OxfordAQA A-Level Chemistry: Equilibrium constant Kp for homogeneous systems (9620) – 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.

Here you practise section 3.1.12 (subsection 3.1.12.1) of the OxfordAQA International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards. Every question here is International A-level only. Unit 4 (Organic 2 and Physical 2) examines this section, and Unit 5 (Practical and synoptic) spans the whole specification. Every outcome in the section is practised: partial pressure from mole fraction, Kp expressions, Kp calculations, and the effects of pressure, temperature and catalysts.

The method is taught in the Kp study guide and summarised in the Kp revision notes. The 9620 chemistry hub links every topic, and the course checklist prints on one sheet. Unless a question says otherwise, each mixture is at constant temperature and every species is a gas.

Questions

1. State what is meant by Kp, and explain why a system described as homogeneous has every species in its Kp expression. [2]

2. Carbon monoxide burns in oxygen: 2CO(g) + O₂(g) ⇌ 2CO₂(g). Write an expression for Kp and deduce its units when pressures are measured in Pa. [2]

3. Nitryl chloride decomposes: 2NO₂Cl(g) ⇌ 2NO₂(g) + Cl₂(g). An equilibrium mixture contains 0.450 mol NO₂Cl, 0.300 mol NO₂ and 0.150 mol Cl₂ at a total pressure of 120 kPa. Calculate the partial pressure of NO₂Cl and of Cl₂. [3]

4. Use your answers to question 3, and p(NO₂) = 40.0 kPa, to calculate Kp for the nitryl chloride equilibrium. Give its units. [3]

5. Propane is dehydrogenated: C₃H₈(g) ⇌ C₃H₆(g) + H₂(g).

(a) 2.00 mol of propane is heated in a sealed vessel. At equilibrium, 0.500 mol of hydrogen is present and the total pressure is 150 kPa. Calculate Kp. [5] (b) At the same temperature, the total pressure is lowered to 30.0 kPa. Show that a mixture in which half of the original propane has reacted is at equilibrium under these conditions. [3]

6. A student writes: “Adding a catalyst to a gas-phase equilibrium increases Kp, because more product is made every second.” Evaluate this statement. [3]

7. For 2CO(g) + O₂(g) ⇌ 2CO₂(g), the forward reaction is exothermic. The temperature is increased at constant total pressure. State and explain the effect on the position of equilibrium and on the value of Kp. [3]

8. cis-But-2-ene and trans-but-2-ene interconvert in the gas phase:

cis-but-2-ene(g) ⇌ trans-but-2-ene(g) Kp = 1.50 at temperature T

(a) Explain why Kp has no units for this equilibrium. [1] (b) 0.800 mol of cis-but-2-ene is sealed in a vessel at temperature T. Calculate the amount of each isomer at equilibrium. [3] (c) The vessel is compressed at temperature T. State and explain the effect on the amount of trans-but-2-ene. [2]

9. Nitrosyl bromide decomposes in a sealed vessel: 2NOBr(g) ⇌ 2NO(g) + Br₂(g). The forward reaction is endothermic. 1.60 mol of NOBr is heated. At equilibrium, 0.400 mol of Br₂ is present and the total pressure is 80.0 kPa.

(a) Calculate the amount of NOBr and of NO at equilibrium. [2] (b) Work out the partial pressure of every gas in the mixture. [2] (c) Calculate the value of Kp, including units. [2] (d) State and explain the effect on Kp of raising the temperature. [2] (e) The total pressure is doubled at constant temperature. Use the Kp expression to explain which way the position of equilibrium moves. [3]

10. Carbon disulfide reacts with hydrogen: CS₂(g) + 4H₂(g) ⇌ CH₄(g) + 2H₂S(g). At a fixed temperature, Kp = 6.25 × 10⁻⁵ kPa⁻². An equilibrium mixture has p(CS₂) = 25.0 kPa, p(H₂) = 40.0 kPa and p(CH₄) = 10.0 kPa.

(a) Calculate p(H₂S). [3] (b) Calculate the total pressure and the mole fraction of H₂S in the mixture. [2] (c) Predict how raising the total pressure would change the equilibrium yield of methane. [1]

Answers

1. Kp is the equilibrium constant calculated from the partial pressures of the gases, for a system at constant temperature [1]. Homogeneous means all species are in the same phase, here all gases, so each has a partial pressure that appears in the expression [1]. [2] Examiner insight: include “at constant temperature” in your definition, because the specification’s own wording of Kp contains it.

2. Kp = p(CO₂)² ÷ (p(CO)² × p(O₂)) [1]. Δn = 2 − 3 = −1, so units Pa⁻¹ [1]. [2] Examiner insight: work the units out from your own expression; if you write it upside down, the units come out as Pa and both answers are wrong.

3. Total amount = 0.450 + 0.300 + 0.150 = 0.900 mol [1]. p(NO₂Cl) = (0.450 ÷ 0.900) × 120 = 60.0 kPa [1]. p(Cl₂) = (0.150 ÷ 0.900) × 120 = 20.0 kPa [1]. [3] Examiner insight: a quick check that 60.0 + 40.0 + 20.0 = 120 kPa catches a wrong total before it carries into question 4.

4. Kp = p(NO₂)² × p(Cl₂) ÷ p(NO₂Cl)² [1] = 40.0² × 20.0 ÷ 60.0² = 32 000 ÷ 3600 = 8.89 [1]; units kPa (Δn = 3 − 2 = +1) [1]. [3] Examiner insight: write the expression in symbols before substituting, so your method is visible even if a value carried from question 3 is wrong.

