Equilibrium and Oxidation-Reduction
CHE-5043-2
Learning Guide
CHE-5043-2
LEARNING GUIDE
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.
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
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
GENERAL INTRODUCTION
OVERVIEW ... 0.12 HOW TO USE THIS LEARNING GUIDE ... 0.12 Learning Activities ... 0.13 Exercises ... 0.13 Self-evaluation Test ... 0.14 Appendices ... 0.14 Materials ... 0.14 CERTIFICATION ... 0.15 INFORMATION FOR DISTANCE EDUCATION STUDENTS ... 0.15 Work Pace ... 0.15 Your Tutor ... 0.15 Homework Assignments ... 0.16 CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION-REDUCTION ... 0.17
LEARNING ACTIVITIES
CHAPTER 1 – EQUILIBRIUM ... 1.1 1.1 EVAPORATION AND DISSOLUTION ... 1.3 Liquid-vapour Equilibrium ... 1.3 Dissolution-crystallization Equilibrium ... 1.8 Experimental Activity 1: Equilibrium Systems ... 1.13 1.2 CHEMICAL REACTIONS ... 1.14 Equilibrium of the Reaction N2O4(g) a 2 NO2(g) ... 1.15 Reaching Equilibrium ... 1.20 Rate and Concentration ... 1.22 Change in Rates of Reaction ... 1.23 Change in Concentrations ... 1.26 The Stationary State: A Special Case ... 1.30 KEY WORDS IN THIS CHAPTER ... 1.34 SUMMARY ... 1.34 REVIEW EXERCISES ... 1.36
CHAPTER 2 – FACTORS AFFECTING CHEMICAL EQUILIBRIUM ... 2.1 2.1 FACTORS AFFECTING EQUILIBRIUM ... 2.3 Concentration ... 2.3 Experimental Activity 2: Disturbing an Equilibrium System ... 2.8
This is a preview of:
- the introduction; and - the first chapter.
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 3.1 ACID IONIZATION CONSTANT Ka ... 3.3 Review on Acids ... 3.3 The pH Scale ... 3.7 The Constant Ka ... 3.12 Experimental Activity 3: Mathematical Expression of Equilibrium and Acid Strength 3.13 Definition ... 3.13 Strong Acids and Weak Acids ... 3.15 Applications ... 3.18 3.2 THE BASE IONIZATION CONSTANT Kb... 3.25 Review on Bases ... 3.25 The Constant Kb ... 3.28 Strong and Weak Bases ... 3.28 Applications ... 3.31 3.3 EQUILIBRIUM OF WATER AND NEUTRALIZATION ... 3.34 The Ion-product Constant for Water ... 3.34 Kwand 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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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 4.1 EQUILIBRIUM OF A CHEMICAL REACTION ... 4.3 The Constant Kc ... 4.4 Definition ... 4.6 Solids and Liquids ... 4.9 The Kcand 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
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 Standard Reduction Potential of the Zn(s)–Zn2+(aq)Half-cell ... 5.31 Standard Reduction Potential of the Ag(s)–Ag+(aq)Half-cell ... 5.34 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 The Zn
(s)–Cu2+
(aq) Cell ... 6.6 Weakening of a Voltaic Cell ... 6.10 6.2 BALANCING OXIDATION-REDUCTION EQUATIONS ... 6.11 Oxidation-Reduction without Ions or Metals ... 6.14 Oxidation Numbers ... 6.15 Half-reactions ... 6.22 Balancing Equations ... 6.25 6.3 APPLICATIONS ... 6.31 Cells and Batteries ... 6.31 Dry Cells ... 6.31 Storage Batteries ... 6.35 Fuel Cells ... 6.38 A Little Bit of History ... 6.41 Corrosion ... 6.42 Corrosion of Iron ... 6.42 Protection Against Corrosion ... 6.47
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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
GENERAL INTRODUCTION
OVERVIEW
Welcome to the course entitled Chemical Reactions 2: Equilibrium and Oxidation- reduction, which is part of the Secondary V Chemistry program. This program comprises the following three courses:
CHE-5041-2 Gases
CHE-5042-2 Chemical Reactions 1: Energy and Chemical Dynamics CHE-5043-2 Chemical Reactions 2: Equilibrium and Oxidation-reduction The three main components of the Chemistry program are related content, the experimental method and the history-technology-society perspective. Whereas the experimental method is developed in the workbook Experimental Activities of Chemistry, the related content and the history-technology-society perspective are covered in the three Learning Guides that complete the three courses which must be taken in sequential order.
