AQA A-Level Chemistry Paper 1, 2021: Question 5

11 marks · Medium difficulty · State/Explain/Numerical

State and explain the effect of pressure on equilibrium yield, calculate the partial pressure of a reactant, determine the expression, value, and units for Kp, and deduce the effect of pressure on Kp.

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Question text

05 This question is about the equilibrium

2 SO2(g) + O2(g) ⇌ 2 SO3(g)

05.1 State and explain the effect, if any, of a decrease in overall pressure on the

equilibrium yield of SO3

[3 marks]

Effect

Explanation

05.2 A 0.460 mol sample of SO2 is mixed with a 0.250 mol sample of O2 in a

sealed container at a constant temperature.

When equilibrium is reached at a pressure of 215 kPa, the mixture contains

0.180 mol of SO3

Calculate the partial pressure, in kPa, of SO2 in this equilibrium mixture.

[4 marks]

Partial pressure of SO2 kPa

05.3 A different mixture of SO2 and O2 reaches equilibrium at a different temperature.

Table 4 shows the partial pressures of the gases at equilibrium.

*14* Table 4

Gas Partial pressure / kPa

SO 1.67 × 102

O 1.02 × 102

SO 1.85 × 102

Give an expression for the equilibrium constant (Kp) for this reaction.

Calculate the value of the equilibrium constant for this reaction and give its units.

[3 marks]

Units

05.4 What is the effect on the value of Kp if the pressure of this equilibrium mixture is

increased at a constant temperature?

2 SO2(g) + O2(g) ⇌ 2 SO3(g)

[1 mark]

*16* Tick ( ) one box.

The value of Kp

increases.

stays the same.

decreases.

Mark scheme

Show the mark scheme

Question Answers Additional comments/Guidelines Mark

M1 decreases yield 1

M2 So equilibrium shifts to side with more moles/molecules or more Allow M2 independent of M1

05.1 moles/molecules on LHS

M3 So equilibrium shifts (to left side) to oppose decrease in Must refer to equilibrium shifting to gain maximum

pressure OR to increase pressure marks

M1 amount SO2 (= 0.46 – 0.18) = 0.28 mol

M2 amount O2 (= 0.25 – 0.09) = 0.16 mol 1

05.2 1

M3 total amount (= 0.28 + 0.16 + 0.18) = 0.62 mol 𝑀1

M4 = x 215

𝑀3

M4 partial pressure of SO2 = 0.28 x 215 = 97(.1) (kPa) 1

0.62

M1 K = (pp SO )2___

p 3 Penalise square brackets in M1 1

(pp SO )2 x pp O

05.3

M2 = 1.2(0) × 10–2 1

M3 = kPa–1

05.4 Stays the same 1

How to answer it

Gas Equilibria, Mole Fractions & Kₚ Calculations

📌 What this question tests

Core Concepts: Le Chatelier's principle applied to gaseous equilibria, stoichiometric ICE table calculations for equilibrium moles, mole fraction and partial pressure calculations, writing Kₚ expressions, calculating Kₚ values with units, and understanding the temperature dependence of equilibrium constants.

Reaction: 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g) | Total Marks: 11

Question 05.1 • 3 Marks

Effect of Decreasing Overall Pressure on Yield

Le Chatelier's Principle & Gaseous Moles

✅ Correct Model Answer

Effect: Decreases yield (of SO₃) [1 mark]
Explanation: Equilibrium shifts to the side with more moles of gas (the left-hand side / LHS) [1 mark]
...in order to oppose the decrease in pressure (or to increase the pressure) [1 mark]

🧠 Exam Technique: 3-Step Le Chatelier

State the direct effect clearly first (decreases).
Compare total gas moles: LHS = 2 + 1 = 3 moles; RHS = 2 moles.
Quote Le Chatelier's rule explicitly: the position of equilibrium shifts to oppose the change. State the direction and the specific change being opposed.

❌ Common Errors & Pitfalls

Vague statements: Writing "shifts to side with more molecules" without stating which side or linking it to opposing the pressure drop.
Leaving out "equilibrium shifts": Mark 3 specifically requires mentioning that the equilibrium shifts to oppose the decrease.

