Question

Given are the bond enthalpies: ${{\varepsilon }_{\left( N=N \right)}}=945kJmo{{l}^{-1}}$ , ${{\varepsilon }_{\left( H-H \right)}}=436kJmo{{l}^{-1}}$ and ${{\varepsilon }_{\left( N-H \right)}}=391kJmo{{l}^{-1}}$ . The enthalpy change of the reaction
\[{{N}_{2\left( g \right)}}+3{{H}_{2\left( g \right)}}\to 2N{{H}_{3\left( g \right)}}\] is
(A) $-93kJmo{{l}^{-1}}$
(B) $-89kJmo{{l}^{-1}}$
(C) $-105kJmo{{l}^{-1}}$
(D) $105kJmo{{l}^{-1}}$

Answer 1

Hint: The general knowledge about bond dissociation energy, enthalpy of formation and Hess’s law, we can solve the given illustration easily.
We have a relationship that relates bond enthalpy and enthalpy of formation; we can solve by that method too.

Complete step by step solution:
Let us firstly understand the given terms;
Bond dissociation enthalpy (or simply bond enthalpy/energy)-
It is a measure of strength of a chemical bond. It can also be defined as the standard enthalpy change when the bond (A-B) is cleaved by homolysis to give fragments of A and B.
Enthalpy of formation (or reaction)-
It is the change in the enthalpy of reaction during the formation of 1 mole of the substance from its constituent elements, with all the contributing substances in their standard states.
Hess’s law-
The law states that the change in enthalpy for a reaction is the same whether the reaction takes place in one step or a series of steps regardless of path by which reaction occurs.
Now, we can describe the required equations in an illustration and solve it as follows,
Given that,
Bond enthalpy of ${{H}_{2}}$ = 436 kJ/mol
Bond enthalpy of ${{N}_{2}}$ = 945 kJ/mol
Bond enthalpy of $N-H$ = 391 kJ/mol
The reaction is described as follows;
\[{{N}_{2\left( g \right)}}+3{{H}_{2\left( g \right)}}\to 2N{{H}_{3\left( g \right)}}\]
This can be divided into simpler steps as,
Step (I)- Dissociation of hydrogen and nitrogen molecules into their respective atoms.
$\begin{align}
& 3{{H}_{2}}\to 6H \\
& {{N}_{2}}\to 2N \\
\end{align}$
Step (II)- Combination of these atoms to form $N{{H}_{3}}$ .
$6H+2N\to 2N{{H}_{3}}$
Since, the first step is an endothermic process thus, $\Delta {{H}_{1}}>0$ . Whereas, the next step releases the energy thus, $\Delta {{H}_{2}}<0$ .
So,
$\begin{align}
& \Delta {{H}_{1}}=\Delta {{H}_{N-N}}+\left( 3\times \Delta {{H}_{H-H}} \right) \\
& \Delta {{H}_{1}}=945+\left( 3\times 436 \right)=2253kJ/mol \\
\end{align}$
And,
$\begin{align}
& \Delta {{H}_{2}}=-2\times 3\times \Delta {{H}_{N-H}} \\
& \Delta {{H}_{2}}=-2346kJ/mol \\
\end{align}$
Thus, applying Hess’s law we get,
$\begin{align}
& \Delta H'=\Delta {{H}_{1}}+\Delta {{H}_{2}} \\
& \Delta H'=2253-2346=-93kJ/mol \\
\end{align}$

Therefore option (A) is correct.

Note: An alternative method to solve the same illustration can be the direct relationship of bond dissociation enthalpy and enthalpy of reaction as stated,
$\Delta {{H}_{reaction}}=\sum{{{\left( B.E \right)}_{reac\tan t}}-\sum{{{\left( B.E \right)}_{product}}}}$ .
Solving by this method will give the same answer.