The values of K_{P_1} and K_{P_2} for the rea | vedclass

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(A) For reaction $(1)$: $X \rightleftharpoons Y + Z$Initial moles: $1, 0, 0$Equilibrium moles: $(1-\alpha), \alpha, \alpha$. Total moles = $1+\alpha$.

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$36 : 1$

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(A) For reaction $(1)$: $X ightleftharpoons Y + Z$Initial moles: $1, 0, 0$Equilibrium moles: $(1-\alpha), \alpha, \alpha$. 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$$K_{P_1} = \frac{P_Y P_Z}{P_X} = \frac{\alpha^2 P_1}{1-\alpha^2}$For reaction $(2)$: $A ightleftharpoons 2B$Initial moles: $1, 0$Equilibrium moles: $(1-\alpha), 2\alpha$. Total moles = $1+\alpha$.$P_A = \frac{1-\alpha}{1+\alpha} P_2, P_B = \frac{2\alpha}{1+\alpha} P_2$$K_{P_2} = \frac{P_B^2}{P_A} = \frac{(2\alpha)^2 P_2^2}{(1+\alpha)^2} 	imes \frac{(1+\alpha)}{(1-\alpha) P_2} = \frac{4\alpha^2 P_2}{1-\alpha^2}$Given $\frac{K_{P_1}}{K_{P_2}} = \frac{9}{1}$.$\frac{\alpha^2 P_1 / (1-\alpha^2)}{4\alpha^2 P_2 / (1-\alpha^2)} = \frac{P_1}{4P_2} = \frac{9}{1}$$\frac{P_1}{P_2} = \frac{36}{1}$.

The values of $K_{P_1}$ and $K_{P_2}$ for the reactions
$X \rightleftharpoons Y + Z$ --- $(1)$
$A \rightleftharpoons 2B$ --- $(2)$
are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio:

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The equilibrium constants $K_{p1}$ and $K_{p2}$ for the reactions $X \rightleftharpoons 2Y$ and $Z \rightleftharpoons P + Q$ respectively are in the ratio of $1 : 9$. If the degree of dissociation of $X$ and $Z$ be equal,then the ratio of total pressures at these equilibria is:

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The values of $K_{P_1}$ and $K_{P_2}$ for the reactions$X \rightleftharpoons Y + Z$ --- $(1)$$A \rightleftharpoons 2B$ --- $(2)$are in the ratio of $9 : 1$. If the degree of dissociation of $X$ and $A$ is equal,then the total pressure at equilibrium for $(1)$ and $(2)$ are in the ratio: