Posted by bobjahng2 on .
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.
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