# Can you approximate the change in entropy of a reaction?

I found a problem in which I am given a hypothetical reaction in which from 2 liquids (reactants) I get 4 moles of gas and 1 mole of a solid substance (products):

$$\ce{A (l) + B (l) -> 4C (g) + D (s)}$$

I am asked about the change in entropy (an approximation of it), so I am curious if there is some kind of approximation for the change in entropy of a reaction based on the change in the amount of substance of gaseous compounds (since I know that the highest influence on entropy is that of gases).

The molar standard entropy of formation ($$S_{f}^{\ominus}$$) for many common substances is documented in tables. You can approximate the overall change in entropy of the reaction ($$\Delta S^{\ominus}_{reaction}$$) with with Hess's law:

$$\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}_{f,molar,products} - \sum S^{\ominus}_{f,molar,reactants}$$

You notice that there is no $$\Delta$$ in front of $$S^{\ominus}_{f,molar}$$, since entropy is an absolute physical quantity. Every system has an absolute entropy.

Don't forget to multiply every $$S^{\ominus}_{f,molar}$$ with the number of moles that you have. This is only an approximation because the molar entropy of a substance depends on Temperature as well, and the reaction likely doesn't take place at standard temperature.

As a sidenote please be aware that the molar entropy of formation for elements is not zero, although the molar enthalpy of formation is!

In such a case, you may simply count the number of gaseous molecules on each side of the arrow. In your reaction there are no gas molecules in the reactants, and 4 in the products. The number of moles of gaseous substances increases. So the qualitative entropy increases.

• You're oversimplifying even this simplified task. – Mithoron Feb 17 '20 at 21:25
• No , I don't require a qualitative approximation ( I mean it is obvious that the entropy will increase ) , my question is numerically if there is a way of giving some sort of approximation (like for example for the enthalpy of vaporization you can use Trouton's constant for any liquid in which there are no hydrogen bonds or other similar interactions). – David Sterlinsky Feb 17 '20 at 21:55