Calcualting Activation Energy using Arrhenius equation and plot

  • Thread starter FaraDazed
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  • #1
FaraDazed
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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.
 

Answers and Replies

  • #2
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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
 
  • #3
FaraDazed
347
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

Right ok, thanks for clearing that up ie J/mol and not just J. Much appreciated, thanks :)
 

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