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Topic: Arrhenius Equation  (Read 4406 times)

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sundrops

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Arrhenius Equation
« on: September 23, 2005, 04:01:45 PM »
I've never liked this equation because everytime I try to solv for it I get a different answer. I'm getting really frusterated - here's hoping that one of you guys can point out my mistake  :D

rate constant = 1.67×102 L mol-1 s-1 at 311°C (584K)
rate constant = 2.32×104 L mol-1 s-1 at 518°C (791K)

so here's what I've been doing:

ln(k2/k1) = Ea/R(1/T1-1/T2)
ln(2.32E4 / 1.67E2) = Ea/8.3145(1/584 - 1/791)
(14.1443)(8.3145) = Ea(4.48106E-4)
Ea = 262505 J/mol

but thats not the answer! what am i doing wrong?!? ???

sundrops

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Re:Arrhenius Equation
« Reply #1 on: September 23, 2005, 04:18:39 PM »
nvmnd i figured it out :)

u need to be REALLY careful with ln and where u put your brackets! :)

thx anyways guys - but it sure feels good to figure it out on my own - even if I
m o my 11th try answering this silly question. :S

sundrops

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Re:Arrhenius Equation
« Reply #2 on: September 23, 2005, 05:13:31 PM »
ok, so I thought that I understood this Arrhenius stuff - but I clearly DON'T since I can't figure out  part 2 of the question.  ::)

so i determined that Ea=91600 J/mol

but now I need to find the rate constank k at 384C (657K)

so heres what I did:

ln(k2/k1) = Ea/R(1/T1-1?T2)
ln(k2/1.67E2) = 11016.898(1/584-1/657)
ln(k2) - ln(1.67E2) = 2.09606
ln(k2) = 7.21399

and that where I'm stuck - how do I extrapolate k2?
or am I even going about it the right way?
any input is appreciated! ;D

 

Offline Donaldson Tan

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Re:Arrhenius Equation
« Reply #3 on: September 23, 2005, 05:39:54 PM »
k = Ae-Ea/RT

since you have calculated the value of Ea, and that you know the rate constants and their corresponding temperatures, just substitute the quantities into Arrhenius' equation to find the value of A.

Once you have the value of A, just substitute it into the Arrhenius equation, the corresponding temperature for K2, to find the value of K2.
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