Equilibrium Revision
Writing equilibrium constant expressions
Answers
What does K tell us?
Le Chatelier’s principle
Question One – Le Chatelier’s Principle
Answers to Question One
Question Two – Le Chatelier’s Principle
Answers to Question Two
Question Three – Le Chatelier’s Principle
Answers to Question 3
Questions – K values
Answers – K values
The moral
Writing equilibrium
constant expressions
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Write the Equilibrium constant expressions
for the following reactions.
1. N2(g) + 3H2(g)  2NH3(g)
2. 2SO2(g) + O2(g)  2SO3(g)
3. 2NO2(g)  N2O4(g)
4. CaCO3(s)  CO2(g) + CaO(s)
5. 2Ag+(aq) + Cu(s)  Cu2+(aq)+ 2Ag(s)
6. HCl (aq) +H2O(l)  H3O+(aq) + Cl-(aq)
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Answers
1
[NH3]2
[N2] x [H2]3
2
[SO3]2
[O2] x [SO2]2
3
[N2O4]
[NO2]2
4
[CO2]
5
[Cu2+]
[Ag+]2
6
[H3O+] [Cl-]
[HCl]
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What does K tell us?
 If K is large, it tells us the position of
equilibrium lies to the right – product favoured
reaction
 If K is small, it tells us the position of
equilibrium lies to the left – reactant favoured
reaction
 If K is 1, it tells us the position of equilibrium
lies in the centre – equal products and
reactants
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Le Chatelier’s principle

1.
2.
3.
4.
If a change is imposed on a system at equilibrium,
the system reacts to try and minimise the change.
An increase in concentration will produce a move in
the direction to reduce the concentration
An increase in pressure (by decreasing the volume)
will produce a move in the direction of the smallest
number of moles of gas
An increase in temperature will produce a movement
in the endothermic direction
A catalyst doesn’t produce a move – but speeds up
both forward and reverse reactions
Question One – Le
Chatelier’s Principle
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In the gas phase equilibrium described by the equation
below, state what would happen to the equilibrium
and give reasons for your answer.
N2O4(g) 
2NO2(g)
Kc = 4.6 x 10-3
1. Some extra N2O4 gas is injected into the equilibrium mixture.
2. Increasing the pressure of the equilibrium mixture.
3. Adding a catalyst.
4. N2O4 gas is colourless and NO2 gas is brown.
Describe any colour change noticed when the pressure
on the equilibrium is decreased.
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Answers to Question One
1. More NO2 would form/Reaction will shift to the right.
Increasing the concentration of the N2O4 will increase the rate of the
forward reaction and cause more products to be formed.
2. More N2O4 will be formed/Reaction will shift to the left.
In this gas phase reaction there are two moles of gaseous products
compared to one mole of gaseous reactants so increasing the
pressure will shift the equilibrium to the side with fewer gaseous
molecules.
3. The catalyst will not affect the position of the equilibrium.
However it will increase the rate of both the forward and reverse
reactions causing it to reach equilibrium much faster.
4. Decreasing the pressure will cause the mixture to be a darker
brown.
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Question Two – Le
Chatelier’s Principle
For the reaction in Q1 [N2O4(g) 
the ∆rH value is 54 kJmol-1.
2NO2(g)]
1. Is this reaction exothermic or endothermic?
2. Discuss what you would observe if an
equilibrium mixture of these two gases was
heated.
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Answers to Question Two
1. Endothermic.
2. As the mixture was heated the colour
would become a darker brown.
Heat will favour the endothermic forward
reaction producing more NO2 gas and
increasing the intensity of the brown colour
in the mixture.
Question Three – Le
Chatelier’s Principle
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In the gas phase equilibrium described by the equation
below, state what would happen to the equilibrium
and give reasons for your answer.
N2(g) + 3H2(g)

2NH3(g)
∆rH = - 92 kJmol-1.
1. The reaction mixture was cooled.
2. The pressure was increased.
3. A catalyst was added.
4. Ammonia was removed.
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Answers to Question 3
1. Cooling the reaction mixture will increase the equilibrium
concentration of NH3 gas. This is because the exothermic
forward reaction will be favoured by the removal of heat.
2. Increasing the pressure will increase the equilibrium
concentration of NH3 gas. This is because the forward
reaction involves a reduction in the number of gaseous
molecules from 4 to 2. This will be favoured by an increase in
pressure.
3. The catalyst will not affect the position of the equilibrium.
However it will increase the rate of both the forward and
reverse reactions causing it to reach equilibrium much faster.
4. Removal of ammonia will cause more ammonia to be
produced. Removing ammonia decreases the rate of the
reverse reaction and so the forward reaction will
predominate.
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Questions – K values
For the equilibrium:
2SO2(g) +
O2(g) 
2SO3(g)
DH -ve
What would happen to the value of the equilibrium
constant, K if;
1. The concentration of SO2 was increased
2. The total pressure was increased by decreasing the
volume of the container
3. The temperature was increased
4. A catalyst was introduced
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Answers – K values
1. No change – all concentrations return to the
position of equilibrium at this temperature
2. No change – changing the volume changes
the concentration of all components – the
ratio will be the same.
3. The K value would decrease as the
concentration of reactants increases and the
concentration of products decreases
4. No change – both forward and reverse
reactions speed up but there is no change in
K
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The moral
 Only a change in temperature affects the
value of K.