Question

In Haber's process, the equivalent weight of nitrogen gas is:

• A. $M$
• B. $\frac{M}{3}$
• C. $\frac{M}{6}$
• D. $3M$

Answer 1

The correct option is C $\frac{M}{6}$

$N_2+3H_2\to 2N{H}_3$
Here, the oxidation state of $N$ changes from 0 to -3 when nitrogen gas becomes ammonia. Since there are two nitrogen atoms in the nitrogen gas molecule, the total change in oxidation number is 6. $\therefore$, the equivalent weight = $\frac{M}{6}$