AQA A-Level Chemistry AS Paper 2, June 2025: Question 18

1 mark · Medium difficulty · Multiple Choice

Calculate the standard enthalpy of formation of benzene from standard enthalpies of combustion.

Practise this question

Question

Question 18 asking: 'The standard enthalpies of combustion for carbon, hydrogen and benzene (C6H6) are -394, -286 and -3273 kJ mol^-1 respectively. What is the standard enthalpy of formation of benzene, in kJ mol^-1?' followed by four multiple choice options: A +51, B -51, C +807, D -807.
Question text

18 The standard enthalpies of combustion for carbon, hydrogen and benzene (C6H6) are

–394, –286 and –3273 kJ mol–1 respectively.

What is the standard enthalpy of formation of benzene, in kJ mol–1?

[1 mark]

A +51

B –51

C +807

D –807

Mark scheme

Show the mark scheme Mark scheme table row for Question 18 showing answer 'A' (+51) worth 1 mark, categorized under assessment objective AO2.

18 A 1 (AO2) +51

How to answer it

Calculating the Standard Enthalpy of Formation of Benzene

SPECIFICATION CHECK

What this question tests

  • Constructing Hess’s Law cycles or using formulaic methods from enthalpy of combustion (ΔcH) data.
  • Writing and balancing the defining equation for the standard enthalpy of formation (ΔfH) of a compound.
  • Correct handling of stoichiometric multipliers (e.g. 6 C and 3 H₂ for C₆H₆).
  • Careful evaluation of algebraic signs when subtracting negative combustion values.
QUESTION 18 • 1 MARK • AO2

Question Breakdown & Direct Answer

Calculate standard enthalpy of formation of benzene from standard enthalpies of combustion

✅ Correct Answer: A (+51)

The standard enthalpy of formation of benzene (C₆H₆) is +51 kJ mol⁻¹.

Applying Hess’s Law with combustion data:

ΔfH = ΣΔcH(reactants) − ΣΔcH(products)

ΔfH = [6(−394) + 3(−286)] − [−3273] = +51 kJ mol⁻¹

📐 Step-by-Step Calculation

Step 1: Write the target formation equation

6C(s) + 3H₂(g) → C₆H₆(l)

Requires 6 moles of C and 3 moles of H₂ gas.

Step 2: Sum the combustion enthalpies of reactants

Carbon: 6 × (−394) = −2364 kJ mol⁻¹

Hydrogen: 3 × (−286) = −858 kJ mol⁻¹

ΣΔcH(reactants) = −2364 + (−858) = −3222 kJ mol⁻¹

Step 3: Identify the combustion enthalpy of products

Benzene: 1 × (−3273) = −3273 kJ mol⁻¹

Step 4: Subtract product from reactants

ΔfH = −3222 − (−3273)

ΔfH = −3222 + 3273 = +51 kJ mol⁻¹

💡 Key Knowledge: Enthalpy Definitions & Cycles

  • Standard enthalpy of formation (ΔfH°): Enthalpy change when 1 mole of a compound is formed from its constituent elements in their standard states under standard conditions.
  • Combustion cycle orientation: In a Hess's cycle involving combustion data, combustion products (6CO₂ + 3H₂O) sit at the bottom. The arrows point downwards from both the elements and the compound toward the combustion products.
Hess's Cycle Mental Layout:
Top Left: 6C(s) + 3H₂(g)  —[ ΔfH (Target) ]→  Top Right: C₆H₆(l)
            ↓ ΣΔcH(reactants)                                       ↓ ΔcH(product)
Bottom: 6CO₂(g) + 3H₂O(l)
Therefore: ΔfH = ΣΔcH(reactants) − ΣΔcH(products)

❌ Common Errors & Distractor Traps

  • Option B (−51): The Inverted Formula Trap
    Students confuse combustion cycles with formation cycles and compute Products − Reactants:
    (−3273) − (−3222) = −51 . Remember: When given combustion data, it is Reactants − Products.
  • Option C (+807) & D (−807): The Diatomic Gas Trap
    Students see the subscript "6" in C₆H₆ and multiply hydrogen's value by 6 instead of 3 (using 6H₂ instead of 3H₂):
    6(−394) + 6(−286) = −4080
    −3273 − (−4080) = +807 (or −4080 − (−3273) = −807 ).

🧠 Exam Technique for Energetics MCQs

  • Write the balanced equation first: Never calculate directly in your head. Writing out 6C + 3H₂ → C₆H₆ takes 3 seconds and guarantees you use the correct stoichiometric coefficient for H₂ (3, not 6).
  • Double-negative check: Subtraction of a negative number means addition: −3222 − (−3273) = −3222 + 3273 = +51 . A quick sign check rules out options B and D immediately if you know the cycle direction.
  • Rule of thumb:
    • Given ΔfH values? Formula is: ΣΔfH(Products) − ΣΔfH(Reactants)
    • Given ΔcH values? Formula is: ΣΔcH(Reactants) − ΣΔcH(Products)

Topics

Physical Chemistry · 3.1.4 Energetics

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.