AQA A-Level Chemistry: Equilibrium constant Kp for homogeneous systems (7405) – 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 cover section 3.1.10 of the AQA AS and A-level Chemistry specification (7404/7405), version 1.2 (July 2026), for AS and A-level exams from June 2016 onwards. The whole section is A-level only, so none of it appears in the AS papers. The A-level assessment overview lists section 3.1.10 under Paper 1, and Paper 3 can test any content. Show your working in full.

Learn the method first in the Kp study guide. Kc-only questions are in the Kc practice set.

Questions

1. Define the terms mole fraction and partial pressure for a gas in a mixture of gases. [2]

2. Ammonia is oxidised in the gas phase: 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g). Write an expression for Kp for this reaction and deduce its units, using kPa. [2]

3. A gas mixture contains 0.240 mol of carbon monoxide, 0.540 mol of hydrogen and 0.0200 mol of methanol vapour at a total pressure of 150 kPa. Calculate the partial pressure of carbon monoxide and the partial pressure of methanol. [3]

4. For N₂(g) + O₂(g) ⇌ 2NO(g) at a high temperature, the equilibrium partial pressures are p(N₂) = 80.0 kPa, p(O₂) = 20.0 kPa and p(NO) = 4.00 kPa. Calculate Kp and state its units, if any. [2]

5. 1.20 mol of sulfur trioxide is heated in a sealed container. It reaches equilibrium:

2SO₃(g) ⇌ 2SO₂(g) + O₂(g)

At equilibrium, 0.400 mol of oxygen is present and the total pressure is 300 kPa. Calculate Kp and give its units. [5]

6. State the effect of adding a catalyst to a gas-phase equilibrium on (i) the time taken to reach equilibrium and (ii) the value of Kp. [2]

7. For a gas-phase reaction, Kp = 1.8 × 10⁻³ kPa at 500 K and Kp = 4.5 × 10⁻² kPa at 700 K. Deduce whether the forward reaction is exothermic or endothermic. Explain your answer. [3]

8. Nitrosyl chloride decomposes: 2NOCl(g) ⇌ 2NO(g) + Cl₂(g). The mixture is at equilibrium. The total pressure is then increased at constant temperature.

Explain, with reference to Kp, what happens to the mole fraction of chlorine and to the value of Kp. [4]

9. In a laboratory study of the Haber process, 1.00 mol of nitrogen and 3.00 mol of hydrogen are sealed in a vessel with an iron catalyst. At equilibrium, 0.400 mol of ammonia is present and the total pressure is 15.0 MPa.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

(a) Calculate the amount, in moles, of nitrogen and of hydrogen at equilibrium. [2] (b) Calculate the partial pressure of each gas at equilibrium. [2] (c) Calculate Kp and give its units. [2] (d) State the effect of the iron catalyst on the value of Kp. [1]

10. Pure carbonyl chloride is heated in a sealed container at constant temperature. It decomposes and reaches equilibrium:

COCl₂(g) ⇌ CO(g) + Cl₂(g) Kp = 40.0 kPa

At equilibrium, p(COCl₂) = 90.0 kPa.

(a) Explain why p(CO) = p(Cl₂) at equilibrium. [1] (b) Calculate p(CO). [2] (c) Calculate the total pressure at equilibrium. [1] (d) Calculate the percentage of the original carbonyl chloride that has decomposed. [2]

11. 1.00 mol of hydrogen sulfide is heated in a sealed container until it reaches equilibrium:

2H₂S(g) ⇌ 2H₂(g) + S₂(g)

The equilibrium mixture contains 0.20 mol of S₂, and the total pressure is 1.5 × 10² kPa. Calculate Kp. Give your answer to an appropriate number of significant figures, with units. [5]

12. Ethanol is made industrially by the hydration of ethene with an acid catalyst:

C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g) ΔH is negative

A mixture of 1.00 mol of ethene and 1.00 mol of steam reaches equilibrium. The equilibrium mixture contains 0.100 mol of ethanol at a total pressure of 6.00 MPa.

