Balancing the Haber process

My favorite reaction is the Haber process, a method for preparing ammonia (NH₃) by reacting nitrogen gas with hydrogen gas:

N_2(g) + H_2(g) → NH_3(g)

This equation shows you what happens in the reaction, but it doesn’t show you how much of each element you need to produce the ammonia. To find out how much of each element you need, you have to balance the equation — make sure that the number of atoms on the left side of the equation equals the number of atoms on the right. You can’t change the subscripts, so you have to put in some coefficients.

In most cases, waiting until the end to balance hydrogen atoms and oxygen atoms is a good idea; balance the other atoms first. So in this example, you need to balance the nitrogen atoms first. You have two nitrogen atoms on the left side of the arrow (reactant side) and only one nitrogen atom on the right side (product side). To balance the nitrogen atoms, use a coefficient of 2 in front of the ammonia on the right. Now you have two nitrogen atoms on the left and two nitrogen atoms on the right:

N_2(g) + H_2(g) → 2NH_3(g)

Next, tackle the hydrogen atoms. You have two hydrogen atoms on the left and six hydrogen atoms on the right (two NH3 molecules, each with three hydrogen atoms, for a total of six hydrogen atoms). So put a 3 in front of the H₂ on the left, giving you the following:

N_2(g) + 3H_2(g) → 2 NH_3(g)

That should do it. Do a check to be sure: You have two nitrogen atoms on the left and two nitrogen atoms on the right. You have six hydrogen atoms on the left (3 × 2 = 6) and six hydrogen atoms on the right (2 × 3 = 6). The equation is balanced.

You can read the equation this way: one nitrogen molecule reacts with three hydrogen molecules to yield two ammonia molecules.

Here’s a tidbit for you: This equation would also balance with coefficients of 2, 6, and 4 instead of 1, 3, and 2. In fact, any multiple of 1, 3, and 2 would balance the equation, but chemists have agreed always to show the lowest whole-number ratio.