Reversible Reactions and Equilibrium Exam Questions
Practice IGCSE Chemistry exam questions on reversible reactions and equilibrium. Covers dynamic equilibrium, Le Chatelier's principle, the effect of temperature/pressure/concentration on equilibrium position, and the Haber process with mark schemes and examiner notes.
Published by IGCSEChemistry.com.my
Chemistry teaching team: K. S. Tan (15+ years teaching IGCSE Chemistry) and Ms Yash (10+ years teaching IGCSE Chemistry) and Ms Kartini (15+ years teaching IGCSE Chemistry).
Mapped to Cambridge IGCSE Chemistry 0620 (2026–2028). Last updated 2026-08-19.
These questions cover reversible reactions and equilibrium. Write full answers before checking.
Question 1 (3 marks, Supplement)
(a) State what is meant by a reversible reaction. (1)
(b) Define dynamic equilibrium. (2)
Mark scheme
(a) A reaction that can go in both the forward and reverse directions [1]
(b) A dynamic equilibrium is reached when the rate of the forward reaction equals the rate of the reverse reaction [1], so the concentrations of reactants and products remain constant (but both reactions are still occurring) [1]
Examiner note: “The reaction has stopped” is wrong — at equilibrium, both forward and reverse reactions continue at equal rates. The word “dynamic” means both reactions are still happening. Concentrations are constant, not equal.
Question 2 (4 marks, Supplement)
The Haber process involves the following equilibrium:
N2(g) + 3H2(g) ⇌ 2NH3(g), delta H = -92 kJ/mol
Using Le Chatelier’s principle, predict and explain the effect on the yield of ammonia of:
(a) Increasing the pressure. (2)
(b) Increasing the temperature. (2)
Mark scheme
(a) Increasing pressure shifts the equilibrium to the right / towards ammonia [1] because there are fewer moles of gas on the right (2 moles) than on the left (4 moles), so the system shifts to reduce pressure [1]
(b) Increasing temperature shifts the equilibrium to the left / away from ammonia [1] because the forward reaction is exothermic, and the system shifts to oppose the temperature increase by favouring the endothermic (reverse) reaction [1]
Examiner note: Le Chatelier’s principle states that if a system at equilibrium is disturbed, it shifts to oppose the change. For pressure: count moles of gas on each side. For temperature: identify which direction is exothermic/endothermic.
Question 3 (3 marks, Supplement)
In the Haber process, the conditions used are 450 degC, 200 atm, and an iron catalyst.
(a) Explain why 450 degC is used even though a lower temperature would give a higher yield. (2)
(b) State the role of the iron catalyst and its effect on the equilibrium position. (1)
Mark scheme
(a) A lower temperature would increase the yield but the rate of reaction would be too slow [1]. 450 degC is a compromise — it gives a reasonable yield at a reasonable rate [1].
(b) The catalyst increases the rate of both forward and reverse reactions equally, so equilibrium is reached faster, but it does not change the equilibrium position or yield [1]
Examiner note: The catalyst is the most commonly misunderstood factor. A catalyst does NOT shift the equilibrium or change the yield — it only speeds up how quickly equilibrium is reached. The compromise between rate and yield is a key concept.
Question 4 (3 marks, Supplement)
The Contact process involves the equilibrium:
2SO2(g) + O2(g) ⇌ 2SO3(g), delta H = -196 kJ/mol
Predict the effect on the position of equilibrium of:
(a) Increasing the concentration of SO2. (1)
(b) Decreasing the pressure. (1)
(c) Decreasing the temperature. (1)
Mark scheme
(a) Equilibrium shifts to the right / towards SO3, increasing yield [1]
(b) Equilibrium shifts to the left / towards SO2 and O2 [1] (the side with more moles of gas: 3 on left vs 2 on right)
(c) Equilibrium shifts to the right / towards SO3 [1] (forward reaction is exothermic, favoured by lower temperature)
Examiner note: For concentration: adding more of a reactant shifts equilibrium towards the products. For pressure: equilibrium shifts to the side with fewer moles of gas. For temperature: lower temperature favours the exothermic direction.
Question 5 (3 marks, Supplement)
Hydrated copper sulfate and anhydrous copper sulfate exist in a reversible reaction:
CuSO4.5H2O(s) ⇌ CuSO4(s) + 5H2O(l) blue crystals ⇌ white powder + water
(a) State what you would observe if water is added to anhydrous copper sulfate. (1)
(b) State what you would observe if hydrated copper sulfate is heated. (1)
(c) Explain why this reaction is used as a test for water. (1)
Mark scheme
(a) The white powder turns blue and the mixture gets hot [1]
(b) The blue crystals turn white and steam/water is given off [1]
(c) The colour change from white to blue is a clear, specific indicator that water is present [1]
Examiner note: The forward reaction (dehydration) is endothermic — heat must be supplied. The reverse reaction (hydration) is exothermic — the mixture gets hot when water is added. This is a classic reversible reaction at IGCSE level.
Question 6 (3 marks, Supplement)
Explain why removing a product from an equilibrium mixture increases the yield of that product.
Mark scheme
- Removing a product decreases its concentration [1]
- By Le Chatelier’s principle, the equilibrium shifts to the right/forward to replace the removed product [1]
- More reactants are converted to products, increasing the yield [1]
Examiner note: In the Haber process, ammonia is cooled and liquefied to remove it from the equilibrium mixture. The unreacted N2 and H2 are recycled over the catalyst. This continuous removal drives the equilibrium forward.
Question 7 (4 marks, Supplement)
A student writes: “Adding a catalyst to a reversible reaction increases the yield of products.”
Evaluate this statement and correct any errors.
Mark scheme
- The statement is incorrect [1]
- A catalyst speeds up both the forward and reverse reactions equally [1]
- It allows equilibrium to be reached more quickly [1]
- But it does not change the position of equilibrium or the yield at equilibrium [1]
Examiner note: This is one of the most common misconceptions. A catalyst affects the rate but NOT the yield. Only changes in temperature, pressure, or concentration can shift the equilibrium position and change the yield.
What to revise if you scored below 5
If Le Chatelier’s principle was unclear, practise applying it to at least three different equilibria. For each one, predict the effect of changing temperature, pressure, and concentration. If the Haber process conditions confused you, memorise: 450 degC (compromise), 200 atm (high pressure for yield), iron catalyst (faster equilibrium, no yield change). Revisit the reversible reactions and equilibrium notes.
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