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

Mar 9, 2017

$\Delta {S}^{\circ} = +$

#### Explanation:

$2 {C}_{3} {H}_{6} \left(g\right) + 9 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(g\right)$

Now $\Delta {S}^{\circ}$ $=$ $\Sigma {S}_{\text{products"-SigmaS_"reactants}}$, where ${S}_{\text{products}}$, 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 ${S}^{\circ}$ 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, $\Delta {S}^{\circ}$, 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.............

$2 {C}_{3} {H}_{6} \left(g\right) + 9 {O}_{2} \left(g\right) \rightarrow 6 C {O}_{2} \left(g\right) + 6 {H}_{2} O \left(l\right)$

How would you anticipate the change in overall entropy?