Question

Equivalent weight of $N_2$ in the reaction
$N_2+3H_2⟶2N{H}_3$ is:

• A. $\frac{28}{6}$
• B. 28
• C. 28/2
• D. $\frac{28}{3}$

Answer 1

The correct option is A $\frac{28}{6}$

The oxidation number of $N$ in $N_2$ is 0.
The oxidation number of $N$ in $N{H}_3$ is -3.
So change in oxidation state for one $N$ atom is 3.
So, for $N_2$ it will be 6.
Hence, equivalent weight $=\frac{M}{6}=\frac{28}{6}$