The following problem is an example of determining the equilibrium partial pressures in a gasphase equilibrium problem. For background information see the documents on gasphase reactions and the general solution of equilibria problems.
What are the equilibrium partial pressures of N_{2}O_{4} and NO_{2} when 0.2 atm of each gas are introduced into a 4.0 L flask at 100^{o}C, (K_{eq} = 11 atm)?
1. First write the balanced chemical reaction: N_{2}O_{4 (g)} 2 NO_{2 (g)}
2. 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
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, N_{2}O_{4 (g)} > 2 NO_{2 (g)} until it reaches equilibrium.
4. For each mol of N_{2}O_{4} that dissociates, 2 moles of NO_{2} 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}  (1.50x) 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.
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} / (1.50x)
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} = 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_{eq}, so the system is at equilibrium.
GasPhase Equilibria Problems  

1. Introduction and Sample Problem 
2. Simple Practice Problems 
3. Advanced Practice Problems 
Chemistry Practice Problems Copyright © 1997 by Science Hypermedia, Inc. 

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