The general mechanism of any catalytic reaction is the following:
X + Y + C -> XYC* (transition state 1) -> XYC1 (intermediate 1) -> XYC** (transition state 2) -> XYC2 (intermediate 2) -> ... -> ZC (catalyst--product intermediate) -> ZC* -> Z + C
That is, the real mechanism may be very complex, and each step should be treated separately. Of course, you should take X + Y + C as reactants, XYC* as a transition state, and XYC1 as a first product. Next, take XYC1 as a reactant, XYC** as a transition state, and XYC2 (or Z + C if no more steps is expected). And so on. Good luck!
On Mon Feb 22 '10 2:44pm, Ronald Conol wrote
>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.