for a each of the following reactions,
a) 2H2(g) + O2(g) ----> 2H2O(g)
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.