Question

A certain reaction has an activation energy of 59.59 kJ/mol. At what Kelvin temperature will the reaction proceed 3.50 times faster than it did at 365 K?

Answer 1

use the Arrhenius two-point equation:

\(\Huge ln(\dfrac{k_1}{k_2})=\dfrac{E_a}{R}(\dfrac{1}{T_2}-\dfrac{1}{T_1})\)

k is the rate constant, which determines the rate, you want \(k_1\)=3.5, and \(k_2\)=1.