If we express a gas-phase equilibria in terms of partial pressures, we obtain the equilibrium constant Kp. For the reaction below:

CO (g) + 3H_{2} (g) = CH_{4} (g) + H_{2}O (g)

The equilibrium-constant expression in terms of partial pressures becomes:

**K _{P} = (P_{CH4} P_{H2O})/(P_{CO} P_{H2})**

In general, the numerical value of Kp is not the same as the value of Kc. From the relationship P V = n R T, we show that

Kp = Kc (RT)^{Δn}

Where: An is the total amount of moles of gaseous product LESS the total amount of moles of gaseous reactant. Consider the following reaction:

2SO_{2} (g) + O_{2} (g) ↔ 2SO_{3} (g)

Reaction constant Kc for the reaction is 2.8 x 10^{2} at 1,000 °C. Calculate Kp for the reaction at this temperature.

We know that: Kp = Kc (RT)^{Δn}

From the equation we see that Δn = 2 – (2 +1) = -1. We can simply substitute the given reaction temperature and the value of R (0.08206 dm^{3}.atm/molK) to obtain Kp.

Thus,

Kp = 2.8 x 10^{2} x (0.08206 x 1,000)^{-1}

Kp = 3.4