One mole of a compound AB reacts with one mol | vedclass

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Question

(D) The reaction is $AB + CD \rightleftharpoons AD + CB$. At $t = 0$,the moles are $1 \ mol$ of $AB$ and $1 \ mol$ of $CD$,with $0 \ mol$ of products.

Vedclass · Accepted Answer

$9$

Vedclass · Answer

(D) The reaction is $AB + CD ightleftharpoons AD + CB$. At $t = 0$,the moles are $1 \ mol$ of $AB$ and $1 \ mol$ of $CD$,with $0 \ mol$ of products. At equilibrium,$\frac{3}{4} \ mol$ of each reactant has reacted. The moles at equilibrium are: $n(AB) = 1 - \frac{3}{4} = \frac{1}{4} \ mol$ $n(CD) = 1 - \frac{3}{4} = \frac{1}{4} \ mol$ $n(AD) = \frac{3}{4} \ mol$ $n(CB) = \frac{3}{4} \ mol$ Since the volume $V$ is constant,the equilibrium constant $K_c$ is given by: $K_c = \frac{[AD][CB]}{[AB][CD]} = \frac{(\frac{3/4}{V})(\frac{3/4}{V})}{(\frac{1/4}{V})(\frac{1/4}{V})} = \frac{3/4 	imes 3/4}{1/4 	imes 1/4} = \frac{9/16}{1/16} = 9$.

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