Question:
The value of ΔfHΘfor NH3 is – 91.8 kJ mol–1. Calculate the enthalpy change for the following reaction :
2NH3(g) → N2(g) + 3H2(g)
Solution:
Enthalpy change of a reaction is calculated as
: Σbond enthalpy of reactants- Σbond enthalpy of products
for the decomposition 2NH3(g) →N2(g) + 3H2(g) ΔrHΘ will be =
– (– 91.8 kJ mol–1 ) = + 91.8 kJ mol–1
for 2 moles of NH3 enthalpy change of the reaction will be ΔrH = (2 X 91.8 ) = 183.6 kJ mol–1 .