Question

The values of $K_{p_1}$ and $K_{p_2}$ for the reactions:

(i) $X\rightleftharpoons Y+Z$

(ii) $A\rightleftharpoons 2B$

They are in the ratio $9:1$. Assuming degree of dissociation of $X$ and $A$ be same, the dissociation pressure at equilibrium (i) and (ii) are in the ratio :

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

Answer 1

Answer: (A)

$36:1$

$X\rightleftharpoons Y+Z$
$1-\alpha$ $\alpha$ $\alpha$

For (i) $K_{p_1}=\frac{n_Y\cdot n_Z}{n_X}\frac{p_1}{{[\sum n]}_1}=\frac{{\alpha }^2}{(1-\alpha )}\cdot \frac{P_1}{(1+\alpha )}$

$A\rightleftharpoons 2B$

$1-\alpha$ $2\alpha$

For (ii) $K_{p_2}=\frac{{\left(n_B\right)}^2}{\left(n_A\right)}\times \frac{P_2}{{[\sum n]}_2}=\frac{4{\alpha }^2}{(1-\alpha )}\cdot \frac{P_2}{(1+\alpha )}$

$\frac{K_{p_1}}{K_{p_2}}=\frac{{\alpha }^2\times P_1}{(1-\alpha )\times (1+\alpha )}\times \frac{(1-\alpha )(1+\alpha )}{4{\alpha }^2\times P_2}$
$\frac{P_1}{P_2}=\frac{4\times 9}{1}=\frac{36}{1}$