calculating half-life

Question

The decomposition of ethylene oxide (CH₂)₂O(g) → CH₄(g) + CO(g) is a first order reaction with a half-life of 58.0 min at 652 K. The activation energy of the reaction is 218 kJ/mol. Calculate the half-life at 617 K.

Answer 1

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

For a first order reaction, t_1/2 = 0.693 / k

solving for k1 @ 652K we have 58.0 min = 0.693 / k1 and k1 = 0.01195 min^-1

Substituting in the Arrhenius equation:

ln (k2/0.01195) = - 218 kJ/mol / 0.008314 kJ/Kmol (1/617 - 1/652)

ln (k2/0.01195) = -26,221 (0.001621 - 0.001534)

ln (k2/0.01195) = -26,221 x 0.000087)

ln (k2/0.01195) = -2.281

(k2/0.01195) = 0.1022

k2 = 0.00122 min^-1

t_1/2 = 0.693 / k2 = 0.693 / 0.00122 min^-1

t_1/2 = 568 minutes @ 617K

(be sure to check the math)