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Chemical Reactions 2:
Equilibrium and Oxidation-Reduction
CHE-5043-2
Learning Guide
CHEMICAL REACTIONS 2:
EQUILIBRIUM AND OXIDATION-REDUCTION
CHE-5043-2
LEARNING 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.
CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION-REDUCTION
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
Chemical Reactions 2 - Table of Contents
TABLE OF CONTENTS
GENERAL INTRODUCTION
This is a preview
of:
n; and
- the introductio
OVERVIEW ...................................................................................................................
r.
- the first chapte
HOW TO USE THIS LEARNING GUIDE .............................................................................
Learning Activities .................................................................................................
Exercises ............................................................................................................
Self-evaluation Test ...............................................................................................
Appendices ...........................................................................................................
Materials ..............................................................................................................
CERTIFICATION .............................................................................................................
INFORMATION FOR DISTANCE EDUCATION STUDENTS ....................................................
Work Pace ............................................................................................................
Your Tutor .............................................................................................................
Homework Assignments ........................................................................................
CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION-REDUCTION .............................
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LEARNING ACTIVITIES
CHAPTER 1 – EQUILIBRIUM ........................................................................................
1.1
1.1 EVAPORATION AND DISSOLUTION ..........................................................................
Liquid-vapour Equilibrium .......................................................................................
1.3
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
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Chemical Reactions 2 - Table of Contents
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
KEY WORDS IN THIS CHAPTER ..................................................................................... 2.54
SUMMARY .................................................................................................................... 2.54
REVIEW EXERCISES ...................................................................................................... 2.56
CHAPTER 3 – EQUIIBRIUM IN ACIDIC AND BASIC SOLUTIONS .....................................
3.1 ACID IONIZATION CONSTANT Ka .............................................................................
3.1
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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Chemical Reactions 2 - Table of Contents
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
KEY WORDS IN THIS CHAPTER ..................................................................................... 3.66
SUMMARY .................................................................................................................... 3.66
REVIEW EXERCISES ...................................................................................................... 3.68
CHAPTER 4 – THE EQUILIBRIUM CONSTANT ................................................................
4.1 EQUILIBRIUM OF A CHEMICAL REACTION ...............................................................
4.1
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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Chemical Reactions 2 - Table of Contents
KEY WORDS IN THIS CHAPTER ..................................................................................... 4.72
SUMMARY .................................................................................................................... 4.72
REVIEW EXERCISES ...................................................................................................... 4.74
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
Half-cell .......................................... 5.31
Standard Reduction Potential of the Zn(s)–Zn2+
(aq)
Standard Reduction Potential of the Ag(s)–Ag+(aq) Half-cell .......................................... 5.34
KEY WORDS IN THIS CHAPTER ..................................................................................... 5.40
SUMMARY .................................................................................................................... 5.40
REVIEW EXERCISES ...................................................................................................... 5.43
CHAPTER 6 – CELLS AND ELECTROCHEMISTRY ..........................................................
6.1
6.1 VOLTAIC CELLS .....................................................................................................
6.3
Experimental Activity 7: Construction of an Electrochemical (Voltaic) Cell ..........
6.5
2+
(aq)
The Zn(s)–Cu
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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Chemical Reactions 2 - Table of Contents
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
KEY WORDS IN THIS CHAPTER ..................................................................................... 6.67
SUMMARY .................................................................................................................... 6.67
REVIEW EXERCISES ...................................................................................................... 6.70
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
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TABLE OF
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ANSWER
KEY
GENERAL INTRODUCTION
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - General Introduction
OVERVIEW
Welcome to the course entitled Chemical Reactions 2: Equilibrium and Oxidationreduction, 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.
1. The terminal objectives and associated intermediate objectives are listed at the beginning of each chapter.
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TABLE OF
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ANSWER
KEY
Chemical Reactions 2 - General Introduction
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.
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TABLE OF
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ANSWER
KEY
Chemical Reactions 2 - General Introduction
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 palecoloured felt pen) to highlight important ideas, a ruler, an eraser, etc.
