The value of ΔfHΘfor NH3 is – 91.8 kJ mol–1.

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 .

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