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
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