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Math Help - Kinda stuck with this one

  1. #1
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    Kinda stuck with this one

    ok here's the problem i'm havin' a problem with getting the right solution

    "When two chemicals, A and B react, 1 gram of A and 2 grams of B yield 2 grams of C and a by-product. The rate at which C is formed is proportional to the amount of A and B present. Sypose that at the outset there is 50 grams of A and 90 grams of B present, and that after 1 hour, 45 grams of C have been formed."

    Show that the amount of C that has been formed after t hours

    What is the maximum of C that can be obtained from this process?

    i'm mainly having trouble with the first section but you need it for the second part, any help with this would be good
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  2. #2
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    Quote Originally Posted by action259
    ok here's the problem i'm havin' a problem with getting the right solution

    "When two chemicals, A and B react, 1 gram of A and 2 grams of B yield 2 grams of C and a by-product. The rate at which C is formed is proportional to the amount of A and B present. Sypose that at the outset there is 50 grams of A and 90 grams of B present, and that after 1 hour, 45 grams of C have been formed."

    Show that the amount of C that has been formed after t hours

    What is the maximum of C that can be obtained from this process?

    i'm mainly having trouble with the first section but you need it for the second part, any help with this would be good
    I'm stumped on how to do the rates at the moment, but the second question can be done without them. According to the ratios we have a reaction:
    A + 2B \to 2C + D
    Thus if we have 90 g of B we can only use 45 g of A to produce 90 g total of C. (The amounts of B and C are in a 1:1 ratio.)

    -Dan
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  3. #3
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    i acctually forgot a part of the question

    Show that the amount of C that has been formed after t hours is

    <br />
y(t) = 900[(10/11)^t-1]/[10(10/11)^t-9]<br />

    sorry i forgot to include that
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by action259
    ok here's the problem i'm havin' a problem with getting the right solution

    "When two chemicals, A and B react, 1 gram of A and 2 grams of B yield 2 grams of C and a by-product. The rate at which C is formed is proportional to the amount of A and B present. Sypose that at the outset there is 50 grams of A and 90 grams of B present, and that after 1 hour, 45 grams of C have been formed."

    Show that the amount of C that has been formed after t hours

    What is the maximum of C that can be obtained from this process?

    i'm mainly having trouble with the first section but you need it for the second part, any help with this would be good
    You have:

    <br />
\frac{dC}{dt}=kAB<br />
,

    and:

    <br />
 \begin{array}{cc}<br />
{A\ =}& {\ \ \ A_0-C/2}\\<br />
{B\ =}& {B_0-C}<br />
\end{array}<br />
,

    where A_0=50 the initial amount of A, and B_0=90 the initial amount of B.

    Thus you have a first order ODE for C to solve, and the rate constant k will be
    determined by the amount of product at t=1 hour.

    RonL
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  5. #5
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    But;

    \frac{dc}{dt} = k(50-c/2)(90-c)

    With c(1) = 45, it doesn't seem to lead to;

    c(t) = 900\frac{(10/11)^{-t}-1}{10(10/11)^{-t}-9}

    It will lead to an equation with e, and it would be impossible to get rid of them to get to c(t)
    Last edited by AnnM; August 27th 2007 at 12:28 PM.
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  6. #6
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    More specifically;

    \int \frac{dc}{(50-c/2)(90-c)} = \int kdt

    1/5\ln(c-100)-1/5\ln(c-90) = kt+Q

    (c-100)^(1/5)/(c-90)^(1/5) = Qe^{kt}

    (c-100)/(c-90) = Qe^{5kt}

    c = (c-90)Qe^{5kt}

    c = cQe^{5kt}-90Qe^{5kt} + 100

    c - cQe^{5kt} = -90Qe^{5kt} + 100

    c(1 - Qe^{5kt}) = -90Qe^{5kt} + 100

    c = \frac{-90Qe^{5kt} + 100}{1 - Qe^{5kt}}

    It clearly doesn't work.
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  7. #7
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    You're clearly not done! Don't give up just before you see it.

    Substitute t = 0 and c = 0 to solve for Q.

    Substitute for Q as above, then substitute t = 1 and c = 45 to solve for k.

    I think you'll be happier in many things in life if you don't quit too soon.
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  8. #8
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    Setting c = 0 and t = 0 gives Q = 10/9, leading to;

    <br />
c(t)= \frac{-100e^{5kt} + 100}{1 - 10e^{5kt}/9}<br />

    Then, using c = 45, t = 1;

    <br />
k = 1/5\ln(11/10)<br />

    <br />
c(t)= \frac{-100e^{\ln(11/10)t} + 100}{1 - 10e^{\ln(11/10)t}/9}<br />

    <br />
c(t)= \frac{-100(11/10)^t + 100}{1 - (10/9)(11/10)^t}<br />

    <br />
c(t)= \frac{-900(11/10)^t + 900}{9 - 10(11/10)^t}<br />

    <br />
 c(t)= 900\frac{-(11/10)^t + 1}{9 - 10(11/10)^t}<br />

    Almost... it's going to be easy from here.
    Last edited by AnnM; August 27th 2007 at 04:51 PM.
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  9. #9
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    Okay, now you're beginning to distress me. Remember how I said you were close before? Well, now it's as if you were driving from New York to Los Angeles. You drive all the way across the country, through the wonderful deciduous forests fo the East, through the intriguing MidWest, across the expansive and delightful Plains States, up the glorious rockies and coniferous forests, through the trecherous deserts, break into the valley and if you go around one more corner you will see the sign that says "Entering Los Angeles". But you don't go around that corner. There is a crazy guy standing by the road and he has a sign that says Los Angeles isn't there any more. You stop, read the sign, turn around and go back home.

    I am not trying to make fun of you. I am trying to get you to see how easily you have managed to discourage yourself at the tiniest sign of trouble. Just in the last few minutes I have seen you do it TWICE! Here's a plan...Stop It! Show just a hair more effort. Go get the answer, don't wait for it to drop out of the sky.

    Note: No fair editing while I am typing. Good work.
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