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I was wondering how a rate-determining step would be identified for an energy-reaction coordinate diagram. The RDS would be the step with the highest activiation energy, but relative to what? Relative to the intermediate that came right before, or relative to the energy of the reactant as a whole?

In the internet, I found following depiction: enter image description here

From this diagram, it becomes clear, that the second activation energy relative to the first intermediate has to be considered, and is therefore the RDS.

However, in my lectures the professor showed us this slide:

enter image description here

This would imply that we measure the overall activation barrier relative to the initial state; is he correct? The professor used following derivation:

enter image description here

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  • $\begingroup$ Your observations do seem to be correct -- the images are in disagreement. The question is how much of the activation energy from the first step can be "rolled over" to the second? If the reactions are closely coupled, and most of the energy of the first step is recovered, then the first case seems to apply. $\endgroup$
    – DrMoishe Pippik
    Commented Jul 7 at 22:31
  • $\begingroup$ But in the simplest case, if we assume a Boltzmann-like distribution of energy - that means higher energies exponentially decay in probability, then yes, one would need much energy to cross the first barrier, and then even more to cross the second one which would imply the second one to be rate-limiting.. $\endgroup$
    – Mäßige
    Commented Jul 8 at 6:37
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    $\begingroup$ That last derivation is in a way trivial. To get to the second transition intermediate you have to go through the first, therefore the rate to the second can't possibly be greater than the rate to the first. However the relevance to rds is when the ratio of the sequential steps is such that all but one can be ignored (as explained in an answer). $\endgroup$
    – Buck Thorn ♦
    Commented Jul 9 at 19:15
  • $\begingroup$ I am sure you are talking of Shri Ludwig Eduard Boltzmann $\endgroup$
    – user150128
    Commented Jul 11 at 14:00

2 Answers 2

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I do not know where your teacher took his definition of a RDS. A rate determining step (RDS) is a step in a reaction mechanism that controls the rate of the overall reaction. It means that if a rate determining step exists, the rate of the overall reaction depends on the rate constant of the rds and not on the rate constant of the other steps. See link to the IUPAC definition at the end.

The RDS is always defined in term of rate constants (k), not rates (v). Actually, most simple kinetic derivations use the steady state approximation and in that case all the elementary steps have the same reaction rate!

As for the error in the derivation from your lecture : the concentration of B is not fixed by only the equilibrium of the first reaction. It obviously depends on the second reaction too! By writing it this way, your teacher implicitly assumed that the rate of the first reverse reaction (B->A) is faster than the rate of the second one (B->C) so that a pre-equilibrium can be attained for A<->B, independently of the other steps. It can happen, but it is not general.

The RDS depends primarily on the ΔG‡ for each step (Gibbs energy of transition state relative to the reactant in the step).

Please note:

  • A RDS cannot always be found, some reaction mechanisms depend on more parameters.
  • The notion of a RDS has been argued about for decades. In complex reactions, it can be quite difficult to identify one and its identity can depend on more parameters (including concentration of intermediates).

Some refs:

  • Atkins and De Paula, 11th eds. (chapter 17).
  • Laidler, Keith J. "Rate controlling step: A necessary or useful concept?." Journal of Chemical Education 65.3 (1988): 250
  • The IUPAC definition (gold book) : https://goldbook.iupac.org/terms/view/R05139
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IUPAC has formal definition for the rate-determining step:

A rate-controlling step can be formally defined on the basis of a control function (or control factor) CF, identified for an elementary reaction having a rate constant $k_i$ by: $$ \text{CF}=\frac{\partial \ln(v)}{\partial \ln(k_i)} $$ where $v$ is the overall rate of reaction.

Followed by

The elementary reaction having the largest control factor exerts the strongest influence on the rate $v$, and a step having a CF much larger than any other step may be said to be rate-controlling.

You can find more in the below references:

  1. J. Chem. Educ. 1988, 65, 3, 250
  2. Solid State Ionics, Volume 117, Issues 1–2, 1999, Pages 123-128
  3. ChemPhysChem, 12: 1413-1418

The third paper has a detailed discussion on why simply using the magnitude of rate constant and/or activation is barrier is not sufficient to define the rate-determining step.

Take a look at Curtin-Hammett principle, where the overall reaction depends independently on kinetic and thermodynamic factors. Transition State Theory (TST) tries to link the two: rate constant (kinetic factor) and activation enrgy (thermodynamic factor), but their dependency is quite obvious. An important missing factor is the transmission coefficient that determines the formation of product from activated complex.
The rate-determining step is the "bottleneck" of the reaction and it is not straightforward to determine what the rds is even from the reaction coordinate diagram.

The derivation used by your lecturer is erroneous as the relation between $C_\ce{A}$ and $C_\ce{B}$ is derived from an equilibrium condition and therefore, is valid only for that specific condition. The actual equation is $$ \frac{C_\ce{A_{eq}}}{C_\ce{B_{eq}}}= \exp\left(-\frac{\Delta G^\circ_\ce{A->B}}{RT}\right) $$
One cannot just assume $\frac{C_\ce{A}}{C_\ce{B}}=\frac{C_\ce{A_{eq}}}{C_\ce{B_{eq}}}$ under all cases.

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  • $\begingroup$ The third paper is not open access, and I am unaware of the copyright issues if I quote directly from the paper. If there are no issues, I can quote the relevant text and attach figures as well. Can someone from the mod team update me on this. $\endgroup$
    – megamonster68
    Commented Jul 11 at 7:13
  • $\begingroup$ Just referencing to the doi/legitimate link, and quoting the relevant text/figures is fine $\endgroup$
    – Safdar Faisal
    Commented Jul 11 at 9:57

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