OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2022: Question 20

8 marks · Hard difficulty · Structured Questions

Calculate the equilibrium constant Kc for the Haber process and explain the compromise conditions, and determine the effect of temperature and pressure on iron oxide reduction equilibria

Practise this question

Question

Chemistry exam question with two parts about chemical equilibria. Part (a) provides the equilibrium equation for the industrial manufacture of ammonia with enthalpy change, initial moles, container volume, and equilibrium amount of ammonia, asking to determine Kc and explain operational conditions versus maximum yield. Part (b) gives an equilibrium equation for the reduction of iron oxide by hydrogen, with two sub-questions asking to determine if the forward reaction is exothermic or endothermic when temperature decreases and Kp decreases, and to evaluate a student's statement regarding the effect of pressure on the equilibrium position.
Question text

20 This question is about equilibria involving hydrogen.

(a)* Hydrogen is used industrially to manufacture ammonia.

The equilibrium is shown below.

N (g) + 3H (g) 2NH (g) ∆H = –92 kJ mol–1 Equilibrium 20.1

22 3

1.20 mol N (g) is mixed with 3.60 mol H (g) in a 8.00 dm3 container.

The mixture is heated to 550 °C with an iron catalyst and allowed to reach equilibrium.

The equilibrium mixture contains 0.160 mol of NH3.

Determine the equilibrium constant Kc for Equilibrium 20.1, and explain why the operational

conditions used by industry may be different from those required for a maximum equilibrium

yield of ammonia. [6]

Additional answer space if required.

(b) In industry, hydrogen is also used to reduce the iron oxide Fe3O4 as shown in

Equilibrium 20.2.

The reaction is carried out at 500 °C.

Fe3O4(s) + 4H2(g) 3Fe(s) + 4H2O(g) Equilibrium 20.2

(i) When the temperature is decreased, the value of Kp decreases.

Determine whether the forward reaction is exothermic or endothermic.

Explain your answer.

… [1]

(ii) Two students are discussing the effect of pressure on the equilibrium position of

Equilibrium 20.2.

Student 1 says:

“There are more moles of products than reactants, so increasing the pressure will shift

the equilibrium to the left hand side.”

Student 2 disagrees.

Determine which student is correct. Justify your answer.

… [1]

Mark scheme

Show the mark scheme Mark scheme detailing answers for question 20. Part (a) is a 6-mark level-of-response question requiring calculation of equilibrium amounts, concentrations, Kc expression with units (dm6 mol-2), and an explanation of temperature, pressure, and catalyst compromises. Parts (b)(i) and (b)(ii) are 1-mark short-answer questions explaining endothermic reaction direction and evaluating student statements on gas moles in equilibria.

AO

Question Answer Marks Guidance

element

20 (a) Please refer to the marking instructions on page 4 of this 6 AO2.4 Indicative scientific points may include:

mark scheme for guidance on how to mark this question. ×4 IGNORE trailing zeroes

Level 3 (5–6 marks)

Uses correct method to calculate Kc AO1.2 Equilibrium amounts

AND explains why most operational condition is different ×2 n(N2): 1.20 – 0.08 = 1.12, n(H2) : 3.60 – 0.24 = 3.36

with few omissions in the explanation. Equilibrium concentrations

1.12 –3

There is a well-developed line of reasoning which is clear [N2] = 8.00 = 0.140 (mol dm )

and logically structured. The information presented is 3.36 –3

relevant and substantiated. [H2] = = 0.420 (mol dm )

8.00

0.160 –3

Level 2 (3–4 marks) [NH3] = = 0.0200 (mol dm )

8.00

Uses correct method to calculate Kc with few errors Equilibrium expression and Kc value wih units

OR [NH ]2

Derives a correct expression for Kc with an attempt at the Kc = 3

[N2] × [H2]

Kc calculation AND explains why an operational condition is 0.02002

different with some omissions. Kc = 3 = 0.0386

0.140 × 0.420

Calculator: 0.03856417851 Units: dm6 mol–2

There is a line of reasoning presented with some structure.

The information presented is relevant and supported by

Explanation for operational differences.

some evidence.

Temperature

Level 1 (1–2 marks) • Low temperature for maximum yield: (∆H –ve \

Derives a correct expression for Kc AND explains why one exothermic)

operational condition is different with some omissions. • High temperature to increase rate

OR

explains why most operational conditions are different Pressure

• High pressure for maximum yield (fewer

There is an attempt at a logical structure with a line of (gaseous) moles/molecules of products)

reasoning. The information is in the most part relevant. • High pressure expensive to generate

OR high pressure is a safety hazard

AO

Question Answer 23 Marks Guidance

element

0 marks Catalyst

No response or no response worthy of credit. • Allows a lower temperature to be used for

maximum yield.

