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Derivation of Eyring Equation

  1. Feb 2, 2007 #1
    Hi,

    In all derivations of the Eyring equation I've seen, all the steps are carried out in detail until the end, when suddenly the rate of the uni-molecular reaction (transition state -> products) is set to kT/h without justification. Why can all transition states be assumed to react at this rate?

    This seems particularly strange since from the viewpoint of a transition state there is no difference between 'reactant' and 'product', but the rate of the reaction (TS -> reactants) is not set equal to this value! What is happening here? What is the justification behind assuming equilibrium (reversibility) between reactants and TS but not TS and products?

    Also, do all reactions with a given reactant and product pass through the same transition state? TS is the saddle point of potential energy and thus the easiest path to take, but should not the actual reaction proceed through a statistical distribution of paths clustered around the one passing through the TS?

    This is actually a question of physical chemistry, but to me it seems to have to do more with quantum physics than chemistry, that's why I'm posting it here. Thanks for the help.

    Molu
     
  2. jcsd
  3. Feb 5, 2007 #2
    Can anyone help me? Thanks.

    Molu
     
  4. Feb 5, 2007 #3

    Gokul43201

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    Have you read the section in Atkins, Phys Chem? You will likely find it a little bit more satisfying (yet, on the whole, a little unsatisfying, due to some handwaving that you'll encounter here and there).

    The kT/h factor in the decay rate constant for AC --> product comes from a vibrational contribution to the partition function of the activated complex, namely [itex](1-exp(h \nu /kT))^{-1} [/itex], where [itex]\nu [/itex] is the frequency of this vibrational mode. One way to help visualize this vibration is to notice that in many simple bimolecular (gas phase) reactions, the PE vs reaction co-ordinate plot actually has a shallow trough at the TS, which can locally be approximated to a harmonic oscillator potential. This weak elastic interaction between the reactant atoms in the TS gives rise to a very low frequency mode. So, in the limit of [itex]h\nu << kT[/itex], the partition function term from this mode approximates to [itex]kT/h \nu [/itex].

    The role of the partition functions is in describing the "equilibrium" constant for [itex]reactants \leftrightharpoons AC [/itex]. And the reason the [itex]\nu[/itex] does not appear in the final expression for the rate is another argument that this rate (the rate at which the AC decays into the product) is itself proportional to the frequency [itex]\nu[/itex] of the vibrational mode along the reaction co-ordinate.

    I'm not doing a very good job of this - I'm making it more hand-wavy than it already is. So, it would be best for you to look into Atkins.

    As for your second question, if I understand it correctly, I believe you are right, since I expect the graph of PE vs reaction co-ordinate describes the ensemble expectation value for the PE of the system at each stage.
     
    Last edited: Feb 5, 2007
  5. Feb 6, 2007 #4
    I find Atkins to be a remarkably dry book and a rather ill-considered investment, but I'll check it. But tell me this, why is it justifiable to assume equilibrium between reactants and TS but not TS and products? Thanks.

    Molu
     
  6. Feb 7, 2007 #5

    Gokul43201

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    I'm not sure what you mean by 'dry', but I've found it attempts to justify more chemistry from first principles (QM, stat mech, group theoretic/symmetry requirements) than any other book I've come across. But this may simply be from my lack of exposure to good books. Do you have a recommendation for an alternative phys chem book?

    The copy of Atkins I have (6th ed, paperback, OUP) was bought for Rs. 750 (but this was nearly 10 years ago). I would never consider that a bad investment.

    Here it is, at the same price: http://www.firstandsecond.com/store/books/info/bookinfo.asp?txtSearch=1557616



    I guess for the same reason it's justifiable to consider the decay of the activated complex into the products and not the reactants (i.e., it works)? If I'd seen something even a little more insightful than that, I'd have mentioned it. Really, despite the reasonable coverage of the breadth of the theory, there's so much hand-waving in the arguments here that the whole thing sits very uncomfortably with me.
     
  7. Feb 7, 2007 #6
    By dry I meant that it's almost encyclopedic, it lacks the voice of a teacher I expect in a textbook. But that's a very personal opinion. In any case, it's not suited to my academic needs.

    I don't think the back-decay of the TS is not considered, otherwise it could not equilibriate with the reactants. It's the back-decay of the products that seems to be ignored. Isn't the labelling of reactant and product arbitrary? How can the theory distinguish between the two so fundamentally?

    Molu
     
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