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= -1,192.24 J/mol
My answer is incorrect, and I would like to know where I went wrong and what the correct answer is.
Temperature Problem - DrBob222, Saturday, February 2, 2013 at 3:05pm
I believe you have reversed the k1/T1 vs k2/T2. Also I think you are reporting too many significant figures
I would set it up as
ln(k2/k1) = (E/R)(1/T1 - 1/T2)
ln(0.235/1.23E-4) = (Ea/8.314)(1/301.15 - 1/352.15)
My answer is approximately 130,000
Answer This Question
Temperature - Q: A reaction rate has a constant of 1.23 x 10-4/s at 28 degrees C...