Calculate the standard entropy change for the following process at 298 K: 2H2(g)+O2(g)⟶2H2O(g)

Question

Calculate the standard entropy change for the following process at 298 K:

2H2(g)+O2(g)⟶2H2O(g)

The standard entropies are as follows: S∘H2(g)=130.57Jmol K, S∘O2(g)=205.0Jmol K, S∘H2O(g)=188.7Jmol K 

Answer 1

∆Sº_rxn = ∑∆Sº_products - ∑∆Sº_reactants

2H₂(g) + O₂(g) ===> 2H₂O(g)

∆Sº_rxn = 2x188.7 - (2x130.57 + 205) = 377.4 - (261.1 + 205) = 377.4 - 466.1

∆Sº_rxn = -88.7 J/mol-K

Note that the sign is negative indicating a decrease in entropy which is consistent with the fact that the reaction goes from 3 moles of gas on the reactant side to 2 moles of gas on the product side.