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I am very confused right now about my approach to this question could somebody help me out with this. I reasoned my answer as given below

Let's assume if a complex reaction is of zero order. It implies its R.D.S. has order 0 ,but R.D.S. is an elementary reaction. So, that particular elementary step of reaction is 0 order.(for the obvious reasons)

So if we take that particular (elementary) step as individual reaction then it will be elementary and will have 0 order.

R.D.S. here implies Rate determining step

But many people whom I shared my doubt with, explained it on basis of Molecularity = Order Rule for elementary step,but it does not sound logical to me and also bit confusing as if we see pseudo first order reaction there molecularity is 2 but order is 1.

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  • $\begingroup$ The concept of rate determining step is outdated, but unfortunately still used. However, if one would use its implications and the order of the RDS is in fact zero, all steps must be of order zero since the RDS is the slowest. Physically, this doesn't make sense. $\endgroup$ Sep 28 '21 at 18:27
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As far as I understand, the assumption that the RDS has a zero-order is false. The RDS has its order equal to its respective molecularity, however, in zero-order reactions, the rate-determining step will not have the initial reactants as parts of the elementary step. Thus, when you calculate the rate constant in terms of the initial reactants using the other steps in the reaction, you will get their orders to be zero. Another thing, zero-order reactions are ALWAYS complex reactions. There are no elementary zero-order reactions.

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    $\begingroup$ I presume RDS refers to rate-determining step. What does ALWAYS refer to, then? $\endgroup$
    – andselisk
    Sep 28 '21 at 17:49

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