TOPIC 7: EQUILIBRIUM SL
7. 1 Dynamic equilibrium
7.1.1 Outline the characteristics of chemical and physical systems in a state of equilibrium.
Many (theoretically ALL) chemical reactions are reversible and do not go to completion (=complete
conversion of reactants into products).
Reversible reactions are reactions in which some of the product formed from the reactants (by the forward
reaction) are converted back into reactants (reversible reaction).
We can recognize reversible reactions in a concentration-time graph of a reactant during a reaction, the
downward curve does not end at the zero line as some of the product are reformed to reactant.
Simulations which can show the above:
 http://www.chm.davidson.edu/chemistryApplets/equilibria/BasicConcepts.html
 http://www.mhhe.com/physsci/chemistry/animations/chang_7e_esp/kim2s2_5.swf
 http://www.physchem.co.za/OB12-che/equilibrium.htm
Characteristics of an equilibrium system
Dynamic means that opposing changes are constantly taking place. In a dynamic equilibrium opposing
changes are going on at the same time and at the same rate. Other characteristics of a dynamic equilibrium
or when a reversible reaction reaches equilibrium:
 dynamic equilibrium is established when rate of forward reaction equals rate of reverse reaction (see
graph (b)); as a result the concentrations of reactants and products become and remain constant as
shown by the horizontal lines in graph (a) below:
(a)
c
o
n
c
e
n
t
r
a
t
I
o
n
[A]0
(b)
r
a
t
e
equilibrium achieved
[A]equilibrium
forward reaction
[B]equilibrium
reverse reaction
time
time
For reaction A  B, equilibrium is
achieved when the concentration
of A and B do not change. WHY?
the reason for the existence of an equilibrium
is in the graph above: the rate of the reverse
reaction becomes equal to the rate of the forward
(in the system above, the reaction was started with
reactants only)
 equilibrium can be reached from both directions: same equilibrium in terms of concentration of reactants
and products is established (provided temperature remains the same) when either starting with the
reactants only or when starting with products only. See graphs below: in both graphs the final
concentrations of all three chemicals should be the same.
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Equilibrium
ICl (g) + Cl2 (g)  ICl3
c
o
n
c
e
n
t
r
a
t
I
o
n
c
o
n
c
e
n
t
r
a
t
I
o
n
ICl
ICl3
Cl2
(a)
ICl
ICl3
Cl2
(b)
In (a) the equilibrium is approached beginning with ICl and Cl2 (reactants) in the ratio 1:1.
In (b) the equilibrium is approached beginning with ICl3 (product). In both graphs, the concentrations at
equilibrium are the same.
Simulations which show that the same equilibrium can be reached from both reactants and products:
 http://www.tutorvista.com/content/chemistry/chemistry-ii/chemical-equilibrium/chemical-equilibriumanimation.php: this one can also be used to show effect of temperature.
 http://www.chm.davidson.edu/ronutt/che115/equkin/equkin.htm
 macroscopic properties remain constant e.g. concentration and/or colour remain the same, no more gas
produced, etc; but two opposing micro-processes are going on simultaneously;
 the system needs to be closed; a closed system is a system which does not exchange any matter or
energy with its surroundings;
 can involve chemical changes (e.g. decomposition of calcium carbonate; dissolving of solutes) and
physical changes (e.g. evaporation of bromine: rate of evaporation equals rate of condensation);
 the time it takes for a reaction to achieve its equilibrium position is determined by the kinetics barrier of
the reaction (= activation energy); this barrier might even prevent the attainment of an equilibrium. This
can be overcome by for instance increasing the temperature or adding a catalyst or both;
7. 2. Position of equilibrium in homogenous systems
7.2.1. Deduce the equilibrium constant expression (Kc) from the equation for a homogeneous reaction.
7.2.2 Deduce the extent of a reaction from the magnitude of the equilibrium constant.
