Chem Course Booklet

Course BookletChem1012 Final E

xam

Note: This Course Booklet is exclusively recommended for this academic term. The structure and content of courses changes each term and this Course Booklet will not be useful in future academic terms.

C h e m i s t r y 1 0 1 2

© Prep101

©Prep101 Chem1012 Exam Booklet

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please don’t copy!

Thanks a lot, Prep101 Important Disclaimer

This Course Booklet is not a replacement for lectures, tutorials, or textbook readings. It is a

supplement focusing on problem‐solving that will reinforce – not replace – what you’re learning

in lectures, tutorials, and readings.

Preface

This Course Booklet contains a summary of the key material tested on Chem1012 exams. We

encourage you to read through it, and write in any additional notes that you think will help you

better understand the material.

The material is divided into 12 chapters. Each chapter begins with a review of important

concepts, puts the subject into context, and ends with practice exam‐style questions.

Within the text of the Course Booklet are small textboxes with the headings: Need to know,

Important Concept and Question Alert. Read these textboxes as they contain important clues as

to what is commonly tested. There are some concepts that are tested every year and we have

tried to outline these for you.


2

Table of Contents Section Page #

DISTRIBUTION OF MARKS ON PREVIOUS EXAMS .............................................................................. 6

PERIODIC TABLE ................................................................................................................................. 7

FORMULA SHEET ................................................................................................................................ 8

I. FUNDAMENTAL CONCEPTS AND STOICHIOMETRY .................................................................... 15

1.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 15

1.2 Chemical Formulas, Names, and Equations ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 16

1.3 Mass and Mole of Substance ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 19

1.4 Determining Molecular Formula ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 20

1.5 Quantitative Relations in Chemical Reactions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 21

1.6 Aqueous Solutions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 23

1.7 Practice Exam Questions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 25

II. ATOMS AND LIGHT ...................................................................................................................... 34

2.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 34

2.2 Basic Terms and the Photoelectric Effect ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 36

2.3 Quantization of Energy ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 38

2.5 Heisenberg’s Uncertainty Principle ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 41

2.6 Quantization and Quantum Numbers ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 41

2.7 Shapes of Atomic Orbitals ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 43

2.8 Periodicity of Atomic Properties‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 43


III. FUNDAMENTALS OF CHEMICAL BONDING ............................................................................... 49

3.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 49


3

3.2 Overview of Bonding ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 50

3.3 Ionic Bonding ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 50

3.4 Covalent Bonding ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 52

3.5 VSEPR: Molecular Geometry ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 54


IV. THE BEHAVIOUR OF GASES ........................................................................................................ 57

4.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 57

4.2 Molecules in Motion ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 58

4.3 The Ideal Gas Equation ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 59

4.4 Applying the Ideal Gas Equation ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 60

4.5 Gas Mixtures ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 61

4.6 Real Gases ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 62


V. SOLIDS, LIQUIDS, SOLUTIONS AND THEIR PROPERTIES ‐ EFFECTS OF INTERMOLECULAR FORCES ............................................................................................................................................. 72

5.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 72

5.2 Types of Intermolecular Forces ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 73

5.3 Forces in Solids ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 73

5.4 Close‐Packed Crystals ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 76

5.5 Phase Diagrams ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 77

5.6 Solutions and Colligative Properties ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 78


VI. KINETICS: MECHANISMS AND RATES OF REACTIONS .............................................................. 88

6.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 88

6.2 Rates of Chemical Reactions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 90

6.3 Concentration and Reaction Rates ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 90


4

6.4 Experimental Kinetics ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 91

6.5 Linking Mechanisms and Rate Laws ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 92

6.6 Reaction Rates and Temperature ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 93

6.7 Catalysis ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 94


VII. THERMOCHEMISTRY ............................................................................................................... 104

7.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 104

7.2 State and Path Functions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 106

7.3 Energy Change ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 107

7.4 The First Law of Thermodynamics ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 110

7.5 Heat Measurement: Calorimetry ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 111

7.6 Heats of Formation ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 112

7.7 Hess’s Law ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 112

7.8 Bond Dissociation Energy ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 112

7.9 Practice Exam Questions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 113

VIII. CHEMICAL EQUILIBRIUM ....................................................................................................... 118

8.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 118

8.2 Dynamic Equilibrium ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 119

8.3 Equilibrium Constant ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 119

8.4 Applications of the Equilibrium Constant ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 121

8.5 Le Chatelier’s Principle ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 123


IX. AQUEOUS ACID‐BASE EQUILIBRIUM ....................................................................................... 132

9.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 132

9.2 Self Ionization of Water ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 133

9.3 Finding pH of Strong Acid/Base ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 134


5

9.4 Weak Acids and Bases ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 134

9.5 Strategy for Solving Acid‐Base Problems: A Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 138


X. APPLICATIONS OF AQUEOUS EQUILIBRIA ................................................................................ 146

10.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 146

10.2 Buffer Solutions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 147

10.3 Acid‐Base Titrations ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 148

10.4 Solubility Equilibria ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 151

10.5 Complexation Equilibria ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 157

10.6 Practice Exam Questions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 158

XI. THERMODYNAMICS ................................................................................................................. 164

11.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 164

11.2 Spontaneity and Entropy ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 166

11.3 Spontaneity and Free Energy ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 167

11.4 Change in Free Energy under Non‐standard Conditions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 168


XII. ELECTROCHEMISTRY ............................................................................................................... 176

12.1 Chapter Summary ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 176

12.2 Recognizing Redox Reactions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 177

12.3 Balancing Redox Reactions ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 179

12.4 Standard Reduction Potentials ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 184

12.5 Galvanic cells ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 187

12.6 Cell Potentials and Free‐Energy Changes ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 189

12.7 Electrolysis and Electrolytic Cells ‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐‐ 192


STUDY TIPS/EXAM WRITING STRATEGIES ...................................................................................... 204


6

Distribution of Marks on Previous Exams

Summary

Topic 2001 spring

2001 winter

2003 spring

2003 winter

2004 spring

2005 spring

2005 winter

2006 spring

% of marks

Stoichiometry 0 3 3 9 18 6 5 2 6%

Atoms and Bonding 0 0 5 5 0 12 3 7 4%

Gases 15 7 7 14 9 5 2 2 8%

Solids, Liquids, Solutions 6 12 12 5 3 7 15 5 8%

Thermochemistry 20 13 13 9 3 4 22 3 11%

Thermodynamics 6 14 14 7 12 10 3 30 12%

Acids, Bases, & Equil. 9 11 11 16 10 6 15 18 12%

Electro‐chemistry 18 14 15 16 12 20 25 23 17%

Question Pattern on Old Exams

In past exams, questions have come from ALL chapters. Therefore, to do well you need to know

all the material. BUT every chapter has its focus. I advise you to understand the principles and

pay special attention to the key points. Work on the sample exam questions on your own and

then compare with solutions.


7

Periodic Table


8

Formula Sheet

Atoms, Molecules, and Compounds

Mass = mA = AA MMn ×

Number of Atoms = n × NA, NA = 6.022 × 1023 atoms/mol

Molarity (M) = solution of volumeTotal

solute of Moles = mol/L or M =

Vn

Molality (m) = solvent of Masssolute of Moles

= mol/kg

Dilutions: MiVi = MfVf

Chemical Reactions and Stoichiometry

Stoichiometric Ratio: MolesB = ⎟⎟⎠

⎞⎜⎜⎝

⎛

A

B

tCoefficientCoefficien MolesA

Percent Yield = 100% ⎟⎠⎞

⎜⎝⎛

amount lTheoreticaamount Actual .


9

Atoms and Light

Universal Wave Equation: λν = c

Photoelectric Effect: Ephoton = hνphoton

Electron Kinetic Energy = Ekinetic (electron) = hv − hv0 = 2

2mu

Hydrogen’s Electron: En = − 2

18 J 1018.2n

−×

Photon Energy: Ephoton = λhc

= hν

Speed: u = m

Ekinetic2; c = 3×108 m/s

Particle in a Box: En = 2

22

mL8n h


10

Behaviour of Gases

Kinetic Energy: KE = ½ mu2

Root‐Mean‐Square Speed: MRTuurms

32 ==

Average Kinetic Energy: A

kinetic 23

NRTE =

Gas Constant, R = 8.314 J mol‐1K‐1 or 0.08206 L atm K−1

Ideal Gas Equation: PV = nRT

Equation for Molar Mass: PV

mRTMM =

Dalton’s Law of Partial Pressures: Ptotal = P1 + P2 + P3 + … + Pn

= (n1 + n2 + n1 + … + nn)RT / V

Mole Fraction (X): Mole fraction of A = XA = total

A

nn

; XA = total

A

PP

Real Gas Equation:

2

⎟⎠⎞

⎜⎝⎛−

−=

Vna

nbVnRTP


11

Thermodynamic Chemistry:

Heat:

q = nCΔT

qsys = − qsurr

qcalorimeter = CcalΔT

Work:

w = Fd

wsys = − PextΔVsys

wsys = − wsurr

Energy:

ΔEsys = qsys + wsys

ΔE = qv (Constant‐Volume calorimeter)

Enthalpy:

ΔH = qp (heat flow in a constant‐pressure process)

ΔH = ΔE + Δ(PV)

∑ Δ=Δ oofpreaction coeff HH (products) − ∑ Δ o

frcoeff H (reactants)

ΔHsolidification = − ΔHfus

ΔHcondensation = −ΔHvap

ΔEvap = ΔHvap − RTvap


12

Spontaneity:

Entropy:

TqS T=Δ

ΔSuniverse = ΔSsystem + ΔSsurroundings

∑ Δ=Δ oo SS preaction coeff (products) − ∑ Δ oSrcoeff (reactants)

Free Energy:

ΔGsys = ΔHsys − TΔSsys

∑= oofpreaction coeff GG ΔΔ (products) − ∑ o

frcoeff GΔ (reactants)

oGΔ = oHΔ − T oSΔ

ΔConcentration:

ΔSreaction = oreactionSΔ − R ln Q

Qlnreactionreaction RTGG +Δ=Δ o

Chemical Equilibrium:

[ ] [ ][ ] [ ]ba

edK

eqeq

eqeqeq

BA

ED=


13

Acid Base Equilibrium:

pH = −log[H3O+]

pOH = −log[OH−]

pKa = −log Ka

pKb = −log Kb

pH + pOH = 14.00

KaKb = Kw

pKa + pKb = 14.00

Applications of Aqueous Equilibria:

pH = pKa + log[ ][ ] ⎟

⎟

⎠

⎞

⎜⎜

⎝

⎛ −

initial

initialHAA

Electrochemistry: °cellE = °

cathodeE ‐ °anodeE

ΔG = ‐nFE, where F = 96,485.34 C mol‐1

eqKnFRTE ln=°

Nernst equation: QnFRTEE ln−°=


14

Kinetics:

First‐Order Rate Law: Rate = k[A] kt=⎟⎟⎠

⎞⎜⎜⎝

⎛[A][A]

ln 0

Half‐Life (t1/2), 1st Order:

kt 2ln

2/1 =

Second‐Order Rate Expression: Rate = k[A]2 kt=−0[A]

1[A]1

Arrhenius Equation: /RTEaAek −= RT

EAln kln a−=


15

I. Fundamental Concepts and Stoichiometry This is a review of fundamental concepts and the structure of matter: measurement in chemistry;

chemical reactions and stoichiometry.

1.1 Chapter Summary

Atom: the fundamental unit of a chemical substance

Molecule: combination of two or more atoms held together in a specific shape by attractive force

Chemical Compound: a substance that contains more than one element

Molecular Formula: a chemical formula that gives the exact number of different atoms of an

element in a molecule

Empirical Formula: contains the smallest set of whole‐number subscripts that give relative

amounts of constituent atoms

Empirical Formula/Molecular Formula massformulaMolecularmassforumlaEmpiricaln = =

mass

mass

MFEF

Mole: the number equal to the number of atoms in exactly 12 g of pure 12C.

Avogadro Constant: NA = 6.022×1023 atoms/mol

Mass Percent Composition: listing of the mass of each element present in 100 g of a compound,

also called the compound’s elemental analysis

Balanced Chemical Equation: describes a chemical reaction in which the amounts of all elements

and of electrical charge are conserved

Stoichiometric Ratio: ratio of Stoichiometric coefficients from a balanced equation

Yield: amount of a product obtained from a reaction

Theoretical Yield: amount of product predicted by Stoichiometry

Actual Yield: amount actually obtained

Percent Yield: the percentage of the theoretical amount that is actually obtained:


⎜⎝⎛

amount lTheoreticaamount Actual Factional Yield: Fractional yield =

yieldltheoreticayieldactual⋅

⋅

Limiting Reactant: the reactant that runs out first, based on the Stoichiometric ratio of all the

reactants


16

1.2 Chemical Formulas, Names, and Equations

1.2.1 Chemical Equations

Gives a description of a chemical reaction. Stoichiometric coefficients give the ratio in which

the reactants and products exist

The products of a chemical reaction have to account for all the atoms present in the

reactants.

1.2.2 Chemical Formulas

A chemical formula describes the composition of a substance by giving the relative number of

atoms of each element.

1.2.3 Empirical Formulas

An empirical formula contains the smallest set of whole‐number subscripts that give relative

amounts of constituent atoms.

1.2.4 Molecular Formulas

A molecular formula is a chemical formula that gives the exact number of different atoms of

an element in a molecule

To get the molecular formula from the empirical you must be given the molecular weight


17

1.2.5 Naming chemical compounds

The following two tables are provided to help you to review some basic rules.

Table 1.1 −Number Prefixes for Chemical Names

Number Prefix Example Name

1 Mon(o)‐* CO Carbon monoxide

2 Di‐ SiO2 Silicon dioxide

3 Tri‐ NO3 Nitrogen trioxide

4 Tetr(a)‐ CCl4 Carbon tetrachloride

5 Pent(a)‐ PCl5 Phosphorus pentachloride

6 Hex(a)‐ SF6 Sulfur hexafluoride

7 Hept(a)‐ IF7 Iodine heptafluoride

* If the numerical prefix ends with the letter “o” or “a” and the name

of the element begins with a vowel, drop the last letter of the prefix.


18

Table 1.2−Common Polyatomic Ions

Formula Name Formula Name

Cations Oxyanions

NH4+ Ammonium NO3

− Nitrate

H3O+ Hydronium NO2

− Nitrite

Hg22+ Mercury (l) SO4

2− Sulfate

SO32− Sulfite

PO43− Phosphate

PO33− Phosphite

Diatomic Anions

OH− Hydroxide MnO4− Permanganate

CN− Cyanide CrO42− Chromate

Cr2O72− Dichromate

Anions with Carbon

CO32− Carbonate ClO4

− Perchlorate

CH3CO2− Acetate ClO3

− Chlorate

C2O42− Oxalate ClO2

− Chorite

ClO− Hypochlorite


19

Guidelines for Naming Oxyanions

1) When an element forms two different oxyanions, the one with fewer oxygen atoms ends in ‐

ite, and the other ends in ‐ate (for example, SO32−, sulfite, and SO4

2−, sulfate).

2) Chlorine, bromine, and iodine each form four different oxyanions that are distinguished by

prefixes and suffixes. The nomenclature of these ions is illustrated for chlorine, but it applies

to bromine and iodine as well: ClO−, hypochlorite; ClO2−, chlorite; ClO3

−, chlorate; and ClO4−,

perchlorate.

3) A polyatomic anion with a charge more negative than −1 may add a hydrogen cation (H+) to

give another anion. These anions are named from the parent anion by adding the word

hydrogen. For example, HCO3− is hydrogen carbonate, HPO4

2− is hydrogen phosphate, and

H2PO4− is dihydrogen phosphate.

1.3 Mass and Mole of Substance

1.3.1 Molecular weight and Formula weight

Molecular weight (MW) is the sum of the atomic weights of all atoms in a molecule of a

substance. Formula weight (FW) is the sum of the atomic weights of all atoms in a formula

unit of a compound.

To convert between grams and moles use the molar mass

To convert between moles and molecules use avogadro’s number


20

1.3.2 Mole

One mole is the number of atoms in exactly 12 g of the pure isotope carbon‐12. This number

of atoms is called Avogadro constant NA = 6.022×1023 atoms/mol

For all substances the molar mass in grams per mole is numerically equal to the formula

weight in atomic mass units.

1.4 Determining Molecular Formula 1.4.1 Mass percentage from formula

The mass percent composition is a listing of the mass of each element present in 100 g of a

compound

%100% x

wholetheofmasswholetheinAofmassMass =

This percent by mass listing is also called the compound’s elemental analysis.

1.4.2 Determining formulas

Finding the empirical formula from the composition, convert the mass of the elements to

moles

Finding molecular formula from empirical formula


21

Important Concept ! Limiting reactant is

completely consumed. Others produced and reacted must be

calculated based on the limiting reactant.

1.5 Quantitative Relations in Chemical Reactions

1.5.1 Quantitative information from Chemical reactions

Equation gives the number of molecules which react to

form product

This is the ratio of number of moles of reactants

required to give ratio of number of moles of product.

The ratio of grams of reactants cannot be directly

related to the grams of products.

To get grams of product from grams of reactant

o Convert grams of reactants to moles of reactants (use molar mass)

o Convert moles of one reactant to moles of other reactants and products (use

stoichiometric ratio)

o Convert moles back to grams for desired product ( use molar mass)

1.5.2 Limiting reagents

When you have two or more reactants in a chemical reaction,

the limiting reactant is the reactant that runs out first, based

on the stoichiometric ratio of all the reactants

The other materials are said to be in excess

Question ALERT! A common question on exams is to

ask the amount(g or mol), concentrations, # of atoms or

molecules based on the balanced equation.


22

Limiting Reagent: Problem Solving Technique:

o Convert all reagent masses to moles

o Divide number of moles of each reagent by its coefficient in the BALANCED equation

o Inspect the numbers obtained; the smallest value is the LIMITING REAGENT and the

moles of products will depend only on the moles of this reactant, all the others being in

excess.

1.5.3 Percentage yield

The amount of a product obtained from a reaction is often reported as a yield

The amount of product predicted by stoichiometry is the theoretical yield, whereas the

amount actually obtained is the actual yield.

The percent yield is the percentage of the theoretical amount that is actually obtained:


⎜⎝⎛

amount lTheoreticaamount Actual Fractional yield =

yieldltheoreticayieldactual⋅

⋅


23

1.6 Aqueous Solutions

A pure substance dissolved in solution is a solute. A substance used to dissolve solutes is called a

solvent. Most of the time, the solvent is a liquid and is present in much larger quantities than any

solutes. When water is the solvent, the solution is said to be aqueous.

1.6.1 Molarity and molality

Molarity is the number of moles of solute (n) divided by the total volume of the solution (V) in

litres.

Molarity (M) = solution of volumeTotal

solute of Moles = mol/L or M =

Vn.

Molality (M) = Moles of solute

( ) of solventMass in kg = mol/kg or m =

nkg

.

1.6.2 Dilutions The quantitative aspects of dilutions can be found from the fact that the number of moles of

solute does not change during a dilution: Molessolute, initial = Molessolute, final

Because Moles = (Molarity) × (Volume), this leads to a simple equation:

MiVi = MfVf


24

1. Cations whose salts are soluble:

NH4+ and Group 1 (Li+, Na+, K+, etc.)

2. Anions whose salts are soluble:

NO3

− ClO4− HSO4

− CH3CO2−

3. Anions whose salts are mainly soluble:

Cl− Br− I− SO4

2− Exceptions:

AgCl AgBr AgI BaSO4

Hg2Cl2 Hg2Br2 Hg2I2 Hg2SO4 PbCl2 PbBr2 PbI2 PbSO4 4. Anions whose salts are mainly precipitates:

CO3

2−, S2–, PO43− except rule #1.

5. Bases are mainly precipitates except rule #1 and Ba(OH)2.

1.6.3 Precipitation Reactions

A precipitation reaction occurs when cations and anions in aqueous solution combine to form

neutral insoluble salts. The following solubility table is very useful.


25

1.7 Practice Exam Questions

Multiple Choice

1. When a 1.00 g sample of a mixture of NaCl and AlCl3 is treated with excess AgNO3 solution,

0.02142 moles of solid AgCl are obtained. What is the percentage by mass of NaCl in the

mixture? (NaCl, 58.44 g∙mol−1; AlCl3, 133.33 g∙mol−1)

A. less than 10%

B. 20%

C. 25%

D. 31%

E. 42%

2. When 900 g of a compound containing C, H and O was burned completely in a stream of pure,

dry oxygen, 0.880 g of CO2 and 0.180 g H2O were obtained. What is the empirical formula of

the compound?

A. CHO2

B. C3H3O

C. C3H5O6

D. CH2O

E. C2H2O


26

3. A 2.4917 g sample of a hydrate of cobalt (II) fluoride, CoF2 ⋅ xH2O, was heated to drive off all of

the water of hydration. The remaining solid weighed 1.4290 g. What is the formula of the

hydrate?

Molar masses, in g mol‐1:H 1.008 O 16.00 F 19.00 Co 58.93

A. CoF2 ⋅ H2O

B. CoF2 ⋅ 2H2O

C. CoF2 ⋅ 3H2O

D. CoF2 ⋅ 4H2O

E. CoF2 ⋅ 5H2O

4. How many oxygen atoms are present in 2.00 millimoles of H3PO4?

A. 4.82 x 1021

B. 1.20 x 1024

C. 6.02 x 1023

D. 1.20 x 1021

5. Combustion analysis of a 15.0 mg sample of acetaminophen produced 8.10 mg of water.

What is the percent hydrogen in acetaminophen?

A. 24.7%

B. 6.04%

C. 12.1%

D. 3.02%

wer: Bolution: 8.10mg H2O x 1mol H2O x 2mol H x 1.008 g = 0.9064mg H


27

6. The hypothetical element torontonium (symbol To) has two stable isotopes:

To‐46 = 46.046 amu, relative abundance = 64.08%

To‐51 = 50.826 amu, relative abundance = 35.92%

What is the average atomic weight of this soon to be discovered element (amu = atomic mass

units)?

A. 47.44 amu

B. 48.44 amu

C. 49.11 amu

D. 47.76 amu

E. 48.50 amu

7. What is the mass of a single bromine molecule?

A. 1.327×10–22 g

B. 2.654×10–22 g

C. 79.90 g

D. 159.8 g

E. none of the above


28

8. Nitrous oxide, N2O and oxygen react to produce NO2 according the equation

2 N2O + 3 O2 ⎯→ 4 NO2

If 6 moles of N2O and 8 moles of O2 react to produce as much NO2 as possible, how many

moles of which reactant will remain at the end of the reaction?

