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 $+$ C $=$ 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):

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

Let $x$ represent the absolute value of the change in concentration of H_2(g). This can also be written more concisely like this: let $x = | Δ [H_{2}] | .$

	H_2(g)	I_2(g)	2 HI_(g)
I	$2.00$	$3.00$	$0$
C	$- x$	$- x$	$+ 2 x$
E	$2.00 - x$	$3.00 - x$	$2 x$

The E value for I_2(g) is known to be 1.30 mol/L, but our table tells us that it is also $x$ subtracted from 3.00 mol/L, therefore we can set them equal:

$1.30 mol/L = 3.00 mol/L - x ⟹ x = 1.70 mol/L .$

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