved. In such cases the coefficients in the balanced equation represent the number of moles of reactants and products producing the associ­ ated enthalpy change. Balanced chemical equations that show the associated enthalpy change in this way are called thermochemical equations.

GIVE IT SOME THOUGHT What information is summarized by the coefficients in a

thermochemical equation?

The enthalpy change accompanying a reac­ tion may also be represented in an enthalpy diagram such as that shown in Figure 5.13(c). Because the combustion of H2(g) is exothermic, the enthalpy of the products in the reaction is lower than the enthalpy of the reactants. The enthalpy of the system is lower after the reac­ tion because energy has been lost in the form of heat released to the surroundings. The reaction of hydrogen with oxygen is highly exothermic (tJ.H is negative and has a large magnitude), and it occurs rapidly once it starts. It can occur with explosive violence, too, as demonstrated by the disastrous explosions of the German airship Hindenburg in 1937 (Figure 5.14 .,.) and the space shuttle Challenger in 1986. The following guidelines are helpful when using thermochemical equations and enthalpy diagrams: .i.

Figure 5 . 1 4 The burning of the hydrogen-filled airship Hindenburg. This photograph was taken only 22 seconds after the first explosion occurred. This tragedy, which occurred in Lakehurst, New jersey, on May 6, 1 937, led to the discontinuation of hydrogen as a buoyant gas in such craft. Modern-day blimps are filled with helium, which is not as buoyant as hydrogen but is not flammable.

1. Enthalpy is an extensive property. The magnitude of tJ.H, therefore, is directly

proportional to the amount of reactant consumed in the process. For the combustion of methane to form carbon dioxide and liquid water, for exam­ ple, 890 kJ of heat is produced when 1 mol of CH4 is burned in a constant­ pressure system:

tJ.H = -890 kJ

[5.18]

Because the combustion of 1 mol of CH4 with 2 mol of 02 releases 890 kJ of heat, the combustion of 2 mol of C� with 4 mol of 02 releases twice as much heat, 1780 kJ. 2. The enthalpy change for a reaction is equal in magnitude, but opposite in sign, to

tJ.H for the reverse reaction. For example, if we could reverse Equation 5.18 so that CH4(g) and 02(g) formed from C02(g) and H20(1), tJ.. H for the process would be +890 kJ:

t.H,

=

-890 kj

.i.

t.Hz =

890 kj

Figure 5 . 1 5 t. H for a reverse reaction. Reversing a reaction changes the sign but not the magnitude of the enthalpy change: t.H2 = -t.H1 .

tJ.H = 890 kJ

[5.19]

When we reverse a reaction, we reverse the roles of the products and the re­ actants. As a result, the reactants in a reaction become the products of the reverse reaction, and so forth. From Equation 5.16, we can see that revers­ ing the products and reactants leads to the same magnitude, but a change in sign for tJ. Hrxn· This relationship is diagrammed for Equations 5.18 and 5.19 in Figure 5.15 ... . 3. The enthalpy change for a reaction depends on the state of the reactants and

products. If the product in the combustion of methane (Equation 5.18) were gaseous H20 instead of liquid H20, tJ.Hrxn would be -802 kJ instead of

5.5 -890 kJ. Less heat would be available for transfer to the surroundings be­ cause the enthalpy of H20(g) is greater than that of H20( I). One way to see this is to imagine that the product is initially liquid water. The liquid water must be converted to water vapor, and the conversion of 2 mol H20(1) to 2 mol H20(g) is an endothermic process that absorbs 88 kJ:

D.H =

+88 kJ

[5.20]

Thus, it is important to specify the states of the reactants and products in thermochemical equations. In addition, we will generally assume that the reactants and products are both at the same temperature, 25 oc, unless otherwise indicated.

- SAMPLE EXERCISE 5.4 1 Relating

!lH to Quantities of Reactants and Products

How much heat is released when 4.50 g of methane gas is burned in a constant­ pressure system? (Use the information given in Equation 5.18.)

SOLUTION Analyze: Our goal is to use a thermochemical equation to calculate the heat produced

when a specific amount of methane gas is combusted. According to Equation 5.18, 890 kJ is released by the system when 1 mol CH4 is burned at constant pressure = -890 kJ).

(llH Plan: Equation 5.18 provides us with a stoichiometric conversion factor: 1 mol

CH4 "' - 890 kJ. Thus, we can convert moles of CH4 to kJ of energy. First, however, we must convert grams of CH4 to moles of CH4. Thus, the conversion sequence is grams CH4 (given) --> moles CH4 --> kJ (unknown to be found).

By adding the atomic weights of C and 4 H, we have 1 mol CH4 = 16.0 g CH4. We can use the appropriate conversion factors to convert grams of CH4 to moles of CH4 to kilojoules:

Solve:

Heat = (4.50 g CH4)

(

1 mo! CH4 16_0 g CH4

)(

-890 kJ 1 mol CH4

)

= -250 kJ

The negative sign indicates that the system released 250 kJ into the surroundings.

- PRACTICE EXERCISE

Hydrogen peroxide can decompose to water and oxygen by the following reaction:

!lH = -196 kJ

Calculate the value of q when 5.00 g of H202(Z) decomposes at constant pressure. Answer: -14.4 kJ

In many situations it is valuable to know the enthalpy change associated with a given chemical process. As we will see in the following sections, AH can be determined directly by experiment or calculated from the known enthalpy changes of other reactions by invoking the first law of thermodynamics.

5 . 5 CALO RIMETRY The value of AH can be determined experimentally by measuring the heat flow accompanying a reaction at constant pressure. Typically, we can determine the magnitude of the heat flow by measuring the magnitude of the temperature change the heat flow produces. The measurement of heat flow is calorimetry; a device used to measure heat flow is a calorimeter.

Heat Capacity and Specific Heat The more heat an object gains, the hotter it gets. All substances change temper­ ature when they are heated, but the magnitude of the temperature change produced by a given quantity of heat varies from substance to substance.

Calorimetry

179

180

C HAPTER 5

Thermochemistry U S I N G E N T HA L P Y AS A G U I D E

f you hold a brick in the air and let it go, it will fall a s the

I force of gravity pulls it toward Earth. A process that is

thermodynamically favored to happen, such as a falling brick, is called a spontaneous process. A spontaneous process can be either fast or slow. Speed is not the issue in thermo­ dynamics. Many chemical processes are thermodynamically fa­ vored, or spontaneous, too. By 11Spontaneous," we do not mean that the reaction will form products without any inter­ vention. That can be the case, but often some energy must be imparted to get the process started. The enthalpy change in a reaction gives one indication as to whether the reaction is like­ ly to be spontaneous. The combustion of H2(g) and 02(g), for example, is a highly exothermic process:

t.. H = -242 kj Hydrogen gas and oxygen gas can exist together in a vol­ ume indefinitely without noticeable reaction occurring, as in Figure 5.13(a). Once initiated, however, energy is rapidly transferred from the system (the reactants) to the surround­ ings. As the reaction proceeds, large amounts of heat are re­ leased, which greatly increases the temperature of the reactants and the products. The system then loses enthalpy by transferring the heat to the surroundings. (Recall that the first law of thermodynamics indicates that the total energy of the system plus the surroundings will not change; energy is conserved.)

II

1.000 g 820(1)

T = 1s.s •c

+ 4.184 J of heat

1r

1.000 g 820(1)

�

T = 14.s •c

Figure 5.16 Specific heat of water. Specific heat ind icates the amount of heat that must be added to one gram of a substance to raise its temperature by 1 K (or 1 "C). Specific heats can vary slightly with temperature, so for precise measurements the temperature is specified. The specific heat of H20(/) at 1 4.5 oc is 4. 1 84 J/g-K; the addition of 4. 1 84 I of heat to 1 g of liquid water at this temperature raises the temperature to 1 5.5 "C. This amount of energy defines the calorie: 1 cal = 4.184 J. .A.

Enthalpy change is not the only consideration in the spontaneity of reactions, however, nor is it a foolproof guide. For example, the melting of ice is an endothermic process:

t..H = +6.01 kj Even though this process is endothermic, it is spontaneous at temperatures above the freezing point of water (0 °C). The re­ verse process, the freezing of water to ice, is spontaneous at temperatures below 0 oc. Thus, we know that ice at room tern­ perature will melt and that water put into a freezer at -20 °C will turn into ice. Both of these processes are spontaneous under different conditions even though they are the reverse of one another. In Chapter 19 we will address the spontaneity of processes more fully. We will see why a process can be sponta­ neous at one temperature, but not at another, as is the case for the conversion of water to ice. Despite these complicating factors, however, you should pay attention to the enthalpy changes in reactions. As a gener­ al observation, when the enthalpy change is large, it is the dominant factor in determining spontaneity. Thus, reactions for which is large and negative tend to be spontaneous. Reactions for which t.. H is large and positive tend to be sponta­ neous only in the reverse direction. The enthalpy of a reaction can be estimated in a number of ways. From these estimates, the likelihood of the reaction being thermodynamically favor­ able can be predicted.

t..H

Related Exercises: 5.45, 5.46

The temperature change experienced by an object when it absorbs a certain amount of heat is determined by its heat capacity, C. The heat capacity of an object is the amount of heat required to raise its temperature by 1 K (or 1 °C). The greater the heat capacity, the greater the heat required to produce a given increase in temperature. For pure substances the heat capacity is usually given for a specified amount of the substance. The heat capacity of one mole of a substance is called its molar heat capacity, Cm· The heat capacity of one gram of a substance is called its specific heat capacity, or merely its specific heat (Figure 5.16 .. ). The specific heat, C,, of a substance can be determined experimentally by mea­ suring the temperature change, !lT, that a known mass, m, of the substance undergoes when it gains or loses a specific quantity of heat, q : Specific heat =

(quantity of heat transferred) --:='--,----,-,--,­ -,-.,(grams of substance)

X

(temperature change)

q C, = m X !lT

[5.21]

For example, 209 J is required to increase the temperature of 50.0 g of water by 1.00 K. Thus, the specific heat of water is c

s

=

209 1 (50.0 g)(l.OO K)

1

= 4. 18 --

g-K

A temperature change in kelvins is equal in magnitude to the temperature change in degrees Celsius: !lT in K = !lT in °C. = (Section 1.4) When the

5.5

Calorimetry

181

TABLE 5.2 • Specific Heats of Some Substances at 298 K Elements Substance

Compounds

Specific Heat (J/g- K)

Substance

1.04 0.90 0.45 0.14

N2{g) Al(s) Fe(s) Hg(l)

H20(/) CH4{g) C02{g) CaC03(s)

Specific Heat

(Jfg-K)

4.18 2.20

0.84 0.82

sample gains heat (positive q), the temperature of the sample increases (positive llT). Rearranging Equation 5.21, we get q = C5

X

m X llT

[5.22]

Thus, we can calculate the quantity of heat that a substance has gained or lost by using its specific heat together with its measured mass and temperature change. The specific heats of several substances are listed in Table 5.2 A.. Notice that the specific heat of liquid water is higher than those of the other substances listed. For example, it is about five times as great as that of aluminum metal. The high specific heat of water affects Earth's climate because it makes the tem­ peratures of the oceans relatively resistant to change. It also is very important in maintaining a constant temperature in our bodies, as we will discuss in the "Chemistry and Life" box later in this chapter.

GIVE IT SOME THOUGHT Which substance in Table 5.2 will undergo the greatest temperature change when the same mass of each substance absorbs the same quantity of heat?

• I Relating Heat, Temperature Change, and Heat Capacity (a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 oc (about room temperature) to near its boiling point, 98 °C? The specific heat of water is 4.18 J/g-K. (b) What is the molar heat capacity of water?

SAMPLE EXERCISE 5.5

SOLUTION Analyze: In part (a) we must find the quantity(b)of heat (q) needed to warm the water, given the mass of water (m}, its temperature

change (LlT}, and its specific heat (C,). In part from its specific heat (heat capacity per gram).

we must calculate the molar heat capacity (heat capacity per mole, C.,) of water

(a) Given C, m, and LlT, we can calculate the quantity of heat, q, using Equation 5.22. (b) We can use the molar mass of water and dimensional analysis to convert from heat capacity per gram to heat capacity per mole.

Plan: Solve:

(a) The water undergoes a temperature change of

Using Equation 5.22, we have

q = C, X

m

X

LlT

= (4.18 J/g-K)(250 g)(76 K) = 7.9

(b) The molar heat capacity is the heat capacity of one mole of substance. Using the atomic weights of hydrogen and oxygen, we have

1 mol H20 = 18.0 g HP

From the specific heat given in part (a}, we have

C.,

- PRACTICE EXERCISE

=

(4.18 )( 18.0 ) J

g-K

g

1 mol

X 104 J

= 75.2 J/mol-K

(a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g-K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 oc. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? Answers: (a) 4.9 X 105 J, (b) 11 K decrease = 11 °C decrease

C HA PTER 5

182

Thermochemistry

/ Thermometer /

. Glass stirrer

/

Cork stopper

---... · Two Styrofoam® cups nested together containing reactants in solution

Figure 5. 1 7 Coffee-cup calorimeter. This simple apparatus is .A.

used to measure heat-accompanying reactions at constant pressure.

Constant-Pressure Calorimetry The techniques and equipment employed in calorimetry depend on the nature of the process being studied. For many reactions, such as those occurring in so­ lution, it is easy to control pressure so that !:J.H is measured directly. (Recall that !:J.H = qp) Although the calorimeters used for highly accurate work are preci­ sion instruments, a very simple "coffee-cup" calorimeter, as shown in Figure 5.17 6 C0 2(g) + 6 HP(l)

t.H = -2803

kJ

Roughly 40% of the energy produced is ultimately used to do work in the form of muscle contractions and nerve cell activi­ ties. The remainder of the energy is released as heat, part of which is used to maintain body temperature. When the body produces too much heat, as in times of heavy physical exer­ tion, it dissipates the excess to the surroundings. Heat is transferred from the body to its surroundings pri­ marily by

radiation, convection, and evaporation. Radiation is the

direct loss of heat from the body to cooler surroundings, much as a hot stovetop radiates heat to its surroundings. Convection is heat loss by virtue of heating air that is in contact with the body. The heated air rises and is replaced with cooler air, and the process continues. Warm clothing, which usually consists of insulating layers of material with "dead air" in between, de­ creases convective heat loss in cold weather. Evaporative cool­ ing occurs when perspiration is generated at the skin surface by the sweat glands. Heat is removed from the body as the perspiration evaporates into the surroundings. Perspiration is predominantly water, so the process involved is the endother­ mic conversion of liquid water into water vapor:

t.H = +44.0

kJ

A Figure 5.1 9 Marathon runner drinking water. Runners must constantly replenish the water in their bodies that is lost through perspiration.

186

C HAPTER 5

Thermochemistry form C02(g) and liquid water can be thought of as occurring in two steps: (1) the combustion of CH4(g) to form C02(g) and gaseous water, H20(g) and (2) the condensation of gaseous water to form liquid water, H20(!). The enthalpy change for the overall process is simply the sum of the enthalpy changes for these two steps: �H = -802 kJ

CH4(g) + 2 02(g) -----4 C02(g) + 2 H20(g) 2 H20(g) -----4 2 H20(I)

(Add)

CH4(g) + 2 02(g) + 2 H20(g) -----4 C02 (g) + 2 H20(I) + 2 H20(g)

�H = - 88 kJ

�H = -890 kJ

The net equation is CH4(g) + 2 Oz(g) -----4 COz(g) + 2 HzO(l)

>.

iJi

-890 kJ

-802 kJ

C02(g) + 2 HzO(g) -88 kJ C02(g) + 2 H20(/)

.A. Figure 5.20 An enthalpy diagram comparing a one-step and a two-step process for a reaction. The enthalpy change of the direct reaction on the left equals the sum of the two steps on the right. That is, ll.H for the overall reaction equals the sum of the ll.H values for the two steps shown.

�H = -890 kJ

To obtain the net equation, the sum of the reactants of the two equations is placed on one side of the arrow and the sum of the products on the other side. Because 2 H20(g) occurs on both sides, it can be canceled like an algebraic quantity that appears on both sides of an equal sign. Figure 5.20

C02{g)

C02{g) -----> CO{g) +

I:!.H1 = -393.5 kJ -I:!.H2 = 283.0 kJ

! Oz{g)

I:!.H3 = -1 10.5 kJ When we add the two equations, C02{g) appears on both sides of the arrow and therefore cancels out. Likewise, 02{g) is eliminated from each side.

!

Comment: It is sometimes useful to add subscripts to the enthalpy changes, as we

have done here, to keep track of the associations between the chemical reactions and their I:!.H values.

- PRACTICE EXERCISE

Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is -393.5 kJ/rnol and that of diamond is -395.4 kJ/mol:

C(graphite)

I:!.H1 = -393.5 kJ

+ 02{g) -----> C02{g)

I:!.H2 = -395.4 kJ

Calculate I:!.H for the conversion of graphite to diamond:

C(graphite)

----->

C(diamond)

Answer: I:!.H3 = + 1.9 kJ •

SAMPLE EXERCISE 5.9 1 Using Three Equations with Hess's law to Calculate

Calculate I:!.H for the reaction 2 C(s)

+ Hz{g)

1:!. H

----->

CzHz{g)

given the following chemical equations and their respective enthalpy changes:

CzHz{g) + l Oz{g) C(s) + 02{g) H2{g) + ! 02 {g)

-----> 2 C02 {g) + H20(1) ----->

C02{g)

I:!.H = -1299.6 kJ

I:!.H =

I:!.H =

-----> H20(l)

-393.5 kJ

-285.8 kJ

SOLUTION Analyze: We are given a chemical equation and asked to calculate its I:!.H using three chemical equations and their associated enthalpy changes. Plan: We will use Hess's law, summing the three equations or their reverses and

multiplying each by an appropriate coefficient so that they add to give the net equa­ tion for the reaction of interest. At the same time, we keep track of the I:!.H values, re­ versing their signs if the reactions are reversed and multiplying them by whatever coefficient is employed in the equation.

Solve: Because the target equation has C2H2 as a product, we turn the first equation

around; the sign of I:!.H is therefore changed. The desired equation has 2 C(s) as a re­ actant, so we multiply the second equation and its I:!.H by 2. Because the target equa­ tion has H2 as a reactant, we keep the third equation as it is. We then add the three equations and their enthalpy changes in accordance with Hess's law: 2 C02{g) + H20(1) -----> C2H2{g) + 2 C(s) + 2 02{g) -----> 2 C02{g) Hz{g) +

l 02{g)

i Oz{g) -----> H20(1)

2 C(s) +

H 2{g)

----->

C2H2{g)

I:!.H = 1299.6 kJ

I:!.H = -787.0 kJ I:!.H = -285.8 kJ

l

I:!.H = 226.8 kJ

When the equations are added, there are 2 COz, 02, and H20 on both sides of the arrow. These are canceled in writing the net equation.

Hess's Law

187

188

C HA PTER 5

Thermochemistry Check: The procedure must be correct because we obtained the correct net equation.

1n cases like this you should go back over the numerical manipulations of the !lH values to ensure that you did not make an inadvertent error with signs.

- PRACTICE EXERCISE

Calculate !lH for the reaction NO(g) + O(g) -> N02(g)

given the following information: !lH = - 198.9 kJ

NO(g) + 03(g) -> NOz(g) + Oz(g) 03(g)

-->

� Oz(g)

!lH = -142.3 kJ

!l H

Oz(g) -> 2 O(g)

Answer: -304.1 k]

Mf2 =

-6

07 kJ

CO(g) + 2 HzO{l) 1 0z(g) +2

1

Mf3 = -283 kJ

_..:..,__:..._ .._ C02(g) + 2 H20(1)

A 5.21 An enthalpy diagram Illustrating Hess's law. Because H is a state function, the enthalpy change for the combustion of 1 mol CH4 is independent of whether the reaction takes place in one or more steps: t.H1 = t.H2 + t.H3.

Figure

= 495.0 kJ

The key point of these examples is that H is a state function, so for a particular set of reactants and products, !:J.H is the same whether the reaction takes place in one step or in a series of steps. For example, consider the reaction of methane (CH4)

and oxygen (02) to form C02 and H20. We can envision the reaction forming C02 directly or with the initial formation of CO, which is then combusted to C02. These two paths are compared in Figure 5.21

CaO(s) + C02{g)

is 178.1 kj. From the values for the standard enthalpies of formation of CaO(s) and C02{g) given in Table 5.3, calculate the stan­ dard enthalpy of formation of CaC03 (s).

SOLUTION Analyze: We need to obtain llHj (CaC03). Plan: We begin by writing the expression for the standard enthalpy change for the reaction: Solve: Inserting the given !l.H� and the known !l.Hj values

j

from Table 5.3 or Appendix C, we have

178.1 kJ = -635.5 kj - 393.5 kj - !l.H (CaC0 3)

Solving for !l.Hj (CaC03) gives

!l.Hj (CaC0 3) = -1207.1 k)/mol

Check: We expect the enthalpy of formation of a stable solid such as calcium carbonate to be negative, as obtained.

5.8

Foods and Fuels

193

- PRACTICE EXERCISE

Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard en­ thalpy of formation of CuO(s):

Answer: -156.1 k)/mol

CuO(s) + H 2(g) -----> Cu(s) + H20(1)

!!.W

=

-129.7 kj

5.8 FOODS AND FUELS Most chemical reactions used for the production of heat are combustion reac­ tions. The energy released when one gram of a material is combusted is often called its Although fuel values represent the heat released in a com­ bustion reaction, fuel values are reported as positive numbers. The fuel value of any food or fuel can be measured by calorimetry.

fuel value.

Foods Most of the energy our bodies need comes from carbohydrates and fats. The forms of carbohydrate known as starch are decomposed in the intestines into glucose, C6H 1206. Glucose is soluble in blood, and in the human body it is known as blood sugar. It is transported by the blood to cells, where it reacts with 02 in a series of steps, eventually producing C02 (g), H20(/), and energy:

t:.H"

= -2803 kJ

The breakdown of carbohydrates is rapid, so their energy is quickly supplied to the body. However, the body stores only a very small amount of carbohydrates. The average fuel value of carbohydrates is 17 kJ/g (4 kcalfg). Like carbohydrates, fats produce C02 and H20 when metabolized and when subjected to combustion in a bomb calorimeter. The reaction of tristearin, Cs7Hno06, a typical fat, is as follows: 2 Cs7H no06(s) + 163 02(gl

-----+

114 C02(g) + 110 H 20(l)

t:.H"

= -75,520 kJ

The body uses the chemical energy from foods to maintain body temperature (see the "Chemistry and Life" box in Section 5.6), to contract muscles, and to construct and repair tissues. Any excess energy is stored as fats. Fats are well suited to serve as the body's energy reserve for at least two reasons: (1) They are insoluble in water, which facilitates storage in the body; and (2) they produce more energy per gram than either proteins or carbohydrates, which makes them efficient energy sources on a mass basis. The average fuel value of fats is 38 kJ/g (9 kcal/g). The metabolism of proteins in the body produces less energy than com­ bustion in a calorimeter because the products are different. Proteins contain nitrogen, which is released in the bomb calorimeter as N 2. In the body this nitrogen ends up mainly as urea, (NH2)2CO. Proteins are used by the body mainly as building materials for organ walls, skin, hair, muscle, and so forth. On average, the metabolism of proteins produces 17 kJ/g (4 kcalfg), the same as for carbohydrates. The fuel values for a variety of common foods are shown in Table 5.4 "'· Labels on packaged foods show the amounts of carbohydrate, fat, and pro­ tein contained in an average serving, as well as the amount of energy sup­ plied by a serving (Figure 5.23 11> ). The amount of energy our bodies require varies considerably depending on such factors as weight, age, and muscu­ lar activity. About 100 kJ per kilogram of body weight per day is required to keep the body functioning at a minimal level. An average 70-kg (154-lb) person expends about 800 kJ/hr when doing light work, such as slow A Figure 5.23 Labels of processed foods walking or light gardening. Strenuous activity, such as running, often re­ quires 2000 kJ/hr or more. When the energy value, or caloric content, of our food exceeds the energy we expend, our body stores the surplus as fat.

showing nutritional Information. Such labels give information about the quantities of different nutrients and the energy value (caloric value) in an average serving.

194

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Thermochemistry

TABLE 5.4 • Compositions and Fuel Values of Some Common Foods Approximate Composition (% by mass) Fat

Carbohydrate Carbohydrate Fat Protein Apples Beer* Bread Cheese Eggs Fudge Green beans Hamburger Milk (whole) Peanuts

Protein

kJ/g

kcal/g (Cal/g)

100 0.4 0.3 9 28 13 2 1.9 22 3.3 26

17 38 17 2.5 1.8 12 20 6.0 18 1 .5 15 3.0 23

4 9 4 0.59 0.42 2.8 4.7 1.4 4.4 0.38 3.6 0.74 5.5

100 100 13 1.2 52 4 0.7 81 7.0 5.0 22

0.5 3 37 10 11 30 4.0 39

Fuel Value

' Beers typically contain 3.5% ethanol, which has fuel value.

GIVE IT SOME THOUGHT Which releases the greatest amount of energy per gram upon metabolism, carbohy­ drates, proteins, or fats?

- SAMPLE EXERCISE 5.13 [ Comparing Fuel Values

A plant such as celery contains carbohydrates in the form of starch and cellulose. These two kinds of carbohydrates have essentially the same fuel values when com­ busted in a bomb calorimeter. When we consume celery, however, our bodies re­ ceive fuel value from the starch only. What can we conclude about the difference between starch and cellulose as foods?

SOLUTION

If cellulose does not provide fuel value, we must conclude that it is not converted in

the body into C02 and H20, as starch is. A slight, but critical, difference in the struc­ tures of starch and cellulose explains why only starch is broken down into glucose in the body. Cellulose passes through without undergoing significant chemical change. It serves as fiber, or roughage, in the diet, but provides no caloric value.

- PRACTICE EXERCISE

The nutritional label on a bottle of canola oil indicates that 10 g of the oil has an ener­ gy value of 86 kcal. A similar label on a bottle of pancake syrup indicates that 60 mL (about 60 g) has an energy value of 200 kcal. Account for the difference. Answer: The oil has a fuel value of 8.6 kcal/ g, whereas the syrup has a fuel value of about 3.3 kcalfg. The higher fuel value for the canola oil arises because the oil is essentially pure fat, whereas the syrup is a solution of sugars (carbohydrates) in water. The oil has a higher fuel value per gram; in addition, the syrup is diluted by water.

•

SAMPLE EXERCISE 5.14 [ Estimating the Fuel Value of a Food from Its Composition

(a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these kinds of substances, estimate the energy value (caloric content) of this serving. (b) A person of average weight uses about 100 Caljmi when running or jogging. How many servings of this cereal provide the energy value requirements for running 3 mi?

SOLUTION (a) Analyze: The energy value of the serving will be the sum of the energy values of the protein, carbohydrates, and fat.

5.8 We are given the masses of the protein, carbohydrates, and fat contained in the combined . We can use the data in Table 5.4 to convert these masses to their energy values, which we can sum to get the total energy value.

Plan:

7 kJ ) + (26 g carbohydrate) ( Solve: (8 g protein) ( 1 g1protem �� ) + . 1 g car o ydrate (2 g fat)

This corresponds to 160 kcal: (650 kJ)

( :� ) 3

1

t

G.:;�J)

= 650 kJ (to two significant figures)

= 160 kcal

Recall that the dietary Calorie is equivalent to 1 kcal. Thus, the serving provides 160 Cal. Here we are faced with the reverse problem, calculating the quantity of food that provides a specific energy value.

(b) Analyze: Plan: The problem statement provides a conversion factor between Calories and

miles. The answer to part (a) provides us with a conversion factor between servings and Calories.

We can use these factors in a straightforward dimensional analysis to deter­ mine the number of servings needed, rounded to the nearest whole number:

Solve:

(

100 Cal Servings = (3 mi) �

)(

1 serving 160 Cal

)

= 2 servings

- PRACTICE EXERCISE

(a) Dry red beans contain 62% carbohydrate, 22% protein, and 1.5% fat. Estimate the fuel value of these beans. (b) During a very light activity, such as reading or watching television, the average adult uses about kJ/min . How many minutes of such activi­ ty can be sustained by the energy provided by a serving of chicken noodle soup con­ taining 13 g protein, 15 g carbohydrate, and 5 g fat? Answers: (a) 15 kJ/g, (b) 95 min

7

Fuels The elemental compositions and fuel values of several common fuels are com­ pared in Table 5.5 T. During the complete combustion of fuels, carbon is con­ verted to C02 and hydrogen is converted to H20, both of which have large negative enthalpies of formation. Consequently, the greater the percentage of carbon and hydrogen in a fuel, the higher its fuel value. Compare, for example, the compositions and fuel values of bituminous coal and wood. The coal has a higher fuel value because of its greater carbon content. In 2005 the United States consumed 1.05 X 10 17 kJ of energy. This value corresponds to an average daily energy consumption per person of 9.6 X 105 kJ which is roughly 100 times greater than the per capita food-energy needs.

TABLE 5.5 • Fuel Values and Compositions of Some Common Fuels Approximate Elemental Composition (mass %)

Wood (pine) Anthracite coal (Pennsylvania) Bituminous coal (Pennsylvania) Charcoal Crude oil (Texas) Gasoline Natural gas Hydrogen

c

H

0

Fuel Value (kJ/g)

50 82

6 1 5 0 12 15 23 100

44 2

18 31 32 34 45 48 49 142

77

100 85 85 70 0

7 0 0 0 0 0

Foods and Fuels

195

196

C HA P T E R 5

Thermochemistry

Renewable Nuclear energy (6.0%) (8.1%}

i

Although the population of the United States is only about 4.5% of the world's population, the U.S. accounts for nearly one-fourth of the world's total energy consumption. Figure 5.24 ..,. illustrates the sources of this energy. Coal, petroleum, and natural gas, which are the world's major sources of en­ ergy, are known as All have formed over millions of years from the decomposition of plants and animals and are being depleted far more rapidly than they are being formed. consists of gaseous hydrocarbons, com­ pounds of hydrogen and carbon. It contains primarily methane (C�), with small amounts of ethane (C2H6), propane (C3Hg), and butane (C,J110). We deter­ mined the fuel value of propane in Sample Exercise is a liquid composed of hundreds of compounds, most of which are hydrocarbons, with the remainder being chiefly organic compounds containing sulfur, nitrogen, or oxygen. which is solid, contains hydrocarbons of high molecular weight as well as compounds containing sulfur, oxygen, or nitrogen. Coal is the most abundant fossil fuel; it constitutes 80% of the fossil fuel reserves of the United States and 90% of those of the world. However, the use of coal presents a num­ ber of problems. Coal is a complex mixture of substances, and it contains com­ ponents that cause air pollution. When coal is combusted, the sulfur it contains is converted mainly to sulfur dioxide, S02, a very troublesome air pollutant. Because coal is a solid, recovery from its underground deposits is expensive and often dangerous. Furthermore, coal deposits are not always close to locations of high-energy use, so there are often substantial shipping costs. One promising way to utilize coal reserves is to use them to produce a mix­ ture of gaseous hydrocarbons called syngas (for syn thesis gas"). In this process, called coal gasification, the coal typically is pulverized and treated with super­ heated steam. Sulfur-containing compounds, water, and carbon dioxide can be removed from the products, leading to a gaseous mixture of CH4, H 2, and CO, all of which have high fuel values:

fossil fuels. Natural gas

5.11. Petroleum

.6.

Coal,

Figure 5.24 Sources of energy

consumed In the United States. In 2005 the United States consumed a total 1 of 1 .05 x 1 0 7 k) of energy.

"

Coal + steam

conversion

complex mixture

purification

mixture of CH4, H;u CO (syngas)

T H E H Y B R I D CAR he hybrid cars now entering the automobile marketplace

T nicely illustrate the convertibility of energy from one

form to another. Hybrid cars run on either gasoline or electric­ ity. The so-called "full hybrids" are cars capable of running at low speeds on a battery -powered electrical engine alone (Figure 5.25 ,. ). The "mild hybrid" cars are best described as electrically assisted gasoline engines. Full hybrid cars are more efficient than the mild hybrid de­ signs but are more costly to produce and require more techno­ logical advances than the mild hybrid versions. The mild hybrids are likely to be more widely produced and sold within the next several years. Let's consider how they operate and some of the interesting thermodynamic considerations they incorporate. Figure 5.26 ,. shows a schematic diagram of the power system for a mild hybrid car. In addition to the 12-volt battery that is standard on conventional autos, the hybrid car typical­ ly carries a 48-volt battery pack. The electrical energy from this battery pack is not employed to move the car directly; an electrical engine capable of doing that, as in the full hybrids, requires from 150 to 300 volts. (The popular Toyota Prius has a battery pack consisting of 228 cells of 1.2 V each, thereby gen­ erating a nominal voltage of 270 V.)

.6. Figure 5.25 A hybrid car. The Lexus GS450h is the first rear-drive full hybrid luxury car, evidence of the widespread movement of hybrid cars into the marketplace.

In the mild hybrid cars the added electrical source is em­ ployed to run various auxiliary devices that would otherwise be run off the gasoline engine, such as water pump, power steering, and air systems. To save on energy, when the hybrid

5.8

Foods and Fuels

197

Because it is gaseous, syngas can be easily transported in pipelines. Additional­ ly, because much of the sulfur in coal is removed during the gasification process, combustion of syngas causes less air pollution than burning coal. For these reasons, the economical conversion of coal and petroleum into "cleaner" fuels such as syngas and hydrogen is a very active area of current research in chemistry and engineering.

Other Energy Sources Nuclear energy is energy that is released in either the splitting or the fusion

(combining) of the nuclei of atoms. Nuclear power is currently used to produce about 22% of the electric power in the United States and comprises about 8% of the total U.S. energy production (Figure 5.24). Nuclear energy is, in principle, free of the polluting emissions that are a major problem in the generation of energy from fossil fuels. However, nuclear power plants produce radioactive waste products, and their use has therefore been fraught with controversy. We will discuss issues related to the production of nuclear energy in Chapter 21 . Fossil fuel and nuclear energy are nonrenewable sources of energy; they are limited resources that we are consuming at a much greater rate than they are being regenerated. Eventually these fuels will be expended, although estimates vary greatly as to when this will occur. Because nonrenewable sources of energy will eventually be used up, a great deal of research is being conducted into sources of renewable energy, energy sources that are essentially inexhaustible. Renewable energy sources include solar energy from the Sun, wind energy har­ nessed by windmills, geothennal energy from the heat stored in the mass of Earth, hydroelectric energy from flowing rivers, and biomass energy from crops, such as trees and corn, and from biological waste matter. Currently, renewable sources provide about 6.0% of the U.S. annual energy consumption, with hydroelectric and biomass sources as the major contributors. Providing our future energy needs will most certainly depend on develop­ ing the technology to harness solar energy with greater efficiency. Solar energy is the world's largest energy source. On a clear day about 1 kJ of solar energy

car comes to a stop, the engine shuts off. It restarts automati­ cally when the driver presses the accelerator. This feature saves fuel that would otherwise be used to keep the engine idling at traffic lights and other stopping situations.

Transmission A Figure 5.26 Schematic diagram of a mild hybrid car. The 48-volt battery pack provides energy for operating several auxiliary functions. It is recharged from the engine and through the braking system.

The idea is that the added electrical system will improve overall fuel efficiency of the car. The added battery, more­ over, is not supposed to need recharging from an external power source. Where, then, can the improved fuel efficiency come from? Clearly, if the battery pack is to continue to oper­ ate auxiliary devices such as the water pump, it must be recharged. We can think of it this way: The source of the volt­ age that the battery develops is a chemical reaction. Recharg­ ing the battery thus represents a conversion of mechanical energy into chemical potential energy. The recharging occurs in part through the agency of an alternator, which runs off the engine and provides a recharging voltage. In the mild hybrid car, the braking system serves as an additional source of mechanical energy for recharging. When the brakes are applied in a conventional car, the car's kinetic energy is con­ verted through the brake pads in the wheels into heat, so no useful work is done. In the hybrid car, some of the car's kinetic energy is used to recharge the battery when the brakes are applied. Thus, kinetic energy that would other­ wise be dissipated as heat is partially converted into useful work. Overall, the mild hybrid cars are expected to yield 10-20% improvements in fuel economy as compared with similar conventional cars.

198

C HA PTER 5

Thermochemistry reaches each square meter of Earth's surface every second. The average solar en­ ergy that falls on only 0.1% of U.S. land area is equivalent to all the energy that this nation currently uses. Harnessing this energy is difficult because it is dilute (it is distributed over a wide area) and it varies with time of day and weather con­ ditions. The effective use of solar energy will depend on the development of some means of storing the collected energy for use at a later time. Any practical means for doing this will almost certainly involve use of an endothermic chemical pro­ cess that can be later reversed to release heat. One such reaction is the following: CH4(g) + HzO(g) + heat � CO(g) + 3 Hz(g) This reaction proceeds in the forward direction at high temperatures, which can be obtained in a solar furnace. The CO and H2 formed in the reaction could then be stored and allowed to react later, with the heat released being put to useful work. A survey taken about 25 years ago at Walt Disney's EPCOT Center revealed that nearly 30% of the visitors expected that solar energy would be the principal source of energy in the United States in the year 2000. The future of solar energy has proven to be a lot like the Sun itself: big and bright but farther away than it seems. Nevertheless, important progress has been made in recent years. Perhaps the most direct way to make use of the Sun's energy is to convert it directly into electricity by use of photovoltaic devices, sometimes called solar cells. The effi­ ciencies of solar energy conversion by use of such devices have increased dra­ matically during the past few years because of intensive research efforts. Photovoltaics are vital to the generation of power for the space station. More sig­ nificant for our Earth-bound concerns, the unit costs of solar panels have been steadily declining, even as their efficiencies have improved dramatically. In 2006 construction was started in southern Portugal on what the builders claim will be the world's biggest solar energy power station. The first module of the station is planned to cover about 150 acres and is capable of generating 11 MW (megawatts) of electrical power-enough for 8000 homes. When fully constructed, the plant is projected to cover 620 acres and supply over 100 MW of power. Several other large solar plants with capacities over 100 MW have also been announced in Australia, Israel, and China.

- SAMPLE INTEGRATIVE EXERCISE J Putting Concepts Together

Trinitroglycerin, C3H5N309 (usually referred to simply as nitroglycerin), has been widely used as an explosive. Alfred Nobel used it to make dynamite in 1 866. Rather surprisingly, it also is used as a medication, to relieve angina (chest pains resulting from partially blocked arteries to the heart) by dilating the blood vessels. The en­ thalpy of decomposition at 1 atm pressure of trinitroglycerin to form nitrogen gas, carbon dioxide gas, liquid water, and oxygen gas at 25 oc is -1541.4 kJ/mol. (a) Write a balanced chemical equation for the decomposition of trinitroglycerin. (b) Calculate the standard heat of formation of trinitroglycerin. (c) A standard dose of trinitro­ glycerin for relief of angina is 0.60 mg. If the sampie is eventually oxidized in the body (not explosively, though!) to nitrogen gas, carbon dioxide gas, and liquid water, what number of calories is released? (d) One common form of trinitroglycerin melts at about 3 oc. From this information and the formula for the substance, would you expect it to be a molecular or ionic compound? Explain. (e) Describe the various con­ versions of forms of energy when trinitroglycerin is used as an explosive to break rockfaces in highway construction.

SOLUTION

(a) The general form of the equation we must balance is C3HsN309(I) --> N2(g) + C02(g) + H20(/) + 02(g) We go about balancing in the usual way. To obtain an even number of nitrogen atoms on the left, we multiply the formula for C3H5N309 by 2. This then gives us 3 mol of N2, 6 mol of C02 and 5 mol of H20. Everything is balanced except for oxygen. We have an odd number of oxygen atoms on the right. We can balance the oxygen by adding mol of 02 on the right:

1

2 C3H5NP9(1)

-->

3 N2(g) + 6 C02(g) + 5 H20(1) +

1 02(g)

Summary and Key Terms

199

We multiply through by 2 to convert all coefficients to whole numbers:

4 C3H5Np9(1)

----->

6 N2(g)

+

12 C02(g)

+

10 Hp(l)

+

02(g)

(At the temperature of the explosion, water is a gas. The rapid expansion of the gaseous products creates the force of an explosion.) (b) The heat of formation is the enthalpy change in the balanced chemical equation: 3 C(s) +

� Nz(g) + � Hz(g) + � Oz(g) -----> C3HsN309(I)

!lHJ =

?

We can obtain the value of !lHJ by using the equation for the heat of decomposition of trinitroglycerin:

4 C3HsNP9(l) -----> 6 N2(g)

+ 12 C02(g) +

10 H 20(l)

+

02(g)

The enthalpy change in this reaction is 4(-1541.4 kJ) = -6165.6 kJ. [We need to multi­ ply by 4 because there are 4 mol of C�sN309(/) in the balanced equation.] This enthalpy change is given by the sum of the heats of formation of the products minus the heats of formation of the reactants, each multiplied by its coefficient in the balanced equation:

-6165.6 kJ = {6!lHj[N2(g)]

+

12!lHJ[C02(g)]

+

10!lHj[H20(1)] + llHJ[02(g)]} - 4llHj(C3HsN309(/)]

!lHJ values for N2(g) and 02(g) are zero, by definition. We look up the values for H20(l) and C02(g) from Table 5.3 and find that -6165.6 kJ = 12(-393.5 kJ) + 10(- 285.8 kJ) - 4!lHJ(C3H5N309(1)) llHJ(C3HsNP9(I)) = -353.6 k)/mol (c) We know that on oxidation 1 mol of C3H5NP9(I) yields 1541.4 kJ. We need to cal­ culate the number of moles of in C3H5N309(1) in 0.60 mg: 1 mol C3H5Np9 1541.4 kJ 0.60 X 10_3 g C3HsN309 = 4.1 X 10_3 kJ 227 g C3HsN309 1 mol CJHsNPg = 4.1 J

The

(

)(

)

(d) Because trinitroglycerin melts below room temperature, we expect that it is a molecular compound. With few exceptions, ionic substances are generally hard, crys­ talline materials that melt at high temperatures. = (Sections 2.5 and 2.6) Also, the molecular formula suggests that it is likely to be a molecular substance. All the ele­ ments of which it is composed are nonmetals. (e) The energy stored in trinitroglycerin is chemical potential energy. When the sub­ stance reacts explosively, it forms substances such as carbon dioxide, water, and ni­ trogen gas, which are of lower potential energy. In the course of the chemical transformation, energy is released in the form of heat; the gaseous reaction products are very hot. This very high heat energy is transferred to the surroundings; the gases expand against the surroundings, which may be solid materials. Work is done in moving the solid materials and imparting kinetic energy to them. For example, a chunk of rock might be impelled upward. It has been given kinetic energy by transfer of energy from the hot, expanding gases. As the rock rises, its kinetic energy is trans­ formed into potential energy. Eventually, it again acquires kinetic energy as it falls to Earth. When it strikes Earth, its kinetic energy is converted largely to thermal energy, though some work may be done on the surroundings as well.

CHAPTER REV I E W SUMMARY AND KEY TERMS Introduction and Section 5.1 is the study of energy and its transformations. In this chapter we have focused on the transformations of energy-especially heat-during chemical reactions. An object can possess energy in two forms: (1) is the energy due to the motion of the object; and is the energy that an object possesses by virtue of its position relative to other objects. An elec­ tron in motion near a proton, for example, has kinetic ener­ gy because of its motion and potential energy because of its electrostatic attraction to the proton. The SI unit of ener-

Thermodynamics thermochemistry, kinetic energy (2) potential energy

2 2 1 J = 1 kg-m /s Another common gy is the energy unit is the ca which was originally defined as the quantity of energy necessary to increase the temperature of 1 g of water by 1 °C: 1 cal = 4.184 J. When we study thermodynamic properties, we define a specific amount of matter as the Everything outside the system is the When we study a chemical reaction, the system is generally the reactants and prod­ ucts. A closed system can exchange energy, but not matter, with the surroundings. Energy can be transferred between the system and the surroundings as work or

joule G): calorie ( l), system. surroundings.

C HA PTER 5

200

Thermochemistry

heat. is the energy expended to move an object against a is the energy that is transferred from a hotter object to a colder one. is the capacity to do work or to transfer heat.

Workforce. Heat

Energy internal energy

Sedlon 5.2 The of a system is the sum of all the kinetic and potential energies of its component

parts. The internal energy of a system can change because of energy transferred between the system and the sur­ roundings. According to the the change in the internal energy of a system, !!.E, is the sum of the heat, q, transferred into or out of the system and the work, w, done on or by the system: !!.E = q + w. Both q and w have a sign that indicates the direction of en­ ergy transfer. When heat is transferred from the surround­ ings to the system, q > 0. Likewise, when the surroundings do work on the system, w > 0. In an process the system absorbs heat from the surroundings; in an process the system releases heat to the sur­ roundings. The internal energy, E, is a The value of any state function depends only on the state or condition of the system and not on the details of how it came to be in that state. The heat, q, and the work, w, are not state functions; their values depend on the particular way in which a system changes its state.

first law of thermodynamics,

exothermic

endothermic state function.

Sections 5.3 and 5.4 When a gas is produced or con­ sumed in a chemical reaction occurring at constant pres­ sure, the system may perform (P-V) against the prevailing pressure. For this reason, we define a new state function called H, which is related to energy: H = E + PV. In systems where only pressure-volume work that is due to gases is involved, the change in the enthalpy of a system, !!.H, equals the heat gained or lost by the system at constant pressure, !!.H = qp. For an endothermic process, !!.H > 0; for an exothermic process, !!.H < 0. Every substance has a char­ acteristic enthalpy. In a chemical process, the is the enthalpy of the products minus the enthalpy of the reactants: !!.H.,,n = H (products) - H (reactants). Enthalpies of reaction follow some simple rules: (1) The enthalpy of reaction is proportional to the amount of reac­ tant that reacts. (2) Reversing a reaction changes the sign of !!.H. (3) The enthalpy of reaction depends on the physical states of the reactants and products.

work

pressure-volume enthalpy,

enthalpy of

reaction

Section 5.5 The amount of heat transferred between the system and the surroundings is measured experimentally by A measures the temperature change accompanying a process. The temperature change of a calorimeter depends on its the amount of heat required to raise its temperature by 1 K. The heat capac­ ity for one mole of a pure substance is called its for one gram of the substance, we use the term

calorimetry. calorimeter heat capacity, molar heat capacity;

Water has a very high specific heat, 4.18 J/g-K. The amount of heat, q, absorbed by a substance is the prod­ uct of its specific heat (C,), its mass, and its temperature change: q = C, X m X !!.T. If a calorimetry experiment is carried out under a constant pressure, the heat transferred provides a direct measure of the enthalpy change of the reaction. Constant-volume calorimetry is carried out in a vessel of fixed volume called a Bomb calorimeters are used to measure the heat evolved in com­ bustion reactions. The heat transferred under constant­ volume conditions is equal to !!.E. However, corrections can be applied to !!.E values to yield enthalpies of combustion.

specific heat.

bomb calorimeter.

Section 5.6 Because enthalpy is a state function, !!.H de­ pends only on the initial and final states of the system. Thus, the enthalpy change of a process is the same whether the process is carried out in one step or in a series of steps. states that if a reaction is carried out in a series of steps, !!.H for the reaction will be equal to the sum of the enthalpy changes for the steps. We can there­ fore calculate !!.H for any process, as long as we can write the process as a series of steps for which f!.H is known.

Hess's law

Section 5.7 The !!.Hf, of a sub­ stance is the enthalpy change for the reaction in which the substance is formed from its constituent elements. The of a reaction, !!.Ho, is the en­ thalpy change when all reactants and products are at 1 atrn pressure and a specific temperature, usually 298 K (25 °C}. Combining these ideas, the !!.H'J, of a substance is the change in enthalpy for the reaction that forms one mole of the substance from its ele­ ments in their most stable form with all reactants and products at 1 atrn pressure and usually 298 K. For any ele­ ment in its most stable state at 298 K and 1 atrn pressure, !!.H 'J = 0. The standard enthalpy change for any reaction can be readily calculated from the standard enthalpies of formation of the reactants and products in the reaction:

enthalpy of formation, standard enthalpy change standard enthalpy of forma­ tion,

!!.H�xn = Ln!!.H'J (products) - Lm!!.H'J (reactants)

Section 5.8 The of a substance is the heat re­ leased when one gram of the substance is combusted. Different types of foods have different fuel values and differing abilities to be stored in the body. The most com­ mon fuels are hydrocarbons that are found as such as and Coal is the most abundant fossil fuel, but the sulfur present in most coals causes air pollution. Coal gasification is one possible way to use existing resources as sources of clean­ er energy in the future. Sources of include solar energy, wind energy, biomass, and hydro­ electric energy. Nuclear power does not utilize fossil fuels but does create serious waste-disposal problems.

fuel value

fuels,

natural gas, petroleum,

coal. fossil

renewable energy

KEY SKILLS • Interconvert energy units. Express the relationships among the quantities q, w, !!.E, and !!.H. Understand their sign conventions, including how the signs of q and !!.H relate to whether a process is exothermic or endothermic. •

Visualizing Concepts

201

• State the first law of thermodynamics. • Understand the concept of a state function and be able to give examples. • Use thermochemical equations to relate the amount of heat energy transferred in reactions at constant pressure (!lH) to the amount of substance involved in the reaction. • Calculate the heat transferred in a process from temperature measurements together with heat capacities or specific heats (calorimetry). • Use Hess's law to determine enthalpy changes for reactions. • Use standard enthalpies of formation to calculate !lH0 for reactions.

KEY EQUATIONS Kinetic energy

[5.1] • !lE = Efinal - Einitial • !lE = q + w (5.5] • w = -P !lV

[5.4]

[5.7]

The work done by an expanding gas at constant pressure

• !lH = !lE + P !lV = qp • q = C,

X

The change in internal energy Relates the change in internal energy to heat and work (the first law of thermodynamics)

m X !lT

[5.10]

Enthalpy change at constant pressure

(5.22]

Heat gained or lost based on specific heat, mass, and temperature change.

• !lH;xn = L n llH j (products) - LmllHj (reactants)

[5.31]

Standard enthalpy change of a reaction

VISUALIZING CONCEPTS 5.1 Imagine a book that is falling from a shelf. At a particu­ lar moment during its fall, the book has a kinetic energy of 13 J and a potential energy with respect to the floor of 72 J. How does the book's kinetic energy and its poten­ tial energy change as it continues to fall? What is its total kinetic energy at the instant just before it strikes the floor? [Section 5.1] 5.2 Consider the accompa­ nying energy diagram. (a) Does this diagram represent an increase or decrease in the internal energy of the system? (b) What sign is given to ll.E for this process? (c) If there is no work associat­ ed with the process, is it exothermic or endother­ mic? [Section 5.2]

Imagine that you are climbing a mountain. (a) Is the dis­ tance you travel to the top a state function? Why or why not? (b) Is the change in elevation between your base camp and the peak a state function? Why or why not? [Section 5.2]

5.5

In the cylinder diagrammed below, a chemical process occurs at constant temperature and pressure. (a) Is the sign of w indicated by this change positive or negative? (b) If the process is endothermic, does the internal ener­ gy of the system within the cylinder increase or decrease during the change and is ll.E positive or negative? [Sections 5.2 and 5.3]

Reactants

w

(ii)

5.4

Products

5.3 The contents of the closed box in each of the following illustrations represent a system, and the arrows show the changes to the system during some process. The

(i)

lengths of the arrows represent the relative magnitudes of q and w (a) Which of these processes is endothermic? (b) For which of these processes, if any, is ll.E < 0? (c) For which process, if any, is there a net gain in inter­ nal energy? [Section 5.2]

w

(iii)

p

p

202

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Thermochemistry

5.6 Imagine a container placed in a tub of water, as depicted in the accompanying diagram. (a) If the contents of the container are the system and 350 K 290 K heat is able to flow through the container walls, what qualitative changes will occur in the temperatures of the system and in its surround­ ings? What is the sign of q associated with each change? From the system's perspective, is the process exo­ thermic or endothermic? (b) If neither the volume nor the pressure of the system changes during the process, how is the change in internal energy related to the change in enthalpy? [Sections 5.2 and 5.3]

chemical equation for the reaction depicted, and predict whether w is positive, negative, or zero. (b) Using data from Appendix C, determine !l.H for the formation of one mole of the product. Why is this enthalpy change called the enthalpy of formation of the involved prod­ uct? [Sections 5.3 and 5.7]

5.9 Consider the two diagrams below. (a) Based on (1), write an equation showing how !l.HA is related to !l.H8 and !l.Hc. How do both diagram (i) and your equation relate to the fact that enthalpy is a state function? (b) Based on (ii), write an equation relating !l.Hz to the other enthalpy changes in the diagram. (c) How do these diagrams re­ late to Hess's law? [Section 5.6]

5.7 Which will release more heat as it cools from 50 oc to

25 °C, 1 kg of water or 1 kg of aluminum? How do you know? [Section 5.5]

5.8 A gas-phase reaction was run in an apparatus designed to maintain a constant pressure. (a) Write a balanced p

z

p

(ii) 0

N

5.10 Does !l.H,xn for the reaction represented by the follow­

ing equation equal the standard enthalpy of formation for CHpH(I)? Why or why not? [Section 5.7]

C(graphite) + 4 H(g) + O(g)

->

CH 30H(l)

EXERC ISES The Nature of Energy 5.11 In what two ways can an object possess energy? How do these two ways differ from one another?

5.12 Suppose you toss a tennis ball upward. (a) Does the ki­ netic energy of the ball increase or decrease as it moves higher? (b) What happens to the potential energy of the ball as it moves higher? (c) If the same amount of energy were imparted to a ball the same size as a tennis ball, but of twice the mass, how high would it go in comparison to the tennis ball? Explain your answers.

5.13 (a) Calculate the kinetic energy in joules of a 45-g golf ball moving at 61 m/s. (b) Convert this energy to calo­ ries. (c) What happens to this energy when the ball lands in a sand trap?

5.14 (a) What is the kinetic energy in joules of an 850-lb

motorcycle moving at 66 mph? (b) By what factor will the kinetic energy change if the speed of the motorcycle is decreased to 33 mph? (c) Where does the kinetic ener­ gy of the motorcycle go when the rider brakes to a stop?

5.15 The use of the British thermal unit (Btu) is common in much engineering work. A Btu is the amount of heat re­ quired to raise the temperature of 1 lb of water by 1 °F. Calculate the number of joules in a Btu.

5.16 A watt is a measure of power (the rate of energy change) equal to 1 Jjs. (a) Calculate the number of joules in a kilowatt-hour. (b) An adult person radiates heat to the surroundings at about the same rate as a 100-watt elec­ tric incandescent lightbulb. What is the total amount of energy in kcal radiated to the surroundings by an adult in 24 hours?

5.17 (a) What is meant by the term system in thermodynamics? (b) What is a closed system? 5.18 In a thermodynamic study a scientist fo­ cuses on the proper- In ties of a solution in an apparatus as illustrated. A solu­ tion is continuously Out flowing into the ap­ paratus at the top and out at the bottom, such that the amount of solution in the apparatus is constant with time. (a) Is the solution in the apparatus a closed system, open system, or isolat­ ed system? Explain your choice. (b) If it is not a closed system, what could be done to make it a closed system?

Exercises 5.19 (a) What is work? (b) How do we determine the amount

203

of work done, given the force associated with the work?

pencil off the top of a desk. (b) A spring is compressed to half its normal length.

5.20 (a) What is heat? (b) Under what conditions is heat

5.22 Identify the force present, and explain whether work is

transferred from one object to another?

5.21 Identify the force present, and explain whether work is being performed

in

the following cases: (a) You lift a

done when (a) a positively charged particle moves in a circle at a fixed distance from a negatively charged par­ ticle; (b) an iron nail is pulled off a magnet.

The First Law of Thermodynamics 5.23 (a) State the first law of thermodynamics. (b) What is meant by the intem111 energy of a system? (c) By what means can the internal energy of a closed system increase?

5.24 (a) Write an equation that expresses the first law of thermo­

dynamics in terms of heat and work. (b) Under what con­ ditions will the quantities q and w be negative numbers?

5.25 Calculate 11£, and determine whether the process is en­

dothermic or exothermic for the following cases: (a) A sys­ tem absorbs 105 kJ of heat from its surroundings while doing 29 kJ of work on the surroundings; (b) q = 1.50 kJ and w = -657 J; (c) the system releases 57.5 kJ of heat while doing 22.5 kJ of work on the surroundings.

(b) What can you say about the values of q and w in each of these cases? (c) What can you say about the relative values of flE for the system (the gas in the cylinder) in the two cases?

5.28 Consider a system consisting of two oppositely charged spheres hanging by strings and separated by a distance r1, as shown in the accompanying illustration. Suppose they are separated to a larger distance r2, by moving them apart along a track. (a) What change, if any, has oc­ curred in the potential energy of the system? (b) What effect, if any, does this process have on the value of !J.E? (c) What can you say about q and w for this process?

5.26 For the following processes, calculate the change in in­ ternal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing 382 J of work on the atmosphere. (b) A 50-g sample of water is cooled from 30 °C to 15 °C, thereby losing approximate­ ly 3140 J of heat. (c) A chemical reaction releases 6.47 kJ of heat and does no work on the surroundings.

+

'2

5.27 A gas is confined to a cylinder fitted with a piston and an elec­ trical heater, as shown in the ac­ companying illustration. Suppose that current is supplied to the heater so that 100 J of energy is added. Consider two different sit­ uations. In case (1) the piston is al­ lowed to move as the energy is added. In case (2) the piston is fixed so that it cannot move. (a) In which case does the gas have the higher temperature after addition of the electrical energy? Explain.

5.29 (a) What is meant by the term statefunction ? (b) Give an

example of a quantity that is a state function and one that is not. (c) Is work a state function? Why or why not?

5.30 Indicate which of the following is independent of the path by which a change occurs: (a) the change in poten­ tial energy when a book is transferred from table to shelf, (b) the heat evolved when a cube of sugar is oxi­ dized to C02(g) and H20(g), (c) the work accomplished in burning a gallon of gasoline.

Enthalpy 5.31 (a) Why is the change in enthalpy usually easier to mea­ sure than the change in internal energy? (b) For a given process at constant pressure, !J.H is negative. Is the process endothermic or exothermic?

5.32 (a) Under what condition will the enthalpy change of a

process equal the amount of heat transferred into or out of the system? (b) During a constant-pressure process the system absorbs heat from the surroundings. Does the enthalpy of the system increase or decrease during the process?

5.33 You are given !J.H for a process that occurs at constant pressure. What additional information do you need to determine !J.E for the process?

5.34 Suppose that the gas-phase reaction 2 NO(g) + 02(g)

----> 2 N02(g) were carried out in a constant-volume con­ tainer at constant temperature. Would the measured heat change represent !J.H or !J.E? lf there is a difference, which quantity is larger for this reaction? Explain.

5.35 A gas is confined to a cylinder under constant atmos­ pheric pressure, as illustrated in Figure 5.3. When the gas undergoes a particular chemical reaction, it releases 79 kJ of heat to its surroundings and does 18 kJ of P-V work on its surroundings. What are the values of fl H and !J. E for this process?

5.36 A gas is confined to a cylinder under constant atmos­ pheric pressure, as illustrated

in

Figure 5.3. When 378 J

204

C HA PTER 5

Thermochemistry

of heat is added to the gas, it expands and does 56 J of work on the surroundings. What are the values of t.H and D. £ for this process?

many kilojoules of heat are released when 50.9 g of CO(g) reacts completely with H2{g) to form CHPH{g) at constant pressure?

5.37 The complete combustion of acetic acid, CH3COOH(I),

5.43 When solutions containing silver ions and chloride ions

to form H20(1) and C02{g) at constant pressure releas­ es 871.7 kJ of heat per mole of CH3COOH. (a) Write a balanced thermochemical equation for this reaction. (b) Draw an enthalpy diagram for the reaction. 5.38 The decomposition of zinc carbonate, ZnC03(s), into zinc oxide, ZnO(s), and C02{g) at constant pressure re­ quires the addition of 71.5 kJ of heat per mole of ZnC03. (a) Write a balanced thermochemical equation for the re­ action. (b) Draw an enthalpy diagram for the reaction.

5.39 Consider the following reaction, which occurs at room temperature and pressure:

t.H = -243.4 kJ Which has the higher enthalpy under these conditions, 2 Cl{g) or Clz(g)?

5.40 Without referring to tables, predict which of the follow­ ing has the higher enthalpy in each case: (a) 1 mol C02(s)

or 1 mol C02{g) at the same temperature, (b) 2 mol of hy­ drogen atoms or 1 mol of H:u (c) 1 mol H2{g) and 0.5 mol 02{g) at 25 oc or 1 mol HzO{g) at 25 oc, (d) 1 mol N2{g) at 100 oc or 1 mol N2{g) at 300 oc.

5.41 Consider the following reaction: 2 Mg(s) + 02{g) -> 2 MgO(s)

t.H = -1204 kJ

(a) Is this reaction exothermic or endothermic? (b) Cal­ culate the amount of heat transferred when 2.4 g of Mg(s) reacts at constant pressure. (c) How many grams of MgO are produced during an enthalpy change of -96.0 kJ? (d) How many kilojoules of heat are absorbed when 7.50 g of MgO(s) is decomposed into Mg(s) and 02{g) at constant pressure?

5.42 Consider the following reaction:

are mixed, silver chloride precipitates:

t.H = -65.5 kJ

Ag+(aq) + cqaq) -> AgCl(s)

(a) Calculate t.H for production of 0.200 mol of AgCl by this reaction. (b) Calculate t.H for the production of 2.50 g of AgCI. (c) Calculate t.H when 0.150 mmol of AgCI dissolves in water.

5.44 At one time, a common means of forming small quanti­ ties of oxygen gas in the laboratory was to heat KC103: 2 KCI03(s) -> 2 KCl(s) + 3 02{g)

t.H = -89.4 kJ

For this reaction, calculate t.H for the formation of (a) 0.632 mol of 02 and (b) 8.57 g of KCI. (c) The decom­ position of KC103 proceeds spontaneously when it is heated . Do you think that the reverse reaction, the for­ mation of KC103 from KCl and 02, is likely to be feasible under ordinary conditions? Explain your answer.

5.45 Consider the combustion of liquid methanol, CH30H(/):

�

CHPH(I) + 02{g) -> C02{g) + 2 H20(I)

t.H = - 726.5 kJ

(a) What is the enthalpy change for the reverse reaction? (b) Balance the forward reaction with whole-number co­ efficients. What is D.H for the reaction represented by this equation? (c) Which is more likely to be thermody­ namically favored, the forward reaction or the reverse reaction? (d) If the reaction were written to produce H20{g) instead of H20(/), would you expect the magni­ tude of t.H to increase, decrease, or stay the same? Explain.

5.46 Consider the decomposition of liquid benzene, C6H6(/), to gaseous acetylene, C2H2{g):

t.H = +90.7 kJ

(a) Is heat absorbed or released in the course of this reac­ tion? (b) Calculate the amount of heat transferred when 45.0 g of CH30H{g) is decomposed by this reaction at constant pressure. (c) For a given sample of CH30H, the enthalpy change on reaction is 25.8 kJ. How many grams of hydrogen gas are produced? What is the value of t.H for the reverse of the previous reaction? (d) How

t.H = +630 kJ

(a) What is the enthalpy change for the reverse reaction? (b) What is t.H for the formation of 1 mol of acetylene?

(c) Which is more likely to be thermodynamically fa­ vored, the forward reaction or the reverse reaction? (d) If C�6{g) were consumed instead of C6H6(1), would you expect the magnitude of t.H to increase, decrease, or stay the same? Explain.

Calorimetry 5.47 (a) What are the units of molar heat capacity? (b) What

are the units of specific heat? (c) If you know the specif­ ic heat of copper, what additional information do you need to calculate the heat capacity of a particular piece of copper pipe?

5.48 Two solid objects, A and B, are placed in boiling water and allowed to come to temperature there. Each is then lifted out and placed in separate beakers containing 1000 g water at 10.0 oc. Object A increases the water temperature by 3.50 oc; B increases the water tempera­ ture by 2.60 oc. (a) Which object has the larger heat ca­ pacity? (b) What can you say about the specific heats of A and B?

5.49 (a) What is the specific heat of liquid water? (b) What is

the molar heat capacity of liquid water? (c) What is the heat capacity of 185 g of liquid water? (d) How many kJ of heat are needed to raise the temperature of 10.00 kg of liquid water from 24.6 oc to 46.2 °C?

5.50 (a) Which substance in Table 5.2 requires the smallest

amount of energy to increase the temperature of 50.0 g of that substance by 10 K? (b) Calculate the energy need­ ed for this temperature change.

5.51 The specific heat of iron metal is 0.450 J/g-K. How many J of heat are necessary to raise the temperature of a 1.05-kg block of iron from 25.0 oc to 88.5 °C?

Exercises 5.52 The specific heat of ethylene glycol is 2.42 J/g-K. How many j of heat are needed to raise the temperature of 62.0 g of ethylene glycol from 13.1 oc to 40.5 °C?

5.53 When a 9.55-g sample of solid sodium hydroxide dis­ solves in 100.0 g of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from 23.6 oc to 47.4 oc. Calculate !lH (in kjjmol NaOH) for the solu­ tion process NaOH(s) ----+ Na+(aq) + OH-(aq) Assume that the specific heat of the solution is the same as that of pure water.

5.54 (a) When a 3.88-g sample of solid ammonium nitrate dissolves in 60.0 g of water in a coffee-cup calorimeter (Figure 5.17), the temperature drops from 23.0 oc to 18.4 °C. Calculate !lH (in k)/mol NH4N03) for the solu­ tion process

+ NH�03(s) ----+ NH4 (aq) + N0 3-(aq) Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?

5.55 A 2.200 -g sample of quinone (C6H402) is burned in a

bomb calorimeter whose total heat capacity is 7.854 kjj"C. The temperature of the calorimeter increas­ es from 23.44 oc to 30.57 oc. What is the heat of combus­ tion per gram of quinone? Per mole of quinone?

205

5.56 A 1.800-g sample of phenol (C�50H) was burned in a

bomb calorimeter whose total heat capacity is 1 1.66 kjj"C. The temperature of the calorimeter plus contents in­ creased from 21.36 oc to 26.37 oc. (a) Write a balanced chemical equation for the bomb calorimeter reaction. (b) What is the heat of combustion per gram of phenol? Per mole of phenol?

5.57 Under constant-volume conditions the heat of combus­

tion of glucose (C6H1P6) is 15.57 k)/g. A 2.500-g sample of glucose is burned in a bomb calorimeter. The temper­ ature of the calorimeter increased from 20.55 oc to 23.25 oc. (a) What is the total heat capacity of the calorimeter? (b) lf the size of the glucose sample had been exactly twice as large, what would the temperature change of the calorimeter have been?

5.58 Under constant-volume conditions the heat of combus­

tion of benzoic acid (C6H5COOH) is 26.38 k)/g. A 1 .640g sample of benzoic acid is burned in a bomb calorimeter. The temperature of the calorimeter increas­ es from 22.25 oc to 27.20 °C. (a) What is the total heat ca­ pacity of the calorimeter? (b) A 1 .320-g sample of a new organic substance is combusted in the same calorimeter. The temperature of the calorimeter increases from 22.14 oc to 26.82 oc. What is the heat of combustion per gram of the new substance? (c) Suppose that in chang­ ing samples, a portion of the water in the calorimeter were lost. In what way, if any, would this change the heat capacity of the calorimeter?

H es s's Law 5.59 What is the connection between Hess's law and the fact

calculate the heat of the reaction

that H is a state function?

5.60 Consider the following hypothetical reactions: A ----+ B B

----+

C

5.63 From the enthalpies of reaction

!lH = +30 kj !lH

=

+60 kj

(a) Use Hess's law to calculate the enthalpy change for the reaction A ----+ C. (b) Construct an enthalpy dia­ gram for substances A, B, and C, and show how Hess's law applies.

5.61 Calculate the enthalpy change for the reaction P406(s) + 2 02(gl ----+ P4010(s)

!lH = -1640.1 kj

P4(s) + 5 02(g) ----+ P4010(s)

!lH = -2940.1 kj

2 HP(g)

=

-680 kj

!lH = +52.3 kj

calculate flH for the reaction of ethylene with F2: C2H4 (g) + 6 F2(g) ----+ 2 CF4(g) + 4 HF(g)

2 NO(g) + 02(g) ----+ 2 N02(g) 2 N20(g) ----+ 2 N2(g) + 02(g)

!lH = + 180.7 kj !lH = -113.1 kj !lH

=

-163.2 kj

use Hess's law to calculate !lH for the reaction

5.62 From the enthalpies of reaction ----+

!lH

2 C(s) + 2 H2(g) ----+ C2H4(g)

N2(g) + 02 (g) ----+ 2 NO (g)

P4(s) + 3 02(g) ----+ P406(s)

3 02(g) ----+ 2 03(g)

!lH = -537 kj

5.64 Given the data

given the following enthalpies of reaction:

2 H2(g) + 02(g)

Hz(g) + F2 (g) ----+ 2 HF(g) C(s) + 2 F2(g) ----+ CF4(g)

!lH

=

-483.6 kj

N20(g) + NOz(g) ----+ 3 NO(g)

!lH = +284.6 kj

Enthalpies of Formation 5.65 (a) What is meant by the term standard conditions, with

reference to enthalpy changes? (b) What is meant by the term enthalpy offormation? (c) What is meant by the term

standard enthalpy offormation?

5.66 (a) Why are tables of standard enthalpies of formation so useful? (b) What is the value of the standard enthalpy of formation of an element in its most stable form? (c) Write the chemical equation for the reaction whose enthalpy change is the standard enthalpy of formation of glucose, C6Hn06(s), llH'( [C6H n061 ·

206

C HA PTER 5

Thermochemistry

5.67 For each of the following compounds, write a balanced

thermochemical equation depicting the formation of one mole of the compound from its elements in their standard states and use Appendix C to obtain the value of llHJ: (a) NH3(g), (b) S02(g), (c) RbCI03(s),

(d) NH�03(s). 5.68 Write balanced equations that describe the formation of

the following compounds from elements in their stan· dard states, and use Appendix C to obtain the values of their standard enthalpies of formation: (a) HBr(g), (b) AgN03(s), (c) Fep3(s), (d) CH3COOH(/).

5.69 The following is known as the thermite reaction [Figure

5.7(b)]:

2 Al(s)

+

Fep3(s) -----> Alp3(s)

+

2 Fe(s)

This highly exothermic reaction is used for welding massive units, such as propellers for large ships. Using standard enthalpies of formation in Appendix C, calcu­ late !lH0 for this reaction. 5.70 Many cigarette lighters contain liquid butane, C4H 10(/). Using standard enthalpies of formation, calculate the quantity of heat produced when 5.00 g of butane is com­ pletely combusted in air under standard conditions. 5.71 Using values from Appendix C, calculate the standard

enthalpy change for each of the following reactions: (a) 2 S02(g) + 02(g) -----> 2 S03(g) (b) Mg(OHh(s) -----> MgO(s) + H20(/) (c) N204(g) + 4 H2(g) -----> N2(g) + 4 H 20(g) (d) SiC14(1) + 2 H 20(/) -----> Si02(s) + 4 HCI(g) 5.72 Using values from Appendix C. calculate the value of llH0 for each of the following reactions: (a) 4 HBr(g) + 02(g) -----> 2 H 20(/) + 2 Br2(/) (b) 2 Na(OH)(s) + S03(g) -----> Na2S04(s) + Hp(g) (c) CH4(g) + 4 Cl2(g) -----> CC14(1) + 4 HCI(g) (d) Fe203(s) + 6 HCI(g) -----> 2 FeCI3(s) + 3 Hp(g) 5.73 Complete combustion of 1 mol of acetone (C3H 60) liber­

ates 1 790 kJ: C3H60(I) + 4 02(g) -----> 3 C0 2(g)

+

3 H 20(/)

flW = - 1 790 kJ Using this information together with data from Appen­ dix C, calculate the enthalpy of formation of acetone.

5.74 Calcium carbide (CaC2) reacts with water to form acety­

lene (C 2H2) and Ca(OHh- From the following enthalpy of reaction data and data in Appendix C, calculate llHJ for CaC2(s): CaC 2(s) + 2 H20(/) -----> Ca(OHh(s) + C2H2(g) flW = -127.2 kJ

5.75 Gasoline is composed primarily of hydrocarbons, in­ cluding many with eight carbon atoms, called octanes.

One of the cleanest-burning octanes is a compound called 2,3,4-trirnethylpentane, which has the following structural formula: CH3 CH3 CH3 I

I

I

H3C-CH-CH-CH-CH3 The complete combustion of one mole of this compound to C02(g) and H20(g) leads to !lH0 = -5064.9 kJ/mol. (a) Write a balanced equation for the combustion of 1 mol of CsH!s(l). (b) Write a balanced equation for the formation of C8H18(1) from its elements. (c) By using the information in this problem and data in Table 5.3, calculate llHJ for 2,3,4-trimethylpentane. 5.76 Naphthalene (C 10H8) is a solid aromatic compound often sold as mothballs. The complete combustion of this substance to yield C02(g) and H 20(/) at 25 oc yields 5154 k)/mol. (a) Write balanced equations for the forma­ tion of naphthalene from the elements and for its com­ bustion. (b) Calculate the standard enthalpy of formation of naphthalene. 5.77 Ethanol (C 2H50H) is currently blended with gasoline as

an automobile fuel. (a) Write a balanced equation for the combustion of liquid ethanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming H20(g) as a product. (c) Calculate the heat produced per liter of ethanol by combustion of ethanol under constant pressure. Ethanol has a density of 0.789 g/mL. (d) Cal­ culate the mass of C02 produced per kJ of heat emitted. 5.78 Methanol (CH30H) is used as a fuel in race cars. (a) Write a balanced equation for the combustion of liq­ uid methanol in air. (b) Calculate the standard enthalpy change for the reaction, assuming H20(g) as a product. (c) Calculate the heat produced by combustion per liter of methanol. Methanol has a density of 0.791 g/mL. (d) Calculate the mass of C02 produced per kJ of heat emitted.

Foods and Fuels 5.79 (a) What is meant by the term fuel value? (b) Which is

a greater source of energy as food, 5 g of fat or 9 g of carbohydrate? 5.80 (a) Why are fats well suited for energy storage in the human body? (b) A particular chip snack food is com­ posed of 12% protein, 14% fat, and the rest carbohy­ drate. What percentage of the calorie content of this food is fat? (c) How many grams of protein provide the same fuel value as 25 g of fat? 5.81 A serving of condensed cream of mushroom soup con­ tains 7 g fat, 9 g carbohydrate, and 1 g protein. Estimate the number of Calories in a serving.

5.82 A pound of plain M&M® candies contains 96 g fat, 320 g

carbohydrate, and 21 g protein. What is the fuel value in kJ in a 42-g (about 1.5 oz) serving? How many Calories does it provide?

5.83 The heat of combustion of fructose, C6H1 206, is

-2812 kJ/mol. If a fresh golden delicious apple weigh­ ing 4.23 oz (120 g) contains 16.0 g of fructose, what caloric content does the fructose contribute to the apple?

5.84 The heat of combustion of ethanol, C2H50H(I), is -1367 kJ/mol. A batch 10.6% ethanol by

tains

of Sauvignon Blanc wine con­ mass. Assuming the density of

Additional Exercises the wine to be 1.0 g/mL, what caloric content does the alcohol (ethanol) in a 6-oz glass of wine (177 mL) have?

5.85 The standard enthalpies of formation of gaseous propyne (C3H4), propylene (C3H6), and propane (C3Hs) are + 185.4, +20.4, and -103.8 k)/mol, respectively. (a) Cal­ culate the heat evolved per mole on combustion of each substance to yield C02(g) and H20(g). (b) Calculate the heat evolved on combustion of 1 kg of each substance.

207

(c) Which is the most efficient fuel in terms of heat evolved per unit mass?

5.86 It is interesting to compare the "fuel value" of a hydro­ carbon in a world where fluorine rather than oxygen is the combustion agent. The enthalpy of formation of CF4(g) is -679.9 k)/mol. Which of the following two re­ actions is the more exothermic? CH4(g) + 2 02(g) CH4(g) + 4 F2(g)

->

C02(g) + 2 H20(g)

->

CF4(g) + 4 HF(g)

ADDITIONAL EXERCISES 5.87 At 20 °C (approximately room temperature) the average velocity of N2 molecules in air is 1050 mph. (a) What is the average speed in m/s? (b) What is the kinetic energy (in J) of an N2 molecule moving at this speed? (c) What is the total kinetic energy of 1 mol of N2 molecules mov­ ing at this speed?

5.88 Suppose an Olympic diver who weighs 52.0 kg executes a straight dive from a 10-m platform. At the apex of the dive, the diver is 10.8 m above the surface of the water. (a) What is the potential energy of the diver at the apex of the dive, relative to the surface of the water? (b) As­ suming that all the potential energy of the diver is con­ verted into kinetic energy at the surface of the water, at what speed in m/s will the diver enter the water? (c) Does the diver do work on entering the water? Explain.

5.89 When a mole of dry ice, C02(s), is converted to C02(g) at atmospheric pressure and -78 °C, the heat absorbed by the system exceeds the increase in internal energy of the C02. Why is this so? What happens to the remaining energy?

5.90 The air bags that provide protection in autos in the event of an accident expand because of a rapid chemical reaction. From the viewpoint of the chemical reactants as the system, what do you expect for the signs of q and w in this process?

[5.91] An aluminum can of a soft drink is placed in a freezer. Later, you find that the can is split open and its contents frozen. Work was done on the can in splitting it open. Where did the energy for this work come from?

[5.92] A sample of gas is contained in a cylinder-and-piston arrangement. It undergoes the change in state shown in the drawing. (a) Assume first that the cylinder and pis­ ton are perfect thermal insulators that do not allow heat to be transferred. What is the value of q for the state

change? What is the sign of w for the state change? What can be said about fl£ for the state change? (b) Now as­ sume that the cylinder and piston are made up of a ther­ mal conductor such as a metal. During the state change, the cylinder gets warmer to the touch. What is the sign of q for the state change in this case? Describe the differ­ ence in the state of the system at the end of the process in the two cases. What can you say about the relative values of fl£?

[5.93] Limestone stalactites and stalagmites are formed in caves by the following reaction:

Ca2+(aq)

+ 2 HC03 -(aq)

->

CaC03(s) + COz(g) + HzO(/) If 1 mol of CaC03 forms at 298 K under 1 atm pressure, the reaction performs 2.47 kJ of P-V work, pushing back the atmosphere as the gaseous C02 forms. At the same time, 38.95 kj of heat is absorbed from the environment. What are the values of flH and of fl£ for this reaction?

[5.94] Consider the systems shown in Figure 5.9. In one case the battery becomes completely discharged by running the current through a heater, and in the other by run­ ning a fan. Both processes occur at constant pressure. In both cases the change in state of the system is the same: The battery goes from being fully charged to being fully discharged. Yet in one case the heat evolved is large, and in the other it is small. Is the enthalpy change the same in the two cases? If not, how can enthalpy be considered a state function? If it is, what can you say about the rela­ tionship between enthalpy change and q in this case, as compared with others that we have considered?

5.95 The enthalpy change for melting ice at 0 oc and constant atmospheric pressure is 6.01 kJ/mol. Calculate the quan­ tity of energy required to melt a moderately large ice­ berg with a mass of 1.25 million metric tons. (A metric ton is 1000 kg.)

5.96 Comparing the energy associated with the rainstorm and that of a conventional explosive gives some idea of the immense amount of energy associated with a storm. (a) The heat of vaporization of water is 44.0 k)/mol. Calculate the quantity of energy released when enough water vapor condenses to form 0.50 inches of rain over an area of one square mile. (b) The energy released when one ton of dynamite explodes is 4.2 x 106 kj. Calculate the number of tons of dynamite needed to provide the energy of the storm in part (a).

5.97 A house is designed to have passive solar energy features. Brickwork incorporated into the interior of the house acts as a heat absorber. Each brick weighs

208

C HA PTER 5

Thermochemistry

approximately 1.8 kg. The specific heat of the brick is 0.85 J/g-K. How many bricks must be incorporated into the interior of the house to provide the same total heat capacity as 1.7 x 103 gal of water? [5.98] A coffee-cup calorimeter of the type shown in Figure 5.17 contains 150.0 g of water at 25.1 oc. A 121.0-g block of copper metal is heated to 100.4 oc by putting it in a beaker of boiling water. The specific heat of Cu(s) is 0.385 J/g-K. The Cu is added to the calorimeter, and after a time the contents of the cup reach a constant tem­ perature of 30.1°C. (a) Determine the amount of heat, in J, lost by the copper block. (b) Determine the amount of heat gained by the water. The specific heat of water is 4.18 J/g-K. (c) The difference between your answers for (a) and (b) is due to heat loss through the Styrofoam® cups and the heat necessary to raise the temperature of the inner wall of the apparatus. The heat capacity of the calorimeter is the amount of heat necessary to raise the temperature of the apparatus (the cups and the stopper) by 1 K. Calculate the heat capacity of the calorimeter in J/K. (d) What would be the final temperature of the sys­ tem if all the heat lost by the copper block were ab­ sorbed by the water in the calorimeter? [5.99] (a) When a 0.235-g sample of benzoic acid is combusted in a bomb calorimeter, the temperature rises 1.642 oc. When a 0.265-g sample of caffeine, CsH wOzN4, is burned, the temperature rises 1.525 °C. Using the value 26.38 kJ/g for the heat of combustion of benzoic acid, calculate the heat of combustion per mole of caffeine at constant volume. (b) Assuming that there is an uncer­ tainty of 0.002 oc in each temperature reading and that the masses of samples are measured to 0.001 g, what is the estimated uncertainty in the value calculated for the heat of combustion per mole of caffeine? 5.100 How many grams of methane [CH4(g)] must be com­ busted to heat 1.00 kg of water from 25.0 oc to 90.0 oc, assuming H20(/) as a product and 100% efficiency in heat transfer? 5.101 Meals-ready-to-eat (MREs) are military meals that can be heated on a flameless heater. The heat is produced by the following reaction: Mg(s) + 2 H 20(l) Mg(OH)z(s) + H 2(g). (a) Calculate the standard en­ thalpy change for this reaction. (b) Calculate the num­ ber of grams of Mg needed for this reaction to release enough energy to increase the temperature of 25 mL of water from 15 oc to 85 oc. 5.102 Burning methane in oxygen can produce three different carbon-containing products: soot (very fine particles of graphite), CO(g), and C02 (g). (a) Write three balanced equations for the reaction of methane gas with oxygen to produce these three products. In each case assume that HzO(l) is the only other product. (b) Determine the standard enthalpies for the reactions in part (a). (c) Why, when the oxygen supply is adequate, is C02(g) the pre­ dominant carbon-containing product of the combustion of methane? 5.103 (a) Calculate the standard enthalpy of formation of gaseous diborane (BzH6) using the following thermo­ chemical information: !lW = -2509.1 kJ 4 B(s) + 3 02(g) 2 B203(s) !lW = -571.7 kJ 2 H 2(g) + 02(g) 2 H 20(/) B2H6(g) + 3 02(g) B 203(s) + 3 H 20(/) !lW = -2147.5 kJ ->

->

->

->

(b) Pentaborane (B5H9) is another boron hydride. What experiment or experiments would you need to perform to yield the data necessary to calculate the heat of for­ mation of B5H9(/)? Explain by writing out and summing any applicable chemical reactions. 5.104 From the following data for three prospective fuels, cal­ culate which could provide the most energy per unit volume: Density at 20 oc (gjcm3)

Fuel

Nitroethane, C2H5N02(1) 1.052 Ethanol, CzHsOH(I) 0.789 Methylhydrazine, CH6N2(1) 0.874

Molar Enthalpy of Combustion kJ/mol

-1368 -1367 -1305

5.105 The hydrocarbons acetylene (C2H2) and benzene (C6H6)

have the same empirical formula. Benzene is an "aro­ matic" hydrocarbon, one that is unusually stable because of its structure. (a) By using the data in Appendix C, determine the standard enthalpy change for the reaction 3 C2H 2(g) C6H6(l). (b) Which has greater enthalpy, 3 mol of acetylene gas or 1 mol of liquid benzene? (c) Determine the fuel value in kJ/g for acetylene and benzene. [5.106] Ammonia (NH3) boils at -33 oc; at this temperature it has a density of 0.81 g/cm3 The enthalpy of formation of NH3(g) is -46.2 kJ/mol, and the enthalpy of vapor­ ization of NH3(/) is 23.2 kJ/mol. Calculate the enthalpy change when 1 L of liquid NH3 is burned in air to give N 2(g) and HzQ(g). How does this compare with !lH for the complete combustion of 1 L of liquid methanol, CHpH(I)? For CH30H(/), the density at 25 oc is 0.792 gjcm3, and t.Hj equals -239 kjjmol. [5.107] Three common hydrocarbons that contain four carbons are listed here, along with their standard enthalpies of formation: ->

Hydrocarbon

Formula

!lHJ (kJ/mol)

1,3-Butadiene 1 -Butene n-Butane

C4H6(g) C4Hs(gl C4H w(gl

111.9 1 .2 -124.7

(a) For each of these substances, calculate the

molar en­ thalpy of combustion to C02(g) and H20(l). (b) Calcu­ late the fuel value in kJ/g for each of these compounds. (c) For each hydrocarbon, determine the percentage of hydrogen by mass. (d) By comparing your answers for parts (b) and (c), propose a relationship between hydro­ gen content and fuel value in hydrocarbons. 5.108 The two common sugars, glucose (C�1206) and su­ crose (C 12H22011 ), are both carbohydrates. Their stan­ dard enthalpies of formation are given in Table 5.3. Using these data, (a) calculate the molar enthalpy of combustion to C02(g) and H20(/) for the two sugars; (b) calculate the enthalpy of combustion per gram of each sugar; (c) determine how your answers to part (b) compare to the average fuel value of carbohydrates dis­ cussed in Section 5.8.

Integrative Exercises 5.109 A 200-lb man decides to add to his exercise routine by

walking up three flights of stairs (45 ft) 20 times per day. He figures that the work required to increase his poten­ tial energy in this way will permit him to eat an extra order of French fries, at 245 Cal, without adding to his weight. Is he correct in this assumption? 5.110 The sun supplies about 1.0 kilowatt of enerp for each square meter of surface area (1.0 kW/m , where a watt = 1 J/s). Plants produce the equivalent of about 0.20 g of sucrose (C 12H2200) per hour per square meter. Assuming that the sucrose is produced as follows, calcu­ late the percentage of sunlight used to produce sucrose.

12 C02(g)

+

11 H20(1)

---+

C12H22011

+

209

12 02(g)

t.H

= 5645 kJ

[5.111] It is estimated that the net amount of carbon dimdde

fixed by photosynthesis on the landmass of Earth is 5.5 X 10 1 6 g/yr of C02. Assume that all this carbon is converted into glucose. (a) Calculate the energy stored by photo­ synthesis on land per year in kJ. (b) Calculate the average rate of conversion of solar energy into plant energy in MW (1 W = 1 J/s). A large nuclear power plant produces about 103 MW. The energy of how many such nuclear power plants is equivalent to the solar energy conversion?

INTEGRATIVE EXERCISES 5.112 Consider the combustion of a single molecule of CH4(g)

forming H20(I) as a product. (a) How much energy, in J, is produced during this reaction? (b) A typical X-ray photon has an energy of 8 keV. How does the energy of combustion compare to the energy of the X-ray photon? 5.113 Consider the dissolving of NaCI in water, illustrated in Figure 4.3. Assume the system consists of 0.1 mol NaCI and 1 L of water. Considering that the NaCI readily dis­ solves in the water and that the ions are strongly stabi­ lized by the water molecules, as shown in the figure, is it safe to conclude that the dissolution of NaCI in water re­ sults in a lower enthalpy for the system? Explain your response. What experimental evidence would you ex­ amine to test this question? 5.114 Consider the following unbalanced oxidation-reduction reactions in aqueous solution: Ag +(aq) + Li(s) Ag(s) + u+(aq) Fe(s) Na +(aq) Fe 2+(aq) Na(s) ---+

+

+

---+

K(s) + HzO(I)

---+

KOH(aq)

+

H2(g)

(a) Balance each of the reactions. (b) By using data in Appendix C, calculate /lH0 for each of the reactions. (c) Based on the values you obtain for /lH0, which of the reactions would you expect to be thermodynamically favored? (That is, which would you expect to be sponta­ neous?) (d) Use the activity series to predict which of these reactions should occur. (Section 4.4) Are these results in accord with your conclusion in part (c) of this problem? [5.115] Consider the following acid-neutralization reactions in­ volving the strong base NaOH(aq): HN03(aq) + NaOH(aq) NaN03(aq) + HzO(I) HCl(aq) + NaOH(aq) NaCl(aq) + H20(/) NH4+(aq) + NaOH(aq) ----> NH3(aq) + Na+(aq) + H20(/) ax>

---+

---+

(a) By using data in Appendix C, calculate !lH0 for each of the reactions. (b) As we saw in Section 4.3, nitric acid and hydrochloric acid are strong acids. Write net ionic equations for the neutralization of these acids. (c) Com­ pare the values of t.H0 for the first two reactions. What can you conclude? (d) In the third equation NH4 +(aq) is acting as an acid. Based on the value of /lH0 for this reac­ tion, do you think it is a strong or a weak acid? Explain. 5.116 Consider two solutions, the first being 50.0 mL of 1.00 M CuS04 and the second 50.0 mL of 2.00 M KOH. When

the two solutions are mixed in a constant-pressure calorimeter, a precipitate forms and the temperature of the mixture rises from 21.5 oc to 27.7 oc. (a) Before mix­ ing, how many grams of Cu are present in the solution of CuS04? (b) Predict the identity of the precipitate in the reaction. (c) Write complete and net ionic equations for the reaction that occurs when the two solutions are mixed. (d) From the calorimetric data, calculate t.H for the reaction that occurs on mixing. Assume that the calorimeter absorbs only a negligible quantity of heat, that the total volume of the solution is 100.0 mL, and that the specific heat and density of the solution after mixing are the same as that of pure water. 5.117 The precipitation reaction between AgN03(aq) and NaCl(aq) proceeds as follows: AgN03(aq) + NaCl(aq) ---- > NaN03(aq) + AgCl(s) (a) By using Appendix C, calculate !lH0 for the net ionic equation of this reaction. (b) What would you expect for the value of t.Ho of the overall molecular equation com­ pared to that for the net ionic equation? Explain. (c) Use the results from (a) and (b ) along with data in Appendix C to determine the value of t.HJ for AgN03(aq). [5.118] A sample of a hydrocarbon is combusted completely in 02(g) to produce 21.83 g C02(g), 4.47 g H20(g), and 311 kJ of heat. (a) What is the mass of the hydrocarbon sample that was combusted? (b) What is the empirical formula of the hydrocarbon? (c) Calculate the value of t.HJ per empirical-formula unit of the hydrocarbon. (d) Do you think that the hydrocarbon is one of those listed in Appendix C? Explain your answer. 5.119 The methane molecule, CH4, has the geometry shown in Figure 2.21. Imagine a hypothetical process in which the methane molecule is "expanded," by simultaneously extending all four C -H bonds to inlinity. We then have the process CH4(g) ---- > C(g) + 4 H(g) (a) Compare this process with the reverse of the reaction that represents the standard enthalpy of formation. (b) Calculate the enthalpy change in each case. Which is the more endothermic process? What accounts for the difference in !lH0 values? (c) Suppose that 3.45 g CH4(g) is reacted with 1.22 g F2(g), forming CF4(g) and HF(g) as sole products. What is the limiting reagent in this reac­ tion? If the reaction occurs at constant pressure, what amount of heat is evolved?

ELECTRONIC STRUCTURE OF ATOMS

T H E GlASS TUBES OF NEON UGHTS contain various gases that can be excited by electricity. Light is produced when electrically excited atoms return to their lowest-energy states.

210

W H AT ' S 6.1 6.2

6.3

6.4

6.5

A H EA D treats matter as waves. We will see that we can describe how an electron exists in atoms by thinking about the electron as a standing wave surrounding the nucleus. The wavefunctions that mathematically describe the electron's position and energy in an atom are called atomic orbitals. The orbitals can be described in a shorthand notation using quantum numbers.

The Wave Nature of Light

We will learn that light (radiant energy, or electromagnetic radiation) has wavelike properties and so is characterized by wavelength,frequency, and speed. Quantized Energy and Photons

We will recognize that many different types of experiments indicate that electromagnetic radiation also has particle-like properties and can be described in terms of plwtons, "particles" of light. Line Spectra and the Bohr Model

We will explore the fact that atoms give off characteristic colors of light (line spectra), when appropriately stimulated. Line spectra provide clues about how electrons are arranged in atoms. Experiments show that electrons exist only at certain energy levels around a nucleus and that energy is involved in moving an electron from one level to another. The Bohr model of the atom pictures the atom as a miniature solar system, with the nucleus of an atom as the "Sun" about which electrons, like planets, orbit. The Wave Behavior of Matter

We recognize that matter also has wavelike properties that are manifested at the atomic scale. Because of the fundamental particle-wave duality of matter, it is impossible to determine simultaneously the exact position and the exact motion of an electron in an atom (Heisenberg's

Uncertainty Principle). Quantum Mechanics and Atomic Orbitals

We will discover that classical mechanics treats matter as particles, while quantum mechanics

WHAT HAPPENS WHEN

6.6 6.7

6.8

6.9

Representations of Orbitals

We will see how to draw pictures of orbitals and interpret graphs of electron density. Many-El ectron Atoms

We will recognize that the energy levels for one electron in an atom are altered when multiple electrons are in an atom. The electron itself has a quantum-mechanical property called spin. The Pauli exclusion principle states that no two electrons in an atom can have the same four quantum numbers (three for the orbital and one for the spin). Therefore, an orbital can hold a maximum of two electrons. Electron Configurations

We will realize that knowing the energies of orbitals as well as some fundamental characteristics of electrons described by Hund's rule allows us to determine how electrons are distributed among various orbitals in an atom (electron configurations). Electron Configurations and the Periodic Table

We will observe that the electron configuration of an atom is related to the location of the element in the periodic table.

someone switches on a neon light?

The electrons in the neon atoms, which are excited to a higher energy by electricity, emit light when they drop back down to a lower energy. The pleasing glow that results is explained by one of the most revolutionary discoveries of the twentieth century, namely, the quantum theory. This theory explains much of the behavior of electrons in atoms. We will see that the behavior of electrons in an atom is quite unlike anything we see in our macroscopic world. In this chapter we will explore the quantum theory and its importance in chemistry. We will begin by looking more closely at the nature of light and how our description of light was changed by the quantum theory. We will explore some of the tools used in quantum mechanics, the "new" physics that had to be developed to describe atoms correctly. We will then use the quantum

211

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C HA PTER 6

Electronic Structure of Atoms theory to describe the arrangements of electrons in atoms-what we call the of atoms. The electronic structure of an atom refers to the number of electrons in an atom as well as the distribution of the electrons around the nucleus and their energies. We will see that the quantum descrip­ tion of the electronic structure of atoms helps us to understand the elegant arrangement of the elements in the periodic table-why, for example, helium and neon are both unreactive gases, whereas sodium and potassium are both soft, reactive metals. In the chapters that follow, we will see how the concepts of quantum theory are used to explain trends in the periodic table and the formation of bonds between atoms.

electronic structure

6. 1 THE WAVE NATURE O F LIGHT

A

Much o f our present understanding o f the electronic structure of atoms has come from analysis of the light either emitted or absorbed by substances. To un­ derstand electronic structure, therefore, we must first learn more about light. The light that we can see with our eyes, visible light, is an example of Because electromagnetic radiation carries energy through space, it is also known as radiant energy. There are many types of elec­ tromagnetic radiation in addition to visible light. These different forms-such as the radio waves that carry music to our radios, the infrared radiation (heat) from a glowing fireplace, and the X-rays used by a dentist-may seem very dif­ ferent from one another, but they all share certain fundamental characteristics. All types of electromagnetic radiation move through a vacuum at a speed 8 of 3.00 X 10 mjs, the speed of light. Ali have wavelike characteristics similar to those of waves that move through water. Water waves are the result of energy imparted to the water, perhaps by the dropping of a stone or the movement of a boat on the water surface (Figure 6.1 ). The mirror system bounced light from the top of Mount Wilson to the top of Mount San Anto­ nio, 22 miles away, where another mirror system bounced the light back to Mount Wilson. If the speed of light was instanta­ neous, or an exact multiple of the turn speed of the rotating mirror, the reflected spot of light would appear exactly super-

•

imposed on the original spot. Michelson was able to change the speed of the rotating mirror and measure small displa­ cements in the position of the reflected spot. The value for the speed of light (in air) based on this experiment is 2.9980 ± 0.0002 X 108 m/s. The main source of error is the distance between the mirrors at the tops of the two mountains, which was measured within a fifth of an inch in 22 miles.

& Figure 6.5 VIew of Mount San Antonio from the top of Mount Wilson. The mountains are 22 miles apart.

SAMPLE EXERCISE 6.1 I Concepts of Wavelength and Frequency

Two electromagnetic waves are represented in the margin. (a) Which wave has the higher frequency? (b) If one wave represents visible light and the other represents in­ frared radiation, which wave is which?

SOLUTION

(a) The lower wave has a longer wavelength (greater distance between peaks). The longer the wavelength, the lower the frequency (v = cf A). Thus, the lower wave has the lower frequency, and the upper wave has the higher frequency. (b) The electromagnetic spectrum (Figure 6.4) indicates that infrared radiation has a longer wavelength than visible light. Thus, the lower wave would be the infrared radiation.

- PRACTICE EXERCISE

If one of the waves in the margin represents blue light and the other red light, which is which? Answer: The expanded visible-light portion of Figure 6.4 tells you that red light has a longer wavelength than blue light. The lower wave has the longer wavelength (lower frequency) and would be the red light.

•

SAMPLE EXERCISE 6.2 I Calculating Frequency from Wavelength

The yellow light given off by a sodium vapor lamp used for public lighting has a wavelength of 589 nm. What is the frequency of this radiation?

SOLUTION

Analyze: We are given the wavelength, A, of the radiation and asked to calculate its frequency, v. Plan: The relationship between the wavelength (which is given) and the frequency (which is the unknown) is given by Equation 6.1. We can solve this equation for v and

6.2

Quantized Energy and Photons

215

then use the values of A and c to obtain a numerical answer. (The speed of light, c, is a fundamental constant whose value is 3.00 X 108 m/s.) Solve: Solving Equation 6.1 for frequency gives = cjA. When we insert the values v for c and A, we note that the units of length in these two quantities are different. We

(

)(

)

can convert the wavelength from nanometers to meters, so the units cancel: 3.00 X loB m/s 589 nm

1 nm 4 5 .09 X 101 s�1 10-9 m = The high frequency is reasonable because of the short wavelength. The units are proper because frequency has units of "per second," or s�1. c

v=A=

Check: - PRACTICE EXERCISE

(a) A laser used in eye surgery to fuse detached retinas produces radiation with a wavelength of 640.0 nm. Calculate the frequency of this radiation. (b) An FM radio station broadcasts electromagnetic radiation at a frequency of 103.4 MHz (megahertz; MHz = 106 s�1 ). Calculate the wavelength of this radiation. The speed of light is 2.998 X loB m/s to four s�cant digits. Answers: (a) 4.688 X 101 s�l, (b) 2.901 m

GIVE IT SOME THOUGHT Our bodies are penetrated by X-rays but not by visible light. Is this because X-rays travel faster than visible light?

6.2 QUANTIZED ENERGY AND PHOTONS Although the wave model of light explains many aspects of its behavior, this model cannot explain several phenomena. Three of these are particularly perti­ nent to our understanding of how electromagnetic radiation and atoms inter­ act: (1) the emission of light from hot objects (referred to as blackbody radiation because the objects studied appear black before heating), (2) the emission of electrons from metal surfaces on which light shines (the photoelectric effect), and (3) the emission of light from electronically excited gas atoms (emission spectra). We examine the first two phenomena here and the third in Section 6.3.

Hot Objects and the Quantization of Energy When solids are heated, they emit radiation, as seen in the red glow of an elec­ tric stove burner and the bright white light of a tungsten lightbuJb. The wave­ length distribution of the radiation depends on temperature; a red-hot object is cooler than a white-hot one (Figure 6.6 �). During the late 1800s, a number of physicists were studying this phenomenon, trying to understand the relation­ ship between the temperature and the intensity and wavelengths of the emitted radiation. The prevailing laws of physics could not account for the observations. 1n 1900 a German physicist named Max Planck (1858-1947) solved the problem by assuming that energy can be either released or absorbed by atoms only in discrete "chunks" of some minimum size. Planck gave the name (meaning "fixed amount") to the smallest quantity of energy that can be emitted or absorbed as electromagnetic radiation. He proposed that the energy, E, of a single quantum equals a constant times the frequency of the radiation: [6.2] E = hv

quantum

The constant h is called and has a value of 6.626 X 10�34 joule-second (J-s). According to Planck's theory, matter is al­ lowed to emit and absorb energy only in whole-number multiples of hv, such as hv, 2hv, 3hv, and so forth. If the quantity of energy emitted by an atom is 3hv,

Planck's constant

for example, we say that three quanta of energy have been emitted (quanta

.a. Figure 6.6 Color as a function of temperature. The color and intensity of the light emitted by a hot object depend on the temperature of the object. The temperature is highest at the center of this pour of molten steel. A5 a result, the light emitted from the center is most intense and of shortest wavelength.

216

C HA PTER 6

Electronic Structure of Atoms

& Figure 6.7 A model for quantized

energy. The potential energy of a person walking up a ramp (a) increases in a uniform, continuous manner, whereas that of a person walking up steps (b) increases in a stepwise, quantized manner.

being the plural of quantum). Because the energy can be released only in specific amounts, we say that the allowed energies are quantized-their values are re­ stricted to certain quantities. Planck's revolutionary proposal that energy is quantized was proved correct, and he was awarded the 1918 Nobel Prize in Physics for his work on the quantum theory. If the notion of quantized energies seems strange, it might be helpful to draw an analogy by comparing a ramp and a staircase (Figure 6.7 shows the relative energies of the hydrogen atom orbitals through n = 3. Each box represents an orbital; orbitals of the same subshell, such as the 2p, are grouped together. When the electron occupies the lowest­ energy orbital (1s), the hydrogen atom is said to be in its ground state. When the

A Figure 6. 18 Orbital energy levels In the hydrogen atom. Each box represents an orbital. Note that all orbitals with the same value for the principal quantum number, n, have the same energy. This is true only in one-electron systems, such as the hydrogen atom.

228

C HA PTER 6

Electronic Structure of Atoms electron occupies any other orbital, the atom is in an excited state. At ordinary temperatures, essentially all hydrogen atoms are in the ground state. The elec­ tron can be excited to a higher-energy orbital by absorption of a photon of appropriate energy.

GIVE IT SOME THOUGHT In Figure 6.18, why is the energy difference between the n = 1 and n = 2 levels so much greater than the energy difference between the n = 2 and n = 3 levels?

- SAMPLE EXERCISE 6.6 I Subshells of the Hydrogen Atom

(a) Without referring to Table 6.2, predict the number of subshells in the fourth shell, that is, for n = 4. (b) Give the label for each of these subshells. (c) How many orbitals are in each of these subshells?

We are given the value of the principal quantum number, n. We need to determine the allowed values of I and m1 for this given value of n and then count the number of orbitals in each subshell.

Analyze and Plan: SOLUTION

There are four subshells in the fourth shell, corresponding to the four possible values of I (0, 1, 2, and 3). These subshells are labeled 4s, 4p, 4d, and 4f The number given in the designa­ tion of a subshell is the principal quantum number, n; the letter designates the value of the angular momentum quantum number, I: for 1 = 0, s; for l = 1, p; for 1 = 2, d; for l = 3,f. There is one 4s orbital (when l = 0, there is only one possible value of m1: 0). There are three 4p orbitals (when I = 1, there are three possible values of m1: 1, 0, and 1 ) . There are five 4d orbitals (when l = 2, there are five allowed values of mr 2, 1, 0, -1, -2). There are seven 4f orbitals (when l = 3, there are seven permitted values of m1: 3, 2, 1, 0, -1, -2, -3).

- PRACTICE EXERCISE

(a) What is the designation for the subshell with n = 5 and 1 = 1? (b) How many or­ bitals are in this subshell? (c) Indicate the values of m1 for each of these orbitals. Answers: (a) 5p; (b) 3; (c) 1, 0, -1

6.6 REPRESENTATIONS OF ORBITALS In our discussion of orbitals so far, we have emphasized their energies. But the wave function also provides information about the electron's location in space when it occupies an orbital. Let's examine the ways that we can picture the or­ bitals. In doing so, we will examine some important aspects of the electron-den­ sity distributions of the orbitals. First, we will look at the three-dimensional shape of the orbital-is it spherical, for example, or does it have directionality? Second, we will examine how the probability density changes as we move on a straight line farther and farther from the nucleus. Finally, we will look at the typical three-dimensional sketches that chemists use in describing the orbitals.

The s Orbitals One representation of the lowest-energy orbital of the hydrogen atom, the ls, is shown in Figure 6.17. This type of drawing, which shows the distribution of electron density around the nucleus, is one of the several ways we use to help us visualize orbitals. The first thing that we notice about the electron density for the ls orbital is that it is spherically symmetric-in other words, the electron den­ sity at a given distance from the nucleus is the same regardless of the direction in which we proceed from the nucleus. All of the other s orbitals (2s, 3s, 4s, and so forth) are spherically symmetric as well. Recall that the l quantum number

6.6

0

1 2 3 4 5 6 7 8 9 10 Distance from the nucleus, r (A)

0

1 2 3 4 5 6 7 8 9 10 Distance from the nucleus, r (A)

(a)

& Figure 6.1 9 Radial probability functions for the ls,

(b) 2s,

and 3s orbitals. These plots show the probability of finding the electron as a function of distance from the nucleus. As n increases, the most likely distance at which to find the electron moves farther from the nucleus, similar to the Bohr model. In the 2s and 3s orbitals the radial probability function drops to zero at certain distances from the nucleus but then rises again. The points at which the probability is zero are called nodes.

for the s orbitals is 0; therefore the m1 quantum number must be 0. Therefore, for each value of n, there is only one s orbital. So what is different about the s orbitals having different n quantum num­ bers? For example, how does the electron-density distribution of the hydrogen atom change when the electron is excited from the 1s orbital to the 2s orbital? To address questions like this, we must look at the radial probability density, that is, the probability that we will find the electron at a specific distance from the nucleus. In Figure 6.19 & we have plotted the radial probability density for the 1s orbital as a function of r, the distance from the nucleus. The resulting curve is the radial probability function for the 1s orbital. (Radial probability functions are described more fully in the "A Closer Look" box in this section.) We see that the probability of finding the electron rises rapidly as we move away from the nucleus, maximizing at a distance of 0.529 A from the nucleus, and then falls off rapidly. Thus, when the electron occupies the 1s orbital, it is most likely to be found 0.529 A from the nucleus*. We still use the probabilistic description, con­ sistent with the uncertainty principle. Notice also that the probability of finding the electron at a distance greater than 3 A from the nucleus is essentially zero. Figure 6.19(b) shows the radial probability function for the 2s orbital of the hydrogen atom. We can see three significant differences between this plot and that for the 1s orbital: (1) There are two separate maxima in the radial probabili­ ty function for the 2s orbital, namely a small peak at about r = 0.5 A and a much larger peak at about r = 3 A; (2) Between these two peaks is a point at which the function goes to zero (at about r = 1 A). An intermediate point at which a prob­ ability function goes to zero is called a node. There is a zero probability of find­ ing the electron at a distance corresponding to a node, even though the electron might be found at shorter or longer distances; (3) The radial probability function for the 2s orbital is significantly broader (more spread out) than that for the 1s orbital. Thus, for the 2s orbital, there is a larger range of distances from the *In tire quantum mechanical model, the most probable distance at which to find the electron in the 1 s orbital-

0.529

A-is identical to tl1e radius of the orbit predicted by Bohr for n

called the Bohr radius.

=

1 . Tl1e dista11ce 0.529 A is ojte11

Representations of Orbitals

0

229

1 2 3 4 5 6 7 8 9 10 Distance from the nucleus, r (A) (c)

230

C HA PTER 6

Electronic Structure of Atoms

3s

nucleus at which there is a high probability of finding the electron than for the ls orbital. This trend continues for the 3s orbital, as shown in Figure 6. 19(c) . Thera­ dial probability function for the 3s orbital has three peaks of increasing size, with the largest peak maximizing even farther from the nucleus (at about r = 7 A) at which it has two nodes and is even more spread out. The radial probability functions in Figure 6.19 tell us that as n increases, there is also an increase in the most likely distance from the nucleus to find the electron. In other words, the size of the orbital increases with increasing n, just as it did in the Bohr model. One widely used method of representing orbitals is to display a boundary surface that encloses some substantial portion, say 90%, of the total electron density for the orbitaL For the s orbitals, these contour representations are spheres. The contour representations of the ls, 2s, and 3s orbitals are shown in Figure 6.20 �- All the orbitals have the same shape, but they differ in size. Although the details of how the electron density varies within the contour rep­ resentation are lost in these representations, this is not a serious disadvantage. For more qualitative discussions, the most important features of orbitals are their relative sizes and their shapes, which are adequately displayed by contour representations .

.A. Figure 6.20 Contour representations of the ls, 2s, and 3s orbitals. The relative radii of the spheres correspond to a 90% probability of finding the electron within each sphere.

How many maxima would you expected to find in the radial probability function for the 4s orbital of the hydrogen atom? How many nodes would you expect in the 4s ra­ dial probability function?

1s

2s

GIVE IT SOME THO UGHT

PROBAB ILITY D EN S I TY AND RA D I AL P R O B A B ILITY F U N C T I O N S

he quantum mechanical description of the hydrogen

T atom requires that we talk about the position of the elec­

tron in the atom in a somewhat unfamiliar way. In classical physics, we can exactly pinpoint the position and velocity of an orbiting object, such as a planet orbiting a star. Under quan­ tum mechanics, however, we must describe the position of the electron in the hydrogen atom in terms of probabilities rather than an exact location-an exact answer would violate the un­ certainty principle, which becomes important when consider­ ing subatomic particles. The information we need about the probability of finding the electron is contained in the wave functions, rp, that are obtained when Schriidinger 's equation is solved. Remember that there are an infinite number of wave functions (orbitals) for the hydrogen atom, but the electron can occupy only one of them at any given time. Here we will discuss briefly how we can use the orbitals to obtain radial probability functions, such as those in Figure 6.19. In Section 6.5 we stated that the square of the wave func­ tion, rp2, gives the probability that the electron is at any one given point in space. Recall that this quantity is called the probability density for the point. For a spherically symmetric s orbital, the value of rf! depends only on the distance from the nucleus, r. Let's consider a straight line outward from the nu­ cleus, as shown in Figure 6.21 �. The probability of finding the electron at distance r from the nucleus along that line is [rp(r)f, where rp(r) is the value of rf! at distance r. Figure 6.23 � shows 2 plots of [rp (r)] as a function of r for the 1s, 2s, and 3s orbitals of the hydrogen atom.

You will notice that the plots in Figure 6.23 look distinctly different from the radial probability functions plotted in Figure 6.19. These two types of plots for the s orbitals are very closely related, but they provide somewhat different informa­ tion. The probability density, [rp(r)f, tells us the probability of finding the electron at a specific point in space that is at dis­ tance r from the nucleus. The radial probability function, which we will denote P(r), tells us the probability of finding the electron at any point that is distance r from the nucleus. In other words, to get P(r) we need to "add up" the probabilities of finding the electron over all the points at distance r from the nucleus. The difference between these descriptions may seem rather subtle, but mathematics provides us with a precise way to connect them.

A sphere

of radius r around the nucleus

A point at distance r from the nucleus

.A. Figure 6.21 Probability at a point. The probability density, ljf(r)2 gives the probability that the electron will be found at a specific point at distance r from the nucleus. The radial probability function, 41jJr 2 1jJ(r)2, gives the probability that the electron will be found at any point distance r from the nucleus-in other words, at any point on the sphere of radius r.

6.6

231

Representations of Orbitals

The p Orbitals The distribution of electron density for a 2p orbital is shown in Figure 6.22(a) T. As we can see from this figure, the electron density is not distributed in a spher­ ically symmetric fashion as in an s z orbital. Instead, the electron density is concentrated in two regions on ei­ ther side of the nucleus, separated by a node at the nucleus. We say that this dumbbell-shaped orbital has two lobes. Recall that we are making X y X no statement of how the electron is moving within the orbital. The only thing Figure 6.22(a) portrays is the averaged distribution of the electron (a) density in a 2p orbital. Beginning with the 11 = 2 shell, each shell has three p orbitals. Recall that the l quantum number for p orbitals is 1. Therefore, the magnetic quantum number m1 can have three possible values: -1, 0, and + 1 . Thus, there are three 2p orbitals, three 3p orbitals, and so forth, corresponding to the three possible values of m1• Each set of p orbitals has the dumbbell shapes shown in Figure 6.22(a) for the 2p orbitals. For each value of 11, the three p orbitals have the same size and shape but differ from one another in spatial orientation. We usually represent p orbitals by drawing the shape and orientation of their wave func­ tions, as shown in Figure 6.22(b ) . It is convenient to label these as the p., Py and Pz orbitals. The letter subscript indicates the Cartesian axis along which the

I

.X,

2s

1s

n = 1, l = 0

n

= 2, I = 0

I I I I

:......- Node I I I I I I

(a)

(b)

3s n = 3, I = 0

I I I

I I I

Nodes � I ,

I I I I I

(c)

As shown in Figure 6.21, the collection of points at dis­ tance r from the nucleus is simply a sphere of radius r. The probability density at every point on that sphere is [op(r)f. To add up all of the individual probability densities requires the use of calculus and is beyond the scope of this text (in the language of calculus "we integrate the probability density over the surface of the sphere"). The result we obtain is easy

y

Px

Py

(b) A Figure 6.22 The p orbitals. (a) Electron-density distribution of a 2p orbital. (b) Contour representations of the three p orbitals. Note that the subscript on the orbital label indicates the axis along which the orbital lies.

H + + e- is called the ionization energy of hydrogen. The experi­ mentally determined value for the ionization energy of hydrogen is 1310 kJ/mol. How does this compare to your calculation?

6.36 For each of the following electronic transitions in the hy­ drogen atom, calculate the energy, frequency, and wave­ length of the associated radiation, and determine whether the radiation is emitted or absorbed during the transition: (a) from n = 4 to n = 1, (b) from n = S to n = 2, (c) from n = 3 to n = 6. Does any of these transi­ tions emit or absorb visible light?

6.37 The visible emission lines observed by Balmer all in­ volved nf = 2. (a) Explain why only the lines with nf = 2 were observed in the visible region of the electro­ magnetic spectrum. (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which n1 = 3, 4, and S-and identify these ]jnes in the emission spectrum shown in Figure 6.13.

6.38 The Lyman series of emission lines of the hydrogen

atom are those for which n1 = 1. (a) Determine the re­ gion of the electromagnetic spectrum in which the lines of the Lyman series are observed. (b) Calculate the wavelengths of the first three lines in the Lyman series-those for which n1 = 2, 3, and 4.

249

6.39 One of the emission lines of the hydrogen atom has a

wavelength of 93.8 nm. (a) In what region of the electro­ magnetic spectrum is this emission found? (b) Deter­ mine the initial and final values of n associated with this emission. 6.40 The hydrogen atom can absorb light of wavelength 2626 nm. (a) In what region of the electromagnetic spectrum is this absorption found? (b) Determine the initial and final values of n associated with this absorption.

6.41 Use the de Broglie relationship to determine the wave­

lengths of the following objects: (a) an 8S-kg person skiing at SO km/hr, (b) a 10.0-g bullet fired at 2SO mjs, (c) a lithi­ um atom moving at 2.5 X 105 mjs, (d) an ozone (03) mol­ ecule in the upper atmosphere moving at SSO mjs. 6.42 Among the elementary subatomic particles of physics is the muon, which decays within a few nanoseconds after formation. The muon has a rest mass 206.8 times that of an electron. Calculate the de Broglie wavelength associated with a muon traveling at a velocity of 8.8S X 105 cmjs.

6.43 Neutron diffraction is an important technique for deter­ mining the structures of molecules. Calculate the veloci­ ty of a neutron needed to achieve a wavelength of 0.9SS A. (Refer to the inside cover for the mass of the neutron). 6.44 The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of 9.38 X 106 m/s. What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

6.45 Using Heisenberg's uncertainty principle, calculate the

uncertainty in the position of (a) a 1.50-mg mosquito moving at a speed of 1.40 m/s if the speed is known to within ±0.01 m/s; (b) a proton moving at a speed of (S.OO ± 0.01) X 104 m/s. (The mass of a proton is given in the table of fundamental constants in the inside cover of the text.) 6.46 Calculate the uncertainty in the position of (a) an elec­ tron moving at a speed of (3.00 ± 0.01) x 105 m/s, (b) a neutron moving at this same speed. {The masses of an electron and a neutron are given in the table of funda­ mental constants in the inside cover of the text.) (c) What are the implications of these calculations to our model of the atom?

Quantum Mechanics and Atomic Orbitals 6.47 (a) Why does the Bohr model of the hydrogen atom vio­ late the uncertainty principle? (b) In what way is the description of the electron using a wave function consis­ tent with de Broglie's hypothesis? (c) What is meant by the term probability density? Given the wave function, how do we find the probability density at a certain point in space?

6.48 (a) According to the Bohr model, an electron in the ground state of a hydrogen atom orbits the nucleus at a

specific radius of 0.53 A. In the quantum mechanical de­ scription of the hydrogen atom, the most probable dis­ tance of the electron from the nucleus is 0.53 A. Why are these two statements different? (b) Why is the use of Schrodinger's wave equation to describe the location of a particle very different from the description obtained from classical physics? (c) In the quantum mechanical description of an electron, what is the physical signifi­ cance of the square of the wave function, lj/?

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C HA PTER 6

Electronic Structure of Atoms

= 4, what are the possible values of I? (b) For 2, what are the possible values of m1? (c) If m1 is 2, what are the possible values for I? 6.50 How many possible values for I and m1 are there when (a) n = 3; (b) n = 5?

6.49 (a) For n I =

6.51 Give the numerical values of n and I corresponding to

each of the following orbital designations: (a) 3p, (b) 2s,

(c) 4f, (d) Sd. 6.52 Give the values for n, I, and mt for (a) each orbital in the 2p subshell, (b) each orbital in the Sd subshell.

6.53 Which of the following represent impossible combina­

tions of n and I: (a) 1p, (b) 4s, (c) Sf, (d) 2d?

6.54 For the table below, write which orbital goes with the

quantum numbers. Don't worry about x, y, z subscripts. If the quantum numbers are not allowed, write "not allowed." n

2

1

mt

Orbital

-1

2p (example)

1

0

0

3

-3

3

2

2 -2

2

0

0 0

4

2

1

5

3

0

6.55 Sketch the shape and orientation of the following types

of orbitals: (a) s, (b) pz, (c) dxy·

6.56 Sketch the shape and orientation of the following types

of orbitals: (a) Px, (b) di', (c) dx' -y'·

6.57 (a) What are the similarities and differences between the

1s and 2s orbitals of the hydrogen atom? (b) In what sense does a 2p orbital have directional character? Com­ pare the "directional" characteristics of the Px and dx'- y' orbitals (that is, in what direction or region of space is the electron density concentrated?). (c) What can you say about the average distance from the nucleus of an electron in a 2s orbital as compared with a 3s orbital? (d) For the hydrogen atom, list the following orbitals in order of increasing energy (that is, most stable ones first): 4f, 6s, 3d, 1s, 2p. 6.58 (a) With reference to Figure 6.19, what is the relationship between the number of nodes in an s orbital and the value of the principal quantum number? (b) Identify the number of nodes; that is, identify places where the elec­ tron density is zero, in the 2px orbital; in the 3s orbital. (c) What information is obtained from the radial proba­ bility functions in Figure 6.19? (d) For the hydrogen atom, list the following orbitals in order of increasing energy: 3s, 2s, 2p, Ss, 4d.

-1

0

Many-Electron Atoms and Electron Configurations 6.59 For a given value of the principal quantum number, 11,

6.65 (a) What are "valence electrons"? (b) What are "core

how do the energies of the s, p, d, andfsubshells vary for (a) hydrogen, (b) a many-electron atom? 6.60 (a) The average distance from the nucleus of a 3s elec­ tron in a chlorine atom is smaller than that for a 3p elec­ tron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a 3s electron from the chlorine atom, as compared with a 2p electron? Explain.

electrons"?(c) What does each box in an orbital diagram represent? (d) What quantity is represented by the direc­ tion (up or down) of the hali-arrows in an orbital diagram? 6.66 For each element, count the number of valence elec­ trons, core electrons, and unpaired electrons in the ground state: (a) carbon, (b) phosphorus, (c) neon.

6.61 (a) What experimental evidence is there for the electron

6.67 Write the condensed electron configurations for the fol­

having a "spin"? (b) Draw an energy-level diagram that shows the relative energetic positions of a 1s orbital and a 2s orbital. Put two electrons in the 1s orbital. (c) Draw an arrow showing the excitation of an electron from the 1s to the 2s orbital. 6.62 (a) State the Pauli exclusion principle in your own words. (b) The Pauli exclusion principle is, in an impor­ tant sense, the key to understanding the periodic table. Explain why. 6.63 What is the maximum number of electrons that can oc­

cupy each of the following subshells: (a) 3p, (b) Sd, (c) 2s,

(d) 4f?

6.64 What is the maximum number of electrons in an atom

that can have the following quantum numbers: (a) n = 2, m, = -t, (b) n = 5, I = 3; (c) n = 4, I = 3, mt = -3; (d) n = 4, I = 1, m1 = 1 ?

lowing atoms, using the appropriate noble-gas core ab­ breviations: (a) Cs, (b) Ni, (c) Se, (d) Cd, (e) U, (f) Pb. 6.68 Write the condensed electron configurations for the fol­ lowing atoms, and indicate how many unpaired elec­ trons each has: (a) Ga, (b) Ca, (c) V, (d) I, (e) Y, (f) Pt, (g) Lu. 6.69 Ions also have electron configurations (Section 7.4).

Cations have fewer valence electrons, and anions have more valence electrons, respectively, than their parent atoms. For example, chloride, Cl", has an electron con­ figuration of 1s22s22p63s23p6, for a total of 18 electrons, compared to 17 for neutral chlorine, the element. Na has an electron configuration of 1s22s22p63s 1, but Na+ has an electron configuration of 1s22s22p6 Write out the electron configurations for (a) F-, (b) I-, (c) 02-, (d) K+, 2 3 (e) Mg +, (f) Al +

Additional Exercises 6.70 In the transition metals (the d-block), the electron con­

figuration of cations is different than what you might expect. Instead of the d electrons being lost first, s elec­ trons are lost first. For example, the electron confi­ guration of iron, Fe, is [Ar]4s23d6; but the electron configuration of Fe2+ is [Ar]3d6; the 4s electrons are eliminated to make the cation. Write out the electron configurations of (a) Zn2+ (b) Pt2+ (c) Cr3+ (d) Ti4+

6.72 Identify the group of elements that corresponds to each of the following generalized electron configurations: (a) [noble gas] ns2np5 (b) [noble gas] ns2(n - 1)d2 (c) [noble gas] 11s2(11 - 1 )d10np1 (d) [noble gas] ns2(n - 2)f6

6.73 What is wrong with the following electron configura­

tions for atoms in their ground states? (a) Is22s23s1, (b) [Ne]2s22p3, (c) [Ne]3s23d5.

6.71 Identify the specific element that corresponds to each

of the following electron configurations: (a) 1s22s2, (b) 1s22s22p4, (c) [Ar]4s13d5, (d) [Kr]5s24d105p4, (e) 1s1.

251

6.74 The following electron configurations represent excited

states. Identify the element, and write its ground-state condensed electron configuration. (a) Is22s23p24p1, (b) [Ar]3d104s14p45s1, (c) [Kr]4d65s25p1.

ADDITIONAL EXERCISES 6.75 Consider the two waves shown here, which we will con­

the energy of a mole of 320-nm photons. (c) Which are more energetic, photons of UV-A racliation or photons of UV-B radiation? (d) The UV-B radiation from the Sun is considered a greater cause of sunburn in humans than is UV-A radiation. ls this observation consistent with your answer to part (c)?

sider to represent two electromagnetic radiations: (a) What is the wavelength of wave A? Of wave B? (b) What is the frequency of wave A? Of wave B? (c) Identify the regions of the electromagnetic spectrum to which waves A and B belong.

6.79 The watt is the derived 51 unit of power, the measure of energy per unit time: 1 W = 1 J/s. A semiconductor laser in a CD player has an output wavelength of 780 nm and a power level of 0.10 mW. How many pho­ tons strike the CD surface during the playing of a CD 69 minutes in length?

6.80 The color wheel (Figure 24.24) is a convenient way to

relate what colors of light are absorbed by a sample and the visible appearance of the sample. lf all visible colors are absorbed by a sample, the sample appears black. If no colors are absorbed by the sample, the sample ap­ pears white. If the sample absorbs red, then what we see is green; such complementary colors are across the wheel from each other. Carrots appear orange because they contain a compound called carotene. Based on the color wheel, what is the possible wavelength range for the light absorbed by carotene?

6.76 Certain elements emit light of a specific wavelength

when they are burned. Historically, chemists used such emission wavelengths to determine whether specific el­ ements were present in a sample. Some characteristic wavelengths for some of the elements are

Ag

328.1 nm

Fe

372.0 nm

Au

267.6 nm

K

404.7 nm

Ba

455.4 nm

Mg 285.2 nm

Ca

422.7 nm

Na

589.6 nm

Cu

324.8 nm

Ni

341.5 nm

(a) Determine which elements emit radiation in the visi­ ble part of the spectrum. (b) Which element emits pho­ tons of highest energy? Of lowest energy? (c) When burned, a sample of an unknown substance is found to emit light of frequency 6.59 x 1014 s-1. Which of these elements is probably in the sample?

6.81 A photocell, such as the one illustrated in Figure 6.8(b ), is a device used to measure the intensity of light. In a cer­ tain experiment, when light of wavelength 630 nm is di­ rected onto the photocell, electrons are emitted at the rate of 2.6 X 10-12 Coulombs/sec. Assume that each photon that impinges on the photocell emits one electron. How many photons per second are striking the photocell? How much energy per second is the photocell absorbing?

6.82 The light-sensitive substance in black-and-white photo­

6.77 In June 2004, the Cassini-Huygens spacecraft began orbit­

ing Saturn and transmitting images to Earth. The closest distance between Saturn and Earth is 746 million miles. What is the minimum amount of time it takes for the transmitted signals to travel from the spacecraft to Earth?

6.78 The rays of the Sun that cause tanning and burning are

in the ultraviolet portion of the electromagnetic spec­ trum. These rays are categorized by wavelength. So­ called UV-A racliation has wavelengths in the range of 320-380 nm, whereas UV-B radiation has wavelengths in the range of 290-320 nm. (a) Calculate the frequency of light that has a wavelength of 320 nm. (b) Calculate

6.83

graphic film is AgBr. Photons provide the energy neces­ sary to transfer an electron from Br- to Ag+ to produce elemental Ag and Br and thereby darken the film. (a) If a minimum energy of 2.00 x 105 J/mol is needed for this process, what is the minimum energy needed from each photon? (b) Calculate the wavelength of the light neces­ sary to provide photons of this energy. (c) Explain why this film can be handled in a darkroom under red light. In an experiment to study the photoelectric effect, a sci­ entist measures the kinetic energy of ejected electrons as a function of the frequency of radiation hitting a metal surface. She obtains the following plot: The point labeled "vo" corresponds to light with a wavelength of 680 nm .

252

C HA PTER 6

Electronic Structure of Atoms

(a) What is the value of v0 in s-1 ? (b) What is the value of the work function of the metal in units of kJ/mol of ejected electrons? (c) What happens when the metal is irradiated with light of frequency less than v0? (d) Note that when the frequency of the light is greater than v0, the plot shows a straight line with a nonzero slope. Why is this the case? (e) Can you determine the slope of the line segment discussed in part (d)? Explain.

Vo

[ 6.91] Consider the discussion of raclial probability functions in

Frequency -

6.84 The series of emission lines of the hydrogen atom for

which nf = 3 is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed. (b) Calculate the wavelengths of the first three lines in the Paschen series-those for which n, = 4, 5, and 6. 6.85 When the spectrum of light from the Sun is examined in high resolution in an experiment similar to that illustrat­ ed in Figure 6.11, dark lines are evident. These are called Fraunhofer lines, after the scientist who studied them ex­ tensively in the early nineteenth century. Altogether, about 25,000 lines have been identified in the solar spec­ trum between 2950 A and 10,000 A. The Fraunhofer lines are attributed to absorption of certain wavelengths of the Sun's "white" light by gaseous elements in the Sun's at­ mosphere. (a) Describe the process that causes absorption of specific wavelengths of light from the solar spectrum. (b) lf a scientist wanted to know which Fraunhofer lines belonged to a given element, say neon, what experiments could she conduct here on Earth to provide data? [6.86] Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as He+ and u 2+. (a) Why is the Bohr model applicable to He + ions but not to neutral He atoms? (b) The ground-state energies of H, He+, and Li 2+ are tabulated as follows: Atom or ion

H

Ground-state energy -2.18 X 10-1 8 J

-8.72 x 10-18 J

and not visible light, suitable for the determination of structure at the atomic level? (b) How fast would an electron have to be moving to have the same wave­ length as these X-rays? [6.88] An electron is accelerated through an electric potential to a kinetic energy of 18.6 keV. What is its characteristic wavelength? [Hint: Recall that the kinetic energy of a moving object is £ = � mv2, where m is the mass of the object and v is the speed of the object.] 6.89 In the television series Star Trek, the transporter beam is a device used to "beam down" people from the Starship Enterprise to another location, such as the surface of a planet. The writers of the show put a "Heisenberg com­ pensator" into the transporter beam mechanism. Ex­ plain why such a compensator would be necessary to get around Heisenberg's uncertainty principle. 6.90 Which of the quantum numbers governs (a) the shape of an orbital, (b) the energy of an orbital, (c) the spin prop­ erties of the electron, (d) the spatial orientation of the orbital?

-1 .96 X 10- 1 7 J

By examining these numbers, propose a relationship be­ tween the ground-state energy of hydrogen-like sys­ tems and the nuclear charge, Z. (c) Use the relationship you derive in part (b) to predict the ground-state energy of the c5+ ion. 6.87 Under appropriate conditions, molybdenum emits X­ rays that have a characteristic wavelength of 0.711 A. These X-rays are used in diffraction experiments to de­ termine the structures of molecules. (a) Why are X-rays,

the "A Closer Look" box in Section 6.6. (a) What is the difference between the probability density as a function of r and the radial probability function as a function of r? (b) What is the significance of the term 47TT2 in the radi­ al probability functions for the s orbitals? (c) Based on Figures 6.19 and 6.22, make sketches of what you think the probability density as a function of r and the radial probability function would look like for the 4s orbital of the hydrogen atom. 6.92 The "magic numbers" in the periodic table are the atom­ ic numbers of elements with high stability (the noble gases): 2, 10, 18, 36, 54, and 86. In terms of allowed val­ ues of orbitals and spin quantum numbers, explain why these electron arrangements correspond to special stability.

[6.93] For non-spherically symmetric orbitals, the contour rep­

resentations (as in Figures 6.23 and 6.24) suggest where nodal planes exist (that is, where the electron density is zero). For example, the Px orbital has a node wherever x = 0. This equation is satisfied by all points on the yz plane, so this plane is called a nodal plane of the Px or­ bital. (a) Determine the nodal plane of the p, orbital. (b) What are the two nodal planes of the dxy orbital? (c) What are the two nodal planes of the di!-y' orbital?

[6.94] As noted in Figure 6.26, the spin of an electron generates

a magnetic field, with spin-up and spin-down electrons having opposite fields. In the absence of a magnetic field, a spin-up and a spin-down electron have the same ener­ gy. (a) Why do you think that the use of a magnet was important in the discovery of electron spin (see the "A Closer Look" box in Section 6.8)? (b) Imagine that the two spinning electrons in Figure 6.26 were placed be­ tween the poles of a horseshoe magnet, with the north pole of the magnet at the top of the figure. Based on what you know about magnets, would you expect the left or right electron in the figure to have the lower energy? (c) A phenomenon called electron spin resonance (ESR) is closely related to nuclear magnetic resonance. In ESR a compound with an unpaired electron is placed in a mag­ netic field, which causes the unpaired electron to have two different energy states analogous to Figure 6.28.

Integrative Exercises

ESR uses microwave radiation to excite the unpaired electron from one state to the other. Based on your read­ ing of the "Chemistry and Life" box in Section 6.8, does an ESR experiment require photons of greater or lesser energy than an NMR experiment? [6.95] The "Chemistry and Life" box in Section 6.8 described the techniques called NMR and MRI. (a) Instruments for obtaining MRI data are typically labeled with a frequen­ cy, such as 600 MHz. Why do you suppose this label is relevant to the experiment? (b) What is the value of !J.E in Figure 6.28 that would correspond to the absorption of a photon of radiation with frequency 450 MHz? (c) In general, the stronger the magnetic field, the greater the

253

information obtained from an NMR or MRJ experiment. Why do you suppose this is the case? 6.96 Suppose that the spin quantum number, m,, could have three allowed values instead of two. How would this af­ fect the number of elements in the first four rows of the periodic table? 6.97 Using only a periodic table as a guide, write the con­ densed electron configurations for the following atoms: (a) Se, (b) Rh, (c) Si, (d) Hg, (e) Hf. 6.98 Scientists have speculated that element 126 might have a moderate stability, allowing it to be synthesized and characterized. Predict what the condensed electron con­ figuration of this element might be.

INTEGRATIVE EXERCISES 6.99 Microwave ovens use microwave radiation to heat

food. The energy of the microwaves is absorbed by water molecules in food, then transferred to other com­ ponents of the food. (a) Suppose that the microwave radiation has a wavelength of 11.2 em. How many pho­ tons are required to heat 200 mL of coffee from 23 'C to 60 'C? (b) Suppose the microwave's power is 900 W (1 Watt = 1 Joule/sec). How long would you have to heat the coffee in part (a)? 6.100 The stratospheric ozone (03) layer helps to protect us from harmful ultraviolet radiation. It does so by absorbing ultraviolet light and falling apart into an 02 molecule and an oxygen atom, a process known as photodissociation. 0 3(g) 02 (g) + O(g) Use the data in Appendix C to calculate the enthalpy change for this reaction. What is the maximum wave­ length a photon can have if it is to possess sufficient en­ ergy to cause this dissociation? In what portion of the spectrum does this wavelength occur? 6.101 The discovery of hafnium, element number 72, provid­ ed a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an ele­ ment number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconi­ um than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group 4B, can be produced as a metal by reduction of solid ZrCL, with molten sodium metal. Write a balanced chemical equation for the reac­ tion. Is this an oxidation-reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, ZrOz, is reacted with chlorine gas in the pres­ ence of carbon. The products of the reaction are ZrCI4 and two gases, C02 and CO in the ratio 1 : 2 Write a bal­ anced chemical equation for the reaction. Starting with a 55.4-g sample of ZrOz, calculate the mass of ZrC4 formed, assuming that Zr02 is the limiting reagent and assuming 100% yield. (d) Using their electron configu­ rations, account for the fact that Zr and Hf form chlo­ rides MCI4 and oxides M02. ----+

6.102 (a) Account for formation of the following series of ox­

ides in terms of the electron configurations of the ele­ ments and the discussion of ionic compounds in Section 2.7: K20, CaO, Sc203, TiOz, V205, Cr03 . (b) Name these oxides. (c) Consider the metal oxides whose enthalpies of formation (in kJ mol-1) are listed here.

Oxide

K20(s)

CaO(s)

Ti02(s)

-363.2

-635.1

-938.7

-1550.6

Calculate the enthalpy changes in the following general reaction for each case: M110.,(s) + H2(g) nM(s) + mH20(g) (You will need to write the balanced equation for each case, then compute !J.H'.) (d) Based on the data given, estimate a value of !l.HJ for Sc203(s). 6.103 The first 25 years of the twentieth century were momen­ tous for the rapid pace of change in scientists' understan­ ding of the nature of matter. (a) How did Rutherford's experiments on the scattering of a particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals that preceded Thomson's work, that the cathode rays are a wave phenomenon? 2 [6.104] The two most common isotopes of uranium are 35U and 238U. (a) Compare the number of protons, the number of electrons, and the number of neutrons in atoms of these two isotopes. (b) Using the periodic table in the front in­ side cover, write the electron configuration for a U atom. (c) Compare your answer to part (b) to the electron con­ figuration given in Figure 6.31. How can you explain any differences between these two electron configu­ rations? (d) 238U undergoes radioactive decay to 234Th. How many protons, electrons, and neutrons are gained or lost by the 238U atom during this process? (e) Exami­ ne the electron configuration for Th in Figure 6.31. Are you surprised by what you find? Explain. 6.105 Imagine sunlight falling on three square areas. One is an inert black material. The second is a photovoltaic cell surface, which converts radiant energy into electricity. The third is an area on a green tree leaf. Oraw diagrams that show the energy conversions in each case, using Figure 5.9 as a model. How are these three examples re­ lated to the idea of sustainable energy sources? ----+

7

OIL PAINTS CONTAIN PIGMENTS, which are usually highly colored salts, suspended in an organic carrier composed of a variety of heavy hydrocarbon molecules. This painting, by the famous French Impressionist Claude Monet

La rue Montorguei/, fete du 30 juin 1878. (1 840-1 926), is entitled

Photo credit: Claude Monet

(1 840--1 926) "Rue Montorgueil in Paris, Festival of 30 June 1 8 78," 1 878. Herve Lewandowski/Reunion des Muses Nationaux/Art Resource, NY.

254

PERIODIC PROPERTIES OF THE ELEMENTS

W H AT ' S 7.1 7.2

7.3

7.4

A H E A D

Development of the Periodic Table

7.5

Effective Nuclear Charge

7.6

We begin our discussion with a brief history of the periodic table. We next explore the many properties of atoms that depend both on the net attraction of the outer electrons to the nucleus and on the average distance of those electrons from the nucleus. The net positive charge of the nucleus experienced by the outer electrons is called the effective nuclear charge. Sizes of Atoms and Ions

We explore the relative sizes (or radii) of atoms and ions, both of which follow trends that are related to their placement in the periodic table. Ionization Energy

We next encounter the energy-called ionization energy-required to remove one or more electrons from an atom. The periodic trends in ionization energy depend upon variations in effective nuclear charge and atomic radii.

7.7 7.8

Electron Affinities

Next we will examine periodic trends in the energy released when an electron is added to an atom. Metals, Non metals, and Metalloids

We recognize that the physical and chemical properties of metals and nonmetals are distinctly different and can be understood from the fundamental characteristics of atoms discussed earlier in the chapter, particularly ionization energy. Metalloids display properties that are intermediate between metals and nonmetals. Group Trends for the Active Metals

We examine some periodic trends in the chemistry of the active metals (groups lA and 2A). Group Trends for Selected N o n metals

We examine some periodic trends in the chemistry of hydrogen, as well as groups 6A, 7A, and 8A.

THE BEAUTY O F AN I M PRESSIO N I ST O I L PAI NTI NG,

such as the Monet

masterpiece shown here, begins with chemistry. Colorful inorganic salts are suspended in various organic media that contain hydrocarbons and other molecular substances. Indeed, the great p ainters had a gift for using compounds of elements that span nearly the entire periodic table. Today the periodic table is still the most significant tool chemists have for organizing and remembering chemical facts. As we saw in Chapter 6, the peri­ odic nature of the table arises from the repeating patterns in the electron con­ figurations of the elements. Elements in the same column of the table contain the same number of electrons in their valence orbitals, the occupied orbitals 4 that hold the electrons involved in bonding. For example, 0 ([He]2s2 2p ) and S ([Ne]3s23p4) are both members of group 6A. The similarity of the electron distribution in their valence s and p orbitals leads to similarities in the proper­ ties of these two elements. When we compare 0 and S, however, it is apparent that they exhibit dif­ ferences as well, not the least of which is that oxygen is a colorless gas at room 255

256

C HA PTER 7

Periodic Properties of the Elements temperature, whereas sulfur is a yellow solid (Figure 7.1 "ii ) ! One of the major differences between atoms of these two elements is their electron configura­ tions: the outermost electrons of 0 are in the second shell, whereas those of S are in the third shell. We will see that electron configurations can be used to explain differences as well as similarities in the properties of elements. In this chapter we explore how some of the important properties of ele­ ments change as we move across a row or down a column of the periodic table. In many cases the trends within a row or column allow us to make pre­ dictions about the physical and chemical properties of the elements.

7. 1 DEVELO P M ENT O F THE PERI O D I C TABLE

.6.

Figure 7. 1 Oxygen and sulfur. Because they are both group 6A elements, oxygen and sulfur have many chemical similarities. They also have many differences, however, including the forms they take at room temperature. Oxygen consists of 02 molecules that appear as a colorless gas (shown here enclosed in a glass container on the left). In contrast, sulfur consists of 58 molecules that form a yellow solid.

.,. Figure 7.2 Discovering the

elements. Periodic table showing the dates of discovery of the elements.

The discovery of the chemical elements has been an ongoing process since an­ cient times (Figure 7.2 T). Certain elements, such as gold, appear in nature in el­ emental form and were thus discovered thousands of years ago. In contrast, some elements, such as technetium, are radioactive and intrinsically unstable. We know about them only because of technology developed during the twenti­ eth century. The majority of the elements, although stable, readily form compounds . Consequently, they are not found in nature in their elemental form. For cen­ turies, therefore, scientists were unaware of their existence. During the early nineteenth century, advances in chemistry made it easier to isolate elements from their compounds. As a result, the number of known elements more than doubled from 31 in 1800 to 63 by 1865. As the number of known elements increased, scientists began to investigate the possibilities of classifying them in useful ways. In 1869, Dmitri Mendeleev in Russia and Lothar Meyer in Germany published nearly identical classifica­ tion schemes. Both scientists noted that similar chemical and physical proper­ ties recur periodically when the elements are arranged in order of increasing atomic weight. Scientists at that time had no knowledge of atomic numbers. Atomic weights, however, generally increase with increasing atomic number, so both Mendeleev and Meyer fortuitously arranged the elements in proper se­ quence. The tables of elements advanced by Mendeleev and Meyer were the forerunners of the modern periodic table. Although Mendeleev and Meyer came to essentially the same conclusion about the periodicity of elemental properties, Mendeleev is given credit for ad­ vancing his ideas more vigorously and stimulating much new work in chemistry. -

H r----

B

Li

Be

K

Ca Sc Ti

r--Na Mg Rb Sr

y

v

AI

c

Si

N p

0 s

F

He Ne

Cl Ar

Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr

Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te

Cs Ba Lu Hf Ta w Re Os

Ir

I

Xe

Pt Au Hg Tl Pb Bi Po At Rn

Fr Ra Lr Rf Db Sg Bh Hs Mt Ds Rg

0 Ancient Times

0 1735-1843

0 1894-1918

O Middle Ages-1700

0 1 843-1886

0 1923-1961

-

0 1 965-

7.2 TABLE 7.1 • Comparison of the Properties of Eka-Silicon Predicted by Mendeleev with the Observed Properties of Germanium

Property

Mendeleev's Predictions for Eka­ Silicon (made in 1871)

Observed Properties of Germanium (discovered in 1886)

Atomic weight Density (g/cm3) Specific heat 0/g-K) Melting point (0C) Color Formula of oxide Density of oxide (g/cm3) Formula of chloride Boiling point of chloride (0C)

72 5.5 0.305 High Dark gray X02 4.7 XC4 A little under 100

72.59 5.35 0.309 947 Grayish white Ge02 4.70 GeC4 84

His insistence that elements with similar characteristics be listed in the same family forced him to leave several blank spaces in his table. For example, both gallium (Ga) and germanium (Ge) were unknown at that time. Mendeleev bold­ ly predicted their existence and properties, referring to them as eka-aluminum ("under" aluminum) and eka-silicon ("under" silicon), respectively, after the ele­ ments under which they appear in the periodic table. When these elements were discovered, their properties closely matched those predicted by Mendeleev, as shown in Table 7.1 .A• . In 1913, two years after Rutherford proposed the nuclear model of the atom (Section 2.2), an English physicist named Henry Moseley (1887-1915) de­ veloped the concept of atomic numbers. Moseley determined the frequencies of X-rays emitted as different elements were bombarded with high-energy elec­ trons. He found that each element produces X-rays of a unique frequency; fur­ thermore, he found that the frequency generally increased as the atomic mass increased. He arranged the X-ray frequencies in order by assigning a unique whole number, called an atomic number, to each element. Moseley correctly identified the atomic number as the number of protons in the nucleus of the atom. (Section 2.3} The concept of atomic number clarified some problems in the periodic table of Moseley's day, which was based on atomic weights. For example, the atomic weight of Ar (atomic number 18) is greater than that of K (atomic number 19), yet the chemical and physical properties of Ar are much more like that of Ne and Kr than they are like Na and Rb. However, when the elements are arranged in order of increasing atomic number, rather than increasing atomic weight, Ar and K appear in their correct places in the table. Moseley's studies also made it possible to identify "holes" in the periodic table, which led to the discovery of other previously unknown elements. c:x:c

c:x:c

GIVE IT SOME THOUGHT Arranging the elements by atomic weight leads to a slightly different order than ar­ ranging them by atomic number. Why does this happen? Can you find an example, other than the case of Ar and K discussed above, where the order of the elements would be different if the elements were arranged in order of increasing atomic weight?

7.2 EFFECTIVE NUCLEAR CHARGE Because electrons are negatively charged, they are attracted to nuclei, which are positively charged. Many of the properties of atoms depend on their elec­ tron configurations and on how strongly their outer electrons are attracted to the nucleus. Coulomb's law tells us that the strength of the interaction

Effective Nuclear Charge

257

258

C HA PTER 7

Periodic Properties of the Elements

T Figure 7.3 Effective nuclear

charge. (a) The effective nuclear charge experienced by the valence electron in sodium depends mostly on the 1 1 + charge of the nucleus and the 1 0- charge of the neon core. If the neon core were totally effective in shielding the valence electron from the nucleus, then the valence electron would experience an effective nuclear charge of 1 +. (b) The 3s electron has some probability of being inside the Ne core. Because of this "penetration," the core is not completely effective in screening the 3s electron from the nucleus. Thus, the effective nuclear charge experienced by the 3s electron is somewhat greater than 1 +.

0

between two electrical charges depends on the magnitudes of the charges and on the distance between them. a:o (Section 2.3} Thus, the force of attrac­ tion between an electron and the nucleus depends on the magnitude of the net nuclear charge acting on the electron and on the average distance between the nucleus and the electron. The force of attraction increases as the nuclear charge increases, and it decreases as the electron moves farther from the nucleus. In a many-electron atom, each electron is simultaneously attracted to the nucleus and repelled by the other electrons. In general, there are so many electron-electron repulsions that we cannot analyze the situation exactly. We can, however, estimate the net attraction of each electron to the nucleus by considering how it interacts with the average environment created by the nucle­ us and the other electrons in the atom. This approach allows us to treat each electron individually as though it were moving in the net electric field created by the nucleus and the electron density of the other electrons. We can view this net electric field as if it results from a single positive charge located at the nucle­ us, called the effective nuclear charge, Zeff· It is important to realize that the effective nuclear charge acting on an elec­ tron in an atom is smaller than the actual nuclear charge because the effective nuclear charge also accounts for the repulsion of the electron by the other elec­ trons in the atom-in other words, Zeff < Z. Let's consider how we can get a sense of the magnitude of Zeff for an electron in an atom. A valence electron in an atom is attracted to the nucleus of the atom and is repelled by the other electrons in the atom. In particular, the electron density that is due to the inner (core) electrons is particularly effective at partially can­ celing the attraction of the valence electron to the nucleus. We say that the inner electrons partially shield or screen the outer electrons from the attraction of the nucleus. We can therefore write a simple relationship between the effective nuclear charge, Zeff, and the number of protons in the nucleus, Z: Zeff = Z - S

[7.1]

The factor S is a positive number called the screening constant. It represents the portion of the nuclear charge that is screened from the valence electron by the other electrons i n the atom. Because the core elec­ trons are most effective at screening a valence electron from the nucleus, the value of S is usually close to the ..----- [Ne] core (10-) number of core electrons in an atom. Electrons in the same valence shell do not screen one another very effectively, Combined effect = 11 - 10 = 1 + but they do affect the value of S slightly (see "A Closer Look" on Effective Nuclear Charge). Let's take a look at Nucleus (11 + ) a Na atom to see what we would expect for the magnitude of Zeff· Sodium (atomic number 11) has a condensed electron configuration of [Ne]3s 1 . The nuclear charge of (a) the atom is 11+, and the Ne inner core consists of ten electrons (1s22s22p6). Very roughly then, we would ex­ pect the 3s valence electron of the Na atom to experience an effective nuclear charge of about 11 - 10 = 1 +, as pictured in a simplified way in Figure 7.3(a)

a

HC03 -( q)

+

H + (aq)

[7.2]

You might be surprised to know that our bodies need an enzyme for such a simple reaction.

