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[From: ] [author: ] [Date: 11-11-19] [Hit: ]
k2/k1 = ratio of rate constants = 3 (if reaction rate triples,ln3 = (Ea/8.1.0986 = (Ea/8.314)(0.1.......
what is the activation energy for a reaction if its rate constant is found to triple when the temperature is raised from 615 k to 629 k ?

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Use the modified Arrhenius equation:

ln(k2/k1) = (Ea/R)(1/T1 - 1/T2)

T1 = initial temp. = 615 K

T2 = final temp. = 629 K

R = 8.314 J/mol K

Ea = activation energy

k2/k1 = ratio of rate constants = 3 (if reaction rate triples, k triples)

ln3 = (Ea/8.314)(1/615 - 1/629)

1.0986 = (Ea/8.314)(0.0000362)

1.0986 = Ea x 0.000004353

1.0986/0.000004353 = Ea = 252,378 J
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