Question

In Haber’s process 30 liters of dihydrogen and 30 liters of dinitrogen were taken for reaction which yields only $50\%$ of the expected product. What will be the composition of gaseous mixture under the above said conditions in the end?

Answer 1

For this we must have the knowledge of reaction occurring between nitrogen and hydrogen in the formation of ammonia using Haber’s process. $50\%$ is the amount of product formed is given to us with respect to initial concentration.

Complete step by step solution:
The formation of ammonia through Haber’s process follows the reaction:
${{\text{N}}_2} + 3{{\text{H}}_2} \rightleftharpoons 2{\text{N}}{{\text{H}}_3}$
One mole of nitrogen needs three moles of hydrogen to form two mole of ammonia. Accordingly 30 liter of nitrogen will require 90 liters of hydrogen, but we have given only 30 liter of hydrogen. So hydrogen will act as a limiting reagent and product formation will occur according to the amount of hydrogen. Now 3 moles of hydrogen forms 2 moles of ammonia, according to this 1 mole of hydrogen will give $\dfrac{2}{3}$ moles of ammonia or 1 liter of hydrogen will give $\dfrac{2}{3}$ liters of ammonia.
So 30 liter of hydrogen will give us $\dfrac{2}{3} \times 30 = 20$ liters of ammonia. $50\%$ Of 20 will be 10 liters. Let us assume that x amount of nitrogen gets dissociated; two moles of ammonia will form. The reaction will follows as:
${{\text{N}}_2} + 3{{\text{H}}_2} \rightleftharpoons 2{\text{N}}{{\text{H}}_3}$
$30{\text{ 30 0}}$
${\text{30 - x 30 - 3x 2x}}$
According to the above equation the volume of ammonia is $2{\text{x}}$. Comparing both the equation we will get,
$2{\text{x = 10 liters}}$
$\Rightarrow {\text{x = 5 liters}}$
The volume of dinitrogen at equilibrium will be
${\text{30}} - {\text{x = 30}} - {\text{5 liters = 25 liters}}$
The volume of dinitrogen at equilibrium will be
${\text{30}} - 3{\text{x = 30}} - {\text{3}} \times {\text{5 liters = 15 liters}}$
. The volume of ammonia at equilibrium will be
${\text{2x = 2}} \times {\text{5 liters = 10 liters}}$

Note:
The Haber’s process is also known as Haber-Bosch process. It is the most common method used for industrial preparation of ammonia. The catalyst required for this reaction is iron oxide and the promoters are aluminum oxide, potassium oxide etc. the reaction is favorable under high temperature and pressure.