# 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:

• [I]nitial concentration
• [C]hange in concentration
• [E]quilibrium concentration

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

## Example

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):

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

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

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

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

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

By substituting 1.70 mol/L for $\displaystyle x$ 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).