AQA A-Level Chemistry Paper 3, 2024: Question 29

✅ Equilibrium & Le Châtelier

• Predict how pressure changes equilibrium position using moles of gas.
• Recognise that a catalyst changes rate, not equilibrium yield.
• Use the sign of ΔH to predict effect of temperature on Kₚ .

🧠 Multiple choice exam skill

• Quickly eliminate statements that confuse rate vs yield.
• Link each option to a specific principle: pressure, temperature, catalyst, comparison with fermentation.

✅ Correct answer (what to write / tick)

D — Increasing pressure (constant temperature) increases the equilibrium yield of ethanol.

This is the only option consistent with Le Châtelier’s principle for gases in this equation.

💡 Key knowledge (why D is correct)

• Count gas moles: left side has 2 mol gas (C₂H₄ + H₂O), right side has 1 mol gas (C₂H₅OH).
• Increasing pressure favours the side with fewer moles of gas.
• So higher pressure shifts equilibrium to the right → higher ethanol yield.

🧠 Exam technique (fast elimination)

• Pressure question? Immediately compare total moles of gas on each side.
• Temperature question? Use ΔH : exothermic means higher T shifts left and Kₚ decreases.
• Catalyst mentioned? Remember: catalyst changes how fast equilibrium is reached, not the position of equilibrium.

❌ Common errors (examiner-style)

• Mixing up rate and yield: stating a catalyst “increases yield” (it doesn’t at equilibrium).
• Forgetting gas mole counting: pressure only shifts equilibrium when the number of gas moles differs.
• Wrong Kₚ trend with temperature: many students think “higher temperature increases Kₚ” automatically—this depends on ΔH .

❌ Why A is wrong

• A catalyst does not change equilibrium position or equilibrium constant.
• It lowers activation energy for both forward and reverse reactions, so equilibrium is reached faster, but the final equilibrium composition is unchanged.

💡 What’s true about the acid catalyst?

• Industrial hydration uses an acid catalyst (e.g. phosphoric acid) to achieve an acceptable rate at workable conditions.
• If asked “why use a catalyst?” the mark is typically for: increases rate / lowers Eₐ.

❌ Why B is wrong (and a common trap)

• Industrial hydration is generally a fast, continuous process compared with fermentation (which is slower and batch-based).
• This statement isn’t supported by the equilibrium/thermo data given; it’s a context distractor.

🧠 Exam technique

• If the stem gives ΔH and a gas-phase equilibrium, the correct option is usually about pressure/temperature/K, not a vague comparison with fermentation.

❌ Why C is wrong

• The reaction is exothermic ( ΔH is negative).
• Increasing temperature shifts equilibrium to the left (endothermic direction).
• Therefore the equilibrium constant Kₚ decreases when temperature increases for an exothermic reaction.

💡 Temperature ↔ K rule

• If forward reaction is exothermic: ↑T → K ↓.
• If forward reaction is endothermic: ↑T → K ↑.

✅ Why D is correct (full-mark logic)

1. Compare gas moles: left = 2, right = 1.
2. ↑Pressure favours fewer gas moles (Le Châtelier).
3. Equilibrium shifts right → more C₂H₅OH formed → higher equilibrium yield.

📐 “Calculation-style” check (no maths, just a scoring routine)

1. Write moles of gas under each side: LHS 2, RHS 1.
2. Circle the smaller number (1 on RHS).
3. Pressure ↑ → shift towards circled side → products.

This quick routine avoids the common mistake of thinking pressure “increases collisions so more product” (rate idea) instead of using equilibrium reasoning.

❌ Misconceptions that cost the mark

• Catalyst = higher yield (incorrect). Catalysts affect kinetics not thermodynamics.
• Temperature always increases K (incorrect). Must link to sign of ΔH .
• Not checking the state symbols: because all species are gases here, pressure effects are relevant and assessed by gas mole counting.

🧠 What distinguishes top responses

• They immediately use: “2 mol gas → 1 mol gas” to justify pressure shift.
• They keep concepts separated: rate (catalyst) vs yield (equilibrium position).
• They connect ΔH = −46 kJ mol⁻¹ to temperature decreasing Kₚ.