The values of ${\mathrm{K}}_{{\mathrm{p}}_1} \mathrm{and} {\mathrm{K}}_{{\mathrm{p}}_2}$ for the reactions

$\mathrm{X} \rightleftharpoons \mathrm{Y} + \mathrm{Z}$          (i)

and $\mathrm{A} \rightleftharpoons 2\mathrm{B}$         (ii)

are in the ratio of 9 : 1. If the degree of dissociation of X and A be equal, then total pressure at equilibrium (i) and (ii) are in the ratio

In equation,        $X \rightleftharpoons Y + Z$

Initial moles         1           0       0

At equilibrium    $\left(1-\alpha \right)$     $\alpha$      $\alpha$

Where $\alpha$ = degree of dissociation

Total number of moles = 1 - $\alpha$ + $\alpha$ + $\alpha$ = (1 + $\alpha$)

$$\begin{gathered} {\mathrm{p}}_{\mathrm{x}} = \left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1 \\ {\mathrm{p}}_{\mathrm{y}} = \left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1 \\ {\mathrm{p}}_{\mathrm{z}} = \left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1 \\ {\mathrm{K}}_{\mathrm{p}1} = \frac{\left[{\mathrm{p}}_{\mathrm{y}}\right] \left[{\mathrm{p}}_{\mathrm{z}}\right]}{\left[{\mathrm{p}}_{\mathrm{x}}\right]}=\frac{ \left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1 \times \left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1}{ \left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_1} \\ = \frac{ {\left(\frac{\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)}^2{\mathrm{p}}_1}{\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)}=\frac{{\mathrm{\alpha }}^2}{(1+\mathrm{\alpha })(1-\mathrm{\alpha })}{\mathrm{P}}_1 \dots \dots \dots \left(\mathrm{i}\right) \end{gathered}$$

For equation,     $A \rightleftharpoons 2B$

Initial moles       1            0

At equilibrium $\left(1-\alpha \right)$      2$\alpha$

Total number of moles at equilibrium   (1 + $\alpha$)

$$\begin{gathered} {\mathrm{p}}_{\mathrm{B}} = \left(\frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_2 \\ {\mathrm{p}}_{\mathrm{B}} = \left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_2 \\ {\mathrm{K}}_{\mathrm{p}2} = \frac{{\left[{\mathrm{p}}_{\mathrm{B}}\right]}^2}{\left[{\mathrm{p}}_{\mathrm{A}}\right]} = \frac{{\left[\left(\frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_2\right]}^2}{\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right){\mathrm{p}}_2} \\ = \frac{{\left(\frac{2\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)}^2{\mathrm{p}}_2}{\left(\frac{1-\mathrm{\alpha }}{1+\mathrm{\alpha }}\right)}=\frac{4{\mathrm{\alpha }}^2}{(1+\mathrm{\alpha })(1-\mathrm{\alpha })}{\mathrm{P}}_2 \end{gathered}$$

Eq. (i) / (ii)

$$\begin{gathered} \frac{{\mathrm{K}}_{\mathrm{p}1}}{{\mathrm{K}}_{\mathrm{p}2}} = \frac{{\mathrm{\alpha }}^2 \times {\mathrm{p}}_1 }{4{\mathrm{\alpha }}^2 \times {\mathrm{p}}_2} \\ \frac{9}{1} = \frac{{\mathrm{p}}_1}{4{\mathrm{p}}_2} \\ \frac{{\mathrm{p}}_1}{{\mathrm{p}}_2} = \frac{36}{1} = 36 :\hspace{0.17em}1 \end{gathered}$$