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Offline NYM

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Arrhenius equation
« on: November 29, 2006, 03:44:36 PM »
Strictly speaking, the  Arrhenius equation can be applied only to gas reactions.

from
http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/eyr-e.htm

Does anybody know why?

Offline tamim83

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Re: Arrhenius equation
« Reply #1 on: November 30, 2006, 12:49:59 PM »
The Arrhenius equation stems from collision theory, which states that in order for a reaction to happen, the reactant molecules/atoms have to collide with one another in a cetain conformation.  This is implying, I suppose, the gas phase because it does not take things like solvent or hydrogen bonding into acount (i.e.-no entropy term is really involved).  The Eyring Equation does take entropy into account so it is better than the arrehenius equation for solution chemistry, which is the bulk of synthetic chemistry. 

Offline NYM

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Re: Arrhenius equation
« Reply #2 on: November 30, 2006, 01:23:26 PM »
The Arrhenius equation stems from collision theory, which states that in order for a reaction to happen, the reactant molecules/atoms have to collide with one another in a cetain conformation.  This is implying, I suppose, the gas phase because it does not take things like solvent or hydrogen bonding into acount (i.e.-no entropy term is really involved).  The Eyring Equation does take entropy into account so it is better than the arrehenius equation for solution chemistry, which is the bulk of synthetic chemistry. 

Ahhh... thanks:)

Just one more question:
On http://en.wikipedia.org/wiki/Eyring_equation it says that k = reaction rate, but it is the reaction rate constant, right?

Offline tamim83

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Re: Arrhenius equation
« Reply #3 on: December 01, 2006, 02:09:32 PM »
Yes it is, I guess someone needs to make an edit  ;D

Glad to help out

Offline NYM

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Re: Arrhenius equation
« Reply #4 on: December 01, 2006, 03:01:26 PM »

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/eyr-e.htm

Hmm, I don't quite understand equation (2).
1) Why is A+B <-> AB* a second order reaction - how to tell? I thought I was determined by experiments, not by theory.
And I suppose that AB* <-> C and AB* <-> A+B (the opposite way of the first one) is a first order reaction, where AB* is the only factor... but how do I know that? Couldn't the rate equation be something like v2 = k[C][AB*]?

Apart from that:

2) I understand that v1 = k1[A] and v-1 = k-1[AB*], and that they at equilibrium equals each other:
v1 = v-1
So far so good.
But how to combine the rate equation in (2)? I've never seen this before.


That's about it for now. I would really appreciate any help, because I don't really "get it". So help me see the light :)

EDIT:
Oh yeah, one more thing:
Equation (4), is it relatively easy to derive this from the statistical mechanics?
I've tried to find it on Google/wiki, but it is nowhere to be seen. I suspect a rather hard derivation, because I've looked for this Eyring equation, and they usually don't mention where (4) is from, other than statistical mechanics.

thanks!
« Last Edit: December 01, 2006, 04:15:26 PM by NYM »

Offline Yggdrasil

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Re: Arrhenius equation
« Reply #5 on: December 01, 2006, 04:23:08 PM »

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/eyr-e.htm

Hmm, I don't quite understand equation (2).
1) Why is A+B <-> AB* a second order reaction - how to tell? I thought I was determined by experiments, not by theory.
And I suppose that AB* <-> C and AB* <-> A+B (the opposite way of the first one) is a first order reaction, where AB* is the only factor... but how do I know that? Couldn't the rate equation be something like v2 = k[C][AB*]?

For elementary reactions (i.e. the mechanism of the reaction proceeds as the equation is written), the stoichiometry of the reaction determines the order of reaction.  Since we are assuming A + B -> AB* is an elementary reaction, the reaction rate will be k1[A].


Quote
2) I understand that v1 = k1[A] and v-1 = k-1[AB*], and that they at equilibrium equals each other:
v1 = v-1
So far so good.
But how to combine the rate equation in (2)? I've never seen this before.

d[AB*]/dt means the change in [AB*] over time.  Basically to get this you subtract the rate of decomposition of AB* (from AB* -> C and AB* -> A + B) from the rate of formation of AB* (from A+B -> AB*).

Quote
Equation (4), is it relatively easy to derive this from the statistical mechanics?
I've tried to find it on Google/wiki, but it is nowhere to be seen. I suspect a rather hard derivation, because I've looked for this Eyring equation, and they usually don't mention where (4) is from, other than statistical mechanics.

I don't remember the exact derivation off the top of my head, but it has to do with the lifetime of molecular vibrations.

Offline NYM

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Re: Arrhenius equation
« Reply #6 on: December 01, 2006, 04:37:59 PM »

http://www.uni-regensburg.de/Fakultaeten/nat_Fak_IV/Organische_Chemie/Didaktik/Keusch/eyr-e.htm

Hmm, I don't quite understand equation (2).
1) Why is A+B <-> AB* a second order reaction - how to tell? I thought I was determined by experiments, not by theory.
And I suppose that AB* <-> C and AB* <-> A+B (the opposite way of the first one) is a first order reaction, where AB* is the only factor... but how do I know that? Couldn't the rate equation be something like v2 = k[C][AB*]?

