The reaction quotient, Q, is the same as the equilibrium constant expression, but for partial pressures or concentrations of the reactants and products before the system reaches equilibrium.
If Q < K then the reaction is proceeding in the forward direction.
If Q > K then the reaction is proceeding in the reverse direction.
Solving Problems:
Use the following steps to solve equilibria problems.
1) Determine if any reactions will occur and identify the species that will exist in equilibrium.
2) D etermine the pre-equilibrium concentrations or partial pressures of the reactants and products that are involved in the equilibrium.
3) Write the balanced chemical reaction and the equilibrium constant expression.
4) Calculate the reation quotient, Q.
5) Compare Q to Keq to determine if the reaction will proceed in the forward or reverse directions to reach equilibrium.
6) Determine the changes in reactant and product concentrations or partial pressures to reach equilibrium.
7) Use the equilibrium constant expression to solve for the unknown, and then calculate the equilibrium concentrations or partial pressures.
Example:
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. Find the pre-equilibrium 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
2. The balanced chemical reaction is: N2O4 (g) <=> 2 NO2 (g)
and the equilibrium constant expression is: Keq = (PNO2)2 / PN2O4
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 pre-equilibrium partial pressures, changes in partial pressures, and equilibrium partial pressures are given in the following table:
| N2O4 | NO2 | |
|---|---|---|
| Po | 1.50 atm | 1.50 atm |
| DP | -x atm | +2x atm |
| Peq | (1.50-x) atm | (1.50+2x) atm |
Where Po are the pre-equilibrium partial pressures, DP 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)
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 = 1.50 + 2(0.717) = 2.93 atm |
Check results: Q = (2.93)2/0.783 = 11. Q = Keq, 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.