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Chemical Reactions 2:Equilibrium and Oxidation-Reduction

CHE-5043-2

Learning Guide

CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION-REDUCTION

CHE-5043-2LEARNING GUIDE

Chemical Reactions 2: Equilibrium and Oxidation-reduction is the third of the three

Learning Guides for the Secondary V Chemistry program, which comprises the

following three courses:

Gases

Chemical Reactions 1: Energy and Chemical Dynamics

Chemical Reactions 2: Equilibrium and Oxidation-reduction

The three Learning Guides are complemented by the workbook entitled Experimental

Activities of Chemistry, which covers the “experimental method” component of the

program.


This Guide was produced by the Société de formation à distance des commissions

scolaires du Québec.

Production Coordinator: Jean-Simon Labrecque (SOFAD)

Mireille Moisan (First edition)

Coordinator: Céline Tremblay (FormaScience)

Authors: Paule Morazain

André Dumas

Illustrators: Gail Weil Brenner (GWB)

Jean-Philippe Morin (JPM)

Content Revisors: André Dumas (French Version)

Céline Tremblay (FormaScience) (French Version)

Stéphanie Belhumeur (English Version)

Layout: I. D. Graphique inc. (Daniel Rémy)

Translators: Claudia de Fulviis and Barbara Chunn

Linguistic Revisors: Barbara Chunn and Claudia de Fulviis

Proofreader: Gabriel Kabis

First Edition: January 2001

June 2006

© Société de formation à distance des commissions scolaires du Québec

All rights for translation and adaptation, in whole or in part, are reserved for all countries.

Any reproduction by mechanical or electronic means, including microreproduction, is

forbidden without the written permission of a duly authorized representative of the Société

de formation à distance des commissions scolaires du Québec.

Legal Deposit – 2000

Bibliothèque et Archives nationales du Québec

Bibliothèque et Archives Canada

ISBN 978-2-89493-193-6

TABLE OF CONTENTS

GENERAL INTRODUCTION

OVERVIEW ................................................................................................................... 0.12

HOW TO USE THIS LEARNING GUIDE ............................................................................. 0.12

Learning Activities ................................................................................................. 0.13

Exercises ............................................................................................................ 0.13

Self-evaluation Test ............................................................................................... 0.14

Appendices ........................................................................................................... 0.14

Materials .............................................................................................................. 0.14

CERTIFICATION ............................................................................................................. 0.15

INFORMATION FOR DISTANCE EDUCATION STUDENTS .................................................... 0.15

Work Pace ............................................................................................................ 0.15

Your Tutor ............................................................................................................. 0.15

Homework Assignments ........................................................................................ 0.16

CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION-REDUCTION ............................. 0.17

LEARNING ACTIVITIES

CHAPTER 1 – EQUILIBRIUM ........................................................................................ 1.1

1.1 EVAPORATION AND DISSOLUTION .......................................................................... 1.3

Liquid-vapour Equilibrium ....................................................................................... 1.3

Dissolution-crystallization Equilibrium ..................................................................... 1.8

Experimental Activity 1: Equilibrium Systems ..................................................... 1.13

1.2 CHEMICAL REACTIONS .......................................................................................... 1.14

Equilibrium of the Reaction N2O4(g) a 2 NO2(g) ....................................................... 1.15

Reaching Equilibrium ............................................................................................. 1.20

Rate and Concentration ................................................................................... 1.22

Change in Rates of Reaction ........................................................................... 1.23

Change in Concentrations ............................................................................... 1.26

The Stationary State: A Special Case ..................................................................... 1.30

KEY WORDS IN THIS CHAPTER ..................................................................................... 1.34

SUMMARY .................................................................................................................... 1.34

REVIEW EXERCISES ...................................................................................................... 1.36

CHAPTER 2 – FACTORS AFFECTING CHEMICAL EQUILIBRIUM ..................................... 2.1

2.1 FACTORS AFFECTING EQUILIBRIUM ........................................................................ 2.3

Concentration ....................................................................................................... 2.3

Experimental Activity 2: Disturbing an Equilibrium System ................................. 2.8

A Concrete Case: Ammonia Production ................................................................... 2.10

Chemical Reactions 2 - Table of Contents

0.5

Temperature ......................................................................................................... 2.13

Catalysts .............................................................................................................. 2.16

2.2 LE CHÂTELIER’S PRINCIPLE ................................................................................... 2.18

Temperature ......................................................................................................... 2.19

Concentration ....................................................................................................... 2.21

Pressure ............................................................................................................... 2.23

2.3 PRACTICAL APPLICATIONS OF CHEMICAL EQUILIBRIUM ........................................... 2.30

Ammonia: A Very Useful Product ............................................................................ 2.30

From Hard Water to Soft Water .............................................................................. 2.38

Iodine Automobile Headlights ................................................................................. 2.41

2.4 EQUILIBRIUM IN NATURE ....................................................................................... 2.44

The Water Cycle .................................................................................................... 2.45

The Carbon Cycle .................................................................................................. 2.50


SUMMARY .................................................................................................................... 2.54


CHAPTER 3 – EQUIIBRIUM IN ACIDIC AND BASIC SOLUTIONS ..................................... 3.1

3.1 ACID IONIZATION CONSTANT Ka ............................................................................. 3.3

Review on Acids .................................................................................................... 3.3

The pH Scale ................................................................................................. 3.7

The Constant Ka .................................................................................................... 3.12

Experimental Activity 3: Mathematical Expression of Equilibrium and Acid Strength 3.13

Definition ........................................................................................................ 3.13

Strong Acids and Weak Acids .......................................................................... 3.15

Applications .................................................................................................... 3.18

3.2 THE BASE IONIZATION CONSTANT Kb ...................................................................... 3.25

Review on Bases .................................................................................................. 3.25

The Constant Kb .................................................................................................... 3.28

Strong and Weak Bases .................................................................................. 3.28

Applications .................................................................................................... 3.31

3.3 EQUILIBRIUM OF WATER AND NEUTRALIZATION ...................................................... 3.34

The Ion-product Constant for Water ........................................................................ 3.34

Kw and Acidic or Basic Solutions ...................................................................... 3.36

Conjugate Acids and Bases ............................................................................. 3.41

Acid-base Neutralization ........................................................................................ 3.44

Strong Acids and Strong Bases ....................................................................... 3.44

Other Cases ................................................................................................... 3.46

Titration of a Strong Acid with a Strong Base .......................................................... 3.48

Experimental Activity 4: Titration of an Acid ..................................................... 3.48


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3.4 ACID-BASE EQUILIBRIUM: CONCRETE EXAMPLES .................................................... 3.51

Acid-base Equilibrium in the Laboratory .................................................................. 3.51

Acid-base Indicators ........................................................................................ 3.52

Buffer Solutions .............................................................................................. 3.54

The Acid-base Equilibrium in Nature ....................................................................... 3.58

The pH of Blood .............................................................................................. 3.58

Sulphur and the pH of Specific Environments ................................................... 3.60

Acid-base Equilibrium at Home ............................................................................... 3.61

Fluoride Toothpaste ........................................................................................ 3.61

The Magic of Baking Powder ............................................................................ 3.62

Fire Extinguishers ............................................................................................ 3.63


SUMMARY .................................................................................................................... 3.66


CHAPTER 4 – THE EQUILIBRIUM CONSTANT ................................................................ 4.1

4.1 EQUILIBRIUM OF A CHEMICAL REACTION ............................................................... 4.3

The Constant Kc ................................................................................................... 4.4

Definition ........................................................................................................ 4.6

Solids and Liquids .......................................................................................... 4.9

The Kc and Disruptions in Equilibrium ..................................................................... 4.12

Changes in Concentration ............................................................................... 4.12

Changes in Pressure ....................................................................................... 4.16

Temperature Changes ..................................................................................... 4.20

4.2 APPLICATIONS ....................................................................................................... 4.22

Calculation of the Equilibrium Constant Kc .............................................................. 4.22

Calculation of the Concentrations at Equilibrium ..................................................... 4.29

Production of Sulphuric Acid .................................................................................. 4.41

4.3 DISSOLUTION-CRYSTALLIZATION EQUILIBRIUM ........................................................ 4.45

The Solubility Product Constant (Ksp) ...................................................................... 4.45

Concrete Examples ............................................................................................... 4.53

Stones and Articular Gout ............................................................................... 4.53

Calcium Carbonate Deposits ........................................................................... 4.54

Caves with Unusual Shapes ............................................................................ 4.55

Precipitation of Salts ............................................................................................. 4.58

4.4 CHEMISTRY, INDUSTRY AND SOCIETY .................................................................... 4.63

A Brief History of Soda Ash Production ................................................................... 4.63

The Solvay Process ............................................................................................... 4.65

Ammonia .............................................................................................................. 4.70


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SUMMARY .................................................................................................................... 4.72


CHAPTER 5 – OXIDATION-REDUCTION ......................................................................... 5.1

5.1 OXIDATION AND REDUCTION .................................................................................. 5.3

How Reactions Proceed ......................................................................................... 5.5

