For proton, ZPE is obviously zero, since there is no vibration!
There is also no electron and no rotation involved in the possible motion of the
proton. Thus, the energy of proton is translational, and E=1.5*RT=0.001416
Hartree, DH=2.5*RT=0.002360 Hartree=6.197 kJ/mol, DG=DH-TDS=-0.01 Hartree. To
convert energy into enthalpy, we add PV, which is RT for an ideal gas. The
entropy can be calculated using the Sackur Tetrode equation. Note that it is
also translational.
You can also find this gaussian output
helpful:
-------------------
-
Thermochemistry -of the proton -------------------
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
E
(Thermal)
CV
S
KCal/Mol
Cal/Mol-Kelvin Cal/Mol-Kelvin
Total
0.889
2.981
26.014
Electronic
0.000
0.000
0.000
Translational
0.889
2.981
26.014
Rotational
0.000
0.000
0.000
Vibrational
0.000
0.000
0.000
Q
Log10(Q)
Ln(Q)
Total Bot
0.397679D+05
4.599533 10.590816
Total V=0
0.397679D+05
4.599533 10.590816
Vib (Bot)
0.100000D+01
0.000000 0.000000
Vib (V=0)
0.100000D+01
0.000000 0.000000
Electronic
0.100000D+01
0.000000 0.000000
Translational
0.397679D+05
4.599533 10.590816
Rotational
0.100000D+01
0.000000 0.00000
Hope
this would help.
Best regards,
On 12 April
2012 08:17, Tymofii Nikolaienko tim_mail*_*
ukr.net <owner-chemistry : ccl.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
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
also>
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