Question

The degree of dissociation of $S{O}_3$ is $\alpha$ at equilibrium pressure $P_0$
$K_{p}for2S{O}_3\left(g\right)\leftrightharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$ is:

• A. $\left[\frac{\left(P_0{\alpha }^3\right)}{2(1-\alpha {)}^3}\right]$
• B. $\left[\frac{\left(P_0{\alpha }^3\right)}{(2+\alpha )(1-\alpha {)}^2}\right]$
• C. $\left[\frac{P_0{\alpha }^2}{2(1-\alpha {)}^2}\right]$
• D. None of these

Answer 1

The correct option is B $\left[\frac{\left(P_0{\alpha }^3\right)}{(2+\alpha )(1-\alpha {)}^2}\right]$

Given,
Degree of dissociation of $S{O}_3$ is $\alpha .$
Equilibrium pressure = $P_0$
The given equilibrium is,
$2S{O}_3\leftrightharpoons 2S{O}_2\left(g\right)+O_2\left(g\right)$
at equilibrium $(2-2\alpha )2\alpha \alpha$
Total moles at equilibrium $=(2-2\alpha )+2\alpha +\alpha =2+\alpha$
$P_{S{O}_3}=\left(\frac{2-2\alpha }{2+\alpha }\right)P_0{P}_{S{O}_2}=\left(\frac{2\alpha }{2+\alpha }\right)P_0{P}_{O_2}=\left(\frac{\alpha }{2+\alpha }\right)P_0$

$K_p$ for the reaction,
$K_p=\frac{{\left[P_{S{O}_2}\right]}^2\left[P_{O_2}\right]}{[P_{S{O}_3}{]}^2}$
On substituting the values
$K_P=\frac{{\left[\left(\frac{2\alpha }{2+\alpha }\right)P_0\right]}^2\left[\left(\frac{\alpha }{2+\alpha }\right)P_0\right]}{{\left[\left(\frac{2-2\alpha }{2+\alpha }\right)P_0\right]}^2}{K}_p=\frac{P_0{\alpha }^3}{(2+\alpha )(1-\alpha {)}^2}$