CCL:G: Thermodynamic Data & Solvation - Calculation Questions:

 Sent to CCL by: Eric Hermes [erichermes!^!]
 Dr. Mielczarek,
 As always, the free energy of a species is defined with respect to some
  choice of reference state, such that for a reaction of the form:
 (1) A + B -> C
 with a standard state free energy of reaction ΔG°, one can write the
 equilibrium constant as:
 (2) Keq = exp[-ΔG°/(kB T)] = a_C / (a_A a_C)
 Where a is the *activity* of a species. In the case of an ideal gas,
 the activity is simply the density divided by the *standard state*
 (3) a = ρ/ρ°
 The numerical value of Keq then depends on choice of reference state,
 but the equilibrium densities do not. Eq. 2 satisfies this requirement.
 One can show that the free energy of a species at an *arbitrary*
 reference state is given by:
 (4) G = G° + kB T ln[ρ/ρ°]
 This is *general*, but you can show that for an ideal gas the
 expression derives from the translational entropy.
 Now, when we are discussing free energy of solvation, the reaction
 (5) A_g -> A_soln
 Which of course brings with it changes to both the enthalpy and the
 entropy. These issues are *orthogonal* to the reference state issue,
 though. The standard state for species in the gas phase is usually
 taken to be 1 bar (occasionally 1 atm is chosen instead), but the
 standard state for species in *solution* is typically chosen to be 1 M,
 which is very different! With these choices of reference state, the
 free energy of solvation ΔG°solv can be used to calculate the
 equilibrium constant like this (using P for pressure to visually
 distinguish from the density ρ):
 (6) Keq = exp[-ΔG°solv/(kB T)] = (ρ_A/(1 M)) / (P_A / (1 bar))
 As before, the numerical value of Keq depends on the choice of
 reference state, but the equilibrium density and pressure do not.
 So, if your goal is to determine the equilibrium properties of a
 solvation process, then you do not need to be concerned about the
 choice of reference state -- you need to be *aware* of it in order to
 calculate the properties directly, but whichever choice you make you
 will get the same answer. On the other hand, if you wish to compare
 your free energies of solvation to experimental values, you should most
 definitely ensure that the value you are calculating uses a reference
 state of 1 M for solution-phase species and 1 bar for gas-phase
 In the case of SMD, by default it will use the *same* reference state
 for the gas-phase and solution-phase species. The actual value of the
 reference state is irrelevant, as ΔGsolv has the same value for any
 choice of reference state so long as the reactants and products have
 the *same* reference state (if you are not convinced, stare at eqs 2,
 3, and 4 in the context of eq 5 until you believe me :) ).
 This means you can arbitrarily say what reference state ΔGsolv is at,
 say 1 bar, for both reactants and products. Then, if you want to
 calculate ΔG°solv (the *experimental* standard state value), you just
 need to use eq 4:
 (7) ΔGsolv = Gsoln - G°gas (we choice a reference state of 1 bar)
 (8) Gsoln = G°soln + kB T ln[(1 bar / (kB T))/1 M] (ideal gas law)
 (9) ΔG°solv = G°soln - G°gas = ΔGsolv - kB T ln[(1
 bar / (kB T))/1 M]
 Note that you would have gotten the *exact same result* if you had
 arbitrarily chosen a reference state of 1 M instead for ΔGsolv (if you
 don't believe me, try it yourself -- start with ΔGsolv = G°soln -
 Now, as to your point about the entropy of solution phase species --
 I'm not sure the paper you are linking is making the claim you are
 saying it does. It is an *unarguable fact* that (at the typical choice
 of standard states) a species in solution has less entropy than the
 same species in the gas phase. This *must* be the case because the
 molecules in an ideal gas are non-interacting, uncorrelated, and
 undergoing ballistic motion, whereas in solution molecules are *caged*
 by the solvent and undergoing diffusive motion, which means they have
 significantly less freedom of motion.
 The paper you link talks about *vibrational* motion, which is going to
 be *significantly* less perturbed by the presence of solvent. The
 *vast* majority of entropy for gas-phase species comes from
 translational motion, not vibrational motion.
 Finally, I'm not sure SMD is well-suited to give a *breakdown* of
 enthalpic and entropic terms to the free energy of solvation. The
 procedure you discuss will get *some* of that breakdown, but ultimately
 several contributions to ΔGsolv are all entangled with one another. 
