I found a sentence in book which states the reverse step of this reaction ( I forgot what was it!) has the faster step as rate determining .

Even Rate determining step-Wikipedia states:

In chemical kinetics, the rate (or velocity) of a reaction mechanism with several steps is "OFTEN" determined by the slowest step .

which I think 'OFTEN' means not 'everytime' so,

if faster step of the reaction is rate determining. Then why it is?


I can provide you an example. The oxidation of formate ion by peroxydisulfate in water solution: $\ce{HCOO^{-} + S_2O_8^{2-} -> CO_2 + 2SO_4^{2-} + H^+}$

has the following mechanism: enter image description here

As you can see, the first step is the slowest, but by using the rate-determining step approximation you wouldn't arrive at the correct rate law which is: $r=k[\ce{HCOO^-}]^{1/2}[\ce{S_2O_8^{2-}}]$. The first reaction is very slow, so most of peroxydisulfate is consumed in the third reaction. The correct rate law can be obtained by applying the steady-state approximation to the two radical species.

(This isn't the true mechanism but it is good for showing the point. A more complex mechanism includes the formation of OH radicals and several chain termination reactions. That’s why the given rate law is valid only for a limited range of reactant concentrations.)

  • $\begingroup$ But how can peroxydisulphate ion be consumed in 3rd step because any co2- will be formed after first step only. I think! $\endgroup$ – DSinghvi Sep 30 '14 at 16:08
  • $\begingroup$ But how can peroxydisulphate ion be consumed in 3rd step because any co2- will be formed after first step only. I think! $\endgroup$ – DSinghvi Oct 7 '14 at 13:48
  • $\begingroup$ During the conversion of the very first peroxydisulfate anion, there wont be any CO2-, but steady state will be reached very fast afterwards. $\endgroup$ – EJC Sep 1 '15 at 16:22
  • $\begingroup$ This is not really a fair test, as reaction 1 is not needed because $\ce{^.SO4^-}$ is produced is reaction 3 independently of reaction 1, i.e. it is probable that as the k quoted is made up of the rate constants one could put $k_1 = 0$ without it having much effect. $\endgroup$ – porphyrin Sep 6 '16 at 20:10

In the past I was also explicitly told that the slowest reaction is "often" the rate determining step, and like you I figured some day I would find a reaction where the fastest step determines the rate. Thinking over it a bit more now, however, I believe we may have misinterpreted what was meant. Rather than looking for an opposite situation, it is likely we were indirectly told that in some cases the rate-determining-step picture simply isn't applicable in the first place.

In general, reactions can't be modelled though such simple kinetic theory; in truth, when a bunch of reactants are brought together, all steps in every possible reaction route matter. It just happens that for several simple but relevant chemical systems there are few possible steps and few accessible reaction routes, and they have such wildly different rate constants that we can approximate by looking only at the slowest step of the fastest route. A more complex reaction with several steps, several side-reactions and similar rate constants will likely be poorly described by a RDS framework, so one can say that the reaction rate is not determined by the slowest step because the reaction rate won't be well determined by any single step.

  • 1
    $\begingroup$ You may also be interested in this past answer of mine. $\endgroup$ – Nicolau Saker Neto Sep 20 '14 at 18:15

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