A. 1/3 mol O2 remains

B. 2/3 mol O2 remains

C. 1/3 mol N2O remains

D. 2/3 mol N2O remains

E. 1.5 mol O2 remains.

9. Iodine is produced from sodium iodate via the sequence of reactions:

IO3‐(aq) + 3 HSO3

‐(aq) → I‐(aq) + 3 SO42‐(aq) + 2 H+(aq)

5 I‐(aq) + IO3‐(aq) + 6 H+(aq) → 3 I2(s) + 3 H2O(l)

How many moles of I2 can be produced from 0.961 moles HSO3‐?

A. 0.247 moles

B. 0.961 moles

C. 0.247 moles

D. 0.192 moles

E. 0.808 moles


29

10. Methane (CH4) reacts with oxygen (O2) to give carbon dioxide (CO2) and water (H2O). If 16

grams of methane and 32 grams of oxygen react (until one of the reactants is completely

consumed) and 11 grams of carbon dioxide are produced, the percent yield of carbon dioxide

in this reaction is

A. 2.5 %

B. 5.0 %

C. 25%

D. 50%

E. 75%

11. The following reactions are used for the determination of dissolved oxygen in water samples.

2MnSO4(aq) + 4NaOH(aq) + O2(aq) 2MnO2(s) + 2Na2SO4(aq) + 2H2O(l)

MnO2(s) + 2H2SO4(aq) + 2NaI(aq) MnSO4(aq) + I2(aq) + Na2SO4(aq) + 2H2O(l)

I2(aq) + 2Na2S2O3(aq) Na2S4O6(aq) + 2NaI(aq)

If 2 millimoles of Na2S2O3(aq) are used in the titration of I2(aq) in the last equation, how many

millimoles of O2(aq) are initially present in the sample?

A. 0.5

B. 1

C. 4

D. 2


30

12. What is the concentration of bromide ion in a solution prepared by adding 40.0mL of water to

125mL of 0.96M aqueous iron(III) bromide?

A. 3.0M

B. 9.0M

C. 0.73M

D. 2.2M

13. 100.0 grams of A when mixed with 80.0 grams of B gives 180.0 grams of C. In an experiment

20.0 g of A is mixed with 10.0 g of B. How many grams of C will be formed?

A. 10.0

B. 20.0

C. 30.0

D. 18.0

E. 22.5

14. What is the molar mass of chlorophyll which contains 2.72% (m/m) magnesium and it is

known that there is one magnesium atom per chlorophyll molecule?

A. 66.1 g/mol

B. 1120 g/mol

C. 272 g/mol

D. 894g/mol

E. 6610g/mol


31

15. What mass of xenon tetrafluoride has the same number of fluorine atoms as 25.0g of oxygen

difluoride ?

A. 1.92g

B. 12.5g

C. 48.0g

D. 96.0g

E. 192g


32

Short Answer:

16. A. How many moles of Al(OH)3 are formed by the reaction of 500.0mL of 0.183M NaOH (aq)

with 10.0g Al(NO3)3. The molar mass of Al(NO3)3 is 213g/mol.

B. Write the balanced equation for the reaction which occurs when calcium carbonate is used

to neutralize a sulphuric acid spill. Include all states.

C. Suppose that a 1.00g sample of a hair bleaching solution labelled “6% hydrogen peroxide”,

H2O2(aq), is titrated with 0.0175M KMnO4 (aq) in acidic solution. If the titration requires

42.8mL of KMnO4(aq) to react with all of the H2O2(aq), calculate the mass percentage of the

H2O2(aq) solution. The titration equation is:

5H2O2(aq) + 2KMnO4(aq) + 6HCl (aq) → 2MnCl2(aq) + 2KCl(aq) + 8H2O(l) +5O2(g)


33

17. In the textile industry, chlorine is used to bleach fabrics. Any of the toxic chlorine that

remains after the bleaching process is destroyed by reacting it with a sodium thiosulfate

solution. Na2S2O3 + 4 Cl2 + 5 H2O (l) = 2 NaHSO4 (aq) + 8 HCl(aq)

27 kg of Na2S2O3 reacts with 10 kg of Cl2 and 50 kg of H2O.

The final concentration of sodium hydrogen sulfate in 56 litres of the final solution is 0.625 M.

What is the percentage yield?

Solution: Na2S2O3 + 4 Cl2 + 5 H2O (l) = 2 NaHSO4 (aq) + 8 HCl(aq) Mass (kg): 27 10 50 Molar mass (g/mol) 158.1 70.9 18.02 kmol: 0.1708 0.141 2.77 ratio: 0.1708 0.03525 0.555 Cl2 is the limiting reactant and the theoretical yield of NaHSO4 =

42

2

20.1414

NaHSOkmol ClCl

× = 0.0705 kmol NaHSO4 = 70.5 mol NaHSO4

The actual yield of NaHSO4 = 56 L×0.625 M = 35 mol


34

II. Atoms and Light

2.1 Chapter Summary

Frequency (ν): the number of wave crests passing a point in space in one second

Wavelength (λ): the distance between successive crests in a wave

Amplitude: the height of the wave which is related to the intensity of the light (in a light wave)

Photoelectric Effect: Ephoton = hνphoton, the energy of a photon of light equals the frequency of the

photon times Planck’s constant (6.626 x 10‐34 J.s).

Dual Nature of Light: light has some properties of waves and some properties of particles.

Excited State: when an atom absorbs a photon, the atom is transformed to a higher energy state,

atoms in excited states subsequently give up their excess energy to return to lower energy states.

Ground State: lowest energy state of an atom, which is its most stable state

Energy of a Hydrogen’s Electron: is restricted to specific values described by the following

equation: En = − 2

18 J 1018.2n

−×

Heisenberg’s Uncertainty Principle: position and momentum (mv) of a particle cannot be known

exactly at the same time.

Particle in a Box: an electron in a 1‐D box of length L the Energy associated with the nth energy

level is: En = 2

22

mL8n h

.

Quantization of Energy: orbitals within atoms are quantized, the electrons that are bound to

these energy levels are also quantized

Principle Quantum Number: the principle quantum number must be a positive integer:

n = 1, 2, 3, ….,8, 9, etc.

Azimuthal Quantum Number: The azimuthal quantum number (l) can be zero or any positive

integer smaller than n: l = 0, 1, 2, …., (n − 1).


35

Magnetic Quantum Number: the magnetic quantum number (ml) can have any positive or

negative integral value between 0 and ±l: ml = 0, ±1, ±2, …., ±l

Spin Orientation Quantum Number: spin orientation is quantized: Electron spin must be

oriented in one of two ways, labelled “up” or “down”. The (ms) indexes this behaviour. The two

possible values of ms are +½ (up) and −½ (down).

Orbital Size: each orbital becomes smaller as nuclear charge increases. As the positive charge of

the nucleus increases, the electrical force exerted by the nucleus on the negatively charged

electrons increases, too, and electrons become more tightly bound. This in turn reduces the

radius of the orbital.

Energy: E = hν, Ekinetic = 2

2mu Wavelength:

Ehc

=λ

Speed: c = 3×108 m/s u = m

Ekinetic2


36

2.2 Basic Terms and the Photoelectric Effect

We need to define the following terms:

Frequency (ν) is the number of wave crests passing a point in space in one second.

Wavelength (λ) is the distance between successive crests in a wave.

Amplitude is the height of the wave which is related to the intensity of the light (in a light wave).

Light waves always move through a vacuum at the same speed roughly 3.00 x 108 m/s

For any wave, its wavelength (in meters) multiplied by its frequency (in inverse seconds) equals

its speed (m/s) i.e. λν = c

The photoelectric effect shows how the energy of light depends on its frequency and intensity as

follows:

1. Below a characteristic threshold frequency, ν0, no electrons are observed, regardless of the

light’s intensity.

2. Above the threshold frequency, the maximum kinetic energy of ejected electrons increases

linearly with the frequency of the light.

3. Above the threshold frequency, the number of emitted electrons increases with the light’s

intensity, but the kinetic energy per electron does not depend on the light’s intensity.

4. All metals exhibit the same pattern, but as Figure 2.1 shows, each metal has a different

threshold frequency.

Figure 2.1

a0ν b0ν

Frequency of light, ν

Kin

etic

ene

rgy

ofej

ecte

del

ectro

ns


37

Question ALERT! A common question is to

ask the energy and wavelength of the photon

or electron.

The photoelectric effect can be summarized by the following equation:

Photon energy = Electron kinetic energy + Binding energy

hv = Ekinetic (electron) + hv0

When a metal surface absorbs a photon, the energy of the

photon is transferred to an electron. Some of this energy is

used to overcome the forces that bind the electron to the metal

(binding energy or work function), and the remainder shows up

as kinetic energy of the ejected electron.


38

2.3 Quantization of Energy

Quantization means that electrons bound to atoms can have only

certain specific energy levels. Niels Bohr showed that the energy of

the electron in a hydrogen‐like system (one electron species) is

restricted to specific values:

En = − 18 2

2

2.18 10 zn

−×J

In which n is the energy level and z is the nuclear charge (z = 1 for hydrogen). The negative sign

means that the energy of the electron in any orbit is lower than it would be if the electron were

at an infinite distance from the nucleus, where the energy is zero. When the electron is in level n

= 1, the hydrogen atom is in its lowest possible energy state, its ground state. Suppose an

electron in level n of an excited hydrogen atom falls back to level m. The energy will be released:

∆E = 18 2 18 2

182 2 2 2

2.18 10 z 2.18 10 z 1 1( ) 2.18 10 ( )n m m n

− −−× ×

− − − = × − J.

This amount of energy is carried away by the emission of a photon: ∆E = hν = hc/λ.

Therefore, 18 2

734 8 2 2 2 2

1 2.18 10 (1) 1 1 1 1( ) 1.097 10 ( )6.626 10 (3 10 / )

E Jhc Js m s m n m nλ

−

−

Δ ×= = − = × −

× × m–1

Or 22 2

1 1 11.097 10 ( )m nλ

−= × − nm–1


39

Worked example 2.1

Consider the Bohr model of the hydrogen atom containing one electron.

A. At what wavelength do you expect the 4→3 transition to emit light in the H‐atom emission

spectrum?

B. One reaction of a chlorofluorocarbon implicated in the destruction of stratospheric ozone is:

FCCl3 + hν → FCCl2 +Cl The bond dissociation energy of the C‐Cl bond is 328kJ/mol. What is the frequency (ν) and

wavelength (λ) of the light necessary to produce the reaction?

Solution:

A. 22 2

1 1 11.097 10 ( )3 4λ

−= × − nm–1

λ = 1875 nm

B. Convert the bond energy from kJ per mol to J per bond first.

E = 328 kJ mol( )=328 × 103 J mol( )

6.022× 1023 mol−1( )= 5.447 × 10−19 J( )

E =hcλ⇒ λ = hc

E=

6.626× 10−34 J ⋅ s( )⋅ 2.998× 108 m s( )5.447 × 10−19 J( )

= 3.647 × 10−7 m( )= 364.7 nm( )

ν =cλ=

2.998 × 108 m s( )3.647 × 10−7 m( )

= 8.220× 1014 s−1 or Hz( )


40

2.4 Properties of Electrons

There are three main properties of electrons:

1. Each electron has the same mass and charge

2. Electrons behave as magnets. The magnetic properties of electrons arise from a

property of spin. All electrons have a spin constant of ±½ .

3. Electrons have wave properties as well as particle properties (matter waves, proposed

by De Broglie). Therefore it can be described similar to light which also share both

wave and particle properties. Table 2.1 summarizes the properties of photons of light

and free electrons.

Table 2.1 Equations for Photons and Free Electrons

Property Photon Equation Electron Equation

Energy E = hν Ekinetic = 2

2mu

Wavelength Ehc

=λ muh

=λ

Speed c = 3×108 m/s u =

mEkinetic2

Important Concept ! The table shows the dual wave/particle nature of

light and matter.


41

2.5 Heisenberg’s Uncertainty Principle

The position and momentum (mv) of a particle cannot be known exactly at the same time.

ΔpΔx π2h

≥

Particle in a Box:

The “Particle in a Box” is a mathematical representation of the energies of an electron. Its theory

is quite complex and although given in course lectures, memorization of it is not necessary. What

is important is that since electrons have wave properties, it can be described using wave

equations as well as the final wave equation solution:

For an electron in a 1‐D box of length L the Energy associated with the nth energy level is:

En = 2

22

8mLn h

.

2.6 Quantization and Quantum Numbers

Since orbitals within atoms are quantized, the electrons that are bound to these energy levels are

also quantized. That is, free electrons have continuous energy states but bound electrons have

quantized energy states.

Each quantized property of a bound electron can be identified using a set of quantum numbers.

The principle quantum number (n) must be a positive integer: n = 1, 2, 3, ….,8, 9, etc.

The angular‐momentum quantum number (l) can be zero or any positive integer smaller than n: l

= 0, 1, 2, …., (n − 1).


42

Historically, orbital shapes have been identified with letters rather than numbers. These letter

designations correspond to the values of l as follows:

Value of l 0 1 2 3 4

Orbital letter s p d f g

The magnetic quantum number (ml) can have any positive or negative integral value between 0

and ±l: ml = 0, ±1, ±2, …., ±l

The spin quantum number (ms) can have either of two values, +½ or −½ , which is independent of

the other three quantum numbers.

A complete description of an atomic orbital requires a set of three quantum numbers, n, l, m and

a complete description of an atomic electron requires a set of four quantum numbers, n, l, ml,

and ms, which must meet all the restrictions summarized in Table 2.2. Any set of quantum

numbers that does not obey these restrictions does not correspond to an orbital and cannot

describe an electron.

Table 2.2 − Restrictions on Quantum Numbers for Atoms

Quantum Number Restrictions Range

N Natural numbers 1, 2, …, ∞

L Whole numbers less than n 0, 1, …, (n − 1)

ml Integers between l and −l −l…−1, 0, +1, …, +l

ms Half‐integers, +1/2 or −1/2 −1/2, +1/2


43

2.7 Shapes of Atomic Orbitals

2.7.1 Orbital Size

Orbitals get larger as the value of n increases.

2.7.2 Orbital Shapes

The quantum number l = 0 corresponds to an s orbital, which is spherical, with radii and number

of nodes that increase as n increases.

The quantum number l = 2 corresponds to a d orbital. A d electron can have any of five values for

ml (−2, −1, 0, +1, and +2), so there are five different orbitals in each set. Each d orbital has two

nodal planes.

2.8 Periodicity of Atomic Properties

2.8.1 Atomic Radii

Atomic size decreases from left to right and increases from top to bottom of the periodic table.

Bohr’s diagram must be used in order to compare sizes of ions.

2.8.2 Ionization Energy

• The minimum amount of energy needed to remove an electron from a neutral atom is called

the first ionization energy (IE1). Variations in ionization energy mirror variations in orbital

stability, because an electron in a less stable orbital is easier to liberate, than one in a more

stable orbital.

• First ionization energy increases from left to right across each row and decreases from top to

bottom of each column of the periodic table.

• Bohr’s diagram must be used in order to compare second or higher order ionization energies.


44

2.8.3 Electronegativity

• The attraction for the electrons in a covalent bond.

• Electronegativity increases from left to right across each row and decreases from top to

bottom of each column of the periodic table.

2.8.4 Electron Affinity

• A neutral atom can add an electron to form an anion.

• The energy change when an electron is added to an atom is called the electron affinity (EA)

• Electron affinity tends to become more negative from left to right across a row of the periodic

table.


45


Multiple Choice

1. What is the total energy of one mole of photons of wavelength 285 nm?

A. 6.98×10−19 J

B. 3.50×10−18 J

C. 3.15×105 J

D. 4.20×105 J

E. 2.11×106 J

2. What is the wavelength of light emitted when an electron in a hydrogen atom undergoes a

transition from the n = 5 to n = 2 level?

A. 663 nm

B. 833 nm

C. 546 nm

D. 521 nm

E. 434 nm

3. The square of the wave function, Ψ2 for an electron in an atom

A. describes the energy of the electron

B. specifies the momentum of the electron.

C. gives the probability of finding the electron in a region of space.

D. is proportional to the velocity of the electron.

E. is inversely proportional to the distance between the electron and the nucleus.


46

4. Which of the following sets of quantum numbers is not possible?

n l ml ms

A. 1 0 0 +1/2

B. 4 0 0 +1/2

C. 3 3 −3 −1/2

D. 2 1 1 −1/2

E. 2 0 0 +1/2

5. In which atom does the 1s orbital have the highest energy?

A. He

B. H

C. Li

D. Be

E. Ne

6. For the n = 5 level of an atom, the number of orbitals having a given value of l is equal to

A. 2l + 1

B. 2n + 1

C. 2ml + 1

D. n + ml

E. l + ml

l.


47

7. What is the ground‐state electronic configuration of the Cu atom?

A. [Ar]3d 94s2

B. [Ar]3d 95s2

C. [Ar]3d 104s1

D. [Kr]4d 105s1

E. [Ar]4d 95s2

8. Which one of the following transitions for the H atom produces radiation of the shortest

wavelength?

A. n = 2 to n = 3

B. n = 3 to n = 2

C. n = 5 to n = 6

D. n = 6 to n = 5

E. n = 2 to n = 1

9. The minimum energy for photoemission of electrons from a potassium surface is 3.69 x 10‐19J.

What is the kinetic energy of electrons emitted from a potassium surface when it is irradiated

by UV light at 300 nm.

A. 7.60 x 10‐20J

B. 3.69 x 10‐19J

C. 2.93 x 10‐19J

D. 6.62 x 10‐19J

E. 1.03 x 10‐18J

6 x 10 –19J ‐ 3.69 x 10‐19J = 2.93 x 10‐19J


48

10. Which of the following electron configuration does not correspond to a ground‐state

configuration of its atom or its ion?

A. S‐2: [Ar]

B. Ti: [Ar]4s23d2

C. Br: [Ar]4s23d104p5

D. Fe+3: [Ar]3d5

E. V+2: [Ar]4s23d1

Short Answer:

11. What is the ground state electron configuration of a) the Cr atom? b) the Cu+ ion?

12. How many unpaired electrons are there in the ground state of: the V atom? The Cr+2 atom?


49

III. Fundamentals of Chemical Bonding

3.1 Chapter Summary

Covalent Bond: electrons as shared between the nuclei, the attractive energy between the nuclei

and electrons exceeds the repulsive energy arising from nuclear‐nuclear and electron‐electron

interactions

Lewis Structures: schematic drawings of molecules

Single Bonds: two shared electrons

Double Bonds: four shared electrons

Triple Bonds: six shared electrons

Formal Charge (FC): the difference between the number of valence electrons in the free atom

and the number of electrons in the Lewis structure:

FC = (Valence electrons of free atom) – (Valence electrons assigned in Lewis structure)

Valence Shell Electron Pair Repulsion, VSEPR: principle of minimizing electron‐electron repulsion

Hybridization: combining atomic orbitals to form a special set of directional orbitals

Alkanes: contain chains of carbon atoms held together by covalent single bonds

Structural Isomers: when two or more compounds have the same molecular formula but

different arrangements of atoms

Steric Number: the sum of an inner atom’s coordination number and the number of its lone pairs

Coordination Number: the number of other atoms to which an atom is bonded

Shape: describes how atomic nuclei, not orbitals, are arranged in space


50

3.2 Overview of Bonding

• Three bonds within molecules (intramolecular force)

• Ionic – electrostatic forces which hold ions together

• Covalent – sharing electrons between atoms

• Metallic – metal nuclei floating in a sea of electrons

• In all chemical bonds electrons are shared or transferred between atoms

3.3 Ionic Bonding

3.3.1 Lewis Symbols and octet rule

• All chemical bonds are formed from a transfer of valence electrons between atoms

• The electrons involved in bonding are called valence electrons

• Valence electrons are found in the incomplete outermost orbital of an atom

• The number of electrons available for bonding are indicated by unpaired dots

• These symbols are called Lewis symbols

• Octet rule: we know that s2p6 is a noble gas configuration. We assume that an atom is stable

when surrounded by 8 electrons (4 electron pairs)

3.3.2 Energetics of ionic bond formation

• The energy required to separate one mole of ions in an ionic lattice into gaseous ions is called

the lattice energy ΔHlattice

• Lattice energy depends on the charge of the ions and the size of the ions

• The force is given by the equation; 1 22

q qF kd

= and the magnitude of the lattice energy

equals 1 2q qU Fd kd

= = .

• The magnitude of a lattice energy (and the melting point or boiling point of an ionic

compound) depends directly on the charge on the ions and inversely on the size of ions.


51

3.3.3 Born Haber Cycle

• Thermodynamic cycle that analyzes lattice energy

3.3.4 Size of Ions

• Ion size is important in predicting lattice energy

• Cations are smaller that their parents

• Anions are larger than their parents

• For ions of the same charge ion size increases down a group

• All members of an isoelectronic series have the same number of electrons

• As nuclear charge increases in an isoelectronic series the ions become smaller

o O2‐> F‐> Na+> Mg2+> Al3+


52

3.4 Covalent Bonding

• Sharing of electrons to each obtain an octet

• Each pair of shared electrons constitutes one chemical bond

3.4.1 Multiple bond

• Possible to share more than one pair between two atoms

• One shared pair = single bond, e.g. H2

• Two shared pair = double bond, e.g. O2

• Three shared pair = triple bond, e.g. N2

• Generally bond distances decrease as we move from a single through double to triple bonds

3.4.2 Bond polarity and Electronegativity

• Sharing of electrons does not imply equal sharing of electrons

• Unequal sharing results in POLAR BONDS

• Difference in Electronegativity is a gauge of bond polarity

• ΔE ~ 0 = non polar covalent

• ΔE ~ 2 = polar covalent

• ΔE ~ 3 = ionic bonds

• Positive end in a polar bond is represented by δ+ and the negative end by δ‐


53

3.4.3 Drawing Lewis structure

• Calculate the total number of valence electrons

• Draw a skeleton structure by connecting the central atom (usually the one closest to the

centre of the periodic table with the least electronegativity except H) with the surrounding

atoms.

• Place the remaining electrons to satisfy the octet rule for the surrounding atoms(except H).

