The big idea: A reaction at dynamic equilibrium can be disturbed by changing the conditions. Le Châtelier's principle says:
If a system at equilibrium is disturbed, the position of equilibrium shifts in the direction that opposes (partly cancels) the change.
So the system always 'pushes back' — if you add something, it gets used up; if you heat it, the shift takes in heat. The job in the exam is to predict the direction of the shift.
Two terms to keep straight: - Position of equilibrium — the balance of products to reactants. A shift right means more product; a shift left means more reactant. - Equilibrium constant, Kc — fixed at a given temperature. Only a temperature change alters Kc; concentration, pressure and a catalyst leave Kc unchanged.
Read the change → predict the shift:Every prediction follows the same logic: name the stress (added/removed species, pressure, temperature), then ask which direction opposes it — that is the way the position moves.
You change… The position shifts to… add a species the side that uses it up raise pressure the side with fewer gas moles raise temperature the endothermic direction
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Add more of a species and the position shifts away from it (to use it up); remove a species and the position shifts towards it (to replace it). This is the most common Paper 1A and Paper 2 prediction.
| Change you make | Which way it shifts | Why |
|---|---|---|
| Add a reactant | shifts to the products (right) | the system removes the added reactant |
| Add a product | shifts to the reactants (left) | the system removes the added product |
| Remove a product | shifts to the products (right) | the system replaces the lost product |
| Remove a reactant | shifts to the reactants (left) | the system replaces the lost reactant |
Worked context — esterification: For CH3COOH + C2H5OH ⇌ ester + H2O, adding more ethanol shifts the position right (more ester forms), and removing the water also shifts it right. Both increase the yield of ester.
Pressure only matters when gases are involved. Increasing the pressure (by shrinking the volume) makes the position shift towards the side with fewer moles of gas, which reduces the pressure again. Count the gas moles on each side first.
| Change you make | Which way it shifts | Why |
|---|---|---|
| Increase pressure (smaller volume) | shifts to the side with fewer gas moles | to reduce the pressure |
| Decrease pressure (larger volume) | shifts to the side with more gas moles | to raise the pressure |
| Equal gas moles on both sides | no shift | neither side relieves the pressure |
Worked example — counting gas moles
For the synthesis of ammonia, N2(g) + 3H2(g) ⇌ 2NH3(g), predict the effect of increasing the pressure on the position of equilibrium.
Solution
- Count the gas moles on each side: left = 1 + 3 = 4 mol; right = 2 mol.
- Increasing the pressure favours the side with fewer gas moles to reduce the pressure — that is the right (2 mol).
- So the position shifts right, forming more ammonia.
Final answer
Shifts right (towards the 2 mol side), increasing the yield of NH3.
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Temperature is the special case: it is the only change that alters Kc. Treat the forward reaction's enthalpy sign as the clue. Raising the temperature adds heat, so the position shifts in the endothermic direction (which absorbs that heat).
| Change you make | Exothermic forward (ΔH < 0) | Endothermic forward (ΔH > 0) |
|---|---|---|
| Raise temperature | shifts to the reactants (left) | shifts to the products (right) |
| Lower temperature | shifts to the products (right) | shifts to the reactants (left) |
| Effect on Kc | Kc decreases | Kc increases |
The temperature shortcut: Write the heat into the equation as if it were a species:
- Exothermic forward → 'reactants ⇌ products + heat'. Adding heat shifts it left. - Endothermic forward → 'reactants + heat ⇌ products'. Adding heat shifts it right.
Then apply the concentration rule to 'heat' just like any other species.
Worked example — deduce exo/endothermic from a colour change
A sealed equilibrium mixture turns a deeper blue when it is heated. Deduce whether the forward reaction (forming the blue species) is exothermic or endothermic.
Solution
- Heating shifts the position towards more of the blue species — so the forward direction is favoured by adding heat.
- A reaction favoured by adding heat is the one that absorbs heat.
- Therefore the forward reaction is endothermic (ΔH > 0).
Final answer
Endothermic — heating drives the position towards the blue product, the heat-absorbing direction.
A catalyst is the odd one out. It speeds up the forward and reverse reactions equally, so equilibrium is reached sooner — but the position is not shifted and Kc is unchanged.
| Change you make | Effect on position | Effect on rate |
|---|---|---|
| Add a catalyst | no shift — position unchanged | speeds up both directions equally |
| Kc unchanged | equilibrium is reached sooner |
How this is tested: R2.3.2 is one of the most reliable marks in the course.
- Paper 1A (MCQ): predict / deduce the effect of one change — most often pressure or temperature — on the position and on K. - Paper 2 / Paper 1B: explain why a named change increases a yield, or deduce exo/endothermic from how heating shifts the position. A 'sketch' part may ask how a concentration changes with time after the disturbance.
The markers reward a clear direction + reason — name the shift and the principle behind it.
Score it cleanly: (1) For pressure, count the gas moles on each side first. (2) For temperature, use the forward ΔH sign and write heat as a species. (3) Say what happens to K: only temperature changes it — concentration, pressure and a catalyst leave it unchanged.
IB-style question — pressure on the SO₃ synthesis (a)
(a) Sulfur trioxide is made by 2SO2(g) + O2(g) ⇌ 2SO3(g). The total pressure is doubled at constant temperature. Deduce the effect on the position of equilibrium and on the value of Kc. [2]
How to score the marks
- Mark 1 — the shift. Gas moles: left = 2 + 1 = 3 mol, right = 2 mol. Raising the pressure shifts the position to the side with fewer gas moles, so it shifts right (more SO3).
- Mark 2 — Kc. Pressure does not change Kc; only a temperature change does. So Kc is unchanged.
Final answer
Position shifts right (towards SO3, the side with fewer gas moles); Kc is unchanged.
IB-style question — explain a yield increase (b)
(b) In the equilibrium 2CrO4²⁻ + 2H⁺ ⇌ Cr2O7²⁻ + H2O, explain why adding acid (more H⁺) makes the solution turn orange (more Cr2O7²⁻). [2]
How to score the marks
- Mark 1 — the disturbance. Adding acid increases [H⁺], a reactant on the left.
- Mark 2 — the shift. By Le Châtelier's principle the position shifts to oppose the increase — to the right, using up the added H⁺ — so more orange Cr2O7²⁻ forms.
Final answer
More H⁺ disturbs the equilibrium; the position shifts right to remove the added H⁺, forming more orange Cr₂O₇²⁻.