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Ronald Conol
aljaosn@yahoo.com

Can we include the energy of the catalyst in the Product and reactant  when dealing with catalytic reaction  to calculate the energy of activation? because i am just confused about the nature of  a catalyst that it does not participate in the chemical reaction.. and if i wont include the energy of the catalyst. the difference between the energy(product,TS and reactant) from free to catalyze reaction is so far ,say -323.88XX hartree/particle.

here is the reaction that I've studied:

1-methylcyclopentadiene + acrylonitrile ------BF3----> 1-methylbi[2,2,1]cyclohexene.

where BF3 acts as my catalyst..


also.. is it possible to get a negative energy of activation ?. because when i employed HF calculation the Ea is quiet reasonable.. but when i refined my result using MP2 calculation i got negative energy of activation (using catalyst). and the result is the almost the same when i used the PCGAMESS and FIREFLY. 6-31G**(d,p) was the basis set that i have used in both HF and MP2 calculation. i have attached my output file so as to prove that i have located the transition state with catalyst.

This message contains the 889 kb attachment [ mp2saddlebf.out ]