Question

Given that ${{N}_{2(g)}}+3{{H}_{2(g)}}\to 2N{{H}_{3(g)}};{{\Delta }_{r}}H=-92kJ$ , the standard molar enthalpy of formation in $kJ\,mo{{l}^{-1}}$ of $N{{H}_{3(g)}}$ is:
A. $-92$
B. $-46$
C. $-92$
D. $-46$

Answer 1

Hint: Haber process is defined as the artificial nitrogen fixation process, which is used for the production of ammonia. In this process, atmospheric nitrogen is converted into ammonia by a reaction with hydrogen using metal catalysts under very high temperature and pressure.

Formula used:${{\Delta }_{f}}H{}^\circ =\dfrac{1}{2}{{\Delta }_{r}}H$
where, ${{\Delta }_{f}}H{}^\circ $ is the standard molar enthalpy of formation and ${{\Delta }_{r}}H$ is the enthalpy of reaction.

Complete step by step answer:
The conversion from atmospheric nitrogen to ammonia is conducted at a pressure above $10MPa$ and a temperature between $400{}^\circ C$ and $500{}^\circ C$ .
In nitrogen, atoms are held together by strong triple bonds and therefore, it is very unreactive.
This reaction is known as Haber’s process in which production of ammonia takes place. In this process, atmospheric nitrogen is converted into ammonia using metal catalysts under high temperature and pressure. It is an artificial nitrogen fixation process.
Here, the reaction is given as follows:
 ${{N}_{2(g)}}+3{{H}_{2(g)}}\to 2N{{H}_{3(g)}}$
 ${{\Delta }_{r}}H=-92kJ$
where, ${{\Delta }_{r}}H$ is the standard molar enthalpy of formation.
 ${{N}_{2(g)}}+{{H}_{2(g)}}\to N{{H}_{3(g)}}$
Now, the standard molar enthalpy of formation $=\dfrac{1}{2}{{\Delta }_{r}}H$
Now, substituting the values in the above expression, we get,
Standard molar enthalpy of $N{{H}_{3}}=\dfrac{-92.4}{2}$
On further solving, we get,
Standard molar enthalpy of $N{{H}_{3}}=-46.2\,kJ\,mo{{l}^{-1}}$

So, the correct answer is Option D.

Note: The standard molar enthalpy for a reaction takes place under standard state conditions at constant temperature.
The Haber process is used to conduct nitrogen into ammonia using metal catalysts at high temperature and pressure. Nitrogen has a very strong triple bond and is very unreactive. Therefore, they require a strong catalyst that accelerate the scission of its triple bond.

The standard molar enthalpy of formation of a compound is defined as the change in enthalpy when one mole of substance is formed from its pure elements under the standard conditions at pressure of $1$ bar at constant temperature.
It is denoted with symbol ${{\Delta }_{f}}H{}^\circ $