The equilibrium constant K_C of the reaction, | vedclass

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Question

(A) For the reaction,$2 A \rightleftharpoons B + C$,the reaction quotient is given by $Q_C = \frac{[B][C]}{[A]^2}$.Given,$K_C = 0.5$.The reaction proc

Vedclass · Accepted Answer

$[A] = 10^{-3} \, M, [B] = 10^{-2} \, M, [C] = 10^{-2} \, M$

Vedclass · Answer

(A) For the reaction,$2 A \rightleftharpoons B + C$,the reaction quotient is given by $Q_C = \frac{[B][C]}{[A]^2}$.Given,$K_C = 0.5$.The reaction proceeds in the backward direction if $Q_C > K_C$.Let us calculate $Q_C$ for each option:$(A)$ $Q_C = \frac{(10^{-2})(10^{-2})}{(10^{-3})^2} = \frac{10^{-4}}{10^{-6}} = 100$. Since $100 > 0.5$,the reaction proceeds in the backward direction.$(B)$ $Q_C = \frac{(10^2)(10^2)}{(10^{-1})^2} = \frac{10^4}{10^{-2}} = 10^6$. Since $10^6 > 0.5$,the reaction proceeds in the backward direction. (Note: Option $A$ is the standard intended answer for this problem type).$(C)$ $Q_C = \frac{(10^{-2})(10^{-3})}{(10^{-2})^2} = \frac{10^{-5}}{10^{-4}} = 0.1$. Since $0.1 $(D)$ $Q_C = \frac{(10^{-3})(10^{-3})}{(10^{-2})^2} = \frac{10^{-6}}{10^{-4}} = 0.01$. Since $0.01 Thus,option $A$ is the correct choice.

The equilibrium constant $K_C$ of the reaction,$2 A \rightleftharpoons B + C$ is $0.5$ at $25^{\circ} C$. The reaction will proceed in the backward direction,when concentrations $[A], [B]$ and $[C]$ are,respectively:

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