posted by Amy on .
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?
Given that the initial rate constant is 0.0160 s-1 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.