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posted by  canuck911 on 9/26/2008 10:46:22 PM  |  status: Live  

Activation Energies --LIFESAVER--

Course Textbook Chapter Problem
General Chemistry N/A N/A
Question Details:
(i'm reposting this because i forgot to show my work)
A reaction rate increases by a factor of 2640 in the presence of a catalyst at 36°C.
The activation energy of the original, uncatalysed, pathway is 56 kJ mol-1.
What is the activation energy of the catalyzed pathway, all other factors being equal?
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what i did for this question

using the arrenhius equation.                        R = 8.3145 gas constant , Temperate also is converted to K

k2/k1 = Ae^(-Ea/RT) / Ae^(-Ea/RT)
A's cancel out (need to find the green)
2640 = e^(-Ea / (R x 309.15K)) / e^(-56000J/mol/ (R x 309.15))

this is where i get lost, so after simplifying the base I'm left with the Ea varible on top and the 2640 on the left. Am i suppose to take natural log of both sides and then bring it over to the left side, I'm really confused about this problem.
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posted by and1 on 9/28/2008 2:41:51 AM  |  status: Live
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canuck911's comment:
"thank you"
Response Details:
Don't use the arrhenius equation twice... rather, use a modified version of the arrhenius equation which accounts for 2 rate constants rather than just 1:



Using this equation should get you the right answer easily.
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posted by Xylene on 9/28/2008 3:02:04 AM  |  status: Live
Asker's Rating: Lifesaver   
canuck911's comment:
"thank you"
Response Details:
Arrhenius equation is
 K = Ae-Ea/RT
Where K is the rate constant, A is the frequency factor, Ea is the activation energy, R is the gas constant, T is the temperature in Kelvin scale
Apply log on both sides
logK = logA + (-Ea/RT)log10e
For two different  temperatures
logK1 = log A - 2.303Ea/RT1 ------1
log K2 = log A -2.303Ea/RT2-------2
equation 1 - equation 2
log(K1/K2) = 2.303Ea/R[1/T2 - 1/T1]
Use the above equation to calculate to your required data 
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