(a) Write an expression for Kp and deduce its units, using MPa. [2] (b) Calculate Kp. [4] (c) The temperature is increased, at the same total pressure. State and explain the effect on the value of Kp and on the equilibrium yield of ethanol. [2]

Answers

1. Mole fraction = moles of that gas ÷ total moles of gas in the mixture [1]. Partial pressure = the pressure the gas would exert if it alone occupied the container; equal to mole fraction × total pressure [1]. [2] Examiner insight: “Total moles of gas” is needed for the first mark; “moles of gas ÷ moles of mixture” with no reference to the total is usually too vague.

2. Kp = p(NO)⁴ × p(H₂O)⁶ / (p(NH₃)⁴ × p(O₂)⁵) [1]. Units: kPa¹⁰ / kPa⁹ = kPa [1]. [2] Examiner insight: The units mark is normally allowed as error carried forward from a wrong expression, so write the units even if you are unsure of the expression.

3. Total moles = 0.240 + 0.540 + 0.0200 = 0.800 mol [1]. p(CO) = (0.240 / 0.800) × 150 = 0.300 × 150 = 45.0 kPa [1]. p(CH₃OH) = (0.0200 / 0.800) × 150 = 0.0250 × 150 = 3.75 kPa [1]. [3] Examiner insight: A wrong total moles loses the first mark, but both partial pressures can still score by error carried forward if the method is right.

4. Kp = p(NO)² / (p(N₂) × p(O₂)) = 4.00² / (80.0 × 20.0) [1] = 0.0100, no units [1]. [2] Examiner insight: Gas moles are equal on both sides, so the units cancel; writing “kPa” here is an error and normally costs the final mark.

5. Equilibrium moles: SO₃ = 1.20 − 0.800 = 0.400 mol; SO₂ = 0.800 mol; O₂ = 0.400 mol [1]. Total = 1.600 mol [1]. Mole fractions 0.250, 0.500, 0.250, so p(SO₃) = 75.0 kPa, p(SO₂) = 150 kPa, p(O₂) = 75.0 kPa [1]. Kp = p(SO₂)² × p(O₂) / p(SO₃)² = 150² × 75.0 / 75.0² [1] = 300 kPa [1]. [5] Examiner insight: The change in SO₃ and SO₂ is twice the moles of O₂ formed; using a 1 : 1 ratio is the commonest wrong step, and later marks are then awarded only as error carried forward.

6. (i) Equilibrium is reached in a shorter time (the catalyst speeds up the forward and reverse reactions equally) [1]. (ii) Kp is unchanged [1]. [2] Examiner insight: Any statement that the catalyst shifts the equilibrium or increases the yield contradicts the answer and usually loses the mark.

7. Kp increases as the temperature increases [1]. So the equilibrium shifts to the right at higher temperature, in the endothermic direction [1]. The forward reaction is endothermic [1]. [3] Examiner insight: The conclusion alone earns only one mark; the link “higher temperature favours the endothermic direction” is needed for the explanation marks.

8. Kp = p(NO)² × p(Cl₂) / p(NOCl)². With p = x × P, Kp = x(NO)² × x(Cl₂) × P / x(NOCl)² [1]. Kp depends only on temperature, so Kp does not change [1]. As P rises, x(NO)² × x(Cl₂) / x(NOCl)² must fall to keep the expression equal to Kp [1]. The equilibrium shifts left, to the side with fewer moles of gas (2 compared with 3), so the mole fraction of chlorine decreases [1]. [4] Examiner insight: A Le Chatelier answer alone (“fewer moles of gas”) misses the Kp marks; “with reference to Kp” needs the P term shown or described.

