The equilibrium constants K_{p_1} and K_{p_2} | vedclass

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(C) For reaction $1$: $X \rightleftharpoons 2Y$. Let initial moles be $1$,degree of dissociation be $\alpha_1$. At equilibrium,moles are $(1-\alpha_1)

Vedclass · Accepted Answer

$1 : 64$

Vedclass · Answer

(C) For reaction $1$: $X \rightleftharpoons 2Y$. Let initial moles be $1$,degree of dissociation be $\alpha_1$. At equilibrium,moles are $(1-\alpha_1)$ and $2\alpha_1$. Total moles $= 1+\alpha_1 \approx 1$. $K_{p_1} = \frac{(2\alpha_1 P_1)^2}{(1-\alpha_1)P_1} \approx 4\alpha_1^2 P_1$.For reaction $2$: $Z \rightleftharpoons P + Q$. Let initial moles be $1$,degree of dissociation be $\alpha_2$. At equilibrium,moles are $(1-\alpha_2)$,$\alpha_2$,and $\alpha_2$. Total moles $= 1+\alpha_2 \approx 1$. $K_{p_2} = \frac{(\alpha_2 P_2)(\alpha_2 P_2)}{(1-\alpha_2)P_2} \approx \alpha_2^2 P_2$.Given $\frac{K_{p_1}}{K_{p_2}} = \frac{1}{4}$ and $\alpha_1 = 2\alpha_2$.Substituting: $\frac{4\alpha_1^2 P_1}{\alpha_2^2 P_2} = \frac{1}{4} \implies \frac{4(2\alpha_2)^2 P_1}{\alpha_2^2 P_2} = \frac{1}{4} \implies \frac{16\alpha_2^2 P_1}{\alpha_2^2 P_2} = \frac{1}{4} \implies \frac{P_1}{P_2} = \frac{1}{64}$.

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