From owner-chemistry@ccl.net Fri Apr 13 03:28:00 2012 From: "steinbrt=-=rci.rutgers.edu" To: CCL Subject: CCL:G: energy for proton Message-Id: <-46687-120413032458-6490-7zAIA6Kqskj/51v0J6jqrQ=-=server.ccl.net> X-Original-From: steinbrt..rci.rutgers.edu Content-Transfer-Encoding: 8bit Content-Type: text/plain;charset=iso-8859-1 Date: Fri, 13 Apr 2012 03:24:49 -0400 (EDT) MIME-Version: 1.0 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 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 From owner-chemistry@ccl.net Fri Apr 13 04:31:01 2012 From: "Alexander Bagaturyants bagaturyants-.-gmail.com" To: CCL Subject: CCL:G: energy for proton Message-Id: <-46688-120413042921-1273-nANl1vHuUK5L3Wevg0uu4Q|a|server.ccl.net> X-Original-From: "Alexander Bagaturyants" Content-Language: en-us Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset="koi8-r" Date: Fri, 13 Apr 2012 12:29:09 +0400 MIME-Version: 1.0 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 From owner-chemistry@ccl.net Fri Apr 13 06:44:01 2012 From: "Pedro Silva pedros%%ufp.edu.pt" To: CCL Subject: CCL:G: energy for proton Message-Id: <-46689-120413063925-18568-M+yFSxZT+S8BcEcrsGC8IQ_-_server.ccl.net> X-Original-From: Pedro Silva Content-Transfer-Encoding: 8bit Content-Type: text/plain; charset=ISO-8859-1 Date: Fri, 13 Apr 2012 11:39:17 +0100 MIME-Version: 1.0 Sent to CCL by: Pedro Silva [pedros[*]ufp.edu.pt] Dear Alexander, Your argument " one can never consider a proton as an ideal particle." applies equally to every molecule, even to noble gases. We all know that the theory involves approximations, so why should one "pick" on the proton as a uniquely problematic case for statistical thermodynamics? I also cannot understand why "One can never consider a mole of protons in a finite volume or a mole of protons at a finite pressure", as it is possible to generate a mole of protons by bombarding H atoms with x-rays with the appropriate wavelength to eject all electrons. That system does not contain A-, and its thermodynamics must be describable in some way :-) On the other hand, if the issue at hand on " "One can never consider a mole of protons in a finite volume" is that of electrostatic repulsion, the same argumnent would also apply to one mole of Na+, H-, or any other charged species. Pedro S. 2012/4/13 Alexander Bagaturyants bagaturyants-.-gmail.com : > > 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>      http://www.ccl.net/cgi-bin/ccl/send_ccl_message>      http://www.ccl.net/cgi-bin/ccl/send_ccl_message>      http://www.ccl.net/chemistry/sub_unsub.shtml>      http://www.ccl.net/spammers.txt> > -- Pedro J. Silva Associate Professor Universidade Fernando Pessoa Porto - Portugal http://homepage.ufp.pt/pedros/science/science.htm http://biochemicalmatters.blogspot.com From owner-chemistry@ccl.net Fri Apr 13 08:18:00 2012 From: "Alex A. Granovsky gran,+,classic.chem.msu.su" To: CCL Subject: CCL:G: energy for proton Message-Id: <-46690-120413081514-22975-AMUsl3SdjogSmVFfiFNoUg::server.ccl.net> X-Original-From: "Alex A. Granovsky" Content-Transfer-Encoding: 7bit Content-Type: text/plain; format=flowed; charset="iso-8859-1"; reply-type=original Date: Fri, 13 Apr 2012 16:14:40 +0400 MIME-Version: 1.0 Sent to CCL by: "Alex A. Granovsky" [gran[a]classic.chem.msu.su] Dear Pedro, Thermodynamics is only applicable to macroscopic systems with additive energies. Charged plasma like gas of protons does not satisfy this requirement due to long-range nature of Coulomb interaction. Kind regards, Alex Granovsky -----Original Message----- > From: Pedro Silva pedros%%ufp.edu.pt Sent: Friday, April 13, 2012 2:39 PM To: Granovsky, Alex, A. Subject: CCL:G: energy for proton Sent to CCL by: Pedro Silva [pedros[*]ufp.edu.pt] Dear Alexander, Your argument " one can never consider a proton as an ideal particle." applies equally to every molecule, even to noble gases. We all know that the theory involves approximations, so why should one "pick" on the proton as a uniquely problematic case for statistical thermodynamics? I also cannot understand why "One can never consider a mole of protons in a finite volume or a mole of protons at a finite pressure", as it is possible to generate a mole of protons by bombarding H atoms with x-rays with the appropriate wavelength to eject all electrons. That system does not contain A-, and its thermodynamics must be describable in some way :-) On the other hand, if the issue at hand on " "One can never consider a mole of protons in a finite volume" is that of electrostatic repulsion, the same argumnent would also apply to one mole of Na+, H-, or any other charged species. Pedro S. 2012/4/13 Alexander Bagaturyants bagaturyants-.