AQA A-Level Chemistry Paper 1, June 2024: Question 4

9 marks · Medium difficulty · State/Explain/Numerical

Calculate equilibrium mole fractions, write Kp expressions, compute Kp with units, and deduce the effects of temperature and volume changes on gaseous equilibria.

Practise this question

Question

Question 4 contains five parts related to gas phase equilibria. Part 04.1 gives the reaction H2O(g) + CO(g) ⇌ H2(g) + CO2(g) (ΔH = -41 kJ mol⁻¹) and asks to calculate the equilibrium mole fraction of H2 given initial moles and equilibrium moles of H2. Part 04.2 asks why Kp for this reaction has no units. Part 04.3 asks whether the amount of H2 increases, decreases, or stays the same when temperature increases. Part 04.4 gives the reaction C2H4(g) + H2O(g) ⇌ CH3CH2OH(g) (ΔH = -45 kJ mol⁻¹) and a table of mole fractions at 6000 kPa (Ethene: 0.645, Steam: 0.323, Ethanol: 0.0321), requiring an expression for Kp, calculation of Kp, and its units. Part 04.5 asks for the effect of an increase in container volume on Kp.
Question text

04 This question is about some gas mixtures at equilibrium.

This reaction can be used to make hydrogen.

H O(g) + CO(g) ⇌ H (g) + CO (g) ΔH = –41 kJ mol–1

22 2

04.1 A mixture of 2.00 mol of H2O(g) and 2.00 mol of CO(g) is allowed to reach equilibrium

at a constant temperature in a 20 dm3 container.

At equilibrium, there are 0.92 mol of H2(g).

Calculate the mole fraction of H2(g) in the equilibrium mixture.

[2 marks]

Mole fraction of H2(g)

04.2 State why the equilibrium constant (Kp) for this reaction has no units.

[1 mark]

04.3 The temperature of the equilibrium mixture formed in Question 04.1 is increased.

How does the amount of H2(g) change when the new position of equilibrium is

reached?

[1 mark]

Tick ( ) one box.

The amount decreases.

The amount does not change.

The amount increases.

Ethanol can be made from ethene and steam.

C H (g) + H O(g) ⇌ CH CH OH(g) ΔH = –45 kJ mol–1

24 2 3 2

Table 2 shows the mole fractions of each of the gases in an equilibrium mixture at

6000 kPa

Table 2

Gas Mole fraction

Ethene 0.645

Steam 0.323

Ethanol 17 0.0321

04.4 Give an expression for Kp for this reaction.

Calculate the value of Kp at 6000 kPa

State the units.

[4 marks]

Kp

Units

04.5 State the effect, if any, of an increase in volume of the container on the value of Kp for

this reaction at a constant temperature.

[1 mark]

Mark scheme

Show the mark scheme Mark scheme for Question 4. 04.1: Total amount of gas = 4.00 mol, mole fraction of H2 = 0.23 (2 marks). 04.2: Amount in moles of products equals reactants so units cancel out (1 mark). 04.3: The amount decreases (1 mark). 04.4: M1 gives Kp = p(CH3CH2OH) / [p(C2H4)p(H2O)]; M2 calculates partial pressures using 6000 kPa; M3 gives Kp = 2.56 × 10⁻⁵ (or 2.57 × 10⁻⁵); M4 gives units kPa⁻¹ (4 marks). 04.5: No effect (1 mark).

Question Answers Additional comments/Guidelines Mark

Total amount of gas at equilibrium = 4.00 mol

04.1

Accept fractions

Mole fraction of H2 = 0.23 (2 x AO2)

Allow equal moles/molecules/particles on both

(The amount in) moles of products is the same as reactants (so 1

04.2 sides (so units cancel)

the units (of partial pressure) cancel out)

(1 x AO1)

04.3 The amount decreases.

(1 x AO1)

𝑝𝑝(𝐶𝐶𝐶𝐶3𝐶𝐶𝐶𝐶2𝑂𝑂𝐶𝐶) M1 allow p or pp for partial pressure; round

M1 Kp = ( 𝐶𝐶 ) 𝑂𝑂) brackets not necessary BUT penalise square

𝑝𝑝 𝐶𝐶2 4 𝑝𝑝(𝐶𝐶2

brackets.

