Homogeneous and Heterogeneous Equilibria

Equilibria are classified according to how many physical phases the reactants and products occupy,

because this directly determines which species must be written into the equilibrium-constant

expression.

Homogeneous equilibria are those in which every reactant and every product exists in the same

phase — most commonly, entirely gaseous, or entirely in one liquid solution. The ammonia synthesis,

$\text{N}_2(g) + 3\text{H}_2(g) \rightleftharpoons 2\text{NH}_3(g)$, is homogeneous because all three

species are gases; its full equilibrium expression, $K_c = [\text{NH}_3]^2 / ([\text{N}_2][\text{H}_2]^3)$,

includes every species exactly as written in the balanced equation, since every one of them has a

concentration that can genuinely vary.

Heterogeneous equilibria are those in which the reactants and products are spread across more than one phase — typically some combination of solid, liquid and gas. The thermal decomposition of

limestone, $\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g)$, is heterogeneous: two

solids and one gas are present. Here a further rule applies — the concentration (or "active mass") of a pure solid or a pure liquid is taken to be constant, because it does not change as the reaction

proceeds (a pure solid's density and hence its molar concentration per unit volume of that solid is

fixed; only the amount of solid changes, not its concentration). These constant terms are absorbed

into the equilibrium constant itself rather than written out explicitly, so the equilibrium expression

for limestone's decomposition is simply

with $\text{CaCO}_3(s)$ and $\text{CaO}(s)$ omitted entirely. This is why, at a fixed temperature, the

equilibrium pressure of $\text{CO}_2$ above a limestone kiln is a fixed number regardless of how much

solid $\text{CaCO}_3$ or $\text{CaO}$ is present — doubling the amount of solid does not double the

equilibrium $\text{CO}_2$ concentration, precisely because the solids' own concentrations never

appear in $K_c$ in the first place. …