-gmail.com : > > 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> > -- Pedro J. Silva Associate Professor Universidade Fernando Pessoa Porto - Portugal http://homepage.ufp.pt/pedros/science/science.htm http://biochemicalmatters.blogspot.comhttp://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt From owner-chemistry@ccl.net Fri Apr 13 08:53:01 2012 From: "Hoa Thi My Pham hoahoa9h ~ gmail.com" To: CCL Subject: CCL:G: find the binding energy Message-Id: <-46691-120413083652-27928-aWrnx/gsYWYGuGjGFRmmFQ(!)server.ccl.net> X-Original-From: "Hoa Thi My Pham" Date: Fri, 13 Apr 2012 08:36:43 -0400 Sent to CCL by: "Hoa Thi My Pham" [hoahoa9h]|[gmail.com] Dear sir, I'm Hoa, Vietnam. I used gaussian to finding binding energy and activation energy, but I can't how to do that. Can you help me? Thanks so much From owner-chemistry@ccl.net Fri Apr 13 09:46:00 2012 From: "Bruce Palfey brupalf(-)umich.edu" To: CCL Subject: CCL:G: energy for proton Message-Id: <-46692-120413094427-17390-w7cKBpxAa3daA/RbQ5k0ow(0)server.ccl.net> X-Original-From: Bruce Palfey Content-Type: multipart/alternative; boundary=Apple-Mail-5--464836113 Date: Fri, 13 Apr 2012 09:44:09 -0400 Mime-Version: 1.0 (Apple Message framework v1084) Sent to CCL by: Bruce Palfey [brupalf+/-umich.edu] --Apple-Mail-5--464836113 Content-Transfer-Encoding: quoted-printable Content-Type: text/plain; charset=us-ascii I think the forest is being missed for the trees. The heterolytic = dissociation energies that are to be calculated are very much analogous = to reduction potentials - thermodynamic parameters referring to a = fictional chemical reaction involving free electrons. In the present = case, a proton would be produced in all the dissociation reactions. Its = fate, whether realistic or fictional, whether involving chemistry = (transfer to a base) or physics (a plasma), would be the same for all = reactions. Its energy would be the same, so it needn't be calculated. = Assume it's zero for convenience. Regardless, the calculated differences = between HA and A- can be compared to gain chemical insight on = reactivities of compounds, just like reduction potentials can be. And = two (fictional?) reactions can be combined after reversing one to give a = net reaction without a dodgy ill-defined species, and an associated = energy could be calculated with the fictional part canceling; it could = possibly be measured. ciao, Bruce Bruce A. Palfey Associate Professor of Biological Chemistry & Associate Director, Program in Chemical Biology Department of Biological Chemistry University of Michigan Medical School 5220E MSRB III 1150 W. Medical Center Drive Ann Arbor, MI 48109-0606 (734) 615-2452 brupalf!=!umich.edu On Apr 13, 2012, at 8:14 AM, Alex A. Granovsky = gran,+,classic.chem.msu.su wrote: >=20 > Sent to CCL by: "Alex A. Granovsky" [gran[a]classic.chem.msu.su] > Dear Pedro, >=20 > Thermodynamics is only applicable to macroscopic systems with additive = energies. > Charged plasma like gas of protons does not satisfy this requirement = due to long-range nature of Coulomb interaction. >=20 > Kind regards, > Alex Granovsky >=20 >=20 >=20 > -----Original Message-----=20 >> From: Pedro Silva pedros%%ufp.edu.pt > Sent: Friday, April 13, 2012 2:39 PM > To: Granovsky, Alex, A. Subject: CCL:G: energy for proton >=20 >=20 > Sent to CCL by: Pedro Silva [pedros[*]ufp.edu.pt] > Dear Alexander, >=20 > Your argument " one can never consider a proton as an ideal particle." > applies equally to every molecule, even to noble gases. We all know > that the theory involves approximations, so why should one "pick" on > the proton as a uniquely problematic case for statistical > thermodynamics? >=20 > I also cannot understand why "One can never consider a mole of protons > in a finite volume or a mole of protons at a finite pressure", as it > is possible to generate a mole of protons by bombarding H atoms with > x-rays with the appropriate wavelength to eject all electrons. That > system does not contain A-, and its thermodynamics must be describable > in some way :-) > On the other hand, if the issue at hand on " "One can never consider a > mole of protons in a finite volume" is that of electrostatic > repulsion, the same argumnent would also apply to one mole of Na+, H-, > or any other charged species. >=20 >=20 > Pedro S. >=20 > 2012/4/13 Alexander Bagaturyants bagaturyants-.