7. 2.3 Apply Le Chatelier’ s principle to predict the qualitative effects of changes of temperature, pressure and concentration
on the
position of equilibrium and on the value of the equilibrium constant.
7. 2. 4 State and explain the effect of a catalyst on an equilibrium reaction.
7. 2. 5 Apply the concepts of kinetics and equilibrium to industrial processes.
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Equilibrium expression
For most of our study we are only considering homogenous systems (a system in which all species,
reactants and products, are in the same phase e.g. systems involving all gases or systems involving
aqueous solutions).
In any reversible system, at any moment in time there are reactant particles and product particles.
The position of the equilibrium refers to the ratio of reactant particles to product particles at the time of
equilibrium.
To indicate this position accurately, chemists use a quantity called equilibrium constant. The constant
indicates how far a reaction has gone to completion; it shows in which direction the equilibrium is or has
moved.
The constant is specific to a given reaction, changes with temperature and can be calculated using the
equilibrium expression for that reaction.
For a given reaction:
aA (aq) +
 cC (aq) +
bB (aq)
dD (aq)
the equilibrium expression becomes:
[C]c [D]d
Kc =
[A]a [B]b
The equilibrium constant and the equilibrium expression is therefore related to the stoichiometric equation
but does not depend on the reaction mechanism (unlike the rate constant, k) or starting concentrations of
either reactants or products.
Kc tells how far a reaction has gone but does not tell us anything about how fast a reaction goes.
Kc is used when the species involved are in aqueous solutions and their concentrations are used. However,
we can also use concentrations for gases and therefore use the Kc.
How does Kc indicate position of equilibrium?
We should really accept that each reaction, including those going to a virtual completion and those
seemingly not starting at all, have a tendency to attain a position of equilibrium and have therefore a K value.
 If Kc 1 than [products] exceed [reactants] at equilibrium (products or forward reaction is favoured –
equilibrium is towards the right);
 if Kc  1 than reaction goes almost to a virtual completion; the reverse process is minimal; the larger Kc
the more spontaneous the forward reaction and the less spontaneous the reverse reaction;
 If Kc  1 than reactants exceed products at equilibrium; small amount of reactant converted to product –
equilibrium to the left;
 if Kc  1 than reaction hardly proceeds and does apparently not occur at all;
The unit of K depend on the equilibrium expression but usually it is not included as we are only interested in
the value of Kc .
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Factors affecting equilibria
As chemists we are able to shift the position of an equilibrium in a reversible reaction to suit our purposes;
we can do this by changing conditions which favour either the forward reaction or the reverse before a new
equilibrium is reached.
Favouring a reaction means that that reaction will suddenly occur at a rate higher than the other reaction;
however, the other will eventually catch up. They will then again both go on at the same rate again and
concentrations remain constant again but at different levels i.e. reactants more and products less or the
other way around.
To predict the effect of a change of a condition or external factor on an equilibrium reaction, Le Chatelier’s
Principle needs to be used.
Le Chatelier’s Principle:
If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one the
components, the system will shift its equilibrium position so as to oppose the effect of the change.
Change in concentration in one of the components after equilibrium has been established:
Use the following simulation to study the effect of a change in amount or concentration on the following
equilibrium system:
C (s) + H2O (g) CO (g) + H2 (g)
Link: http://www.chm.davidson.edu/vce/equilibria/LeChatelier.html
Observations:

