For the reaction system A_{(g)} rightleftharp | vedclass

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

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

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(C) The equilibrium constant expression for the reaction $A_{(g)} \rightleftharpoons B_{(g)} + 2C_{(g)}$ is given by: $K_c = \frac{[B][C]^2}{[A]}$.Assuming the temperature remains constant,$K_c$ remains constant.Let the initial concentrations be $[A]_1, [B]_1, [C]_1$ and the new concentrations be $[A]_2, [B]_2, [C]_2$.Since $K_c = \frac{[B]_1[C]_1^2}{[A]_1} = \frac{[B]_2[C]_2^2}{[A]_2}$,and given $[B]_2 = 2[B]_1$,we have $\frac{[B]_1[C]_1^2}{[A]_1} = \frac{2[B]_1[C]_2^2}{[A]_2}$.Assuming the volume of the system is constant and the reaction reaches a new equilibrium,the relationship simplifies to $[C]_2^2 = \frac{[C]_1^2}{2}$ (assuming $[A]$ remains relatively constant or considering the ratio change).Therefore,$[C]_2 = \frac{[C]_1}{\sqrt{2}}$.

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

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