Mark Scheme Note: Mark 2 is independent of Mark 1. However, to achieve full marks, the answer must explicitly link the shift to the left with opposing the decrease in pressure.

Question 05.2 • 4 Marks

Calculating Partial Pressure of SO₂ at Equilibrium

ICE Table & Mole Fraction Method

📐 Step-by-Step Calculation

Given: Initial n(SO₂) = 0.460 mol, Initial n(O₂) = 0.250 mol, Equil n(SO₃) = 0.180 mol, Total Pressure (P) = 215 kPa.

Substance	2 SO₂(g)	O₂(g)	2 SO₃(g)
Initial / mol	0.460	0.250	0
Change / mol	-0.180 (2:2 ratio)	-0.090 (1:2 ratio)	+0.180
Equilibrium / mol	0.280	0.160	0.180

Step 1: Equilibrium moles of SO₂
n(SO₂) = 0.460 - 0.180 = 0.280 mol [1 mark]
Step 2: Equilibrium moles of O₂
Reacting ratio of SO₃ : O₂ is 2 : 1.
Moles of O₂ reacted = 0.180 / 2 = 0.090 mol.
n(O₂) = 0.250 - 0.090 = 0.160 mol [1 mark]
Step 3: Total equilibrium moles
Total moles = 0.280 + 0.160 + 0.180 = 0.620 mol [1 mark]
Step 4: Partial pressure of SO₂
Mole fraction of SO₂ = 0.280 / 0.620
pp(SO₂) = (0.280 / 0.620) × 215 kPa = 97.1 kPa (or 97 kPa) [1 mark]

❌ Common Errors

Stoichiometry blunder: Subtracting 0.180 mol from O₂ instead of 0.090 mol. Remember the stoichiometry is 1 mol O₂ per 2 mol SO₃ formed!
Forgetting SO₃ in total moles: Summing only the reactants and omitting the 0.180 mol of product.

💡 Key Formulae

Mole fraction (x) = (Moles of gas) / (Total moles of gas)
Partial pressure (p) = Mole fraction × Total pressure

Question 05.3 • 3 Marks

Kₚ Expression, Value & Units

Equilibrium Constant Expression & Evaluation

✅ 1. Expression for Kₚ

Kₚ = (p(SO₃))² / [ (p(SO₂))² × p(O₂) ]

Also accepted: (pp SO₃)² / ((pp SO₂)² × pp O₂)

❌ Critical Trap: Square Brackets

NEVER use square brackets [ ] in a Kₚ expression! Square brackets represent molar concentrations (mol dm⁻³), which applies to K_c, not Kₚ. Use round brackets with 'p' or 'pp', e.g. (pSO₃)².

📐 2. Calculation of Value and Determination of Units

Calculation:

Substitute values from Table 4:

p(SO₃) = 1.85 × 10² kPa
p(SO₂) = 1.67 × 10² kPa
p(O₂) = 1.02 × 10² kPa

Kₚ = (1.85 × 10²)² / [ (1.67 × 10²)² × (1.02 × 10²) ]
Kₚ = 34225 / (27889 × 102)
Kₚ = 34225 / 2844678 = 1.20 × 10⁻² (or 0.0120) [1 mark]

Units Derivation:

Units = (kPa)² / [ (kPa)² × kPa ] = 1 / kPa = kPa⁻¹ [1 mark]

Always derive units directly by cancelling terms from your expression.

Question 05.4 • 1 Mark

Effect of Pressure on the Value of Kₚ

Temperature vs. Pressure Dependency

✅ Correct Answer

Tick: stays the same [1 mark]

💡 Golden Rule of Equilibrium Constants

ONLY temperature changes the value of Kₚ (or K_c).

Changes in total pressure, volume, or concentration shift the position of equilibrium to keep the value of Kₚ constant. Catalysts speed up both rates equally and also have no effect on Kₚ.

Topics

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

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 1, 2021. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.