• Put the remaining electrons on the central atom, use multiple bonds to minimize formal

charges.

FC = [# valence electrons] – [# non‐bonded electrons] – 1/2 [bonded electrons]

3.4.4 Resonance

• Some molecules not described by Lewis structure

• Structures with multiple bonds can have similar structures with the multiple bonds between

different pairs of atoms

• Common examples are O3, NO3‐, SO4

2‐, NO2

3.4.5 Exceptions to the Octet Rule

• Three exceptions to this rule

o Molecules with an odd number of electrons ClO2, NO and NO2

o Molecules in which one atom has less than an octet BF3, BeF2

o Molecules in which one atom has more than an octet atoms across the 3rd period

PCl5, SF4, AsF6‐, and ICl4

‐


54

Question ALERT! Predict shapes and estimate the bond

angles

3.5 VSEPR: Molecular Geometry

• VSEPR Theory – is a simple model for predicting molecular geometries, it is based on the idea

that the valence shell electron pairs are arranged symmetrically about an atom to minimize

electron pair repulsion

• Valence shell electrons are the bonding electrons and the lone pairs

• Lone pair‐lone pair repulsion is greatest

• Bonding pair‐bonding pair repulsion is least

• Lone pair‐bonding pair repulsion is intermediate

• The molecular geometry takes the one with least

number of 90o lone pair‐lone pair repulsion: Electron Arrangement

class Hybrid orbital

used by the cetral

atom

Molecular geometry

AX2 sp linear

AX3

AX2E

sp2

trigonal planar

angular (bent)

AX4

AX3E

AX2E2

sp3

tetrahedral

trigonal pyramidal

angular (bent)

AX5

AX4E

AX3E2

AX2E3

sp3d

trigonal bipyramidal

seesaw

T‐shaped

Linear

AX6

AX5E

AX4E2

sp3d2

octahedral

square pyramidal

square planar


55


Multiple Choice

1. For the Lewis structure below, the formal charge on Cl, N, and O, respectively are:

A. ‐1, 0, +1

B. ‐1, +1, 0

C. 0, ‐1, +1

D. 0, +1, ‐1

E. 0, 0, 0

2. What is the shape of ICl2‐?

A. linear

B. angular

C. trigonal planar

D. T‐shaped

E. Trigonal bipyramidal

3. Which of the following molecules is polar?

A. PCl5

B. BF3

C. CS2

D. SO2

E. XeF2

Cl N O


56

4. What type of orbitals best describe the bond between oxygen atom 1 and 2 in the

peroxyacetic acid?

A. sp2‐sp2

B. sp3‐sp3

C. sp2‐sp3

D. sp‐sp

E. sp2‐sp

5. Estimate the C‐O‐O bond angle in peroxyacetylnitrate. The structural formula is given below:

A. 90o

B. <120o but >109.5o

C. <109.5o

D. 180o

E. 120o

6. Which of the following species will contribute resonance structures?

A) BF4–

B) NO2–

C) OF2

D) PCl3

E) N2F4

Ans: B) NO2–

Solution:

CH3 C

O

O OH1 2

CH3 C

O

O O N

O

O


57

IV. The Behaviour of Gases

4.1 Chapter Summary

Average Kinetic Energy ( kineticE ): found by adding all the individual molecular energies and

dividing by the total number of molecules. The average kinetic energy depends on the

temperature of the gas: A

kinetic 23

NRTE =

Ideal Gas: one for which both the volume of molecules and the forces among the molecules are

so small that they have no effect on the behavior of the gas, PV = nRT

Units of Pressure: 1 atm = 760 mm Hg = 760 torr = 101.325 kPa = 1.01325 bar

Dalton’s Law of Partial Pressures: for a mixture of n ideal gases in a container, the total pressure

of gas in the container is the sum of pressures that each gas exerts alone,

Ptotal = P1 + P2 + P3 + … + Pn

Mole Fraction (X): fraction of the number of moles of all substances in the mixture,

Mole fraction of A = XA = total

A

total

A

PP

nn

=

Real Gas Equation:

2

⎟⎠⎞

⎜⎝⎛−

−=

Vna

nbVnRTP

where nb is a volume correction factor and an2/V2 is a pressure correction factor.

Gases differ from solids or liquids in the fact that they have much more range of motion due to

the fact that they are not greatly attracted to one another. There’s much more disorder, or

entropy, in a mole of gas than there is in a

mole of either a liquid or solid. This

freedom of motion allows gas molecules to

expand and compress quite readily; in fact,

gases will expand to fill any volume they

are contained in.


58

Important Concept ! Root mean square speed is T and M related but

average kinetic energy is related to T only.

4.2 Molecules in Motion

The kinetic energy of a molecule of a gas is related to its speed: KE = ½ mu2

At any given temperature, ALL gases have the same molecular kinetic energy distribution. Also,

for any given gas, the kinetic energy increases as temperature increases.

4.2.1 Average Kinetic Energy

More molecules have the most probable kinetic energy than any other value. By proportion,

however, very few molecules actually have the most probable kinetic energy. This is because any

sample of gas contains many molecules that move with many different kinetic energies. The

average kinetic energy ( kineticE ) per molecule can be found by adding all the individual molecular

energies and dividing by the total number of molecules. The average kinetic energy depends on

the temperature of the gas: A

kinetic 23

NRTE =

If you want to have the average kinetic energy per mole

of gas molecules simply leave out Avogadro’s number.

4.2.2 Root Mean Square Speed

The average kinetic energy of a gas particle is represented by 1/2mv2. v2 means the average of

the squares of the particle velocities. The square root of it is thus called

root mean square speed: vrms = kineticA

2 2 3 32

RT RTEm m N M

= × =

(M =mNA is the molar mass of the gas in kg/mol NOT g/mol !)


59

4.2.3 Graham’s law of effusion

The relative rates of effusion of two gases at the same temperature and pressure are given by the

inverse ratio of the square roots of the masses of the gas particles:

2

1

12

MRates of effusion for gasRates of effusion for gas M

=

4.3 The Ideal Gas Equation

There are two extremely important assumptions about an ideal gas:

1. The volume occupied by the molecules of an ideal gas is negligible compared with the volume

of its container.

2. The energies generated by forces among ideal gas molecules are negligible compared with

molecular kinetic energies.

The Ideal Gas Equation describes the macroscopic behavior of an ideal gas and is given by:

PV = nRT

It is useful to know the conversion factor of pressure:

1 atm = 760 mm Hg = 760 torr = 101.325 kPa = 1.01325 bar


60

Question ALERT! Understand the relationship based on PV=nRT and link them with different graphs, either a linear or inverse

plot.

4.4 Applying the Ideal Gas Equation

4.4.1 Variations on the Gas Equation

During chemical and physical transformations, any of the four

variables in the ideal gas equation (P, V, n, T) may change, and

any of them may remain constant. A good strategy for

organizing gas calculations is to determine which variables do

not change and rearrange the gas equation to group them all on

the right. Here are some examples:

For a fixed amount at constant temperature, PV = nRT = constant

For a fixed amount at constant volume, VnR

TP= = constant

For a fixed amount at constant pressure, P

nRTV= = constant

For a fixed volume at constant temperature, VRT

nP= = constant

For a fixed pressure at constant temperature, P

RTnV= = constant

4.4.2 Determination of Molar Mass

The ideal gas equation can be combined with the mole‐mass relation to find molar mass of an

unknown gas: PV = nRT (Ideal gas equation) and n = MM

m (Mole‐mass relation)

will give: PV

mRTMM =


61

4.5 Gas Mixtures

4.5.1 Dalton’s Law of Partial Pressures

For a mixture of ideal gases in a container, the total pressure of gas in the container is the sum of

pressures that each gas exerts alone.

Ptotal = P1 + P2 + P3 + … + Pn = (n1 + n2 + n1 + … + nn)RT / V

Since RT/V is the same for each gas in the container, the pressure of an ideal gas mixture is

determined by the total number of gas particles, not by the identity of each component in the

mixture.

4.5.2 Describing Gas Mixtures

The way of stating composition in fractional terms is the mole fraction (X):

Mole fraction of A = XA = total

A

total

A

PP

nn

=

Which gives PA = XAPtotal.


62

Important Concept ! 1. Why is the pressure correction

added? (the attractions make the real gas P smaller)

2. Why is the volume correction subtracted? (particles of a real gas take up some space)

3. a and b tend to be greater for larger molecules.

4. A real gas would behave like an ideal one at high temperature and low pressure.

4.6 Real Gases

Experiments have shown that the ideal gas law describes

the behavior of a real gas quite well at moderate

pressures and temperatures but not so well at high

pressures and low temperatures. At these latter

conditions, the two assumptions(see 4.3. no

interparticle interactions and zero volume for the gas

particle) that are the basis of the ideal gas equation

break down and we have a REAL gas that must be

described using a REAL gas equation:

( )2

2

anP V nb nRTV

⎛ ⎞+ − =⎜ ⎟

⎝ ⎠

where b is a volume correction factor because the volume of a real gas at high pressure is larger

than predicted by the ideal gas law and a is a pressure correction factor because the attractive

forces between the particles become more important at high pressures.

Values of some van der Waals constants are listed in Table below:

GAS a (L2 atm mol−2) b (L mol−1)

He 0.034 0.0237 Ar 1.35 0.0322 CO2 3.59 0.0427 C2H4 4.53 0.0571 H2O 5.46 0.0305 NH2 4.17 0.0371 Cl2 6.49 0.0562


63


Multiple Choice

1. Which graph represents the plot of pressure vs. temperature in degrees Celsius for an ideal

gas at constant volume and number of moles?

A.

B.

C.

D.

E.

2. Which graph represents the plot of the product of pressure and volume vs. volume for an

ideal gas at constant temperature and number of moles?

A.

B.

C.

D.

E.


64

3. Which of the following is true of an ideal gas:

A.

B.

C.

D.

E.

the origin


65

4. Which of the following gases would have a density of 1.39 g/L at 15oC and 780 mmHg?

A. Ar

B. N2

C. CO2

D. O2

E. none of above

5. Which of the following statements regarding the kinetic‐molecular theory of ideal gases is

incorrect?

A. gas molecules collide with each other randomly

B. the size of gas molecules is negligible

C. at the same temperature, the average kinetic energy is the same for different gases

D. attractive and repulsive forces among gas molecules are negligible

E. at the same temperature, the kinetic energy of all the molecules is the same for the same

gas

Answer: E

6. The average speed of He gas at an unknown temperature is found to be 1200 m/s. What is

the average speed of H2O gas at this same temperature?

A. 1200 m/s

B. 267 m/s

C. 400 m/s

D. 2550 m/s

E. 566 m/s


66

7. Consider three identical flasks at room temperature and pressure. Flask A contains NH3 gas,

flask B contains NO2 gas, and flask C contains N2 gas. Which contains the largest mass of

molecules?

A. flask A

B. flask B

C. flask C

D. all are the same

E. not enough information to solve this problem

8. Which of the following gases is likely to behave the least like an ideal gas at room

temperature and pressure and which is most likely to behave as an ideal gas?

i. methane ii. ethane iii. diethyl ether iv. carbon dioxide v. hydrogen

A. (iv) is least likely to behave as an ideal gas and (v) most likely

B. (v) is least likely to behave as an ideal gas and (i) most likely

C. (iii) is least likely to behave as an ideal gas and (v) most likely

D. (i) is least likely to behave as an ideal gas and (iv) most likely

E. (iii) is least likely to behave as an ideal gas and (i) most likely

9. A mixture of helium and argon contains 3 moles of He for every 2 moles of Ar. The partial

pressure of Ar is

A. 2/3 of the total pressure

B. 1/3 of the total pressure

C. 3/5 of the total pressure

D. 1/2 of the total pressure

E. 2/5 of the total pressure


67

10. The root‐mean‐square speed of a nitrogen molecule at 1000°C in units of m/s is

A. more than 1000

B. between 900 and 1000

C. between 500 and 900

D. between 300 and 500

E. less than 300

11. Which of the following is not a unit of pressure?

A. Torr

B. mm Hg

C. N/m

D. Pa

E. bar

12. The “ideal gas”

A. is a good model for gas behaviour at high pressure and high temperature.

B. is a good model for gas behaviour at high pressure and low temperature.

C. is a good model for gas behaviour at low pressures and high temperature.

D. is a good model for gas behaviour at low pressure and low temperature.

E. has a pressure that varies inversely with its volume at different temperatures.

each other much, making the gas behave in an ideal fashion.


68

13. Consider two 1 L flasks, one containing oxygen, the other containing methane (CH4), each at

room temperature and pressure. Which of the following statements is NOT true regarding

these gases?

A. The gases in each flask have the same density

B. Each flask contains the same number of molecules

C. The gases in each flask have the same average kinetic energy

D. The molecular kinetic energy distributions are different

E. The root mean square speeds are different.

ass.

14. In a 2.00 L flask at 300 Co , 3.00 g of chlorine and 4.82 g of fluorine reacted completely to form

an unknown gaseous product. At the end of the reaction, the total pressure in the flask was

2.00atm. The product of the reaction has to be

A. FCl

B. FCl3

C. ClF3

D. F2Cl6

E. Cl2F6


69

15. A vessel of unknown volume contains an ideal gas at 0.980 atm. When a portion of the gas is

withdrawn and adjusted to a pressure of 1.00 atm at the same temperature, it is found to

occupy a volume of 35.0 cm3. It is also found that the pressure of the gas remaining in the

vessel has dropped to 0.855 atm. What is the volume of the vessel?

A. 1.17 L

B. 0.410 L

C. 0.360 L

D. 0.280 L

E. 4.09 L


70

Short Answer: 16. A mixture of Ar and N2 gases have a density of 1.413 g/L at STP. What is the mole fraction of

each gas? 17. 15.0 moles of a gas in a 2.00 L container exert a pressure of 151 atm at 25.0oC. Is this an ideal

gas?


71

18. A sample of butane gas C4H10 of unknown mass is contained in a vessel of unknown volume at 26.8°C and a pressure of 560 mmHg. To this vessel 8.68 g of Neon gas are added in such a way that no butane is lost from the vessel. The final total pressure in vessel is 168 kPa at the same temperature 26.8°C. What is the volume of the vessel and what is the mass of butane gas?

19. A constant volume vessel contains 12.5 g of a gas at 21oC. If the pressure is to remain

constant as the temperature is raised to 210oC, how many grams of gas must be released?

Solution: PV1 = nRT1 PV2 = nRT2

2 2

1 1

273 210V 273 21V T

T+

= =+

= 1.643

V2 = 1.643 V1 This V2 is the total volume at 210oC. The amount of gas remaining in the container at 210oC

would be: 2

12.5 gV

(V1) = 1

12.51.643

gV

(V1) = 7.61 g

12.5 – 7.6 = 4.9 g of the gas must be released.


72

V. Solids, Liquids, Solutions and Their Properties ‐ Effects of Intermolecular Forces

5.1 Chapter Summary

Dispersion Forces: describe the attraction between the negatively charged electron cloud of one

molecule and the positively charged nuclei of neighbouring molecules

Dipolar Forces: describe the attraction between the negatively charged end of a polar molecule

and the positively charged ends of neighbouring polar molecules

Hydrogen Bonding: involves lone pairs of electrons on an electronegative atom of one molecule

and a polar bond to hydrogen in another molecule, they are confined to molecules that contain

O, N, and F atoms.

Polarizability: the ease with which the electron clouds are distorted

Molecular Solid: molecules in molecular solids are held in place by the types of forces: dispersion

forces, dipolar interactions, and/or hydrogen bonds

Metallic Solid: atoms in metallic solids are held in place by delocalised bonding

Network Solid: contains an array of covalent bonds linking every atom to its neighbours

Ionic Solid: contains cations and anions, attracted to one another by coulombic interactions

Close‐Packed Structure: a solid is most stable when each atom, molecule, or ion has as many

neighbours as possible, thus maximizing intermolecular attractions

Phase Diagrams: show how the stability of a phase depends on temperature and pressure, it is a

P‐T graph that shows the ranges of temperature and pressure over which each phase is stable


73

5.2 Types of Intermolecular Forces

force Strength Characteristics

Ion‐dipole Moderate (10‐50 kJ/mol) Occurs between ions and polar solvents

Dipole‐dipole Weak (3 ‐ 4 kJ/mol) Occurs between polar molecules

London dispersion Weak (1 ‐ 10 kJ/mol) Occurs between all molecules; strength

depends on size, polarizability

Hydrogen bond Moderate (10‐40 kJ/mol) Occurs between molecules with O‐H, N‐H, and

F‐H bonds

5.3 Forces in Solids

The molecules (or atoms, for noble gases) of a molecular solid are held in place by the

intermolecular forces: dispersion forces, dipolar interactions, and/or hydrogen bonds. The

atoms of a metallic solid are held in place by delocalised bonding. A network solid contains an

array of covalent bonds linking every atom to its neighbours. An ionic solid contains cations and

anions, attracted to one another by ionic bonds.

Solid type Molecular Molecular Molecular Metallic Network Ionic Attractive Forces

Dispersion Dispersion + dipolar

Dispersion + dipolar + H bonding

Delocalized Bonding

Covalent Ionic bonds

Energy (kJ / mol)

0.05 – 40 5 – 25 10 – 40 75 – 1000 150 – 500 400 – 4000

Example Ar HCl H2O Cu SiO2 NaCl Melting Point (K)

84 158 273 1357 1983 1074

Molecular Picture


74

Question ALERT! A common question is to ask the order of melting

and boiling points.

Many properties of liquids (viscosity, surface tension, melting point, and boiling point) and solids

(melting point) are related to the intermolecular forces. It is important to recognize the relative

importance of different types of intermolecular forces and to

compare the relative strength of different kinds of forces.

Strategy to find the order of melting or boiling points of

different substances:

Eg. Arrange the following in the decreasing order of their

melting points NaCl, KCl, HCl, HF, F2, Ne, MgO.

Step 1. Group substances into two: group 1 with high mp/bp and group 2 with low mp/bp.

Covalent network (diamond and quartz) and ionic compounds (made of metal and nonmetal) are

in group 1 because they all have high mp/bp. Molecular compounds (made of nonmetal and

nonmetal) are in group 2 because they all have low mp/bp.

NaCl, KCl, and MgO are in group 1 and HCl, HF, F2, and Ne are in group 2.

Step 2. Compare the relative mp/bp for ionic compounds in group 1 based on the following two

rules in order:

Rule 1. Higher the charge of the ions stronger the ionic bond and higher the mp/bp;

Rule 2. Larger the size of the ions weaker the ionic bond and lower the mp/bp.

MgO has charged particles Mg2+ and O2– and it has the highest mp/bp based on the rule 1. K+ is

bigger than Na+ and NaCl has higher mp/bp than KCl based on the rule 2.

Therefore, the increasing order of mp/bp for this group is: KCl<NaCl<MgO.


75

Step 3. Compare the relative mp/bp for molecular compounds in group 2 based on the following

three rules in order

Rule 1. A compound with hydrogen bond (containing O‐H, N‐H, or F‐H bond) would have a

relative high mp/bp as long as the molar masses are not very different

Rule 2. The compound with relatively high molar mass would usually have higher mp/bp because

of the relatively greater London forces due to higher polarizability)

Rule 3. The polar molecule will have relatively higher intermolecular forces than the nonpolar

one when they have similar molar masses.

HF has the highest mp/bp in group 2 based on the rule 1 because it has hydrogen bond.

Ne has the lowest mp/bp in group 2 based on the rule 2 because it has the lowest molar mass

and weakest London forces.

HCl has relatively higher mp/bp than F2 based on the rule 3 because they have similar molar mass

but HCl has dipole forces in addition to London forces.

The decreasing order of their melting points is: MgO> NaCl> KCl> HF > HCl > F2> Ne.


76

Important Concept! Understand how to derive this table

5.4 Close‐Packed Crystals

A solid is most stable when each atom, molecule, or ion has as many neighbours as possible, thus

maximizing intermolecular attractions. An arrangement that accomplishes this is called a close‐

packed structure.

Summary of the Three Kinds of Packing for Spheres

Unit cell Primitive cubic Body centered cubic Face centered cubic

Structure Simple cubic Body centered cubic Cubic closest

packed

Stacking pattern a‐a‐a‐a‐a‐ a‐b‐a‐b‐ a‐b‐c‐a‐b‐c‐

Coordination number 6 8 12

Atom(s) per unit cell 1 2 4

Atomic radius (r) and

length of unit cell (a)

2r = a 4 3r a= 4 2r a=

Packing efficiency

calculation

334 52%

3r aπ÷ =

3342 68%

3r aπ

× ÷ =3

344 74%3r aπ

× ÷ =

Density

calculation

3

A

MM aN

÷ 32A

MM aN

× ÷ 34A

MM aN

× ÷

Examples Po only Na, Fe,

and 14 others

Ag, Cu,

and 16 others


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5.5 Phase Diagrams

Phase diagrams show how the stability of a phase depends on temperature and pressure. That is,

a phase diagram is a map of the pressure‐temperature world showing the phase behavior of a

substance.

The phase diagram for water, shown in Figure below, illustrates the features for a familiar

substance. The figure shows that liquid water and solid ice coexist at the normal freezing point, T

= 273.15 K at P = 1.00 atm. Liquid water and water vapor coexist at the normal boiling point, T =

373.15 K and P = 1.00 atm. The triple point of water occurs at T = 273.16K and P = 0.0060 atm.

The figure shows that when P is lower than 0.0060 atm, there is no temperature at which water is

stable as a liquid. At low pressure, ice sublimes but it does not melt.

The dashed lines on the phase diagram for water show two paths that involve phase changes for

water. The horizontal dashed line shows what happens as the temperature is changed at

constant pressure of 1 atm and the dashed vertical line shows what happens as the pressure on

water is reduced at constant temperature of 298 K. The ice/water line bends to the left,

indicating that the melting point is decreased as the pressure is increased. Notice that the

liquid/gas boundary line abruptly ends at the critical point, where the two phases blend into each

other to form a supercritical fluid that is neither true liquid nor true gas. No distinct physical

phase change occurs on going beyond the critical point. For water, the critical temperature Tc =

374.4oC and the critical pressure Pc = 217.7 atm.

fp (273.15 K) bp (373.15 K)

tp (0.0060 atm, 273.16 K)

298 K

Temperature, T

Pres

sure

, P 1 atm

Solid Liquid

Vapor

The phase diagram for water. Arrows indicate the triple point (tp), normal freezing point (fp), and normal boiling point (bp).


