Consider a reaction with an activation energy of 11090.1 J/mol run at 298.15 K and at...
Consider a reaction with an activation energy of 11090.1 J/mol run at 298.15 K and at 2228 K. How many times larger is the rate constant, k , for the reaction at 2228 K than at 298 K?
Homework Answers
Activation energy = 11090.1 J/mol
Temperature T1 = 298.15
T2 = 2228 K
rate constant k1 . k2
ln (k2 / k1) = Ea / R [1 / T1 - 1 / T2]
ln (k2 / k1) = 11090.1 / 8.314 [1/298.15 - 1/2228]
k2 / k1 = 48.2
number of times = 48 times
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