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9701 · 23.4

Gibbs free energy change, ΔG — practice questions

Practice and worked examples for 9701 Gibbs free energy change, ΔG. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

The decomposition of calcium carbonate, CaCO₃(s) → CaO(s) + CO₂(g), has a standard enthalpy change of reaction, ΔH°, of +178 kJ mol⁻¹ and a standard entropy change, ΔS°, of +161 J K⁻¹ mol⁻¹. Calculate the minimum temperature at which this reaction becomes spontaneous under standard pressure.

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  1. Identify the condition for spontaneity: The reaction becomes spontaneous when ΔG° ≤ 0. The minimum temperature for spontaneity is when ΔG° = 0.
  2. State the Gibbs equation: ΔG° = ΔH° - TΔS°
  3. Set ΔG° to zero: 0 = ΔH° - TΔS°
  4. Rearrange for T: T = ΔH° / ΔS°
  5. Convert units: ΔS° is in J K⁻¹ mol⁻¹, but ΔH° is in kJ mol⁻¹. We must convert ΔS° to kJ K⁻¹ mol⁻¹. ΔS° = 161 J K⁻¹ mol⁻¹ / 1000 = 0.161 kJ K⁻¹ mol⁻¹
  6. Substitute values and solve: T = (+178 kJ mol⁻¹) / (+0.161 kJ K⁻¹ mol⁻¹) T = 1105.6 K
  7. Final Answer: The reaction becomes spontaneous above 1106 K (to 4 s.f.).

Worked example 2

The standard electrode potential, E°cell, for the reaction Zn(s) + Cu²⁺(aq) → Zn²⁺(aq) + Cu(s) is +1.10 V. Calculate the standard Gibbs free energy change, ΔG°, for this reaction at 298 K. (Faraday constant, F = 96500 C mol⁻¹).

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  1. Identify the relevant equation: ΔG° = -nFE°
  2. Determine 'n': The half-equations are Zn → Zn²⁺ + 2e⁻ and Cu²⁺ + 2e⁻ → Cu. The number of moles of electrons transferred, n, is 2.
  3. Substitute the values: ΔG° = -(2 mol) × (96500 C mol⁻¹) × (+1.10 V) Note: 1 Volt = 1 Joule per Coulomb (J C⁻¹) ΔG° = -2 × 96500 × 1.10 J ΔG° = -212300 J
  4. Convert to standard units (kJ mol⁻¹): ΔG° = -212300 J / 1000 ΔG° = -212.3 kJ mol⁻¹
  5. Final Answer: The standard Gibbs free energy change is -212 kJ mol⁻¹ (to 3 s.f.). The negative value confirms the reaction is spontaneous, as expected from the positive E°cell.
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