The values of K_p_1 and K_p_2 for the… - NEET UG 2008 Chemistry PYQ

NEET UG 2008 — Chemistry Organic Chemistry

2008mcqeasy

The values of $K_{p_1}$ and $K_{p_2}$ for the reactions $X \rightleftharpoons Y+Z$ ...(i) and $A \rightleftharpoons 2 B$ ...(ii) are in ratio of $9: 1$ . If degree of dissociation of $X$ and $A$ be equal, then total pressure at equilibrium (i) and (ii) are in the ratio

Official previous-year question

Held on 30 Apr 2008 · Verified 9 Jul 2026.

Options

A
$3: 1$
B
$1: 9$
C
$36: 1$
D
$1: 1$

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Solution

From equation. $\begin{array}{lcc} & X & \rightleftharpoons & Y & + & Z \\ \text {Initial mole } & 10 & & 0 & & 0 \\ \text {mole at equilibrium } & (1-a) & & a & & a \end{array}$ $\begin{aligned} & K_{p_1}=\frac{p_Y \times p_z}{p_X} \\ & =\frac{\left[\frac{\alpha \times p_1}{1+\alpha}\right]\left[\frac{\alpha \times p_1}{1+\alpha}\right]}{\left[\frac{1-\alpha}{1+\alpha}\right] p_1} \\ & K_{P_1}=\frac{\alpha^2 p_1}{1-\alpha^2} \end{aligned}$ From equation $\begin{array}{lcc} & A & \rightleftharpoons & 2 B \\ \text {Initial mole } & 10 & & 0 \\ \text {mole at equilibrium } & (1-a) & & 2 a \end{array}$ $\begin{aligned} & K_{p_2}=\frac{\left[\frac{2 \alpha}{1+\alpha} \cdot p_2\right]^2}{\left[\frac{1-\alpha}{1+\alpha}\right] p_2}=\frac{4 \alpha^2 p_2}{1-\alpha^2} \end{aligned}$ From Eqs (i) and (ii) $\frac{K_{P_1}}{K_{P_2}}=\frac{p_1}{4 p_2}$ $\begin{array}{ll} \Rightarrow & \frac{9}{1}=\frac{p_1}{4 p_2} \\ \Rightarrow & \therefore \frac{p_1}{p_2}=\frac{36}{1} \end{array}$

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