Calculate S^o for a each of the following reactions,

a) 2H₂(g) + O₂(g) ----> 2H₂O(g)

Answer

S°_rxn = 2S°(H₂O(g)) - [2S°(H₂(g)) + S°(O₂(g))]
S°_rxn = 2(189 J mol^-1 K^-1) - [2(130.6 J mol^-1 K^-1) + (205 J mol^-1 K^-1)]
S°_rxn = (378 J mol^-1 K^-1) - [(466.2 J mol^-1 K^-1)]
S°_rxn = -88.2 J mol^-1 K^-1

We notice that the calculated value is negative. Just as we predicted on the basis of comparing the relative order/disorder of the reactants and products. In this case we concluded S was likely to be negative becaue there are more moles of gas on the reactants side compared to the product side.