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Abhisar

  • one year ago

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

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  1. Abhisar
    • one year ago
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    How to derive this \(\sf \Large \frac{d}{dt}(ln K)=\frac{E_a}{RT^2}\)

  2. anonymous
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    lol thanks! XD

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