AQA A-Level Chemistry Paper 1, 2019: Question 7

10 marks · Medium difficulty · State/Explain/Numerical

Calculate equilibrium quantities, determine mole fractions, write the expression for Kp, find total pressure, and deduce enthalpy and modified equilibrium constants for the decomposition of sulfur trioxide.

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Question

Question 7 covers the equilibrium 2SO3(g) ⇌ 2SO2(g) + O2(g). Part 07.1 asks for the mass of O2 at equilibrium given 6.08 g of SO2 is formed. Part 07.2 provides a table with amounts at equilibrium at 1050 K (SO3: 0.320 mol, SO2: 1.20 mol, O2: 0.600 mol) and Kp = 7.62 x 10^5 Pa, asking for mole fractions, the Kp expression, and total pressure. Part 07.3 asks to explain why the forward reaction is endothermic given Kp at 1050 K and 500 K. Part 07.4 asks to calculate Kp and its units at 500 K for the halved equation SO3(g) ⇌ SO2(g) + 1/2 O2(g).
Question text

07 Sulfur trioxide decomposes on heating to form an equilibrium mixture containing

sulfur dioxide and oxygen.

2SO3(g) ⇌ 2SO2(g) + O2(g)

07.1 A sample of sulfur trioxide was heated and allowed to reach equilibrium at a given

temperature.

The equilibrium mixture contained 6.08 g of sulfur dioxide.

Calculate the mass, in g, of oxygen gas in the equilibrium mixture.

[2 marks]

16 Mass g

07.2 A different mass of sulfur trioxide was heated and allowed to reach equilibrium

at 1050 K

2SO3(g) ⇌ 2SO2(g) + O2(g)

The amounts of each substance in the equilibrium mixture are shown in Table 4.

Table 4

Substance Amount at equilibrium / mol

sulfur trioxide 0.320

sulfur dioxide 1.20

oxygen 0.600

For this reaction at 1050 K the equilibrium constant, K = 7.62 x 105 Pa

p

Calculate the mole fraction of each substance at equilibrium.

Give the expression for the equilibrium constant, Kp

Calculate the total pressure, in Pa, of this equilibrium mixture.

[4 marks]

Mole fraction SO3

Mole fraction SO2

Mole fraction O2

Kp

Total pressure17 Pa

07.3 For this reaction at 1050 K the equilibrium constant, K = 7.62 x 105 Pa

p

For this reaction at 500 K the equilibrium constant, K = 3.94 x 104 Pa

p

*16* Explain how this information can be used to deduce that the forward reaction is

endothermic.

[2 marks]

07.4 Use data from Question 07.3 to calculate the value of Kp, at 500 K, for the equilibrium

represented by this equation.

Deduce the units of Kp

SO3(g) ⇌ SO2(g) + O2(g)

[2 marks]

Kp

Units

Mark scheme

Show the mark scheme Mark scheme for Question 07: 07.1 gives 1 mark for finding moles of SO2 = 0.0949 and moles of O2 = 0.0474, and 1 mark for mass of O2 = 1.52 g. 07.2 gives 1 mark for mole fractions (SO3 = 0.15, SO2 = 0.57, O2 = 0.28), 1 mark for Kp expression using partial pressures, 1 mark for rearranging for total pressure P, and 1 mark for P = 1.91 x 10^5 Pa. 07.3 awards 2 marks for stating Kp is higher at higher temperature and deducing that the equilibrium shifts to the right favoring products. 07.4 gives 1 mark for value 198.5 (or 198) by taking the square root, and 1 mark for units Pa^(1/2) or Pa^0.5.

