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I’ve heard many conflicting points on this single question alone. Even the usually reliable sources I run to are pretty ambiguous on the matter. So now I’m here, after 1 and a half weeks of confusion.

As I understand it, there is usually a faster elementary step and a slower elementary step. The slower one is called the “rate-determining step.” My intuition tells me that the actual overall reaction rate (the reaction rate for the whole process) should be higher than the RDS, since the faster step, although it may be magnitudes faster, still takes up time. Thus, it is only the lower limit of the overall reaction rate.

However, I’ve heard plenty of times that the RDS is the overall rate, and when I ask why, the answer is always the same analogy of the motorway with two toll booths. I get the analogy, but I’m not exactly sure if it’s a complete analogy to how a reaction works. No reaction is instantaneous, and this should show especially if the fast step isn’t much faster than the slow step. Yet, I still hear this all over the place, and this wouldn’t be the first time I was wrong about chemistry.

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    $\begingroup$ I am not sure if this shall be a comment or an answer. Are you thinking about the duration of a reaction at molecular scale? It is certainly true that the duration of the overall event is the sum of two or more (extremely short) time intervals. But this is not dn/dt for a large ensemble. You can think of a Baker making 100 breads daily and a delivery boy able to deliver 200 breads. Of course it takes someone more than 1 day to be delivered with his ordered bread. The difference would be noticeable about the first day only. The same here, except that events take times in the order ~ 10^-13 s $\endgroup$
    – Alchimista
    Dec 15 '21 at 8:47
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    $\begingroup$ Do you mean overall rate or rate constant?. The overall rate has to be the appearance of product; in $A-B-C$ then $dC/dt$. The overall rate constant is harder to define, some sort of average but perhaps the half-life of the rise-time of the product would do. You could calculate this in limiting cases for the ABC scheme to satisfy yourself as to how this depends on rate constants. $\endgroup$
    – porphyrin
    Dec 15 '21 at 14:48
  • $\begingroup$ Is not an complete analogy an oxymoron? It would be an equivalent, not an analogy. $\endgroup$
    – Poutnik
    Jan 14 at 10:12
  • $\begingroup$ The overall rate can be temporarily slower than RDS rate, if the system is not in the steady state yet, so the rate of the fast reaction after RDS has not reached RDS rate yet. Typical case would be generation of uranium decay chain products from purified uranium. $\endgroup$
    – Poutnik
    Jan 14 at 11:34
  • $\begingroup$ The rate determining step approximation is only a just well enough approximation if there is one step, and exactly only one step, which is many magnitudes slower than every other possible step. A better approximation for the rate of a reaction is the energy span model. $\endgroup$ Jan 15 at 13:46
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The step called rate determining because the rate over all steps is limited by this single step.

Say the reaction would be

$$\ce{A -> B -> C}$$

with $\ce{A -> B}$ as rate determining. Because $\ce{B -> C}$ only can happen when $\ce{A -> B}$ takes place and must await the formation of at least one intermediate molecule $\ce{B}$ (in an analogy, can't process this item until it actually arrives on the conveyor), there is no faster advancement than $\ce{A -> B}$.

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    $\begingroup$ OP is probably aware of this. The question is why the reaction isn't slower. Eventually it is, for the first 10^-13 s or something like that. The point is that reaction kinetics deals with the transformation of about 10^20 or more particles, and includes the time needed for them to find each others and effectively react. It has little to do with the duration of single events or a sequence of them in the situation when they actually occur. $\endgroup$
    – Alchimista
    Dec 15 '21 at 9:29
  • $\begingroup$ @Alchimista So you think OP points to rate of the elementary step (contact by collision $\ce{->}$ detachment [if it is a substitution]) vs. observed rate of a diffusion controlled reaction? $\endgroup$
    – Buttonwood
    Dec 15 '21 at 10:11
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    $\begingroup$ The question seems to be formulated having in mind how long it takes for a reaction at molecular scale to occur, and that is obviously the sum of the various steps duration.. It seems to me the only way to give sense to the question. Let us see if OP will comment. $\endgroup$
    – Alchimista
    Dec 15 '21 at 10:17
  • $\begingroup$ @Paulemic A reason for my referring to diffusion control are surveys by the Mayr group, Munich like this one. $\endgroup$
    – Buttonwood
    Dec 15 '21 at 10:59
  • $\begingroup$ @alchimista Your interpretation is a bit closer to what I was trying to express. I’m just going off of my intuition at a molecular scale to justify my conception. $\endgroup$
    – Paulemic
    Dec 15 '21 at 16:11

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