The calculated ΔS°rxn for the combustion of ammonia to nitrogen dioxide and water vapor is 112.5 J/ (mol K)
NH3=192.5
ΔsO2=205.1
ΔsNO2=240.1
H2O=188.8
4NH3+7O2→4NO2+6H2O
Reaction entropy: The entropy of the production of the products and reactants can be used to compute the entropy change of a reaction. It equates to:
Hence, rxn=iviSoi
The Third Law of Thermodynamics explains why all of the formation values are positive.
From the information provided, we learn:
Consequently, ΔS°rxn = 6(188.8) + 4(240.1) + 4(192.5) + 7(205.1)
= 112.5 J/ (mol K)
a single molecule (0) is the sign of deltaS0. NH3 should therefore have a coefficient of one.
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