CCL:G: Thermodynamic Data & Solvation - Calculation Questions:
- From: MIELCZAREK Detlev Conrad
- Subject: CCL:G: Thermodynamic Data & Solvation - Calculation
- Date: Fri, 7 Apr 2017 06:41:58 +0000
Sent to CCL by: MIELCZAREK Detlev Conrad [detlev-conrad.mielczarek * ifpen.fr]
Dear Uwe, many thanks for the explanation and the reference/paper.
Dear Eric, many thanks for your very comprehensive response - with a detailed
explanation how and why the terms are chosen - possibly the best I have seen
Hopefully I won't have any further questions after sitting down to properly read
your response and the linked paper.
Once again, many thanks for your two comprehensive responses.
Detlev Conrad Mielczarek
Scientific Visitor/Post Doctorant
IFP Energies nouvelles
detlev-conrad.mielczarek - - ifpen.fr
> From: owner-chemistry+detlev-conrad.mielczarek==ifpen.fr|ccl.net [mailto:owner-chemistry+detlev-conrad.mielczarek==ifpen.fr|ccl.net] On
Behalf Of Eric Hermes erichermes**gmail.com
Sent: 06 April 2017 18:57
To: MIELCZAREK Detlev Conrad
Subject: CCL:G: Thermodynamic Data & Solvation - Calculation Questions:
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*
(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 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*
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.
On Thu, 2017-04-06 at 06:24 +0000, MIELCZAREK Detlev Conrad detlev-
> Sent to CCL by: MIELCZAREK Detlev Conrad [detlev-
> conrad.mielczarek]^[ifpen.fr] 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
> (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
> 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: 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
> 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
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