Chemistry [Fifth Edition] 0534011845, 9780534011840

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CHEMISTRY FIFTH EDITION

MORTIMER

0 2

He Heliunn III

A

IV

A

V A

A

VI

6

7

Boron

Carbon

Nitrogen

Oxygen

1081

12 Oil

0067

159994

5

VII

9

N

B 14

A

13

00260 10

F

Ne

Fluorine

Neon

18

998403 17

16

4

20 179 18

''

S AlumirVn^ < II

B

26 981544


28

29

30

31

Ni

Cu

Zn

Nickel

Copper

Zinc

S^e Gallium

Gernnanium

58 69

63 546

65 38

69 72

46

47

48

49

06

4^

CI Chlorine

Argon

35453

39 948

36

.33

of

«ate ial%n©r»t»

Se

Br

Kr

Arsenic

Selenium

Bromine

Krypton

72 59

74.9216

7896

79 904

83 80

50

51

52

53

54

Pd

Ag

Cd

In

Sn

Sb

Te

Palladium

Silver

Cadmium

Indium

Tin

Antimony

Tellurium

Iodine

Xenon

106.42

107868

112 41

114 82

11869

12760

126 9045

131.29

121

75

79

80

81

82

83

84

85

86

Au

Hg

Tl

Pb

Po

At

Rn

Platinum

Gold

Mercury

Thallium

Lead

Bi Bismuth

Polonium

Astatine

Radon

195 08

196 9665

200 59

204 383

2072

208 9804

(209)''

(210)'

(222)»

78

metals

63

64

65

66

67

68

69

nonmetals

70

71

Eu

Gd

Tb

Dy

Ho

Er

Tm

Yb

Lu

Europium

Gadolinium

Terbium

Dysprosium

Holmium

Erbium

Thulium

Ytterbium

Lutetium

151.96

15725

158 9254

162 50

164 9304

167 26

168 9342

173 04

174

967

95

96

97

98

99

100

101

102

103

An-.

Cm

6k

Cf

Es

Fm

Md

No

Americium

Curium

Berkelium

Californium

Einsteinium

Fermium

Mendelevium

Nobelium

Lr Lawrencium

(243)^

(247)^

(247r

(251)"

(252)"

(257)^

(258)-

(259)"

(260)"

Digitized by tlie Internet

:

in

Arch ive

2014

https://archive.org/cIetails/chemistryOOmort

Chemistry

E

S

T

R

Y

Fifth Edition

Charles E.

Mortimer Muhlenberg College

Wadswon. P„..„„, Co.,,„,

.

Be,„„„, Ca„r„™,

.

. 0,™,„„

„r

Wad»„„,

,

nc.

To: J.S.M. and C.E.M.

ISBN D-SaM-DllfiM-S A

study guide has been specially designed to help students master the concepts presented Order from your bookstore.

in this textbook.

Cover photos:

Pamela S. Roberson (first and third photos from top) tiller, Peter Arnold, Inc. (second photo) Werner H. The Image Bank West/Gabe Palmer (fourth photo)

M

Chemistry Editor: Jack Carey Produetion Editor: Hal Humphrey Designer: Adriane Bosworth Technical Illustrators : J & R Art Services Photo Researchers : Roberta Spieckerman Associates

©

1983 by Wadsworth, Inc.

reserved.

©

1979, 1975 by Litton Educational Publishmg, Inc. All rights

No part of this book may be reproduced,

stored in a retrieval system, or transcribed,

any form or by any means, electronic, mechanical, photocopying, recording, or otherwise, without the prior written permission of the publisher, Wadsworth Publishing Company, Belmont, California 94002, a division of Wadsworth, Inc.

in

Printed in the United States of America

23456789

10—87

86 85 84 83

Library of Congress Cataloging

in

Publication Data

Mortimer, Charles E. Chemistry.

(Wadsworth

series in chemistry)

Includes index. 1.

Chemistry.

I.

Title.

JT ^Series. 82-11133

PREFACE

Sometimes it seems as though every product that is advertised on television is said new and improved. I hope that this trite phrase describes this edition, but at the same time, I beheve that the character of the text that has made it successful through the previous four editions has been retained. This book was written to

to be

explain chemistry, not just present chemical facts. Consequently, each concept

continues to be explained as fully as

is

necessary for understanding, simplified

where necessary, but never distorted. Probably the most prominent change made editioi) is the alteration in the

to the

understanding of

all

in the

preparation of the

sequence of topics. Stoichiometry, which

chemical concepts,

is

is

introduced early (Chapter

a result, the use of stoichiometric principles can be

fifth

central 2).

As

expanded and reinforced

throughout the entire course. Furthermore, this early placement facilitates the design of a coordinated laboratory program (for which a section on "Reactions in Solution" has been introduced). Thermochemistry (Chapter 3) follows stoichiometry, underscoring the fact that chemistry

is

a science that

is

concerned with both energy and matter and that

both are amenable to quantitative treatment. The early discussion of thermochemistry prepares the way for the use of energy concepts (such as lattice energy,

and bond energy) in the development of later topics. new chapter on Reactions in Aqueous Solution (Chapter 11) is introduced the chapter on Solutions (Chapter 10), which it logically follows. The discuss-

ionization energy,

A after

ion of these reactions, which constitute a high proportion of all chemical reactions

groundwork for later discussions (notably, ionic equilibria, and bases, electrochemistry, and descriptive chemistry). Electrochemistry (Chapter 18) is postponed until after Thermodynamics (Chapter 17) and Equilibrium (Chapters 13 through 16) have been discussed. In this way, the principles of thermodynamics (particularly Gibbs free energy) and investigated, lays the

acids

equilibrium (notably, equilibrium expressions) can be used to develop electro-

chemical concepts (electromotive force, electrode potentials, the Nernst equation).

Oxidation numbers and oxidation-reduction reactions are discussed earlier

The

than

in the

in

Chapter

1

1,

formal treatment of electrochemistry.

descriptive chemistry of the Nonmetals (Chapters 19 through 22) appears

later in the

book

A discussion

so that

it

continues to follow the treatment of electrochemistry.

if it is to be more than superficial, demands an understanding of electrochemical concepts electrolysis and electrode potentials. The chemistry of oxygen and hydrogen is now incorporated into the treatment of the nonmetals rather than appear as a separate chapter (old Chapter 8). The descriptive chemistry of the nonmetals is followed by a chapter on Metals (Chapter 23) and industrial aspects are emphasized in both.

of descriptive chemistry,

—

on Organic Chemistry (Chapter 26) together with a new chapter on Biochemistry (Chapter 27), which logically follows, are placed at the end of the book. The chapter on Nuclear Chemistry (Chapter 25) has been moved forward to accommodate this placement. Several chapters have been divided to make the text more flexible and to facihtate the preparation of a course outline Stoichiometry (Chapter 2) and Thermo-

The

chapter"

:

chemistry (Chapter

now

covered

in a

are

3)

now

Bond (Chapter 5) is Bond (Chapters 6 and 7); appear separately; and the

separate chapters; the Ionic

separate chapter from the Covalent

Kinetics (Chapter 12)

and Equilibrium (Chapter

13)

descriptive chemistry of the Nonmetals has been divided into four parts (Chapters 19 through 22).

Many

changes have been made

in the text itself in the interest

of keeping

it

up-to-date and improving the clarity and usefulness of the presentation. For

example, sections on air pollution, the corrosion of iron, and industrial uses of the nonmetals have been added.

A number of features appear in given at the end of each chapter.

of the concepts covered

Key terms

are listed

this edition to help the student.

They provide

Summaries are

the student with a quick check-list

in the chapter.

and defined

at the

end of each chapter. Students

will find

these glossaries useful as an aid in studying the material of the chapter, as a quick

reference for future work,

follow these

and as a help

in solving the

chapter-end problems that

Important new terms continue to be set where they are first introduced and defined.

lists.

point in the text

Boxes are used

color type at the

in

to set off step-by-step directions for the solution of basic types

of problems. Students will find this boxed material useful for

and also for reference

in later

initial

assignments

work.

Chapter-end problems, totaling about 1200, are grouped according to type.

Many

of these problems are new. Answers for approximately half the problems,

those that are color-keyed, are given in the appendix. are

marked with

Examples, designed to

throughout the

The more

difficult

problems

asterisks.

text.

illustrate

how

to solve chemical problems, are used

The number of these examples has been

increased, particularly

in the early chapters.

Photographs, of scientists and subjects of chemical interest, are used to enliven

They accompany the figures, which augment and amplify the discussion. Notes on mathematical operations appear in the appendix. These notes include discussions on the use of exponents, scientific notation, common and natural the text.

logarithms, and the quadratic formula.

The following supplemetary items are available study guide, solutions manual, answer booklet, instructor's manual with test items, and transparency masters. I sincerely thank the following persons for their comments and suggestions: :

David L. Adams, North Shore Community College John E. Bauman Jr., University of Missouri Paul A. Barks, North Hennepin Community College Neil R. Coley, Chabot College John DeKorte, Northern Arizona University Geoffry Davies, Northeastern University Phil Davis, University of Tennessee

Lawrence Epstein, University of Pittsburgh Des Moines Area Community College Peter J. Hansen, Northwestern College

Patrick Garvey,

Preface

Larry C. Hall, Vanderbilt University

David W. Herlocker, Western Maryland College Delwin Johnson, St. Louis Community College at Forest Park George B. Kauffman, California State University, Fresno Robert P. Lang, Quincy College Lester R. Morss, Argonne National Laboratory John Maurer, University of Wyoming William McCurdy, .Ohio State University Robert C. Melucci, Community College of Philadelphia

Lucy

T. Pryde, Southwestern College

Fred H. Redmore, Highland Community College Lewis Radonovich, University of North Dakota Roland R. Roskos, University of Wisconsin Larry Thompson, University of Minnesota

'

James A. Weiss, Penn State University, Scranton Campus

The Wadsworth tion of this book.

I

staff

have been congenial, helpful, and

appreciate their efforts and express

efficient in the

my gratitude. My

produc-

thanks go

Chemistry Editor; Hal Humphrey, Production Editor; Adriane Bosworth, Designer; and Harriet Serenkin. Suggestions for the improvement of this edition will be welcomed. especially to: Jack Carey,

Charles E. Mortimer

Preface

vil

BRIEF CONTENTS

1

Introduction

2

Stoichiometry

23

3

Thermochemistry

51

4

Atomic Structure

73

5

Properties of

6

The Covalent Bond

7

Molecular Geometry Molecular Orbitals

155

8

Gases

178

9

Liquids and Solids

214

10

Solutions

246

11

Reactions

12

Chemical Kinetics

305

13

Chemical Equilibrium

336

14

Theories

15

Ionic Equilibrium, Part

I

371

16

Ionic Equilibrium, Part

II

412

17

Elements

1

Atoms and the

Ionic

Bond

135 ;

in

of

of

114

Aqueous Solution

275

Acids and Bases

354

Chemical Thermodynamics

429 450

18

Electrochemistry

19

The Nonmetals, Part

I:

20

The Nonmetals, Part

II:

21

The Nonmetals, Part

22

The Nonmetals, Part

23

Metals and Metallurgy

587

24

Complex Compounds

627

25

Nuclear Chemistry

648

26

Organic Chemistry

675

27

Biochemistry

711

Appendices

732

Index

759

viii

Hydrogen and the Halogens

492

A Elements

515

III:

The Group V A Elements

540

IV:

Carbon, Silicon, Boron, and the Noble Gases

565

The Group

VI

DETAILED CONTENTS

Preface

1

Introduction

1

1.1

5

1.3

The Development of Modern Chemistry Elements, Compounds, and Mixtures The Metric System

1.4

Significant Figures

1.5

Chemical Calculations

1.2

Summary

/

Key Terms

1

8 11

14 ,

Problems

20

2

Stoichiometry

23

2.1

23

2.6

Dalton's Atomic Theory Atomic Weights Formulas The Mole Derivation of Formulas Percentage Composition of Compounds

2.7

Chemical Equations

36

2.8

Problems Based on Chemical Equations

2.9

Stoichiometry of Reactions

2.2 2.3

2.4 2.5

Summary

in

Key Terms

24 25

27 30 34

38

43

Solution

Problems

46

Thermochemistry

3

51

3.1

Energy Measurement

3.2

Temperature and Heat

52

3.3

Calorimetry

54

3.4

Thermochemical Equations

56

3.5

The Law of Hess

59

3.6

Enthalpies of Formation

61

3.7

Bond Energies

Summary

65

Key Terms

/

Problems

68

Atomic Structure

4 4.

51

1

4.2

The Electron The Proton

73 •

"

73 75

ix

4.5

4.6

Isotopes

4.7

Atomic Weights

80

4.8

Electromagnetic Radiation

83

4.9

4.11

Atomic Spectra Atomic Number and Wave Mechanics

4.12

Quantum Numbers

4.13

Orbital Filling and Hund's Rule

4.4

4.10

4.14

_^

77

-

78

'

79

'

.

the Periodic

Law

and

87

93

97 100

Types of Elements /

107

Filled Subshells

Half-filled

4.16

84 91

Electronic Structures of the Elements

4.15

Summary

107

Key Terms

Atoms and

/

Problems

108

Bond

114

5

Properties of

5.1

Atomic

5.2

Ionization Energies

116

5.3

Electron Affinities

119

5.4

The Ionic Bond

120

the Ionic

114

Sizes

5.5

Lattice Energies

122

5.6

Types of Ions

125

5.7

Ionic Radius

128

5.8

Nomenclature of Ionic Compounds

Summary

/

Key Terms

/

130

Problems

132

6

The Covalent Bond

6.1

Covalent Bonding

135

6.2

Formal Charge

137

6.3

Lewis Structures

140

6.4

Resonance

144

6.5

Transition between Ionic and Covalent Bonding

146

6.6

Electronegativity

149

6.7

Nomenclature of Covalent Binary Compounds Summary / Key Terms / Problems

151

7

Molecular Geometry Molecular Orbitals

155

7.1

Exceptions to the Octet Rule

7.2

Electron-Pair Repulsions and Molecular

7.3

Hybrid Orbitals

7.4

Molecular Orbitals

7.5

Molecular Orbitals

7.6

pn-dn Bonding

135

;

Summary

X

76

The Neutron The Nuclear Atom Atomic Symbols

4.3

8

Gases

8.1

Pressure

8.2

Boyle's

Contents

/

152

155

Geometry

156 163 165

in

Polyatomic Species

172

174

Key Terms

/

Problems

175 178

Law

178

180

Law

182

8.3

Charles'

8.4

Amontons' Law

8.5

Ideal

8.6

Kinetic Theory of Gases

184

-

Gas Law

185

190

Gas Law from

the Kinetic Theory

8.7

Derivation of the Ideal

8.8

Gay-Lussac

8.9

Stoichiometry and

8.10

Dalton's

8.11

Molecular Speeds

200

8.12

Graham's Law of Effusion

202

8.13

Real Gases

204

8.14

Liquefaction of Gases

Law

Summary

191

Law

of Combining Volumes and Avogadro's Principle s

Gas Volumes

of Partial Pressures

/

Key Terms

192 195

198

207 /

Problems

208

9

Liquids and Solids

214

9.1

Intermolecular Forces of Attraction

214

9.2

The Hydrogen Bond The Liquid State

216

9.3

9.4

Evaporation

220

9.5

Vapor Pressure

221

9.6

Boilmg Point

222

9.7

Enthalpy of Vaporization

223

9.8

224

9.10

The Freezing Point Vapor Pressure of a Phase Diagrams

9.11

Types of Crystalline Solids

229

9.12

Crystals

231

9.13

X-ray Diffraction of Crystals

235

9.14

Crystal Structure of Metals

237

9.15

Ionic Crystals

238

9.16

Defect Structures

240

9.9

Summary

/

219

225

Solid

Key Terms

226

/

Problems

241

Solutions

246

10.1

Nature of Solutions

246

10.2

The Solution Process

247

10.3

Hydrated Ions

248

10.4

Enthalpy of Solution

250

10.5

Effect of

10.6

Concentrations of Solutions

252

10.7

257

10.9

Vapor Pressure of Solutions Boiling Point and Freezing Point of Solutions Osmosis

10.10

Distillation

266

10.11

Solutions of Electrolytes

268

10.12

Interionic Attractions in Solution

268

Summary

270

10

10.8

Temperature and Pressure on Solubility

/

Key Terms

,'

Problems

251

261

264

Contents

xi

11

Reactions

in

275

11.1

Metathesis Reactions

275

11.2

Oxidation Numbers

279

11.3

Oxidation-Reduction Reactions

282

11.4

Arrhenius Acids and Bases

289

11.5

Acidic and Basic Oxides

11.6

Nomenclature

11.7

Volumetric Analysis

11.8

Equivalent Weights and

298

Summary

300

/

ot~

291

Acids and Salts

'

293 295

Normal Solutions Key Terms / Problems

Chemical Kinetics

305

12.1

Reaction Rates

305

12.2

Concentrations and Reaction Rates

308

12.3

Single-Step Reactions

311

12.4

Rate Equations for Single-Step Reactions

316

12.5

Reaction Mechanisms

318

12.6

Rate Equations and Temperature

322

12.7

Catalysts

327

12

Summary

/

Key Terms

/

Problems

332

Chemical Equilibrium

336

13.1

Reversible Reactions and Chemical Equilibrium

336

13.2

Equilibrium Constants

13.3

Equilibrium Constants Expressed

13.4

Le Chatelier's Principle

13

Summary

/

Key Terms

339 in

Pressures

344 348

/

Problems

Acids and Bases

351

354

14

Theories

14.1

14.2

The Arrhenius Concept The Bronsted-Lowry Concept

355

14.3

Strength of Bronsted Acids and Bases

356

14.4

Acid Strength and Molecular Structure

359

14.5

The Lewis Concept

363

14.6

Solvent Systems

366

of

Summary 15

xii

Aqueous Solution

/

Key Terms

Ionic Equilibrium, Part

/

354

Problems

I

368

371

15.1

Weak

Electrolytes

371

15.2

The Ionization of Water

378

15.3

pU

380

15.4

Indicators

384

15.5

The Common-Ion

15.6

Buffers

387

15.7

Polyprotic Acids

393

15.8

Ions That Function as Acids and Bases

397

Contents

Effect

386

15.9

Acid-Base Titrations

Summary

/'

402

Key Terms

16

Ionic Equilibrium, Part

16.1

The

16.2

Precipitation

16.3

Precipitation of Sulfides

Problems

408 412

11

Solubility Product

Equilibria Involving

16.5

Amphoterism

Summary Elements

/

of

Law

412

and the Solubility Product

16.4

17

/

415

419

Complex Ions

422 425

Key Terms

/

Problems

427

Chemical Thermodynamics

429

of Thermodynamics

429

17.1

First

17.2

Enthalpy

431

17.3

Second Law of Thermodynamics

434

17.4

Gibbs Free Energy

436

17.5

Standard Free Energies

439

17.6

Absolute Entropies

440

17.7

Gibbs Free Energy and Equilibrium

443

Summary

446

/

Key Terms

/

Problems

18

Electrochemistry

450

18.1

Metallic Conduction

450

18.2

Electrolytic

18.3

Electrolysis

453

18.4

Stoichiometry of Electrolysis

459

18.5

Voltaic Cells

462

18.6

Electromotive Force

464

18.7

Electrode Potentials

466

18.8

Gibbs Free Energy Change and Electromotive Force

18.9

Effect of Concentration

18.10

Concentration Cells

479

18.11

Electrode Potentials and Electrolysis

480

18.12

481

18.13

The Corrosion of Iron Some Commercial Voltaic

18.14

Fuel Cells

Conduction

Summary

451

471

475

on Cell Potentials

482

Cells

484

Key Terms

,

485

Problems

Hydrogen and the Halogens

19

The Nonmetals, Part Hydrogen

492

19.1

Occurrence and Properties of Hydrogen

492

19.2

Industrial Production of

19.3

Hydrogen from Displacement Reactions Reactions of Hydrogen Industrial Uses of Hydrogen

19.4 19.5

I:

Hydrogen

The Halogens 19.6

492

Properties of the Halogens

493

494 495 497 -

498

498

Contents

xlil

19.7

Occurrence and Industrial Prepaiation of the Halogens

500

19.8

Laboratory Preparation of the Ha'ogens

502

19.9

502

19.11

The Interhalogen Compounds The Hydrogen Halides The Metal Halides

19.12

Oxyacids of the Halogens

19.13

Industrial Uses of the

19.10

Summary

/

II:

505

507

Halogens

Key Terms

The Nonmetals, Part

20

504

513

Problems

/

The Group

Group VI

512

'

A

A Elements

VI

515

Elements

515

20.1

Properties of the

20.2

Occurrence and Industrial Production of Oxygen

516

20.3

Laboratory Preparation of Oxygen

517

20.4

Reactions of Oxygen

518

20.5

Industrial Uses of

20.6

Ozone

20.7

Air Pollution

20.8

AUotropic Modifications of

20.9

Occurrence and Industrial Preparation of

Oxygen

521

521

522 S. Se,

and Te

524 S, Se,

and Te

Hydrogen Compounds of S, Se, and Te The 4 + Oxidation State of S, Se, and Te 20.12 The 6+ Oxidation State of S, Se, and Te 20.13 Electrode Potential Diagrams for S 20. 14 Industrial Uses of S, Se, and Te Summary / Key Terms / Problems 20.10

527

20.

529

1 1

21

The Nonmetals, Part

III:

531

536 536 537

The Group V A Elements

Group V

A

540

Elements

21.1

Properties of the

21.2

The Nitrogen Cycle

21.3

Occurrence and Preparation of the Group

21.4

Nitrides and Phosphides

545

21.5

546

21.9

Hydrogen Compounds Halogen Compounds Oxides and Oxyacids of Nitrogen Oxides and Oxyacids of Phosphorus Oxides and Oxyacids of As, Sb, and

21.10

Industrial Uses of the

21.6 21.7 21.8

Summary 22

/

540 543

IV:

VA

Elements

/

544

549 551

556 Bi

Group V A Elements

Key Terms

The Nonmetals, Part Noble Gases Carbon and

xiv

526

Problems

560 561 562

Carbon, Silicon, Boron, and the 565

Silicon

Group IV

565

A

22.1

Properties of the

22.2 22.3

Occurrence and Preparation of Carbon and Silicon Carbides and Silicides

22.4

Oxides and Oxyacids of

571

22.5

Sulfur, Halogen,

575

Contents

Elements

C and Si and Nitrogen Compounds of Carbon

565 567

570

Boron

576

22.6

Properties of the

22.7

Boron

22.8

Compounds

Group

III

A Elements

576 577

of Boron

578

The Noble Gases 22.9

581

Properties of the Noble Gases

Summary

Key Terms

/

/

582

Problems

585

23

Metals and Metallurgy

587

23.1

The Metallic Bond

587

23.2

Semiconductors

589

23.3

Physical Properties of Metals

591

23.4

Natural Occurrence of Metals

594

Preliminary Treatment of Ores

23.5

Metallurgy

23.6

Metallurgy: Reduction

23.7

Metallurgy: Refining

The 23.9 The 23.10 The 23.11 The 23.12 The 23.13 The 23.8

Group Group

:

595 598 603

A Metals II A Metals

605

1

608

Transition Metals

612

Lanthanides

617

Group III A Metals of Group IV A Summary / Key Terms Metals of

619 621

Problems

623

24

Complex Compounds

627

24.1

Structure

627

24.2

Labile and Inert Complexes

632

24.3

Nomenclature

633

24.4

Isomerism

24.5

The Bonding

634

Summary

in

637

Complexes

Key Terms

645

Problems

25

Nuclear Chemistry

648

25.1

The Nucleus

648

25.2

Radioactivity

651

25.3

Rate of Radioactive Decay

655

25.4

Radioactive Disintegration Series

661

25.5

Nuclear Bombardment Reactions

662

25.6

Nuclear Fission and Fusion

666

25.7

Uses of Radioactive Nuclides

669

Summary

/

Key Terms

26

Organic Chemistry

26.1

TheAlkanes TheAlkenes TheAlkynes

26.2 26.3

/

Problems

67

675

675 .

-•

679 681

Contents

XV

26.4

Aromatic Hydrocarbons

26.5

Reactions of the Hydrocarbons

685

26.6

Alcohols and Ethers

690

26.7

Carbonyl Compounds

26.8

Carboxylic Acids and Esters

26.9

Amines and Amides

26.10

Polymers

682

-

695

'

698 701 •

Summary

/

Key Terms

703

706

Problems

/

27

Biochemistry

711

27.1

Proteins

711

27.2

Carbohydrates

716

27.3

Fats and Oils

719

27.4

Nucleic Acids

720

27.5

Enzymes

724

27.6

Metabolism

Summary

726 /

Key Terms

Problems

/

728

Appendices

A

International System of Units (SI)

B

Values of

C

Notes on Mathematical Operations

D

Logarithms

E F

Standard Electrode Potentials

G

Answers

xvi

Contents

to Selected

Factors

733

734 740

Equilibrium Constants

Index

732

Some Constants and Conversion

at 25

at 25

C

C

Numerical Problems

742

744 746

759

CHAPTER

INTRODUCTION

1

Chemistry

may be defined as the science that is concerned

with the characterization,

composition, and transformations of matter. This definition, however, adequate. The interplay between the branches of

boundaries between them to be so indistinct that out a

field

and say

"this

is

modern

it is

far

from

almost impossible to stake

Not only do

chemistry.""

is

science causes the

the interests of scientific

but concepts and methods find universal application. Moreover,

fields overlap,

convey the spirit of chemistry, for it, like all science, is a growing enterprise, not an accumulation of knowledge. It is self-generating; the very nature of each new chemical concept stimulates fresh observation and experimentation that lead to progressive refinement as well as to the development of other concepts, in the light of scientific grow th, it is not surprising that a given this definition fails to vital,

human-imposed boundaries. somewhat vague, understanding of

scientific pursuit frequently crosses artificial,

Nevertheless, there

is

a

common,

if

province of chemistry, and we must return to our preliminary definition; a

the

fuller

this book unfolds. Chemistry is concerned with and the structure of substances and with the forces that hold the

understanding should emerge as the composition

structures together. The physical properties of substances are studied since they provide clues for structural determinations, serve as a basis for identification and classification,

and indicate possible uses

for specific materials.

The focus of

probably the chemical reaction. The interest of chemistry extends to every conceivable aspect of these transformations and includes such

chemistry, however,

is

considerations as detailed descriptions of how and at w hat rates reactions proceed, the conditions required to bring about desired changes

changes, the energy changes that

of substances that occur

in

and the quantitative mass

accompany chemical

and

to prevent undesired

reactions, the syntheses

nature and of those that have no natural counterparts, relations

between the materials involved

in

chemical

changes.

1.1

The Development Modern

of

Modern Chemistry

chemistry, which emerged late in the eighteenth century, took hundreds

of years to develop. The story of five

1.

its

development can be divided roughly into

periods

Practical arts

(

—

to

600 B.C.).

The production of metals from

facture of pottery, brewing, baking,

ores, the

manu-

and the preparation of medicines, dyes, and

1

:

drugs are ancient

Archaeological evidence proves that the inhabitants of

arts.

ancient Egypt and they developed are

Mesopotamia were not known.

skilled in these crafts, but

how and when

which are chemical processes, became highly developed during enipiiicaL that is, based on practical experience alone without reference to underlying chemical principles. The Egyptian metalworkers knew how to obtain copper by heating malachite ore with charcoal. They did not know, nor did they seek to know, why the process worked These

arts,

this period.

The development, however, was

and what actually occurred

in the fire.

Greek (600 B.C. to 300 B.C.). The philosophical aspect (or theoretical aspect) of chemistry began in classical Greece about 600 B.C. The foundation of Greek science was the search for principles through which an understanding of nature 2.

could be obtained.

Two

theories of the

Greeks became very important

in the

centuries that followed a.

A

concept that

(earth, air, fire,

all

terrestial

and water)

in

substances are

composed of four elements

various proportions originated with Greek

philosophers of this period.

A theory that matter consists of separate and distinct units was proposed by Leucippus and extended by Democritus in the

b.

called atoms fifth

century

B.C.

Plato proposed that the atoms of one element differ in shape from the atoms

of another. Furthermore, he believed that atoms of one element could be changed (or transmuted) into

atoms of another by changing the shape of the atoms.

The concept of transmutation is also found in Aristotle's theories. Aristotle (who did not believe in the existence of atoms) proposed that the elements, and therefore all substances, are composed of the same primary matter and differ only

in the

forms that

this

To Aristotle,

primary matter assumes.

the

form included

not only the shape but also the qualities (such as color and hardness) that distinguish one substance

from

others.

He proposed

that changes in

form constantly

occur in nature and that all material things (animate and inanimate) grow and develop from immature forms to adult forms. (Throughout the middle ages, it

was believed

that minerals could

grow and

that

mines would be replenished after

minerals were removed from them.) 3. Alchemy (300 B.C. to 1650 A.D.). The philosophical tradition of ancient Greece and the craft tradition of ancient Egypt met in Alexandria, Egypt (the city founded by Alexander the Great in 331 B.C.), and alchemy was the result of the union.

The

early alchemists used Egyptian techniques for the handling of materials to

investigate theories concerned with the nature of matter.

known works on chemical topics) apparatus and descriptions of many laboratory dria (the oldest

Books written

in

Alexan-

contain diagrams of chemical operations (for example, dis-

and sublimation). The philosphical content of alchemy incorporated elements of astrology and mysticism into the theories of the earlier Greeks. A dominant interest of the alchemists was the transmutation of base metals, such as iron and lead, into the noble metal, gold. They believed that a metal could be changed by changing its qualities (particularly its color) and that such changes occur in nature that metals strive for the perfection represented by gold. Furthermore, the alchemists believed that these changes could be brought about by means of a very small tillation, crystallization,

—

amount of a powerful transmuting agent civilization (including

Chapter

1

Introduction

Egypt

in

(later called the

philosophers stone).

Arabs conquered the centers of Hellenistic 640 a.d.), and alchemy passed into their hands.

In the seventh century a.d., the

The Alchemist, painted by the Flemish

artist

David Teniers

in

1648. Fisher Scientific

Company.

Greek texts were translated into Arabic and served as the foundation for the work of Arab alchemists. The Arabs called the philospher's stone aliksir (which was later corrupted into elixir). Arab alchemists believed that this substance could not only ennoble metals by transmuting them into gold but also could ennoble life by curing all diseases. For centuries afterward, the two principal goals of alchemy were the transmutation of base metals into gold and the discovery of an elixir

of life that could make

man immortal by

preventing death.

alchemy was gradually introduced into Europe by the translation of Arabic works into Latin. Most of the translations were made in Spain w here, after the Islamic conquest in the eighth century, a rich Moorish culture was established and flourished. A school of iatrochcmistry, a branch of alchemy concerned with medicine, flourished in the sixteenth and seventeenth centuries. On the whole, however, European alchemists added little that was new to alchemical theory. Their work is important because they preserved the large body of chemical data that they received from the past, added to it, and passed it on to later chemists. Alchemy lasted until the seventeenth century. Gradually the theories and attitudes of the alchemists began to be questioned. The work of Robert Boyle, who published The Sceptical Chymist in 1661, is noteworthy. Although Boyle believed that the transmutation of base metals into gold might be possible, he severely criticized alchemical thought. Boyle emphasized that chemical theory should be derived from experimental evidence. In the twelfth

4.

and thirteenth

Phlogiston (1650 to 1790).

centuries,

Throughout most of

the eighteenth century,

the

phlogiston theory dominated chemistry. This theory, which was later shown to

1.1

The Development

of

Modern Chemistry

be erroneous, was principally the work of Georg Ernst Stahl. Phlogiston (a "fire principle") was assumed to be a constituent of any substance that could undergo

combustion.

Upon combustion, a substance was thought to lose its phlogiston and be reduced to a simpler form. Air was believed to function in a combustion by carrying off the phlogiston as it was released. Whereas we would think of the combustion of wood

following terms:

in the

wood + oxygen

gas (from

air)

—ashes +

oxygen-containing gases

according to the phlogiston theory,

wood

—

ashes

+

phlogiston (removed by

air)

compound composed of ashes and phlowere thought to be rich in phlogiston. The phlogiston theory interpreted calcination in a similar way. The formation of a metal oxide (called a calx) by heating a metal in air is called a calcination: Wood,

therefore,

was beheved

to be a

giston. Readily combustible materials

+

metal

oxygen gas (from

air)

—

calx (metal oxide)

According to the phlogiston theory, a metal is assumed to be a compound composed of a calx and phlogiston. Calcination, therefore, was thought to be the loss of phlogiston by a metal: metal

—

calx +

phlogiston (removed by

air)

The phlogiston theory was extended to explain many other chemical phenomena. The preparation of certain metals, for example, can be accomplished by heating the metal oxide with carbon: calx (metal oxide)

+

carbon

—

metal + carbon monoxide

gas

In a process of this type, the carbon (supposedly rich in phlogiston)

was thought

to replace the phlogiston lost by calcination:

calx

4-

phlogiston (from carbon)

—

metal

One difficulty inherent in the phlogiston theory was never adequately explained. When wood burns, it supposedly loses phlogiston and the resulting ashes weigh less

than the original piece of wood.

On

the other hand, in calcination, the loss

accompanied by an increase in weight since the calx (a metal oxide) weighs more than the original metal. The adherents of the phlogiston theory recognized this problem, but throughout most of the eighteenth century the importance of weighing and measuring was not realized.

of phlogiston

5.

is

Modern chemistry (1790— ).

teenth century

is

Lavoisier deliberately set

He relied on the results of quantitative experimentation (he used the chemical balance extensively) to arrive at his explanations of a number of chemical

chemistry. Antoine Lavoisier, 1743-1794. Smithsonian Institution.

4

The work of Antoine Lavoisier in the late eighmodern chemistry. out to overthrow the phlogiston theory and revolutionize

generally regarded as the beginning of

phenomena.

Chapter

1

Introduction

The law of conservation of mass

states that there

is

no detectable change

in

mass

during the course of a chemical reaction. In other words, the total mass of

all

materials entering into a chemical reaction equals the total mass of all the products

of the reaction. This law

is

implicit in earlier work, but Lavoisier stated

it

explicitly

and used it as the cornerstone of his science. To Lavoisier, therefore, the phlogiston theory was impossible. The roles that gases play in reactions proved to be a stumbling block to the development of chemical theory. When the law of conservation of mass is applied to a combustion or to a calcination, the masses of the gases used or produced in these reactions must be taken into account. The correct interpretation of these processes, therefore, had to wait until chemists identified the gases involved and developed methods to handle and measure gases. Lavoisier drew upon the results of other scientists' work with gases to explain these reactions. In interpreting chemical phenomena, Lavoisier used the modern definitions of elements and compounds (see Section 1.2). The phlogiston theory regarded a metal as a compound composed of a calx and phlogiston. Lavoisier showed that a metal is an element and that the corresponding calx is a compound composed of the metal and oxygen from the air. In his book Traite Elcmentaire de Chimie [Elementary Treatise on Chemistry), pubhshed in 1789, Lavoisier used essentially modern terminology. The present-day language of chemistry is based on the system of nomenclature that Lavoisier helped to devise.

The achievements of scientists since the 790s are described throughout this book. More has been learned about chemistry in the two centuries following Lavoisier than in the twenty centuries preceding him. Chemistry has gradually 1

developed

five principal

classification

is

branches (these divisions, however, are arbitrary and the

subject to criticism):

The chemistry of most of the compounds of carbon. was assumed that these compounds could only be obtained from plant or animal life or derived from other compounds that had been obtained from living material. a.

Organic chemistry.

At one time

b.

it

The chemistry of all the elements except carbon. Some compounds (for example, carbon dioxide and the carbonates)

Inorganic chemistry.

simple carbon

are traditionally classified as inorganic c.

Analytical chemistry.

The

compounds.

identification of the composition, both qualitative

and quantitative, of substances. d.

e.

1.2

The study of the physical and chemical transformations.

Physical chemistry.

slruclure of matter Biochemistry.

The chemistry of

living systems,

principles that underlie the

both plant and animal.

Elements, Compounds, and Mixtures is composed, may be defined as anything and has mass. Mass is a measure of quantity of matter. A body that is not being acted upon by some external force has a tendency to remain at rest or, when it is in motion, to continue in uniform motion in the same direction. This property is known as inertia. The mass of a body is proportional to the inertia

Matter, the material of which the universe that occupies space

of the body.

1.2

Elements, Compounds, and Mixtures

body is not. Weight is the on a body the weight of a given body varies with the distance of that body from the center of the earth. The weight of a body is directly proportional to its mass as well as to the earth's gravitational attraction. At any given place, therefore, two objects of equal mass

The mass of

body

a

is

invariable; the weight of a

gravitational force of attraction exerted by the earth

;

have equal weights.

The ancient Greeks originated the concept that all matter is composed of a number of simple substances called elements. The Greeks assumed that all terrestrial matter is derived from four elements: earth, air, fire, and water. Since heavenly bodies were thought to be perfect and unchangeable, celestial matter was assumed to be composed of a different element, the ether, which later came to be known as quintessence (from Latin, meaning^/?/; element). This Greek limited

theory dominated scientific thought for centuries. In 1661, Robert Boyle proposed an essentially

book The Sceptical Chymist

in his

:

'T

modern

definition of an element

now mean by Elements

.

.

.

certain Primitive

and Simple, or perfectly unmingled bodies; which not being made of any other bodies, or of one another, are the Ingredients of which all those call'd perfectly mixt Bodies are immediately compounded, and into which they are ultimately resolved."

Boyle

made no attempt

to identify specific substances as elements.

He

did,

however, emphasize that the proof of the existence of elements, as well as the identification of them, rested

on chemical experimentation.

Boyle's concept of a chemical element was firmly established by Antoine

Lavoisier in the following century. Lavoisier accepted a substance as an element if it

could not be decomposed into simpler substances. Furthermore, he showed

compound

that a

is

produced by the union of elements. Lavoisier correctly and several

identified 23 elements (although he incorrectly included light, heat,

simple

compounds

in his

list).

known. Of these, 85 have been isolated from natural sources; the remainder have been prepared by nuclear reactions

At

the present time, 106 elements are

(Section 25.5).

Each element

is

assigned a one- or two-letter chemical symbol that has been

decided upon by international agreement. Whereas the differ

from one language

to another, the

symbol does

name of an element may

not. Nitrogen, for example,

and stickstoff in German, but the symbol for nitrogen any language. These symbols are listed in the table of the elements that appears inside the back cover of this book. Most of the symbols correspond closely to the English names of the elements. Some of them, however, do not. The symbols for some of the elements have been assigned on the basis of their Latin names; these elements are listed in Table 1.1. The symbol for tungsten, W, is derived from the German name for the element, is

called azoto in Italian

is

N

in

wolfram.

The 15 most abundant elements in the earth's crust, bodies of water, and atmosphere are listed in Table 1.2. This classification includes those parts of the universe from which we can obtain the elements. The earth consists of a core (which is probably composed of iron and nickel) surrounded successively by a mantle and a thin crust. The crust is about 20 to 40 miles thick and constitutes only about T'^ of the earth's mass. If the entire earth were considered, a list different from that of Table .2 would result, and the most abundant element would be iron. On the other hand, the most abundant element in the universe as a whole is hydrogen, which is thought to constitute about 75''o of the total mass of the universe. 1

Chapter

1

Introduction

Table Latin

Symbols

1.1

of

elements derived from

names

English

Name

Latin

Name

Symbol

antimony

stibium

Sb

copper

cuprum

Cu

gold

aurum

Au

iron

ferrum

Fe

lead

plumbum

Pb

mercury

hydrargyrum

Hg

potassium

kalium

K

silver

argentum

Ag

sodium

natrium

Na

tin

stannum

Sn

Table 1.2 Abundance and atmosphere)

Rank

of the elements (earth's crust, bodies of water, j

Percent by Mass

Symbol

Element

1

oxygen

O

49.2

2

silicon

Si

25.7

3

aluminum

Al

7.5

4

iron

Fe

4.7

5

calcium

Ca

3.4

6

sodium

Na

2.6

7

potassium

K

2.4

8

magnesium

Mg

1.9

9

hydrogen

H

0.9

10

titanium

Ti

0.6

11

chlorine

CI

02

12

phosphorus

P

0.1

13

manganese

0.1

14

carbon

Mn C

15

sulfur

S

0.05

all

0.09

0.56

others

Whether an element finds wide commercial use depends not only upon its abundance but also upon its accessibility. Some familiar elements (such as copper, tin, and lead) are not particularly abundant but are found in nature in deposits of ores from which they can be obtained readily. Other elements that are more abundant (such as titanium, rubidium, and zirconium) are not widely used because either their ores are widespread in nature or the extraction of the elements from their ores

is

difficult

Compounds

or expensive.

are substances that are

The law of definite

composed of two or more elements

in

proposed by Joseph Proust in 1 799) states that a pure compound always consists of the same elements combined in the same proportion by mass. The compound water, for example, is

fixed proportions.

proportions

1.2

(first

Elements, Compounds, and Mixtures

always formed from the elements hydrogen and oxygen in the proportion 11.19% hydrogen to 88.81% oxygen. Over ten thousand inorganic compounds are known,

and over one million organic compounds have been either synthesized or isolated from natural sources. Compounds have properties that are different from the properties of the elements of which they are composed.

An element

or a

compound

is

called a pure suhsfanc;

All other kinds of matter

are mixtures. Mixtures consist of two or more pure substances and have variable compositions. The properties of a mixture depend upon the composition of the_

mixture and the properties of the pure substances that form the mixture. There A heterogeneous mixture is not uniform throughout

are two types of mixtures.

A

but consists of parts that are physically distinct. sand, for example,

is

a heterogeneous mixture.

uniform throughout and

and a

is

A

sample containing iron and homogeneous mixture appears

usually called a solution. Air, salt dissolved in water,

silver-gold alloy are examples of a gaseous, a liquid,

and a soHd

solution,

respectively.

The

classification of matter

is

summarized

in

Figure

see that the only type of heterogeneous matter

is

1.1.

From

the figure

we

the heterogeneous mixture.

The classification homogeneous matter, however, includes homogeneous mixtures and pure substances (elements and compounds). A physically distinct portion of matter that is uniform throughout in comHomogeneous materials consist of only position and properties is called a pii; one phase. Heterogeneous materials consist of more than one phase. The phases of heterogeneous mixtures have distinct boundaries and are usually easily -.

.

discernible.

In the heterogeneous mixture granite, for example,

it

is

possible to identify

pink feldspar crystals, colorless quartz crystals, and shiny black mica crystals.

When

the

number of phases

in

a sample

is

being determined,

all

portions of the

same kind are counted as a single phase. Granite, therefore, is said to consist of three phases. The relative proportions of the three phases of granite may vary from sample to sample. Figure 1.1 notes that both types of mixtures can be separated into their components by physical means, but that compounds can be separated into

their

constituent elements only by chemical means. Changes in state (such as the melting

of a solid and the vaporization of a liquid), as well as changes in shape or state the production of distillation)

may

new chemical

— changes

that do not involve means (such as filtration and components of a mixture, but a substance

of subdivision, are examples of physical changes

species. Physical

be used to separate the

was not present in the original mixture is never produced by these means. Chemical changes, on the other hand, are transformations in which substances that

are converted into other substances.

1.3

The Metric System The metric system of measurement

is

used

in all scientific studies.

treaty signed in 1875, metric conventions are estabhshed

As

a resuh of a

and modified when

From time to time, an international group, and Measures, meets to ratify improvements in the m.etric system. The currently approved International System of Units {Le Systeme International d'Unites, officially abbreviated '^?) is a modernization and simplification of an old system that developed from one proposed by the French necessary by international agreement. the General Conference of Weights

8

Chapter

1

Introduction

MATTER

HETEROGENEOUS by physical

MIXTURES

means

(variable

^

HOMOGENEOUS MATTER

into

composition)

HOMOGENEOUS MIXTURES

PURE SUBSTANCES

by physical

(SOLUTIONS)

means

(variable

into

(fixed composition)

composition)

COMPOUNDS

Figure

by chemical

means

ELEMENTS

^ into

Classification of matter

1.1

Academy

of Science in 1790. Lavoisier was a

member of

the

committee that

formulated the original system.

The International System is founded on seven base units and two supplementary Table .3 and the appendix). The selection of primar>' standards for the

units (see

1

base units is

is

arbitrary.

For example, the primary standard of mass, the kilogram,

defined as the mass of a cylinder of platinum-irridium alloy that

is

kept at the

International Bureau of Weights and Measures at Sevres, France. the years, the primary standards for

new standards appeared

Table

1.3

Base

units

some base

to be superior to old ones.

and supplementary

units of the International

Measurement Base

units

of Units

Symbol

Unit

meter

m

mass

l^ilogram

kg

time

second

s

electric current

ampere

temperature

Icelvin

A K

of

substance

luminous intensity units

System

length

amount

Supplementary

units

Throughout have been changed when

mole

mol

candela

cd

plane angle

radian

rad

solid angle

steradlan

sr

1.3

The Metric System

:

Factor

Abbreviation

Prefix

tera-

T-

giga-

G-

mega-

M-

kilo-

k-

hecto-

h-

deka-

da-

deci-

d-

1

000 000 000 000 X or 10'^ 1

000 000 000 X or 10' '

1

000 000 X or 10" 1

000 X or 10^ 100 X or 10^

centi-

c-

milli-

m-

10 X or 10 0.1

X

0.01

X

or 10^' or 10"or 10"^

X

0.001

0.000 001 X

or 10""

n-

0.000 000 001 X

or 10-*^

pico-

P-

0.000 000 000 001 X

or 10

femto-

f-

0.000 000 000 000 001 X

or

atto-

a-

0.000 000 000 000 000 001 X or 10"'*

micro-

nano-

10^"

Multiples or fractions of base units are indicated by the use of prefixes (see

Table

.4).

1

The base

unit of length, the meter (m),

the distances between

The kilometer

is

adding the prefix

1

km =

A

cities.

1

usually not used to record

is more convenient. name for this unit is obtained by to the name of the base unit:

equal to 1000 meters, and the kilo-

(which means 1000 x

)

m

1000

The centimeter (cm) is a smaller unit than X and centimeter is 0.01 meter

0.01

is

larger unit, the kilometer (km),

the meter.

The

prefix centi-

means

1

,

cm =

0.01

m

Note that the name for the base unit for mass, the kilogram, contains a prefix. The names of other units of mass are obtained by substituting other prefixes for the prefix kilo-. The name of no other base unit contains a prefix. Other SI units, called derived units, are obtained from the base units by algebraic combination. Examples are the SI unit for volume, which is the cubic meter (abbreviated m^) and the SI unit for velocity, which is the meter per second (abbreviated m/s or

Some is

m

s~^).

derived units are given special names. The SI unit for force, for example,

the newton, N. This unit

kg), length (the meter, m),

gives a

1

mass of

N =

1

1

is

derived from the base units for mass (the kilogram,

and time

rules

Some

1

s).

The newton

is

the force that

m/s'^ (see Section 3.1):

the International System has been developed were defined prior to this time do not conform to SI The use of some of these units is, however, permitted.

units that

and are not SI

Chapter

1

kgm/s"

The current terminology of since 1960.

(the second,

kg an acceleration of

Introduction

units.

The liter,

may

for example,

be used

in

which

is

addition to the

defined as

cubic decimeter (and hence

1

official SI unit

is 1000 cm-'), of volume, the cubic meter. Certain

other units that are not a part of SI are to be retained for a limited period of time.

The standard atmosphere (atm, use of

a unit of pressure) falls into this category.

other units that are outside the International System

still

is

The

discouraged.

For example, the International Committee of Weights and Measures considers preferable to avoid the use of the calorie as an energy unit. Not all scientists have adopted SI units, but use of the system appears to be growing. Strict adherence to the International System, however, poses a problem since it eliminates some units that previously have been used widely. Since much of the data found in the chemical literature has been recorded in units that are not SI units, one must be familiar with both the old and the new units. it

1.4 Significant Figures

Every measurement

is

uncertain to

some extent. Suppose, for example, that we If we use a platform balance, we can deter-

wish to measure the mass of an object.

mine the mass

to the nearest 0.

1

g.

An

analytical balance,

capable of giving results correct to the nearest 0.0001

on the other hand,

The exactness, or

g.

is

precision,

of the measurement depends upon the limitations of the measuring device and the skill with which

The record

it is

precision of a

The

it.

used.

measurement

digits in a properly

is

indicated by the

number of

recorded measurement are

figures used to

sifjnificant figures.

These figures include all those that are known with certainty plus one more, which is an estimate. Suppose that a platform balance is used and the mass of an object is determined to be 12.3 g. The chances are slight that the actual mass of the object is exactly 12.3 g, no more nor less. We arc sure of the first two figures (the and the 2); we known that the mass is greater than 12 g. The third figure (the 3), however, is 1

somewhat

inexact.

At

best,

to either 12.2 g or 12.4

12.33

.

.

.

g,

the value

g.

it

If.

tells

us that the true

mass lies closer mass were

for example, the actual

would be correctly recorded

in either

to 12.3 g

than

12.28 ... g or

case as 12.3 g lo three

significant figures.

measurement, we indicate a value which is incorrect and misleading. This value indicates that the actual mass is between 12.29 g and 12.31 g. We have, however, no idea of the magnitude of the integer of the second decimal place since we have determined the value only to the nearest 0. g. The zero does not indicate that the second decimal place is unknown or undetermined. Rather, it should be interpreted in the same way that any other figure is (see, however, in

If,

our example, we add a zero

to the

containing Jour significant figures (12.30

g),

1

rule digit

1

that follows). Since the uncertainty in the

should be the

On

measurement

lies in

the

3, this

last significant figure reported.

the other hand,

we have no

right to

drop a zero

if

it is

significant.

A

value

of 12.0 g that has been determined to the precision indicated should be recorded that way. It is incorrect to record 12. g for this measurement since 12. g indicates a precision of only two significant figures instead of the three significant figures

of the measurement.

The following

rules can be used to determine the proper

figures to be recorded for a

number of significant

measurement.

1.4

Significant Figures

11

:

:

Modern analytical balances capable of giving results to the nearest 0.1 mg. Left: A mechanical single-pan balance. Right: An electronic, digital read-out balance that can be Interfaced with other equipment. Sauter Division of Mettler Instrument Corporation.

Zeros used

1.

to locate the

distance between two points

be expressed as 0.03 3

cm =

0.03

m since

decimal point are not significant. Suppose that the is measured as 3 cm. This measurement could also 1

cm

is

0.01

m

m

Both values, however, contain only one

significant figure.

The zeros

in the

second

value, since they merely serve to locate the decimal point, are not significant.

The

precision of a measurement cannot be increased by changing units. Zeros that arise as a part of a measurement are significant. The number 0.0005030

has four significant figures. The zeros after 5 are significant. Those preceding the

numeral

5 are

not significant since they have been added only to locate the

decimal point. Occasionally,

it

is

difficult to interpret the

number of

significant figures in a

value that contains zeros, such as 600. Are the zeros significant, or do they merely serve to locate the decimal point? This type of scientific

notation (see Appendix C.2).

of 10 employed; the

first

problem can be avoided by using is located by the power

The decimal point

part of the term contains only significant figures.

The

value 600, therefore, can be expressed in any of the following ways depending

upon how

precisely the

6.00 X 10' 6.0 X 10^

6 X 10^ 2.

measurement has been made

(three significant figures)

(two significant figures) (one significant figure)

Certain values, such as those that arise from the definition of terms, are exact. definition, there are exactly 1000 ml in 1 liter. The value 1000

For example, by

may be considered to have an infinite number of significant figures (zeros) following the decimal point.

12

Chapter

1

Introduction

Values obtained by counting

may also

be exact. The

H2

molecule, for example,

Other counts, however, are inexact. The population of the world, for example, is estimated and is not derived from an

contains exactly 2 atoms, not

2.1

or

2.3.

actual count.

At times, the answer to a calculation contains more figures than are significant. The following rules should be used to round off such a value to the correct number 3.

of digits. a.

If the figure following the last

unwanted 3.6247 b.

figures are discarded

is

number

and the

number

is less

is left

than

5, all

the

unchanged:

3.62 to three significant figures

If the figure following the last

or 5 with other digits following

unwanted

to be retained

last

it,

number

to

be retained

the last figure

is

is

greater than

increased by

1

5,

and the

figures are discarded

7.5647

is

7.565 to four significant figures

6.2501

is

6.3 to

two

significant figures

and there are only increased by 1 if it is an odd number or left unchanged if it is an even number. In a case of this type, the last figure of the rounded-ofl" value is always an even number. Zero is considered to be an even number: If the figure following the last figure to be retained

c.

zeros following the

5,

two

the 5

is

discarded and the

3.250

is

3.2 to

7.635

is

7.64 to three significant figures

8.105

is

8.10 to three significant figures

The

5

significant figures

idea behind this procedure, which

many

is

last figure is

is

arbitrary,

is

that

on the average as

values will be increased as are decreased.

The number of significant figures in the answer to a calculation depends upon numbers of significant figures in the values used in the calculation. Consider the following problem. If we place 2.38 g of salt in a container that has a mass of 52.2 g, what will be the mass of the container plus salt? Simple addition gives 54.58 g. But we cannot know the mass of the two together any more precisely than we know the mass of one alone. The result must be rounded off to the nearest the

0.1 g,

which gives 54.6

g.

The result of an addition or subtraction should he reported to the same number of decimal places as that of the term with the least number of decimal places. The answer for the addition

4.

161.032 5.6

32.4524 199.0844

should be reported as 199.1 since the number 5.6 has only one digit following the

decimal point.

1.4

Significant Figures

13

rounded off to the same number of significant figures as is possessed by the least precise term used in the calculation. The

The answer

5.

to

a multiplication or division

'

is

of the multiplication

result

=

152.06 X 0.24

36.4944

should be reported as 36 since the

least precise

term

in the calculation

is

0.24

(two significant figures).

1.5

Chemical Calculations Units should be indicated as an integral part of all measurements. sense to say that the length of an object

m,

5.0

What does

is 5.0.

making

makes little mean 5.0 cm,

It

:

problem solving and reduce the

5.0 ft? Careful use of units will simplify

probability of

this value

errors.

employed in a calculation should undergo the same mathematical operations as the numbers. In any calculation, the units that appear in both numerator and denominator are canceled, and those remaining appear as a part of the answer. If the answer does not have the unit sought, a mistake has been made in the way that the calculation has been set up. Many problems may be solved by the use of one or more conversion factors. A factor of this type is derived from an equality and is designed to convert a measurement from one unit into another. Suppose, for example, that we wish to

The

unit labels that are included in the terms

calculate the that

we need

2.54

If

we

cm =

problem

is

in 5.00 inches (in.).

cm

1.00

in.

The conversion

factor

derived from the exact relation

1.00 in.

divide both sides of this equation by 1.00

2.54

The

number of centimeters to solve the

=

in.,

we obtain

1

factor (2.54 cm/1 .00 in.)

is

equal to

1

since the

numerator and the denominator

are equivalent.

Our problem can be ?

cm =

5.00

stated in the following

in.

Multiplication by the conversion factor that

Since the factor

way:

is

equal to

1

,

this

we derived

solves the

problem

operation does not change the value of the given

quantity. Notice that the inch labels cancel, which leaves the answer in the desired unit, centimeters.

A

second conversion factor can be derived from the relation

2.54

14

Chapter

cm =

1

1.00

in.

Introduction

by dividing both sides of the equation by 2.54 cm: 1.00 in.

"

2.54

cm

This factor, which

also equal to

is

is

1,

the reciprocal of the factor previously

derived and can be used to convert centimeters to inches. For example, the

number of

?

m.

=

inches that equals 20.0

—

/l.OO in.\ 20.0

gfrr

=

cm

can be found

way:

in the following

7.87 m.

\2.54Grr\J

A

two

single equality that relates

can be used to derive two

units, therefore,

conversion factors. The factors are reciprocals of one another. In the solution to a problem, the correct factor to use

one that

the

is

will lead to the cancellation

of the unit, or units, that must be eliminated. Notice that this unit should be

denominator of the factor. some problems requires the use of several factors. If we wish to find the number of centimeters in 0.750 ft. we can state the problem in the following way:

found

in the

The

solution to

?cm = Since 1.00 is

0.750

ft

=

ft

12.0

in.,

of course equal to

we

derive the conversion factor (12.0

in. /l.OO ft),

which

Multiplication by this factor converts feet into inches

1.

but does not complete the solution: 12.0 in.

?cm = 0.750A ,

The

factor needed to convert inches into centimeters

is

(2.54

cm

1

.00 in.),

and

thus

?cm =

0.750

ft-

i?:?^f21''™) = \

Example

\\

.00

.00

Iff.

22.9cn,

J

1.1

Verne had used SI units, what title would he have given his book Twenty Thousand Leagues under the Sea'] Express the answer in the SI unit that will give the smallest number that is greater than One league is 3.45 miles; mile is 1609 m. If Jules

1

1

.

Solution First

we convert

leagues into meters.

The conversion

is

accomplished by the use

of two factors derived from the data given:

?m =

20,000

kagCe

P y

1

-"^^

"^ V

leagtle

)\

^ \

jaite

1

1 1

,000,000

m=

1.1!

x 10«

m

J

1.5

Chemical Calculations

:

Notice that the factors successively convert leagues to miles and then miles to A given factor converts the units in the denominator of the factor to the

meters.

units in the numerator of the factor.

we change

the units of the answer

from the base unit meter to the SI From Table 1.4 we note that a megameter (Mm) is 10^ meters and a gigameter (Gm) is 10'' meters. The magnitude of our answer (10^) is between the two. In order to get an answer that is greater than 1, we convert to megameters: Next,

unit that will satisfy the requirement stated in the problem.

?

Mm

=

1.11

X

10**

bt(^-^*^

1

=

1.11

X 10-

One Hundred and Eleven Megameters under

or.

of the earth

is

only 6.37

Mm

Mm

=

111

Mm

the Sea. (Notice that the radius

!)

Factors can be derived from percentages. Consider, for example, the percentages used to express the composition of the alloy used to five-cent piece.

The "nickel"

is

actually 75.0% copper

Six factors, counting reciprocals, can be derived

Since a percentage

is

the

number of

from

make

and 25.0%

the

American

nickel,

by mass.

this data.

parts per hundred,

it

is

convenient to use

one hundred mass units of the alloy in the derivation of the factors. In 100.0 g of alloy there would be 75.0 g of copper and 25.0 g of nickel. Thus,

exactly

1.

100.0 g alloy

=

75.0 g

Cu

2.

100.0 g alloy

=

25.0 g

Ni

=

25.0

75.0

3.

Each of

gCu

gNi

these relations will yield

—

two factors one the inverse of the other. The problem can be derived from that relation

factor required for the solution of a that involves the pertinent units.

Example

1.2

How many grams of nickel must be added

to 50.0 g of copper to

make

the coinage

Cu (in

the

alloy previously described?

Solution

The problem ?

To

g Ni

=

is

stated in the following

50.0 g

Cu

solve the problem,

inator) to g

factor;

?

it is

g Ni

Chapter

1

=

Ni

(in

way

we need

a conversion factor that relates g

denom-

the numerator). Relation 3 given previously can yield such a

(25.0 g Ni/75.0 g Cu).

The

solution

is

/25 0 g Ni\ 50.0

Introduction

^^z^] =

16.7 g

Ni

Example

1.3

Sterling silver

is

grams of sterling

an alloy consisting of 92.5% Ag and 7.5% Cu. How many silver can be made from 3.00 kg of pure silver?

kilo-

Solution

The problem ?

kg

is

sterling

way

stated in the following

=

3.00 kg

Ag

that we need a factor in which the unit in the denominator is kg Ag. can derive the desired factor from the percentage of silver in sterling silver. Since sterling silver is 92.5% Ag by mass, 100.0 kg of sterling silver contains It is clear

We

kg of

92.5

silver:

100.0 kg sterling

The

=

92.5 kg

Ag

we need, therefore, is (100.0 kg sterling/92.5 kg Ag). Notice that the kg Ag appears in the denominator of this factor and will cancel the unit of the

factor

label

given quantity

?

kg sterhng

=

.

kg

00.0 kg^ /1"^^^^I^^^^ .

3.00

.

sterlingX

(

Frequently, information

is

= )

^

'^^'"'•"S

given in the form of ratios.

The

cost per unit item,

and the number of items per unit mass are examples. The word per implies division, and the number in the denominator is 1 (exactly) unless specified otherwise. A speed of 50 kilometers per hour is 50 km/1 hr. The numerator and denominator of such a ratio are equivalent: the distance traveled per unit time,

50

km =

hr

1

These ratios may be used, therefore, as conversion factors either in the form in which they are given (50 km/1 hr) or in the inverted form (1 hr 50 km). If a ratio is desired as the answer to a problem, the calculation is set up by using the numerator of the ratio as the quantity desired and the denominator of the ratio as the quantity given. If, for example, we wish to find a rate of travel in kilometers per hour, we state the problem in the following way: ?

km =

Example What

is

1

hr

1.4

the speed of a car (in

km

hr)

if it

travels 16

km

in 13

min?

Solution

The problem

is

stated

by writing

1.5

Chemical Calculations

17

?

km =

hr

1

The only information a conversion factor 16

km =

13

that

is

given in the problem and that can be used to derive

is

min

Since the answer is to be expressed in km, the factor that has this unit in the numerator is the one that should be used (16 km/ 13 min). The unit in the denominator of the factor (min), however, will not cancel the unit of the given quantity (hr). This cancellation can be brought about by using another factor, one derived from

=

60 min

The

1

hr

factor derived

from

this equality

is

(60 min/1 hr) since the unit in the

denom-

nator (hr) will cancel the unit of the given quantity. Use of both factors gives

The

car traveled at the rate of 74 km/hr.

Density

is

one type of ratio that is frequently used in chemistry. The density is the mass per unit volume of that substance:

of a substance

Density

mass — volume

=

density

is

;

usually expressed in g/cm^. For pure liquids or solutions, the units

usually employed are g/ml. Since one

1000 ml,

cm^ equals

1

Example

m^.

ml,

and g/cm^

liter is is

1000 cm^ and one

liter

contains

equivalent to g/ml.

1.5

The mass of 10"^

1

the earth

What

is

the

is

5.976 x

mean

10^"^

kg and the volume of the earth

is

1.083 x

density of the earth?

Solution

The problem can be solved by finding

the

number of grams contained

in

one

cubic centimeter: ?

g

=

1

cm^

Since the volume

must be changed 1

cm =

Chapter

1

10"~

is

given in the problem in terms of m^, the cm^ in our set-up By taking the third power of both sides of the equation

to m^.

m

Introduction

we

derive 1

cm^

=

which we use to convert the cm^ units

?g =

states that

=

1.083 X 10-'

5.976 x 10'^ kg

and the factor we derive from

The 1

1^

1

solution

kg

=

The mean 1.00

is

=

kg:

10'* kg^

10^'^^ g:

^/lO-^fTr^\/5.976 X I0'*kg\/I03g\

density of the earth

is

5.518 g/cm^. (In comparison, the density of water

g/cm^)

1.6

The mean

density of the

g.

in

10^ g

Example

10'^

J(^L083 X

used next to find the mass

completed by converting kg into

1000 g

,

is

this relation is

^/10"''Hr^\/5.976 X

_

?g -

our set-up to m-

cnf^l

1

The problem

9

in

What

is

the

moon

is

3.341 g/cm^

and the mass of the

moon is 7.350

x

volume of the moon?

Solution Density relates mass and volume. to find the volume.

Icm^ = The (1

?cm^ =

*

state the

We

are given the

mass of the moon and asked

problem as

7.350 x 10"' g

factor that

cm^/3.341

We

g).

we

use to solve the problem

the reciprocal of the density

is

In this way, the g units will cancel:

7.350 X

10"

^(^/^Y^)

=

2-200 x

10" cm^*

Cancellation will not be indicated in future examples.

1.5

Chemical Calculations

The Conversion Factor Method of Problem Solving 1

.

State the problem. Write the unit in which the answer should be expressed,

an equal

sign,

and that quantity given

in the

problem that

will lead to

a

solution. 2.

Derive a conversion factor

same as the

mation given 3.

in

which the unit

unit of the given quantity. in the

The

factor

If this unit is

the

unit.

Write the conversion factor after the given quantity (written

When this multiplication

the answer will be expressed in the unit in the

is

be derived from infor-

problem or from the definition of a

indicate multiplication. Cancel units.

4.

denominator

in the

may

in step 1) to

is

performed,

numerator of the

factor.

not the one sought, additional conversion factors must be

employed. The unit in the

in the denominator of each factor should cancel the unit numerator of the preceding factor.

5.

Continue the process

6.

Perform the mathematical operations indicated and obtain the answer.

uncanceled unit

until the only

is

the desired unit.

Summary The

topics that have been discussed in this chapter are

The development of chemistry from its roots in the practical arts of ancient civilizations and the theories of

1.

the ancient Greeks.

The classification of matter into pure substances (elements and compounds) and mixtures. Since chemistry is the study of the composition, properties, and transforma-

4.

The abundance of the elements.

5.

The metric system of measurement.

The use of significant figures to indicate the precision of measurements. 6.

2.

tions of matter, this classification 3.

is

of central importance.

The assignment of chemical symbols

7.

A

method of calculating

that

employs conversion

factors. 8.

Calculations involving percentages, rates, and densities.

to the elements.

Key Terms Some of the more important terms introduced in this chapter are listed below. Definitions for terms not included in this list may be located in the text by use of the index.

C

(Section 1.2) A one- or two-letter abbreviation a-^Mgned by international agreement to each element.

ChemistPi (Introduction) The science that is concerned with the characterization, composition, and transformations of matter. Cbi

20

^i.^

L.

(Section 1.2)

A

pure substance that

Chapter

1

is

com-

Introduction

posed of two or more elements in fixed proportions and that can be chemically decomposed into these elements. Conversion factor (Section 1.5) A ratio in which the numerator and denominator are equivalent quantities expressed in different units. A conversion factor is equal to and is used in a calculation to convert the units of a 1

measurement

into other units.

Densitj' (Section 1.5)

Mass

Element (Section

A

decomposed

1.2)

per unit volume.

pure substance that cannot be

into simpler substances.

Law

There is no de1.1) mass during the course of a chemical

of conservation of mass (Section

change

tectable

in

reaction.

physically

portion of composition and

distinct

uniform throughout

is

in

properties.

A

comLaw .2) pound always consists of the same elements combined in the same proportions by mass. of definite proportions (Section

Mass

A

Pliase (Section 1.2)

matter that

(Section

A

.2)

1

Matter (Section

1

pure

measure of quantity of matter.

has mass.

A

decimal system of mea-

Mixture (Section 1.2) A sample of matter that consists of two or more pure substances, does not have a fixed composition, and may be decomposed into its components

by physical means.

known

with certainty

an estimate.

A mixture of two or more pure uniform throughout (homogeneous).

Solution (Section 1.2)

substances that

scientific studies.

is

measurement

measurement. These

figures include all those that are

plus one more, which

Metric system (Section 1.3) surement that is used in all

Digits in a

Significant figures (Section 1.4)

that indicate the precision of the

Anything that occupies space and

1.2)

SI unit (Section 1.3) A unit that is used in the InternaSystem of Units (Le Systenie International d' Unites).

tional

is

.Substance (Section 1.2) An element or a compound. Substances have fixed compositions and properties.

Weight (Section 1.2) The gravitational force of by the earth on a body.

attrac-

tion exerted

Problems* Compare and contrast: (a) law of conservation of mass, law of definite proportions ;(b) mixture, compound; mixture, homogeneous mixture; (c) heterogeneous (d) physical changes, chemical changes; (e) organic chem1.1

biochemistry.

istry,

not an SI

1

which is defined as 10" '° m, nanometers are equal to

unit (A),

How many

unit, (a)

How many

picometers are equal to A'.' (c) The atom is 0.99 A. What is the distance nanometers and in picometers?

A'.' (b)

1

radius of the chlorine in

Give the names of the cicmenls for which the symbols are: (a) Sr, (b) Sb, (c) Al, (d) Au. (e) Ag, (f) Si. (g) Hg, Na, 0) Ne, (k) Ca, (I) C d. (h) He, 1.2

The angstrom

1.9 is

1.10

How many meters tall

One hand

is

is

a horse that stands 15

4.00 inches, and

inch

1

is

hands?

2.54 cm.

(t)

Give the symbols for the following elements: (a) tin, titanium, (c) phosphorus, (d) potassium, (e) copper, cobalt, (g) iron, (h) iodine, (i) chlorine, (j) chromium,

1.11 One furlong is defined as one-eighth of a mile. How many kilometers are there in a six-furlong race? The following relations are exact: inch = 2.54 cm. 12 inches = foot, 5280 feet = mile. Give answer to three signifi-

(k)

magnesium,

cant figures.

1.3 (b)

manganese, (m) lithium,

(I)

Determine the number of significant

1.4

of the following: (e)

0.03040,

(a) 500.0. (b)

500.

(n) lead.

figures in each

(c) 0.05.

(d) 10.072,

6,(XX).(g) 6003. (h) 0.6542.

(f)

+

(d) 5.0

12.34

-I-

+

1.234,(b) 123.4 12.34.(c) 6.524

0.005, (e) 16.0 x

18.75 x 0.375,

(f)

-

5.624.

