AQA A-Level Chemistry AS Paper 1, June 2022: Question 7

13 marks · Medium difficulty · State/Explain/Numerical

Apply Le Chatelier's principle to interpret equilibrium yield curves and calculate equilibrium amounts and reverse reaction constants using Kc.

Practise this question

Question

Question 7 consists of five parts on gaseous equilibria. Figure 1 displays a graph of percentage yield versus pressure with two downward-sloping curves: one at T = 300 K (higher yield) and one at T = 600 K (lower yield). Parts 7.1 to 7.3 ask candidates to explain why the reaction is exothermic, deduce whether moles of gas increase or decrease, and state why a catalyst has no effect on yield. Parts 7.4 and 7.5 present the reaction 3H2(g) + N2(g) ⇌ 2NH3(g) with Kc = 0.118 mol^-2 dm^6, asking for the moles of ammonia at equilibrium in a 0.150 dm^3 vessel, followed by the value and units of Kc for the reverse dissociation reaction.
Question text

07 This question is about gaseous equilibria.

Figure 1 shows the effect of pressure on the percentage yield of a reaction at

equilibrium at two different temperatures.

Figure 1

07.1 Explain how Figure 1 shows that the forward reaction in this equilibrium is

exothermic.

[2 marks]

07.2 State whether the forward reaction in this equilibrium results in an increase, decrease

or no change in the amount, in moles, of gas.

Explain your answer.

[3 marks]

Tick ( ) one box.

increase

decrease

no change

Explanation

07.3 Explain why using a catalyst has no effect on the percentage yield.

[1 mark]

Hydrogen and nitrogen react to form ammonia.

3H2(g) + N2(g) ⇌ 2NH3(g)

At 745 K, the equilibrium constant, K = 0.118 mol–2 dm6

c

07.4 At 745 K, 0.150 dm3 of an equilibrium mixture contains 0.0285 mol of hydrogen and

0.0870 mol of nitrogen.

Calculate the amount, in moles, of ammonia present in this equilibrium mixture.

[5 marks]

Amount of ammonia mol

07.5 Calculate the value, at 745 K, for the equilibrium constant Kc for this dissociation of

ammonia to give hydrogen and nitrogen.

State the units.

2NH3(g) ⇌ 3H2(g) + N2(g)

[2 marks]

Value

Units

Mark scheme

Show the mark scheme Mark scheme for Question 7. 07.1 awards 2 marks for noting lower yield at higher temperature, so equilibrium shifts left/backwards in the endothermic direction. 07.2 awards 3 marks for selecting 'increase', stating yield decreases at higher pressure, and explaining equilibrium shifts left to fewer moles of gas. 07.3 awards 1 mark for stating that a catalyst increases the rate of forward and reverse reactions equally. 07.4 awards 5 marks for the Kc expression, calculating concentrations, finding [NH3]^2 = 4.69 x 10^-4, [NH3] = 0.0217 mol dm^-3, and moles of NH3 = 3.25 x 10^-3 mol. 07.5 awards 2 marks for Kc = 1/0.118 = 8.47 and units mol^2 dm^-6.

Question Marking guidance Additional Comments/Guidelines Mark

Allow converse arguments 1

Lower yield at higher temperature Higher yield at lower temperature

07.1 So, equilibrium has shifted backwards/left/in the endothermic direction So, equilibrium has shifted forwards/right/in the (2 x AO3)

to oppose the increase in temperature exothermic direction to oppose the decrease in

temperature

Ignore reference to pressure

Increase 1

Lower yield at higher pressure Higher yield at lower pressure 1

07.2 So, equilibrium has shifted backwards/to the side with fewest number So, equilibrium has shifted forwards/to the side with 1

of moles to oppose the increase in pressure highest number of moles to oppose the decrease in (3 x AO3)

pressure

Ignore reference to temperature

Increases the rate of forwards and reverse reactions equally – HEMISTRY – – 1

07.3

(AO1)

[𝑁𝐻 ]2 M1: Kc expression 1

20 3

𝐾𝑐 = 3

[𝐻2] [𝑁2]

[𝑁𝐻 ]2

0.118 = 3 M2: converts moles to concentration; divides mole

[0.0285⁄ ] [0.0870⁄ ] quantities by 0.150

0.150 0.150

[𝑁𝐻 ]2

0.118 = 3

[0.190] [0.580]

07.4

2 −4 M3: calculation of [NH3]

[𝑁𝐻3] = 4.69 × 10

[ ] −3 M4 = √𝑀3

𝑁𝐻3 = 0.0217 𝑚𝑜𝑙 𝑑𝑚

−3 M5 = M4 x 0.150 (allow ecf on an incorrect volume (5 x AO2)

𝑛(𝑁𝐻3) = 0.0217 ×0.150 = 3.25 × 10 𝑚𝑜𝑙

used in M2)

If Kc upside down then can still score 4

1 Allow 8.45 – 8.5 1

𝐾𝑐(= ) = 8.47

0.118

07.5

Units – mol2 dm–6 1

(2 x AO2)

How to answer it

Gaseous Equilibria, Le Chatelier's Principle & Kc Calculations

📋 What this question tests

This question assesses your mastery of Le Chatelier's Principle (interpreting yield curves against changes in temperature and pressure), the kinetic role of a catalyst in reversible reactions, multi-step Kc calculations involving volume conversions, and determining the equilibrium constant and units for a reverse reaction.

