I came across the following sentence in a publication$^1$: "At this point we have to remember that the computed MEP path [sic] represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic." Further, they write that, "This path is accessed because after the transition state the molecule is 'ballistic'."

A minimum energy path (MEP) is calculated "step-wise", taking the previous geometry as a starting point, and the program follows the gradient. In classical terms, this might be to place a ball on a tilted plane, releasing it, and stopping it after a given distance, and repeat.

When they say that, in a real reaction, the molecules will have kinetic energy and be ballistic, I think they mean that the potential energy surface (PES) may be explored in directions slightly different than the gradient. This can be compared to releasing a ball on the top of a changing surface and allowing it to roll freely, where the ball will not follow the gradient when its velocity is high.

I cannot find a decent explanation of this anywhere, and it would be nice to get some input.

$^1$ De Vico, Liu, Krogh, and Lindh, 2007. J. Phys. Chem. A 111:8013-8019. doi:10.1021/jp074063g. Cited passages are on p. 8016, Section 3.3.


I think the description you lay out in your question has it basically right.

MEPs, IRCs, etc. all assume that the geometric rearrangements that occur over the course of a reaction strictly follow the gradient uphill from reactants to transition state, and then downhill from TS to products (or intermediates). Energetically, this provides the "path of least resistance."

In the paper you cited, it looks like the MEP would require some atoms in the system to "take a right-angle turn." Quoting a broader excerpt around the first sentence you include (emphasis added):

As can be seen in Figure 3, after TS‘ S$_0$ the MEP takes a turn on the PES:  the O−O‘ distance increasing and the torsion around the O−C−C‘−O‘ dihedron nearly stopping, while the increase of the C−C‘ distance comes into action.
At this point we have to remember that the computed MEP path represents the reaction in the absence of any kinetic energy, while in reality the true reaction path, to some degree, will be ballistic.
What would be the consequences if the reaction continued along the torsional mode? Section 3.2 already suggested that this is of interest—the torsional mode is the principal internal coordinate of the T$_1$ MEP.

Nuclei have mass (obviously), and thus have inertia. While Newton's relevance at the atomic scale can be fuzzy at times, classical treatment of atomic motion has sufficient validity that acceleration of nuclei can be considered to require a suitable accompanying force (viz., PES gradient). If the system has enough "inertia" and the PES gradient provides too small an acceleration to overcome it, the progress of the reaction can absolutely move away from the MEP.

Alternatively, at least one additional mechanism can introduce deviations from the MEP: molecules at finite temperature are always vibrating. Thus, even in a case where the key internal molecular coordinates (bond lengths, angles, dihedrals) for a given reaction strictly follow the IRC, the "uninvolved" vibrational motion is still occurring, and so the system is actually always oscillating around the IRC in a high-dimensional sense. If the energy of the system is high enough, these oscillations have the potential to be of sufficient magnitude to "bump" the reaction coordinate away from the gradient-following pathway.

Rotational motion may also contribute to this, but I would think in a more subtle way than vibrational motion.

For further reading: Steven Bachrach of the Computational Organic Chemistry blog has an entire tagged category, "Dynamics," dedicated to this phenomenon. In no particular order, some posts of possible survey value include:

The entirety of Chapter 8 of his book is also dedicated to this topic.

Addendum: FWIW, as a chemical engineer, the (imperfect) analogy that I like to use is to liken the MEP to the 'reversible process' of thermodynamics:

  1. Start at the beginning state
  2. Take an infinitesimal step
  3. Let the system infinitesimally relax
  4. Repeat 2 & 3 until you reach the end state

As I (very approximately) understand it, irreversibility comes from changing the state of a system "too much, too fast", leading to net generation of entropy as the system relaxes back to the reversible pathway, and it's "kinetic energy" in a qualitative, generalized sense that allows the system to depart from the reversible path.


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