Experimental Activity 5: Displacement of Metals ............................................. 5.5

Half-reactions and the Complete Reaction ............................................................. 5.12

Spontaneous Redox Reactions .............................................................................. 5.15

5.2 THE DANIELL CELL: PRECURSOR OF THE MODERN CELL ........................................ 5.22

Experimental Activity 6: Potential Difference Between Two Metals .................... 5.28

5.3 STANDARD REDUCTION POTENTIAL ........................................................................ 5.29

Standard Hydrogen Half-cell ................................................................................... 5.30

Standard Reduction Potential of the Zn(s)–Zn2+(aq) Half-cell .......................................... 5.31

Standard Reduction Potential of the Ag(s)–Ag+(aq) Half-cell .......................................... 5.34


SUMMARY .................................................................................................................... 5.40


CHAPTER 6 – CELLS AND ELECTROCHEMISTRY .......................................................... 6.1

6.1 VOLTAIC CELLS ..................................................................................................... 6.3

Experimental Activity 7: Construction of an Electrochemical (Voltaic) Cell .......... 6.5

The Zn(s)–Cu2+(aq) Cell .............................................................................................. 6.6

Weakening of a Voltaic Cell .................................................................................... 6.10

6.2 BALANCING OXIDATION-REDUCTION EQUATIONS ..................................................... 6.11

Oxidation-Reduction without Ions or Metals ............................................................ 6.14

Oxidation Numbers ................................................................................................ 6.15

Half-reactions ....................................................................................................... 6.22

Balancing Equations .............................................................................................. 6.25

6.3 APPLICATIONS ....................................................................................................... 6.31

Cells and Batteries ............................................................................................... 6.31

Dry Cells ........................................................................................................ 6.31

Storage Batteries ............................................................................................ 6.35

Fuel Cells ....................................................................................................... 6.38

A Little Bit of History ............................................................................................. 6.41

Corrosion ............................................................................................................. 6.42

Corrosion of Iron ............................................................................................. 6.42

Protection Against Corrosion ........................................................................... 6.47


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Electrolysis ........................................................................................................... 6.52

Electroplating .................................................................................................. 6.53

Aluminum Production ............................................................................................ 6.56

A Short History of Aluminum ........................................................................... 6.60

Magnesium Production .......................................................................................... 6.61

Production from Seawater ............................................................................... 6.62

Production from Mining Tailings ....................................................................... 6.64


SUMMARY .................................................................................................................... 6.67


CONCLUSION

SELF-EVALUATION TEST ................................................................................................ C.5

ANSWER KEY

CHAPTER 1 – EQUILIBRIUM ................................................................................... C.25

CHAPTER 2 – FACTORS AFFECTING CHEMICAL EQUILIBRIUM ................................... C.32

CHAPTER 3 – EQUILIBRIUM IN ACIDIC AND BASIC SOLUTIONS ................................ C.44

CHAPTER 4 – THE EQUILIBRIUM CONSTANT ........................................................... C.72

CHAPTER 5 – OXIDATION-REDUCTION ..................................................................... C.106

CHAPTER 6 – CELLS AND ELECTROCHEMISTRY ...................................................... C.116

ANSWER KEY FOR THE SELF-EVALUATION TEST ...................................................... C.139

APPENDIX A – THE INTERNATIONAL SYSTEM OF UNITS (SI) ............................................. C.155

Symbols of Quantity and Their Units ................................................................. C.155

Multiples and Submultiples of SI Units ............................................................. C.155

APPENDIX B – SCIENTIFIC NOTATION ............................................................................. C.156

APPENDIX C – THE LAW OF EXPONENTS ........................................................................ C.157

APPENDIX D – LOGARITHMS ......................................................................................... C.158

APPENDIX E – SOLVING SECOND-ORDER EQUATIONS ..................................................... C.163

APPENDIX F – BALANCING EQUATIONS .......................................................................... C.165

APPENDIX G – LIST OF FIGURES .................................................................................... C.168

BIBLIOGRAPHY ............................................................................................................. C.171

GLOSSARY ................................................................................................................. C.173


0.9

GENERAL INTRODUCTION

OVERVIEW

Welcome to the course entitled Chemical Reactions 2: Equilibrium and Oxidation-

reduction, which is part of the Secondary V Chemistry program. This program

comprises the following three courses:

CHE-5041-2 Gases

CHE-5042-2 Chemical Reactions 1: Energy and Chemical Dynamics

CHE-5043-2 Chemical Reactions 2: Equilibrium and Oxidation-reduction

The three main components of the Chemistry program are related content, the

experimental method and the history-technology-society perspective. Whereas the

experimental method is developed in the workbook Experimental Activities of

Chemistry, the related content and the history-technology-society perspective are

covered in the three Learning Guides that complete the three courses which must

be taken in sequential order.

Chemical Reactions 2: Equilibrium and Oxidation-reduction is the third in the set of

three Learning Guides. It is divided into six chapters, corresponding to the three

terminal objectives of the program.1 This Guide is to be used in conjunction with the

workbook Experimental Activities of Chemistry. You will find references to the

appropriate sections of the Workbook throughout the Guide.

The course Chemical Reactions 2: Equilibrium and Oxidatio-reduction will help you

gain a better understanding of chemical equilibrium and oxidation-reduction

reactions, together with the related technical applications, social changes and

environmental consequences.

HOW TO USE THIS LEARNING GUIDE

This Guide is the main work tool for this course and has been designed to meet the

specific needs of adult students enrolled in individualized learning programs or distance

education courses.

Each chapter covers a certain number of themes, using explanations, tables,

illustrations and exercises designed to help you to master the different program

objectives. A list of key words, a summary and review exercises are included at the

end of each chapter.

Chemical Reactions 2 - General Introduction

0.12

1. The terminal objectives and associated intermediate objectives are listed at the beginning of each chapter.

The conclusion contains a summary covering all the courses in the program along

with a self-evaluation test. It also includes an Answer Key for the self-evaluation test,

for the exercises in each chapter and for the review exercises. A glossary with definitions

of the key words, a bibliography, appendices and an index are also provided in the

conclusion. You may wish to consult the books and publications in the bibliography

for further information on the topics covered in this course.

Learning Activities

The Guide contains theoretical sections as well as practical activities in the form of

exercises. The exercises come with an Answer Key.

Start by skimming through each part of the Guide to familiarize yourself with the

content and the main headings. Then read the theory carefully:

– Highlight the important points.

– Make notes in the margins.

– Look up new words in the dictionary.

– Summarize important passages in your own words, in your notebook.

– Study the diagrams carefully.

– Write down questions relating to ideas you don’t understand.

Exercises

The exercises come with an Answer Key, which is located in the coloured section at

the end of the Guide.

• Do all the exercises.

• Read the instructions and questions carefully before writing your answers.

• Do all the exercises to the best of your ability without looking at the Answer Key.

Reread the questions and your answers, and revise your answers, if necessary. Then

check your answers against the Answer Key and try to understand any mistakes

you made.

• Complete each chapter before doing the corresponding review exercises. Doing these

exercises without referring to the lesson you have just completed is a better way

of preparing for the final examination.


0.13

Self-evaluation Test

The self-evaluation test is a step that prepares you for the final evaluation. You must

complete your study of the course before attempting to do it. Reread your notebook

and the definitions of the key words in the chapters. Make sure you understand how

they relate to the course objectives listed at the beginning of each chapter. Then do

the self-evaluation test without referring to the main body of the Guide or the Answer

Key. Compare your answers with those in the Answer Key and review any areas you

had difficulty with.

Appendices

The appendices contain a review of some concepts you should be familiar with before

beginning the course. The complete list of appendices appears in the table of contents.

Materials

Have all the materials you will need close at hand:

• Learning material: this Guide and a notebook where you will summarize important

concepts relating to the objectives (listed in the introduction of each chapter). You

will also need to use your periodic table and the workbook Experimental Activities

of Chemistry.

• Reference material: a dictionary.

• Miscellaneous material: a calculator, a pencil for writing your answers and notes

in your Guide, a coloured pen for correcting your answers, a highlighter (or a pale-

coloured felt pen) to highlight important ideas, a ruler, an eraser, etc.


0.14

CERTIFICATION

To earn credits for this course, you must obtain at least 60% on the final examination

which will be held in an adult education centre.

The evaluation for the course Chemical Reactions 2: Equilibrium and Oxidation-

reduction is divided into two separate parts.

Part I consists of a two-hour written examination made up of multiple-choice, short-

answer and essay-type questions. This part is worth 70% of your final mark and deals

with the objectives covered in this Guide. You may use a calculator.

Part II is designed exclusively to evaluate the experimental method. It will be held

in the laboratory during a 120-minute session. This part is worth 30% of your final

mark and deals with the course objectives covered in Section C of Experimental Activities

of Chemistry.

INFORMATION FOR DISTANCE EDUCATION STUDENTS

Work Pace

Here are some tips for organizing your work:

• Draw up a study timetable that takes into account your personality and needs, as

well as your family, work and other obligations.

• Try to study a few hours each week. You should break up your study time into several

one- or two-hour sessions.

