0
$\begingroup$

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.

$\endgroup$
5
  • $\begingroup$ The concept of rate determining step is outdated, but unfortunately still used.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. $\endgroup$ Commented Sep 28, 2021 at 18:27
  • $\begingroup$ Photolytic reactions $\ce{A ->[light]B}$ have typically molecularity 1 and order 0. $\endgroup$
    – Poutnik
    Commented Jun 17, 2022 at 6:03
  • $\begingroup$ Aren't they first order in photons, the forgotten reagent? What happens at very low pressures? I encountered one zero order reaction that I attributed to product catalysis but never figured out the details. $\endgroup$
    – jimchmst
    Commented Jun 17, 2022 at 18:23
  • $\begingroup$ The combustion of wax in a burning candle is zeroth-order. Has it anything to do with a rate determining step ? $\endgroup$
    – Maurice
    Commented Oct 15, 2022 at 19:03
  • $\begingroup$ The rate of burning is related to the diffusion of wax in the wick definitely rate determining. It is similar to turning the gas valve on your stove. $\endgroup$
    – jimchmst
    Commented Oct 15, 2022 at 21:52

1 Answer 1

1
$\begingroup$

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.

$\endgroup$
2
  • 2
    $\begingroup$ I presume RDS refers to rate-determining step. What does ALWAYS refer to, then? $\endgroup$
    – andselisk
    Commented Sep 28, 2021 at 17:49
  • $\begingroup$ When a zero order rate is observed it means that the reaction is not a single step reaction and either two mechanisms are involved, or a reactant involved in the rate determining step is being throttled, or possibly there is product catalysis. Such an event invites further investigation. Perhaps someone can tell me what the rate determining step has been replaced by? $\endgroup$
    – jimchmst
    Commented Oct 15, 2022 at 22:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.