Question

The equilibrium constants $K_{p_1}$ and $K_{p_2}$ for the reactions, $X\rightleftharpoons 2Y$ and $Z\rightleftharpoons P+Q$, respectively are in the ratio of $1:9$. If the degree of dissociation of $X$ and $Z$ be equal then the ratio of total pressures at the equilibrium is:

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

Answer 1

The correct option is B $1:36$

The given reaction is :

$X\rightleftharpoons 2Y$

Initial pressure : $A$ $0$

At eq. : $A(1-\alpha )$ $2\alpha A$

Now, Total pressure at equilibrium, $P_1=A(1+\alpha )\to \left(I\right)$

Now, $K_{P_1}=\frac{{\left(p_Y\right)}^2}{px}=\frac{{\left(2\alpha A\right)}^2}{A(1-\alpha )}$

$\Rightarrow K_{P_1}=\frac{4{\alpha }^{2}A}{(1-\alpha )}$

$\Rightarrow K_{P_1}=\frac{4{\alpha }^2}{(1-\alpha )}.\frac{P_1}{(1+\alpha )}$ (from (I))

$\Rightarrow K_{P_1}=\frac{4{\alpha }^2{P}_1}{(1-{\alpha }^2)}\to \left(II\right)$

$\left(II\right)$ Now, in reaction,

$Z\rightleftharpoons P+Q$

Initial pressure : $B$ $0$ $0$

At eqm : $B(1-\alpha )$ $B\alpha$ $B\alpha$ (let degree of dissociation $=\alpha$)

Total pressure at equilibrium, $P_2=B(1+\alpha )\to \left(II\right)$

Now, ${Kp}_2=\frac{p_p.p_Q}{pz}=\frac{B\alpha .B\alpha }{B(1-\alpha )}$

$\Rightarrow {Kp}_2=\frac{{B\alpha }^2}{(1-\alpha )}$

$\Rightarrow {Kp}_2=\frac{{\alpha }^2}{(1-\alpha )}.\frac{P_2}{(1+\alpha )}$ (from (III))

$\Rightarrow {Kp}_2=\frac{P_2{\alpha }^2}{(1-{\alpha }^2)}\left(IV\right)$

Given, ${Kp}_1/{Kp}_2=1/9\Rightarrow \frac{4{\alpha }^2{P}_1}{P_2{\alpha }^2}=\frac{1}{9}\Rightarrow \frac{P_1}{P_2}=\frac{1}{36}$

Option (B) is correct.