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5.6 Solutions and Colligative Properties

5.6.1 vapour pressure of a pure substance.

Experiments showed that the pressure P, enthalpy of vaporization, �Hvap, and temperature T are

related, P = A exp (‐ �Hvap / R T) or ln P = – �Hvap / (RT) + C

Where A and C are constants characteristic of a given liquid. This is known as the Clausius‐

Clapeyron equation, which allows us to estimate the vapor pressure at another temperature.

5.6.2 Vapour pressure lowering of solutions: Raoult’s Law

For a solution containing a nonvolatile solute:

The vapour pressure of the solution, Psoln = Psolv Xsolv

The vapour pressure lowering, ∆Psoln = Psolv Xsolute

A solution of an ionic compound usually requires a van’t hoff factor (i) to count for the partial

dissociation of ionic compounds.

For a solution containing a volatile solute:

The total vapour pressure of the solution is

Ptotal = PA + PB = PoA XA + P

oB XB

5.6.3 Boiling point elevation and freezing point depression of solutions

The boiling point elevation, ∆Tb = Kb m i

The freezing point depression, ∆Tf = Kf m i

5.6.4 Osmosis and osmotic pressure

Osmosis (the migration of solvent molecules through a semipermeable membrane) occurs when

solvent and solution are separated by the membrane. Π = iMRT


79


Multiple Choice

1. Consider the following molecules, all with molecular formula C6H14:

1

2

3

4

Which has the lowest boiling point?

A. All have the same boiling point because they have the same molecular formula.

B. 1

C. 4

D. 2

E. 3

CH3CH2CH2CH2CH2CH3

CH3CH2CH2CHCH3

CH3

CH3CH2CHCH2CH3

CH3

CH3CH2CCH3

CH3

CH3


80

2. The quantity ln (Pvap) is plotted as a function of I/T for both water and acetone. Which ONE of

the following is TRUE?

A. The plots are steeply rising curves because vapour pressure increases with

temperature.

B. The plots are straight lines with the same slopes and y‐intercepts.

C. The plots are straight lines and the slope of the plot for water is steeper than

for acetone.

D. The plots are straight lines and the slope of the plot for acetone is steeper than

for water.

E. The plots are straight lines with the same slopes but different y‐intercepts.

3. Which of the following substances should have the lowest boiling point?

A. Na2S

B. HF

C. NH3

D. N2

E. H2O


81

4. Which substance in each pair is more likely to boil at higher temperature?

A. CH3CH2CH2CH2CH3 or CH3CH2CH2CH3

B. CH3OCH3 or H2O

C. He or Xe

D. CH3OH or CF4

Solution: (a) CH3CH2CH2 CH2CH3

5. Which ONE of the following statements about the solid‐liquid‐vapour phase diagram shown

below is FALSE?

A. Point B represents a melting point for the substance.

B. C represents the triple point.

C. The liquid phase boils at point D.

D. K represents the normal boiling point for this substance.

E. There are different phases present at point B than at point M.

D

M

K C

B

Solid

liquid

Vapour

Pres

sure

(atm

)

Temperature (K)

1.0


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6. Which ONE of the following decreases as the strength of the attractive intermolecular forces

increases?

A. The heat of vaporization

B. The normal boiling point

C. The extent of deviations from the ideal gas law

D. The sublimation temperature of the solid

E. The vapor pressure of a liquid

7. The density of a solid phase of a substance is 0.90 g/cm3 and the density of the liquid phase is

1.0 g/cm3. The phase diagram for such a substance suggests that an increase in

pressure .

A. lowers the freezing point

B. raises the freezing point

C. lowers the boiling point

D. raises the triple point

E. lowers the triple point

8. When a nonvolatile solute is added to a volatile solvent, the solution vapour pressure

__________, the boiling point ___________, the freezing point ____________, and the

osmotic pressure across a semipermeable membrane ___________.

A. decreases, increases, decreases, decreases

B. increases, increases, decreases, increases

C. increases, decreases, increases, decreases

D. decreases, increases, decreases, increases

E. decreases, decreases, increases, decreases


83

9. Thyroxine is an important hormone that controls the rate of metabolism in the body. At

25°C, an osmotic pressure of 109 mmHg is recorded for a 10.00 mL solution containing 45.5

mg of thyroxine. What is the molar mass of thyroxine?

A. 5.90×105 g/mol

B. 7.77×102 g/mol

C. 2.33×103 g/mol

D. 2.85×102 g/mol

E. 3.76×103 g/mol

10. The phase diagram for a certain substance is shown below. At which of the following values

of T and P is the substance a pure liquid?

A. T = 8°C, P = 1 atm

B. T = 10°C, P = 0.5 atm

C. T = 70°C, P = 1.2 atm

D. T = 80°C, P = 1atm

E. T = 10°C, P = 1atm

Answer: C

Solution: On the diagram, the left portion

is solid, the central portion is liquid and

10

70

0.5

0

1

T (°C)

P (a

tm)

−50


84

Short Answer:

11. In each of the following groups of substances, pick the one that has the given property. Justify

each answer.

1) highest boiling point: HCl, Ar, F2

2) highest freezing point: H2O, NaCl, HF

3) lowest vapor pressure at 25oC: Cl2, Br2, I2

4) lowest freezing point: N2, CO, CO2

5) lowest boiling point: CH4, C2H6, C3H8

6) highest boiling point: HF, HCl, HBr

7) lowest vapor pressure at 25oC: C3H8, (CH3)2CO, C3H7OH


85

12. 90.0 g of a protein are dissolved in 2.00 L of aqueous solution. The osmotic pressure of the

solution at 25 Co is 12.7 torr.

A. What is the van’t Hoff factor for a normal protein?

B. Calculate the molarity of the protein solution?

C. Calculate the molar mass of the protein (in units of g/mol).


86

13. Liquids A and B, form ideal solutions. A solution of A and B is prepared at 300 K in which the

mole fraction of A is 0.256. In the vapor phase, in equilibrium with this solution, the mole

fraction of A is 0.318 and the total pressure is 673 torr.

A. Calculate the partial pressures of A and B in the vapor phase.

B. Calculate the vapor pressures of pure A and pure B at 300 K.

C. Provide, in terms of intermolecular interactions, three reasons that may explain the

difference between the vapor pressures of pure A and pure B.


87

14. The radius of a silver atom is 144 pm and the density of crystalline silver is 10.6 g/cm3. What

type of cubic close‐packed crystal structure should crystalline silver have?

3 = 2(107.8682g/6.022x1023)/10.6 g/cm3 a = 3.23x10–8 cm = 323 pm From the radius formula: 4r = a (the right triangle has one leg of a, another leg of a, and the hypotenuse of 4r) r = = x323 pm/4 = 140 pm This doesn’t matches the radius given.


88

VI. Kinetics: Mechanisms and Rates of Reactions

6.1 Chapter Summary

Reaction Mechanism: the exact molecular pathway that starting materials follow on their way to

becoming products

Rate‐Determining Step: the slowest elementary step in a mechanism

Rate Law: links the rate of a reaction with the concentrations of the reactants through a rate

constant (k)

First‐Order Rate Law: Rate = k[A], where A is a reactant in the overall reaction, this can be

converted to kt=⎟⎟⎠

⎞⎜⎜⎝

⎛[A][A]

ln 0

Half‐Life (t1/2), 1st Order: when half the original concentration has been consumed,

[A] = 0.5[A]0, k

t 2ln2/1 =

Second‐Order Rate Expression: Rate = k[A]2, this can be converted to kt=−0[A]

1[A]1

Bimolecular elementary reaction: A + B → products Elementary rate = k[A][B]

Unimolecular elementary reaction: C → products Elementary rate = k[C]

First Step is Rate‐Determining: the predicted rate law for the overall reaction is the rate

expression for that first step

Activation Energy (Ea): energy barrier, the minimum energy that must be supplied before the

reaction can occur


89

Activated Complex: the molecular arrangement at the point of highest energy along the energy

level diagram

Temperature and Rate: the rate of reaction increases with temperature.

Arrhenius Equation: /RTEaAek −= or RT

EAln kln a−=

Catalyst: functions by changing the mechanism of a reaction in a manner that lowers activation

energy barriers, it is not part of the overall stoichiometry of the reaction and is not consumed as

the reaction proceeds


90

Question ALERT! A common question is to derive the rate law and

calculate k.

6.2 Rates of Chemical Reactions

A Macroscopic View: Concentration Changes

In order to measure the rate of a reaction, the concentration of a particular compound is

monitored as a function of time. For a reaction of the form

a A + b B → d D + e E, the rate expressions are related as follows:

Reaction rate = ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−

ΔtΔ[A]

a1

= ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛−

ΔtΔ[B]

b1

= ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

ΔtΔ[D]

d1

= ⎟⎠⎞

⎜⎝⎛⎟⎠⎞

⎜⎝⎛

ΔtΔ[E]

e1

6.3 Concentration and Reaction Rates

Rate Laws

A rate law links the rate of a reaction with the concentrations

of the reactants through a rate constant (k): Rate = k[A]y[B]z

These exponents (y and z) are called the orders of the reaction.

The sum of the exponents is known as the overall order of the

reaction. A rate law must always be determined by conducting

experiments; it can never be derived from the stoichiometry of

the overall chemical reaction.


91

Important Concept! Understand how to derive this table and apply the concepts to solve problems. Notice that the differential equation is used to deal with rates and the integrated equation is used to solve time.

6.4 Experimental Kinetics Order 0th order First order Second order Rate law Rate = k Rate = k[A] Rate = k[A]2 Unit of k Ms–1 s–1 M–1s–1 [ ] vs time [A] = –kt +[A]o ln [A] = –kt +

ln [A]o 1 1

[ ] [ ]o

ktA A

= +

Half life [ ]2

oAk

ln 2k

1

[ ]ok A

Linear plot [A]

t

ln [A] t

[A]

t


92

6.5 Linking Mechanisms and Rate Laws

6.5.1 Reaction Mechanism

Each reaction consists of a sequence of molecular events called a reaction mechanism.

A reaction mechanism is the exact molecular pathway that starting materials follow on their way

to becoming products.

6.5.2 The Rate‐Determining Step

The slowest elementary step in a mechanism is called the rate‐determining step. The rate‐

determining step governs the rate of the overall chemical reaction because no net chemical

reaction can proceed faster than its slowest step.

The mechanism of any chemical reaction has the following characteristics:

1) The sum of the individual steps in the mechanism must give the overall balanced chemical

equation. Sometimes a step may occur more than once in the mechanism.

2) The reaction mechanisms must be consistent with the experimental rate law.

Bimolecular elementary reaction:

A + B → products Elementary rate = k[A][B]

Unimolecular elementary reaction:

C → products Elementary rate = k[C]


93

6.6 Reaction Rates and Temperature

6.6.1 Activation Energy

In almost all chemical reactions, the molecules must overcome an energy barrier before the

starting materials can become products. This energy barrier, the minimum energy that must be

supplied before the reaction can occur, is called the activation energy (Ea) of the chemical

reaction.

6.6.2 The Arrhenius Equation

The value of the rate constant for a particular reaction depends on the activation energy of the

reaction, the temperature of the system, and how often a collision occurs in which the atoms are

in the required orientation. All of these factors can be summarized in a single equation, called

the Arrhenius equation: /RTEaAek −=

Qualitatively, this means that as Ea gets larger, the rate constant gets smaller, and that as T gets

larger, the rate constant gets larger. This equation can be rewritten in another form, which is

easier to graph:

RTE

Aln kln a−=

This leads to a linear graph: plotting ln k along the y‐axis versus 1/T along the x‐axis gives a

straight line with a slope of –Ea/R and an intercept of ln A. Thus A and Ea can be evaluated after k

has been measured at several different temperatures.


94

6.7 Catalysis

Another way to make a reaction go faster is to add a substance called a catalyst. A catalyst

functions by changing the mechanism of a reaction in a manner that lowers activation energy

barriers. Although the catalyst changes the mechanism of a reaction, it is not part of the overall

stoichiometry of the reaction and is not consumed as the reaction proceeds.


95


Multiple Choice

1. For the overall chemical reaction shown below,

2H2S(g) + O2(g) → 2S(s) + 2H2O(g)

Which one of the following statements can you rightly assume?

A. The reaction is third‐order overall

B. The rate law is: rate= k[H2S]2 [O2].

C. The reaction is second‐order with respect to H2S.

D. The rate law is: rate=k[H2O]2 /{[H2S]

2 [O2]}.

E. The rate law cannot be determined from the reaction stoichiometry.

2. The rate of consumption of iodide, I– (aq), in the reaction:

S2O82–

(aq) + 3I– (aq) → I3

– (aq) + 2SO4

2– (aq)

is 5.40(mol I–)•L–1•s–1. What is the rate of formation of SO42–

(aq)?

A. 1.80 (mol SO42–)•L–1•s–1

B. 3.60 (mol SO42–)•L–1•s–1

C. 5.40 (mol SO42–)•L–1•s–1

D. 8.10 (mol SO42–)•L–1•s–1

E. 10.8 (mol SO42–)•L–1•s–1


96

3. The following data were obtained for the reaction of NO with O2. Concentrations are in

molecules/L and rates are in molecules/L⋅s.

[NO]0 [O2]0 Initial Rate

1×1021 1×1021 2.0×1019

2×1021 1×1021 8.0×1019

3×1021 1×1021 18.0×1019

1×1021 2×1021 4.0×1019

1×1021 3×1021 6.0×1019

Which of the following is the correct rate law?

A. Rate = k[NO][O2]

B. Rate = k[NO][O2]2

C. Rate = k[NO]2[O2]

D. Rate = k[NO]2

E. Rate = k[NO]2[O2]2


97

4. The rate constant k depends on

A. the concentration of the reactants and the temperature.

B. the nature of the reactants and the order of the reaction

C. the nature of the reactants and the temperature

D. the order of the reaction and the concentration of the reactants

E. the order of the reaction and the temperature

5. If the reaction

2 HI (g) ⎯→ H2 (g) + I2 (g)

is second order, which of the following plots would be linear?

A. 1/[HI] vs. time

B. 1/[HI]2 vs. time

C. ln[HI] vs. time

D. −[ ]dt

d HIvs. time.

E. None of the above would be linear.


98

6. The reaction 2 H2O2 ⎯→ 2 H2O + O2 has the following mechanism.

H2O2 + I− ⎯→ H2O + IO

−

H2O2 + IO− ⎯→ H2O + O2 + I

−

What is the catalyst in the reaction?

A. H2O

B. I−

C. H2O2

D. IO−


Use the potential energy diagram below to answer the following two questions.

E

Reaction coordinate

Pote

ntia

l Ene

rgy

A

B D

C


99

7. Which letter in the above diagram represents the activation energy for the forward reaction,

i.e for the uncatalyzed conversion of reactants into products?

A. A

B. B

C. D

D. E

E. None of the above.

8. Which letter represents the change in potential energy for the overall reaction?

A. A

B. B

C. D

D. E



100

9. The rate law for a reaction is rate = k[A]2[B]. Which of the following mechanisms supports

this rate law?

i) A + B E (fast)

E + B ⎯→ C + D (slow)

ii) A + B E (fast)

E + A ⎯→ C + D (slow)

iii) A + A ⎯→ E (slow)

E + B ⎯→ C + D (fast)

A. i

B. ii

C. iii

D. two of these

E. none of these.

⎯→ ←⎯

⎯→ ←⎯


101

Short Answer:

10. Rate constants for a certain reaction are 3.88×10−5 s−1 at 27oC and 5.98×107 s−1 at 727oC

A. What are the activation energy and pre‐exponential factor for the reaction?

B. Suppose that the reaction is catalyzed at 227oC and the rate constant is measured to be

4.40×107 s−1. Assuming the catalytic mechanism is entirely due to lowering of activation

energy, how much has the transition state energy been lowered?

C. A t what temperature does the uncatalyzed reaction proceed at the same rate as the

catalyzed reaction at 227oC?


102

11. For the reaction

2 A + B ⎯→ Products

The following mechanism is proposed:

A + B M Forward rate constant: k1 Backward rate constant: k−1

A + M ⎯→ Products Forward rate constant: k2

A. Assuming that the second step is the rate‐determining step and the first step is a fast

equilibrium step, determine the rate law. Represent the rate constant in terms of k1, k−1,

and k2.

B. Using the steady state approximation, determine the rate law.

C. Under what conditions are the rate laws above the same?

⎯→ ←⎯


103

12. In experiments investigating the thermal decomposition of azomethane in a flask of fixed

volume, CH3NNCH3 (g) ⎯→ C2H6 (g) + N2 (g)

The following data were obtained:

Exp. Initial [CH3NNCH3] Initial Rate

1 5.13×10−2 M 2.8×10−6 M/s

2 2.05×10−1 M 1.1×10−5 M/s

A. Determine the rate law and the value of the rate constant.

B. In Exp. 1 above, how long would it take for 90% of the original azomethane to

decompose?

C. In Exp. 1 above, what would be the rate of decomposition of CH3NNCH3 once 90% of the

original azomethane had decomposed?


104

VII. Thermochemistry

7.1 Chapter Summary

Thermodynamic System: anything that we want to describe and study by itself

Surroundings: everything else that is not the system

Boundary: separates a system from its surroundings across which matter and/or energy can be

transferred

Open System: exchanges matter and energy with its surroundings

Closed System: exchanges energy but not matter with its surroundings

Isolated System: exchanges neither matter nor energy with the surroundings

State: the conditions that describe a system

State Variables: conditions that must be specified to establish the state of a system, pressure (P),

volume (V), temperature (T), and amounts of substances (n)

Physical Change of State: some of the state variables changes, but the chemical composition of

the system stays the same

Chemical Change of State: the amounts of reactants and products change

Chemical Reaction: some bonds break, and new bonds form

State Function: a property that depends only on the state of the system

Path Function: a property that depends on how a change takes place

Kinetic Energy: the energy associated with motion.

Potential Energy: stored energy

Molar Heat Capacity: the amount of heat needed to raise the temperature of 1 mol of substance

by 1 Kelvin (1K)

Work (w): energy used to move an object against an opposing force, w= Fd

First Law of Thermodynamics: ΔEsys = qsys + wsys

Calorimeter: a device that measures heat flow


105

Exothermic Process: if the chemicals release heat, this heat gain raises the temperature of the

calorimeter

Endothermic Process: if the chemicals absorb heat, this heat loss lowers the temperature of the

calorimeter, qcalorimeter = CcalΔT, q = nCΔE

Enthalpy: a thermodynamic quantity whose change equals the heat flow at constant pressure,

ΔH = ΔE + Δ(PV)

Standard Enthalpy of Formation ( ofHΔ ): enthalpy change accompanying the formation of one

mole of a chemical substance from pure elements in their most stable forms under standard

conditions

Hess’s Law: the enthalpy change for any overall process is equal to the sum of enthalpy changes

for any set of steps that leads from the reactants to the products

Molar Heat of Solution: measures net energy flow that occurs as substance dissolves

Molar Heat of Vaporization: the heat needed to vaporize one mole of a substance at its normal

boiling point

Molar Heat of Fusion: heat needed to melt one mole of a substance at its normal melting point

Sublimation: a phase change in which a solid converts directly to a vapor without passing

through the liquid phase, ΔEvap = ΔHvap − RTvap

Bond Dissociation Energy: a standard enthalpy change, ΔHo, for the corresponding bond‐

breaking reaction.


106

T I P! State function is like a

vector and path function is a scalar.

7.2 State and Path Functions

Open System: exchanges matter and energy with its surroundings.

Closed System: exchanges energy but not matter with its surroundings.

Isolated System: exchanges neither matter nor energy with the surroundings.

A change in energy depends on the difference between the initial conditions and final conditions

(changes in temperature, pressure, etc.), but it does not depend on how the change of state is

accomplished.

State Function: a property that depends only on the state of the

system.

Path Function: a property that depends on how a change takes

place.

Energy Matter

Open

Energy Matter Energy Matter

Closed Isolated

ΔE = q + w = 0


107

7.3 Energy Change

Breaking a bond always consumes energy and making a bond always releases energy. The

balance between two processes results in two classes of reactions, which are illustrated in the

Figure below.

When the energy released by bond forming is greater than the energy consumed by bond

breading, there is a net release of chemical energy by the system.

When the energy needed for bond breaking is greater than the energy released by bond

forming, the chemical reaction must be driven by energy absorbed from the surroundings.

(a) Energy‐releasing process (b) Energy‐absorbing process

Schematic diagrams showing energy flows that accompany reactions. (a) When a reaction

releases energy, the surroundings gain energy. For these reactions, change in energy (ΔE) is

negative for the system and positive for the surroundings. (b) When the reaction absorbs energy,

the surroundings lose energy. For these reactions, ΔE is positive for the system and negative for

the surroundings.

When a system gains energy, the surroundings lose the same amount of energy, and vice versa.

Consequently, the sign of ΔE for a system is always different from the sign of ΔE for its

surroundings.

If a system gains energy, ΔE is positive for the system and negative for the surroundings.

If a system loses energy, ΔE is negative for the system and positive for the surroundings.

Surroundings E increases +ΔE Energy flow

Reactants System −ΔE E decreases Products

Surroundings E decreases −ΔE Energy flow

E increases System Products

+ΔE

Reactants


108

q = nCΔT

q = mcΔT

q = nΔH

7.3.1 Energy and Heat

When a system’s energy changes, there must be an exchange of energy with the surroundings.

Heat (q): one way to exchange energy is by a flow of Heat. Flows are changes in energy, so they

are measured in joules (J). Temperature changes often accompany heat flows.

Changes in Temperature (ΔT): although temperature does not have energy units, we can use

temperature changes to calculate energy changes. Experiments show that a change in an

object’s temperature (ΔT) depends on four factors:

1. ΔT depends on q, the amount of heat transferred.

2. ΔT depends on the direction in which heat is transferred. If a system absorbs heat, ΔT will be

positive; but if a system releases heat, ΔT will be negative.

3. ΔT depends inversely on the amount of material.

4. ΔT depends on the identity of the material.

Molar Heat Capacity (C; unit, J mol−1K−1) is the amount of heat needed to

raise the temperature of 1 mol of the substance by 1 Kelvin (1K). Using C, all

four factors can be expressed in one equation that describes the temperature change resulting

from a heat transfer:

Specific Heat Capacity (c; unit, J g−1K−1) is the amount of heat needed to

raise the temperature of 1 g of the substance by 1 Kelvin (1K).

