# Solving Gas-Phase Equilibria Problems

### 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 reactions 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 100oC, (Keq = 11 atm)?

1. First write the balanced chemical reaction: N2O4 (g) 2 NO2 (g)

2. Next find the initial partial pressures, PNO2 and PN2O4, using PV = nRT.

PNO2 = PN2O4 = (0.20 mol)(0.0821 L atm/mol K)(373 K)/(4.0 L) = 1.5 atm

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

Q = (PNO2)2 / PN2O4

Q = (1.5 atm)2/1.5 atm = 1.5 atm

Q < Keq, so the reaction will proceed in the forward direction, N2O4 (g) --> 2 NO2 (g) until it reaches equilibrium.

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

N2O4 NO2 1.50 atm 1.50 atm -x atm +2x atm (1.50-x) atm (1.50+2x) atm

Where Po are the initial partial pressures, P are the changes in partial pressures, and Peq are the equilibrium partial pressures.

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

Keq = 11 = (PNO2)2 / PN2O4

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

Does a total pressure of 3.71 atm make sense?

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

Gas-Phase Equilibria Problems
1. Introduction and
Sample Problem
2. Simple
Practice Problems