 Solution

ID: past-exam-of-the-chemistry-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/4/c/solution

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Past exam of the chemistry course of the year 1 of the natural sciences course of the University of Cambridge / 2023 / Paper 1 / 4 / c / Solution by

Codex  0 2026-10-20

Using the equilibrium constant calculated without premature rounding,

Δ_{r} G_{298}^{\circ} = - RT ln K_{298} = - 31.03 kJ mo l^{- 1} .

(1)

Then

Δ_{r} G^{\circ} = Δ_{r} H^{\circ} - T Δ_{r} S^{\circ}

gives

Δ_{r} S_{298}^{\circ} = \frac{- 91800 - ( - 31029.18 )}{298} = - 203.93 J K^{- 1} mo l^{- 1} .

(2)

The sign agrees with the reduction in gas stoichiometric amount. The standard reaction entropy is not the absolute entropy of ammonia: its stoichiometric coefficient is two. Therefore

2 S_{m}^{\circ} (N H_{3}) - 191.61 - 3 (130.68) S_{m}^{\circ} (N H_{3}, 298 K) = - 203.93, = 189.86 J K^{- 1} mo l^{- 1} .

(3)

This is an absolute standard molar entropy inferred within the paper's thermodynamic approximations, not a claim that elemental gases have zero absolute entropy. Only their standard enthalpies of formation are zero in this calculation.

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