• Do your best to stick to your study timetable.

Your Tutor

Your tutor is the person who will give you any help you need throughout this course.

He or she will answer your questions and correct and comment on your homework

assignments.

Don’t hesitate to contact your tutor if you are having difficulty with the theory or the

exercises, or if you need some words of encouragement to help you get through this

course. Write down your questions and get in touch with your tutor during his or

her available hours. The letter included with this Guide or that you will receive shortly

tells you when and how to contact your tutor.


0.15

Your tutor will assist you in your work and provide you with the advice, constructive

criticism and support that will help you succeed in this course.

Homework Assignments

In this course, you will have to do three homework assignments: the first after

completing Chapter 2, the second after completing Chapter 4, and the third after

completing Chapter 6. Each homework assignment also contains questions on the

experimental method you studied in Experimental Activities of Chemistry.

These assignments will show your tutor whether you understand the subject matter

and are ready to go on to the next part of the course. If your tutor feels you are not

ready to move on, he or she will indicate this on your homework assignment, providing

comments and suggestions to help you get back on track. It is important that you

read these corrections and comments carefully.

The homework assignments are similar to the examination. Since the exam will be

supervised and you will not be able to use your course notes, the best way to prepare

for it is to do your homework assignments without referring to the Learning Guide

and to take note of your tutor’s corrections so that you can make any necessary

adjustments.

Remember not to send in the next assignment until you have received the corrections

for the previous one.


0.16


In the first course of the Secondary V Chemistry program, students learn about gas-

related phenomena and the energy balance of chemical reactions. The second

course in the program examines energy transfers from a broader perspective and also

deals with reaction rates and the factors affecting those rates.

This course, Chemical Reactions 2: Equilibrium and Oxidation-reduction, deals with

two types of chemical reactions in greater depth. In Chapter 1, students have the chance

to become familiar with the conditions that characterize a system at equilibrium and

to analyze the equilibrium of reversible reactions. Chapter 2 deals with what

happens to a system at equilibrium when it is disrupted by external factors.

After completing this qualitative analysis of equilibrium in the first two chapters, in

the third and fourth chapters students go on to study the quantitative aspects of

chemical equilibrium, which include the calculation of equilibrium constants.

The last two chapters delve into oxidation-reduction reactions, a chemical reaction

that involves a transfer of electrons. In this part of the Guide, students will also learn

how electrochemical cells work. This material represents the final stepping stones

in completing the Secondary V Chemistry program.

As in the first two Guides, a table of contents diagram at the beginning of each chapter

shows you how the chapter fits into the course as a whole. The content of the chapter

you are about to begin is in bold type and in larger characters, whereas the content

of completed chapters is in italics. For example, the table of contents diagram for

Chapter 2 is reproduced on the next page. The section for Chapter 2 is in bold type

and the content of Chapter 1 is in italics and smaller type. You will find this diagram

a very useful tool as you go through the course.

Enjoy your work and good luck!


0.17

1. Chemical EquilibriumLiquid-vapour equilibriumDissolution-crystallization

equilibrium Chemical equilibriumReaching equilibriumStationary state

2. Factors Affecting Chemical EquilibriumTypes of disturbances:

concentrationtemperaturecatalysts

Le Châtelier’s principleApplicationsEquilibrium in nature

6. Cells and ElectrochemistryVoltaic cellsBalancing oxidation-

reduction equationsCells and batteriesCorrosionElectrolysis

CH

EMICAL REACTIONS

2:

EQU

ILIBRIUM

AND OXIDATION-REDUCTIO

N

5. Oxidation-ReductionOxidation and reductionDaniell cellStandard reduction potential

4. The Equilibrium ConstantChemical equilibriumApplicationsDissolution-crystallization equilibrium Chemistry, industry and society

3. Equilibrium in Acidic and Basic SolutionsConstants:

Ka, Kb and Kw

NeutralizationTitration

Examples of acid-base equilibrium


0.18

CHAPTER 1

EQUILIBRIUM

GWB

Terminal Objective 1

To analyze qualitatively the state of equilibrium of a system.

Intermediate Objectives

1.1 To state the three conditions that characterize a system in a state of equilibrium.

1.2 To describe a vapour-liquid equilibrium and an equilibrium involving solutions.

1.3 To verify, through experimentation, whether or not a system is in equilibrium.

1.4 To associate equilibrium with the reversibility of reactions.

1.5 To interpret curves illustrating forward and reverse reactions over time.

Chemical Reactions 2 - Chapter 1: Equilibrium

1. Chemical EquilibriumLiquid-vapour equilibrium

Dissolution-crystallization

equilibrium

Chemical equilibrium

Reaching equilibrium

Stationary state

2. Factors Affecting Chemical Equilibrium

Types of disturbances:

concentration

temperature

catalysts

Le Châtelier’s principle

Applications

Equilibrium in nature

6. Cells and Electrochemistry

Voltaic cells

Balancing oxidation-

reduction equations

Cells and batteries

Corrosion

Electrolysis

CH

EMICAL REACTIONS

2:

EQU

ILIBRIUM

AND OXIDATION-REDUCTI

ON

5. Oxidation-Reduction

Oxidation and reduction

Daniell cell

Standard reduction potential

4. The Equilibrium Constant

Chemical equilibrium

Applications

Dissolution-crystallization equilibrium

Chemistry, industry and society

3. Equilibrium in Acidic and Basic Solutions

Constants:

Ka, Kb and Kw

Neutralization

Titration

Examples of acid-base equilibrium

A tightrope walker balancing on the high wire, a vase on a table and a canoe floating

on water are a few examples of equilibrium. Although some of these situations may

be viewed as precarious, the tightrope artist and the objects are stable and immobile.

They are said to be in static equilibrium. In a chemistry context, the state of

equilibrium is less apparent and, although the term “equilibrium” has the same general

meaning, several key differences have to be considered. That is because we are dealing

with dynamic equilibrium and reversibility, and reaction rates are involved.

In this chapter, we will study a variety of situations in order to shed light on the state

of equilibrium. First we will look at evaporation and dissolution to determine under

what conditions these processes can reach equilibrium. After that, we will look at

the characteristics of a reversible chemical reaction at equilibrium.

1.1 EVAPORATION AND DISSOLUTION

The word “equilibrium” makes one think of stability. In chemistry, a system at

equilibrium1 is stable in that no apparent changes are occurring. Periodic checks should

not turn up any signs of change. To help you better understand this concept, let’s

consider the phenomenon of evaporation. Under certain conditions, equilibrium may

be reached between a liquid and its vapour form.

LIQUID-VAPOUR EQUILIBRIUM

Say there are two glasses of water on a counter in a room where the temperature is

kept constant. One of the two glasses has an airtight lid. Each of the glasses and its

contents represents a system. If no one touches the glasses and we check them twice

a day, what do you think will happen?

Figure 1.1 - Evaporation

Water evaporates gradually from the first glass, whereas the liquid in the glass with a lid remains at the same level.


1.3

1. The word “system” denotes the equipment, setup, substances and all other elements involved in an experiment.

From experience, you probably know that the water level in the air-tight sealed glass

will not change as time passes, whereas the water in the other glass will evaporate

slowly until it is all gone. You are right!

Exercise 1.1

In your view, can either of the two glasses described above be considered a systemat equilibrium? Explain.

Let’s now look at what is happening at the molecular level. The explanation has to

do with the kinetic model of matter that was discussed in earlier courses. In a glass

of water, the molecules are close together and are drawn to one another with a certain

amount of force. However, the force of attraction is insufficient to keep the molecules

in a fixed position like in a solid. The molecules of liquid move around and bump

into one another; some of them acquire enough speed (kinetic energy) to leave the

surface of the liquid. These molecules enter the gaseous phase, and we say that some

of the liquid has evaporated.

In the open system, the water molecules leave the liquid gradually, and the level of

the liquid in the glass decreases. If the system is exposed to the air for a few days,

all of the water will evaporate.

Evaporation also takes place in the glass with an airtight lid; however, the vapour is

trapped in the container. The drops of water that collect on the sides of the container

show that there is condensation, which is a sign of evaporation. However, very little

water evaporates and the level in the glass stays the same. We can conclude from

this that, in a closed system, the process of evaporation stabilizes quickly.

In an open system, evaporation can continue until all of the water has disappeared.

By contrast, in a closed system at room temperature, evaporation takes place, but it

appears to stop after a while. This is only the ways things seem, however.


1.4

?

Figure 1.2 - Open system and closed system

a) In an open system, a state of equilibrium is not feasible. The system is constantly changing because molecules of vapour are escaping to the air.

b) In a closed system, evaporation is the dominant process initially. After a certain time,molecules in the gaseous phase return to the liquid phase. The arrows indicate

the movement of the molecules between the two phases.

c) After a while, the system shows no apparent signs of change. It has reached a state ofequilibrium, since an equal number of molecules are moving in both directions between the liquid phase and the gaseous phase. The arrows of equal length illustrate this phenomenon.

d) Evaporation and condensation are two opposite processes. The dotted line represents the evaporation rate and the solid line the condensation rate.

