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

Table of contents
- Solution b

Solution

Let

ρ_{i} = N_{i} / V

be a molecular number density and define the volume-independent molecular partition function density

f_{i} = q_{i} / V

. Then

q_{i} / N_{i} = f_{i} / ρ_{i}

. Also define the reaction reference-energy difference

Δ ε_{0} = ε^{0, C} - ε^{0, A} - ε^{0, B} .

Substitution into the equilibrium chemical-potential relation gives

0 \frac{ρ _{C}}{ρ _{A} ρ _{B}} = - k_{B} T ln \frac{f _{C}}{ρ _{C}} + k_{B} T ln \frac{f _{A}}{ρ _{A}} + k_{B} T ln \frac{f _{B}}{ρ _{B}} + Δ ε_{0}, = \frac{f _{C}}{f _{A} f _{B}} exp (- \frac{Δ ε _{0}}{k _{B} T}) .

An equilibrium constant expressed using concentration activities

a_{i} = ρ_{i} / c^{\circ}

is dimensionless:

K_{c} = \frac{a _{C}}{a _{A} a _{B}} = \frac{ρ _{C} c ^{\circ}}{ρ _{A} ρ _{B}} = \frac{f _{C} c ^{\circ}}{f _{A} f _{B}} e^{- Δ ε_{0} / (k_{B} T)} .

Here every

f_{i}

has units

m^{- 3}

, and

c^{\circ}

is the chosen standard concentration, initially one molecule per cubic metre. The factor

c^{\circ}

is essential: without it the association quotient has units of volume.

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