-gmail.com > : >>=20 >> 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. >>=20 >> 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. >>=20 >> One cannot consider an equilibrium like AH -> A(-) + H(+), and the >> equilibrium constant for this reaction makes no sense. >>=20 >> Sorry for these trivial explanations. >>=20 >> 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. >>=20 >> Best regards >> Alexander >>=20 >>=20 >>> -----Original Message----- >>> From: owner-chemistry+sasha=3D=3Dphotonics.ru-*-ccl.net = [mailto:owner- >>> chemistry+sasha=3D=3Dphotonics.ru-*-ccl.net] On Behalf Of steinbrt=3D-= >>> =3Drci.rutgers.edu >>> Sent: 13 April, 2012 11:25 >>> To: Alexander Bagaturyants >>> Subject: CCL:G: energy for proton >>>=20 >>>=20 >>> Sent to CCL by: steinbrt!A!rci.rutgers.edu Dear CCLers, >>>=20 >>> 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. >>>=20 >>> 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). >>>=20 >>> 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. >>>=20 >>> 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. >>>=20 >>> 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) >>>=20 >>> If I am wrong in any of the above, I would be happy to be corrected = :-) >>>=20 >>> Kind Regards, >>>=20 >>> Thomas >>>=20 >>> 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=3D=3Dphotonics.ru=3D=3D=3Dccl.net = [mailto:owner- >>> >> chemistry+sasha=3D=3Dphotonics.ru=3D=3D=3Dccl.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. >>> >> >>> >> =3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3DOriginal message = text=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D 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=3DNoRaman 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=3D 0.000000 >>> >> (Hartree/Particle) >>> >> Thermal correction to Energy=3D 0.001416 >>> >> Thermal correction to Enthalpy=3D 0.002360 >>> >> Thermal correction to Gibbs Free Energy=3D -0.010000 >>> >> Sum of electronic and zero-point Energies=3D = 0.000000 >>> >> Sum of electronic and thermal Energies=3D = 0.001416 >>> >> Sum of electronic and thermal Enthalpies=3D = 0.002360 >>> >> Sum of electronic and thermal Free Energies=3D = -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{=3D}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=3Dultrafine >>> >> > >>> >> > 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=3D 0.000000 (Hartree/Particle) Thermal >>> >> > correction to Energy=3D 0.001416 Thermal correction to = Enthalpy=3D >>> >> > 0.002360 Thermal correction to Gibbs Free Energy=3D -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 =3D 298.15 K, while "0.002360" (" Thermal correction to >>> >> > Enthalpy") equals 3/2*k*T + k*T since the enthalpy is H =3D U + = P*v >>> >> > while P*v =3D 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=3D=3Dphotonics.ru|,|ccl.net >>> >> >>> [mailto:owner- >>> >> >>> chemistry+sasha=3D=3Dphotonics.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=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3DEnd of >>> >> >>> original message text=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D> To = recover the email address >>> of >>> >> the author of the message, please >>> >> change the strange characters on the top line to the =3D=3D=3D = sign. You >>> >> can >>> >> also> >>> > >>> > >>>=20 >>>=20 >>> 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> >>=20 >=20 >=20 >=20 > --=20 > Pedro J. Silva > Associate Professor > Universidade Fernando Pessoa > Porto - Portugal > http://homepage.ufp.pt/pedros/science/science.htm > = http://biochemicalmatters.blogspot.comhttp://www.ccl.net/cgi-bin/ccl/send_= ccl_messagehttp://www.ccl.net/chemistry/sub_unsub.shtmlhttp://www.ccl.net/= spammers.txt >=20 >=20 > -=3D This is automatically added to each message by the mailing script = =3D- > To recover the email address of the author of the message, please = change>=20>=20>=20=>=20>=20Conferences: = http://server.ccl.net/chemistry/announcements/conferences/ >=20>=20>=20>=20 >=20 >=20 >=20 --Apple-Mail-5--464836113 Content-Transfer-Encoding: quoted-printable Content-Type: text/html; charset=us-ascii Bruce A. Palfey
Associate = Professor of Biological Chemistry &
Associate Director, Program in Chemical = Biology
Department of Biological Chemistry
University of Michigan = Medical School
5220E MSRB III
1150 W. Medical Center Drive
Ann = Arbor, MI 48109-0606
(734) 615-2452
brupalf!=!umich.edu
On Apr 13, 2012, at 8:14 AM, Alex A. Granovsky = gran,+,classic.chem.msu.su wrote:


