chemistry

The given reaction,
cyclopropane propene (isomerization),
follows first-order kinetics. The half-live was found to be 4.57 min at 796 K and rate constant was found to be 0.0328 s-1 at 850 K. Calculate the activation energy (in kJ/mol) for this reaction. Round your answer to 3 significant figures.

asked by bobjahng2
  1. No guarantees, but here goes:

    rate constant = C*e^(-Ea/RT)

    A half life of 4.57 min = 274 s corresponds to a rate constant of ln2/274 = 2.53*10^-3 s^-1

    Ea is the activation energy

    0.0328 = C*e^(-Ea/850R)
    0.00253 = C*e^(-Ea/796R)

    12.96 = e^[-(Ea/R)(1/850 - 1/796)]

    Solve for Ea. R = 8.317 J/mole*K

    2.562 = (-Ea/R)*(-7.98*10^-5 K^-1)
    Ea/R = 3.21*10^4 K
    Ea= 2.67*10^5 J/mole
    = 267 kJ/mole

    posted by drwls

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