• Reducing fuel expense OR increasing rate

(b) (i) Equilibrium (position) shifts to the left (as T is decreased) 1 AO1.2 ALLOW ‘favours backward reaction’

AND Implies shift to left

(forward) reaction is endothermic

ALLOW ‘shifts in exothermic direction’ BUT only if

(forward) reaction stated as endothermic

(ii) Student 2 is correct 1 AO3.2 ALLOW AW that suggests student 2 is correct

AND

same number of gas particles/ gas(eous) molecules/moles

of gas on each side (of equation)

Total 8

How to answer it

Equilibria Involving Hydrogen (Haber Process & Reduction)

What this question tests

This multi-part question assesses your ability to calculate equilibrium constants (Kc), construct equilibrium expressions with correct units, apply Le Chatelier's principle to industrial trade-offs (compromise conditions), interpret enthalpy changes relative to temperature changes, and evaluate the effect of pressure based on gaseous mole ratios.

Question 20 (a) — 6 Marks

Calculating Kc and Explaining Industrial Conditions

📐 Step-by-Step Calculation (Kc)

  1. Find equilibrium moles of reactants:
    Ratio is 1 : 3 : 2 for N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
    Equilibrium moles of NH₃ = 0.160 mol.
    Since 2 moles of NH₃ form from 1 mole of N₂, the moles of N₂ reacted = 0.160 / 2 = 0.080 mol.
    Equilibrium n(N₂) = 1.20 - 0.08 = 1.12 mol .
    Moles of H₂ reacted = 3 × 0.080 = 0.240 mol.
    Equilibrium n(H₂) = 3.60 - 0.24 = 3.36 mol .
  2. Convert to equilibrium concentrations:
    Container volume = 8.00 dm³.
    [N₂] = 1.12 / 8.00 = 0.140 mol dm⁻³
    [H₂] = 3.36 / 8.00 = 0.420 mol dm⁻³
    [NH₃] = 0.160 / 8.00 = 0.0200 mol dm⁻³
  3. Construct Kc expression and calculate:
    Kc = [NH₃]² / ([N₂] × [H₂]³)
    Kc = (0.0200)² / (0.140 × (0.420)³)
    Kc = 0.000400 / 0.0103823 = 0.0386 (to 3 sig figs)
  4. Determine Units:
    Units = (mol dm⁻³)² / (mol dm⁻³ × (mol dm⁻³)³)
    Units = 1 / (mol² dm⁻⁶) = dm⁶ mol⁻²

💡 Industrial Compromise Conditions

  • Temperature Trade-off: The forward reaction is exothermic (ΔH = -92 kJ mol⁻¹). A low temperature maximises equilibrium yield, but the rate becomes too slow. A higher temperature is used industrially to increase the rate of reaction.
  • Pressure Trade-off: Fewer moles of gas on the product side (2 moles vs 4 moles total on reactant side). A high pressure maximises yield and rate, but high pressures require expensive reinforced equipment and carry high safety risks.
  • Catalyst: An iron catalyst allows a lower temperature to be used while maintaining a viable rate, saving fuel expenses and improving overall efficiency.

❌ Common Errors

  • Forgetting to divide equilibrium moles by the volume (8.00 dm³) to find concentrations.
  • Using initial moles instead of equilibrium moles in the Kc expression.
  • Inverting the Kc expression (putting reactants on top).
  • Incorrectly cancelling units to give dm⁻⁶ mol² instead of dm⁶ mol⁻² .

🧠 Level of Response Guidance

This is a Level of Response question (6 marks total). To achieve Level 3 (5–6 marks), you must correctly calculate the numeric value of Kc with units and thoroughly explain why industrial operational conditions (temperature and pressure) differ from theoretical maximum yield conditions due to rate and economic/safety constraints.

Question 20 (b)(i) — 1 Mark

Enthalpy and Temperature Shifts

✅ Correct Answer

The forward reaction is endothermic.

Explanation: When temperature is decreased, Kp decreases, meaning the equilibrium position shifts to the left (favours the backward reaction). Decreasing temperature always shifts equilibrium in the exothermic direction. Therefore, if decreasing temperature shifts it left, the reverse reaction is exothermic, meaning the forward reaction must be endothermic.

Mark Scheme Note: Award 1 mark for stating the forward reaction is endothermic based on the shift to the left when temperature decreases.
Question 20 (b)(ii) — 1 Mark

Pressure Effect on Equilibrium 20.2

✅ Correct Answer & Justification

Student 2 is correct.

Justification: Look at the gaseous moles in the equation: Fe₃O₄(s) + 4H₂(g) ⇌ 3Fe(s) + 4H₂O(g) .
There are 4 moles of gas on the left (4 H₂) and 4 moles of gas on the right (4 H₂O). Note that Fe₃O₄ and Fe are solids (s) and do not count toward gaseous moles! Because there is an equal number of gaseous molecules on both sides, changing pressure has no effect on the position of equilibrium.

❌ Common Student Traps

Students often blindly count total species or forget to check state symbols ( (s) vs (g) ), incorrectly treating solid iron or iron oxide as moles of gas.

Mark Scheme Note: Award 1 mark for identifying Student 2 and stating that there is the same number of gas particles/molecules on each side of the equation.

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 2022. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.