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Rate law

  1. Mar 22, 2015 #1
    The rate law is given by R=k[A]^m[ B]^n where m,n needn't be equal to stoichiometric coefficients....but we use r=k[A]^a[ B]^b where a,b are stoichiometric coefficients when we use ratio rate of forward to rate of backward reaction in getting value of equilibrium constant K for any given reaction at equilibrium.....why is it different in both cases...does it mean all equilibrium reactions are elementary reactions? But that isn't possible as many intermediate steps are equilibrium reactions in the mechanism of complex reactions...
     
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  3. Mar 22, 2015 #2

    Borek

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    Staff: Mentor

    No. It is kind of a coincidence that it works this way.
     
  4. Mar 22, 2015 #3
    Well, it's a coincidence for all reactions in equilibrium sounds peculiar....
    Like I said a complex mechanism too can be in equillibrium....in that case the rate law equation should have different coefficients....but all problems we solve in equillibrium chapter have stoichiometric coefficients only
    t
     
  5. Mar 22, 2015 #4

    Borek

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    Equilibrium can be derived from the thermodynamics, then it is clear why it follows the reaction stoichiometry. As you have correctly said, reaction kinetics doesn't follow stoichiometry - or rather, it sometimes follows stoichiometry of the elementary reactions (which are not necessarily the same as the forward and/or backward reaction). For such simple cases kinetic approach yields the same formula for the equilibrium - no coincidence here. Generalization of that approach yields the same result as the thermodynamics, but is a coincidence in the sense that the generalization is incorrect.
     
  6. Mar 22, 2015 #5

    epenguin

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    We have discussed this quite often here including quite recently, so briefly - you can derive the equilibrium laws from quite general thermodynamic arguments (found in all University level starting textbooks of physical chemistry) and so whatever mechamism and rate law, the equilibrium law has to be obeyed. So given a rate law fornthenforward reaction you can deduce that for the back reaction.

    I think you could not come up with a mechanism that made sense in both directions for which you didn't get the right equilibrium law, but you are welcome to try - it might ba the best way to convince yourself and understand.

    There is a physical principle you have to incorporate into this - that is that the path of the reverse reaction is the same as the forward reaction in reverse - you can't go back another way. If you could then indeed you don't generally get the ordinary equilibrium equations. But classical and quantum mechanics say that this 'principle of microscopic reversibility' aka 'detailed balance' is true.
     
    Last edited: Mar 22, 2015
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