The Haber’s process for producing ammonia from N_{2} and H_{2} is an important industrial example of a reaction that does not go to completion.

It establishes an equilibrium state where all the three species are present.

**Example: **Suppose we place 1.00 mol of N_{2} and 3.00 mol of H_{2} in a reaction vessel at a temperature of 450 °C and pressure of 10.0 atm. The Haber’s reaction is:

N_{2} (g) + 3H_{2} (g) ↔ 2NH_{3} (g)

What is the composition of the equilibrium mixture if it contains 0.08 mol of NH_{3}?

**Solution: **Using the information given, a table can be created in terms of “x”, the change in the number of moles of one reactant as shown in the table below:

The equilibrium amount of NH_{3} is equal to 0.08 mol. Therefore, 2x = 0.08 mol and then x = 0.04 mol of NH_{3}.

- Equilibrium amount of N
_{2}= 1.00 – 0.04 = 0.96 mol of N_{2} - Equilibrium amount of H
_{2}= 3.00 – (3 x 0.04) = 2.88 mol o H_{2} - Equilibrium amount of NH
_{3}= 2 x = 0.08 mol of NH_{3}

There is far more H_{2} and N_{2} present at equilibrium than NH_{3}. This will not change, no matter how long the reaction is left. Being at equilibrium does not mean that the amount of reactants equals the amount of product.