5. (a) Change row from the 1 : 1 : 1 ratio: 0.500 mol of propane reacts, giving 0.500 mol C₃H₆ and 0.500 mol H₂ [1]. Equilibrium: C₃H₈ 1.50 mol, total = 2.50 mol [1]. Partial pressures: p(C₃H₈) = (1.50 ÷ 2.50) × 150 = 90.0 kPa; p(C₃H₆) = p(H₂) = (0.500 ÷ 2.50) × 150 = 30.0 kPa [1]. Kp = p(C₃H₆) × p(H₂) ÷ p(C₃H₈) = 30.0 × 30.0 ÷ 90.0 [1] = 10.0 kPa [1]. (b) Per 1 mol of propane at the start: 0.500 mol C₃H₈, 0.500 mol C₃H₆, 0.500 mol H₂; total 1.500 mol, so each mole fraction is 1/3 [1]. Each partial pressure = (1/3) × 30.0 = 10.0 kPa [1]. 10.0 × 10.0 ÷ 10.0 = 10.0 kPa, which equals Kp, so the mixture is at equilibrium [1]. [8] Examiner insight: on a “show that”, finish by comparing your value with Kp in words; a bare number with no conclusion does not show anything.

6. The statement is wrong about Kp: a catalyst does not change the value of Kp [1]. It raises the rates of the forward and reverse reactions by the same factor [1], so the system arrives at equilibrium in less time while the equilibrium partial pressures, and so the yield, are unchanged [1]. [3] Examiner insight: “Evaluate” needs a judgement as well as reasons; agreeing that “more product per second” means a higher yield gains nothing.

7. The position of equilibrium moves to the left [1], because raising the temperature favours the endothermic direction, which is the reverse reaction [1]. The value of Kp decreases [1]. [3] Examiner insight: name the endothermic direction (or say the reverse reaction absorbs heat); “to oppose the change” on its own does not explain which way the equilibrium moves.

8. (a) There is 1 mole of gas on each side, so the pressure units cancel [1]. (b) Let y mol of cis convert: Kp = y ÷ (0.800 − y) = 1.50 [1]. y = 1.20 − 1.50y, so y = 0.480 [1]. Equilibrium: 0.480 mol trans and 0.320 mol cis [1]. (c) No change in the amount of trans-but-2-ene [1]. The total pressure cancels from the Kp expression because both sides have equal moles of gas, so compression does not shift the position [1]. [6] Examiner insight: in part (b) you can work in moles because Δn = 0; say so, or show that the total pressure cancels, so the method is clear.

9. (a) 0.400 mol Br₂ forms from 0.800 mol NOBr, which also gives 0.800 mol NO [1]. NOBr left = 1.60 − 0.800 = 0.800 mol; NO = 0.800 mol [1]. (b) Total = 0.800 + 0.800 + 0.400 = 2.00 mol [1]. p(NOBr) = p(NO) = (0.800 ÷ 2.00) × 80.0 = 32.0 kPa; p(Br₂) = (0.400 ÷ 2.00) × 80.0 = 16.0 kPa [1]. (c) Kp = p(NO)² × p(Br₂) ÷ p(NOBr)² = 32.0² × 16.0 ÷ 32.0² [1] = 16.0 kPa [1]. (d) Kp increases [1], because the equilibrium moves in the endothermic, forward direction, raising the partial pressures of the products relative to the reactant [1]. (e) Writing each p as (mole fraction × P) gives Kp = (mole-fraction terms) × P, because Δn = +1 [1]. If P doubles with the mole fractions unchanged, the right-hand side becomes 32.0 kPa, larger than Kp [1]. The mole fractions of NO and Br₂ must fall and that of NOBr must rise, so the position moves to the left, the side with fewer moles of gas [1]. [11] Examiner insight: in part (e), a correct Le Chatelier answer alone may not gain full credit when the question asks you to use the Kp expression; show where the factor of P comes from.

10. (a) p(H₂S)² = Kp × p(CS₂) × p(H₂)⁴ ÷ p(CH₄) [1] = 6.25 × 10⁻⁵ × 25.0 × 40.0⁴ ÷ 10.0 = 400 [1]. p(H₂S) = 20.0 kPa [1]. (b) Total pressure = 25.0 + 40.0 + 10.0 + 20.0 = 95.0 kPa [1]. Mole fraction of H₂S = 20.0 ÷ 95.0 = 0.211 [1]. (c) The yield of methane increases, because the position moves right, to the side with fewer moles of gas (3 rather than 5) [1]. [6] Examiner insight: forgetting the square root in part (a) gives 400 kPa, a value larger than the whole mixture; a sense check on size would catch it.

Where marks are usually lost

• Totalling the starting amounts rather than the equilibrium amounts.
• Applying a 1 : 1 change when the equation’s ratio is 2 : 2 : 1 or 1 : 4.
• Leaving a power off a partial pressure, such as p(H₂)⁴.
• Units with the wrong sign when Δn is negative, or units given when Δn = 0.
• Mixing Pa and kPa within one substitution.
• No closing comparison with Kp on a “show that” question.
• Stating that a pressure change alters Kp, or that a catalyst raises the yield.
• Temperature answers that never name the endothermic or exothermic direction.
• Missing the square root when the unknown appears squared.

Next steps

• Go over weak points with the Kp revision notes
• Go back to the full explanations in the Kp study guide
• Revisit AS-level Kc work with the AS equilibria question set
• Move on to other topics from the 9620 chemistry hub, tracking them on the course checklist
• Try all free 10-minute diagnostics
• Book a free trial class

Official syllabus

OxfordAQA, International AS and A-level Chemistry (9620) specification, Version 5.3, for International AS and A-level exams May/June 2020 onwards: section 3.1.12 Equilibrium constant Kp for homogeneous systems (International A2), subsection 3.1.12.1.