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Homework Help: Calcualting Activation Energy using Arrhenius equation and plot

  1. Feb 18, 2014 #1
    This is part of a lab report so not sure if the template agrees with what I am asking but basically using the Arrhenius equation in the form of [itex]ln(k)=ln(A)+(\frac{-E_a}{R})(\frac{1}{T})[/itex] I plotted a straight line graph of ln(k) vs 1/T and found the gradient to be -5525.

    Its the units that are confusing me so not sure if it is as straight forward as what I have done below.
    [tex]
    gradient=\frac{-E_a}{R} \\
    (gradient)(R)=-E_a \\
    (-5525)(8.3145)=-E_a \\
    -45937.6=-Ea \\
    ∴ E_a=45937.6 Joules
    [/tex]

    any help is really appreciated, thanks.
     
  2. jcsd
  3. Feb 18, 2014 #2
    Yes, you did it correctly. The E/RT is dimensionless, so the gradient has units of 1/T, and the activation energy is in J/mole.

    Chet
     
  4. Feb 18, 2014 #3
    Right ok, thanks for clearing that up ie J/mol and not just J. Much appreciated, thanks :)
     
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