Remember: The
only thing which
changes the value of
Kc for a particular
reaction is the
temperature.
Industrial Application of
Le Chatelier’s Principle
Goal:
a- Maximize the yield by choosing conditions
that will cause the equilibrium to move to the right.
(Considering the equilibrium)
b- Maximize the rate of the reaction (Considering the kinetics)
c- Sometimes these two criteria work against each other.
d- Chemists must choose the best compromise between them.
The Haber process:
N2(g) + 3H2(g)  2NH3(g)
H = -93 KJ mol-1
Applying Le Chatelier’s principle, we can therefore consider the optimum
conditions for this reaction:

Concentration: the reactants nitrogen and hydrogen are supplied in the
molar ration 1:3 in accordance with their stoichiometry in the equation. The
product ammonia is removed as it forms, thus helping to pull the
equilibrium to the right and increasing the yield.

Pressure: as the forward reaction involves a decrease in the number of
molecules, it will be favoured by a high pressure. The usual pressure
used in the Haber process is about 200 atmospheres.

Temperature: as the forward reaction is exothermic, it will be favoured by
a lower temperature. However, too low a temperature would cause the
reaction to be uneconomically slow and so a moderate temperature of
about 450C is used.

Catalyst: although a catalyst will not increase the yield of ammonia, it will
speed up the rate of production and so help to compensate for the
1
moderate temperature used. A catalyst of finely divided iron is used, with
small amounts of aluminum and magnesium oxides added to improve
its activity.
In questions on industrial processes
involving chemical equilibria, the
main focus of your answer should be
on the reasons for the optimum
conditions chosen, which should be
explained using the concepts of
equilibrium and kinetics. This is more
important than remembering too
many details about specific
conditions for different reactions.
2
The Contact process: Production of sulfuric
acid, H2SO4
The production of sulfuric acid involves series of three simple reactions:

The combustion of sulfur
S(s) + O2(g)  SO2(g)

The oxidation of sulfur dioxide
2SO2(g) + O2(g)  2SO3(g)
H=-196 KJ mol-1

The combination of sulfur trioxide with water. This usually takes place
indirectly by first absorbing the sulfur trioxide into a flowing solution of
concentrated sulfuric acid and then allowing the product of this reaction
to react with water. This avoids the problem caused by the violent
nature of the direct reaction between sulfur trioxide and water:
SO3(g) + H2SO4(l)  H2S2O7(l) + H2O(l)  2H2SO4(aq)
The overall rate of the process depends on the second reaction above, the
oxidation of sulfur dioxide. So applying the Le Chatelier’s principle to this step,
we can predict the conditions that will most favour the formation of product.
These are summarized in the table below.
2SO2(g) + O2(g)  2SO3(g) H=-196 KJ mol-1
Pressure
Influence of reaction
Conditions used
Forward reaction involves
2 atm (this gives a very high
reduction in the number of
yield, so still higher pressure is
molecules of gas: high
not needed)
pressure will favour the
product.
Temperature
Forward reaction is
450C
exothermic: low temperature
will increase the yield, but
decrease the rate.
Catalyst
Increases the rate of reaction
Vanadium (V) oxide, V2O5
3
4
Examples:
1- In the Haber process for the synthesis of ammonia, what effects does the
catalyst have?
Rate of formation of NH3(g)
Amount of NH3(g) formed
Increases
Increases
Increases
Decreases
Increases
No change
No change
Increases
2- 2SO2(g) + O2(g)  2SO3(g) H=-196 KJ mol-1
According to the above information, what temperature and pressure
conditions produce the greatest amount of SO3?
Temperature
Pressure
Low
Low
Low
High
High
High
High
Low
3- Predict how you would expect the value for Kc for the Haber process to
change as the temperature is increased. Explain the significance of this in
terms of the reaction yield.
4- Consider the following equilibrium reaction:
2SO2(g) + O2(g)  2SO3(g)
H=-196 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)
5
b- the pressure increases
(2)
5- The table below gives information about the percentage yield of ammonia
obtained in the Haber process under different conditions.
Pressure/atmosphere
TemperatureC
200
300
400
500
10
50.7
14.7
3.9
1.2
100
81.7
52.5
25.2
10.6
200
89.1
66.7
38.8
18.3
300
89.9
71.1
47.1
24.4
400
94.6
79.7
55.4
31.9
600
95.4
84.2
65.2
42.3
a- From the table, identify which combination of temperature and pressure gives
the highest yield of ammonia.
(1)
b- The equation for the main reaction in Haber process is:
N2(g) + 3H2(g)  2NH3(g)
H = -93 KJ mol-1
Use the information to state and explain the effect on the yield of ammonia of:
i)
increasing the pressure
(2)
ii)
increasing the temperature.
(2)
c- In practice, typical conditions used in the Haber process are a temperature of
500C and a pressure of 200 atmospheres. Explain why these conditions are
used rather than those that given the highest yield.
(2)
6
d- Write the equilibrium constant expression, Kc, for the production of ammonia.
(1)
Calculating the equilibrium constant from initial and
equilibrium concentrations (Using ICE Table)
1- Write the balanced equation
2- Under the equation, write in the values of the concentrations of each
component using three rows: initial, change, and equilibrium

