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Topic: rate determining step  (Read 5801 times)

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Offline AhmedEzatAlzawalaty

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rate determining step
« on: March 14, 2007, 05:26:48 PM »
why the rate of reaction is determined by the rate determining step i think it may be the rate of the combination of all reaction steps not only one step.?

Offline Yggdrasil

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Re: rate determining step
« Reply #1 on: March 16, 2007, 03:50:18 AM »
If one step is significantly slower than the other steps in a reaction mechanism, then a common assumption people make is that the rate of that slow step will be the rate of the overall reaction.  This is a very crude assumption and you are correct that the real expression for the reaction rate is a combination of the rates of all reaction steps.

Offline AhmedEzatAlzawalaty

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Re: rate determining step
« Reply #2 on: March 16, 2007, 06:57:49 AM »
look at this reaction
                           2NO2+F2=2NO2F
the mechanism is that NO2+F2=NO2F+F
                                NO2+F=NO2F   rate=k.[NO2][F2]
could i know how?

Offline ARGOS++

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Re: rate determining step
« Reply #3 on: March 24, 2007, 05:11:36 PM »
Dear ahmedezat,

In one point the answer from ggdrasil is correct:
    At the absolute final glance, the Reaction-Rate is ALL times THE complete
    combination
of ALL possible and impossible reaction-steps.
But his remaining picture is at least wrong coloured, and NEVER “crude” in any case, and sense, too!
Think about: You have to distinguish between two situations, a consecution of reaction-steps,
and one, or more, “sidecar”-reactions !!

But what’s the difference, and why is it important?:
   A.)  In a consecutionall” has to go through ONE and ONLY one AND the same Pipe,
         what means that it has only as consequence:  Even the faster as the fastest has
         to wait for the slowest ever! 
         That’s why it is so extraordinary heavy to catch/detect a fast intermediate, because
         its concentration is all times very, very low!
   B.)  In the case of side-reactions-steps you as a chemist are “claimed”, as the ratio of the
         different reactions-steps will tell you, how much of which product you will earn!
         So, you as a Chemist need all your imagination to get a higher, and even higher, yield
         of the product of your/other interest! Some times, only 2% side-reactions may flag
         the end of all your endeavours,  — so be warned, even if it is anyway “one” of the
         most interesting “jobs” on Earth!
         (But the opposite can also be the absolute “accolade” for a Chemist, to get only,
         but 0.3% of that what is of interest!)

As a conclusion you may extract: In Chemistry “the slowest” is nearly all times
of "most interest", even, if not every times identical founded.

The “Thumb” may tell you: As lower as the “Hill of Activation” is, as faster the
corresponding reaction-step may be.

A normal Chemistry-Book will explain you which of your reaction-steps above will be
the slowest.

So I hope You will all times take not the crude colours.
Good Luck!
                      ARGOS++



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