OCR A-Level Chemistry Periodic table, elements and physical chemistry (01), June 2025: Question 6

1 mark · Medium difficulty · Multiple Choice

Calculate the C–H bond enthalpy in methane using the given enthalpy change of combustion and bond enthalpies.

Practise this question

Question

Question 6: Methane reacts with oxygen according to the equation CH4(g) + 2O2(g) -> CO2(g) + 2H2O(g) with an enthalpy change of reaction of -730 kJ per mole. A table shows bond enthalpies in kJ per mole: O-H is +464, O=O is +498, and C=O is +805. Four multiple-choice options are provided for the bond enthalpy of the C-H bond: A (+109), B (+367), C (+435), and D (+568).

Mark scheme

Show the mark scheme Mark scheme for question 6 showing the correct answer is C, with 1 mark awarded.

How to answer it

OCR A-Level Chemistry • Module 3: Enthalpy & Bond Energies

Bond Enthalpy Calculation from Enthalpy of Combustion

What this question tests

This question evaluates your ability to rearrange the mean bond enthalpy expression to determine an unknown single bond enthalpy from an overall enthalpy change of reaction (ΔrH):

  • Counting covalent bonds per molecule using molecular structures (CH₄, O₂, CO₂, H₂O).
  • Applying stoichiometric molar ratios to bond totals.
  • Using the fundamental enthalpy relationship: ΔrH = Σ(bonds broken) − Σ(bonds made) .
  • Correctly tracking positive and negative signs during algebraic rearrangement.

Question 6 (Multiple Choice)

Enthalpy of reaction: CH₄(g) + 2O₂(g) → CO₂(g) + 2H₂O(g)  |  ΔrH = −730 kJ mol⁻¹

✅ Correct Answer

C : +435 kJ mol⁻¹

Award [1] mark for identifying key C.

💡 Key Knowledge

  • Bond breaking is endothermic (+ΔH), occurring in reactants.
  • Bond making is exothermic (−ΔH), occurring in products.
  • Formula: ΔrH = Σ(Reactant bonds) − Σ(Product bonds)
  • Every bond enthalpy value listed is endothermic and thus has a positive (+) sign.

📐 Step-by-Step Calculation

1 Tally bonds in reactants (bonds broken):

  • CH₄ contains 4 × (C−H) bonds = 4 × E(C−H)
  • 2O₂ contains 2 × (O=O) bonds = 2 × (+498) = +996 kJ mol⁻¹
  • Σ(bonds broken) = 4 × E(C−H) + 996

2 Tally bonds in products (bonds formed):

  • CO₂ contains 2 × (C=O) double bonds = 2 × (+805) = +1610 kJ mol⁻¹
  • 2H₂O contains 2 × 2 = 4 × (O−H) single bonds = 4 × (+464) = +1856 kJ mol⁻¹
  • Σ(bonds made) = 1610 + 1856 = +3466 kJ mol⁻¹

3 Substitute into the reaction enthalpy formula:

ΔrH = Σ(bonds broken) − Σ(bonds made)
−730 = [4 × E(C−H) + 996] − 3466

4 Rearrange algebraically to solve for E(C−H):

−730 = 4 × E(C−H) − 2470
4 × E(C−H) = 2470 − 730
4 × E(C−H) = +1740 kJ mol⁻¹
E(C−H) = 1740 ÷ 4 = +435 kJ mol⁻¹

🧠 Exam Technique & Examiner Insight

  • Draw out structural formulae: Sketches like O=C=O and H−O−H ensure you do not miss double bonds or count only one bond per molecule.
  • Check your answer is sensible: Typical covalent single bond enthalpies (C−H, C−C, C−O) fall in the range of 300–500 kJ mol⁻¹. A value of +435 kJ mol⁻¹ is chemically realistic, whereas +109 kJ mol⁻¹ (Option A) is unusually low.
  • Keep track of the negative sign: The reaction is exothermic, so ΔrH is −730 , not +730 .

❌ Common Calculation Traps

  • Forgetting 2 moles of H₂O: Water has two O−H bonds, meaning 2H₂O has 4 O−H bonds. Counting only 2 O−H bonds leads to an incorrect answer.
  • Confusing Hess cycles with bond enthalpies: Students often incorrectly use Products − Reactants (which applies only to enthalpies of formation, ΔfH). For bond enthalpies, it must always be Reactants (broken) − Products (made) .
  • Dividing incorrectly at the end: Forgetting to divide by 4 at the final step yields +1740 kJ mol⁻¹ (the energy of all 4 C−H bonds together).

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), June 2025. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.