For the reaction : N₂(g) + 3H₂(g) ? 2NH₃(g)

Equilibrium constant K_c = $\frac{{\left[{\mathbf{N}\mathbf{H}}_3\mathbf{}\right]}^2}{\left[{\mathbf{N}}_2\mathbf{}\right]{\left[{\mathbf{H}}_3\mathbf{}\right]}^3\mathbf{ }}$

Some reactions are written below in Column I and their equilibrium constants in terms of K_c are written in Column II. Match the following reactions with the corresponding equilibrium constant.

Column I (Reaction)	Column II (Equilibrium constant)
(i) 2N2(g) + 6H2(g) ⇌ 4NH3(g)	(a) 2Kc
(ii) 2NH3(g) ⇌N2(g) + 3H2(g)	(b) Kc
(iii) N2(g) +  2H2(g) ⇌ NH3(g)	(c)
	(d) K2c

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)}$