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In the RASPA2 manual, https://raw.githubusercontent.com/numat/RASPA2/master/Docs/raspa.pdf

page 106,

ExternalTemperature 450.0

ExternalPressure 101300.

...

MoleculeName propene

LnPartitionFunction 87.1384


In version 2.0.37 manual, page 106

PartitionFunction 6.977909e37


looks to me the  LnPartitionFunction was obtained by taking natural log on the previous version (exp(87.1384)~ 6.97...e37)


However, if I compare

https://github.com/numat/RASPA2/blob/master/examples/Tutorial/ReactionEnsembleAmmonia/simulation.input

N2

LnPartitionFunction 208.188


In version 2.0.37, examples/Tutorial/ReactionEnsembleAmmonia/simulation.input

N2 is

PartitionFunction 3.81253e4


that is not associated with exp, namely exp(208.188) >> 3.8..e4

I think in the N2 example, the atomization energy was used in the new partition function after version 2.0.37. (https://github.com/iRASPA/RASPA2/blob/master/ChangeLog.md, "Updated partition factors for ReactionEnsembleAmmonia example")


With/without atomization energies (from https://cccbdb.nist.gov/atomize2.asp, Propene about 3395 kJ/mol experimental atomization energies), I cannot obtain the 6.977909e37 for propene.


May I know how the propene partition function value was obtained?  Thank you very much

Hi,

  This is related to https://forums.iraspa.org/index.php?topic=1227.0
  In RASPA2, if I understand correctly, https://pure.tudelft.nl/ws/portalfiles/portal/71296377/Thesis_Ahmadreza_Rahbari.pdf, A73, including an exponential factor exp(D_e/(kT)).

  If the system includes a charged species, e.g., OH^-, will the atomization energy be calculated from negatively charged atom, e.g., OH^- -> O^- and H?

  From what I understand, the atomization energy is for setting up zero point, as long as the references (e.g., neutral atoms for neutral molecules) are consistent, the calculations should be fine. But, I know very little about the implementation of RASPA2 (go through the source code is quite complicated), does OH^- -> O^- and H OK for simulating charged molecules? Or should I use a different approach?

Thank you very much

 

From https://github.com/iRASPA/RASPA2/blob/master/examples/Tutorial/ReactionEnsembleAmmonia/simulation.input

"

Component 2 MoleculeName NH3 

 

MoleculeDefinition Local 

 

LnPartitionFunction 253.69"


From

https://github.com/iRASPA/RASPA2/issues/4 and
https://github.com/iRASPA/RASPA2/files/3927153/RASPA.Ammonia.Synthesis.Partition.Function.pdf

Table 2,
"NH3 1.8118E+02 1.9951E+02 1.1422E+00 4020.5 1.66002E+08 1.498133E+08"

The partition function seems to be ~ 1*10^8. I calculated the partition function of NH3 using quantum chemistry software and got ~ 10^8 value.


I am unable to relate 253.69 to ~1*10^8, for example, from https://www.rapidtables.com/calc/math/Ln_Calc.html

ln(1.498133e8) gives me ~ 18.82.

May I know how the partition function and LnPartitionFunction are related? Thanks