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.

Corrected Temperature Problem - DrBob222, Saturday, February 2, 2013 at 3:37pm
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)
Since we are using at least four places here, I used -0.69315 for ln0.5 on the left and I used -0.003354 for 3957.180*(1/298.15) but neither of those change the answer significantly. The next line contains a larger error.
-0.6932 = 3,957.180(1/T2) - 13.2565
13.187T2 = 3,957.180
T2 = 300.08 K = 26.98 degrees C

-0.6932 = (3957.18/T2) - 13.2565
13.2565-0.6392 = 3957.18/T2
12.617 T2 = 3957.18
T2 = 313.6
If you use the -0.69315 for ln0.5 and you use the 0.003354 from my work above, T2 is about a degree higher or 314.58 K = about 41 C. After I post this response I'll look to see that it's ok; if not I'll post a P.S.

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