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)

10 L ammonia, 25 L nitrogen, 15 L hydrogen

Answer 2

The correct answer is Option D) 10 L ammonia, 25 L nitrogen, 15 L hydrogen

In Haber process 30L of dihydrogen and 30L dinitrogen were taken for reaction which yielded only 50% of the expected product.

The Haber process is the technique used to make ammonia. Ammonia is created when hydrogen and nitrogen react. Extremely exothermic in nature, this reaction produces heat as it proceeds. The result is

$N_2 + 3H_2 \rightarrow 2NH_3$
$_{10L}^{1V} \, \, \, \, _{30L}^{3V} \, \, \, \, \, _{20L}^{2V}$

Two moles of ammonia are produced when one mole of nitrogen combines with three moles of hydrogen.
As only 50% of the expected product is formed,
hence only 10 L of $NH_3$ is formed.
the composition of gaseous mixture under the aforesaid condition in the end will be-
$H_2 = 30 - 15 = 15 \, L$
$N_2 = 30 - 5 = 25 \, L$
$NH_3 = 10 \, L$

If the reaction contains a coefficient, we must subtract the assumed quantity multiplied by the reactant's coefficient from the supplied amount of the reactant to get the amount of reactant that is still present in the reaction.

**Discover more from this chapter: **Equilibrium