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 H_2(g) + I_2(g) ⇌ 2 HI_(g), 2.00 mol of H_2(g) and 3.00 mol of I_2(g) are placed in a 1.00 L container. Calculate the other two equilibrium concentrations if I_2(g) has an equilibrium concentration of 1.30 mol/L.

Here is the I calculation for H_2(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 H_2(g). This can also be written more concisely like this: let $\displaystyle x = \left \lvert \Delta{} \left [ \text{H}_{2} \right ] \right \rvert$ .

	H_2(g)	I_2(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 I_2(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 H_2(g) and HI_(g), we can easily find their concentrations at equilibrium as well (0.30 mol/L and 3.40 mol/L respectively).