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Equilibrium

Question
CBSEENCH11006412

The dissociation of phosgene is represented as follows:

COCl subscript 2 left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space CO left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis

When the mixture of these three gases is compressed at constant temperature, what happens to:
(i) the amount of CO in the mixture.
(ii) the partial pressure of COCl2
(iii) the equilibrium constant for the reaction?

Solution
COCl subscript 2 left parenthesis straight g right parenthesis space space rightwards harpoon over leftwards harpoon space space space space CO left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis
(i) The forward reaction is accompanied by the increase of pressure. If the pressure of the equilibrium mixture is increased at constant temperature then according to Le-Chatelier’s principle, the equilibrium will shift in the backward direction i.e. the amount of CO in the mixture will decrease.
(ii) As the equilibrium shifts in the backward direction, the partial pressure of COCl2will increase.
(iii) For the reaction
   COCl subscript 2 left parenthesis straight g right parenthesis space rightwards harpoon over leftwards harpoon space space CO left parenthesis straight g right parenthesis space plus space Cl subscript 2 left parenthesis straight g right parenthesis
straight K space equals space fraction numerator open square brackets CO close square brackets space open square brackets Cl subscript 2 close square brackets over denominator open square brackets COCl subscript 2 close square brackets end fraction
As [CO] and [Cl2] decrease and [COCl2] increases, the value of equilibrium constant K will decrease.