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Kinetic Theory and Activation Energy

  1. Nov 9, 2011 #1
    For a reaction to occur, we assume that two particles collide, have sufficient energy to react, and are oriented appropriately....or so I've read. The first one is straight forward. The second two I have questions about.

    For convenience, let's say A + B-> AB is what we're looking at. This reaction could have a recorded activation energy, presumably something determined experimentally at thermal equilibrium. Since both particles have a velocity distribution, a certain fraction of collisions will have sufficient energy to react. However, what if we think of A* + B -> AB? We have a bit of extra energy so I'm guessing we'd need a third body or we'd see a temperature increase, but let's ignore that. Is the activation energy still valid? I realize that the Ea tabulated is for an energy distribution that accounts for a certain fraction of A*, but let's say we have an overabundance of A* particles that are still at thermal equilibrium. Does a unit of energy in the form of electrical potential equal that "pound for pound" with that of heat/velocity? Is there any theory/method to figure out an appropriate Ea for the second process?

    In regards to the orientation, I assume this has something to do with the proper orbitals overlapping (can you tell I'm not a chemist yet?). So this last criterion is something that would always average out into the value of Ea if everything is isotropic, no? So in a weak electric field, this would change a bit even if no excited states were being generated? Also, does the influence of A and B on each other with no external field cause the particles/molecules to reorient themselves to encourage/discourage the reaction? Or is it just a strict probability of the molecules being at the right orientation when they collide?

    Thank you for any help. I hope I posted this in an ok location.
     
  2. jcsd
  3. Nov 11, 2011 #2
    I don't really know how to answer your second question, but ill take a stab at your first.

    I believe you can add as much energy to the system as you want, and it will just add up with the rest and give you a higher probability of the reaction occuring. I'm pretty sure that Ea is a constant, and doesn't depend on the energy in the system.
     
  4. Nov 17, 2011 #3
    elegysix,

    You would seem to agree with what I'm finding online. However, this doesn't seem to sit well with me. Let's say we are still considering A + B -> AB, where A and B are both in a ground state. Now let's say that they are at some temperature that gives them a translational energy of 2 doodads each. Now let's say A is now A*, which has a potential energy of 1 doodad (sorry for the silly name, trying to be abstract here). We conclude that we have a potential 3 doodads versus 2 doodads available in a collision, hence the reaction will proceed faster in the second case.

    However, let's say we have all A* travelling with an energy of only 1 doodad. Now the available energy in the collision is identical. So based on our discussion, we conclude that the systems in both examples react at the same rate, meaning electronic energy and translation add up apples for apples.

    This doesn't feel right. The electronic configuration would surely affect the reaction rate in a more complex way, no? And faster moving particles would also have less time to react. I'm not a chemist, but the longer you have an overlapping wave function, the better the odds of a reaction, no?
     
  5. Nov 23, 2011 #4
    I think I get what you're asking -
    in the rate equation from basic chem, the rate is proportional to the temperature, the rate constant k, and the concentrations.
    I suspect that the rate constant k depends on the electronic configuration.
    I don't know how/if k is derived somewhere, but you've got a starting place.
     
  6. Nov 23, 2011 #5
    Hmmm....so it sounds like you are saying R = k*exp(Ea/KT) and that k will be different for the two reactions, but presumably not Ea? Perhaps I can dig out some old chem books and see how each is derived. Thanks.
     
  7. Nov 23, 2011 #6
    I was referring to the other form: [itex]R =k*T*[A]^{m}*^{n} [/itex]

    but I'm sure it affects both
     
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