Change in States while the temperature is kept constant


109

w = Fd

wsys = − wsurr

wsys = − PextΔVsys

7.3.2 Energy and Work

Work (w): energy used to move an object against an opposing force. For an

automobile to move, work must be done to overcome the forces of friction

and gravity. The amount of work depends on the magnitude of the force that must be overcome

and the amount of movement or displacement.

Expansion Work: A gas in a container with movable walls can expand,

and this expansion results in a movement against the force exerted by

the walls. Thus, work is done in moving the walls of the container.

Expansion Work is expressed in terms of the opposing pressure and

the change of volume of the system:

Every energy transfer has a direction. When a system does work on the

surroundings, the system transfers energy to the surroundings. Thus w is

positive for the surroundings and negative for the system in an expression:


110

ΔEsys = qsys + wsys

qsys = − qsurr

wsys = − wsurr

Question ALERT! Apply the first law to solve different

terms

7.4 The First Law of Thermodynamics

A closed system can exchange energy with its

surroundings in two ways:

• Energy can be transferred as heat

• Energy can be transferred as work

Because energy must be conserved, the energy change of the system is linked to the flow of heat

and work:

This equation summarizes the observation that the change of energy of a system equals the heat

transferred into or out of the system plus the work transferred into or out of the system. The

above equation applies to any process and is called the First Law of Thermodynamics.

Energy transfer is directional, and for this reason it is essential to keep track of the signs

associated with heat and work.

When heat flows into a system, qsys is positive and the energy of the system increases.

When heat flows out of a system, qsys is negative and the energy of the system decreases.

When work is done on a system, wsys is positive and the energy of the system increases.

When a system does work on the surroundings, wsys is negative and the energy of the system

decreases.

Furthermore, whatever the system loses, the surroundings gain, and vice

versa. Whenever a system undergoes any change in energy, the

surroundings undergo an equal and opposite change in energy, so the total

energy of the universe remains unchanged.


111

7.5 Heat Measurement: Calorimetry

In a calorimetry experiment, the heat flow resulting from a process is determined by measuring

the temperature change of the calorimeter. Then q can be related to energy change through the

first law of thermodynamics.

Because heat and work are path functions, however, the conditions of the experiment must be

specified.

For a constant‐volume calorimeter, ΔVsys = 0, so there is no expansion work:

wv = 0. Thus: ΔE = qv (Constant‐Volume calorimeter)

From the equation defining enthalpy (H = E + PV) , we can relate enthalpy changes (ΔH) to the

heat flow in a constant‐pressure process:

qp = ΔE – wsys = ΔE − (−PextΔVsys) = ΔE + PextΔVsys = ΔE+ Δ(PV) = ΔH

Virtually all aqueous solution chemistry occurs at constant pressure.

For solids and liquids, volume changes during chemical reactions are very small:

Δ(PV)condensed phases ≈ 0.

Thus enthalpy changes and energy changes are essentially equal for processes that involve only

liquids and solids. Pressure and/or volume change significantly when gases are produced or

consumed during a reaction. Consequently, the enthalpy change is likely to be different from the

energy change for any chemical reaction involving gaseous reagents.

The following four step procedures should be followed


112

reaction fnH HΔ = Δ∑o o (products) − fn HΔ∑ o (reactants)

reaction DHΔ =∑o (reactants) − D∑ (products)

7.6 Heats of Formation

A formation reaction produces 1 mol of a chemical substance from the elements in their most stable

forms

The standard enthalpy of formation ( ofHΔ ) is the enthalpy change accompanying the formation reaction

under standard conditions, which are defined to be 298 K and 1 atm pressure.

The reaction enthalpy for any chemical reaction can be found from the standard enthalpies of

formation for all the reactants and products:

7.7 Hess’s Law

Enthalpy is a state function, and the change of any state function is independent of the path of

the reaction. Thus, the enthalpy change for any overall process is equal to the sum of enthalpy

changes for any set of steps that leads from the reactants to the products. This is known as Hess’s

law.

7.8 Bond Dissociation Energy

A bond dissociation energy is a standard enthalpy change, , for the corresponding bond‐breaking

reaction.

For the reaction H − Y → H + Y oHΔ = D = bond dissociation energy

We can calculate an approximate standard enthalpy change for a reaction by using the D values:


113


Multiple Choice

1. Suppose you add 55 J of heat to a system, let it do 10 J of expansion work, then return the

system to its initial state by cooling and compression. Which statement is true for this

process?

A. ΔH < ΔE

B. The work done in compressing the system must exactly equal the work done by the

system in the expansion step.

C. ΔH = 45 J

D. The change in the internal energy for this process is zero.


2. A 40.2 g sample of a metal is heated to 99.3 Co and then placed in a calorimeter containing

120.0 g of water (c = 4.18 J/g Co ) at 21.8 Co . The final temperature of the water is 24.5 Co .

Which metal was used?

A. Gold (c = 25.4 J/mol Co )

B. Iron (c = 25.1 J/mol Co )

C. Copper (24.4 J/mol Co )

D. Lead (c = 29.0 J/mol Co )

E. Graphite (c = 8.53 J/mol Co )

= 25.1 J/

mol (1 mol/55.845 g) = 0.45 J/g


114

3. Under conditions of constant pressure, how much heat is required to convert 100.0 g

H2O (l) at 25°C to H2O (g) at 125°C ?

Specific heat capacity (H2O (l)) = 4.18 J/g°C;

Specific heat capacity (H2O (g)) = 1.84 J/g°C;

ΔHvap (100°C) = 40.7 kJ/mol−1.

A. 36kJ

B. 262kJ

C. 257kJ

D. 108kJ

E. 236kJ

+ 4.6kJ = 262k

J

4. For a particular process, q = 20 kJ and w = 15 kJ. Which statement is true?

A. Heat flows from the system to the surroundings.

B. The system does work on the surroundings.

C. ΔE = 35 kJ

D. All of the above are true.

E. None of the above are true.


115

5. You place 5 ice cubes at 0°C, each of mass 10.0 g in an insulated glass. You then add 300.0 g

of water at 25°C to the ice. Assuming the glass does not absorb or release heat, what will be

the final temperature of the contents of the glass? Heat of fusion of ice = 6.02 kJ/mol;

specific heat capacity of water = 4.18 J °C−1 g−1.

A. 10.0°C

B. 11.7°C

C. 14.0°C

D. 21.4°C

E. 32.8°C

6. Two metals of equal mass with different heat capacities are subjected to the same amount of

heat. Which undergoes the smallest change in temperature?

A. The metal with the higher heat capacity.

B. The metal with the lower heat capacity

C. Both undergo the same change in temperature.

D. The temperature change cannot be determined because the initial temperature of the

metals are needed.

E. The temperature change cannot be determined because the heat capacities of the metals

and the amount of head added are needed.

Answer: A Solution: q = mcΔT. The larger the heat capacity the smaller the change in temperature will be because q and m are kept constant.


116

Short Answer:

7. Consider the following data:

ΔH (kJ)

Ca (s) + 2 C (graphite) ⎯→ CaC2 (s) −62.8

Ca (s) + 21 O2 (g) ⎯→ CaO (s) −635.5

CaO (s) + H2O (l) ⎯→ Ca(OH)2 (aq) −653.1

C2H2 (g) + 25 O2 (g) ⎯→ 2 CO2 + H2O (l) −1300.0

C (graphite) + O2 (g) ⎯→ CO2 (g) −393.51

Use Hess’s law to find the change in enthalpy at 25 Co for the following equation:

CaC2 (s) + 2 H2O (l) ⎯→ Ca(OH)2 (aq) + C2H2 (g)


117

8. Suppose a calorimeter composed of a strong metal container with a movable piston

surrounded by one litre of liquid water is used to study the burning of propane gas, C3H8, in

oxygen.

A. When 1.00 g of propane is burned, the observed temperature change of the calorimeter

system is +3.37 Co . Assuming that the heat of reaction is used to raise the temperature

of the metal container and surrounding water only, estimate the heat capacity of the

metal container. rH ×Δ = −2200 kJ/mol.

B. Calculate the molar enthalpy of burning butane gas, C4H10, in oxygen if burning 1.00g of

butane in the same calorimeter system leads to a temperature rise of 3.67 Co .


118

VIII. Chemical Equilibrium

8.1 Chapter Summary

Equilibrium Constant: [ ] [ ][ ] [ ]ba

edK

eqeq

eqeqeq

BA

ED=

Keq applied only at equilibrium

Keq is independent of initial conditions

Keq is related to the stoichiometry of the balanced net reaction

Pure Liquid and Solid: the concentrations of pure liquids or solids are always equal to their

standard concentrations, therefore division by standard concentration results in a value of 1

Large Keq: indicates that the reaction goes virtually to completion

Vapor Pressure: equilibrium pressure

Le Châtelier’s Principle: when a change is imposed on a system at equilibrium, the system will

react in the direction that reduces the amount of change

Catalysts: does not affect the equilibrium constant, it allows a reaction to reach equilibrium more

rapidly, but it does not alter the equilibrium position

Temperature: is the only variable that causes a change in the value of Keq, an increase in

temperature always shifts the equilibrium position in the endothermic direction, and a decrease

in temperature always shifts the equilibrium position in the exothermic direction


119

8.2 Dynamic Equilibrium

The double arrow implies the process is dynamic and reversible

Rate of forward reaction = rate of reverse reaction

Concentrations of all species are constant

8.3 Equilibrium Constant

8.3.1 Definition

For a reaction: aA + bB ↔ cC + dD

The equilibrium constant expression is: Kc = [ ] [ ][ ] [ ]

c d

a b

C DA B

or Kp = c d

c da b

a b

p pp p

Kc is based on the molarities of the reaction components at equilibrium

Kp is based on the partial pressure (atm) of the reaction components at equilibrium

Values of Kc and Kp are dimensionless.

Use ideal gas equation Kp = Kc (RT)Δn where Δn is the difference between the total number

of moles of gases on the right side and the left side of the equation and R equals 0.08206

L atm mol−1K−1.

Pure solids, pure liquids, and solvents are not included in the expression.

Values of K are related to the balanced equation and temperature only (not concentrations,

not pressures, and not catalysts).


120

8.3.2 Equilibrium constants and reaction equations

If the reaction equation is written in reverse, the new K value is the reciprocal.

If the reaction equation is doubled, the K value is squared.

If the reaction equations are added, the K value for the new equation is the product of the

original ones.

If the two equations are subtracted, the K value for the new equation is the quotient of the

original ones.

8.3.3 Equilibrium constants and temperature

The equilibrium constant of an exothermic reaction decreases with increasing temperature,

whereas the equilibrium constant of an endothermic reaction increases with increasing

temperature. This can be derived qualitatively from Le Chatelier’s Principle. It can also be

quantitatively derived from the standard free energy of the reaction:

Step 1. Calculate ΔGo at the new temperature: ΔGo = ΔHo – T ΔSo

Step 2. Calculate K from the equation: ΔGo = – RT ln K


121

8.4 Applications of the Equilibrium Constant

8.4.1 Judging the extent of reaction

Large K more products: Small K more reactants

K>>1 the products dominate and reactions go essentially to completion.

K<<1 the reactants dominate and the reaction does not occur to any significant extent.

• The size of K and the rate of the reaction are not directly related.

8.4.2 Predicting the direction of reaction

If Q > K, net reaction goes from left to right.

If Q < K, net reaction goes from right to left.

If Q = K, no net reaction occurs; the system is at equilibrium.


122

Question ALERT! A common question is related to equilibrium

calculations.

8.4.3 Solving equilibrium problems

This is a very common problem on the exam. To solve this kind problem, follow the steps

summarized:

Step 1. Write the balanced equation. This is directly related to K.

Don’t alter it even though sometimes it may go from the right side

to the left.

Step 2. Under the balanced equation, list the following:

i) The initial concentration ← the concentrations can have any values to start the

reaction

ii) The change in concentration ← all concentrations must follow the stoichiometric

ratio of the balanced equation.

iii) The equilibrium concentration ← all concentrations must follow the equilibrium

constant expression. Write the equation.

Step 3. Solve the equation set up in step 2 iii).

The key step is to analyze the ICE table in step 2. You can always find one concentration or define

x as the concentration of one of the substances and others can be derived by using the

stoichiometry of the reaction.


123

8.5 Le Chatelier’s Principle

Le Chatelier’s principle provides us with the means for making qualitative predictions about

changes in chemical equilibria.

CHANGES Equilibrium will shift to K will

[reactant] increases The right not change

[product] increases The left not change

increasing the pressure by reducing the

volume of the container

The direction having less

number of moles of gases

not change

increasing the temperature The endothermic

direction

increase if the

forward is endo

addition of solids, liquids, catalyst, or

noble gases without changing the

volume

No shift not change


124


Multiple Choice

1. Given: P4(s) + 6Cl2(g) ↔ 4PCl3(l) K

The equilibrium constant for the reaction below is

2PCl3(l) ↔ 3Cl2(g) + ½ P4(s)

A. 1/K

B. K1/2

C. –K1/2

D. 1/K2

E. 1/K1/2

2. What is the proper expression for Kc for the reaction

CaCO3(s) → CaO(s) + CO2(g) ?

A. Kc = [CO2 ]

B. Kc = [CO2 ] [CaO] / [CaCO3 ]

C. Kc = [CaCO3 ] / ([CO2 ] [CaO])

D. Kc = [CO2 ] [CaO] [CaCO3 ]

E. Kc = 1 / [CO2 ]


125

3. Phosgene is a deadly gas, which decompose at elevated temperature according to the

following equilibrium:

COCl2 (g) ←→ CO (g) + Cl2 (g) K = 8.3×10−4 at 360°C.

What is the equilibrium pressure of CO if a reaction vessel containing 6.3 atm of phosgene is

heated to 360°C and allowed to come to equilibrium?

A. 7.2×10−2 atm

B. 5.2×10−2 atm

C. 5.2×10−3 atm

D. 8.3×10−4 atm

E. 1.3×10−4 atm

4. The value of the equilibrium constant K depends on

i. the temperature of the system.

ii. the nature of the reactants and products.

iii. The concentration of the reactants.

iv. The concentration of the products.

A. i, ii

B. ii, iii

C. iii, iv

D. It is dependent on three of the above

E. none of these

Answer: A

Solution: K is related to the nature of the reaction equation and temperature only.


126

5. A sample of nitrosyl bromide decomposes according to the following equation:

2NOBr (g) ⇔ 2NO (g) + Br2 (g)

An equilibrium mixture in a 5.00 L vessel at 100 Co was found to contain 3.22 g NOBr, 3.08 g

NO and 4.19 g Br2. The value of equilibrium constant (Kp ) for the reaction as written is:

A. 1.97

B. 0.56

C. 3.83

D. 0.353

E. 1.40

6. At 200°C, nitrogen oxide reacts with oxygen to form nitrogen dioxide as follows:

2 NO + O2 ⇔ 2 NO2, K = 3×106

To a reaction vessel was introduced 0.010 atm NO, 0.10 atm NO2, and 0.010 atm O2. This

system is:

A. At equilibrium.

B. Not at equilibrium and must proceed from left to right to reach equilibrium.

C. Not at equilibrium and must proceed from right to left to reach equilibrium.

D. Not at equilibrium but insufficient information is given to predict which direction the

reaction must go to reach equilibrium.



127

7. Which of the following statements concerning the reaction quotient, Q, and equilibrium

constant, K, is false?

A. Q may be equal to zero

B. Q may be larger than K

C. Q may be smaller than K

D. The value of Q changes as the reaction proceeds towards equilibrium

E. One of the statements above (a, b, c, d) is false

8. For the reaction 3H2(g) + N2(g) ↔ 2NH3(g), what is the relationship between Kc and Kp at

temperature T?

A. Kc = Kp

B. Kp = Kc(RT)

C. Kc = Kp(RT)2

D. Kc = Kp1/2

E. none of these


128

9. The temperature in an automobile engine during ignition is approximately 1000K. At this

temperature, N2(g) and O2(g) from the air react in the engine to form NO(g):

N2(g) + O2(g) 2NO(g)

The standard free energy of formation, ΔGof of nitric oxide NO at 1000K is 78 kJ/mol.

Calculate the equilibrium constant for the above reaction at 1000K.

A. 1.8 x 10‐19

B. 0.98

C. 4.3 x 10‐13

D. 7.1 x 10 –9

E. 8.4 x 10‐5

10. When 0.1 moles of methylamine is dissolved in 500mL of water, the following hydrolysis

reaction occurs:

CH3NH2 (aq) ↔ CH3NH3+ (aq) + OH– (aq)

The hydroxide concentration is found to be 8.6 x 10–3M. What is the value of Keq for this

reaction?

A. 4.3 x 10–2

B. 7.4 x 10–4

C. 7.4 x 10–5

D. 3.7 x 10–4

E. 7.4 x 10–3


129

11. Consider a warm (30oC) 355mL can of soda, under 2.5atm CO2 pressure. What volume of CO2

will be released from the soda in order to reach equilibrium after opening where the pressure

of CO2 = 3.25 x 10–4 atm and the total pressure is 758mm Hg (KH = 1.6 x 10

–2 M/atm)?

A. 3.5 mL

B. 3.5 L

C. 35 mL

D. 0.35 L

E. 100 mL

0.35L

12. For the hypothetical reactions 1 and 2 below, K1 = 103 and K2 = 10

–5.

1. A + 2B ↔ AB2

2. A + B ↔ AB

3. AB2 ↔ AB + B

What is the value of K for reaction 3?

A. 10‐8

B. 10‐2

C. 102

D. 108

E. none of these

Answer: A Solution: the equation is reversed, K is the reciprocal: AB2↔ A + 2B K = 10–3

A + B ↔ AB K= 10–5 The equations are added, K’s are multiplied AB2 ↔ AB + B K = 10–8


130

Short Answer:

13. An equal number of moles of gases A2 and B2 were placed in an evacuated 2.00 L flask at

25°C and allowed to reach equilibrium according to the reaction.

A2 (g) + B2 (g) ↔ 2 AB (g) Kc = 2.0×10−4.

At equilibrium, the total pressure inside the flask is 3.20 atm.

A. What was the initial concentration of A2?

B. What is the equilibrium concentration of AB?


131

14. Equal volumes of sulphur dioxide and oxygen gases are introduced into a 4.00 L vessel

and establish the following equilibrium: 2SO2(g) + O2 (g) 2SO3

At equilibrium, 4.00 moles of sulphur trioxide gas are present at a pressure of 164 atm and a

temperature of 527oC.

A. How many moles of sulphur dioxide were introduced?

B. If at equilibrium a quantity of sulphur dioxide is added, we find that at the new

equilibrium there are twice as many moles of sulphur trioxide as there are of oxygen. How

many moles of sulphur dioxide were added?

⎯→←⎯


132

IX. Aqueous Acid‐Base Equilibrium

9.1 Chapter Summary

Strong Electrolytes: ionize completely, almost 100% become ions.

Weak Electrolytes: ionize partially, only small amount of ions present in the mixture.

Arrhenius Acid: a species that produces hydrogen ions in aqueous solution.

Arrhenius Base: a species that produces hydroxide ions in aqueous solution.

BrØnsted‐Lowry Acid: a species that donates a proton is an acid

BrØnsted‐Lowry Base: a species that accepts a proton is a base

Lewis Acid: an electron‐pair acceptor.

Lewis Base: an electron‐pair donor.

Amphiprotic Species: a chemical species that can both donate and accept protons

Water Equilibrium Constant (Kw): Kw = [H3O+][OH−] = 1.00×10−14 (at 298 K)

pH Scales: pH = −log[H3O+] pOH = −log[OH−] pKa = −log Ka

pKb = −log Kb pH + pOH = 14.00 KaKb = Kw pKa + pKb = pKw = 14.00 (25oC)

Conjugate Acid‐Base Pair: any proton donor and species generated by removing one of its

protons


133

9.2 Self Ionization of Water

9.2.1 The Proton in water

Water can act as both acid and base depending on what it reacts with

2 H2O (l) ↔ OH− (aq) + H3O+ (aq) (auto ionization of water)

Keq = Kw = [H3O+][OH−] = 1.00×10−14 (at 298 K)

The equilibrium constant for this reaction is the water equilibrium constant (Kw)

or ionic product of water

[H3O+]eq = [OH

−]eq = 1.00×10−7 M (pure water at 298 K)

9.2.2 The pH scale

pH = – log [H3O]+ [H3O

+] = 10–pH

pOH = –log[OH–] [OH–] = 10–pOH

In pure water at 25oC: pH = pOH = 7.00

Acidic = [H3O+]> 1.0 x 10–7 M, so pH < 7

Basic = [OH–] > 1.0 x 10–7 M, so pH > 7

In aqueous solution at 25oC: pH + pOH = pKw = 14 or [H3O+] [OH–] = 10–14

9.2.3 Other “p” scales

pKa = –log Ka; pKb = –log Kb; pKw = –log Kw.

The smaller the pKa value, the greater the Ka value, the stronger the acid will be.

The smaller the pKb value, the greater the Kb value, the stronger the base will be.

pH + pOH = 14


134

9.3 Finding pH of Strong Acid/Base

The following strong acids will ionize completely in water:

HNO3, H2SO4, HClO3, HClO4, HCl, HBr, HI.

Note that H2SO4 is strong for its first step dissociation and the second step has equilibrium.

The following strong bases will ionize completely in water:

Group IA bases: LiOH, NaOH, KOH, … and Ba(OH)2.

Based on the complete ionization, the [H3O+] and [OH–] can be calculated directly from the

balanced equation. e.g. 0.1 M Ba(OH)2 → Ba2+ + 2 OH–

[OH–] = 0.2 M, pOH = 0.70.

If the concentrations are extremely low, such as <10‐6, then autoionization of water needs to

be taken into account. See the practice exam questions.

9.4 Weak Acids and Bases

Weak acids and bases are partially ionized and their equilibrium constants are associated with

the equilibrium equations. Their percentage ionization = the ionized part/initial total.

9.4.1 Weak acid

The Larger the Ka the stronger the acid and the smaller the pKa, the stronger the acid

Use ICE table to analyze the equilibrium concentration

HA + H2O ↔ H2O+ + A–

Initial , M ca ‐ 0 0

Change, M –x ‐ +x +x

Equilibrium, M ca – x ‐ X x

2

3[ ] ( )a a aa

xK x H O K c xc x

+= = = −−

, assume x to be 0 to calculate the first

approximated x value. Use iteration to get the accurate value of x if necessary.


