Hi! I am having a lot of trouble with these two problems:
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?
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.
Q:Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C and 0.235 at 79 degrees C. Determine the activation barrier for the reaction.
A: My work thus far:
ln(1.23 x 10-4/0.235)=(Ea/8.314)(1/301.15-1/352.15)
Ea = 13,0553.145
2,723 J/mol = 0.002723 kJ/mol
title changed from: Can someone tell me what I am doing wrong with these chemistry problems? to more descriptive