Question

For the gaseous reactions:
$A\left(g\right)\rightleftharpoons 3B\left(g\right):K_{P_1}andP\left(g\right)\rightleftharpoons 2Q\left(g\right)+R\left(g\right);K_{P_2}$
Occuring in two different vessels at the same temperature, the degree of dissociation of A and P are same when the total equilibrium pressure of gases in both vessels are same. If $K_{P_1}:K_{P_2}$ is $X:4,$ then value of "X" is

Answer 1

$A\left(g\right)\rightleftharpoons 3B\left(g\right);K_{P_1}$
$101-\alpha 3\alpha Atequilibrium{K}_{P_1}=\frac{(3\alpha {)}^3}{(1-\alpha )}\frac{P^2}{(1+2\alpha {)}^2}P\left(g\right)\rightleftharpoons 2Q\left(g\right)+R\left(g\right);K_{P_2}$
$1-\alpha 2\alpha \alpha atequilibrium{K}_{P_2}=\frac{(2\alpha {)}^2\alpha }{(1-\alpha )}\frac{P^2}{(1+2\alpha {)}^2}\frac{K_{P_1}}{K_{P_2}}=\frac{X}{4}=\frac{\frac{3^3{\alpha }^3{P}^2}{(1-\alpha )(1+2a{)}^2}}{\frac{4{\alpha }^3{P}^2}{(1-\alpha )(1+2\alpha {)}^2}}\therefore X=27$