Equilibrium is reached when the rates are equal.

Let’s take a closer look at the graph in Figure 1.2d). In the beginning (t = 0), there is

only evaporation. The molecules of liquid that have enough energy to free themselves

from the liquid’s attractive forces convert to the gaseous phase. The rate of

condensation is zero because the air contains almost no water vapour at this time.

At t = 1, some molecules in the gaseous phase return to the liquid phase as a result

of their random movements. The process whereby gaseous molecules return to the

liquid phase is called condensation. At this point in time, evaporation is still

predominant, because the evaporation rate exceeds the condensation rate.

Equilibrium: liquid a vapour

Rat

e

Time0 1 2 3 4 5 6

Evaporation rate

Condensation rate


1.5

a) b) c)

d)

After a while, that is, at t = 4, the two curves come together. The rate of evaporation

is equal to the rate of condensation at this point. There are as many molecules escaping

from the liquid phase as there are molecules returning to it. The level of the liquid

remains constant. We say that the two phases are in equilibrium. This liquid-vapourequilibrium is represented by the following equation:

liquid a vapourH2O(l) a H2O(g)

The use of two arrows pointing in opposite directions in the equation comes from a

combination of the following two equations:

H2O(l) → H2O(g) evaporationH2O(l) ← H2O(g) condensation

In the equilibrium equation, the two arrows point in opposite directions to show the

continual back-and-forth movement of the molecules escaping from the liquid

phase (→) to the gaseous phase and those going from the gaseous phase to the liquid

phase (←). The two arrows indicate that the reaction is reversible, that is, the two

processes are taking place in opposite directions. At equilibrium, the processes of

evaporation and condensation occur at the same speed, hence the use of opposite

arrows of equal length. The right arrow (→) indicates the forward reaction and the

left arrow (←) the reverse reaction (condensation).

The double arrow indicates that the system is in dynamic equilibrium. The word

“dynamic” means that even if no change is evident, there is movement, in this case,

the back-and-forth movement of the molecules between two phases. In order for

dynamic liquid-vapour equilibrium to occur, the system must be closed so that no

matter can escape from it. In other words, there must be no exchange of matter between

the system and the surroundings.

Exercise 1.2

a) Name an important characteristic of a closed system.

b) What distinguishes an open system from a closed system?


1.6

?

c) What do the two opposing arrows mean in the equation H2O(l) a H2O(g)?

d) A system at equilibrium shows no apparent change, and looks as though nothing

were happening in it. However, we say that the system H2O(l) a H2O(g) is at dynamic

equilibrium. What does this mean?

e) In the equation H2O(l) a H2O(g), why are the arrows of equal length?

f) State an essential pre-condition for dynamic equilibrium to exist in a system.

In summary, three macroscopic2 conditions characterize a liquid-vapour equilibrium

system.

• The level of the liquid remains constant: there is no apparent change.

• The liquid and vapour are both present: both states must be present in order forequilibrium to exist between the liquid and its vapour form.

• The system is closed: there must be no exchange of matter or energy between thesystem and the surroundings.

At the molecular level, liquid-vapour equilibrium is a reversible process that proceeds

at the same rate in both directions. Now let’s look at another type of system, that is,

an aqueous solution obtained by dissolving a solid in water. What form will

equilibrium take in this case? How is this situation similar to liquid-vapour

equilibrium?


1.7

2. “Macroscopic” means that which can be seen with the naked eye.

DISSOLUTION-CRYSTALLIZATION EQUILIBRIUM3

Consider two beakers half-full of water at room temperature. We add a handful of

table salt to the first beaker and only a teaspoonful of salt to the second. We then

mix the two solutions for several minutes. What do you think will happen?

You will probably say, based on experience, that the salt will be completely dissolved

in the beaker to which we added a teaspoon of salt whereas there will still be some

undissolved salt in the other beaker. In the second case, what tells us we are dealing

with an equilibrium system? Certain clues allow us to conclude this, for example,

the amount of salt at the bottom of the beaker is constant, that is, no change is evident,

and both the solid and liquid phases are present. Hence equilibrium can be reached.

But can the system be considered closed? Does an exchange take place between the

system and the surroundings?

The gaseous phase does not play an important role in our example. The system consists

of a beaker and the liquid and solid it contains. There is no lid on the beaker. If we

observe the system for a few hours only, we can assume that there will be no change

in the level of the solution, that is, evaporation is negligible and no exchange of matter

occurs between the system and the surroundings. There is no exchange of energy

between the system and the surroundings either, since the water was initially at room

temperature. Given these conditions, we can consider the system to be closed when

analyzing the equilibrium between the undissolved and dissolved solute.

Now, let’s see what is happening at the molecular level. When we add a handful of

salt (NaCl) to a beaker of water at room temperature and we stir the solution, after

a while we can see that there is still some salt at the bottom of the beaker. The solutionis saturated, meaning that it contains a maximum amount of dissolved salt. We can

see that the amount of salt at the bottom of the beaker remains constant. This is an

indication that the dissolved ions and the undissolved salt crystals at the bottom of

the beaker have reached dynamic equilibrium (Figure 1.3). The following equation

represents this situation:

NaCl(s) a Na+(aq) + Cl–

(aq)

The double arrow tells us that the process is reversible, that the system has reached

equilibrium and that the rates of the opposing processes are equal. Let’s now examine

in detail the characteristics of a system in dissolution-crystallization equilibrium.


1.8

3. In some textbooks, the equivalent expression “equilibrium involving dissolution of solids” is used.

When we add salt to water, dissolution (→) begins as the NaCl dissociates into Na+(aq)

and Cl–(aq) ions. After a while, the continually moving dissolved ions start to accumulate

in the solution, and some of them become reattached to the solid NaCl crystals, through

the process of crystallization (←). When the rate of crystallization (←) equals the

rate of rate of dissolution (→), the number of ions in solution no longer increases

and we say that the solution is saturated.

Figure 1.3 - Solution at equilibrium

a) A large quantity of salt is added to a container of water.Initially, the Na+ and Cl– ions leave the crystals and go into solution. A single arrow is shown,

since only the process of dissolution is taking place. Crystallization has not yet started.

b) After a while, dissolution continues but some ions become reattached to the crystals as theymove about in the solution. At this point, crystallization is slower than dissolution.

c) At equilibrium, the solution is saturated and the rate of crystallization equals the rate of dissolution. The arrows are of equal length. The amount of

dissolved solid is constant, as is the amount of ions in solution.

The up arrow represents ions dissociating. The down arrow represents the ions returning to the crystal.

a) b) c)


1.9

Exercise 1.3

a) Based on what you have learned about evaporation, list three macroscopicconditions indicating that a system has reached a dissolution-crystallizationequilibrium.

b) We pour some sugar into a glass of water. We stir the solution, and, after a fewseconds, the sugar disappears completely. Is the sugar water at dissolution-crystallization equilibrium? Explain your answer.

c) We add some more sugar and, after stirring the solution well, we note that an excessof undissolved sugar settles at the bottom of the glass. The system is at equilibrium.Describe the equilibrium reached by the sugar by considering what happens atthe molecular level.

d) In the equation sugar(s) a sugar(aq), what is the meaning of the two opposing arrowsand why are they the same length?


1.10

?

Exercise 1.4

In the example given in the text, some salt (NaCl) is dissolved in a glass of water.

a) A system can be considered to be at equilibrium only if it is closed. What conditionmust apply in order for the glass of saltwater to be considered a closed system?

b) Macroscopically speaking, what indicates whether or not the system is atequilibrium?

Take a Little Shot … and Keep Your Balance4

When we drink alcoholic beverages, the stomach wall absorbs about 20% of the alcohol and 80% reaches

the wall of the small intestine. The alcohol enters the tiny blood vessels in these walls and dissolves in the

blood; it then circulates through the veins of the body until it reaches the lungs, where gas exchange occurs.

The lungs are made up of a multitude of small air sacs called “alveoli,” which have very thin walls containing

numerous capillaries. The alveoli contain inhaled air, and it is through their walls that gas diffusion occurs,

in two directions: oxygen from the air enters the blood, and CO2 leaves the blood and is expelled to the air.


1.11

4. A light bulb indicates additional information: this information is not part of the course and will not be coveredin the final examination.

?

Respiratory system

When dissolved alcohol circulates in the blood, a part of this alcohol becomes gaseous in the lungs, like

CO2. With each inhalation, equilibrium is established very rapidly between the blood and the air so that, upon

exhalation, the concentration of alcohol in the exhaled breath is representative of the concentration of alcohol

in the blood. The higher the blood alcohol content, the higher the concentration of alcohol in the exhaled

breath. However, the concentrations differ. There is about 2 000 times more alcohol in the blood than in the

exhaled breath. The Breathalyzer test is based on this difference. This device used by police officers precisely

measures the concentration of alcohol in exhaled breath and computes the amount of alcohol in the driver’s

blood. If the result exceeds 0.08%, that is, 0.08 g of alcohol per 100 mL of blood, the driver’s license will

be suspended on the spot for a period of 15 days for a first offense, and 30 days for a repeat offense.