Chemical Reactions 2: Equilibrium and Oxidation-reductionis the third in the set of three Learning Guides. It is divided into six chapters, corresponding to the three terminal objectives of the program.1This 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-reductionwill 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.
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1. The terminal objectives and associated intermediate objectives are listed at the beginning of each chapter.
The conclusion contains a summary covering all the courses in the program along with a self-evaluation test. It also includes an Answer Key for the self-evaluation test, for the exercises in each chapter and for the review exercises. A glossary with definitions of the key words, a bibliography, appendices and an index are also provided in the conclusion. You may wish to consult the books and publications in the bibliography for further information on the topics covered in this course.
Learning Activities
The Guide contains theoretical sections as well as practical activities in the form of exercises. The exercises come with an Answer Key.
Start by skimming through each part of the Guide to familiarize yourself with the content and the main headings. Then read the theory carefully:
– Highlight the important points.
– Make notes in the margins.
– Look up new words in the dictionary.
– Summarize important passages in your own words, in your notebook.
– Study the diagrams carefully.
– Write down questions relating to ideas you don’t understand.
Exercises
The exercises come with an Answer Key, which is located in the coloured section at the end of the Guide.
• Do all the exercises.
• Read the instructions and questions carefully before writing your answers.
• Do all the exercises to the best of your ability without looking at the Answer Key.
Reread the questions and your answers, and revise your answers, if necessary. Then check your answers against the Answer Key and try to understand any mistakes you made.
• Complete each chapter before doing the corresponding review exercises. Doing these exercises without referring to the lesson you have just completed is a better way of preparing for the final examination.
Self-evaluation Test
The self-evaluation test is a step that prepares you for the final evaluation. You must complete your study of the course before attempting to do it. Reread your notebook and the definitions of the key words in the chapters. Make sure you understand how they relate to the course objectives listed at the beginning of each chapter. Then do the self-evaluation test without referring to the main body of the Guide or the Answer Key. Compare your answers with those in the Answer Key and review any areas you had difficulty with.
Appendices
The appendices contain a review of some concepts you should be familiar with before beginning the course. The complete list of appendices appears in the table of contents.
Materials
Have all the materials you will need close at hand:
• Learning material: this Guide and a notebook where you will summarize important concepts relating to the objectives (listed in the introduction of each chapter). You will also need to use your periodic table and the workbook Experimental Activities of Chemistry.
• Reference material: a dictionary.
• Miscellaneous material: a calculator, a pencil for writing your answers and notes in your Guide, a coloured pen for correcting your answers, a highlighter (or a pale- coloured felt pen) to highlight important ideas, a ruler, an eraser, etc.
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CERTIFICATION
To earn credits for this course, you must obtain at least 60% on the final examination which will be held in an adult education centre.
The evaluation for the course Chemical Reactions 2: Equilibrium and Oxidation- reductionis divided into two separate parts.
Part Iconsists of a two-hour written examinationmade up of multiple-choice, short- answer and essay-type questions. This part is worth 70%of your final mark and deals with the objectives covered in this Guide. You may use a calculator.
Part II is designed exclusively to evaluate the experimental method. It will be held in the laboratory during a 120-minute session. This part is worth 30%of your final mark and deals with the course objectives covered in Section C of Experimental Activities of Chemistry.
INFORMATION FOR DISTANCE EDUCATION STUDENTS
Work Pace
Here are some tips for organizing your work:
• Draw up a study timetable that takes into account your personality and needs, as well as your family, work and other obligations.
• Try to study a few hours each week. You should break up your study time into several one- or two-hour sessions.
• Do your best to stick to your study timetable.
Your Tutor
Your tutor is the person who will give you any help you need throughout this course.
He or she will answer your questions and correct and comment on your homework assignments.
Don’t hesitate to contact your tutor if you are having difficulty with the theory or the exercises, or if you need some words of encouragement to help you get through this course. Write down your questions and get in touch with your tutor during his or her available hours. The letter included with this Guide or that you will receive shortly tells you when and how to contact your tutor.
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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CHEMICAL REACTIONS 2: EQUILIBRIUM AND OXIDATION- REDUCTION
In the first course of the Secondary V Chemistryprogram, students learn about gas- related phenomena and the energy balance of chemical reactions. The second course in the program examines energy transfers from a broader perspective and also deals with reaction rates and the factors affecting those rates.
