The decomposition of ethylene oxide is a fist order reaction with a half life of 58.0 min at 652k. The activation energy of the reaction is 218 kj/mol. Calculate the half life at 623k

Answer 1

For a 1st order reaction, t_1/2 = 0.693 / k

58.0 min = 0.693 / k

k = 0.0119 min^-1 @ 652K

Now we can use the Arrhenius equation to find k @ 623K

ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

k1 = 0.0119

k2 = ?

T1 = 652K

T2 = 623K

Ea = 218 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/Kmol (convert to keep units consistent with those of Ea)

ln (k2/0.0119) = -(218 / 0.008314) (1/623 - 1/652)

ln (k2/0.0119) = -26,221 (0.001605 - 0.001534)

ln (k2/0.0119) = -26,221 x 0.000071

ln (k2/0.0119) = -1.86

ln k2 - ln 0.0119 = -1.86

ln k2 + 4.43 = -1.86

ln k2 = -6.29

k2 = 0.00185 @ 623K

t_1/2 = 0.693 / k

t_1/2 = 0.693 / 0.00185

t_1/2 = 375 min @ 623K

NOTE: Be sure to check ALL of the math. This is the correct approach to the problem, but math errors can creep in.