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Two-Step Mechanism

  1. Aug 26, 2008 #1
    Hi,

    So if we have a two-step mechanism that goes something like this:

    1) A + B -> C
    2) C -> D
    -------------
    A + B -> D

    and the second step is the slow step, the reaction rate would be r = k2[C] correct? Well, seeing as reactants A and B are zero order, varying their concentrations wouldn't affect the rate. However, one of the questions I encountered said if you ADD A or B or DECREASE A or B, the rate subsequently increases and decreases respectively. I understand the reasoning for this since more A or B would drive the reaction to the right and thus increasing the concentration of intermediate C, which is important in determining the rate. But really, what's the difference between varying the concentration of A or B and adding A or B; if we add A, are we not effectively increasing its concentration?

    I gotta second question kind of related to this. We have a reaction A + B -> C with equilibrium constant: Kc = [C]/[A]. Now, if we add reactant A, BUT at the same concentration as the equilibrium concentration, then according to the equation, the reaction does not shift to the right, correct?

    Thanks.
     
  2. jcsd
  3. Aug 26, 2008 #2
    if A and B are solids, then, adding either of them to the reacting solution would increase their concentration, and shift the equilibrium more to the right. you are increasing the number of moles of A or B, but keeping the volume fixed.

    if A and B are solutions, then the concentration does not change. if you add 10 cm3 of 1 M HCl to 30cm3 of 1 M HCl, then you get 40cm3 1 M HCl. the concentration does not change. there is no change in reaction rate.


    if the concentration of A at equilibrium is x M, and the concentration of the additional A you add is also x M, then the concentration of A in the solution does not change.
     
  4. Aug 26, 2008 #3
    The thing is though, A and B are not part of the rate-determining step so technically even increasing either of their concentrations should not affect the rate 9ince they are zero order no?
     
  5. Aug 26, 2008 #4

    Ygggdrasil

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    2015 Award

    The concentration dependence on A and B are not zero order as they affect the concentration of [C]. In other words, [C] is a function of [A] and [ B]. For example, if you assume that step 1 reaches equilibrium much more quickly than a turnover of step 2, then [C] = K1[A][ B], where K1 = k1/k-1 is the equilibrium constant for reaction 1. In this case, your rate equation becomes:

    rate = k2K1[A][ B]

    and, the reaction is first order in both A and B.
     
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