How To Calculate Equilibrium Pressures

Chemical equilibrium describes the "steady state" of a reversible reaction, where the reactants turn into the products and vice-versa at the same rate.

Chemists characterize this state using the equilibrium constant, K_p, and you can use the full expression for this constant to determine the equilibrium pressure for either the reactants or the products. The process requires a little bit of algebra, but at the end, you get a straightforward equation you can use to find the equilibrium pressure for your reaction.

The process for finding the equilibrium pressure depends strongly on how chemists describe an equilibrium state for a reaction. For a generic gas-phase reaction:

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

The general formula for the equilibrium constant is:

\(K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\)

Here, the Ps refer to the partial pressures of the gases A, B, C and D (indicated by subscripts). You can remember this formula easily as "products over reactants," but this simple form is only valid if each of the components of the reaction has 1 as its stoichiometric coefficient (a, b, c and d in the above equation). Remember that this formula is only valid at equilibrium.

In some cases, you might see the equilibrium constant called K_c, and the equation for it written in terms of molar concentrations, and the relationship between these two is:

\(K_p = K_c (RT)^{∆n}\)

In this formula, ∆n is the change in the number of moles of gas from the reactants to the products.

The key to solving this equation to find the equilibrium pressure is introducing the quantity x, which corresponds to the change from the initial pressures to the equilibrium pressure. Imagine you know the pressure of the reactants but don't know what it will be at equilibrium. In that case, you can simply write each equilibrium pressure as (initial pressure = x).

Since the products in this case start with a partial pressure of 0 (because they aren't present), their equilibrium pressure will be equal to x. Provided you also know the equilibrium constant itself, you can use this information to find the partial pressures you don't know.

Call the initial pressure P_i and note that both of the reactants must be at the same pressure. Also, to simplify the equation this will be written with the coefficients a, b, c and d as equal to 1. So you can write the equation as:

\(\begin{aligned}\)
\(K_p &= \frac{x^2}{(P_i – x) (P_i – x)}\)
\(&=\frac{x^2}{(P_i – x)^2}\)
\(\end{aligned}\)

Rearranging this equation for x allows you to find the difference between the initial pressure and the equilibrium pressure for your reactants, and the value for the products too. After taking the square root of both sides, multiplying by the denominator of the fraction, and rearranging a little more, the equation for x is:

\(x = \frac{\sqrt{K_p}P_i}{1 + \sqrt{K_p}}\)

This becomes an equilibrium partial pressure formula when you note that the equilibrium pressure of your reactants is P_i – x, and for your products it is simply x.

Working through an example will cement the approach and help you get more comfortable with using the equation. Consider the reaction used to produce chloromethane:

\(\text{CH}_3\text{OH(g)} + \text{HCl(g)} \rightleftharpoons \text{CH}_3\text{Cl(g)} + \text{H}_2\text{O(g)}\)

With K_p = 5,900. The reactants have an initial pressure (in atmospheres, atm) of P_i = 0.75 atm. Insert these values into the formula and run through the calculations to find the partial pressures:

\(\begin{aligned}\)
\(x &= \frac{\sqrt{K_p}P_i}{1 + \sqrt{K_p}}\)
\(&= \frac{\sqrt{5900}× 0.75 \text{ atm}}{1 + \sqrt{5900}}\)
\(&= 0.74 \text{ atm}\)
\(\end{aligned}\)

This is the value for the equilibrium pressures of the products, and for the reactants, all you need to do is subtract this from the initial value P_i to find the result. Here, 0.75 atm – 0.74 atm = 0.01 atm.