At certain temperature $28\mathrm{gm}$ of ${\mathrm{N}}_2$ and $6\mathrm{gm}$ of ${\mathrm{H}}_2$ are taken in a 2 lit vessel in Haber's process if $17\mathrm{gm}$ of ${\mathrm{NH}}_3$ is formed at equilibrium, the concentration of ${\mathrm{N}}_2$ at equilibrium is

 At equilibrium the reaction will be:

Suppose the value of x is 0.5.

At equilibrium the amount of $N_2$ will be = 1 – 0.5 = 0.5

The amount of $H_2$ will be = 3 – 1.5 = 1.5

The amount of $N{H}_3$ will be = $2\times 0.5=1$

The concentration of ${\mathrm{N}}_2$ at equilibrium is lower than the concentration of ${\mathrm{H}}_2$.

Therefore, the correct option is (C).