OCR A-Level Chemistry AS Breadth in chemistry (01), June 2023: Question 24

9 marks · Medium difficulty · Calculations

Explain homogeneous equilibrium, apply Le Chatelier's principle to changes in pressure, temperature and catalyst for the sulfur dioxide oxidation, calculate the equilibrium constant Kc from given equilibrium concentrations, and explain the use of an excess of oxygen industrially.

Practise this question

Question

A structured chemistry question about the reversible reaction 2SO2(g) + O2(g) right-left-arrow 2SO3(g) with enthalpy change -197 kJ mol^-1. Part (a) asks why Equilibrium 24.1 is a homogeneous equilibrium. Part (b) asks to explain the effect of pressure, temperature and a catalyst on the yield of SO3 using Le Chatelier's principle. Part (c)(i) provides a table of equilibrium concentrations for SO2, O2 and SO3 and asks to write the Kc expression and calculate its numerical value. Part (c)(ii) asks why an excess of O2 is used industrially.
Question text

24 The reaction between sulfur dioxide, SO2(g) and oxygen, O2(g), to form sulfur trioxide, SO3(g), is

a key step in the industrial manufacture of sulfuric acid.

This is a reversible reaction, shown in Equilibrium 24.1:

2SO (g) + O (g) 2SO (g) ΔH = –197 kJ mol–1 Equilibrium 24.1

22 3

(a) Why is Equilibrium 24.1 a homogeneous equilibrium?

… [1]

(b) Le Chatelier’s principle can be used to predict how different conditions affect the equilibrium

position in Equilibrium 24.1.

Explain how changing pressure, temperature and using a catalyst affect the equilibrium yield

of SO3.

In your answer, use le Chatelier’s principle and other chemical concepts, where appropriate.

… [5]

(c) A mixture of SO2(g) and O2(g) is allowed to reach equilibrium at a constant temperature.

The equilibrium concentrations are shown in the table.

Equilibrium concentration

Substance –3

/ mol dm

SO (g) 3.0 × 10–3

O (g) 3.5 × 10–3

SO (g) 5.0 × 10–2

(i) Write the expression for Kc and calculate the numerical value for Kc in Equilibrium 24.1

at this constant temperature.

Give your answer to an appropriate number of significant figures and in standard

form.

K = … dm3 mol–1 [2]

c

(ii) In the industrial production of SO3, an excess of O2(g) is used rather than a 2:1

proportion of SO2(g) to O2(g) which would match the stochiometry in Equilibrium 24.1.

Suggest, in terms of equilibrium, why an excess of O2(g) is used industrially.

… [1]

Mark scheme

Show the mark scheme The mark scheme for question 24 giving answers for parts a, b, c(i) and c(ii). Part (a) awards 1 mark for stating all reaction species have the same state or phase. Part (b) awards 5 marks for explaining pressure, temperature and catalyst effects using Le Chatelier's principle. Part (c)(i) awards 2 marks for the correct Kc expression and calculation resulting in 7.9 x 10^4. Part (c)(ii) awards 1 mark for stating the equilibrium shifts to the right.

24 (a) 1 AO1.1 ALLOW SO2 AND O2 AND SO3 for all species

All reaction species have same state/phase OR reactants and products are gases

OR OR the molecules are all gases

Reactants AND products has same state ✓

IGNORE reactants and catalyst have same state

(b) Throughout, 5 FULL ANNOTATIONS MUST BE USED

ALLOW suitable alternatives for right-hand side, ----------------------------------------------------------------

e.g. towards SO3/products

OR forward direction

--------------------------------------------------------------

Pressure 2 marks

Increased pressure shifts equilibrium to right ORA for reverse reaction

OR favours the right e.g. decreased pressure shifts equilibrium to left

OR increases yield (of SO3) ✓ AO1.1

Right-hand side has fewer (gaseous) moles For moles, ALLOW molecules/particles

OR 3 (gaseous) moles → 2 (gaseous) moles ✓ AO1.2

Temperature 2 marks

Increased temperature shifts equilibrium to left ORA for reverse reaction

OR favours the left e.g. decreased temperature shifts equilibrium to right

OR decreases yield (of SO3) ✓ AO1.1

(Forward) reaction is exothermic/∆H is negative ALLOW reverse reaction is endothermic

OR (Forward) reaction gives out heat ✓ AO1.2 /∆H is positive/takes in heat

Catalyst 1 mark

No shift in equilibrium ALLOW rates of forward and reverse reaction

OR no effect on yield (of SO3) ✓ AO1.2 increase by same amount

IGNORE ‘no increase in yield’

Yield could still decrease

(c) (i) FIRST CHECK THE ANSWER ON ANSWER LINE 18 2 AO2.6 IF there is an alternative answer, check for any ECF

IF answer = 7.9 104 award 2 marks 2 credit possible using working below.

