Le Châtelier's principle (topic 7.9) predicts which way a disturbed equilibrium shifts. On the free-response section, though, "it shifts to relieve the stress" earns nothing. Readers want the reason in terms of Q and K. This page shows how to turn every Le Châtelier prediction into a Q vs K justification, and why that comparison is the real reason the system shifts.
The idea
At equilibrium, Q = K. A disturbance breaks that equality in one of two ways:
- Concentration, pressure or volume changes move Q; K stays the same.
- A temperature change moves K; Q stays the same at that instant.
Either way, Q ≠ K afterward, and the rule from topic 7.3 takes over: Q < K means net forward, Q > K means net reverse, until Q = K again.
Concentration changes
Worked example. N2O4(g) ⇌ 2 NO2(g), Kc = 4.6 × 10−3, is at equilibrium with [N2O4] = 0.0400 M and [NO2] = 0.0136 M. NO2 is added to make [NO2] = 0.0300 M. Justify the shift.
1. New Q: Q = (0.0300)2 / 0.0400 = 2.3 × 10−2.
2. Compare: 2.3 × 10−2 > 4.6 × 10−3, so Q > K.
3. Justification: "Adding NO2 increases Q above K, so the net reaction proceeds in reverse, forming N2O4, until Q again equals K."
You often do not need numbers. Adding a product raises the top of Q, so Q > K; removing a reactant lowers the bottom, so again Q > K. Adding a reactant or removing a product makes Q < K.
Volume changes
Changing the volume changes every gas concentration by the same factor. Whether Q changes depends on the exponents. For N2O4 ⇌ 2 NO2, doubling the volume halves both concentrations:
Qnew = (½[NO2])2 / (½[N2O4]) = ¼/½ × Qold = ½ Qold
So Q falls below K and the net reaction goes forward, toward the side with more gas molecules, exactly as Le Châtelier predicts. When the gas coefficients are equal on both sides, the factors cancel, Q is unchanged and there is no shift.
Worked example. N2(g) + 3 H2(g) ⇌ 2 NH3(g) is at equilibrium. The volume is halved. By what factor does Q change, and which way does the system shift?
1. Every concentration doubles. Top: (2)2 = 4. Bottom: (2)(2)3 = 16.
2. Qnew = 4/16 × Qold = 0.25 Qold, so Q < K.
3. The net reaction goes forward, making NH3 (2 mol gas from 4), until Q = K.
An inert gas added at constant volume changes no concentration in Q, so Q = K still and nothing shifts.
Temperature changes
Heating or cooling does not change the concentrations at the moment it happens, so Q is unchanged. What changes is K. For an endothermic forward reaction, K increases with temperature; for an exothermic one, K decreases.
Worked example. N2O4(g) ⇌ 2 NO2(g) is endothermic. An equilibrium mixture is heated. Justify the color change (NO2 is brown).
"The forward reaction is endothermic, so raising the temperature increases K. The concentrations have not yet changed, so Q is now less than the new K. The net reaction proceeds forward, producing more brown NO2, until Q equals the new K."
What earns the point, and what does not
| Answer | Credit? | Why |
|---|---|---|
| "Adding H2 makes Q < K, so the net reaction proceeds forward until Q = K." | Yes | Names the change in Q and compares it with K |
| "Adding H2 increases K." | No | K does not change with concentration |
| "The forward rate increases, so it shifts right." | Usually no | Rate language alone; the question asks for Q vs K |
| "It shifts right to relieve the stress." | No | Restates the principle without reasoning |
| "Heating increases K for this endothermic reaction, so Q < K and it shifts forward." | Yes | Correctly says K, not Q, changes |