A reaction has a rate constant of 0.364 min^-1 at 90.0 C degrees and a rate constant of 0.879 min ^-1 at 430 C degrees.

Question

Calculate the energy of activation of this reaction in kJ/mol

Answer 1

Arrhenius equation:

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

k1 = 0.364

k2 = 0.879

T1 = 90 C + 273 = 363K

T2 = 430 C + 273 = 703K

R = 8.314 J / Kmol

Ea = ?

ln (0.879 / 0.364) = - Ea / 8.314 (1/703 - 1/363)

ln 2.415 = - Ea / 8.314 x (0.001422 - 0.002755)

ln 2.415 = - Ea / 8.314 x -0.001333

0.8817 = 0.001333 Ea / 8.314

Ea = 5499 J / mol

Ea = 5.50 kJ / mol

(be sure to check the math)