How does the entropy change in the reaction #2C_3H_6(g) + 9O_2(g) -> 6CO_2(g) + 6H_2O(g)#?

1 Answer
Mar 9, 2017

Answer:

#DeltaS^@=+#

Explanation:

#2C_3H_6(g) + 9O_2(g) rarr 6CO_2(g) + 6H_2O(g)#

Now #DeltaS^@# #=# #SigmaS_"products"-SigmaS_"reactants"#, where #S_"products"#, etc, are measured under standard conditions.

There are 12 moles of GASEOUS product, and 11 moles of GASEOUS reactant in the equation as written; and of course we must assume standard conditions.

We need values for #S^@# for products and reactants, before we make the summation quantitatively. However, because there is ONE more mole of GASEOUS product than GASEOUS reactant, we would predict the enthalpy change, #DeltaS^@#, would be positive, given that a gaseous product has a more positive entropy, a greater statistical probability for disorder, than a liquid one.

For the reaction.............

#2C_3H_6(g) + 9O_2(g) rarr 6CO_2(g) + 6H_2O(l)#

How would you anticipate the change in overall entropy?