To find $K$ you need every equilibrium concentration, but experiments usually measure only one. The coefficients of the balanced equation link the others to it, using the same mole ratios as in stoichiometry. An ICE table keeps track.
Key idea
Make a table with three rows: Initial, Change and Equilibrium. Put the known change for one species in the C row, scale it by the coefficients for every other species (minus for reactants, plus for products), add down the columns, and put the E row into $K$.
Why it works
The amounts react in the ratio of the coefficients, just as they do in a reaction that goes to completion. The only difference is that the reaction stops at equilibrium instead of when a reactant runs out. Concentrations in one container change in the same ratio as moles, so the C row is just a mole ratio.
Worked examples
Worked example: One measured amount
A flask starts with $0.400\ \text{M}$ $\mathrm{SO_2}$ and $0.200\ \text{M}$ $\mathrm{O_2}$. At equilibrium, $[\mathrm{SO_3}] = 0.300\ \text{M}$ in $\mathrm{2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)}$. Find $K_c$.
- I: $\mathrm{SO_2}$ 0.400, $\mathrm{O_2}$ 0.200, $\mathrm{SO_3}$ 0. C: $\mathrm{SO_3}$ $+0.300$, so $\mathrm{SO_2}$ $-0.300$ (same coefficient) and $\mathrm{O_2}$ $-0.150$ (half as much).
- E: $[\mathrm{SO_2}] = 0.100$, $[\mathrm{O_2}] = 0.050$, $[\mathrm{SO_3}] = 0.300\ \text{M}$.
- $$K_c = \dfrac{(0.300)^2}{(0.100)^2(0.050)} = \dfrac{0.0900}{5.0 \times 10^{-4}} = 180$$
Answer: $K_c = 1.8 \times 10^{2}$.
| $\mathrm{SO_2}$ | $\mathrm{O_2}$ | $\mathrm{SO_3}$ | |
|---|---|---|---|
| I | 0.400 | 0.200 | 0 |
| C | −0.300 | −0.150 | +0.300 |
| E | 0.100 | 0.050 | 0.300 |
Worked example: Equal starting amounts
$\mathrm{H_2}$ and $\mathrm{I_2}$, each $1.00\ \text{M}$, react in a sealed flask: $\mathrm{H_2 + I_2 \rightleftharpoons 2HI}$. At equilibrium $[\mathrm{HI}] = 1.57\ \text{M}$. Find $K_c$.
- HI rose by 1.57, so $\mathrm{H_2}$ and $\mathrm{I_2}$ each fell by half of that: $0.785\ \text{M}$.
- E: $[\mathrm{H_2}] = [\mathrm{I_2}] = 1.00 - 0.785 = 0.215\ \text{M}$.
- $K_c = \dfrac{(1.57)^2}{(0.215)^2} = 53.3$.
Answer: $K_c = 53.3$.
Two habits prevent most errors. First, work in molarities: if you are given moles and a volume, divide before you fill in the table (for a reaction in one container, moles also work in the C row, but $K_c$ needs molarities). Second, make the C row from a single unknown and the coefficients, then check that the E row has no negative entries.
Common mistake
Putting the starting concentrations into $K$, or using the same change for every species. $K$ needs equilibrium values, and the changes follow the coefficients: when 2 $\mathrm{SO_3}$ form, only 1 $\mathrm{O_2}$ is used. Check: the C row should be proportional to the coefficients, with opposite signs on the two sides.