An increase in the amount, and therefore concentration, of hydrogen, which is a product in this reaction,
results in an increase in the amount or concentration of the reactants.

A decrease in the amount of hydrogen results in an increase in carbon dioxide which is the other product.

An increase in the amount of water produces an increase in the amount of product.
Explanation:
In all three cases, the system responds to compensate the change imposed, i.e.:


when hydrogen is increased, the forward reaction is favoured to compensate for the increase by using up
hydrogen at a faster rate increasing the amount of product
when hydrogen is decreased the reverse reaction is favoured to compensate by making more hydrogen
On a more complex level, an increase in [reactant] results in a lowering of the ratio of product over reactant
(the equilibrium expression gives a value lower than the equilibrium constant; this value can be called the
reaction quotient) so the equilibrium shifts to the right (the forward reaction is increased) to decrease the
[reactant] and to increase [products] to increase the value of reaction quotient back to the value of the
equilibrium constant at that temperature.
Pressure:
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Equilibrium
The table below shows how the percentage of ammonia in the equilibrium mixture changes as the pressure
is changed on the following system:
3H2 (g)

+ N2 (g)
Total pressure (atm)
Percentage of NH3 at equilibrium at 723K
2NH3 (g)
1
0.24
2
9.5
100
16.2
200
25.3
Conclusion:
When the pressure is increased the equilibrium shifts to the right.
Explanation:
If pressure is increased, the equilibrium shifts to the side with the fewest number of molecules to
reduce the pressure. Increased pressure pushes the particles in the system closer together, this can be
opposed by favouring the reaction which produces fewer particles reducing the pressure in the system. At
the new equilibrium both reactions will go faster.
Decrease in pressure favours the reaction (or the equilibrium shifts to the side) which produces the
larger number of particles. At the new equilibrium both reactions will go slower.
The pressure in a closed system can be changed by changing the volume of the reacting vessel or by
adding an unreactive gas like helium to the reacting mixture – this increases the pressure as it adds more
particles to the same volume.
Catalyst:
A catalyst does not affect the equilibrium constant or the position of the equilibrium. It only shortens the time
it takes for the system to reach its equilibrium as it lowers the activation energy with the same amount for
both the reverse and the forward reaction. Catalysts speed up the rates of both forward and reverse
reactions EQUALLY!!!
Temperature:
Use the following simulation to study the effect of a change in temperature on the following equilibrium
system:
C (s) + H2O (g) CO (g) + H2 (g)
ΔHrxn = + 131.3 kJ mole-1
http://www.chm.davidson.edu/vce/equilibria/Temperature.html
Alternatively, consider the data in the table below:
3H2 (g) + N2 (g 
2NH3 (g)
ΔHrxn = - 92 kJ mole-1
T (K)
400
500
600
700
800
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N2O4 (g) 
2NO2 (g)
ΔHrxn = + 57 kJ mole-1
Kp
1.0 x 102
1.6 x 10-1
3.1 x 10-3
6.3 x 10-5
7.9 x10-6
T (K)
200
300
400
500
600
Kp
1.9 x 10-6
1.7 x 10-1
5.1 x 10
1.5 x 103
1.4 x 104
5 hours
2SO2 (g) + O2 (g) 
2SO3 (g)
ΔHrxn = - 197 kJ mole-1
T (K)
600
700
800
900
1000
Kp
3.2 x 103
2.0 x 102
3.1 x 10
6.3
20
Equilibrium
Conclusion:

If temperature is increased the equilibrium shifts to the side which is endothermic; the
endothermic reaction is favoured as it uses up the extra heat.
This can decrease or increase the Kc.


If the forward reaction is endothermic than Kc increases as the yield of the product increases;

if the reverse reaction is endothermic than Kc will decrease as the yield of product decreases.
Decrease in temperature favours the exothermic reaction as the system will release heat to replace
the heat removed.
Application of equilibrium principle: Haber Process + Contact Process
Haber Process:
3H2 (g)
+ N2 (g)
Contact Process: 2SO2 (g)
+ O2 (g)


2NH3 (g)
2SO3 (g)
In both processes the position of the chemical equilibrium is moved to the right by the same conditions.
This is the case because in both reversible systems, the forward reaction is exothermic and produces
fewer number of moles.
So the conditions needed are:

low temperatures; however, in reality they do not do the reaction at very low temperatures because,
although it would give a high yield, it would be very slow;

high pressure; although the higher the pressure the higher the yield, the process is not carried out at the
highest pressure because that would make the process too expensive as vessel and pipes would need to
be made a lot thicker to withstand very high pressures.

