Edexcel A-Level Chemistry AS Paper 2, June 2025: Question 3

✅ Correct Answer

-3975 kJ mol⁻¹

(Acceptable range: -3950 to -4010 kJ mol⁻¹)

💡 Key Knowledge

Each additional -CH₂- group increases the combustion enthalpy by a roughly constant amount (~650–655 kJ mol⁻¹):

• Ethanol − Methanol = −1367 − (−726) = −641 kJ mol⁻¹
• Propan-1-ol − Ethanol = −2021 − (−1367) = −654 kJ mol⁻¹
• Butan-1-ol − Propan-1-ol = −2676 − (−2021) = −655 kJ mol⁻¹

Hexan-1-ol has two extra -CH₂- units compared to butan-1-ol:

-2676 + 2(-650) ≈ -3976 kJ mol⁻¹

❌ Common Errors

• Adding only one -CH₂- step: Calculating for pentan-1-ol (5 carbons) instead of hexan-1-ol (6 carbons).
• Positive sign: Giving +3975 kJ mol⁻¹. Combustion is always exothermic, so positive signs are strictly rejected.

✅ Acceptable Final Answers

• -3710 kJ mol⁻¹ (3 s.f.)
• -3709 kJ mol⁻¹ (4 s.f.)
• -3708.7 kJ mol⁻¹

(Negative sign is required for the final mark!)

🧠 Exam Technique: Securing 4/4

• Use mass of water for q: Always use 200 g (water), NEVER the alcohol mass, in q = mcΔT .
• Avoid premature rounding: If you round moles to 0.006 (1 s.f.), you will lose the final accuracy mark. Keep the full fraction 0.80 / 130 in your calculator.
• Check your sign: Combustion produces heat (exothermic), so Δ_cH must carry a negative (−) sign.

❌ Common Traps to Avoid

• Forgetting the negative sign: An answer of +3710 kJ mol⁻¹ scores 3/4 marks (loses M4).
• Adding 273 to ΔT: Temperature change is identical in °C and K. Adding 273 ( 273 + 27.3 = 300.3 ) causes the loss of M3.
• Omitting unit conversion: Forgetting to divide by 1000 gives -3 708 700 kJ mol⁻¹ , losing the final mark if written as kJ mol⁻¹.

✅ Model Answer (2 Marks)

1. The experimental value of Δ_cH will be less negative / less exothermic (or data book value is more negative).

2. Because heat is lost / escapes to the surroundings / calorimeter.

🧠 Exam Technique: Word Precision

Be extremely careful when describing enthalpy values that are negative numbers:

• Say "less negative" or "less exothermic".
• Do NOT write "the value is smaller/lower" without context, as mathematically −2000 is greater than −3700, leading to ambiguity.
• Link cause directly to the specific fault given: the prompt specifies "calorimeter is not insulated", so the reason must be heat loss.

❌ Examiner Pitfalls

• Irrelevant reasons: Mentioning "incomplete combustion" or "specific heat capacity is an approximation" scores 0 for the explanation mark here because the question specifically prompts about lack of insulation.
• Vague descriptions: "Energy lost" or "energy dissipates" without specifying to the surroundings or to the calorimeter loses the second mark.
• Saying "more positive": Strictly not allowed by the mark scheme.

✅ Model Answer (2 Marks)

1. Effect on temperature: The temperature recorded by the thermometer will be higher (as the bottom of the beaker is hotter than the average temperature without stirring) / the recorded temperature change ( ΔT ) would be larger.  [1 Mark]

2. Effect on calculated enthalpy: Therefore, the calculated Δ_cH will be more negative / more exothermic / larger in magnitude.  [1 Mark]

💡 Key Knowledge

The spirit burner flame is directly under the base of the beaker. Convection distributes heat, but without stirring, a localised thermal gradient forms:

Base of beaker (where thermometer rests) >> Average temperature of water

A higher recorded ΔT leads to an artificially high calculated heat value q , making Δ_cH more exothermic (more negative).

❌ Common Errors

• Claiming reaction is incomplete: Writing that alcohol "did not react completely if not stirred" is physically incorrect; stirring affects the water, not the flame.
• Vague statements: Stating merely that "there is a temperature difference" or "uneven heat" without stating the recorded reading will be higher.
• Ambiguous enthalpy effect: Stating just "enthalpy change would be bigger" without clarifying that it is bigger in magnitude, more exothermic, or more negative.