Equilibrium Concentration Example Problem

Equilibrium Concentration Example Problem This example problem demonstrates how to calculate the equilibrium concentrations from initial conditions and the reactions equilibrium constant. This equilibrium constant example concerns a reaction with a small equilibrium constant. Problem: 0.50 moles of N2 gas is mixed with 0.86 moles of O2 gas in a 2.00 L tank at 2000 K. The two gasses react to form nitric oxide gas by the reactionN2(g) O2(g) ↔ 2 NO(g).What are the equilibrium concentrations of each gas?Given: K 4.1 x 10-4 at 2000 K Solution: Step 1 -Â  Find initial concentrations: [N2]o 0.50 mol/2.00 L [N2]o 0.25 M [O2]o 0.86 mol/2.00 L [O2]o 0.43 M [NO]o 0 M Step 2 -Â  Find equilibrium concentrations using assumptions about K: The equilibrium constant K is the ratio of products to reactants. If K is a very small number, you would expect there to be more reactants than products. In this case, K 4.1 x 10-4 is a small number. In fact, the ratio indicates there are 2439 times more reactants than products. We can assume very little N2 and O2 will react to form NO. If the amount of N2 and O2 used is X, then only 2X of NO will form. This means at equilibrium, the concentrations would be [N2] [N2]o - X 0.25 M - X[O2] [O2]o - X 0.43 M - X[NO] 2XIf we assume X is negligible compared to the concentrations of the reactants, we can ignore their effects on the concentration[N2] 0.25 M - 0 0.25 M[O2] 0.43 M - 0 0.43 MSubstitute these values in the expression for the equilibrium constantK [NO]2/[N2][O2]4.1 x 10-4 [2X]2/(0.25)(0.43)4.1 x 10-4 4X2/0.10754.41 x 10-5 4X21.10 x 10-5 X23.32 x 10-3 XSubstitute X into the equilibrium concentration expressions[N2] 0.25 M[O2] 0.43 M[NO] 2X 6.64 x 10-3 MStep 3 - Test your assumption:When you make assumptions, you should test your assumption and check your answer. This assumption is valid for values of X within 5% of the concentrations of the reactants.Is X less than 5% of 0.25 M?Yes - it is 1.33% of 0.25 MIs X less than 5% of 0.43 MYes - it is 0.7% of 0.43 MPlug your answer back into the equilibrium constant equationK [NO]2/[N2][O2]K (6.64 x 10-3 M)2/(0.25 M)(0.43 M)K 4.1 x 10-4The value of K agrees with the value given at the beginning of the problem. The assumption is proven valid. If the value of X was greater than 5% of the concentration, then the quadratic equation would have to be used as in this example problem. Answer: The equilibrium concentrations of the reaction are[N2] 0.25 M[O2] 0.43 M[NO] 6.64 x 10-3 M