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Math Help - Separable Equations Application

  1. #1
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    Separable Equations Application

    In an elementary chemical reaction, single molecules of two reactants A and B form a molecule of the product C: A + B -> C. The law of mass action states that the rate of reaction is proportional to the product of the concentrations of A and B: d[C]/dt = k[A][B]. Thus, if the initial concentrations are [A] = a moles/L and [B] = b moles/L and we write x= [C], then we have dx/dt= k(a-x)(b-x).

    (a) Assuming that a does not equal b, find x as a function of t. Use the fact that the initial concentration of C is 0.

    (b) Find x(t) assuming that a = b. How does this expression for x(t) simplify if it is known that [C] = (1/2)a after 20 seconds?
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  2. #2
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    What ideas have you had so far?
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  3. #3
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    I separated my variables. But I don't know how to go about integrating it.
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  4. #4
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    Well, show me what you've gotten so far.
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  5. #5
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    dx/((a-x)(b-x)) = k dt
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  6. #6
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    The integral on the right is quite straight-forward, right? I'm assuming it's the LHS that is a bit more difficult. The usual technique to integrate the LHS would be partial fractions. What does that give you?
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  7. #7
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    [a-abe^(akt-bkt)]/[1-(a/b)e^(akt-bkt)]
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  8. #8
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    So, there are two checks you need to perform (which you should ALWAYS do with every single DE you ever solve):

    1. Check that your solution satisfies the initial conditions.
    2. Check that your solution satisfies the original DE.

    Does your solution satisfy those two criteria?
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  9. #9
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    I'm not sure if it does. I don't think so.
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  10. #10
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    I don't think so, either. Can you show all your steps from the separation of variables to your final solution, including plugging in the initial condition? Your answer is quite close to the correct answer, so I imagine it's just a small algebra error somewhere.
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  11. #11
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    My friend just told me that answer. I just don't know how he got it..
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  12. #12
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    Oh, I see.

    Do you remember the method of partial fractions for integrating?
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  13. #13
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    A/(a-x) + B/(b-x) = C
    Ab-Ax+Ba-Bx=0, right?
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  14. #14
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    Close, but not quite. You have the fraction

    \displaystyle\frac{1}{(a-x)(b-x)}

    as the integrand on the LHS of your separated DE. You would like to write this as two separate fractions. So, pretend you can and see if it works:

    \dfrac{1}{(a-x)(b-x)}=\dfrac{A}{a-x}+\dfrac{B}{b-x}\quad\Rightarrow\quad 1=A(b-x)+B(a-x).

    Did you see how that last step was done? It's just a common denominator on the RHS, and then canceling the denominators of both sides.

    Now what?
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  15. #15
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    We distribute the A and B?
    Ab-Ax+Ba-Bx=1
    (A+B)=0
    (Ab-Ba)=1
    Right?
    Last edited by CluelessSenior; February 18th 2011 at 06:05 PM.
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