For elementary reactions (i.e. the mechanism of the reaction proceeds as the equation is written), the stoichiometry of the reaction determines the order of reaction.  Since we are assuming A + B -> AB* is an elementary reaction, the reaction rate will be k1[A].


Quote
2) I understand that v1 = k1[A] and v-1 = k-1[AB*], and that they at equilibrium equals each other:
v1 = v-1
So far so good.
But how to combine the rate equation in (2)? I've never seen this before.

d[AB*]/dt means the change in [AB*] over time.  Basically to get this you subtract the rate of decomposition of AB* (from AB* -> C and AB* -> A + B) from the rate of formation of AB* (from A+B -> AB*).

Quote
Equation (4), is it relatively easy to derive this from the statistical mechanics?
I've tried to find it on Google/wiki, but it is nowhere to be seen. I suspect a rather hard derivation, because I've looked for this Eyring equation, and they usually don't mention where (4) is from, other than statistical mechanics.

I don't remember the exact derivation off the top of my head, but it has to do with the lifetime of molecular vibrations.

That makes sense, thank you for clarifying all this.
I haven't dealt with statistical mechnics yet, but from what I've seen on Wiki, it is quite messy (especially when you havent' dealt with statistics and probability in general!) :p
I would give you another Scoobie snack, but

Sorry, you can't repeat a karma action without waiting 24 hours.

:)

Offline Yggdrasil

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Re: Arrhenius equation
« Reply #7 on: December 01, 2006, 04:44:56 PM »
Statistical mechanics may seem somewhat messy, but it is a really elegant and integral part of physical chemistry.  Whereas quantum mechanics focuses solely on the microscale and thermodynamics considers only the macroscale, statistical mechanics allows you to relate whats happening at the microscale to the macroscopic properties of the ensemble.  Very cool indeed.

Offline NYM

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Re: Arrhenius equation
« Reply #8 on: December 04, 2006, 04:01:58 AM »
Sorry to bother you again, but does anybody know how to derive equation (7)?
Or maybe a link?
I've looked on Google, but can't seem to find anything.

Offline Yggdrasil

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Re: Arrhenius equation
« Reply #9 on: December 04, 2006, 04:56:23 PM »
The overall rate of the reaction is the rate of formation of C (d[C]/dt).  This is just the forward rate of the second reaction in the mechanism:

d[C]/dt = k2[AB*]

From the previous equations, we have shown that the concentration of [AB*] = K*[A] where K* is the thermodynamic equilibrium constant for the first reaction in the mechanism.

Therefore, d[C]/dt = k2 K* [A]

From equation (4) we substitute the expression for k2 to get the equation in (6).

Now, consider the overal process A+B -> C.  The rate of this reaction would be written:

v = k[A]nm

(since A+B -> C is not an elementary reaction, we cannot know a priori the order of reaction with respect to A and B)

Since in (6), we derived that the overall rate is:

v = (kbT/h) K* [A]

Comparing the two equations shows that k = (kbT/h) K*, n = 1, and m = 1.

Offline NYM

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Re: Arrhenius equation
« Reply #10 on: December 05, 2006, 02:24:43 AM »
The overall rate of the reaction is the rate of formation of C (d[C]/dt).  This is just the forward rate of the second reaction in the mechanism:

d[C]/dt = k2[AB*]

From the previous equations, we have shown that the concentration of [AB*] = K*[A] where K* is the thermodynamic equilibrium constant for the first reaction in the mechanism.

Therefore, d[C]/dt = k2 K* [A]

From equation (4) we substitute the expression for k2 to get the equation in (6).

Now, consider the overal process A+B -> C.  The rate of this reaction would be written:

v = k[A]nm

(since A+B -> C is not an elementary reaction, we cannot know a priori the order of reaction with respect to A and B)

Since in (6), we derived that the overall rate is:

v = (kbT/h) K* [A]

Comparing the two equations shows that k = (kbT/h) K*, n = 1, and m = 1.


Arhhh sorry! I meant equation (4) - the one with statistical mechanics. They look very alike, and I was in a hurry, so I must have misread is. I actually did understand equation (7) (yeah, yeah, don't mock me!)...

If you don't wan't to, it's okay, you've helped med a lot already. Sorry for the confusion.


Offline NYM

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Re: Arrhenius equation
« Reply #11 on: December 11, 2006, 03:17:31 AM »

Arhhh sorry! I meant equation (4) - the one with statistical mechanics. They look very alike, and I was in a hurry, so I must have misread is. I actually did understand equation (7) (yeah, yeah, don't mock me!)...

If you don't wan't to, it's okay, you've helped med a lot already. Sorry for the confusion.

I found a book. It looks rather difficult.
I think I'll just ignore it in my assignment :)

Offline ajsammut

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Re: Arrhenius equation
« Reply #12 on: May 16, 2007, 04:38:31 AM »
What are the units of Ea in the Arrhenius equation? Are they kJ/mol or J/mol? 10ks for help

Offline Yggdrasil

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Re: Arrhenius equation
« Reply #13 on: May 16, 2007, 06:08:33 AM »
You can use any units of energy per mole.  You just have to make sure the units of RT match.  For example, if you use R = 8.314 J/(mol*K), then your Ea should be in J/mol.  If you use R = 0.08206 L*atm/(mol*K) then your Ea should be in units of L*atm.

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