OCR A-Level Chemistry AS Depth in chemistry (02), June 2018: Question 4

12 marks · Medium difficulty · Structured Questions

Calculate the energy released from converting 5.10 tonnes of ammonia, write an equilibrium constant expression, and predict the conditions for a maximum yield of nitrogen monoxide with LE Chatelier's principle and compromise factors.

Practise this question

Question

A structured multi-part chemistry question about the industrial oxidation of ammonia. Part (a)(i) requires completing an enthalpy profile diagram with activation energy and enthalpy change for the exothermic reaction. Part (a)(ii) asks to calculate the energy released when 5.10 tonnes of ammonia are converted, giving the answer in standard form and appropriate significant figures. Part (b) asks for the expression for the equilibrium constant Kc. Part (c) requires predicting temperature and pressure conditions for maximum yield using Le Chatelier's principle and discussing compromise factors.
Question text

4 The reaction of ammonia, NH3, with oxygen to form nitrogen monoxide, NO, is an important

industrial process.

The equation for this reaction is shown in equilibrium 4.1 below.

4NH (g) + 5O (g) 4NO(g) + 6H O(g) ∆H = –905 kJ mol–1 Equilibrium 4.1

32 2

(a) The forward reaction in equilibrium 4.1 converts NH3 into NO.

(i) Complete the enthalpy profile diagram for this reaction.

On your diagram:

• Label the activation energy, Ea

• Label the enthalpy change of reaction, ∆H

• Include the formulae of the reactants and products.

Enthalpy

Progress of reaction

[2]

(ii) 5.10 tonnes of NH3 are converted into NO.

Calculate the energy released, in kJ, for this conversion.

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

energy released = … kJ [4]

(b) Write an expression for the equilibrium constant, Kc, in equilibrium 4.1.

[1]

(c) Predict the conditions of temperature and pressure for a maximum equilibrium yield of

nitrogen monoxide in equilibrium 4.1.

• Explain your prediction in terms of le Chatelier’s principle.

• State and explain how these conditions could be changed to achieve a compromise

between equilibrium yield, rate and other operational factors.

… [5]

Mark scheme

Show the mark scheme The mark scheme provides a completed enthalpy profile diagram showing reactants higher than products, activation energy Ea, and enthalpy change delta H. It details the step-by-step calculation for finding the energy released (6.79 x 10^7 kJ) and shows the Kc expression with square brackets. Finally, it lists marking points for equilibrium conditions including temperature, pressure, optimum conditions, rate considerations, and industrial operational factors.

Question Answer Marks Guidance

4 (a) (i) 2 ANNOTATE ANSWER WITH TICKS AND

CROSSES ETC

IGNORE state symbols

ALLOW 1 mark for a correctly labelled endothermic

diagram

Ea ALLOW no arrowhead or arrowheads at both

end of Ea line.

Ea line must reach maximum (or near to

maximum) on curve

Reactants, products and Ea

Reactants on LHS 4NH + 5O For Ea, ALLOW AE OR AE

3(g) 2(g)

AND

Products on RHS 4NO(g) + 6H2O(g)

AND ∆H DO NOT ALLOW –∆H

Activation energy correctly labelled / E DO NOT ALLOW double headed arrow on ∆H

a

ALLOW ∆H arrow even with small gap at the

∆H top and bottom, i.e. line does not quite reach

∆H labelled with product below reactant reactant or product line.

AND

Arrow downwards ALLOW –905 for ∆H

H032/02 Mark scheme June 2018

(ii) FIRST CHECK ON ANSWER LINE 14 4 IGNORE (-) SIGN

If answer = 6.79 107 (kJ) award 4 marks Throughout: IGNORE trailing zeroes in intermediate

If answer = 2.72 108 (kJ) award 3 marks (no ÷ 4) working,

e.g. For n(NH ) ALLOW 3 105 for 3.00 105

-------------------------------------------------------------------------- 3

n(NH3) --------------------------------------------------------------

= 5.1 x 106 = 3.00 105 (mol)

Stoichiometry and ∆H

1 mol NH3 releases 905 OR 226.25 (kJ)

Energy released

(3.00 105) 905 OR 67875000 (kJ)

ALLOW ECF from incorrect n(NH3) OR 905/4

ALLOW 3 SF up to calc value correctly rounded.

Value will depend on intermediate rounding

Final answer to 3SF AND standard form Common Errors

71.09 109 (x 4 instead of ÷ 4) 3 marks

= 6.79 10 (kJ)

2.72 108 (no ÷ 4) 3 marks

standard form AND 3 SF required

6.79 101 (no tonnes g) 3 marks

(b) 1

[NO(g)]4 [H O(g)]6 Square brackets required

(Kc = ) [NH (g)]4 [O (g)]5

IGNORE state symbols

H032/02 Mark scheme June 2018

4 (c) EQUILIBRIUM CONDITIONS 5 ANNOTATE ANSWER WITH TICKS AND

CROSSES ETC

Temperature: 1 mark

(Forward) reaction is exothermic/ΔH is negative

OR (Forward) reaction gives out heat

Pressure: 1 mark ALLOW reverse arguments

Left-hand side has fewer (gaseous) moles

OR 9 (gaseous) moles form 10 (gaseous) moles

OPTIMUM EQUILIBRIUM CONDITIONS: 1 mark

(for maximum yield of NO)