1.0625/505.

Perform the following calculations and record the answer to the proper number of significant figures: (a) 6.50 X 10"' - 5.603 x 10 ^ (b) 6.50 x 10' + 5.603 x 1.6

10\ 0.03,

(c) (5.5 (f)

X 10

')'. (d)

1

1.12

(3.52 x 25) 91.75, (e) 13.6

+

156.2/0.62.

A

day

in

many hours 1.13

Perform the following calculations and report the answer to the proper number of significant figures: (a) 1.5

123.4

1

1

is

A

is

Venus

is

l.OI

x

How many

this?

10' seconds long.

tun consists of four hogsheads, one hogshead one butt is 126 gallons, one gallon is

0.500 butt,

3.785

one hter

liters,

is

1

.00

dm^.

How many m

1

nanoseconds are

in 10

milliseconds?

(d)

1

.'

kilometer? (c)

(b)

How many

How many

tera-

'

are equal

to 1.00 tun?

The mean distance of the earth from the sun is 10* km, which is defined as one astronomical unit (au). The mean radius of the orbit of the moon around the earth is 0.002570 au. The mean radius of the earth at the equator is 6378 km. The distance from the 1.14

1.496 X

earth to the

moon

is

the equivalent of

around the circumference of the earth 1.7 (a) How many centimeters are in How many kilograms are in milligram

How

earth days?

how many

at the

trips

equator?

Convert the measurements of the following recipe cup sugar, \ cup butter, 2 eggs (which cannot be converted), teaspoon vanilla, 1| cup flour, \ tablespoon baking powder, { teaspoon salt, and \ cup milk. Given: pint = 473.2 ml, pint = 2 cups, cup = 16 tablespoons, tablespoon =

1.15

(for plain layer cake) into milliliters:

1

1

meters are

in

100 micrometers?

The liter is many liters are

1.8

meters are

*

The more

in

1

defined as in

1

1

cubic decimeter,

cubic meter?

(b)

(a)

How many

problems are marked with

1

1

1

1

3 teaspoons.

liter?

difficult

How cubic

asterisks.

The appendix contains answers

to color-keyed problems.

Problems

21

1.16

If the

metric system

is

adopted for everyday use,

plete revolution,

in

1

1

:

;

24 karat (also written carat, but abbreis made of 14 k gold. What percentage of the metal is gold? (b) A gold alloy used in dental work contains 92% Au. How is this alloy rated in terms

Pure gold

1.17

is

A

viated K). (a)

ring

at at

this

speed in m/s?

make a comwhat is the circumference of the earth the equator (in m)? (c) What is the radius of the earth the equator (in m)?

kilograms (not pounds). Given the following: 1 inch = 2.54 cm, 1 pint = 473.2 ml, and pound = 453.6 g, what percentage increase does each of the following changes represent'.' (a) fabric: 1.00 meter 1 .00 liter instead of instead of .00 yard (b) milk 1.00 quart ;(c) meat: 1.00 kilogram instead of 2.00 pound.

and meat

is

takes 24.0 hours for the earth to

(b)

(not quarts),

What

rate of 1039 miles per hour, (a)

fabric will be sold in meters (not yards), milk in liters

Since

it

Under certain conditions, the speed of a hydrogen molecule is 1.84 x 10^ m/s. In one second, however, the molecule undergoes 1 .40 x 10' "collisions with other mol-

1.26

ecules.

On

the average,

tween collisions

(a) in

what

is

the distance traveled be-

meters? (b)

in

micrometers?

The density of diamond is 3.51 g/cm^. What is the volume of a 1.00 carat diamond? (One carat is 0.200 g.)

1.27

of karats of gold?

The density of diamond is 3.51 g/cm^ and of graphite Both substances are pure carbon. What volume would 10.0 g of carbon occupy (a) in the form of diamond? (b) in the form of graphite?

1.28

A

form of white gold called platinum gold is 60.0% Au and 40.0% Pt. (a) How many grams of platinum must be used to make 0.500 kg of alloy? (b) How many grams of the alloy can be made from 0.500 kg of platinum?

1.18

is

1

How many

1.19

grams of zinc must be used with 1.95 kg make a type of brass that is 35.0% Zn and

of copper to 65.0",,

Cu?

200-pound person contains 3.6 g of iron. What the percentage of iron in the body? One pound is 453.6

is

g.

A white brass consists of 60.0% copper, 15.0% nickel, and 25.0% zinc, (a) How many grams of the alloy can be made from 1500 g of Cu, 360 g of Ni, and 500 g of Zn? (b) How many grams of each of the pure metals would be

*1.21

left

over?

.22

1

.

1

in

1

1

1.23

The speed

Convert

this

1.24

feet

=

that

22

1

on a highway

1

.0

g of seawater contains

1

:

1

inch

=

2.54 cm, 12 inches

1

foot,

mile.)

The equator of Mars is 2.13 Mars about The earth

=

x

10'*

its

axis

km is

long.

The

0.240 km/s

How many hours constitute a day on Mars?

around its axis with a velocity such a person standing on the equator is moving at a rotates

Chapter

1

Introduction

very long tube with a cross-sectional area of

is

1.00 g/cm^.)

A

quarter-pound stick of butter measures li^ inch by 1^ inch by 4jl inch, (a) What is the density of butter in g/cm^?(b) Will butter float or sink in water (density = 1.00 g/cm^)?

sity

is

A

of water

*1.32

at the equator. .25

to estimates,

cm^ is filled with mercury to a height of 76.0 cm. At what height would water stand in this tube if it were filled with a mass of water equal to that of the mercury? (The density of mercury is 13.60 g/cm^ and the density

55 miles per hour.

limit

rotational velocity of

1

According

1.00

value to kilometers per hour. (The following

relations are exact

5280

*1.30

'1.31

508 miles per hour. What is this speed One knot is inch = cm/s? (The following relations are exact: 2.54 cm, 12 inches = foot, 5280 feet = mile.)

1

.29

4.0 pg of Au. If the total mass of the oceans is 1.6 x 10'^ Tg, how many grams of gold are present in the

oceans of the earth?

A

1.20

2.22 g/cm^.

The mass of Mars is 6.414 x Mars is 3.966 g/cm^. What

of

10'* is

Tg and

the den-

the radius of

Mars?

Express the answer in the SI unit that will give the smallest number greater than 1.

The density of Venus is 0.9524 times and the volume of Venus is 0.8578 times The mass of earth is 5.975 x 10^^ g. What Venus?

1.33

that of earth, that of earth. is

the

mass of

2.1

STOICHIOMETRY

CH APTE

2

North Whitehead, the philosopher and mathematician, wrote, "All it grows towards perfection becomes mathematical in its ideas." Modern chemistry began when Lavoisier and chemists of his time recognized the importance of careful measurement and began to ask questions that could be answered quantitatively. Stoichiometry (derived from the Greek stoicheion, meaning "element," and metron, meaning "to measure") is the branch of chemistry that deals with the quantitative relationships between elements and compounds in chemical reactions. The atomic theory of matter is basic to this study. Alfred

science as

Dalton's Atomic Theory Credit for the this

prevailed is

first

atomic theory

is

usually given to the ancient Greeks, but

may have had its origins in even among the Greeks. Aristotle (fourth

concept

earlier civilizations.

Two

theories

century b.c.) believed that matter

continuous and, hence, hypothetically can be divided endlessly into smaller

and smaller

particles.

The atomic theory of Leucippus and Democritus

(fifth

century B.C.) held that the subdivision of matter would ultimately yield atoms,

which could not be further divided. The word atom is derived from the Greek word atonws, which means "uncut" or "indivisible." The theories of the ancient Greeks were based on abstract thought, not on planned experimentation. For approximately two thousand years the atomic theory remained mere speculation. The existence of atoms was accepted by Robert Boyle in his book The Sceptical Chymist (1661) and by Isaac Newton in his books Principia {16^1) and Opticks {\104). John Dalton, however, proposed an atomic theory, which he developed in the years 1803 to 1808, that is a landmark in the history of chemistry.

Many scientists of the time believed that all matter consists of atoms, but Dalton went further. Dalton made the atomic theory quantitative by showing that it is possible to determine the relative masses of the atoms of different elements. The principal postulates of Dalton's theory are 1.

Elements are composed of extremely small particles called atoms. All atoms

of the same element are alike, and atoms of different elements are different. 2.

The separation of atoms and the union of atoms occur in chemical reactions. no atom is created or destroyed, and no atom of one element converted into an atom of another element.

In these reactions, is

23

John Dalton, 1766-1844 Smithsonian Institution.

A

3.

chemical

more elements

compound

is

the result of the combination of

atoms of two or

a simple numerical ratio.

in

atoms of a given element have equal atomic masses. elements consist of several types of atoms that differ in mass (see the discussion of isotopes in Section 4.6). All atoms of the same element, however, react chemically in the same way. We can, therefore, work with Dalton's theory by using an average mass for the atoms of each element. In most calculations, no mistake is made by proceeding as if an element consists of only one type of atom with an average mass. Dalton derived the quantitative aspects of his theory from the laws of chemical change. His second postulate accounts for the law of conservation of mass, which states that there is no detectable change in mass during the course of a chemical reaction. Since chemical reactions consist of the separating and joining of atoms and since atoms are neither created nor destroyed in these processes, the total mass of all the materials entering into a chemical reaction must equal the total mass Dalton believed that

Today we know

that

all

many

of all the products of the reaction.

The

third postulate of Dalton's theory explains the law of definite proportions,

pure compound always contains the same elements combined same proportions by mass. Since a given compound is the result of the combination of atoms of two or more elements in a fixed ratio, the proportions by mass of the elements present in the compound are also fixed. On the basis of his theory, DaUon proposed a third law of chemical combination, the law of multiple proportions. This law states that when two elements, A and B, form more than one compound, the amounts of A that are combined in these compounds with a fixed amount of B are in a small whole-number ratio. This law follows from Dalton's view that the atoms in a compound are combined in a fixed proportion. For example, carbon and oxygen form two compounds: carbon dioxide and carbon monoxide. In carbon dioxide, two atoms of oxygen are combined with one atom of carbon, and in carbon monoxide, one atom of oxygen is combined with one atom of carbon. When the two compounds are compared, therefore, the masses of oxygen that combine with a fixed mass of carbon stand in the ratio of 2: 1. The experimental verification of the law of multiple proportions was strong support for Dalton's theory.

which

states that a

in the

2.2

Atomic Weights

A

very important aspect of Dalton's work is his attempt to determine the relative masses of atoms. Water is a compound that consists of 88.8% oxygen and 1.2% hydrogen by mass. Dalton incorrectly assumed that one atom of oxygen is com1

bined with one atom of hydrogen in water. On this basis, the mass of a single oxygen atom and the mass of a single hydrogen atom would stand in the ratio of 88.8 to 1.2, which is approximately 8 to 1. The arbitrary assignment of a mass of to the hydrogen atom would give a relative mass of approximately 8 to the oxygen atom.* 1

1

The formulation that Dalton used for water is incorrect. Actually, one atom is combined with two atoms of hydrogen. The oxygen atom, therefore, has a m.ass that is approximately 8 times the mass of two hydrogen atoms. If one

of oxygen

*

Since the figures that Dalton used for the percent composition of water were very inaccurate, Dalton mass of 7 for the oxygen atom.

actually proposed a relative

Chapter 2

Stoichiometry

hydrogen atom is assigned a mass of 1 two hydrogen atoms would have a combined mass of 2. Thus on this scale, the oxygen atom has a relative mass of approximately ,

8 times 2, or 16.

The relative mass of the carbon atom can be derived in the following way. Carbon monoxide consists of approximately 4 parts by mass of oxygen and 3 parts by mass of carbon. One oxygen atom is combined with one carbon atom in this compound, and therefore the masses of the oxygen atom and the carbon atom must stand in the ratio of approximately 4 to 3. Based on the value of 16 that we derived for the oxygen atom, the approximate relative mass of the carbon atom is 12.

Notice that two types of information are needed to apply Dalton's method

combining ratio by mass of the elements in a compound and the combining ratio by numbers of atoms of these elements. The atomic combining ratios were not available to Dalton and the chemists of his time. Many years passed before a way around this difficulty was discovered and a correct set of relative atomic masses was developed (see Section 4.7). At the present time, they are determined by the use of instruments known as mass spectrometers (see Section 4.6), rather than by means of chemical analysis. Even though Dalton made errors in assigning relative atomic masses, he must be given credit for introducing the concept and recognizing its importance. The relative masses of atoms serves as the cornerstone of chemical stoichiometry. These values are called atomic weights, a term that is not literally correct (since it refers to masses, not weights), but a term that is sanctioned by long usage. Any relative atomic weight scale must be based on the arbitrary assignment of a value to one atom that is chosen as a standard. Dalton used the hydrogen atom In later years, chemists used naturally as his standard and assigned it a value of occurring oxygen as the standard and set its atomic weight equal to exactly 16. Today, a particular type of carbon atom, the carbon- 12 atom, is employed as the standard and assigned a mass of exactly 12 (see Section 4.7). An element may occur in nature as a mixture of various types of atoms that have identical chemical properties but that dilfer slightly in mass. With very few exceptions, a mixture of this type has a constant composition. The atomic weight of such an element is an average value that takes into account the masses of these types of atoms and their relative abundances in nature (see Section 4.7). Several types of carbon atoms occur in nature. The carbon- 12 atom, which is employed as the standard for atomic weights, is the most abundant one. When the percentages and masses of all the types of carbon atoms are taken into account, the average relative mass for naturally occurring carbon is 12.011, which is recorded as the atomic weight of carbon. About three-quarters of the atomic weights of the elements are average values that take into account the several types of atoms that make up the clement. The remainder give the relative mass of a single type of atom. A table of modern atomic weights appears inside the back cover of this book. successfully: the

1

.3

.

Formulas The chemical symbols

that are assigned to the elements (Section

1

.2)

are used to

compounds. The formula are two atoms of hydrogen for every

write formulas that describe the atomic composition of

of water is HjO. which indicates that there one atom of oxygen in the compound. The subscripts of the formula indicate the relative number of atoms of each type that are combined. If a symbol carries no

2.3

Formulas

1 is assumed. The formula of sulfuric acid is H2SO4, which combining ratio of this compound is two atoms of hydrogen to one atom of sulfur to four atoms of oxygen. A molecule is a particle formed from two or more atoms. Some (but not all) compounds occur in molecular form. In these cases, the formula gives the number of atoms of each type in a single molecule of the compound. Formulas of this type are sometimes called molecular formulas. Both water and sulfuric acid are molecular in nature, and both H2O and H2SO4 are molecular formulas. The molecular formula of hydrogen peroxide, HjOj, indicates that there are two atoms of hydrogen and two atoms of oxygen in a molecule of hydrogen peroxide. Notice that the ratio of hydrogen atoms to oxygen atoms (2 to 2) is not the simplest whole-number ratio (which is 1 to 1). A formula that is written using

subscript, the

number

indicates that the

Q

ci-

Na+

whole-number ratio is called the simplest formula, or empirical The molecular formula of hydrogen peroxide is H2O2; the empirical formula is HO. At times the data at hand are sufficient to derive only an empirical the simplest

Figure

2.1

Sodium chloride

IsiiiiissEsUK

crystal lattice

formula.

For some molecular compounds, the molecular and empirical formulas are examples are HjO, H2SO4, CO2, and NH3. For many molecular compounds, however, the molecular and empirical formulas are different. The identical;

molecular formulas

N2H4

QH,

B3N3H,

correspond to the empirical formulas

CH

BNH2

NH,

Notice that the atomic ratio for an empirical formula can be obtained by reducing the atomic ratio of the molecular formula to the lowest possible set of

whole

numbers.

Some compounds is made up

example,

(Chapter

5).

are not

composed of molecules. Sodium

—

of ions

chloride,

for

particles that bear positive or negative charges

The sodium and chloride ions, which are derived from sodium and A sodium chloride crystal is composed of large

chlorine atoms, form a crystal.

numbers of

these ions,

which are held together by the attractions between the

dissimilar charges of the ions (see Figure 2.1).

In the crystal, there is one sodium ion (Na"^) for every one chloride ion (CI ). The formula of the compound is NaCl. The formula does not describe a molecule nor does it indicate that the ions are paired, since no ion m the crystal can be

considered as belonging exclusively to another. Rather, the formula gives the

atoms of each type required to produce the compound. The formula NaCl, therefore, is an empirical formula. The formulas of ionic compounds are derived from the formulas of their ions.

simplest ratio of

Smce

a crystal

is

electrically neutral, the total

charge of the positive ions must

equal the total charge of the negative ions. Consider barium chloride. The formula

of the barium ion

is

Ba^"^,

and the formula of the chloride ion

is

CP. The two

ions

Ba^^

and

2 CI"

form

BaCl,

The arrangement of ions in a BaCl2 crystal is different from that in the NaCl The ions in each of these crystals are present in the ratio indicated by their formulas {one Ba'^ to two CI" in BaCl2, and one Na"" to one C\ in NaCl).

crystal.

26

Chapter 2

Stolchiometry

2.4

The formulas of ionic compounds such as BaClj and NaCl are empirical The formulas of a few ionic compounds, however, can be reduced to simpler terms. Sodium peroxide is such a compound. In sodium peroxide, two sodium ions (Na"^) are present for every one peroxide ion (02~): formulas.

2Na-

and

This formula can be reduced to peroxide.

Na,0,

form

What do we

is

the empirical formula of

sodium

It seems odd to call Na-)02 a an ionic compound and molecules of Na202

formula Na202'?

call the

molecular formula since

NaO, which

NajOj

is

this reason, some chemists prefer to call all molecular formulas The problem encountered with the formula of sodium peroxide is not common. The formulas of most ionic compounds are empirical formulas, and

do not

exist.

For

true fornuilas.

Table 2.1 Elements that occur in nature as diatomic molecules

'

Element

Formula

hydrogen

H2

nitrogen

N2

oxygen

O2

fluorine

F2

chlorine

CI2

bromine

Brj

iodine

I2

the atomic ratios they indicate cannot be reduced.

Formulas are also used to designate the atomic composition of the molecules which some elements are composed. A number of elements occur in nature as of diatomic molecules molecules that contain two atoms joined together. These elements, together with their molecular formulas, are listed in Table 2.1. Some elements are composed of molecules that are formed from more than two atoms. Sulfur molecules, for example, consist of eight atoms and have the molecular formula Sg. The molecular formula of the phosphorus molecules is P4. The empirical formulas of all elements, of course, consist of only the symbols for the

—

elements.

The Mole The atomic weight of fluorine is 19.0 and of hydrogen is 1.0, which means that an atom of fluorine is 19 times heavier than an atom of hydrogen.* If we take 100 fluorine atoms and 100 hydrogen atoms, the mass of the collection of fluorine atoms will be 19 times the mass of the collection of hydrogen atoms. The masses of any two samples of fluorine and hydrogen that contain the same number of atoms

will

stand in the ratio of 19.0 to

1.0,

which

is

the ratio of their atomic

1.0

g of hydrogen, w hich are

weights.

Now

suppose that we take 19.0 g of fluorine and

values in grams nutherically equal to the atomic weights of the elements. Since

must sample of any element that has a the atomic weight of the element will contain

the masses of the samples stand in the ratio of 19.0 to 1.0, the samples

contain the same

number of atoms.

mass

In fact, a

in grams numerically equal to same number of atoms. This number is called Avofjadro's number, named

this

gadro,

who

first

in

honor of Amcdeo Avo-

interpreted the behavior of gases in chemical reactions in terms

number of reacting molecules (see Section 8.8). The value of Avogadro's number has been experimentally determined; to six significant figures it is 6.02205 X 10^^. The amount of a substance that contains Avogadro's number of elementary units is called a mole (abbreviated mol), which is an SI base unit. The mole is defined as the amount of substance that contains as many elementary entities as there are atoms in exactly 12 g of carbon- 12. of the

*

In order to simplify the discussion, following the decimal point.

we have rounded

off these

atomic weights to the

2.4

first

figure

The Mole

27

:

Thus, a sample of an element that has a mass in grams numerically equal to the atomic weight of the element is a mole of atoms of the element and contains Avogadro's number of atoms. The atomic weight of beryllium, for example, is 9.0121 8. A 9.01218 g sample of beryllium is a mole of beryllium atoms and contains 6.02205 X 10^^ beryllium atoms (Avogadro's number).

The atomic weights used to solve a problem should be expressed to the proper number of significant figures. The data given in the statement of a problem determine how precise the answer to the problem should be. The atomic weights used in the solution should be expressed to the number of significant figures that reflects this precision.

Example

2.1

How many

moles of aluminum are there

in 125 g

of Al?

Solution Notice that the answer should be expressed to three significant figures. state the ?

problem

Next,

we

We is

atomic weight of Al

mol Al =

we

125 g Al

derive a conversion factor to solve the problem.

figures the

1

=

mol Al

First,

way:

in the following

is

To

three significant

27.0; therefore,

27.0 g Al

use the conversion factor that has the unit g Al in the denominator since this

the unit that

?

mol Al =

must be eliminated: ^

125 g Al

=

(

4.63

mol Al

V27.0 g Al/

Example

2.2

How many

grams of gold constitute 0.2500 mol of Au?

Solution

The answer should be expressed ?

To

g

Au =

0.2500 m.ol

to four significant figures.

The

problem

Au

is

197.0; therefore,

mol Au = 197.0 g Au must be eliminated in this problem must have this unit in the denominator:

unit that

factor

?

state the

Au

four significant figures the atomic weight of 1

We

g

Au =

0.2500 mol

Au f.^^'^f,^" ^ = y mol Au y 1

Chapter 2

Stoichiometry

is

49.25 g

nwl Au, and

Au

the conversion

Example

2.3

How many carbon atoms are there in a 1.000 carat diamond? Diamond carbon and one carat is exactly 0.2 g.

is

pure

Solution g, which arises from the definition of the carat, is exact, it does not limit the number of significant figures in the answer. The precision of the answer is limited by the value 1.000 carat (four significant figures). The problem

Since the value 0.2

is

?

We

atoms

C =

0.2000 g

C

derive a conversion factor from the atomic weight of

C

(to four significant

figures):

1

molC =

12.01

with the unit g

?

atoms

At

C

C =

in the

0.2000

mol

C =

this factor

atoms

C

this unit will cancel:

gc(jL^) would give an answer expressed

(to four significant figures),

6.022 x

with the unit mol

?

denominator so that

this point, the calculation

Avogadro's number 1

gC

10^-'

in the

atoms

we derive

in

mol C. Using from

a conversion factor

C

denominator so

that

it

will cancel.

Muhiplication by

completes the solution:

C =

molC\

1

0.2000 g

C

12.01

=

/6.022 x

gCyV

10^

'

atoms

C

ImolC

10" atoms C

1.003 X

What about compounds? The formula weight of a substance is the sum of the atomic weights of all the atoms in the formula of the substance. The formula weight of H2O, for example, can be calculated as follows: 2(atomic weight H)

=

=

2.0

O=

16.0

HjO =

18.0

2(1.0)

atomic weight formula weight

The formula weight of BaCl2

is

atomic weight Ba 2(atomic weight CI)

=

2(35.5)

formula weight BaCl, If the

formula

in

= = =

137.3

71.0

208.3

question pertains to a molecular substance and is a molecular may also be called a molecular weight.

formula, the corresponding formula weight

2.4

The Mole

29

A

molecular weight

a molecule.

is

the

sum

of the atomic weights of the atoms that constitute

The formula weight of H2O

stance since the formula

is

also the molecular weight of the sub-

is

a description of the composition of the water molecule.

is not a molecular weight BaClj is an ionic compound and molecules of BaCl2 do not exist. How to distinguish between molecular and ionic substances is a topic that will be discussed

In the case of BaClj, however, the formula weight since

in later chapters.

A

mole consists of Avogadro's number of

substance consists of Avogadro's

entities.

A

mole of a molecular

number of molecules. For

these substances, a

sample that has a mass in grams numerically equal to the molecular weight

is

a

mole of the substance and contains Avogadro's number of molecules. A mole of HjO, therefore, has a mass of 18.0 g and contains 6.02 x 10^^ HjO molecules. Since there are two atoms of

of

H2O

molecules (18.0

H

and one atom of O

g) contains two moles of

H

in a

molecule of H2O, a mole

atoms

(2.0 g)

and one mole of

O atoms (16.0 g). When

mole designation is used, the type of entity being measured must be mole of H atoms contains 6.02 x 10^^ H atoms and to three significant figures has a mass of 1.01 g; a mole of H2 molecules contains 6.02 x 10^^ H2 molecules and has a mass of 2.02 g. For fluorine, specified.

the

A

1

mol F =

1

mol F2

6.02 X

=

10" F atoms =

19.0 g fluorine

6.02 X 10^^ Fj molecules

=

38.0 g fluorine

What about ionic substances? The designation "1 mol BaCl2" means that the sample contains Avogadro's number of BaCl2 formula units the entity specified. One mole of BaClj. therefore, has a mass of 208.3 g, the formula weight of BaCl2. In reality, one mole of BaClj contains

—

1

molBa^^ =

2molCr =

6.02 x

2(6.02 x

10" Ba^^

ions

=

137.3 g barium*

Cr

ions

=

2(35.5) g

10")

CI"

=

71 .0 g chlorine

which together make up 1

mol BaClj

2.5 Derivation of

=

6.02 x

10" BaCl,

units

=

208.3 g BaCl2

Formulas

Data from the chemical analysis of a compound are used

to derive the empirical

formula of the compound. The analysis gives the proportions by mass of the elements that make up the compound. The simplest or empirical formula indicates the atomic proportions of the

various types that

make up

the

compound

— the

relative

numbers of atoms of

compound.

Since a mole of atoms of one element contains the same number of atoms as mole of atoms of any other element, the ratio by moles is the same as the ratio by atoms. The number of moles of each element present in a sample of the compound is readily obtained from the mass of each element present. The simplest

a

*

For

all

practical purposes, an ion has the

Chapter 2

Stolchiometry

same mass as

the

atom from which

it is

derived.

Derivation of Empirical Formulas If the

1.

data are given in terms of percentage composition, base the calcu-

on a 100.0 g sample of the compound. In this instance, the number of grams of each element present in the sample will be numerically equal to the percentage of that element present in the compound. There is no need to find percentages if the data are given in terms of the number of grams of lation

each element present in a sample.

Convert the number of grams of each element present in the sample to number of moles of atoms of each element. The conversion factors needed are derived from the fact that 1 mol of atoms of an element (numerator) is an atomic weight in grams (denominator). 2.

the

Divide each of the values obtained in step 2 by the smallest value. If

3.

number obtained

every

way

in this

number by the same simple integer

not a whole number multiply each

is

in

such a way that whole numbers

will

result.

A ratio by moles of atoms is the same as a ratio by atoms. The whole numbers obtained in step 3 are the subscripts of the empirical formula.

4.

whole-number

ratio

by moles (which is the same as the ratio by atoms) is used The procedure is illustrated in the examples that

to write the empirical formula.

follow.

Example What is

2.4

the empirical formula of a

compound

that contains

43.6% P and 56.4%

O?

Solution

We

assume, for convenience, that we have a sample that has a mass of 100.0

On

the basis of the percentage composition, this sample contains 43.6 g

56.4

g.

P and

gO.

method illustrated in Example 2.1 to find the number of moles atoms in these quantities. To three significant figures, the atomic weight of P is 3 .0 and of O is 6.0 Next,

we use

the

of P atoms and

O

1

?

mol P

=

43.6 g

1

P (^

?

The

mol

O=

56.4 g

O

3j"^"gp ^

=

(^j^^^)

=

'

41

mol P

3.53

mol

O

by atoms is the same as the ratio by moles of atoms. There are, therefore, atoms of P for every 3.53 atoms of O in the compound. We need, however, the simplest whole-number ratio in order to write the formula. By dividmg the two ^jlues^ by th e smaller value, we get ratio

1.41

2.5

Derivation of Formulas

41

1

for P,

for

still

2.50

1.41

do not have

a

each of these values by the empirical formula

Example

=

O,

1.41

We

'

3.53

=1 .00

ratio, but we can get one by multiplying Hence, the simplest whole-number ratio is 2 to 5, and

whole-number 2.

is

P2O5.

2.5

and kola nuts, is a stimulant for the central sample of pure caffeine contains 0.624 g C, 0.065 g H, 0.364 g N, and 0.208 g O. What is the empirical formula of caffeine? Caffeine, which occurs in coffee, tea,

nervous system.

A

1

.261 g

Solution

The

of a chemical analysis are usually reported

results

in

terms of percentages.

However, any mass ratio can be converted into a mole ratio and, in this form, used to derive an empirical formula. There is no need to convert the data given in this example into percentages. We calculate the number of moles of each element present

?

mol

C =

in the

0.624 g

sample:

C

p"^"g

=

0 0520 mol

=

0.065 mol

C

(^

-

?

?

?

mol

H =

0.065 g

H

|

.

\

1

-0

H

g

I

H

y

mol

N =

0.364 g

N

=

0.0260 mol

mol

O=

0.208 g

O ^^?^H = goy

0.0130 mol

) 1

(

N O

V16.0

Division of each of these values by the smallest value (0.0130) gives the ratio

4 mol

C

:

5

mol

H

:

2

mol

N

:

1

and the empirical formula of caffeine,

The molecular formula of formula

if

Example

a

mol

O

therefore,

is

C4H5N2O.

compound can be derived from compound is known.

the empirical

the molecular weight of the

2.6

What is the molecular formula of the oxide of phosphorus that has the empirical formula P2O5 (derived in Example 2.4) if the molecular weight of this compound is 284?

Chapter 2

Stoichiometry

Solution

The value obtained by adding the atomic weights indicated by the empirical formula P2O5 is 142. If we divide this formula weight into the actual molecular weight, we get 284

^

^ "

142

There is

are, therefore, twice as

many atoms of each kind

indicated by the empirical formula.

Example

present in a molecule as

The molecular formula

is

P40io-

2.7

The molecular weight of

C4H5N2O. What

is

cafieine

is

194 and the empirical formula of caffeine

is

the molecular formula of caffeine?

Solution

The formula weight indicated by C4H5N2O

is

97. Since the

twice this value, the molecular formula of caffeine

Example

molecular weight

is

CgH,oN4.02.

2.8

Glucose, a simple sugar, utilized

is

by

is

6.73% H, and

53.3",,

human blood and tissue fluids and is compound contains 4().()"„ C, molecular weight of 80.2. What is the molecular

a constituent of

source of energy. The

cells as a principal

O and has a

1

formula of glucose?

Solution

problem is to calculate the number of moles mole of glucose. We first determine the number of grams of each element in a mole of glucose 80.2 g). Since the compound contains 40.0% C, there are 40.0 g of C in 100 g of glucose, and we use the factor (40.0 g

The most convenient way of each element present

to solve the

in

a

( 1

C/ 1 00 g glucose): 9 ?

g

r C =

1

1

molII glucose

\ In like

'>

.g

80.2 g gluco se p-,

/^l

manner, the number of grams of

H =

1

mol mo

I

lucose gucosei

g

O =

1

mol glucose

we determine

the

40.0 g r

C

=

,

W

gi^^^^g y

mol glucose

72.1 g

C

and of O can be found: 6.73 g

yOO

jOjgglucoseW 1

Next,

H

^O-^ g glu^ose j

?

V

mol glucose /ylOO g glucose/

1

/ \

H

\

^^^IgH

g glucosey

53.3

go

„A

100 e glucose /

number of moles of atoms

that each of these values

represents

2.5

Derivation of Formulas

?

mol

C

72.1

1

gC

mol

{ 12.0 ?mol

H

12.1

?molO

C 6.00

gC5

mol

C

gH

12.0

mol

H

O

6.00

mol

O

96.0 g

These values are the number of moles of atoms of each element present in a mole of glucose molecules. They are also the number of atoms of each type present in a molecule of glucose.

The problem may the analytical data

The molecular formula,

also be solved by

(it is

CHjO) and

first

therefore,

is

CgHjjO^.

determining the empirical formula from

then using the molecular weight to derive the

molecular formula.

2.6

Percentage Composition of Compounds The percentage composition of a compound is readily calculated from the formula of the compound. The subscripts of the formula give the number of moles of each element in a mole of the compound. From this information and from the atomic weights of the elements, we can obtain the number of grams of each element contained in a mole of the compound. The percentage of a given element is 100 times the mass of the element divided by the mass of a mole of the compound. The process is illustrated in Example 2.9.

Example What

is

2.9

the percentage of

Fe

FcjOj calculated

in

to four significant figures?

Solution

One mole 2

of Fe203 contains

mol Fe = 2(55.8) g Fe mol O = 3(16.0) gO

3

= =

1 1 1

.6

g Fe

48.0 g

O

159.6 g

The sum of tage of Fe in

the masses, 159.6 g,

FcjO^

111.6g Fe 159.6

gFe^Os

is

the

mass of one mole of Fe203. The percen-

is

X 100%

=

69.92% Fein Fe^Oj

The percentage composition of a compound is frequently determined by chemical analysis. These data can then be used to find the empirical formula of

Chapter 2

Stoichiometry

compound. Example 2.10

the

is

an

illustration of a

method used

for the analysis

of organic compounds.

Example

2.10

is a compound that contains carbon, hydrogen, and nitrogen. If a 2.50 g sample of nicotine is burned in oxygen, 6.78 g of CO2, .94 g of HjO, and 0.210 g of N2 are the products of the combustion. What is the percentage composition

Nicotine

1

of nicotine?

Solution

Note

that

we

will

work

to three significant figures.

We first calculate

of each element present in the 2.50 g sample of nicotine. formed 6.78 g of CO 2- We ask, therefore, ?

g

C =

6.78 g

The carbon

the quantity

in the

sample

CO2

The conversion factor that we use to solve this problem is the fraction that we mol CO, (44.0 g COj) would use to find the percentage of C in CO,. Since contains mol C (12.0 g C), 1

1

12.0

We

gC =

44.0

gCO^

derive the conversion factor (12.0 g C/44.0 g COj):

7gC = 678ga),(^-J^)=l.85gC The same procedure is used to find the number of grams of hydrogen sample of nicotine. The hydrogen of the nicotine formed 1.94 g H2O. In of H2O (18.0 g) there are 2 mol of H atoms (2.02 g). Therefore,

^gH=1.94gH,o(^i^)

in the 1

mol

= 02,8gH

In a combustion such as the one described, the nitrogen does not combine with oxygen but is evolved as N2. Hence, the sample contained 0.432 g N. The quantity of each element present in the 2.50 g sample is used to determine the percentage composition of nicotine: ' '^^ ^ ^ X 2.50 g nicotine

0.218

gH

100%

=

74.0%

C

in nicotine

X 100%

=

8.72%

H

in

X 100%

=

17.3%

N

in nicotine

nicotine

2.50 g nicotine

0.432 g

N

2.50 g nicotine

These data can be used to find the empirical formula of nicotine, which

2.6

Percentage Composition

of

is

C5H7N.

Compounds

Some

simple stoichiometric problems can be solved by use of the proportions

derived from formulas.

Example

2.11

AgjS, occurs as the mineral argentite, which is an ore of silver. grams of silver are theoretically obtainable from 250.0 g of an impure 70.00% AgjS?

Silver sulfide,

How many ore that

is

Solution

The problem can be ?

g

Ag =

stated as follows

250.0 g ore

we take 100 g of the ore, we will get 70.00 g Ag^S since the ore is 70.00% Ag2S. Notice that the number 100 is exact (it arises from the definition of percent); the If

number 70.00 70.00

and

not. Therefore,

is

gAgjS =

100 g ore

the factor (70.00 g

g

Ag =

Ag^S/lOO g '70.00 g

250.0 g ore

ore) can be derived:

AgjS^

100 g ore

The g ore labels cancel, and at this point we have an answer in g Ag2S. The solution to the problem can be completed by use of the same factor that would be used to find the percentage of Ag in Ag2S. From the formula AgjS, we derive 2

mol Ag

=

1

2(107.9)gAg 215.8

gAg =

mol AgjS

=

247.9

247.9

gAg^S

gAg^S

Therefore,

.,

,

A, =

:50.0 g ore

(Z5^5^)(ii!liA* \

2.7

100 g ore

/

\

247.9 g

Ag ,S

)

.

,

/

Chemical Equations Chemical equations are representations of reactions in terms of the symbols and formulas of the elements and compounds involved. The reactants are indicated on the left and the products on the right. An arrow is used instead of the customary equal sign of the algebraic equation it may be considered as an abbrevia;

tion for the

word

yields.

Chemical equations report the results of experimentation. One of the goals of is the discovery and development of principles that make it possible to

chemistry

Chapter 2

Stoichiometry

predict the products of chemical reactions; careful attention will be given to any such generalizations. All too often, however, the products of a particular set of reactants must be memorized, and any prediction is subject to modification if

experiment dictates. What may appear reasonable on paper is not necessarily what occurs in the laboratory. The first step in writing a chemical equation is to ascertain the products of the reaction in question. Carbon disulfide, CSj, reacts with chlorine, CI2, to produce carbon tetrachloride, CCI4, and disulfur dichloride, S2C12. To represent thjs,

we

write

CS2 + CI2

—

This equation

is

CCI4

+

S2CI2

not quantitatively correct because

it

violates the law of con-

must be as many atoms of each element, combined or uncombined, indicated on the left side of an equation as there are on the right side. Although one carbon atom and two sulfur atoms are indicated on both the left side and the right side of the equation, two chlorine atoms (one CU molecule) appear on the left and six chlorine atoms appear on the right. The equation can servation of mass. There

be balanced* by indicating that three molecules of chlorine should be used for the reaction. Thus,

CS2

+

3CI2

—

CCI4 + S2CI2

The simplest types of chemical equations are balanced by as the following examples will illustrate.

trial

and

error,

When

steam is passed over hot iron, hydrogen gas and an oxide of iron that has the formula Fe304 are produced. Thus,

Fe

+ H2O

—Fe304 +

H,

It is tempting to substitute another oxide of iron, FeO, for Fe304, since this would immediately produce a balanced equation. Such an equation, however, would be without value; experiment indicates that Fe304, not FeO, is a product

of the reaction. Balancing an equation

is

never accomplished by altering the

formulas of the products of the reaction. In the equation for the reaction of iron

and steam, three atoms of Fe and four molecules of H2O are needed the iron and oxygen atoms required for the formation of Fe304:

3Fe + 4H2O The equation

—

Fe304 + H2

now balanced

is

to provide

except for hydrogen, which

may

be balanced as

follows:

3Fe + 4H2O In

much

the

—Fe304 + 4H2

same manner we can balance an equation for the complete comThe products of this reaction are carbon

bustion of ethane (CjHf,) in oxygen. dioxide and water:

C2H6 + O2

* Strictly

—

CO2 + H2O

speaking, the expression

is

not an "equation" until

it is

balanced.

2.7

Chemical Equations

37

To balance the two carbon atoms of CjHf,, the production of two molecules of CO, must be indicated, and the six hydrogen atoms of C2H6 require that three

HjO

molecules of

C2H6 + O2

be produced

—

CO2 +

2

H2O

3

Only the oxygen remains unbalanced; there are seven atoms of oxygen on the right and only two on the left. In order to get seven atoms of oxygen on the left, we would have to take 3^, or ^, molecules of O2:

CjHg + i02

—

2CO2 + 3H2O

Customarily, equations are written with whole number coefficients. the entire equation by two,

2C2H6 + 7O2

2.8

—

we

By multiplying

get

4CO2 + 6H2O

Problems Based on Chemical Equations

A

chemical equation can be interpreted

m

several different ways. Consider, for

example, the equation

—

2H2 + O2

On

2H2O shows that hydrogen

the simplest level this equation

produce water. 2 molecules

On

the atomic-molecular level,

Hj +

1

molecule

O2

—

-

it

reacts with

oxygen

to

states that

2 molecules

H2O

can also be read as

It

2

+

mol H2

since

1

1

mol O2

mol of Hj,

1

—

2

mol

mol of O2, and

H2O 1

mol of

H2O

all

contain the same number

of molecules (Avogadro's number).

The

last

interpretation

is

the

one that enables us to solve stoichiometric

problems. The coefficients of the chemical equation give the ratios, by moles, in 1

which the substances react and are produced. Since 2 mol of H2 react with

mol of O2,

10

H2 produce ofH20.

of

Example

2

mol of H2 would require 5 mol of O2 for reaction. Since 2 mol mol of H2O, the reaction of 10 mol of H, would produce 10 mol

2.12

Determine the number of moles of O2 that are required C2H6 according to the equation

of

2C2H6 + 7O2 Chapter 2

—4CO2 + 6H2O

Stoichiometry

to react with 5.00

mol

Solution

The problem can be

?mol02 =

stated as follows

mol C2H6

5.00

The stoichiometric equation 2

mol CjHg =

From

relationship derived

is (7

?

coefficients of the chemical

mol O2

7

we can derive the conversion factor that we need to solve must have the units mol CiHf, in the denominator, mol O2/2 mol CjHf,)- The solution is

this relationship,

the problem. Since it

from the

is

mol O2 -

this ratio

mol C2H6

5.00

Jmol C2H6

=17.5 mol O2

2

Example

2.13

Chlorine can be prepared by the reaction

+ 4HC1

MnO.,

—MnCl2 +

CI2

+ 2H2O

How many grams of HCl are required to react many grams of CI 2 are produced by the reaction? (a)

with 25.0 g of

Mn02?

(b)

How

Solution

The problem

is

?gHCl = 25.0gMnO2 The stoichiometric

ratio to

pressed in moles. Therefore,

weight of

Mn02

is

be derived from the chemical equation will be exwe convert g MnOj into niol MnOi The molecular -

86.9:

/imolMnOjX

^ ^ -^""°"°^^"»4 86.9gMnoj

r,

The chemical equation 1

mol

Mn02 =

4 mol

gives the relation

HCl

from which we derive the conversion factor

(4

mol HCl/1 mol Mn02):

Mn02\

/

4mol HCl

At

/I

Tx^,

o

mol

\

would give the number of moles of HCl required. mol HCl into g HCl to get our answer. The molecular

this point, the calculation

We must,

therefore, convert

weight of

HCl

g

36.5:

HCl =

25.0 g

^ M„0,

=

42.0 g

HCl

TT^,

o

7

is

-.cr.

/I mol

Mn02\/ 4mol HCl

\ /36.5 g

HC1\

( 8,,^Mnoj (rmdMnO;j (otCI j

2.8

Problems Based on Chemical Equations

The same procedure is used to solve Ihis problem. Grams of Mn02 MnOj. The mole relation from the chemical equation

(b)

is

converted

into moles of 1

is

MnO, =

mol

mol CI2

1

number of moles of CI2 produced.

used to find the

converted into grams of CI2

7

CI, g ^ '

=

2

=

Example

M„0,

25.0 g B

Finally,

moles of CI2

is

:

r'-^o'MnOA/ y^^ g g Mn02; V

1

mol CI.

VTLOgClA

Mn02/ V mol 1

CI2/

20.4 g CI2

2.14

The amount of carbon monoxide

in a

sample of a gas can be determined by the

reaction

+ 5CO

I2O5

—

*

I2

+ 5CO2

sample liberates 0.192 g of

If a gas

Ij,

how many grams

of

CO

were present

in

the sample?

Solution

The

by moles, of the two substances of

relation,

chemical equation. 5

mol

We also

CO =

know 28.0 g

mol

CO =

1

mol

I2

=

is

obtained from the

mol

1

need to

1

interest

It is

254 g

CO

I,

Conversion factors derived from these three relations are needed to solve the problem. The solution is ^

g ^

CO =

8CO\

0,192

f^"'°'COy28-0 ^ tuC-^'M ^ -V254gl2;V mollj /VI molCOy

^

1

some problems, quantities are given for two or more reactants. Suppose, we are asked how much H2O can be prepared from 2 mol of H2 and 2 mol of O2. The chemical equation In

for example, that

2H2 + O2

—

*

2H2O

mol of mol of H2 and

states that 2

ever, 2

can be used. point,

1

Chapter 2

When

mol of O2

all

will react with only

Stoichiometry

mol of Oj

the

will

1

mol of O2. In the problem, how-

More O2 has been supplied than H2 has been consumed, the reaction will stop. At this have been used and J mol of O2 will remain unreacted.

2

are given.

The amount of

supplied limits the reaction and determines

be formed. Hydrogen, therefore,

will

is

how much H2O

called the limiting reactant.

Whenever the quantities of two or more reactants are given in a problem, we must determine which one limits the reaction before the problem can be solved.

Example

2.15

How many 5.00 3

moles of H2 can theoretically be prepared from 4.00 mol of Fe and

mol of H2O? The chemical equation Fe

+

4

—

H,0

for the reaction

is

Fe304 + 4 H2

Solution

The

step

first

the

problem

4.00 3

mol Fe

in

terms of moles specifies

4.00 mol Fe, which

=

3

mol

Fe.

The amount of Fe given

in

is

1.33

mol Fe

times the

amount

4 mol

specifies

which

is

determine which reactant limits the reaction. The chemical

to

is

equation read

specified in the chemical equation.

H2O

and the amount given

in the

The chemical equation is 5.00 mol H2O,

problem

is

5.00

H2O _ H2O "

mol

4 mol

amount

The H2O, therefore, limits amount has been supplied .25 times the amount of H2O specified .33). Since (1 .25 is a smaller number than in the equation has been supplied, only .25 times the amount of Fe specified in the equation can react. The remainder will be left over. The problem is solved times the

specified in the chemical equation.

the extent of the reaction since a smaller proportionate 1

1

1

on the basis of the

Example

H2O

supplied (the limiting reactant):

2.16

How many grams of N2F4 can theoretically be prepared from 4.00 g of and 14.0 g of F2? The chemical equation for the reaction is

2NH3 + 5F2

NH3

— N2F4 + 6HF

Solution

The first step is to determine which reactant limits the reaction. We find the number of moles of each reactant present before reaction. The molecular weight of

NH3

is

17.0

and of F2

is

38.0:

2.8

Problems Based on Chemical Equations

41

NH3 =

?

mol

?

mol ¥2

=

4.00 g

14.0 g

NH3

F2

(yy^^l^)

3g"^"gp' )

=

=

0-368

0-235

mol

NH3

mol

(^

If

mol

2

We

we read

the chemical equation in terms of moles,

NH3 =

compare

of the

number of moles supplied with mol of NH3, which is

these quantities.

The problem

NH,

amount

0.368 mol F2

stated in the relation derived

from the chemical equation. The

is

0.368 mol F2 5

that

mol

the

specified 0.235

2 mol

5

we note

=

0.0736

mol F2

of the amount given in the relation from the chemical equation. The F2, therefore, is the limiting reactant since a smaller proportionate amount has been supplied

Calculations 1.

Based on Chemical Equations

State the problem. Indicate the substance sought (using grams as the

desired unit), an equal sign, and the mass of the substance given (in grams).

Enter the factor that will convert the mass of the substance given into moles of substance given. The conversion factor is derived from the fact that 1 mol of a substance (numerator) is a molecular weight in grams 2.

(denominator). 3.

Enter the conversion factor that

is

derived from the coefficients of the

I

chemical equation and that relates the number of moles of substance sought (numerator) to the number of moles of substance given (denominator). 4.

Enter the factor that

will

convert the

number of moles of substance

sought to grams of substance sought. The molecular weight of the substance sought in grams (numerator) is 1 mol of the substance sought (denominator). 5.

Carry out the mathematical operations indicated

All units should cancel except If more than one quantity 1

.

Calculate the

ceding step 2.

is

to obtain the answer.

grams of the substance sought.

given in the problem:

number of moles of each of

the given reactants (see pre-

2).

Divide each of these values by the coefficient of the chemical equation

that pertains to the substance being considered. 3.

The

I

number obtained in step 2 pertains to the reactant that of the reaction. Use the quantity of this reactant to solve the way previously outhned.

smallest

limits the extent

the

problem

in

j

42

Chapter 2

Stoichiometry

is a smaller number than amount of F2 supplied

0.118).

(0.0736 the

?

g

The

N2F4 =

5

is

=

104:

is

= 104gN2F4

mol N2F4

The solution

It is

mol N2F4

1

The molecular weight of N2F4 1

solved on the basis of

employed and the quantity of N2F4

derived from the chemical equation.

mol F2

is

mol F2

between the quantity of

relation

produced

0.368

The problem

to the

problem

is

Frequently, the quantity of a product actually obtained from a reaction

than the

amount

calculated.

It

may

that part of the reactants react in a

or that not

of the product

all

of product that

would predict

.

is

is

is

less

be that part of the reactants do not react, or

way

different

The

recovered.

from that desired

(side reactions),

percent yield relates the

actually obtained (the actual yield) to the

amount

amount

that theory

(the theoretical yield):

.

.

percent yield

actual yield

=

.

.

.

.

,

x 100%

.

theoretical yield

Example

2.17

If 4.80 g of

what

is

N2F4

is

obtained from the experiment described

in

Example

2.16,

the percent yield?

Solution

The theoretical yield is the result of the calculation of Example 2.16, N2F4. The actual yield is 4.80 g N2F4. Therefore, the percent yield is

!.9

Stoichiometry of Reactions

Many

in

7.65 g

Solution

chemical reactions are carried out

in

aqueous solutions. The quantities

of reactants for a reaction of this type are usually stated in terms of solution

2.9

Stoichiometry ot Reactions

in

Solution

43

concentrations. tion

is

The

The amount of

molaritj,

M,

of a solution

a solute) dissolved in one

one

a substance dissolved in a given

volume of

solu-

called the concentration of the solution.

liter.

of solute

liter

is

the

number of moles of

a substance (called

of solution. Notice that the definition

The value stated as the molarity of that would be present in exactly one

is

based on

a solution pertains to the liter

amount

of the solution. If a sample of

the solution is less (or more) than one liter, the number of moles of solute in the sample is proportionately less (or more) than the numerical value of the molarity. Notice also that the definition of molarity is based on one liter of solution and not on one liter of solvent (which is usually water). Because the definition is made in this way, it is relatively simple to determine the number of moles of

measured volume of a solution on the basis of the molar concentration One liter (which is 1000 ml) of a 3.0 solution contains 3.0 mol of solute, one-half liter (which is 500 ml) contains 1.5 mol of solute, one-quarter liter (which is 250 ml) contains 0.75 mol of solute, and so on. solute in a

M

of the solution.

Example

2.18

How many

AgN03

moles of

are present in 25.0 ml of 0.600

M AgNOj solution?

Solution

The problem ?

mol

is

AgN03 =

25.0

ml

Since the concentration of

0.600 mol

AgNOj =

AgNOj

AgNOj

1000 ml

soFn

in the solution

AgNOa

is

0.600

M,

sol'n

from which we derive a conversion factor to solve the problem

mol

AgNOi =

=

Example

25.0

ml

AgNOj

1000 ml 0.0150 mol

AgNOj

2.19

How many solution of

grams of

NaOH

are required to prepare 0.250

NaOH?

Solution

The solution 0.300 mol

Chapter 2

AgN03 AgNOj sol'n

0.600 mol

soFn

is

0.300

M, and

NaOH =

Stoichiometry

1

liter

therefore

NaOH

sol'n

liter

of a 0.300

M

We find the number of moles of NaOH required ?

NaOH =

mol

0.250

liter

NaOH

"""^

soln ( \1

=

NaOH

NaOH =

40.0 g

1

g ^

NaOH =

is

40.0. Therefore,

0.0750 mol

is

NaOH K^"^^!?^^" = mol NaOH )

•

3.00 g ^

NaOH

/

2.20

of 0.750 M NaOH are required M H2SO4 according to the following equation?

How many 0.150

soFn

NaOH

mol

'

Example

NaOH

to three significant figures

The number of grams of NaOH needed ?

liter

NaOH

0.0750 mol

The formula weight of

to prepare 0.250 liter of solution:

milliliters

HjSO^ +

2

NaOH

—

to react with 50.0

ml of

+ 2H2O

* Na2S04

Solution

We find ?

number of moles of H2SO4

mol H2SO4

From 2

the

=

50.0 ml

=

0.00750 mol

the equation,

NaOH =

mol

we 1

sample:

in the

H,S04 soln

0.150 mol H2SO4 ,nrv.-'Tir^r^--l^ y 1000 ml H2SO4 sol n /

H2SO4

see that

mol H2SO4

Therefore, -)

?

Finally,

of

NaOH =

mol

we

find the

0.00750 mol

H,S04( -JIL"L^^£|^

)

=

0.0150 mol

mol H2S04y

volume of 0.750

M NaOH

NaOH

solution that contains 0.0150

mol

NaOH: ?

ml

NaOH soln = =

0.0150 mol 20.0 ml

NaOH

NaOH

The problem could have been solved

.

1000 ml

NaOH sol'n NaOH

0.750 mol

sofn

in

one step as

illustrated in the following

example.

2.9

Stoichiometry of Reactions

in

Solution

45

Example

A

2.21

soda mint tablet contains

of 0.138

NaHCOj

of

NaHCOj

as

an antacid. One tablet requires 34.5 ml Determine the number of grams

M HCl solution for complete reaction. that

one

NaHCOj + HCl

tablet contains:

—

NaCl + H^O + CO^

Solution

7

g

NaHCO> =

34 5 ml

=

0.400 g

The

first

,

,

V

HCl HCl sol n j(

0.138 mol

/

HCl soPn (

,^00

1

mol 1

NaHCOa

mol HCl

Y )\

84.0 g 1

mol

NaHCOj N NaHCOaJ

NaHCOj

factor (derived

from the molarity of the HCl solution) is used to find in the sample of HCl solution. The second factor

number of moles of HCl

the

(derived from the coefficients of the chemical equation) converts this

moles of last

HCl

into the

number of moles of

NaHCOj

NaHCOj)

from the formula weight of grams of NaHCOj.

factor (derived

NaHCOa

into

number of it. The

that will react with

converts moles of

Additional discussion of solutions and examples that pertain to them

is

found

in Section 10.6.

Summary The 1.

topics that have been discussed in this chapter are

Dalton's atomic theory, which provided the founda-

tion for chemical stoichiometry. 2.

The assignment of atomic weights

to the elements, the

basis of chemical stoichiometry.

centage composition of

8.

An

9.

The

The types of chemical formulas that are used scribe compounds, and their interpretations.

to de-

Avogadro's number of entities that permits calculations involving realistic quantities of elements and compounds. unit consisting of

The assignment of molecular weights and formula

the calculation of the theoretical quantity of

— or

a substance

produced by

—

method to determine the limiting when amounts of two or more reactants are given

chemical reaction, the problem,

(b) a

and

(c)

the calculation of the percent

yield of a chemical process. 10.

5.

(a)

a reactant needed for

in

the use of these

solution of problems based on chemical equations,

reactant

The mole, a

compounds and

calculations.

introduction to the chemical equation.

including

3.

4.

some simple

values in

The use of molarity

substance

to express the concentration of a

in solution.

weights. 11. 6.

How

7.

The

46

formulas can be derived from experimental data. derivation from chemical formulas of the per-

Chapter 2

Stoichiometry

Calculations that involve reactions that take place in

solution.

Key Terms Some

of the more important terms introduced in this

chapter are Usted below. Definitions for terms not

cluded

may be

in this list

in-

located in the text by use of

the index.

Actual yield (Section 2.8) The amount of product actually obtained from a chemical reaction.

Atom

(Section 2.1)

The

smallest particle of an element

that retains the properties of the element.

as exactly

1

The number of

number (Section 2.4) one mole; 6.02205 x 10".

entities

Chemical equation (Section 2.7) A representation of a chemical reaction in terms of the symbols and formulas of the elements and compounds involved. Concentration (Section 2.9) The amount of a substance dissolved in a given quantity of solution or solvent.

Diatomic molecule (Section of two atoms.

that

ratio of

is

A

2.3)

A

molecule consisting

formula for a com-

whole-number

written using the simplest

atoms present

in the

compound

liter

a sub-

of solution.

The amount of substance that consame number of elementary entities as there are atoms in exactly 12g of carbon-12; a collection of Avogadro's number of units.

.Mole (Section 2.4)

A

also called the

;

simplest formula.

Formula weight (Section 2.4) The sum weights of the atoms in a formula.

of

the

atomic

chemical formula for number and type of

a molecular substance that gives the

each atom present

in a

molecule of the substance.

Molecular weight (Section 2.4) The sum of the atomic weights of the atoms that constitute a molecule.

Molecule (Section more atoms.

2.3)

A

particle

formed from two or

Percent yield (Section 2.8) 100% times the actual yield divided by the theoretical yield.

The

St()ichiometr\ (Introduction)

Empirical formula (Section 2.3)

pound

The number of moles of

stance (called a solute) dissolved in one

.Molecular iornuila (Section 2.3)

2.

.\vogadro*s in

Molaritj' (Section 2.9)

tains the

Atomic weight (Section 2.2) The average mass of the atoms of an element, relative to the mass of a carbon- 12

atom taken

The reactant that, based on the chemical equation, is supplied in the smallest stoichiometric amount and hence limits the quantity of product that can be obtained from a chemical reaction.

Limiting reactant (Section 2.8)

ships between the elements

quantitative relation-

and compounds

in

chemical

reactions.

nieorctical yield (Section 2.8)

The maximum amount of

product that can be obtained from a chemical reaction, as calculated by use of stoichiometric theory on the basis of the chemical equation for the reaction.

A particle made up of an atom or a group of atoms that bears either a positive or a negative

Ion (Section 2.3)

charge.

Problems* Dalton's Theory, Atomic Weight 2.1

proportions.

How

Dalton's theory account for 2.2

mass and the law of do they differ'.' How does them?

State the law of conservation of

definite

Compare and

contrast the law of definite proportions

atomic mass to the carbon atom on the basis of the mass of the hydrogen atom being set equal to .00. 1

2.5

List the seven

elements that occur

in

nature as di-

atomic molecules.

and the law of multiple proportions. Use the compounds NO and NO2 in your discussion.

why

2.3

Explain

2.4

Methane has

Show how

*

The more

relative

atomic weights have no

the formula

CH4

and

is

The Mole, Avogadro's Number units.

75.0% carbon.

these data can be used to assign a relative

difficult

problems are marked with

asterisks.

2.6

How many moles and how many molecules are present

in 50.0

g of (a) H,,(b) H20,(c)

The appendix contains answers

H^SO^?

to color-keyed problems.

Problems

47

How many

2.7

atoms are present

each of the samples

in

Only one type of aluminum atom occurs four significant figures, what is the mass of one Al atom?

2.8

in nature.

To

(in

10"

is

international prototype of the kilogram

The

2.10

grams)

of an element has a mass of 9.786 x the atomic weight of the element?

One atom g. What

2.9

is

a

90.000% platinum and 10.000% iridium, (a) How many moles of Pt and how many moles of Ir are present in the cylinder? (b) How many atoms of

cylinder of an alloy that

is

each kind are present? 2.11 One ounce (avdp) is 28.350 g. (a) How many moles and how many atoms of Au are present in .0000 ounce of Au? (b) If gold sells for $650.00 an ounce, how many atoms can you buy for a dollar? 1

The

*2.12

distance from the earth to the sun

km. Suppose

10**

that the

larged into spheres 1.000

were arranged in a reach to the sun? *2.13 sists

Pure gold

is

line

atoms

cm

in

1.000

is

1.496 x

mol were en-

diameter. If these spheres

If a 14.0

of 14.0 parts by mass of

of Cu, how many every one atom of

in

touchmg one another, would they

24 karat.

compound

Cu atoms Au?

karat gold alloy con-

Au and

mass

10.0 parts by

are present in the alloy for

4

for

Determine the molecular formulas of the compounds which the following empirical formulas and molecular

HBS,, 227.81; (b) NaSO,. 174.10; V3S4,281.06;(d) NaPOj, 815.69;(e) CH,, 56.11. (a)

Determine the molecular formulas of the compounds for which the following empirical formulas and molecular weights pertain: (a) COS, 60.07; (b) B,H4, 232.3.1; (c) S,N, 56.25 ;(d) NSF, 195.20; (e) PNCU", 579.43.

2.15

1

What

2.16 is

1.40%

2.17

the empirical formula of a

is

compound

is

that

11.53% B. and 81.07% F?

Li,

H

A

was burned

A sample of a compound that contains only C, H, and S was burned in oxygen, and 15.84 g of CO2, 3.24 g of H,0, and 5.77 g of SO, were obtained, (a) How many moles of C atoms, moles of H atoms, and moles of S atoms did the sample contain? (b) What is the empirical formula of the compound ?(c) What was the mass of the sample that was burned?

2.25

A

sample of /)-aminobenzoic acid (PABA, a in sunscreen products) was burned in oxygen and 17.1 g of CO^, 3.50 g of H,0, and 0.777 g of N2 were obtained. The compound contains carbon, hydrogen, nitrogen, and oxygen, (a) How many moles of C atoms, moles of H atoms, and moles of N atoms did the sample contain'.' (b) What mass of C, H, and N did the sample contain? (c) Based on the mass of the original sample, what mass of O did the sample contain? (d) How many moles of O atoms did the sample contain ?(e) What is the empirical formula of PABA?

*2.26

7.61 g

compound used

Upon

heating 7.50 g of a hydrate

is 74.05% C, 7.46"o H, 9.86"„ O, the empirical formula of quinine?

Putrescine, a product of decaying flesh, is 54.50",, C, 13.72% H, and 31.78% N. What is the empirical formula of putrescine?

2.28

Upon

heating 6.45 g of a hydrate

a vacuum, the water was driven

drous CuSO^ remained. What formula CuS04-.vH,0? *2.29

A

Hydroxyl apatite, an important constituent of bones 39.895% Ca, 18.498"., P, 41.406",, O, and 0.20r'o H. What is the empirical formula of hydroxyl teeth, is

apatite? is

60.00"„ C. 4.48"^ H, and 35.52",, O.

What

the empirical formula of aspirin?

2.21

L-Dopa,

disease,

What 2.22

is

a

drug used

54.82",,

C.

in the

treatment of Parkinson's

5.62",,

32.46",,

O.

The molecular weight of citric acid is 192.13 and the is 37,51",, C, 58.29",, O, and 4.20",, H. What

the molecular formula of citric acid?

2.23

48

The molecular weight of saccharin

2.30

2.31

The process

To

183.18 and the

Chapter 2

in the

yields 17.19 g of AgCl.

three significant figures,

Stoichiometry

To

is

in-

What

vanadium?

.25

2.34

2.35

what percentage of nickel

nickel?

BaCOj,

is

barium?

To four significant figures, what percentage of zircon,

ZrSi04,

1

is

four significant figures, what percentage of the

mineral witherite,

is

zirconium?

What mass

of zinc

is

theoretically obtainable

kg of sphalerite ore that

is

75. 0"^

from

ZnS?

What mass of copper is theoretically obtainable from kg of chalcocite ore that is 25.0% Cu^S?

How many

retically

2.36 is

.v

Percentage Composition

10.0

compound is

the value of

the empirical formula of the chloride of

2.33

H, 7.10",, N. and the empirical formula of L-Dopa? is

in

6.29 g sample of a compound that contains vanais dissolved in water. The addition of

soluble in water.

2.32

Aspirin

is

CuSO^-.vHjO

and 4.82 g of anhy-

a water-soluble silver salt precipitates AgCl, which

2.19

is

off"

dium and chlorine

carbonyl, NilCOlj,

2.20

in a

and 8.63%

2.18

and

CoCUwHjO

CoCU-.vH20?

is

Quinine

What

N.

S,

the molecular formula of saccharin?

vacuum, the water was driven off" and 4.09 g of anhydrous C0CI2 remained. What is the value of .v in the formula

weights pertain: (c)

is

sample of a compound that contains only C and in oxygen and 9.24 g of CO, and 3.15 g of HjO were obtained, (a) How many moles of C atoms and how many moles of H atoms did the sample contain? (b) What is the empirical formula of the compound? (c) What was the mass of the sample that was burned?

2.24

2.27

Formulas 2.1

C, 2.75% H, 26.20% O, 17.50%

45.90"/„

is

and 7.65% N. What

described in Problem 2.6?

grams of xenon and of fluorine are theo-

needed to make

How many

theoretically

1

.000 g of

XeF4?

grams of lithium and of nitrogen are needed to make 5.000 g of LijN?

A

sample of a compound that contains only in oxygen, and 5.28 g of CO, and 2.70 g of H2O were obtained. What is the percentage composition of the compound ? 2.37

C

and

1.74 g

H

was burned

is a compound that contains carbon, hydrogen, and oxygen. The combustion of a 9.50 g sample of the compound yields 29.20 g of CO2 and 10.18 g of HiO. What is the percentage composition of the

*2.38 Cholesterol

compound? The mineral hematite is FejOj. Hematite ore conunwanted material, called gangue, in addition to FejOj. If 1.000 kg of ore contains 0.5920 kg of Fe, what percentage of the ore is FcjO,?

2.46 Pure, dry

NO

gas can be

made by

the following

reaction

3KNO2 + KNO3 + How many make

4NO + 2K2Cr04

Cr203

grams of each of the reactants are needed

2.50 g of

2.47 Determine the number of grams of HI that produced by adding 3.50 g of PI3 to excess water:

PI3

*2.39

+

H2O

3

to

NO?

3

HI

will

be

+ H3PO3

tains

compounds are an undesirable The amount of sulfur in an oil can be determined by oxidizing the S to sulfate, SO4

2.48 A 13.38 g sample of a material that contains some As40e, requires 5.330 g of I2 for reaction according to the chemical equation

*2.40 Sulfur-containing

component of some

As^O^ + 4I2 + 4 H2O

oils.

2

AS2O5

4-

8

HI

,

and precipitating the sulfate ion as barium sulfate, BaSOj, which can be collected, dried, and weighed. From an 8.25 g sample of an oil. 0.929 g of BaSO^ was obtained. What is

the percentage of sulfur in the oil?

What mass of As40e, reacted with the 1 2 What percentage of the sample is AS4O,,'.'

(a) (b)

percentage of the sample 2.49

A 6.55

is

supplied? (c)

What

As?

g sample of a mixture of Na2S03 and Na2S04 in water and heated with solid sulfur. The

was dissolved

Na2S04 does

not react, but the

Na2S03

reacts as follows:

Chemical Equations

Na2S03 + S

NajSjOa

Balance the following chemical equations:

2.41 (a)

VPs +

(b)

B2O3 +

(c)

Bi

-I-

O2

1.23 g of S dissolved to form Na^SiO,. centage of the original mixture was Na2SOj?

and

V3O3 + H,0 B4C + CO

H2

C—

(c)

NO2 + H2O AI2S3 + H2O SiCU + Si

(d)

(NH4)2Cr20,

(e)

Ca3N2 + H2O

(a)

(b)

CS2 +

Balance the following chemical equations:

2.42

HNO3 + NO - Al(OH)3 + H2S

—

Gasohol

of F, and

2F2

N2 + H2O + Cr203

NH3

2.52

is

the chemical equation for the

(C2H(,0)

in

combustion of ethyl alcohol

O2. (The products of the reaction are

CO

and H2O.)

be

-f-

can be made from equation for the

NHj? The

grams of OF, can be made from NaOH ? The equation is

1.60 g

.60 g of

1

Si2CU

Ca(OH)2 +

NH4SCN

per-

NH4SCN + H2S

2NH3

How many

2.51

a mixture of gasoline and ethyl alcohol, (a) Write the chemical equation for the combustion of octane (CgHig, a component of gasoline) in O2. (The products of the reaction are CO 2 and H2O.) (b) Write 2.43

grams of

5.00 g of CS, and 4.00 g of reaction is

Ca(OH)2 + H2C2 CaCj + H2O BaS04 + HNO3 Ba(N03)2 + H2SO4

(d) (e)

How many

2.50

BiiOj

What

2

OF2

NaOH

+ 2NaF

-I-

HjO

Determine the number of grams of B2H,, that can 3.204 g of NaBH4 and 5.424"g of BF3 by

made from

the following reaction:

3NaBH4 + 4BF3 2.53

—

3NaBF4 +

^^z^f,

Determine the number of grams of SF4 that can be 4.00 g of SCI2 and 2.(X) g of NaF by the

made from

following reaction

3SCI2 + 4

NaF

SF4 + S,Cl2 + 4NaCl

Problems Based on Chemical Equations 2.54 (a)

Determine the number of grams of be made from 100.0 g of P40,o:

2.44

P4O10 2.45

Use

+ 6H2O

H3PO4

that can

—•4H3PO4

N2O

If 3.50 g of OPlNHjIs were isolated, what was the percentage yield?

NaN3

-I-

NaOH + NH3

determine the number of grams of NaNHj and of that are required to make 5.00 g of NaNj. (b) How

many grams

of

NH3

OP(NH2)3 + 3NH4CI

(b)

2.55 (a) (a) to

grams of OP(NH2)3 should be se7.(X) g of OPCI3 and 5.00 g of

OPCI3 + 6NH3

the equation

2NaNH2 + N2O

How many

cured from the reaction of NH3? The equation is

are produced?

How many

grams of Ti metal are required to The equation for the reaction

react with 3.513 g of TiC^? is

3TiCl4

+

Ti

4riCl3

Problems

49

How many

(b)

grams of TiCl, should be produced by

How many

2.61

the reaction?

5.000

3.000 g of TiClj are isolated as the product of the reaction, what is the percentage yield?

