$(i) X (g) ⇌ Y (g) + z (g) K_{p_{1}} = 3 (ii) A (g) ⇌ 2 B (g) K_{p_{2}} = 1$
If the degree of dissociation and initial concentration of both the reactants X(g) and A(g) are equal, then the ratio of the total pressure at equilibrium $(\frac{p_{1}}{p_{2}})$ is equal to x : 1. The value of x is ___________(Nearest integer)

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answer is 12.

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Detailed Solution

$\begin{array}{ccc} (i) X_{(g)} & ⇌ Y_{(g)} + Z_{(g)} & K_{p_{1}} = 3 \\ 1 & 0 & 0 \\ 1 - x & x & x \end{array}$
Total moles at equillibrium = 1 + x
$\begin{array}{l} K_{p} = \frac{\frac{x}{1 + x} \cdot P_{1} \times \frac{x}{1 + x} \cdot P_{1}}{\frac{1 - x}{1 + x}, P_{1}} \\ K_{P} = \frac{P_{1} \cdot x^{2}}{1 - x^{2}} \end{array}$
$\begin{array}{l} (ii) A_{(g)} ⇌ 2 B_{(g)} K_{P_{2}} = 1 \\ 1 0 \\ 1 - x 2 x \\ K_{P_{2}} = \frac{\frac{2 x}{1 + x} \cdot P_{2} \times \frac{2 x}{1 + x} \cdot P_{2}}{\frac{1 - x}{1 + x} \cdot P_{2}} K_{P_{2}} = P_{2} \cdot \frac{4 x^{2}}{1 - x^{2}} \end{array}$
$\frac{{K_{P}}_{1}}{K_{p_{2}}} = \frac{3}{1} = \frac{P_{1} \cdot x^{2} / 1 - x^{2}}{P_{2} \cdot 4 x^{2} / 1 - x^{2}} \Rightarrow \frac{P_{1}}{P_{2}} = 12$