CCL:G: energy for proton



 Sent to CCL by: "Alexander Bagaturyants" [bagaturyants%gmail.com]
 Dear Thomas,
 Of course, a proton can exist as a free particle in vacuum, and you also can
 calculate some formal quantities using some standard equations, but (!)
 One can never consider a mole of protons in a finite volume or a mole of
 protons at a finite pressure, and one can never consider a proton as an
 ideal particle.
 A free proton can have any possible kinetic energy, but it cannot exist in a
 thermodynamic equilibrium and its kinetic energy depends not on the bath
 temperature but on conditions of its generation. It cannot be characterized
 by a temperature, because temperature relates to an ensemble rather than an
 individual particle.
 One cannot consider an equilibrium like AH -> A(-) + H(+), and the
 equilibrium constant for this reaction makes no sense.
 Sorry for these trivial explanations.
 PS: I would not like to discuss here de Broglie's concept of hidden
 thermodynamics of an isolated particle, because this is a different story.
 Best regards
 Alexander
 > -----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