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ID: past-exam-of-the-thermal-and-statistical-physics-course-of-the-university-of-cambridge/2021/section-b/4/d/solution

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Past exam of the thermal and statistical physics course of the University of Cambridge / 2021 / Section B / 4 / d / Solution by

Codex  0 2026-10-18

Introduce an extent of reaction

ξ

increasing in the dissociation direction. The allowed particle-number changes conserve both carbon and oxygen atoms:

d N_{c} = - 2 d ξ, d N_{CO} = 2 d ξ, d N_{O_{2}} = d ξ .

(1)

At fixed temperature and volume, the resulting Helmholtz free energy change is

d F = (- 2 μ_{c} + 2 μ_{CO} + μ_{O_{2}}) d ξ .

(2)

At an interior chemical equilibrium, both forward and backward changes are allowed, so the coefficient must vanish:

2 μ_{C O_{2}} = 2 μ_{CO} + μ_{O_{2}} .

(3)

Equality of the three individual chemical potentials would be wrong: a reaction exchanges species in its stoichiometric proportions rather than molecule for molecule. Stability is also explicit here, since

\frac{d ^{2} F}{d ξ ^{2}} = k_{B} T (\frac{4}{N _{c}} + \frac{4}{N _{CO}} + \frac{1}{N _{O_{2}}}) > 0

(4)

when all three species are present. Thus the equilibrium composition minimizes the free energy along the conserved-composition constraint.

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