A certain reaction has an activation energy of 34.04 kJ/mol. At what Kelvin temperature will the reaction proceed 5.50 times faster than it did at 315 K?

Answer 1

Arrhenius equation: ln (k2/k1) = -Ea/R (1/T2 - 1/T1)

The rate of the reaction is directly related to the rate constant (1st order) so we can set k1 = 1 and then k2 = 5.50 to correspond to an increase in rate of 5.50 times.

k1 = 1

k2 = 5.50

T1 = 315K

T2 = ?

Ea = 34.04 kJ/mol

R = 8.314 J/Kmol = 0.008314 kJ/mol

Solving for T2, we have...

ln (5.50/1) = -34.04 / 0.008314 (1/T2 - 1/315)

1.70 = -4094 (1/T2 - 0.00317)

1.70 = -4094 /T2 + 12.98

11.28 = 4094 / T2

T2 = 363K