OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2017: Question 20

10 marks · Hard difficulty · Structured Questions

Explain the effects of pressure and temperature on the yield of hydrogen in an equilibrium reaction, write an expression and units for Kc, and calculate the equilibrium amount of SO3 in a sulfur dioxide and oxygen equilibrium mixture.

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Question text

20 This question is about equilibrium reactions.

(a) Hydrogen gas is manufactured by the chemical industry using the reaction of methane and

steam. This is a reversible reaction, shown in equilibrium 20.1 below.

equilibrium 20.1 CH (g) + H O(g) 3H (g) + CO(g) ∆H = +210 kJ mol–1

42 2

Explain, in terms of le Chatelier’s principle, the conditions of pressure and temperature

for a maximum yield of hydrogen from equilibrium 20.1, and explain why the operational

conditions used by the chemical industry may be different.

… [4]

(b) A chemist investigates the equilibrium reaction between sulfur dioxide, oxygen, and sulfur

trioxide, shown below.

2SO2(g) + O2(g) 2SO3(g)

• The chemist mixes together SO2 and O2 with a catalyst.

• The chemist compresses the gas mixture to a volume of 400 cm3.

• The mixture is heated to a constant temperature and is allowed to reach equilibrium

without changing the total gas volume.

The equilibrium mixture contains 0.0540 mol SO2 and 0.0270 mol O2.

At the temperature used, the numerical value for K is 3.045 × 104 dm3 mol–1.

(i) Write the expression for Kc and the units of Kc for this equilibrium.

[2]

(ii) Determine the amount, in mol, of SO3 in the equilibrium mixture at this temperature.

Give your final answer to an appropriate number of significant figures.

Show all your working.

equilibrium amount of SO3 = … mol [4]

Mark scheme

Show the mark scheme

Question Answer Marks Guidance

20 (a) Conditions ANNOTATE ANSWER WITH TICKS AND

Low/decreased pressure 4 CROSSES ETC

AND high/increased temperature

Pressure:

Right-hand/product side has more (gaseous) DO NOT ALLOW more atoms on right-hand

moles/molecules side OR fewer atoms on left-hand side.

OR left-hand side/reactant side has fewer (gaseous) DO NOT ALLOW incorrect shift direction

moles/molecules

Temperature:

(Forward) reaction is endothermic / takes in heat

OR reverse reaction is exothermic / gives out heat

Low pressure gives a slow rate ORA

High temperature uses a large amount of energy/fuel IGNORE ‘expensive’

IGNORE use of catalyst

(b) (i) 2

[SO ]2 IGNORE state symbols in K expression,

3 c

(Kc = ) [SO ]2 [O ] even if wrong.

3 –1 For units, ALLOW mol–1 dm3

Units: dm mol

DO NOT ALLOW dm3/mol

NOTE: If Kc upside down, units become

mol dm–3 by ECF

No other ECF allowed for units.

(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE 4 FULL ANNOTATIONS NEEDED

IF answer = 2.45, Award 4 marks. IF there is an alternative answer, check to

-------------------------------------------------------------------- see if there is any ECF credit possible using

Equilibrium concentrations (moles 2.5) 1 MARK working below

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

SO = 0.135 (mol dm–3)

AND O = 0.0675 (mol dm–3)

Calculation of [SO3(g)] 2 MARKS ALLOW ECF from incorrect concentrations of

SO2 and/or O2

[SO ] = (K [SO ]2 O )

3 c 2 2

OR ( (3.045 104) 0.1352 0.0675) ALLOW ECF from incorrect [SO ]

= 6.12039291 (mol dm–3) ALLOW 3 SF, 6.12, up to calculator value of

Answer scores both [SO3] marks automatically 6.12039291 correctly rounded.

Common errors

37.5 1 mark

3 No for [SO ]2 and no scaling by 1/2.5

Calculation of n(SO3) in 400 cm 1 MARK 3

n(SO3) = 6.12039291/2.5 = 2.45 (mol) 15.0 2 marks

No for [SO ]2

3SF required (Appropriate number)

0.619 3 marks

Use of mol of SO2 and O2

1.55 2 marks

No conc used and Use of mol of SO2 and O2

Total 11

How to answer it

Equilibrium Reactions & Chemical Kinetics Study Guide

What this question tests

This question assesses your mastery of dynamic equilibria, Le Chatelier's principle, writing equilibrium constant expressions with correct units, and executing multi-step equilibrium concentration/amount calculations using Kc values.

