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

 Sent to CCL by: Andreas Klamt []
 Hi together,
 it is surely true that the most fundamental way to separate H and TS is
 doing a temperature derivative. But that requires that the free energy
 has the correct temperature dependence. Please note that the 2nd
 derivatives  in solution, if calculate from a continuum solvation model,
 are not the true vibrational frequencies. It is pure fiction to assume
 that in this way you would end up the correct enthalpy and entropy.
 In COSMO-RS we have a consistent, although not completely ab initio (no
 practically usable solvation model is completely ab initio!), scheme for
 temperature and mixture dependent free energies, and thus for H and TS.
 Best regards
 Am 07.04.2017 um 09:12 schrieb adon cumi adonmage-
 > Sent to CCL by: adon cumi [adonmage ..]
 > 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
 > The JAGUAR program from SCHRODINGER Inc. It provide a tool via
 > jaguar-->single point energy-->vibrational frequencies. in the tab
 > below you should be able to increment the temperature to do your
 > "dG/dT" !
 > 2017-04-06 17:08 UTC+02:00, Eric Hermes erichermes**
 > <owner-chemistry^_^>:
 >> 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*
 >> density:
 >> (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
 >> becomes:
 >> (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
 >> 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
 >> species.
 >> 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
 >> 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
 >> 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
 >> (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 -
 >> Ggas).
 >> ---
 >> 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
 >> 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
 >> 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
 >> method.
 >> 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,
 >>> 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
 >>> molecular dynamics simulations etc.)
 >>> Solvation models are normally parametrised for dGsolv - so this
 >>> 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
 >>> structure.)
 >>> (I guess there is the case of stabilisation and complexes, such as
 >>> are reported for water. However this is currently beyond the scope
 >>> my work.)
 >>> The topic of solvation has been discussed previously on the CCL
 >>> And there is the book "Essentials of Computational Chemistry
 >>> and Models" from Professor Cramer with a section on phase
 change (the
 >>> source of my confusion).
 >>> Specifically, the discussion concerning the energy change related
 >>> the state conversion causes me some grief.
 >>> On the one hand, the CCL responses read as if this should be
 >>> 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
 >>> 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
 >>> agree with the published values in the SMD paper:
 >>> /doi/abs/10.1021/jp810292n (Supplementary Data)
 >>> In addition, a regular computational chemistry calculation sees
 >>> little (virtually no) difference in the entropy between the gaseous
 >>> and solvated phase. This would agree with the CCL-linked paper here
 >>> 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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