2.62

If

(c)

grams of KIO3 are needed to prepare

of a 0.1000

liter

How many

M solution of KIO3? NaOH

grams of

are needed to prepare

How many grams of NaNj should be secured from the reaction of 3.50 g of NaNHj and 3.50 g of

M solution of NaOH? 2.63 How many of 3.00 M H3PO4 are required to react with 28.8 ml of 5.00 M KOH? The equation for

NaNO,? The

the reaction

2.56 (a)

equation

is

3NaNH2 + NaNOj If

(b)

1

NaNj

.20 g of

are isolated,

what

is

the percentage

A

which converts the oxides to sulfates. Barium sulfate, BaS04, precipitates from the solution, but sodium sulfate, Na2S04, is soluble and remains in solution. The BaSO^ is collected by filtration and is found to weigh 3.43 g when dried. What percentage of the original sample of mixed oxides is BaO?

A

H3PO4

is

4-

KOH

3

How many

2.64

10.50 g sample of a mixture of calcium carbonate, sulfate, CaS04, is heated to decom-

K3PO4

tion for the reaction

5

How many

2.66

grams of

pose the carbonate:

reaction

CaO + CO,

percentage of the original mixture

A 9.90 g sample of a mixture of CaC03 and heated and the compounds decompose:

CaO

CaCOj

2NaHC03

—

is

2.60

How many

50

CaCl, +

grams of

I,

H,0

are required to react with is

I,

2NaI +

"^^a^S^O^

How many grams of Na2C03 are present in an impure sample of the compound if 35.0 ml of 0.250

2.68 (a)

HCl

* Na2C03 + CO2 + H2O

are required to react with it?

reaction

grams of H2SO4 are needed

M solution of H2SO4?

Chapter 2

to prepare

Stoichiometry

The equation

for the

is

Na2C03 (b)

Solution

375 ml of a 6.00

for the

M

+ CO 2

CaCOj? in

are needed to react

HCl? The equation

M Na,S,03? The equation

2Na,S,03 +

NaHCOj

If the

material

Reactions

HCl

45.0 ml of 0.500

The decomposition of the sample yielded 2.86 g of CO, and 0.900 g of HjO. What percentage of the original mixture

CaO

solid

CaCO,?

*2.59 is

2

How many

2.67

M

+ MnCl2 + KCl + 4H,0

is

CaO +

is

HCl

5FeCl3

with 50.0 ml of 0.600

The CO, gas escapes and the CaS04 is not decomposed by heating. The final mass of the sample is 7,64 g. What

8

M

KMn04 are needed FeClj? The equation is

50

1

M

ml of 0.250

+ KMn04 +

FeCl2

M AgN03 are needed Na,Cr04? The equa-

—- Ag2Cr04 + 2NaN03

How many milliliters of 0.

to react with 15.0

+ 3H2O

is

Na2Cr04 + 2AgN03 2.65

M

ml of 0.750

CaCOj, and calcium

CaCOj—

of 0.500

milliliters

to react with 25.0

mixture of sodium oxide, NajO, and barium oxide, BaO, that weighs 5.00 g is dissolved in water. This solution is then treated with dilute sulfuric acid, H2SO4,

*2.58

milliliters

NaN, + 3NaOH + NH3

yield? '2.57

0.250 hter of a 1.50

2

HCl

2NaCl + CO, + H,0

sample weighed 1.25

is

Na2C03?

g,

what percentage of the

CHAPTER

THERMOCHEMISTRY

3

In the course of a chemical reaction, energy

is

either liberated or absorbed.

Calculations relating to these energy changes are as important as those concerned

with the masses of reacting substances. Thermochemistry

is

the study of the heat

released or absorbed

by chemical and physical changes. In succeeding chapters, calculations involving these energy changes will be frequently encountered. In this chapter, this

.1

type of calculation will be introduced.

Energy Measurement It is

If

common

to think of force as the application of physical strength

the effects of friction are neglected, a

body

in

motion remains

constant velocity, and a body at rest stays at rest

(its

velocity

bodies are pushed, there will be a change in their velocities. velocity per unit time

is

is

— as pushing.

in

motion

at a

zero). If these

The

increase in

called the acceleration.

m/s. Suppose, for example, that we have a body moving at a velocity of Assume that this body is acted on by a constant force, that is, given a sustained second it may be push. The body will move faster and faster. At the end of 1

1

moving at the rate of 2 m/s. At the end of 2 s, its speed may be 3 m/s. If the body picks up speed at the rate of one meter per second in a second, its acceleration is

said to be

A

1

m/s^.

gram body an acceleration of m/s" is not so large as body the same acceleration. The magnitude of proportional to the mass of the body (m) as well as to the

force that gives a one

1

a force that gives a one kilogram a force (F), therefore,

is

acceleration [a) that the force produces:

F ^ ma The

(3.1)

SI unit of force

is

called the

newton (symbol, N) and

is

derived from the

base units of mass (the kilogram), length (the meter), and time (the second):

F = ma 1

1

N = N =

(1

1

kg)(I m/s^)

kgm/s^

Work (W)

is

defined as the force times the distance through which the force

acts(t/):

51

W=

Fd

(3.2)

In the International System, the unit of is

defined as the

work done when a

work

is

force of one

the joule (symbol,

newton

J).

The

joule

acts through a distance

of one meter

W

= Fd

IJ

= = =

(1

N)(l m)

1

N-m

1

kgm^/s-

may

Energy

be defined as the capacity to do work. There are

many forms of

energy, such as heat energy, electrical energy, and chemical energy.

form of energy destroyed.

The

is

converted into another form, energy

SI unit of work, the joule,

ments, including heat measurements.

The

(1818-1889), a student of John Dalton, of

3.2

work always produces

the

is

the unit used for

unit

is

named

in

who demonstrated

same quantity of

When one

neither created nor

is

ail

energy measure-

honor of James Joule that a given quantity

heat.

Temperature and Heat Most is

liquids

expand as the temperature

increases.

The mercury thermometer

designed to use the expansion of mercury to measure temperature. The ther-

mometer

consists of a small bulb sealed to a tube that has a

a capillary tube). the

mercury

is

The bulb and

evacuated, and the upper end of the tube

temperature increases, the mercury expands and

The

narrow bore

(called

part of the tube contain mercury, the space above is

sealed.

When

the

rises in the capillary tube.

named for Anders Celsius, a Swedish astronemployed in scientific studies and is part of the International System. The scale is based on the assignment of 0 C to the normal freezing point of water and 100 C to the normal boiling point of water. When a thermometer is placed in a mixture of ice and water, the mercury will stand at a height that is marked on the tube as 0 C. When the thermometer is placed in boiling water under standard atmospheric pressure, the mercury will rise to a position that is marked lOO'C. The tube is marked between these two fi.xed points to indicate 100 equal divisions, each of which represents one degree. The thermometer is calibrated below 0 C and above 100 C by marking off degrees of the same size. The Celsius scale was formerly called the ccntioradc scale, derived from the Latin words centum (a hundred) and yradus (a degree). On the Fahrenheit temperature scale (named for G. Daniel Fahrenheit, a German instrument maker) the normal freezing point of water is 32 F and the normal boiling point of water is 212 F. Since there are 100 Celsius degrees and 180 (212 minus 32) Fahrenheit degrees between these two fixed points, 5 Celsius omer,

Celsius temperature scale,

is

degrees equal 9 Fahrenheit degrees.

The Fahrenheit temperature scale is not used in scientific work. Conversion of a temperature from the Fahrenheit scale (/p) to the Celsius scale (/() can be accomplished 1.

in the

following way:

Subtract 32 from the Fahrenheit reading.

Fahrenheit degrees the temperature

Chapter 3

Thermochemistry

is

The value obtained

tells

how many

above the freezing point of water.

Celsius (centigrade)

Fahrenheit

scale

scale

normal boiling point

^oo°c

of

water

212°F

100°C = 180°F-

0°C

I


' (or brightness) of the radiation is proportional to the square of the

2.

Figure 4.6 amplitude,

Wavelength, a, of

a

/,

and

wave

amplitude, a^. 3.

vacuum,

In a

all

waves, regardless of wavelength, travel at the same speed,

2.9979 X 10^ m/s. This speed 4.

The frequency of

is

called the speed of

the radiation, v (nu),

the

is

li

4d''>4f^ 5s-

4/)*'

6.r

The notations for lanthanum (Z = 57) and the lanthanides (Z = 58 to 71) pose 4/° Ss'^ 5p^ 5d^ 6s^. One might expect The notation for 5 7La ends 4/^ 55^ 5p^ 5d^ 65^. that the notation for the next element, sgCe, would end

a problem.

.

.

.

.

Instead, the differentiating 5d electron

added

subshell so that the notation for jgCe ends

notations

end

.

.

.

4f' 5s' 5^" 5d^ 6s-

(for

.

.

for .

^jLa

4f-

,gPr),

5.s'^

.

.

.

.

.

falls

5p^

back into the 4f Subsequent

5c/° 65^.

4f^ 5s-

5/j^ 5c/°

65^

(for

6oNd), and so on.

Example

4.6

Write the electronic notation for the electronic configuration of tungsten (Z

=

74).

Solution first

period:

\s'

second period 2s' 2p^ :

third period

:

3^^ 3/;^

fourth period: 4s' 3d^^ 4p^ fifth

5s' 4d''' 5p''

period:

sixth period:

Upon

6s' 4/^'* 5d'^

rearrangement, the notation

Is' 2s' 2/?"

is

3s' 3p'' 3d"' 4s' 4p'' 4d''' 4f'^ 5s' 5p^ 5d* 6s'

The aufbau order cannot be used to interpret processes that involve the of electrons (ionizations). The configuration of the iron atom, Fe, is

loss

Is' 2s' 2/7^ 3s' 3/7" 3d'' 4s',

106

Chapter 4

Atomic Structure

and

that of the

Fe"^ ion

is

Is' 2s' 2p'' 3s' 3p'> 3d^.

Ionization, therefore, results in the loss of the 4s electrons even though 3d elec-

added by the aufbau method. The Fe atom has 26 protons in and 26 electrons. The Fe^^ ion has 26 protons in the nucleus but only 24 electrons. The order of orbital energies is diflferent in the atom and the ion. In general, the first electrons lost in an ionization are those with the highest value of n and /. Electron notations, therefore, should be written by increasing value of n and not by the hypothetical order of filling. trons are the last the nucleus

L15 Half-filled

and

Filled

Subshells

In Table 4.8 the correct electronic configurations of the elements are listed.

The

configurations predicted by the aufbau procedure are confirmed by spectral and

magnetic studies for most elements. There are a few, however, that exhibit

slight

from the standard pattern. In certain instances, it is possible to explain variations on the basis of the stability of a filled or half-filled subshell. these The predicted configuration for the 3d and 4s subshells in the chromium atom (Z, 24) is 3d* 4.v', whereas the experimentally derived configuration is 3^/- 4.v'. Presumably the stability gained by having one unpaired electron in each of the five 3d orbitals (a half-filled subshell) accounts for the fact that the 3d^ 4.s' configuration is the one observed. The existence of a half-filled subshell also accounts for the fact that the configuration for the 4d and 5,v subshells of molybdenum (Z, 42) is 4d^ 5s^ rather than the predicted 4d* 5s^. variations

The

stability

gadolinium (Z,

of a half-filled 4/ subshell 64).

is

evident in the configuration of

The configuration predicted by

inner-transition element ends

.

.

.

4/^

5.v^ 5/?^'

5d"

aufbau method for this The accepted structure of

the

6s^.

4/'^ Ss'^

5p^ 5d^ 6s~. which contains a half-filled 4/ subshell and one 5d electron. For copper (Z, 29), the predicted configuration for the last two subshells is 3d^ 4s^, whereas the experimentally derived structure is 3d^° 4,v'. The explana-

gadolinium, however, ends

.

.

.

tion for this deviation lies in the stability of the 3(/'" 4i

from the completed

3(/

subshell. Silver (Z, 47)

'

configuration that results

and gold (Z.

79) also

have con-

d subshells instead of the (ri — \)d'* ns'^ configuthe rations predicted. In the case of palladium (Z. 46), two electrons are involved only case with a difference of more than one electron. The predicted configuration for the last two subshells of palladium is 4d*^ 5s' the observed configuration is

figurations with completely filled

—

;

4

principle (Section 4.11)

It

is

impossible to

determine, simultaneously, the exact position and exact

momentum

of an electron.

spins.

charge (Section 4.2) 1.6022 x 10" C. The magnitude of the charge on the proton and electron; the proton has a unit positive charge and the electron a unit negative linit

is

drawn

teristic I'. riiKl

in

A substance that behavior that is characof substances that contain unpaired electrons. ce (Section 4.13)

;

inio a

magnetic

(Section 4.10)

field,

Those elements

that are arranged

horizontal rows in the periodic table. i

properties

(Section 4.10)

of the

elements

periodic

physical

functions

of

atomic number.

A quantum

i'iioicn (Section 4.8)

sumber,

]'

//

The values of « are

of radiant energy.

(Section 4.12)

the energy shell of the electron to

Indicates

which the value pertains.

positive integers;

1,

V alence electrons (Section 4.13)

A

110

A

electrons found in

\\a\e

function,

Chapter 4

Atomic Structure

i//

(Section

4.11)

A

solution

to

the

Schrodinger wave equation; it describes an orbital. The square of the wave function, i/'", at any point is proportional to the electron charge density or the probability of finding the electron at that point. Wavelength.

).

2, 3

subatomic particle that has a

atom of an

family element.

similar points I'roton (Section 4.2)

The

the outermost shell in the ground state of an

The chemical and are

charge.

radiation.

(Section 4.8) The distance between two on two successive waves of electromagnetic

Problems* Nuclear Atom, Atomic Symbols 4.1

J.

J.

Thomson determined

the ratio of charge to

Why

was the method he used

mass of an electron

{I'lm).

Complete

4.11

Which

4.2

field?

positive ion

Why?

The proton

4.3

10"'^

or

(a)

cm and

is

a

deflected

is

K

Ne^

(b)

Ne^

or

1.67 x

10

(a)

What

19

41

40

90

Protons

Neutrons

Electrons

126

40 82

78 46 66

46

55

Ne-^

"^g.

A

Mn

an electric

in

Pb Xe

thought to have a radius of 1.30 x

mass of

132

is

g/cm'? Assume the proton sphere. V, is given by

the density of the proton in to be spherical.

more

Z

Symbol

unable to yield either value separately?

the following table:

48

The volume of a

V = f 7tr \ where r is the radius of the sphere, basketball has a radius of 12.0 cm. What would be

the formula

A

(b)

mass of a basketball if proton? Could you lift it? the

it

Isotopes, Atomic Weights

had the same density as a

Describe the three types of rays emitted by radio-

and

contrast:

(a)

isotope,

isobar;

atomic mass, mass number; (c) nucleon, neutron; atomic number, atomic weight.

(b)

4.4

Compare

4.12

(d)

active substances that occur in nature.

Each

4.13

alpha particle scattering experiments. Compare the number of wide-angle deflections observed for a Cu foil with the number observed for an Au foil of the same thickness.

Rutherford used several

4.5

metal

The approximate radius of a

*4.6

foils

nucleus,

his

in

/,

is

atom

(n) listed

A

given by

= /l"^(1.3 x 10"'-* cm), where A is the mass number of the nucleus. The atomic radius of the 13AI atom is approximately 143 pm. If the diameter of an ,]A1 atom were l.(K)km (which is 0.621 mile), what would be the diameter of its nucleus (in cm)? the formula r

Use the data given

*4.7

in

Problem 4.6 to calculate the

percentage of the total volume of the aluminum atom

occupied by the nucleus. The volume of a sphere,

that

is

V.

given by the formula

is

V =

^nr^, where r

is

the radius

(a)

capital letter in the following

beside

26p, 28n 24p, 30n

C

23 p, 27 n

Classify the

stands for an

it:

B

them

list

number of protons (p) and neutrons

that contains the

D E F

atoms

24 26

p,

23

p,

p,

26 n 27 n 30 n

into groups of isotopes, (b) Classify

into groups of isobars,

(c) Write the atomic symbol, complete with atomic number and mass number, for each atom.

Which is larger: an atomic mass unit based on the mass of the '^F atom set at exactly 19, or the current standard? Only one isotope of fluorine occurs in nature,

4.14

19p

of the sphere.

The element vanadium consists of two isotopes: I3V, which has an atomic mass of 49.9472 u, and 23V. which has an atomic mass of 50.9440 u. The atomic weight of vanadium is 50.9415. What is the percent abundance of each of the two isotopes? 4.15

(a) Describe .the composition of the 'f.Ba atom, Give the .symbol that designates the atom that contains 83 protons and 126 neutrons. 4.8

'

(b)

Describe the composition of the 'vsPt atom, (b) Give the symbol that designates the atom that contains 30 protons and 34 neutrons 4.9 (a)

4.10

Complete the following

Symbol P

Z 15

A

37

Neutrons

Electrons

37

85 140

Fe^^

80

82

30 35

difficult

of two

Lithium occurs

in

isotopes:

nature as a mixture of iLi atoms

(mass. 6.015 u) and jLi atoms (mass, 7.016

122

Ce

consists

fsRe, which has an atomic mass of 186.956 u. The atomic weight of rhenium is 186.207. What is the percent abundance of each of the two isotopes? 4.17

31

Hg

The more

Protons

The element rhenium

'75Re, which has an atomic mass of 184.953 u, and '

48

Ti

*

table:

4.16

44 problems are marked with

36

asterisks.

weight of lithium

is

6.941

.

What

is

u).

the percent

The atomic abundance

of each of the two isotopes? 4.18 If an element consists of 60.10"„ of atoms with a mass of 68.926 u and 39.90"„ of atoms with a mass of 70.925 u, what is the atomic weight of the element?

The appendix contains answers

to color-keyed problems.

Problems

111

an element consists of 90.51°o of atoms with a mass of 19.992 u, 0.27% of atoms with a mass of 20.994 u, and 9.22% of atoms with a mass of 21.990 u, what is the

4.19 If

Compare and

contrast: (a) wavelength, frequency;

wavelength, amplitude

spectrum;

ground

(d)

;

(c)

state,

spectrum, continuous excited state; (e) photons,

Which

radiation

is

more

energetic: (a) infrared ra(c)

a

radiowave or a microwave?

A compound

maximum

used in sunscreens, /7-aminobenozic absorbs ultraviolet radiation with

PABA)

acid (called

absorption occurring at 265 nm.

What

The speed of

light,

frequency?

corresponding

(a)

10**

What

4.23

the

is

c,

is

m/s.

with

light, f, is

6.63 X

quantum

the frequency and energy per

is

nm

red light with a wavelength of 700

light

and

of:

violet

(b)

wavelength of 400 nm? The speed of 3.00 X 10^ m/s and Planck's constant, /;, is a

10-^-*J

s.

and energy per quantum of a 1 .00 pm and (b) microwave with a wavelength of 0.100 cm? The speed of is Ught, c. is 3.00 X 10* m/s and Planck's constant.

What

4.24 (a)

a

10

ray with a wavelength of

'•'J-

What

4.25

the frequency

is

gamma

6.63 X

is

s.

the wavelength

and energy per quantum

a radiowave with a frequency of 9.00 x

(a)

lO'/s

of:

and

an X ray with a frequency of 1.20 x 10"*/s? The speed of light, c. is 3.00 x 10* m/s and Planck's constant is6.63 X 10~^*J s.

(b)

How many

4.26

of

photons are in a 1.00 x 10 '" J signal with a wavelength of 500 nm? The speed of 10** m/s and Planck's constant, /;, is is 3.00 X

light

light, c,

6.63 X 4.27

10'"

Voyager

J 1

sent back pictures of Saturn

How

reach the earth?

The speed of

and 1.000 mile

1609 m.

is

long did

from a

dis-

it

take a signal to

light, c. is

3.00 x 10* m/s

4.30

irradiated by light.

It

requires a photon with a

minimum

energy of 7.58 x 10" J to eject an electron from silver. What frequency and wavelength of light correspond to this value? The speed of light, c. is 3.00 x 10* m/s and Planck's constant, /?, is 6.63 x 10"^'* J s.

The

According to Bohr, what

is

the source of the light

What is the wavelength of the spectral line that corresponds to an electron transition from the /? = 4 level to the n = 1 level in the hydrogen atom?

4.31

What is the wavelength of the spectral line that corresponds to an electron transition from the n = 4 level

4.32

to the

=

12

The

3 level in the

hydrogen atom?

hydrogen in the visible region correspond to electron transitions to the h = 2 level from spectral lines of

higher levels.

sponds 4.34

A

What

to the 410.2

handbook

the electron transition that corre-

is

nm

spectral line?

identifies the persistent spectral lines

of

hydrogen as having wavelengths of 121.6 nm, 486.1 nm, and 656.3 nm. (a) In what region of the electromagnetic spectrum does each occur? (b) Identify the electron transition that corresponds to each line.

The Pfund series of spectral lines of hydrogen occurs wavelengths of from 2.279 ^/m to 7.459 /im. What are the corresponding electron transitions?

*4.35 at

*4.36

to remove the most from an isolated atom in its ground

The amount of energy required

loosely held electron state

called the

is

first

ionization energy of the element

under consideration, (a) Calculate the frequency of the hydrogen spectral line that corresponds to an electron transition from « = ooton = 1. (b) Calculate the energy of this transition. Planck's constant is 6.626 x 10"^"* J What is the first ionization energy of hydrogen

(c)

s.

in

The Bohr theory successfully interprets the spectra of species that have only one electron. For the Li^"^ ion, the two lines of lowest energy that correspond to electron level have wavelengths of 13.50 nm transitions to the/; = and 11.39 nm. (a) What is the value of K in the relation-

*4.37

1

The

photoelectric effect consists of the emission of electrons from the surface of a metal when the metal is

*4.29

s.

kJ/mol?

s.

tance of 8.0 X 10" miles.

*4.28

6.63 x 10"^'* J

Atomic Spectra

4.33

3.00 X

is

emitted by a substance in a spectroscope?

diation or microwaves, (b) yellow light or blue light,

4.22

frequency and corresponding wave-

line

light ray.

4.21

minimum

are the

Planck's constant,

Electromagnetic Radiation

(b)

how much energy (c) What

required to release theelectron from the metal?

length of light required to release an electron from potassium? The speed of light, c, is 3.00 x 10* m/s and

atomic weight of the element?

4.20

to the^ejected electron as kinetic energy, is

photoelectric effect consists of the emission of

electrons from the surface of a metal

when

the metal

is

400 nm falls on the surface of potassium metal, electrons, each with a kinetic energy of 1.38 x 10 " J, are ejected, (a) What is the energy of a 400 nm photon ?(b) If .38 x 10 J of the energy of the incident photon is imparted irradiated by light. If light with a wavelength of

=

ship: V

A'[(l

//;,-)

-

(

1 //!^^

)]

? (b)

of the next line in the series?

What is the wavelength What are the wave-

(c)

lengths of the two spectral lines that correspond to transitions

from the n =

3

and n = 4

levels to the

ii

=

2 level?

Based on the answer to Problem 4.37a and the in Example 4.3, derive an equation relating in the equation v = K[(\/nf) - (l/"o)] and the atomic number, Z, of the one-electron species under consideration, (b) Predict the value of A' for He^.

*4.38 (a)

equation used the value of A'

Periodic

Law

1

112

Chapter 4

Atomic Structure

4.39

Compare

the elements of (a) a period

and

(b) a

group.

What change

4.40

What

4.41

make

did Moseley

in

Mendeleev's pe-

have

is

the origin of

X

=

/

many

law?

riodic

as

1

one of

their

electrons have m,

quantum numbers?

= +

as one of their

1

numbers? (c) How many electrons have of their quantum numbers?

rays?

Moseley found that the frequency, v, of a characof the X-ray spectrum of an element is related to the atomic number, Z, of the element by the formula ^'V = a(Z - b) where a is approximately 5.00 x 10^/^s and b is approximately .00. What is the atomic number of an element for which the corresponding line in the X-ray spectrum occurs at a wavelength of 0. 164nm?

/«,

how

(b)

quantum

= +2

as one

*4.42

teristic line

1

What

Electronic Configurations 4.54

Give the orbital diagram for the electronic configuraand the corresponding electronic notation.

tion of 2bFe 4.55 State

Hund's

How have magnetic measurements

the element?

is

4.56 Identify the

What

*4.43

rule.

helped to establish this rule?

is

the wavelength of the line in the X-ray spec-

trum of 24Cr that conforms Problem 4.42?

the formula given in

to

3p\{b) 3s- 3/?^ 3d^ 4s\ 3d'°4s\(e) 4s^4p*'.

(a) 3s-

4.57 Identify the

Quantum Numbers

atoms that have the following ground-

state electronic configurations in their outer shell or shells:

atoms

(c) 3.s- 3/7*'

45-. (d) 3s^ 3p^

that have the following ground-

state electronic configurations in their outer shell or shells: (a) 5s- 5/7^ (b) 4s^ 4p'' 4^/'" 5s\ (c) 4^^ 4p'' 4^/'" 4/ 5s^ '

4.44 List the four

and

fies,

quantum numbers,

state the values that

=

4.45 Describe the « bitals,

and

moving

at

may

each

level in

what each

iden-

assume.

is

-V

65-. (d)

4.58

terms of sublevels, or-

electrons.

What

*4.46 (a)

4

tell

5,v- Sp""

5d'

6.s-.(e) 5.v-

Which of the atoms

listed in

5/)*" 6.v-.

Problems 4.56 and 4.57

are paramagnetic? 4.59 Write the notations for the ground-state electronic

the de Broglie wavelength of an electron

one-tenth the speed of light?

What

(b)

the

is

de Broglie wavelength of a 70.0 kg person jogging at the rate of 2.70 m/s? The mass of an electron is 9.11 x 10"^** g, the speed of light is 3.00 x 10" m s, Planck's constant is 6.63 x 10^^'* J s, and 1 J is kg m"^ s^. 1

configurations of the following atoms:

^^Sm.

(c)

jgNi. (e) -,Lu.

(d)

4.60 State the

atoms

the

number of unpaired electrons in each of Problem 4.59. Which are paramagnetic?

listed in

Write the notations for the ground-slate electronic

4.61

configurations of the following atoms:

Sketch boundary-surface diagrams for

4.47

id

2p.

\s,

and

j^Sr. (b) soSn,

(a)

«oHg.

(f)

sa^e, (b) yj'J-

(a)

,3V.(d).,^W,(e) ,4Xe.(l) -oYb.(g) ^„Zr."

(c)

orbitals.

4.62 Slate the

What

4.48

is

Bohr radius, ci„. in terms of wave mechanics?

the interpretation of the

terms of the Bohr theory and

in

the

atoms

4.63

Give the values for all four quantum numbers for each electron in the ground state of the neon atom. Use positive values of m, and first.

4.49

an atom may be characterized by a set of four quantum numbers. For each of the following parts, tell how many different sets of quantum numbers are possible such that each set contains all of the values

Each electron

4.50

in

number of unpaired electrons in each of Problem 4.61 Which are paramagnetic?

listed in

.

Write the notations for the ground-state electronic

configurations of the

«,Pb-'.

(c)

,,Sc^'.

(d)

following

,^Ct'\

ions: (e)

,,S-

(a) .

47Ag*,

(f)

(b)

.,,r.

number of unpaired electrons in each of the Which are paramagnetic?

4.64 State the

ions listed in Problem 4.63.

how

4.65 Describe

the ground-stale electronic configura-

0;

compare to one another and how the ground-slate electronic configurations of the elements of a group are related.

an atom may be characterized by a set of four quantum numbers. For each of the following parts, tell how many diflerent sets of quantum numbers are possible such that each set contains all of the values

Table 4.8. List the elements of the first six periods that have configurations that deviate from those predicted by the aufbau method. In which of these cases can the deviation be ascribed to the presence of a half-filled subshell?

(d)

/!

(f) //

/

=

n

(b)

/

=

(c)

=

(a) 4,

4;{f) n

/;

/

/;

;

Each electron

n

/i

/

listed: (c)

(a)

/

4.51

=

=3; 3, 3; = 3, = 2. ni, = -2; (e) » = 3, = 3, = 0, w, = + 2 (g) = 3, / =

listed:

=

=

3,

/

w,

0,

= =

2;

4.66

1

in

= 0; « = 4, 1=3. m, = 0; = 4,1 = 3.(g)

(b)

/

(d) /;

=

/)

//

= =

=

4.

/

4,

1=2:

0.

=

/;;,

(e)

/;

the

+?>:

a filled subshell? 4.67 Classify

4,

4.

electronic

configurations

given

in

each of the following elements as a noble

gas. a representative element, a transition element, or an

ground state of ,-Rb: (a) how many electrons 0 as one of their quantum numbers':* (b) how many electrons have />7, = 0 as one of their quantum numbers'.' (c) how many electrons have /»/, = — as one /

Examine

=

4.52 In the

have

tions of the elements of a period

=

inner-transition element. Also slate whether the element is

(e)

a metal or a nonmetal: (a) Na, (b) N.

Nd.

(()

(c)

Ni, (d) In,

Ne.

1

of their

quantum numbers?

4.53 In the

ground

state of ^^Xe: (a)

how many

electrons

Problems

113

CHAPTER

PROPERTIES OF ATOMS

AND THE

IONIC

BOND

Chemical bonds, which form when atoms combine, are the result of changes electron distribution. There are three fundamental types of bonding:

in

when electrons are transferred from one type The atoms of one of the reacting elements lose electrons and become positively charged ions. The atoms of the other reactant gain electrons and become negatively charged ions. The electrostatic (plus-minus) attraction Ionic bonding (Chapter 5) results

1.

of atom to another.

between the oppositely charged ions holds them In covalent bonding (Chapters 6

2.

and

in a crystal.

7) electrons are

shared, not transferred.

A single covalent bond consists of a pair of electrons shared cules are

made up of atoms

Metallic bonding (Chapter 23)

3.

arranged

5.1

found

is

in

in a three-dimensional structure.

move throughout

free to

by two atoms. Mole-

covalently bonded to each other.

the structure

metals and alloys. The metal atoms are The outer electrons of these atoms are

and are responsible

for binding

it

together.

Atomic Sizes

How

an atom reacts depends upon many

configuration are the most important.

importance. Determination of this

size,

Nuclear charge and electron of the atom is also of however, is a problem. The wave theory factors.

The

effective size

beyond a region of high density, the electron cloud of an atom thins out gradually and ends only at infinity. We cannot isolate and measure a single

predicts that

atom.

measure in several ways the distance between the bonded together. AUKnic radii are derived from these bond distances.* The CI CI bond distance in the CI, molecule, for example, is 198 pm.^ One-half of this value, 99 pm, is assumed to be the atomic radius of CI chlorine. In turn, the atomic radius of CI (99 pm) can be subtracted from the C bond distance (176 pm) to derive the atomic radius of C (77 pm). The effective It is

possible, however, to

two atoms

nuclei of

that are

—

—

*

The bond

*

Atomic dimensions have been recorded

distances o( single covalent bonds are employed. See Section 6.1. in

angstrom units

in the past (1

International System, these dimensions are recorded in nanometers (i

pm =

10

1.98

A =

114

'

"

m).

0.198

The CI

nm =

-

CI

198

bond distance

pm =

1.98

is

x 10"'"

m

(1

nm =

A =

10

'"

m). In the

10"' m) or picometers

Cs

Rb

inner-transition

elements I

Na

transition

Br

transition

transition

elements

elements

elennents period 3

period 4

period 2

period 5

period 6

H 36

18

10

86

54

Atomic number Figure 5.1

size

Atomic

of an atom

radii of ttie

may

elements

in

picometers

(1

pm =

10"'^ m)

vary slightly from bond to bond when the atom

is

bonded

to

other types of atoms. Such variations, however, are usually less than a few pico-

Data derived in this way, therefore, can be used in making comparisons. Atomic radius is plotted against atomic number in Figure 5.1. Values for the

meters.

noble gases are not available.

1.

Two

trends should be noted.

Within a group of the periodic table, an increase

observed from top to bottom. Values for the group

in I

A

atomic radius elements

(Li,

is

generally

Na, K, Rb,

and Cs) and the group VII A elements (F, CI, Br, and I) are labeled in Figure 5.1. The increase in atomic radius within each group is clearly evident. As we move from one atom to the next down a group, an additional electron level is employed, and an increase in atomic size follows. There is, however, also an increase in the number of protons in the nucleus. The resulting increase in nuclear charge is a factor that lends to decrease atomic size. The full nuclear charge, however, is shielded from the outer electrons by electrons that lay between them and the nucleus. The number of shielding electrons increases from atom to atom in a group along with the increase in nuclear charge. As a result, the effective nuclear char 2Br(g) is +224 kJ/mol Br,. The lattice energy of RbBr is —666 kJ/mol.

5.24 Consider the lattice energies of

Which member of each of the following

you predict

Rb

compound?

Name the following: (a) MnS04, (b) Mg3(P04)2, PbCO,,(d) HgCl,,(e) Na,0,,(f) AKfSO^),."

5.40 (c)

Give fomuilas for the oxides, chlorides, and nitrides of sodium, magnesium, and aluminum.

5.26

134

Chapter 5

Properties of

5.41 (c)

Atoms and

Name

SnF,,(d)

the Ionic

the following: (a) Ca(CI04),. (b)

KMnO^.ie) FeP04,

Bond

(f)

Hg^l,.

ColNOj),,

CHAPTER

THE COVALENT BOND

6

In the last chapter

we

discussed the formation and properties of ionic compounds.

bond (in which electrons are shared by the bonded atoms) will be introduced. In addition, we will consider bonds that have a character In this chapter, the covalent

intermediate between the purely ionic and the purely covalent.

1

Covalent Bonding

When atoms

of nonmetals interact, molecules are formed which are held together by covalent bonds. Since these atoms are similar in their attraction for electrons (identical when two atoms of the same element are considered), electron transfer does not occur; instead, electrons are shared. of electrons (with opposite spins) that

is

A

covalent bond consists of a pair

shared by two atoms.

As an example, consider the bond formed by two hydrogen atoms. An inatom has a single electron that is symmetrically distributed

dividual hydrogen

around the nucleus in a \s orbital. When two hydrogen atoms form a covalent bond, the atomic orbitals overlap in such a way that the electron clouds reinforce each other in the region between the nuclei, and there is an increased probability of finding an electron in this region. According to the Pauli exclusion principle, the two electrons of the bond must have opposite spins. The strength of the covalent bond comes from the attraction of the positively charged nuclei for the

bond (see Figure 6.1). The hydrogen molecule can be represented by

negative cloud of the

—

the symbol H:H or H H. Although the electrons belong to the molecule as a whole, each hydrogen atom can be considered to have the noble-gas configuration of helium (two electrons in the A7 = level). This consideration is based on the premise that both shared electrons contribute to the stable configuration of each hydrogen atom. The formula, H2, describes a discrete unit a molecule and hydrogen gas consists of a collection of such molecules. There are no molecules in strictly ionic materials. The formula Na2Cl2 is incorrect because sodium chloride is an ionic compound and the simplest ratio of ions in a crystal of sodium chloride is to a molecule of formula Na2Cl2 does not exist. For covalent materials, however, a formula such as H2O2 can be correct this formula describes a molecule containing two hydrogen atoms and two oxygen atoms. The hydrogen molecule can be described as being diatomic (containing two 1

—

—

1

1

;

atoms). Certain other elements also exist as diatomic molecules.

any group VII

A element,

for example, has seven valence electrons.

An atom of By the forma-

135

Figure

6.1

Representation

the electron distribution

hydrogen molecule

in

a

of

tion of a covalent

bond between two of

these atoms, each

atom

attains

an octet

configuration characteristic of the noble gases. Thus, fluorine gas consists of F2

molecules :F-

+

-F:

—

:f:f:

Only the electrons between the two atoms are shared and form

a part of the

covalent bond (although the molecular orbital theory considers that electrons affect the bonding

More (group

V

— see Section

than one covalent bond

A) has

five

of the

may form between two

atoms.

A

nitrogen

atom

valence electrons:

+ -N:

:NIn the molecule,

all

7.4).

Nj,

six electrons are

—

:n:::n:

shared in three covalent bonds (usually

called a triple bond). Notice that, as a result of this formulation, each of the

nitrogen atoms can be considered to have an octet of electrons. exist as diatomic molecules are H2, F,, CI2, Brj, N2, and O2. (Oxygen is a special case and will be discussed in Section 7.4). These elements are always indicated in this way in chemical equations. The electron-dot formulas we have been using are called valence-bond structures or Lewis structures, named after Gilbert N. Lewis who proposed this theory of covalent bonding in 1916 (more recent theories of covalent bonding are discussed in Chapter 7). The Lewis theory emphasizes the attainment of noble-gas configurations on the part of atoms in covalent molecules. Since the number of valence electrons is the same as the group number for the nonmetals, one might predict that VII A elements, such as CI, would form one covalent bond to attain a stable octet VI A elements, such as O and S, two covalent bonds V A elements, such as N and P, three covalent bonds and IV A elements, such as C, four covalent bonds. These predictions are borne out in many compounds containing only simple covalent bonds:

Nonmetallic elements that

I2,

;

Gilbert N. Lewis, 1875-1946.

Photographer Johan

Hagemeyer Bancroft :

Library, courtesy

;

;

AlP Neils Bot)r Library

H- +

-Cl:

—

H:C1: hydrogen chloride

2H- + -6:

—

H H:0: water

3H- + -N-

—

H H:N:H ammonia

4H- +

-C-

—

H H:C:H

H methane

Notice that

in these molecules,

complete n

=

1

each hydrogen atom can be considered to have a

shell; the other

atoms have

The covalent bonding of compounds can

characteristic noble-gas octets.

also be indicated

dash represents one bond, a pair of electrons:

136

Chapter 6

The Covalent Bond

by dashes; each

.2

H H :C1— P— Ci:

^Cl—

H— C— C^H

Cl:

H

:Cl: phosphorous

dichlorine oxide

H

ethane

trichloride

The following are examples of molecules

+

:o:

+ :o:

:c:

—

that contain double

and

triple

bonds:

(or:0=C=o:)

:o::c::o: carbon dioxide

2H- +

+

-C:

H H — H:C::C:H + 2H— +

:C-

f

f

(or

H— C=C— H)

(or

H—C=C— H)

ethylene

H- + -Ci + iC-

H:C:::C:H

-H

acetylene

Notice that

in

compound the number of covalent bonds on each atom number predicted.

each

agrees with the

A method for drawing Lewis structures is given in Section 6.3. Before the method is presented, however, the concept of formal charge must be introduced since the method relies upon the use of formal charge to check the validity of its results.

Formal Charge In the formation of certain covalent bonds, hoih of the shared electrons are

furnished by one of the bonded atoms. For example, in the reaction of

with a proton (a hydrogen pair of the nitrogen

atom

atom of

stripped of

NH3

is

its

electron), the

ammonia

unshared electron

used to form a new covalent bond:

H

H:N:H +

H

H:N:H H

A bond

formed in this way is frequently called a '"coordinate covalent" bond, probably unwise to do so. Labeling a specific bond as a "coordinate covalent" bond implies that it is different from other covalent bonds and has little justification. All electrons are alike no matter what their source. All the bonds in NH4 are identical. It is impossible to distinguish between them. Notice, however, that the number of covalent bonds on the N atom of NH4 but

it

is

does not agree with the number predicted in the preceding section. Since a nitrogen atom has five valence electrons (group V A), it would be expected to satisfy the octet principle through the formation of three covalent bonds. This prediction is

correct for

NH3;

it

is

not correct for

NH4.

An

answer to the question may be obtained by calculating the formal charges of the atoms in NH4 The formal charge of an atom in a molecule is calculated in the following way. The number of valence electrons that an A family atom has .

6.2

Formal Charge

137

is

equal to

atom

its

group number.

+ {group

valence electrons were removed from the

If all the

would have

in question, the resulting ion

a positive charge equal to

no.)

This would be the charge on the atom if it had no valence electrons associated it. In a molecule, however, the atom has valence electrons associated with

with it

— shared

atom

in

bonds) and, in some cases, unshared. If the electron pair

(as covalent

of each covalent bond

divided equally between the two atoms that

it

bonds, the

question will get one electron for each covalent bond that

it

has

is

charge for each bond). In addition, the atom will have a

unshared electron that sources

bonds)

+

(no.

The formal charge of formal charge Since the electrons,

its

1

—

charge for each

has in the molecule. The total negative charge from these

= +

unshared e~

the

)'\

bonded atom,

(group no.)

—

therefore,

—

(no bonds)

N atom (a group V A atom) in NH4 formal charge

formal charge

H

1

(a

is

— [(/7 stalli/.alion (Section 9.8) The enthalpy change associated with the conversion of a given quantity of a liquid (usually one mole or one gram) into a solid at

sure at which a substance can simultaneously exist as a

a specified temperature.

that will reproduce the crystal

Enthalpy of fusion (Section 9.8) The energy required to melt a given quantity of a solid (usually one mole or one gram) at a specified temperature.

dimensions.

Enthalpy of vapori/.ation (Sections 9.4 and 9.7) The energy required to vaporize a given quantity of a liquid (usually one mole or one gram) at a specified temperature.

242

Chapters

Liquids and Solids

solid, a liquid,

Unit

cell

and a gas

in

The

(Section 9.12)

Vapor pressure (Section

equilibrium with each other.

9.5)

smallest part of a crystal

when

repeated in three

The pressure of vapor

in

equilibrium with a pure liquid or a pure solid at a given temperature. Viscosity (Section 9.3) to flow.

A

property of liquids; resistance

Problems Intermolecular Attractive Forces

London forces and what types of molecular substances do each exist? Which type is stronger in most molecular substances where both exist? Describe the difference between

9.1

Diagram the structures of the following molecules and explam how the water solubility of each compound is enhancedby hydrogen bonding: (a) NH3, (b) HjN OH, 9.13

dipole-dipole forces. In

—

(c)

H3COH,

Although

9.14

(such as

Give an explanation for the following,

9.2

moment of OF2

(a)

The dipole

D

but the dipole moment of BeFj is zero, (b) The dipole moment of PF3 is 1.03 D but the dipole moment of BFj is zero, (c) The dipole moment of SF4

is

D

0.63

is

0.30

but the dipole

moment

of SnF^

is

zero.

HXO.

(d)

there

NaHS04)

corresponding normal

exceptions,

are

are

more

O

CH3— C— CH3 mixtures have higher boiling points than either pure

component.

The normal

9.16

positions?

why chloroform,

HCCI3, and acetone,

zero?

could measurement of the dipole moment of the trigonal bipyramidal molecule PCI2F3 help to determine whether the CI atoms occupy axial or equatorial

(Na2S04). Offer an explana-

salts

9.15 Offer an explanation as to

Which molecules (not ions) given as examples in Table 7.1 would you expect to have a dipole moment of

How

salts

tion for ihis generalization.

9.3

9.4

most acid

soluble in water than the

boiling point of the

compound

ethylene

diamine, H2NCH2CH2NH2, is 17 C and that of propyl amine, CH3CH2CH2NH2, is 49 C. The molecules, however, are similar in size and molecular weight. What 1

The dipole moment of PF3 is moment. Explain.

9.5

1.03

D, whereas PF5

has no dipole

The dipole moment of

9.6

NH3

(1.49

D)

is

greater than

NF3 (0.24 D). On the other hand, the dipole moment of PH3 (0.55 D) is less than that of PF3 (1.03 D).

reason can you give for the difference

in boiling

point?

that of

The Liquid State

Explain these results.

Explain why the dipole moment of SCO is 0.72 D, whereas the dipole moment of COj is zero. Would CS, have a dipole moment?

9.7

Table 6.1 the electronegativity of C is given as 2.6 and that of O is given as 3.4. On the other hand, in Table 9.1 the dipole moment of CO is listed as only 0.12 D (the dipole-dipole forces of CO are negligible). Draw the Lewis structure of CO and offer an explanation for the low dipole moment of CO. 9.8 In

Explain

9.9

why

order given: F2

and

the following melting points

(-233X), CI, (-103

fall in

the

(-7

C),

C), Brj

I2 (113.5°C).

Consider the following molecules, each of which with the C atom as the central atom: CH4, CH3CI, CH,Cl2, CHCI3, ecu. In which of these tetrahedral

compounds would dipole-dipole forces state? In what order would you expect of the compounds to fall?

how and why each

of the following

gives an indication of the strength of the intermolecular

forces of attraction of a substance (a) critical temperature, :

tension,

(b)

surface

(e)

enthalpy of vaporization,

9.18 Explain,

curve, 9.19

using a

pressure, (d) vapor normal boiling point.

(c) viscosity, (f)

Maxwell-Boltzmann distribution liquid becomes cool.

why an evaporating

What

is

an equilibrium state? Describe the condition

when an evaporating

that exists

liquid

is

placed

in

a

container.

Why

does the boiling point of a hquid vary with is the normal boiling point? Use the curves of Figure 9.7 to estimate the boiling point of diethyl ether, ethyl alcohol, and water at a pressure of 0.50 atm.

9.20

pressure?

9.10 is

9.17 Briefly explain

What

Use the data of Table 8.3 to estimate the boiling (a) 0.010 atm and (b) 0.025 atm.

exist in the liquid

9.21

the boihng points

point of water at

Phase Diagrams The Hydrogen Bond Use the following data to draw a rough phase diagram for hydrogen: Normal melting point, 14.01 K; normal boiling point, 20.38 K; triple point, 13.95 K, 7 X 10"^ atm; critical point, 33.3 K, 12.8 atm; vapor x 10"^ atm. pressure of solid at 10 K, 9.22

of hydrogen bonding on the properties of the following hydrides falls in the order given: HjO > 9.11

The

effect

HE > NH3.

Explain this observation.

The compound reaction of KF and

9.12

structure of the

*

The more

HF,

difficult

KHF, HE in

can be prepared from the water solution. Explain the

ion.

problems are marked with

asterisks.

1

9.23 Use the following data to draw a rough phase diagram for krypton: Normal boiling point, — 152'C;

The appendix contains answers

to color-keyed problems.

Problems

243

normal melting point, -157"C;

triple poijit,

-169°C,

0.175 atm; critical point, -63'C, 54,2 atm; vapor pressure of solid at - 199 C, 1.3 x 10'^ atm. Which has the higher density at a pressure of

1

atm: solid Kr or liquid

LiH or H,,

Li or H2,(b)

(a)

(6)1^2 or

Li or LiH. (d)

(c)

or Clj,

HCl?

Crystals

Kr?

Neon

a face-centered cubic

diagram for water. Describe the phase changes that occur, and the approximate pressures at which they occur, when the pressure on a

9.36

H2O

system is gradually increased (a) at a constant temperature of — I 'C, (b) at a constant temperature of 50 C, (c) at a constant temperature of — 50'C.

9.37 Barium crystallizes in a body-centered cubic lattice, and the edge of the unit cell is 502 pm. The atomic weight of barium is 1 37. What is the density of crystalline barium ?

and describe the phase changes and the approximate temperatures at which they occur, when water is heated from — lO'C to 110°C (a) under a pressure of 1 x 10~^ atm, (b) undera pressure of 0.5 atm. (c) under a pressure of 1.1 atm.

and the edge 336 pm. The density of Po is 9.20 g/cm^, and the atomic weight of Po is 210. What type of cubic unit cell does Po form?

9.24 Figure 9.9

is

the phase

9.25 Refer to Figure 9.9

that occur,

Polonium

9.38

of the unit

Gold

(a) at a constant temperature constant temperature of 0 C.

increased (b) at a

— 60'C,

of

An

lattice,

and describe the phase changes that occur, and the approximate temperatures at which they occur, when carbon dioxide is heated (a) at a constant pressure of 2.0 atm,

(b) at a

constant pressure of 6.0 atm.

how and why

the slope of the melting-

point curve of a phase diagram for a substance depends

upon

and

the relative densities of the solid

sometimes used to purify solids. is heated, and the pure crystalline product condenses on a cold surface. Is it possible to purify ice by sublimation? What conditions would have to be employed?

The impure

crystallizes in a cubic system, is

element crystallizes

An

9.41

cell is

286 pm. The density

7.92 g/cm^. Find the atomic weight of

is

element crystallizes

a

in

face-centered

cubic

556 pm. The density 1.55 g/cm^. Find the atomic weight of

and the edge of the unit

of the element the element.

a body-centered cubic

in

and the edge of the unit

is

cell is

liquid forms

of the substance. 9.29 Sublimation

crystallizes in a cubic system,

of the element the element.

lattice,

9.28 Briefly explain

lattice,

cell is

the unit cell

9.40

9.27 Refer to Figure 9.10

in

and the edge of 407 pm. The density of Au is 19.4 g/cm^, and the atomic weight of Au is 197. What type of cubic unit cell does Au form?

9.39

and describe the phase changes that occur and the approximate pressures at which they occur when the pressure on a CO, system is gradually 9.26 Refer to Figure 9.10

crystallizes

and the edge of the unit cell is 452 pm. The atomic weight of neon is 20.2. What is the density of crystalline neon?

is

9.42

Tungsten

The

density of

of

W

184.

is

body-centered cubic lattice. and the atomic weight the length of an edge of the unit cell?

crystallizes in a

W

What

is is

19.4 g/cm^,

material

9.43

Palladium crystallizes

The

density of

of Pd 9.44

is

106.

Pd

What

Potassium

is is

in a face-centered

cubic

lattice.

and the atomic weight the length of an edge of a unit cell? 12.0 g/cm^,

crystallizes

in

a

body-centered

cubic

and the length of an edge of the unit cell is 533 pm. The atomic weight of K is 39.1, and the density of K is 0.858 g/cm^. Use these data to calculate Avogadro's number. lattice,

Types

of Crystalline Solids

9.30 List the types of crystals;

each

is

composed and

tell

the particles of which

the kinds of forces that hold

them 9.45 Silver crystallizes

together.

What type of forces must be overcome in order to melt crystals of the following: (a) Si, (b) Ba, (c) F,, (d) BaF,. (e) BF3. (f) PF,. 9.31

Ag 9.46

9.32

What

type of forces must be overcome in order to

melt crystals of the following: (d) Ba, (e) BaBr,, (f) BaO?

(a)

O,,

(b)

Br,,

(c)

Br,0,

(107.7

a face-centered cubic lattice,

g).

Copper

crystallizes in a face-centered cubic lattice,

and the edge of a

unit cell

is

361

pm. Calculate the atomic

radius of Cu. 9.47

Which substance of each of the following pairs would you expect to have the higher melting point: (a) CIF or BrF, (b) BrCl or CI .(c) CsBr or BrCl, (d) Cs or Br,, (e) C (diamond) or Cl,?'why? 9.33

in

and the edge of a unit cell is 407.7 pm. Calculate the dimensions of a cube that would contain one mole of

Chromium

crystallizes

and the edge of a unit the atomic radius of Cr. lattice,

in

a

cell is

body-centered cubic 287.5 pm. Calculate

,

9.48

Molybdenum has an atomic

radius of 136

crystallizes in a body-centered cubic lattice.

9.34

Which substance of each of

would you expect (a)

Sr or CI,,

(d)

SiCU or SCI4

9.35

(b) ,

following pairs

(c)

SiCU or

SiC (carborundum) or SiCU?

Which substance of each of to

is

and the

length of an edge of a unit cell?

have the higher melting point:

SrCl, or SiCU,

(e)

would you expect

244

to

the

pm

What

the

SiBr^,

Why?

following pairs

have the higher melting point:

Chapter 9

Liquids and Solids

X-Ray

Diffraction of Crystals

X rays with a wavelength of 35 pm, a first-order reflection was obtained 9.49 In the dilTraction of a crystal using 1

at

an angle of 11.2

.

What

The

diffracted planes?

the distance between the

is

sine of 11.2

is

0.1942.

The

lattice.

of a crystal using X rays with a wavelength of 94 pm, a first-order reflection was obtained What is the distance between the at an angle of 25.9 diffracted planes? The sine of 25.9 is 0.4368.

9.50 In the diffraction

sodium chloride Pb"^ ion and a

*9.59 (a) Lead(II) sulfide crystalhzes in the

S^" ion

shortest distance between a

What

(b)

What

297 pm.

is

PbS

unit cell of

1

is

is

the length of the edge of a

shown in Figure 9.21? PbS?

similar to that

the density, in g/cm^, of

.

In the X-ray diffraction of a set of crystal planes for

9.51

Cadmium

*9.60 (a)

The

lattice.

sulfide crystallizes in the zinc blende

shortest distance between a

253 pm.

What

Cd"*

ion

and

the length of the edge of a

which d is 248 pm, a first-order reflection was obtained What is the wavelength of the X rays at an angle of 8.21 that were employed? The sine of 8.21 is 0.1428.

a S^" ion

of a crystal using X rays with a wavelength of 179 pm, a first-order reflection is found at an angle of 21.0\ What is the wavelength of X rays that show the same reflection at an angle of 25.0 ? The sine of 21.0' is 0.3584, and the sine of 25.0 is 0.4226.

In a given crystal, the distance between the centers of a cation and a neighboring anion is approximately equal to the sum of the ionic radii of the two ions. Some

is

is

shown

unit cell similar to that

Figure 9.21?

in

(b)

What

.

9.52 In the diffraction

At what angle would a

9.53

first-order reflection be ob-

served in the X-ray diffraction of a set of crystal planes for

if the X rays used have a wavelength angle would a second-order reflection

pm

which d is 300

pm? At what

of 154

is

CdS?

the density of

9.61

Na^, 95 pm; K"^, 133 pm; Ca^*, 99 pm; pm; Ni'^, 69 pm; Ag + 126 pm; CI", 181 pm; Br", 195 pm; O'", 140 pm; S^", 184 pm. Refer ionic radii are:

Ba2\

to

135

,

Equation 9.7

in

Section 9.15 and arrange the following

(each of which crystallizes in a sodium-chloride type lattice) in

value

decreasing order of lattice energy (most negative

first):

AgCl, BaO, CaS, KCl, NaBr. and NiS.

be observed?

At what angle would a first-order reflection be in the X-ray diffraction of a set of crystal planes for which d is 349 pm if the X rays used have a wavelength of 194 pm? At what angle would a second-order reflection

Defect Structures

be observed'?

crystals.

9.54

observed

9.62 List

*9.63

and describe

Cadmium

oxide,

types of defects

the

CdO,

crystallizes

in

found

the

in

sodium

Cd^^ and four O'" ions Figure 9.21). The compound, however,

chloride structure, with four

per unit Ionic Crystals 9.55 (a)

is

How many

shown in the diagrammed in

ions of each type are

cesium chloride crystal The density of cesium chloride is 3.99 g/cm^. What is the length of the edge of the unit cell shown in the figure'.' (c) Use the equations given in Section 9.12 to determine the shortest distance between a Cs^ ion and a CI" ion. unit cell of the

Figure 9.21?

9.56 (a)

(b)

How many

9.21 in a

The

(b)

ions of each type are

sodium chloride

unit cell of the

shown

in the

crystal depicted in Figure

density of silver chloride, which crystallizes

sodium chloride

lattice, is

length of the edge of the

AgCl

5.57

g/cm\ What

unit cell of the type

is

the

shown

Use the equations given in Section 9.12 determine the shortest distance between a Ag^ ion

in the figure? (c)

to

and a CI 9.57 (a)

ion.

How many

ions of each type are

shown

unit cell of the zinc sulfide crystal depicted in

9.21?

(b)

The

stallizes in

density of copper

a zinc blende

lattice, is

the length of the edge of the

shown

in the figure? (c)

(I)

Use

CuCl

in the

Figure

chloride, which cry-

4.14 g

cm \ What

unit cell

the equations given in Section

9.12 to determine the shortest distance between a ion

and

*9.58 (a)

a CI

The

chloride lattice. ion

cell

What

is

mates CdOo.995. The defect occurs because some cation Cd atoms instead of Cd^ * ions and an equivalent number of anion positions are vacant, (a) What percent of the anion sites are vacant? (b) If the edge of a unit cell is 469.5 pm, what would be positions in the crystal are occupied by

the density of a perfect crystal? (c) What is the density of the nonstoichiometric crystal? The atomic weight of Cd is 112.40 and of O is 16.00. *9.64 Iron(II) oxide crystalhzes

the cesium

shortest distance between a Tl*

in

the

sodium chloride

ions per unit Fe"* and four O" cell (see Figure 9.21). The crystals, however, are always deficient in iron. Some cation sites are vacant and some cation sites contain Fe^^ ions instead of Fe^^ ions, but the combination is such that the structure is electrically neutral. The formula usually approximates Feo.930. (a) What is the ratio of Fe^* ions to Fe^* ions in the crystal? (b) What percentage of the cation sites are vacant? structure, with four

O^"

Hint: consider a crystal that contains 100 *9.65 (a)

9.21

is

The edge of

563.8

pm

Use these data of

NaCl

the

NaCl

unit cell

shown

and the density of NaCl

is

ions.

in

Figure

2.165

g/cm^

to calculate the apparent molecular weight

to four significant figures,

(b)

The

diflerence

between the value calculated from crystal data and the actual molecular weight of

crystallizes in

is 333 pm. What is the length of the of TlCl similar to that shown in Figure the density, in g,cm\ of TlCl?

and a CI" ion

9.21? (b)

Cu^

ion.

ThalUum(I) chloride

edge of a unit

is

of the type

cell (see

usually nonstoichiometric. with a formula that approxi-

NaCl

type of lattice defect in which

of

Na*

(58.44)

Na atoms

ions in the crystal and an equal

ions are missing from lattice positions.

your answer to part sites that are

(a),

ascribed to a

is

number number of CI"

replace a

On

the basis of

calculate the percentage of anion

vacant.

Problems

245

:

CHAPTER

SOLUTIONS

Solutions are

homogeneous mixtures. They

their physical state; gaseous, liquid,

and

are usually classified according to

solid solutions

can be prepared. Dalton's

law of partial pressures describes the behavior of gaseous solutions, of which air

Not

is

all

common

most

the

is

silver

copper dissolved

example. Certain alloys are solid solutions; coinage in silver,

and brass

alloys are solid solutions, however.

is

a solid solution of zinc in copper.

Some

are heterogeneous mixtures,

and some are intermetallic compounds. Liquid solutions are the most and are probably the most important to the chemist.

10.1

common

Nature of Solutions The component of a solution that is present the solxcni. and all other components are

in greatest

quantity

is

usually called

called solutes. This terminology

is

sometimes convenient to designate a component as the solvent even though it is present in only small amounts. At other times, the assignment of the terms solute and solvent has little significance (for example, in describing gaseous solutions). loose and arbitrary.

It is

Certain pairs of substances will dissolve in each other in

Complete and some

miscibility

given solvent.

temperature

amount of

is

characteristic of the

components of

pairs of

however, there

is

The the

sohibilit>

liquid

solid solutions.

is

proportions.

For most materials,

of a substance

in a particular solvent at a specified

and produce

of the solute that

will dissolve in a definite

a stable system.

For a given solution, the amount of solute dissolved given

all

gaseous solutions

the substance that will dissolve in a

maximum amount

the solvent

and

all

amount of

a limit on the

is

components of

in

an amount of solvent

the concenlration of the solute. Solutions containing a relatively

low

concentration of solute are called dilute solutions; those of relatively high concentrations are called concentrated solutions. If

an excess of solute (more than

of a liquid solvent, an equilibrium

will

is

normally dissolve)

is

added

to a quantity

established between the pure solute

and the

dissolved solute

solutCp^r,

The pure

solute

soluted„,„|,,d

may

be a solid, liquid, or gas. At equilibrium in such a system,

the rate at which the pure solute dissolves equals the rate at which the dissolved solute

246

comes out of

solution.

The concentration of

the dissolved solute, there-

fore, is a constant.

concentration

is

A

solution of this type

and

called a saturated solution,

is

its

the solubility of the solute in question.

That such dynamic equilibria

exist has

been shown experimentally.

If

small

crystals of a solid solute are placed in contact with a saturated solution of the solute, the crystals are observed to change in size and shape. Throughout this experiment, however, the concentration of the saturated solution does not change, nor does the quantity of excess solute decrease or increase.

An

unsaturated solution has a lower concentration of solute than a saturated On the other hand, it is sometimes possible to prepare a supersaturated

solution. solution,

one

solution.

A

in

which the concentration of solute

supersaturated solution, however,

amount of pure

solute

is

added

to

it,

is

is

higher than that of a saturated

metastable, and

the solute that

is

in excess

if

a very small

of that needed

to saturate the solution will precipitate.

).2

The Solution Process London

forces are the only intermolecular forces between nonpolar covalent

molecules.

On

the other hand, the intermolecular attractions between polar

covalent molecules are due to dipole-dipole forces as well as to In substances in

which there

is

London

forces.

hydrogen bonding the intermolecular forces are

unusually strong.

Nonpolar substances and polar substances are generally insoluble in one Carbon tetrachloride (a nonpolar substance) is insoluble in water (a polar substance). The attraction of one water molecule for another water molecule is much greater than an attraction between a carbon tetrachloride molecule and a water molecule. Hence, carbon tetrachloride molecules are "squeezed out," and these two substances form a two-liquid-layer system. Iodine, a nonpolar material, is soluble in carbon tetrachloride. The attractions between 1 2 molecules in solid iodine are approximately of the same type and magnitude as those between CCI4 molecules in pure carbon tetrachloride. Hence, significant iodine carbon tetrachloride attractions are possible, and iodine molecules can mix with carbon tetrachloride molecules. The resulting solution another.

is

a

random molecular mixture. CH jOH, like

water, consists of polar molecules that are

Methyl alcohol,

highly associated. In both pure liquids the molecules are attracted to one another

through hydrogen bonding:

H O H~0 H— O H— C— H H— C— H H~C— H

H

H

H— O

•

•

•

H— O H

H

•

•

•

H— H

H

Methyl alcohol and water are miscible in all proportions. In solutions of methyl alcohol in water, CH3OH and H^O molecules are associated through hydrogen bonding:

H— O H

•

•

•

H— O H— H H— C— H •

•

•

H 10.2

The Solution Process

0O3 ^

Solution of an ionic crystal

in

Methyl alcohol does not dissolve

ni

Figure 10.1

water

nonpolar solvents. The strong intermolecular overcome unless the solvent molecules can tbrm attractions of equal, or almost equal, strength with the methyl alcohol attractions of pure methyl alcohol are not

molecules.

and nonpolar

In general, polar materials dissolve only in polar solvents,

substances are soluble in nonpolar solvents. This "like dissolves like."

atoms

that

make up

in all liquids.

Network

crystals (the

diamond,

the

first

rule of solubility:

for example), in

which the

the crystal are held together by covalent bonds, are insoluble

This crystalline structure

solution process.

is

Any

far too stable to

is

down by a approach the

be broken

potential solute-solvent attractions cannot

strength of the covalent bonding of the crystal.

Polar liquids (water,

in particular)

can function as solvents for

compounds. The ions of the solute are solvent molecules

electrostatically attracted

many

ionic

by the polar

— negative ions by the positive poles of the solvent molecules,

positive ions by the negative poles of the solvent molecules. These ion-dipole

attractions can be relatively strong.

Figure 10.1 diagrams the solution of an ionic crystal center of the crystal are attracted equally in

all

in water.

The

ions in the

directions by the oppositely charged

The electrostatic attractions on the ions of the surface of the however, are unbalanced. Water molecules are attracted to these surface ions, the positive ends of the water molecules to the anions and the negative ends of the water molecules to the cations. The ion dipole attractions formed in this ions of the crystal.

crystal,

,v.

way allow

the ions to escape

from the

dissolved ions are hydrated and

crystal

and

move through

drift into the liquid phase.

The

the solution surrounded by a

sheath of water molecules. All ions are hydrated in water solution.

10.3 Hydrated Ions in water solution by means of attractions between and the hydrogen atoms of the water molecule. In some cases (the sulfate for example) these attractions may be one or more hydrogen bonds:

Negative ions are hydrated the ion ion,

248

Chapter 10

Solutions

H

H

O—

O

O

H—

O—

o

o

H—

H

H

by means of attractions between the ion and the unshared electron pairs of the oxygen atom of the water molecule. These attractions are strong. In many cases, each cation is hydrated by a definite number of H2O molecules Positive ions are hydrated

H

H

H—

P

H

O

O

H

H

H

Be.

H

Additional water molecules form hydrogen bonds with those molecules that are

bonded are

to the cation or anion.

more

What

These outer layers of water molecules, however,

loosely held. factors lead to the formation of strong interactions between the ion

and

water molecules?

4

H

O atoms of the HjO molecule.

1.

Ions with high charges strongly attract the

2.

Small ions are more effective than large ones because the charge

or

is

concentrated in small ions.

.\

more highly r^^^

,

A

number of covalent compounds of metals produce hydrated ions in water The compounds of beryllium, for example, are covalent when pure. The same factor that is largely responsible for the covalent character of the compounds of beryllium (a high ratio of ionic charge to ion size) also leads to

solution.

the formation of very stable hydrated ions:

BeCl2(s)

+ 4H2O

Be(H20)r(aq) + 2Cr(aq)

bond always liberates energy, and the breaking of a bond The energy released by a hypothetical process in which formed from gaseous ions is called the enthalpy of hydration

The formation of

^ways

— a

requires energy.

hydrated ions are

-rri^i ,

^

--n

/

of the ions. For example.

K^(g)

The

size

+ Cr(g)

—

K*(aq)

^H =

+ Cr(aq)

-684.1 kJ

of the enthalpy of hydration depends upon the concentration of the no distinction is made (as in the example previously given), it is

final solution. If

assumed that the enthalpy change pertains

to a process in

which the ions are

hydrated to the greatest possible extent. This high degree of hydration would occur only if the solution were very dilute. The corresponding A// values are called enthalpies of hydration at infinite dilution.

10.3

Hydrated Ions

249

:

:

:

The value of an enthalpy of hydration the attractions between the ions

A

and

gives an indication of the strength of

the water molecules that hydrate them.

large negative value (which indicates a large quantity of energy evolved)

shows

that the ions are strongly hydrated.

Frequently, hydrated ions remain in the crystalline solids that are obtained by the evaporation of

FeClj-eH^O BeCl2-4H20

aqueous solutions of

Thus,

Fe(H.O)^+ and CI" ions BelH^O);^ and CI" ions ofZniH^O)!^ and S04(H20)-" ions of CulHjO^ and SO^lHiO)^" ions

consists of consists of

ZnS04-7H20 consists CuSO^-SH.O consists Water molecules can

also occur in crystalline hydrates

in the crystal structure is

salts.

by taking

lattice positions

without associating with any specific ion (BaCl2'2H20

an example) or by occupying

silicates called zeolites are

interstices (holes)

of a crystal structure (the hydrous

examples).

10.4 Enthalpy of Solution

The enthalpy change solvent

is

associated with the process in which a solute dissolves in a

The value of an enthalpy of

called the enthalpy of solution.

(given in kJ/mol of solute) depends

upon

solution

the concentration of the final solution

in the same way that an enthalpy of hydration does. Unless otherwise noted, an enthalpy of solution is considered to apply to the preparation of a solution

that

is

The enthalpy of

infinitely dilute.

solution

is

virtually constant for all the

dilute solutions of a given solute-solvent pair.

The enthalpy change observed when

a solution

is

prepared

the net result

is

of the energy required to break apart certain chemical bonds or attractions (solutesolute

and solvent-solvent) and the energy released by the formation of new ones

(solute

KCl

solvent).

The enthalpy of

in water, for

solution for the preparation of a solution of

example, can be considered to be the

sum

of two enthalpy

changes

1.

The energy required

to break apart the

KCl

crystal lattice

and form gaseous

ions

KCl(s)

2.

—K^(g) +

A//=+701.2kJ

Cl"(g)

The enthalpy of hydration of KCl, which

is

the energy released

when the gaseous

ions are hydrated

K^(g)

+

CI

(g)

—*K^(aq) + Cr(aq)

This enthalpy change

is

actually the

sum

AH =

of two enthalpy changes: the energy

required to break the hydrogen bonds between the energy released

when

two

250

Chapter 10

Solutions

the water molecules

more energy

It is,

and

however,

elTects separately.

In this example, the overall process positive because

some of

these water molecules hydrate the ions.

difficult to investigate these

is

-684.1 kJ

is

is

endothermic. The enthalpy of solution

required in step

1

than

is

released in step 2:

K^(aq)

KCl(s)

+

(,

The

— —

Mn- ^ 4

H3ASO4

can be brought into material balance by the addition and 8 to the left side. In the second partial equation, lOHjO must be added to the left side to make up the needed 10 oxygens. If we stopped at this point, there would be 20 hydrogen atoms on the left and 12 on the right. Therefore, 8 H"^ must be added to the right: 2.

of 4

first

HjO

partial equation

to the right side

H ^ + MnO;

8

IOH2O + AS4O6 3.

To

— Mn- + HjO —4H3ASO4 + 8H ^

+

balance the net charges, electrons are added

+ 8H^ + Mn04'

5e"

10H,O + As40^ 4.