This course, Chemical Reactions 2: Equilibrium and Oxidation-reduction, deals with two types of chemical reactions in greater depth. In Chapter 1, students have the chance to become familiar with the conditions that characterize a system at equilibrium and to analyze the equilibrium of reversible reactions. Chapter 2 deals with what happens to a system at equilibrium when it is disrupted by external factors.
After completing this qualitative analysis of equilibrium in the first two chapters, in the third and fourth chapters students go on to study the quantitative aspects of chemical equilibrium, which include the calculation of equilibrium constants.
The last two chapters delve into oxidation-reduction reactions, a chemical reaction that involves a transfer of electrons. In this part of the Guide, students will also learn how electrochemical cells work. This material represents the final stepping stones in completing the Secondary V Chemistry program.
As in the first two Guides, a table of contents diagram at the beginning of each chapter shows you how the chapter fits into the course as a whole. The content of the chapter you are about to begin is in bold type and in larger characters, whereas the content of completed chapters is in italics. For example, the table of contents diagram for Chapter 2 is reproduced on the next page. The section for Chapter 2 is in bold type and the content of Chapter 1 is in italics and smaller type. You will find this diagram a very useful tool as you go through the course.
Enjoy your work and good luck!
1. Chemical Equilibrium Liquid-vapour equilibrium Dissolution-cr ystallization
equilibrium Chemical equilibrium Reaching equilibrium Stationar y state
2. Factors Affecting Chemical Equilibrium Types of disturbances:
concentration temperature catalysts Le Châtelier’s principle Applications Equilibrium in nature 6. Cells and Electrochemistry
Voltaic cells Balancing oxidation-
reduction equations Cells and batteries Corrosion Electrolysis
CH
EMICAL
REACTIO NS
2:
EQ UIL
IBR IUM
ANDOXIDATION-REDUCTION
5. Oxidation-Reduction Oxidation and reduction Daniell cell
Standard reduction potential
4. The Equilibrium Constant Chemical equilibrium Applications
Dissolution-cr ystallization equilibrium Chemistr y, industr y and society
3. Equilibrium in Acidic and Basic Solutions Constants:
Ka, Kband Kw Neutralization Titration Examples of acid-base equilibrium
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CHAPTER 1
EQUILIBRIUM
GWB
Terminal Objective 1
To analyze qualitatively the state of equilibrium of a system.
Intermediate Objectives
1.1 To state the three conditions that characterize a system in a state of equilibrium.
1.2 To describe a vapour-liquid equilibrium and an equilibrium involving solutions.
1.3 To verify, through experimentation, whether or not a system is in equilibrium.
1.4 To associate equilibrium with the reversibility of reactions.
1.5 To interpret curves illustrating forward and reverse reactions over time.
1. Chemical Equilibrium Liquid-vapour equilibrium Dissolution-crystallization
equilibrium Chemical equilibrium Reaching equilibrium Stationary state
2. Factors Affecting Chemical Equilibrium Types of disturbances:
concentration temperature catalysts Le Châtelier’s principle Applications Equilibrium in nature 6. Cells and Electrochemistry
Voltaic cells Balancing oxidation-
reduction equations Cells and batteries Corrosion Electrolysis
CHEMI
CAL REACTIO NS
2:
EQ UIL
IBR IUM
AND OXIDATION-RED UCTIO
N
5. Oxidation-Reduction Oxidation and reduction Daniell cell
Standard reduction potential
4. The Equilibrium Constant Chemical equilibrium Applications
Dissolution-crystallization equilibrium Chemistry, industry and society
3. Equilibrium in Acidic and Basic Solutions Constants:
Ka,K
band K
w
Neutralization Titration Examples of acid-base equilibrium
A tightrope walker balancing on the high wire, a vase on a table and a canoe floating on water are a few examples of equilibrium. Although some of these situations may be viewed as precarious, the tightrope artist and the objects are stable and immobile.
They are said to be in static equilibrium. In a chemistry context, the state of equilibrium is less apparent and, although the term “equilibrium” has the same general meaning, several key differences have to be considered. That is because we are dealing with dynamic equilibrium and reversibility, and reaction rates are involved.
In this chapter, we will study a variety of situations in order to shed light on the state of equilibrium. First we will look at evaporation and dissolution to determine under what conditions these processes can reach equilibrium. After that, we will look at the characteristics of a reversible chemical reaction at equilibrium.
1.1 EVAPORATION AND DISSOLUTION
The word “equilibrium” makes one think of stability. In chemistry, a system at equilibrium1is 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.
