[Schrödinger]

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?

[curiouscat]

Can you reproduce the context / paragraph?

[Schrödinger]

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.

[curiouscat]

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.

[Schrödinger]

But aren't 3N-5 modes calculated inclusive of that? The only vibrational mode corresponds to the imaginary one, does it not?

[curiouscat]

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.

[Schrödinger]

Hmm... So, doesn't look like the simple 3N-5 is useful here.

[curiouscat]

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.

[Schrödinger]

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.

[curiouscat]

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.

[Schrödinger]

That would be Chemical Kinetics by Keith Laidler. It doesn't say 2 per se. But that's what it implies

[curiouscat]

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?

[Schrödinger]

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

[fledarmus]

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.

[curiouscat]

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.