The standard enthalpy of formation of NH3 is -46.0 kJ/mol. If the enthalpy of formation of H2 from its atoms is kJ/mol and that of N2 is -712 kJ/mol, the average bond enthalpy of N−H bond in NH3 is: kJ/mol +352 kJ/mol +1056 kJ/mol -102 kJ/mol

Enthalpy of formation of NH3 = -46 kJ/mole
N2 + 3H2 → 2NH3; ΔHr = -46 x 2 kJ
Bond breaking is endothermic and bond formation is exothermic
Assuming x is the bond energy of N−H bond (kJ/mol)
712 + (3 × 436) - 3x = -46 × 2
712 + 1308 - 3x = -92
2020 - 3x = -92
3x = 2112
x = 704 kJ/mol
Hence, the average bond enthalpy of N-H bond is 704 kJ/mol. However, this value is not among the options provided. Let's re-examine the calculation.

Let's use the given information and apply Hess's Law:
The formation of 2 moles of NH3 can be represented as:
1/2 N2(g) + 3/2 H2(g) → NH3(g) ΔHf = -46 kJ/mol (given)
This means that for 1 mole of NH3, the enthalpy of formation is -46 kJ/mol.

The bond energies involved are:

• N≡N bond energy in N2: 712 kJ/mol (given, this is the energy required to break the triple bond)
• H−H bond energy in H2: Let's denote this as y (kJ/mol) (given)
• N−H bond energy in NH3: Let's denote this as x (kJ/mol) (to be calculated)

Applying Hess's Law:
Energy required to break bonds - Energy released in forming bonds = Enthalpy change of reaction
1/2(712) + 3/2(y) - 3x = -46
356 + 1.5y - 3x = -46

We don't have a value for y (the H-H bond energy). The problem statement seems to be incomplete or contains inconsistent data, rendering a precise solution based on the provided information impossible. The calculation above demonstrates the correct approach, but requires additional data to arrive at a numerical answer consistent with the given options.