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).