AQA A-Level Chemistry AS Paper 2, June 2025: Question 7

10 marks · Medium difficulty · State/Explain/Numerical

Define enthalpy change, calculate the mass of ethanol burned using calorimetry data, show the bond enthalpy expression for combustion products of an alkane, and determine its molecular formula.

Practise this question

Question

Question 07 is divided into four parts about combustion enthalpies. Part 07.1 asks for the definition of enthalpy change for 1 mark. Part 07.2 gives the enthalpy of combustion of ethanol as -1371 kJ/mol and asks to calculate the mass of ethanol burned to heat 55.0 g of water by 46.2 °C (c = 4.18 J K^-1 g^-1) for 5 marks. Part 07.3 displays a combustion equation for an alkane CnH2n+2 and Table 2 with mean bond enthalpies for C=O (805 kJ/mol) and O-H (463 kJ/mol), asking to show that the bond enthalpies for the products sum to 2536n + 926 (1 mark). Part 07.4 provides the reactants' bond enthalpy expression (1916n + 724) and an enthalpy of combustion of -3302 kJ/mol to calculate n and deduce the alkane's molecular formula (3 marks).
Question text

07 This question is about enthalpies of combustion.

07.1 Define enthalpy change.

[1 mark]

07.2 The enthalpy of combustion of ethanol is –1371 kJ mol–1

A sample of ethanol is burned completely in air to raise the temperature of

55.0 g of water by 46.2 °C

Calculate the mass of ethanol in the sample.

Assume there is no heat loss.

Specific heat capacity of water, c = 4.18 J K–1 g–1

[5 marks]

Mass of ethanol g

An alkane burns completely in oxygen. The equation for this reaction is shown.

3n + 1

CnH2n+2 + O2 → n CO2 + (n+1)H2O

Table 2 gives some mean bond enthalpies.

Table 2

Bond C=O O–H

Bond enthalpy / kJ mol–1 805 463

07.3 Use Table 2 to show that an expression for the sum of the bond enthalpies for all the

bonds in the products is 2536n + 926

[1 mark]

07.4 An expression for the sum of the bond enthalpies for all the bonds in the reactants is

1916n + 724

A calculated value for the enthalpy of combustion of an alkane is –3302 kJ mol–1

Use this enthalpy change and the expressions to calculate a value of n for this alkane.

Use your value of n to deduce the molecular formula of the alkane.

[3 marks]

Value of n

Molecular formula

Mark scheme

Show the mark scheme Mark scheme for Question 07. 07.1: heat (energy) change at constant pressure (1 mark). 07.2: M1 uses q = mcΔT = 55 × 4.18 × 46.2; M2 = 10621 J or 10.621 kJ; M3 defines amount = q/ΔH; M4 calculates moles = 0.007747 mol; M5 calculates mass = 0.007747 × 46.0 = 0.356 g (5 marks). 07.3: shows 2n(805) + 2(n+1)463 = 2536n + 926 (1 mark). 07.4: M1 sets up -3302 = (1916n + 724) - (2536n + 926); M2 solves for n = 5; M3 gives formula C5H12 (3 marks).

Question Marking guidance Additional Comments/Guidelines Mark

change in heat content at constant pressure

07.1 heat (energy) change at constant pressure

ignore reference to standard states (1 x AO1)

M1 q (= mcΔT) = 55 × 4.18 × 46.2 Ignore minus sign in M1/2

Allow ECF from M1 to M2

M2 = 10621 J or 10.621 kJ

Allow ECF from M2 to M4

q 5

M3 amount of ethanol = (–) Allow ECF from M3 to M4

07.2 ∆H (1 x AO1,

Allow ECF from M4 to M5 4 x AO3)

M2 in kJ

M4 = (= 0.007747) 356 g scores 4 marks

1371

Use of ΔT as (46.2+273) gives 73.4 kJ and 2.46 g

M5 mass ethanol = M4 × 46.0 = 0.356 (g) (min 2sf) – scores 4 marks

2n(805) + 2(n+1)463 Required brackets must be used correctly

OR 2n(805) + (2n+2)463 Accept 2 x n x 805 + 2 x (n+1) x 463

07.3 – AS CHEMISTRY – 7404/2 –

OR 1610n + 926n + 926 (1 x AO2)

(= 2536n + 926)

M1 Energy change = (1916n + 724) – (2536n + 926) No ECF from M1 to M2

or –3302 = (1916n + 724) – (2536n + 926) Could allow M1, M2 and/or M3 for correct answers

with no working

or 620n = 3100 3

07.4

Allow ECF from M2 to M3 using nearest integer (3 x AO3)

M2 n = 5 (must be an integer)

(but must be a positive integer)

M3 C5H12

Answer for M3 is dependent on their value for M2

How to answer it

Enthalpy of Combustion & Mean Bond Enthalpies

📋 What this question tests

This question assesses fundamental thermochemistry knowledge and multi-step quantitative calculation skills:

  • Recalling the exact definition of enthalpy change.
  • Performing calorimetry calculations using q = mcΔT and rearranging with ΔcH to determine the mass of fuel burned.
  • Deriving algebraic expressions for sum of bond enthalpies in combustion products.
  • Applying ΔH = Σ(bonds broken) - Σ(bonds formed) to solve for unknown chain length n and determining molecular formulas.
Part 07.1 • 1 Mark

Definition of Enthalpy Change

Define enthalpy change.

✅ Correct Answer

Heat (energy) change at constant pressure.

