Question

A certain reaction has an activation energy of 67.72 kJ/mol. At what Kelvin temperature will the reaction proceed 4.00 times faster than it did at 353 K?

Answer 1

T1 = 363 K
T2 = ??
R = 8.3145 J/mol K
K1 = 1
K2 = 4
Ea = 67.72 x 1000 = 67720 J/mol

T2 =1 / (1/363) – (8.3145 / 67720) x ln(4 / 1)

T2 = 1/0.00258461502
T2 = 386.90 K