Question

The values of $K_{P_1}$ and $K_{P_2}$ for the reactions
$X\rightleftharpoons Y+Z......\left(1\right)$ and
$A\rightleftharpoons 2B.....\left(2\right)$
are in ratio of $9:1$. If degree of dissociation of $X$ and $A$ be equal, then total pressure at equilibrium $\left(1\right)$ and $\left(2\right)$ are in the ratio.

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

Answer 1

The correct option is B $36:1$

$X\rightleftharpoons Y+Z{K}_{P1}$ $A\rightleftharpoons 2B{K}_{P2}$

$t=0$ $a$ $0$ $0$ $a$ $0$

$t=t_{eq}$ $a(1-\alpha )$ $a\alpha$ $a\alpha$ $a(1-\alpha )$ $2a\alpha$

total pressure $\alpha$ no. of moles $P_2$ total pressure in

Given $\frac{K_{P1}}{K_{P2}}=9$ case (2)

$P_1$ total pressure in case $1$ $K_{P2}=\frac{{\left(\frac{2a\alpha }{a(1+\alpha )}{P}_2\right)}^2}{\frac{a(1-\alpha )}{a(1+\alpha )}{P}_2}\Rightarrow P_2.\frac{{4\alpha }^2}{1-{\alpha }^2}$

$K_{P1}=\frac{\frac{a\alpha }{a(1+\alpha )}{P}_1\frac{a\alpha }{a(1+\alpha )}{P}_1}{\frac{a(1-\alpha )}{a(1+\alpha )}{P}_1}\Rightarrow \frac{{\alpha }^2{P}_1}{1-{\alpha }^2}$

$\frac{K_{P1}}{K_{P2}}=\frac{a}{1}$

$\frac{\frac{{\alpha }^2{P}_1}{1-{\alpha }^2}}{\frac{4{\alpha }^2{P}_2}{1-{\alpha }^2}}=a/1\Rightarrow \frac{P_1}{4P_2}=\frac{9}{1}$

$\Rightarrow \frac{P_1}{P_2}=\frac{36}{1}$