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Integrating the differential rate equation

  1. Apr 23, 2012 #1
    Heres the differential rate equation for a 0 order reaction in chemistry:
    [itex]Rate = {{-d[A]} / {dt}} = k[/itex]
    which can be rearranged to this:
    [itex]-d[A] = dt k[/itex]
    and when you integrate this you get the integrated rate equation but I don't understand how this works. The site I'm reading says you integrate both sides of the equation and get this:
    http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord_z5.gif
    is that supposed to be a definite integral or something? Anyhow, after integrating they get this:
    http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord_z3.gif
    I don't get it. Firstly [A] and t are variables so shouldn't they be raised a power and get divided by 2? In other words, shouldn't they become [itex][A]^{2} / 2[/itex] and [itex]t^{2} / 2[/itex]? Secondly, whats going on with the d. d means the change in something but I've never seen it been integrated before so I don't know what happens to it. Why did it disappear?

    The 1st order equation has me equally confused:
    http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1_ip1.gif
    integratin that, they get this:
    http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/Grafik/ord1-3.gif
    has does d[A]/[A] become ln[A]?
     
    Last edited: Apr 23, 2012
  2. jcsd
  3. Apr 23, 2012 #2
    Any time there are limits of integration it is a definite integral.

    You need to remember back to your Calc I/II days. The d means an infitessimal change, you covered this when going over Riemann sums in Calc I. In more general terms, the d term lets you know for what term you are integrating with respect to. dx means you are integrating with respect to x and all other variables are held constant.

    ##\int x \ dx = \frac{x^2}{2} + C## because the variable you are integrating with respect you is x. However, ##\int y \ dx = xy + C## because y is held constant.

    So, in this problem we have ##\displaystyle \int_{[A]_0}^{[A]} d[A] = -k \int_{t_o}^{t} dt## We are simply integrating the constant 1 in both of these equations, so the integral yields ##[A]## and t respectively, evaluated at the proper limits of integration.

    ##\displaystyle \int_{[A]_0}^{[A]} d[A] = -k \int_{t_o}^{t} dt## yields ##[A] - [A]_0 = -k (t - t_0)##. In these types of problems, ##t_0## is usually 0 so the equation simplifies to ##[A] = -kt + C##. I believe the + C term is the initial amount of the concentration of A, we never wrote it like this, we kept the ##[A]_0## term in the equation.

    Remember ##\displaystyle \int \frac{1}{x} dx = ln|x| + C##.
     
  4. Apr 24, 2012 #3

    HallsofIvy

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    [tex]\int_{[A]_0}^{[A]} \frac{d[A]}{[A]}[/tex]
    is just bad notation. Although it is, unfortunately, often done you should NOT use the same symbol for the "variable of integration" and the limits of integration. Better would be
    [tex]\int_{[A]_0}^{[A]} \frac{dt}{t}[/tex]
     
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