There are missing information in the question. The bonds enthalpy are:
N=N: 409 kJ/mol
N-H: 388 kJ/mol
O=O: 498 kJ/mol
N=O: 630 kJ/mol
O-H: 463 kJ/mol
Answer:
+508 kJ/mol
Explanation:
The formation of a bond is endothermic, so the enthalpy is positive, and the break is exothermic, so the enthalpy is negative. The enthalpy of the reactants must be negative, thus, the change in enthalpy can be calculated as:
ΔH = ∑n*H products - ∑n*H reactants, where n is the number of bonds.
N₂H₂ has 1 N=N bond and 2 N-H bonds; O₂ has 1 O=O bond; NO has 1 N=O bond; and H₂O has 2 O-H bonds, so:
ΔH = (4*1*630 + 2*2*463) - [2*(1*409 + 2*388) + 3*1*498)
ΔH = +508 kJ/mol
