Heterogeneous Systems

In a heterogeneous system the species are in more than one phase. When writing K_c, solids are left out of the expression (and so are pure liquids that are not part of a solution).The reason is that the concentration of a pure solid is fixed by its density and molar mass. Adding more of the solid does not change its concentration, so it is effectively constant and is included in the value of K_c.Example 1.CaCO₃(s) ⇌ CaO(s) + CO₂(g)K_c = [CO₂]    units: mol dm⁻³At a given temperature, the equilibrium concentration of CO₂ above the solids is fixed, whatever the amounts of CaCO₃ and CaO present.Example 2.C(s) + H₂O(g) ⇌ CO(g) + H₂(g)K_c = [CO][H₂] / [H₂O]Units: (mol dm⁻³)² / mol dm⁻³ = mol dm⁻³. Here the water is a gas, so it is included.Example 3. For a solid in equilibrium with its ions in solution:Ag⁺(aq) + Fe²⁺(aq) ⇌ Ag(s) + Fe³⁺(aq)K_c = [Fe³⁺] / [Ag⁺][Fe²⁺]    units: dm³ mol⁻¹Consequence: adding more of a solid does not shift the position of a heterogeneous equilibrium, because it does not appear in K_c. Always check the state symbols before writing the expression.