Calcualting Activation Energy using Arrhenius equation and plot

[FaraDazed]

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 $ln(k)=ln(A)+(\frac{-E_a}{R})(\frac{1}{T})$ 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.
$$gradient=\frac{-E_a}{R} \\ (gradient)(R)=-E_a \\ (-5525)(8.3145)=-E_a \\ -45937.6=-Ea \\ ∴ E_a=45937.6 Joules$$

any help is really appreciated, thanks.

 

[Chestermiller]

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

 

[FaraDazed]

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 :)