 Solution

ID: past-exam-of-the-chemistry-a-course-of-the-university-of-cambridge/2023/paper-2/a8/b/solution

μ_{AB} = μ_{A} + μ_{B}

. Using the number-concentration form gives

0 = Δ ε_{0} - k_{B} T ln [\frac{f _{A} f _{B}}{f _{AB}} \frac{c _{AB}}{c _{A} c _{B}}], Δ ε_{0} = ε^{0, A} + ε^{0, B} - ε^{0, AB} .

Here

Δ ε_{0}

is the reaction energy between the chosen molecular reference states. Rearranging,

\frac{c _{A} c _{B}}{c _{AB}} = \frac{f _{A} f _{B}}{f _{AB}} e^{- Δ ε_{0} / (k_{B} T)} .

K_{c} = \frac{( c _{A} / c ^{\circ} ) ( c _{B} / c ^{\circ} )}{c _{AB} / c ^{\circ}} = \frac{c _{A} c _{B}}{c _{AB} c ^{\circ}} .

Thus

K_{c} = \frac{f _{A} f _{B}}{f _{AB}} \frac{1}{c ^{\circ}} e^{- Δ ε_{0} / (k_{B} T)} .

c^{\circ}

has the same units as

c_{i}

. It converts concentration ratios to ideal thermodynamic activities; changing it changes the numerical dimensionless

K_{c}

for this dissociation, which creates an extra particle. The unnormalized quotient

c_{A} c_{B} / c_{AB}

instead has concentration units.^[ref]

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