Le Chatelier's Principle

Hope these revision cards help everyone understand this a bit more! 

Please comment and rate!

Thanks :)

HideShow resource information

Principle

The equilibirum shifts either left or right to oppose a change.

1 of 6

Example equation

This will be the equation I shall use to explain;

A + B <--> 2C + D

Just pretend that the <--> is the symbol for a reversible reaction.

2 of 6

Temperature- Exothermic (Positive)

A + B <--> 2C + D
Exothermic reaction 

If I increase the temperature of this reaction, the equilibrium will shift to the left hand side to oppose the change; therefore there will be more reactant formed.

If I decrease the temperature of this reaction, the equilibrium will shift to the right hand side; therefore more product will be formed.

There is no way of calculating this, you just have to remember that if temp is increased the equilibrium shifts to the left and if the temp is decreased the equilibrium shifts to the right.

3 of 6

Temperature- Endothermic (Negative)

A + B <--> 2C + D
Endothermic reaction

If I increase the temperature of the reaction the equilibrium will shift to the right hand side, to oppose the change, therefore more product is made.

If I decrease the temperature the equilibrium will shift to the left hand side, therefore more reactant is made.

Again for this reaction there is no way of calculating which way the equilibrium will shift, you just have to learn it! 

4 of 6

Pressure

A + B <--> 2C + D

If I increase the pressure of the system, the equilibrium will shift to oppose the change so it will shift to the side with the fewer moles, in this reaction the equilibrium will shift to the left hand side; therefore more reactant is made. 

It moves to the left because of the fewer moles: Each letter represents one mole. So A + B = 2 moles but 2C + D = 3 moles.

If I decrease the pressure the equilibrium will shift to the side with more moles, the right hand side in this equation; therefore more product is made.

5 of 6

Concentration

A + B <--> 2C + D

If I increase the concentration of A the equilibrium would shift to the right hand side to try and remove the extra particles of A; therefore more product.

If I decrease the concentration of A the equilibrium would shift to replace the lost particles of A so the equilibrium would shift to the left hand side; therefore less product is made.

Even if I remove/decrease the amount of B, the equilibrium would shift in the same way as it did for A. It would shift the opposite way if I was increasing or decreasing the concentration of C or D.

6 of 6

Comments

No comments have yet been made

Similar Chemistry resources:

See all Chemistry resources »See all Electrolysis resources »