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Deriving the rate laws for first and second order reactions

  1. Dec 12, 2005 #1
    So, the average rate for a reaction of type A --> product is given by [tex]\text{rate} = -\frac{\Delta A}{\Delta t}[/tex]. Also, [tex]\text{rate} = k \cdot \text{A}[/tex].

    The instantaneous rate for a reaction of that type is [tex]\lim_{\Delta t\rightarrow\0} -\frac{\Delta A}{\Delta t} = -\frac{dA}{dt}[/tex].

    Setting the instantaneous rate for a reaction equal to the second equation, there is [tex]-\frac{dA}{dt} = k \cdot \text{A}[/tex].

    Well, this is a very friendly separable differential equation. I get [tex]\ln A = -kt + C[/tex].

    How do I get [tex]\ln{\frac{A}{A_o}} = -kt[/tex] from my derivation? Definite integration?
     
  2. jcsd
  3. Dec 12, 2005 #2


    At time =0 C=Ao
     
  4. Dec 12, 2005 #3

    GCT

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    yes, definite integration, you'll merely have one term on the right since initial time is considered zero. Also remember to always derive the equation for a particular reaction....if you're gonna write the rate equation in terms of a reactant/product with a unique coefficient, they'll be multiples to account for.
     
  5. Dec 12, 2005 #4
    Thanks! It's like every intro chem book always uses the exact phrase, "Using calculus, we can derive..."

    I mean at the intro chem level most students haven't taken calculus, so to show the derivation is unnecessary. I was just curious :)
     
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