Equilibrium and Le Chatelier's principle in QCE Chemistry, explained

Published 2026-05-26 · Updated 2026-05-26

Equilibrium and Le Chatelier's principle in QCE Chemistry, explained

Chemical equilibrium is the first big topic of QCE Chemistry Unit 3 (Equilibrium, Acids and Redox Reactions), and it is the idea the rest of the unit is built on. Acids, bases, buffers and solubility are all equilibrium problems wearing different clothes. Get equilibrium right early and the back half of Unit 3 becomes much easier.

It is also the topic students most often half-understand. Le Chatelier's principle is easy to recite and easy to misapply. This guide covers what dynamic equilibrium actually is, how each disturbance shifts a system and why, the difference between Kc and Q, and the mistakes that quietly cost marks in the IA1 Data Test and the external exam.

What "dynamic equilibrium" really means

A reversible reaction can run forwards and backwards. In a closed system, it reaches dynamic equilibrium when the forward and reverse reactions are happening at the same rate.

The word that matters is dynamic. At equilibrium the reaction has not stopped. Both reactions are still going, molecule for molecule, but because they go at equal rates the concentrations of reactants and products stay constant. Nothing you can measure changes, yet everything is still moving.

Two conditions are needed:

A useful mental model: at equilibrium the amounts are constant but not equal. A system can sit at equilibrium with 90% products and 10% reactants, or the reverse. "Equilibrium" says the rates match, not that the amounts match.

Le Chatelier's principle

If a system at equilibrium is disturbed, it shifts to partially oppose the disturbance and reach a new equilibrium.

The key word is partially. The system never fully cancels the change; it only reduces it. Here is how each disturbance plays out.

Concentration

Add more of a substance and the system shifts to consume it; remove a substance and the system shifts to replace it.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g):

Important: changing a concentration shifts the position of equilibrium, but it does not change Kc. The system moves to restore the same ratio.

Pressure (gases only)

Increasing pressure by decreasing the volume shifts the system toward the side with fewer moles of gas, because fewer gas molecules exert less pressure.

For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), the left has 4 moles of gas and the right has 2. Increasing pressure shifts right.

Two traps:

Temperature

This is the only disturbance that actually changes Kc. Treat heat as a reagent.

So temperature is special: it does not just move the position, it moves the equilibrium constant itself.

Catalysts

A catalyst speeds up the forward and reverse reactions equally. It lowers the activation energy for both directions, so the system reaches equilibrium faster, but the position does not move and Kc is unchanged. A catalyst changes when, not where.

Disturbance Position shifts? Kc changes?
Add/remove a substance Yes No
Change pressure (volume) Yes (toward fewer gas moles) No
Change temperature Yes Yes
Add a catalyst No No

Kc and Q: the maths of equilibrium

The equilibrium constant Kc is the ratio of product to reactant concentrations at equilibrium, each raised to its coefficient. For aA + bB ⇌ cC + dD:

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

Rules that catch students out:

The reaction quotient Q uses the exact same expression, but with the concentrations at any moment, not necessarily at equilibrium. Comparing Q with Kc tells you which way a system will move:

In QCE you are expected to calculate Kc from equilibrium concentrations, and to use an ICE (Initial, Change, Equilibrium) table to find unknown concentrations. The coefficient is where sign and ratio errors creep in, so write the balanced equation above the table every time.

Solubility: equilibrium in disguise

The last piece of the equilibrium strand is the solubility product Ksp for sparingly soluble salts. A salt like AgCl sitting in contact with its saturated solution is at equilibrium:

AgCl(s) ⇌ Ag⁺(aq) + Cl⁻(aq), so Ksp = [Ag⁺][Cl⁻]

Same logic as Kc: the solid is left out, and comparing the ionic product with Ksp predicts whether a precipitate forms (ionic product > Ksp means it precipitates). The common ion effect, where adding a shared ion suppresses solubility, is just Le Chatelier applied to a dissolving equilibrium. If you see solubility as equilibrium, you have already learned it.

Common mistakes that cost marks

  1. Saying a catalyst shifts the equilibrium. It speeds both directions equally; the position does not move.
  2. Claiming the reaction stops at equilibrium. The rates are equal, not zero. It is dynamic.
  3. Saying any concentration or pressure change alters Kc. Only temperature changes Kc.
  4. Including solids and liquids in Kc or Ksp. Only gases and aqueous species appear.
  5. Counting all moles for a pressure change. Only moles of gas matter.
  6. Forgetting "partially". Le Chatelier shifts oppose the change but never fully undo it.

How to prepare

Equilibrium rewards reasoning, not memorising. For each scenario, name the disturbance, state which way the system shifts, and justify it with the principle. Then practise the calculations until writing a Kc expression and running an ICE table is automatic, because the exam will pair a conceptual "explain the shift" with a numerical "calculate Kc".

The fastest way to find your gaps is specific feedback on practice answers, since the trap here is usually a small reasoning slip you cannot see yourself. Avocado is an AI-powered Chemistry tutor built specifically for the QCE syllabus, so you can work through Unit 3 equilibrium questions, attempt Kc and ICE-table calculations, and get specific feedback on exactly where your Le Chatelier reasoning went wrong.

Frequently asked questions

What is dynamic equilibrium? The state in a closed system at constant temperature where the forward and reverse reactions occur at the same rate, so concentrations stay constant even though both reactions are still happening.

What is Le Chatelier's principle? If a system at equilibrium is disturbed, it shifts to partially oppose the disturbance and settle at a new equilibrium.

Does a catalyst change the position of equilibrium? No. A catalyst speeds up the forward and reverse reactions equally, so the system reaches equilibrium faster but at the same position, with the same Kc.

What changes the value of Kc? Only temperature. Concentration and pressure changes shift the position but leave Kc the same.

What is the difference between Kc and Q? They use the same expression. Kc uses equilibrium concentrations; Q uses the concentrations at any moment. Comparing Q with Kc tells you which direction the system will move.

Content aligned to the QCAA Chemistry General Senior Syllabus, Unit 3 (Equilibrium, Acids and Redox Reactions). Always confirm current syllabus detail with your teacher and the QCAA website.