Question
Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K .
Calculate the activation energy of a reaction, whose rate constant doubles on raising the temperature from 300 K to 600 K .
A. 3.45 ~kJ / mol
( k_2 k_1 )= E_a 2.303 R [ 1 T_1 - 1 T_2 ] or, ( 2 k k )= E_a 2.303 8.3 [600-300] 300 600 E_a 3.45 ~kJ / mol
Related: Chemistry — Chemical Kinetics · All PYQ Banks