Chemical Forums
Chemistry Forums for Students => Physical Chemistry Forum => Topic started by: Schrödinger on December 07, 2012, 02:29:43 PM
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Hey guys!
I'm going through Transition state theory, and there's this part on vibrational motions of the transition state that I couldn't quite understand.
As far as I know :
Since a bond consists of only 2 atoms, it is linear. Hence, the total number of degrees of freedom for this system would be 3N = 3*2 = 6. Subtracting 3 for the net translation of the TS and 2 for the rotational motions possible, one should end up with 1 degree of freedom for the vibrational mode. i.e., this bind that links the 2 atoms of the TS has 1 vibrational degree of freedom. Yet, the text says 2. Why is this so?
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Can you reproduce the context / paragraph?
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Well, actually this is from my class notes. When I cross-checked with the text, I found a factor difference of 2. Anyway, I got to reading up a bit more and I found partition functions and statistical mechanics being used to get symmetry numbers, blah blah blah. What I wanted to know was if it was as simple as what I've written. i.e., being able to explain in terms of 3N-6 (or 3N-5, depending) rule.
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Well, actually this is from my class notes. When I cross-checked with the text, I found a factor difference of 2. Anyway, I got to reading up a bit more and I found partition functions and statistical mechanics being used to get symmetry numbers, blah blah blah. What I wanted to know was if it was as simple as what I've written. i.e., being able to explain in terms of 3N-6 (or 3N-5, depending) rule.
If you are asking about degrees of freedom yes. 3N-6 in general but 3N-5 for linear molecules.
Note that the TS will have one imaginary mode corresponding to the reaction coordinate at the saddle point.
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But aren't 3N-5 modes calculated inclusive of that? The only vibrational mode corresponds to the imaginary one, does it not?
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But aren't 3N-5 modes calculated inclusive of that? The only vibrational mode corresponds to the imaginary one, does it not?
Right. In a diatomic it does. There's no other modes besides the imaginary in the TS of a diatomic.
AFAIK.
In a poly-atomic molecule there will be the imaginary mode and several other real ones.
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Hmm... So, doesn't look like the simple 3N-5 is useful here.
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Hmm... So, doesn't look like the simple 3N-5 is useful here.
Why not? A diatomic that is not a TS will yet have 3N-5.
The TS just steals a vibration.
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No no, I was referring to the TS situation.
Any normal system obeys 3N-6. When it comes to TS, we should probably look for something better because there are restrictions being placed on the atoms/groups in the transition state.
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Subtracting 3 for the net translation of the TS and 2 for the rotational motions possible, one should end up with 1 degree of freedom for the vibrational mode. i.e., this bind that links the 2 atoms of the TS has 1 vibrational degree of freedom. Yet, the text says 2. Why is this so?
Which text is it that says 2 degrees of freedom? I am still curious. Something's not right here. I'm probably mis-understanding your question.
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That would be Chemical Kinetics by Keith Laidler. It doesn't say 2 per se. But that's what it implies
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That would be Chemical Kinetics by Keith Laidler. It doesn't say 2 per se. But that's what it implies
Online preview:
http://books.google.com/books?id=Dw-eLpmuFMIC&printsec=frontcover#v=onepage&q&f=false
Could you point the relevant extract?
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Around pages 95-96. Once again, I'd like to remind you that it doesn't say this per se. Partition functions have been used, but according to what our prof told us, I tried to correlate the 2 equations and ended with a factor of 2
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Yes, oddly enough I saw the same reference to two degrees of vibrational freedom here:
http://www.transtutors.com/physics-homework-help/thermal-physics/degree-of-freedom.aspx (http://www.transtutors.com/physics-homework-help/thermal-physics/degree-of-freedom.aspx)
No idea why they would say that, however.
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Around pages 95-96. Once again, I'd like to remind you that it doesn't say this per se. Partition functions have been used, but according to what our prof told us, I tried to correlate the 2 equations and ended with a factor of 2
I think it's your "correlation" step where the bug lies. But of course, that's only my hunch. I read through those pages and didn't find any (unless I missed it!) indication of the extra mode.
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@curiouscat : Yes, I'm pretty sure the correlation's at fault. Since my lecturer said so without giving the reason, I wanted to be able to understand it in an easier way. He did say statistical mechanics needs to be applied. But since he didn't teach us that part, I tried to make the best out of it with my pre-existing knowledge.
The main reason I got hooked to this explanation is because he specifically said "a bond has 2 vibrational degrees of freedom"