Question

In Haber process $30\,L$ of dihydrogen and $30\,L$ of dinitrogen were taken for reaction which yielded only $50\%$ of the expected product. What will be the composition of gaseous mixture under the aforesaid condition in the end?

• A. 20 L ammonia, 10 L nitrogen, 30 L hydrogen
• B. 20 L ammonia, 25 L nitrogen, 15 L hydrogen
• C. 20 L ammonia, 20 L nitrogen, 20 L hydrogen
• D. 10 L ammonia, 25 L nitrogen, 15 L hydrogen

Answer 1

Answer: (D)

(d) $N_2 + 3H_2 \rightarrow 2NH_3$
$_{10L}^{1V} \, \, \, \, _{30L}^{3V} \, \, \, \, \, _{20L}^{2V}$
As only 50% of the expected product is formed,
hence only 10 L of $NH_3$ is formed.
Thus, for the production of 10 L of $ NH_3 , 5 L \, of \, N_2$ and 15 L of $H_2$ are used and composition of gaseous mixture under the aforesaid condition in the end is
$H_2 = 30 - 15 = 15 \, L$
$N_2 = 30 - 5 = 25 \, L$
$NH_3 = 10 \, L$