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.
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Figure 1.2 - Open system and closed system
a) In an open system, a state of equilibrium is not feasible. The system is constantly changing because molecules of vapour are escaping to the air.
b) In a closed system, evaporation is the dominant process initially. After a certain time, molecules in the gaseous phase return to the liquid phase. The arrows indicate
the movement of the molecules between the two phases.
c) After a while, the system shows no apparent signs of change. It has reached a state 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.
Equilibrium: liquid avapour
Rate
Time
0123456 Evaporation rate
Condensation rate
a) b) c)
d)
After a while, that is, at t = 4, the two curves come together. The rate of evaporation is equal to the rate of condensation at this point. There are as many molecules escaping from the liquid phase as there are molecules returning to it. The level of the liquid remains constant. We say that the two phases are in equilibrium. This liquid-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) evaporation H2O(l) ← H2O(g) condensation
In the equilibrium equation, the two arrows point in opposite directions to show the continual back-and-forth movement of the molecules escaping from the liquid phase (→) to the gaseous phase and those going from the gaseous phase to the liquid phase (←). The two arrows indicate that the reaction is reversible, that is, the two processes are taking place in opposite directions. At equilibrium, the processes of evaporation and condensation occur at the same speed, hence the use of opposite arrows of equal length. The right arrow (→) indicates the forward reaction and the left arrow (←) the reverse reaction (condensation).
The double arrow indicates that the system is in dynamic equilibrium. The word
“dynamic” means that even if no change is evident, there is movement, in this case, the back-and-forth movement of the molecules between two phases. In order for dynamic liquid-vapour equilibrium to occur, the system must be closed so that no matter can escape from it. In other words, there must be no exchange of matter between the system and the surroundings.
Exercise 1.2
a) Name an important characteristic of a closed system.
b) What distinguishes an open system from a closed system?
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c) What do the two opposing arrows mean in the equation H2O(l)aH2O(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)aH2O(g) is at dynamic equilibrium. What does this mean?
e) In the equation H2O(l)aH2O(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 macroscopic2conditions 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?
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.
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3. In some textbooks, the equivalent expression “equilibrium involving dissolution of solids” is used.
When we add salt to water, dissolution (→) begins as the NaCl dissociates into Na+(aq) and Cl–(aq)ions. After a while, the continually moving dissolved ions start to accumulate in the solution, and some of them become reattached to the solid NaCl crystals, through the process of crystallization (←). When the rate of crystallization (←) equals the rate of rate of dissolution (→), the number of ions in solution no longer increases and we say that the solution is saturated.
Figure 1.3 - Solution at equilibrium
a) A large quantity of salt is added to a container of water.
Initially, the Na+and Cl–ions leave the crystals and go into solution. A single arrow is shown, since only the process of dissolution is taking place. Crystallization has not yet started.
b) After a while, dissolution continues but some ions become reattached to the crystals as 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.
The up arrow represents ions dissociating.
The down arrow represents the ions returning to the crystal.
a) b) c)
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 dissolution- crystallization 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)asugar(aq), what is the meaning of the two opposing arrows and why are they the same length?
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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 CO2leaves 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
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Respiratory system
When dissolved alcohol circulates in the blood, a part of this alcohol becomes gaseous in the lungs, like CO2. With each inhalation, equilibrium is established very rapidly between the blood and the air so that, upon exhalation, the concentration of alcohol in the exhaled breath is representative of the concentration of alcohol in the blood. The higher the blood alcohol content, the higher the concentration of alcohol in the exhaled breath. However, the concentrations differ. There is about 2 000 times more alcohol in the blood than in the exhaled breath. The Breathalyzer test is based on this difference. This device used by police officers precisely measures the concentration of alcohol in exhaled breath and computes the amount of alcohol in the driver’s blood. If the result exceeds 0.08%, that is, 0.08 g of alcohol per 100 mL of blood, the driver’s license will be suspended on the spot for a period of 15 days for a first offense, and 30 days for a repeat offense.
Equilibrium in the alveoli
Alveolus
Capillary at the outset
Capillary in equilibrium
Ethanol Ethanol
Alveolus
1.12
We have outlined the conditions necessary for a system to be at either liquid-vapour equilibrium or dissolution-crystallization equilibrium. In the following experimental activity, you will study two new situations that will allow you to learn more about the characteristics of an equilibrium system.
Experimental Activity 1: Equilibrium Systems
In this activity, you will study various systems and determine under what conditions they are at equilibrium.
In the first part, you will examine the reaction produced by dissolving an effervescent salt in water, which, once dissolved, produces carbon dioxide (CO2) gas. You will be required to compare the properties of the reaction in an open system versus a closed system.
In the second part, you will have a chance to verify that the following reaction is reversible:
CaCl2(aq) + Na2SO4(aq) a 2 NaCl(aq) + CaSO4(s)
Allow about 30 minutes for all of the steps in the experiment. All of the information you will need to carry out this activity is given in Section C of the workbook Experimental Activities of Chemistry. Enjoy your work!
The experimental activity you have just completed allowed you to observe that, in the closed system, part of the CO2was 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 CO2dissolved in the water and the CO2above 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)
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 H2react with one mole of N2to form two moles of NH3, and that 92.2 kJ of heat is released when two moles of NH3are formed. We assume that the reaction is complete and that there is no reactant left at the end.
Irreversible reactions generally keep proceeding until one of the reactants has been completely depleted. There are, however, many reversible reactions too. The following analogy will give you a better understanding of reversible chemical reactions.
Suppose that a leaky boat is one-quarter full of water and that we are bailing out water as quickly as it is entering the boat. Equilibrium has been established and the water level remains stable, since the rate at which the water is entering the boat equals the rate at which it is being removed. Likewise, if the boat is half-full and we remove the same amount of water as is entering the boat, the state of equilibrium will continue.
Nonetheless, the boat is still half-full of water. If the boat is three-quarters full and we are again able to remove water as fast as water is entering the boat, the state of equilibrium will continue, but the boat will still be three-quarters full of water.
1.14
Figure 1.4 - Precarious situation or equilibrium?
The boat will stay afloat as long as the amount of water entering it is counterbalanced by the water being bailed out of it. The water level in the boat depends on how long it took for equilibrium to be reached. Fortunately for our “balanced” friend, the water level is not rising!
We can apply this analogy to a chemical reaction at equilibrium. At equilibrium, the rate of the opposing reactions is the same but the quantities of reactants and products are not necessarily equal. As in our analogy, at equilibrium we could have a boat that is one-quarter full, one-third full, half full, three-quarters full, and so on.
To conclude this analogy, let’s say that the sinking of the boat would correspond to a chemical reaction in which equilibrium cannot be reached. This is the case for irreversible chemical reactions in which the reactants are converted entirely into products. These reactions stop by themselves when there are no reactants left. For example, when we burn propane gas in the open air, the reaction produces carbon dioxide (CO2) and water (H2O). The reaction is forward and irreversible. The single arrow in the equation denotes this.
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O(g)
Let’s now see how a reversible chemical reaction proceeds.
EQUILIBRIUM OF THE REACTION N
2O
4(g) aa 2 NO
2(g)
At a macroscopic level, the chemical reaction N2O4(g)a2 NO2(g) is an interesting one.
N2O4is colourless, whereas NO2is 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.
GWB
To produce this reaction, suppose that we add some dinitrogen tetraoxide (N2O4), a colourless gas, to a container in which we have first created a vacuum. We heat the container in a bath of boiling water and keep the temperature at 100°C. The average kinetic energy of the molecules in the system will be constant at that point. Figure 1.5 shows what happens while the N2O4 is being heated and equilibrium is being established.
Figure 1.5 - Reaching equilibrium in the reaction N
2O
4(g) aa 2 NO
2(g)
At t = 0, the container filled with the colourless gas N2O4is 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 N2O4starts 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 NO2is 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, N2O4and NO2, are present by analyzing a gas sample from the system.
t= 0 t = 1 t= 2 t= 3 t = 4
Time
Forward reaction: ↑; reverse reaction: ↓; N2O4: ; NO2:
1.16
Forward reaction reverse reaction
Now let’s consider the opposite reaction (Figure 1.6). We immerse a one-L container of pure NO2in 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→N2O4is 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 NO2are present by analyzing a gas sample from the container.
Figure 1.6 - Equilibrium 2 NO
2(g)aa N
2O
4(g)
At t= 0, the container is immersed in an ice-water bath (0°C). As time goes by, the reddish brown colour turns lighter, indicating a decrease in the concentration of NO2(t = 1, t = 2).
Equilibrium is reached when the intensity of the colour stabilizes (t = 3, t = 4).
Note that the forward reaction (↑) is faster in the beginning (length of the vertical arrow).
