Question

So I was wondering how to derive an differential form of Arrhenius Equation.

Answer 1

How to derive this

\(\sf \Large \frac{d}{dt}(ln K)=\frac{E_a}{RT^2}\)

Answer 2

This comes from taking the empirically observed Arrhenius equation:

\[k = Ae^{-E_a/RT}\]
then rearranging it to get a linear plot by using a natural log:
\[\ln k = -\frac{E_a}{R} \frac{1}{T} + \ln A\]
By linear plot, that means we plot things in a very peculiar way, https://en.wikipedia.org/wiki/Arrhenius_plot#/media/File:Arrhenius_plot_with_break_in_y-axis_to_show_intercept.svg

So that if we look at it through the lens of \(y=mx+b\) and plot \(\frac{1}{T} = x\) and \(\ln k = y\) we have slope \(\frac{-E_a}{R}=m\) and y-intercept \(\ln A = b\).

From here, we recognize from calculus that the slope is the derivative, so:

\[\frac{d}{dx}(y) = m\]
Plugging in everything,
\[\frac{d}{d (1/T)} \ln k = \frac{-E_a}{R}\]

Then we use the chain rule on the derivative to make the funky derivative look nicer:

\[\frac{d (\ln k)}{d (1/T)} = \frac{d (\ln k)}{d T} \frac{d T}{d (1/T)}\]

To evaluate this funky thing, we just recognize that:
\[\frac{d T}{d (1/T)} = \frac{1}{\left(\frac{d (1/T)}{d T}\right)} \]

It's simple enough to take this derivative now:

\[\frac{d}{dT} T^{-1} = -T^{-2}\]

Push all this back together and we get:

\[\frac{d}{dT} \ln k = \frac{E_a}{RT^2}\]

Now what's the point of doing all this? Well remember that plot I showed, this only allows the Arrhenius equation to have linear slopes, that is to say constant slope. However since the formula is empirically discovered, this simply allows us to push it a little further to fit more data to a similar thing and is sometimes used as the definition of activation energy.

Answer 3

Wow!!
That's beautiful D:

I mean i know the part up to comparing the differentiated equation with straight line equation but never figured it could be further beautified to something like this......A vulcan salute for you c:

Thanks a bunch !

Answer 4

lol thanks! XD