The gaseous reaction, ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3,$ takes place at ${450}^0\mathrm{C}$. 1 mole of Nitrogen and 2 mole of Hydrogen are mixed in a 1 litre vessel and 1 mole of Ammonia is formed at equilibrium. Then ${\mathrm{K}}_{\mathrm{c}}$ for the above reaction is

$\mathrm{V}=1 \mathrm{lit};$

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

1 mole of ${\mathrm{N}}_2$ gives 2 moles of ${\mathrm{NH}}_3$

'${\mathrm{X}}_1$' moles of ${\mathrm{N}}_2$ gives 1 mole of ${\mathrm{NH}}_3$

Number of moles of ${\mathrm{N}}_2$ reacted = 0.5

1 mole of ${\mathrm{N}}_2$ reacts with 3 moles of ${\mathrm{H}}_2$

0.5 moles of ${\mathrm{N}}_2$ reacts with '${\mathrm{X}}_2$' moles of ${\mathrm{H}}_2$

Number of moles of ${\mathrm{H}}_2$ reacted = 1.5

Stoichiometry
Initial moles	1	2		0
Moles at equilibrium	(1 - 0.5)	(2 - 1.5)		1
Equilibrium concentration	0.5	0.5		1

$$\begin{gathered} {\mathrm{K}}_{\mathrm{c}}=\frac{{\left[{\mathrm{NH}}_3\right]}^2}{\left[{\mathrm{N}}_2\right]{\left[{\mathrm{H}}_2\right]}^3}; \\ {\mathrm{K}}_{\mathrm{c}}=\frac{1^2}{\left(0.5\right){\left(0.5\right)}^3}=16; \end{gathered}$$