AQA A Level Chemistry
AQA A Level Chemistry 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc Exam Questions
Free AQA A Level Chemistry exam questions for 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc, with a mark scheme for every pack. Covers dynamic equilibrium, Le Chatelier's principle, industrial compromise conditions, the equilibrium constant Kc and its calculations.
3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc exam questions
1 question pack of real past exam questions, each with its mark scheme. Do not stop after a few: answer as many as you can without notes, then mark every answer with the mark scheme. The more questions you do, the more quickly you will spot how examiners ask about each idea.
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About AQA A Level Chemistry 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc
This topic introduces reversible reactions and how they behave once a dynamic equilibrium is reached. On the AQA A Level course you need to apply Le Chatelier's principle to changes in concentration, pressure and temperature, explain why catalysts do not change the position of equilibrium, and discuss compromise conditions in industrial processes such as ammonia, methanol and ethanol production. You also write expressions for Kc, calculate Kc and equilibrium amounts, and know that only temperature changes the value of Kc. It links to kinetics, Kp in 3.1.10 and acid dissociation constants in 3.1.12. Examiners favour Kc calculations from initial amounts and explanations of yield versus rate.
What you need to know
- Dynamic equilibrium In a closed system, the forward and reverse reactions occur at the same rate, so the concentrations of reactants and products stay constant.
- Le Chatelier's principle If a factor affecting an equilibrium is changed, the position of equilibrium shifts to oppose the change. Higher pressure favours the side with fewer gas molecules; higher temperature favours the endothermic direction.
- Catalysts A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached faster but the position of equilibrium and the yield do not change.
- Industrial compromise Conditions balance yield, rate, cost and safety. A lower temperature may favour an exothermic forward reaction but makes the rate too slow; very high pressure increases yield but is expensive and hazardous.
- Kc expression For aA + bB ⇌ cC + dD, Kc = [C]^c[D]^d / [A]^a[B]^b using equilibrium concentrations. Work out units by cancelling mol dm-3 terms; some Kc values have no units.
- Kc calculations Set up initial, change and equilibrium amounts using the mole ratio, divide by volume to get concentrations, then substitute. Only temperature changes Kc: if the forward reaction is exothermic, raising temperature lowers Kc.
Common mistakes in 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc
- Saying a catalyst shifts the equilibrium or increases the yield. It only increases the rate at which equilibrium is reached.
- Stating that changing concentration or pressure changes Kc. Only temperature changes the value of Kc.
- Substituting initial amounts instead of equilibrium amounts into the Kc expression.
- Forgetting to divide moles by volume when the volume is not 1 dm3 and the moles do not cancel.
- Giving a direction of shift without the reason. Say what the system opposes, for example 'fewer moles of gas on the right, so the position shifts right to reduce the pressure'.
Try three 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc exam-style questions
1. State and explain the effect of increasing the total pressure on the position of equilibrium for N2(g) + 3H2(g) ⇌ 2NH3(g). (2 marks)
Show the mark scheme answer
The position of equilibrium shifts to the right; there are fewer moles of gas on the right (2 compared with 4), so this opposes the increase in pressure.
2. 0.500 mol of H2 and 0.500 mol of I2 were sealed in a 1.00 dm3 flask. At equilibrium, 0.780 mol of HI was present. H2(g) + I2(g) ⇌ 2HI(g). Calculate Kc. (3 marks)
Show the mark scheme answer
Moles of H2 and I2 reacting = 0.390, so 0.110 mol of each remains; Kc = [HI]^2 / [H2][I2] = 0.780^2 / (0.110 x 0.110); Kc = 50.3 (no units).
3. The forward reaction of an equilibrium is exothermic. State and explain what happens to the value of Kc when the temperature is increased. (2 marks)
Show the mark scheme answer
Kc decreases; the equilibrium shifts in the endothermic (reverse) direction to absorb heat, so the ratio of products to reactants falls.
3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc FAQs
What does AQA A Level Chemistry 3.1.6 Chemical Equilibria cover?
It covers dynamic equilibrium, Le Chatelier's principle, the effect of catalysts, industrial compromise conditions, and writing and calculating the equilibrium constant Kc.
Which paper is 3.1.6 Chemical Equilibria examined in?
It can be tested in both Paper 1 and Paper 2, and Paper 3 can include it as well.
How do I set out a Kc calculation?
Write the balanced equation, then a table of initial moles, change and equilibrium moles using the mole ratio. Convert to concentrations if needed, substitute into the Kc expression and work out the units by cancelling.
