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Rate constant and Kinetic model

  1. Mar 13, 2014 #1
    Hi everyone,
    I have a question on the rate constant of a kinetic model(like the one we use in describing chemical reaction). Suppose we have the following model between two states A and B:
    A--(a)-->B
    A<--(b)--B
    where a, b are the rate constant for the transition. So the corresponding differential equation will be
    [itex]\frac{dA}{dt}=-aA+bB[/itex]
    [itex]\frac{dB}{dt}=-bB+aA[/itex]
    If now I add an intermediate state C, so that if we just measure the amount of A and B, the two models gives basically the same description (i.e. the overall rate constant from A to C then to B is equivalent to a and the same for reverse direction)
    A--(a1)-->C--(a2)-->B
    A<--(b1)--C<--(b2)--B

    What is the functional relationship between the rate constant a1,a2 and a and similarly for b1, b2 and b? I am confused on how to find out the functional form. Is it just simply
    [itex]\frac{1}{a}=\frac{1}{a1}+\frac{1}{a2}[/itex]?
     
  2. jcsd
  3. Mar 13, 2014 #2

    pasmith

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    Homework Helper

    There is no functional relationship. What this system gives you is
    [tex]
    \frac{dA}{dt} = -a_1A + b_1 C, \\
    \frac{dB}{dt} = -b_2 B + a_2 C, \\
    \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.
    [/tex]

    There is no way to eliminate [itex]C[/itex] from this system in order to end up with a system involving only [itex]A[/itex] and [itex]B[/itex]. The complication comes from the fact that you also have [itex]A \to C \to A[/itex] and [itex]B \to C \to B[/itex], so not all [itex]A[/itex] turns to [itex]B[/itex] before turning back to [itex]A[/itex] and vice-versa.
     
  4. Mar 13, 2014 #3
    Maybe I change my question, is there any 'effective' rate constant between A and B in the new system
    [tex]
    \frac{dA}{dt} = -a_1A + b_1 C, \\
    \frac{dB}{dt} = -b_2 B + a_2 C, \\
    \frac{dC}{dt} = a_1 A + b_2 B - (a_2 + b_1) C.
    [/tex] ?
     
  5. Mar 13, 2014 #4
    If C is short-lived so that it is at quasi steady state, then

    [tex]C=\frac{a_1 A + b_2 B}{(a_2 + b_1)}[/tex]

    If we substitute this into the other differential equations, we get:

    [tex]\frac{dA}{dt}=\frac{-a_1a_2A+b_1b_2B}{(a_2 + b_1)}=-a_3A+b_3B[/tex]
    [tex]\frac{dB}{dt}=\frac{a_1a_2A-b_1b_2B}{(a_2 + b_1)}=a_3A-b_3B[/tex]

    Chet
     
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