Abhisar
  • Abhisar
So I was wondering how to derive an differential form of Arrhenius Equation.
Chemistry
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Abhisar
  • Abhisar
How to derive this \(\sf \Large \frac{d}{dt}(ln K)=\frac{E_a}{RT^2}\)
anonymous
  • 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.
Abhisar
  • 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|

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
lol thanks! XD

Looking for something else?

Not the answer you are looking for? Search for more explanations.