In the absence of carbonic

anhydrase, however, the C02 produced in cells when they are oxidizing glucose or other fuels in vigorous exercise would be

cleared out much too slowly. About 20% of the C02 produced by cell metabolism binds to hemoglobin and is carried to the

lungs, where it is expelled. About 70% of the C02 produced is converted to bicarbonate ion through the action of carbonic

.6. Figure 7.9 Ridding the body of carbon dioxide. Illustration of the flow of C02 from tissues into blood vessels and eventually into the lungs. About 20% of the C02 binds to hemoglobin and is released in the lungs. About 70% is converted by carbonic anhydrase into HC03- ion, which remains in the blood plasma until the reverse reaction releases C02 into the lungs. Small amounts of C02 simply dissolve in the blood plasma and are released in the lungs. 2 2+ ferent? Both occur as 2+ ions, but Zn + is smaller than Cd . 2+ 2+ is 0.88 A; that of Cd is 1.09 A. Can this

The radius of Zn

anhydrase. When the C02 has been converted into bicarbon­ ate ion, it diffuses into the blood plasma and eventually is

difference be the cause of such a dramatic reversal of biologi­

These

factor, it is very important. In the carbonic anhydrase enzyme 2+ the Zn ion is found electrostatically bonded to atoms on the 2+ protein, as shown in Figure 7.10 T. It turns out that Cd binds 2+ in this same place preferentially over Zn , thus displacing it. 2+ 2+ When Cd is present instead of zn , however, the reaction of 2 C02 with water is not facilitated. More seriously, Cd + inhibits

passed into the lungs in the reverse of Equation

processes are illustrated

7.2.

in Figure 7.9 I> . In the absence of zinc,

the carbonic anhydrase would be inactive, and serious imbal­ ances would result in the amount of C02 present in blood.

Zinc is also found in several other enzymes, including

some found in the liver and kidneys. It is obviously essential to life. By contrast, cadmium, zinc's neighbor in group

28,

is

extremely toxic to humans. But why are two elements so dif-

cal properties? The answer is that while size is not the only

reactions that are essential to the kidneys' functioning.

Related Exercises: 7.30, 7.91, and 7.92

..,. Figure 7.1 0 A zinc-containing enzyme. The enzyme called carbonic anhydrase (left) catalyzes the reaction between C02 and water to form HC03-. The ribbon represents the folding of the protein chain. The "active site" of the enzyme (represented by the ball-and-stick model) is where the reaction occurs. (H atoms have been excluded from this model for clarity.) The red sphere represents the oxygen of a water molecule that is bound to the zinc ion (gold sphere) at the center of the active site. The water molecule is replaced by C02 in the reaction. The bonds coming off the five-member rings attach the active site to the protein (nitrogen and carbon atoms are represented by blue and black spheres, respectively).

needed to remove the first electron from a neutral atom. For example, the first ionization energy for the sodium atom is the energy required for the process Na(g)

�

Na+(g) + e-

[7.3]

The second ionization energy, l2o is the energy needed to remove the second electron, and so forth, for successive removals of additional electrons. Thus, I2 for the sodium atom is the energy associated with the process 2 Na+(g) � Na +(g) + e[7.4] The greater the ionization energy, the more difficult it is to remove an electron.

C HA PTER 7

266

Periodic Properties of the Elements

GIVE IT SOME THOUGHT

Light can be used to ionize atoms and ions, as in Equations 7.3 and 7.4. Which of the two processes, [7.3] or [7.4], would require shorter wavelength radiation?

Variations in Successive Ionization Energies Ionization energies for the elements sodium through argon are listed in Table 7.2 T. Notice that the values for a given element increase as successive electrons are removed: h < [z < !3, and so forth. This trend exists because with each successive removal, an electron is being pulled away from an increasingly more positive ion, requiring increasingly more energy.

TABLE 7.2 • Successive Values of Ionization Energies, I, for the Elements Sodium through Argon (kJ/mol) Element

Na Mg

11

495

738

12

1817

p

1012

1907

Cl

1251

Si

s

Ar

786

1000 1521

I•

Is

16

I7

(inner-shell electrons)

578

AI

I3

1577

3232

2252

3357

2298 2666

2914

4964

3822

5159

3931

4556

7004

5771

7238

6542

9362 8781

27,107 11,018

11,995

A second important feature shown in Table 7.2 is the sharp increase in ion­ ization energy that occurs when an inner-shell electron is removed. For exam­ ple, consider silicon, whose electron configuration is ls22s22p63s23p2 or [Ne]3s23p2 The ionization energies increase steadily from 786 kJ/mol to 4356 kJ/mol for the loss of the four electrons in the outer 3s and 3p subshells. Removal of the fifth electron, which comes from the 2p subshell, requires a great deal more energy: 16,091 kJ/mol. The large increase occurs because the 2p electron is much more likely to be found close to the nucleus than are the four n = 3 electrons, and therefore the 2p electron experiences a much greater effec­ tive nuclear charge than do the 3s and 3p electrons.

GIVE IT SOME THOUGHT

Which would you expect to be greater, 11 for a boron atom or lz for a carbon atom?

IIIIIIIIIIIIIU 111111111111111

Every element exhibits a large increase in ionization energy when electrons are removed from its noble-gas core. This observation supports the idea that only the outermost electrons, those beyond the noble-gas core, are involved in the sharing and transfer of electrons that give rise to chemical bonding and re­ actions. The inner electrons are too tightly bound to the nucleus to be lost from the atom or even shared with another atom.

- SAMPLE EXERCISE 7.5 J Trends in Ionization Energy

Three elements are indicated in the periodic table in the margin. Based on their loca­ tions, predict the one with the largest second ionization energy.

7.4

Ionization Energy

267

SOLUTION Analyze and Plan: The locations of the elements in the periodic table allow us to

predict the electron configurations. The greatest ionization energies involve removal of core electrons. Thus, we should look first for an element with only one electron in the outermost occupied shell.

Solve: The element in group 1A (Na), indicated by the red box, has only one valence electron. The second ionization energy of this element is associated, therefore, with the removal of a core electron. The other elements indicated, S (green box) and Ca (blue box), have two or more valence electrons. Thus, Na should have the largest sec­ ond ionization energy. Check: If we consult a chemistry handbook, we find the following values for the second ionization energies (12) of the respective elements: Ca (1,145 kJ/mol) < S (2,252 kJ/mol) < Na (4,562 k)/mol).

- PRACTICE EXERCISE Which will have the greater third ionization energy, Ca or S? Answer: Ca

Periodic Trends in Fi rst Ionization Energies We have seen that the ionization energy for a given element increases as we remove succes­ sive electrons. What trends do we observe in ionization energy as we move from one ele­ ment to another in the periodic table? Figure 7.11 1> shows a graph of It versus atomic num­ ber for the first 54 elements. The important trends are as follows: 1. Within each row (period) of the table, It generally increases with increasing atomic number. The alkali metals show the lowest ionization energy in each row, and the noble gases show the highest. There are slight irregularities in this trend that we will discuss shortly. 2. Within each column (group) of the table, the ionization energy generally decreases with increasing atomic number. For exam­ ple, the ionization energies of the noble gases follow the order He > Ne > AI > Kr > Xe.

2500

He I

I I I I I I I I 1500 - �-I I I I I I 1000 I I I - I I I

2000

500

Li

Ne

Na

K

Rb

0 �----�--��--10 20 40 0 50 30 Atomic number

3. The s- and p-block elements show a larger range of values of It than do the transition-metal elements. Generally, the ionization energies of the transi­ tion metals increase slowly as we proceed from left to right in a period. The /-block metals, which are not shown in Figure 7.11, also show only a small variation in the values of It·

The periodic trends in the first ionization energies of the s- and p-block ele­ ments are further illustrated in Figure 7.12 T. In general, smaller atoms have higher ionization energies. The same factors that influence atomic size also influence ionization energies. The energy needed to remove an electron from the outermost occupied shell depends on both the effective nuclear charge and the average distance of the electron from the nu­ cleus. Either increasing the effective nuclear charge or decreasing the distance from the nucleus increases the attraction between the electron and the nucleus.

.6. Figure 7. 1 1 First Ionization energy versus atomic number. The red dots mark the beginning of a period (alkali metals), the blue dots mark the end of a period (noble gases), and the black dots indicate s- and p·block elements, while green dots are used to represent the transition metals.

268

C HA PTER 7

Periodic Properties of the Elements

As this attraction increases, it becomes more difficult to remove the electron and, thus, the ionization energy increases. As we move across a --.....__ period, there is both an increase in effective nuclear charge and a decrease in atomic radius, causing the ion--.....__ . -------- -...._ ization energy to increase. As we move ;-----....__ ------down a column, however, the atomic radius increases, while the effective nu­ clear charge increases rather gradually. Thus, the attraction between the nucle­ us and the electron decreases, causing the ionization energy to decrease. The irregularities within a given row are somewhat subtler but still read­ ily explained. For example, the decrease in ionization energy from beryllium ([He]2s2) to boron ([He]2s 22p\ shown in Figures 7.11 and 7.12, occurs because the third valence electron of B must occupy the 2p subshell, which is empty for Be. Recall that, as we discussed earlier, the 2p subshell is at a higher energy than the 2s subshell (Figure 6.23). The decrease in ionization energy when mov­ ing from nitrogen ([He]2s22p3) to oxygen ([He]2s22p4) .6. Figure 7. 1 2 Trends In first is because of the repulsion of paired electrons in the p4 Ionization energy. First ionization configuration, as shown in Figure 7.13 ... . Remember that energies for the s- and p-block elements according to Hund's rule, each electron in the p3 configuration resides in a differ­ in the first six periods. The ionization energy generally increases from left to ent p orbital, which minimizes the electron-electron repulsion among the three right and decreases from top to bottom. 2p electrons. o:o (Section 6.8)

-----1---_ _

-------

-------

The ionization energy of astatine has not been determined.

I 1111I

- SAMPLE EXERCISE 7.6 I Periodic Trends in Ionization Energy

111111I

SOLUTION

2P H

Oxygen

2P

Nitrogen

.6. Figure 7. 1 3 2p orbital filling In nitrogen and oxygen. The presence of a fourth electron in the 2p orbitals of oxygen leads to an extra repulsion associated with putting two electrons in a single orbital. This repulsion is responsible for the lower first ionization energy of oxygen.

Referring to a periodic table, arrange the following atoms in order of increasing first ionization energy: Ne, Na, P, Ar, K. Analyze and Plan: We are given the chemical symbols for five elements. To rank them according to increasing first ionization energy, we need to locate each element in the periodic table. We can then use their relative positions and the trends in first ionization energies to predict their order. Solve: Ionization energy increases as we move left to right across a row. It decreases as we move from the top of a group to the bottom. Because Na, P, and Ar are in the same row of the periodic table, we expect It to vary in the order Na < P < Ar. Because Ne is above Ar in group SA, we expect Ne to have the greater first ion­ ization energy: Ar < Ne. Similarly, K is the alkali metal directly below Na in group lA, and so we expect It for K to be less than that of Na: K < Na. From these observations, we conclude that the ionization energies follow the order K < Na < P < Ar < Ne Check: The values shown in Figure 7.12 confirm this prediction. - PRACTICE EXERCISE Which has the lowest first ionization energy, B, AI, first ionization energy? Answer: A! lowest, C highest

C, or Si? Which has the highest

Electron Configurations of Ions When electrons are removed from an atom to form a cation, they are always re­ moved first from the occupied orbitals having the largest principal quantum

7.4 number,

n.

For example, when one electron is removed from a lithium atom

(ls22s1 ), it is the 2s1 electron that is removed: Li (ls22s 1 ) => Li+ (ls2)

+ e2 2 Likewise, when two electrons are removed from Fe ([Ar]3d64s ), the 4s electrons are the ones removed: Fe ([Ar]3d64s 2) => Fe 2+ ([Ar]3d6) + 2e3 If an additional electron is removed, forming Fe +, it now comes from a 3d orbital because all the orbitals with n = 4 are empty: Fe 2+ ([Ar]3d6) => Fe 3+ ([Ar]3d5) + e-

It may seem odd that the 4s electrons are removed before the 3d electrons in forming transition-metal cations. After all, in writing electron configurations, we added the 4s electrons before the 3d ones. In writing electron configurations for atoms, however, we are going through an imaginary process in which we move through the periodic table from one element to another. In doing so, we are adding both an electron to an orbital and a proton to the nucleus to change the identity of the element. In ionization, we do not reverse this process because no protons are being removed. If there is more than one occupied subshell for a given value of n the elec­ trons are first removed from the orbital with the highest value of I. For example a tin atom loses its Sp electrons before it loses its Ss electrons: Sn ([Kr]4d 10Ss 2Sp 2) => Sn2+ ([Kr]4d 10Ss 2) + 2 e - => Sn4+ ([Kr]4d10) + 4e­ When electrons are added to an atom to form an anion, they are added to the empty or partially filled orbital having the lowest value of n. For example, when an electron is added to a fluorine atom to form the F- ion, the electron goes into the one remaining vacancy in the 2p subshell : F (ls22s22p5) + e - => F - (ls22s22p 6)

GIVE IT SOME THOUGHT Would Cr3+ and V 2+ have the same or different electron configurations?

- SAMPLE EXERCISE 7.7 J Electron Configurations of Ions Write the electron configuration for (a) Ca2+ (b) Co3+, and (c) s2-. SOLUTION Analyze and Plan: We are asked to write electron configurations for three ions. To do so, we first write the electron configuration of the parent atom. We then remove electrons to form cations or add electrons to form anions. Electrons are first removed from the orbitals having the highest value of n. They are added to the empty or par­ tially filled orbitals having the lowest value of n.

Solve:

(a) Calcium (atomic number 20) has the electron configuration

Ca: [Ar]4s 2 To form a 2+ ion, the two outer electrons must be removed, giving an ion that is iso­ electronic with Ar: Ca 2+: [Ar] (b) Cobalt (atomic number 27) has the electron configuration

Co: [Ar]3d74s 2 To form a 3+ ion, three electrons must be removed. As discussed in the text preced­ ing this Sample Exercise, the 4s electrons are removed before the 3d electrons. Conse­ quently, the electron configuration for Co3+ is Co3+: [Ar]3d6

Ionization Energy

269

270

C HA PTER 7

Periodic Properties of the Elements (c) Sulfur (atomic number

16) has the electron configuration 5: [Ne]3s 2 3p4

To form a 2- ion, two electrons must be added. There is room for two additional elec­ trons in the 3p orbitals. Thus, the 52- electron configuration is 52-: [Ne]3s 2 3p6 = [Ar]

Comment: Remember that many of the common ions of the s- and p-block elements, such as Ca2+ and 52-, have the same number of electrons as the closest noble gas. = (Section 2.7) - PRACTICE EXERCISE Write the electron configuration for (a) Ga 3+, (b) Cr 3+, and (c) Br-. Answers: (a) [Ar]3d 10, (b) [Ar]3d3, (c) [Ar]3d 104s 24p6 = [Kr]

7.5 ELECTRON AFFINITIES The first ionization energy of an atom is a measure of the energy change associ­ ated with removing an electron from the atom to form a positively charged ion. For example, the first ionization energy of Cl(g), 1251 kJ/mol, is the energy change associated with the process Ionization energy: Cl(g) � Cl+ (g) + e!:J.E = 1251 kJ/mol [7.5] [Ne]3s 23p4

The positive value of the ionization energy means that energy must be put into the atom to remove the electron. In addition, most atoms can gain electrons to form negatively charged ions. The energy change that occurs when an electron is added to a gaseous atom is called the electron affinity because it measures the attraction, or affinity, of the atom for the added electron. For most atoms, energy is released when an elec­ tron is added. For example, the addition of an electron to a chlorine atom is ac­ companied by an energy change of -349 kJ/mol, the negative sign indicating that energy is released during the process. We therefore say that the electron affinity of Cl is -349 kJ/mol:*

Electron affinity: Cl{g) + e-

�

Cl-{g) [Ne]3s 23p6

!1£ = -349 kJ/mol

[7.6]

It is important to understand the difference between ionization energy and electron affinity: Ionization energy measures the ease with which an atom loses an electron, whereas electron affinity measures the ease with which an atom gains an electron. The greater the attraction between a given atom and an added electron, the more negative the atom's electron affinity will be. For some elements, such as the noble gases, the electron affinity has a positive value, meaning that the anion is higher in energy than are the separated atom and electron:

!:J.E > 0 [Ne]3s 23p6

[7.7]

[Ne]3s 23p64s 1

The fact that the electron affinity is a positive number means that an electron will not attach itself to an Ar atom; the Ar- ion is unstable and does not form. �'Two sign conventions are used for electron affinity. In most introductory texts, including this one, the ther­ modynamic sign convention is used: a negative sign indicates that the addition of an electron is an exother­ mic process, as in the electron affinity given for chlorine, -349 kjjmol. Historically, however, electron affinity has been defined as the energy released when an electron is added to a gaseous atom or ion. Because 349 k)jmol is released when an electron is added to Cl(g), the electron affinity by this convention would be +349 kjfmol.

7.6

Metals, Nonmetals, and Metalloids

Figure 7.14 � shows the electron affinities for the s- and p-block 'if -73 elements in the first five rows of the periodic table. Notice that the trends in electron affinity as we proceed through the periodic table Be Li are not as evident as they were for ionization energy. The halogens, -60 > 0 which are one electron shy of a filled p subshell, have the most-neg­ Na Mg ative electron affinities. By gaining an electron, a halogen atom -53 > 0 forms a stable negative ion that has a noble-gas configuration K Ca (Equation 7.6). The addition of an electron to a noble gas, however, -48 -2 requires that the electron reside in a higher-energy subshell that is Rb Sr empty in the neutral atom (Equation 7.7). Because occupying a -47 - 5 higher-energy subshell i s energetically very unfavorable, the eleclA 2A tron affinity is highly positive. The electron affinities of Be and Mg are positive for the same reason; the added electron would reside in a previously empty p subshell that is higher in energy. The electron affinities of the group SA elements (N, P, As, Sb) are also inter­ esting. Because these elements have half-filled p subshells, the added electron must be put in an orbital that is already occupied, resulting in larger electron-electron repulsions. Consequently, these elements have electron affini­ ties that are either positive (N) or less negative than their neighbors to the left (P, As, Sb ). Recall that we saw a discontinuity in the regular periodic trends for first ionization energy in Section 7.4 for the same reason. Electron affinities do not change greatly as we move down a group. For ex­ ample, consider the electron affinities of the halogens (Figure 7.14). For F, the added electron goes into a 2p orbital, for Cl a 3p orbital, for Br a 4p orbital, and so forth. As we proceed from F to I, therefore, the average distance between the added electron and the nucleus steadily increases, causing the electron-nucleus attraction to decrease. However, the orbital that holds the outermost electron is increasingly spread out, so that as we proceed from F to I, the electron-electron repulsions are also reduced. As a result, the reduction in the electron-nucleus attraction is counterbalanced by the reduction in electron-electron repulsions.

271

'He"" >0

0 B c N F Ne -27 -122 > 0 - 141 -328 > 0

AI p Ar Cl Si s -43 -134 -72 -200 -349 > 0 Ge As Se Br Kr Ga -30 - 1 19 -78 - 195 -325 > 0

I Xe 1n Sn Sb Te -30 -107 - 103 -190 -295 > 0 3A 4A SA 6A 7A 8A

Figure 7. 14 Electron affinity. Electron affinities in k]/mol for the s- and p-block elements in the first five rows of the periodic table. The more negative the electron affinity, the greater the attraction of the atom for an electron. An electron affinity > 0 indicates that the negative ion is higher in energy than the separated atom and electron. .A.

GIVE IT SOME THOUGHT Suppose you were asked for a value for the first ionization energy of a CC(g) ion. What is the relationship between this quantity and the electron affinity of Cl(g)?

T Figure 7. 1 5 Metals, metalloids,

and nonmetals. The majority of elements are metals. Metallic character increases from right to left across a period and also increases from top to bottom in a group.

7.6 M ETALS, NONM ETALS , AND METALLOI D S Atomic radii, ionization energies, and electron affinities are proper­ ties of individual atoms. With the exception of the noble gases, however, none of the elements exists in nature as individual atoms. To get a broader under­ standing of the properties of ele­ ments, we must also examine periodic trends in properties that involve large collections of atoms. The elements can be broadly grouped into the categories of metals, nonmetals, and metal­ loids. (Section 2.5) This classi­ fication is shown in Figure 7.15 �. Roughly three-quarters of the elements are metals, situated in the left and middle portions of the table. The nonmetals are located ceo

lA I

[I] 1

H

Increasing metallic character 2A 2

� � Li Be

nr-u Na 19 K

37 Rb

Mg 20 Ca 38

s,

55

56

87 F•

88 Ra

_g._�

3A 13

3B 3 21 Sc

39 y

71 Lu

1 1103 L•

48 4

n

22

40 z, 72

HI

104 Rf

5B 5

v

23

41 Nb 73 Ta

105 Db

68 6 24

c.

42 Mo 74 w

106 Sg

7B 7

25 Mn 43 Tc

75

Re

107 Bh

8

26 Fe

44

Ru 76 Os

108 Hs

8B 9

27 Co

45 Rh

77 Ir

109 Mt

5

8

10

28

N;

46

Pd 78 Pt

110 Ds

1B II

2B 12

47 Ag

48 Cd

Ill Rg

112

29 Cu

79 Au

13 AI

4A 14 6

c

14

s;

6A 16

7A 17

16

s

0

51 Sb

52 Te

53 I

115

116

SA 15 7 N

IS

p

30 Zn

31 Ga

32 Ge

33 As

80 Hg

81 Tl

82 Pb

83

49 In

113

50 Sn

114

a;

8 0

34 Se

84 Po

9 F

17

35 B•

85 At

8A 18

r--z He 10 Ne 18

M

36 K•

54 Xe

Rn 86

118

�

C HA PTER 7

272

Periodic Properties of the Elements

TABLE 7.3 • Characteristic Properties of Metals and Nonmetals

Metals

Nonmetals

Have a shiny luster; various colors, although most are silvery Solids are malleable and ductile Good conductors of heat and electricity Most metal oxides are ionic solids that are basic

Do not have a luster; various colors Solids are usually brittle; some are hard, some are soft Poor conductors of heat and electricity Most nonmetal oxides are molecular substances that form acidic solutions Tend to form anions or oxyanions in aqueous solution

Tend to form cations in aqueous solution

at the top right comer, and the metalloids lie between the metals and non­ metals. Hydrogen, which is located at the top left corner, is a nonmetal. This is why we set off hydrogen from the remaining group 1A elements in Figure 7.15 by inserting a space between the H box and the Li box. Some of the distinguish­ ing properties of metals and nonmetals are summarized in Table 7.3 .a.. The more an element exhibits the physical and chemical properties of met­ als, the greater its metallic character. As indicated in Figure 7.15, metallic char­ acter generally increases as we proceed down a column of the periodic table and increases as we proceed from right to left in a row. Let's now examine the close relationships that exist between electron configurations and the proper­ ties of metals, nonmetals, and metalloids.

Metals

.A. Figure 7. 1 6 The luster of metals. Metallic objects are readily recognized by

their characteristic shiny luster.

Most metallic elements exhibit the shiny luster that we associate with metals (Figure 7.16 ... ). Metals conduct heat and electricity. In general they are mal­ leable (can be pounded into thin sheets) and ductile (can be drawn into wires). All are solids at room temperature except mercury (melting point = -39 oc), which is a liquid. Two metals melt at slightly above room temperature, cesium at 28.4 oc and gallium at 29.8 oc. At the other extreme, many metals melt at very high temperatures. For example, chromium melts at 1900 °C.

Metals tend to have low ionization energies and therefore tend to form positive ions relatively easily. As a result, metals are oxidized (lose electrons) when they un­

2500

He

'3 E

....._

g

>, ""

�

c: 0 ·.o "' N

·;:; .9

Ar

1500

\

1000

�

.� "'

dergo chemical reactions. This fact is illustrated in Figure 7. 1 7 'Y, where the first ionization energies of the metals, nonmetals, and metalloids are compared. Among the fundamental atomic properties (ra­ dius, electron configuration, electron affinity, and so forth), the first ionization energy is the best indicator of whether an element will be­ have as a metal or a nonmetal. The relative ease of oxidation of common metals was discussed earlier, in Section 4.4. As we noted there, many metals are oxidized by a variety of common substances, including 02 and acids. Kr

500

Li

Na

K

Xe

I I _ 1 ___________ _ T I I I I I I

Rb

Q L------J 0

10

20 30 Atomic number

40

50

..,. Figure 7. 17 A comparison of the first Ionization energies of metals vs. nonmetals. Values of first ionization energy for metals are markedly lower than those of nonmetals. The red circles correspond to metallic elements, the blue squares to nonmetals, and the black triangles to metalloids. The dashed line at 925 k)/mol separates the metals from the nonmetals.

7.6

Metals, Nonmetals, and Metalloids

lA

EJ

7A

3A

2A

4A

z Na + Mg + K+

SA

6A

N3 - oz-

u+

Transition metals

2 Ca + Sc3+ Ti4+

2

2 cu+ y5 + Mn2 + Fe + C o + Ni2 + Zn2+ Cr3+ 3 Cu2+ y4+ Mn4 + Fe + Co3 +

Rb+ Sr 2+

2 Pd + Ag+ Cd2+

cs + Ba2+

2 Au + Hg�+ Pt + Au3+ Hgz+

p3 -

Al3 +

Figure 7.18 "' shows the oxidation states of some representative ions of both metals and nonmetals. As we noted in Section 2.7, the charge on any alkali metal ion is always 1 +, and that on any alkaline earth metal is always 2+ in their compounds. For atoms belonging to either of these groups, the outer s electrons are easily lost, yielding a noble-gas electron configuration. For metals belonging to groups with partially occupied p orbitals (Groups 3A-7A), the ob­ served cations are formed either by losing only the outer p electrons (such as 2 Sn +) or the outer s and p electrons (such as Sn4+). The charge on transition­ metal ions does not follow an obvious pattern. One of the characteristic features of the transition metals is their ability to form more than one positive ion. For example, iron may be 2+ in some compounds and 3+ in others.

GIVE IT SOME THOUGHT Based on periodic trends discussed in this chapter, can you see a general relationship between the trends in metallic character and those for ionization energy?

Compounds of metals with nonmetals tend to be ionic substances. For example, most metal oxides and halides are ionic solids. To illustrate, the reaction be­ tween nickel metal and oxygen produces nickel oxide, an ionic solid containing 2 2 Ni + and 0 - ions: ------>

2 NiO(s)

[7.8]

The oxides are particularly important because of the great abundance of oxy­ gen in our environment. Most metal oxides are basic. Those that dissolve in water react to form metal hydroxides, as in the following examples: Metal oxide + water Na20(s) + H 20(!) CaO(s)

+

H20(!)

------>

metal hydroxide

------>

2 NaOH(aq)

[7.9]

------>

Ca(OHh(aq)

[ 7.10]

The basicity of metal oxides is due to the oxide ion, which reacts with water according to the net ionic equation [7.11]

p-

N 0 B

L

Cl-

E

Se2- Br-

G A s E s

Sn2+ Sb3 + T 2 e Sn4+ Sb5+ 2+ + Bi3 Pb Pb4+ Bi5+

"" Figure 7.18 Representative oxidation states. Oxidation states found in ionic compounds, including some examples of higher oxidation states adopted by p-block metals. Notice that the steplike line that divides metals from nonmetals also separates cations from anions.

2 Ni(s) + 02(g)

sz-

�

SA

r

-

273

274

C HA PTER 7

Periodic Properties of the Elements

II> Figure 7.1 9 Metal oxides react with acids. (a) Nickel oxide (NiO), nitric acid (HN03), and water. (b) NiO is insoluble in water, but reacts with HN03 to give a green solution of the salt Ni(N03)2.

NiO (a)

(b)

Metal oxides also demonstrate their basicity by reacting with acids to form a salt plus water, as illustrated in Figure 7.19 .6.: Metal oxide + acid NiO(s) + 2 HN03 (aq)

-->

-->

salt + water Ni(N0 3h(aq) + HzO(l)

[7.12]

1n contrast, we will soon see that nonmetal oxides are acidic, dissolving in

water to form acidic solutions and reacting with bases to form salts.

I Metal Oxides - SAMPLE EXERCISE (a) Would you expect scandium oxide to be a solid, liquid, or gas at room tempera­ ture? (b) Write the balanced chemical equation for the reaction of scandium oxide with nitric acid.

7.8

SOLUTION Analyze and Plan: We are asked about one physical property of scandium oxide­ its state at room temperature-and one chemical property-how it reacts with nitric acid. Solve: (a) Because scandium oxide is the oxide of a metal, we would expect it to be an ionic solid. Indeed it is, with the very high melting point of 2485 oc. (b) In its compounds, scandium has a 3+ charge, Sc 3+; the oxide ion is 0 2-. Conse­ quently, the formula of scandium oxide is Sc203. Metal oxides tend to be basic and therefore to react with acids to form a salt plus water. In this case the salt is scandium nitrate, Sc(N03)J. The balanced chemical equation is

T Figure 7.20 The diversity of

nonmetals. Nonmetallic elements are diverse in their appearances. Shown here are (cloc kwise from top left) sulfur, bromine, phosphorus, iodine, and carbon.

- PRACTICE EXERCISE

Write the balanced chemical equation for the reaction between copper(II) oxide and sulfuric acid. Answer: CuO(s) + H 2S04(aq) ---> CuS04(aq) + H 20(!)

Nonmetals Nonmetals vary greatly in appearance (Figure 7.20

acid

C02(g) + H20(!)

------>

H2C0 3 (aq)

[7.14]

P40w(s) + 6 H20(!)

------>

4 H 3P04(aq)

[7.15]

The reaction of carbon dioxide with water (Figure 7.21 T) accounts for the acidity of carbonated water and, to some extent, rainwater. Because sulfur is present in oil and coal, combustion of these common fuels produces sulfur dioxide and sulfur trioxide. These substances dissolve in water to produce acid rain, a major pollution problem in many parts of the world. Like acids, most nonmetal oxides dissolve in basic solutions to form a salt plus water: Nonmetal oxide + base

------>

salt + water [7.16]

GIVE IT SOME THOUGHT

A compound ACI3 (A is an element) has a melting point of - 112 oc. Would you ex­

pect the compound to be a molecular or ionic substance? If you were told that ele­ ment A was either scandium (Sc) or phosphorus (P), which do you think would be a more likely choice?

HzSe03(aq) (It does not matter that Se02 is a solid and C02 is a gas under ambient conditions; the point is that both are water-soluble nonmetal oxides.) (b) The reaction with sodium hydroxide is like the reaction summarized by Equation

7.16:

Se0 (s) 2

+ 2 NaOH(aq)

---->

Na Se03(aq) 2

+

HzO(I)

- PRACTICE EXERCISE

Write the balanced chemical equation for the reaction of solid tetraphosphorus hex­ oxide with water. Answer: P406(s) + 6 HzO(l) ----> 4 H 3P03 (aq)

Meta l loids

& Figure 7.22 Elemental silicon. Silicon is an example of a metalloid. Although it looks metallic, silicon is brittle and is a poor thermal and electrical conductor as compared to metals. Large crystals of silicon are sliced into thin wafers for use in integrated circuits.

T Figure 7.23 Alkali metals. Sodium and the other alkali metals are soft enough to be cut with a knife. The shiny metallic surface quickly tarnishes as the metal reacts with oxygen in the air.

Metalloids have properties intermediate between those of metals and those of nonmetals. They may have some characteristic metallic properties but lack oth­ ers. For example, silicon looks like a metal (Figure 7.22 ..,.), but it is brittle rather than malleable and is a much poorer conductor of heat and electricity than are metals. Compounds of metalloids can have characteristics of the compounds of metals or nonmetals, depending on the specific compound. Several of the metalloids, most notably silicon, are electrical semiconduc­ tors and are the principal elements used in the manufacture of integrated cir­ cuits and computer chips. One of the reasons metalloids such as silicon can be used for integrated circuits is the fact that their electrical conductivity is inter­ mediate between that of metals and nonmetals. Very pure silicon is an electrical insulator, but its conductivity can be dramatically increased with the addition of specific impurities (dopants) . This modification provides a mechanism for controlling the electrical conductivity by controlling the chemical composition. We will return to this point in Chapter 12.

7.7 GROUP TREN D S FOR THE ACTIVE METALS Our discussion of atomic radius, ionization energy, electron affinity, and metal­ lic character gives some idea of the way the periodic table can be used to orga­ nize and remember facts. As we have seen, elements in a group possess general similarities. However, trends also exist as we move through a group. In this sec­ tion we will use the periodic table and our knowledge of electron configura­ tions to examine the chemistry of the alkali metals (group lA) and the alkaline earth metals (group 2A).

Group 1 A: The Alkali Metals

The alkali metals are soft metallic solids (Figure 7.23 ... ). All have characteristic metallic properties such as a silvery, metallic luster and high thermal and elec­ trical conductivities. The name alkali comes from an Arabic word meaning "ashes." Many compounds of sodium and potassium, two alkali metals, were isolated from wood ashes by early chemists.

7.7 TABLE 7.4 • Some Properties of the Alkali Metals

Element

Electron Configuration

Melting Point (°C)

Density (g/cm3)

Atomic Radius

tJ.H =

4 HF(aq) + 02(g) SiF4(g) + Oz(g)

758.9 kJ

[7.31]

tJ.H = -704.0 kJ

[7.32]

-

As a result, fluorine gas is difficult and dangerous to use in the laboratory, requiring specialized equipment. Chlorine is the most industrially useful of the halogens. In 2005, total pro­ duction was 22 billion pounds, making it the eighth most produced chemical in the United States. Unlike fluorine, chlorine reacts slowly with water to form rel­ atively stable aqueous solutions of HCl and HOC! (hypochlorous acid): Clz(g) + H20(1) .A. Figure 7.35 Elemental halogens. All three of these elements-from left to right, iodine (12), bromine (Br2), and chlorine (CI2)-exist as diatomic molecules.

---->

[7.33]

HCl(aq) + HOCl(aq)

Chlorine is often added to drinking water and swimming pools, where the HOCl(aq) that is generated serves as a disinfectant. The halogens react directly with most metals to form ionic halides. The halo­ gens also react with hydrogen to form gaseous hydrogen halide compounds: Hz(g) + Xz

---->

2 HX(g)

[7.34]

These compounds are all very soluble in water and dissolve to form the hydro­ halic acids. As we discussed in Section 4.3, HCl(aq), HBr(aq), and HI(aq) are strong acids, whereas HF(aq) is a weak acid.

GIVE IT SOME THOUGHT Can you use data in Table 7.7 to provide estimates for the atomic radius and first ion­ ization energy of an astatine atom?

Group 8A: The Noble Gases The group 8A elements, known as the noble gases, are all nonmetals that are gases at room temperature. They are all monntomic (that is, they consist of single atoms rather than molecules). Some physical properties of the noble-gas elements are listed in Table 7.8T. The high radioactivity of radon (Rn, atomic number 86) has limited the study of its reaction chemistry and some of its properties. The noble gases have completely filled s and p subshells. All elements of group 8A have large first ionization energies, and we see the expected decrease as we move down the column. Because the noble gases possess such stable elec­ tron configurations, they are exceptionally unreactive. In fact, until the early TABLE 7.8 • Some Properties of the Noble Gases Element Helium Neon Argon Krypton Xenon Radon

Electron Configuration

1 s2 [He)2sz2p6 [Ne)3s23p6 [Ar]3d104s24p6 [Kr]4d105s25p6 [Xe]4/45d106s26p6

Boiling Point (K)

Density (g/L)

27.1 87.3 120 165 211

0.18 0.90 1.78 3.75 5.90 9.73

4 .2

Atomic Radius• CAl 0.32 0.69 0.97 1.10 1.30 1.45

I1

(kJ/mol)

2372 2081 1521 1351 1170 1037

*Only the heaviest of the noble-gas elements form chemical compounds. Thus, the atomic radii for the lighter noble-gas elements are estimated values.

7.8

Group Trends for Selected Nonmetals

285

1960s the elements were called the inert gases because they were thought to be in­ capable of forming chemical compounds. In 1962, Neil Bartlett at the University of British Columbia reasoned that the ionization energy of Xe might be low enough to allow it to form compounds. In order for this to happen, Xe would have to react with a substance with an extremely high ability to remove elec­ trons from other substances, such as fluorine. Bartlett synthesized the first noble-gas compound by combining Xe with the fluorine-containing compound PtF6. Xenon also reacts directly with F2(g) to form the molecular compounds XeFz, XeF4, and XeF6 (Figure 7.36 1>). Krypton has a higher 11 value than xenon and is therefore less reactive. 1n fact, only a single stable compound of krypton is known, KrF2. In 2000, Finnish scientists reported the first neutral molecule that contains argon, the HArP molecule, which is stable only at low temperatures.

.;. Figure 7.36 A compound of xenon. - SAMPLE INTEGRATIVE EXERCISE I Putting Concepts Together The element bismuth (Bi, atomic number 83) is the heaviest member of group SA. A salt of the element, bismuth subsalicylate, is the active ingredient in Pepto-Bismol®, an over-the-counter medication for gastric distress. (a) The covalent atomic radii of thallium (Tl) and lead (Pb) are 1.48 A and 1.47 A, respectively. Using these values and those in Figure 7.7, predict the covalent atomic radius of the element bismuth (Bi). Explain your answer. (b) What accounts for the general increase in atomic radius going down the group SA elements? (c) Another major use of bismuth has been as an ingredient in low-melting metal alloys, such as those used in fire sprinkler systems and in typesetting. The element itself is a brittle white crystalline solid. How do these characteristics fit with the fact that bismuth is in the same periodic group with such nonmetallic elements as nitro­ gen and phosphorus? (d) Bi203 is a basic oxide. Write a balanced chemical equation for its reaction with di­ lute nitric acid. If 6.77 g of Bi203 is dissolved in dilute acidic solution to make 0.500 L 3 of solution, what is the molarity of the solution of Bi + ion? 2 (e) 09Bi is the heaviest stable isotope of any element. How many protons and neu­ trons are present in this nucleus? 3 (f) The density of Bi at 2S oc is 9.808 g/cm . How many Bi atoms are present in a cube of the element that is S.OO em on each edge? How many moles of the element are present? SOLUTION (a) Note that there is a gradual decrease in radius of the elements in Groups 3A-SA as we proceed across the fifth period, that is, in the series In-Sn-Sb. Therefore, it is reasonable to expect a decrease of about 0.02 A as we move from Pb to Bi, leading to an estimate of 1.4S A. The tabulated value is 1.46 A. (b) The general increase in radius with increasing atomic number in the group SA el­ ements occurs because additional shells of electrons are being added, with corre­ sponding increases in nuclear charge. The core electrons in each case large!y shield the outermost electrons from the nucleus, so the effective nuclear charge does not vary greatly as we go to higher atomic numbers. However, the principal quantum number, 11, of the outermost electrons steadily increases, with a corresponding in­ crease in orbital radius. (c) The contrast between the properties of bismuth and those of nitrogen and phos­ phorus illustrates the general rule that there is a trend toward increased metallic character as we move down in a given group . Bismuth, in fact, is a metal. The in­ creased metallic character occurs because the outermost electrons are more readily lost in bonding, a trend that is consistent with its lower ionization energy. (d) Following the procedures described in Section 4.2 for writing molecular and net ionic equations, we have the following:

Molecular equation: Net io11ic equatio11:

Bi203(s) + 6 HN03(aq) --> 2 Bi(N03h(aq) + 3 Hp(l ) 3 Bi203(s) + 6 H +(aq) --> 2 Bi +(aq) + 3 H20(l)

In the net ionic equation, nitric acid is a strong acid and Bi(N03)3 is a soluble salt, so we need show only the reaction of the solid with the hydrogen ion forming the 3 Bi +(aq) ion and water. To calculate the concentration of the solution, we proceed as follows (Section 4.5): 3 6.77 g Bi 203 1 mol Bi203 2 mol Bi3+ O.OS81 mol Bi + = 0.0S81 M -:-:: -:=-:::--' '-:-"- X 466.0 g Bi203 X 1 mol BiP3 O.SOO L soln L soln

Crystals of XeF4, which is one of the very few compounds that contain a group 8A element.

286

C HA PTER 7

Periodic Properties of the Elements (e) We can proceed as in Section 2.3. Bismuth is element 83; there are therefore 83 protons in the nucleus. Because the atomic mass number is 209, there are 209 - 83 = 126 neutrons in the nucleus. (f) We; noceed as in Sections 1.4 and 3.4: the volume of the cube is (5.00)3 cm3 = 125 em . Then we have 9.808 g Bi 1 mol Bi -X 125 cm3 Bi X --= 5.87 mol Bi 209.0 g Bi 1 em 3 23 6.022 X 10 atom Bi = 3.54 X 1024 atoms Bi 5.87 mol Bi X 1 mo1 Bi

CHAPTER REV I E W SUMMARY AND KEY TERMS Introduction and Section 7.1 The periodic table was first developed by Mendeleev and Meyer on the basis of the similarity in chemical and physical properties exhibit­ ed by certain elements. Moseley established that each ele­ ment has a unique atomic number, which added more order to the periodic table. We now recognize that ele­ ments in the same column of the periodic table have the same number of electrons in their valence orbitals. This similarity in valence electronic structure leads to the sim­ ilarities among elements in the same group. The differ­ ences among elements in the same group arise because their valence orbitals are in different shells. Section 7.2 Many properties of atoms are due to the av­ erage distance of the outer electrons from the nucleus and to the effective nuclear charge experienced by these elec­ trons. The core electrons are very effective in screening the outer electrons from the full charge of the nucleus, whereas electrons in the same shell do not screen each other effectively. As a result, the effective nuclear charge experienced by valence electrons increases as we move left to right across a period. Section 7.3 The size of an atom can be gauged by its bonding atomic radius, based on measurements of the distances separating atoms in their chemical compounds. In general, atomic radii increase as we go down a column in the periodic table and decrease as we proceed left to right across a row. Cations are smaller than their parent atoms; anions are larger than their parent atoms. For ions of the same charge, size increases going down a column of the peri­ odic table. An isoelectronic series is a series of ions that has the same number of electrons. For such a series, size decreases with increasing nuclear charge as the electrons are attracted more strongly to the nucleus. Section 7.4 The first ionization energy of an atom is the minimum energy needed to remove an electron from the atom in the gas phase, forming a cation. The second ionization energy is the energy needed to remove a sec­ ond electron, and so forth. Ionization energies show a

sharp increase after all the valence electrons have been re­ moved, because of the much higher effective nuclear charge experienced by the core electrons. The first ioniza­ tion energies of the elements show periodic trends that are opposite those seen for atomic radii, with smaller atoms having higher first ionization energies. Thus, first ionization energies decrease as we go down a column and increase as we proceed left to right across a row. We can write electron configurations for ions by first writing the electron configuration of the neutral atom and then removing or adding the appropriate number of electrons. Electrons are removed first from the orbitals with the largest value of n. If there are two valence or­ bitals with the same value of n (such as 4s and 4p), then the electrons are lost first from the orbital with a higher value of l (in this case, 4p). Electrons are added to orbitals in the reverse order. Section 7.5 The electron affinity of an element is the energy change upon adding an electron to an atom in the gas phase, forming an anion. A negative electron affinity means that the anion is stable; a positive electron affinity means that the anion is not stable relative to the separat­ ed atom and electron, in which case its exact value cannot be measured. In general, electron affinities become more negative as we proceed from left to right across the peri­ odic table. The halogens have the most-negative electron affinities. The electron affinities of the noble gases are positive because the added electron would have to occu­ py a new, higher-energy subshell. Section 7.6 The elements can be categorized as metals, nonmetals, and metalloids. Most elements are metals; they occupy the left side and the middle of the periodic table. Nonmetals appear in the upper-right section of the table. Metalloids occupy a narrow band between the met­ als and nonmetals. The tendency of an element to exhibit the properties of metals, called the metallic character, in­ creases as we proceed down a column and decreases as we proceed from left to right across a row. Metals have a characteristic luster, and they are good conductors of heat and electricity. When metals react with

Key Equations nonmetals, the metal atoms are oxidized to cations and ionic substances are generally formed. Most metal oxides are basic; they react with acids to form salts and water. Nonmetals lack metallic luster and are generally poor conductors of heat and electricity. Several are gases at room temperature. Compounds composed entirely of nonmetals are generally molecular. Nonmetals usually form anions in their reactions with metals. Nonmetal ox­ ides are acidic; they react with bases to form salts and water. Metalloids have properties that are intermediate between those of metals and nonmetals. Section 7.7 The periodic properties of the elements can help us understand the properties of groups of the repre­ sentative elements. The alkali metals (group 1A) are soft metals with low densities and low melting points. They have the lowest ionization energies of the elements. As a result, they are very reactive toward nonmetals, easily losing their outer s electron to form 1 + ions. The alkaline earth metals (group 2A) are harder and more dense and have higher melting points than the alkali metals. They are also very reactive toward nonmetals, although not as reactive as the alkali metals. The alkaline earth metals readily lose their two outer s electrons to form 2+ ions. Both alkali and alkaline earth metals react with hydrogen to form ionic substances that contain the hydride ion, H-.

287

Section 7.8 Hydrogen is a nonmetal with properties that are distinct from any of the groups of the periodic table. It forms molecular compounds with other non­ metals, such as oxygen and the halogens. Oxygen and sulfur are the most important elements in group 6A. Oxygen is usually found as a diatomic molecule, 02. Ozone, 03, is an important allotrope of oxygen. Oxygen has a strong tendency to gain electrons from other elements, thus oxidizing them. In combina­ tion with metals, oxygen is usually found as the oxide ion, 0 2-, although salts of the peroxide ion, 022-, and superoxide ion, 02-, are sometimes formed. Elemental sulfur is most commonly found as S8 molecules. In com­ bination with metals, it is most often found as the sul­ fide ion, s 2-. The halogens (group 7A) are nonmetals that exist as diatomic molecules. The halogens have the most negative electron affinities of the elements. Thus their chemistry is dominated by a tendency to form 1 - ions, especially in reactions with metals. The noble gases (group 8A) are nonmetals that exist as monatomic gases. They are very unreactive because they have completely filled s and p subshells. Only the heaviest noble gases are known to form compounds, and they do so only with very active nonmetals, such as fluorine.

KEY SKILLS • Understand the meaning of effective nuclear charge, z.ff, and how Zeff depends upon nuclear charge and electron configuration. • Use the periodic table to predict the trends in atomic radii, ionic radii, ionization energy, and electron affinity. • Understand how the radius of an atom changes upon losing electrons to form a cation or gaining electrons to form an anion. • Understand how the ionization energy changes as we remove successive electrons. Recognize the jump in ionization energy that occurs when the ionization corresponds to removing a core electron. • Be able to write the electron configurations of ions. • Understand how irregularities in the periodic trends for electron affinity can be related to electron configuration. • Recognize the differences in chemical and physical properties of metals and nonmetals, including the basicity of metal oxides and the acidity of nonmetal oxides. • Understand how the atomic properties, such as ionization energy and electron configuration, are related to the chemical reactivity and physical properties of the alkali and alkaline earth metals (groups 1A and 2A). • Be able to write balanced equations for the reactions of the group lA and 2A metals with water, oxygen, hydrogen, and the halogens. • Understand and recognize the unique characteristics of hydrogen. • Understand how the atomic properties (such as ionization energy, electron configuration, and electron affinity) of group 6A, 7A, and 8A elements are related to their chemical reactivity and physical properties.

KEY EQUATIONS 0

Zeff = Z - S

[7.1]

Estimating effective nuclear charge

288

C HA PTER 7

Periodic Properties of the Elements

VISUALIZING CONCEPTS

7.1

We can draw an analogy between the attraction of an electron to a nucleus and seeing a lightbulb-in essence, the more nuclear charge the electron "sees," the greater the attraction. (a) Within this analogy, discuss how the shielding by core electrons is analogous to putting a frosted-glass lampshade between the Lightbulb and your eyes, as shown in the illustration. (b) Explain how we could mimic moving to the right in a row of the peri­ odic table by changing the wattage of the lightbulb. (c) How would you change the wattage of the bulb and/ or the frosted glass to mimic the effect of moving down a column of the periodic table? [Section 7.2]

I

Light bulb

""'

\ Observer

7.4

111111111111111�

electron transfer, an elec­ tron is transferred from one atom or molecule to another (We will talk about electron transfer extensively in Chapter 20.) A simple electron transfer reaction is

7.5 In the chemical process called

Frosted glass

A(g) + A(g)

• 7.3 Consider the A2X4 molecule depicted below, where A

and X are elements. The A -A bond length in this mol­ ecule is d1, and the four A - X bond lengths are each dz. (a) In terms of d1 and dz, how could you define the bonding atomic radii of atoms A and X? (b) In terms of d1 and dz, what would you predict for the X - X bond length of an X2 molecule? [Section 7.3]

----+

A+(g) + A-(g)

In terms of the ionization energy and electron affinity of atom A, what is the energy change for this reaction? For a representative nonmetal such as chlorine, is this process exothermic? For a representative metal such as sodium, is this process exothermic? [Sections 7.4 and 7.5]

7.2 Fluorine has atomic number 9. If we represent the ra­ dius of a fluorine atom with the billiard ball illustrated here, would the analogy be more appropriate for the bonding or nonbonding atomic radius? If we used the same billiard ball to illustrate the concept of fluorine's bonding atomic radius, would we overestimate or underestimate the bonding atomic radius? Explain. [Section 7.3]

Make a simple sketch of the shape of the main part of the periodic table, as shown. (a) Ignoring H and He, write a single straight arrow from the element with the smallest bonding atomic radius to the element with the largest. (b) Ignoring H and He, write a single straight arrow from the element with the smallest first ionization energy to the element with the largest. (c) What signifi­ cant observation can you make from the arrows you drew in parts (a) and (b )? [Sections 7.3 and 7.4]

7.6

An element X reacts with F2(g) to form the molecular product shown below. (a) Write a balanced equation for this reaction (do not worry about the phases for X and the product). (b) Do you think that X is a metal or non­ metal? Explain. [Section 7.6]

Exercises

289

EXERCISES Periodic Ta ble; Effective Nuclear Charge 7.7

Why did Mendeleev leave blanks in his early version of the periodic table? How did he predict the properties

7.12 (a)

7.8

The prefix

a many-electron atom? (b) Which experiences a greater

eka- comes from the Sanskrit word for one.

effective nuclear charge in a Be atom, the 1s electrons or

the 2s electrons? Explain.

Mendeleev used this prefix to indicate that the un­ known element was one place away from the known el­

ement that followed the prefix. For example,

eka-silicon,

7.13

which we now call germanium, is one element below silicon. Mendeleev also predicted the existence of

manganese, until

1937

is

7.2, what

Mendeleev called

7.9 In Chapter 1

0.00

iron, which accounts for less than

5%

7.2),

7.14

7.15

(c) Why

1.00

and valence electrons contribute

to the screening constant?

(b) What values do you

Which will experience the greater effective nuclear charge, the electrons in the

n

= 3 shell in Ar or the n = 3

shell in Kr? Which will be closer to the nucleus? Explain.

7.16

atomic weight?

What is meant by the term

6.12+, respectively. (a) What

the steady increase in Zeff that occurs upon moving left to right across a period?

sured. Why was this important to Mendeleev's formula­

7.11 (a)

and

method of approximation more accurately accounts for

atomic weights of many elements were accurately mea­

more closely related to atomic number than they are to

4.29+

estimate for Zeff using Slater's rules? (c) Which ap­ proach gives a more accurate estimate of Zeff? (d) Which

discovery?

are chemical and physical properties of the elements

Detailed calculations show that the value of Zeff for Si

and Cl atoms is

0.00

7.10 (a) During the period from about 1800 to about 1865, the

emitted from an element in his experiments?

increase in Zeff that occurs upon moving down a group?

trons contribute

abundance how do you account for its relatively late

(b) What property of the atom

(b) What values do you

most electron in both Si and Cl by assuming core elec­

of Earth's crust,

did Moseley associate with the wavelength of X-rays

and valence electrons contribute

value do you estimate for Zeff experienced by the outer­

whereas

has been known since prehistoric times. Given silicon's

tion of the periodic table?

1.00

to the screening constant?

Z ff

we see that silicon is not among the elements that have been known since ancient times (Figure

3.49+, respectively. (a) What

either method of approximation account for the gradual

we learned that silicon is the second most

1 .6). Yet

and

estimate for Zeff using Slater's rules? (c) Which ap­ proach gives a more accurate estimate of e ? (d) Does

eka-manganese?

than one-fourth of the mass of the crust (Figure

2.51 +

most electron in both Na and K by assuming core elec­

do we now call the element

abundant element in Earth's crust, accounting for more

and K atoms is

trons contribute

radioactive and

does not occur in nature. Based on the periodic table shown in Figure

Detailed calculations show that the value of Zeff for Na

value do you estimate for Zeff experienced by the outer­

eka­

which was not experimentally confirmed because this element

How is the concept of effective nuclear charge used

to simplify the numerous electron-electron repulsions in

of the elements that belonged in those blanks?

Arrange the following atoms

the n

effective nuclear charge?

in order

of increasing ef­

fective nuclear charge experienced by the electrons in

= 3 electron shell: K, Mg, P, Rh, and Ti. Explain the

basis for your order.

(b) How does the effective nuclear charge experienced

by the valence electrons of an atom vary going from left to right across a period of the periodic table?

Atomic and Ionic Radii 7.17 (a)

Because an exact outer boundary cannot be mea­

sured or even calculated for an atom, how are atomic

radii determined?

(b) What is the difference between a (c) For a

bonding radius and a nonbonding radius?

7.20

Based on the radii presented in Figure

7.21

Estimate the As- I bond length from the data in Figure

given element, which one is larger?

7.18 (a)

Why does the quantum mechanical description of

many-electron atoms make it difficult to define a precise atomic radius?

(b) When nonbonded atoms come up

against one another, what determines how closely the

nuclear centers can approach?

7.19

7.22

The distance between W atoms in tungsten metal is 2.74 A.

What is the atomic radius of a tungsten atom in this envi­ ronment? (This radius is called the

metallic radius.)

7.7,

predict the

and compare your value to the experimental As- I

bond length in arsenic triiodide, Asi3, 2.55 A.

The experimental Bi- I bond length in bismuth triio­

2.81 A. Based on this value 7.7, predict the atomic radius of Bi.

dide, Bil3, is

Figure

7.23

7.7,

distance between Si atoms in solid silicon.

and data in

How do the sizes of atoms change as we move

(a) from left to right across a row in the periodic table, (b) from top to bottom in a group in the periodic table? (c) Arrange the following atoms in order of increasing

atomic radius: F, P, S, As.

290

C HA PTER 7

7.24 (a)

Periodic Properties of the Elements

Among the nonmetallic elements, the change in

is smaller than the change in moving one row up or

the following ions identify the neutral atom that has the

(b) Arrange the fol­

same number of electrons and determine if this atom has the same electron configuration.

Al, Ge, Ga.

does not exist explain why: (d) zn2+, (e) Sn4+

Using only the periodic table, arrange each set of atoms

(a) Ca, Mg, Be; (b) Ga, Br,

7.33

tribute

(a) Ba, Ca, Na; (b) Sn, Sb,

(d) For isoelectronic ions, how are effective nuclear

sponding neutral atoms? (b) Why are monatomic an­

charge and ionic radius related?

7.34

[ij] � Wltich sphere represents a metal and which a nonmetal? Explain.

7.30

ions. (c) Repeat this calculation using Slater's rules to esti­

For isoelectronic ions

how are effective nuclear charge and ionic radius related?

7.35

Consider S, Cl, and K and their most common ions.

(a) List the atoms in order of increasing size. (b) List the (c) Explain any differ­

ions in order of increasing size.

ences in the orders of the atomic and ionic sizes.

7.36

For each of the following sets of atoms and ions, arrange 2 2 the members in order of increasing size: (a) Se -, Te -, Se; (b) Co3+, Fe 2+, Fe 3+; (c) Ca, Ti 4+, Sc 3+; (d) Be 2+ , Na+, Ne.

7.37

For each of the following statements, provide an explana­ 2 2 2 tion: (a) 0 - is larger than 0; (b) s - is larger than 0 -; 2 2 (c) S - is larger than K+; (d) K+ is larger than Ca +

7.38

1n the ionic compounds LiF, NaCl, KBr, and Rbi, the

measured cation-anion distances are

2.01 A (Li-F), 2.82 A

(Na-Cl),

(Rb-I), respectively.

3.30 A

(K-Br), and

3.67 A

(a) Predict the cation-anion distance using the values of ionic radii given in Figure 7.8. (b) Is the agreement be­ tween the prediction and the experiment perfect? If not, why not? (c) What estimates of the cation-anion distance

2 Wltich one represents Ca, which Ca +, and which Mg 2+? What is an isoelectronic series?

S. (d)

mate the screening constant,

Consider the following spheres:

7.31 (a)

1.00 and valence electrons contribute noth­ Zeff for these two

ing to the screening constant, S, calculate

Consider a reaction represented by the following spheres: Products

+ (a) Which ion is (b) Use Equation 7.1 and assuming that core elec­

trons contribute

Explain the following variations in atomic or ionic radii: (a) l- > I > I+, (b)Ca 2+ > Mg 2+ > Be 2+ , 2 (c) Fe > Fe + > Fe 3+

Reactants

Consider the isoelectronic ions o- and K smaller?

down a column in the periodic table?

7.29

(c) Repeat this calculation

using Slater's rules to estimate the screening constant, S.

7.27 (a) Why are monatomic cations smaller than their corre­ (c) Why does the size of ions increase as one proceeds

1.00 and valence electrons con­ 0.00 to the screening constant, S, calculate Zeff for

the 2p electrons in both ions.

As; (c) AI, Be, Si.

ions larger than their corresponding neutral atoms?

(a) Which (b) Using Equation 7.1 and assuming that

core electrons contribute

Using only the periodic table, arrange each set of atoms in order of increasing radius:

Consider the isoelectronic ions F- and Na+ ion is smaller?

Ge; (c) AI, Tl, Si.

7.28

If such an atom (a) CC (b) Sc 3+, (c) Fe2+,

lowing atoms in order of increasing atomic radius: Si,

in order of increasing radius:

7.26

Some ions do not have a corresponding neutral atom that has the same electron configuration. For each of

down. Explain these observations.

7.25

7.32

atomic radius in moving one place left or right in a row

(b) Which neutral

would you obtain for these four compounds using

bonding atomic radii? Are these estimates as accurate as the

atom is isoelectronic with each of the following ions: 4+ 3+ 2 At , Ti , Br-, Sn +.

estimates using ionic radii?

Ionization Energies; Electron Affinities 7.39

Write equations that show the processes that describe the first, second, and third ionization energies of a boron atom.

7.40

Write equations that show the process for two ionization energies of

7.41

tin and (b)

ties?

the fourth ioniza­

ergy? Which has the smallest?

7.44 (a)

(b) What is the trend in first ionization energies as one this trend compare with the trend in atomic radii?

(b) The difference between the third and

tion energies of titanium. Why?

(c) Why does Li have a

much larger second ionization energy than Be?

elements? Explain how

moves across the fourth period from K to Kr? How does

Why does Li have a larger first ionization energy

fourth ionization energies of scandium is much larger

7A

this trend relates to the variation in atomic radii.

atom always greater than its first ionization energy?

than the difference between the third and fourth ioniza­

What is the trend in first ionization energies as one

proceeds down the group

(b) Why does F have a larger first ionization energy

than N a?

(b) Which ele­

ment in the periodic table has the largest ionization en­

than 0? (c) Why is the second ionization energy of an

7.42 (a)

(a) What is the general relationship between the size of an atom and its first ionization energy?

(a) the first

tion energy of titanium. _ (a) Why are ionization energies always positive quanti­

_ _ _

7.43

7.45

Based on their positions in the periodic table, predict which atom of the following pairs will have the larger first ionization energy:

(d) S, Ge; (e) Sn, Te.

(a) Cl, Ar; (b) Be, Ca; (c ) K, Co;

Exercises

7.46 For each of the following pairs, indicate which element has the larger first ionization energy: (a) Ti, Ba; (b) Ag, Cu; (c) Ge, Cl; (d) Pb, Sb. (In each case use electron con­ figuration and effective nuclear charge to explain your answer.)

7.47 Write the electron configurations for the following ions: 2 (a) !n3+, (b) Sb3+, (c) Te -, (d) Te 6+, (e) Hg 2+, (f) Rh 3+ 7.48 Write electron configurations for the following ions, and determine which have noble-gas configurations: 2 (a) Cr3+, (b) N 3-, (c) Sc 3+ , (d) Cu +, (e) Tl +, (f) Au+ 2 7.49 Write the electron configuration for (a) the Ni + ion and 2 (b) the Sn + ion. How many unpaired electrons does each contain?

7.50 Identify the element whose ions have the following elec­

tron configurations: (a) a 2+ ion with [Ar]3tf', (b) a 1 + 0 2 ion with [Xe]4f14Sd1 6s How many unpaired electrons does each ion contain?

7.51 The first ionization energy of Ar and the electron affini­ ty of Ar are both positive values. What is the signifi­ cance of the positive value in each case?

291

7.52 The electron affinity of lithium is a negative value,

whereas the electron affinity of beryllium is a positive value. Use electron configurations to account for this observation.

7.53 While the electron affinity of bromine is a negative quantity, it is positive for Kr. Use the electron configura­ tions of the two elements to explain the difference.

7.54 What is the relationship between the ionization energy

of an anion with a 1 - charge such as F- and the electron affinity of the neutral atom, F?

7.55 Consider the first ionization energy of neon and the electron affinity of fluorine. (a) Write equations, includ­ ing electron configurations, for each process. (b) These two quantities will have opposite signs. Which will be positive, and which will be negative? (c) Would you ex­ pect the magnitudes of these two quantities to be equal? If not, which one would you expect to be larger? Explain your answer.

7.56 Write an equation for the process that corresponds to the electron affinity of the Mg + ion. Also write the electron configurations of the species involved. What is the magnitude of the energy change in the process? [Hint: The answer is in Table 7.2.]

Properties of Metals and Nonmetals 7.57 How are metallic character and first ionization energy related?

7.58 Arrange the following pure solid elements in order of increasing electrical conductivity: Ge, Ca, S, and Si. Ex­ plain the reasoning you used.

7.59 If we look at groups 3A through SA, we see two metal­

loids for groups 4A (Si, Ge) and SA (As, Sb), but only one metalloid in group 3A (B). To maintain a regular geometric pattern one might expect that aluminum would also be a metalloid, giving group 3A two metal­ loids. What can you say about the metallic character of aluminum with respect to its neighbors based on its first ionization energy?

7.60 For each of the following pairs, which element will have

the greater metallic character: (a) Li or Be, (b) Li or Na, (c) Sn or P, (d) Al or B?

7.61 Predict whether each of the following oxides is ionic or molecular: SOz, MgO, LizO, PzOs, Yz03, NzO, and Xe03. Explain the reasons for your choices.

7.62 Some metal oxides, such as Sc20:y do not react with pure water, but they do react when the solution becomes either acidic or basic. Do you expect Sc203 to react when the so­ lution becomes acidic or when it becomes basic? Write a balanced chemical equation to support your answer.

7.63 (a) What is meant by the terms acidic oxide and basic oxide? (b) How can we predict whether an oxide will be acidic or basic, based on its composition?

7.64 Arrange the following oxides in order of increasing acidity: COz, CaO, AlzQ3, S03, SiOz, and PzOs.

7.65 Chlorine reacts with oxygen to form Cl207. (a) What is the name of this product (see Table 2.6)? (b) Write a bal­ anced equation for the formation of Cl207(1) from the el­ ements. (c) Under usual conditions, Cl207 is a colorless liquid with a boiling point of 81 °C. Is this boiling point expected or surprising? (d) Would you expect Cl207 to be more reactive toward H+(aq) or OH-(aq)? Explain.

[7.66] An element X reacts with oxygen to form X02 and with chlorine to form XC4. X02 is a white solid that melts at high temperatures (above 1000 °C). Under usual condi­ tions, XC4 is a colorless liquid with a boiling point of 58 °C. (a) XC4 reacts with water to form X02 and anoth­ er product. What is the likely identity of the other product? (b) Do you think that element X is a metal, nonmetal, or metalloid? Explain. (c) By using a source­ book such as the CRC Handbook of Chemistry and Physics, try to determine the identity of element X.

7.67 Write balanced equations for the following reactions:

(a) barium oxide with water, (b) iron(II) oxide with per­ chloric acid, (c) sulfur trioxide with water, (d) carbon dioxide with aqueous sodium hydroxide.

7.68 Write balanced equations for the following reactions: (a) potassium oxide with water, (b) diphosphorus triox­ ide with water, (c) chromium(III) oxide with dilute hy­ drochloric acid, (d) selenium dioxide with aqueous potassium hydroxide.

292

C HA PTER 7

Periodic Properties of the Elements

Group Trends in Metals and Nonmetals 7.69

Compare the elements sodium and magnesium with re­

7.76

with hydrogen

tion,

halogens-for

(b) most common ionic charge, (c) first ionization energy, (d) reactivity toward water, (e) atomic radius.

7.70

7.7, the alkali metals react

form hydrides

example,

and

fluorine-to

react with

form

halides.

Compare the roles of hydrogen and the halogen in these reactions. In what sense are the forms of hydrogen and

(a) Compare the electron configurations and atomic

halogen in the products alike?

the reaction of hydrogen with calcium. What are the

spects are their electronic configurations similar? Ac­

(b) As with rubidium, silver is most commonly found as

the 1 + ion, Ag +. However, silver is far less reactive. Ex­

similarities among the products of these reactions?

7.77

tion, (b) most common ionic charge, (c) first ionization

(d) reactivity toward water, (e) electron affinity, (f) atomic radius. Account for the differences between

(a) Why is calcium generally more reactive than magne­

energy,

(b) Why is calcium generally less reactive than

potassium?

the two elements.

(a) Why is cesium more reactive toward water than is lithium?

Compare the elements fluorine and chlorine with re­ spect to the following properties: (a) electron configura­

plain these observations.

sium?

(b) Write balanced equa­

tions for the reaction of fluorine with calcium and for

7.7) of rubidium and silver. In what re­

count for the difference in radii of the two elements.

7.72

to

Account for the differences between the two elements. radii (see Figure

7.71

(a) As described in Section

spect to the following properties: (a) electron configura­

(b) One of the alkali metals reacts with oxygen

7.78

Little is known about the properties of astatine, At, because of its rarity and high radioactivity. Neverthe­

to form a solid white substance. When this substance is

less, it is possible for us to make many predictions about

dissolved in water, the solution gives a positive test for

its properties. (a) Do you expect the element to be a gas,

hydrogen peroxide, H202. When the solution

liquid, or solid at room temperature? Explain.

is tested in

a burner flame, a lilac-purple flame is produced. What is the likely identity of the metal?

(b) What

is the chemical formula of the compound it forms

(c) Write a balanced

with Na?

chemical equation for reaction of the white substance

7.79

with water.

7.73

Write a balanced equation for the reaction that occurs in

gases. The term rare gases was dropped after it was dis­

each of the following cases: (a) Potassium metal burns in an atmosphere of chlorine gas. added to water.

(d)

atmosphere. (a) Why was the term inert gases dropped?

(b) What discovery triggered this change in name? (c) What name is applied to the group now?

Sodium metal is reacted with

molten sulfur.

7.74

covered that argon accounts for roughly 1% of Earth's

(b) Strontium oxide is

(c) A fresh surface of lithium metal is ex­

posed to oxygen gas.

Write a balanced equation for the reaction that occurs in

7.80

(b) Stontium is added to water. (c) Sodium reacts (d) Calcium reacts with iodine.

with oxygen.

7.75

(a)

If

7.81

dioxygen.

(b) Xenon reacts with fluorine. (Write three (c) Sulfur reacts with hydrogen gas.

different equations.)

(Li-Ne) in order of increasing first ionization energy,

(d) Fluorine reacts with water.

(b) If we now

arrange the elements of the third period (Na-Ar) in

Write a balanced equation for the reaction that occurs in each of the following cases: (a) Ozone decomposes to

we arrange the elements of the second period

where would hydrogen fit into this series?

Why does xenon react with fluorine, whereas neon does not?

each of the following cases: (a) Cesium is added to water.

Until the early 1960s the group 8A elements were called the inert gases; before that they were called the rare

7.82

Write a balanced equation for the reaction that occurs in

order of increasing first ionization energy, where would

each of the following cases: (a) Chlorine reacts with

lithium fit into this series? (c) Are these series consistent

water.

with the assignment of hydrogen as a nonmetal and

hydrogen gas.

lithium as a metal?

rine reacts with magnesium metal.

(b) Barium metal is heated in an atmosphere of (c) Lithium reacts with sulfur. (d) Fluo­

ADDITI ONAL EXERC ISES 7.83

Consider the stable elements through lead

(Z

=

82). In

how many instances are the atomic weights of the ele­ ments in the reverse order relative to the atomic num­ bers of cases?

7.84

the elements? What is the explanation for these

(a) Which will have the lower energy, a tron in an As atom?

4s

or a 4p elec­

(b) How can we use the concept of

effective nuclear charge to explain your answer to part (a)?

7.85 (a) If the core electrons were totally effective at shielding the valence electrons and the valence electrons provided

no shielding for each other, what would be the effective nuclear charge acting on the 3s and 3p valence electrons

(b) Repeat these calculations using Slater's rules. (c) Detailed calculations indicate that the effective nu­

in P?

clear charge is 5.6+ for the 3s electrons and 4.9+ for the 3p electrons. Why are the values for the 3s and 3p elec­ trons different?

(d) If you remove a single electron from

a P atom, which orbital will it come from? Explain.

Additional Exercises 7.86 Nearly all the mass of an atom is in the nucleus, which has a very small radius. When atoms bond together (for example, two fluorine atoms in F2), why is the distance

separating the nuclei so much larger than the radii of the

nuclei?

(7.87] Consider the change in effective nuclear charge experi­ enced by a

2p

electron as we proceed from C to N.

293

cube in the figure (the distance from the center of an

atom at one comer to the center of an atom at a neigh­ boring comer). (c) The experimental density of SrO is 3 Given your answer to part (b), what is the

5.10 g/cm

number of formula units of SrO that are contained in the cube in the figure? (We will examine structures like those in the figure more closely in Chapter

11.)

(a) Based on a simple model in which core electrons

screen the valence electrons completely and valence electrons do not screen other valence electrons, what do

you predict lor the change in Zeff from C to N? (b) What

change do you predict using Slater's rules? (c) The actu­ al change in z.11 from C to N is 0.70+. Which approach to estimating Zeu is more accurate? (d) The change in Zeff from N to 0 is smaller than that from C to N. Can you provide an explanation for this observation?

7.88 As we move across a period of the periodic table, why do the sizes of the transition elements change more gradually than those of the representative elements?

7.89 In the series of group SA hydrides, of general formula MH3, the measured bond distances are P-H, 1.419 A; As-H, 1.519 A; Sb - H, 1.707 A. (a) Compare these values with those estimated by use of the atomic radii in

Figure 7.7. (b) Explain the steady increase in M- H bond distance in this series

in terms of the electronic

configurations of the M atoms.

7.94 Explain the variation in ionization energies of carbon, as displayed in the following graph:

7.90 It is possible to produce compounds of the form

GeCIH3, GeC1 Hz, and GeC13H. What values do you 2 predict for the Ge - H and Ge-Cl bond lengths in these compounds?

7.91 Note from the following table that the increase in atom­

ic radius in moving from Zr to Hf is smaller than in

moving from effect.

Y

to La. Suggest an explanation for this

Atomic Radii (A) Sc

1.44

Ti

1.69

HI

1.62

y La

1.36

Zr

1 .48

Ionization number

1.50

7.95 Do you agree with the following statement? "A negative 7.92 The "Chemistry and Life" box on ionic size in Section 7.3 compares the ionic radii of Zn2+ and Cd2+ (a) The 2+

ion of which other element seems the most obvious one

to compare to Zn2+ and Cd2+? (b) With reference to Figure

2.24,

is the element in part (a) essential for life?

(c) Estimate the ionic radius of the 2+ ion of the element

in part (a). Explain any assumptions you have made. (d) Would you expect the 2+ ion of the element in part

(a) to be physiologically more similar to Zn2+ or to Cd2+?

(e) Use a sourcebook or a Web search to determine

whether the element in part (a) is toxic to humans.

(7.93] The ionic substance strontium oxide, SrO, forms from the direct reaction of strontium metal with molecular

oxygen. The arrangement of the ions in solid SrO is

analogous to that

in solid NaCl

(see Figure

2.23) and is

shown here. (a) Write a balanced equation for the forma­

tion of SrO(s) from the elements. (b) Based on the ionic

radii in Figure

7.8,

predict the length of the side of the

value for the electron affinity of an atom occurs when the outermost electrons incompletely shield one another

from the nucleus." If not, change it to make it more near­ ly correct in your view. Apply either the statement as given or your revised statement to explain why the elec­

tron affinity of bromine is neighbor Kr is > 0.

-325 kJ/mol and

that for its

7.96 Use orbital diagrams to illustrate what happens when

an oxygen atom gains two electrons. Why is it extreme­ ly difficult to add a third electron to the atom?

(7.97] Use electron configurations to explain the following ob­

servations: (a) The first ionization energy of phosphorus

is greater than that of sulfur. (b) The electron affinity of

nitrogen is lower (less negative) than those of both

carbon and oxygen. (c) The second ionization energy of oxygen is greater than the first ionization energy of

fluorine. (d) The third ionization energy of manganese is

greater than those of both chromium and iron.

294

C HA PTER 7

Periodic Properties of the Elements

7.98 The following table gives the electron affinities, in kJ/mol, for the group 1B and group 2B metals: (a) Why are the electron affinities of the group 2B elements greater than zero? (b) Why do the electron affinities of the group 1B elements become more negative as we move down the group? [Hint: Examine the trends in the electron affinity of other groups as we proceed down the periodic table.] Zn

Cu -119

>0

Ag - 1 26

Cd >0

Au -223

Hg >0

7.99 Hydrogen is an unusual element because it behaves in some ways like the alkali metal elements and in other ways like a nonmetal. Its properties can be explained in part by its electron configuration and by the values for its ionization energy and electron affinity. (a) Explain why the electron affinity of hydrogen is much closer to the values for the alkali elements than for the halogens. (b) Is the following statement true? "Hydrogen has the smallest bonding atomic radius of any element that forms chemical compounds." If not, correct it. If it is, ex­ plain in terms of electron configurations. (c) Explain why the ionization energy of hydrogen is closer to the values for the halogens than for the alkali metals. [7 .100] The first ionization energy of the oxygen molecule is the energy required for the following process:

02(gl

-----+

02+(g) + e-

The energy needed for this process is 1175 kJ/mol, very similar to the first ionization energy of Xe. Would you expect 02 to react with F2? If so, suggest a product or products of this reaction.

7.101 Based on your reading of this chapter, arrange the fol­ lowing in order of increasing melting point: K, Br:u Mg, and 02. Explain the factors that determine the order. 7.102 The element strontium is used in a variety of industrial processes. It is not an extremely hazardous substance, but low levels of strontium ingestion could affect the health of children. Radioactive strontium is very haz­ ardous; it was a by-product of nuclear weapons testing and was found widely distributed following nuclear tests. Calcium is quite common in the environment, in­ cluding food products, and is frequently present in drinking water. Discuss the similarities and differences between calcium and strontium, and indicate how and why strontium might be expected to accompany calci­ um in water supplies, uptake by plants, and so on. [7.103] There are certain similarities in properties that exist be­ tween the first member of any periodic family and the element located below it and to the right in the periodic table. For example, in some ways Li resembles Mg, Be resembles Al, and so forth. This observation is called the diagonal relationship. Using what we have learned in this chapter, offer a possible explanation for this rela­ tionship. [7.104] A historian discovers a nineteenth-century notebook in which some observations, dated 1822, on a substance thought to be a new element, were recorded. Here are some of the data recorded in the notebook: Ductile, silver­ white, metallic looking. Softer than lead. Unaffected by water. Stable in air. Melting point: 153 oc Density: 7.3 g/ cm3. Electrical conductivity: 20% that of copper. Hardness: About 1% as hard as iron. When 4.20 g of the unknown is heated in an excess of oxygen, 5.08 g of a white solid is formed. The solid could be sublimed by heating to over 800 oc. (a) Using information in the text and a handbook of chemistry, and making allowances for possible variations in numbers from current values, iden­ tify the element reported. (b) Write a balanced chemical equation for the reaction with oxygen. (c) Judging from Figure 7.2, might this nineteenth-century investigator have been the first to discover a new element?

INTEGRATIVE EXERCI SES [7.105] Moseley established the concept of atomic number by studying X-rays emitted by the elements. The X-rays emitted by some of the elements have the following wavelengths:

Element

Wavelength (A)

Ne

14.610

Ca

3.358

Zn

1 .435

Zr

0.786

Sn

0.491

(a) Calculate the frequency, v, of the X-rays emitted by each of the elements, in Hz. (b) Using graph paper (or

suitable computer software), plot the square root of v versus the atomic number of the element. What do you observe about the plot? (c) Explain how the plot in part (b) allowed Moseley to predict the existence of undis­ covered elements. (d) Use the result from part (b) to pre­ dict the X-ray wavelength emitted by iron. (e) A particular element emits X-rays with a wavelength of 0.980 A. What element do you think it is? [7.106] (a) Write the electron configuration for Li, and estimate the effective nuclear charge experienced by the valence electron. (b) The energy of an electron in a one-electron atom or ion equals (-2.18 X 10-lB J)

(�:)

where Z is

the nuclear charge and n is the principal quantum num­ ber of the electron. Estimate the first ionization energy of Li. (c) Compare the result of your calculation with the

Integrative Exercises value reported in table 7.4, and explain the difference.

(d) What value of the effective nuclear charge gives the

proper value for the ionization energy? Does this agree

with your explanation in (c)?

[7.107] One way to measure ionization energies is photoelec­ tron spectroscopy

(PES), a technique based on the

295

The mass of the mixture of MgO and magnesium ni­

tride after burning is 0.470 g. Water is added to the cru­ cible, further reaction occurs, and the crucible is heated

to dryness until the final product is 0.486 g of MgO.

What was the mass percentage of magnesium nitride in

the mixture obtained after the initial burning? (d) Mag­

(Section 6.2) In PES, monochro­

nesium nitride can also be formed by reaction of the

to be emitted. The kinetic energy of the emitted elec­

balanced equation for this reaction. If a 6.3-g Mg ribbon

photoelectric effect.

ooo

matic light is directed onto a sample, causing electrons

metal with ammonia at high temperature. Write a

the photons and the kinetic energy of the electrons cor­

reacts with 2.57 g NH3{g) and the reaction goes to com­ pletion, which component is the limiting reactant?

(that is, the ionization energy). Suppose that a PES ex­ periment is performed in which mercury vapor is irradi­

nitride is -461.08 kJ/mol. Calculate the standard en­

trons is measured. The difference between the energy of

responds to the energy needed to remove the electrons

ated with ultraviolet light of wavelength 58.4 nm.

(a) What is the energy of a photon of this light, in eV? (b) Write an equation that shows the process corre­ sponding to the first ionization energy of Hg. (c) The ki­

netic energy of the emitted electrons is measured to be

10.75 eV. What is the first ionization energy of Hg, in kJ/mol? (d) With reference to Figure 7.11, determine

which of the halogen elements has a first ionization en­

ergy closest to that of mercury.

7.108 Consider the gas-phase transfer of an electron from a sodium atom to a chlorine atom: Na{g)

+

Cl{g)

---->

Na+{g)

+

Cl-{g)

(a) Write this reaction as the sum of two reactions, one

that relates to an ionization energy and one that relates

What mass of H2{g) is formed in the reaction? (e) The standard enthalpy of formation of solid magnesium thalpy change for the reaction between magnesium

metal and ammonia gas.

7.110 (a) The experimental Bi- Br bond length in bismuth tri­

A.

bromide, BiBr3, is 2.63 Based on this value and the data in Figure 7.7, predict the atomic ractius of Bi.

(b) Bismuth tribromide is soluble in acidic solution. It is formed by treating solid bismuth(III) oxide with aque­ ous hydrobromic acid. Write a balanced chemical equa­

tion for this reaction. (c) While bismuth(III) oxide is soluble in acidic solutions, it is insoluble in basic solu­

tions such as NaOH(aq). Based on these properties, is

bismuth characterized as a metallic, metalloid, or non­

metallic element? (d) Treating bismuth with fluorine gas

forms BiF5. Use the electron configuration of Bi to ex­

plain the formation of a compound with this formula­

to an electron affinity. (b) Use the result from part (a),

tion. (e) While it is possible to form BiF5 in the manner

thalpy of the above reaction. Is the reaction exothermic or endothermic? (c) The reaction between sodium metal and chlorine gas is highly exothermic and produces

for the other halogens. Explain why the pentahalide might form with fluorine, but not with the other halo­

data in this chapter, and Hess's law to calculate the en­

NaCl(s), whose structure was discussed in Section 2.7.

Comment on this observation relative to the calculated

enthalpy for the aforementioned gas-phase reaction.

[7.109] When magnesium metal is burned in air (Figure 3.5),

just described, pentahalides of bismuth are not known

gens. How does the behavior of bismuth relate to the fact that xenon reacts with fluorine to form compounds,

but not with the other halogens?

7.111 Potassium superoxide, KOz, is often used in oxygen

two products are produced. One is magnesium oxide,

masks (such as those used by firefighters) because K02 reacts with C02 to release molecular oxygen. Experi­ ments indicate that 2 mol of K02(s) react with each mole

water is added to magnesium nitride, it reacts to form

and 02{g). Write a balanced equation for the reaction be­

MgO. The other is the product of the reaction of M g with molecular nitrogen, magnesium nitride. When

magnesium oxide and ammonia gas. (a) Based on the charge of the nitride ion (Table 2.5), predict the formula

of magnesium nitride. (b) Write a balanced equation for the reaction of magnesium nitride with water. What is the driving force for this reaction? (c) In an experiment a

piece of magnesium ribbon is burned in air in a crucible.

of C02{g). (a) The products of the reaction are K2C03(s)

tween K02(s) and COz(g) . (b) Indicate the oxidation

number for each atom involved in the reaction in part (a). What elements are being oxidized and reduced?

(c) What mass of K02(s) is needed to consume 18.0 g

C02(g)? What mass of 02{g) is produced during this

reaction?

BASIC CONCEPTS OF CHEMICAL BONDING

THE WHITE CLIFFS OF DOVER in southeastern England are made up largely of chalk, a porous form of limestone. The mineral calcite, with a composition of CaC03, is the predominant chemical substance in both chalk and limestone. Much of Earth's

calcite is produced by marine organisms, which combine Ca2+ and C03 2- ions to form shells of CaC03. The presence of CO ions in oceans can be traced to dissolved C0 2 from the atmosphere.

296

/­

W H AT ' S 8.1

A H E A D

We can broadly characterize chemical bonds into three types: ionic, covalent, and metallic. In evaluating bonding, Lewis symbols provide a useful shorthand notation for keeping track of the valence electrons in atoms and ions.

8.2

8.3

8.4

8.5

patterns within molecules. In addition to the octet rule, we will see that the concept of formal charge can be used to identify the most favorable Lewis structure.

Chemical Bonds, Lewis Symbols, and the Octet Rule

8.6

Ionic Bonding

We will observe that in ionic substances the atoms are held together by the electrostatic attractions between ions of opposite charge. We will study the energetics of formation of ionic substances and describe the lattice energy of these substances.

8.7

Covalent Bonding

We also recognize that the atoms in molecular substances are held together by the sharing of one or more electron pairs between atoms. In general the electrons are shared in such a way that each atom attains an octet of electrons. Bond Polarity and Electronegativity

We define electronegativity as the ability of an atom in a compound to attract electrons to itself. In general, electron pairs will be shared unequally between atoms of differing electronegativity, leading to polar covalent bonds.

8.8

Resonance Structu res

We observe that in some cases more than one equivalent Lewis structure can be drawn for a molecule or polyatomic ion. The actual structure in such cases is a blend of two or more contributing Lewis structures, called resonance structures. Exceptions to the Octet Rule

We recognize that the octet rule is more of a guideline than an inviolate rule. Exceptions to the octet rule include molecules with an odd number of electrons; molecules where large differences in electronegativity prevent an atom from completing its octet; and, most commonly, molecules where an element from the third period or lower attains more than an octet of electrons. Strengths of Covalent Bonds

We observe that bond strengths vary with the number of shared electron pairs as well as other factors. We can use average bond enthalpy values to estimate the enthalpies of reactions in cases where thermodynamic data such as heats of formation are unavailable.

Drawing Lewis Structu res

We will see that Lewis structures are a simple yet powerful way of predicting the covalent bonding

CALC I U M CARBO NATE, CaC03, IS O N E O F TH E MOST I NTERE STI NG A N D VERSAT I L E C O M P O U N D S

on the planet. It accounts for

roughly 4% of Earth's crust and is the major component of rocks such as limestone and marble. The white cliffs of Dover in sou theastern England are one of the most famous natural formations of CaC03. The cliffs consist almost entirely of a porous form of limestone called chalk. Unlike most inorganic substances, CaC03 is widely used by living organ­ isms and is found in objects such as seashells, coral, eggshells, and pearls. Calcium carbonate plays a role in the complex chemistry associated with the greenhouse effect because it is formed in the oceans through the reaction be­ tween calcium ions and dissolved carbon dioxide. When limestone is heated to elevated temperatures, CaC03 decomposes into solid CaO (the important in­ dustrial chemical called quicklime) and gaseous C02• Calcium carbonate undergoes the reactions that it does because the atoms in CaC03 are held 297

298

C HA PTER 8

Basic Concepts of Chemical Bonding

T Figure 8.1 Chemical bonds.

Examples of substances in which (a) ionic, (b) covalent, and (c) metallic bonds are found.

Magnesium oxide

Potassium dichromate Nickel(II) oxide (a) Sulfur

Sucrose

Bromine (b)

Magnesium

together by a combination of two different types of bonds. The carbon and oxygen atoms that make up the carbonate ion share valence electrons, result­ ing in the formation of covalent bonds. The oppositely charged Ca2+ and CO /­ ions are held together by electrostatic attractions, which are called ionic bonds. The properties of substances are determined in large part by the chemical bonds that hold their atoms together. What determines the type of bonding in each substance? How do the characteristics of these bonds give rise to differ­ ent physical and chemical properties? The keys to answering the first question are found in the electronic structures of the atoms involved, which we dis­ cussed in Chapters 6 and 7. In this chapter and the next, we will examine the relationship between the electronic structures of atoms and the chemical bonds they form. We will also see how the properties of ionic and covalent substances arise from the distributions of electronic charge within atoms, ions, and molecules.

8. 1 CHEMICAL BONDS, LEWIS SYM B OLS , AND THE OCTET RULE Whenever two atoms or ions are strongly attached to each other, we say there is a chemical bond between them. There are three general types of chemical bonds: ionic, covalent, and metallic. Figure 8.1 ... shows examples of substances in which we find each of these types of attractive forces. The term ionic bond refers to electrostatic forces that exist between ions of opposite charge. Ions may be formed from atoms by the transfer of one or more electrons from one atom to another. Ionic substances generally result from the interaction of metals on the left side of the periodic table with nonmetals on the right side (excluding the noble gases, group 8A). Ionic bonding will be dis­ cussed in Section 8.2. A covalent bond results from the sharing of electrons between two atoms. The most familiar examples of covalent bonding are seen in the interactions of nonmetallic elements with one another. We devote much of this chapter and the next to describing and understanding covalent bonds. Metallic bonds are found in metals, such as copper, iron, and aluminum. Each atom in a metal is bonded to several neighboring atoms. The bonding electrons are relatively free to move throughout the three-dimensional struc­ ture of the metal. Metallic bonds give rise to such typical metallic properties as high electrical conductivity and luster. We will examine these bonds in Chapter 23.

Lewis Symbols The electrons involved in chemical bonding are the valence electrons, which, for most atoms, are those residing in the outermost occupied shell of an atom. CXX> (Section 6.8) The American chemist G. N. Lewis (1875-1946) suggested a simple way of showing the valence electrons in an atom and tracking them in the course of bond formation, using what are now known as Lewis electron-dot symbols, or merely Lewis symbols. The Lewis symbol for an element consists of the chemical symbol for the element plus a dot for each valence electron. Sulfur, for example, has the elec­ 4 tron configuration [Ne]3s23p ; its Lewis symbol therefore shows six valence electrons:

·�· Copper

Gold (c)

The dots are placed on the four sides of the atomic symbol: the top, the bottom, and the left and right sides. Each side can accommodate up to two electrons. All four sides of the symbol are equivalent, which means that the choice of which sides accommodate the fifth and sixth electrons is arbitrary.

8.2

TABLE 8.1 • Lewis Symbols

Ionic Bonding

299

Element

Electron Configuration

Lewis Symbol

Element

Electron Configuration

Lewis Symbol

Li

[He]2s 1

Li·

Na

[Ne]3s 1

Na ·

Be

[He]2s2

·Be ·

Mg

[Ne]3s2

· Mg ·

B

[He]2s 22pl

·B·

AI

[Ne]3s23p1

·AI·

c

[He]2s 22pz

·¢·

Si

[Ne]3s23p2

· �i ·

N

[He]2s22p3

[Ne]3s23p3

. �:

0

[He]2s 22p4

[Ne]3s23p4 [Ne]3s23p5

· ¢.1:

Ne

[He]2s22p6

Cl

:$:

F

[He]2s 22p5

.�: :¢: .f. :

p

[Ne]3s23p6

:�r :

s

Ar

:r-;(e :

The electron configurations and Lewis symbols for the representative ele­ ments of the second and third rows of the periodic table are shown in Table 8.1 A. Notice that the number of valence electrons in any representative element is the same as the group number of the element. For example, the Lewis symbols for oxygen and sulfur, members of group 6A, both show six dots.

GIVE IT SOME THOUGHT Are all three of the following Lewis symbols for Cl correct?

:¢)· :(;.1: : lists the lattice energies of NaCl and other ionic compounds. All are large positive values, indicating that the ions are strongly attracted to one another in these solids. The energy released by the

TABLE 8.2 • Lattice Energies for Some Ionic Compounds Lattice Energy Lattice Energy c ompound

NaCI(s)

a HJ [NaCI(s)] = -411 kJ

------>

! Cl2(g)

Na(g)

------>

Cl(g)

aHJ [Na(g)] = 108 kJ

a HJ [Cl(g)J

= 122 kJ

------>

Cl(g) + e-

Na+(g) + e------>

Cl-(g)

aH = 11 (Na) = 496 kJ aH = E(Cl) = -349 kJ

------>

aH =

NaCl(s)

-a Hiattice

=?

-411 kJ

=

Solving for

a Hiattke

108 kJ + 122 kJ + 496 kj - 349 kJ - aHiattke

a Hiatticc' = 108 kJ + 122 kJ + 496 kJ - 349 kj + 411 kJ

= 788 kJ

Thus, the lattice energy of NaCl is 788 kJ/mol.

Related Exercises: 8.26, 8.27, and 8.28

E(Cl) Na)

[8.7]

[8.8]

[8.9]

Finally, we combine the gaseous sodium and chloride ions to form solid sodium chloride. Because this process is just the reverse of the lattice energy (breaking a solid into gaseous ions), the enthalpy change is the negative of the lattice energy, the quantity that we want to determine: Na+(g) + Cl-(g)

+ft(Na) + E(Cl) - aHiattke

[8.6]

Both of these processes are endothermic; energy is required to generate gaseous sodium and chlorine atoms. In the next two steps we remove the electron from Na(g) to form Na+(g) and then add the electron to Cl(g) to form Cl-(g). The enthalpy changes for these processes equal the first ionization energy of Na, /1 (Na), and the electron affinity of Cl, denoted E(Cl), respectively: = (Sections 7.4, 7.5) Na(g)

a HJ [NaCl(s)] = aHJ [Na(g)] + aHJ [Cl(g)]

[8.5]

The indirect route consists of five steps, shown by the green arrows in Figure 8.4. First, we generate gaseous atoms of sodium by vaporizing sodium metal. Then we form gaseous atoms of chlorine by breaking the bonds in the Cl2 molecules. The enthalpy changes for these processes are avail­ able to us as enthalpies of formation (Appendix C): Na(s)

The sum of the five steps in the indirect path (green arows) gives us NaCl(s) from Na(s) and ! Cl2(g) . Thus, from Hess's law we know that the sum of the enthalpy changes for these five steps equals that for the direct path, indicated by the red arrow, Equation 8.5:

[8.10]

Na+(g) + eng)

�

Na(g) + Cl(g) (g)] (g)]

+

1 2 Cl2(g)

Na(s) +

� Clz(g)

Na(g)

0 "' z 0

>, bD

0: @

Q)

E"'

_, I

aH; [NaCl(s)]

0 "' z 0

r

�

j

NaCl(s)

.a. Figure 8.4 The Born-Haber cycle. This representation shows

the energetic relationships in the formation of ionic solids from the elements. By Hess's law, the enthalpy of formation of NaCI(s) from elemental sodium and chlorine (Equation 8.5) is equal to the sum of the energies of several individual steps (Equations 8.6 through 8.1 0).

largest value of n. Thus, in forming ions, transition metals lose the valence-shell s electrons first, then as many d electrons as are required to reach the charge of the ion.

2 Let's consider Fe, which has the electron configuration [Ar]3d64s In forming 2 the Fe + ion, the two 4s electrons are lost, leading to an [Ar]3d6 configuration. 3 Removal of an additional electron gives the Fe + ion, whose electron configura­ tion is [Ar]3d5.

8.3

305

Covalent Bonding

GIVE IT SOME THOUGHT Which element forms a 2 + ion that has the electron configuration [Kr]4d8?

8.3 COVALENT BONDING Ionic substances possess several characteristic properties . They are usually brit­ tle substances with high melting points. They are usually crystalline. Further­ more, ionic crystals often can be cleaved; that is, they break apart along smooth, flat surfaces. These characteristics result from electrostatic forces that maintain the ions in a rigid, well-defined, three-dimensional arrangement such as that shown in Figure 8.3. The vast majority of chemical substances do not have the characteristics of ionic materials. Most of the substances with which we come into daily con­ tact-such as water-tend to be gases, liquids, or solids with low melting points. Many, such as gasoline, vaporize readily. Many are pliable in their solid forms-for example, plastic bags and paraffin. For the very large class of substances that do not behave like ionic sub­ stances, we need a different model for the bonding between atoms. G. N. Lewis reasoned that atoms might acquire a noble-gas electron configuration by shar­ ing electrons with other atoms. As we noted in Section 8.1, a chemical bond formed by sharing a pair of electrons is called a covalent bond. The hydrogen molecule, H2, provides the simplest example of a covalent bond. When two hydrogen atoms are close to each other, electrostatic interac­ tions occur between them. The two positively charged nuclei repel each other, the two negatively charged electrons repel each other, and the nuclei and elec­ trons attract each other, as shown in Figure 8.5 � . Because the H2 molecule ex­ ists as a stable entity, the attractive forces must exceed the repulsive ones. Let's take a closer look at the attractive forces that hold this molecule together. By using quantum mechanical methods analogous to those employed for atoms cx:o (Section 6.5), it is possible to calculate the distribution of electron density in molecules. Such a calculation for H2 shows that the attractions be­ tween the nuclei and the electrons cause electron density to concentrate be­ tween the nuclei, as shown in Figure 8.5(b). As a result, the overall electrostatic interactions are attractive. Thus, the atoms in H2 are held together principally because the two nuclei are electrostatically attracted to the concentration of negative charge between them. In essence, the shared pair of electrons in any covalent bond acts as a kind of "glue" to bind atoms together.

GIVE IT SOME THOUGHT If a Hz molecule is ionized to form Hz+, it will change the bond strength. Based on the simple description of covalent bonding given above, would you expect the H bond in Hz+ to be weaker or stronger than the H -H bond in Hz?

H-

Lewis Structures The formation of covalent bonds can be represented using Lewis symbols for the constituent atoms. The formation of the H2 molecule from two H atoms, for example, can be represented as H· + · H ---+

@@_)

In this way, each hydrogen atom acquires a second electron, achieving the sta­ ble, two-electron, noble-gas electron configuration of helium. The formation of a bond between two Cl atoms to give a Cl2 molecule can be represented in a similar way: :

¢)

·+ ·

¢)

:

-

�

/

Attraction

8

/

Electron

j "">.

8/ Repulsion """8/ Nucleus "">.

1

""8/

/

(a)

+

+

(b)

"" Figure 8.5 The covalent bond In H2. (a) The attractions and repulsions among electrons and nuclei in the hydrogen molecule. (b) Electron distribution in the Hz molecule. The concentration of electron density between the nuclei leads to a net attractive force that constitutes the covalent bond holding the molecule together.

306

C HA PTER 8

Basic Concepts of Chemical Bonding By sharing the bonding electron pair, each chlorine atom has eight electrons (an octet) in its valence shell. It thus achieves the noble-gas electron configura­ tion of argon. The structures shown here for H2 and Cl2 are called Lewis structures (or Lewis electron-dot structures). In writing Lewis structures, we usually show each electron pair shared between atoms as a line and the unshared electron pairs as dots. Written this way, the Lewis structures for H2 and Cl2 are H-H For the nonmetals, the number of valence electrons in a neutral atom is the same as the group number. Therefore, one might predict that 7A elements, such as F, would form one covalent bond to achieve an octet; 6A elements, such as 0, would form two covalent bonds; SA elements, such as N, would form three co­ valent bonds; and 4A elements, such as C, would form four covalent bonds. These predictions are borne out in many compounds. For example, consider the simple hydrogen compounds of the nonmetals of the second row of the periodic table: H H-f. :

H-0:

H-N-H

�

�

I

H-C-H

�

Thus, the Lewis model succeeds in accounting for the compositions of many compounds of nonmetals, in which covalent bonding predominates. - SAMPLE EXERCISE

8.3 I Lewis Structure of a Compound

Given the Lewis symbols for the elements nitrogen and fluorine shown in Table 8.1, predict the formula of the stable binary compound (a compound composed of two el­ ements) formed when nitrogen reacts with fluorine, and draw its Lewis structure. SOLUTION

Analyze: The Lewis symbols for nitrogen and fluorine reveal that nitrogen has five valence electrons and fluorine has seven. Plan: We need to find a combination of the two elements that results in an octet of

electrons around each atom in the compound. Nitrogen requires three additional elec­ trons to complete its octet, whereas fluorine requires only one. Sharing a pair of electrons between one N atom and one F atom will result in an octet of electrons for fluorine but not for nitrogen. We therefore need to figure out a way to get two more electrons for the N atom. Nitrogen must share a pair of electrons with three fluorine atoms to complete its octet. Thus, the Lewis structure for the resulting compound, NF3, is

Solve:

·

N

·

+ 3

·

f. :

�

:il:f..J :il: :p : ··

··

:il-f..J -il: .. .. I :F:

The Lewis structure in the center shows that each atom is surrounded by an octet of electrons. Once you are accustomed to thinking of each line in a Lewis struc­ ture as representing two electrons, you can just as easily use the structure on the right to check for octets.

Check:

- PRACTICE EXERCISE

Compare the Lewis symbol for neon with the Lewis structure for methane, CH . In what important way are the electron arrangements about neon and carbon alike?4 In what important respect are they different? Answer: Both atoms have an octet of electrons about them. However, the electrons about neon are unshared electron pairs, whereas those about carbon are shared with four hydrogen atoms.

8.4

Bond Polarity and Electronegativity

Multiple Bonds The sharing of a pair of electrons constitutes a single covalent bond, generally referred to simply as a single bond. In many molecules, atoms attain complete octets by sharing more than one pair of electrons. When two electron pairs are shared, two lines are drawn, representing a double bond. In carbon dioxide, for example, bonding occurs between carbon, with four valence electrons, and oxy­ gen, with six:

:6: + ·¢· + :¢: ------> q::c ::q

(or

Q=C=Q)

As the diagram shows, each oxygen acquires an octet of electrons by sharing two electron pairs with carbon. Carbon, on the other hand, acquires an octet of elec­ trons by sharing two electron pairs in each of the two bonds it forms with oxygen. A triple bond corresponds to the sharing of three pairs of electrons, such as in the N2 molecule:

:N·

+

·N: ------>

: N : : : N:

(or : N- N :)

Because each nitrogen atom possesses five electrons in its valence shell, three electron pairs must be shared to achieve the octet configuration. The properties of N2 are in complete accord with its Lewis structure. Nitro­ gen is a diatomic gas with exceptionally low reactivity that results from the very stable nitrogen-nitrogen bond. Study of the structure of N2 reveals that the nitrogen atoms are separated by only 1.10 A. The short N -N bond dis­ tance is a result of the triple bond between the atoms. From structure studies of many different substances in which nitrogen atoms share one or two electron pairs, we have learned that the average distance between bonded nitrogen atoms varies with the number of shared electron pairs : N-N 1.47 A

N=N 1.24 A

N == N

uo A

As a general rule, the distance between bonded atoms decreases as the number of shared electron pairs increases. The distance between the nuclei of the atoms involved in a bond is called the bond length for the bond. We first encountered bond lengths in Section 7.3 in our discussion of atomic radii, and we will discuss them further in Section 8.8.

GIVE IT SOME THOUGHT

A,

The C - O bond length in carbon monoxide, CO, is 1.13 whereas the C - O bond length in C02 is 1 .24 Without drawing a Lewis structure, do you think that carbon monoxide has a single, double, or triple bond between the C and 0 atoms?

A.

8.4 BOND P O LARITY AND ELECTRON EGAT IVITY When two identical atoms bond, as in Clz or Hz., the electron pairs must be shared equally. In highly ionic compounds, on the other hand, such as NaCI, there is relatively little sharing of electrons, which means that NaCI is best de­ scribed as composed of Na+ and Cl- ions. The 3s electron of the Na atom is, in effect, transferred completely to chlorine. The bonds that are found in most substances fall somewhere between these extremes. The concept of bond polarity helps describe the sharing of electrons be­ tween atoms. A nonpolar covalent bond is one in which the electrons are shared equally between two atoms, as in the Cl2 and N2 examples we just cited.

307

308

C HA PTER

8

Basic Concepts of Chemical Bonding

In

a

polar covalent bond,

one of the atoms exerts a greater attraction for the

bonding electrons than the other. If the difference in relative ability to attract electrons is large enough, an ionic bond is formed.

Electro negativity We use a quantity called electronegativity to estimate whether a given bond will be nonpolar covalent, polar covalent, or ionic. as the ability of an atom

Electronegativity is defined

in a molecule to attract electrons to itself. The greater an

atom's electronegativity, the greater is its ability to attract electrons to itself. The electronegativity of an atom in a molecule is related to its ionization energy and electron affinity, which are properties of isolated atoms. The

ionization energy

measures how strongly a gaseous atom holds on to its electrons, = (Section while the

electron affinity is

7.4)

a measure of how strongly an atom attracts addi­

tional electrons. = (Section 7.5) An atom with a very negative electron affinity

T Figure 8.6 Electronegatlvltles of

and high ionization energy will both attract electrons from other atoms and

the elements. Electronegativity generally increases from left to right across a period and decreases from top to bottom down a group.

resist having its electrons attracted away; it will be highly electronegative. Numerical estimates of electronegativity can be based on a variety of prop­ erties, not just ionization energy and electron affinity. The American chemist Linus Pauling

(1901-1994)

developed the first and most widely used elec­

tronegativity scale; he based his scale on thermochemical data. Figure

8.6 .,.

shows Pauling's electronegativity values for many of the elements. The values are unitless. Fluorine, the most electronegative element, has an electronegativity of

4.0.

The least electronegative element,

cesium, has an electronegativity of >- 4

increase in electronegativity from left to

"'

� 2

>LI

The values for all

Within each period there is generally a steady

:�

J

0.7.

other elements lie between these two extremes.

0

�

right; that is, from the most metallic to the most nonmetallic elements. With 1A

some

exceptions

(especially

within the transition metals), elec­ 2A

tronegativity decreases with increas­ 3B

3.0-4.0 2.0-2.9 1.5-1.9 < 1 .5

4B

ing atomic number in any one group. SB

This is what we might expect because 6B

7B�

1B

we know that ionization energies tend

to

decrease

with

increasing

atomic number in a group and electron 2B

affinities do not change very much. You do 3A 4A

not need to memorize numerical values for SA 6A

electronegativity. Instead, you should know the periodic trends so that you can predict which of two 7A

elements is more electronegative.

GIVE IT SOME THOUGHT How does the electronegativity of an element differ from its electron affinity?

Electronegativity and Bond Pola rity We can use the difference in electronegativity between two atoms to gauge the polarity of the bonding between them. Consider these three fluorine-containing compounds:

Compound Electronegativity difference Type of bond

F2

HF

4.0 - 4.0 = 0

4.0 - 2.1 = 1.9

Nonpolar covalent

Polar covalent

LiF 4.0 - 1.0 = 3.0

Ionic

8.4

Bond Polarity and Electronegativity

• HF

LiF

.A. Figure 8.7 Electron density distribution. This computer-generated rendering shows the calculated electron-density distribution on the surface of the F2, HF, and LiF molecules. The regions of relatively low electron density (net positive charge) appear blue, those of relatively high electron density (net negative charge) appear red, and regions that are close to electrically neutral appear green.

In F2 the electrons are shared equally between the fluorine atoms, and thus the covalent bond is nonpolar. In general, a nonpolar covalent bond results when the electronegativities of the bonded atoms are equal. In HF the fluorine atom has a greater electronegativity than the hydrogen atom, with the result that the sharing of electrons is unequal-the bond is polar. In general, a polar covalent bond results when the atoms differ in electronega­ tivity. In HF the more electronegative fluorine atom attracts electron density away from the less electronegative hydrogen atom, leaving a partial positive charge on the hydrogen atom and a partial negative charge on the fluorine atom. We can represent this charge distribution as 8+

5-

H-F

The 8 + and 8- (read "delta plus" and "delta minus") symbolize the partial positive and negative charges, respectively. In LiF the electronegativity difference is very large, meaning that the elec­ tron density is shifted far toward F. The resultant bond is therefore most accu­ rately described as ionic. This shift of electron density toward the more electronegative atom can be seen in the results of calculations of electron densi­ ty distributions. For the three species in our example, the calculated electron density distributions are shown in Figure 8.7 .a.. The regions of space that have relatively higher electron density are shown in red, and those with a relatively lower electron density are shown in blue. You can see that in F2 the distribution is symmetrical, in HF it is clearly shifted toward fluorine, and in LiF the shift is even greater. These examples illustrate, therefore, that the greater the difference in

electronegativity between two atoms, the more polar their bond. GIVE IT SOME THOUGHT

Based on differences in electronegativity, how would you characterize the bonding in silicon nitride, Si3N4? Would you expect the bonds between Si and N to be nonpolar, polar covalent, or ionic?

- SAMPLE EXERCISE

8.4 J

Bond Polarity

In each case, which bond is more polar: (a) B-Cl or C - CI, (b) P - F or P - C!? Indicate in each case which atom has the partial negative charge. SOLUTION Analyze: We are asked to determine relative bond polarities, given nothing but the atoms involved in the bonds. Plan: Because we are not asked for quantitative answers, we can use the periodic table and our knowledge of electronegativity trends to answer the question .

309

310

C HA PTER 8

Basic Concepts of Chemical Bonding Solve: (a) The chlorine atom is common to both bonds. Therefore, the analysis reduces to a comparison of the electronegativities of B and C. Because boron is to the left of carbon in the periodic table, we predict that boron has the lower electronegativity. Chlorine, being on the right side of the table, has a higher electronegativity. The more polar bond will be the one between the atoms having the lowest electronegativity (boron) and the highest electronegativity (chlorine). Consequently, the B - Cl bond is more polar; the chlorine atom carries the partial negative charge because it has a higher electronegativity. (b) In this example phosphorus is common to both bonds, and the analysis reduces to a comparison of the electronegativities of F and Cl. Because fluorine is above chlo­ rine in the periodic table, it should be more electronegative and will form the more polar bond with P. The higher electronegativity of fluorine means that it will carry the partial negative charge. Check: (a) Using Figure 8.6: The difference in the electronegativities of chlorine and boron is 3.0 - 2.0 = 1.0; the difference between chlorine and carbon is 3.0 - 2.5 = 0.5. Hence the B-Cl bond is more polar, as we had predicted. (b) Using Figure 8.6: The difference in the electronegativities of chlorine and phosphorus is 3.0 - 2.1 = 0.9; the difference between fluorine and phosphorus is 4.0 - 2.1 = 1.9. Hence the P-F bond is more polar, as we had predicted. - PRACTICE EXERCISE

Which of the following bonds is most polar: S-CI, S-Br, Se-C!, or Se- Br?

Answer: Se-C!

Dipole Moments The difference in electronegativity between H and F leads to a polar covalent bond in the HF molecule. As a consequence, there is a concentration of negative charge on the more electronegative F atom, leaving the less electronegative H atom at the positive end of the molecule. A molecule such as HF, in which the centers of positive and negative charge do not coincide, is said to be a polar molecule. Thus, we describe both bonds and entire molecules as being polar and nonpolar. We can indicate the polarity of the HF molecule in two ways: H

6-

H-F

Q+

Q-

.A. Figure 8.8 Dipole and dipole moment. When charges of equal magnitude and opposite sign Q+ and Q- are separated by a distance r, a dipole is produced. The size of the dipole is given by the dipole moment, J.L, which is the product of the c harge separated and the distance of separation between the charge centers: J.L = Qr.

or

H-F

Recall from the preceding subsection that "ll+ " and "ll-" indicate the partial positive and negative charges on the H and F atoms. In the notation on the right, the arrow denotes the shift in electron density toward the fluorine atom. The crossed end of the arrow can be thought of as a plus sign that designates the positive end of the molecule. Polarity helps determine many of the properties of substances that we ob­ serve at the macroscopic level, in the laboratory and in everyday life. Polar mol­ ecules align themselves with respect to one another, with the negative end of one molecule and the positive end of another attracting each other. Polar mole­ cules are likewise attracted to ions. The negative end of a polar molecule is at­ tracted to a positive ion, and the positive end is attracted to a negative ion. These interactions account for many properties of liquids, solids, and solutions, as you will see in Chapters 11, 12, and 13. How can we quantify the polarity of a molecule? Whenever a distance sep­ arates two electrical charges of equal magnitude but opposite sign, a dipole is established. The quantitative measure of the magnitude of a dipole is called its dipole moment, denoted J.L. If a distance r separates two equal and opposite charges Q+ and Q-, the magnitude of the dipole moment is the product of Q and r (Figure 8.8 .,.. ) : J.L = Qr

[8.11]

8.4

Bond Polarity and Electronegativity

311

The dipole moment increases as the magnitude of charge that is separated increases and as the distance between the charges increases. For a nonpolar molecule, such as Fz, the dipole moment is zero because there is no charge separation. GIVE

IT

S O M E

THOUGHT

The molecules chlorine monofluoride, CIF, and iodine monofluoride, IF, are exam­ ples of interhalogen compounds---

Cl - CH 3{g) + H -Cl{g)

ilH,xn = ?

[8.13]

Our two-step procedure is outlined in Figure 8.14 .A.. We note that in the course of this reaction the following bonds are broken and made:

Bonds broken: 1 mol C- H, 1 mol Cl- Cl Bonds made: 1 mol C- Cl, 1 mol H- Cl We first supply enough energy to break the C- H and Cl-Cl bonds, which will raise the enthalpy of the system. We then form the C -Cl and H -Cl bonds, which will release energy and lower the enthalpy of the system. By using Equation 8.12 and the data in Table 8.4, we estimate the enthalpy of the reaction as il Hrxn = [D(C -H) + D(Cl-Cl) - [D(C- Cl) + D(H-Cl)] = (413 kJ + 242 kJ) - (328 kJ + 431 kJ) = -104 kJ

The reaction is exothermic because the bonds in the products (especially the H -Cl bond) are stronger than the bonds in the reactants (especially the Cl-Cl bond). We usually use bond enthalpies to estimate il Hrxn only if we do not have the needed LlH J values readily at hand. For the above reaction, we cannot calculate ilH,xn from LlHJ values and Hess's law because the value of LlHJ for CH 3Cl{g) is not given in Appendix C. If we obtain the value of LlH J for CH3Cl{g) from another source (such as the CRC Handbook of Chemistry and Physics) and use Equation 5.31, we find that ilH,xn = -99.8 kJ for the reaction in Equation 8.13. Thus, the use of average bond enthalpies provides a reason­ ably accurate estimate of the actual reaction enthalpy change. It is important to remember that bond enthalpies are derived for gaseous molecules and that they are often averaged values. Nonetheless, average bond enthalpies are useful for estimating reaction enthalpies quickly, especially for gas-phase reactions.

328

C HA PTER 8

Basic Concepts of Chemical Bonding

Chemis

Put to Work

EXP L O S I V E S A N D ALFRED N O B E L

E bonds. Perhaps the most graphic illustration of this fact is

normous amounts of energy can be stored in chemical

shown here (resonance structures are not shown for clarity) . TNT contains the six-membered ring characteristic of benzene.

seen in certain molecular substances that are used as explosives. Our discussion of bond enthalpies allows us to examine more closely some of the properties of such explosive substances. An explosive must have the following characteristics: (1) It must decompose very exothermically; (2) the products of its decomposition must be gaseous, so that a tremendous gas pressure accompanies the decomposition; (3) its decom­ position must occur very rapidly; and (4) it must be stable enough so that it can be detonated predictably. The combina­ tion of the first three effects leads to the violent evolution of heat and gases. To give the most exothermic reaction, an explosive should have weak chemical bonds and should decompose into mole­ cules with very strong bonds. Looking at bond enthalpies (Table 8.4), the N==N, C == O, and C = O bonds are among the strongest. Not surprisingly, explosives are usually designed to produce the gaseous products N2{g), CO{g), and C02{g). Water vapor is nearly always produced as well. Many common explosives are organic molecules that contain nitro (N02) or nitrate (N03) groups attached to a carbon skeleton. The Lewis structures of two of the most familiar explosives, nitroglycerin and trinitrotoluene (TNT), are

Nitroglycerin

Trinitrotoluene (TNT)

- SAMPLE EXERCISE

8.12 [ Using Average Bond Enthalpies

Using Table 8.4, estimate 6.H for the following reaction (where we explicitly show the bonds involved in the reactants and products): H

H

I

I

H- - -H{g) +

T T H H

�

� Oz{g)

2 O =C=O(g) + 3 H-0-H(g)

SOLUTION Analyze: We are asked to estimate the enthalpy change for a chemical process by using average bond enthalpies for the bonds that are broken in the reactants and formed in the products. Plan: Among the reactants, we must break six C - H bonds and a C - C bond in C2H6; we also break 02 bonds. Among the products, we form four C = O bonds (two in each COz) and six 0 - H bonds (two in each HzO). Solve: Using Equation 8.12 and data from Table 8.4, we have

�

6.H

= = = =

6D(C -H) + D(C -C) + � 0(02) - 4D(C=O) - 60(0-H) (495 kJ) - (4(799 kJ) + 6(463 kJ))

6(413 kJ) + 348 kJ + 4558 kJ - 5974 kJ -1416 kJ

�

Check: This estimate can be compared with the value of -1428 kj calculated from more accurate thermochemical data; the agreement is good. - PRACTICE EXERCISE Using Table 8.4, estimate

6.H for the reaction

H-N-N-H(g)

I

H

Answer: -86 kJ

I

H

---+

N - N(g) + 2 H-H{g)

8.8

Strengths of Covalent Bonds

329

Nitroglycerin is a pale yellow, oily liquid. It is highly

shock-sensitive: Merely shaking the liquid can cause its explo­ sive decomposition into nitrogen, carbon dioxide, water, and oxygen gases: 4 C:;HsN309(l)

---->

6 N2(g) + 12 C02(g) + 10 HzO(g) + 02(g) The large bond enthalpies of the N2 molecules (941 kJ/mol), C02 molecules (2 X 799 kJ/mol), and water mol­ ecules (2 x 463 k)/mol) make this reaction enormously exothermic. Nitroglycerin is an exceptionally unstable explo­ sive because it is in nearly perfect explosive balance: With the exception of a small amount of 02(g) produced, the only prod­ ucts are N:u C02, and H20. Note also that, unlike combustion reactions (Section 3.2), explosions are entirely self-contained. No other reagent, such as 02(g), is needed for the explosive decomposition. Because nitroglycerin is so unstable, it is difficult to use as a controllable explosive. The Swedish inventor Alfred Nobel (Figure 8.15 �) found that mixing nitroglycerin with an ab­ sorbent solid material such as diatomaceous earth or cellulose gives a solid explosive (dynamite) that is much safer than liq­ uid nitroglycerin.

Related Exercises: 8.93 and 8.94

"' Figure 8. 1 5 Alfred Nobel (1833-1896), the Swedish Inventor of dynamite. By many accounts Nobel's discovery that nitroglycerin could be made more stable by absorbing it onto cellulose was an accident. This discovery made Nobel a very wealthy man. He was also a complex and lonely man, however, who never married, was frequently ill, and suffered from chronic depression. He had invented the most powerful military explosive to date, but he strongly supported international peace movements. His will stated that his fortune be used to establish prizes awarding those who "have conferred the greatest benefit on mankind," including the promotion of peace and "fraternity between nations." The Nobel Prize is probably the most coveted award that a scientist, writer, or peace advocate can receive.

Bond Enthalpy and Bond Length Just as we can define an average bond enthalpy, we can also define an average bond length for a number of common bond types. Some of these are listed in Table 8.5 "'· Of particular interest is the relationship among bond enthalpy, bond length, and the number of bonds between the atoms. For example, we can use data in Tables 8.4 and 8.5 to compare the bond lengths and bond enthalpies of carbon-carbon single, double, and triple bonds:

c-c 1.54 A

C=C 1 .34 A

C == C 1 .20 A

348 k)jmol

614 k)/mol

839 k)/mol

As the number of bonds between the carbon atoms increases, the bond enthalpy increases and the bond length decreases; that is, the carbon atoms are held more closely and more tightly together. In general, as the number of bonds between two

atoms increases, the bond grows shorter and stronger.

TABLE 8.5 • Average Bond Lengths for Some Single, Double, and Triple Bonds Bond

Bond Length (;\)

Bond

Bond Length (;\)

c-c

1.54 1.34 1.20 1.43 1.38 1.16

N-N N=N N == N N-0 N=O

1.47 1.24 1.10 1.36 1.22

0-0 0=0

1.48 1.21

C=C C==C C-N C=N C == N

c-o C=O C == O

1.43 1.23 1.13

330

C HA PTER 8

Basic Concepts of Chemical Bonding

- SAMPLE INTEGRATIVE EXERCISE

I Putting Concepts Together

Phosgene, a substance used in poisonous gas warfare duirng World War I, is so named because it was first prepared by the action of sunlight on a mixture of carbon monoxide and chlorine gases. Its name comes from the Greek words phos (light) and genes (born of). Phosgene has the following elemental composition: 12.14% C, 16.17% 0, and 71.69% Cl by mass. Its molar mass is 98.9 gjmol. (a) Determine the molecular formula of this compound. (b) Draw three Lewis structures for the molecule that satisfy the octet rule for each atom. (The Cl and 0 atoms bond to C.) (c) Using formal charges, determine which Lewis structure is the most important one. (d) Using average bond enthalpies, estimate llH for the formation of gaseous phos­ gene from CO(g) and Cl2(g). SOLUTION (a) The empirical formula of phosgene can be determined from its elemental compo­ sition. exx> (Section 3.5) Assuming 100 g of the compound and calculating the number of moles of C, 0, and Cl in this sample, we have

( ( (

0)

(12.14 g C) (16.17 g

1 mo! C

) ) )=

-

= 1.011 mol C

1 moJ O 16_00 g 0

= 1.011 mol O

12. 01 g C

1 mo! Cl (71.69 g Cl) --35.45 g CI

2.022 mol Cl

The ratio of the number of moles of each element, obtained by dividing each number of moles by the smallest quantity, indicates that there is one C and one 0 for each two Cl in the empirical formula, COCI2• The molar mass of the empirical formula is 12.01 + 16.00 + 2(35.45) = 98.91 g/mol, the same as the molar mass of the molecule. Thus, COCI2 is the molecu­ lar formula.

(b) Carbon has four valence electrons, oxygen has six, and chlorine has seven, giving 4 + 6 + 2(7) = 24 electrons for the Lewis structures. Drawing a Lewis structure with all single bonds does not give the central carbon atom an octet. Using multiple bonds, three structures satisfy the octet rule: ·.a ·. .. 1 .. : o- (g) The lattice energy of K20(s) is 2238 kJ/mol. Use these data along with data in Appendix C and Figure 7.12 to calculate the "second electron affinity" of oxygen, corre­ sponding to the reaction 2 o-(g) + e- - o -(g)

Integrative Exercises 8.100 The reaction of indium, In, with sulfur leads to three bina­ ry compounds, which we will assume to be purely ionic. The three compounds have the following properties: Compound

Mass % In

Melting Point (•C)

A B

87 7 78.2 70.5

653 692 1050

c

(a) Determine the empirical formulas of compounds A, B, and C. (b) Give the oxidation state of In in each of the three compounds. (c) Write the electron configuration for the In ion in each compound. Do any of these config­ urations correspond to a noble-gas configuration? (d) In which compound is the ionic radius of In expected to be smallest? Explain. (e) The melting point of ionic com­ pounds often correlates with the lattice energy. Explain the trends in the melting points of compounds A, B, and C in these terms. [8.101] One scale for electronegativity is based on the concept that the electronegativity of any atom is proportional to the ionization energy of the atom minus its electron affinity: electronegativity = k(IE - EA), where k is a proportionality constant. (a) How does this definition explain why the electronegativity of F is greater than that of Cl even though Cl has the greater electron affini­ ty? (b) Why are both ionization energy and electron affinity relevant to the notion of electronegativity? (c) By using data in Chapter 7, determine the value of k that would lead to an electronegativity of 4.0 for F under this definition. (d) Use your result from part (c) to determine the electronegativities of Cl and 0 using this scale. Do these values follow the trend shown in Figure 8.6? 8.102 The compound chloral hydrate, known in detective sto­ ries as knockout drops, is composed of 14.52% C, 1 .83% H, 64.30% Cl, and 19.35% 0 by mass and has a molar mass of 165.4 g/mol. (a) What is the empirical for­ mula of this substance? (b) What is the molecular for­ mula of this substance? (c) Draw the Lewis structure of the molecule, assuming that the Cl atoms bond to a sin­ gle C atom and that there are a C - C bond and two c-o bonds in the compound. 8.103 Barium azide is 62.04% Ba and 37.96% N. Each azide ion has a net charge of 1-. (a) Determine the chemical for­ mula of the azide ion. (b) Write three resonance struc­ tures for the azide ion. (c) Which structure is most important? (d) Predict the bond lengths in the ion. 8.104 Acetylene (C2H2) and nitrogen (N2) both contain a triple bond, but they differ greatly in their chemical proper­ ties. (a) Write the Lewis structures for the two sub­ stances. (b) By referring to Appendix C, look up the enthalpies of formation of acetylene and nitrogen and compare their reactivities. (c) Write balanced chemical equations for the complete oxidation of N2 to form N205(g) and of acetylene to form C02(g) and H20(g). (d) Calculate the enthalpy of oxidation per mole of N2 and C2H2 (the enthalpy of formation of N205(g) is 11.30 kJ/mol). How do these comparative values relate to your response to part (b)? Both N2 and C2H2 possess triple bonds with quite high bond enthalpies (Table 8.4). What aspect of chemical bonding in these molecules or

339

in the oxidation products seems to account for the dif­ ference in chemical reactivities? [8.105] Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of 69.6% S and 30.4% N. Measurements of its molecular mass yield a value of 184.3 g mol-1 . The compound oc­ casionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond dis­ tances between the atoms in the ring. (Note: The S-S distance in the 58 ring is 2.05A .) (d) The enthalpy of for­ mation of the compound is estimated to be 480 kJ mol-1 . t!.HJ of S(g) is 222.8 kJ mol-1 Estimate the average bond enthalpy in the compound.

[8.106] A common form of elemental phosphorus is the tetrahe­ dral P4 molecule, where all four phosphorus atoms are equivalent:

At room temperature phosphorus is a solid. (a) Do you think there are any unshared pairs of electrons in the P4 molecule? (b) How many P - P bonds are there in the molecule? (c) Can you draw a Lewis structure for a lin­ ear P4 molecule that satisfies the octet rule? (d) Using formal charges, what can you say about the stability of the linear molecule vs. that of the tetrahedral molecule? [8.107] Consider benzene (C6H6) in the gas phase. (a) Write the reaction for breaking all the bonds in C6H6(g), and use data in Appendix C to determine the enthalpy change for this reaction. (b) Write a reaction that corresponds to breaking all the carbon--. Therefore, an AB2 molecule must be either linear (bond angle = 180°) or bent (bond angle # 180°). For ex­ ample, C02 is linear, and 502 is bent. For AB3 molecules, the two most common 000

(c) A Figure 9.1 Tetrahedral geometry. (a) A tetrahedron is an object with four faces and four vertices. Each face is an equilateral triangle. (b) The geometry of the CCI4 molecule. Each C - CI bond in the molecule points toward a vertex of a tetrahedron. All of the C-CI bonds are the same length, and all of the CI-C-CI bond angles are the same. This type of drawing of CCI4 is called a ball-and-stick model. (c) A representation of CCI4, called a space-filling model. It shows the relative sizes of the atoms, but the geometry is somewhat harder to see.

9.1

Molecular Shapes

343

AB2

Bent

Linear

..,. Figure 9.2 Shapes of AB2 and AB3 Trigonal planar

Trig onal pyramidal

T-shaped

molecules. Top: AB2 molecules can be either linear or bent. Bottom: Three possible shapes for AB3 molecules.

shapes place the B atoms at the comers of an equilateral triangle. If the A atom T Figure 9.3 Shapes of ABn lies in the same plane as the B atoms, the shape is called trigonal planar. If the A molecules. For molecules whose atom lies above the plane of the B atoms, the shape is called trigonal pyramidal formula is of the general form ABn. there (a pyramid with an equilateral triangle as its base). For example, S03 is trigonal are five fundamental shapes. planar, and NF3 is trigonal pyramidal. Some AB3 molecules, such as CIF3, exhibit the more unusual T shape shown in Figure 9.2. The shape of any particular ABn molecule can usually be derived from one of the five basic geometric structures shown in Figure 9.3 11>. Starting with a tetrahedron, for example, we can remove atoms successively from the comers as shown in Figure 9.4 T. When an atom is removed Linear Tetrahedral Trigonal planar from one corner of the tetrahedron, the remaining fragment has a trigonal-pyramidal geometry, such as that found for NF3. When two atoms are removed, a bent geometry results. Why do so many ABn molecules have shapes related to the basic structures in Figure 9.3, and can we predict these shapes? When A is a representative element (one of the elements from the s block or p block of the periodic table), we can answer these questions by using the valence-shell electron-pair repulsion (VSEPR) model. Trigonal bipyramidal Octahedral Although the name is rather imposing, the model is quite simple. It has useful predictive capabilities, as we will see in Section 9. 2.

Tetrahedral

Trigonal pyramidal

Bent

..,. Figure 9.4 Derivatives from the ABn geometries. Additional molecular shapes can be obtained by removing corner atoms from the basic shapes shown in Figure 9.3. Here we begin with a tetrahedron and successively remove corners, producing first a trigonal-pyramidal geometry and then a bent geometry, each having ideal bond angles of 1 09.5°. Molecular shape is meaningful only when there are at least three atoms. If there are only two, they must be arranged next to each other, and no special name is given to describe the molecule.

344

CHA PTER 9

Molecular Geometry and Bonding Theories GIVE IT SOME THOUGHT One of the common shapes for AB4 molecules is square planar: All five atoms lie in the same plane, the B atoms lie at the corners of a square, and the A atom is at the center of the square. Which of the shapes in Figure 9.3 could lead to a square-planar geome­ try upon the removal of one or more atoms?

9.2 THE VSEPR M O DEL (a) Two balloons adopt a linear arrangement.

(b) Three balloons adopt a trigonal-planar arrangement.

Imagine tying two identical balloons together at their ends. As shown in Figure 9.5(a) ..,., the balloons naturally orient themselves to point away from each other; that is, they try to "get out of each other's way" as much as possible. If we add a third balloon, the balloons orient themselves toward the vertices of an equilateral triangle as shown in Figure 9.5(b). If we add a fourth balloon, they adopt a tetrahedral shape [Figure 9.5(c)]. We see that an optimum geome­ try exists for each number of balloons. In some ways the electrons in molecules behave like the balloons shown in Figure 9.5. We have seen that a single covalent bond is formed between two atoms when a pair of electrons occupies the space between the atoms. = (Section 8.3) A bonding pair of electrons thus defines a region in which the electrons will most likely be found. We will refer to such a region as an electron domain. Likewise, a nonbonding pair (or lone pair) of electrons defines an electron domain that is located principally on one atom. For example, the Lewis structure of NH3 has four electron domains around the central nitrogen atom (three bonding pairs and one nonbonding pair): /Nonbonding pair

H-N-H

. Bond mg prurs .

;;� \ �

Each multiple bond in a molecule also constitutes a single electron domain. Thus, the following resonance structure for 03 has three electron domains around the central oxygen atom (a single bond, a double bond, and a nonbond­ ing pair of electrons): (c) Four balloons adopt a tetrahedral arrangement.

& Figure 9.5 A balloon analogy for

electron domains. Balloons tied together at their ends naturally adopt their lowest-energy arrangement.

:Q-O=Q general, each nonbonding pair, single bond, or multiple bond produces an electron domain around the central atom. In

GIVE IT SOME THOUGHT An AB3 molecule has the resonance structure

Does this Lewis structure follow the octet rule? How many electron domains are there around the A atom?

The VSEPR model is based on the idea that electron domains are negative­ ly charged and therefore repel one another. Like the balloons in Figure 9.5, elec­ tron domains try to stay out of one another's way. The best arrangement ofa given

number of electron domains is the one that minimizes the repulsions among them.

In fact, the analogy between electron domains and balloons is so close that the same preferred geometries are found in both cases. Like the balloons in

9.2

TABLE 9.1 • Electron-Domain Geometries as a Function of the Number of Electron Domains Number of Electron Domains

Arrangement of Electron Domains

Electron-Domain Predicted Bond Angles Geometry

180° 180°

2

Linear

3

Trigonal planar

4

Tetrahedral

10gs

5

Trigonal bipyramidal

120° goo

6

Octahedral

goo

120°

Figure 9.5, two electron domains are arranged linearly, three domains are arranged in a trigonal-planar fashion, and four are arranged tetrahedrally. These arrangements, together with those for five electron domains (trigonal bipyrami­ dal) and six electron domains (octahedral), are summarized in Table 9.1 £. If you compare the geometries in Table 9.1 with those in Figure 9.3, you will see that they are the same. The shapes of different AB0 molecules or ions depend on the num­ ber of electron domains surrounding the central A atom.

The arrangement of electron domains about the central atom of an AB, molecule or ion is called its electron-domain geometry. In contrast, the molecular geometry is the arrangement of only the atoms in a molecule or ion­ any nonbonding pairs are not part of the description of the molecular geometry. In the VSEPR model, we predict the electron-domain geometry. From knowing how many domains are due to nonbonding pairs, we can then predict the mol­ ecular geometry of a molecule or ion from its electron-domain geometry.

The VSEPR Model

345

346

CHA PTER

9

Molecular Geometry and Bonding Theories

/ 1\

/I / I / I // I

H-N-H Lewis

I

H structure

\

'

\

\

' H�N "-.... , - - - - ·H \ I - - - -"'>,'

;z-

J.r - Electron-domain geometry (tetrahedral) \

Molecular geometry (trigonal pyramidal)

A

Figure 9.6 The molecular geometry of NH3. The geometry is predicted by first drawing the Lewis structure, then using the VSEPR model to determine the electron-domain geometry, and finally focusing on the atoms themselves to describe the molecular geometry.

When all the electron domains in a molecule arise from bonds, the molecu­ lar geometry is identical to the electron-domain geometry. When, however, one or more of the domains involve nonbonding pairs of electrons, we must re­ member to ignore those domains when predicting molecular shape. Consider the NH3 molecule, which has four electron domains around the nitrogen atom (Figure 9.6 £). We know from Table 9.1 that the repulsions among four electron domains are minimized when the domains point toward the vertices of a tetra­ hedron, so the electron-domain geometry of NH3 is tetrahedral. We know from the Lewis structure of NH3 that one of the electron domains is due to a non­ bonding pair of electrons, which will occupy one of the four vertices of the tetrahedron. Hence the molecular geometry of NH3 is trigonal pyramidal, as shown in Figure 9.6. Notice that the tetrahedral arrangement of the four elec­ tron domains leads us to predict the trigonal-pyramidal molecular geometry. We can generalize the steps we follow in using the VSEPR model to predict the shapes of molecules or ions: 1.

2. 3.

Draw the Lewis structure of the molecule or ion, and count the total number of electron domains around the central atom. Each nonbonding electron pair, each single bond, each double bond, and each triple bond counts as an electron domain. Determine the electron-domain geometry by arranging the electron domains about the central atom so that the repulsions among them are minimized, as shown in Table 9.1. Use the arrangement of the bonded atoms to determine the molecular geometry.

Figure 9.6 shows how these steps are applied to predict the geometry of the NH3 molecule. Because the trigonal-pyramidal molecular geometry is based on tetrahedral electron-domain geometry, the ideal bond angles are 109.5°. As we will soon see, bond angles deviate from the ideal angles when the surrounding atoms and electron domains are not identical. Let's apply these steps to determine the shape of the C02 molecule. We first draw its Lewis structure, which reveals two electron domains (two double bonds) around the central carbon:

Q=C=Q Two electron domains will arrange themselves to give a linear electron­ domain geometry (Table 9.1). Because neither domain is a nonbonding pair of electrons, the molecular geometry is also linear, and the 0-C-0 bond angle is 180°. Table 9.2 � summarizes the possible molecular geometries when an AB" molecule has four or fewer electron domains about A. These geometries are important because they include all the commonly occurring shapes found for molecules or ions that obey the octet rule.

9.2

The VSEPR Model

347

TABLE 9.2 • Electron-Domain Geometries and Molecular Shapes for Molecules with Two, Three, and Four Electron Domains around the Central Atom Number of Electron Domains

ElectronDomain Geometry

2

Bonding Domains

Nonbonding Domains

2

0

Molecular Geometry

Linear

3

Linear

3

Q=C=Q :ii:

0

:.� :"

Trigonal planar

I

B

':t

Trigonal planar

2

1

Bent

4

4

Example

H I

0

Tetrahedral

[·t�N�:r H/J< H H

Tetrahedral

3

N

1

H/J H "H Trigonal pyramidal

2

2

Bent

0.

" H/J H

- SAMPLE EXERCISE 9.1 J Using the VSEPR Model Use the VSEPR model to predict the molecular geometry of (a) 03, (b) SnC13-. SOLUTION Analyze: We are given the molecular formulas of a molecule and a polyatomic ion, both conforming to the general formula AB, and both having a central atom from the p block of the periodic table. Plan: To predict the molecular geometries of these species, we first draw their Lewis structures and then count the number of electron domains around the central atom. The number of electron domains gives the electron-domain geometry. We then obtain the molecular geometry from the arrangement of the domains that are due to bonds.

348

C HA PTER

9

Molecular Geometry and Bonding Theories

Solve:

:Q-i:i=Q ------. Q=O-Q:

(a) We can draw two resonance structures for 03: Because of resonance, the bonds between the central 0 atom and the outer 0 atoms are of equal length. In both resonance structures the central 0 atom is bonded to the two outer 0 atoms and has one nonbonding pair. Thus, there are three electron domains about the central 0 atoms. (Remember that a double bond counts as a single electron domain.) The arrange­ ment of three electron domains is trigonal planar (Table 9.1). Two of the domains are from bonds, and one is due to a non­ bonding pair. So, the molecule has a bent shape with an ideal bond angle of 120° (Table 9.2).

o

/0:,._

""'o

As this example illustrates, when a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular geometry.

[:C:i-s�-ci:]..

(b) The Lewis structure for the SnC13- ion is

The central Sn atom is bonded to the three Cl atoms and has one nonbonding pair. Therefore, the Sn atom has four electron domains around it. The resulting electron-domain geometry is tetrahedral (Table 9.1) with one of the corners occupied by a nonbonding pair of electrons. The molecular geometry is therefore trigonal pyramidal (Table 9.2), like that of NH3.

..

I : Figure 9.45 The effect of 2J-2p Interactions. When the 2s and 2p

Increasing 2s-2p interaction

orbitals interact, the "ls MO falls in energy and the "lp MO rises in energy. For 02, F2, and Ne2, the interaction is small, and the "lp MO remains below the "lp MOs, as in Figure 9.43. For B2, C2, and N2, the 2s-2p interaction is great enough that the

4 S2 (g)

(a) With respect to electronic structure, which element in the second row of the peri­ odic table is most similar to sulfur? (b) Use the VSEPR model to predict the S-S-S bond angles in S8 and the hybridi2ation at S in S8. (c) Use MO theory to predict the sulfur-sulfur bond order in S2. Is the molecule expected to be diamagnetic or para­ magnetic? (d) Use average bond enthalpies (Table 8.4) to estimate the enthalpy change for the reaction just described. Is the reaction exothermic or endothermic? SOLUTION 2 (a) Sulfur is a group 6A element with an [Ne)3s 3p4 electron configuration. It is ex­ pected to be most similar electronically to oxygen (electron configuration, 2 [He)2s 2p4), which is immediately above it in the periodic table. (b) The Lewis struc­ ture of Ss is

=s-s=

. ./ :s I :s

. ."

"'-·· s: I s: /•

:�-�:

There is a single bond between each pair of S atoms and two nonbonding elec­ tron pairs on each S atom. Thus, we see four electron domains around each S atom, and we would expect a tetrahedral electron-domain geometry corresponding to sp3 hybridization. = (Sections 9.2, 9.5) Because of the nonbonding pairs, we would expect the S-S-S angles to be somewhat less than 109°, the tetrahedral angle. Experimentally, the S-S-S angle in S8 is 108°, in good agreement with this predic­ tion. Interestingly, if S8 were a planar ring (like a stop sign), it would have S-S-S angles of 135°. Instead, the S8 ring puckers to accommodate the smaller angles dictat­ ed by sp3 hybridization. (c) The MOs of 52 are entirely analogous to those of 02, al­ though the MOs for 52 are constructed from the 3s and 3p atomic orbitals of sulfur. Further, 52 has the same number of valence electrons as 02. Thus, by analogy to our discussion of 02, we would expect 52 to have a bond order of 2 (a double bond) and to be paramagnetic with two unpaired electrons in the 7T3p molecular orbitals of 52. = (Section 9 .8) (d) We are considering the reaction in which an 58 molecule falls apart into four 52 molecules. From parts (b) and (c), we see that S8 has S-S single bonds and 52 has S = S double bonds. During the course of the reaction, therefore, we are breaking eight S-S single bonds and forming four S=S double bonds. We can estimate the enthalpy of the reaction by using Equation 8.12 and the average bond enthaipies in Table 8.4:

tlH,xn

= 8 D(S -S)

-

4 D(S =S) = 8(266 kJ)

-

4(418 kJ) = +456 kJ

Because tlH,xn > 0, the reaction is endothermic. = (Section 5.4) The very positive value of tlH,xn suggests that high temperatures are required to cause the reaction to occur.

CH APT ER REVI E W SUMMARY AND KEY TERM S

The three-dimensional shapes and sizes of molecules are determined by their bond angles and bond lengths. Molecules with a central atom A surrounded by n atoms B, denoted ABn, adopt a number of different geometric shapes, depending on the value of n and on the particular atoms involved. In the Introduction and Section 9.1

overwhelming majority of cases, these geometries are re­ lated to five basic shapes (linear, trigonal pyramidal, tetrahedral, trigonal bipyramidal, and octahedral). Section 9.2 The valence-shell electron-pair repulsion (VSEPR) model rationalizes molecular geometries based

382

C HA PTER 9

Molecular Geometry and Bonding Theories

on the repulsions between electron domains, which are regions about a central atom in which electrons are likely to be found. Bonding pairs of electrons, which are those involved in making bonds, and nonbonding pairs of electrons, also called lone pairs, both create electron do­ mains around an atom. According to the VSEPR model, electron domains orient themselves to minimize electro­ static repulsions; that is, they remain as far apart as possi­ ble. Electron domains from nonbonding pairs exert slightly greater repulsions than those from bonding pairs, which leads to certain preferred positions for nonbond­ ing pairs and to the departure of bond angles from ideal­ ized values. Electron domains from multiple bonds exert slightly greater repulsions than those from single bonds. The arrangement of electron domains around a central atom is called the electron-domain geometry; the arrangement of atoms is called the molecular geometry. Section 9.3 The dipole moment of a polyatomic mole­ cule depends on the vector sum of the dipole moments associated with the individual bonds, called the bond dipoles. Certain molecular shapes, such as linear AB2 and trigonal planar AB3, assure that the bond dipoles cancel, producing a nonpolar molecule, which is one whose dipole moment is zero. In other shapes, such as bent AB2 and trigonal pyramidal AB 3, the bond dipoles do not cancel and the molecule will be polar (that is, it will have a nonzero dipole moment). Section 9.4 Valence-bond theory is an extension of Lewis's notion of electron-pair bonds. In valence-bond the­ ory, covalent bonds are formed when atomic orbitals on neighboring atoms overlap one another. The overlap re­ gion is a favorable one for the two electrons because of their attraction to two nuclei. The greater the overlap between two orbitals, the stronger will be the bond that is formed. Section 9.5 To extend the ideas of valence-bond theory to polyatomic molecules, we must envision mixing s, p, and sometimes d orbitals to form hybrid orbitals. The process of hybridization leads to hybrid atomic orbitals that have a large lobe directed to overlap with orbitals on another atom to make a bond. Hybrid orbitals can also accommodate nonbonding pairs. A particular mode of hybridization can be associated with each of the five common electron­ 2 domain geometries (linear = sp; trigonal planar = sp ; 3 3 tetrahedral = sp ; trigonal bipyramidal = sp d; and 2 octahedral = sp 3d ). Section 9.6 Covalent bonds in which the electron den­ sity lies along the line connecting the atoms (the internu­ clear axis) are called sigma (u) bonds. Bonds can also be formed from the sideways overlap of p orbitals. Such a bond is called a pi ( TT) bond. A double bond, such as that in C2H4, consists of one u bond and one 7T bond; a triple bond, such as that in C2H2, consists of one u and two 7T bonds. The formation of a 7T bond requires that molecules adopt a specific orientation; the two CH2 groups in C2H4, for example, must lie in the same plane. As a result, the presence of 7T bonds introduces rigidity into molecules.

In molecules that have multiple bonds and more than one resonance structure, such as C6H6, the 7T bonds are delocalized; that is, the 7T bonds are spread among sever­ al atoms. Section 9.7 Molecular orbital theory is another model used to describe the bonding in molecules. In this model the electrons exist in allowed energy states called molecular orbitals (MOs). These orbitals can be spread among all the atoms of a molecule. Like an atomic orbital, a molecular orbital has a definite energy and can hold two electrons of opposite spin. The combination of two atomic orbitals leads to the formation of two MOs, one at lower energy, and one at higher energy relative to the en­ ergy of the atomic orbitals. The lower-energy MO concen­ trates charge density in the region between the nuclei and is called a bonding molecular orbital. The higher-energy MO excludes electrons from the region between the nu­ clei and is called an antibonding molecular orbital. Occupation of bonding MOs favors bond formation, whereas occupation of antibonding MOs is unfavorable. The bonding and antibonding MOs formed by the combi­ nation of s orbitals are sigma (u) molecular orbitals; like u bonds, they lie on the internuclear axis. The combination of atomic orbitals and the relative energies of the molecular orbitals are shown by an energy-level (or molecular orbital) diagram. When the appropriate number of electrons are put into the MOs, we can calculate the bond order of a bond, which is half the difference between the number of electrons in bond­ ing MOs and the number of electrons in antibonding MOs. A bond order of 1 corresponds to a single bond, and so forth. Bond orders can be fractional numbers. Section 9.8 Electrons in core orbitals do not contribute to the bonding between atoms, so a molecular orbital de­ scription usually needs to consider only electrons in the outermost electron subshells. In order to describe the MOs of second-row homonuclear diatomic molecules, we need to consider the MOs that can form by the combi­ nation of p orbitals. The p orbitals that point directly at one another can form u bonding and u* antibonding MOs. The p orbitals that are oriented perpendicular to the internuclear axis combine to form pi (TT) molecular orbitals. In diatomic molecules the 7T molecular orbitals occur as pairs of degenerate (same energy) bonding and antibonding MOs. The u2p bonding MO is expected to be lower in energy than the 7T2p bonding MOs because of larger orbital overlap. This ordering is reversed in B2, C:u and N2 because of interaction between the 2s and 2p atomic orbitals of different atoms. The molecular orbital description of second-row di­ atomic molecules leads to bond orders in accord with the Lewis structures of these molecules. Further, the model predicts correctly that 02 should exhibit paramagnetism, an attraction of a molecule by a magnetic field due to un­ paired electrons. Those molecules in which all the elec­ trons are paired exhibit diamagnetism, a weak repulsion from a magnetic field.

Visualizing Concepts

383

KEY SKILLS • • • • • • •

Be able to describe the three-dimensional shapes of molecules using the VSEPR model. Determine whether a molecule is polar or nonpolar based on its geometry and the individual bond dipole moments. Be able to identify the hybridization state of atoms in molecules. Be able to sketch how orbitals overlap to form sigma (u) and pi (1T) bonds. Be able to explain the concept of bonding and antibonding orbitals. Be able to draw molecular orbital energy-level diagrams and place electrons into them to obtain the bond orders and electron configurations of diatomic molecules using molecular orbital theory. Understand the relationships among bond order, bond strength, and bond length.

KEY EQUATIONS •

Bond order = t (no. of bonding electrons - no. of antibonding electrons)

VISUALIZING CONCEPTS 9.1 A certain AB4 molecule has a "seesaw" shape:

9.5 The plot below shows the potential energy of two Cl atoms as a function of the distance between them.

(a) To

what does an energy of zero correspond in this dia­

gram? (b) According to the valence-bond model, why does the energy decrease as the Cl atoms move from a

large separation to a smaller one? (c) What is the signifi­ cance of the Cl-Cl distance at the minimum point in the

plot? (d) Why does the energy rise at Cl-Cl distances less than that at the minimum point in the plot? (e) How

From which of the fundamental geometries shown in Figure 9.3 could you remove one or more atoms to cre­

can you estimate the bond strength of the Cl-Cl bond

from the plot? [Section 9.4]

ate a molecule having this seesaw shape? [Section 9.1]

9.2

(a) If the three balloons shown on the right are all the same size, what angle is formed be­

tween the red one and the

green one? (b) If additional air is added to the blue balloon so that it gets larger, what hap­

pens to the angle between the red and green balloons?

(c) What aspect of the VSEPR

model is illustrated by part {b)? [Section 9.2]

9.3

An

AB5 molecule adopts the geometry shown below.

(a) What is the name of this geometry? (b) Do you think

there are any nonbonding elec­

tron pairs on atom A? Why or why not?

(c) Suppose the atoms

B are halogen atoms. Can you determine tmiquely to which

9.6 Shown below are three pairs of hybrid orbitals,

with each set at a characteristic angle. For each pair,

group in the periodic table atom

determine the type or types of hybridization that

A belongs? [Section 9.2]

9.4 The molecule shown here is

difluoromethane

(CH2F2),

- . (a) Based on the

which is used as a refrigerant called R 32

structure, how many electron domains surround the C atom in this molecule? (b) Would the molecule have a nonzero dipole moment? (c)

If

the molecule is polar, in what direction will the overall dipole

moment vector point in the

molecule? [Sections 9.2 and 9.3]

Cl-Cl distance ------+-

could lead to hybrid orbitals at the specified angle. [Section 9.5]

384

CHA PTER 9

Molecular Geometry and Bonding Theories

9.7 The orbital diagram below presents the final step in the formation of hybrid orbitals by a silicon atom. What type of hybrid orbitals is produced in this hybridiza­ tion? [Section 9.5]

ITJ I 3s

(b)

(a)

3p

9.8 Consider the hydrocarbon drawn below. (a) What is the hybridization at each carbon atom in the molecule? (b) How many u bonds are there in the molecule? (c) How many 7T bonds? [Section 9.6]

H H H I

I

I

H I

H-C=C-c-c -c-C-H I

H

I

H

9.9 For each of the following contour representations of molecular orbitals, identify (i) the atomic orbitals (s or p) used to construct the MO, (ii) the type of MO (u or 7r), and (iii) whether the MO is bonding or antibonding. [Sections 9.7 and 9.8]

(c) 9.10 The diagram below shows the highest occupied MOs of a neutral molecule CX, where element X is in the same row of the periodic table as C. (a) Based on the number of electrons, can you determine the identity of X? (b) Would the molecule be diamagnetic or paramagnet­ ic? (c) Consider the 7Tzp MOs of the molecule. Would you expect them to have a greater atomic orbital contribu­ tion from C, have a greater atomic orbital contribution from X, or be an equal mixture of atomic orbitals from the two atoms? [Section 9.8]

EXERCISES Molecular Shapes; the VSEPR Model 9.11 An AB2 molecule is described as linear, and the A- B bond length is known. (a) Does this information com­ pletely describe the geometry of the molecule? (b) Can you tell how many nonbonding pairs of electrons are around the A atom from this information? 9.12 (a) Methane (CH4) and the perchlorate ion (Cl04-) are both described as tetrahedral. What does this indicate about their bond angles? (b) The NH3 molecule is trigo­ nal pyrantidal, while BF3 is trigonal planar. Which of these molecules is flat? 9.13 (a) What is meant by the term electron domain? (b) Ex­ plain in what way electron domains behave like the bal­ loons in Figure 9.5. Why do they do so? 9.14 (a) How does one determine the number of electron domains in a molecule or ion? (b) What is the difference between a bonding electron domain and a nonbonding

electron domain?

