Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2022 / Paper 1 / 4 / b / i / Solution

For red-to-yellow conversion, subtract the red values from the yellow values:

Δ_{r} H^{\circ} = (- 90.2) - (- 90.7) = + 0.5 kJ mo l^{- 1}, Δ_{r} S^{\circ} = 73.0 - 72.0 = + 1.0 J K^{- 1} mo l^{- 1} .

Hence

Δ_{r} G^{\circ} (T) = 500 - T J mo l^{- 1}, Δ_{r} G^{\circ} (298 K) = + 202 J mo l^{- 1} ≃ + 0.20 kJ mo l^{- 1} .

The positive value makes the red form the lower-Gibbs free energy form at 298 K within the supplied bulk-phase model. The calculation uses 298 K as the conventional reference temperature; the paper supplies no separate numerical temperature for this subpart. Keeping enthalpy and entropy in the same energy units is essential.

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