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(C) The equilibrium constant expression for the reaction $A_{(s)} \rightleftharpoons 2B_{(g)} + 3C_{(g)}$ is given by $K_c = [B]^2 [C]^3$.Since $A$ is

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$\frac{1}{2\sqrt{2}}$ times its original concentration

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(C) The equilibrium constant expression for the reaction $A_{(s)} ightleftharpoons 2B_{(g)} + 3C_{(g)}$ is given by $K_c = [B]^2 [C]^3$.Since $A$ is a solid,its activity is taken as $1$.For a constant temperature,$K_c$ remains constant.Let the initial concentrations be $[B]_1$ and $[C]_1$. Then $K_c = [B]_1^2 [C]_1^3$.If the new concentration of $C$ is $[C]_2 = 2[C]_1$,let the new concentration of $B$ be $[B]_2$.Then $K_c = [B]_2^2 [C]_2^3 = [B]_2^2 (2[C]_1)^3 = [B]_2^2 	imes 8[C]_1^3$.Equating the two expressions for $K_c$:$[B]_1^2 [C]_1^3 = [B]_2^2 	imes 8[C]_1^3$.$[B]_1^2 = 8[B]_2^2$.$[B]_2^2 = \frac{[B]_1^2}{8}$.$[B]_2 = \frac{[B]_1}{\sqrt{8}} = \frac{[B]_1}{2\sqrt{2}}$.Thus,the concentration of $B$ becomes $\frac{1}{2\sqrt{2}}$ times its original concentration.

For the equilibrium system $A_{(s)} \rightleftharpoons 2B_{(g)} + 3C_{(g)}$,if the concentration of $C$ is doubled at equilibrium,then the concentration of $B$ at equilibrium will become ...

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