Equilibrium in the alveoli

Alveolus

Capillary atthe outset

Capillary inequilibrium

Ethanol Ethanol

Alveolus


1.12

We have outlined the conditions necessary for a system to be at either liquid-vapour

equilibrium or dissolution-crystallization equilibrium. In the following experimental

activity, you will study two new situations that will allow you to learn more about

the characteristics of an equilibrium system.

Experimental Activity 1: Equilibrium Systems

In this activity, you will study various systems and determine

under what conditions they are at equilibrium.

In the first part, you will examine the reaction produced by

dissolving an effervescent salt in water, which, once dissolved,

produces carbon dioxide (CO2) gas. You will be required to

compare the properties of the reaction in an open system

versus a closed system.

In the second part, you will have a chance to verify that the following reaction is

reversible:

CaCl2(aq) + Na2SO4(aq) a 2 NaCl(aq) + CaSO4(s)

Allow about 30 minutes for all of the steps in the experiment. All of the information

you will need to carry out this activity is given in Section C of the workbook

Experimental Activities of Chemistry. Enjoy your work!

The experimental activity you have just completed allowed you to observe that, in

the closed system, part of the CO2 was dissolved in water and the other part was trapped

between the cap and the surface of the water. The evidence was that, when you opened

the bottle, a large quantity of bubbles spontaneously appeared throughout the solution

and then escaped from the bottle. From this observation, it can be deduced that the

CO2 dissolved in the water and the CO2 above the water’s surface were in equilibrium

before the bottle was opened. This situation is represented by the following equation.

CO2(aq) a CO2(g)

In the second part of the activity, you were able to note that the reaction is reversible:

CaCl2(aq) + Na2SO4(aq) a 2 NaCl(aq) + CaSO4(s)


1.13

The system was in a state of equilibrium and you were able to determine that both

products and reactants were present, that there was no exchange of matter with the

surroundings and that the system showed no apparent signs of change.

Dynamic equilibrium can therefore exist between a chemical reaction and the reverse

reaction. The process is similar to that involved in liquid-vapour equilibrium and

dissolution-crystallization equilibrium, described earlier in the chapter. In the

following section, we will look at the equilibrium of reversible reactions, including

the state of chemical equilibrium and the processes involved in reaching equilibrium.

1.2 CHEMICAL REACTIONS

Very often, when we write the equation for a chemical reaction, we assume that all

the reactants have been converted into products, and that by the end of the reaction

at least one of the reactants has been completely used up. For example, the equation

for the following reaction:

3 H2(g) + N2(g) → 2 NH3(g) + 92.2 kJ

tells us that precisely three moles of H2 react with one mole of N2 to form two moles

of NH3, and that 92.2 kJ of heat is released when two moles of NH3 are formed. We

assume that the reaction is complete and that there is no reactant left at the end.

Irreversible reactions generally keep proceeding until one of the reactants has been

completely depleted. There are, however, many reversible reactions too. The following

analogy will give you a better understanding of reversible chemical reactions.

Suppose that a leaky boat is one-quarter full of water and that we are bailing out

water as quickly as it is entering the boat. Equilibrium has been established and the

water level remains stable, since the rate at which the water is entering the boat equals

the rate at which it is being removed. Likewise, if the boat is half-full and we remove

the same amount of water as is entering the boat, the state of equilibrium will continue.

Nonetheless, the boat is still half-full of water. If the boat is three-quarters full and

we are again able to remove water as fast as water is entering the boat, the state of

equilibrium will continue, but the boat will still be three-quarters full of water.


1.14

Figure 1.4 - Precarious situation or equilibrium?

The boat will stay afloat as long as the amount of water entering it is counterbalanced by the water being bailed out of it. The water level in the boat depends on how long it took

for equilibrium to be reached. Fortunately for our “balanced” friend, the water level is not rising!

We can apply this analogy to a chemical reaction at equilibrium. At equilibrium, the

rate of the opposing reactions is the same but the quantities of reactants and products

are not necessarily equal. As in our analogy, at equilibrium we could have a boat that

is one-quarter full, one-third full, half full, three-quarters full, and so on.

To conclude this analogy, let’s say that the sinking of the boat would correspond to

a chemical reaction in which equilibrium cannot be reached. This is the case for

irreversible chemical reactions in which the reactants are converted entirely into

products. These reactions stop by themselves when there are no reactants left. For

example, when we burn propane gas in the open air, the reaction produces carbon

dioxide (CO2) and water (H2O). The reaction is forward and irreversible. The single

arrow in the equation denotes this.

C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)

Let’s now see how a reversible chemical reaction proceeds.

EQUILIBRIUM OF THE REACTION N2O4(g)aa 2 NO2(g)

At a macroscopic level, the chemical reaction N2O4(g) a 2 NO2(g) is an interesting one.

N2O4 is colourless, whereas NO2 is reddish brown. The gradual change in colour that

can be observed in the reactor indicates that a reaction is taking place. In chemistry,

it is relatively rare to be able to observe the evolution of a chemical reaction to

equilibrium based on a change in colour.


1.15

GWB

To produce this reaction, suppose that we add some dinitrogen tetraoxide (N2O4), a

colourless gas, to a container in which we have first created a vacuum. We heat the

container in a bath of boiling water and keep the temperature at 100°C. The average

kinetic energy of the molecules in the system will be constant at that point. Figure 1.5

shows what happens while the N2O4 is being heated and equilibrium is being

established.

Figure 1.5 - Reaching equilibrium in the reaction N2O4(g) aa 2 NO2(g)

At t = 0, the container filled with the colourless gas N2O4 is immersed in boiling water (100°C). A reddish brown colour gradually appears, signalling the presence of NO2. The intensity of the colour increases for a while (t = 1, t = 2) and then stabilizes (t = 3, t = 4). At this point,

the colour of the mixture is a dark reddish brown, but not as dark as pure NO2. Equilibrium is reached when there is no further change in the colour. Note that the

forward reaction (↑) is fast at first and then slows down (length of the vertical arrow).

As soon as the container is immersed in hot water (t = 0), the N2O4 starts to decompose

to NO2 (↑). A reddish brown colour appears; it is pale at first, and gradually becomes

more intense. This colour indicates that nitrogen dioxide (NO2) is present in the system,

and we can say that the forward reaction N2O4 → 2 NO2 is occurring. As the reaction

proceeds, the reddish brown colour becomes more intense but it stabilizes after a

while (t = 4). There is no further change in the colour then and the system is at

equilibrium. At that point, we can verify that the two gases, N2O4 and NO2, are present

by analyzing a gas sample from the system.

t = 0 t = 1 t = 2 t = 3 t = 4Time

Forward reaction: ↑; reverse reaction: ↓; N2O4 : ; NO2 :


1.16

Forward reactionreverse reaction

Now let’s consider the opposite reaction (Figure 1.6). We immerse a one-L container

of pure NO2 in an ice bath (0°C). At first, the gas is a very dark reddish brown. We

can see the colour gradually decrease in intensity: it turns a lighter colour. This change

shows that the reaction 2 NO2 → N2O4 is proceeding. After a while, there is no further

change in the colour of the gas. We can prove that at this point the two gases N2O4

and NO2 are present by analyzing a gas sample from the container.

Figure 1.6 - Equilibrium 2 NO2(g)aa N2O4(g)

At t= 0, the container is immersed in an ice-water bath (0°C). As time goes by, the reddish brown colour turns lighter, indicating a decrease in the concentration of NO2 (t = 1, t = 2).

Equilibrium is reached when the intensity of the colour stabilizes (t = 3, t = 4). Note that the forward reaction (↑) is faster in the beginning (length of the vertical arrow).

In both cases, we can see that the system has reached equilibrium. When no further

change is observed in the colour of the gaseous mixture, we can conclude that the

reaction N2O4 a 2 NO2 is at equilibrium. Colour is a macroscopic characteristic of

the system.

In a reversible reaction, not all the reactants are converted into products. Dynamic

equilibrium is reached between the two opposing reactions, which means that reactants

and products can both be present at the same time. The preceding description clearly

shows the reversible nature of the reaction. For example, when we heat N2O4, NO2

is formed and, conversely, when NO2 is cooled, N2O4 is formed. In Figure 1.5, the

colourless gas N2O4 was not completely converted since the colour of pure NO2 is

much darker than that observed at the end. Likewise, in Figure 1.6 the reddish brown

NO2 was not completely converted into colourless N2O4, since the gaseous mixture

was still slightly coloured at the end.

t = 0 t = 1 t = 2 t = 3 t = 4Time

Forward reaction: : ↑; reverse reaction: : ↓; N2O4 : ; NO2 :


1.17

Forward

reaction

reverse

reaction

When the system is at equilibrium, the equation for the reaction is:

N2O4(g) a 2 NO2(g)

The above equation sums up the two equations below:

N2O4(g) → 2 NO2(g) forward reactionN2O4(g) ← 2 NO2(g) reverse reaction

As before, the double arrow shows that the system is in dynamic equilibrium and

that the rate of decomposition of the N2O4 into NO2 equals the rate at which N2O4 is

formed from NO2. At equilibrium, the two gases are present in constant but not

necessarily equal quantities. For example, an equilibrium system can contain

0.4 mole of N2O4 and 1.2 moles of NO2.

