Effect of concentration on equilibrium

What is the effect of concentration on equilibrium?

As mention in an earlier article, the equilibrium constant of a reaction, K, is given by:

K=e^{-\frac{\Delta_rG^o}{RT}}\; \; \; \; \; \; \; \; 17

Eq17 shows that the value of K for a particular reaction only varies with temperature and is independent of the concentration of reaction species. With that in mind, how does concentration affect the position of chemical equilibrium of a reaction? Consider the hydrolysis of ethyl ethanoate,

CH_3COOC_2H_5(l)+H_2O(l)\rightleftharpoons CH_3COOH(l)+C_2H_5OH(l)

the equilibrium constant for the reaction is given by

K=\frac{[CH_3COOH][C_2H_5OH]}{[CH_3COOC_2H_5][H_2O]}

When the reaction reaches dynamic equilibrium at a particular temperature, the ratio of the products and the reactants is a constant. If we increase the concentration of ethyl ethanoate at this stage, the excess ethyl ethanoate reacts with water to give more products to maintain the constant value of K, thereby reducing the concentration of ethyl ethanoate and shifting the position of the equilibrium to the right to attain a new dynamic equilibrium. This is consistent with Le Chatelier’s principle.

If we now increase the concentration of ethanoic acid or decrease the concentration of ethyl ethanoate, the position of the equilibrium will shift to the left to minimise the change and maintain the same value of K.

We can also explain the above observation using chemical kinetics, where, at equilibrium, the rate of the forward reaction is equal to the rate of the reverse reaction. If we increase the concentration of ethyl ethanoate, the greater number of ethyl ethanoate molecules leads to a higher frequency of collisions between ethyl ethanoate and water molecules, which increases the rate of the forward reaction relative to the reverse reaction. As a result, the position of equilibrium shifts to the right to maintain the value of K.

 

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