-----Original Message-----
From: owner-chemistry+sasha==photonics.ru-*-ccl.net [mailto:owner-
chemistry+sasha==photonics.ru-*-ccl.net] On Behalf Of steinbrt=-
=rci.rutgers.edu
Sent: 13 April, 2012 11:25
To: Alexander Bagaturyants
Subject: CCL:G: energy for proton
Sent to CCL by: steinbrt!A!rci.rutgers.edu Dear CCLers,
I found the most recent comments on this by Alexander suprising. This
is a particularly interesting discussion thread and I hope someone else
will comment more on this.
Why would thermodynamic parameters for a proton not make sense? As a
free particle in vacuum it exists, and for any particle with a mass I
can compute ideal values for its enthalpy, entropy and temperature
(well, for one particle, kinetic energy at least).
Moving from a description of a single particle in vacuum to a mol of
(ideal) particles at standard state is then at least conceptually
possible.
The fact that a mol of H+ does not exist any more than a mol of Ac-
doesn't change the fact that I can use it as a reference in computing
thermodynamic properties of the reaction HAc -> H+ + Ac- from the
properties of its constituent molecules.
Furthermore, why can a deprotonation reaction not be described by
thermodynamics? The fragmentation of e.g. HF -> F- + H+ in vacuum will
have an equilibrium that lies exceptionally far to the left, but at
high temperature some fragmentation would occur (disregarding the
alternative radical cleavage for now, which would be more likely but
shouldn't stop the heterolytic cleavage from happening)
If I am wrong in any of the above, I would be happy to be corrected :-)
Kind Regards,
Thomas
On Thu, April 12, 2012 1:17 pm, Alexander Bagaturyants
sasha#photonics.ru
wrote:
>
> Sent to CCL by: "Alexander Bagaturyants" [sasha .. photonics.ru]
A
> good comment by Prof. Sukumar!
> However, it seems that it was not too straightforward, because some
> misleading comments still appear and appear.
> The matter of fact is that neither enthalpy, nor entropy, nor
> temperature of a free proton makes physical sense. One cannot
> construct a (thermodynamic) ensemble of free protons. A reaction in
> which a proton is detached from a molecule can proceed only under
> nonequilibrium conditions, it is a dynamic rather than thermodynamic
> process.
> That is, it senseless to calculate formally any thermodynamic
function
> of a free (individual) proton.
> Hope this will make things a little bit more clear.
> Best regards
> Alexander
>
>> -----Original Message-----
>> From: owner-chemistry+sasha==photonics.ru===ccl.net [mailto:owner-
>> chemistry+sasha==photonics.ru===ccl.net] On Behalf Of N. Sukumar
>> nagams(a)rpi.edu
>> Sent: 12 April, 2012 15:27
>> To: Alexander Bagaturyants
>> Subject: CCL:G: energy for proton
>>
>>
>> Sent to CCL by: "N. Sukumar" [nagams~~rpi.edu] "Why not
perform the
>> calculation on proton?"
>>
>> This is an interesting philosophical/pedagogical question. My answer
>> would be: because for many students (and others), the output from a
>> computer is the end of the problem, not the beginning of the
question!
>> If the computation is used as an aid to understand the chemistry,
>> well and good. But many people these days will not believe a
>> numerical answer unless it is produced by a calculator or a
computer.
>> And they may see no need to question those numbers/output any
further.
>>
>> N. Sukumar
>> Rensselaer Exploratory Center for Cheminformatics Research Professor
>> of Chemistry Shiv Nadar University
>> --------------------------
>> "When you get exactly the opposite result to what you predict, you
>> know it is right, because there is no bias." -- David Nutt,
Imperial
>> College, London.
>>
>> ==============Original message text=============== On Thu, 12 Apr
>> 2012
>> 4:13:44 EDT "Peeter Burk peeter.burk^ut.ee" wrote:
>>
>>
>> Sent to CCL by: Peeter Burk [peeter.burk * ut.ee] Why not perform
the
>> calculation on prorton? With Gaussian 09 (if I remember correctly,
>> then with g03 you had to use Freq=NoRaman to get the same results)
>> you will
>> get:
>>
>> -------------------
>> - Thermochemistry -
>> -------------------
>> Temperature 298.150 Kelvin. Pressure 1.00000 Atm.
>> Atom 1 has atomic number 1 and mass 1.00783
>> Molecular mass: 1.00783 amu.
>> Zero-point vibrational energy 0.0 (Joules/Mol)
>> 0.00000 (Kcal/Mol)
>> Vibrational temperatures:
>> (Kelvin)
>>
>> Zero-point correction= 0.000000
>> (Hartree/Particle)
>> Thermal correction to Energy= 0.001416
>> Thermal correction to Enthalpy= 0.002360
>> Thermal correction to Gibbs Free Energy= -0.010000
>> Sum of electronic and zero-point Energies= 0.000000
>> Sum of electronic and thermal Energies= 0.001416
>> Sum of electronic and thermal Enthalpies= 0.002360
>> Sum of electronic and thermal Free Energies= -0.010000
>>
>>
>> Peeter Burk
>> University of Tartu
>>
>> On 04/12/2012 10:17 AM, Tymofii Nikolaienko tim_mail*_*ukr.net
wrote:
>> >
>> > Sent to CCL by: Tymofii Nikolaienko [tim_mail{=}ukr.net] Yes, ZPE
>> > is zero.
