The values of K_p for the reactions,X rightle | vedclass

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(A) For reaction $(i)$: $X \rightleftharpoons Y + Z$Initial moles: $1, 0, 0$At equilibrium: $(1-\alpha), \alpha, \alpha$Total moles $= 1+\alpha$$P_X =

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$36 : 1$

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(A) For reaction $(i)$: $X ightleftharpoons Y + Z$Initial moles: $1, 0, 0$At equilibrium: $(1-\alpha), \alpha, \alpha$Total moles $= 1+\alpha$$P_X = \frac{1-\alpha}{1+\alpha} P_1, P_Y = \frac{\alpha}{1+\alpha} P_1, P_Z = \frac{\alpha}{1+\alpha} P_1$$K_{p1} = \frac{P_Y P_Z}{P_X} = \frac{\alpha^2 P_1}{1-\alpha^2}$For reaction $(ii)$: $A ightleftharpoons 2B$Initial moles: $1, 0$At equilibrium: $(1-\alpha), 2\alpha$Total moles $= 1+\alpha$$P_A = \frac{1-\alpha}{1+\alpha} P_2, P_B = \frac{2\alpha}{1+\alpha} P_2$$K_{p2} = \frac{P_B^2}{P_A} = \frac{4\alpha^2 P_2}{1-\alpha^2}$Given $\frac{K_{p1}}{K_{p2}} = \frac{9}{1}$$\frac{\alpha^2 P_1}{1-\alpha^2} 	imes \frac{1-\alpha^2}{4\alpha^2 P_2} = \frac{9}{1}$$\frac{P_1}{4P_2} = 9 \implies \frac{P_1}{P_2} = 36$Thus,the ratio is $36 : 1$.

The values of $K_p$ for the reactions,
$X \rightleftharpoons Y + Z$ $...(i)$
$A \rightleftharpoons 2B$ $...(ii)$
are in the ratio $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the ratio of total pressure at equilibrium for $(i)$ and $(ii)$ is:

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The values of $K_p$ for the reactions,$X \rightleftharpoons Y + Z$ $...(i)$$A \rightleftharpoons 2B$ $...(ii)$are in the ratio $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the ratio of total pressure at equilibrium for $(i)$ and $(ii)$ is: