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Topic: Vibrational degrees of freedom of a bond  (Read 7277 times)

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Offline Schrödinger

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Vibrational degrees of freedom of a bond
« on: December 07, 2012, 02:29:43 PM »
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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Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #1 on: December 07, 2012, 02:44:03 PM »
Can you reproduce the context / paragraph?

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #2 on: December 07, 2012, 02:47:45 PM »
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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Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #3 on: December 07, 2012, 02:51:25 PM »
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.

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #4 on: December 07, 2012, 02:56:12 PM »
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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Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #5 on: December 07, 2012, 03:01:23 PM »
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.

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #6 on: December 07, 2012, 03:03:06 PM »
Hmm... So, doesn't look like the simple 3N-5 is useful here.
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Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #7 on: December 07, 2012, 03:05:53 PM »
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.

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #8 on: December 07, 2012, 03:09:36 PM »
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.
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
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Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #9 on: December 08, 2012, 02:39:28 AM »
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.

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #10 on: December 08, 2012, 07:39:12 AM »
That would be Chemical Kinetics by Keith Laidler. It doesn't say 2 per se. But that's what it implies
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #11 on: December 08, 2012, 08:05:49 AM »
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?

Offline Schrödinger

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Re: Vibrational degrees of freedom of a bond
« Reply #12 on: December 08, 2012, 10:54:23 AM »
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
"Destiny is not a matter of chance; but a matter of choice. It is not a thing to be waited for; it is a thing to be achieved."
- William Jennings Bryan

Offline fledarmus

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Re: Vibrational degrees of freedom of a bond
« Reply #13 on: December 08, 2012, 03:53:38 PM »
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

No idea why they would say that, however.

Offline curiouscat

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Re: Vibrational degrees of freedom of a bond
« Reply #14 on: December 09, 2012, 09:08:32 AM »
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.
« Last Edit: December 09, 2012, 09:35:11 AM by curiouscat »

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