Question

the standard enthalpy of formation of NH3 is _46.0 kj/mol.if the enthalpy of formation of H2 from its atom is _436 kj/mol and that of N2 is _712 kj/mol,the average bond enthalpy of N-H bond in NH3 is:

Answer 1

The balanced chemical equation for formation of NH3 would be as follows:
​12N2(g) + 32H2(g)-->NH3(g) 
The energy required to break required bonds for the formation of product is calculated as:
1/2 mole of N2 will require 712 kJ·mol-1/2  = 356 kJ/mol 
3/2 mole of H2 will require 3×  436 kJ·mol-1/2 = 654 kJ/mol 
Total energy require to break the bonds = (356 + 654) kJ/mol = 1010 kJ/mol
total energy of reaction = (total energy of bonds broken) - (total energy of bonds formed)  
 -46.3 kJ/mol =1010) kJ/mol - 3(N-H)
3(N-H) = 1010 kJ/mol + 46.3kJ/mol  = 1056.3 kJ/mol 
(N-H) =  1056.3 kJ/mol /3 = 352.1kJ/mol