When we have a chemical reaction that has reached a state of dynamic equilibrium, there will be one particular set of equilibrium concentrations of reactants and products that is described by the position of equilibrium.
If we change the conditions of the reaction or system (such as the initial concentrations of reactants and products, temperature, pressure) and allow it to re-establish equilibrium, there will be a new position of equilibrium. A different set of equilibrium concentrations of reactants and products.
By the late 1880s, Henri Le Chatelier had studied enough equilibrium reactions to state that …
when an external change is made to a system in dynamic equilibrium, the system responds to minimise the effect of the change
Not only can we use Le Chatelier’s principle to explain how changing the conditions of a system at equilibrium affects the position of equilibrium, we can use it to manipulate reversible reactions to favour a particular product. This is exceedingly useful in the chemical industry.
Consider a bottle of sparkling water. It is sparkling because there is carbon dioxide dissolved in the water under considerable pressure.
Some of the dissolved carbon dioxide reacts with water to form hydrogen carbonate ions.
CO2(aq) + H2O(l) ⇌ HCO3–(aq) + H+(aq)
Once the reactions have been allowed to establish equilibrium at a particular pressure and temperature, the position of equilibrium describes the concentrations of CO2(aq), HCO3–(aq), H+(aq) as well as the partial pressure of CO2(g) present in the bottle, our closed system.
What happens when we remove the cap from the bottle, count to 5 and then replace it? Clearly the pressure in the system is reduced and some of the CO2(g) escapes. The system frantically tries to minimise the effect of this change in the external conditions by increasing the rate of both back reactions.
After some time, a new equilibrium will be established where the rate of each forward reaction equals the rate of each back reaction. There is a new position of equilibrium in which the concentrations / partial pressures of all the substances is lower than it was previously, before we opened the bottle.
Let’s look at the effect of changing the concentration, pressure and temperature on the position of equilibrium with some specific examples.
- Changing the concentration
Iron(III) ions react with thiocyanate ions to form an iron thiocyanate complex, [Fe(SCN)]2+ in a reversible reaction. We can estimate the concentration of [Fe(SCN)]2+ by the colour of the mixture.
When equilibrium is established, the rate of the forward reaction equals the rate of the back reaction and the concentration of reactants and products is constant (but not necessarily equal).
What happens if we add more Fe3+(aq) to the equilibrium mixture?
This is the equivalent of increasing the concentration of Fe3+(aq) in the system.
- Le Chatelier’s principle states that the system will work to minimise / oppose this change
- the position of equilibrium shifts to the right, temporarily increasing the rate of the forward reaction to use up the excess Fe3+ (this will also use up SCN– and produce more [Fe(SCN)]2+)
- the mixture will look a deeper, darker red colour
- a new position of equilibrium is established with higher concentrations of Fe3+ and [Fe(SCN)]2+ but a lower concentration of SCN– than in the original equilibrium mixture
Language is crucial to picking up the marks in exams. This is often a long answer question and so many students tie themselves up in knots trying to explain Le Chatelier’s principle 😳. You should use mark schemes for your exam board to write some ‘perfect’ answers using the exact same wording that they use to describe a shift in the position of equilibrium and defining key terms, and learn them!
2. Changing the pressure
The pressure of a gas is proportional to the amount of gas (in terms of molecules of gas) present in a reaction system.
Imagine that we place a mixture of nitrogen, hydrogen and ammonia in a gas syringe at constant temperature and seal the nozzle. After a period of time, the mixture will establish equilibrium.
N2(g) + 3H2(g) ⇌ 2NH3(g)
What happens if we push the plunger in?
This will increase the pressure in the system.
- according to Le Chatelier’s principle, the system will move to minimise / oppose the change
- the position of equilibrium shifts to the side with fewer moles of gas (reducing the pressure in the syringe), in this case, it shifts to the right
reactants (4 moles) ⇌ products (2 moles)
- a new position of equilibrium is established in which there are fewer moles of hydrogen and nitrogen, and more moles of ammonia than in the original equilibrium mixture
3. Changing the temperature
Nitrogen(IV) oxide is a brown gas that exists in equilibrium with colourless dinitrogen tetroxide.
2NO2(g) ⇌ N2O4(g) ΔrH⦵ = -57 kJ mol-1 (always quoted for the forward reaction)
brown colourless
ΔrH⦵ tells us that the forward reaction is exothermic (gives out heat) and the back reaction is endothermic (absorbs heat).
What happens if we place the sealed gas syringe in hot water?
- increasing the temperature of the system causes the position of equilibrium to shift to the left in the endothermic direction to oppose the change (endothermic reactions absorb energy from their surroundings, hence using up some of the excess heat)
- the mixture becomes a deeper brown colour
What happens if we place the sealed gas syringe in iced water?
- decreasing the temperature of the system causes the position of equilibrium to shift to the right in the exothermic direction to oppose the change (exothermic reactions produce heat)
- the mixture becomes paler in colour
If we were trying to produce N2O4 on an industrial scale we would ideally have our gaseous mixture at a very low temperature so that the position of equilibrium was shifted to the far right BUT cold temperatures mean we have a very slow rate of reaction, which is not economical …
If we were trying to produce NO2 on a an industrial scale we would want a high temperature so that the position of equilibrium was shifted to the far left BUT high temperatures cost energy, lead to high pressures and there are safety implications …
We always have to compromise in real life!