4

The

first

—

*

Mn-+ + 4H2O

—4H3ASO4 + 8H^ +

8^-

must be multiplied through by 8 and the second by same number of electrons are lost in the oxidation partial equation

partial equation

5 so that the

as are gained in the reduction partial equation

40f-

—8Mn^+ + 32H2O 5OH2O + 5AS4O6 —2OH3ASO4 + 40H^ +

+ 64H^ + 8Mn04-

40e-

When

these two partial equations are added, water molecules and hydrogen must be canceled as well as electrons. It is poor form to leave an equation with 64 H"" on the left and 40^1"*^ on the right: 5.

ions

24H^ + I8H2O + 5AS4O6 + 8Mn04 286

Chapter

11

Reactions

in

Aqueous Solution

—2OH3ASO4 + 8Mn^

+

Equations for reactions that take place

manner are the

slightly different

from those

in alkaline solution are

balanced

in a

that occur in acidic solution. All the steps

same except the second one;

reactions that occur in alkaline solution.

cannot be used to balance equations for As an example, consider the reaction

MnO^ + N,

Mn04 + N2H4

that takes place in alkaline solution:

1.

The equation

—

MnO^ N2H4

—

is

divided into two partial equations:

MnO, N2

,

For reactions occurring in alkaline solution, OH " and HjO are used to balance oxygen and hydrogen. For each oxygen that is needed, 2 OH" ions are added to the side of the partial equation that is deficient, and one HjO molecule is added to the opposite side. For each hydrogen that is needed, one H^O molecule is added to the side that is deficient, and one OH~ ion is added to the opposite side. In the first partial equation, the right side is deficient by 2 oxygen atoms. We to the right side and 2 H,0 to the left: add, therefore, 4 OH

2.

2H2O + Mn04

—

»

MnO, + 40H-

In order to bring the second partial equation into material balance, we must add four hydrogen atoms to the right side. For each hydrogen atom needed, we add one HjO to the side deficient in hydrogen and one OH" to the opposite side. In the present case we add 4H2O to the right side and 4 OH" to the left to make up the four hydrogen atoms needed on the right:

40H^ + N,H4 3.

—Nj + 4H,0

Electrons are added to effect charge balances:

3f"

+

2

HP

+ MnO^

40H + N2H4 4.

MnO^ + 40H"

—N, + 4H,0 +

Ae'

The lowest common multiple of

equation

is

3 and 4 is 12. Therefore, the first partial 4 and the second by 3 so that the number of by through multiplied

electrons gained equals the

12e"

number

+ 8H2O + 4Mn04

120H- + 3N2H4

lost:

—4MnO, + 160H"

—

3N2 + I2H2O +

12^-

5. Addition of these partial equations, with cancellation of molecules as well as electrons, gives the final equation

4Mn04 + 3N2H4 As

OH"

ions and

HjO

—*4Mn02 + 3N2 + 4H2O + 40H-

a final example, consider Lhe following skeleton equation for a reaction in

alkaline solution:

11.3

Oxidation-Reduction Reactions

:

+ Br"

BrOj"

Br,

In this reaction the

same substance, Br 2,

is

both oxidized and reduced. Such

reactions are called disproportionations or auto-oxidation-reduction reactions:

Br2

1.

Br^

120H" +

2.

Br,

Br, 3.

4.

120H- + 2e' +

Br,

120H

Br2

\0e5.

+

+

Br2

SBi-j

120H~ + 6Br2 When

— —2Br" — — — —

2Br03

6H2O

2Br03- +

»

2Br-

2Br03 + 6H2O +

lOe"

2Br

— — 2Br03 + 6H2O + —2Br03 + + 6H2O lOe"

*

lOBr

lOBr"

the ion-electron

method

is

applied to a disproportionation reaction,

the coefficients of the resulting equation usually are divisible by

number this

some common

one reactant was used in both partial equations. The coefficients of equation are all divisible by 2 and should be reduced to the lowest possible since

terms

60H- +

3Br2

—

3H2O

BrOa + SBr" +

The Ion-Electron Method

for Balancing Oxidation-Reduction Equations

1.

Divide the equation into two skeleton partial equations.

2.

Balance the atoms that change their oxidation numbers in each partial

equation. 3.

Balance the

O and H atoms in each partial equation.

For reaction in acid solution: For each O atom that is needed, add one

a.

i.

H2O

to the side of the

partial equation that is deficient in oxygen. ^ where needed to bring the hydrogen into balance. ii. Add

H

For reactions in alkaline For each O atom that

b.

solution is

i.

the partial equation that

:

needed, add two

OH"

ions to the side of

O, and add one

is

deficient in

is

needed, add one

H2O

to the

opposite side. ii.

For each

H

atom

partial equation that

that is

H2O

to the side of the

H, and add one

deficient in

OH"

ion to the

opposite side. 4.

To each

on the 5.

left

partial equation,

If necessary,

make

the

Add

in the

Chapter

11

such a way that the net charge

on the

right side.

lost in

one

partial equation equal the

the partial equations. In the addition, cancel terms

Reactions

in

number

other partial equation.

sides of the final equation.

288

in

multiply one or both partial equations by numbers that will

number of electrons

of electrons gained 6.

add electrons

side of the equation equals the net charge

Aqueous

Solution

common

to

both

Most oxidation-reduction equations may be balanced by the ion-electron method, which is especially convenient for electrochemical reactions and reactions of ions in water solution. However, several misconceptions that can arise must be pointed out. Half reactions cannot occur alone, and partial equations do not represent complete chemical changes. Even in electrochemical cells, where the two half reactions take place

at different electrodes, the

two half reactions always

occur simultaneously.

Whereas the

partial equations

probably represent an overall,

way an oxidation-reduction

view of the

the same reaction in a beaker may not take place method should not be interpreted as necessarily giving

cell,

by which a reaction occurs. reaction

It is, at

if

in this

way

the correct

4+

—^80^ + 6+

5+

SOj- + CIO3

3

at all. The mechanism

times, difficult to recognize whether a given

a legitimate example of an electron-exchange reaction.

is

not detailed,

reaction occurs in an electrochemical

The

reaction

+

CIO2-

made to take place in an electroand can be balanced by the ion-electron method. However, this reaction has been shown to proceed by direct oxygen exchange (from CIO3 to SOj") and not by electron exchange.

looks hke an electron-exchange reaction, can be

chemical

4

cell,

Arrhenius Acids and Bases The

and bases that are

several concepts of acids

Chapter

14.

The Arrhenius concept of

in current use are the topic

acids and bases, the oldest of these,

is

of

pre-

sented in this section.

An ions,

acid

defined as a substance that dissociates in water to produce

is

which are sometimes shown as H'^(aq)

H— O:

+

H— Cl:(g)

HCl

HjO^

For example,

H— O—

(aq)

+

:Cl:

(aq)

H

H Pure

ions.

gas consists of covalent molecules. In water, the H^ (which is nothing a proton) of the HCl molecule is strongly attracted by an electron

more than

atom of a H2O molecule. The transfer of the proton to the HjO molecule produces what is called a hydronium ion (HaO"^) and leaves behind a pair of the

Cr

O

ion.

in water solution, which is indicated by the symbol (aq) of the ion. This symbolism does not give the number formula placed behind the of water molecules associated with each ion. The number in most cases is not known and in many cases is variable. The H^ ion, however, is a special case. The positive charge of the H ^ ion, the proton, is not shielded by any electrons and in comparison to other ions is extremely small. The H* ion, therefore, is strongly attracted to an electron pair of a HjO molecule. There is, however, evidence that the H3O* ion has three additional water

Every ion

is

hydrated

an ion that has the formula H9O4 Other evidence supports the idea that several types of hydrated ions exist simultaneously in water solution. Some chemists, therefore, prefer to represent the hydrated proton as

molecules associated with

it

in

H+(aq). The process in which

HCl molecules dissolve

.

in

water would be indicated

11.4

Arrhenius Acids and Bases

289

:

HCl(g)

—

H + (aq) +

Cr(aq),

In the Arrhenius system, a base

OH",

is

a substance that contains hydroxide ions,

or dissolves in water to produce hydrated hydroxide ions, OH"(aq):

NaOH(s) Ca(OH)2(s)

The only

—Na —

+ OH"(aq)

+ iaq)

Ca2 + (aq) + 20H-(aq)

soluble metal hydroxides are those of the group

Ba(OH)2, Sr(OH)2, and Ca(OH)2 of group

II

I

A

elements and

A. Insoluble hydroxides, however,

react as bases with acids.

The

reaction of an acid and a base

thesis reaction in

which water

HjO + laq) + OH may

which

is

is

called a neutralization,

which

produced. The net ionic equation

is

a meta-

is

—^IHjO

(aq)

also be written

— H2O

H^(aq) + OH-(aq)

Ionic equations for two neutralization reactions are

Ba^ + (aq)

+ 20H-(aq) + 2H^(aq) + 2Cr(aq)

—

Ba^^(aq)

+ 3H

Fe(OH)3(s)

+ (aq)

+ SNOjlaq)

—

+

2

Cr(aq) +

2

H.O

Fe^^(aq) + SNOjtaq) + 3H2O

The barium chloride (BaClj) and

iron(III) nitrate [Fe(N03)3] produced by these which are ionic compounds with cations derived from bases and anions derived from acids. Acids are classified as strong or weak depending upon the extent of their dissociation in water (see Table 11.3). A strong acid is 100% dissociated in water solution. The common strong acids are HCl, HBr, HI, HNO3, H2SO4 (first H"^ dissociation only), HCIO4, and HCIO3. Other common acids are weak acids, which are less than 100% dissociated in water solution. Acetic acid (HC2H3O2), for example, is a weak acid

reactions are called

:.h

H2O + HC2H302(1) The

reversible

arrow (^)

both directions. In a

HC2H3O2

of the

^ is

H30^(aq) + C2H302(aq)

used in

equation to show that reactions occur

this

M solution of acetic acid, a balance

1

100%

ionic

one

an aqueous solution of ammonia,

is

pure. There are a few molecular

H2O

NH3(g) +

In this reaction, the to

form the

in

achieved with 0.4%

dissociated into ions.

All soluble metal hydroxides are strong bases. After

when

is

—

weak

all,

these substances are

bases.

The most common

NH3:

NH;(aq) + OH-(aq)

ammonia molecule

ammonium

accepts a proton from the water molecule

ion and the hydroxide ion.

The

reaction, however,

is

not complete.

Acids that can lose only one proton per molecule (such as HCl,

and 290

HNO3)

Chapter

11

are called monoprotic acids.

Reactions

in

Aqueous

Solution

Some

acids can lose

HC2H3O2,

more than one

proton per molecule; they are called polyprotic acids. for example, can lose two protons:

H2S04(1)

+ H^O

—

HS04-(aq)

+ H^O

^HjO

Only one proton

is

A molecule of sulfuric acid,

HjO^laq) + HS04(aq) + iaq)

neutralized

+ SOr(aq)

if

mol of H2SO4

1

is

reacted with

1

mol of

NaOH: H2S04(aq)

The

salt

obtained,

NaHS04,

is

NaHS04(aq) + H2O called

an acid

salt

because

it

contains an acid

mol of H2SO4 is reacted with 2 mol of NaOH, both neutralized. The normal salt. Na2S04, is obtained:

hydrogen. If are

—

+ NaOH(aq)

1

H2S04(aq)

The acid

salt

+ 2NaOH(aq)

—

can be reacted with

NaHS04(aq) + NaOH(aq) The products of

Na2S04(aq)

NaOH

—

to

acid hydrogens

+ lU^O

produce the normal

Na2S04(aq) +

salt:

H2O

the neutralization of a polyprotic acid, therefore,

the quantities of acid

Phosphoric acid, H3PO4, has three acid hydrogens, and three obtained from

NaH2P04

5 Acidic

depend upon

and base employed. salts

can be

it:

Na2HP04

Na3P04

and Basic Oxides

The oxides of metals are called basic oxides. The oxides of the group I A metals and those of Ca, Sr, and Ba dissolve in water to produce hydroxides. All these oxides are ionic. When one of them dissolves in water, it is the oxide ion that reacts with the water

O^

(aq)

The oxides

+ H2O

—20H-(aq)

(as well as the

hydroxides) of other metals are insoluble in water.

Nevertheless, metal oxides and hydroxides are chemically related. heated,

most hydroxides are converted

Mg(OH)2(s)

—

When

into oxides:

MgO(s) + H20(g)

Metal oxides, as well as hydroxides, can be neutralized by acids:

MgO(s) + 2H^(aq) Mg(OH)2(s)

+ 2H^(aq)

The insoluble Fe203 to

—Mg^^(aq) + H2O —Mg'^(aq) + 2H2O

will react

with acids, even though

it

will

not react with water

produce a hydroxide: 11.5

Acidic and Basic Oxides

291

::

FcjO^ls)

:

—2Fe^

+ 6H^(aq)

+ (aq)

Most of the oxides of the nonmetals

+ 3H2O

are acidic Qxides.

Many of them

react with

water to produce oxyacids:

H2O

—

CUO, + H2O

*

6H2O

—

CI2O +

2HOC1

— 2HCIO4 N2O5 + H2O —2HNO3

P4O10 +

SO3 + SO2 +

4H3PO4

H2O— H2SO4 H20^ H2SO3

CO2 + H2O

^

H2CO3

two cases, both the oxides [S02(g) and C02(g)] and the acids [H2S03(aq) and H2C03(aq)] exist in the solutions. For some nonmetal oxides, there are no

In the last

corresponding acids.

Oxides of nonmetals will neutralize bases. The same products are obtained from the reaction of an acidic oxide as are obtained from the reaction of the corresponding acid '

H2S03(aq) S02(g)

+ 20H (aqj^SO^

(aq)

+ 2H2O

+ 20H

(aq)

+ H2O

(aq)

—

SO5

Mortar consists of lime [Ca(OH)2], sand [Si02], and water. The

initial

hardening

of mortar occurs when the mortar dries out. Over a long period of time, however, the mortar sets by absorbing C02(g)

Ca(OH)2(s)

Some

+

C02(g)

from the

air

and forming insoluble CaCOs

—CaCOjis) + H2O

oxides have both acidic and basic properties (for example.

BeO and

AI2O3). They are called amphoteric oxides and are formed principally by the

elements

in the center

of the periodic table, near to the borderline between the

metals and the nonmetals: Al203(s) Al203(s)

+ 20H

—2Al^ + H2O — AUOH);

+ 6H^(aq)

(aq)

+

3

+ (aq)

2

(aq)

aluminate ion

ZnO(s) ZnO(s)

+ 2H + (aq)

+ 20H'(aq) + H2O

—*Zn^

—

+ (aq)

+ H2O

Zn(OH)^-(aq) zincate ion

Acidic and basic oxides react directly, and industrial significance. In the

(limestone)

is

used as a

292

Chapter

11

Reactions

in

manufacture of pig iron

of these reactions are of (see Section 23.6),

CaC03

The CaC03 decomposes into CaO and CO2 at the blast furnace. The CaO, a basic oxide, reacts with SiOj,

flux.

high temperatures of the

an acidic oxide present the unwanted Si02

many

in the iron ore, to

Aqueous

Solution

form

slag

(CaSi03) and thereby remove

CaC03(s) In the is

+

Si02(s)

—

open hearth process

often Uned with

CaO

or

+

CaSiOjfl)

COaCg)

for the manufacture of steel

MgO

(basic oxides) to

from pig iron, the hearth remove the oxides of siUcon,

phosphorus, and sulfur (acidic oxides), which are present

in the pig iron as

impurities.

made from various combinations of acidic and basic oxides. Ordinary made from lime (CaCOa), soda (NajCOj), and sihca (SiOj). The corresponding basic oxides are CaO and Na20, the acidic oxide is SiOj, and the Glass

is

soft glass

is

product

a mixture of sodium and calcium

is

made

Substitutions are sometimes oxide,

is

substituted for a part of the SiOj a borosilicate (Pyrex) glass

for lenses).

.6

(light

as a part of the basic-oxide component produces flint glass (used The use of some basic oxides produces colored glasses; for example, green), CrjOj (dark green), and CoO (blue).

Some common

1.

salts

Aqueous

of

Acids and Salts

acids are listed in Table 11.3. Rules for

naming

these

compounds

derived from them follow.

named by combined with hydrogen. are used followed by the word acid:

solutions o{ binary compounds that function as acids are

modifying the root of the name of the element that

The

an acidic produced.

PbO

Nomenclature

and the

is

,

The use of

FeO

silicates.

for these oxides. If boric oxide,

prefix hydro-

and the

Table11.3

suffix -ic

Some common

is

acids

Binary

ACIDS Compounds Polyprotic Acids

Monoprotic Acids

HF*

hydrofluoric acid

HCI

hydrochloric acid

HBr

hydrobromic acid

HI

hydroiodic acid

H,S*

hydrosulfuric acid

Ternary Compounds

Monoprotic Acids

*

•*

Weak

Polyprotic Acids

HNOj

nitric

H2SO4**

sulfuric acid

HNO2*

nitrous acid

HjSOj*

sulfurous acid

HCIO4

perchloric acid

H3PO4*

phosphoric acid

HCIO3

chloric acid

H2CO3*

carbonic acid

HCIO,*

chlorous acid

HOCI*

hypochlorous acid

H3BO3*

boric acid

HC2H3O 2*

acetic acid

acid

acid.

The second dissociation

is

weak.

11.6

Nomenclature

of

Acids and Salts

293

:

HCl forms H2S forms 2.

Fe(OH)2

:

)

:

)

hydrochloric acid hydrosulfuric acid

Metal hydroxides are niimed

Mg(OH)2

3.

:

manner described

in Section 5.8:

magnesium hydroxide

is

hydroxide or ferrous hydroxide

iron(II)

is

in the

Salts of bifiary acids are themselves binary

the customary -ide ending.

They

are

compounds, and

named according

names have

their

to the rules given in Section

5.8.

Ternary acids are composed of three elements.

4.

compound

the

a.

If

is

called

When oxygen is the third element,

an oxyacid.

an element forms only one oxyacid, the acid is named by changing the -ic and adding the word acid:

ending of the name of the element to

H3BO3 b.

two common oxyacids of the same element, the ending -ous is naming the oxyacid of the element in its lower oxidation state; the

If there are

used

in

ending

used to denote the higher oxidation state (see Table

-ic is

HNO2 HNO3 c.

boric acid

is

is

nitrous acid

is

nitric acid

There are a few

series

1

1.3):

of oxyacids for which two names are not enough. in Table 11.3. The prefix hypo- is

See the names of the oxyacids of chlorine

added

to the

name

of an -ous acid to indicate an oxidation state of the central

element lower than that of the -ous acid

HCIO2

HOCl The

chlorous acid (CI has an oxidation number of 3 + hypochlorous acid (CI has an oxidation number of

is

is

prefix per-

is

added

of the central atom that

HCIO3 HCIO4

to the is

name

of an

-ic

)

1

+

acid to indicate an oxidation state

higher than that of the

-ic

acid

number of 5 + number of 7 +

IS

chloric acid (CI has an oxidation

is

perchloric acid (CI has an oxidation

)

The names of the anions of normal salts are derived from the names of the acids from which the salts are obtained. The -ic ending is changed to -ate. The -ous ending is changed to -ite. Prefixes, if any, are retained 5.

S04~ (from sulfuric acid) is the sulfate ion OCl~ (from hypochlorous acid) is the hypochlorite The name of the

name

is

Fe(C104.)3

Chapter

11

if

obtained by combining the

of the anion

NaN02

294

the salt itself

ion

sodium is

nitrite

iron(III) perchlorate or ferric perchlorate

Reactions

in

Aqueous

Solution

name of

the cation with

Figure 11.1 An acid-base titration, (a) The buret contains the standard solution. The

unknown solution and the flask, (b) Solution

indicator are placed is

Eq jivalence point changes color.

to the flask.

in

added from the buret is

reached when

the indicator

6.

In

naming the anion of an acid salt, the number of acid hydrogens must be indicated. The prefix mono- is usually omitted

retained by

the anion

H2PO4 HP04~ PO4" The

is

the dihydrogen phosphate ion

is

the hydrogen phosphate ion

(the

prefix bi-

anion of the normal

may be

used

in

salt) is the

place of the

phosphate ion

word hydrogen

in the

name

of the anion

of an acid salt derived from a diprotic acid:

HCO3 HSO3

7

is

is

hydrogen carbonate ion or the bicarbonate ion

the

the hydrogen sulfite ion or the bisulfite ion

Volumetric Analysis

A is

is one that relies on the measurement of the volume of a an exactly known concentration. A procedure called a titration

volumetric analysis

solution that has

employed

(see

Figure

11.1).

called a standard solution, until the reaction

called a buret.

solution to be

unknown

is

is

In a titration, a solution of

complete. The standard solution

The buret

is

withdrawn

known

concentration,

added to a measured volume of an unknown solution fitted

in

controlled amounts.

solution, or a weighed

is

placed

in

a graduated tube

with a stopcock at the lower end to permit the

mass of a

A

measured volume of the

unknown dissolved in water, is substance known as an indicator.

solid

placed in a flask together with a few drops of a

The standard solution from the buret is slowly added

to the flask until the indicator

11.7

Volumetric Analysis

295

:

changes color. Throughout the addition, the contents of the flask are kept well mixed by swirling. At the equivalence point, as shown by the color change of the indicator, equivalent amounts of the two reactants have been used. The volume of standard solution employed

is read from the buret. Three types of volumetric analyses are in use. They are based on precipitation reactions, acid-base neutralizations, and oxidation-reduction reactions. These three types are illustrated in the examples that follow.

Example

11.5

effluent from a manufacturing process is analyzed for CI" content. A 10.00 g sample of the waste water requires 30.20 ml of a 0.1050 AgNOj solution for

An

M

What is

reaction.

the percent CI

waste water? The equation for the reaction

in the

is

Cr(aq) + AgN03(aq)

—

AgCl(s)

+

N03"(aq)

Solution First

we

number of moles of AgN03

^ = 30.20 .^.^ mo AgNO, ^ ,

?

find the

.

/0.1050 mol

,

,

ml so n

^

= From 1

and

=

mol CI

1

V

mol

we

sample

7gC|- =

/

AgN03

see that

AgNOj

since the atomic weight of CI

in the

AgNOjX ~

lOOOmlsoln

3.171 X 10"^ mol

the chemical equation,

that have been used:

in the

following

3.171

X

is

35.45 g/mol,

we can

find the

mass of CI"

way /

IO^'molAgNO,(

1

mol CI"

\/35.45

gCr\

,_^^,^^^pj (^^^^-j

= 0.1124gCr The mass percent of CI "

in the

sample

is

\ 10.00 g sample/

Example

11.6

A

25.00 g sample of vinegar, which contains acetic acid (HC2H3O2), requires NaOH solution for neutralization. What is the mass percent 37.50 ml of 0.4600

M

of acetic acid

in the

vinegar? The equation

NaOH(aq) + HC2H302(aq)

296

Chapter

11

Reactions

in

Aqueous

—

*

Solution

is

NaC2H302(aq) + H2O

Solution

The number of moles of NaOH employed can be found 0 ?

I

way:

/O ^eOO mol NaOH\ 1-7 37.50 ml solr n 1000 ml soFn V /

r^u = NaOH XT

mol

in the following

1

=

10"2 mol

1.725 X

NaOH

Since the equation shows 1

mol HC2H3O2 =

1

mol

NaOH

and since the molecular weight of

?gHC2H302 =

HC2H3O2

60.05 g/mol,

is

10- mol NaOH ([^^S^^^^^]

1.725 X

\

=

1.036 g

The mass percent of

mol

NaOH

/ \

1

mol HC2H3O2 /

HC2H3O2

HC2H3O2

/1.036gHC2H3O2

1

in the vinegar

sample

100% = 4.144% HC2H3O2

is

in

vinegar

25.00 g vinegar

Example 11.7

A 0.4308

g sample of iron ore

Fe^"^ state. This solution

is

is

dissolved in acid and the iron converted into the

reacted with a solution of potassium permanganate.

M

The reaction requires 27.35 ml of 0.02496 KMn04 The chemical equation is

solution.

What

is

the

mass

percent of iron in the ore?

8H+ + 5Fe^+ + MnO:

—-5Fe^* + Mn^^ + 4H2O

Solution First

?

we

find the

mol

number of moles of

KMn04 = =

KMn04 consumed

in the reaction

—

27.35 ml soln

/0.02496 mol KMn04^ .^^^ sol 1000 ml Y

6.827 X 10

mol

,

KMn04

Since the equation shows 5

mol Fe^^ =

1

mol

KMn04

and since the atomic weight of iron

?

g ^ Fe

=

=

6.827 x lO--* mol

0.1906

is

55.85 g/mol,

KMn04

mol Fe

/

5

\\

mol

.

^

\ /55.85 g Fe

„

KMn04y

\

1

mol Fe

gFe

11.7

Volumetric Analysis

297

:

This

is

the quantity of iron found in the 0.4308 g sample of ore. Therefore,

and

11.8 Equivalent Weights

Normal Solutions way that was used in the on the basis of the mole and molar solutions. There is, however, another method, one based on what are called equivalents and normal solutions. The definition of an equivalent depends upon the type of reaction being considered. The definition, however, is always framed so that one equivalent of a given reactant will react with exactly one equivalent of another. Two types of reactions for which equivalents are defined are neutralization reactions and oxidation-reduction reactions. The mass of one equivalent of a compound is called an equivalent weight. In general,

All volumetric analysis problems can be solved in the last section,

=

equivalent weight

molecular weight (11.1)

a

where the value of a depends upon the type of reaction considered. 1.

For neutralization

reactions, equivalent weights are based

H"^(aq) ion reacts with one

amount of

OH"(aq)

ion.

One

on the

fact that

equivalent weight of an acid

is

one the

one mole of H^(aq) ions, and one equivalent weight of a base is the amount of the base that supplies one mole of OH"(aq) ions. The value of (/ in Equation 11.1, therefore, is the number of moles of H^(aq) supplied by one mole of the acid or the number of moles of OH~(aq) supplied by one mole of the base in the reaction considered. 2.

the acid that supplies

For oxidation-reduction

reactions, equivalent weights are based

on oxidation

number changes. In an oxidation-reduction reaction, the increase in oxidation number of one element must equal the decrease in oxidation number of another. An equivalent is defined in terms of an oxidation number change of one unit, and a in Equation 11.1 is the total change in oxidation number (either up or down) that the atoms in a formula unit of the compound undergo in the reaction under consideration. The equivalent weight of KMn04 for a reaction in which the Mn04 ion IS reduced to the Mn'^ ion (a change in oxidation number of 5 units) is the formula weight of KMn04 divided by 5. The equivalent weight of KMnO^. for a reaction in which the MnO^ ion is reduced to Mn02 (a change in oxidation number of

3 units) is the

The

formula weight of

normality, N, of a solution

solute dissolved in

one

liter

KMn04 divided by is

the

3.

number of gram equivalent weights of

of solution. The normality of a solution and

its

molarity are related

N ^ aM

298

Chapter

11

Reactions

(11.2)

in

Aqueous Solution

The number of equivalents of A in a sample of a solution of A, e^, can be obtained by multiplying the volume of the sample, (in liters), by the normality (which is the number of equivalents of in one liter of solution) of the solution,

A

= VkNa

eA

By

design,

Cf^

(K^

—

e-^,

in liters)

(11.3)

and therefore (11-4)

^^aA^a^^^b^b Since a

volume term appears on both

be used to express V\ and

same

sides of

Equation

1 1 .4,

any volume unit can

provided that both volumes are expressed

in the

unit.

Example 11.8 (a)

What is the normality of a solution of H2SO4 if 50.00 ml of the solution

37.52 (b)

ml of 0.1492

What

is

N NaOH

requires

for complete neutralization?

the molarity of the solution?

Solution

V^N^ = V^N^

(a)

(b)

(50.00 ml)A^A

=

(37.52 ml)(0.1492 A^)

N/,

=

0.1120

Since

1

mol of H2SO4

is

2 equivalents, a

=

2.

Therefore

N = aM 0.

1

120 equiv/liter

=

(2

M

=

0.05600 mol/liter

equiv/mol)A/

Example 11.9

A

0.4308 g sample of iron ore is dissolved in acid and the iron converted to the Fe^'*^ state. The solution is reacted with a solution of potassium permanganate. The reaction, in which Fe'^ is oxidized to Fe-^"", requires 27.35 ml of 0.1248

N

KMn04. What

is

the

in the

mass percent of iron

ore?

Solution This problem solve

it

is

the

same

as that given in

Example

by using equivalents and normality.

KMn04

We find

Here, however, we will number of equivalents of

11.7.

the

used

1 1

.8

Equivalent Weights and Normal Solutions

299

— = =

liter )(0. 1248 equiv/liter) 3.413 X 10"^ equiv

(0.02735

The number

of equivalents of

alents of iron

KMn04

used

number of The equivalent weight of iron,

to 3

+

).

as the

number

of equiv-

the iron increases by one unit (from

In the reaction, the oxidation

2+

same

the

is

ore sample.

in the

therefore,

is

same

the

as the atomic

weight, 55.85:

?gFe =

3.413 X

10^ 3

r^,

equiv Fe

Fe — ^— Fe

/^55.85 g

yl equiv

The mass percent of Fe 0.1906 g Fe

in the ore

=

0.1906

sample, therefore,

100% = 44.24% Fe

gFe

/ is

in ore

0.4308 g ore

Summary The 1.

topics that have been discussed in this chapter are

Aqueous metathesis

reactions,

which occur because of

the forination of a precipitate, gas, or a

weak

The assignment of oxidation numbers compounds. 2.

3.

5.

The Arrhenius concept of acids and

6.

Acidic and basic oxides.

7.

Nomenclature of acids and

8.

Volumetric analyses based on precipitation, neutral-

electrolyte.

to the

atoms

in

ization

bases.

salts.

and oxidation-reduction

reactions,

and the

stoi-

chiometric calculations involved in these procedures.

O.xidation and reduction.

How to balance oxidation-reduction equations by the oxidation-number method and by the ion-electron method.

9. The use of equivalent weights and normality in chiometric calculations.

4.

stoi-

Key Terms Some

of the more important terms introduced in this chapter are listed below. Definitions for terms not included in this list may be located in the text by use of the

Base (Section

pound

1

1

.4)

acts with water to

A covalent compound of hydrogen water to produce H^(aq) ions (or

.\cid (Section 11.4)

that dissociates

in

Arrhenius system, a comproduce OH'(aq) ions.

the

An

Basic oxidu (Section 11.5)

index.

H3O+

In

that dissociates in water to

oxide of a metal that

Disproportionaf ion (Section 11.3)

substance

ions).

is

re-

form a base.

both

oxidized

A

and

reaction in which a

reduced;

an

auto-

oxidation-reduction reaction. ,\d(iic oxicfi (Section

reacts with water to .Acid salt (Section

1

1 1

An

.5)

form an

1.4)

A

oxide of a nonmetal that Equivalent wci'

A =

log

A

b).

plot of log

is A:

in the

form of an equation for a straight Hne

against (l/T)

is

a straight line with a slope of

— E^/2.303R

and a j-intercept of log^ (Figure 12.10). If values of k are deterfor the mined at several temperatures and the data plotted in this manner, reaction can be calculated from the slope of the curve and A can be obtained by taking the antilogarithm of the _v-intercept.

The curve shown 2N205(g)

The

—

in

4N02(g) + 03(g)

slope of the curve

^

Figure 12.10 pertains to the first-order reaction:

-5350

—5350 K, from which we obtain

is

K

2.30/?

E,

The

= =

(5350 K)(2.30)[8.31 J/(K mol)] 102,000 J/mol

v-intercept

is

=

13.51

A =

3.2 X

log^

13.51,

and therefore

lO^-'/s

The values of E^ and A for

324

two

= 102kJ/mol

for a reaction can also be

found from the rate constants is and at Tj is k2, then

different temperatures. If the rate constant at

Chapter 12

Chemical Kinetics

-

23§Rf,

(12.6)

= ''^^ '

1303^

(12.7)

=

^""^'^

^""^^

and

''^'^

Subtraction of Equation 12.7 from Equation 12.6 gives

Since logx

—

log v

is

log(x/>'),

(12.9)

K^)=2:^(7,-7;) or

'°«ft)

=

2:l^(%jf)

If this

equation

ship

obtained

is

=

2.303.

The uses of the

solved for the energy of activation, E„, the following relation-

is

(^),„g(^) last

two equations are

illustrated in the following examples.

Example 12.2 For the reaction

2NOCl(g)

—

the rate equation rate of

The (mol

2NO(g) + is

production of CI,

rate constant, •

s) at

400 K.

Cl2(g)

A:,

is

What

= A[NOCl]-

2.6 x is

10'^ liter/(mol

•

s)

at

300

K

and 4.9 x 10

* hter/

the energy of activation, £„, for the reaction?

Solution Let

T,

=

300

K

=

400

K 12.6

Rate Equations and Temperature

.= 2.6 X 10"^ liter/(mol-,s)

= R

is

4.9 X 10"* liter/(mol

8.31 J/(K

mol),

•

s)

and substitution into Equation

=

2.30/?

=

.1 0 J/(K mol)](^ 2.30[8.31

= = =

[19.1J/(K mol)](1200K)log(1.88 x 10*

12.11 gives

log

w«

„-,/^

(300K)(4O0K)\

[4.9

^^^_3^^^j ^"^ 2.6

=

98,000 J/mol

98 kJ/mol

12.3

Given the data found

in

Example

12.2, find the

value of k at 500 K.

Solution Let

K = 500 K

r,

=

^1

=4.9

A,

= unknown

400

=

* liter/(mol s)

X 10

9.8 X

10* J/mol

use Equation 12.10:

log

kj

230R\ 9.8 X 10* J/mol

~ = =

/500

K -

(5.13 X

10^ K)(5.00 X

10"*/K)

2.57

antilog2.57

=

3.7 x

10'

or A-2

326

= = =

(3.7 X

10-)Ai

(3.7

X 102)[4.9 X 10"* liter/(mol

0.18

liter/

Chapter 12

K\

2.30 [8.31 J/(K mol)] \(400 K)(500 K) /

Therefore,

= ^ ki

400

(mol

s)

Chemical Kinetics

s)]

s)

\

x 10-Miter/(mol s)j

(22,900 J/mol)(4.28)

Example

We

x 10-Miter/(mol-

T is exponential, a small change in T causes Any change in the rate constant is, of course, reflected

Since the relation between k and a relatively large change in k. in the rate

of the reaction. For the reaction considered

in

Examples

12.2

and

12.3,

a 100° increase in the temperature causes the following effects:

300

400

K to 400 K K to 500 K

The marked

rate increases 18,800 times rate increases 367 times

of a temperature increase

is obvious. Notice, however, that the low temperatures than at high temperatures. The energies of activation of many reactions range from 60 kJ/mol to 250 kJ/mol, values which are on the same scale as bond energies. For a 10 rise in temperature, from 300 K to 310 K, the rate of reaction varies with the energy of activation in the following way:

effect

reaction rate

.7

is

more

affected

at

=

60 kJ/mol

rate increases about 2 times

=

250 kJ/mol

rate increases about 25 times

Catalysts

A catalyst

is

a substance that increases the rate of a chemical reaction without

being used up in the reaction.

A

catalyst

may

be recovered unchanged at the end

Oxygen can be prepared by heating potassium chlorate (KCIO3) by itself. Or, a small amount of manganese dioxide (MnO,) can be used as a catalyst for this reaction. When Mn02 is present, the reaction is much more rapid of the reaction.

and the decomposition of KCIO3 takes place

at a satisfactory rate at a lower

temperature

2KC103(s) "^""^^ 2KC1(1)

The

catalyst

is

+

302(g)

written over the arrow in the chemical equation since a catalyst

does not affect the overall stoichiometry of the reaction. The

Mn02 may

be

recovered unchanged at the conclusion of the reaction.

The mere presence of

A

a catalyst does not cause the effect

on the reaction

catalyzed reaction takes place by a pathway, or mechanism, that

is

rate.

different

from the one that the uncatalyzed reaction follows. Suppose, for example, that an uncatalyzed reaction occurs by collisions between X and Y molecules: 1

X+ Y

—XY

The catalyzed reaction might follow 1.

2.

x + c XC + Y

where

C

is

mechanism consisting of

—XC

— XY

+ C is used in the first step and regenand over again. Consequently, only

the catalyst. Notice that the catalyst

erated in the second.

a small

a two-step

amount of a

It is,

therefore, used over

catalyst

is

needed to do the job.

12.7

Catalysts

327

:

:

Reaction coordinate

Reaction coordinate

Uncatalyzed reaction

Catalyzed reaction

Figure 12.11

energy diagrams

Potential

A catalyst,

for a reaction in tiie

of

a catalyst

works by opening a new path by which the reaction can

therefore,

The catalyzed path has

take place.

absence and presence

a lower overall energy of activation than the

uncatalyzed path does (see Figure 12.11), which accounts for the more rapid reaction rate.

1.

Two

additional points can be derived from Figure 12.1

The enthalpy change.

A//, for the catalyzed reaction

is

the

same

1

as the

A//

for

the uncatalyzed reaction. 2.

For

that

it

same effect on the reverse reaction The energy of activation for the reverse reaction, the same extent that energy of activation for the

reversible reactions, the catalyst has the

has on the forward reaction.

lowered by the catalyst to r is forward reaction, E^,j, is lowered.

A Jiomogeneous catalyst is present in the same phase as the reactants. An example of homogeneous catalysis in the gas phase is the effect of chlorine gas on the decomposition of dinitrogen oxide gas. Dinitrogen oxide, N2O, is relatively unreactive at room temperature, but at temperatures near 600°C it decomposes according to the equation

2N20(g)— 2N2(g) + The uncatalyzed

reaction

02(g) is

thought to occur by means of a complex mechanism

that includes the following steps:

1.

Through

energy to

N20(g) 2.

collisions

split

between

N2O

molecules,

—N2(g) +

The oxygen atoms

+ N20(g)

Cfiapter 12

N2O

molecules gain enough

0(g)

are very reactive.

molecules:

0(g)

some

apart

—

N2(g)

Chennical Kinetics

+

02(g)

They

readily react with other

N2O

The

N2 and O2 The O atom

products of the reaction are

final

.

is

a reaction inter-

mediate and not a final product. The energy of activation for the uncatalyzed reaction is about 240 kJ/mol.

The reaction is catalyzed by a trace of chlorine gas. The path that has been proposed for the catalyzed reaction consists of the following steps: 1.

At the temperature of the decomposition, and particularly some chlorine molecules dissociate into chlorine atoms

in the

presence of

light,

Cl^lg)

2.

—

2Cl(g)

The chlorine atoms

N20(g) + Cl(g) 3.

readily react with

—

N2(g)

+

NjO

molecules:

ClO(g)

The decomposition of the unstable CIO molecules follows

2C10(g)-^Cl2(g) + 02(g) Notice that the catalyst (CI2

is returned to its original state in the last step. The products of the catalyzed reaction (2 N2 and O2) are the same as those of the uncatalyzed reaction. CI and CIO are not products because they are used in )

final

steps that

foHow

lower than

The energy of activation about 140 kJ/mol, which is considerably

the ones in which they are produced.

for the reaction catalyzed

by chlorine

is

for the uncatalyzed reaction (240 kJ/mol).

and catalyst are present in different on the surface of the catalyst in these

In heterogeneous catalysis the reactants phases. Reactant molecules are adsorbed processes,

and the reaction takes place on

that surface. Adsorption

which molecules adhere to the surface of a in

solid.

is

a process

Charcoal, for example,

is

in

used

gas masks as an adsorbent for noxious gases. In ordinary physical adsorption, the

molecules are held to the surface by

Heterogeneous

catalysis,

London

forces.

however, usually takes place through chemical ad-

) in which the adsorbed molecules are held to the surface by bonds that are similar in strength to those in chemical compounds. When these bonds form, the chemisorbed molecules undergo changes in the arrangement of their electrons. Some bonds of the molecules may be stretched and weakened, and

sorptio n (or chemisorption

some cases, even broken. Hydrogen molecules, for example, are believed to be adsorbed as hydrogen atoms on the surface of platinum, palladium, nickel, and other metals. The chemisorbed layer of molecules or atoms, therefore, functions in

as a reaction intermediate in a surface catalyzed reaction.

The decomposition of the gold-catalyzed

1.

NjO

is

decomposition

catalyzed by gold. is

diagrammed

in

A

proposed mechanism for

Figure 12.12. The steps are

Molecules of N20(g) are chemisorbed on the surface of the gold:

N20(g)

—

N20(on Au)

The bond between the O atom and the adjacent N atom of a N2O molecule is bond breaks and an weakened when the O atom bonds to the gold. This

2.

N—O

N2 molecule

leaves

N20(onAu)

—N2(g) +

O(on Au)

" -

12.7

Catalysts

329

,

Au

Au Proposed mode

Figure 12.12

3.

Two O atoms on

of

N,0 on Au

the surface of the gold

combine

of

decomposition

to

form an O2 molecule which

enters the gas phase:

0(on Au) + 0(on Au) The energy of activation which

02(g)

for the gold-catalyzed decomposition

is

about 120 kJ/mol,

lower than £„ for either the uncatalyzed decomposition (240 kJ/mol) or the chlorine-catalyzed decomposition (140 kJ/mol). is

The second

step of the

mechanism of the gold-catalyzed decomposition is The rate of this step is proportional to the

believed to be the rate-determining step.

fraction of the gold surface that holds chemisorbed

the surface

is

covered, step 2

is

faster than

occupied. This fraction, however,

N20(g).

If the pressure

is

rate

N2O

molecules. If one-half

only one-quarter of the surface

is first

is

is

directly proportional to the pressure of

low, the fraction of surface covered will be low.

rate of the reaction, therefore,

and the decomposition

is

if

The

proportional to the concentration of N20(g),

order:

= AIN2O]

At high pressures of N^O, the surface of the gold becomes completely covered; Under these conditions, the reaction becomes zero order, is equal to that is, the rate is unatTected by changes in the concentration of N20(g): the fraction

rate

=

1

.

k

The gold surface is holding all the N2O molecules that it can, and the pressure of N20(g) is high enough to keep the surface saturated. Small changes in the pressure of N20(g) do not cause the chemisorbed NjO molecules to decompose any more slowly or rapidly.

The

and arrangement of atoms on the surface of a catalyst determine its activity. Lattice defects and irregularities are thought to be active sites for catalysis. The surface of some catalysts can be changed by the addition of substances called promoters, which enhance the catalytic activity. In the synelectronic structure

thesis of

ammonia

Cfiapter 12

Cfiemical Kinetics

A cross-sectional view

of a catalytic converter

used

in automobiles. Engine exhaust, which converter and forced to pass through dual beds of catalytic beads before exiting at the bottom and to the left. Air is inducted into the chamber between the catalytic beds. The beads contain Pt. Pd, and Rh and are designed to catalyze the oxidation of CO and hydrocarbons to CO, and the transformation of the oxides of nitrogen into N, and O,. General Motors Corporation.

enters on the right,

+

N2(g)

is

3H2(g)

an iron catalyst are

conducted

added to

is

—

to the top of the

2NH3(g)

made more

effective

when

traces of potassium or

vanadium

it.

Catalytic poisons are substances that inhibit the activity of catalysts. For example, small amounts of arsenic destroy the power of platinum to catalyze the preparation of sulfur trioxide from sulfur dioxide:

2S02(g)

+ 02{g)-^2S03(g)

Presumably, platinum arsenide forms on the surface of the platinum and destroys its

catalytic activity.

Catalysts are generally highly specific in their activity. In

some cases a given

substance will catalyze the synthesis of one set of products from certain reagents,

whereas another substance

will catalyze the synthesis

of completely different

products from the same reactants. In these cases, both reactions are possible and the products obtained are those that are produced most rapidly. Carbon monoxide and hydrogen can be made to yield a wide variety of products depending upon the catalyst

employed and the conditions of the

reaction.

CO

and Hj produce mixtures of hydrocarbons. One hydrocarbon produced, for example, is methane, CH4: If a

cobalt or nickel catalyst

CO(g)

On

is

+ 3H2(g)^CH4(g) +

the other hand,

methanol

is

used,

H20(g)

the product of the reaction of

a mixture of zinc and chromic oxides

CO(g)

+

2H2(g)

is

employed as a

CO

and H2 when

catalyst:

CH30H(g)

The catalytic converter installed on car mufflers is a recent application of surface Carbon monoxide and hydrocarbons from unburned fuel are present

catalysis.

automobile engine exhaust and are serious air pollutants. In the converter, and additional air are passed over a catalyst that consists of metal oxides. The CO and hydrocarbons are converted into CO2 and HjO, which are in

the exhaust gases

12.7

Catalysts

331

relatively harmless and are released to the atmosphere. Since the catalyst is poisoned by lead, unleaded gasohne must be used in automobiles equipped with

catalytic converters.

Many catalysts

industrial processes

known

depend upon

complicated substances catalyze cell synthesis.

The

large

life,

to

man. These extremely

processes such as digestion, respiration, and

number of complex chemical

body, and are necessary for

body because of

life

catalytic procedures, but the natural

more important

as enzymes are even

can occur

reactions that occur in the

at the relatively

low temperature of the

Thousands of enzymes are known, and each serves a specific function. Research into the structure and action of enzymes may lead to a better understanding of the causes of disease and the mechanism of the action of enzymes.

growth.

Summary The 1.

topics that have been discussed in this chapter are

How

reaction rates are expressed.

2. Rate equations, which give the quantitative relationship between reaction rate and reactant concentrations.

3. The collision theory, which accounts for the rate of a chemical reaction on the basis of effective collisions between the reacting molecules, and the transition state theory, which describes a step of a chemical reaction on the basis of the attainment of a transition state arrangement of the reacting molecules.

The molecularity of the steps of reaction mechanisms and the corresponding rate equations. 4.

5. Reaction mechanisms, which may consist of one step or of several and which describe the ways in which reactions occur on an atomic, molecular, or ionic level.

6. The effect of a temperature change on reaction rate; the Arrhenius equation. 7.

Catalysts and

how

they function.

Key Terms Some of the more important terms introduced in this chapter are listed below. Definitions for terms not included in this list

may

be located in the text by use of the index.

An unstable arrangemoment in the course

Activated complex (Section 12.3)

ment of atoms

that exists only for a

of a chemical reaction, also called a transition

with temperature and energy of activation.

reaction.

for a reaction in which, after

an

are repeated over and over again. is

a reactant in

the second of these steps

332

is

A

mechanism step, two steps

multistep

initiation

A

product of the first of the second, and a product of

a reactant in the

first.

Chapter 12

Chemical Kinetics

the rates

reactions.

A

theory that describes

reactions in terms of collisions between reacting particles

(atoms, molecules, or ions). KtTective collision (Section 12.3)

Chain mechanism (Section 12.5)

The study of

Chemical kinetics (Introduction)

and mechanisms of chemical

Collision theory (Section 12.3)

A

substance that increases the rate of a chemical reaction without being used up in the

these steps

by bonds that are similar in strength to chemical bonds, and in the process the adsorbed molecules become activated.

An equation that dehow the rate constant for a chemical reaction varies

C atalyst (Section 12.7)

reactant molecules are held to the surface of the catalyst

state.

Arriienius equation (Section 12.6)

scribes

Chemical adsorption (Section 12.7) A process through which solid, heterogeneous catalysts work. The adsorbed

particles (atoms, molecules,

A collision

between two

or ions) that results in a

reaction. f'juTgy

of activation (Section

12.3)

The

difference

in

energy between the potential energy of the reactants of a reaction and the potential energy of the activated complex.

Enzyme

A

(Section 12.7)

natural catalyst that

is

effective

a biochemical process (such as digestion, respiration,

in

and

cell synthesis).

Heterogeneous catalyst (Section 12.7) A catalyst that present in a different phase from that of the reactants.

Homogeneous

catalyst (Section 12.7)

present in the

same phase

A

is

catalyst that

is

A

Reaction intermediate (Section 12.3)

substance that

is

produced and used in the course of a chemical reaction and is, therefore, neither a reactant nor a product of the reaction.

as the reactants.

The number of reacting par(atoms, molecules, or ions) that participate in a single step of a reaction mechanism. A step may be iininsoit cular, bimolecular, or termolecuhir depending upon whether one, Molecularity (Section 12.4) ticles

two, or three particles react in

Rate equation (Section 12.2) A mathematical equation that relates the rate of a chemical reaction to the concentrations of reactants.

it.

Order of a chemical reaction (Section 12.2) The sum of the exponents of the concentration terms in the rate equation for the reaction.

Reaction mechanism (Section 12.5) The detailed description of the way a reaction occurs based on the behavior of

may

atoms, molecules, or ions;

more than one

include

step.

Reaction rate (Section 12.1) The rate at which a reaction proceeds, expressed in terms of the increase in the concentration of a product per unit time or the decrease in the concentration of a reactant per unit time; value changes

during the course of the reaction. Rate constant (Section 12.2) stant in a rate equation.

The proportionality conTransition

Rate-determininj; step (Section 12.5) a multistep reaction

how

mechanism

;

the

The slowest

step in

one that determines

state

theopi

(Section

A

12.3)

theory that

assumes that reacting particles must assume a specific arrangement, called a transition state, before they can form the products of the reaction.

fast the overall reaction proceeds.

Problems* Reaction Rates, Rate Equations

The

12.1

2

rate equation for the reaction

NO(g) + 2

H2(g)—*

N,(g)

+

2

[A]

[B]

Rate of formation of

(mol/liter)

(mol/liter)

(mol/liter-

0.03

0.03

0.3 X

0.06

0.06

1.2

0.06

0.09

2.7 X

H,0(g)

is second order in NO(g) and first order in Hjig). (a) Write an equation for the rate of appearance of Njlg). (b) If

would

(a)

an equation for the rate of disappearance of NO(g). Would k in this equation have the same numerical value as k in the equation of part

(b)

concentrations are expressed

in

mol

liter,

what

units

the rate constant, k, have'.' (c) Write

What What

12.4

The

expressed

C

s)

lO""^

X 10"* *

10

the rate equation for the reaction?

is is

the numerical value of the rate constant,

A

rate equation for the reaction in the

B

-(-

kl

C

is

form

(a)? 12.2

For a reaction

in

which

A and B form C, the following

rate of disappearance of

A =

k\^\Y

data were obtained from three experiments:

The value of the [A] (mol/hter)

[B] (mol liter)

0.30

0.15

(a) (b)

12.3

Rate of formation of (mol/liter-

7.0 X

0.60

0.30

2.8 X

0.30

0.30

1.4

What What

is is

and [A]

C

s)

is 0.

What

100 (units unspecified)

are the units of k

the rate of the reaction (in mol/liter(a)

zero order in A,

(b) first

s) if

order in A,

and

the reaction

(c)

is:

second order

A?

in

IQ-^

The rate equation for the reaction A -» B -I- C is expressed in terms of the concentration of A only The rate

X 10"^

the numerical value of the rate constant, in

rate constant, k.

0.050 mol/liter.

10"*

the rate equation for the reaction?

For a reaction

=

A-?

which A and B form C, the followfrom three experiments:

12.5

.

of disappearance of A is 0.0080 mol/liter s when [A] = 0.20 mol/liter. Calculate the value of k if the reaction is: (a) zero order in A, (b) first order in A, (c) second order in

A.

ing data were obtained

*

The more

difficult

problems are marked with

asterisks.

The appendix contains answers

to color-keyed problems.

Problems

333

:

For

12.6

2NO(g) +

and

is

second order

third order overall.

reaction of a mixture of 0.02 Cliig) in a

1

liter

container with

NO(g),

in

first

the rate of the reaction

(a)

half the

The

reversible;

is

NO + NO, -^2 NO, Assume

that [NO3] becomes constant after the reaction has been going on for a while (that is, NO3 is used as fast as it is produced). Show that the mechanism leads to the observed rate equation.

According to the collision theory, a first-order decomposition (A products) proceeds by the following

*12.14

A

single-step reaction

NO,Cl(g)

+ NO(g) E^j

is

28.9 kJ

and

41.8 kJ.

is

Draw

A

4-

A

NO,(g) + ONCl(g)

-f

A common

12.8

and serious mistake

is

rate equation for reaction can be derived

assume that the from the balanced

to

chemical equation for the reaction by using the coefficients of the chemical equation as exponents in the rate expression. Why cannot rate equations be derived in this way?

Why

12.9

are

some

collisions

between

molecules

of

reactants not effective?

In step two A molecules collide, energy is transferred, and one molecule, marked A*, attains a high-energy state.

A to

How well

would you expect

A* can cause

subsequent collision of 2).

Some A*

form the products

the process to

molecules, however, decompose

rate-determining step. Derive

in the

a rate equation from this

mechanism by assuming

that

[A*] becomes constant after the reaction has been going on for a while (that is. A* is used as fast as it is produced).

Remember

much

smaller

The following mechanism has been postulated

for the

that since step 3

is

slow,

is

A:,-

12.10

the synthesis of perbromates?

,

reverse (step

than

The synthesis of perbromates has only recently been accomplished. The best preparation involves the oxidation of bromates in alkaline solution by fluorine. What reason can you give to account for the difficulty encountered in

\+ h

'

products 1

and A// on the diagram.

,,

A

A*

A

A

a

potential energy diagram for the reaction. Indicate E^j,

E^

^N.O,

steps

Cljfg) in a 0.5 liter container. 12.7

is

NO2 + NO3

NO2 + NO

order in

Compare the initial rate of mol NO(g) and 0.02 mol

NO(g) has been consumed, (b) the rate of the reaction when half the CUig) has been consumed, (c) the rate of the reaction when two-thirds of NO{g) has been consumed, (d) the initial rate of a mixture of 0.04 mol NO(g) and 0.02 mol CKfg) in a 1 liter container, (e) the initial rate of a mixture of 0.02 mol NO(g) and 0.02 mol

when

A-

N2O5

20NCl(g)

Cl^ig)-

the rate equation Cljlg),

The suggested mechanism

the reaction

The

*12.15

rate equation for the reaction

-30,

2O3

has been determined experimentally as

perbromates to function as oxidizing agents? rate of disappearance of

O = 3

[03]^ k

[O2]

Reaction Mechanisms, Catalysis

decomposition of ozone. Define the following: (a) activated complex, energy of activation, (c) reaction order, (d) catalyst, chemisorption, (f) rate-determining step, (g) reaction

12.11 (b)

(e)

intermediate, 12.12

The

(h)

+

H^lg)

at temperatures

first

zero-order reaction.

reaction

21Cl(g)

Klg) +

above 200

C

2HCl(g)

is first

order

H, and

first

*12.13

in ICl.

For the reaction

N,05(g)

+ NO(g)

^

O2 +

O

O +

O3

The in

Suggest a mechanism of two steps with the step rate-detennining to account for the rate equation.

order

O3

rate equation

^l^'3[N20 53[NO] A-,[NO,]

334

O3 20,

third step of the

Assume that

mechanism

is

the rate-determining

the concentration of

or A,.

Show how

the suggested

mechanism

leads to the

observed rate expression.

3N02(g)

N^O^

O

after the reaction has

A',

is

rate of disappearance of

A,

O becomesconstant been going on for a while (that is, O is used as fast as it is produced) and find an expression for [O]. Remember that A' 3 is much smaller than either step.

For the mechanism outlined in problem 12.15, the proposed activation energies for the three steps are 100 kJ, 4 kJ, and 24 kJ. The value of A// for the overall reaction is —285 kJ. Draw a potential energy diagram for the 12.16

The

O2 +

Chapter 12

+

A:3[NO]

Chemical Kinetics

reaction.

*12.17

The

K

rate expression for the reaction

1400 and 0.659 liter/mol s at of activation for the reaction?

2N02(g)

2NO(g) + 02(g) second order

+

NO(g)

in

02(g)

^

in OjCg).

The

What is the energy

is

^

The

first

K

energy of activation for the reaction?

N03(g)

N03(g)-^N0(g) +

The

12.23

02(g)

first

+ NO(g)-^^2N02(g)

N03(g)

500 K.

reaction: HI(g) + CH3l(g) CH4(g) -I- 12(g) order in each of the reactants and second order overall. The rate constant is 1.7 x 10"-^ liter/mol s at 430 and 9.6 x 10" ^ hter/mol s at 450 K. What is the

12.22

NO(g) and first order following mechanism has been suggested: is

1

at

reaction:

order

600

K

and

1.6 x

C2H4(g)

CjHsCUg)

C2H5CI. The

in

rate constant

10" "/s at 650 K.

is

What

+

HCl(g)

3.5 x is

is

10~*/s

the energy

of activation for the reaction?

The

third step of the

Assume

mechanism

is

the rate-determining

that the concentration of

NO

3 becomes conhas been going on for a while (that is, NO3 is used as fast as it is produced). Remember that A- 3 is much smaller than either fcj or A^2- Show that this mechanism leads to the observed rate equation.

step.

stant after the reaction

12.18

The reaction

12.24 For the reaction: NO,Cl(g) -1- NO(g) -> NO,(g) -IONCl(g), A is 8.3 X 10** and is 28.9 kJ/mol. The rate equation is first order in NOjCl and first order in NO. What is the rate constant. A:, at 500 K?

For the reaction: NO(g) + N,0(g) -* N02(g) -IA is 2.5 X 10" and is 209" kJ/mol. The rate equation is first order in NO and first order in N2O. What

12.25

N2(g),

is

CHM) +

CHjCKg) + HCl(g)

Cl2(g)

proceeds by a chain mechanism. The chain propagators are CI atoms and CH3 radicals, and it is believed that free

H atoms are not involved. Write a series of equations showing the mechanism and identify the chain-initiating, chain-sustaining, 12.19

Use

and chain-terminating

potential energy diagrams to explain

how

a

Does a catalyst affect the value of AH for the reaction? Can a catalyst slow a reaction down? Can found that

affects only the

forward reaction

of a reversible reaction? Explain each of your answers. 12.20

a

What

is

and which each

the difference between a heterogeneous

homogeneous

catalyst? Describe the

way

in

-f-

1

k for the reaction conducted

The

at

730

K?

NO(g) + N20(g)

-> NO.ig) + ^iig) each of the reactants and second order overall. The energy of activation of the reaction is 209 kJ/mol, and k is 0.77 liter/mol s at 950 K. What is the value of A' for a reaction conducted at 1000 K? is

catalyst functions.

a catalyst be

The reaction: C,H4(g) C2H6(g) is first H.ig) order in C2H4, first order in H,, and second order overall. The energy of activation for the reaction is 81 kJ/mol and A: is 1.3 X 10"Miter/mol sat700K. Whatisthevalueof 12.27

steps.

K?

the rate constant, k, at 1000

12.26

first

12.28

reaction:

order

The

in

reaction:

C2H5Br(g)

^

C2H4(g) + HBr(g)

is

order in CiHjBr. The rate constant is 2.0 x 10" '/s at 650 K, and the energy of activation is 226 kJ/mol. At what temperature is the rate constant 6.0 x 10" '/s? first

type of catalyst functions. 12.29

What

is

the energy of activation of a reaction that

increases ten-fold in rate

from 300

Rate Equations and Temperature

K

to

310

when

the temperature

is

increased

K?

At 400 K a certain reaction is 50.0" „ complete in 1.50 min. At 430 K the same reaction is 50.0% complete in 0.50 min. Calculate the activation energy for the reaction.

12.30 12.21

The

order in

reaction:

NO. The

2NO(g)

rate constant

is

+

02(g) is second 0.143 liter/mol s at

N2(g)

Problems

335

:

CHAPTER

CHEMICAL EQUILIBRIUM

13

Under

suitable conditions, nitrogen

N2(g)

On

+

and hydrogen

react to

form ammonia:

—2NH3(g)

3H,(g)

the other hand,

ammonia decomposes

at

high temperatures to yield nitrogen

and hydrogen

2NH3(g) The

—

reaction

N,(g)

is

N2(g)+

3UM

reversible,

and the equation

for the reaction

can be written

+ 3H2(g)-=-2NH3(g)

The double arrow (^)

indicates that the equation can be read in either direction.

All reversible processes tend to attain a state of equilibrium. For a reversible

chemical reaction, an equilibrium state chemical reaction

is

is

attained

when

the rate at which a

proceeding equals the rate at which the reverse reaction

is

proceeding. Equilibrium systems that involve reversible chemical reactions are the topic of this chapter.

13.1

Reversible Reactions and Chemical Equilibrium Consider a hypothetical reversible reaction A2(g)

+ B2(g)^2AB(g)

This equation

may

decompose Suppose

to

A2 and B2 are mixed, sample of pure AB, on the other hand, will

be read either forward or backward. If

they will react to produce

AB.

A

form A, and Bj.

we place a mixture of A2 and B2 in a container. They will react AB, and their concentrations will gradually decrease as the forward occurs (see Figure 13.1). Since the concentrations of A2 and Bj decrease, that

to produce

reaction

the rate of the forward reaction decreases.

At the start of the experiment, the reverse reaction cannot occur since there no AB present. As soon as the forward reaction produces some AB, however, the reverse reaction begins. The reverse reaction starts slowly (since the concentration of AB is low) and gradually picks up speed. is

336

Time Curves showing changes in concentrations of materials with time 2 AB. Equilibrium is attained at time + Bj

Figure 13.1

equilibrium

is

An

established.

Iwo^pposing tendencies

two

rates are equal.

equilibrium condition

same

AB

rate that

made

is

is

are equal.

At equilibrium, the concentrations of concentration of

are being

and the rate of the At Jhis point, chemical one in which the rates of

passes, the rate of the forward reaction decreases

As time

reverse reaction increases until the

at the

for

^

the reaction Aj

it is

the substances are constant.

all

AB

The

produced by the forward reaction used by the reverse reaction. In like manner, A2 and B2

constant since

is

(by the reverse reaction) at the

same

rate that they are being used

important to note that the concentrations are constant because the rates of the opposing reactions are equal and noi because all activity has stopped. Data for the experiment are plotted in Figure 13.1. (by the forward reaction).

It

is

EquiUbrium is attained at time If we assume that the forward and reverse reactions occur by simple one-step mechanisms, the rate of the forward reaction is rate^

=J(f\A2][^2]

and the rate of thejreverse/' reaction

is

f

At equilibrium, these two

rates are equal,

rate^

=

rate^

^/[A2][B2]

=

^.[AB]^

and therefore

This equation can be rearranged to

kj

_

[AB]^

13

1

Reversible Reactions and Chemical Equilibrium

337

The

rate constant of the forward.jeaction, kf, divided

reverse reaction, k^,

is

by the rate constant of the

equal to a third constant, which

is

called the equilibrium

K:

constant,

(13.1)

Therefore,

_[AB]^ [A2][B,]

The numerical value of K varies with temperature. There are an unlimited number of possible equilibrium systems for this reaction. The concentrations, however, of Aj, 63, and will,

when expressed

AB

in the

for

any system

in

equilibrium at a given temperature

preceding manner, equal the same value of K.

In general, for any reversible reaction

H'W + .tX

^

vY + zZ

(13.2)

at equihbrium,

j.=ixflgi By convention,

,13.3,

the concentration terms for the materials

chemical equation are written

in the

on the

right of the

numerator of the expression for the equi-

hbrium constant. If the

equation

yY + zZ

^

is

written in the reverse form,

vrW + A-X

the expression for the equilibrium constant (which

^

we

will indicate as K')

is

[W]»-[X][Y]^[Z]=

Notice that A"

is

the reciprocal of K:

1

In our derivation, we assumed that the forward and reverse reactions occurred by mechanisms each consisting of a single step. Does the law of chemical equilibrium hold for reactions that occur by mechanisms of more than one step? The answer to the question is that it does. Consider the reaction

2N02Cl(g) for

338

^

2N02(g) +

Cl,(g)

which the expression for the equilibrium constant

Chapter 13

Chemical Equilibrium

is

_ [NO,]^[Cl,] [NO^Cl]^ '

The reaction

is

believed to occur by

means of a mechanism

consisting of two

i

^u^/e-^^f

steps:

N02C1^N02

1.

2.

-

1

^

NO.Cl +

CI

+

4^ NO,

C1

A.

,,

4^^/'^

+ Ci;

i

,v^r.JLL

"'2

.,|

Civ*!.

the steps.

be reversible. of the

^

When

is

is

oti-'

^'""^ •

'

c^'^

^i-,

"

'

^/.^

f

c a^y

reversible,

equilibrium

mechanism must be

each step of the mechanism must also established for the overall reaction, each step

,

^

p -

Since the overall reaction

(3r

1'^^

Symbols for the rate constants appear above and below the arrows in these equations. The symbol k is used for the rate constant of the forward reaction, and k' is used for the reverse reaction. The subscripts of these symbols designate

ff-^->

S-r:\

>

\

5

'

''-C

\

in equilibrium. Therefore,

_k, _ [NO,][Cl] [NO2CI]

k\

and

_ [NO,] [CI,]

_/c, ^

[NO,Cl][Cl]

k\

The product of these expressions

K K _

^1^2

^ [N023[C1] [NO,] [CI,] _ [NO,]^[Cl,] [NO,Cl]

/t'lA:,

which

is

the

is

same

[N02C1][C1]

[NO,Cl]^

as the expression for the equilibrium constant that

we derived

from the equation for the overall change. In this case, the equilibrium constant for the overall change is the product of the equilibrium constants of

directly

each of the steps

2 Equilibrium Constants

For the reaction H2(g)

+ I,(g)^2HI(g)

the equilibrium constant at 425

;f

=

JMU!-

=

C

is

54.8

[HJ[1J 13.2

Equilibrium Constants

339

]

The numerical value of K must be determined experimentally. of the materials

If the

concentrations

mol/Hter) present in any equilibrium mixture at 425°C are

(in

substituted into the expression for the equilibrium constant, the result will equal 54.8. If

any value other than 54.8

is

obtained, the mixture

The equilibrium condition may be approached from

is

not in equilibrium.

That is, an equilibrium mixture can be obtained by mixing hydrogen and iodine, by allowing pure hydrogen iodide to dissociate, pr by mixing all three materials. The magnitude of the value of the equilibrium constant gives an indication of the position of equilibrium. Recall that the concentration terms of substances on the right of the equation are written in the numerator of the expression for the either direction.

equilibrium constant. For the reaction

CO(g) + Cl^ig)

—

COCl2(g)

at 100°C,

rcoci,!

From

this relatively large

value of K,

we conclude

that equilibrium concentrations

CO

and CI2 are small and that the synthesis of COCI2 is virtually complete. In other words, the reaction to the right is fairly complete at equilibrium. For the reaction of

+

N2(g)

at

02(g)

—

2NO(g)

2000°C,

A-

We

=

,

^

conclude from

and O2

=

4.08 X lO--^

this small value

at equilibrium.

The

of

K

that

reaction to the

Equilibria between substances in two or equilibria.

The concentration of a pure

NO

is

largely dissociated into

left is fairly

N2

complete.

more phases

are called heterogeneous

solid or a pure liquid

is

constant

when

the

temperature and pressure are constant. For any heterogeneous equilibrium, therefore, the values of the concentrations of solids or liquids involved are included in

the value of

A^,

and concentration terms

for these substances

do not appear

in the

expression for the equilibrium constant.

For example, CaC03(s)

for the reaction

^

CaO(s)

+ C02(g)

the values for the concentrations of A',

and the expression

K=

CaO

and

CaCOj

for the equilibrium constant

are included in the value of

is

[CO2]

Hence, at any fixed temperature the equilibrium concentration of CO2 over a mixture of the solids is a definite value. The equilibrium constant for the reaction

340

Chapter 13

Chemical Equilibrium

+ 4H20(g)

3Fe(s)

is

^

Fe304(s)

+ 4H2(g)

expressed in the following terms:

Some

facts dealing with expressions for

equiHbrium constants may be sum-

marized as follows:

The concentration terms

on the right side of the numerator of the expression for K. The concentration terms for substances that appear on the left side are written in the 1.

for substances that appear

chemical equation are written

in the

denominator. 2.

No

of

K includes

3.

The value of K for a given equilibrium

concentration terms are included for pure solids or pure liquids. The value these terms. is

constant

if

the equilibrium temperature

does not change. At a different temperature, the value of

K is

different.

K for a given equilibrium indicates the position of equilibK indicates that the reaction to the right is fairly complete. A small value of K indicates that the reaction to the left is fairly complete. A value of K that is neither very large nor very small indicates an intermediate situation.* 4.

The magnitude of

rium.

A

large value of

The following examples librium constants

Example

may be

illustrate

how some

simple problems involving equi-

solved.

13.1

For the reaction

N204{g)^2NO,(g) the concentrations of the substances present in an equilibrium mixture at 25

C

are

[N2O4] = 4.27 X 10'^ mol/Hter

[NO2] =

What

is

1.41

X 10"^ mol/liter

the value of

K for this temperature?

To be strictly accurate, equations for equilibrium constants should be written in terms of activities concentrations and pressures up to rather than concentrations or pressures (see Section 17.7). At low *

a few atmospheres, however, concentraiions

may

be used with reasonable accuracy.

13.2

Equilibrium Constants

:

Solution

[NO3? [N2O,] X 10~^ mol/liter)2

_ ~

(1.41

=

4.66 X 10"

(4.27 X

Example

10"^ mol/liter) 3

mol/liter

13.2

At 500 K, 1.00 mol of ONCI(g) is introduced rium, the ONCl(g) is 9.0% dissociated

20NCl(g)^2NO(g) + Calculate the value of

K for

into a

1

liter

container.

At

equilib-

Cl2(g)

the equilibrium at 500 K.