Sent to CCL by: "Alex A. Granovsky" = [gran[a]classic.chem.msu.su]
Dear Pedro,

Thermodynamics is = only applicable to macroscopic systems with additive = energies.
Charged plasma like gas of protons does not satisfy this = requirement due to long-range nature of Coulomb interaction.

Kind = regards,
Alex Granovsky



-----Original Message----- =
From: Pedro Silva = pedros%%ufp.edu.pt
Sent: Friday, April 13, 2012 2:39 = PM
To: Granovsky, Alex, A. Subject: CCL:G: energy for = proton


Sent to CCL by: Pedro Silva = [pedros[*]ufp.edu.pt]
Dear Alexander,

Your argument " one can = never consider a proton as an ideal particle."
applies equally to = every molecule, even to noble gases. We all know
that the theory = involves approximations, so why should one "pick" on
the proton as a = uniquely problematic case for statistical
thermodynamics?

I = also cannot understand why "One can never consider a mole of = protons
in a finite volume or a mole of protons at a finite = pressure", as  it
is possible to generate a mole of protons by = bombarding H atoms with
x-rays with the appropriate wavelength to = eject all electrons. That
system does not contain A-, and its = thermodynamics must be describable
in some way :-)
On the other = hand, if the issue at hand on " "One can never consider a
mole of = protons in a finite volume" is that of electrostatic
repulsion, the = same argumnent would also apply to one mole of Na+, H-,
or any other = charged species.


Pedro S.

2012/4/13 Alexander = Bagaturyants bagaturyants-.-gmail.com
&= lt;owner-chemistry|*|ccl.net>:

