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Follow the cycle.

Six original Cambridge-style questions on Hess's law: the statement, formation and combustion cycles, and enthalpy changes that cannot be measured directly.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Draw the cycle, then follow the arrows.

01
Recall
[1 marks]

State Hess's law.

  • The total enthalpy change for a reaction is independent of the route taken, provided the initial and final conditions are the same. ✓
02
Calculation
[3 marks]

Draw a cycle and calculate ΔHf of methane using ΔHc: C −394, H2 −286, CH4 −890 kJ mol−1.

  • Cycle: C + 2H2 → CH4 across the top; both sides go down to CO2 + 2H2O. ✓
  • ΔHf + (−890) = −394 + 2(−286) = −966. ✓
  • ΔHf = −76 kJ mol−1. ✓
03
Calculation
[2 marks]

Calculate ΔH for 2NO2(g) → N2O4(g), given ΔHf(NO2) = +33 and ΔHf(N2O4) = +9 kJ mol−1.

  • ΔH = ΣΔHf(products) − ΣΔHf(reactants) = 9 − 2 × 33. ✓
  • ΔH = −57 kJ mol−1. ✓
04
Calculation
[2 marks]

Calculate ΔH for C2H4(g) + H2O(l) → C2H5OH(l) using ΔHc: C2H4 −1411, C2H5OH −1367 kJ mol−1.

  • Water does not burn, so its ΔHc is zero. ✓
  • ΔH = ΣΔHc(reactants) − ΣΔHc(products) = −1411 − (−1367) = −44 kJ mol−1. ✓
05
Analysis
[3 marks]

Anhydrous copper(II) sulfate dissolves with ΔH1 = −66.5 kJ mol−1; hydrated CuSO4·5H2O dissolves with ΔH2 = +11.7 kJ mol−1. Calculate ΔH for CuSO4(s) + 5H2O(l) → CuSO4·5H2O(s) and explain why it cannot be measured directly.

  • Both solids give the same solution, CuSO4(aq). ✓
  • ΔH = ΔH1 − ΔH2 = −66.5 − 11.7 = −78.2 kJ mol−1. ✓
  • Adding exactly 5 mol of water to the solid cannot be controlled; the solid would partly dissolve and the heat could not be measured cleanly. ✓
06
Explain
[2 marks]

Explain why the enthalpy change of formation of carbon monoxide must be found using Hess's law.

  • Burning carbon in a limited supply of oxygen always gives a mixture of CO and CO2, never pure CO. ✓
  • So ΔHf(CO) is found indirectly from ΔHc(C) and ΔHc(CO). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.