135

9.4.2 Weak Base

The Larger the Kb the stronger the base and the smaller the pKb, the stronger the base

Using ICE table to analyze the equilibrium concentration or

2

[ ] ( )b b bb

xK x OH K c xc x

−= = = −−

, assume x to be 0 to calculate the first approximated x

value. Use iteration to get the accurate value of x if necessary.

9.4.3 Conjugate acid‐base pair

Conjugate acid‐base pair differs by only one proton

The stronger the acid the weaker the conjugate base

For conjugate acid and base pair:

Ka x Kb = Kw or pKa+ pKb = pKw = 14

B + H2O ↔ HB+ + OH–

Initial , M cb ‐ 0 0

Change, M –x ‐ +x +x

Equilibrium, M cb–x ‐ x x


136

Question ALERT! A common question is

related to the pH calculation of various

solutions.

9.4.4 Acid – base reactions

The acid base reaction is actually the proton transfer between two pairs of acids and bases:

Acid1 + base2 ↔ base1 + acid2

The reaction favors the direction from strong acid1 and strong

base2 to weak base1 and weak acid2.

From this rule, we can do two things:

a) Predict the direction of the reaction, or compare the

equilibrium constant, etc.

b) Compare the relative strengths of the acids or bases in the reaction system if the direction of

the reaction is given.

Worked example:

What would be the major reaction between H2PO4– and HCO3

– ? The equilibrium constant for the

given reaction is greater than 1 or less than 1? Give a brief explanation.

(H3PO4: pKa1 = 2.12; pKa2 = 7.21; pKa3 = 12.32; H2CO3: pKa1 = 6.37; pKa2 = 10.25)

Solution:

Both of H2PO4– and HCO3

– can behave as acids and bases.

As acids: H2PO4– has pKa = 7.21 and HCO3

– has pKa = 10.25. H2PO4– is a stronger acid than HCO3

–

As bases: H2PO4– has pKb = 14 – 2.12 = 11.88 and HCO3

– has pKa = 14 – 6.37 = 7.63. H2PO4– is a

weaker base than HCO3– .

The major reaction should have H2PO4– to react as an acid and HCO3

– as a base.

H2PO4– + HCO3

– ↔ HPO42– + H2CO3

A1 B2 B1 A2


137

This is an acid‐base reaction, in which pKa for the acid H2PO4– (A1) = 7.21 and pKa for the acid

H2CO3 (A2) = 6.37. H2PO4– is a weaker acid than H2CO3 and HPO4

2– is a stronger base than HCO3– .

The reaction is running from weak acid 1 and weak base 2 to stronger base 1 and stronger acid 2.

It favours the reverse direction and K value is less than 1.

9.4.5 Acid – base properties of salt

Most soluble salts are strong electrolytes and will ionize completely.

The ions produced may be acidic, basic, or neutral, depending on their different properties.

Salts made from Nature of solution

Strong acid + strong base Neutral

Strong acid + Weak base Acidic

Weak acid + strong base Basic

Weak acid + Weak base Neutral, acidic, or basic

depending on the relative

strength


138

9.5 Strategy for Solving Acid‐Base Problems: A Summary

Step 1. Calculate the number of moles of all the species in the solution by using the formula:

n = M L = (molar concentration) x (volume in litres).

The rule is to break strong electrolytes to ions and keep the weak ones unchanged.

Step 2. If there are acid‐base reactions, write the balanced equation and determine the number

of moles of each of the species after the reaction based on the stoichiometric ratios.

If there is no reaction, go to step 3.

Step 3. If there is only one acid or base, use the formula 3[ ] ( )a aH O K c x+ = − for the weak

acid, [ ] ( )b bOH K c x− = − for the weak base, and [H3O+] or [OH–] directly for the strong

acid or base. If there are conjugate acid and base together in the system, use the formula

base logacidapH pK ⎛ ⎞= + ⎜ ⎟

⎝ ⎠ to calculate pH. The advantage to use this formula is that the

number of moles of the conjugate acid and base can be directly used whereas the acid

and base formulas require their molar concentrations.


139


Multiple Choice

1. The conjugate base of OH– is

A. H2O

B. O–

C. H+

D. OH–

E. O2–

2. Given:

HClO, pKa = 7.53

CH2ClCOOH, pKa = 2.85

H2CO3, pKa1 = 6.37 and pKa2 = 10.25

List the conjugate bases, ClO–, CH2ClCOO–, and CO3

2–, in order of decreasing basicity.

A. ClO–> CO32–> CH2ClCOO

–

B. CO32–> ClO–> CH2ClCOO

–

C. CH2ClCOO–> ClO–> CO3

2–

D. CO32–> CH2ClCOO

–> ClO–


140

3. Consider the following monoprotic acids

I. Acetic acid (CH3COOH) Ka = 1.8×10−5

II. Formic acid (HCO2H) Ka = 1.8×10−4

III. Hypobromous acid (HOBr) Ka = 2.4×10−9

IV. Nitrous acid (HNO2) Ka = 5.1×10−4

V. Phenol (C6H5OH) Ka = 1.0×10−10

Which of the following aqueous solutions will have the LOWEST pH?

A. 0.10 M CH3COONa

B. 0.10 M HCO2Na

C. 0.10 M NaOBr

D. 0.10 M NaNO2

E. 0.10 M C6H5ONa

4. A 245.0 mL sample of 0.15 M Ba(OH)2 is added to 438.0 mL of 0.2 M HNO3. What is the pH of

the resulting solution?

A. 0.021

B. 1.68

C. 12.32

D. 0.70

E. 13.30

Ba2+ + 2 OH− There will be 0.15 M x 0.245 L x 2 = 0.0735 mol OH− HNO3 is a strong acid: HNO3 + H2O → H3O+ + NO3

− There will be 0.2 M x 0.438 L = 0.0876 mol H3O+


141

5. What is the pOH of 1.0×10−8 M solution of NaOH?

A. 7.0

B. 8.0

C. 6.0

D. slightly less than 7.0

E. slightly more than 7.0

6. Consider two solutions, A and B. [H3O+] in solution A is 500 times greater than in solution B.

What is the difference in pH values between the two solutions?

A. 1.3

B. 2.7

C. 3.1

D. 6.2

E. cannot be determined from the given.

7. Given three separate solutions containing equal concentrations of formic acid (Ka = 1.7×10−4),

phenol (Ka = 1.3×10−10), and acetic acid (Ka = 1.8×10−5) in water. Select the response below

that has the acid solutions arranged in order of increasing pH.

A. formic < phenol < acetic

B. formic < acetic < phenol

C. acetic < formic < phenol

D. phenol < acetic < formic

E. No response is correct


142

8. At 0oC the ion product constant of water Kw = 1.2 x 10−15. What is the pH of pure water at

0oC?

A. 7.00

B. 6.88

C. 7.56

D. 7.46

E. none of these

9. A solution of total volume 0.50 L was prepared by the addition of 0.10 moles of KF to

sufficient water. From the following, select the major species and the pH of the solution.

A. K+, HF; HO−, H2O, pH = 8.23

B. K+, F−, H+, H2O, pH = 5.76

C. K+, F−, H2O, pH = 8.23

D. K+, F−, H2O, pH = 5.76

E. None of these

would be:, in w

hich Kb = Kw/Ka.

10. When ammonia dissolves in water, the new major species present are:

A. H2O; NH2–;H+

B. H2O; NH3

C. H2O; NH4+;OH–

D. H2O; NH2–; H+

E. H2O; N3–;H+


143

11. Which of the following is the conjugate base of the dihydrogenphosphate ion ?

A. H3PO4

B. NaH2PO4

C. OH–

D. HPO42–

E. PO43–

12. A 1.35g sample of ammonium chloride is added to sufficient water to give 150.0 mL of

solution whose pH is less than 7.0. Several dissociation reactions occur. Which of the

following goes essentially to completion?

A. H2O H+ + OH–

B. NH4+ NH3 + H

+

C. NH3 NH2– + H+

D. NH4Cl NH4+ + Cl–

E. NH4+ + H2O H3O

+ + NH3

13. Calculate [H+] in a 0.010M solution of HCN, Ka = 6.2 x 10–10.

A. 1.0 x 10–7M

B. 2.5 x 10–6M

C. 3.6 x 10–3M

D. 6.2 x 10–10M

E. none of these is close


144

14. For a 0.080M aqueous solution of weak base B, pKb =9, which of the following is true?

A. [B] ≅ 0.04M

B. [B] = [OH−]

C. [BH+] = [OH−]

D. [B] = [BH+]

Short Answer:

15. a) A solution of sodium cyanide labeled “0.200M NaCN (aq)” has a pH =11.31.

Calculate the Ka of HCN.

b) The pH of an aqueous solution of ammonium cyanide is less than 7, greater than 7, or

undetermined. Give a brief explanation.


145

16. Consider the following acids:

HW (pKa = 2), HX (pKa = 6), HY (pKa = 10), and HZ (pKa = 20)

a) The conjugate bases of which of the above acids would react almost completely with water

to form hydroxide ion?a, and thus the weakest acid.

b) If you want to convert HX to X−, which of the other bases W−, Y−, Z− can you use?

Answer: Y−, Z−


146

X. Applications of Aqueous Equilibria

10.1 Chapter Summary

Buffer Solution: contains both a weak acid and its conjugate base as major species in solution,

pH = pKa + logBA

⎛ ⎞⎜ ⎟⎝ ⎠

Buffer Capacity: the amount of added H3O+ or OH− the buffer solution can tolerate without

exceeding a specified pH range

pH Titration Curve: a plot of the pH of the solution as a function of the volume of added titrant.

Indicator: a weak organic acid that contains a highly delocalized π‐bonding network, at pH < pKa

of the weak organic acid, the indicator is in its protonated form and at pH > pKa of the weak

organic acid, the indicator gives up its proton and is converted to its conjugate base

Equivalence Point: the point at which stoichiometrically equivalent quantities of acid and base

have been mixed together.

End Point: the point at which the indicator changes color. Although the equivalence point is not

necessarily the same as the end point, careful selection of the indicator will ensure that the error

is negligible.

Titration of Weak Acid with Strong Base: at the midpoint of a titration of a weak acid by a strong

base, the pH of the solution equals the pKa of the weak acid

Solubility Product (Ksp): describes a solubility equilibrium from the perspective of a salt dissolving

in water. This is an equilibrium constant and has only one value for a given solid at a given

temperature.

Solubility: describes an equilibrium position. It has different units, such as g/mL or mol/L. The

solubility varies according to the concentrations of the ions present in the solution. Common ion

effect may change the solubility significantly. However, in all cases the product of the ion

concentrations must satisfy the Ksp expression.


147

Question ALERT! A common question is related to the pH calculation of various

titration curves.

10.2 Buffer Solutions

10.2.1 Characteristics of a buffer solution

A solution which resists change in pH when a small amount of

strong acid or base is added is called a buffer.

It contains a weak acid and its conjugate base

Ratio between acid and its conjugate is between 0.1 and 10.

The total concentration is not too low

10.2.2 Buffer capacity

This is the amount of acid or base neutralized by the buffer before there is a significant change in

pH.

A buffer solution works because there are a significant amount of acid and base coexisting in the

solution. If a small amount of strong base is added, the acid component of the suffer solution

neutralizes the added base, making the pH relatively stable; if a small amount of strong acid is

added, the base component of the buffer solution neutralizes the added acid, making the pH

relatively stable. Thus, Buffer capacity depends on the composition of buffer:

(i) the total amount of the conjugate acid‐base pair present;

(ii) the ratio of the conjugate acid‐base pair (It would be better if the ratio is close to 1).

The Effective range of buffer is pH = pKa ± 1.


148

10.2.3 Making a buffer

Add conjugate pairs to water NH3/NH4Br or CH3COOH/CH3COONa

Add strong base into weak acid ( partial neutralization)

Add strong acid into weak base ( partial neutralization)

It is always best to choose the pKa of the weak acid close to the pH of the buffer required in order

to make a buffer with high capacity.

Henderson‐ Hasselbalch equation is used to calculate the pH: base logacidapH pK ⎛ ⎞= + ⎜ ⎟

⎝ ⎠

10.3 Acid‐Base Titrations

10.3.1 Titration

In an acid‐base titration a solution containing a known concentration of base is slowly added to

an acid until the acid is completely neutralized. Alternatively, a known concentration of acid is

slowly added to a basic solution until the base is completely neutralized.

• The equivalence point is the point at which a stoichiometrically equivalent amount of base

and acid are used

• This situation can be identified by noting an appropriate (pH‐induced) color change in an

indicator that is added to the solution. Or a pH meter can be used to measure the pH of the

solution as a function of added base

• A graph or plot of the pH as a function of added titrant (e.g. base solution) is called a

titration curve


149

10.3.2 Titration curve

• A titration curve can help to figure out when the equivalence point occurs, and also the value

of the acid or base dissociation constant (i.e. Ka, or Kb). The titration curve can also help

identify what type of indicator would be most useful in following the acid‐base neutralization

reaction

• The end point is where the indicator changes color. Careful selection of the indicator will

make the end point very close to the equivalence point.

Strong Acid ‐ Strong Base Titration

• A strong acid ionizes completely in solution. Likewise, a strong base.

• Titration curve exhibits a large change in pH in the immediate vicinity of the equivalence

point

• pH = 7 at the equivalence point

Weak Acid – Strong Base Titration

• The strength of a weak acid has a significant effect on the shape of its pH curve. The vertical

region that surrounds the equivalence point becomes shorter as the acid becomes weaker.

• pH is > 7 at the equivalence point because the anion of a weak acid is a base and

[OH–] = ( )b bK c x− , where Kb = kw/Ka and cb is the concentration of the conjugate base of

the weak acid being titrated. The weaker the acid, the greater the pH value at the equivalence

point.

• In the region midway to the equivalence point, pH rises slowly due to the buffering action of

the weak acid‐conjugate base mixture and pH = pKa for the weak acid exactly halfway to the

equivalence point.

• Beyond the equivalence point, the pH is determined by the concentration of OH– from the

excess NaOH, which is the same as the curve for the strong acid‐strong base titration.


150

Weak Base ‐ Strong Acid Titration

• The strength of a weak base has a significant effect on the shape of its pH curve. The vertical

region that surrounds the equivalence point becomes shorter as the base becomes weaker.

pH is < 7 at the equivalence point because it is a solution of the conjugate acid of the weak

base and [H3O+] = ( )a aK c x− , where Ka = kw/Kb and ca is the concentration of the conjugate

acid of the weak base being titrated. The weaker the base, the smaller the pH value at the

equivalence point.

• In the region midway to the equivalence point, pH changes slowly due to the buffering action

of the weak base‐conjugate acid mixture and pH = pKa for the conjgate acid exactly halfway

to the equivalence point.

• Beyond the equivalence point, the pH is determined by the concentration of H3O+ from the

excess strong acid, which is the same as the curve for the strong base – strong acid titration.

10.3.3 Indicators

• An acid‐base titration can be performed without measuring the pH of the solution, provided

that an indicator is present that gives a colour change when the titration reaches the

stoichiometric point.

• The pH of a solution changes during titration. As the pH passes the value of pKIn, the indicator

changes colour. A good indicator changes colour very near the pH of the stoichiometric point.

• Because this pH depends on the pKa of the substance being titrated, different titrations

require different indicators.

• Because the pH changes rapidly near the stoichiometric point, an indicator is suitable if the

pH at the stoichiometric point is within one unit of pKIn: pHstoichiometric point = pKIn ± 1


151

10.4 Solubility Equilibria

10.4.1 Precipitation Equilibria

The solubility product (Ksp) describes solubility equilibrium from the perspective of a salt

dissolving in water. When the solubility equilibrium is reached, the solution is saturated and no

more solid can dissolve into the solution.

For example, Ca3(PO4)2 (s) ↔ 3 Ca2+ (aq) + 2 PO43–(aq)

Ksp = [Ca2+]3[PO43–]2.

The equilibrium constant is called solubility product because the concentration of the solid on the

left side is not included in the Ksp expression.

10.4.2 Calculating Ksp from solubility

Worked example 10.1

A saturated solution of Ca(OH)2 prepared by dissolving solid Ca(OH)2 in water has its pH = 12.324.

Calculate Ksp for Ca(OH)2.

Solution:

Step 1. Write the balanced equilibrium equation:

Ca(OH)2 (s) ↔ Ca2+ (aq) + 2 OH–(aq)

[ ] at equilibrium,

(M)

0.010545 0.02109

Step 2. Determine concentrations of all species:

From pOH = 14 – PH = 1.676, [OH−] = 0.02109 M and [Ca2+] = 0.010545 M

Ksp = [Ca2+][OH–]2 = (0.010545)(0.02109)2 = 4.69x10–6


152

T I P! Do not multiply by 2

because OH− and Ca2+ are from different sources.

In this example, [Ca2+] and [OH–] are in the ratio of 1:2 because the saturated solution is prepared

by dissolution of Ca(OH)2 in pure water. If the saturated solution is prepared by mixing solutions

of Ca(NO3)2 and NaOH, there will be no separate restrictions on [Ca2+] and [OH–] as the following

example shows.

Worked example 10.2

A saturated solution of Ca(OH)2 prepared by mixing solutions of Ca(NO3)2 and NaOH has [Ca2+] =

0.002168 M and pH = 12.6676. Calculate Ksp for Ca(OH)2.

Solution:


Ca(OH)2 (s) ↔ Ca2+ (aq) + 2 OH–(aq)

[ ] at equilibrium,

(M)

0.002168 0.04652

Step 2. Determine concentrations of all species:

From pOH = 14 – pH = 1.3324, [OH−] = 0.04652 M

and [Ca2+] = 0.002168 M

10.4.3 Calculating solubility from Ksp

It is very important to distinguish between the solubility and its solubility product. The solubility

product is an equilibrium constant and has only one value for a given solid at a given

temperature. Solubility varies according to the concentration of the common ion present in the

solution. However, in all cases the product of the ion concentrations must satisfy the Ksp

expression.


153

Worked example 10.3

Calculate the molar solubility of PbI2 in pure water at 25oC (given Ksp = 8.5x10–9)

Solution:


PbI2 (s) ↔ Pb2+ (aq) + 2 I–(aq)

[ ] at equilibrium,

(M)

x 2x

Step 2. Determine concentrations of all species: let x be the molar solubility of PbI2

[Pb2+] = x M and [I−] = 2x M

Ksp = [Pb2+][I–]2 = (x)(2x)2 = 8.5x10–9

x = [Pb2+] = molar solubility = 1.29x10–3 M

Note that the number 2 appears twice in the expression (x)(2x)2. The exponent 2 is required

because of the equilibrium equation, Ksp = [Pb2+][I–]2. The coefficient 2 in 2x is required because

each mole of PbI2 that dissolves produces 2 mol of I–.


154

Worked example 10.4

Calculate the molar solubility of PbI2 in 0.100 M NaI at 25oC (given Ksp = 8.5x10–9)

Solution:

Step 1. NaI is a strong electrolyte, which will ionize completely:

NaI → Na+ + I–

(0.1 M) 0.1 M 0.1 M

Write the balanced equilibrium equation :

PbI2 (s) ↔ Pb2+ (aq) + 2 I–(aq)

[ ] at equilibrium,

(M)

x 0.1 + 2x

Step 2. Determine concentrations of all species: let x be the molar solubility of PbI2

[Pb2+] = x M and [I−] = 0.1 + 2x M (I− comes from two sources: 0.1 M is from the complete

ionization of NaI and 2x is from the solid PbI2. Note that 0.1 M should not be multiplied by 2

because it is from a different source.)

Ksp = [Pb2+][I–]2 = (x)(0.1+2x)2 = 8.5x10–9 (2x is negligible compared to 0.1)


Note that the calculated solubility of PbI2 is much less than that in pure water (about 1 500) due

to the common ion effect (I–), which shifts the equilibrium position in the direction that reduces

this added concentration.


155

10.4.4 Determining whether a precipitate forms

The forward direction of the solubility equilibrium is the dissolution of the solid and the reverse

direction is the formation of the precipitate. In order to determine whether a precipitate will

form or not, we need to compare the reaction quotient (Q) with the equilibrium constant (K):

If Q > K, the reaction proceeds in the forward direction. The solution is not saturated yet. The

solid will dissolve and no precipitate forms;

If Q = K, the reaction is just at equilibrium and the solution is just saturated. There will be no

precipitate produced;

If Q < K, the reaction proceeds in the reverse direction, the precipitate will form.


156

Worked example 10.5

Will a precipitate form when 30.0 mL of 0.001M Pb(NO3)2 and 20 mL of 0.001 M NaI are mixed?

(Ksp for PbI2 = 8.5x10–9)

Solution:

Step 1. Calculate the molar concentrations of all the ions in the solution by using c1V1=c2V2:

Pb(NO3)I2 → Pb2+ (aq) + 2 NO3–(aq)

M

In the solution

30x0.001/50 6x10–4 1.2 x10–3

NaI ↔ Na+ (aq) + I–(aq)

M

In the solution

20x0.001/50 4x10–4 4 x10–4

Step 2. Write the balanced equilibrium equation and show the concentrations of the ions present

in the mixture in order to calculate the Q value:

PbI2 (s) ↔ Pb2+ (aq) + 2 I–(aq)

[ ] at equilibrium,

(M)

6x10–4 4x10–4

Q = [Pb2+][I–]2 = (6x10–4)( 4x10–4)2 = 9.6x10–11 < 8.5x10–9 = K

(2x is negligible compared to 0.1)


Notice the differences between two equations: Q = [Pb2+][I–]2 and K = [Pb2+][I–]2.

The concentrations used to calculate Q can be any values but the concentrations used to

calculate K must be at equilibrium.


157

10.5 Complexation Equilibria

Metal cations in aqueous solution can form bonds to anions or neutral molecules that have lone

pairs of electrons. This leads to formation of complex ions and to chemical equilibria involving

complexation. The complexation equilibrium between Ag+ and NH3 is an example:

Ag+ (aq) + 2 NH3 (aq) ↔Ag(NH3)2+ Kf =

[ ]3 2

23

( )Ag NH

Ag NH

+

+

⎡ ⎤⎣ ⎦⎡ ⎤⎣ ⎦

The treatment of complexation equilibria is more complicated than treatments for acid‐base or

solubility, both because the stoichiometries of complexes are not obvious and because complex

formation and dissociation frequently involve multiple steps.