Exercise 1.5

What signs indicate that the reaction N2O4(g) a 2 NO2 has reached equilibrium?

Exercise 1.6

Decide whether the following statements are true or false. Modify the false ones in

order to make them true.

a) The symbol a indicates that at equilibrium there are as many moles of NO2 asthere are moles of N2O4.

b) At equilibrium, the system shows no apparent sign of change.

c) In any chemical equilibrium, both the products and the reactants are present.


1.18

?

?

Let’s see what happens when we inject 1 mole of N2O4 into a 1-L container. The

temperature is raised to and held at 120°C. When equilibrium is reached, the colour

of the gaseous mixture is stable and there are 0.4 mole of N2O4 and 1.2 moles of NO2

in the container. The total number of moles is greater than at the beginning. How is

this possible? Let’s examine the equation for the reaction and the process by which

equilibrium is established.

N2O4(g) a 2 NO2(g)

Initially 1.0 mol 0,0 molConversion of system – 0.6 mol + 1.2 molAt equilibrium 0.4 mol 1.2 mol

The coefficients in the equation show that 1 molecule of N2O4 decomposes to produce

2 molecules of NO2. We know that initially we have 1 mole of N2O4; however, only

0.4 mole is left at equilibrium. This means that 0.6 mole (1.0 – 0.4 = 0.6) of N2O4 has

decomposed to produce 1.2 moles of NO2, as shown in the small box. The second

line of text beneath the equation represents this conversion: on one side, we subtract

0.6 mole of decomposed N2O4 and on the other, we add 1.2 moles of NO2. The last

line beneath the equation gives the quantities at equilibrium; the quantities are not

equal but they remain constant.

Remember that the process of equilibrium is always dynamic. The two reactions occur

simultaneously and at the same speed. Thus, each time one molecule of N2O4

decomposes, another is formed by the fusion of two NO2 molecules.

Exercise 1.7

A technician adds 138 g of NO2 to a container. The gas is heated to a given

temperature and held there. The colour of the gas turns lighter, indicating that N2O4

has been formed. After a while, the colour stabilizes and equilibrium is reached. We

take a sample and calculate that 1.20 moles of NO2 are left in the system.

a) Determine the number of moles of NO2 that were added to the system.


1.19

Conversion:

N2O4(g) → 2 NO2(g)

1,0 mol 2,0 mol

0.6 mol → ?,0 mol

0.6 mol →

Æ 1.2 mol

?

b) How many moles of NO2 were converted into N2O4 in the time it took the systemto reach equilibrium?

c) Complete the lines beneath the equation to determine the quantities of each ofthe gases present at equilibrium.

N2O4(g) a 2 NO2(g)

Initially ________ ________

Conversion of system ________ ________

At equilibrium ________ ________

In summary, a reversible chemical reaction produces a system that is in dynamic

equilibrium. Macroscopically speaking, dynamic equilibrium has the following

characteristics.

• The system shows no apparent signs of change (the colour and other characteristics

are stable).

• All the reactants and the products are present.

• The system is closed; no matter is exchanged between the system and the

surroundings, and the temperature is constant (no transfer of energy).

When the reactants come together, it takes a certain amount of time for equilibrium

to be established. We have presented an overview of the process that precedes

equilibrium. We will now study this process in more depth, by seeing how the rates

of the opposing reactions change between the time the reactants are brought

together and the time equilibrium is established. Remember that the rates of the

opposing reactions are equal at equilibrium.

REACHING EQUILIBRIUM

In an equilibrium system, the rates of the forward and reverse reactions are the same.

However, before equilibrium is reached, one of the two reactions is faster than the

other. Before we examine how reaction rates change over time, let’s look at how the

rate of a reaction is expressed.


1.20

Rate and Concentration

The previous chemistry course5 dealt with the rate of chemical reactions, and more

specifically, the factors that influence the reaction rate. Remember that the rate of

a reaction corresponds to the speed with which the reactants disappear and is generally

expressed as a change in concentration (∆C) over time (∆t), with concentration

expressed in moles per litre (mol/L). We have:

∆Cv = ––––

∆t

For example, a rate of 0.4 mole per litre per second is written as v = 0.4 mol/L•s.

According to the collision theory, a chemical reaction can occur only when the reactants

collide with sufficient kinetic energy (speed). A small number of these collisions give

rise to the reaction and are called effective collisions. The greater the concentration

of the reactants, the greater the number of effective collisions and the faster the reaction.

Chemists have shown that the speed of a reaction is proportional to the concentration

of reactants, raised to the power corresponding to their respective coefficients in the

balanced equation. For example, the rate of formation of PCl5(g) depends on the

concentration of the reactants PCl3(g) and Cl2(g).

PCl3(g) + Cl2(g) a PCl5(g)

The rate of formation of PCl5(g) is expressed mathematically as follows:

vPCl5= k [PCl3] [Cl2]

where k is the constant of proportionality for the reaction. The expressions [PCl3]

and [Cl2] are read “concentration of PCl3” and “concentration of Cl2” and are

expressed in moles per litre (mol/L).

Since the reaction is reversible, we can express the rate of the reverse reaction in the

same way, that is, the rate at which the products disappear to form the reactants.

The rate of decomposition of PCl5 is therefore written as:

vPCl3 + Cl2= k [PCl5]

where k is the constant of proportionality and [PCl5] is the concentration of PCl5.


1.21

5. Blondin, André, Chemical Reactions 1: Energy and Chemical Dynamics (Chemistry, Secondary V), Learning Guideproduced by SOFAD

Below are other examples for writing the rate at which the products of a reaction

are formed. In all cases, the expression represents the rate of the forward reaction.

3 H2(g) + N2(g) a 2 NH3(g) vNH3= k [H2]

3 [N2]

2 H2(g) + O2(g) a 2 H2O(g) vH2O = k [H2]2 [O2]

N2O4(g) a 2 NO2(g) vNO2= k [N2O4]

For the first reaction, the concentration of H2 is assigned the exponent 3, since the

coefficient of H2 is 3 in the balanced equation. Following the same rule, we have [H2]2

in the expression for the rate of the second reaction. Thus, the rate of a reaction is

proportional to the concentration of the reactants, raised to the power corresponding

to their respective coefficients in the balanced equation.

All of the above reactions are reversible. The rate of the reverse reactions for the above

equations is expressed according to the same rule. This rate is proportional to the

concentrations, each raised to the power corresponding to its coefficient in the equation.

Therefore:

vH2 + N2= k [NH3]

2

vH2 + O2= k [H2O]2

vN2O4= k [NO2]

2

To simplify matters, we generally write vf for the rate of the forward reaction—the

reaction which is read from left to right in the equation—and vr for the rate of the

reverse reaction—the reaction that is read from right to left in the equation.

Therefore:

3 H2(g) + N2(g) a 2 NH3(g) vf = k [H2]3 [N2]

vr = k [NH3]2

In the next chapter, we will use this method of expressing the rate of reaction to

determine the concentration of the substances in equilibrium systems. It is important

to remember that, at equilibrium, the rate at which the products (vf) are formed equals

the rate at which the reactants (vr) are formed.


1.22

Exercise 1.8

For each of the following reactions, write the equation representing the rate of

formation of the products (forward reaction) and the equation representing the rate

of formation of the reactants (reverse reaction).

Reaction Rate of forward reaction Rate of reverse reaction

CO(g) + H2O(g) a CO2(g) + H2(g)

H2(g) + Cl2(g) a 2 HCl(g)

2 NO(g) + O2(g) a 2 NO2(g)

4 NH3(g) + 5 O2(g) a 4 NO(g) + 6 H2O(g)

Note that the value of the constant of proportionality varies from one equation to

the next, and according to the temperature. The symbol k stands for “constant.”

Change in Rates of Reaction

Let’s go back to the system consisting of N2O4 and NO2. The rate of the forward reaction

(vf) is a function of the N2O4 concentration, whereas the rate of the reverse reaction (vr)

is expressed using the NO2 concentration. We therefore have:

N2O4(g) a 2 NO2(g) vf = k [N2O4] and vr = k [NO2]2

The graph in Figure 1.7 shows the simultaneous change in the rates of the forward

and reverse reactions of the system when equilibrium is being established.


1.23

?

Figure 1.7 - N2O4(g)aa 2 NO2(g) system

a) Conversion of N2O4 into NO2 until equilibrium is reached

b) Change in rates of reaction in N2O4 a 2 NO2

vf: rate of forward reaction; vr: rate of reverse reaction

The forward reaction proceeds rapidly at first, since the concentration of N2O4 is high; the reversereaction is slow due to the low concentration of NO2. The rate of the forward reaction decreases

gradually, whereas that of the reverse reaction speeds up. At equilibrium, the rate of the forward reaction equals that of the reverse reaction.

