N_{2(g)} + 3H_{2(g)} ⇌ 2NH_{3(g)}

At the beginning of an experiment, there were 2.1 mol of nitrogen gas, and 6.9 mol of hydrogen gas. The reaction was allowed to proceed to equilibrium in a 10L container, at which point 1.2 mol of N_{2} was remaining. What is the value of K, assuming fixed temperature?

The first step is to note the concentration of each item individually at equilibrium. If 12 mol of N_{2} is remaining then 0.9 mol must have been converted. Therefore 3 x 0.9 mol of H_{2} must also have been converted, leaving 0.3 mol of H_{2} (for every one mole of nitrogen gas, 3 moles of hydrogen gas is converted, as per the chemical equation). In addition, there must now be 2 x 0.9 mol of NH_{3}.

We now know the concentrations of each substance. The next trick is to note that, when dealing with gases, you must account for the size of the container, as it is very rarely a simple 1L in exams. You must do this because concentration is proportional to pressure, and the size of a container influences the pressure.

Once the 10L has been taken into account, K can be calculated:

$K = \frac{1.2^{0.1} 0.3^{0.3}}{1.8^{0.2}}$ = 0.6310

Note that the indices have been adjusted (divided by 10) to account for the 10L container.