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Topic: How to get activation energy from the rate law?  (Read 5723 times)

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Offline Marija

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How to get activation energy from the rate law?
« on: July 16, 2009, 12:34:51 PM »
I've been doing some kinetics and a problem has occured to me. I've tried to solve it myself, but without any success. So I would be really thankful if one of you could help me with it.

It's a problem regarding the simplified mechanism of the pyrolysis of ethanal. These are the given reactions and data:
CH3CHO  :rarrow: CH3 (radical) + HCO (radical) rate constant: k1 activation energy: 358 kJ/mol
CH3 (rad.) + CH3CHO  :rarrow: CH4 + CH3CO (rad.)  k2 act. E: 8 kJ/mol
CH3CO (rad.)  :rarrow: CH3 (rad.) + CO   k3 act. E: 59 kJ/mol
HCO (rad.)  :rarrow: H (rad.) + CO   k4 act. E: 65 kJ/mol
H (rad.) + CH3CHO  :rarrow: H2 + CH3CHO (rad.)   k5 act. E: 15 kJ/mol
2CH3 (rad.)  :rarrow: C2H6   k6 act. E: 0 kJ/mol

You have to find 2 pathways for the dissociation of ethanal. This was easy:
1) CH3CHO  :rarrow: CH4 + CO
2) 2CH3CHO  :rarrow: C2H6 + 2CO + H2

Then, you have to find the order of reaction with respect to ethanal for each pathway. Using the steady state aproximation for intermediates and writing down the rate laws I got the right answer that the first reaction is 3/2 order with respect to ethanal and the second is 1 order with respect to ethanal. (The rate laws are attached!)
And finnally, the main problem - you have to calculate the activation energy for both reactions. Since I have the solutions I somehow managed to find the right answer, but I'm not sure about it/ I don't understand it fully. First equation: taking the natural logarithm of the rate constants which stand before ethanal in the rate law for methane gives (according to Arrhenius): Ea= Ea2 + 1/2 Ea1 - 1/2 Ea6 which then gives the right answer (187 kJ/mol).  For the second equation I took the natural logarithm of k1 (which stands before ethanal in the rate law for ethane) and (according to Arrhenius): Ea= Ea1 which giver the right answer (358 kJ/mol). However, these answers confuse me - why do we take only the constant which stands before ethanal in rate laws and why don't we take the sum of all constants (also those which are included in the rate laws for CO and H2)?
I would realy appreciate it if someone colud clear this up for me, because this is the first problem where I had to calculate activation E from the mechanism and I don't know how to do it (I just know how to get it from rate constants at different T by using Arrhenius).
Thanks in advance! ;D
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Offline Hunt

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Re: How to get activation energy from the rate law?
« Reply #1 on: July 24, 2009, 07:34:57 AM »
Just write the mechanism apply the SS and find the expression , for e.g.

Rate = k2.k1/24/ k1/26 [A] = kobs [A]   where [A] is the molarity of ethanal.

The differential form of the Arrhenius equation :

dLnkobs/ dT = Eobs / RT2

dLnk2/ dT + 1/2 dLnk4/ dT - 1/2 dLnk6/ dT=Eobs / RT2

Eobs = E2 + 1/2 E4 - 1/2 E6

I hope this helps





Offline Marija

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Re: How to get activation energy from the rate law?
« Reply #2 on: July 28, 2009, 12:48:07 PM »
Thanks. ;D
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