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The standard enthalpy of formation of $N{{H}_{3}}$is $-46kJmo{{l}^{-1}}$. If the enthalpy of formation of ${{H}_{2}}$from its atoms is $-436kJmo{{l}^{-1}}$and that of ${{N}_{2}}$is $-712kJmo{{l}^{-1}}$ the average bond enthalpy of N-N bond in $N{{H}_{3}}$is
(A) $+1056kJmo{{l}^{-1}}$
(B) $-1102kJmo{{l}^{-1}}$
(C) $-964kJmo{{l}^{-1}}$
(D) \[+352kJmo{{l}^{-1}}\]

Last updated date: 17th Jun 2024
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Hint: The amount of energy needed to be supplied to break a chemical bond between two species is known as Bond Dissociation Enthalpy.

Complete step by step solution:
-We will begin solving this question by writing the chemical equation for the question-
$\dfrac{1}{2}{{N}_{2}}+\dfrac{3}{2}N{{H}_{3}}\to N{{H}_{3}}$
-The change in enthalpy during the formation of one mole of the substance from its constituent elements, with all substances in their standard states is known as the standard enthalpy of change or standard enthalpy of formation. This is represented by $\Delta H{}^\circ $.
$\Delta H{}^\circ =-\Delta H{{{}^\circ }_{f}}(N{{H}_{3}})=-(-46kJmo{{l}^{-1}})=46kJmo{{l}^{-1}}$
Also, $\Delta H{}^\circ =3\Delta {{H}_{N-N}}+\dfrac{1}{2}\Delta {{H}_{N\equiv N}}+\dfrac{3}{2}\Delta {{H}_{H-H}}...(i)$
-According to the question,
\[\Delta {{H}_{N\equiv N}}=712kJmo{{l}^{-1}}\]
$\Delta {{H}_{H-H}}=-436kJmo{{l}^{-1}}$
$\Delta H{}^\circ =46kJmo{{l}^{-1}}$
-Inserting the above values in equation (i), we will get
$46=3\Delta {{H}_{N-H}}+\dfrac{1}{2}(-712)+\dfrac{3}{2}(-436)$
$\Rightarrow \Delta {{H}_{N-H}}=\dfrac{1}{3}[1056]=+352kJmo{{l}^{-1}}$

So, the correct answer is option D.

Note: -Bond dissociation enthalpy gives the strength of the chemical bond between any two species. It is generally measured as enthalpy change at standard conditions, which is 298K.
-The bond dissociation energy of a chemical bond is frequently defined as the enthalpy change occurring through homolytic fission of the bonds at absolute zero (0K).
-For diatomic molecules, the bond dissociation enthalpy is equal to the value of bond energy.
-The strongest bond dissociation enthalpy is found to exist between the bonds between silica and fluorine. The bond dissociation energy required to break the first bond between silicon and fluorine in s silicon tetrafluoride molecule is estimated to be $166 kcalmo{{l}^{-1}}$. The high energy is due to the difference in electronegativities of the silicon and fluorine atoms.
-When considering neutral compounds, the carbon-oxygen bond in carbon monoxide is said to have the highest strength with a bond dissociation energy $257 kcalmo{{l}^{-1}}$.
-The carbon-carbon bond in the ethyne molecule is found to have a relatively high bond dissociation energy $160 kcalmo{{l}^{-1}}$.
-The weakest bond dissociation energy is said to exist between the atoms or molecules with covalent bonds.