Question

For the reaction:
$AB\left(g\right)\rightleftharpoons A\left(g\right)+B\left(g\right)$
$AB$ is $33$% dissociated at a total pressure of $p$. Then $\frac{p}{K_p}$ will be equal to :

Answer 1

Let $P_i=$ initial pressure of AB. Also, let x be the change in pressure of AB to reach equilibrium. The equilibrium pressure of AB will be
$P_i-x$. The equilibrium pressures of A and B will be x and x respectively.
The total pressure is $p=(P_i-x)+x+x$
$p=(P_i+x)$.....(1)
AB is 33% dissociated at a total pressure of p
So $x=\frac{\text{33}}{\text{100}}\times P_i=0.33P_i$....(2)
Substitute this in equation (1)
$p=(P_i+0.33P_i)$
$p=1.33P_i$
$P_i=\frac{p}{1.33}$
Substitute this in equation (2)
$x=0.33P_i$
$x=0.33\times \frac{p}{1.33}$
$x=0.248p$
The equilibrium pressures of A and B will be 0.248 p and 0.248 p respectively.
The equilibrium pressure of AB will be $P_i-x=\frac{p}{1.33}-0.248p=0.504p$
$K_p=\frac{P_A{P}_B}{P_{AB}}$
$K_p=\frac{0.248p\times 0.248p}{0.504p}$
$K_p=\frac{0.248p\times 0.248p}{0.504p}$
$K_p=0.122p$
$\frac{p}{K_p}=\frac{1}{0.122}=8$