9.15 How many nonbonding electron pairs are there in each of the following molecules: (a) (CH3)zS; (b) HCN; (c) H2Cz; (d) CH3F? 9.16 Describe the characteristic electron-domain geometry of each of the following numbers of electron domains about a central atom: (a) 3, (b) 4, (c) 5, (d) 6. 9.17 What is the difference between the electron-domain geometry and the molecular geometry of a molecule? Use the water molecule as an example in your discussion. 9.18 An AB3 molecule is described as having a trigonal­ bipyramidal electron-domain geometry. How many nonbonding domains are on atom A? Explain.

9.19 Give the electron-domain and molecular geometries of a molecule that has the following electron domains on its central atom: (a) four bonding domains and no non­ bonding domains, (b) three bonding domains and two nonbonding domains, (c) five bonding domains and one nonbonding domain, (e) four bonding domains and two nonbonding domains. 9.20 What are the electron-domain and molecular geome­ tries of a molecule that has the following electron do­ mains on its central atom? (a) three bonding domains and no nonbonding domains, (b) three bonding do­ mains and one nonbonding domain, (c) two bonding domains and two nonbonding domains. 9.21 Give the electron-domain and molecular geometries for 2 the following molecules and ions: (a) HCN, (b) S03 -, (c) SF4, (d) PF6-, (e) NH3Cl+, (f) N3 -. 9.22 Draw the Lewis structure for each of the following mol­ ecules or ions, and predict their electron-domain and molecular geometries: (a) PF3, (b) CH3 +, (c) BrF3, (d) Cl04- (e) XeFz, (f) BrOz-. 9.23 The figure that follows shows ball-and-stick drawings of three possible shapes of an AF3 molecule. (a) For each shape, give the electron-domain geometry on which the molecular geometry is based. (b) For each shape, how many nonbonding electron domains are there on atom A? (c) Which of the following elements will lead to an AF3 molecule with the shape in (ii): Li, B, N, AI, P, Cl? (d) Name an element A that is expected to lead to the AF3 structure shown in (iii) . Explain your reasoning.

Exercises

385

9.26 Give approximate values for the indicated bond angles in the foUowing molecules: I

(iii)

(ii)

(i)

9.24 The figure that follows contains ball-and-stick drawings of three possible shapes of an AF4 molecule.

(a) For each

shape, give the electron-domain geometry on which the

n .. .. (a) H-0-N=O .. .. V 2 (b)

4 H 3 . ;- l H - C - C =O

H

many nonbonding electron domains are there on atom

(c) Which of the foUowing elements will lead to an

AF4 molecule with the shape in (iii): Be, C, S, Se, Si, Xe?

(d) Name an element A that is expected to lead to the

AF4 structure shown in (i).

6

7

H

./). . .. I I

molecular geometry is based. (b) For each shape, how

A?

H

s;- 1 ./.;'\

(c) H-t'!"-Q-H

(d) H-

1 n

r

.{

C=N:

H

H

9.27 Predict the trend in the F(axial) - A -F(equatorial) bond angle in the following AFn molecules: PF5, SF4, and ClF3.

9.28 The three species NH2- NH3, and NH4 + have H -N -H bond angles of 105", 107", and 109", respec­ tively. Explain this variation in bond angles.

9.29 (a) Explain why BrF4- is square planar, whereas BF4- is tetrahedral. (b) Water, H20, is a bent molecule. Predict the shape of the molecular ion formed from the water

(ii)

(i)

molecule if you were able to remove four electrons 4 to make (H20) +.

(iii)

9.25 Give the approximate values for the indicated bond an­ gles in the following molecules: I

n .. .. (a) H-0-Cl-0: .. � .. I :o:

H

(b)

angles: C!Oz- and NOz-. Predict the bond angle in each case.

(b) Explain why the XeF2 molecule is linear and

not bent.

5

n

(c) H-C - C-H

4

I n H-C -0-H .. �I

9.30 (a) Explain why the following ions have different bond

(d)

:o: H 6t> ll . H-C -0-C-H

H

.

1 -.s

V I 7

H

Polarity of Polyatomic Molecules 9.31 (a) Does SCI2 have a dipole moment? If so, in which di­ rection does the net dipole point? (b) Does BeC12 have a dipole moment? If so, in which direction does the net

9.37 Dichloroethylene (C2H2Cl2) has three forms (isomers), each of which is a different substance. (a) Draw Lewis

9.32 (a) The PH3 molecule is polar. How does this offer experi­

carbon carbon double bond. (b) Which of these isomers has a zero dipole moment? (c) How many isomeric

dipole point?

mental proof that the molecule cannot be planar?

(b)

It

turns out that ozone, 03, has a small dipole moment. How is this possible, given that all the atoms are the same?

9.33 (a) Consider the AF3 molecules in Exercise 9.23. Which of these will have a nonzero dipole moment? Explain.

structures of the three isomers, all of which have a

forms can chloroethylene, C2H3Cl, have? Would they be expected to have dipole moments?

9.38 Dichlorobenzene, C�4Ciz, exists in three forms (isomers), called

ortlw, meta, and para:

(b) Which of the AF4 molecules in Exercise 9.24 will

Jv � u U

have a zero dipole moment?

9.34 (a) What conditions must be met if a molecule with polar bonds is nonpolar? (b) What geometries will give nonpo­

cl

lar molecules for ABz, AB3, and AB4 geometries?

9.35 Predict whether each of the following molecules is polar or nonpolar: (a) IF, (b) CS2, (c) 503, (d) PCJ3, (e) SF , 6 (f) IFs .

9.36 Predict whether each of the following molecules is polar or nonpolar: (a) CC4, (b) NH3, (c) SF4, (d) XeF I II 3 c c H/ �c/ "'-� . . c- -H I II I 1

electrons in ethyl acetate? (c) How many of the valence

electrons are used to make

(c) What is the total number of u

bonds in the entire molecule, and what is the total num­

(b) What are the hy­

bridizations of the orbitals on the two oxygens and

the nitrogen atom, and what are the approximate bond

molecule or ion will exhibit delocalized 7r bonding? (c) Is the 7r bond in N02- localized or delocalized?

9.58 (a) Write a single Lewis structure for SO:v and determine the hybridization at the S atom. (b) Are there other equiv­ alent Lewis structures for the molecule? (c) Would you expect S03 to exhibit delocalized 7r bonding? Explain.

Molecular Orbitals 9.59 (a) What is the difference between hybrid orbitals and molecular orbitals? (b) How many electrons can be

placed into each MO of a molecule? (c) Can antibonding molecular orbitals have electrons in them?

9.60 (a) If you combine two atomic orbitals on two different atoms to make a new orbital, is this a hybrid orbital or a

molecular orbital?

bitals on

(b) If you combine two atomic or­

one atom to make a new orbital, is this a hybrid

orbital or a molecular orbital? (c) Does the Pauli exclu­ sion principle (Section 6.7) apply to MOs? Explain.

9.61 Consider the H2+ ion. (a) Sketch the molecular orbitals of the ion, and draw its energy-level diagram. (b) How many electrons are there in the H z+ ion? (c) Draw the elec­ tron configuration of the ion in terms of its MOs.

(d) What is the bond order in H2 +7 (e) Suppose that the ion is excit­

ed by light so that an electron moves from a lower-energy to a higher-energy MO. Would you expect the excited­

state H2 + ion to be stable or to fall apart? Explain.

9.62 (a) Sketch the molecular orbitals of the H2- ion, and draw its energy-level diagram. (b) Write the electron

configuration of the ion in terms of its MOs. (c) Calculate

the bond order in H2-

(d) Suppose that the ion is excited

by light, so that an electron moves from a lower-energy to a higher-energy molecular orbital. Would you expect the excited-state Hz- ion to be stable? Explain.

9.63 Draw a picture that shows all three atom and all three

2p orbitals on one

2p orbitals on another atom. (a) Imag­

ine the atoms coming close together to bond. How many

u bonds can the two sets of 2p orbitals make with

each

other? (b) How many 7r bonds can the two sets of

2p

orbitals make with each other? (c) How many antibond­

ing orbitals, and of what type, can be made from the two sets of

2p orbitals?

9.64 (a) What is the probability of finding an electron on the internuclear axis if the electron occupies a 7r molecular orbital?

(b) For a homonuclear diatomic molecule, what 7T2p

similarities and differences are there between the

2px atomic orbitals and the Trzp MO 2py atomic orbitals? (c) Why are the Tr2p MOs lower in energy than the Tr�p MOs? MO made from the made from the

9.65 (a) What are the relationships among bond order, bond length, and bond energy? (b) According to molecular or­

bital theory, would either Be2 or Be2+ be expected to

exist? Explain.

peroxide ion, o,Z-, has a su.peroxide ion, 02-. (b) The

9.66 Explain the following: (a) The longer bond length than the

magnetic properties of B2 are consistent with the Trzp MOs 2 uzp MO. (c) The 02 +

being lower in energy than the

ion has a stronger 0---0 bond than 02 itself.

9.67 (a) What does the term

diamagnetism

mean? (b) How

does a diamagnetic substance respond to a magnetic field?

(c) Which of the following ions would you expect 2 to be diamagnetic: N,Z-, 02 -, Be/+, c,-?

9.68 (a) What does the term

paramagnetism

mean?

(b) How

can one determine experimentally whether a substance is paramagnetic?

(c) Which of the following ions would

you expect to be paramagnetic:

o,+, N,

2 -, Liz+,

o,Z-?

For those ions that are paramagnetic, determine the number of unpaired electrons.

388

C HA PTER 9

Molecular Geometry and Bonding Theories

9.69

Using Figures 9.37 and 9.45 as guides, draw the molecu­ lar orbital electron configuration for (a) B2+, (b) Li2+, (c) N2 +, (d) Nel+ In each case indicate whether the ad­ dition of an electron to the ion would increase or de­ crease the bond order of the species.

9.70

If we assume that the energy-level diagrams for homo­ nuclear diatomic molecules shown in Figure 9.42 can be applied to heteronuclear diatomic molecules and ions, predict the bond order and magnetic behavior of (a) co+, (b) NO-, (c) OF+, (d) NeF+

9.71

Determine the electron configurations for CN+, CN, and CN-. (a) Which species has the strongest C-N bond? (b) Which species, if any, has unpaired electrons?

9.72

(a) The nitric oxide molecule, NO, readily loses one elec­ tron to form the NO+ ion. Why is this consistent with the electronic structure of NO? (b) Predict the order of the N - 0 bond strengths in NO, NO+, and NO-, and describe the magnetic properties of each. (c) With what neutral homonuclear diatomic molecules are the NO+ and NO- ions isoelectronic (same number of electrons)?

[9.73]

Consider the molecular orbitals of the P2 molecule. Assume that the MOs of diatomics from the third row of the periodic table are analogous to those from the sec­ ond row. (a) Which valence atomic orbitals of P are used

to construct the MOs of P2? (b) The figure below shows a sketch of one of the MOs for P2. What is the label for this MO? (c) For the P2 molecule, how many electrons occupy the MO in the figure? (d) Is P2 expected to be diamagnetic or paramagnetic? Explain.

[9.74]

The iodine bromide molecule, !Br, is an interhalogen com­ pound. Assume that the molecular orbitals of !Br are analogous to the homonuclear diatomic molecule F2. (a) Which valence atomic orbitals of I and of Br are used to construct the MOs of !Br? (b) What is the bond order of the !Br molecule? (c) One of the valence MOs of !Br is sketched below. Why are the atomic orbital contribu­ tions to this MO different in size? (d) What is the label for the MO? (e) For the IBr molecule, how many elec­ trons occupy the MO?

ADDITI ONAL EXERC ISES

9.75

(a) What is the physical basis for the VSEPR model? (b) When applying the VSEPR model, we count a dou­ ble or triple bond as a single electron domain. Why is this justified?

9.76

The molecules SiF4, SF4, and XeF4 have molecular for­ mulas of the type AF4, but the molecules have different molecular geometries. Predict the shape of each mole­ cule, and explain why the shapes differ.

[9.77]

The vertices of a tetrahedron correspond to four alter­ nating corners of a cube. By using analytical geometry, demonstrate that the angle made by connecting two of the vertices to a point at the center of the cube is 109.5°, the characteristic angle for tetrahedral molecules.

9.78

Consider the molecule PF4CJ. (a) Draw a Lewis structure for the molecule, and predict its electron-domain geome­ try. (b) Which would you expect to take up more space, a P-F bond or a P-C! bond? Explain. (c) Predict the molecular geometry of PF 4CI. How did your answer for part (b) influence your answer here in part (c)? (d) Would you expect the molecule to distort from its ideal electron-domain geometry? If so, how would it distort?

9.79

From their Lewis structures, determine the number of a and 7T bonds in each of the following molecules or ions: (a) CO� (b) thiocyanate ion, NCS-; (c) formaldehyde, H2CO; (d) formic acid, HCOOH, which has one H and two 0 atoms attached to C.

9.80

The lactic acid molecule, CH3CH(OH)COOH, gives sour milk its unpleasant, sour taste. (a) Draw the Lewis structure for the molecule, assuming that carbon always forms four bonds in its stable compounds. (b) How many 7T and how many a bonds are in the molecule? (c) Which CO bond is shortest in the molecule? (d) What

is the hybridization of atomic orbitals around each car­ bon atom associated with that short bond? (e) What are the approximate bond angles around each carbon atom in the molecule?

9.81

The PF3 molecule has a dipole moment of 1.03 D, but BF3 has a dipole moment of zero. How can you explain the difference?

9.82

There are two compounds of the formula Pt(NH3hCli NH3

I

Cl-Pt-CI

I

NH3

Cl

I

Cl-Pt-NH3

I

NH3

The compound on the right, cisplatin, is used in cancer therapy. The compound on the left, transplatin, is inef­ fective for cancer therapy. Both compounds have a square-planar geometry. (a) Which compound has a nonzero dipole moment? (b) The reason cisplatin is a good anticancer drug is that it binds tightly to DNA. Cancer cells are rapidly dividing, producing a lot of DNA. Consequently cisplatin kills cancer cells at a faster rate than normal cells. However, since normal cells also are making DNA, cisplatin also attacks healthy cells, which leads to unpleasant side effects. The way both molecules bind to DNA involves the Cl- ions leaving the Pt ion, to be replaced by two nitrogens in DNA. Draw a picture in which a long vertical line represents a piece of DNA. Draw the Pt(NH3h fragments of cisplatin and transplatin with the proper shape. Also draw them attaching to your DNA line. Can you explain from your drawing why the shape of the cisplatin causes it to bind to DNA more effectively than transplatin?

Additional Exercises [9.83] The 0-H bond lengths in the water molecule (H20) are 0. 6 and the H - 0 - H angle is 104.5°. The di­ pole moment of the water molecule is 1.85 D. (a) In what directions do the bond dipoles of the 0 - H bonds point? In what direction does the dipole moment vector of the water molecule point? (b) Calculate the magni­ tude of the bond dipole of the 0- H bonds. (Note: You will need to use vector addition to do this.) (c) Compare your answer from part (b) to the dipole moments of the hydrogen halides (Table 8.3). Is your answer in accord with the relative electronegativity of oxygen?

9 A,

389

(a) Predict the bond angles around each of the carbon atoms, and sketch the molecule. (b) Compare the bond lengths to the average bond lengths listed in Table 8.5. Can you explain any differences? [9.89] The sketches below show the atomic orbital wave func­ tions (with phases) used to construct some of the MOs of a homonuclear diatomic molecule. For each sketch, de­ termine the MO that will result from mixing the atomic orbital wave functions as drawn. Use the same labels for the MOs as in the "Closer Look" box on phases.

[9.84] The reaction of three molecules of fluorine gas with a Xe atom produces the substance xenon hexafluoride, XeF6: Xe(g) + 3 F2(g)

--->

XeF6(s)

(a) Draw a Lewis structure for XeF6. (b) If you try to use

the VSEPR model to predict the molecular geometry of XeF6, you run into a problem. What is it? (c) What could you do to resolve the difficulty in part (b)? (d) Suggest a hybridization scheme for the Xe atom in XeF6. (e) The molecule IF7 has a pentagonal-bipyramidal structure (five equatorial fluorine atoms at the vertices of a regu­ lar pentagon and two axial fluorine atoms). Based on the structure of IF7, suggest a structure for XeF6.

(a)

(b)

[9.85] The Lewis structure for allene is

Make a sketch of the structure of this molecule that is analogous to Figure addition, answer the follow­ ing three questions: (a) Is the molecule planar? (b) Does it have a nonzero dipole moment? (c) Would the bonding in allene be described as delocalized? Explain.

9.27. In

[9.86] The azide ion, N3 -, is linear with two N -N bonds of equal length, 1.16 A. (a) Draw a Lewis structure for the azide ion. (b) With reference to Table 8.5, is the observed N -N bond length consistent with your Lewis struc­ ture? (c) What hybridization scheme would you expect at each of the nitrogen atoms in N3 -? (d) Show which hybridized and unhybridized orbitals are involved in the formation of a and rr bonds in N3 -. (e) It is often ob­ served that a bonds that involve an sp hybrid orbital are 3 shorter than those that involve only sp 2 or sp hybrid or­ bitals. Can you propose a reason for this? Is this obser­ vation applicable to the observed bond lengths in N3-? [9.87]

In ozone, 03, the two oxygen atoms on the ends of the molecule are equivalent to one another. (a) What is the best choice of hybridization scheme for the atoms of ozone? (b) For one of the resonance forms of ozone, which of the orbitals are used to make bonds and which are used to hold nonbonding pairs of electrons? (c) Which of the orbitals can be used to delocalize the rr electrons? (d) How many electrons are delocalized in the rr system of ozone?

9.88 Butadiene, C4H6, is a planar molecule that has the fol­ lowing carbon-carbon bond lengths: H2C LJ4 A CH 1 4s A CH !.J4A CH2 .

(c) [9.90] The cyclopentadienide ion has the formula C5H5 -. The ion consists of a regular pentagon of C atoms, each bonded to two C neighbors, with a hydrogen atom bonded to each C atom. All the atoms lie in the same plane. (a) Draw a Lewis structure for the ion. According to your structure, do all five C atoms have the same hybridiza­ tion? Explain. (b) Chemists generally view this ion as having sp2 hybridization at each C atom. Is that view consistent with your answer to part (a)? (c) Your Lewis structure should show one nonbonding pair of electrons. Under the assumption of part (b), in what type of orbital must this nonbonding pair reside? (d) Are there reso­ nance structures equivalent to the Lewis structure you drew in part (a)? If so, how many? (e) The ion is often drawn as a pentagon enclosing a circle. Is this represen­ tation consistent with your answer to part (d)? Explain. (f) Both benzene and the cyclopentadienide ion are often described as systems containing six rr electrons. What do you think is meant by this description? 9.91 Write the electron configuration for the first excited state for N2-that is, the state with the highest-energy elec­ tron moved to the next available energy level. (a) Is the nitrogen in its first excited state diamagnetic or para­ magnetic? (b) Is the N-N bond strength in the first ex­ cited state stronger or weaker compared to that in the ground state? Explain.

9.47

9.92 Figure shows how the magnetic properties of a compound can be measured experimentally. When such measurements are made, the sample is generally cov­ ered by an atmosphere of pure nitrogen gas rather than air. Why do you suppose this is done? 9.93 Aw dyes are organic dyes that are used for many applica­ tions, such as the coloring of fabrics. Many azo dyes are derivatives of the organic substance azobenzene, C1 2H10N2.

390

C HA PTER 9

Molecular Geometry and Bonding Theories

A closely related substance is hydrazobenzene, C12H12N2. The Lewis structures of these two substances are

@-N=N--© @-y -y-@ H

Azobenzene

H

Hydrazobenzene

(Recall the shorthand notation used for benzene.) (a) What is the hybridization at the N atom in each of the substances? (b) How many unhybridized atomic orbitals are there on the N and the C atoms in each of the substances? (c) Predict the N -N - C angles in each of the substances. (d) Azobenzene is said to have greater delocalization of its ., electrons than hydra­ zobenzene. Discuss this statement in light of your an­ swers to (a) and (b). (e) All the atoms of azobenzene lie in one plane, whereas those of hydrazobenzene do not. Is this observation consistent with the statement in part (d)? (f) Azobenzene is an intense red-orange color, whereas hydrazobenzene is nearly colorless. Which molecule would be a better one to use in a solar energy conversion device? (See the "Chemistry Put to Work" box for more information about solar cells.) [9.94] (a) Using only the valence atomic orbitals of a hydrogen atom and a fluorine atom, how many MOs would you expect for the HF molecule? (b) How many of the MOs from part (a) would be occupied by electrons? (c) Do you think the MO diagram shown in Figure 9.49 could be used to describe the MOs of the HF molecule? Why or

why not? (d) It turns out that the difference in energies between the valence atomic orbitals of H and F are suffi­ ciently different that we can neglect the interaction of the 1s orbital of hydrogen with the 2s orbital of fluorine. The 1s orbital of hydrogen will mix only with one 2p orbital of fluorine. Draw pictures showing the proper orienta­ tion of all three 2p orbitals on F interacting with a 1s or­ bital on H. Which of the 2p orbitals can actually make a bond with a 1s orbital, assuming that the atoms lie on the z-axis? (e) In the most accepted picture of HF, all the other atomic orbitals on fluorine move over at the same energy into the molecular orbital energy level diagram for HF. These are called "nonbonding orbitals." Sketch the energy level diagram for HF using this information, and calculate the bond order. (Nonbonding electrons do not contribute to bond order.) (f) Look at the Lewis struc­ ture for HF. Where are the nonbonding electrons? [9.95] Carbon monoxide, CO, is isoelectronic to N2. (a) Draw a Lewis structure for CO that satisfies the octet rule. (b) Assume that the diagram in Figure 9.49 can be used to describe the MOs of CO. What is the predicted bond order for CO? Is this answer in accord with the Lewis structure you drew in part (a)? (c) Experimentally, it is found that the highest-energy electrons in CO reside in a 0'-type MO. Is that observation consistent with Figure 9.49? If not, what modification needs to be made to the di­ agram? How does this modification relate to Figure 9.45? (d) Would you expect the 1r2p MOs of CO to have equal atomic orbital contributions from the C and 0 atoms? If not, which atom would have the greater contribution?

INTEGRATIVE EXERCISES 9.96 A compound composed of 2.1% H, 29.8% N, and 68.1% 0 has a molar mass of approximately 50 g/mol. (a) What is the molecular formula of the compound? (b) What is its Lewis structure if H is bonded to 0? (c) What is the geom­ etry of the molecule? (d) What is the hybridization of the orbitals around the N atom? (e) How many 0' and how many ., bonds are there in the molecule? 9.97 Sulfur tetrafluoride (SF4) reacts slowly with 02 to form sulfur tetrafluoride monoxide (OSF4) according to the following unbalanced reaction: SF4(g) + 02(gl -> OSF4(g) The 0 atom and the four F atoms in OSF4 are bonded to a central S atom. (a) Balance the equation. (b) Write a Lewis structure of OSF4 in which the formal charges of all atoms are zero. (c) Use average bond enthalpies (Table 8.4) to estimate the enthalpy of the reaction. Is it endothermic or exothermic? (d) Determine the electron­ domain geometry of OSF4, and write two possible mole­ cular geometries for the molecule based on this electron-domain geometry. (e) Which of the molecular geometries in part (d) is more likely to be observed for the molecule? Explain. [9.98] The phosphorus trihalides (PX3) show the following vari­ ation in the bond angle X -P-X: PF3, 96.3°; PCI3, 100.3°; PBr3, 101.0°; Pl3, 102.0°. The trend is generally at­ tributed to the change in the electronegativity of the halo­ gen. (a) Assuming that all electron domains are the same size, what value of the X- P- X angle is predicted by

the VSEPR model? (b) What is the general trend in the X-P-X angle as the electronegativity increases? (c) Using the VSEPR model, explain the observed trend in X -P-X angle as the electronegativity of X changes. (d) Based on your answer to part (c), predict the struc­ ture of PBrCI4. [9.99] The molecule 2-butene, C4H8, can undergo a geometric change called cis-trans isomerization: H3C H

cis-2-butene

"'C=C

/

H

/ "'CH3

trans-2-butene

As discussed in the "Chemistry and Life" box on the chemistry of vision, such transformations can be in­ duced by light and are the key to human vision. (a) What is the hybridization at the two central carbon atoms of 2-butene? (b) The isomerization occurs by rota­ tion about the central C - C bond. With reference to Figure 9.32, explain why the ., bond between the two central carbon atoms is destroyed halfway through the rotation from cis- to trans-2-butene. (c) Based on average bond enthalpies (Table 8.4 ) , how much energy per mole­ cule must be supplied to break the C - C ., bond? (d) What is the longest wavelength of light that will pro­ vide photons of sufficient energy to break the C -C ., bond and cause the isomerization? (e) Is the wavelength in your answer to part (d) in the visible portion of the

Integrative Exercises electromagnetic spectrum? Comment on the importance of this result for human vision. 9.100 (a) Compare the bond enthalpies (fable 8.4) of the car­ bon-- 6 C(g) + 6 H(g) Compare the value to that obtained by using ilHJ data given in Appendix C and Hess's law. To what do you at­ tribute the large discrepancy in the two values? 9.102 For both atoms and molecules, ionization energies (Section 7.4) are related to the energies of orbitals: The lower the energy of the orbital, the greater the ioruzation energy. The first ioruzation energy of a molecule is therefore a measure of the energy of the highest occu­ pied molecular orbital (HOMO). See the "Chemistry Put to Work" box on Orbitals and Energy. The first ioniza­ tion energies of several diatomic molecules are given in electron-volts in the following table: Molecule

11

(eV)

Hz

15.4

Nz

15.6

Oz

12.1

Fz

15.7

(a) Convert these ioruzation energies to kJ/mol. (b) On the same plot, graph It for the H, N, 0, and F atoms (Figure 7.11) and 11 for the molecules listed. (c) Do the ionization energies of the molecules follow the same periodic trends as the ionization energies of the atoms? (d) Use molecular orbital energy-level diagrams to explain the trends in the ionization energies of the molecules. [9.103] Many compounds of the transition-metal elements con­ tain direct bonds between metal atoms. We will assume that the z-axis is defined as the metal-metal bond axis. (a) Which of the 3d orbitals (Figure 6.24) can be used to make a u bond between metal atoms? (b) Sketch the scale, now known as the Kelvin scale. On this scale 0 K, which is , , called absolute zero, equals -273.15 °C. (Section 1.4) In terms , 10 of the Kelvin scale, Charles's law can be stated as follows: The vol­

.i.

Figure 1 0.8 An Illustration of the

0 -300

, ,

,,''I t

-200

- 273 °C

- 100 100 0 Temperature ("C)

.i. Figure 1 0.9 Graph based on Charles's law. At constant pressure, the volume of an enclosed gas increases as the temperature increases. The dashed line is an extrapolation to temperatures at which the substance is no longer a gas.

oc:c

200

300

ume ofa fixed amount ofgas maintained at constant pressure is directly proportional to its absolute temperature. Thus, doubling the absolute temperature, say from 200 K to 400 K, causes the gas volume to double. Mathematically, Charles's law takes the following form:

V

= constant x T

or

v T

= constant

[10.3]

The value of the constant depends on the pressure and amount of gas.

GIVE IT SOME THOUGHT Does the volume of a fixed quantity of gas decrease to half its original value when the temperature is lowered from 100 oc to 50 °C?

The Quantity-Volume Relationship: Avogadro's Law As we add gas to a balloon, the balloon expands. The volume of a gas is affected by pressure, temperature, and the amount of gas. The relationship between the quantity of a gas and its volume follows from the work of Joseph Louis Gay-Lussac (1778-1823) and Amedeo Avogadro (1776-1856).

10.3 Observation

One volume oxygen

Two volumes hydrogen

Two volumes water vapor

The Gas Laws

401



2 Cl02(g) + 2 NaCl(s)

If you allow 10.0 g of NaC102 to react with 2.00 L of chlorine gas at a pressure of 1 .50 atm at 21 oc, how many grams of Cl02 can be prepared? 10.120 Natural gas is very abundant in many Middle Eastern oil fields. However, the costs of shipping the gas to mar­ kets in other parts of the world are high because it is necessary to liquefy the gas, which is mainly methane and thus has a boiling point at atmospheric pressure of -164 oc. One possible strategy is to oxidize the methane to methanol, CH30H, which has a boiling point of 65 oc and can therefore be shipped more readily. Suppose 3 that 10.7 x 109 ft of methane at atmospheric pressure and 25 oc are oxidized to methanol. (a) What volume of methanol is formed il the density of CH30H is 0.791 g/mL? (b) Write balanced chemical equations for the oxidations of methane and methanol to C02(g) and H20(/). Calculate the total enthalpy change for complete combustion of the 10.7 x 109 ft:f of methane described

435

above and for complete combustion of the equivalent amount of methanol, as calculated in part (a). (c) Methane, when liquefied, has a density of 0.466 g/mL; the density of methanol at 25 oc is 0.791 g/mL. Compare the enthalpy change upon com­ bustion of a unit volume of liquid methane and liquid methanol. From the standpoint of energy production, which substance has the higher enthalpy of combustion per unit volume? [10.121] Gaseous iodine pentafluoride, IF5, can be prepared by the reaction of solid iodine and gaseous fluorine: I2(s) + 5 �(g)

---->

2 IF5(g)

A 5.00-L flask containing 10.0 g I2 is charged with 10.0 g F2, and the reaction proceeds until one of the reagents is completely consumed. After the reaction is complete, the temperature in the flask is 125 oc. (a) What is the partial pressure of IF5 in the flask? (b) What is the mole fraction of IF5 in the flask? [10.122] A 6.53-g sample of a mixture of magnesium carbonate and calcium carbonate is treated with excess hydrochlo­ ric acid. The resulting reaction produces 1.72 L of carbon dioxide gas at 28 oc and 743 torr pressure. (a) Write bal­ anced chemical equations for the reactions that occur between hydrochloric acid and each component of the mixture. (b) Calculate the total number of moles of car­ bon dioxide that forms from these reactions. (c) Assum­ ing that the reactions are complete, calculate the percentage by mass of magnesium carbonate in the mixture.

INTERMOLECULAR FORCES, LIQUIDS, AND SOLIDS

THE WATER IN HOT SPRINGS is at least 5 - l 0 "C warmer than the mean annual air temperature where they are located. There are over 1 000 hot springs in the United States, the majority in the West. Some hot springs are popular for bathing, while others are far too hot for people to enter them safely.

436

W H AT ' S

A H E A D

11.1 A Molecular Comparison of Gases,

Liquids,

and Solids

We begin with a brief comparison of solids, Hquids, and gases from a molecular perspective. This comparison reveals the important roles that temperature and intemzolecular forces play in determining the physical state of a substance.

1 1.2

11.3

11.4

I ntermolecular Forces

1 1.5

11.6

We then examine the main kinds of intermolecular forces that occur within and between substances:

ion-dipoleforces, dipole-dipoleforces, London dispersion forces, and hydrogen bonds. Some Properties of Liquids

We will learn that the nature and strength of the intermolecular forces between molecules are largely responsible for many properties of liquids, including their viscosity, which is a measure of a liquid's resistance to flow, and surface tension, which is a measure of a liquid's resistance to increasing its surface area. Phase Changes

We will explore phase changes the transitions of matter between the gaseous, liquid, and solid states--and their associated energies. -

1 1.7

11.8

Vapor Pressure

We will examine the dynamic equilibrium that exists between a liquid and its gaseous state and introduce the idea of vapor pressure. A liquid boils when its vapor pressure equals the pressure acting on the surface of the liquid. Phase Diagrams

We will recognize that in a phase diagram the equilibria among the gaseous, liquid, and solid phases are displayed graphically. Structures of Solids

We then observe that the orderly arrangements of molecules or ions in three dimensions characterize crystalline solids. We also examine how the structure of a crystalline solid can be conveyed in terms of its unit cell and how simple molecules and ions are most efficiently arranged in three dimensions. Bonding in Solids

We learn that solids can be characterized and classified according to the attractive forces between their component atoms, molecules, or ions. We examine four such classes: molecular solids, covalent-network solids, ionic solids, and metallic solids.

S ITTI NG IN A HOT SPRING ON A SNOWY

day is not something

many of us have experienced. If we were in the hot spring shown in the chapter-opening photo, however, we would be surrounded simultaneously by all three phases of water-gas, liquid, and solid. The water vaporor humidity-in the air, the water in the hot spring, and the surrounding snow are all forms of the same substance, H20. They all have the same chemical properties. Their physical properties differ greatly, however, because the physical properties of a substance depend on its physical state. In Chapter 10 we discussed the gaseous state in some detail. In this chapter we turn our attention to the physical properties of liquids and solids and to the phase changes that occur between the three states of matter. Many of the substances that we will consider in this chapter are molecular. In fact, virtually all substances that are liquids at room temperature are molec­ ular substances. The intramolecular forces within molecules that give rise to covalent bonding influence molecular shape, bond energies, and many aspects of chemical behavior. The physical properties of molecular liquids and solids, however, are due largely to intermolecular forces, the forces that exist between molecules. We learned in Section 10.9 that attractions between gas molecules lead to deviations from ideal-gas behavior. But how do these intermolecular attractions arise? By understanding the nature and strength of intermolecular forces, we can begin to relate the composition and structure of molecules to their physical properties.

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TABLE 1 1 .1 • Some Characteristic Properties of the States of Matter Gas

Assumes both the volume and shape of its container Is compressible Flows readily Diffusion within a gas occurs rapidly

Liquid

Assumes the shape of the portion of the container it occupies Does not expand to fill container Is virtually incompressible Flows readily Diffusion within a liquid occurs slowly

Solid

Retains its own shape and volume Is virtually incompressible Does not flow Diffusion within a solid occurs extremely slowly

1 1 . 1 A M OLECULAR C O M PARI SON OF GAS ES , LIQU I D S , A N D S O L I D S Some o f the characteristic properties of gases, liquids, and solids are listed in Table 11.1 &. These properties can be understood in terms of the energy of mo­ tion (kinetic energy) of the particles (atoms, molecules, or ions) of each state, compared to the energy of the intermolecular interactions between those parti­ cles. As we learned from the kinetic-molecular theory of gases in Chapter 10, the average kinetic energy, which is related to the particle's average speed, is proportional to the absolute temperature. Gases consist of a collection of widely separated particles in constant, chaotic motion. The average energy of the attractions between the particles is much smaller than their average kinetic energy. The lack of strong attractive forces between particles allows a gas to expand to fill its container. In liquids the intermolecular attractive forces are stronger than in gases and are strong enough to hold particles close together. Thus, liquids are much denser and far less compressible than gases. Unlike gases, liquids have a defi­ nite volume, independent of the size and shape of their container. The attractive forces in liquids are not strong enough, however, to keep the particles from moving past one another. Thus, any liquid can be poured, and it assumes the shape of whatever portion of its container it occupies. In solids the intermolecular attractive forces are strong enough to hold par­ ticles close together and to lock them virtually in place. Solids, like liquids, are not very compressible because the particles have little free space between them. Because the particles in a solid or liquid are fairly close together compared with those of a gas, we often refer to solids and liquids as condensed phases. Often the particles of a solid take up positions in a highly regular three­ dimensional pattern. Solids that possess highly ordered three-dimensional structures are said to be crystalline. Because the particles of a solid are not free to undergo long-range movement, solids are rigid. Keep in mind, however, that the units that form the solid, whether ions or molecules, possess thermal energy and vibrate in place. This vibrational motion increases in amplitude as a solid is heated. In fact, the energy may increase to the point that the solid either melts or sublimes. Figure 11.1 � compares the three states of matter. The particles that com­ pose the substance can be individual atoms, as in Ar; molecules, as in H20; or ions, as in NaCI. The state of a substance depends largely on the balance between the kinetic energies of the particles and the interparticle energies of attraction. The kinetic energies, which depend on temperature, tend to keep the particles apart and moving. The interparticle attractions tend to draw the particles together.

11.2

Cool or increase pressure

-

Cool

Heat

Liquid

Gas Total disorder; much empty space; particles have complete freedom of motion; particles far apart

439

Figure 1 1 .22 Illustration of the equilibrium vapor pressure over a liquid. In (a) we imagine that no molecules of the liquid are in the gas phase initially; there is zero vapor pressure in the vessel. In (b), after equilibrium is reached, the rate at which molecules leave the surface equals the rate at which gas molecules return to the liquid phase. These equal rates produce a stable vapor pressure that does not change as long as the temperature remains constant.

"'

-3

�e

Liquid ethanol Liquid before any evaporation

At equilibrium, molecules enter and leave liquid at the same rate .

(a)

(b)

I

Explaining Vapor Pressure on the Molecu lar Level

:

The molecules of a liquid move at various speeds. Figure

Lower temperature 1

1 1 .23 (Section 11 .2) The triple point of H 20 (0.0098 oc and 0.00603 atm) is at much lower pres­ sure than that of C02 ( - 56 .4 °C and 5.11 atm). For C02 to exist as a liquid, the pressure must exceed 5.11 atm. Consequently, solid C02 does not melt but

a three-phase system. In this generic diagram, the substance being investigated can exist as a solid, a liquid, or a gas, depending on pressure and temperature. Beyond the critical point (C), the distinction between liquid and gas is lost, and the substance is a supercritical fluid.

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Intermolecular Forces, Liquids, and Solids

II> Figure 1 1 .27 Phase diagrams of H20 and C02• The axes are not drawn

to scale in either case. In (a), for water, note the triple point T at 0.0098 oc and 0.00603 atm, the normal melting (or freezing) point of 0 oc at 1 atm, the normal boiling point of 1 00 oc at 1 atm, and the critical point C (374.4 oc and 21 7.7 atm). In (b), for carbon dio xide, note the triple point T at -56.4 oc and 5.1 1 atm, the normal sublimation point of -78.5 oc at 1 atm, and the critical point C (3 1 .1 oc and 73.0 atm).

c

218 atm

�

�

""

1 atm 0.00603 atm

C02(s)

� " "' "'

5.11 � atm

I I

Ice (solid)

73 atm

""

Water vapor (gas) 100 0.0098 Temperature (0C}

1 atm

37

- 78.5

- 56.4

31.1

Temperature COC) (b)

(a)

sublimes when heated at 1 atm. Thus, C02 does not have a normal melting point; instead, it has a normal sublimation point, -78.5 oc. Because C02 sublimes rather than melts as it absorbs energy at ordinary pressures, solid C02 (dry ice) is a convenient coolant. For water (ice) to sublime, however, its vapor pressure must be below 0.00603 atm. Food is "freeze-dried" by placing frozen food in a low-pressure chamber (below 0.00603 atm) so that the ice in it sublimes. - SAMPLE EXERCISE 1 1 .6

[ Interpreting a Phase Diagram

Referring to Figure 11.28 ..,., describe any changes in the phases present when H20 is (a) kept at 0 oc while the pressure is increased from that at point 1 to that at point 5 (vertical line), (b) kept at 1.00 atm while the temperature is increased from that at point 6 to that at point 9 (horizontal line). SOLUTION Analyze: We are asked to use the phase diagram provided to deduce what phase changes might occur when specific pressure and temperature changes are brought about. Plan: Trace the path indicated on the phase diagram, and note what phases and phase changes occur. Solve:

Temperature .t.

Figure 1 1 .28 Phase diagram of

H20.

(a) At point 1, H20 exists totally as a vapor. At point 2 a solid-vapor equilibrium exists. Above that pressure, at point 3, all the H20 is converted to a solid. At point 4 some of the solid melts and equilibrium between solid and liquid is achieved. At still higher pressures all the H20 melts, so only the liquid phase is present at point 5 . (b) At point 6 the H20 exists entirely as a solid. When the temperature reaches point 4, the solid begins to melt and equilibrium exists between the solid and liquid phases. At an even higher temperature, point 7, the solid has been converted entirely to a liquid. A liquid-vapor equilibrium exists at point 8. Upon further heating to point 9, the H20 is converted entirely to the vapor phase. Check: The indicated phases and phase changes are consistent with our knowledge of the properties of water. - PRACTICE EXERCISE Using Figure 11.27(b}, describe what happens when the following changes are made to a C02 sample: (a) Pressure increases from 1 atm to 60 atm at a constant tempera­ ture of -60 °C. (b) Temperature increases from -60 oc to -20 oc at a constant pres­ sure of 60 atm. Answers: (a) C02{g) --> C02(s); (b) C02(s) --> C02(1)

1 1 .7 STRUCT URES O F S OLIDS Throughout the remainder o f this chapter w e will focus o n how the proper­ ties of solids relate to their structures and bonding. Solids can be either crys­ talline or amorphous (noncrystalline). In a crystalline solid the atoms, ions, or molecules are ordered in well-defined three-dimensional arrangements.

11.7

(a) Pyrite (fool's gold)

(b ) Fluorite

Structures of Solids

(c) Amethyst

These solids usually have flat surfaces, or faces, that make definite angles with one another. The orderly stacks of particles that produce these faces also cause the solids to have highly regular shapes (Figure 11.29 .A). Quartz and diamond are crystalline solids. An amorphous solid (from the Greek words for "without form") is a solid in which particles have no orderly structure. These solids lack well-defined faces and shapes. Many amorphous solids are mixtures of particles that do not stack together well. Most others are composed of large, complicated molecules. Familiar amorphous solids include rubber and glass. Quartz (Si02) is a crystalline solid with a three-dimensional structure like that shown in Figure 11 .30(a) �. When quartz melts (at about 1600 oq, it be­ comes a viscous, tacky liquid. Although the silicon-oxygen network remains largely intact, many Si -0 bonds are broken, and the rigid order of the quartz is lost. If the melt is rapidly cooled, the atoms are unable to return to an orderly arrangement. As a result, an amorphous solid known either as quartz glass or as silica glass results [Figure 11.30(b)].

(a) Crystalline Si02

.A Figure 1 1 .29 Crystalline solids. Crystalline solids come in a variety of forms and colors: (a) pyrite (fool's gold), (b) fluorite, (c) amethyst.

� Figure 1 1 .30 Schematic comparisons of (a) crystalline SI02 (quartz) and (b) amorphous SI02 (quartz glass). The structures are actually three-dimensional and not planar as drawn. The two-dimensional unit shown as the basic building block of the structure (silicon and three oxygens) actually has four oxygens, the fourth coming out of the plane of the paper and capable of bonding to other silicon atoms. The actual three-dimensional building block is shown.

(b) Amorphous Si02

Two-dimensional unit

459

Actual three-dimensional unit

460

C HAPTER 1 1

Intermolecular Forces, Liquids, and Solids Because the particles of an amorphous solid lack any long-range order, intermolecular forces vary in strength throughout a sample. Thus, amorphous solids do not melt at specific temperatures. Instead, they soften over a tempera­ ture range as intermolecular forces of various strengths are overcome. A crys­ talline solid, in contrast, melts at a specific temperature.

+----+-----+

+ i + i + i +

+----+----+

+ i + : + +

�----� +

+

+

+ +

+ + +

+ + +

.A. Figure 1 1 .31 A two-dimensional analog of a lattice and Its unit cell. The wallpaper design shows a characteristic repeat pattern. Each dashed blue square denotes a unit cell of the pattern. The unit cell could equally well be selected with red figures at the corners.

Lattice point

Lattice point

.A. Figure 1 1 .32 Part of a simple

crystal lattice and Its associated unit cell. A lattice is an array of points that define the positions of particles in a crystalline solid. Each lattice point represents an identical environment in the solid. The points here are shown connected by lines to help convey the three-dimensional character of the lattice and to help us see the unit cell.

(; I V E I T S O M E T H O U (; H T What is the general difference in the melting behaviors of crystalline and amorphous solids?

U nit Cells The characteristic order of crystalline solids allows us to convey a picture of an entire crystal by looking at only a small part of it. We can think of the solid as being built by stacking together identical building blocks, much as stacking rows of individual "identical" bricks forms a brick wall. The repeating unit of a solid, the crystalline "brick," is known as the unit c ell . A simple two-dimension­ al example appears in the sheet of wallpaper shown in Figure 11.31 ""i. There are several ways of choosing a unit cell, but the choice is usually the smallest unit cell that shows clearly the symmetry characteristic of the entire pattern. A crystalline solid can be represented by a three-dimensional array of points called a crystal lattice. Each point in the lattice is called a lattice point, and it represents an identical environment within the solid. The crystal lattice is, in effect, an abstract scaffolding for the crystal structure. We can imagine forming the entire crystal structure by arranging the contents of the unit cell re­ peatedly on the crystal lattice. In the simplest case the crystal structure would consist of identical atoms, and each atom would be centered on a lattice point. This is the case for most metals. Figure 11.32 ""i shows a crystal lattice and its associated unit cell. In general, unit cells are parallelepipeds (six-sided figures whose faces are parallelograms). Each unit cell can be described by the lengths of the edges of the cell and by the angles between these edges. Seven basic types of unit cells can describe the lat­ tices of all crystalline compounds. The simplest of these is the cubic unit cell, in which all the sides are equal in length and all the angles are 90° . Three kinds of cubic unit cells are illustrated in Figure 11.33 T. When lattice points are at the corners only, the unit cell is called primitive cubic (or simple cubic). When a lattice point also occurs at the center of the unit cell, the cell is body-centered cubic. When the cell has lattice points at the center of each face, as well as at each corner, it is face-centered cubic. The simplest crystal structures are cubic unit cells with only one atom cen­ tered at each lattice point. Most metals have such structures. Nickel, for exam­ ple, has a face-centered cubic unit cell, whereas sodium has a body-centered cubic one. Figure 11.34 I> shows how atoms fill the cubic unit cells. Notice that

Primitive cubic

Body-centered cubic

Face-centered cubic

.A. Figure 1 1 .33 The three types of unit cells found In cubic lattices. For clarity, the corner spheres are red and the body-centered and face-centered ones are yellow. Each sphere represents a lattice point (an identical environment in the solid).

11.7 1 atom at center

k atom at 8 comers

Structures of Solids

461

1 atom at 6 faces

! atom at

8 comers

cubic

the atoms on the comers and faces do not lie wholly within the unit cell. Instead, these atoms are shared between unit cells. As an example, let's look at the prim­ itive cubic structure in Figure 11 .34. In an actual solid, this primitive cubic struc­ ture has other primitive cubic unit cells next to it in all directions, on top of it, and underneath it. If you look at any one comer of the primitive cubic unit cell, you will see that is shared by 8 unit cells. Therefore, in an individual primitive cubic unit cell, each corner contains only one-eighth of an atom. Because a cube has eight comers, each primitive cubic unit cell has a total of 1/8 X 8 = 1 atom. Similarly, each body-centered cubic unit cell shown in Figure 11.34 contains two atoms (1/8 X 8 = 1 from the corners, and 1 totally inside the cube). Two unit cells share equally atoms that are on the faces of a face-centered cubic unit cell so that only one-half of the atom belongs to each unit cell. Therefore, the total number of atoms in the face-centered cubic unit cell shown in Figure 11.34 is four (that is, 1/8 X 8 = 1 from the corners and 1/2 X 6 = 3 from the faces). Table 11.6 � summarizes the fraction of an atom that occupies a unit cell when atoms are shared between unit cells.

.A. Figure 1 1 .34 Space-filling view of cubic unit cells- Only the portion of each atom that belongs to the unit cell is shown.

TABLE 1 L6 • Fraction of an Atom That Occupies a Unit Cell for Various Positions in the Unit Cell Position in Unit Cell Center Face Edge

Corner

Fraction in Unit Cell I 2 I 4 I 8

GIVE IT SOME THOUGHT If you know the unit cell dimensions for a solid, the number of atoms per unit cell, and the mass of the atoms, show how you can calculate the density of the solid.

The Crystal Structure of Sodium Chloride In the crystal structure of NaCl (Figure 11.35 T) we can center either the Na+ ions or the Cl- ions on the lattice points of a face-centered cubic unit cell. Thus, we can describe the structure as being face-centered cubic.

(a )

(b)

T Figure 1 1 .35 Two ways of defining the unit cell of NaCt A representation of an NaCI crystal lattice can show either (a) Cl- ions (green spheres) or (b) Na+ ions (purple spheres) at the lattice points of the unit cell. In both cases, the red lines define the unit cell. Both of these choices for the unit cell are acceptable; both have the same volume, and in both cases identical points are arranged in a face-centered cubic fashion.

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Intermolecular Forces, Liquids, and Solids

In

.,. Figure 1 1 .36 Relative size of Ions In an NaCI unit cell. As in Figure 1 1 .35, purple represents Na+ ions and green represents Cl- ions. Only portions of most of the ions lie within the boundaries of the single unit cell.

Figure

11.35

the Na+ and Cl- ions have been moved

apart so the symmetry of the structure can be seen more clear­ ly. In this representation no attention is paid to the relative sizes of the ions. The representation in Figure

11.36 �,

on the

other hand, shows the relative sizes of the ions and how they fill the unit cell. Notice that other unit cells share the particles at corners, edges, and faces. The total cation-to-anion ratio of a unit cell must be the same as that for the entire crystaL Therefore, within the unit + cell of NaCI there must be an equal number of Na and Cl­ 2+ ions. Similarly, the unit cell for CaC12 would have one Ca for every two CC and so forth.

- SAMPLE EXERCISE

11.7 I Determining the Contents of a Unit Cell

Determine the net number of Na+ and o- ions in the NaC! unit cell (Figure 11.36).

SOLUTION

We must sum the various contributing elements to determine the number of Na+ and Cl- ions within the unit cell. To find the total number of ions of each type, we must identify the different locations within the unit cell and determine the fraction of the ion that lies within the unit cell boundaries. Solve: There is one-fourth of an Na+ on each edge, Na+, (! Na+ per edge )(12 edges) = 3 Na+ a whole Na+ in the center of the cube (refer also to (1 Na+ per center)(1 center) = 1 Na+ Figure 11.35), one-eighth of a o- on each comer, and one-half of a Cr on each face. Thus, we have Cl-: (� Cl- per corner )(8 corners) = 1 Cl­ the following: (i o- per face ) Section 9.8, "A Closer Look: Phas­ es in Atomic and Molecular Orbitals"). The most effective diffraction of light occurs when the wavelength of the light and the width of the slits are similar in magnitude. The spacing of the layers of atoms in solid crystals is usu­ ally about 2-20 A. The wavelengths of X-rays are also in this range. Thus, a crystal can serve as an effective diffraction grat­ ing for X-rays. X-ray diffraction results from the scattering of X-rays by a regular arrangement of atoms, molecules, or ions. Much of what we know about crystal structures has been ob­ tained from studies of X-ray diffraction by crystals, a tech­ nique known as X-ray crystallography. Figure 11.38 "' depicts the diffraction of a beam of X-rays as it passes through a crys­ tal. The diffracted X-rays were formerly detected by photo­ graphic film. Today, crystallographers use an array detector, a device analogous to that used in digital cameras, to capture and measure the intensities of the diffracted rays. The diffrac­ tion pattern of spots on the detector in Figure 11.38 depends on the particular arrangement of atoms in the crystal. Thus, different types of crystals give rise to different diffraction

patterns. In 1913 the English scientists William and Lawrence Bragg (father and son) determined for the first time how the spacing of layers in crystals leads to different X-ray diffraction patterns. By measuring the intensities of the diffracted beams and the angles at which they are diffracted, it is possible to reason backward to the structure that must have given rise to the pattern. One of the most famous X-ray diffraction pat­ terns is the one for crystals of the genetic material DNA (Figure 11.39 "'), first obtained in the early 1950s. Working from X-ray diffraction data obtained by Rosalind Franklin, Franklin, Maurice Wilkins, James Watson, and Francis Crick determined the double-helix structure of DNA, one of the most important discoveries in molecular biology. For this achievement, Watson, Crick, and Wilkins were awarded the Nobel Prize in Physiology or Medicine in 1962. Franklin died in 1958, at age 37, from cancer; Nobel Prizes are awarded only to the living (and can be shared by three people at the most). Today X-ray crystallography is used extensively to deter­ mine the structures of molecules in crystals. The instruments used to measure X-ray diffraction, known as X-ray diffrac­ tometers, are now computer-controlled, making the collection of diffraction data highly automated. The diffraction pattern of a crystal can be determined very accurately and quickly (sometimes in a matter of hours), even though thousands of diffraction points are measured. Computer programs are then used to analyze the diffraction data and determine the arrangement and structure of the molecules in the crystal.

Related Exercises: 11.95, 11.96

... Figure 1 1 .38 Diffraction of X-rays by a crystal. In X-ray crystallography, an X-ray beam is diffracted by a crystal. The diffraction pattern can be recorded as spots where the diffracted X-rays strike a detector, which records the positions and intensities of the spots.

screen

X-ray detector II> Figure 1 1 .39 The X-ray diffraction photograph of one form of crystalline DNA. This photograph was taken in the early 1 950s. From the pattern of dark spots, the double-helical shape of the DNA molecule was deduced.

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Intermolecular Forces, Liquids, and Solids The properties of molecular solids depend on the strengths of

©

Benzene Melting point ('C) 5 Boiling point ('C) 80

the forces that exist between molecules and on the abilities of the molecules to pack efficiently in three dimensions. Benzene (C6H6), for example, is a highly symmetrical planar molecule. o:x:> (Section

8.6)

It has a higher melting point than toluene, a compound in

Toluene -95

Phenol

which one of the hydrogen atoms of benzene has been replaced by

111

182

43

a CH3 group (Figure 11.40

(e) The critical temperature and pressure of C52

temperatures is especially notable. The higher values for C52 arise from the greater London dispersion attractions between the C52 molecules compared with C02. These greater attractions are due to the larger size of the sulfur compared to oxygen and therefore its greater polarizability. (f) The density would be lower at the higher temperature. Density decreases with in­ creasing temperature because the molecules possess higher kinetic energies. Their more energetic movements result in larger average distances between molecules, which translate into lower densities.

CHAPTER REVIEW SUM MARY AND KEY TERMS Introduction and Section 1 1 . 1 Substances that are

molecular motion and to force the particles to occupy

gases or liquids at room temperature are usually com­

specific locations in a three-dimensional arrangement.

posed of molecules. In gases the intermolecular attractive forces are negligible compared to the kinetic energies of

Section 1 1 .2 Three types of intermolecular forces exist between neutral molecules: dipole-dipole forces, London

and undergo constant, chaotic motion. In liquids the

dispersion forces, and hydrogen bonding. Ion-dipole forces are important in solutions in which ionic com­

the molecules; thus, the molecules are widely separated

intermolecular forces are strong enough to keep the mol­ ecules

in close proximity; nevertheless, the molecules are

pounds are dissolved in polar solvents. London dispersion

free to move with respect to one another. In solids the in­

forces operate between all molecules (and atoms, for atom­

terparticle attractive forces are strong enough to restrain

ic substances such as He, Ne, AI, and so forth). The relative

470

CHAPTER

11

Intermolecular Forces, Liquids, and Solids

strengths of the dipole-dipole and dispersion forces de­ pend on the polarity, polarizability, size, and shape of the molecule. Dipole-dipole forces increase in strength with increasing polarity. Dispersion forces increase in strength with increasing molecular weight, although molecular shape is also an important factor. Hydrogen bonding occurs in compounds containing 0-H, N-H, and F-H bonds. Hydrogen bonds are generally stronger than dipole-dipole or dispersion forces. Sedlon 1 1 . 3 The stronger the intermolecular forces, the greater is the viscosity, or resistance to flow, of a liq­ uid. The surface tension of a liquid also increases as inter­ molecular forces increase in strength. Surface tension is a measure of the tendency of a liquid to maintain a mini­ mum surface area. The adhesion of a liquid to the walls of a narrow tube and the cohesion of the liquid account for capillary action and the formation of a meniscus at the surface of a liquid.

The normal boiling point is the temperature at which the vapor pressure equals 1 atm. Section 1 1 .6 The equilibria between the solid, liquid, and gas phases of a substance as a function of temperature and pressure are displayed on a phase diagram. Aline indi­ cates equilibria between any two phases. The line through the melting point usually slopes slightly to the right as pressure increases, because the solid is usually more dense than the liquid. The melting point at 1 atrn is the normal melting point. The point on the diagram at which all three phases coexist in equilibrium is called the triple point.

Section 1 1 .4 A substance may exist in more than one state of matter, or phase. Phase changes are transforma­ tions from one phase to another. Changes of a solid to liq­ uid (melting), solid to gas (sublimation), and liquid to gas (vaporization) are all endothermic processes. Thus, the heat of fusion (melting), the heat of sublimation, and the heat of vaporization are all positive quantities. The re­ verse processes are exothermic. A gas cannot be liquefied by application of pressure if the temperature is above its critical temperature. The pressure required to liquefy a gas at its critical temperature is called the critical pressure.

Section 1 1.7 In a crystalline solid, particles are arranged in a regularly repeating pattern. An amorphous solid is one whose particles show no such order. The es­ sential structural features of a crystalline solid can be rep­ resented by its unit cell, the smallest part of the crystal that can, by simple displacement, reproduce the three-di­ mensional structure. The three-dimensional structures of a crystal can also be represented by its crystal lattice. The points in a crystal lattice represent positions in the struc­ ture where there are identical environments. The simplest unit cells are cubic. There are three kinds of cubic unit cells: primitive cubic, body-centered cubic, and face­ centered cubic. Many solids have a close-packed struc­ ture in which spherical particles are arranged so as to leave the minimal amount of empty space. Two closely related forms of close packing, cubic close packing and hexagonal close packing, are possible. In both, each sphere has a coordination number of 12.

Section 1 1 .5 The vapor pressure of a liquid indicates the tendency of the liquid to evaporate. The vapor pres­ sure is the partial pressure of the vapor when it is in dynamic equilibrium with the liquid. At equilibrium the rate of transfer of molecules from the liquid to the vapor equals the rate of transfer from the vapor to the liquid. The higher the vapor pressure of a liquid, the more readi­ ly it evaporates and the more volatile it is. Vapor pressure increases nonlinearly with temperature. Boiling occurs when the vapor pressure equals the external pressure.

Section 1 1 .8 The properties of solids depend both on the arrangements of particles and on the attractive forces be­ tween them. Molecular solids, which consist of atoms or molecules held together by intermolecular forces, are soft and low melting. Covalent-network solids, which consist of atoms held together by covalent bonds that extend throughout the solid, are hard and high melting. Ionic solids are hard and brittle and have high melting points. Metallic solids, which consist of metal cations held togeth­ er by a sea of electrons, exhibit a wide range of properties.

KEY S KILLS •

Understand and be able to describe the intermolecular attractive interactions (ion-dipole, dipole-dipole, London dispersion, hydrogen bonding) that exist between molecules or ions, and be able to compare the relative strengths of intermolecular attractions in substances based on their molecular structure, or physical properties.

•

Understand the concept of polarizability.

•

Understand the concepts of viscosity and surface tension in liquids.

•

Know the names of the various phase changes for a pure substance.

•

Interpret heating curves and be able to calculate quantities related to temperature and enthalpies of phase changes.

•

Define critical pressure, critical temperature, vapor pressure, normal boiling point, normal melting point, critical point, triple point.

•

Be able to interpret and sketch phase diagrams; know how water's phase diagram differs from most other sub­ stances, and why.

471

Visualizing Concepts

• Know the difference between crystalline and amorphous solids, and be able to explain the differences between

primitive cubic, body-centered cubic, and face-centered cubic unit cells. • Classify solids based on their bonding/intermolecular forces and understand how difference in bonding relates to

physical properties.

VISUALIZING CONCEPTS 11.1 Does the following diagram best describe a crystalline

-.---, 800 .-------------------

solid, liquid, or gas? Explain. [Section 11.1]

700

0:0 :r: 600 e

.§.

[ �

500

" 400

� "'

>

11.2 (a) What kind of intermolecular attractive force is

300 200 100

shown in each of the following cases? (b) Predict which two interactions are stronger than the other two. [Section 11.2]

30

50

Temperature (0C)

11.5 The following molecules have the same molecular for­ (a)

H F

•

H F

(b)

(c)

.

mula (C 3H80), yet they have different normal boiling points, as shown. Rationalize the difference in boiling points. [Sections 11.2 and 11.5]

u

(d)

(a) Propanol 97.2°C

11.3 The molecular models of glycerol and 1-propanol are given here.

(b) Ethyl methyl ether 10.8°C

11.6 The phase diagram of a hypothetical substance is shown below.

(a) Glycerol

(b) 1-propanol

Do you expect the viscosity of glycerol to be larger or smaller than that of 1-propanol? Explain. [Section 11.3]

11.4 Using the following graph of CS2 data, determine (a) the approximate vapor pressure of cs2 at 30 °C, (b) the tem­ perature at which the vapor pressure equals 300 torr, (c) the normal boiling point of CS2. [Section 11.5]

I

1.0

� � iil 0.5

�

p...

OL______L______L_____�------� 300 0 100 200 400 Temperature (K)

(a) Estimate the normal boiling point and freezing

point of the substance. (b) What is the physical state of the substance under the following conditions? (i) T = 150 K, P = 0.2 atrn (ii) T = 100 K, P = 0.8 atrn (iii) T = 300 K, P = 1.0 atm (c) What is the triple point of the substance? [Section 11.6]

472

CHAPTER

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Intermolecular Forces, Liquids, and Solids

11.7 Niobium(II) oxide crystallizes in the following cubic unit cell.

11.8 (a) What kind o f packing arrangement is seen i n the ac­

companying photo? (b) What is the coordination num­ ber of each orange in the interior of the stack? (c) If each orange represents an argon atom, what category of solid is represented? [Sections 11.7 and 11.8]

• Oxygen �Niobium

(a) How many niobium atoms and how many oxygen atoms are within the unit cell?

(b) What is the empirical formula of niobium oxide? (c) Is this a molecular, covalent-network, or ionic solid? [Sections 11.7 and 11.8]

EXERC ISES Molecular Comparis o n s of Gases, Liquids, and Solids 11.9 List the three states of matter in order of (a) increasing molecular disorder and (b) increasing intermolecular

attractions. (c) Which state of matter is most easily compressed?

11.10 (a) How does the average kinetic energy of molecules

compare with the average energy of attraction between molecules in solids, liquids, and gases? (b) Why does increasing the temperature cause a solid substance to change in succession from a solid to a liquid to a gas? (c) What happens to a gas if you put it under extremely high pressure?

11.11 If you mix olive oil with water, the olive oil will float on

room temperature. (a) Is the density of olive oil more or less than 1.00 g/cm3? (b) The density of olive oil in its liquid phase does vary with temperature. Do you think olive oil would be more dense or less dense at higher temperatures? Explain.

11.12 Benzoic acid, C6H5 COOH, melts at 122 oc. The density in the liquid state at 130 oc is 1.08 g/cm3. The density of solid benzoic acid at 15 oc is 1.266 g/cm3 . (a) 1n which of

these two states is the average distance between mole­ cules greater? (b) Explain the difference in densities at the two temperatures in terms of the relative kinetic en­ ergies of the molecules.

top of the water. The density of water is 1.00 g/cm3 at

Intermolecular Forces 11.13 Which type of intermolecular attractive force operates between (a) all molecules, (b) polar molecules, (c) the

hydrogen atom of a polar bond and a nearby small elec­ tronegative atom?

atomic weight 84, boils at 120.9 K, whereas Cl 2, molecu­ lar weight about 71, boils at 238 K. (d) Acetone boils at 56 oc, whereas 2-methylpropane boils at -12 oc.

T

H3

11.14 Based on what you have learned about intermolecular

forces, would you say that matter is fundamentally at­ tracted or repulsed by other matter?

CH3-CH-CH3 Acetone

2-Methylpropane

11.15 Describe the intermolecular forces that must be over­ come to convert each of the following from a liquid or solid to a gas: (a) lz, (b) CH3 CHzOH, (c) H Se. 2 11.16 What type of intermolecular force accounts for the fol­ lowing differences in each case? (a) CH30H boils at 65 oc, CH3 SH boils at 6 oc. (b) Xe is liquid at atmos­ pheric pressure and 120 K, whereas Ar is a gas. (c) Kr,

11.17 (a) What is meant by the term polarizability? (b) Which of

the following atoms would you expect to be most polar­ izable: N, P, As, Sb? Explain. (c) Put the following mole­ cules in order of increasing polarizability: GeCl4, CH4, SiC4, SiH4, and GeBr4• (d) Predict the order of boiling points of the substances in part (c).

Exercises

True or false: (a) The more polarizable the molecules, the stronger the dispersion forces between them. (b) The boiling points of the noble gases decrease as you go down the column in the periodic table. (c) In general, the smaller the molecule, the stronger the dispersion forces. (d) All other factors being the same, dispersion forces between molecules increase with the number of electrons in the molecules. 11.19 Which member of the following pairs has the larger London dispersion forces: (a) H20 or H2S, (b) C02 or CO, (c) SiH4 or GeH4? 11.20 Which member of the following pairs has the stronger intermolecular dispersion forces: (a) Br2 or O:u (b) CH3CH2CH2CH2SH or CH3CH2CH2CH2CH2SH, (c) CH3CH2CH2CI or (CH3hCHCI? 11.21 Butane and 2-methylpropane, whose space-filling mod­ els are shown, are both nonpolar and have the same molecular formula, yet butane has the higher boiling point ( 0 5 oc compared to -11.7 °C}. Explain. 11.18

-

11.22

.

(b) 2-Methylpropane (a) Butane Propyl alcohol (CH3CH2CH20H) and isopropyl alcohol ((CH3hCHOH]. whose space-filling models are shown, have boiling points of 97.2 oc and 82.5 °C, respectively. Explain why the boiling point of propyl alcohol is higher, even though both have the molecular formula of C3HsO.

(a) Propyl alcohol

(b) Isopropyl alcohol

473

What atoms must a molecule contain to participate in hydrogen bonding with other molecules of the same kind? (b) Which of the following molecules can form hydrogen bonds with other molecules of the same kind: CH3F, CH3NH2, CH30H, CH3Br? 11.24 Rationalize the difference in boiling points between the members of the following pairs of substances: (a) HF (20 oq and HCI (-85 oq (b) CHCI3 (61 oq and CHBr3 (150 °C}, (c) Br2 (59 °C) and ICI (97 °C). 11.25 Ethylene glycol (HOCH2CH20H), the major substance in antifreeze, has a normal boiling point of 198 oc. By comparison, ethyl alcohol (CH3CH20H) boils at 78 oc at atmospheric pressure. Ethylene glycol dimethyl ether (CH30CH2CH20CH3) has a normal boiling point of 83 oc, and ethyl methyl ether (CH3CH20CH3) has a normal boiling point of 11 oc. (a) Explain why replace­ ment of a hydrogen on the oxygen by CH3 generally re­ sults in a lower boiling point. (b) What are the major factors responsible for the difference in boiling points of the two ethers? 11.26 Identify the types of intermolecular forces present in each of the following substances, and select the sub­ stance in each pair that has the higher boiling point: (a) C6H14 or CsHJs, (b) C3Hs or CH30CH3, (c) HOOH or HSSH, (d) NH2NH2 or CH3CH3. 11.27 Look up and compare the normal boiling points and normal melting points of H20 and H2S. (a) Based on these physical properties, which substance has stronger intermolecular forces? What kind of intermolecular forces exist for each molecule? (b) Predict whether solid H2S is more or less dense than liquid H2S. How does this compare to H20? Explain. (c) Water has an unusual­ ly high specific heat. Is this related to its intermolecular forces? Explain. 11.28 The following quote about ammonia (NH3) is from a textbook of inorganic chemistry: "It is estimated that 26% of the hydrogen bonding in NH3 breaks down on melting, 7% on warming from the melting to the boiling point, and the final 67% on transfer to the gas phase at the boiling point." From the standpoint of the kinetic energy of the molecules, explain (a) why there is a decrease of hydrogen-bonding energy on melting and (b) why most of the loss in hydrogen bonding occurs in the transition from the liquid to the vapor state.

11.23 (a)

Viscosity and Surface Te nsion

Explain why surface tension and viscosity decrease with increasing temperature. (b) Why do substances with high surface tensions also tend to have high viscosities? 11.30 (a) Distinguish between adhesive forces and cohesive forces. (b) What adhesive and cohesive forces are in­ volved when a paper towel absorbs water? (c) Explain the cause for the U-shaped meniscus formed when water is in a glass tube. _ 11.31 Explain the following observations: (a) The surface ten­ sion of CHBr3 is greater than that of CHC!3. (b) As tern11.29 (a)

___

perature increases, oil flows faster through a narrow tube. (c) Raindrops that collect on a waxed automobile hood take on a nearly spherical shape. (d) Oil droplets that collect on a waxed automobile hood take on a flat shape. 11.32 Hydrazine (H2NNH2), hydrogen peroxide (HOOH), and water (H20) all have exceptionally high surface ten­ sions compared with other substances of comparable molecular weights. (a) Draw the Lewis structures for these three compounds. (b) What structural property do these substances have in common, and how might that account for the high surface tensions?

474

CHAPTER

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Intermolecular Forces, Liquids, and Solids

Phase Changes 11.33 Name the phase transition in each of the following situ­ ations, and indicate whether it is exothermic or en­ dothermic: (a) When ice is heated, it turns to water. (b) Wet clothes dry on a warm summer day. (c) Frost appears on a window on a cold winter day. (d) Droplets of water appear on a cold glass of beer.

11.34 Name the phase transition in each of the following situ­ ations, and indicate whether it is exothermic or en­ dothermic: (a) Bromine vapor turns to bromine liquid as it is cooled. (b) Crystals of iodine disappear from an evaporating dish as they stand in a fume hood. (c) Rub­ bing alcohol in an open container slowly disappears. (d) Molten lava from a volcano turns into solid rock.

11.35 Explain why the heat of fusion of any substance is gen­ erally lower than its heat of vaporization.

11.36 Ethyl chloride (C2H5Cl) boils at 12 oc. When liquid C2H5Cl under pressure is sprayed on a room-temperature (25 oq surface in air, the surface is cooled considerably. (a) What does this observation tell us about the specific heat of C2H5Cl(g) as compared with C2H5Cl(/)? (b) As­ sume that the heat lost by the surface is gained by ethyl chloride. What enthalpies must you consider if you were to calculate the final temperature of the surface?

11.37 For many years drinking water has been cooled in hot climates by evaporating it from the surfaces of canvas bags or porous clay pots. How many grams of water can be cooled from 35 oc to 20 oc by the evaporation of 60 g of water? (The heat of vaporization of water in this tem­ perature range is 2.4 kJ/g. The specific heat of water is 4.18 J/g-K.)

11.38 Compounds like CC12F2 are known as chlorofluorocar­ bons, or CFCs. These compounds were once widely used as refrigerants but are now being replaced by com­ pounds that are believed to be less harmful to the envi­ ronment. The heat of vaporization of CC12F2 is 289 J/g. What mass of this substance must evaporate to freeze 200 g of water initially at 15 'C? (The heat of fusion of water is 334 J/g; the specific heat of water is 4.18 J/g-K.)

11.39 Ethanol (C2H50H) melts at -114 'C and boils at 78 'C.

Its density is 0.789 g/mL. The enthalpy of fusion of ethanol is 5.02 kJ/mol, and its enthalpy of vaporization is 38.56 kJ/mol. The specific heats of solid and liquid ethanol are 0.97 J/g-K and 2.3 J/g-K, respectively. (a) How much heat is required to convert 25.0 g of ethanol at 25 oc to the vapor phase at 78 °C? (b) How much heat is required to convert 5.00 L of ethanol at -140 oc to the vapor phase at 78 'C?

11.40 The fluorocarbon compound C2Cl3F3 has a normal boil­ ing point of 47.6 'C. The specific heats of C2Cl3F3(1) and C2Cl#3(g) are 0.91 J/g-K and 0.67 J/g-K, respectively. The heat of vaporization for the compound is 27.49 kJ/mol. Calculate the heat required to convert 50.0 g of C2Cl3F3 from a liquid at 10.00 'C to a gas at 85.00 'C.

11.41 (a) What is the significance of the critical pressure of a substance? (b) What happens to the critical temperature of a series of compounds as the force of attraction be­ tween molecules increases? (c) Which of the substances listed in Table 11.5 can be liquefied at the temperature of liquid nitrogen ( -196 'C)?

11.42 The critical temperatures (K) and pressures (atm) of a series of halogenated methanes are as follows:

Compound Critical Temperature

471

385

302

227

Critical Pressure

43.5

40.6

38.2

37.0

(a) List the intermolecular forces that occur for each compound. (b) Predict the order of increasing intermol­ ecular attraction, from least to most, for this series of compounds. (c) Predict the critical temperature and pressure for CC14 based on the trends in this table. Look up the experimentally determined critical temperatures and pressures for CC14, using a source such as the CRC Handbook of ChemislnJ and Physics, and suggest a reason for any discrepancies.

Vapor Pressure and Boiling Point 11.43 Explain how each of the following affects the vapor pressure of a liquid: (a) volume of the liquid, (b) surface area, (c) intermolecular attractive forces, (d) tempera­ ture, (e) density of the liquid. 11.44 A liquid that has an equilibrium vapor pressure of 130

torr at 25 'C is placed into a 1-L vessel like that shown in Figure 11.22. What is the pressure difference shown on the manometer, and what is the composition of the gas in the vessel, under each of the following conditions: (a) Two hundred mL of the liquid is introduced into the vessel and frozen at the bottom. The vessel is evacuated and sealed, and the liquid is allowed to warm to 25 'C. (b) Two hundred milliliters of the liquid is added to the vessel at 25 'C under atmospheric pressure, and after a

few minutes the vessel is closed off. (c) A few mL of the liquid is introduced into the vessel at 25 'C while it has a pressure of 1 atm of air in it, without allowing any of the air to escape. After a few minutes a few drops of ___ liquid remain in the vessel.

11.45 (a) Place the following substances in order of increasing

volatility: C�, CBr4, CH2Clz, CH3Cl, CHBr3, and CHzBr2. Explain. (b) How do the boiling points vary through this series?

11.46 True or false: (a) CBr4 is more volatile than CC14. (b) CBr4 has a higher boiling point than CC4.

(c) CBr4 has weaker intermolecular forces than CC14.

(d) CBr4 has a higher vapor pressure at the same temperature than CC14.

Exercises

Two pans of water are on different burners of a stove. One pan of water is boiling vigorously, while the other is boiling gently. What can be said about the tempera­ ture of the water in the two pans? (b) A large container of water and a small one are at the same temperature. What can be said about the relative vapor pressures of the water in the two containers? 11.48 Explain the following observations: (a) Water evapo­ rates more quickly on a hot, dry day than on a hot, humid day. (b) It takes longer to boil water for tea at high altitudes than at lower altitudes. 11.49 Using the vapor-pressure curves in Figure 11 .24, (a) esti­ mate the boiling point of ethanol at an external pressure of 200 torr; (b) estimate the external pressure at which ethanol will boil at 60 °C; (c) estimate the boiling point of diethyl ether at 400 torr; (d) estimate the external pressure at which diethyl ether will boil at 40 oc. 11.47 (a)

11.50

475

Appendix B lists the vapor pressure of water at various external pressures. (a) Plot the data in Appendix B, vapor pressure (torr) vs. temperature (0C). From your plot, estimate the vapor pressure of water at body temperature, 37 oc. (b) Explain the significance of the data point at 760.0 torr, 100 °C. (c) A city at an altitude of 5000 ft above sea level has a barometric pressure of 633 torr. To what temper­ ature would you have to heat water to boil it in this city? (d) A city at an altitude of 500 ft below sea level would have a barometric pressure of 774 torr. To what temperature would you have to heat water to boil it in this city? (e) For the two cities in parts (c) and (d), compare the average kinetic energies of the water molecules at their boiling points. Are the kinetic energies the same or different? Explain.

Phase Diagrams

What is the significance of the critical point in a phase diagram? (b) Why does the line that separates the gas and liquid phases end at the critical point? 11.52 (a) What is the significance of the triple point in a phase diagram? (b) Could you measure the triple point of water by measuring the temperature in a vessel in which water vapor, liquid water, and ice are in equilibri­ um under one atmosphere of air? Explain. 11.53 Refer to Figure 11.27(a), and describe all the phase changes that would occur in each of the following cases: (a) Water vapor originally at 0.005 atm and -0.5 oc is slowly compressed at constant temperature until the final pressure is 20 atm. (b) Water originally at 100.0 oc and 0.50 atm is cooled at constant pressure until the temperature is -10 oc. 11.51 (a)

Refer to Figure 11.27(b), and describe the phase changes (and the temperatures at which they occur) when C02 is heated from -80 oc to -20 oc at (a) a constant pressure of 3 atm, (b) a constant pressure of 6 atm. 11.55 Sketch a generic phase diagram for a substance that has a more dense solid phase than a liquid phase. Label all regions, lines, and points. 11.56 The normal melting and boiling points of 02 are -218 oc and - 183 oc respectively. Its triple point is at -219 OC and 1.14 torr, and its critical point is at -119 oc and 49.8 atm. (a) Sketch the phase diagram for 02, showing the four points given and indicating the area in which each phase is stable. (b) Will 02(s) float on 02(1)? Explain. (c) As it is heated, will solid 02 sublime or melt under a pressure of 1 atrn? 11.54

Structures of Solids

Draw a picture that represents a crystalline solid at the atomic level. (b) Now draw a picture that represents an amorphous solid at the atomic level. 3 11.58 Amorphous silica has a density of about 2.2 gjcm , whereas the density of crystalline quartz is 2.65 g/cm3 Account for this difference in densities. 11.59 Tausonite, a mineral composed of Sr, 0, and Ti, has the cubic unit cell shown in the drawing. (a) What is the chemical formula of this mineral? (b) It is easy to see that Ti is coordinated by six oxygen atoms, because the Ti atom is located at the center of the unit cell. To see the full coordination environment of the other ions, we have to consider neighboring unit cells. How many oxygens are coordinated to strontium?

11.57 (a)

Strontium

. Oxygen Ql

11.60

Titanium

Rutile is a mineral composed of Ti and 0. Its unit cell, shown in the drawing, contains Ti atoms at each corner and a Ti atom at the center of the cell. Four 0 atoms are

476

CHAPTER

11

Intermolecular Forces, Liquids, and Solids

on the opposite faces of the cell, and two are entire!y within the cell.

(a) What

Exercise 11.61. If each

mineral? (b) What is the nature of the bonding that holds the solid together? (c) What is the coordination number of each atom?

AI

atom has a radius of 1.43

A,

what is the length of a side of the unit cell? (d) Calculate

is the chemical formula of this

the density of aluminum metaL

11.63

An element crystallizes in a body-centered cubic lattice.

The edge of the unit cell is 2.86 A, and the density of the crystal is 7.92 g/cm3 Calculate the atomic weight of the element.

11.64 Oxygen

KCl has the same structure as NaCl. The length of the 3 unit cell is 628 pm. The density of KCl is 1.984 g/cm , and its formula mass is 74.55 amu. Using this informa­

Titanium

tion, calculate Avogadro's number.

11.65

NaF has the same structure as NaCI. (a) Use ionic radii from Chapter 7 to estimate the length of the unit cell edge for NaF. (b) Use the unit cell size calculated in part (a) to estimate the density of NaF.

11.61

Iridium crystallizes in a face-centered cubic unit cell that

has an edge length of 3.833 A. The atom in the center of

11.66

What is the coordination number of (a) Na+ in the NaCl 2 structure, Figure 11.35; (b) Zn + in the ZnS unit cell, 2 Figure 11.42(b); (c) Ca + in the CaF2 unit cell, Figure

11.67

Clausthalite is a mineral composed of lead selenide

the face is in contact with the comer atoms, as shown in

the drawing. (a) Calculate the atomic radius of an iridi­ um atom. (b) Calculate the density of iridium metaL

11.42(c)?

(PbSe). The mineral adopts a NaCl-type structure. The 3 density of PbSe at 25 'C is 8.27 gjcm Calculate the length of an edge of the PbSe unit cell.

11.68

A particular form of cinnabar (HgS) adopts the zinc blende structure, Figure 11.42(b). The length of the unit cell side is 5.852

11.62

structure

(a)

(face-centered

cubic

cell,

Figure

11.34).

of the unit cell side in this mineral is 6.085

A. What ac­

counts for the larger unit cell length in tiemmanite?

(b) What is the coordination number of each aluminum

(c)

Calculate the density of HgS in

a solid phase with the zinc blende structure. The length

How many aluminum atoms are in a unit cell?

atom?

A. (a)

this form. (b) The mineral tiemmanite (HgSe) also forms

Aluminum metal crystallizes in a cubic close-packed

(c) Which of the two substances has the higher density?

Assume that the aluminum atoms can be

How do you account for the difference in densities?

represented as spheres, as shown in the drawing for

Bonding in Solids 11.69

It is possible to change the temperature and pressure of

11.74

(a)

(c) high melting point and poor electrical (e) charged

melting point;

conductivity; (d) network of covalent bonds;

Silicon is the fundamental component of integrated

Is

Si

a molecular, metallic, ionic, or covalent-network solid?

_ __

11.75

dissolves in water to form a conducting solution. Which

silica to form?

type of solid (Table 11.7) might the substance be?

What kinds of attractive forces exist between particles in

(a)

11.76

molecular crystals, (b) covalent-network crystals,

(melting point, 2677 'C),

(e) benzene, (f) !2.

(a) (d)

CaC0 3, (b) Pt, (c) Zr02

is a nonconductor of electricity and is insoluble

in water. Which type of solid (Table 11.7) might this substance be?

Indicate the type of crystal (molecular, metallic, cova­ lent-network, or ionic) each of the following would

You are given a white substance that sublimes at 3000 "C; the solid

(c) ionic crystals, (d) metallic crystals?

form upon solidification:

particles throughout the solid. _ A white substance melts with some decomposition at 730 'C. As a solid, it does not conduct electricity, but it

(b) Silica is Si02. What type of solid would you expect

11.73

high mobility of elec­

atoms? (b) Is the solid argon a "covalent network

circuits. Si has the same structure as diamond.

11.72

(a)

trons throughout the solid; (b) softness, relatively low

What intermolecular forces exist between argon

solid"? Why or why not?

11.70 (a)

11.71

Which type (or types) of crystalline solid is character­ ized by each of the following:

a vessel containing argon gas so that the gas solidifies.

11.77

For each of the following pairs of substances, predict which will have the higher melting point and indicate

table sugar (C12H22011),

why:

11.78

(a) Ar, Xe; (b) SiOz, COz; (c)

KBr, Brz; (d) C6 Cl iY C6H 6·

For each of the following pairs of substances, predict

Covalent bonding occurs in both molecular and cova­

which will have the higher melting point, and indicate

lent-network solids. Why do these two kinds of solids

why:

differ so greatly in their hardness and melting points?

(a) HF, HCl; (b) (d) LiF, MgF2.

C (graphite), CH4; (c) KCl, Clz;

Additional Exercises

HN-H·· ·O N -N�1't--�N··· -, N-< ·HH-NH N F

ADDITIO NAL EXERCISES 11.79 As the intermolecular attractive forces between mole­ cules increase in magnitude, do you expect each of the following to increase or decrease in magnitude? (a) vapor pressure, (b) heat of vaporization, (c) boiling point, (d) freezing point, (e) viscosity, (f) surface tension, (g) critical temperature. 11.80 Suppose you have two colorless molecular liquids, one boiling at - 84 oc, the other at 34 oc, and both at atmos­ pheric pressure. Which of the following statements is correct? For those that are not correct, modify the state­ ment so that it is correct. (a) The higher-boiling liquid has greater total intermolecular forces than the other. (b) The lower boiling liquid must consist of nonpolar molecules. (c) The lower-boiling liquid has a lower molecular weight than the higher-boiling liquid. (d) The two liquids have identical vapor pressures at their nor­ mal boiling points. (e) At 34 •c both liquids have vapor pressures of 760 Hg. 11.81 Two isomers of the planar compound 1,2-dichloro­ ethylene are shown here, along with their melting and boiling points.

rnm

477

� �

�

O..

Cytosine

Guanine

You can see that AT base pairs are held together by two hydrogen bonds, and the GC base pairs are held togeth­ er by three hydrogen bonds. Which base pair is more stable to heating?

11.85 Ethylene glycol [CH2(0H)CH2(0H)] is the major com­ ponent of antifreeze. It is a slightly viscous liquid, not very volatile at room temperature, with a boiling point of 198 oc. Pentane (C5Hn), which has about the same molecular weight, is a nonviscous liquid that is highly volatile at room temperature and whose boiling point is 36.1 oc. Explain the differences in the physical proper­ ties of the two substances.

11.86 Using the following list of normal boiling points for a

-80.5

-49.8

series of hydrocarbons, estimate the normal boiling point for octane, C8H18: propane (C3H8, -42.1 oq, bu­ tane (C4H10, -0.5 •q, pentane (CsHu, 36.1 °C), hexane (C6H14• 68.7 •q, heptane (C7H16• 98.4 °C). Explain the trend in the boiling points.

60.3

47.5

[11.87] Notice in Figure 11.21 that there is a pressure-reduction

cis isomer Melting point (0C) Boiling point (°C)

trans isomer

(a) Which of the two isomers will have the stronger

dipole-dipole forces? Is this prediction borne out by the data presented here? (b) Based on the data presented here, which isomer packs more efficiently in the solid phase? 11.82 1n dichloromethane, CH2Cl2 (J.t � 1.60 D), the disper­ sion force contribution to the intermolecular attractive forces is about five times larger than the dipole-dipole contribution. Would you expect the relative importance of the two kinds of intermolecular attractive forces to differ (a) in dibromomethane (J.t � 1 .43 D), (b) in difluo­ romethane (J.t � 1.93 D)? Explain. 11.83 When an atom or group of atoms is substituted for an H atom in benzene (C,;I-16), the boiling point changes. Explain the order of the following boiling points: C,;I-16 (80 •q, C,;I-IsCI (132 oq, C6HsBr (156 oq, C6HsOH (182 oq. 11.84 The DNA double helix (Figure 25.40) at the atomic level looks like a twisted ladder, where the "rungs" of the lad­ der consist of molecules that are hydrogen-bonded together. Sugar and phosphate groups make up the sides of the ladder. Shown are the structures of the adenine­ thymine (AT) "base pair" and the guanine-cytosine (GC) base pair:

CH3 O···H-NH N �N-H··· N"=N K-1't----, N-< ar( 0

!sug

Thymine

�

Adenine

valve in the line just before the supercritical C02 and dis­ solved caffeine enter the separator. Suggest an explana­ tion for the function of this valve in the overall process.

11.88 (a) When you exercise vigorously, you sweat. How does this help your body cool? (b) A flask of water is connected to a vacuum pump. A few moments after the pump is turned on, the water begins to boil. After a few minutes, the water begins to freeze. Explain why these processes occur.

[11.89] The following table gives the vapor pressure of hexaflu­ orobenzene (C6F6) as a function of temperature:

Temperature (K)

Vapor Pressure (torr)

280.0 300.0 320.0 330.0 340.0

32.42 92.47 225.1 334.4 482.9

(a) By plotting these data in a suitable fashion, determine whether the Clausius-Clapeyron equation is obeyed. lf it is obeyed, use your plot to determine llHv•p for C6F6· (b) Use these data to determine the boiling point of the compound. [11.90] Suppose the vapor pressure of a substance is measured at two different temperatures. (a) By using the Clausius­ Clapeyron equation, Equation 11.1, derive the following relationship between the vapor pressures, P1 and P2, and the absolute temperatures at which they were measured, T1 and T2:

478

CHAPTER

11

Intermolecular Forces, Liquids, and Solids related to the relative sizes of the cation and anion and

(b) Gasoline is a mixture of hydrocarbons, a major com­ ponent of which is octane, CH3CH2CH2CH2CH2CH2

the overall stoichiometry. Li20 crystallizes with a struc­

CH2CH 3 . Octane has a vapor pressure of 13.95 torr at

ture called the antifluorite structure. This structure is

25 'C and a vapor pressure of 144.78 torr at 75 'C. Use

identical to the fluorite structure shown in Figure 11.42(c),

these data and the equation in part (a) to calculate the

but the cations and anions have been switched.

heat of vaporization of octane.

(c)

in part (a) and the data given in part (b), calculate the

(b) As the cation radius increases, would you expect the

normal boiling point of octane. Compare your answer to

coordination number to increase or decrease (assuming

the one you obtained from Exercise 11.86.

(d) Calculate

the anion size does not change)?

the vapor pressure of octane at -30 'C.

[11.91]

(a) In

Li20, what are the coordination numbers for each ion?

By using the equation

(c)

Why do you

think

the antifluorite structure becomes unstable for A20 (A

The following data present the temperatures at which certain vapor pressures are achieved for dichloro­ methane (CH2Cl2) and methyl iodide (CH 3I):

= Li, Na, K, Rb, Cs) compounds of the heavier alkali

metal ions?

11.94

Gold crystallizes in a face-centered cubic unit cell that

has an edge length of 4.078 A. The atom in the center of

in contact with the corner atoms, as shown in (a) Calculate the appar­

Vapor Pressure (torr):

10.0

40.0

100.0

400.0

the drawing for Exercise 11.61.

T for CH2Cl2 ('C):

- 43.3

-22.3

-6.3

24.1

the density of gold metal.

T for CH 3I ('C):

-45.8

- 24.2

-7.0

25.3

the face is

ent radius of a gold atom in this structure. (b) Calculate

11.95 (a)

Explain why X-rays can be used to measure atomic

distances in crystals but visible light cannot. (b) Why

(a) Which of the

two substances is expected to have the

greater dipole-dipole forces? Which is expected to have

can't CaCI2 have the same crystal structure as NaCI?

11.96 In

(b)

diffraction

CRC

drawing an appropriate graph.

11.92

The elements xenon and gold both have solid-state

and

relationship

(A), the

angle at

and the distance

(d)

is given by

nA

=

2d sin 9. (a)

X-rays from

spacing in the crystal, assuming

n = 1 (first-order dif­

fraction). (b) Repeat the calculation of part (a) but for the

n = 2 case (second-order diffraction). 3 [11.97] (a) The density of diamond [Figure 11.41(a)] is 3.5 g/cm , 3 and that of graphite [Figure 11.41(b)] is 2.3 g/cm . Based

structures with face-centered cubic unit cells, yet Xe

on the structure of buckminsterfullerene (Figure 11.43),

melts at -112 'C and gold melts at 1064 'C. Account for

what would you expect its density to be relative to these

these greatly different melting points.

11.93 In Chapter

(9),

William

the

con. Using the Bragg equation, calculate the interplanar

for the two substances in order for this phenomenon to Substantiate your answer for part (c) by

that

diffracted at an angle of 14.22 degrees by crystalline sili­

volatility of these two substances changes as the tem­

(d)

diffraction,

a copper X-ray tube that have a wavelength of 1.54 A are

The order of

perature is increased. What quantity must be different occur?

X-ray

determined

between the layers of atoms in the crystal that cause the

would you expect to have the higher boiling point?

Handbook of Chemistry and Physics. (c)

of

Bragg

which the radiation is diffracted

Which compound

Check your answer in a reference book such as the

study

among the wavelength of the radiation

swers, explain why it is difficult to predict which com­ pound would be more volatile.

their

Lawrence

the greater London dispersion forces? Based on your an­

other forms of carbon? (b) X-ray diffraction studies of

7 we saw that Li reacts with oxygen to form

buckminsterfullerene show that it has a face-centered

lithium oxide, Li20, while the larger alkali metals react

cubic lattice of

c 60

molecules. The length of a side of

with oxygen to form peroxides (such as Na20:u K20:u

the unit cell is 14.2 A. Calculate the density of buckmin­

etc.). The origin of this curious pattern of reactivity is

sterfullerene.

INTEGRAT IVE EXERCISES 11.98

Spinel is a mineral that contains 37.9% AI, 17.1% Mg, and 3 45.0% 0, by mass, and has a density of 3.57 g/cm

which has about the same molecular weight. What

The unit cell is cubic, with an edge length of 809 pm. How

relate to the difference in normal boiling points for

accounts for this difference? How does your answer these two substances?

many atoms of each type are in the unit cell?

11.99 (a)

At the molecular level, what factor is responsible

11.100

Acetone, (CH3hCO, is widely used as an industrial

(a)

for the steady increase in viscosity with increasing mol­

solvent.

ecular weight in the hydrocarbon series shown in

molecule, and predict the geometry around each car­

Draw the Lewis structure for the acetone

Table 11.4? (b) Although the viscosity varies over a fac­

bon atom. (b) Is the acetone molecule polar or non­

tor of more than two in the series from hexane to

polar?

(c) What kinds of intermolecular attractive forces

nonane, the surface tension at 25 'C increases by only

exist

between

about 20% in the same series. How do you account for

CH3CH2CH20H, has a molecular weight that is very

this?

similar to that of acetone, yet acetone boils at 56.5 'C

(c) n-Oc:rl alcohol, CH3(CH2);,0H, has a viscosity X 10- kg m- ls- 1 , much higher than nonane,

of 1.01

acetone

molecules?

(d)

1-Propanol,

and 1-propanol boils at 97.2 'C. Explain the difference.

Integrative Exercises 11.101 The table shown here lists the molar heats of vaporiza­ tion for several organic compounds. Use specific exam­ ples from this list to illustrate how the heat of vaporization varies with (a) molar mass, (b) molecular shape, (c) molecular polarity, (d) hydrogen-bonding in­ teractions. Explain these comparisons in terms of the na­ ture of the intermolecular forces at work. (You may find it helpful to draw out the structural formula for each compound.)

Compound CH3CH2CH3 CH3CH2CH2CH2CH3 CH3CHBrCH3 CH3COCH3 CH3CH2CH2Br CH3CH2CH20H

Heat of Vaporization (kJ/mol)

19.0 27.6 31.8 32.0 33.6 47.3

11.102 Liquid butane, C4H10, is stored in cylinders, to be used as a fuel. The normal boiling point of butane is listed as - 0.5 'C. (a) Suppose the tank is standing in the sun and reaches a temperature of 35 'C. Would you expect the pressure in the tank to be greater or less than atmos­ pheric pressure? How does the pressure within the tank depend on how much liquid butane is in it? (b) Suppose the valve to the tank is opened and a few liters of butane are allowed to escape rapidly. What do you expect would happen to the temperature of the remaining liquid butane in the tank? Explain. (c) How much heat must be added to vaporize 250 g of butane if its heat of vaporization is 21.3 kJ/mol? What volume does this much butane occupy at 755 torr and 35 'C?

[11.103] Using information in Appendices B and C, calculate the

minimum number of grams of propane, C3H8(g), that must be combusted to provide the energy necessary to convert 5.50 kg of ice at -20 'C to liquid water at 75 'C.

11.104

479

In a certain type of nuclear reactor, liquid sodium metal is employed as a circulating coolant in a closed system, protected from contact with air or water. Much like the coolant that circulates in an automobile engine, the liq­ uid sodium carries heat from the hot reactor core to heat exchangers. (a) What properties of the liquid sodium are of special importance in this application? (b) The viscos­ ity of liquid sodium varies with temperature as follows:

Temperature ('C)

Viscosity (kg m -1 s -1)

100 200 300 600

7.05 4.50 3.45 2.10

X 10-4 X 10-4 X 10-4 X 10-4

What forces within the liquid sodium are likely to be the major contributors to the viscosity? Why does viscosity decrease with increasing temperature? 11.105 The vapor pressure of a volatile liquid can be deter­ mined by slowly bubbling a known volume of gas through it at a known temperature and pressure. In an experiment, 5.00 L of N2 gas is passed through 7.2146 g of liquid benzene, C�6, at 26.0 'C. The liquid remain­ ing after the experiment weighs 5.1493 g. Assuming that the gas becomes saturated with benzene vapor and that the total gas volume and temperature remain constant, what is the vapor pressure of the benzene in torr? 11.106 The relative humidity of air equals the ratio of the par­ tial pressure of water in the air to the equilibrium vapor pressure of water at the same temperature, times 100%. If the relative humidity of the air is 58% and its temper­ ature is 68 'F, how many molecules of water are present in a room measuring 12 ft x 10 ft X 8 ft? 11.107 Use a reference source such as the CRC Handbook of Chemistry and Physics to compare the melting and boil­ ing points of the following pairs of inorganic sub­ stances: (a) W and WF6, (b) S02 and SF4, (c) Si02 and SiCI4. Account for the major differences observed in terms of likely structures and bonding.

MODERN MATERIALS

HERE, THERE, AND EVERYWHERE. The element silicon

is the primary component of

computer processor chips and commercial solar panels.

480

W H AT ' S

A H EA D

12.1 Classes of Materials

We will characterize materials according to their predominant bonding interactions and their ability to conduct electricity. Ionic solids have localized electrons and are insulators. Covalent-network solids also have localized electrons, but in some cases conductivity can result when electrons are excited or impurities are added. Hence they are called semiconductors. The valence electrons in metallic solids are collectively shared, resulting in high electrical conductivity.

12.2 Electronic Structure of Materials

We see that the behavior of electrons in bulk solids can be explained by the rules of molecular orbital theory. The large number of atoms in a solid leads to bands of allowed energy instead of discrete molecular orbitals. The electrical and optical properties of materials depend upon the filling of these bands.

12.3 Semiconductors

We next study semiconductors, where there is an energy gap between the filled and empty bands called the band gap. The size of the band gap can be controlled by manipulating the orbital overlap and bond polarity. The conductivity of semiconductors can be controlled by intentionally adding controlled amounts of impurities called dopants. A variety of important technological devices can be built from semiconductors, including computer chips, solar cells, and light-emitting diodes (LEDs).

12.4 Ceramics

We investigate a class of materials called ceramics, which are inorganic solids, such as oxides, nitrides, carbides, and silicates, that are highly resistant to heat, corrosion, and wear. Many ceramics are stronger and lighter than metals, but their applications are somewhat limited by their brittle nature.

12.5 Superconductors

We examine materials with the remarkable ability to conduct electricity with no resistance when they are cooled to low temperature. Another unusual property of superconductors is that they expel magnetic fields.

12.6 Polymers and Plastics Next we investigate polymers-long chainlike

molecules where the motif of a small molecule is repeated many times over. We will see how the shapes, interactions, and bonding between polymer chains affect the physical properties.

12.7 Biomaterials

We recognize that materials used in biomedical applications must meet special requirements to be compatible with living organisms. We will see several examples of biomaterials and how they work in such applications as heart valves, artificial tissues, and vascular grafts.

12.8 Liquid Crystals

We will learn that instead of passing directly from the solid to the liquid phase when heated, some substances pass through an intermediate liquid crystalline phase that has some of the structural order of a solid and some of the freedom of motion possessed by liquids. Liquid crystals are widely used in electronic displays.

12.9 Nanomaterials

We will discuss how physical and chemical properties of bulk materials change when their crystals become very small. These effects begin to occur when materials have sizes on the order of 1 -100 nm. Materials that take advantage of these effects are collectively called nanomaterinls.

THE MODERN WORLD IS FULL OF "STUFF"-materials that are used to make computers, compact discs, cell phones, contact lenses, skis, furniture, and a host of other things. Chemists have contributed to the discovery and development of new materials by inventing entirely new substances and by developing the means for processing naturally occurring materials to form fibers, films, coatings, adhesives, and substances with special electrical, magnetic, or optical properties. in this chapter we will discuss classes of materials, their properties, and their applications in modern society. Our aim is to show how we can understand the physical or chemical properties of materials by applying the principles discussed in earlier chapters.

48 1

482

C HA PTER 12

Modern Materials This chapter demonstrates the important point that the observable, macro­ scopic properties of materials are the result of atomic- and molecular-level structures and processes.

1 2 . 1 CLAS SES OF MATERIALS When we say "material," we generally mean a substance or mixture of sub­ stances that is held together by strong chemical bonds throughout the whole sample-in other words, covalent-network solids, ionic solids, and metallic solids.

cxx:>

(Section 11 .8) The electrical, mechanical, optical, and magnetic prop­

erties of these substances find numerous technological uses. Because the types of materials and their applications are so diverse, we need a system for classify­ ing them. One possibility is to classify materials according to the type of bonding that holds the atoms together: ionic, covalent, or metallic. Another possible organization would be to classify materials according to their electrical conductivity: insulators, semiconductors, or conductors. Because bonding and conductivity both depend upon the behavior of the electrons, these two classi­ fication schemes are related. In ionic solids the valence electrons are localized predominantly on the an­ ions. The localized nature of the electrons leads to insulating behavior. In metallic solids the valence electrons are delocalized and shared collectively. The delocalized nature of the valence electrons makes it possible for the elec­ trons to move freely throughout the sample and is responsible for the high electrical and thermal conductivity of metals. In covalent network solids the electrons are localized in covalent bonds. This localization limits the conduc­ tivity, but in many such materials conductivity can result if light, heat, or an electrical field excites some of the electrons. Therefore, many covalent-network solids are classified as semiconductors because they have conductivities that are intermediate between metals and insulators.

In this chapter we will discuss

semiconductors and their applications in some detail. We will also consider ce­ ramics, which are an important class of materials that are normally insulators . We will defer a detailed discussion of the largest class of conductors, metals, to Chapter

23.

Examples of functional and structural materials also exist among molecular solids.

In these materials the intermolecular forces we learned about in Chapter

11-London

dispersion forces, dipole--dipole interactions,

and hydrogen

bonds-play an important role. Because intermolecular forces are relatively weak, molecular materials are often softer than materials with extended net­ works of chemical bonds. Hence, molecular materials are often referred to as "soft" materials, while extended solids are referred to as "hard" materials. Two important types of soft materials are investigated in this chapter: polymers and liquid crystals. Both are examples of molecular solids in which the intermolec­ ular forces give rise to unusual properties that are the basis of important tech­ nological applications .

1 2.2 ELECTRO N I C S TRUCTURE OF M ATERIALS To understand the relationship between conductivity and bonding, let's consider how the atomic orbitals interact in a crystalline solid.

In a crystal there are numer­

ous atoms, each with its own atomic orbitals. These atomic orbitals overlap and

combine to give numerous molecular orbitals. Nevertheless, the bonding con­ cepts that apply to a simple diatomic molecule also apply to a crystal with a gi­ gantic number of atoms. Let's briefly review some of the basic rules of molecular orbital theory.

cxx:>

(Section

9.7)

First, the atomic orbitals combine in a molecule

to make molecular orbitals that can extend over the entire molecule. Second, a given molecular orbital can contain up to two electrons, depending on the energy level of the molecular orbital and how many electrons the molecule possesses.

12.2

Electronic Structure of Materials Most antibonding

\

\ \

\

\

\

Most bonding

2

4

Many

Number of interacting atoms

Finally, the number of molecular orbitals in a molecule is equal to the number of atomic orbitals that combine to make the molecular orbitals. The electronic structures of bulk solids and small molecules have many simi­ larities as well as some important differences. To illustrate this point, consider the following thought experiment. How does the molecular orbital diagram for a chain of hydrogen atoms change as we increase the length of the chain (Figure 12.1 A)? For one hydrogen atom there is only a single half-filled 1s orbital. For two atoms the molecular orbital diagram corresponds to the molecular orbital dia­ gram of an H2 molecule that we considered in detail in Chapter 9. oco (Section 9.7) For a four-atom chain there are now four molecular orbitals. These range from the lowest energy orbital, where the orbital interactions are completely bonding (0 nodal planes), to the highest energy orbital, where all of the orbital interactions are antibonding (a nodal plane between each pair of atoms). As the length of the chain increases, the number of molecular orbitals also in­ creases. Fortunately, we can apply a few general concepts regardless of the length of the chain. First, the lowest energy molecular orbitals are always the most bonding, while the highest energy molecular orbitals are the most antibonding. In the middle, the bonding and antibonding interactions tend to balance out. Second, because each hydrogen atom has only one atomic orbital, the number of molecular orbitals is equal to the number of hydrogen atoms in the chain. Given the fact that each hydrogen atom has one electron, this means that half of the molecular orbitals will be fully occupied while the other half will be empty, regardless of chain length.* If the length of the chain becomes very large, there are so many molecular or­ bitals that the energy separation between them becomes vanishingly small, which leads to the formation of a continuous band of energy states (Figure 12.1). For the bulk solids we will discuss later in this chapter, the number of atoms is very large. Consequently, their electronic structures will consist of bands, as shown for the very long chain of hydrogen atoms, rather than discrete molecular orbitals. The electronic structures of most materials are more complicated be­ cause we have to consider more than one type of atomic orbital on each atom. As each type of orbital can give rise to its own band, the electronic structure of a solid will consist of a series of bands. Consequently, the electronic structure of a bulk solid is referred to as a band structure. 'This is only strictly truefor chains with an even number of atoms.

483



-->

u

=

=

\I \\ \\ \\ II

I I I I

II I I II

I.-1 -.I 1-2.l___- ,�, C

p 1 \ Eg 11111 '�I I C2 ' ' \ I I I \ I I I I I \I

s

=

55

/,

eV

I

'-�.---1 1--.- .l---1 ,I : 'DOls

1 / 1 ' C

I I I I I I I I I I

I f I

Valence band C (d1amond) Large overlap, large Eg

C

2

2p

(b )

r

Conduction band

/'-r-----.----c-'\ \ \ I �I I �:-;;-'----' .\ JL L_�,...-;;-' ·---'tl /

�I (\ E,Tov \/:'j:qfI 1

Si

\

\

',

I

I

,

Valence band

11

I

I

Silicon Smaller overlap, smaller Eg

Semiconductors

12.3

487

Conduction band

Conduction band

I r,'/, oo: II ;

>,

w

� "' "

'

Ge 4p

Ge 4s

'

I

I ..

II

\

I

\ \

\

I

I

\

�r.l ,

Valence band

\

\

\

\

\

\,',l ' 1 \

Ge 4p

,' 'ill Ge 4s

I

1

I I I I I

I

I 1

1 '- - -J I ----,- ----; \I I I

I I I I I I I I I I

l,-_,-1 1--,------ll-, / Ga 4p

1n1

1111' \ �I I Ga 4s l \

II

I

Eg

=

Ge

1 1

\ \I I

1

\I

�

\\ ,---, ,------, ._ .L._ -,J- _J \IJL As 4P /1

1 .43 eV

\I \I

I I I

I

_

lf1Il // /� 1

1

1

As 4s

Valence band GaAs

.A. Figure

- SAMPLE EXERCISE

12.1

Comparison of Semiconductor Band Gaps I Qualitative

Will GaP have a larger or smaller band gap than ZnS? Will it have a larger or smaller band gap than GaN? SOLUTION

Analyze: The size of the band gap depends upon the vertical and horizontal posi­ tions of the elements in the periodic table. The band gap will increase when either of the following conditions is met: ( 1 ) The elements are located higher up in the peri­ odic table, where enhanced orbital overlap leads to a larger splitting between bond­ ing and antibonding orbitals: or (2) The horizontal separation between the elements increases, whkh leads to an increase in the electronegativity difference and the bond polarity. Plan: We must look at the periodic table and compare the relative positions of the el­ ements in each case. Solve: Gallium is in the fourth period and group 3A. Its electron configuration is [Ar]3d104s 24p1 . Phosphorus is in the third period and group SA. Its electron configu­ 2 ration is [Ne]3s 3p3 . Zinc and sulfur are in the same periods as gallium and phospho­ rus, respectively. However, zinc, in group 2B, is one element to the left of gallium and sullur in group SA, is one element to the right of phosphorus. Thus we would expect the electronegativity difference to be larger for ZnS, which should result in ZnS hav­ ing a larger band gap than GaP. For both GaP and GaN the more electropositive element is gallium. So we need only compare the positions of the more electronegative elements, P and N. Nitrogen is located above phosphorus in group SA. Therefore, based on increased orbital over­ lap, we would expect GaN to have a larger band gap than GaP. Additionally, nitrogen is more electronegative than phosphorus, which also should result in a larger band gap for GaP. Check: Looking in Table 12.2 we see that the band gap of GaP is 2.26 eV. ZnS and GaN are not listed in the table, but external references show the band gaps for these compounds to be 3.6 eV for ZnS and 3.4 eV for GaN. - PRACTICE EXERCISE Will ZnSe have a larger or smaller band gap than ZnS?

Answer: Because zinc is common to both compounds and selenium is below sulfur in the periodic table, the band gap of ZnSe will be smaller than ZnS.

12.6 The relationship between bond polarity and band gap. In germanium the bonding is purely covalent. In gallium arsenide the difference in electronegativity introduces polarity into the bonds. The gallium atoms are less electronegative than germanium, which is reflected in an upward shift of the energies of the gallium atomic orbitals. The arsenic atoms are more electronegative than germanium, which is reflected in a downward shift of the energies of the arsenic atomic orbitals. The introduction of bond polarity increases the band gap from 0.67 eV for Ge to 1 .43 eV for GaAs.

488

>.

Q) c:: �

i:J'

C HA PTER

D

12

Modern Materials

D D DVB Ova EJv CB

Intrinsic

CB

n-Type

Semiconductor Doping

CB

p-Type

of impurity atoms to a material is known as doping. Consider what happens B

A Figure 1 2.7 The addition of controlled small amounts of Impurities (doping) to a semiconductor changes the electronic properties of the material. Left: A pure, intrinsic semiconductor has a filled valence band and an empty conduction band (ideally). Middle: The addition of a dopant atom that has more valence electrons than the host atom adds electrons to the conduction band (i.e., phosphorus doped into silicon). The resulting material is an n·type semiconductor. Right: The addition of a dopant atom that has fewer valence electrons than the host atom leads to fewer electrons in the valence band or more holes in the valence band (i.e., aluminum doped into silicon). The resulting material is a p·type semiconductor.

The electrical conductivity of a semiconductor is influenced by the presence of small amounts of impurity atoms. The process of adding controlled amounts when a few phosphorus atoms (known as dopants) replace silicon atoms in a silicon crystal. Phosphorus has five valence electrons; silicon has only four. Therefore, the extra electrons that come with the phosphorus atoms are forced to occupy the conduction band because the valence band is already completely filled (Figure

12.7 �. middle). The resulting material is called an n-type semicon­

ductor, "n" signifying that the number of negatively charged electrons in the conduction band has increased. These extra electrons can move very easily in the conduction band. Thus, just a few parts per million (ppm) of phosphorus in silicon can increase silicon's intrinsic conductivity by a factor of a million! The dramatic change in conductivity in response to the addition of a trace amount of a dopant means that extreme care must be taken to control the impurities in semiconductors. We will return to this point later. It is also possible to dope semiconductors with atoms that have fewer valence electrons than the host material. Consider what happens when a few aluminum atoms, in group 3A, replace silicon atoms in a silicon crystal. Aluminum has only three valence electrons compared to silicon's four. Thus, there are electron vacancies, known as

holes, in the valence band when silicon is

doped with aluminum. Since the negatively charged electron is not there, the hole can be thought of as having a positive charge. When an adjacent electron jumps into the hole, the electron leaves behind a new hole. Thus, the positive hole moves about in the lattice like a particle itself. This movement is analogous to watching people changing seats in a classroom; you can watch the people (electrons) move about the seats (atoms), or you can watch the empty seats (holes) "move." Thus, holes can conduct, too, and a material like this is called a

p-type

semiconductor, "p" signifying that the number of positive holes in the

material has increased. Like n-type conductivity, only parts-per-million levels of a p-type dopant can lead to a rnillionfold increase in the conductivity of the semiconductor-but in this case, the holes in the valence band are doing the conduction (Figure

12.7, right). The junction of an n-type semiconductor with a

p-type semiconductor forms the basis for diodes, transistors, solar cells, and other devices.

- SAMPLE EXERCISE

12.2 [ Identifying Types of Semiconductors

Which of the following elements, if doped into silicon, would yield an n-type semi­ conductor? Ga; As; C.

SOLUTION

Analyze: An n-type semiconductor means that the dopant atoms must have more valence electrons than the host material. Silicon is the host material in this case.

Plan: We must look at the periodic table and determine the number of valence elec­ trons associated with Si, Ga, As, and C. The elements with more valence electrons than silicon are the ones that will produce an n-type material upon doping.

Solve: Si is in column 4A, and so has four valence electrons. Ga is in column 3A, and so has three valence electrons. As is in column SA, and so has five valence electrons; C is in column 4A, and so has four valence electrons. Therefore, As, if doped into sil­ icon, would yield an n-type semiconductor. - PRACTICE EXERCISE Suggest an element that could be used to dope silicon to yield a p-type material. 3A. Boron and aluminum are both good choices-both are in group 3A. In the semiconductor industry boron and aluminum are commonly used dopants for silicon.

Answer: Because Si is in group 4A, we need to pick an element in group

12.3

.6. Figure

12.8 Photograph of a Pentium 4 processor computer chip. On this scale, mostly what you can see is the connections between transistors.

Semiconductors

.6. Figure 12.9 Making silicon chips. IBM's clean room for the fabrication of 300 mm silicon wafers, in East Fishkill, New York. Tom Way/Courtesy of International Business Machines Corporation. Unauthorized use not permitted.

The Sil icon Chip Silicon is the most important commercial semiconductor. It is also one of the most abundant elements on Earth. = (Section 1.2) The semiconductor industry that is responsible for the circuitry in computers, cell phones, and a host of other devices relies on silicon wafers called "chips," on which complex patterns of semicon­ ductors, insulators, and metal wires are assembled (Figure 12.8 .6.). Although the details of integrated circuit design are beyond the scope of this book, we can say that the process of making chips relies mostly on chemistry. Silicon is the semiconductor of choice because it is abundant and cheap in its raw form (obtained from sand), it can be made 99.999999999% pure in highly spe­ cialized facilities called "clean rooms" (Figure 12.9.6.), and it can be grown into enormous crystals that are nearly atomically perfect. If you recall that as little as parts-per-million level of impurities can change the conductivity of silicon by a mil­ lionfold, it becomes clear why clean rooms are necessary. In addition, the individ­ ual units that make up integrated circuits are about 50 nm wide, which makes them similar in size to an inclividual virus particle and quite a bit smaller than orclinary dust particles. Thus, to an integrated circuit, a dust particle looks like a boulder. Silicon has some other important advantages over competing semiconduc­ tors. It is nontoxic (compared to GaAs, its closest competitor), and its surface can be chemically protected by Si02o the natural product of its reaction with air. Si02 is an excellent insulator, and its atoms are in excellent registry with the underly­ ing silicon substrate, which means it is easy to grow crystalline layers of Si02 on top of Si. The basic unit of the integrated circuit, the transistor, requires a metal/insulator "gate" between a semiconductor "source" and a semiconductor "drain." Electrons move from the source to the drain when voltage is applied to the gate. That silicon's naturally occurring oxide is a very chemically stable insu­ lator provides a huge benefit to silicon technology. Solar Energy Conversion The magnitude of the band gap for semiconductors is -50 kJ/mol to -300 kJ/mol (Table 12.2)-just into the range of the bond-dissociation energies for single chemical bonds-and corresponds to photon energies of infrared, visible, and low-energy ultraviolet light. oc:c (Section 6.2) Thus, if you shine light of the appropriate wavelength on a semiconductor, you will promote electrons from bonding molecular orbitals in the valence band to antibonding molecular orbitals in the conduction band. The promotion of an electron into the empty conduction band allows it to move freely through the crystal.

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THE TRAN S I STOR

T small electric charge, a transistor controls the flow of a sub­ he transistor is the heart of integrated circuits. By applying a

stantially larger current. Therefore, the transistor controls infor­ mation flow in the form of electrons; it also amplifies the signal. One common type of transistor is the MOSFET, which stands for metal-oxide-semiconductor field-effect tra115istor. The de­ sign of a MOSFET is shown in Figure 12.10 � . A piece of silicon is p-doped to form the substrate, and then patches of n-type doping are created and connected to metal wires. The back of the sub­ strate is also connected to a metal wire. The n-type regions are the source and drain. They are separated by a p-type channel, which is about 65 nm long in a Pentium 4 processor. Deposited on top of the channel is the gate, an insulating oxide-traditionally SiOz, but it can also be silicon nitride, Si3N4- or a mixture of these two. In the Pentium 4 chip, the oxide layer is only about 2 nm, or 20 atoms, thick. A metal contact is made to this oxide gate. When a tiny positive charge is applied to the gate, electrons from the n-type source are drawn into the channel and flow to the drain. The larger the charge applied, the more the channel is "opened" and the greater the number of electrons that flow. The reverse situation, in which the source and drain are p-type and the channel is n-type, can also be made. There are many other transistor designs. The size of a MOSFET is defined by the gate channel length, which defines the distance electrons must travel to get from the source to the drain. Using Si02 as the gate oxide, the smallest circuit elements in commercial use are 65 nm wide. This means that a typical integrated circuit board holds in ex­ cess of 65 million transistors over an area the size of your finger­ nail. The electrical properties of SiOz prevent further reduction in the size of the gate channel. To continue shrinking the size of transistors scientists have turned to hafnium oxide, HfOz, as the gate oxide of choice. This materials advance has enabled chip designers to reduce the gate channel length to 45 nm, resulting

.A. Figure 12. 10 Design of a basic MOSFET. The substrate is greenish blue, the source and drain are pale gray, the channel is blue, and the gate oxide is pale brown. The blue lines and gray cylinders represent metallic contacts. A small positive voltage applied to the gate creates a positive channel through which electrons (red arrows) from the source are drawn, and go into the drain. in a twofold increase in the number of transistors that can be manufactured per unit area. The patterning of 300-mm commercial silicon wafers, which are about the size of a medium pizza, results from a series of chemical reactions that put down layers of thin films of semiconductor, insulator, and metal in patterns that are defined by a series of masks. Figure 12.11 T shows that the thickness of the layers varies from about 100 nm to 3000 nm. In the re­ search laboratory, film thicknesses have been made as thin as several atoms.

Related Exercises: 12.19, 12.20, 12.83

The First Steps in Making a Transistor: 1. Wafer is oxidized

Silicon dioxide

2. Oxidized wafer is covered

with photoresist

5. Oxide now unprotected by photoresist is etched away in hydrofluoric acid

3. Wafer is exposed to UV

light through a photomask

�l l �

4. Exposed photoresist is dissolved in developer solution

Ultraviolet radiation

Mask

6. The rest of the photoresist is removed; wafer is now ready for doping

..,. Figure 1 2. 1 1 Schematic Illustration of the process of photolithography. Typical film

thicknesses are 200 - 600 l"m for the silicon substrate; 80 nm for the silicon dioxide layer; and 0.3 - 2.5 l"m for the photoresist, which is a polymer that degrades when irradiated with ultraviolet light so it will be chemically dissolved by organic solvents in the "developer" solution. The mask protects parts of the photoresist from light and provides patterning to the surface on the -1 00-nm scale. Commercial silicon chips can have as many as 30 layers. Other forms of lithography utilize means other than light to pattern the surface (for example, ion-beam lithography and electron-beam lithography use beams of ions or electrons to 11Write" on the surface).

12.3

The same can be said for the hole that is left behind in the valence band. Hence, the Photon material becomes more conductive when irradiated by photons whose energy is greater than the band gap. This property, known as photoconductivity, is essential for many solar energy conversion applications. A solar cell is a semiconductor device Conduction that converts photons from the sun into band electricity. The heart of a solar cell is a junc­ tion between an n-type semiconductor and a p-type semiconductor (Figure 12.12,.. ). When a photon is absorbed in the region near the junction, an electron is excited from the valence band to the conduction band, creating a hole at the same time. The pho­ toexcited electron moves toward the n-type semiconductor and the hole moves toward Valence the p-type semiconductor. In this way the band energy of the photon is partially converted into an electrical current (Figure 12.13,.. ). The efficiency of a solar cell depends n-Type upon a number of factors including the semiconductor band gap. If the energy of the photon is smaller than the band gap, the photon will not be absorbed and its energy can­ not be harvested. On the other hand, if the energy of the photon is larger than the band gap energy, the excess energy of the photon is converted into thermal energy rather than electrical energy. Therefore, a tradeoff in the size of the band gap is necessary. For a solar cell made out of a single material, the maximum ef­ ficiency occurs for a material with a band gap of approximately 1.3 eV, where the conversion of optical energy into electrical energy is theoretically predicted to be as high as 31%. In the laboratory single crystal silicon solar cells have been pro­ duced with efficiencies of approximately 24%, while the efficiencies of commer­ cial silicon solar cells tend to fall closer to 15%. Power generation using solar cells has many attractive aspects. Sunlight is re­ newable, abundant, free, and widely distributed. There are no greenhouse gas emissions. The fundamental limitation is the cost of producing solar cells. Assuming a 20-year system lifetime, the present cost of electricity generated from solar cells is in the range of $0.25-$0.65 per kilowatt hour (kWh). In comparison, electricity generated from a coal-based power plant is closer to $0.04-$0.06 per kWh. This cost comparison does not take into account the long-term environ­ mental costs of burning coal, which are significant. Burning coal releases a num­ ber of undesirable pollutants into the atmosphere including sulfur dioxide and nitrogen oxides, which are responsible for acid rain oco (Section 18.4}, as well as fly ash and other particulates. The sheer volume of coal that is burned means that even elements that are present in on!y trace amounts, such as mercury, arsenic, and uranium, are released in fairly large quantities over time. In the most advanced coal-fired power plants, modern technologies are employed that significantly reduce the emission of many pollutants. However, it is not possible to remove carbon dioxide, the main product of coal burning. It is estimated that a 500 MW coal-fired power plant, which is large enough to power a city of 140,000, releases 3.7 million tons of C02 every year. Given the concerns about global warming and climate change ooc (Section 18.4}, this re­ lease of C02 will continue to be a major concern associated with generating electricity from coal. Driven in large part by environmental concerns, many governments have initiated programs to encourage development of solar power, which has led to an almost exponential growth in the market for solar cells. In 2005 the market exceeded $10 billion.

Semiconductors

491

Conduction band

Valence band

p-Type semiconductor p-n Junction

.l Figure 12.12 Absorption of light at a p-n junction. This figure illustrates

the process by which light is absorbed at the junction between a p-type and an n-type semiconductor. First, a photon is absorbed in the junction exciting an electron from the valence band into the conduction band creating an electron­ hole pair (this process is marked with a 1 ). Next the electron (e-) is attracted to the n-type semiconductor and the hole (h+) toward the p-type semiconductor (this process is marked with a 2). In this way the energy of the photon can be converted into electrical energy.

.l

Figure

1 2. 1 3 Electricity from

Solar panels, made of silicon, are used both as an energy source and as architectural elements in this apartment building in southern California. sunlight.

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Semiconductor Light-Emitting Diodes

� 7

p-Type Junction n-Type

Light

A Figure 1 2.1 4

Light emitting

Top: The heart of a light emitting diode is a p-n junction where an applied voltage drives electrons and holes to meet. Bottom: The color of light emitted depends upon the band gap of the semiconductor used to form the p-n junction. For display technology red, green, and blue are the most important colors because all other colors can be made by mixing th ese colors . diodes.

Light-emitting diodes (LEDs) are used as displays in clocks; brake lights, tum signals, and dashboards of some cars; as traffic lights; and in many other places. There is a growing interest to replace incandescent and fluorescent lights with white LEDs because of their efficiency and long lifetimes, which are approximately 25 times longer than incandescent lights and roughly twice as long as fluorescent lights. Their efficiency is currently 2-3 times greater than incandescent lights and competitive with fluorescent lights. The mechanism by which LEDs work is the opposite of the mechanism operating in solar energy conversion devices. In an LED, a small voltage is applied to a semiconductor device that, like a solar cell, has a junction between an n-type semiconductor and a p-type semiconductor. The junction forms a p-n diode, in which electrons can flow only one way. When a voltage is applied, electrons in the conduction band from the n-side are forced to the junction where they meet holes from the p-side. The electron falls into the empty hole, and its energy is converted into light whose photons have energy equal to the band gap (Figure 12.14 .,.. ) . In this way electrical energy is converted into optical energy. Because the wavelength of light that is emitted is inversely proportional to the band gap of the semiconductor, the color of light produced by the LED can be controlled by appropriate choice of semiconductor. Most red LEDs are made of a 9 mixture of GaP and GaAs. The band gap of GaP is 2.26 eV (3.62 x 10-1 J), which corresponds to a green photon with a wavelength of 549 nm. cr.o (Section 6.1) On the other hand, GaAs has a band gap of 1.43 eV (2.29 X 10- 1 9 J), which corre­ sponds to an infrared photon with a wavelength of 867 nm. By forming solid solutions of these two compounds, with stoichiometries of GaP1 -xAsx, the band gap can be adjusted to any intermediate value. Thus, GaP1 -xAsx is the solid solu­ tion of choice for red, orange, and yellow LEDs. Green LEDs are made from mix­ tures of GaP and AlP (E g = 2.43 eV). Blue LEDs, made from GaN (E g = 3.4 eV) and InN (E g = 2 .4 eV), are beginning to appear in consumer products as well. White light can be produced from LEDs using different approaches. In some cases, light is combined from blue, green, and red LEDs. More commonly a blue LED is coated with a material that converts some of the blue light into yellow light. In either case the combined colors appear white to the eye. - SAMPLE EXERCISE

12.3 [ Designing an LED

Green light-emitting diodes can be made from a solid solution of GaP and AlP. These two compounds have band gaps that are 2.26 eV and 2.43 eV, respectively. If we assume that the band gap of a Ga1-xAlxP solid solution varies linearly from GaP to AlP, what composition would be needed for the emitted light to have a wavelength of 520 nm? SOLUTION Analyze: The wavelength of the emitted light has an energy that is nearly equal to the band gap of the semiconductor. The band gap depends upon the composition. Plan: First we must convert the desired wavelength of emitted light, 520 nm, to an energy in eV. Next we must estimate the value of x that gives a band gap of this size. Solve: The wavelength of light can be determined from Equation 6.3:

=A= 520 x 10-" m 1 eV E = (3.82 X 10-19 J) X = 2.38 eV 1.602 X 10 19 J E

he

(6.626 x 10-34 J-s)(3.00 x 10 8 m/s)

The band gap changes linearly from 2.26 eV for GaP to 2.43 for AlP. Therefore, we can estimate the band gap of any composition in the Gal-xAlxP solid solution from the following expression:

E

=

2.26 eV

+

x(2.43 eV

-

2.26 eV)

=

(2.26 + 0.17x) eV

12.4 We can solve for x by rearranging the equation and inserting the desired value of the band gap: E - 2.26 2.38 - 2.26 = 0.71 X = --- = 0.17 0.17 Thus, the desired composition would be G Ti(OCH2CH3l4(s) + 2 H2{g)

[12.1]

The alkoxide product, Ti(OCH2CH3)4, is dissolved in an appropriate alcohol solvent. Water is then added, and it reacts with the alkoxide to form Ti-OH groups and regenerate ethanol. Ti(OCH2CH3l4(soln) + 4 H20(1) --> Ti(OHJ4(s)

+

4 CH3CH20H(I)

[12.2]

Even though the ethanol is simply regenerated, the initial reaction with ethanol is important because the direct reaction of Ti(s) with H20(1) leads to a complex mixture of titanium oxides and titanium hydroxides. The intermediate forma­ tion of Ti(OC2H5)4(s) ensures that a uniform suspension of Ti(OH)4 will be formed. The Ti(OH)4 is present at this stage as a sol, a suspension of extremely small particles. The acidity or basicity of the sol is adjusted to split out water from between two of the Ti -OH bonds. (HO)JTi- 0- H(s) + H - 0- Ti(OH)J(s) --> (HO)JTi- 0 - Ti(OHh(s) + H20(I)

[12.3]

This is an example of a condensation reaction. We will learn more about con­ densation reactions when discussing polymers. Condensation also occurs at some of the other OH groups bonded to the central titanium atom, producing a

12.5

Superconductors

495

three-dimensional network. The resultant material, called a gel, is a porous, interconnected network of extremely small particles with the consistency of gelatin. When this material is heated carefully at 200 °C to 500 °C, all the liquid is removed, and the gel is converted to a finely divided metal oxide powder with particles in the range of 3 to 100 nm in diameter. These particles might be used in an application where small, uniform particle size is required, such as the Ti02 solar cells discussed in Chapter 9 (Section 9.8: "Chemistry Put to Work: Orbitals and Energy"), or they might be used as the starting point in the sinter­ ing process. Figure 12.15 11> shows Si02 particles, formed into remarkably uni­ form spheres by a precipitation process similar to the sol-gel process.

1 2 . 5 SUPERC O N D U CTORS Even metals are not infinitely conductive; there is some resistance to electron .i. Figure 1 2.1 5 Uniformly sized flow due to the vibrations of atoms and the presence of impurities and defects. spheres of amorphous silica, 5102• These are formed by precipitation from a However, in 1911 the Dutch physicist H. Kamerlingh Onnes discovered methanol solution of Si(OCH3)4 upon that when mercury is cooled below 4.2 K, it loses all resistance to the flow of addition of water and ammonia. The an electrical current. Since that discovery, scientists have found that many average diameter is 550 nm. substances exhibit this "frictionless" flow of electrons, known as supercon­ ductivity. Substances that exhibit superconductivity do so only when cooled below a particular temperature called the superconducting transition temperature, T,. The observed values of T, are gen­ TABLE 1 2.4 • Superconducting Materials: erally very low. Table 12.4 11> lists numerous superconducting Dates of Discovery and Transition Temperatures materials, the year of their discovery, and their transition tem­ T, (K) Substance Discovery Date perature. Some are notable for their relatively high T0 and oth­ ers for the mere fact that a material of that composition would 1911 4.0 Hg superconduct at all. It is surprising to see that the materials 1.5 1933 NbO with the highest transition temperatures are ceramics rather 1941 16.1 NbN than metals. Until a few decades ago most scientists working Nb3Sn 18.0 1954 in the area of superconductivity concentrated their studies on Nb3Ge 22.3 1973 metallic elements and compounds. BaPbl-xBix03 13 1975 Superconductivity has tremendous economic potential. If 1986 Laz-xBa.Cu04 35 1987 95 YBazCu307 electrical power lines or the conductors in a variety of electrical 1988 95 Bi2Sr3_.ca.Cu20s+y devices were capable of conducting current without resistance, 125 1988 TlzBazCazCu30 10 enormous amounts of energy could be saved. Unfortunately, 133 1993 HgBazCazCu30s+x this advantage is offset by the need to cool the superconduct­ 1993 138 Hgo.sTlo.zBazCazCu308.33 ing wires to a temperature below T, (Figure 12.16 T). However, 40 1995 Cs3C60 superconducting power lines have the advantage that they 39 MgB2 2001 can carry 2-5 times more current than a conventional power ... Figure 1 2.1 6 Superconducting power transmission cable. The

/

Pipe coating

/

Electrical insulation

/ Superconductor

superconducting wires surround a hollow core through which liquid nitrogen flows, cooling the superconductor to the zero resistance state. An electrical insulator surrounds the superconducting wires.

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.6. Figure 1 2. 1 7 Magnetic levitation. A small permanent magnet is levitated by its interaction with a ceramic superconductor that is cooled to liquid nitrogen temperature, 77 K. The magnet floats in space because the superconductor excludes magnetic field lines, a property known as the Meissner effect.

.6. Figure 1 2. 18 Superconductlon In action. High speed superconductive magnetic levitation (maglev) train. Shown is the ML002 experimental train being tested in Japan.

transmission cable in the same volume of space. This characteristic makes them attractive alternatives in crowded urban areas where power distribution demands are pushing the capacity of the system, but the population density makes it very expensive to dig up the streets to install new lines. Prototype un­ derground superconducting transmission lines installed in the United States and Japan have demonstrated the potential of this technology. In addition to their remarkable electrical properties, superconducting ma­ terials exhibit a property called the Meissner effect (Figure 12.17 .6.}, in which they exclude all magnetic fields from their volume. The Meissner effect is being explored for high-speed trains that are magnetically levitated "maglev" (Figure 12.18 .6.}. Because superconductivity appears in most materials only at very low temperatures, applications of this phenomenon to date have been limited. One important use of superconductors is in the windings of the large magnets that form the magnetic fields needed in magnetic resonance imaging (MRl} instruments, used in medical imaging (Figure 12.19 T). The magnet

II> Figure 1 2. 19 A magnetic resonance Imaging (MRI) machine used In medical diagnosis. The

magnetic field needed for the procedure is generated by current flowing in superconducting wires, which must be kept below their superconducting transition temperature, T,, of 1 8 K. This low temperature requires liquid helium as a coolant.

12.5

Superconductors

497

windings, typically formed from Nb3Sn, must be kept cooled with liquid heli­ um, whlch boils at about 4 K. The cost of liquid helium is a significant factor in the cost of using an MRI instrument. Ceramic Su perconductors Before 1986 the hlghest value observed for T, was about 22 K for the compound Nb3Ge (Table 12.4). Whlle most of the known superconductors at that time were elements and compounds containing only metals, superconductivity at a re­ spectable T, of 13 K had been found a decade earlier in an oxide material, BaBi1-xPbx03 by A. W. Sleight and coworkers at DuPont in Wilmington, Delaware. In 1986, working on a structurally related oxide system, J. G. Bednarz and K. A. Milller, at the IBM research laboratories in Zurich, Switzerland, discov­ ered superconductivity above 30 K in a ceramic oxide containing lanthanum, bar­ ium, and copper. That discovery, for whlch Bednarz and Muller received the Nobel Prize in Physics in 1987, set off a flurry of research activity all over the world. Before the end of 1987 scientists had verified the onset of superconductiv­ ity at 95 K in yttrium-barium-copper oxide, YBa2Cup7. The highest temperature observed to date for onset of zero resistance at 1 atm pressure is 138 K, which was achleved in another complex copper oxide, Hgo.sTlo.2Ba2Ca2Cu30s.33· The discovery of so-called high-temperature (high-T,) superconductivity is of great significance. Because maintaining extremely low temperatures is costly, many applications of superconductivity will become feasible only with the development of usable hlgh-temperature superconductors. Liquid nitrogen is the favored coolant because of its low cost (it is cheaper than milk), but it can cool objects to only 77 K. The only readily available safe coolant at tempera­ tures below 77 K is liquid helium, whlch costs as much as fine wine. Thus, many practical applications of superconductors, such as the power transmis­ sion cables discussed in the previous section, will become commercially viable only if T, is well above 77 K. What is the mechanism of superconductivity in the materials with unex­ pectedly high values of T,? The answer to thls question is still being vigo­ rously debated. One of the most widely studied ceramic superconductors is YBa2Cu307, whose structure is shown in Figure 12.20 �. The copper-<Jxygen planes present in YBaCu307 (marked in transparent blue) are present in all of the high-T, oxide superconductors containing copper and oxygen in Table 12.4, as well as many others. Extensive work has gone into studying these materials. Thls work indicates that the copper-<Jxygen planes are responsible for the su­ perconductivity, as well as the relatively hlgh metallic-type conductivity ob­ served at temperatures above T0 where the electrical conductivity parallel to 4 the copper-<Jxygen planes is 10 times greater than in the perpendicular direc­ 2 tion. The Cu + ions have an [Ar]3JI electron configuration with a single elec­ tron in the 3dx'- t/ orbital. -[CH=CH],Note that all atoms originally present in acetylene end up in the polyacetylene product. (c) We can use the ideal-gas equation as follows: PV = nRT

(1.00 atrn)(5.00 L) = n(0.08206 L atrn/K mol)(298 K) n = 0.204 mol Acetylene has a molar mass of 26.0 g/mol; therefore, the mass of 0.204 mol is (0.204 mol)(26.0 g/mol) = 5.32 g acetylene Note that from the answer to part (b), all the atoms in acetylene go into polyacety­ lene. Due to conservation of mass, then, the mass of polyacetylene produced must also be 5.32 g, if we assume 100% yield. (d) Let's consider the case for n = 1. We note that the reactant side of the equation in part (b) has one C == C triple bond and two C- H single bonds. The product side of the equation in part (b) has one C = C double bond, one C-C single bond (to link to the adjacent monomer), and two C- H single bonds. Therefore, we are breaking one C==C triple bond and are forming one C=C double bond and one C- C single bond. Accordingly, the enthalpy change for polyacetylene formation is:

!lH,xn

= (C==C triple bond enthalpy) - (C=C double bond enthalpy)

!lH,xn

= (839 k)/mol) - (614 k)/mol) - (348 k)/mol)

- (C-C single bond enthalpy) = - 123 k)/mol

Because aH is a negative number, the reaction releases heat, and is exothermic.

CH A PTER REVIEW SUM MARY AND KEY TERMS Section 1 2. 1 In this chapter we consider the classifica­ tion of materials according to their bonding (ionic, metal­ lic, or covalent) and electrical conductivity (insulators, metals, or semiconductors). Ionic solids are typically insulators. Metallic solids are typically good conductors of electricity. Covalent-network solids can be either semi­ conductors or insulators.

Section 1 2.2 In extended solids, continuous bands rather than discrete molecular orbitals are the best way to picture electron energy levels. Metals have a partially filled band, which makes it easy for electrons to move between filled and empty molecular orbitals in the solid. Semi­ conductors and insulators have an energy gap known as the band gap that separates the filled valence band and

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In such materials electrons

Section 1 2.6 Polymers are molecules of high molecular

have to be excited over the band gap for the material to

mass formed by joining together large numbers of small

conduct electricity.

molecules, called monomers. In an addition polymeriza­

Section 1 2.3 Elemental semiconductors contain a sin­ gle element while compound semiconductors contain

existing

the empty conduction band.

two or more elements.

In a typical semiconductor the va­

lence band is made up of bonding molecular orbitals and the conduction band of antibonding molecular orbitals. The size of the band gap increases (a) as the bond distance decreases, due to increased orbital overlap, and

(b) as the

difference in electronegativity between the two elements increases, due to increased bond polarity.

Doping semiconductors changes their ability to con­ duct electricity by orders of magnitude. An n-type semi­ conductor is one that is doped so that there are excess electrons in the conduction band; a p-type semiconductor is one that is doped so that there are excess holes in the valence band. The reasons for silicon's preeminence in electronics are its abundance, low cost, and ability to be grown into highly pure crystals, as well as the ability of its native oxide to grow on it in perfect registry with the atoms un­ derneath. The design of the transistor, the heart of inte­ grated circuitry, relies on the junction between n-type and p-type silicon. Solar cells are also based on a junction between an n-type and a p-type material. In a solar cell, photons with energies larger than its band gap energy can be absorbed near the p-n junction leading to the conver­ sion from optical energy to electrical energy. Light emitting diodes (LEOs) operate like a solar cell in reverse. Electrons in the conduction band from the n-type semiconductor are forced to the junction where they meet holes from the p-type semiconductor. The electron falls into the empty hole and its energy is converted into

tion reaction, the molecules form new linkages by opening 7T

bonds. Polyethylene forms, for example, when

the carbon-carbon double bonds of ethylene open up. In a

condensation polymerization reaction, the monomers are joined by splitting out a small molecule from between

them. The various kinds of nylon are formed, for example, by removing a water molecule from between an amine and

a carboxylic acid. A polymer formed from two different monomers is called a copolymer.

Plastics are materials that can be formed into various shapes, usually by the application of heat and pressure.

Thermoplastic polymers can be reshaped, perhaps through heating, in contrast to thermosetting plastics, which are formed into objects through an irreversible chemical process and cannot readily be reshaped. An elastomer is a material that exhibits elastic behavior; that is, it returns to its original shape following stretching or bending. Polymers are largely amorphous, but some materials

possess a degree of crystallinity. For a given chemical

composition, the crystallinity depends on the molecular mass and the degree of branching along the main polymer

chain. High-density polyethylene, for example, with little side-chain branching and a high molecular mass, has a higher degree of crystallinity than low-density polyethyl­ ene, which has a lower molecular mass and a relatively high degree of branching. Polymer properties are also strongly affected by cross-linking, in which short chains of atoms connect the long polymer chains. Rubber is cross­ linked by short chains of sulfur atoms i n the process called

vulcanization. Section 1 2.7

A

biomaterial is any material that has a

light whose photons have energy equal to the band gap.

biomedical application. Biomaterials are typically in con­

Section 1 2.4 Ceramics are inorgani c solids with gener­

patible, which means that they are not toxic, nor do they

ally high thermal stability, usually formed through three­ dimensional network bonding. The bonding in ceramics may be either covalent or ionic, and ceramics may be crys­ talline or amorphous. The sol-gel process plays an impor­ tant role in the processing of commercial ceramics. The process begins with the formation of a sol, a suspension of very small, uniformly sized particles, which is then trans­ formed to a gel, where the particles form an interconnect­ ed network. Finally, the small particles are compressed and heated at high temperature. They coalesce through a

tact with body tissues and fluids. They must be biocom­ cause an inflammatory response. They must meet physi­ cal requirements, such as long-term reliability, strength, and flexibility or hardness, depending on the application. They must also meet chemical requirements of nonreac­ tivity in the biological environment, or of biodegradabili­ ty. Biomaterials are commonly polymers with special properties matched to the application.

Section 1 2.8

A liquid crystal is a substance that exhibits

one or more ordered phases at a temperature above the

process known as sintering.

melting point of the solid. In a nematic liquid crystalline phase the molecules are aligned along a common direc­

Section 1 2.5 A superconductor is a material that is

tion, but the ends of the molecules are not lined up.

capable of conducting an electrical current without any ap­

smectic liquid crystalline phase the ends of the molecules

In a

parent resistance when cooled below its superconducting

are lined up so that the molecules form sheets. Nematic

transition temperature, Tc. Since discovery of the phe­ nomenon in 1911, many elements and compounds have

and smectic liquid crystalline phases are generally com­ posed of molecules with fairly rigid, elongated shapes,

been found to be superconducting when cooled to very

with polar groups along the molecules to help retain rela­

low temperatures. Superconducting ceramics such as

tive alignment through dipole-dipole interactions. The

YBa2Cu307 are capable of superconductivity at tempera­ tures higher than that for any nonceramic superconductor.

cules that align as in nematic liquid crystalline phases, but

cholesteric liquid crystalline phase is composed of mole­

Visualizing Concepts with each molecule twisted with respect to to form a helical structure.

its neighbors,

Section 1 2.9 Nanotechnology is the manipulation of

energy

becomes size-dependent.

519

Metal nanoparticles

have different chemical and physical properties in the 1-100-nm size range. Gold, for example, is more reactive and no longer has a golden color. Carbon nanotubes

matter, and fabrication of devices, on the 1-100 nm scale.

are sheets of graphite rolled up, and they can behave

The llllusual properties that semiconductors, metals, and

as either semiconductors or metals depending on how

carbon have on the size scale of 1-100 nm are explored.

the sheet was rolled. Applications of these

Quantum dots are semiconductor particles with diame­

are being

ters of 1-10 nm. In this size range the material's band gap

developed

now

nanomaterials

for imaging,

electronics,

and medicine.

KEY SKILLS • Understand the relationship between the type of bonding that holds atoms together in a material and the electrical conductivity of the material. • Understand how molecular orbitals combine to form bands and be able to differentiate the electronic band struc­ tures of metals, semiconductors, and insulators. • Be able to recognize elemental and compolllld semiconductors from empirical formulas. Be able to use the period­ ic table to qualitatively compare the band gap energies of semiconductors. • Understand how n-type and p-type doping can be used to control the conductivity of semiconductors and how the most important semiconductor devices, the silicon chip, LEDs, and solar cells, operate. • Understand how forming solid solutions can control the band gaps of semiconductors. • Be familiar with the bonding and properties of ceramics. • Understand how the electrical and magnetic properties of superconductors differ from ordinary conductors. Understand the differences between conventional superconductors and high Tc superconductors.

• Understand how polymers are formed from monomers. Be able to recognize the features of a molecule that allow it to react to form a polymer. Understand the differences between addition polymerization and condensation polymerization. • Understand how the interactions between polymer chains impact the physical properties of polymers. • Be familiar with the characteristics and applications of biomaterials. • Understand how the molecular arrangements characteristic of nematic, smectic, and cholesteric liquid crystals dif­ fer from ordinary liquids and from each other. Be able to recognize the features of a molecule that favor formation of a liquid crystalline phase. • Understand how the properties of bulk semiconductors and metals change as the size of the crystals decreases into the nanometer length scale. • Be familiar with the structures and unique properties of carbon nanotubes.

VISUALIZING CONCEPTS 12.1 Based on the following band structures, which material is a metal? [Section 12.2]

E

B D D

Material A

Material B

12.2 The band structures below both belong to semiconduc­ tors. One is AlAs and the other GaAs. Which is which? [Section 12.3]

E

D D

Material X

D

D

Material y

C HA P T E R

520

12

Modern Materials

12.3 Shown below are cartoons of two different polymers. Based on these cartoons, which polymer would you ex­ pect to be denser? Which one would have the higher melting point? [Section 12.6]

conductor, and a superconductor. Which plot corre­ sponds to which material? (Hint: Remember in a semi­ conductor thermal energy is needed to excite electrons from the valence band to the conduction band.) [Sections 12.1, 12.3, and 12.5]

8

]

y

.s. .c

6 X 10-5 ,--.,.-----;--,

4 X 10-5

:� 2 X 10-5 (a)

(b )

12.4 Which picture best represents molecules that are in a liq­ uid crystalline phase? [Section 12.8]

�

§

Q

X 10+0 '----...!.----'!...------'4 30 20 10 0 0 Temperature, (K)

1 X 10+3 ,..-----,

] 1 X 10+ 1 0

; 1 X 10- 1 (a)

·:;: · t-''....,----1--- 1 1 X 10-3 1----1--·;;; ;;; Q)

(b )

12.5 Consider the molecules shown below. (a) Which of these would be most likely to form an addition polymer? (b) Which would be most likely to form a condensation polymer? (c) Which would be most likely to form a liq­ uid crystal? [Sections 12.6 and 12.8]

() (i)

� (ii)

H

eN

I

0

II

H N-C-C-OH

,

I

H (iii)

12.6 The three plots to the right show the conductivity as a function of temperature for three different types of mate­ rials: a semiconductor with a small band gap, a metallic

� 1 X 10-5 '----...!....----'--' 100 150 200 250 300

§

t -t� -;-,___ 1 _.. E

Temperature, (K)

§

-L__,-----,

2 X 10-6 '--,----

! 1 X 10-6

___j

I

/

.£ 8 X 10-7 ·E ·� 4 X 10-7 & 0 X lO+O '--"""'---------' 0 50 100 150 200 250 300 Temperature, (K)

EXERC ISES

Classes of Materials and Electronic Structure of Materials 12.7 Classify each of the following materials as metal, semi­ conductor, or insulator: (a) GaN (b) B (c) ZnO (d) Pb

12.10 Repeat exercise 12.9 for a linear chain of eight hydrogen atoms.

12.9 The molecular orbital diagrams for linear chains of hydro­ gen atoms with two and four atoms in the chain are shown in Figure 12.1. Construct a molecular orbital diagram for a chain containing six hydrogen atoms and use it to answer the following questions. (a) How many molecular orbitals are there in the diagram? (b) How many nodes are there in the lowest energy molecular orbital? (c) How many nodes are there in the highest energy molecular orbital? (d) How many nodes are there in the highest occupied molecular orbital (HOMO)? (e) How many nodes are there in the lowest unoccupied molecular orbital (LUMO)?

12.12 State whether each statement is true or false, and why. (a) A typical band gap energy for an insulator is 400 kJ/mol. (b) The conduction band is higher in energy than the valence band. (c) Electrons can conduct well if they are in a filled va­ lence band. (d) Holes refer to empty atomic sites in a solid crystal.

12.8 Classify each of the following materials as metal, semi­ conductor, or insulator: (a) lnAs (b) MgO (c) HgS (d) Sn

12.11 State whether each statement is true or false, and why. (a) Semiconductors have a larger band gap than insulators. (b) Doping a semiconductor makes it more conductive. (c) Metals have delocalized electrons. (d) Most metal oxides are insulators.

Exercises

521

Semiconductors 12.13 For each of the following pairs of semiconductors, which one will have the larger band gap: (a) CdS or CdTe (b) GaN or InP (c) GaAs or InAs? 12.14 For each of the following pairs of semiconductors, which one will have the larger band gap: (a) InP or InAs (b) Ge or AlP (c) Agl or CdTe? 12.15 lf you want to dope GaAs to make an n-type semicon­ ductor with an element to replace Ga, which element(s) would you pick?

12.16 If you want to dope GaAs to make a p-type semiconduc­ tor with an element to replace As, which element(s) would you pick?

12.17 What advantages does silicon have over other semicon­

12.22 Cadmium telluride is an important material for solar cells. (a) What is the band gap of CdTe? (b) What wave­ length of light would a photon of this energy corre­ spond to? (c) Draw a vertical line at this wavelength in the figure shown with exercise 12.21, which shows the light output of the Sun as a function of wavelength.

(d) With respect to silicon, does CdTe absorb a larger or smaller portion of the solar spectrum? 12.23 The semiconductor GaP has a band gap of 2.2 eV. Green LEDs are made from pure GaP. What wavelength of light would be emitted from an LED made from GaP?

