Edexcel International A Level Chemistry: Chemical equilibria (YCH11) – 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.

This set covers Topic 13: Chemical Equilibria of the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry specification (YCH11), Issue 1, September 2017, outcomes 13.1 to 13.9. It is IA2 content, taught in Unit 4: Rates, Equilibria and Further Organic Chemistry. New to the topic? Read the study guide and revision notes first. Then log gaps on the chemistry checklist, use the Edexcel course hub or run a free diagnostic.

Take R = 8.31 J K⁻¹ mol⁻¹ where needed.

Questions

1. For each equilibrium, write an expression for Kc and give its units, if any.

(a) 2CH₄(g) ⇌ C₂H₂(g) + 3H₂(g) [2] (b) 2Fe³⁺(aq) + Cu(s) ⇌ 2Fe²⁺(aq) + Cu²⁺(aq) [2] (c) Ca(OH)₂(s) ⇌ Ca²⁺(aq) + 2OH⁻(aq) [2]

2. A flask contains 0.300 mol ethyl ethanoate, 0.500 mol water and dilute hydrochloric acid containing 0.0100 mol HCl as a catalyst. It is sealed and left until this equilibrium is reached:

CH₃COOC₂H₅(l) + H₂O(l) ⇌ CH₃COOH(l) + C₂H₅OH(l)

All the acid in the equilibrium mixture is then neutralised by 68.50 cm³ of 2.00 mol dm⁻³ sodium hydroxide. Ignore the water added with the catalyst.

(a) Calculate the amount, in mol, of ethanoic acid at equilibrium. [2] (b) Calculate Kc for the hydrolysis. [4]

3. 0.500 mol of antimony(V) chloride vapour is heated in a sealed vessel. It sets up the equilibrium

SbCl₅(g) ⇌ SbCl₃(g) + Cl₂(g)

At equilibrium, the vessel contains 0.150 mol Cl₂ and the total pressure is 1.95 atm.

(a) Write an expression for Kp. [1] (b) Calculate the value of Kp and give its units. [4]

4. Solid ammonium chloride is heated in an evacuated sealed container at a fixed temperature:

NH₄Cl(s) ⇌ NH₃(g) + HCl(g)

At equilibrium, with some solid still present, the total pressure is 0.40 atm.

(a) Calculate Kp at this temperature. [2] (b) Extra ammonia is pumped in at the same temperature. When equilibrium is re-established, with solid still present, p(NH₃) = 0.32 atm. Calculate p(HCl) and the new total pressure. [2]

5. Ammonia is shaken with water and trichloromethane, two liquids that do not mix, until equilibrium is reached. A 10.0 cm³ portion of the aqueous layer is neutralised by 9.60 cm³ of 0.500 mol dm⁻³ hydrochloric acid. A 25.0 cm³ portion of the trichloromethane layer is neutralised by 25.0 cm³ of 0.0200 mol dm⁻³ hydrochloric acid.

NH₃ + HCl → NH₄Cl

Calculate the equilibrium constant K = [NH₃(aq)] / [NH₃(trichloromethane)]. [4]

6. Methanal can be made by dehydrogenating methanol vapour:

CH₃OH(g) ⇌ HCHO(g) + H₂(g)

The forward reaction is endothermic. For each change, state the effect on the equilibrium yield of methanal and on the value of Kp.

(a) The total pressure is increased at constant temperature. [2] (b) A catalyst is added. [2] (c) The temperature is raised. [2]

7. This question uses the equilibrium mixture from question 3, where p(SbCl₅) = 1.05 atm, p(SbCl₃) = p(Cl₂) = 0.450 atm and Kp = 0.193 atm. The volume of the vessel is suddenly halved at constant temperature.

(a) Calculate the value of p(SbCl₃) × p(Cl₂) / p(SbCl₅) immediately after the volume is halved. [2] (b) Use your answer to explain the direction in which the equilibrium moves, and state the effect on Kp. [2]

8. Three reactions have these equilibrium constants at 298 K:

Reaction	K
P	3 × 10²²
Q	2 × 10⁻³¹
R	0.6

(a) Calculate ΔStotal for reaction P at 298 K. [1] (b) State what you would expect to see for each reaction if the reactants were mixed and left to reach equilibrium. [2] (c) Reaction P shows no visible change after a day at room temperature. Suggest why. [1]

9. Use these data for CH₃OH(g) ⇌ HCHO(g) + H₂(g): ΔH = +92.4 kJ mol⁻¹ and ΔSsystem = +109.7 J K⁻¹ mol⁻¹. Assume both values do not change with temperature.

