Question

When $1mole$ of $A\left(g\right)$ is introduced in a closed rigid $1liter$ vessel maintained at a 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 $\frac{[C{]}_q}{[B{]}_{eq}}=\frac{1}{5}$

• A. $4$
• B. $0.5$
• C. $2$
• D. $0.16$

Answer 1

The correct option is D $0.16$

	A	B	C
Initial moles	1	0	0
Equilibrium moles	$1-x$	$x+y$	$x-y$

	C	D	B
Initial moles	0	0	0
Equilibrium moles	$x-y$	$y$	$x+y$

Total number of moles $=1-x+x+y+x-y+y=2$
$1+x+y=2$
$x+y=1$
$x=1-y$......(1)
$\frac{\left[C\right]}{\left[B\right]}=\frac{1}{5}$
$\frac{x-y}{x+y}=\frac{1}{5}$
$x-y=0.2$......(2)
Substitute equation (1) in equation (2)
$1-y-y=0.2$
$2y=0.8$
$y=0.4$......(3)
Substitute equation (3) in equation (1)
$x=1-0.4=0.6$
$K_{C1}=\frac{(x+y)(x-y)}{(1-x)}=\frac{1}{5(1-x)}$
$K_{C2}=\frac{y(x+y)}{(x-y)}=\frac{y}{5}$
$\frac{K_{C2}}{K_{C1}}=\frac{\frac{y}{5}}{\frac{1}{5(1-x)}}=y(1-x)=0.4\times 0.4=0.16$