I have a seemingly easy question about performing an Arrhenius fit to the equation
$$y = A \times \exp \left( -\frac{E_A}{RT} \right)$$
I can either fit this in the exponential form using a non-linear fitting algorithm, or I can transform my temperature values to $1/T$ and my Y-Values to $\log(Y)$ and fit the data with a linear equation.
I then just need to set $$A = \exp(\text{Intercept})$$ and $$E_A = -\text{Slope} \times R$$
I would presume these should give the same result for the same data, however when I actually try it out, I get rather different results for $A$ and $E_A$.
The reason for this is probably, that for the exponential form, the higher values contribute more to the sum of squares than the smaller y-values. In the linear form, all y-values are roughly in the same order of magnitude, thus avoiding this "weighting" effect.
But the question now is: Which is the correct answer? Should I use the exponential fit or the linear approach?
If you want to try it out (X values are in kelvin, Y-Values are conductivity values)
Temperature / K
253.15
263.15
273.15
283.15
293.15
303.15
313.15
323.15
333.15
Conductivity / S/cm
2.70763399971192E-4
4.06886505509869E-4
5.63162560690061E-4
7.38270829563633E-4
9.40304202004938E-4
0.00115908392908651
0.0013418776248027
0.00154927476612532
0.00173264667362589
From a linear fit I get:
E_A = 16150.00143
A = 0.656834781
From an exponential fit I get:
E_A = 14235.1261
A = 0.307851979
So there is quite a large difference in the results between the two methods.