--------------------------------------------------------------------– -------------------------------------------------------------

Kc expression

[SO ]2 (5.0 10–2)2

(Kc = ) [SO ]2 [O ] OR –3 2 –3 Square brackets required for Kc expression

22 (3.0 10 ) (3.5 10 )

OR 79365 … ✓ ALLOW ECF to 2 SF and standard form

ONLY from inverted K expression → 1.3 10–5

c

Answer to 2 SF and in standard form

4 [SO ]2

Kc = 7.9 10 ✓ 3

DO NOT ALLOW [SO ]2 + [O ] = 0.71 (no marks)

IGNORE attempts at units

(ii) Equilibrium shifts to the right/towards products/SO3 ✓ 1 AO3.1 ALLOW equilibrium favours the right

How to answer it

Industrial Equilibria and Kc Calculations

OCR AS Level Chemistry • Physical Chemistry

What this question tests

This question assesses your understanding of chemical equilibria, specifically homogeneous vs heterogeneous systems, applying Le Chatelier's principle to predict the effects of temperature, pressure, and catalysts on equilibrium yield, writing Kc expressions, performing complex equilibrium Kc calculations using standard form and significant figures, and explaining industrial trade-offs (using excess reactants).

Part (a): Homogeneous Equilibria

Why is Equilibrium 24.1 a homogeneous equilibrium? [1 mark]

✅ Correct Answer

All reaction species (reactants and products) are in the same state/phase (all are gases: SO₂₍₀₉₎, O₂₍₉₎, and SO₃₍₉₎).

❌ Common Errors

Students often lose this mark by stating "reactants and catalysts are in the same state" (catalysts are not part of the equilibrium species) or vaguely stating "they are all liquids/gases" without specifying that it applies to all species in the equation.

Part (b): Le Chatelier's Principle

Explain how changing pressure, temperature and using a catalyst affect the equilibrium yield of SO₃. [5 marks]

💡 Key Knowledge: Breaking Down the 5 Marks

  • Pressure (2 marks): Increased pressure shifts equilibrium to the right / increases SO₃ yield because the right-hand side has fewer gaseous moles (3 moles on left → 2 moles on right).
  • Temperature (2 marks): Increased temperature shifts equilibrium to the left / decreases SO₃ yield because the forward reaction is exothermic (ΔH is negative).
  • Catalyst (1 mark): No shift in equilibrium position / no effect on yield, but increases the rate at which equilibrium is reached.

🧠 Exam Technique

Always structure your answer clearly by addressing each variable (Pressure, Temperature, Catalyst) in turn. Use precise terminology: state the direction of shift (right/left), link it to the moles/enthalpy change, and explicitly state the outcome on the yield of SO₃.

Mark scheme requirement: Full annotations must be used. Mentioning ORA (opposite-refers-to-alternative) is acceptable if discussing decreases in pressure/temperature.

Part (c)(i): Equilibrium Constants (Kc)

Write the expression for Kc and calculate the numerical value for Kc in Equilibrium 24.1. [2 marks]

📐 Step-by-Step Calculation

Step 1: Write the Kc expression
Kc = [SO₃]² / ([SO₂]² [O₂])

Step 2: Substitute the equilibrium values
[SO₃] = 5.0 × 10⁻²
[SO₂] = 3.0 × 10⁻³
[O₂] = 3.5 × 10⁻³

Step 3: Calculate the numerical value
Kc = (5.0 × 10⁻²)² / ((3.0 × 10⁻³)² × (3.5 × 10⁻³))
Kc = 79365...

Step 4: Format to 2 significant figures and standard form
7.9 × 10⁴ (Units: dm³ mol⁻¹, though ignored by mark scheme here).

❌ Common Calculation Traps

  • Forgetting square brackets in the Kc expression will cost you the first mark.
  • Incorrect stoichiometry powers: Forgetting to square [SO₃] and [SO₂] .
  • Significant figures: Leaving the answer as 79365 or 79000 instead of strict standard form to 2 SF ( 7.9 × 10⁴ ).

Part (c)(ii) Industrial Application of Equilibria

Suggest, in terms of equilibrium, why an excess of O₂ is used industrially. [1 mark]

✅ Correct Answer

Equilibrium shifts to the right (towards products / favours SO₃ production), OR increases the conversion/yield of the more expensive reactant (SO₂).

🧠 Industrial Context

Sulfur dioxide (SO₂) is derived from burning sulfur and is more costly/hazardous. Oxygen (O₂) is cheap and readily available from the air. Pushing equilibrium to the right by flooding the system with excess O₂ ensures maximum conversion of the limiting reactant (SO₂).

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2023. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.