158


Multiple Choice

1. The pH of a saturated solution of a slightly soluble metal hydroxide, M(OH)2 is 9.53 at 25oC.

Its Ksp at this temperature is

A. 1.15 x 10–9

B. 5.76 x 10–10

C. 1.95 x 10–14

D. 2.89 x 10–10

E. 3.39 x 10–5

2. Given that the following Solubility Products (Ksp)

A. Co(OH)2 1.3 x 10–15

B. Ag2CrO4 1.1 x 10–12

C. PbCO3 7.4 x 10–14

D. AgCl 1.8 x 10–10

E. AgBr 5.0 x 10–13

Which compound is the most soluble (in moles/litre)?


159

3. If 10mL of a 1.0 x 10‐4M solution of a strong acid were added to 100mL each of one solution

containing 1.8 x 10‐5M HCl and a second solution containing 1.0 M acetic acid (Ka =1.8 x 10‐5)

plus 1.0 M sodium acetate, it is expected that the:

A. pH of both solutions would remain unchanged.

B. change in pH would be very large in both solutions.

C. change in pH would be the same in both solutions.

D. change in pH would be larger in solution containing acetic acid and sodium acetate.

E. change in pH would be larger in the HCl solution.

4. One liter of a solution contains 0.500 moles each of HA and NaA. If 0.0100 moles of HCl is

added to it, what would be the pH of the resulting solution assuming that its volume remains

constant? (Ka for HA is 1.800×10−4)

A. 0.300

B. 2.000

C. 3.727

D. 3.745

E. 3.762

5. How much solid NaCN must be dissolved in 1.0L of a 0.5 M HCN solution to produce a

solution with pH 7.0? (Ka = 6.2×10−10 for HCN.)

A. 0.0034 g

B. 11 g

C. 160 g

D. 24 g

E. 0.15 g


160

6. Which of the following will NOT make a buffered solution?

A. 100 mL of 0.1 M Na2CO3 and 50 mL of 0.1 M HCl

B. 100 mL of 0.1 M NaHCO3 and 25 mL of 0.2 M HCl

C. 100 mL of 0.1 M Na2CO3 and 75 mL of 0.2 M HCl

D. 50 mL of 0.2 M Na2CO3 and 5 mL of 1.0 M HCl

E. 100 mL of 0.1 M NaHCO3 and 50 mL of 0.2 M HCl

7. Which of the following solutions is not a buffer?

1. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M NaOH


3. 50 mL of 0.1 M acetic acid mixed with 25 mL of 0.05 M sodium acetate


5. 50 mL of 0.1 M sodium acetate mixed with 25 mL of 0.05 M NaOH

A. none of the above

B. 5 only

C. 1, 2 and 5

D. 3, 4 and 5

E. 2, 4 and 5


161

8. Ksp for AgCl and AgBr are 1.8×10−10 and 3.3×10−13 respectively. Solid AgNO3 is added

dropwise to a solution which is 0.0010 M each in NaCl and NaBr. What % of the bromide ions

remain in solution just before AgCl begins to precipitate?

A. 0.18 %

B. 0.018 %

C. 0.0018 %

D. less than 0.00010 %

E. more than 1.00 %

0.0 = 0.18 %.

Short Answer

9. A mixture of the solids AgCl and CaCO3 sits at the bottom of a flask in equilibrium with their

ions in aqueous solution. What should you do in order to separate the AgCl from the CaCO3 so

that you can keep the AgCl (silver !!). Explain your procedures briefly.

You may use the following reagents: AgNO3(aq), HCl(aq), NaOH(aq), Ca(NO3)2(aq).

(Ksp for AgCl = 1.8×10−10, Ksp for CaCO3 = 5.0×10−9)


162

10. How many moles of NH3 are needed to prepare 100 mL of an ammonia solution that will have

a pH of 11.11? Kb (NH3) = 1.82×10−5.

11. A solution is 0.200 M in both Ni2+ and Fe3+. Solubility product constants for Ni(OH)2 and

Fe(OH)3 are 6.0×10−16 and 3.0×10−39, respectively.

A. What must be the pH of the solution in order to precipitate 99.99% of the Fe3+ as

Fe(OH)3?


163

B. Show that no Ni(OH)2 will have precipitated at the pH calculated in part a).

C. At what pH will Ni(OH)2 first begin to precipitate?

D. What is the concentration of Fe3+ when Ni(OH)2 first begins to precipitate?


164

XI. Thermodynamics


Second Law of Thermodynamics: any spontaneous process increases the disorder of the universe

Entropy: the state function that provides a quantitative measure of disorder and is symbolized S,

TqS T=Δ

Entropy Change of the Universe: total entropy change,


Reaction Entropies: reaction pnS SΔ = Δ∑o o (products) − rn SΔ∑ o (reactants)

Reaction Enthalpies: reaction p fnH HΔ = Δ∑o o (products) − r fn HΔ∑ o (reactants)

Free Energy (G): a state function whose change for the system predicts spontaneity and is

defined by, Free energy = G = H − TS

Change in Free Energy: ΔGsys = ΔHsys − TΔSsys, ΔGsys is negative for all spontaneous processes

under conditions of constant temperature and pressure

Standard Molar Free Energy of Formation ( ofGΔ ): the change of free energy when one mole of

that substance is formed from elements in their standard states,

reaction p fnG GΔ = Δ∑o o (products) − r fn GΔ∑ o (reactants)

Entropy Change under Non‐Standard Conditions: can be expressed in terms of the standard

entropy change and Q:

ΔSreaction = oreactionSΔ − R ln Q



165

Influence of Temperature on Spontaneity: oTGΔ = oHΔ − T oSΔ

oHΔ oSΔ oGΔ (high T) oGΔ (low T) Spontaneity

− + − − All T

+ − + + No T

+ + − + High T

− − + − Low T


166

surrHS

T−Δ

Δ =


Important Concept! It involves two factors: 1. The enthalpy change for the reaction 2. The constant temperature of the surroundings.

11.2 Spontaneity and Entropy

Every process has a preferred direction, which is referred to in thermodynamics as the

spontaneous direction. A process follows its spontaneous direction unless acted on by some

external intervention.

11.2.1 Entropy and the Second Law of thermodynamics

Entropy: the state function that provides a quantitative measure of disorder and is symbolized S

The Second Law of Thermodynamics provides a clear‐cut criterion of spontaneity:

the total entropy of the universe increases in spontaneous processes,

that is,

If ΔSuniverse > 0, the reaction is spontaneous.

If ΔSuniverse < 0, the reaction is nonspontaneous.

If ΔSuniverse = 0, the reaction mixture is at equilibrium.

To determine ΔSuniverse, we need ΔSsystem and ΔSsurroundings.

For a reaction that occurs at constant pressure,

ΔSsurr can be calculated according to the

equation:

• Because ΔH is an energy term, the units of ΔS are energy/temperature, or J/K.

• If the reaction is exothermic, ΔH is negative, but since heat flows into the surroundings, ΔSsurr

is positive.

• At low temperature, there is little disorder, so the addition of given amount of heat increases

the disorder of the surroundings significantly.

• At high temperature, in contrast, there is considerable disorder, so adding the same amount

of heat increases total disorder of the surroundings by a smaller amount.


167

11.2.2 Standard Reaction Entropies

ΔSsystem is the entropy change of the reaction, ΔSreaction, is which can be calculated from standard

molar entropies in exactly the same way as reaction enthalpies can be calculated from standard

heats of formation.

11.3 Spontaneity and Free Energy

Free Energy (G): a state function whose change for the system

predicts spontaneity and is defined by the equation below:

The definition of free energy gives an equation that relates the

change in free energy for any system to other thermodynamic

changes:

ΔGsys is negative for all spontaneous processes under conditions of constant temperature and

pressure.

Because free energy is a state function, its values can be tabulated for use in chemical

calculations.

The Standard Free Energy of Formation ( ofGΔ ) of a substance is the change of free energy when

one mole of that substance is formed from its stable elements in their standard states. To

calculate the free energy change for any chemical reaction:

reaction no oS SΔ = Δ∑ (products) − n oSΔ∑ (reactants)

Free energy = G = H − TS

ΔGsys = ΔHsys − TΔSsys

reaction p rn ( ) n ( tan )o o of fG G products G reac tsΔ = Δ − Δ∑ ∑


168

11.4 Change in Free Energy under Non‐standard Conditions

Standard State Conditions: Solids, liquids, and gases are in pure form at 1 atm pressure; solutes

are at 1 M concentration at a specified temperature, usually at 25oC.

Many chemical reactions occur at different concentrations, pressures, and temperatures. To use

GΔ as a measure of spontaneity under nonstandard conditions, we must calculate ∆G under

nonstandard‐state conditions.

11.4.1 ∆G at different temperatures

Changing the temperature of the system is one way to influence the spontaneity of a reaction.

The equation for ΔG has two parts, ΔH and TΔS, which can work together or in opposition:

This equation is valid at any temperature as long as the temperature is constant. However,

application of the equation at a temperature different from 298 K requires the appropriate values

of ΔHreaction and ΔSreaction at the corresponding temperature. The temperature effect on the

entropy change for a reaction is usually small enough that we can consider ΔSreaction to be

independent of temperature. ΔHreaction also does not change rapidly with temperature. As a

result, free energy changes at temperatures other than 298 K can be estimated by assuming that

standard enthalpies and entropies at 298 K also apply at any other temperature.

reaction, reaction reactionTG H T SΔ ≅ Δ − Δo o o


169


The effects of temperature on spontaneity are summarized in the table below:

Influence of Temperature on Spontaneity

oHΔ oSΔ oGΔ (high T) oGΔ (low T) Spontaneity

− + − − All T

+ − + + No T

+ + − + High T

− − + − Low T

11.4.2 ∆G under different concentrations and pressures

where Q is the reaction quotient.

For reactions that involve gases, Qp should be used because the standard state for gases is 1 atm

pressure.

For reactions that involve solutes in solution, Qc should be used because the standard state for

solutes is 1 M concentration.

For reactions that involve both gases and solutes in solution, Q contains partial pressures of gases

and molar concentrations of solutes.


170

11.4.3 Comparison of different methods to calculate ∆G

Methods to calculate ∆G Comments

reaction p rn ( ) n ( tan )o o of fG G products G reac tsΔ = Δ − Δ∑ ∑ ∆Go under standard conditions at 25oC

when ∆Gfo ‘s are given.

∆Go = – RT ln K ∆Go under standard conditions at the given

temperature when K is given.

Additivity rule similar to those for finding ∆H using

Hess’s law.

∆Go under the same condition as the

reactions given.

∆G = ∆H – T ∆S ∆G at different temperatures. Note that ∆H

and ∆S are temperature insensitive.

∆G = ∆Go + RT ln Q ∆G at nonstandard conditions.

Note that ∆G is temperature is sensitive.


171


Multiple Choice

1. For the process CHCl3 (s) → CHCl3 (l), oHΔ = 9.2 kJ/mol and oSΔ = 43.9 J/mol⋅K. What is the

melting point of chloroform?

A. −63 Co

B. 210 Co

C. 5 Co

D. 63 Co

E. −5 Co

2. Calculate oGΔ for H2O (g) + ½ O2 (g) ↔ H2O2 (g) at 600 K using the following data:

H2 (g) + O2 (g) ↔ H2O2 (g) Kp = 2.3×106 at 600 K

2 H2 (g) + O2 (g) ↔ 2 H2O (g) Kp = 1.8×1037 at 600 K

A. +140 kJ

B. −220 kJ

C. −290 kJ

D. −350 kJ

E. +290 kJ


172

3. A mixture of hydrogen and chlorine does not react until it is exposed to ultraviolet light from

a burning magnesium strip. Then the following reaction occurs very rapidly:

H2 (g) + Cl2 (g) ⎯→ 2 HCl (g) ΔG = −45.54 kJ

ΔH = −44.12 kJ

ΔS = −4.76 J/K

Which statement best explains this behavior?

A. The reactants are thermodynamically more stable than the products.

B. The reaction has a small equilibrium constant.

C. The ultraviolet light raises the temperature of the system and makes the reaction more

favorable.

D. The negative value for ΔS slows down the reaction.

E. The reaction is spontaneous, but the reactants are kinetically stable.

Answer: E

Solution: Once the reaction has started, it will go by itself. This means that it is spontaneous. Spontaneity has nothing to do with the rate. In this case, the rate is slow under normal temperature.


173

Short Answer:

4. Fe3O4, magnetite and Fe2O3, haematite, are two prevalent oxides of iron. Given the following

thermodynamic data at 298 K:

Fe2O3 (s) Fe3O4(s) O2 (g)

ΔHf° (kJ/mol) −824.2 −1118.4 0

S° (J/K mol) 87.4 146.4 205.1

A. Write a balanced reaction equation for the thermal decomposition of one mole of Fe2O3

into Fe3O4 and O2.

B. Calculate ΔG° for the thermal decomposition of one mole of Fe2O3.

C. Is the decomposition reaction thermodynamically spontaneous under a partial O2 (g)

pressure of 0.010 atm? Explain your reasoning.


174

5. A beaker of water is put under an air‐tight bell jar that is kept at 10°C. The air is removed

and, after the system is allowed to come to equilibrium, a quantity of water remains in the

beaker.

ΔHf° (kJ mol−1) S°(J mol−1 K−1)

H2O (l) −286 70

H2O (g) −242 189

A. Calculate ΔH° and ΔS° for the vaporization of one mole of water.

B. The standard entropy, ΔS°, of vaporization of water is greater than that of a “typical”

liquid. Give a possible explanation for this in one sentence only.


175

C. Determine the equilibrium vapor pressure of water at 10°C. Assuming that ΔH° and ΔS°

are temperature independent for the vaporization of water


176

XII. Electrochemistry


Oxidation number: the assumed charge

Oxidation: the loss of electrons from a substance

Reduction: the gain of electrons by a substance

Balancing redox Equations: the key to balancing complicated redox equations is to balance

electrons as well as atoms

Indirect Electron Transfer: spontaneous redox reactions can occur by indirect electron transfer,

species involved in the redox chemistry are not allowed to come into direct contact with one

another

Salt Bridges: serve as separators that allow ion migration without free mixing of solutions

Electrodes: some redox reactions take place at the surfaces of metal plates, where electrons are

gained and lost by metal atoms and ions

Anode: the electrode where oxidation occurs

Cathode: the electrode where reduction occurs

Galvanic Cells: use redox reactions to generate electrical current, Galvanic cells find wide usage

in the form of batteries.

Cell Potential: the difference in electrical potential between the two electrodes, when a reaction

is multiplied by any integer, its cell potential remains unchanged.

Molar Free Energy Change: the product of the electrical potential, the number of electrons

transferred, and the Faraday constant: ΔG = –nFE where F = 96,485.34 C mol–1

Standard Potential and Equilibrium Constant: –RT ln Keq = nFE° ⇒ eqKnFRTE ln=°

Nernst Equation: QnFRTEE ln−°=


177

12.2 Recognizing Redox Reactions

12.2.1 Oxidation‐Reduction Reactions

In a redox reaction, one species loses electrons while another species gains electrons.

Oxidation is the loss of electrons from a substance.

Reduction is the gain of electrons by a substance.

(OIL RIG) (LEO says GER)

A chemical species that loses electrons is oxidized, and a chemical species that gains electrons is

reduced. Remember also that electrons are conserved in all chemical processes. That is,

electrons can be transferred from one species to another, but they are neither created nor

destroyed. Thus it is impossible for oxidation to occur without reduction also occurring, because

when one species gains electrons, another species must lose electrons. Oxidation and reduction

always occur together.

The oxidizing agent makes others to be oxidized but itself is reduced;

the reducing agent makes others to be reduced but itself is oxidized.

12.2.2 Oxidation Numbers

In a redox reaction, electron transfer causes some of the atoms to change their oxidation

numbers. Thus redox reactions can be identified by noting changes in oxidation numbers.

Oxidation increases the oxidation number of an element by losing its electrons; reduction

decreases the oxidation number by gaining electrons.

There are 5 basic guidelines for assigning oxidation numbers:

1. An atom in its elemental state has an oxidation number of 0

2. A simple ion has its oxidation number equal to its charge; an ion from IA group is +1; an ion

from IIA group is +2; F is –1 in all compounds.

3. Hydrogen is usually +1; it is –1 when bonded to a metal, such as NaH, CaH2.

4. Oxygen is usually –2 but it can be others.

5. The sum of the oxidation numbers is equal to the net charge.


178

T I P! Assign x for the last one in order to calculate the oxidation number

Worked example 12.1

Assign oxidation numbers to each atom in the following substances:

A. Fe3O4

B. NH4NO3

C. H2O2

D. OF2

E. Na2Cr2O7 f) Fe

Solution:

A. 3x + 4(–2) = 0; x = 8/3 note that oxidation number can be fractions.

B. Break it to the ions: NH4+ and NO3

– and calculate the oxidation numbers separately

x + 4(1) = 1; x = – 3; y + 3(–2) = –1; y = +5

C. Based on the rules, H is assigned as +1 and leave O as x; 2(1) + 2x = 0; x = –1

D. Based on the rules, F is assigned as –1 and leave O as x; x + 2(–1) = 0; x = + 2

E. Break it to the ions: Na+ and Cr2O7–2 and calculate the oxidation numbers separately

Na is +1 based on the rule #2; assign x for Cr and the equation is: 2x + 7(–2) = –2; x = 6

F. Fe is 0 based on the rule #1.

Oxidation

number

a

)

X –2 b) x +1 y –2 c) +1 x d) x –1

Formula Fe3 O4 N H4 N O3 H2 O2 O F2


179

12.3 Balancing Redox Reactions

The key to balancing complicated redox equations is to balance electrons as well as atoms.

Because electrons do not appear in chemical formulas or balanced net reactions, the number of

electrons transferred in a redox reaction often is not obvious. To balance complicated redox

reactions, therefore, we need a procedure that shows the electrons involved in the oxidation and

the reduction.


180

12.3.1 Balancing Redox Molecular Reactions Using the Oxidation Number Method

Worked example 12.2

Balance the equation using the oxidation number method:

KMnO4 + H2O2 + H2SO4 → MnSO4 + O2

Solution:

Step 1. Assign oxidation numbers to each atom in the equation:

Step 2. From the changes of the oxidation numbers, find the electrons gained or lost, multiply the

number restricted due to the formula:

Mn changes from (+7) in KMnO4 to (+2) in MnSO4: Mn gains 5 electrons;

O changes from (–1) in H2O2 to (0) in O2: O loses 1x2 electrons (multiply 1 by 2 due to the

restriction of two O in the formula H2O2).

Step 3. Multiply by the appropriate number to make the number of electrons gained equal to the

number of electrons lost.

The least common multiple of 5 and 2 is 10. It requires 2 Mn and 10 O.

2 KMnO4 + 5 H2O2 + H2SO4 → 2 MnSO4 + 5 O2

Step 4. Balance the other spectators except H and O.

2 KMnO4 + 5 H2O2 + 3 H2SO4 → 2 MnSO4 + 5 O2 + K2SO4

Step 5. Balance H by using H2O.

2 KMnO4 + 5 H2O2 + 3 H2SO4 → 2 MnSO4 + 5 O2 + K2SO4 + 8 H2O

Step 6. Use O to check that both sides are balanced.

Left side: 8 + 10 + 12 Right side: 8 + 10 + 4 + 8

Both sides are equal and the equation is balanced.


181

12.3.2 Balancing Redox ionic Reactions Using the half‐Reaction Method

This is done in 4 steps (5 steps for basic solution), which must be followed in the order listed:

Step 1: Break the unbalanced equation into half‐reactions by inspection.

Step 2: Balance each half‐reaction, following the four‐step procedure.

A. Balance all elements except O and H.

B. Balance O using H2O.

C. Balance H using H+.

D. Balance the charge using electrons

Step 3: Multiply the half‐reactions by integers that will lead to cancellation of electrons.

Step 4: Add the half‐reactions, and cancel identical species.

Step 5: If the reaction takes place in basic solution, add the appropriate number of OH – to both

sides of the equation to neutralize the H+, and cancel identical H2O if necessary.


182

Worked example 12.3

Balance the equation in acidic solution:

Cr2O7 2– + C2H5OH → Cr3+ + CO2

Solution:

Step 1. Write the half‐reactions:

Cr2O7 2– → Cr3+

C2H5OH → CO2

Step 2: Balance half‐reactions:

a) Balance all elements except O and H.

Cr2O7 2– → 2 Cr3+

C2H5OH → 2 CO2

b) Balance O using H2O.

Cr2O7 2– → 2 Cr3+ + 7 H2O

C2H5OH + 3 H2O → 2 CO2

c) Balance H using H+.

14 H+ + Cr2O7 2– → 2 Cr3+ + 7 H2O

C2H5OH + 3 H2O → 2 CO2 + 12 H+

d) Balance the charge using electrons

14 H+ + Cr2O7 2– + 6 e → 2 Cr3+ + 7 H2O (gaining electrons is reduction)

C2H5OH + 3 H2O → 2 CO2 + 12 H+ + 12 e (losing electrons is oxidation)


28 H+ + 2 Cr2O7 2– + 12 e → 4 Cr3+ + 14 H2O

C2H5OH + 3 H2O → 2 CO2 + 12 H+ + 12 e


16 H+ + 2 Cr2O7 2– + C2H5OH → 4 Cr3+ + 11 H2O + 2 CO2


183

Worked example 12.4

Balance the equation in basic solution:

Ag + CN – + O2 → Ag(CN)2 –

Solution:

Step 1. Write the half‐reactions:

Ag + CN – → Ag(CN)2 –

O2 →

Step 2: Balance half‐reactions:

a) Balance all elements except O and H.

Ag + 2 CN – → Ag(CN)2 –

O2 →

b) Balance O using H2O.

Ag + 2 CN – → Ag(CN)2 –

O2 → 2 H2O

c) Balance H using H+.

Ag + 2 CN – → Ag(CN)2 –

4 H+ + O2 → 2 H2O

d) Balance the charge using electrons

Ag + 2 CN – → Ag(CN)2 – + e (losing electrons is oxidation)

4 H+ + O2 + 4 e → 2 H2O (gaining electrons is reduction)


4 Ag + 8 CN – → 4 Ag(CN)2 – + 4 e

4 H+ + O2 + 4 e → 2 H2O


4 Ag + 8 CN – + 4 H+ + O2 → 4 Ag(CN)2 – + 2 H2O


184

Step 5: Add 4 OH – to both sides of the equation to neutralize the 4 H+ on the left, and cancel 2

H2O on both sides.

