Posted by Amy on Friday, February 1, 2013 at 6:50pm.
Q: The activation energy of a certain reaction is 32.9 kJ/mol At 25 degrees C the rate constant is 0.0160 units s1 units. At what temperature in degrees Celsius would this reaction go twice as fast?
My work:
ln(0.0160/0.032) = (32.9 x 103/8.314)(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. In addition, I need this answer for a followup question.

Temperature and Rate  Devron, Friday, February 1, 2013 at 7:07pm
Where did you get 0.032 to plug in for K2?

Temperature and Rate  DrBob222, Friday, February 1, 2013 at 9:31pm
Thank you for posting your work.
I didn't work the problem but it appears to me that the error is in Ea. That's 32.9 kJ/mol which is 32,900 J/mol and that's the number that goes in for Ea (not 32.9E3). 0.032 is twice 0.0160 and that's ok. By the way, since the rate doubles you know the temperature is higher by approximately 10 C. Do you remember that from class; i.e., the rate (approximately) doubles for every 10 C rise in temperature.

Temperature and Rate  Amy, Saturday, February 2, 2013 at 2:05am
DrBob222, when you divide 32,900 by 8.314, you still get 3,957.180.
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