catalyst: Fe in the Haber Process and vanadium oxide in the Contact Process
In reality, the Haber process for instance is carried out at 450 C and 250 atmospheres.
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IB questions
1. (M07)
2. (M07)
3. (N06)
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4. (N06)
5. (M06)
6. (M06)
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7. (N05)
8. (N05)
9. (M05)
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10. (M05)
11. (N04)
12. (N04)
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13. (M04)
14. (M04)
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15. Which statement is true about chemical reactions at equilibrium?
A.
B.
C.
D.
The forward and backward reaction proceed at equal rates.
The forward and backward reactions have stopped.
The concentrations of the reactants and products are equal.
The forward reaction is exothermic.
16. Chemical equilibrium is referred to as dynamic because, at equilibrium, the
A.
B.
C.
D.
Equilibrium constant changes.
Reactants and products keep reacting.
Rates of the forward and backward reactions change.
Concentrations of the reactants and products continue to change.
17. What is the equilibrium expression for the reaction N2 + 3H2 
2NH3 ?
[NH3]
A.
Kc =
[H2]a [N2]
2 [NH3]
B.
Kc =
[H2]a [N2]
2[NH3]
C.
Kc =
[H2]a 3[N2]
[NH3]2
D.
Kc =
[H2]3 [N2]
18.
For a reaction which goes to completion, the equilibrium constant, Kc , is:
A.  1
B.  1
C. = 1
D. = 0
19. The equilibrium constant for the reaction below is 1.0 x 10-14 at 25 C and 2.1 x 10-14 at 35 C. What can
be concluded from this information?
2H2O (l)  H3O+ (aq) + OH- (aq)
A. [H3O+ ] decreases as the temperature is raised.
B. [H3O+ ] is greater than [OH- ] at 35 C.
C. Water is a stronger electrolyte at 25 C.
D. The ionization of water is endothermic
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Equilibrium
20. Ethanol is manufactured from ethene using the reaction below:
C2H4 (g) + H2O (g)
 C2H5OH (l)
H = - 46 kJ
Which conditions favour the highest yield of ethanol?
A.
B.
C.
D.
High pressure and low temperature
High pressure and high temperature
Low pressure and low temperature
Low pressure and high temperature
PAPER 2
1. (M07) Ammonia is produced by the Haber process according to the following reaction.
N2 (g) 3H2 (g) 2NH3 (g)
H is negative.
(a)
State the equilibrium constant expression for the above reaction.
(b)
Predict, giving a reason, the effect on the position of equilibrium when the pressure in the reaction
vessel is increased.
[2]
(c)
State and explain the effect on the value of Kc when the temperature is increased.
(d)
Explain why a catalyst has no effect on the position of equilibrium.
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[1]
[2]
[1]
Equilibrium
2. (M06)
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3. (N05)
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4.
(N04)
5. (M02) Ammonia is made on a large scale by the Haber process. The main reaction occurring is
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Equilibrium
3H2 (g)
+ N2 (g)

H = -92 kJ mol-1
2NH3 (g)
(a) State two characteristics of a reversible reaction at equilibrium.
[2]
(b) This reaction is described as homogeneous. State what is meant by the term homogeneous.
[1]
(c) Write the equilibrium constant expression for the reaction.
[2]
(d) When nitrogen and hydrogen are mixed together at room temperature and atmospheric
pressure the reaction is very slow. In industry, typical values of pressure and temperature used are
250 atmospheres and 450 C.
(i) State the effects on both the rate of reaction and the value of the equilibrium constant of
increasing the temperature.
[2]
(ii) State the effects on both the rate of reaction and the value of the equilibrium constant of
increasing the pressure.
[2]
(iii) Suggest why a pressure of 1000 atmospheres is not used.
[1]
(e) Name the catalyst used in the Haber process. State and explain its effect on the value of the
equilibrium constant.
[3]
(f) Use the collision theory to explain the effect of increasing the temperature on the rate of reaction
between nitrogen and hydrogen.
[3]
(g) A mixture of nitrogen and hydrogen is left at 450 C and 250 atmospheres until equilibrium is
reached. Use Le Chatelier.s principle to state and explain what will happen to the position of
equilibrium when
(i) some of the ammonia is removed;
(ii) the pressure is increased.
[2]
[2]
6. (M04) Consider the following equilibrium reaction.
2SO2 (g)
+ O2 (g)

2SO3 (g)
H = -198 kJ mol-1
Using Le Chatelier’s Principle, state and explain what will happen to the position of equilibrium if
(a) the temperature increases.
[2]
(b) the pressure increases.
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Equilibrium
Mark scheme
Paper 1
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
Paper 2
1. (M06)
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2. (N05)
3. (N04)
3. (M02)
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5. (M04)
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Equilibrium
6. (M03)
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