4 Ag + 8 CN – + 2 H2O + O2 → 4 Ag(CN)2 – + 4 OH–

12.4 Standard Reduction Potentials

12.4.1 Definition of Standard Reduction Potentials

Here is a summary of the definitions and conventions for working with electrochemical

potentials:

1. Standard conditions for electrochemical half‐reactions are the same as those in

thermodynamics: 25 °C (298.15 K), 1 M for solutes, and 1 atm for gases.

2. Standard reduction potentials (Eo)are assigned to half‐reactions written as reduction

processes (gaining electron reaction).

3. The half‐reaction under standard conditions, written in either directions (2 H+ + 2e → H2 or

H2 → 2 H+ + 2e ), is assigned an arbitrary reference potential of zero volts.

4. Whenever the direction of a half‐reaction is reversed, the sign of Eo must be reversed.

5. The value of Eo is not changed when a half‐reaction is multiplied.

12.4.2 Meaning of Standard Reduction Potentials

1. It is a measure of the tendency to have a reduction reaction; the higher the reduction

potential, the greater the tendency for the reduction reaction.

2. It is a measure of the ability to gain electrons; the higher the reduction potential, the greater

the tendency of gaining electrons.

3. It is a measure of the relative strength of the oxidizing agent; the higher the reduction

potential, the stronger the oxidizing agent will be.


185

12.4.3. Predicting Spontaneous Redox Reactions

The relative values of standard reduction potentials would enable us to arrange oxidizing or

reducing agents in order of increasing strength, and this would permit us to predict the

spontaneity of redox reactions.

Worked example 12.4

Predict whether the reaction can occur under standard state conditions:

2 Ag+ + Zn → 2 Ag + Zn2+

Solution:

Method 1. Compare relative strength of the oxidizing and reducing agents

Ag+ + e ↔ Ag Eredo = 0.80 V

Zn2+ + 2e ↔ Zn Eredo = –0.76 V

Ag+ is a stronger oxidizing agent than Zn2+ and Zn is a stronger reducing agent than Ag.

The reaction will proceed spontaneously from the stronger oxidizing agent (Ag+) and reducing

agent (Zn ) to form the weaker reducing agent (Ag) and oxidizing agent (Zn2+) direction.

The reaction (2 Ag+ + Zn → 2 Ag + Zn2+) is spontaneous in the forward direction.

Method 2. Use the value of Eo for the reaction.

Reduction: [ Ag+ + e → Ag ] x 2 Eredo = 0.80 V

Oxidation: Zn → Zn2+ + 2e Eoxo = –(–0.76) V

Overall reaction: 2 Ag+ + Zn → 2 Ag + Zn2+ Eo = 1.56 V

Because Eo is positive, the reaction is a spontaneous one under the standard conditions.


186

Remember that:

(i) Eo for oxidation of Zn is the negative of the standard reduction potential (–0.76 V )

(ii) We do not multiply the Eo value for reduction of Ag+ ( 0.80 V) by a factor of 2 as mentioned in

section 12.4.1.

Method 3. From the metal reactivity series, Zn is more reactive than Ag and the single

displacement reaction ( 2 Ag+ + Zn → 2 Ag + Zn2+) will take place spontaneously.


187

12.5 Galvanic cells

When a redox reaction occurs in solution, the electrons are transferred directly, and no useful

work can be obtained from the chemical energy involved in the reaction. However, if the

oxidizing agent is physically separated form the reducing agent so that the electrons must flow

through a wire from one to the other, the chemical energy can produce useful work. A galvanic

cell is a device in which chemical energy is transformed to electrical energy.

12.5.1 Description of a Galvanic Cell

The description will include the cell reaction, the cell potential, and the physical setup of the cell.

Worked example 12.5

Describe a galvanic cell based on the following half‐reactions:

Solution:

Step 1. Determine the cell potential:

A cell will always run spontaneously in the direction that produces a positive cell potential. Thus,

the half‐reaction involving Fe with lower reduction potential must be reversed, since this choice

leads to a positive cell potential:

Fe → Fe2+ + 2e Eo = –(–0.44) V

MnO4– + 8 H+ + 5e → Mn2+ + 4 H2O Eo = 1.51 V

The cell potential is Ecell = –(–0.44) V + 1.51 V = 1.95 V.


188

Step 2. Describe the reactions at the electrodes and the cell reaction.

Because we have determined the directions of half‐reactions, it is easy to know:

The electron‐losing reaction is oxidation at the anode: [ Fe → Fe2+ + 2e ] x 5

The electron‐gaining reaction is reduction at the cathode:

[ MnO4– + 8 H+ + 5e → Mn2++ 4 H2O ] x 2

The overall cell reaction is: 2 MnO4– + 16 H+ + 5 Fe → 2 Mn2++ 8 H2O + 5 Fe

2+

Step 3. The shorthand notation for the cell

Fe(s) │ Fe2+(aq) ║ MnO4– (aq), Mn2+(aq)│Pt

The left side is the anode (negative electrode) and the right side is the cathode ( positive

electrode). The electrons flow from Fe to MnO4– , or from the anode to the cathode through the

external wire, and ions flow from one compartment to the other to keep the net charge zero, as

is always the case.

12.5.2 Shorthand Notation for Galvanic Cells

It is convenient to use a shorthand notation for representing the cell. For a cell reaction

Sn2+ + 2 Ag+ → 2 Ag + Sn4+

We can write the following expression:

Pt(s) │ Sn2+(aq), Sn4+(aq) ║Ag+(aq) │Ag

A. A single vertical line represents a phase boundary but different solutes are separated by a

comma

B. A double vertical line denotes a salt bridge

C. The electrodes are written on the extreme left (anode) and on the extreme right (cathode)

D. The reactants in each half‐cell are written first, followed by the products.

With these arbitrary conventions, electrons move through the external circuit from left to right

(from anode to cathode or from negative to positive ).


189

12.5.3 General Rules for Anode and Cathode

Anode: Is where oxidation occurs Cathode: Is where reduction occurs

Is where electrons are produced Is where electrons are consumed

Is what anions migrate toward Is what cations migrate toward

Has a negative sign Has a positive sign

12.6 Cell Potentials and Free‐Energy Changes

12.6.1 Cell Potential and Work

Electrons flow through the external circuit from the anode (– electrode) to the cathode ( +

electrode). The work that can be accomplished when electrons are transferred depends on the

“push” (the electromotive force ) behind the electrons. This driving force (emf) is defined in

terms of a potential difference ( in volts), or the cell potential.

emf = cell potential ( V ) = work ( J ) / charge ( C ) or w = – q E (1)

Work is done by the galvanic cell ( the system ). Thus work flowing out of the system is indicated

by a minus sign. When a cell produces a current, the cell potential is positive, and the current can

be used to do work. Thus the cell potential E and the work w have opposite sign.

12.6.2 Free Energy and Electrochemistry A spontaneous electrochemical reaction generates a positive cell potential. Thermodynamically,

the spontaneity of a redox reaction is determined by its free energy change: A spontaneous

reaction has a negative ΔG. Thus, a reaction that has a negative change in free energy generates

a positive cell potential:

ΔGreaction < 0: Spontaneous redox reaction

Ecell > 0: Spontaneous redox reaction


190

12.6.3 Cell Potential and Free Energy

The linkage between free energy and cell potentials can be made quantitative. The more negative

the value of ΔG° for a reaction, the more positive its standard cell potential.

The amount of charge provided by one mole of electrons is called the Faraday constant (F),

(96,485.34 C mol‐1 or 96500 C/mol e) and we have q = n F, in which n is the number of moles of

electrons transferred during the process. Substitute the q into the equation (1):

w = – n F E (2)

For a process carried out at constant temperature and pressure, the change in free energy equals

the maximum useful work obtainable from that process: w = ΔG

Substitute this into equation (2):

ΔG = – nFE (3)

Remember that the number of electrons transferred is not explicitly stated in a net redox

reaction. This means that any overall redox reaction must be broken down into its balanced half‐

reactions to determine n, the ratio between the number of electrons transferred and the

stoichiometric coefficients for the chemical reagents. The equation (3) above applies under all

conditions, including standard conditions.

At standard conditions, ΔG = ΔG° and E = E°: ΔGo = – nFEo (3)


191

12.6.4 Cell Potentials and Chemical Equilibrium

Combining the equations:

ΔG° = ‐nFE° and ΔG° = ‐RT ln Keq

and grouping the constants gives a new equation that links standard potentials directly to

equilibrium constants:

‐RT ln Keq = nFE° ⇒ eqKnFRTE ln=° or eqK

nE logV0592.0

⎟⎠⎞

⎜⎝⎛=° at 25oC

12.6.5 The Nernst Equation

The equation that links ΔG° with free energy changes under nonstandard conditions is:

ΔG = ΔG° + RT ln Q

Here, Q is the reaction quotient. We can see how concentration and pressure affect cell

potentials by substituting for ΔG and ΔG° using ΔG = ‐nFE and ΔG° = ‐nFE°.

‐nFE = ‐nFE° + RT ln Q

Dividing both sides by (‐nF) gives the Nernst equation: QnFRTEE ln−°=

The Nernst equation is used to convert between standard cell potentials and potentials of

electrochemical cells operating under nonstandard concentration conditions. When

measurements are made at standard temperature, 298 K, the numerical value calculated earlier

can replace RT/F:

298KTwhenlogV 0.0592=⎟

⎠⎞

⎜⎝⎛−°= Q

nEE


192

12.7 Electrolysis and Electrolytic Cells

12.7.1 Electrolysis

The equation that links ΔG° with free energy changes under nonstandard conditions is:

ΔG = ΔG° + RT ln Q

A galvanic cell produces current when an oxidation‐reduction reaction proceeds spontaneously.

An electrolytic cell uses electrical energy to produce chemical change. The process of electrolysis

involves forcing a current through a cell to produce a chemical change for which the cell potential

is negative: that is, electrical work causes an otherwise nonspontaneous chemical reaction to

occur. The process of using an electric current to bring about chemical change is called

electrolysis.

12.7.2 Electrolysis of water

Hydrogen and oxygen combine spontaneously to form water and that the accompanying

decrease in free energy can be used to run a fuel cell to produce electricity. The reverse

nonspontaneous process can be forced by electrolysis, which takes place in an electrolytic cell. An

electrolytic cell has two electrodes that dip into an electrolyte and are connected to an external

battery. The battery pushes electrons from its negative terminal into the cathode to force the

reduction reaction to take place and pulls electrons out of the anode to force the oxidation

reaction to take place.

Anode reaction: 2 H2O → O2 + 4 H+ + 4e Eox

o = – 1.23 V

Cathode reaction: 4 H2O + 4e → 2 H2 + 4 OH– Ered

o = – 0.83 V

Net reaction: 6 H2O → 2 H2 + O2 + 4 H+ + 4OH– Eo = – 2.06 V

or 2 H2O → 2 H2 + O2 Eo = – 2.06 V

(4 H+ + 4OH– → 4 H2O)


193

Note that these potentials assume an anode chamber with 1 M H+ and a cathode chamber with 1

M OH–. In pure water, where [H+] = [ OH– ] =10–7 M, the potential for the overall process is –1.23

V.

12.7.3 Electrolysis of aqueous sodium chloride

In the electrolysis of aqueous sodium chloride, the battery pushes electrons into the cathode and

the cathode half‐reaction might be either the reduction of Na+ to sodium metal or the reduction

of water to hydrogen gas:

Na+(aq) + e → Na(s) Eredo = – 2.71 V

2 H2O(l) + 2e → H2(g) + 2 OH–(aq) Ered

o = – 0.83 V

Because the standard reduction potential is much less negative for the reduction of water than

for the reduction of Na+, water is reduced preferentially and bubbles of hydrogen gas are

produced at the cathode. Notice that the solution at the cathode will be basic due to the

production of OH– and the anion OH– will migrate to the anode which is connected to the +

terminal of the external battery.

The positive terminal of the battery pulls electrons out of the anode to force the oxidation

reaction to take place. If there are different species at the anode, which one will be oxidized first?

It would be easier to know the following qualitative rules:

A. If the electrode is made of a non‐inert metal, such as Cu or Fe, the metal will lose the electron

to be oxidized first;

B. If the electrode is inert, the following is a list of increasing difficulties in transferring electrons

from the species in the solution to the atoms on the electrode:

I– < Br– < Cl– < H2O


194

The Cl– is oxidized preferentially and bubbles of chlorine gas are produced at the anode.

The electrode reactions and overall cell reaction for electrolysis of aqueous sodium chloride are

Anode (oxidation): 2 Cl– (aq) → Cl2(g) + 2 e Eoxo = – 1.36 V

Cathode (reduction): 2 H2O(l) + 2e → H2 (g) + 2 OH–(aq) Ered

o = – 0.83 V

Overall cell reaction: 2 Cl– (aq) + 2 H2O(l) → Cl2(g) + H2 (g) + 2 OH–(aq) Eo = – 2.19 V

12.7.4 Quantitative aspects of Electrolysis

Current

and

time

↔

Quantity of

charge in

coulombs

↔

Moles of

electrons

↔

Moles of

substance

oxidized or

reduced

↔

Grams of

substances

oxidized or

reduced

The above table shows how the quantities of substances produced or consumed in electrolysis

are related to the quantity of electrical charge that is used. The same relationships can also be

applied to galvanic cells. In other words, electrons can be thought of as reagents in electrolysis

reaction.


195

Worked example 12.5

Determine the mass of copper that is plated out when a current of 10.0 A is passed for 30.0

minutes through a solution containing Cu2+.

Solution:

Step 1. Calculate the charge in coulombs: Q = IΔt = 10 A (30x60 s) = 18000 C ( note that 1 A = 1

C/s)

Step 2. Calculate the moles of electrons by using the conversion factor of 1 F = 96500 C/1 mol e.

18000 C x 1

96500mol e

C = 0.187 mol e

This means that 0.187 mol of electrons flowed into the Cu2+ solution.

Step 3. Calculate the moles of copper produced

From the balanced reduction half reaction: Cu2+ + 2e → Cu.

1 mol Cu2+ requires 2 mol electrons to produce 1 mol Cu.

0.187 mol e x 12mol Cumol e

= 0.0935 mol Cu

Step 4. Calculate the mass of copper produced

0.0935 mol Cu x 63.546

1g

mol = 5.94 g Cu


196


Multiple Choice

1. A strip of copper is placed in a 1 M solution of copper nitrate and a strip of silver is placed in a

1 M solution of silver nitrate. The two metal strips are connected to a voltmeter by wires and

a salt bridge connects the solutions. The following standard reduction potentials apply:

Ag+ (aq) + e− ⎯→ Ag (s) oE = +0.80 V

Cu2+ (aq) + 2e− ⎯→ Cu (s) oE = +0.34 V

When the voltmeter is removed and the two electrodes are connected by a wire, which of the

following does not take place?

A. Electrons flow in the external circuit from the copper electrode to the silver electrode.

B. The silver electrode increases in mass as the cell operates.

C. There is a net general movement of silver ions through the salt bridge to the copper half‐

cell.

D. Negative ions pass through the salt bridge from the silver half‐cell to the copper half‐cell.

E. Some positive copper ions pass through the salt bridge from the copper half‐cell to the

silver half‐cell.


197

For the following 3 questions, consider an electrochemical cell with a cobalt electrode

immersed in 1.0 M Co2+ and a lead electrode immersed in 1.0 M Pb2+.

Co2+ + 2e− ⎯→ Co oE = −0.28 V

Pb2+ + 2e− ⎯→ Pb oE = −0.13 V

2. Calculate oE for this cell

A. −0.15 V

B. 0.15 V

C. −0.41 V

D. 0.41 V

E. none of these

3. Which of the electrodes is the cathode?

A. The cobalt electrode

B. The lead electrode

4. If [Co2+]0 is 0.0010 M and [Pb2+]0 is 0.10 M, calculate E.

A. 0.15 V

B. 0.091 V

C. 0.27 V

D. 0.21 V

E. none of these

Answer: D

Solution: Write the balanced cell reaction:


198

5. The following oxidation‐reduction reaction occurs in basic solution:

CrO42− + Fe(OH)2 ⎯→ Fe(OH)3 + Cr(OH)4

−

Complete and balance the above equation. What is the sum of the reaction coefficients in

the balanced equation?

A. 10

B. 11

C. 13

D. 15

E. 18


199

6. Which of the following is the strongest oxidizing agent?

MnO4− + 4 H+ + 3 e− ⎯→ MnO2 + 2 H2O E° = 1.68 V

I2 + 2 e− ⎯→ 2 I− E° = 0.54 V

Zn2+ + 2 e− ⎯→ Zn E° = −0.76 V

A. MnO4−

B. I2

C. Zn2+

D. Zn

E. MnO2

The following 5 questions refer to the galvanic cell shown below which uses inert platinum

electrodes and operates at a temperature of 25°C.

Br2(aq) + 2 e− ⎯→ 2 Br− (aq) E° = 1.09 V

I2 (aq) + 2 e− ⎯→ 2 I− (aq) E° = 0.54 V

[Br−] = 0.20 M [Br2] = 0.20 M

[I−] = 10.0 M [I2] = 0.10 M

Salt Bridge

V

Pt Pt


200

7. If the galvanic cell operates at standard state conditions, which species will be oxidized and

which species will be reduced?

A. Br− is oxidized and I− is reduced.

B. I− is oxidized and Br2 is reduced.

C. Br2 is oxidized and I2 is reduced.

D. I2 is oxidized and H+ is reduced.

E. H2O is oxidized and Br2 is reduced.

Answer: B

8. In which direction will electrons flow in the external circuit?

A. Left to right

B. Right to left

C. No current flows since the cell is at equilibrium.

D. No current flows since the concentration of Br2 is exactly the same as that of Br−.

9. What is the value of E◦cell?

A. 0.33V

B. 0.44V

C. 0.55V

D. 0.66V

E. 0.77V


201

10. What is the value of Ecell of the galvanic cell shown in the diagram?

A. 0.33V

B. 0.44V

C. 0.55V

D. 0.66V

E. 0.77V

11. Indicate the direction of ion migration through the salt bridge if it contains concentrated

KNO3 solution.

A. K+ migrates to right and NO3– migrates to left

B. NO3– migrates to right and K+ migrates to left

C. Both K+ and NO3– migrate to riht

D. Both K+ and NO3– migrate to left

E. No ion migrates through the salt bridge since the cell is at equilibrium.

12. For the galvanic cell,

Cu (s) | Cu2+ (aq, 1.0 M) | | Ag+ (aq, 1.0 M) | Ag (s)

Which of the following does NOT take place?

A. The silver electrode increases in mass as the cell operates.

B. The cell, as described above, would be operating at standard state conditions.

C. The cell potential, E would be less than E° if the concentration of Cu2+ was greater than 1.0

M.

D. Electrons flow in the external circuit from the copper electrode to the silver electrode.

E. The concentration of Cu2+ decreases as the cell operates.


202

Short Answer:

13. A galvanic cell contains a silver electrode and a copper electrode.

Ag+ + e‐ ↔ Ag(s) Eo = +0.799

Cu2+ + 2e‐ ↔ Cu(s) Eo = 0.337

A. Write the half‐reactions that occur at the anode and the cathode and identify the

oxidizing and reducing agents.

B. Calculate the standard cell potential, ∆Eo, and the equilibrium constant K for the cell at

standard conditions.


203

C. What concentration of Ag+ will produce a voltage of 0.500 V if [Cu2+] = 0.100 M?

2Ag+ + Cu 2Ag + Cu2+

E = E° – [ ]2

2

Ag

][log0592.0+

+Cun

E = .500 V, E° = .462 V, and n = 2

–0.038 = [ ]2

2

Ag

][log2

0592.0+

+Cu

log 0.1 – 2log [Ag+] = 2x(–0.038)/0.0592 = –1.284 log[Ag+] = 0.142 [Ag+] = 1.39 M


204

Study Tips/Exam Writing Strategies

This course covers a wide variety of fundamental concepts and principles in Chemistry. Students

always find it hard to put all these together especially when they are preparing for the final exam.

This booklet is designed to help you spend your valuable time effectively and to ensure that it is a

source of the information that you need for learning, skill development, and test preparation.

The chapter summaries at the beginning of this Study Guide are a simple way to study the basic

concepts covered in this course in a short amount of time. Read through the detailed contents

and do all practice problems on your own, then look at the solutions provided. This way will make

your learning successful. You learn things by doing them. Doing practice questions contained

within this booklet is a good way to help you acquire both conceptual understanding and

problem‐solving skills for the exam.

The amount of information you will be expected to learn can sometimes seem overwhelming. It is

essential to recognize those concepts and skills that are particularly important. The following

table outlines some important concepts and skills you must master for the exam.


205

Chapter Test items Preparation focus 1 Limiting reactant stoichilmetry

Solution stoichiometry Empirical and molecular formulas

Stoichiometry calculations based on the balanced equations

2 Quantum numbers and atomic structures Atom and light, photoelectric effect Periodicity

Understand the meaning of quantum numbers. The nature of the particle wave.

3 Lewis structures VSEPR model

Connect the type of molecule with the shape.

4 Speed, kinetic energy, and temperature Ideal gas equations Real gas concepts

Understand the basic formulas and pay attention to the unit conversion.

5 Comparison of intermolecular forces Phase diagram Colligative properties

Determine the relative strength of different attractive forces

6 Rate law derived from a tabulated data Match the rate law with different mechanisms

The table summarizes different kinetics and worked examples help you understand the mechanisms

7 First law Calorimetry, Hess’s law, standard enthalpy of formation

Understand the signs for energy, heat, and work.

8 Calculate the equilibrium concentrations and the equilibrium constant

Use ICE table to analyse the concentrations at equilibrium

9 Calculate pH in different systems. It may be strong acid or base; weak acid or base; mixtures of acid or base.

The three step analyzing procedures are designed for you to work on different systems.

10 Buffer calculation Acid‐base titration curve analysis Solubility equilibrium calculation

The worked examples guide you step by step to deal with different situations

11 Spontaneity and ∆S,∆G,∆H relationships Conceptual questions

Understand the concepts among different terms

12 Describe the galvanic cells Nernst equation

Understand the cell mechanism Relate to the reduction potentials

A central theme of this booklet is a systematic approach to problem solving. I hope the material is useful in helping you understand chemistry better and achieve higher on your exams.

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