12.24 The first LEDs were made from GaAs, which has a band gap of 1 .43 eV. What wavelength of light would be emit­

12.18 Why is it important for Si crystals to be 99.999999999% pure, as opposed to 99% pure, for silicon chips?

ted from an LED made from GaAs? What region of the electromagnetic spectrum does this light correspond to: UV, Visible, or IR?

12.19 What material is traditionally used to make the gate in a

12.25 GaAs and GaP make solid solutions that have the same

ductors for use in integrated circuits?

MOSFET transistor? What material is used in the next generation transistors?

12.20 Look up the diameter of a silicon atom, in A. The chan­ nel length in a Pentium 4 processor chip is 65 nm long. How many silicon atoms does this correspond to?

12.21 Silicon has a band gap of 1.1 eV at room temperature. (a) What wavelength of light would a photon of this energy correspond to? (b) Draw a vertical line at this wavelength in the figure shown, which shows the light output of the sun as a function of wavelength. Does sili­ con absorb all, none or a portion of the visible light that comes from the sun?

crystal structure as the parent materials, with As and P randomly distributed through the crystal. GaPxASt-x exists for any value of x. lf we assume that the band gap

varies linearly with composition between x = 0 and x = 1, estimate the band gap for GaP05As05. What wave­ length of light does this correspond to?

12.26 Red light-emitting diodes are made from GaAs and GaP solid solutions, GaPxASt -x (see exercise 12.25). The origi­ nal red LEDs emitted light with a wavelength of 660 nm . If we assume that the band gap varies linearly with com­ position between x = 0 and x = 1, estimate the composi­ tion (the value of x) that is used in these LEDs.

outside atmosphere

Wavelength, m

I 2.6 X 10-6 10-6

3 X 10-6

Ceramics 12.27 Metals, such as AI or Fe, and many plastics are recyclable.

With the exception of many glasses, such as bottle glass, ce­ ramic materials in general are not recyclable. What charac­ teristics of ceramics make them less readily recyclable?

12.28 It is desirable to construct automobiles out of light­ weight materials to maximize fuel economy. All of the ceramics listed in Table 12.3 are less dense than steel. Why do you think ceramic materials are not more wide­ ly used in the construction of automobiles?

12.29 Why is the formation of very small, uniformly sized and shaped particles important for many applications of ce­ ramic rnaterials?

12.30 Describe the general chemical steps in a sol-gel process, beginning with Zr(s) and CH3CH20H(/). Indicate whether

each step is an oxidation-reduction reaction (refer to Sec­ tion 4.4), condensation reaction, or other process.

12.31 The hardnesses of several substances according to a scale known as the Knoop value are as follows:

Substance

Knoop Value

Ag CaC03 MgO Soda-lime glass Cr ZrB2 AI203 TaC

60 135 370 530 935 1550 2100 2000

522

C HA P T E R

12

Modern Materials

Which of the materials in this list would you classify as a ceramic? What were your criteria for making this clas­ sification? Does classification as a ceramic correlate with Knoop hardness? If you think it does, is hardness alone a sufficient criterion to determine whether a substance is a ceramic? Explain. 12.32 Silicon carbide, SiC, has the three-dimensional structure shown in the figure.

(a) Name another compound that has the same structure. (b) Would you expect the bond­ ing in SiC to be predominantly ionic, metallic, or covalent? (c) How do the bonding and structure of SiC lead to its high thermal stability (to 2700 °C) and exceptional hardness?

Superconductors 12.33 To what does the term superconductivity refer? Why might superconductive materials be of value? 12.34 What are the differences in the electrical and magnetic properties of an excellent metallic conductor of electricity (such as silver) and a superconducting substance (such as Nbpn) below its superconducting transition temperature? 12.35 The following graph shows the resistivity (a measure of electrical resistance) of MgB2 as a function of tempera­ ture in the region from about 4 K to 100 K. What is the significance of the sharp drop in resistivity below 40 K?

100

��

-

-

-

-

-

0

-

0

I

I

20

I

I

I

40

60

I 80

I

100

12.36 (a) What is the superconducting transition temperature, T,? (b) The discovery by Muller and Bednarz of super­ conductivity in a copper oxide ceramic at 35 K set off a frantic scramble among physicists and chemists to find materials that exhibit superconductivity at higher tem­ peratures. What is the significance of achieving T, val­ ues above 77 K? 12.37 Explain how the Meissner effect can be used to levitate trains. What can the tracks and train wheels be made of, and which one would be more likely to be cooled? 12.38 The group 4B metal nitrides (TiN, ZrN, and HfN) as well as the group SB metal nitrides (VN, NbN, and TaN) are all superconductors at low temperature. Niobium(III) nitride, which has the highest T0 superconducts below 16.1 K. All of these compounds have crystal structures that are analogous to the sodium chloride structure. Scandium nitride also adopts the sodium chloride struc­ ture, but it is not a superconductor. (a) At room tempera­ ture will NbN be a metallic conductor or an insulator? (b) At room temperature will SeN be a metallic conductor or an insulator? (c) Why do you think the properties of SeN are so different than the group 4B and SB metal nitrides? (Hint: Consider the electron configuration of the metal cation.)

Temperature, (K)

Polymers 12.39 What is a monomer? Give three examples of monomers, taken from the examples given in this chapter. 12.40 The structure of decane is shown in Practice Exercise 12.5. Decane is not considered a polymer, whereas poly­ ethylene is. What is the distinction? 12.41

An ester is a compound formed by a condensation reac­ tion between a carboxylic acid and an alcohol. Use the index to find the discussion of esters in Chapter 25, and give an example of a reaction forming an ester. How might this kind of reaction be extended to form a poly­ mer (a polyester)?

12.42 Write a chemical equation for formation of a polymer via a condensation reaction from the monomers succinic acid (HOOCCH2CH2COOH) and ethylenediamine (H2NCH2CH2NHz).

12.43 Draw the structure of the monomer(s) employed to form each of the following polymers shown in Table 12.5 (a) polyvinyl chloride, (b) nylon 6,6, (c) polyethyl­ ene terephthalate. 12.44 Write the chemical equation that represents the forma­ tion of (a) polychloroprene from chloroprene CH2=CH-C=CH2

I

Cl (Polychloroprene is used in highway-pavement seals, expansion joints, conveyor belts, and wire and cable jackets.); (b) polyacrylonitrile from acrylonitrile CH2=CH

I

CN (Polyacrylonitrile is used in home furnishings, craft yarns, clothing, and many other items.)

Exercises

U.45 The nylon Nomex® , a condensation polymer, has the

'@( '@

In this structure, R represents - H, -CH 3, or another group of atoms. Draw the general structure for a polyamino acid polymer formed by condensation poly­ merization of the molecule shown here.

following structure:

fo o II

C

II

'

NH

NH

1

"

Draw the structures of the two monomers that yield ® Nomex .

12.46 Proteins are polymers formed by condensation reactions of amino acids, which have the general structure H

I

R

I

0

II

523

U.47 What molecular features make a polymer flexible? Explain how cross-linking affects the chemical and physical properties of the polymer.

12.48 What molecular structural features cause high-density polyethylene to be denser than low-density polyethylene?

12.49

Are high molecular masses and a high degree of crys­ tallini ty always desirable properties of a polymer? Explain.

12.50 Briefly describe each of the following: (a) elastomer, (b) thermoplastic, (c) thermosetting plastic, (d) plasticizer.

H-N-C-C-0-H

I

H

Bio materials U.51 Neoprene is a polymer of chlorobu tadiene. CH2=C-C=CH2

� d

l

The polymer can be used to form flexible tubing that is resistant to chemical attack from a variety of chemical reagents. Suppose it is proposed to use neoprene tubing as a coating for the wires running to the heart from an implanted pacemaker. What questions would you ask to determine whether it might be suitable for such an application?

12.52 On the basis of the structure shown in Table 12.5 for polystyrene and polyurethane, which of these two class­ es of polymer would you expect to form the most effec­ tive interface with biological systems? Explain.

12.53 Patients who receive vascular grafts formed from poly­

® mer material such as Dacron are required to take anti­ coagulation drugs on a continuing basis to prevent blood clots. Why? What advances in such vascular im­ plants are needed to make this precaution unnecessary?

12.54 Several years ago a biomedical company produced and marketed a new, efficient heart valve implant. It was later withdrawn from the market, however, because pa­ tients using it suffered from severe loss of red blood cells. Describe what properties of the valve could have been responsible for this result.

12.55 Skin cells from the body do not differentiate when they are simply placed in a tissue culture medium; that is, they do not organize into the structure of skin, with dif­ ferent layers and different cell types. What is needed to cause such differentiation to occur? Indicate the most important requirements on any material used.

12.56 If you were going to attempt to grow skin cells in a

medium that affords an appropriate scaffolding for the cells and you had only two fabrics available, one made from polystyrene and the other from polyethyl­ eneterephthalate (Table 12.5), which would you choose for your experiments? Explain.

-------

Liquid Crystals

12.57 In what ways are a nematic liquid crystalline phase and an ordinary liquid the same, and in what ways do they differ? 12.58 In contrast to ordinary liquids, liquid crystals are said to possess "order." What does this mean?

12.59 Describe what is occurring at the molecular level as a substance passes from the solid to the nematic liquid crystalline to the isotropic (normal) liquid phase upon heating.

12.60 What observations made by Reinitzer on cholesteryl benzoate suggested that this substance possesses a liq­ uid crystalline phase?

12.61 The molecules shown in Figure 12.39 possess polar groups (that is, groupings of atoms that give rise to sizable

dipole moments within the molecules). How might the presence of polar groups enhance the tendency toward liquid crystal formation?

12.62 Liquid crystalline phases tend to be more viscous than the isotropic, or normal, liquid phase of the same sub­ stance. Why?

12.63 The smectic liquid crystalline phase can be said to be more highly ordered than the nematic. In what sense is this true?

12.64 One of the more effective liquid crystalline substances employed in LCDs is this molecule.

)=H2-CH"

CH3(CH,},CH=CH-CH

/CH2-cH,,

CH-cH

\:H;r-CH/

CH-C ;;;;;N

'cH,...H..C /

524

C HA P T E R

12

Modern Materials

By comparing this structure with the structural formu­ las and models shown in Figure

12.39, describe the fea­

12.66 It often happens that a substance possessing a smectic liquid crystalline phase just above the melting point

tures of the molecule that promote its liquid crystalline

passes into a nematic liquid crystalline phase at a higher

behavior.

temperature. Account for this type of behavior in terms of the ideas developed in Chapter 11 relating molecular

12.65 Describe how a cholesteric liquid crystal phase differs

energies to temperature.

from a nematic phase.

Nanomaterials 12.67 Explain why "bands" may not be the most accurate

12.70 True or false:

description of bonding in a solid when the solid has

1f you want a semiconductor that emits blue light, you

nanoscale dimensions.

could either use a material that has a band gap corre­

12.68 CdS has a band gap of 2.4 eV. 1f large crystals of CdS are

sponding to the energy of a blue photon or you could

illuminated with ultraviolet light they emit light equal to the band gap energy. light?

(b) Would

(a)

What color is the emitted

appropriately sized CdS quantum dots

be able to emit blue light?

use a material that has a smaller band gap, but make a

(c) What about red light?

nanoparticle out of it of the right size.

12.71 Gold is a face-centered cubic structure that has a unit cell edge length of 4.08 A (Figure

12.69 True or false:

(a)

The band gap of a semiconductor decreases as the particle size decreases, in the 1-1 0-nm range.

(b) The light that is emitted from a semiconductor,

11.34). How many gold 20 nrn in diameter?

atoms are there in a sphere that is

Recall that the volume of a sphere is

� 1rr3

12.72 Cadmium telluride, CdTe, takes the zinc blende struc­ ture (Section

upon external stimulation, is longer and longer in

11.8) with a unit cell edge length of 6.49 A.

There are four cadmium atoms and four tellurium

wavelength as the particle size of the semiconductor

atoms per unit cell. How many of each type of atom are

decreases.

there in a cubic crystal with an edge length of

5 nrn?

A D D ITI ONAL EXERC ISES 12.73 One major difference i n the behavior of semiconductors

12.77 Ceramics are generally brittle, subject to crack failure,

In contrast, plastics are

and metals is that semiconductors increase their con­

and stable to high temperatures.

ductivity as you heat them (up to a point), but the con­

generally deformable under stress and have limited ther­

ductivity of a metal decreases as you heat it. Suggest an

mal stability. Discuss these differences in terms of the structures and bonding in the two classes of materials.

explanation.

12.74 What properties of the typical nematic liquid crystalline molecule are likely to cause it to reorient when it is

12.78 A watch with a liquid crystal display (LCD) does not function properly when it is exposed to low tempera­

placed in an electrical field that is perpendicular to the direction of orientation of the molecules? ® is a polymer formed by the polymerization of

12.75 Teflon

F2C =CF2. Draw the structure of a section of this poly­

tures during a trip to Antarctica. Explain why the LCD might not function well at low temperature.

12.79 The temperature range over which a liquid possesses liquid crystalline behavior is rather narrow (for exam­

mer. What type of polymerization reaction is required to

t i::l

ples, see Figure

form it?

12.76 Classify each of the following as a ceramic, polymer, or

teryl benzoate is warmed to well above its liquid crys­

liquid crystal.

talline range and then cooled. On cooling, the sample

CH

(a)

CH2-

COOCH3

unexpectedly remains clear until it reaches a tempera­

t� j

ture just below the melting point, at which time it solidi­

(b) LiNb03

n

fies. What explanation can you give for this behavior?

12.81 Hydrogen bonding between polyamide chains plays an important role

CH

(c) SiC

12.39). Why?

12.80 Suppose that a liquid crystalline material such as choles­

(d)

such as nylon

determining the properties of a nylon Draw the structural for­

mulas for two adjacent chains of nylon

i

CH3

in

6,6 (Table 12.5).

6,6, and

show

where hydrogen-bonding interactions could occur be­

n

tween them.

[12.82[

A particular liquid crystalline substance has the phase diagram shown in the figure. By analogy with the phase diagram for a non-liquid crystalline substance (Section

11.6), identify the phase present in each area.

Integrative Exercises

525

12.83 In fabricating microelectronics circuits, a ceramic con­

ductor such as TiSi2 is employed to connect various re­ gions of a transistor with the outside world, notably aluminum wires. The TiSi2 is deposited as a thin film via chemical vapor deposition, in which TiCl4(g) and SiH4(g) are reacted at the Si surface. Write a balanced equation for the reaction, assuming that the other products are H2 and HCl. Why might TiSi2 behave better as a conducting interconnect on Si than on a metal such as Cu?

p

T

INTEG RATIVE EXERCIS ES (12.84] Employing the bond enthalpy values listed in Table 8.4,

estimate the molar enthalpy change occurring upon (a) polymerization of ethylene, (b) formation of nylon 6,6, (c) formation of polyethylene terephthalate (PET). (12.85] Although polyethylene can twist and turn in random

ways, as illustrated in Figure 12.24, the most stable form is a linear one with the carbon backbone oriented as shown in the figure below: H H '

H H '

H H '

H H '

H H '

H H '

In

what respects is the electron configuration of copper different from that of the other two metallic elements in this compound? 12.87 A

sample of the superconducting oxide HgBa2Ca2Cu30s+x is found to contain 14.99% oxygen by mass. (a) Assuming that all other elements are present in the ratios represented by the formula, what is the value of x in the formula? (b) Which of the metallic elements in the compound is (are) most likely to have noninteger average charges? Explain your answer. (c) Which of the metallic ions in the substance is likely to have the largest ionic ra­ dius? Which will have the smallest?

12.5, which bonds have the lowest average bond enthalpy? (b) In the thermal conversion of polyvinyl chloride to diamond, at high pressure, which bonds are most likely to rupture first on heating the material? (c) Employing the values of average bond enthalpy in Table 8.4, estimate the overall enthalpy change lor converting PVC to diamond.

12.88 (a) In polyvinyl chloride shown on Table

The solid wedges in the figure indicate bonds from carbon that come out of the plane of the page; the dotted wedges indicate bonds that lie behind the plane of the page. (a) What is the hybridization of orbitals at each carbon atom? What angles do you expect between the bonds? (b) Now imagine that the polymer is polypropylene rather than polyethylene. Draw structures for polypropylene in which (i) the CH3 groups all lie on the same side of the plane of the paper (this form is called isotactic polypropylene); (ii) the CH3 groups lie on alternating sides of the plane (syndiotactic polypropylene); or (iii) the CH3 groups are random­ ly distributed on either side (atactic polypropylene). Which of these forms would you expect to have the highest crystallinity and melting point, and which the lowest? Explain in terms of intermolecular inter­ actions and molecular shapes. (c) Polypropylene fibers have been employed in athlet­ ic wear. The product is said to be superior to cotton or polyester clothing in wicking moisture away from the body through the fabric to the outside. Ex­ plain the difference between polypropylene and polyester or cotton (which has many -OH groups along the molecular chain), in terms of intermolecu­ lar interactions with water. 12.86 In the superconducting ceramic YBa2Cu307, what is the

average oxidation state of copper, assuming that Y and Ba are in their expected oxidation states? Yttrium can be replaced with a rare-earth element such as La, and Ba can be replaced with other similar elements without lun­ damentaJly changing the superconducting properties of the material. However, general replacement of copper by any other element leads to a loss of superconductivity.

12.89 Consider para-azoxyanisole, which is a nematic liquid

crystal in the temperature range of

21 oc to 47 °C:

--©- �--©0

CH30

N=

OCH3

(a) Write out the Lewis structure lor this molecule, show­ ing all lone-pair electrons as well as bonds. (b) Describe the hybrid orbitals employed by each of the two nitro­ gens. What bond angles do you anticipate about the nitrogen atom that is bonded to oxygen? (c) Replacing one of the -OCH3 groups in para-azoxyanisole by a -CH2CH2CH2CH3 group causes the melting point of the substance to drop; the liquid crystal range changes to 19 oc to 76 oc . Explain why this substitution produces the observed changes in properties. (d) How would you expect the density of para-azoxyanisole to change upon melting at 117 °C? Upon passing from the nematic to the isotropic liquid state at 137 °C? Explain.

12.3) with unit cell edge length of 5.43 A and eight atoms per unit cell. 3 (a) How many silicon atoms are there in 1 cm of mater­ 3 ial? (b) Suppose you dope that 1-cm sample of silicon with 1 ppm of phosphorus that will increase the conduc­ tivity by a factor of a million. How many milligrams of phosphorus are required?

[12.90] Silicon has diamond structure (Figure

PROPERTIES OF SOLUTIONS

A TROPICAL BEACH. Ocean water is a complex aqueous solution of many dissolved substances of which sodium chloride is highest in concentration.

Dana Edmunds/PacificStock. cam

526

W H AT ' S

A H E A D

13.1 The Solution Process We begin by considering what happens at a molecular level when a substance dissolves, paying particular attention to the role of intermolecular forces in the process. Two important aspects of the solution process are the changes in energy and the changes in how particles are distributed in space as a result of the solution process. 13.2 Saturated Solutions and Solubi lity We will see that in saturated solutions the dissolved and undissolved solutes are in equilibrium. The amount of solute in a saturated solution defines its solubility, the extent to which a particular solute dissolves in a particular solvent. 13.3 Factors Affecting Solubility We next consider the major factors affecting solubility. The nature of the solute and solvent determines the kinds of intermolecular forces between and within them and strongly influences solubility. Temperature also influences solubility: Most solids are more soluble in water at higher temperatures, whereas gases are less soluble in water at higher temperatures. In addition, the solubility of gases increases with increasing pressure.

13.4 Ways of Expressing Concentration We observe that many physical properties of solutions depend on their concentration, and we examine several common ways of expressing concentration. 13.5 Colligative Properties We observe that solutes affect the properties of solutions. The physical properties of solutions that depend only on concentration and not on the identity of the solute are called colligative properties. These properties include the extent to which the solute lowers the vapor pressure, increases the boiling point, and decreases the freezing point of the solvent. The osmotic pressure of a solution is also a colligative property. 13.6 Colloids We close the chapter by investigating colloids, mixtures in which particles larger than typical molecular sizes are dispersed in another component.

ON A WARM DAY AT TH E BEACH, we are unlikely to consider chemistry while we enjoy the water and warmth of the sun. But as we breathe the air, swim in the water, and walk on the sand, we are experiencing the three states of matter. In Chapter 10 and 11, we explored the properties of gases, liquids, and solids. Most of the discussion focused on pure substances. However, the matter that we encounter in our daily lives, such as air, seawater, and sand, is usually composed of mixtures. In this chapter we examine mixtures, although we limit ourselves to those that are homogeneous. As we have noted in earlier chapters, a homogeneous mixture is called a solution. oco (Sections 1.2 and 4.1 ) When we think of solutions, we usually first think about liquids, such as a solution of salt in water, like the seawater shown in this chapter's opening photo. Sterling silver, which is used in jewelry, is also a solution-a homoge­ neous distribution of about 7% copper in silver. Sterling silver is an example of

527

528

C HA PTER 1 3

Properties of Solutions a solid solution. Numerous examples of solutions abound in the world around us-some solid, some liquid, and some gas. For example, the air we breathe is a solution of several gases; brass is a solid solution of zinc in copper; and the fluids that run through our bodies are solutions that carry a variety of essen­ tial nutrients, salts, and other materials. Each of the substances in a solution is called a component of the solution. As we saw in Chapter 4, the solvent is normally the component present in the greatest amount. Other components are called solutes. Because liquid solutions are the most common, we will focus our attention on them in this chapter. Our primary goal is to examine the physical properties of solutions, comparing them with the properties of their components. We will be particularly con­ cerned with aqueous solutions, which contain water as the solvent and a gas, liquid, or solid as a solute. 1 3 . 1 THE S O LUTION P RO CESS

A solution is formed when one substance disperses uniformly throughout another. As we noted in the introduction, solutions may be gases, liquids, or solids. Each of these possibilities is listed in Table 13.1 T. The ability of substances to form solutions depends on two general factors: (1) the types of intermolecular interactions involved in the solution process, and (2) the natural tendency of substances to spread into larger volumes when not restrained in some way. We begin our discussion of the solution process by examining the role of intermolecular interactions. The Effect of I ntermolecu lar Forces

Any of the various kinds of intermolecular forces that we discussed in Chapter 11 can operate between solute and solvent particles in a solution. Ion-dipole forces, for example, dominate in solutions of ionic substances in water. Disper­ sion forces, on the other hand, dominate when a nonpolar substance such as CJI14 dissolves in another nonpolar one like CC14. Indeed, a major factor deter­ mining whether a solution forms is the relative strengths of intermolecular forces between and among the solute and solvent particles. That is, the extent to which one substance is able to dissolve in another depends on the relative mag­ nitudes of the solute-solvent, solute-solute, and solvent-solvent interactions involved in the solution process. Solutions form when the magnitudes of the attractive forces between solute and solvent particles are comparable to or greater than those that exist between the solute particles themselves or between the solvent particles themselves. For example, the ionic substance NaCI dissolves readily in water because the attractive interactions between the ions and the polar H20 molecules (solute­ solvent interactions) overcome the attraction between the ions in the solid NaCI

TABLE 1 3.1

• Examples of Solutions

State of Solution

State of Solvent

State of Solute

Example

Gas Liquid Liquid Liquid Solid Solid Solid

Gas Liquid Liquid Liquid Solid Solid Solid

Gas Gas Liquid Solid Gas Liquid Solid

Air Oxygen in water Alcohol in water Salt in water Hydrogen in palladium Mercury in silver Silver in gold

13.1

(b)

(a) .t.

The Solution Process

529

(c)

Figure 1 3.1

Dissolution of an Ionic solid In water. (a) A crystal of the ionic solid is hydrated by water molecules, with the oxygen atoms of the water molecules oriented toward the cations (purple) and the hydrogens oriented toward the anions (green). (b, c) As the solid dissolves, the individual ions are removed from the solid surface and become completely separate hydrated species in solution.

(solute-solute interactions) and between H20 molecules in the solvent (solvent­ solvent interactions). Let's examine this solution process more closely, paying attention to these various attractive forces. When NaCI is added to water (Figure 13.l .t.), the water molecules orient themselves on the surface of the NaCI crystals. The positive end of the water di­ pole is oriented toward the Cl- ions, and the negative end of the water dipole is oriented toward the Na+ ions. The ion-dipole attractions between the ions and water molecules are strong enough to pull the ions from their positions in the crystal. Once separated from the crystal, the Na+ and Cl- ions are surrounded by water molecules, as shown in Figure 13.1(b and c) and Figure 13.2 1l>. We learned in Section 4.1 that interactions such as this between solute and solvent mole­ cules are known as solvation. When the solvent is water, the interactions are also referred to as hydration. In addition to the solvent-solute interactions (the ion-dipole attractions be­ tween H20 molecules and the Na+ and Cl- ions) and the solute-solute interac­ tions (between the Na+ and Cl - ions in the solid), we must consider one other interaction: the solvent-solvent interaction (in this case the hydrogen-bonding attractions between H20 molecules). In forming the solution, the water mole­ cules must make room for the hydrated Na+ and Cl - ions in their midst, causing some water molecules to move apart. GIVE IT

SOME THOUGHT

Why doesn't NaCl dissolve in nonpolar solvents such as hexane, C6H14?

Energy Changes and Solution Formation We can analyze the roles played by the solute-solvent, solute-solute, and solvent-solvent interactions by examining the energy changes associated with each. Let's continue to analyze the process of dissolving NaCl in water as an ex­ ample of how energy considerations provide insight into the solution process. We have observed that sodium chloride dissolves in water because the water molecules have a strong enough attraction for the Na+ and CC ions to overcome the attraction of these two ions for one another in the crystal. In addi­ tion, water molecules must separate from one another to form spaces in the solvent that the Na+ and Cl- ions will occupy. Thus, we can think of the over­ all energetics of solution formation as having three components-associated

.t. Figure 1 3.2 Hydrated Na +and Cl­ lons. The negative ends of the water dipoles point toward the positive ion, and the positive ends point toward the negative ion.

530

C HA P T E R

13

Properties of Solutions

II> Figure 1 3. 3 Enthalpy contributions to t.H..,In· The enthalpy changes t.H1 and t.H2 represent endothermic processes, requiring an input of energy, whereas t.H3 represents an exothermic process.

• t.H1 : Separation of solute molecules

• llHi Separation of solvent molecules

+

t.H3: Formation of solute-solvent interactions

with breaking the solute-solute and the solvent-solvent interactions and forming the solute-solvent interactions-as illustrated schematically in Figure 13.3 •. The overall enthalpy change in forming a solution, t.H501n, is the sum of three terms associated with these three processes: [13.1]

Regardless of the particular solute being considered, separation of the solute particles from one another requires an input of energy to overcome their attractive interactions. The process is therefore endothermic (t.H1 > 0). Separa­ tion of solvent molecules to accommodate the solute also always requires ener­ gy (t.H2 > 0). The third component, which arises from the attractive interactions between solute and solvent, is always exothermic (t.H3 < 0). As shown in Figure 13.411>, the three enthalpy terms in Equation 13.1 can be added together to give either a negative or a positive sum, depending on the relative magnitudes of the terms. Thus, the formation of a solution can be either exothermic or endothermic. For example, when magnesium sulfate, MgS04, is added to water, the resultant solution gets quite warm: t.Hsoin = -91.2 kJ/mol. In contrast, the dissolution of ammonium nitrate (NH4N03) is endothermic: t.Hsoln = 26.4 kJ/mol. These particular substances have been used to make the instant heat packs and ice packs that are used to treat athletic injuries (Figure 13.5 II>). The packs consist of a pouch of water and a dry chemical, MgS04 for hot packs and NH4N03 for cold packs. When the pack is squeezed, the seal sep­ arating the solid from the water is broken and a solution forms, either increas­ ing or decreasing the temperature. In Chapter 5 we learned that the enthalpy change in a process can provide information about the extent to which a process will occur. ceo (Section 5 .4) Processes that are exothermic tend to proceed spontaneously. A solution will not form if t.Hsoln is too endothermic. The solvent-solute interaction must be

13.1

531

0).

separated solute particles

.___J c_ _._ -----, _

The Solution Process

Net endothermic process

(b) Endothermic

(a) Exothermic

strong enough to make t.. H3 comparable in magnitude to t.. H1

+ t.. H2. This fact

explains why ionic solutes such as NaCI do not dissolve in nonpolar liquids such as gasoline. The nonpolar hydrocarbon molecules of the gasoline would experience only weak attractive interactions with the ions, and these interac­ tions would not compensate for the energies required to separate the ions from one another.

ceo

By similar reasoning, a polar liquid such as water does not form solutions with a nonpolar liquid such as octane (C8H18). The water molecules experience strong hydrogen-bonding interactions with one another.

(Section

11.2)

These attractive forces must be overcome to disperse the water molecules throughout the nonpolar liquid. The energy required to separate the H20 mol­ ecules is not recovered in the form of attractive interactions between H20 and C8H18 molecules.

G IVE

IT

SOME

THOUGHT

Label the following processes as exothermic or endothermic: (a) forming solvent­ solute interactions, (b) breaking solvent-solvent interactions.

Solution Formation, Spontaneity, and Entropy When carbon tetrachloride (CC14) and hexane (C6H14) are mixed, they dissolve in one another in all proportions. Both substances are nonpolar, and they have similar boiling points

(77 oc

for CC14 and

69 oc for

C�14). It is therefore rea­

sonable to suppose that the magnitudes of the attractive forces (London disper­ sion forces) among molecules in the two substances and in their solution are comparable. When the two substances are mixed, dissolving occurs sponta­ neously; that is, it occurs without any extra input of energy from outside the system. Processes that occur spontaneously involve two distinct factors. The most obvious factor is energy, which we have used to analyze the dissolving of NaCl in water. The other factor is the distribution of each component into a larger volume--the tendency in nature for substances to mix and spread out into larger volumes.

"" Figure 1 3.5 Endothermic dissolution. Ammonium nitrate instant ice packs are often used to treat athletic injuries. To activate the pack, the container is kneaded. The kneading breaks an interior seal separating solid NH4N03 from water. The heat of solution of NH4N03 is positive, so the temperature of the solution decreases.

532

C HA P T E R

13

Properties o f Solutions We see the influence of energy all around us. If you let go of a book, it falls to the floor because of gravity. At its initial height, the book has a high­ er potential energy than it has when it is on the floor. Unless the book is

500 rnL 500 rnL

restrained, it falls; and as it does, potential energy is converted into kinetic energy. When the book strikes the floor, the kinetic energy is converted large­ ly into heat energy, which is dispersed throughout the surroundings. The book has lost energy to its surroundings in this process. This fact leads us to the first basic principle identifying spontaneous processes and the

Processes in which the energy content of the system decreases tend to occur spontaneously. Spontaneous processes tend to be exothermic. = (Section 5.4, "Strategies in Chemistry: Using Enthalpy as a Guide")

direction they take:

(a)

Change tends to occur in the direction that leads to a lower energy or lower enthalpy for the system. Some spontaneous processes, however, do not result in lower energy for a system, and even some endothermic processes occur spontaneously. For

-1000 rnL CCI4 +

C6H14

example, NH4N03 readily dissolves in water, even though the solution process is endothermic. Processes such as this are characterized by a more dispersed state of one or more components, resulting in an overall increase in the randomness of the system. In this example the densely ordered solid, + NH4N03, is dispersed throughout the solution as separated NH4 and N03ions. The mixing of CC14 and C6H14 provides another simple example.

(b)

A Figure 1 3.6 Increasing randomneu In a solution proceu. A homogeneous solution of CCI4 and C6H14 forms when a barrier separating the two liquids is removed. Each CCI4 molecule of the solution in (b) is more dispersed in space than it was in the left compartment in (a), and each C6H14 molecule in (b) is more dispersed than it was in the right compartment in (a).

Suppose that we could suddenly remove a barrier that separates

500 mL

of CC4 from 500 mL of C6H14, as shown in Figure 1 3 . 6 (a) � . Before the barrier is removed, each liquid occupies a volume of 500 mL. All the CCI4 molecules are in the the

500

500 mL to the left of the barrier, and

all the C6H14 molecules are in

mL to the right. When equilibrium has been established after the

barrier has been removed, the two liquids together occupy a volume of about 1000 mL. Formation of a homogeneous solution has increased the degree of dispersal, or randomness, because the molecules of each substance are now mixed and distributed in a volume twice as large as that which they occupied individually before mixing. The degree of randomness in the system, some­ times referred to as disorder, is given by a thermodynamic quantity called

Processes occur­ ring at a constant temperature in which the randomness or dispersal in space (entropy) of the system increases tend to occur spontaneously. entropy. This example illustrates our second basic principle:

When molecules of different types are brought together, mixing-hence an increased dispersal-occurs spontaneously unless the molecules are restrained by sufficiently strong intermolecular forces or by physical barriers. Thus, gases spontaneously mix and expand unless restrained by their containers; in this case intermolecular forces are too weak to restrain the molecules. However, because strong bonds hold sodium and chloride ions together, sodium chloride does not spontaneously dissolve in gasoline. We will discuss spontaneous processes again in Chapter

19.

At that time

we will consider the balance between the tendencies toward lower enthalpy and toward increased entropy in greater detail. For the moment, we need to be aware that the solution process involves two factors: a change in enthalpy and a change in entropy. In most cases the formation of solutions is favored by the in­

crease in entropy that accompanies mixing. Consequently, a solution will form un­ less solute-solute or solvent-solvent interactions are too strong relative to the solute-solvent interactions.

GIVE

I T

SOME

THOUGHT

Silver chloride, AgCI, is essentially insoluble in water. Would you expect a significant change in the entropy of the system when 10 g of AgCI is added to 500 mL of water?

13.1

(a)

Solution Formation and Chemical Reactions

In all our discussions of solutions, we must be careful to distinguish the physi­ cal process of solution formation from chemical reactions that lead to a solu­ tion. For example, nickel metal is dissolved on contact with hydrochloric acid solution because the following chemical reaction occurs: +

2 HCl(aq) � NiC)z(aq) + H2(g)

[13.2]

In this instance the chemical form of the substance being dissolved is changed from Ni to NiCI2. If the solution is evaporated to dryness, NiC12 6 H20(s), not Ni(s), is recovered (Figure 13.7� ). When NaCI(s) is dissolved in water, on the other

"" Figure 1 3.7 The nickel-add reaction Is not a simple dissolution. (a) Nickel metal and hydrochloric acid. (b) Nickel reacts slowly with hydrochloric acid, forming NiCI2(aq) and H2(g). (c) NiCI2 • 6 H20 is obtained when the solution from (b) is evaporated to dryness. Because the residue left after evaporation is chemically different from either reactant, we know what takes place is a chemical reaction rather than merely a solution process.

•

hand, no chemical reaction occurs. If the solution is evaporated to dryness, NaCI is recovered. Our focus throughout this chapter is on solutions from which the solute can be recovered unchanged from the solution.

HYDRATE S

F obtained by evaporation of water from aqueous solutions.

requently, hydrated ions remain in crystalline salts that are

Common examples include FeCI3 • 6 H20 [iron(ill) chloride he­ xahydrate] and Cu504 • 5 H20 [copper(JI) sulfate pentahydrate]. The FeCI3 • 6 H20 consists of Fe(H20)63+ and o- ions; the Cu504 • 5 H20 consists of Cu(H20) /+ and S04(H2of- ions. Water molecules can also occur in positions in the crystal lat­ tice that are not specifically associated with either a cation or an anion. BaCJ2 • 2 H20 (barium chloride dihydrate) is an ex­ ample. Compounds such as FeCI3 • 6 H20, Cu504 • 5 H20, and BaCI2 • 2 H20, which contain a salt and water combined in definite proportions, are known as hydrates. The water asso­ ciated with them is called water of hydration. Figure 13.8 � shows an example of a hydrate and the corresponding anhy­ drous (water-free) substance.

Related Exercises: 13.4, 13.107

533

(c)

(b)

Ni(s)

The Solution Process

"" Figure 1 3.8 Hydrates and anhydrous salts. Hydrated cobalt(ll) chloride, CoCI2 • 6 H20 (left), and anhydrous CoCI2 (right).

534

C HA P T E R

13

Properties o f Solutions

- SAMPLE EXERCISE 1 3. 1

I Assessing Entropy Change

In the process illustrated below, water vapor reacts with excess solid sodium sulfate to form the hydrated form of the salt. The chemical reaction is Na2S04(s)

+ 10 H20(g)

----+

Na2S04 · 10 H20(s)

Essentially all of the water vapor in the closed container is consumed in this reaction. 1f we consider our system to consist initially of Na2504(s) and 10 H20(g), (a) does the system become more or less ordered in this process, and (b) does the entropy of the system increase or decrease?

SOLUTION Analyze: We are asked to determine whether the reaction of water vapor with the solid salt causes the system to become more or less dispersed, or random, and to de­ termine whether the process results in a higher or lower entropy for the system.

Plan: We need to examine the initial and final states and judge whether the process has made the system more or less dispersed. Depending on our answer to that ques­ tion, we can determine whether the entropy has increased or decreased.

Solve: (a) In the course of forming the hydrate of Na2S04(s), the water vapor moves from the vapor state, in which it is dispersed throughout the entire volume of the con­ tainer, to the solid state, where it is confined to the Na2S04 · 10 H20(s) lattice. This means that the water vapor becomes less dispersed (more ordered, or less random). (b) When a system becomes less dispersed, or more ordered, its entropy is decreased.

- PRACTICE EXERCISE

Does the entropy of the system increase or decrease when the stopcock is opened to allow mixing of the two gases in this apparatus?

Answer: The entropy increases because each gas eventually becomes dispersed in twice the volume it originally occupied.

1 3.2 SATURATED S O LUTIONS AND S O LU B I LI TY As a solid solute begins to dissolve in a solvent, the concentration of solute par­

.i. Figure 1 3.9 Dynamic equilibrium

In a saturated solution. In a solution in which excess ionic solute is present, ions on the surface of the solute are continuously passing into the solution as hydrated species, while hydrated ions from the solution are deposited on the surfaces of the solute. At equilibrium in a saturated solution, the two processes occur at equal rates .

ticles in solution increases, thus increasing the chances of the solute particles colliding with the surface of the solid (Figure

13.9 �).

Because of such a colli­

sion, the solute particle may become reattached to the solid. This process, which is the opposite of the solution process, is called crystallization. Thus, two opposing processes occur in a solution in contact with undissolved solute. This situation is represented in a chemical equation by use of two half arrows: Solute

+

solvent

� solution crystallize

[13.3]

13.3

(a) A seed crystal of NaCH3COO being added to the supersaturated solution.

(b) Excess NaCH3COO crystallizes from the solution.

(c) The solution arrives at saturation.

When the rates of these opposing processes become equal, there is no further net increase in the amount of solute in solution. A dynamic equilibrium is es­ tablished similar to the one between evaporation and condensation discussed in Section 11.5. A solution that is in equilibrium with undissolved solute is saturated. Addi­ tional solute will not dissolve if added to a saturated solution. The amount of solute needed to form a saturated solution in a given quantity of solvent is known as the solubility of that solute. That is, the solubility is the maximum amount

of solute that will dissolve in a given amount of solvent at a specified temperature, given that excess solute is present. For example, the solubility of NaCl in water at 0 °C is 35.7 g per 100 mL of water. This is the maximum amount of NaCl that can be dis­

solved in water to give a stable equilibrium solution at that temperature. If we dissolve less solute than that needed to form a saturated solution, the so­ lution is unsaturated. Thus, a solution containing only 10.0 g of NaCl per 100 mL of water at 0 oc is unsaturated because it has the capacity to dissolve more solute. Under suitable conditions it is sometimes possible to form solutions that contain a greater amount of solute than that needed to form a saturated solution. Such solutions are supersaturated. For example, considerably more sodium ac­ etate (NaCH3COO) can dissolve in water at high temperatures than at low tem­ peratures. When a saturated solution of sodium acetate is made at a high temperature and then slowly cooled, all of the solute may remain dissolved even though the solubility decreases as the temperature decreases. Because the solute in a supersaturated solution is present in a concentration higher than the equi­ librium concentration, supersaturated solutions are unstable. Supersaturated solutions result for much the same reason as supercooled liquids (Section 11.4). For crystallization to occur, the molecules or ions of solute must arrange them­ selves properly to form crystals. The addition of a small crystal of the solute (a seed crystal) provides a template for crystallization of the excess solute, lead­ ing to a saturated solution in contact with excess solid (Figure 13.10�).

G IVE IT SOME THOUGHT Is a supersaturated solution of sodium acetate a stable equilibrium solution?

13 .3 FACTORS AFFECTING SOLUBIL ITY The extent to which one substance dissolves in another depends on the nature of both the solute and the solvent. (Section 13.1) It also depends on temperature and, at least for gases, on pressure. Let's consider these factors more closely. c:xx>

Factors Affecting Solubility

� Figure 1 3.1 0 Sodium acetate readily forms a supersaturated solution In water.

535

536

C HA PTER 1 3

Properties of Solutions

Sol ute-Solvent Interactions TABLE 1 3.2 • Solubilities of Gases in Water at 20 •c, with 1 atm G as

Pressure Gas Nz

co Oz

Ar Kr

Solubility (M)

0.69 1.04 1.38 1.50 2.79

X

3 3

-

10

-

X 10 3 X 10 -

10 3 3 X 10-

X

-

One factor determining solubility is the natural tendency of substances to mix (the tendency of systems to move toward a more dispersed, or random, state). ooo (Section 13.2) If this were the only factor involved, however, we would expect all substances to be completely soluble in one another. This is clearly not the case. So what other factors are involved? As we saw in Section 13.1, the rel­ ative forces of attraction among the solute and solvent molecules also play very important roles in the solution process. Although the tendency toward dispersal and the various interactions among solute and solvent particles are all involved in determining the solubilities, con­ siderable insight can often be gained by focusing on the interaction between the solute and solvent. The data in Table 13.2 ..,. show, for example, that the solubil­ ities of various simple gases in water increase with increasing molecular mass or polarity. The attractive forces between the gas and solvent molecules are mainly of the London dispersion type, which increase with increasing size and mass of the gas molecules. ooo (Section 11.2) Thus, the data indicate that the solubilities of gases in water increase as the attraction between the solute (gas) and solvent (water) increases. In general, when other factors are comparable, the stronger the

attractions are between solute and solvent molecules, the greater the solubility.

Acetone

Hexane Is Insoluble In water. Hexane is the top layer because it is less dense than water.

Because of favorable dipole-dipole attractions between solvent molecules and solute molecules, polar liquids tend to dissolve readily in polar solvents. Water is both polar and able to form hydrogen bonds. ooo (Section 11.2) Thus, polar molecules, especially those that can form hydrogen bonds with water mole­ cules, tend to be soluble in water. For example, acetone, a polar molecule with the structural formula shown in the margin, mixes in all proportions with water. Acetone has a strongly polar C = O bond and pairs of nonbonding elec­ trons on the 0 atom that can form hydrogen bonds with water. Pairs of liquids such as acetone and water that mix in all proportions are miscible, whereas those that do not dissolve in one another are immiscible. Gasoline, which is a mixture of hydrocarbons, is immiscible with water. Hydro­ carbons are nonpolar substances because of several factors: The C - C bonds are nonpolar, the C - H bonds are nearly nonpolar, and the shapes of the mol­ ecules are symmetrical enough to cancel much of the weak C -H bond dipoles. The attraction between the polar water molecules and the nonpolar hy­ drocarbon molecules is not sufficiently strong to allow the formation of a solu­ tion. Nonpolar liquids tend to be insoluble in polar liquids. As a result, hexane (C6H14) does not dissolve in water, as the photo in the margin shows. The series of compounds in Table 13.3 T demonstrates that polar liquids tend to dissolve in other polar liquids and nonpolar liquids in nonpolar ones. These organic compounds all contain the OH group attached to a C atom. Organic com­ pounds with this molecular feature are called alcohols. The 0- H bond is polar and is able to form hydrogen bonds. For example, CH3CH20H molecules can form hydrogen bonds with water molecules as well as with each other (Figure 13.11 .,.). As a result, the solute-solute, solvent-solvent, and solute-solvent forces

TABLE 1 3.3 • Solubilities of Some Alcohols in Water and in Hexane•

Alcohol CH30H (methanol) CH3CHzOH (ethanol) CH3CH2CHzOH (propanol) CH3CH2CH2CH20H (butanol) CH3CHzCH2CHzCHzOH (pentanol) CH3CHzCHzCHzCHzCHzOH (hexanol)

Solubility in H20 00 00 00

0.11 0.030 0.0058

Solubility in C6H14

0.12 00 00 00

00 00

*Expressed in mol alcohol/100 g solvent at 20 oc. The infinity symbol (oo) indicates that the alcohol is completely miscible with the solvent.

13.3

Factors Affecting Solubility

537

are not greatly different within a mixture of CH3CH20H

and

H20.

No major

change occurs in the environments of the molecules as they are mixed. Therefore, the increased dispersal (entropy) of the two components into a larger combined volume as they mix plays a significant role in the formation of the solution. Ethanol (CH3CH20H), therefore, is com­ pletely miscible with water. The number of carbon atoms in an alcohol affects its solubility in water. As the length of the carbon chain increases, the polar OH group becomes an ever

(a )

smaller part of the molecule, and the

molecule behaves more like a hydrocarbon. The solubility of the alcohol in water decreases correspondingly. On the other hand, the solubility of the alco­ hol in a nonpolar solvent like hexane (C6H14) increases as the nonpolar hydro­ carbon chain increases in length. One way to enhance the solubility of a substance in water is to increase the number of polar groups it contains. For example, increasing the number of OH groups along a carbon chain of a solute increases the extent of hydrogen bond­ ing between that solute and water, thereby increasing solubility. Glucose (C6H1206) has five OH groups on a six-carbon framework, which makes the molecule very soluble in water

(83

g clissolves in

The glucose molecule is shown in Figure

13.12 ,..

100 mL of water

at

17.5 °C).

HYDROGEN B O N D I NG AND AQUEOUS SOLU B I LITY

The presence of OH groups capable of hydrogen bonding with water enhances the aqueous solubility of organic molecules.

Cydohexane, C6Hl2• which has no polar OH groupsi is essentially insoluble in water. &

Figure

1 3. 1 2 Structure and solubility.

Glucose, CoHn06, has five OH groups and is highly soluble in water.

& Figure 1 3. 1 1 Hydrogen-bonding Interactions. (a) Between two ethanol molecules and (b) between an ethanol molecule and a water molecule.

538

C HA P T E R

13

Properties o f Solutions

� ,iliiji;j I!!!I �II Ij ft �II!I lII �l!lfi!!J!Ir�l!'i � .,,., ij iQ ',I!IJ!I •��; ,ijiiJI �[II,.-I

-----;F;-_;A T- A N D WAT ER-S 0 L U B LE V I TA M I N S

Vsolubilities in different parts of the human body. Vitamins

itamins have unique chemical structures that affect their

B and C are water soluble, for example, whereas vitamins A, D,

E, and K are soluble in nonpolar solvents and in the fatty tissue

of the body (which is nonpolar). Because of their water solubil­ ity, vitamins B and C are not stored to any appreciable extent in the body, and so foods containing these vitamins should be included in the daily diet. In contrast, the fat-soluble vitamins are stored in sufficient quantities to keep vitamin-deficiency diseases from appearing even after a person has subsisted for a long period on a vitamin-deficient diet. The different solubility patterns of the water-soluble vit­ amins and the fat-soluble ones can be rationalized in terms of

the structures of the molecules. The chemical structures of vi­ tamin A (retinol) and of vitamin C (ascorbic acid) are shown in Figure 13.13 T. Notice that the vitamin A molecule is an al­ cohol with a very long carbon chain. Because the OH group is such a small part of the molecule, the molecule resembles the long-chain alcohols listed in Table 13.3. This vitamin is nearly nonpolar. In contrast, the vitamin C molecule is small­ er and has more OH groups that can form hydrogen bonds with water. It is somewhat like glucose, which was discussed earlier. It is a more polar substance.

Related Exercises: 13.7, 13.44

Vitamin C

Vitamin A (a)

(b)

.6. Figure 1 3. 1 3 Vitamins A and C. (a) The molecular structure of vitamin A, a fat-soluble vitamin. The

molecule is composed largely of carbon-carbon and carbon-hydrogen bonds, so it is nearly nonpolar. (b) The molecular structure of vitamin C, a water-soluble vitamin. Notice the OH groups and the other oxygen atoms in the molecule, which can interact with water molecules by hydrogen bonding.

Examination of different combinations of solvents and solutes such as those considered in the preceding paragraphs has led to an important general­

Substances with similar intermolecular attractive forces tend to be soluble in one another. This generalization is often simply stated as "like dissolves like."

ization:

Nonpolar substances are more likely to be soluble in nonpolar solvents; ionic and polar solutes are more likely to be soluble in polar solvents. Network solids such as diamond and quartz are not soluble in either polar or nonpolar solvents because of the strong bonding forces within the solid.

GIVE IT SOME THOUGHT Suppose the hydrogens on the OH groups in glucose (Figure 13.12) were replaced with methyl groups, CH3. Would you expect the water solubility of the resulting mol­ ecule to be higher than, lower than, or about the same as the solubility of glucose?

13.3

- SAMPLE EXERCISE 1 3.2

Factors Affecting Solubility

539

I Predicting Solubility Patterns

Predict whether each of the following substances is more likely to dissolve in the nonpolar solvent carbon tetrachloride (CC14) or in water: C7H16, Na2S04, HCl, and 12. SOLUTION Analyze: We are given two solvents, one that is nonpolar (CCI4) and the other that is polar (HzO), and asked to determine which will be the best solvent for each solute listed. Plan: By examining the formulas of the solutes, we can predict whether they are ionic or molecular. For those that are molecular, we can predict whether they are polar or nonpolar. We can then apply the idea that the nonpolar solvent will be best for the nonpolar solutes, whereas the polar solvent will be best for the ionic and polar solutes. Solve: C7H16 is a hydrocarbon, so it is molecular and nonpolar. Na2S04, a compound containing a metal and nonmetals, is ionic. HCl, a diatomic molecule containing two nonmetals that differ in electronegativity, is polar. lz, a diatomic molecule with atoms of equal electronegativity, is nonpolar. We would therefore predict that C7H16 and 12 (the nonpolar solutes) would be more soluble in the nonpolar CCI4 than in polar H20, whereas water would be the better solvent for Na2S04 and HCI (the ionic and polar covalent solutes). - PRACTICE EXERCISE

Arrange the following substances in order of increasing solubility in water: H

H

H

H

H

t t t t t � � � � �

H- - - - - -H

H

H

H

H

H

t t t t t � � � � �

H- - - - - -OH

� � � � � � � � � �

HO-C-C-C-C-C-OH

H

H

H

H

H

t t t t t � � � � �

H- - - - - -CI

Answer: C5H12 < C5H11Cl < C5H110H < C5H10(0H)z (in order of increasing pola­ rity and hydrogen-bonding ability)

(a)

Pressure Effects The solubilities of solids and liquids are not appreciably affected by pressure, whereas the solubility of a gas in any solvent is increased as the pressure over the solvent increases. We can understand the effect of pressure on the solubility of a gas by considering the dynamic equilibrium illustrated in Figure 13.14 �. Suppose that we have a gaseous substance distributed between the gas and so­ lution phases. When equilibrium is established, the rate at which gas molecules enter the solution equals the rate at which solute molecules escape from the so­ lution to enter the gas phase. The small arrows in Figure 13.14(a) represent the rates of these opposing processes. Now suppose that we exert added pressure on the piston and compress the gas above the solution, as shown in Figure 13.14(b). If we reduced the volume to half its original value, the pressure of the gas would increase to about twice its original value. The rate at which gas mol­ ecules strike the surface to enter the solution phase would therefore increase. As a result, the solubility of the gas in the solution would increase until equilib­ rium is again established; that is, solubility increases until the rate at which gas molecules enter the solution equals the rate at which solute molecules escape from the solution. Thus, the solubility of the gas increases in direct proportion to its

partial pressure above the solution.

(b) .II. Figure 1 3. 14 Effect of pressure on gas solubility. When the pressure is increased, as in (b), the rate at which gas molecules enter the solution increases. The concentration of solute molecules at equilibrium increases in proportion to the pressure.

C HAPTER 1 3

540

Properties of Solutions The relationship between pressure and the solubility of a gas is expressed by a simple equation known as Henry's law: [13.4]

& Figure 1 3. 1 5 Solubility decreases as prenure decreases. C02 bubbles out of solution when a carbonated beverage is opened, because the C02 partial pressure above the solution is reduced.

Here, Sg is the solubility of the gas in the solution phase (usually expressed as molarity), P8 is the partial pressure of the gas over the solution, and k is a pro­ portionality constant known as the Henry's law constant. The Henry's law con­ stant is different for each solute-solvent pair. It also varies with temperature. As an example, the solubility of N2 gas in water at 25 oc and 0.78 atm pres­ sure is 5.3 X 10-4 M. The Henry's law constant for N2 in water at 25 °C is thus given by (5.3 X 10-4 moljL)/0.78 atm = 6.8 X 10-4 moi/L-atm. If the partial pressure of N2 is doubled, Henry's law predicts that the solubility in water at 25 oc will also double, to 1.36 x 10-3 M. Bottlers use the effect of pressure on solubility in producing carbonated beverages such as beer and many soft drinks. These are bottled under a carbon dioxide pressure greater than 1 atm. When the bottles are opened to the air, the partial pressure of C02 above the solution decreases. Hence, the solubility of C02 decreases, and C02(g) escapes from the solution as bubbles (Figure 13.15 �). - SAMPLE EXERCISE 1 3.3

[

A

Henry's Law Calculation

Calculate the concentration of C02 in a soft drink that is bottled with a partial pressure of C02 of 4.0 atrn over the liquid at •c. The Henry's law constant for C02 in water 2 at this temperature is 3.1 x 0- mol/L-atm.

1

25

SOLUTION

Analyze: We are given the partial pressure of C02, Pc0,, and the Henry's law con­ stant, k, and asked to calculate the concentration of C02 in the solution. Plan: With the information given, we can use Henry's Jaw, Equation late the solubility,

Sco,.

13.4, to calcu­

BLOOD GASES AND DEEP-SEA DIVING

B

ecause the solubility of gases increases with increasing

pressure, divers who breathe compressed air (Figure

13.161>)

must be concerned about the solubility of gases in

their blood. Although the gases are not very soluble at sea

C02 poisoning occurs. At excessive concentrations in the

body, carbon dioxide acts as a neurotoxin, interfering with nerve conduction and transmission.

Related Exercises: 13.55, 13.56, 13.105

level, their solubilities can become appreciable at deep levels where their partial pressures are greater. Thus, deep-sea divers must ascend slowly to prevent dissolved gases from

being released rapidly from blood and other fluids in the

body. These bubbles affect nerve impulses and give rise to the affliction known as decompression sickness, or "the bends,"

which is painful and potentially fatal. Nitrogen is the main problem because it has the highest partial pressure in air and

because it can be removed only through the respiratory sys­ tem. Oxygen, in contrast, is consumed in metabolism.

Deep-sea divers sometimes substitute helium for nitrogen

in the

air that they breathe, because helium has a much lower

solubility in biological fluids than N2. For example, divers

working at a depth of 100 ft experience a pressure of about 4 atm. At this pressure a mixture of 95% helium and 5% oxy­ gen will give an oxygen partial pressure of about 0.2 atrn, which is the partial pressure of oxygen in normal air at 1 atm. If the oxygen partial pressure becomes too great, the urge to

breathe is reduced, C02 is not removed from the body, and

&

Figure 1 3. 1 6 Solubility Increases as pressure Increases. Divers who use compressed gases m ust be concerned about the solubility of the gases in their blood. Doug Perrine/PacificStock.com

13.3

Solve: Sco,

� kPco, � (3.1

2 X 10- moljL-atm)(4.0 atm)

�

0.12 moljL

Factors Affecting Solubility

� 0 . 12 M

Check: The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of C02 and the value of Henry's constant. - PRACTICE EXERCISE

Calculate the concentration of C02 in a soft drink after the bottle is opened and equi­ 4 librates at 25 oc under a C02 partial pressure of 3.0 X 10- atm. X 10...,; M

Answer: 9.3

Temperature Effects

The solubility of most solid solutes in water increases as the temperature of the solution increases. Figure 13.17T shows this effect for several ionic substances in water. There are exceptions to this rule, however, as seen for Ce2(S04)3, whose solubil­ ity curve slopes downward with increasing temperature.

In contrast to solid solutes, the solubility of gases in water decreases with in­ creasing temperature (Figure 13.18 T). If a glass of cold tap water is warmed, you can see bubbles of air on the inside of the glass. Similarly, carbonated beverages go flat as they are allowed to warm. As the temperature of the solution increas­ es, the solubility of C02 decreases, and C02(g) escapes from the solution. The decreased solubility of 02 in water as temperature increases is one of the effects of thermal pollutwn of lakes and streams. The effect is particularly seri­ ous in deep lakes because warm water is less dense than cold water. It therefore

tends to remain on top of cold water, at the surface. This situation impedes the dissolving of oxygen into the deeper layers, thus stifling the respiration of all aquatic life needing oxygen. Fish may suffocate and die under these conditions.

GIVE IT SOME THOUGHT Why do bubbles form on the inside wall of a cooking pot when water is heated on the stove, even though the temperature is well below the boiling point of water?

CH4 2.0

I

� .!:>

" 1.0 0 rJl

t �

He

0 Temperature (0C} & Figure 1 3 . 1 7 Solubilities of several Ionic compounds In water as a function of temperature.

10

20 30 Temperature CC)

40

& Figure 1 3. 18 Variation of gas solubility with temperature. Note that solubilities are in units of millimoles per liter (mmoljl), for a constant total pressure of 1 atm in the gas phase.

50

541

542

C HAPTER 1 3

Properties of Solutions 1 3 .4 WAYS O F EXP RESSING CONCENTRATI O N

The concentration of a solution can b e expressed either qualitatively or quan­ titatively. The terms dilute and concentrated are used to describe a solution qualitatively. A solution with a relatively small concentration of solute is said to be dilute; one with a large concentration is said to be concentrated. We use several different ways to express concentration in quantitative terms, and we examine four of these in this section: mass percentage, mole fraction, molarity, and molality. Mass Percentage, ppm, and ppb

One of the simplest quantitative expressions of concentration is the percentage of a component in a solution, given by Mass % of component

=

mass of component in soln total mass of soln

X 100

mass

[13.5]

where we have abbreviated "solution" as "soln." Percent means per hundred. Thus, a solution of hydrochloric acid that is 36% HCl by mass contains 36 g of HCl for each 100 g of solution. We often express the concentrations of very dilute solution in parts per million (ppm), or parts per billion (pfb). These quantities are similar to mass percentage but use 106 (a million) or 10 (a billion), respectively, in place of 100 as a multiplier for the ratio of the mass of solute to the mass of solution. Thus, parts per million is defined as ppm of component

=

mass of component in soln total mass of soln

x

106

[13.6]

A solution whose solute concentration is 1 ppm contains 1 g of solute for each million (106) grams of solution or, equivalently, 1 mg of solute per kilogram of solution. Because the density of water is 1 g/mL, 1 kg of a dilute aqueous solu­ tion will have a volume very close to 1 L. Thus, 1 ppm also corresponds to 1 mg of solute per liter of aqueous solution. The acceptable maximum concentrations of toxic or carcinogenic sub­ stances in the environment are often expressed in ppm or ppb. For example, the maximum allowable concentration of arsenic in drinking water in the United States is 0.010 ppm; that is, 0.010 mg of arsenic per liter of water. This concen­ tration corresponds to 10 ppb. GIVE IT SOME THOUGHT A solution of 502 in water contains 0.00023 g of 502 per liter of solution. What is the concentration of 502 in ppm? In ppb?

• SAMPLE EXERCISE 13.4 I Calculation of Mass-Related Concentrations (a) A solution is made by dissolving 13.5 g of glucose (C6H 1206) in 0.1 00 kg of water. What is the mass percentage of solute in this solution? (b) A 2.5-g sample of ground­ 2 2 water was found to contain 5.4 �J-g of Zn +. What is the concentration of zn + in parts per million? SOLUTION

(a) Analyze: We are given the number of grams of solute (13.5 g) and the number of grams of solvent (0.100 kg = 100 g). From this we must calculate the mass percent­ age of solute. Plan: We can calculate the mass percentage by using Equation 13.5. The mass of the solution is the sum of the mass of solute (glucose) and the mass of solvent (water).

13.4

Solve: Mass % of glucose

=

mass glucose mass soln

13.5 g

x 100 = 13 5 g

+ 100 g

Ways of Expressing Concentration

x 100 = 11.9%

Comment: The mass percentage of water in this solution is (100 - 11 .9)% = 88.1 %.

(b) Analyze: In this case we are given the number of micrograms of solute. Because 1 !-'g is 1 X 10.. ,; g, 5.4J.'g = 5.4 X 10..,; g.

Plan: We calculate the parts per million using Equation 13.6. 5.4 X 10-6 g mass of solute x 106 = Solve: ppm = x 106 = 2.5 g mass of soln

- PRACTICE EXERCISE

2.2 ppm

(a) Calculate the mass percentage of NaCI in a solution containing 1.50 g of NaCI in 50.0 g of water. (b) A commercial bleaching solution contains 3.62 mass % sodium hypochlorite, NaOCI. What is the mass of NaOCl in a bottle containing 2.50 kg of bleaching solution? Answers: (a) 2.91 %, (b) 90.5 g of NaOCl

Mole Fraction, Molarity, and Molality Concentration expressions are often based on the number of moles of one or more components of the solution. The three most commonly used are mole fraction, molarity, and molality. Recall from Section 10.6 that the mole fraction of a component of a solution is given by Mole fraction of component =

moles of component tota1 mo1es of a11 components

[13.7]

The symbol X is commonly used for mole fraction, with a subscript to indicate the component of interest. For example, the mole fraction of HCI in a hydro­ chloric acid solution is represented as XHCI · Thus, a solution containing 1.00 mol of HCI (36.5 g) and 8.00 mol of water (144 g) has a mole fraction of HCI of XHcl = (1.00 mol)/(1.00 mol + 8.00 mol) = 0.111. Mole fractions have no units because the units in the numerator and the denominator cancel. The sum of the mole fractions of all components of a solution must equal 1 . Thus, in the aque­ ous HCI solution, XH,o = 1 .000 - 0.111 = 0.889. Mole fractions are very useful when dealing with gases as we saw in Section 10.6 but have limited use when dealing with liquid solutions. Recall from Section 4.5 that the molarity (M) of a solute in a solution is de­ fined as moles solute Molarity = [13.8] liters soln For example, if you dissolve 0.500 mol of Na2C03 in enough water to form 0.250 L of solution, then the solution has a concentration of (0.500 mol)/(0.250 L) = 2.00 M in Na2C03. Molarity is especially useful for relating the volume of a solution to the quantity of solute it contains, as we saw in our cliscussions of titrations. cx:c (Section 4.6) The molality of a solution, denoted m, is a unit that we have not encoun­ tered in previous chapters. This concentration unit equals the number of moles of solute per kilogram of solvent: Molality =

moles of solute kilograms of solvent

[l3_9]

Thus, if you form a solution by mixing 0.200 mol of NaOH (8.00 g) and 0.500 kg of water (500 g), the concentration of the solution is (0.200 mol)/(0.500 kg) = 0.400 m (that is, 0.400 molal) in NaOH.

543

544

C HA P T E R 1 3

Properties of Solutions The definitions of molarity and molality are similar enough that they can be easily confused. Molarity depends on the volume of solution, whereas molali­ ty depends on the

mass of solvent.

When water is the solvent, the molality and

molarity of dilute solutions are numerically about the same because 1 kg of sol­ vent is nearly the same as 1 kg of solution, and 1 kg of the solution has a volume of about

1 L.

The molality of a given solution does not vary with temperature because masses do not vary with temperature. Molarity, however, changes with temper­ ature because the expansion or contraction of the solution changes its volume. Thus molality is often the concentration unit of choice when a solution is to be used over a range of temperatures.

GIVE IT SOME THOUGHT If an aqueous solution is very dilute, will the molality of the solution be (a) greater than its molarity, (b) nearly the same as its molarity, or (c) smaller than its molarity?

I Calculation of Molality A solution is made by dissolving 4.35 g glucose (C 6H 1206) in 25.0 mL of water at 25 oc. Calculate the molality of glucose in the solution. Water has a density of 1.00 g/mL.

- SAMPLE EXERCISE 1 3.5

SOLUTION

We are asked to calculate a molality. To do this, we must determine the number of moles of solute (glucose) and the number of kilograms of solvent (water). Plan: We use the molar mass of C6H 1 206 to convert grams to moles. We use the density of water to convert mililliters to kilo­ grams. The molality equals the number of moles of solute divided by the number of kilograms of solvent (Equation 13.9). Solve: Use the molar mass of glucose, 180.2 g/mL, to convert grams to moles:

Analyze:

Because water has a density of 1.00 g/mL, the mass of the solvent is

(25.0 mL)(1.00 g/mL)

=

25.0 g

=

0.0250 kg

Finally, use Equation 13.9 to obtain the molality: - PRACTICE EXERCISE

What is the molality of a solution made by dissolving 36.5 g of naphthalene (C 10H8) in 425 g of toluene (C7H8)?

Answer: 0.670 m

Conversion of Concentration U n its Sometimes the concentration of a given solution needs to be known

in several

different concentration units. It is possible to interconvert concentration units as shown in Sample Exercises

13.6 and 13.7.

I Calculation of Mole Fraction and Molality An aqueous solution of hydrochloric acid contains 36% HCl by mass. (a) Calculate the mole fraction of HCl in the solution. (b) Calculate the molality of HCl in the solution.

- SAMPLE EXERCISE 1 3.6

SOLUTION

We are asked to calculate the concentration of the solute, HCI, in two related concentration units, given only the per­ centage by mass of the solute in the solution. Analyze:

13.4

Ways of Expressing Concentration

545

Plan: In converting concentration units based on the mass or moles of solute and solvent (mass percentage, mole fraction, and molality), it is useful to assume a certain total mass of solution. Let's assume that there is exactly 100 g of solution. Because the solution is 36% HCl, it contains 36 g of HCl and (100 - 36) g = 64 g of H20. We must convert grams of solute (HCl) to moles to calculate either mole fraction or molality. We must convert grams of solvent (H20) to moles to calculate mole fractions, and to kilograms to calculate molality. 1 mol HCI Solve: (a) To calculate the mole fraction of = 0.99 mol HCl Moles HCl = (36 g HCl) 36.5 g HCl HCl, we convert the masses of HCI and H20 to moles and then use Equation 13.7: 1 mol H20 = 3.6 mol H20 Moles H20 = (64 g H20) 18 g H20 moles HCl � = 0.99 XHCI = = = 0_22 moles H 20 + moles HCI 3.6 + 0.99 4.6

( (

(b) To calculate the molality of HCl in the solution, we use Equation 13.9. We calculat­ ed the number of moles of HCI in part (a), and the mass of solvent is 64 g = 0.064 kg:

Molality of HCI =

0.99 mol HCl 0_064 kg H20

) )

=

15 m

- PRACTICE EXERCISE A commercial bleach solution contains 3.62 mass % NaOCI in water. Calculate (a) the mole fraction and (b) the molality of NaOCI in the solution. Answers: (a) 9.00 x 10-3, (b) 0.505 m.

To interconvert molality and molarity, we need to know the density of the solution. Figure 13.19 T outlines the calculation of the molarity and molality of a solution from the mass of solute and the mass of solvent. The mass of the so­ lution is the sum of masses of the solvent and solute. The volume of the solu­ tion can be calculated from its mass and density.

Molality (mol/kg solvent) Mass of solvent

Mass of solute

Moles of solute

Molarity (moi/L solution)

A Figure 1 3.1 9 Calculating molality and molarity. This diagram summarizes the calculation of molality and molarity from the mass of the solute, the mass of the solvent, and the density of the solution.

546

C HA P T E R

13

Properties o f Solutions

I Calculation of Molarity Using the Density of the Solution solution with a density of 0.876 g/mL contains 5.0 g of toluene (C7H8) and 225 g of benzene. Calculate the molarity of the solution.

• SAMPLE EXERCISE 1 3.7 A

SOLUTION

Our goal is to calculate the molarity of a solution, given the masses of solute (5.0 g) and solvent (225 g) and the density of the solution (0.876 g/mL). Plan: The molarity of a solution is the number of moles of solute divided by the number of liters of solution (Equation 13.8). The number of moles of solute (C7H8) is calculated from the number of grams of solute and its molar mass. The volume of the solu­ tion is obtained from the mass of the solution (mass of solute + mass of solvent = 5.0 g + 225 g = 230 g) and its density. 1 mol C7H8 Solve: The number of moles of solute is Moles C7Ha = (5.0 g C7Ha) 92 g C7H8 = 0.054 mol The density of the solution is used to con­ 1 L vert the mass of the solution to its volume: = 263 mL Milliliters soln = (230 g) 08 g Molarity is moles of solute per liter of . = moles C7Ha = 0.054 mol C7Ha 1000 mL soln = 0·21 M solution: Molanty liter soln 263 mL soln 1 L soln Check: The magnitude of our answer is reasonable. Rounding moles to 0.05 and (0.05 mol)/(0.25 L) = 0.2 M liters to 0.25 gives a molarity of Analyze:

(

)

( �) ) (

(

)(

)

The units for our answer (mol/L) are correct, and the answer, 0.21 M, has two significant figures, corresponding to the number of significant figures in the mass of solute (2). Comment: Because the mass of the solvent (0.225 kg) and the volume of the solution (0.263 L) are similar in magnitude, the molarity and molality are also similar in magnitude:

(0.054 mol C7H8)/(0.225 kg solvent)

=

0.24 m

- PRACTICE EXERCISE

A solution containing equal masses of glycerol (C3H803) and water has a density of 1.10 g/mL. Calculate (a) the molality of glycerol, (b) the mole fraction of glycerol, (c) the molarity of glycerol in the solution.

Answers: (a) 10.9 m, (b) Xc,H,o, = 0.163, (c) 5.97 M

1 3 . 5 C OLLI GATIVE P ROPERTIES Some physical properties o f solutions differ i n important ways from those of

the pure solvent. For example, pure water freezes at 0 °C, but aqueous solutions

freeze at lower temperatures. Ethylene glycol is added to the water in radiators of cars as an antifreeze to lower the freezing point of the solution. It also raises the boiling point of the solution above that of pure water, making it possible to operate the engine at a higher temperature. The lowering of the freezing point and the raising of the boiling point are physical properties of solutions that depend on the quantity (concentration) but

kind or identity of the solute particles. Such properties are called colligative pro p erties (Colligative means "depending on the collection"; col­

not the

.

ligative properties depend on the collective effect of the number of solute parti­ cles.) In addition to the decrease in freezing point and the increase in boiling point, vapor-pressure reduction and osmotic pressure are colligative proper­

ties. As we examine each of these, notice how the concentration of the solute af­ fects the property relative to that of the pure solvent.

Lowering the Vapor Pressure We learned in Section 11.5 that a liquid in a closed container will establish equilib­ rium with its vapor. When that equilibrium is reached, the pressure exerted by the vapor is called the

vapor pressure. A substance that has no measurable vapor

pressure is nonvolatile, whereas one that exhibits a vapor pressure is volatile.

13.5

Colligative Properties

547

When we compare the vapor pressures of various solvents with those of their solutions, we find that adding a nonvolatile solute to a solvent always lowers the vapor pressure. This effect is illustrated in Figure 13.20 .,.. The extent to which a nonvolatile solute lowers the vapor pressure is proportional to its concentration. This relationship is expressed by Raoult's law, which states that the partial pressure exerted by solvent vapor above a solution, P1v equals the product of the mole fraction of the solvent in the solution, XA, times the vapor pressure of the pure solvent, P'}.. :

[13.10] For example, the vapor pressure of water is 17.5 torr at 20 'C. Imagine holding the temperature constant while adding glucose (C6H1206) to the water so that the resulting solution has XH,o = 0.800 and Xc6H12o6 = 0.200. According to Equation 13.10, the vapor pressure of water over the solution will be 80.0% of that of pure water: Jlc.i,o

=

(0.800)(17.5 torr)

=

Solvent alone

Solvent + solute

(a)

(b)

.l Figure 1 3 .20 Vapor-pressure lowering. The vapor pressure over a solution formed by a volatile solvent and a nonvolatile solute (b) is lower than that of the solvent alone (a). The extent of the decrease in the vapor pressure upon addition of the solute depends on the concentration of the solute.

14.0 torr

In other words, the presence of the nonvolatile solute lowers the vapor pressure of the volatile solvent by 17.5 torr 14.0 torr = 3.5 torr. Raoult's law predicts that when we increase the mole fraction of non­ volatile solute particles in a solution, the vapor pressure over the solution will be reduced. In fact, the reduction in vapor pressure depends on the total con­ centration of solute particles, regardless of whether they are molecules or ions. Remember that vapor-pressure lowering is a colligative property, so it depends on the concentration of solute particles and not on their kind. -

G IVE IT SOME THOUGHT Adding 1 mol of NaCl to 1 kg of water reduces the vapor pressure of water more than adding 1 mol of C6H1206. Explain.

• SAMPLE EXERCISE I Calculation of Vapor-Pressure lowering Glycerin (C3Hs03) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25 'C. Calculate the vapor pressure at 25 'C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25 'C is 23.8 torr (Appendix B), and its density is 1.00 g/mL.

13.8

SOLUTION Analyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.

(

)(

)

Plan: We can use Raoult's law (Equation 13.10) to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, XA, is the ratio of the number of moles of solvent (H20) to total moles of solution (moles C3H803 + moles H20). Solve: To calculate the mole fraction of water in the solution, we must deter­ mine the number of moles of C3H803 and H20:

We now use Raoult's law to calculate the vapor pressure of water for the solution:

1.26 g C3Hs03 1 mol C3Hs03 (50.0 mL C3H803) 1 92_1 g C3H803 mL C3H803 1.00 g Hz0 1 mol H20 Moles H20 = (500.0 mL H zO) = 27.8 mol 1 mL HzO 18_0 g HzO mol H20 27.8 = 0_976 XH,o = mol H20 + mol C3Hs03 27.8 + 0.684 Moles C�8o3 =

(

)(

)

=

0.684 mol

(0.976)(23.8 torr) = 23.2 torr 0.6 torr relative to that of pure water.

l1i,o = XH,o Pf'r,o =

The vapor pressure of the solution has been lowered by - PRACTICE EXERCISE

The vapor pressure of pure water at 110 'C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110 'C. Assuming that Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution?

Answer: 0.290

548

C HA P T E R

13

Properties o f Solutions

An ideal gas obeys the ideal-gas equation (Section 10.4), and an ideal solu­ tion obeys Raoult's law. Real solutions best approximate ideal behavior when the solute concentration is low and when the solute and solvent have similar molecular sizes and similar types of intermolecular attractions. Many solutions do not obey Raoult's law exactly: They are not ideal solu­ tions. If the intermolecular forces between solvent and solute are weaker than those between solvent and solvent and between solute and solute, then the vapor pressure tends to be greater than predicted by Raoult's law. Conversely, when the interactions between solute and solvent are exceptionally strong, as might be the case when hydrogen bonding exists, the vapor pressure is lower than Raoult's law predicts. Although you should be aware that these departures from ideal solution occur, we will ignore them for the remainder of this chapter.

GIVE IT SOME THOUGHT In Raoult's law, PA

= XAP'A, what are the meanings of PA

and PA?

IDEAL SOLUTIONS WITH TWO OR MORE VOLATILE COMPONENTS

S Gasoline, for example, is a complex solution containing olutions sometimes have two or more volatile components.

several volatile substances. To gain some understanding of such mixtures, consider an ideal solution containing two com­ ponents, A and B. The partial pressures of A and B vapors above the solution are given by Raoult's law: PA = XAPA and Ps = XePl\ The total vapor pressure over the solution is the sum of the partial pressures of each volatile component: Ptota l = PA

+

Pa = XAP'A

+

XaPl\

Consider, for example, a mixture of benzene (C6H 6) and toluene (C7H8) containing 1.0 mol of benzene and 2.0 mol of toluene (Xben = 0.33 and Xtol = 0.67). At 20 oc the vapor pressures of the pure substances are Benzene: Pben = 75 torr Toluene: P loi = 22 torr Thus, the partial pressures of benzene and toluene above the solution are ll,.,n = (0.33)(75 torr) = 25 torr P101 = (0.67)(22 torr) = 15 torr The total vapor pressure is Ptotal = 25 torr + 15 torr = 40 torr The vapor, therefore, is richer in benzene, the more volatile component. The mole fraction of benzene in the vapor is given by the ratio of its vapor pressure to the total pressure (Equation 10.15): . Fben 25 torr = 0 63 = Xben m vapor = . 40 torr Ptotal Although benzene constitutes only 33% of the molecules in the solution, it makes up 63% of the molecules in the vapor. When ideal solutions are in equilibrium with their vapor, the more volatile component of the mixture will be relatively

richer in the vapor. This fact forms the basis of dstillation, i a technique used to separate (or partially separate) mixtures containing volatile components. Distillation is the procedure by which a moonshiner obtains whiskey using a still and by which petrochemical plants achieve the separation of crude petroleum into gasoline, diesel fuel, lubricating oil, and so forth (Figure 13.21T). Distillation is also used routinely on a small scale in the laboratory. A specially designed fractional distillation apparatus can achieve in a single operation a de­ gree of separation that would be equivalent to several succes­ sive simple distillations.

Related Exercises: 13.63, 13.64

"' Figure 1 3.21 Separating volatile components. In an industrial distillation tower, such as the ones shown here, the components of a volatile organic mixture are separated according to boiling-point range.

13.5

Colligative Properties

549

Boi l ing-Point Elevation In Sections 11.5 and 11.6 we examined the vapor pres­ sures of pure substances and how they can be used to Liquid construct phase diagrams. How will the phase diagram of a solution, and hence its boiling and freezing points, differ from those of the pure solvent? The addition of a nonvolatile solute lowers the vapor pressure of the solution. Thus, as shown in Figure 13.22 �, the vapor­ pressure curve of the solution (blue line) will be shifted downward relative to the vapor-pressure curve of the Boiling point of solution pure liquid (black line); at any given temperature Freezing point 1 the vapor pressure of the solution is lower than that of solution Tnple pomt 11 . . t Boiling point of the pure liquid. Recall that the normal boiling point 1 F reezmg porn 1 of solution 1 of solvent of a liquid is the temperature at which its vapor pres­ of solvent sure equals 1 atm. = (Section 11 .5) At the normal Temperature boiling point of the pure liquid, the vapor pressure of the solution will be less than 1 atm (Figure 13.22). Therefore, a higher temperature is required to attain a vapor pressure of 1 atm. .a. Figure 1 3.22 Phase diagrams for a pure solvent and for a solution of a Thus, the boiling point of the solution is higher than that of the pure liquid. nonvolatile solute. The vapor pressure The increase in boiling point relative to that of the pure solvent, t:J.Tb, is a of the solid solvent is unaffected by the positive quantity obtained by subtracting the boiling point of the pure solvent presence of solute if the solid freezes out from the boiling point of the solution. The value of t:J.Tb is directly proportional without containing a significant concentration of solute, as is usually to the concentration of the solution expressed by its molality, m: •

t:J.Tb

=

Kbm

•

Vv

[13.11]

the case.

The magnitude of Kb, which is called the molal boiling-point-elevation constant, depends only on the solvent. Some typical values for several common solvents are given in Table 13.4 \'. Because solutions generally do not behave ideally, however, the constants listed in Table 13.4 serve well only for solutions that are not too concentrated. For water, Kb is 0.51 'C/m; therefore, a 1 m aqueous solution of sucrose or any other aqueous solution that is 1 m in nonvolatile solute particles will boil 0.51 'C higher than pure water. The boiling-point elevation is proportional to the concentration of solute particles, regardless of whether the particles are molecules or ions. When NaCI dissolves in water, 2 mol of solute particles (1 mol of Na+ and 1 mol of Cl-) are formed for each mole of NaCl that dissolves. Therefore, a 1 m aqueous solution ofNaCI is 1 m in Na+ and 1 m in CC making it 2 m in total solute particles. As a result, the boiling-point elevation of a 1 m aqueous solution of NaCI is approximately (2 m)(0.51 'C/m) = 1 'C, twice as large as a 1 m solution of a nonelectrolyte such as sucrose. Thus, to properly predict the effect of a particular solute on the boiling point (or any other colliga­ tive property), it is important to know whether the solute is an electrolyte or a nonelectrolyte. = (Sections 4.1 and 4.3)

G IVE IT SOME THOUGHT An unknown solute dissolved in water causes the boiling point to increase by 0.51 'C. Does this mean that the concentration of the solute is 1.0 m? TABLE 1 3.4 • Molal Boiling-Point-Elevation and Freezing-Point-Depression Constants

Solvent

Normal Boiling Point ('C)

Kb ('Cjm)

Normal Freezing Point ('C)

Kt ('Cjm)

Water, H20 Benzene, C6H6 Ethanol, C2HsOH Carbon tetrachloride, CCI4 Chloroform, CHCl3

100.0 80.1 78.4 76.8 61.2

0.51 2.53 1.22 5.02 3.63

0.0 5.5 -114.6 -22.3 -63.5

1.86 5.12 1.99 29.8 4.68

550

C HA P T E R

13

Properties of Solutions

Freezing-Point Depression When a solution freezes, crystals of pure solvent usually separate out; the solute molecules are not normally soluble in the solid phase of the solvent. When aqueous solutions are partially frozen, for example, the solid that sepa­ rates out is almost always pure ice. As a result, the part of the phase diagram in Figure

13.22 that represents the vapor pressure of the solid is the same as that

for the pure liquid. The vapor-pressure curves for the liquid and solid phases meet at the triple point.

cx:o

(Section

11.6) In Figure 13.22 we see that the triple

point of the solution must be at a lower temperature than that in the pure liquid because the solution has a lower vapor pressure than the pure liquid. The freezing point of a solution is the temperature at which the first crystals of pure solvent begin to form in equilibrium with the solution. Recall from Section

11.6

that the line representing the solid-liquid equilibrium rises nearly

vertically from the triple point. Because the triple-point temperature of the solu­

the freezing point of the solution is lower tlwn that ofthe pure liquid. The decrease in freezing point, 6. T1, is a positive quan­

tion is lower than that of the pure liquid,

tity obtained by subtracting the freezing point of the solution from the freezing point of the pure solvent. Like the boiling-point elevation, 6. Tf is directly proportional to the molality

of the solute:

[13.12]

The values of Kf, the molal freezing-point-depression constant, for several common solvents are given in Table 13.4. For water, K1 is 1.86 °C/m; therefore, a

1 m aqueous solution of sucrose

Antifreeze being added to an automobile radlatlor. Antifreeze consists of an aqueous solution of ethylene glycol, CH2(0H)CH2(0H).

1 m in m NaCl) will freeze 1.86 °C lower than

or any other aqueous solution that is

nonvolatile solute particles (such as 0.5

pure water. The freezing-point lowering caused by solutes explains the use of antifreeze in cars and the use of calcium chloride (CaC12) to melt ice on roads during winter.

- SAMPLE EXERCISE

13.9 1 Calculation of Boiling-Point Elevation

and Freezing-Point Lowering Automotive antifreeze consists of ethylene glycol, CH2(0H)CH2(0H), a nonvolatile nonelectrolyte. Calculate the boiling point and freezing point of a 25.0 mass % solu­ tion of ethylene glycol in water. SOLUTION

Analyze: We are given that a solution contains 25.0 mass % of a nonvolatile, nonelec­ trolyte solute and asked to calculate the boiling and freezing points of the solution. To do this, we need to calculate the boiling-point elevation and freezing-point depression. Plan: To calculate the boiling-point elevation and the freezing-point depression using Equations 13.11 and 12, we must express the concentration of the solution as molality. Let's assume for convenience that we have 1000 g of solution. Because the solution is 25.0 mass % ethylene glycol, the masses of ethylene glycol and water in the solution are 250 and 750 g, respectively. Using these quantities, we can calculate the molality of the solution, which we use with the molal boiling-point-elevation and freezing-point­ depression constants (Table 13.4) to calculate LlTb and LlTt. We add LlTb to the boiling point and subtract Ll T1 from the freezing point of the solvent to obtain the boiling point and freezing point of the solution. Solve: The molality of the solution is calculated as follows: 250 g CzH602 1 mol C2H 602 1000 g HzO moles C2H 602 . Molality = = kilograms HzO 750 g HzO 62.1 g C2H 602 1 kg HzO = 5.37 m We can now use Equations 13.11 and 13.12 to calculate the changes in the boiling and freezing points: LlTb = Kbm = (0.51 °C/m)(5.37 m) = 2.7 °C LlTt = Ktm = (1.86 °C/m)(5.37 m) = 10.0 °C

(

)(

)(

)

13.5

Hence, the boiling and freezing points of the solution are Boiling point

=

(normal bp of solvent)

= 100.0

oc + 2.7 oc

+ �Tb

= 102.7 oc

- � Tt oc - 10.0 oc = -10.0 oc

Freezing point = (normal fp of solvent) = 0.0

Comment: Notice that the solution is a liquid over a larger temperature range than the pure solvent. - PRACTICE EXERCISE

Calculate the freezing point of a solution containing 0.600 kg of CHC13 and 42.0 g of eucalyptol (C10H180), a fragrant substance found in the leaves of eucalyptus trees. (See Table 13.4.)

Answer: -65.6 oc

I Solutions List the following aqueous solutions in order of their expected freezing point:

- SAMPLE EXERCISE

13.10

Freezing-Point Depression in Aqueous

CaClz, 0.15 111 NaCI, 0.10 111 HCI, 0.050 m CH3COOH, 0.10 m C12H22011.

0.050 111

SOLUTION

We must order five aqueous solutions according to expected freezing points, based on molalities and the solute formulas. Analyze:

Plan: The lowest freezing point will correspond to the solution with the greatest con­ centration of solute particles. To determine the total concentration of solute particles in each case, we must determine whether the substance is a nonelectrolyte or an elec­ trolyte and consider the number of ions formed when an electrolyte ionizes. Solve: CaC12, NaCI, and HCl are strong electrolytes, CH3COOH (acetic acid) is a weak electrolyte, and C12H22011 is a nonelectrolyte. The molality of each solution in total particles is as follows: 2 0.050 m CaC12 => 0.050 m in Ca + and 0.10 m in Cl- => 0.15 m in particles 0.15 m NaCI => 0.15 m Na+ and 0.15 m in Cl- => 0.30 m in particles 0.10 m HCI => 0.10 m H+ and 0.10 m in Cl- => 0.20 m in particles 0.050 m CH3COOH => weak electrolyte => between 0.050 m and 0.10 m in particles 0.10 m C12H22011 => nonelectrolyte => 0.10 m in particles

Because the freezing points depend on the total molality of particles in solution, the expected ordering is 0.15 m NaCI (lowest freezing point), 0.10 m HCl, 0.050 m CaC12, 0.10 m C12H2p11, and 0.050 m CH3COOH (highest freezing point). - PRACTICE EXERCISE

Which of the following solutes will produce the largest increase in boiling point upon addition to 1 kg of water: 1 mol of Co(N03)z, 2 mol of KCI, 3 mol of ethylene glycol (C,H60,)? Answer: 2 mol of KCI because it contains the highest concentration of particles, 2 m K+ and 2 m Cl-, giving 4 m in all

Osmosis Certain materials, including many membranes in biological systems and syn­ thetic substances such as cellophane, are semipermeable. When in contact with a solution, they allow some molecules to pass through their network of tiny pores but not others. Most importantly, semipermeable membranes generally allow small solvent molecules such as water to pass through but block larger solute molecules or ions. This selectivity gives rise to some interesting and im­ portant applications.

Colligative Properties

551

552

C HA P T E R 1 3

Properties o f Solutions Consider a situation in which only solvent molecules are able to pass through a membrane. If such a membrane is placed between two solutions of different concentration, solvent molecules move in both directions through the membrane. The concentration of solvent is higher in the solution containing less solute, however, so the rate with which solvent passes from the less concentrat­ ed (lower solute concentration) to the more concentrated solution (higher solute concentration) is greater than the rate in the opposite direction. Thus, there is a net movement of solvent molecules from the less concentrated solution into the more concentrated one. In this process, called osmosis, the net movement of sol­ vent is always toward the solution with the higher solute concentration.

Osmosis is illustrated in Figure 13.23 T. Let's begin with two solutions of

different concentration separated by a semipermeable membrane. Because the solution on the left is more concentrated than the one on the right, there is a net movement of solvent through the membrane from right to left, as if the solutions

OSMOSIS

I n osmosis, the net movement of solvent i s always toward the solution with the higher solute concentration. There is a net movement of solvent through the semipermeable membrane, as if the solutions were driven to attain equal concentrations. The difference in liquid level, and thus pressure, eventually causes theflow to cease. Applying pressure to the arm with the higher liquid level can also halt the flow. Applied pressure,

II, stops net move­

ment of solvent.

Net movement of solvent from the pure solvent or a solution with low solute concentration to a solution with high solute concentration.

&

Figure

1 3.23 Osmosis,

Osmosis stops when the column of a solution on the left becomes high enough to exert sufficient pressure at the membrane to counter the net movement of solvent. At this point the solution on the left has become more dilute, but there stil1 exists a difference in concentrations between the two solutions.

Applied pressure on the left arm of the apparatus stops net movement of solvent from the right side of the semipermeable membrane. This applied pressure is the osmotic pressure of the solution.

13.5

Colligative Properties

553

were driven to attain equal concentrations. As a result, the liquid levels in the two arms become unequal. Eventually, the pressure difference resulting from the unequal heights of the liquid in the two arms becomes so large that the net flow of solvent ceases, as shown in the center panel. Alternatively, we may apply pressure to the left arm of the apparatus, as shown in the panel on the right, to halt the net flow of solvent. The pressure required to prevent osmosis by pure solvent is the osmotic pressure, ll of the solution. The osmotic pressure obeys a law similar in form to the ideal-gas law, nv = nRT where V is the volume of the solution, n is the number of moles of solute, R is the ideal-gas constant, and T is the temperature on the Kelvin scale. From this equation, we can write ll =

(�)RT

=

[13.13]

MRT

where M is the molarity of the solution. If two solutions of identical osmotic pressure are separated by a semiper­ meable membrane, no osmosis will occur. The two solutions are isotonic. If one solution is of lower osmotic pressure, it is hypotonic with respect to the more concentrated solution. The more concentrated solution is hypertonic with re­ spect to the dilute solution.

0.5

0.20

GIVE IT SOME THOUGHT Of two KBr solutions, one to the other?

m

and the other

m, which is

hypotonic with respect

Osmosis plays a very important role in living sys­ tems. The membranes of red blood cells, for example, are semipermeable. Placing a red blood cell in a solu­ tion that is hypertonic relative to the intracellular solu­ tion (the solution within the cells) causes water to move out of the cell, as shown in Figure 13.24(a) �. This causes the cell to shrivel, a process called crenation. Placing the cell in a solution that is hypotonic relative to the intracellular fluid causes water to move into the cell, as illustrated in Figure 13.24(b). This may (a) Crenation cause the cell to rupture, a process called hemolysis. People who need body fluids or nutrients replaced but cannot be fed orally are given solutions by intravenous (IV) infusion, which feeds nutrients directly into the veins. To prevent crenation or hemolysis of red blood cells, the IV solu­ tions must be isotonic with the intracellular fluids of the cells.

- SAMPLE EXERCISE

13.11

25 oc. What molarity of glucose

I Calculations Involving Osmotic Pressure

The average osmotic pressure of blood is 7.7 attn at (C6Hl206) will be isotonic with blood? SOLUTION

25

Analyze: We are asked to calculate the concentration of glucose in water that would be isotonic with blood, given that the osmotic pressure of blood at oc is 7.7 atm. Plan: Because we are given the osmotic pressure and temperature, we can solve for the concentration, using Equation Solve: II =

M =

MRT

_I!_ = RT

13.13.

attn 0.31 L-attn (0.0821 mo 8 )( 9 2 K) l-K 7.7

=

M

(b) Hemolysis

"" Figure

1 3.24 Osmosis through red blood cell wall. The blue arrows represent the net movement of water molecules.

554

C HA P T E R 1 3

Properties o f Solutions

Comment: In clinical situations the concentrations of solutions are generally ex­ pressed as mass percentages. The mass percentage of a 0.31 M solution of glucose is 5.3%. The concentration of NaCI that is isotonic with blood is 0.16 M, because NaCl ionizes to form two particles, Na+ and Cl- (a 0.155 M solution of NaCI is 0.310 M in particles). A 0.16 M solution of NaCI is 0.9 mass % in NaCI. This kind of solution is known as a physiological saline solution. - PRACTICE EXERCISE

What is the osmotic pressure at 20 oc of a 0.0020 M sucrose (C12H22011 ) solution? Answer: 0.048 atm, or 37 torr

C O L L I GATIVE P R O P E RTIES OF ELECTROLYTE S O L UT I O N S

T concentration of solute particles, regardless of whether

he colligative properties of solutions depend on the total

the particles are ions or molecules. Thus, we would expect a 0.100 m solution of NaCl to have a freezing-point depression of (0.200 m)(1.86 oC/m) = 0.372 oc because it is 0.100 m in Na+ (aq) and 0.100 m in o- (aq). The measured freezing-point depression is only 0.348 oc, however, and the situation is similar for other strong electrolytes. A 0.100 m solution of KCI, for example, freezes at -0.344 °C. The difference between the expected and observed colliga­ tive properties for strong electrolytes is due to electrostatic at­ tractions between ions. As the ions move about in solution, ions of opposite charge collide and "stick together" for brief moments. While they are together, they behave as a single par­ ticle called an ion pair (Figure 13.25 'f'). The number of indepen­ dent particles is thereby reduced, causing a reduction in the freezing-point depression (as well as in the boiling-point eleva­ tion, the vapor-pressure reduction, and the osmotic pressure).

One measure of the extent to which electrolytes dissociate is the van't Hofffactor, i. This factor is the ratio of the actual value of a colligative property to the value calculated, assum­ ing the substance to be a nonelectrolyte. Using the freezing­ point depression, for example, we have

L = !:l.Tf (calculated for nonelectrolyte) � TJ (measured)

.

[13.14]

The ideal value of i can be determined for a salt from the num­ ber of ions per formula unit. For N aCI, for example, the ideal van't Hoff factor is 2 because NaCI consists of one Na+ and one o- per formula unit; for K2S04 it is 3 because K2S04 con­ sists of two K+ and one so/-. In the absence of any informa­ tion about the actual value of i for a solution, we will use the ideal value in calculations. Table 13.5 'f' gives the observed van't Hoff factors for sev­ eral substances at different dilutions. Two trends are evident in these data. First, dilution affects the value of i for electro­ lytes; the more dilute the solution, the more closely i ap­ proaches the ideal or limiting value. Thus, the extent of ion pairing in electrolyte solutions decreases upon dilution. Second, the lower the charges on the ions, the less i departs from the limiting value because the extent of ion pairing decreases as the ionic charges decrease. Both trends are consis­ tent with simple electrostatics: The force of interaction between charged particles decreases as their separation in­ creases and as their charges decrease.

Related Exercises: 13.79, 13.80, 13.99, 13.102

•

Concentration Compound

A Figure 1 3.25 lon pairing and colllgatlve properties. A solution of NaCI contains not only separated Na +(aq) and cnaq) ions but ion pairs as well. lon pairing becomes more prevalent as the solution concentration increases and has an effect on all the colligative properties of the solution.

Sucrose NaCI K2S04 MgS04

0.100 m

1.00 1.87 2.32 1.21

0.0100

1.00 1.94 2.70 1.53

m 0.00100 m

1.00 1.97 2.84 1.82

Limiting Value

1.00 2.00 3.00 2.00

13.5

Colligative Properties

555

There are many interesting biological examples of osmosis. A cucumber placed in concentrated brine loses water via osmosis and shrivels into a pickle. If a carrot that has become limp because of water loss to the atmosphere is placed in water, the water moves into the carrot through osmosis, making it firm once again. People who eat a lot of salty food retain water in tissue cells and intercellular space because of osmosis. The resultant swelling or puffiness is called edema. Water moves from soil into plant roots and subsequently into the upper portions of the plant at least in part because of osmosis. Bacteria on salted meat or canclied fruit lose water through osmosis, shrivel, and die-thus preserving the food. The movement of a substance from an area where its concentration is high to an area where it is low is spontaneous. Biological cells transport water and other select materials through their membranes, permitting nutrients to enter and waste materials to exit. In some cases substances must be moved across the cell membrane from an area of low concentration to one of high concentration. This movement-called active transport-is not spontaneous, so cells must ex­ pend energy to do it.

G IVE IT SOME THOUGHT Is the osmotic pressure of a 0.10 M solution of NaCI greater than, less than, or equal to that of a 0.10 M solution of KBr? Determination of Molar Mass The colligative properties of solutions provide a useful means of experimental­ ly determining molar mass. Any of the four colligative properties can be used, as shown in Sample Exercises 13.12 and 13.13.

- SAMPLE EXERCISE

13.12 I Molar Mass from Freezing-Point Depression

A solution of an unknown nonvolatile nonelectrolyte was prepared by dissolving 0.250 g of the substance in 40.0 g of CCI4. The boiling point of the resultant solution was 0.357 oc higher than that of the pure solvent. Calculate the molar mass of the solute. SOLUTION Analyze:

CCI4, 11 Tb

Our goal is to calculate the molar mass of a solute based on knowledge of the boiling-point elevation of its solution in = 0.357 °C, and the masses of solute and solvent. Table 13.4 gives Kb for the solvent (CC4), Kb = 5.02 °C/m.

Plan: We can use Equation 13.11, 11Tb = Kbm, to calculate the molality of the solution. Then we can use molality and the quanti­ ty of solvent (40.0 g CC14) to calculate the number of moles of solute. Finally, the molar mass of the solute equals the number of grams per mole, so we divide the number of grams of solute (0.250 g) by the number of moles we have just calculated. Solve: From Equation

13.11 we have

Thus, the solution contains 0.0711 mol of solute per kilogram of solvent. The solution was prepared using 40.0 g = 0.0400 kg of solvent (CCI4). The number of moles of solute in the solution is therefore The molar mass of the solute is the number of grams per mole of the substance: - PRACTICE EXERCISE

.

0.357 oc

11 Tb

Molality = K; = = 0.0711 m 5.02 oc;m

(

(0.0400 kg CCI4) 0.0711

mol solute kg CCI4

)

=

2.84

3

x w-

mol solute

0.250 g Molar mass = 88.0 g/mol 2.84 X 10_3 mol =

Camphor (C10H160) melts at 179.8 oc, and it has a particularly large freezing-point-depression constant, Kt = 40.0 °Cfm. When 0.186 g of an organic substance of unknown molar mass is dissolved in 22.01 g of liquid camphor, the freezing point of the mixture is found to be 176.7 oc. What is the molar mass of the solute? Answer: 110 g/mol

556

C HA P T E R 1 3

Properties o f Solutions

I Molar Mass from Osmotic Pressure The osmotic pressure of an aqueous solution of a certain protein was measured to determine the protein's molar mass. The solution contained 3.50 mg of protein dissolved in sufficient water to form 5.00 mL of solution. The osmotic pressure of the solution at 25 oc was found to be 1.54 torr. Treating the protein as a nonelectrolyte, calculate its molar mass.

• SAMPLE EXERCISE

13.13

SOLUTION

Our goal is to calculate the molar mass of a high-molecular-mass protein, based on its osmotic pressure and a knowledge of the mass of protein and solution volume. Analyze:

Plan: The temperature (T = 25 oq and osmotic pressure (II = 1 .54 torr) are given, and we know the value of R so we can use Equation 13.13 to calculate the molarity of the solution, M . 1n doing so, we must convert temperature from °C to K and the osmotic pressure from torr to atm. We then use the molarity and the volume of the solution (5.00 mL) to determine the number of moles of solute. Finally, we obtain the molar mass by dividing the mass of the solute (3.50 mg) by the number of moles of solute. Solve:

(

Solving Equation 13.13 for molarity gives . Molanty

=

=

II

RT

(

)

Because the volume of the solution is 5.00 ml of protein must be Moles

=

(8.28

x

)

1 atm (1.54 torr) 760 torr:...._ -'--::-- _7___ -,-""' L-atm 0.0821 (298 K) mol-K =

=

5.00 X

10-5 mol/1)(5.00 x 10-3 L)

8 28

X

10_5

mol L

10-3 L, the number

=

4.14

X

of moles

10-7 mol

The molar mass is the number of grams per mole of the substance. The sample has a mass of 3.50 mg = 3.50 X 10-3 g. The molar mass is the number of grams divided by the number of moles: Molar mass

=

grams 1 mo es

-

=

3.50 x 10-3 g 4.14 x 10-7 mol

=

8.45

x

103 g/mol

Comment: Because small pressures can be measured easily and accurately, osmotic pressure measurements provide a useful way to determine the molar masses of large molecules. - PRACTICE EXERCISE

A sample of 2.05 g of polystyrene of uniform polymer chain length was dissolved in

enough toluene to form 0.100 L of solution. The osmotic pressure of this solution was found to be 1.21 kPa at 25 oc. Calculate the molar mass of the polystyrene. Answer: 4.20 x 104 g/mol

1 3.6 COLLO I D S When finely divided clay particles are dispersed throughout water, they even­ tually settle out of the water because of gravity. The dispersed clay particles are much larger than most molecules and consist of many thousands or even mil­ lions of atoms. In contrast, the dispersed particles of a solution are of molecular size. Between these extremes lie dispersed particles that are larger than typical molecules, but not so large that the components of the mixture separate under the influence of gravity. These intermediate types of dispersions or suspensions are called colloidal dispersions, or simply colloids. Colloids form the dividing line between solutions and heterogeneous mixtures. Like solutions, colloids can be gases, liquids, or solids. Examples of each are listed in Table 13.6 �.

13.6

TABLE

Colloids

557

1 3.6 • Types of Colloids

Phase of Colloid

Dispersing (solutelike) Substance

Dispersed (solventlike) Substance Colloid Type

Gas Gas Gas Liquid Liquid Liquid Solid Solid Solid

Gas Gas Gas Liquid Liquid Liquid Solid Solid Solid

Gas Liquid Solid Gas Liquid Solid Gas Liquid Solid

Aerosol Aerosol Foam Emulsion Sol Solid foam Solid emulsion Solid sol

The size of the dispersed particles is used to classify a mixture as a colloid. Colloid particles range in diameter from approximately 5 to 1000 nrn. Solute particles are smaller. The colloid particle may consist of many atoms, ions, or molecules, or it may even be a single giant molecule. The hemoglobin molecule, for example, which carries oxygen in blood, has molecular dimensions of 65 A x 55 A x 50 A and a molar mass of 64,500 g/mol. Although colloid particles may be so small that the dispersion appears uni­ form even under a microscope, they are large enough to scatter light very effec­ tively. Consequently, most colloids appear cloudy or opaque unless they are very dilute. (Homogenized milk is a colloid.) Furthermore, because they scatter light, a light beam can be seen as it passes through a colloidal suspension, as shown in Figure 13.26 I> . This scattering of light by colloidal particles, known as the Tyndall effect, makes it possible to see the light beam of an automobile on a dusty dirt road or the sunlight corning through a forest canopy [Figure 13.27(a) "'] . Not all wavelengths are scattered to the same extent. As a result, brilliant red sunsets are seen when the sun is near the horizon and the air con­ tains dust, smoke, or other particles of colloidal size [Figure 13.27(b)].

Example

None (all are solutions) Fog Smoke Whipped cream Milk Paint Marshmallow

Butter Ruby glass

&

Figure 1 3.26 Tyndall effect In the laboratory. The glass on the left contains a colloidal suspension; that on the right contains a solution. The path of the beam through the colloidal suspension is visible because the light is scattered by the colloidal particles. Light is not scattered by the individual solute molecules in the solution.

Hydrophilic and Hydrophobic Colloids The most important colloids are those in which the dispersing medium is water. These colloids may be hydrophilic (water loving) or hydrophobic (water fearing). Hydrophilic colloids are most like the solutions that we have previously examined. In the human body the extremely large molecules that make up such important substances as enzymes and antibodies are kept in sus­ pension by interaction with surrounding water molecules. The molecules fold in such a way that the hydrophobic groups are away from the water molecules, CH4(g) + H2(g) + CO(g)

If the initial pressure of (CH3)zO is 135 torr, what is its pressure after 1420 s?

Answer:

51 torr

Equation 14.13 can be used to verify whether a reaction is first order and to determine its rate constant. This equation has the form of the general equation for a straight line, y = mx + b, in which m is the slope and b is the y-intercept of the line (Appendix A.4):

[]

tn A 1

I

=

-

[]

II I

m·x +

y

Methyl isonitrile

k - 1 + ln A 0

b

For a first-order reaction, therefore, a graph of ln[A]1 versus time gives a straight line with a slope of -k and a y-intercept of ln[A]0. A reaction that is not first order will not yield a straight line. As an example, consider the conversion of methyl isonitrile (CH3NC) to acetonitrile (CH3CN) (Figure 14.6 �). Because experiments show that the reac­ tion is first order, we can write the rate equation: ln[CH3NC]1 =

-kt +

Acetonitrile

ln[CH3NC]0

Figure 14.7(a) Y shows how the pressure of methyl isonitrile varies with time as it rearranges in the gas phase at 198.9 °C. We can use pressure as a unit of concentration for a gas because from the ideal-gas law the pressure is directly proportional to the number of moles per unit volume. Figure 14.7(b) shows a plot of the natural logarithm of the pressure versus time, a plot that yields a straight line. The slope of this line is -5.1 X 10-5 s-1 (You should verify this for yourself, remembering that your result may vary slightly from ours because of inaccuracies associated with reading the graph.) Because the slope of the line equals -k, the rate constant for this reaction equals 5.1 X 10-5 s- 1

A Figure 14.6 A first-order reaction. The transformation of methyl isonitrile (CH3NC) to acetonitrile (CH3CN) is a first-order process. Methyl isonitrile and acetonitrile are isomers, molecules that have the same atoms arranged differently. This reaction is called an isomerization reaction.

GIVE IT SOME THOUGHT

What do the y-intercepts in Figure 14.7(a) and (b) represent?

'§" 140

£

u z 100

£ u �

�

"



N2(g) + O(g)

(slow) (fast) (a) Write the equation for the overall reaction. (b) Write the rate law for the overall reaction. N20(g) + O(g)

--->

N2(g) + 02(g)

SOLUTION Analyze: Given a multistep mechanism with the relative speeds of the steps, we are asked to write the overall reaction and the rate law for that overall reaction. Plan: (a) Find the overall reaction by adding the elementary steps and eliminating

the intermediates. (b) The rate law for the overall reaction will be that of the slow, rate-determining step. (a) Adding the two elementary reactions gives

Solve:

2 N20(g) + O(g) -----> 2 N2(g) + 02(g) + O(g)

Omitting the intermediate, O(g), which occurs on both sides of the equation, gives the overall reaction: 2 N20(g)

--->

2 N2(g) + 02(g)

The rate law for the overall reaction is just the rate law for the slow, rate­ determining elementary reaction. Because that slow step is a unimolecular elemen­ tary reaction, the rate law is first order: (b)

Rate =

- PRACTICE EXERCISE

k[N20)

Ozone reacts with nitrogen dioxide to produce dinitrogen pentoxide and oxygen: 03(g) + 2 N02(g)

--->

The reaction is believed to occur in two steps: 03(g) + N02(gl

N03(g) + N02(gl

--->

--->

N20 s (g) + 02(g) N03(g) + 02(gl N20s(g)

The experimental rate law is rate = k[03][N02]. What can you say about the relative rates of the two steps of the mechanism? Answer: Because the rate law conforms to the molecularity of the first step, that must be the rate-determining step. The second step must be much faster than the first one.

Mechanisms with a Fast Initial Step

It is less straightforward to derive the rate law for a mechanism in which an inter­ mediate is a reactant in the rate-determining step. This situation arises in multistep mechanisms when the first step is fast and therefore not the rate-determining one.

14.6

Let's consider one example: the gas-phase reaction of nitric oxide (NO) with bromme (Brz). [14.24] 2 NO(g) + Br2(g) � 2 NOBr(g) The experimentally determined rate law for this reaction is second order in NO and first order in Br2: Rate = k[NOf[Brz] [14.25] We seek a reaction mechanism that is consistent with this rate law. One possi­ bility is that the reaction occurs in a single termolecular step: NO(g) + NO(g) + Br2(g) � 2 NOBr(g) Rate = k[NOf[Br2] [14.26] As noted in Practice Exercise 14.13, this does not seem likely because termolec­ ular processes are so rare. Let's consider an alternative mechanism that does not invoke a termolecu­ lar step: Step 1: NO(g)

+

Step 2: N0Br2(g)

kl Br2(g) � NOBr2(g)

+

(fast)

NO(g) � 2 NOBr(g) (slow)

this mechanism, step 1 actually involves two processes: a forward reaction and its reverse. Because step 2 is the slow, rate-determining step, the rate law for that step governs the rate of the overall reaction: Rate = k[NOBr2][NO] [14.27] However, NOBr2 is an intermediate generated in step 1. Intermediates are usu­ ally unstable molecules that have a low, unknown concentration. Thus, our rate law depends on the unknown concentration of an intermediate. Fortunately, with the aid of some assumptions, we can express the concen­ tration of the intermediate (N0Br2) in terms of the concentrations of the start­ ing reactants (NO and Br2). We first assume that NOBr2 is intrinsically unstable and that it does not accumulate to a significant extent in the reaction mixture. There are two ways for NOBr2 to be consumed once it is formed: It can either react with NO to form NOBr or reform NO and Br2 . The first of these possibili­ ties is step 2, a slow process. The second is the reverse of step 1, a unirnolecular process: NOBr2(g) � NO(g) + Br2(g) [14.28] Because step 2 is slow, we assume that most of the NOBr2 falls apart according to Equation 14.28. Thus, we have both the forward and reverse reactions of step 1 occurring much faster than step 2. Because the forward and reverse processes of step 1 occur rapidly with respect to the reaction in step 2, they establish an equilibrium. We have seen examples of dynamic equilibrium before, in the equilibrium between a liquid and its vapor cxx:> (Section 11.5) and between a solid solute and its solution. (Section 13.3) As in any dynamic equilibrium, the rates of the forward and reverse reactions are equal. Thus, we can equate the rate expression for the forward reaction in step 1 with the rate expression for the reverse reaction: k1 [NO][Brz] k-1 [NOBr2] 1n

cxx:>

Rate of forward reaction

Rate of reverse reaction

Solving for [N0Br2], we have k [NOBrz] = l [NO][Brz] k-1

Reaction Mechanisms

603

604

C HAPTER 1 4

Chemical Kinetics Substituting this relationship into the rate law for the rate-determining step (Equation 14.27), we have

This result is consistent with the experimental rate law (Equation 14.25). The ex­ perimental rate constant, k, equals k2k1/L1. This mechanism, which involves only unimolecular and bimolecular processes, is far more probable than the single termolecular step (Equation 14.26). In general, whenever a fast step precedes a slow one, we can solve for the con­

centration of an intermediate by assuming that an equilibrium is established in the fast step.

I with a Fast Initial Stepalso produces a rate law con­ Show that the following mechanism for Equation

•

Deriving the Rate Law for a Mechanism

SAMPLE EXERCISE 14.15

sistent with the experimentally observed one:

14.24

(fast equilibrium)

SOLUTION Analyze: We are given a mechanism with a fast initial step and asked to write the rate law for the overall reaction. Plan: The rate law of the slow elementary step in a mechanism determines the rate

law for the overall reaction. Thus, we first write the rate law based on the molecular­ ity of the slow step. In this case the slow step involves the intermediate N202 as a re­ actant. Experimental rate laws, however, do not contain the concentrations of intermediates; instead they are expressed in terms of the concentrations of starting substances. Thus, we must relate the concentration of N202 to the concentration of NO by assuming that an equilibrium is established in the first step. The second step is rate determining, so the overall rate is Rate = k2[NP2HBr2l We solve for the concentration of the intermediate N202 by assuming that an equilib­ rium is established in step 1; thus, the rates of the forward and reverse reactions in step 1 are equal: k1 [NOf = k-1 [N202J Solving for the concentration of the intermediate, N202, gives

Solve:

:�1 [N0]2[Br2l = k [NOj2[Br2l

Substituting this expression into the rate expression gives Rate = k2

Thus, this mechanism also yields a rate law consistent with the experimental one.

- PRACTICE EXERCISE

The first step of a mechanism involving the reaction of bromine is

(

Br2(g)

)

k,

� 2 �c_,

Br(g) (fast, equilibrium)

What is the expression relating the concentration of Br(g) to that of Br2(g)? 1/2 k Answer: [Br] = �1 [Br2l

14.7

Catalysis

14.7 CATALYSIS A catalyst is a substance that changes the speed of a chemical reaction without

undergoing a permanent chemical change itself in the process. Catalysts are very common; most reactions in the body, the atmosphere, and the oceans occur with the help of catalysts. Much industrial chemical research is devoted to the search for new and more effective catalysts for reactions of commercial importance. Extensive research efforts also are devoted to finding means of in­ hlbiting or removing certain catalysts that promote undesirable reactions, such as those that corrode metals, age our bodies, and cause tooth decay. Homogeneous Catalysis A catalyst that is present in the same phase as the reacting molecules is called a

homogeneous catalyst. Examples abound both in solution and in the gas phase. Consider, for example, the decomposition of aqueous hydrogen perox­ ide, H202(aq), into water and oxygen:

2 H 202 (aq) � 2 H20(/)

+

02(g)

[14.29]

In the absence of a catalyst, thls reaction occurs extremely slowly.

Many different substances are capable of catalyzing the reaction represented by Equation 14.29, including bromide ion, Br-(aq), as shown in Figure 14.19T.

T Figure 14.19 Effect of catalyst. (H20 molecules and Na+ ions are omitted from the molecular art for clarity.)

HOMOG E N EOUS CATALYS I S

A catalyst that is present in the same phase as the reacting molecules is a homogeneous catalyst.

in the absence of a catalyst, H202(aq) decomposes very slowly.

Shortly after the addition of a small amount of NaBr(aq) to H202(aq), the solution turns brown because Brz is generated (Equation 14.30). The buildup of Br2 leads to rapid evolution of 02 according to Equation 1 4.31 .

After all of the H202 has decomposed, a colorless solution of NaBr(aq) remains. Thus, NaBr has catalyzed the reaction even though it is not consumed during the reaction.

605

606

C HAPTER 1 4

Chemical Kinetics The bromide ion reacts with hydrogen peroxide in acidic solution, forming aque­ ous bromine and water: (14.30] The brown color observed in the middle photograph of Figure 14.19 indicates the formation of Br2(aq). If this were the complete reaction, bromide ion would not be a catalyst, because it undergoes chemical change during the reaction. However, hydrogen peroxide also reacts with the Br2(aq ) generated in Equation 14.30: (14.31] The bubbling evident in Figure 14.19(b) is due to the formation of 0 2(g). The sum of Equations 14.30 and 14.31 is just Equation 14.29: 2 HzOz(aq) -----+ 2 HzO(I) +

2 HP + Oz + 2 Br- + 2 H+

Reaction pathway .A.

Figure 14.20

Energy profiles for uncatalyzed and catalyzed reactions. The energy profiles for the uncatalyzed decomposition of hydrogen peroxide and for the reaction as catalyzed by Br­ are compared. The catalyzed reaction involves two successive steps, each of which has a lower activation energy than the uncatalyzed reaction. Notice that the energies of reactants and products are unchanged by the catalyst.

Oz(g)

When H202 has been completely decomposed, we are left with a colorless solution of Br-(aq), as seen in the photo­ graph on the right in Figure 14.19. Bromide ion, therefore, is indeed a catalyst of the reaction because it speeds the overall reaction without itself undergoing any net change. It is added at the start of the reaction, reacts, and then reforms at the end. In contrast, Br2 is an intermediate because it is first formed (Equation 14.30) and then con­ sumed (Equation 14.31). Neither the catalyst nor the intermediate appears in the chemical equation for the overall reaction. Notice, however, that the catalyst is there at the

start of the reaction, whereas the intermediate is formed during the course of the reaction.

On the basis of the Arrhenius equation (Equation 14.19), the rate constant (k) is determined by the activation energy (Ea) and the frequen­ cy factor (A). A catalyst may affect the rate of reaction by altering the value of either Ea or A. The most dramatic catalytic effects come from lowering Ea. As a general rule, a catalyst lowers the overall activation energy for a chemical reaction. A catalyst usually lowers the overall activation energy for a reaction by providing a different mechanism for the reaction. In the decomposition of hy­ drogen peroxide, for example, two successive reactions of H202, with bromide and then with bromine, take place. Because these two reactions together serve as a catalytic pathway for hydrogen peroxide decomposition, both of them must have significantly lower activation energies than the uncatalyzed decomposi­ tion, as shown schematically in Figure 14.20 �-

GIVE IT SOME THOUGHT

How does a catalyst increase the rate of a reaction? Heterogeneous Catalysis A heterogeneous catalyst exists in a different phase from the reactant mole­ cules, usually as a solid in contact with either gaseous reactants or with reac­ tants in a liquid solution. Many industrially important reactions are catalyzed by the surfaces of solids. For example, hydrocarbon molecules are rearranged to form gasoline with the aid of what are called "cracking" catalysts (see the "Chemistry Put to Work" box in Section 25.3). Heterogeneous catalysts are often composed of metals or metal oxides. Because the catalyzed reaction oc­ curs on the surface, special methods are often used to prepare catalysts so that they have very large surface areas. The initial step in heterogeneous catalysis is usually adsorption of reac­ tants. Adsorption refers to the binding of molecules to a surface, whereas

14.7

absorption refers to the uptake of molecules into the interior of another sub­ stance. (Section 13.6) Adsorption occurs because the atoms or ions at the surface of a solid are extremely reactive. Unlike their counterparts in the interi­ or of the substance, surface atoms and ions have unused bonding capacity. This unused bonding capability may be used to bond molecules from the gas or so­ lution phase to the surface of the solid. The reaction of hydrogen gas with ethylene gas to form ethane gas pro­ vides an example of heterogeneous catalysis: OX>

Ethylene

Ethane

[14.32]

Even though this reaction is exothermic, it occurs very slowly in the absence of a catalyst. In the presence of a finely powdered metal, however, such as nickel, palladium, or platinum, the reaction occurs rather easily at room temperature. The mechanism by which the reaction occurs is diagrammed in Figure 14.21 T. Both ethylene and hydrogen are adsorbed on the metal surface [Figure 14.21(a)]. Upon adsorption the H- H bond of H2 breaks, leaving two H atoms that are bonded to the metal surface, as shown in Figure 14.21(b). The hydrogen atoms are relatively free to move about the surface. When a hydrogen encoun­ ters an adsorbed ethylene molecule, it can form a O" bond to one of the carbon atoms, effectively destroying the C -C 7r bond and leaving an ethyl group (C2H5) bonded to the surface via a metal-to-carbon O" bond [Figure 14.21(c)]. This O" bond is relatively weak, so when the other carbon atom also encounters a hydrogen atom, a sixth C- H O" bond is readily formed and an ethane mole­ cule is released from the metal surface [Figure 14.21(d)]. The site is ready to ad­ sorb another ethylene molecule and thus begin the cycle again.

Hydrogen

I

. . . . � . . . . . . . � . . . . . . . . . � . • • • • • • • •

• • • • • • • • • • • • • • • • (a )

(c)

(b)

(d)

A Figure 14.21 Mechanism for reaction of ethylene with hydrogen on a catalytic surface. (a) The hyd rogen and ethylene are adsorbed at the metal surface. (b) The H - H bond is broken to give adsorbed hydrogen atoms. (c) These migrate to the adsorbed ethylene and bond to the carbon atoms. (d) As C - H bonds are formed, the adsorption of the molecule to the metal surface is decreased and ethane is released.

Catalysis

607

608

C HAPTER

Chemis

14

Chemical Kinetics

Put to Work

CATALYTIC CO NVERTERS

H against urban air pollution. Two components of automo­ eterogeneous catalysis plays a major role in the fight

bile exhausts that help form photochemical smog are nitrogen oxides and unburned hydrocarbons of various types (Section 18.4). ln addition, automobile exhausts may contain consider­ able quantities of carbon monoxide. Even with the most care­ ful attention to engine design, it is impossible under normal driving conditions to reduce the quantity of these pollutants to an acceptable level in the exhaust gases. It is therefore nec­ essary to remove them from the exhaust before they are vent­ ed to the air. This removal is accomplished in the catalytic

converter. The catalytic converter, which is part of the exhaust sys­ tem, must perform two distinct functions: (1) oxidation of CO and unburned hydrocarbons (CxHy) to carbon dioxide and water, and (2) reduction of nitrogen oxides to nitrogen gas: CO, CxH y � C02 + HzO NO, N02 ----> Nz

These two functions require two distinctly different catalysts, so the development of a successful catalyst system is a diffi­ cult challenge. The catalysts must be effective over a wide range of operating temperatures. They must continue to be ac­ tive despite the fact that various components of the exhaust can block the active sites of the catalyst. And the catalysts must be sufficiently rugged to withstand exhaust gas turbu­ lence and the mechanical shocks of driving under various conditions for thousands of miles. Catalysts that promote the combustion of CO and hydro­ carbons are, in general, the transition-metal oxides and the noble metals, such as platinum. A mixture of two different metal oxides, CuO and Cr203, might be used, for example. These materials are supported on a structure (Figure 14.22 .,.) that allows the best possible contact between the flowing ex­ haust gas and the catalyst surface. A honeycomb structure made from alumina (Al20 3) and impregnated with the cata­ lyst is employed. Such catalysts operate by first adsorbing oxygen gas, also present in the exhaust gas. This adsorption weakens the 0-0 bond in 02, so that oxygen atoms are available for reaction with adsorbed CO to form COz. Hydro­ carbon oxidation probably proceeds somewhat similarly, with

the hydrocarbons first being adsorbed followed by rupture of a C -H bond. The most effective catalysts for reduction of NO to yield N2 and 02 are transition-metal oxides and noble metals, the same kinds of materials that catalyze the oxidation of CO and hydrocarbons. The catalysts that are most effective in one reac­ tion, however, are usually much Jess effective in the other. It is therefore necessary to have two different catalytic components. Catalytic converters are remarkably efficient heteroge­ neous catalysts. The automotive exhaust gases are in contact with the catalyst for only 100 to 400 ms. ln this very short time, 96% of the hydrocarbons and CO is converted to C02 and H20, and the emission of nitrogen oxides is reduced by 76%. There are costs as well as benefits associated with the use of catalytic converters. Some of the metals used in the convert­ ers are very expensive. Catalytic converters currently account for about 35% of the platinum, 65% of the palladium, and 95% of the rhodium used annually. All of these metals, which come mainly from Russia and South Africa, are far more expensive than gold.

Related Exercises: 14.56, 14.75, and 14.76

& Figure 14.22 Cross section of a catalytic converter. Automobiles are equipped with catalytic converters, which are part of their exhaust systems. The exhaust gases contain CO, NO, NOz, and unburned hydrocarbons that pass over surfaces impregnated with catalysts. The catalysts promote the conversion of the exhaust gases into C02, H20, and N2.

GIVE IT SOME THOUGHT How does a homogeneous catalyst compare with a heterogeneous one regarding the ease of recovery of the catalyst from the reaction mixture?

Enzymes Many of the most interesting and important examples of catalysis involve re­ actions within living systems. The human body is characterized by an ex­ tremely complex system of interrelated chemical reactions. All these reactions must occur at carefully controlled rates to maintain life. A large number of

14.7

marvelously efficient biological catalysts known as enzymes are necessary for many of these reactions to occur at suitable rates. Most enzymes are large protein molecules with molecular weights ranging from about 10,000 to about 1 million amu. They are very selective in the reactions that they catalyze, and some are absolutely specific, operating for only one substance in only one reaction. The decomposition of hydrogen peroxide, for example, is an important biological process. Because hydrogen peroxide is strongly oxidizing, it can be physiologically harmful. For this reason, the blood and livers of mammals contain an en­ zyme, catalase, which catalyzes the decomposition of hydrogen peroxide into water and oxygen (Equation 14.29). Figure 14.23 � shows the dramatic acceleration of this chemical reaction by the catalase in beef liver. Although an enzyme is a large molecule, the reaction is cat­ alyzed at a very specific location in the enzyme, called the active site. The substances that undergo reaction at this site are called substrates. The lock-and-key model, illustrated in Figure 14.24 T, provides a simple ex­ planation for the specificity of an enzyme. The substrate is pictured as fitting neatly into a special place on the enzyme (the active site), much like a specific key fits into a lock. The active site is created by coiling and folding of the long protein molecule to form a space, something like a pocket, into which the sub­ strate molecule fits. Figure 14.25 � shows a model of the enzyme lysozyme with and without a bound substrate molecule. The combination of the enzyme and the substrate is called the enzyme­ substrate complex. Although Figure 14.24 shows both the active site and its complementary substrate as having rigid shapes, the active site is often fairly flexible. Thus, the active site may change shape as it binds the substrate. The binding between the substrate and the active site involves intermolecular forces such as dipole-dipole attractions, hydrogen bonds, and London disper­ sion forces. = (Section 11.2) As the substrate molecules enter the active site, they are somehow acti­ vated, so that they are capable of extremely rapid reaction. This activation may result from the withdrawal or donation of electron density at a particular bond by the enzyme. In addition, in the process of fitting into the active site, the substrate molecule may be distorted and thus made more reactive. Once the reaction occurs, the products then depart, allowing another substrate mol­ ecule to enter.

Catalysis

609

.A. Figure 14.23 Effect of an enzyme. Ground-up beef liver causes hydrogen peroxide to decompose rapidly into water and oxygen. The decomposition is catalyzed by the enzyme catalase. Grinding the liver breaks open the cells, so that the reaction takes place more rapid ly. The frothing is due to escape of oxygen gas from the reaction mixture.

(a) Substrate

Enzyme

Products

Enzyme-substrate complex

Enzyme

.t. Figure 14.24 The lock-and-key model for enzyme action. The correct substrate is recognized by its ability to fit the active site of the enzyme, forming the enzyme­ substrate complex. After the reaction of the substrate is complete, the products separate from the enzyme.

A Figure 14.25

(b)

Molecular model of an enzyme. (a) A molecular model of the enzyme lysozyme. Note the characteristic cleft, which is the location of the active site. (b) Lysozyme with a bound substrate molecule.

610

C HAPTER

Chemis

14

Chemical Kinetics

N I T RO G E N F IXAT I O N AND N I T R O G E NA S E

and Li e

itrogen is one of the most essential elements i n living or­

N ganisms. It is found in many compounds that are vital to

life, including proteins, nucleic acids, vitamins, and hor­ mones. Plants use very simple nitrogen-containing com­ pounds, especially NH3, NH 4+, and N03 -, as starting materials from which such complex, biologically necessary compounds are formed. Animals are unable to synthesize the complex nitrogen compounds they require from the simple substances used by plants. Instead, they rely on more compli­ cated precursors present in vitamin- and protein-rich foods. Nitrogen is continually cycling through this biological arena in various forms, as shown in the simplified nitrogen cycle in Figure 14.26 T. For example, certain microorganisms

convert the nitrogen in animal waste and dead plants and ani­ mals into molecular nitrogen, Nz(g), which returns to the at­ mosphere. For the food chain to be sustained, there must be a means of reincorporating this atmospheric N2 in a form that plants can utilize. The process of converting N2 into corn­ pounds that plants can use is called nitrogenfixation. Fixing ni­ trogen is difficult; N2 is an exceptionally unreactive molecule, in large part because of its very strong N==N triple bond. = (Section 8.3) Some fixed nitrogen results from the action of lightning on the atmosphere, and some is produced industrial­ ly using a process we will discuss in Chapter 15. About 60% of fixed nitrogen, however, is a consequence of the action of a re­ markable and complex enzyme called nitrogenase. This enzyme ZnC12(aq) + H2{g) Suppose you are asked to study the kinetics of this reac­ tion by monitoring the rate of production of H2(g). (a) By using a reaction flask, a manometer, and any other common laboratory equipment, design an experi­ mental apparatus that would allow you to monitor the partial pressure of H2{g) produced as a function of time. (b) Explain how you would use the apparatus to deter­ mine the rate law of the reaction. (c) Explain how you would use the apparatus to determine the reaction order for [H+] for the reaction. (d) How could you use the ap­ paratus to determine the activation energy of the reac­ tion? (e) Explain how you would use the apparatus to determine the effects of changing the form of Zn(s) from metal strips to granules. The gas-phase reaction of NO with F2 to form NOF and F has an activation energy of Ea = 6.3 kJ/mol and a

Integrative Exercises frequency factor of A = 6.0 X 108 M-1 s-1. The reaction is bel.ieved to be bimolecular: NO(g) + l':l(g) --+ NOF(g) + F(g) (a) Calculate the rate constant at 100 •c. (b) Draw the

Lewis structures for the NO and the NOF molecules, given that the chemical formula for NOF is misleading because the nitrogen atom is actually the central atom in the molecule. (c) Predict the structure for the NOF mole­ cule. (d) Draw a possible transition state for the forma­ tion of NOF, using dashed l.ines to indicate the weak bonds that are beginning to form. (e) Suggest a reason for the low activation energy for the reaction. 14.108 The mechanism for the oxidation of HBr by 02 to form 2 H20 and Br2 is shown in Exercise 14.70. (a) Calculate

the overall standard enthalpy change for the reaction process. (b) HBr does not react with 02 at a measurable rate at room temperature under ordinary conditions. What can you infer from this about the magnitude of the activation energy for the rate-determining step? (c) Draw a plausible Lewis structure for the intermedi­ ate HOOBr. To what familiar compound of hydrogen and oxygen does it appear similar?

14.109 Enzymes, the catalysts of biological systems, are high

molecular weight protein materials. The active site of the enzyme is formed by the three-dimensional arrange­ ment of the protein in solution. When heated in solu­ tion, proteins undergo denaturation, a process in which the three-dimensional structure of the protein unravels or at least partly does so. The accompanying graph shows the variation with temperature of the activity of a typical enzyme. The activity increases with temperature to a point above the usual operating region for the en­ zyme, then decl.ines rapidly with further temperature increases. What role does denaturation play in deter­ mining the shape of this curve? How does your expla­ nation fit in with the lock-and-key model of enzyme action?

625

addition of Mn2+(aq), according to the following mechanism: Ce3+(aq) + Mn3+(aq) Ce4+(aq) + Mn2+(aq) 3+(aq) + Mn4+(aq) 4 + Mn3+(aq) --+ Ce Ce +(aq) 4 Mn +(aq) + Tl+(aq) --+ Mn2+(aq) + Tl3+(aq) --+

(a) Write the rate law for the uncatalyzed reaction. (b) What is unusual about the uncatalyzed reaction? Why might it be a slow reaction? (c) The rate for the cat­ alyzed reaction is first order in [Ce 4+] and first order in 2 [Mn +]. Based on this rate law, which of the steps in the catalyzed mechanism is rate determining? (d) Use the available oxidation states of Mn to comment on its spe­ cial suitability to catalyze this reaction. [14.111] The rates of many atmospheric reactions are accelerated by the absorption of light by one of the reactants. For ex­ ample, consider the reaction between methane and chlo­ rine to produce methyl chloride and hydrogen chloride:

Reaction 1: CH4(g)

+

Clz(g)

--+

CH3Cl(g) + HCI(g)

This reaction is very slow in the absence of light. How­ ever, Cl2(g) can absorb light to form Cl atoms: Reaction 2: Cl2(g) + hv --+ 2 Cl(g) Once the Cl atoms are generated, they can catalyze the reaction of CH4 and Cl2, according to the following pro­ posed mechanism: Reaction 3: CH 4(g) + Cl(g) --+ CH3(g) + HCI(g) Reaction 4: CH 3(g) + Cl2(g) --+ CH Cl(g) + Cl(g) 3 The enthalpy changes and activation energies for these two reactions are tabulated as follows: E. (kJ/mol)

Reaction

ll.H�an (kJ/mol)

3

+4

17

4

-109

4

(a) By using the bond enthalpy for Cl2 (Table 8.4), deter­ mine the longest wavelength of light that is energetic enough to cause reaction 2 to occur. ln which portion of the electromagnetic spectrum is this light found? (b) By using the data tabulated here, sketch a quantitative en­ ergy profile for the catalyzed reaction represented by re­ actions 3 and 4. (c) By using bond enthalpies, estimate where the reactants, CH4(g) + C12(g), should be placed on your diagram in part (b). Use this result to estimate the value of E. for the reaction CH4(g) + Cl2(g) CH 3(g) + HCI(g) + Cl(g). (d) The species Cl(g) and CH3(g) in reactions 3 and 4 are radicals, that is, atoms or molecules with unpaired electrons. Draw a Lewis struc­ ture of CH:v and verify that it is a radical. (e) The se­ quence of reactions 3 and 4 comprise a radical chain mechanism. Why do you think this is called a "chain re­ action"? Propose a reaction that will terminate the chain reaction. --+

Temperature ("C)

100

[14.110) Metals often form several cations with different charges.

Cerium, for exam}'le, forms Ce3+ and Ce4+ ions, and thallium forms Tl and Tl 3+ ions. Cerium and thallium ions react as follows: 3 2 Ce 4+(aq) + Tl+(aq) --+ 2 Ce +(aq) + Tl3+(aq) This reaction is very slow and is thought to occur in a single elementary step. The reaction is catalyzed by the

CHEMICAL EQUILIBRIUM

TRAFFIC ENTERING AND LEAVING San Francisco over the Golden Gate Bridge in the early morning.

626

W H AT ' S

A H E A D

15.1 The Concept of Equilibrium

15.5 Calculating Equilibrium Constants

15.2 The Equilibrium Constant We then define the equilibrium constant and learn how to write equilibrium-constant expressions for

15.6 Applications of Equilibrium Constants

We begin by examining the concept of equilibrium.

homogeneous reactions.

15.3 I nterpreting and Working with Equilibrium Constants

We also learn how to interpret the magnitude of an equilibrium constant and how to determine the way in which its value is affected when the chemical equation is reversed or changed in some other fashion. 15.4 Heterogeneous Equili bria

We then learn how to write equilibrium-constant expressions for heterogeneous reactions.

We will see that the value of an equilibrium constant can be calculated using equilibrium concentrations of reactants and products. We will also see that equilibrium constants can be used to predict the equilibrium concentrations of reactants and products and to determine the direction in which a reaction mixture must proceed to achieve equilibrium.

15.7 Le Chatelier's Principle

We conclude the chapter with a discussion of Le Chiite/ier's principle, which predicts how a system at equilibrium responds to changes in concentration, volume, pressure, and temperature.

TO BE I N EQU ILIBRIUM IS TO BE IN A STATE OF BALANCE. A tug of war in which the two sides are pulling with equal force so that the rope does not move is an example of a static equilibrium, one in which an object is at rest.

Equilibria can also be dynamic, as illustrated in the chapter-opening photograph, which shows cars traveling in both directions over a bridge that serves as an entranceway to a city. If the rate at which cars leave the city equals the rate at which they enter, the two opposing processes are in balance, and the net number of cars in the city is constant. We have already encountered several instances of dynamic equilibrium. For example, the vapor above a liquid is in equilibrium with the liquid phase. ooo (Section 11 .5) The rate at which molecules escape from the liquid into the gas phase equals the rate at which molecules in the gas phase strike the sur­ face and become part of the liquid. Similarly, in a saturated solution of sodium chloride the solid sodium chloride is in equilibrium with the ions dispersed in water. ooo (Section 13.2) The rate at which ions leave the solid surface equals the rate at which other ions are removed from the liquid to become part of the solid. Both of these examples involve a pair of opposing processes. At equilib­ rium these opposing processes are occurring at the same rate. 627

628

C HA PTER

15

Chemical Equilibrium In this chapter we will consider yet another type of dynamic equilibrium,

occurs when opposing re­ actions are proceeding at equal rates: The rate at which the products are formed from one involving chemical reactions. Chemical equilibrium

the reactants equals the rate at which the reactants are formed from the products. As a result, concentrations cease to change, making the reaction appear to be stopped. How fast a reaction reaches equilibrium is a matter of kinetics. Chemical equilibria are involved in a great many natural phenomena, and they play important roles in many industrial processes.

In this and the next two

chapters, we will explore chemical equilibrium in some detail. Here we will learn how to express the equilibrium position of a reaction in quantitative terms. We will also study the factors that determine the relative concentrations of reac­ tants and products in equilibrium mixtures.

1 5 . 1 THE CONCEPT O F EQU I L I B RI U M Let's examine a simple reaction to see how it reaches an equilibrium ture

state-a mix­

of reactants and products whose concentrations no longer change with time.

We begin with brown. Figure

N20-v a colorless substance that dissociates to form N02, which is 15.1 11> shows a sample of frozen N204 inside a sealed tube resting

in a beaker. Because the chemical reaction occurs in a closed system, the reaction will eventually reach equilibrium.

N204 vaporizes as it is warmed above its boiling point (21 .2 °C), Nz04 gas dissociates into N02 gas (Figure 15 .1 ) . Eventually, even though there is still N204 in

The solid

and the gas turns progressively darker as the colorless brown

the tube, the color stops getting darker because the system reaches equilibrium. We are left with an

equilibrium mixture of N204 and N02 in which the concentra­

tions of the gases no longer change as time passes.

reversible. N204 can N02, and N02 can react to form N204. This situation is represent­

The equilibrium mixture results because the reaction is react to form

ed by writing the equation for the reaction with two half arrows pointing in both directions: cx:o (Section 4 . }

1 Nz04(g)

�

Colorless

2 NOz(g)

[15.1]

Brown

We can analyze this equilibrium using our knowledge of kinetics. Let's call the

N204 to form N02 the forward reaction and the reaction of N02 N204 the reverse reaction. In this case both the forward reaction and the

decomposition of to re-form

reverse reaction are elementary reactions. As we learned in Section

14.6, the rate

laws for elementary reactions can be written from their chemical equations:

Forward reaction: Reverse reaction:

N204(g) � 2 NOz(g) 2 NOz(g) � N204(g)

= k,[NOzf

Rate! = Rate,

kf[N204]

[15.2] [15.3]

where kJ and k, are the rate constants for the forward and reverse reactions, respectively. At equilibrium the rate at which products are produced from reac­ tants equals the rate at which reactants are produced from products:

kf[Nz04]

k,[NOz]2

Forward reaction

Reverse reaction

[15.4]

Rearranging this equation gives

[NOzf kf = k; = a constant [15.5] [Nz04] As shown in Equation 15.5, the quotient of two constants, such as kf and k,, is it­

self a constant. Thus, at equilibrium the ratio of the concentration terms involving

N204

and

N02

equals a constant. fYVe will consider this constant, called the

equilibrium constant, in Section 15 .2.) It makes no difference whether we start with

N204 or with N02, or even with some mixture of the two. At equilibrium the

ratio equals a specific value. Thus, there is an important constraint on the propor­ tions of

N204 and N02 at equilibrium.

15.1

The Concept of Equilibrium

629

ESTAB L I S H I N G EQU I LI B RI U M

The condition in which the concentrations of all reactants and products in a closed system cease to change with time is called chemical equilibrium.

As N204 is warmed above its boiling point, it starts to dissociate into brown N02 gas.

Frozen Nz04 is nearly colorless.

A Figure 15.1

The N204(g)

= 2

Eventually the color stops changing as N204(g) and NO,(g) reach concentrations at which they are interconverting at the same rate. The two gases are in equilibrium.

N02(g) equilibrium.

Once equilibrium is established, the concentrations of N204 and N02 no longer change, as shown in Figure 15.2(a) 'f'. If the composition of the equilibri­ um mixture remains constant with time, it does not mean, however, that N204 and N02 stop reacting. On the contrary, the equilibrium is dynamic-some N204 is still converting to NO: and some N02 is still converting to N204. At equilibrium, however, the two processes occur at the same rate, as shown in Figure 15.2(b) . u

and so the reaction will proceed from right to left, forming more 503.

Qp

Qp Kp,

Calculating Equilibrium Concentrations

Chemists frequently need to calculate the amounts of reactants and products present at equilibrium. Our approach in solving problems of this type is simi­ lar to the one we used for evaluating equilibrium constants: We tabulate the initial concentrations or partial pressures, the changes therein, and the final equilibrium concentrations or partial pressures. Usually we end up using the equilibrium-constant expression to derive an equation that must be solved for an unknown quantity, as demonstrated in Sample Exercise 15.11.

646

C HA PTER 15

Chemical Equilibrium

I For the Haber process, N2(g) + 3 H2(g) � 2 NH3(g), Kp = 1.45 x 10-5 at 500 'C. In an equilibrium mixture of the three gases at 500 'C, the partial pressure of H2 is 0.928 atm and that of N2 is 0.432 atm. What is the partial pressure of NH3 in this equilibri­ um mixture?

• SAMPLE EXERCISE

15.1 1 Calculating Equilibrium Concentrations

SOLUTION

We are given an equilibrium constant, Kp, and the equilibrium partial pressures of two of the three substances in the equation (N2 and H2), and we are asked to calculate the equilibrium partial pressure for the third substance (NH3). Plan: We can set Kp equal to the equilibrium-constant expression and substitute in the partial pressures that we know. Then we can solve for the only unknown in the equation. Analyze:

Solve: We tabulate the equilibrium pressures as follows:

Because we do not know the equilibrium pressure of NH3, we represent it with a variable, x. At equilibrium the pressures must satisfy the equilibrium-constant expression:

Equilibrium pressure (ahn)

0.432

We now rearrange the equation to solve for x:

(PNH/ = 1.45 X 10-5 KP = --= FN,(FH/ (0.432)(0.928)3 x2 = (1.45 X 10-5)(0.432)(0.928) 3 = 5.01 X 10-6

We can always check our answer by using it to recalculate the value of the equilibrium constant:

Kp =

Comment:

X

0.928

x2

x=

V5.01 x 10-6 = 2.24 x w-3 atm = fNH, 32 (2.24 x w- ) (0.432)(0.928) 3

= 1.45

x w-5

- PRACTICE EXERCISE

At 500 K the reaction PC15(g) � PC13(g) + Cl2(g) has Kp = 0.497. In an equilibrium mixture at 500 K, the partial pressure of PC15 is 0.860 atm and that of PC13 is 0.350 atm. What is the partial pressure of Cl2 in the equilibrium mixture?

Answer: 1.22 atm

In many situations we will know the value of the equilibrium constant and the initial amounts of all species. We must then solve for the equilibrium amounts. Solving this type of problem usually entails treating the change in concentration as a variable. The stoichiometry of the reaction gives us the rela­ tionship between the changes in the amounts of all the reactants and products, as illustrated in Sample Exercise 15.12. I A 1.000-L flask is filled with 1.000 mol of H2 and 2.000 mol of I2 at 448 'C. The value of the equilibrium constant K, for the reaction

• SAMPLE EXERCISE

15.12 Calculating Equilibrium Concentrations from Initial Concentrations

H2(g) + I2(g) � 2 HI(g) at 448 'C is 50.5. What are the equilibrium concentrations of H2, I2, and HI in moles per liter? SOLUTION

We are given the volume of a container, an equilibrium constant, and starting amounts of reactants in the container and are asked to calculate the equilibrium concentrations of all species. Plan: In this case we are not given any of the equilibrium concentrations. We must develop some relationships that relate the ini­ tial concentrations to those at equilibrium. The procedure is similar in many regards to that outlined in Sample Exercise 15.9, where we calculated an equilibrium constant using initial concentrations.

Analyze:

Solve: First, we note the initial concen­ trations of H2 and I2 in the 1.000-L flask: Second, we construct a table in which we tabulate the initial concentrations:

[H2] = 1 .000 M and [I2] = 2.000 M H2(g) + Initial

Change

Equilibrium

1.000 M

2 HI(g) 2.000 M

OM

15.6

Third, we use the stoichiometry of the reaction to determine the changes in concentration that occur as the re­ action proceeds to equilibrium. The concentrations of H2 and l2 will de­ crease as equilibrium is established and that of HI will increase. Let's rep­ resent the change in concentration of H2 by the variable x. The balanced chemical equation tells us the rela­ tionship between the changes in the concentrations of the three gases:

Fourth, we use the initial concentra­ tions and the changes in concentra­ tions, as dictated by stoichiometry, to express the equilibrium concentra­ tions. With all our entries, our table now looks like this:

Fifth, we substitute the equilibrium concentrations into the equilibrium­ constant expression and solve for the unknown, x:

If you have an equation-solving cal­ culator, you can solve this equation directly for x. If not, expand this ex­ pression to obtain a quadratic equa­ tion in x:

Solving the quadratic equation (Ap­ pendix A.3) leads to two solutions for x: When we substitute x = 2.323 into the expressions for the equilibrium concentrations, we find negative con­ centrations of H2 and 12. Because a negative concentration is not chemi­ cally meaningful, we reject this solu­ tion. We then use x = 0.935 to find the equilibrium concentrations: Check: We can check our solution by putting these numbers into the equi­ librium-constant expression to assure that we correctly calculate the equi­ librium constant:

Applications of Equilibrium Constants

647

For each x mol of H2 that reacts, x mol of 12 are consumed and 2x mol of HI are produced:

+ Initial

2 HI(g)

1.000M

2.000 M

-x

-x

Change

OM +2x

Equilibrium

2 HI(g)

+ Initial

1.000 M

2.000 M

-x

-x

+2x

(1.000 - x) M

(2.000 - x) M

2x M

Change Equilibrium

c

K

OM

[Hif (2x)2 = [H2][12] = (1.000 - x)(2.000 - x) = 50 5 ·

--

4x2 = 50.5(x2 - 3.000x + 2.000) 46.5x2 - 151.5x + 101.0 = 0 X

_

-

-(-1s1.s)

±

Vl(-151.5)2 - 4(46.5)(10I.o) 2(46.5)

_

- 2.323 or 0.935

[H2J = 1.000 - x = 0.065 M [I:J = 2.000 - x = 1.065 M [HI] = 2x = 1.87 M

c K

(1.87)2

= [H2] [I2J = (0.065)(1.065) = Sl [Hif

Comment: Whenever you use a quadratic equation to solve an equilibrium problem, one of the solutions will not be chemically meaningful and should be rejected. - PRACTICE EXERCISE

For the equilibrium PCls(g) = PCl3(g) + Cl2(g), the equilibrium constant Kp has the value 0.497 at 500 K. A gas cylinder at 500 K is charged with PC15(g) at an initial pressure of 1.66 atm. What are the equilibrium pressures of PCl5, PCl3, and Cl2 at this temperature? Answer: PpCJ5 = 0.967 atrn; Ppe13 = PCJ, = 0.693 atm

648

CHAPTER 15

Chemical Equilibrium 1 5.7 LE CHATELIER'S P RINCI PLE

80 60 Percent NH3 at equilibrium 40 20 0

A Figure 1 5.1 0

Effect of temperature and pressure on the percentage of NH3 In an equilibrium mixture of N:z., H:z., and NH3• Each mixture was produced by starting with a 3 : 1 molar

mixture of H2 and N2. The yield of NH3 is greatest at the lowest temperature and at the highest pressure.

When Haber developed his process for making ammonia from N2 and H2, he sought the factors that might be varied to increase the yield of NH3. Using the values of the equilibrium constant at various temperatures, he calculated the equilib­ rium amounts of NH3 formed under a variety of conditions. Some of Haber's results are shown in Figure 15.10 0 [15.23] Co(H20)62+(aq) + 4 Cqaq) ;:::::::': CoCI/-(aq) + 6 H20(I) Pale pink

Deep blue

2 2 The formation of CoCI4 - from Co(H20)6 + is an endothermic process. We will discuss the significance of this enthalpy change shortly. Because Co(H20)62+ is pink and CoCI/- is blue, the position of this equilibrium is readily apparent from the color of the solution. Figure 15.14(left) '+' shows a room-temperature so­ 2 lution of CoC12 in HCI(aq). Both Co(H20)6 + and CoCI/- are present in signifi­ cant amounts in the solution; the violet color results from the presence of both the pink and blue ions. When the solution is heated [Figure 15.14(middle)], it be­ comes intensely blue in color, indicating that the equilibrium has shifted to form more CoCl/-. Cooling the solution, as in Figure 15.14(right), leads to a pink solu­ tion, indicating that the equilibrium has shifted to produce more Co(H20)62+. How can we explain the dependence of this equilibrium on temperature? We can deduce the rules for the temperature dependence of the equilibrium constant by applying Le Chatelier's principle. A simple way to do this is to treat heat as if it were a chemical reagent. In an endothermic (heat-absorbing) reaction we can consider heat as a reactant, whereas in an exothennic (heat-releasing) re­ action we can consider heat as a product. ;:::::::':

Endothermic:

Reactants + heat

products

Exothermic:

Reactants ;:::::::': products + heat

the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothennic reaction or a product to an exothennic reaction. The equi­ librium shifts in the direction that consumes the excess reactant (or product), namely heat. When

Le Chatelier's Principle

651

652

C HA PTER 15

Chemical Equilibrium

E F F ECT OF TEM PERATU R E C H AN G ES

Almost every equilibrium constant changes in value as the temperature changes. In an endothermic reaction, such as the one shown, heat is absorbed as reactants are converted to products. Increasing the temperature causes the equilibrium to shift to the right and K to increase. Lowering the temperature shifts the equilibrium in the direction that produces heat, to the left, decreasing K. Co(H20)62+ (aq)+ 4 Cqaq) === CoCll-(aq)+ 6 H20(1)

At room temperature both the pink Co(H 20)62+ and blue CoCii - ions are present in significant amounts, giving a violet color to the solution.

