AQA A-Level Chemistry Paper 1, 2023: Question 4

11 marks · Medium difficulty · State/Explain/Numerical

Calculate mole fractions, partial pressure, and total pressure using Kp for the decomposition of nitrogen dioxide, and predict the effects of changing volume and temperature on equilibrium.

Practise this question

Question

Multi-part question based on the equilibrium 2NO2(g) ⇌ 2NO(g) + O2(g) with ΔH = +113 kJ mol⁻¹. Question 04.1 gives 0.317 mol of NO2 initially and 0.120 mol of O2 at equilibrium, asking to calculate the mole fraction of each substance. Question 04.2 asks for the partial pressure of NO2 at a total pressure of 120 kPa. Question 04.3 provides a table of mole fractions (NO2: 0.310, NO: 0.460, O2: 0.230) and Kp = 59.7 kPa, asking for the Kp expression and the total pressure in kPa. Question 04.4 asks for the effect of compression on the yield of O2 and Kp. Question 04.5 asks for an explanation of why the equilibrium yield of O2 decreases at lower temperature.
Question text

04 Nitrogen dioxide decomposes at a high temperature.

2 NO (g) ⇌ 2 NO(g) + O (g) ΔH = +113 kJ mol–1

04.1 A 0.317 mol sample of nitrogen dioxide is placed in a sealed flask and heated at a

constant temperature until equilibrium is reached.

At equilibrium, the flask contains 0.120 mol of oxygen.

Calculate the mole fraction of each substance at equilibrium.

[3 marks]

Mole fraction of NO2

Mole fraction of NO

Mole fraction of O2

04.2 The total pressure in the flask in Question 04.1 is 120 kPa at equilibrium.

Calculate the partial pressure, in kPa, of NO2

If you were unable to answer Question 04.1 you should assume that the mole fraction

of NO2 is 0.380. This is not the correct answer.

[1 mark]

Partial pressure8 kPa

04.3 Table 1 shows the mole fractions of the three gases in a different equilibrium mixture.

2 NO (g) ⇌ 2 NO(g) + O (g) ΔH = +113 kJ mol–1

*07* Table 1

Gas Mole fraction

NO2 0.310

NO 0.460

O2 0.230

For this equilibrium mixture, Kp = 59.7 kPa

Give an expression for Kp for this reaction.

Use your expression and the data in Table 1 to calculate the total pressure, in kPa, in

the flask.

[3 marks]

Kp

Total pressure9 kPa

04.4 The equilibrium mixture in Question 04.3 is compressed into a smaller volume.

Deduce the effect, if any, of this change on the equilibrium yield of oxygen and on the

value of Kp

*08* [2 marks]

Effect on yield of oxygen

Effect on Kp

04.5 The equilibrium mixture in Question 04.3 is allowed to reach equilibrium at a lower

temperature.

Explain why the equilibrium yield of oxygen decreases.

[2 marks]

Mark scheme

Show the mark scheme Mark scheme for Question 04. 04.1 gives mole fractions: NO2 = 0.176, NO = 0.549, O2 = 0.275 (3 marks). 04.2 gives partial pressure = 21.1 kPa (1 mark). 04.3 gives Kp expression Kp = p(NO)^2 p(O2) / p(NO2)^2, rearranged formula for total pressure, and final answer 117.9 kPa or 118 kPa (3 marks). 04.4 awards 1 mark for yield decreasing and 1 mark for no change in Kp. 04.5 awards 2 marks for stating the reaction is endothermic (or reverse is exothermic) and equilibrium shifts in the reverse direction to raise the temperature.

Question Answers Additional comments/Guidelines Mark

M1 0.176 Allow answers to 2 significant figures 3

04.1 M2 0.549

(3 x AO2)

M3 0.275

21.1 (kPa) Allow answer to question 04.1 x 120 and answer 1

in kPa (AO2)

04.2 Allow 21.6 (kPa)

Answer using given value of 0.380 mol =

45.6(kPa)

M1 K = p(NO)2p(O )

p 2

p(NO )2

2 Do not allow square brackets

04.3 M2 = K mol frac (NO )2 OR 59.7 ×(0.31)2

p x 2 (1 x AO1,

mol frac (NO)2x mol frac (O ) (0.46)2 × (0.23) Rearrangement

22 x AO2)

M3 = 117.9 (kPa) or 118 (kPa)

Decrease

04.4 (1 x AO1,

No change

1 x AO3)

M1 Reaction is endothermic OR exothermic in backwards direction

04.5

M2 Equilibrium shifts/moves in backwards direction/to the left (2 x AO3)

to raise the temp/oppose the decrease in temp

How to answer it

Gas Phase Equilibria: Mole Fractions, Kₚ & Le Chatelier's Principle

What this question tests

This multi-step question assesses key Physical Chemistry concepts in homogeneous gas equilibria:

  • Constructing an equilibrium mole table (ICE method: Initial, Change, Equilibrium) using stoichiometric reacting ratios.
  • Calculating mole fractions and converting mole fractions to partial pressures using Dalton's law ( p = x × P_total ).
  • Writing standard expressions for the equilibrium constant Kₚ without invalid notation (no square brackets!).
  • Algebraic rearrangement of the Kₚ expression to solve for total pressure.
  • Predicting and explaining the effect of pressure and temperature changes on equilibrium yield and the numerical value of Kₚ .
Reaction: 2NO₂(g) ⇌ 2NO(g) + O₂(g)   ΔH = +113 kJ mol⁻¹
Question 04.1 • 3 Marks

Equilibrium Mole Fractions

Calculate the mole fraction of each substance at equilibrium given 0.317 mol initial NO₂ and 0.120 mol O₂ at equilibrium.

