How to Calculate Equilibrium Pressures

By Robert Schrader; Updated April 25, 2017
Calculating the pressures of gases at equilibrium can be simple.

As you read your chemistry textbook, you may notice that some reactions are written with arrows that point in both directions. This signifies that a reaction is reversible--that the reaction's products can re-react with one another and re-form the reactants. The point at which a reaction occurs at the same rate in both directions is known as equilibrium. When gases react at equilibrium, it's possible to calculate their pressures using a number known as the equilibrium constant, which is different for each reaction.

Set up expressions for the equilibrium pressures of your reactants and products, keeping in mind that both reactants (and both products) will have equal pressures and that all gases in the reaction system change by the same amount as the reaction proceeds. Assign this change the variable "x." If, for example, you're trying to calculate equilibrium pressures in a system of fluoromethane, CH3F, which reacts according to the equilibrium CH3OH + HF <--> CH3F + H2O (where all reactants and products are in gas phase), and you know that the initial pressures of CH3OH and HF were 0.5 atmospheres (atm), you can set equilibrium pressures for the reactants equal to "0.5 - x"--initial pressure minus change--and products equal to "x"--the change" itself, since they had no pressure before the reaction began (they didn't exist).

Set your equilibrium constant equal to the product of your products' equilibrium pressures over the product of your reactants' equilibrium. For the example--assuming that the reaction has an equilibrium constant, Kp, of 8.1 x 10^3--write this expression as follows: Kp = [CH3F][H2O]/[CH3OH][H2F] = (x)(x)/(.5-x)(.5-x) = x^2/(.5-x)^2 = 8.1 x 10^3 = 8,100.

Simplify your equation by taking the square root of both sides. For the example, this would be sqrt(x^2/(.5-x)^2) = sqrt (8,100), or x/(5-x) = 90.

Solve your equation for x. First, multiply both sides by (.5 - x) to get rid of the denominator, as follows: x/(.5 - x) = x and 90(.5 - x) = (90 x .5) - (90x) = 45 - 90x. Notice that x = 45 - 90x and add 90x to both sides to see that 91x = 45, or x = 45/91 = 0.495.

Insert the value of x into your expressions to calculate the equilibrium pressure of your reactants and products. For your reactants, you expressed equilibrium pressure as .5 -x. So, the pressures of HF and CH3OH at equilibrium are equal to 0.5 - 0.495, or .005 atm. The pressures of the products CH3F and H2O are equal to x, or .495 atm.

About the Author

Robert Schrader is a writer, photographer, world traveler and creator of the award-winning blog Leave Your Daily Hell. When he's not out globetrotting, you can find him in beautiful Austin, TX, where he lives with his partner.