&

Figure 1 5.14

Heating the solution shifts the equilibrium to the right, forming ' more blue CoCI. -.

Cooling the solution shifts the equilibrium to the left, toward pink Co(H20)62+ .

Temperature and

(The reaction shown is Co(H z 0) 62+(aq) + 4 Cl-(aq) = CoCI/-(aq) + 6 H20(/).)

equilibrium.

GIVE IT SOME THOUGHT

Use Le Chatelier's principle to explain why the equilibrium vapor pressure of a liq­ uid increases with increasing temperature. In an endothermic reaction, such as Equation 15.23, heat is absorbed as reac­ tants are converted to products. Thus, increasing the temperature causes the equilibrium to shift to the right, in the direction of products, and K increases. For Equation 15 .23, increasing the temperature leads to the formation of more 2 CoC14 -, as observed in Figure 15.14(b). In an exothermic reaction the opposite occurs. Heat is absorbed as products are converted to reactants; therefore the equilibrium shifts to the left and K decreases. We can summarize these results as follows:

Endothermic:

Increasing T results in an increase in K.

Exothermic:

Increasing T results in a decrease in K.

15.7

Cooling a reaction has the opposite effect. As we lower the temperature, the equilibrium shifts to the side that produces heat. Thus, cooling an endothermic reaction shifts the equilibrium to the left, decreasing K . We observed this effect in Figure 15.14(c). Cooling an exothermic reaction shifts the equilibrium to the right, increasing K . - SAMPLE EXERCISE 1 5.1 3

Le Chatelier's Principle to Predict Shifts inUsing Equilibrium

Consider the equilibrium

N204(g) :;:::::::= 2 N02(g)

!lW

=

58.0 kJ

In which direction will the equilibrium shift when (a) N204 is added, (b) N02 is removed, (c) the total pressure is increased by addition of N2(g), (d) the volume is increased, (e) the temperature is decreased? SOLUTION Analyze: We are given a series of changes to be made to a system at equilibrium and are asked to predict what effect each change will have on the position of the equilibrium. Plan: Le Chatelier's principle can be used to determine the effects of each of these changes. Solve:

(a) The system will adjust to decrease the concentration of the added N204, so the equilibrium shifts to the right, in the direction of products. (b) The system will adjust to the removal of N02 by shifting to the side that produces more N02; thus, the equilibrium shifts to the right. (c) Adding N2 will increase the total pressure of the system, but N2 is not involved in the reaction. The partial pressures of N02 and N204 are therefore unchanged, and there is no shift in the position of the equilibrium. (d) If the volume is increased, the system will shift in the direction that occupies a larger volume (more gas molecules); thus, the equilibrium shifts to the right. (This is the opposite of the effect observed in Figure 15.13, where the volume was decreased .) (e) The reaction is endothermic, so we can imagine heat as a reagent on the reactant side of the equation. Decreasing the temperature will shift the equilibrium in the di­ rection that produces heat, so the equilibrium shifts to the left, toward the formation of more N204. Note that only this last change also affects the value of the equilibrium constant, K. - PRACTICE EXERCISE For the reaction

!lW

= 87.9 kJ

in which direction will the equilibrium shift when (a) Cl2(g) is removed, (b) the tem­ perature is decreased, (c) the volume of the reaction system is increased, (d) PCI3(g) is added? Answers: (a) right, (b) left, (c) right, (d) left

- SAMPLE EXERCISE 1 5.14

I

Predicting the Effect of Temperature on K

(a) Using the standard heat of formation data in Appendix C, determine the stan­

dard enthalpy change for the reaction N2(g)

+ 3 H2(g) :;:::::::= 2 NH3(g)

(b) Determine how the equilibrium constant for this reaction should change with temperature. SOLUTION

Analyze: We are asked to determine the standard enthalpy change of a reaction and how the equilibrium constant for the reaction varies with temperature. Plan: (a) We can use standard enthalpies of formation to calculate !lH' for the reac­ tion. (b) We can then use Le Chatelier's principle to determine what effect tempera­ ture will have on the equilibrium constant.

Le Chatelier's Principle

653

C HA PTER 15

654

Chemical Equilibrium Solve:

TABLE 1 5.2

• Variation in KP for the

Equilibrium N2 + 3 H2 :;=== 2 NH3 as a Function of Temperature Temperature (•C)

4.34 1.64 4.51 1 .45 5.38 2.25

300 400 450 500 550 600

X

X

X

X

X

X

10-J 10-4 10-5 10-5 10-6 10-6

(a) Recall that the standard enthalpy change for a reaction is given by the sum of the standard molar enthalpies of formation of the products, each multiplied by its coeffi­ cient in the balanced chemical equation, less the same quantities for the reactants. At 25 •c, llHJ for NH3(g) is -46.19 kJ/mol. The flH'j values for H2(g) and N2(g) are zero by definition because the enthalpies of formation of the elements in their normal states at 25 •c are defined as zero (Section 5.7). Because 2 mol of NH3 is formed, the total enthalpy change is

(2 mol)(-46.19 kJ/mol) - 0

=

-92.38 kJ

(b) Because the reaction in the forward direction is exothermic, we can consider heat

a product of the reaction. An increase in temperature causes the reaction to shift in the direction of less NH3 and more N2 and H2. This effect is seen in the values for Kp presented in Table 15.2 ... . Notice that Kp changes markedly with changes in tempera­ ture and that it is larger at lower temperatures. Comment: The fact that Kp for the formation of NH3 from N2 and H2 decreases with increasing temperature is a matter of great practical importance. To form NH3 at a reasonable rate requires higher temperatures. At higher temperatures, however, the equilibrium constant is smaller, and so the percentage conversion to NH3 is smaller. To compensate for this, higher pressures are needed because high pressure favors NH3 formation. - PRACTICE EXERCISE

Using the thermodynamic data in Appendix C, determine the enthalpy change for the reaction 2 POCl3(g) :;=== 2 PCl3(g) + 02(g) Use this result to determine how the equilibrium constant for the reaction should change with temperature. Answer: flH• = 508.3 kJ; the equilibrium constant will increase with increasing temperature

r----------- - ---------,

6'o Et

�1

E,

_.___ _ _

A

Reaction pathway

.A. Figure 15.15

Effect

of a catalyst

At equilibrium for the hypothetical reaction A = B, the forward reaction rate, r1, equals the reverse reaction rate, 'r· The violet curve represents the path over the transition state in the absence of a catalyst. A catalyst lowers the energy of the transition state, as shown by the green curve. Thus, the activation energy is lowered for both the forward and the reverse reactions. As a result, the rates of forward and reverse reactions in the catalyzed reaction are increased. on equilibrium.

j

The Effect of Catalysts What happens if we add a catalyst to a chemical system that is at equilibrium? As shown in Figure 15.15 ..,., a catalyst lowers the acti­ vation barrier between the reactants and products. The activation energy of the forward reaction is lowered to the same extent as that for the reverse reaction. The catalyst thereby increases the rates of both the forward and reverse reactions. As a result, a catalyst increas­

es the rate at which equilibrium is achieved, but it does not change the com­ position of the equilibrium mixture. The value of the equilibrium

constant for a reaction is not affected by the presence of a catalyst. The rate at which a reaction approaches equilibrium is an im­ portant practical consideration. As an example, let's again consider the synthesis of ammonia from N2 and H2. In designing a process B for ammonia synthesis, Haber had to deal with a rapid decrease in the equilibrium constant with increasing temperature, as shown in Table 15.2. At temperatures sufficiently high to give a satisfactory reaction rate, the amount of ammonia formed was too small. The solution to this dilemma was to develop a catalyst that would produce a reasonably rapid approach to equilibrium at a sufficiently low temperature, so that the equilibrium constant was still reasonably large. The development of a suitable catalyst thus became the focus of Haber's research efforts. After trying different substances to see which would be most effective, Haber finally settled on iron mixed with metal oxides. Variants of the original catalyst formulations are still used. These catalysts make it possible to obtain a reasonably rapid approach to equilibrium at temperatures around 400 •c to 500 •c and with gas pressures of 200 to 600 atm. The high pressures are needed to obtain a satisfactory degree of conversion at equilibrium. You can see from

15.7

Figure 15.10 that if an improved catalyst could be found-one that would lead to sufficiently rapid reaction at temperatures lower than 400 oc to 500 °C-it would be possible to obtain the same degree of equilibrium conversion at much lower pressures. This would result in great savings in the cost of equipment for ammonia synthesis. In view of the growing need for nitrogen as fertilizer, the fixation of nitrogen is a process of ever-increasing importance. G IVE IT SOME THOUGHT Does the addition of a catalyst have any effect on the position of an equilibrium?

- SAMPLE INTEGRATIVE EXERCISE

I

Putting Concepts Together

At temperatures near 800 °C, steam passed over hot coke (a form of carbon obtained from coal) reacts to form CO and H2: C(s) + H 20(g) = CO(g) + H 2(g) The mixture of gases that results is an important industrial fuel called water gas. (a) At 800 oc the equilibrium constant for this reaction is Kp = 14.1. What are the equilibri­ um partial pressures of H 20, CO, and H2 in the equilibrium mixture at this tempe­ rature if we start with solid carbon and 0.100 mol of HP n i a 1.00-L vessel? (b) What is the minimum amount of carbon required to achieve equilibrium under these condi­ tions? (c) What is the total pressure in the vessel at equilibrium? (d) At 25 oc the value of Kp for this reaction is 1.7 x 10-21 . Is the reaction exothermic or endothermic? (e) To produce the maximum amount of CO and H 2 at equilibrium, should the pressure of the system be increased or decreased? SOLUTION

(a) To determine the equilibrium partial pressures, we use the ideal gas equation, first determining the starting partial pressure of hydrogen.

�,o

=

n H,oRT

-v-

=

(0.100 mol)(0.0821 L-atm/mol-K)(1073 K) l.OO L

=

8.81 atm

We then construct a table of starting partial pressures and their changes as equilibri­ um is achieved: C(s)

H 20(g)

CO(g)

8.81 atm

0 atm

O atm

-x

+x

+x

8.81 - x atm

x atm

x atrn

+

Initial Change Equilibrium

+

H 2(g)

There are no entries in the table under C(s) because the reactant, being a solid, does not appear in the equilibrium-constant expression. Substituting the equilibrium par­ tial pressures of the other species into the equilibrium-constant expression for the re­ action gives

KP =

o� , Pc �,0

= � = 14.1 (8.81 - x)

Multiplying through by the denominator gives a quadratic equation in x: x2

x2 + 14.1x - 124.22

=

=

(14.1)(8.81 - x) 0

Solving this equation for x using the quadratic formula yields x = 6.14 atm. Hence, the equilibrium partial pressures are Pco = x = 6.14 atm, �2 = x = 6.14 atm, and PH o = (8.81 - x) = 2.67 atm. (b), Part (a) shows that x = 6.14 atrn of H20 must react for the system to achieve equilibrium. We can use the ideal-gas equation to convert this partial pressure into a mole amount.

n

=

(6.14 atrn)( l . OO L) PV = 0.0697 mol = RT (0.0821 L-atrn/mol-K)(1073 K)

Le Chatelier's Principle

655

656

C HA PTER 15

Chemical Equilibrium

Thus, 0.0697 mol of H20 and the same amount of C must react to achieve equi­ librium. As a result, there must be at least 0.0697 mol of C (0.836 g C) present among the reactants at the start of the reaction. (c) The total pressure in the vessel at equilibrium is simply the sum of the equilibri­ um partial pressures: Ptotal = ll-!,o + Pea + ll-!, = 2.67 atm + 6.14 atm + 6.14 atm = 14.95 atm (d) In discussing Le Chi\telier's principle, we saw that endothermic reactions exhibit an increase in KP with increasing temperature. Because the equilibrium constant for this reaction increases as temperature increases, the reaction must be endothermic. From the enthalpies of formation given in Appendix C, we can verify our prediction by calculating the enthalpy change for the reaction, b.W = b.HJ(CO) + b.HJ (H2) - b.W(C) - b.HJ(H 20) = +131.3 kJ. The positive sign for b.Ho indicates that the reaction is endothermic. (e) According to Le Cha.telier's principle, a decrease in the pressure causes a gaseous equilibrium to shift toward the side of the equation with the greater number of moles of gas. In this case there are two moles of gas on the product side and only one on the reactant side. Therefore, the pressure should be reduced to maximize the yield of the CO and H2.

Chemis

Put to Work

CONTROLLING N I T R I C OXIDE E M I S S I O N S

T teresting example of the practical importance of changes he formation of NO from N2 and 02 provides another in­

in the equilibrium constant and reaction rate with tempera­ ture. The equilibrium equation and the standard enthalpy change for the reaction are

[15.24] b.W = 90.4 kJ The reaction is endothermic; that is, heat is absorbed when NO is formed from the elements. By applying Le Chate­ lier's principle, we deduce that an increase in temperature will shift the equilibrium in the direction of more NO. The equilib­ rium constant Kp for formation of 1 mol of NO from the ele­ ments at 300 K is only about 10�15. In contrast, at a much higher temperature of about 2400 K the equilibrium constant is 1013 times as large, about 0.05. The manner in which Kp for Equation 15.24 varies with temperature is shown in Figure 15.16 �. This graph helps to explain why NO is a pollution prob­ lem. In the cylinder of a modem high-compression auto en­ gine, the temperatures during the fuel-burning part of the cycle may be approximately 2400 K. Also, there is a fairly large excess of air in the cylinder. These conditions favor the forma­ tion of some NO. After the combustion, however, the gases are quickly cooled. As the temperature drops, the equilibrium in Equation 15.24 shifts strongly to the left (that is, in the direc­ tion of N2 and 02). The lower temperatures also mean that the rate of the reaction is decreased, however, so the NO formed at high temperatures is essentially "frozen" in that form as the gas cools. The gases exhausting from the cylinder are still quite hot, perhaps 1200 K. At this temperature, as shown in Figure 15.16, the equilibrium constant for formation of NO is much smaller. However, the rate of conversion of NO to N2 and 02 is too slow to permit much loss of NO before the gases are cooled still further.

As discussed in the "Chemistry Put to Work" box in Section 14.7, one of the goals of automotive catalytic convert­ ers is to achieve the rapid conversion of NO to N2 and 02 at the temperature of the exhaust gas. Some catalysts for this re­ action have been developed that are reasonably effective under the grueling conditions found in automotive exhaust systems. Nevertheless, scientists and engineers are continual­ ly searching for new materials that provide even more effec­ tive catalysis of the decomposition of nitrogen oxides.

1

Kp

l�t

x 10�5

l

Cylinder temperature Exhaust gas during combustion ---r------- l 1 x 10�10 _____ temperature

I

1 X 10�Js

_ _

+

0

1000

2000

Temperature (K)

.A. Figure 15.16 Equilibrium and temperature. The graph shows how the equilibrium constant for the reaction ! N2(g) + ! 02(g) = NO(g) varies as a function of temperature. The equilibrium constant increases with increasing temperature because the reaction is endothermic. It is necessary to use a log scale for Kp because the values vary over such a large range.

Key Skills

657

CHAPTER REVIEW

SUM MARY AND KEY TERMS Introduction and Section 1 5.1

A chemical

reaction

can achleve a state in which the forward and reverse processes are occurring at the same rate. This condition is

Section 1 5.4 Equilibria for which all substances are in homogeneous equilibria; in heterogeneous equilibria two or more phases are pre­ the same phase are called

chemical equilibrium, and it results in the forma­

sent. The concentrations of pure solids and liquids are left

tion of an equilibrium mixture of the reactants and prod­

out of the equilibrium-constant expression for a heteroge­

ucts of the reaction. The composition of an equilibrium

neous equilibrium.

called

mixture does not change with time.

Section 1 5.5 If the concentrations of all species in an

Section 1 5.2 An equilibrium that is used throughout this

N2(g) + 3 H2(g) � 2 NH3(g).

chapter is the reaction

This reaction is the basis of the

Haber process for the pro­

duction of ammonia. The relationship between the concen­ trations of the reactants and products of a system at

equilibrium are known, the equilibrium-constant expres­ sion can be used to calculate the value of the equilibrium constant. The changes in the concentrations of reactants and products on the way to achieving equilibrium are governed by the stoichiometry of the reaction.

equilibrium is given by the law of mass action. For an equi­

Section 1 5.6 The reaction quotient, Q, is found by

librium equation of the form

substituting reactant and product concentrations or par­

the

a A + b B � d D + e E,

equilibrium-constant expression is written as

tial pressures at any point during a reaction into the equilibrium-constant expression. If the system is at equi­ librium, Q =

K.

If Q #

equilibrium. When Q < where

Kc

is a constant called the

equilibrium constant.

When the equilibrium system of interest consists of gases, it is often convenient to express the concentrations of re­ actants and products in terms of gas pressures:

K

P

(Po)d(PE)' "-:(PA)"(Ps)b

equilibrium by forming more products (the reaction

moves from left to right); when Q >

K,

the reaction will

proceed from right to left. Knowing the value of K makes it possible to calculate the equilibrium amounts of reac­ tants and products, often by the solution of an equation in which the unknown is the change

'-'--' " = -'--"

Kc and Kp are related by the expression Kp

K , however, the system is not at K, the reaction will move toward

in

a partial pressure

or concentration.

=

" Kc(RT) "·

Section 1 5.7 Le Chatelier's principle states that if a sys­

tem at equilibrium is disturbed, the equilibrium will shift to

minimize the disturbing influence. By this principle, if a

Section 1 5.3 The value of the equilibrium constant

reactant or product

changes with temperature.

the equilibrium will shift to consume the added substance.

A large value of Kc indicates

is

added to a system at equilibrium,

that the equilibrium mixture contains more products

The effects of removing reactants or products and of chang­

than reactants and therefore lies toward the product side

ing the pressure or volume of a reaction can be similarly de­

of the equation.

A small

value for the equilibrium con­

stant means that the equilibrium mixture contains less

duced. For example, if the volume of the system is reduced,

the equilibrium will shift in the direction that decreased the

products than reactants and therefore lies toward the re­

number of gas molecules. The enthalpy change for a reac­

actant side. The equilibrium-constant expression and the

tion indicates how an increase in temperature affects the

equilibrium constant of the reverse of a reaction are the

equilibrium: For an endothermic reaction, an increase in

reciprocals of those of the forward reaction. If a reaction

temperature shifts the equilibrium to the right; for an

is the sum of two or more reactions, its equilibrium con­

exothermic reaction, a temperature increase shifts the equi­

stant will be the product of the equilibrium constants for

librium to the left. Catalysts affect the speed at which

the individual reactions.

equilibrium is reached but do not affect the magnitude of K.

KEY SKILLS • •

•

Understand what is meant by chemical equilibrium and how it relates to reaction rates. Write the equilibrium-constant expression for any reaction. Relate Kc and

Kp.

• Relate the magnitude of an equilibrium constant to the relative amounts of reactants and products present in an equilibrium mixture. •

Manipulate the equilibrium constant to reflect changes in the chemical equation.

C HA PTER 15

658 •

Chemical Equilibrium

Write the equilibrium-constant expression for a heterogeneous reaction.

•

Calculate an equilibrium constant from concentration measurements.

•

Calculate equilibrium concentrations given the equilibrium constant and all but one equilibrium concentration.

•

•

•

Predict the direction of a reaction given the equilibrium constant and the concentrations of reactants and products. Calculate equilibrium concentrations given the equilibrium constant and the starting concentrations. Understand how changing the concentrations, volume, or temperature of a system at equilibrium affects the equi­ librium position.

KEY EQUATIONS

[D] d [E] ' [A]"[B] b

•

Kc

= --

•

Kp

=

(Pd(PE)' (PA)"(Ps)b

[15.11]

The equilibrium-constant expression in terms of equilibrium partial pressures

•

Kp

=

Kc(RT)I!."

[15.14]

Relating the equilibrium constant based on pressures to the equilibrium constant based on concentration

[15.8]

The equilibrium-constant expression for a general reaction of the type a A + b B :;::::::=: d D + e E; the concentrations are equilibrium concentrations only

The reaction quotient. The concentrations are for any time during a reaction. If the concentrations are equilibrium concentrations, then Q, = K,.

VIS UALI Z I N G C O N C EPTS 15.1 (a) Based on the following energy profile, predict whether kt > k, or kt < k,. (b) Using Equation 15.5,

predict whether the equilibrium constant for the process is greater than 1 or Jess than 1. [Section 15.1]

>, LL_

� ] .§

t'!'

£

Reactants

Products Reaction pathway

15.2 The following diagrams represent a hypothetical reac­ tion A ----> B, with A represented by red spheres and B

represented by blue spheres. The sequence from left to right represents the system as time passes. Do the dia­ grams indicate that the system reaches an equilibrium state? Explain. [Sections 15.1 and 15.2]

15.3 The following diagram represents an equilibrium mix­

ture produced for a reaction of the type A + X � AX. lf the volume is 1 L, is K greater or smaller than 1? [Section 15.2]

15.4 The following diagram represents a reaction shown

going to completion. (a) Letting A = red spheres and B = blue spheres, write a balanced equation for the re­ action. (b) Write the equilibrium-constant expression for the reaction. (c) Assuming that all of the molecules are in the gas phase, calculate lln, the change in the number of gas molecules that accompanies the reaction. (d) How can you calculate Kp if you know Kc at a particular tem­ perature? [Section 15.2]

Key Equations

659

15.5 Ethene (C2�l reacts with halogens (X2) by the follow­

ing reaction: C2H4(g) + X2(g) � C2H4X2(g) The following figures represent the concentrations at equi­ librium at the same temperature when X2 is Cl2 (green), Br2 (brown), and I2 (purple). List the equilibria from small­ est to largest equilibrium constant. [Section 15.3]

15.8 The following diagram

(a)

represents the equilib­ rium state for the reac­ tion A2(gl + 2 B(g) � 2 AB(g). (a) Assuming the volume is 1 L, calcu­ late the equilibrium con­ stant, K0 for the reaction. (b) If the volume of the equilibrium mixture is decreased, will the num­ ber of AB molecules increase or decrease? [Sections 15.5 and 15.7]

(b)

15.9 The following diagrams represent equilibrium mixtures for the reaction A2 + B � A + AB at (1) 300 K and (2) 500 K. The A atoms are red, and the B atoms are blue.

Is the reaction exothermic or endothermic? [Section 15.7]

(c) 15.6 The reaction A2 + B2 � 2 AB has an equilibrium con­ stant Kc = 1.5. The following rliagrams represent reaction

mixtures containing A2 molecules (red), B2 molecules (blue), and AB molecules. (a) Which reaction mixture is at equilibrium? (b) For those mixtures that are not at equilib­ rium, how will the reaction proceed to reach equilibrium? [Sections 15.5 and 15.6]

(a)

(b)

15.10 The following graph represents the yield of the compound

AB at equilibrium in the reaction A (g) + B(g) ----> AB(g).

(i)

(ii)

(iii)

15.7 The reaction A2(g) + B(g) � A(g) + AB(g) has an equilibrium constant of Kp = 2. The accompanying

diagram shows a mixture containing A atoms (red), A2 molecules, and AB molecules (red and blue). How many B atoms should be added to the diagram to illus­ trate an equilibrium mixture? [Section 15.6]

T (a) Is this reaction exothermic or endothermic? (b) Is P = x greater or smaller than P = y? [Section 15.7]

660

C HA PTER 15

Chemical Equilibrium

EXERC I S E S Equilibrium ; The Equilibrium C o nstant 15.11 Suppose that the gas-phase reactions A ---> B and B ---> A are both elementary processes with rate con­ z 1 stants of 3.8 x w- s- and 3.1 x 10-1 s-1, respectively. (a) What is the value of the equilibrium constant for the equilibrium A(g) :;= B(g)? (b) Which is greater at equilibrium, the partial pressure of A or the partial pres­ sure of B? Explain.

15.12 Consider the reaction A + B :;= C + D. Assume that both the forward reaction and the reverse reaction are elementary processes and that the value of the equilibri­ um constant is very large. (a) Which species predomi­ nate at equilibrium, reactants or products? (b) Which reaction has the larger rate constant, the forward or the reverse? Explain. 15.13 Write the expression for K, for the following reactions. In each case indicate whether the reaction is homoge­ neous or heterogeneous. (a) 3 NO(g) :;= N20(g) + NOz(g) (b) CH4(g) + 2 HzS(g) :;= CS2(g) + 4 Hz(g) (c) Ni(C0)4(g) :;= Ni(s) + 4 CO(g) (d) HF(aq) :;= H+(aq) + F-(aq) z (e) 2 Ag(s) + Zn +(aq) :;= 2 Ag+(aq) + Zn(s)

15.14 Write the expressions for K, for the following reactions. In each case indicate whether the reaction is homoge­ neous or heterogeneous. (a) 2 03(g) :;= 3 Oz(g) (b) Ti(s) + 2 Clz(g) :;= TiCl4(1) (c) 2 CzH4(g) + 2 HzO(g) :;= 2 CzH6(g) + Oz(g) (d) C(s) + 2 H2(g) :;= CH4(g) (e) 4 HCl(aq) + Oz(g) :;= 2 HzO(I) + 2 Cl2(g)

15.15 When the following reactions come to equilibrium, does the equilibrium mixture contain mostly reactants or mostly products? (a) Nz(g) + Oz(g) :;= 2 NO(g); K, = 1.5 x 10-10 (b) 2 SOz(g) + 02(g) :;= 2 S03(g); Kp = 2.5 x 109

15.16 Which of the following reactions lies to the right, favor­ ing the formation of products, and which lies to the left, favoring formation of reactants? z (a) 2 NO(g) + Oz(g) :;= 2 NOz(g); Kp = 5.0 X 101 (b) 2 HBr(g) :;= H2(g) + Brz(g); Kc = 5.8 X 10- 18 15.17 If K, = 0.042 for PCl3(g) + Cl2(g) :;= PCl5(g) at 500 K, what is the value of Kp for this reaction at this temperature?

15.18 Calculate K, at 303 K for S02(g) + Cl2(g) :;= SOzCl2(g) if Kp = 34.5 at this temperature. 15.19 The equilibrium constant for the reaction

2 NO(g) + Brz(g) :;= 2 NOBr(g)

is K, = 1.3 X 10-z at 1000 K. (a) Calculate K, for 2 NOBr(g) :;= 2 NO(g) + Brz(g). (b) At this tempera­ ture does the equilibrium favor NO and Brz, or does it favor NOBr?

15.20 Consider the following equilibrium:

2 Hz(g) + S2(g) :;= 2 HzS(g) K, = 1.08 X 10 7 at 700 oc

(a) Calculate Kp. (b) Does the equilibrium mixture con­ tain mostly Hz and Sz or mostly HzS?

15.21 At 1000 K, Kp

(a)

= 1.85 for the reaction

SOz(g) +

i Oz(g) :;= S03(g) i

What is the value of Kp for the reaction S03(g) :;= SOz(g) + Oz(g)? (b) What is the value of Kp for the reaction 2 SOz(g) + Oz(g) :;= 2 S03(g)? (c) What is the value of K, for the reaction in part (b)?

15.22 Consider the following Kp = 0.0752 at 480 oe:

(a)

equilibrium,

for

2 Clz(g) + 2 HzO(g) :;= 4 HCl(g) + Oz(g)

which

What is the value of KP for the reaction 4 HCl(g) + 02(g) :;= 2 Clz(g) + 2 HzO(g)? (b) What is the value of Kp for the reaction Clz(g) + HzO(g) :;= 2 HCl(g) + Oz(g)? (c) What is the value of K, for the reaction in part (b)?

i

15.23 The following equilibria were attained at 823 K: CoO(s) + Hz(g) :;= Co(s) + HzO(g) K,

CoO(s) + CO(g) :;= Co(s) + C02(g)

K,

=

67

= 490

Based on these equilibria, calculate the equilibrium con­ stant for Hz(g) + COz(g) :;= CO (g) + HzO(g) at 823 K.

15.24 Consider the equilibrium

Nz(g) + Oz(g) + Brz(g) :;= 2 NOBr(g)

Calculate the equilibrium constant Kp for this reaction, given the following information (at 298 K):

2 NO(g) + Br2(g) :;= 2 NOBr(g) K, = 2.0 2 NO(g) :;= N2(g) + Oz(g) K, = 2.1 X 1030

15.25 Mercury(!) oxide decomposes into elemental mercury and elemental oxygen: 2 HgzO(s) :;= 4 Hg(l) + Oz(g) (a) Write the equilibrium-constant expression for this reaction in terms of partial pressures. (b) Explain why we normally exclude pure solids and liquids from equilibrium-constant expressions. 15.26 Consider the equilibrium NazO(s) + SOz(g) :;= NazS03(s). (a) Write the equilibrium-constant expres­ sion for this reaction in terms of partial pressures. (b) Why doesn't the concentration of NazO appear in the equilibrium-constant expression?

Calculating Equi librium Cons tants 15.27 Gaseous hydrogen iodide is placed in a closed container at 425 oc, where it partially decomposes to hydrogen and iodine: 2 Hl(g) :;= Hz(g) + I2(g). At equilibrium it is

found that [HI] = 3.53 X 10- M, [Hz] = 4.79 X 10- M, 4 and [I2] = 4.79 X 10- M. What is the value of K, at this temperature?

3

4

Exercises 15.28 Methanol (CH30H) is produced commercially by the catalyzed reaction of carbon monoxide and hydrogen: CO(g) + 2 H2(g) = CH30H(g). An equilibrium mix­ ture in a 2.00-L vessel is found to contain 0.0406 mol CH30H, 0.170 mol CO, and 0.302 mol H2 at 500 K. Cal­ culate K, at this temperature. 15.29 The equilibrium 2 NO(g) + Cl2(g) = 2 NOCI(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures of 0.095 atm, 0.171 atm, and 0.28 atm for NO, Cl2, and NOCl, respectively. Calculate Kp for this reaction at 500 K. 15.30 Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: PC13 + Cl2(g) = PC1 (g). A gas vessel is charged with a mixture of 5 PCI3(g) and Cl2(g), which is allowed to equilibrate at 450 K. At equilibrium the partial pressures of the three gases are PpCI, = 0.124 atm, Po, = 0.157atm, and Prc1, = 1.30 atm. (a) What is the value of Kp at this temperature? (b) Does the equilibrium favor reactants or products? 15.31 A mixture of 0.10 mol of NO, 0.050 mol of H2o and 0.10 mol of H20 is placed in a 1.0-L vessel at 300 K. The following equilibrium is established: 2 NO(g) + 2 H2(g) = N2(g)

661

15.32 A mixture of 1.374 g of H2 and 70.31 g of Br2 is heated in a 2.00-L vessel at 700 K. These substances react as follows: H2(g) + Br2(g) = 2 HBr(g) At equilibrium the vessel is found to contain 0.566 g of H2. (a) Calculate the equilibrium concentrations of H2, Br2, and HBr. (b) Calculate K,.

15.33 A mixture of 0.2000 mol of C02, 0.1000 mol of H2o and 0.1600 mol of H20 is placed in a 2. 000-L vessel. The fol­ lowing equilibrium is established at 500 K: C02(g)

+ H2(g) = CO (g) + H20(g)

(a) Calculate the initial partial pressures of C02o H2,

and H20. (b) At equilibrium I'r!,o = 3.51 atm. Calculate the equilibrium partial pressures of C02o H2, and CO. (c) Calculate Kp for the reaction.

15.34 A flask is charged with 1.500 atm of N204(g) and 1.00 atm N02(g) at 25 'C, and the following equilibrium is achieved: N204(g) = 2 N02(g) After equilibrium is reached, the partial pressure of N02 is 0.512 atm. (a) What is the equilibrium partial pressure of N204? (b) Calculate the value of Kp for the reaction.

+ 2 H20(g)

At equilibrium [NO] = 0.062 M. (a) Calculate the equilib­ rium concentrations of H2o N2o and H20. (b) Calculate K,.

Applications of Equilibrium Constants 15.35 (a) How does a reaction quotient differ from an equilib­ rium constant? (b) If Q, < K,, in which direction will a reaction proceed in order to reach equilibrium? (c) What condition must be satisfied so that Q, = K,? 15.36 (a) How is a reaction quotient used to determine whether a system is at equilibrium? (b) If Q, > K,, how must the reaction proceed to reach equilibrium? (c) At the start of a certain reaction, only reactants are present; no products have been formed. What is the value of Q, at this point in the reaction? 15.37 At 100 'C the equilibrium constant for the reaction COCI2(g) = CO(g) + Cl2(g) has the value K, = 2.19 x 10-lo Are the following mixtures of COCl2o CO, and Cl2 at 100 'C at equilibrium? If not, indicate the direc­ tion that the reaction must proceed to achieve equilibrium. 3 (a) [COC12] = 2.00 x 10- M, [CO] = 3.3 X 10-6 M, 2 [CliJ = 6.62 X 10-6 M; (b) [COCI2] = 4.50 X 10- M, [CO] = 1.1 X 10-7M, [Cl2l = 2.25 X 10--.; M; (c) [COC12J = 0.0100 M, [CO] = [C12J = 1.48 X 10--.; M

15.38 As shown in Table 15.2, Kp for the equilibrium N2(g) + 3 H2(g) = 2 NH3(g)

is 4.51 x 10-5 at 450 'C. For each of the mixtures listed here, indicate whether the mixture is at equilibrium at 450 'C. If it is not at equilibrium, indicate the direction (toward product or toward reactants) in which the mix­ ture must shift to achieve equilibrium. (a) 98 atm NH3, 45 atm N2, 55 atm H2 (b) 57 atm NH3, 143 atm N2o no H2 (c) 13 atm NH3, 27 atm N2, 82 atm H2.

15.39 At 100 'C, K, = 0.078 for the reaction In an

502C12(g) = 502(g)

+ C12(g)

equilibrium mixture of the three gases, the concen­ trations of 50202 and 502 are 0.108 M and 0.052 M, re­ spectively. What is the partial pressure of Cl2 in the equilibrium mixture?

15.40 At 900 K the following reaction has Kp 2 502(g)

=

0.345:

+ 02(g) = 2 503(g)

In an equilibrium mixture the partial pressures of 502 and 02 are 0.135 atm and 0.455 atm, respectively. What is the equilibrium partial pressure of 503 in the mixture?

15.41 (a) At 1285 'C the equilibrium constant for the reaction 3 Br2(g) = 2 Br(g) is K, = 1.04 X 10- A 0.200-L vessel containing an equilibrium mixture of the gases has 0.245 g Br2(g) in it. What is the mass of Br(g) in the vessel? (b)Forthe reactio nH2(g) + l2(g) = 2 HI(g), K, = 55.3 at 700 K. In a 2.00-L flask containing an equilibrium mix­ ture of the three gases, there are 0.056 g H2 and 4.36 g !2. What is the mass of HI in the flask?

15.42 (a) At 800 K the equilibrium constant for I2(g) = 2 I(g) is K, = 3.1 x lo-s If an e�uilibrium mixture in a 10.0-L vessel contains 2.67 x 10- g of I(g), how many grams of 12 are in the mixture? (b) For 2 502(g) + 02(g) = 2 503(g), Kp = 3.0 X 104 at 700 K. In a 2.00-L vessel the equilibrium mixture con­ tains 1.17 g of 503 and 0.105 g of 02. How many grams of 502 are in the vessel?

662

C HAPTER 15

Chemical Equilibrium

2000 oc the equilibrium constant for the reaction 2 NO(g) = N2(g) + 02(g) is Kc = 2.4 X 103. If the initial concentration of NO is 0.200 M, what are the equilibrium concentrations of NO,

15.43 At

N;t, and 02?

15.44 For the equilibrium Br2(gl + Cl2(gl

= 2 BrCI(g)

at 400 K, Kc = 7.0. If 0.25 mol of Br2 and 0.25 mol of CJ2 are introduced into a 1.0-L container at 400 K, what will be the equilibrium concentrations of Br0 Cl2, and BrCl?

= 0.416 for the equilibrium 2 NOBr(g) = 2 NO(g) + Br2(g)

15.45 At 373 K, KP

If the pressures of NOBr(g) and NO(g) are equal, what is the equilibrium pressure of Br2(g)?

15.46 At

218 °C, Kc = 1.2 x 10-4 for the equilibrium NH4HS(s) = NH3(g) + H2S(g)

Calculate the equilibrium concentrations of NH3 and H2S if a sample of solid NH�S is placed in a closed ves­ sel and decomposes until equilibrium is reached.

15.47 Consider the reaction Ca504(s)

= Ca2+(aq) +

At 25 oc the equilibrium constant is Kc = 2.4 x 10-5 for this reaction . (a) If excess CaS04(s) is mixed with water at 25 oc to produce a saturated solution of CaS04, what 2 2 are the equilibrium concentrations of Ca + and 504 -? (b) If the resulting solution has a volume of 3.0 L, what is the minimum mass of CaS04(s) needed to achieve equilibrium?

15.48 At 80 °C, Kc

= 1.87

X

PH3BCl3(s)

10-3 for the reaction = PH3(g) + BCl3(g)

(a) Calculate the equilibrium concentrations of PH3 and BCl3 if a solid sample of PH3BC13 is placed in a closed vessel and decomposes until equilibrium is reached. (b) If the flask has a volume of 0.500 L, what is the mini­ mum mass of PH3BC13(s) that must be added to the flask to achieve equilibrium?

15.49 For the reaction 12 + Br2(g) = 2 IBr(g), Kc = 280 at 150 oc. Suppose that 0.500 mol IBr in a 1.00-L flask is al­ lowed to reach equilibrium at 150 oc. What are the equi­ librium concentrations of IBr, I;t, and Br2?

15.50 At 25 oc the reaction

= Ca2+(aq) + Crol-(aq) has an equilibrium constant Kc = 7.1 X 10-4. What are 2 CaCr04(g)

2 the equilibrium concentrations of Ca + and Cr04 saturated solution of CaCr04?

sol-(aq)

in a

Le Chatel ier's Principle 15.51 Consider the following equilibrium, for which

2 S02(g) +

02(gl

= 2 S03(g)

!lH
06), are present in many fruits as well as in certain vegetables, such as rhubarb and tomatoes. Acids and bases are important in numerous chemical processes that occur around us-from industrial processes to biological ones, from reactions in the laboratory to those in our environment. The time required for a metal object immersed in water to corrode, the ability of an aquatic environment to sup­ port fish and plant life, the fate of pollutants washed out of the air by rain, and even the rates of reactions that maintain our lives all critically depend upon the acidity or basicity of solutions. Indeed, an enormous amount of chemistry can be understood in terms of acid-base reactions. 667

668

C HA PTER 1 6

Acid-Base Equilibria We have encountered acids and bases many times in earlier discussions. For example, a portion of Chapter 4 focused on their reactions. But what makes a substance behave as an acid or as a base? In this chapter we reexamine acids and bases, taking a closer look at how they are identified and characterized. In doing so, we will consider their behavior both in terms of their structure and bonding and in terms of the chemical equilibria in which they participate.

1 6 . 1 ACIDS AND BASES: A B RIEF REVI EW From the earliest days of experimental chemistry, scientists have recognized acids and bases by their characteristic properties. Acids have a sour taste and cause certain dyes to change color (for example, litmus turns red on contact with acids). Indeed, the word acid comes from the Latin word acidus, meaning sour or tart. Bases, in contrast, have a bitter taste and feel slippery (soap is a good example). The word base comes from an old English meaning of the word, which is "to bring low." (We still use the word debase in this sense, meaning to lower the value of something.) When bases are added to acids, they lower the amount of acid. Indeed, when acids and bases are mixed in certain proportions, their characteristic properties disappear altogether. = (Section 4.3) Historically, chemists have sought to relate the properties of acids and bases to their compositions and molecular structures. By 1830 it was evident that all acids contain hydrogen but not all hydrogen-containing substances are acids. During the 1880s, the Swedish chemist Svante Arrhenius (1859-1927) linked acid behavior with the presence of H + ions and base behavior with the presence of OH - ions in aqueous solution. Arrhenius defined acids as substances that produce H+ ions in water and bases as substances that produce OH - ions in water. Indeed, the properties of aqueous solutions of acids, such as sour taste, are due to H + (aq), whereas the properties of aqueous solutions of bases are due to OH-(aq). Over time the Ar­ rhenius concept of acids and bases came to be stated in the following way: • •

An acid is a substance that, when dissolved in water, increases the concen­ tration of H+ ions. A base is a substance that, when dissolved in water, increases the concentra­ tion of OH- ions.

Hydrogen chloride is an Arrhenius acid. Hydrogen chloride gas is highly soluble in water because of its chemical reaction with water, which produces hydrated H + and Cl- ions: HCl(g)

H,O H +(aq) + Cqaq)

[16.1]

The aqueous solution of HCl is known as hydrochloric acid. Concentrated hy­ drochloric acid is about 37% HCl by mass and is 12 M in HCI. Sodium hydrox­ ide, on the other hand, is an Arrhenius base. Because NaOH is an ionic compound, it dissociates into Na + and OH- ions when it dissolves in water, thereby releasing OH - ions into the solution.

GIVE IT SOME THOUGHT What two ions are central to the Arrhenius definitions of acids and bases?

1 6.2 B R0NSTED-LOWRY ACI D S AND BASES The Arrhenius concept of acids and bases, while useful, has limitations. For one thing, it is restricted to aqueous solutions. In 1923 the Danish chemist Johannes Bronsted (1879- 1947) and the English chemist Thomas Lowry (1874-1936) in­ dependently proposed a more general definition of acids and bases. Their con­ cept is based on the fact that acid-base reactions involve the transfer ofH + ions from

one substance to another.

16.2

Bmnsted-Lowry Acids and Bases

669

The H + lon in Water

In Equation 16.1 hydrogen chloride is shown ionizing in water to form H+(aq).

An H+ ion is simply a proton with no surrounding valence electron. This small, positively charged particle interacts strongly with the nonbonding electron pairs of water molecules to form hydrated hydrogen ions. For example, the interaction of a proton with one water molecule forms the hydronium ion, H 30+(aq): [16.2]

The formation of hydronium ions is one of the complex features of the in­ teraction of the H+ ion with liquid water. In fact, the H 30+ ion can form hydro­ gen bonds to additional H20 molecules to generate larger clusters of hydrated hydrogen ions, such as H502+ and H904+ (Figure 16.1 .,. ). Chemists use H+(aq) and H 30+ (aq) interchangeably to represent the same thing-namely the hydrated proton that is responsible for the characteristic properties of aqueous solutions of acids. We often use the H+(aq) ion for simplic­ ity and convenience, as we did in Equation 16.1. The H30+ (aq) ion, however, more closely represents reality. H Proton-Transfer Reactions

When we closely examine the reaction that occurs when HCl dissolves in water, we find that the HCl molecule actually transfers an H+ ion (a proton) to a water molecule as depicted in Figure 16.2 .,.. Thus, we can represent the reaction as occurring between an HCl molecule and a water molecule to form hydronium and chloride ions:

H

"-o ·:

/ ·..

•

•

An acid is a substance (molecule or ion) that donates a proton to another substance. A base is a substance that accepts a proton.

---->

. .. :CJ: .

+

[ ]

+ H I H-N-H I H

..

H

(b)

.t. Figure 16.1 Hydrated hydronium Lewis structures and molecular models for H502 + and H904+. There is good experimental evidence for the existence of both these species.

Ions.

H

+

H+

[16.4]

This reaction can occur in the gas phase. The hazy film that forms on the windows of general chemistry laboratories and on glassware in the lab is largely solid NH4Cl formed by the gas-phase reaction of HCl and NH3 (Figure 16.3 T).

H

H-0: I H H9o4+

Thus, when HCl dissolves in water (Equation 16.3), HCl acts as a Bronsted­ Lowry acid (it donates a proton to H20), and H20 acts as a Bronsted-Lowry base (it accepts a proton from HCl). Because the emphasis in the Bmnsted-Lowry concept is on proton transfer, the concept also applies to reactions that do not occur in aqueous solution. In the reaction between HCl and NH3, for example, a proton is transferred from the acid HCI to the base NH3 : H .. I :Cl-H + :N-H .. I H

'cf

H

I I:I

[16.3] The polar H20 molecule promotes the ionization of acids in water solution by accepting a proton to form H 30+. Bmnsted and Lowry proposed definitions of acids and bases in terms of their ability to transfer protons:

H+

:c:( · · ".

+

H

.t. Figure 16.2 A proton·transfer When a proton is transferred from HCI to H20, HCI acts as the Bnmsted-Lowry acid and H20 acts as the Bnmsted- Lowry base. reaction.

670

C HA PTER 16

Acid-Base Equilibria Let's consider another example that compares the relationship between the Arrhenius definitions and the Bmnsted-Lowry definitions of acids and bases­ an aqueous solution of ammonia, in which the following equilibrium occurs:

NH3(aq) + HzO(l) � NH4+(aq) + OH-(aq)

.6. Figure 16.3 A gas-phase add-base reaction. The HCI(g) escaping from concentrated hydrochloric acid and the NH 3(g) escaping from aqueous ammonia (here labeled ammonium hydroxide) combine to form a white fog of NH4CI(s).

[16.5]

Ammonia is an Arrhenius base because adding it to water leads to an increase in the concentration of OH-(aq). It is a Bnmsted-Lowry base because it accepts a proton from H20. The H20 molecule in Equation 16.5 acts as a Bmnsted- Lowry acid because it donates a proton to the NH3 molecule. An acid and a base always work together to transfer a proton. In other words, a substance can function as an acid only if another substance simultaneously be­ haves as a base. To be a Bnmsted-Lowry acid, a molecule or ion must have a hydro­ gen atom that it can lose as an H+ ion. To be a Bmnsted-Lowry base, a molecule or ion must have a nonbonding pair of electrons that it can use to bind the H+ ion. Some substances can act as an acid in one reaction and as a base in another. For example, H20 is a Bmnsted-Lowry base in its reaction with HCl (Equation 16.3) and a Bmnsted-Lowry acid in its reaction with NH3 (Equation 16.5). A sub­ stance that is capable of acting as either an acid or a base is called amphiprotic. An amphiprotic substance acts as a base when combined with something more strongly acidic than itself and as an acid when combined with something more strongly basic than itself.

GIVE IT SOME THOUGHT In the forward reaction, which substance acts as the Bmnsted-Lowry base:

HS04-(aq) + NH 3(aq)

;;===:

sol-(aq) + NH 4+(aq)?

Conjugate Acid-Base Pairs

In any acid-base equilibrium both the forward reaction (to the right) and the re­

verse reaction (to the left) involve proton transfers. For example, consider the reac­ tion of an acid, which we will denote HX, with water:

[16.6] In the forward reaction HX donates a proton to H20. Therefore, HX is the Bmnsted-Lowry acid, and H20 is the Bmnsted-Lowry base. In the reverse reaction the H3o+ ion donates a proton to the x- ion, so H3o+ is the acid and x­ is the base. When the acid HX donates a proton, it leaves behind a substance, x-, which can act as a base. Likewise, when H20 acts as a base, it generates H30+, which can act as an acid. An acid and a base such as HX and x- that differ only in the presence or absence of a proton are called a conjugate acid-base pair.* Every acid has a conjugate base, formed by removing a proton from the acid. For example, OH- is the conjugate base of H20, and x- is the conjugate base of HX. Similarly, every base has associated with it a conjugate acid, formed by adding a proton to the base. Thus, H30+ is the conjugate acid of H20, and HX is the conjugate acid of x-. In any acid-base (proton-transfer) reaction we can identify two sets of con­ jugate acid-base pairs. For example, consider the reaction between nitrous acid (HNOz) and water: remove H...

Acid

Base

Conjugate

base

add H +

*The word conjugate means "joined together as a pair. "

Conjugate

J

aCJ d

[16.7]

Bmnsted-Lowry Acids and Bases

16.2

Likewise, for the reaction between NH3 and H20 (Equation 16.5), we have add H+

[16.8] Base

- SAMPLE EXERCISE

Acid

I

Conj':l g ate actd remove H

Conjugate base

l

+

16.1 I Identifying Conjugate Acids and Bases

(a) What is the conjugate base of each of the following acids: HCl04, H25, PH 4+,

HC03-? (b) What is SO/-, HzO, HC03-?

the conjugate acid of each of the following bases:

CN-,

SOLUTION Analyze: We are asked to give the conjugate base for each of a series of species and to give the conjugate acid for each of another series of species. Plan: The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve: (a) HCI04 less one proton (H+) is Cl0 4-. The other conjugate bases are HS-, PH3, and col-. (b) eN - plus one proton (H+) is HCN. The other conjugate acids are H504-, H30+, and H2C03. Notice that the hydrogen carbonate ion (HC03-) is arnphiprotic. It can act as

either an acid or a base.

- PRACTICE EXERCISE

3-, CO.

Write the formula for the conjugate acid of each of the following: HS03 -, F-, P04

Answers: H2503, HF, HPO/-, HCO+

- SAMPLE EXERCISE

16.2 I Writing Equations for Proton-Transfer Reactions

The hydrogen sulfite ion (H503-) is amphiprotic. (a) Write an equation for the reac­ tion of HS03- with water, in which the ion acts as an acid. (b) Write an equation for the reaction of HS03- with water, in which the ion acts as a base. In both cases iden­ tify the conjugate acid-base pairs. SOLUTION Analyze and Plan: We are asked to write two equations representing reactions be­ tween HS03- and water, one in which HS03- should donate a proton to water, there­ by acting as a Bmnsted-Lowry acid, and one in which HS03- should accept a proton from water, thereby acting as a base. We are also asked to identify the conjugate pairs in each equation. Solve:

(a) The conjugate pairs in this equation are HS03- (acid) and and H20 (base) and H3o+ (conjugate acid).

2 503 - (conjugate base);

(b) The conjugate pairs in this equation are H20 (acid) and H503- (base) and H2503 (conjugate acid).

OH- (conjugate base), and

- PRACTICE EXERCISE When lithium oxide (LizO) is dissolved in water, the solution turns basic from the 2 reaction of the oxide ion (0 -) with water. Write the reaction that occurs, and identi­ fy the conjugate acid-base pairs.

Answer: 022-(aq) + H20(/) the base 0

----->

OH-(aq) + OH-(aq). OH- is the conjugate acid of -. OH- is also the conjugate base of the acid H20.

671

672

C HA PTER 16

Acid-Base Equilibria Relative Strengths of Acids and Bases

10 % { E

ionized in

H20

ACID

HCl

H2S04 HN03 Hp+ (aq)

H2P04-

Figure 16.4

gate base; the stronger a base, the weaker is its conjugate acid. Thus, if we know something about the strength of an acid (its ability to donate protons), we also know something about the strength of its conjugate base (its ability to accept protons). The inverse relationship between the strengths of acids and the strengths of their conjugate bases is illustrated in Figure 16.4

Pol-

� ..0 "So

OH-

OH oz-

Hz

H-

z

CH4

CH3-

:.:1 bO '"

&

HS04N0 3-

NH4 +

HC03 -

I

Some acids are better proton donors than others; likewise, some bases are bet­ ter proton acceptors than others. If we arrange acids in order of their ability to donate a proton, we find that the more easily a substance gives up a proton, the less easily its conjugate base accepts a proton. Similarly, the more easily a base accepts a proton, the BASE less easily its conjugate acid gives up a proton. In cr other words, the stronger an acid, the weaker is its conju­

00

§

b "'

}10 %

protonated in H20

Relative strengths of some conjugate acid-base

The two members of each pair are listed opposite each other in the two columns. The acids decrease in strength from top to bottom, whereas their conjugate bases increase in strength from top to bottom. pairs.

2. A weak acid only partially dissociates in aqueous so­

lution and therefore exists in the solution as a mix­ ture of acid molecules and their constituent ions. The conjugate base of a weak acid shows a slight ability to remove protons from water. (The conjugate

base of a weak acid is a weak base.) negligible acidity, such as CH4,

3. A substance with

contains hydrogen but does not demonstrate any acidic behavior in water. Its conjugate base is a strong base, reacting completely with water, ab­ stracting protons to form OH- ions.

GIVE IT SOME THOUGHT Using the three categories above, specify the strength of HN03 and the strength of its conjugate base, N03-.

We can think of proton-transfer reactions as being governed by the relative abilities of two bases to abstract protons. For example, consider the proton transfer that occurs when an acid HX dissolves in water:

[16.9] If H20 (the base in the forward reaction) is a stronger base than x- (the conju­ gate base of HX), then H20 will abstract the proton from HX to produce H30+ and x-. As a result, the equilibrium will lie to the right. This describes the behavior of a strong acid in water. For example, when HCl dissolves in water, the solution consists almost entirely of H3o + and Cl- ions with a negligible concentration of HCl molecules.

[16.10] H20 is a stronger base than Cl- (Figure 16.4), so H20 acquires the proton to become the hydronium ion.

16.3

When x- is a stronger base than H20, the equilibrium will lie to the left. This situation occurs when HX is a weak acid. For example, an aqueous solu­ tion of acetic acid (CH3COOH) consists mainly of CH3COOH molecules with only a relatively few H30+ and CH3Coo- ions. CH3COOH(aq) + H20(1) :;::::::': H30 +(aq) + CH3COO-(aq) [16.11] CH3Coo- is a stronger base than H20 (Figure 16.4) and therefore abstracts the proton from H30 + From these examples, we conclude that in every acid-base reaction the position of the equilibrium favors transfer of the proton from the stronger acid to the stronger base to form the weaker acid and the weaker base. As a result, the equilibrium mix­ ture contains more of the weaker acid and weaker base and less of the stronger acid and stronger base. • SAMPLE EXERCISE

16.3 Predicting the Position of a I Proton-Transfer Equilibrium

For the following proton-transfer reaction, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left (that is, K, < 1 ) or to the right (K, > 1):

HS0 4-(aq) + C03 2-(aq) = so/-(aq) + HCo3-(aq)

SOLUTION Analyze: We are asked to predict whether the equilibrium shown lies to the right, fa­ voring products, or to the left, favoring reactants. Plan: This is a proton-transfer reaction, and the position of the equilibrium will favor the proton going to the stronger of two bases. The two bases in the equation are col-, the base in the forward reaction as written, and so. 2-, the conjugate base of HS04-. We can find the relative positions of these two bases in Figure 16.4 to deter­ mine which is the stronger base. Solve: COl- appears lower in the right-hand column in Figure 16.4 and is therefore 2 a stronger base than so. 2-. C03 -, therefore, will get the proton preferentially to become HC03-, while SO/- will remain mostly unprotonated. The resulting equi­ librium will lie to the right, favoring products (that is, K, > 1).

HS04-(aq) Acid

+

C03 2-(aq) = 504 2-(aq) Base

Conjugate base

+

HC03-(aq)

Conjugate acid

K, > 1

Comment: Of the two acids in the equation, HS04 - and HC03 -, the stronger one gives up a proton more readily while the weaker one tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base. - PRACTICE EXERCISE For each of the following reactions, use Figure 16.4 to predict whether the equilibrium lies predominantly to the left or to the right:

2 (a) HP0 4 -(aq) + H20(/) = H2P0 4-(aq) + OH-(aq) (b) NH4 +(aq) + OH-(aq) = NH3(aq) + H20(/) Answers: (a) left, (b) right

1 6.3 THE AUTO IONIZATION OF WATER One of the most important chemical properties of water is its ability to act as either a Bmnsted acid or a Bmnsted base, depending on the circumstances. In the presence of an acid, water acts as a proton acceptor; in the presence of a base, water acts as a proton donor. In fact, one water molecule can donate a proton to another water molecule: [16.12]

We call this process the autoionization of water. No individual molecule re­ mains ionized for long; the reactions are extremely rapid in both directions.

The Autoionization of Water

673

674

C HA PTER 1 6

Acid-Base Equilibria At room temperature only about two out of every 109 molecules are ionized at any given instant. Thus, pure water consists almost entirely of H20 molecules and is an extremely poor conductor of electricity. Nevertheless, the autoioniza­ tion of water is very important, as we will soon see. The lon Product of Water Because the autoionization of water (Equation 16.12) is an equilibrium process, we can write the following equilibrium-constant expression for it:

[16 . 13] The term [H20] is excluded from the equilibrium-constant expression because we exclude the concentrations of pure solids and liquids. = (Section 15.4) Because this equilibrium-constant expression refers specifically to the autoioniza­ tion of water, we use the symbol Kw to denote the equilibrium constant, which 4 we call the ion-product constant for water. At 25 °C, Kw equals 1.0 x 10-1 . Thus, we have

[16.14] Because we use H+(aq) and H 30+(aq) interchangeably to represent the hy­ drated proton, the autoionization reaction for water can also be written as

[16.15] Likewise, the expression for Kw can be written in terms of either H 30- or H+ , and Kw has the same value in either case: Kw = [H 30+][0H-] = [H+][OH-] = 1.0

X 10-14

(at 25

oq

[16.16]

This equilibrium-constant expression and the value of Kw at 25 °C are extremely important, and you should commit them to memory. What makes Equation 16.16 particularly useful is that it is applicable to pure water and to any aqueous solution. Although the equilibrium between H+(aq) and OH-(aq) as well as other ionic equilibria are affected somewhat by the presence of additional ions in solution, it is customary to ignore these ionic effects except in work requiring exceptional accuracy. Thus, Equation 16.16 is taken to be valid for any dilute aqueous solution, and it can be used to calculate either [H+] (if [OH-] is known) or [OH-] (if [H+] is known). A solution in which [H+] = [OH-] is said to be neutral. In most solutions H+ and OH- concentrations are not equal. As the concentration of one of these ions increases, the concentration of the other must decrease, so that the product of their concentrations equals 1.0 X 10-H In acidic solutions [H+] exceeds [OH-]. In basic solutions [OH-] exceeds [H+]. - SAMPLE EXERCISE 1 6.4 [

Calculating [H+] for Pure Water

Calculate the values of [H+] and [OH-] in a neutral solution at 25 oc. SOLUTION Analyze: We are asked to determine the concentrations of H+ and OH- ions in a neutral solution at 25 'C. Plan: We will use Equation 16.16 and the fact that, by definition, [H+] = [OH-] in a neutral solution. Solve: We will represent the concentration of [H+] and [OH-] in neutral solution with x. This gives [H+][oH-]

=

(x)(x)

=

1.0

x2 = 1.0 x 10-14

x

10-14

x = 1.0 X 10-7 M = [H+] = [OH-] In an acid solution [H+] is greater than 1.0 x 10- 7 M; in a basic solution [H+] is less than 1.0 X 10-7 M.

16.4

The pH Scale

675

- PRACTICE EXERCISE Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: (a) [H+] � 4 x 10-9 M; (b) [OH-] � 1 x 10-7 M; 3 (c) [OH-] � 7 X 10-1 M. Answers: (a) basic, (b) neutral, (c) acidic

- SAMPLE EXERCISE

16.5 I Calculating [H+] from [OHl

Calculate the concentration of H+(aq) in (a) a solution in which [OHl is 0.010 M, (b) a solution in which [OH-] is Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 oc.

1.8

x

10-9 M.

SOLUTION Analyze: We are asked to calculate the hydronium ion concentration in an aqueous solution where the hydroxide concentration is known. Plan: We can use the equilibrium-constant expression for the autoionization of water and the value of Kw to solve for each un­ known concentration. Solve:

[H+][oH-] � 1.0 X 10-14

(a) Using Equation 16.16, we have:

[H+] �

This solution is basic because

[OW] > [H+]

(b) In this instance

[H"']

This solution is acidic because

[H+] > [OW]

�

(1.0

X

(1.0 X 10-14 ) [OHl

10-14 )

[OH l

�

�

1.0

X 10-14 � 1.0 0.010

1.0 X 10-14 1.8 X 10-'�

- PRACTICE EXERCISE

Calculate the concentration of OH-(aq) in a solution in which (a) (a) 5 X 10-'� M, (b) 1.0 X 10-7 M, (c) 1.0 X 10-B M

Answers:

[H"']

�

2 X

�

5.6

10� M; (b) [H+]

The molar concentration of H+(aq) in an aqueous solution is usually very small. For convenience, we therefore usually express [H+] in terms of pH, which is the negative logarithm in base 10 of [H+].• [16.17] pH � -log[H+] If you need to review the use of logs, see Appendix A. We can use Equation 16.17 to calculate the pH of a neutral solution at 25 oc (that is, one in which [H+] 1 .0 X 10-7 M): pH � -log(l.O X 10-7) � -(-7.00) � 7.00 The pH of a neutral solution is 7.00 at 25 °C. Notice that the pH is reported with two decimal places. We do so because only the numbers to the right of the dec­ imal point are the significant figures in a logarithm. Because our original value for the concentration (1.0 x 10-7 M) has two significant figures, the corre­ sponding pH has two decimal places (7.00). What happens to the pH of a solution as we make the solution acidic? An acidic solution is one in which [H+] > 1.0 X 10-7 M. Because of the negative sign in Equation 16.17, the pH decreases as [H+] increases. For example, the pH of an acidic solution in which [H+] 1.0 x 10-3 M is pH � -log(l.O X 10-3) � -(-3.00) � 3.00 �

At 25 oc the pH of an acidic solution is less than 7.00.

'Because [W] and [H30+1 are used interchangeably, you might see pH defined as -Log[HJO+/.

10-1 2 M

X 10� M

1 6.4 THE pH S CALE

�

X

�

[OH-]; (c) [H+]

�

100

X

[OH-].

676

C HA PTER 1 6

Acid-Base Equilibria

TABLE 16.1

• Relationships among [H +], [OH -], and pH at 25 •c

Solution Type

[H + ] (M)

[OH - ] (M)

pH Value

Acidic Neutral Basic

>l.O x w-7 =l.O x to-7 1.0 x 10-7 M . Suppose [OH -] = 2.0 X 10- M . We can use Equation + 16.16 to calculate [H ] for this solution, and Equation 16.17 to calculate the pH: [H +]

K 10 = __ w_ = . [OH-]

X 10-14

2.0 X 10-

pH = -log(5.0 x

w-12)

3

= 5.0 X 10-12 M

= 1 1 .30

At 25 •c the pH of a basic solution is greater than 7.00. The relationships among [H +] , [OH-], and pH are summarized in Table 16.1 A and in Figure 16.5 T. The pH values characteristic of several familiar solutions are shown in Figure 16.5. Notice that a change in [H+] by a factor of 10 causes the pH to change by 1. Thus, a solution of pH 6 has 10 times the concentration of H +(aq) as a solution of pH 7.

GIVE IT SOME THOUGHT (a) What is the significance of pH = 7? (b) How does the pH change as OH- is added to the solution?

You might think that when [H +] is very small, as it is for some of the exam­ ples shown in Figure 16.5, it would be unimportant. Nothing is further from the truth. If [H +] is part of a kinetic rate law, then changing its concentration will

.,. Figure 16.5 H + concentrations and pH values of some common substances at 25 •c. The pH of a

solution can be estimated using the benchmark concentrations of H+ and OH- corresponding to whole-number pH values.

Gastric juice - - - - - Lemon juice - - - - - Cola, vinegar - - - - -

Wine - - - - - - - - - Tomatoes - - - - - - Banana - - - - - - - - Black coffee - - - - - Rain - - - - - - - - - Saliva - - - - - - - - Milk - - - - - - - - - -

Human blood, tears Egg white, seawater Baking soda - - - - - Borax - - - - - - - - -

.�

"'"' ..0 0

�

::E

Milk of magnesia - - Lime water - - - - - -

[H + ] (M)

pH

pOH

[OW] (M)

1 (1 x 10- �

0.0

14.0

1x 10- 14

1.0

13.0

1x 1o- 13

1x 10 - 1

I

1 X10-2

2.0

12.0

1X 10- 12

1x10-3

3.0

11.0

1 Xl0-ll

1X 10-4

4.0

10.0

1X 10-10

1 x1o-5

5.0

9.0

1X 10- 9

1 x 10-6

6.0

8.0

1x 10-8

1 >

H30+(aq)

+

N03-(aq) (complete ionization) [16.21]

We have not used equilibrium arrows for Equation 16.21 because the reaction lies entirely to the right, the side with the ions. (Section 4. 1) As noted in Section 16.3, we use H30+(aq) and H+(aq) interchangeably to represent the hy­ drated proton in water. Thus, we often simplify the equations for the ionization reactions of acids as follows: orx>

HN03(aq)

-----'>

H+(aq) + N03-(aq)

In an aqueous solution of a strong acid, the acid is normally the only significant source of H+ ions.• As a result, calculating the pH of a solution of a strong monoprotic acid is straightforward because [H +] equals the origi­ nal concentration of acid. In a 0.20 M solution of HN03(aq), for example, [H +] = [N03-] = 0.20 M. The situation with the diprotic acid H2S04 is more complex, as we will see in Section 16.6. *if the concentration of the acid is 10-6 M or less, we also need to consider H + ions that result from the autoionization of H20. Normally, the concentration of H + from H20 is so small that it can be neglected.

Strong Acids and Bases

679

680

CHA PTER 16

Acid-Base Equilibria

16.8 I Calculating the pH of a Strong Acid

- SAMPLE EXERCISE

What is the pH of a 0.040 M solution of HC104? SOLUTION

Analyze and Plan: Because HC104 is a strong acid, it is completely ionized, giving [H+] = [Cl04 -] = 0.040 M. Solve: The pH of the solution is given by pH = -log(0.040) = 1.40. 2 Check: Because [H+] lies between 1 X 10- and 1 X 10-1, the pH will be between 2.0 and 1.0. Our calculated pH falls within the estimated range. Furthermore, because the concentration has two significant figures, the pH has two decimal places. - PRACTICE EXERCISE An aqueous solution of HN03 has a pH of 2.34. What is the concentration of the acid?

Answer: 0.0046 M Strong Bases

There are relatively few common strong bases. The most common soluble strong bases are the ionic hydroxides of the alkali metals (group lA) and the heavier alkaline earth metals (group 2A), such as NaOH, KOH, and Ca(OH)2. These compounds completely dissociate into ions in aqueous solution. Thus, a solution labeled 0.30 M NaOH consists of 0.30 M Na+(aq) and 0.30 M OH-(aq); there is essentially no undissociated NaOH. - SAMPLE EXERCISE

16.9 I Calculating the pH of a Strong Base

What is the pH of (a) a O.Q28 M solution of NaOH, (b) a 0.0011 M solution of Ca(OH)z? SOLUTION Analyze: We are asked to calculate the pH of two solutions of strong bases. Plan: We can calculate each pH by either of two equivalent methods. First, we could use Equation 16.16 to calculate [H+] and then use Equation 16.17 to calculate the pH. Alternatively, we could use [OH-] to calculate pOH and then use Equation 16.20 to calculate the pH. Solve:

(a) NaOH dissociates in water to give one OH- ion per formula unit. Therefore, the

OH- concentration for the solution in (a) equals the stated concentration of NaOH, namely 0.028 M.

Method 1: [H+] =

Method 2:

l.O X

4 10-!

O.Q28

pOH

=

= 3 . 57 X 1 0-13 M

-log(0.028)

=

1.55

pH = -log(3.57 X 10-13) = 12.45

pH

=

14.00 - pOH

=

12.45

(b) Ca(OH)z is a strong base that dissociates in water to give two OH- ions per for­

mula unit. Thus, the concentration of OH-(aq) for the solution in part (b) is 2 X (0.0011 M) = 0.0022 M.

Method 1: [H+] =

Method 2:

1 .0

X

10-14

0.0022

= 4.55 X 10-12 M

pOH = -log(0.0022) = 2.66

pH = -log(4.55 X 10-12) = 11.34

pH = 14.00 - pOH = 11 .34

- PRACTICE EXERCISE What is the concentration of a solution of (a) KOH for which the pH is 11.89; (b) Ca(OH)z for which the pH is 11 .68? Answers: (a) 7.8 x 10-3 M , (b) 2.4 x 10-3 M

16.6

Although all the hydroxides of the alkali metals (group 1A) are strong elec­ trolytes, LiOH, RbOH, and CsOH are not commonly encountered in the labora­ tory. The hydroxides of the heavier alkaline earth metals, Ca(OHh, Sr(OHh, and Ba(OHh, are also strong electrolytes. They have limited solubilities, how­ ever, so they are used only when high solubility is not critical. Strongly basic solutions are also created by certain substances that react with water to form OH-(aq). The most common of these contain the oxide ion. Ionic metal oxides, especially Na20 and CaO, are often used in industry when a strong base is needed. The 0 2- reacts with water to form OH-, leaving virtu­ ally no 02- remaining in the solution: [16.22] 2 OH-(aq) 02- (aq) + H20( I) ----

Thus, a solution formed by dissolving 0.010 mol of Na20(s) in enough water to form 1.0 L of solution will have [OH-] = 0.020 M and a pH of 12.30. GIVE IT SOME THOUGHT The CH3- ion is the conjugate base of CH4, and CH4 shows no evidence of being an acid in water. What happens when CH3- is added to water?

1 6.6 WEAK AC IDS Most acidic substances are weak acids and are therefore only partially ionized in aqueous solution. We can use the equilibrium constant for the ionization re­ action to express the extent to which a weak acid ionizes. If we represent a gen­ eral weak acid as HA, we can write the equation for its ionization reaction in either of the following ways, depending on whether the hydrated proton is rep­ resented as H30+(aq) or H+(aq): HA(aq) + H20(1) :;::::::': H30+(aq) + A-(aq)

or

[16.23] [16.24]

Because H20 is the solvent, it is omitted from the equilibrium-constant expres­ sion. (Section 15.4) Thus, we can write the equilibrium-constant expression as either CX10

As we did for the ion-product constant for the autoionization of water, we change the subscript on this equilibrium constant to indicate the type of equa­ tion to which it corresponds. Ka

=

[H30+][A-] or Ka [HA]

=

[HAl [H+][A-]

[16.25]

The subscript a on Ka denotes that it is an equilibrium constant for the ioniza­ tion of an acid, so Ka is called the acid-dissociation constant. Table 16.2T shows the names, structures, and Ka values for several weak acids. Appendix D provides a more complete list. Many weak acids are organic compounds composed entirely of carbon, hydrogen, and oxygen. These com­ pounds usually contain some hydrogen atoms bonded to carbon atoms and some bonded to oxygen atoms . In almost all cases the hydrogen atoms bonded to carbon do not ionize in water; instead, the acidic behavior of these com­ pounds is due to the hydrogen atoms attached to oxygen atoms. The magnitude of Ka indicates the tendency of the acid to ionize in water: The larger the value ofK3, the stronger the acid. Hydrofluoric acid (HF), for example, is the strongest acid listed in Table 16.2, and phenol (HOC6H5) is the weakest. Notice that Ka is typically less than 10-3 •

Weak Acids

681

682

C HA PTER 16

Acid-Base Equilibria

TABLE 1 6 .2 • Some Weak Acids in Water at 25 •c Acid

Structural Formula•

Conjugate Base

Equilibrium Reaction

Hydrofluoric (HF)

H-F

F-

HF(aq) + H 20(1) � Hp+(aq) + F-(aq)

6.8 X 10-4

Nitrous (HN02)

H-0-N=O

N02-

HN02(aq) + H20(/) � Hp+(aq) + N0 2- (aq)

4.5 X 10-4

C6H 5coo-

C6H 5COOH(aq) + H 20(1) � H p+(aq) + C6H 5COO-(aq)

6.3 X 10-5

CH 3coo-

CH 3COOH(aq) + H20(/) � H p+(aq) + CH 3COO-(aq)

1.8 X 10-

-@ 0

Ka

0

Benzoic (C6H5COOH)

H-0-CII

Acetic (CH3COOH)

H-0-C-C-H

Hypochlorous (HCIO)

H- O-C!

oo-

HClO(aq) + H 20(1) � Hp +(aq) + CIO-(aq)

3.0 X 10-8

Hydrocyanic (HCN)

H-C =N

CN-

HCN(aq) + H 20(1) � H 30+(aq) + CN-(aq)

4.9 X 10-lO

Phenol (HOC6H5)

H-o

C6H 50-

HOC6H5(aq) + HzO(I) � Hp+(aq) + C6H50-(aq)

1.3 X 1 0-IO

0

II

H

I

I

H

+ The proton that ionizes is shown in blue.

--@

5

Calculating K0 from pH In order to calculate either the Ka value for a weak acid or the pH of its solu­ tions, we will use many of the skills for solving equilibrium problems that we developed in Section 15.5. In many cases the small magnitude of Ka allows us to use approximations to simplify the problem. In doing these calculations, it is important to realize that proton-transfer reactions are generally very rapid. As a result, the measured or calculated pH for a weak acid always represents an equilibrium condition. - SAMPLE EXERCISE 1 6.1 0 J A student prepared a 0.10 M solution of formic acid (HCOOH) and measured its pH. The pH at 25 •c was found to be 2.38. Calculate Ka for formic acid at this temperature.

Calculating K0 from Measured pH

SOLUTION Analyze: We are given the molar concentration of an aqueous solution of weak acid and the pH of the solution, and we are asked to determine the value of Ka for the acid. Plan: Although we are dealing specifically with the ionization of a weak acid, this problem is very similar to the equilibrium prob­ lems we encountered in Chapter 15. We can solve this problem using the method first outlined in Sample Exercise 15.9, starting with the chemical reaction and a tabulation of initial and equilibrium concentrations. Solve: The first step in solving any equilibrium problem is to write the equation for the equilibrium reac­ tion. The ionization of formic acid can be written as follows:

HCOOH(aq) � H+(aq) + HCOO-(aq)

The equilibrium- Kb, the ion will cause the solution to be acidic. If Kb > Ka, the solution will be basic. cx:o

A Cation's Abil ity to React with Water

Polyatomic cations whose formulas contain one or more protons can be consid­ ered the conjugate acids of weak bases. NH4+, for example, is the conjugate acid of the weak base NH3. Thus, NH4 + is a weak acid and will donate a proton to water, producing hydronium ions and thereby lowering the pH: NH4+(aq)

+

H20( I) :;::::::=: NH3(aq)

+

H30+(aq)

[16.44]

-rhese rules apply to what are called normal salts. These salts contain no ionizable protons on the anion. The pH of an acid salt (such as NaHC03 or NaH,P04) is affected by the hydrolysis of the anion and by its acid dissocation, i as shown in Sample Exercise 16.19.

Acid-Base Properties of Salt Solutions

16.9

697

Most metal ions can also react with water to decrease the pH of an aqueous solution. The mechanism by which metal ions produce acidic solutions is de­ scribed in Section 16.11. However, ions of alkali metals and of the heavier alka­ line earth metals do not react with water and therefore do not affect pH. Note that these exceptions are the cations found in the strong bases. <XX> (Section 16.5) G IVE IT S O M E THOUGHT Which of the following cations has no effect on the pH of a solution: K+, Fe2+, or A13+ ?

Combined Effect of Cation and Anion in Solution

If an aqueous salt solution contains an anion that does not react with water and a cation that does not react with water, we expect the pH to be neutral. If the so­ lution contains an anion that reacts with water to produce hydroxide and a cation that does not react with water, we expect the pH to be basic. If the solu­ tion contains a cation that reacts with water to produce hydronium and an anion that does not react with water, we expect the pH to be acidic. Finally, a so­ lution may contain an anion and a cation both capable of reacting with water. In this case both hydroxide and hydronium will be produced. Whether the solu­ tion is basic, neutral, or acidic will depend upon the relative abilities of the ions to react with water. To summarize: anion that is the conjugate base of a strong acid, for example, Br-, will not affect the pH of a solution. (It will be a spectator ion in acid-base chemistry.) 2. An anion that is the conjugate base of a weak acid, for example, CN-, will cause an increase in pH. 3. A cation that is the conjugate acid of a weak base, for example, CH3NH3+, will cause a decrease in pH. 4. The cations of group 1A and heavier members of group 2A (Ca2+, Sr2+, and Ba 2+) will not affect pH. These are the cations of the strong Arrhenius bases. (They will be spectator ions in acid-base chemistry.) 5. Other metal ions will cause a decrease in pH. 6. When a solution contains both the conjugate base of a weak acid and the conjugate acid of a weak base, the ion with the larger equilibrium constant, Ka or Kb, will have the greater influence on the pH. 1. An

Figure 16.11 T demonstrates the influence of several salts on pH.

.. Figure 16.1 1 Salt solutions can be neutral, addle, or basic. These three

solutions contain the acid-base indicator bromthymol blue. (a) The NaCI solution is neutral (pH = 7.0); (b) the NH4CI solution is acidic (pH = 3.5); (c) the NaCIO solution is basic (pH = 9.5).

(a)

(b)

(c)

698

CHAPTER 16

Acid-Base Equilibria - SAMPLE EXERCISE 16.18

1 Determining Whether Salt Solutions Are Acidic, Basic, or Neutral

Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: (a) Ba(CH3COO):u (b) NH4Cl, (c) CH�H3Br, (d) KN03, (e) Al(Cl04)J. SOLUTION Analyze: We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral. Plan: We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the pH. Solve:

2

(a) Tilis solution contains barium ions and acetate ions. The cation, Ba +, is an ion of one of the heavy alkaline earth metals and will therefore not affect the pH (summary point 4).

The anion, CH3Coo-, is the conjugate base of the weak acid CH3COOH and will hy­ drolyze to produce OH- ions, thereby making the solution basic (summary point 2). (b) This solution contains NH 4+ and o- ions. NH 4+ is the conjugate acid of a weak base (NH3) and is therefore acidic (summary point 3). o- is the conjugate base of a strong acid (HCl) and therefore has no influence on the pH of the solution (summary point 1). Because the solution contains an ion that is acidic (NH4+) and one that has no influence on pH (Cq, the solution of NH4Cl will be acidic. (c) This solution contains CH3NH3 + and Br- ions. CH3NH3+ is the conjugate acid of a weak base (CH�H:u an amine) and is therefore acidic (summary point 3). Br- is the conjugate base of a strong acid (HBr) and is therefore pH-neutral (summary point 1). Because the solution contains one ion that is acidic and one that is neutral, the solution of CH3NH3Br will be acidic. (d) This solution contains the K+ ion, which is a cation of group 1A, and the N03- ion, which is the conjugate base of the strong acid HN03. Neither of the ions will react with water to any appreciable extent (summary points 1 and 4), making the solution neutral. 3 3 (e) This solution contains Al + and Cl0 4- ions. Cations, such as Al +, that are not in groups 1A or 2A are acidic (summary point 5). The Cl04- ion is the conjugate base of a strong acid (HC104) and therefore does not affect pH (summary point 1). Thus, the solution of Al(Cl04)3 will be acidic. - PRACTICE EXERCISE

In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) 0.010 M solution: (a) NaN03, or Fe(N03)3; (b) KBr, or KBrO; (c) CH3NH3Cl, or BaC12, (d) NH4NO:u or NH4N03. Answers: (a) Fe(N03)3, (b) KBr, (c) CH�H3Cl, (d) NH4N03

- SAMPLE EXERCISE 16.19

1 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic

Predict whether the salt Na2HP04 will form an acidic solution or a basic solution on dissolving in water. SOLUTION Analyze: We are asked to predict whether a solution of Na2HP04 will be acidic or basic. This substance is an ionic compound composed of Na+ and HPO/- ions. Plan: We need to evaluate each ion, predicting whether each is acidic or basic. Because Na+ is a cation of group 1A, we know that it has no influence on pH. It is merely a spectator ion in acid-base chemistry. Thus, our analysis of whether the 2 solution is acidic or basic must focus on the behavior of the HP04 - ion. We need 2 to consider the fact that HP04 - can act as either an acid or a base. 3 [16.45] HPO/-(aq) :;==! H+(aq) + P0 4 -(aq) 2 [16.46] HP04 -(aq) + HzO :;==! HzP0 4-(aq) + OH-(aq) The reaction with the larger equilibrium constant will determine whether the solu­ tion is acidic or basic. 3 Solve: The value of Ka for Equation 16.45, as shown in Table 16.3, is 4.2 X 10-1 We must calculate the value of Ku for Equation 16.46 from the value of Ka for its conjugate acid, H2P04-. We make use of the relationship shown in Equation 16.40.

Ka X Ku = Kw

16.10

We want to know Kb for the base acid H PO•-=

2

Acid-Base Behavior and Chemical Structure

HP042-, knowing the value of K, for the conjugate

Kb(HP042l X K.(H 2P04-)

=

Kw

=

1.0 X 10-14

Because K, for H 2P04- is 6.2 X 10--a (Table 16.3), we calculate Kb for HPOl- to be 1.6 x 10-7. This is more than 105 times larger than K, for HPO/-; thus, the reac­ tion shown in Equation 16.46 predominates over that in Equation 16.45, and the solution will be basic. - PRACTICE EXERCISE Predict whether the dipotassium salt of citric acid or basic solution in water (see Table 16.3 for data). Answer: acidic

(K2HC6Hs07) will form an acidic

1 6 . 1 0 ACID-BASE BEHAVIO R AND CHEM ICAL STRUCTURE When a substance is dissolved in water, it may behave as an acid, behave as a base, or exhibit no acid-base properties. How does the chemical structure of a substance determine which of these behaviors is exhibited by the substance? For example, why do some substances that contain OH groups behave as bases, releasing OH- ions into solution, whereas others behave as acids, ionizing to release H+ ions? Why are some acids stronger than others? In this section we will discuss briefly the effects of chemical structure on acid-base behavior. Factors That Affect Acid Strength

A molecule containing H will transfer a proton only if the H-X bond is polar­ ized in the following way: +--+

H-X

In ionic hydrides, such as NaH, the reverse is true; the H atom possesses a neg­ ative charge and behaves as a proton acceptor. Essentially nonpolar H- X bonds, such as the H -C bond in CH4, produce neither acidic nor basic aque­ ous solutions. A second factor that helps determine whether a molecule containing an H-X bond will donate a proton is the strength of the bond. Very strong bonds are less easily dissociated than weaker ones. This factor is important, for exam­ ple, in the case of the hydrogen halides. The H- F bond is the most polar H- X bond. You therefore might expect that HF would be a very strong acid if the first factor were all that mattered. However, HF has the highest bond strength among the hydrogen halides, as seen in Table 8.4. As a result, HF is a weak acid, whereas all the other hydrogen halides are strong acids in water. A third factor that affects the ease with which a hydrogen atom ionizes from HX is the stability of the conjugate base, x-. In general, the greater the sta­ bility of the conjugate base, the stronger is the acid. The strength of an acid is often a combination of all three factors: (1) the polarity of the H- X bond, (2) the strength of the H-X bond, and (3) the stability of the conjugate base, x-. Binary Acids

In general, the H-X bond strength is the most important factor determining

acid strength among the binary acids (those containing hydrogen and just one other element) in which X is in the same group in the periodic table. The strength of an H-X bond tends to decrease as the element X increases in size. As a re­ sult, the bond strength decreases and the acidity increases down a group. Thus, HCl is a stronger acid than HF, and H2S is a stronger acid than H20.

699

700

CHAPTER 16

Acid-Base Equilibria GROUP 4A CH4 Period 2 No acid or base properties Period 3

SiH4 No acid or base properties

SA

6A

NH3

HzO

HF

Weak base

---

Weak acid

7A

PH3

H2S

HCl

Weak base

Weak acid

Strong acid

Increasing acid s Increasing base streng!h

.& Figure 1 6. 1 2 Trends In acid-base properties of binary hydrides. The acidity of the

binary compounds of hydrogen and nonmetals increases moving left to right across a period and moving top to bottom down a group.

Bond strengths change less moving across a row in the periodic table than they do down a group. As a result, bond polarity is the major factor determin­ ing acidity for binary acids in the same row. Thus, acidity increases as the elec­ tronegativity of the element X increases, as it generally does moving from left to right in a row. For example, the acidity of the second-row elements varies in the following order: CH4 < NH3 Ka2· (d) How does the osmotic pressure of a 0.050 M solution of HCl compare qualitatively with that of a 0.050 M solution of H�03? Explain. SOLUTION We will use what we have learned about molecular structure and its impact on acidic behavior to answer part (a). We will then use stoichiometry and the relationship be­ tween pH and [H+] to answer parts (b) and (c). Finally, we will consider acid strength in order to compare the colligative properties of the two solutions in part (d).

ooo

(a) Acids have polar H- X bonds. From Figure 8.6 we see that the electronegativity

of H is 2.1 and that of P is also 2.1. Because the two elements have the same elec­ tronegativity, the H- P bond is nonpolar. (Section 8.4) Thus, this H cannot be acidic. The other two H atoms, however, are bonded to 0, which has an electronega­ tivity of 3.5. The H -0 bonds are therefore polar, with H having a partial positive charge. These two H atoms are consequently acidic. (b) The chemical equation for the neutralization reaction is

ooo

H�03(aq) + 2 NaOH(aq)

--->

Na2HP03(aq) + 2 H20(!)

From the definition of molarity, M = moljL, we see that moles = M X L. (Section 4.5) Thus, the number of moles of NaOH added to the solution is 3 (0.0233 L)(0.102 mol/L) = 2.38 X 10- mol NaOH. The balanced equation indicates that 2 mol of NaOH is consumed for each mole of H3P03. Thus, the number of moles of H3P03 in the sample is

(

1 mol H3P03 3 (2.38 x 10- mol NaOH) 2 mol NaOH

)

=

1.19

x

3 10- mol H3P03

3 The concentration of the H3P03 solution, therefore, equals (1.19 x 10- mol)/ (0.0250 L) = 0.0476 M. (c) From the pH of the solution, 1.59, we can calculate [H+] at equilibrium. [H+]

=

antilog(-1.59)

=

5 10- 1 . 9

=

0.026 M (two significant figures)

Because Ka1 >> Ka:v the vast majority of the ions in solution are from the first ioniza­ tion step of the acid. H3P03(aq)

:;===='

H+(aq) + H2P03-(aq)

Because one H2P03- ion forms for each H+ ion formed, the equilibrium concentrations of H+ and H 2P03- are equal: [H+] = [H 2P03 -] = 0.026 M. The equilibrium con­ centration of H3P03 equals the initial concentration minus the amount that ionizes to form H+ and H2P03- : [H3P03] = 0.0476 M - 0.026 M = 0.022 M (two signifi­ cant figures). These results can be tabulated as follows: + Initial

0.0476 M

0

0

Change

-0.026 M

+0.026 M

+0.026 M

0.022 M

0.026 M

0.026 M

Equilibrium

Summary and Key Terms

707

The percent ionization is Percent ionization

[H+l equHibdum

0.026 M X 100% = X 100% = 55% = [H PO ] 0.0476 M J Jn ; ;tiol

The first acid-dissociation constant is a K l =

-

[H'][H2P03 ] [H3P03]

=

(0.026)(0.026) 0.022

=

0.03 l

(d) Osmotic pressure is a colligative property and depends on the total concentration of particles solution. 000 (Section 13.5) Because HCl is a strong acid, a 0.050 M solution will contain 0.050 M H+(aq) and 0.050 M Cl-(aq), or a total of 0.100 mol/L of particles. Because H3P03 is a weak acid, it ionizes to a lesser extent than HCI, and, hence, there are fewer particles the H3P03 solution. As a result, the H3P03 solution will have the lower osmotic pressure.

in

in

CHAPTER REVIEW

SUMMARY AN D KEY TERM S

Section 16.1 Acids and bases were first recognized by the properties of their aqueous solutions. For example, acids tum litmus red, whereas bases turn litmus blue. Ar­ rhenius recognized that the properties of acidic solutions are due to H+(aq) ions and those of basic solutions are due to OH-(aq) ions. Section 16.2 The Brensted-Lowry concept of acids ar1d bases is more general than the Arrhenius concept ar1d em­ phasizes the transfer of a proton (H+) from ar1 acid to a base. The H+ ion, which is merely a proton with no sur­ rounding valence electrons, is strongly bound to water. For this reason, the hydronium ion, H30+(aq), is often used to represent the predominar1t form of H+ in water in­ stead of the simpler H+(aq). A Brensted-Lowry acid is a substance that donates a proton to ar1other substance; a Bronsted-Lowry base is a substar1ce that accepts a proton from another substance. Water is ar1 amphiprotic substar1ce, one that can function as either a Brensted-Lowry acid or base, depending on the substance with which it reacts. The conjugate base of a Brensted-Lowry acid is the species that remains when a proton is removed from the acid. The conjugate acid of a Brensted-Lowry base is the species formed by adding a proton to the base. Together, ar1 acid and its conjugate base (or a base and its conjugate acid) are called a conjugate acid-base pair. The acid-base strengths of conjugate acid-base pairs are related: The stronger an acid, the weaker is its conju­ gate base; the weaker an acid, the stronger is its conjugate base. In every acid-base reaction, the position of the equi­ librium favors the transfer of the proton from the stronger acid to the stronger base. Section 1 6.3 Water ionizes to a slight degree, forming H+(aq) and OH-(aq). The extent of this autoionization is expressed by the ion-product constant for water:

Kw = [H+][oH-) = 1.0 X 10-1 4 (25 °C). This relationship describes both pure water and aqueous solutions. The Kw expression indicates that the product of [H+) and [OH-) is a constant. Thus, as [H+) increases, [OH-) decreases. Acidic solutions are those that contain more H+(aq) tharJ OH-(aq): basic solutions contain more OH-(aq) than H+(aq).

Section 1 6.4 The concentration of H+(aq) can be ex­ pressed in terms of pH: pH = -log[H+]. At 25 °C the pH of a neutral solution is 7.00, whereas the pH of an acidic solution is below 7.00, and the pH of a basic solution is above 7.00. The pX notation is also used to represent the negative log of other small quar1tities, as in pOH and pKw· The pH of a solution can be measured using a pH meter, or it can be estimated using acid-base indicators.

Section 16.5 Strong acids are strong electrolytes, ioniz­ ing completely in aqueous solution. The common strong acids are HCI, HBr, HI, HN03, HCl03, HC104, and H2S04. The conjugate bases of strong acids have negligi­ ble basicity. Common strong bases are the ionic hydroxides of alkali metals and the heavy alkaline earth metals. The cations of these metals have negligible acidity. Section 1 6.6 Weak acids are weak electrolytes; only some of the molecules exist in solution in ionized form. The extent of ionization is expressed by the acid-dissociation constant, K., which is the equilibrium constar1t for the reaction HA(aq) � H +(aq) + A-(aq), which car1 also be written HA(aq) + H20(1) � Hp+(aq) + A-(aq). The larger the value of K., the stronger is the acid. For solutions of the same concentration, a stronger acid also has a larger percent ionization. The concentration of a weak acid ar1d its Ka value can be used to calculate the pH of a solution. Polyprotic acids, such as H2S03, have more than one ionizable proton. These acids have acid-dissociation constants that decrease in magnitude in the order

708

CHAPTER 1 6

Acid-Base Equilibria Section 1 6. 1 0 The tendency of a substance to show

Ka 1 > Ka2 > Ka3· Because nearly all the H+(aq) in a polyprotic acid solution comes from the first dissociation step, the pH can usually be estimated satisfactorily by considering only Kal·

acidic or basic characteristics in water can be correlated with its chemical structure. Acid character requires the presence of a highly polar H- X bond. Acidity is also fa­ vored when the H - X bond is weak and when the x- ion is very stable. For oxyacids with the same number of OH groups and the same number of 0 atoms, acid strength increases with increasing electronegativity of the central atom. For oxyacids with the same central atom, acid strength in­ creases as the number of oxygen atoms attached to the central atom increases. The structures of carboxylic acids, which are organic acids containing the COOH group, also help us to understand their acidity.

Sections 1 6.7 and 1 6.8 Weak bases include NH3, amines, and the anions of weak acids. The extent to which

a weak base reacts with water to generate the correspond­ ing conjugate acid and OH - is measured by the base­ dissociation constant, Kb. This is the equilibrium constant for the reaction B(aq) + H20(1) � HB+(aq) + OH-(aq), where B is the base. The relationship between the strength of an acid and the strength of its conjugate base is expressed quan­ titatively by the equation Ka X Kb = Kw, where Ka and Kb are dissociation constants for conjugate acid-base pairs.

Section 1 6. 1 1 The Lewis concept of acids and bases emphasizes the shared electron pair rather than the pro­ ton. A Lewis acid is an electron-pair acceptor, and a Lewis base is an electron-pair donor. The Lewis concept is more general than the Bnmsted-Lowry concept because it can apply to cases in which the acid is some substance other than H+ The Lewis concept helps to explain why many hydrated metal cations form acidic aqueous solutions. The acidity of these cations generally increases as their charge increases and as the size of the metal ion decreases.

Section 1 6.9 The acid-base properties of salts can be as­ cribed to the behavior of their respective cations and an­ ions. The reaction of ions with water, with a resultant change in pH, is called hydrolysis. The cations of the alka­ li metals and the alkaline earth metals and the anions of strong acids do not undergo hydrolysis. They are always spectator ions in acid-base chemistry.

KEY S KILLS •

•

•

Understand the nature of the hydrated proton, represented as either H+(aq) or H30+(aq). Define and identify Arrhenius acids and bases. Define and identify Bronsted-Lowry acids and bases, and identify conjugate acid-base pairs.

• Relate the strength of an acid to the strength of its conjugate base. • Understand how the equilibrium position of a proton transfer reaction relates the strengths of the acids and bases involved. Describe the autoionization of water and understand how [H30+] and [OH -] are related. •

•

Calculate the pH of a solution given [H30+] or [OH -].

•

Calculate the pH of a strong acid or strong base given its concentration.

•

Calculate the pH of a weak acid or weak base or its percent ionization given its concentration and Ka or Kb.

•

Predict whether an aqueous solution of a salt will be acidic, basic, or neutral.

•

•

Calculate Ka or Kb for a weak acid or weak base given its concentration and the pH of the solution. Calculate Kb for a weak base given Ka of its conjugate acid, and silnilarly calculate Ka from Kb.

•

Predict the relative strength of a series of acids from their molecular structures.

•

Define and identify Lewis acids and bases.

KEY EQUATIONS • Kw = [H30+][0H-] = [H+][OH-] =

•

pH = -log[H+]

• pOH = -log[OW]

[16.17] [16.18]

1.0 X 1 0-14

[16.16]

The ion product of water at 25 °C Definition of pH Definition of pOH

Visualizing Concepts •

pH

+ pOH

=

14.00

709

Relationship between pH and pOH

[16.20]

The acid dissociation constant for a weak acid, HA Percent ionization of a weak acid

[16.27]

=

[BH+][OW] [B]

•

Kb

•

K, X Kb

=

Kw

[16.34]

The base dissociation constant for a weak base, B

[16.40]

The relationship between the acid and base disso­ ciation constants of a conjugate acid-base pair

VISUAL IZING CONCEPTS 16.1 (a) Identify the Brensted-Lowry acid and the Brensted­ Lowry base in the following reaction:

=H

=N

(b) Which acid would have the smallest acid-dissociation constant, K,? (c) Which solution would have the highest pH? [Sections 16.5 and 16.6]

=X

(b) Identify the Lewis acid and the Lewis base in the re­ action. [Sections 16.2 and 16.11] 16.2 The following diagrams represent aqueous solutions of two monoprotic acids, HA (A = X or Y). The water molecules have been omitted for clarity. (a) Which is the stronger acid, HX or HY? (b) Which is the stronger base, x- or y-? (c) If you mix equal concentrations of HX and NaY, will the equilibrium HX(aq) + y-(aq) = HY(aq) + x-(aq) lie mostly to the right (Kc > 1) or to the left (Kc < 1)? [Section 16.2]

16.4

In which of the following cases is the approximation that the equilibrium concentration of H+(aq) is small relative to the initial concentration of HA likely to be most valid: (a) initial [HA] = 0.100 M and K, = 1.0 X 10... ;;, (b) ini­ tial [HA] = 0.100 M and K, = 1.0 X 10-4, (c) initial 3 [HA] = 0.100 M and K, = 1.0 x 10- ? [Section 16.6]

16.5 (a) Which of these three lines represents the effect of concentration on the percent ionization of a weak acid? (b) Explain in qualitative terms why the curve you choose has the shape it does. [Section 16.6]

= HA A

c 16.3 The following diagrams represent aqueous solutions of three acids, HX, HY, and HZ. The water molecules have been omitted for clarity, and the hydrated proton is rep­ resented as a simple sphere rather than as a hydronium ion. (a) Which of the acids is a strong acid? Explain.

0 �------� 0 Acid concentration

710

CHAPTER

16

Acid-Base Equilibria

16.6 Refer to the diagrams accompanying Exercise 16.3. (a) Rank the anions, x-, Y -, and z-, in order of increasing basicity. (b) Which of the ions would have the largest base-dissociation constant, Kb? [Sections 16.2 and 16.8]

16.10 In this model of acetylsalicylic acid (aspirin), identify the carboxyl group in the molecule. [Section 16.10]

16.7 (a) Draw the Lewis structure for the following molecule and explain why it is able to act as a base. (b) To what class of organic compounds does this substance belong? (See the color key in Exercise 16.1.) [Section 16.7]

•

• 16.8 The following diagram represents an aqueous solution formed by dissolving a sodium salt of a weak acid in water. The diagram shows only the Na + ions, the x- ions, and the HX molecules. What ion is missing from the diagram? If the drawing is completed by drawing all the ions, how many of the missing ions should be shown? [Section 16.9]

•• • 0 • 0

• I

• I /.

•

"

•

16.11 Rank the following acids in order of increasing acidity: CH3COOH, CH2CICOOH, CHC12COOH, CC13COOH, CF3COOH. [Section 16.10] 16.12 (a) The following diagram represents the reaction of PC14+ with o-. Draw the Lewis structures for the reac­ tants and products, and identify the Lewis acid and the Lewis base in the reaction.

0

+

16.9 (a) What kinds of acids are represented by the following molecular models? (b) Indicate how the acidity of each molecule is affected by increasing the electronegativity of the atom X, and explain the origin of the effect. [Section 16.10]

(a)

l

•

(b) The following reaction represents the acidity of a hy­ drated cation. How does the equilibrium constant for the reaction change as the charge of the cation increases? [Section 16.11]

+

(b)

+

..�·

EXERC I S ES Arrhenius and Br0nsted-Lowry Acids and Bases 16.13 Although HCl and HzS04 have very different properties as pure substances, their aqueous solutions possess many common properties. List some general properties of these solutions, and explain their common behavior in terms of the species present. 16.14 Although pure NaOH and NH3 have very different properties, their aqueous solutions possess many com­ mon properties. List some general properties of these solutions, and explain their common behavior in terms of the species present. 16.15 (a) What is the difference between the Arrhenius and the Bmnsted-Lowry definitions of an acid? (b) NH3(g) and HCI(g) react to form the ionic solid NH4Cl(s) (Figure 16.3). Which substance is the Bmnsted-Lowry acid in this reaction? Which is the Bnmsted-Lowry base?

16.16 (a) What is the difference between the Arrhenius and the Bmnsted-Lowry definitions of a base? (b) When ammo­ nia is dissolved in water, it behaves both as an Arrhenius base and as a Bnmsted-Lowry base. Explain. 16.17 (a) Give the conjugate base of the following Bmnsted­ Lowry acids: (i) HI03, (ii) NH4+. (b) Give the conjugate acid of the following Bmnsted-Lowry bases: (i) 02-, (ii) H 2P04-. 16.18 (a) Give the conjugate base of the following Bmnsted­ Lowry acids: (i) C6H5COOH. (ii) HPO l-. (b) Give the conjugate acid of the following Bmnsted-Lowry bases: (i) col-, (ii) CzH5NHz. 16.19 Designate the Bremsted-Lowry acid and the Bremsted­ Lowry base on the left side of each of the following

Exercises equations, and also designate the conjugate acid and conjugate base on the right side: (a) NH4+(aq) + CN -(aq) � HCN(aq) + NH3(aq) (b) (CH3)JN(aq) + H 20(/) � (CH3)JNH+(aq) + OH-(aq)

3

(c) HCH0 2(aq) + P04 -(aq) �

CH02-(aq) + HPO/-(aq)

16.20 Designate the Bmnsted-Lowry acid and the Bmnsted­ Lowry base on the left side of each equation, and also designate the conjugate acid and conjugate base on the right side. (a) HBrO(aq) + H 20(/) � H30+(aq) + BrO-(aq) (b) HS04-(aq) + HC03 -(aq) �

SO/-(aq) + H 2C03(aq)

(c) HS03-(aq) + H30+(aq) � H 2S03(aq) + H 20(/)

16.21 (a) The hydrogen oxalate ion (HC 204l is amphiprotic. Write a balanced chemical equation showing how it acts as an acid toward water and another equation showing how it acts as a base toward water. (b) What is the conju­ gate acid of HC 204-? What is its conjugate base? 16.22 (a) Write an equation for the reaction in which H 2C6H;>0 5-(aq) acts as a base in H20(/). (b) Write an equation for the reaction in which H2C,;H;>0 -(aq) acts 5 as an acid in H20(/). (c) What is the conjugate acid of H 2C6H;>Os-? What is its conjugate base? 16.23 Label each of the following as being a strong base, a weak base, or a species with negligible basicity. In each case write the formula of its conjugate acid, and indicate whether the conjugate acid is a strong acid, a weak acid,

or a species with negligible acidity: (a) (b) HC03 -, (c) 02-, (d) CC (e) NH3.

711

CH3Coo-,

16.24 Label each of the following as being a strong acid, a weak acid, or a species with negligible acidity. In each case write the formula of its conjugate base, and indicate whether the conjugate base is a strong base, a weak base, or a species with negligible basicity: (a) HN02, (b) H2S04, (c) HPO/-, (d) CH4, (e) CH 3NH 3+ (an ion related to NH4 +). 16.25 (a) Which of the following is the stronger Bmnsted­ Lowry acid, HErO or HBr? (b) Which is the stronger Bnm­ sted-Lowry base, F- or Cr""'? Briefly explain your choices. 16.26 (a) Which of the following is the stronger Bmnsted­ Lowry acid, HN03 or HN02? (b) Which is the stronger Bmnsted-Lowry base, NH3 or H20? Briefly explain your choices. 16.27 Predict the products of the following acid-base reac­ tions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) 02-(aq) + H 20(/) � (b) CH3COOH(aq) + HS-(aq) � (c) N02-(aq) + H 20(/) � 16.28 Predict the products of the following acid-base reac­ tions, and predict whether the equilibrium lies to the left or to the right of the equation: (a) NH 4 +(aq) + OH-(aq) � (b) CH3coo-(aq) + H30+(aq) � (c) HC03-(aq) + F-(aq) �

Autoionization of Water 16.29 (a) What does the term autoioniwtion mean? (b) Explain why pure water is a poor conductor of electricity. (c) You are told that an aqueous solution is acidic. What does this statement mean?

16.32 Calculate [OH-] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) [H+] = 0.0045 M; (b) [H+] = 1.5 X 10-9 M; (c) a solution in which [H+] is 10 times greater than [OH].

16.30 (a) Write a chemical equation that illustrates the au­ toionization of water. (b) Write the expression for the ion-product constant for water, Kw. Why is [H 20] absent from this expression? (c) A solution is described as basic. What does this statement mean?

16.33 At the freezing point of water (0 °C), Kw = 1.2 X 10-IS Calculate [H1 and [OH-] for a neutral solution at this temperature.

16.31 Calculate [H+] for each of the following solutions, and indicate whether the solution is acidic, basic, or neutral: (a) [OH-] = 0.00045 M; (b) [OH] = 8.8 X 10-9 M; (c) a solution in which [OH-] is 100 times greater than [H+].

16.34 Deuterium oxide (D20, where D is deuterium, the hydrogen-2 isotope) has an ion-product constant, Kw, of 8.9 x w-16 at 20 oc. Calculate [D+] and [OD-] for pure (neutral) D20 at this temperature.

The pH Scale 16.35 By what factor does [H1 change for a pH change of (a) 2.00 units, (b) 0.50 units? 16.36 Consider two solutions, solution A and solution B. [H+] in solution A is 500 times greater than that in solution B. What is the difference in the pH values of the two solutions?

16.37 (a) If NaOH is added to water, how does [H+] change? How does pH change? (b) Use the pH values in Figure 16.5 to estimate the pH of a solution with [H+] = 0.0006 M. Is the solution acidic or basic? (c) If the pH of a solution is 5.2, first estimate and then cal­ culate the molar concentrations of H+(aq) and OH-(aq) in the solution.

712

CHAPTER

16

Acid-Base Equilibria

16.38 (a) If HN03 is added to water, how does [OH-] change?

How does pH change? (b) Use the pH values in Figure 16.5 to estimate the pH of a solution with [OH-] = 0.014 M. Is the solution acidic or basic? (c) If pH = 6.6, first estimate and then calculate the molar concentrations of H+(aq) and OH-(aq) in the solution.

16.39 Complete the following table by calculating the missing

entries and indicating whether the solution is acidic or basic.

+

[H ]

16.40 Complete the following table by calculating the missing

entries. In each case indicate whether the solution is acidic or basic.

pH

pOH

[H + ]

11 .25 6.02

4.4 X 10-4 M

Acidic or OH-(aq)

pH

pOH

basic?

3 7.5 X 10- M 3.6 X 10-to M

Acidic or basic?

[OW]

3 8.5 X 10- M

16.41 The average pH of normal arterial blood is 7.40. At nor­ mal body temperature (37 °C), Kw = 2. 4 X 10-14 Calcu­

late [H+], [OH-], and pOH for blood at this temperature.

8.25 5.70

16.42 Carbon dioxide in the atmosphere dissolves in raindrops

to produce carbonic acid (H2C03), causing the pH of clean, unpolluted rain to range from about 5.2 to 5.6. What are the ranges of [H+] and [OH-] in the raindrops?

Stron g Acids and Bases 16.43 (a) What is a strong acid? (b) A solution is labeled

0.500 M HCI. What is [H+] for the solution? (c) Which of the following are strong acids: HF, HCI, HBr, HI?

16.44 (a) What is a strong base? (b) A solution is labeled

0.035 M Sr(OH)2. What is [OH-] for the solution? (c) Is the following statement true or false? Because Mg(OHh is not very soluble, it cannot be a strong base. Explain.

16.45 Calculate the pH of each of the following strong acid

3 solutions: (a) 8.5 X 10- M HBr, (b) 1.52 g of HN03 in 575 mL of solution, (c) 5.00 mL of 0.250 M HC104 diluted to 50.0 mL, (d) a solution formed by mixing 10.0 mL of 0.100 M HBr with 20.0 mL of 0.200 M HCI.

16.46 Calculate the pH of each of the following strong acid

solutions: (a) 0.00135 M HN03, (b) 0.425 g of HC104 in 2.00 L of solution, (c) 5.00 mL of 1.00 M HCl diluted to 0.500 L, (d) a mixture formed by adding 50.0 mL of 0.020 M HCl to 150 mL of 0.010 M HI.

16.47 Calculate [OH-] and pH for

3

(a) 1.5 x 10- M Sr(OHh, (b) 2.250 g of LiOH in 250.0 mL of solution, (c) 1.00 mL

of 0.175 M NaOH diluted to 2.00 L, (d) a solution formed by adding 5.00 mL of 0.105 M KOH to 15.0 mL of 2 9.5 X 10- M Ca(OH) 2. 16.48 Calculate [OH-] and pH for each of the following strong

base solutions: (a) 0.082 M KOH, (b) 1.065 g of KOH in 500.0 mL of solution, (c) 10.0 mL of 0.0105 M Ca(OHh diluted to 500.0 mL, (d) a solution formed by mixing 10.0 mL of O.D15 M Ba(OHh with 40.0 mL of 3 7.5 X 10- M NaOH.

16.49 Calculate the concentration of an aqueous solution of

NaOH that has a pH of 11 .50.

16.50 Calculate the concentration of an aqueous solution of

Ca(OHh that has a pH of 12.05.

Weak Acids 16.51 Write the chemical equation and the Ka expression for

the ionization of each of the following acids in aqueous solution. First show the reaction with H+(aq) as a prod­ uct and then with the hydronium ion: (a) HBr02, (b) C2H5COOH.

16.52 Write the chemical equation and the

Ka expression

for the acid dissociation of each of the following acids in aqueous solution. First show the reaction with H+(aq) as a product and then with the hydronium ion: (a) C6H5COOH, (b) HC03-.

16.53 Lactic acid (CH3CH(OH)COOH) has one acidic hydro­

gen. A 0.10 M solution of lactic acid has a pH of 2.44. Calculate Ka.

16.54 Phenylacetic acid (C6H5CH2COOH) is one of the sub­

stances that accumulates in the blood of people with phenylketonuria, an inherited disorder that can cause mental retardation or even death. A 0.085 M solution of C6H5CH2COOH has a pH of 2.68. Calculate the Ka value for this acid.

713

Exercises 16.55 A 0.100 M solution of chloroacetic acid (CICH2COOH) is

11.0%

ionized. Using this information, calculate [ClCH2COOl, [H+], [CICH2COOH)], and K, for chloroacetic acid.

0.100 M solution of bromoacetic acid (BrCH2COOH) is 13.2% ionized. Calculate [H+], [BrCH2COOl, and [BrCH2COOH].

16.56 A

16.57 A particular sample of vinegar has a pH of

2.90.

If acetic acid is the only acid that vinegar contains 5 (K, = 1.8 X 10- ), calculate the concentration of acetic acid in the vinegar.

6.8 X 10-4) must be pre­ 0.200 L to form a solution with a pH of 3.25?

16.58 How many moles of HF (K, = sent in

16.59 The

acid-dissociation constant for benzoic acid s (C6HsCOOH) is 6.3 x 10- Calculate the equilibrium concentrations of H30+, C6H5Coo-, and C6HsCOOH in the solution if the initial concentration of C6H5COOH is 0.050 M.

16.60 The acid-dissociation constant for hypochlorous acid

(HClO) is 3.0 x 10-s. Calculate the concentrations of H3o+, CIO-, and HCIO at equilibrium if the initial con­ centration of HCIO is 0.0090 M.

16.61 Calculate the pH of each of the following solutions (K, and Kb values are given in Appendix 0): (a) 0.095 M propionic acid (C2H5COOH), (b) 0.100 M hydrogen chromate ion (HCr04 -), (c) 0.120 M pyridine (CsHsN). 16.62 Determine the pH of each of the following solutions (K, and Kb values are given in Appendix 0): (a) 0.095 M hypochlorous acid, (b) 0.0085 M phenol, (c) 0 . 095 M hydroxylamine.

16.63 Saccharin, a sugar substitute, is a weak acid with pK, = 2.32 at follows:

25 oc.

HNC7H4S03(aq) What is the pH of a

16.64 The active ingredient in aspirin is acetylsalicylic acid

(HC9H�4), a monoprotic acid with K, = 3.3 x 10-4 at 25 oc. What is the pH of a solution obtained by dissolving two extra-strength aspirin tablets, containing 500 mg of acetylsalicylic acid each, in 250 mL of water?

16.65 Calculate the percent ionization of hydrazoic acid (HN3)

in solutions of each of the following concentrations (K, is given in Appendix 0): (a) 0.400 M, (b) 0.100 M,

(c) 0.0400 M.

16.66 Calculate the percent ionization of propionic acid

(C2H5COOH) in solutions of each of the following con­ centrations (K, is given in Appendix 0): (a) 0.250 M,

(b) 0.0800 M, (c) 0.0200 M.

[16.67] Show that for a weak acid, the percent ionization should

vary as the inverse square root of the acid concentration.

[16.68] For solutions of a weak acid, a graph of pH versus the

log of the initial acid concentration should be a straight line. What is the magnitude of the slope of that line?

[16.69] Citric acid, which is present in citrus fruits, is a triprotic

acid (Table 16.3). Calculate the pH and the citrate ion 3 (C6H507 -) concentration for a 0.050 M solution of citric acid. Explain any approximations or assumptions that you make in your calculations.

[16.70] Tartaric acid is found in many fruits, including grapes,

and is partially responsible for the dry texture of certain wines. Calculate the pH and the tartarate ion 2 (C4H406 -) concentration for a 0.250 M solution of tartaric acid, for which the acid-dissociation constants are listed in Table 16.3. Explain any approximations or assumptions that you make in your calculation.

It ionizes in aqueous solution as

�

H+(aq) + NC7H4S03 -(aq)

0.10 M

solution of this substance?

Weak Bases 16.71 What is the essential structural feature of all Brensted­ Lowry bases?

16.72 What are two kinds of molecules or ions that commonly function as weak bases?

0.550 M solution of hypobromite ion (BrO-; Kb = 4.0 x 10-6). What is the pH of this solution?

16.77 Ephedrine, a central nervous system stimulant, is used

16.73 Write the chemical equation and the Kb expression for

the ionization of each of the following bases in aqueous solution: (a) dimethylamine, (CH3hNH; (b) carbonate 2 ion, C03 -; (c) formate ion, CH02 -.

16.74 Write the chemical equation and the Kb expression for

the reaction of each of the following bases with water: (a) propylamine, C3H7NH2; (b) monohydrogen phos­ phate ion, HPO/-; (c) benzoate ion, C6H5C02 -.

16.75 Calculate the molar concentration of OH- ions in a

0.075 M solution of ethylamine (C2HsNH� 6.4 x 10-4) . Calculate the pH of this solution.

16.76 Calculate the molar concentration of OH - ions in a

Kb =

in nasal sprays as a decongestant. This compound is a weak organic base:

C10H1sON(aq) + H20(/) A

0.035 M

� C10H1sONH+(aq) +

OH-(aq)

solution of ephedrine has a pH of

11.33.

What are the e�ilibrium concentrations of C1oH 1 50N, CwH 1 sONH , and OH-? (b) Calculate Kb for ephedrine.

(a)

16.78 Codeine (C18H21N03) is a weak organic base. A

5.0 X 10-3M solution of codeine has a pH of 9.95. Calcu­ late the value of Kb for this substance. What is the pKb

for this base?

714

CHAPTER 1 6

Acid-Base Equilibria

The Ka-Kb Relationship ; Acid-Base Properties of Salts 16.79 Although the acid-dissociation constant for phenol

(C6H50H) is listed in Appendix D, the base-dissociation constant for the phenolate ion (C�50-) is not. (a) Ex­ plain why it is not necessary to list both Ka for phenol and Kb for the phenolate ion. (b) Calculate Kb for the phenolate ion. (c) Is the phenolate ion a weaker or stronger base than ammortia? 16.80 We can calculate Kb for the carbonate ion if we know the Ka values of carbonic acid (H2C03). (a) Is Ka 1 or Ka2 of carbonic acid used to calculate Kb for the carbonate ion? Explain. (b) Calculate Kb for the carbonate ion. (c) Is the carbonate ion a weaker or stronger base than ammortia? 16.81 (a) Given that Ka for acetic acid is 1.8 x

10-s and that for

hypochlorous acid is 3.0 X 10-6, which is the stronger acid? (b) Which is the stronger base, the acetate ion or the hypochlorite ion? (c) Calculate Kb values for CH3COO- and CIO-. 5 16.82 (a) Given that Kb for ammortia is 1.8 X 10- and that for hydroxylamine is 1 . 1 x 10-8, which is the stronger base? (b) Which is the stronger acid, the ammortium ion or the hydroxylammonium ion? (c) Calculate Ka values for NH4+ and H3NOH+ 16.83 Using data from Appendix D, calculate [OH-] and pH

for each of the following solutions: (a) 0.10 M NaCN, (b) 0.080 M Na2C03, (c) a mixture that is 0.10 M in NaN02 and 0.20 M in Ca(N02)z .

16.85 Predict whether aqueous solutions of the following

compounds are acidic, basic, or neutral: (a) NH4Br, (b) FeCI3, (c) Na2C03, (d) KCI04, (e) NaHCz04.

16.86 Predict whether aqueous solutions of the following sub­

stances are acidic, basic, or neutral: (a) CrBr3, (b) Lii, (c) K3P04, (d) [CH3NH3]Cl, (e) KHS04.

16.87 An unknown salt is either NaF, NaCl, or NaOCI. When

0.050 mol of the salt is dissolved in water to form 0.500 L of solution, the pH of the solution is 8.08. What is the identity of the salt?

16.88 An unknown salt is either KBr, NH4Cl, KCN, or K2C03.

If a 0.100 M solution of the salt is neutral, what is the identity of the salt?

16.89 Sorbic acid (C5H,.COOH) is a weak monoprotic acid

with Ka = 1. 7 X 10-5. Its salt (potassium sorbate) is added to cheese to inhibit the formation of mold. What is the pH of a solution contairting 11.25 g of potassium sorbate in 1 .75 L of solution?

16.90 Trisodium phosphate (Na3P04) is available in hardware

stores as TSP and is used as a cleaning agent. The label on a box of TSP warns that the substance is very basic (caustic or alkaline). What is the pH of a solution con­ taining 35.0 g of TSP in a liter of solution?

16.84 Using data from Appendix D, calculate [OH-] and pH

for each of the following solutions: (a) 0.105 M NaF, (b) 0.035 M Na2S, (c) a mixture that is 0.045 M in CH3C00Na and 0.055 M in (CH3COO)zBa.

Acid-Base Character and Chemical Structure 16.91 How does the acid strength of an oxyacid depend on

(a) the electronegativity of the central atom; (b) the num­

ber of nonprotonated oxygen atoms in the molecule? 16.92

(a) How does the strength of an acid vary with the polar­ ity and strength of the H- X bond? (b) How does the acidity of the binary acid of an element vary as a function of the electronegativity of the element? How does this re­ late to the position of the element in the periodic table? (a) HN03 is a stronger acid than HNOz; (b) H2S is a stronger acid than H20; (c) H2S04 is a stronger acid than HS04-; (d) H2S04 is a stronger acid than H2Se04; (e) CC13COOH is a stronger acid than CH3COOH.

16.95 Based on their compositions and structures and on

conjugate acid-base relationships, select the stronger base in each of the following pairs: (a) Bro- or CIO-, (b) Bro­ 2 or BrOz-, (c) HP04 - or HzP04-.

16.96 Based on their compositions and structures and on con­

jugate acid-base relationships, select the stronger base in each of the following pairs: (a) N03- or N02-, (b) P04 3- or As043-, (c) HC03- or CO l-.

16.93 Explain the following observations:

16.94 Explain the following observations: (a) HCI is a stronger

acid than H2S; (b) H�04 is a stronger acid than H�sO� (c) HBr03 is a stronger acid than HBrOz; (d) H2C204 is a stronger acid than HC204-; (e) benzoic acid (C�5COOH) is a stronger acid than phenol (�50H).

16.97 lndicate whether each of the following statements is

true or false. For each statement that is false, correct the statement to make it true. (a) 1n general, the acidity of binary acids increases from left to right in a given row of the periodic table. (b) 1n a series of acids that have the same central atom, acid strength increases with the number of hydrogen atoms bonded to the central atom. (c) Hydrotelluric acid (H2Te) is a stronger acid than H2S because Te is more electronegative than S.

Additional Exercises 16.98 Indicate whether each of the following statements is true or false. For each statement that is false, correct the statement to make it true. (a) Acid strength in a series of H- X molecules increases with increasing size of X. (b) For acids of the same general structure

715

but differing electronegativities of the central atoms, acid strength decreases with increasing electroneg­ ativity of the central atom. (c) The strongest acid known is HF because fluorine is the most electronega­ tive element.

Lewis Acids and Bases 16.99 If a substance is an Arrhenius base, is it necessarily a Bmnsted-Lowry base? Is it necessarily a Lewis base? Explain. 16.100 If a substance is a Lewis acid, is it necessarily a Bmnsted-Lowry acid? Is it necessarily an Arrhenius acid? Explain. 16.101 Identify the Lewis acid and Lewis base among the reactants in each of the following reactions: (a) Fe(Cl04)3(s) + 6 HzO(I) = 3 Fe(Hz0)6 +(aq) + 3 Cl04-(aq) (b) CN-(aq) + H20(1) = HCN(aq) + OW(aq) (c) (CH3)3N(g) + BF3(g) = (CH3)3NBF3(s) (d) HIO(Iq) + NH2-(lq) = NH 3(1q) + IO-(Iq) (lq denotes liquid ammonia as solvent)

16.102 Identify the Lewis acid and Lewis base in each of the fol­ lowing reactions: (a) HNOz(aq) + OH-(aq) = NOz-(aq) + HzO(I) (b) FeBr3(s) + Br-(aq) = FeBr4-(aq) 2 2 (c) Zn +(aq) + 4 NH 3(aq) = Zn(NH 3)4 +(aq) (d) S02(g) + H20(1) = H2S0 3(aq) 16.103 Predict which member of each pair produces the more 2 2 acidic aqueous solution: (a) K+ or Cu +, (b) Fe + or 3 3 3 Fe +, (c) Al + or Ga +. Explain. 16.104 Which member of each pair produces the more acidic aqueous solution: (a) ZnBr2 or CdCI2, (b) CuCI or Cu(N03)z, (c) Ca(N03)z or NiBrz? Explain.

ADDITI ONAL EXERC ISES 16.105 In your own words, define or explain (a) Kw, (c) pOH, (d) pKb.

(b)

K.,

16.106 Indicate whether each of the following statements is cor­ rect or incorrect. For those that are incorrect, explain why they are wrong. (a) Every Bnmsted-Lowry acid is also a Lewis acid. (b) Every Lewis acid is also a Bnmsted-Lowry acid. (c) Conjugate acids of weak bases produce more acidic solutions than conjugate acids of strong bases. (d) K+ ion is acidic in water because it causes hydrat­ ing water molecules to become more acidic. (e) The percent ionization of a weak acid in water in­ creases as the concentration of acid decreases.

Using data from Appendix D, calculate the equilibrium constant for the reaction of citric acid with methyl­ amine, if only the first proton of the citric acid (K,1) is important in the neutralization reaction.

16.109 Hemoglobin plays a part in a series of equilibria involving protonation-deprotonation and oxygenation-deoxygena­ tion. The overall reaction is approximately as follows: HbH +(aq) + 02(aq) = Hb02(aq) + H+(aq) where Hb stands for hemoglobin, and Hb02 for oxyhe­ moglobin. (a) The concentration of 02 is higher in the lungs and lower in the tissues. What effect does high [02] have on the position of this equilibrium? (b) The normal pH of blood is 7.4. Is the blood acidic, basic, or neutral? (c) If the blood pH is lowered by the presence of large amounts of acidic metabolism products, a con­ dition known as acidosis results. What effect does low­ ering blood pH have on the ability of hemoglobin to transport Oz?

16.107 Predict whether the equilibrium lies to the right or to the left in the following reactions: (a) NH4+(aq) + Pol-(aq) = 2 NH3(aq) + HP04 -(aq) (The ammonium ion is a stronger acid than the hydrogen phosphate ion.) (b) CH 3COOH(aq) + CN-(aq) = CH 3COO-(aq) + HCN(aq) (The cyanide ion is a stronger base than the acetate ion.)

[16.110] Calculate the pH of a solution made by adding 2.50 g of lithium oxide (Li20) to enough water to make 1.500 L of solution.

16.108 The odor of fish is due primarily to amines, especially methylamine (CH3NHz). Fish is often served with a wedge of lemon, which contains citric acid. The amine and the acid react forming a product with no odor, there­ by making the less-than-fresh fish more appetizing.

16.111 Which of the following solutions has the higher pH? (a) a 0.1 M solution of a strong acid or a 0.1 M solution of a weak acid, (b) a 0.1 M solution of an acid with Ka = 2 X 10-3 or one with Ka = 8 X 10--.;, (c) a 0.1 M solution of a base with pKb = 4.5 or one with pKb = 6.5.

716

C HAPTER 1 6

Acid-Base Equilibria

[16.112] What is the pH of a solution that is 2.5

x

NaOH? Does your answer make sense?

10-9 M in

[16.119] The amino acid glycine (HzN -CHz -COOH) can participate in the following equilibria in water: H2N-CH2-COOH + H20 :;==="' HzN-CH2-coo- + H3o+

16.113 Caproic acid (C5HnCOOH) is found in small amounts

in coconut and palm oils and is used in making artificial flavors. A saturated solution of the acid contains 11 g/L and has a pH of 2.94. Calculate Ka for the acid.

16.115 Butyric acid is responsible for the foul smell of rancid

=

4.3

x

10-

3

H2N-CH2-COOH + H20 :;==="' �3N-CH2-COOH + OH- Kb = 6.0 x 10-5 (a) Use the values of Ka and Kb to estimate the equilibri­ um constant for the intramolecular proton transfer to form a zwitterion:

[16.114] A hypothetical acid H2X is both a strong acid and a

diprotic acid. (a) Calculate the pH of a 0.050 M solution of H2X, assuming that only one proton ionizes per acid mol­ ecule. (b) Calculate the pH of the solution from part (a), now assuming that both protons of each acid molecule completely ionize. (c) In an experiment it is observed that the pH of a 0.050 M solution of H2X is 1.27. Comment on the relative acid strengths of H2X and HX-. (d) Would a solution of the salt NaHX be acidic, basic, or neutral? Explain.

Ka

H2N-CH2-COOH

:;==="'

+H3N -CH2-COO­

What assumptions did you need to make? (b) What is the pH of a 0.050 M aqueous solution of glycine? (c) What would be the predominant form of glycine in a solution with pH 13? With pH 1? 16.120 The structural formula for acetic acid is shown in Table

16.2. Replacing hydrogen atoms on the carbon with chlorine atoms causes an increase in acidity, as follows:

butter. The pKa of butyric acid is 4.84. (a) Calculate the pKb for the butyrate ion. (b) Calculate the pH of a 0.050 M solution of butyric acid. (c) Calculate the pH of a 0.050 M solution of sodium butyrate.

Acid

Formula

Ka (25 •q

Acetic

CH3COOH

increasing acidity (decreasing pH): (i) NH�03, {il) NaN03, (iii) CH3COONH4, (iv) NaF, (v) CH3COONa. 2 [16.117] What are the concentrations of H+, H2P0 4-, HP04 -, 3 and P04 - in a 0.0250 M solution of H3P04?

Chloroacetic

CH2ClCOOH

1 .8 X 10-5

Dichloroacetic

CHC12COOH

Trichloroacetic

CCl3COOH

[16.118] Many moderately large organic molecules containing

Using Lewis structures as the basis of your discussion, explain the observed trend in acidities in the series. Cal­ culate the pH of a 0.010 M solution of each acid.

16.116 Arrange the following 0.10 M solutions in order of

basic nitrogen atoms are not very soluble in water as neutral molecules, but they are frequently much more soluble as their acid salts. Assuming that pH in the stomach is 2.5, indicate whether each of the following compounds would be present in the stomach as the neutral base or in the protonated form: nicotine, Kb = 7 X 10-7; caffeine, Kb = 4 X 10- 14; strychnine, Kb = 1 X 10---i;; quinine, Kb = 1.1 X 10-6

1.4 X 10-

3

2

3.3 X 10-

2 X 10-l

INTEGRATIVE EXERCI SES

(b) What volume of C02 at 25 •c and 1.0 atrn is dis­ solved in a 20.0-L bucket of today's rainwater?

16.121 Calculate the number of H+(aq) ions i n 1.0 mL o f pure

water at 25 •c. 16.122 How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 10.0 L of a solution that has a pH of 2.05? 16.123 The volume of an adult's stomach ranges from about 50 mL when empty to 1 L when full. If the stomach vol­ ume is 400 mL and its contents have a pH of 2, how many moles of H+ does the stomach contain? Assuming that all the H+ comes from HCl, how many grams of sodium hy­ drogen carbonate will totally neutralize the stomach acid? 16.124 Atmospheric C02 levels have risen by nearly 20% over the past 40 years from 315 ppm to 380 ppm. (a) Given that the average pH of clean, unpolluted rain today is 5.4, determine the pH of unpolluted rain 40 years ago. Assume that carbonic acid (H2C03) formed by the reac­ tion of C02 and water is the only factor influencing pH. C02{g) + H20(/) :;==="' H2C03(aq)

[16.125]

In many reactions the addition of AlC13 produces the same effect as the addition of H+. (a) Draw a Lewis structure for AJC13 in which no atoms carry formal charges, and determine its structure using the VSEPR method. (b) What characteristic is notable about the structure in part (a) that helps us understand the acidic character of AlC13? (c) Predict the result of the reaction between AIC13 and NH3 in a solvent that does not par­ ticipate as a reactant. (d) Which acid-base theory is most suitable for discussing the sintilarities between A1Cl3 and H+?

[16.126] What is the boiling point of a 0.10 M solution of NaHS04 if the solution has a density of 1.002 g/mL?

[16.127] Cocaine is a weak organic base whose molecular formu­

la is C17H21N04. An aqueous solution of cocaine was found to have a pH of 8.53 and an osmotic pressure of 52.7 torr at 15 •c. Calculate Kb for cocaine.

Integrative Exercises

[16.128[ The iodate ion is reduced by sulfite according to the fol­ lowing reaction: 103 -(aq)

2

+ 3 503

-(aq) ---> qaq)

+ 3 so/-(aq)

The rate of this reaction is found to be first order in 103 -, first order in SO l-, and first order in H+ (a) Write the rate law for the reaction. (b) By what factor will the rate of the reaction change if the pH is lowered from 5.00 to 3.50? Does the reaction proceed faster or slower at the lower pH? (c) By using the concepts discussed in Section 14.6, explain how the reaction can be pH-dependent even though H + does not appear in the overall reaction.

[16.129] (a) Using dissociation constants from Appendix D, de­ termine the value for the equilibrium constant for each of the following reactions. (Remember that when reac­ tions are added, the corresponding equilibrium con­ stants are multiplied.) (i) HC03-(aq) + OH-(aq) = Col-(aq) + H 20( /) (ii) NH 4+(aq) + Co l-(aq) = NH3(aq) + HC03 -(aq) (b) We usually use single arrows for reactions when the forward reaction is appreciable (K much greater than 1) or when products escape from the system, so that equi­ librium is never established. If we follow this conven­ tion, which of these equilibria might be written with a single arrow?

717

[16.130] Lactic acid, CH3CH(OH)COOH, received its name be­ cause it is present in sour milk as a product of bacterial action. It is also responsible for the soreness in muscles after vigorous exercise. (a) The pK0 of lactic acid is 3.85. Compare this with the value for propionic acid (CH3CH2COOH, pKa = 4.89), and explain the differ­ ence. (b) Calculate the lactate ion concentration in a 0.050 M solution of lactic acid. (c) When a solution of sodium lactate, CH3CH(OH)COONa, is mixed with an aqueous copper(ll) solution, it is possible to obtain a solid salt of copper(II) lactate as a blue-green hydrate, (CH3CH(OH)C00)2Cu · xH20. Elemental analysis of the solid tells us that the solid is 22.9% Cu and 26.0% C by mass. What is the value for x in the formula for the hydrate? (d) The acid-dissociation constant for the 2 Cu +(aq) ion is 1.0 x 10-8 Based on this value and the acid-dissociation constant of lactic acid, predict whether a solution of copper(II) lactate will be acidic, basic, or neutral. Explain your answer.

ADDITIONAL ASPECTS OF AQUEOUS EQUILIBRIA

CANARY SPRING, which is part of Mammoth Hot Springs in Yellowstone National Park.

718

W H AT ' S 17.1

A H EA D

The Common-ion Effect

We begin by considering a specific example of Le Chatelier's principle known as the

1 7.5

common-ion effect.

17.2

17.3

17.4

Buffered Solutions

We then consider the composition of buffered solutions, or buffers, and learn how they resist pH change upon the addition of small amounts of a strong acid or a strong base. Acid-Base Titrations

We continue by examining acid-base titration in detail, and we explore how to determine pH at any point in an acid-base titration.

17.6

17.7

Factors That Affect Solubility

We investigate some of the factors that affect solubility, including the common-ion effect and the effect of acids. Precipitation and Separation of Ions

Continuing the discussion of solubility equilibria, we learn how to precipitate ions selectively. Qualitative Analysis

for Metallic Elements

We conclude the chapter with an explanation of how the principles of solubility and complexation equilibria can be used to identify ions qualitatively in solution.

Solubility Equilibria

Next, we learn how to use equilibrium constants known as solubility-product constants to determine to what extent a sparingly soluble salt will dissolve in water.

WATER IS THE MOST COMMON AND MOST IMPORTANT SOLVENT ON EARTH. In a sense, it is the solvent of life. It is difficult to imagine how living matter in all its complexity could exist with any liquid other than water as the solvent. Water occupies its position of importance because of its abundance and its exceptional ability to dissolve a wide variety of substances. For example, the chapter-opening photograph shows a hot spring; this water contains a high concentration of ions (especially Mg2+, Ca2+, Fe 2+, C03 2-, and SO/-). The ions are dissolved as the hot water, initially underground, passes through various rocks on its way to the surface and dissolves minerals in the rocks. When the solution reaches the surface and cools, the minerals deposit and make the terracelike formations seen in the photograph. The various aqueous solutions encountered in nature typically contain many solutes. For example, the aqueous solutions in hot springs and oceans, as well as those in biological fluids, contain a variety of dissolved ions and molecules. Consequently, many equilibria can occur simultaneously in these solutions. 719

720

C HA PTE R

17

Additional Aspects o f Aqueous Equilibria In this chapter we take a step toward understanding such complex solu­ tions by looking first at further applications of acid-base equilibria. The idea is to consider not only solutions in which there is a single solute but also those containing a mixture of solutes. We then broaden our discussion to include two additional types of aqueous equilibria: those involving slightly soluble salts and those involving the formation of metal complexes in solution. For the most part, the discussions and calculations in this chapter are an extension of those in Chapters 15 and 16.

1 7 . 1 THE C O M M ON - I O N EFFECT In Chapter 16 we examined the equilibrium concentrations of ions in solutions containing a weak acid or a weak base. We now consider solutions that contain a weak acid, such as acetic acid (CH3COOH), and a soluble salt of that acid, such as sodium acetate (CH3COONa). Notice that these solutions contain two substances that share a common ion CH3Coo-. It is instructive to view these so­ lutions from the perspective of Le Chatelier's principle. = (Section 15.7) Sodi­ um acetate is a soluble ionic compound and is therefore a strong electrolyte. = (Section 4.1) Consequently, it dissociates completely in aqueous solution to form Na+ and CH3Coo- ions: CH3COONa(aq)

-----4

Na+ (aq) + CH3COO-(aq)

In contrast, CH3COOH is a weak electrolyte that ionizes as follows: CH3COOH(aq)

==

H + (a q) + CH3COO-(aq)

[17.1]

The CH3Coo- from CH3C00Na causes this equilibrium to shift to the left, thereby decreasing the equilibrium concentration of H + (a q) . CH3COOH(aq)

==

H + (a q) + CH3COO-(aq)

Addition of CH 3coo- shifts equilibrium, reducing [H+]

In other words, the presence of the added acetate ion causes the acetic acid to ion­ ize less than it normally would.

Whenever a weak electrolyte and a strong electrolyte contain a common ion, the weak electrolyte ionizes less than it would ifit were alone in solution. We call this observation the common-ion effect. Sample Exercises 17.1 and 17.2 illustrate how equilibrium

concentrations may be calculated when a solution contains a mixture of a weak electrolyte and a strong electrolyte that have a common ion. The procedures are similar to those encountered for weak acids and weak bases in Chapter 16.

• SAMPLE EXERCISE I Calculating the pH When a Common lon Is Involved What is the pH of a solution made by adding 0.30 mol of acetic acid and 0.30 mol of sodium acetate to enough water to make 1.0 L of solution?

17.1

SOLUTION Analyze: We are asked to determine the pH of a solution of a weak electrolyte (CH3COOH) and a strong electrolyte

(CH3COONa) that share a common ion, CH3Coo-.

In any problem in which we must determine the pH of a solution containing a mixture of solutes, it is helpful to proceed by a series of logi­ cal steps:

Plan:

1. Consider which solutes are strong electrolytes and which are weak electrolytes, and identify the major species in solution.

2. Identify the important equilibrium that is the source of H+ and therefore deter­

mines pH. Tabulate the concentrations of ions involved in the equilibrium. 4. Use the equilibrium-constant expression to calculate [H+] and then pH.

3.

17.1

The Common-Ion Effect

721

Solve: First, because CH3COOH is a weak electrolyte and CH3C00Na is a strong electrolyte, the major species in the solution are CH3COOH (a weak acid), Na+ (which is neither acidic nor basic and is therefore a spectator in the acid-base chemistry), and CH3coo- (which is the conjugate base of CH3COOH). Second, [H+] and, therefore, the pH are controlled by the dissociation equilibrium of CH3COOH: (We have written the equilibrium using H +(aq) rather than H30 +(aq) but both representations of the hydrated hydrogen ion are equally valid.) Third, we tabulate the initial and equilibrium concentrations as we did in solving other equilibrium prob­ lems in Chapters 15 and 16:

Initial

0.30 M

0

Change

-x M

+x M

+x M

(0.30 - x) M

xM

(0.30 + x) M

Equilibrium

0.30 M

The equilibrium concentration of CH3coo- (the common ion) is the initial concentration that is due to CH3C00Na (0.30 M) plus the change in concentration (x) that is due to the ionization of CH3COOH. Now we can use the equilibrium­ constant expression: (The dissociation constant for CH3COOH at 25 oc is from Appendix D; addition of CH3COONa does not change the value of this constant.) Substituting the equilibrium-constant concentrations from our table into the equilibrium expression gives

Because Ka is small, we assume that x is small compared to the original concen­ trations of CH3COOH and CH3coo­ (0.30 M each). Thus, we can ignore the very small x relative to 0.30 M, giving The resulting value of x is indeed small relative to 0.30, justifying the approximation made in simplifying the problem.

K0 =

18

•

X

10_5 =

=

1.8

X

10-5

X =

1.8

X

10-5 M

K0

pH

=

=

-log(l.8 X

x(0.30 + x) 0.30 - X

x(0.30) 0.30

= [H +]

10-5)

=

4.74

Finally, we calculate the pH from the equilibrium concentration of H +(aq): Comment: In Section 16.6 we calculated that a 0.30 M solution of CH3COOH has a pH of 2.64, corresponding to [H +] = 2.3 X 10-3 M. Thus, the addition of CH3C00Na has substantially decreased [H +], as we would expect from Le Chatelier 's principle. - PRACTICE EXERCISE Calculate the pH of a solution containing 0.085 M nitrous acid (HN02; K0 = 4.5 X

Answer: 3.42

10-4) and 0.10 M potassium nitrite (KN02).

722

CHAPTER

17

Additional Aspects of Aqueous Equilibria

• SAMPLE EXERCISE 1 7.2 I Calculating lon Concentrations When a Common lon Is Involved Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCI. SOLUTION Analyze: We are asked to determine the concentration of p- and the pH in a solution containing the weak acid HF and the strong

acid HCI. In this case the common ion is H+

Plan: We can again use the four steps outlined in Sample Exercise 17.1. Solve: Because HF is a weak acid and HCl is a strong acid, the major species in solution are HF, H+, and o-. The CC, which is the conjugate base of a strong acid, is merely a spectator ion in any acid-base chemistry. The problem asks for [F-], which is formed by ionization of HF. Thus, the important equilibrium is

The common ion in this problem is the hydrogen (or hydronium) ion. Now we can tabulate the initial and equilibrium concentrations of each species involved in this equilibrium:

Initial

0.20 M

0.10 M

0

Change

-x M

+x M

+x M

(0.20 - x) M

(0.10 + x) M

xM

Equilibrium The equilibrium constant for the ionization of HF, from Appendix D, is 6.8 x 10-4. Substituting the equilibrium-constant concentrations into the equilibrium expression gives If we assume that x is small relative to 0.10 or 0.20 M, this expression simpli­ fies to

+

HF(aq)

Ka

=

6.8 X 10_4

(0.10)(x) 0.20

=

x=

=

[H+][p-j [HF]

=

(0.10 + x)(x) 0.20 - X

6. 8 X 10-4 0 20 3 · (6.8 X 10-4) = 1.4 X 10- M = [F-] 0.10

This p- concentration is substantially smaller than it would be in a 0.20 M solution of HF with no added HCI. The common ion, H+, suppresses the ionization of HF. The concentration of H+(aq) is

[H+] = (0.10 + x) M

Thus,

pH = 1.00

�

0.10 M

Comment: Notice that for all practical purposes, [H+] is due entirely to the HCI; the HF makes a negligible contribution by

comparison.

- PRACTICE EXERCISE

Calculate the formate ion concentration and pH of a solution that is 0.050 M in formic acid (HCOOH; Ka = 1.8 in HN03. 5 Answer: [HCOO-] = 9.0 x 10- ; pH = 1.00

x

10-4) and 0.10 M

Sample Exercises 17.1 and 17.2 both involve weak acids. The ionization of a weak base is also decreased by the addition of a common ion. For example, the addition of NH4+ (as from the strong electrolyte NH4Cl) causes the base-disso­ ciation equilibrium of NH3 to shift to the left, decreasing the equilibrium con­ centration of OH- and lowering the pH: NH3(aq) + H 20(l) :;::::::=: NH 4+(aq) + OH-(aq) Addition of NH4+ shifts equilibrium, reducing [OH-]

[17.2]

17.2

Buffered Solutions

723

G IVE IT SOME THOUGHT A mixture of 0.10 mol of NH4CI and 0.12 mol of NH3 is added to enough water to

make 1.0 L of solution. (a) What are the initial concentrations of the major species in the solution? (b) Which of the ions in this solution is a spectator ion in any acid-base chemistry occurring in the solution? (c) What equilibrium reaction determines [OH-] and therefore the pH of the solution?

1 7.2 BUFFERED SOLUTIONS Solutions such as those discussed in Section 17.1, which contain a weak conjugate acid-base pair, can resist drastic changes in pH upon the addition of small amounts of strong acid or strong base. These solutions are called buffered solutions (or merely buffers). Human blood, for example, is a complex aqueous mixture with a pH buffered at about 7.4 (see the "Chemistry and Life" box near the end of this section). Much of the chemical behavior of seawater is determined by its pH, buffered at about 8.1 to 8.3 near the surface. Buffered solutions find many impor­ tant applications in the laboratory and in medicine (Figure 1 7. 1 � ). Composition and Action of Buffered Solutions

buffer resists changes in pH because it contains both an acid to neutralize OH- ions and a base to neutralize H+ ions. The acid and base that make up the buffer, however, must not consume each other through a neutralization reac­ tion. These requirements are fulfilled by a weak acid-base conjugate pair such as CH3COOH -CH3Coo- or NH4 +-NH 3. Thus, buffers are often prepared by mixing a weak acid or a weak base with a salt of that acid or base. The CH3COOH-CH3Coo- buffer can be prepared, for example, by adding CH3COONa to a solution of CH3COOH. The NH4+ - NH3 buffer can be pre­ pared by adding NH4CI to a solution of NH3• By choosing appropriate compo­ nents and adjusting their relative concentrations, we can buffer a solution at virtually any pH. A

G IVE IT S O M E THOUGHT Which of the following conjugate acid-base pairs will not function as a buffer: C2H5COOH and C2H5coo-; HC03- and co /-; HN03 and N03- ? Explain.

To understand better how a buffer works, let's consider a buffer composed of a weak acid (HX) and one of its salts (MX, where M+ could be Na+, K+, or another cation). The acid-dissociation equilibrium in this buffered solution involves both the acid and its conjugate base: [17.3] The corresponding acid-dissociation-constant expression is [17.4] Solving this expression for [H+], we have [H +]

=

Ka

[HX] [x-]

[17.5]

We see from this expression that [H+], and thus the pH, is determined by two factors: the value of Ka for the weak-acid component of the buffer and the ratio of the concentrations of the conjugate acid-base pair, [HX]/[X-].

"" Figure 1 7.1 Buffer solutions.

Prepackaged buffer solutions and ingredients for making up buffer solutions of predetermined pH can be purchased.

724

C HA PTE R 1 7

I> Figure 1 7.2

Additional Aspects of Aqueous Equilibria

Buffer action. When a small portion of OW is added to a buffer consisting of a mixture of the weak acid HF and its conjugate base (left), the OW reacts with the H F, decreasing [H F] and increasing [F-] in the buffer. Conversely, when a small portion of H+ is added to the buffer (right), the H+ reacts with the F-, decreasing [F-] and increasing [H F] in the buffer. Because pH depends on the ratio of F- to HF, the resulting pH change is small.

Buffer with equal concentrations of weak acid and its conjugate base

Buffer after addition of OH-

------

-----

HF

F-

Buffer after addition of H+

- - - - ,----,,--- - - - -

F-

HF

-----

------

HF

F-

H+ + F - HF

>

If OH- ions are added to the buffered solution, they react with the acid component of the buffer to produce water and x-: HX(aq) added base

[17.6]

weak acid in buffer

This reaction causes [HX] to decrease and [X-] to increase. As long as the amounts of HX and x- in the buffer are large compared to the amount of OH­ added, however, the ratio [HX]/ [X-] does not change much, and thus the change in pH is small. A specific example of such a buffer, the HF /F- buffer, is shown in Figure 17.2 .6.. If H + ions are added, they react with the base component of the buffer: H+(aq) + added base

x-(aq)

� HX (aq)

[17.7]

weak acid in buffer

This reaction can also be represented using H30+: H30+(aq) + x-(aq) � HX(aq) + H 20( 1) Using either equation, we see that the reaction causes [X-] to decrease and [HX] to increase. As long as the change in the ratio [HX]/ [X-] is small, the change in pH will be small. Figure 17.2 shows a buffer consisting of equal concentrations of hydrofluo­ ric acid and fluoride ion (center). The addition of OH- (left) reduces [HF] and increases [F-]. The addition of H+ (right) reduces [F-] and increases [HF].

G IVE IT SOME THOUGHT (a) What happens when NaOH is added to a buffer composed of CH3COOH and CH3coo - ? (b) What happens when HCl is added to this buffer? Calculating the pH of a Buffer Because conjugate acid-base pairs share a common ion, we can use the same procedures to calculate the pH of a buffer that we used to treat the common­ ion effect (see Sample Exercise 17. 1). However, we can sometimes take an al­ ternate approach that is based on an equation derived from Equation 17.5.

Buffered Solutions

17.2

(

)

725

Taking the negative log of both sides of Equation 17.5, we have -log[H+] = -log Ka

[HX] [X-]

= -logKa - log

[HX] [X-]

Because -log[H+] = pH and -logKa = pK., we have pH = pKa - log

In general,

[HX] [X-]

= pKa

pH = pK. + log

+ log [HX] [X-]

[17.8]

[base]

[17.9]

[acid]

where [acid] and [base] refer to the equilibrium concentrations of the conjugate [base] = [acid], pH = pK•. Equation 1 7.9 is known as the Henderson-Hasselbalch equation. Biolo­ gists, biochemists, and others who work frequently with buffers often use this equation to calculate the pH of buffers. In doing equilibrium calculations, we have seen that we can normally neglect the amounts of the acid and base of the buffer that ionize. Therefore, we can usually use the starting concentrations of the acid and base components of the buffer directly in Equation 17.9.

acid-base pair. Note that when

- SAMPLE EXERCISE 17.3

I Calculating the pH of a Buffer

�

What is the pH of a buffer that is 0.12 M in lactic acid [CH3C OH)COOH, or HC3H503] and [CH3CH(OH)COONa or NaC3H503]? For lactic acid, Ka � 1.4 X 10 .

0.10 M

in sodium lactate

SOLUTION Analyze: We are asked to calculate the pH of a buffer containing lactic acid HC3H503 and its conjugate base, the lactate ion (C3H s03-). Plan: We will first determine the pH using the method described in Section 17.1. Because HC3H503 is a weak electrolyte and NaC3H503 is a strong electrolyte, the major species in solution are HC�503, Na+, and C�H503 -. The Na + ion is a spectator ion. The HC3H 503-C3H 503- conjugate acid-base pair determines [H+] and thus pH; [H ] can be determined using the acid­ dissociation equilibrium of lactic acid. Solve: The initial and equilibrium concentrations of the species involved in this equilibrium are

Initial

0.12 M

0

-x M

+x M

(0.12 - x) M

xM

Change Equilibrium

0.1 0 M +x M (0.10

The equilibrium concentrations are go­ verned by the equilibrium expression: Because Ka is small and a common ion is present, we expect x to be small rel­ ative to either 0.12 or 0.10 M. Thus, our equation can be simplified to give Solving for x gives a value that justi­ fies our approximation:

Ka

�

[H+] pH

Alternatively, we could have used the Henderson-Hasselbalch equation to calculate pH directly: - PRACTICE EXERCISE

X

x(0.10) 10-4 ; 0.12

;x;

X

1.4

(�:��)(1.4

10-4) ; 1.7

X

10-4 M

� -log(1 7 X 10-4) � 3.77 .

(

pH ; pK. + log

[base]

[acid]

0.10 ) ) ; 3.85 + ( 0.12 log

; 3.85 + (-0.08) ; 3.77

Calculate the pH of a buffer composed of 0.12 M benzoic acid and 0.20 M sodium benzoate. (Refer to Appendix D.)

