3.1.10 - Equilibrium Constant Kp for Homogeneous Systems

From Total Pressure to Partial Pressure

In a mixture of gases, each gas contributes part of the total pressure. That matters because a gaseous equilibrium is described by the pressure exerted by each gas at equilibrium, not just by the overall pressure in the container.

If a gas A has mole fraction $x_A$, then:

So the route is always the same: work out total moles, find the mole fraction of the gas you want, then multiply by the total pressure. This is why balanced, careful arithmetic matters before you even think about $K_p$.

Suppose an equilibrium mixture contains $0.60\,\mathrm{mol} \ce{H_2}$, $0.20\,\mathrm{mol} \ce{N_2}$, and $0.20\,\mathrm{mol} \ce{NH_3}$ at a total pressure of $250\,\mathrm{kPa}$. The total moles of gas are $1.00\,\mathrm{mol}$, so the mole fraction of hydrogen is $0.60$. Therefore:

You would repeat the same process for the other gases until you had the full set of equilibrium partial pressures.

Once you can turn total pressure into separate partial pressures, you have the quantities needed to build the equilibrium constant for the gaseous system.

Building the Kp Expression

$K_p$ applies to a homogeneous equilibrium in the gas phase. Here, homogeneous means every species in the equilibrium mixture is in the same phase, so for this topic all the species involved are gases.

The structure of a $K_p$ expression follows the balanced equation exactly. Products go on the top line, reactants go on the bottom line, and each partial pressure is raised to the power of its stoichiometric coefficient.

For the Haber equilibrium,

the expression is:

The powers matter. They come from the balanced equation, not from the numerical values you substitute. A very common mistake is to forget the square on $\ce{NH_3}$ or the cube on $\ce{H_2}$.

Units depend on the overall powers of pressure. If partial pressures are in $\mathrm{kPa}$, then the Haber $K_p$ above has units:

For some equilibria, the total power of pressure is the same on both sides, so the units cancel. That happens for:

because there are two moles of gas on each side overall.

Using Kp in Calculations

There are two common calculation routes. In one, the equilibrium partial pressures are given directly and you substitute them into the expression. In the other, you are given equilibrium amounts and a total pressure, so you must find the partial pressures first and then substitute.

Always check one word carefully: equilibrium. $K_p$ is built from equilibrium values only. Initial amounts are not used unless you are first taken to the equilibrium amounts.

For the equilibrium

suppose the equilibrium partial pressures are $p(\ce{H_2})=40\,\mathrm{kPa}$, $p(\ce{I_2})=40\,\mathrm{kPa}$, and $p(\ce{HI})=240\,\mathrm{kPa}$.

Here the units cancel, so the final answer is simply $36$. Because $K_p$ is greater than $1$, the equilibrium mixture contains relatively more products than reactants, although reactants are still present because the reaction is reversible.