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. $(ii)$

$\frac{K_{\mathrm{pl}}}{K_{\mathrm{p} 2}}=\frac{\alpha^{2} \times p_{1}}{4 \alpha^{2} \times p_{2}}$

$\frac{9}{1}=\frac{p_{1}}{4 p_{2}}$

$\frac{p_{1}}{p_{2}}=\frac{36}{1}=36: 1$