Part (a) — Le Chatelier's Principle & Industrial Trade-offs

Applying Equilibrium Laws to Industrial Processes

✅ Correct Answer Requirements

Conditions: Low pressure AND high temperature.
Pressure reasoning: Right side has more gaseous moles (4 vs 2), so low pressure shifts equilibrium to the right.
Temperature reasoning: Forward reaction is endothermic (positive ΔH), so high temperature shifts equilibrium to the right.
Industrial compromise: Low pressure gives an unacceptably slow reaction rate; high temperature requires immense energy/fuel costs.

❌ Common Student Errors

Confusing equilibrium yield shifts with reaction kinetics (e.g., claiming temperature increases rate without linking it to equilibrium position).
Vague statements like "it is too expensive" instead of specifying why (large amount of energy/fuel required).
Getting moles confused: counting total atoms instead of gaseous molecules when explaining pressure effects.

🧠 Exam Technique & Examiner Tips

To score all 4 marks here, you must structure your answer in two clear blocks: (1) Maximising Yield using Le Chatelier's principle with explicit references to moles and enthalpy changes, and (2) Industrial Realities explaining economic or kinetic constraints (rate vs. yield compromise).

Part (b)(i) — Kc Expressions and Units

Writing Equilibrium Constants

✅ Correct Answer

Kc Expression: [SO₃]² / ([SO₂]² [SO₂]) or [SO₃]² / ([SO₂]² [O₂])

Units of Kc: dm³ mol⁻¹ (or mol⁻¹ dm³ )

💡 Key Knowledge

Products go on the numerator, reactants on the denominator.
Stochiometric balancing numbers become powers.
To find units, substitute concentration units ( mol dm⁻³ ) into the expression and cancel them out.

Part (b)(ii) — Advanced Kc Calculation

Determining Equilibrium Amount (mol)

📐 Step-by-Step Calculation Guide

Convert moles to concentrations: The mixture volume is 400 cm³ ( 0.400 dm³ ).
[SO₂] = 0.0540 mol / 0.400 dm³ = 0.135 mol dm⁻³
[O₂] = 0.0270 mol / 0.400 dm³ = 0.0675 mol dm⁻³
Rearrange the Kc expression for [SO₃]:
Kc = [SO₃]² / ([SO₂]² [O₂]) ⟹ [SO₃]² = Kc × [SO₂]² × [O₂]
Substitute values and square root:
[SO₃]² = 3.045 × 10⁴ × (0.135)² × 0.0675 = 37.457...
[SO₃] = √(37.457...) = 6.12039 mol dm⁻³
Scale concentration back to moles in 400 cm³:
Moles = Concentration × Volume (in dm³)
n(SO₃) = 6.12039 × 0.400 = 2.448... mol
Apply Significant Figures:
Data is given to 3 sf, so round final answer to 3 sf: 2.45 mol.

❌ Calculation Traps to Avoid

Forgetting volume scaling: Working directly with moles instead of concentrations in the Kc expression loses marks instantly.
Missing the square root: Forgetting to square root [SO₃]² leads to common alternative incorrect answers like 37.5.
Rounding too early: Always keep full calculator numbers through intermediate steps and round only at the very end.

🧠 Top-Level Examiner Note

Full marks (4/4) are awarded for the correct final answer of 2.45 . ECF (Error Carried Forward) is applied generously if your intermediate concentration conversions or algebraic rearrangements contained minor slips, provided your method logic remains sound.

Mark Scheme Allocation: 1 mark for concentrations / 2 marks for calculating [SO₃] / 1 mark for final moles scaled to volume and given to appropriate 3 sf.

Topics

Module 5: Physical chemistry and transition elements · 5.1 Rates, equilibrium and pH

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), June 2017. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.