Time0 1 2 3 4 5 6 7 8

Rat

e

N2O4 → 2 NO2 (forward reaction)

Start of equilibrium

N2O4 ← 2 NO2 (reverse reaction)

N2O4 a 2 NO2 (equilibrium)

vf0

vf1

vf2

vr2

vr1

vr0

t = 0 t = 1 t = 2 t = 3 t = 4Time

Forward reaction: ↑; reverse reaction: ↓; N2O4 : ; NO2 :


1.24

Let’s look more closely at Figure 1.7. At t = 0, when the container is immersed in hot

water, it contains only molecules of the colourless gas N2O4. The concentration of

N2O4 is high and, consequently, the molecules collide frequently. The length of the

arrow (↑) in the first container indicates that the forward reaction is predominant.

The graph shows that the rate of the forward reaction is very high (vf0).

At t = 1, the rate of the reverse reaction is considerable, but still slower than the rate

of the forward reaction. The rate of decomposition of the N2O4 (vf1) is even faster than

the rate at which it is being formed (vr1).

At t = 2, the rate of the forward reaction has decreased considerably (vf2) because the

concentration of N2O4 is much lower than it was at the beginning. By contrast, the rate

of the reverse reaction has increased (vr2) because the concentration of NO2 is higher.

At t = 3 and t = 4, equilibrium is reached. The meeting point of the two curves represents

the start of the state of equilibrium. From this point on, the rate of the forward reaction

equals that of the reverse reaction. This portion of the curve is horizontal and represents

a constant rate.

Exercise 1.9

Decide whether the following statements are true or false. Modify the false ones inorder to make them true.

a) At equilibrium, the rate of the forward reaction equals that of the reverse reaction.

b) At equilibrium, the concentration of the reactants always equals that of the products.

Exercise 1.10

The following questions refer to Figure 1.7.

a) What do the two curves on this graph represent?


1.25

?

?

b) The rate of disappearance of the N2O4 is fastest at the beginning of the reaction.Explain why.

c) The rate of regeneration of N2O4 is low at the beginning of the reaction. Explainwhy.

d) From t = 3 on, the curve for the forward reaction and that for the reverse reactionare superimposed. What does this mean?

Change in Concentrations

The graph in Figure 1.7 shows that the rate of the forward reaction decreases over

time and the rate of the reverse reaction increases. At equilibrium, the two rates are

equal. This information, however, does not tell us up to what point the reactants have

combined to form the products.

In general, chemists and chemical manufacturers are interested in the yield from a

reaction, that is, the quantity of products that it can provide. It is therefore important

to know the concentrations that can be obtained from a reaction. For example, a high

concentration of reactants indicates that production conditions are mediocre. As in

the previous course, we will study the change in concentration over time. For this,

we need only take samples periodically and analyze them to determine the

concentration of substances. Based on the data collected, we can draw the graphs of

the concentration as a function of time. In practice, the curves can be useful to chemists

in establishing optimal production conditions.

The graphs in the following three figures show the change in the concentrations of

the reactants and products for three different chemical reactions. In more concrete


1.26

terms, we might speak of the speed with which the reactants disappear and the speed

with which the products appear. The first graph (Figure 1.8) shows the change in the

concentration of the substances in a one-L container to which 1 mole of H2 and 1 mole

of I2 have been added and which is held at a temperature of 440°C. The forward reaction

is:

H2 + I2 → 2 HI vf = k [H2] [I2]

At the beginning of the reaction, the concentration of the reactants decreases

rapidly, since the rate of the reaction is high and the collisions between the molecules

are frequent. As the reaction proceeds, the concentration of the reactants decreases

and eventually stabilizes at 0.22 mol/L. On the product side, [HI], which was zero at

the start, gradually increases and eventually stabilizes at 1.6 mol/L. Then, all the

concentrations remain stable, indicating that the system has reached equilibrium and

the rates of the two reactions are the same. Note that the final concentration of HI

is higher than that of H2 and I2. When this happens, chemists say that the products

are favoured.

Figure 1.8 - Change in concentrations in the reaction H2(g) + I2(g) a 2 HI(g) at 440°C

Equilibrium is reached when the concentrations have stabilized. Under the conditions in whichthe reaction was carried out, [H2] = [I2] = 0.22 mol/L and [HI] = 1.6 mol/L.

Let’s look at another example. We add 1 mole of CO2 and 1 mole of H2 to a one-L

container and the temperature is kept at 550°C. The graph in Figure 1.9 shows the

change in the concentration of the reactants and products. At equilibrium,

[CO2] = [H2] = 0.73 mol/L, and [CO] = [H2O] = 0.27 mol/L. Note that there is a larger

quantity of reactants than products. Hence, the reactants are favoured.

Con

cent

ration

(m

ol/L

)

Time

[H2] and [I2]

Reactants

[HI]

Product

2.0

1.61.5

1.0

0.5

0.22

0


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Figure 1.9 - Change in concentrations in the reaction CO2(g) + H2(g) a CO(g) + H2O(g)

at 550°C

Equilibrium is reached when the concentrations of the reactants and the products have stabilized.At this point, the curves in the graph form plateaus.

Finally, the graph in Figure 1.10 shows the change in the concentration of the

substances in a one-L container to which 0.4 mole of SO2 and 0.3 mole of O2 were

previously added, and which was held at 450°C. Note the difference in the

concentrations of the substances at equilibrium. The products are favoured.

Figure 1.10 - Change in concentrations in the reaction 2 SO2(g) + O2(g) a 2 SO3(g)

The dotted vertical line indicates the point at which the system reaches equilibrium. At this point, the concentrations of SO2, O2 and SO3 are constant.

Con

cent

ration

(m

ol/L

)

Time

[SO2] Reactant

[SO3] Product

[O2] Reactant

0.400

0.300

0.200

0.100

0

Con

cent

ration

(m

ol/L

)

Time

[CO] and [H2O]

Products

[CO2] and [H2]

Reactants

1.00

0.73

0.50

0.27

0


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The three graphs we have just looked at show the behaviour of systems involving

reversible chemical reactions. We have seen that these systems evolve toward a dynamic

equilibrium in which the two opposing reactions continue to take place. Conversely,

an irreversible reaction moves in one direction only and the process stops when one

of the reactants has been used up. The graph in Figure 1.11 can help you better

understand this difference; it shows the change in concentrations over time for an

irreversible chemical reaction. You are already familiar with this type of graph from

the previous course.

Figure 1.11 - Change in an irreversible chemical reaction

A irreversible reaction can proceed as long as there are reactants. The process stops completely when the concentration of at least one of the reactants is zero.

Exercise 1.11

a) Explain the curves in the graph in Figure 1.8.


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b) What distinguishes the chemical reaction represented by the graph in Figure 1.8from that represented by the graph in Figure 1.11? Explain your answer.

A reversible chemical reaction is initially characterized by rapid conversion of the

reactants into products, and by slow conversion of the products into reactants. As

the reaction proceeds, however, the rate of the forward reaction decreases and that

of the reverse reaction increases. When the reaction has reached a state of equilibrium,

the two rates are equal and the concentration of all the substances remains constant.

By contrast, in an irreversible chemical reaction, called a complete reaction, there

is no reverse reaction. The forward reaction proceeds until the reactants are used

up, unless it is interrupted sooner by an external factor.

Let’s now look at an interesting case: a stable system where one of the conditions

necessary for equilibrium is not met. This is referred to as a stationary state. But

watch out! Appearances can be deceiving!

THE STATIONARY STATE: A SPECIAL CASE

Macroscopically speaking, a system at equilibrium has the following characteristics:

there is no exchange of matter or energy between the system and the surroundings

(closed), both reactants and products are present and the concentration of all the

substances remains constant, so there is no sign of change in the system.

Let’s study the flame of a Bunsen burner (Figure 1.12) or of a gas stove. We see a

steady flame that shows no apparent signs of change. Its colour, height and the quantity

of heat it releases are constant. Are the Bunsen burner and gas stove examples of

equilibrium systems? No, because they are not closed systems and they continuously

exchange matter and energy with the surroundings. They require a constant supply

of fuel and oxygen and the products of combustion escape to the atmosphere. In

addition, the reaction is not reversible.


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A steady flame is a stationary system. It shows no apparent change, and reactants

and products are both present; however, it is not an equilibrium system, since there

is a continuous exchange of matter and energy with the surroundings.

Figure 1.12 - Stationary system

A Bunsen burner constitutes a stationary system. For combustion to occur, the burner mustreceive a continuous supply of propane and oxygen from the outside; the reaction is not reversible.

Exercise 1.12

State whether the following situations constitute an open or a closed system.

Explain your answer. (Assume that the temperature is constant or that there is no

exchange of energy between the system and the surroundings.)

a) A wood fire in a slow-combustion stove.

C3H8 (propane gas)

CO2 + H2O

O2 O2


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b) A frozen dish in vacuum packaging.

c) A bottle of carbonated water that is opened and then re-capped.

d) A deodorant stick placed on a shelf in the bathroom.

