Author Topic: How the partition function for propene in the RASPA manual was obtained?  (Read 251 times)

w_wonder

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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

David Dubbeldam

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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.