Past exam of the chemistry A course of the University of Cambridge / 2024 / Paper 2 / A6 / a

Solution

Treat each species as a dilute classical ideal gas: molecules of the same species are indistinguishable, while different species are distinguishable. Let

N_{i}

i

q_{i} (T, V)

its molecular partition function with energies measured from a chosen internal reference, and

ε^{0, i}

that reference energy per molecule relative to a common chemical energy convention. Then the mixture's canonical partition function factorizes:

Q = \prod_{i} \frac{q _{i}^{N_{i}}}{N _{i} !} exp [- β N_{i} ε^{0, i}], β = \frac{1}{k _{B} T} .

The

N_{i}!

corrects the overcounting of indistinguishable particles. The exponential restores the species-dependent reference energies that matter for reactions. Thus the Helmholtz free energy is

A = - k_{B} T \sum_{i} [N_{i} ln q_{i} - ln (N_{i}!)] + \sum_{i} N_{i} ε^{0, i} .

ln (N_{i}!) ≃ N_{i} ln N_{i} - N_{i}

, and differentiating at fixed

T, V, N_{j \neq = i}

yields

μ_{i} = (\frac{\partial A}{\partial N _{i}})_{T, V, N_{j \neq = i}} = - k_{B} T ln \frac{q _{i}}{N _{i}} + ε^{0, i} .

q_{i}

does not depend on the number of molecules, so no additional

\partial q_{i} / \partial N_{i}

term appears. These are chemical potentials per molecule; multiply by the Avogadro constant for a molar chemical potential. The result assumes the thermodynamic, nondegenerate-gas limit.

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