Homework Help: Temperature and Rate

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 s-1 units. At what temperature in degrees Celsius would this reaction go twice as fast?

My work:
ln(0.0160/0.032) = (32.9 x 10-3/8.314)(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 follow-up 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 32.9E-3). 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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