# Which rate is faster: one step with Ea or two steps each with Ea/2?

I have been struggling with the following problem posed at a prior lab meeting. The answer seems trivial to me, so can you tell me if I'm doing this wrong? Thanks!

Consider the following two scenarios. They represent a small portion of the same catalytic cycle.

1. There is a reaction step of that cycle wherein $A\rightarrow P$ with an activation energy $E_{a}$.
2. Upon further inspection, the above reaction is actually $A\rightarrow I\rightarrow P$, where each of the two steps has an activation barrier of $E_{a}/2$ such that their sum is equal to the barrier height in the first scenario. Refer to the schematic here for a depiction if needed.

The question is: which scenario is faster? My thought process is as follows.

First, we need to define which rate we are talking about. I am going to assume the rate of production of product, $dP/dt$, is compared. I need information about pre-factors, but I'll have them all equal an arbitrary value of 1 for the time being. I have set $[A]_{t=0}=1$, $[P]_{t=0}=0$, and $[I]_{t=0}=0$. I now write out my ODEs and solve them. I've assumed $E_{a}=RT$ for demonstration purposes.

Case 1:

Case 2:

Comparison of Rates:

My Issue:

Can the question not be definitively answered? It seems like the answer (whether case 1 is faster than case 2) is that it depends on what conversion you're running to since the answer changes at a certain point. However, this was asked in the context of catalysis. I've posed this problem to multiple people and they all brought up the point that they don't usually think about time or conversion in this regard. They just want to know which one of these two cases will be faster when looking at two potential energy profiles. Isn't that too naive, or am I thinking about this all wrong? At full disclosure, this is mildly related to a prior question I posed in the comments located here.

Edit: To emphasize, if I change the pre-factors or the value of $E_{a}$, the order of which case is faster will also change. So, the answer to the question in the title is that "it depends".

• You've answered your own question by plotting the rates and concentrations. The single step case is always faster to reach a given point in the reaction (say 50% complete) but at some points the instantaneous rate of the two step process is higher. – bon Dec 17 '16 at 18:44
• Yes, I agree. Thank you for the confirmation. Just to note for others that visit this post, that trend changes if I change the parameters. It's very easy to get the two-step case to be faster if you change the pre-factors or value of $E_{a}$. So, the end conclusion is that you can't a priori say which one is faster without further information. – Argon Dec 17 '16 at 20:23

Instead of looking at $\mathrm{d}[\ce{P}]/\mathrm{d}t$ against time $t$, a better idea might be to plot $[\ce{P}]_t$ against $t$, which you already did. By eyeballing the graphs I'm fairly sure that $[\ce{P}]_t$ in case 2 is always below $[\ce{P}]_t$ in case 1.
In org chem, people probably aren't too interested in what the absolute rate of the reaction at time $t$ is, they are probably more interested in how long the reaction will take to run. Because of the maths, the reaction will never hit 100%, so you can't use that as a benchmark, but you can use e.g. 95% complete (i.e. $[\ce{P}]_t = 0.95$) as a benchmark.