## Abhisar one year ago So I was wondering how to derive an differential form of Arrhenius Equation.

1. Abhisar

How to derive this $$\sf \Large \frac{d}{dt}(ln K)=\frac{E_a}{RT^2}$$

2. anonymous

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.

3. Abhisar

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 !|dw:1441389492893:dw|

4. anonymous

lol thanks! XD