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

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.

Get our expert's

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions.

A community for students.

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

Chemistry
See more answers at brainly.com
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.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this and thousands of other questions

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

Not the answer you are looking for?

Search for more explanations.

Ask your own question

Other answers:

lol thanks! XD

Not the answer you are looking for?

Search for more explanations.

Ask your own question