Q.For the reaction

[!FORMULA] $SrCO_3(s) \rightleftharpoons SrO(s) + CO_2(g),$

the value of the equilibrium constant $K_P = 2.2 \times 10^{-4}$ at 1002 K. Calculate $K_C$ for the reaction.

In a heterogeneous equilibrium, any pure solid or pure liquid present is dropped from the equilibrium-constant expression, because a pure solid or liquid 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 active mass (concentration) is itself a fixed constant that can simply be absorbed into $K_C$ rather than written out as a variable term.

For $CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$, writing out the full expression $K_C=\dfrac{[CaO(s)][CO_2(g)]}{[CaCO_3(s)]}$ and then folding the two constant solid-concentration terms into $K_C$ itself leaves

so the equilibrium constant depends only on the pressure or concentration of the gaseous $CO_2$, regardless of how much solid $CaCO_3$ or $CaO$ happens to be present. Exactly the same logic drops a pure liquid's concentration too: for $CO_2(g)+H_2O(l)\rightleftharpoons H^+(aq)+HCO_3^-(aq)$, with $H_2O(l)$ a pure liquid, $K_C=\dfrac{[H^+(aq)][HCO_3^-(aq)]}{[CO_2(g)]}$. …