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TABLE OF
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ANSWER
KEY
Chemical Reactions 2 - General Introduction
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 Oxidationreduction is divided into two separate parts.
Part I consists of a two-hour written examination made up of multiple-choice, shortanswer 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.
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TABLE OF
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ANSWER
KEY
Chemical Reactions 2 - General Introduction
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.
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TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - General Introduction
CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATIONREDUCTION
In the first course of the Secondary V Chemistry program, students learn about gasrelated 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!
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ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - General Introduction
2. Factors Affecting Chemical Equilibrium
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CH
E
ION
UM
UC
T
RI
5. Oxidation-Reduction
Oxidation and reduction
Daniell cell
Standard reduction potential
2:
ILIB
EQU
6. Cells and Electrochemistry
Voltaic cells
Balancing oxidationreduction equations
Cells and batteries
Corrosion
Electrolysis
M
AL R EACTIO
IC
N
S
1. Chemical Equilibrium
Liquid-vapour equilibrium
Dissolution-crystallization
equilibrium
Chemical equilibrium
Reaching equilibrium
Stationary state
ED
AN
D OXI DATION-R
Types of disturbances:
concentration
temperature
catalysts
Le Châtelier’s principle
Applications
Equilibrium in nature
3. Equilibrium in Acidic and Basic Solutions
Constants:
Ka, Kb and Kw
Neutralization
Titration
Examples of acid-base equilibrium
4. The Equilibrium Constant
Chemical equilibrium
Applications
Dissolution-crystallization equilibrium
Chemistry, industry and society
TABLE OF
CONTENTS
ANSWER
KEY
CHAPTER 1
EQUILIBRIUM
B
GW
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
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.
1. Chemical Equilibrium
ION
UM
UC
T
CH
E
RI
5. Oxidation-Reduction
Oxidation and reduction
Daniell cell
Standard reduction potential
2. Factors Affecting Chemical Equilibrium
Types of disturbances:
concentration
temperature
catalysts
Le Châtelier’s principle
Applications
Equilibrium in nature
2:
ILIB
EQU
6. Cells and Electrochemistry
Voltaic cells
Balancing oxidationreduction equations
Cells and batteries
Corrosion
Electrolysis
M
AL R EACTIO
IC
N
S
Liquid-vapour equilibrium
Dissolution-crystallization
equilibrium
Chemical equilibrium
Reaching equilibrium
Stationary state
ED
AN
D OXI DATION-R
3. Equilibrium in Acidic and Basic Solutions
Constants:
K a, Kb and Kw
Neutralization
Titration
Examples of acid-base equilibrium
4. The Equilibrium Constant
Chemical equilibrium
Applications
Dissolution-crystallization equilibrium
Chemistry, industry and society
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: 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. The word “system” denotes the equipment, setup, substances and all other elements involved in an experiment.
1.3
TABLE OF
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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 system
at 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
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
Figure 1.2 - Open system and closed system
a)
b)
d)
c)
Rate
Evaporation rate
Equilibrium: liquid a vapour
Condensation rate
0
1
2
3
4
5
6
Time
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 of
equilibrium, 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.
1.5
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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-vapour
equilibrium is represented by the following equation:
liquid a vapour
H2O(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)
H2O(l) ← H2O(g)
evaporation
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
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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 for
equilibrium to exist between the liquid and its vapour form.
• The system is closed: there must be no exchange of matter or energy between the
system 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?
2. “Macroscopic” means that which can be seen with the naked eye.
1.7
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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 solution
is 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.
3. In some textbooks, the equivalent expression “equilibrium involving dissolution of solids” is used.
1.8
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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)
b)
c)
The up arrow represents ions dissociating.
The down arrow represents the ions returning to the crystal.
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 they
move 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.
