It is know that the entropy of a system can be calculated using the formula (1):

\begin{equation} \label{eq:entropy_Boltz} S = {k_B}\ln \Omega\tag1 \end{equation}

where $k_B$ is the Boltzmann constant, $\Omega$ is the number of microstates of the system and $S$ is its entropy. This equation is one of the fundamental equations of Statistical Mechanics. Since $\Omega$ represents the number of microstates of the system, the more microstates, the greater the disorder of the system and therefore, the greater the entropy. One of the problems in using equation \eqref{eq:entropy_Boltz} is the need to know the number of microstates of the system.

Instead of expression \eqref{eq:entropy_Boltz}, Oganov and Valle [1] proposed an expression for the calculation of a quasi-entropy representing the structure disorder for a crystal:

\begin{equation} \label{eq:quasi-entropy} S_{str} = - \sum\limits_A {\frac{{{N_A}}}{N}}\left\langle\ln (1 - {F_{{A_i}{A_j}}})\right\rangle \end{equation} where $A$ represents the chemical species in the structure, $N_A$ is the number of atoms of the chemical species $A$, $N$ is the total number of atoms, and $F_{{A_i}{A_j}}$ is the distance between the fingerprints of sites $i$ and $j$ of the chemical species $A$. The fingerprints of each structure can be calculated as

\begin{equation} \label{eq:fingerprints} {F_{{A_i}B}} = - \sum\limits_{{B_j}} {\left\{ {\frac{{\delta (R - {R_{ij}})}}{{4\pi {R_{ij}}^2({N_B}/V)\Delta }}} \right\}} - 1 \end{equation} where $R$ is a parameter representing the maximum distance between atoms, $R_{ij}$ is the distance between the sites $i$ and $j$, $N_B$ is the number of atoms of the chemical species $B$, $V$ is the volume of the structure, and $\Delta$ is another parameter associated with the distance between pairs of structures [1].

Using this for molecules imply you have to create a pseudo crystal with your system.

On the other hand, the entropy associated with of bit string can be calculated using the Shannon Entropy. A modification of it was proposed by Grenville J. Croll and called BiEntropy [2]. In this case, it is necessary to use binary fingerprints of molecules.

The problem

My systems consist on randomly decorate a nanostructure with organic groups like $\ce{-OH}$ and $\ce{-COOH}$, then select one (using the entropy as criteria) and run the calculations with it (here is an example of the type of system I am talking about).

Is there a way to calculate the conformational (disorder) entropy of a molecule?


[1] A.R. Oganov, M. Valle, How to quantify energy landscapes of solids, J. Chem. Phys. 130 (2009) 104504

[2] G.J. Croll, BiEntropy - The Approximate Entropy of a Finite Binary String, arXiv:1305.0954 [cs.OH].


1 Answer 1


tldr, Extensively sample the potential energy surface of the molecule, then Boltzmann-weight the relative energies. The CREST tool is useful.

Nice to see this question as I'm writing / editing a paper on the conformational entropy of molecules.

There's already a question about the translational and rotational entropy of molecules. For those, you optimize a geometry, then calculate from the molecular mass and moments of inertia.

Similarly, one gets the vibrational entropy under the rigid rotor approximation by calculating the vibrational modes from the Hessian and thus the vibrational partition function.

These procedures, though, start from one conformer geometry. For any molecule with conformational degrees of freedom, there are likely to be multiple minima.

In principal, you could run molecular dynamics (even ab initio MD) for a long time, collect all the geometries, and Boltzmann weight the relative energies. Unfortunately, there's no guarantee that "a long time" is enough to capture all the thermally-accessible minima.

There are a few more practical methods at the moment:

The latter will directly calculate conformational entropies of the resulting ensemble through Boltzmann weighting and degeneracies.

For example, you might get output like this:

T /K                                  :   298.15
E lowest                              :   -43.34438
ensemble average energy (kcal)        :    0.028
ensemble entropy (J/mol K, cal/mol K) :   13.834    3.306
ensemble free energy (kcal/mol)       :   -0.986
population of lowest in %             :   79.407
 number of unique conformers for further calc            3
   1       0.000
   2       0.006
   3       2.879

CREST is time-consuming, and in principal, you should also use QM minimization and re-ranking relative energies.

  • 1
    $\begingroup$ Good answer. Something that always bothered me was people choosing the quicker option over the better one. $\endgroup$
    – B. Kelly
    Commented May 30, 2020 at 18:03
  • $\begingroup$ @CharlieCrown - I'm not aware of many people calculating conformer entropies - even with MD. I'd be curious for references - I think I know all of them coming from the QM community. $\endgroup$ Commented May 30, 2020 at 18:13
  • 2
    $\begingroup$ Mine was just a general comment somewhat in defense of using the more time consuming Grimme software. Just a general pet peeve really and not anything against your answer. $\endgroup$
    – B. Kelly
    Commented May 30, 2020 at 18:17
  • $\begingroup$ In my experience free energy calculations are much more common than entropy calculations with MD and for a good reason: it is straightforward to obtain entropies if you have a free energy profile/difference and a reasonable estimate of the average energy (very easy with MD). $\endgroup$
    – Godzilla
    Commented May 30, 2020 at 20:34
  • $\begingroup$ Thanks for your answers and comments. My problem is that I have to randomly decorate the nanostructure and then select one structure as representative. Til now, I generated 10.000 (for eachone of the 5 functionalization concentrations) and then calculated the entropy using the Oganov approach (I am testing using BiEntropy). As there are a lot of structures, using time consuming (even being better methods) is out of question :( $\endgroup$
    – Camps
    Commented May 30, 2020 at 23:01

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .