**Gaseous Equilibria**

The concentrations of gases in a mixture usually are expressed in terms of the partial pressure of the gases. Since the partial pressure of an ideal gas is proportional to its concentration, c, the expression for the equilibrium constant for the reaction,

a A (g) + b B (g) ↔ l L(g) + m M (g) may be written in the form

K_{P} = [P^{l}_{L} x P^{m}_{M}] / [P^{a}_{A} x P^{b}_{B}]

when K_{P}, is the equilibrium constant in terms of partial pressure and the partial pressures are the equilibrium values K_{P} and K_{C} are not always identical. This is because the partial pressures are proportional to, but not equal to, gas concentration expressed in mol L^{-1}.

**Example:** The gases SO_{2}, O_{2} and SO_{3} were allowed to come to equilibrium in a closed vessel under certain conditions of temperature and pressure. The partial pressures of the gases were P_{SO2} = 0.050 atm, P_{O2} = 0.025 atm and P_{SO3} = 1.00 atm. Find the values of K_{P}, for the following equilibria:

(a) SO_{2} (g) + ½ O_{2} (g) ↔ SO_{3} (g)

(b) 2 SO_{2} (g) + O_{2} (g) ↔ 2 SO_{3} (g)

Solution: For, (a); K_{P} = P_{SO3} / (P_{SO2} + P_{O2}) = 1.00 / (0.050 x 0.025^{1/2}) = 126.58

For, (b); K_{P} = P^{2}_{SO3} / (P^{2}_{SO2} + P_{O2}) = 1.00^{2} / (0.050^{2} x 0.025) = 16×10^{3}

It can be seen that the value of a equilibrium constant for the same reaction may be different when the mole ratios of the reactants and products are shown differently in balanced equation.