J.R. S. answered 06/23/21
Ph.D. University Professor with 10+ years Tutoring Experience
For this, you can use the Arrhenius equation:
ln (k2/k1) = -Ea/R (1/T2- 1/T1)
k1 =1.35x102 s-1
k2 = ?
Ea = 55.5 kJ/mol
R = 8.314 J/Kmol = 0.008314 kJ/Kmol
T1 = 25 + 273 = 298K
T2 = 35 + 273 = 308K
ln (k2/1.35x102) = -55.5 / 0.008314 (1/308 - 1/298)
ln (k2/1.35x102) = -6675 (0.00325 - 0.00336)
ln (k2/1.35x102) = 0.734
k2 (k@35ºC) = 2.81x102 s-1 (check the math to be sure)