Posted by Amy on .
Q: A reaction rate has a constant of 1.23 x 104/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 104/0.235)=(Ea/8.314)(1/301.151/352.15)
7.551=(Ea/8.314)(0.0048090535)
7.551=(Ea/8.314)(0.0048090535))
15,702.808=(Ea/8.314)
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,
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(0.235/1.23E4) = (Ea/8.314)(1/301.15  1/352.15)
My answer is approximately 130,000