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Activation energy

  1. Feb 11, 2009 #1
    First off, an introduction.
    I am a AP Chemistry student (though not a very good one, to be honest) in Texas.
    My homework has an equation for activation energy that I have to solve, and I'm having a bit of trouble with it.

    CH3I + C2H5ONa --> CH3OC2H5+ NaI
    the following data has been obtained
    Temperature (K)......K (M-1 s-1)

    273....................... 5.60 x 10-5
    279....................... 11.8 x 10-5
    285....................... 24.5 x 10-5
    291....................... 48.8 x 10-5
    297....................... 100. x 10-5

    That is the information I have been given.


    ln((5.6 x 10-5)/(11.8 x 10-5)) = Ea/8.31 (1/279 - 1/273)


    -.74533 = Ea/8.31 (-7.877 x 10-5)
    -6.1937 = Ea(-7.877 x 10-5)

    Ea = 78.6 KJ

    That's what I got, but I'm doubtful about how well I did it.


    Anyway, I guess I'm just asking someone to check my work.
    [EDIT] Whoops, I must've put the numbers in the calculator wrong, it was 78.6 KJ, not 7.86 of anything
     
    Last edited: Feb 11, 2009
  2. jcsd
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