4. Using a catalyst
Catalysts increase the rate of a reaction by lowering the activation energy, working equally on the rate of both the forward and back reactions. We can use a catalyst to achieve equilibrium more quickly but it would not affect the position of equilibrium.
Practice questions
- The Contact process is used to manufacture sulphur trioxide:
2SO2(g) + O2(g) ⇌ 2SO3(g) ΔrH⦵ = -197 kJ mol-1
Which of the following changes in the conditions of the reaction would move the position of equilibrium to the right?
(a) increasing the temperature
(b) increasing the concentration of oxygen
(c) decreasing the pressure
(d) adding a catalyst
2. The Bosch process is used to produce carbon monoxide and hydrogen:
C(s) + H2O(g) ⇌ CO(g) + H2(g) ΔrH⦵ = +131 kJ mol-1
Using le Chatelier’s principle, determine the conditions of temperature and pressure that could be used to obtain the maximum yield.
3. Under certain conditions, nitrogen and oxygen react to form nitrogen(II) oxide:
N2(g) + O2(g) ⇌ 2NO(g)
(a) A student states that when equilibrium is achieved, the rate of the forward reaction equals the rate of the back reaction so the concentrations of nitrogen, oxygen and nitrogen(II) oxide will also be equal. Comment on this statement.
(b) Nitrogen(II) oxide can also be made from ammonia:
4NH3(g) + 5O2(g) ⇌ 4NO(g) + 6H2O(g)
What will be the effect, if any, of increasing the pressure on the yield of nitrogen(II) oxide? Use le Chatelier’s principle to help answer the question.
4. Ammonia is made in the Haber process at a temperature of 450°C, 200 atm pressure and an iron catalyst:
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔrH⦵ = -92 kJ mol-1
It is suggested that the process be run at 600°C and 250 atm pressure with the iron catalyst. Evaluate these changes in terms of their effect on the yield of ammonia, the rate and economics of the reaction.
5. Antimony(III) chloride can be made by dissolving antimony oxychloride in hydrochloric acid with a concentration of 6 mol dm-3.
SbCl3(aq) + H2O(l) ⇌ SbClO(s) + 2HCl(aq)
(a) Suggest why the solution is prepared using 6 mol dm-3 hydrochloric acid.
(b) What would be observed if
(i) the solution was diluted?
(ii) solid sodium hydroxide was added?
6. When excess barium sulphate is shaken with water, a saturated solution quickly forms and the mixture reaches equilibrium.
BaSO4(s) ⇌ Ba2+(aq) + SO42-(aq)
Sketch a graph to show how the concentration of Ba2+(aq) changes when excess solid barium sulphate is shaken with water and the equilibrium established.
Answers
- (a) the forward reaction is exothermic so an increase in temperature would shift the equilibrium position to the left (in the endothermic direction)
(b) increasing the concentration of oxygen would shift the equilibrium position to the right as the reaction moves to oppose the change in conditions
(c) decreasing the pressure would would shift the equilibrium position to the left to the side with more moles of gas (three moles of reactants versus two moles of product)
(d) adding a catalyst has no effect on the equilibrium position, only the rate at which equilibrium is reached
2. To obtain the maximum yield of carbon monoxide and hydrogen we need to shift the position of equilibrium to the right in the endothermic direction which would be achieved by running the process at a high temperature. There are more moles of gas on the product side so lowering the pressure would shift the equilibrium position to the right (but this would also decrease the rate of the reaction).
3. (a) The first statement is correct, the second is incorrect. At equilibrium, the concentrations of reactants and products remains constant but not necessarily equal.
(b) There are 9 moles gas on the left side and 10 moles of gas on the right. Increasing the pressure will shift the position of equilibrium to the left to oppose the change to the side with fewer moles of gas, decreasing the yield of NO.
4. A higher temperature will shift the position of equilibrium to the left in the endothermic direction, decreasing the yield of ammonia, however it would increase the rate of the reaction.
A higher pressure will shift the position of equilibrium to the right to the side with fewer moles of gas increasing the yield of ammonia and would also increase the rate of the reaction (although it is possible that increasing the pressure has no further effect on the rate if the active sites of the catalyst are already full).
Running an industrial process at a higher temperature would incur higher energy costs. High pressure incurs higher running and equipment costs. It would seem that the engineer’s suggestions would not lead to a more viable process overall.
5. (a) A very high concentration of HCl will give enough HCl to react with the SbClO and as it will be in excess, it causes the position of equilibrium to shift to the left to oppose the change (according to le Chatelier’s principle) so that there is a greater concentration of SbCl3 in solution at equilibrium.
(b) (i) if the solution was diluted no change would be observed as the concentration of both reactants and products is reduced equally (water is already in excess when there are substances in solution).
(ii) NaOH will react with the HCl, neutralising it. The decrease in HCl concentration causes position of equilibrium to shift to the right and SbClO would be seen precipitating out of solution.
6.
