Equilibrium - Le Chatelier's Principle.

In this post, I willl be discussing about Le Chatelier's Principle and its application in equilibria.

Le Chatelier's Princple states that if there is a disturbance to the equilibrium, the equilibrium will respond in a manner which will attempt to restore the original conditions of the equilibrium.

This definition is worded in an interesting manner. It does not claim that the disturbance is removed, but it does imply that the equilibrium will attempt to do something to regain the original condition.

I will be using Haber Process for this discussion:

N₂ + 3H₂ <-> 2NH₂

Enthlpy change of reaction is exothermic.
Catalysis used: Fe

Now, how does the various disturbance affect the equilibrium position?

1) Increasing either H₂ or N₂

At an equilibrium, when reactants are added will cause the equilibrium to be disturbed. The equilibrum responds to the increase by having the equilibrium position to shift to the right so as to attempt to restore the original conditions.

The forward reaction is hence favoured and although, there will be a decrease in the amount of the reactants, at the new equilibrium the amount of the reactants will still be higher than at the old equilibrium. Needless to say, the amount of products increases too.

2) Increasing Temperature

Since the forward reaction of the above equilibrium is exothermic, increasing temperature (changing it from T₁ to T₂) would cause the equilibrium to respond to this disturbance, since the original temperature which equilibrium exists in has been disturbed.

Therefore, the the equilibrium position will shift to the left and the backward reaction is thus favoured. More reactants and less products at the new equilibrium.

Although, this shift favours the endothermic backward reaction, it appears that heat is taken in and thus temperature decreases. However, that is not true. The temperature which the new equilibrium is at would be T₂.

3) Increasing Pressure

Increasing the pressure of the equilibrium, will also cause the equilibrium to respond to this disturbance. As there are more gaseous reactants than gaseous products, increasing pressure causes the total pressures of the reactants to increase more than the increase for the reactants. Hence, it causes the equilibrium position to shift to the right, favouring the forward reaction. Therefore, at the new equilibrium, there are more products and less of reactants.

In addition, like changing of temperature, changing of pressure does not mean that the equilibrium reduces the increased pressure. At the new equilibrium, we actually observed the equilibrium existing at the new higher pressure. Thus, if initially the equilibrium was at 2 atm and we increase the pressure to 5 atm. At the new equilibrium, the pressure will be 5 atm.

Lastly, in order for the equilibrium to be disturbed by changing the pressure. Two criteria must be satisfied. (1) There must be gases present in the equilibrium system. (2) The total amount of gaseous reactants must not be the same the total amount of gaseous reactants.

4) Adding of catalyst

A catalyst works to speed the rate both forward and backward reactions equally. Hence, the equilibrium is attained faster. The equilibrium composition remains the same hence there is no change to the equilibrium position.

5) Adding an inert gas (such as He).

When an inert gas is added, the total pressure of the reaction system increases. However, the individual partial pressures of N₂, H₂ and NH₃ remain the same.

This is unlike increasing the total pressure of the system. In this situation, the indivdual partial pressures of N₂, H₂ and NH₃ increase; which results in the forward reaction to be favoured.

Hence, when the inert gas is added, the equilibrium position remains the same. Neither the forward nor the backward reaction is favoured.

How do we make an equilibrium to cease from being an equilibrium?

A) We could remove the products of the reaction. Thus, continuously, the equilibrium is being disturbed. The equilibrium position will continue to shift to the right and the forward reaction is favoured until all the reactants are used up. This understanding explains why equivolumes of 1 moldm^-3 of CH₃COOH and 1 moldm^-3 of HCl required same volume of NaOH for neutralisation, despite CH₃COOH is the weaker acid.

B) Alternatively, we could add so much of one reactant such that it causes the equilibrium position to shift to the right; this results in the other reactant to be used up. This can be a possible explanation to why when we have an insoluble salt and add it into a infinitely large amount of water, the salt actually dissolves.

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Article written by Kwok YL 2007. (Edited in June 2009)

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