Question Answers Additional Comments/Guidelines Mark

Moles SO2 eqbm (=6.08/64.1 = 0.0949) so moles O2 eqbm = 0.0474 Allow 0.0475 1

07.1

Mass of oxygen (= 0.0474 x 32(.0)) = 1.52 g Allow M1 x 32 1

M1: Mole fraction SO3 = 0.15 Accept fractions for M1 1

Mole fraction SO2 = 0.57

Mole fraction O2 = 0.28

M2: K = (pSO )2 x (pO ) ( = (λSO )2 P2 x (λO ) P ) Do not accept [ ] 1

p 2 2 2 2 λ = mole fraction

(pSO )2 (λSO )2 P2

07.2

M3: P = K x (λSO )2 or K x (0.15)2

p 3 p M3 is for rearrangement with or without numbers 1

(λSO )2 x (λO ) (0.57)2 x (0.28)

22 If incorrect rearrangement allow correct M1 and

M2 only

M4 P = 1.91 x 105 (Pa) Allow range 1.88 x 105 to 1.94 x 105

M1 Kp is higher at higher temperature or converse 1

07.3 M2 At higher temperature more dissociation occurs / more products M2: Allow converse arguments 1

are formed / equilibrium shifts to the right/forward direction M2 dependent on M1.

(√3.94 x 104 Pa) = 198.5 Allow 198 – 198.5 (answer is 198.49) 1

07.4

Pa1/2 or Pa0.5 If √7.62 x 105 = 873 then lose M1 but allow M2 1

How to answer it

Gas Equilibria, Mole Fractions & Kp Calculations

📌 What this question tests

This question assesses core physical chemistry skills for gas-phase equilibria:

  • Using stoichiometric molar ratios to calculate reacting quantities and masses.
  • Writing expressions for Kp using partial pressures (and strictly avoiding square brackets).
  • Calculating mole fractions, substituting partial pressures into a Kp expression, and algebraically rearranging to solve for total pressure ( P ).
  • Relating changes in equilibrium constant ( Kp ) with temperature to deduce reaction enthalpy using Le Chatelier's principle.
  • Determining the modified value and units of Kp when stoichiometric coefficients are halved.

Part 07.1: Stoichiometric Mole & Mass Calculation

Equilibrium reaction: 2SO₃(g) ⇌ 2SO₂(g) + O₂(g) [2 marks]

📐 Step-by-Step Calculation

  1. Calculate moles of SO₂ formed:
    Mr(SO₂) = 32.1 + 2(16.0) = 64.1 g mol⁻¹
    Moles SO₂ = 6.08 / 64.1 = 0.09485 mol (Mark 1)
  2. Use mole ratio for O₂:
    Reaction ratio is 2 SO₂ : 1 O₂
    Moles O₂ = 0.09485 / 2 = 0.04743 mol
  3. Calculate mass of O₂:
    Mr(O₂) = 32.0 g mol⁻¹
    Mass = 0.04743 × 32.0 = 1.52 g (Mark 2)

✅ Correct Answer

Mass of oxygen: 1.52 g

Mark Breakdown:
• 1 mark: Finding moles of SO₂ and dividing by 2 to get moles of O₂ ( 0.0474 to 0.0475 mol ).
• 1 mark: Correct mass of O₂ ( 1.52 g ). Allow ecf from M1 × 32.

❌ Common Traps

  • Wrong Molar Mass for Oxygen: Multiplying by 16 instead of 32 for molecular oxygen (O₂).
  • Missing the 2:1 Ratio: Assuming moles of O₂ equals moles of SO₂. Look closely at the balanced equation!

