Deriving the rate laws for first and second order reactions

  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. At time =0 C=Ao
     
  4. GCT

    GCT 1,769
    Science Advisor
    Homework Helper

    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. 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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