# Catalytic energy profiles (is this Wikipedia image misleading?) [duplicate]

I am looking at the image below from the Wikipedia page for energy profiles in chemistry. I do not understand why the "$E_a$ (with catalyst)" is labeled as it is. I would argue that is not the activation barrier with the catalyst. Shouldn't all of the individual steps need to be considered to get the rate of production of product (or, if one step has an especially high barrier, then that step can be assumed to be rate-limiting and that step's $E_a$ is the barrier with the catalyst)? I suppose in other words, if I do $r = A\exp(-E_a/k_{B}T)$ where $E_{a}$ is the activation energy with catalyst, that will not get me the rate, right? For the record, I have taken grad-level courses in chemistry, but I got myself into a position of questioning everything I once learned and here I am.

## marked as duplicate by Ivan Neretin, airhuff, Klaus-Dieter Warzecha, Todd Minehardt, ronApr 13 '17 at 15:26

However, there is still some truth in labelling ‘$E_\mathrm{a}$ with catalyst’ as such. Even though you have much smaller steps you will still need to add that labelled energy in net terms to overcome the barrier. The principal difference is that it need not be added all at once but can be added stepwise. You could even say that the total activation energy (i.e. adding up all the individual $E_\mathrm{a}$) be even higher, but you regain some of that from the intermediate reactions.
• @Zhe Thank you for your replies. I am going to have to convince myself of this. I ask because the question that motivated me to even submit the original question was the following. Is the rate the same between: Case 1) the reaction proceeds in one step with a barrier height that goes to $E_{a}$ compared to the reactant; Case 2) That same reaction is actually two sequential steps, where the highest of the two peaks reaches a height $E_{a}$ compared to the reactant. Everything else is identical. From your answer, it seems the answer is yes, the two cases have the same rate. – Argon Dec 14 '16 at 4:40