Question

A certain reaction has an activation energy of 48.97 kJ/mol. At what Kelvin temperature will the reaction proceed 6.50 times faster than it did at 295 K?

Answer 1

Given:

T1 = 295 oC

=(295+273)K

= 568 K

K2/K1 = 6.5

Ea = 48.97 KJ/mol

= 48970 J/mol

use:

ln(K2/K1) = (Ea/R)*(1/T1 - 1/T2)

ln(6.5) = (48970.0/8.314)*(1/568 - 1/T2)

1.8718 = 5890.065*(1/568 - 1/T2)

T2 = 693 K

Answer: 693 K