Le Chatelier’s principle says an equilibrium shifts to undo a disturbance. With Q and K you can see why, and predict the direction every time: a disturbance either changes $Q$ (and the reaction runs until $Q = K$ again) or changes $K$ itself.
Key idea
Adding or removing a species, or changing the volume, changes $Q$ but not $K$. The reaction then runs in the direction that brings $Q$ back to $K$. Only a change in temperature changes $K$. Adding an inert gas at constant volume changes nothing. A catalyst changes nothing either.
Why it works
Halving the volume doubles every gas concentration. Each concentration in $Q$ is raised to its coefficient, so $Q$ is multiplied by $2^{\Delta n}$. For $\mathrm{N_2 + 3H_2 \rightleftharpoons 2NH_3}$, $\Delta n = -2$, so $Q$ drops to $\tfrac{1}{4}$ of $K$ and the reaction moves forward, toward fewer moles of gas. An inert gas raises the total pressure but leaves each reacting gas's concentration (and partial pressure) unchanged, so $Q$ stays equal to $K$. Temperature is different: $K$ itself depends on it. For an exothermic reaction, $K$ gets smaller as $T$ rises.
Worked examples
Worked example: Compressing a gas equilibrium
$\mathrm{N_2O_4(g) \rightleftharpoons 2NO_2(g)}$ is at equilibrium. The volume is suddenly halved. Which way does it shift?
- Every concentration doubles. $Q = \dfrac{[\mathrm{NO_2}]^2}{[\mathrm{N_2O_4}]}$ becomes $\dfrac{(2)^2}{2} = 2$ times $K$.
- $Q > K$: the reaction runs in reverse, toward $\mathrm{N_2O_4}$, the side with fewer moles of gas.
Answer: Toward $\mathrm{N_2O_4}$.
Worked example: Heating an exothermic reaction
$\mathrm{N_2(g) + 3H_2(g) \rightleftharpoons 2NH_3(g)}$ has $\Delta H^\circ = -91.8\ \text{kJ}$. What does heating do to $K$ and to the amount of ammonia?
- The forward reaction releases heat. Treat heat as a product: adding heat pushes the reaction backward.
- In terms of $K$: for an exothermic reaction, $K$ decreases as $T$ rises (this follows from ΔG° = −RT ln K). With a smaller $K$, the current $Q$ is now too large, so ammonia breaks down.
Answer: $K$ falls and the mixture holds less ammonia. Industry still runs at about 400–450 °C because the reaction is too slow when cold: a compromise between yield and rate.
| Change | Q or K? | Effect |
|---|---|---|
| Add a reactant | $Q$ falls | shifts forward |
| Remove a product | $Q$ falls | shifts forward |
| Halve the volume | $Q \times 2^{\Delta n}$ | toward fewer moles of gas |
| Add inert gas, fixed V | no change | none |
| Raise T, exothermic | $K$ falls | shifts in reverse |
| Raise T, endothermic | $K$ rises | shifts forward |
| Add a catalyst | no change | none (equilibrium is reached sooner) |
Common mistake
Thinking a change in concentration changes $K$, or that an inert gas at constant volume shifts the equilibrium. Only temperature changes $K$. Check: ask what happens to each concentration in $Q$; if none changes, nothing shifts.