Solution

By considering exactly 1 liter, we simplify the problem since the number of moles of is the same as the concentration of the gas (in mol/liter).

a gas

Since the

ONCl

is

number of moles

We

9.0%

dissociated,

=

dissociated

0.090(1.00 mol)

=

0.090 mol

ONCl

must subtract this quantity from the number of moles of The concentration of ONCl at equilibrium, therefore, is

ONCl

present

initially.

[ONCl] = The ONCl

-

1.00 mol/liter

0.090 mol/liter

that dissociates produces

NO

=

0.91 mol/hter

and CI 2-

We

can derive the amounts

of these substances produced from the coefficients of the chemical equation: 2

ONCl

^

2

NO

+

0.090 mol

Since 2 mol of

0.090 mol of

ONCl

NO. The

CI2 0.045 mol

produces 2 mol of

of CI 2, therefore, 0.090 mol of ONCl

0.91 mol/hter

[NO] =

0.090 mol/liter

=

0.045 mol/liter

[CI 2]

Therefore,

_ [NO]^[Cl2] [ONCl]^

342

_ ~

(0.090 mol/liter)^(0.045 mol/liter)

=

4.4 X

(0.91 mol/hter)^

Chapter 13

0.090 mol of

10"* mol/liter

Chemical Equilibrium

ONCl

ONCl

will

produce

produces only 1 mol will produce 0.045 mol of CI 2. The equilibrium

concentrations are

[ONCl] =

NO,

equation shows that 2 mol of

Example 13.3

K for

HI equilibrium

the

425°C

is

54.8:

+ l2(g)^2HI(g)

H2(g)

A

at

quantity of HI(g)

placed in a 1.00 liter container and allowed to come to What are the concentrations of H2(g) and 12(g) in equilibrium

is

equilibrium at 425°C.

with 0.50 mol/liter of HI(g)?

Solution

The concentrations of H2(g) and 12(g) must be equal since they are produced equal amounts by the decomposition of HI(g). Therefore, let

=

[H2]

We

[la]

= ^

are told that the equilibrium concentration of

=

[HI]

We

in

HI

is

0.50 mol/hter

substitute these values into the equation for the equilibrium constant

and

solve for x:

(0.50 mol/liter)2 y-

— =

54.8x^

X

54.8

=

0.25 moP/liter^

=

0.00456 mol^/liter^

=

0.068 mol/liter

The equilibrium concentrations are

[H2]

=

[HI]

-

0.50 mol/liter

[I2]

=

0.068 mol/liter

Example 13.4 For the reaction H2(g)

K 1

at

is

+ C02(g)

^

H20(g)

+ CO(g)

mol of Hj and 0.0100 mol of CO2 are mixed in a 750 C, what are the concentrations of all substances present

0.771 at 750'C. If 0.0100

liter

container at

equilibrium?

Solution

mol of H2 reacts with x mol of CO2 out of the of H2O and .Y mol of CO will be produced. Since

If

1

.Y

liter,

amounts supplied, .y mol contamer has a volume of

total

the

at equilibrium the concentrations are (in mol/liter) 13.2

Equilibrium Constants

343

+

H2Cg)

-

(0.0100

^

C02(g) (0.0100

x)

-

Hpig) + CO(g) X

X

A)

[H,Q][CO] _ - [H.][CO,] (0.0100 If

we

-

A)(0.0100

-

0.771 x)

find the square root of both sides of this equation,

X

-

(0.0100

=

0.878

=

0.00468

we

get

X)

X

Therefore, at equilibrium

[H2]

= [CO2] = (0.0100 - A) = 0.0053 mol/Hter = 5.3 X 10~^ mol/Hter

[H.O] = [CO] = X = 0.00468

-

4.68 X

10^

13.3 Equilibrium Constants in

mol/liter

mol/liter

Expressed

Pressures

The

partial pressure of a gas

is

a measure of

its

stants for reactions involving gases, therefore, partial pressures of the reacting gases.

An

concentration. Equilibrium con-

may

be written

in

terms of the

equilibrium constant of this type

is

given the designation Kp.

For the calcium carbonate equilibrium

CaC03(s)

^

CaO(s)

+ CO,(g)

the equilibrium constant in terms of partial pressures

=

is

PC02

For the equilibrium N,(g)

The Kp

+ 3H2(g)^2NH3(g)

is

There

is

a simple relation between the

Kp

for a reaction

constant derived from concentrations. Consider the reaction

344

Chapter 13

Chemical Equilibrium

and

the equilibrium

H'W + -yX

^

>'Y

+ rZ

(13.4)

these materials are gases,

If all

Assume

that each of the gases follows the ideal gas law

PV = nRT Then

the partial pressure of any gas, p,

The concentration of a gas

in mol/liter

is

is

equal to n/V. Therefore, for gas

p^=[^]RT (P.r = If

we

W (13.6)

V^TiRTr

(13.7)

substitute expressions such as Equation 13.7 for the partial pressure

terms in the expression for Kp (Equation

13.5),

we

get

lYY(RTylZy(RTf

~

[W]"'(i?7r[X]^(/?rf

[W]1X] The

fractional term in the last equation

Kp If

equal to K:

= K{RT)^

we read

wW in

is

+

(13.9)

the chemical equation for the reaction

A-X

^>'Y

+ zZ

(13.4)

molar quantities,

y +

z

= number

w + X = number

We let

An equal

read from

An =

of moles of gases on the right of moles of gases on the

the change in the

left

number of moles of gases when

the equation

is

left to right

{y

+

z)

-

(vv

+

.y)

=

+.v

+

r

- w -

x

(13.10)

Therefore,

K.^K{RTf"

'

13.3

Equilibrium Constants Expressed

(13.11)

in

Pressures

345

Partial 'pressures are expressed

moles per

in

liter,

R

is

0.08206

Jn atmospheres, concentrations are expressed atm/(K mol), T is the absolute temperature in

liter

•

K.

For the reaction PCl5(g)

A/7

+

is

1

^

+

PCl3(g)

C\,(g)

Therefore,

.

- K(RTr' For the reaction

A/7

c\M

+

co(g)

= —

—

coci^ig)

Therefore,

1.

K=K(RT)-'

K=

or

For the reaction

+ l2(g)^2HI(g)

H2(g) A/7

=

0.

Therefore,

Kp = K{RTf

Example At

1

13.5

100 K, the equilibrium constant for the reaction

2S03(g) is

= K

or

—

2SO,(g)

+

What

is

0.0271 mol/liter.

03(g)

Kp

at this

temperature?

Solution

Two

moles of S03(g) produce a

mol of gases. Therefore,

+1

A/7

= KiRTr' = (0.0271 mol/hter) = 2.45 atm

[[0.0821

j!e 13.6

What

total of 3

is

K for the

Chapter 13

reaction

Chemical Equilibrium

liter

•

atm/(K

•

mol)](1100 K)}

+ 3H2(g)^2NH3(g)

N2(g) at 500'

C

if

Kp

is

1

.50 x

Vatm^

10

at this temperature?

Solution There are 4 mol of gases indicated on the

left

of the chemical equation and 2 mol

of gases indicated on the right. Therefore,

-2

^n =

The temperature

r=

773

is

500 C, which

is

K

Therefore,

Kp

=

K{RT)'^

(1.50 X

10-Vatm')

=

K{[0.0821

(1.50 X

lO-^atm^)

= ^^TTT; (63.5 liter

^

K-

•

•

atm/(K

mol)](773 K)}-^

i^ atm/mol)

10"Vatm^)(4.03 x 10^

(1.50 X

=

liter

liter^

•

atm^mol")

6.04 X 10"^ Uter^/moP

Example 13.7 For the reaction C(s)

+

C02(g)

167.5

Kj, is

atm

at

—

2CO(g)

1000 C.

What

is

CO(g) 0.100 atm?

the partial pressure of

system in which the partial pressure of C02(g)

is

in

an equilibrium

Solution

=

167.5

atm

=

167.5

atm

iPco?

=

16.8

atm^

Pco

=

4.10

atm

K = "

PC02

{Pcof 0.100 atm

Example 13.8 Kp

for the equilibrium

FeO(s)

+ CO(g)

^

Fe(s)

+

C02(g)

13.3

Equilibrium Constants Expressed

in

Pressures

:

at

lOOO'C

0.403. If CO(g), at a pressure of 1.000 atm,

is

and excess FeO(s) are

placed in a container at 1000 C, what are the pressures of CO(g) and C02(g) equilibrium

when

attained?

is

Solution Let of

.V

equal the partial pressure of

COo

equal to the decrease in

FeO(s)

+

CO(g)

(

-

i

.000

Pco,

^

x)

^

COj when

CO the pressure of CO

produced for every

is

1

^

mol of

Fe(s)

equilibrium

is

attained. Since

used, the partial pressure of

1

mol

CO2

is

+ CO^fg) x atm

atm

0.403

Pco

X atm

-

(1.000

0.403

.x)atm

X 1.000

=

pf-Q^

=

- x ^ pco =

0.287 atm

0-713

atm

13.4 Le Chatelier's Principle

What happens

an equilibrium system if an experimental condition (such as is changed? The effects of such changes were sum1884 by Henri Le Chatelier. Le Chateher's principle states that a to

temperature or pressure)

marized system

in

in

equilibrium reacts to a stress in a

establishes a

new equilibrium

state.

way

that counteracts the stress

This important generahzation

is

and

very simple

to apply:

1.

Concentration changes.

equilibrium will shift

If the

way

was added. Suppose

stance that

H2(g)

in a

+

12(g)

—

concentration of a substance

is

increased, the

that will decrease the concentration of the subthat

we have a system

in

equihbrium

2HI(g)

and we increase the concentration of H2 by adding more H2 to the system. The equilibrium is upset, and the system will react in a way that will decrease the concentration of H2- It can do that by using up some of the H2 (and some I, as well) to form more HI. When equilibrium is established again, the concentration of HI will be higher than it was initially. The position of equilibrium is said to have shifted to the right. If the concentration of HI is increased by adding HI to the system, the position of equilibrium

will shift to the left. In this

Henri Le Chatelier, 1850-1936.

Smithsonian

Institution.

equilibrium

is

than they were

initially.

Removal of one of

the substances

the position of equilibrium to shift.

348

Chapter 13

way some HI will be used up. When Hj and I2 will be higher

established again, the concentrations of

Chemical Equilibrium

from an equilibrium system If,

for example,

will also

cause

HI could be removed,

the

would shift to the right. Additional HI would be produced and 1 2 would decrease. and the concentrations of By the continuous removal of a product it is possible to drive some reversible reactions "to completion," Complete conversion of CaCOjls) into CaO(s): position of equilibrium

CaC03(s)

^

CaO(s)

+

COjig)

can be accomplished by removing the 2.

Pressure changes.

gas as fast as

it is

produced.

Le Chatelier's principle may also be used

to

make

qualita-

predictions of the effect of pressure changes on systems in equilibrium.

tive

Consider the

and

CO,

effect

of a pressure increase on an equilibrium mixture of SO2, O2,

SO 3

2S02(g) In the

+ 02(g)^2S03(g)

forward reaction two gas molecules (2SO3) are produced by the disap(2 50, + O2). Two gas molecules do not exert

pearance of three gas molecules

as high a pressure as three gas molecules.

mixture

is

When

the pressure

on an equilibrium

increased (or the volume of the system decreased), the position of equi-

way

librium shifts to the right. In this

the system counteracts the change. Alter-

natively, decreasing the pressure (or increasing the

of this equilibrium to shift to the

For reactions

in

which An

=

volume) causes the position

left.

0,

pressure changes have no effect on the position

of equilibria. Equilibria involving the systems

H2(g)

N2(g) H2(g)

+ l2(g)^2HI(g) + 02(g)^2NO(g)

+ C02(g)^ H20(g) +

CO(g)

are not influenced by changing the pressure since there total

volume

in

is

no

difference in the

either the forward or reverse direction for any one of these

reactions.

For a system that involves only liquids and solids, the effect of pressure on the is slight and may usually be ignored for ordinary changes in pressure. Large pressure changes, however, can significantly alter such equilibria; and at times, even slight changes in such equilibria are of interest. For

position of equilibrium

example, the position of equilibrium

H20(s)^

H20(l)

forced to the right by an increase in pressure because a given quantity of water the solid state (its density occupies a smaller volume in the liquid state than is

m

is

higher in the liquid state).

Pressure changes affect equilibria involving gases to a much greater degree. For example, a high pressure would favor the production of a high yield of

ammonia from

the equilibrium

N2(g)+ 3H2(g)^2NH3(g) is of practical importance as an aid in determining favorable reaction conditions for the production of a desired substance.

Hence, Le Chatelier's principle

13.4

Le Chatelier's Principle

349

For heterogeneous equilibria the effect of pressure is predicted by counting number of moles of gas indicated on each side of the equation. For example,

the

the position of equilibrium

+ 4H20(g)

3Fe(s)

is

virtually unaffected

on each 3.

^

Fe304(s)

+

4H2(g)

by pressure because there are four moles of gas indicated

side of the equation.

Temperature changes.

In order to predict the effect of a temperature

on a system m equilibrium, reaction must be known. At of ammonia

the thermochemical equation for the synthesis

is

N2(g)+ 3H2(g)^2NH3(g) Since A//

change accompanies the

the nature of the heat effect that

A//=-92.4kJ

We

negative, the reaction to the right evolves heat.

is

can write the

equation to indicate heat as a product:

+

N2(g)

3

H^ig)

^

NH3(g)

2

The forward reaction

+

92.4 kJ

exothermic, and the reverse reaction

is

heat. If heat

is

added

(the

temperature of the system

equilibrium will shift to the

mixture in

is

left

— the direction

in

is

raised), the position

is

which heat

absorbed.

is

is

The highest

evolved.

temperatures. Unfortunately,

if

yields of

NH3

the temperature

will

is

of

If the

— the direction

cooled, the position of equilibrium will shift to the right

which heat

endo thermic.

and the reverse reaction uses

In other words, the forward reaction produces heat,

be obtained at the lowest

too low, the reaction will be

extremely slow. High pressures and temperatures around 500

C

are employed

commercial process. Consider the reaction

in the

CO.ig) Since

AH

+ is

41.1 kJ

H^lg)

^

positive, the

+ CO,(g) +

AH=

CO(g) + H^Olg) forward reaction

H2(g)

^

+41.1 kJ

We

endothermic.

is

write the equation

CO(g) + H20(g)

Increasing the temperature always favors the endothermic change, and decreasing the temperature always favors the exothermic change. In this case, the reaction is

forced to the right by an increase in temperature.

causes the position of equilibrium to shift to the

The numerical value of changed.

The

A

decrease in temperature

left.

the equilibrium constant changes

reaction of

CO2 and

is

when

shifted to the right

the temperature

by an increase

is

in

temperature. The concentrations of the substances on the right are increased by this shift.

The terms

for the concentrations of these substances

numerator of the expression for A'. As the temperature the value of K increases. For this reaction, at 700 C, K

K= 4.

is

appear

in the

increased, therefore,

=

0.63,

no

effect

and

at 1000°C,

1.66.

Addition of a catalyst.

The presence of

a catalyst has

on the position

of a chemical equilibrium since a catalyst affects the rates of the forward and reverse reactions equally (see Section 12.7).

A

system to attain equilibrium more rapidly than

350

Chapter 13

Chemical Equilibrium

it

catalyst will, however, cause a

otherwise would.

Summary topics that have been discussed in this chapter are

The

Reversible reactions

1.

and chemical equilibrium;

Equilibrium constants expressed in partial pressures, and their relation to constants expressed in molar concentrations, A^'s or K^'s 3.

A'p's,

the

derivation of equilibrium constants.

4.

The meaning of an equilibrium constant and

2.

its

use in

problem solving.

Le Chatelier's principle applied to chemical equilibria change in conditions on a system in

the effects of a

equilibrium.

Key Terms Some of

more important terms introduced

the

in

this

chapter are listed below. Definitions for terms not included in this list may be located in the text by use of the index.

Chemical equilibrium (Section rate of a reversible reaction in in the

A state in which the 13.1 one direction equals the rate

I

k'terogeneoiLs equilibrium (Section

between substances

state

Equilibrium constant (Section 13.1)

A

constant for an

equilibrium system equal to a fraction in which the nuis

use of molar concentrations is given the symbol A' or A'^; one written in terms of the partial pressures of reacting gases is given the symbol K^.

)

other direction.

merator

equation, each raised to a power equal to its coefficient in the balanced chemical equation. A value obtained by the

in

1

3.2

An equilibrium

)

more than one phase.

Homogeneous equilibrium (Section 13.2) An equilibrium between substances in the same phase.

state

the product of the concentrations of the sub-

on the right of the chemical equation, each raised to a power equal to its coefficient in the balanced chemical equation, and the denominator is the product of the concentrations of the substances on the left of the chemical stances

Le Chatelier's principle (Section 13.4)

A

system

librium reacts to a change in conditions in a tends to offset the change and establish a

in equi-

way

that

new equilibrium

state.

Problems* Chemical Equilibrium, Le Chatelier's Principle For each of the following reactions, write an expres-

13.1

^ ^ ^ ^

(a)

2

CO,(g)

(b)

2

Pb304(s)

(c)

2

Ag,0(s)

(d)

CH^(g)

(e)

C(s)

2

CO(g) + 0,(g) 6 PbO(s)

4 Ag(s)

+ 2 H,S(g) + C02(g)

2

+

13.4

*

The more

+

H2(g)

difficult

—

0.08 at 400

C

and

2 S02(g)

CS,(g)

0.41 at

600 C.

Is the

reaction as

For the equilibrium

0,(g)

+

4 H,(g)

CO(g) + H,0(g)

problems are marked with

+

02(g)

—

2

S03(g)

A'pisl.Ox 10^'atmat700 Kand 1.1 x lO^tmatSOOK. reaction as written exothermic or endothermic?

Is the

CO(g)

For the equilibrium

C02(g)

is

+ 0,(g)

13.2 Indicate in which direction each of the equilibria given in problem 13.1 will shift with an increase in pressure. 13.3

K

written exothermic or endothermic?

sion for the equilibrium constant, A':

asterisks.

+ B(g):^C(g) is exothermic as would an increase in temperature have on the numerical value of A"? (b) How would the numerical values of A' and A'p vary with an increase in

13.5

The

reaction A(g)

written, (a)

What

effect

total pressure?

The appendix contains answers

to color-keyed problems.

Problems

351

:

:

The reaction A(g) + B(g)^C(g) is exothermic as Assume that an equihbrium system is established. How would the equilibrium concentration of C(g) change

13.6

equilibrium

established, 0.0040

is

:

mol of S03(g) is present

written.

an increase in pressure, (c) the addition of A(g), (d) the addhion of a catalyst, (e) the removal of B (g )? How would the numerical value of A' change with (f) an increase in tempera ture,(g) an with

an increase

(a)

in

temperature,

increase in pressure, (h) the addition of a catalyst,

(i)

the

addition of A(g)?

which each of the following

13.7 State the direction in

(a)

+

2 SOjIg)

0,(g)

2

SOjlg) (exothermic)

What

are the equilibrium concentrations of S02(g)

and 0,(g)?(b) What at

1000

is

the value of

K for

the equilibrium

K?

13.12 If 0.025

mol of COCl2(g)

is

container at 400 C, 16.0% of the when equilibrium is established

equilibrium systems would be shifted upon the application of the stress listed after the equation (a)

02(g)^2S03(g)

2S02(g) +

(b)

COCl2(g)

^

CO(g) +

Calculate the value of

K

placed in a one-hter

COCK

is

dissociated

Cl2(g)

for the equilibrium at

400 C.

decrease temperature

+ CO,(g)

C(s)

(b)

13.13

CO(g) (endothermic)

2

increase temperature (c)

N,04(g)^

(d)

CO(g) + H^Oig)

CO(g) +

^

at

CO,(g)

+

^ ^

Br.lg)

+

4 H,0(g)

+

0,(g) add a catalyst

2

13.14

FcjO^fs)

+

4 H2(g)

10.2 liter-/mol-.

equilibrium with

at

1800 K,

NO(g)

CaO(s) + CO,(g) CaCOjIs) remove CO 2(g)

is

in

A

N2(g)

13.15

increase the concentration of Hjig)

H2(g)

concentration of

at

a

concentration

of

+

02(g)

8.36 x

is

10^ What concentration of

equilibrium with N2(g) at a concentration

of 0.0500 mol/liter 0.0500 mol/hter?

+ 3H2(g)^2NH3(g)

N^ig)

and

What

CO (g) at a concentration

For the equilibrium

2NO(g)^

S03(g)

CH30H(g)

in

of 0.020 mol/liter

+

^

is

0.020 mol/hter?

2 SO,(g)

(i)

is

2 NOBr(g) =F=i 2 NO(g)

add Fe(s)

(h)

)

decrease total pressure

3 Fe(s)

(g)

A

225 C,

CH 30H(g

HM)

decrease total pressure (f)

2 H2(g)

2NO,(g)

increase total pressure

(e)

For the equilibrium

and 02(g)

at

a

concentration

of

For the equilibrium

Br2(g)

+ Cl2(g)^2BrCl(g)

400 K,

A

Problems Based on Equilibrium Constants at

What

13.8

is

the value of the equilibrium constant.

the following system at 395

H,(g)

+

12(g)

A',

for

C:

what

^2HI(g)

13.16

=

0.0064 mol/liter. [I,]

7.0. If

0.060 mol of Br2(g) and 0.060 mol

is

the concentration of BrCl(g)

when equilibrium

is

established?

The equilibrium concentrations [H,]

is

of Cl2(g)are introduced into a one-liter container at 400 K,

=

At 425 C.

A

is

54.8 for the equilibrium

for such a system are:

0.0016 mol/liter, [HI]

=

H2(g)

+ l2(g)^2HI(g)

0.0250 mol/liter. 13.9 Solid

NH4HS

container at 24'C.

NH4HS(s)

^

was introduced

When

0.614 atm.

What

At 250 C.

NH3 for

is

0.1 10

a one-liter container.

and HjS taken together) was the equilibrium at 24 C?

mol of PCl5(g) was introduced mto Equilibrium was established:

PCl5(g)^PCl3(g) + CU(g) concentration of PCl3(g) was the equilibrium concentrations of Cl2(g) and PCl5(g)'? (b) What is the value of A'

At

equilibrium,

0.050 mol/liter. at

(a)

If 1.000 mol of H2(g), 1.000 mol of ^(g), and 1.000 mol of HI(g) are placed in one-liter container at 425'C, what are the concentrations of all gases present at

equilibrium?

NHjig) + H,S(g)

the total pressure (of

13.10

into an evacuated

equilibrium was established,

the

What were

13.17

1.82 x

10"- for the equilibrium

12(g)

Assume that an equilibrium is established at 425'C by adding only HI(g) to the reaction flask, (a) What are the concentrations of H2(g) and 12(g) in equihbrium with 0.0100 mol/hter of HI(g)? (b) What was the initial concentration of HI(g) before equilibrium was established? (c) What percent of the HI(g) added is dissociated at equilibrium? 13.18

For the equilibrium

A

mixture of 0.0080 mol of S02(g) and 0.0056 mol of 02(g) is placed in a one-liter container at 1(K)0 K. When

352

A is

2HI(g)^H2(g) +

250 C?

13.11

At 425°C,

Chapter 13

Chemical Equilibrium

2IBr(g)^l2(g) +

Br2(g)

10'^ at 150'C. If 0.0300 mol of IBr(g) is is 8.5 X introduced into a one-liter container, what is the concen-

K

tration of this substance after equilibrium

3Fe{s)

+ 4H,0(g)

^

CHjOHlg)

established?

is

For the equilibrium

13.19

For the equilibrium

13.24

at 275 C,

A?

+

FcjO^ls)

4H,(g)

at

900 C, A' is 5.1. If 0.050 mol of H^Oig) and excess solid Fe are placed in a one-liter container, what is the concentration of H2(g) when equilibrium is e.stablished at 900'C?

Ap

What

(b)

is

1.14 X

are

A

mixture consisting of 1.000 mol of CO(g| and mol of H,0(g) is placed in a 10.00 liter container at 800 K. At equiirbrium, 0.665 mol of CO^lg) and 0.665 mol

What in

of Hjtg) are present:

CO(g) (a)

What

gases?(b)

What

At 585

is

the

is

K?

K and a

A

of

1

is

with

CaCOj and CaO

solid

at

is

this

For the equilibrium

+ 0,(g)^2N0(g) 10"^ at 2400 K. The partial pressure of N^ig)

2.5 X

NO(g) when equilibrium

13.27

established at 2400

is

K?

For the equilibrium

N,(g)

0NCl(g)^2NO(g) +

+ COjlg)

atm, NOCl(g)

.00

56.4/0 dissociated

2

0.220 atm for the equilibrium

0.50 atm, and the partial pressure of Oilg) is 0.50 atm in a mixture of these two gases. What is the partial pressure

of

total pressure

the value of

is

What

the value of ATat 800 K?(c)

is

four

is

the concentration of CO^Ig), in mol/liter, that

is

N.lg)

all

What

for the following equilibrium

temperature?

H,(g)

are the equilibrium concentrations of

value of Ap at 800 13.21

+

CO^lg)

is

CaO(s)

equilibrium

13.26

^

+ Hpig)

Ap

2H2(g)^CH30H(g)

At 800 K, Ap

13.25

1.000

2 H,(g)

10^ atm'. (a)

A and

CaCOjis) 13.20

CO(g) +

275 C:

CO(g) +

at

^

+ 3H,(g)^2NH3(g)

CU(g) 350 C, Ap is 7.73 x 10 ^/atm-. If the partial pressure of Njig) is 9.4 atm and the partial pressure of H,(g) is 28.0 atm in an equilibrium mixture at 350 C, what is the partial pressure of NHjIg) ' What is the total pressure? What is the mole fraction of NHjlg) present? at

Assume

that 1.00

mol of ONCl(g) was present before

How many

sociation, (a)

dis-

moles of ONCl(g), NO(g). and

Clilg) are present at equilibrium

number of moles of gas present

What

(b)

is

equilibrium?

at

the total (c)

What

are the equilibrium partial pressures of the three gases? (d)

What At

13.22

is

the numerical value of A'p at 585

K

100 C,

4.57 X

is

10"

liter,

mol

for

the

equilibrium

CO(g) + Cl2(g)

^

13.28

For the equilibrium

K?

PCljfg)^

are (a)

COCl.tg)

K and

^

temperature, A'p is 2.25 atm. A quantity of introduced into an evacuated flask at the refer-

at a given

PCUlg) COCl,(g) for the equilibrium

(b)

CO(g) +

is

ence temperature.

When

13.23

sociated into PCl^lg) and CKfg)?

Cl,(g)

temperature?

*13.29

480'C for the equilibrium

889 liter/mol at 480 C. (a) What is the value of at this temperature? (b) Calculate the values of A' and

2

CUg) +

2

H,0(g)

^

in this

are the

What mole

percent of

system at equilibrium?

At a given temperature and

are Pf^^o^

at

What

a total pressure of

N30^(g)^2NO,(g)

4HCUg) + 0,(g)^2Cl,(g) + 2H,0(g) is

established, the

1.00 atm, the partial pressures of an equilibrium mixture

For the equilibrium

K

is

0.25 atm.

is

equilibrium partial pressures of PCl3(g) and CUlg)? What was the initial pressure of PCl^lg) before any of it had disPClslg) has dissociated

at this

equilibrium

pressure of PCl5(g|

partial

What

+ CU(g)

PCl.,(g)

at this A,,

=

0.50

atm and

temperature?

to 2.00

atm and

/7no,

(b) If

=

0.50 atm.

(a)

the total pressure

What is

is

A'p

increased

is constant, what are components of an equilibrium quadratic formula must be used

the temperature

the partial pressures of the

4 HCl(g)

+

0:(g)

mixture? Note that the to solve this problem.

Problems

^

353

.

CHAPTER

THEORIES OF ACffiS

AND BASES

14

X' :

^ ^ IV

^_

,

Throughout the history of chemistry various acid-base concepts have been proposed and used. In this chapter four concepts in current use are reviewed. Each of the definitions can be apphed with advantage in appropriate circum-

'A

"Ty-^

^

-I/

r

r.i't.-i

u)

1

1.

stances. In a given situation the chemist uses the concept that best suits the purpose.

-

tiiM ',,->

is

Uv.£

Xhe

earhest criteria for the characterization of acids and bases were the ex-

perimentally observed properties of aqueous solutions.

An

acid

was defined

as a substance that in water solution tastes sour, turns litmus red, neutralizes bases,

and so on.

A

substance was a base

if its

aqueous solution

tastes bitter,

turns litmus blue, neutralizes acids, and so on. Concurrent with the development

of generalizations concerning the structure of matter, scientists searched for a correlation between acidic

and basic properties and the structure of compounds

that exhibit these properties.

14.1

\

.

_j

H

•

jp^^'a.. Cf S

*stem. The remainder is called the surroundings. A mixture of chemical compounds, for example, can constitute a system. The container and everything else around the system make to the

that

is

up what

A

is

called the surroundings.

system

forms

is

assumed

to have

an

energy, E, which includes all possible Important contributions to the internal

ir.«.;mal

of energy attributable to the system.

429

energy of a system include the attractions and repulsions between the atoms, molecules, ions, and subatomic particles that energies of

According system

of

all

the system

and the

kinetic

law of thermodynamics, the internal energy of an isolated

to the first

The actual value of E for any system is not known and cannot Thermodynamics, however, is concerned only with changes in

constant.

is

be calculated.

and these changes can be measured.

internal energy,

The

make up

parts.

its

state

of a system can be defined by specifying the values of properties

such as temperature, pressure, and composition. The internal energy of a system

depends upon the

of the system and not upon

state

that state. Internal energy

is

how

the system arrived at

therefore called a state function. Consider a sample

1 liter at 100 K and 1 atm pressure (state At 200 K and 0.5 atm (state B), the sample occupies a volume of 4 liters. According to the first law, the internal energy of the system in state A, E/^, is

of an ideal gas that occupies a volume of A).

a constant, as It is

B.

the internal energy of the system in state B, E^.

is

follows that the diff'erence in the internal energies of the two states, Af,

and

also a constant It

makes no

is

-

is

£, If the .

energy of the system

+ If the H',

or, indeed,

whether the

in

an

initial state

and that the

internal energy

system absorbs heat from the surroundings,

will

now

q,

the internal

be

q

system

now uses some of its

internal energy to

do work on the surroundings,

the internal energy of the system in the final state, E^, will be

Ej^

It

state

Ef^

Suppose that we have a system of the system

and

heated before the pressure change,

whether the heating is done after the pressure change, total change is brought about in several steps:

A£ =

A

independent of the path taken between state

is

difference whether the gas

is

—

Ej-

=

Ei

+

Ej

=

q

— w

AE =

q

- w

q

— w

(17.1)

important to keep

in

mind

the conventions regarding the signs of these

quantities:

q, positive

=

heat absorbed by the system

negative

=

heat evolved by the system

q, M',

H',

positive

= work done

by the system

negative

= work done

on the system

The values of q and w involved in changing a system from an initial state to a depend upon the way in which the change is carried out. The value

final state

iq — H'), however, is a constant, equal to A^", for the change no matter how it brought about. If a system undergoes a change in which the internal energy of the system remains constant, the work done by the system equals the heat absorbed by the system.

of is

430

Chapter 17

Elements

of

Chemical Thermodynamics

/'

2 Enthalpy

For ordinary chemical reactions, the work term generally

consequence

arises as a

of pressure-volume changes. The work done against the pressure of the atmosphere the system expands in the course of the reaction is an example of pressurevolume work. The term PV has the dimensions of work. Pressure, which is force per unit area, may be expressed in newtons per square meter (N/m'). If volume if

is

expressed in cubic meters (m^), the product

(N/m^Km^) =

PV =

PK

is

Nm

is a joule) is a unit of work, since work is defined as newton) times distance (the meter). In like manner, liter atmospheres are units of work. If the pressure is held constant, the work done in expansion is to from

The newton- meter (which force (the

•

w = P(V^ -

No

V^)

=

PAV

(17.2)

pressure-volume work can be done by a process carried out at constant

volume, and

AE =

q

w =

0.

Thus, at constant volume the equation

— w

becomes

AE = where

(1V.3)

q^

t/y is

the heat absorbed by the system at constant volume.

Processes carried out at constant pressure are far

more common

in

chemistry

than those conducted at constant volume. If we restrict our attention to pressurevolume work, the work done in constant pressure processes '\s P AV. Thus, at constant pressure the equation

AE —

q

— w

becomes

AE ^ q^- P AV Or, by rearranging,

qf

=

AE+PAV the heat absorbed by the system at constant pressure. thermodynamic function enthalpy, H, is defined by the equation

where qp

The

H

(17.4)

is

= E + PV

(17.5)

Therefore,

q,

= AH

(17.6)

'

17.2

Enthalpy

431

The heat absorbed by a change

and

the system

The

reaction conducted at constant pressure

enthalpy. Enthalpy, like internal energy,

in

independent of the manner

is

Hess

validity of the law of

When Section

bomb

a

3.3),

calorimeter

the heat effect

is

are run at constant pressure.

change

in internal

energy

on

rests

The

is

equal to the

a function of the state of

which the

state

was achieved.

this fact (see Section 3.5).

make

used to

is

measured

in

is

a calorimetric determination (see

at constant

volume. Ordinarily, reactions

relationship between change in enthalpy

and

used to convert heats of reaction at constant volume

is

= ^E) to heats of reaction at constant pressure {qp — AH). The conversion made by considering the change in volume of the products. The changes in

(^v is

volumes of liquids and For reactions involving

solids are so small that they are neglected.

the

Let us say that

is

volume changes may be significant. volume of gaseous reactants, is the total volume

gases, however,

the total

is the number of moles of gaseous reactants, is number of moles of gaseous products, and the pressure and temperature

of gaseous products,

the

are

constant:

PV^ = n^RT and PV^ = n^RT Thus,

P AF = PV^ - PV^ = n^RT - n^RT

= (An)RT

(17.7)

Since

AH = AE+PAV

(17.8)

then,

AH = AE + where An

is

the

[An)RT

(17.9)

number of moles of gaseous products minus

the

number of moles

of gaseous reactants.

we must express the value of atmospheres are units of energy. atm/(K mol), may be converted to J/(K mol)

In order to solve problems using this equation,

R

in

appropriate units.

We

have noted that

The value of

liter

i?, 0.082056 liter by use of factors derived from the relations •

•

1

atm = 1.01325 x

1

liter

=

1

J

=

1

1

X 10"^

10'

N/m^

m^

Nm

Thus,

atm/1.01325 x 10^ con^A X ,^-,liter 10 8.2056

K

mol

\

1

atm

8.3143 J/(K mol)

432

Chapter 17

Elements

of

Chemical Thermodynamics

N/m^

3.

Various types of calculations involving enthalpy changes are a topic of Chapter list of standard enthalpies of formation is found in Table 3.1

A

Example

17.1

The heat of combustion calorimeter at 25"C and

volume of CH4(g) is measured in a found to be -885,389 J/mol. What is the Mil

at constant is

bomb

Solution

For the reaction

—

CH4(g) + 202(g) CO^ig) + 2H20(1) A« = 1 - (2 + 1) = -2

^E=

-885,389

J

Therefore,

AH = AE + = = -

{An)RT -885,389 J + (-2mol)[8.3143 J/(K mol)](298.2 K) -885,389 J -4959 J -890,348 J = -890.348 kj

Example 17.2 Calculate

OF2(g)

AH° and AE"

+

H^OCg)

for the reaction

—

+ 2HF(g)

02(g)

The standard enthalpies of formation are OF2(g), +23.0kJ/mol; H20(g), —241.8 kJ/mol; and HF(g), -268.6 kJ/mol.

Solution

The standard enthalpies of formation are used

to calculate

AH

for the reaction

(see Section 3.6):

AH° = 2A//}(HF) - [AHjiOF,) + A//>(H20)] = 2( - 268.6 kJ) - [( + 23.dkJ) + (-241.8 kJ)] = -537.2 kJ + 218.8 kJ = -318.4 kJ This value of

A£- -

AH

AH

is

used to find

AE

.

For the reaction An

= +1

mol,

-(An)RT

= -318.4 = -318.4 = -318.4

kJ kJ kJ

-

(1 mol) 2476 J

2.5 kJ

[(8.31

J/(K mol)] (298 K)

= -320.9

kJ

17.2

Enthalpy

Example

17.3

—

For the reaction B.Hgig)

+ 303(g)

—

BPjis) + 3H20(1)

for the reaction, (b) Determine the value is — 2143.2 kJ. (a) Calculate A// of the standard enthalpy of formation of BjHJ^g). For 8203(5), AH} = - 1264.0 kJ/mol and for H20(l), AH} = -285.9 kJ/mol.

A^"

Solution (a)

An = — 4. Therefore,

AH° = AE' + (An)RT = -2143.2 kJ + (-4mol)[8.314 = -2143.2 kJ - 9.9 kJ

=

17.3

-2153.1 kJ A//=

(b)

= A//>(BP3) + 3A//^(H20) - A//}(B2H6)

-2153.1 kJ

= (-

A//}(B2H6)

= +31.4

Second Law The

first

changes

x 10"^ kJ/(K mol)](298 K)

of

1264.0 kJ)

+ 3(-285.9

kJ)

- AHjiB^Hf,)

kJ

Thermodynamics

law of thermodynamics puts only one restriction on chemical or physical first law, however, provides no basis

— energy must be conserved. The

for determining

whether a proposed change

will

be spontaneous. The second law

of thermodynamics establishes criteria for making

The thermodynamic function

may

is

this

dition

is

is

One

one, entropy

is:

is

affected.

The final state is more random and hence more probable than the initial state. The random motion of the gas molecules has produced a more disordered condition. The fact that the gases mix spontaneously is not surprising; one would have predicted it from experience. Indeed, it would be surprising if the reverse were to be observed a gaseous This change represents an increase

434

be regarded as a

is

neither the internal energy nor the enthalpy of the system

Figure 17.1 Spontaneous mixing of two gases

may

statement of the second law of thermodynamics

accompanied by an increase in entropy. As an example of a spontaneous change, consider the mixing of two ideal gases. The two gases, which are under the same pressure, are placed in bulbs that are joined by a stopcock (see Figure 17.1). When the stopcock is opened, the gases spontaneously mix until each is evenly distributed throughout the entire apparatus. Why did this spontaneous change occur? The first law cannot help us answer this question. Throughout the mixing, the volume, total pressure, and temperature remain constant. Since the gases are ideal, no intermolecular forces exist, and every spontaneous change

mixing

A

said to have a high entropy. Since a disordered con-

more probable than an ordered

probability function.

after

important prediction.

central to the second law. Entropy

be interpreted as a measure of the randomness, or disorder, of a system.

highly disordered system

before mixing

entropy. S,

in

entropy.

—

mixture spontaneously separating into two pure gases, each occupying one of the bulbs.

Chapter 17

Elements

of

Chemical Thermodynamics

For a given substance the

solid, crystalline state

(most ordered); the gaseous state

and the liquid

is

is

the state of lowest entropy

the state of highest entropy (most random);

intermediate between the other two. Hence,

when a substance entropy increases. The reverse changes, crystallization and condensation, are changes in which the entropy of the substance decreases. Why, then, should a substance spontaneously freeze at temperatures below its state

is

either melts or vaporizes,

its

melting point, since this change represents a decrease in the entropy of the substance? All the entropy effects that result

from the proposed change must be conmix by the process previously described, there is no exchange of matter or energy between the isolated system in which the change occurs and its surroundings. The only entropy effect is an increase in the entropy

When two

sidered.

ideal gases

of the isolated system

Usually, however, a chemical reaction or a physical such a way that the system is not isolated from its surroundings. The total change in entropy is equal to the sum of the change in the entropy of the system {AS^^^^^J and the change in entropy of the surroundings

change

is

conducted

urroundings'

A5',o,ai

=

itself.

in

•

A^jy,,^^

(17.10)

-H AS^y^^oyndings

When a liquid freezes, the enthalpy of fusion is evolved by the liquid and absorbed by the surroundings. This energy increases the random motion of the surrounding molecules and therefore increases the entropy of the surroundings. The spontaneous freezing of a

liquid at a temperature

below the melting point

occurs, therefore, because the decrease in entropy of the liquid (AS^y^i^^,)

than offset by the increase in entropy of the surroundings there

is

more

so that

a net increase in entropy.

is

The

(AS'^urroundings)

total

taneity.

change

When

in

entropy should always be considered to determine spon-

a substance melts, the entropy increases, but this effect alone does

is spontaneous. The entropy of must also be considered, and spontaneity is indicated only if the total entropy of system and surroundings taken together increases. The data of Table 17.1 pertain to the freezing of water. The meaning of the units of AS values will be discussed in later sections. For the moment, let us be concerned only with the numerical values listed in the table. At - 1°C the change is spontaneous; AS^„^^^ is positive. At +rC, however, AS^„^^^ is negative, and

not determine whether or not the transformation the surroundings

not a spontaneous change. On the other hand, the reverse change, spontaneous at -I- C (the signs of all AS values would be reversed). At O'C, the melting point, AS^„^^^ is zero, which means that neither freezing nor melting is spontaneous. At this temperature a water-ice system would be in equilibrium, and no net change would be observed. Note, however, that freezing freezing

is

melting,

is

T

I

Table

1 7.1

Entropy changes for the transformation HjOO) -» liOCs) at

1

athf

Temperature

AS^jsi^m

(°C)

[J/(Kmol)]

(J/(Kmol)]

[J/(Kmol)]

1

-22.13

+ 22.05

-0.08

0

-21.99

-1

-21.85

+ 21.99 + 21.93

+ 0.08

+

17.3

0

Second Law

of

Thermodynamics

;

or meltmg can be made to occur at 0 C by removing or adding heat, but neither change will occur spontaneously. Thus, the A5,o,3, of a postulated change may be used as a criterion for whether the change will occur spontaneously. The entropy of the universe is steadily increasing as spontaneous changes occur. Rudolf Clausius summarized the first and second laws of thermodynamics as: "The energy of the universe is constant the entropy of the universe tends toward a maximum.'' Entropy, like internal energy and enthalpy, is a state function. The entropy, or randomness, of a system in a given state is a definite value, and hence, AS for a change from one state to another is a definite value depending only on the initial and final states and not on the path between them. It must be emphasized that, whereas thermodynamic concepts can be used to determine what changes are possible, thermodynamics has nothing to say about the rapidity of change. Some thermodynamically favored changes occur very slowly. Although reactions between carbon and oxygen, as well as between hydrogen and oxygen, at 25 "C and 1 atm pressure are definitely predicted by theory, mixtures of carbon and oxygen and mixtures of hydrogen and oxygen can be kept for prolonged periods without significant reaction; such reactions are generally initiated by suitable means. Thermodynamics can authoritatively indicate postulated changes that will not occur and need not be attempted, and it can tell us how to alter the conditions of a presumably unfavored reaction in such a manner that the reaction will be thermodynamically possible.

17.4 Gibbs Free Energy

The type of change of primary interest to the chemist is, of course, the chemical reaction. The ASj^^oundings for ^ reaction conducted at constant temperature and pressure may be calculated by means of the equation

'^'^surroundings

AH

~

(17.11)

t-

where A// is the enthalpy change of the reaction and Tis the absolute temperature. The change in the entropy of the surroundings is brought about by the heat transferred into or out of the surroundings because of the enthalpy change of the reaction. Since heat evolved by the reaction is absorbed by the surroundings (and vice versa), the sign of A// must be reversed. Hence, the larger the value of — AH, the more disorder created in the surroundings and the larger the value of ^"^surroundings

On

•

the other hand, the change in the entropy of the surroundings

is

inversely

proportional to the absolute temperature at which the change takes place.

A

given quantity of heat added to the surroundings at a low temperature (where the

randomness

is

relatively

low

initially) will

create a larger diff"erence in the disorder

of the surroundings than the same quantity of heat added at a high temperature

(where the randomness

measured

in units

In the last section

'^'^tolal

436

Chapter 17

is

relatively high to begin with).

of J/K.

we noted

that

'^'^syslem ~^ ^"^surroundings

Elements

of

Chemical Thermodynamics

Entropy

is

therefore

If

-^H|T

is substituted for A5,„,,„„„di„g3 and if the symbol ^S (without a subused to indicate the entropy change of the system, the following equation obtained

script) is

is

A'^'.otal

=

jr

Multiplication by

rA5,

Is

IF

\

/V96,500Cy

F

reaction

Cu'^ + 2e~

is

-> Cu(s),

and therefore

2

F

plate out 63.5 g

of Cu(s):

?

g

Cu =

Example

0.00466 F

/^Ml^^ j

0.

148 g Cu(s)

18.3

What volume of O^lg) at STP is liberated at the anode in the electrolysis of CUSO4 described in Example 18.2? (b) If 100 ml of 1.00 M CuSO^ is employed

(a)

in the cell,

that there

the

what is the H (aq) concentration at the end of the electrolysis? Assume is no volume change for the solution during the experiment and that

anode reaction

2H2O

is

4H^(aq) +

O^ig)

+ 4e-

Solution (a)

Four faradays produce 22.4

? liter

Chapter 18

02(g)

liter

/22.4

=

0.00466 F

=

0.0261

Electrochemistry

liter

0,(g)\ ^ '

\

I

liter

of O^ig) at STP:

02(g)

(b)

Four faradays also produce 4 mol of H^(aq):

?molH-(aq) = 0.00466 F'-"^^'"''"^'^ F 1

=

0.00466 mol

H + (aq)

The small contribution of H'^(aq) from the ionization of water may be ignored, may assume that there are 0.00466 mol H^(aq) in 100 ml of solution:

and we o ?

+ = .AAA mol H*(aq) 1000 ml ,

.

,

,

solution

,

0.00466 mol

'

H + (aq) ^

100 ml solution

= The solution

is

H + (aq)

therefore 0.0466

In Figure 18.4,

through one

0.0466 mol

M

in

hydrogen

ion.

two electrolytic cells are set up in series. Electricity passes and then through the other before returning to the current

cell first

source. If silver nitrate

Ag^ + e-

—

is

electrolyzed in one of the cells, the cathode reaction

is

Ag(s)

By weighing this electrode one can determine the quantity of silver plated before and out and hence the number of coulombs that have passed through the cell. One faraday would plate out 107.868 g of silver. One coulomb, therefore, is equivalent and metallic

silver

is

plated out on the electrode used.

after the electrolysis,

to

Figure 18.4

Silver coulometer

in

series with a cell for electrolysis

18.4

Stoichiometry

of Electrolysis

461

-

(107.868 g Ag)/(96,485 C)

lO'^gAg/C

1.1180 x

The same number of coulombs pass through both cells in a given time when these cells are arranged in series. The number of coulombs used in an electrolysis, therefore, can be determined

by the addition,

in series,

of this

silver

coulometer

to the circuit of the experimental cell.

Example (a)

What mass of copper

time that in

18.4 plated out in the electrolysis of

is

takes to deposit 1.00 g of

it

seri^ with the

Ag

in a silver

CUSO4 cell? (b) If a current of

1

.00

CUSO4

in the

coulometer that

is

same

arranged

A is used, how many minutes

are required to plate out this quantity of copper?

Solution (a)

From the electrode gAg:

reactions

we

see that 2

F

deposit 63.5 g

Cu and

F

1

deposits

107.9

(b)

?min =

1.00

107.9

=

14.9

C \ / s\/l min gAg/ VI C/\ 60s

96,500

gAg

1

min

18.5 Voltaic Cells

(

X

'

'

^

y^.n

)

A cell that is used as a source of electrical energy is called a voltaic cell or a galvanic Alessandro Volta (1800) or Luigi Galvani (1780),

cell after

who

first

experimented

with the conversion of chemical energy into electrical energy.

The

reaction between metallic zinc and copper(II) ions in solution

of a spontaneous change Zn(s)

+

Cu^^(aq)

in

is

illustrative

which electrons are transferred:

—Zn^^(aq) +

The exact mechanism by which

Cu(s)

electron transfer occurs

is

not known.

We

may,

however, represent the above reaction as a combination of two half-reactions:

Zn^^(aq) + le~

Zn(s)

2e'

+

Cu^^(aq)

—

* Cu(s)

In a voltaic cell these half-reactions are

made

to occur at different electrodes so

that the transfer of electrons takes place through the external electrical circuit

rather than directly between zinc metal

462

Chapter 18

Electrochemistry

and copper(II)

ions.

anode

The Daniell

Figure 18.5

The

cathode

cell

diagrammed

is designed to make use of this reaction The half-cell on the left contains a zinc metal electrode and ZnS04 solution. The half-cell on the right consists of a copper metal electrode in a solution of CUSO4. The half-cells are separated by a porous

to

cell

produce an

in

Figure 18.5

electric current.

1^'^

'^

mechanical mixing of the solutions but permits the passage of ions under the influence of the flow of electricity. A cell of this type is partition that prevents the

called a Daniell cell.

When

and copper electrodes are joined by a wire, electrons flow from copper electrode. At the zinc electrode the zinc metal is oxidized to zinc ions. This electrode is the anode, and the electrons that are the product of the oxidation leave the cell from this pole (see Table 18.1). The electrons travel the external circuit to the copper electrode where they are used in the reduction of copper(II) ions to metallic copper. The copper thus produced plates out on the electrode. The copper electrode is the cathode. Here, the electrons enter the cell and reduction occurs. Since electrons are produced at the zinc electrode, this anode is designated as the negative pole. Electrons travel from the negative pole to the positive pole in the external circuit of any voltaic cell when the cell is operating. The cathode, the zinc

the zinc electrode to the

where electrons are used Within the it is

cell

the

in the electrode reaction,

movement of ions completes

is

/ l

-

.

-

"

therefore the positive pole.

the electric circuit.

At

first

glance,

surprising that anions, which are negatively charged, should travel toward the negative electrode. Conversely, cations, which carry a positive

an anode that

is

charge, travel

toward the cathode, which

is

the positive pole.

Careful consideration of the electrode reactions provides the answer to this

problem. At the anode, zinc ions are being produced and electrons left behind metal. At all times the electncal neutrality of the solution is maintained. In

in the

the solution

surrounding the electrode there must be as

much

negative charge

18.5

Voltaic Cells

463

from cHiions as there is positive charge from cations. Hence, S04~ ions move toward the anode to neutralize the effect of the Zn- ions that are being produced. At the same time, zinc ions move away from the anode toward the cathode. At the cathode, electrons are being used to reduce Cu^ ions to copper metal. While the Cu^"^ ions are being discharged,

more Cu^"^

ions

move

into the region sur-

rounding the cathode to take the place of the ions being removed. If this did not occur, a surplus of SO4" ions would build up around the cathode. The porous partition is added to prevent mechanical mixing of the solutions of the

would be transferred

move

into contact with the zinc metal electrode,

directly rather than

through the

circuit. In the

Cu~^ away from the zinc electrode. would work if a solution of an electrolyte other than ZnS04

normal operation of the ions

came

Cu'"^ ions

half-cells. If

electrons

cell, this

"short circuit" does not occur because the

in a direction

Actually this

cell

anode compartment, and if a metal other than copper were used The substitutes, however, must be chosen so that the electrolyte in the anode compartment does not react with the zinc electrode and the cathode does not react with Cu" ions. were used

in the

for the cathode.

18.6 Electromotive Force If

1

cell

M

ZnS04 and M CUSO4 solutions may be represented by the notation

Zn(s)|Zn2^(l M)|Cu^ in

are

1

which the

+ (l

listed in the

is

phase boundaries. By convention, the sub-

listed first.

The other materials of

the cell are then

order that one would encounter them leading from the anode to the

The composition of the cathode

cathode.

in the Daniell cell, the

M)|Cu(s)

vertical lines represent

stance forming the anode

employed

is

given

last.

produced by a voltaic cell as a result of the electromotive force (emf of the cell, which is measured in volts. The greater the tendency for the cell reaction to occur, the higher the emf of the cell. The emf of a given cell, howElectric current

is

)

ever, also

A

which state

depends upon the concentrations of the substances used to make the

standard emf, S" \ pertains to the electromotive force of a reactants

all

and products are present

of a solid or a liquid

is,

— that

is,

in

solution

is

cell.

25'C,

in

The standard or pure liquid itself. The

standard

of course, the pure solid

standard state of a gas or a substance activity

in their

cell, at

states.

a defined state of ideal unit

corrections are applied for deviations from ideality caused by

intermolecular and interionic attractions. For our discussion we shall make the assumption that the activity of ions may be represented by their molar concentrations and the activity of gases by their pressures in atmospheres. Hence, according concentrations to this approximation, a standard cell would contain ions at 1

M

and gases

at

1

atm

be indicated only If the cell

emf of

pressures. In the cell notations that follow, concentrations will

if

they deviate from standard.

a cell

is

to be

used as a reliable measure of the tendency for the must be the maximum value obtainable for the

reaction to occur, the voltage

particular cell under consideration. If there

is an appreciable flow of electricity during measurement, the voltage measured, S\ will be reduced because of the

internal resistance of the

cell.

In addition,

when

the cell delivers current, the

electrode reactions produce concentration changes that reduce the voltage.

Chapter 18

Electrochemistry

The emf of

must be measured with no appreciable flow of is accomplished by the use of a potentiometer. The circuit of a potentiometer mcludes a current source of variable voltage and a means of measuring this voltage. The cell being studied is connected

electricity

to the

a cell, therefore,

through the

This measurement

cell.

potentiometer circuit

in

such a way that the emf of the

emf of the potentiometer current

cell is

opposed by the

source.

If the emf of the cell is larger than that of the potentiometer, electrons will flow in the normal direction for a spontaneously discharging cell of that type. On the other hand, if the emf of the potentiometer current source is larger than

that of the cell, electrons will flow in the opposite direction, thus causing the cell

When

reaction to be reversed. flow. This voltage

is

the

the reversible

two emf's are exactly balanced, no electrons emf of the cell. The emf of a standard Daniell

1.10 V.

cell is

Faraday's laws apply to the

cell reactions of voltaic, as well as electrolytic, precaution must be observed, however. Electricity is generated by the simultaneous oxidation and reduction half-reactions that occur at the anode and

One

cells.

if the cell is to deliver current. Two faradays of electricity will be produced, therefore, by the oxidation of 1 mol of zinc at the anode together with the reduction of mol of Cu^^ ions at the cathode. The partial equations

cathode, respectively. Both must occur

1

Zn

anode: cathode:

when read

in

2e~

+ Cu^^

-Zn^^ + 2e'

— Cu

terms of moles, represent the flow of two times Avogadro's number

of electrons or the production of 2 F of electricity.

The quantity of

oHhe

electrical energy, in joules,

produced by a

quantity of electricity delivered, in coulombs, and the

cell is

the product

emf of

the

cell, in

The electrical energy produced by the reaction between metal and mol of copper(II) ions may be calculated as follows:

volts (see Section 18.1). 1

mol of zinc

1

2(96,500 C)( 1. 10 V)

One

=

212,000

J

volt coulomb is a joule. The emf used in the preceding

=

212 kJ

•

standard Daniell

cell

calculation

is

the reversible

and hence the maximum voltage

emf

{S") of the

for this cell. Therefore,

maximum work that can be obtained from the The maximum net work* that can be obtained from a chemical reaction conducted at a constant temperature and pressure is a measure of the decrease in the Gibbs free energy (see Section 17.4) of the system. the value secured (212 kJ)

operation of this type of

is

the

cell.

Hence, (18.1)

*

Some

the

reactions proceed with an increase in volume, and the system must do work to expand against in order to maintain a constant pressure. The energy for this pressure-volume work

atmosphere

not available for any other purpose; it must be expended in this way if the reaction is to occur at constant pressure. Pressure-volume work is not included in the potentiometric measurement of the electrical work of any cell. Net work (or available work) is work other than pressure-volume work. is

18.6

Electromotive Force

is the number of moles of electrons transferred in the reaction (or the number of faradays produced), F is the value of the faraday in appropriate units, and (f is the emf in volts. If F is expressed as 96,487 C, AG is obtained in joules.

where n

A change in free energy derived

from a standard emf, S"', is given the symbol AG°. change of a reaction is a measure of the tendency of the reaction to occur. If work must be done on a system to bring about a change, the change is not spontaneous. At constant temperature and pressure a spontaneous change is one from which net work can be obtained. Hence, for any spontaneous reaction

The

free energy

AG

the free energy of the system decreases;

only

if

S

negative. Since

is

AG = —nFS,

and serve as a source of

positive will the cell reaction be spontaneous

is

electrical energy.

18.7 Electrode Potentials In the

same way

may

that a cell reaction

be regarded as the sum of two half-

reactions, the

emf of a cell may be thought of as

However,

impossible to determine the absolute value of the potential of a single

half-cell.

it is

A

relative scale has

the

sum of two half-cell

potentials.

been established by assigning a value of zero to the

voltage of a standard reference half-cell and expressing

all

half-cell potentials

relative to this reference electrode.

The

reference half-cell used

of hydrogen gas,

at

1

atm

is

the standard hydrogen electrode, which consists

pressure, bubbling over a platinum electrode (coated

with finely divided platinum to increase solution containing

H^(aq)

its

surface) that

at unit activity. In

shown connected by means of a

is

immersed

in

an acid

Figure 18.6 a standard hydrogen bridge to a standard

Cu^*/Cu

electrode

is

electrode.

A salt bridge is a tube filled with a concentrated solution of a salt (usually

salt

KCl), which conducts the current between the half-cells but prevents the mixing of the solutions of the

half-cells.

The

cell

may be diagrammed

of Figure 18.6

as

PtlHjIH + IICu^ + ICu

A

double bar indicates a

copper electrode

The

cell

is

The hydrogen electrode is the anode, the emf of the cell is 0.34 V. be the sum of the half-cell potential for the oxida-

salt bridge.

the cathode, the

emf is considered to we shall

tion half reaction (which

give the symbol Sl^)

for the reduction half reaction (which

we

and the

half-cell potential

shall indicate as #red)-

For the

cell

of

Figure 18.6,

n2^2\\^+2e-

anode: cathode:

le'

+ Cu-+

—

*

Since the hydrogen electrode "

=

0.00

V

= +0.34 V

Cu

arbitrarily assigned a potential of zero, the entire

The value +0.34 V is Cu^'^/Cu electrode. Notice that electrode potentials are given for reduction half-reactions. If the symbol S" (without a subscript) is used for an electrode potential, S'^^^ is understood. If a cell is constructed from a standard hydrogen electrode and a standard cell

^

is

Sl^

emf

is

ascribed to the standard Cu^'^/Cu electrode.

called the standard electrode potential of the

Zn^'^/Zn electrode, the zinc electrode Thus,

Chapter 18

Electrochemistry

v

is

the anode,

and the emf of the

cell is

0.76 V.

—Zn^^+le" + 2H^ — H2 Zn

anode:

le'

cathode:

Cx=+0.76V

Cd =

0.00

V

The value +0.76 V

it

corresponds

to

is

a reduction

is sometimes called an oxidation potential, since an oxidation half-reaction. An electrode potential, however,

potential.

To

obtain the electrode potential of the

Zn^^/Zn

couple,

we must

change the sign of the oxidation potential so that the potential corresponds to the reverse half-reaction, a reduction

Ced ^ "0.76 V

2e-

+ Zn^^

It is

not necessary to use a

Zn

cell

containing a standard hydrogen electrode to

obtain a standard electrode potential. For example, the standard potential of the

Ni~^/Ni electrode may be determined from the

cell

Ni|Ni^ + ||Cu^ + |Cu

The emf of

this cell

+ Cu^+

Ni

—

is

Ni^^

The standard electrode 2eIf

we

+ Cu^^

+ Cu

^ieii

= +0.59 V

potential of the Cu'"^/Cu electrode has been determined:

— Cu

Ced = +0.34 V

subtract the Cu^'^/Cu half-reaction from the

half-cell potential

Ni

0.59 V, and the nickel electrode functions as the anode:

from the

—Ni^++2e-

cell

emf,

cell

reaction

and subtract the

we obtain

= +0.25V

.

18.7

Electrode Potentials

467

" ""

Table 18.2

Standard'telectrode potentials at

25°^

Half-Reaction

+

LI

Ba"

^

Ca"

*

1

e

Ca

-He

Na

Na

Ba

Mg^ ^

4-

2e"

Mg

Al'^

-1-

3e"

Al

O

LI HiO

H,

Zn^* + 2e

Zn

Cr^* *

Fe'

Cd"* Ni"

*

1

-I-

3e

-H

2e

Fe

-1-

2e"

Cd

-h

2e

-h

2e

Sn

-1-

2e"

Pb

"

""^

Pb^^ *i

— .662 — 0.82806 — 0.7628 — 0.744 — 0.4402 — 0.4029 — 0.250 1

-H

2

OH

Cr

-1-

Ni ^

Sn^

— 2.714 — 2.363

~

O

-t

^

\iL

I

'

/.'i'

'

;.

.

f

therefore

- -0.25 V

Ni

Standard electrode potentials are listed in Table 18.2, and a more complete list found in the appendix. The table is constructed with the most positive electrode potential (greatest tendency for reduction) at the bottom. Hence, if a pair of is

I '

—

+ Ni^+

2e-

is

electrodes

of the

is

combined

cell will

be that

to

make

a voltaic cell, the reduction half-reaction (cathode)

listed for the electrode that

stands lower in the table, and

the oxidation half-reaction (anode) will be the reverse of that

shown

for the

electrode that stands higher in the table.

For example, consider

from standard

The

table entries for these electrodes are

+

2e-

^

Ni^^

Ag+ +

468

a cell constructed

electrodes.

Chapter 18

e-

= -0.250 V

Ni

^Ag

Electrochemistry

S;,^

= +0.799 V

Ni'^^^/Ni

and Ag^/Ag

Of

the two ions, the Ag"^ ion shows the greater tendency for reduction. The Ag''/Ag electrode is therefore the cathode, and the Ni^ + /Ni electrode is the anode. The half-reaction that takes place at an anode is an oxidation, and the halfcell potential is an oxidation potential. The sign of the table entry for the Ni^ + /Ni half-cell, therefore, must be reversed to give an

cathode:

The

cell

Ni

—Ni^^ ^ + 2Ag+ —lAg Ni

anode:

le'

reaction and cell

+ 2Ag+

—

2e'

il^

= +0.799 V

emf may be obtained by

+

Ni-+

2

= +0.250V

Ag

=

Sl^^,

addition

+1.049

V

Notice that the half-reaction for the reduction of Ag"^ must be multiplied by 2 before the addition so that the electrons lost

cancel.

The S°

for the

Ag^/Ag

and gained

electrode, however,

in the half-reactions will

not multiplied by

is

2.

The

magnitude of an electrode potential depends upon the temperature and the concentrations of materials used in the construction of the half-cell. These variables are fixed for standard electrode potentials. Indication of the stoichiometry of the cell

reaction does not imply that a concentration change has been made.

Actually the half-reactions implied by the half-cell potentials are

2H^ + Ni W2 + 2Ag+ Ni + 2Ag+

— — —

+

+

+ 2Ag + 2Ag

Ni^^

Notice, however, that and H"^ ions cancel.

Ni^

in the

addition of these half-reactions, the

H, molecules

Electrode potentials are also useful for the evaluation of oxidation-reduction reactions that take place outside of electrochemical cells.

about an oxidation and

An

oxidizing agent

is

a substance that brings

A

strong oxidizing agent, therefore, has a high positive reduction potential,

The strongest oxidizing agent given given in the table

F2(g)

+

2e--

Table 18.2

is

process

is

itself

reduced.

F2(g) since the highest

S',^^.

S'^^^

is

—

2F-(aq)

The best oxidizing agents CrjOy" in acid.

A

in

in the

reducing agent

is

Ced = +2.87

V

listed in the table are

itself

oxidized

in

Fj,

Mn04

in acid, CI2,

bringing about a reduction.

reducing agent, therefore, has a high, positive oxidation potential.

A

and

strong

Remember

that Sl^ values are obtained by changing the signs of the table values; the corresponding oxidation half-reactions are derived by reversing the partial equations shown. The strongest reducing agent given in Table 18.2 is Li metal since the

highest Sl^ derived

Li{s)

The

—

from the table values

Li^(aq) + e'

S'l,

=

is

+3.045

V

best reducing agents given in the table are the active metals Li, K, Ba, Ca,

and Na.

18.7

Electrode Potentials

469

Whether or not a proposed reaction will be spontaneous with all substances present at unit activity can be determined by use of electrode potentials. spontaneous reaction is indicated only if the emf of the reaction is positive.

A

Example

18.5

Use electrode potentials are spontaneous with

Ch(g)

(a)

(b) 2

Ag(s)

whether the following proposed reactions

to determine

——

all

substances present at unit activity:

+ 2r(aq) + 2 H " (aq)

+ I^ls) Ag + (aq) + H^ig)

2Cl-(aq) 2

Solution In the proposed reaction, the CI2

(a)

for this half-reaction)

and the I"

is

is

reduced to Cl~ (and we need an

oxidized to I2 (and

we need an

(f^ed

Sl^ for this

half-reaction):

+

2e-

Cl2(g)

2r(aq)

Cyg) + 2r(aq) Since the overall

—2Cr(aq) — + —2Cr(aq) + l2(s)

emf is

(b) In this reaction,

2Ag(s)

The

reaction

is

l2(s)

positive, the reaction

Ag

2H + (aq) + 2^" 2Ag(s) + 2H + (aq)

= + 1.360 V = -0.536 V emf = +0.824 V

Cd

2e-

is

oxidized

{S^„^

— —2Ag^(aq) +

—

spontaneous.

is

needed) and

+

(S^^^ needed):

SI,

H2(g)

not spontaneous as written.

and H2) would be spontaneous (emf

reduced

= -0.799 V 0.000 V S°,i = emf = -0.799 V

le'

H2(g)

2Ag^(aq)

is

The

= +0.799

There are several factors that must be kept

reverse reaction (between

Ag^

V).

in

mind when using

a table of

electrode potentials to predict the course of a chemical reaction. Because

changes with changes

in concentration,

many presumably unfavored

S

reactions

can be made to occur by altering the concentrations of the reacting species. In addition, some theoretically favored reactions proceed at such a slow rate that they are of no practical consequence.

Correct use of the table also demands that given element be considered before

making

all

pertinent half-reactions of a

a prediction.

On

the basis of the

half-reactions

7>e-

+ Fe^^

2e-

+

2

H+

^ ^

Fe

S',,^

= -0.036 V

H,

Cd

=

0.000

V

one might predict that the products of the reaction of iron with H"^ would be hydrogen gas and Fe^"^ ions (emf for the complete reaction, +0.036 V). The oxidation state iron(II), however, lies between metallic iron and the oxidation state

470

Chapter 18

Electrochemistry

Once an

iron(III).

further oxidation

.

+

e-

^

Fe^+

atom has lost two electrons and becomes an Fe^"" ion, opposed, as may be seen from the reverse of the following:

iron is

=

Fe^^

Thus, the reaction yields Fe^

^

V

+0.771

ions only. This fact could have been predicted by

an examination of the half-reaction

2^-

+

Fe^+

^

Fe

= -0.440 V

S',^^

The

(fo, for the production of Fe^"^ ions from the reaction of iron metal and H"^ ions ( + 0.440 V) is greater than that for the production of Fe^^ ions + 0.036 V), and hence the former is favored. We may summarize the electrode potentials for iron and its ions as follows:

(

„

3+

Pe^+

V ^ 2+ Fe^

+0.77

V „

-0.440

>

Fe

*

V

-0.036

The preceding predictions are immediately evident from ber that oxidition

is

this

diagram,

if

we remem-

the reverse of the relation corresponding to an electrode

potential.

Occasionally an oxidation state of an element portionation (auto oxidation-reduction, see Section for

copper and

its

may

ions

„

Cu 2+

1

is 1

unstable toward dispro-

.3).

The

electrode potentials

be summarized as follows: +0.153

V

„

Cu

*

+

+0.521

V „ *

Cu

+ 0.337 V

From

this

we

see that the

Cu^

ion

is

are disproportionate to copper metal

2Cu + (aq)

—

Cu(s)

+

not a very stable one. In water,

and Cu^^

Cu^

ions

ions:

Cu^^(aq)

The emf for this reaction is +0.521 - 0.153 = +0.368 V. Species instable toward disproportionation may be readily recognized by the fact that the electrode potential for the reduction to the next lower oxidation state is more positive than the electrode potential for the couple with the next higher oxidation state.

inspection of the diagram for iron

and

its

ions

shows

that the Fe""^ ion

is

An

stable

toward such disproportionation.

Gibbs Free Energy Change and Electromotive Force The

reversible

energy for the

emf of

a

cell, CI > Br > I, and the oxidizing power of the halogens decreases in the same order. Electronegativity.

3.

and

fluorine

is

the

Bond energy

Bond energy.

4.

size of the

CU

decreases from

halogen atom makes

and

to Br, to

h

since the increasing

bond between atoms of the X, molecule. The bond dissociation energy of the F2 molecule is, however, unusually low and out of line in comparison to the other values. The reason for this relatively low value is not completely understood but is ascribed to the efl'ect of the nonbonding electrons in the Fj molecule. A repulsion between it

easier

easier to break the

the

the highly dense electron clouds of the small fluorine

atoms is believed to weaken bond and lower the energy required to break it. The bond formed between fluorine and an element other than itself is always stronger than the bonds formed by any of the other halogens with the same element. The bond energies of the hydrogen halides, for example, are HF, 565 kJ/ mol; HCI, 431 kJ mol; HBr. 364 kJ mol; and HI. 297 kJ mol. The high order the

of chemical reactivity of F, in reactions with other nonmetals therefore, of the

with the high bond energies of the Electrode potentials.

