# 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 more information see the document describing the general solution of equilibria problems.

What are the equilibrium partial pressures of N_{2}O_{4} and NO_{2} when 0.2 mol of each gas are mixed in a 4.0 L flask at 100^{o}C, (K = 11 atm)?

First write the balanced chemical reaction: N_{2}O_{4 (g)} 2 NO_{2 (g)}

Next find the initial 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

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, so the reaction will proceed in the forward direction, N_{2}O_{4 (g)} --> 2 NO_{2 (g)} until it reaches equilibrium.

For each mol of P_{N2O4} that dissociates, 2 moles of P_{NO2(g)} will form. The changes and equilibrium partial pressures are given in the following table:

| N_{2}O_{4} | NO_{2} |

P_{o} | 1.50 atm | 1.50 atm |

P | -x atm | +2x atm |

P_{eq} | (1.50-x) atm | (1.50+2x) atm |

Where **P**_{o} are the initial partial pressures, **P** are the changes in partial pressures, and **P**_{eq} are the equilibrium partial pressures.

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:

4x^{2} + 17x - 14.25 =0

Find x using the quadratic equation:

x = 0.717

P_{N2O4} = 1.50 - 0.717 = 0.783 atm

P_{NO2(g)} = 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 = K, so the system is at equilibrium.

Copyright © 2000 by Brian M. Tissue, all rights reserved.