III. Gas-Phase Equilibria | ||||
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III-1. Introduction and Background | III-2. Sample Problem | III-3. List of Problems | III-4. Simple Problems | III-5. Advanced Problems |
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 equilibria 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. Find the pre-equilibrium 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
2. The balanced chemical reaction is: N_{2}O_{4 (g)} 2 NO_{2 (g)}
and the equilibrium constant expression is: K_{eq} = (P_{NO2})^{2} / P_{N2O4}
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 pre-equilibrium partial pressures, changes in partial pressures, 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 pre-equilibrium 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} / (1.50-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} = 1.50 + 2(0.717) = 2.93 atm |
Check results: Q = (2.93)^{2}/0.783 = 11. Q = K_{eq}, so the system is at equilibrium.
Does a total pressure of 3.71 atm at equilibrium make sense? Think about the total pressure before the system was at equilibrium, and the direction that we said the reaction would proceed to reach equilibrium.
Equilibrium Practice Problems | |
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