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(A) For reaction $1$: $X \rightleftharpoons 2Y$. Let initial moles be $1$,degree of dissociation be $\alpha$. At equilibrium: $X = 1-\alpha$,$Y = 2\al

Vedclass · Accepted Answer

$1 : 36$

Vedclass · Answer

(A) For reaction $1$: $X \rightleftharpoons 2Y$. Let initial moles be $1$,degree of dissociation be $\alpha$. At equilibrium: $X = 1-\alpha$,$Y = 2\alpha$. Total moles $= 1+\alpha$. $K_{p1} = \frac{(2\alpha/(1+\alpha))^2 P_1}{(1-\alpha)/(1+\alpha)} = \frac{4\alpha^2 P_1}{1-\alpha^2}$.For reaction $2$: $Z \rightleftharpoons P + Q$. Let initial moles be $1$,degree of dissociation be $\alpha$. At equilibrium: $Z = 1-\alpha$,$P = \alpha$,$Q = \alpha$. Total moles $= 1+\alpha$. $K_{p2} = \frac{(\alpha/(1+\alpha))^2 P_2}{(1-\alpha)/(1+\alpha)} = \frac{\alpha^2 P_2}{1-\alpha^2}$.Given $\frac{K_{p1}}{K_{p2}} = \frac{1}{9}$.$\frac{4\alpha^2 P_1 / (1-\alpha^2)}{\alpha^2 P_2 / (1-\alpha^2)} = \frac{1}{9} \implies \frac{4P_1}{P_2} = \frac{1}{9}$.Therefore,$\frac{P_1}{P_2} = \frac{1}{36}$.

For the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$,the equilibrium constants $K_p$ and $K_q$ are in the ratio $1:9$. If the degree of dissociation of $X$ and $Z$ is the same,then the ratio of their total pressures is:

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