Question

When one mole of $P\left(g\right)$ is introduced in one litre closed rigid vessel maintained at constant temperature, the following equilibrium is established.
$P\left(g\right)\rightleftharpoons Q\left(g\right)+R\left(g\right)⟹K\left(C_1\right)$
$R\left(g\right)\rightleftharpoons S\left(g\right)+Q\left(g\right)⟹K\left(C_2\right)$
If the final equilibrium pressure is $2$ times of the initial pressure, then the value of $\frac{K\left(C_2\right)}{K\left(C_1\right)}$is ________.

[Ratio of the equilibrium concentration of $Q$ and $R$ is $5$]

Answer 1

For the first equilibrium :

	P	Q	R
Initial Moles	1	0	0
Equilibrium Moles	$1-x$	$x+y$	$x-y$

For the second equilibrium :

	P	Q	R
Initial Moles	$x$	0	$x$
Equilibrium Moles	$x-y$	$y$	$x+y$

$\frac{\left[Q\right]}{\left[R\right]}=5$

$\frac{x+y}{x-y}=5$
$x+y=5x-5y$
$x=1.5y$......(1)
As the final equilibrium pressure is $2$ times of the initial pressure, the final equilibrium number of moles is twice the initial number of moles.
Total number of moles $=1-x+x+y+x-y+y=1+x+y=2$
$1+1.5y+y=2$
$2.5y=1$
$y=0.4$

$K\left(C_1\right)=\frac{\left[Q\right]\left[R\right]}{\left[P\right]}$

$K\left(C_2\right)=\frac{\left[S\right]\left[Q\right]}{\left[R\right]}$

$\frac{K\left(C_2\right)}{K\left(C_1\right)}=\frac{\left[S\right]\left[Q\right]\left[P\right]}{\left[R\right]\left[Q\right]\left[R\right]}=\frac{\left[S\right]\left[P\right]}{[R{]}^2}$

Substitute values in the above expression.

$\frac{K\left(C_2\right)}{K\left(C_1\right)}=\frac{y(1-x)}{(x-y{)}^2}$

Substitute $x=1.5y$ in the above expression.

$\frac{K\left(C_2\right)}{K\left(C_1\right)}=\frac{y(1-1.5y)}{\frac{y^2}{4}}=4\left(\frac{1-1.5y}{y}\right)$

Substitute $y=0.4$ in the above expression.

$\frac{K\left(C_2\right)}{K\left(C_1\right)}=4\left(\frac{1-1.5\left(0.4\right)}{0.4}\right)=4$