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=3D=3Dphotonics.ru-*-ccl.net = [mailto:owner-
chemistry+sasha=3D=3Dphotonics.ru-*-ccl.net] On Behalf Of = steinbrt=3D-
=3Drci.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=3D=3Dphotonics.ru=3D=3D=3Dccl.net = [mailto:owner-
>> = chemistry+sasha=3D=3Dphotonics.ru=3D=3D=3Dccl.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.
>>
>> =3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3DOriginal message text=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D= =3D=3D 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=3DNoRaman 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)
>> =             &n= bsp;           &nbs= p;            = 0.00000 (Kcal/Mol)
>>   Vibrational = temperatures:
>> =            (Kelvin)=
>>
>>   Zero-point = correction=3D =             &n= bsp;           &nbs= p; 0.000000
>> = (Hartree/Particle)
>>   Thermal = correction to Energy=3D =             &n= bsp;      0.001416
>> =   Thermal correction to Enthalpy=3D =             &n= bsp;    0.002360
>>   Thermal = correction to Gibbs Free Energy=3D =        -0.010000
>> =   Sum of electronic and zero-point Energies=3D =             &n= bsp;0.000000
>>   Sum of = electronic and thermal Energies=3D =             &n= bsp;   0.001416
>>   Sum of = electronic and thermal Enthalpies=3D =             &n= bsp; 0.002360
>>   Sum of = electronic and thermal Free Energies=3D =           -0.010000
<= /blockquote>
>>
>>
>> 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{=3D}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=3Dultrafine
>> = >
>> > 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=3D 0.000000 (Hartree/Particle) = Thermal
>> > correction to Energy=3D 0.001416 Thermal = correction to Enthalpy=3D
>> > 0.002360 Thermal = correction to Gibbs Free Energy=3D = -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 =3D 298.15 K, = while "0.002360" (" Thermal correction = to
>> > Enthalpy") equals 3/2*k*T + k*T since the = enthalpy is H =3D U + P*v
>> > while P*v =3D 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
<= /blockquote>
>> 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=3D=3Dphotonics.ru|,|ccl.net
>> = >>> [mailto:owner-
>> >>> = chemistry+sasha=3D=3Dphotonics.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
>> >>> also=Conferences:>
>> = >>>
>> 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=3D=3D=3D=3D=3D=3D= =3D=3D=3D=3D=3DEnd of
>> >>> original = message text=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D=3D> To recover the email = address
of
>> the author of the = message, please
>> change the strange = characters on the top line to the =3D=3D=3D 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>




--
Pedro J. = Silva
Associate Professor
Universidade Fernando Pessoa
Porto - = Portugal
http://homepage= .ufp.pt/pedros/science/science.htm
http://biochemicalmatters.blogsp= ot.comhttp://www.ccl.net/cgi-bin/ccl/send_ccl_messagehttp://www.ccl.net/ch= emistry/sub_unsub.shtmlhttp://www.ccl.net/spammers.txt


-=3D = This is automatically added to each message by the mailing script = =3D-
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
look up the X-Original-From: line in the mail = header.
=     http://www.ccl.net/cgi-bin/ccl/send_ccl_message =     http://www.ccl.net/cgi-bin/ccl/send_ccl_message
Before posting, check wait time at: = http://www.ccl.net

Search = Messages: http://www.ccl.net/chemistry/searchccl/index.shtml

If = your mail bounces from CCL with 5.7.1 error, check:
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= --Apple-Mail-5--464836113-- From owner-chemistry@ccl.net Fri Apr 13 16:50:00 2012 From: "Prof. Dr. N. Sekar nethi.sekar(-)gmail.com" To: CCL Subject: CCL: G09: TDDFT adiabatic excitation energy Message-Id: <-46693-120412231933-21320-tZy5oQnUiPuFIxSF0Biq8A=server.ccl.net> X-Original-From: "Prof. Dr. N. Sekar" Content-Type: multipart/alternative; boundary=14dae9340ef76db63004bd86f00e Date: Fri, 13 Apr 2012 08:49:23 +0530 MIME-Version: 1.0 Sent to CCL by: "Prof. Dr. N. Sekar" [nethi.sekar]=[gmail.com] --14dae9340ef76db63004bd86f00e Content-Type: text/plain; charset=ISO-8859-1 Dear John McKelvey, You will be able to see the vertical excitation energy, which is nothing but what you are looking for, from the output file of TDDFT On Fri, Apr 13, 2012 at 6:10 AM, John McKelvey jmmckel],[gmail.com < owner-chemistry|,|ccl.net> wrote: > > Sent to CCL by: John McKelvey [jmmckel],[gmail.com] > One might some papers on this by Stefan Grimme by Googling him and > looking at his pub list > > . > > John McKelvey > > On Thu, Apr 12, 2012 at 1:17 PM, Vera Cathrine > vera.cathrine]-[yahoo.com wrote: > > > > Sent to CCL by: "Vera Cathrine" [vera.cathrine+/-yahoo.com] > > Hello Everyone, > > > > I want to calculate the adiabatic energy difference between first > excited state and the ground state for my system. I would like to use > DFT/BLYP method and TDDFT/BLYP method respectively for optimization of my > geometries for these states. However, I am not sure how to calculate the > gap between these state from out file. What are the keywords which I have > to look for in the out files? > > I would thank you for your help in advance. > > Best regards, > > Vera> > > > > > > -- > John McKelvey > 10819 Middleford Pl > Ft Wayne, IN 46818 > 260-489-2160 > jmmckel a gmail.com> > > -- Thanks and regards Prof. Dr. N. Sekar CCol FSDC Co-ordinator, UGC-CAS and Professor in Tinctorial Chemistry Dyestuff Technology Department Institute of Chemical Technology (formerly UDCT) Matunga, Mumbai-400019 Mob +91-9867958452 n.sekar|,|ictmumbai.edu.in sekarnm|,|rediffmail.com drnsekar4562000|,|yahoo.co.in website: http://ictmumbai.edu.in/Fac_FacDetails.aspx?fidno=116 --14dae9340ef76db63004bd86f00e Content-Type: text/html; charset=ISO-8859-1 Content-Transfer-Encoding: quoted-printable
Dear John McKelvey,
You will be able to see the vertical exc= itation energy, which is nothing but what you are looking for, from the out= put file of TDDFT

