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New Eley-Rideal families for RMG-database #63
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Great! |
for the dangling double falls flat, I don't think that follows the octet rule either. I think it would have to be something like this:
where 3 would be a hydrogen or something? |
I edited your original post above, to add some numbers, as I was getting confused by the counting. Now maybe the other comments need editing so they correspond . Also, reading Franklin's first comment on reaction 1 had me (or him) confused if 1 is itself an X (as I now drew) or was adsorbed to an X. |
Thanks, that helps.
which, with the bond order increase, does preserve the octet rule.
where |
I don't like those as much. My original meaning behind that one was to get benzene fully adsorbed to the surface and then hydrogenated and desorbed. |
Look for things by one of Norskov's students. An oral talk at AIChE a year or two ago. Probably published by now. |
I can't figure out how to open up an issue on the RMG-database so I just put it here instead
Now that Eley-Rideal is a little added, we came across some other families to potentially add to help us deal with bidentate/multidentate species.
Have just added:
1 EleyRideal_H_addition_multiple_bond
where
4=3
can be a double, triple, or quadruple bond, and 2 is an H atom.Some potential families to add are:
2
where
2
is a radical, probably an H.Would be good to have
3
where 3 is a radical, probably a hydrogen or something?
Probably less likely than number 2, above.
If you had that starting point, it would probably just go like this instead:
4
where
2=3
can be a double or triple bond. 4 is probably an HProbably a similar story to number 3. It's overall 3rd order (two surface sites and a gas phase species)
5
where
2=3
can be a double or triple bond. 4 is probably an HProbably a similar story to number 3. It's overall 3rd order (two surface sites and a gas phase species)
6
3 is a radical, abstracts 2.
eg. 1 is *CH2, 3 is CH3, 2 is H, you get methane and *=CH2
Seems reasonable. Not unlike H abstraction in gas phase.
Would be good to have
7
Not going to work.
because you have to break a double bond (1=2),
and 2 ends up unhappy.
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