The degree of dissociation of ${\mathrm{SO}}_3$ is $\alpha$ at equilibrium pressure ${\mathrm{P}}_0$.

$K_p\text{ for }2{\mathrm{SO}}_3\left(g\right)\rightleftharpoons 2{\mathrm{SO}}_2\left(g\right)+{\mathrm{O}}_2\left(g\right)$ is:

In the following reaction we have,

$2{\mathrm{SO}}_3\rightleftharpoons 2{\mathrm{SO}}_2+{\mathrm{O}}_2$

$a(1-\alpha )a\alpha \frac{a\alpha }{2}$

$\begin{array}{l}{n}_T=a\left(1+\frac{\alpha }{2}\right)\\ X_{{\mathrm{SO}}_3}=\frac{1-\alpha }{1+\frac{\alpha }{2}};X_{{\mathrm{SO}}_2}=\frac{\alpha }{1+\frac{\alpha }{2}}\\ X_{{\mathrm{O}}_2}=\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\end{array}$

In the following reaction we have,

$K_P=\frac{{\left(\frac{\alpha }{1+\frac{\alpha }{2}}\right)}^2{P}_0^2\times \left(\frac{\frac{\alpha }{2}}{1+\frac{\alpha }{2}}\right)\times P_0}{\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right)\times P_0^2}$

$K_P=\frac{P_0{\alpha }^3}{(2+\alpha )(1-\alpha {)}^2}$