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
Partition functions are not part of RASPA but are input. So you will need to compute them. There are many different ways of doing that, and they do differ in results.
For some pointers:
https://homepage.tudelft.nl/v9k6y/thesis-Rahbari.pdf
Page 262 and following, and see Table A5.