X_{2(g)} + Y_{2(g)} rightleftharpoons 2Z_{(g) | vedclass

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(B) The equilibrium reaction is $X_{2(g)} + Y_{2(g)} \rightleftharpoons 2Z_{(g)}$. The equilibrium constant $K_{C}$ is calculated as: $K_{C} = \frac{[

Vedclass · Accepted Answer

$15$

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(B) The equilibrium reaction is $X_{2(g)} + Y_{2(g)} ightleftharpoons 2Z_{(g)}$. The equilibrium constant $K_{C}$ is calculated as: $K_{C} = \frac{[Z]^2}{[X_2][Y_2]} = \frac{(9/1)^2}{(3/1) 	imes (3/1)} = \frac{81}{9} = 9$. When $10 \ mol$ of $Z_{(g)}$ is added,the total moles of $Z$ become $9 + 10 = 19 \ mol$. The reaction shifts backward to re-establish equilibrium. Let $x$ moles of $X_2$ and $Y_2$ be formed. Equilibrium moles: $X_2 = 3+x$,$Y_2 = 3+x$,$Z = 19-2x$. $K_{C} = \frac{(19-2x)^2}{(3+x)(3+x)} = 9$. Taking the square root on both sides: $\frac{19-2x}{3+x} = 3$. $19-2x = 9+3x$. $10 = 5x \Rightarrow x = 2$. Moles of $Z$ at new equilibrium $= 19 - 2(2) = 15 \ mol$.

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