The value of K꜀ is 64 at 800 K for the reaction
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The value of K꜀ for the following reaction is:
NH₃(g) ⇌ ½ N₂(g) + ³⁄₂ H₂(g)

A(g) ->B(g) + $\frac{1}{2}$ (g)

Initial moles             n                         0             0

Eqb. moles               n(1 – a)   $\frac{n\alpha }{2}$           n_a            

total moles =   $n\left(1+\frac{\alpha }{2}\right)$

Eqb. pressure   $\frac{\left(1-\alpha \right)p}{1+\frac{\alpha }{2}}$ $\frac{\alpha p}{1+\frac{\alpha }{2}}$ $\frac{\left(\frac{\alpha }{2}\right)p}{1+\frac{\alpha }{2}}$

$\frac{K_p=\frac{\alpha p}{\left(1+\frac{\alpha }{2}\right)}\times {\left[\frac{\alpha p}{\left(2+\alpha \right)}\right]}^{\frac{1}{2}}}{\frac{\left(1+\alpha \right)p}{1+\frac{\alpha }{2}}}$

$K_p=\frac{{\alpha }^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{p}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{{\left(2+\alpha \right)}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\left(1-\alpha \right)}$