Edexcel A-Level Chemistry Paper 1, June 2024: Question 5

📐 Step-by-Step Calculation

1. Bonds Broken (Reactants):
   Ethanol (CH₃CH₂OH):
   • 5 × (C−H) = 5 × 413 = 2065 kJ mol⁻¹
   • 1 × (C−C) = 1 × 347 = 347 kJ mol⁻¹
   • 1 × (C−O) = 1 × 358 = 358 kJ mol⁻¹
   • 1 × (O−H) = 1 × 464 = 464 kJ mol⁻¹
   Oxygen (3 × O=O):
   • 3 × 498 = 1494 kJ mol⁻¹
   Total Broken = +4728 kJ mol⁻¹
2. Bonds Formed (Products):
   Water (3 × H₂O):
   • 3 × 2 × (O−H) = 6 × 464 = 2784 kJ mol⁻¹
   Carbon Dioxide (2 × CO₂):
   • 2 × 2 × (C=O) = 4 × 805 = 3220 kJ mol⁻¹
   Total Formed = −6004 kJ mol⁻¹
3. Enthalpy Change:
   ΔH = Σ(broken) − Σ(formed)
   ΔH = +4728 − 6004 = −1276 kJ mol⁻¹

❌ Common Errors

• Forgetting bonds in ethanol: Overlooking the single C−C bond or mistaking ethanol for having 6 identical C−H bonds instead of 5 C−H and 1 O−H.
• Missing the stoichiometric multipliers: Forgetting that 2CO₂ contains 4 C=O bonds (not 2), and 3H₂O contains 6 O−H bonds (not 3).
• Sign inversion: Writing +1276 kJ mol⁻¹. Bond breaking is endothermic (+), bond forming is exothermic (−). Combustion must have a negative ΔH!

✅ Key Elements of the Cycle

• Bottom Box: 2C(s) + 3H₂(g) + 3½O₂(g) (or C(graphite)). State symbols are required!
• Arrow Direction: Both arrows must point UP from the elements in standard states to the compounds.
• Labels: Left arrow: ΔfH°[C₂H₅OH(l)] or −277. Right arrow: 3×ΔfH°[H₂O(l)] + 2×ΔfH°[CO₂(g)] or −1646.

🧠 Exam Technique

By definition, standard enthalpy of formation (ΔfH°) is the formation of 1 mole of a compound from its elements. Therefore, the cycle arrows always point away from the elements.

Note: Elements in their standard states like O₂(g) have ΔfH° = 0, which is why only C₂H₅OH needs forming on the left side (from 2C + 3H₂ + ½O₂), while the remaining 3O₂ passes through unchanged.

✅ Two Acceptable Reasons (1 mark each)

1. Mean Bond Enthalpies: The bond enthalpy calculation uses mean (average) bond enthalpies taken from many different chemical environments, which are not the exact bond enthalpies for these specific molecules.
2. Physical States (Gaseous vs. Liquid): Bond enthalpies apply only to substances in the gas state, but ethanol and water are liquids under standard conditions (energy is released during condensation: H₂O(g) → H₂O(l)).

❌ Examiner Pitfalls

• Vague answers: Stating simply "experimental error" or "heat loss" receives 0 marks. Neither calculation was an experimental calorimetry procedure!
• Standard conditions: Do not just state "conditions were not standard" without mentioning that water and ethanol are liquids, not gases.

✅ Correct Half-Equations

Oxidation half-equation:

C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻

Reduction half-equation:

4H⁺ + 4e⁻ + O₂ → 2H₂O

(Multiples like 12H⁺ + 12e⁻ + 3O₂ → 6H₂O are also accepted)

🧠 How to Balance the Ethanol Half-Equation

1. Balance Carbons: C₂H₅OH → 2CO₂
2. Balance Oxygens using H₂O: Reactant has 1 'O', products have 4 'O'. Add 3H₂O to LHS:
   C₂H₅OH + 3H₂O → 2CO₂
3. Balance Hydrogens using H⁺: LHS has 6 + 6 = 12 'H'. Add 12H⁺ to RHS:
   C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺
4. Balance Charge using electrons: Add 12e⁻ to RHS:
   C₂H₅OH + 3H₂O → 2CO₂ + 12H⁺ + 12e⁻