Le Chatelier's principle

In a nutshell

Many physical changes can occur during a reversible equilibrium reaction. Le Chatelier's principle is how the equilibrium changes position to adjust to any changes such as temperature, concentration and pressure.

Le Chatelier's principle

Le Chatelier's principle is defined as any change in temperature, concentration or pressure, the equilibrium will move to help counteract this change.

If the concentration of the reactants increases, the equilibrium will shift to the right to remove any excess reactant concentration. This means the reactant concentration will start decreasing as the reactant will be getting used up to form the product. The equilibrium shifting to the right means the forward reaction is favoured.

If the concentration of the product increases, the equilibrium will shift to the left to remove any excess product concentration. This means the product concentration will start decreasing as the product is getting used up to form the reactants. The equilibrium shifting to the left means the backward reaction is favoured.

Example

$H_2 (g) + O_2 (g) ⇌ H_2O (g)$

Hydrogen and oxygen react to form water in this reversible reaction. If you increase the concentration of water, the equilibrium will shift to the left in order to decrease this extra concentration of water.

Rules for Le Chatelier's principle

Concentration

Increasing or decreasing the concentration causes the equilibrium position to shift to counteract the change.

Example

$2NO(g)+O_2(g)⇌2NO_2(g)$

Nitric oxide and oxygen react to form nitrogen dioxide in this reversible reaction. If you increase the reactant concentration ( $NO$ or/and $O_2$ ), the equilibrium position will shift to the right to counteract this change.

If you increase the product concentration ( $NO_2$ ), the equilibrium position will shift to the left to counteract this change. The opposite effect will occur if you decrease the concentration.

Pressure

Increasing or decreasing the pressure causes the equilibrium position to shift to counteract the change. If you increase the pressure, the equilibrium position will shift to the side with less gaseous molecules. If you decrease the pressure, the equilibrium position will shift to the side with more gaseous molecules.

Example

$2NO(g)+O_2(g)⇌2NO_2(g)$

If you increase the pressure, the equilibrium position will shift to the right as there are only two gaseous moles on the right. This is less than the three gaseous moles on the left. If you decrease the pressure, the equilibrium position will shift to the left as there are more gaseous moles on the left.

Temperature

Increasing or decreasing the temperature causes the equilibrium position to shift to counteract the change. If a reaction is exothermic, it will have a negative enthalpy change. If a reaction is endothermic, it will have a positive enthalpy change.

Increasing the temperature will favour the endothermic reaction. If you increase the temperature, heat is being added so the equilibrium position will shift to decrease the excess heat. Endothermic reactions use the increased heat to react. This is why the endothermic reaction is favoured so it can use up all this excess heat. Therefore, temperature will be reduced.

Decreasing the temperature will favour the exothermic reaction. If you decrease the temperature, heat is being removed so the equilibrium position will shift to decrease the excess heat. Exothermic reactions release heat during the reaction. This is why the exothermic reaction is favoured so it can release heat to increase the temperature.

Example

$2 NO(g) + O_2(g) ⇌ 2 NO_2(g)$

$Enthalpy\>change,\> △H = -110 \>kJ\>mol^{-1}$

Nitric oxide and oxygen react to form nitrogen dioxide in this reversible reaction. A negative enthalpy change indicates the forward reaction is exothermic. Decreasing the temperature will favour the exothermic reaction. This causes the equilibrium to shift to the right to counteract this change.

Increasing the temperature will favour the endothermic reaction. This causes the equilibrium to shift to the left to counteract this change.

Choosing conditions in industry

In real life, scientists have to choose the optimal conditions for a reaction. This means they have to take into account the temperature, concentrations and pressures of the reaction. They want a reaction with low energy costs and lots of products. They do this by using the Le Chatelier's principle.

Low temperature is great for reducing energy costs. However, the temperature can't be too low or else the reaction will be too slow. You want a fast enough reaction with a compromised low temperature.

High pressures increase the rate of reaction. If it is too high, unwanted side reactions can also occur. This ultimately decreases the product yield. High pressure conditions are not ideal as it is expensive. Stronger equipment is needed to control high pressure.

Example

$C_2H_4 (g) + H_2O (g) ⇌ C_2H_5OH (g)$

$△H = -46\> kJ/mol$

A real life example is the production of ethanol steam from ethene and water.

Decreasing the temperature shifts the equilibrium position to the right. Therefore, lower temperature gives more ethanol. The ideal temperature for this reaction is $300\degree C$ . This is a compromise between having a low temperature and a fast enough reaction.

Increasing the pressure will shift the equilibrium position to the right. To compromise between the expense and the product yield, $60-70 \>atmosphere$ is used.