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Homework Help: Lowering of activation barrier

  1. Jul 18, 2007 #1

    It is stated that changing the medium - water to ethanol f.ex - the experimentally measured rate enhancement is 10^6 fold which leads to a lowering of the activation barrier of \approx 9 kcal/mol.
    I don't quite understand which parameters I am missing in order to calculate that? Any help on which equations could be used to deduced this would be very much appreciated.

    Thanks in advance

    best regards,
  2. jcsd
  3. Jul 18, 2007 #2


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    What exactly do you want to calculate - just the relationship between reaction rate and activation energy, or the relationship between solvent and activation energy?

    I'm not sure how easily you can calculate the change in activation energy (given a change in solvent), but if you can do that, you can then determine the effect on the rate constant through the Arrhenius equation (and then, the effect on the reaction rate from the rate equation). The latter two equations should be covered in any standard textbook.

    The change in activation comes from the difference in stabilization of the reactants and the transition state, by the solvent. For instance, a TS with a greatly increased charge density (compared to the reactant) will experience much larger charge-dipole interactions with a polar medium. A polar solvent thus stabilizes the TS more than the reactant and, as a result, increases the activation energy for the forward reaction.
  4. Jul 18, 2007 #3


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    To add to what Gokul stated:

    http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=JCPSA6000084000009004894000001&idtype=cvips&gifs=yes [Broken]
    Last edited by a moderator: May 3, 2017
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