5.

the

same order

halogen

is

the result,

low bond energy of the F2 molecule (energy required), coupled

new bonds (energy

The values

for the

X, X'

released).

electrode potentials

as the electronegativity values. Fluorine

— iodine

is

the

most

fall

in

reactive

the least. Since electrode potentials refer to processes that

aqueous solution and since fluorine reacts with water, the Fj F~ electrode potential is obtained by calculation rather than by direct measurement.

occur

in

|^^b!^9^ Some

properties of the halogens

Property

F,

color

Br,

CI,

pale yellow

yellow-

red-

violet-

green

brown

black

melting point (°C)

-218

-101

-7

boiling point (°C)

- 188

-35

+ 59

atomic radius (pm) ionic radius, first

X (pm)

ionization energy (kj/mol)

bond energy (kJ/mol) standard electrode potential 2e" + X, ;^=^ 2X

72

99

114

133

136

181

195

216

1.68 X 10^ 4.0

electronegativity

+ 113 + 183

1.14 X 10^

1.25 X 10^ 3.2

3.0

1.00 X 10^ 2.7

155

243

193

151

+ 2.87

+ 1.36

+ 1.07

+ 0.54

(V)

19.6

Properties of the Halogens

499

The

relative oxidizing ability of the

halogens

may

be observed

in

displacement

and iodine from their their salts; and bromine and iodine from salts; chlorine can displace bromine iodides: iodine from can displace reactions. Thus, fluorine can displace chlorine, bromine,

F,(g)

CU(g)

+

2 NaCl(s)

+ 2Br'(aq)

Br.(l)

— + —2Cr(aq) + 2 NaF(s)

+ 2r(aq)

2Br-(aq)

+

CI ,(g)

Br,(H ^(s)

Since fluorine actively oxidizes water (producing Oj), displacement reactions

cannot be run

involving

19.7

Occurrence and of the Halogens The

in

water solution.

Industrial Preparation

principal natural sources of the halogens are listed in Table 19.3.

are too reactive to occur in elemental form.

they occur

is

The most common

The halogens

state in

which

as halide ions.

The halogens

are produced commercially in the following ways:

Fluorine. Fluorine must be prepared by an electrochemical process because no suitable chemical agent is sufficiently powerful to oxidize fluoride ion to 1.

fluorine.

Furthermore, since the oxidation of water

is

easier to accomplish than

must be carried out under

the oxidation of the fluoride ion, the electrolysis

anhydrous conditions. In actual practice, a solution of potassium fluoride in anhydrous hydrogen fluoride is electrolyzed. Pure HF does not conduct electric current; the KF reacts with HF to produce ions (K^ and HF2 that can act as charge carriers. The HF7 ion is formed from a F~ ion hydrogen-bonded to a HF molecule (F H"*F"); this hydrogen bond is so strong that the H atom is exactly midway between the two F atoms. The hydrogen fluoride used in the electrolysis is commercially derived from fluorospar, CaF, (see Section 19.10): )

—

2KHF,(1)

'^'"'"''''y'' ,

H,(g)

+

F,(g)

+ 2KF(1)

heat

Table 19.3

Occurrence

of the halogeni^

Percent

of

Element

Earth's Crust

fluorine

6.5 X 10"'

Occurrence

CaF,

(fluorospar), Na,AIF(, (cryolite),

Ca5('P04),F (fluorapatite)

chlorine

5.5 X

10"^

bromine

1.6 X

10"^

Br" (sea water, underground brines, solid salt beds)

iodine

3.0 X

10~*

r

Cl~ (sea water and underground brines)

NaCI (rock

salt)

(oil-well brines,

sea water)

NalO,, Na O4 (impurities saltpeter,

500

Chapter 19

The Nonmetals, Part

I:

NaNOj)

Hydrogen and the Halogens

in

Chilean

Cells used for the electrolysis of

sodium cfiloride brine. Hooker Chemical Company.

2.

Chlorine.

The

principal industrial source of chlorine

is

the electrolysis of

aqueous sodium chloride, from which sodium hydroxide and hydrogen are also products:

2Na^(aq) + 2Cr(aq) +

2H,0

^'^^'^^'y^'^

H,(g)

+

Cl,(g)

+ 2Na^(aq) + 20H

(aq)

Chlorine is also obtained, as a by-product, from the industrial processes in which the reactive metals Na, Ca, and Mg are prepared. In each of these processes an anhydrous molten chloride is electrolyzed and CU gas is produced at the anode. For example, '''"'-"'y^'^

2NaCl(l) 3.

Bromine.

2Na(l)

Bromine

is

+

CU(g)

commercially prepared by the oxidation of the bromide

ion of salt brines or sea water with chlorine as the oxidizing agent:

Cl.lg)

+

2Br-(aq)

—

2Cr(aq) +

19.7

Br,(l)

Occurrence and

Industrial Preparation of thie

Halogens

liberated Bvj is removed from the solution by a stream of air and, in subsequent steps, collected from the air and purified.

The

4.

Iodine.

found

In the United States the principal source of iodine

Free iodine

in oil-well brines.

Cl2(g)

In addition, iodine

Chilean

-ICrtaq) +

+ 2r(aq)

nitrates.

2I03(aq)

is

bisulfite is

—

+ SHSOjIaq)

in

used to reduce the iodate ion:

+ 5SO^"(aq) + 3H + (aq) + H^O

1,(5)

Halogens

the exception of fluorine (which

halogens are usually prepared

free

I^is)

commercially obtained from the iodate impurity found

Sodium

19.8 Laboratory Preparation of the

With

the iodide ion

is

obtained by chlorine displacement:

is

must be prepared electrochemically), the

in the

laboratory by the action of oxidizing

agents on aqueous solutions of the hydrogen halides or on solutions containing the

sodium halides and

From

sulfuric acid.

a table of standard electrode potentials

we can

get an

approximate idea

of what oxidizing agents will satisfactorily oxidize a given halide ion. Thus, any couple with a standard electrode potential

the chloride ion

= +1.36 V) as (^^ed — +0.54 V).

(tfred

and the iodide ion

more

C

with

all

only +1.23 V, is

MnOi

used (rather than

HCl

at unit activity)

and

if

->

Mn^^

concentrated

if

the reaction

KMn04, KjCrjO,, PbO^. and Mn02

actual practice,

MnO,

capable of oxidizing the chloride ion

is

should oxidize = +107 V)

materials in their standard

Thus, even though the standard electrode potential of

states.

HCl

+ 1.36 V

bromide ion (4ed

Recall, however, that standard electrode

potentials are listed for half-reactions at 25

is

positive than

well as the

is

heated. In

are frequently used to

prepare the free halogens from halide ions:

MnO,(s) +

4H + (aq) +

2Cr(aq)

2Mn04(aq) + 16H + (aq) + 10Br~(aq) Cr20r(aq) + 14H^(aq) + 6r(aq)

19.9

Mn^^(aq) +

Cl2(g)

—2Mn-^(aq) + 2Cr^ + (aq) +

+ 2H2O

5 Br2(l)

3 l2(s)

+

SHp

+ 7H,0.

The Interhalogen Compounds Some

of the reactions of the halogens are summarized in Table 19.4.

The halogens

produce a number of interhalogen compounds. All the compounds with the formula XX' (such as BrCl) are known except IF. Four XX3 compounds have been prepared (CIF3, BrFj, ICI3, and IF3), and three XXj compounds are known (CIF5, BrFj, and IF5). The only XX, compound that has been made is IF7. With the exception of ICI3, all the molecules for which /; of the formula XX^ is greater than are halogen fluorides in which fluorine atoms (the smallest and most electronegative of all the halogen atoms) surround a CI, Br, or I atom. The react with each other to

1

stability of these

502

Chapter 19

compounds

The Nonmetals. Part

I:

increases as the size of the central

Hydrogen and the Halogens

atom

increases.

Some reactions of the

Table 19.4

liaiogens (Xj

=

F^, CI2, Bfj, or Ij)

General Reaction

+ 2M

nX^

2

Remarks

MX„

F,, CI, with practically all metals;

Rr '2 1 lA/ith oil i-"2' "Villi all y A2

4-

H n

ov^or,t IIOUIG r\r\V^\ck CAUcpi

rr,^t.,lf rrieiais

rlA

with excess P; similar reactions with r^o, OU, dllLI 01

5X2 + 2P

2PX5

with excess X,. but not with

I,'

SbF,, SbCl5, ASF5, ASCI5, and BiF, may be similarly prepared

+ 2S

X, X,

-)-

H,0

2X2 + 2H,0

+ H2S X2 + CO

X2

X2 X2

+ so. + 2X'"

X2

+

X2

S2X2

+ X

— '

with CI2, Br2

+ HOX

4H^ + 4X' + O2 2HX + S

not with Fj F, rapidly; Clj. Br2 slowly in sunlight

COX,

CI,, Br,

S0,X2

F2.CI2

X',

+ 2X"

2 XX'

F2

>

CI2

>

>

Br2

I,

formation of the interhalogen (all except IF)

compounds

Hence, neither BrF, nor CIF7 has been prepared, although IF7 is known; CIF5 readily decomposes into CIF3 and Fj, whereas BrF, and IF, are stable at temperatures above 400 C.

The

compounds have received conatom in each of these Figure 19.1). The central atom of

structures of the higher interhalogen

siderable attention because the bonding of the central

molecules violates the octet principle (see

each in its

XX3 molecule has three bonding pairs and two nonbonding pairs of electrons valence shell, and the molecules, therefore, are T-shaped. The XX5 molecules

are square pyramidal since each of the central atoms has five bonding pairs and

one nonbonding pair of electrons in its valence shell. The nonbonding electron pairs of the central atoms of these two types of molecules introduce some distortion. The I atom of the IF7 molecule has seven bonding pairs of electrons in its valence shell. The IF7 molecule is pentagonal bipyramidal.

19.9

The Interhalogen Compounds

— 19.10

The Hydrogen Halides Each of the hydrogen hahdes may be prepared by the

direct reaction of

hydrogen

The

reactions

with the corresponding free halogen:

U2 + X2 The vigor

—2HX

of the reaction decreases marlcedly from fluorine to iodine.

serve as important industrial sources of HCl,

HF and HCl can be prepared by

Both

HBr, and HI.

the action of warm concentrated sulfuric

CaF2 and NaCl. Both

acid on the corresponding natural halide,

reactions serve

as important industrial sources of the gases:

CaF.is)

+

H,S04(1)

NaCl(s)

+

HjSO^Il)

—

CaSO^ls)

+ 2HF(g)

—*NaHS04(s) +

All hydrogen halides are colorless gases at the other hand,

is

room temperature;

sulfuric acid,

on

a high-boiling liquid. Thus, the foregoing reactions are examples

method

of a general

HCl(g)

for the preparation of a volatile acid

from

its salts

by means

of a nonvolatile acid. At higher temperatures (about 500 C), further reaction

occurs between

NaCl(s)

NaHS04

and NaCl:

+ NaHS04(l)

—

HCl(g)

+ Na2S04(s)

Hydrogen bromide and hydrogen iodide cannot be made by

the action of

concentrated sulfuric acid on bromides and iodides because hot, concentrated sulfuric acid oxidizes these anions to the free halogens.

ions are easier to oxidize than the fluoride

2NaBr(s)

+ 2H2S04(1)

Since the iodide ion

bromide

ion,

is

—*

Br,(g)

+

and chloride

SO.ig)

The bromide and

iodide

ions:

+ Na2S04(s) + 2H20(g)

a stronger reducing agent (more easily oxidized) than the

S and HjS, as well as SO,, are obtained as reduction products

from the reaction of Nal with hot concentrated sulfuric acid. Pure HBr or HI can be obtained by the action of phosphoric acid on or Nal; phosphoric acid

is

an essentially nonvolatile acid and

is

a

NaBr

poor oxidizing

agent:

NaBr(s)

+

H3P04(1)

Nal(s)

+

H3P04(1)

The hydrogen

* HBr(g) + NaH2P04(s)

—

halides

HI(g)

may

+ NaH2P04(s)

be prepared by the reaction of water on the ap-

propriate phosphorus trihalide:

PX3 + 3H2O

—

3HX(g) + H3P03(aq)

Convenient laboratory preparations of HBr and HI have been developed in which red phosphorus, bromine or iodine, and a limited amount of water are employed and in which no attempt is made to isolate the phosphorus trihalide intermediate.

Hydrogen fluoride molecules associate with each other through hydrogen bonding. The vapor consists of aggregates up to (HF)^, at temperatures near the 504

Chapter 19

The Nonmetals. Part

I:

Hydrogen and the Halogens

boiling point (19. 4°C) but

is

highly associated at higher temperatures. Gaseous

less

HCl, HBr, and HI consist of single molecules. Liquid HF and solid HF are more highly hydrogen bonded than gaseous HF, and the boiling point and melting point of HF are abnormally high in comparison with those of the other hydrogen halides.

All hydrogen halides are very soluble in water; water solutions are called hydrohalic acids. Aqueous HI, for example, is called hydroiodic acid. The bond is stronger than any other H X bond; HF is a weak acid in water solution,

H—

whereas HCl, HBr, and HI are completely dissociated:

HF(aq)

The F"

^

H^(aq)

+

F-(aq)

ions from this dissociation are largely associated with

F-(aq)

+

HF(aq)

HF

Concentrated

^

HF^

solutions are

Hydrofluoric acid reacts with

+ 6HF(aq)

When warmed,

more

strongly ionic than dilute solutions and

SiO^lg)

For

+ 4HF(aq)

this reason,

silica,

—

HFj H,F3 and .

,

SiO^, and glass, which

—2H^(aq) +

the reaction

molecules:

(aq)

contain high concentrations of ions of the type

Si02(s)

HF

SiF^-(aq)

higher.

made from

is

silica:

+ 2H,0

is

SiFJg)

4-

2H20(g)

hydrofluoric acid must be stored

in

wax or

plastic containers

instead of glass bottles.

The

is an example of a large group of complex ions Halo complexes are formed by most metals (with the notable exceptions of the group I A, group II A, and lanthanide metals) and with some nonmetals (for example, BF4 The formulas of these complex ions are most commonly of the types (MX4)''''" where n is the and (MX^)"' oxidation number of the central atom of the complex.

fluorosilicate ion, SiF^

formed by the halide

,

ions.

).

19.11

The Metal Halides Metal halides can be prepared by direct interaction of the elements, by the

re-

actions of the hydrogen halides with hydroxides or oxides, and by the reactions

of the hydrogen halides with carbonates:

K2C03(s)

+

2HF(1)

—

2KF(s) + C02(g) +

HjO

bonding in metal halides varies widely as do the physical compounds. A metal that has a low ionization energy generally forms halides that are highly ionic and consequently have high melting and

The character of

the

properties of these

boiling points.

On

the other hand, metals that have comparatively high ionization

energies react, particularly with bromine and iodine, to form halides in which the bonding has a high degree of covalent character. These compounds have comparatively low melting points and boiling points. 19.11

The Metal Halides

505

KCI

00

O 13

CD

>

CaCI 2 50

TiCU 0 Figure 19.2

Equivalent conductances

In general, the halides of the

of

I

some molten

A

chlorides near the melting point

metals, the

A

II

metals (with the exception of

and most of the inner-transition metals are largely ionic; halides of the remaining metals are covalent to a varying degree. Within a series of halides of metals of the same period, ionic character decreases from compound to compound as the oxidation number of the metal increases and the size of the cation undergoes a concomitant decrease. The ionic character of a compound is reflected in its ability to conduct electric current as measured by its conductance. The equivalent conductances of the molten chlorides of some fourth-period metals are plotted in Figure 19.2. The "cations'" of the compounds (K*, Ca^^, Sc^"^, and Ti"^"*") are isoelectronic. Potassium chloride is a completely ionic solid, and molten KCI has the highest conductance of the four compounds. Titanium(IV) chloride, TiCl4, is a covalent liquid and is nonconducting. If a metal exhibits more than one oxidation state, the halides of the highest oxidation state are the most covalent. Thus, the second member of each of the Be),

following pairs

is

a highly covalent, volatile liquid (melting points are given in

parentheses): SnCl2 (246 C),

SnCU (-33

C); PbClj (SOl C),

PbCU

(-15°C);

SbCl3(73.4 C), SbCl,(2.8 C). Since fluorine is the most electronegative halogen, fluorides are the most ionic of the halides; ionic character decreases in the order; fluoride

bromide >

iodide.

The

halides of

aluminum

>

chloride

>

are an excellent example of the

and the

relationship between ionic or covalent character

size

of the halide ion.

an ionic substance. Aluminum chloride is semicovalent and crystallizes in a layer lattice in which electrically neutral layers are held together by London forces. Aluminum bromide and aluminum iodide are essentially covalent the crystals consist of AUBr^ and AUIe molecules, respectively. The water solubility of fluorides is considerably difi"erent from that of the chlorides, bromides, and iodides. The fluorides of lithium, the group II A metals, and the lanthanides are only slightly soluble, whereas the other halides of these

Aluminum

fluoride

is

;

metals are relatively soluble.

Most form

chlorides, bromides,

slightly soluble

cury(I), lead(II), copper(I),

The

and

Chapter 19

AgCl

The Nonmetals, Part

I:

in water.

The

cations that

these halide ions include silver(I), mer-

thallium(I).

insolubility of the silver salts of CI",

test for these halide ions;

506

and iodides are soluble

compounds with

is

white,

Br

AgBr

,

and

is

I

is

the basis of a

cream, and Agl

Hydrogen and the Halogens

is

common

yellow.

The

Oxyhalogen acids

Table 19.5

Name

Oxidation State of tfie

Halogen

Formula

Name

Acid

of

of

Acid

of Anion Derived from Acid

1

HOF

+

1

HOBr

HOCI

+ 5+ 7+

HOI

1

HCIO,

3

HCIO-;

HBrO"3

HIOS

HCIOi

HBrOi

fHIO,

1

y |JL/M d

1

u u o doiu

hypofialite ion

halous acid

halite ion

halic acid

halate ion

perhalic acid

perfialate ion

Ih,io, strong acids.

'

silver halide precipitates

may

be formed by the addition of a solution of

nitrate to a solution containing the appropriate halide ion. Silver iodide

silver is

in-

ammonia however. AgCl readily dissolves to form the AgiNHjjj complex ion and AgBr dissolves with difficulty. Silver fluoride is soluble. Precipitates of MgF2 or CaF2 are usually used to confirm the presence of the fluoride soluble in excess

;

ion in a solution. If the iodide ion in an aqueous solution is oxidized to iodine (the usual procedure employs chlorine), the 1 2 can be extracted by cyclohexane, which forms in cyclohexane is violet a two-liquid-layer system with water. The solution of colored. In the corresponding test for the bromide ion, the Br2-cyclohexane

solution

is

brown.

19.12 Oxyacids of the Halogens ..

The oxyacids

The only oxyacid of fluorine a thermally unstable com-

of the halogens arc listed in Table 19.5.

that has been prepared

is

HOF,

hypofluorous acid,

pound. The acids of chlorine and compounds.

their salts are the

most important of these

Lewis formulas for the oxychlorine acids are shown in Figure 19.3; removal of H"" from each of these structures gives the electronic formula of the corresponding anion. However, Lewis structures, in which each CI and O atom has a

show

valence-electron octet, do not

that the

CI— O

bonds

in these

compounds

can have a considerable amount of double-bond character due to pn-dn bonding (see Section 7.6).

H

— 0—

Cl:

H

— 0—

Ci:

H— 0—

© CI

:0:

—0:

H

..

..

..

"

1® — 0— CI— 0: ..

1

..

1

:0:

:0:

:0: hypochlorous

chlorous

chloric

perchloric

acid

acid

acid

acid

Figure 19.3

Lewis formulas

for the

oxychlorine acids

19.12

Oxyacids

of the

Halogens

507

Cl

0

0

o hypochlorite,

CIO'

chlorite,

CIO2'

0

chlorate,

CIO3

perchlorate,

CIO4'

Structures of the oxychlorine anions

Figure 19.4

The oxybromine and oxyiodine anions have configurations similar to the analogous oxychlorine anions shown in Figure 19.4. The HjIO^ molecule is octahedral with five OH groups and one O atom in the six positions surrounding the central I atom. The standard

electrode potentials for Cl,, Br,,

I2,

and the compounds of these

in Figure 19.5. The number shown over each arrow is the S\^^, in volts, for the reduction of the species on the left to that on the right. An oxidation potential for a transformation

elements in acidic and alkaline solution are summarized

from

right to left

may, of course, be obtained by changing the sign of (f ^ed-

Perchloric acid (HCIO4), perbromic acid (HBr04), and the halic acids

HBrOs, and HIO3)

(HCIO3,

are strong acids, but the remaining oxyacids are incompletely

dissociated in water solution and exist in solution largely in molecular form.

Hence, molecular formulas are shown

in

Figure 19.5 for the weak acids in acidic

solution. In general, acid strength increases with increasing

Much

of the chemistry of these compounds

is

oxygen content.

effectively correlated

by means

of these electrode potential diagrams. Remember, however, that standard electrode potentials refer to reductions that take place at 25 in their (f ^

standard

states.

C

emf tells nothing about some reactions for which

values. Furthermore, a cell

mation

to

which

it

with

all

substances present

Concentration changes and temperature changes

applies;

alter

the speed of the transforthe cell emfs are positive

occur so slowly that they are of no practical importance. According to S,^^ values,

OH

the oxidation of water (in acid) and of the ion (in base) should be readily accomplished by many oxyhalogen compounds. These reactions, however, occur

so slowly that

One of

it is

possible to observe

(^red

compounds in solution. compounds is their values are positive. The (fred

compounds

are stronger oxidizing agents

all

the oxyhalogen

the outstanding characteristics of the oxyhalogen

ability to function as oxidizing agents; all the

values also

show

in acidic solution

that, in general, these

than in alkaline solution.

Many of these oxyhalogen compounds are unstable toward disproportionation. Such materials are readily identified from the diagrams of Figure 19.5. For example,

in alkaline solution:

cio-^:^ci,^tl^ciSince

+

1.36

is

more

40H- +

Cl,

+ 20H- +

Cl,

2e"

Chapter 19

CI2

positive than +0.40, Cl, will disproportionate;

—

2C10" + 2H,0 + —2Cr —CIO" + Cr + H2O

The Nonmetals, Part

le-

I:

^;ed 6,„,f

Hydrogen and the Halogens

= -0.40 V = +1.36 V = +0.96 V

ACIDIC SOLUTION +

cor

CIO3-

1.47

il65^HOCI^^^, -±12^ CI"

HCIO.

4\

+

+ 1.49

1.43

+ > BrOa"

Br04~

1.52

+ 1.50

"""^'^^

+1.60 ^

,^0 > HOBr V.

>

,

^

+1.07^ >

Br2

Br" '

t

+1.33 I

+ H5IO6

^^'^^

>

1,20

T

+ 1.13 I03-

,

+1.45

HOI

>h

+0.54 ^

> ,r

+0.99

ALKALINE SOLUTION + 0.63 ^0-36

CO,

^

cib3-^^CI0a--^:^CI0--±°^CI.-±lgg^CI

1

+ 0.89

+ 0.50

+ 0.61 + 0 54

> BrOa-

BrOr

>BrO"

+0.45

^

>Br2

+1.07 ^ „

>

Br

+ 0.76

i

+ 0.49

t

+ 0.26

> I03-

HalOi-

Figure 19.5 ttieir

Standard electrode potential diagrams

for ctilorine.

bromine, iodine, and

compounds

The

HOI, HCIO,. and CIO3 should ^,ed values indicate that HOCl, HOBr. in acidic solution. In alkaline solution all species should dis-

disproportionate

proportionate except the halide ions.

1.

CIO4 BrO^ BrOj and H ,10g^.

Hypohalous acids and hypohalites.

.

.

,

The hypohalous

acids

(HOX)

are the weak-

exist in solution but cannot be prepared in pure form.

halogen o.xyacids. They Each of the three halogens is

est

slightly soluble in

water and reacts to produce a low

concentration of the corresponding hypohalous acid:

X2 + H,0

^

H

"(aq)

+ X

(aq)

+ HOX(aq)

All standard potentials for these reactions are negative, and the reactions proceed to a very limited extent. Thus, at 25 C saturated solutions of the halogens contain the following

HOI, 6 X

10

HOX "

concentrations:

HOCl,

3 x

10

'

M; HOBr,

x 10"^

1

M;

M.

19.12

Oxyacids

of ttie

Halogens

509

The X2-H2O

may be AgjO

reactions

increased, by the addition of

ion and

precipitate the

4 HgO(s)

HgO

to the reaction mixture.

HOX

These oxides

remove H"^(aq):

—*2AgX(s) + 2HOX(aq)

2X2 + AgjOls) + H2O

2X2 +

driven to the right, and the yields of or

+ H2O

—

HgX2

Each of the three halogens dissolves

3

•

HgO(s) +

2

HOX(aq)

in alkaline solution to

produce a hypohalite

ion and a halide ion:

X2 + 20H~(aq)

—*X~(aq) + XO-(aq) + H2O

The emf for each of these

reactions

is

positive,

and each reaction

is

rapid.

However, and

the hypohalite ions disproportionate in alkaline solution to the halate ions

halide ions. Fortunately the disproportionation reactions of

C10~ and BrO"

are

slow at temperatures around O'C, so that these ions can be prepared by

this

method. The hypoiodite

low

ion,

however, disproportionates rapidly even

at

temperatures. Solutions of sodium hypochlorite, used commercially in cotton bleaching (for

example, Clorox), are prepared by electrolyzing cold sodium chloride solutions such as the ones used

preparation of chlorine (see Section

in the

19.7). In this

process, however, the products of the electrolysis are not kept separate. Rather, the electrolyte

vigorously mixed so that the chlorine produced at the anode

is

produced

reacts with the hydroxide ion

anode: 2^" CI2

cathode:

—CU + + 2H2O —*20H- + H, + 20H- —»ocr + cr + H2O CP

2

cathode:

at the

2e~

The

overall equation for the entire process

CI-

+ H.o

''"'''^''y'"

ocr

,

is

+ h.

cold

The hypohalous

acids and hypohalites are good oxidizing agents, particularly The compounds decompose not only by disproportionation liberation of from the solutions. The reactions in which

in acidic solution.

but also by the is

2.

liberated are slow, however,

Halous acids and

This it

decomposes

The only known halous acid

halites.

compound cannot be rapidly.

and are catalyzed by metal

isolated in pure form,

Chlorous acid

is

weak

chlorous acid. The disproportionation of

HCIO2. The

chlorous acid (HCIO,).

and even

acid, but

HOCl

is

salts.

it is

in

aqueous solution

stronger than hypo-

in acidic solution

electrode potentials indicate, however, that

it

does not yield

should be possible

to make the chlorite ion by the disproportionation of CIO" in alkaline solution. The disproportionation of ClO^ into CIO3 and CI", however, is more favorable, and CIO J cannot be made from CIO".

Chlorites are comparatively stable in alkaline solution and may be prepared from CIO2 gas. Chlorine dioxide, a very reactive odd-electron molecule, is prepared by the reduction of chlorates in aqueous solution using sulfur dioxide gas as the

reducing agent:

2C103"(aq)

510

Chapter 19

+

S02(g)

—

The Nonmetals, Part

CI02(g)

I:

+ SOr(aq)

Hydrogen and the Halogens

The

chlorite ion is prepared by the reaction of CIO2 gas with an alkahne solution 2' in alkaline solution) of sodium peroxide (which forms the hydroperoxide ion

HO

2C102(g)

+ HOjlaq) + OH-(aq)

—

Solid chlorites are dangerous chemicals.

+

2C102"(aq)

02(g)

+ H2O

They detonate when heated and are

explosive in contact with combustible material. 3.

We

and halates.

Halic acids

have mentioned the disproportionation reactions

of the hypohalite ions:

3XO" Thus

if

— XO3 + 2X-

a free halogen

is

added

to a hot, concentrated solution of alkali, the

corresponding halide and halate ions are produced rather than the halide and hypohalite ions:

5X-(aq) + X03(aq) + BHjO

3X2 + 60H-(aq) The chlorates

are commercially prepared by the electrolysis of hot, concentrated

solutions of chlorides (instead of the cold solutions used for the electrochemical

preparation of the hypochlorites). The electrolyte the chlorine produced at the

anode

stirred vigorously so that

is

reacts with the hydroxide ion that

is

a product

of the reduction at the cathode:

2e~

cathode:

3CI2 If

— CU + + IH^O — 20H + H2 + 60H —SCI + 2CI

anode:

le'

"

ClO; + 3H2O

the three foregoing equations are

been multiplied through by

CI-

3,

added

after the first

the equation

two equations have each

for the overall process

is

obtained:

+3H2 + 3H20-^^q^feci03 ^ hot

The chlorate

from the concentrated solution employed as the elecAlthough chlorates are generally water soluble, they are much

crystallizes

trolyte of the cell.

than the corresponding chlorides. Solutions of a halic acid can be prepared by adding sulfuric acid to a solution

less soluble

of the barium

Ba^*(aq)

salt

of the acid

+ 2X03(aq) + 2H"(aq) + SO^Iaq)

—

BaS04(s)

+ 2H^(aq) + 2X03"(aq)

be isolated from aqueous solutions because of decomposition. Iodic acid. HIO3. however, can be obtained as a white solid; this acid is generally prepared by oxidizing iodine with concentrated nitric acid.

Pure

HBrOi

or

HCIO3 cannot

All halic acids are strong.

dizing agents.

The

The

halates, as well as the halic acids, are strong oxi-

reactions of chlorates with easily oxidized materials

may

be

explosive.

Chlorates decompose upon heating in a variety of ways. At high temperatures, and particularly in the presence of a catalyst, chlorates decompose into chlorides and oxygen

19.12

Oxyacids

of the

Halogens

511

heal

2 KC103(s)

+

* 2KCl(s)

MnO,

302(g)

At more moderate temperatures, in and chlorides:

the absence of a catalyst, the decomposition

yields perchlorates

Iwai

4 KC103(s)

4.

3

KC104(s)

+

KCl(s)

made by

Perchlorate salts are

Perhalic acids and perhalates.

the controlled

thermal decomposition of a chlorate or by the electrolysis of a cold solution of a chlorate.

The

free acid, a clear

mixture of a perchlorate talline

salt

hygroscopic liquid,

hydrates of perchloric acid are solid

Perchloric acid

may

isomorphous

is

is

a strong acid

react violently

be prepared by

distilling a

A number of crysknown. The compound HCIO4 HjO is of

interest; the lattice positions of the crystal are

and the

may

with concentrated sulfuric acid.

(of the

and

same

occupied by HjO"^ and CIO4 ions,

crystalline structure) with

NH4CIO4. HCIO4

a strong oxidizing agent. Concentrated

when heated with organic substances;

the reactions are

frequently explosive.

Perbromates are prepared by the oxidation of bromates

in alkaline solution

using Ft as the oxidizing agent. Several periodic acids have been prepared; the

most

common

one

is

HjIO^,. Periodates are

means of CU in alkaline lOg", and IO5" have been obtained. usually by

Uses

19.13 Industrial

of the

The most important

made by

the oxidation of iodates,

solution. Salts of the anions

IO4

,

I2O9",

Halogens

industrial uses of the halogens

and halogen compounds

are

The most important fluorides produced commercially are synthetic and the fluorocarbons. Cryolite, NajAlFg, is used as a supporting electrolyte in the electrolysis of molten AUO, for the production of aluminum 1.

Fluorine.

cryolite

The fluorocarbons

(Hall process).

Freons

(for

example,

CCUFj)

are noted for their chemical inertness.

Polytetrafluoroethylene (polymerized

polymer that

The

are used as refrigerants and aerosol propellants.

F2C=CF2.

Teflon)

is

a solid fluorocarbon

highly resistant to chemical attack. Liquid fluorocarbons are

is

used as chemically resistant lubricants.

The

principal use of elemental fluorine

is

in the

separation of "^^'U (the isotope

of uranium that undergoes atomic fission) from natural uranium (which ^^^U).

Uranium metal (which contains both

isotopes)

is

is

mostly

converted into

UF^

(which sublimes at 56 C). Since ^-^'UF^ has a lower molecular weight than -^^UFg, the --^-^UFg vapor effuses through a porous barrier more rapidly than the ^^^UFg does and a separation of the two isotopes is effected by this means. Minor but well publicized uses of fluorides (principally NaF), are their addition to water supplies 2.

Chlorine.

A

and toothpaste

to prevent tooth decay.

number of chlorine-containing compounds are produced compounds are organic compounds that are made or hydrogen chloride. They are used, for example, as plastics.

large

commercially. Most of these

by use of chlorine

512

Chapter 19

The Nonmetals,

Part

I:

Hydrogen and the Halogens

solvents, pesticides, herbicides, pharmaceuticals, refrigerants,

quantities of

HCl

and

dyes. Large

are produced to be used not only in the synthesis of organic

petroleum technology, metallurgy, metal cleaning (to metals), food processing, and in the manufacture of inorganic chlorides. Chlorine is used in the manufticture of paper, rayon, products,

but also

in

remove metal oxides from

hydrogen chloride, bromine, iodine, sodium hypochlorite, and metal chlorides, in the disinfection of water, and in the bleaching of textiles. 3.

The

Bromine.

principal use of bromine, at present,

(CH^BrCHjBr) which

of ethylene dibromide tetraethyl

is

swept out of the engine

is

is

manufacture

in the

used with the anti-knock agent

bromine to combustion temperatures the exhaust. The importance of this use of bromine

lead in leaded gasolines. Ethylene dibromide supplies

convert the lead into PbBr2 which

and

is

in

is

volatile at cylinder

declining since antipollution laws forbid the use of leaded gasoline in

Bromine-containing organic compounds are used as intermediates syntheses, dyes, pharmaceuticals, fumigants,

bromides are used 4.

cars.

and fire-proofing agents. Inorganic

medicine, in bleaching, and in photography (AgBr).

in

Iodine and

Iodine.

new

in industrial

compounds

its

are not used as extensively as the other

halogens and halides. Significant uses include the production of pharmaceuticals, dyes,

and

silver iodide (for

photography).

Summary The

topics that have been discussed in this chapter are

physical properties of hydrogen.

1.

The occurrence and

2.

The industrial production of hydrogen.

6.

7.

which hydrogen

3.

Displacement reactions

4.

The

5.

Commercial uses of hydrogen.

in

The halogens:

dustrial

Compounds

is

liberated.

properties,

occurrence,

in-

of the halogens: hydrogen halides, metal and their salts, electrode-potential

oxyacids

halides,

reactions of hydrogen; the hydrides.

physical

and laboratory preparations.

diagrams. 8.

Industrial uses of the halogens

and

their

compounds.

Key Terms Some

of the more important terms introduced in this chapter are listed below. Definitions for terms not included in this list may be located in the text by use of the index.

reaction

(Section

19.3)

A

reaction

in

which one element (or group of elements) displaces another element (or group of elements) from a compound

I,

19.6)

A

group VII

.A

element;

F.

or At.

compound (Section 19.9) composed of two different halogens.

Interhalogen is

Cracking (Section 19.2) A process in which high molecular weight hydrocarbons are broken down into lower molecular weight compounds: used in petroleum refining. Displacement

Halogen (Section CI, Br,

A compound

that

Steam reformer process (Section 19.2) An industrial process used to prepare hydrogen by the reaction of steam with a hydrocarbon. \V ater

gas (Section 19.2)

and hydrogen produced steam and coke.

Key Terms

A mixture of carbon

monoxide

industrially by the reaction of

513

Problems* The Halogens

Hydrogen

What

19.1

ways

are the principal

chemical equations to show how Cl2(g) CI (aq) by use of (a) Mn02(s), (b) PbOjIs), (c) Mn04(aq), (d) Cr202"(aq). (e) Can analogous reactions be used to prepare fluorine, bromine, or iodine? Justify your answer.

19.13 Write

which hydrogen which

in

may be prepared from

nature? Discuss three methods by hydrogen is prepared from water industrially. occurs

in

Write chemical equations for the preparation of hydrogen from (a) Na(s) and HjO, (b) Fe(s) and steam, + (c) Zn(s) and H (aq), (d) Zn(s) and OH"(aq), (e) C(s) and steam, (f) CH4(g) and steam, (g) CaHjls) and HjO. 19.2

19.14 Write chemical equations to

iS3 Write chemical equations for the reactions of hydrogen with (a) Na(s), (b) Ca(s), (c) Cl,(g). (d) N,(g), (e) Cu,0(s), (f) CO(g), (g) WO,{s).

Describe the properties and structures of

19.4

hydrides,

(b)

interstitial

hydrides,

(g)

19.16

with

19.5 How do the physical properties of hydrogen reflect the nature of the London forces between Hj molecules?

What accounts for the catalytic activity of Pd, and Ni in many reactions that involve hydrogen? 19.6

What mass

19.7

(a) 10.0

g of Ca(s),

is

(b) 10.0

required to liberate

1

(b)

Zn, and

(c)

or SrCU.

Table

19.21

NH,(g).

(d)

Compare your

electrolyte

Which (b)

is

larger:

stirred,

the acid strength of

(a)

the melting point of F2 or of

(d)

CK,

(c)

hot,

stirred,

HF

or of

the boiling

19.23 Write chemical equations for the reactions used to identify each of the halide ions in

enthalpy of sublimation of Na(s) the first ionization energy of Na(g) kJ/mol. The bond energy of Hjig) is +435

19.24

How many

aqueous solution.

faradays of electricity are needed to

produce 1.000 kg of CKfg) by the molten NaCl?

kJ/mol, and the first electron affinity of H(g) is -73 kJ/mol. Compare the value you get with the lattice energy of NaCl(s), which is -789 kJ/mol.

What mass

the

HF or of HCl, (d) the first ionization energy of F2 or of I,, (e) the bond energy of F, or of CI,, (f) the bond energy of CI, or of I,?

— 57.3 kJ/mol. The is + 108 kJ/mol, and

19.12 (a)

with

solution

point of

values with

19.11 Use the following data to calculate the lattice energy of NaH(s): The enthalpy of formation of NaH(s) is

+496

Write chemical equations for the electrolysis of molten NaCl, (b) cold NaCl solution, (c) cold

NaCl

those Hsted in Table 3.1.

is

geometry of the interhalogen and XX',.

concentrated NaCl solution with the electrolyte (e) cold NaClOj solution.

Sn

19.22

(c)

XX 5,

(a) dry,

HCl,

HiOlg).

BeF, or BeClj.

compounds XX3,

3.2 to cal-

(b)

(c)

19.20 Discuss the molecular

culate the standard enthalpy of formation of (a) HF(g),

in

compounds, would have the higher electrical

the following pairs of that

conductivity in the molten state. Explain the basis for your prediction in each case, (a) FeCl, or FeCla, (b) RbCl

.000 g of hydrogen from excess acid.

Use the bond energies found

From each of compound

select the

and calculate the mass of 22.4 liters of air at STP. (c) If a 22.4-liter balloon were filled with hydrogen at STP, the difference between the mass of 22.4 liters of air and the mass of 22.4 liters of hydrogen would be the approximate value of the liftmg power of the balloon. Calculate this value. Approximately how many times its own mass can a sample of hydrogen lift? (a) Al,

from NaCl and NaBr and H2SO4

g

What is the mass of 22.4 liters of H^tg) at STP? Assume that air is 22.0% Ojlg) and 78.0°o N,(g)

mass of

show why concentrated HF(aq)

HCl(g) can be prepared is it that the reaction of cannot be used to prepare HBr(g)? 19.19

the

HF

strongly ionic than dilute HF(aq).

H2SO4, why

19.8 (a)

19.10

more

19.18 Since

ofCaHjis)?

19.9 Calculate

Write chemical equations for the reactions of SiOj. (b) Na2C03, (c) KF, (d) CaO.

(a)

19.17 Write equations to

Pt,

of hydrogen can be obtained by the

reaction of excess water with

(b)

Write chemical equations for the reactions of CI2 (b) Zn, (a) H2, (c) P, (e) HjS, (d) S. (f) CO, SO2, (h) r(aq), (i) cold H^O.

19.15

complex hydrides,

(c)

to prepare

(a)

with (a) saltlike

covalent hydrides.

(d)

show how

from CaF,, (b) CI 2 from NaCl, (c) Br, from sea water, (d) 1 2 from NaI03, (e) HBr from PBrj. the following;

electrolysis of dry,

How many

grams of Cl2(g) are produced in the takes to prepare .000 kg of NaOH by the electrolysis of an aqueous solution of NaCl? 19.25

same time

of hydrogen would theoretically

that

it

1

be required to reduce 1.00 kg of WOjfs) to yield W(s)? (b)

What volume would

this

mass of

H ,(g) occupy at STP?

19.26

How many

hour by the if

The appendix contains answers

514

to color-keyed

1000

A

of electricity are supplied?

problems

Chapter 19

The Nonmetals, Part

I:

Cl,(g) are produced in one an aqueous solution of NaCl

grams of

electrolysis of

Hydrogen and the Halogens

THE NONMETALS, PART II: THE GROUP VI A ELEMENTS

CHAPTER

20 A includes oxygen, sulfur, selenium, tellurium, and polonium. Oxygen most important and abundant of the group. Since the chemistry of oxygen is different from that of the other members, oxygen is considered separately before the others. Polonium is the product of the radioactive disintegration of radium. The most abundant isotope of polonium. -'"Po, has a half-life of only Group VI

is

the

138.7 days.

20.1

Properties of the Group VI

The

A Elements

group VI A elements are listed in Table 20. two electrons short of a noble-gas structure. Hence, these elements

electronic configurations of the

Each element

is

1

attain a noble-gas electronic configuration in the formation of ionic

compounds

by accepting two electrons per atom

2Na^ The elements

acquire

also

rsV-

noble-gas configurations

through covalent-bond

formation

H:Se:

H Certain properties of the group VI A elements are summarized in Table 20.2. Each member of the group is a less active nonmetal than the halogen of its period. The electronegatives of the elements decrease, in the expected manner, with increasing atomic number. Oxygen is the second most electronegative element

^Table

20.1

Elements

Electronic configurations of the group VI

Z

i

6

A elemenR

3s

3p

3d

4s

4p

4

4d

4f

5s

5p

5d

6s

6p

10

2

4

1

0

8

2

2

4

S

16

2

2

6

2

4

Se

34

2

2

6

2

6

10

2

Te

52

2

2

6

2

6

10

2

6

10

Po

84

2

2

6

2

6

10

2

6

10

'

14

515

2

4

2

6

Some

Table 20.2

properties of the group Vi

A elements .

Oxygen

Property

Sulfur

Selenium

....

.J

Tellurium 1

silver-white

color

colorless

yellow

red to

molecular formula

O2

Sg rings

Seg rings Se„ cfiains

melting point (°C)

-218.4

119

217

452

boiling point (°C)

-182.9

444.6

688

1390

black

74

104

117

137

140

184

198

221

1312

1004

946

870

atomic radius (pm)

(2-

ionic radius first

(pm)

ion)

Te„ chains

ionization energy

(kJ/mol) electronegativity

bond energy

2.1

(single

bonds) (kJ/mol) S'° for

2.6

2.6

3.4

138

213

184

138

+ 1.23

+ 0.14

-0.40

-0.72

reduction of

element

H,X

to

in

acid

solution (V)

(fluorine

first);

is

sulfur

is

about as electronegative as iodine. Thus, the oxides

of most metals are ionic, whereas the sulfides, selenides, and tellurides of only the

most

A and II A metals) are truly ionic compounds. A elements are predominantly nonmetallic in chemical behavior;

active metals (such as the

The group VI

I

however, metallic characteristics appear

The trend

in increasing metallic

in the heavier

members of

the group.

character parallels, as expected, increasing atomic

number, increasing atomic radius, and decreasing ionization potential. Polonium the most metallic member of the group; it appears to be capable of form.ing cations that exist in aqueous solution, and the 2— state of polonium (in H2P0, for example) is unstable. Whereas tellurium is essentially nonmetallic in character, unstable salts of tellurium with anions of strong acids have been reported. The ordinary form of tellurium is metallic. Selenium exists in both metallic and nonis

metallic crystalline modifications. Sulfur, selenium, and tellurium exist in positive oxidation states in compounds which they are combined with more electronegative elements (such as oxygen and the halogens). Oxygen is considered to have a positive oxidation number only in the few compounds that it forms with fluorine. For sulfur, selenium, and tellurium, the oxidation states of 4+ and 6+- are particularly important. The electrode potentials listed in Table 20.2 give an idea of the strength of the group VI A elements as oxidizing agents. Oxygen is a strong oxidizing agent, but there is a striking decrease in this property from oxygen to tellurium. In fact, HjTe and H ,Se are better reducing agents than hydrogen. Compare the 6 values listed in Table 20.2 with those given for the halogens in Table 19.2. in

20.2

Occurrence and of

Industrial Production

Oxygen

is the most abundant element (see Table 1.2). Free oxygen makes up about 21.0% by volume or 23. 2°,, by mass of the atmosphere. Most minerals

Oxygen

516

Chapter 20

The Nonmetals, Part

II:

The Group

VI

A Elements

Composition

Table 20.3

of dry air

Percent by Volunne

Substance N2

78.00

O2

20.95

Ar

0.93

CO, Ne

He

Percent by Volume

Substance

CH4

2 X 10

*

X 10'*

Kr

1

N,0

5 X 10"'

0.03

H2

5 X 10"*

0.0018

Xe

8 X 10"*^

0.0005

O3

1

X 10"*

contain combined oxygen. Silica, Si02, is a common ingredient of many minerals and the chief constituent of sand. Sihcon is second to oxygen in the order of natural abundance because of the widespread occurrence of silica. Other oxygencontaining minerals are oxides, sulfates, and carbonates. Oxygen is a constituent of the compounds that make up plant and animal matter. The human body is

more than 60% oxygen. Three isotopes of oxygen occur in nature: '"O (99.759",,), '^O (0.204"„). and »^0 (0.037%). The isotopes "*0, ''O, ''*0, and -°0 are artificial and unstable. The principal commercial source of oxygen is the atmosphere. Air is a mixture. The composition of air varies with altitude and to a lesser extent with location. The analysis of air is made after water and solid particles (such as dust and spores) have been removed. Some of the components of air are listed in Table 20.3. The percentages given (by volume) are for clean, dry air at sea level. Over 99",, of the oxygen produced industrially is obtained from the liquefaction and fractional distillation of air. In the process, filtered, dry air from which the CO2 has been removed is liquefied by compression and cooling. When the air is allowed to warm, nitrogen (boiling point, - 196 C) boils away from the oxygen (boiling point, -183 C). The noble gases are obtained from the nitrogen and oxygen fractions by repeated

A

amount

small

distillations

and other separation techniques. is produced

of very pure but relatively expensive oxygen

commercially by the electrolysis of water:

2H,0^i^^^^fe2H,(g) +

20.3 Laboratory Preparation of

Oxygen

is

0,(g)

Oxygen

usually prepared in the laboratory by the thermal decomposition of

certain oxygen-containing

compounds. The following are used:

The oxides of silver (Ag^O), mercury (HgO), to give oxygen gas and the free metal: heating on and gold (Au,0,) decompose

1.

Oxides

of

2HgO(s) 2.

metals

is

low

reactivity.

—2Hg(l| +

Peroxides.

Ot",

of

0,(g)

Oxygen and -1

2

the oxide ion are produced

when

the peroxide ion,

heated:

:0:0:

2 :0:

2

-

+ O,

20.3

Laboratory Preparation

of

Oxygen

517

Thus,

2Na202(s) 2Ba02(s)

—2Na20(s) +

—

O.Jg)

2BaO(s) + O^Cg)

Oxygen can be obtained from the temperature. The other product

is

40H "(aq) +

2H,0

205-(aq) +

sodium peroxide and water at room an aqueous solution of sodium hydroxide:

reaction of

©.(g)

Certain other compounds release all or part of 3. Nitrates and chlorates. oxygen upon heating. Nitrates of the I A metals form nitrites: 2

—

NaN03(l)

2

NaNO.lD +

Potassium chlorate loses

all its

their

O.Jg)

oxygen; a catalyst (Mn02) this decomposition:

generally used to

is

lower the temperature required for

2KC103(s)

—2KCl(s) +

20.4 Reactions of

The

302(g)

Oxygen

reactions of oxygen are often

more

sluggish than

would be predicted from

the fact that oxygen has a high electronegativity (3.4), second in this property

only to fluorine

oxygen

is

(4.0).

The reason

for this slowness

is

that the

bond energy of

high (494 kJ/mol); therefore, reactions that require the oxygen-to-oxygen

broken occur only at high temperatures. Many of these reactions are exothermic and produce sufficient heat to sustain themselves after having once been initiated by external heating. Whether self-sustaining or not, most oxygen reactions occur at temperatures considerably higher than room

bond

to be

relatively highly

temperature.

Oxygen forms

four different anions: the superoxide, peroxide, oxide,

and

ozonide ions. Molecular orbital diagrams for oxygen, the superoxide ion, and the peroxide ion are given in Figure 20. 1 The superoxide ion. 02" can be considered ,

.

from the addition of one electron to the n*2p orbital of the O2 molecule, which reduces the number of unpaired electrons to and the bond order to I5. The iixisic ioii, Oj", contains two more electrons (in the n*2p orbitals) than the O2 molecule hence, the bond order is reduced to 1 and the ion is diamagnetic. to arise

1

'i

,

;

The oxide

ion,

O'",

is

isoelectronic with

neon and

is

diamagnetic. The ozonide

paramagnetic with one unpaired electron and is produced by reactions of ozone, O ,, with hydroxides of K, Rb, and Cs (see Section 20.6). All metals except the less-reactive metals (for example, Ag and Au) react with oxygen. Oxides of all metals are known but some must be made indirectly. The most reactive metals of group I A (and those with largest atomic radii) Cs, Rb, and K react with oxygen at atmospheric pressure to produce superoxides. For example, ion.Oj,

is

—

—

Cs(s)

518

+

Chapter 20

02(g)

—

Cs02(s)

The Nonmetals, Part

II

;

The Group

VI

A Elements

o

o

o

a*2p

a*2p

a*2p

n

[

1

n'2p

nj

1

TT*2p

m

n

in n2p

1 I

1 [

TT*2p

n

n

Tr2p

'ir2p

®

T a2p

ct2p

® ®

®

CT*2S

a*2s

ct2s

a2s

O2 oxygen bond order, 2

Oi

or

superoxide ion

peroxide ion

ff2p

(t'2s

® bond

unpaired electrons, 2

(j2s

Molecular orbital energy-level diagrams

Figure 20.1

bond

order,

unpaired electrons,

for

1

order,

1

unpaired electrons, 0

oxygen, the superoxide

Ion,

and

the peroxide ion

Sodium peroxide

+

2Na(s)

produced by the reaction of sodium with oxygen

is

02(g)

—

Na^Ojis)

Lithium metal forms an ordinary oxide with O2 rather than a peroxide or a superoxide because the small Li^ ion cannot form a stable lattice with the larger

O2

~

or

O2

4Li(s)

ions:

+

02(g)

—

2Li20(s)

Generally, oxides form at

much

higher temperatures than either peroxides or

The ordinary oxides of Na, K, Rb, and Cs can be obtained by

superoxides.

heating oxygen with an excess of the metal.

With the exception of barium (which

reacts with

oxygen

to yield

barium

peroxide), the remaining metals generally produce normal oxides in their reactions

with oxygen

2Mg(s) 4Al(s)

+

+

02(g)

302(g)

—2MgO(s) —

2Al203(s)

20.4

Reactions

of

Oxygen

519

Analogous reactions can be written for the preparation of CaO, CuO, ZnO, PbO, and other oxides. The reaction of mercury and oxygen is reversible: 2Hg(l)

+ 02(g)^2HgO(s)

For metals that have more than one electrovalence number, the oxide produced generally depends upon the quantity of oxygen, the quantity of the metal, and the reaction conditions. Thus, the reaction of iron and oxygen can be made to yield FeO (low pressure of oxygen, temperature above 600 C), Fe304 (finely divided iron, heated in air at 500 'C), or FcjOj (iron heated in air at temperatures above 500 C). Hydrated Fe203 is iron rust. Except for the noble gases and the group VII A elements, all nonmetals react with oxygen. Oxides of the halogens and those of the heavier members of the noble-gas family have been prepared by indirect means. The reaction of oxygen with hydrogen produces water. The product of the reaction of carbon with oxygen depends upon the proportion of carbon to oxygen employed:

+ 02(g)—*2CO(g)

2C(s)

+

C(s)

02(g)

C02(g)

In like manner, the product of the reaction of

upon whether phosphorus

phosphorus and oxygen depends

reacted in a limited oxygen supply (P4O6) or in

is

excess oxygen (P4O10). Sulfur reacts to produce SO,:

S(s)

+

02(g)

—

S02(g)

The reaction of nitrogen with oxygen requires extremely high temperatures. The following reaction occurs in a high-energy electric arc:

+

N2(g)

02(g)

—

2

NO(g)

Additional oxides of sulfur (for example, SO3) and nitrogen (for example,

NO2

and N2O5) are prepared by means other than the direct combination of the Lower oxides can be reacted with oxygen to produce higher oxides.

elements.

For example,

2Cu20(s) + 02(g)

—*4CuO(s)

2CO(g) + 02(g)

2C02(g)

Most

reactions of

be obtained directly.

if

—

compounds with oxygen yield the same products that would make up the compounds were reacted

the individual elements that

Thus,

2H2S(g) + 302(g) CS2(1)

+ 302(g)

2C2H2(g) + 502(g) C2H,0(1) + 302(g)

The

H20(g) + 2S02(g)

2

C02(g)

+

2

SO 2(g)

4C02(g) + 2H20(g) C02(g)

2

+ 3H20(g)

reaction of zinc sulfide with oxygen illustrates a metallurgical process

as roasting.

520

— — — —

Chapter 20

Many

sulfide ores are subjected to this

The Nonmetals, Part

II

:

The Group

VI

A Elements

procedure

known

(see Section 23.5):

2ZnS(s)

+

2ZnO(s) + 2S02(g)

302(g)

The products of the reaction of a hydrocarbon with oxygen depend upon the amount of oxygen suppHed. Thus, when natural gas (methane, CH4) is burned in air,

H20(g),

C(s),

CO(g), and C02(g) are produced by the oxidation. hydrogen and oxygen form a compound called hydrogen

In addition to water,

H2O2, which is colorless liquid that boils at 150.2 C and -0.4r C. Hydrogen peroxide can be made by treating peroxides with

peroxide,

+ 2H^(aq) + SOr(aq)

Ba02(s)

barium

In this preparation the

—

sulfate,

H202(aq) which

+

BaS04(s)

removed by

insoluble, can be

is

freezes at

acids:

filtration.

The peroxide linkage 820^" or ion

is

H2O2 and

(

— O—O—

)

exists in covalent

O2 ion. For example, [OjSdoSOj]^", contains this linkage

addition to

the

ions in

The S^Og"

(see Section 20.12).

under suitable conditions; the ion with water serves as a commercial preparation of hydrogen

produced by the

reaction of this

compounds and

the peroxydisulfate ion, written

electrolysis of sulfuric acid

peroxide:

S20r(aq) + 2H2O

—

Hydrogen peroxide

is

* H202(aq)

+ 2HS04(aq) One

a weak, diprotic acid in water solution.

or two sodium

hydrogens can be neutralized by sodium hydroxide to produce hydroperoxide (NaHO,) or sodium peroxide (Na202). In the laboratory. H2O2 either

is

used as an oxidizing or reducing agent.

20.5 Industrial

Uses

of

Oxygen

Most of the commercial

uses of oxygen stem from

its

ability to

support combustion

oxygen or oxygen-enriched air and sustain life. In many and speed of reaction and intensity the increases air in place of atmospheric principal uses of oxygen are: The yields. improves and thereby lowers costs applications, the use of

1.

Production of

2.

Processing and fabrication of metals

3.

Production of oxygen-containing compounds such as sodium peroxide and

organic

20.6

steel

compounds

4.

Oxidizer for rocket fuels

5.

The oxyacetylene torch

6.

Biological treatment of waste water

7.

Life support systems in medicine, in air

and space

travel,

and

in

submarines

Ozone The is

The

more than one form m and the forms are called ailotropes.

existence of an element in

called allotropy,

the

same

physical state

A number

of elements

20.6

Ozone

fuel

used

to propel this

Delta space vehicle was oxidized by liquid oxygen,

nasa.

and phosphorus. Oxygen

exhibit allotropy, for example, carbon, sulfur, in a

triatomic form, ozone, in addition to the

common

exists

diatomic modification.

The ozone molecule is diamagnetic and has an angular structure. Both oxygento-oxygen bonds have the same length (128 pm), which is intermediate between the double-bond distance (110 pm) and the single-bond distance (148 pm). The molecule

may

be represented as a resonance hybrid: ..

Ozone

a pale blue gas with a characteristic

is

H times that of O2 melting point

.

—

is

boiling point of ozone

193'

slightly

C.

It is

atoms and the combination of an

0 +

odor; predictably,

The normal

Ozone is produced by passing a The reaction proceeds through the

—O

iO,

..®

©

0,— O3

11 2"

atom with

its

density

is

C, and the normal is

Oj.

discharge through oxygen gas.

dissociation of an

O

—

soluble in water than

silent electric

- +247

A//

more

is

O, molecule into oxygen O2 molecule:

a second

kJ

A//=-105kJ

released in the second step, in which a new bond is formed, is not compensate for the energy required by the first step, in which a bond broken. Hence, the overall reaction for the preparation of ozone is endothermic:

The energy

sufficient to is

|0,

—

Ozone

is

^H} = +142

O3

highly reactive;

it

is

kJ explosive at temperatures above 300

the presence of substances that catalyze

its

decomposition. Ozone

C

will react

many substances at temperatures that are not high enough to produce with O2 The higher reactivity of O3 in comparison to O2 is consistent

or

in

with

reaction

with the

higher energy content of O3.

20.7 Air Pollution Several oxides, found in air in variable amounts, are air pollutants.

technological civilization at

an ever-increasing

rate.

Modern

introducing foreign substances into the atmosphere

is

The

principal air pollutants in terms of quantities

present are the following:

1.

Carbon monoxide

is

produced by the incomplete combustion of

automobile's internal-combustion engine

The mass of CO produced by

this

source

is

is

fuels.

The

the principal source of this pollutant.

about equal to half the mass of gasoline

consumed.

Carbon monoxide is toxic because it combines with the hemoglobin of the blood and prevents the hemoglobin from carrying oxygen to the body tissues (see Section 24. In other ways, however, CO is not very reactive. Reaction with O2 of the air to form CO2 does occur, but very slowly. 1

2.

).

Oxides ofsulfur (SOj and SO3) result from the combustion of coal, metallurgical and petroleum combustion and refining. The major source is the

processes,

522

Chapter 20

The Nonmetals, Part

II

:

The Group

VI

A Elements

combustion of coal (which contains from 0.5%

The roasting of source of SO2 pollution: 2 PbS(s)

+

3

02(g)

—

+

2 PbO(s)

Sulfur dioxide from these sources

The SO3 forms acid.

to

3.0%

These substances

are,

is

of

also a significant

2 S02(g)

is

SO3 by O2 in the air. and SO2 forms sulfurous

slowly oxidized to

sulfuric acid with atmospheric water,

The oxides of

S) in the generation

sulfide ores (see Section 23.5)

electricity.

of course, extremely corrosive.

some ways

the most serious air pollutants. They and are a serious health threat. They damage plant life, corrode metals, and erode marble and limestone. Ancient monuments (such as the Parthenon in Athens) which have stood for centuries are crumbling sulfur are in

are toxic, cause respiratory ailments,

because of the pollution of modern civilization.

Oxides of nitrogen (NO and NO2) are produced from the N2 and O, of the high temperatures characteristic of some combustions. Significant amounts of NO are formed in the combustions carried out in automobile engines and in electric generating plants. Nitrogen dioxide is formed in the air by the 3.

air at the

oxidation of

NO.

Small amounts of

NO and NO2

ordinarily occur in air

of the nitrogen cycle. Nitrogen dioxide

The

is

and form

a

minor part

considerably more toxic than

NO.

gases, however, usually occur at relatively

low concentrations so that the direct effect of these pollutants is not serious. The significance of these oxides lies in the role that they play in the formation of other, more serious pollutants (see

Hydrocarbons, below).

Hydrocarbons are compounds that contain carbon and hydrogen. They are found in petroleum, natural gas, and coal. The compounds are released into the atmosphere by evaporation, petroleum refining, and incomplete combustion of fuels. The unburned hydrocarbons in automobile exhaust constitute a major source of this type of contamination. A few hydrocarbons are carcinogenic (cancer-producing). The principal danger associated with hydrocarbon pollution, however, lies in the pollutants that are produced from hydrocarbons in the air. Nitrogen dioxide decomposes 4.

in sunlight to give

N02(g) These

—

O

NO(g) + 0(g)

O atoms react

0(g)

+

Ozone

with

O2

to

produce ozone, O3:

03(g)

02(g)

is

atoms:

highly reactive and reacts with

oxygen-containing organic compounds.

some hydrocarbons

Since the process

is

initiated

to

produce

by sunlight,

the products are sometimes called

These substances are toxic and very irritating to the eyes, skin, and respiratory tract. They cause extensive crop damage and the deterioration of materials. They constitute what is called photochemical smog. 5.

Small particles suspended

pollution.

The

particles

from about 0.01 /S0

^SsOi-

+ 0.51

+ 0.08

t

ALKALINE SOLUTION -0.66

1

-0.59

Figure 20.5

Electrode potential diagrams for sulfur and

its

compounds

values given

in volts)

Like selenium, tellurium

used

is

in the

facture of glass, ceramics, alloys,

vulcanization of rubber and in the

manu-

and enamel pigments.

Summary The

topics that have been discussed in this chapter are

The

2.

Oxygen: occurrence,

Sulfur, selenium,

and

properties of the group VI elements.

1.

7.

and tellurium:

allotropes. occurrence,

industrial preparation.

Compounds of sulfur, selenium, and tellurium: hydrogen compounds, compounds with the group Vi A element in a 4+ oxidation state, compounds with the group VI A element in a 6+ oxidation state, and electrode potential diagrams of S.

8.

industrial

and laboratory prep-

arations. 3.

Reactions of oxygen: oxides, superoxides, peroxides,

and ozonides. 4.

industrial uses of oxygen.

9.

5.

Ozone, an ailotrope of oxygen.

6.

Air pollution.

Industrial uses of sulfur, selenium,

their

and tellurium and

compounds.

Key Terms Some of the more important terms introduced in this chapter are listed below Definitions for terms not included in this list may be located in the text by use of the index.

Frasch process (Section 20.9) A process in which molten sulfur is obtained from underground deposits.

.

Allotropes (Section 20.6)

element

in the

Two or more

same physical

forms of the same

Peroxy acid (Section 20.12) An acid that contains a O O somewhere in the molecule. peroxide group

manu-

Air pollutants Photoctiemical pollutants (Section 20.7) produced by a sequence of reactions that is initiated by

state.

Contact process (Section 20.12)

A

process for the

which SO, is catalytically oxidized to SO^. the SO3 vapor dissolved in H2SO4, and the resulting H2S2O7 diluted with water to give H2SO4. facture of sulfuric acid

in

(

— — —

)

sunlight.

Key Terms

537

,

Problems* *20.12The standard enthalpy of formation of ozone, O3, is -I- 142 kJ/mol. The bond dissociation energy of 02(g) is -1-494 kJ/mol. What is the average bond energy of the two bonds in ozone?

Oxygen forms in which oxygen occurs oxygen produced industrially?

List the

20.1

How 20.2

is

in

nature.

chemical equations for the preparation from (a) HgO(s), (b) Na,0,(s) and HjO,

Write

of oxygen

Selenium, and Tellurium

Sulfur,

NaNO.,(s).(d) KC103(s).(e) H,0.

(c)

20.13 Describe the changes in sulfur that occur as the

Write chemical equations for the reactions of oxygen with (a) K(s|. (b) Na(s). (c) Li(s), (d) Mg(s), (e) Hg(l), (f) Ba(s),(g) C(s), (h) S(s),(i) Cu,0(s). (j) P4(s). 20.3

temperature

is

increased.

20.14 Describe the Frasch process for

mining elementary

sulfur.

Write chemical equations for the complete comin oxygen of (a) C4H,o.(b) CsHjjS, (c) C3H8O, ZnS,(e) PbS.

20.4

bustion

20.15 Write chemical equations for the reactions of sulfur

(d)

with (f)

20.5 The products of the combustion of a hydrocarbon in oxygen depend upon the amount of oxygen supplied. Write chemical equations for the reactions of methane, CH4(g), with oxygen that yield (a) C(s), (b) CO(g),

(f)

the oxides of the

I

A elements with

Ci,H,,0,i.

(b)

(b)

of the elements),

the products of their reactions with water.

(e)

Draw

O,.

(b)

electrons

Cu.

(d)

H2SO4 Zn," (e)

with

ZnS,

make each of the following and H2S (not by direct union

(b)

H2S03,(c) Na2S203,(d) NaHS04,

H2S2O7.

20.18 Write equations for the reactions of S02(g) with (a)

O2

(e)

OH

,

O,". State the number of unpaired bond order of each.

(c)

and the

that contain the dioxygenyl ion,

20.19

O2

(aq),(f)

SOf

Draw Lewis

S4OS

(e)

.

H.O,

(Pt catalyst) (b) CKlg), (c)

(aq)

(d)

C103"(aq),

and H2O.

and describe the geo,(c) SOj ,(d) S2O5", H2S20,,(g) H2S2O8. structures for

metric structure of (a) Si

known. Draw molecular-orbital energy-level diagrams for the dioxygenyl ion, and the superoxide ion, 07. Compare the two ions as to (a) bond order, (b) number of

,(f)

,(b)

SO5

an equation for the reaction of water with (b) SO,.(c) SO,.(d) Al, 83,(6) Se03, Te03.(g) H2S03,(h) H2S2O7.

20.20 Write

unpaired electrons. 20.9

(c)

molecular-orbital energy-level diagrams for

are

(to

F2,

O2

Compounds

20.8

(e)

that start with elemental S (a)

(d)

(a)

(d) Fe,

Fe263.'

elements in regard to (a) their physical state, their melting point, (c) the nature of the bonding,

20.7

S05"(aq),

(c)

NaN03,

that could be used to

those of

A

the VI

S-"(aq),

(b)

HNO3.

20.17 Write a sequence of equations representing reactions

Compare

20.6

O,,

20.16 Write equations for the reactions of (a)

CO,(g).

(c)

(a)

Cl2.(g)

(a)

Because the oxygen of H,02 can be either oxidized O,) or reduced (to HiO), hydrogen peroxide can

(f)

CH3C(NH,)S,

20.21

function as a reducing agent or as an oxidizing agent.

Write chemical equations for the reactions of 02(g)

with(a) H2S,(b) H2Te,(c) PbS,{d)

Na^SOj.

Using the ion-electron method, write balanced chemical 20.22 Write equations for the reactions of

equations for the following reactions of HjOi (a) the oxidation of PbStoPbSO^in acid solution. (b) the oxidation of

Cr(OH ),

of

Mn04

Ag,0

to

to

to

Ag

CrOj

Mn-*

in

(a)

20.23 Write

alkaline solution, (c) the reduction

in acid solution, (d)

balanced chemical equations proportionation reaction of (a) S (to

the reduction of

in alkaline solution, (b)

in alkaline solution.

Draw the resonance forms of the ozone molecule, O3. What is the bond order? What is the shape of the

'20.24

molecule?

SO2

The standard enthalpy of formation of H,0(1)

-285.9 kJ/mol and of 03(g) is 142.3 kJ/mol. What is change when one mole of H^Od) is prepared from (a) H2(g) and 0,(g).(b) Hjig) and 63(g)? In general, how do enthalpy changes for the reactions of ozone compare with those of oxygen'.' -I-

The more

538

ditficult

problems are marked with

asterisks.

Chapter 20

According (in

dis-

to standard electrode potentials, both

acid solution) and

20.25 Explain

can

SOf"

(in alkaline solution)

Part

II

:

why OF4 cannot be prepared

be.

The appendix contains answers

The Nonmetals,

(to

the

and 8203") S and SO2) in acid

should disproportionate. These reactions, however, are slow. What products should be obtained in each of these disproporlionation reactions? Note that it is necessary to take all possibilities given in Figure 20.5 into account.

is

the enthalpy

*

820^"

for

solution.

20.10

20.11

HCl(aq) with

Na2S03.{b) Na^S.Cc) Na2S203.

The Group

to color-keyed problems.

VI

A Elements

but SF4

20.26

What

prefixes per-

is

the difference in

and peroxy- as applied

20.27 Describe

a

laboratory

meaning between the in the naming of acids?

test

for

(a)

(aq),

(b)SOr(aq).(c) SOr(aq),(d) Sp^Caq). 20.28 Chlorosulfonic acid,

HOSOjCl.

is

the product of

SO3 and HCl. The peroxysulfuric acids (H2SO5 and H2S20g) can be prepared by the reaction of one mole of hydrogen peroxide, H2O2, with either the reaction of

one or two moles of chlorosulfonic acid. Write chemical equations for the reactions. 20.29 Describe the geometric shapes of the

H,S.