On Fri, Apr 13, 2= 012 at 6:10 AM, John McKelvey jmmckel],[gmail.= com <ow= ner-chemistry|,|ccl.net> wrote:

Sent to CCL by: John McKelvey [jmmckel],[gmail.com]
One might =A0some papers on this by Stefan Grimme by Googling him and
looking at his pub list

.

John McKelvey

On Thu, Apr 12, 2012 at 1:17 PM, Vera Cathrine
vera.cathrine]-[yahoo.com <owner-chemistry a ccl.ne= t> wrote:
>
> Sent to CCL by: "Vera =A0Cathrine" [vera.cathrine+/-yahoo.com]
> Hello Everyone,
>
> I want to calculate the adiabatic energy difference between first exci= ted state and the ground state for =A0my system. I would like to use DFT/BL= YP method and TDDFT/BLYP method respectively for optimization of my geometr= ies for these states. However, I am not sure how to calculate the gap betwe= en these state from out file. What are the keywords which I have to look fo= r in the out files?
> I would thank you for your help in advance.
> Best regards,
> Vera> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_m= essage> =A0 =A0 =A0http://www.ccl.net/cgi-bin/ccl/send_ccl_messag= e> =A0 =A0 =A0http://www.ccl.net/chemistry/sub_unsub.shtml> = =A0 =A0 =A0ht= tp://www.ccl.net/spammers.txt>
>



--
John McKelvey
10819 Middleford Pl
Ft Wayne, IN 46818
260-489-2160
jmmckel a gmail.com



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--
Thanks= and regards

Prof. Dr. N. Sekar=A0=A0 CCol FSDC
Co-ordinator, UGC= -CAS and Professor in Tinctorial Chemistry
Dyestuff Technology Departmen= t
Institute of Chemical Technology (formerly UDCT)
Matunga, Mumbai-400019<= br>
Mob +91-9867958452
n.sekar|,|ictmumbai.edu.in
sekarnm|,|rediffmail.com
drnsekar45= 62000|,|yahoo.co.in
=A0

--14dae9340ef76db63004bd86f00e-- From owner-chemistry@ccl.net Fri Apr 13 18:19:00 2012 From: "errol lewars elewars###trentu.ca" To: CCL Subject: CCL:G: find the binding energy Message-Id: <-46694-120413133202-14522-+aERVdFRz22jSjpso8EpKQ|-|server.ccl.net> X-Original-From: errol lewars Content-Transfer-Encoding: 7bit Content-Type: text/plain; charset=ISO-8859-1; format=flowed Date: Fri, 13 Apr 2012 13:32:09 -0400 MIME-Version: 1.0 Sent to CCL by: errol lewars [elewars=-=trentu.ca] 2012 April 13 Binding energy is E(products) - E(reactants). Activation E is E(transition state) - E(reactants). The reliability of your quantities depends on the method used and how accurate you need the results to be. Since you are evidently just beginning calculations like these, you should read some introductions to computational chemistry and relevant papers in journals. E. Lewars Trent University == Hoa Thi My Pham hoahoa9h ~ gmail.com wrote: > Sent to CCL by: "Hoa Thi My Pham" [hoahoa9h]|[gmail.com] > Dear sir, I'm Hoa, Vietnam. I used gaussian to finding binding energy and > activation energy, but I can't how to do that. Can you help me? Thanks so much> > >