1.9
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
Exercise 1.3
a) Based on what you have learned about evaporation, list three macroscopic
conditions indicating that a system has reached a dissolution-crystallization
equilibrium.
b) We pour some sugar into a glass of water. We stir the solution, and, after a few
seconds, the sugar disappears completely. Is the sugar water at dissolutioncrystallization equilibrium? Explain your answer.
c) We add some more sugar and, after stirring the solution well, we note that an excess
of 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 at
the molecular level.
d) In the equation sugar(s) a sugar(aq), what is the meaning of the two opposing arrows
and why are they the same length?
1.10
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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 condition
must 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 at
equilibrium?
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.
4. A light bulb indicates additional information: this information is not part of the course and will not be covered
in the final examination.
1.11
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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
Ethanol
Capillary at
the outset
1.12
Alveolus
Ethanol
Capillary in
equilibrium
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
Figure 1.4 - Precarious situation or equilibrium?
B
GW
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) a 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
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
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) a 2 NO2(g)
Forward reaction
reverse reaction
t=0
Forward reaction: ↑;
t=1
reverse reaction: ↓;
t=2
Time
N2O4 :
t=3
;
t=4
NO2 :
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.
1.16
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
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) a N2O4(g)
Forward
reaction
reverse
reaction
t=0
t=1
Forward reaction: : ↑; reverse reaction: : ↓;
t=2
Time
N 2O4 :
t=3
;
t=4
NO2 :
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.
1.17
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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)
N2O4(g) ← 2 NO2(g)
forward reaction
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 as
there 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
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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)
Initially
Conversion of system
At equilibrium
1.0 mol
– 0.6 mol
0.4 mol
a
2 NO2(g)
0,0 mol
+ 1.2 mol
1.2 mol
Conversion:
N2O4(g) → 2 NO2(g)
1,0 mol
2,0 mol
0.6 mol → ?,0 mol
0.6 mol →
Æ
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
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
b) How many moles of NO2 were converted into N2O4 in the time it took the system
to reach equilibrium?
c) Complete the lines beneath the equation to determine the quantities of each of
the 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
TABLE OF
CONTENTS
ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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:
∆
∆
∆C
v = ––––
∆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:
vPCl = k [PCl3] [Cl2]
5
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:
vPCl
3 + Cl2
= k [PCl5]
where k is the constant of proportionality and [PCl5] is the concentration of PCl5.
5. Blondin, André, Chemical Reactions 1: Energy and Chemical Dynamics (Chemistry, Secondary V), Learning Guide
produced by SOFAD
1.21
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
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)
vNH = k [H2]3 [N2]
2 H2(g) + O2(g) a 2 H2O(g)
vH
N2O4(g) a 2 NO2(g)
vNO = k [N2O4]
3
2O
= k [H2]2 [O2]
2
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:
vH
= k [NH3]2
vH
= k [H2O]2
2 + N2
2 + O2
vN
2O4
= 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.
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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
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TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
Figure 1.7 - N2O4(g) a 2 NO2(g) system
a) Conversion of N2O4 into NO2 until equilibrium is reached
t=0
t=1
Forward reaction: ↑;
t=2
Time
reverse reaction: ↓;
N2O4 :
t=3
;
t=4
NO2 :
b) Change in rates of reaction in N2O4 a 2 NO2
N2O4 → 2 NO2 (forward reaction)
vf
0
Start of equilibrium
vf
Rate
1
N2O4 a 2 NO2 (equilibrium)
vf
2
vr
2
N2O4 ← 2 NO2 (reverse reaction)
vr
1
vr
0
0
1
2
3
4
5
Time
6
7
8
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 reverse
reaction 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.
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TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
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 (vf ).
0
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 (vf ) is even faster than
1
the rate at which it is being formed (vr ).
1
At t = 2, the rate of the forward reaction has decreased considerably (vf ) because the
2
concentration of N2O4 is much lower than it was at the beginning. By contrast, the rate
of the reverse reaction has increased (vr ) because the concentration of NO2 is higher.
2
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 in
order 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?
