A certain reaction has an activation energy of 66.58 kJ/mol. At what Kelvin temperature will the reaction proceed 4.50 times faster than it did at 289 K

Answer 1

given:
Ea= 66.58 kj/mol=66,580 j/mol
T1=298.15k
K2=4.5 (K1)

find: T2

solution:
ln (k2/k1)= Ea/R (1/T1-1/T2)
ln (4.5 k1/k1) =Ea/R ( 1/298.15k-1/T2)

T2=316 K

Hope that helps 👍