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anonymous

  • one year ago

In the Thrall model, the reaction rate (with respect to time) is proportional to the amount x of the new substance at time t , and the reaction rate is also proportional to a - x where a is the original amount of the first substance. a. Why does this mean that dx/dt = K x (a - x) for some positive constant K ? b. Differentiate with respect to x your expression for dx/dt.

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  1. anonymous
    • one year ago
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    I think I can probably handle b, but what do you think these variables mean? given a catalytic reaction. What variables would represent "first substance"? "the new substance"? K ? etc..

  2. anonymous
    • one year ago
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    Here's a bit more of that question ... if it helps at all. "R. M. Thrall and his University of Michigan colleagues (Report No. 40241-R-7, University of Michigan, 1967) gave the following crisp description of autocatalytic reaction of one substance into a new substance: "An autocatalytic reaction progresses in such a way that the first substance catalyzes its own formation."

  3. hartnn
    • one year ago
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    we won't need a variable to represent 'first substance', the reaction rate depends on the amount of 1st substance, which is already represented by 'a'. same goes for new substance, amount of new substance is represented by 'x' K is just a constant of proportionality! such constants come on all variation problems... ex: if x is proportional to y, x= ky, k= constant of proportionality.

  4. Michele_Laino
    • one year ago
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    I think that the differential equation: dx/dt = K x (a - x) is right!

  5. anonymous
    • one year ago
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    Is there any reason why I should assume here that a=x when t=0 ?

  6. Michele_Laino
    • one year ago
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    a is the initial concentration

  7. hartnn
    • one year ago
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    Direct variation: If \(x\) is \(\text{directly proportional to} ~ y,\) then we say x/y is a constant. this constant is the constant of proportionality = k (say) so x/y = k, or x =ky in your problem, reaction rate(dx/dt) is proportional to x and (a-x) hence, dx/dt = K x(a-x)

  8. Michele_Laino
    • one year ago
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    that is a second order process, which involves two substances

  9. hartnn
    • one year ago
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    a is the amount of 1st substance x is the amount of new substance at t=0, the reaction didn't happen yet, so at t = 0, x = 0

  10. anonymous
    • one year ago
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    awesome.. that's the piece had me confused..

  11. anonymous
    • one year ago
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    the equation makes sense now.

  12. hartnn
    • one year ago
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    had some doubt about part b. what you found in part a is dx/dt already! because is the reaction 'rate' Then how could we find dx/dt by differentiating the expression w.r.t x.... won't that give you 2nd derivative? d^2x/ dt^2

  13. anonymous
    • one year ago
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    well that was another thing I was wondering... if a 2nd differential was going to come into play here.

  14. hartnn
    • one year ago
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    oh, they are asking to differentiate the expression for dx/dt ! ok, got it :P

  15. hartnn
    • one year ago
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    yes, the answer for part b will indeed be d^2x/dt^2

  16. hartnn
    • one year ago
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    the wording were confusing! "for dx/dt" >>> i read it as 'to get dx/dt'

  17. anonymous
    • one year ago
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    oh right on.. I see .. differentiate the dx/dt .. all good.

  18. anonymous
    • one year ago
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    well if it confuses you, it must be bad ... lol.

  19. anonymous
    • one year ago
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    alright.. thanks again you guys.. I got this.. :)

  20. hartnn
    • one year ago
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    welcome ^_^

  21. Preetha
    • one year ago
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    Hey Hugh, give one of the Qualified Helpers a rating so they can get their share of OwlBucks!

  22. anonymous
    • one year ago
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    right on, sorry.. I usually always do that * eventually :)

  23. Preetha
    • one year ago
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    Thanks.

  24. hartnn
    • one year ago
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    Thanks hugh! much appreciated :)

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