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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.

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  • $\begingroup$ $EA \to$ 'Activation energy'? $\endgroup$ – HDE 226868 Nov 30 '14 at 23:08
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    $\begingroup$ Yes normally $E_A$ - I will edit latex $\endgroup$ – tom Nov 30 '14 at 23:41
  • $\begingroup$ I would expect you get the same results with both methods, provided you properly weight your data points by 1/σ², where σ in the uncertainty, and take care of transforming the uncertainties when you transform the data. $\endgroup$ – Edgar Bonet Dec 2 '14 at 16:34
  • $\begingroup$ Hi Jens. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Sep 25 '15 at 6:18
  • $\begingroup$ @Qmechanic Well, since there wasn't any 'homework' involved and I'm used to remove unnecessary tags, this tag seemed (and for me still seems) pointless. But I understand that there are policies, however blurry in this case. Please feel free to add the tag again. I just don't think it fits. $\endgroup$ – Jens Sep 25 '15 at 6:31
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Using the linearization procedure has become so common in experimental chemical kinetics that practitioners have taken to using it to define the activation energy for a reaction. That is the activation energy is defined to be (-R) times the slope of a plot of ln(k) vs. (1/T )

Thus, you should go with the linearized version because it is the common practice in the field. However, if I had to fit the equation without an apriori preference, I would use both methods, calculate the errors of each one, and choose the one with the smaller errors (in the least squares sense). There is no correct answer, it is matter of choice to what part of the data you want to give more weight (in the linear format you give more weight to, or a better fit for, the smaller values). For instance, if you had them available, you could consider if the relative errors for the measured conductivity are larger for the smaller or for the larger values, and choose the wighting based on that heuristics. If you do not have those, it is very limited what you can do in terms of good statistics. Also, I would add to the data the point (0,0), which is the only one theoretical value that you do know for sure.

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  • $\begingroup$ Thank you for your answer. I think you are correct, that in such a case sticking to the custom routines is the best way to go, when no clear answer could be given (following Chris's argumentation in his answer). One thing: Do you have a source for the quote you have given? Thank you. $\endgroup$ – Jens Dec 1 '14 at 18:21
  • $\begingroup$ sure, actually, it the same you gave in your question :) en.wikipedia.org/wiki/Arrhenius_equation $\endgroup$ – Wolphram jonny Dec 1 '14 at 18:24
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The sign of a good fit is that the residuals have the same distribution as your model for the errors. Usually the assumption that goes into fitting methods is that the errors are normally distributed. That is, given perfect inputs $x_i$, and an ideal relation $y_i = f(x_i)$, you will measure $y_i + \epsilon_i$, where $\epsilon_i$ are distributed normally.

So if you suspect a fit is not good (and it's great that you pay attention to these details!) a good first step is to plot the data over the fit, and to plot the residuals as well. For your data, the linear and exponential fits are plotted below.

linear fit and residuals exponential fit and residuals

Neither set of residuals looks particularly bad in terms of the distribution. It's not like you have one or two high points and everything else is low.1

You can also check by plotting the distributions of residuals for the two methods, as I've done below. The dashed line is the best-fit normal distribution. However, without a lot more points, it's hard to visually identify non-gaussianity. There are various statistical tests you can perform to quantitatively analyze the distribution of residuals, but that's getting to be a lot of worry over not-too-bad data.

linear residual distribution exponential residual distribution

In this particular case I'm not sure how worried I'd be. If you are only interested in extracting $E_A$, for instance, the disagreement is only about $6\%$, which is often pretty good for things that involve exponential dependence. If you still want some way to determine which fit to use, you can always think about whether your method for measuring conductivity should have normal or log-normal measurement errors.


1Actually, there is definitely a trend. One usually hopes for residuals that look independently drawn from the same distribution -- independent of where the point is on the $x$-axis. But the middle residuals are systematically higher than the ends. This is something to consider, but it's beyond the scope of what you asked.

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  • $\begingroup$ Thank you very much for the work you put into it. You are right about looking at the residuals is a good idea, and one can't really come to a conclusion in this case. As an accepted answer I will however go with wolprhram's answer, since I guess he is correct about using adapted procedures. I would love to combine both of your answers though, since the your arguments and his conclusion is basically the go to solution I guess. Thanks again. $\endgroup$ – Jens Dec 1 '14 at 18:19

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