Using the table of entropy values above, calculate So
for a each of the following reactions,
b) 2C2H6(g) + 7O2(g) ---->
6H2O(g) + 4CO2(g)
Answer
We notice that the calculated value is very negative. Just as
we should predict on the basis of comparing the relative order/disorder of the
reactants and products. There are 4 moles of product gases and 9 moles of reactant
gases. Also there are 6 moles of products in the liquid phase and no reactants
in the liquid phase. The products are much more ordered compared to the reactants.
We must conclude the S of the reaction
is negative.