Arrhenius activation energy Ea is approximately equal to delta H double dagger?
they differ by RT
Can't work out the equation...
For 1st order rxn, k = kBT/h*exp(-dG**/RT)
as dG** = dH** - TdS**
so k = kBT/h* exp(-(dH** - TdS**)/RT)
k = kBT/h* exp((-dH** + TdS**)/RT)
k = kBT/h* exp(-dH**/RT)* exp(dS**/R)
Comparing with k = A exp(-Ea/RT) (this is eqn 1)
A= kBT/h* exp(dS**/R), which is consistent with their eqn. 5 :A = (c kBT/h) exp(dS**/R)
Ea=dH**, which is not....
I am not sure the derivation is right... Really differ by RT?