Question

$A\rightleftharpoons 2B,K_p$ and $C\rightleftharpoons D+E,K_p^{\prime }$. If degrees of dissociation of $A$ and $C$ are same and $K_p=2K_p^{\prime }$, then the ratio of total pressure $\frac{p}{p{}^{\prime }}$ is:

• A. $1/2$
• B. $1/3$
• C. $1/4$
• D. $2$

Answer 1

Answer: (A)

$1/2$

$\left(I\right)$ The given reaction is ;-

$A\rightleftharpoons 2B$

Initial pressure : $P$ $0$ (let)

At eqm : $P(1-2)$ $2\alpha P$ (given degree of dissociation $=\alpha$)

Now, Total pressure, $p=(P+Px)$

Now, $K_P={\left(p_B\right)}^2/p_A$

$=\frac{{\left(2\alpha P\right)}^2}{P(1-2)}$

$\Rightarrow K_P=\frac{{4\alpha }^{2}P}{(1-\alpha )}\to \left(I\right)$

$\left(II\right)$ Given reaction is :-

$C\rightleftharpoons D+E$

Initial pressure : $P^1$ $0$ $0$ (let)

At eqm : $P^{{}^{\prime }}(1-\alpha )$ ${\alpha P}^{{}^{\prime }}$ ${\alpha P}^{{}^{\prime }}$ (given)

Total pressure, $p^{{}^{\prime }}=p^{{}^{\prime }}(1+\alpha )$

Now, $K_{P^{{}^{\prime }}}=\frac{p_D.p_6}{p_C}$

$=\frac{{\alpha p}^{{}^{\prime }}.{\alpha p}^{{}^{\prime }}}{P^{{}^{\prime }}(1-\alpha )}$

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

Now, given $K_P={2K}^{{}^{\prime }}p$

$\Rightarrow \frac{{\alpha }^2{P}^{{}^{\prime }}}{(1-\alpha )}=\frac{2\times 4{\alpha }^{2}P}{(1-\alpha )}$

$\Rightarrow \frac{P}{p^{{}^{\prime }}}=\frac{1}{8}$