For a reaction at equilibrium $\mathrm{A}\left(\mathrm{g}\right)\rightleftharpoons \mathrm{B}\left(\mathrm{g}\right)+\frac{1}{2}\mathrm{C}\left(\mathrm{g}\right)$ the relation between dissociation constant (K),
degree of dissociation ($\mathrm{\alpha }$) and equilibrium pressure (p) is given by :

$\begin{array}{ccc}\mathbf{A}\left(\mathrm{g}\right)& \rightleftharpoons & \mathrm{B}\left(\mathrm{g}\right)+\frac{1}{2}\mathrm{C}\left(\mathrm{g}\right)\\ \text{ Initial : }{\mathbf{P}}_{\mathrm{i}}& \mathbf{0}& 0\\ \text{ At eq.: }{\mathrm{P}}_{\mathrm{i}}(1-\mathrm{\alpha })& {\mathrm{P}}_{\mathrm{i}}\cdot \mathrm{\alpha }& {\mathrm{P}}_{\mathrm{i}}\frac{\mathrm{\alpha }}{2}\end{array}$

$$\begin{gathered} \text{ Now, equilibrium pressure (p), } \\ P=P_i\times \left(1+\frac{\alpha }{2}\right) \\ \therefore {\mathrm{P}}_{\mathrm{A}}=\left(\frac{1-\alpha }{1+\frac{\alpha }{2}}\right)\mathrm{P} \\ {\mathrm{P}}_{\mathrm{B}}=\left(\frac{\mathrm{\alpha }}{1+\frac{\mathrm{\alpha }}{2}}\right)\mathrm{P} \\ {\mathrm{P}}_{\mathrm{C}}=\left(\frac{\frac{\mathrm{\alpha }}{2}}{1+\frac{\mathrm{\alpha }}{2}}\right)\mathrm{P} \\ \begin{array}{r}\therefore \mathrm{K}=\frac{{\mathrm{P}}_{\mathrm{c}}^{\frac{1}{2}}\times {\mathrm{P}}_{\mathrm{B}}}{{\mathrm{P}}_{\mathrm{A}}}\\ \mathrm{K}=\frac{{\mathrm{\alpha }}^{\frac{3}{2}}{\mathrm{p}}^{\frac{1}{2}}}{(2+\mathrm{\alpha }{)}^{\frac{1}{2}}(1-\mathrm{\alpha })}\end{array} \end{gathered}$$