Question

Bond dissociation enthalpies of ${{H}_{2}}(g)$ and ${{N}_{2}}(g)$ are $436.0\text{ kJmo}{{\text{l}}^{-1}}$ and $941.8\text{ kJmo}{{\text{l}}^{-1}}$, enthalpy of formation of $N{{H}_{3}}(g)$ is $-46\text{ kJmo}{{\text{l}}^{-1}}$. What is enthalpy of atomization of $N{{H}_{3}}(g)$ ? What is the average bond enthalpy of $N-H$ bond?

Answer 1

While calculation; keep in mind that the enthalpy of formation is considered for only one mole of the substance formed while enthalpy of atomization is the enthalpy change in which total separation of all atoms of the substance occurs.

Complete step by step solution:
Given that,
The bond dissociation enthalpy of ${{H}_{2}}$ is $436.0\text{ kJmo}{{\text{l}}^{-1}}$ and that of ${{N}_{2}}$ is $941.4\text{ kJmo}{{\text{l}}^{-1}}$.
The enthalpy of formation of $N{{H}_{3}}$ is $-46\text{ kJmo}{{\text{l}}^{-1}}$.
The reaction can be shown as follows:
${{N}_{2}}(g)+{{H}_{2}}(g)\to 2N{{H}_{3}}(g)$
We can see that, two moles of ammonia (i.e. $N{{H}_{3}}$) is formed during the reaction and it is important to note that, the standard enthalpy of formation ($\Delta {{H}_{f}}$) is the change of enthalpy during the formation of one mole of the substance from its constituent elements. But there is formation of two moles of ammonia.
So, the enthalpy of formation of $N{{H}_{3}}$ will be $=2\times -46\text{ kJmo}{{\text{l}}^{-1}}$.
The enthalpy formation of a reaction equals to the difference between the bond dissociation enthalpy of the reactants ($B{{E}_{\text{Reactants}}}$) and the products ($B{{E}_{\text{Products}}}$).
$\Delta {{H}_{f}}=B{{E}_{\text{Reactants}}}-B{{E}_{\text{Products}}}$
So, here the reactants are the one mole of ${{N}_{2}}$ and three moles of ${{H}_{2}}$ and the product is $N{{H}_{3}}$ which has six $N-H$. Now, by placing the values we will get:
$\implies 2\times -46=(941.8+3\times 436)-6X$, where X represents the bond energy of $N-H$ bond.
Then, $-92=2249.8-6X$
So, $X=390.3$ $390.3\text{ kJ/mol}$
We should know that ammonia has one atom of nitrogen and three atoms of hydrogen atoms and heat of atomization refers to the enthalpy change in which total separation of all atoms occurs.
Thus, the enthalpy of atomization of ammonia is given by $=3\times \text{BE of N-H}$ (as it has three $N-H$ bonds).
So, enthalpy of atomization will be $=2\times 390.3=1170.9\text{ kJ/mol}$.

Hence, the enthalpy of atomization of $N{{H}_{3}}(g)$ is $1170.9\text{ kJ/mol}$ and the average bond enthalpy of $N-H$ bond is $390.3\text{ kJ/mol}$.

Note: Possible mistake could be, you may get confused with the number of moles involved while calculating enthalpy of formation and number of atoms present in a molecule. The enthalpy of formation is considered for only one mole of the substance formed so; you must check how many moles are formed.