How Does Changing the Medium Lower the Activation Barrier?

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SUMMARY

Changing the medium from water to ethanol results in a rate enhancement of 10^6 fold and a reduction of the activation barrier by approximately 9 kcal/mol. The relationship between reaction rate and activation energy can be analyzed using the Arrhenius equation, which connects the rate constant to the activation energy. The solvent's effect on the transition state (TS) stabilization compared to the reactants is crucial in understanding this phenomenon. Charge-dipole interactions in polar solvents significantly influence the activation energy for reactions.

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  • Understanding of the Arrhenius equation
  • Familiarity with reaction kinetics
  • Knowledge of transition state theory
  • Basic principles of solvent effects in chemical reactions
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  • Study the Arrhenius equation in detail
  • Research transition state theory and its implications
  • Explore the role of solvent polarity in reaction mechanisms
  • Examine experimental methods for measuring activation energy changes
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Chemists, chemical engineers, and students studying reaction kinetics and solvent effects on chemical reactions will benefit from this discussion.

greisen
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Hi,

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,
 
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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.
 
To add to what Gokul stated:

http://scitation.aip.org/getabs/servlet/GetabsServlet?prog=normal&id=JCPSA6000084000009004894000001&idtype=cvips&gifs=yes
 
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