The values of $k_{p_{1}}$ and $K_{p_{2}}$ for the reactions $X ⇌ Y + Z . . . (i)$ and $A ⇌ 2 B . . . . (ii)$ are in the ratio 9 : l. If degree of dissociation of X and A be equal, then total pressure at equilibrium (1) and (2) are in the ratio:

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3:1

36:1

1:1

1:9

answer is B.

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Detailed Solution

$K_{P_{1}} = \frac{n_{Y} \cdot n_{Z}}{n_{X}} {[\frac{P_{1}}{Σ_{n_{1}}}]}^{1} K_{P_{2}} = \frac{{(n_{B})}^{2}}{n_{A}} \times {[\frac{P_{2}}{{Σn}_{2}}]}^{1}$
for $\underset{1 - α}{\underset{1}{X}} ⇌ \underset{α}{\underset{0}{Y}} + \underset{α}{\underset{0}{Z}} for \underset{1 - α}{\underset{1}{A}} ⇌ \underset{2 α}{\underset{0}{2 B}}$
$\begin{array}{l} ∴ \frac{K_{P_{1}}}{K_{P_{2}}} = \frac{P_{1}}{P_{2}} \times \frac{n_{Y} \cdot n_{Z}}{n_{X}} \times \frac{(n_{A})}{{(n_{B})}^{2}} \times [\frac{{Σn}_{2}}{{Σn}_{1}}] \\ 9 = \frac{P_{1}}{P_{2}} \times \frac{α \cdot α}{1 - α} \times \frac{(1 - α)}{(2 α)^{2}} \times \frac{(1 + α)}{(1 + α)} \\ ∴ \frac{P_{1}}{P_{2}} = 4 \times 9 = 36 \end{array}$