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A2 · Practice questions · Gibbs free energy change

Balance ΔH against TΔS.

Six original Cambridge-style questions on Gibbs free energy and feasibility.

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.
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01
Calculation
[3 marks]

For CaCO3(s) → CaO(s) + CO2(g), ΔH = +178 kJ mol−1 and ΔS = +161 J K−1 mol−1. Calculate ΔG at 298 K and state whether the reaction is feasible.

  • ΔS = 0.161 kJ K−1 mol−1. ✓
  • ΔG = 178 − 298 × 0.161 = 178 − 48.0 = +130 kJ mol−1. ✓
  • ΔG is positive, so it is not feasible at 298 K. ✓
02
Calculation
[2 marks]

Calculate the minimum temperature at which calcium carbonate decomposes, using the data in the previous question.

  • At the switch-over ΔG = 0, so T = ΔH / ΔS. ✓
  • T = 178 / 0.161 = 1106 K (1110 K to 3 s.f.); feasible above this temperature. ✓
03
Calculation
[3 marks]

For N2(g) + 3H2(g) → 2NH3(g), ΔH = −92 kJ mol−1 and ΔS = −199 J K−1 mol−1. Calculate ΔG at 700 K and comment on feasibility.

  • ΔG = −92 − 700 × (−0.199) = −92 + 139.3. ✓
  • ΔG = +47.3 kJ mol−1. ✓
  • Positive, so not feasible at 700 K (it is feasible only below 462 K); the industrial process relies on high pressure and the equilibrium being reached quickly with a catalyst. ✓
04
Calculation
[3 marks]

For N2O4(g) → 2NO2(g), ΔH = +57 kJ mol−1 and ΔS = +176 J K−1 mol−1. Calculate ΔG at 298 K and the temperature above which the reaction is feasible.

  • ΔG = 57 − 298 × 0.176 = 57 − 52.4 = +4.55 kJ mol−1: not feasible at 298 K. ✓
  • T = 57 / 0.176 = 324 K; feasible above 324 K. ✓
05
Explain
[2 marks]

Ammonium nitrate dissolves in water spontaneously even though the process is endothermic. Explain using the Gibbs equation.

  • Dissolving the solid gives ions dispersed through the solution, so ΔS is positive. ✓
  • At room temperature TΔS is larger than ΔH, so ΔG = ΔH − TΔS is negative and dissolving is feasible. ✓
06
Analysis
[3 marks]

A reaction has ΔH negative and ΔS negative. Explain how its feasibility depends on temperature, and why a feasible reaction might still not be observed.

  • At low T, −TΔS is small, so the negative ΔH makes ΔG negative: feasible. ✓
  • At high T, −TΔS becomes large and positive, so ΔG becomes positive above T = ΔH/ΔS. ✓
  • ΔG gives no information about rate; a high activation energy can make a feasible reaction too slow to observe. ✓

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.