Concentration at equilibrium

An ICE Table can be used to find the concentrations of all aqueous and gaseous reactants and products when a chemical reaction achieves equilibrium. It is a method of organizing stoichiometric calculations, and its letters stand for the following:

All three are measured in mol/L, and they are related by I +\displaystyle + C =\displaystyle = E.


In the reaction H2(g) + I2(g) ⇌ 2 HI(g), 2.00 mol of H2(g) and 3.00 mol of I2(g) are placed in a 1.00 L container. Calculate the other two equilibrium concentrations if I2(g) has an equilibrium concentration of 1.30 mol/L.

Here is the I calculation for H2(g) (you don’t need to show all of them):

c=nV=2.00mol1.00L=2.00mol/L\displaystyle c = \frac{n}{V} = \frac{2.00 \, \text{mol}}{1.00 \, \text{L}} = 2.00 \, \text{mol/L}.

Let x\displaystyle x represent the absolute value of the change in concentration of H2(g). This can also be written more concisely like this: let x=Δ[H2]\displaystyle x = \left \lvert \Delta{} \left [ \text{H}_{2} \right ] \right \rvert.

H2(g) I2(g) 2 HI(g)
I 2.00\displaystyle 2.00 3.00\displaystyle 3.00 0\displaystyle 0
C x\displaystyle - x x\displaystyle - x +2x\displaystyle + 2 x
E 2.00x\displaystyle 2.00 − x 3.00x\displaystyle 3.00 − x 2x\displaystyle 2 x

The E value for I2(g) is known to be 1.30 mol/L, but our table tells us that it is also x\displaystyle x subtracted from 3.00 mol/L, therefore we can set them equal:

1.30mol/L=3.00mol/Lxx=1.70mol/L\displaystyle 1.30 \, \text{mol/L} = 3.00 \, \text{mol/L} - x \qquad \Rightarrow \qquad x = 1.70 \, \text{mol/L}.

By substituting 1.70 mol/L for x\displaystyle x into the E expressions for H2(g) and HI(g), we can easily find their concentrations at equilibrium as well (0.30 mol/L and 3.40 mol/L respectively).