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So often I see total energies of transition states corrected for zero-point vibrational energy which always confuses me. Zero point energy is the lowest energy that a ground state minimum energy configuration can have at 0K in a vacuum. Transition states are indeed not minima, can never be isolated, have one imaginary mode of vibration, etc.

So why do people constantly factor in zero-point vibrational energy to the total energy for a transition state? Obviously this is done to have 'comparable' minima and TS energetics (if you are applying ZPE to the minima) but is this even physical? If zero-point correcting the TS, should the zero-point energy of the imaginary mode be included in the summation or excluded (or should both quantities be provided)?

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  • $\begingroup$ Transition states are much more real than you think. Read chemistry.stackexchange.com/a/16800/186. They can not be isolated, but they can be observed. The 1999 Nobel Prize in Chemistry was awarded for such observations. $\endgroup$
    – Wildcat
    Sep 29, 2014 at 14:08
  • $\begingroup$ @Wildcat I've omitted the word "observed". $\endgroup$ Sep 29, 2014 at 14:13
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    $\begingroup$ So, basically, the answer is, yes, you simply omit the normal mode with an imaginary frequency while calculating ZPE for TS. The mode with imaginary frequency is treated either as a very loose vibration with frequency that tends to zero or even as a translation. And in both cases it won't contribute to zero point vibrational energy. $\endgroup$
    – Wildcat
    Sep 29, 2014 at 14:37
  • $\begingroup$ @Wildcat Right but including ZPVE with a TS goes against its very definition. I think omitting the imaginary mode is mathematically convenient and it just 'sounds good' but in the end applying ZPVE to a TS is completely unphysical. $\endgroup$ Sep 30, 2014 at 11:32
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    $\begingroup$ Which quantities? Reaction rates, first of all. And here I can tell you only one thing: for unimolecular reaction rates RRKM theory in which contributions from normal-mode vibrations (and rotations) are included in the model works better than its predecessor, RRK theory. $\endgroup$
    – Wildcat
    Sep 30, 2014 at 12:00

1 Answer 1

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At the transition state, you have exactly one vibration mode with imaginary frequency (which corresponds locally to the reaction coordinate), and $3N-1$ other “normal” vibration modes. In the transition state theory, you separate out the reaction coordinate (consider it infinitely slow compared to the other movements). Thus, when you consider the energy of the transition state, you have to include the contributions of ground state energy of all the other vibration modes.

The more rigorous way of looking at it is to use statistical physics, and to consider the partition function of the transition state. See here or there for such treatments.

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  • $\begingroup$ So what you're saying is, we just throw out vibrational modes until we get a 'ground state' molecule? (I say modes because this concept can be applied to higher order saddle points) $\endgroup$ Oct 23, 2013 at 12:06
  • $\begingroup$ @LordStryker you calculate the ZPE by summing over all modes that have an imaginary frequency (i.e. correspond to negative eigenvalues of the Hessian matrix). For a true transition state, this should be all modes but one. $\endgroup$
    – F'x
    Oct 23, 2013 at 12:41
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    $\begingroup$ The only thing I'm still unclear about is what consequences are there to throwing away vibrational modes? Now you'd be comparing the relative energy of a TS with X-1 number of modes to say an intermediate with X number of modes. $\endgroup$ Oct 23, 2013 at 12:50
  • $\begingroup$ @LordStryker that's normal. If you want to know why, read the statistical physics treatment of transition state theory, and you'll see why! $\endgroup$
    – F'x
    Oct 23, 2013 at 12:56
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    $\begingroup$ @LordStryker TST has much worse flaws than ignoring the imaginary mode. For saddle points of higher order, TST is not even applicable anymore.// Rotational and translational modes are also projected out, so you end up with $3N-7$ or $3N-6$ (for linear molecules). $\endgroup$ Sep 30, 2014 at 11:29

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