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Rate law equation

  1. Aug 13, 2010 #1
    When determining a rate law equation you find the individual orders via experimental data, thus you don't assume the coefficients in front of the reactants are their order in the rate law equation. aA + bB -> cC | Rate law equation is not necessarily, rate = k[A]^a^b

    But how come when given the "slow" step for a multi-step process, I assume the coefficient is the order for the reactant? aA + bB -> cC | Rate law equation is, rate = k[A]^a^b
     
    Last edited: Aug 13, 2010
  2. jcsd
  3. Aug 15, 2010 #2
    In a multi-step process, the "rate-liming" (slow) step is the only one that matters (approximately) for determining the overall rate. The slow step creates a bottleneck, the other fast steps either before or after the bottleneck, are having no effect on the overall process.
     
  4. Aug 15, 2010 #3
    Thanks for the reply. That part I fully understand. But what is confusing me is why the coefficient is the power (or order) the concentration is raised to in the rate equation.
     
  5. Aug 16, 2010 #4

    Borek

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    Staff: Mentor

    Honestly I don't understand what you are asking about. Please try to reword your question.

    As far as I can tell order of the slow step is determined experimentally as well.

    --
    methods
     
  6. Aug 16, 2010 #5
    Ah, I see, sorry.
    When you have a single indivisible reaction it just comes down to statistics.
    If your reaction looks like A + 2B --> C, you require the interaction of one A molecule and two B molecules, simultaneously. The probability of that occuring is A times B times B or
    [tex]
    R \propto A\times B^2
    [/tex]
     
  7. Aug 16, 2010 #6

    alxm

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    Well, actually it almost never happens that reactions occur that way; it's extremely improbable to have three molecules colliding with each other simultaneously. In such a case, you would typically have some short lived intermediate, so that:
    A + B <-> AB
    AB + B --> C

    If both these are first-order, the rate of formation of AB is proportional to [A] and the rate of formation of C is proportional to [AB], so the overall rate is proportional to [A]^2.
     
  8. Aug 16, 2010 #7
    I'm reviewing my old chemistry stuff and the textbooks I read just assumes that the rate law equation is based off of the coefficients of the rate-determining step rather than experimental data.

    Example:

    2A + 3B > 4C
    4C + D > 3E
    _______________
    2A + 3B + D > 3E

    If the rate-determing step is 2A + 3B -> 4C
    Then the rate law = kA2B3
     
  9. Aug 17, 2010 #8

    Borek

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    How do you know what is the rate determining step? From the experiment... so the reaction equation is already based on the experimental results, no wonder it is exact and can be used to calculate the reaction order.

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  10. Aug 17, 2010 #9
    Often you can predict it theoretically. Consider bimolecular and unimolecular nucleophilic substitution reactions in organic chemistry.
     
  11. Aug 17, 2010 #10
    Ohhh, you're right. If you know the rate determining step then you have experimental data. So then it is correct to assume the coefficient is the exponent to which the concentration is raised, given it is the rate determining step?
     
  12. Aug 17, 2010 #11

    Borek

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    A little bit convoluted, but sounds about right.

    --
     
  13. Aug 17, 2010 #12

    alxm

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    You can't really predict the rate-limiting step theoretically, short of doing an explicit quantum-chemical calculation. In fact, it's difficult to predict any of the steps. Most of the time you're doing educated guesses based on chemical intuition and similar known mechanisms. Determining the reaction order is relatively simple to do experimentally (plot concentrations versus time basically), and also the transition state energy (same experiment, but vary the temperature so you can determine dS). Depending on your interest, this is often good enough - as one professor told me as an undergrad - it doesn't really matter if you have two or fifty intermediates, if they're not rate-limiting.

    On the other hand, if your main interest is in catalysis, and the details of how the reaction occurs, it's tricky. The only way to "directly" observe the reaction is to use ultrafast laser spectroscopy, which is difficult and still a fairly rare procedure. There's a whole host of experimental techniques to try to glean information though, looking for bireactions; reacting with analogues, using isotope labeling of certain atoms to see where they end up, using various reactants and tricks to 'trap' the reaction at some intermediate that you can characterize, looking for kinetic isotope effects (e.g. a deuterium bond will break more slowly than a hydrogen bond since it's heavier and slower), and so on.

    Then there's pure theory (what I do), which is to test the plausible mechanisms and use quantum chemical methods to calculate the theoretical transition-state energies for the various methods. QC methods aren't yet accurate enough (in most cases) to correctly predict kinetics, but they're at least accurate enough to give an idea of which mechanisms are more likely and which ones are certainly impossible.
     
  14. Aug 18, 2010 #13

    DrDu

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    I wanted to mention RRKM theory (Rice Ramsberger Kassel Marcus) which explains that even the order of an unimolecular reaction can change with the energy available to the molecule.
     
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