$56\mathrm{gm}$ of nitrogen and $4\mathrm{gm}$ of hydrogen are taken in a closed vessel of 5 litre capacity and suitable conditions are provided for the reaction. When equilibrium is reached it is found that $0.5$ moles of nitrogen is used up. The equilibrium concentration of ammonia is ...mole/lit

${\mathrm{N}}_{2( \mathrm{g})}+3{\mathrm{H}}_{2( \mathrm{g})}\leftrightharpoons 2{\mathrm{NH}}_{3( \mathrm{g})}$

Given that when the equilibrium is reached, it is found that a half mole of ${\mathrm{N}}_2$ is used up, i.e.,

Amount of ${\mathrm{N}}_2$ used $=0.5$ moles

$\therefore$ Amount of ${\mathrm{NH}}_3$ produced by 1 mole of ${\mathrm{N}}_2=2$ moles

$\Rightarrow$ Amount of ${\mathrm{NH}}_3$ produced by $0.5$ mole of ${\mathrm{N}}_2=1$ mole

Volume of vessel $=5 \mathrm{L}$

Equilibrium concentration of ${\mathrm{NH}}_3=\frac{\text{ No. of moles }}{\text{ Volume }}=\frac{1}{5}=0.2$moles

Hence the correct option is (D) $0.2$.