The equilibrium expression can be evaluated for any mixture, not only one at equilibrium. The result is called the reaction quotient, $Q$. Comparing it with $K$ tells you which way the reaction will go.
Key idea
$Q$ has the same form as $K$ but uses the current concentrations. If $Q < K$, the reaction goes forward (toward products). If $Q > K$, it goes in reverse (toward reactants). If $Q = K$, it is at equilibrium.
Why it works
A reaction always moves toward equilibrium, where $Q = K$. If $Q$ is too small, there is too little product compared with reactant; making more product raises the top of the fraction and lowers the bottom, so $Q$ climbs to $K$. If $Q$ is too big, the reverse reaction brings it down.
Worked examples
Worked example: Q below K
For $\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$, $K_c = 4.6 \times 10^{-3}$ at $298\ \text{K}$. A flask holds $0.10\ \text{M}$ $\mathrm{N_2O_4}$ and $0.010\ \text{M}$ $\mathrm{NO_2}$. Which way does it go?
- $$Q = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]} = \dfrac{(0.010)^2}{0.10} = 1.0 \times 10^{-3}$$
- $Q < K$: there is too little $\mathrm{NO_2}$. The reaction goes forward, making more $\mathrm{NO_2}$.
Answer: Forward, toward $\mathrm{NO_2}$.
Worked example: Q above K
For $\mathrm{H_2(g) + I_2(g) \rightleftharpoons 2HI(g)}$, $K_c = 54.3$ at $703\ \text{K}$. A mixture has $[\mathrm{HI}] = 0.50\ \text{M}$ and $[\mathrm{H_2}] = [\mathrm{I_2}] = 0.050\ \text{M}$. Which way does it go?
- $$Q = \dfrac{(0.50)^2}{(0.050)(0.050)} = \dfrac{0.25}{0.0025} = 100$$
- $Q > K$: too much HI. The reverse reaction runs, turning some HI into $\mathrm{H_2}$ and $\mathrm{I_2}$.
Answer: In reverse, toward $\mathrm{H_2}$ and $\mathrm{I_2}$.
| Comparison | Too much of… | Reaction goes |
|---|---|---|
| $Q < K$ | reactants | forward (to the right) |
| $Q = K$ | — | nowhere: at equilibrium |
| $Q > K$ | products | in reverse (to the left) |
Worked example: A solid in the mixture
For $\mathrm{CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)}$ at a certain temperature, $K_p = 0.25$. A sealed flask holds both solids and $\mathrm{CO_2}$ at $0.40\ \text{atm}$. Which way does it go?
- Solids are left out, so $Q = P_{\mathrm{CO_2}} = 0.40$.
- $Q > K_p$: too much $\mathrm{CO_2}$. Some of it reacts with CaO to re-form $\mathrm{CaCO_3}$ until the pressure falls to 0.25 atm.
Answer: In reverse, toward $\mathrm{CaCO_3}$.
Common mistake
Saying a reaction with $Q < K$ shifts toward reactants. It is the other way: small $Q$ means not enough products yet. Check with the extreme case: a flask with only reactants has $Q = 0$, and it obviously reacts forward.
$Q$ is also the bridge to free energy: ΔG = ΔG° + RT ln Q is negative exactly when $Q < K$.