OCR A-Level Chemistry AS Breadth in chemistry (01), June 2019: Question 24

10 marks · Medium difficulty · Calculations

Apply Le Chatelier's principle to ammonia synthesis, calculate Kc from equilibrium concentrations, state standard conditions, and calculate standard enthalpy change of formation for NO.

Practise this question

Question

A multi-part chemistry question about ammonia and nitric acid production. Part (a) provides the equilibrium equation N2(g) + 3H2(g) <=> 2NH3(g) with delta H = -92 kJ mol-1, followed by a request to explain conditions using Le Chatelier's principle (4 marks) and a table of equilibrium concentrations to calculate Kc in standard form and appropriate significant figures (2 marks). Part (b) provides the equation 4NH3(g) + 5O2(g) -> 4NO(g) + 6H2O(g) with delta H = -908 kJ mol-1, a table of standard enthalpy changes of formation, a request to state standard temperature and pressure conditions (1 mark), and a calculation of the standard enthalpy change of formation for NO(g) (3 marks).
Question text

24 This question is about ammonia, NH3.

(a) In industry, ammonia is made from nitrogen and hydrogen. This is a reversible reaction, as

shown in equilibrium 24.1 below.

N (g) + 3H (g) 2NH (g) ∆H = −92 kJ mol−1 Equilibrium 24.1

22 3

(i) Explain how le Chatelier’s principle can be used to predict the conditions of temperature

and pressure for a maximum equilibrium yield of ammonia.

… [4]

(ii) Using certain conditions, equilibrium 24.1 has the equilibrium concentrations in the

table.

Equilibrium concentration

Substance −3

/ mol dm

N2(g) 1.25

H2(g) 2.75

NH3(g) 0.862

Calculate the numerical value for Kc for equilibrium 24.1 under these conditions.

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

(b) Ammonia is used to make nitric acid. The first stage of the reaction is shown below.

4NH (g) + 5O (g) → 4NO(g) + 6H O(g) ∆H = −908 kJ mol−1

32 2

Standard enthalpy changes of formation, ∆ H o, are given in the table.

f

Substance ∆ H o / kJ mol−1

f

NH3(g) −46

O2(g) 0

Kc = … [2]

H2O(g) −242

(i) State the conditions of temperature and pressure used for standard enthalpy

measurements.

Temperature …

Pressure …

[1]

(ii) Calculate the standard enthalpy change of formation for NO(g).

Give your answer to a whole number.

∆ H o for NO(g) = … kJ mol−1 [3]

f

Mark scheme

Show the mark scheme The mark scheme details points for part (a)(i) explaining pressure and temperature effects using Le Chatelier's principle (4 marks), part (a)(ii) showing the Kc expression, substitution, and final answer in standard form to 3 SF (2 marks), part (b)(i) stating 298 K / 25 °C and 100 kPa (1 mark), and part (b)(ii) detailing the use of formation enthalpy values, correct subtraction using delta H, and division by 4 to find +90 kJ mol-1 (3 marks).

AO

Question Answer Marks Guidance

element

24 (a) (i) 4 FULL ANNOTATIONS MUST BE USED

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

ALLOW suitable alternatives for right-hand side,

e.g.: towards NH3/products

OR forward direction

OR increases yield

Pressure:

Right-hand side has fewer (gaseous) moles

OR 4 (gaseous) moles form 2 (gaseous) moles AO1.2 For moles, ALLOW molecules/particles

High pressure AO2.1

Temperature:

(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

Low temperature AO2.1 ORA for reverse reaction

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

IF answer = 2.86 × 10–2 award 2 marks ×2 ECF credit possible using working below.

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

Kc expression

[NH ]2 0.8622 ALLOW calculated value 0.02858291 correctly

(Kc = ) 3 OR 3 rounded to 3 or more SF for 1st marking point

[N2] [H2] 1.25 × 2.75

OR 0.02858 …

ALLOW ECF to 3 SF and standard form

ONLY from inverted K expression → 3.50 × 101

Answer to 3 SF and in standard form c

K = 2.86 × 10–2

c

[NH ]2

DO NOT ALLOW 3 = 0.0337 (no marks)

[N2] + [H2]

IGNORE attempts at units

AO

12 element

(b) (i) 298 K/25ºC 1 AO1.1 ALLOW ‘a stated temperature’

AND To accept that other standard temperatures can be

100 kPa used and 298 should strictly be added as ∆H o

ALLOW 1 × 105 Pa, 101 kPa, 1.01 × 105 Pa,

1 atm, 1 bar

(ii) FIRST, CHECK THE ANSWER ON ANSWER LINE 3 AO2.6 FULL ANNOTATIONS MUST BE USED

IF answer = (+)90 (kJ mol–1) award 3 marks

×3 ALLOW ECF if common errors not seen

IF answer = –90 (kJ mol–1) award 2 marks

IF answer = (+)360 (kJ mol–1) award 2 marks

IF ∆H of –908 has NOT been used,

--------------------------------------------------------------------– ONLY award 1st mark

Use of ∆fH values and balancing numbers -----------------------------------------------------------------

± (4 × –46) OR ± 184 COMMON ERRORS

AND

± (6 × –242) OR ± 1452 seen anywhere 1 mark

Incorrect signs(s) AND missing ÷4

Correct subtraction using ∆H = –908 ±2544 from ± ( 184 + 1452 + 908)