📐 Step-by-Step Calculation

  1. Identify stoichiometric changes:
    Ratio is 2 NO₂ : 2 NO : 1 O₂ .
    Since 0.120 mol of O₂ formed:
    • Moles of NO formed = 2 × 0.120 = +0.240 mol
    • Moles of NO₂ reacted = 2 × 0.120 = -0.240 mol
  2. Determine equilibrium moles:
    n(NO₂) = 0.317 - 0.240 = 0.077 mol
    n(NO) = 0.240 mol
    n(O₂) = 0.120 mol
  3. Calculate total moles:
    n(total) = 0.077 + 0.240 + 0.120 = 0.437 mol
  4. Calculate mole fractions (x = n / n_total):
    x(NO₂) = 0.077 / 0.437 = 0.176
    x(NO) = 0.240 / 0.437 = 0.549
    x(O₂) = 0.120 / 0.437 = 0.275

✅ Correct Answers

Mole fraction of NO₂ = 0.176 (or 0.18)

Mole fraction of NO = 0.549 (or 0.55)

Mole fraction of O₂ = 0.275 (or 0.27 / 0.28)

Award 1 mark for each correct mole fraction (allow 2 significant figures).

❌ Common Errors

  • Forgetting that 2 moles of NO₂ react for every 1 mole of O₂ produced, simply subtracting 0.120 from 0.317.
  • Dividing by the initial moles (0.317) instead of finding the new total moles at equilibrium (0.437).
Question 04.2 • 1 Mark

Partial Pressure Calculation

Total pressure is 120 kPa. Calculate the partial pressure of NO₂.

📐 Calculation

Apply Dalton's law of partial pressure:

p(NO₂) = mole fraction × P_total

p(NO₂) = 0.176 × 120 kPa = 21.1 kPa

(Using 0.18 gives 21.6 kPa; using fallback 0.380 gives 45.6 kPa)

🧠 Exam Technique & Guidance

Error-Carried-Forward (ECF): If you miscalculated part 04.1, you can use your own calculated value × 120.

The exam provided a fallback value ( 0.380 ). If used, 0.380 × 120 = 45.6 kPa scores the mark, ensuring you aren't penalised twice!

1 mark for 21.1 kPa (or 21.6 kPa, or 45.6 kPa from fallback).
Question 04.3 • 3 Marks

Expression for Kₚ & Calculating Total Pressure

Give the Kₚ expression and use Table 1 data (Kₚ = 59.7 kPa) to calculate total pressure.

💡 Kₚ Expression Rule

Write the expression using small 'p' for partial pressure:

Kₚ = [p(NO)]² × p(O₂) / [p(NO₂)]²

⚠️ NEVER use square brackets like [NO]! Square brackets denote concentration in mol dm⁻³ (for K꜀). Using them forfeits mark 1 automatically.

📐 Step-by-Step Pressure Calculation

  1. Express partial pressures in terms of total pressure P:
    p(NO) = 0.460 × P
    p(O₂) = 0.230 × P
    p(NO₂) = 0.310 × P
  2. Substitute into Kₚ:
    Kₚ = (0.460P)² × (0.230P) / (0.310P)²
    Kₚ = [ (0.460)² × 0.230 / (0.310)² ] × [ P³ / P² ]
    Kₚ = [ 0.048668 / 0.0961 ] × P = 0.50643 × P
  3. Rearrange to solve for P:
    P = Kₚ × (0.310)² / [ (0.460)² × 0.230 ]
    P = 59.7 / 0.50643 = 117.9 kPa (or 118 kPa)
M1: Correct Kₚ expression
M2: Correct rearrangement for total pressure
M3: Final answer = 117.9 (or 118) kPa
Question 04.4 • 2 Marks

Effect of Compression (Increased Pressure)

Mixture compressed into a smaller volume. Deduce effect on yield of oxygen and value of Kₚ.

✅ Correct Deductions

Effect on yield of oxygen: Decrease

Effect on value of Kₚ: No change

1 mark for each correct deduction.

🧠 Chemical Rationale

  • Smaller volume = higher total pressure: By Le Chatelier's principle, the equilibrium shifts to the side with fewer moles of gas to oppose the increase in pressure. Left side has 2 moles of gas; right side has 3 moles (2 + 1). Shift is to the left, decreasing O₂ yield.
  • Constant Kₚ: The equilibrium constant Kₚ is only affected by temperature. Changes in pressure or concentration shift equilibrium positions but do not change the numerical value of Kₚ.
Question 04.5 • 2 Marks

Effect of Lowering Temperature

Explain why the equilibrium yield of oxygen decreases at a lower temperature.

✅ Required Examiner Points

  • Mark 1: The forward reaction is endothermic (ΔH = +113 kJ mol⁻¹) OR the backward reaction is exothermic.
  • Mark 2: The equilibrium shifts to the left / in the backwards direction to oppose the decrease in temperature / raise the temperature (by releasing heat).
2 marks total (1 mark for thermodynamic classification, 1 mark for direction of shift linked to Le Chatelier).

❌ Common Student Pitfalls

  • Failing to state the direction of enthalpy: Simply saying "it shifts left because temperature decreases" loses Mark 1. You must explicitly link the sign of ΔH (+113 kJ mol⁻¹) to being endothermic.
  • Confusing yield with rate: Stating that "molecules have less kinetic energy so fewer successful collisions occur" explains rate, not the thermodynamic shift in equilibrium position.

Topics

Physical Chemistry · 3.1.10 Equilibrium Constant Kp · 3.1.6 Chemical Equilibria, Le Chatelier's Principle and Kc · 3.1.2 Amount of Substance

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.