7.10 Q and Le Châtelier

Q Chases K

Perturb equilibrium with concentration or temperature buttons and watch Q(t) and K(t) trace on a log timeline. Concentration jolts Q, temperature jolts K, Q relaxes to K.

Q vs K3 ReactionsConcentration → QTemperature → K
Topic 7.10

Reaction Quotient and Le Châtelier’s Principle

Explain the relationships between Q, K, and the direction in which a reversible reaction will proceed to reach equilibrium.

EK 7.10.A.1 gives the mechanism behind Le Châtelier's principle: a disturbance to a system at equilibrium causes Q to differ from K, thereby taking the system out of equilibrium. The system responds by bringing Q back into agreement with K, establishing a new equilibrium state.

This reframes 7.9 from a memorized table into a calculation you can actually do:

  • Q < K → too few products relative to equilibrium → net forward reaction until Q rises to K.
  • Q > K → too many products → net reverse reaction until Q falls to K.
  • Q = K → at equilibrium, no net change.

Why this is the better tool. Le Châtelier's principle is a qualitative heuristic with edge cases that trip people up. The Q-vs-K analysis is exact and handles all of them:

  • Adding an inert gas at constant volume: partial pressures are unchanged, so Q is unchanged, so no shift. The reasoning is immediate.
  • Adding a solid: solids do not appear in Q, so Q is unchanged, so no shift.
  • Diluting an aqueous equilibrium: every concentration drops by the same factor, but the exponents differ between numerator and denominator, so Q changes by a predictable factor. Compute it and you know the direction.

The one thing Q cannot explain is a temperature change — because temperature changes K itself rather than Q. That is why 7.9 treats temperature separately.

Procedure: compute Q with the current values, compare to K, and state the direction of net reaction.

Key points

  • Q < K → forward. Q > K → reverse. Q = K → equilibrium.
  • Every stress except temperature works by changing Q while K stays fixed.
  • Temperature changes K itself, which is why it is the exception.
  • Q analysis handles inert gases, added solids, and dilution correctly and automatically.

Equations

  • on the exam sheetSame form as Kc; evaluated at any moment.

Common mistakes

  • Using Q to predict a temperature effect. Temperature changes K, not Q.
  • Forgetting to exclude solids from Q.
  • Assuming dilution always shifts one way. The direction depends on which side has more dissolved particles.
  • Comparing Q to K without recomputing after a volume change. Changing V changes every concentration.

Worked example

For 2 SO₂(g) + O₂(g) ⇌ 2 SO₃(g), Kc = 4.2 at some temperature. A mixture contains [SO₂] = 0.50 M, [O₂] = 0.30 M, [SO₃] = 1.2 M. (a) Is it at equilibrium? (b) If not, which way will it proceed?

(a) Compute Q:
Qc = [SO₃]² / ([SO₂]²[O₂]) = (1.2)² / [(0.50)²(0.30)]
Qc = 1.44 / (0.25 × 0.30) = 1.44 / 0.075 = 19.2

Q = 19.2 vs K = 4.2, so Q > K. The system is not at equilibrium.

(b) Direction: Q is too large, meaning the numerator (products) is too big relative to the denominator. To decrease Q toward K the system must consume SO₃ and produce SO₂ and O₂, so the net reaction proceeds in the reverse direction (toward reactants).

Kinetic restatement: the reverse rate currently exceeds the forward rate, producing a net conversion of products to reactants until the two rates equalize.

Full notes for topic 7.10 →