Pawan Prajapati
Last Activity: 3 Years ago
Bond dissociation enthalpy/energy is defined as the amount of energy required to dissociate a particular bond present in a compound. It is done via homolytic fracture which produces the radical species as a result.
N2+BondEnergy→2N
Complete step by step solution:
The chemical reaction for the formation of Ammonia (NH3
) can be given as:
12N2+13H2→NH3
We have been given in the question that the:
Standard enthalpy of formation of NH3
(ΔfH
) = 46.0 kJ/mol.
Enthalpy of formation or Bond enthalpy of H2
from its atoms (ΔHH2
)= -436 kJ/mol
Enthalpy of formation or Bond enthalpy of N2
from its atoms (ΔHN2
) = -712 kJ/mol
We know that,
Δ Hrxn=∑Bond enthalpy ofproduct-∑Bond enthalpyofreactants
ΔfH=[12ΔHN2+32ΔHH2]−3[ΔHN−H]
Putting the values in above equation, we get,
−46=[12×712+32×436]−3[ΔHN−H]
After rearranging,
3[ΔHN−H]=1010+46
[ΔHN−H]=10533
[ΔHN−H]=+352kJmol−1
Therefore, the average bond enthalpy of N-H bond in NH3 is +352kJmol−1
.
So, the correct option is (c).
Note: Enthalpy of formation or bond formation enthalpy is given in the question is negative but the reaction mechanism we have taken in the answer involves bond breaking. Hence, the values need to be taken as positive.
