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Math Help - chemical reaction....

  1. #1
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    chemical reaction....

    ok im not sure how to set this problem up


    two chemicals A and B are combined to form a chemical C. the rate or velocity of the reaction is proportional to the product of the instantaneous amounts of A and B not converted to chemical C. initially there are 40 grams of A and 50 grams of B and for each gram of B, 2 grams of A is used. it is observed that 10 grams of C is formed in 5 minutes. ho much is formed in 20 minutes? what is the limiting amount of C after a long time?

    i know im suppose to be using this formula...



    i also know that X(5)=10, the relationship between A and B is 2 to 1 or 2X to X, in jus not sure how to get the alpha or beta....thanks in advance.
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  2. #2
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    ok my book has the setup...



    im not sure how it came up with the 120 and the 150?
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  3. #3
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    Quote Originally Posted by slapmaxwell1 View Post
    ok my book has the setup...



    im not sure how it came up with the 120 and the 150?
    You are correct that the above post is wrong, I had a serious brain cramp!

    \displaystyle \alpha=\frac{a(M+N)}{M}, \text{ and } \beta=\frac{b(M+N)}{N}

    Where a is the intial amount of A
    Where b is the intial amount of B
    M is the proportion of A and N is the proportion of B

     \displaystyle \alpha=\frac{a(M+N)}{M}=\frac{40(2+1)}{2}

     \displaystyle \beta=\frac{b(M+N)}{N}=\frac{50(2+1)}{1}

    Now we get

    \displaystyle \left( \frac{40(2+1)}{2}-X\right) \left( \frac{50(2+1)-X}{1}\right)=\frac{1}{2}(120-2X)(150-X)

    Again sorry for the brain cramp!
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    You are correct that the above post is wrong, I had a serious brain cramp!

    \displaystyle \alpha=\frac{a(M+N)}{M}, \text{ and } \beta=\frac{b(M+N)}{N}

    Where a is the intial amount of A
    Where b is the intial amount of B
    M is the proportion of A and N is the proportion of B

     \displaystyle \alpha=\frac{a(M+N)}{M}=\frac{40(2+1)}{2}

     \displaystyle \beta=\frac{b(M+N)}{N}=\frac{50(2+1)}{1}

    Now we get

    \displaystyle \left( \frac{40(2+1)}{2}-X\right) \left( \frac{50(2+1)-X}{1}\right)=\frac{1}{2}(120-2X)(150-X)

    Again sorry for the brain cramp!
    ok so what happens to the 1/2? thanks in advance!
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  5. #5
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    Quote Originally Posted by slapmaxwell1 View Post
    ok so what happens to the 1/2? thanks in advance!
    Don't forget the k

    So you get

    \displaystyle \frac{dX}{dt}=k\left( \frac{1}{2}\right)(120-2X)(150-X)= \hat{k}(120-2X)(150-X)

    Since k is determined by the X(5)=10

    We can relabel and use a different arbitrary constant.
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