Question
The standard enthalpy of formation of $$N{H_3}$$ is $$ - 46.0\,kJ\,mo{l^{ - 1}}.$$ If the enthalpy of formation of $${H_2}$$ from its atoms is $$ - 436\,kJ\,mo{l^{ - 1}}$$ and that of $${N_2}$$ is $$ - 712\,kJ\,mo{l^{ - 1}},$$ the average bond enthalpy of $$N - H$$ bond in $$N{H_3}$$ is
A.
$$ - 964\,kJ\,mo{l^{ - 1}}$$
B.
$$ + 352\,kJ\,mo{l^{ - 1}}$$
C.
$$ + 1056\,kJ\,mo{l^{ - 1}}$$
D.
$$ - 1102\,kJ\,mo{l^{ - 1}}$$
Answer :
$$ + 352\,kJ\,mo{l^{ - 1}}$$
Solution :
$${N_2} + 3{H_2} \to 2N{H_3}\,\,\,\,\Delta H = 2 \times - 46.0\,kJ\,mo{l^{ - 1}}$$
Let $$x$$ be the bond enthalpy of $$N - H$$ bond then
[Note : Enthalpy of formation or bond formation enthalpy is given which is negative but the given reaction involves bond breaking hence values should be taken as positive. ]
$$\eqalign{
& \Delta H = \sum {{\text{Bond energies of products}} - \sum {{\text{Bond energies of reactants}}} } \cr
& 2 \times - 46 = 712 + 3 \times \left( {436} \right) - 6x; - 92 = 2020 - 6x \cr
& 6x = 2020 + 92 \Rightarrow 6x = 2112 \Rightarrow x = + 352\,kJ/mol \cr} $$