Expressing the Equilibrium Constant of a Gas in Terms of Pressure

Equilibrium Constants for Gases

Up to this point, we have been discussing equilibrium constants in terms of concentration. For gas-specific reactions, however, we can also express the equilibrium constant in terms of the partial pressures of the gases involved. Take the general gas-phase reaction:

$aA(g)+bB(g)\rightleftharpoons cC(g)+dD(g)$

Our equilibrium constant in terms of partial pressures, designated K_P, is given as:

$K_P=\frac{P^c_CP^d_D}{P^a_AP^b_B}$

Note that this expression is extremely similar to K_C, the equilibrium expression written in terms of concentrations. In order to prevent confusion, do not use brackets ([ ]), when writing K_P expressions.

K_P and the Ideal Gas Law

The reason we are allowed to write a K expression in terms of partial pressures for gases can be found by looking at the ideal gas law. Recall that the ideal gas law is given by:

$PV=nRT$

Re-writing this expression in terms of P, we have:

$P=\frac{n}{V}RT$

Note that in order for K to be constant, temperature must be constant as well. Therefore, the term RT is a constant in the above expression. As for n/V (moles per unit volume) this is simply a measure of concentration. Pressure is directly proportional to concentration, so we are justified in our use of K_P.

Lastly, there is a very important equation that relates K_P and K_C. It is given as follows:

$K_{p} = K_{c}(RT) ^{ \Delta n}$

In this expression, $\Delta n$ is a measure of the change in number of moles of gas in the reaction. For instance, if a reaction produces three moles of gas, and consumes two moles of gas, then $\Delta n=(3-2)=1$.

Liquefied gas

Inside this tank, propane is compressed into a liquid, which is in equilibrium with its gaseous headspace. The internal pressure of the gaseous propane is a function of temperature.