Introduction

The following problem is an example of determining the equilibrium partial pressures in a gas-phase equilibrium problem. For background information see the documents on gas-phase equilibria and the general solution of equilibria problems.

Sample Problem

What are the equilibrium partial pressures of N₂O₄ and NO₂ when 0.2 atm of each gas are introduced into a 4.0 L flask at 100^oC, (K_eq = 11 atm)?

1. Find the pre-equilibrium partial pressures, P_NO2 and P_N2O4, using PV = nRT.

P_NO2 = P_N2O4 = (0.20 mol)(0.0821 L atm/mol K)(373 K)/(4.0 L) = 1.5 atm

2. The balanced chemical reaction is: N₂O_{4 (g)} 2 NO_{2 (g)}

and the equilibrium constant expression is: K_eq = (P_NO2)² / P_N2O4

3. Now calculate the reaction quotient, Q, to determine the direction in which the reaction will proceed to reach equilibrium.

Q = (P_NO2)² / P_N2O4

Q = (1.5 atm)²/1.5 atm = 1.5 atm

Q < K_eq, so the reaction will proceed in the forward direction, N₂O_{4 (g)} 2 NO_{2 (g)} until it reaches equilibrium.

4. For each mol of N₂O₄ that dissociates, 2 moles of NO₂ will form. The pre-equilibrium partial pressures, changes in partial pressures, and equilibrium partial pressures are given in the following table:

	N2O4	NO2
Po	1.50 atm	1.50 atm
P	-x atm	+2x atm
Peq	(1.50-x) atm	(1.50+2x) atm

Where P_o are the pre-equilibrium partial pressures, P are the changes in partial pressures, and P_eq are the equilibrium partial pressures.

5. We can now calculate the equilibrium partial pressures using the equilibrium constant expression:

K_eq = 11 = (P_NO2)² / P_N2O4

11 = (1.50+2x)² / (1.50-x)

Rearranging gives:

4x² + 17x - 14.25 =0

Find x using the quadratic equation:

x = 0.717

PN2O4 = 1.50 - 0.717 = 0.783 atm

PNO2 = 1.50 + 2(0.717) = 2.93 atm

Check results: Q = (2.93)²/0.783 = 11. Q = K_eq, so the system is at equilibrium.

Does a total pressure of 3.71 atm at equilibrium make sense? Think about the total pressure before the system was at equilibrium, and the direction that we said the reaction would proceed to reach equilibrium.