4 × ∆fH(NO) ±728 from ± ( 184 + 1452 – 908)

= (4 × –46) – (6 × –242) – 908 ±2176 from ± (–184 + 1452 + 908)

= –184 + 1452 – 908 –360 from – (–184 + 1452 – 908)

= (+)360 (kJ mol–1)

2 marks

Calculation of ∆fH(NO) formation by ÷4 Incorrect signs(s)

360 –1 ±636 from ± ( 184 + 1452 + 908) = ± 2544÷4

∆fH(NO) = = (+)90 (kJ mol ) ±182 from ± ( 184 + 1452 – 908) = ± 728÷4

±544 from ± (–184 + 1452 + 908) = ± 2176÷4

–90 from – (–184 + 1452 – 908) = – 360÷4

Total 10

How to answer it

Equilibrium and Enthalpy Changes Study Guide

What this question tests

This question assesses your understanding of chemical equilibria, Le Chatelier's principle, equilibrium constant calculations (Kc), standard conditions, and thermodynamic calculations using standard enthalpy changes of formation (ΔfH°). You must demonstrate precise data handling, correct unit application, and strict adherence to significant figure rules.

Question 24 (a)(i)

Applying Le Chatelier's Principle

✅ Correct Answer

Pressure: High pressure. The right-hand side has fewer moles of gas (4 moles forming 2 moles), so the equilibrium shifts right to oppose the increase in pressure.

Temperature: Low temperature. The forward reaction is exothermic (ΔH is negative / gives out heat), so the equilibrium shifts right to oppose the decrease in temperature by releasing heat.

🧠 Exam Technique

Structure your answer into two distinct parts: Pressure and Temperature. For each factor, explicitly state:

  1. The condition needed (High pressure / Low temperature).
  2. The chemical reasoning (compare gas moles; state enthalpy sign/exothermic nature).
  3. The direction of shift and its effect on yield.
🎯 Mark breakdown: 4 marks available (1 mark for pressure reasoning, 1 mark for high pressure condition, 1 mark for temperature reasoning, 1 mark for low temperature condition).
Question 24 (a)(ii)

Calculating the Equilibrium Constant (Kc)

📐 Step-by-Step Calculation

  1. Write the Kc expression: Kc = [NH₃]² / ([N₂][H₂]³)
  2. Substitute equilibrium concentrations: Kc = (0.862)² / ((1.25) × (2.75)³)
  3. Calculate raw value: Kc = 0.743044 / (1.25 × 20.796875) = 0.0285829...
  4. Format output: Convert to standard form and 3 significant figures: 2.86 × 10⁻²

❌ Common Errors & Traps

  • Power errors: Forgetting to raise concentrations to the power of their stoichiometric balancing numbers (e.g., forgetting cubed on [H₂] or squared on [NH₃]).
  • Formatting failures: Failing to report the final answer in standard form or rounding to incorrect significant figures will lose method/accuracy marks.
🎯 Mark breakdown: 2 marks available (1 mark for correct Kc expression with correct substitution, 1 mark for correct final answer given to 3 SF in standard form).
Question 24 (b)(i)

Standard Conditions for Enthalpy Measurements

💡 Key Knowledge

Temperature: 298 K (or 25 °C)

Pressure: 100 kPa (or 1 atm / 101 kPa / 100,000 Pa)

❌ Common Errors

Writing vague terms like "room temperature" or "normal pressure" without specifying numerical values will result in zero marks.

🎯 Mark breakdown: 1 mark available for stating both correct standard values.
Question 24 (b)(ii)

Standard Enthalpy Change of Formation Calculation

📐 Step-by-Step Calculation

  1. Recall the thermodynamic cycle formula: ΔH = Σ ΔfH(products) - Σ ΔfH(reactants)
  2. Account for balancing numbers from equation:
    Reactants: 4NH₃ (4 × -46 = -184) + 5O₂ (5 × 0 = 0) = -184 kJ mol⁻¹
    Products: 6H₂O (6 × -242 = -1452) + 4NO (4 × x)
  3. Set up the equation: -908 = [(-1452) + 4x] - [-184]
  4. Rearrange and solve for 4x: 4x = -908 + 1452 - 184 → 4x = +360
  5. Divide by 4 for one mole of NO: x = 360 / 4 = +90 kJ mol⁻¹

❌ Common Calculation Traps

  • Sign inversion: Getting confused with minus signs during subtraction of negative enthalpy values (e.g., missing the double negative when subtracting reactant values).
  • Omitting division: Forgetting to divide by the stoichiometric coefficient of NO (4) at the final step.
  • Units and signs: Failing to include the explicit positive sign ( +90 ) can lose marks depending on strict mark scheme enforcement; always show signs for enthalpy changes.
🎯 Mark breakdown: 3 marks available (1 mark for correct use of ΔfH values and stoichiometry, 1 mark for correct algebraic rearrangement/subtraction yielding 360, 1 mark for final division and correct whole number answer +90).

Topics

Module 3: Periodic table and energy · Module 5: Physical chemistry and transition elements · 3.2 Physical chemistry · 5.1 Rates, equilibrium and pH · 5.2 Energy

Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.