Question

The rate constant of a first-order reaction is 3.20 × 10−4 s−1 at 350.°C. If the activation energy is 135 kJ/mol, calculate the temperature at which its rate constant is 9.15 × 10−4 s−1.

Answer 1

use the arrhenious equation

ln K2/K1 = Ea / R [1/T1 - 1/T2]

where

K1 = 3.20 × 10^−4 s−1, T1 = 273+350 = 623K

K2 = 9.15 × 10^−4 s−1 T2 = we have to calculate

Ea activation energy given as 135 kJ/mol K convert in to joules =135000 J

R = constant = 8.314 J/mol K

plug in these values in above equation

ln(9.15 × 10^−4 s−1 / 3.20 × 10^−4 s−1) = 135000 / 8.314 [ 1 / 623 - 1/T2]

1.0506 = 16237.67 [ 0.001605 -1/T2]

[ 0.001605 -1/T2] = 1.0506 / 16237.67 = 0.0000647

0.001605 - 0.0000647 = 1 /T2

1/ T2 = 0.00154

T2 = 1/0.00154 = 649.35 kelvin

T2 = 649.35 - 273 = 376 ºC

T2 = 376 ºC