SO,.{c) S03,(d) (g)SF4,(h) SF,.(i) H^TeO,,.

(a)

(b)

20.30 Explain (b)

H2SO4

is

why

(a)

HjTe

SO^ is

.le)

S,Oi

following: ,(f)

SOl

a stronger acid than

,

H,S,

a stronger acid than Hf,TeO(,.

Problems

539

CHAPTER

THE NONMETALS, PART III: THE GROUP V A ELEMENTS

Group V A

includes nitrogen, phosphorus, arsenic, antimony,

show

Collectively, these elements

by either the group VI

21.1

A

elements or the group VII

Properties of the Group V Within any

A

and bismuth.

a wider range of properties than

A

exhibited

is

elements.

A Elements

family of the periodic classification, metallic character increases

(and nonmetallic character decreases) with increasing atomic number, atomic weight, and atomic size. This trend first

is

particularly striking in

group

V

A. The

ionization energies of the elements in the group, which are listed in Table 21.1,

decrease from values typical of a nonmetal (N) to those characteristic of a metal

Nitrogen and phosphorus are generally regarded as nonmetals, arsenic and antimony as semimetals or metalloids, and bismuth as a metal. The electronic configurations of the elements are listed in Table 21.2. Each element has three electrons less than the noble gas of its period, and the formation of trinegative ions might be expected. Nitrogen forms the nitride ion, N"*", in combination with certain reactive metals, and phosphorus forms the phosphide ion, P'^". less readily. The remaining elements of the group (As. Sb. and Bi), however, are more metallic than N and P and have no tendency to form comparable

(Bi).

anions.

The teristic

loss

of electrons and consequent formation of cations, which

of metals,

is

is

charac-

observed for the heavier members of the group. High ionization

energies prohibit the loss of all five valence electrons by any element. Consequently, 5

+

ions

do not

exist,

and the

5

+

oxidation state

is

attained only through covalent

bonding. In addition, most of the compounds in which the group

V A

elements

appear in the 3 + oxidation state are covalent. Antimony and bismuth, however, can form (^/"'.s' ions. Sb'^^ and Bi^^, through loss of the p electrons of their valence levels.

The 3+

The compounds Sb2(S04)3. BiFj. and Bi(C104)3-5H20

are ionic.

ions of antimony and bismuth react with water to form antimonyl and

bismuthyl ions (SbO^ and BiO"^). as well as hydra ted forms of these ions (for

example.

BKOH)! ):

Bi-'"(aq)

+

H,0^ BiO^(aq)

+ 2H^(aq)

Nitrogen, phosphorus, and arsenic do not form simple cations.

The oxides of

the

group

V A

elements become

the metallic character of the element increases.

540

less acidic

Thus N2O3.

and more basic as ^jO^,. and AS4O6

Table 21.1

Some

properties of the group V

Property

A elements

Nitrogen

color

Phosphorus

colorless

molecular formula

-210

melting point (°C)

gray metallic, yellow

gray metallic, yel low

gray metallic

black

Pj (white)

As„ (metallic)

Sb„ (metallic)

Bi„

P„ (black)

AS4 (yellow)

Sb4 (yellow)

44.1 (white)

814 (36 atm)

630.5 (metallic)

white, red,

N,

Bismuth

Antimony

Arsenic

271

(metallic)

-

boiling point (°C)

195.8

280

633 (sublimes)

74

110

121

atomic radius (pm) ionic radius (pm) first

140 (N'")

1325

1560 152

141

92

185 (P^")

108 (Bi^

(Sb-*"^)

+ )

ionization

energy (kJ/mol)

1399

electronegativity

FTable 21.2

2.2

Electronic configurations of

ELEMENT

965

1061

3.0

tlie

group V

A

2.0

2.1

elements

Is

N

772

830

2,2

3d

4s

4p

3

4d

2

3

P

15

2

2

6

2

3

As

33

2

2

6

2

6

10

2

Sb

51

2

2

6

2

6

10

2

6

10

Bi

83

2

2

6

2

6

10

2

6

10

5p

14

2

3

2

6

5d

6s

6p

10

2

3

are acidic oxides they dissolve in water to form acids, and they dissolve in solutions ;

of alkalies to form salts of these acids. The compound Sb^Of, is amphoteric; it will dissolve in hydrochloric acid as well as in sodium hydroxide. The comparable oxide of bismuth is strictly basic; BijO, is not soluble in alkalies, but the compound will dissolve in acids to

produce bismuth

salts.

which the elements exhibit a 5+ oxidation state are acidic, but the acidity declines markedly from N,05 to Bi205. In addition, the stability of the 5+ oxidation state decreases with increasing atomic number; B]20f is All the oxides in

extremely unstable and has never been prepared

in a

pure

state.

of the properties of nitrogen are anomalous in comparison to those of the other V A elements. This departure is characteristic of the first members of the groups of the periodic classification. Free nitrogen is surprisingly unreactive,

Many

partly because of the great strength of the

bonding

in the

N2

molecule:

:N=N: According to the molecular orbital theory, two n bonds and one rr bond join the atoms of a N, molecule, and the bond order is 3. The energy required to dissociate molecular N2 into atoms is very high (941 kJ mol). Since nitrogen has no d orbitals in its valence level {n = 2), the maximum number of covalent bonds formed by nitrogen is four (for example, in NH4). In

21.1

Properties of the Group V

A Elements

541

the valence levels of the other

may be many as

utilized in covalent six

V A

empty d orbitals which and Bi form as PCI5, PClg, AsFj, SbCl^, and

elements, there are

bond formation. Hence,

covalent bonds in such species as

P, As, Sb,

BiClf"-

For the group as a whole, the

common. The importance and

3

—

,

stability

3

+ and 5+ oxidation states are most 5 + and 3 — states decline from the ,

of the

lighter to the heavier elements. Nitrogen,

however, appears

in

every oxidation

from 3 — to 5 + Nitrogen also has a tendency toward the formation of multiple bonds (for example, in the cyanide ion, C=N~). The other V A elements do not form n bonds with p orbitals, but some multiple bond character can arise in the compounds of these elements (particularly those of P) from pn-dn bonding. Phosphorus, arsenic, and antimony occur in allotropic modifications. There are three important forms of phosphorus: white, red, and black. White phosphorus, a waxy solid, is obtained by condensing phosphorus vapor. Crystals of white phosphorus are formed from P4 molecules (see Figure 21.1) in which each phosphorus atom has an unshared pair of electrons and completes its octet by forming single covalent bonds with the other three phosphorus atoms of the state

.

molecule.

White phosphorus is soluble in a number of nonpolar solvents (for example, benzene and carbon disulfide). In such solutions, in liquid white phosphorus, and in phosphorus vapor, the element exists as P4 molecules. At temperatures above 800 C a slight dissociation of the P4 molecules of the vapor into Pj molecules is observed; these latter molecules are assumed to have a structure similar to that of the N2 molecule. White phosphorus is the most reactive form of the element

and

is

stored under water to protect

it

from atmospheric oxygen with which

it

heating white phosphorus to about 250

C

spontaneously reacts.

Red phosphorus may be prepared by

which many phosphorus atoms are joined in a network, but the details of the structure of red phosphorus are not known. Red phosphorus is not soluble in common solvents and is considerably less reactive than the white variety. It does not react with oxygen at room in the

absence of air.

It is

a polymeric material in

temperature.

common allotrope, is made by subjecting the element by a slow crystallization of liquid white phosphorus in the presence of mercury as a catalyst and a seed of black phosphorus. Crystalline black phosphorus consists of layers of phosphorus atoms covalently joined into a network (see Figure 21.2). The distance between P atoms of adjacent layers is much greater than the distance between P atoms of the same layer since the Black phosphorus, a

less

to very high pressures or

Figure 21.2

Structure of a layer of

Chapter 21

The Nonmetals, Part

tfie

III:

black phosphorus crystal

The Group V A Elements

P atoms of is

bonded to one another and the layers weak London forces. Hence, black phosphorus

a given layer are covalently

are held together by comparatively a flaky material

much

like graphite

(which also has a layer-type crystal, see

Section 22.2), and like graphite, black phosphorus

an electrical conductor. form of the element. Arsenic and antimony exist in soft, yellow, nonmetallic modifications which are thought to be formed from tetrahedral AS4 and Sb4 molecules analogous to the P4 molecules of white phosphorus. These yellow forms may be obtained by the rapid condensation of vapors and are soluble in carbon disulfide. They are unstable and are readily converted into stable, gray, metallic modifications. Bismuth commonly occurs as a light gray metal with a reddish cast; the element

Black phosphorus

does not exist

is

the least soluble

and

The

in other modifications.

antimony, and bismuth are comparatively luster.

metallic modifications of arsenic,

soft

and

brittle

and have

a metallic

Their crystalline structures are similar to the structure of black phosphorus,

and they are

21.2

is

least reactive

electrical conductors.

The Nitrogen Cycle is constantly being removed from the atmosphere and returned atmosphere by several natural and artificial processes. These processes, taken together, constitute what is called the nitrogen cycle. Nitrogen is a constituent element of all plant and animal protein. Since nitrogen is a comparatively unreactive element, the cells of living systems cannot directly

In nature, nitrogen to the

assimilate the nitrogen of the air to use in the synthesis of proteins.

of the

air,

however,

is

The nitrogen

converted by several nitrogen-fixation processes into com-

pounds

that can be used by plants. These nitrogen-fixation processes constitute

the

part of the nitrogen cycle.

to

first

During storms, lightning form nitrogen oxide: N2(g)

flashes cause

some nitrogen and oxygen of

the air

+ 0,(g)-^2N0(g)

Nitrogen dioxide

is

produced by the reaction of

NO

with additional

O2 from

the

air:

2NO(g) + 02(g) The NO2

—

reacts with water to

3N02(g) + H20(l)

The

2N02(g)

nitric acid is

form

nitric acid:

—*2HN03(1) +

washed

to the earth

NO(g)

where

it

forms nitrates

in the soil,

which

can be used by plants as nutrients.

Certam

soil bacteria, as well as nitrogen-fixing bacteria in the

of leguminous plants (such as peas, beans, and alfalfa)

fix

root nodules

atmospheric nitrogen

that plants can assimilate. Fertilizers are used to augment the ammonia, fixed nitrogen in the soil. Nitrogen-containing fertilizers are made from the Hancr process, nitrogen-fixation a by commercially which is itself produced into

compounds

process

N2(g)

+ 3H2(g)-2NH3(g) 21.2

The Nitrogen Cycle

543

— In the second stage of the nitrogen cycle, plants use the fixed nitrogen in the soil to

used to

make plant protein. The plant protein make animal protein. Indeed, humans

the ingestion of both plant

eaten by animals and

in turn

is

obtain their fixed nitrogen from

and animal protein.

In the third part of the nitrogen cycle, the cycle

is

completed. The decay of the

waste products of animal metabolism and the death and decay of plants and

animals liberates nitrogen as an end product. The N^, therefore, the

21.3

is

returned to

air.

Occurrence and Preparation Group V A Elements The

of the

group

principal natural sources of the

VA

elements are

listed in

Table

21.3.

Nitrogen, like oxygen (see Section 20.2) and the noble gases (see Section 22.9),

produced commercially by the fractionation of liquid air. Nitrogen from this is usually employed when the gas is needed in the laboratory. On occasion, however, small amounts for laboratory use may be obtained by heating an aqueous solution saturated with ammonium chloride and sodium is

source, in cylinders,

nitrate,

NH;(aq) + N02(aq)

^Njig)

+ lU^O

or by heating either sodium azide or barium azide,

2NaN3(s)

—- 2Na(l) + 3N3(g)

Phosphorus is the only member of group V A that does not occur in nature uncombined element. It is prepared industrially by heating a mixture of phosphate rock, sand, and coke in an electric furnace:

as an

Percent

Element

of

Occurrence

Earth's crust

0.0046 (0.03 including atmosptiere)

N2 (atmosphere)

phosphorus

0.12

CajtPOj), (phosphate rock), Ca,(P04)jF and Ca5(P04) ,CI

arsenic

5 X 10

nitrogen

NaNO,

(Chilean saltpeter)

(apatite)

-4

FeAsS(arsenopyrite), AS4S4

As,S, (orpiment), As^O^,

(realgar),

antimony

:

Cu, Pb, Co,

Ni,

SbjSj

5 X 10

native As:

(arsenolite)

(stibnite),

Sb Pb, Ag, and Hg

tite)

bismuth

1

X 10

- 5

;

native

Chapter 21

The Nonmetals, Part

III:

ores

;

of

Au

SbiOj, (senarmonores of Cu,

in

81283 (bismuthinite), Bi,0, (bismite) native Bi in ores of Cu, Pb, Sn, Co, Ni, Ag, and Au ,

544

in

Zn, Sn, Ag, and

The Group V A Elements

;

Mining phosphate rock, W. R Grace & Co.

2Ca3(P04)2(s)

+

P40,o(g)

6SiO,(s)

+

lOC(s)

—

eCaSiOjll) + P40,o(g)

—

P4(g)

+ lOCO(g)

The calcium

silicate is withdrawn as a molten slag from the bottom of the furnace, and the product gases are passed through water, which condenses the phosphorus vapor into a white solid. Arsenic, antimony, and bismuth are obtained by carbon reduction of their

oxides at elevated temperatures:

As406(s)

An

+ 6C(s)^As4(g) + 6CO(g)

important industrial source of the oxides

is

the flue dust obtained

from the

processes used in the production of certain metals, notably copper and lead. In addition, the oxides are obtained by roasting the sulfide ores of the elements in air; for

example,

2Sb2S3(s)

+

90,(g)

—Sb40^(g) +

6SO,(g)

arsenic, antimony, and bismuth all occur as native ores, only the deposits of native bismuth are sufficiently large to be of commercial importance.

Although

21.4 Nitrides

and Phosphides

Elementary nitrogen reacts with a number of metals

at high

temperatures to form

ionic nitrides, high-melting, white, crystalline solids that contain the N^" ion. The group II A metals, cadmium, and zinc form ionic nitrides with the formula

21.4

Nitrides

and Phosphides

545

M3N2

(where

M

is

Mg, Ca,

Be,

Sr, Ba,

+ 6H2O

Ca3N2(s)

—

Cd, or Zn) and lithium forms LijN. Ionic

ammonia and hydroxides:

nitrides react with water to yield

3Ca^^(aq)

+ 60H-(aq) + 2NH3(g)

made at elevated temperatures from many transition powdered form, and nitrogen or ammonia. A crystal of an interstitial nitride (VN, Fe4N, W^N, and TiN are examples) consists of metal atoms arranged in a lattice with nitrogen atoms occupying lattice holes (the interstices). These substances, therefore, frequently deviate from exact stoichiometry. They resemble metals and are hard, extremely high melting, good electrical conductors, and Interstitial nitrides are

metals, in

chemically unreactive.

Covalent nitrides include such compounds as S4N4, P3N5, Si3N4. Sn3N4, BN, and AIN. Some of these compounds are molecular in form. Others, such as BN and AIN, are substances in which a large number of atoms of the two elements are covalently bonded together into a network crystal. Both BN and AIN are made by reacting the elements at high temperatures. Two C atoms taken together have the same number of valence electrons (8) as one B atom (3 valence electrons) and one N atom (5 valence electrons) combined. The compound BN, therefore, may be considered to be isoelectronic with carbon. Indeed, BN is known in two crystalline modifications, one resembling graphite and another extremely hard form resembling diamond. Many metals react with white phosphorus to form phosphides. The group II A elements form phosphides with the formula M3P2 (where is Be. Mg. Ca, Sr. or Ba). lithium forms Li3P and sodium forms NajP. These compounds readily react with water to form phosphine. For example,

M

Ca3P,(s)

—

+

The phosphides of

the

3Ca-^(aq)

group HI

A

+ 60H

(aq)

+

PH3(g)

elements (such as BP, AlP. and GaP) form

covalent network crystals similar to silicon, and like silicon these substances are

semiconductors.

Many

phosphides of the transition metals are known (FeP,

FctP, C02P, RuP, and OsP, are examples). These substances are gray-black, semimetallic crystals that are electrical conductors.

The

reactions of metals with arsenic, antimony, and, to a lesser extent, bismuth

yield arsenides, stibnides, sively

more

difficult to

and bismuthides. These compounds become progres-

prepare as the atomic number of the group

V A

element

increases.

21.5

Hydrogen Compounds The group V A elements all form hydrogen compounds, the most important of which is ammonia, NH3. Large quantities of ammonia are commercially prepared by the direct union of the elements (Haber process): N2(g)

+ 3H2(g)^2NH3(g)

Ammonia prepared

is

the only hydrogen

directly.

1000 atm), at 400

546

Chapter 21

The to

reaction

is

550 C, and

The Nonmetals. Part

III

;

of the V A elements that can be conducted under high pressures (from 100 to

compound in the

presence of a catalyst.

The Group V A Elements

One

catalyst, so

employed, consists of of K2O and AI2O3.

and Fe304 containing small amounts

finely divided iron

Smaller quantities of

ammonia

are produced as a by-product in the

manu-

Ammonia was formerly cyanamide, CaNCN, with

facture of coke by the destructive distillation of coal.

produced commercially by the reaction of calcium steam under pressure

CaNCN(s) + 3H,0(g)

—

*

+ 2NH3(g)

CaCOafs)

However, the Haber process has

method

largely displaced this

source of ammonia, and calcium cyanamide

is

produced

as a commercial

chiefly as a fertilizer

and as a raw material in the manufacture of certain nitrogen-containing organic compounds. Calcium cyanamide is produced in a two-step process. Calcium carbide, CaC2, is made by the reaction of CaO and coke in an electric furnace: CaO(s)

+

3C(s)

—

CaC.fs) + CO(g)

and the calcium carbide is reacted with 1000 C to produce calcium cyanamide: CaC2(s)

+

N,(g)

—

CaNCN(s) +

ammonia

In the laboratory,

NaOH

or Ca(OH),,

conveniently prepared by the hydrolysis of

is

ammonium

salt

with a strong alkali,

either dry or in solution:

—

NH;(aq) + OH-(aq)

approximately

C(s)

by heating an

nitrides (see Section 21.4) or

such as

relatively pure nitrogen at

H^O

NHjtg) +

The ammonia molecule,

N

H

H

H is

trigonal pyramidal with the nitrogen

atom

at the

apex; this

compound

is

associated through hydrogen bonding in the liquid and solid states.

Aqueous

solutions of

NHjlaq) +

H,0

ammonia

^

In solution or as a dry gas.

NH3(g) + HCl(g)

The ammonium Nitrogen

is

ion

NH;(aq) + OH'(aq)

ammonia

reacts with acids to

is

—2N2(g) +

a mixture of

nitric oxide,

ammonium

salts:

tetrahcdral.

formed when ammonia

However, when

produce

—NH^CKs)

4NH3(g) + 302(g)

1000 C,

are alkaline:

NO,

4NH3{g) + 502(g)

is

is

burned

pure oxygen

in

6H20(g)

ammonia and

air

is

passed over platinum gauze at

produced:

—4NO(g) +

6H20(g)

21.5

Hydrogen Compounds

547

H This catalyzed oxidation of

H

NH3

is

:

a part of the Ostwald process for the

manu-

facture of nitric acid (see Section 21.7).

N2H4, may be considered atom by a NHj group:

Hydrazine,

ment of a

H

to be derived

from

NH3

by the replace-

—

H— N—N— H

H

is a liquid, may be prepared by oxidizing NH3 with NaOCl. Hydrazine is less basic than NH3 but does form cations in which one proton or two protons are bonded to the free electron pairs of the molecule. For example.

The compound, which

H

H

H— N— N— H

H

H— N— N

CI

H

H

H

2cr

H

Hydrazine is a strong reducing agent and has found some use in rocket fuels. Hydroxylamine, NH^OH. is another compound which, like hydrazine, may be considered to be derived from the NH3 molecule:

H— N— O— H is a weaker base than NH3, but salts containing may be prepared. Hydrazoic acid, HNj, is another hydrogen compound of nitrogen. The structure of hydrazoic acid may be represented as a resonance hybrid

Like hydrazine, hydroxylamine the

[NHjOH]^

ion

©

..

e

H— N=N=N:

..

e ®

H— N"N=N:

—

N N bond distances of the molecule are not the same; the distance from the central N to the N bearing the H atom (124 pm) is longer than the other (113 pm). Hydrazoic acid may be made by reacting hydrazine (which forms the The two

[N2H5]*

ion in acidic solution) with nitrous acid

N,H5^(aq)

+ HNO,(aq)

—

HN3(aq) +

(HNO2):

H + (aq) + 2H2O

may be obtained by distillation of the water Hydrazoic acid is a weak acid. The heavy metal salts of the acid, such as lead azide, Pb(N3)2, explode upon being struck and are used in detonation caps. Phosphine (PH3) is a very poisonous, colorless gas that is prepared by the hydrolysis of phosphides or by the reaction of white phosphorus with concen-

The

free acid, a low-boiling liquid,

solution.

trated solutions of alkalies:

P4(s)

+ 30H

(aq)

+ 3H2O

PH3(g)

3H2PO;(aq)

+

hypo[ilu>splulc ion

The PH3 molecule however, the

548

Chapter 21

is

pyramidal, similar to the

compound

is

NH3

molecule. Unlike

NH3,

not associated by hydrogen bonding in the liquid state.

The Nonmetals, Part

III:

The Group V A Elements

Phosphine

is

much

less basic

decompose component

than ammonia. Phosphonium compounds, such

made from dry PH3(g) and

as PH4l(s) which can be

low temperatures, or

at relatively

in

HI(g), are unstable.

aqueous solution,

They

to yield the

gases.

Arsine (AsH,), stibine (SbHj), and bismuthine (BiH,) are extremely poisonous gases that may be produced by the hydrolysis of arsenides, stibnides, and bis-

muthides (for example, NajAs, ZujSbj, and Mg3Bi2). The yields of the hydrogen compounds become poorer with increasing molecular weight. Very poor yields of bismuthine are obtained by this method. 1 he stability of the hydrogen com-

pounds declines in the series from NH3 to BiH3. Bismuthine is very unstable and decomposes to the elements at room temperature. Arsine and stibine may be similarly decomposed by warming. Arsine, stibine, and bismuthine have no basic properties and do not form salts with acids.

21.6

Halogen Compounds The most important halides of the V A elements are the trihalides (for example, NF3) and the pentahalides (for example, PF5). All four binary trihalides of each

V A elements have been made,

but the tribromide and triiodide of nitrogen form of ammoniates (NBr3-6NH3 and Nl3-.vNH3). The nitrogen trihalides are prepared by the halogenation of ammonia gas (NF3, NBr3), of an ammonium salt in acidic solution (NCI3, NBrj), or of concentrated aqueous ammonia (NI3). Each of the trihalides of P, As, Sb. and Bi is prepared

of the

can be isolated only

in the

by direct halogenation of the

V A

element using a stoichiometric excess of the

VA

element to prevent pentahalide formation. Bismuth trifiuoride is an ionic compound, but the other trihalides are covalent. In the gaseous state the covalent trihalides exist as trigonal pyramidal molecules (see Figure 21.3). This molecular form persists in the liquid state and in all of the solids except Aslj, Sblj,

Nitrogen trifiuoride

is

and

Bilj.

which

of nitrogen are explosively unstable.

trihalides

undergo hydrolysis:

+ 3H.O

NH3(g) + 3HOCl(aq)

PCl3(l)

+ 3H,0

H3P03(aq) + 3H'(aq) + 3C1

AsCl3(l)

+ 3H,0

H3As03(aq) + 3H*(aq) + 3C1

SbCl3(s)

+

H2O

SbOCl(s)

BiCl3(s)

+

H,0

BiOCl(s)

is

more

CI appear in 3

1

-

state.

The

(aq) (aq)

+ 2H"(aq) + 2Cr(aq) + 2H*(aq) +

electronegative than CI,

- and

hydrolysis reactions, the 1

The

NCl3(l)

Nitrogen

a

crystallize in covalent layer lattices.

a very stable colorless gas. whereas the other trihalides

+ oxidation V A element

and

20

in the

(aq)

hydrolysis of

states, respectively.

appears

metallic character of the

VA

in a 3

+

NCI 3,

N

and

In each of the other

oxidation state and CI

in

elements increases with increasing

atomic number, and Sb and Bi occur as 0x0 cations (SbO* and BiO") in the hydrolysis products of SbClj and BiCl3. The pentahalide series is not so complete as the trihalide series. Since N has no d orbitals in its valence level, N can form no more than four covalent bonds.

21.6

Halogen Compounds

Figure 21.3

Molecular

structure of the covalent trihalides of

group V A

elements

549

Pentahalides of N, therefore, do not

exist. All

pentahalides of phosphorus are

known with the exception of the pentaiodide; presumably there is not sufficient room around a phosphorus atom to accommodate five large iodine atoms. In addition,

AsFj, SbFj, BiFj. and SbCl; have been prepared.

The pentahalides may be prepared by

direct reaction of the elements using

an

excess of the halogen and by the reaction of the halogen with the trihalide:

PCl3(l)

+

^

Cl2(g)

The preceding

reaction

PClsfs)

reversible. In the gas

is

phase the pentahalides dissociate

to varying degrees. Figure 21.4

gaseous PCI

Structure of the 5

molecule

The pentahalides

and liquid SbCl5 consists of such molecules. Solid PCI 5 and PBr,, however, form ionic lattices composed of PCI4 and PClg respectively. Apparently it is impossible to pack six bromine and PBr^ and Br atoms around a phosphorus atom since PBv^ does not form. The cations are tetrahedral and the PC1(~ ion is octahedral (see Figure 21.^). The phosphorus pentahalides undergo hydrolysis in two steps. For example, are trigonal bipyramidal molecules in the gaseous

state (see Figure 21.4).

The

crystal lattice of

,

PCl5(s)

POCljd) +

+ H2O

3H2O

The phosphoryl lysis

— —

halides,

POCl3(l)

+ 2H^(aq) + 2Cr(aq)

H3P04(aq) + 3H^(aq) + 3Cr(aq)

POX3

(X

=

F, CI, or Br) can be prepared

of the appropriate pentahalide in a limited

amount

by the hydro-

of water or by the reaction

of the trihalide with oxygen:

2PCl3(l)

+

0,(g)

—

2POCl3(l)

Molecules of the phosphoryl halides have a PX3 grouping arranged as a trigonal pyramid (see Figure 21.3) with an oxygen atom bonded to the phosphorus atom, thus forming a distorted tetrahedron.

A number of mixed trihalides (for example, NF^Cl, PFBr^, and SbBrI,) and mixed pentahalides (for example, PCI2F3, PCIF4, and SbCl3F2) have been prepared. In addition, halides are known that conform to the general formula E2X4: N2F4, P2CI4, P2I4, iind AS2I4. These compounds have molecular structures similar to the structure of hydrazine,

N2H4.

Isomers are substances that have the same molecular formula but differ in the the constituent atoms are arranged into molecules. Dinitrogen difluoride exists in two isomeric forms

way

F

\

N=N

N=N \

\

/

F

F

trans

F cis

The double bond, composed of

a a bond and a n bond, between the two nitrogen atoms prevents free rotation about the nitrogen-nitrogen axis. In the cis isomer both fluorine atoms are on the same side of the double-bonded nitrogen atoms, whereas in the trans isomer the fluorine atoms are on opposite sides. Both mole-

cules are planar.

21.7 Oxides

and Oxyacids

Oxides are known 1.

Then-

Nitrogen

of

for every oxidation state of nitrogen

—

1

-I-

to

5-1-

:

Dinitrogen oxide (also called nitrous oxide), N2O,

oxidation state.

prepared by gently heating molten

NH^NOjd)

from

ammonium

is

nitrate:

Npig) + 2HP(g)

It is a colorless gas and is relatively unreactive. At temperatures around 500 C. however, dinitrogen oxide decomposes to nitrogen and oxygen; hence, N2O

supports combustion. Molecules of of the

compound may

0

Dinitrogen oxide

is

its

..

©

..

:N=N=0:

commonly

produces when breathed

and because of

N2O

and the electronic structure

are linear,

be represented as a resonance hybrid:

in

— :n=N— ©

..

e

O:

called "laughing gas" because of the effect

small amounts.

solubility in cream,

it is

The gas

is

it

used as a general anesthetic,

the gas used to charge

whipped cream

aerosol cans.

The 2-1- oxidation state. Nitrogen oxide (also called nitric oxide), NO, prepared by the direct reaction of the elements at high temperatures: 2.

N2(g)

The

+

be

02(g)— 2 NO(g)

reaction

NO

may

is

endothermic (A//

=

-1-90.4 kJ mol).

but even at 3000

C

the yield

only approximately 4"o. In a successful preparation the hot gases from back the reaction must be rapidly cooled to prevent the decomposition of into nitrogen and oxygen. By this reaction, atmospheric nitrogen is fixed during

of

is

NO

lightning storms. This reaction also serves as the basis of the arc process of nitrogen

which an electric arc is used to provide the high temperatures necessary for the direct combination of nitrogen and oxygen. As a commercial source of NO, the arc process has been supplanted by the catalytic oxidation of ammonia from the Haber process (see Section 21.5). fixation in

21.7

Oxides and Oxyacids

of

Nitrogen

:

The NO molecule contains an odd number of electrons, which means that one electron must be unpaired for this reason, NO is paramagnetic. The electronic structure of the molecule may be represented by the resonance forms e ® ;

:n=o:-^:N--6: The

structure, however,

best described by molecular orbital theory, which

is

bond order of 2j to the molecule and indicates that the odd electron is in a n* orbital. The loss of an electron from NO produces the nitrosonium ion, NO"*". Since the electron is lost from an antibonding orbital, the NO"^ ion has a bond order of 3, and the bond distance in NO^ (106 pm) is shorter than the bond distance in the NO molecule (114 pm). Ionic compounds of NO^ are known (for example, NO^ HSO;, NO^ CIO4, and NO^ BF;). Whereas odd-electron molecules are generally very reactive and highly colored, nitric oxide is only moderately reactive and is a colorless gas (condensing to a blue liquid and blue solid at low temperatures). In addition, NO shows little assigns a

N2O2

tendency to associate into

oxygen

reacts instantly with

—

2NO(g) + 02(g)

at

molecules by electron pairing. Nitrogen oxide

room temperature

to

form nitrogen dioxide:

2N02(g)

The 3+ oxidation state. Dinitrogen trioxide, NjOj, forms as a blue liquid when an equimolar mixture of nitric oxide and nitrogen dioxide is cooled to 3.

-20 C

:

NO(g) + N02(g)

—

NPsd)

The compound is unstable under ordinary conditions and decomposes into NO and NO,. Both NO and NO, are odd-electron molecules; N2O3 is formed by and the N2O3 molecule

electron pairing,

Dinitrogen trioxide

aqueous 4.

The 4

N2O4,

alkali to -I-

is

is

thought to contain a

the anhydride of nitrous acid,

produce the

oxidation state.

nitrite ion,

N— N

HNO2, and

bond.

dissolves in

N02'-

Nitrogen dioxide, NO,, and dinitrogen tetroxide,

exist in equilibrium:

2NO2

^

N2O4

Nitrogen dioxide consists of odd-electron molecules, is paramagnetic, and is brown in color. The dimer, in which the electrons are paired, is diamagnetic and colorless. In the solid state the oxide is colorless and consists of pure N2O4.

The liquid is yellow in color and consists of a dilute solution of NO2 in N2O4. As the temperature is raised, the gas contains more and more NO2 and becomes deeper and deeper brown in color. At 135 C approximately 99"„ of the mixture is NO2. Nitrogen dioxide molecules are angular: .

N

The

©

®:q.

.o'

structure of

N2O4

.

N o. is

©

.0:°

N

N -q.

.0.

thought to be planar with two

N— N bond. 552

Chapter

21

The Nonmetals, Part

III

:

The Group V A Elements

.0.

NO2

.0 units joined by a

Nitrogen dioxide

The

S+

nitric acid

produced by the reaction of

oxidation state. ;

is

nitric oxide with oxygen. In conveniently prepared by heating lead nitrate:

—2PbO(s) +

2Pb(N03)2(s) 5.

is

compound

the laboratory the

4N02(g) + O^Cg)

Dinitrogen pentoxide, NjOj,

is

the acid anhydride of

N2O5 may be prepared from this acid by dehydration using phosphorus

(V) oxide:

4HN03(g) + The compound

P40,o(s)

2N,05(g)

a colorless, crystalline material that sublimes at 32.5 C.

is

vapor consists of

—4HP03(s) +

The

N2O5

molecules which are thought to be planar with the two nitrogen atoms joined through an oxygen atom (O2NONO2).

The

electronic structure of the molecule

may be

represented as a resonance

hybrid of

:0:

:o:

J ®N e. /

II

N® \ / \ .0.

.0.

and other equivalent structures bonds. The

compound

is

..e

.0.

that have different arrangements of the double

unstable

in the

vapor

and decomposes according

state

to the equation

2N20.,(g)

—

4N02(g) + 02(g)

composed of nitronium, NOj and nitrate, NO3 ions. two ions in solutions in anhydrous sulfuric acid, nitric acid, and phosphoric acid. The nitrate ion is triangular planar. The nitronium ion is linear; it is isoelectronic with CO2 and may be considered as a nitrogen dioxide molecule minus the odd electron. The ion is probably a Crystals of

N2O5

The compound

is

are

.

,

dissociated into these

reaction intermediate in certain reactions of nitric acid in the presence of sulfuric

compounds have been prepared

acid (nitrations). Ionic nitronium

(for

example,

NOj^ClO^-, N02'BF4", N02^PF6").

The most important oxyacid of nitrogen is nitric acid, HNO3, in which nitrogen an oxidation number of 5 + Commercially, nitric acid is produced by

exhibits

.

the Ostwald process. Nitric oxide from the catalytic oxidation of

ammonia

is

reacted with oxygen to form nitrogen dioxide. This gas, together with excess

oxygen,

is

passed into a tower where

3N02(g) +

The

H2O

—2H*(aq) +

excess oxygen converts the

as before. In this cyclic into nitric acid.

manner

The product

NO

2N03-(aq) into

NO.; is

warm

from

by

water:

+ NO(g)

NO2

this

then reacts with water

eventually completely converted

of the Ostwald process

as concentrated nitric acid;

it

reacts with

the nitric oxide

known Pure

it

is

more concentrated

about 70°„

solutions

HNO3

may

and

is

be prepared

distillation.

nitric acid is a colorless liquid that boils at

the laboratory by heating

sodium

83'C.

It

may

be prepared

in

nitrate with concentrated sulfuric acid:

21.7

Oxides and Oxyacids

of Nitrogen

553

N

:

—

H 2804(1)

NaNOsCs) +

-

NaHS04(s) + HNOjfg)

This preparation (from Chilean saltpeter)

is

a

minor commercial source of the

acid.

The

HNO3

molecule

is

planar and

may

be represented as a resonance hybrid:

-q:®

•q.

®^ H— O— N

—

'

..

H— O—

-

®

is

and

a strong acid

is

almost completely dissociated

Most salts of nitric acid, which are The nitrate ion is triangular planar:

solution.

water.

.. e :o:

Nitric acid

aqueous

e

:o:

N®

.0.°

N®

.0:°

°:q.

a powerful oxidizing agent;

is

in

called nitrates, are very soluble in

:o:

N® q.

"

.p-

.q.e Nitric acid

/

it

°:q.

oxidizes

.0

most nonmetals (generally and all metals with the

to oxides or oxyacids of their highest oxidation state)

exception of a few of the noble metals.

and copper, that do not as HCl, dissolve in nitric

react to yield

Many

unreactive metals, such as silver

hydrogen with nonoxidizing

acids, such

acid.

hydrogen is almost never obtained; instead, a variety compounds, in which nitrogen is in a lower oxidation state, is produced (see Table 21 .4). The product to which HNO3 is reduced depends upon the concentration of the acid, the temperature, and the nature of the material In nitric acid oxidations,

of nitrogen-containing

A

being oxidized.

product oxides

in

many

mixture of products

cases

usually obtained, but the principal

is

NO when dilute HNO3

is

when concentrated

HNO3

is

is

employed and the nitrogen(IV)

used.

Dilute:

3Cu(s)

+ 8H + (aq) +

2 N03~(aq)

—

3

Cu^ ^(aq) + 2NO(g) +

4H2O

Concentrated

+ 4H^(aq) +

Cu(s)

Table 21.4

2 N03"(aq)

—

Cu-^(aq)

Standard electrode potentials

for

+ 2NO:(g) + 2H,0

reductions of the nitrate ion

Standard Half Reaction

+ 2H+ + + 10 + 2e' + 3H+ + 3e" + 4 + 8e~ + 10 H +2 e'

no;

8e"

NO3

*

lOe"

554

Chapter 21

+

6

+

2

^=

no; NO3"

no;

NOJ

The Nonmetals, Part

;j=±

III

:

Electrode Potential

NO2 + H2O NH; + 3 HjO HNO2 + H,0

+ 0.80 V + 0.88 V + 0.94 V

NO + 2 H^O N2O + 5 HjO N2 + 6 H2O

+ 0.96 V + 1.12 V

The Group V A Elements

+ 1.25 V

In

some

instances, however, strong reducing agents are

pure compounds of nitrogen

lower oxidation

in

of zinc and dilute nitric acid yields

The Table

21.4)

show

to

produce almost

as the reduction product of

the

^"^^^

solutions of

HCl

The oxyacid

HNO3.)

2

M,

has

nitric acid

upon

the H"^(aq)

experimentally observed

is

more oxidizing power than

little

of corresponding concentration.

of nitrogen in which nitrogen has an oxidation

HNOj. The compound

nitrous acid,

(particularly

values to be strongly dependent

This concentration dependence

below a concentration of

is

number

of 3

+

unstable toward disproportionation

when warmed):

—

3HN02(aq) As

known

(For example, the reaction

half-reactions for the reduction of the nitrate ion in acid solution (see

concentration.

is

NH3

states.

H^(aq)

HNO2

a result, pure

+

N03-(aq)

+ 2N0(g) + H^O

has never been prepared. Instead, aqueous solutions of

the acid are usually prepared by adding a strong acid (such as HCl) to a cold

aqueous solution of

H^(aq)

a nitrite (such as

+ N02(aq)

—

Such solutions are used

NaNO,):

* HN02(aq)

directly, without

ratory procedures that require

attempting to isolate

HNO^. Aqueous

solutions of

HNO2,

in labo-

HNO2 may

also be

prepared by adding an equimolar mixture of NO(g) and N02(g) to water:

NO(g) + N02(g) + H2O

The

reaction

is

N2O3, The acid is The electronic

acid,

is

^

2HN02(aq)

exothermic and reversible. Note that the acid anhydride of nitrous prepared by the reaction of NO(g) and N02(g) at -20 C. weak and may function as an oxidizing agent or a reducing agent. structure

may

be represented as:

N .

/

•o. /

.O'

•

H Nitrites are prepared

NO(g)

by adding NO(g) and N02(g) to solutions of

+ N02(g) + 20H

A

(aq)

—

alkalies:

2N02(aq) + H2O

metals are formed when the nitrates are heated. They

The

nitrites of the

may

also be prepared by heating the nitrate with a reducing agent, such as lead,

I

iron or coke:

NaN03(l) +

The

nitrite ion

C(s)

is

—

NaN02(l) + CO(g)

angular:

N

N .

//

•o.

\ .0: .

21.7

Oxides and Oxyacids

of Nitrogen

555

ACIDIC SOLUTION

+ 0.96

+094

NO3

+100

HNOj +

..^ +159. ^ +1.77,> NO >N20-'

N2

^ +0.94^ >

Ng

.,

^

+0.27.

> NHj

1,12

ALKALINE SOLUTION

-0

15

r" +00^

^,r^

NO3

^0 46

m^NO2

^,^+0 NO

76

-

N2O

-0.7 3

NH3

t

+ 0.10

Figure 21.6 Electrode potential diagrams for nitrogen and given in volts)

some

of its

compounds

(values

and some of its compounds appear in The compounds of nitrogen shown in the diagrams are much stronger

Electrode potential diagrams for

Figure 21.6.

NO3' compound

oxidizing agents in acidic solution than in alkaline solution. In fact, the ion

is

a very

of nitrogen

is

in alkaline solution.

Conversely, a given

easier to oxidize in alkaline solution than in acidic solution with the

NO

which cannot be oxidized. The potentials also indicate that NO and NO3" in acid, whereas alkaline solution the N02~ ion is stable toward such disproportionation.

exception of

HNO2 in

weak oxidant

is

21.8 Oxides

3"

unstable toward disproportionation to

and Oxyacids

of

The two important oxides

3+

(P4O6) and

Phosphorus

of phosphorus contain phosphorus in oxidation states

5+

(P40io)- Phosphorus(III) oxide, P^Pe^ is frequently called phosphorus trioxide, a name which dates from a time when only the empirical of

formula of the compound, P2O3, was known. The is

in

compound

is

a colorless

C and is the principal

product when white phosphorus burned in a limited supply of air. The structure of the P40f, molecule (shown Figure 21.7) is based on a P4 tetrahedron (see Figure 21.1 and has an O atom

substance that melts at 23.9

)

inserted in each of the six edges of the tetrahedron.

Phosphorus(V) oxide, P4O10, the empirical formula, P2O5).

is

often called

The oxide

is

phosphorus pentoxide (based on

the product of the combustion of

in an excess of oxygen. It is a white powder that sublimes at 360 C. The molecular structure of P4O10 (see Figure 21.7) can be derived from the structure of P40f, by adding an extra O atom to each P atom.

white phosphorus

Phosphorus(V) oxide has a great affinity for water and is a very eflFective Many different phosphoric acids may be prepared by the addition of water to P40,o- The most important acid of phosphorus in the 5+ state.

drying agent.

556

Chapter

21

The Nonmetals, Part

III:

The Group V A Elements

orthophosphoric acid (usually called phosphoric acid) hydration of the oxide:

6H2O

P40,o +

results

from the complete

—*4H3P04

Phosphoric acid is obtained commercially by rock with sulfuric acid:

this

means or by

phosphate

treating

P4O6

Ca3(P04)2(s)

The compound

+ 3H2S04(1)^2H3P04(1) + is

a colorless solid but

electronic structure of

H ,P04 may ..

is

3CaS04(s)

generally sold as an 85"„ solution.

The

be represented as

e

:o:

:o:

H— O— P^O-H

O— P— O—

H

or

:0:

:0:

H

H

—

P O bond has double-bond character from pn-dn bonding. The molecule and the ions derived from it are tetrahedral (Figure 21.8). Phosphoric acid is a weak, triprotic acid without effective oxidizing power:

since the

H3PO4

H.,P04(aq)

H2P04:(aq)

HPOa"(aq)

^ ^ ^

^

P

Figure 21.7

H*(aq)

+ H,P04(aq)

H'(aq)

+ HPOr(aq)

H "(aq) + PO^

Structures of P4O,,

(aq)

series of salts may be derived from H3PO4 (the sodium salts, for example, NaH2P04, Na2HP04. and Na3P04). The product of a given neutralization depends upon the stoichiometric ratio of H3PO4 to alkali.

Three are:

Phosphates are important ingredients of commercial fertilizers. Phosphate is too insoluble in water to be used directly for this purpose. The more

T

rock

soluble calcium dihydrogen phosphate

however, and

may

Ca3(P04)2(s)

The mixture

of

ingredient,

fertilizer

.

be obtained by treatment of phosphate rock with an acid:

+ 2H,S04(1)

—

Ca(H2P04),(s)

Ca(H2P04)2 and CaS04

higher yield of the dihydrogen phosphate

ployed

satisfactory

a

is

called superphosphate fertilizer.

is is

+ 2CaS04(s)

obtained

if

phosphoric acid

is

A

em-

in the reaction instead of sulfuric acid:

Ca3(P04),(s)

+ 4H3P04(1)

—

Q'

3Ca(H2P04)2(s)

/

Since nitrates are also important constituents of

fertilizers,

by treatment of phosphate rock with

is

Ca3(P04)2(s)

nitric acid

the mixture obtained

o

a highly eflfective fertilizer:

+ 4HN03(1) ^Ca(H2P04)2(s) +

O ^o

2Ca(N03)2(s)

P^Of"

P

Condensed phosphoric acids have more than one P atom per molecule. The members of one group of condensed phosphoric acids, the polyphosphoric acids, conform to the general formula

H„ + 2Pn03n+

1

where «

is

2 to 10.

21.8

Examples are

Oxides and Oxyacids

of

Figure 21.8

and

Phosphorus

structures of

PO^

p,or 557

H

o

o

O— P~0— P— O H H4P2O7,

o

o

H— O— P—O— P—

H

O

O

H

H

diphosphoric acid

The polyphosphoric

PO4

o

'

HjPjOio,

O H

O

O

H

H

triphosphoric acid

and polyphosphates have chain structures based on O atoms that are common to adjacent

acids

tetrahedra which are joined through

tetrahedra (Figure 21.8).

The metaphosphoric acids constitute another group of condensed phosphoric These compounds have the general formula H„P„03„ where n is 3 to 7.

acids.

Some

of the metaphosphoric acids are cyclic. For example:

O

H

o

O \

H— O— P— O— P^O—

00

P

00

H— O— P— O P^O—

II

P

H"0 H 3P3O0

,

o

'

/ \ / \

O

O— H

O H4P40

irimetaphosphoric acid

1

1

,

O

telrametaphosphoric acid

In addition, there are high-molecular-weight, long-chain metaphosphoric acids which are always obtained as complex mixtures that are assigned the formula (HPO3),,. The molecules of these mixtures are based on long chains of

O

— P— O— OH units joined in such a

way

that each

P atom

is

tetrahedrally

bonded

to four

O

atoms, but the complete structures are very complicated and involve phosphorus-

oxygen units that link two chains together. The condensed phosphoric acids may be obtained by the controlled addition of water to P40,o- For example, P^Oiots)

+ 4H2O

2H4P,07(S) diphosphoric acid

P40io(s)

+ 2H,0

—-

H4P40i,(s) tetrametaphosphoric acid

The dehydration of H3PO4 by heating

H4P207(1)

2H3P04(1)

also yields condensed acids.

+ H,0(g)

diphosphoric acid

(HPO3UI)

«H3P04(1)

+

/!H,0(g)

metaphosphoric acid mixture

558

Chapter 21

The Nonmetals,

Part

III

:

The Group V A Elements

For example,

On

standing in water, all condensed phosphoric acids revert to H,P04. The oxyacid in which phosphorus has an oxidation number of 3+ is phosphorous acid, H3PO3. Note that the -oiis ending of the name of the acid differs from the -us ending of the name of the element. Phosphorous acid can be made

by adding F4P(, to cold water, P406(s)

+ 6H2O

4H3P03(aq)

Condensed phosphorous acids

Even though the H3PO3 molecule is a weak, diprotic acid and is

also exist.

contains three hydrogen atoms, phosphorous acid

probably better formulated as H2(HP03):

H,(HP03)(aq)^H^(aq) + H(HP03)-(aq) H(HP03)

(aq)

^

H^(aq) + HPOi~(aq)

The sodium salts NaHjPOj and Na^HPOj are known, but it is impossible to prepare Na3P03. Phosphorus has an oxidation number of + in hypophosphorous acid, H3P02. 1

Solutions of salts of the acid

may

be prepared by boiling white phosphorus with

solutions of alkalies:

P4(s)

+ 30H

(aq)

+ 3H2O

—

PH3(g)

+ 3H2P02(aq)

which is a colorless crystalline material, may be obtained by treating a solution of barium hypophosphite with sulfuric acid:

The

acid,

Ba^^laq)

+ 2H2P02"(aq) + 2H^(aq) + SOilaq)

— BaSO^Is)

+

2

H3P02(aq)

removing the precipitated BaSO^ by filtration, and evaporating the solution. Hypophosphorous acid is a weak monoprotic acid. The formula of the compound, therefore, is sometimes written H(H2P02):

H(H2P02)(aq)

^

H"(aq) + H2P02(aq)

The number of protons released by phosphoric (three), phosphorous (two), and hypophosphorous (one) acids may be explained on the basis of the molecular structures of these compounds:

0 H

O P— O-H

O 1

H

O

o

H— 0"^P— 0"H

H^O P— H

H phosphorous acid

hypophosphorous acid

phosphoric acid

hydrogen atoms bonded to oxygen atoms are acidic: those bonded to phosphorus atoms do not dissociate as H*(aq) in water. In each of these molecules the P— O bonds have pn-dn double-bond character. Electrode potential diagrams for phosphorus and some of its compounds the appear in Figure 21.9. The most striking feature of the diagrams is that all

Only

the

21.10

Industrial

Uses

of the

Group V A Elements

559

ACIDIC SOLUTION -0.16 -0.28

-0.50

> H3PO3

H3PO4

-0.51

'

> HsPOs

>

-0.06 P4-

^ >

PH3

-0.28

f

I

ALKALINE SOLUTION -1.18

f

^JL31

I

Electrode potential diagrams for phosphorus and

Figure 21.9

some

of

compounds

its

potentials are negative

Thus none of

—

in contrast to the potentials of the nitrogen

diagram.

good oxidizing agent, particularly in alkaline solution. Rather, H3PO3, H3PO2, and the salts of these acids have strong reducing properties; the acids are readily oxidized to H3PO4 and the anions are readily oxidized to PO4". With the exception of HPO3' in alkaline solution, the substances

is

a

PH3

species of intermediate oxidation state disproportionate;

all

is

one of the

products of each disproportionation. 21.9 Oxides

and Oxyacids

When arsenic,

of As, Sb,

and

antimony, or bismuth

is

Bi

heated

in air, a 3

+

oxide

is

formed AS4O5, :

Sb40(,, or Bi203. These oxides serve as an excellent example of the change in metallic character that table.

The

therefore,

and

its

lightest

is

the

is

observed within a group of elements of the periodic

element of a group

most

is

oxide, therefore,

is

the

most nonmetallic, and its oxide, a group is the most metallic,

The heaviest member of the most basic.

acidic.

The oxide of the lightest element of the three, AS4O5, is the most acidic of Aqueous solutions of As40f, are acidic and thought to contain arsenious acid, H3ASO3. When As40(, dissolves in aqueous alkalies, arsenites (salts of AsO|~ and other forms of this anion) form. The oxide is exclusively acidic and 1.

the three.

will 2.

not dissolve

The oxide of

in

aqueous

acids.

the intermediate element of the three, Sb40(,,

has both acidic and basic properties). The oxide it

solutions (to give salts of

The oxide of

It will

in

water or

Chapter 21

salt,

in

aqueous

amphoteric

group V A elements. Bi(OH)3 precipitates.

The Nonmetals, Part

III

:

(it

SbO^ and )

also

or Sb^^).

alkalies,

but

it

is

exclusively basic.

will dissolve in acids

BiO"^ or Bi^^). Bismuth(III) hydroxide, Bi(OH)3,

true hydroxide of the

of a Bi^"^

SbO^

the heaviest element of the three. Bi203,

not dissolve

(to give salts of

560

is

not dissolve in water, but

will dissolve in alkaline solutions (to give antimonites, salts of

in acidic 3.

will

When

OH

The Group V A Elements

(aq)

is

added

is

the only

to a solution

The molecular structures of the 5 + oxides of arsenic and antimony are not known, and empirical formulas (As^O, and Sb^O,) are employed.

accurately

These oxides are prepared by the action of concentrated HNO3 on the elements or the 3+ oxides, followed by the dehydration of the products of these reactions (2H3As04-H20 and SbjOj vv H2O). The 5+ oxides are exclusively acidic. Orthoarsenic acid (H3ASO4, a triprotic acid) forms when AsjO, is dissolved in water. Arsenates (salts of ASO4") or antimonates (salts of Sb(OH)6) can be prepared by dissolving As,©; or SbjO, in aqueous alkalies. Sodium antimonate,

NaSb(OH)6, used as a

is

one of the

test for

least soluble

sodium

salts,

and

its

formation

often

is

Na^

Bismuth(V) oxide has never been prepared in the pure state; it is unstable readily loses oxygen. The red-brown product obtained by the action of strong oxidizing agents (such as CU, OCl", and S2OI") on a suspension of Bi^Oj in an alkaline solution is thought to be impure BijO,.

and

Uses of the Group V A Elements

21.10 Industrial

The most important industrial uses of the group V

A elements and their compounds

are:

1.

Nitrogen.

The manufacture

of

ammonia

elemental nitrogen. Smaller amounts of

constitutes the principal

use of

are used in the production of calcium

cyanamid (CaCNj). Because is relatively unreactive, gaseous nitrogen is used as an inert atmosphere in place of air for chemical and metallurgical processes that must be run in the absence of oxygen. The gas is used in food processing and packaging (coffee, for example) to prevent the spoilage and deterioration that

is

brought about by exposure to atmospheric oxygen. Liquid nitrogen has in cryogenic work (low-temperature work) to avoid the danger

replaced liquid air

associated with contact between the oxygen of the air and combustible materials.

Liquid nitrogen

is

used for the preparation and transportation of frozen food

as well as the transportation of perishable food.

The metal electrical

nitrides are high-melting, very hard, chemically unreactive,

conductors, and their uses

reflect these properties.

They are used

and

in the

fabrication of refractory materials (heat-resistant materials), abrasives, grinding

and cutting tools, and semiconductors. The major industrial use of ammonia is the manufacture of nitric acid. Large amounts of ammonia are converted into various ammonium salts, used principally as fertilizers, and ammonia itself is used directly as a fertilizer. The compound is used to make hydrazine (H2NNH2. a component of rocket fuels) and urea (H2NCONH2, a fertilizer and an ingredient in the manufacture of plastic resins). Ammonia is employed in the Srhn- process (for the manufacture of sodium carbonate), in petroleum, paper, rubber, and textile technology, and in the manufacture of dyes, drugs, and explosives. The major use of nitric acid is in the production of nitrates, which are principally used in fertilizers

organic

and

compounds produce

explosives. a

number

The

reactions of nitric acid with certain

of commercial explosives (nitroglycerine,

and trinitrotoluene are examples). Nitric acid has a large number minor applications; all the nitrogen compounds of commerce are produced from nitric acid and or ammonia.

nitrocellulose,

of

21 .10

Industrial

Uses

of the

Group V A Elements

2.

Most elemental phosphorus

Phosphorus.

oxide, phosphoric acids,

and the

make matches, warfare

agents,

is

make phosphorus(V) Some phosphorus is used to

used to

salts of these acids.

and rodent poisons. Metal phosphides, and and copper. Some phosphides (GaP, BP, AlP, and InP) are semiconductors. Phorphorus(V) oxide is used to make phosphoric acids, phosphates, and phosphorus

itself,

are used as alloying ingredients in the metallurgy of steel

flame-retardant materials. is

The oxide

a desiccant (drying agent),

and

is

functions as a catalyst for

used as a dehydrating agent

some reactions, some organic

in

reactions.

Phosphoric acid

The

a fertilizer.

acid

is is

used

in the

manufacture of

as directly as

fertilizers as well

used in phosphatizing, a process that produces a corrosion-

on iron objects and is applied prior to painting the objects. used in polishing aluminum objects, in electropolishing steel articles, and in several ways in food technology. Phosphates are used as fertilizers and in food processing, drugs, detergents, scouring powders, and toothpastes. Ammonium phosphate functions as a flame retardant and is used in textile technology. resistant coating

Phosphoric acid

is

These elements are not used in the amounts Arsenic, Antimony, and Bismuth. nor to the extent that nitrogen and phosphorus are. Arsenic, antimony, and bismuth are used in the production of a wide range of alloys. Either antimony

3.

or bismuth

is

a usual

component of

metal alloys expand upon freezing.

Molten type-

the alloys used as type metal.

When

they are used to

make

type for printing,

and bismuth and electrical fuses. Arsenic is used in lead and copper alloys as a hardener and arsenic compounds find use as insecticides, rodent poisons, and weed killers. An important the casts obtained have sharp edges. Low-melting alloys of antimony

are used to

use of

all

make

safety plugs for boilers, fire sprinklers, solders,

three elements

is

the fabrication of semiconductors.

Summary The

topics that have been discussed in this chapter are

oxyacids diagrams. oxides,

The group V A elements: their physical properties, allotropic forms, occurrence, and industrial preparation.

and

their

electrode-potential

salts,

1.

2.

The nitrogen

4.

Industrial uses of the

group V

A

elements and their

compounds.

cycle.

Compounds of the group V A elements: nitrides, phosphides, hydrogen compounds, halogen compounds.

3.

Key Terms Some

of the

more important terms introduced

in

Cyanamid process

this

in this list

may

be located

in the text

by use of the index.

Chapter 21

The Nonmetals,

21.5)

in

Arc process (Section 21 .7) A nitrogen-fixation process in which nitrogen oxide, NO, is produced by the reaction of nitrogen and oxygen in an electric arc.

562

(Section

A

which calcium cyanamid. CaNCN. calcium carbide, CaC,, and nitrogen.

chapter are listed below. Definitions for terms not found

Haber process (Sections 21.2 and

nitrogen-fixation

21.5)

fixation process for the preparation of

nitrogen and hydrogen.

Part

III

:

The Group V A Elements

prepared from

is

A

nitrogen-

ammonia from

Isomers (Section 21.6) Substances that have the same molecular formula but differ in the way the constituents are arranged into a molecule.

Nitrogen cycle (Section 21 .2) A group of natural and processes by which nitrogen is constantly being removed from the atmosphere and returned to it.

artificial

Nitrogen fixation (Section 21.2) A process in which is converted into a nitrogen-containing

Ostwald process (Section 21.7) A commercial process for the manufacture of nitric acid ammonia is catalytically oxidized to NO, the is reacted with O, to form NO,, and the NO2 is reacted with water to form HNO3. ;

NO

Superphosphate

(Section 21.8)

fertilizer

Ca(H2P04)2 and CaSO^ used by the reaction of sulfuric

A

mixture of

and produced acid and phosphate rock. as a fertilizer

elemental nitrogen

compound.

Problems* Discuss how the properties of the elements of group V and their compounds change with increasing atomic number.

21

.1

A

Write all the equations that you can for the reactions of elementary nitrogen. Explain the low chemical reactivity of N,. 21.2

21.3

Why

is

crops sown

it

advantageous

in a

21.4 In 1892

for a

farmer to rotate the

Lord Rayleigh observed

that the density of

The normal boiling point of the compound ethylene diamme, H2NCH,CH,NH,. is 17 C and that of propyl amine. CH3CH ,CH,NH," is 49 C. The molecules, however, are similar in size and molecular weight. What reason can you give for the difference in boiling point? 21.5

1

the comparative reactivities, solubilities, and electrical conductivities of the three allotropic forms of phosphorus in terms of molecular structure. 21.6 Discuss

Write chemical equations for the separate reactions of each of the following with H,0. with aqueous NaOH, and with aqueous Hcf: (a) P/)„. (b) Sb/)„. (c) BiPj. 21.7

21.8

What

(e)

NH3(g) + 02(g)

(f)

NH3(g)

(g)

NH3(g) + V(s)

— ^ ^—

are the characteristics of the three types of

(h)

NH3(1) + Na(s)

•

NH3(aq) + H^(aq) NH3(aq) + HjO + CO,(aq) NH3(g) + 0,(g)

(c)

(d)

Describe,

using drawings, the structures of these

Why

is

and properties of

the boiling point of

NH,

high

in

21.15 State the products of the thermal decomposition of (a)

NH4NO3.

(e)

NaNj.

21.16 Write

NH4NO2,

equations

that produces

H,

with (c)

in

(a)

HCl

NF3 and

Pb(N03)2.

(c)

the

for

;

(b) the

NH3

in

NaNOj,

oxidation-

with

OCl"

reaction of

solution that produces

NH4F

(d)

following

the reaction of

NjH^ and CL

the electrolysis of

duces

anhydrous

NO2

[NHjOHjCl;

HF

that pro-

Hj.

21.17 Draw electronic structures for orthophosphoric, phosphorous, and hypophosphorous acids. Tell how these

in

explain

the

when each

number of H'^(aq) compounds

of these

dissociated is

may

21.19 Write

(i)

all

acids

be obtained by the hydrolysis of P40,o.

an equation

following with (d)

dissolved

water.

21.18 Write equations for the preparations of

SbHj?

The appendix contains answers

(b)

reduction reactions:

comparison to the boiling points of PH,. AsH,, and

*

as the source of nitrogen.

21.14 (a) List the oxides of nitrogen, (b) Write a chemical equation for the preparation of each of them, (c) Diagram the electronic structure of each oxide.

that

21.10 Discuss the preparation, structure,

ammonia.

N,

an equation for the reaction of HNO3 with Cu.(b) Zn,(c) P^Oio.W NH3, (e) Ca(OH)2.

per mole

ite.

with

21.13 Write (a)

structures

substances.

-^^^^

HCl(g)

-f

nitric acid starting

Boron nitride, BN. exists in two crystalline modifications one resembling diamond and the other graph-

—

of

21.12 Write a series of equations for the preparation of

nitrides? 21.9

reactions

•

NH3(aq) + Ag^(aq)

(b)

HiO, CO,,

and O,) was different from the density of nitrogen prepared from NH4CI and NaNO,. On the basis of this observation, Rayleigh and William Ramsey isolated argon (1894). Assume that air is l.(X)"„ Ar, 78.00",, N,, and 21 .00",, O, (by volume). How should the densities of the two "'nitrogen" samples compare?

following

the

for

(a)

given field?

nitrogen prepared from air (by the removal of

Write equations

21.11

ammonia

CaNCN,

(e)

water:

NCI3,

each of the Li3N, (b) AIN, (c) Ca3P2, PCI,, (g) PCI5, (h) H4P2O7,

for the reaction of (a) (f)

NO,.

to color-keyed problems.

Problems

563

:

What is the acid anhydride of each of H2N.O,, (b) HNO,, (c) HNO3, H3P36,, (I) H3PO3, (g) H3ASO4

21 .20 (a) (e)

Diagram the resonance forms of (c)NO. (d) NO3, (e)HN03,(f) HN3. 21.22 Write

(e)

(a)

NO2

H,NNH„

(b)

H,NOH.

(c)

.(b)

As,

(b)

(a)

NajAs,

NOj,

(f)

PH4I.

As,Os,

NH3,

structures

for cis-

and trans-N2F2-

the shape of each of the NH,-, (c) PCi;, (d) PCI 6,

for (c)

the

NO,

SbClj,

the commercial processes used manufacture of (a) P^, (b) NH3, (c) H3PO4. (d) (e) superphosphate fertilizer, (() AS4.

HNO3,

(()Sb(OH)^. 21.27 Describe

NaNO,,

reactions (e)

HCl with (d) NH3,

of

O,

with

with

Sb^Sj.

(d)

As^O,,.

Chapter 21

(a)

NH4Cl,(b) P4,

21.29 Describe

the reactions of water with

MgjBi.,

following: (e)

(b)

21.28 Write equations for the reactions of

equations for P4, (c) PCI3, (d)

(b)

Draw Lewis

21.26 Describe

(NH4)2C63.

21.24 Write equations

564

21.25

H4P2O7,

(a)

equations for the reactions of

21.23 Write (a)

(d)

.'

21.21

(a)

the following

(e)

(a)

P4. (b)

the

POCI3,

III

:

aqueous As40e,(d) Sb,©,.

the

NaOH

shape of the following molecules: (c)

P40^.

SbClj.

The Nonmetals. Part

(c)

in

The Group V A Elements

(d)

P40,o.

CHAPTER

THE NONMETALS, PART IV: CARBON, SILICON, BORON, AND THE NOBLE GASES

22

In this chapter, the nonmetals of group IV A (carbon and siHcon), group III A (boron alone), and group 0 (the noble gases) are considered. These discussions complete the survey of the nonmetals that was begun in Chapter 19.

CARBON AND SILICON Carbon,

silicon,

germanium,

and lead make up group IV A. The compounds compounds of any other element with

tin,

of carbon are more numerous than the

the possible exception of hydrogen. In fact, approximately ten

contain carbon are

known

for every

compound

that does not.

compounds that The chemistry of

the compt)unds of carbon (most of which also contain hydrogen)

of

22.1

.'.p,.....^

i\ (see

^

Chapter

Properties of the Group IV

The

transition

is

the subject

26).

A Elements

from nonmetallic character

to metallic character with increasing

atomic number that is exhibited by the elements of group V A is also evident in the chemistry of the IV A elements. Carbon is strictly a nonmetal (although graphite is an electrical conductor). Silicon is essentially a nonmetal in its chemical behavior, but

Germanium

is

its

electrical

a semimetal;

and physical properties are those of its

properties are

more

a semimelal.

metallic than nonmetallic.

Tin and lead are truly metallic, although some vestiges of nonmetallic character remain (the oxides and hydroxides of tin and lead, for example, are amphoteric). Carbon exists in network crystals with the atoms of the crystal held together by covalent bonds (see Section 22.2). A large amount of energy is required to rupture some or all of these bonds in fusion or vaporization. Carbon, therefore, has the highest melting point and boiling point of the family (see Table 22. ). 1

The

heaviest

member

of the family, lead, exists

in a typical metallic lattice.

The two

forms of the intervening members show a transition between the extremes displayed by carbon and lead, and this accounts for the trend in melting points and boiling points of the elements (see Table 22.1).

crystalline

The crystalline forms of silicon and germanium are similar to the diamond. The bonds are not so strong as those in the diamond, however, and silicon and

565

,

^^^^Bable 22.1 Some

properties of the group IV

U

Prop6rty melting point (°C) hnilinn nnint (°C\

i

A eEements

Germanium

Silicon

b U Ti

Lead

Tin

3570

1420

959

4827

2355

2700

2360

1755

77

117

122

141

154

1090 2350 4620 6220

782 1570 3230 4350

782 1530 3290 4390

atomic radius (pm)

Oil

f

ionization energy (kJ/mol) first

second third

fourth

electronegativity *

714 1

*HJU

2940 3800

2.0

1.9

2.6

704 1

A^^n

3090 4060

2.0

2.3

Diamond

germanium

may

are semiconductors (the

diamond

is

not). Silicon

and germanium

be used to prepare impurity semiconductors (see Section 23.2) which are

employed

The

Although one modification of form of tin is metallic.

in transistors.

structure, the principal

has a diamond-type

tin

electronic configurations of the elements are listed in Table 22.2.

With

the possible exception of carbon, the assumption of a noble-gas configuration

through the formation of a negativities of the

The

4— ion by electron gain is not A elements are generally low

group IV

observed.

ionization energies of the elements (see Table 22.1)

required for the removal of

The

Table

(see

show

electro-

22.1).

that the energy

four valence electrons from any given element

all

extremely high. Consequently, simple

4+

ions of group IV

A

is

elements are un-

known. Germanium, tin, and lead appear to be able to form d^°s'^ ions by the loss of two electrons. Of these 2 ions, however, only some of the compounds of Pb'"^ (such as PbF2 and PbCl2) are ionic, the compounds of Ge^^ and Sn*"^ are predominantly covalent, and those of Ge^"^ are relatively unstable toward disproportionation to the 4-1- and 0 states. In the majority of their compounds, the IV A elements are covalently bonded. Through the formation of four covalent bonds per atom, a IV A element attains -I-

the electronic configuration of the noble gas of

AB4

are tetrahedral. All the IV

a few

compounds of

A

this type are

they, with the exception of

its

period.

Compounds

of the type

elements can form such compounds, but only

known

for lead (PbF4,

PbC^, and PbH^) and

PbF4, are thermally unstable.

bonds saturates the However, the other members of the group have empty d orbitals available in their valence levels and can form species in which the atom of the IV A element exhibits a covalence greater than four. The ions SiF^ ~ GeCl^ " SnBr^ ~ Sn(OH)g~, Pb(OH)^-, and PbCl^" are octahedral. The most important way in which carbon differs from the remaining elements of group IV A (as well as from all other elements) is the pronounced ability of carbon to form compounds in which many carbon atoms are bonded to each other in chains or rings. This property, called catenation, is exhibited by other elements near carbon in the periodic classification (such as boron, nitrogen, phosphorus, sulfur, oxygen, silicon, germanium, and tin) but to a much lesser extent than carbon; this property of carbon accounts for the large number of organic compounds. In the case of carbon, the formation of four covalent

valence

level.

,

566

Chapter 22

The Nonmetals, Part

IV:

Carbon, Silicon, Boron, and the Noble Gases

,

In group IV A the tendency for self-linkage diminishes markedly with increasing atomic number. The hydrides, which have the general formula E„H2„ + 2 (where E is a group IV A element), illustrate this trend. There appears to be no limit to the number of carbon atoms that can bond together to form chains, and a very large number of hydrocarbons are known. For the other IV A elements, the most

complex hydrides that have been prepared are Si6Hi4, Ge9H2o, SnjHg, and PbH4. The carbon-carbon single bond energy (347 kJ/mol) is much greater than that of the silicon-silicon bond (226 kJ/mol), the germanium-germanium bond (188 kJ/mol), or the tin-tin bond (151 kJ/mol). In addition, the C C bond is about as strong as any bond that carbon forms with any other element. Typical bond energies for carbon bonds are: C C, 347 kJ/mol; H, 414 kJ/mol; and C CI, 326 kJ/mol. In 335 kJ/mol; C Si bond is much weaker than the bonds that silicon forms comparison, the Si Si, 226 kJ/mol with other elements. Some bond energies for silicon bonds are Si Silicon, thereSi CI. 391 kJ/mol. Si— O, 368 kJ/mol: Si H. 328 kJ/mol; and to itself elements than to bond bond to other fore, has more of a tendency to multiple form ability to is its pronounced characteristic of carbon Another bonds with itself and with other nonmetals. Groupings such as

—

—

—

—

:

^C=C^, / \

— C=C—

are frequently encountered.