 What I mean by that is if you simply perform two *single point*
 calculations on the same species, one with and one without SMD
 correction, the difference in the potential energies between those two
 calculations will include a mixture of enthalpic and entropic effects. 
 Doing the full thermodynamic calculations (i.e. doing a frequency
 calculation and reading the thermodynamic data printed in the Gaussian
 output file) will give you a *full* estimate for ΔGsolv. But, in both
 calculations (with and without SMD) the calculated enthalpic and
 entropic contributions are arising from the same set of approximations
 -- harmonic oscillator for vibration, rigid rotor for rotation, and
 *ideal gas* for translation.
 This works because the *potential energy* difference between the
 systems is parameterized such that the free energy difference
 calculated by Gaussian is a good approximatino of the total free energy
 of solvation. Since the enthalpy and entropy calculated by Gaussian for
 both calculations are using the same approximations, they will have
 very similar (but not identical!) values. All of those complicated
 factors such as the loss of translational entropy due to solvation are
 baked into the *electronic energy* that is calculated by the SMD
 The most *robust* way of separating enthalpic and entropic
 contributions would be to calculate the full value of the free energy
 at a variety of different temperatures, then numerically evaluating
 dG/dT to find the entropy. However, SMD as implemented in Gaussian is
 not temperature-dependent, so you cannot actually do this. There are
 other continuum solvation models developed by Cramer and Truhlar which
 do have temperature-dependence, which would allow you to do this, but
 as far as I can tell they are not included in Gaussian.
 I hope this answers all of the questions that you had. Please let me
 know if anything I said was unclear or you want additional assistance.
 Eric Hermes
 On Thu, 2017-04-06 at 06:24 +0000, MIELCZAREK Detlev Conrad detlev- wrote:
 > Sent to CCL by: MIELCZAREK Detlev Conrad [detlev-
 > conrad.mielczarek]^[]
 > Dear CCL, a question on thermodynamic data & solvation from me, maybe
 > you can help me.
 > So, the basic problem for me is, that I am calculating/want to
 > calculate thermodynamic data (Hf, S - hence also dG) in solvation,
 > using implicit solvation models, SMD with a COSMO cavity to be
 > specific. For my application, these should be accurate enough. (So no
 > molecular dynamics simulations etc.)
 > Solvation models are normally parametrised for dGsolv - so this value
 > can be extracted from the quantum chemistry calculation as the
 > difference in the calculated Gibbs Free Enthalpy.
 > Hf can calculated easily in the gas phase, and a re-optimisation of
 > the structure with solvation should capture the majority of the
 > impact of solvation on the enthalpy. (Which is dominated by molecular
 > structure.)
 > (I guess there is the case of stabilisation and complexes, such as
 > are reported for water. However this is currently beyond the scope of
 > my work.)
 > The topic of solvation has been discussed previously on the CCL here:
 > And there is the book "Essentials of Computational Chemistry Theories
 > and Models" from Professor Cramer with a section on phase change (the
 > source of my confusion).
 > Specifically, the discussion concerning the energy change related to
 > the state conversion causes me some grief.
 > On the one hand, the CCL responses read as if this should be applied
 > in the case of any phase change, but then others suggest this is
 > applicable only if the process is a second order reaction and thus
 > the total number of mols changes? - The latter view seems to agree
 > with the book...
 > So if I have compound A in both the gas and liquid phase (from a
 > quantum chemistry calculation), do I need to account for the phase
 > change/change of state or not? Or is it something that can be
 > included in the parametrisation of the solvation model/the quantum
 > chemistry code already?
 > Just to add more confusion to the topic: I have trialled a commercial
 > product which gives the Gibbs Enthalpy of Solvation in kcal/mol for
 > mol/L concentrations and using a very low end/fast functional, it
 > gives values similar to when a correction term is added... on the
 > other hand, where available, the calculated values without correction
 > agree with the published values in the SMD paper:
 > /doi/abs/10.1021/jp810292n (Supplementary Data)
 > In addition, a regular computational chemistry calculation sees very
 > little (virtually no) difference in the entropy between the gaseous
 > and solvated phase. This would agree with the CCL-linked paper here h
 > ttp:// . But this would clash
 > with the common expectation that entropy in the liquid phase is
 > reduced...
 > Hence, I would highly appreciate if someone knowledgeable in the
 > field of solvation could guide me onto the correct track.
 > Detlev Conrad Mielczarek
 > Scientific Visitor/Post Doctorant
 > IFP Energies nouvelles
 > France
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