OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), November 2020: Question 6

1 mark · Medium difficulty · Multiple Choice

Calculate the standard enthalpy of combustion of C7H8 using given enthalpies of formation values.

Practise this question

Question

Multiple choice question 6 shows the combustion equation for C7H8(l) + 9O2(g) -> 7CO2(g) + 4H2O(l) and a table of enthalpies of formation for C7H8(l), CO2(g), and H2O(l). Four multiple-choice options A, B, C, and D are given representing possible enthalpy of combustion values.
Question text

6 The equation for the combustion of C7H8 is shown in the following equation.

C7H8(l) + 9O2(g) 7CO2(g) + 4H2O(l)

Enthalpy changes of formation are shown in the table.

Substance C7H8(l) CO2(g) H2O(l)

∆ H / kJ mol−1 +12 −394 −286

f

Calculate the enthalpy of combustion, in kJ mol−1, for the hydrocarbon C H .

A −3914

B −692

C +692

D +3914

Your answer [1]

Mark scheme

Show the mark scheme The mark scheme indicates that the correct answer for question 6 is option A.

6 A 1 2.2

How to answer it

Calculating Enthalpy of Combustion from Enthalpies of Formation

What this question tests

This question assesses your understanding of thermodynamic cycles, specifically how to apply standard enthalpies of formation (ΔfH) to calculate the standard enthalpy of combustion (ΔcH) for a hydrocarbon using a balanced chemical equation.

Question 6 Analysis

Multiple Choice Question (1 Mark)

✅ Correct Answer

A (value: -3914 kJ mol⁻¹ )

Mark Awarded: 1 / 1

💡 Key Knowledge

  • Enthalpy of Formation (ΔfH): Enthalpy change when 1 mole of a compound is formed from its elements under standard conditions.
  • Enthalpy of Combustion (ΔcH): Enthalpy change when 1 mole of a substance is burned completely in oxygen.
  • Elements in standard states: Oxygen O₂(g) has an enthalpy of formation of 0 kJ mol⁻¹ .

🧠 Exam Technique

  • Always construct or recall the formula using enthalpies of formation:
    ΔH = ΣΔfH(products) - ΣΔfH(reactants)
  • Don't forget to multiply each enthalpy of formation value by its balancing stoichiometric coefficient from the equation!

❌ Common Errors

  • Reversing the product and reactant subtraction order (e.g., reactants minus products), leading to an incorrect positive sign ( +3914 ).
  • Forgetting to multiply the individual values by the balancing numbers ( 7 for CO₂ and 4 for H₂O ).
  • Omitting oxygen because its value wasn't given in the table (remember: elements have a formation enthalpy of zero).

📐 Step-by-Step Calculation

Given equation: C₇H₈(l) + 9O₂(g) → 7CO₂(g) + 4H₂O(l)

  1. Identify the formula: ΔH = ΣΔfH(products) - ΣΔfH(reactants)
  2. Calculate products total:
    (7 × ΔfH[CO₂]) + (4 × ΔfH[H₂O])
    = (7 × -394) + (4 × -286)
    = -2758 + (-1144) = -3902 kJ mol⁻¹
  3. Calculate reactants total:
    (1 × ΔfH[C₇H₈]) + (9 × ΔfH[O₂])
    = (1 × +12) + (9 × 0) = +12 kJ mol⁻¹
  4. Perform final subtraction (Products - Reactants):
    -3902 - (+12) = -3914 kJ mol⁻¹

Topics

Module 3: Periodic table and energy · 3.2 Physical chemistry

Question and mark scheme from the OCR A-Level Chemistry examination, Periodic table, elements and physical chemistry (01), November 2020. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.