Asked by Jessica
The activation energy (E*) for a reaction is 195 KJ. If the k for this reaction is 3.46 x 10-5/min at 35oC, what will the temperature be in OF when k is 1.5 x 10-3/min?
Answers
Answered by
DrBob222
ln(k2/k1) = Ea(1/T1 - 1/T2)/R
ln(1.5E-3/3.46E-5) = 195000(1/308 - 1/T2)/8.314
Solve for T2.
ln(1.5E-3/3.46E-5) = 195000(1/308 - 1/T2)/8.314
Solve for T2.
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