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I was solving some questions related chemical kinetics and encountered one question with multiple correct answer , the question is :

For a complex (multistep) reaction,

(a) the overall rate of reaction is the rate of slowest step.

(b) the overall molecularity has no significance.

(c) molecularity and order may or may not be same.

(d) the number of rate determining steps may be more than one.

The question have all the four options as correct answer. There's no doubt about first three options, but about the fourth one I want to ask , How a complex reaction have more than one rate determining steps? And if it has then how the overall order and molecularity of the reaction may be given as for complex reation we look at the slowest step and with the help of slowest step we decide the order and molecularity . Then with multiple RDS how it is possible ?

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    $\begingroup$ I am not really sure, but maybe its possible that two steps are equally or similarly slow? You can identify a step as rate determining only when its rate is appreciably slower than the other steps. $\endgroup$
    – S R Maiti
    Jul 22, 2021 at 2:53
  • $\begingroup$ @S R Maiti So we can take any of the slowest step to determine the order and molecularity ? $\endgroup$
    – Curiousminded
    Jul 22, 2021 at 3:02
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    $\begingroup$ No, I think if there are multiple steps which are slow, then you would have a very complicated rate equation. Sometimes you can derive these by steady state approximation. Other times it might not even be possible to write a rate law. For example the decomposition of HBr has a very weird looking rate law with 5/2 exponents and a denominator etc. The order would be undefined in that case. And molecularity is not even defined except for elementary reactions, so there is no question of molecularity for multi-step processes. $\endgroup$
    – S R Maiti
    Jul 22, 2021 at 3:05
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    $\begingroup$ There is always one step that will be the slowest even if the difference is miniscule. By definition there can only be one slowest step. I think someone above mentioned that there could be a case where two steps have identical velocities but it's unlikely. $\endgroup$
    – CsmcBurrito
    Jul 22, 2021 at 4:31
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    $\begingroup$ For a sufficiently complicated overall reaction which hides a large interconnected reaction network, the concept of a "rate determining step" may effectively not exist (at least not in any way that we might be familiar with from more simple cases), such that sentence (d) kind of loses meaning. I briefly talk about it in this past answer and the second half of this one. $\endgroup$
    – Nicolau Saker Neto
    Jul 22, 2021 at 4:38

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