Given reaction (all substances in standard state):
N2(g) + 3 H2(g) -> 2 NH3(g) ; Delta H= -92.4 kJ mol^-1
This enthalpy change is for the formation of 2 moles of NH₃(g) from its elements in their standard states.
By definition, the standard enthalpy of formation of NH₃(g) is for the reaction producing 1 mole of NH₃(g):
N2(g) + 3/2H2(g) -> NH3(g)
Since enthalpy is an extensive property, divide the given ΔH° by 2:
NH3(g)= \frac{-92.4}{2} = -46.2 kJ mol^-1
NH3(g) = -46.2 kJ mol^-1