0.0321×6000 192.6

M2 = = 4

(0.645 x 6000)×(0.323×6000) 3870 x 1938 M2 allow correct use of 6000 to calculate partial

04.4

pressure (4 x AO2)

–5 M3 Allow 2.57 × 10–5

M3 Kp = 2.56 × 10

M4 kPa–1

04.5 no effect 1

(1 x AO1)

How to answer it

Gas Phase Equilibria: Mole Fractions, Kp, and Le Chatelier’s Principle

AQA A-Level Chemistry • Physical Chemistry: Equilibrium Constants

📋 What this question tests

  • Setting up and completing equilibrium mole tables (ICE tables) for gaseous systems.
  • Calculating mole fractions: mole fraction = moles of gas / total moles of gas .
  • Writing expressions for the gaseous equilibrium constant, Kp, using partial pressures.
  • Determining the numerical value and correct units of Kp from mole fractions and total pressure.
  • Applying Le Chatelier’s principle to predict shifts when temperature changes.
  • Understanding the fundamental rule: only temperature changes the value of Kp (volume/pressure changes do not).
Question 04.1 • 2 Marks

Equilibrium Mole Fraction Calculation

H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)

📐 Step-by-Step Calculation

  1. Determine equilibrium moles using reacting ratios (1:1:1:1):
    Species H₂O(g) CO(g) H₂(g) CO₂(g)
    Initial 2.00 2.00 0.00 0.00
    Change −0.92 −0.92 +0.92 +0.92
    Equilibrium 1.08 mol 1.08 mol 0.92 mol 0.92 mol
  2. Calculate total gaseous moles at equilibrium:
    Total moles = 1.08 + 1.08 + 0.92 + 0.92 = 4.00 mol
  3. Calculate mole fraction of H₂(g):
    Mole fraction = 0.92 / 4.00 = 0.23

❌ Common Errors & Traps

  • Dividing by volume: The question states the container is 20 dm³. Many students mistakenly divided moles by 20 to find concentrations. Mole fraction does not depend on volume.
  • Assuming total moles = initial moles: While total moles happens to remain 4.00 here because Δn = 0 (2 moles reactants → 2 moles products), always write out the ICE table to avoid careless errors.
  • Giving units: Mole fractions are dimensionless ratios; never add units like mol or mol dm⁻³ .

✅ Final Answer

Mole fraction of H₂(g) = 0.23 (or equivalent fraction 23/100)

Mark Scheme Breakdown:
• M1: Total amount of gas at equilibrium = 4.00 mol [AO2]
• M2: Mole fraction of H₂ = 0.23 (accept fractions) [AO2]
Question 04.2 • 1 Mark

Units of Kp

Explaining why Kp is dimensionless

💡 Key Knowledge

The units of Kp depend on the powers of partial pressure in the numerator versus the denominator:

Units = (pressure) × (pressure) / [ (pressure) × (pressure) ] = no units

Because there are equal numbers of moles of gaseous products and gaseous reactants, the pressure terms fully cancel out.

🧠 Exam Technique

  • Always refer to both sides of the equation having the same number of moles / molecules.
  • Explicitly mention that the units of partial pressure cancel out.

✅ Accepted Answer

There are the same number of moles (or molecules/particles) of gas on both sides, so the units of partial pressure cancel out.

