[youngtreeest]

I was confused by the difference btw Ea and delta G double dagger.

Through fitting different k at different temp by Arrenhius equation k=Aexp(-Ea/RT), we can fit and obtain Ea.

But when a textbook mentions the energy profile, it always refers it to delta G double dagger. I been doing a experimental and theoretic study on a hydroxyl radical reacting with a siloxane chemical. I found the theoretic delta G double dagger at 25 C is different from the Ea that I experimentally obtained from fitting different k at different temp through Arrenhius equation.

Am I comparing apple to apple?

Thank you very much if you can help me figure this out!

[Irlanur]

I think this helps:
http://www.cup.uni-muenchen.de/oc/zipse/arrhenius-and-eyring-equations.html

[youngtreeest]

I think this helps:
http://www.cup.uni-muenchen.de/oc/zipse/arrhenius-and-eyring-equations.html

Thank you very much! Very helpful!

So for the first order rxn, Arrhenius activation energy Ea is approximately equal to delta H double dagger?

In addition, the website is saying "...The third part of equation (2) contains the activation free enthalpy dG** of the reaction which,..."
activation free energy dG** instead of "the activation free enthalpy dG** right?

[Irlanur]

In German "g=h-Ts" refers to "freie Enthalpie", which I think is called "Gibbs free Energy" in English. "f=u-Ts" stands for "freie Energie" or "Helmholtz free Energy". correct me if I'm wrong.

[Irlanur]

Arrhenius activation energy Ea is approximately equal to delta H double dagger?

they differ by RT

[youngtreeest]

In German "g=h-Ts" refers to "freie Enthalpie", which I think is called "Gibbs free Energy" in English. "f=u-Ts" stands for "freie Energie" or "Helmholtz free Energy". correct me if I'm wrong.

I will ask my German colleague about it.

Thank you!

[youngtreeest]

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?