Computing the rotational entropy in RRHO approximation

Question

I am implementing the computation of Gibbs energies by their molecular parameters (structure, frequencies) in my program Chemcraft (the main task is the possibility to cut small frequencies which produce big errors in the vibrational entropy). The rotational entropy of nonlinear molecule is computed in RRHO approximation as follows:

$$S_\text{rot} = R\left(\ln\left(\frac{2^\frac{9}{2}\pi^\frac{7}{2}(kt)^\frac{3}{2}(I_{xx}I_{yy}I_{zz})^\frac{1}{2}}{h^3\sigma}\right)+\frac{3}{2}\right)$$

Here sigma is the symmetry number, depending on the point group. I don’t understand what this value means. At one textbook I found that $\sigma$ is $2$ for $C_2$ symmetry and $1$ for $C_i$ or $C_s$ symmetry; $5$ for $C_5$, $C_{5v}$, $C_{5h}$, $S_{10}$ symmetry and $10$ for $D_5$, $D_{5d}$, $D_{5h}$ symmetry. Is this true?

I find these formulas slightly strange. For example, if we have a benzene molecule with $D_{6h}$ symmetry, we can compute its entropy; but if we alter a coordinate of one atom with $0.000001$Å, the symmetry lowers to $C_1$ and the entropy becomes different (I verified this with Gaussian). Don’t you find this strange?

I like to talk about physics on physics.stackexchange.com, and I have a feeling that the thermodynamics is rather a philosophy, not a science (this is related to the $S=k\ln(W)$ formula where it is not fully clear what means the $W$).

So, my question is: maybe it will be useful if I add the option “Do not use the symmetry number” in Chemcraft? This option will compute the rotational entropy without the $\sigma$.

Answer 1

The symmetry number is the order (number of symmetry elements) of the rotational subgroup of the molecular point group. I.e. the rotational subgroup of $C_{2v}$ is $C_2$, the rotational subgroup of $D_{6h}$ is $D_6$, etc. This is what comes out of the stat. mech. treatment within the Rigid-Rotor Harmonic-Oscillator (RRHO) approximation, although the above formula for $S_\text{rot}$ involves additional assumptions, as the rotational energy levels for a general molecule cannot be condensed into a closed formula for the partition function.

Just to follow up on the illustrative example given by wzkchem5, the RRHO expression for the vibrational entropy breaks down when a frequency goes to zero. In the harmonic approximation, the vibrational entropy goes to infinity when the frequency goes to zero, and contributions from frequencies less than, say, $\pu{100 cm-1}$ are likely severely overestimated. If the motion in reality corresponds to an internal rotation, then the limiting entropy for a free rotor is $R$, and there are suggestions for how to interpolate between these limits, for example work by Stefan Grimme.

Answer 2

The argument "a slightly perturbed benzene molecule has an entropy that differs a lot from the entropy of the equilibrium structure" does not hold, because entropies are only defined for equilibrium geometries (when they are calculated by e.g. the harmonic approximation) or phase space regions with non-zero measure (when they are calculated by phase space integration). The latter case is self-evident; the former case exists because when we talk about the entropy of an equilibrium structure, we are actually talking about the entropy in the basin of attraction around that equilibrium structure, i.e. the entropy of the molecule when it is free to wander around in the local PES basin but not beyond. Now consider the entropy of a non-equilibrium structure. How would you define it? If you can define it at all, you have to use the "basin of attraction" definition, but then it is the same as the entropy of the equilibrium geometry. This is along the same line as the famous rule that RRHO thermochemical properties can only be obtained by frequency calculations at equilibrium structures. Thus, if you expect that the exact converged structure has $D_{6h}$ symmetry, then you should use the symmetry number appropriate for $D_{6h}$ even if your (not fully converged) optimized structure has $C_1$ symmetry; in this case, the entropy at that $C_1$ structure is not only non-calculable, but not even defined (unless you trivially define it as the entropy of the $D_{6h}$ structure).

Nevertheless, it is possible to make the argument more rigorous by the following way. Suppose you study the equilibrium structures of a series of hexa-substituted benzenes, with increasing tendency of second-order Jahn-Teller symmetry breaking into the bond-alternating $D_{3h}$ structure. Then you'll find that the point group of the molecule changes abruptly from $D_{6h}$ to $D_{3h}$ somewhere in the series, where the $D_{6h}$ now becomes a saddle point. The symmetry number thus suddenly reduces from 12 to 6, such that the entropy of each of the $D_{3h}$ structures will experience a finite jump from the trend line, similar as what you said. This is indicative of a breakdown of the RRHO approximation (since near the point where the $D_{6h}$ changes from a minimum to a saddle point, one of the vibrational frequencies is extremely close to 0, therefore the associating vibration mode is extremely anharmonic). I believe no single-structure based anharmonic correction scheme will remove this finite jump, either, and you probably have to treat that vibrational mode by e.g. the WKB approximation, in order to properly describe the entropy of such systems.