5.3.4Additional lessons
Le Chatelier’s Principle — IB Chemistry SL Revision Notes & Topical Questions
5 min read 5 parts Lesson 76 of 98
Part 1 of 5The Principle
Statement
- If a change is applied to a system at equilibrium, the position shifts to oppose the change and partly cancel its effect.
Shift right
- More products form at the expense of reactants, so the concentration of products rises.
Shift left
- More reactants form at the expense of products, so the concentration of reactants rises.
Re-establishing equilibrium
- After the shift the system settles at a new equilibrium with the same K, unless the temperature changed.
Part 2 of 5Concentration and Pressure
Changes in concentration or pressure shift the position but do not change K.
Concentration and Pressure
Add reactant
- Equilibrium shifts right to use up some of the extra reactant, giving more product.
Remove product
- Equilibrium shifts right to replace the removed product.
Add product or remove reactant
- Equilibrium shifts left to reduce the added product or replace the removed reactant.
Raise pressure
- Equilibrium shifts towards the side with fewer moles of gas, lowering the pressure again.
Lower pressure
- Equilibrium shifts towards the side with more moles of gas, raising the pressure again.
Equal moles of gas
- Changing pressure has no effect on the position when both sides have the same number of gas moles.
Part 3 of 5Temperature, Catalysts and Industrial Compromise
1
Raising the temperature favours the endothermic direction, which absorbs the extra heat.
2
Lowering the temperature favours the exothermic direction, which releases heat to oppose the cooling.
3
Unlike concentration and pressure changes, a temperature change alters the value of K.
4
A catalyst speeds up the forward and reverse reactions equally, so it only shortens the time needed to reach equilibrium.
5
In industry a compromise is used: low temperature gives a better yield for an exothermic reaction but is too slow, so a moderate temperature and catalyst are chosen.
Part 4 of 5Worked Example
Question For the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹, predict and explain the effect on the position of equilibrium and on K of (a) increasing the pressure and (b) increasing the temperature.
- Count gas moles4 mol on the left (1 + 3) and 2 mol on the right, so the right side has fewer moles of gas.
- (a) Higher pressure shifts the position right towards fewer gas moles;
- K is unchanged because temperature is constant.
- (b) The forward reaction is exothermic, so higher temperature shifts the position left, which absorbs heat;
- K decreases.
Answer
- (a) More NH₃ forms and K is unchanged.
- (b) Less NH₃ forms at equilibrium and K decreases.
Part 5 of 5Exam Checklist & Tips
5.3.4 – Key Points for the Exam
- Le Chatelier's principle A system at equilibrium responds to a change by shifting the position to oppose that change.
- Yield The amount of product obtained at equilibrium compared with the maximum possible.
- Catalyst A substance that speeds up a reaction without being used up and does not alter the position of equilibrium.
- Compromise conditions Temperature and pressure chosen to balance a reasonable yield against a reasonable rate and cost.
Common Mistakes
- Saying that a catalyst moves the position of equilibrium to the right.
- Applying a pressure argument to a reaction with no gases or equal gas moles on both sides.
- Stating that K changes whenever the position of equilibrium shifts; K only changes with temperature.
- Explaining a shift as 'to make the reaction go faster' rather than 'to oppose the change'.
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3 & IGCSE —
Kc
- : Kc! (ΔH < 0) Kc. (ΔH > 0) Kc.
- : ( ).
- : Kc .