Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I have literature values for the standard enthalpy and Gibbs energy changes of formation at 298 K, $\Delta H_\text{f}^o$ and $\Delta G_\text{f}^o$, for product in the ideal gas state. Since $G=H-TS$ and by definition, the reactions are carried out at constant temperature I can solve for $\Delta S_\text{f}^o$ via:

$$\Delta S_\text{f}^o=\frac{\Delta H_\text{f}^o-\Delta G_\text{f}^o}{T}$$

I've done this for three hexanes, n-hexane, 2-methylpentane, and 2,3-dimethylbutane. Below are the results:

hexanes thermochemistry values

I guess the sign makes sense to me. It says that all three molecules represent more order than their reactants (C and H2). What is strange to me is the order of the standard entropies of formation. Since these are values in the ideal gas state, entropy effects due to intermolecular forces should not be included, correct? Hence, I would have thought that n-hexane would be the most "ordered" of the three molecules. Surely a linear molecule is less likely (requires higher order) than a branched molecule. What have I missed?

share|cite|improve this question
up vote 1 down vote accepted

I can't find any references to support this, but I would guess that because the linear molecule is more flexible than the branched molecules it has a higher entropy associated with the number of arrangements of the molecule.

See for some comments on this, though this is just some site I found with Google so caveat emptor.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.