This example problem demonstrates how to calculate the equilibrium concentrations from initial conditions and the reaction's equilibrium constant. This equilibrium constant example concerns a reaction with a "small" equilibrium constant.**Problem:**

0.50 moles of N_{2} gas is mixed with 0.86 moles of O_{2} gas in a 2.00 L tank at 2000 K. The two gasses react to form nitric oxide gas by the reaction

N_{2}(g) + O_{2}(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** - Find initial concentrations

[N_{2}]_{o} = 0.50 mol/2.00 L

[N_{2}]_{o} = 0.25 M

[O_{2}]_{o} = 0.86 mol/2.00 L

[O_{2}]_{o} = 0.43 M

[NO]_{o} = 0 M**Step 2** - 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 N_{2} and O_{2} will react to form NO. If the amount of N_{2} and O_{2} used is X, then only 2X of NO will form.

This means at equilibrium, the concentrations would be

[N_{2}] = [N_{2}]_{o} - X = 0.25 M - X

[O_{2}] = [O_{2}]_{o} - X = 0.43 M - X

[NO] = 2X

If we assume X is negligible compared to the concentrations of the reactants, we can ignore their effects on the concentration

[N_{2}] = 0.25 M - 0 = 0.25 M

[O_{2}] = 0.43 M - 0 = 0.43 M

Substitute these values in the expression for the equilibrium constant

K = [NO]^{2}/[N_{2}][O_{2}]

4.1 x 10^{-4} = [2X]^{2}/(0.25)(0.43)

4.1 x 10^{-4} = 4X^{2}/0.1075

4.41 x 10^{-5} = 4X^{2}

1.10 x 10^{-5} = X^{2}

3.32 x 10^{-3} = X

Substitute X into the equilibrium concentration expressions

[N_{2}] = 0.25 M

[O_{2}] = 0.43 M

[NO] = 2X = 6.64 x 10^{-3} M**Step 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 M

Is X less than 5% of 0.43 M

Yes - it is 0.7% of 0.43 M

Plug your answer back into the equilibrium constant equation

K = [NO]^{2}/[N_{2}][O_{2}]

K = (6.64 x 10^{-3} M)^{2}/(0.25 M)(0.43 M)

K = 4.1 x 10^{-4}

The 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

[N_{2}] = 0.25 M

[O_{2}] = 0.43 M

[NO] = 6.64 x 10^{-3} M