Answer:

4.42

+

x) M

726

C HA PTE R 1 7

Additional Aspects of Aqueous Equilibria

- SAMPLE EXERCISE 1 7.4

I Preparing a Buffer

How many moles of NH4Cl must be added to 2.0 L of 0.10 M NH3 to form a buffer whose pH is 9.00? (Assume that the addition of NH4Cl does not change the volume of the solution.) SOLUTION Analyze: Here we are asked to determine the amount of NH4+ ion required to prepare a buffer of a specific pH. Plan: The major species in the solu-

tion will be NH/, CC and NH3 . Of these, the Cl- ion is a spectator (it is the conjugate base of a strong acid). Thus, the NH4+-NH 3 conjugate acid-base pair will determine the pH of the buffer solution. The equilibrium relationship between NH4+ and NH3 is given by the base­ dissociation constant for NH3 :

Kb

=

[NH4 +][OH-] [NH3 ]

=

1 . 8 X 10-s

The key to this exercise is to use this Kb expression to calculate [NH4+]. Solve: We obtain [OH-] from the given pH:

pOH

=

14.00 - pH

=

14.00 - 9.00

=

(1.8 x 10-5)

=

5.00

and so Because Kb is small and the common ion NH4+ is present, the equilibrium concentration of NH 3 will essentially equal its initial concentration:

[NH 3]

We now use the expression for Kb to calculate [NH4+]:

[NH4+]

Thus, for the solution to have pH = 9.00, [NH4+] must equal 0.18 M. The number of moles of N�Cl needed to produce this concentration is given by the product of the volume of the solu­ tion and its molarity:

(2.0 L)(0.18 mol NH4CljL)

=

0.10 M

=

Kb

[NH 3] [OH-]

=

(0.10 M) (1.0 x 10-s M)

=

0.18 M

0.36 mol NH4Cl

Comment: Because NH4 + and NH3 are a conjugate acid-base pair, we could use the Henderson-Hasselbalch equation (Equation

17.9) to solve this problem. To do so requires first using Equation 16.41 to calculate pKa for NH4+ from the value of pKb for NH3 . We suggest you try this approach to convince yourself that you can use the Henderson-Hasselbalch equation for buffers for which you are given Kb for the conjugate base rather than Ka for the conjugate acid. - PRACTICE EXERCISE

Calculate the concentration of sodium benzoate that must be present in a 0.20 M solution of benzoic acid (Ct;H5COOH) to produce a pH of4.00. Answer: 0.13 M

Buffer Capacity and pH Range Two important characteristics of a buffer are its capacity and its effective pH range. Buffer capacity is the amount of acid or base the buffer can neutralize before the pH begins to change to an appreciable degree. The buffer capacity depends on the amount of acid and base from which the buffer is made. The pH of the buffer depends on the Ka for the acid and on the relative concentrations of the acid and base that comprise the buffer. According to Equation 17.5, for ex­ ample, [H+] for a 1-L solution that is 1 M in CH3COOH and 1 M in CH3COONa will be the same as for a 1-L solution that is 0.1 M in CH3COOH and 0.1 M in CH3C00Na. The first solution has a greater buffering capacity, however, be­ cause it contains more CH3COOH and CH3Coo -. The greater the amounts of the conjugate acid-base pair, the more resistant is the ratio of their concentra­ tions, and hence the pH, is to change.

17.2

Buffered Solutions

The pH range of any buffer is the pH range over which the buffer acts effec­ tively. Buffers most effectively resist a change in pH in either direction when the concentrations of weak acid and conjugate base are about the same. From Equation 17.9 we see that when the concentrations of weak acid and conjugate base are equal, pH = pK This relationship gives the optimal pH of any buffer. Thus, we usually try to select a buffer whose acid form has a pK, close to the de­ sired pH. In practice, we find that if the concentration of one component of the buffer is more than 10 times the concentration of the other component, the buffering action is poor. Because log 10 = 1, buffers usually have a usable range within :!: 1 pH unit of pK. (that is, a range of pH = pK. :t 1 ) . •.

G IVE IT SOME THOUGHT What is the optimal pH buffered by a solution containing CH3COOH and CH3COONa? (K, for CH3COOH is 1.8 x 10-s. )

Addition of Strong Acids or Bases to Buffers Let's now consider in a more quantitative way the response of a buffered solu­ tion to the addition of a strong acid or base. In solving these problems, it is im­ portant to understand that reactions between strong acids and weak bases proceed essentially to completion, as do those between strong bases and weak acids. Thus, as long as we do not exceed the buffering capacity of the buffer, we can assume that the strong acid or strong base is completely consumed by reac­ tion with the buffer. Consider a buffer that contains a weak acid HX and its conjugate base x - . When a strong acid is added to this buffer, the added H+ is consumed by x - to produce HX; thus, [HX] increases and [X-] decreases. (See Equation 17.7.) When a strong base is added to the buffer, the added OH- is consumed by HX to pro­ duce x-; in this case [HX] decreases and [X-] increases. (See Equation 17.6.) These two situations are summarized in Figure 17.2. To calculate how the pH of the buffer responds to the addition of a strong acid or a strong base, we follow the strategy outlined in Figure 17.3 T: 1. Consider the acid-base neutralization reaction, and determine its effect on [HX] and [X-]. This step of the procedure is a stoichiometry calculation. 2. Use K, and the new concentrations of [HX] and [X-] from step 1 to calculate [H+]. This second step of the procedure is a standard equilibrium calculation and is most easily done using the Henderson-Hasselbalch equation.

The complete procedure is illustrated in Sample Exercise

17.5.

Neutralization

Use K,, [HX], and [X-] to calculate [H+] Neutralization HX + OH � x- + HzO

----- Stoichiometry calculation -----+-!+---- Equilibrium calculation

.A

_j

Figure 1 7.3 Calculation of the pH of a buffer after the addition of acid or base. First consider how the

neutralization reaction between the added strong acid or strong base and the buffer affects the composition of the buffer (stoichiometry calculation). Then calculate the pH of the remaining buffer (equilibrium calculation). As long as the amount of added acid or base does not exceed the buffer capacity, the Henderson-Hasselbalch equation, Equation 1 7.9, can be used for the equilibrium calculation.

727

C HAPTER

728

17

Additional Aspects of Aqueous Equilibria

- SAMPLE EXERCISE 17.5

I Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol CH3COOH and 0.300 mol CH 3C00Na to enough water to make 1.00 L of solution. The pH of the buffer is 4.74 (Sample Exercise 17.1). (a) Calculate the pH of this solution after 0.020 mol of NaOH is added. (b) For comparison, calculate the pH that would result if 0.020 mol of NaOH were added to 1.00 L of pure water (neglect any volume changes). SOLUTION Analyze: We are asked to determine the pH of a buffer after addition of a small

amount of strong base and to compare the pH change to the pH that would result if we were to add the same amount of strong base to pure water. Plan: (a) Solving this problem involves the two steps outlined in Figure 17.3. Thus, we must first do a stoichiometry calculation to determine how the added OH- reacts with the buffer and affects its composition. Then we can use the resultant composi­ tion of the buffer and either the Henderson-Hasselbalch equation or the equilibrium­ constant expression for the buffer to determine the pH.

Stoichiometry Calculation: The OH- provided by NaOH reacts with CH3COOH, the weak acid component of the buffer. Prior to this neutralization reac­ tion, there are 0.300 mol each of CH3COOH and CH3Coo-. Neutralizing the 0.020 mol OH- requires 0.020 mol of CH3COOH. Consequently, the amount of CH3COOH decreases by 0.020 mol, and the amount of the product of the neutra­ lization, CH3Coo-, increases by 0.020 mol. We can create a table to see how the com­ position of the buffer changes as a result of its reaction with OH-: Solve:

Buffer before addition Addition Buffer after addition 0.280 M CH3COOH 0.320 M CH3COONa pH = 4.80

4.74 7Q' Q'

0.020 mol

0.280 mol

0

-

0.300 mol

-

0.320 mol

-

CH3COOH(aq) �

H+(aq)

+ CH3COO-(aq)

Using the quantities of CH3COOH and CH3Coo- remaining in the buffer, we can determine the pH using the Henderson-Hasselbalch equation. =

4.74 + log

0.320 molj1.00 L 0.280 mol/1.00 L

=

4.80

Comment Notice that we could have used mole amounts in place of concentrations in the Henderson-Hasselbalch equation and gotten the same result. The volumes of the acid and base are equal and cancel. If 0.020 mol of H+ was added to the buffer, we would proceed in a similar way to calculate the resulting pH of the buffer. In this case the pH decreases by 0.06 units, giving pH = 4.68, as shown in the figure in the margin.

0.300 M CH3COOH 0.300 M CH3C00Na pH

0

-

Equilibrium Calculation: We now tum our attention to the equilibrium that will determine the pH of the buffer, namely the ionization of acetic acid.

pH

Buffer

0.300 mol

(b) To determine the pH of a solution made by adding 0.020 mol of NaOH to 1.00 L of pure water, we can first determine pOH using Equation 16.18 and subtracting from 14. pH = 14 - (-log 0.020) = 12.30

·� 0

�b ·/ &.

0.320 M CH 3COOH 0.280 M CH3COONa pH = 4.68

Note that although the small amount of NaOH changes the pH of water significantly, the pH of the buffer changes very little.

- PRACTICE EXERCISE

Determine (a) the pH of the original buffer described in Sample Exercise 17.5 after the addition of 0.020 mol HCl and (b) the pH of the solution that would result from the addition of 0.020 mol HCl to 1.00 L of pure water. Answers: (a) 4.68, (b) 1 .70

17.2

Chemis

and Li e

Buffered Solutions

729

BLOOD AS A B UF FERED SOLUTION

M systems are extremely sensitive to pH. Many of the

any of the chemical reactions that occur in living

enzymes that catalyze important biochemical reactions, for example, are effective only within a narrow pH range. For this reason the human body maintains a remarkably in­ tricate system of buffers, both within tissue cells and in the fluids that transport cells. Blood, the fluid that transports oxygen to all parts of the body (Figure 17.4 � ), is one of the most prominent examples of the importance of buffers in living beings. Human blood is slightly basic with a normal pH of 7.35 to 7.45. Any deviation from this normal pH range can have extremely disruptive effects on the stability of cell mem­ branes, the structures of proteins, and the activities of en­ zymes. Death may result if the blood pH falls below 6.8 or rises above 7.8. When the pH falls below 7.35, the condition is called acidosis; when it rises above 7.45, the condition is called alkalosis. Acidosis is the more common tendency be­ cause ordinary metabolism generates several acids within the body. The major buffer system that is used to control the pH of blood is the carbonic acid-bicarbonate buffer system. Carbonic acid (H2C03) and bicarbonate ion (HC03l are a conjugate acid-base pair. In addition, carbonic acid can decompose into carbon dioxide gas and water. The important equilibria in this buffer system are

H+(aq) + HC03-(aq)

�

H2C03(aq) � H20(/) + C02{g)

(17.10]

Several aspects of these equilibria are notable. First, although carbonic acid is a diprotic acid, the carbonate ion (COll is unimportant in this system. Second, one of the components of this equilibrium, CO;u is a gas, which provides a mechanism for the body to adjust the equilibria. Removal of C02 via exhalation shifts the equilibria to the right, consum­ ing H+ ions. Third, the buffer system in blood operates at a pH of 7.4, which is fairly far removed from the pKal value of H2C03 (6.1 at physiological temperatures). For the buffer to have a pH of 7.4, the ratio [base]/[acid] must have a value of about 20. In normal blood plasma the concentrations of HC0 3- and H2C03 are about 0.024 M and 0.0012 M, respec­ tively. Consequently, the buffer has a high capacity to neutral­ ize additional acid, but only a low capacity to neutralize additional base. The principal organs that regulate the pH of the carbonic acid-bicarbonate buffer system are the lungs and kidneys. Some of the receptors in the brain are sensitive to the concen­ trations of H+ and C02 in bodily fluids. When the concentra­ tion of C02 rises, the equilibria in Equation 17.10 shift to the left, which leads to the formation of more H+. The receptors trigger a reflex to breathe faster and deeper, increasing the rate of elimination of C02 from the lungs and shifting the equilib­ ria back to the right. The kidneys absorb or release H+ and HC03-; much of the excess acid leaves the body in urine, which normally has a pH of 5.0 to 7.0.

& Figure 1 7.4 Red blood cells. A scanning electromicrograph of a group of red blood cells traveling through a small branch of an artery. Blood is a buffered solution whose pH is maintained between 7.35 and 7.45.

The regulation of the pH of blood plasma relates directly to the effective transport of 02 to bodily tissues. The protein hemoglobin, which is found in red blood cells, carries oxygen. Hemoglobin (Hb) reversibly binds both H+ and 02. These two substances compete for the Hb, which can be represented ap­ proximately by the following equilibrium: HbH+ + 02 � Hb02

+

H+

[17.11]

Oxygen enters the blood through the lungs, where it pass­ es into the red blood cells and binds to Hb. When the blood reaches tissue in which the concentration of 02 is low, the equilibrium in Equation 17.11 shifts to the left and 02 is re­ leased. An increase in H+ ion concentration (decrease in blood pH) also shifts this equilibrium to the left, as does increasing temperature. During periods of strenuous exertion, three factors work together to ensure the delivery of 02 to active tissues: (1) As 02 is consumed, the equilibrium in Equation 17.11 shifts to the left according to Le Chatelier's principle. (2) Exertion raises the temperature of the body, also shifting the equilibrium to the left. (3) Large amounts of C02 are produced by metabo­ lism, which shifts the equilibrium in Equation 17.10 to the left, thus decreasing the pH. Other acids, such as lactic acid, are also produced during strenuous exertion as tissues become starved for oxygen. The decrease in pH shifts the hemoglobin equilibrium to the left, delivering more 02. In addition, the de­ crease in pH stimulates an increase in the rate of breathing, which furnishes more 02 and eliminates C02. Without this elaborate arrangement, the 02 in tissues would be rapicUy depleted, making further activity impossible. Related Exercises: 1 7.29 and

17.90

730

C HA PTE R 1 7

Additional Aspects of Aqueous Equilibria

/ Buret containing NaOH(aq) I

1

I

••

�

.A.

Figure 1 7.5 Measuring pH during a A typical setup for using a pH

titration.

meter to measure data for a titration curve. In this case a standard solution of NaOH (the titrant) is added by buret to a solution of HCI. The HCI solution is stirred during the titration to ensure uniform composition.

1 7.3 ACID-BA S E TITRATIONS In Section 4.6 w e briefly described titrations. In an acid-base titration, a solu­ tion containing a known concentration of base is slowly added to an acid (or the acid is added to the base). Acid-base indicators can be used to signal the equivalence point of a titration (the point at which stoichiometrically equiv­ alent quantities of acid and base have been brought together). Alternatively, a pH meter can be used to monitor the progress of the reaction producing a pH titration curve, a graph of the pH as a function of the volume of the added titrant. The shape of the titration curve makes it possible to determine the equivalence point in the titration. The titration curve can also be used to select suitable indicators and to determine the Ka of the weak acid or the Kb of the weak base being titrated. A typical apparatus for measuring pH during a titration is illustrated in Figure 17.5 ... . The titrant is added to the solution from a buret, and the pH is continually monitored using a pH meter. To understand why titration curves have certain characteristic shapes, we will examine the curves for three kinds of titrations: (1) strong acid-strong base, (2) weak acid-strong base, and (3) polyprotic acid-strong base. We will also briefly consider how these curves relate to those involving weak bases.

G IVE IT SOME THOUGHT For the setup shown in Figure 17.5, will pH increase or decrease as titrant is added?

Strong Acid-Strong Base Titrations The titration curve produced when a strong base is added to a strong acid has the general shape shown in Figure 17.6 �. This curve depicts the pH change that occurs as 0.100 M NaOH is added to 50.0 mL of 0.100 M HCI. The pH can be cal­ culated at various stages of the titration. To help understand these calculations, we can divide the curve into four regions:

The initial pH (initial acid):

1.

The pH of the solution before the addition of any base is determined by the initial concentration of the strong acid. For a solu­ tion of 0.100 M HCI, [H+] = 0.100 M, and hence pH = -log(O.lOO) = 1.000. Thus, the initial pH is low.

2.

Between the initial pH and the equivalence point (remaining acid):

As NaOH is added, the pH increases slowly at first and then rapidly in the vicinity of the equivalence point. The pH of the solution before the equivalence point is determined by the concentration of acid that has not yet been neutralized. This calculation is illustrated in Sample Exercise 17.6(a).

3.

The equivalence point: At the equivalence point an equal number of moles of the NaOH and HCI have reacted, leaving only a solution of their salt, NaCI. The pH of the solution is 7.00 because the cation of a strong base (in this case Na+) and the anion of a strong acid (in this case Cl-) do not hydrolyze and therefore have no appreciable effect on pH. cx:o (Section 16.9)

4.

After the equivalence point (excess base):

The pH of the solution after the equivalence point is determined by the concentration of the excess NaOH in the solution. This calculation is illustrated in Sample Exercise 17.6(b).

17.3

Acid-Base Titrations

731

14 r----r---, 13

Initial acid (HCI)

Remaining acid (HCI + NaCI)

Equivalence point (NaCI)

Excess base (NaCI + NaOH)

.A Figure 1 7.6 Adding a strong base to a strong add. The pH curve for titration of 50.0 ml of a 0.1 00 M solution of a strong acid with a 0.1 00 M solution of a strong base. In this case the acid is HCI and the base is NaOH. The pH starts out at a low value characteristic of the acid and then increases as base is added, rising rapidly at the equivalence point. Both phenolphthalein and methyl red change color at the equivalence point. (For clarity, water molecules have been omitted from the molecular art.) - SAMPLE EXERCISE 1 7.6

I Calculating pH for a Strong Acid-Strong Base Titration

Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution: (a) 49.0 mL, (b) 51.0 mL. SOLUTION Analyze: We are asked to calculate the pH at two points in the titration of a strong acid with a strong base. The first point is just before the equivalence point, so we expect the pH to be determined by the small amount of strong acid that has not yet been neu­ tralized. The second point is just after the equivalence point, so we expect this pH to be determined by the small amount of excess strong base. Plan: (a) As the NaOH solution is added to the HCI solution, H+(aq) reacts with OH-(aq) to form H20. Both Na+ and o- are spectator ions, having negligible effect on the pH. To determine the pH of the solution, we must first determine how many moles of H+ were originally present and how many moles of OH- were added. We can then calculate how many moles of each ion re­ main after the neutralization reaction. To calculate [H+], and hence pH, we must also remember that the volume of the solution increases as we add titrant, thus diluting the concentration of all solutes present. Solve: The number of moles of H+ in the original HCl solution is given by the product of the volume of the solu­ tion (50.0 mL = 0.0500 L) and its mo­ larity (0.100 M):

(0.0500 L soln)

(

� )

0 100 mo H+ · 1 L so

=

3 5.00 X 10- mol H+

732

CHAPTER 1 7

Additional Aspects o f Aqueous Equilibria

Likewise, the number of moles of OH- in 49.0 mL of 0.100 M NaOH is Because we have not yet reached the equivalence point, there are more moles of H+ present than OH-. Each mole of OH- will react with one mole of H+ Using the convention intro­ duced in Sample Exercise 17.5, During the course of the titration, the volume of the reaction mixture in­ creases as the NaOH solution is added to the HCl solution. Thus, at this point in the titration, the total volume of the solutions is (We assume that the total volume is the sum of the volumes of the acid and base solutions.) Thus, the concen­ tration of H+(aq) is The corresponding pH equals

(0.0490 L soln)

(

� )=

0· 100 mol 1 L so

Addition

50.0 mL + 49.0 mL

=

99.0 mL

3 10- mol OH+

=

=

0 4.90 x 10-3 mol

liters soln

0

-

0.0990 L

3 0 . 10 X 10- mol 0.09900 L

moles H+(aq)

3 -log(l.O X 10- )

x

3 0.10 X 10- mol

After addition

=

4.90

5.00 X 10-3 mol

Before addition

[H +]

H-

=

3 1.0 X 10- M

3.00

Plan: (b) We proceed in the same way as we did in part (a), except we are now past the equivalence point and have more OH- in

the solution than H+. As before, the initial number of moles of each reactant is determined from their volumes and concentra­ tions. The reactant present in smaller stoichiometric amount (the limiting reactant) is consumed completely, leaving an excess of hydroxide ion. Solve:

+

Before addition

5.00

x

10-3 mol

After addition In this case the total volume of the so­ lution is

50.0 mL + 51.0 mL

Hence, the concentration of OH-(aq) in the solution is

[OH-]

Thus, the pOH of the solution equals

pOH

and the pH equals

pH

=

=

=

0 =

101.0 mL

moles OH-(aq) .

hters soln

3 -log(l.O X 10- )

14.00 - pOH

=

= =

0.10

=

x

0

3 10- mol 3 0.10 x 10- mol

5.10

Addition

0.1010 L

x 10-3 mol 0.1010 L

=

-

1.0 X 10-3 M

3.00

14.00 - 3.00

=

11.00

- PRACTICE EXERCISE

Calculate the pH when the following quantities of 0.100 M HN03 have been added to 25.0 mL of 0.100 M KOH solution: (a) 24.9 mL, (b) 25.1 mL. Answers: (a) 10.30, (b) 3.70

Optimally, an indicator would change color at the equivalence point in a titration. In practice, however, that is unnecessary. The pH changes very rapid­ ly near the equivalence point, and in this region merely a drop of titrant can change the pH by several units. Thus, an indicator beginning and ending its color change anywhere on this rapid-rise portion of the titration curve will give a sufficiently accurate measure of the volume of titrant needed to reach the equivalence point. The point in a titration where the indicator changes color is called the end point to distinguish it from the actual equivalence point that it closely approximates.

17.3

Acid-Base Titrations

733

�Tilis is strictly true onlyfor very dilute solutions. The values of equilibrium constants are somewhat altered when the total concentration of ionc i substances in water s i increased. However, we will ignore these effects, which are taken into consideration only for work that requires exceptional accuracy.

.6. Figure 1 7.14 Relationships between solubility and K,.. The solubility of any compound in grams per liter can be converted to molar solubility. The molar solubility can be used to determine the concentrations of ions in solution. The concentration of ions can be used to calculate Ksp· The steps can be reversed, and solubility calculated from Ksp ·

740

CHAPTER 1 7

Additional Aspects o f Aqueous Equilibria

- PRACTICE EXERCISE A saturated solution of Mg(OHh in contact with undissolved solid is prepared at 25 oc. The pH of the solution is found to be 10.17. Assuming that Mg(OHh dissociates completely in water and that there are no other simultaneous equilibria involving 2 the Mg + or OH- ions in the solution, calculate K,p for this compound. 2 Answer: 1.6 X 10-1

- SAMPLE EXERCISE 1 7.1 1 I Calculating Solubility from K,p The Ksp for CaF2 is 3.9 X 10-l l at 25 oc. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility, calculate the solubility of CaF2 in grams per liter. SOLUTION Analyze: We are given Ksp for CaF2 and are asked to determine solubility. Recall that the solubility of a substance is the quantity that can dissolve in solvent, whereas the solubility-product constant, K,p, is an equilibrium constant. Plan: We can approach this problem by using our standard techniques for solving equilibrium problems. We write the chemical equation for the dissolution process and set up a table of the initial and equilibrium concentrations. We then use the equilibrium­ constant expression. In this case we know K,p, and so we solve for the concentrations of the ions in solution. Solve: Assume initially that none of the salt has dissolved, and then allow

x moles/liter of CaF2 to dissociate com­

pletely when equilibrium is achieved.

+ Initial

-

Change

-

Equilibrium The stoichiometry of the equilibrium dictates that 2x moles/liter of F- are produced for each x moles/liter of CaF2 that dissolve. We now use the ex­ pression for Ksp and substitute the equilibrium concentrations to solve for the value of x: (Remember that Vy = y113; to calcu­ late the cube root of a number, you can use the yx function on your calcu­ lator, with x = Thus, the molar sol­ ubility of CaF2 is 2 . 1 x 10-4 mol/L. The mass of CaF2 that dissolves in water to form a liter of solution is

t.J

K,p = X =

(

2 [Ca +][p-f

�3 /3.9 X410

=

!I

(x)(2x)2 = 4x3 =

4 2.1 x 10- mol CaF2 1 L soIn

2.1 X

)(

=

3.9

0

0

+x M xM

+2x M 2x M

x 10-11

4 10- M

78.1 g CaF2 1 moI CaF2

)

=

1.6

X 10-2 g CaFz/L soln

Check: We expect a small number for the solubility of a slightly soluble salt. If we reverse the calculation, we should be able 2 to recalculate the solubility product: Ksp = (2.1 x 10-4)(4.2 x 10-4) = 3.7 x 10-1 1 , close to the starting value for K,p, 3.9 X 10-ll Comment: Because F- is the anion of a weak acid, you might expect that the hydrolysis of the ion would affect the solubility of CaF2. The basicity of p- is so small (Kb = 1.5 X 10- 1 1), however, that the hydrolysis occurs to only a slight extent and does not significantly influence the solubility. The reported solubility is 0.017 g/L at 25 oc, in good agreement with our calculation. - PRACTICE EXERCISE The Ksp for LaF3 is 2 x 10-1 9 What is the solubility of LaF3 in water in moles per liter? Answer: 9 X 10-6 mol/L

17.5

Factors That Affect Solubility

741

LIMITATIONS O F S O LUBILITY PRO DUCTS

Ksp sometimes

T deviate appreciably from those found experimentally. he concentrations of ions calculated from

In part, these deviations are due to electrostatic interactions between ions in solution, which can lead to ion pairs. (See Section 13.5, "A Closer Look: Colligative Properties of Electrolyte Solutions.") These interactions increase in magni­ tude both as the concentrations of the ions increase and as their charges increase. The solubility calculated from tends to be low unless it is corrected to account for these interactions between ions. Chemists have developed procedures for correcting for these "ionic-strength" or "ionic-activity" effects, and these procedures are examined in more advanced chemistry courses. As an example of the effect of these interi­ onic interactions, consider CaC03 (calcite), whose solubility 9 product, = 4.5 x 10- , gives a calculated solubility of 6.7 x 10-5 moljL. Making corrections for the interionic interactions in the solution yields a higher solubility, 7.3 X 10-5 mol/L. The reported solubility, however, is twice as high (1.4 X 10-4 moljL), so there must be one or more additional factors involved.

Ksp

K,P

Ksp

Another common source of error in calculating ion concen­ trations from is ignoring other equilibria that occur simulta­ neously in the solution. It is possible, for example, that acid-base equilibria take place simultaneously with solubility equilibria. In particular, both basic anions and cations with high charge-to­ size ratios undergo hydrolysis reactions that can measurably in­ crease the solubilities of their salts. For example, CaC03 contains 4 the basic carbonate ion = 1.8 X 10- ), which hydrolyzes in water: Co l-(aq) + HzO(l) :;==:: HC03-(aq) + OH-(aq). If we consider both the effect of the interionic interactions in the solu­ tion and the effect of the simultaneous solubili and hydrolysis equilibria, we calculate a solubility of 1.4 X 10 moljL, in agree­ ment with the measured value. Finally, we generally assume that ionic compounds disso­ ciate completely into their component ions when they dissolve. This assumption is not always valid. When MgF2 dissolves, for 2 example, it yields not only Mg + and F- ions but also MgF+ ions in solution. Thus, we see that calculating solubility using can be more complicated than it first appears and it requires considerable knowledge of the equilibria occurring in solution.

(Kb

jY

Ksp

17.5 FACTORS THAT AFFECT SOLUB ILITY The solubility of a substance is affected by temperature as well as by the pres­ ence of other solutes. The presence of an acid, for example, can have a major in­ fluence on the solubility of a substance. In Section 17.4 we considered the dissolving of ionic compounds in pure water. In this section we examine three factors that affect the solubility of ionic compounds: (1) the presence of com­ mon ions, (2) the pH of the solution, and (3) the presence of complexing agents. We will also examine the phenomenon of amphoterism, which is related to the effects of both pH and complexing agents. Common-ion Effect

The presence of either Ca 2+(aq) or F- (aq) in a solution reduces the solubility of CaF2, shifting the solubility equilibrium of CaF2 to the left. CaF2(s) � Ca2+ (aq) 2 F- (aq)

+

2 Addition of Ca + or p- shifts equilibrium, reducing solubility

ceo

This reduction in solubility is another application of the common-ion effect. (Section 17.1) In general, the solubility of a slightly soluble salt is decreased by the presence of a second solute that furnishes a common ion. Figure 17.15� shows how the solubility of CaF2 decreases as NaF is added to the solution. Sample Exercise 17.12 shows how the K,P can be used to calculate the solubility of a slightly soluble salt in the presence of a common ion.

0

0.05 0.10 0.15 0.20 Concentration of NaF (moi/L)

"' Figure 1 7.15 Common-ion effect.

The way in which NaF concentration affects the solubility of CaF2 demonstrates the common-ion effect. Notice that the CaF2 solubility is on a logarithmic scale.

742

CHAPTER 1 7

Additional Aspects o f Aqueous Equilibria

• SAMPLE EXERCISE 1 7.12

I Calculating the Effect of a Common lon on Solubility

Calculate the molar solubility of CaF2 at 25 'C in a solution that is (a)

0.010 M in Ca(N03):u (b) 0.010 M in NaF.

SOLUTION

Analyze: We are asked to determine the solubility of CaF2 in the presence of two strong electrolytes, each of which contains an ion 2 common to CaF2. In (a) the common ion is Ca +, and N03- is a spectator ion. In (b) the common ion is F-, and Na + is a spectator ion. Plan: Because the slightly soluble compound is CaF2, we need to use the K,p for this compound, which is available in Appendix D: 2 K,p = [Ca +] [p-f = 3.9 X 10-11

The value ofKsp is unchanged lly the presence ofadditional solutes. Because of the common-ion effect, however, the solubility of the salt will decrease in the presence of common ions. We can again use our standard equilibrium techniques of starting with the equa­ tion for CaF2 dissolution, setting up a table of initial and equilibrium concentrations, and using the K,p expression to determine the concentration of the ion that comes only from CaF2. Solve: (a) In this instance the initial 2 concentration of Ca + is 0.010 M because of the dissolved Ca(N03)2:

+ -

Initial Change

-

Equilibrium

-

0.010 M +x M (0.010 + x) M

0 +2x M 2x M

Substituting into the solubility-product expression gives This would be a messy problem to solve exactly, but fortunately it is possible to simplify matters greatly. Even without the common-ion effect, the solubility of CaF2 is very small (2.1 x 10-4 M). Thus, we assume that 2 the 0.010 M concentration of Ca + from Ca(N03h is very much greater than the small additional concentration re­ sulting from the solubility of CaF:b that is, x is small compared to 0.010 M, and 0.010 + x � 0.010. We then have

3.9 X 10-11

=

(0.010)(2xf 3.9 x w-11 = 9.8 X 10-10 x2 = 4(0.010) x = Y9.8 x 10 10 = 3.1 x w-s M

The very small value for x validates the simplifying assumption we have made. Our calculation indicates that solid CaF2 dissolves per liter of the 0.010 M Ca(N03h solution.

3.1 X 10-s mol of

(b) In this case the common ion is F-, and at equilibrium we have Assuming that 2x is small compared to (that is, 0.010 + 2x "' 0.010), we have

0.010 M

Thus, 3.9

3.9 X 10-11 X

x(0.010)2 3.9 X 10-11 = (0.010)2 =

=

3.9 X 10-7 M

X 10-7 mol of solid CaF2 should dissolve per liter of 0.010 M NaF solution.

Comment: The molar solubility of CaF2 in pure water is 2.1 x 10-4 M (Sample Exercise 17.11). By comparison, our calculations 2 above show that the solubility of CaF2 in the presence of 0.010 M Ca + is 3.1 x 10-5 M, and in the presence of 0.010 M p- ion it 2 is 3.9 x 10-7 M. Thus, the addition of either Ca + or F- to a solution of CaF2 decreases the solubility. However, the effect ofF- on 2 the solubili is more pronounced than that of Ca + because [F-] appears to the second power in the Ksp expression for CaF0 whereas Ca + appears to the first power.

�

- PRACTICE EXERCISE The value for K,p for manganese(II) hydroxide, Mn(OH)2, is that contains 0.020 M NaOH.

Answer:

4.0

x

1.6 X 10-13

Calculate the molar solubility of Mn(OHh in a solution

10-10 M

Solubil ity and pH The pH of a solution will affect the solubility of any substance whose anion is basic. Consider Mg(OH)z, for example, for which the solubility equilibrium is 2 Mg(OHh(s) � Mg +(aq) + 2 OH-(aq) Ksp = 1.8 X 10- 1 1 (17.17]

17.5

Factors That Affect Solubility

743

A saturated

solution of Mg(OHh has a calculated pH of 10.52 and contains [Mg2+] = 1.7 X 10-4 M. Now suppose that solid Mg(OHh is equilibrated with a solution buffered at a more acidic pH of 9.0. The pOH, therefore, is 5.0, so [OH-] = 1.0 X 10-5 . Inserting this value for [OH-] into the solubility-product expression, we have K,p = [Mg 2+][oH-f = 1.8 X 10-11

[Mg 2+](1.0 X 10-5) 2

[Mg 2+]

=

=

1.8 X 10-11

1.8 X 10-11

(1.0 X 10-5) 2

=

0.18 M

Thus, Mg(OHh dissolves in the solution until [Mg 2+] = 0.18 M. It is apparent that Mg(OHh is quite soluble in this solution. If the concentration of OH­ were reduced even further by making the solution more acidic, the Mg2+ con­ centration would have to increase to maintain the equilibrium condition. Thus, a sample of Mg(OHh will dissolve completely if sufficient acid is added (Figure 17.16 T).

T Figure

1 7.16 Dissolution of a

precipitate In add. A white precipitate of Mg(OH)z(s) in contact with its saturated solution is in the test tube on the left. The dropper poised above the solution surface contains hydrochloric acid. (The anions accompanying the acid have been omitted to simplify the art.)

SOLUBILITY AND pH

The solubility of a substance whose anion is basic will be affected to some extent by the pH of the solution. The solubility of Mg(OH)z greatly increases as the acidity of the solution increases.

Mg2 +(aq) + 2 H20 (l)

2 H+(aq) + Mg(OHh(s) A precipitate of Mg(OH)z(s)

The precipitate dissolves upon addition of acid.

A sample of Mg(OH), will dissolve completely if sufficient acid is added.

744

CHAPTER 1 7

Additional Aspects of Aqueous Equilibria

Chemistr and Li e

S I N K H OLES

A stone,

principal cause of sinkholes is the dissolution of lime­ which is calcium carbonate, by groundwater. Although CaC03 has a relatively small solubility-product con­ stant, it is quite soluble in the presence of acid. 2 2 CaC03(s) � Ca +(aq) + C03 -(aq) Ksp = 4.5 X 10-9

The sudden formation of large sinkholes can pose a serious threat to life and property. The existence of deep sinkholes also increases the risk of contamination of the aquifer.

Rainwater is naturally acidic, with a pH range of 5 to 6, and can become more acidic when it comes into contact with decaying plant matter. Because carbonate ion is the conjugate base of the weak acid, hydrogen carbonate ion (HC03-), it readily com­ bines with hydrogen ion. COl-(aq)

+ H+(aq)

-----> HC03 -(aq)

The consumption of carbonate ion shifts the dis­ solution equilibrium to the right, thus increasing the solubility of CaC03. This can have profound conse­ quences in areas where the terrain consists of porous calcium carbonate bedrock covered by a relatively thin layer of clay and/ or topsoil. As acidic water percolates through and gradually dissolves the lime­ stone, it creates underground voids. A sinkhole re­ sults when the overlying ground can no longer be supported by the remaining bedrock and col­ lapses into the underground cavity [Figure 17.17 � ].

-" Figure 1 7.1 7 Sinkhole formation. An underground void develops as limestone, CaC03(s), dissolves. Collapse of the overlying ground into an underground cavity causes sinkhole formation. The large sinkhole shown here occured in Orlando, Florida and destroyed several buildings and part of a highway.

The solubility of almost any ionic compound is affected if the solution is made sufficiently acidic or basic. The effects are very noticeable, however, only when one or both ions involved are at least moderately acidic or basic. The metal hydroxides, such as Mg(OHh, are examples of compounds containing a strongly basic ion, the hydroxide ion. In general, ifa compound contains a basic anion (that is, the anion of a weak acid), its solubility will increase as the solution becomes more acidic. As we have seen, the solubility of Mg(OHh greatly increases as the acidity of the solution increases. The solubility of CaF2 increases as the solution becomes more acidic, too, be­ cause the p- ion is a weak base; it is the conjugate base of the weak acid HF. As a result, the solubility equilibrium of CaF 2 is shifted to the right as the concen­ tration of p- ions is reduced by protonation to form HF. Thus, the solution process can be understood in terms of two consecutive reactions: CaF2(s) � Ca2+ (aq) F-(aq )

+ 2 F-(aq)

+ H+ (aq) � HF(aq)

[17.18] [17.19]

The equation for the overall process is CaF2(s) 5.0

3.0 pH

1.0

-" Figure 1 7. 1 8 The effect of pH on the solubility of CaF2. The solubility

increases as the solution becomes more acidic (lower pH). Notice that the vertical scale has been multiplied by 1 03 .

+ 2 H+(aq) � Ca2+ (aq) + 2 HF(aq)

[17.20]

Figure 17.18 ..,. shows how the solubility of CaF2 changes with pH. 2 Other salts that contain basic anions, such as col-, P0 43-, CN-, or s -, behave similarly. These examples illustrate a general rule: The solubility of slight­ ly soluble salts containing basic anions increases as [H +] increases (as pH is lowered). The more basic the anion, the more the solubility is influenced by pH. Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes.

17.5

- SAMPLE EXERCISE 17.13

I Predicting the Effect of Acid on Solubility

Which of the following substances will be more soluble in acidic solution than in basic solution: (a) Ni(OHh(s), (b) CaC03(s), (c) BaF2(s), (d) AgCl(s)? SOLUTION Analyze: The problem lists four sparingly soluble salts, and we are asked to deter­ mine which will be more soluble at low pH than at high pH. Plan: Ionic compounds that dissociate to produce a basic anion will be more soluble in acid solution. Solve:

(a) Ni(OHh(s) will be more soluble in acidic solution because of the basicity of OH-; the H+ ion reacts with the OH- ion, forming water. Ni(OHh(s) = Ni2+(aq) + 2 OH-(aq)

2 OH-(aq) + 2 H+(aq) = 2 H20(/) Overall:

Ni(OHh(s) + 2 H+(aq) = Ni2+(aq) + 2 H 20(/)

(b) Similarly, CaC03(s) dissolves in acid solutions because C032- is a basic anion. 2 2 CaC03(s) = Ca +(aq) + C03 -(aq) col-(aq) + 2 H+(aq) = H 2C03(aq) H 2C03(aq) = C0 2(g) + H 20(/)

Overall:

CaC03(s) + 2 H+(aq) = Ca2+(aq) + C02(g) + H 20(/)

The reaction between col- and H+ occurs in a stepwise fashion, first forming HC03 -. H2C03 forms in appreciable amounts only when the concentration of H+ is sufficiently high. (c) The solubility of BaF2 is also enhanced by lowering the pH, because F- is a basic anion. BaF2(s) = Ba2+(aq) + 2 F-(aq)

2 F-(aq) + 2 H+(aq) = 2 HF(aq) Overall:

BaF2(s) + 2 H+(aq) = Ba2+(aq) + 2 HF(aq)

(d) The solubility of AgCl is unaffected by changes in pH because Cl- is the anion of a strong acid and therefore has negligible basicity. - PRACTICE EXERCISE

�

Write the net ionic equation for the reaction of the following copper(II) compounds with acid: (a) CuS, (b) Cu N3)2. Answers: (a) CuS(s) + H (aq) = Cu2+(aq) + HS-(aq) (b) Cu(N3h(s) + 2 H+(aq) = Cu2+(aq) + 2 HN3(aq)

Formation of Complex Ions A characteristic property of metal ions is their ability to act as Lewis acids, or electron-pair acceptors, toward water molecules, which act as Lewis bases, or electron-pair donors. c:co (Section 16.11) Lewis bases other than water can also interact with metal ions, particularly with transition-metal ions. Such interac­ tions can dramatically affect the solubility of a metal salt. AgCI, for example, which has K,P = 1.8 X 10-10, will dissolve in the presence of aqueous ammonia because Ag+ interacts with the Lewis base NH3, as shown in Figure 17.19T. This process can be viewed as the sum of two reactions, the dissolution of AgCI and the Lewis acid-base interaction between Ag + and NH3. AgCl(s) Ag +(aq)

Overall: AgCl(s)

+ 2 NH3(aq) + 2 NH3(aq)

;;:::::::!

Ag+(aq) + Cqaq)

[17.21]

;;:::::::! Ag(NH3h +(aq)

[17.22]

;;:::::::! Ag(NH3h+(aq) + Cqaq)

[17.23]

Factors That Affect Solubility

745

746

CHAPTER 1 7

Additional Aspects of Aqueous Equilibria

FO RMATION OF COMPLEX IONS

Lewis bases can interact with metal ions, particularly with transition-metal ions, which can dramatically affect the solubility of a metal salt. AgCl, for example, will dissolve in the presence of aqueous ammonia because Ag + interacts with the Lewis base NH3.

-+

AgCI(s) A saturated solution of AgCI in contact with solid AgCI.

_..

+

Ag(NH3)z+(aq)

When concentrated ammonia is added, Ag+ ions are consumed in the formation of the complex ion Ag(NH3h+. The AgCI solid is being dissolved by the addition of NH3.

+ Cqaq)

Removal of Ag+ ions from the solution shifts the dissolution equilibrium to the right, causing AgCI to dissolve. Addition of sufficient ammonia results in complete dissolution of the AgCI solid.

Figure 1 7. 1 9 Using NH3(oq) to dissolve AgCI(s).

The presence of NH3 drives the reaction, the dissolution of AgCI, to the right as Ag+(aq) is consumed to form Ag(NH 3)2+ For a Lewis base such as NH3 to increase the solubility of a metal salt, it must be able to interact more strongly with the metal ion than water does. The NH3 must displace solvating H20 molecules (Sections 13.1 and 16.11) in order to form Ag(NH 3)2+ : [1 7.24]

17.5

Chemis

and Li e

ooth enamel consists mainly o f a mineral called hydrox­

the body. Tooth cavities are caused when acids dissolve tooth enamel.

+ 8 H+(aq)

----->

10 Ca2+(aq)

+ 6 HPO/-(aq) + 2 H 20(/)

The resultant Ca2+ and HPO/- ions diffuse out of the tooth enamel and are washed away by saliva. The acids that attack the hydroxyapatite are formed by the action of specific bacteria on sugars and other carbohydrates present in the plaque adhering to the teeth. Fluoride ion, present in drinking water, toothpaste, and other sources, can react with hydroxyapatite to form fluoro­ apatite, Ca10(P04)6F2. This mineral, in which F- has replaced

OH-, is much more resistant to attack by acids because the flu­ oride ion is a much weaker Bmnsted-Lowry base than the hydroxide ion. Because the fluoride ion is so effective in preventing cavi­ ties, it is added to the public water supply in many places to give a concentration of 1 mg/L (1 ppm). The compound added may be NaF or NazSiF6. NazSiF6 reacts with water to release fluoride ions by the following reaction: SiF62-(aq)

+ 2 H20(/)

f

=

[ Ag(NH3h+] [Ag +] [NH3f

=

Related Exercise: 17.110

1.7 X 107

[ 17.25]

The equilibrium constant for this kind of reaction is called a formation constant, K1. The formation constants for several complex ions are listed in Table 17.1 T.

TABLE 17.1

• Formation Constants for Some Metal Complex Ions in Water at 25 •c

Complex Ion

Kr

Equilibrium Equation

Ag(NH3h+

1.7 X 107

Ag(CNh-

1 X 1021

CdBr/-

10 1 5 X 103

Cr(OH)4 -

8

Co(SCN)/-

1 X 10

Cu(NH3)/+

5 X 10 12

Cu(CN)/-

1 X 1025

Ni(NH3)62+ 4 Fe(CN)6 3 Fe(CN)6 -

1.2 X 109 3 1 X 10 5

+ 2 NH3(aq) = Ag(NH3)z+(aq) Ag+(aq) + 2 CN-(aq) = Ag(CNh -(aq) 3 Ag+(aq) + 2 S 20l-(aq) = Ag(S 203)z -(aq) 2 2 Cd +(aq) + 4 Br-(aq) = CdBr4 -(aq) 3 Cr +(aq) + 4 OH-(aq) = Cr(OH)4-(aq) Co2+(aq) + 4 SCN-(aq) = Co(SCN)/-(aq) Cu2+(aq) + 4 NH3(aq) = Cu(NH3)/+(aq) 2 Cu2+(aq) + 4 CN-(aq) = Cu(CN)4 -(aq) 2 2 +(aq) Ni(NH3)6 Ni +(aq) + 6 NH3(aq) = 4 2 Fe +(aq) + 6 CN-(aq) = Fe(CN)6 -(aq) 3 3 Fe +(aq) + 6 CN-(aq) = Fe(CN)6 -(aq)

3 Ag(Sz03)z -

2.9 X

3

X 1029 3

1 X 1042

Ag+(aq)

----->

6 F-(aq) + 4 H+(aq) + Si0 2(s)

About 80% of all toothpastes now sold in the United States contain fluoride compounds, usually at the level of 0.1% fluoride by mass. The most common compounds in toothpastes are sodium fluoride (NaF), sodium monofluorophosphate (Na2P03F), and stannous fluoride (SnF2).

An assembly of a metal ion and the Lewis bases bonded to it, such as Ag(NH3)z+, is called a complex ion. The stability of a complex ion in aqueous solution can be judged by the size of the equilibrium constant for its formation from the hydrated metal ion. For example, the equilibrium constant for forma­ tion of Ag(NH 3)z+ (Equation 17.24) is 1.7 X 107:

K

747

TOOTH D ECAY A N D F L U O RI DAT I O N

T yapatite, Ca10(P04)6(0Hh. It is the hardest substance in Ca 10(P04)6(0Hh(s)

Factors That Affect Solubility

748

CHAPTER 1 7

Additional Aspects of Aqueous Equilibria • SAMPLE EXERCISE 17.14 I Evaluating an Equilibrium Involving a Complex ion Calculate the concentration of Ag + present in solution at equilibrium when concen­ trated ammonia is added to a 0.010 M solution of AgN03 to give an equilibrium con­ centration of [NH3] = 0.20 M. Neglect the small volume change that occurs when NH3 is added. SOLUTION Analyze: When NH3(aq) is added to Ag+(aq), a reaction occurs forming Ag(NH3h+ as shown in Equation 17.22. We are asked to determine what concentration of Ag+(aq) will remain uncombined when the NH3 concentration is brought to 0.20 M in a solution originally 0.010 M in AgN03.

Plan: We first assume that the AgN03 is completely dissociated, giving 0.10 M Ag+. Because K1 for the formation of Ag(NH3h+ is quite large, we assume that essentially all the Ag+ is then converted to Ag(NH3h+ and approach the problem as though we are concerned with the dissociation of Ag(NH3h+ rather than its formation. To facil­ itate this approach, we will need to reverse the equation to represent the formation of Ag + and NH3 from Ag(NH3h+ and also make the corresponding change to the equilibrium constant. :;====:

Ag +(aq) + 2 NH3(aq) 1 ...!._ = ---- = 5.9 X 10-B Kt 1.7 X 107

Ag(NH3h +(aq)

0.010 M initially, then [Ag(NH3h+l will be 0.010 M following addition of the NH3. We now construct a table to solve this equilibrium problem. Note that the NH3 concentration given in the problem is an equilibrium concentration rather than an initial concentration. Solve: If [Ag+] is

+ Initial Change Equilibrium

0.010 M -x M 0.010 - x M

OM +x M xM

0.20 M

Because the concentration of Ag + is very small, we can ignore x in comparison with 0.010. Thus, 0.010 - x "" 0.010 M. Substituting these values into the equilibrium­ constant expression for the dissociation of Ag(NH3h +, we obtain

(x)(0.20j2 _..c.,- = --- = 5.9 X 10_8 0.010

[Ag+] [NH3 f

__: ::__

[Ag(NH3h+]

Solving for x, we obtain x = 1.5 X 10-B M = [Ag +]. Thus, formation of the Ag(NH3h+ complex drastically reduces the concentration of free Ag+ ion in solution. - PRACTICE EXERCISE

3 Calculate [Cr +] in equilibrium with Cr(OH)4- when 0.010 mol of Cr(N03h is dis­ solved in a liter of solution buffered at pH 10.0.

Answer:

1 X 10-16 M

The general rule is that the solubility of metal salts increases in the presence of suitable Lewis bases, such as NH 3, CN-, or OH-, if the metal forms a com­ plex with the base. The ability of metal ions to form complexes is an extremely important aspect of their chemistry. In Chapter 24 we will take a much closer look at complex ions. In that chapter and others we will see applications of complex ions to areas such as biochemistry, metallurgy, and photography. Amphoterism Some metal oxides and hydroxides that are relatively insoluble in neutral water dissolve in strongly acidic and strongly basic solutions. These substances are soluble in strong acids and bases because they themselves are capable of behaving as either an acid or base; they are amphoteric oxides and hydroxides.

Factors That Affect Solubility

17.5

AM PHOTERISM

Metal oxides and hydroxides that are relatively insoluble in neutral water, but dissolve in both strongly acidic and strongly basic solutions, are said to be amphoteric. Their behavior results from the formation ofcomplex anions containing several hydroxides bound to the metal ion.

--+

Al(Hz0)63+(aq)

An aqueous solution of Al3+.

--+

Al(H20)J(OH)J (s) As NaOH is added, a precipitate of Al(OH)J forms.

Al(H20)z(OH)4-(aq)

The precipitate then dissolves as more NaOH is added, demonstrating the amphoterism of the Al(OH)3.

.II. Figure 1 7.20 Amphoterism.

2 2 Amphoteric oxides and hydroxides include those of Al3+, Cr3+, Zn +, and Sn +. Notice that the term amphoteric is applied to the behavior of insoluble oxides and hydroxides that can be made to dissolve in either acidic or basic solutions. The similar term amphiprotic, which we encountered in Section 16.2, relates more generally to any molecule or ion that can either gain or lose a proton. Amphoteric species dissolve in acidic solutions because they contain basic anions. What makes amphoteric oxides and hydroxides special, though, is that they also dissolve in strongly basic solutions (Figure 17.20 ..t.). This behavior re­ sults from the formation of complex anions containing several (typically four) hydroxides bound to the metal ion. [17.26] Amphoterism is often explained by the behavior of the water molecules that surround the metal ion and that are bonded to it by Lewis acid-base interac­ tions. oao (Section 16.11) For example, Al3+(aq) is more accurately represented as Al(H20)63+(aq) because six water molecules are bonded to the Al3+ in aqueous solution. Recall from Section 16.11 that this hydrated ion is a weak acid.

749

750

C HA PTE R 1 7

Additional Aspects of Aqueous Equilibria As a strong base is added, Al(H20)63+ loses protons in a stepwise fashion, eventu­ ally forming neutral and water-insoluble Al(H20) 3(0H) 3. This substance then dis­ solves upon removal of an additional proton to form the anion Al(H 20h(OH)4-. The reactions that occur are as follows: Al(H20)63+(aq) 2 Al(H20)s(OH) +(aq)

+ OH-(aq) � Al(H20)s(OH)2+(aq) + H20(!)

Al(H20)4(0Hh+(aq)

+ OH-(aq) � Al(H20h(OH)J(s) + H20(!)

Al(H20h(OH)J(s)

+ OH-(aq) � Al(H20)4(0Hh+(aq) + H20(!) + OH-(aq) � Al(H20h(OH)4-(aq) + H20(I)

Removing additional protons is possible, but each successive reaction occurs less readily than the one before. As the charge on the ion becomes more negative, it becomes increasingly difficult to remove a positively charged proton. Addition of an acid reverses these reactions. The proton adds in a stepwise fashion to convert the OH- groups to H20, eventually re-forming Al(H20)63+ The common prac­ tice is to simplify the equations for these reactions by excluding the bound H20 molecules. Thus, we usually write Al3+ instead of Al(H20)63+, Al(OHh instead of Al(H20h(OHh, Al(OH)4- instead of Al(H20h(OH)4 -, and so forth. The extent to which an insoluble metal hydroxide reacts with either acid or base varies with the particular metal ion involved. Many metal hydroxides­ such as Ca(OHh, Fe(OHh, and Fe(OH)J-are capable of dissolving in acidic so­ lution but do not react with excess base. These hydroxides are not amphoteric. The purification of aluminum ore in the manufacture of aluminum metal provides an interesting application of the property of amphoterism. As we have seen, Al(OHh is amphoteric, whereas Fe(OHh is not. Aluminum occurs in large quantities as the ore bauxite, which is essentially Al203 with additional water molecules. The ore is contaminated with Fe203 as an impurity. When bauxite is added to a strongly basic solution, the Al203 dissolves because the aluminum forms complex ions, such as Al(OH)4-. The Fe203 impurity, however, is not amphoteric and remains as a solid. The solution is filtered, getting rid of the iron impurity. Aluminum hydroxide is then precipitated by addition of acid. The purified hydroxide receives further treatment and eventually yields alu­ minum metal. 000 (Section 23.3)

GIVE IT SOME THOUGHT What kind of behavior characterizes an amphoteric oxide or an amphoteric hydroxide?

1 7.6 PRECIPITATI O N AND SEPARAT I O N OF IONS Equilibrium can b e achieved starting with the substances on either side o f a 2 2 chemical equation. The equilibrium among BaS04(s), Ba + (aq), and S04 -(aq) (Equation 17.15) can be achieved starting with solid BaS04. It can also be 2 2 reached starting with solutions of salts containing Ba + and S04 -, say BaC12 and Na2S04. When these two solutions are mixed, BaS04 will precipitate if the 2 2 product of the initial ion concentrations, Q = [Ba +][S04 -], is greater than Ksp· The use of the reaction quotient, Q, to determine the direction in which a reaction must proceed to reach equilibrium was discussed earlier. (Section 15.6) The possible relationships between Q and Ksp are summarized as follows: ere

K,P, precipitation occurs until Q

= Ksp·

•

If Q

>

•

If Q

•

=

If Q < K,P, solid dissolves until Q = Ksp·

K,p, equilibrium exists (saturated solution).

17.6

- SAMPLE EXERCISE 17.15

Precipitation and Separation of Ions

751

I Predicting Whether a Precipitate Will Form

Will a precipitate form when 0.10 L of 8.0 X 10-3 M Pb(N03h is added to 0.40 L of 5.0 x 10-3 M Na2S04? SOLUTION Analyze:

The problem asks us to determine whether a precipitate will form when two salt solutions are combined.

We should determine the concentrations of all ions immediately upon mixing of the solutions and compare the value of the reaction quotient, Q, to the solubility-product constant, Ksp, for any potentially insoluble product. The possible metathesis prod­ ucts are PbS04 and NaN03. Sodium salts are quite soluble; PbS04 has a Ksp of 6.3 x 10-7 (Appendix D), however, and will preci­ pitate if the Pb 2+and S04 2- ion concentrations are high enough for Q to exceed K,p for the salt. Plan:

Solve: When the two solutions are mixed, the total volume becomes 0.10 L + 0.40 L 0.50 L. The num­ ber of moles of Pb 2+ in 0.10 L of 8.0 X 10-3 M Pb(N03h is The concentration of Pb2+ in the 0.50-L mixture is therefore

m l (0.10 L) 8.0 x 10-3

The number of moles of SO/- in 0.40 L of 5.0 x 10-3 M Na2S04 is

m (0.40 L) 5.0 X 10-3

Therefore, [SO/-] in the 0.50-L mix­ ture is

[SO42-]

We then have

Q =

=

[Pb2'")

(

=

(

�)

8.0 X 10-4 mol 0.50 L

=

t)

=

=

2.0 X 10-3 mol 0.50 L

8.0 x 10-4 mol

1.6 X 10-3 M

=

2.0 X 10-3 mol

=

4.0 X 10-3 M

[Pb 2+][SO/l = (1.6 X 10-3)(4.0 X 10-3) = 6.4 X 10-6

Because Q > Ksp, PbS04 will precipitate. - PRACTICE EXERCISE

Will a precipitate form when 0.050 L of 2.0 x 10-2 M NaF is mixed with 0.010 L of 1.0 X 10-2 M Ca(N03h? Answer: Yes, CaF2 precipitates because Q = 4.6 X 10-8 is larger than Ksp = 3.9 X 10-11

Selective Precipitation of Ions Ions can be separated from each other based on the solubilities of their salts. 2 Consider a solution containing both Ag + and Cu + . If HCl is added to the 2 solution, AgCl (K,P = 1.8 X 10-10) precipitates, while Cu + remains in solu­ tion because CuC12 is soluble. Separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or a few of the ions is called

selective precipitation. - SAMPLE EXERCISE 1 7.16

I Calculating lon Concentrations for Precipitation

A solution contains 1.0 x 10-2 M Ag + and 2.0 x 10-2 M Pb2+ When o- is added to the solution, both AgCI (Ksp = 1.8 x 10-10) and PbCI2 (Ksp = 1.7 x 10-5) precipitate from the solution. What concentration of o- is necessary to begin the precipitation of each salt? Wh1Ch salt precipitates first? SOLUTION

We are asked to determine the concentration of o- necessary to begin the precipitation from a solution containing Ag+ and Pb2+ ions, and to predict which metal chloride will begin to precipitate first.

Analyze:

Plan: We are given Ksp values for the two possible precipitates. Using these and the metal ion concentrations, we can calculate what concentration of o- ion would be necessary to begin precipitation of each. The salt requiring the lower o- ion concen­ tration will precipitate first. Solve:

For AgCI we have

Because [Ag+] = 1.0 X 10-2 M, the greatest concentration of o- that can be present without causing precipita­ tion of AgCI can be calculated from the Ksp expression:

K,r

=

[Ag+ncn

=

1.8 x 10-10

10-2ncn = 1.s x 10-10 1.8 x 10-10 = 1.8 x 10-s M 1.0 x 10-2

K,r = r1.o rcn

=

x

752

C HA PTE R 1 7

Additional Aspects of Aqueous Equilibria

Any CJ- in excess of this very small concentration will cause AgCl to pre­ cipitate from solution. Proceeding similarly for PbC12, we have

Ksp = [Pb2+] [Cn2 = 1.7 X 10-5

r2.o x 10-2Hcn2 = 1.7 x 10-5 1.7 X 10-5 = 8.5 X 10-4 2.0 X 10-2 rcn = 8.s x 10-4 = 2.9 x 10-2 M [Cn2 =

V

Thus, a concentration of CJ- in excess of 2.9 X 10-2 M will cause PbC12 to precipitate. Comparing the concentrations of o- required to precipitate each salt, we see that as o- is added to the solution, AgCI will precipitate first because it requires a much smaller concentration of CJ-. Thus, Ag + can be separated from Pb 2+ by slowly adding Cl- so [Cn is between 1.8 X 10-B M and 2.9 X 10-2 M. - PRACTICE EXERCISE

A solution consists of 0.050 M Mg 2+ and 0.020 M Cu2+ Which ion will precipitate first as OH- is added to the solution? What con­ centration of OH- is necessary to begin the precipitation of each cation? [K,p = 1.8 x 10-11 for Mg(OH)z, and K,p = 4.8 x 10-20 for Cu(OH)z.] Answer: Cu(OH)z precipitates first. Cu(OH)z begins to precipitate when [OH-] exceeds 1.5 x 10-9 M; Mg(OH)z begins to precip­ itate when [OH-] exceeds 1.9 X 10-5 M.

Sulfide ion is often used to separate metal ions because the solubilities of sulfide salts s�an a wide range and depend greatly on the pH of the solution. 2 2 Cu + and Zn , for example, can be separated by bubbling H2S gas through an 37 acidified solution. Because CuS (K,p = 6 x 10- ) is less soluble than ZnS 25 (Ksp = 2 X 10- ), CuS precipitates from an acidified solution (pH = 1) while ZnS does not (Figure 17.21 "' ): [17.27]

SElECTIVE PRECIPITATION OF IONS

Separation of ions in an aqueous solution by using a reagent that forms a precipitate with one or a few of the ions is called selective precipitation.

Remove CuS and increase pH

Solution containing Zn2+(aq) and Cu2+ (aq).

& Figure 1 7.21 Selective precipitation.

When H2S is added to a solution whose pH exceeds 0.6, CuS precipitates.

After CuS is removed, the pH is increased allowing ZnS to precipitate.

17.7

Qualitative Analysis for Metallic Elements

753

The CuS can be separated from the zn2+ solution by filtration. The CuS can then be dissolved by using a high concentration of H +, shifting the equilibrium shown in Equation 17.27 to the left.

G IVE IT SOME THOUGHT What experimental conditions will leave the smallest concentration of Cu2+ ions in solution according to Equation 17.27?

1 7.7 QUALITATIVE ANALYSIS FOR METALLIC ELEMENTS In this chapter we have seen several examples of equilibria involving metal ions in aqueous solu­ tion. In this final section we look briefly at how solubility equilibria and complex-ion formation At3+ Fe2+ Fe3+ Co2+, Ni2+ eel+ Zn2+ Mn2+ can be used to detect the presence of particular iBa2+, Ca2+, �1 metal ions in solution. Before the development of modern analytical instrumentation, it was neces­ Na+, K+, NH/ sary to analyze mixtures of metals in a sample by so-called wet chemical methods. For example, a Add 6 M HCI metallic sample that might contain several metal­ Precipitate lic elements was dissolved in a concentrated acid solution. This solution was then tested in a sys­ Group 1-Insoluble tematic way for the presence of various metal ions. chlorides: AgCI, Remaining cations Qualitative analysis determines only the Hg2Cl:z, PbCI2 presence or absence of a particular metal ion, whereas quantitative analysis determines how F======l Add H2S, Precipitate much of a given substance is present. Wet meth­ 0.2 M HCI ods of qualitative analysis have become less im­ portant as a means of analysis. They are Group 2-Acid-insoluble frequently used in general chemistry laboratory Remaining cations sulfides: CuS, Bi2� CdS, programs, however, to illustrate equilibria, to PbS, HgS, A5z� � SnSz teach the properties of common metal ions in so­ F==-----==-1 Add (NH4)zS, lution, and to develop laboratory skills. Typically, Precipitate pH 8 such analyses proceed in three stages: {I) The ions are separated into broad groups on the basis of Group 3-Base-insoluble sulfides solubility properties. (2) The individual ions with­ Remaining cations and hydroxides: Al(OHlJ, Fe(OH)J­ in each group are then separated by selectively Cr(OH)J- ZnS, NiS, CoS, MnS dissolving members in the group. (3) The ions are then identified by means of specific tests. Precipitate A scheme in general use divides the common cations into five groups, as shown in Figure 17.22 �. The order of addition of reagents is im­ Group 4-insoluble Group 5-Alkali metal phosphates: Ba3(P04):z, portant. The most selective separations-those ions and ammonium ion: Ca (P04):z, MgN�04 3 Na+, K+, NH/ that involve the smallest number of ions-are carried out first. The reactions that are used must "' Figure 1 7.22 Qualitative analysis. proceed so far toward completion that any concentration of cations remaining in A flowchart showing the separation of the solution is too small to interfere with subsequent tests. Let's take a closer look cations into groups as a part of a at each of these five groups of cations, briefly examining the logic used in this common scheme for identifying cations. qualitative analysis scheme. + + + 2 1. Insoluble chlorides: Of the common metal ions, only Ag , Hg/ , and Pb

.

form insoluble chlorides. When dilute HCl is added to a mixture of cations, therefore, only AgCI, Hg2Cl2, and PbCI2 will precipitate, leaving the other cations in solution. The absence of a precipitate indicates that the starting solution contains no Ag +, Hg 22+, or Pb 2+ .

754

C HA PTE R 17

Additional Aspects of Aqueous Equilibria 2.

Acid-insoluble sulfides: After any insoluble chlorides have been removed, the remaining solution, now acidic, is treated with H2S. Only the most in­ soluble metal sulfides-CuS, Bi2S3, CdS, PbS, HgS, As2S3, Sb2S3, and SnS2can precipitate. (Note the very small values of Ksp for some of these sulfides in Appendix D.) Those metal ions whose sulfides are somewhat more soluble--for example, ZnS or NiS-remain in solution.

3.

Base-insoluble sulfides and hydroxides: After the solution is filtered to remove any acid-insoluble sulfides, the remaining solution is made slightly basic, 2 and (NH4hS is added. In basic solutions the concentration of 5 - is higher than in acidic solutions. Thus, the ion products for many of the more soluble sulfides are made to exceed their Ksp values and precipitation 2 occurs. The metal ions precipitated at this stage are AJ3+, Cr 3+, Fe 3+, zn +, 2 2 2 3 3 3 Ni +, Co +, and Mn +. (Actually, the AJ +, Fe +, and Cr + ions do not form insoluble sulfides; instead they are precipitated as insoluble hydro­ xides at the same time.)

4.

Insoluble phosphates: At this point the solution contains only metal ions from periodic table groups lA and 2A. Adding (NH4hHP04 to a basic solu­ 2 2 2 2 tion precipitates the group 2A elements Mg +, Ca +, Sr +, and Ba + because these metals form insoluble phosphates.

5.

The alkali metal ions and NH4+: The ions that remain after removing the in­ soluble phosphates form a small group. We can test the original solution for each ion individually. A flame test can be used to determine the pres­ ence of K+, for example, because the flame turns a characteristic violet color if K+ is present. GIVE IT SOME THOUGHT

If a precipitate forms upon addition of HCI to an aqueous solution, what conclusions can you draw about the contents of the solution?

Additional separation and testing is necessary to determine which ions are present within each of the groups. Consider, for example, the ions of the insolu­ ble chloride group. The precipitate containing the metal chlorides is boiled in water. The PbC12 is relatively soluble in hot water, whereas AgCl and Hg2Cl2 are not. The hot solution is filtered, and a solution of Na2Cr04 is added to the 2 filtrate. If Pb + is present, a yellow precipitate of PbCr04 forms. The test for Ag+ consists of treating the metal chloride precipitate with dilute ammonia. Only Ag+ forms an ammonia complex. If AgCI is present in the precipitate, it will dissolve in the ammonia solution. [17.28] After treatment with ammonia, the solution is filtered and the filtrate made acidic by adding nitric acid. The nitric acid removes ammonia from solution by forming NH4+, thus releasing Ag+, which re-forms the AgCI precipitate. Ag(NH 3h+(aq)

+

Cr-(aq)

+ 2 H+(aq)

;;::::::= AgCI(s) + 2 NH4+ (aq)

[17.29]

The analyses for individual ions in the acid-insoluble and base-insoluble sulfides are a bit more complex, but the same general principles are involved. The detailed procedures for carrying out such analyses are given in many labo­ ratory manuals.

Qualitative Analysis for Metallic Elements

17.7

- SAMPLE INTEGRATIVE EXERCISE

I Putting Concepts Together

A sample of 1.25 L of HCl gas at 21 •c and 0.950 atm is bubbled through 0.500 L of 0.150 M NH 3 solution. Calculate the pH of the resulting solution assuming that all the HCl dissolves and that the volume of the solution remains 0.500 L. SOLUTION

The number of moles of HCl gas is calculated from the ideal-gas law.

n=

PV

RT

=

(0.950 atm)(1.25 L) = 0·0492 mol HCI (0.0821 L-atmjmol-K)(294 K)

The number of moles of NH 3 in the solution is given by the product of the volume of the solution and its concentration. Moles NH3

=

(0.500 L)(0.150 mol NH 3/L)

=

0.0750 mol NH 3

The acid HCI and base NH3 react, transferring a proton from HCI to NH3, pro­ ducing NH4 + and Cl- ions. HCl(g) + NH 3(aq) --+ NH4+(aq) + cqaq) To determine the pH of the solution, we first calculate the amount of each reactant and each product present at the completion of the reaction. HCI(g) Before addition

0.0492 mol

O mol

Addition

O mol

O mol

0.0492 mol

0.0492 mol

0.0750 mol

After addition

O mol

0.0258 mol

Thus, the reaction produces a solution containing a mixture of NH3, NH4+, and o-. The NH3 is a weak base (Kb 1.8 x 10-5), NH4+ is its conjugate acid, and o- is neither acidic nor basic. Consequently, the pH depends on [NH3] and [NH4+]. =

[NH3 ] =

0.0258 mol NH 3 = 0.0516 M 0.500 L soln

[NH4 +] =

0.0492 mol NH4 + = 0.0984 M O.SOO L soln

We can calculate the pH using either Kb for NH3 or Ka for NH4+ Using the Kb expres­ sion, we have

Initial

-

0.0516 M

Change

-

-x M (0.0516 - x) M

Equilibrium

Kb

=

[NH4+][oH-] [NH 3]

=

-

0.0984 M

0

+x M

+x M

(0.0984 + x) M

xM

(0.0984 + x)(x) (0.0984)x "' (0.0516 - x) 0.0516

=

1.8 X 1O_5

(0.0516)(1.8 X 10-5) = 9.4 X 10-6 M 0.0984 Hence, pOH = -log(9.4 X 10-6) = 5.03 and pH = 14.00 - pOH = 14.00 - 5.03 = 8.97. X

= [OH-] =

755

756

C HA PTE R 17

Additional Aspects of Aqueous Equilibria

C H APTER RE V IEW

S U MMARY AND KEY TERMS

In this chapter we have considered sever­ al types of important equilibria that occur in aqueous so­ lution. Our primary emphasis has been on acid-base equilibria in solutions containing two or more solutes and on solubility equilibria. The dissociation of a weak acid or weak base is repressed by the presence of a strong electrolyte that provides an ion common to the equilibri­ um. This phenomenon is called the common-ion effect. Section 1 7. 1

A particularly important type of acid-base mixture is that of a weak conjugate acid-base pair. Such mixtures function as buffered solutions (buffers). Addi­ tion of small amounts of a strong acid or a strong base to a buffered solution causes only small changes in pH because the buffer reacts with the added acid or base. (Strong acid-strong base, strong acid-weak base, and weak acid-strong base reactions proceed essentially to comple­ tion.) Buffered solutions are usually prepared from a weak acid and a salt of that acid or from a weak base and a salt of that base. Two important characteristics of a buffered solution are its buffer capacity and its pH. The pH can be calculated using Ka or Kb. The relationship between pH, pK0, and the concentrations of an acid and its conjugate base can be expressed by the Henderson-Hasselbalch . [base] equahon: pH = pK0 + log . [acid] Section 1 7.2

Section 1 7.3 The plot of the pH of an acid (or base) as a function of the volume of added base (or acid) is called a pH titration curve. Titration curves aid in selecting a proper pH indicator for an acid-base titration. The titra­ tion curve of a strong acid-strong base titration exhibits a large change in pH in the immediate vicinity of the equivalence point; at the equivalence point for this titra­ tion, pH = 7. For strong acid-weak base or weak acid-strong base titrations, the pH change in the vicinity of the equivalence point is not as large. Furthermore, the pH at the equivalence point is not 7 in either of these cases. Rather, it is the pH of the salt solution that results from the neutralization reaction. It is possible to calculate the pH at any point of the titration curve by first consid­ ering the effects of the reaction between the acid and base on solution concentrations and then examining equilibria involving remaining solute species. Section 1 7.4 The equilibrium between a solid com­ pound and its ions in solution provides an example of heterogeneous equilibrium. The solubility-product

(or simply the solubility product), Ksp, is an equilibrium constant that expresses quantitatively the ex­ tent to which the compound dissolves. The Ksp can be used to calculate the solubility of an ionic compound, and the solubility can be used to calculate Ksp· constant

Section 1 7.5 Several experimental factors, including temperature, affect the solubilities of ionic compounds in water. The solubility of a slightly soluble ionic compound is decreased by the presence of a second solute that fur­ nishes a common ion (the common-ion effect). The solu­ bility of compounds containing basic anions increases as the solution is made more acidic (as pH decreases). Salts with anions of negligible basicity (the anions of strong acids) are unaffected by pH changes. The solubility of metal salts is also affected by the pres­ ence of certain Lewis bases that react with metal ions to form stable complex ions. Complex-ion formation in aqueous solution involves the displacement by Lewis bases (such as NH3 and CN-) of water molecules attached to the metal ion. The extent to which such complex forma­ tion occurs is expressed quantitatively by the formation constant for the complex ion. Amphoteric oxides and hydroxides are those that are only slightly soluble in water but dissolve on addition of either acid or base. Acid-base reactions involving the OH- or H20 groups bound to the metal ions give rise to the amphoterism. Section 1 7.6 Comparison of the ion product, Q, with the value of Ksp can be used to judge whether a precipi­ tate will form when solutions are mixed or whether a slightly soluble salt will dissolve under various con­ ditions. Precipitates form when Q > Ksp· Ions can be separated from each other based on the solubilities of their salts. Section 1 7.7 Metallic elements vary a great deal in the solubilities of their salts, in their acid-base behavior, and in their tendencies to form complex ions. These differ­ ences can be used to separate and detect the presence of metal ions in mixtures. Qualitative analysis determines the presence or absence of species in a sample, whereas quantitative analysis determines how much of each species is present. The qualitative analysis of metal ions in solution can be carried out by separating the ions into groups on the basis of precipitation reactions and then analyzing each group for individual metal ions.

Visualizing Concepts

757

KEY SKILLS •

Describe the common-ion effect.

•

Explain how a buffer functions.

•

Calculate the pH of a buffer solution.

•

Calculate the pH of a buffer after the addition of small amounts of a strong acid or a strong base.

•

Calculate the pH at any point in an acid-base titration of a strong acid and strong base.

•

Calculate the pH at any point in a titration of a weak acid with a strong base or a weak base with a strong acid.

•

Understand the differences between the titration curves for a strong acid-strong base titration and those when either the acid or base is weak.

•

Calculate K,P from molar solubility and molar solubility from Ksp·

•

Calculate molar solubility in the presence of a common ion.

•

Predict the effect of pH on solubility.

•

Predict whether a precipitate will form when solutions are mixed by comparing Q and Ksp ·

•

Calculate the ion concentrations required to begin precipitation.

•

Explain the effect of complex-ion formation on solubility.

KEY EQUATIONS •

pH

=

a

pK + log

[base] [acid]

[17.9]

The Henderson-Hasselbalch equation, used to calculate the pH of a buffer from the concentrations of a conjugate acid-base pair

VISUALIZING CONCEPTS 17.1

The following boxes represent aqueous solutions con­ taining a weak acid, HX, and its conjugate base, x-. Water molecules and cations are not shown. Which so­ lution has the highest pH? Explain. [Section 17.1] = HX

17.2

The beaker on the right contains 0.1 M acetic acid solu­ tion with methyl orange as an indicator. The beaker on the left contains a mixture of 0.1 M acetic acid and 0.1 M sodium acetate with methyl orange. (a) Using Figure

16.7, estimate the pH of each solution, and explain the difference. (b) Which solution is better able to main­ tain its pH when small amounts of NaOH are added? Explain. [Sections 17.1 and 17.2]

758

CHA PTER

17

Additional Aspects o f Aqueous Equilibria

17.3 A buffer contains a weak acid, HX, and its conjugate base. The weak acid has a pKa of 4.5, and the buffer solution has a pH of 4.3. Without doing a calculation, predict whether [HX] = [X-], [HX] > [X-], or [HX] < [X-]. Explain. [Section 17.2] 17.4 The drawing on the left represents a buffer composed of equal concentrations of a weak acid, HX, and its conju­ gate base, x-. The heights of the columns are propor­ tional to the concentrations of the components of the buffer. (a) Which of the three drawings, (1), (2), or (3), represents the buffer after the addition of a strong acid? (b) Which of the three represents the buffer after the addition of a strong base? (c) Which of the three repre­ sents a situation that cannot arise from the addition of either an acid or a base? [Section 1 7.2]

17.7 Equal volumes of two acids are titrated with 0.10 M NaOH resulting in the two titration curves shown in the following figure. (a) Which curve corresponds to the more concentrated acid solution? (b) Which cor­ responds to the acid with the largest Ka? Explain. [Section 17.3]

12

:a

10

8 6 4

10

HX x-

HX x-

HX x-

HX x-

(1)

(2)

(3)

17.5 The following drawings represent solutions at various stages of the titration of a weak acid, HA, with NaOH. (The Na+ ions and water molecules have been omit­ ted for clarity. ) To which of the following regions of the titration curve does each drawing correspond: (a) before addition of NaOH, (b) after addition of NaOH but before equivalence point, (c) at equivalence point, (d) after equivalence point? [Section 17.3]

• = OW

= HA

(ii)

(i)

(iii)

(iv)

17.6 Match the following descriptions of titration curves with the diagrams: (a) strong acid added to strong base, (b) strong base added to weak acid, (c) strong base added to strong acid, (d) strong base added to poly­ protic acid. [Section 17.3]

mL titrant

mL titrant

mL titrant

(i)

(ii)

(iii)

(iv)

50

17.8 The following drawings represent saturated solutions of three ionic compounds of silver-AgX, AgY, and AgZ. (Na+ cations, which might also be present for charge balance, are not shown.) Which compound has the smallest K,p? [Section 17.4] =

AgX

x-, y-, or z-

AgY

AgZ

17.9 The figures below represent the ions in a saturated aqueous solution of the slightly soluble ionic com­ pound MX: MX(s) :;====' M 2+(aq) + X 2-(aq). (Only the M 2+ and X2- ions are shown.) (a) Which figure repre­ sents a solution prepared by dissolving MX in water? (b) Which figure represents a solution prepared by dis­ solving MX in a solution containing Na2X? (c) If xz- is a basic anion, which figure represents a saturated solu­ tion with the lowest pH? (d) If you were to calculate the Ksp for MX, would you get the same value in each of the three scenarios? Why or why not? [Sections 17.4 and 17.5]

xz ---. • Mz+-. • • • • • (a)

mL titrant

20 30 40 mL NaOH

D (b)

� t_;__ij (c)

17.10 The following graphs represent the behavior of BaC03 under different circumstances. In each case the vertical axis indicates the solubility of the BaC03 and the

Exercises

horizontal axis represents the concentration of some other reagent. (a) Which graph represents what happens to the solubility of BaC0 3 as HN03 is added? (b) Which graph represents what happens to the BaC03 solubility as Na2C03 is added? (c) Which represents what hap­ pens to the BaC03 solubility as NaN03 is added? [Section 17.5]

Cone 17.11

Cone

17.12

759

Three cations, Ni2+, Cu2+, and Ag+, are separated using two different precipitating agents. Based on Figure 17.22, what two precipitating agents could be used? Using these agents, indicate which of the cations is A, which is B, and which is C. [Section 17.7] Cation A removed

Mixture of cations A,B,C

Cations B,C

Cation B removed

Cation A

Cone

What is the name given to the kind of behavior demon­ strated by a metal hydroxide in this graph? [Section 17.5]

Cation A •

Cation B e

Cation C •

pH EXERCISES Common-Ion Effect 17.13

(a) What is the common-ion effect? (b) Give an example of a salt that can decrease the ionization of HN02 in solution.

17.14

(a) Consider the equilibrium B(aq) + H 20(/) = HB+(aq) + OH-(aq). Using Le Chatelier's principle, explain the effect of the presence of a salt of HB+ on the ionization of B. (b) Give an example of a salt that can decrease the ionization of NH3 in solution.

17.15

Use information from Appendix D to calculate the pH of (a) a solution that is 0.060 M in potassium propionate (C2H5COOK or KC3H50 2) and 0.085 M in propionic acid (C2H5COOH or HC3 H502); (b) a solution that is 0.075 M in trimethylamine, (CH3hN, and 0.10 M in trimethylam­ monium chloride, (CH3hNHCl; (c) a solution that is made by mixing 50.0 mL of 0.15 M acetic acid and 50.0 mL of 0.20 M sodium acetate.

17.16

17.17

17.18

Use information from Appendix D to calculate the pH of (a) a solution that is 0.150 M in sodium formate (HCOONa) and 0.200 M in formic acid (HCOOH); (b) a solution that is 0.210 M in pyridine (C5H5N) and 0.350 M in pyridinium chloride (C5H5NHC1); (c) a solu­ tion that is made by combining 125 mL of 0.050 M hy­ drofluoric acid with 50.0 mL of 0.10 M sodium fluoride. (a) Calculate the percent ionization of 0.0075 M butanoic acid (Ka = 1.5 X 10-5). (b) Calculate the percent ioniza­ tion of 0.0075 M butanoic acid in a solution containing 0.085 M sodium butanoate. (a) Calculate th=f ercent ionization of 0.085 M lactic acid

1.4 X 1 0 ) (b) Calculate the percent ionization of 0.095 M lactic acid in a solution containing 0.0075 M sodium lactate. (Ka

=

.

Buffers 17.19

17.20

Explain why a mixture of CH3COOH and CH3C00Na can act as a buffer while a mixture of HCl and NaCl cannot. Explain why a mixture formed by mixing 100 mL of 0.100 M CH3COOH and 50 mL of 0.100 M NaOH will act as a buffer.

17.21

(a) Calculate the pH of a buffer that is 0.12 M in lactic acid and 0.11 M in sodium lactate. (b) Calculate the pH of a buffer formed by mixing 85 mL of 0.13 M lactic acid with 95 mL of 0.15 M sodium lactate.

17.22

(a) Calculate the pH of a buffer that is 0.105 M in NaHC03 and 0.125 M in Na2C03 . (b) Calculate the pH of a solution formed by mixing 65 mL of 0.20 M NaHC03 with 75 mL of 0.15 M Na2C03 .

760

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17

Additional Aspects o f Aqueous Equilibria

17.23 A buffer is prepared by adding 20.0 g of acetic acid

(CH3COOH) and 20.0 g of sodium acetate (CH3C00Na) to enough water to form 2.00 L of solution. (a) Deter­ mine the pH of the buffer. (b) Write the complete ionic equation for the reaction that occurs when a few drops of hydrochloric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of sodium hydroxide solution are added to the buffer.

17.24 A buffer is prepared by adding 7.00 g of ammonia (NH3)

and 20.0 g of ammonium chloride (NH4CI) to enough water to form 2.50 L of solution. (a) What is the pH of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equa­ tion for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

17.25 How many moles of sodium hypobromite (NaBrO)

should be added to 1.00 L of 0.050 M hypobromous acid (HBrO) to form a buffer solution of pH 9.15? Assume that no volume change occurs when the NaBrO is added.

17.26 How many grams of sodium lactate [CH3CH(OH)COONa

or NaC�503] should be added to 1.00 L of 0.150 M lactic acid [CH3CH(OH)COOH or HC3H503] to form a buffer solution with pH 4.00? Assume that no volume change oc­ curs when the sodium lactate is added.

17.27 A buffer solution contains 0.10 mol of acetic acid and

0.13 mol of sodium acetate in 1 .00 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the

addition of 0.02 mol of KOH? (c) What is the pH of the buffer after the addition of 0.02 mol of HN03? 17.28 A buffer solution contains 0.10 mol of propionic acid

(C2H5COOH) and 0.13 mol of sodium propionate (C2H5COONa) in 1.50 L. (a) What is the pH of this buffer? (b) What is the pH of the buffer after the addi­ tion of 0.01 mol of NaOH? (c) What is the pH of the buffer after the addition of O.Dl mol of HI?

17.29 (a) What is the ratio of HC03- to H2C03 in blood of

pH 7.4? (b) What is the ratio of HC03- to H2C03 in an exhausted marathon runner whose blood pH is 7.1? 2 17.30 A buffer, consisting of H2P0 4- and HP04 -, helps con­ trol the pH of physiological fluids. Many carbonated soft drinks also use this buffer system. What is the pH of a soft drink in which the major buffer ingredients are 6.5 g of NaH2P04 and 8.0 g of Na2HP04 per 355 mL of solution? 17.31 You have to prepare a pH 3.50 buffer, and you have the

following 0.10 M solutions available: HCOOH, CH3COOH, H�04, HCOONa, CH3COONa, and NaH2P04. Which solutions would you use? How many milliliters of each solution would you use to make ap­ proximately a liter of the buffer?

17.32 You have to prepare a pH 4.80 buffer, and you have the

following 0.10 M solutions available: formic acid, sodi­ um formate, propionic acid, sodium propionate, phos­ phoric acid, and sodium dihydrogen phosphate. Which solutions would you use? How many milliliters of each solution would you use to make approximately a liter of the buffer?

Acid-Base Titrations 17.33 The accompanying graph shows the titration curves for two monoprotic acids. (a) Which curve is that of a

strong acid? (b) What is the approximate pH at the equivalence point of each titration? (c) How do the original concentrations of the two acids compare if 40.0 mL of each is titrated to the equivalence point with 0.100 M base? 14 r-�-r----r--r-,

strong base differ from titration of a weak, monoprotic acid with a strong base with respect to the following: (a) quantity of base required to reach the equivalence point, (b) pH at the beginning of the titration, (c) pH at the equivalence point, (d) pH after addition of a slight excess of base, (e) choice of indicator for determining the equivalence point?

17.35 Predict whether the equivalence point of each of the

following titrations is below, above, or at pH 7: (a) NaHC03 titrated with NaOH, (b) NH3 titrated with HCI, (c) KOH titrated with HBr.

12 10 pH

17.34 How does titration of a strong, monoprotic acid with a

8

17.36 Predict whether the equivalence point of each of the fol­ lowing titrations is below, above, or at pH 7: (a) formic

6

acid titrated with NaOH, (b) calcium hydroxide titrated with perchloric acid, (c) pyridine titrated with nitric acid.

4

M in concentration, are titrated with 0.100 M NaOH. The pH at the equiv­ alence point for HX is 8.8, and that for HY is 7.9. (a) Which is the weaker acid? (b) Which indicators in Figure 16.7 could be used to titrate each of these acids?

17.37 Two monoprotic acids, both 0.100

10 20 30 40 50 60 mL NaOH

Exercises 17.38

17.39

17.40

17.41

Assume that 30.0 mL of a 0.10 M solution of a weak base B that accepts one proton is titrated with a 0.10 M solu­ tion of the monoprotic strong acid HX. (a) How many moles of HX have been added at the equivalence point? (b) What is the predominant form of B at the equi­ valence point? (c) What factor determines the pH at the equivalence point? (d) Which indicator, phenol­ phthalein or methyl red, is likely to be the better choice for this titration? How many milliliters of 0.0850 M NaOH are required to titrate each of the following solutions to the equivalence point: (a) 40.0 mL of 0.0900 M HN03, (b) 35.0 mL of 0.0850 M CH3COOH, (c) 50.0 mL of a solution that con­ tains 1.85 g of HCl per liter? How many milliliters of 0.105 M HCl are needed to titrate each of the following solutions to the equivalence point: (a) 45.0 mL of 0.0950 M NaOH, (b) 22.5 mL of 0.118 M NH:v (c) 125.0 mL of a solution that contains 1.35 g of NaOH per liter? A 20.0-mL sample of 0.200 M HBr solution is titrated with 0.200 M NaOH solution. Calculate the pH of the solution after the following volumes of base have been added: (a) 15.0 mL, (b) 19.9 mL, (c) 20.0 mL, (d) 20.1 mL, (e) 35.0 mL.

761

17.42

A 30.0-mL sample of 0.150 M KOH is titrated with 0.125 M HCI04 solution. Calculate the pH after the fol­ lowing volumes of acid have been added: (a) 30.0 mL, (b) 35.0 mL, (c) 36.0 mL, (d) 37.0 mL, (e) 40.0 mL.

17.43

A 35.0-mL sample of 0.150 M acetic acid (CH3COOH) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added: (a) 0 mL, (b) 17.5 mL, (c) 34.5 mL, (d) 35.0 mL, (e) 35.5 mL, (f) 50.0 mL.

17.44

Consider the titration of 30.0 mL of 0.030 M NH3 with 0.025 M HCI. Calculate the pH after the following vol­ umes of titrant have been added: (a) 0 mL, (b) 10.0 mL, (c) 20.0 mL, (d) 35.0 mL, (e) 36.0 mL, (f) 37.0 mL.

17.45

Calculate the pH at the equivalence point for titrating 0.200 M solutions of each of the following bases with 0.200 M HBr: (a) sodium hydroxide (NaOH), (b) hy­ droxylamine (NH20H), (c) aniline (C6HsNH2l·

17.46

Calculate the pH at the equivalence point in titrating 0.100 M solutions of each of the following with 0.080 M NaOH: (a) hydrobromic acid (HBr), (b) lactic acid [CH 3CH(OH)COOH], (c) sodium hydrogen chromate (NaHCr04).

Solubility Equilibria and Factors Affecting Solubility 17.47

17.48

17.49

17.50

17.51

(a) Why is the concentration of undissolved solid not explicitly included in the expression for the solubility­ product constant? (b) Write the expression for the solubility-product constant for each of the following strong electrolytes: Agl, SrS04, Fe(OH)z, and Hg2Br2. (a) Explain the difference between solubility and solubility-product constant. (b) Write the expression for the solubility-product constant for each of the following ionic compounds: MnCO:v Hg(OH)z, and Cu3(P04)z. (a) If the molar solubility of CaF 2 at 35 oc is 1.24 X 10-3 moljL, what is Ksp at this temperature? (b) It is found that 1.1 x 10-2 g of SrF2 dissolves per 100 mL of aqueous solution at 25 oc. Calculate the solubility product for SrF2. (c) The Ksp of Ba(I03)z at 25 oc is 6.0 X 10-IO What is the molar solubility of Ba(I03)z? (a) The molar solubility of PbBr2 at 25 oc is

1.0 X 10-2 moljL. Calculate Ksp. (b) If 0.0490 g of Ag!03 dissolves per liter of solution, calculate the solubility­ product constant. (c) Using the appropriate K,p value from Appendix 0, calculate the solubility of Cu(OH)z in grams per liter of solution. A 1.00-L solution saturated at 25 oc with calcium ox­ alate (CaC204) contains 0.0061 g of CaC 204. Calculate the solubility-product constant for this salt at 25 oe,

17.52

17.53

17.54

A 1.00-L solution saturated at 25 oc with lead(II) iodide contains 0.54 g of Pbl2. Calculate the solubility-product constant for this salt at 25 oc. Using Appendix 0, calculate the molar solubility of AgBr in (a) pure water, (b) 3.0 X 10-2 M AgN03 solu­ tion, (c) 0.10 M NaBr solution. Calculate the solubility of LaF3 in grams per liter in pure water, (b) 0.010 M KF solution, (c) 0.050 M LaC13 solution. (a)

17.55

Calculate the solubility of Mn(OHh in grams per liter when buffered at pH (a) 7.0, (b) 9.5, (c) 11.8.

17.56

Calculate the molar solubility of Fe(OH)z when buffered at pH (a) 8.0, (b) 10.0, (c) 12.0.

17.57

Which of the following salts will be substantially more soluble in acidic solution than in pure water: (a) ZnC03, (b) ZnS, (c) Bil3, (d) AgCN, (e) Ba3(P04)z?

17.58

For each of the following slightly soluble salts, write the net ionic equation, if any, for reaction with acid: (a) MnS, (b) PbF2, (c) AuCI3, (d) Hg2C204, (e) CuBr.

17.59

From the value of Kr listed in Table 17.1, calculate the concentration of Cu2+ in 1.0 L of a solution that contains a total of 1 X 10-3 mol of copper(II) ion and that is 0.10 M in NH3 .

762

CHA PTER

17

Additional Aspects o f Aqueous Equilibria

17.60 To what final concentration of NH3 must a solution

be adjusted to just dissolve 0.020 mol of NiC 204 (Ksp = 4 x 10-10) in 1.0 L of solution? (Hint: You can ne­ 2 glect the hydrolysis of C 204 - because the solution will be quite basic.)

17.62 Using the value of Ksp for Ag2S, Ka 1 and Ka2 for H2S, and

K1 = 1.1 x 105 for AgC12-, calculate the equilibrium constant for the following reaction:

Ag2S(s)

+ 4 Cr-(aq) + 2 H+(aq) �

2 AgCI2 -(aq)

+ H2S(aq)

17.61 By using the values of K,p for Agl and Kt for Ag(CNh-,

calculate the equilibrium constant for the reaction Agl(s)

+ 2 CW(aq) �

Ag(CNh -(aq)

+ r-(aq)

Precipitation; Qualitative Analysis 17.63

(a) Will Ca(OHh precipitate from solution if the pH of a 0.050 M solution of CaCI2 is adjusted to 8.0? (b) Will Ag2S04 precipitate when 100 mL of 0.050 M AgN03 is 2 mixed with 10 mL of 5.0 x 10- M Na2S04 solution?

17.64

(a) Will Co(OHh precipitate from solution if the pH of a

Which metal ions discussed in Section 17.7 are possibly present? Which are definitely absent within the limits of these tests? 17.70 An unknown solid is entirely soluble in water. On addi­

tion of dilute HCl, a precipitate forms. After the precipi­ tate is filtered off, the pH is adjusted to about 1 and H2S is bubbled in; a precipitate again forms. After filtering off this precipitate, the pH is adjusted to 8 and H2S is again added; no precipitate forms. No precipitate forms upon addition of (NH4hHP04. The remaining solution shows a yellow color in a flame test. Based on these ob­ servations, which of the following compounds might be present, which are definitely present, and which are def­ initely absent: CdS, Pb(N03):u HgO, ZnS04, Cd(N03)u and Na2S04?

0.020 M solution of Co(N03h is adjusted to 8.5? (b) Will Ag!03 precipitate when 20 mL of 0.010 M AgN03 is mixed with 10 mL of O.D15 M Na!03? (Ksp of Ag!03 is 3.1 X 10-8.)

17.65 Calculate the minimum pH needed to precipitate

2 Mn(OHh so completely that the concentration of Mn + is less than 1 ,.,_g per liter [1 part per billion (ppb)].

17.66 Suppose that a 10-mL sample of a solution is to be tested

for Cl- ion by addition of 1 drop (0.2 mL) of 0.10 M AgN03. What is the minimum number of grams of Cr that must be present for AgCl(s) to form?

4 solution contains 2.0 X 10- M Ag+ and 3 2 10- M Pb + If Nal is added, will Agl (Ksp = 8.3 X 10-17) or Pbl2 (Ksp = 7.9 X 10-9) precipi­ tate first? Specify the concentration of r- needed to begin precipitation.

17.67 A

1.5

x

17.68 A solution of Na2S04 is added dropwise to a solution

2 2 that is 0.010 M in Ba + and 0.010 M in Sr + (a) What concentration of so/- is necessary to begin precipita­ tion? (Neglect volume changes. BaS04: K,p = 1.1 X 10-10; SrS04: K,p = 3.2 x 10-7.) (b) Which cation r.recipi­ tates first? (c) What is the concentration of S04 - when the second cation begins to precipitate?

17.69 A solution containing an unknown number of metal

ions is treated with dilute HCl; no precipitate forms. The pH is adjusted to about 1, and H2S is bubbled through. Again, no precipitate forms. The pH of the solution is then adjusted to about 8. Again, H2S is bubbled through. This time a precipitate forms. The filtrate from this solu­ tion is treated with (NH4hHP04. No precipitate forms.

17.71 In the course of various qualitative analysis procedures,

2 the following mixtures are encountered: (a) zn + and 2 2 Cd +, (b) Cr(OH)J and Fe(OH)J, (c) Mg + and K+, 2 (d) Ag + and Mn + Suggest how each mixture might be separated.

17.72 Suggest how the cations in each of the following solution

2 mixtures can be separated: (a) Na+ and Cd +, (b) Cu2+ and Mg 2+, (c) Pb 2+ and Al3+, (d) Ag+ and Hg 2+

17.73

(a) Precipitation of the group 4 cations (Figure 17.22) re­

quires a basic medium. Why is this so? (b) What is the most significant difference between the sulfides precipi­ tated in group 2 and those precipitated in group 3? (c) Suggest a procedure that would serve to redissolve the group 3 cations following their precipitation.

17.74 A student who is in a great hurry to finish his laboratory

work decides that his qualitative analysis unknown contains a metal ion from the insoluble phosphate group, group 4 (Figure 17.22). He therefore tests his sample directly with (NH4hHP04, skipping earlier tests for the metal ions in groups 1, 2, and 3. He observes a precipitate and concludes that a metal ion from group 4 is indeed present. Why is this possibly an erroneous conclusion?

Additional Exercises

763

ADD ITIONAL EXERCI SES 17.75

Derive an equation similar to the Henderson­ Hasselbalch equation relating the pOH of a buffer to the pK0 of its base component.

17.76

Benzenesulfonic acid is a monoprotic acid with pK, = 2.25. Calculate the pH of a buffer composed of 0.150 M benzenesulfonic acid and 0.125 M sodium benzensulfonate.

17.77

Furoic acid (HCsH303) has a Ka value of 6.76 X 10-4 at oc. Calculate the pH at 25 oc of (a) a solution formed by adding 25.0 g of furoic acid and 30.0 g of sodium furoate (NaC5H303) to enough water to form 0.250 L of solution; (b) a solution formed by mixing 30.0 mL of 0.250 M HC5H303 and 20.0 mL of 0.22 M NaC5H303 and diluting the total volume to 125 mL; (c) a solution pre­ pared by adding 50.0 mL of 1.65 M NaOH solution to 0.500 L of 0.0850 M HC5H303. 25

17.78

The acid-base indicator bromcresol green is a weak acid. The yellow acid and blue base forms of the indi­ cator are present in equal concentrations in a solution when the pH is 4.68. What is the pKa for bromcresol green?

17.79

Equal quantities of 0.010 M solutions of an acid HA and a base B are mixed. The pH of the resulting solution is 9.2. (a) Write the equilibrium equation and equilibrium­ constant expression for the reaction between HA and B. (b) If Ka for HA is 8.0 x 10-5, what is the value of the equilibrium constant for the reaction between HA and B? (c) What is the value of K0 for B?

17.80

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make 1.00 L of solution. Buffer A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCI. (d) Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCI. (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

17.81

A biochemist needs 750 mL of an acetic acid-sodium acetate buffer with pH 4.50. Solid sodium acetate (CH3COONa) and glacial acetic acid (CH3COOH) are available. Glacial acetic acid is 99% CH3COOH by mass and has a density of 1.05 gfmL. If the buffer is to be 0.15 M in CH3COOH, how many grams of CH3C00Na and how many milliliters of glacial acetic acid must be used?

17.82

A sample of 0.2140 g of an unknown monoprotic acid was dissolved in 25.0 mL of water and titrated with 0.0950 M NaOH. The acid required 27.4 mL of base to

reach the equivalence point. (a) What is the molar mass of the acid? (b) After 15.0 mL of base had been added in the titration, the pH was found to be 6.50. What is the Ka for the unknown acid? 17.83

Show that the pH at the halfway point of a titration of a weak acid with a strong base (where the volume of added base is half of that needed to reach the equiva­ lence point) is equal to pK. for the acid.

17.84

Potassium hydrogen phthalate, often abbreviated KHP, can be obtained in high purity and is used to determine the concentrations of solutions of strong bases. Strong bases react with the hydrogen phthalate ion as follows: 2 HP-(aq) + OH-(aq) ----> H20(1) + P -(aq) The molar mass of KHP is 204.2 g/mol and Ka for the HP- ion is 3.1 x 10-6 . (a) If a titration experiment begins with 0.4885 g of KHP and has a final volume of about 100 mL, which indicator from Figure 16.7 would be most appropriate? (b) If the titration required 38.55 mL of NaOH solution to reach the end point, what is the concentration of the NaOH solution?

17.85

If 40.00 mL of 0.100 M Na2C03 is titrated with 0.100 M HCI, calculate (a) the pH at the start of the titration; (b) the volume of HCl required to reach the first equiv­ alence point and the predominant species present at this point; (c) the volume of HCl required to reach the second equivalence point and the predominant species present at this point; (d) the pH at the second equiva­ lence point.

17.86

A hypothetical weak acid, HA, was combined with NaOH in the following proportions: 0.20 mol of HA, 0.080 mol of NaOH. The mixture was diluted to a total volume of 1.0 L, and the pH measured. (a) If pH = 4.80, what is the pKa of the acid? (b) How many additional moles of NaOH should be added to the solution to in­ crease the pH to 5.00?

[17.87]

What is the pH of a solution made by mixing 0.30 mol NaOH, 0.25 mol Na2HP04, and 0.20 mol H3P04 with water and diluting to 1.00 L?

[17.88]

Suppose you want to do a physiological experiment that calls for a pH 6.5 buffer. You find that the organism with which you are working is not sensitive to the weak acid H2X (Kal = 2 X 10-2; Ka2 = 5.0 X 10-7) or its sodium salts. You have available a 1.0 M solution of this acid and a 1.0 M solution of NaOH. How much of the NaOH solution should be added to 1.0 L of the acid to give a buffer at pH 6.50? (Ignore any volume change.)

[17.89]

How many microliters of 1.000 M NaOH solution must be added to 25.00 mL of a 0.1000 M solution of lactic acid [CH3CH(OH)COOH or HC3H503] to produce a buffer with pH = 3 . 75?

764 17.90

17.91

17.92

17.93

17.94 [17.95]

CHA PTER

17

Additional Aspects o f Aqueous Equilibria

A person suffering from anxiety begins breathing rapid­ ly and as a result suffers alkalosis, an increase in blood pH. (a) Using Equation 17.10, explain how rapid breath­ ing can cause the pH of blood to increase. (b) One cure for this problem is breathing in a paper bag. Why does this procedure lower blood pH? For each pair of compounds, use Ksp values to determine which has the greater molar solubility: (a) CdS or CuS, (b) PbC03 or BaCr04, (c) Ni(OHh or NiC03, (d) Agi or Ag2S04. Describe the solubility of CaC03 in each of the follow­ ing solutions compared to its solubility in water: (a) in 0.10 M NaCI solution; (b) in 0.10 M Ca(N03h solution; (c) 0.10 M Na2C03; (d) 0. 1 0 M HCl solution. (Answer same, less soluble, or more soluble.) Tooth enamel is composed of hydroxyapatite, whose sim­ plest formula is Ca5(P04h0H, and whose corresponding Ksp = 6.8 X 10-27. As discussed in the "Chemistry and Life" box in Section 17.5, fluoride in fluorinated water or in toothpaste reacts with hydroxyapatite to form fluoroa­ patite, Ca5(P04hF, whose Ksp = 1 .0 X 10__,;o. (a) Write the expression for the solubility-constant for hydroxy­ apatite and for fluoroapatite. (b) Calculate the molar solubility of each of these compounds. Calculate the solubility of Mg(OHh in 0.50 M NH4Cl. Seawater contains 0.13% magnesium by mass, and has a density of 1.025 g/mL. What fraction of the magnesium can be removed by adding a stoichiometric quantity of CaO (that is, one mole of CaO for each mole of Mg2+)?

The solubility-product constant for barium perman­ ganate, Ba(Mn04)u is 2.5 x 10-10 Assume that solid Ba(Mn04h is in equilibrium with a solution of KMn04. What concentration of KMn04 is required to establish a concentration of 2.0 x 10-8 M for the Ba2+ ion in solution? 2 2 17.97 Calculate the ratio of [Ca +] to [Fe +] in a lake in which the water is in equilibrium with deposits of both CaC03 and FeC03. Assume that the water is slightly basic and that the hydrolysis of the carbonate ion can therefore be ignored. [17.98] The solubility products of PbS04 and SrS04 are 6.3 X 10-7 and 3.2 x 10-7, respectively. What are the values of [5042-], [Pb2+], and [Sr2+] in a solution at equilibrium with both substances? 2 [17.99] What pH buffer solution is needed to give a Mg + con­ centration of 3.0 x 10-2 M in equilibrium with solid magnesium oxalate? 2 [17.100] The value of Ks for Mg3(As04)2 is 2.1 X 10- 0 p The As043- ion is derived from the weak acid H�s04 (pKa1 = 2.22; pKa2 = 6.98; pKa3 = 11.50). When asked to calculate the molar solubility of Mg3(As04h in water, a student used the Ksp expression and assumed that [Mg2+] = 1.5[As043-]. Why was this a mistake? [17.101] The solubility product for Zn(OHh is 3.0 x 10-1 6 The formation constant for the hydroxo complex, Zn(OH)42-, is 4.6 X 10-17. What concentration of OH­ is required to dissolve O.Q15 mol of Zn(OHh in a liter of solution? 17.96

INTEGRATIVE EXERCISES

Write the net ionic equation for the reaction that oc­ curs when a solution of hydrochloric acid (HCl) is mixed with a solution of sodium formate (NaCH02). (b) Calculate the equilibrium constant for this reaction. (c) Calculate the equilibrium concentrations of Na+, CC H+, CH02-, and HCH02 when 50.0 mL of 0.15 M HCl is mixed with 50.0 mL of 0.15 M NaCH02. (a) A 0.1044-g sample of an unknown monoprotic acid requires 22.10 mL of 0.0500 M NaOH to reach the end point. What is the molecular weight of the unknown? (b) As the acid is titrated, the pH of the solution after the addition of 11.05 mL of the base is 4.89. What is the Ka for the acid? (c) Using Appendix D, suggest the identity of the acid. Do both the molecular weight and Ka value agree with your choice? A sample of 7.5 L of NH3 gas at 22 oc and 735 torr is bub­ bled into a 0.50-L solution of 0.40 M HCI. Assuming that all the NH3 dissolves and that the volume of the solution remains 0.50 L, calculate the pH of the resulting solution.

17.102 (a)

17.103

17.104

17.105

Aspirin has the structural formula

At bod§ temperature (37 °C), Ka for aspirin equals 3 X 10- . If two aspirin tablets, each having a mass of 325 mg, are dissolved in a full stomach whose volume is 1 L and whose pH is 2, what percent of the aspirin is in the form of neutral molecules? 17.106

What is the pH at 25 oc of water saturated with C02 at a partial pressure of 1.10 atm? The Henry's law constant for C02 at 25 °C is 3.1 X 10-2 mol/L-atrn. The C02 is an acidic oxide, reacting with H20 to form H2C03.

Integrative Exercises 17.107

Excess Ca(OH)z is shaken with water to produce a satu­ rated solution. The solution is filtered, and a 50.00-mL sample titrated with HCl requires 11.23 mL of 0.0983 M HCl to reach the end point. Calculate K,p for Ca(OH)z. Compare your result with that in Appendix D. Do you think the solution was kept at 25 °C?

17.108

The osmotic pressure of a saturated solution of stron­ tium sulfate at 25 oc is 21 torr. What is the solubility product of this salt at 25 °C?

17.109

A concentration of 10- 100 parts per billion (by mass) of Ag + is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the Ag+ can cause adverse health effects. One way to main-

765

tain an appropriate concentration of Ag + is to add a slightly soluble salt to the pool. Using Ksp values from Aprendix D, calculate the equilibrium concentration of Ag in parts per billion that would exist in equilibrium with (a) AgCl, (b) AgBr, (c) Agl. [17.110)

Fluoridation of drinking water is employed in many places to aid in the prevention of dental caries. Typically the p- ion concentration is adjusted to about 1 ppb. Some water supplies are also "hard"; that is, they contain cer­ 2 tain cations such as Ca + that interfere with the action of 2 soap. Consider a case where the concentration of Ca + is 8 ppb. Could a precipitate of CaF2 form under these con­ ditions? (Make any necessary approximations.)

CHEMISTRY OF THE ENVIRONMENT

PHOTOGRAPH OF

EARTH

taken from Space looking south along the Red Sea.

766

W H AT ' S 18.1

18.2

18.3

18.4

A H EA D many of these minor constituents, including those that cause acid rain and photochemical smog, have been increased by human activities. Carbon dioxide is an important minor constituent of the atmosphere because it acts as a "greenhouse" gas that warms Earth's atmosphere.

Earth's Atmosphere

We begin this chapter with a general look at Earth's atmosphere. We will consider the temperature profile of the atmosphere, its pressure profile, and its chemical composition. Outer Regions of the Atmosphere

We next examine the outer (or upper) regions of the atmosphere, where the pressure is very low. These regions absorb a great deal of high-energy radiation from the Sun through photoionization and photodissociation reactions. By filtering out high­ energy radiation, these processes make it possible for life as we know it to exist on Earth. Ozone in the U pper Atmosphere

We will see that ozone in the stratosphere acts as a filter of high-energy ultraviolet light. Human activities have contributed to the depletion of the ozone layer by introducing into the stratosphere chemicals that perturb the natural cycle of ozone formation and decomposition. Notable among these chemicals are chlorofluorocarbons.

1 997,

REPRESENTATIVES OF

18.6

1 8.7

Chemistry of the Troposphere We complete our discussion of the atmosphere by examining its innermost region, the troposphere, which is the region in which we live. Many minor constituents of the troposphere affect air quality and the acidity of rainwater. The concentrations of

IN

18.5

1 30

The World Ocean

We next turn our attention to our water environment. Almost all the water on Earth is in the world ocean. Seawater contains many salts that participate in the global cycling of the elements and nutrients. Freshwater

Our discussion of water concludes with a look at freshwater. Processing seawater to obtain freshwater is energy intensive. We count on freshwater sources to supply most of our needs, but these sources often require treatment to render them usable. Green Chem istry

We conclude our discussion of the environment by examining green chemistry, which seeks to eliminate pollution at its source. Green chemistry is an international initiative to make industrial products, processes, and chemical reactions compatible with a sustainable society and environment.

NATIONS MET IN KYOTO, JAPAN,

to discuss the impact of human activities on global warming. Out of that meeting came an initiative to work toward a global treaty that would, among other things, spell out actions to be taken to reduce emissions of gases that cause global warming. 1n 2001 in Bonn, Germany, 178 nations signed a treaty based on the so-called Kyoto Protocols. in 2007, the United Nations intergovernmental Panel on Climate Change issued a report declaring that human activity has "very likely" been the driving force for the warming trend observed worldwide over the past 50 years. These efforts to address environmental concerns at the international level indicate that many of the most urgent environmental problems are global in nature. As technology has advanced and the world human population has increased, we have put new and greater stresses on our environment. Paradoxically, the very technology that can cause pollution also provides the tools to help us understand and manage our environment in a beneficial way. 767

768

CHAPTER 1 8

Chemistry of the Environment Chemistry is often at the heart of these issues. The economic growth of both developed and developing nations depends critically on chemical processes that range from treatment of water supplies to industrial processes. Some of these processes produce products or by-products that are harmful to the environment. We are now in a position to apply the principles we have learned in earlier chapters to an understanding of how our environment operates and how human activities affect it. In this chapter we consider some aspects of the chem­ istry of our environment, focusing on Earth's atmosphere and water. Both the air and water of our planet make life as we know it possible. To understand and maintain the environment in which we live, we must understand how human­ made and natural chemical compounds interact on land and in the sea and sky. Our daily decisions as consumers mirror those of the leaders meeting in Bonn and similar international meetings: We must weigh the costs versus the benefits of our actions. Unfortunately, the environmental impacts of our decisions are often very subtle and not immediately evident. 1 8 . 1 EART H ' S ATM O S P H E RE

T Figure 1 8.1 Temperature and pressure In the atmosphere.

(a) Temperature variations in the atmosphere at altitudes below 1 1 0 km. (b) Variations in atmospheric pressure with altitude. At 80 km the pressure is approximately 0.01 torr.

Because most of us have never been very far from Earth's surface, we often take for granted the many ways in which the atmosphere determines the environ­ ment in which we live. In this section we will examine some of the important characteristics of our planet's atmosphere. The temperature of the atmosphere varies in a complex manner as a func­ tion of altitude, as shown in Figure 18.1(a)
Chemistry of the Troposphere

.A. Figure 18.10

779

Carbon monoxide

Kerosene lamps and stoves have warning labels concerning use in enclosed spaces, such as an indoor room. Incomplete combustion can produce colorless, odorless carbon monoxide, CO, which is toxic. warnings.

Nitrogen oxides are primary components of smog, a phenomenon with which city dwellers are all too familiar. The term smog refers to a particularly unpleas­ ant condition of pollution in certain urban environments that occurs when weather conditions produce a relatively stagnant air mass. The smog made fa­ mous by Los Angeles, but now common in many other urban areas as well, is more accurately described as photochemical smog because photochemical processes play a major role in its formation (Figure 18.11 � ). The majority of nitrogen oxide emissions (about 50%) comes from cars, buses, and other forms of transportation. Nitric oxide, NO, forms in small quantities in the cylinders of internal combustion engines by the direct combi­ nation of nitrogen and oxygen: Nz(g) + 02(g) :;:::::::': 2 NO (g)

t.H = 180.8 kJ

[18.11]

As noted in the "Chemistry Put to Work" box in Section 15.7, the equilibrium constant K for this reaction increases from about 10- IS at 300 K (near room tem­ perature) to about 0.05 at 2400 K (approximately the temperature in the cylin­ .A. Figure 18. 1 1 Photochemical smog. Smog is produced largely by the action der of an engine during combustion). Thus, the reaction is more favorable at of sunlight on automobile exhaust gases. higher temperatures. In fact, some NO is formed in any high-temperature com­ bustion. As a result, electrical power plants are also major contributors to nitro­ gen oxide pollution. Before the installation of pollution-control devices on automobiles, typical emission levels of NOx were 4 grams per mile (g/mi). (The x is either 1 or 2 because both NO and N02 are formed, although NO TABLE 18.5 National Tailpipe Emission Standards predominates.) Starting in 2004, the auto emission standards for NOx call for a phased-in reduction to 0.07 g/mi Year Hydrocarbons (gjmi) Nitrogen Oxides g ( jmi) -'-=-"---' by 2009. Table 18.5 � summarizes the federal standards 1975 1.5 (0.9) for hydrocarbons and NOx emissions since 1975 as well 3.1 (2.0) 1980 0.41 (0.41) 2.0 (1.0) as the more restrictive standards enforced in California. 0.41 (0.41) 1985 1.0 (0.4) In air, nitric oxide (NO) is rapidly oxidized to nitro­ 1.0 (0.4) 1990 0.41 (0.41) gen dioxide (NOz):

•

-

2 NO(g) + 02(g) :;:::::::': 2 N02(g)

t.H = -113.1 kJ

[18.12]

1995 2004

-

0.41 (0.25) 0.075 (0.05)

"California standards in parentheses

-

-

0.4 (0.4) 0.07 (0.05)

­

780

CHAPTER 1 8

Chemistry of the Environment The equilibrium constant for this reaction decreases from about 101 2 at 300 K to about w-s at 2400 K. The photodissociation of N02 initiates the reactions associated with photochemical smog. The dissociation of N02 into NO and 0 requires 304 kJ/mol, which corresponds to a photon wavelength of 393 nm. In sunlight, therefore, N02 undergoes dissociation to NO and 0: N02(g)

+ hv � NO(g) + O(g)

[18.13]

The atomic oxygen formed undergoes several possible reactions, one of which gives ozone, as described earlier: O(g)

+ 02 + M(g) � 03(g) + M*(g)

[18.14]

Ozone is a key component of photochemical smog. Although it is an essential UV screen in the upper atmosphere, ozone is an undesirable pollutant in the troposphere. It is extremely reactive and toxic, and breathing air that contains appreciable amounts of ozone can be especially dangerous for asthma sufferers, exercisers, and the elderly. We therefore have two ozone problems: excessive amounts in many urban environments, where it is harmful, and depletion in the stratosphere, where it is vital. In addition to nitrogen oxides and carbon monoxide, an automobile engine also emits unburned hydrocarbons as pollutants. These organic compounds, which are composed entirely of carbon and hydrogen, are the principal compo­ nents of gasoline (Section 25.3) and are major ingredients of smog. A typical en­ gine without effective emission controls emits about 10 to 15 grams of these compounds per mile. Current standards require that hydrocarbon emissions be less than 0.075 grams per mile. Hydrocarbons are also emitted naturally from living organisms (see "A Closer Look" box later in this section). Reduction or elimination of smog requires that the essential ingredients for its formation be removed from automobile exhaust. Catalytic converters are de­ signed to drastically reduce the levels of NOx and hydrocarbons, two of the major ingredients of smog. = (Section 14.7, "Chemistry Put to Work: Catalytic Converters")

GIVE IT SOME THOUGHT What photochemical reaction involving nitrogen oxides initiates the formation of photochemical smog?

Water Vapor, Carbon Dioxide, and Cli mate We have seen how the atmosphere makes life as we know it possible on Earth by screening out harmful short-wavelength radiation. In addition, the atmos­ phere is essential in maintaining a reasonably uniform and moderate tempera­ ture on the surface of the planet. The two atmospheric components of greatest importance in maintaining Earth's surface temperature are carbon dioxide and water. Without them, the average surface temperature of Earth would be 254 K, a temperature too low to sustain life. Earth is in overall thermal balance with its surroundings. This means that Earth radiates energy into space at a rate equal to the rate at which it absorbs ener­ gy from the Sun. The Sun has a surface temperature of about 6000 K. As seen from outer space, Earth is relatively cold (254 K). The temperature of an object deter­ mines the distribution of wavelengths in the radiation it emits. = (Section 6.2) Why does Earth, viewed from outside its atmosphere, appear so much colder than the temperature we usually experience at its surface? The troposphere is transparent to visible light but not to infrared radiation. Figure 18.12 11> shows the distribution of radiation from Earth's surface and the wavelengths absorbed by atmospheric water vapor and carbon dioxide. According to the graph, these atmospheric gases absorb much of the outgoing radiation from Earth's surface. In doing so, they help to maintain a livable uniform temperature at the surface by holding in, as it were, the infrared radiation, that we feel as heat.

18.4

Chemistry of the Troposphere

Wavelengths absorbed by C02 Infrared

II

� II

Wavelengths absorbed by H20

.s

0: 0 •jj "'

'B

�

.

-

Wavelengths emitted by Earth"s surface

781



--->

CaC03(s) Mg (OHh(s)

Ion exchange is a typical household method for water softening. In this procedure the hard water is passed through a bed of an ion-exchange resin: plastic beads with covalently bound anion groups such as -coo- or -503-. These nega­ tively charged groups have Na+ ions attached to balance their charges. The Ca2+ and other cations in the hard water are at­ tracted to the anionic groups and displace the lower-charged Na+ into the water. Thus, one type of ion is exchanged for an­ other. To maintain charge balance, 2 Na+ enter the water for each Ca2+ removed. If we represent the resin with its anionic site as R-Coo-, we can write the equation for the process as follows:

2 Na(R-COO)(s) + Ca2+(aq)

:;====:

Ca(R-COOh(s)

+ 2 Na+(aq)

Water softened in this way contains an increased concen­ tration of Na+. Although Na+ does not form precipitates or cause other problems associated with hard-water cations, in­ dividuals concerned about their sodium intake, such as those who have high blood pressure (hypertension), should avoid drinking water softened in this way. When all the available Na+ have been displaced from the ion-exchange resin, the resin is regenerated by flushing it with a concentrated solution of NaCI. Homeowners can do this by charging their units with large amounts of NaCl(s), which can be purchased at most grocery stores. The high concentration of Na+ forces the equilibrium shown in the earlier equation to shift to the left, causing the Na+ to displace the hard-water cations, which are flushed down the drain.

1 8.7 G REEN C H E M ISTRY

The planet on which we live is, to a large extent, a closed system, one that ex­ changes energy but not matter with its environment. If humankind is to thrive in the future, all the processes we carry out should be in balance with Earth's natural processes and physical resources. This goal requires that no toxic mate­ rials are released to the environment, that our needs are met with renewable re­ sources, and that we consume the least possible amount of energy. Although the chemical industry is but a small part of human activity, chemical processes are involved in nearly all aspects of modem life. Chemistry is therefore at the heart of efforts to accomplish these goals. Green chemistry is an initiative that promotes the design and application of chemical products and processes that are compatible with human health and that preserve the environment. Some of the major principles that govern green chemistry are the following: •

The potential hazards of substances should be a major consideration in the design of chemical processes. It is better to prevent waste than to treat or clean it up after it has been created.

18.7 • •

•

The methods employed in synthesizing substances should generate as little waste product as possible. Chemical processes should be designed to be as energy efficient as possi­ ble, avoiding high temperatures and pressures. Thus, catalysts should be sought that facilitate reactions under mild reaction conditions. The raw materials for chemical processes should be renewable whenever technically and economically feasible.

To illustrate how green chemistry works in practice, consider the manufacture of styrene, a very important building block in the formation of many polymers, in­ cluding the expanded polystyrene packages used to pack eggs and restaurant takeout food. The global demand for styrene is more than 25 million metric tons per year. For many years styrene has been produced in a two-step process. In the first step, benzene and ethylene react in the presence of a solid catalyst, forming ethyl benzene. In the second step, the ethyl benzene is mixed with high tempera­ ture steam and passed over an iron oxide catalyst at 625 °C yielding styrene:

©+ H2C=CH2

Acid catalyst

Benzene

�CH2CH3

Iron oxide catalyst -H,

�CH=CH2 Styrene

Ethyl benzene

This process has several shortcomings. One is that both benzene, which is formed from crude oil, and ethylene, which is formed from natural gas, are high-priced starting materials for a product that should be a low-priced com­ modity. From an environmental point of view, the high toxicity of benzene and the high energy cost overall for the process are both undesirable. Researchers at Exelus Inc. have demonstrated a process that bypasses some of these shortcomings. The two-step process is replaced by a one-step process in which toluene is reacted with methanol at 425 °C over a special catalyst: Base catalyst -H2 -H20

Toluene

=C C 2 H H � Styrene

The new process saves money because toluene and methanol are less expensive than benzene and ethylene, and the reaction conditions require much lower energy input. Adclitional benefits are that the methanol could be produced from biomass, and that benzene is replaced by less toxic toluene. The hydrogen formed in the reaction can be recycled as a source of energy. This example demonstrates how finding the right catalyst was key in discovering the new process. Let's consider some other examples in which green chemistry can operate to improve environmental quality.

Solvents and Reagents A major area of concern in chemical processes is the use of volatile organic com­ pounds as solvents for reactions. Generally, the solvent in which a reaction is run is not consumed in the reaction, and there are unavoidable releases of sol­ vent into the atmosphere even in the most carefully controlled processes. Fur­ ther, the solvent may be toxic or may decompose to some extent during the reaction, thus creating waste products. The use of supercritical fluids (Section 11.4: "Chemistry Put to Work: Supercritical Fluid Extraction") represents a way to replace conventional solvents with other reagents. Recall that a supercritical fluid is an unusual state of matter that has properties of both a gas and a liquid. (Section 11.4) Water and carbon dioxide are the two most popular choices as supercritical fluid solvents. One recently developed industrial process, c:x:o

Green Chemistry

789

790

CHAPTER 1 8

Chemistry of the Environment for example, replaces chlorofluorocarbon solvents with liquid or supercritical C02 in the production of polytetrafluoroethylene ([CF2CF2 Jn, sold as Teflon ®). The chlorofluorocarbon solvents have harmful effects on Earth's ozone layer (Section 18.3). Though C02 is a greenhouse gas, no new C02 need be manufac­ tured for use as a supercritical fluid solvent. As a further example, para-xylene is oxidized to form terephthalic acid, which in tum is used to make polyethylene terephthalate (PET) plastic and polyester fiber (Section 12.6, Table 12.5):

CH3

-©-

CH3 + 3 0 2

190'C, 20 atm C ata1yst

para-Xylene

�

11 �C-OH 11 + 2 H 20

0 HO -C

0

Terephthalic acid

This commercial process requires pressurization and a relatively high tem­ perature. Oxygen is the oxidizing agent, and acetic acid (CH3COOH) is the sol­ vent. An alternative route employs supercritical water as the solvent {Table 11.5) and hydrogen peroxide as the oxidant. This alternative process has several potential advantages, most particularly the elimination of acetic acid as solvent and the use of a different oxidizing agent. Whether it can successfully replace the existing commercial process, however, depends on many factors, which will require further study.

G IVE IT SOME THOUGHT We noted earlier that increasing carbon dioxide levels contribute to global warming, which seems like a bad thing, but now we are saying that using carbon dioxide in industrial processes is a good thing for the environment. Explain this seeming contradiction.

Other Processes

.6. Figure 1 8.22 Green chemistry for your clothes. This dry-cleaning

apparatus employs liquid C02 as the solvent.

Many processes that are important in modem society use chemicals not found in nature. Let's briefly examine two of these processes, dry cleaning and the coating of automotive bodies to prevent corrosion, and consider alternatives being developed to reduce harmful environmental impacts. Dry cleaning of clothing typically uses a chlorinated organic solvent such as tetrachloroethylene (Cl2C = CC12), which may cause cancer in humans. The widespread use of this and related solvents in dry cleaning, metal cleaning, and other industrial processes has contaminated groundwater in some areas. Alter­ native dry-cleaning methods that employ liquid C02- along with special clean­ ing agents, are being successfully commercialized (Figure 18.22 ._) . The metal bodies of cars are coated extensively during manufacture to pre­ vent corrosion. One of the key steps is the electrodeposition of a layer of metal ions that creates an interface between the vehicle body and the polymeric coat­ ings that serve as the undercoat for painting. In the past, lead was the metal of choice for inclusion in the electrodeposition mixture. Because lead is highly toxic, however, its use in other paints and coatings has been virtually eliminat­ ed, and a relatively nontoxic yttrium hydroxide alternative to lead has been developed as an automotive coating (Figure 18.23 � ). When this coating is sub­ sequently heated, the hydroxide is converted to the oxide, producing an insolu­ ble cerarniclike coating. = (Section 12.4) The Chal lenges of Water Purification Access to clean water is essential to the workings of a stable, thriving society. We have seen in the previous section that disinfection of water is an important step in water treatment for human consumption. Water disinfection is one of the greatest

18.7 public health innovations in human his­ tory. It has dramatically decreased the in­ cidences of waterborne bacterial diseases such as cholera and typhus. But this great benefit comes at a price. In 1974 scientists in both Europe and the United States discovered that chlori­ nation of water produces a group of by­ products that had previously gone undetected. These by-products are called trihalomethanes (THMs) because all have a single carbon atom and three halogen atoms: CHCI3, CHC12Br, CHC!Brz, and CHBr3. These and many other chlorine­ and bromine-containing organic sub­ stances are produced by the reaction of aqueous chlorine with organic materials present in nearly all natural waters, as well as with substances that are by-products of human activity. Recall that chlorine dissolves in water to form HClO, which is the active oxidizing agent oco (Section 7.8): Clz(g) + HzO(l)

----+

.a. Figure 18.23 Green chemistry for your car. An automobile body receives a corrosion-protection coating containing yttrium in place of lead.

HC10(aq) + H+(aq) + Cqaq)

[18.16]

HClO in tum reacts with organic substances to form the THMs. Bromine enters through the reaction of HClO with dissolved bromide ion: HOC1(aq) + Br-(aq)

----+

HBrO(aq) + Cqaq)

[18.17]

HBrO(aq) halogenates organic substances analogously to HClO(aq). Some THMs and other halogenated organic substances are suspected car­ cinogens; others interfere with the body's endocrine system. As a result, the World Health Organization and the Environmental Protection Agency (EPA) have placed concentration limits of 80 J.Lg/L (80 ppb) on the total quantity of such substances in drinking water. The goal is to reduce the levels of THMs and related substances in the drinking water supply while preserving the antibacte­ rial effectiveness of the water treatment. In some cases simply lowering the con­ centration of chlorine may provide adequate disinfection while reducing the concentrations of THMs formed. Alternative oxidizing agents, such as ozone (03) or chlorine dioxide (CI02), produce less of the halogenated substances, but they have their own disadvantages. For example, each is capable of oxidiz­ ing aqueous bromide, as shown here for ozone: 03(aq) + Br-(aq) + H 20(l)

----+

HBrO(aq) + 2 03(aq)

----+

Green Chemistry

HBrO(aq) + 02(aq) + OH-(aq)

[18.18]

Br03-(aq) + 2 02(aq) + H +(aq)

[18.19]

As we have seen, HBrO(aq) is capable of reacting with dissolved organic sub­ stances to form halogenated organic compounds. Furthermore, bromate ion has been shown to cause cancer in animal tests. There seem to be no completely satisfactory alternatives to chlorination or ozonation at present. The risks of cancer from THMs and related substances in municipal water are very low, however, compared to the risks of cholera, typhus, and gastrointestinal disorders from untreated water. When the water supply is cleaner to begin with, less disinfectant is needed; thus, the danger of contami­ nation through disinfection is reduced. Once the THMs are formed, their concen­ trations in the water supply can be reduced by aeration because the THMs are more volatile than water. Alternatively, they can be removed by adsorption onto activated charcoal or other adsorbents.

791

792

CHA PTER 1 8

Chemistry of the Environment - SAMPLE INTEGRATIVE EXERCISE

I Putting Concepts Together

(a) Acids from acid rain or other sources are no threat to lakes in areas where the rock is limestone (calcium carbonate), which can neutralize the excess acid. Where the rock is granite, however, no such neutralization occurs. How does the limestone neu­ tralize the acid? (b) Acidic water can be treated with basic substances to increase the pH, although such a procedure is usually only a temporary cure. Calculate the mini­ mum mass of lime, CaO, needed to adjust the pH of a small lake (V = 4 X 109 L) from 5.0 to 6.5. Why might more lime be needed?

SOLUTION Analyze: We need to remember what a neutralization reaction is, and calculate the amount of a substance needed to effect a certain change in pH.

Plan: For (a), we need to think about how acid can react with calcium carbonate,

which evidently does not happen with acid and granite. For (b), we need to think about what reaction would happen with acid and CaO and do stoichiometric calcu­ lations. From the proposed change in pH, we can calculate the change in proton con­ centration needed and then figure out how much CaO it would take to do the reaction.

Solve:

(a) The carbonate ion, which is the anion of a weak acid, is basic. = (Sections 16.2

and 16.7) Thus, the carbonate ion, col-, reacts with H+(aq). If the concentration of H+(aq) is small, the major product is the bicarbonate ion, HC0 3 -. If the concentration of H+(aq) is higher, however, H2C03 forms and decomposes to C02 and H20. = (Section 4.3) (b) The initial and final concentrations of H+(aq) in the lake are obtained from their pH values:

[H+hnitial

=

10-5·0

=

1 X 10-5 M

and

[H+Jfinal

=

10-6.5

=

3 X 10-7 M

Using the volume of the lake, we can calculate the number of moles of H+(aq) at both pH values: (1 (3

X X

10-S moljL)(4.0 10-7moljL)(4.0

X X

109 L)

=4X

104 mol

109 L) = 1 X 10 3 mol

Hence, the change in the amount of H+(aq) is 4 X 104 mol - 1 x 103 mol "' 4 X 104 mol. Let's assume that all the acid in the lake is completely ionized, so that only the free H+(aq) measured by the pH needs to be neutralized. We will need to neutralize at least that much acid, although there may be a great deal more acid in the lake than that. The oxide ion of CaO is very basic. = (Section 16.5) 1n the neutralization reac­ 2 tion 1 mol of 0 - reacts with 2 mol of H+ to form H20. Thus, 4 X 104 mol of H+ re­ quires the following number of grams of CaO:

(4 x 104 molH+)

(

1 mo1Ca0 2 molH+

)(

56.1 g CaO 1 mol C a0

)=

1

x

106 g cao

This amounts to slightly more than a ton of CaO. That would not be very costly because CaO is an inexpensive base, selling for less than $100 per ton when purchased in large quantities. The amount of CaO calculated above, however, is the very minimum amount needed because there are likely to be weak acids in the water that must also be neutralized. This liming procedure has been used to adjust the pH of some small lakes to bring their pH into the range necessary for fish to live. The lake in our example would be about a half mile long, a half mile wide, and have an average depth of 20 ft.

C H APTER RE V IEW

SUMMARY AND KEY TERMS

Sections 1 8.1 and 18.2 In these sections we examined the physical and chemical properties of Earth's atmosphere. The complex temperature variations in the atmosphere give rise to four regions, each with characteristic proper­ ties. The lowest of these regions, the troposphere, extends

from Earth's surface up to an altitude of about 12 km. Above the troposphere, in order of increasing altitude, are the stratosphere, mesosphere, and thermosphere. In the upper reaches of the atmosphere, only the simplest che­ mical species can survive the bombardment of highly

Key Skills energetic particles and radiation from the Sun. The aver­ age molecular weight of the atmosphere at high eleva­ tions is lower than that at Earth's surface because the lightest atoms and molecules diffuse upward and also be­ cause of photodissociation, which is the breaking of bonds in molecules because of the absorption of light. Absorption of radiation may also lead to the formation of ions via photoionization. Section 18.3 Ozone is produced in the upper atmos­ phere from the reaction of atomic oxygen with 02• Ozone is itself decomposed by absorption of a photon or by reaction with an active species such as Cl. Chlorofluorocarbons can undergo photodissociation in the stratosphere, intro­ ducing atomic chlorine, which is capable of catalytically destroying ozone. A marked reduction in the ozone level in the upper atmosphere would have serious adverse conse­ quences because the ozone layer filters out certain wave­ lengths of ultraviolet light that are not removed by any other atmospheric component. Section 1 8.4 In the troposphere the chemistry of trace

atmospheric components is of major importance. Many of these minor components are pollutants. Sulfur dioxide is one of the more noxious and prevalent examples. It is oxi­ dized in air to form sulfur trioxide, which, upon dissolving in water, forms sulfuric acid. The oxides of sulfur are major contributors to acid rain. One method of preventing the escape of S02 from industrial operations is to react the S02 with CaO to form calcium sulfite (CaS03). Carbon monoxide (CO) is found in high concentra­ tions in automobile engine exhaust and in cigarette smoke. CO is a health hazard because it can form a strong bond with hemoglobin and thus reduce the capacity for blood to transfer oxygen from the lungs. Photochemical smog is a complex mixture of compo­ nents in which both nitrogen oxides and ozone play im­ portant roles. Smog components are generated mainly in automobile engines, and smog control consists largely of controlling auto emissions. Carbon dioxide and water vapor are the major compo­ nents of the atmosphere that strongly absorb infrared radi­ ation. C02 and H20 are therefore critical in maintaining Earth's temperature. The concentrations of C02 and other so-called greenhouse gases in the atmosphere are thus im­ portant in determining worldwide climate. Because of the

793

extensive combustion of fossil fuels (coal, oil, and natural gas), the concentration of carbon dioxide in the atmos­ phere is steadily increasing. Section 18.5 Seawater contains about 3.5% by mass of dissolved salts and is described as having a salinity of 35.

Seawater's density and salinity vary with depth. Because most of the world's water is in the oceans, humans may eventually look to the seas for freshwater. Desalination is the removal of dissolved salts from seawater or brackish water to make it fit for human consumption. Desalination may be accomplished by distillation or by reverse osmosis. Section 1 8.6 Freshwater contains many dissolved sub­ stances, including dissolved oxygen, which is necessary for fish and other aquatic life. Substances that are decom­ posed by bacteria are said to be biodegradable. Because the oxidation of biodegradable substances by aerobic bac­ teria consumes dissolved oxygen, these substances are called oxygen-demanding wastes. The presence of an ex­ cess amount of oxygen-demanding wastes in water can sufficiently deplete the dissolved oxygen to kill fish and produce offensive odors. Plant nutrients can contribute to the problem by stimulating the growth of plants that be­ come oxygen-demanding wastes when they die. The water available from freshwater sources may re­ quire treatment before it can be used domestically. The several steps generally used in municipal water treatment include coarse filtration, sedimentation, sand filtration, aeration, sterilization, and sometimes water softening. Water softening is required when the water contains ions such as Mg 2+ and Ca2+, which react with soap to form soap scum. Water containing such ions is called hard water. The lime-soda process, which involves adding CaO and Na2C03 to hard water, is sometimes used for large-scale municipal water softening. Individual homes usually rely on ion exchange, a process by which hard­ water ions are exchanged for Na+ ions. Section 18.7 The green chemistry initiative promotes

the design and application of chemical products and processes that are compatible with human health and that preserve the environment. The areas in which the princi­ ples of green chemistry can operate to improve environ­ mental quality include choices of solvents and reagents for chemical reactions, development of alternative processes, and improvements in existing systems and practices.

KEY SKILLS • •

•

•

•

•

•

Describe the regions of Earth's atmosphere in terms of how temperature varies with altitude. Describe the composition of the atmosphere in terms of the major components in dry air at sea level. Calculate concentrations of gases in parts per million (ppm). Describe the processes of photodissociation and photoionization and their role in the upper atmosphere. Use bond energies and ionization energies to calculate the minimum frequency or maximum wavelength needed to cause photodissociation or photoionization. Explain the role of ozone in the upper atmosphere. Explain how chlorofluorocarbons (CFCs) are involved in depleting the ozone layer.

794

CHA PTER 1 8

Chemistry of the Environment

•

Describe the origins and behavior of sulfur oxides, carbon monoxide, and nitrogen oxides as air pollutants, including the generation of acid rain and photochemical smog.

•

Describe how water and carbon dioxide in the atmosphere affect atmospheric temperature via the greenhouse effect.

•

Explain what is meant by the salinity of water and describe the process of reverse osmosis as a means of desalination.

• List the major cations, anions, and gases present in natural waters and describe the relationship between dissolved oxygen and water quality. • List the main steps involved in treating water for domestic uses. •

Describe the main goals of green chemistry.

VIS UALIZING CONCEPTS K) and 1 atm pressure, one mole of an ideal gas occupies 22.4 L. = (Section 10.4) (a) Looking back at Figure 18.1, do you predict that 1 mole of an ideal gas in the middle of the stratosphere would occupy a greater or smaller volume than 22.4 L? (b) Looking at Figure 18.1(a), we see that the tempera­ ture is lower at 85 km altitude than at 50 km. Does this mean that one mole of an ideal gas would occupy less volume at 85 km than at 50 km? Explain. [Section 18.1]

18.1 At room temperature (298

18.2 Molecules in the upper atmosphere tend to contain dou­

Add hard water to top of column

-.Special plastic bead

ble and triple bonds rather than single bonds. Suggest an explanation. [Section 18.2] 18.3 Why does ozone concentration in the atmosphere vary

as a function of altitude (see Figure 18.4)? [Section 18.3] 18.4 You are working with an artist who has been commis­

sioned to make a sculpture for a big city in the eastern United States. The artist is wondering what material to use to make her sculpture, because she has heard that acid rain in the eastern U.S. might destroy it over time. You take samples of granite, marble, bronze, and other materi­ als, and place them outdoors for a long time in the big city. You periodically examine the appearance and measure the mass of the samples. (a) What observations would lead you to conclude that one, or more, of the materials were well-suited for the sculpture? (b) What chemical process (or processes) is (are) the most likely responsible for any observed changes in the materials? [Section 18.4] 18.5 How does carbon dioxide interact with the world ocean?

[Section 18.5] 18.6 The following picture represents an ion-exchange column,

2 in which water containing "hard" ions, such as Ca +, is added to the top of the column, and water containing "soft" ions such as Na+ come out the bottom. Explain what is happening in the column. [Section 18.6]

18.7 Describe the basic goals of green chemistry. [Section 18.7] 18.8 One mystery in environmental science is the imbalance

in the "carbon dioxide budget." Considering only human activities, scientists have estimated that 1.6 bil­ lion metric tons of C02 is added to the atmosphere every year because of deforestation (plants use COz, and fewer plants will leave more C02 in the atmos­ phere). Another 5.5 billion tons per year is put into the atmosphere because of burning fossil fuels. It is further estimated (again, considering only human activities) that the atmosphere actually takes up about 3.3 billion tons of C02 per year, while the oceans take up 2 billion tons per year, leaving about 1.8 billion tons of C02 per year unaccounted for. This "missing" C02 is assumed to be taken up by the "land." What do you think might be happening?

EXERCISES Earth's Atmosphere 18.9 (a) What is the primary basis for the division of the at­

18.10 (a) How are the boundaries between the regions of the

mosphere into different regions? (b) Name the regions of the atmosphere, indicating the altitude interval for each one.

atmosphere determined? (b) Explain why the strato­ sphere, which is more than 20 miles thick, has a smaller total mass than the troposphere, which is less than 10 miles thick.

Exercises 18.11

Air pollution in the Mexico City metropolitan area is among the worst in the world. The concentration of ozone in Mexico City has been measured at 441 ppb (0.441 ppm). Mexico City sits at an altitude of 7400 feet, which means its atmospheric pressure is only 0.67 atm. Calculate the partial pressure of ozone at 441 ppb if the atmospheric pressure is 0.67 atm.

18.12

From the data in Table 18.1, calculate the partial pres­ sures of carbon dioxide and argon when the total atmos­ pheric pressure is 96.5 kPa.

795

The average concentration of carbon monoxide in air in an Ohio city in 2006 was 3.5 ppm. Calculate the number of CO molecules in 1.0 L of this air at a pressure of 755 torr and a temperature of 18 oc. 18.14 (a) From the data in Table 18.1, what is the concentration of neon in the atmosphere in ppm? (b) What is the con­ centration of neon in the atmosphere in molecules per L, assuming an atmospheric pressure of 733 torr and a temperature of 292 K? 18.13

The Upper Atmosphere; Ozone The dissociation energy of a carbon-bromine bond is typically about 210 kJ/mol. What is the maximum wavelength of photons that can cause C- Br bond dissociation? 18.16 In CF3CI the C -CI bond-dissociation energy is 339 kJ/mol. In CCI4 the C - CI bond-dissociation ener­ gy is 293 kJ/mol. What is the range of wavelengths of photons that can cause C -Cl bond rupture in one mol­ ecule but not in the other?

18.15

18.17

18.18

(a) Distinguish between photodissociation and photoioni­ zation. (b) Use the energy requirements of these two processes to explain why photodissociation of oxygen is more important than photoionization of oxygen at altitudes below about 90 km. Why is the photodissociation of N2 in the atmosphere relatively unimportant compared with the photodisso­ ciation of 02?

What is a hydrofluorocarbon? Why are these com­ pounds potentially less harmful to the ozone layer than CFCs? 18.20 Draw the Lewis structure for the chlorofluorocarbon CFC-11, CFCI3 . What chemical characteristics of this substance allow it to effectively deplete stratospheric ozone? 18.19

(a) Why is the fluorine present in chlorofluorocarbons not a major contributor to depletion of the ozone layer? (b) What are the chemical forms in which chlorine exists in the stratosphere following cleavage of the carbon­ chlorine bond? 18.22 Would you expect the substance CFBr to be effective in 3 depleting the ozone layer, assuming that it is present in the stratosphere? Explain.

18.21

Chemistry of the Troposphere For each of the following gases, make a list of known or possible naturally occurring sources: (a) CH4, (b) 502, (c) NO, (d) CO. 18.24 Why is rainwater naturally acidic, even in the absence of polluting gases such as 502? 18.23

(a) Write a chemical equation that describes the attack of acid rain on limestone, CaC0 3. (b) If a limestone sculp­ ture were treated to form a surface layer of calcium sul­ fate, would this help to slow down the effects of acid rain? Explain. 18.26 The first stage in corrosion of iron upon exposure to air 2 is oxidation to Fe + (a) Write a balanced chemical equa­ tion to show the reaction of iron with oxygen and pro­ tons from acid rain. (b) Would you expect the same sort of reaction to occur with a silver surface? Explain. 18.25

18.27

Alcohol-based fuels for automobiles lead to the produc­ tion of formaldehyde (CH20) in exhaust gases. Formal­ dehyde undergoes photodissociation, which contributes to photochemical smog: CH20 + hv CHO + H The maximum wavelength of light that can cause this reaction is 335 nrn. (a) In what part of the electro­ magnetic spectrum is light with this wavelength found? ----->

(b) What is the maximum strength of a bond, in kJ/mol, that can be broken by absorption of a photon of 335-nm light? (c) Compare your answer from part (b) to the ap­ propriate value from Table 8.4. What do you conclude about the C- H bond energy in formaldehyde? (d) Write out the formaldehyde photodissociation reac­ tion, showing Lewis-dot structures. 18.28 An important reaction in the formation of photochemi­ cal smog is the photodissociation of NOi N02 + hv NO(g) + O(g) The maximum wavelength of light that can cause this reaction is 420 nm. (a) In what part of the electromagnet­ ic spectrum is light with this wavelength found? (b) What is the maximum strength of a bond, in kJ/mol, that can be broken by absorption of a photon of 420-nm light? (c) Write out the photodissociation reaction show­ ing Lewis-dot structures. ----->

Explain why increasing concentrations of C02 in the at­ mosphere affect the quantity of energy leaving Earth but do not affect the quantity entering from the Sun. 18.30 (a) With respect to absorption of radiant energy, what distinguishes a greenhouse gas from a nongreenhouse gas? (b) CH4 is a greenhouse gas, but Ar is not. How might the molecular structure of CH4 explain why it is a greenhouse gas? 18.29

796

CHAPTER 1 8

Chemistry of the Environment

The World Ocean Assume the density of seawater is 1.03 g/mL and that your gold recovery process is 50% efficient.

18.31 What is the molarity of Na+ in a solution of NaCl whose

salinity is 5.6 if the solution has a density of 1.03 g/mL?

18.32 Phosphorus is present in seawater to the extent of

0.07 ppm by mass. If the phosphorus is present as 3 phosphate, P04 -, calculate the corresponding molar concentration of phosphate in seawater.

18.35 Suppose that one wishes to use reverse osmosis to reduce

the salt content of brackish water containing 0.22 M total salt concentration to a value of 0.01 M, thus rendering it usable for human consumption. What is the minimum pressure that needs to be applied in the permeators (Figure 18.16) to achieve this goal, assuming that the op­ eration occurs at 298 K? (Hint: Refer to Section 13.5.)

18.33 A first-stage recovery of magnesium from seawater is

precipitation of Mg(OHh with CaO:

Mg2+(aq) + CaO(s) + HzO(I)

--->

Mg(OHh(s) + Ca2+(aq)

What mass of CaO, in grams, is needed to precipitate 1000 lb of Mg(OHh?

18.36 Assume that a portable reverse-osmosis apparatus such

as that shown in Figure 18.17 operates on seawater, whose concentrations of constituent ions are listed in Table 18.6, and that the desalinated water output has an effective molarity of about 0.02 M. What minimum pres­ sure must be applied by hand pumping at 297 K to cause reverse osmosis to occur? (Hint: Refer to Section 13.5.)

18.34 Gold is found in seawater at very low levels, about

0.05 ppb by mass. Assuming that gold is worth about $800 per troy ounce, how many liters of seawater would you have to process to obtain $1,000,000 worth of gold?

Freshwater 18.37 List the common products formed when an organic ma­

terial containing the elements carbon, hydrogen, oxy­ gen, sulfur, and nitrogen decomposes (a) under aerobic conditions, (b) under anaerobic conditions.

18.38 (a) Explain why the concentration of dissolved oxygen

in freshwater is an important indicator of the quality of the water. (b) How is the solubility of oxygen in water affected by increasing temperature?

magnesium ions are removed in water treatment by the addition of slaked lime, Ca(OHh.

18.42

(a) Which of the following ionic species could be, 2 responsible for hardness in a water supply: Ca +, K+, 2 2 Mg +, Fe +, Na+ ? (b) What properties of an ion deter­ mine whether it will contribute to water hardness?

18.43 How many moles of Ca(OHh and Na2C03 should

18.39 The organic anion

{-@-

be added to soften 1200 L of water in which 4 [Ca2+] = 5.0 X 1 0- M and [HC03-] = 7.0 X 1 0-4 M ?

H H3C-(CH2)9-

18.41 Write a balanced chemical equation to describe how

so,­

CH3

is found in most detergents. Assume that the anion un­ dergoes aerobic decomposition in the following manner: --->

2 C1sH29S03-(aq) + 51 0 (aq) 2 36 COz(aq) + 28 HzO(/) + 2 H+(aq) + 2 Sol-(aq) What is the total mass of 0 required to biodegrade 1.0 g 2 of this substance?

18.40 The average daily mass of 02 taken up by sewage dis­

charged in the United States is 59 g per person. How many liters of water at 9 ppm 0 are totally depleted of 2 oxygen in 1 day by a population of 120,000 people?

18.44 The concentration of Ca2+ in a particular water supply is

5.7 x 10-3 M. The concentration of bicarbonate ion, HC03-, in the same water is 1 .7 x 10-3 M. What masses of Ca(OHh and Na C03 must be added to 5.0 X 107 L of this water to 2reduce the level of Ca2+ to 20% of its original level?

18.45 Ferrous sulfate (FeS04) is often used as a coagulant in

water purification. The iron(II) salt is dissolved in the water to be purified, then oxidized to the iron(III) state by dissolved oxygen, at which time gelatinous Fe(OH)J forms, assuming the pH is above approximately 6. Write balanced chemical equations for the oxidation of Fe2+ 3 to Fe + by dissolved oxygen, and for the formation of 3 Fe(OH)J(s) by reaction of Fe +(aq) with HC03 -(aq).

18.46 What properties make a substance a good coagulant for

water purification?

Green Chemistry 18.47 One of the principles of green chemistry is that it is better

to use as few steps as possible in making new chemicals. How does this principle relate to energy efficiency?

18.48 Discuss how catalysts can make processes more energy

efficient.

797

Additional Exercises 18.49

The Baeyer-Villiger reaction is a classic organic oxida­ tion reaction for converting ketones to lactones, as in this reaction:

18.50

The reaction shown here was performed with an iridi­ um catalyst, both in supercritical C0 2 (scC02) and in the chlorinated solvent CH2CI2. The kinetic data for the re­ action in both solvents are plotted in the graph. Why is this a good example of a green chemical reaction? scC02 ... Clliral lr-cat

Ketone

HJC:

..

H

PhxNPh H

3-Chloroperbenzoic acid

Lactone

7 6 Time lhl

3-Chlorobenzoic acid

The reaction is used in the manufacture of plastics and pharmaceuticals. The reactant 3-chloroperbenzoic acid is somewhat shock sensitive, however, and prone to explode. Also, 3-chlorobenzoic acid is a waste product. An alter­ native process being developed uses hydrogen peroxide and a catalyst consisting of tin deposited within a solid support. The catalyst is readily recovered from the reac­ tion mixture. (a) What would you expect to be the other product of oxidation of the ketone to lactone by hydro­ gen peroxide? (b) What principles of green chemistry are addressed by use of the proposed process?

lr-catal)·st

10 22

23

24

ADDITIONAL EXERCI SES 18.51

18.52

18.53

18.54

18.55

18.56

A friend of yours has seen each of the following items in newspaper articles and would like an explanation: (a) acid rain, (b) greenhouse gas, (c) photochemical smog, (d) ozone depletion. Give a brief explanation of each term, and identify one or two of the chemicals as­ sociated with each. Suppose that on another planet the atmosphere consists of 17% Kr, 38% CH4, and 45% 02. What is the average molar mass at the surface? What is the average molar mass at an altitude at which all the 02 is photodissociated?

