AQA A-Level Chemistry AS Paper 1, November 2021: Question 7

Mark scheme

Show the mark scheme

Question Marking guidance Additional Comments/Guidelines Mark

Enthalpy change when one mole of a substance burns completely in Allow heat energy change / allow fully combust 1

oxygen

07.1

With all substances in their standard states (at stated temperature 1

and 100kPa)

q = m c ∆T = 150 × 4.18 × 13.9 = 8715.3 J 1

0.497 1

n(propan-1-ol)= = 0.00828 mol

60.0

07.2

8.7153 -1 1

∆H = − = −1050 kJ mol M3 = – M1×10–3/M2

0.00828

Minimum of 2 sf needed

Must be negative

Incomplete combustion Evaporation of fuel 1

07.3 Experiment not completed under standard

conditions

How to answer it

Enthalpy of Combustion Study Guide

What this question tests

This question assesses your understanding of thermochemistry definitions, calorimetric calculations involving heat capacity, and the evaluation of experimental errors in combustion experiments. Key skills include applying q = mcΔT, handling units (Joules to kiloJoules), finding moles, assigning correct signs to enthalpy changes, and identifying practical limitations.

Part 07.1

Defining Standard Enthalpy of Combustion

State the meaning of the term standard enthalpy of combustion. [2 marks]

✅ Correct Answer

The enthalpy change when one mole of a substance is burned completely in oxygen, with all substances in their standard states (at a stated temperature and 100 kPa).

💡 Key Knowledge

Mark 1: Awarded for mentioning "one mole of a substance" burning completely in oxygen (or fully combusting).
Mark 2: Awarded for stating that all substances are in their standard states under standard conditions.

Part 07.2

Calorimetry Calculation

Calculate a value, in kJ mol⁻¹, for the enthalpy of combustion of propan-1-ol in this experiment. [3 marks]

📐 Step-by-Step Calculation

Step 1: Calculate heat energy released (q)
q = m × c × ΔT
m = 150 g , c = 4.18 J K⁻¹ g⁻¹ , ΔT = 35.1 - 21.2 = 13.9 K
q = 150 × 4.18 × 13.9 = 8715.3 J (or 8.7153 kJ)
Step 2: Calculate moles of propan-1-ol burned
Moles = mass / Mᵣ = 0.497 / 60.0 = 0.008283 mol
Step 3: Calculate ΔH per mole and apply sign
ΔH = - (q in kJ / moles) = -8.7153 / 0.008283 = -1052 kJ mol⁻¹
Rounding to appropriate sig fig (minimum 2 sf, matching 13.9 and 0.497 gives 3 sf): -1050 kJ mol⁻¹

❌ Common Errors & Traps

Missing negative sign: Combustion is exothermic, so ΔH must be negative. Omitting the minus sign loses M3.
Unit conversion failures: Forgetting to divide q by 1000 to convert Joules to kiloJoules before dividing by moles.
Using mass of alcohol instead of water: Using 0.497 g for m in q = mcΔT instead of the mass of water (150 g).

Mark Scheme Breakdown:
• M1: Correct calculation of q (8715.3 J or 8.715 kJ)
• M2: Correct calculation of moles (0.00828 mol)
• M3: Correct evaluation of ΔH with a negative sign and appropriate units (-1050 kJ mol⁻¹)

Part 07.3

Evaluating Experimental Limitations

Give one possible reason for this, other than heat loss. [1 mark]

✅ Correct Answer

Any one of the following:

Incomplete combustion
Evaporation of the fuel / alcohol
Experiment not completed under standard conditions

🧠 Exam Technique

Read the stem carefully! The question explicitly states "other than heat loss". Writing heat loss will score zero. Keep your answer concise and scientifically precise—mentioning "incomplete combustion" immediately secures the mark.

Topics

Physical Chemistry · Required Practicals · 3.1.4 Energetics · Required Practical 2: Measurement of an enthalpy change