Initial represent the concentration originally placed in the flask.

Unless stated otherwise, we assume the initial product concentration is
zero.

Change represent the amount that react to reach equilibrium

Equilibrium is the concentration present in the equilibrium mixture.

Equilibrium concentration = initial concentration  change in
concentration.
Example:
1- The equilibrium constant Kc for the reaction
SO3(g) + NO(g)  NO2(g) + SO2(g)
was found to be 6.78 at specific temperature. If the initial concentrations of NO
and SO3 were both 0.03 mol dm-3, what would be the equilibrium concentration of
each component?
Answers:
[SO3] = 0.0083 mol dm-3,
[NO] = 0.0083 mol dm-3,
[NO2] = 0.022 mol dm-3,
[SO2] = 0.022 mol dm-3
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Calculating the equilibrium concentrations when Kc is
very small

When Kc is very small (less than 10-3), it represent a reaction in which the
forward reaction has hardly processed and the equilibrium mixture
consists almost entirely of reactants.

In this case, the change in reactant concentrations is close to zero.

Therefore, [reactants]initial  [reactants]equilibrium
2- The thermal decomposition of water has a very small value of Kc. At
1000C, Kc = 7.3 x 10-18 for the reaction
2H2O(g)  2H2(g) + O2(g)
A reaction is set up at this temperature with an initial H2O concentration of
0.10 mol dm-3. Calculate the H2 concentration at equilibrium.
Answer: [H2]eqm = 5.3 x 10-7 mol dm-3
Manipulating Kc for different reaction equations
Effect on equilibrium
Effect on Kc
expression
Reversing the reaction
Inverts the expression
1/Kc or Kc-1
Doubling the reaction
Squares the expression
Kc2
Square root the
Kc
coefficients
Halving the reaction
coefficient
Adding together two
reactions
expression
Multiples the two
Kcx Kc
expressions
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Example:
3- If the equilibrium constant for the reaction
2HI(g)  H2(g) + I2(g)
is 0.04 at a certain temperature, what would be the value of the equilibrium
constant for the following reaction at the same temperature?
1
1
H2(g) + I2(g)  HI(g)
2
2
Solution
Kc =
[H 2 ][I 2 ]
[HI]
K c =
[HI]
=
[H 2 ]1/ 2[I 2 ]1/ 2
1
Kc
As the reaction is reversed and halved, the value of Kc becomes
So, Kc =
4- (a)
(
-1
(K c
1
) = 5.0
0.04
The equation for the decomposition of hydrogen iodide is:
2HI(g)  H2(g) + I2(g)
H = + 52 KJ
predict and explain the effect on the position of equilibrium of:
i)
increasing the pressure, at constant temperature
(2)
ii)
increasing the temperature, at constant pressure
(2)
iii)
adding catalyst, at constant temperature and pressure
(2)
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(b)
deduce the expression for Kc for the forward reaction (1)
(c)
the equilibrium formed during this reaction was investigated in two
experiments carried out at different temperatures. The results are shown in the
table below.
Experiment
Initial concentration
Equilibrium concentration
(mol dm-3)
(mol dm-3)
[HI]
[H2]
[I2]
1
0.06
0.00
0.00
2
0.00
0.04
0.04
i)
[HI]
[H2]
[I2]
0.01
0.04
for each experiment, deduce the concentrations of the other species
present at equilibrium. Calculate the values of Kc for the forward
reaction for each experiment.
ii)
(6)
Use the two calculated values of Kc to deduce which of the two
experiments was carried out at the higher temperature, and explain
your choice. (if you were not able to calculate the values of K c in
