Equilibrium shifts to oppose a change in conditions
Definition
Position of equilibrium
The relative amounts of reactants and products in a reaction mixture at equilibrium.
Definition
Le Chatelier's principle
When the conditions of a system at equilibrium are changed, the position of equilibrium shifts to oppose the change.
- When an equilibrium lies to the right, the mixture contains mostly products.
- When it lies to the left, the mixture contains mostly reactants.
- Changing the concentration, temperature or pressure disturbs the equilibrium.
- The system responds by shifting in the direction that opposes the change, until a new equilibrium is reached.
- At the new equilibrium, the forward and reverse rates are equal again, but the amounts of each substance are different.
Key Idea
To predict a shift, ask what the change did and which direction would undo some of it.
Adding or removing a substance shifts the equilibrium
- Adding more reactant shifts the equilibrium to the right, forming more product.
- Removing product as it forms also shifts the equilibrium to the right.
- Adding more product shifts the equilibrium to the left, forming more reactant.
- Removing a reactant shifts the equilibrium to the left, to replace some of it.
Example
- In 2CrO42−(aq)+2H+(aq)⇌Cr2O72−(aq)+H2O(l)2\text{CrO}_4^{2-}\text{(aq)} + 2\text{H}^+\text{(aq)} \rightleftharpoons \text{Cr}_2\text{O}_7^{2-}\text{(aq)} + \text{H}_2\text{O(l)}2CrO42−(aq)+2H+(aq)⇌Cr2O72−(aq)+H2O(l), adding acid turns the yellow solution orange as the equilibrium shifts right.
- Adding alkali removes H+\text{H}^+H+ ions, so the equilibrium shifts left and the solution turns yellow again.
Raising temperature favours the endothermic direction
- If the forward reaction is exothermic, the reverse reaction is endothermic by the same amount.
- Raising the temperature shifts the equilibrium in the endothermic direction, which takes in energy and so opposes the rise.
- Lowering the temperature shifts the equilibrium in the exothermic direction, which releases energy.
- So for an exothermic forward reaction, a lower temperature gives more product at equilibrium.
- For an endothermic forward reaction, a higher temperature gives more product.
Example
In 2NO2(g)⇌N2O4(g)2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}2NO2(g)⇌N2O4(g) the forward reaction is exothermic, so warming the mixture shifts it left and the colour darkens as more brown NO2\text{NO}_2NO2 forms.
Raising pressure favours fewer gas molecules
- To compare the sides, add up the balancing numbers of the gases on each side of the equation.
- Raising the pressure shifts the equilibrium to the side with fewer gas molecules, which reduces the pressure.
- Lowering the pressure shifts it to the side with more gas molecules.
- In 2NO2(g)⇌N2O4(g)2\text{NO}_2\text{(g)} \rightleftharpoons \text{N}_2\text{O}_4\text{(g)}2NO2(g)⇌N2O4(g), there are 2 gas molecules on the left and 1 on the right, so high pressure shifts the equilibrium to the right.
- If both sides have the same number of gas molecules, as in H2(g)+I2(g)⇌2HI(g)\text{H}_2\text{(g)} + \text{I}_2\text{(g)} \rightleftharpoons 2\text{HI(g)}H2(g)+I2(g)⇌2HI(g), pressure does not change the position.
Common Mistake
Do not count solids, liquids or dissolved substances when comparing the two sides, because pressure only affects the gases.
Choosing conditions to give the most product
- First find which side the desired product is on and whether its formation is exothermic or endothermic.
- Choose a low temperature if the forward reaction is exothermic, or a high temperature if it is endothermic.
- Choose a high pressure if the product side has fewer gas molecules, or a low pressure if it has more.
- Keep the reactant concentrations high and remove the product as it forms.
- A catalyst speeds up the forward and reverse reactions equally, so it does not change the position of equilibrium.
- Industry often uses compromise conditions that also give a fast enough rate at a sensible cost, and these are covered with industrial processes.
Example
- Suggest conditions to make as much methanol as possible: CO(g)+2H2(g)⇌CH3OH(g)\text{CO(g)} + 2\text{H}_2\text{(g)} \rightleftharpoons \text{CH}_3\text{OH(g)}CO(g)+2H2(g)⇌CH3OH(g), where the forward reaction is exothermic.
- The forward reaction is exothermic, so a low temperature shifts the equilibrium to the right.
- There are 3 gas molecules on the left and 1 on the right, so a high pressure shifts the equilibrium to the right.
- Use a low temperature, a high pressure and plenty of carbon monoxide and hydrogen, and remove the methanol as it forms.
Self review
- What does it mean when an equilibrium lies to the left?
- What happens to the position of equilibrium when a product is removed?
- Which direction does a rise in temperature favour?
- In CH4(g)+H2O(g)⇌CO(g)+3H2(g)\text{CH}_4\text{(g)} + \text{H}_2\text{O(g)} \rightleftharpoons \text{CO(g)} + 3\text{H}_2\text{(g)}CH4(g)+H2O(g)⇌CO(g)+3H2(g), the forward reaction is endothermic; which temperature and pressure give the most hydrogen?
- Why does a catalyst not change the position of equilibrium?
