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.
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:
|Po||1.50 atm||1.50 atm|
|P||-x atm||+2x atm|
|Peq||(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)
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.