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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 N2O4 and NO2 when 0.2 mol of each gas are mixed in a 4.0 L flask at 100oC, (K = 11 atm)?

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

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

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

For each mol of PN2O4 that dissociates, 2 moles of PNO2(g) 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.

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(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.

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