Part 07.2: Mole Fractions, Kp Expression & Total Pressure

Given: amounts at 1050 K and Kp = 7.62 × 10⁵ Pa [4 marks]

📐 Step-by-Step Calculation

Step 1: Total moles and mole fractions (M1)

Total moles = 0.320 + 1.20 + 0.600 = 2.120 mol
x(SO₃) = 0.320 / 2.120 = 0.151 (or 0.15)
x(SO₂) = 1.20 / 2.120 = 0.566 (or 0.57)
x(O₂) = 0.600 / 2.120 = 0.283 (or 0.28)

Step 2: Partial pressures in terms of total pressure P (M3)

p(X) = mole fraction × P
Kp = [(0.566 P)² × (0.283 P)] / (0.151 P)²
The P² terms cancel out!
Kp = [(0.566)² / (0.151)²] × 0.283 × P
Kp = [0.320 / 0.0228] × 0.283 × P = 3.97 × P

Step 3: Solve for total pressure P (M4)

P = Kp × (xSO₃)² / [(xSO₂)² × xO₂]
P = 7.62 × 10⁵ / 3.97 = 1.91 × 10⁵ Pa

✅ Correct Answers

Mole fractions:
SO₃: 0.15 (or 4/26.5)
SO₂: 0.57 (or 15/26.5)
O₂: 0.28 (or 7.5/26.5)

Kp expression:
Kp = (pSO₂)² × pO₂ / (pSO₃)²

Total pressure: 1.91 × 10⁵ Pa (allow range 1.88 × 10⁵ to 1.94 × 10⁵ Pa )

🧠 Exam Technique: Notation Alert

Never use square brackets [ ] in a Kp expression!

Square brackets denote molar concentration (mol dm⁻³), which invalidates the expression in AQA exams. Always write p(SO₂) , pSO₂ , or P(SO₂) .

❌ Common Algebra Slip

Students often forget to square the total pressure term P along with the mole fraction. Because both numerator and denominator have squared terms ( P² and P² ), they cancel out nicely, leaving just one P term from pO₂ .

Part 07.3: Deducing Reaction Enthalpy from Kp

Data: Kp at 1050 K = 7.62 × 10⁵ Pa | Kp at 500 K = 3.94 × 10⁴ Pa [2 marks]

💡 Key Knowledge

Temperature is the ONLY factor that changes the value of Kp.

  • If temperature increases and Kp increases, the equilibrium has shifted to the right (favouring products).
  • By Le Chatelier's principle, an increase in temperature favours the endothermic direction (absorbing the added thermal energy).

✅ Model Answer (2 Marks)

Mark 1: As temperature increases (from 500 K to 1050 K), the value of Kp increases (from 3.94 × 10⁴ to 7.62 × 10⁵ Pa).
(or converse: Kp is lower at lower temperature)

Mark 2: Therefore, at higher temperatures, more dissociation occurs / the equilibrium shifts to the right / forward direction (which is endothermic).
Note: Mark 2 is dependent on Mark 1.

Part 07.4: Modifying the Equilibrium Expression & Units

New equation: SO₃(g) ⇌ SO₂(g) + ½O₂(g) at 500 K [2 marks]

💡 Key Rule: Changing Stoichiometry

When the stoichiometric coefficients of a chemical equation are multiplied by a factor n :

Kp' = (Kp)ⁿ

Here, all coefficients are halved ( n = ½ ):

Kp' = (pSO₂ × (pO₂)^½) / pSO₃ = √(Kp)

📐 Calculation & Units

  1. Calculate value of Kp':
    Use data at 500 K: Kp = 3.94 × 10⁴ Pa
    Kp' = √(3.94 × 10⁴) = 198.5 (or 198 ) (Mark 1)
  2. Deduce the units:
    Units = [Pa × Pa^½] / [Pa] = Pa^½ (or Pa⁰˙⁵ ) (Mark 2)

❌ Common Misconception

Halving Kp instead of taking the square root: Students frequently write 3.94 × 10⁴ / 2 = 1.97 × 10⁴ . Stoichiometric factors become exponents in equilibrium expressions, not multipliers!

Selecting the wrong temperature: The question asks for the value at 500 K. If you used the value for 1050 K ( √(7.62 × 10⁵) = 873 ), you lose Mark 1, but Mark 2 can still be awarded for the correct units ( Pa¹ᐟ² ).

Topics

Physical Chemistry · 3.1.2 Amount of Substance · 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, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.