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The standard enthalpy of formation of NH3 is 46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is -436 kJ/mol and that of N2 is -712 kJ/mol, the average bond enthalpy of N — H bond in NH3 is: a) -1102 kJ/mol b) -964 kJ/mol c) +352 kJ/mol d) +1056 kJ/mol

The standard enthalpy of formation of NH3
is 46.0 kJ/mol. If the enthalpy of formation of H2
from its atoms is -436 kJ/mol and that of N2
is -712 kJ/mol, the average bond enthalpy of N — H bond in NH3
is:
a) -1102 kJ/mol
b) -964 kJ/mol
c) +352 kJ/mol
d) +1056 kJ/mol

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 9723 Points
21 days 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.

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