AQA A-Level Chemistry Paper 3, 2019: Question 8

1 mark · Easy difficulty · Multiple Choice

Calculate the enthalpy change for the oxidation of ammonia to nitrogen dioxide using Hess's law.

Practise this question

Question

Multiple choice question showing two chemical equations with enthalpy changes: 4NH3(g) + 5O2(g) to 4NO(g) + 6H2O(g) with Delta H = -909 kJ mol^-1, and 2NO(g) + O2(g) to 2NO2(g) with Delta H = -115 kJ mol^-1. Students are asked to calculate the enthalpy change in kJ mol^-1 for 4NH3(g) + 7O2(g) to 4NO2(g) + 6H2O(g), with options A: -679, B: -794, C: -1024, and D: -1139.
Question text

08 Nitrogen dioxide is produced from ammonia and air as shown in these equations

4 NH (g) + 5O (g) → 4NO(g) + 6 H O(g) ΔH = –909 kJ mol–1

32 2

2 NO(g) + O (g) → 2 NO (g) ΔH = –115 kJ mol–1

What is the enthalpy change (in kJ mol–1) for the following reaction?

4NH3(g) + 7O2(g) → 4NO2(g) + 6H2O(g)

[1 mark]

A –679

B –794

C –1024

D –1139

Mark scheme

Show the mark scheme Mark scheme table row showing question number 8 with correct answer D and 1 mark.

8 D 1

How to answer it

Hess's Law: Combining Enthalpy Changes

📋 What this question tests

This question assesses your ability to apply Hess's Law to determine an unknown reaction enthalpy from given thermochemical equations. Key skills include: balancing intermediate species, scaling stoichiometric coefficients, and tracking negative signs during algebraic addition.

Question 08

Calculating ΔH for the Oxidation of Ammonia to Nitrogen Dioxide

Multiple Choice (1 Mark)

✅ Correct Answer

D  (−1139 kJ mol⁻¹)

Mark Scheme Breakdown:
• 1 mark for selecting option D.

💡 Key Knowledge

  • Hess's Law: Total enthalpy change is independent of the route taken.
  • If a chemical equation is multiplied by a factor n , its enthalpy change ΔH must also be multiplied by n .
  • When reactions are added together, intermediate species that appear equally on both reactant and product sides cancel out.

📐 Step-by-Step Solution

1. Identify the Target Equation:

4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g)   ΔH = ?

2. Inspect the Given Equations:

(1) 4NH₃(g) + 5O₂(g) → 4NO(g) + 6H₂O(g)   ΔH₁ = −909 kJ mol⁻¹
(2) 2NO(g) + O₂(g) → 2NO₂(g)   ΔH₂ = −115 kJ mol⁻¹

3. Manipulate the Equations:

  • Equation (1): Contains the required 4NH₃(g) as reactants and 6H₂O(g) as products. Keep as is:
    ΔH = −909 kJ mol⁻¹
  • Equation (2): Contains 2NO₂(g) as a product, but our target needs 4NO₂(g) . Multiply equation (2) by 2:
    2 × [2NO(g) + O₂(g) → 2NO₂(g)]
    4NO(g) + 2O₂(g) → 4NO₂(g)
    ΔH = 2 × (−115) = −230 kJ mol⁻¹

4. Add the Two Equations Together:

[4NH₃ + 5O₂] + [4NO + 2O₂] → [4NO + 6H₂O] + [4NO₂]
Cancelling 4NO(g) on both sides:
4NH₃(g) + 7O₂(g) → 4NO₂(g) + 6H₂O(g)

5. Sum the Enthalpy Changes:

ΔHtarget = ΔH₁ + 2(ΔH₂)
ΔHtarget = (−909) + 2(−115)
ΔHtarget = −909 − 230 = −1139 kJ mol⁻¹

🧠 Exam Technique

  • Target Tracker: Look at unique species in the target. NO₂ only appears in Equation 2. Because the target requires 4 moles of NO₂ and Equation 2 has 2 moles, multiplying Equation 2 by 2 is mandatory.
  • Check intermediate cancellation: Notice that 4NO produced in (1) is completely consumed in 2 × (2), confirming the reaction pathway works.

❌ Distractor Breakdown & Common Errors

  • Option C (−1024): Simply added (−909) + (−115) without multiplying Equation 2 by 2.
  • Option A (−679): Subtracted instead of adding: −909 − (−230) = −679.
  • Option B (−794): Forgot to multiply by 2 AND subtracted: −909 − (−115) = −794.
  • All three incorrect options correspond to typical student arithmetic and stoichiometric slip-ups!

Topics

Physical Chemistry · 3.1.4 Energetics

Question and mark scheme from the AQA A-Level Chemistry examination, Paper 3, 2019. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.