Monday
April 21, 2014

Homework Help: Corrected Temperature Problem

Posted by Amy on Saturday, February 2, 2013 at 10:13am.

Q: The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?

My work:
ln(0.0160/0.032) = (32,900/8.314)(1/T2 - 1/298.15)
-0.6931 = 3,957.180(1/T2 - 0.00335)
-0.6932 = 3,957.180(1/T2) - 13.2565
13.187T2 = 3,957.180
T2 = 300.08 K = 26.98 degrees C

The answer above is not correct, and I would like to know where I went wrong. Previously, I divided 32.9 x 10-3 by 8.314. I corrected that, but I still get 3,957.180. In addition, I need this answer for a follow-up question.

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