Question

Calculate the entropy change for the rusting of iron according to the reaction :

4 Fe (s) + 3 O₂ (g)→ 2 Fe₂O₃ (s), Δ H° = –1648 kJ mol^–1

Given that the standard entropies of Fe, O₂ and Fe₂O₃ are 27.3, 205.0 and 87.4 J K^–1 mol^–1 respectively. Will the reaction be spontaneous at room temperature (25°C) ? Justify your answer with appropriate calculations.

Answer 1

Δ_rS° =∑ S° (Products) ∑S° (Reactants) = 2 S° (Fe₂O₃ ) – [4 S° (Fe) + 3 S° (O₂ )]

= 2 × 87.4 – [4 × 27.3 + 3 × 205.0] J K^–1 mol^–1 = – 549.4 J K^–1 mol^–1

This is the entropy change of the reaction, i.e., system (ΔS_system)

Now, Δ_rG° = Δ_rH° – TΔ_rS°

= – 1648000 J mol^–1 – 298 K × –549.4 J K^–1 mol^–1)

= – 1648000 + 163721 J K^–1 mol^–1 = – 1484279 J K^–1 mol^–1

As ΔG° is –ve, the reaction is spontaneous.