In both cases, we can see that the system has reached equilibrium. When no further change is observed in the colour of the gaseous mixture, we can conclude that the reaction N2O4a2 NO2is 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, N2O4is formed. In Figure 1.5, the colourless gas N2O4was not completely converted since the colour of pure NO2is much darker than that observed at the end. Likewise, in Figure 1.6 the reddish brown NO2was not completely converted into colourless N2O4, since the gaseous mixture was still slightly coloured at the end.
t= 0 t = 1 t= 2 t= 3 t = 4
Time
Forward reaction: : ↑; reverse reaction: : ↓; N2O4: ; NO2:
Forward reaction reverse reaction
When the system is at equilibrium, the equation for the reaction is:
N2O4(g) a2 NO2(g)
The above equation sums up the two equations below:
N2O4(g) →2 NO2(g) forward reaction N2O4(g) ←2 NO2(g) reverse reaction
As before, the double arrow shows that the system is in dynamic equilibrium and that the rate of decomposition of the N2O4 into NO2equals the rate at which N2O4is 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 N2O4and 1.2 moles of NO2.
Exercise 1.5
What signs indicate that the reaction N2O4(g)a2 NO2has 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 NO2as 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
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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 N2O4and 1.2 moles of NO2 in the container. The total number of moles is greater than at the beginning. How is this possible? Let’s examine the equation for the reaction and the process by which equilibrium is established.
N2O4(g) a 2 NO2(g)
Initially 1.0 mol 0,0mol
Conversion of system – 0.6 mol + 1.2 mol
At equilibrium 0.4 mol 1.2 mol
The coefficients in the equation show that 1 molecule of N2O4decomposes 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 N2O4has 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 N2O4and 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 NO2are left in the system.
a) Determine the number of moles of NO2that were added to the system.
Conversion: N2O4(g) → 2 NO2(g) 1,0mol 2,0mol 0.6 mol → ?,0mol 0.6 mol →
Æ 1.2 mol
?
b) How many moles of NO2were converted into N2O4in 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
Rate and Concentration
The previous chemistry course5dealt 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:
vPCl5 = k [PCl3] [Cl2]
where k is the constant of proportionality for the reaction. The expressions [PCl3] and [Cl2] are read “concentration of PCl3” and “concentration of Cl2” and are expressed in moles per litre (mol/L).
Since the reaction is reversible, we can express the rate of the reverse reaction in the same way, that is, the rate at which the products disappear to form the reactants.
The rate of decomposition of PCl5is therefore written as:
vPCl3 + 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
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)a2 NH3(g) v
NH3 = k [H2]3[N2] 2 H2(g) + O2(g) a 2 H2O(g) v
H2O = k [H2]2[O2] N2O4(g) a 2 NO2(g) v
NO2 = k [N2O4]
For the first reaction, the concentration of H2is assigned the exponent 3, since the coefficient of H2is 3 in the balanced equation. Following the same rule, we have [H2]2 in the expression for the rate of the second reaction. Thus, the rate of a reaction is proportional to the concentration of the reactants, raised to the power corresponding to their respective coefficients in the balanced equation.
All of the above reactions are reversible. The rate of the reverse reactions for the above equations is expressed according to the same rule. This rate is proportional to the concentrations, each raised to the power corresponding to its coefficient in the equation.
Therefore:
vH2 + N2 = k [NH3]2 vH2 + O2 = k [H2O]2 vN2O4 = k [NO2]2
To simplify matters, we generally write vf for the rate of the forward reaction—the reaction which is read from left to right in the equation—and vrfor the rate of the reverse reaction—the reaction that is read from right to left in the equation.
Therefore:
3 H2(g) + N2(g) a 2 NH3(g) vf= k [H2]3[N2] vr= k [NH3]2
In the next chapter, we will use this method of expressing the rate of reaction to determine the concentration of the substances in equilibrium systems. It is important to remember that, at equilibrium, the rate at which the products (vf) are formed equals the rate at which the reactants (vr) are formed.
1.22
Exercise 1.8
For each of the following reactions, write the equation representing the rate of formation of the products (forward reaction) and the equation representing the rate of formation of the reactants (reverse reaction).
Reaction Rate of forward reaction Rate of reverse reaction
CO(g)+ H2O(g)aCO2(g)+ H2(g) H2(g)+ Cl2(g) a2 HCl(g) 2 NO(g)+ O2(g)a2 NO2(g)
4 NH3(g)+ 5 O2(g)a4 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 N2O4and 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.
