${\mathrm{N}}_2\left(\mathrm{g}\right)+3{\mathrm{H}}_2\left(\mathrm{g}\right)\rightleftharpoons 2{\mathrm{NH}}_3\left(\mathrm{g}\right)$  For the reaction initially the mole ratio was 1 : 3 of N₂, : H₂. At equilibrium 50% of each has
reacted. If the equilibrium pressure is -Q the partial pressure of NH₃, at equilibrium is

               ${\mathrm{N}}_2 + 3{\mathrm{H}}_2 \rightleftharpoons 2{\mathrm{NH}}_3$ 

t=0 ;       a         3a            0

t_eq. ;  (a)(1−α)   3a(1−α)    2aα

Given 50% reacted α=1/2 , n=a-a$\alpha$+3a-3a$\alpha$+2a$\alpha$=2a(2-$\alpha$)

${\mathrm{P}}_{{\mathrm{NH}}_3}$​=(mole fraction)×(total pressure)

${\mathrm{P}}_{{\mathrm{NH}}_3}=\frac{2\mathrm{a\alpha }}{2\mathrm{a}(2-\mathrm{\alpha })}\mathrm{P}=\frac{\mathrm{P}}{3}$