>> > However, if considering temperatures higher than 0 K, we can NOT
>> > neglect the kinetic energy of the proton, since its thermal
avarage
>> is
>> > 3 * kT / 2 !
>> >
>> > It is easy to demonstrate if you run the following for example
with
>> > H
>> atom:
>> >
>> > # opt freq b3lyp/aug-cc-pVQZ int=ultrafine
>> >
>> > H atom
>> >
>> > 0 2
>> > H 0.0 0.0 0.0
>> >
>> > And than you read in the output file:
>> > ...
>> > - Thermochemistry -
>> > -------------------
>> > Temperature 298.150 Kelvin. Pressure 1.00000 Atm.
>> > ...
>> > Zero-point correction= 0.000000 (Hartree/Particle) Thermal
>> > correction to Energy= 0.001416 Thermal correction to Enthalpy=
>> > 0.002360 Thermal correction to Gibbs Free Energy= -0.010654
>> >
>> > These thermal corrections would be just that same for the proton
>> since
>> > when calculating thermochemistry Gaussian assumes ground electron
>> > state only (so no electronic degrees of freedom contribute to
>> > thermal corrections; see
>> > http://www.gaussian.com/g_whitepap/thermo.htm ).> Note
that
>> > "0.001416" (the "Thermal correction to
Energy") equals 3/2*k*T for
>> > T = 298.15 K, while "0.002360" (" Thermal
correction to
>> > Enthalpy") equals 3/2*k*T + k*T since the enthalpy is H = U +
P*v
>> > while P*v = k*T for ideal gas - the model for calculating
>> > thermochemistry Gaussian assumes (where v is the gas volume per
>> > particle). To obtain Gibbs free energy use the -T*s term where s
is
>> > the entropy of ideal gas per particle at given temperature.
>> >
>> > Yours sincerely
>> > Tymofii Nikolaienko
>> >
>> >
>> > 12.04.2012 8:30, Alexander Bagaturyants bagaturyants-.-gmail.com
>> wrote:
>> >> Sent to CCL by: "Alexander Bagaturyants"
>> >> [bagaturyants_-_gmail.com] Dear Arturo, Proton has no internal
>> >> degrees of freedom; therefore, its energy is zero, if we
neglect
its kinetic energy.
>> >> Naturally, the kinetic energy (of a free proton) can take on
any
>> >> value, so that we may speak about so-called dissociation
threshold.
>> >> A piece of advice: when you consider chemistry, you should not
>> >> sometimes forget about physics.
>> >> Best regards
>> >> Alexander
>> >>
>> >>> -----Original Message-----
>> >>> From: owner-chemistry+sasha==photonics.ru|,|ccl.net
>> >>> [mailto:owner-
>> >>> chemistry+sasha==photonics.ru|,|ccl.net] On Behalf Of
Arturo
>> >>> chemistry+Espinosa
>> >>> artuesp|*|um.es
>> >>> Sent: 11 April, 2012 21:12
>> >>> To: Alexander Bagaturyants
>> >>> Subject: CCL: energy for proton
>> >>>
>> >>>
>> >>> Sent to CCL by: Arturo Espinosa [artuesp(_)um.es] Dear CCL
users:
>> >>>
>> >>> I am trying to compute ZPE-corrected dissociation energies
for
>> >>> some particular bonds, in order to correlate these values
with
>> >>> other properties computed at the same level (starting
from,
let's
>> >>> say,
>> >>> B3LYP- D/def2-TZVP). My problem (perhaps a bit stupid)
comes
when
>> >>> dealing with heterolytic dissociations of a A-H bond to
give A-
>> >>> (anion) and H+ (a proton). Moreover I am intending to
compare
>> >>> this dissociation with the other possible heterolytic
>> >>> dissociation and even with the homolytic one. Calculation
of the
>> >>> A-H and A- species is straighforward (no matter what level
of
>> >>> calculation), but the problem is what value (in atomic
>> >>> units) should I assign to the H+ species. No QC
calculation is
>> >>> possible as there are no electrons. I recognize that I am
a bit
>> lost.
>> >>> Suggestions are wellcome.
>> >>> Thank you in advance and best regards,
>> >>> Arturo> To recover the email address of the author of
the
>> >>> Arturo> message,
>> >>> please
>> >>> change the strange characters on the top line to the |,|
sign.
>> >>> You can
>> >>> alsohttp://www.ccl.net/chemistry/sub_unsub.shtmlConferences:>;
>> >>>
>> http://server.ccl.net/chemistry/announcements/conferences/http-:-//www
>> >>> .ccl.net/cgi-
>> bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/su
>> >>> b_unsub.shtmlhttp://www.ccl.net/spammers.txt===========End of
>> >>> original message text===========> To recover the email
address
of
>> the author of the message, please
>> change the strange characters on the top line to the === sign. You
>> can
>> also>
>
>
Dr. Thomas Steinbrecher
formerly at the
BioMaps Institute
Rutgers University
610 Taylor Rd.
Piscataway, NJ 08854> To recover the email address of the author of
the
message, please
change the strange characters on the top line to the -*- sign. You can
also>