I've seen some controversy on this question while doing a brief search. For example, this SE answer quotes Wikipedia and says that the RDS is the step with the largest $E_a$. However, this UC Davis page says that

In the potential energy profile, the rate-determining step is the reaction step with the highest energy of transition state.

My textbooks don't explicitly mention whether the RDS is the step with the largest $E_a$ or the step with the highest energy transition state. When I was doing a past exam, one of the questions was on this topic, and the correct answer assumed that the RDS is the step with the highest energy transition state. However, I've always thought that the RDS is the slowest step, and the slowest step has the highest $E_a$. So which viewpoint is actually correct?

  • $\begingroup$ Are you doing Chemistry Olympiad? $\endgroup$ – Yunfei Ma Apr 16 '16 at 0:22
  • $\begingroup$ Anslyn & Dougherty p 361: "The step of the reaction whose rate determines the observed rate of product formation is called the rate determining step (rds). It is commonly the chemical step that involves the highest energy transition state." (emphasis mine) $\endgroup$ – orthocresol Apr 16 '16 at 0:23
  • $\begingroup$ The operative word in the Anslyn & Dougherty quote is commonly. Dogmatic statements in chemistry always bother me. More often than not there are exceptions. $\endgroup$ – MaxW Apr 16 '16 at 0:31
  • $\begingroup$ But really the previous question that you linked has the answer in a comment. I don't have the time to write up an answer but the summary is that neither of your proposed approaches is complete. The reference is J. Chem. Educ., 1981, 58, 32 cc. @MaxW $\endgroup$ – orthocresol Apr 16 '16 at 0:32
  • $\begingroup$ @orthocresol I can't read the article you linked because it says I do not have a subscription to the publication. Can you post a summary of (or quote) what it says? $\endgroup$ – carbenoid Apr 16 '16 at 1:30

Yes, the rate determining step is the largest energy difference between any starting material or intermediate on a potential energy diagram and any transition state that comes after it. That transition state will then be the rate-determining step of a given reaction.

The transition state with highest absolute energy may not necessarily correspond to the rate determining step, because if it is transforming from another high energy state, then little additional energy is required to reach this (low $E_a$).

  • $\begingroup$ See the Anslyn & Dougherty quote pointed out orthocresol above. Is and is commonly are very different. There are no exceptions to is. $\endgroup$ – MaxW Apr 16 '16 at 0:40
  • $\begingroup$ @MaxW I saw it, and I don't see any disagreement. "Commonly," "not necessarily"-- neither are absolute statements. $\endgroup$ – ringo Apr 16 '16 at 0:50
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    $\begingroup$ Also as a cool side note, Anslyn is currently my professor for organic chemistry. He is an immensely smart guy. $\endgroup$ – ringo Apr 16 '16 at 0:53
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    $\begingroup$ You said "Yes, the rate determining step is the largest energy difference..." I think it should be "Yes, the rate determining step is commonly the largest energy difference..." $\endgroup$ – MaxW Apr 16 '16 at 0:53
  • $\begingroup$ That is how the RDS is determined though. It does not have to include the highest energy transition state if there is an intermediate following it with a higher following $E_a$. I think that is where it is important to qualify the statement with "commonly." $\endgroup$ – ringo Apr 16 '16 at 1:21

I couldn't fit this as a comment, so I'll post it as an "answer." This are just my ponderings.

For everybody, this is the problem: (it's not that clear, but the vertical axis is Gibbs Free Energy). enter image description here

The correct answer according to the problem source is III-IV, even though the first and second transition states have higher $E_a$ s from their respective "valleys."

The most relevant equation here is $k = Ae^{-E_a/RT}$. This is a high school level competition, so this is purely a conceptual problem involving that equation. As we all know, as the activation energy of an elementary step increases, the rate of that step slows down as the fraction of molecules with the requisite kinetic energy decreases. ($e^{E_a/RT}$ ) is the fraction factor: the fraction of particles that have equal to or greater than $E_a$.) Fraction of what, though? I believe that's where the answer lies.

My reasoning in favor of III-IV this: even though $E_a$ for III-IV may not be as high as I-II or II-III, already only a small fraction of molecules out of the total even made it to state III, due to the high potential energy at state III. Thus, when we apply the fraction factor again to that fraction via the Arrhenius equation, the final fraction to of particles that have enough energy to progress from III-IV is indeed the smallest.

To put it mathematically (and this is a purely mathematical argument, not chemical), the fraction of the particles that have enough energy to progress from III-IV is the fraction of total particles that are at state III, multiplied by the fraction of those particles at state III that have the energy to go from there to IV. Multiplying the two fraction factors, adding the exponents, the net "minimum energy" is the sum of the energies involved in getting to III and the $E_a$ of the III-IV reaction, giving the "absolute energy" of the transition state between III and IV, which is the highest.

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    $\begingroup$ But for any reaction, the fraction of particles that pass through the last step is always the smallest since it needs to have passed through all the previous steps (and each step multiplies another "fraction factor" onto it), right? Also, if II-III has the largest $E_a$, that step would be slowest since a small fraction of the particles that reach II have enough energy to go to III. But of the particles that make it to III, a larger fraction of them have enough energy to go to IV, so III-IV progresses faster. So wouldn't the RDS be II-III since a reaction can't be faster than the slowest step? $\endgroup$ – carbenoid Apr 16 '16 at 13:42
  • $\begingroup$ @Yunfei Ma Does the energy difference of I to III-IV transition state matters? In order for reactants to go to products they must posses kinetic energy greater than the III-IV (with respect to I) transition state energy. So why we don't take into account this? $\endgroup$ – ado sar Jan 27 at 14:46

From the Wade textbook on kinetics and activation energy: the rate-determining step, is the step which leads to the transition state with the highest overall energy.

If there is an early step with a low transition state, the intermediate will accumulate (the forward and reverse reaction of the first step achieve equilibrium), and the overall reaction will depend on the later step with the higher transition state.


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