Question

In the 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 above condition in the end?
A. 20 L ammonia, 20 L nitrogen, 20 L hydrogen
B. 10 L ammonia, 25 L nitrogen, 15 L hydrogen
C. 20 L ammonia, 10 L nitrogen, 30 L hydrogen
D. 20 L ammonia, 25 L nitrogen, 15 L hydrogen

Answer 1

The concept of stoichiometry is to be used. Form the equation of Haber's process and analyse the concentrations before and at the time of equilibrium. Form the necessary equation with the condition given in the question and solve.

Complete answer:
In order to answer the question, we need to be clear with the concepts of stoichiometry. Before that, we know that in Haber's Process, nitrogen mixes with hydrogen to yield ammonia gas and the reaction is represented as ${{N}_{2}}+{{H}_{2}}\rightleftharpoons N{{H}_{3}}$ . However this equation is not balanced and if we use an unbalanced equation, then we will get the answer wrong. Let us try to balance the equation. If we add $3{{H}_{2}}$ on the reactant side and $2N{{H}_{3}}$ on the product side, then we can observe that the atoms of each element on both sides are equal, which in turn balances the reaction. The final equation is ${{N}_{2}}+3{{H}_{2}}\rightleftharpoons 2N{{H}_{3}}$ Now, we proceed with the question.
Initially, we have 30 L of ${{N}_{2}}$ and 30L of ${{H}_{2}}$ and no concentration of ammonia, as the reaction is not started yet. Now, we will see the concentrations at the time when equilibrium is reached. Let ‘x’ be the volume that is consumed during the reaction. Now we will write the concentrations based on the number of moles of the substances. The amount of nitrogen remaining is 30-x as there is one mole. The amount of hydrogen left is 30-3x as there are 3 moles of hydrogen present. And on the product side we have 2x the amount of ammonia left because there are 2 moles of it present.
However, according to the question we have,$2x=(\dfrac{50}{100}\times 20\,)litre$ which gives us x as 5L.
So, the volume of nitrogen left is 30-5= 25 litre.
The volume of hydrogen left is $30-(3\times 5)$ which is 15 litre.
The volume of ammonia left is $2\times 5$ , which is 10 litres.
So, we get option B as the correct answer for the question.

Note: It is to be noted that temperature plays a role in the equilibrium . If the temperature were increased, the equilibrium would have shifted to the left, based on Le Chatelier's principle and then we will get different volumes of product and reactant left at equilibrium.