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Le Chatelier's principle is a Higher tier topic in the AQA GCSE Chemistry specification. It allows you to predict how the position of equilibrium will shift when conditions are changed. Mastering this principle is essential for understanding industrial processes such as the Haber process and for answering questions about reversible reactions at equilibrium.
Le Chatelier's principle states:
If a system at equilibrium is subjected to a change in conditions, the equilibrium position will shift to counteract (oppose) the change.
In other words, the system adjusts to partially undo the change that was made.
| Change Made | System Response |
|---|---|
| Increase in concentration of a substance | Equilibrium shifts to reduce the concentration of that substance |
| Decrease in concentration of a substance | Equilibrium shifts to increase the concentration of that substance |
| Increase in temperature | Equilibrium shifts in the endothermic direction (to absorb the extra heat) |
| Decrease in temperature | Equilibrium shifts in the exothermic direction (to release heat) |
| Increase in pressure | Equilibrium shifts towards the side with fewer moles of gas |
| Decrease in pressure | Equilibrium shifts towards the side with more moles of gas |
Exam Tip: [H] Le Chatelier's principle is all about the system opposing the change. The word "counteract" or "oppose" is essential. If the temperature is increased, the equilibrium shifts to reduce the temperature — i.e. in the endothermic direction. If pressure is increased, the equilibrium shifts to reduce the pressure — i.e. towards fewer gas molecules.
Consider the reaction: A + B <=> C + D
| Change | Equilibrium Shifts | Effect on Yield of Products |
|---|---|---|
| Increase concentration of A | To the right (towards products) | More C and D produced |
| Decrease concentration of A | To the left (towards reactants) | Less C and D produced |
| Increase concentration of C | To the left (towards reactants) | More A and B produced |
| Remove product D | To the right (towards products) | More C and D produced (to replace D) |
graph TD
A[Increase concentration of reactant] --> B[System opposes: uses up the extra reactant]
B --> C[Forward reaction speeds up]
C --> D[Equilibrium shifts to the RIGHT]
D --> E[More product formed]
F[Increase concentration of product] --> G[System opposes: uses up the extra product]
G --> H[Reverse reaction speeds up]
H --> I[Equilibrium shifts to the LEFT]
I --> J[More reactant formed]
In industry, removing a product as it forms shifts the equilibrium to the right, producing more product. For example, in the Haber process, ammonia is cooled and liquefied to remove it from the equilibrium mixture, driving the reaction forward to produce more ammonia.
Exam Tip: [H] If a question asks you to explain how to increase the yield of a product, consider: (1) increasing the concentration of reactants, (2) removing the product as it forms, (3) changing temperature or pressure (see below). Use Le Chatelier's principle to justify each answer.
The effect of temperature depends on whether the forward reaction is exothermic or endothermic.
Consider: A + B <=> C + D (forward reaction is exothermic)
| Change | Equilibrium Shifts | Effect on Yield of Products | Effect on Rate |
|---|---|---|---|
| Increase temperature | To the left (endothermic direction) | Less product formed | Rate increases |
| Decrease temperature | To the right (exothermic direction) | More product formed | Rate decreases |
For a reaction where the forward reaction is endothermic, the effects are reversed:
| Change | Equilibrium Shifts | Effect on Yield of Products |
|---|---|---|
| Increase temperature | To the right (endothermic direction) | More product formed |
| Decrease temperature | To the left (exothermic direction) | Less product formed |
In industry, there is often a compromise between yield and rate:
A compromise temperature is used to balance these two factors.
The effect of pressure depends on the number of moles of gas on each side of the equation.
Consider: N2(g) + 3H2(g) <=> 2NH3(g)
Left side: 1 + 3 = 4 moles of gas Right side: 2 moles of gas
| Change | Equilibrium Shifts | Explanation |
|---|---|---|
| Increase pressure | Towards the side with fewer moles of gas (right, towards products) | System opposes the increase by reducing the number of gas molecules |
| Decrease pressure | Towards the side with more moles of gas (left, towards reactants) | System opposes the decrease by increasing the number of gas molecules |
graph TD
A["N2 + 3H2 <=> 2NH3"] --> B[Left: 4 moles of gas]
A --> C[Right: 2 moles of gas]
D[Increase pressure] --> E[Equilibrium shifts to side with fewer gas moles]
E --> F[Shifts RIGHT - more NH3]
G[Decrease pressure] --> H[Equilibrium shifts to side with more gas moles]
H --> I[Shifts LEFT - less NH3]
If the number of moles of gas is the same on both sides, changing the pressure has no effect on the position of equilibrium.
Example: H2(g) + I2(g) <=> 2HI(g) — 2 moles on each side, so pressure changes have no effect.
Exam Tip: [H] Always count the moles of GAS on each side. Solids and liquids are not affected by pressure changes. If a substance is (s) or (l), do not include it in your count. Only (g) substances matter for pressure effects.
A catalyst does NOT change the position of equilibrium.
| Property | Effect of Catalyst |
|---|---|
| Rate of forward reaction | Increases |
| Rate of reverse reaction | Increases by the same amount |
| Position of equilibrium | Unchanged |
| Yield of product | Unchanged |
| Time to reach equilibrium | Decreased (equilibrium reached faster) |
A catalyst increases the rate of both the forward and reverse reactions equally. The equilibrium position remains the same, but it is reached more quickly.
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