9. (a) 0.400 mol NH₃ forms, so 0.200 mol N₂ and 0.600 mol H₂ react [1]. N₂ = 0.800 mol; H₂ = 2.40 mol [1]. (b) Total = 0.800 + 2.40 + 0.400 = 3.60 mol [1]. p(N₂) = 3.33 MPa, p(H₂) = 10.0 MPa, p(NH₃) = 1.67 MPa [1]. (c) Kp = p(NH₃)² / (p(N₂) × p(H₂)³) = 1.667² / (3.333 × 10.0³) [1] = 8.33 × 10⁻⁴ MPa⁻² [1]. (d) No effect on Kp (it only lets equilibrium be reached faster) [1]. [7] Examiner insight: Rounding the partial pressures to 3 s.f. before (c) gives 8.38 × 10⁻⁴, which can lose the accuracy mark; carry full calculator values through.

10. (a) One mole of CO and one mole of Cl₂ form for every mole of COCl₂ that decomposes, and there was none of either at the start, so their moles (and so their partial pressures) are equal [1]. (b) Kp = p(CO) × p(Cl₂) / p(COCl₂), so p(CO)² = 40.0 × 90.0 = 3600 [1]; p(CO) = 60.0 kPa [1]. (c) Total = 90.0 + 60.0 + 60.0 = 210 kPa [1]. (d) Partial pressure is proportional to moles in the same mixture, so moles decomposed : moles remaining = 60.0 : 90.0 [1]. Percentage decomposed = 60.0 / (60.0 + 90.0) × 100 = 40.0% [1]. [6] Examiner insight: In (d), dividing 60.0 by 90.0 (the amount remaining, not the original) is a common slip; the proportionality idea can still earn the first mark.

11. Equilibrium moles: S₂ = 0.20 mol, H₂ = 0.40 mol, H₂S = 1.00 − 0.40 = 0.60 mol [1]. Total = 1.20 mol [1]. p(H₂S) = 75 kPa, p(H₂) = 50 kPa, p(S₂) = 25 kPa [1]. Kp = p(H₂)² × p(S₂) / p(H₂S)² = 50² × 25 / 75² [1] = 11 kPa (2 s.f., as the data are given to 2 s.f.) [1]. [5] Examiner insight: The final mark needs both the 2 s.f. answer and the unit; 11.1 kPa is over-precise for 2 s.f. data and may not be credited.

12. (a) Kp = p(C₂H₅OH) / (p(C₂H₄) × p(H₂O)) [1]. Units: MPa / MPa² = MPa⁻¹ [1]. (b) Equilibrium moles: C₂H₄ = 0.900 mol, H₂O = 0.900 mol, C₂H₅OH = 0.100 mol [1]. Total = 1.900 mol [1]. p(C₂H₄) = p(H₂O) = (0.900 / 1.900) × 6.00 = 2.842 MPa; p(C₂H₅OH) = (0.100 / 1.900) × 6.00 = 0.3158 MPa [1]. Kp = 0.3158 / (2.842 × 2.842) = 0.0391 MPa⁻¹ [1]. (c) The forward reaction is exothermic, so a higher temperature shifts the equilibrium left, in the endothermic direction [1]. Kp decreases and the equilibrium yield of ethanol decreases [1]. [8] Examiner insight: In (c), “Kp decreases” without the reason (exothermic forward reaction, endothermic direction favoured) usually scores only the second mark.

Where marks are usually lost

• Using a 1 : 1 ratio in the moles table when the equation has different coefficients (Questions 5, 9 and 11).
• Calculating mole fractions from initial moles instead of equilibrium moles.
• Units: giving units when gas moles are equal, or getting the sign of the power wrong.
• Rounding intermediate partial pressures and losing the final accuracy mark.
• Quoting more significant figures than the least accurate data allow.
• Saying an increase in pressure or a catalyst changes Kp.
• Temperature explanations that never name the endothermic or exothermic direction.

Next steps

• Recheck formulas in the Kp revision notes.
• Reread the worked examples in the Kp study guide.
• Practise Kc and Le Chatelier questions in the Kc practice set.
• See every topic on the AQA A-Level Chemistry hub and track progress on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

AQA AS and A-level Chemistry specification (7404/7405), version 1.2, July 2026, for AS and A-level exams June 2016 onwards, published by AQA. Section 3.1.10: Equilibrium constant Kp for homogeneous systems (A-level only).