Equilibrium constants for heterogeneous equilibrium

For a heterogeneous equilibrium, the equilibrium-constant expression can be simplified further, because a pure solid or pure liquid does not have a variable concentration the way a gas or a dissolved species does.

Consider $CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)$. Writing the equilibrium constant in the usual way would give $K_C = \dfrac{[CaO(s)][CO_2(g)]}{[CaCO_3(s)]}$. But a pure solid, unlike a gas, does not expand to fill its container -- it always has the same number of moles per litre of its own volume at a given temperature, so its concentration (active mass) is itself a constant. Since both $[CaCO_3(s)]$ and $[CaO(s)]$ are constants, they can be absorbed into the equilibrium constant itself, leaving

So the equilibrium constant for this reaction depends only on the concentration (or partial pressure) of the gaseous carbon dioxide, not at all on how much solid calcium carbonate or calcium oxide happens to be present. The same reasoning applies to a pure liquid, whose active mass likewise stays constant at a given temperature -- so pure-liquid concentration terms can also be dropped from an equilibrium-constant expression. For example, in

since $H_2O(l)$ is a pure liquid, its concentration is excluded and $K_C$ is written

Worked example: writing Kp and Kc for three reactions.

1. $2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)$: both sides fully gaseous, so $K_C = \dfrac{[SO_3]^2}{[SO_2]^2[O_2]}$ and $K_P = \dfrac{p_{SO_3}^2}{p_{SO_2}^2\,p_{O_2}}$.

2. $2CO(g) \rightleftharpoons CO_2(g) + C(s)$: heterogeneous, with pure solid carbon dropped from the expression, so $K_C = \dfrac{[CO_2]}{[CO]^2}$ and $K_P = \dfrac{p_{CO_2}}{p_{CO}^2}$.

3. $Ag_2O(s) + 2NH_3(aq) \rightleftharpoons 2AgNO_3(aq) + H_2O(l)$: with both the pure solid $Ag_2O(s)$ and the pure liquid $H_2O(l)$ dropped, $K_C = \dfrac{[AgNO_3]^2}{[NH_3]^2}$. …