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A reaction has a rate constant of 0.393 at 291 K and 1.41 at 345 K....

A reaction has a rate constant of 0.393 at 291 K and 1.41 at 345 K. Calculate the activation energy of this reaction in kJ/mol.

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Answer #1

K1   = 0.393

T1   = 291K

K2   = 1.41

T2   = 345K

logK2/K1   = Ea/2.303R [1/T1-1/T2]

log1.41/0.393 = Ea/2.303*8.314 [1/291 -1/345]

0.5548           = Ea/19.147 (0.00343-0.00289)

0.5548*19.147    = Ea *0.00054

Ea                      = 0.5548*19.147/0.00054   = 19671.77J/mole   = 19.672KJ/mole >>>>>answer

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