— C^N,

,

No

other IV

A

C=0, /

and

\=S /

element uses p orbitals to form n

bonds.

Only

the truly nonmetallic

members of

the family, carbon

and

silicon, are

treated in the sections that follow.

22.2

Occurrence and Preparation Carbon and Silicon

of

Carbon constitutes approximately 0.03% of the earth's crust. In addition, the atmosphere contains 0.03% CO2 by volume and carbon is also an important constituent of all plant and animal matter. The allotropes of carbon, diamond and graphite, as well as impure forms of the element such as coal occur in nature. In combined form the element occurs in compounds with hydrogen (which are called hydrocarbons and are found in natural gas and petroleum), in the atmosphere as CO2, and in carbonate minerals such as limestone (CaCOj), dolomite (CaCOj •

MgCOj),

siderite

(FeCOj) witherite (BaCOj), and malachite (CuC03-Cu(OH)2).

22.2

Occurrence and Preparation

of

Carbon and

Silicon

567

Figure 22.3

In

Resonance forms

for a

fragment

diamond each carbon atom

is

of a grapfiite layer

bonded through

sp^ hybrid orbitals to four

other carbon atoms arranged tetrahedrally (see Figure 22.1). Strong bonds hold

network crystal together. Furthermore, all valence electrons of each carbon are paired in bonding orbitals the valence level of each carbon atom can hold no more than eight electrons. Thus the diamond is extremely hard, high melting, stable, and a nonconductor of electricity. Whereas the diamond is a colorless, transparent material with a high refracthis

—

atom

is a soft, black solid with a slight metallic luster. The graphite composed of layers formed from hexagonal rings of carbon atoms (see Figure 22.2). The layers are held together by relatively weak London forces; the distance from carbon atom to carbon atom in adjacent planes is 335 pm as compared with a distance of 141.5 pm between bonded carbon atoms of a plane.

graphite

tivity,

crystal

Since

is

it is

slippery

easy for the layers to shde over one another, graphite

feel. It is less

is

soft

and has a

dense than diamond.

The nature of the bonding in the layers of the graphite crystal accounts for some of the properties of this substance. Each carbon atom is bonded to three other carbon atoms, and all the bonds are perfectly equivalent. The C C bond distance in graphite (141.5 pm) compared with that in the diamond (154 pm)

—

suggests that a degree of multiple bonding exists in the former. Graphite

represented as a resonance hybrid (see Figure 22.3) in which each

bond

bond.

568

Cfiapter 22

The Nonmetals, Part

IV:

Carbon, Silicon, Boron, and

ttie

Noble Gases

may is

be

a 1^

Each

C atom

in graphite

forms a bonds with three other

The

use of sp^ hybrid orbitals.

structure, therefore,

is

C atoms

through the

planar with the three a

bonds of each atom directed to the corners of equilateral triangles. This bonding accounts for three of the four valence electrons of each C atom; the fourth, a p electron, If

is

not involved in

bond formation

cr

(see

Figure 22.4a).

only two adjacent atoms in a molecule had this electron arrangement, the

p electrons would pair to form a localized n bond, and thus the atoms would be joined by a double bond. However, in graphite each C atom has an additional p electron, and the resonance forms depict the possibility of forming conventional double bonds in three ways. Since each p orbital overlaps with more than one other p orbital, an extended n bonding system that encompasses the entire structure forms (see Figure 22.4b). The electrons in this n bonding system are not localized between two atoms but are free to move throughout the entire layer. Hence, graphite has a metallic luster and is an electrical conductor. The additional

conductivity

is

fairly large in a direction parallel to the layers

but

is

small in a

direction perpendicular to the planes of the crystal. Silicon, which constitutes approximately 28% of the earth's crust, is the second most abundant element (oxygen is first). The element does not occur free in nature rather, it is found as silicon dioxide (sometimes called silica) and in an enormous

variety of silicate minerals. Silicon

is

prepared by the reduction of silicon dioxide by coke at high temper-

atures in an electric furnace

Si02(l)

If a larger

+

2 C(s)

—

Si(l)

+

quantity of carbon

2

is

CO(g)

employed,

silicon carbide (SiC,

"carborundum") is produced rather than silicon. The only known modification of silicon has a structure Crystalline silicon silicon are

is

a gray, lustrous solid that

not so strong as those

in

is

is

similar to

a semiconductor.

is

called

diamond.

The bonds

in

diamond, and the bonding electrons are not

may

be thermally excited to a energetically close to the valence band (see Sec-

so firmly localized. Evidently, in silicon, electrons

conductance band that

which

tion 23.2).

Very pure silicon, which is used in transistors, is prepared by a series of steps. First, impure silicon is reacted with chlorine to produce SiC^. The tetrachloride, a volatile liquid, is purified by fractional distillation and then reduced by hydrogen to elementary silicon. This product is further purified by zone refining (see Figure and 23.7). In this process a short section of one end of a silicon rod is melted, this melted zone is caused to move slowly along the rod to the other end by the

22.2

Occurrence and Preparation

of

Carbon and

Silicon

569

H movement of the

heater.

Pure

silicon crystalHzes

from the melt, and the impurities

are swept along in the melted zone to one end of the rod which

sawed

is

subsequently

and discarded.

off

22.3 Carbides

and

Silicides

A large number of carbides are known. These compounds may be made by heating the

appropriate

metal

or

oxide

its

with

carbon,

carbon monoxide, or a

hydrocarbon.

The

saltlike carbides are

made up

of metal cations together with anions that

and II A metals, as well as Cu"^, Ag"^, Au"*", Zn^"^, and Cd^"^, form carbides that are sometimes called acetylides because they contain the acetyUde ion, C\~ which has the structure contain carbon alone. The

I

A

,

[:C^C:]^-

Upon

hydrolysis, acetylides yield acetylene,

C2H2:

H— C=C^H For example,

+ 2H2O

CaC^is)

Calcium carbide

CaNCN(see

is

—

Ca(OH)2(s)

used

in the

+ C2H2(g)

commercial production of calcium cyanamid,

Section 21.5).

Beryllium carbide, Be^C, and aluminum carbide, AI4C3, contain the methanide ion, C'^

,

which upon hydrolysis

yields

methane, CH^.:

H

H— C— H For example,

AUCsls)

+ I2H2O

—

4Al(OH)3(s) + 3CH4(g)

formed by the transition metals and consist of metallic crystals with carbon atoms in the holes between the metal atoms of the crystal structure (the interstices). Examples of this type of carbide include TiC, TaC, WjC, VC, and MojC. Because interstitial carbides are very hard, high-melting, and chemically unreactive, they are used in the fabrication of cutting tools. They resemble metals in appearance and electrical conductivity. The bonding in the covalent carbides SiC and Be4C is completely covalent. These compounds are very hard, chemically unreactive, and do not melt even at Interstitial carbides are

high temperatures. Because of these properties, they are used as abrasives the place of industrial diamonds). Silicon carbide (carborundum) in

an

Chapter 22

The Nonmetals. Part

IV:

Carbon, Silicon, Boron, and the Noble Gases

and

electric furnace.

The SiC

(in

produced by

C

the reaction of Si02

570

is

crystal consists of a

diamond-like tetrahedral network formed from alternating Si and C atoms. Boron carbide, B4C, which is harder than SiC, is made by the reduction of B2O3 by carbon in an electric furnace.

molten metals, and in many of these instances, produced (Mg2Si, CaSiz, LijSi, and FeSi are examples). Although probably none of the silicides is truly ionic, some of them react with water to produce hydrogen-silicon compounds that are called silicon SiHcon dissolves

definite

almost

in

compounds

all

called silicides are

hydrides or silanes.

The silanes are compounds compounds in which n equals hydrocarbons that conform to

conform to the general formula Si„H2„ + 2; known. They resemble structurally the the general formula C„H2„ + 2 and are called the alkanes (see Figure 22.5). Although there is presumably no limit to the number of C atoms that can join together to form alkanes, the number of Si atoms that can bond together to form silanes appears to be limited because of the comparative weakness of the Si Si bond (see Section 22.1). In addition, the alkanes are much less reactive than the silanes, which are spontaneously flammable in air: that

1

to 6 are

—

2Si2H6(g)

+

— C bonds

The C

—

702(g)

4 Si02(s)

alkanes are

in the

+ 6H20(g) all

hydrocarbons bonds between C atoms. Silicon bonds are unknown. The hydrocarbons single bonds, but there are

(acetylene for example) which contain multiple

hydrides that contain multiple

Si to Si

are discussed in Chapter 26. (b)

Arrangement

Figure 22.5

22.4 Oxides and Oxyacids

of

C and

atonns

Si

(b)

Carbon monoxide is formed by the combustion of carbon oxygen at high temperatures (approximately 1000 C): 2C(s)

It is

also

in a limited

supply of

in (a)

of

CH4 and SiH4 and

C,H„ and SuH^. (Larger

spheres, C or spheres, H.)

Si

;

smaller

+ 02(g)—*2CO(g) CO2 and C

produced by the reaction of

constituent of water gas (see Section

19.2).

at high

temperatures and

Carbon monoxide

is

is

a

isoelectronic

with nitrogen:

©:C=0:® and has two n bonds and one a bond joining the atoms. Carbon monoxide burns in air: 2

CO(g)

+

—

02(g)

Since this reaction

is

2

C02(g)

AH = - 283 kj

highly exothermic, carbon

monoxide can be used

as a fuel.

The compound reacts with the halogens in sunlight to produce the carbonyl halides (COX2) and with sulfur vapor at high temperatures to form carbonyl sulfide (COS). Carbon monoxide high temperatures,

it

is

used as a reducing agent in metallurgical processes. At many metal oxides to yield the free metal and

reacts with

CO2: FeO(s)

+ CO(g)

—

Fed)

+ C02(g)

22.4

Oxides and Oxyacids

of

C and

Si

571

The catalyzed

reactions of carbon

monoxide and hydrogen are commercially

important for the production of hydrocarbons and methanol (see Section 26.6). Certain transition metals and transition metal salts react with CO to produce in

C atom

which the

CO

molecule uses the unshared electron pair of the

bond formation. Examples of metal carbonyls include Ni(CO)4,

for

a

tetrahedral molecule; Fe(CO)5, a trigonal bipyramidal molecule; and CrlCO)^,

an octahedral molecule. Carbon monoxide is poisonous because it combines with the iron atom of the hemoglobin of the blood, thus preventing the hemoglobin from combining with oxygen (see Section 24.1 for a more complete discussion).

formed by the complete combustion of carbon or compounds It is also produced by the reaction of carbonates or hydrogen carbonates with acids,

Carbon dioxide

is

of carbon (notably the hydrocarbons).

+ 2H^(aq)

CaC03(s)

—^Ca^

+ Caq)

+

C02(g)

+ H2O

and by heating carbonates,

CaC03(s)

—CaO(s) +

The molecule

is

linear

C02(g)

and nonpolar:

:6=C=-6: In the oxygen-carbon-dioxide cycle in nature, oxygen is removed from the air and CO2 is added to the air by the respiration of animals, the combustion of fuels, and the decay of organic matter. The reverse occurs in photosynthesis. In the formation of carbohydrates by plants, CO2 is removed from the air and O2 is released to the air. The energy for photosynthesis is supplied by sunlight and the process is catalyzed by the green coloring matter of plants, chlorophyll. The geochemical equilibrium between CO,, H2O, and limestone, CaC03, also serves to control the level of

CaC03(s)

CO2

in the air:

+ C02(g) + H20^Ca^^(aq) + 2HC03-(aq)

At the present time, the rate at which CO, is being added to the atmosphere exceeds the rate at which it is being removed by other processes. The concentration of CO2 in the atmosphere, therefore, is steadily increasing. The concentration today is about 10% higher than it was a century ago. Carbon dioxide is moderately soluble in water. It is the acid anhydride of carbonic acid, H2CO3. The acid, however, has never been obtained in the pure state.

than is

Solutions of

1%

CO2

in

of the dissolved

water consist mainly of dissolved

CO2

a weak, diprotic acid,

is

in the

form of

and equations

for

H2CO3

its

CO2

molecules

—

less

molecules. Carbonic acid

ionizations are probably best

written

C02(aq)

+ H2O

HC03-(aq)

Two

series

^ ^

H^(aq)

+ HC03(aq)

H^(aq)

+ COr(aq)

the hydrogen carbonates, such as

NaHCOj

structure of the carbonate ion has been discussed in

Figure

572

Na2C03 and and Ca(HC03)2 The Sections 7.1 and 7.6 (see

of salts are formed: the normal carbonates, such as

CaCOj, and 7.18).

Chapter 22

The Nonmetals. Part

IV;

Carbon, Silicon, Boron, and the Noble Gases

Mining trona ore in Green River, Wyoming. The ore is the principal source of sodium carbonate, which is used in the manufacture of glass, soap, detergents, paper, and other chemicals. Allied Corporation.

The only

well-characterized oxide of silicon

is

SiOj. In contrast to the oxides

of carbon, which are volatile molecular species held together by London forces in the solid state, Si02 forms very stable, nonvolatile, three-dimensional network

~ 1700 C). One of the three crystal modifications of SiOj be considered to be derived from the diamond lattice with silicon atoms replacing carbon atoms and an oxygen atom midway between each

crystals (melting point,

has a lattice that

may

pair of silicon atoms.

The chemistry of

silicon

linkage. Silicon dioxide

is

dominated by compounds that contain the

is

the product of the reaction of the elements;

it

Si is

— also

produced by reaction of the spontaneously flammable silicon hydrides with air. Hydrous SiOj is the product of the hydrolysis of many silicon compounds. Silicon dioxide occurs in several forms in nature; among them are sand, flint, agate, jasper, onyx,

and quartz.

Silicon dioxide isolated.

is

an acidic oxide; however, no acids of

The oxide does not

silicon

have ever been

react directly with water; acidification of a water

solution of a soluble silicate yields only hydrous SiOj. Silicates

may be made by

heating metal oxides or metal carbonates with SiOj. Certain silicates of the I A metals (with a molar ratio of silicon dioxide to metal oxide of not more than 2 to 1)

are water soluble.

A

large

number of

silicates

of various types occurs

in nature.

The

basic unit

of all silicates is the tetrahedral Si04 group (see Figure 22.6). The simple ion SiOl" occurs in certain minerals (zircon, ZrSi04, is an example):

Quartz crystals. U.S. Geological Survey.

:0:

:0— Si— O: :0:

22 4

Oxides and Oxyacids

of

C and

Si

573

Figure 22.6

Schematic representation

of the

arrangement

of

atoms

in

the silicate ions

The four bond pairs of the Si atom are arranged in a tetrahedral manner. The other silicate anions contain more than one SiOj. tetrahedron joined together by bridge oxygen atoms (that is, oxygen atoms that are shared by two tetrahedra). The mineral thortveitite (ScjSi^Ov) contains the Si207~ ion, which is formed by two Si04 tetrahedra joined by a single bridge oxygen atom (see Figure 22.6):

:0:

:o:

:0— Si— O— Si :o:

574

Chapter 22

The Nonmetals, Part

IV:

O:

:o:

Carbon. Silicon, Boron, and the Noble Gases

Cyclic silicate anions are known in which tetrahedra are joined in a circle with each tetrahedron sharing two bridge oxygen atoms. The SijOg" anion in bentonite (BaTiSijOg) consists of three Si04 tetrahedra joined in a circle, and in the SieOjg" ion found in beryl (BcjAljSigOig) there are six (see Figure 22.6).

oxygen atoms of each Si04 tetrahedron are used as bridge atoms, a The anion (Si205")„ occurs in talc, Mg3(Si205)(OH)2.

If three

sheetlike anion results.

Because of the layer structure, this material feels slippery. Occasionally, aluminum atoms take the places of some of the silicon atoms in certain anions. The hypothetical AIO4 tetrahedron would have a 5— charge; consequently, such substitutions increase the negative charge of the anion. Muscovite, KAl3Si30io(OH)2 contains a sheetlike alumino-silicate anion with one-fourth of the silicon atoms of the (81205" )„ structure replaced by aluminum atoms. If all four oxygen atoms of each Si04 tetrahedron are used as bridge atoms, the three-dimensional, diamond-like network crystal of Si02 results. If some of the Si atoms of this SiOj structure are replaced with Al atoms, an aluminosilicate anion is produced. Since Si atoms (each with four valence electrons) are replaced by aluminum atoms (each with three valence electrons), additional electrons (which provide the charge on the anion) are required to satisfy the bonding requirements of the structure. Framework aluminosilicates. such as the feldspars and zeolites, are of this type.

Glass

is

carbonates.

a mixture of silicates

Common

Special glasses

may

be

(such as AI2O3, B2O3,

mixture

made from

made by fusing Si02 with metal oxides and is made from Na2C03, CaCOs, and SiOz-

soda-lime glass

made by

and basic oxides complex aluminosilicate

the addition of other acidic

PbO, and K2O). Cement

is

a

limestone (CaC03) and clay (H4Al2Si209).

and Nitrogen Carbon

22.5 Sulfur, Halogen,

Compounds

of

disulfide. CS2, is a volatile liquid prepared commercially by heating carbon and sulfur together in an electric furnace. The structure of the molecule

Carbon is

similar to that of

CO2

:S=C='s: good solvent for nonpolar substances such as waxes and grease. commercially Its use, however, is limited by its toxicity and flammability. It is used tetrachloride. carbon of in the manufacture of rayon and in the production Carbon tetrachloride is made commercially by heating carbon disulfide with

The compound

is

a

chlorine:

CS2(g)

+

3Cl2(g)

—

CCUg) +

S2Cl2(g)

Carbon tetrachloride, which is a liquid under ordinary conditions, is not flammable fire and is a good solvent for many nonpolar materials. It was formerly used in extinguishers and in dry cleaning, but these uses are

now

illegal

because of the

compound. CF4, which can be obtained by the fluorination of almost chlorine-fluorine any carbon-containing compound, is a very stable gas. Mixed They are -freons." the compounds of carbon (in particular CCI2F2) are called

toxicity of the

Carbon

tetrafluroride.

refrigerants. very stable, odorless, nontoxic gases that are used principally as

22.5

Sulfur, Halogen,

and Nitrogen Compounds

of

Carbon

575

— Carbon tetrabromide, CBr4, and carbon

tetraiodide, CI4, are solids.

They

are

thermally unstable, presumably because of the difficulty of the carbon atom in

accommodating four large atoms around itself. Other halogen-containing compounds of carbon are discussed in Section 26.5. There are many compounds in which carbon is bonded to nitrogen. Hydrogen cyanide, HCN, is commercially prepared by the catalyzed reaction of methane (CH4), ammonia, and oxygen at high temperatures:

+

2 CH^lg)

2

NH3(g) +

3 02(g)

—

2

HCN(g) +

HMg)

6

The compound is a highly poisonous, low-boiling liquid (boiling point, 26.5"C) and is used in the manufacture of plastics. Hydrogen cyanide readily dissolves in water to produce solutions of hydrocyanic acid, a weak acid. Cyanides, such as sodium cyanide (NaCN), are produced and CO: by neutralization of this acid. The cyanide ion is isoelectronic with

[:C=N:]The ion has a strong tendency to form covalent complexes with metal cations such as Ag(CN)2 Cd(CN)r, Ni(CN)r, Hg(CN)r. Fe(CN)r, Fe(CN)r, and ,

Cr(CN)r. Mild oxidation of the cyanide ion in aqueous solution produces the cyanate OCN " The corresponding acid, cyanic acid HOCN), is not stable in aqueous solution and decomposes to CO2 and NH3. An analogous ion, the thiocyanate

ion,

ion,

{

.

SCN",

is

prepared by melting a mixture of a

I

A

cyanide with sulfur.

BORON the group III A elements (boron, aluminum, gallium, indium, and thallium), boron alone is a nonmetal.

Of

22.6 Properties of the

Group

III

A Elements

The electronic configurations of the elements are listed in Table 22.3 and properties of the elements are given in Table 22.4. The boron atom is much smaller than those of the other elements of the group. This difference accounts for the sharp dis-

between the nonmetallic boron and the other group members, which are metallic. In general, metals have atomic radii that are greater than 120 pm and nonmetals have atomic radii that are less than 120 pm. The atomic radii of Ga, In, and Tl are influenced by the electronic inner-building of elements that immediately precede them in the periodic table (particularly in the case of Tl, which follows the lanthanides). Atomic radius, therefore, does not rapidly and regularly increase with increasing atomic number in the series from Al to Tl. tinction in properties

None

of the

III

A elements shows the slightest tendency to form simple anions.

The most important oxidation

would seem reasonable The first three ionization energies of boron are relatively high because of the small size of the boron atom. As a result, the B^^ ion is never formed. The energy required to remove from the

576

Chapter 22

«,s'

state of this

group

is

3

-t-

,

as

np^ electron configuration of the valence level.

The Nonmetals, Part

IV:

Carbon, Silicon. Boron, and the Noble Gases

Table 22.3

Electronic configurations of

—

-

B

W"

III

A elements

-

Z

Element

group

tlie

-

1

s

J5

3s

ZjJ

-

3p

3d

4s

4p

5

2

2

1

Al

13

2

2

6

2

1

Ga

31

2

2

6

2

6

10

2

1

In

49

2

2

6

2

6

10

2

6

10

Tl

81

2

2

6

2

6

10

2

6

10

Some

"table 22.4

properties of the group

ill

5p

14

2

1

2

6

5d

6s

6p

10

2

1

A elements Aluminum

Boron

Property

4^

4a'

Gallium

Indium

Thallium

1

melting point {°C)

2300

659

30

155

304

boiling point (°C)

2550

2500

2070

2100

1457

80

125

125

150

155

52

62

81

95

801

579

2422 3657

1814

579 1968 2953

560 1814 2692

589 1959 2866

atomic radius (pm) ionic radius,

ionization

M'" (pm)

energy (kJ/mol)

first

second third

electrode potential.

results

when

P

The value of P depends upon This conclusion of a d^ ion there

is

is

the metal ion

;

A„

is

different for each complex.

also valid for complexes o(d^, d^,

and

only one possible configuration:

six electrons

d'' ions.

For a complex paired in the

and two unpaired electrons in the orbitals. Likewise, complexes of d'^ and d^° ions exist in only one configuration. Values of A„ can be obtained from studies of the absorption spectra of complexes. In the complex [TiiHiO)^]^^, there is only one electron to be accommoorbitals. In the ground state this electron occupies a dated in either the ?2g or orbitals

Cg orbital is possible when the energy The absorption of light by the complex can bring about such excitations. The wavelength of light absorbed most strongly of by the [Ti(H20)5]^"^ ion is approximately 490 nm, which corresponds to a about 243 kj/mol. The single absorption band of this complex spreads out over a considerable portion of the visible spectrum. Most of the red and violet light,

orbital. Excitation of the electron to

required for this transition, A„,

however,

is

not absorbed, and

is

an

supplied.

this causes the red-violet color of the

interpretation of the absorption spectra of complexes with

d electron is considerably more complicated, since of d electrons are then possible. In general, for a given metal ion of one set In

of ligands by another causes a change

and

the

complex.

more than one more than two arrangements

The

many

orbitals, A^,

which gives

in the

rise to different

instances a striking color change

is

the replacement

energy difference between

light-absorption properties.

observed when the ligands of a com-

plex are replaced by other ligands. Ligands may be arranged in a spcctrochemical series according to the magnitude

of A„ they bring about.

complexes,

it

From

the experimental study of the spectra of

has been found that the order

is

generally the

same

for the

many

complexes

all of the transition elements in their common oxidation states with only occasional inversions of order between ligands that stand near to one another on

of

the

list.

I


• liquid solid at 4 x 10"* atm at

5.5

50 C:

9.28 Increasing the pressure

0 C:

on a solid-liquid equilibrium

system will cause the equilibrium to shift to the more dense form. If solid is the more dense form, the substance

CO2

will freeze (curve slants to the right); if liquid

58.0 g

forces

9.26 (a) at -60 C: vapor -» solid at 4.0 atm, (b) at vapor -> liquid at 34 atm

500K

1.25 times faster than

liter, (b)

is

boiling point

at 0.1 atm,(c) at

(1 - .v) mol ^.O* undissocialed, mol NO, produced; therefore, 0.62 mol N2O4 and 0.76 mol no",, (c) .V^,,o. = 0.45. A\o; = 0.55, (d) P^.q, = 0.45 atm, /Jno. = 0.55 atm

at

—

molecules.

9.24 (a) at

2.V

s

HjO

solid

279

molecule and a F" F H -F"

which the vapor pressure of a liquid equals the pressure of the surroundings. The temperature at which the vapor pressure of a liquid equals 1 atm is called the normal boiling point. At 0.50 atm, the boiling points are: diethyl ether, 15 C; ethyl alcohol. 60 C: water, 80.3 C.

*8.54 (a) 1.38 mol, (b)

m

held together by hydrogen bonding:

ion consists of an

hydrogen bonding

ml

8.58

HF

ion

9.20

lOH^OlD

.Yco 2.45 g CO.

point.

The HF2

of each are similar. The H atoms and electron pairs of the -NH, groups, however, can enter into hydrogen bonding. Since H2NCH2CH2NH, has nvo - NH, groups,

mol hydrocarbon, 0.8125 mol O^. 0.500 mol €02,0.625 mol H^O. (b) 2 (hydrocarbon). nO^.StCOj). 8C0,(g) + 10 (H,0), (c) 2C4H,„(g) + 130,(g)

(c)

polarities

same order; 1 2 is London forces,

9.16 Since the molecules are similar, the

3CHjg) +

H, Oil)

(b) 1.00 liter

=

forces increase in the

9.12

with

8.45 (a) 0.125

=

bond

a dipole.

The anion of an acid salt (HSO4. for example) contains a H atom with which it can hydrogen bond

8.42 0.507 liter gas

8.49 (a)

virtually nonpolar),

and has the highest melting

4.70gCaH2

4 AKOHljIs),

CO2 and

9.14

(c) 1.03

8.40 (a) Al4C3(s)

The London

is

in

polarities cancel

the largest molecule, has the strongest

ml

10'

(S^C

bond

and negative charge both fall in SCO, the two bonds are

the center of the molecule). In

9.9

10Na(s) + 6NjO(g)

molecules, the

6H,0(g),

5.71 g/litcr

8-47 Pn;0

in these

not identical

'

(b)

CSj, and

4NO(g) +

NO

8.28 0^50 liter Cl^, 1.50 liter

6NaOH(s) +

at the

molecule is linear: Both double bonds are the same

S=C=S.

liter

in

8.36 (a)

P atom

S=C=0, 0=C=0,

Each

9.7

CH4,

8.24 30.0 hter

8.31

nonbonding electron

trigonal pyramidal with a

the

pair.

g/mol

34.1

is

(c)

apex of the pyramid; PF5 is trigonal hipyramidal and does not have a nonbonding pair

8.21

PF3 on

PF3 is trigonal SF4 is irregular

linear, (b)

BF3 is triangular planar, tetrahedral, SnF4 is tetrahedral

8.19 0.474

higher.

is

150ml,(c) 1.50 x 10' ml

10^ balloons

1.00 X

8.5

(b)

temperature

because the

1.77 X 10"^" liter/molecule, (b) 0.0478%

8.74 (a) 8.3 (a)

of molecular volume,

effect

0.611, 1.31

(c)

mol

dense form, the substance

8.64 0.900 g liter

left).

8.68 (a) 0.514 liter, 0.0321 liter, (b) the first caused prin-

9.31

cipally by the effect of intermolecular forces, the second

(e)

Appendix G

Answers

to

(a)

network,

London,

(f)

(b)

will

metallic,

London and

Color-Keyed Problems

is

the

more

melt (curve slants to the

(c)

London,

(d)

ionic,

dipole-dipole

749

9.33 (a)

10.47

-

BrF than

10.49

122 g/mol

BrF. Electronegativity difference is larger for for CIF; dipole-dipole forces are stronger. Electron cloud of BrF is larger than that of CIF; London forces are stronger, (b) BrCl. Stronger London forces in BrCl. Dipole-dipole forces exist between BrCl molecules but not between CI, molecules, (c) CsBr. Ionic forces are stronger than London or dipole-dipole forces, (d) Cs. Metallic bonding is stronger than London attractions, (e) C (diamond). Network bonding is stronger than

London

forces.

9.38

simple cubic

cell,

9.45 a

gC3H5(OH)3

10.54 27

10.56 62.0

g/mol

10.59 2.91

atm

6.70 X

10-^

g/mol 1 1

.0

cm

atm

9.53

14.9

C

-0.242

10.69

cm on

cube 2.169

9.51

9.49

Nj, 1.49 x

(b)

10.67 2.67

pm

pm 348 pm 70.8 pm

9.46

N2, 0.800 atm; O,. 0.200 atm, 0^,7.97 X 10 " ^ (c) -0.0024 C

';

10.65 27.1

9.40 55.8 9.42 316

10

10.63 (a) 0.0106 atm, (b)

atom/unit

1

C

*10.52 (a)

10.61

g/cm^

9.36 1.45

1.75

an edge

Chapter

128

11.1

11

ZnS(s)

(a)

+ 2H^ + 2Cr

+

H,S(g)

Zn-* + 2Cr.

2Na^ + COl~ + Sr-^ + 2C,H30,SrC03(s) 2C,H307. + 2Na^ * N.R., (c) Sn-+ + 2Cr + 2nh; + SOi (d) Mg-^ + 2NO3 +Ba-* + 2 OH Mg(OH),(s) 3Br + Ba-^ + 2NO3. (e) 3Na" + PO^ + 3H^ H3PO4 3Na^ + 3 Br (b)

30.9

.

-t-

9.55 (a)

one Cs^

pm,

*9.58 (a) 385 9.61

(b)

Cr

one

ion,

ion, (b)

412 pm,

(c)

357

pm

6.98 g/cm^

NiS, BaO, CaS, NaBr, AgCl,

g/cm\

*9.63 (a) 0.5"„, (b) 8.241

*9.65 (a) 58.41 g/mol, (b)

(c)

KCl

-t-

8.235 g/cm^

11.6 (a) 64-, (b) 6-H, (g) 3

0.08%

+

,

(h) 5

11.8 (a)

Chapter 10 10.2 (a)

3-h.

(g) 3-t-,(h) 5

CH3OH,

(b)

NaCl,

5

(c)

4-1-,

1+.

(e)

4

+

,

(f)

2-,

(d) 5-^,

(e)

4

+

.

(f)

5-f-,

(d)

,

(b)

3

+

,

+

11.10 Oxidized (reducing agent): (a) Zn. (b) SbCl3,

CH,F

(c)

+

(c)

+

Mg. (d) NO, (e) H, Reduced (oxidizmg agent): {a)CK, (b)ReCU.

(c)

10.5 (a)

10.8

Fe^\(b)

Li + ,(c)

F

,

(d)

Sn'^,

(e)

Al^^,(f)

Mg^"

(d)

-47kJ/mol

—785 kJ/mol; energy required to separate water molecules from each other and energy released when the water molecules hydrate the ions

10.10

*10.15 10.16

10'- mol CO2, 2.12

X

10.12 4.81

m =

100 M/(100.v

CH3OH,

0.294;

-

A

gCO,

(a)2H,0 + 4Mn04 + 3CIO2 4MnO, 3CIO4 + 40H", 2Cr^^ + 3S + 7H2O, (b) 8H^ + Cr,0? + 3H2S (c) 6H,0 + P"4 4^ lOHOCl 4H3PO4 + -t-

locr + iOH\ 3Cu + 8H+ + 2NO3 (e) PbO, + 4HI Pbl,

H,0, 0.706

11.15 (a) 10.19

71.4%Cf,Hi206

10.21

7.01 g

(b)

33.7 ml cone

14

-I-

-l-

-I-

HBr

—

-I-

3.35

10.29 25.5 10.31

M HCl ml cone

0.316

10.35 0.264

M HNO3 atm

3H,0 (c)

g/mol

-f-

11.21

Answers

Br

to

(a) 8

0H

Color-Keyed Problems

+S'

,

+ 2H2O,

-f

4H,0 + 2Re6, + 6Cr + 6H +

C2H4(OH)2

Appendix G

- 3H3ASO4 + ——*2Se HSO4 +

Br03"

(e)

mum-boiling

750

3H3ASO3

{d)2H,Se03 + H,S

10.42 (a) 0.434 atm, (b) negative, (c) evolved, (d) maxi-

10.45 333 g

-I-

-i-

H3PO4

10.36 0.514 10.39 70

As40^ +

5Bi''+

10.25 (a) 5.51 A/, (b) 6.93 in 10.27

2H,0

6H2O + 4ASH3 + 24Ag^

5 Bi03 2Mn04 2Mn-+ + 7H,0, 2rslO + 4NO3 3N,04 + 2H2O, (c) 4H^ * 2Mn--' + (d) \\H* + 2Mn04 + 5HCN + sV 5ICN + 8H,0, 3Zn-+ 3Zn 4- 2H,Mo04 (e) 12H^ 2Mo^-' + 8H2O cr + 3I2 + 3H2O, 11.17 (a) 6H' 4-CIO3 4Zn-* + NH; + (b) 10 H* + 4Zn + NO3 (b)

cone HBr,

3Cu-" + 2NO + 4H2O,

L

24Ag + 24H\

KOH

10.23 (a) 50.6 g

CuCK.

11.12

(d)

v)

(c)

0,,(e)W03

3

CI,

+4I2

*

2HRe04

-I-

SOr + 4H,0 + 8I

,

H20 + 3CN- + 2Mn04: 3CNO- + 2Mn02 + 20H-, (c) 2H2O + 4Au + 8CN" +0, 4Au(CN)7 + (b)

SiOr

+ 20H-

Si

—

+ 2H2.

Br"

CO(NH2)2 + 3 OBr

CO2 + N2 + 3

— —

Chapter 13 13.2 (a) left,(b) left,(c) left,(d) left,(e) left

13.6 (a) decrease, (b) increase, (c) increase, (d)

PbS + 4H2O2 PbS04 + 4H2O. +2Cr(OH)3 + 3H20, (b) 40H * 2CrOr +8H2O, (c) 6H^ + 2Mn04" + SH^Oj 2Mn^^ + 5O2 + 8H2O, * 2Ag + 02 + H20 (d) Ag20 + H202 11.26 (a)

bromic

11.32 (a)

acid,

hydrobrotnic acid,

hydrogen sulfate, hydrogen carbonate, (i)

potassium

(b)

sodium

(d)

(e) (i)

potassium sulfite, (g) sodium (h) sodium dihydrogen phosphate,

69.5%

11.42

6.00

KHCSH4O4

6, (c)

1

(c)

I

=

0.895% NaCl

6H2O,

3N, + 2Br 1

6, (e)

1

2.

1

14

(f)

M H2SO4, 2.00 M H3PO4

HCl. 3.00

(b)

Chapter 12 12.2 (a)

rate of formation

(mol

liter

of

A.-[A][B],

(b)

0.016

mol/liter,

=

=

liter^/mol^

1.78

system:

)

(e)

,

(b)

H3ASO4,

HAsOj

(b)

,

PH4%

(c)

NOj

(c)

HC2H3O2,

,

(d)

S^'

12.12 step

0.100

=

0.100 mol,/(liter s), rate = 0.100 mol (liter = s, rate = 0.0050 mol (liter s), (c) k

=

k

14.11

s).

(b)

0.100

liter

(mol

-

s),

=

rate

0.00025 mol

(liter

s)

12.6 (a) 3 16 of the original rate, (b) zero, (c) 2 27 of the original rate, (d) 4 times the original rate, (e) 8 times the

1:

*12.14

[A*]

=

ICl

f

HCl + HI;

H,

+ CH4

(A'l

CI

HI +

CH3 +

—CHjCl

,

,

,

,

,

,

HBr is a strong acid, H2S is a weak acid, and AsHj not acidic. The acid strength of the hydrides of the

14.20

A-2)[A]: therefore, rate

CI; chain terminating:

CH3 +

step 2;

I2

12.18 chain initiating: CI2

CI

HOCl +

(a)

H,0 + NH, HCO3 + OH" NH3 + H .

HCl +

ICI

HS

OH

14.17 (a) yes,(b) no.(c) no,(d) yes

k

(d)

HPO;

original rate

(b)

=

0.75 atm, initial pressure of PCI 5

H,0 + OCl", NH3 + OH H2O + COi (c) H2 + NH,(d) 14.16 (a) HjO", H3PO4, HCN, H2O, NH3, OH CN H2PO; H2O (b) NH2

12.4 (a)

10" atm

x

(b) 6.71

(b) A'

to

H2ASO3

14.8 (a)

s)

•

[CO2] =

.00 atm, 75" „ dissociated

14.6 (a)

C =

mol/liter,

)

M KMn04

0.0150

mol/liter

acid (NH4CI) reacts with base form solvent (2NH3) and salt (NaCl). Bronsted: acid, (NH4 reacts with base2 (NH, to form conjugate base, (NH3) and acid2 (NH3)— NH3 is amphiprotic. Lewis: a nucleophilic displacement of NH3 by NH, on NH4%

N KMn04,

0.0750

0.60 mol/liter,

atm

=

(NaNH2)

K2Cr20-,

effect,

Chapter 14

one

;V

10

0.563 mol'/liter-, 8.79 X 10 %atm^

14.4 Solvent

0.1200

effect,

mol'liter, (b) 3.94, (c) 3.94

2.19 X

13.28

5, (d)

no no

= 2.362

mol/Hter, [HI]

[CO] = [H2O] = 0.0335

13.26 0.024

i\

11.55 (a) 90.0 g,(b) 11.57 (a)

= [l2] = 0.319

[H2]

13.24 (a)

.V/

11.53 0.409

10""^ mol/liter

13.22 (a)

11.49 (a) 1/4, (b) 11.51

13.14 5.5 X

Mg(OH)2

3N2H4 + 2Br03 24.0",, N2H4

=

0.050 mol/liter, [PCI5]

10""^ mol/liter

[H,] = 0.0665

11.46 (a) (b)

=

13.12 7.6 X

13.20 (a)

11.44 (a) 0.0448 g NaCl.{b)

(h)

13.18 0.025 mol/liter

N203.(c) SO,.(d) B,03,(e) A\,0^.

M

11.40 41.28''o

effect,

effect

(f)

'K2O

11.38 0.3858

no

(g)

0.042 mol/liter

13.16

11.35 (a) Ci,07,(b)

ZnO,

no

decrease,

(f)

13.10 (a) [CI2] (b)

copper(Il) nitrate

(f)

decrease,

13.8 61

bromate, potassium

(e)

nitrite,

endothermic

13.3

+ 2 HjO

Br

s)

kJ/mol

12.30 52.3

——

1

(c)

s)

K

12.28 668

40H" +2Cr(OH)3 + 3BrO* 2CrOa + + 5H2O 11.23 (a) OH' + 5HCIO2 4CIO2 +C1 + 3H,0, (b) 8 OH + 8 MnO^ + 8 MnOi" + lO; + 4 H2O, (c) 40H + 2H2O + P4 2HPOr + 2PH3, (d) OH' + SbHj + 3H2O Sb(0H)4 + 3H,, (e)

(e)

10^ liter/(mol

12.26 4.7 X 10'^ liter/(mol

40H~, (d) H2O + 3

266kJ/mol

12.21

12.24 7.9 X

HCl,

2C1

*

=

(A.A^ A-2)[A]

—

is

elements of a period increases from

2C1; chain propagating:

CH3 +

CI2

Clj, 2CH3

CH3CI +

to right (with

H3P04,(b) H3As04,(c) H2SO4, HjCOj.Ce) HBr

14.22 (a)

C2Ht,,

(d)

Appendix G

left

increasing electronegativity).

Answers

to

Color-Keyed Problems

751

14.23 (a) P'

FefCO);;

(c)

Mn(CO)J

;(e) acid:

,

CF

.

NH2

displaced by (c)

nucleophilic;

(d)

OH^

nucleophilic:

base:

displaced by

H

,(b) electro-

nucleophilic:

16.1

2.0 X

16.4

2.1

HjO

GeS

displaced by Ge,

O'

,(f)

electrophilic:

Br^ displaced by FeBr3,(g) electrophilic: CH3 displaced by AICI3, (h) nucleophilic: I displaced by OH"

16.5 1.6 X 10"' mol .CuCOj/liter 10"-* mol Ag^COj/liter

1,5

15.4

2.3 X

10^5

15.5

7.3 X

10'^

[Na+]=0.16

[C^Or] =

0.18

15.15

1.0

M

16.14

1.2

16.17

0.18

16.19

ion product (1.8 x

X

1.5

M

[NH;] = 2.7 [NH3] = 0.15 M (b)

(a)[H+]

[OH

"]

=

(c) 1.9

=

10'' A/,[N3]

X 10"' A/,

=

[OH

0.13

]

=

M,

0.10

M, [OH ] = 5.0 x 10" '^M, M,[H^] = 2.5 x 10"'^ A/

0.020

0.040

15.27

3.1

X

10

15.29

7.3 X

10

lO"'"* A/,(b) 4.8 x

10"' A/,

M

*

17.8

-66.58 kJ

10"'

5.5 X

15.38

0.70

15.40

4.85

15.41

0.045

15.44

[NH;]/[NH3] =

M M 0.56

15.46 (a)

[H+] = [HCOj'l =

[CO', "]

=

4.8 X

15.48 (a) 4.9 X

8.18

15.52

15.57

10" " 5

-

A/, (b) 7.4 x

10

752

-

1366.8 kj,

kJ

+103.0 J/K.(b)

- 587.6

-207.0

kJ

17.18

For BF3, AG = + 12.09 kJ; for BCI3, AG kJ BCI3 will hydrolyze at 25 C. BF3 will not.

kJ,

kJ.

HjSlg) and H20(l)

;

-48.39

kJ, yes

17.23 (a)

-101.01 kJ,(b) - 120.6 J/K,

17.25 (a)

+38.3

k.l,

no.

17.28

239.7 K, -33.4

17.30

+

17.32

208 J/(K

17.34

2.438 J/(K

17.35

-

17.36

5.12 X

-

(c)

-

120.6

J/K

11.8 kJ, yes

C

mol

18.25 kJ •

(b)

mol) •

mol).

diamond

1095.02 kJ 10"''

17.37 0.443

A/

10"^

17.39

2.07

17.41

+38.5 kJ

17.43

+27

kJ

Chapter 18

5.82

6.2 X

(b)

A/

18.7 0.570 g Ni

15.61 (a) 3.92, (b) 8.46, (c) 12.16

15.63

no

184.79 kJ; mol

(b) 3.92

M

4.3 X

A'jp,

+267.4

2.82

15.55 0.59

*15.59

10

X 10~*

M. 10"" M, [CO,] = 0.034 M. 1.2

smaller than

17.16

17.21

15.36

10"

10"' A/

- 5459.55

- 266. 19

15.35 (a) 3.80, (b) 0.46"o

15.51

A/,

(minimum)

17.6

^ mol HCIO2

6.57-8.35

X

= 0.070

M

- 1364.3 kJ/mol, - 277.9 kJ mol

17.14 (a)

1.9

^]

Af

17.3 (a)

M,

(c)

x 10"^ A/,(d) 4.6 x 10"'^

15.50

[Ba'

Chapter 17

15.25 (a) 3.5 X

15.32

A/,

10"- mol AgCl/liter, 10"^ mol AgBr/liter, X 10"' mol Agl/liter

17.10

1.3

= 0.30

X

15.23 (a) 4.14. (b) 1.08. (c) 10.52. (d) 12.62

(c)

]

16.28 (a) 2.4 X

M

10'

15.19

15.21

x

0.086 A/ (minimum)

(b) 1.4

0.10

1.3

M

10"*

9.5 X

16.26

M

[H^] =

15.16

[HN3] =

M

16.12

16.24 2.3 X

10"^ A/,(b) 0.72°o

X

X

A/, [CI X \0-^

2.1

16.22 0.23 A/

*

15.7 4.4 X 10

15.12

than

10"'^ mol Ni(OH)2

16.8 3.3 X

16.10

lower

*

X 10

15.1

(a) 1.8

10"'* A/

X

Chapter 15

15.9

10""*

base:

HF, base F

N02^ displaced by H^,

philic:

displaced by

Chapter 16

NOj.(e) F"

acid: Ag"^, base:

14.28 (a) nucleophilic:

(e)

,(d)

(b) acid: H"^, SH NH3; (d) acid: H"",

CSj, base:

14.26 (a) acid:

*

NHj.Cc) SiOf

.(b)

18.9 (a)

10

(b)

Appendix G

Answers

to

Pb-^ +

0.558 g PbOj.

Color-Keyed Problems

2H2O (c)

46.6

PbO,(s)

mm

+ 4H^ +

le'

=

18.11

min

28.6

A

18.13 (a) 780 C.(b) 0.867

333

18.17

9.22

(g)

18.20 (a) 5.00 liter CI,, (b) 0.893

18.22

M

Mg +

+2.227 V,(b)

Sn^^

-

Mg2" +

+0.8240 V.

18.29

-

1.789

(c)

the

Sn,

H

;

:

:

Cr^"

;(b)

Au^

Ga Cr Zn

(c)

;

;

;

Pb02, (d)

HF(1)

Au ^ Cr

Hjig)

» 2Ag + 02 + 2H\ H,02 + 2Ag' Ag + Fe^\ (d) no. (ejHjSOj + 2H,S SS + 3H2O

18.37 (a) In^ will disproportionate, (b) In^ '

3 In

2 In

+

3H,,2In + 3a,

'

In 6 H + 2 In 2In'" + 6C1 '

.

+

(b)

(d)

Pb^^ + SOr -45.0 kJ

M8.47 (b)

Cl,(g),

Cl,(g)

(d)

- -* 2 in'

+

2H^(aq) + + 6HF(aq) SiFriaq) + 2H2O, SiOjfs) + 4HF(aq)

19.16 (a) SiOjls)

+ 2H2O, * 2F COr(aq) + 2HF(aq) CO^Ig) + H,0. KHFjd), (c) KF(s) + HF(1) (b)

no

(d)

PbS04,(c) +0.233 V,

CaO(s) + 2HF(aq)

19.19 (a)

Pt(s)|H,(g)|OH

(a)

Nad) +

2NaCl(l)

2Cr(aq) + Br^d), + 2Br-(aq) 2I03(aq) + SHSOjIaq) 1,(8) + 5SOl (aq) 3H+(aq) + HjO, 3HBr(g) + H3P03(aq) (e) PBr,(l) + 2Hfi

(c)

(c) yes,

(aq)||Pb-Maq)lPb(s),

18.45 (a) Pb(s)|PbS04(s)|S0i

"

heat

2KF(1),

-

F,(g)

heat

1.208 V,(b) no,(c) yes, Ti^*

18.43 (a) +1..397 V,(b) yes,(c)

2HF(g) + CaS04(s), electrolysis

SiF4(g)

-

18.40 (a)

critical

H,

KHF,(s). 2KHF:(1)

eiectroivsis (b)

no

18.39 (a) no.(b) yes.(c)

0.958 g

(b)

+ H2SO4I

—

+ KF(s)

Ag^ + Fe-"

(d)

and

;

18.34 (a) no, (b)

in-^

3H,0(g)

electrode

Cr^Ml' (c)

H,,

19.14 (a) CaF,(s)

;

* W(s)

point, boiling point,

805.8 kJ'mol

19.11

V

PbS04; Cd'" HOCl, H " Ce* "

—CHjOHd). +

The low melting

19.7 (a) 0.503 g

Cl^/CP

18.32 (a) '

3

CO(g) + 2H2(g) WOjfs) + 3H2(g)

18.27 (a) Pt(s)|l,(s)|r(aq)||Cr(aq)|Cl,(g)|Pt(s), (b)

Cljig)

temperature reflect the weak nature of the London forces between H, molecules. The forces are weak because the electron cloud of the molecule is relatively small.

the Sn^'^/Sn electrode

(c)

Njlg)

19.5

5.77 hr

18.25 (a)

H2(g) =?=^ 2 NHjIg). Cu,0(s) + H,(g) 2Cu(s) + H,0(g).

(f)

CI2

liter

:'lCl(g),

H,(g)

(d) (e)

gC

18.15

+ +

(c)

+0.2943 V.(b) -56.79 kJ.(c) - 121.7

18.52 (a)

-509.5 kJ.(b) -43.5 kJ mol

K

J

(aq)

+

+ H,0

CaF,(s)

FeCK.(b) RbCI.(c) BeFj

19.22 (a) HCI.(b) Cl,.(c)

(aq)llH '(aq)|H,(g)|Pt(s),

+0.828 V. -79.9 kJ

18.49 (a)

— —

19.24 28.2

F

19.26

gCli

1.^24

HF.(d) F,,(e) CI,

.(f)

Clj

Chapter 20

+125.2kJ mol, (b) + 1.76 V, ilworeiically. H^O,, Co'' 5,05 ,03, and F,

18.54 (a) (c)

20.2 (a)

,

10-'

18.55 7.1

X

18.57

X 10

18.59

1.0

-0.284

'*

V

20;

(c)

2NaN03(l)

(d)2KC10,(s)

+

Mg'*

Ni''

+

= -0.13

^

mcreased by

V. no.(b)

=

(b)

+0.05 V, yes

0.(X)891 V.

decreased by

(d)

V

18.70 (a)

(e)

+0.010 V,(b) [Sn-*] =

1..^7 .W,

(b)

+

0,(g),

0,(g).

(c)

+0.175

—

CH4(g) + 02(g) 2CH4(g) + 302(g) CH4(g) + 202(g)

(c)

H, + 2H*(0.0250

Ml

V

Chapter 19

2NaH(s), (a)2Na(s) + Hjig) CaHjIs), + H2(g)

Ca(s)

Appendix G

* C(s)

+

2

H20(g),

*2CO(g) + 4H20(g),

+ 2H20(g) PbS04 + 4H2O, 20.9 (a) PbS + 4H2O2 (b) 40H' + 2Cr(OH)3 + 3H2O2 2CrOr + 8H2O, -2Mn^^ + (c) 6H^ + 2Mn04" +5H2O2 5O2 + 8H2O, 2 Ag + O2 + H2O (d) Ag20 + H2O,

+0.136 V.

H2 + 2H'(5.00iV/)

(b)

2KCl(s)

0,(g),

SCO^ + lOH^O, (a)2C4H,o + 130, 5CO2 + 6H2O + SO2, C,H,,S + 9O2 6CO, + 8H,0. 2C3H80 + 90, 2ZnO + 2SO,, 2ZnS + 2O2 2 PbO + 2 SO, 2 PbS + 3 O,

20.5 (a)

.V/

(b)

19.3

2NaNO,(l) +

MnO,

+

(aq)

(e)2H,0^iH^!l£!H:,2H,(g) + 0,(g)

(c)

[Pb-"] = 0.63

Ojig),

20.4

18.68 (a)

18.69 (a)

+

40H

Ni,

M

18.66 (a)

0.00891

2Hg(l)

+ 2H,0

(b)

(aq)

heat

heal

18.61 (a) +2.157 V,(b) Mg (c) the Ni^ ' Ni electrode

18.64 2.04

2HgO(s)

Answers

to

Color-Keyed Problems

C02(g)

753

-285.9

20.11 (a)

-

(b)

kJ,

(b)

(c)

+

20.15 (a) S(s)

S^iaq). + * S^Oriaq), + SOr(aq) FeS(s), + Fe(s) + 2F,(g) SF^ig),

(b) S(s) (c) S(s)

(d) S(s) (e) S(s)

+

S(s)

+

S(s)

——

—-

+

SO^ig)

S(s)

(b)

—

+ 0,(g) SOjig). H^SO^d). + H,0 H^(aq) + HSO" (aq) + OH-(aq) 2SO,(g)

(d)

SOjIg)

* NH4Cl(s), NH3(g) + HCl(g) 2NH3(g) + 2V(s)^2VN(s) + 3 H2(g), (h) 2NH3(1) + 2Na(s) 2Na+ + 2NH2~ + H2(g) (g)

+ 8H^(aq) + 2N03(aq) 3Cu-^(aq) + 2NO(g) + 4H2O,

21.13 (a) 3Cu(s)

(b)

(c)

(d)

H,0 +

(e)

H^S^O^d)

H2S04(1)

CHjCfNHOS + H,0

20.20 (a)

+ 9H + (aq) + NOjfaq)

CHjClNH,)© + H,S, H^SOj, (b) SO, + H2O * H,S04, (c) SO3 + H2O 2Al(OH)3 + 3H2S, (d) A1,S3 + 6H2O H,Se04. (e) SeOj + H,0 Te03 + 3 H2O HfeTeO^, H2SO4 + H2O,, (g) H,S05 + H20 (h) H,S207 + H2O 2H2SO4

(c)

SOf

SO,

SO^

in

Na(s)

+

in acid solution

goes to

alkaline solution goes to

(c)

2NH3(aq) + OCl

its

valence

level. its

The

O

is

+

2N02(g) + 5H2(g) + 2H^(aq)

NH4F +

2HF(1)

limited to four

valence level has only four orbitals.

acidification produces H2S(g), (b) acidification produces S02(g), (c) BaS04(s) precipitates when a solution of BaCl, IS added, (d) acidification produces S(s) and S02(g)

20.27 (a)

NF3(g) +

+ 3H,0

3H2(g)

+ + NH3(g), (b) 2AlN(s) + 3H,0 AUOjIs) + 2NH3(g), 3 Ca^ ' (aq) + (c) Ca3P2(s) + 6 H2O 60H-(aq) + 2PH3(g), (d) CaNCN(s) + 3H,0 CaC03(s) + 2NH3(g), (e) NCl3(l) + 3H2O NH3(g) + 3HOCl(aq). (f) PCljO) + 3H2O H3P03(aq) + 3H^(aq) + 3Cr(aq) POCl3(l) + 2H^(aq) + (g) PCU(s) + H2O * H3P04(aq) + 2C1 (aq), PCl5(s) + 4H,0 5H^(aq) + 5C1 (aq), (h) H4P207(1) + H2O 2H3P04(aq), (i) 3N02(g) + H2O 2H^(aq) + 2N03(aq) + NO(g) 3

SO4" and 840^". SO|^ and 8,03"

atom

N2H4(1)

(aq)

+ H2O,

21.19 (a) Li3N(s)

The S atom in SF4 has four bonding electron pairs and one nonbonding electron pair (a total of five electron electron pairs since

02(g),

N2(g)

2NH30H + (aq) + 2H2O,

20.25

pairs) in

+ 2H2O

(e)

(b)

—

(aq)

*20.24

2N205(s),

(c)

CI (aq)

(aq) + 2H + (aq) S02(g) + H2O, * H2S(g), + 2H"(aq) S(s) + S02(g) + HjO SjOriaq) + 2H^(aq)

S-

NH4(aq), Ca-*(aq)

N20(g) + H2O, (b) N2(g) + H2O, PbO(s) + N02(g) + 02(g), (d) NaN02(s) +

21.16 (a)

—

(b)

4HP03(s) +

4HN03(l) + P40io(s) H"(aq) + NH3(aq) 2H^(aq) + Ca(OH)2(s)

(f)

20.22 (a)

4Zn^*(aq) +

21.15 (a)

+

S03(g)

4Zn(s)

NH3(g) + 3H2O,

HS04(aq). (e)

—

(f)

FeS(s),

+ 2H + (aq) Fe'+laq) + H,S(g), SO,(g). + 0,(g) H,S03(aq), SO.(gl + H,0 (c) HjSOjiaq) + 2 0H (aq) SO^laq) + H^O, * S20r(aq), S(s) + SOr(aq) FeS(s)

+ 3H2O

+

'

2NH3(aq) + Ag^(aq) - Ag(NH3)2"(aq), (b) NH3(aq) + H + (aq) NH;(aq), (c) NH3(aq) + H,0 + C02(aq) NH;(aq) + HC03(aq), (d) 4NH3(g) + 302(g) 2N2(g) + 6H20(g), (e) 4NH3{g) + 502(g) *4NO(g) + 6H20(g),

2S(s)

20.17 (a) Fe(s)

N.R., BijOjfs) N.R., Bi203(s) + 6H+(aq)

21.11 (a)

SF^ig).

3F2(g)

(aq)

2Bi^*{aq)

^S^CI^d), + C^g) (g) S(s) + 4H"(aq) + 4N03 (aq) 4N02(g) + 2H2O (f)

—+- H2O

Bi203(s)

OH

SOjig),

O^lg) S--(aq)

+

N.R., Sb406(s)

4Sb02"(aq) + H2O, Sb406(s) + 4H^(aq) + 4Cl-(aq) 4SbOCl(s) + 2H2O,

+299.5 kJ/mol

*20.12

+ H2O

Sb406(s)

40H'(aq)

more exothermic

333.3 kJ, they are

0H

(aq)

N,0, (b) N,03, P40„(g) AS2O5

21.20 (a) («)

3Li^(aq)

(c)

N2O,,

(d)

P4O,,,, (e) P4O10,

H2NNH3'^(aq), H2NNH,(1) + H^(aq) H3NNH-3^(aq), H,NNH2(1) + 2H + (aq) H3NOH^(aq), (b) H2NOH(s) + H^(aq) HNO,(aq). (c) N67(aq) + H-'(aq) (d) NH"3(aq) + H^(aq) * NH;(aq). » 2NH;(aq) + (e) (NH4)2C03(s) + 2H^(aq) C02(g) + HjO 21.22 (a)

20.29 (a) angular, (d)

trigonal

(g)

irregular

(b)

angular,

pyramidal, tetrahedral.

(e)

(c)

triangular

tetrahedral.

(h)

octahedral,

(f)

planar.

tetrahedral.

(i)

octahedral

—

21

21.4

At STP, the density of pure N, is 1.250 "N2 from air" is 1.257 g/liter

g, liter,

the

21.23 (a) 4As(s)

density of

21.7 (a) P406(s)

F406(s) P406(s) P406(s)

754

+ 6H2O

(b) P4(s)

4H3P03(aq),

P40,o(s), (c) 2PCl3(l)

*4HPO-; (aq) + 2H2O. + 80H-(aq) 4H2P03(aq), + 40H-(aq) + 2H2O + H + (aq) * N.R.,

—

Appendix G

Answers

+

to

+ 30,(g)

302(g)

+

0,(g)

(d)

2NOig) + 02(g)

(e)

2Sb2S3(s)

Color-Keyed Problems

+

902(g)

As40e,(s),

P406(s), P4(s)

+

502(g)

2POCl3(l),

2N02(g), * Sb40fi(s)

+ 6S02(g)

"

3

0H

(b) (c)

(d) (e) (f)

+

NajAsIs)

21.24 (a)

—

AsjOsls) + 3H2O MgjBi^ls) + 6H2O As406(s) + 6H2O

—

AsHjfg) + 3Na^{aq) +

water the least soluble of all sulfates, (c) The principal anions found in sea water are CO3 HCO3", SO4", CI", and Br". The Na"" and Mg^"" salts of all of these anions ,

2H3As04(aq),

+

2BiH3(g)

3

Mg(OH)2(s),

23.16 (a)

octahedral,

(e) trigonal

bipyramidal,

NH;(aq) + OH-(aq)

21.28 (a)

+ 30H

(b) P4(s)

(aq)

(f)

Al,03,

(b)

and

3V2O5 +

2Ta

Ta205 + 10 Na 5Na20, (d) ThO, + 2Ca Th + 2CaO. (e) WO3 + 2 Al + AI2O3

—

octahedral

+ H2O, PH3(g) +

NH3(g)

+ 3H2O

U+

UO3 + 2A1

eV + 5A1,03,

2A1

21.26 (a) trigonal pyramidal, (b) angular, (c) letrahedral, (d)

sulfides of Pb, Ni,

Bi are very insoluble in water.

SbOCl(s) + 2 H ^ (aq) + 2 Cr (aq), + H + (aq) + r(aq)

PH3(g)

The

are very soluble in water, (d)

4H3As03(aq),

+ H 2O

SbCl3(s)

PH^Ks)

—

H^O

3

(aq),

(c)

+

W

i^CaO(s) + CO,(g). Ca^''(aq) + 2 6h (aq), + H2O Mg'^(aq) + 20H"(aq) Mg(OH)2(s). Mg(OH),(s) + 2H + (aq) + 2Cr(aq) Mg'^(aq) + 2Cl-(aq) + 2H2O, CaC03(s)

23.18

CaO(s)

3H2P02"(aq), As406(s) + (d) Sb205(s) +

120H (aq) *4AsO-] (aq) + 6H,0. 2Na + (aq) + 20H (aq) + 5 HjO

(c)

—

2NaSb(OH)6(s)

MgCl2(l)

^i^^g^

+

Mg(l)

Cl2(g)

Chapter 22 23.20 (a) Leaching of the carbonate ore

(c)

(e)

sodium hydrogen carbonate,

reduced by blowing air through the molten material and impure, blister Cu is obtained. This impure Cu is refined by electrolysis. The impure Cu is made the anode of the cell and pure Cu plates out on the cathode.

sodium carbonate

(h)

—

(d)

C0Cl2(g), CO(g) + Cl2(g) CO(g) + S(s) COS(g), 2 C02(g), 2 CO(g) + 02(g) CO(g) + FcO(s) * Fe( + COjlg),

(e)

4CO(g) +

22.6 (a) (b) (c)

—

22.10

1

—

+ H2O + OCN

2£'

23.23 (a)

(b)

CN

PbS(s)

PbS(s)

—

—

3

0,(g)

+ CO(g), 2PbO(s) +

2S02(g),

PbS04(s).

3Pb(i) + S02(g), 2Pb(l) + 2S02(g)

—

N R., 2NaH,(b) Na + N2 2Na + H2 Na,02,(d) 2Na + CI2 * 2NaCl, 2Na + O2 4 NajP, Na2S. (I) 2 Na + P4 (e) 2 Na + S *Na2C,,(h) 2Na + 2H2O (g) 2Na + 2C 2NaOH + H2, 2NaNH2 + H, (i) 2Na + 2NH3

—

1

BF; MgB,. (e) BF, + F 28 + Mg B(OH)3. (f) 8,03 + 3H,0 B(OH)4. (g) B(OH)3 + OH HBO, + H ,0. (i) 2 B(OH)3 (h) B(OH)3 8,03 + 3H,0. 2LiBH4 2LiH + B2H,, ())

—

B^Hio

Xe + F,

22.17 (a)

Xe +

+ 2HF. 22.19 (a)

3

(e)

XeF,.

(b)

Xe + 2 F, + H,0

—-

Cs is the largest I A metal atom, Cs is the atom it is most easy to remove one electron (smallest first ionization energy). The Li^ ion, however, is the smallest I A metal cation and more energy is released when Li^ is hydrated in water solution than when any

XeFfe

+ 3H2O

from which

XeF^.

other

XeOF^

XeFe,. (d) XcF,,

F,

SXe

A cation

is

hydrated. Standard electrode potentials

occur

in

water solution. The relathe hydration

amount of energy released by Li^ ion makes up for the relatively

tively large

+

of the of energy required for the ionization of

6Mn04 + 18H". 0.120 iW, 3.59

I

refer to processes that

XeOj + 6HF

5Xe03 + 6Mn^^ + QHjO

—

23.25 Since

—

(b)

+

Pb(l)

23.24 (a)

(d)

(c)

—-

(c)

2BN,

22.15

C(s)

+ 2 0,(g) + 2PbO(s) + PbS04(s)

.

PbO(s) + C02(g),

PbC03(s)

2PbS(s)

PbS(s)

+ 20H-,

Pb + 2 0H + H2O + PbO PbO + CN" Pb + OCN ^28 + 3 MgO. 22.13 (a) B,03 + 3Mg - 2B + 6HBr. (c) 2 B + N, (b) 2BBr3 + 3H,

le

+

PbO(s)

)

Ni(CO)4(g)

Ni(s)

sulfuric

acid gives a solution of

see Section 24.3, (g)

by dilute

CUSO4, from which the Cu is obtained by electrolysis, (b) The CuS ore is concentrated by flotation and smelted to CU2S (matte). The CU2S is

hydrogen cyanide, (b) hydrocyanic acid, potassium cyanide, (d) potassium cyanate, potassium thiocyanate, (() pentacarbonyliron(O),

22.5 (a)

gXeOj

large

amount

Li.

Ca3N2, CaH2,(b) 3Ca + N2 Ca + H2 CaCl2, 2CaO.(d) Ca + CI2 2Ca + O, * 2Ca3P2, * CaS. (f) 6Ca + P4 (e) Ca + S Ca(OH)2 + CaC2 (h) Ca + 2 H2O (g) Ca + 2 C Ca(NH2)2 + H2 H2.(i) Ca + 2NH3 23.27 (a)

The London forces of He are weaker than those of H2. Whereas both He and Hj have two electrons, the electron cloud of He is smaller and less polarizable than 22.20

—

—

(c)

.

that of H,.

4Al (c) 2A1 (d) 2A1 (e) 2Al (f) 2A1

*23.7 2.4

(b)

cm

Au, and Ag are relatively unreactive elements, BaSO^. SrS04. and PbS04 are very insoluble in

23.9 (a) Pt, (b)

Appendix G

Answers

to

—

2A1 + 3CI2 * 2 AICI3, 2AI2O3, + 30, AI2S3, + 3S 2A1N, + N2 2 A1(0H)4 + 3H2, + 2 0H" + 6H2O 2AP^ + 3H2 + 6H^

23.30 (a)

Chapter 23

Color-Keyed Problems

—

755

The Al^^ The S^"

23.33

ion has a high charge and hydrolyzes ion also hydrolyzes.

readily.

hydrolyzed

AUSj

is

completely

in water.

6H2O

—

(9)

—

—

+ 2H2O +

23.35 (a) PbO^ls)

Pb(OH)^(aq), (b) 3Pb02(s) (c)

2

0H'(aq)

—

(c)

ammonium

(d)

(e) (f)

Pb304(s) 2 PbO(s)

+ O^ig), + Ojig),

KFe[Fe(CN),].(b) Fe[Fe(CN)6], Cu2[Fe(CN)6],(d) K2Fe[Fe(CN),]

(b)

[Co(NH3),(S04)]N03, [Mn(CO)5(NCS)],

(c)

[Pt(NH3)3Cl][Pt(NH3)Cl3],

(d)

[Co(en)2(H,0)Br]Br2-H20

24.13 (a)

Pb(OH)3 + H2O + OH (aq) N.R., (e) SnS(s) + S^ (aq) (f) SnS.ls) + S''(aq) SnS^faq), PbCl3-(aq). (g) PbCl,(s) + CI (aq) SnFg ^aq) (h) SnPjs) + 2F"(aq)

(aq),

24.15 Let

a

= NH3. There

[Pta4][PtCl6]

2Fe +

3 CI, 2FeCl3, 2Cr + 3 CI, * ZnCl^, + CI, 2Cr,03. (b) 3Fe + 20, - Fe304, 4Cr + 3 0^ 2ZnO. 2Zn + O, * CrS. Zn + S ZnS, (c) Fe + S FeS. Cr + S 2CrN, N.R., 2Cr + N, (d) Fe + N, Zn + N.R., (e) 3Fe + 4H2O Fe304 + 4H,, 2Cr + < ZnO + H,, 3H,0 Cr,03 + 3H2,Zn + H,b » Fe^"- +H,, Cr + 2H^ (f) Fe + 2H+ * Cr-^ + * Zn^^ + H^. H,, Zn + 2H^ * N R., 2Cr + 60H + 6H2O (g) Fe + OH 2Cr(OH)r + 3H,, Zn + 20H + 2H,0 Zn(OH)r + H,

2CrCl3, Zn

pentachloroaquoferrate(III),

tetraamminecopper(II) tetrachloroplatinate(II), nitritopentaammineiridium(III) chloride, hexaamminecobalt(III) tetracyanonickelate(II)

24.11 (a)

PbO(s)

23.36 (a)

potassium tetracyanonickelate(O), potassium tetracyanonickelate(II),

(c)

2 PbOjis)

(d)

(b)

24.9 (a)

• SnOj, - SnCl4,(b) Sn + O2 Sn + 2CI2 Sn^^ + H2, SnSj. (d) Sn + 2H'^ + 2S Sn(OH)3 + H2,(«) 3Sn + + OH" + 2H2O - 3Sn02 + 4NO + 2H2O 4H" + 4NO3

23.34 (a)

(c)

(e)

2Ap-'(aq) + 3S^-(aq) + 2Al(OH)3(s) + 3H,S(g)

(c)Sn (e) Sn

K,[Rh(H,0)CU].(b) [Co(NH3)4(S04)]N03, Na3[ReO,(CN)4],(d) [Co(NH3)2(en),]Cl2, K4[Ni(CN)4],(f) K4[Ni(CN)4], [Cu(NH3)4]3[CrCle]2

24.7 (a)

one

and

is

of

one stereoisomer of For [Pta3Cl][PtaCl5].

[Pta3Cl3][PtaCl3] there are two:

—

CI

/CI Pt

cr

CI

CI