$X \rightleftharpoons Y + Z$ $...(i)$
$A \rightleftharpoons 2B$ $...(ii)$
are in the ratio $9 : 1.$ If degree of dissociation of $X$ and $A$ be equal, then total pressure at equilibrium $(i)$ and $(ii)$ are in the ratio
Initial moles $\quad 0\quad \quad 0 \; \quad 0$
At equil. $\;\;\;\;(1-\alpha) \quad \alpha \quad \alpha$
where, $\alpha=$ degree of dissociation
Total number of moles
$=1-\alpha+\alpha+\alpha=(1+\alpha)$
$P_{x}=\left(\frac{1-\alpha}{1+\alpha}\right) P_{1}$
$P_{y}=\left(\frac{\alpha}{1+\alpha}\right) P_{1}$
$P_{z}=\frac{\alpha}{1-\alpha} P_{1}$
$K_{p 1}=\frac{[p y][p z]}{[p x]}=\frac{\left(\frac{a}{1+\alpha}\right) p 1 \times\left(\frac{a}{1+\alpha}\right) p 1}{\left(\frac{1-\alpha}{1+\alpha}\right) p 1}$
$=\frac{\left(\frac{\alpha}{1+\alpha}\right)^{2} p 1}{\left(\frac{1-\alpha}{1+\alpha}\right)} \quad \cdots(i)$
For equation, $A \rightleftharpoons 2 B$
Initial moles $\;\;1 \;\;\;\quad 0$
At equil. $\;\;(1-\alpha) \;\;\;2 \alpha$
Total number of moles at equilibrium $=(1+\alpha)$ $p_{B}=\left(\frac{2 \alpha}{1+\alpha}\right) p_{2}$
$p_{A}=\left(\frac{1-\alpha}{1+\alpha}\right) p_{2}$
$K_{p 2}=\frac{\left[p_{B}\right]^{2}}{\left[p_{A}\right]}$$=\frac{\left[\left(\frac{2 a}{1+\alpha}\right) p_{2}\right]^{2}}{\left(\frac{1-\alpha}{1+\alpha}\right)} \quad \cdots (i i)$
Eq. $(i)$ divide by Eq. $(ii)$
$\frac{K_{\mathrm{pl}}}{K_{\mathrm{p} 2}}=\frac{\alpha^{2} \times p_{1}}{4 \alpha^{2} \times p_{2}}$
$\frac{9}{1}=\frac{p_{1}}{4 p_{2}}$
$\frac{p_{1}}{p_{2}}=\frac{36}{1}=36: 1$
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