The value of Δ_fH^Θfor NH₃ is – 91.8 kJ mol^–1. Calculate the enthalpy change for the following reaction :

2NH₃(g) → N₂(g) + 3H₂(g)

o Enthalpy change of a reaction is calculated as
: Σbond enthalpy of reactants- Σbond enthalpy of products

o The standard molar enthalpy of formation (for one mole) Δ_fH^Θ is given for NH₃ = – 91.8 kJ mol^–1 which is associated with the reverse exothermic reaction N₂(g) + 3H₂(g) → 2NH₃(g) i.e. heat released information of ammonia is – 91.8 kJ mol ^–1.

Hence, for the decomposition 2NH₃(g) →N₂(g) + 3H₂(g) Δ_rH^Θ will be =

- (– 91.8 kJ mol^–1 ) = + 91.8 kJ mol^–1 .

And in this case for 2 moles of NH₃ enthalpy change of the reaction will be Δ_rH = (2 X 91.8 ) = 183.6 kJ mol^–1 .