(a) Calculate ΔSsurroundings and ΔStotal at 700 K and at 900 K. [3] (b) Calculate the equilibrium constant, K, at each temperature using ΔStotal = R lnK. [2] (c) Explain, in terms of entropy changes, why K is larger at the higher temperature. [2]

10. Steam reacts with hot iron in a closed vessel:

3Fe(s) + 4H₂O(g) ⇌ Fe₃O₄(s) + 4H₂(g)

The forward reaction is exothermic. At one temperature, with both solids present, the equilibrium partial pressures are p(H₂) = 0.90 atm and p(H₂O) = 0.60 atm.

(a) State whether this is a homogeneous or heterogeneous equilibrium, and write an expression for Kp. [2] (b) Calculate Kp and state its units, if any. [2] (c) The volume of the vessel is halved at constant temperature. State and explain the effect on the composition of the gas mixture. [2] (d) The temperature is raised. State the effect on Kp and on the ratio p(H₂) / p(H₂O). [2] (e) At a higher temperature, Kp = 2.40 and the total equilibrium pressure is 3.00 atm. Calculate p(H₂). [2]

Answers

1. (a) Kc = [C₂H₂][H₂]³ / [CH₄]² [1]; units (mol dm⁻³)⁴ / (mol dm⁻³)² = mol² dm⁻⁶ [1]. (b) Kc = [Fe²⁺]²[Cu²⁺] / [Fe³⁺]², with solid copper left out [1]; units mol dm⁻³ [1]. (c) Kc = [Ca²⁺][OH⁻]² [1]; units mol³ dm⁻⁹ [1]. Examiner insight: Show the cancelling that gives each unit; a bare unit with no working gives the examiner nothing to check if the expression itself has slipped.

2. (a) n(NaOH) = 0.06850 dm³ × 2.00 mol dm⁻³ = 0.137 mol [1]. Subtract the 0.0100 mol HCl: n(CH₃COOH) = 0.127 mol [1]. (b) Ethanol formed equals ethanoic acid formed, 0.127 mol [1]. Ester left = 0.300 − 0.127 = 0.173 mol; water left = 0.500 − 0.127 = 0.373 mol [1]. Kc = [CH₃COOH][C₂H₅OH] / ([CH₃COOC₂H₅][H₂O]), and V cancels because there are two species on each side [1]. Kc = (0.127 × 0.127) / (0.173 × 0.373) = 0.250, no units [1]. Examiner insight: Write Kc for the reaction exactly as the question gives it; the inverted (esterification) expression gives about 4.00, which answers a different question.

3. (a) Kp = p(SbCl₃) × p(Cl₂) / p(SbCl₅) [1] (b) Equilibrium moles: SbCl₅ = 0.500 − 0.150 = 0.350; SbCl₃ = Cl₂ = 0.150; total = 0.650 mol [1]. Partial pressures: p(SbCl₅) = (0.350 / 0.650) × 1.95 = 1.05 atm; p(SbCl₃) = p(Cl₂) = (0.150 / 0.650) × 1.95 = 0.450 atm [1]. Kp = 0.450² / 1.05 = 0.193 [1] atm [1]. Examiner insight: Set out the total moles and each mole fraction × total pressure on separate lines, so that every stage of the method can be seen and credited.

4. (a) Equal amounts of the two gases form, so p(NH₃) = p(HCl) = 0.40 / 2 = 0.20 atm [1]. Kp = 0.20 × 0.20 = 0.040 atm² [1]. (b) p(HCl) = 0.040 / 0.32 = 0.125 atm [1]. Total pressure = 0.32 + 0.125 = 0.445 atm [1]. Examiner insight: Say that Kp is unchanged at fixed temperature; it justifies part (b).

5. n(NH₃) in 10.0 cm³ of aqueous layer = 9.60 × 10⁻³ dm³ × 0.500 mol dm⁻³ = 4.80 × 10⁻³ mol [1]. [NH₃(aq)] = 4.80 × 10⁻³ / 0.0100 = 0.480 mol dm⁻³ [1]. n(NH₃) in 25.0 cm³ of organic layer = 0.0250 × 0.0200 = 5.00 × 10⁻⁴ mol, so the trichloromethane layer has [NH₃] = 0.0200 mol dm⁻³ [1]. K = 0.480 / 0.0200 = 24.0, no units [1]. Examiner insight: Divide each amount by the volume of its own sample, not the acid volume; mixing these up is the usual way this calculation goes wrong.