section (C)(i), assume that the values are 0.1 for experiment 1 and 0.2
for experiment 2. Although these are not the correct values.
(2)
10
11
12
Introducing ICE table
1- Hydrogen and carbon dioxide react as shown in the equation below.
H2(g) + CO2(g)  H2O(g) + CO(g)
For this reaction the values of Kc with different temperature are
Temperature /K
Kc
500
7.76 x 10-3
700
1.23 x 10-1
900
6.01 x 10-1
Which statement for the reaction is correct?
A
The forward reaction is endothermic
B
H2O(g) and CO(g) are more stable than H2(g) and CO2(g)
C
The reaction goes almost to completion at high temperatures
D The reverse reaction is favoured by high temperature
2- In the reaction below
N2(g) + 3H2(g)  2NH3(g)
H = -92 KJ
Which of the following changes will increase the amount of ammonia at
equilibrium?
I
Increasing the pressure
II
Increasing the temperature
III
Adding a catalyst
A
I only
B
II only
C
I and II only
D
II and III only
3- Which of the factors below affect the equilibrium vapour pressure of a
liquid in a container?
I
Temperature
II
Surface of the liquid
III
Volume of the container
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A
I only
B
I and II only
C
II and III only
D
I, II and III
1- The position of equilibrium in a reversible reaction is shifted to the right until it
reaches equilibrium again. Which statement must be true for the reaction when
the new position of equilibrium is reached?
A. The rate of the forward reaction is greater than the rate of the reverse reaction
B. The concentrations of reactants and products do not change
C. The concentrations of reactants and products are equal
D. The value of Kc is greater than 1
2- Which change will shift the position of equilibrium to the right in this reaction?
N2(g)+3H2(g)  2NH3(g)
ΔH = −92kJ
A. Increasing the temperature
B. Decreasing the pressure
C. Adding a catalyst
D. Removing ammonia from the equilibrium mixture
3- What is the effect of adding a catalyst to a reaction mixture at equilibrium?
A. It decreases the activation energy of the forward reaction and increases the
activation energy of the reverse reaction
B. It decreases both the activation energy and the enthalpy change of the forward
reaction
C. It decreases the activation energies of both forward and reverse reactions
D. It decreases the activation energies and enthalpy changes of both forward and
reverse reactions
4- 10.0 cm3 of liquid bromine is placed in an empty 100 cm3 bottle, which is then
sealed and left to reach equilibrium at room temperature. What happens
first?
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A. The rate of evaporation is greater than the rate of condensation
B. The rate of condensation is greater than the rate of evaporation
C. The rate of evaporation is equal to the rate of condensation
D. There is no evaporation or condensation
5- Which statement(s) about the following reaction at 100 !C is/are correct?
N2(g) + 3H2(g)  2NH3(g)
I.
Every collision between N2(g) and H2(g) molecules is expected to produce
NH3(g)
II.
This reaction must involve a collision between one N2(g) and three H2(g)
molecules.
A. I only
B. II only
C. Both I and II
D. Neither I nor II
6- The reaction
2NO2(g)  N2O4 (g)
is exothermic. Which of the following could be used to shift the equilibrium to
the right?
I. Increasing the pressure
II. Increasing the temperature
A. I only
B. II only
C. Both I and II
D. Neither I nor II
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