1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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 :)
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Calcualting Activation Energy using Arrhenius equation and plot
  1. Arrhenius equation (Replies: 5)

Loading...