Coefficients in the Equilibrium Equation

It is very important to realize that the coefficients tell how the molecules react, not the amount or concentration of either reactants or products. For example, the equation

2 HI (g) H2(g) + I2(g)

does not mean:

It does mean that everytime 2 molecules of HI decompose you will get 1 molecule of  H2 and 1 molecule of I2 formed.

It is easy to verify that this is true by running the computer simulation.  You will notice that there is almost no equilibrium condition where you will get a ratio of 2 : 1 : 1 between HI : H2 : I2. 

In fact, if you start with:

[[H2] = 1.0 M, [I2] = 1.0 M, and HI] = 2.0 M at T = 400 oC

you will end up with concentrations of:

[H2] = 0.504 M,  [I2] = 0.504 M, and [HI] = 2.99 M

which is a ratio of:

HI : H2 : I2 of 2.99 : 0.504 : 0.504 which is approximately 6 : 1 : 1

There is no need for the H2 or I2 to be equal either. You can use any amount you want (within limitations such as the physical capacity of the container). Try running the simulation with:

[H2] = 0.1 M, [I2] = 0.5 M, and [HI] = 0.4 M at T = 356 °C

to see what happens. Check if you have the right concept by answering these questions.

1. The ratio of HI : H2 : I2 at equilibrium when the starting concentrations of [HI] = 0.4 M, [H2] = 0.1 M, and [I2] = 0.5 M and T = 356 °C is:
a) You can't do this!   The coefficients show that you'd have to have 0.2 M H2 and I2 for the reaction to be able to occur.
b) The ratio will always be 2 : 1 : 1 because that's what the balanced equation indicates
c) The ratio at equilibrium will be 4 : 1 : 5, just like the starting amounts.
d) The ratio at equilibrium will be 33 : 1 : 25

2. If you were investigating the following equilibrium equation:

N2(g) + 3 H2 (g) 2 NH3 (g)

and placed 0.1 mole of N2 (g), and 0.3 mole of H2 (g) in a 1 L container, then when equilibrium was reached, how much NH3 (g) would you have?
a) 0.2 mole
b) There isn't enough information to answer this question.