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Biological mathematics - trimerisations - law of mass action

  1. Aug 5, 2011 #1
    1. The problem statement, all variables and given/known data

    In the real world trimolecular reactions are rare, although trimerizations are not. Consider
    the following trimerization reaction in which three monomers of A combine to form the
    trimer C,

    A + A [itex]\Updownarrow[/itex] B, where k1 and k-1 are the forward and reverse rate constants respectively.
    and,
    A + B [itex]\Updownarrow[/itex] C. where k2 and k-2 are the forward and reverse rate constants respectively.

    (a) Use the law of mass action to find the rate of production of the trimer C.
    (b) Suppose k−1 >> k−2, k2A. Use the appropriate quasi-steady-state approximation to find the rates of production of A and C, and show that the rate of production of C is proportional to [A]3. Explain in words why this is so.

    2. Relevant equations

    law of mass action, quasi-steady-state approximation

    3. The attempt at a solution

    let c = concentration of C, a = concentration of A, b = concentration of B
    using law of mass action:
    dc/dt = k2ab - k-2c

    Is this correct??

    Also I don't know how to go about part b? don't really understand the quasi-steady state for this system.
     
  2. jcsd
  3. Aug 5, 2011 #2
    Let:

    d[A]/dt = rate in - rate out
    rate in = 2 k-1 + k-2 [C]
    rate out = k1 [A]2 + k2 [A]

    d/dt = rate in - rate out
    rate in = k1 [A]2 + k-2 [C]
    rate out = k-1 + k2 [A]

    d[C]/dt = rate in - rate out
    rate in = k2 [A]
    rate out = k-2 [C]

    where [X] denotes the concentration of X.

    You're correct (for that last part).

    Now all that's left is the calculus that you require to solve this. The exponential function is your friend. Make sure you don't forget any free constants that may arise from integration.

    After solving the system, eradicate all the terms with k−2, k2 A in it. That's what it means by letting k−1 >> k−2, k2 A.

    I'm not sure if there's a typo here. The production of C is proportional to [A]3 when the forward reactions are much stronger than the backwards reactions, or when

    k1 and k2 >> k-1 and k-2

    This is because you can just substitute A+A=B into B in A+B=C to make A+A+A=C. Good luck separating the system btw.

    I've no idea why it changes my uppercase B in brackets to lowercase b.

    Edit: Oops. X)
     
    Last edited: Aug 5, 2011
  4. Aug 5, 2011 #3
    how come in your d[A]/dt and d/dt expressions they involve derivatives but in d[C]/dt it does not??
    Ok. I will work on that. thank you.
     
  5. Aug 5, 2011 #4
    Oh snap. I screwed up. :(

    I'm sincerely sorry for that. Read the edited post. Hopefully that's correct now.
     
    Last edited: Aug 5, 2011
  6. Aug 5, 2011 #5
    haha all good. that makes more sense. Just wondering though should the rate out for d[A]/dt be 2k1[A]2 + k2[A]?
     
  7. Aug 5, 2011 #6
    I'm pretty sure it's k1 [A]2, since the molecules colliding is what causes that part to be squared. I'm not too sure about the rate out having the coefficient 2 k-1, but I have a hunch that it is.

    http://en.wikipedia.org/wiki/Rate_equation
     
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