6. (a) Yield of methanal decreases, because the equilibrium moves to the side with fewer gas moles (1 mol on the left, 2 mol on the right) [1]. Kp is unchanged [1]. (b) Yield unchanged (equilibrium reached sooner) [1]. Kp unchanged [1]. (c) Kp increases because the forward reaction is endothermic [1]. Yield of methanal increases [1]. Examiner insight: State the yield effect and the Kp effect separately.

7. (a) New partial pressures: p(SbCl₅) = 2.10 atm, p(SbCl₃) = p(Cl₂) = 0.900 atm [1]. Value = 0.900² / 2.10 = 0.386 atm [1]. (b) 0.386 atm is greater than Kp, so the fraction must fall: SbCl₃ and Cl₂ combine to form SbCl₅, and the equilibrium moves to the left [1]. Kp is unchanged because the temperature has not changed [1]. Examiner insight: Compare your calculated value with Kp; a Le Chatelier answer alone ignores the calculation.

8. (a) ΔStotal = R lnK = 8.31 × ln(3 × 10²²) = +430 J K⁻¹ mol⁻¹ [1]. (b) P goes essentially to completion and Q hardly reacts at all [1]; R gives an equilibrium mixture with significant amounts of both reactants and products [1]. (c) The activation energy is high, so the rate is very slow: P is kinetically stable even though it is thermodynamically favoured [1]. Examiner insight: In part (c) name rate or activation energy; “K is large” only restates the data.

9. (a) ΔSsurroundings = −ΔH / T, with ΔH = +92 400 J mol⁻¹ [1]. At 700 K: ΔSsurroundings = −132.0 J K⁻¹ mol⁻¹ and ΔStotal = 109.7 − 132.0 = −22.3 J K⁻¹ mol⁻¹ [1]. At 900 K: ΔSsurroundings = −102.7 J K⁻¹ mol⁻¹ and ΔStotal = 109.7 − 102.7 = +7.0 J K⁻¹ mol⁻¹ [1]. (b) At 700 K: K = e^(−22.3 / 8.31) = 0.068 [1]. At 900 K: K = e^(7.0 / 8.31) = 2.3 [1]. (c) As T rises, −ΔH / T becomes less negative, so ΔSsurroundings is smaller in size while ΔSsystem stays about the same [1]. ΔStotal therefore increases, and since ΔStotal = R lnK, K increases [1]. Examiner insight: Convert ΔH to J mol⁻¹ before dividing; kJ makes every later value wrong by a factor of 1000.

10. (a) Heterogeneous (solids and gases present) [1]. Kp = p(H₂)⁴ / p(H₂O)⁴ [1]. (b) Kp = (0.90 / 0.60)⁴ = 5.06 [1]; no units [1]. (c) Both partial pressures double, so the fraction (2p(H₂))⁴ / (2p(H₂O))⁴ is still equal to Kp [1]. There are equal gas moles on each side, so the composition does not change [1]. (d) Kp decreases because the forward reaction is exothermic [1]. The ratio p(H₂) / p(H₂O) falls [1]. (e) p(H₂) / p(H₂O) = 2.40^(1/4) = 1.245 [1]. p(H₂) = 3.00 × 1.245 / (1 + 1.245) = 1.66 atm [1]. Examiner insight: Show the fourth root as its own step so the method is visible even if the final pressure slips.

Where marks are usually lost

• Writing solids into Kc or Kp, or putting aqueous ions into Kp.
• Substituting moles into Kc when the indices above and below do not balance.
• Forgetting to allow for the acid catalyst in the ester titration.
• Finding mole fractions but not multiplying by the total pressure.
• Stating that pressure or a catalyst changes Kp.
• Explaining a temperature effect without saying that K changes.
• Dividing ΔH in kJ by T and adding the result to ΔSsystem in J.

Next steps

• Review the chemical equilibria revision notes.
• Reread weak sections of the chemical equilibria study guide.
• Visit the course hub.
• Mark outcomes 13.1 to 13.9 as done on the printable checklist.
• Try all free 10-minute diagnostics.
• Book a free trial class.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Chemistry (XCH11/YCH11) specification, Issue 1, September 2017, Topic 13: Chemical Equilibria (outcomes 13.1 to 13.9), in Unit 4: Rates, Equilibria and Further Organic Chemistry. Published by Pearson Education Limited.