(b) Propose a mechanism by which the presence of halons in the stratosphere could lead to the depletion of stratospheric ozone. 18.57

It is estimated that the lifetime for HFCs in the strato­ sphere is 2-7 years. If HFCs have such long lifetimes, why are they being used to replace CFCs?

[18.58]

The hydroxyl radical, OH, is formed at low altitudes via the reaction of excited oxygen atoms with water: O'(g)

If an average 03 molecule "lives" only 100-200 seconds in the stratosphere before undergoing dissociation, how can 03 offer any protection from ultraviolet radiation?

2 OH(g)

Once produced, the hydroxyl radical is very reac­ tive. Explain why each of the following series of re­ actions affects the pollution in the troposphere: (b) OH + N02 HN03 (c) OH + CO + 02 C02 + OOH OOH + NO OH + N0 2 (d) OH + CH4 H20 + CH3 CH3 + 02 OOCH3 OOCH3 + NO OCH3 + N0 2 ------+

------+

------+

What properties of CFCs make them ideal for various commercial applications but also make them a long­ term problem in the stratosphere? are fluorocarbons that contain bromine, such as CBrF3. They are used extensively as foaming agents for fighting fires. Like CFCs, halons are very unreactive and ultimately can diffuse into the stratosphere. (a) Based on the data in Table 8.4, would you expect photo­ dissociation of Br atoms to occur in the stratosphere?

------+

(Hint: It has one unpaired electron.)

Show how Equations 18.7 and 18.9 can be added to give Equation 18.10. (You may need to multiply one of the re­ actions by a factor to have them add properly.)

Halons

+ H20(g)

(a) Write the Lewis structure for the hydroxyl radical.

------+

------+

------+

18.59

Explain, using Le Chi\telier's principle, why the equilib­ rium constant for the formation of NO from N2 and 02 increases with increasing temperature, whereas the equilibrium constant for the formation of N0 2 from NO and 02 decreases with increasing temperature.

798

CHAPTER 1 8

Chemistry of the Environment

18.60 The affinity of carbon monoxide for hemoglobin is

about 210 times that of 02. Assume a person is inhaling air that contains 125 ppm of CO. If all the hemoglobin leaving the lungs carries either oxygen or CO, calculate the fraction in the form of carboxyhemoglobin.

18.61 Natural gas consists primarily of methane, CH4(g).

(a) Write a balanced chemical equation for the complete combustion of methane to produce C02(g) as the only carbon-containing product. (b) Write a balanced chemi­ cal equation for the incomplete combustion of methane to produce CO{g) as the only carbon-containing prod­ uct. (c) At 25 oc and 1.0 atm pressure, what is the mini­ mum quantity of dry air needed to combust 1.0 L of CH4(g) completely to C02(g)?

18.65

18.66

18.62 One of the possible consequences of global warming is

an increase in the temperature of ocean water. The oceans serve as a "sink" for C02 by dissolving large amounts of it. (a) How would the solubility of C02 in the oceans be affected by an increase in the temperature of the water? (b) Discuss the implications of your an­ swer to part (a) for the problem of global warming.

169 watts per square meter. The rate of energy radiated from Earth's surface averages 390 watts per square meter. Comparing these numbers, one might expect that the planet would cool quickly, yet it does not. Why not?

[18.67]

18.63 The rate of solar energy striking Earth averages

18.64 The solar power striking Earth every day averages

169 watts per square meter. The peak electrical power usage in New York City is 12,000 megawatts. Considering

[18.68]

that present technology for solar energy conversion is only about 10% efficient, from how many square meters of land must sunlight be collected in order to provide this peak power? (For comparison, the total area of the city is 2 830 krn .) Write balanced chemical equations for each of the following reactions: (a) The nitric oxide molecule under­ goes photodissociation in the upper atmosphere. (b) The nitric oxide molecule undergoes photoionization in the upper atmosphere. (c) Nitric oxide undergoes oxi­ dation by ozone in the stratosphere. (d) Nitrogen dioxide dissolves in water to form nitric acid and nitric oxide. (a) Explain why Mg(OHh precipitates when COl- ion 2 is added to a solution containing Mg + (b) Will Mg(OHh precipitate when 4.0 g of Na2C03 is added to 2 1.00 L of a solution containing 125 ppm of Mg +? It has been pointed out that there may be increased amounts of NO in the troposphere as compared with the past because of massive use of nitrogen-containing corn­ pounds in fertilizers. Assuming that NO can eventually diffuse into the stratosphere, how might it affect the conditions of life on Earth? Using the index to this text, look up the chemistry of nitrogen oxides. What chemical pathways might NO in the troposphere follow? As of the writing of this text, EPA standards limit atmos­ pheric ozone levels in urban environments to 84 ppb. How many moles of ozone would there be in the air above Los Angeles County (area about 4000 square miles; consider a height of 10 rn above the ground) if ozone was at this concentration?

INTEGRATIVE EXERCI S E S 18.69 The estimated average concentration of N02 in air in the

United States in 2006 was 0.016 ppm. (a) Calculate the partial pressure of the N02 in a sample of this air when the atmospheric pressure is 755 torr (99.1 kPa). (b) How many molecules of N02 are present under these condi­ tions at 20 oc in a room that measures 15 X 14 X 8 ft?

1986 an electrical power plant in Taylorsville, Geor­ gia, burned 8,376,726 tons of coal, a national record at that time. (a) Assuming that the coal was 83% carbon and 2.5% sulfur and that combustion was complete, cal­ culate the number of tons of carbon dioxide and sulfur dioxide produced by the plant during the year. (b) If 55% of the 502 could be removed by reaction with pow­ dered CaO to form CaS03, how many tons of CaS03 would be produced?

[18.70] In

18.71 The water supply for a midwestern city contains the fol­

lowing impurities: coarse sand, finely divided particu­ lates, nitrate ion, trihalomethanes, dissolved phosphorus in the form of phosphates, potentially harmful bacterial strains, dissolved organic substances. Which of the fol­ lowing processes or agents, if any, is effective in removing each of these impurities: coarse sand filtration, activated carbon filtration, aeration, ozonization, precipitation with aluminum hydroxide?

18.72 The concentration of H20 in the stratosphere is about

5 ppm. It undergoes photodissociation as follows: H20{g)

-----+

H(g) + OH(g)

(a) Write out the Lewis-dot structures for both products

and reactant. (b) Using Table 8.4, calculate the wavelength required to cause this dissociation. (c) The hydroxyl radicals, OH, can react with ozone, giving the following reactions: OH(g)

+ 03{g)

H02{g) + O(g)

-----+ ----->

H02{gl

+ 02{gl

OH(g) + 02{g)

What overall reaction results from these two elementary reactions? What is the catalyst in the overall reaction? Explain. 18.73 Biorernediation is the process by which bacteria repair their

environment in response, for example, to an oil spill. The efficiency of bacteria for "eating" hydrocarbons depends on the amount of oxygen in the system, pH, temperature, and many other factors. In a certain oil spill, hydrocarbons from the oil disappeared with a first-order rate constant of 2 X 10-6 s- I How many days did it take for the hydrocar­ bons to decrease to 10% of their initial value?

18.74 The standard enthalpies of formation of CIO and Cl02

are 101 and 102 kJ/rnol, respectively. Using these data and the thermodynamic data in Appendix C, calculate the overall enthalpy change for each step in the following catalytic cycle: ClO{g) + 03{g) Cl02{gl + O(g)

-----+ -----+

Cl02{g) + 02{gl ClO{g) + 02{gl

Integrative Exercises What is the enthalpy change for the overall reaction that results from these two steps? 18.75

[18.76]

The main reason that distillation is a costly method for purifying water is the high energy required to heat and vaporize water. (a) Using the density, specific heat, and heat of vaporization of water from Appendix B, calcu­ late the amount of energy required to vaporize 1.00 gal of water beginning with water at 20 oc. (b) If the energy is provided by electricity costing $0.085/kWh, calculate its cost. (c) If distilled water sells in a grocery store for $1.26 per gal, what percentage of the sales price is repre­ sented by the cost of the energy? A reaction that contributes to the depletion of ozone in the stratosphere is the direct reaction of oxygen atoms with ozone:

18.78

18.77

Nitrogen dioxide (N02) is the only important gaseous species in the lower atmosphere that absorbs visible light. (a) Write the Lewis structure(s) for N02. (b) How does this structure account for the fact that N02 dirner­ izes to form N204? Based on what you can find about this dirnerization reaction in the text, would you expect to find the N02 that forms in an urban environment to be in the form of dimer? Explain. (c) What would you expect as products, if any, for the reaction of N02 with CO? (d) Would you expect N02 generated in an urban environment to migrate to the stratosphere? Explain.

The following data was collected for the destruction of 03 by H (0 3 + H 02 + OH) at very low concentrations: ->

Experiment

2 3 (a)

[03l, M

[H], M

Initial Rate, M/s

5.17 x 10-33 2.59 x 1 0-33 5.19 x 10-33

2 3.22 x 1o- 6 2 3.25 x 10_ 6 6.46 x 10-26

1.88 x 10-1 4 9.44 X 10-1 5 3.77 X 10-1 4

Write the rate law for the reaction.

(b) Calculate the rate constant.

18.79

The degradation of CF3CH2F (an HFC) by OH radicals in the troposphere is first order in each reactant and has a rate constant of k = 1 . 6 X 108 M-1 s-1 at 4 oc. If the tropospheric concentrations of OH and CF3CH2F are 8.1 X 105 and 6.3 X 108 molecules cm-3, respectively, what is the rate of reaction at this temperature in M/s?

[18.80]

The Henry's law constant for C02 in water at 25 oc 2 is 3.1 x 10- M atm-1 (a) What is the solubility of C02 in water at this temperature if the solution is in contact with air at normal atmospheric pressure? (b) Assume that all of this C02 is in the form of H2C03 produced by the reaction between C02 and H20:

--

2 02(g) O(g) + 03(g) At 298 K the rate constant for this reaction is 4.8 x 105 M-1 s-1 (a) Based on the units of the rate con­ stant, write the likely rate law for this reaction. (b) Would you expect this reaction to occur via a single elementary process? Explain why or why not. (c) From the magni­ tude of the rate constant, would you expect the activa­ tion energy of this reaction to be large or small? Explain. (d) Use tl.HJ values from Appendix C to estimate the en­ thalpy change for this reaction. Would this reaction raise or lower the temperature of the stratosphere?

799

C02(aq) + H20(/) -- H2C03(aq) What is the pH of this solution? 2 [18.81] If the pH of a 1.0-in. rainfall over 1500 mi is 3.5, how many kilograms of H2S04 are present, assuming that it is the only acid contributing to the pH? [18.82] The precipitation of AI(OHh (Ksp = 1.3 x 10-33) is sometimes used to purify water. (a) Estimate the pH at which precipitation of AI(OHh will begin if 5.0 lb of AJ2(S04h is added to 2000 gal of water. (b) Approxi­ mately how many pounds of CaO must be added to the water to achieve this pH?

CHEMICAL THERMODYNAMICS

THE ORANGE EXTERNAL TAN K under the Space Shuttle holds tanks of hydrogen and oxygen. The reaction of hydrogen and oxygen producing water propels the Space Shuttle into orbit. Once the mixture is ignited, the reaction proceeds spontaneously to completion, generating heat and light. The reverse process, formation of hydrogen and oxygen from water, does not occur of

800

its own accord.

W H AT ' S 19. 1

19.2

A H EA D

Spontaneous Processes

In this chapter we will see why changes that occur in nature have a directional character. They move spontaneously in one direction but not in the reverse direction. Entropy and the Second Law of Thermodyna mics

We will discuss entropy, which is a thermodynamic state function that is important in determining whether a process is spontaneous. The second law of thermodynamics tells us that in any spontaneous process the entropy of the universe (system plus surroundings) increases. 19.3

19.4

19.5

19.6

The Molecular Interpretation of Entropy

On a molecular level, we will see that an entropy increase is associated with an increase in the number of accessible microstates. Entropy change can also be thought of as a measure of a system's randomness at a stated temperature. The third law of thermodynamics states that the entropy of a perfect crystalline solid at 0 K is zero.

19.7

Gibbs Free Energy

We next encounter another thermodynamic state function, free energy (or Gibbsfree energy), whkh is a measure of how far removed the system is from equilibrium. Free energy change measures the maximum amount of useful work obtainable from a process and yields information on the direction in which a chemical reaction will proceed spontaneously. Free Energy and Te mperature

We will consider how the relationship among free-energy change, enthalpy change, and entropy change provides insight into how temperature affects the spontaneity of a process. Free Energy and the Equilibrium Constant

Finally, we will consider how the standard free-energy change for a chemical reaction can be used to calculate the equilibrium constant for the process.

Entropy Changes in Chemical Reactions

Using tabulated standard molar entropies, we can calculate the standard entropy changes for systems undergoing reaction.

THE EN ERGY REQU IRED TO PROPEL THE SPACE SH UTTLE INTO SPACE is obtained from two solid-fuel booster rockets and a rocket engine that relies on the combustion of hydrogen and oxygen to form water. The hydrogen and oxygen are stored as liquids at very low temperatures in tanks mounted below the Space Shuttle. As the hydrogen and oxygen vapors are ingnited, they react very rapidly and virtually completely, producing enormous quantities of water vapor and heat. Two of the most important questions chemists ask when designing and using chemical reactions are "How fast is the reaction?" and "How far does it proceed?" The first question is addressed by the study of chemical kinetics, which we discussed in Chapter 14. The second question involves the equilibrium constant, which was the focus of Chapter 15. In Chapter 14 we learned that the rates of chemical reactions are con­ trolled largely by a factor related to energy, namely the activation energy of the reaction. cx:o (Section 14.5) In general, the lower the activation energy, the faster a reaction proceeds. In Chapter 15 we saw that equilibrium depends on 801

802

CHAPTER 1 9

Chemical Thermodynamics the rates of the forward and reverse reactions: Equilibrium is reached when the opposing reactions occur at equal rates. = (Section 15.1) Because reaction rates are closely tied to energy, it is logical that equilibrium also depends in some way on energy. In this chapter we will explore the connection between energy and the extent of a reaction. Doing so requires us to take a deeper look at chemical thermodynamics, the area of chemistry that deals with energy relationships. We first encountered thermodynamics in Chapter 5, where we discussed the nature of energy, the first law of thermodynamics, and the concept of enthalpy. Recall that the enthalpy change is the heat transferred between the system and its surroundings during a constant-pressure process. = (Section 5.3) Now we will see that reactions involve not only changes in enthalpy but also changes in entropy-another important thermodynamic quantity. Our discussion of entropy will lead us to the second law of thermodynamics, which provides insight into why physical and chemical changes tend to favor one direction over another. We drop a brick, for example, and it falls to the ground. We do not expect bricks to spontaneously rise from the ground to our outstretched hand. We light a candle, and it burns down. We do not expect a half-consumed candle to regenerate itself spontaneously, even if we have kept all the gases produced when the candle burned. Thermodynamics helps us understand the significance of this directional character of processes, whether they are exothermic or endothermic. 1 9 . 1 S P O N TANEO U S P ROCESSES

A

B

1 atm

Evacuated (a) Spontaneous

Not spontaneous

A

B

0.5 aim

0.5 atm

(b)

.A.

Figure 1 9. 1 Spontaneous

expansion of an Ideal gas Into an evacuated space. In (a) flask 8 holds

an ideal gas at 1 atm pressure and flask A is evacuated. In (b) the stopcock connecting the flasks has been opened. The ideal gas expands to occupy both flasks A and 8 at a pressure of 0.5 atm. The reverse process-all the gas molecules moving back into flask 8-is not spontaneous.

The first law of thermodynamics states that energy is conserved. = (Section 5.2) In other words, energy is neither created nor destroyed in any process, whether it is the falling of a brick, the burning of a candle, or the melting of an ice cube. Energy can be transferred between a system and the surroundings or can be converted from one form to another, but the total energy remains constant. We expressed the first law of thermodynamics mathematically as ilE = q + w, where ilE is the change in the internal energy of a system, q is the heat absorbed by the system from the surroundings, and w is the work done on the system by the surroundings. The first law helps us balance the books, so to speak, on the heat transferred between a system and its surroundings and the work done by a particular process or reaction. But the first law does not address another important feature of reactions-the extent to which they occur. As we noted in the introduction, our ex­ perience tells us that physical and chemical processes have a directional character. For instance, sodium metal and chlorine gas combine readily to form sodium chloride, which we also know as table salt. We never find table salt decomposing of its own accord to form sodium and chlorine. (Have you ever smelled chlorine gas in the kitchen or seen sodium metal on your table salt?) In both processes­ the formation of sodium chloride from sodium and chlorine and the decomposi­ tion of sodium chloride into sodium and chlorine-energy is conserved, as it must be according to the first law of thermodynamics. Yet one process occurs, and the other does not. A process that occurs of its own accord without any ongoing outside intervention is said to be spontaneous. A spontaneous process is one that proceeds on its own without any outside assistance. A spontaneous process occurs in a definite direction. Imagine you were to see a video clip in which a brick rises from the ground. You would conclude that the video is running in reverse--bricks do not magically rise from the ground! A brick falling is a spontaneous process, whereas the reverse process is nonspontaneous. A gas will expand into a vacuum as shown in Figure 19.1 ..,., but the process will never reverse itself. The expansion of the gas is spontaneous. Likewise, a nail left out in the weather will rust (Figure 19.2 � ) . In this process the iron in the nail reacts with oxygen from the air to form an iron oxide. We would never ex­ pect the rusty nail to reverse this process and become shiny. The rusting process

19.1

Spontaneous Processes

803

is spontaneous, whereas the reverse process is nonspontaneous. There are countless other examples we could cite that illustrate the same idea: Processes

that are spontaneous in one direction are nonspontaneous in the opposite direction.

Experimental conditions, such as temperature and pressure, are often impor­ tant in determining whether a process is spontaneous. Consider, for example, the melting of ice. When the temperature of the surroundings is above 0 °C at ordi­ nary atmospheric pressures, ice melts spontaneously and the reverse process­ liquid water turning into ice--is not spontaneous. However, when the surround­ ings are below 0 °C, the opposite is true. Liquid water converts into ice sponta­ neously, and the conversion of ice into water is not spontaneous (Figure 19.3 T). What happens at T = 0 °C, the normal melting point of water, when the flask of Figure 19.3 contains both water and ice? At the normal melting point of a substance, the solid and liquid phases are in equilibrium. =(Section 11.5) At this temperature the two phases are interconverting at the same rate and there is no preferred direction for the process. It is important to realize that the fact that a process is spontaneous does not necessarily mean that it will occur at an observable rate. A chemical reaction is spontaneous if it occurs on its own accord, regardless of its speed. A sponta­ neous reaction can be very fast, as in the case of acid-base neutralization, or very slow, as in the rusting of iron. Thermodynamics can tell us the direction and extent of a reaction but tells us nothing about the speed of the reaction.

Spontaneous

&

Figure 19.2

Non­ spontaneous

A spontaneous

process. Elemental iron in the shiny nail in the top photograph spontaneously combines with H20 and 02 in the surrounding air to form a layer of rust­ Fe203--on the nail surface.

Figure 19.3 Spontaneity can depend on the temperature. At T > 0 oc ice melts spontaneously to liquid water. At T < 0 oc the reverse process, water freezing to ice, is spontaneous. At T = 0 oc the two states are in equilibrium.

&

- SAMPLE EXERCISE 19.1

I Identifying Spontaneous Processes

Predict whether the following processes are spontaneous as described, spontaneous in the reverse direction, or in equilibrium: (a) When a piece of metal heated to 150 oc is added to water at 40 oc, the water gets hotter. (b) Water at room temperature decomposes into H2(g) and 02(g), (c) Benzene vapor, C6H6(g), at a pressure of 1 atrn condenses to liquid benzene at the normal boiling point of benzene, 80.1 °C. SOLUTION Analyze: We are asked to judge whether each process will proceed spontaneously in

the direction indicated, in the reverse direction, or in neither direction.

Plan: We need to think about whether each process is consistent with our experience

about the natural direction of events or whether we expect the reverse process to occur.

Solve: (a) This process is spontaneous. Whenever two objects at different tempera­ tures are brought into contact, heat is transferred from the hotter object to the colder one. Thus, heat is transferred from the hot metal to the cooler water. The final tern­ perature, after the metal and water achieve the same temperature (thermal equilibri­ um), will be somewhere between the initial temperatures of the metal and the water. (b) Experience tells us that this process is not spontaneous-we certainly have never seen hydrogen and oxygen gases spontaneously bubbling up out of water! Rather, the

804

CHAPTER 1 9

Chemical Thermodynamics reverse process-the reaction of H2 and 02 to form H20-is spontaneous. (c) By defin­ ition, the normal boiling point is the temperature at which a vapor at 1 atrn is in equi­ librium with its liquid. Thus, this is an equilibrium situation. If the temperature were below 80.1 oc, condensation would be spontaneous. - PRACTICE EXERCISE

Under 1 atrn pressure C02(s) sublimes at -78 °C. Is the transformation of C02(s) to C02(g) a spontaneous process at -100 °C and 1 atrn pressure? Answer: No, the reverse process is spontaneous at this temperature.

Seeking a Criterion for Spo ntaneity A marble rolling down an incline or a brick falling from your hand loses energy. The loss of energy is a common feature of spontaneous change in mechanical sys­ tems. During the 1870s Marcellin Bertholet (1827-1907), a famous chemist of that era, suggested that the direction of spontaneous changes in chemical systems was also determined by the loss of energy. He proposed that all spontaneous chemical and physical changes were exothermic. It takes only a few moments, however, to find exceptions to this generalization. For example, the melting of ice at room temperature is spontaneous even though melting is an endothermic process. Similarly, many spontaneous solution processes, such as the dissolving of NH4N03, are endothermic, as we discovered in Section 13.1. We conclude therefore that, although the majority of spontaneous reactions are exothermic, there are spontaneous endothermic ones as well. Clearly, some other factor must be at work in determining the natural direction of processes. What is this factor? To understand why certain processes are spontaneous, we need to consider more closely the ways in which the state of a system can change. Recall that quantities such as temperature, internal energy, and enthalpy are state functions, properties that define a state and do not depend on how we reach that state. (Section 5.2) The heat transferred between a system and its surroundings, q, and the work done by or on the system, w, are not state functions. The values of q and w depend on the specific path taken from one state to another. One of the keys to understanding spontaneity is distinguishing between reversible and irreversible paths between states. coo

G IVE IT SOME THOUGHT If a process is nonspontaneous, does that mean the process cannot occur under any circumstances?

Reversible and Irreversible Processes 1n 1824 a 28-year-old French engineer named Sadi Camot (1796-1832) published an analysis of the factors that determine how efficiently a steam engine can con­ vert heat to work. Camot considered what an ideal engine, one with the highest possible efficiency, would be like. He observed that it is impossible to convert the energy content of a fuel completely to work because a significant amount of heat is always lost to the surroundings. Camot's analysis gave insight into how to build better, more efficient engines, and it was one of the earliest studies in what has developed into the discipline of thermodynamics. About 40 years later, Rudolph Clausius (1822-1888), a German physicist, extended Carnot's work in an important way. Clausius concluded that a special significance could be ascribed to the ratio of the heat delivered to an ideal engine and the temperature at which it is delivered, q/T. He was so convinced of the importance of this ratio that he gave it a special name, entropy. He deliberately selected the name to sound like energy to emphasize his belief that the impor­ tance of entropy was comparable to that of energy.

19.1

Spontaneous Processes

805

An ideal engine, one with the maximum efficiency, operates under an ideal set of conditions in which all the processes are reversible. In a reversible process, a system is changed in such a way that the system and surroundings can be restored to their original state by exactly reversing the change. In other words, we can completely restore the system to its original condition with no net change to either the system or its surroundings. An irreversible process is one that can­ not simply be reversed to restore the system and its surroundings to their original states. What Carnot discovered is that the amount of work we can extract from any spontaneous process depends on the manner in which the process is carried out. A reversible change produces the maximum amount of work that can be achieved by the system on the surroundings (wrev = Wmax ) .

G IVE IT S O M E THOUGHT If you evaporate water and then condense it, have you necessarily performed a reversible process?

Let's examine some examples of reversible and irreversible processes. When two objects at different temperatures are in contact, heat will flow spontaneously from the hotter object to the colder one. Because it is impossible to make heat flow in the opposite direction, the flow of heat is irreversible. Given these facts, can we imagine any conditions under which heat transfer can be made reversible? Consider two objects or a system and its surroundings that are at essentially the same temperature, with just an infinitesimal difference (an extremely small tem­ perature difference, aT) to make heat flow in the desired direction (Figure 19.4 �) . We can then reverse the direction of heat flow by making an infinitesimal change of temperature in the opposite direction. Reversible processes are those that reverse di­

rection whenever an infinitesii1Ull change is made in some property of the system. Now consider another example, the expansion of an ideal gas at constant tem­ perature. A constant-temperature process such as this is said to be isothermal. To keep the example simple, consider the gas in the cylinder-and-piston arrange­ ment shown in Figure 19.5 -.-. When the partition is removed, the gas expands spontaneously to fill the evacuated space. Because the gas is expanding into a vac­ uum with no external pressure, it does no P-V work on the surroundings (w = 0). CXlO (Section 5.3) We can use the piston to compress the gas back to its original state, but doing so requires that the surroundings do work on the system (w > 0). That is, reversing the process has produced a change in the surroundings as energy is used to do work on the system. The fact that the system and the sur­ roundings are not both returned to their original conditions indicates that the process is irreversible. Movable partition

Vacuum

(a)

(b)

(c)

.6. F igure 19.5 An Irreversible process. Restoring the system to its original state after an irreversible process changes the surroundings. In (a) the gas is confined to the right half of the cylinder by a partition. When the partition is removed (b), the gas spontaneously (irreversibly) expands to fill the whole cylinder. No work is done by the system during this expansion. In (c) we can use the piston to compress the gas back to its original state. Doing so requires that the surroundings do work on the system, which changes the surroundings forever.

Surroundings (a)

Surroundings (b)

.6. F igure 1 9.4 Reversible flow of heat. Heat can flow reversibly between a system and its surroundings if the two have only an infinitesimally small difference in temperature, � T. The direction of heat flow can be changed by increasing or decreasing the temperature of the system by � T. (a) Increasing the temperature of the system by � T causes heat to flow from the system to the surroundings. (b) Decreasing the temperature of the system by � T causes heat to flow from the surroundings into the system.

806

CHAPTER 1 9

Chemical Thermodynamics What might a reversible, isothermal expansion of an ideal gas be like? It will occur only if the external pressure acting on the piston exactly balances the pressure exerted by the gas. Under these conditions, the piston will not move unless the external pressure is reduced infinitely slowly, allowing the pressure of the confined gas to readjust to maintain a balance in the two pres­ sures. This gradual, infinitely slow process in which the external pressure and internal pressure are always in equilibrium is reversible. If we reverse the process and compress the gas in the same infinitely slow manner, we can re­ turn the gas to its original volume. The complete cycle of expansion and com­ pression in this hypothetical process, moreover, is accomplished without any net change to the surroundings. Because real processes can at best only approximate the slow, ever-in­ equilibrium change associated with reversible processes, all real processes are irreversible. Further, the reverse of any spontaneous process is a nonsponta­ neous process. A nonspontaneous process can occur only if the surroundings do work on the system. Thus, any spontaneous process is irreversible. Even if we return the system to the original condition, the surroundings will have changed.

1 9.2 ENTROPY AN D T H E S E C O N D LAW OF T HERMO DYNA M I C S We are now closer to understanding spontaneity because we know that any spontaneous process is irreversible. But how can we use this idea to make pre­ dictions about the spontaneity of an unfamiliar process? Understanding spon­ taneity requires us to examine the thermodynamic quantity called entropy. Entropy has been variously associated with the extent of randomness in a system or with the extent to which energy is distributed or dispersed among the vari­ ous motions of the molecules of the system. In fact, entropy is a multifaceted concept whose interpretations are not so quickly summarized by a simple defi­ nition. In this section we consider how we can relate entropy changes to heat transfer and temperature. Our analysis will bring us to a profound statement about spontaneity that we call the second law of thermodynamics. In Section 19.3 we examine the molecular significance of entropy. Entropy Change

The entropy, S, of a system is a state function just like the internal energy, E, and enthalpy, H. As with these other quantities, the value of S is a characteristic of the state of a system. = (Section 5.2) Thus, the change in entropy, tJ.S, in a sys­ tem depends only on the initial and final states of the system and not on the path taken from one state to the other: tJ.S = Sfinal - Sinitial

[19.1]

For the special case of an isothermal process, tJ.S is equal to the heat that would be transferred if the process were reversible, qrev' divided by the temper­ ature at which the process occurs: tJ.S =

qrev

T

(constant T)

[19.2]

Because S is a state function, we can use Equation 19.2 to calculate tJ.S for any isothermal process, not just those that are reversible. If a change between two states is irreversible, we calculate tJ.S by using a reversible path between the states.

19.2

Entropy and the Second Law of Thermodynamics

807

G IVE IT SOME THOUGHT How do we reconcile the fact that S is a state function but that tJ.S depends on q, which is not a state function?

,iS for Phase Changes

The melting of a substance at its melting point and the vaporization of a sub­ stance at its boiling point are isothermal processes. Consider the melting of ice. At 1 atrn pressure, ice and liquid water are in equilibrium with each other at 0 °C. Imagine that we melt one mole of ice at 0 °C, 1 atrn to form one mole of liquid water at 0 °C, 1 atm. We can achieve this change by adding a certain amount of heat to the system from the surroundings: q = LlHfusion· Now imag­ ine that we carry out the change by adding the heat infinitely slowly, raising the temperature of the surroundings only infinitesimally above 0 °C. When we make the change in this fashion, the process is reversible. We can reverse the process simply by infinitely slowly removing the same amount of heat, LlHfusion' from the system, using immediate surroundings that are infinitesimally below 0 °C. Thus, qrev = LlHfusion and T = 0 °C = 273 K. The enthalpy of fusion for H20 is LlHfusion = 6.01 kJ/mol. (The melting is an endothermic process, and so the sign of llH is positive.) Thus, we can use Equation 19.2 to calculate LlSfusion for melting one mole of ice at 273 K: LlSfusion

=

qrev

llHfusion

T = --T -

=

3 (1 mol)(6.01 X 10 J/mol) 273 K

=

22 .O l K

Notice that the units for llS, J/K, are energy divided by absolute temperature, as we expect from Equation 19.2.

- SAMPLE EXERCISE 19.2

I Calculating tJ.S for a Phase Change

The element mercury, Hg, is a silvery liquid at room temperature. The normal freezing point of mercury is -38.9 °C, and its molar enthalpy of fusion is tJ.Hfusion = 2.29 k)/mol. What is the entropy change of the system when 50.0 g of Hg(l) freezes at the normal freezing point? SOLUTION Analyze: We first recognize that freezing is an exothem1ic process; heat is transferred from the system to the surroundings when a liquid freezes (q < 0). The enthalpy of fusion is tJ.H for the melting process. Because freezing is the reverse of melting, the en­

(

)(

thalpy change that accompanies the freezing of 1 mol of Hg is -tJ.Hfusion

=

Plan: We can use - tJ.Hfusion and the atomic weight of Hg to calculate q for freezing 50.0 g of Hg:

1 mol Hg q = (50.0 g Hg) 200.59 g Hg

-2.29 kj 1 mol Hg

We can use this value of q as q,,v in Equation 19.2. We must first, how­ ever, convert the temperature to K:

-38.9 oc = (-38.9 + 273.15) K

=

)(lkJI)

-2.29 k)/mol.

1000

= -571 j

234.3 K

Solve: We can now calculate the

-571 j tJ.Ssys = q,.v = = -2.44 J/K value of tJ.Ssys T 234.3 K Check: The entropy change is negative because heat flows from the system, making qrev negative.

Comment: The procedure we have used here can be used to calculate tJ.S for other isothermal phase changes, such as the vapor­ ization of a liquid at its boiling point.

- PRACTICE EXERCISE

The normal boiling point of ethanol, C2H50H, is 78.3 oc, and its molar enthalpy of vaporization is 38.56 k)/mol. What is the change in entropy in the system when 68.3 g of C2H50H(g) at 1 atrn condenses to liquid at the normal boiling point? Answer: -163 J/K

808

CHAPTER

19

Chemical Thermodynamics T H E ENTROPY CHANGE WHEN A GAS EXPA N D S I S O T H E RMALLY

I out, or more random, the system's entropy increases. Thus,

n general, we will see that if a system becomes more spread

we expect the spontaneous expansion of a gas to result in an increase in entropy. To illustrate how the entropy change asso­ ciated with an expanding gas can be calculated, consider the expansion of an ideal gas that is initially constrained by a pis­ ton, as in Figure 19.5(c). If the gas undergoes a reversible isothermal expansion, the work done on the surroundings by the moving piston can be calculated with the aid of calculus:

Wrev = -nRT In Yt2 V

In this equation, n is the number of moles of gas, R is the gas constant, T is the absolute temperature, V1 is the initial volume, and V2 is the final volume. Notice that if V2 > Vv as it must be in our expansion, then Wrev < 0, meaning that the expanding gas does work on the surroundings. One of the characteristics of an ideal gas is that its inter­ nal energy depends only on temperature, not on pressure. Thus, when an ideal gas expands at a constant tempera­ ture, tlE = 0. Because tlE = qrev + Wrev = 0, we see that

qrev = -w,ev = nRT ln(V2fVt)· Then, using Equation 19.2, we can calculate the entropy change in the system:

tlSsys

=

qrev y

=

nRT in �

Vt ___ _

T

= nR In �

[19.3]

V1

For 1.00 L of an ideal gas at 1.00 atrn and 0 •c, we can calculate the number of moles, n = 4.46 x 10-2 mol. The gas constant, R, can be expressed in units of J/mol-K, 8.314 J/mol-K (Table 10.2). Thus, for the expansion of the gas from 1.00 L to 2.00 L, we have flSsys

(

J

= (4.46 X 10-2 mol) 8.314 -= 0.26 J/K

-

mo 1-K

)(

2.00 L

In - 1.00 L

)

In Section 19.3 we will see that this increase in entropy is a measure of the increased randomness of the molecules be­ cause of the expansion.

Related Exercises: 19.27 and 19.28

The Second Law of Thermodynam ics The key idea of the first law of thermodynamics is that energy is conserved in any process. Thus, the quantity of energy lost by a system equals the quantity gained by its surroundings. = (Section 5.1) We will see, however, that entropy is different because it actually increases in any spontaneous process. Thus, the sum of the entropy change of the system and surroundings for any spontaneous process is always greater than zero. Entropy change is like a signpost indicating whether a process is spontaneous. Let's illustrate this generalization by again considering the melting of ice, designating the ice and water as our system. Let's calculate the entropy change of the system and the entropy change of the surroundings when a mole of ice (a piece roughly the size of an ordinary ice cube) melts in the palm of your hand. The process is not reversible because the system and surroundings are at different temperatures. Nevertheless, because !!. S is a state function, the entropy change of the system is the same regardless of whether the process is reversible or irreversible. We calculated the entropy change of the system just before Sample Exercise 19.2: (1mol)(6.01 X 103 J/mol) J qrev = 22.0 !!. Ssys = T = K 273 K The surroundings immediately in contact with the ice are your hand, which we will assume is at body temperature, 37 •c = 310 K. The heat lost by your hand is equal in magnitude to the heat gained by the ice but has the opposite sign, -6.01 X 103 J/mol. Hence the entropy change of the surroundings is (1 mol)( -6.01 X 103 J/mol) J qrev fl Ssurr = T = = -19.4 K 310 K Thus, the total entropy change is positive:

19.3

The Molecular Interpretation of Entropy

If the temperature of the surroundings were not 310 K but rather some temper­ ature infinitesimally above 273 K, the melting would be reversible instead of irreversible. In that case the entropy change of the surroundings would equal -22.0 J/K and aStotal would be zero. In general, any irreversible process results in an overall increase in entropy, whereas a reversible process results in no overall change in entropy. This gener­ al statement is known as the second law of thermodynamics. The sum of the entropy of a system plus the entropy of the surroundings is everything there is, and so we refer to the total entropy change as the entropy change of the uni­ verse, asuniv· We can therefore state the second law of thermodynamics in terms of the following equations:

Reversible process: Irreversible process:

asuniv = aSsys + aSsurr = 0 aSuniv = aSsys + aSsurr > 0

[19.4]

All real processes that occur of their own accord are irreversible (with reversible processes being a useful idealization) . These processes are also spontaneous. Thus, the total entropy of the universe increases in any spontaneous process. This profound generalization is yet another way of expressing the second law of thermodynamics.

G IVE IT SOME THOUGHT

The rusting of iron is accompanied by a decrease in the entropy of the system (the iron and oxygen). What can we conclude about the entropy change of the surroundings?

The second law of thermodynamics tells us the essential character of any spontaneous change-it is always accompanied by an overall increase in en­ tropy. We can, in fact, use this criterion to predict whether processes will be spontaneous. Before beginning to use the second law to predict spontaneity, however, we will find it useful to explore further the meaning of entropy from a molecular perspective. Throughout most of the remainder of this chapter, we will focus mainly on the systems we encounter rather than on their surroundings. To simplify the notation, we will usually refer to the entropy change of the system merely as as rather than explicitly indicating assys·

1 9.3 THE MO LECULAR INTERP RETATION OF ENTROPY As chemists, we are interested in molecules. What does entropy have to do with them and their transformations? What molecular property does entropy reflect? Ludwig Boltzmann (1844-1906) gave conceptual meaning to entropy. To understand Boltzmann's contribution, we need to examine the ways in which molecules can store energy. Molecular Motions and Energy When a substance is heated, the motion of its molecules increases. When we studied the kinetic-molecular theory of gases, we found that the average kinet­ ic energy of the molecules of an ideal gas is directly proportional to the absolute temperature of the gas. ooo (Section 10.7) That means the higher the tempera­ ture, the faster the molecules are moving and the more kinetic energy they pos­ sess. Moreover, hotter systems have a broader distribution of molecular speeds, as you can see by referring back to Figure 10.18. The particles of an ideal gas, however, are merely idealized points with no volume and no bonds, points that we visualize as flitting around through space. Real molecules can undergo more complex kinds of motions.

809

810

CHAPTER 1 9

Chemical Thermodynamics

/ L----- Vibrations -------' .i. Figure 1 9.6 VIbrational and rotational motions In a water molecule. Vibrational motions in the

molecule involve periodic displacements of the atoms with respect to one another. Rotational motions involve the spinning of a molecule about an axis .

L---- Rotation ----'

Molecules can undergo three kinds of motion. The entire molecule can move in one direction, as in the motions of the particles of an ideal gas or the motions of larger objects, such as a baseball being thrown around a baseball field. We call such movement translational motion. The molecules in a gas have more freedom of translational motion than those in a liquid, which in tum have more translational freedom than the molecules of a solid. A molecule may also undergo vibrational motion, in which the atoms in the molecule move periodically toward and away from one another, much as a tuning fork vibrates about its equilibrium shape. In addition, molecules may possess rotational motion, as though they were spinning like tops. Figure 19.6 .i. shows the vibrationa! motions and one of the rotationa! motions possible for the water molecule. These different forms of motion are ways in which a molecule can store energy, and we refer to the various forms collectively as the "motional energy" of the molecule.

G IVE IT SOME THOUGHT What kinds of motion can a molecule undergo that a single atom cannot?

Boltzmann's Equation and Microstates The science of thermodynamics developed as a means of describing the proper­ ties of matter in our macroscopic world without regard to the microscopic structure of the matter. In fact, thermodynamics was a well-developed field be­ fore the modern views of atomic and molecular structure were even known. The thermodynamic properties of water, for example, addressed the behavior of bulk water (or ice or water vapor) as a substance, without considering any specific properties of individual H20 molecules. To connect the microscopic and macroscopic descriptions of matter, scien­ tists have developed the field of statistical thermodynamics, which uses the tools of statistics and probability to provide the link between the microscopic and macroscopic worlds. Here we will show how entropy, which is a property of bulk matter, can be connected to the behavior of atoms and molecules. Because the mathematics of statistical thermodynamics is quite complex, our discussion will be largely conceptual. Let's begin by considering one mole of an ideal gas in a particular thermody­ namic state, which we can define by specifying the temperature, T, and volume, V, of the gas. (Remember that the energy, E, of an ideal gas depends only on its temperature and that by fixing the values of n, T, and V, we also fix the value of the pressure, P.) What is happening to our ideal gas sample at the microscopic level, and how does what is going on at the microscopic level relate to the en­ tropy of the sample? To address these questions, we need to consider both the po­ sitions of the gas molecules and their individual kinetic energies, which depend on the speeds of the molecules. In our discussion of the kinetic-molecular theory, we considered the gas molecules to be in constant motion within the entire volume of the container. We also saw that the speeds of the gas molecules follow a well-defined distribution at a given temperature, such as that shown in Figure 10.18. cx:o (Section 10.7)

19.3

The Molecular Interpretation of Entropy

Now imagine that we could take a "snapshot" of the positions and speeds of all of the molecules at a given instant. That particular set of 6 x 10 23 positions and energies of the individual gas molecules is what we call a microstate of the thermodynamic system. A mi­ crostate is a single possible arrangement of the positions and kinetic energies of the gas molecules when the gas is in a specific thermody­ namic state. We could envision continuing to take snapshots of our system to see other possible microstates. In fact, as you no doubt see, there would be such a staggeringly large number of microstates that taking individual snapshots of all of them is not feasible. Because we are examining such a large number of particles, however, we can use the tools of statistics and probability to determine the total number of microstates for the thermodynamic state. (That is where the statistical part of statistical thermodynamics comes in.) Each thermodynamic state has a characteristic number of microstates associated with it, and we will use the symbol W for that number. The connection between the number of microstates of a system, W, and its entropy, 5, is expressed in a beautifully simple equation developed by Boltzmann: S = k ln W [19.5 ] In this equation, k is Boltzmann's constant, 1.38 X 10-23 J/K. Thus, entropy is a

measure of how many microstates are associated with a particular macroscopic state. Equation 19.5 appears on Boltzmann's gravestone (Figure 19.7 ,.. ) .

GIVE IT SOME THOUGHT What is the entropy of a system that has only a single microstate? The entropy change accompanying any process is

Wfinal !J.S = k In Wfinal - k In Winitial = k I n -ltVirutial

[19.6]

Thus, any change in the system that leads to an increase in the number of micro­ states leads to a positive value of !J.S: Entropy increases with the number of microstates

of the system.

Let's briefly consider two simple changes to our ideal-gas sample and see how the entropy changes in each case. First, suppose we increase the volume of the system, which is analogous to allowing the gas to expand isothermally. A greater volume means that there are a greater number of positions available to the gas atoms. Thus, there will be a greater number of microstates for the system after the increase in volume. The entropy therefore increases as the volume in­ creases, as we saw in the "A Closer Look" box in Section 19.2. Second, suppose we keep the volume fixed but increase the temperature. How will this change af­ fect the entropy of the system? Recall the distribution of molecular speeds pre­ sented in Figure 10.18. An increase in temperature increases the average (rms) speed of the molecules and broadens the distribution of speeds. Hence, the mole­ cules have a greater number of possible kinetic energies, and the number of mi­ crostates will once again increase. The entropy of the system will therefore increase with increasing temperature. If we consider real molecules instead of ideal-gas particles, we must also con­ sider the different amounts of vibrational and rotational energies the molecules have in addition to their kinetic energies. A collection of real molecules therefore has a greater number of microstates available than does the same number of ideal-gas particles. In general, the number of microstates available to a system increas­

es with an increase in volume, an increase in temperature, or an increase in the number of molecules because any of these changes increases the possible positions and energies of the molecules of the system.

k

lo�

811

\

.A. Figure 19.7 Ludwig Boltzmann's gravestone. Boltzmann's gravestone in Vienna is inscribed with his famous relationship between the entropy of a state and the number of available microstates. (In Boltzmann's time, "log" was used to represent the natural logarithm.)

CHAPTER 19

812

Chemical Thermodynamics

ENTROPY AND PROBABI LITY

T plore the idea of the microstates associated with a partic­

he game of poker is sometimes used as an analogy to ex­

such as that depicted in Figure 19.1. When the stopcock is opened, the gas molecules are less constrained, and there are more possible arrangements for them (more microstates) in the larger volume. The various microstates are depicted in a schematic way in Figure 19.8't'. In this figure, we make no attempt to describe the motion of the particles, focusing instead only on their locations. The spreading of the molecules over the larger volume represents movement to the more probable state. When we use the terms randomness and disorder to de­ scribe entropy, we have to be careful not to carry an aesthetic sense of what we mean. What we must remember is that the fundamental connection to entropy is not tied directly with randomness, disorder, or energy dispersal, but with the num­ ber of available microstates.

ular state. There are about 2.6 million different five-card poker hands that can be dealt, and each of these hands can be viewed as a possible "microstate" for the hand dealt to any one player in a game. Table 19.1 't' shows two poker hands. The probability that a particular hand will contain five specific cards is the same regardless of which five cards are specified. Thus, there is an equal probability of dealing either of the hands shown in Table 19.1. However, the first hand, a royal flush (the ten through ace of a single suit), strikes us as much more highly ordered than the second hand, a "nothing." The reason for this is clear if we compare the number of five­ card arrangements that correspond to a royal flush to the number corresponding to a nothing: only four hands (microstates) for a royal flush but more than 1.3 million for a nothing hand. The nothing state has a higher probability of being dealt from a shuffled deck than the royal-flush state because there are so many more arrangements of cards that correspond to the nothing state. In other words, the value of W in Boltzmann's equa­ (a) (b) tion (Equation 19.5) is much greater for a nothing than for a royal flush. This example teaches us that there is a connection be­ .6. Figure 1 9.8 Probability and the locations of gas molecules. The two molecules are colored red and blue to keep track of them. (a) Before the stopcock is tween probability and entropy. opened, both molecules are in the right-hand flask. (b) After the stopcock is opened, The entropy of any system has a natural there are four possible arrangements of the two molecules. Only one of the four tendency to increase, because increased en­ arrangements corresponds to both molecules being in the right-hand flask. The tropy represents a movement toward a state greater number of possible arrangements corresponds to greater disorder in the of higher probability. Let's use this reasoning system. In general, the probability that the molecules will stay in the original flask to explain the isothermal expansion of a gas, is { !)", where n is the number of molecules.

Hand

State

l l l liJ I

Royal flush

"Nothing"

Number of Hands that Lead to This State

4

1,302,540

Chemists use several different ways to describe an increase in the number of microstates and therefore an increase in the entropy for a system. Each of these ways seeks to capture a sense of the increased freedom of motion that causes molecules to spread out if not restrained by physical barriers or chemical bonds.

19.3

The Molecular Interpretation of Entropy

Some say the increase in entropy represents an increase in the

randomness or disorder of the system. Others liken an increase in entropy to an increased dispersion (spreading out) of energy

because there is an increase in the number of ways the posi­ tions and energies of the molecules can be distributed throughout the system. Each of these descriptions (random­ ness, disorder, and energy dispersal) is conceptually help­ ful if applied correctly. Indeed, you will find it useful to keep these descriptions in mind as you evaluate entropy changes.

813

....

......

Making Qual itative Predictions About � 5

It is usually not difficult to construct a mental picture to estimate qualitatively how the entropy of a system changes during a simple process. In most instances, an increase in the number of microstates, and hence an increase in entropy, parallels an in­ crease in 1. temperature

2. volume

3. number of independently moving particles

Thus, we can usually make qualitative predictions about entropy changes by focusing on these factors. For example, when water vaporizes, the molecules spread out into a larger volume. Because they occupy a larger space, there is an increase in their freedom of motion, giving rise to more accessible microstates and hence to an increase in entropy. Consider the melting of ice. The rigid structure of the water molecules, shown in Figure 19.9 .6., restricts motion to only tiny vibrations throughout the crystal. In contrast, the molecules in liquid water are free to move about with respect to one another (translation) and to tumble around (rotation) as well as vibrate. During melting, therefore, the number of accessible microstates increa­ ses and so does the entropy. When an ionic solid, such as KCI, dissolves in water, a mixture of water and ions replaces the pure solid and pure water (Figure 19.10 T). The ions now move in a larger volume and possess more motional energy than in the rigid solid. We have to be careful, however, because water molecules are held around the ions as water of hydration. coo (Section 13.1) These water molecules have less motional energy than before because they are now con­ fined to the immediate environment of the ions. The greater the charge of an ion, the greater are the ion-dipole attractions that hold the ion and the water together, thereby restricting mo­ tions. Thus, even though the solution process is normally accompanied by a net increase in entropy, the dissolving of salts with highly charged ions can result in a net decrease in entropy.

� Figure 19.10 Dissolving an Ionic

solid In water. The ions become more spread out and random in their motions, but the water molecules that hydrate the ions become less random.

.6. Figure 19.9 Structure of Ice. The intermolecular attractions in the three­ dimensional lattice restrict the molecules to vibrational motion only.

814

CHAPTER 19

Chemical Thermodynamics The same ideas apply to systems involving chemical reactions. Con­ sider the reaction between nitric oxide gas and oxygen gas to form nitro­ gen dioxide gas: [19.7]

2 NO(g) + 02(g) .6.

Figure 1 9. 1 1

Entropy change for

In this case the reaction results in a decrease in the number of molecules­ three molecules of gaseous reactants form two molecules of gaseous products (Figure 19.11 H20(g) Ag+(aq) + Cqaq) ----> AgCl(s) 4 Fe(s) + 3 02(g) ----> 2 Fe203(s) N2(g) + 02(g) ----> 2 NO(g)

SOLUTION Analyze: We are given four equations and asked to predict the sign of chemical reaction.

6.5 for each

Plan: The sign of 6.5 will be positive if there is an increase in temperature, an in­ crease in the volume in which the molecules move, or an increase in the number of gas particles in the reaction. The question states that the temperature is constant. Thus, we need to evaluate each equation with the other two factors in mind. Solve:

(a) The evaporation of a liquid is accompanied by a large increase in volume. One mole of water (18 g) occupies about 18 mL as a liquid and if it could exist as a gas at STP it would occupy 22.4 L. Because the molecules are distributed throughout a much larger volume in the gaseous state than in the liquid state, an increase in mo­ tional freedom accompanies vaporization. Therefore, 6.5 is positive. (b) In this process the ions, which are free to move throughout the volume of the solution, form a solid in which they are confined to a smaller volume and restricted to more highly constrained positions. Thus, 6.5 is negative. (c) The particles of a solid are confined to specific locations and have fewer ways to move (fewer microstates) than do the molecules of a gas. Because 02 gas is converted into part of the solid product Fe203, 6.5 is negative. (d) The number of moles of gases is the same on both sides of the equation, and so the entropy change will be small. The sign of 6.5 is impossible to predict based on our discussions thus far, but we can predict that 6.5 will be close to zero. - PRACTICE EXERCISE

Indicate whether each of the following processes produces an increase or decrease in the entropy of the system: (a) C02(s) ----> C02(g) (b) CaO(s) + C02(gl ----> CaC03(s) (c) HCl(g) + NH3(g) ----> NH4Cl(s) (d) 2 S02(g) + 02(g) ----> 2 S03(g)

Answers: (a) increase, (b) decrease, (c) decrease, (d) decrease

19.3

- SAMPLE EXERCISE

The Molecular Interpretation of Entropy

19.4 1 Predicting Which Sample of Matter Has the Higher Entropy

Choose the sample of matter that has greater entropy in each pair, and explain your choice: (a) 1 mol of NaCl(s) or 1 mol of HCl{g) at 25 'C, (b) 2 mol of HCI{g) or 1 mol of HCl{g) at 25 'C, (c) 1 mol of HCl{g) or 1 mol of Ar{g) at 298 K. SOLUTION

We need to select the system in each pair that has the greater entropy. To do this, we examine the state of the system and the complexity of the mole­ cules it contains. Solve: (a) Gaseous HCl has the higher entropy because gases have more available motions than solids. (b) The sample containing 2 mol of HCI has twice the number of molecules as the sample containing 1 mol. Thus, the 2-mol sample has twice the num­ ber of microstates and twice the entropy when they are at the same pressure. (c) The HCl sample has the higher entropy because the HCl molecule is capable of storing en­ ergy in more ways than is Ar. HCI molecules can rotate and vibrate; Ar atoms cannot. Analyze: Plan:

- PRACTICE EXERCISE

Choose the substance with the greater entropy in each case: (a) 1 mol of H2(g) at STP or 1 mol of H2(g) at 100 'C and 0.5 atrn, (b) 1 mol of H20(s) at 0 'C or 1 mol of H20(1) at 25 'C, (c) 1 mol of H2{g) at STP or 1 mol ofS02{g) at STP, (d) 1 mol of N204{g) at STP or 2 mol of N02 {g) at STP. Answers: (a) 1 mol of H2(g) at 100 'C and 0.5 atm, (b) 1 mol of H20(/) at 25 'C, (c) 1 mol of S02{g) at STP, (d) 2 mol of N02{g) at STP

Chemis

and Li e

ENTROPY A N D LIFE

T patterns of form and color. Both plant systems and animal he ginkgo leaf shown in Figure 19.12(a)� reveals beautiful

systems, including those of humans, are incredibly complex structures in which a host of substances come together in orga­ nized ways to form cells, tissue, organ systems, and so on. These various components must operate in synchrony for the organism as a whole to be viable. lf even one key system strays far from its optimal state, the organism as a whole may die. To make a living system from its component molecules­ such as a ginkgo leaf from sugar molecules, cellulose mole­ cules, and the other substances present in the leaf-requires a very large reduction in entropy. It would seem, then, that liv­ ing systems might violate the second law of thermodynamics. They seem spontaneously to become more, not less, organized as they develop. To get the full picture, however, we must take into account the surroundings. We know that a system can move toward lower entropy if we do work on it. (That is, if we supply energy to the system in a very specific way.) When we do work on a gas, for exam­ ple, by compressing it isothermally, the entropy of the gas is lowered. The energy for the work done is provided by the sur­ roundings, and in the process the net entropy change in the universe is positive. The striking thing about living systems is that they are or­ ganized to recruit energy from their surroundings sponta­ neously. Some single-celled organisms, called autotrophs, capture energy from sunlight and store it in molecules such as sugars and fats [Figure 19.12(b)]. Others, called heterotrophs, absorb food molecules from their surroundings and then break down the molecules to provide needed energy. Whatev­ er their mode of existence, however, living systems gain their order at the expense of the surroundings. Each cell exists at the expense of an increase in the entropy of the universe.

.a.

Figure 19.12 Entropy and life. (a) This ginkgo leaf represents a highly organized living system. (b) Cyanobacteria absorb light energy and utilize it to synthesize the substances needed for growth.

815

816

CHAPTER 1 9

Chemical Thermodynamics

I> Figure 1 9. 1 3 A perfectly ordered crystalline solid at and above 0 K.

At absolute zero (left), all lattice units are in their lattice sites, devoid of thermal motion. As the temperature rises above 0 K (right), the atoms or molecules gain energy and their vibrational motion increases.

Increasing temperature

T=OK 5=0

T>OK 5>0

The Third Law of Thermodynamics If we decrease the thermal energy of a system by lowering the temperature, the energy stored in translational, vibrational, and rotational forms of motion de­ creases. As less energy is stored, the entropy of the system decreases. If we keep lowering the temperature, do we reach a state in which these motions are essen­ tially shut down, a point described by a single microstate? This question is addressed by the third law of thermodynamics, which states that the entropy ofa pure crystalline substance at absolute zero is zero: 5(0 K) = 0. Figure 19.13 .1i. shows schematically a pure crystalline solid. At absolute zero all the units of the lattice have no thermal motion. There is, therefore, only one microstate. As a result 5 = k In W = k In 1 = 0. As the temperature is increased from absolute zero, the atoms or molecules in the crystal gain energy in the form of vibrational motion about their lattice positions. Thus, the degrees of freedom of the crystal increase. The entropy of the lattice therefore increases with temper­ ature because vibrational motion causes the atoms or molecules to have a greater number of accessible microstates. What happens to the entropy of the substance as we continue to heat it? Figure 19.14 K

0

0 •

•

Equilibrium mixture

(Q = K,!lG

- SAMPLE EXERCISE

=

"

r

Pure NH3

0)

19.6 1 Calculating Free-Energy Change from !lH',

T,

and !lS'

Calculate the standard free energy change for the formation of NO(g) from N2{g) and 02{g) at 298 K: N2{g) + 02{g) -----> 2 NO{g) given that !lH' = 180.7 kj and !lS' = 24.7 )/K. Is the reaction spontaneous under these circumstances? SOLUTION

We are asked to calculate !lG' for the indicated reaction (given !lH', !lS', and T) and to predict whether the reaction is spontaneous under standard conditions at 298 K. Plan: To calculate !lG', we use Equation 19.12, !lG' = !lH' - T!lS'. To determine whether the reaction is spontaneous under standard conditions, we look at the sign of !lG'. Analyze:

Solve:

!lG'

=

= =

=

!lH' - T!lS'

180.7 kj - (298 K){24.7 J/K) 180.7 kj - 7.4 kj 173.3 kj

( 101:)J )

Because !lG' is positive, the reaction is not spontaneous under standard conditions at 298 K. Comment: Notice that we had to convert the units of the T !lS' term to kJ so that they could be added to the !lH' term, whose units are kJ. - PRACTICE EXERCISE

A particular reaction has !lH'

=

=

821

K) and the NH 3 decomposes spontaneously into N 2 and H 2. Both of these spontaneous processes are "downhill" in free energy. At equilibrium (center), Q K and the free energy is at a minimum (!lG = 0).

Spontaneous

•

Gibbs Free Energy

24.6 kJ and !lS' 132 J/K at 298 K. Calculate !lG'. Is the reaction spontaneous under these conditions? Answer: !lG' = -14.7 kJ; the reaction is spontaneous.

822

CHAPTER 1 9

Chemical Thermodynamics

TABLE 1 9.3

• Conventions Used in Establishing Standard Free Energies

State of Matter

Standard State

Solid

Pure solid

Liquid

Pure liquid

Solution

1 atm pressure 1 M concentration

Elements

Standard free energy

Gas

of formalion of an element in its standard state is defined as zero

Standard Free Energy of Formation Free energy is a state function, like enthalpy. We can tabulate standard free energies of formation for substances, just as we can tabulate standard enthalpies of formation. cxx:> (Section 5.7) It is important to remember that standard values for these functions imply a particular set of conditions, or standard states. The stan­ dard state for gaseous substances is 1 atm pressure. For solid substances, the standard state is the pure solid; for liquids, the pure liquid. For substances in solu­ tion, the standard state is normally a concentration of 1 M. (In very accurate work it may be necessary to make certain corrections, but we need not worry about these.) The temperature usually chosen for purposes of tabulating data is 25 °C, but we will calculate 6.G0 at other temperatures as well. Just as for the standard heats of formation, the free energies of elements in their standard states are set to zero. This arbitrary choice of a reference point has no effect on the quantity in which we are interested, namely, the difference in free energy between reactants and products. The rules about standard states are summarized in Table 19.3 3 C02(g) + 4 H20(/) aH' = -2220 kJ (a)

Without using data from Appendix C, predict whether aG' for this reaction is more

negative or less negative than aH'. (b) Use data from Appendix C to calculate the standard free-energy change for the reaction at 298 K. Is your prediction from part (a) correct? SOLUTION

In part (a) we must predict the value for aG' relative to that for aHo on the basis of the balanced equation for the reaction. ln part (b) we must calculate the value for aG' and compare with our qualitative prediction. Analyze:

Plan: The free-energy change incorporates both the change in enthalpy and the change in entropy for the reaction (Equation 19.11), so under standard conditions:

aG'

=

aH' - TaS'

To determine whether a co is more negative or less negative than aH', we need to determine the sign of the term T aS'. T is the absolute temperature, 298 K, so it is a positive number. We can predict the sign of aS' by looking at the reaction.

Gibbs Free Energy

823

824

CHAPTER 19

Chemical Thermodynamics Solve:

(a) We see that the reactants consist of six molecules of gas, and the products consist of three molecules of gas and four molecules of liquid. Thus, the number of molecules of gas has decreased significantly during the reaction. By using the general rules we discussed in Section 19.3, we would expect a decrease in the number of gas molecules to lead to a decrease in the entropy of the system-the products have fewer accessible microstates than the reactants. We therefore expect tl.S" and therefore T tl.Sa to be neg­ ative numbers. Because we are subtracting T tl.S", which is a negative number, we would predict that tl.G" is less negative than tl.H0• (b) Using Equation 19.14 and values from Appendix C, we can calculate the value of {).Go tl.Ga = 3tl.GJ[C02(g)] =

+ 4tl.GJ[H20(1) ]

- tl.GJ[C3H s(g) ] - S tl.GJ[Oz(g)]

3 mol(-394.4 k)/mol) + 4 mol(-237.13 k)/mol) -

1 mol( -23.47 k)/mol) - 5 mol(O kJ/mol)

=

-2108 kJ

Notice that we have been careful to use the value of tl.GJ for H20(1), as in the calcu­ lation of tl.H values, the phases of the reactants and products are important. As we predicted, tl.G" is less negative than tl.H0 because of the decrease in entropy during the reaction. - PRACTICE EXERCISE

Consider the combustion of propane to form C02(g) and H20(g) at 298 K: C3H8(g) + 5 02(g) ----> 3 C02(g) + 4 H20(g). Would you expect tl.G0 to be more negative or less negative than tl.H0? Answer: more negative

1 9 .6 F REE ENERGY AND TEMP ERATURE We have seen that tabulations of l!l.GJ, such as those in Appendix C, make it possible to calculate !!J.G0 for reactions at the standard temperature of 25• C. However, we are often interested in examining reactions at other temperatures. How is the change in free energy affected by the change in temperature? Let's look again at Equation 19.11:

!!J.G

=

!!J.H - T!!J.S

=

!!J.H

Enthalpy term

+

( -T!!J.S) Entropy term

Notice that we have written the expression for !!J.G as a sum of two contribu­ tions, an enthalpy term, !!J.H, and an entropy term, T !!J.S. Because the value of -T !!J.S depends directly on the absolute temperature T, !!J.G will vary with tem­ perature. T is a positive number at all temperatures other than absolute zero. We know that the enthalpy term, !!J.H, can be positive or negative. The entropy term, -T !!J.S, can also be positive or negative. When !!J.S is positive, which means that the final state has greater randomness (a greater number of micro­ states) than the initial state, the term -T !!J.S is negative. When !!J.S is negative, the term -T !!J.S is positive. The sign of !!J.G, which tells us whether a process is spontaneous, will de­ pend on the signs and magnitudes of !!J.H and -T!!J.S. When both !!J.H and -T !!J.S are negative, !!J.G will always be negative and the process will be sponta­ neous at all temperatures. Likewise, when both !!J.H and -T !!J.S are positive, !!J.G will always be positive and the process will be nonspontaneous at all tempera­ tures {the reverse process will be spontaneous at all temperatures). When !!J.H and -T!!J.S have opposite signs, however, the sign of !!J.G will depend on the magnitudes of these two terms. In these instances temperature is an important consideration. Generally, !!J.H and !!J.S change very little with temperature. -

19.6

Free Energy and Temperature

However, the value of T directly affects the magnitude of -T ilS. As the temper­ ature increases, the magnitude of the term -T ilS increases and it will become relatively more important in determining the sign and magnitude of .:lG. For example, let's consider once more the melting of ice to liquid water at 1 atm pressure:

This process is endothermic, which means that .:lH is positive. We also know that the entropy increases during this process, so .:lS is positive and -T .:lS is negative. At temperatures below 0 oc (273 K) the magnitude of .:lH is greater than that of -TilS. Hence, the positive enthalpy term dominates, leading to a positive value for .:lG. The positive value of .:lG means that the melting of ice is not spontaneous at T < 0 oc; rather, the reverse process, the freezing of liquid water into ice, is spontaneous at these temperatures. What happens at temperatures greater than 0 °C? As the temperature in­ creases, so does the magnitude of the entropy term -T ilS. When T > 0 oc, the magnitude of -TilS is greater than the magnitude of .:lH. At these tempera­ tures the negative entropy term dominates, which leads to a negative value for .:lG. The negative value of .:lG tells us that the melting of ice is spontaneous at T > 0 oc. At the normal melting point of water, T = 0 oc, the two phases are in equilibrium. Recall that .:lG = 0 at equilibrium; at T = 0 °C, .:lH and -T.:lS, are equal in magnitude and opposite in sign, so they cancel one another and give .:lG = 0. GIVE IT SOME THOUGHT

The normal boiling point of benzene is 80 oc. At 100 oc and 1 atm, which term is greater for the vaporization of benzene, ilH or T ilS? The possible situations for the relative signs of .:lH and .:lS are given in Table 19.4 T, along with examples of each. By applying the concepts we have developed for predicting entropy changes, we often can predict how .:lG will change with temperature. Our discussion of the temperature dependence of .:lG is also relevant to standard free-energy changes. As we saw earlier in Equation 19.12, we can cal­ culate .:lGo from .:lHo and T .:l$0: .:lG0 = .:lH0 - T 6.5°. We can readily calculate the values of .:lHo and .:lSo at 298 K from the data tabulated in Appendix C. If we assume that the values of .:lH0 and .:l$0 do not change with temperature, we can then use Equation 19.12 to estimate the value of .:lGo at temperatures other than 298 K.

TABLE 19.4 ilH

ilS

• Effect of Temperature on the Spontaneity of Reactions

-TilS

ilG = ilH - TilS

+ +

+

+

Reaction Characteristics

Example

Spontaneous at all temperatures

2 03(g) ----> 3 02(g)

Nonspontaneous at all temperatures

3 02(g) ----> 2 03(g) H 20(/) ----> H 20(s)

+

+

+

+ or -

Spontaneous at low T; nonspontaneous at high T

+ or -

Spontaneous at high T; nonspontaneous at low T

H20(s) ----> H 20(/)

825

826

CHAPTER 1 9

Chemical Thermodynamics • SAMPLE EXERCISE 19.9

the Effect of Temperature 1 onDetermining Spontaneity

The Haber process for the production of ammonia involves the equilibrium N2(gl + 3 H 2(gl � 2 NH3(g)

Assume that !J.H' and !J.S' for this reaction do not change with temperature. (a) Pre­ dict the direction in which !J.G' for this reaction changes with increasing tempera­ ture. (b) Calculate the values of !J.G' for the reaction at 25 'C and 500 'C. SOLUTION Analyze: In part (a) we are asked to predict the direction in which !J.G' for the am­ monia synthesis reaction changes as temperature increases. In part (b) we need to de­ termine !J.G' for the reaction at two different temperatures. Plan: In part (a) we can make this prediction by determining the sign of !J.S for the reaction and then using that information to analyze Equation 19.12. In part (b) we need to calculate !J.H' and !J.S' for the reaction by using the data in Appendix C. We can then use Equation 19.12 to calculate !J.G'. Solve:

(a) Equation 19.12 tells us that !J.G' is the sum of the enthalpy term !J.H' and the en­ tropy term -T!J.S'. The temperature dependence of !J.G' comes from the entropy term. We expect !J.S' for this reaction to be negative because the number of molecules of gas is smaller in the products. Because !J.S' is negative, the term -T !J.S' is positive and grows larger with increasing temperature. As a result, !J.G' becomes less nega­ tive (or more positive) with increasing temperature. Thus, the driving force for the production of NH3 becomes smaller with increasing temperature. (b) We calculated the value of !J.H' in Sample Exercise 15.14, and the value of !J.S' was determined in Sample Exercise 19.5: !J.H' = -92.38 kJ and !J.S' = - 198.3 J/K. If we assume that these values do not change with temperature, we can calculate !J.G' at any temperature by using Equation 19.12. At T = 298 K we have:

tJ.G' = -92.38 kJ - (298 K)(-198.3 J/KJ =

-92.38 kJ + 59.1 kJ

At T = 500 + 273 = 773 K we have

=

(

-33.3 kJ

!J.c' = -92.38 kJ - (773 KJ -198.3 =

-92.38 kJ + 153 kJ

Notice that we have been careful to convert added to !J.H', which has units of kJ.

.

=

-

61 kJ

C�ok� ) 1

k)C�:� ) 1

T !J.S' into units of kJ so that it can be

Comment: Increasing the temperature from 298 K to 773 K changes !J.G' from -33.3 kJ to +61 kJ Of course, the result at 773 K depends on the assumption that !J.H' and !J.S' do not change with temperature. In fact, these values do change slightly with tempera­ ture. Nevertheless, the result at 773 K should be a reasonable approximation. The posi­ tive increase in !J.G' with increasing T agrees with our prediction in part (a) of this exercise. Our result indicates that a mixture of N2(g), H2(g), and NH3(g), each present at a partial pressure of 1 atrn, will react spontaneously at 298 K to form more NH3(g). In contrast, at 773 K the positive value of !J.G' tells us that the reverse reaction is sponta­ neous. Thus, when the mixture of three gases, each at a partial pressure of 1 atrn, is heated to 773 K, some of the NH3(g) spontaneously decomposes into N2(g) and H2(g). - PRACTICE EXERCISE

---->

(a) Using standard enthalpies of formation and standard entropies in Appendix C, calculate !J.H' and !J.S' at 298 K for the following reaction: 2 S02(g) + 02(g) 2 S03(g). (b) Using the values obtained in part (a), estimate !J.G' at 400 K. Answers: (a) !J.H' = - 196.6 k}, !J.S' = -189.6 }/K; (b) !J.G' = -120.8 kj

1 9 .7 F REE ENERGY AND THE EQU I L I BRIUM CONSTANT In Section 19.5 we saw a special relationship between D.G and equilibrium: For a system at equilibrium, D.G = 0. We have also seen how we can use tabulated thermodynamic data, such as those in Appendix C, to calculate values of the standard free-energy change, D.G'. In this final section of this chapter, we will

Free Energy and the Equilibrium Constant

19.7

827

learn two more ways in which we can use free energy as a powerful tool in our analysis of chemical reactions. First, we will learn how to use the value of a co to calculate the value of llG under nonstandard conditions. Second, we will see how we can directly relate the value of a co for a reaction to the value of the equilibriurn constant for the reaction. The set of standard conditions for which aco values pertain are given in Table 19.3. Most chemical reactions occur under nonstandard conditions. For any chemical process the general relationship between the standard free-energy change, llG0, and the free-energy change under any other conditions, ac, is given by the expression llG = a co + RT In Q [19.16] In this equation R is the ideal-gas constant, 8.314 J/mol-K, Tis the absolute tem­ perature; and Q is the reaction quotient that corresponds to the particular reac­ tion mixture of interest. = (Section 15.6) Recall that the expression for Q is identical to the equilibrium-constant expression except that the reactants and products need not necessarily be at equilibrium. Under standard conditions the concentrations of all the reactants and prod­ ucts are equal to 1. Thus, under standard conditions Q = 1 and therefore In Q = 0. We see that Equation 19.16 therefore reduces to llG = a co under standard conditions, as it should.

I Relating aG to a Phase Change at Equilibrium As we saw in Section 11.5, the normal boiling point is the temperature at which a pure liquid is in equilibrium with its vapor at a pressure of 1 atm. (a) Write the chemical equation that defines the normal boiling point of liquid carbon tetrachloride, CCI4(1). (b) What is the value of a co for the equilibrium in part (a)? (c) Use thermodynamic data in Appendix C and Equation 19.12 to es­ timate the normal boiling point of CCI4.

- SAMPLE EXERCISE 19.10

SOLUTION Analyze: (a) We must write a chemical equation that describes the physical equilibrium between liquid and gaseous CCI4 at the normal boiling point. (b) We must determine the value of ac o for CCI4, in equilibrium with its vapor at the normal boiling point. (c) We must estimate the normal boiling point of CC4, based on available thermodynamic data. Plan: (a) The chemical equation will merely show the change of state of CCI4 from liquid to solid. (b) We need to analyze Equation 19.16 at equilibrium (aG = 0). (c) We can use Equation 19.12 to calculate T when aG = 0. Solve: (a) The normal boiling point of CCI4 is the temperature at which pure liquid CCI4 is in equilibrium with its vapor at a pressure of 1 atm:

(b) At equilibrium aG = 0. ln any normal boiling-point equilibrium both the liquid and the vapor are in their standard states (Table 19.2). Consequently, Q = 1, In Q = 0, and aG = a co for this process. Thus, we conclude that a co = 0 for the equilibrium involved in the normal boiling point of any liquid. We would also find that a co = 0 for the equilibria relevant to normal melting points and normal sublimation points of solids.

(c) Combining Equation 19.12 with the result from part (b), we see that the equality at the normal boiling point, Tb, of CC14(1) or any other pure liquid is Solving the equation for Tb, we obtain Strictly speaking, we would need the values of aHo and aso for the equilibrium between co.(l) and CCI4(g) at the normal boiling point to do this calcula­ tion. However, we can estimate the boiling point by using the values of aHo and aso for CCI4 at 298 K, which we can obtain from the data in Appendix C and Equations 5.31 and 19.8: Notice that, as expected, the process is endothermic (llH > 0) and produces a gas in which energy can be more spread out (AS > 0). We can now use these val­ ues to estimate Tb for CCI4(1):

aw

=

(1 mol)(-106.7 kJ/mol) - (1 mol)(-139.3 kJ/mol)

=

+32.6 kJ

aso = (1 mol)(309.4 Jjmol-K) - (1 mol)(214.4 J/mol-K) = +95.0 J/K

�

b=

aHo a so

=

(

32.6 kJ 95.0 J/K

)(

1000 J 1 kJ

)

= 343 K = 70o C

Note also that we have used the conversion factor between J and kJ to make sure that the units of aHo and AS0 match.

828

CHAPTER 1 9

Chemical Thermodynamics

Check: The experimental normal boiling point of CCI4(1) is 76.5 oc. The small deviation of our estimate from the experimental value is due to the assumption that !l.H0 and !l.S0 do not change with temperature. - PRACTICE EXERCISE

Use data in Appendix C to estimate the normal boiling point, in K, for elemental bromine, Br2(1). (The experimental value is given in Table 11.3.)

Answer: 330 K

When the concentrations of reactants and products are nonstandard, we must calculate the value of Q to determine the value of d. G. We illustrate how this is done in Sample Exercise 19.11. At this stage in our discussion, it becomes important to note the units associated with Q in Equation 19.16. The convention used for standard states imposes itself into the way Q is expressed: In Equation 19.16 the concentrations of gases are always expressed by their partial pressures in atmospheres, and solutes are expressed by their concentrations in molarity. • SAMPLE EXERCISE 19. 1 1

I Nonstandard Conditions

Calculating the Free-Energy Change under

We will continue to explore the Haber process for the synthesis of ammonia:

N2(g) + 3 H2(g) � 2 NH 3(g)

Calculate !l.G at 298 K for a reaction mixture that consists of 1.0 atm N2, 3.0 atm H;t, and 0.50 atm NH3. SOLUTION Analyze: We are asked to calculate !l.G under nonstandard conditions. Plan: We can use Equation 19.16 to calculate fl. G. Doing so requires that we calculate the value of the reaction quotient Q for the specified partial pressures of the gases and evaluate !l.G0, using a table of standard free energies of formation. Solve: Solving for the reaction quotient gives: 2 2 FN H_ ,_ = Q = __ FN, Prr/ (1.0)(3.0)3

� = 9.3 X 10-3

In Sample Exercise 19.9 we calculated !l.G0 = -33.3 kJ for this reaction. We will have to change the units of this quantity in applying Equation 19.16, however. For the units in Equation 19.16 to work out, we will use kJjmol as our units for !l.G0, where "per mole" means "per mole of the reaction as written." Thus, !l.G0 = -33.3 kJ/mol implies per 1 mol of N2, per 3 mol of H2, and per 2 mol of NH3. We can now use Equation 19.16 to calculate !l.G for these nonstandard conditions:

!l.G = !l.Go + RT ln Q

3

= (-33.3 kJ/mol) + (8.314 J/moi-K)(298 K)(1 kJ/1000 J) ln(9.3 x 10- ) = (-33.3 kJ/mol) + (-11.6 kJ/mol) = -44.9 kJ/mol Comment: We see that !l.G becomes more negative, changing from -33.3 kJ/mol to -44.9 kJ/mol, as the pressures of N;t, H2, and NH3 are changed from 1.0 atm each (standard conditions, !l.G0) to 1.0 atm, 3.0 atm, and 0.50 atm, respectively. The larger negative value for !l.G indicates a larger "driving force" to produce NH3. We would have made the same prediction based on Le Chatelier's principle. = (Section 15.7) Relative to standard conditions, we have increased the pressure of a reactant (H2) and decreased the pressure of the product (NH3). Le Chiitelier's prin­ ciple predicts that both of these changes should shift the reaction more to the product side, thereby forming more NH3. - PRACTICE EXERCISE

Calculate !l.G at 298 K for the reaction of nitrogen and hydrogen to form ammonia if the reaction mixture consists of 0.50 atm N;t, 0.75 atm H;t, and 2.0 atm NH3. Answer: -26.0 kJ/mol

19.7

Free Energy and the Equilibrium Constant

We can now use Equation 19.16 to derive the relationship between .:leo and the equilibrium constant, K. At equilibrium .:le = 0. Further, recall that the re­ action quotient, Q, equals the equilibrium constant, K, when the system is at equilibrium. Thus, at equilibrium, Equation 19.16 transforms as follows:

.:le = .:leo + RT ln Q 0 = .:leo + RT ln K .:leo = -RT ln K

[19.17]

Equation 19.17 also allows us to calculate the value of K if we know the value of .:le0• If we solve the equation for K, we obtain K = e-i'.Go(RT

[19.18]

As we pointed out in discussing Equation 19.16, some care is necessary in the choice of units. Thus, in Equations 19.17 and 19.18 we again express .:leo in kJ/mol. For the reactants and products in the equilibrium-constant expres­ sion, we use the following conventions: Gas pressures are given in atm; solution concentrations are given in moles per liter (molarity); and solids, liquids, and solvents do not appear in the expression. = (Section 15.4) Thus, for gas-phase reactions the equilibrium constant is Kp, whereas for reactions in solution it is Kc. =(Section 15.2) From Equation 19.17 we can see that if .:leo is negative, then In K must be positive. A positive value for In K means K > 1. Therefore, the more negative .:leo is, the larger the equilibrium constant, K. Conversely, if .:leo is positive, then In K is negative, which means that K < 1. Table 19.5 1> summarizes these conclu­ sions by comparing .:leo and K for both positive and negative values of .:leo.

829

TABLE 19.5 • Relationship between .lG0 and K at 298 K tl.G0, kJ/mol

+200 +100 +50 +10 +1.0 0 -1.0 -10 -50 -100 -200

K 3 8.7 x 10- 6 3.0 x 10-1 8 1.7 X 10--'J 2 1.8 x 106.7 x 10-1 1.0 1.5 5.7 5.8 3.4 1.1

X 101 X 108 X 1017 X 1035

I Calculating an Equilibrium Constant from tl.G0 Use standard free energies of formation to calculate the equilibrium constant, K, at 25 oc for the reaction involved in the Haber process: N2(g) + 3 H2(g) = 2 NH3(g)

- SAMPLE EXERCISE 19.12

The standard free-energy change for this reaction was calculated in Sample Exercise 19.9: tl.G0

=

-33.3 kJ/mol

=

-

33,300 J/mol.

SOLUTION Analyze: We are asked to calculate K for a reaction, given tl.G0• Plan: We can use Equation 19.18 to evaluate the equilibrium constant, which in this case takes the form

In this expression the gas pressures are expressed in atmospheres. (Remember that we use kJ/mol as the units of tl.G0 when using Equations 19.16, 19.17, or 19.18.) Solve: Solving Equation 19.17 for the exponent -tl.G0/RT, we have

-tl.Go RT

-( -33,300 J/mol) (8.314 J/moi-K)(298K)

= 13 4 '

We insert this value into Equation 19.18 to obtain K: Comment: This is a large equilibrium constant, which indicates that the product, NH3, is greatly favored in the equilibrium mix­ ture at 25 oc. The equilibrium constants for temperatures in the range of 300 °C to 600 °C, given in Table 15.2, are much smaller than the value at 25 oc. Clearly, a low-temperature equilibrium favors the production of ammonia more than a high-temperature one. Nevertheless, the Haber process is carried out at high temperatures because the reaction is extremely slow at room temperature. Remember: Thermodynamics can tell us the direction and extent of a reaction, but tells us nothing about the rate at which it will occur. If a catalyst were found that would permit the reaction to proceed at a rapid rate at room temperature, high pressures would not be needed to force the equilibrium toward NH3. - PRACTICE EXERCISE Use data from Appendix C to calculate the standard free-energy change, tl.G0, and the equilibrium constant, K, at 298 K for the re­ action Hz(g) + Brz(l) = 2 HBr(g). Answer: tl.G0 = -106.4 kJ/mol, K = 4 X 1018

830

CHAPTER 1 9

Chemical Thermodynamics

Chemistr and Li e

D RI V I N G N O N S P O N TA N E O U S REACT I O N S

M ber that are central to living systems, are nonsponta­

any desirable chemical reactions, incluc:ling a large num­

neous as written. For example, consider the extraction of copper metal from the mineral chalcocite, which contains Cu2S. The decomposition of Cu2S to its elements is nonspontaneous: Cu2S(s)

--->

2 Cu(s)

+

S(s)

LlG o = +86.2 kJ

Because LlGo is very positive, we cannot obtain Cu(s) directly via this reaction. Instead, we must find some way to 11do work" on the reaction to force it to occur as we wish. We can do this by coupling the reaction to another one so that the overall reaction is spontaneous. For example, we can envision the S(s) reacting with 02(gl to form S02(g): S(s)

+ 02(g)

--->

S02(g)

UG0 = -300.4 kJ

By coupling these reactions, we can extract much of the cop­ per metal via a spontaneous reaction: Cu2S(s)

+ 02(g)

--->

2 Cu(s)

+ S02(g)

UG0 = (+86.2 kJ) + (-300.4 kJ) = -214.2 kJ In essence, we have used the spontaneous reaction of S(s) with 02(gl to provide the free energy needed to extract the copper metal from the mineral. Biological systems employ the same principle of using spontaneous reactions to drive nonspontaneous ones. Many of the biochemical reactions that are essential for the forma­ tion and maintenance of highly ordered biological structures are not spontaneous. These necessary reactions are made to

occur by coupling them with spontaneous reactions that release energy. The metabolism of food is the usual source of the free energy needed to do the work of maintaining biological systems. For example, complete oxidation of the sugar glucose, C6H1206, to C02 and H20 yields substantial free energy: C6H n06(s) + 6 02(g)

--->

6 C02(g) + 6 H20(l)

UG0 = -2880 kJ

This energy can be used to drive nonspontaneous reactions in the body. However, a means is necessary to transport the ener­ gy released by glucose metabolism to the reactions that re­ quire energy. One way, shown in Figure 19.20 T, involves the interconversion of adenosine triphosphate (ATP) and adeno­ sine diphosphate (ADP), molecules that are related to the building blocks of nucleic acids. The conversion of ATP to ADP releases free energy (LlG0 = -30.5 kJ) that can be used to drive other reactions. In the human body the metabolism of glucose occurs via a complex series of reactions, most of which release free ener­ gy. The free energy released during these steps is used in part to reconvert lower-energy ADP back to higher-energy ATP. Thus, the ATP-ADP interconversions are used to store energy during metabolism and to release it as needed to drive non­ spontaneous reactions in the body. If you take a course in bio­ chemistry, you will have the opportunity to learn more about the remarkable sequence of reactions used to transport free energy throughout the human body.

Related Exercises: 19.94 and 19.95

Glucose (C6H1206)

ree energy released by ATP converts simple molecules to ore complex ones Simpler molecules A Figure 19.20 Free energy and cell metabolism. This schematic representation shows part of the free-energy changes that occur in cell metabolism. The oxidation of glucose to C02 and H20 produces free energy that is then used to convert ADP into the more energetic ATP. The ATP is then used, as needed, as an energy source to convert simple molecules into more complex cell constituents. When it releases its stored free energy, ATP is converted back into ADP.

19.7

- SAMPLE INTEGRATIVE EXERCISE

Free Energy and the Equilibrium Constant

I Putting Concepts Together

Consider the simple salts NaCl(s) and AgCl(s). We will examine the equilibria in which these salts dissolve in water to form aqueous solutions of ions: NaCl(s) = Na+(aq) + Cl-(aq) AgCl(s) = Ag+(aq) + Cqaq) (a) Calculate the value of llG0 at 298 K for each of the preceding reactions. (b) The two values from part (a) are very different. Is this difference primarily due to the en­ thalpy term or the entropy term of the standard free-energy change? (c) Use the values of a co to calculate the K'l' values for the two salts at 298 K. (d) Sodium chlo­ ride is considered a soluble salt, whereas silver chloride is considered insoluble. Are these descriptions consistent with the answers to part (c)? (e) How will aGo for the solution process of these salts change with increasing T? What effect should this change have on the solubility of the salts? SOLUTION (a) We will use Equation 19.14 along with llGJ values from Appendix C to calculate the llG�oln values for each equilibrium. (As we did in Section 13.1, we use the sub­ script "soln" to indicate that these are thermodynamic quantities for the formation of a solution.) We find

llG�oJn(NaCl) = (-261.9 kJ/mol) + (-131.2 kJ/mol) - (-384.0 kJ/mol) = -9.1 kJ/mol llG�oJn(AgCl) = ( +77.11 kJ/mol) + ( -131.2 kJ/mol) - ( -109.70 kJ/mol) = +55.6 kJ/mol

(b) We can write llG�oln as the sum of an enthalpy term, llH�oln• and an entropy term, -TllS�oln' llG�1n = llH�1n + (-TllS�ln ). We can calculate the values of llH�oln and llS�oln by using Equations 5.31 and 19.8. We can then calculate -T llS�oln at T = 298K. All these calculations are now familiar to us. The results are summa­ rized in the following table: Salt

dH�oln

d S �oln

-T8 S�oln

NaCl AgCl

+3.6 kJ/mol +65.7 kJ/mol

+43.2 J/mol-K +34.3 J/mol-K

-12.9 kJ/mol -10.2 kJ/mol

The entropy terms for the solution of the two salts are very similar. That seems sensi­ ble because each solution process should lead to a similar increase in randomness as the salt dissolves, forming hydrated ions. cxc (Section 13.1) In contrast, we see a very large difference in the enthalpy term for the solution of the two salts. The difference in the values of llG�oln is dominated by the difference in the values of llH�oln · (c) The solubility product, K,P, is the equilibrium constant for the solution process. cxc (Section 17.4) As such, we can relate Ks directly to llG�1n by using Equation 19.18: p Ksp

=

e-6.G�InfRT

We can calculate the Ksp values in the same way we applied Equation 19.18 in Sample Exercise 19.12. We use the llG�oln values we obtained in part (a), remembering to con­ vert them from kJ/mol to J/mol:

37 3 NaCl: Ksp = [Na+(aq)][Cqaq)] = e-(-9100)/[(8. 14)(298)) = e+ . = 40 3 AgCl: Ksp = [Ag+(aq)][Cqaq)] = e-(+55,600)/1(8. 14)(298)1

=

e-22.4 = 1 .9

x

10-10

The value calculated for the K� of AgCl is very close to that listed in Appendix D. (d) A soluble salt is one that d1ssolves appreciably in water. cxc (Section 4.2) The Ksp value for NaCl is greater than 1, indicating that NaCl dissolves to a great extent. The K,p value for AgCl is very small, indicating that very little dissolves in water. Silver chlo­ ride should indeed be considered an insoluble salt. (e) As we expect, the solution process has a positive value of llS for both salts (see the table in part b). As such, the entropy term of the free-energy change, -T llS�olno is nega­ tive. If we assume that llH�1n and llS�1n do not change much with temperature, then an increase in T will serve to make llG�In more negative. Thus, the driving force for disso­ lution of the salts will increase with increasing T, and we therefore expect the solubility of the salts to increase with increasing T. In Figure 13.17 we see that the solubility of NaCl (and that of nearly any salt) increases with increasing temperature. cxc (Section 13.3)

831

832

CHAPTER 19

Chemical Thermodynamics

C H APTER RE V IEW

SUMMARY AND KEY TERMS

Section 1 9. 1 Most reactions and chemical processes have an inherent directionality: They are spontaneous in

one direction and nonspontaneous in the reverse direc­ tion. The spontaneity of a process is related to the thermo­ dynamic path the system takes from the initial state to the final state. In a reversible process, both the system and its surroundings can be restored to their original state by exactly reversing the change. In an irreversible process the system cannnot return to its original state without there being a permanent change in the surroundings. Any spontaneous process is irreversible. A process that occurs at a constant temperature is said to be isothermal.

Section 1 9.2 The spontaneous nature of processes is re­ lated to a thermodynamic state function called entropy, denoted S. For a process that occurs at constant tempera­ ture, the entropy change of the system is given by the heat absorbed by the system along a reversible path, divided by the temperature: 6-S = qrevfT. The way entropy con­ trols the spontaneity of processes is given by the second law of thermodynamics, which governs the change in the entropy of the universe, 6-Suniv = 6-Ssys + 6-Ssurr· The second law states that in a reversible process 6-Suniv = 0; in an irreversible (spontaneous) process 6-Suniv > 0. Entropy values are usually expressed in units of joules per kelvin, J/K. Section 1 9.3 Molecules can undergo three kinds of mo­ tion: In translational motion the entire molecule moves in space. Molecules can also undergo vibrational motion,

in which the atoms of the molecule move toward and away from one another in periodic fashion, and rotational motion, in which the entire molecule spins like a top. A particular combination of motions and locations of the atoms and molecules of a system at a particular in­ stant is called a microstate. Entropy is a measure of the number of microstates, W, over which the energy of the system is distributed: S = k In W. The number of avail­ able microstates, and therefore the entropy, increases with an increase in volume, temperature, or motion of molecules because any of these changes increases the possible motions and locations of the molecules. As a re­ sult, entropy generally increases when liquids or solu­ tions are formed from solids, gases are formed from either solids or liquids, or the number of molecules of gas increases during a chemical reaction. The third law of

thermodynamics states that the entropy of a pure crys­

talline solid at 0 K is zero.

Section 1 9.4 The third law allows us to assign entropy

values for substances at different temperatures. Under standard conditions the entropy of a mole of a substance is called its standard molar entropy, denoted S0• From tabulated values of S0, we can calculate the entropy change for any process under standard conditions. For an isothermal process, the entropy change in the surround­ ings is equal to -6-H/T. Section 1 9.5 The Gibbs free energy (or just free energy), G, is a thermodynamic state function that

combines the two state functions enthalpy and entropy: G = H - TS. For processes that occur at constant tem­ perature, 6-G = 6-H - T 6-S. For a process occurring at constant temperature and pressure, the sign of 6-G relates to the spontaneity of the process. When 6-G is negative, the process is spontaneous. When 6-G is positive, the process is nonspontaneous but the reverse process is spontaneous. At equilibrium the process is reversible and 6-G is zero. The free energy is also a measure of the maxi­ mum useful work that can be performed by a system in a spontaneous process. The standard free-energy change, 6-Go, for any process can be calculated from tabulations of standard free energies of formation, 6-G 'j, which are defined in a fashion analogous to standard enthalpies of formation, 6-Hf. The value of 6-G 'j for a pure element in its standard state is defined to be zero. Sections 1 9.6 and 1 9. 7 The values of 6-H and 6-S gen­ erally do not vary much with temperature. Therefore, the dependence of 6-G with temperature is governed mainly by the value of T in the expression 6-G = 6-H - T 6-S. The entropy term -T 6-S has the greater effect on the tempera­ ture dependence of 6-G and, hence, on the spontaneity of the process. For example, a process for which 6-H > 0 and 6-S > 0, such as the melting of ice, can be nonsponta­ neous (6-G > 0) at low temperatures and spontaneous (6-G < 0) at higher temperatures. Under nonstandard conditions 6-G is related to 6.G0 and the value of the reac­ tion quotient, Q: 6-G = 6.G0 + RT In Q. At equilibrium (6-G = 0, Q = K), 6.G0 = -RT In K. Thus, the standard free-energy change is directly related to the equilibrium constant for the reaction. This relationship expresses the temperature dependence of equilibrium constants.

KEY SKILLS •

•

•

Understand the meaning of spontaneous process, reversible process, irreversible process, and isothermal process. State the second law of thermodynamics. Describe the kinds of molecular motion that a molecule can possess.

833

Visualizing Concepts Explain how the entropy of a system is related to the number of accessible microstates. Predict the sign of t.. S for physical and chemical processes. • State the third law of thermodynamics. • Calculate standard entropy changes for a system from standard molar entropies. • Calculate entropy changes in the surroundings for isothermal processes. • Calculate the Gibbs free energy from the enthalpy change and entropy change at a given temperature. • Use free energy changes to predict whether reactions are spontaneous. • Calculate standard free energy changes using standard free energies of formation. • Predict the effect of temperature on spontaneity given t.. H and t.. S . • Calculate t.. G under nonstandard conditions. • Relate t.. G o and equilibrium constant. •

•

KEY EQUATIONS 0

•

il.S

s=

=

• il. S0

qrev

T

(constant T)

k in w [19.5] = L nS0(products)

•

-il.Hsys il. Ssurr = ---T

•

t.. G = t.. H - Tt.. S

•

il.G0 =

•

il.G = - wmax

•

t..G = il.G0 + RT In Q

•

t.. G o = -RT ln K

Relating entropy change to the heat absorbed or released in a reversible process

[19.2]

The second law of thermodynamics (spontaneous process)

[19.4]

L m5°(reactants)

-

Relating entropy to the number of microstates [19.8]

The entropy change of the surroundings for a process at constant temperature and pressure

[19.9]

Calculating the Gibbs free-energy change from enthalpy and entropy changes at constant temperature

[19.11]

L nil.GJ(products)

Calculating the standard entropy change from standard molar entropies

-

L mil.GJ(reactants)

[19.14]

Calculating the standard free-energy change from standard free energies of formation Relating the free-energy change to the maximum work a process can perform.

[19.15] [19.16]

[19.17]

Calculating free-energy change under nonstandard conditions. Relating the standard free-energy change and the equilibrium constant.

VISUALIZING CONCEPTS 19.1 Two different gases occupy two separate bulbs. Consid­

er the process that occurs when the stopcock separating the gases is opened, assuming the gases behave ideally. (a) Draw the final (equilibrium) state. (b) Predict the signs of t.H and t.S for the process. (c) Is the process that occurs when the stopcock is opened a reversible one? (d) How does the process affect the entropy of the surroundings? [Sections 19.1 and 19.2]

the sign of t.G? (c) If energy can flow in and out of the system to maintain a constant temperature during the process, what can you say about the entropy change of the surroundings as a result of this process? [Sections 19.2 and 19.5]

• •

19.2 (a) What are the signs of t.S and t.H for the process de­

picted to the right? (b) How might temperature affect

•

• • •

• • •

834

CHAPTER 1 9

Chemical Thermodynamics

19.3 Predict the sign of tl.S accompanying this reaction.

Explain your choice. [Section 19.3]

temperature. (a) At what temperature is the system at equilibrium? (b) In what temperature range is the reac­ tion spontaneous? (c) Is tl.H positive or negative? (d) Is tl.S positive or negative? [Sections 19.5 and 19.6]

19.4 The diagram below shows the variation in entropy with

temperature for a substance that is a gas at the highest temperature shown. (a) What processes correspond to the entropy increases along the vertical lines labeled 1 and 2 in this diagram? (b) Why is the entropy change for 2 larger than that for 1? [Section 19.3]

Ternperature 19.7 Consider a reaction A2(g) + B2(g) ;;===: 2 AB(g), with

atoms of A shown in red and atoms of B shown in blue. (a) If Kc = 1, which system is at equilibrium? (b) What is the sign of tl.G for any process in which the contents of a reaction vessel move to equilibrium? (c) Rank the boxes in order of increasing magnitude of tl.G for the reaction. [Sections 19.5 and 19.7]

Temperature (K) 19.5 The diagram below shows how tl.H (red line) and T tl.S

(blue line) change with temperature for a hypothetical reaction. (a) What is the significance of the point at 300 K, where tl.H and T tl.S are equal? (b) In what temperature range is this reaction spontaneous? [Section 19.6]

ru m I ..-.1 (1)

. ,, . (2)

,

(3)

19.8 The diagram below shows how the free energy, G, changes

during a hypothetical reaction A(g) + B(g) ----> AB(g). On the left are pure reactants, each at 1 atrn, and on the right is the pure product, also at 1 atm. (a) What is the significance of the minimum in the plot? (b) What does the quantity x, shown on the right side of the diagram, represent? [Section 19.7]

300 K Temperature 19.6 Consider the accompanying diagram, which repre­

sents how tl.G for a hypothetical reaction responds to

Progress of reaction

EXERCISES Spontaneous Processes 19.9 Which of the following processes are spontaneous, and

which are nonspontaneous: (a) the ripening of a banana, (b) dissolution of sugar in a cup of hot coffee, (c) the re­ action of nitrogen atoms to form N2 molecules at 25 'C

and 1 atrn, (d) lightning, (e) formation of CH4 and 02 molecules from C02 and H20 at room temperature and 1 atm of pressure?

Exercises

processes that occur in the world around us reversible in nature? Explain.

19.10 Which of the following processes are spontaneous:

(a) the melting of ice cubes at 10 oc and 1 atm pressure; (b) separating a mixture of Nz and 02 into two separate samples, one that is pure N2 and one that is pure 02; (c) alignment of iron filings in a magnetic field; (d) the reaction of sodium metal with chlorine gas to form sodi­ um chloride; (e) the dissolution of HCl(g) in water to form concentrated hydrochloric acid?

19.11 (a) Give two examples of endothermic processes that are

spontaneous. (b) Give an example of a process that is spontaneous at one temperature but nonspontaneous at a different temperature.

19.16 (a) What is meant by calling a process

19.17 Consider a process in which an ideal gas changes from

state 1 to state 2 in such a way that its temperature changes from 300 K to 200 K. (a) Describe how this change might be carried out while keeping the volume of the gas constant. (b) Describe how it might be carried out while keeping the pressure of the gas constant. (c) Does the change in t.E depend on the particular path­ way taken to carry out this change of state? Explain.

when placed in a large, closed, dry vessel:

2 and back to state 1. (a) What is the relationship between the value of t.E for going from state 1 to state 2 to that for going from state 2 back to state 1? (b) Without further information, can you conclude anything about the amount of heat transferred to the system as it goes from state 1 to state 2 as com­ pared to that upon going from state 2 back to state 1? (c) Suppose the changes in state are reversible processes. Can you conclude anything about the work done by the system upon going from state 1 to state 2 as compared to that upon going from state 2 back to state 1?

19.18 A system goes from state 1 to state

19.13 Consider the vaporization of liquid water to steam at a

pressure of 1 atm. (a) Is this process endothermic or exothermic? (b) In what temperature range is it a spon­ taneous process? (c) In what temperature range is it a nonspontaneous process? (d) At what temperature are the two phases in equilibrium?

19.14 The normal freezing point of 1-propanol (C3H80) is

-127 oc. (a) Is the freezing of 1-propanol an endother­ mic or exothermic process? (b) In what temperature range is the freezing of 1-propanol a spontaneous process? (c) In what temperature range is it a nonspon­ taneous process? (d) Is there any temperature at which liquid and solid 1-propanol are in equilibrium? Explain.

19.15 (a) What is special about a

reversible process? (b) Sup­

irreversible?

(b) After an irreversible process the system is restored to its original state. What can be said about the condition of the surroundings after the system is restored to its original state? (c) Under what conditions will the con­ densation of a liquid be an irreversible process?

19.12 The crystalline hydrate Cd(N03)z · 4H20(s) loses water

Cd(N03)z · 4HzO(s) -----> Cd(N03)z(s) + 4HzO(g) This process occurs even though it is endothermic; that is, t.H is positive. Is thls process an exception to Bertholet's generalization? Explain.

835

19.19 Consider a system consisting of an ice cube. (a) Under

what conditions can the ice cube melt reversibly? (b) If the ice cube melts reversibly, is t.E zero for the process? Explain. 19.20 Consider what happens when a sample of the explosive

TNT (Section 8.8: "Chemistry Put to Work: Explosives and Alfred Nobel") is detonated. (a) Is the detonation a spontaneous process? (b) What is the sign of q for this process? (c) Can you determine whether w is positive, negative, or zero for the process? Explain. (d) Can you determine the sign of t.E for the process? Explain.

pose a reversible process is reversed, restoring the sys­ tem to its original state. What can be said about the surroundings after the process is reversed? (c) Under what circumstances will the vaporization of water to steam be a reversible process? (d) Are any of the

Entropy and the Second Law of Thermodynamics 19.21 (a) How can we calculate t.S for an isothermal process?

(b) Does t.S for a process depend on the path taken from the initial to the final state of the system? Explain.

19.22 Suppose we vaporize a mole of liquid water at 25

oc

and another mole of water at 100 oe, (a) Assuming that the enthalpy of vaporization of water does not change much between 25 oc and 100 °C, which process in­ volves the larger change in entropy? (b) Does the en­ tropy change in either process depend on whether we carry out the process reversibly or not? Explain.

19.23 The normal boiling point of methanol (CH30H) is 64.7

oc,

and its molar enthalpy of vaporization is t.Hvap = 71.8 kJ/mol. (a) When CH30H(I) boils at its normal boiling point, does its entropy increase or decrease?

(b) Calculate the value of is vaporized at 64.7 oe,

t.S when

1.00 mol of CH30H(I)

19.24 The element cesium (Cs) freezes at 28.4 oc, and its molar

enthalpy of fusion is t.Hfus = 2.09 kJ/mol. (a) When molten cesium solidifies to Cs(s) at its normal melting point, is t.S positive or negative? (b) Calculate the value of t.S when 15.0 g of Cs(l) solidifies at 28.4 oe ,

19.25 (a) Express the second law of thermodynamics in

words. (b) If the entropy of the system increases during a reversible process, what can you say about the entropy change of the surroundings? (c) In a cer­ tain spontaneous process the system undergoes an entropy change, t.S = 42 J/K. What can you conclude about dSsurr?

836

CHAPTER 1 9

Chemical Thermodynamics

Express the second law of thermodynamics as a mathematical equation. (b) In a particular spontaneous process the entropy of the system decreases. What can you conclude about the sign and magnitude of C.Ssun? (c) During a certain reversible process, the surroundings undergo an entropy change, C.Ssun = -78 J/K. What is the entropy change of the system for this process?

19.26 (a)

19.27

19.28

The volume of 0.100 mol of helium gas at 27 'C is increased isothermally from 2.00 L to 5.00 L. Assu­ ming the gas to be ideal, calculate the entropy change for the process. The pressure on 0.850 mol of neon gas is increased from 1.25 atm to 2.75 atm at 100 'C. Assuming the gas to be ideal, calculate C.S for this process.

The Molecular Interpretation of Entropy 19.29

19.30

How would each of the following changes affect the number of microstates available to a system: (a) increase in temperature, (b) decrease in volume, (c) change of state from liquid to gas? (a) Using the heat of vaporization in Appendix B, calcu­ late the entropy change for the vaporization of water at 25 'C and at 100 'C. (b) From your knowledge of mi­ crostates and the structure of liquid water, explain the difference in these two values. What do you expect for the sign of C.S in a chemical reaction in which two moles of gaseous reactants are converted to three moles of gaseous products? (b) For which of the processes in Exercise 19.9 does the entropy of the system increase? (a) In a chemical reaction two gases combine to form a solid. What do you expect for the sign of C.S? (b) How does the entropy of the system change in the processes described in Exercise 19.1 0?

19.37

19.38

19.31 (a)

19.32

19.33 19.34

How does the entropy of the system change when (a) a solid melts, (b) a gas liquefies, (c) a solid sublimes? How does the entropy of the system change when (a) the temperature of the system increases, (b) the vol­ ume of a gas increases, (c) equal volumes of ethanol and water are mixed to form a solution. State the third law of thermodynamics. (b) Distin­ guish between translational motion, vibrational motion, and rotational motion of a molecule. (c) lllustrate these three kinds of motion with sketches for the HCl molecule. (a) The energy of a gas is increased by heating it. Using C02 as an example, illustrate the different ways in which additional energy can be distributed among the molecules of the gas. (b) You are told that the number of microstates for a system increases. What does this tell you about the entropy of the system?

19.39

19.40

19.41

19.35 (a)

19.36

(a) Using Figure 19.14 as a model, sketch how the en­ tropy of water changes as it is heated from -50 'C to 110 'C at sea level. Show the temperatures at which there are vertical increases in entropy. (b) Which process has the larger entropy change: melting ice or boiling water? Explain. Propanol (C3H;DH) melts at -126.5 'C and boils at 97.4 'C. Draw a qualitative sketch of how the entropy changes as propanol vapor at 150 'C and 1 atm is cooled to solid propanol at -150 'C and 1 atm. For each of the following pairs, choose the substance with the higher entropy per mole at a given tempera­ ture: (a) Ar(/) or Ar(g), (b) He(g) at 3 atm pressure or He(g) at 1.5 atm pressure, (c) 1 mol of Ne(g) in 15.0 L or 1 mol of Ne(g) in 1.50 L, (d) C02(g) or C02(s). For each of the following pairs, indicate which substance possesses the larger standard entropy: (a) 1 mol of P4(g) at 300 'C. 0.01 atm, or 1 mol of As4(g) at 300 'C. 0.01 atm; (b) 1 mol of H20(g) at 100 'C, 1 atm, or 1 mol of H20(/) at 100 'C, 1 atm; (c) 0.5 mol of N2(g) at 298 K, 20-L volume, or 0.5 mol CH4(g) at 298 K, 20-L volume; (d) 100 g Na2S04(s) at 30 'C or 100 g Na2S04(aq) at 30 'C. Predict the sign of the entropy change of the system for each of the following reactions: (a) 2 S0 2(g) + 02(g) -----> 2 S03(g) (b) Ba(OH)z(s) BaO(s) + H 20(g) (c) CO(g) + 2 H2(g) -----> CHpH(l) (d) FeC12(s) + H 2(g) -----> Fe(s) + 2 HCI(g) Predict the sign of C.Ssys for each of the following process­ es: (a) Gaseous Ar is liquefied at 80 K. (b) Gaseous N204 dissociates to form gaseous N02. (c) Solid potassium re­ acts with gaseous 02 to form solid potassium superox:ide, K02 . (d) Lead bromide precipitates upon mixing Pb(N03)z(aq) and KBr(aq). ----->

19.42

Entropy Changes in Chemical Reactions each of the following pairs, which compound would you expect to have the higher standard molar entropy: (a) CzHz(g) or CzH6(g); (b) C02(g) or CO(g)? Explain. Cyclopropane and propylene isomers both have the for­ mula C3H6. Based on the molecular structures shown, which of these isomers would you expect to have the higher standard molar entropy at 25 'C?

19.43 In

19.44

H

H H

'c/

H

�\�

/

" H

Cyclopropane

H

H H

'c

/ "'==C '/

/H '­

H

Propylene

H

Exercises 19.45 Use Appendix C to compare the standard entropies at 25 oc for the following pairs of substances: (a) Sc(s) and Sc(g); (b) NH3(g) and NH3(aq); (c) 1 mol P4(g) and 2 mol P2(g); (d) C(graphite) and C(diamond). 1n each case

explain the difference in the entropy values.

19.46 Using Appendix C, compare the standard entropies at 25 oc for the following pairs of substances: (a) CuO(s) and Cu20(s); (b) 1 mol N204(g) and 2 mol N02(g); (c) Si02(s) and C02(g); (d) CO(g) and C02(g). For each

pair, explain the difference in the entropy values.

[19.47] The standard entropies at 298 K for certain of the group 4A elements are as follows: C(s, diamond) = 2.43 J/mol-K; Si(s) = 18.81 J/mol-K; Ge(s) = 31 .09 J/mol-K; and Sn(s) = 51.18 J/mol-K.

All but Sn have the diamond structure. How do you account for the trend in the so values? [19.48] Three of the forms of elemental carbon are graphite, dia­ mond, and buckminsterfullerene. The entropies at 298 K for graphite and diamond are listed in Appendix C.

837

(a) Account for the difference in the s o values of graphite and diamond in light of their structures (Figure 11.41). (b) What would you expect for the so value of buckminsterfullerene (Figure 11.43) relative to the val­ ues for graphite and diamond? Explain.

19.49 Using so values from Appendix C, calculate 6.S0 values

for the following reactions. 1n each case account for the sign of 6.S0• (a) C2H4(g) + H2(g) ---> C2H6(g) (b) N204(g) ---> 2 N02(g) (c) Be(OHh(s) ---> BeO(s) + H20(g) (d) 2 CH30H(g) + 3 02(g) ---> 2 C02(g) + 4 H20(g)

19.50 Calculate t;.So values for the following reactions by using tabulated so values from Appendix C. 1n each case explain the sign of t;.S0• (a) N2H4(g) + H2(g) ---> 2 NH3(g) (b) K(s) + 02(g) ---> K02(s) (c) Mg(OHh(s) + 2 HCl(g) ---> MgCI2(s) + 2 H20(/) (d) CO(g) + 2 H2(g) ---> CH30H(g)

Gibbs Free Energy 19.51 (a) For a process that occurs at constant temperature,

express the change in Gibbs free energy in terms of changes in the enthalpy and entropy of the system. (b) For a certain process that occurs at constant T and P, the value of t;.G is positive. What can you conclude? (c) What is the relationship between t;.G for a process and the rate at which it occurs? 19.52 (a) What is the meaning of the standard free-energy change, 6.G0, as compared with t;.G? (b) For any process that occurs at constant temperature and pressure, what is the significance of t;.G = 0? (c) For a certain process, t;.G is large and negative. Does this mean that the process necessarily occurs rapidly? 19.53 For a certain chemical reaction, t;.Ho = -35.4 kJ and t;.So = -85.5 J/K. (a) Is the reaction exothermic or en­ dothermic? (b) Does the reaction lead to an increase or

decrease in the randomness or disorder of the system? (c) Calculate t;.Go for the reaction at 298 K. (d) Is the re­ action spontaneous at 298 K under standard conditions? 19.54 A certain reaction has t;.H0 = -19.5 kJ and !;.5° = +42.7 J/K. (a) Is the reaction exothermic or endo­ thermic? (b) Does the reaction lead to an increase or decrease in the randomness or disorder of the system? (c) Calculate !;.Go for the reaction at 298 K. (d) Is the re­ action spontaneous at 298 K under standard conditions?

19.55 Using data in Appendix C, calculate t;.H0, t;.S0, and t;.G0 at 298 K for each of the following reactions. In each case show that t;.Go = t;.Ho - Tt;.S0• (a) H2(g) + F2(g) ---> 2 HF(g) (b) C(s, graphite) + 2 Cl2(g) ---> CCI4(g) (c) 2 PCI3(g) + 02(g) ---> 2 POCl3(g) (d) 2 CH30H(g) + H2(g) ---> C2H6(g) + 2 H20(g)

19.56 Use data in Appendix C to calculate t;.H0, !;.S0, and !;.Go at 25 oc for each of the following reactions. 1n each case show that t;.Go = t;.Ho - T!;.S0• (a) 2 Cr(s) + 3 Br2(g) ---> 2 CrBr3(s) (b) BaC03(s) ---> BaO(s) + C02(g) (c) 2 P(s) + 10 HF(g) ---> 2 PF5(g) + 5 H2(g) (d) K(s) + 02(g) ---> K02(s) 19.57 Using data from Appendix C, calculate !;.G0 for the

following reactions. lndicate whether each reaction is spontaneous under standard conditions. (a) 2 S02(g) + 02(g) ---> 2 S03(g) (b) N02(g) + N20(g) ---> 3 NO(g) (c) 6 Cl2(g) + 2 Fe203(s) 4 FeCI3(s) + 3 02(g) (d) S02(g) + 2 H2(g) ---> S(s) + 2 H20(g) --->

19.58 Using data from Appendix C, calculate the change in

Gibbs free energy for each of the following reactions. 1n each case indicate whether the reaction is sponta­ neous under standard conditions. (a) H2(g) + Cl2(g) 2 HCI(g) (b) MgC12(s) + H20(/) ---> MgO(s) + 2 HCl(g) (c) 2 NH3(g) ---> N2H4(g) + H2(g) (d) 2 NOCl(g) ---> 2 NO(g) + Cl#) --->

19.59 Cyclohexane (C6Hn) is a liquid hydrocarbon at room temperature. (a) Write a balanced equation for the

combustion of C6H1 2(/) to form C02(g) and H20(/).

(b) Without using thermochemical data, predict whether !;.G0 for this reaction is more negative or less negative than 6.H0•

838

CHAPTER 1 9

Chemical Thermodynamics

19.60 Sulfur dioxide reacts with strontium oxide as follows: S02{g) + SrO(s) -----> SrS03(s)

(a) Without using thermochemical data, predict whether �Go for this reaction is more negative or less negative than �H0• (b) If you had only standard enthalpy data for this reaction, how would you go about making a rough estimate of the value of �Go at 298 K, using data from Appendix C on other substances?

19.61 Classify each of the following reactions as one of the four possible types summarized in Table 19.4: (a) N2{g) + 3 F2{g) -----> 2 NF3{g) �w = -249 kJ; �so = -278 J/K (b) N2{g) + 3 Cl2{g) -----> 2 NCl3{g) �W = 460 kJ; �so = -275 J/K (c) N2F4{g) -----> 2 N�(g) �W = 85 kJ; �so = 198 J/K 19.62 From the values given for �H0 and �S0, calculate �G0 for each of the following reactions at 298 K. If the reac­ tion is not spontaneous under standard conditions at 298 K, at what temperature (if any) would the reaction become spontaneous? (a) 2 PbS(s) + 3 02(g) -----> 2 PbO(s) + 2 S02(g) �w = -844 kJ; uso = -165 J/K (b) 2 POCl3{g) -----> 2 PCI3{g) + 02{g) � H o = 572 kJ; �So = 179 J/K

19.63 A particular reaction is spontaneous at 450 K. The enthalpy change for the reaction is +34.5 kj. What can you conclude about the sign and magnitude of �S for the reaction?

19.64 A certain reaction is nonspontaneous at -25 oc. The en­ tropy change for the reaction is 95 J/K. What can you conclude about the sign and magnitude of �H?

19.65 For a particular reaction, �H = -32 kj and �S = -98 J/K. Assume that � H and �S do not vary with tem­ perature. (a) At what temperature will the reaction have �G = 0? (b) If T is increased from that in part (a), will the reaction be spontaneous or nonspontaneous?

19.66 Reactions in which a substance decomposes by losing CO are called decarbonylation reactions. The decarbony­ lation of acetic acid proceeds as follows: CH3COOH(I) -----> CH30H{g) + CO{g) By using data from Appendix C, calculate the minimum temperature at which this process will be spontaneous under standard conditions. Assume that �Ho and �so do not vary with temperature.

19.67 Consider the following reaction between oxides of nitrogen: N02{g) + N20{g) -----> 3 NO{g) (a) Use data in Appendix C to predict how �Go for the reaction varies with increasing temperature. (b) Calcu­ late �Go at 800 K, assuming that � H o and �so do not change with temperature. Under standard conditions is the reaction spontaneous at 800 K? (c) Calculate �Go at 1000 K. Is the reaction spontaneous under standard conditions at this temperature?

19.68 Methanol (CHPH) can be made by the controlled oxi­ dation of methane: CH4{g) +

! 02{g) -----> CHPH{g)

(a) Use data in Appendix C to calculate �Ho and �so for this reaction. (b) How is �Go for the reaction expect­ ed to vary with increasing temperature? (c) Calculate �Go at 298 K. Under standard conditions, is the reaction spontaneous at this temperature? (d) Is there a tempera­ ture at which the reaction would be at equilibrium under standard conditions and that is low enough so that the compounds involved are likely to be stable?

[19.69] (a) Use data in Appendix C to estimate the boiling point of benzene, C6H6(1). (b) Use a reference source, such as

the CRC Handbook of Chemistry and Physics, to find the experimental boiling point of benzene. How do you ex­ plain any deviation between your answer in part (a) and the experimental value?

[19.70] (a) Using data in Appendix C, estimate the temperature at which the free-energy change for the transformation from I2(s) to I2{g) is zero. What assumptions must you make in arriving at this estimate? (b) Use a reference source, such as WebElements (www.webelements.com), to find the experimental melting and boiling points of I2. (c) Which of the values in part (b) is closer to the value you obtained in part (a)? Can you explain why this is so?

19.71 Acetylene gas, C2H2(g), is used in welding. (a) Write a balanced equation for the combustion of acetylene gas to C02(g) and HP(I). (b) How much heat is produced in burning 1 mol of C2H2 under standard conditions if both reactants and products are brought to 298 K? (c) What is the maximum amount of useful work that can be accom­ plished under standard conditions by this reaction?

19.72 (a) How much heat is produced in burning 1 mol of ethane (C2H6) under standard conditions if reactants and products are brought to 298 K and H20(1) is formed? (b) What is the maximum amount of useful work that can be accomplished under standard condi­ tions by this system?

Free Energy and Equilibrium 19.73 Explain qualitatively how �G changes for each of the fol­ lowing reactions as the partial pressure of 02 is increased: (a) 2 CO(g) + 02(g) -----> 2 C02{g) (b) 2 H202(I) -----> 2 H20(/) + 02{g) (c) 2 KC103(s) -----> 2 KCl(s) + 3 02(g)

19.74 Indicate whether �G increases, decreases, or does not change when the partial pressure of H2 is increased in each of the following reactions:

(a) N2{g) + 3 H2{g) -----> 2 NH3{g) (b) 2 HBr(g) -----> H2{g) + Br2{g) (c) 2 H2{g) + C2H2{g) -----> C2H6{g)

19.75 Consider the reaction 2 N02(g) -----> N204(g). (a) Using data from Appendix C, calculate �Go at 298 K. (b) Cal­ culate � G at 298 K if the partial pressures of N02 and N204 are 0.40 atrn and 1.60 atrn, respectively.

Additional Exercises 19.76 Consider the reaction 6 H 2{g) + P4{g) -----> 4 PH 3{g). (a) Using data from Appendix C, calculate dG0 at 298 K. (b) Calculate dG at 298 K if the reaction mixture consists of 8.0 atm of H2, 0.050 atm of P4, and 0.22 atm of PH3 . 19.77 Use data from Appendix C to calculate the equilibrium constant, K, at 298 K for each of the following reactions: (a) Hz{g) + lz{g) = 2 HI{g) (b) C2H50H{g) = C2H4{g) + H 20{g) (c) 3 CzHz{g) = C6H 6{g) 19.78 Write the equilibrium-constant expression and calculate

the value of the equilibrium constant for each of the fol­ lowing reactions at 298 K, using data from Appendix C: (a) NaHC03(s) = NaOH(s) + C02{g) (b) 2 HBr{g) + Cl2{g) = 2 HCl{g) + Br2{g) (c) 2 SOz{g) + Oz{g) = 2 S03{g)

19.79 Consider the decomposition of barium carbonate:

BaC0 3(s) = BaO(s) + C02{g) Using data from Appendix C, calculate the equilibrium pressure of C02 at (a) 298 K and (b) 1100 K.

839

19.80 Consider the following reaction:

PbC03(s) = PbO(s) + C02{g) Using data in Appendix C, calculate the equilibrium pres­ sure of C02 in the system at (a) 400 oc and (b) 180 oc. 19.81 The value of Ka for nitrous acid (HN02) at 25 oc is given in Appendix D. (a) Write the chemical equation for the equi­ librium that corresponds to K•. (b) By using the value of K,.

calculate dG0 for the dissociation of nitrous acid in aque­ ous solution. (c) What is the value of dG at equilibrium? (d) What is the value of dG when [H+] = 5.0 x w-2 M, [N02-] = 6.0 X 10-4 M, and [HN02] = 0.20 M? 19.82 The Kb for methylamine (CH3NH2) at 25 oc is given in Appendix D. (a) Write the chemical equation for the equi­ librium that corresponds to Kb· (b) By using the value of Kb, calculate dG0 for the equilibrium in part (a). (c) What is the value of dG at equilibrium? (d) What is the value of dG when [CH 3NH 3"'] = [H+] = 1.5 X 10-8 M, [CH 3NH 3+j =5.5 X 10-4 M, and [CH 3NH2] = 0.120 M?

ADDITI ONAL EXERC ISES 19.83 Indicate whether each of the following statements is true or false. If it is false, correct it. (a) The feasibility

19.84

19.85

19.86

[19.87]

of manufacturing NH 3 from N 2 and H2 depends entirely on the value of dH for the process N2{g) + 3 H 2{g) -----> 2 NH 3{g). (b) The reaction of Na(s) with Cl2{g) to form NaCl(s) is a spontaneous process. (c) A spontaneous process can in principle be conducted reversibly. (d) Spontaneous processes in gen­ eral require that work be done to force them to proceed. (e) Spontaneous processes are those that are exothermic and that lead to a higher degree of order in the system. For each of the following processes, indicate whether the signs of dS and dH are expected to be positive, neg­ ative, or about zero. (a) A solid sublimes. (b) The tem­ perature of a sample of Co(s) is lowered from 60 oc to 25 oc. (c) Ethyl alcohol evaporates from a beaker. (d) A diatomic molecule dissociates into atoms. (e) A piece of charcoal is combusted to form COz{g) and HzO{g). 2 MgO(s) is highly The reaction 2 Mg(s) + 02{g) spontaneous and has a negative value for d5°. The sec­ ond law of thermodynamics states that in any sponta­ neous process there is always an increase in the entropy of the universe. Is there an inconsistency between the above reaction and the second Jaw? Ammonium nitrate dissolves spontaneously and en­ dothermally in water at room temperature. What can you deduce about the sign of dS for this solution process? Trouton's rule states that for many liquids at their normal boiling points, the standard molar entropy of vaporiza­ tion is about 88 J/moi-K. (a) Estimate the normal boiling point of bromine, Br2, by determining dH�ap for Br2 using data from Appendix C. Assume that dH�ap re­ mains constant with temperature and that Trouton's rule holds. (b) Look up the normal boiling point of Br2 in a chemistry handbook or at the WebElements web site (www.webelements.com). ----->

[19.88] For the majority of the compounds listed in Appendix

C, the value of dGJ is more positive (or less negative) than the value of dHJ. (a) Explain this observation, using NH3{g), CCL,(/), and KN03(s) as examples. (b) An exception to this observation is CO{g). Explain the trend in the dHJ and dGJ values for this molecule. 19.89 Consider the following three reactions: (i) Ti(s) + 2 Cl2{g) TiCl4{g) (ii) C2H 6 {g) + 7 Cl2{g) -----> 2 CCl4{g) + 6 HCl{g) BaC03(s) (iii) BaO(s) + C0 2{g) (a) For each of the reactions, use data in Appendix C to calculate 6H0, 6G0, and d5° at 25 oc. (b) Which of these reactions are spontaneous under standard condi­ tions at 25 oc? (c) For each of the reactions, predict the manner in which the change in free energy varies with an increase in temperature. 19.90 Using the data in Appendix C and given the pressures listed, calculate dG for each of the following reactions: (a) Nz{g) + 3 Hz{g) 2 NH 3{g) �2 = 2.6 atm, PH, = 5.9 atm, PNH, = 1.2 atm (b) 2 N2H4{g) + 2 N02{g) 3 N 2{g) + 4 H20{g) �2H4 = PNo, = 5.0 X 10-2 atrn, P,'J, = 0.5 atrn, PH,O = 0.3 atm (c) NzH4{g) Nz{g) + 2 Hz{g) �,H, = 0.5 atm, �2 = 1.5 atrn, f\.!2 = 2.5 atm 19.91 (a) For each of the following reactions, predict the sign of dH0 and 65° and discuss briefly how these factors de­ termine the magnitude of K. (b) Based on your general chemical knowledge, predict which of these reactions will have K > 0. (c) In each case indicate whether K should increase or decrease with increasing temperature. (i) 2 Mg(s) + 02{g) = 2 MgO(s) (ii) 2 KI(s) = 2 K{g) + 12{g) (iii) Na 2{g) = 2 Na{g) (iv) 2 Vz05(s) = 4 V(s) + 5 Oz{g) ----->

----->

----->

----->

----->

840

C HAPTER 19

Chemical Thermodynamics

19.92 Acetic acid can be manufactured by combining methanol

with carbon monoxide, an example of a carbonylation reaction: CH 30H(I) + CO{g) CH3COOH(I) (a) Calculate the equilibrium constant for the reaction at 25 •c . (b) Industrially, this reaction is run at tempera­ tures above 25 •c . Will an increase in temperature pro­ duce an increase or decrease in the mole fraction of acetic acid at equilibrium? Why are elevated temperatures used? (c) At what temperature will this reaction have an equilibrium constant equal to 1? (You may assume that flH• and llS• are temperature independent, and you may ignore any phase changes that might occur.) 19.93 The oxidation of glucose (C6H 1 206) in body tissue pro­ duces C02 and H20. In contrast, anaerobic decom­ position, which occurs during fermentation, produces ethanol (CzHsOH) and C02. (a) Using data given in Appendix C. compare the equilibrium constants for the following reactions: C� u06(s) + 6 Oz(g) :o===: 6 COz(g) + 6 H20(/) C� u06(s) :o===: 2 C2H50H{l) + 2 C0 2{g) (b) Compare the maximum work that can be obtained from these processes under standard conditions. [19.94) The conversion of natural gas, which is mostly methane, into products that contain two or more carbon atoms, such as ethane (C2H6), is a very important industrial chemical process. In principle, methane can be convert­ ed into ethane and hydrogen: 2 CH 4{g) C2H6{g) + H 2{g) In practice, this reaction is carried out in the presence of oxygen: C2H6{g) + H20{g) 2 CH 4{g) + � Oz{g) --->

--->

--->

(a) Using the data in Appendix C. calculate K for these reactions at 25 •c and 500 •c . (b) Is the difference in

flG• for the two reactions due primarily to the enthalpy term {llH) or the entropy term (- T llS)? (c) Explain how the preceding reactions are an example of driving a non­ spontaneous reaction, as discussed in the "Chemistry and Life" box in Section 19.7. (d) The reaction of CH4 and 02 to form CzH6 and H20 must be carried out care­ fully to avoid a competing reaction. What is the most likely competing reaction? [19.95) Cells use the hydrolysis of adenosine triphosphate {ATP) as a source of energy (Figure 19.20). The conversion of ATP to ADP has a standard free-€nergy change of -30.5 kJ/mol. If all the free energy from the metabolism of glucose, C�u06(s) + 6 0z{g) ---> 6 C02{g) + 6 H20(1) goes into the conversion of ADP to ATP, how many moles of ATP can be produced for each mole of glucose?

[19.96) The potassium-ion concentration in blood plasma is about 3 5.0 X 10- M, whereas the concentration in muscle-cell fluid is much greater (0.15 M). The plasma and intracellu­

lar fluid are separated by the cell membrane, which we as­ sume is permeable only to K+ (a) What is llG for the transfer of 1 mol of K+ from blood plasma to the cellular fluid at body temperature 37 •c? (b) What is the minimum amount of work that must be used to transfer this K+? [19.97) The relationship between the temperature of a reaction, its standard enthalpy change, and the equilibrium constant at that temperature can be expressed as the following lin­ ear equation: - fl H• In K = � + constant (a) Explain how this equation can be used to determine

flH• experimentally from the equilibrium constants at several different temperatures. (b) Derive the preceding equation using relationships given in this chapter. To what is the constant equal? [19.98) One way to derive Equation 19.3 depends on the obser­ vation that at constant T the number of ways, W, of ar­ ranging m ideal-gas particles in a volume V is proportional to the volume raised to the m power:

W ex

V"'

Use this relationship and Boltzmann's relationship between entropy and number of arrangements (Equation 19.5) to de­ rive the equation for the entropy change for the isothermal expansion or compression of 11 moles of an ideal gas. [19.99) About 86% of the world's electrical energy is produced by using steam turbines, a form of heat engine. In his analysis of an ideal heat engine, Sadi Carnot concluded that the maximum possible efficiency is defined by the total work that could be done by the engine, divided by the quantity of heat available to do the work (for exam­ ple from hot steam produced by combustion of a fuel such as coal or methane). This efficiency is given by the ratio (Th;gh - 1iowl/Th;gh' where Thigh is the temperature of the heat going into the engine and T10w is that of the heat leaving the engine. (a) What is the maximum possi­ ble efficiency of a heat engine operating between an input temperature of 700 K and an exit temperature of 288 K? (b) Why is it important that electrical power plants be located near bodies of relatively cool water? (c) Under what conditions could a heat engine operate at or near 100% efficiency? (d) It is often said that if the energy of combustion of a fuel such as methane were cap­ tured in an electrical fuel cell instead of by burning the fuel in a heat engine, a greater fraction of the energy could be put to useful work. Make a qualitative drawing like that in Figure 5.10 that illustrates the fact that in prin­ ciple the fuel cell route will produce more useful work than the heat engine route from combustion of methane.

I N TEGRATIVE EXERC ISES Substance

Normal Boiling Point (•C)

flHvap

Acetone, (CH3)zCO Dimethyl ether, (CH3)zO Ethanol, C2H50H Octane, C8H 18 Pyridine, C5H5N

56.1 -24.8 78.4 125.6 115.3

29.1 21.5 38.6 34.4 35.1

19.100 Most liquids follow Trouton's rule, which states that the

molar entropy of vaporization lies in the range of 88 ± 5 J/mol-K. The normal boiling points and en­ thalpies of vaporization of several organic liquids are as follows:

(kJ/mol)

Integrative Exercises (a) Calculate li. Svap for each of the liquids. Do all of the liquids obey Trouton's rule? (b) With reference to inter­ molecular forces (Section 11.2), can you explain any ex­ ceptions to the rule? (c) Would you expect water to obey Trouton's rule? By using data in Appendix B, check the accuracy of your conclusion. (d) Chlorobenzene (C6HsCl) boils at 131.8 •c. Use Trouton's rule to esti­ mate II. Hvap for this substance.

19.101 Consider the polymerization of ethylene to polyethylene. aoo (Section 12.6) (a) What would you predict for the sign of the entropy change during polymerization (li.SpoJy)? Explain your reasoning. (b) The polymerization of ethyl­ ene is a spontaneous process at room temperature. What can you conclude about the enthalpy change during poly­ merization (li.Hpo1y)? (c) Use average bond enthalpies (Table 8.4) to estimate the value of li.Hpoly per ethylene monomer added. (d) Polyethylene is an addition polymcr. By comparison, Nylon 66 is a condensation polymcr. How would you expect LI.Spoly for a condensation polymer to compare to that for an addition polymer? Explain. 19.102 In chemical kinetics the entropy of activation is the en­ tropy change for the process in which the reactants reach the activated complex. The entropy of activation for bimolecular processes is usually negative. Explain this observation with reference to Figure 14.15. 19.103 The following processes were all discussed in Chapter 18, "Chemistry of the Environment." Estimate whether the entropy of the system increases or decreases during each process: (a) photodissociation of 02(g), (b) forma­ tion of ozone from oxygen molecules and oxygen atoms, (c) diffusion of CFCs into the stratosphere, (d) desalina­ tion of water by reverse osmosis. 19.104 Carbon disulfide (CS2) is a toxic, highly flammable sub­ stance. The following thermodynamic data are available for CS2(1) and CS2(g) at 298 K:

CS2(I) CS2(g)

li.HJ (kJ/mol)

li.GJ (kJ/mol)

89.7 117.4

65.3 67.2

(a) Draw the Lewis structure of the molecule. What do you predict for the bond order of the c-s bonds? (b) Use the VSEPR method to predict the structure of the C� molecule. (c) Liquid CS2 burns in 02 with a blue flame, forming C02(g) and S02(g). Write a balanced equation for this reaction. (d) Using the data in the preceding table and in Appendix C, calculate aH• and ac• for the reaction in part (c). Is the reaction exothermic? Is it spontaneous at 298 K? (e) Use the data in the preceding table to calculate as• at 298 K for the vaporization of C�(l). Is the sign of as• as you would expect for a vaporization? (f) Using data in the preceding table and your answer to part (e), estimate the boiling point of C�(l). Do you predict that the substance will be a liquid or a gas at 298 K and 1 atm? [19.105] The following data compare the standard enthalpies and free energies of formation of some crystalline ionic substances and aqueous solutions of the substances:

841

Substance

li.HJ (kJ/mol)

li.GJ (kJ/mol)

AgN03(s) AgN03(aq) MgS04(s) MgS04(aq)

-124.4 -101.7 -1283.7 -1374.8

-33.4 -34.2 -1169.6 -1198.4

(a) Write the formation reaction for AgN03(s). Based on this reaction, do you expect the entropy of the system to increase or decrease upon the formation of AgN03(s)? (b) Use li.HJ and li.GJ of AgN03(s) to determine the en­ tropy change upon formation of the substance. Is your answer consistent with your reasoning in part (a)? (c) Is dissolving AgN03 in water an exothermic or endother­ mic process? What about dissolving MgS04 in water? (d) For both AgN03 and Mg504, use the data to calcu­ late the entropy change when the solid is dissolved in water. (e) Discuss the results from part (d) with refer­ ence to material presented in this chapter and in the sec­ ond "Closer Look" box in Section 13.5. [19.106] Consider the following equilibrium: N204(g) :;::::: 2 N02(g) Thermodynamic data on these gases are given in Appendix C. You may assume that li.Ho and as• do not vary with temperature. (a) At what temperature will an equilibrium mixture contain equal amounts of the two gases? (b) At what temperature will an equilibrium mix­ ture of 1 atm total pressure contain twice as much N02 as N204? (c) At what temperature will an equilibrium mix­ ture of 10 atm total pressure contain twice as much N02 as N204? (d) Rationalize the results from parts (b) and (c) by using Le Chatelier's principle. cco (Section 15.7)

[19.107] The reaction S02(g) + 2 H2S(g)

:;:::::

3 S(s) + 2 H20(g)

is the basis of a suggested method for removal of 502 from power-plant stack gases. The standard free energy of each substance is given in Appendix C. (a) What is the equilibrium constant for the reaction at 298 K? (b) In principle, is this reaction a feasible method of removing S02? (c) If Pso, = �,s and the vapor pressure of water is 25 torr, calculate the equilibrium 502 pressure in the system at 298 K. (d) Would you expect the process to be more or less effective at higher temperatures?

19.108 When most elastomeric polymers (e.g., a rubber band) are stretched, the molecules become more ordered, as illustrated here:

Suppose you stretch a rubber band. (a) Do you expect the entropy of the system to increase or decrease? (b) If the rubber band were stretched isothermally, would heat need to be absorbed or emitted to maintain con­ stant temperature?

2D

A VARIETY OF BATIERIES of different sizes, composition, and voltages.

842

ELECTROCHEMISTRY

W H AT ' S 20.1

A H EA D

Oxidation States and Oxidation-Reduction Reactions

20.5

We briefly review oxidation states and oxidation­

reduction (redox) reactions.

20.2

20.3

20.4

Balancing Oxidation-Reduction Equations

20.6

We learn how to balance redox equations using the method of half-reactions. Voltaic Cells

We next consider voltaic cells, which produce electricity from spontaneous redox reactions. In these cells, solid electrodes serve as the surfaces at which oxidation and reduction reactions take place. The electrode where oxidation occurs is the anode, and the electrode where reduction occurs is the cathode. Cell EMF under Standard Conditions

We see that one of the important characteristics of a voltaic cell is its emf, or voltage, which is the difference in the electrical potentials at the two electrodes. Electrode potentials are tabulated for reduction half-reactions under standard conditions (standard reduction potentials). They are used to calculate cell voltages, to determine the relative strengths of oxidizing agents and reducing agents, and to predict whether particular redox reactions are spontaneous.

20.7

20.8

20.9

Free Energy and Redox Reactions

We relate the Gibbs free energy, ll.G0, to the emf of an electrochemical cell. Cell EMF under Nonstandard Conditions

We can calculate the voltage of cells that run under nonstandard conditions by using standard voltages and the Nernst equation. Batteries and Fuel Cells

We describe batteries and fuel cells, which are commercially important energy sources based on electrochemical reactions. Corrosion

Next, we discuss corrosion, a spontaneous electrochemical process involving metals. Electrolysis

Finally, we focus on nonspontaneous redox reactions, examining electrolytic cells, which use electricity to perform chemical reactions.

WE ARE SURROUN DED BY AN AMAZI NG ARRAY of portable electronic gadgets including cell phones, portable music players, laptop computers, and gaming devices. In the absence of batteries, however, all of today' s wirelessly connected electronic gadgetry would be nothing more than extra weight. Thus, a variety of batteries of different sizes, compositions, and voltages have been developed, as shown in the chapter-opening photograph. Considerable research is in progress to develop new batteries with more power, faster recharging ability, lighter weight, or cheaper price. For example, in an effort to develop a battery that lasts long enough for all-day computing, battery makers are trying to increase the capacity of batteries without increasing their mass. At the heart of such development is the utilization of oxidation-reduction (redox) reactions, which are the chemical reactions that power batteries. Redox reactions are among the most common and important chemical re­ actions. They are involved in the operation of batteries and in a wide variety of important natural processes, including the rusting of iron, the browning of 843

844

CHAPT E R 20

Electrochemistry foods, and the respiration of animals. As we discussed in Chapter 4, oxidation refers to the Joss of electrons, and reduction refers to the gain of electrons. (Section 4.4) Thus, oxidation-reduction reactions occur when electrons are transferred from an atom that is oxidized to an atom that is reduced. The examples of redox reactions that we have cited in this introduction are all spontaneous processes. We can also use electrical energy to make certain nonspontaneous redox reactions occur. A common example is electroplating, in which layers of one metal are coated onto another by the application of a voltage. Electrochemistry is the study of the relationships between electricity and chemical reactions. It includes the study of both spontaneous and nonspontaneous processes. c:x:c

20. 1 OXI DATION STATES AND OXIDAT I O N ­ RED U C T I O N REACTIONS How d o w e determine whether a given chemical reaction i s an oxidation­ reduction reaction? We can do so by keeping track of the oxidation numbers (oxidation states) of all the elements involved in the reaction. (Section 4.4) This procedure identifies whether any elements are changing oxidation state. For example, consider the reaction that occurs when zinc metal is added to a strong acid (Figure 20.1 T): c:x:c

Zn(s)

+

2 H +(aq) --> zn2+(aq)

+ H2(g)

[20.1]

The reaction proceeds spontaneously and produces energy in the form of heat. The chemical equation for this reaction can be written as

[20.2]

I> Figure 20.1 Redox reaction In addle solution. The addition of zinc

metal to hydrochloric acid leads to a spontaneous oxidation-reduction reaction: Zinc metal is oxidized to Zn 2+(aq), and H+(aq) is reduced to H 2 (g), which produces the vigorous bubbling.

Zn(s)

2 HCI(aq)

ZnCl2(aq)

20.1

Oxidation States and Oxidation-Reduction Reactions

By writing the oxidation number of each element above or below the equation, we can see how the oxidation numbers change. The oxidation number of Zn changes from 0 to +2, and that of H changes from + 1 to 0. Thus, tills is an oxidation-reduction reaction. Electrons are transferred from zinc atoms to hydrogen ions, and therefore Zn is oxidized and H+ is reduced. In a reaction such as Equation 20.2, a clear transfer of electrons occurs. Zinc 2 loses electrons as Zn(s) is converted to Zn +(aq), and hydrogen gains electrons as H +(aq) is turned into H2(g). In some reactions, however, the oxidation num­ bers change, but we cannot say that any substance literally gains or loses elec­ trons. For example, consider the combustion of hydrogen gas:

[20.3]

In tills reaction, hydrogen is oxidized from the 0 to the + 1 oxidation state and oxygen is reduced from the 0 to the -2 oxidation state. Therefore, Equation 20.3 is an oxidation-reduction reaction. Water is not an ionic substance, however, and so there is not a complete transfer of electrons from hydrogen to oxygen as water is formed. Using oxidation states, therefore, is a convenient form of "bookkeeping," but you should not generally equate the oxidation state of an atom with its actual charge in a chemical compound. ooo (Section 8.5 "A Closer Look: Oxidation Numbers, Formal Charges, and Actual Partial Charges") GIVE IT SOME THOUGHT

What are the oxidation numbers, or states, of the elements in the nitrite ion, N02-? In any redox reaction, both oxidation and reduction must occur. In other words, if one substance is oxidized, then another must be reduced: The elec­ trons formally have to go from one place to another. The substance that makes it possible for another substance to be oxidized is called either the oxidizing agent or the oxidant. The oxidizing agent removes electrons from another substance by acquiring them itself; thus, the oxidizing agent is itself reduced. Similarly, a reducing agent, or reductant, is a substance that gives up electrons, thereby causing another substance to be reduced. The reducing agent is there­ fore oxidized in the process. In Equation 20.2, H +(aq) is the oxidizing agent and Zn(s) is the reducing agent.

- SAMPLE EXERCISE 20.1

I Identifying Oxidizing and Reducing Agents

The nickel-cadmium (nicad) battery, a rechargeable "dry cell" used in battery-operated devices, uses the following redox reaction to generate electricity: Cd(s) + Ni02(s) + 2 H 20(1) Cd(OHh(s) + Ni(OHh(s) Identify the substances that are oxidized and reduced, and indicate which is the oxidizing agent and which is the reducing agent. ->

SOLUTION

We are given a redox equation and asked to identify the substance oxi­ dized and the substance reduced and to label one as the oxidizing agent and the other as the reducing agent. Plan: First, we assign oxidation states, or numbers, to all the atoms in the reaction and determine the elements that are changing oxidation state. Second, we apply the definitions of oxidation and reduction.

Analyze:

Solve:

® e9 e@

Cd(s) + Ni02(s) + 2H20(1)

�

@9� @9e

->

Cd(OH s) + Ni(OHh( s)

C d increases i n oxidation state from 0 to +2, and N i decreases from + 4 to +2.

845

846

CHAPTER 20

Electrochemistry Because the Cd atom increases in oxidation state, it is oxidized (loses electrons) and therefore serves as the reducing agent. The Ni atom decreases in oxidation state as Ni02 is converted into Ni(OH)z. Thus, Ni02 is reduced (gains electrons) and there­ fore serves as the oxidizing agent. Comment: A common mnemonic for remembering oxidation and reduction is "LEO the lion says GER": losing electrons is oxidation; gaining electrons is reduction. - PRACTICE EXERCISE

Identify the oxidizing and reducing agents in the oxidation-reduction reaction 2 H20(1) + Al(s) + Mn04-(aq)

--->

Al(OH)4 -(aq) + Mn02(s)

Answer: Al(s) is the reducing agent; Mn04-(aq) is the oxidizing agent.

20.2 BALANC I N G OXI D ATION­ RED U CT I O N EQUATIONS Whenever w e balance a chemical equation, w e must obey the law of conserva­ tion of mass: The amount of each element must be the same on both sides of the equation. (Atoms are neither created nor destroyed in any chemical reaction.) As we balance oxidation-reduction reactions, there is an additional require­ ment: The gains and losses of electrons must be balanced. In other words, if a substance loses a certain number of electrons during a reaction, then another substance must gain that same number of electrons. (Electrons are neither creat­ ed nor destroyed in any chemical reaction.) In many simple chemical reactions, such as Equation 20.2, balancing the electrons is handled "automatically"; w e can balance the equation without ex­ plicitly considering the transfer of electrons. Many redox reactions are more complex than Equation 20.2, however, and cannot be balanced easily without taking into account the number of electrons lost and gained in the course of the reaction. In this section we examine the method of half reactions, which is a sys­ tematic procedure for balancing redox equations. Ha lf-Reactions Although oxidation and reduction must take place simultaneously, it is often convenient to consider them as separate processes. For example, the oxidation 2 of Sn + by Fe3+ : 4 Sn2+ (aq) + 2 Fe3+ (aq) -----> Sn +(aq) + 2 Fe2+ (aq) can be considered to consist of two processes: (1) the oxidation of Sn2+ (Equation 20.4) and (2) the reduction of Fe3+ (Equation 20.5):

Oxidation: Reduction:

Sn2+ (aq) 3 + 2 Fe (aq) + 2 e-

----->

----->

4 Sn +(aq) + 2 e2 Fe 2+(aq)

[20.4] [20.5]

Notice that electrons are shown as products in the oxidation process, whereas electrons are shown as reactants in the reduction process. Equations that show either oxidation or reduction alone, such as Equations 20.4 and 20.5, are called half-reactions. In the overall redox reaction, the number of electrons lost in the oxidation half-reaction must equal the number of electrons gained in the reduction half-reaction. When this condition is met and each half-reaction is balanced, the electrons on the two sides cancel when the two half­ reactions are added to give the overall balanced oxidation-reduction equation. Ba lancing Equations by the Method of Half-Reactions The use of half-reactions provides a general method for balancing oxidation­ reduction equations. We usually begin with a "skeleton" ionic equation that shows only the substances undergoing oxidation and reduction. In such cases,

Balancing Oxidation-Reduction Equations

20.2

847

we usually do not need to assign oxidation numbers unless we are unsure whether the reaction actually involves oxidation-reduction. We will find that H+ (for acidic solutions), OH- (for basic solutions), and H20 are often involved as reactants or products in redox reactions. Unless H +, OH-, or H20 are being oxidized or reduced, they do not appear in the skeleton equation. Their pres­ ence, however, can be deduced during the course of balancing the equation. For balancing a redox reaction that occurs in acidic aqueous solution, the procedure is as follows: 1. Divide the equation into two half-reactions, one for oxidation and the other

for reduction.

2. Balance each half-reaction. (a) First, balance the elements other than H and 0.

(a)

(b) Next, balance the 0 atoms by adding H20 as needed.

--...,._--

(c) Then, balance the H atoms by adding H + as needed. (d) Finally, balance the charge by adding e- as needed.

This specific sequence is important, and it is summarized in the diagram in the margin. At this point, you can check whether the number of electrons in each half-reaction corresponds to the changes in oxidation state. 3. Multiply the half-reactions by integers, if necessary, so that the number of electrons lost in one half-reaction equals the number of electrons gained in the other. 4. Add the two half-reactions and, if possible, simplify by canceling species appearing on both sides of the combined equation. 5. Check to make sure that atoms and charges are balanced.

(b)

(c)

---..----

--...,.----

(d)

As an example, let's consider the reaction between permanganate ion (Mn04-) and oxalate ion (C 20/-) in acidic aqueous solution. When Mn04- is added to an acidified solution of C20l-, the deep purple color of the Mn04- ion fades, as illustrated in Figure 20.2 ..- . Bubbles of C02 form, and the solution takes on the pale pink color of Mn2+. We can therefore write the skeleton equation as Mn04-(aq) + C20l-(aq) � Mn2+(aq) + C02(aq) [20.6] Experiments show that H + is consumed and H20 is produced in the reaction. We will see that their involvement in the reaction is deduced in the course of balancing the equation. To complete and balance this equation, we first write the two half-reactions (step 1). One half-reaction must have Mn on both sides of the arrow, and the other must have C on both sides of the arrow: Mn04-(aq) � Mn2+(aq) c2o/-(aq) � C02(gl summarizes the relationships among the anode, the cathode, the chemical process occurring in a voltaic cell, the di­ rection of migration of ions in solution, and the motion of electrons between electrodes in the external circuit. Notice in particular that

in any voltaic cell the electrons flow from the anode through the external circuit to the cathode. Because the negatively charged electrons flow from the anode to the cathode, the anode in a voltaic cell is labeled with a negative sign and the cathode with a positive sign; we can envision the electrons as being attracted to the positive cathode from the negative anode through the external circuit. - SAMPLE EXERCISE

Anode compartment Oxidation occurs

20.4 I Describing a Voltaic Cell

The oxidation-reduction reaction

Cr20/-(aq) + 14 H+(aq) + 6 qaq) ----> 2 Cr3+(aq) + 3 I2(s) + 7 H20(1) is spontaneous. A solution containing K2Cr207 and H2S04 is poured into one beaker, and a solution of KI is poured into another. A salt bridge is used to join the beakers. A metallic conductor that will not react with either solution (such as platinum foil) is suspended in each solution, and the two conductors are connected with wires through a voltmeter or some other device to detect an electric current. The resultant voltaic cell generates an electric current. Indicate the reaction occurring at the anode, the reaction at the cathode, the direction of electron migration, the direction of ion migration, and the signs of the electrodes. SOLUTION Analyze: We are given the equation for a spontaneous reaction that takes place in a voltaic cell and a description of how the cell is constructed. We are asked to write the half-reactions occurring at the anode and at the cathode, as well as the directions of electron and ion movements and the signs assigned to the electrodes. Plan: Our first step is to divide the chemical equation into half-reactions so that we can identify the oxidation and the reduction processes. We then use the definitions of anode and cathode and the other terminology summarized in Figure 20.6. Solve: In one half-reaction, Cr20l-(aq) is converted into Cr3+(aq). Starting with these ions and then completing and balancing the half-reaction, we have

Cr20/-(aq) + 14 H+(aq) + 6 e- ----> 2 Cr3+(aq) + 7 H20(1) In the other half-reaction, I-(aq) is converted to I2(s): 6 qaq) ----> 3 I2(s) + 6 eNow we can use the summary in Figure 20.6 to help us describe the voltaic cell. The first half-reaction is the reduction process (electrons on the reactant side of the equation). By definition, the reduction process occurs at the cathode. The second half­ reaction is the oxidation process (electrons on the product side of the equation), which occurs at the anode. The r- ions are the source of electrons, and the Cr20/- ions accept the electrons. Hence, the electrons flow through the external circuit from the electrode immersed in the KI solution (the anode) to the electrode immersed in the K2Cr207-H2504 solu­ tion (the cathode). The electrodes themselves do not react in any way; they merely provide a means of transferring electrons from or to the solutions. The cations move through the solutions toward the cathode, and the anions move toward the anode. The anode (from which the electrons move) is the negative electrode, and the cathode (toward which the electrons move) is the positive electrode.

