# Activation energy

1. Feb 11, 2009

### Sir_Viking

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