At 550 K,the K_c for the following reaction | vedclass

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(A) The equilibrium constant expression for the reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ is $K_c = \frac{[Z]}{[X][Y]}$.Given that $K_c

Vedclass · Accepted Answer

$2 	imes 10^{-4}$

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(A) The equilibrium constant expression for the reaction $X_{(g)} + Y_{(g)} ightleftharpoons Z_{(g)}$ is $K_c = \frac{[Z]}{[X][Y]}$.Given that $K_c = 10^4 \ L \ mol^{-1}$ and at equilibrium $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$.From the given relations,we have $[X] = \frac{1}{2}[Z]$ and $[Y] = [Z]$.Substituting these values into the $K_c$ expression:$10^4 = \frac{[Z]}{(\frac{1}{2}[Z])([Z])}$$10^4 = \frac{[Z]}{\frac{1}{2}[Z]^2} = \frac{2}{[Z]}$$[Z] = \frac{2}{10^4} = 2 	imes 10^{-4} \ mol \ L^{-1}$.

At $550 \ K$,the $K_c$ for the following reaction is $10^4 \ L \ mol^{-1}$: $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$. At equilibrium,it was observed that $[X] = \frac{1}{2}[Y] = \frac{1}{2}[Z]$. What is the value of $[Z]$ (in $mol \ L^{-1}$) at equilibrium?

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