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  1. #1
    Senior Member euclid2's Avatar
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    chemistry

    -\frac{\delta[A]}{\deltat}=r=k[A] How would i manipulate this using integral calculus to yield the expression In \frac{([A]_o)}{([A])}= kt (This is a question asked out of curiosity)
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  2. #2
    o_O
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    For a first-order reaction, the differential rate is given by: r = -\frac{d[A]}{dt} = k[A]

    To get its integrated form, separate the variables:
    \begin{aligned}-\frac{d[A]}{[A]} & = k dt \\ -{\color{red}\int}\frac{d[A]}{[A]} & = {\color{red}\int} k dt \\ \ln [A] & = -kt + C\end{aligned}

    Normally, we're given an initial concentration, i.e. [A]_0 \ \ @ \ \ t=0

    This enables us to solve for the constant: \ln [A]_0 = -k(0) + C \ \Leftrightarrow \ C = \ln [A]_0

    I'm sure you can take it from here.
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  3. #3
    Senior Member euclid2's Avatar
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    Quote Originally Posted by o_O View Post
    For a first-order reaction, the differential rate is given by: r = -\frac{d[A]}{dt} = k[A]

    To get its integrated form, separate the variables:
    \begin{aligned}-\frac{d[A]}{[A]} & = k dt \\ -{\color{red}\int}\frac{d[A]}{[A]} & = {\color{red}\int} k dt \\ \ln [A] & = -kt + C\end{aligned}

    Normally, we're given an initial concentration, i.e. [A]_0 \ \ @ \ \ t=0

    This enables us to solve for the constant: \ln [A]_0 = -k(0) + C \ \Leftrightarrow \ C = \ln [A]_0

    I'm sure you can take it from here.
    Thanks very much. For some reason I was expecting a reply from you!

    I suppose we can assume then after one half life we can simplify further to;

     In  2 = kt_\frac{1}{2}
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    o_O
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    This is only applicable to first-order reactions though as I'm sure you will learn soon enough.
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  5. #5
    Senior Member euclid2's Avatar
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    Quote Originally Posted by o_O View Post
    This is only applicable to first-order reactions though as I'm sure you will learn soon enough.
    Something to look forward to! Ahh the wonders of chemistry
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