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 pressure at these equilibria is

Question

Dr. Abubacker · Accepted Answer

(a)
$X\ \ \ \ \  ightleftharpoons \ \ \ \ \ 2Y$
1 0
At equilibrium $1 - x$ $2x$
Total no. of moles = $1 + x$
$K_{p1} = \frac{\left\lbrack \frac{2x}{1 + x}.p_{1} ightbrack^{2}}{\left( \frac{1 - x}{1 + x} ight).P_{1}} = \frac{4x^{2}}{(1 - x)}\ \frac{p_{1}}{(1 + x)}$
$Z\ \ \ \ \ \ \ \ \ \ \  ightleftharpoons \ \ \ \ \ \ \ \ \ P\  + \ \ \ \mathcal{Q}$
1 0 0
At equilibrium $1 - x$ $x$ $x$
Similarly, $K_{p2} = \frac{x^{2}}{(1 - x)}\ \frac{p_{2}}{(1 + x)}$
$\frac{K_{p1}}{K_{p2}} = \frac{1}{9}$
So, $\frac{4 	imes p_{1}}{p_{2}} = \frac{1}{9}$
$\Rightarrow$ $\frac{p_{1}}{p_{2}} = \frac{1}{36}$

Q. 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 pressure at these equilibria is

Explanation

Reviewed by Dr. Abubacker

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