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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. Explain
why.
d) From t = 3 on, the curve for the forward reaction and that for the reverse reaction
are 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
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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
Concentration (mol/L)
2.0
[HI]
1.6
1.5
Product
1.0
0.5
[H2] and [I2]
0.22
Reactants
0
Time
Equilibrium is reached when the concentrations have stabilized. Under the conditions in which
the 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.
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TABLE OF
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Chemical Reactions 2 - Chapter 1: Equilibrium
Figure 1.9 - Change in concentrations in the reaction CO2(g) + H2(g) a CO(g) + H2O(g)
at 550°C
Concentration (mol/L)
1.00
[CO2] and [H2]
0.73
Reactants
0.50
[CO] and [H2O]
0.27
Products
0
Time
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)
Concentration (mol/L)
0.400
0.300
0.200
0.100
[SO3] Product
[O2] Reactant
[SO2] Reactant
0
Time
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.
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ANSWER
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Chemical Reactions 2 - Chapter 1: Equilibrium
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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ANSWER
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Chemical Reactions 2 - Chapter 1: Equilibrium
b) What distinguishes the chemical reaction represented by the graph in Figure 1.8
from 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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ANSWER
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TABLE OF
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Chemical Reactions 2 - Chapter 1: Equilibrium
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
CO2 + H2O
O2
O2
C3H8 (propane gas)
A Bunsen burner constitutes a stationary system. For combustion to occur, the burner must
receive 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.
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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 water
level in the reservoir remains constant thanks to precipitation and the river that
flows into it.
e) Some sugar is poured into a glass. The mixture is stirred and the sugar dissolves.
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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
and products
D) Reversible reaction
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 liquidvapour 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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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
KEY WORDS IN THIS CHAPTER
Closed system
Complete reaction
Dissolution-crystallization
equilibrium
Dynamic equilibrium
Forward reaction
Kinetic energy
Liquid-vapour equilibrium
Open system
Reverse reaction
Reversible reaction
SUMMARY
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 between
the system and the surroundings.
• The products and the reactants (or phases) between which equilibrium is
established are present at the same time.
• The system shows no sign of change, that is, the quantities of the substances (and
phases) present are constant. The colour and any other observable characteristics
remain 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
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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 reaction
equals 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.
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
REVIEW EXERCISES
EXERCISE DE SYNTHÈSE
?
Exercise 1.15
From a macroscopic standpoint, name the three conditions that must exist for a system
to 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 equilibrium
when there is no undissolved salt at the bottom of the container.
?
Exercise 1.17
State whether the following situations represent equilibrium systems. If you answer
no for a situation, give at least one reason to support your answer.
a) A lit welding torch.
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ANSWER
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Chemical Reactions 2 - Chapter 1: Equilibrium
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 solution
has been mixed thoroughly for a few minutes, it is left to stand and a deposit of solid
sugar forms at the bottom of the glass. Based on these observations, can you say that
this 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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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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 than
the dissolved iodine molecules return to the solid phase.
C) The iodine molecules in solution crystallize faster than the molecules of solid iodine
go into solution.
D) The iodine molecules in solution crystallize at the same rate as the molecules of
solid 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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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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.
1.39
ANSWER
KEY
TABLE OF
CONTENTS
Chemical Reactions 2 - Chapter 1: Equilibrium
?
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? Explain
your 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.
2.0
Concentration (mol/L)
?
[HI] = 1.564
1.6
1.5
Product
1.0
0.5
[H2] = [I2] = 0.218
0.22
Reactants
0
Time
1.40
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ANSWER
KEY
Chemical Reactions 2 - Chapter 1: Equilibrium
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 over
time.
c) If we use 0.5 mole of I2 instead of 1.0 mole as initially stated, the graph will be
different. The following graph shows the change in the concentration of the
substances when we start with 1.0 mole of H2(g) and 0.5 mole of I2(g). What allows
us to say that this reaction is at equilibrium?
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ANSWER
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Chemical Reactions 2 - Chapter 1: Equilibrium
d) Draw a rough graph of the reaction rate of the reactants and the products over
time.
?
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 reverse
reactions.
c) Determine the concentration of each substance in the equilibrium system.
1.42