I would like to ask how we can see entropy as an intrinsic property of molecules, namely as something depending also on their geometry (and not only as gross disorder). Is this what is called the "shape entropy" or there's no relation between the two concepts? Which are the parameters that determine this "geometric" or conformational entropy?

EDIT: The answer written by Buck Thorn provides an interesting mathematical insight of entropy considering the conformations or "geometries" available to molecules in thermal equilibrium.

Is there also a qualitative analysis of this phenomenon (in terms of the possible conformations playing a role in entropy - the parameters I was referring to in the question)?

  • $\begingroup$ Maybe it's just me but I can't tell if your question is even valid. Even if it is I doubt it would actually be about chemistry. $\endgroup$
    – Mithoron
    Nov 19, 2019 at 17:31
  • $\begingroup$ @Mithoron I ask you kindly to re-evaluate this; my chemistry teacher mentioned it, and I would not know which other field this could be related to $\endgroup$ Nov 19, 2019 at 17:34
  • $\begingroup$ @Mithoron my teacher talked about entropy in stretching and movement of the bonds, rotations and so on... But I would like to delve deeper into this $\endgroup$ Nov 19, 2019 at 17:36
  • $\begingroup$ Well, you'd need to completely rewrite it. I have a feeling you'd need to learn a bit more to properly word it, though :( $\endgroup$
    – Mithoron
    Nov 19, 2019 at 17:44
  • $\begingroup$ @Mithoron yeah, probably, English is not my native language. If you did sense something from my confused words, could you pls suggest an edit? $\endgroup$ Nov 19, 2019 at 17:48

2 Answers 2


As an example consider an idealised/simplified model of butane, which has trans and gauche configurations as shown in the figure. We shall suppose these have different energies due to interactions between the protons on carbon $1$ and $4$ as bond C2 to C3 rotates.


We take the energy of the trans state at $0^\text{o}$ to be zero, $E_0 = 0$. The energy of the other two conformations, measured relative to this, have minima at $120^\text{o}$ and $240^\text{o}$ and are $E_{120}$ and $E_{240}$.

The partition function is the sum of the statistical weights of the energy levels. The statistical weight of a state $i$ with energy $E_i$ is the Boltzmann factor $\displaystyle e^{-E_i/k_BT}$, multiplied by the degeneracy of that state $g_i$. The partition function is therefore $Z=\sum_i g_ie^{-E_i/k_BT}$.

The partition function for our model butane is therefore $\displaystyle Z = 1 + e^{-E_{120}/ k_BT} + e^{-E_{240}/ k_BT}$. By symmetry, the energy of the $120$ and $240$ states are the same and the probability of being in either is $\displaystyle 2e^{-E_g / k_BT}/Z$ where $E_g$ is an abbreviation for $E_{120}$ and $E_{240}$. These two levels are accidentally degenerate.

Once the partition function has been found then the entropy $S$ can be calculated as can the internal energy $U$ and other thermodynamic quantities; $\displaystyle U=RT^2\left(\frac{d\ln(Z)}{dT}\right)_V;\; S=R\ln(Z) +\frac{U}{T}$.

If you calculate the conformational entropy you will find it is zero at low temperatures, because all molecules are in the lowest trans state, and rises to 3R at high temperatures.

Try not to think of entropy as disorder, it is generally more useful to think of it as the number of ways of placing the molecule in its many possible energy states.

  • $\begingroup$ Really thank you for the clear and detailed explanation! I've just a doubt, and I don't know if the answer might already be in the sentence "The energy of the other two conformations [... ] have minima at 120° and 240°". Well my question is: can we theoretically talk about infinitely many possible conformations, some of which are only the most probable? Is this the meaning of minima of energy? Again thanks a lot $\endgroup$ Nov 20, 2019 at 18:39
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    $\begingroup$ yes you can do this if you want, and may be necessary for some problems, then of course you will have to find an infinite sum which may or may not be easy. $\endgroup$
    – porphyrin
    Nov 21, 2019 at 11:30

You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium.

For molecules in an ideal gas the molar entropy is $$S=R\ln z +RT\left(\frac{\partial z}{\partial T} \right) _V - R \ln N_A + R$$ Here $z$ is the molecular partition function $$z=\sum_i e^{\epsilon_i/kT}$$ where the summation extends over all states available to a molecule and $\epsilon _i$ is the energy of state $i$ (states degenerate in energy are counted separately).

If you can identify all the possible conformations ("geometries") available to the molecule and their respective possible energy states, you can use the formula to compute the entropy.

Qualitatively speaking, if you increase the number of conformations you increase the entropy, since the summation in the partition function now extends over more states.

You might invoke Boltzmann's equation $$p_i=\frac{e^{-\epsilon_i/kT}}{z}$$ to compute the probability of finding a particular molecule in state i.


McQuarrie and Simon, Statistical Thermodynamics, University Science Books (1999).


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