OCR A-Level Chemistry AS Breadth in chemistry (01), June 2018: Question 11
1 mark · Medium difficulty · Multiple Choice
Calculate the standard enthalpy change of combustion of aluminium sulfide using standard enthalpy changes of formation.
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Question text
11 The equation for the reaction of aluminium sulfide, Al2S3, with oxygen is shown below.
2Al2S3(s) + 9O2(g) → 2Al2O3(s) + 6SO2(g)
The table shows standard enthalpy changes of formation, ∆ H o.
f
Substance Al2S3(s) O2(g) Al2O3(s) SO2(g)
∆ H o / kJ mol−1 −723.8 0 −1675.7 −296.8
f
What is the standard enthalpy change of combustion of Al S (s), in kJ mol−1?
A −3684.6
B −1842.3
C +1842.3
D +3684.6
Your answer [1]
Mark scheme
Show the mark scheme
11 B 1
How to answer it
Calculating Enthalpy Change of Combustion from Enthalpies of Formation
What this question tests
This question assesses your understanding of thermodynamic definitions, specifically standard enthalpy changes of formation (ΔfHꭒ) and standard enthalpy changes of combustion. It tests your ability to apply Hess's Law using a cycle or formula involving enthalpy of formation data, manipulate stoichiometry from a balanced chemical equation, and properly scale values for 1 mole of substance.
Exam Breakdown & Solution
✅ Correct Answer: B (-1842.3 kJ mol⁻¹)
The correct option is B. The enthalpy change of combustion represents the reaction of 1 mole of a substance in excess oxygen. Because the given equation reacts 2 moles of Al₂S₃, the final calculated enthalpy change for the equation must be divided by 2.
💡 Key Knowledge
- Enthalpy of Formation (ΔfHꭒ): Enthalpy change when 1 mole of a compound is formed from its elements in their standard states.
- Enthalpy of Combustion (ΔcHꭒ): Enthalpy change when 1 mole of a substance reacts completely in oxygen under standard conditions.
- Elements in standard states: O₂(g) has an enthalpy of formation of exactly 0 kJ mol⁻¹.
📐 Step-by-Step Calculation
- Recall the formula using enthalpies of formation:
ΔH = Σ ΔfHꭒ (products) − Σ ΔfHꭒ (reactants) - Substitute values from the balanced equation:
Equation: 2Al₂S₃ + 9O₂ → 2Al₂O₃ + 6SO₂
Products: (2 × −1675.7) + (6 × −296.8)
= −3351.4 + (−1780.8) = −5132.2 kJ mol⁻¹ - Calculate reactants:
Reactants: (2 × −723.8) + (9 × 0)
= −1447.6 kJ mol⁻¹ - Find ΔH for the written equation:
ΔH = (−5132.2) − (−1447.6) = −3684.6 kJ mol⁻¹ - Adjust for 1 mole of fuel (Enthalpy of Combustion):
The equation shows 2 moles of Al₂S₃ reacting.
ΔcHꭒ = −3684.6 ÷ 2 = −1842.3 kJ mol⁻¹
❌ Common Errors & Examiner Traps
- Forgetting to divide by 2 (Distractor A): Many students calculate −3684.6 kJ mol⁻¹ and immediately select option A. Remember that enthalpy of combustion is defined per 1 mole of the substance burned.
- Sign errors (Distractor C/D): Subtracting reactants in the wrong order or mishandling negative numbers leads to positive enthalpy values. Combustion reactions are exothermic, so the value must be negative.
- Ignoring stoichiometry multipliers: Forgetting to multiply individual formation values by the stoichiometric balancing numbers (2, 6, and 9) in the equation.
🧠 Exam Technique & Top-Level Insights
In multiple-choice calculation questions, examiners deliberately include intermediate values (such as the total for 2 moles, found in option A) as distractors to catch students who rush. Always double-check what the question is asking for: per mole of reaction versus per mole of a specific reactant.
Topics
Module 3: Periodic table and energy · 3.2 Physical chemistry
Question and mark scheme from the OCR A-Level Chemistry examination, AS Breadth in chemistry (01), June 2018. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.