= Solution
At <chemical equilibrium>, $\mu_{\rm AB}=\mu_{\rm A}+\mu_{\rm B}$. Using the number-concentration form gives
$$
0=\Delta\varepsilon_0-k_BT\ln\left[
\frac{f_{\rm A}f_{\rm B}}{f_{\rm AB}}
\frac{c_{\rm AB}}{c_{\rm A}c_{\rm B}}\right],
\qquad
\Delta\varepsilon_0=\varepsilon^{0,\rm A}+\varepsilon^{0,\rm B}-\varepsilon^{0,\rm AB}.
$$
Here $\Delta\varepsilon_0$ is the reaction energy between the chosen molecular reference states. Rearranging,
$$
\frac{c_{\rm A}c_{\rm B}}{c_{\rm AB}}
=\frac{f_{\rm A}f_{\rm B}}{f_{\rm AB}}e^{-\Delta\varepsilon_0/(k_BT)}.
$$
Define the <dimensionless concentration equilibrium constant> by
$$
K_c=\frac{(c_{\rm A}/c^\circ)(c_{\rm B}/c^\circ)}{c_{\rm AB}/c^\circ}
=\frac{c_{\rm A}c_{\rm B}}{c_{\rm AB}c^\circ}.
$$
Thus
$$
\boxed{K_c=\frac{f_{\rm A}f_{\rm B}}{f_{\rm AB}}
\frac1{c^\circ}e^{-\Delta\varepsilon_0/(k_BT)}}.
$$
The <standard concentration> $c^\circ$ 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 $c_{\rm A}c_{\rm B}/c_{\rm AB}$ instead has concentration units.<IUPAC Compendium of Chemical Terminology (Gold Book) by International Union of Pure and Applied Chemistry>[entries equilibrium constant and standard equilibrium constant: distinguish concentration quotients from the thermodynamic dimensionless standard constant]{ref}
Back to article page
