OCR A Level Chemistry A: Chemical equilibrium (H432) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions practise section 3.2.3, Chemical equilibrium, of the OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017. Together they test every learning outcome, 3.2.3(a) to (g), from Module 3, Periodic table and energy. H432 is the full A level. The last two questions are extended problems drawing on several outcomes at once.

Revise with the Chemical equilibrium study guide and the Chemical equilibrium revision notes. Other Module 3 topics are on the OCR chemistry hub; tick topics off on the printable checklist. The free diagnostics cover other topics as well.

All data is given in the questions. Give Kc without units, as 3.2.3 does not require them.

Questions

1. A reversible reaction is left in a sealed flask at constant temperature until it reaches dynamic equilibrium. State two features of the mixture at dynamic equilibrium. [2]

2. Nitrosyl bromide decomposes in a reversible reaction. The forward reaction is endothermic.

2NOBr(g) ⇌ 2NO(g) + Br₂(g)

(a) Predict and explain the effect on the position of equilibrium of increasing the total pressure. [2] (b) Predict and explain the effect on the amount of Br₂ at equilibrium of increasing the temperature. [2]

3. Ethanol is made industrially by reacting ethene with steam over a phosphoric acid catalyst. Explain how the catalyst affects the forward and reverse rates, the position of equilibrium, and the amount of ethanol made per hour. [3]

4. When iron(III) ions and thiocyanate ions are mixed, a deep red complex forms.

Fe³⁺(aq) + SCN⁻(aq) ⇌ [FeSCN]²⁺(aq) pale yellow, colourless ⇌ deep red

(a) A few drops of concentrated potassium thiocyanate solution are added to a pale red equilibrium mixture. Describe and explain what you would see. [2] (b) Sodium hydroxide solution is added to a second sample. It removes Fe³⁺(aq) as a precipitate of iron(III) hydroxide. Predict and explain the change in the red colour of the solution. [2]

5. Cobalt(II) ions take part in this equilibrium in solution:

[Co(H₂O)₆]²⁺(aq) + 4Cl⁻(aq) ⇌ [CoCl₄]²⁻(aq) + 6H₂O(l) pink ⇌ blue

You are given a violet solution of this equilibrium mixture. Describe how you would find out whether the forward reaction is exothermic or endothermic. Include how you would interpret your observations. [4]

6. Write the expression for Kc for each equilibrium.

(a) 4HCl(g) + O₂(g) ⇌ 2Cl₂(g) + 2H₂O(g) [1] (b) C₂H₆(g) ⇌ C₂H₄(g) + H₂(g) [1]

7. For the equilibrium in question 2, a mixture at one temperature contains [NOBr] = 0.0640 mol dm⁻³, [NO] = 0.0180 mol dm⁻³ and [Br₂] = 0.00900 mol dm⁻³. Calculate Kc. Give your answer to 3 significant figures. [2]

8. For the equilibrium in question 6(a), Kc = 0.400 at a certain temperature. An equilibrium mixture at this temperature contains [HCl] = 0.200 mol dm⁻³, [O₂] = 0.100 mol dm⁻³ and [H₂O] = 0.0500 mol dm⁻³. Calculate the equilibrium concentration of Cl₂. [3]

9. Three homogeneous reactions, J, K and L, have the values of Kc shown at 298 K.

Reaction	Kc
J	4 × 10⁻¹⁶
K	0.9
L	6 × 10¹¹

For each reaction, estimate the position of equilibrium and describe the composition of the equilibrium mixture. [3]

10. Ethanol is made by the hydration of ethene. The forward reaction is exothermic.

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)

(a) Explain why a moderate temperature is used, rather than a low or a very high one. [3] (b) Explain why a raised pressure is used, and why the pressure is not made as high as possible. [3] (c) Only a small fraction of the ethene reacts in each pass through the reactor. Suggest how the plant keeps waste of ethene low. [1] (d) At one temperature, an equilibrium mixture contains [C₂H₄] = 0.0800 mol dm⁻³, [H₂O] = 0.0950 mol dm⁻³ and [C₂H₅OH] = 0.0120 mol dm⁻³. Write the Kc expression and calculate Kc. [2] (e) Use your answer to (d) to estimate the position of equilibrium at this temperature. [1]

11. Butane can be converted into its isomer, methylpropane. Both are gases under the conditions used.

CH₃CH₂CH₂CH₃(g) ⇌ (CH₃)₃CH(g)

(a) Butane is sealed in a flask with a catalyst. Explain why the concentrations of both gases eventually stop changing although both reactions continue. [2] (b) At equilibrium, [butane] = 0.0500 mol dm⁻³ and [methylpropane] = 0.0900 mol dm⁻³. Write the Kc expression and calculate Kc. [2] (c) Use your value of Kc to state which isomer is present in the larger amount at equilibrium. [1] (d) A second flask with no catalyst is set up at the same temperature with the same starting amount of butane. Compare the second flask with the first in terms of the time taken to reach equilibrium, the equilibrium composition and the value of Kc. [3] (e) The flask is compressed to a smaller volume at constant temperature. State and explain what happens to the position of equilibrium. [1]

Answers

1. Forward and reverse reactions take place at the same rate [1]. The concentration of each substance in the flask stays constant [1]. [2] Examiner insight: “The concentrations are equal” is a common wrong answer and gains no credit; the word needed is constant, or “do not change”.

2. (a) There are 2 gas moles on the left and 3 on the right [1]. The position moves left, because forming fewer gas molecules lowers the pressure [1]. (b) The forward reaction is endothermic, so the position shifts to the right to absorb the added heat [1]. The amount of Br₂ increases [1]. Examiner insight: Each prediction needs its reason; a bare “shifts left” or “more Br₂” usually earns only one of the two marks.

