# Why does transition state loses its ability to vibrate? (Transition State Theory)

In transition state theory, (according to my textbook) it is assumed that the the transition state loses its structure, and the ability to vibrate and rotate. If the transition state cannot vibrate, then my questions are

• how does the previous bonds break and the newer bonds form?

• Isn't the instablity and high potential energy of transition state due to the extra vibration of the atoms in the molecules?

my book states:

"The transition state does not represent a real molecule, it is impossible to isolate a transition state. However it is assumed to possess properties common to real molecule, such as molecular weight, intermediate distances, a definite enthalpy, definite composition but loses its structure, and the ability to rotate and vibrate. The transition state may either return to the initial reactant or it proceeds to form products."

• I am suspicious that your book makes this claim (exact quote would be helpful here). It would be more accurate to say that the transition state is a structure where the vibrations and rotations of the species will, in one direction, lead to product and, in the other direction, lead to the reactant. – Zhe Jan 25 at 13:56
• Well the "transition state" is, by definition, a state where all the bonds are at their highest energy level. It implies that if any bond changes even a little bit, it would no longer be the "highest energy level". – SteffX Jan 25 at 13:59
• Thx for the reply! I will quote the exact statement from my book. – HoneyChem Jan 25 at 13:59
• One way to think about it is that the lifetime of the transition state is shorter than the fastest bond vibration or rotation. In order to break a bond, that bond is stretched out to beyond its maximum length. If the stretching mode has time to vibrate back to a shorter bond length, the energy goes back down. – Andrew Jan 25 at 14:34

Only some of the statements you quote are true, certainly a transition state cannot be isolated since it lasts for less than a picosecond. In fact there is hardly any direct measurement of transition states, such as a spectrum or time profile or any structure. It is not that they are too fleeting to measure but that they occur very infrequently in any reaction so their population is always vanishingly small. This is a real experimental challenge.

What we do know is that the transition state is just like any other molecule with bonds vibrating/rotating normally except for the bond(s) that are involved in reaction. They stretch and break then form new ones as the transition state is crossed. This pathway is the lowest point on the trajectory from reactants to products and is usually drawn at the top of the reaction energy profile as a 'saddle point'. (When at the saddle point looking at right angles to the reaction path the energy increases, looking forwards or backwards along the reaction path the energy decreases). As the new bond has lots of energy it is initially vibrationally hot and rapidly loses energy either to solvent or into other vibrational modes in the product molecule or both.

In a complex molecule you can imagine the reaction starting as the species approach one another. The energy starts to increase as the atoms that are going to react approach one another, but collisions with solvent can randomly knock them off course, or further on course for a while, or vibrations in the molecule may randomly add or remove some energy and so hinder or accelerate the reaction. One can then imagine the species behaving somewhat as if they are diffusing along the reaction path, sometimes going forwards but sometimes being hindered. At any point, including the saddle point, they can diffuse around a bit and either return to reactants or proceed to products.

Even in simple atom-diatom reactions in the gas phase, such as $$\ce{H + Cl_2}$$ vibrations in the diatomic molecule can inhibit reaction even when there is enough total energy to react; sometimes the energy is just in the wrong place and cannot aid the reaction. This can be the case because the diatomic is both rotating and vibrating and reaction will depend on the angle which it has to the line of approach ( atom to diatom) at the separation at which reaction starts.

• Thnx for helping – HoneyChem Jan 26 at 1:00

I think there is some confusion between removing a single degree of freedom (the vibration that changes the length of the bond that gets broken) and - in the text quoted in the question - freezing all degrees of vibrational freedom. I also don't understand the part about the transition state having no structure at all.

According to this paper's introduction, the transition state looks like the reactant except for a set of bond lengths of the vibrational mode that is frozen:

Transition state theory is a statistical mechanical theory of chemical reaction rates that may be derived from two fundamental assumptions. First one defines a reaction coordinate $$s$$ leading from reactants (negative $$s$$) to products (positive $$s$$) and a (generalized) transition state as a system partway between reactants and products with a fixed value for $$s$$ (thus the transition state is a system with one less degree of freedom than the reactants).

The hand-waving reason for doing this is that the degrees of freedom kept affect the entropy of the transition state (and can treated like a system at equilibrium), while the degree of freedom associated with making and breaking bonds has to be treated in a special way (it has to be frozen to be at the maximal energy). So if you think of the energy landscape around the transition state as a saddle point, the removed degree of freedom is along the coordinate that shows a maximum, and the remaining degrees of freedom are at a minimum.

Just in case you are wondering about the second assumption:

The second assumption is that any system passing through the transition state does so only once (before the next collision or before it is stabilized or thermalized as reactant or product). These assumptions may be called the local-equilibrium and no-recrossing assumptions.

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To put it simply, transition states do not exist. They are artificial constructs. Thus, any attempts to work with them outside transition state theory are technically incorrect.

Molecules are quantum systems. We do not have a clear theory for interaction of quantum systems, only for isolated quantum system evolving in time, and even that is quite wonky. Thus any models about molecule interactions are very, very up in the air. It is actually very surprising that ANY of our models for chemical reactions work, since most of our models are quasi-classical in some sense.

Isn't the instablity and high potential energy of transition state due to the extra vibration of the atoms in the molecules?

No. Vibrations contribute very little to energy of any molecule. In fact, trying to ascribe high energy of transition state to something specific to molecules like vibration or strain energy is kinda questionable.

TS has high energy because it is chosen (defined) this way. It is, by definition, the highest point of the reaction path. The causes of the form of the reaction path are not considered in transition state theory. Actually, there are reactions that do not have a barrier and thus a transition state per se.

• Your last two paragraphs make it sound like the energy of a transition state is arbitrary, defined as higher than reactants/products. If that is what you meant to say, that isn't correct. If it isn't what you meant, can you please rephrase to clarify? – buckminst Jan 26 at 4:11
• @buckminst to some extent. I rephrased a bit, is it more understandable now? – permeakra Jan 26 at 9:54