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Differential v. Integral Rate Laws; Kinetics

  1. Apr 22, 2010 #1
    I'm nearing the end of AP Chemistry, and in review for the big test I came across my old friend, Chemical Kinetics. Also being near the end of Precalculus, the "differential" and "integral" laws piqued my interest. Remembering that I have not taken any Calculus lessons, can someone help explain the mathematics involved in deriving the formulae we use in kinetics?
  2. jcsd
  3. Apr 23, 2010 #2


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    Well, that's a kind of tough order, since it's all about calculus, really. But here's an attempt at explaining the basic calculus behind it all:

    Chemical kinetics is all about the rate at which things react, i.e. the rate at which molecules are formed. So you're dealing with finding out:
    [tex]\frac{d[A]}{dt}[/tex], i.e. the rate of change of the concentration of A with time.

    Now, typically, the rate at which the concentration changes will depend on the concentration itself. More molecules of A means more reactions taking place, so in general, you can write:

    [tex]-\frac{d[A]}{dt} = k[A]^n[/tex]

    Which says that the concentration of A decreases over time in proportion to the concentration of A (itself), raised to the power of n. Where n is called the order of the reaction. Chemical reactions are most often first or second order.

    (Although for a zeroth-order reaction n=0 then [tex][A]^n = 1[/tex], meaning in that one case the rate at which [A] changes does not depend on [A].)

    So with the exception of zeroth-order, you have an equation with a function ([A], the concentration of A is a function of time) which depends on its own derivative ([tex]\frac{d[A]}{dt}[/tex]). That's termed a differential equation, hence a 'differential rate law'.

    So this tells us how the [A] changes with time. But we're not usually interested in that. What you're usually interested in is [A] itself. What's [A] at time t? To get that, you have to solve the differential equation. Taking as an example the rate law for a first-order reaction:
    [tex]-\frac{d[A]}{dt} = k[A][/tex]
    The solution to this differential equation (the second-simplest D.E. possible) is, if you haven't learned it yet:
    [tex][A] = [A]_0 e^{-kt}[/tex]

    Where [tex][A]_0[/tex] is the concentration at t = 0. Since the differential equation only tells us how [A] changes with time, obviously we have to know its value at the starting point to be able to say anything about what happens later.

    The solved equation is called the integrated rate law.

    Now, it can get trickier: The rate for [X] might depend on the concentration of more than one substance, (e.g. [Y]) and the concentration [Y] might depend on [X]. That's called a system of coupled differential equations (and you'll probably how to solve them eventually). A nifty property of those is that, if the conditions are right, they can oscillate wildly (chaotically) between different values before reaching equilibrium. An example is the http://www.youtube.com/watch?v=QdscFBvdTLg". This is also seen in nature where you have 'cycles' where the populations of predators and prey oscillate. (predator-prey models are a very popular example of coupled diff equations)
    Last edited by a moderator: Apr 25, 2017
  4. Apr 23, 2010 #3
    Although this is a differential equation, it is a very simple one to solve. In fact, the other simple rate laws (k[A]2 or k[A]3) are easy to solve as well. It only gets difficult when you include [A] and

    In order to solve this just integrate (still beyond precalculus level though):

    Rearrange [tex]-\frac{d[A]}{dt} = k[A][/tex]

    [tex]\frac{d[A]}{[A]} = -k[dt][/tex]

    Then integrate both sides to get

    ln[A] - ln [A0] = -kt

    which rearranges to

    [tex][A] = [A]_0 e^{-kt}[/tex]
    Last edited: Apr 23, 2010
  5. Apr 23, 2010 #4

    Char. Limit

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    Except for the fact that you just solved a seperable differential equation right there.

    EDIT: Crossposted with the edit and didn't even read the box once I started posting... sorry Cesium.
  6. Apr 23, 2010 #5
    Many thanks, all - this is very helpful. When I learn a concept, I fail at remembering it unless I feel the topic has been logically proved in my own mind. Understanding a bit more about the mathematics behind kinetics seems to "complete" the concepts thereof.
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