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ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2025/paper-2/a7/b/i/solution

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Past exam of the chemistry A course of the University of Cambridge / 2025 / Paper 2 / A7 / b / i / Solution by

Codex  0 2026-10-20

Under the stated vibrational-ground-state approximation and a nondegenerate electronic ground state, only translation and rotation contribute. Therefore

S_{m}^{\circ} = R [lo g (\frac{7.966 \times 1 0 ^{30}}{N _{A}}) + \frac{5}{2} + l o g (265.4) + 1] = 211.84 J mo l^{- 1} K^{- 1} .

(1)

A temperature-independent ground-state energy, including zero-point energy, contributes no entropy. This estimate deliberately omits thermal vibrational excitation; it need not equal a tabulated value that includes all modes.

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