Low temperature AND low pressure

RATE: 1 mark

Low temperature/pressure gives a slow rate/slower reaction Answer MUST relate temp/pressure to rate /

so high temperatures / higher pressure needed to increase frequency of collisions

rate OR frequency of collisions

INDUSTRIAL CONDITIONS / OPERATIONAL FACTORS: 1 ALLOW Temperature / pressure not too high

mark because yield reduced

High pressure provides a safety risk

OR IGNORE stated temperatures and pressures

Higher temperatures increase energy costs / reduce yield /

shift equilibrium to left IGNORE catalyst

OR

(High) pressure is expensive (to generate) / uses a lot of

energy

Total 12

How to answer it

Industrial Ammonia Oxidation Equilibrium

OCR AS Level Chemistry • Physical Chemistry & Energetics

What this question tests

This question assesses your understanding of energetic profiles (enthalpy profile diagrams), stoichiometric calculations involving mass conversions and enthalpy changes, writing equilibrium constant expressions (Kc), and applying Le Chatelier's principle to balance yield, rate, and economic factors in industrial chemical processes.

Part (a)(i): Enthalpy Profile Diagram

Complete the enthalpy profile diagram for the exothermic reaction.

✅ Correct Answer Requirements

  • Reactants written on the left higher energy line: 4NH₃(g) + 5O₂(g)
  • Products written on the right lower energy line: 4NO(g) + 6H₂O(g)
  • An arched curve showing a peak representing activation energy ( Eₐ )
  • A downward arrow for enthalpy change ( ΔH ) starting from the reactant level down to the product level.

🧠 Exam Technique & Mark Scheme

  • 2 Marks total: One mark for correct relative energy levels and formulae of reactants/products, and one mark for labelling Eₐ and ΔH correctly.
  • Make sure the ΔH arrow points downwards because the reaction is exothermic ( ΔH = -905 kJ mol⁻¹ ). Never use a double-headed arrow for ΔH .
Available marks: [2]

Part (a)(ii): Enthalpy Calculation

Calculate the energy released, in kJ, when 5.10 tonnes of NH₃ are converted into NO. Give your answer in standard form and to 3 significant figures.

📐 Step-by-Step Calculation

  1. Convert tonnes to grams / moles:
    Mass of NH₃ = 5.10 tonnes = 5.10 × 10⁶ g.
    Molar mass of NH₃ = 14.0 + (3 × 1.0) = 17.0 g mol⁻¹.
    Moles of NH₃ ( n ) = (5.10 × 10⁶) / 17.0 = 3.00 × 10⁵ mol .
  2. Use stoichiometry from the balanced equation:
    The equation shows 4 mol of NH₃ release 905 kJ .
    Therefore, 1 mol releases 905 / 4 = 226.25 kJ.
  3. Calculate total energy released:
    Energy = 3.00 × 10⁵ mol × (905 / 4) kJ mol⁻1 = 67,875,000 kJ = 6.79 × 10⁷ kJ .

❌ Common Calculation Traps

  • Stoichiometry ratio error: Multiplying by 4 instead of dividing by 4 gives 1.09 × 10⁹ (3 marks max).
  • Forgetting to divide by 4 entirely: Gives 2.72 × 10⁸ (3 marks).
  • Incorrect mass conversion: Forgetting that 1 tonne = 10⁶ g.
  • Formatting: Failing to provide the final answer in standard form and 3 significant figures loses the final accuracy mark.
Available marks: [4]

Part (b): Equilibrium Constant Expression

Write an expression for the equilibrium constant, Kc , for equilibrium 4.1.

✅ Correct Answer

Kc = [NO]⁴[H₂O]⁶ / [NH₃]⁴[O₂]⁵

💡 Key Knowledge

  • Products go on the numerator (top), reactants on the denominator (bottom).
  • Stoichiometric balancing numbers become the powers (indices).
  • Square brackets [ ] denote equilibrium concentrations. State symbols are omitted from Kc expressions.
Available marks: [1]

Part (c): Le Chatelier's Principle & Compromise Conditions

Predict the conditions of temperature and pressure for a maximum equilibrium yield of nitrogen monoxide, and explain trade-offs.

💡 Core Principles

  • Temperature: Forward reaction is exothermic ( ΔH = -905 kJ mol⁻¹ ). Low temperature shifts equilibrium to the right (maximum yield).
  • Pressure: LHS has 4 + 5 = 9 moles of gas; RHS has 4 + 6 = 10 moles of gas. Low pressure shifts equilibrium to the right towards more moles of gas.

🧠 Industrial Compromise & Rate Factors

  • Rate conflict: Low temperature and low pressure drastically decrease the rate of reaction (fewer successful collisions per second).
  • Compromise conditions: Higher temperatures and pressures are used industrially to ensure a commercially viable reaction rate, despite slightly reducing equilibrium yield.
  • Economic / Operational factors: Very high pressures are expensive to generate and maintain due to safety risks and heavy-duty plant costs.
Available marks: [5]

Topics

Module 2: Foundations in chemistry · Module 5: Physical chemistry and transition elements · Module 3: Periodic table and energy · 2.1 Atoms and reactions · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, AS Depth in chemistry (02), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.