Question

When 1 mole of A(g) is introduced in a closed rigid 1 litre vessel maintained at constant temperature the following equilibria are established

$A\left(g\right)\rightleftharpoons B\left(g\right)+C\left(g\right):K_{C_1}$
$C\left(g\right)\rightleftharpoons D\left(g\right)+B\left(g\right):K_{C_2}$

The pressure at equilibrium is twice the initial pressure. Calculate the value of $\frac{K_{C_2}}{K_{C_1}}$ if $\frac{[C{]}_{eq}}{[B{]}_{eq}}=\frac{1}{5}$.

Answer 1

$\underset{P_0-x}{A}\rightleftharpoons \underset{x+y}{B}+\underset{x-y}{C}$

$\underset{x-y}{C}\rightleftharpoons \underset{y}{D}+\underset{y+x}{B}$

Given $\frac{x-y}{x+y}=\frac{1}{5}\&2P_0=P_0-x+x+y+x-y+y$

So $x+y=P_0$

$x+y=P_0,\frac{x-y}{x+y}=\frac{1}{5}$

$\Rightarrow x-y=\frac{P_0}{5}$

$x=\frac{3P_0}{5},y=\frac{2P_0}{5}$

$\frac{K{c}_2}{K{c}_1}=\frac{(y+x)\left(y\right)}{(x-y)}\times \frac{(P_0-x)}{(x+y)(x-y)}$

$\Rightarrow \frac{y(P_0-x)}{(x-y{)}^2}=\frac{y(P_0-x)}{(\frac{P_0}{5}{)}^2}$

$\Rightarrow \frac{\left(\frac{2P_0}{5}\right)\left(\frac{2P_0}{5}\right)}{(\frac{P_0}{5}{)}^2}=4$