Mark Scheme Breakdown:
• 1 mark: (The amount in) moles of products is the same as reactants (so the units cancel out). [AO1]
Question 04.3 • 1 Mark

Effect of Temperature on Equilibrium

H₂O(g) + CO(g) ⇌ H₂(g) + CO₂(g)    ΔH = −41 kJ mol⁻¹

💡 Key Knowledge

  • The forward reaction has a negative enthalpy change ( ΔH = −41 kJ mol⁻¹ ), meaning it is exothermic.
  • By Le Chatelier's Principle, an increase in temperature favours the endothermic direction (the reverse reaction) to absorb added thermal energy.
  • The position of equilibrium shifts to the left.

❌ Common Errors

  • Confusing rate with equilibrium position (thinking higher temperature always means more products formed).
  • Misreading the sign of ΔH (negative means exothermic, not endothermic).

✅ Correct Box to Tick

☑ The amount decreases.

Mark Scheme Breakdown:
• 1 mark: The amount decreases. [AO1]
Question 04.4 • 4 Marks

Kp Expression, Calculation & Units

C₂H₄(g) + H₂O(g) ⇌ CH₃CH₂OH(g)    (Total Pressure = 6000 kPa)

📐 Step-by-Step Calculation

  1. Write the Kp expression:
    Use p(species) notation. Square brackets are penalised!
    Kp = p(CH₃CH₂OH) / [ p(C₂H₄) × p(H₂O) ]
  2. Calculate partial pressures:
    partial pressure = mole fraction × total pressure
    • p(CH₃CH₂OH) = 0.0321 × 6000 = 192.6 kPa
    • p(C₂H₄) = 0.645 × 6000 = 3870 kPa
    • p(H₂O) = 0.323 × 6000 = 1938 kPa
  3. Substitute into Kp:
    Kp = 192.6 / (3870 × 1938) = 192.6 / 7 500 060
    Kp = 2.57 × 10⁻⁵ (or 2.56 × 10⁻⁵)
  4. Deduce units:
    Units = kPa / (kPa × kPa) = kPa⁻¹

🧠 Exam Technique & Examiner Tips

  • Brackets matter: In Kp expressions, never write [CH₃CH₂OH] . Square brackets represent concentration in mol dm⁻³. Always write p(CH₃CH₂OH) or pp(CH₃CH₂OH) .
  • Keep the pressure unit consistent: The total pressure is in kPa, so partial pressures are in kPa, making the units kPa⁻¹. If you convert to Pa (6 000 000 Pa), your value will be 2.57 × 10⁻⁸ and units Pa⁻¹. It is much safer to stay in kPa.
  • Significant figures: The given mole fractions have 3 significant figures, so round your final numerical value to 3 s.f. (2.57 × 10⁻⁵).

✅ Final Answers

• Expression: Kp = p(CH₃CH₂OH) / ( p(C₂H₄) × p(H₂O) )

• Kp Value: 2.57 × 10⁻⁵ (allow 2.56 × 10⁻⁵)

• Units: kPa⁻¹

Mark Scheme Breakdown:
• M1: Expression with partial pressures (penalise square brackets) [AO2]
• M2: Correct use of 6000 to find partial pressures [AO2]
• M3: Correct numerical answer: 2.56 × 10⁻⁵ or 2.57 × 10⁻⁵ [AO2]
• M4: Units = kPa⁻¹ [AO2]
Question 04.5 • 1 Mark

Effect of Volume / Pressure on Kp

Constant temperature condition

💡 The Golden Rule of Equilibrium Constants

The equilibrium constant Kp is dependent ONLY on temperature.

Increasing the volume of the container decreases the total pressure. While this will cause the position of equilibrium to shift towards the side with more gaseous moles (to the left), the value of Kp remains completely unchanged.

❌ The #1 Student Pitfall

Students frequently confuse the position of equilibrium with the equilibrium constant (Kp).

  • Position of equilibrium: Shifts to the left.
  • Value of Kp: No effect / Unchanged.

✅ Correct Answer

No effect (or unchanged / stays the same)

Mark Scheme Breakdown:
• 1 mark: no effect [AO1]

Topics

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

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.