Why only structures corresponding to valleys on PES are taken into account?

Question

I have had to ask this question again because it was marked as a duplicate but it wasn't. Here it goes..

Suppose we want to estimate $G_f^\circ$ of a molecule. If it has rotational conformers, right way to calculate $G$ should include a Boltzmann average of them. We will do it as:

$$\large \langle G_f^\circ \rangle = \frac{\sum_{i}G_f(i)e^{-G_f(i)/k_BT}}{\sum_{i}e^{^{-G_f(i)/k_BT}}}$$ where the sum goes over the conformers.

Question:

Why $\langle G_f^\circ \rangle$ is estimated as the sum of $G_f(i)$ of only most stable conformers? (corresponding to minima on PES)

Example

Suppose the following rotational conformers:

Consider the $y-axis$ to be gibbs free energy, we would estimate $\langle G_f^\circ \rangle$ as the sum of structures corresponding to minima (ie 60º, 180º), but there are non-minimum structures around 180º with less energy than the minima in 60º. Why aren't them considered?

EDIT

Someone could argue that a conformer in 150º won't exist because there will be a driving force to 180º, but if we take the relative energies $\frac{G_f(150º)}{G_f(60º)}$ we get that 150º must have almost the same population as 60º. There might be some criteria I am missing or maybe a conceptual error.

Any edit will be welcome.

Answer 1

The vast majority of thermodynamics we consider is at equilibrium, we is the issue here.

At equilibrium we will only find our S states at the bottom of the valleys (or minima) populated. This this because they have now force acting on them to drive them to another configuration.

An important relationship to remember with potential energy surfaces is: $$ F = -\frac{dE}{dx} $$ Force is the negative derivative of potential energy with respect to some coordinate. This means that the driving force is given by the slope of the PES and is pushing down the PES. Then a molecule with a $\phi = 150^\circ$ experiences a rotational force pushing it towards the minimum at $\phi = 180^\circ$.

Answer:

The result is that any molecule not at a minimum will be quickly driven towards the nearest minimum, so only the minima will be occupied and need to be summed over.

Side Note - Maxima

Maxima (the peaks) also experience zero force. However this is only true at the exact peak itself. Any point infinitesimally far away from the maxima experiences a force pushing it down the hill into a minima.

Thermal fluctuations and collisions with other molecules will continually slightly nudge the molecules around on the potential energy surface. At maxima this will push any molecules there off the peaks and down into minima. In minima, the force will bring the molecules back into the minima.

Answer 2

You are right that this is an approximation. It is, however, a much better approximation than you might think. The assumption is that each of the wells are shaped approximately the same. Furthermore, because the population of a state decreases exponentially with energy, it is really the approximation that each of the wells is shaped the same near the minimum.

It is well known that the smaller the displacements from the minimum, the more harmonic the motion will be. So, a measure of how much error is introduced by this approximation is the difference in harmonic frequency of the torsion in each of these wells. And, in the case that the frequencies in each of these wells is the same, the equation you have written is exact within the harmonic approximation. If the true (experimental) frequencies are identical, then the equation you have written is exact.

To be extra clear, for identically shaped wells, all deviations from the minimum energy structure will cancel out because when you do the Boltzmann averaging in each well, the population above any minimum will follow a Boltzmann distribution. Thus, the contributions from these deviations will be very minimal, and depend on there being different shapes of the well.

Probably the simplest way to include the shapes of the different wells in a calculation like this would be to construct the PES using some kind of ab initio method and then do a dynamics simulation on this surface to get the free energies this way.

Answer 3

The Boltzmann distribution doesn't only take into account the most stable conformers. It can take the energy of any conformer (relative to the energy of the most stable conformer).

I assume that you're calculating the Gibbs energy using $\Delta G = - RT \text{ ln}K$. In this equation, $K$ represents the equilibrium constant between two local minima, i.e. between anti and gauche. So, when you calculate the Gibbs energy in this way, you're actually calculating the Gibbs energy for the rotation between two conformers (you can sort of imagine this as a reaction).