Question

In Haber’s process $30$ litres of dihydrogen and $30$ litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture under these conditions in the end?
(a) $10$ litres ammonia, $25$ litres nitrogen, $15$ litres hydrogen
(b) $20$ litres ammonia, $10$ litres nitrogen, $30$ litres hydrogen
(c) $20$ litres ammonia, $25$ litres nitrogen, $15$ litres hydrogen
(d) $20$ litres ammonia, $20$ litres nitrogen, $20$ litres hydrogen

Answer 1

This reaction involves the knowledge of the law of constant composition for you to be able to solve this question easily. Also, remember that the volume occupied by $1$ mole of a gas at STP is $22.4$ litres.

Complete Solution :
Let us first look at the law of constant proportion, a law crucial to the solution of this question.
In chemistry, the law of definite proportion, sometimes called Proust's law, or law of constant composition states that a given chemical compound always contains its component elements in fixed ratio and does not depend on its source and method of preparation. For example, in a nitrogen dioxide ($N{{O}_{2}}$) molecule, the ratio of the number of nitrogen and oxygen atoms is $1 : 2$ but the mass ratio is $14 : 32$ (or $7 : 16$).

- Now, let us look at the Haber’s process of Ammonia production and observe the reaction that takes place in the same.
The Haber Process combines nitrogen from the air with hydrogen derived mainly from natural gas (methane) into ammonia. The reaction is reversible and the production of ammonia is exothermic.
${{N}_{2}}\,+\,3{{H}_{2}}\,\xrightarrow[KOH]{Fe}\,\,\,2N{{H}_{3}}$
The catalyst is actually slightly more complicated than pure iron. It has potassium hydroxide added to it as a promoter - a substance that increases its efficiency.

- Now, let us apply this knowledge to the given question.
$\begin{matrix} {{N}_{2}} \\\ 1\text{ volume} \\\ 10\text{ litres} \\\ \end{matrix}\text{ }\begin{matrix} \+ \\\ {} \\\ {} \\\ \end{matrix}\text{ }\begin{matrix} 3{{H}_{2}} \\\ \text{3 volume} \\\ 30\text{ litres} \\\ \end{matrix}\text{ }\begin{matrix} \to \\\ {} \\\ {} \\\ \end{matrix}\text{ }\begin{matrix} 2N{{H}_{3}} \\\ \text{2 volume} \\\ \text{20 litres} \\\ \end{matrix}$

It is given that only 50% of the expected product is formed, hence only $10$ litres of $N{{H}_{3}}$ is formed.
${{N}_{2}}\operatorname{used} = 5\operatorname{litres}\operatorname{left} = 30 - 5 = 25\operatorname{litres}$
${{H}_{2}}\operatorname{used} = 15\operatorname{litres}\operatorname{left} = 30 - 15 = 15\operatorname{litres}$
So, the correct answer is “Option A”.

Note: Keep the expected product fraction in mind while solving the given question so as to ensure that you obtain the correct number of products evolved as a result of this procedure. Also, be very wary of not getting confused between the laws of constant and multiple proportions.