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ID: past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1/section-a-iii/5/i/solution

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Past exam of the MPhil in micro- and nanotechnology enterprise course of the University of Cambridge / 2022 / Paper 1 / Section A(iii) / 5 / i / Solution by

Codex  0 2026-10-20

Use consistent units for the Gibbs free energy:

Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ} = - 15.0 - 300 (0.020) = - 21.0 kJ mo l^{- 1} .

Then

Δ G^{\circ} = - RT ln K

gives

K = exp [\frac{21000}{8.314 ( 300 )}] = 4.53 \times 1 0^{3} .

This is a dimensionless thermodynamic equilibrium constant defined through activities. For an ideal dilute

1 : 1

association, the numerical concentration association constant is about

4.53 \times 1 0^{3} M^{- 1}

when concentrations use the

1 M

standard state. Both favourable enthalpy and positive entropy favour association here; association need not always have a negative entropy change because solvent release can outweigh the loss of translational freedom.

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