Arrhenius Question
posted by Amy .
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?
Given that the initial rate constant is 0.0160 s1 at an initial temperature of 25 degrees C , what would the rate constant be at a temperature of 130 degrees C for the same reaction described in Part A?

I managed to be given the formula: ln(k2/0.0160) = (32900/0.08206)(1/298/15  1/403.15) but when I answered it, I got the wrong answer.

Note that there are two questions here. The answer I gave you yesterday was for the second one. Did you report that answer for the first or the second one? Next, you simply copied what I gave you. Show your work, including the answer, and I will try to find the error.
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