December 26, 2024, 08:03:24 AM
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Topic: Getting rate second rate of reaction from firs  (Read 1751 times)

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Offline dzsdgfhjkl

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Getting rate second rate of reaction from firs
« on: February 13, 2024, 01:54:55 AM »


I attempted the "traditional" way of doing it. But the answer I received was nowhere near the answers given. I tried some YouTuber's (Useless) way of doing it and it was basically the same as the start point. At this point I am starting to unlearn everything I have so far and just confuse myself. Not covered in class, in the textbook or seemingly anywhere online (correctly at least)

Original attempt:
ln[k2/k1] = -Ea/R * (1/T2 - 1/T1)
k2 = k1*e^(-Ea/R * (1/T2 - 1/T1))
k2 = (1.35*10^2)*e^(-91000/8.314 * (1/348.15 - 1/298.15))
k2 = k1*e^(-10945.4) * (-4.82*10^-4))
k2 = k1*e^(5.27)
k2 = k1*194.86
k2 = 26306.6

This answer isn't anywhere on the sheet

Tried an ALTERNATIVE approach from online

ln[k2/k1] = -Ea/R * (1/T2 - 1/T1)/(T1*T2)
k2 = k1*e^(-91000/8.314 * ((1/348.15 - 1/298.15)/(298.15*348.15))
k2 = k1*e^(-10945.4 * ((-4.82*10^-4)/(103800.9225))
k2 = k1*e^(-10945.4 * (-4.64E-9)
k2 = k1*e^5.079E-5
k2 = k1*1.000050794
k2 = 135.0068572


Also none of the answers and hardly any different

Incredibly frustrating as I have almost 100% on this assignment SO FAR and if I do one thing wrong I have to do it ALL OVER again. So guessing isn't an option and it would likely glean nothing

Offline Borek

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Re: Getting rate second rate of reaction from firs
« Reply #1 on: February 13, 2024, 03:24:32 AM »
There is an obvious problem with what you copied from the online source - looks like an incorrect attempt at adding two fractions by finding a common denominator:

[tex]\frac 1 {T_2} - \frac 1 {T_1} = \frac {T_1} {T_2\times T_1} - \frac {T_2} {T_1 \times T_2} = \frac {T_1 - T_2} {T_2\times T_1}[/tex]

That being said, I see nothing wrong with your own calculations, 26300 is what I got too.
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Offline Aldebaran

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Re: Getting rate second rate of reaction from firs
« Reply #2 on: February 13, 2024, 04:25:12 AM »
Same answer here too, 26300.

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