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(D) For the reversible first-order reaction $X \rightleftharpoons Y$,the equilibrium constant $K_{eq}$ is given by:$K_{eq} = \frac{K_{f}}{K_{b}} = \fr

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(D) For the reversible first-order reaction $X \rightleftharpoons Y$,the equilibrium constant $K_{eq}$ is given by:$K_{eq} = \frac{K_{f}}{K_{b}} = \frac{[Y]_{eq}}{[X]_{eq}}$Given $K_{f} = 2 \ s^{-1}$ and $K_{b} = 1 \ s^{-1}$,we have:$K_{eq} = \frac{2}{1} = 2$Therefore,$\frac{[Y]_{eq}}{[X]_{eq}} = 2$,which implies $[Y]_{eq} = 2[X]_{eq}$.This means that at equilibrium,the concentration of $Y$ is twice the concentration of $X$. As the reaction proceeds,the concentration of $X$ decreases from its initial value $[X]_{0}$ to $[X]_{eq}$,and the concentration of $Y$ increases from $0$ to $[Y]_{eq}$.Since $[Y]_{eq} = 2[X]_{eq}$,the final concentration of $Y$ must be higher than the final concentration of $X$. Plot $(d)$ correctly shows $[Y]_{eq} > [X]_{eq}$ at equilibrium.

List-$I$ (Reaction)	List-$II$ $(K_p)$
$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ at $298 \ K$	$4.0 \times 10^{24}$
$2 SO_{2(g)} + O_{2(g)} \rightleftharpoons 2 SO_{3(g)}$ at $700 \ K$	$3.0 \times 10^{4}$
$N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)}$ at $298 \ K$	$0.98$
$N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)}$ at $500 \ K$	$1700$

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