*From*: Eric Hermes <erichermes*o*gmail.com>*Subject*: CCL: Entropy of a Bimolecular System*Date*: Tue, 07 Feb 2017 15:22:24 -0600

Sent to CCL by: Eric Hermes [erichermes[#]gmail.com] EC, You are on the right path. First, it is important to understand how free energies of gas-phase species are calculated. I would suggest reading the relevant chapters of McQuarrie's Statistical Mechanics (5 through 8 in my version) or Hill's Introduction to Statistical Thermodynamics (5 through 9). The free energy of a gas-phase molecule has contributions from the electronic degrees of freedom, translation, rotation, and vibration. Typically, the free energy from electronic degrees of freedom is assumed to be the potential energy plus an entropic term arising from multiplicity. This comes from the assumption that excited electronic states are thermally inaccessible, but this assumption breaks down if the system has low-lying excited states or is metallic. The translational free energy is typically calculated using the particle-in-a-box picture and employing the ideal gas approximation. In your case, this is where the confusion arises. If you treat the bimolecular system as a single system with only three degrees of translational freedom, you are going to massively underestimate the amount of entropy. Instead, you must consider them as two separate non- interacting systems with three degrees of translational freedom each. The rotational free energy is typically calculated by employing the rigid rotor approximation. If you treat your bimolecular system as a single system, you will also be calculating this value incorrectly, as a system composed of two non-bonded molecules is not a rigid rotor. The vibrational free energy is typically calculated by employing the harmonic oscillator approximation. For this, the second derivative matrix of the energy (the Hessian) is calculated and diagonalized. The eigenvalues of the Hessian correspond to vibrational frequencies and the eigenvectors the corresponding normal modes. The Hessian is 3N dimensional, and since there are only 3N-6 vibrational modes (3N-5 if the molecule is linear), the 6 lowest frequency modes are typically discarded -- these should correspond to some linear combination of translational and rotational modes. If you treat your bimolecular system as a single system, you will likely have an additional 6 low frequency modes corresponding to additional rotations or intermolecular translations. In summary, make sure you are accounting for each species degrees of freedom correctly: 1 electronic, 3 translational, 3 rotational (2 if linear), and 3N-6 vibrational (3N-5 if linear) modes per molecule. Also, remember that the resulting free energy is the free energy at the standard state, typically 298 K and 1 atm. Eric On Tue, 2017-02-07 at 11:48 -0600, Ernest Chamot echamota/chamotlabs.com wrote: > Hi All, > > I seem to have argued myself into a state of confusion: I guess I > just don’t really understand the entropy of a bimolecular system. > > I can calculate the enthalpy of a molecule with any number of > methods, and so long as I also do an IR or frequency calculation, I > can also get the entropy, and ultimately the free energy of the > molecule. So if I am considering the equilibrium of a dissociation > reaction, I can get the heat of reaction by modeling all three > species, and subtracting the enthalpy of the reactant from the sum of > the enthalpies of the products. But how do I calculate the free > energy of reaction? > > I can’t just add up the individual free energies, can I? Isn’t the > entropy of the pair of product molecules different from just the sum > of the two individual entropies? Since there are two separate > molecules in the same frame of reference, there should be an > additional 6 degrees of freedom for the second molecule, even at > infinite separation. Or do these all have a correspondence with a > vibrational mode in the original reactant molecule? Doesn’t there > need to be an additional term or factor: ln(2), or angular momentum, > or something? > > (I’m interested in the overall reaction, not with the two product > molecules still bound together in some intermediate complex. > Otherwise I could just model that.) > > Thanks for any help. > > EC > > > Ernest Chamot > Chamot Labs, Inc. > http://www.chamotlabs.com > >