Cathode compartment Reduction occurs

.A. Figure 20.6 A summary of the terminology used to describe voltaic cells. Oxidation occurs at the anode;

reduction occurs at the cathode. The electrons flow spontaneously from the negative anode to the positive cathode. The movement of ions in solution completes the electrical circuit. Anions move toward the anode, whereas cations move toward the cathode. The cell compartments can be separated by either a porous glass barrier (as in Figure 20.4) or by a salt bridge (as in Figure 20.5).

854

CHAPTER 20

Electrochemistry - PRACTICE EXERCISE

The two half-reactions in a voltaic cell are

T

Figure 20.7

Atomic-level depletion 2 of the Zn(s)-Cu + (aq) reaction. The

water molecules and anions in the solution are not shown. (a) A Cu 2+ ion comes in contact with the surface of the Zn strip and gains two electrons from a Zn atom; the Cu 2+ ion is reduced, and the Zn atom is oxidized. (b) The resulting Zn 2+ ion enters the solution, and the Cu atom remains deposited on the strip. Atoms in Zn strip

Cu2+ ions in solution

(a)

Figure 20.8 Atomic-level depletion of the voltaic cell In Figure 20.5. At

T

the anode a Zn atom loses two electrons and becomes a Zn 2+ ion; the Zn atom is oxidized. The electrons travel through the external circuit to the cathode. At the cathode a Cu 2+ ion gains the two electrons, forming a Cu atom; the Cu2+ ion is reduced. Ions migrate through the porous barrier to maintain charge balance between the compartments.

Zn(s)

/

Zn(s) -----> zn2+(aq) + 2 eCJ03 -(aq) + 6 H+(aq) + 6 e- -----> cqaq) + 3 H 20(/)

(a) Indicate which reaction occurs at the anode and which at the cathode. (b) Which electrode is consumed in the cell reaction? (c) Which electrode is positive?

Answers: (a) The first reaction occurs at the anode and the second reaction at the cath­ ode. (b) The anode (Zn) is consumed in the cell reaction. (c) The cathode is positive.

A Molecular View of Electrode Processes To better understand the relationship between voltaic cells and sponta­ neous redox reactions, let's look at what happens at the atomic or mole­ cular level. The actual processes in­ volved in the transfer of electrons are quite complex; nevertheless, we can learn much by examining these processes in a simplified way. Consider the spontaneous reac(b) tion between Zn(s) and Cu2+(aq), in which Zn(s) is oxidized to Zn2+ (aq) and Cu2+(aq) is reduced to Cu(s), as shown in Figure 20.3. Figure 20.7 .,.. shows a schematic diagram of these processes at the atomic level. We can envision a Cu2+ ion coming into contact with the strip of Zn metal, as in Figure 20.7(a). Two elec­ trons are transferred from a Zn atom to the Cu2+ ion, leading to a Zn2+ ion and a Cu atom. The zn2+ ion migrates away into the aqueous solution while the Cu atom remains deposited on the metal strip [Figure 20.7(b)]. As the reaction pro­ ceeds, we produce more and more Cu(s) and deplete the Cu2+(aq). GIVE IT SOME THOUGHT What happens to the surface Zn atoms as they lose electrons?

The voltaic cell in Figure 20.5 is also based on the oxidation of Zn(s) and the reduction of 2+ Cu (aq). In this case, however, the electrons are not transferred directly from one reacting species to the other. As shown in Figure 20.8 .,.. , a Zn atom at the surface of the anode "loses" two electrons and becomes a Zn2+ (aq) ion in the anode com­ partment. We envision the two electrons traveling from the anode through the wire to the cathode. At the surface of the cathode, a Cu2+ ion from the solution gains the two electrons to form a Cu atom, which is deposited on the cathode. The redox reaction between Zn and Cu2+ is spontaneous regardless of whether they react directly or in the separate compartments of a voltaic cell. In each case the overall reaction is the same-only the path by which the electrons are transferred from the Zn atom to a Cu2+ ion is different.

20.4

Cell EMF under Standard Conditions

855

20.4 CELL E M F U N D ER STANDARD C O N D ITIONS Why d o electrons transfer spontaneously 2 from a Zn atom to a Cu + ion, either di­ rectly as in the reaction of Figure

20.3 or

through an external circuit as in the volta­

Anode

l

ic cell of Figure 20.4? In a simple sense, we

11-

can compare the electron flow to the flow of water in a waterfall (Figure

20.9 �).

Water flows spontaneously over a water­

l!_

fall because of a difference in potential en­ ergy between the top of the falls and the stream below. o:o (Section 5.1)

�

In a similar

fashion, electrons flow from the anode of a voltaic cell to the cathode because of a dif­ ference in potential energy. The potential energy of electrons is higher in the anode

+

Cathode

than in the cathode, and they sponta­ neously flow through an external circuit from the anode to the cathode. That is, electrons flow spontaneously toward the electrode with the more positive electrical potential. The difference in potential energy per

potential difference) between two electrodes is measured in (V) is the potential difference required to impart 1 joule 0) of energy to a charge of 1 coulomb (C). electrical charge (the

units of volts. One volt

1v = Recall that one electron has a charge of

1l c

1 .60

X 10-19 C.

o:o

(Section 2.2)

The potential difference between the two electrodes of a voltaic cell pro­ vides the driving force that pushes electrons through the external circuit. There­ fore, we call this potential difference the electromotive ("causing electron

motion") force, or emf. The emf of a cell, denoted Ecell• is also called the cell potential. Because Ecell is measured in volts, we often refer to it as the cell voltage.

For any cell reaction that proceeds spontaneously, such as that in a voltaic cell, the cell potential will be

positive.

The emf of a particular voltaic cell depends on the specific reactions that occur at the cathode and anode, the concentrations of reactants and products, and the temperature, which we will assume to be 25 °C unless otherwise noted. In this

oc under standard conditions. 19.5 that standard conditions include 1 M concentrations for reactants and products in solution and 1 atm pressure for those that are gases (Table 19.3). Under standard conditions the emf is called the standard emf, or the standard cell potential, and is denoted E�u· For the Zn-Cu voltaic cell in Figure 20.5, for example, the standard cell potential at 25 oc is + 1.10 V. section we will focus on cells that are operated at 25

Recall from Section

2 Zn(s) + Cu +(aq,

Recall that the superscript

GIVE

2

1 M) --> zn +(aq, 1 M) + Cu(s)

E�ell = + 1 .10 V

o indicates standard-state conditions.

o:o

(Section 5.7)

IT SOME THOUGHT

If a standard cell potential is +0.85 V at 25 oc, is the redox reaction of the cell

spontaneous?

.i. Figure 20.9 Water analogy for electron flow. The flow of electrons

from the anode to the cathode of a voltaic cell can be likened to the flow of water over a waterfall. Water flows over the waterfall because its potential energy is lower at the bottom of the falls than at the top. Likewise, if there is an electrical connection between the anode and cathode of a voltaic cell, electrons flow from the anode to the cathode to lower their potential energy.

856

CHAPTER 20

Electrochemistry Standard Reduction (Half-Cel l) Potentials The emf, or cell potential, of a voltaic cell, E�eu, depends on the particular cathode and anode half-cells involved. We could, in principle, tabulate the standard cell potentials for all possible cathode/ anode combinations. However, it is not neces­ sary to undertake this arduous task. Rather, we can assign a standard potential to each individual half-cell and then use these half-cell potentials to determine E�ell· The cell potential is the difference between two electrode potentials, one as­ sociated with the cathode and the other associated with the anode. By conven­ tion, the potential associated with each electrode is chosen to be the potential for reduction to occur at that electrode. Thus, standard electrode potentials are tabulated for reduction reactions; they are standard reduction potentials, de­ noted E�ed · The cell potential, E�eu, is given by the standard reduction potential of the cathode reaction, E�ed (cathode), minus the standard reduction potential of the anode reaction, E�ed (anode): E�ell = E�ed (cathode)

-

E�ed (anode)

[20.8]

For all spontaneous reactions at standard conditions, E�ell > 0. Because every voltaic cell involves two half-cells, it is not possible to mea­ sure the standard reduction potential of a half-reaction directly. If we assign a standard reduction potential to a certain reference half-reaction, however, we can then determine the standard reduction potentials of other half-reactions rel­ ative to that reference. The reference half-reaction is the reduction of H +(aq) to H2 (g) under standard conditions, which is assigned a standard reduction po­ tential of exactly 0 V. E�ed

=

OV

[20.9]

An electrode designed to produce this half-reaction is called a standard hydrogen electrode (SHE), or the normal hydrogen electrode (NHE). An SHE consists of a platinum wire connected to a piece of platinum foil covered with finely divided platinum that serves as an inert surface for the reaction. The elec­ trode is encased in a glass tube so that hydrogen gas under standard conditions (1 atrn) can bubble over the platinum, and the solution contains H +(aq) under standard (1 M) conditions (Figure 20.10 T).

�L

To external circuit

-1 atm H2(g)

H2 molecule

\

Reduction ...---

Oxidation

(a) .6.

Figure 20.10

(b) The standard hydrogen electrode (SHE) used as a reference

(a) An SHE consists of an electrode with finely divided Pt in contact with H2(g) at 1 atm pressure and an acidic solution with [H+] = 1 M. (b) Molecular depiction of the processes that occur at the SHE. When the SHE is the cathode of a cell, two H+ ions each accept an electron from the Pt electrode and are reduced to H atoms. The H atoms bond together to form H2• When the SHE is the anode of a cell, the reverse process occurs: An H2 molecule at the electrode surface loses two electrons and is oxidized to H+ The H+ ions in solution are hydrated. electrode.

20.4

Figure 20.11 � shows a voltaic cell using an SHE and a standard electrode. The sponta­ neous reaction is the one shown in Figure 20.1, namely, the oxidation of Zn and the reduction of H +: Zn2+ (aq) + Hz(g) Zn(s) + 2 H+(aq )

Anode compartment

E�ell = E�ed (cathode) - E�ed (anode) +0.76 V = 0 V - E�ct (anode)

=

E�ed (anode)

-0.76 V

.6.

.......-

Figure 20.11

We write the reaction as a reduction even though it is "running in reverse" as an oxidation in the cell in Figure 20.11. Whenever we assign an electrical potential to a

half-reaction, we write the reaction as a reduction. The standard reduction potentials for other half-reactions can be estab­ lished from other cell potentials in a fashion analogous to that used for 2 the Zn +;zn half-reaction. Table 20.1 T lists some standard reduction potentials; a more complete list is found in Appendix E. These standard reduction poten­ tials, often called half-cell potentials, can be combined to calculate the emfs of a large variety of voltaic cells.

TABLE 20.1

• Standard Reduction Potentials in Water at 25 oc

(V)

+2.87 +1 .51 +1.36 +1.33 +1.23 +1 .06 +0.96 +0.80 +0.77 +0.68 +0.59 +0.54 +0.40 +0.34 0 [defined] -0.28 -0.44 -0.76 -0.83 -1.66 -2.71 -3.05

Cathode compartment (standard hydrogen electrode)

A voltaic cell using a standard hydrogen electrode.

Thus, a standard reduction potential of -0.76 V can be assigned to the reduction 2 of Zn + to Zn. 2 Zn +(aq, 1 M) + 2 e- --+ Zn(s) E�cd = -0.76 V

Potential

857

Zn anode

--+

2 Notice that the Zn +/Zn electrode is the anode and the SHE is the cathode and that the cell voltage is +0.76 V. By using the defined stan­ dard reduction potential of H + (E�ed = 0) and Equation 20.8, we can determine the stan­ 2 dard reduction potential for the Zn +/Zn half­ reaction:

Cell EMF under Standard Conditions

Reduction Half-Reaction F2(g) + 2 e- ----> 2 F-(aq) 2 Mn04-(aq) + 8 H+(aq) + 5 e- ----> Mn +(aq) + 4 H20(/) Cl2(g) + 2 e- ----> 2 cqaq) 2 Cr20 7 -(aq) + 14 H+(aq) + 6 e- ----> 2 Cr 3+(aq) + 7 H20(/) 02(g) + 4 H+(aq) + 4 e- ----> 2 H20(/) Br2(1) + 2 e- ----> 2 Br-(aq) N03-(aq) + 4 H+(aq) + 3 e- ----> NO(g) + 2 H20(/) Ag+(aq) + e- ----> Ag(s) 2 Fe 3+(aq) + e- ----> Fe +(aq) 02(g) + 2 H+(aq) + 2 e- ----> H202(aq) Mn04-(aq) + 2 H20(/) + 3 e- ----> Mn02(s) + 4 OH-(aq) 1 2 (s) + 2 e- ----> 2 qaq) 02(g) + 2 H20(/) + 4 e- ----> 4 OH-(aq) 2 Cu +(aq) + 2 e- ----> Cu(s) 2 H+(aq) + 2 e- ----> H2(g) 2 Ni +(aq) + 2 e- ----> Ni(s) 2 Fe +(aq) + 2 e- ----> Fe(s) 2 Zn +(aq) + 2 e- ----> Zn(s) 2 H20(/) + 2 e- ----> H2(g) + 2 OH-(aq) Al3+(aq) + 3 e- ----> Al(s) Na+(aq) + e- ----> Na(s) u+(aq) + e- ----> Li(s)

858

CHAPTER 20

Electrochemistry G IVE IT SOME THOUGHT For the half-reaction Cl2(g) + 2 e - ---> 2 Cqaq), what are the standard conditions for the reactant and product?

Because electrical potential measures potential energy per electrical charge, standard reduction potentials are intensive properties. ceo (Section 1.3). In other words, if we increased the amount of substances in a redox reaction, we would increase both the energy and charges involved, but the ratio of energy Qoules) to electrical charge (coulombs) would remain constant (V = J/C). Thus,

changing the stoichiometric coefficient in a half-reaction does not affect the value of the standard reduction potential. For example, E�ed for the reduction of 10 mol Zn2+ is

the same as that for the reduction of 1 mol Zn2+: 10 Zn2+ (a q, 1 M) + 20 e- -----> 10 Zn(s)

E�ed = - 0.76 V

- SAMPLE EXERCISE I Calculating f�ed from f�ell For the Zn-Cu 2+ voltaic cell shown in Figure 20.5, we have

20.5

Zn(s) + Cu2+(aq, 1 M) ---> Zn2+(aq, 1 M) + Cu(s) £�11 = 1.10 V 2 Given that the standard reduction potential of Zn + to Zn(s) is -0.76 V, calculate the E�ect for the reduction of Cu2+ to Cu: Cu2+(aq, 1 M) + 2 eSOLUTION

--->

Cu(s)

Analyze: We are given £�11 and E�ed for Zn2+ and asked to calculate E�ed for Cu2+ Plan: 1n the voltaic cell, Zn is oxidized and is therefore the anode. Thus, the given E�ect for Zn2+ is E�ed (anode). Because Cu2+ is reduced, it is in the cathode half-cell. Thus, the unknown reduction potential for Cu2+ is E�ed (cathode). Knowing E�ell and E�ect (anode), we can use Equation 20.8 to solve for E�ect (cathode). Solve: E�ell

1.10 V E�ect (cathode)

=

=

=

E�ed (cathode) - E�cd (anode) E�ect (cathode) - (-0.76 V )

1.10 V - 0.76 V

=

0.34 V

Check: This standard reduction potential agrees with the one listed in Table 20.1. Comment: The standard reduction potential for Cu2+ can be represented as Ecu2+ = 0.34 V, and that for Zn2+ as Ezn2+ = -0.76 V. The subscript identifies the ion that is reduced in the reduction half-reaction. - PRACTICE EXERCISE

A voltaic cell is based on the half-reactions ln+(aq)

--->

ln3+(aq) + 2 e -

Br2(1) + 2 e- ---> 2 Br -(aq) The standard emf for this cell is 1.46 V. Using the data in Table 20.1, calculate E�ed for the reduction of 1n3+ to 1n+ Answer: -0.40 V

- SAMPLE EXERCISE

20.6 I Calculating f�eu from f�ed

Using the standard reduction potentials listed in Table 20.1, calculate the standard emf for the voltaic cell described in Sample Exercise 20.4, which is based on the reaction Cr2072-(aq) + 14 H+(aq) + 6 l-(aq)

--->

2 Cr3+(aq) + 3 I2(s) + 7 H 20(1)

SOLUTION Analyze: We are given the equation for a redox reaction and asked to use data in Table 20.1 to calculate the standard emf (standard potential) for the associated volta­ ic cell.

20.4

Cell EMF under Standard Conditions

Plan: Our first step is to identify the half-reactions that occur at the cathode and the anode, which we did in Sample Exercise 20.4. Then we can use data from Table 20.1 and Equation 20.8 to calculate the standard emf. Solve: The half-reactions are Cathode: Cr2072-(aq) + 14 H+(aq) + 6 e- -> 2 Cr3+(aq) + 7 H 20(1)

6 r-(aq) -> 3 I 2(s) + 6 e-

Anode:

According to Table 20.1, the standard reduction potential for the reduction of Cr2ol­ to Cr 3+ is +1.33 V, and the standard reduction potential for the reduction of !2 to !­ (the reverse of the oxidation half-reaction) is +0.54 V. We then use these values in Equation 20.8. E�ell

More positive

= E�ed (cathode) - E�ed (anode) = 1.33 V - 0.54 V = 0.79 V

E�ell

- PRACTICE EXERCISE

2 Al(s) + 3 ! 2(s) -> 2 Al3+(aq) + 6 r-(aq) 2.20 V

j

E�ed(anode) Anode (oxidation)

Using data in Table 20.1, calculate the standard emf for a cell that employs the follow­ ing overall cell reaction: =

1

Cathode (reduction) ';..(rn�odo)

Although we must multiply the iodide half-reaction at the anode by 3 to obtain a bal­ anced equation for the reaction, the value of E�ed is not multiplied by 3. As we have noted, the standard reduction potential is an intensive property, so it is independent of the specific stoichiometric coefficients. Check: The cell potential, 0.79 V, is a positive number. As noted earlier, a voltaic cell must have a positive emf in order to operate.

Answer: 0.54 V - ( -1.66 V)

859

More negative

For each of the half-cells in a voltaic cell, the standard reduction potential provides a measure of the driving force for reduction to occur: The more positive

A Figure 20.12

the value ofE�ed' the greater the drivingforce for reduction under standard conditions.

potential measures the difference in the standard reduction potentials of the cathode and the anode reactions: E�ell = E�ed (cathode) - E:'ed (anode). In a voltaic cell the cathode reaction is always the one that has the more positive (or less negative) value for E:'.,ct.

In any voltaic cell operating under standard conditions, the reaction at the cath­ ode has a more positive value of E�ed than does the reaction at the anode. Equation 20.8, which indicates that the standard cell potential is the differ­ ence between the standard reduction potential of the cathode and the standard reduction potential of the anode, is illustrated graphically in Figure 20.12 �. The standard reduction potentials of the cathode and anode are shown on a scale. The more positive E�ed value identifies the cathode, and the difference be­ tween the two standard reduction potentials is the standard cell potential. Figure 20.13 � shows the specific values of E�ed for the two half-reactions in the Zn-Cu voltaic cell illustrated in Figure 20.5. GIVE IT SOME THOUGHT

Is the following statement true or false? The smaller the difference is between the standard reduction potentials of the cathode and anode, the smaller the driving force for the overall redox reaction. - SAMPLE EXERCISE

20.7 1 Calculating Determining Half-Reactions at Electrodes and Cell EMF

A voltaic cell is based on the following two standard half-reactions: Cd 2+(aq) + 2 e- -> Cd(s) Sn2+(aq) + 2 e- -> Sn(s) By using the data in Appendix E, determine (a) the half-reactions that occur at the cathode and the anode, and (b) the standard cell potential. SOLUTION

We have to look up E�ed for two half-reactions and use these values to pre­ dict the cathode and anode of the cell and to calculate its standard cell potential, E�ell·

Analyze:

Standard cell potential of a voltaic cell. The cell

More positive

+0.34

� o� "' E�ell

-0.76

1

Cu2 + + 2 e- -----7 Cu Cathode = ( +0.34) - ( -0.76)

= +1.10 V

1---'---:::-:--­ 2 Anode Zn -----7 zn + + 2 e

A Figure 20.13

Half-cell potentials.

The half-cell potentials for the voltaic cell in Figure 20.5, diagrammed in the style of Figure 20.12.

860

CHAPTER 20

Electrochemistry Plan: The cathode will have the reduction with the more positive E�ed value. The anode will have the less positive E�ed· To write the half-reaction at the anode, we re­ verse the half-reaction written for the reduction, so that the half-reaction is written as an oxidation. Solve: (a) According to Appendix E, E�ed(Cd2+/Cd{ = -0.403 V and E�ct(Sn2+/Sn) = -0.136 V. The standard reduction potential for Sn + is more positive (less negative) than that for Cd 2+; hence, the reduction of Sn2+ is the reaction that occurs at the cathode.

Cathode: The anode reaction therefore is the loss of electrons by Cd. Cd(s) ----> Cd 2+(aq) + 2 e -

Anode:

(b) The cell potential is given by Equation 20.8. E�ell

=

E:'ect (cathode) - E�ed (anode)

=

(-0.136 V) - (-0.403 V)

=

0.267 V

Notice that it is unimportant that the E�ed values of both half-reactions are negative; the negative values merely indicate how these reductions compare to the reference reaction, the reduction of H+(aq). Check: The cell potential is positive, as it must be for a voltaic cell. - PRACTICE EXERCISE

A voltaic cell is based on a Co2+;co half-cell and an AgCl/Ag half-cell. (a) What half-reaction occurs at the anode? (b) What is the standard cell potential? Answers: (a) Co ----> Co2+ + 2 e-; (b) +0.499 V

Strengths of Oxidizing and Reducing Agents

We have thus far used the tabulation of standard reduction potentials to exam­ ine voltaic cells. We can also use E:ed values to understand aqueous reaction chemistry. Recall, for example, the reaction between Zn(s) and Cu2+(aq) shown in Figure 20.3. 2 2 Zn(s) + Cu +(aq) ----> zn + (aq) + Cu(s) 2 Zinc metal is oxidized, and Cu + (aq) is reduced in this reaction. These sub­

stances are in direct contact, however, so we are not producing usable electrical work; the direct contact essentially "short-circuits" the cell. Nevertheless, the driving force for the reaction is the same as that in a voltaic cell, as in Figure 20.5. Because the E:ed value for the reduction of Cu2+(0.34 V) is more positive than the E:ed value for the reduction of Zn2+( -0.76 V), the reduction of Cu 2+(aq) by Zn(s) is a spontaneous process. We can generalize the relationship between the value of E:ed and the spon­ taneity of redox reactions: The more positive the E�ed value for a half-reaction, the

greater the tendency for the reactant of the half-reaction to be reduced and, therefore, to oxidize another species. In Table 20.1, for example, F2 is the most easily reduced species, so it is the strongest oxidizing agent listed. Fz(g) + 2 e-

---->

2 F-(aq)

E:ed

= 2.87 V

Among the most frequently used oxidizing agents are the halogens, O:u and oxyanions such as Mn04-, Cr2ol-, and N03-, whose central atoms have high positive oxidation states. According to Table 20.1, all these species undergo reduction with large positive values of E�ed· Lithium ion (Li+) is the most difficult species to reduce and is therefore the poorest oxidizing agent:

Li + (aq)

+ e-

---->

Li(s)

E�ed = -3.05 V

Because u+ is so difficult to reduce, the reverse reaction, the oxidation of Li(s) to u+(aq), is a highly favorable reaction. The half-reaction with the smallest reduction potential is most easily reversed as an oxidation. Thus, lithium metal has a great ten­ dency to transfer electrons to other species. In water, Li is the strongest reducing

20.4

Cell EMF under Standard Conditions

861

agent among the substances listed in Table 20.1. Be­ Strongest Most post·tive va1ues of £0red cause Table 20.1 lists half-reactions as reductions, oxidizing agent --.:::_ only the substances on the reactant side of these + 2 eequations can serve as oxidizing agents; only those on the product side can serve as reducing agents. Commonly used reducing agents include H2 and the active metals such as the alkali metals and the al­ kaline earth metals. Other metals whose cations have negative E�ed values-Zn and Fe, for example-are also used as reducing agents. Solutions of reducing agents are difficult to store for extended periods be­ cause of the ubiquitous presence of 02, a good oxi­ dizing agent. For example, developer solutions used in film photography are mild reducing agents; they have only a limited shelf life because they are readily oxidized by 02 from the air. The list of E�ed values in Table 20.1 orders the ability of substances to act as oxidizing or reducing agents and is summarized in Figure 20.14 �. The sub­ stances that are most readily reduced (strong oxidiz­ ing agents) are the reactants on the top left of the Li+(aq) + e.. t-- Strongest ILifs)l...__. table. Their products, on the top right of the table, c_ _j ._ are oxidized with difficulty (weak reducing agents). reducing Most negative values of E�d agent The substances on the bottom left of the table are re­ duced with difficulty, but their products are readily .A. Figure 20.14 Relative strengths of oxidized. This inverse relationship between oxidizing and reducing strength is oxidizing agents. The standard similar to the inverse relationship between the strengths of conjugate acids and reduction potentials, �ed• listed in Table bases. = (Section 16.2 and Figure 16.4) 20.1 are related to the ability of substances to serve as oxidizing or To help you remember the relationships between the strengths of oxidizing reducing agents. Species on the left side and reducing agents, recall the very exothermic reaction between sodium metal of the half-reactions can act as oxidizing and chlorine gas to form sodium chloride (Figure 8.2). In this reaction Cl2(g) is agents, and those on the right side can reduced (it serves as a strong oxidizing agent) and Na(s) is oxidized (it serves as act as reducing agents. As E;',d becomes + a strong reducing agent) . The products of this reaction-Na and Cl- ions-are more positive, the species on the left become stronger and stronger oxidizing very weak oxidizing and reducing agents, respectively. agents. As f�d becomes more negative, the species on the right become stronger and stronger reducing agents. - SAMPLE EXERCISE Determining the Relative Strengths of

�

_ _ _ _ _ _ _ _ _

20.8 1

Oxidizing Agents

Using Table 20.1, rank the following ions in order of increasing strength as oxidizing agents: N03-(aq), Ag+(aq), Cr20 /- (aq). SOLUTION Analyze: We are given several ions and asked to rank their abilities to act as oxidiz­ ing agents. Plan: The more readily an ion is reduced (the more positive its E�ed value), the stronger it is as an oxidizing agent. Solve: From Table 20.1, we have

E�ed = + 0.96 V N03-(aq) + 4 H+(aq) + 3 e- ----> NO(g) + 2 H20(/) E�ed = + 0.80 V Ag+(aq) + e- ----> Ag(s) E�ed = + 1.33 V Cr2o/-(aq) + 14 H+(aq) + 6 e- ----> 2 Cr3+(aq) + 7 H20( /) 2 2 Because the standard reduction potential of Cr207 - is the most positive, Cr207 - is the 2 strongest oxidizing agent of the three. The rank order is Ag + < N03- < Cr207 -. - PRACTICE EXERCISE

Using Table 20.1, rank the following species from the strongest to the weakest reduc­ ing agent: qaq), Fe(s), Al(s) . Answer: Al(s) > Fe(s) > r(aq)

_ _ _

862

CHAPTER 20

Electrochemistry

20.5 F REE ENERGY AND RED OX REACTIONS We have observed that voltaic cells use redox reactions that proceed spontaneously. Any reaction that can occur in a voltaic cell to produce a positive emf must be spontaneous. Consequently, it is possible to decide whether a redox reaction will be spontaneous by using half-cell potentials to calculate the emf associated with it. The following discussion will pertain to general redox reactions, not just reac­ tions in voltaic cells. Thus, we will make Equation 20.8 more general by writing it as E0 = E�ed (reduction process) - E�ed (oxidation process)

[20.10]

In modifying Equation 20.8, we have dropped the subscript "cell" to indicate that the calculated emf does not necessarily refer to a voltaic cell. Similarly, we have generalized the standard reduction potentials on the right side of the equation by referring to the reduction and oxidation processes, rather than the cathode and the anode. We can now make a general statement about the spontaneity of a reac­ tion and its associated emf, E: A positive value of E indicates a spontaneous process; a negative value ofE indicates a nonspontaneous one. We will use E to represent the emf under nonstandard conditions and Eo to indicate the standard emf. - SAMPLE EXERCISE 20.9

I Spontaneous or Not?

Using standard reduction potentials in Table 20.1, determine whether the following reactions are spontaneous under standard conditions. 2 (a) Cu(s) + 2 H +(aq) -----> Cu +(aq) + H 2(g) (b) Cl2(g) + 2 qaq) -----> 2 Cqaq) + I 2(s) SOLUTION Analyze: We are given two equations and must determine whether each is spontaneous. Plan: To determine whether a redox reaction is spontaneous under standard conditions, we first need to write its reduction and oxidation half-reactions. We can then use the standard reduction potentials and Equation 20.10 to calculate the standard emf, £0, for the reaction. If a reaction is spontaneous, its standard emf must be a positive number. Solve:

2 (a) In this reaction Cu is oxidized to Cu + and H+ is reduced to H2. The corresponding half-reactions and associated standard reduction potentials are Notice that for the oxidation, we use the standard reduction potential from Table 20.1 for the re­ 2 duction of Cu + to Cu. We now calculate P by using Equation 20.10: Because P is negative, the reaction is not sponta­ neous in the direction written. Copper metal does not react with acids in this fashion. The reverse re­ action, however, s i spontaneous and would have an P of +0.34V.

Reduction: Oxidation: P

=

=

In this case

£�00 (reduction process)

2 Cu +(aq)

+

Reduction: Oxidation: Eo

=

Cu(s)

(0 V) - (0.34 V)

2 Cu + can be reduced by H2. (b) We follow a procedure analogous to that in (a):

2 H+(aq) + 2 e- ----->

H 2(g)

=

----->

----->

-

H2(g) Cu2+(aq)

£�00

=

=

OV +0.34 V

£�00 (oxidation process)

-0.34 V

Cu(s) + 2 H+(aq)

Cl2(g) + 2 e- -----> 2 Cqaq) 2 r-(aq)

(1.36 V) - (0.54 V)

£�00

+ 2 e-

=

----->

I2(s)

£0

=

+0.34 V

E�cd

+ 2 e- E�ed

=

=

+ 1.36 V +0.54 V

+0.82 V

Because the value of P is positive, this reaction is spontaneous and could be used to build a voltaic cell. - PRACTICE EXERCISE

Using the standard reduction potentials listed in Appendix E, determine which of the following reactions are spontaneous under standard conditions: 2 (a) I2(s) + 5 Cu +(aq) + 6 Hp(l) -----> 2 !03 -(aq) + 5 Cu(s) + 12 H+(aq) 2 (b) Hg +(aq) + 2 qaq) -----> Hg(l) + I2(s) (c) H2S03(aq) + 2 Mn(s) + 4 H+(aq) -----> S(s) + 2 Mn2+(aq) + 3 Hp(l)

Answer: Reactions (b ) and (c) are spontaneous.

20.5

Free Energy and Redox Reactions

863

We can use standard reduction potentials to understand the activity series of metals. = (Section 4.4) Recall that any metal in the activity series will be ox­ idized by the ions of any metal below it. We can now recognize the origin of this rule based on standard reduction potentials. The activity series, tabulated in Table 4.5, consists of the oxidation reactions of the metals, ordered from the strongest reducing agent at the top to the weakest reducing agent at the bottom. (Thus, the ordering is "flipped over" relative to that in Table 20.1.) For example, nickel lies above silver in the activity series. We therefore expect nickel to dis­ place silver, according to the net reaction Ni(s) + 2 Ag+(aq) ----+ Ni2+(aq) + 2 Ag(s) In this reaction Ni is oxidized and Ag+ is reduced. Therefore, using data from Table 20.1, the standard emf for the reaction is Eo = E�ed

(Ag+/Ag) - E�ed (Ni2+/Ni)

= ( +0.80 V) - (-0.28 V) = + 1.08 V

The positive value of fO indicates that the displacement of silver by nickel is a spontaneous process. Remember that although we multiply the silver half­ reaction by 2, the reduction potential is not multiplied. GIVE IT SOME THOUGHT Based on Table 4.5, which is the stronger reducing agent, Hg(l) or Pb(s)?

EMF and � G

The change in the Gibbs free energy, llG, is a measure of the spontaneity of a process that occurs at constant temperature and pressure. = (Section 19.5) The emf, E, of a redox reaction also indicates whether the reaction is sponta­ neous. The relationship between emf and the free-energy change is llG =

-nFE

[20.11]

In this equation n is a positive number without units that represents the number of electrons transferred in the reaction. The constant F is called Faraday's constant, named after Michael Faraday (Figure 20.15 II>). Faraday's constant is the quantity of electrical charge on one mole of electrons. This quantity of charge is called a faraday (F).

1 F = 96,485 Cfmol = 96,485 J/V-mol The units of llG calculated by using Equation 20.11 are J/mol; as in Equation 19.16, we use "per mole" to mean "per mole of reaction as written." = (Section 19.7) Both n and F are positive numbers. Thus, a positive value of E in Equation 20.11 leads to a negative value of llG. Remember: A positive value ofE and a neg­ ative value of LlG both indicate that a reaction is spontaneous. When the reactants and products are all in their standard states, Equation 20.11 can be modified to relate LlG0 and E0• [20.12] Equation 20.12 is very important. It relates the standard emf, P, of an elec­ trochemical reaction to the standard free-energy change, LlG0, for the reaction. Because llGo is related to the equilibrium constant, K, for a reaction by the ex­ pression llGo = -RT In K = (Section 19.7), we can relate the standard emf to the equilibrium constant for the reaction.

"' Figure 20.15

Michael Faraday.

Faraday (1 791 -1 867) was born in England, a child of a poor blacksmith. At the age of 1 4 he was apprenticed to a bookbinder who gave Faraday time to read and to attend lectures. In 1 81 2 he became an assistant in Humphry Davy's laboratory at the Royal Institution. He succeeded Davy as the most famous and influential scientist in England, making an amazing number of important discoveries, including his formation of the quantitative relationships between electrical current and the extent of chemical reaction in electrochemical cells.

864

CHAPTER 20

- SAMPLE EXERCISE

Electrochemistry

20.10 I Determining LlGo and K

(a) Use the standard reduction potentials listed in Table 20.1 to calculate the standard free-energy change, um constant, K, at 298 K for the reaction 4 Ag(s) + 02(g) + 4 H+(aq)

----->

LlG0, and the equilibri­

4 Ag +(aq) + 2 H20(1)

! 02(g) + 2 H+(aq) -----> 2 Ag+(aq)

(b) Suppose the reaction in part (a) was written 2 Ag(s) + What are the values of E0,

+ H20(1)

LlG0, and K when the reaction is written in this way?

SOLUTION Analyze: We are asked to determine

LlG0 and K for a redox reaction, using standard reduction potentials.

Plan: We use the data in Table 20.1 and Equation 20.10 to determine Eo for the reaction and then use Eo in Equation 20.12 to cal­

culate

LlG0• We will then use Equation 19.17, LlG0

Solve:

(a) We first calculate E" by breaking the equation into two half-reactions, as we did in Sample Exercise 20.9, and then obtain E�cd values from Table 20.1 (or Appendix E):

= -

RT In K, to calculate K.

Reduction: Oxidation:

02(g) + 4 H+(aq) + 4 e-

----->

4 Ag(s)

----->

2 H 0(1) 2 4 Ag+(aq) + 4 e-

E;;,d E�d

= +1.23 V = +0.80 V

Even though the second half-reaction has 4 Ag, we use the E�ed value directly from Table 20.1 because emf is an intensive property. Using Equation 20.10, we have The half-reactions show the transfer of four electrons. Thus, for tltis reaction = 4. We now use Equation 20.12 to calculate LlG0:

n

E" = (1.23 V) LlG0 = -nFE"

= -(4)(96,485 J/V-mol)(+0.43) V

The positive value of E" leads to a negative value of Now we need to calculate the equilibrium constant, K, using LlG0 = - In K. Because LlGo is a large negative number, which means the reaction is thermodynami­ cally very favorable, we expect K to be large.

RT

(0.80 V) = 0.43 V

LlG0

= -1.7 x 105 Jfmol = -170 kJ/mol

LlG0• =

-RT In K

-1.7 x 105 Jfmol = -(8.314 J/K mol) (298 K) ln K In K =

-1.7

X

105 J/mol

-(8.314 J/K mol)(298 K)

In K = 69

2 K = 9 X 10 9

K is indeed very large! This means that we expect silver metal to oxidize in acidic aqueous environments, in air, to Ag+.

Notice that the voltage calculated for the reaction was 0.43 V, which is easy to measure. Directly measuring such a large equi­ librium constant by measuring reactant and product concentrations at equilibrium, on the other hand, would be very difficult.

!·

(b) The overall equation is the same as that in part (a), multiplied by The half­ reactions are The values of E;;,d are the same as they were in part (a); they are not changed by multiplying the half-reactions by Thus, E" has the same value as in part (a):

!.

!

Notice, though, that the value of has changed to n = 2, which is the value in part (a). Thus, LlG' is half as large as in part (a).

n

Now we can calculate K as before:

E"

! 02(g)

+ 2 H+(aq) + 2 e-

----->

H20( I)

E;;,d

= +0.80 V

= +0.43 v

LlGo

= -(2)(96,485 J/V-mol)(+0.43 V) = -83 k J/mol

-8.3

x

104 J/mol = -(8.314 J/K mol)(298 K) In K K = 4 X 1014

is an quantity, so multiplying a chemical equation by a certain factor will not affect the value of E0• Multiplying an equation will change the value of however, and hence the value of LlG0• The change in free energy, in units of J/mol of reaction as written, is an quantity. The equilibrium constant is also an extensive quantity.

Comment:

Eo

Reduction: Oxidation:

intensive

extensive

n,

- PRACTICE EXERCISE For the reaction 3 Ni2+(aq) + 2 Cr(OH)J(s) + 10 OH-(aq)

----->

(a) What is the value of n? (b) Use the data in Appendix E to calculate

Answers: (a) 6, (b)

1 + 87 kJ/mol, (c) K = 6 x 10- 6

3 Ni(s) + 2 Cr0 2-(aq) + 8 H20(/) 4 = 298 K.

LlG0• (c) Calculate K at T

20.6

Cell EMF under Nonstandard Conditions

20.6 CELL EMF UNDER NON STAN DARD CONDITIONS We have seen how to calculate the emf of a cell when the reactants and products are under standard conditions. As a voltaic cell is discharged, however, the reactants of the reaction are consumed, and the products are generated, so the concentrations of these substances change. The emf progressively drops until E = 0, at which point we say the cell is "dead." At that point the concentrations of the reactants and prod­ ucts cease to change; they are at equilibrium. In this section we will examine how the cell emf depends on the concentrations of the reactants and products of the cell reaction. The emf generated under nonstandard conditions can be calculated by using an equation first derived by Walther Nernst (1864-1941), a German chemist who established many of the theoretical foundations of electrochemistry. The Nernst Equation

The dependence of the cell emf on concentration can be obtained from the depen­ dence of the free-energy change on concentration. oco (Section 19.7) Recall that the free-energy change, 6.G, is related to the standard free-energy change, 6.G0: 6.G = 6.G 0 + RT In Q [20.13] The quantity Q is the reaction quotient, which has the form of the equilibrium­ constant expression except that the concentrations are those that exist in the re­ action mixture at a given moment. oco (Section 15.6) Substituting 6-G = -nFE (Equation 20.11) into Equation 20.13 gives -nFE = -nFEo + RT In Q Solving this equation for E gives the Nernst equation: RT ln Q E = £0 [20.14] nF This equation is customarily expressed in terms of common (base 10) loga­ rithms, which is related to natural logarithms by a factor of 2.303: 2.303 RT log Q E = £0 [20.15] nF At T = 298 K the quantity 2.303 RT/F equals 0.0592, with units of volts (V), and so the equation simplifies to 0 0592 V (T = 298 K) log Q - · [20.16] n We can use this equation to find the emf produced by a cell under nonstandard conditions or to determine the concentration of a reactant or product by mea­ suring the emf of the cell. To show how Equation 20.16 might be used, consider the following reac­ tion, which we discussed earlier: Zn(s) + Cu2+(aq) � Zn2+(aq) + Cu(s) In this case n = 2 (two electrons are transferred from Zn to Cu2+), and the stan­ dard emf is + 1.10 V. Thus, at 298 K the Nernst equation gives [Zn2+] 0.0592 v E = 1.10 V log [20.17] 2 [Cu2+] Recall that pure solids are excluded from the expression for Q. oco (Section 15.6) According to Equation 20.17, the emf increases as [Cu2+] increases and as [Zn2+] decreases. For example, when [Cu2+] is 5.0 M and [Zn2+] is 0.050 M, we have

E

=

£0

--- --

E

=

0.0592 V 0.050 1.10 V - --- log (5.0 2

=

1.10 v -

0.0592 v ( -2.00) 2

=

)

1.16 v

865

866

CHAPTER 20

Electrochemistry Thus, increasing the concentration of the reactant (Cu2+) and decreasing the concentration of the product (Zn2+ ) relative to standard conditions increases the emf of the cell (E = + 1.16 V) relative to standard conditions (E0 = + 1.10 V). The Nernst equation helps us understand why the emf of a voltaic cell drops as the cell discharges. As reactants are converted to products, the value of Q increases, so the value of E decreases, eventually reaching E = 0. Because t.G = nFE (Equation 20.11), it follows that t.G = 0 when E = 0. Recall that a system is at equilibrium when t.G = 0. (Section 19.7) Thus, when E = 0, the cell reaction has reached equilibrium, and no net reaction is occurring. In general, increasing the concentration of reactants or decreasing the con­ centration of products increases the driving force for the reaction, resulting in a higher emf. Conversely, decreasing the concentration of reactants or increasing the concentration of products causes the emf to decrease. -

c:xx:>

• SAMPLE EXERCISE

20.11

Cell EMF under Nonstandard Conditions I Voltaic

Calculate the emf at 298 K generated by the cell described in Sample Exercise 20.4 when [CrzOl-J = 2.0 M, [H+] = 1.0 M, [r-] = 1.0 M, and [Cr 3+] = 1.0 X 10-5 M. 2 Crz07 -(aq) + 14 H+(aq) + 6 l-(aq) -> 2 Cr 3+(aq) + 3 I2(s) + 7 H 20(!) SOLUTION Analyze: We are given a chemical equation for a voltaic cell and the concentrations of reactants and products under which it operates. We are asked to calculate the emf of the cell under these nonstandard conditions. Plan: To calculate the emf of a cell under nonstandard conditions, we use the Nernst equation in the form of Equation 20.16. Solve: We first calculate E" for the cell from standard reduction potentials (Table 20.1 or Appendix E). The standard emf for this reaction was calculated in Sample Exercise 20.6: E" = 0.79 V. As you will see if you refer back to that exercise, the balanced equation shows six electrons transferred from reducing agent to oxidizing agent, so n = 6. The reaction quotient, Q, is 2 [Cr 3+f (1 . 0 x 10-s ) _ = = 5 .0 X 10 11 Q= [Cr2o/-][H+] 14 [r-]6 (2.0)(1.0) 14 (1.0)6

Using Equation 20.16, we have E = 0.79 V =

v = 0.79 v 0.79

0·0592 V log(5.0 X 10-11) 6 0 0592 ' ( -10.30) 6 = 0.89 v + 0.10 -

v

-

v

Check: This result is qualitatively what we expect: Because the concentration of Cr20/- (a reactant) is greater than 1 M and the concentration of Cr3+ (a product) is less than 1 M, the emf is greater than E". Q is about 10-10, so log Q is about -10. Thus, the correction to E" is about 0.06 x (10)/6, which is 0.1, in agreement with the more detailed calculation. - PRACTICE EXERCISE

Calculate the emf generated by the cell described in the practice exercise accompany­ ing Sample Exercise 20.6 when [Al 3+] = 4.0 X 10-3 M and [r-] = 0.010 M. Answer: E = +2.36 V

• SAMPLE EXERCISE I Calculating Concentrations in a Voltaic Cell 2 If the voltage of a Zn-H+ cell (like that in Figure 20.11) is 0.45 V at 25 °C when [Zn +] = 1.0 M and Jl-r, = 1.0 atrn, what is the con­ centration of H+?

20.12

SOLUTION Analyze: We are given a description of a voltaic cell, its emf, and the concentrations of all reactants and products except H+, which we are asked to calculate.

20. 6

Cell EMF under Nonstandard Conditions

867

20.1

Plan: First, we write the equation for the cell reaction and use standard reduction potentials from Table to calculate Eo for the reaction. After determining the value of from our reaction equation, we solve the Nernst equation for Q. Finally, we use the equation for the cell reaction to write an expression for Q that contains [H+] to determine [H+]. 2 Solve: The cell reaction is H2 (gl H+(aq) --> Zn +(aq) Zn(s)

n

The standard emf is

+ +2 - E�ed 0 - (-0.76 +0.76 n=2 0.45 0.76 0.0592 v Q Q (0. 76 v - 0.45 v>(o � ) 10. 5 .05 2 v Q = 1010. 5 = 3 X 101 0 [Zn2+]%, (1.0)(1.0) Q = � = [H+j2 = 3 X 1010 1 0 = 3 X 10-11 =3 X-·1010 = \h X 10-11 = 6 10--;; M + 16.4) [H ] 20.11 0.542 V? £" = E�ed (reduction) =

Because each Zn atom loses two electrons, Using Equation

20.16,

we can solve for

Q:

V

log

Q has the form of the equilibrium constant for the reaction Solving for

V

=

(oxidation)

V) =

V

-- log V-2

=

=

[H+f

[H+], we have

[H+]

X

Comment: A voltaic cell whose cell reaction involves H+ can be used to measure voltaic cell with a voltmeter calibrated to read pH directly. cx:c (Section - PRACTICE EXERCISE

or pH. A pH meter is a specially designed

What is the pH of the solution in the cathode compartment of the cell pictured in Figure anode compartment is and cell emf is Answer: pH = (using data from Appendix E to obtain £" to three significant figures)

4.23 0.10 M,

when %2

Concentration Cells In each of the voltaic cells that we have looked at thus far, the reactive species at the anode has been different from the one at the cathode. Cell emf depends on concentration, however, so a voltaic cell can be constructed using the same species in both the anode and cathode compartments as long as the concentra­ tions are different. A cell based solely on the emf generated because of a differ­ ence in a concentration is called a concentration cell. A diagram of a concentration cell is shown in Figure 20.16(a) 1'. One compartment consists of a strip of nickel metal immersed in a 1 .00 M solution 2 of Ni +(aq). The other compartment also has an Ni(s) electrode, but it is

Ni cathode

[Ni2+] =

1.00 10-3M x

[Ni2 + ] =

1.00 M

(a)

.A 2Figure 20.16

2 Concentration cell based on the Ni + -Ni cell reaction.

(b)

In (a) the concentrations of Ni +(aq) in the two compartments are unequal, and the cell generates an electrical current. The cell 2 operates until the concentrations of Ni +(aq) in the two compartments become equal, (b) at which point the cell has reached equilibrium and is "dead."

= 1.0

atm, [Zn2+] in the

868

CHAPTER 20

Electrochemistry

3 2 immersed in a 1.00 X 10- M solution of Ni +(aq). The two compartments are connected by a salt bridge and by an external wire running through a volt­ meter. The half-cell reactions are the reverse of each other: Ni(s ) � Ni 2+(aq) + 2 e­

Anode: Cathode:

E�ed

Ni2+ (aq) + 2 e- � Ni(s )

-0.28 v

E�ed = -0.28 V

-

-

Although the standard emf for this cell is zero, E�ell = E �ect (cathode) ( -0.28 V) = 0 V, the cell operates under nonstandard conditions because the concentration of Ni 2+(aq) is different in the two compartments. In fact, the cell will operate until the concentrations of Ni 2+ in both compartments are equal. Oxidation of Ni(s) occurs in the half­ cell containing the more dilute solution, thereby increasing the concentration of Ni 2+(aq). It is therefore the anode compartment of the cell. Reduction of 2 Ni +(aq) occurs in the half-cell containing the more concentrated solution, thereby decreasing the concentration of Ni 2+(aq), making it the cathode compartment. The overall cell reaction is therefore

E�ect (anode) = ( -0.28 V)

2 Ni(s) � Ni +(aq, dilute) + 2 e-

Anode: Cathode:

2 Ni + (aq, concentrated) + 2 e- � Ni(s)

Overall:

Chemistr and Li e

2 2 Ni + (aq, concentrated) � Ni + (aq, dilute)

H EARTB EATS A N D E L E C T RO CARD I O G RA P H Y

T ty. In a typical day an adult's heart pumps more than he human heart i s a marvel o f efficiency and dependabili­

7000 L of blood through the circulatory system, usually with no maintenance required beyond a sensible diet and lifestyle. We generally think of the heart as a mechanical device, a muscle that circulates blood via regularly spaced muscular contractions. However, more than two centuries ago, two pio­ neers in electricity, Luigi Galvani (1729-1787) and Alessandro Volta (1745--1827), discovered that the contractions of the heart are controlled by electrical phenomena, as are nerve impulses throughout the body. The pulses of electricity that cause the heart to beat result from a remarkable combination of electro­ chemistry and the properties of semipermeable membranes. = (Section 13.5) Cell walls are membranes with variable permeability with respect to a number of physiologically important ions (espe­ 2 cially Na+, K+, and Ca +). The concentrations of these ions are different for the fluids inside the cells (the intracellular fluid, or ICF) and outside the cells (the extracellularfluid, or ECF). In car­ diac muscle cells, for example, the concentrations of K+ in the ICF and ECF are typically about 135 millimolar (mM) and 4 mM, respectively. For Na+, however, the concentration differ­ ence between the ICF and ECF is opposite that for K+; typical­ ly, [Na +JrcF = 10 mM and [Na+ ]ECF = 145 mM.

The cell membrane is initially permeable to K+ ions, but is 2 much less so to Na+ and Ca +. The difference in concentration of K+ ions between the ICF and ECF generates a concentration cell: Even though the same ions are present on both sides of the membrane, there is a potential difference between the two fluids that we can calculate using the Nemst equation with E" = 0. At physiological temperature (37 "C) the potential in millivolts for moving K+ from the ECF to the ICF is

E

=

£0

_

2.30 RT

nF

log

[K+lJcF

(--135 ) = -94

[K+]ECF

= 0 - (61.5 mV) log

mM

4 mM

mV

In essence, the interior of the cell and the ECF together serve as a voltaic cell. The negative sign for the potential indicates that work is required to move K + into the intracellular fluid. Changes in the relative concentrations of the ions in the ECF and ICF lead to changes in the emf of the voltaic cell. The cells of the heart that govern the rate of heart contraction are called the pacemaker cells. The membranes of the cells regu­ late the concentrations of ions in the ICF, allowing them to change in a systematic way. The concentration changes cause the emf to change in a cyclic fashion, as shown in Figure 20.17�.

20.6

Cell EMF under Nonstandard Conditions

869

We can calculate the emf of a concentration cell by using the Nemst equation. For this particular cell, we see that n = 2. The expression for the reaction quotient for the overall reaction is Q = [Ni2+ldilute/[Ni2+lconcentrated· Thus, the emf at 298 K is

E = £0 _

- O

_

_

0.0592 V n

log Q

0.0592 V [Ni2+Jdilute Iog · + 2 [NI2 ]concentrated

3 - V Iog 1.00 X 10- M - 0.0592 2 1.00 M -

-

+0.0888 V

This concentration cell generates an emf of nearly 0.09 V even though E0 = 0. The difference in concentration provides the driving force for the cell. When the concentrations in the two compartments become the same, Q = 1 and E = 0. The idea of generating a potential by a difference in concentration is the basis for the operation of pH meters (Figure 16.6). It is also a critical aspect in bi­ ology. For example, nerve cells in the brain generate a voltage across the cell membrane by having different concentrations of ions on either side of the mem­ brane. The regulation of the heartbeat in mammals, as discussed in the "Chem­ istry and Life" box in this section, is another example of the importance of electrochemistry to living organisms.

Time -----+-

Figure 20.17

Variation of the electrical potential caused by changes of ion concentrations in the pacemaker cells of the heart.

£ Figure 20.18 A typical electrocardiogram. An electrocardiogram (ECG) printout records the electrical events monitored by electrodes attached to the body surface. The horizontal axis is time, and the vertical displacement is the emf.

The emf cycle determines the rate at which the heart beats. If the pacemaker cells malfunction because of disease or injury, an artificial pacemaker can be surgically implanted . The arti­ ficial pacemaker is a small battery that generates the electrical pulses needed to trigger the contractions of the heart. During the late 1800s scientists discovered that the elec­ trical impulses that cause the contraction of the heart muscle are strong enough to be detected at the surface of the body.

This observation formed the basis for electrocardiography, noninvasive monitoring of the heart by using a complex array of electrodes on the skin to measure voltage changes during heartbeats. A typical electrocardiogram is shown in Figure 20.18 £. It is quite striking that, although the heart's major function is the mechanical pumping of blood, it is most easily monitored by using the electrical impulses generated by tiny voltaic cells.

£

lon concentration and emf In the human

heart.

870

CHAPTER 20

Electrochemistry

• SAMPLE EXERCISE I Determining pH Using a Concentration Cell A voltaic cell is constructed with two hydrogen electrodes. Electrode 1 has l'ti, = 1.00 atrn and an unknown concentration of H+(aq). Electrode 2 is a standard hydrogen electrode ([H+] = 1.00 M, l'ti, = 1.00 atm). At 298 K the measured cell voltage is 0.211 V, and the electrical current is observed to flow from electrode 1 through the external circuit to electrode 2. Calculate [H+] for the solution at electrode 1. What is its pH?

20.13

SOLUTION Analyze: We are given the voltage of a concentration cell and the direction in which the current flows. We also have the concen­ trations of all reactants and products except for [H+] in half-cell 1, which is our unknown. Plan: We can use the Nernst equation to determine Q and then use Q to calculate the unknown concentration. Because this is a concentration cell, E�n = 0 V. Solve: Using the Nernst equation, we have

0.211 V

=0

0.0592 v

- --- log Q 2

log Q = -(0.211 v)

( 0.05�2 v ) =

-7.13

Q = 10-7.13 = 7.4 X 10-8 Because electrons flow from electrode 1 to electrode 2, electrode 1 is the anode of the cell and electrode 2 is the cathode. The electrode reactions are therefore as follows, with the concentration of H+(aq) in elec­ trode 1 represented with the unknown x:

H 2(g, 1.00 atm) -----> 2 H+(aq, x M) + 2 e-

Electrode 1: Electrode 2:

2 H+(aq; 1.00 M) + 2 e-

----->

H 2(g, 1.00 atm)

E�ed = O V E�ed = O V

[H+ (electrode 1)f l'ti, (electrode 2) -- --' -, ---------Q = -----: '--[H+ (electrode 2)f l'ti, (electrode 1)

Thus,

x2(1.00) = x2 = 7.4 X 10-8 (1 .00)2(1.00) X=

\h.4 X 10-8 = 2.7 X 10-4

=

2.7 X 10-4 M

At electrode 1, therefore,

[H+ ]

and the pH of the solution is

pH = -log[H+] = -log(2.7 X 10-4) = 3.57

Comment: The concentration of H+ at electrode 1 is lower than that in electrode 2, which is why electrode 1 is the anode of the cell: The oxidation of H2 to H+(aq) increases [H+] at electrode 1. - PRACTICE EXERCISE

=

A concentration cell is constructed with two Zn(s)-Zn2+(aq) half-cells. The first half-cell has [Zn2+] 1.35 M, and the second half­ cell has [Zn2+] = 3 75 X 10-4 M. (a) Which half-cell is the anode of the cell? (b) What is the emf of the cell? Answers: (a) the second half-cell, (b) 0.105 V

.

20.7 BATTERIES AND FUEL CELLS A battery is a portable, self-contained electrochemical power source that con­ sists of one or more voltaic cells. For example, the common 1.5-V batteries used to power flashlights and many consumer electronic devices are single voltaic cells. Greater voltages can be achieved by using multiple voltaic cells in a single battery, as is the case in 12-V automotive batteries. When cells are connected in series (with the cathode of one attached to the anode of another), the battery produces a voltage that is the sum of the emfs of the individual cells. Higher emfs can also be achieved by using multiple batteries in series (Figure 20.19 � ) . The electrodes of batteries are marked following the convention of Figure 20.6-­ the cathode is labeled with a plus sign and the anode with a minus sign. Although any spontaneous redox reaction can serve as the basis for a volta­ ic cell, making a commercial battery that has specific performance characteris­ tics can require considerable ingenuity. The substances that are oxidized at the

20.7

Batteries and Fuel Cells

871

anode and reduced by the cathode determine the emf of a battery, and the us­ able life of the battery depends on the quantities of these substances packaged in the battery. Usually a barrier analogous to the porous barrier of Figure 20.6 separates the anode and the cathode compartments. Different applications require batteries with different properties. The bat­ tery required to start a car, for example, must be capable of delivering a large electrical current for a short time period. The battery that powers a heart pace­ maker, on the other hand, must be very small and capable of delivering a small but steady current over an extended time period. Some batteries are primary cells, meaning that they cannot be recharged. A primary cell must be discarded or recycled after its emf drops to zero. A secondary cell can be recharged from an external power source after its emf has dropped. In this section we will briefly discuss some common batteries. As we do so, notice how the principles we have discussed so far in this chapter help us un­ derstand these important sources of portable electrical energy. Lead-Acid Battery A 12-V lead-acid automotive battery consists of six voltaic cells in series, each producing 2 V. The cathode of each cell consists of lead dioxide (Pb02) packed on a metal grid. The anode of each cell is composed of lead. Both electrodes are im­ mersed in sulfuric acid. The electrode reactions that occur during discharge are

Cathode: Anode:

Pb02(s) + HS04-(aq) + 3 H+ (aq) + 2 e-

Pb(s) + HS04-(aq)

------>

------>

PbS04(s)

"' Figure 20.19

Combining batteries.

When batteries are connected in series, as in most flashlights, the total emf is the sum of the individual emfs.

+ 2 H20(I)

PbS04(s) + H+(aq) + 2 e-

The standard cell potential can be obtained from the standard reduction potentials in Appendix E: E�ell = E�d (cathode) - E�ed (anode) = (+1.685 V) - (-0.356 V) = +2.041 V The reactants Pb and Pb02 serve as the electrodes. Because the reactants are solids, there is no need to separate the cell into anode and cathode compartments; the Pb and Pb02 cannot come into direct physical con­ tact unless one electrode plate touches another. To keep the electrodes from touching, wood or glass-fiber spacers are placed between them (Figure 20.20 � ) . Using a reaction whose reactants and products are solids has another benefit. Because solids are excluded from the reaction quotient Q, the rel­ ative amounts of Pb(s), Pb02(s), and PbS04(s) have no effect on the emf of the lead storage battery, helping the battery maintain a relatively constant emf during discharge. The emf does vary somewhat with use because the concentration of H2S04 varies with the extent of cell discharge. As the equation for the overall cell reaction indicates, H2S04 is consumed during the discharge. One advantage of a lead-acid battery is that it can be recharged. Dur­ ing recharging, an external source of energy is used to reverse the direc­ tion of the overall cell reaction, regenerating Pb(s) and Pb02(s).

l

//

H2S04

electrolyte

\�'\"\Lead

Lead grid filled with spongy lead (anode)

"' Figure 20.20

In an automobile the alternator, driven by the engine, provides the energy necessary for recharging the battery. Recharging is possible because PbS04 formed during discharge adheres to the electrodes. As the external source forces electrons from one electrode to another, the PbS04 is converted to Pb at one electrode and to Pb02 at the other.

A 12-V

grid filled with Pb02 (cathode)

automotive

Each anode/cathode pair of electrodes in this schematic cutaway produces a potential of about 2 V. Six pairs of electrodes are connected in series, producing the desired battery voltage. lead-add battery.

872

CHAPTER 20

Electrochemistry Alkaline Battery

.A. Figure 20.21 Cutaway view of a miniature alkaline battery.

The most common primary (nonrechargeable) battery is the alkaline battery. 1 More than 10 0 alkaline batteries are produced annually. The anode of this bat­ tery consists of powdered zinc metal immobilized in a gel in contact with a con­ centrated solution of KOH (hence the name alkaline battery) . The cathode is a mixture of Mn02(s) and graphite, separated from the anode by a porous fabric. The battery is sealed in a steel can to reduce the risk of leakage of the concentrat­ ed KOH. A schematic view of an alkaline battery is shown in Figure 20.21 2 MnO(OH)(s) + 2 OH-(a q)

Anode:

Zn(s) + 2 OH-(a q)

-----> Zn(OHh(s) + 2 e-

The emf of an alkaline battery is 1.55 V at room temperature. The alkaline bat­ tery provides far superior performance over the older "dry cells" that were also based on Mn02 and Zn as the electrochemically active substances. N ickel-Cad mium, Nickel- Metal-Hyd ride, and Lithium-ion Batteries The tremendous growth in high-power-demand portable electronic devices, such as cell phones, notebook computers, and video recorders, has increased the demand for lightweight, readily recharged batteries. One of the most com­ mon rechargeable batteries is the nickel-cadmium (nicad) battery. During dis­ charge, cadmium metal is oxidized at the anode of the battery while nickel oxyhydroxide [NiO(OH)(s)] is reduced at the cathode.

Cathode: 2 NiO(OH)(s) + 2 H20(1) + 2 e- -----> 2 Ni(OHh(s) + 2 OH-(a q)

Anode:

Cd(s) + 2 OH-(aq)

----->

Cd(OHh(s) + 2 e-

As in the lead-acid battery, the solid reaction products adhere to the electrodes, which permits the electrode reactions to be reversed during charging. A single nicad voltaic cell has an emf of 1.30 V. Nicad battery packs typically contain three or more cells in series to produce the higher emfs needed by most elec­ tronic devices. There are drawbacks to nickel-cadmium batteries. Cadmium is a toxic heavy metal. Its use increases the weight of batteries and provides an environ­ mental hazard-roughly 1.5 billion nickel-cadmium batteries are produced an­ nually, and these must eventually be recycled as they lose their ability to be recharged. Some of these problems have been alleviated by the development of nickel-metal-hydride (NiMH) batteries. The cathode reaction of NiMH batter­ ies is the same as that for nickel-cadmium batteries, but the anode reaction is very different. The anode consists of a metal alloy, such as ZrNiv that has the ability to absorb hydrogen atoms (we will discuss alloys in Section 23.6). Dur­ ing the oxidation at the anode, the hydrogen atoms lose electrons, and the re­ sultant H+ ions react with OH- ions to form H20, a process that is reversed during charging. The current generation of hybrid gas-electric automobiles, which are powered by both a gasoline engine and an electric motor, use NiMH batteries to store the electrical power. The batteries are recharged by the electric motor while braking. Due to the robustness of the batteries toward discharge and recharge, the batteries can last up to 8 years. The newest rechargeable battery to receive large use in consumer electronic devices is the lithium-ion (Li-ion) battery. You will find this battery in cell phones and laptop computers. Because lithium is a very light element, Li-ion batteries achieve a greater energy density-the amount of energy stored per unit mass-than nickel-based batteries. The technology of Li-ion batteries is very dif­ ferent from that of the other batteries we have described here; it is based on the ability of u+ ions to be inserted into and removed from certain layered solids.

20.7

Batteries and Fuel Cells

873

For example, u+ ions can be inserted reversibly into layers of graphite (Figure 11.41). In most commercial cells, one electrode is graphite or some other car­ bon-based material, and the other is usually made of lithium cobalt oxide (LiCo02). When the cell is charged, cobalt ions are oxidized, and u + ions mi­ grate into the graphite. During discharge, when the battery is producing elec­ tricity for use, the u+ ions spontaneously migrate from the graphite anode to the cathode, enabling electrons to flow through the external circuit. An Li-ion battery produces a maximum voltage of 3.7 V, considerably higher than typical 1.5-V alkaline batteries. Hydrogen Fuel Cells The thermal energy released by the combustion of fuels can be converted to electrical energy. The heat may convert water to steam, which drives a turbine that in turn drives a generator. Typically, a maximum of only 40% of the energy from combustion is converted to electricity; the remainder is lost as heat. The direct production of electricity from fuels by a voltaic cell could, in principle, yield a higher rate of conversion of the chemical energy of the reaction. Voltaic cells that perform this conversion using conventional fuels, such as H2 and CH4, are called fuel cells. Strictly speaking, fuel cells are not batteries because they are not self-contained systems-the fuel must be continuously supplied to generate electricity. The most promising fuel-cell system involves the reaction of H2(g) and 02(g) to form H20(!) as the only product. These cells can generate electricity twice as efficiently as the best internal combustion engine. Under acidic condi­ tions, the electrode reactions are

Cathode: Anode:

02(g) + 4 H+ + 4 e- � 2 H20(!) 2 H2(g) � 4 H+ + 4 e-

Overall: The standard emf of an H2-02 fuel cell is + 1 .23 V, reflecting the large driving force for the reaction of H2 and 02 to form H20. In this fuel cell (known as the PEM fuel cell, for "proton-exchange mem­ brane"), the anode and cathode are separated by a thin polymer membrane that is permeable to protons but not to electrons. The polymer membrane, therefore, acts as a salt bridge. The electrodes are typically made from graphite. A PEM cell operates at a temperature of around 80 °C. At this low temperature the elec­ trochemical reactions would normally occur very slowly, and so a thin layer of platinum on each electrode catalyzes the reactions. Under basic conditions the electrode reactions in the hydrogen fuel cell are

Cathode: Anode:

-

4 e- + 02(g) + 2 H20(1) � 4 OH (aq)

Figure 20.22 A low-temperature H,-02 fuel cell. The porous membrane

T

allows the H+ ions generated by the oxidation of H 2 at the anode to migrate to the cathode, where H20 is formed.

2 H2(g) + 4 OH-(a q) � 4 H20(!) + 4 e2 Hz(g) + Oz(g) � 2 HzO(l)

NASA has used this basic hydrogen fuel cell as the energy source for its spacecraft. Liquid hydrogen and oxygen are stored as fuel, and water, the product of the reaction, is drunk by the spacecraft crew. A schematic drawing of a low-temperature Hz-02 fuel cell is shown in Figure 20.22 �. This technology is the basis for fuel cell-powered vehi­ cles that are under study by major automobile manufacturers. Currently a great deal of research is going into improving fuel cells. Much effort is being directed toward developing fuel cells that use conventional fuels such as hydrocarbons and alcohols, which are not as difficult to handle and distribute as hydrogen gas.

H2 exhaust

- o2 inlet

membrane

874

CHAPTER 20

Chemis

Electrochemistry

Put

to

�rk

D I RECT M ET HANOL FUEL C ELLS

Wa clean and efficient alternative to gasoline-powered in­

hile the hydrogen fuel cell has been widely proposed as

ternal combustion engines, liquid methanol, CH30H, is a far more attractive fuel to store and transport than hydrogen gas. Furthermore, methanol is a clean-burning liquid, and its use would require only minor modifications to existing engines and to fuel-delivery infrastructure. One of the intriguing aspects of methanol as a fuel is that manufacturing it could make use of carbon dioxide, a source of global warming. coo (Section 18.4) Methanol can be made by combining C02 and H:z, although the process is presently costly. Imagine, though, that the synthesis can be improved and that the C02 used in the synthesis is captured from ex­ haust gases from power plants or even directly from the at­ mosphere. 1n such cases, the C02 released by subsequently burning the methanol would be cancelled by the carbon diox­ ide captured to make it. Thus, the process would be carbon neutral, meaning that it would not increase the concentration

of C02 in the atmosphere. The prospect of a liquid fuel that could replace conventional fuels without contributing to the greenhouse effect has spurred considerable research to reduce the cost of methanol synthesis and to develop and improve methanol fuel cell technology. A direct methanol fuel cell has been developed that is sim­ ilar to the PEM hydrogen fuel cell. The reactions in the cell are

Cathode: � 02(g) Anode: Overall:

+ 6 H+ + 6 e-

-----> 3 H20(g)

CHPH(l) + HzO(g) -----> COz(g) + 6 H+ + 6 eCH30H(g) + � Oz (g) -----> COz(g) + 2 HzO(g)

The direct methanol fuel cell is currently too expensive to be used in passenger cars because of the quantity of platinum catalyst it requires to operate. Nevertheless, small methanol fuel cells could appear in mobile devices such as computers or cell phones in the near future.

20.8 C O RROSION Batteries are examples of how spontaneous redox reactions can be used pro­ ductively. In this section we will examine the undesirable redox reactions that lead to the corrosion of metals. Corrosion reactions are spontaneous redox reac­ tions in which a metal is attacked by some substance in its environment and converted to an unwanted compound. For nearly all metals, oxidation is a thermodynamically favorable process in air at room temperature. When the oxidation process is not inhibited in some way, it can be very destructive to whatever object is made from the metal. Oxidation can form an insulating protective oxide layer, however, that prevents further reaction of the underlying metal. Based on the standard reduction po­ tential for Al3+, for example, we would expect aluminum metal to be very read­ ily oxidized. The many aluminum soft-drink and beer cans that litter the environment are ample evidence, however, that aluminum undergoes only very slow chemical corrosion. The exceptional stability of this active metal in air is due to the formation of a thin protective coat of oxide-a hydrated form of Al203-on the surface of the metal. The oxide coat is impermeable to 02 or H20 and so protects the underlying metal from further corrosion. Magnesium metal is similarly protected. Some metal alloys, such as stainless steel, likewise form protective impervious oxide coats. The semiconductor silicon, as we saw in Chapter 12, also readily forms a protective Si02 coating that is important to its use in electronic circuits. Corrosion of I ro n

.A. Figure 20.23 Corrosion. The corrosion of iron is an electrochemical process of great economic importance. The annual cost of metallic corrosion in the United States is estimated to be $70 billion.

The rusting of iron (Figure 20.23 -Ecell· For example, if a nonspontaneous process has Ecell = -0.9 V, then the external potential Eext must be greater than 0.9 V in order for the process to occur. When an external potential Eext is applied to a cell, the surroundings are doing work on the system. The amount of work performed is given by w =

nFEext

1W

=

[20.20]

Unlike Equation 20.19, there is no minus sign in Equation 20.20. The work cal­ culated in Equation 20.20 will be a positive number because the surroundings are doing work on the system. The quantity n in Equation 20.20 is the number of moles of electrons forced into the system by the external potential. The prod­ uct n x F is the total electrical charge supplied to the system by the external source of electricity. Electrical work can be expressed in energy units of watts times time. The watt (W) is a unit of electrical power (that is, the rate of energy expenditure). 1 J/s

Thus, a watt-second is a joule. The unit employed by electric utilities is the kilowatt-hour (kWh), which equals 3.6 x 106 }. 1 kWh = (1000 W)(1 ru>

(3��s ) (��) = 3.6

X 106 J

[20.21]

is

---+

880

CHAPTER 20

Electrochemistry Using these considerations, we can calculate the maximum work obtain­ able from the voltaic cells and the minimum work required to bring about de­ sired electrolysis reactions.

Calculating Energy in Kilowatt-hours

I Calculate the number of kilowatt-hours of electricity required to produce 1.0 x 103 kg of aluminum by electrolysis of Al3+ if the applied voltage is 4.50 V. We areforgiven the mass ofAI produced from Al3+ and the applied voltage and asked to calculate the energy, in kilowatt­ hours, required the reduction. AI, we can20.calculate number ofis moles AI and inthencoulombs the number required to obtain4.5that mass. WeFromcanthethenmassuseofEquation 20, w = first thewhere the totalof charge and of coulombs is the applied potential, 0 V. 1000 First, we need to calculate the g Al )( ( )( )( 96,485 C 3 mol 1 mol eAl Coulombs = (1.00 X 103 kg Al) 1 kg AI 27.0 g AI 1 mol AI 1 mole e- ) number of coulombs required: = 1. 0 7 101° C ) We can now calculate w. doing so, we Kilowatt-hours = (1.07 X 101° C)(4.50 ( 1 CJ- )( 3.61 kWh must apply several conversion factors, i n ­ X 10 J cluding Equation kilowatt-hours which givesandthejoules: con­ version between = 1.34 X 104 kWh This quantity of energy does not include the energy used to mine, transport, and aluminum process theorealuminum ore, and to keep the electrolysis bath molten during electrolysis. A typical electrolytic cell used to reduce to aluminum metal is only 40%toefficient, withkg 60% of the electrical energy being dissipated as heat. It therefore requires approximately 33 kWh of electricity produce of aluminum. The aluminum industry consumes about 2% of the electrical energy generated in the United States. Because this energy is used mainly to reduce aluminum, recycling this metal saves large quantities of energy.

- SAMPLE EXERCISE

20.15

SOLUTION

Analyze: Plan:

nFEexu

Solve:

nF

Eext

nF,

X

In

1

V)

20.21,

V

6

Comment:

1

- PRACTICE EXERCISE

Calculate of kilowatt-hours of electricity applied emf11.the0isnumber 5.kWh00 Assume that the process is 100%required efficient.to produce 1.00 kg of Mg from electrolysis of molten MgC12 if the Answer:

V.

- SAMPLE INTEGRATIVE EXERCISE

I Putting Concepts Together

The Ksp at 298 K for iron(II) fluoride is 2.4 X 0- (a) Write a half-reaction that gives the likely products of thereduction two-electron reduction of+ (aq) FeFto2(s)calculate in water.the(b)standard Use the Ksp 2 value and the standard potential of Fe re­ duction potential for the half-reaction in part (a). (c) Rationalize the difference in the 2 reduction potential for the half-reaction in part (a) with that for Fe + (aq). 1

6

SOLUTION

are going to have to combine we know about equilibrium con­ stants andWe electrochemistry to obtain reductionwhatpotentials. For (a)electrons we needandtowrite determine whichreaction ion, Fefor2+ FeF or F-,+is2 more likely to be re­ duced by the overall 2 e- ? For (b) we need to write the K5p reaction and manipulate it to get Eo for the reaction in (a). For (c) we need to see what we get for (a) and (a)predict Iron(II)where fluoridetwois anelectrons ionic substance that consists of .FeWe2+ can't and F-ions. Weadding are asked toelectrons could be added to FeF envision 2 2 to theWeF-therefore ions topredict form Fthe-, sohalf-reaction it seems likely that we could reduce the Fethe2+ ions to Fe(s). FeF2(s) + 2 e- ---> Fe(s) + 2 F-(aq) (b) The K,P value refers to the following equilibrium =(Section 17.4): FeF2(s) � Fe2+(aq) + 2 F-(aq) K,p [Fe2+][F-j2 2.4 X 10-6 2 We also asked the standard reduction andwere standard voltagetoareuselisted in Appendix E: potential of Fe +, whose half-reaction Fe2+(aq) + 2 e- Fe(s) = -0.440 Analyze: Plan:

--->

2

(b).

Solve:

=

--->

£"

=

V

Summary and Key Terms

881

Recallcanthataddaccording to Hess'squantities law, we canlikeaddt.Hreactions to get the one we want and we thermodynamic and t.G to solve for the enthalpy oraddfreetheenergy of the reaction we want. (Section 5.6) this case notice that if we Ksp reaction to the standard reduction half-reaction for Fe 2+, we get the half­ reaction we want: FeF2(s) ----> Fe2+(aq) + 2 F-(aq) 1. 2 2. Fe +(aq) 2 e- ----> Fe(s) Overall: 3. FeF2(s) 2 e ----> Fe(s) + 2 F (aq) Reaction 3 is still a half-reaction, so we do see the free electrons. If we knew D.G0 for reactions 1 and 2, we could add them to get D.G0 for reaction 3. Recallknowthat weforcan relate1;D.G0 to Eo by t.Go � -nFEo and to K by t.Go � -RT ln K. We reaction it is Ksp· We know Eo for reaction 2. Therefore we can calcu­ late t.Go for reactions 1 and 2: Reaction t.Go � -RT ln K � -(8.314 J/K mol)(298 K) ln(2.4 10__,;) � 3.21 104 )/mol Reaction D.G0 -nF£0 -(2 mol)(96,485 Cjmol)(-0.440 J/C) 8.49 X 104 J (RecallThen, thatD.G0 1 voltforisreaction 1 joule3,perthecoulomb. ) is 3.21 X 104 J (for one mole of FeF ) + one we want, 2 4 J 1.17 105 J. We can convert this to Eo easily from the relationship 8.D.G049 � 10-nf£0: 1.17 105 J -(2 mol)(96,485 Cjmol) £" £" � -0.606 J/C � -0.606 The standard reduction potential for FeF2of( -0.FeF606is theislessmorefavorable negativeprocess. than thatWhen for FeFeF2+ is(-0.reduced, 440 wetelling us that the reduction 2 2 only reduce the Fe + ions but also break up the ionic solid. Be­ 2 additionalnotenergy cause than thethisreduction of Fe 2+ must be overcome, the reduction of FeF2 is less favorable In

exx>

+

+

K

1:

2:

X

�

x

�

�

x

�

X

X

�

V.

(c)

V)

V),

CHAPTER REVIEW

SUM MARY AN D KEY TERMS

Introduction and Section 20.1 In this chapter we have focused on electrochemistry, the branch of chem­ istry that relates electricity and chemical reactions. Elec­ trochemistry involves oxidation-reduction reactions, also called redox reactions. These reactions involve a change in the oxidation state of one or more elements. In every oxi­ dation-reduction reaction one substance is oxidized (its oxidation state, or number, increases) and one substance is reduced (its oxidation state, or number, decreases). The substance that is oxidized is referred to as a reducing agent, or reductant, because it causes the reduction of some other substance. Similarly, the substance that is re­ duced is referred to as an oxidizing agent, or oxidant, be­ cause it causes the oxidation of some other substance. Section 20.2 An oxidization-reduction reaction can be balanced by dividing the reaction into two half-reactions, one for oxidation and one for reduction. A half-reaction is a balanced chemical equation that includes electrons. In oxidation half-reactions the electrons are on the product

(right) side of the reaction; we can envision that these electrons are transferred from a substance when it is oxi­ dized. In reduction half-reactions the electrons are on the reactant (left) side of the reaction. Each half-reaction is balanced separately, and the two are brought together with proper coefficients to balance the electrons on each side of the equation. Section 20.3 A voltaic (or galvanic) cell uses a sponta­ neous oxidation-reduction reaction to generate electricity. In a voltaic cell the oxidation and reduction half-reactions often occur in separate compartments. Each compart­ ment has a solid surface called an electrode, where the half-reaction occurs. The electrode where oxidation oc­ curs is called the anode; reduction occurs at the cathode. The electrons released at the anode flow through an ex­ ternal circuit (where they do electrical work) to the cath­ ode. Electrical neutrality in the solution is maintained by the migration of ions between the two compartments through a device such as a salt bridge.

882

CHAPTER 20

Electrochemistry

Section 20.4 A voltaic cell generates an electromotive force (emf) that moves the electrons from the anode to the

cathode through the external circuit. The origin of emf is a difference in the electrical potential energy of the two elec­ trodes in the cell. The emf of a cell is called its cell potential, EcetV and is measured in volts. The cell potential under standard conditions is called the standard emf, or the standard cell potential, and is denoted E �ell · A standard reduction potential, E�ed' can be assigned for an individual half-reaction. This is achieved by compar­ ing the potential of the half-reaction to that of the standard hydrogen electrode (SHE), which is defined to have E�ed = 0 V and is based on the following half-reaction: 2 H+ (aq, 1 M) + 2 e- � H2{g, 1 atm) E�ed = 0 V The standard cell potential of a voltaic cell is the differ­ ence between the standard reduction potentials of the half-reactions that occur at the cathode and the anode: E�ell = E�ed (cathode) - E�ed (anode). The value of E�ell is positive for a voltaic cell. For a reduction half-reaction, E�ed is a measure of the tendency of the reduction to occur; the more positive the value for E�ed' the greater the tendency of the substance to be reduced. Thus, E�ed provides a measure of the oxidiz­ ing strength of a substance. Fluorine (F2) has the most pos­ itive value for E�ed and is the strongest oxidizing agent. Substances that are strong oxidizing agents produce prod­ ucts that are weak reducing agents, and vice versa. Section 20.5 The emf, E, is related to the change in the Gibbs free energy, .:lG: .:lG = -nFE, where n is the num­ ber of electrons transferred during the redox process and F is Faraday's constant, defined as the quantity of electrical charge on one mole of electrons. This amount of charge is 1 faraday (F): 1 F = 96,485 C/mol. Because E is related to .:lG, the sign of E indicates whether a redox process is spontaneous: E > 0 indicates a spontaneous process, and E < 0 indicates a nonspontaneous one. Because .:lG is also related to the equilibrium constant for a reaction (.:lG = - RT In K), we can relate E to K as well. Section 20.6 The emf of a redox reaction varies with temperature and with the concentrations of reactants and products. The Nernst equation relates the emf under nonstandard conditions to the standard emf and the reac­ tion quotient Q: E = Eo - (RTfnF) In Q = P - (0.0592/n) log Q The factor 0.0592 is valid when T = 298 K. A concen­ tration cell is a voltaic cell in which the same half-reaction occurs at both the anode and cathode but with different concentrations of reactants in each compartment.

At equilibrium, Q = K and E = 0. The standard emf is therefore related to the equilibrium constant. Section 20.7 A battery is a self-contained electrochemi­ cal power source that contains one or more voltaic cells. Batteries are based on a variety of different redox reactions. Several common batteries were discussed. The lead-acid battery, the nickel-cadmium battery, the nickel-metal­ hydride battery, and the lithium-ion battery are examples of rechargeable batteries. The common alkaline dry cell is not rechargeable. Fuel cells are voltaic cells that utilize redox reactions in which reactants such as H2 have to be continuously supplied to the cell to generate voltage. Section 20.8 Electrochemical principles help us under­ stand corrosion, undesirable redox reactions in which a metal is attacked by some substance in its environment. The corrosion of iron into rust is caused by the presence of water and oxygen, and it is accelerated by the presence of electrolytes, such as road salt. The protection of a metal by putting it in contact with another metal that more readily undergoes oxidation is called cathodic protection. Galva­ nized iron, for example, is coated with a thin layer of zinc; because zinc is oxidized more readily than iron, the zinc serves as a sacrificial anode in the redox reaction. Section 20.9 An electrolysis reaction, which is carried out in an electrolytic cell, employs an external source of electricity to drive a nonspontaneous electrochemical re­ action. The negative terminal of the external source is connected to the cathode of the cell, and the positive ter­ minal to the anode. The current-carrying medium within an electrolytic cell may be either a molten salt or an elec­ trolyte solution. The products of electrolysis can general­ ly be predicted by comparing the reduction potentials associated with possible oxidation and reduction processes. The electrodes in an electrolytic cell can be ac­ tive, meaning that the electrode can be involved in the electrolysis reaction. Active electrodes are important in electroplating and in metallurgical processes. The quantity of substances formed during electrolysis can be calculated by considering the number of electrons involved in the redox reaction and the amount of electrical charge that passes into the cell. The maximum amount of electrical work produced by a voltaic cell is given by the product of the total charge delivered, nF, and the emf, E: Wmax = -nFE. The work performed in an electrolysis is given by w = nFEexv where Eext is the applied external po­ tential. The watt is a unit of power: 1 W = 1 J/s. Electrical work is often measured in kilowatt-hours.

KEY S KILLS • • • •

Identify oxidation, reduction, oxidizing agent, and reducing agent in a chemical equation. Complete and balance redox equations using the method of half-reactions. Sketch a voltaic cell and identify its cathode, anode, and the directions that electrons and ions move. Calculate standard emfs (cell potentials), E�ell , from standard reduction potentials.

Visualizing Concepts

883

Use reduction potentials to predict whether a redox reaction is spontaneous. Relate E�ell to 6.G0 and equilibrium constants. • Calculate emf under nonstandard conditions. • Describe the reactions in electrolytic cells. • Relate amounts of products and reactants in redox reactions to electrical charge.

• •

KEY EQUATIONS •

E�ell = E�d (cathode) - E�ed (anode)

•

[20.11) -nFE 0·0592 V log Q (at 298 K) E=Pn

•

[20.8)

Relating standard emf to standard reduction potentials of the reduction (cathode) and oxidation (anode) half reactions Relating free energy change and emf

[20.16)

The Nernst equation, expressing the effect of concentration on cell potential

llG =

VISUALIZING CONCEPTS 20.1

20.2

Inacid-base the Brensted-Lowry conceptas ofproton-transfer acids and bases, reactions are viewed reac­ tions. The stronger the acid, the weaker is its conj u gate base. In what ways are redox reactions analogous? [Sections 20.1 and 20.2] Consider the reaction insolution Figurecontained 20.3. Describe what would happen if the cadmium(II) sulfate and theandmetal the solution silver nitrate the was metalzinc, was(b)copper. [Sectioncontained 20.3] The diagram below represents a molecular view process occurring at an electrode in a voltaic cell. of a

(a)

(d)

20.5

(a)

20.3

What additionsemf?must(b)you makeelectrode to the cell for it toasgen­ erate a standard Which functions the cathode? (c) Which direction do electrons move through theunder externalstandard circuit?conditions? What vol[Sections tage will20.the3 andcell20.gener­ ate 4] Where on Figure 20. 1 4 would you fi n d the chemical species that (b) the20.chemical species that isis the theeasiest easiesttotooxidize, reduce?and [Section 4] For the generic reaction A(aq) + B(aq) ----> A-(aq) + B+(aq) for which Eo is a positive number, answer theisfollowi ng questions: What being oxidi zed, and what is being reduced? (b) If you made a voltaic cell out of this reaction, what half-reaction wouldwould be occurring at theatcathode, and what half-reaction be occurring the anode? (c) Which half-reaction from (b) is higher in potential energy?is the sign of the free energy change for there­ What action? [Sections 20.4 and 20.5] Consider thethehalflinesreaction Ag+(aq) diagram + e- ----> Ag(s). Which of in the following indicates how the reduction potential varies with the concen­ tration of Ag+? (b) What is the value of E,ed when log[Ag+] = 0? [Section 20.6]

20.6

(a)

(a)

(d)

Does theelectrode processtherepresent oxidation or(c)reduction? Is the anode or cathode? Why are the atoms in the electrode represented by larger spheres than the ions in the solution? [Section 20.3] Assume that you want to construct a voltaic cell that uses the following half reactions: A2+(aq) + 2 e- ----> A(s) E�ed -0.10 V B2+(aq) + 2 e- ----> B(s) E�d -1.10 V You with the incomplete whichbegin the electrodes are immersedcellinpictured water. below, in (a) (b)

20.4

20.7

(a)

=

=

,-----'�}---.. Voltmeter I

I

20.8

20.9 20.10

log[Ag+] Draw a generic picture of aafuel cell. What is theofmaithen difference between it and battery, regardless redox reactions that occur inside? [Section 20.7] How does oxidation? a zinc coatingSection on iron unwanted 20.8protect ] the iron from [ You may haveBased heardonthatwhat"antioxidants" are good for your health. you have learned in this chapter, [Sections what 20.1 anddo20.you2] deduce an "antioxidant" is?

884

CHAPTER 20

Electrochemistry

EXERCISES Oxidation-Reduction Reactions 20.11 (a) What is meant by the term oxidation? (b) On which side of an oxidation half-reaction do the electrons ap­ pear? (c) What is meant by the term oxidant? (d) What is meant by the term oxidizing agent? 20.12 (a) What is meant by the term reduction? (b) On which side of a reduction half-reaction do the electrons ap­ pear? (c) What is meant by the term reductant? (d) What is meant by the term reducing agent? 20.13 Indicate whether each of the following statements is true or false: (a) If something is oxidized, it is formally losing electrons. 3 (b) For the reaction Fe +(aq) + Co2+(aq) ---. 3 3 Fe 2+(aq) + Co +(aq), Fe +(aq) is the reducing agent and Co2+(aq) is the oxidizing agent. (c) If there are no changes in the oxidation state of the reactants or products of a particular reaction, that reaction is not a redox reaction. 20.14 Indicate whether each of the following statements is true or false: (a) If something is reduced, it is formally losing electrons. (b) A reducing agent gets oxidized as it reacts. (c) Oxidizing agents can convert CO into C02 .

20.15 In each of the following balanced oxidation-reduction equations, identify those elements that undergo changes in oxidation number and indicate the magnitude of the change in each case. (a) I 205(s) + 5 CO(g) ---. I 2 (s) + 5 C0 2(g) (b) 2 Hg2+(aq) + N2H 4(aq) ---. 2 Hg(I) + N2(g) + 4 H+(aq) (c) 3 H 2S(aq) + 2 H+(aq) + 2 N03-(aq) ---. 3 S(s) + 2 NO(g) + 4 HzO{l) (d) Ba2+(aq) + 2 OH -(aq) + HzOz(aq)

+ 2 Cl02(aq) ---. Ba(C!Oz)2(s) + 2 HzO(I) + 02(g)

20.16 Indicate whether the following balanced equations in­ volve oxidation-reduction. If they do, identify the ele­ ments that undergo changes in oxidation number.

(a) PBr3(I) + 3 HzQ(I) ---. H 3P0 3 (aq) + 3 HBr(aq) (b) Nai(aq) + 3 HOCI(aq) ---. Nai03(aq) + 3 HCI(aq) (c) 3 S02 (g) + 2 HN03(aq) + 2 H 20{l) ---. 3 H 2504(aq) + 2 NO(g) (d) 2 H 2504(aq) + 2 NaBr(s) ---. Brz{l) + SOz(g) + Na2S04(aq) + 2 HzO{l)

Balancing Oxidation-Reduction Reactions 20.17 At 900 oc titanium tetrachloride vapor reacts with molten magnesium metal to form solid titanium metal and molten magnesium chloride. (a) Write a balanced equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance is the reductant, and which is the oxidant? 20.18 Hydrazine (N2H4) and dinitrogen tetroxide (N204) form a self-igniting mixture that has been used as a rocket propellant. The reaction products are N2 and H20. (a) Write a balanced chemical equation for this reaction. (b) What is being oxidized, and what is being reduced? (c) Which substance serves as the reducing agent, and which as the oxidizing agent? 20.19 Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxida­ tion or a reduction. (a) Sn2+(aq) ---. Sn4\aq) (acidic or basic solution) (b) Ti0 2(s) ---. Ti 2+(aq) (acidic solution) (c) Cl03-(aq) ---. Cqaq) (acidic solution) (d) N2(g) ---. NH 4+(aq) (acidic solution) (e) OH-(aq) ---. 0 2(g) (basic solution) (f) 50 32-(aq) ---- soi-(aq) (basic solution) (g) N2(g) ---. NH3(g) (basic solution) 20.20 Complete and balance the following half-reactions. In each case indicate whether the half-reaction is an oxida­ tion or a reduction. 3 (a) Mo +(aq) ---. Mo(s) (acidic or basic solution) (b) H 2S03(aq) ---. SOi-(aq) (acidic solution)

(c) N03-(aq) --- . NO(g) (acidic solution) (d) (e) (f) (g)

0 2(g) ---. HzO(I) (acidic solution) Mn2+(aq) ---. Mn02(s) (basic solution) Cr(OH)J(s) ---. Croi-(aq) (basic solution) 0 2(g) ---. HzO(I) (basic solution)

20.21 Complete and balance the following equations, and identify the oxidizing and reducing agents: 3 (a) Crpl-(aq) + r-(aq) ---. Cr +(aq) + I03-(aq) (acidic solution) (b) Mn04-(aq) + CH30H(aq) ---. Mn2+(aq) + HC02H(aq) (acidic solution) (c) lz{s) + OCqaq) ---. I0 3-(aq) + Cqaq) (acidic solution) (d) As203(s) + N03-(aq) ---. H 3As04(aq) + N20 3(aq) (acidic solution) (e) Mn0 4-(aq) + Br-(aq) ---. Mn02(s) + Br03-(aq) (basic solution) (f) Pb(OH)4 2-(aq) + CIO-(aq) ---. PbOz(s) + cqaq) (basic solution) 20.22 Complete and balance the following equations, and identify the oxidizing and reducing agents. Recall that the 0 atoms in hydrogen peroxide, HzOz, have an atyp­ ical oxidation state. = (Table 2.5) (a) N0 2-(aq) + Cr20l- (aq) ---. 3 Cr + (aq) + N03-(aq) (acidic solution) (aq) HN0 (b) S(s) + ---. H 2S0 3(aq) + NzO(g) 3 (acidic solution)

885

Exercises

N02-(aq) Al(s) -----> NH4+(aq) Al02-(aq) (basic solution) (f) H 202(aq) Cl02(aq) -----> Cl02-(aq) 02(g) (basic solution)

Cr20l-(aq) CH30H(aq) -----> HC02H(aq) + Cr3+(aq) 2 (acidic solution) (d) Mn0 4-(aq) Cr(aq) -----> Mn +(aq) Cl2(aq) (acidic solution) +

(c)

+

+

+

+

+

(e)

+

Voltaic Cells

What20.are3 andthe Figure similarities and(b) differences between Figure 20. 4 ? Why are Na+ ions drawn into the cathode compartment as the voltaic cell shown in Figure 20.5 operates? (a) What is the role of the porous glass disc shown in the Figure 20.4? (b) WhyasdotheN0 3- ionscell migrate anode compartment voltaic shown inintoFigure 20.5 operates? Astructed. voltaic One cell similar to that shown in consists Figure 20.of5aissilver con­ electrode compartment stripironplaced inplaced a solution of AgN0of3, FeCl2. and theTheother has ancell strip in a solution overall reaction is Fe(s) 2 Ag+(aq) -----> Fe2+(aq) 2 Ag(s) (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two elec­ trode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the

20.23 (a)

20.24

20.25

(e)

20.26

----->

+

+

+

electrodes. electronsorflow elec­ trode to the ironDoelectrode, from from the ironthetosilver the silver? (f) In which directions do the cations and anions mi­ grate through the solution? Astructed. voltaic One cell similar to that shown in Figure 20.of5 anis con­ electrode compartment consists alu­ minum strip placed in a solution of Al(N0 )J, and the. 3 other has a nickel strip placed i n a solution of NiS0 4 The overall cell reaction is 2 2 Al(s) 3 Ni +(aq) 2 Al3+(aq) 3 Ni(s) (a) What is being oxidized, and what is being reduced? (b) Write the half-reactions that occur in the two elec­ trode compartments. (c) Which electrode is the anode, and which is the cathode? (d) Indicate the signs of the electrodes. Do electrons flow orfromfromthethealuminum electrode to the nickel electrode, nickel to the aluminum? (f) In which directions do the cations migrate AIandis anions not coated withthrough its oxide.the solution? Assume the +

(e)

Cell EMF under Standard Conditions

Whatis does the termof theelectromotive force mean? What the definition volt? (c) What does the term cell potential mean? (a) Which electrode of a voltaic cell, the cathode or the anode, corresponds to are thethehigher potential energypoten­ for the electrons? (b) What units for electrical tial? How(c) What does isthisspecial unit about relate atostandard energycellexpressed joules? potential?in (a) Write the half-reaction that occurs at a hydrogen elec­ trode in acidic aqueouscell.solution whenis itstandard serves asabout the cathode of a voltaic (b) What (c) What the role ofthethestandard platinumhydrogen foil in aelectrode? standard hydrogen electrode? (a) Write the half-reaction that occurs at a hydrogen electrode inofacidic aqueous solution when it serves as the anode a voltaic cell. (b) The platinum electrode standard hydrogen electrode is specially prepared to(c)in aSketch have aalarge surface area. Why is this important? standard hydrogen electrode. (a) What is a standard reduction potential? (b) What is the standard reduction potential of a standard hydrogen electrode? (a Why is it impossible to measure the standard reduc­ tion potential reduction of a single potential half-reaction? (b) Describe how the standard of a half-reaction can be determined. A voltaic cell that uses the reaction Tl3+(aq) 2 Cr +(aq) -----> Tl+(aq) 2 Cr3+(aq) has a measured standard cell potential of 1.19 V.

20.27 (a)

20.28

20.29

is

20.30

20.31

20.32

20.33

20.34

+

2

+

+

2

+

+

+

E,

(c)

20.35

+

+

+

+

20.36

+

E,

----->

+

+

+

+

)

+

(a) Write the two half-cell reactions. (b) By using data from Appendix determine E�ed for the reduction of Tl3+(aq) tocathode, Tl+(aq)and indicate Sketch the anode and thevoltaic directioncelofl, label electrontheflow. A voltaic cell that uses the reaction PdC142-(aq) Cd(s) -----> Pd(s) 4 Cr(aq) Cd +(aq has a measured standard cell potential of 1.03 V. (a) Write the two half-cell reactions. (b) By using data from Appendix determine E�cd for the reaction in­ volving Pd. Sketch the voltaic cell, label the anode and cathode, and indicate the direction of electron flow. Using standard reduction potentials (Appendi x E), calcu­ late the standard emf for each of the following reactions: (a) Cl2(g) 2 qaq) -----> 2 cqaq) 12(s) (b) Ni( s) + 2 Ce4+(aq) -----> Ni 2+(aq) + 2 Ce3+(aq) (c) Fe(s) 2 Fe3+(aq) -----> 3 Fe2+(aq) (d) 2 Al3+(aq) 3 Ca(s) -----> 2 Al(s ) 3 Ca2+(aq) Usingeachdataof thein Appendix calculate the standard emf for following reactions: 2 H+(aq) 2 F-(aq) (a) H2(g) F2(g) (b) Cu 2+(aq) + Ca(s) -----> Cu(s) Ca2+(aq) 2 (c) 3 Fe +(aq) -----> Fe(s) 2 Fe3+(aq) (d) Hgl+(aq) 2 Cu+(aq) -----> 2 Hg(l) 2 Cu2+(aq) The standard reduction potentials of the following half­ reactions are given in Appendix Ag+(aq) + e- -----> Ag(s) Cu22+(aq) 2 e- -----> Cu(s) Ni +(aq) + 2 e- -----> Ni(s) Cr3+(aq) 3 e- -----> Cr(s) E, . (c)

(b)

+

20.37

+

E:

+

+

)

886

CHAPTER 20

Electrochemistry

(a) Determine which combination of these half-cell reac­ tions leads to the cell reaction with the largest positive cell emf, and calculate the value. (b) Determine which combi­ nation of these half-cell reactions leads to the cell reaction with the smallest positive cell emf, and calculate the value. 20.38 Given the following half-reactions and associated stan­ dard reduction potentials: 4 Br-(aq) AuBr4-(aq) e- -----> Au(s)

+3 + +2

3 Eu +(aq) IO-(aq)

+

H 20(/)

+

=

-0.858 E�ed = -0.43 +2 E�ed +0.49 E�ed -0. 1 4 E�ed

e- -----> Eu2+(aq)

V

V

e- -----> r-(aq)

OH-(aq) =

V

=

V

(a) Write the cell reaction for the combination of these

half-cell reactions that leads to the largest positive cell emf, and calculate the value. (b) Write the cell reaction for the combination of half-cell reactions that leads to the smallest positive cell emf, and calculate that value.

1M

20.39 A solution of Cu(N03)z is placed in a beaker with a strip of Cu metal. A 1 M solution of SnS04 is placed in a second beaker with a strip of Sn metal. A salt bridge connects the two beakers, and wires to a voltmeter link the two metal electrodes. (a) Which electrode serves as the anode, and which as the cathode? (b) Which elec­ trode gains mass and which loses mass as the cell reac­ tion proceeds? (c) Write the equation for the overall cell reaction. (d) What is the emf generated by the cell under standard conditions? 20.40 A voltaic cell consists of a strip of cadmium metal in a solution of Cd(N03)z in one beaker, and in the other beaker a platinum electrode is immersed in a NaCJ solu­ tion, with Cl2 gas bubbled around the electrode. A salt bridge connects the two beakers. (a) Which electrode serves as the anode, and which as the cathode? (b) Does the Cd electrode gain or lose mass as the cell reaction proceeds? (c) Write the equation for the overall cell reac­ tion. (d) What is the emf generated by the cell under standard conditions?

Strengths of Oxidizing and Reducing Agents 20.41 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger re­ ducing agent: (a) Fe(s) or Mg(s) (b) Ca(s) or Al(s) (c) H2(g, acidic solution) or H2S(g) (d) H2S03(aq) or H2C204(aq) 20.42 From each of the following pairs of substances, use data in Appendix E to choose the one that is the stronger oxi­ dizing agent: (a) Clz(g) or Brz(l) (b) zn2+(aq) or Cd 2+(aq) (c) Br03-(aq) or I03 -(aq) (d) HzOz(aq) or 03(g) 20.43 By using the data in Appendix E, determine whether each of the following substances is likely to serve as an oxidant or a reductant: (a) Cl2(g), (b) Mn0 4-(aq, acidic solution), (c) Ba(s), (d) Zn(s). 20.44 Is each of the following substances likely to serve as an 3 oxidant or a reductant: (a) Ce +(aq), (b) Ca(s), (c) CJ03 -(aq), (d) N205(g)?

20.45 (a) Assuming standard conditions, arrange the following in order of increasing strength as oxidizing agents in acidic solution: Cr20l-, H20z, Cu2+, Clz, 02. (b) Arrange the following in order of increasing strength as reducing agents in acidic solution: Zn, r-, Sn2+, H20z, Al. 20.46 Based on the data in Appendix E, (a) which of the follow­ ing is the strongest oxidizing agent and which is the weak­ 4 est in acidic solution: Ce +, Brz, H20z, Zn? (b) Which of the following is the strongest reducing agent, and which is the weakest in acidic solution: F -, Zn, N2H5+, Iz, NO? 20.47 The standard reduction potential for the reduction of 3 Eu +(aq) to Eu2+(aq) is -0.43 V. Using Appendix E, which of the following substances is capable of reducing 3 Eu +(aq) to Eu2+(aq) under standard conditions: AI, Co, HzOz, NzHs +, HzCz04? 20.48 The standard reduction potential for the reduction of Ru04-(aq) to Ru042-(aq) is V. By using Appendix E, which of the following substances can oxidize Ruol-(aq) to Ru04-(aq) under standard conditions: 2 2 Brz(l), Br03-(aq), Mn +(aq), Oz(g), Sn +(aq)?

+0.59

Free Energy and Redox Reactions 20.49 Given the following reduction half-reactions: 3 Fe +(aq) e- -----> Fe 2+(aq)

+

5206 2-(aq) N20(g)

+4 + +

VOz+(aq)

H+(aq)

2 H+(aq)

+ +

2 H+(aq)

=

+0.77 E�ed +0. 6 0 + E�ed -1. 7 7 + E�ed +1. 0 0 E�ed

2 e - -----> 2 H 2S03(aq) 2 e - -----> N2(g)

+

V

=

V

=

V

H 20(/)

e - -----> V0 2+(aq)

=

HzO(l) V

(a) Write balanced chemical equations for the oxidation of Fe2+(aq) by Sz06 2-(aq), by NzO(aq), and by VOz+(aq). (b) Calculate 6.G0 for each reaction at K. (c) Calculate the equilibrium constant K for each reaction at

298

298 K.

20.50 For each of the following reactions, write a balanced equation, calculate the standard emf, calculate 6-Go at and calculate the equilibrium constant K at (a) Aqueous iodide ion is oxidized to I2(s) by Hg 22+(aq). (b) In acidic solution, copper(!) ion is oxidized to cop­ per(II) ion by nitrate ion. (c) In basic solution, Cr(OH)J(s) is oxidized to Crol-(aq) by ClO-(aq).

298 K,

298 K.

887

Exercises 20.51 If the equilibrium constant for a two-electron redox re­ action at K is x calculate the correspond­ ing .lG' and E�ell·

1.5 10-4,

298

20.52 If the equilibrium constant for a one-electron redox reac­ tion at K is X calculate the corresponding .lG' and E�ell·

8.7 104,

298

20.53 Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at K: (a) Fe(s) Ni 2+(aq) ---> Fe 2+(aq) + Ni(s) 2 H2(g) (b) Co(s) 2H+(aq) ---> Co +(aq) H+(aq) ---> Br-(aq) Mn0 4-(aq) (c) 2 Br2(1) H20(1) 2 Mn +(aq)

10

298

+ +

+ + 16 +8

+2

+5

20.54 Using the standard reduction potentials listed in Appendix E, calculate the equilibrium constant for each of the following reactions at K: Ag(s) (a) Cu(s) 2 Ag+(aq) ---> Cu2+(aq) 4 (b) Ce +(aq) Bi(s) H20(1) ---> 3 H+(aq) BiO+(aq) 3 Ce +(aq) 3 (c) N2Hs+(aq) + 4 Fe(CN)6 -(aq) ---> 4 4 Fe(CN)6 -(aq) 5 H+(aq) N2(g)

+

3

298

+

+

+2

+2

+

+

+

+0.177 298 3?

20.55 A cell has a standard emf of V at K. What is the value of the equilibrium constant for the cell reac­ = = = tion (a) if n 1? (b) if n 2? (c) if n

298

+0.17

20.56 At K a cell reaction has a standard emf of V. The equilibrium constant for the cell reaction is X l OS. What is the value of n for the cell reaction?

5.5

Cell EMF under Nonstandard Conditions 20.57 (a) Under what circumstances is the Nernst equation ap­ plicable? (b) What is the numerical value of the reaction quotient, Q, under standard conditions? (c) What hap­ pens to the emf of a cell if the concentrations of the reac­ tants are increased? 20.58 (a) A voltaic cell is constructed with all reactants and products in their standard states. Will this condition hold as the cell operates? Explain. (b) Can the Nernst equation be used at temperatures other than room tem­ perature? Explain. (c) What happens to the emf of a cell if the concentrations of the products are increased? 20.59 What is the effect on the emf of the cell shown in Figure which has the overall reaction Zn(s) H+(aq) ---> Zn2+(aq) H2(g), for each of the following changes? (a) The pressure of the H2 gas is increased in the cathode compartment. (b) Zinc nitrate is added to the anode compartment. (c) Sodium hydrox­ ide is added to the cathode compartment, decreasing [H1. (d) The surface area of the anode is doubled.

20.11, +2

+

20.60 A voltaic cell utilizes the following reaction: 3 Al(s) Ag+(aq) ---> Al +(aq) Ag(s) What is the effect on the cell emf of each of the following changes? (a) Water is added to the anode compartment, diluting the solution. (b) The size of the aluminum elec­ trode is increased. (c) A solution of AgN03 is added to the cathode compartment, increasing the quantity of Ag+ but not changing its concentration. (d) HCI is added to the AgN03 solution, precipitating some of the Ag+ as AgCl.

+3

+3

20.61 A voltaic cell is constructed that uses the following reac­ tion and operates at K: 2 2 Ni(s) Zn(s) Ni +(aq) ---> Zn +(aq)

+

298

+

(a) What is the emf of this cell under standard conditions?

(b) What is the emf of this cell when [Ni2+] = and [Zn2+] = (c) What is the emf of the cell when 2+] = 2 and [Zn 1 = [Ni

M? 0.0.210000 M, 298 3 +

3.00 M

0.900 M?

20.62 A voltaic cell utilizes the following reaction and oper­ ates at K: 3 3 Cr +(aq) Ce 4+(aq) Cr(s) ---> Ce +(aq)

3

+

(a) What is the emf of this cell under standard conditions?

4 (b) What is the emf of this cell when [Ce +] = 3 3 [Ce +] = and [Cr +] = (c) What is the emf of the cell when [Ce 41 = [Ce3+] = 3 and [Cr +] =

0.10 M, 2.5 M?

3.0 M, 1.75 M,

0.0.10010M,M?

20.63 A voltaic cell utilizes the following reaction: 3 4 Fe 2+(aq) H 20(1) 02(g) 4 H+(aq) ---> 4 Fe +(aq) (a) What is the emf of this cell under standard conditions? (b) What is the emf of this cell when [Fe 21 = 3 Po, = atm, and the pH of the so­ [Fe +] = lution in the cathode compartment is 3.50?

+

+2

+

0.010 M,

1.3 M,

0.50

20.64 A voltaic cell utilizes the following reaction: 3 2 2 H+(aq) H2(g) ---> Fe +(aq) 2 Fe +(aq) (a) What is the emf of this cell under standard conditions? (b) What is the emf for this cell when [Fe 3+] = %, = atm, [Fe 2+] = and the pH in both compartments is

2

+

0.85

+

2.50 M,

5.00? 0.0010 M,

20.65 A voltaic cell is constructed with two zn2+-Zn electrodes. The two cell compartments have [Zn2+] = and [Zn2+] = x respectively. (a) Which elec­ trode is the anode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Zn2+] will increase, decrease, or stay the same as the cell operates.

1.8 M

1.00 10-2 M,

20.66 A voltaic cell is constructed with two silver-silver chlo­ ride electrodes, each of which is based on the following half-reaction:

+

+

AgCl(s) e- ---> Ag(s) cqaq) The two cell compartments have [Cn = and [Cll = respectively. (a) Which electrode is the cathode of the cell? (b) What is the standard emf of the cell? (c) What is the cell emf for the concentrations given? (d) For each electrode, predict whether [Cll will increase, decrease, or stay the same as the cell operates.

2.55 M,

0.0150 M

20.11

20.67 The cell in Figure could be used to provide a mea­ sure of the pH in the cathode compartment. Calculate the pH of the cathode compartment solution if the cell emf at K is measured to be V when [Zn2+] = and %, = atm.

2980.30 M

0.90

+0.684

888

CHAPTER 20

Electrochemistry

20.68 A voltaic cell is constructed that is based on the follow­ ing reaction:

Sn2+ (aq)

+

Pb(s) --> Sn(s)

+ Pb 2+(aq)

2 (a) If the concentration of Sn + in the cathode com­ partment is M and the cell generates an emf of what is the concentration of Pb 2+ in the anode compartment? (b) If the anode compartment contains [SO/-) = M in equilibrium with PbS04(s), what is the K,P of PbS04?

1.00

+0.22 V,

1.00

Batteries and Fuel Cells 20.69 (a) What happens to the emf of a battery as it is used? Why does this happen? (b) The AA-size and D-size al­ kaline batteries are both batteries that are based on the same electrode reactions. What is the major differ­ ence between the two batteries? What performance fea­ ture is most affected by this difference?

20.74 Mercuric oxide dry-cell batteries are often used where a high energy density is required, such as in watches and cameras. The two half-cell reactions that occur in the battery are

20.70 (a) Suggest an explanation for why liquid water is needed in an alkaline battery. (b) What is the advantage of using highly concentrated or solid reactants in a voltaic cell?

(a) Write the overall cell reaction. The value of E�ed for the cathode reaction is The overall cell po­ tential is Assuming that both half-cells operate under standard conditions, what is the standard reduc­ tion potential for the anode reaction? (c) Why is the po­ tential of the anode reaction different than would be expected if the reaction occurred in an acidic medium?

1.5-V

402

20.71 During a period of discharge of a lead-acid battery, g of Pb from the anode is converted into PbS04(s). What mass of Pb02(s) is reduced at the cathode during this same period?

20.72 During the discharge of an alkaline battery, g of Zn are consumed at the anode of the battery. What mass of Mn02 is reduced at the cathode during this discharge?

4.50

20.73 Heart pacemakers are often powered by lithium-silver chromate "button" batteries. The overall cell reaction is:

2 Li(s)

+

Ag 2Cr04(s) --> Li 2Cr04(s)

+2

Ag(s)

(a) Lithium metal is the reactant at one of the electrodes of the battery. Is it the anode or the cathode? (b) Choose

the two half-reactions from Appendix E that most closely approximate the reactions that occur in the battery. What standard emf would be generated by a voltaic cell based on these half-reactions? (c) The battery generates an emf of How close is this value to the one calculated in part (b)? (d) Calculate the emf that would be generat­ ed at body temperature, °C. How does this compare to the emf you calculated in part (b)?

+3.5 V.

37

HgO(s)

+

+ 2 e- --> Hg(l) + 2 0H-(aq) + 2 OH-(aq) --> + +2 (b) +0.098 V. + 1.35 V. Hz0(1)

Zn(s)

ZnO(s)

Hz0(1)

e­

20.75 (a) Suppose that an alkaline battery was manufactured using cadmium metal rather than zinc. What effect would this have on the cell emf? (b) What environmen­ tal advantage is provided by the use of nickel-metal­ hydride batteries over the nickel-cadmium batteries? 20.76 (a) The nonrechargeable lithium batteries used for ph­ tography use lithium metal as the anode. What advan­ tages might be realized by using lithium rather than zinc, cadmium, lead, or nickel? (b) The rechargeable lithium­ ion battery does not use lithium metal as an electrode ma­ terial. Nevertheless, it still has a substantial advantage over nickel-based batteries. Suggest an explanation. 20.77 The hydrogen-oxygen fuel cell has a standard emf of What advantages and disadvantages are there to using this device as a source of power, compared to a alkaline battery? 20.78 (a) What is the difference between a battery and a fuel cell? (b) Can the "fuel" of a fuel cell be a solid? Explain.

1.23 V. 1.55-V

Corrosion 20.79 (a) Write the anode and cathode reactions that cause the corrosion of iron metal to aqueous iron(II). (b) Write the balanced half-reactions involved in the air oxidation of 2 Fe + (aq) to Fez03 • H20.

3

20.80 (a) Based on standard reduction potentials, would you expect copper metal to oxidize under standard condi­ tions in the presence of oxygen and hydrogen ions? (b) When the Statue of Liberty was refurbished, Teflon spacers were placed between the iron skeleton and the copper metal on the surface of the statue. What role do these spacers play? 20.81 (a) Magnesium metal is used as a sacrificial anode to protect underground pipes from corrosion. Why is the magnesium referred to as a "sacrificial anode"? (b) Looking in Appendix E, suggest what metal the un­ derground pipes could be made from in order for mag­ nesium to be successful as a sacrificial anode.

20.82 An iron object is plated with a coating of cobalt to pro­ tect against corrosion. Does the cobalt protect iron by cathodic protection? Explain. 20.83 A plumber's handbook states that you should not con­ nect a brass pipe directly to a galvanized steel pipe be­ cause electrochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. Brass is a mixture of copper and zinc. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations. 20.84 A plumber's handbook states that you should not con­ nect a copper pipe directly to a steel pipe because elec­ trochemical reactions between the two metals will cause corrosion. The handbook recommends you use, instead, an insulating fitting to connect them. What spontaneous redox reaction(s) might cause the corrosion? Justify your answer with standard emf calculations.

Additional Exercises

889

Electrolysis; Electrical Wo rk 20.85 (a) What is electrolysis? (b) Are electrolysis reactions ther­

modynamically spontaneous? Explain. (c) What process occurs at the anode in the electrolysis of molten NaCI? 20.86 (a) What is an electrolytic cell? (b) The negative terminal of a voltage source is connected to an electrode of an electrolytic cell. Is the electrode the anode or the cathode of the cell? Explain. (c) The electrolysis of water is often done with a small amount of sulfuric acid added to the water. What is the role of the sulfuric acid?

20.87 (a) A Cr3+(aq) solution is electrolyzed, using a current of

7.(b)60What A. What mass of Cr(s) is plated out after 2.00 days? amperage is required to plate out 0.250 mol Cr from a Cr3+ solution in a period of 8.00 h?

20.88 Metallic magnesium can be made by the electrolysis of molten MgClz. (a) What mass of Mg is formed by pass­

4.55

3.50

ing a current of A through molten MgCl:u for days? (b) How many minutes are needed to plate out g Mg from molten MgC12, using A of current?

10.00

3.50

20.89 A voltaic cell is based on the reaction

Sn(s )

+ I2(s)

----+

Sn2+(aq)

+ 2 qaq)

Under standard conditions, what is the maximum elec­ trical work, in joules, that the cell can accomplish if g of Sn is consumed? 20.90 Consider the voltaic cell illustrated in Figure which is based on the cell reaction Zn(s) Cu2+(aq) ----+ Zn2+(aq) Cu(s) Under standard conditions, what is the maximum elec­ trical work, in joules, that the cell can accomplish if g of copper is plated out?

75.0

20.5,

+

+

50.0

20.91 (a) Calculate the mass of Li formed by electrolysis of

7.5 10

4 molten LiCI by a current of X A flowing for a pe­ riod of h. Assume the electrolytic cell is efficient. (b) What is the energy requirement for this electrolysis per mole of Li formed if the applied emf is V? 20.92 Elemental calcium is produced by the electrolysis of molten CaC12 . (a) What mass of calcium can be pro­ duced by this process if a current of x A is ap­ plied for h? Assume that the electrolytic cell is efficient. (b) What is the total energy requirement for this electrolysis if the applied emf is V?

24

48

85% +7.5 7.5 103 68% +5.00

ADDITI ONAL EXERCI SES

disproportionation reaction is an oxidation-reduction reaction in which the same substance is oxidized and reduced. Complete and balance the following dispro­ portionation reactions: (a) Ni+(aq) ----> Ni 2+(aq) Ni(s) (acidic solution) (b) Mno/-(aq) Mn04 -(aq) Mn0 2(s) (acidic solution)

20.93 A

+

----+

+

(acidic solution)

(d) CI2(aq) ----> cqaq)

+ CIO-(aq) (basic solution)

20.94 This oxidation-reduction reaction in acidic solution is

spontaneous: Fe2+(aq) Mn04- (aq)

5

+

+ 8 H+(aq) --2 --> + Mn +(aq) + 4 H20(1)

5 Fe 3+(aq)

A solution containing KMn04 and H2S04 is poured into one beaker, and a solution of FeS04 is poured into anoth­ er. A salt bridge is used to join the beakers. A platinum foil is placed in each solution, and a wire that passes through a voltmeter connects the two solutions. (a) Sketch the cell, indicating the anode and the cathode, the direction of electron movement through the external circuit, and the direction of ion migrations through the solutions. (b) Sketch the process that occurs at the atomic level at the surface of the anode. (c) Calculate the emf of the cell under standard conditions. (d) Calculate the emf of the cell at K when the concentrations are the following: pH = [Fe2"] = [Mn04l = 3 [Fe +] X [Mn2"]

298 =2.0.05, 10-4 M, 0.10 M,= 0.001 M.

1.50 M,

20.95 A common shorthand way to represent a voltaic cell is

to list its components as follows: anodelanode solutionllcathode solutionlcathode

A double vertical line represents a salt bridge or a porous barrier. A single vertical line represents a change in phase, such as from solid to solution. (a) Write the half-reactions and overall cell reaction represented by FeiFe 2+IIAg+lAg; sketch the cell. (b) Write the half-reactions and overall cell reaction represented by ZniZn2+IIH +IHz; sketch the cell. (c) Using the notation just described, represent a cell based on the following reaction: Cl0 3-(aq) Cu(s) H+(aq) ----> HzO(I) Cu2+(aq) cqaq) Pt is used as an inert electrode in contact with the Cl03and Cl-. Sketch the cell. 20.96 Predict whether the following reactions will be sponta­ neous in acidic solution under standard conditions: (a) oxidation of Sn to Sn2+ by 12 (to form n, (b) reduc­ tion of Ni2+ to Ni by ,- (to form 12), (c) reduction of Ce4+ to Ce 3+ by H202, (d) reduction of Cu 2+ to Cu by Sn2+ (to form Sn4+). [20.97] Gold exists in two common positive oxidation states, and The standard reduction potentials for these oxi­ dation states are Au+(aq) e- ----> Au(s) E�ed V Au3+ (aq) e- ----> Au(s) E�ed = V (a) Can you use these data to explain why gold does not tarnish in the air? (b) Suggest several substances that should be strong enough oxidizing agents to oxidize gold metal. (c) Miners obtain gold by soaking gold­ containing ores in an aqueous solution of sodium cyanide. A very soluble complex ion of gold forms in the aqueous solution because of the redox reaction Au(s) NaCN(aq) H 20(1) Oz(g) ----> Na [Au(CN)z](aq) + NaOH(aq)

+3

+6

+3

+3

+1

+3.

+ +3

4

+8

= +1.69 +1.50

4

+2

+

4

890

CHAPTER 20

Electrochemistry

What is being oxidized, and what is being reduced, in this reaction? (d) Gold miners then react the basic aqueous product solution from part (c) with Zn dust to get gold metal. Write a balanced redox reaction for this process. What is being oxidized, and what is being reduced? 20.98 Two important characteristics of voltaic cells are their cell potential and the total charge that they can deliver. Which of these characteristics depends on the amount of reactants in the cell, and which one depends on their concentration? 20.99 A voltaic cell is constructed from an Ni2+(aq)-Ni(s) half­ cell and an Ag +(aq)-Ag(s) half-cell. The initial concentra­ z tion of Ni2+(aq) in the Ni +_Ni half-cell is [Ni2+] = M. The initial cell voltage is + V. (a) By using data in Table calculate the standard emf of this voltaic cell. (b) Will the concentration of NiZ+(aq) increase or decrease as the cell operates? (c) What is the initial concentration of Ag+(aq) in the Ag+-Ag half-cell? [20.100] A voltaic cell is constructed that uses the following half­ cell reactions:

0.0100

1.12

20.1,

+ +2 298

e- ----> Cu(s)

Cu+(aq) lz (s)

e- ---->

2

r-(aq)

The cell is operated at K with [Cu+] = M and [I-] = M. (a) Determine E for the cell at these con­ centrations. (b) Which electrode is the anode of the cell? (c) Is the answer to part (b) the same as it would be if the cell were operated under standard conditions? (d) If [Cu+] was equal to M, at what concentration of r­ would the cell have zero potential? 20.101 Derive an equation that directly relates the standard emf of a redox reaction to its equilibrium constant. 20.102 Using data from Appendix E, calculate the equilibrium constant for the disproportionation of the copper(!) ion at room temperature: Cu+(aq) ----> Cu2+(aq) Cu(s) 20.103 (a) Write the reactions for the discharge and charge of a nickel-cadmium rechargeable battery. (b) Given the fol­ lowing reduction potentials, calculate the standard emf of the cell: Cd(OH)z(s) e- ----> Cd(s) OH-(aq)

3.5

0.25

0.15 2

+

+2

NiO(OH)(s)

+

HzO(l)

+2

+

E�ed =

e- ----> Ni(OH)z(s)

+

E�ed =

-0.76 +0.49

V

OH-(aq) V

(c) A typical nicad voltaic cell generates an emf of

+ 1.30

V. Why is there a difference between this value and the one you calculated in part (b)? (d) Calculate the equilibrium constant for the overall nicad reaction based on this typical emf value.

20.104 The capacity of batteries such as the typical AA alkaline battery is expressed in units of milliamp-hours (mAh).

An "AA" alkaline battery yields a nominal capacity of mAh. (a) What quantity of interest to the consumer is being expressed by the units of mAh? (b) The starting voltage of a fresh alkaline battery is V. The voltage decreases during discharge and is V when the bat­ tery has delivered its rated capacity. If we assume that the voltage declines linearly as current is withdrawn, es­ timate the total maximum electrical work the battery could perform during discharge.

2850

1.55 0.80

20.105

If you were going to apply a small potential to a steel ship resting in the water as a means of inhibiting corrosion, would you apply a negative or a positive charge? Explain.

20.106 The following quotation is taken from an article dealing with corrosion of electronic materials: "Sulfur dioxide, its acidic oxidation products, and moisture are well es­ tablished as the principal causes of outdoor corrosion of many metals." Using Ni as an example, explain why the factors cited affect the rate of corrosion. Write chemical equations to illustrate your points. (Note: NiO(s) is solu­ ble in acidic solution.) [20.107] (a) How many coulombs are required to plate a layer of chromium metal mm thick on an auto bumper with a total area of m2 from a solution containing CrO/-? The density of chromium metal is g/cm 3 . (b) What current flow is required for this electroplating if the bumper is to be plated in s? (c) If the external source has an emf of V and the electrolytic cell is effi­ cient, how much electrical power is expended to electrc:r plate the bumper?

0.32

0.25

7.20

10.0

+6.0

65%

20.108 (a) What is the maximum amount of work that a 6-V lead-acid battery of a golf cart can accomplish if it is rated at A-h? (b) List some of the reasons why this amount of work is never realized.

300

[20.109] Some years ago a unique proposal was made to raise the Titanic. The plan involved placing pontoons within the ship using a surface-controlled submarine-type vessel. The pontoons would contain cathodes and would be filled with hydrogen gas formed by the electrolysis of water. It has been estimated that it would require about X mol of Hz to provide the buoyancy to lift the ship (}. Chern. Educ., Vol. (a) How many coulombs of electrical charge would be required? (b) What is the minimum voltage required to generate Hz and Oz if the pressure on the gases at the depth of the wreckage (2 mi) is atm? (c) What is the minimum electrical energy required to raise the Titanic by electroly­ sis? (d) What is the minimum cost of the electrical energy required to generate the necessary Hz if the electricity costs cents per kilowatt-hour to generate at the site?

7 108

50, 1973, 61).

300

85

INTEGRATIVE EXERCI S E S 20.110 Two wires from a battery are tested with a piece of filter paper moistened with NaCl solution containing phe­ nolphthalein, an acid-base indicator that is colorless in acid and pink in base. When the wires touch the paper about an inch apart, the rightmost wire produces a pink coloration on the filter paper and the leftmost produces none. Which wire is connected to the positive terminal of the battery? Explain.

20.111 The Haber process is the principal industrial route for converting nitrogen into ammonia: Nz(g) Hz(g) ----> NH 3(g) (a) What is being oxiclized, and what is being reduced? (b) Using the thermodynamic data in Appendix C, cal­ culate the equilibrium constant for the process at room temperature. (c) Calculate the standard emf of the Haber process at room temperature.

+3

2

Integrative Exercises

(1.00

[20.112] In a galvanic cell the cathode is an Ag + M)/ Ag(s) half-cell. The anode is a standard hydrogen electrode immersed in a buffer solution containing 0.10 M benzoic acid (C6H5COOH) and M sodium benzoate (C6H5COOrJa+ ). The measured cell voltage is V. What is the pK, of benzoic acid? 20.113 Consider the general oxidation of a species A in solu­ tion: A ---+ A+ + e -. The term "oxidation potential" is sometimes used to describe the ease with which species A is oxidized-the easier a species is to oxidize, the greater its oxidation potential. (a) What is the relation­ ship between the standard oxidation potential of A and the standard reduction potential of A+? (b) Which of the metals listed in Table 4.5 has the highest standard oxida­ tion potential? Which has the lowest? (c) For a series of substances, the trend in oxidation potential is often re­ lated to the trend in the first ionization energy. Explain why this relationship makes sense. [20.114] Gold metal dissolves in aqua regia, a mixture of concen­ trated hydrochloric acid and concentrated nitric acid. The standard reduction potentials Au 3+(aq) 3 e- ---+ Au(s) E�ec1 = V AuCl4 -(aq) 3 e - ---+ Au(s) + 4 Cqaq)

0.050

(a) Use data in Appendix C to calculate ilH0 and 115° for the above reaction. We will assume that these values do not change appreciably with temperature. (b) Based on the values from part (a), what trend would you expect for the magnitude of ilG for the above reaction as the tem­ perature increases? (c) What is the significance of the change in the magnitude of ilG with temperature with respect to the utility of hydrogen as a fuel (recall Equation (d) Based on the analysis here, would it be more efficient to use the combustion method or the fuel-cell method to generate electrical energy from hydrogen?

1.030

+ +

E�cd

=

+1.498 1.002

+

V

are important in gold chemistry. (a) Use half-reactions to write a balanced equation for the reaction of Au and ni­ tric acid to produce Au 3+ and NO(g), and calculate the standard emf of this reaction. Is this reaction sponta­ neous? (b) Use half-reactions to write a balanced equa­ tion for the reaction of Au and hydrochloric acid to produce AuC14-(aq) and H2(g), and calculate the stan­ dard emf of this reaction. Is this reaction spontaneous? (c) Use half-reactions to write a balanced equation for the reaction of Au and aqua regia to produce AuCI4-(aq) and NO(g), and calculate the standard emf of this reac­ tion. Is this reaction spontaneous under standard condi­ tions? (d) Use the Nemst equation to explain why aqua regia made from concentrated hydrochloric and nitric acids is able to dissolve gold.

20.115 A voltaic cell is based on Ag+(aq)/Ag(s) and Fe 3+(aq)jFe 2+(aq) half-cells. (a) What is the standard emf of the cell? (b) Which reaction occu rs at the cathode, and which at the anode of the cell? (c) Use so values in Appendix C and the relationship between cell potential and free-energy change to predict whether the standard cell potential increases or decreases when the tempera­ ture is raised above oc. 20.116 Hydrogen gas has the potential as a clean fuel in reac­ tion with oxygen. The relevant reaction is H2(g) + 02(g) ---+ 2 H20(/)

25

2

Consider two possible ways of utilizing this reaction as an electrical energy source: (i) Hydrogen and oxygen gases are combusted and used to d rive a generator, much as coal is currently used in the electric power industry; (ii) hydrogen and oxygen gases are used to generate elec­ tricity directly by using fuel cells that operate at oc.

85

891

19.15)?

20.117 Cytochrome, a complicated molecule that we will repre­ sent as CyFe 2+, reacts with the air we breathe to supply energy required to synthesize adenosine triphosphate (ATP). The body uses ATP as an energy source to drive other reactions. (Section 19.7) At pH the following re­ duction potentials pertain to this oxidation of CyFe 2+:

7.0

02 (g)

+

4 H+(aq)

+

4 e- ---+ 2 H20(/)

2 CyFe 3+(aq) + e - ---+ CyFe +(aq)

E� = E�ed

=

+0.82 +0.22

V V

(a) What is ilG for the oxidation of CyFe 2+ by air? (b) If the synthesis of mol of ATP from adenosine diphos­ phate (ADP) requires a ilG of 37.7 kJ, how many moles of ATP are synthesized per mole of 02?

1.00

[20.118] The standard potential for the reduction of AgSCN(s) is

+0.0895

v.

AgSCN(s)

+

e - ---+ Ag(s) + SCN-(aq)

Using this value and the electrode potential for Ag +(aq), calculate the Ksp for AgSCN. [20.119] The K,p value for PbS(s) is x -28 By using this value together with an electrode potential from Appendix E, determine the value of the standard reduc­ tion potential for the reaction PbS(s) + 2 e - ---+ Pb(s) + 5 2-(aq)

8.0 10

16.6)

[20.120] A pH meter (Figure employs a voltaic cell for which the cell potential is very sensitive to pH. A simple (but impractical) pH meter can be constructed by using two hydrogen electrodes: one standard hydrogen electrode (Figure and a hydrogen electrode (with atm pressure of H2 gas) dipped into the solution of unknown pH. The two half-cells are connected by a salt b ridge or porous glass disk. (a) Sketch the cell described above. (b) Write the half-cell reactions for the cell, and calculate the standard emf. (c) What is the pH of the solution in the half-cell that has the standard hydrogen electrode? (d) What is the cell emf when the pH of the unknown so­ lution is (e) How precise would a voltmeter have to be in order to detect a change in the pH of pH units?

20.10)

1

5.0?

0.01

[20.121] A student designs an ammeter (a device that measures electrical current) that is based on the electrolysis of water into hydrogen and oxygen gases. When electrical current of unknown magnitude is run through the de­ vice for min, mL of water-saturated Hz(g) is collected. The temperature of the system is oc, and the atmospheric pressure is torr. What is the magni­ tude of the current in amperes?

2.00

12.3

768

25.5

NUCLEAR CHEMISTRY

THE SUN'S ENERGY COMES FROM NUCLEAR reactions of hydrogen atoms in its core.

892

W H AT ' S 21.1

A H EA D

Radioactivity

21.2

In this chapter webywillequations learn howanalogous to describeto nuclear reactions chemical equations, in which the nuclearare charges and masses of reactants and products in decay balance. Radioactive nuclei most commonly by emission of alplul, beta, or gamma radiations. We recognize that nuclear stability is determined largely by the neutron-to-proton ratio. For stable nucleir thisAllratio increases with increasi ng atomic number. nuclei with or more protons areseries radioactive. Heavy nuclei gain stability by a of nuclear disintegrations leading to stable nuclei. We study nuclear transmutations, which are nuclear reactions induced byaccelerated bombardment of aparticle, nucleus bysucha neutron or an charged as an alpha particle. We alsokinetic learnprocesses that radioisotope decays are first­ order that exhibit characteristic half-lives. decays ofof ancient radioisotopes canandbe used togeological determineThe the ages artifacts formations. Patterns of Nuclear Stability

21.6

=

21.7

84

21.3

21.4

21.5

Nuclear Transmutations

Rates of Radioactive Decay

Detection of Radioactivity

We see thatisthedetected radiation emitted bycounters a radioactive substance with Geiger and scintillation counters.

Energy Changes in Nuclear Reactions

We recognize that energy changes in nuclear reactions are related to mass changes via Einstein's famous equation, £ mc2 . The binding energies of nuclei are reflected in the difference betweenofthethemasses of nuclei andthey the are sum of the masses nucleons of which composed.

21.8

21.9

Nuclear Power: Fission

We learnsplits that inunder a nuclear fission reaction a heavy nucleus nuclear bombardment to of form two or more product nuclei, with release energy. type ofpower nuclearplants. reaction is the energy source forThisnuclear Nuclear Power: Fusion

We observe that nuclearfusion results from the fusion ofone.two light nuclei to form a more stable, heavier Radiation in the Environment and Living Systems

We discover that naturally occuring radioisotopes bathe our planet, and us, with low levels of radiation. The radiation emitted in nuclear reactions hascan, the potential to cause cell damage in animals that in lead to cancer and other illnesses. of radiation, on the other hand,Appropriate can be usedusemedically to diagnose and treat cancer. tum,

THE SUN IS THE MAJOR ENERGY SOURCE FOR OUR PLAN ET AND IS ESSENTIAL TO LIFE ON EARTH. Sunlight provides approximately 1 000 watts (1 watt

=

1 J/s) of power per square meter at

Earth's surface. Sunlight is used by plants for photosynthesis, the process that produces food for plants and oxygen that most life on Earth needs to survive. What makes the Sun shine? Hydrogen fuses in the Sun's core to make helium, releasing tremendous amounts of energy in the process. This hydrogen fusion reaction is an example of a

nuclear reaction. A nuclear reaction involves changes in the nucleus of an atom. Nuclear chemistry is the study of nuclear reactions, with an emphasis on their uses in chemistry and their effects on biological systems. Nuclear chemistry affects our lives in many ways, particularly in energy and medical applications. For example, radioactive elements can be used as therapeutic and diagnostic tools.

893

894

CHAPTER 2 1

Nuclear Chemistry

.A. Figure 2 1 . 1 Radioisotope scanning. A gamma ray detector scan of a normal human heart, obtai ned following intravenous injection of the radioisotope thallium-20 1, a gamma emitter. The donut-shaped pink and red area represents uptake of the radioisotope by healthy heart muscles.

Radiation therapy is a common form of cancer therapy in which gamma rays, emitted from a radioactive substance such as cobalt-60, are directed to cancerous tumors to destroy them. In the widely used thallium "stress test," a compound containing thallium-201 (another radioactive substance that emits gamma rays) is injected intravenously into the patient. A scanner detects the gamma emis­ sions from the thallium that has concentrated in the heart muscle and reveals the condition of the heart and surrounding arteries (Figure 21.1 "11 ) . Radioactivity is also used to help determine the mechanisms of chemical reactions, to trace the movement of atoms in biological systems and the environment, and to date important historical artifacts. Nuclear reactions are also used to generate electricity. Roughly 10% of the total electricity generated in the world comes from nuclear power plants. The percentage of electricity generated from nuclear power varies widely among countries: 80% in France, 50% in Sweden, 20% in the United States, less than 5% each in India and China, and essentially zero in Italy (Figure 21.2 ") The use of nuclear energy for power generation and the disposal of nuclear wastes from power plants are controversial social and political issues. It is imperative, there­ fore, that as a citizen with a stake in these matters, you have some understanding of nuclear reactions and the uses of radioactive substances. -

2 1 . 1 RAD I OACTIVITY To understand nuclear reactions, we must review and develop some ideas intro­ duced in Section 2.3. First, recall that two types of subatomic particle reside in the nucleus: the proton and the neutron. We will refer to these particles as nucleons.

I> Figure 21.2 Nuclear energy usage for electricity, 2004. This bar graph

shows, for selected countries, the percentage of electricity supplied by nuclear power.

Italy China Pakistan India Brazil Netherlands Mexico South Africa Argentina Romania Canada Russia United Kingdom United States Spain Finland Japan Czech Republic Germany Hungary Republic of Korea Slovenia Armenia Switzerland Bulgaria Ukraine Sweden Belgium Slovakia Lithuania France

�

�8 II

0

10

20

30

40 Percent

50

60

70

80

21.1

Recall also that all atoms of a given element have the same number of protons; this number is the element's atomic number. The atoms of a given element can have different numbers of neutrons, however, so they can have different mass numbers; the mass number is the total number of nucleons in the nucleus. Atoms with the same atomic number but different mass numbers are known as isotopes. The different isotopes of an element are distinguished by their mass numbers. For example, the three naturally occurring isotopes of uranium are uranium-234, uranium-235, and uranium-238, where the numerical suffixes represent the mass numbers. These isotopes are also labeled, using chemical symbols, as 2�U, 2��U, and 2��U. The superscript is the mass number; the sub­ script is the atomic number. Different isotopes have different natural abundances. For example, 99.3% of naturally occurring uranium is uranium-238, 0.7% is uranium-235, and only a trace is uranium-234. Different nuclei also have different stabilities. Indeed, the nuclear properties of an atom depend on the number of protons and neu­ trons in its nucleus. A nuclide is a nucleus with a specified number of protons and neutrons. Nuclei that are radioactive are called radionuclides, and atoms containing these nuclei are called radioisotopes. Nuclear Equations

Most nuclei found in nature are stable and remain intact indefinitely. Radionu­ clides, however, are unstable and spontaneously emit particles and electromag­ netic radiation. Emission of radiation is one of the ways in which an unstable nucleus is transformed into a more stable one with less energy. The emitted ra­ diation is the carrier of the excess energy. Uranium-238, for example, is radioac­ tive, undergoing a nuclear reaction in which helium-4 nuclei are spontaneously emitted. The helium-4 particles are known as alpha (a) particles, and a stream of these particles is called alpha radiation. When a uraniurn-238 nucleus loses an alpha particle, the remaining fragment has an atomic number of 90 and a mass number of 234. If you look at the periodic table, you will find that the element with atomic number 90 is Th, thorium. Therefore, the products of uranium-238 decomposition are an alpha particle and a thorium-234 nucleus. We represent this reaction by the following nuclear equation: [21.1] When a nucleus spontaneously decomposes in this way, it is said to have de­ cayed, or to have undergone radioactive decay. Because an alpha particle is involved in this reaction, scientists also describe the process as alpha decay. GIVE IT SOME THOUGHT

What change in the mass number of a nucleus occurs when the nucleus emits an alpha particle? In Equation 21.1 the sum of the mass numbers is the same on both sides of the equation (238 = 234 + 4). Likewise, the sum of the atomic numbers on both sides of the equation is equal (92 = 90 + 2). Mass numbers and atomic numbers must be balanced in all nuclear equations. The radioactive properties of the nucleus are independent of the chemical state of the atom. In writing nuclear equations, therefore, we are not concerned with the chemical form (element or compound) of the atom in which the nucle­ us resides.

Radioactivity

895

896

CHAPTER 2 1

Nuclear Chemistry I Predicting the Product of a Nuclear Reaction What product is formed when radium-226 undergoes alpha decay?

- SAMPLE EXERCISE 21.1

SOLUTION

We are asked to determine the nucleus that results when radium-226 loses an alpha particle. Analyze: Plan:

We can best do this by writing a balanced nuclear reaction for the process.

The periodic table shows that radium has an atomic number of 88. The complete chemical symbol for radium-226 is therefore ��Ra. An alpha particle is a helium-4 nucleus, and so its symbol is iHe (sometimes written as ia). The alpha particle is a product of the nuclear reaction, and so the equation is of the form Solve:

2g�Ra -> �X + iHe

where A is the mass number of the product nucleus and Z is its atomic number. Mass numbers and atomic numbers must balance, so 226 = A + 4 and 88 Hence, A = 222

= z

+2

and

z = 86

Again, from the periodic table, the element with Z = 86 is radon (Rn). The prod­ uct, therefore, is ��Rn, and the nuclear equation is ��Ra -. 2�Rn + iHe - PRACTICE EXERCISE

Which element undergoes alpha decay to form lead-208? Answer: 2�o

Types of Radioactive Decay

The three most common kinds of radioactive decay are alpha (a), beta ({3), and gamma (-y) radiation. (Section 2.2) Table 21.1 T summarizes some of the im­ portant properties of these kinds of radiation. As we have just discussed, alpha radiation consists of a stream of helium-4 nuclei known as alpha particles, which we denote as iHe or ia. Beta radiation consists of streams of beta ({3) particles, whkh are high­ speed electrons emitted by an unstable nucleus. Beta particles are represented in nuclear equations by the symbol ?e or sometimes ?13 . The superscript zero in­ dicates that the mass of the electron is exceedingly small compared to the mass of a nucleon. The subscript -1 represents the negative charge of the particle, which is opposite that of the proton. Iodine-131 is an isotope that undergoes decay by beta emission: cxx:>

[21.2] TABLE 2 1 . 1

• Properties of Alpha, Beta, and Gamma Radiation

Type of Radiation

Property

Charge Mass Relative penetrating power Nature of radiation

2+

6.64 X 10-24 g 1 iHe nuclei

19.11 X 10-28 g 100 Electrons

y

0 0 10,000 High-energy photons

21.1

You can see from Equation 21.2 that beta decay causes the atomic number to increase from 53 to 54. Therefore, beta emission is equivalent to the conversion of a neutron ( bn) to a proton ( jp or jH), thereby increasing the atomic number by 1 : [21.3] Just because an electron is ejected from the nucleus, however, we should not think that the nucleus is composed of these particles, any more than we consid­ er a match to be composed of sparks simply because it gives them off when struck. The electron comes into being only when the nucleus undergoes a nuclear reaction. Gamma radiation (or gamma rays) consists of high-energy photons (that is, electromagnetic radiation of very short wavelength). It changes neither the atomic number nor the mass number of a nucleus and is represented as 81' , or merely y. Gamma radiation usually accompanies other radioactive emission because it represents the energy lost when the remaining nucleons reorganize into more stable arrangements. Generally, the gamma rays are not shown when writing nuclear equations. Two other types of radioactive decay are positron emission and electron capture. A positron is a particle that has the same mass as an electron, but an opposite charge.* The positron is represented as �e. The isotope carbon-11 de­ cays by positron emission: J�c � l�B + ?e [21.4] Positron emission causes the atomic number to decrease from 6 to 5. The emis­ sion of a positron has the effect of converting a proton to a neutron, thereby de­ creasing the atomic number of the nucleus by 1: [21.5] lp � bn + �e Electron capture is the capture by the nucleus of an electron from the elec­ tron cloud surrounding the nucleus. Rubidium-81 undergoes decay in this fashion, as shown in Equation 21.6: ��Rb + -?e (orbital electron) � �!Kr [21.6] Because the electron is consumed rather than formed in the process, it is shown on the reactant side of the equation. Electron capture, like positron emission, has the effect of converting a proton to a neutron: [21.7] Table 21.2 � summarizes the symbols used to represent the various elementary particles commonly encountered in nuclear reactions. GIVE IT SOME THOUGHT

Which of the particles listed in Table 21.2 result in no change in nuclear charge when emitted in nuclear decay? - SAMPLE EXERCISE 21 .2

I Writing Nuclear Equations

Write nuclear equations for thedecays following processes: mercury-201 undergoes elec­ tron capture; thorium-231 to form protactinium-231. must and writeproducts balancedare nuclear charges ofWereactants equal. equations in which the masses and can begin writingin thetheproblem. complete chemical symbols for the nuclei and decay We particles that arebygiven (a)

(b)

SOLUTION

Analyze: Plan:

*Tize positron has a very short life because it gamma rays: �e + -?e ------+ 2 8y.

is

annil1ilated when it collides with an electron, producing

Radioactivity

897

TABLE 2 1 .2 • Common Particles in Radioactive Decay and Nuclear Transformations Particle

Symbol

Neutron Proton Electron Alpha particle Beta particle Positron

On ] y � Y Y

Hor ]p _e He or � -ee or Y/3 -

a

898

CHAPTER 2 1

Nuclear Chemistry Solve:

(a) The information given in the question can be summarized as

2i)hHg + �e -----> �X The mass numbers must have the same sum on both sides of the equation: 201 + 0 = A Thus, the product nucleus must have a mass number of 201. Similarly, balancing the atomic numbers gives 80 - 1 = z Thus, the atomic number of the product nucleus must be 79, which identifies it as gold (Au): zgdHg + _£e -----> 29§Au

(b) In this case we must determine what type of particle is emitted in the course of the radioactive decay: �6Th -----> �jPa + �X =

=

=

=

From 231 231 + A and 90 91 + Z, we deduce A 0 and Z -1. According to Table 21.2, the particle with these characteristics is the beta particle (electron). We therefore write - PRACTICE EXERCISE

Write a balanced nuclear equation for the reaction in which oxygen-15 undergoes positron emission. Answer: 1 go -----> 1�N + �e

2 1 .2 PATTERNS OF N U C LEAR S TA B ILITY The stability of a particular nucleus depends o n a variety of factors. No single rule allows us to predict whether a particular nucleus is radioactive and how it might decay. However, several empirical observations will help you predict the stability of a nucleus.

Neutron-to-Proton Ratio Because like charges repel each other, it may seem surprising that a large number of protons can reside within the small volume of the nucleus. At close distances, however, a strong force of attraction, called the nuclear force, exists between nu­ cleons. Neutrons are intimately involved in this attractive force. All nuclei other than l H contain neutrons. As the number of protons in the nucleus increases, there is an ever-greater need for neutrons to counteract the effect of proton­ proton repulsions. Stable nuclei with low atomic numbers (up to about 20) have approximately equal numbers of neutrons and protons. For nuclei with higher atomic numbers, the number of neutrons exceeds the number of protons. Indeed, the number of neutrons necessary to create a stable nucleus increases more rapid­ ly than the number of protons, as shown in Figure 21.3 11>. Thus, the neutron-to­ proton ratios of stable nuclei increase with increasing atomic number. The colored band in Figure 21.3 is the area within which all stable nuclei are found. This area is known as the belt of stability. The belt of stability ends at element 83 (bismuth). All nuclei with 84 or more protons (atomic number 2: 84) are radioactive. For example, all isotopes of uranium, atomic number 92, are radioactive. G IVE IT SOME THOUGHT

Using Figure 21.3, estimate the optimal number of neutrons for a nucleus containing 70 protons.

21.2

Patterns of Nuclear Stability

899

130 (1.5:1 ratio)2�Hg 120 110 t:.i . �: : . · _j 100 ·�· Belt of ;.i : 90 stability � :::' · I!! 80 g (1.4:1 ratw). 120 Sn �:r:: r:· · 70 :J·: 60 E (1.25:1 ratio);:&Zr � �� z 50 :.r: 40 1:1 neutron-to­ 30 proton ratio

t

•

50

" "' "

�

0 � "' .0 "

10 20 30 40 50 60 70 80 Number of protons The belt of stability. The number of neutrons is graphed versus the number of protons for stable nuclei. As the atomic number increases, the neutron-to-proton ratio of the stable nuclei increases. The stable nuclei are located in the shaded area of the graph known as the belt of stability. The majority of radioactive nuclei occur outside this belt.

.A Figure 21 .3

The type of radioactive decay that a particular radionuclide undergoes de­ pends largely on how its neutron-to-proton ratio compares to those of nearby nuclei within the belt of stability. We can envision three general situations: 1.

2.

3.

Loss of _ye

Nuclei above the belt of stability (high neutron-to-proton ratios). These neu­ tron-rich nuclei can lower their ratio and move toward the belt of stability by emitting a beta particle. Beta emission decreases the number of neu­ trans and increases the number of protons in a nucleus, as shown in Equation 21.3.

Loss of �ecapture or electron

Nuclei below the belt of stability (low neutron-to-proton ratios). These proton­ rich nuclei can increase their ratio by either positron emission or electron capture. Both kinds of decay increase the number of neutrons and de­ crease the number of protons, as shown in Equations 21.5 and 21.7. Positron emission is more common than electron capture among the lighter nuclei. However, electron capture becomes increasingly common as nuclear charge increases. 2: 84. These heavy nuclei, which lie beyond the upper right edge of the band of stability, tend to undergo alpha emission. Emission of an alpha particle decreases both the number of neutrons and the number of protons by 2, moving the nucleus diagonally toward the belt of stability.

Nuclei with atomic numbers

These three situations are summarized in Figure 21.4 �.

A

Number of protons Figure 2 1 .4

-

Proton and neutron changes In nuclear processes.

The graph shows the results of alpha emission