Question

The standard enthalpy of formation of$N{{H}_{3}}$ is -46.0$kJmo{{l}^{-1}}$ . If the enthalpy of formation of ${{H}_{2}}$ from its atoms is -436$kJ mol{{l}^{-1}}$ and that of \[{{N}_{2}}\] is -712$kJmo{{l}^{-1}}$ , the average bond enthalpy of N-H bond in $N{{H}_{3}}$ is:
A. -964$kJmo{{l}^{-1}}$
B. +352$kJmo{{l}^{-1}}$
C. +1056$kJmo{{l}^{-1}}$
D. -1102$kJmo{{l}^{-1}}$

Answer 1

Hint: The enthalpy for the formation of any compound is a relative value and it is measured by measuring the sums of all the bond energies of the reactants and products separately and then subtracting the bond energies of the products from the reactants.

Complete step by step solution:
-The formation of any compound from the pure elements is possible only when the energy is released during that process. It means that the process should be exothermic for it to go in a forward direction.
-To form a bond with a new atom or molecule, the original bonds are broken which require energy to do so and the new bonds are made which release the energy. The energy released in making new bonds should exceed that in breaking old bonds.
-Thus we see that a single step of any reaction can be both exothermic and endothermic but the complete process needs to be exothermic so that the compound formed remains stable.
-The amount of energy that is lost or gained during the formation of any compound is represented as a change in the enthalpy. The reaction is favourable when the enthalpy of the reactants is more than that of the products.
-This means that the change in enthalpy should be negative in order to make the reaction exothermic and should be positive in order to make the reaction endothermic.
-Enthalpy is found by the knowledge of the bond energies of the molecules. If we are to find the enthalpy of $N{{H}_{3}}$ we need to know the bond energies of both nitrogen and hydrogen molecules and also the bond energy between N-H bonds.
-The formula for the change in the enthalpy of a reaction can be shown as
\[\Delta H=\sum{B{{E}_{reactants}}-\sum{B{{E}_{products}}}}\]
-In the question, we are given the standard enthalpy of formation of$N{{H}_{3}}$ to be -46.0$kJmo{{l}^{-1}}$. Also, the enthalpy of formation of ${{H}_{2}}$ from its atoms is -436$kJ mol{{l}^{-1}}$ and that of \[{{N}_{2}}\] is -712$kJmo{{l}^{-1}}$. We have to find the bond enthalpy of the N-H bond which occurs in the product.
-The reaction of nitrogen and hydrogen to form ammonia can be shown as
\[{{N}_{2}}+3{{H}_{2}}\to 2N{{H}_{3}}\]
-We see that nitrogen is simply broken down into 2 N atoms which are needed in the product formation. So 1 mole nitrogen is completely needed. 1 mole of hydrogen releases 2 hydrogen atoms. The bond energy for that is given. If we need only 1 atom, the bond energy will be half of it. Here we need 3 atoms. So the bond energy required for that will be $\dfrac{3}{2}xB{{E}_{{{H}_{2}}}}$
-Thus putting all the values in the formula for change in enthalpy, we get
$-46=712+\dfrac{3}{2}x\left( 436 \right)-B{{E}_{product}}$
The bond energy of the product is nothing but three times the average bond enthalpy of N-H bond in $N{{H}_{3}}$ as there are 3 such bonds in ammonia. It can be shown as
$-46=712+\dfrac{3}{2}x\left( 436 \right)-3x$
-Solving this we get the value of x to be +352. The negative values of the bond energies has been removed as they are the bond energies of formation of hydrogen and nitrogen. These bonds are to be broken and so they are taken as negative of the formation energy values.

Therefore the correct option is B.

Note: The standard enthalpy of formation of a pure element which occurs naturally in its original form is considered to be zero. This is the reference value. Enthalpy for any other compound is measured from this value itself. For any compound formation, energy is released and so the enthalpy can never be zero for any compound.