Part 07.1

Explaining Exothermic Equilibrium from Graphical Data

2 Marks • Assessment Objective: AO3

✅ Model Answer

  • Mark 1: At higher temperature (600 K), the percentage yield is lower (or yield is higher at 300 K).
  • Mark 2: According to Le Chatelier's principle, an increase in temperature shifts the equilibrium in the endothermic direction to oppose the increase. Since the yield decreases, the reverse reaction is endothermic, meaning the forward reaction must be exothermic.

🧠 Exam Technique

Always structure Le Chatelier graph explanations in two clear steps:

  1. State the direct graphical observation comparing values (e.g. yield decreases from 300 K to 600 K at any fixed pressure).
  2. Apply the principle explicitly: state which direction the position of equilibrium shifts and why (endothermic opposes heat addition).
Examiner Insight: Ignore any reference to pressure in this part. Candidates often lose marks by simply stating "higher temperature favours endothermic" without linking it directly to the evidence from the graph (that yield is lower at 600 K).
Part 07.2

Deducing Mole Changes from Pressure Variation

3 Marks • Assessment Objective: AO3

✅ Model Answer

Box ticked: [✓] Increase (1 Mark)

Explanation:

  • Mark 2: As pressure increases, the percentage yield decreases (or yield is higher at lower pressure).
  • Mark 3: Increasing pressure shifts the equilibrium in the direction of fewer gaseous moles to oppose the increase. Since the yield of product decreases, the backward direction has fewer moles, meaning the forward reaction results in an increase in moles of gas.

❌ Common Errors

  • Ticking "decrease" due to confusing the direction of shift with the effect of the forward reaction.
  • Writing "equilibrium moves to the right" without linking to how the graph shows a drop in yield.
  • Forgetting to explicitly mention "moles of gas" (stating just "fewer particles" or "fewer molecules").
Marking breakdown: 1 mark for ticking "increase", 1 mark for stating that yield decreases as pressure increases, 1 mark for explaining that equilibrium shifts backwards / to the side with fewer moles of gas to oppose the increase in pressure.
Part 07.3

Effect of a Catalyst on Equilibrium Yield

1 Mark • Assessment Objective: AO1

✅ Model Answer

A catalyst increases the rate of both the forward and reverse reactions equally.

💡 Key Knowledge

A catalyst lowers the activation energy (Ea) by providing an alternative pathway. Because it lowers Ea for both the forward and reverse reactions by the same extent, both rates increase by the identical factor.

Therefore, equilibrium is reached faster, but the position of equilibrium and percentage yield remain unchanged.

Part 07.4

5-Mark Equilibrium Calculation (Haber Process Kc)

Equation: 3 H₂(g) + N₂(g) ⇌ 2 NH₃(g)

📐 Step-by-Step Calculation

  1. Step 1: Write the Kc expression (Mark 1)
    Kc = [NH₃]² / ([H₂]³[N₂])
  2. Step 2: Convert moles to equilibrium concentrations in mol dm⁻³ (Mark 2)
    Given volume = 0.150 dm³:
    [H₂] = 0.0285 / 0.150 = 0.190 mol dm⁻³
    [N₂] = 0.0870 / 0.150 = 0.580 mol dm⁻³
  3. Step 3: Substitute values into the expression to solve for [NH₃]² (Mark 3)
    0.118 = [NH₃]² / ((0.190)³ × (0.580))
    (0.190)³ × 0.580 = 0.006859 × 0.580 = 3.9782 × 10⁻³
    [NH₃]² = 0.118 × 3.9782 × 10⁻³ = 4.694 × 10⁻⁴ mol² dm⁻⁶
  4. Step 4: Take the square root to find concentration [NH₃] (Mark 4)
    [NH₃] = √(4.694 × 10⁻⁴) = 0.02167 mol dm⁻³
  5. Step 5: Convert concentration back to moles of NH₃ (Mark 5)
    n(NH₃) = [NH₃] × Volume = 0.02167 × 0.150 = 3.25 × 10⁻³ mol (allow 3.25 × 10⁻³ to 3.26 × 10⁻³)

❌ Common Traps in this Calculation

  • Forgetting to divide by volume: Because the powers in numerator and denominator are not equal (2 vs 4), the volume does NOT cancel out!
  • Forgetting to cube [H₂]: The stoichiometric coefficient of 3 becomes an exponent of 3 in the expression.
  • Stopping at concentration: The question specifically asked for amount, in moles. Multiplying back by 0.150 dm³ is essential for the final mark.

🧠 Error-Carried-Forward (ECF) Note

If you set up Kc inverted (reactants over products), the mark scheme allows up to 4 out of 5 marks if the subsequent algebraic manipulation is followed correctly.

Part 07.5

Equilibrium Constant for the Reverse Reaction

2 Marks • Assessment Objective: AO2

✅ Model Answer

Equation: 2 NH₃(g) ⇌ 3 H₂(g) + N₂(g)

  • Value: Kc' = 1 / 0.118 = 8.47 (allow 8.45 to 8.5) (1 Mark)
  • Units: mol² dm⁻⁶ (1 Mark)

💡 Key Knowledge: Inverting Reactions

When an equilibrium reaction is reversed:

  • The new equilibrium constant is the mathematical reciprocal of the original:
    Kc(reverse) = 1 / Kc(forward)
  • The units are also completely inverted:
    Original units = mol⁻² dm⁶
    New units = 1 / (mol⁻² dm⁶) = mol² dm⁻⁶

Topics

Physical Chemistry · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 1, June 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.