Edexcel A-Level Chemistry Paper 1, June 2025: Question 4

8 marks · Medium difficulty · Calculations

Calculate the equilibrium constant Kp and its units for the industrial hydration of ethene, and explain the effects of pressure and temperature on the equilibrium and Kp.

Practise this question

Question

Question 4 gives the equation for the industrial production of ethanol: C2H4(g) + H2O(g) ⇌ C2H5OH(g) with delta H = -45.0 kJ mol^-1. A mixture of 90.0 mol ethene and 54.0 mol steam is heated to 500 K at a total pressure of 60.0 atm, resulting in 5% conversion of ethene to ethanol. Part (a) asks to calculate Kp with expression and units (5 marks). Part (b)(i) asks to explain the effect on yield of increasing pressure to 70.0 atm (2 marks). Part (b)(ii) is a multiple-choice question asking which factor changes the value of Kp (1 mark).

Mark scheme

Show the mark scheme Mark scheme for Question 4: (a) 5 marks awarded for equilibrium moles (85.5, 49.5, 4.5), mole fractions (0.6129, 0.3548, 0.0323), partial pressures (36.77, 21.29, 1.94 atm), expression for Kp = p(C2H5OH) / (p(C2H4)p(H2O)), and final calculated value 2.47 x 10^-3 to 2.5 x 10^-3 atm^-1. (b)(i) 2 marks for stating increased pressure increases ethanol yield because the equilibrium shifts to the RHS with fewer moles/molecules of gas. (b)(ii) 1 mark for C: decreasing the temperature.

How to answer it

Gas Equilibria: Kp & Le Chatelier's Principle

Edexcel A-Level Chemistry • Physical Chemistry • Equilibria

📋 What this question tests

  • ICE Table & Percentage Yield: Determining equilibrium moles when conversion percentage is specified.
  • Gas Calculations: Converting moles to mole fractions and determining equilibrium partial pressures using total pressure.
  • Kp Expression & Units: Writing gaseous equilibrium expressions using round brackets and partial pressure symbols ( p ), followed by dimensional analysis for units.
  • Le Chatelier's Principle: Predicting and justifying shifts in equilibrium position caused by pressure alterations.
  • Equilibrium Constant Invariance: Recognizing that temperature is the only factor that alters the numerical value of Kp.

Part (a) — Calculating Kp and Deducing Units

5 Marks

📐 Step-by-Step Calculation

Reaction: C₂H₄(g) + H₂O(g) ⇌ C₂H₅OH(g)  |  Total Pressure (Ptotal) = 60.0 atm

Step 1: Calculate reacting moles and equilibrium moles
5% of ethene is converted: 5% × 90.0 mol = 4.50 mol converted.
Stoichiometry is 1 : 1 : 1.

Substance C₂H₄(g) H₂O(g) C₂H₅OH(g)
Initial moles 90.0 54.0 0.0
Change in moles −4.5 −4.5 +4.5
Equilibrium moles 85.5 49.5 4.5

Total moles at equilibrium: 85.5 + 49.5 + 4.5 = 139.5 mol

Step 2: Calculate mole fractions (x)
• x(C₂H₄) = 85.5 / 139.5 = 0.6129
• x(H₂O) = 49.5 / 139.5 = 0.3548
• x(C₂H₅OH) = 4.5 / 139.5 = 0.03226

Step 3: Calculate partial pressures (p = x × Ptotal)
• p(C₂H₄) = 0.6129 × 60.0 = 36.77 atm
• p(H₂O) = 0.3548 × 60.0 = 21.29 atm
• p(C₂H₅OH) = 0.03226 × 60.0 = 1.935 (or 1.94) atm

Step 4: Write Kp expression
Kp = p(C₂H₅OH) / [p(C₂H₄) × p(H₂O)]

Step 5: Substitute values and deduce units
• Value: Kp = 1.935 / (36.77 × 21.29) = 2.47 × 10⁻³ (or 2.48 × 10⁻³ with intermediate rounding; allow 0.0025)
• Units: atm / (atm × atm) = atm⁻¹

✅ Mark Breakdown

  • M1: Equilibrium moles: 85.5 (C₂H₄), 49.5 (H₂O), 4.5 (C₂H₅OH).
  • M2: Mole fractions: 0.6129, 0.3548, 0.0323.
  • M3: Partial pressures: 36.77, 21.29, 1.94 atm.
  • M4: Correct expression using partial pressures ( p ).
  • M5: Final value (2.47 × 10⁻³ to 2.5 × 10⁻³) AND units (atm⁻¹).
Note: Transfer of Error (TE) is allowed throughout if arithmetic slips occur early on.

❌ Common Errors to Avoid

  • Using Square Brackets: Writing [C₂H₅OH] instead of p(C₂H₅OH) instantly forfeits M4. Square brackets denote concentration (mol dm⁻³), not partial pressure!
  • Forgetting to subtract reacted moles: Leaving H₂O at 54.0 mol instead of subtracting 4.5 mol.
  • Omitting Units: Forgetting units or writing kPa⁻¹ when given data is in atm .
  • Premature Rounding: Rounding mole fractions to only 1 or 2 sig figs early on skews the final value.

Part (b)(i) — Effect of Increased Pressure on Yield

2 Marks

✅ Model Answer

Increasing the pressure will increase the yield of ethanol (1 mark).

This is because the equilibrium position shifts to the right-hand side / products side, as there are fewer moles of gas on the RHS (1 mole of gas) compared to the LHS (2 moles of gas) (1 mark).

🧠 Exam Technique: 2-Step Pressure Template

Always structure Le Chatelier pressure answers in two distinct sentences:

  1. State the effect on yield: "Yield of [product] increases / decreases."
  2. Quote the gas mole comparison: "Position of equilibrium shifts to the side with fewer moles/molecules of gas to oppose the increase in pressure."
Crucial: Always specify moles of gas, not just "moles".

Part (b)(ii) — Factor That Changes the Value of Kp

1 Mark

✅ Correct Answer

C — decreasing the temperature (1 mark)

💡 Golden Rule of Equilibrium Constants

  • Only temperature changes Kc and Kp.
  • Catalysts: Increase forward and reverse reaction rates equally; no effect on equilibrium position or Kp.
  • Pressure & Volume: Shift the position of equilibrium to keep the ratio equal to Kp, but the numerical value of Kp remains completely unchanged.
  • Since ΔH = −45.0 kJ mol⁻¹ (exothermic), decreasing temperature shifts equilibrium to the right, which actually increases Kp.

Topics

Physical Chemistry · Topic 11: Equilibrium II · Topic 10: Equilibrium I

Question and mark scheme from the Edexcel A-Level Chemistry examination, Paper 1, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.