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

• From: Andreas Klamt <klamt]|[cosmologic.de>
• Subject: CCL:G: Thermodynamic Data & Solvation - Calculation Questions:
• Date: Sat, 8 Apr 2017 10:44:37 +0200

``` Sent to CCL by: Andreas Klamt [klamt-.-cosmologic.de]
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
Andreas
> HI TO ALL!
>
>
> 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" !
>
> BEST REGARDS,
>
> 2017-04-06 17:08 UTC+02:00, Eric Hermes erichermes**gmail.com
> <owner-chemistry^_^ccl.net>:
>> Sent to CCL by: Eric Hermes [erichermes!^!gmail.com]
>> 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
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
>> 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
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 -
>> 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
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
>> 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.
>>
>> ---
>>
>> 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-
>>> Sent to CCL by: MIELCZAREK Detlev Conrad [detlev-
>>> 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:
>>> http://www.ccl.net/chemistry/resources/messages/2011/12/01.001-dir/
>>> http://www.ccl.net/chemistry/resources/messages/2011/10/06.005-dir/
>>> http://www.ccl.net/chemistry/resources/messages/2014/05/01.004-dir/
>>> 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
>>>
>>> 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: http://pubs.acs.org
>>> /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://pubs.acs.org/doi/abs/10.1021/jp205508z . 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.
>>>
>>> Scientific Visitor/Post Doctorant
>>> IFP Energies nouvelles
>>> France
>>>
>>> www.ifpenergiesnouvelles.fr
>>>
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--
--------------------------------------------------
Prof. Dr. Andreas Klamt
CEO / Geschäftsführer
COSMOlogic GmbH & Co. KG
Imbacher Weg 46
D-51379 Leverkusen, Germany
phone  	+49-2171-731681
fax    	+49-2171-731689
e-mail 	klamt_+_cosmologic.de
web    	www.cosmologic.de
[University address:      Inst. of Physical and
Theoretical Chemistry, University of Regensburg]
HRA 20653 Amtsgericht Koeln, GF: Prof. Dr. Andreas Klamt
Komplementaer: COSMOlogic Verwaltungs GmbH
HRB 49501 Amtsgericht Koeln, GF: Prof. Dr. Andreas Klamt
```