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When we calculate rate of reaction pfor a complex reaction, if the rate determining step have intermediates, we calculate the rate of reaction using the step which is at equilibrium (and doesn't have any intermediate).

For example:-

Let the given reaction be a complex reaction$$\ce {A +B->C}$$ If this reaction takes place as follows:- $$\ce{A +B<=> P + Q}.....(fast)$$ $$\ce{P +Q-> C}.......(slow)$$ So here rate determining step is second one. But as it consists of intermediate we say that the first elementary reaction is at equilibrium and solve.

But why do we assume that the first reaction is at equilibrium? Maybe it is still not at equilibrium. Also what if they can never reach equilibrium because of the conditions? Then how can we calculate rate of reaction?

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    $\begingroup$ Be sure you have searched this site and libretexts.org for related keywords, as this and similar topics are evergreens and are asked regularly. // One of many links: chem.libretexts.org/Courses/Purdue/… $\endgroup$
    – Poutnik
    Commented Jul 26, 2023 at 8:27
  • $\begingroup$ I read the suggested article and it helped me. But I still didn't get what if there is no equilibrium in elementary reaction? $\endgroup$
    – NOTE Book
    Commented Jul 26, 2023 at 9:04
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    $\begingroup$ There is many more. Do a proper review before asking. As example, searching and review search results for site:chemistry.stackexchange.com OR site:chem.libretexts.org reaction kinetics fast slow step or for similar keywords. // Then enumerate what you learnt and know what you do not understand yet. No explicitly listed effort = no effort, as nobody can distinguish the cases. $\endgroup$
    – Poutnik
    Commented Jul 26, 2023 at 9:22
  • $\begingroup$ I read many articles before. If I found my answer I wouldn't post the question at first place. $\endgroup$
    – NOTE Book
    Commented Jul 26, 2023 at 12:48
  • $\begingroup$ You should include in questions a short summary of what you already know or found (with eventual quotes/links), so others will not write what you are already aware of, or what you should know if you searched about it. If nothing was found, write at least what you have searched for. $\endgroup$
    – Poutnik
    Commented Jul 26, 2023 at 13:45

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But why do we assume that the first reaction is at equilibrium?

It makes it easier to solve.

Maybe it is still not at equilibrium.

At the very beginning of the reaction, it will be far from equilibrium. So the model (fast equilibrium between reactants and intermediates) will not be a good description of the initial phase of the reaction.

Also what if they can never reach equilibrium because of the conditions?

Technically, no matter what the conditions, it does not reach equilibrium until all reactions are at equilibrium. However, in some cases, it is a pretty good approximation for most of the reaction time.

Then how can we calculate rate of reaction?

You can always break time into discrete steps and do a numeric approximation (step-wise integration). Their might be some special cases where you can get a closed-form integral, but that is probably rare.

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