Exercise 1.13

State whether the following situations represent dynamic equilibrium, a stationary

state or neither. Explain your answer.

a) A cat whose weight remains constant.

b) N2(g) + 3 H2(g) a 2 NH3(g)

c) A fire in a propane gas fireplace.

d) The Daniel Johnson dam in Manicouagan and the reservoir behind it. The waterlevel in the reservoir remains constant thanks to precipitation and the river thatflows into it.

e) Some sugar is poured into a glass. The mixture is stirred and the sugar dissolves.


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Exercise 1.14

Among the following expressions, choose the ones that apply to an equilibrium system.

A) An unchanging amount of matter B) Forward reaction and reverse reaction

C) Equal amounts of reactants D) Reversible reactionand products

E) Stationary state F) Constant concentrations

G) Continuous supply of reactants H) Open system

I) Dynamic equilibrium

This first chapter describes the characteristics shared by systems involving liquid-

vapour equilibrium, dissolution-crystallization equilibrium and chemical

equilibrium. Macroscopically speaking, an equilibrium state is characterized by

three essential conditions: the system shows no apparent signs of change, the

substances or the phases between which equilibrium is established are present

and the system is closed, that is, there is no exchange of matter or energy between

the system and the surroundings.

The rest of this Guide is devoted to equilibrium in chemical reactions. What

happens when an equilibrium system is disturbed by an external factor, for

example, by adding a reactant to the system? The system remains open as long

as the disturbance lasts. How does the system react when the disturbance ceases?

Is equilibrium still possible? The next chapter will provide answers to these

questions.


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Closed system Complete reaction

Dissolution-crystallization Dynamic equilibriumequilibrium

Forward reaction

Kinetic energy

Liquid-vapour equilibrium

Open system

Reverse reaction Reversible reaction

Liquid-vapour equilibrium, dissolution-crystallization equilibrium and chemical

equilibrium all present similar characteristics. Macroscopically speaking, three

conditions are essential for a system to be in equilibrium:

• The system is closed, that is, there is no exchange of matter or energy betweenthe system and the surroundings.

• The products and the reactants (or phases) between which equilibrium isestablished are present at the same time.

• The system shows no sign of change, that is, the quantities of the substances (andphases) present are constant. The colour and any other observable characteristicsremain constant.

In the equation for a reaction, equilibrium is represented by two opposite arrows of

equal length (a).

liquid a vapour solute(s) a solute(aq) reactants a products

In liquid-vapour equilibrium, the gas molecules return to the liquid state at the same

speed as molecules leave the liquid phase to enter the gaseous phase. The level of

the liquid in the container remains constant. In dissolution-crystallization equilibrium,

the solution must be saturated and in contact with the solid solute at the bottom of


1.34

KEY WORDS IN THIS CHAPTER

SUMMARY

the container. In a chemical reaction at equilibrium, the concentrations of the products

and reactants remain constant.

At the molecular level, equilibrium is a dynamic phenomenon in which two opposing

processes occur simultaneously and at the same rate. The rate of the forward reactionequals the rate of the reverse reaction. In other words, the molecules of reactants

are converted into products at the same rate as the molecules of products are converted

into reactants. Equilibrium is therefore a dynamic state, since the two reactions keep

going. This is possible only with reversible processes. If the system is left alone and

is not disturbed, the reactants will never be completely converted into products.

It takes time for a system to reach equilibrium. During the period before equilibrium,

the rates of the forward and reverse reactions are not equal; they depend on the

concentrations of the substances that are present. Mathematically, the rate of a reaction

is proportional to the concentration of the reactants raised to the power corresponding

to their respective coefficients in the balanced equation. For example, in the

equation:

3 H2(g) + N2(g) a 2 NH3(g)

vf = kf [H2]3 [N2] and vr = kr [NH3]

2

where kf and kr are the constants of proportionality for the forward and reverse

reactions.

A stationary system is not at equilibrium even though it shows no apparent signs of

change. It is an open system and the reaction involved is not reversible.


1.35

EXERCISE DE SYNTHÈSE

Exercise 1.15

From a macroscopic standpoint, name the three conditions that must exist for a systemto be at equilibrium.

Exercise 1.16

Among the following statements, select those that are true. If a statement is false,

explain why.

a) Systems involving reversible chemical reactions are always at equilibrium.

b) A system at equilibrium shows no apparent signs of change.

c) A saltwater solution can be said to be in dissolution-crystallization equilibriumwhen there is no undissolved salt at the bottom of the container.

Exercise 1.17

State whether the following situations represent equilibrium systems. If you answerno for a situation, give at least one reason to support your answer.

a) A lit welding torch.


1.36

REVIEW EXERCISES

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b) H2SO4(aq) + Zn(s) → ZnSO4(aq) + H2(g)

c) CaCO3(s) a CaO(s) + CO2(g)

d) An open bottle of soft drink.

Exercise 1.18

Some sugar is stirred into a glass of water at room temperature. After the solutionhas been mixed thoroughly for a few minutes, it is left to stand and a deposit of solidsugar forms at the bottom of the glass. Based on these observations, can you say thatthis system is at equilibrium?

Exercise 1.19

Choose the term in the following list that best describes each of the equations below.

Liquid-vapour equilibrium, dissolution-crystallization equilibrium,chemical equilibrium, reverse reaction or forward reaction.

a) CaCl2(aq) + Na2SO4(aq) a CaSO4(s) + 2 NaCl(aq) ____________________________

b) H2O(l) a H2O(g) ____________________________

c) NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) ____________________________

d) C6H12O6(s) a C6H12O6(aq) ____________________________

e) H2O(l) ← H2O(g) ____________________________


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Exercise 1.20

We mix some iodine (I2) crystals with some alcohol. After stirring the solution for a

few minutes, we let it stand. The intensity of the reddish brown colour does not change

and the deposit of iodine at the bottom of the container remains unchanged. Choose

the statement below that best describes the behaviour of the iodine molecules.

A) The iodine molecules in solution are immobile.

B) The iodine molecules go from the solid phase to the dissolved phase faster thanthe dissolved iodine molecules return to the solid phase.

C) The iodine molecules in solution crystallize faster than the molecules of solid iodinego into solution.

D) The iodine molecules in solution crystallize at the same rate as the molecules ofsolid iodine go into solution.

Exercise 1.21

When a system is at equilibrium, no visible change occurs. Explain how this

condition is manifested in the following case. We place 1 mole of N2O4(g) in a sealed

container which we then heat to 100°C. The temperature is kept constant and, after

a while, the system is at equilibrium.

N2O4(g) a 2 NO2(g)


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Exercise 1.22

Which of the following situations represent reversible processes?

A) A puddle of water evaporating B) 2 NO2(g) a N2O4(g)

C) Wood burning in a fireplace D) A closed bottle of carbonated water

E) A gas stove F) H2(g) + I2(g) a 2 HI(g)

Exercise 1.23

Which of the following characteristics can be associated with wood burning in a

fireplace?

A) Reversible reaction B) Stationary state C) Forward reaction

D) Dynamic equilibrium E) Constant temperature F) Open system

G) Constant concentrations

Exercise 1.24

A vacuum is created in a container equipped with a tap. Two colourless gases are

then added to it. The tap is closed and the system is left at room temperature for a

few hours. No change is observed. Is the system at equilibrium? Explain your answer.


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Exercise 1.25

Two colourless solutions are mixed in a flask, which is then air-tight sealed with a

stopper. As the reaction proceeds, the solution turns blue and bubbles escape from

the liquid. After a few minutes, the colour of the solution stabilizes, no further change

can be observed and the solution is at room temperature. The flask is then opened.

Gas escapes and the blue colour becomes more intense.

Can we say that the system was at equilibrium before the flask was opened? Explainyour answer.

Exercise 1.26

A vacuum is created in a container before 1.0 mole of H2(g) and 1.0 mole of I2(g) are

added to it. The mixture is heated to 440°C and held at this temperature. The equation

for this reaction is:

H2(g) + I2(g) a 2 HI(g)

The graph below shows the change in the concentrations of the substances during

the reaction.

Con

cent

ration

(m

ol/L

)

Time

[H2] = [I2] = 0.218

Reactants

[HI] = 1.564

Product

2.0

1.61.5

1.0

0.5

0.22

0


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a) What allows us to say that the reaction has reached equilibrium?

b) Draw a rough graph of the rate of reaction of the reactants and the products overtime.

c) If we use 0.5 mole of I2 instead of 1.0 mole as initially stated, the graph will bedifferent. The following graph shows the change in the concentration of thesubstances when we start with 1.0 mole of H2(g) and 0.5 mole of I2(g). What allowsus to say that this reaction is at equilibrium?


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d) Draw a rough graph of the reaction rate of the reactants and the products overtime.

Exercise 1.27

A vacuum is created in a 1-L container before 1.0 mole of H2(g) and 1.0 mole of I2(g)

are added to it. The mixture is heated to 440°C and held at this temperature. At

equilibrium, there are 1.564 moles of HI in the system.

a) Write the equation for this reaction.

b) Write the mathematical expressions for the rates of the forward and reversereactions.

c) Determine the concentration of each substance in the equilibrium system.


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