- #1

Sir_Viking

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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.

the following data has been obtained

Temperature (K)...K (M

273...... 5.60 x 10

279...... 11.8 x 10

285...... 24.5 x 10

291...... 48.8 x 10

297...... 100. x 10

That is the information I have been given.

ln((5.6 x 10

-.74533 = Ea/8.31 (-7.877 x 10

-6.1937 = Ea(-7.877 x 10

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

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.

**CH**_{3}I + C_{2}H_{5}ONa --> CH_{3}OC_{2}H_{5}+ NaIthe 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

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