Transition state theory tells us that the rate constant of an elementary step is $$k_\mathrm{r} = \frac{\kappa k_\mathrm{B}T}{h}\exp\left\{\frac{- \Delta G^\ddagger}{RT}\right\},$$ where $\Delta G^\ddagger$ is the activation free energy of the step, i.e. the difference between the Gibbs free energy of the transition state and the reactant state.

Then the rate law would be $$r = k_\mathrm{r}[A]^a[B]^b[C]^c\dots,$$ where $A$, $B$, $C$, ... are all the reactants that come together in that elementary step. So far so good.

But I have come across reactions where changing the concentration of a reactant (by a large amount) in a multi-step reaction seems to change the rate determining step of the whole reaction.

The explanation offered was that changing concentration changed the Gibbs free energy of the reactant or intermediate states (as Gibbs free energy depends on the concentration in solution). The states which contain the species (whose concentration is changed) are affected whereas others are not, so the Gibbs free energies of activation are changed, which means the rate constants are changed, which can change the rate determining step.

But if concentrations can affect $k_\mathrm{r}$ then it seems like the rate law cannot be written like that.

Does this sound right, or am I overthinking this?

  • 2
    $\begingroup$ If the rate constant depends on the concentration of something, it is not a rate constant. $\endgroup$
    – Maurice
    Commented Dec 2, 2020 at 22:04
  • $\begingroup$ Then how would you define rate constant where the rate determining step can change if the concentration of one component is changed to a large extent? $\endgroup$
    – S R Maiti
    Commented Dec 2, 2020 at 22:09
  • $\begingroup$ TST is a crude approximation. It works only for elementary steps, only for the defined potential hyper surface. If you change the concentrations, you change the surface, the approximation breaks. P.S.: Please don't use mhchem for non-chemistry items. $\endgroup$ Commented May 13, 2021 at 17:40
  • $\begingroup$ @Martin-マーチン Thank you. So what happens to the rate constant then? It changes, right? $\endgroup$
    – S R Maiti
    Commented May 13, 2021 at 17:51
  • $\begingroup$ If with "change" you mean it was once that and became that, then: no. If with "change" you mean it is completely different and has absolutely no relationship with anything else, then: yes. That includes the cases, where that rate appears to be equal. I think I need to stress that again: TST only ever works for a single, well defined potential hyper surface. $\endgroup$ Commented May 13, 2021 at 17:57

1 Answer 1


It seems possible that you mix the terms. There is the reaction rate constant -- in your equation $k_r$, and there is the rate of reaction -- in your equation $r$ and sometimes expressed by $v$.

Depending on the order of a reaction, the overall rate of reaction may, but need not depend on the concentration of the reactants available. There are reaction types known by experience to proceed typically one kinetic order (e.g., heterogeneous hydrogenation); however, reading the stoichiometric reaction equation often does not provide sufficient information and the rate law has to be determined experimentally.

Examples of kinetic reaction order are zero, one, two, three, but fractional orders are known, too. It is possible to perform reactions which are of second order ($v = k_r[\ce{A}][\ce{B}]$) in a pseudo-first order regime if the concentration of reagent A is so much much higher, than the one of [B], that you detect after consumption of B still much of A. This however assumes that the reaction still proceeds by the same pathway, thus requires experimental evidence to be supported.

This contrasts to $k_r$, which is independent from the concentration of the reagents, however depends on the temperature $T$ (keyword Boltzmann, and even more, Arrhenius equation).

  • $\begingroup$ If the Gibbs free energy depends on the concentration of the reagents, and $\ce{k_r}$ depends on the Gibbs free energy difference, then surely $\ce{k_r}$ can depend on the concentration? $\endgroup$
    – S R Maiti
    Commented Dec 6, 2020 at 14:49

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