3. The catalyst speeds up the forward reaction and the reverse reaction equally [1]. The position of equilibrium is unchanged, so the equilibrium yield of ethanol is the same [1]. Equilibrium is reached sooner, so more ethanol is made per hour [1]. [3] Examiner insight: Writing that the catalyst moves the position or improves the yield contradicts the creditworthy points, so keep the ideas of yield and rate apart.

4. (a) The solution becomes a deeper red [1]. The added SCN⁻ shifts the position to the right, forming more [FeSCN]²⁺ [1]. (b) The red colour fades [1]. Removing Fe³⁺ shifts the position to the left to replace some of it, so less [FeSCN]²⁺ is present [1]. Examiner insight: Describe what you would see, a colour, as well as the shift; an answer that only says “equilibrium moves right” misses the observation mark.

5. Put equal volumes of the violet solution into three test tubes, keeping one at room temperature as a control [1]. Place one tube in a hot water bath and one in an ice bath for a few minutes [1]. Compare the colours of the heated and cooled tubes with the control [1]. If the hot tube turns blue (and the cold one pink), the forward reaction is endothermic; if the hot tube turns pink, it is exothermic [1]. [4] Examiner insight: A conclusion needs to link the colour seen at the higher temperature to the direction favoured; stating “it is endothermic” with no reasoning from the observation may not gain the final mark.

6. (a) Kc = [Cl₂]²[H₂O]² / ([HCl]⁴[O₂]) [1] (b) Kc = [C₂H₄][H₂] / [C₂H₆] [1] Examiner insight: Every power must match the balancing number; [HCl] written without the power 4 makes the expression wrong and loses the mark.

7. Kc = [NO]²[Br₂] / [NOBr]² = (0.0180² × 0.00900) / 0.0640² [1] = 2.916 × 10⁻⁶ / 4.096 × 10⁻³ = 7.12 × 10⁻⁴ [1]. [2] Examiner insight: Show the substituted expression first, so your method is visible to the examiner even if the arithmetic slips.

8. [Cl₂]² = Kc × [HCl]⁴[O₂] / [H₂O]² [1] = 0.400 × 0.200⁴ × 0.100 / 0.0500² = 0.400 × 1.60 × 10⁻⁴ / 2.50 × 10⁻³ = 0.0256 [1] [Cl₂] = √0.0256 = 0.160 mol dm⁻³ [1] [3] Examiner insight: Stopping at 0.0256 gives [Cl₂]², not [Cl₂]; the final mark needs the square root and the unit of concentration.

9. J: Kc is much less than 1, so the position is far to the left; the mixture is almost all reactants [1]. K: Kc is close to 1, so there are similar amounts of reactants and products [1]. L: Kc is much greater than 1, so the position is far to the right; the mixture is almost all products [1]. [3] Examiner insight: A qualitative estimate is what is asked for; link each value to a position, rather than trying to calculate percentages.

10. (a) The forward reaction is exothermic, so the equilibrium yield of ethanol is greater at low temperature [1]. But at a low temperature the rate would be too slow [1]. A moderate temperature is a compromise that gives an acceptable yield at an acceptable rate; a very high temperature would cut the yield too much [1]. (b) There are 2 gas moles on the left and 1 on the right, so a higher pressure shifts the position right and raises the yield (it also increases the rate) [1]. Very high pressures need expensive, thick-walled plant and use a lot of energy for compression [1]. They also increase the safety risk, so a compromise pressure is chosen [1]. (c) Unreacted ethene (and steam) is separated from the ethanol and recycled through the reactor [1]. (d) Kc = [C₂H₅OH] / ([C₂H₄][H₂O]) = 0.0120 / (0.0800 × 0.0950) [1] = 0.0120 / 7.60 × 10⁻³ = 1.58 [1] (e) Kc is close to 1, so neither side is strongly favoured; reactants and product are present in comparable amounts [1]. Examiner insight: In compromise answers, the yield argument and the rate (or cost and safety) argument are separate marking points, so state both explicitly rather than writing “it is a compromise” alone.

11. (a) The forward rate falls as butane is used up and the reverse rate rises as methylpropane forms, until the two rates are equal [1]. Each isomer then forms as fast as it is used up, so concentrations stay constant [1]. (b) Kc = [methylpropane] / [butane] = 0.0900 / 0.0500 [1] = 1.80 [1] (c) Kc is greater than 1, so methylpropane is present in the larger amount [1]. (d) The flask without a catalyst takes longer to reach equilibrium [1]. The equilibrium composition is the same in both flasks [1]. Kc is the same, because the temperature is the same [1]. (e) It does not move: with 1 gas mole on each side, neither direction lowers the pressure [1]. Examiner insight: In part (d) a catalyst comparison needs all three ideas, time, composition and Kc, as separate statements; a single sentence saying “the catalyst makes no difference” earns little.

Where marks are usually lost

• Writing “equal concentrations” in place of “constant concentrations” for dynamic equilibrium.
• Predicting a shift without the reason, such as fewer gas moles or the endothermic direction.
• Claiming that a catalyst increases the yield or speeds up only the forward reaction.
• Describing a practical without a control, or giving a conclusion that does not follow from the colour seen.
• Missing a power in a Kc expression, or putting reactants on top.
• Answering an industrial compromise question with yield arguments only.

Next steps

• Recap with the Chemical equilibrium revision notes.
• Re-read any outcome you found hard in the Chemical equilibrium study guide.
• Browse the OCR chemistry hub.
• Track your progress on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

OCR A Level Chemistry A (H432) specification, Version 3.1 (May 2026), for first assessment in 2017, section 3.2.3 Chemical equilibrium. Published by OCR.