Reaction Rates Problem

Question

A certain reaction has an activation energy of 58.63KJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 357 K?

Answer 1

Given Ea, T1 and 2 rates (rate 1 and rate 2 which is 6.50 x rate 1).

Asked to find T2

You can use the Arrhenius equation in the form of ln (k1/k2) = Ea/R (1/T2 - 1/T1) and we can use relative values for k1 and k2 which essentially represents rates as they are rate constants. So, we can set k1 = 1 and k2 = 6.50.

ln (1/6.5) = 58630 J/mol / 8.314 J/mol-K (1/T2 - 1/357)

-1.872 = 7052 (1/T2 - 0.00280)

-1.872 = 7052/T2 - 19.75

T2 = 394K (assuming my math is correct)