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Homework Help: DE Chemical Combinations

  1. Mar 6, 2010 #1
    Two chemicals, A and B, are combined, forming chemical C. The rate of the reaction is jointly proportional to the amounts of A and B not yet converted to C. Initially, there are 50 grams of A and 80 grams of B, and, during the reaction, for each two grams of A used up in the conversion, there are three grams of B used up. An experiment shows that 100 grams of C are produced in the first 10 minutes. After a long period of time, how much of A and B remains, and how much of C has been reproduced?

    dx/dt = k*(50-[tex]\frac{2}{5}[/tex]*x)*(80-[tex]\frac{3}{5}[/tex]*x)


    After separation and solving for partial fractions, I obtain:

    [tex]\int\frac{1}{10-2*x}[/tex] - [tex]\frac{3}{2}[/tex][tex]\int\frac{1}{16-3*x}[/tex] = k*t+c

    Which then yields:

    [tex]\frac{16-3*x}{10-2*x}[/tex] = C*e[tex]^{2*k*t}[/tex]

    C=8/5

    k=[tex]\frac{ln(71/76)}{20}[/tex]

    However, something is wrong with my final equation solved for x(t) due to x(0) doesn't = 0 and x(10) doesn't = 100.
     
  2. jcsd
  3. Mar 6, 2010 #2

    Dick

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    x is the amount of chemical C, right? I think you making more than one mistake. You've got a (50-2x/5) going in and a (10-2*x) coming out in the partial fractions. Those aren't proportional to each other. (50-2x/5)=(250-2x)/5=(125-x)*(2/5).
     
  4. Mar 6, 2010 #3

    epenguin

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    You might as well get that sorted out, but an odd or trick question it seems to me, with superfluous information. Are you sure it is about kinetics despite the kinetic info in it? :confused: The equlibibrium is all towards product since there is no mention of back reaction in the kinetic laws given. 50g of A reacting in a 2:3 ratio with B will after a long time give you 125g of C leaving 0 A and 5g of B it seems to me.
     
  5. Mar 6, 2010 #4

    Dick

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    Good point. I wasn't taking the long view.
     
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