if the rate constant for a reaction tripled when the temperature rises from 3.00×10^2 k to 3.10 ×10^2 k.what is the activation energy of the reaction?
rate constant k2 = 3 k1
temperature T1 = 300 K
T2 = 310 K
ln (k2 / k1) = Ea / R [1 / T1 - 1/ T2]
ln (3) = Ea / 8.314 x 10^-3 [1/ 300 - 1/310]
Ea = 84.9 4 kJ /mol
activation energy = 85.0 kJ /mol
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