devarshi-dt-logo

Question:

What is the activation energy (KJ/mol) for a reaction if its rate constant doubles when the temperature is raised from 300K to 400K?

68.8

6.88

3.44

34.4

Solution:

Correct option is B. 6.88
log(K2/K1) = Ea/2.3R(1/T1 - 1/T2)
log(2/1) = Ea/2.3 * 8.31(1/300 - 1/400)
0.3 * 2.3 * 8.31 * 1000/120000 = Ea
Ea = 6.88 kJ