Equlibrium
HL and SL
7.1 Dynamic Equilibrium
Many chemical reactions go to completion i.e. they finish when
one or other of the reactants has been used up.
However, many other reactions do not go to completion and are
reversible.
e.g yellow chromate(VI) ions turn orange when acid is added as
dichromate(VI) ions are formed:
2CrO42- + 2H+  Cr2O72- + H2O
If sodium hydroxide is added to the orange solution, the reaction is
reversed and yellow chromate(VI) ions are formed:
Cr2O72- + 2OH-  2CrO42- + H2O
This can be summarised as:
2CrO42- + 2H+ Ý
Cr2O72- + H2O
Ý is used to show that the reaction is reversible.
The reaction from left to right is known as the forward
reaction.
The reaction from right to left is known as the backward or
reverse reaction.
Since the reaction continues in both directions, it is said to
be dynamic.
When the rate of the forward reaction is equal to the rate of
the backward reaction the chemical equilibrium has been
reached.
At chemical equilibrium, there are reactants and products
present at all times and, the concentration of the reactants
and products remains constant (not the same as each other)
so there is no apparent change occurring.
A state of dynamic equilibrium can only be reached in a
closed system. A closed system is one in which neither
matter nor energy can be lost or gained.
The position of equilibrium can be reached either
starting with the reactants or starting with the products.
7.2 The position of equilibrium
The value of the equilibrium constant, Kc gives an
indication of whether the equlibrium position is to the
right hand side (mostly products) or to the left hand
side (mostly reactants).
 high value (>> 1) for an equilibrium constant means
that there is a high theoretical yield of products and
the reaction almost goes to completion.
 small value (<<1) indicate that there are more
reactants than products and the reaction barely
proceeds.
Expressions for Kc
For homogenous equilibria i.e. where all reactants are
in the same phase (state).
• Kc – expressed in terms of concentrations measured
in moldm-3
Expressions for Kc can be derived from a balanced
chemical equation.
For a reaction
aA + bB
cC + dD
The equilibrium constant, Kc is given by
Kc =
[C]c[D]d
[A]a[B]b
where [X] = concentration of X in moldm-3
The units for Kc must be derived from the expression
for Kc, this is done in the same way as we derived
units for the rate constant, k.
Handy hint: 1/moldm-3 = mol-1dm3
Write an expression for Kc for the following equilibria
and state the units.
1. 2SO2 + O2
Kc =
2SO3
]2
[SO3
[SO2]2[O2]
2. 3H2 + N2
Kc =
[NH3]2
[H2]3[N2]
Click here to try quizzes.
Units:
concentration2
concentration3
= 1/concentration
= 1/moldm-3
= mol-1dm3
2NH3
Units:
concentration2
concentration4
= 1/concentration2
= 1/mol2dm-6
= mol-2dm6
Changing Conditions
We will consider the effect on an equilibrium of:
• change in concentration of reactants and products,
• increase or decrease in temperature,
• changing pressure (only affects reactions involving gases)
and,
• the addition of a catalyst.
To help:
Le Chatelier’s principle can help us to predict how
an equilibrium responds to a change in conditions.
A system at equilibrium will
react to oppose any change
imposed upon it.
Change of concentration of reactant or products
If the concentration of any species in an equilibrium mixture
is changed, then the concentration of the other substances
must also change to re-establish the equilibrium.
Addition of reactant / removal of a product – causes
equilibrium to move to right and more product is obtained.
Removal of a reactant / addition of a product – causes
equilibrium to move to the left and more reactant is
obtained.
Change in total pressure
Only has an effect on gaseous equilibria with differing
number of moles of gas on either side of the equation. So:
CH4(g) + H2O(g) Ý
3H2(g) + CO(g)
But not:
2HI(g) Ý
H2(g) + I2(g)
An increase in pressure causes the equilibrium to displace to
the side with the fewer moles of gas. Reducing the number
of moles reduces the pressure.
The reverse is true if there is a reduction in pressure.
Change in temperature
A change in temperature alters the rate of the forward and
backward reaction by different amounts so the position of the
equilibrium alters. However, we can still use Le Chatelier’s
principle.
Exothermic forward reaction – heat energy is released. An
increase in temperature will be opposed which means the reaction
moves in the endothermic direction i.e. to the left and there are
fewer products. However, as the temperature is higher
equilibrium is reached faster.
The reverse applies if temperature is decreased.
Endothermic forward reaction – heat energy is absorbed. An
increase in temperature will be opposed which means the
reaction moves in the endothermic direction i.e. to the right and
there are more products. However, as the temperature is higher
equilibrium is reached faster.
The reverse applies if temperature is decreased.
Now complete the table below:
Enthalpy
change
Change in
temperature
Exothermic
Increase
Exothermic
Decrease
Endothermic Increase
Endothermic Decrease
Displacement Yield of
of equilibrium product
Rate
Addition of a catalyst
Has no effect on composition of equilibrium mixture. A catalyst
increases the rate of the forward and backward reactions by the
same amount.
The equilibrium position is reached faster but the composition is
unchanged.
Importance of equilibria:
e.g. the Haber process
N2(g) + 3H2(g)
2NH3(g)
DH = - 92 kJmol-1
Aim: to obtain highest possible yield, in shortest possible time
for the lowest possible cost.
Pressure: 4 moles of reactant  2 moles of product so applying
Le Chatelier’s principle, a high pressure favours the formation
of ammonia. It also increases the rate. 2.0 x 104 kPa is used.
Higher pressures are uneconomic.
2.0 x 104 kPa is a good compromise between yield and cost.
Temperature: The forward reaction is exothermic so applying
Le Chatelier’s principle, if we increase the temperature the
yield of ammonia will decrease as the reaction moves in the
endothermic direction.
So the best equilibrium yield is obtained at a low temperature.
However, at low temperatures the rate is low.
A compromise is needed between rate and yield. The usual
operating temperature is 650 – 720 K.
Use of a catalyst: The catalyst used is iron, with potassium
hydroxide added to promote its activity.
The catalyst increases the rate of the forward and backward
reaction by the same extent.
e.g. The Contact Process (manufacture of sulphuric acid)
Essential reaction is the formation of sulphur trioxide from
sulphur dioxide:
2SO2(g) + O2(g) Ý 2SO3(g)
DH = -192 kJmol-1
What conditions would you choose?
High yield favoured by a high pressure and a low temperature.
High rate favoured by high pressure, high temperature and a
catalyst.
Plants operate at around 450 °C with catalyst of vanadium(V)
oxide, V2O5.
Although in theory a high pressure is required, in practice only 1
- 2 atmospheres is used as this gives a 98 % yield.