The standard enthalpy of formation of NH 3 is - 46.0 kJ/mol. If bond enthalpy of H 2 is - 436 kJ/mol and that of N 2 is - 712 kJ/mol , the average bond enthalpy of N - H bond in NH 3 is :

Step-by-step solution

For The Reaction 1 2 N 2 g + 3 2 H 2 g → NH 3 g ; Δ H f ∘ = - 4 6 KJ mol - 1 Δ H f ∘ = - 4 6 = ∑ B.E - ∑ B.E Reactants Product ∑ B.E = ∑ B.E + 4 6 Product Reactants = 1 2 7 1 2 + 3 2 4 3 6 + 4 6 = 356 + 654 + 46 = 1056 Averge B.E for N-H Bond = 1 0 5 6 3 = 3 5 2 KJ mol - 1