Award [1 mark] for both "heat (energy) change" and "constant pressure".

💡 Key Knowledge

  • Enthalpy ( H ) is defined specifically at constant pressure.
  • Alternative acceptable wording: "Change in heat content at constant pressure".

❌ Common Errors

  • Forgetting to state "at constant pressure" (results in 0 marks).
  • Stating "energy change at constant temperature" (confusing conditions).

🧠 Exam Technique

Standard conditions (100 kPa, 298 K) are not required here since the question asked for enthalpy change, not standard enthalpy change. However, mentioning standard conditions is ignored and will not penalise you.

Part 07.2 • 5 Marks

Reverse Calorimetry Calculation: Mass of Ethanol

Calculate the mass of ethanol burned to raise 55.0 g of water by 46.2 °C.

📐 Step-by-Step Calculation

  1. Calculate heat absorbed by the water (q):
    q = m × c × ΔT
    q = 55.0 g × 4.18 J K⁻¹ g⁻¹ × 46.2 °C = 10621.14 J
    [M1] Correct substitution: 55 × 4.18 × 46.2
    [M2] 10621 J or 10.621 kJ
  2. Calculate amount of ethanol burned (moles):
    Convert q to kJ: 10.621 kJ
    n(ethanol) = q / |ΔcH| = 10.621 kJ / 1371 kJ mol⁻¹ = 0.007747 mol
    [M3] Stating or using amount = q / ΔH
    [M4] 10.621 / 1371 = 0.007747 mol
  3. Calculate mass of ethanol:
    Mr(C₂H₅OH) = (2 × 12.0) + (6 × 1.0) + 16.0 = 46.0 g mol⁻¹
    mass = n × Mr = 0.007747 mol × 46.0 g mol⁻¹ = 0.356 g
    [M5] 0.356 g (accept min 2 sig figs, e.g. 0.36 g )

❌ Common Errors & Lost Marks

  • Unit conversion fail: Forgetting to divide q by 1000 before dividing by 1371 kJ mol⁻¹. This yields 356 g (caps at 4 marks).
  • Temperature trap: Adding 273 to the temperature rise (treating 46.2 as a temperature rather than a change ΔT). This gives 73.4 kJ and 2.46 g (caps at 4 marks).
  • Wrong mass in q: Using mass of ethanol instead of mass of water (55.0 g) in q = mcΔT .

🧠 Exam Technique

  • Error-Carried-Forward (ECF): ECF applies consecutively from M1 through to M5. Always show full workings!
  • Signs: Combustion is exothermic (ΔH is negative), but heat released q is a magnitude. Minus signs are ignored in steps M1/M2.
Part 07.3 • 1 Mark

Algebraic Expression for Bond Enthalpies in Products

Show that the sum of bond enthalpies for products is 2536n + 926

💡 Products Structure Breakdown

Equation shows products: n CO₂ + (n + 1) H₂O

  • Each CO₂ (O=C=O) has two C=O bonds → Total C=O bonds = 2n
  • Each H₂O (H-O-H) has two O-H bonds → Total O-H bonds = 2(n + 1)

✅ Correct Answer / Working

= 2n(C=O) + 2(n + 1)(O-H)

= 2n(805) + 2(n + 1)(463)

= 1610n + (2n + 2)(463)

= 1610n + 926n + 926 = 2536n + 926

Award [1 mark] for correctly writing out the unsimplified or intermediate algebraic expression with correct brackets.

❌ Common Errors

  • Forgetting that each molecule has two bonds (e.g. writing n(805) + (n+1)(463) ).
  • Missing brackets: writing 2n + 1 × 463 instead of 2(n + 1) × 463 .
Part 07.4 • 3 Marks

Deducing Value of n and Molecular Formula

Using Reactants = 1916n + 724 and ΔcH = -3302 kJ mol⁻¹, find n and the alkane formula.

📐 Step-by-Step Calculation

  1. Set up enthalpy change formula:
    ΔH = Σ(bonds broken in reactants) - Σ(bonds formed in products)
    -3302 = (1916n + 724) - (2536n + 926)
    [M1] Correct expression equating ΔH to reactants minus products.
  2. Simplify and solve for n:
    -3302 = (1916n - 2536n) + (724 - 926)
    -3302 = -620n - 202
    -3100 = -620n
    n = 3100 / 620 = 5
    [M2] n = 5 (must be an integer).
  3. Deduce molecular formula:
    General formula for alkane: CnH2n+2
    For n = 5 : C₅H₍₂ₓ₅₊₂₎ = C₅H₁₂ (pentane)
    [M3] C₅H₁₂ (dependent on positive integer from M2).

❌ Common Errors

  • Sign reversal: Writing Products - Reactants instead of Reactants - Products for bond enthalpies, resulting in a negative value for n.
  • Sign of ΔH: Forgetting that combustion is exothermic and using +3302 instead of -3302 .
  • Bracket expansion error: Failing to distribute the negative sign: -(2536n + 926) becomes -2536n - 926 , not -2536n + 926 .

🧠 Exam Technique

  • Remember: B-B-O-F → Bonds Broken (reactants) minus Bonds Formed (products).
  • n must always be a whole positive number for an alkane. If you get a fraction or negative, re-check your signs immediately!

Topics

Physical Chemistry · Organic Chemistry · 3.1.4 Energetics · 3.1.2 Amount of Substance · 3.3.2 Alkanes

Question and mark scheme from the AQA A-Level Chemistry examination, AS Paper 2, June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.