Question

A first order reaction is 50 % complete in 50 minutes at 300 K and the same reaction is again 50 % complete in 25 minutes at 350 K. Calculate activation energy of the reaction.

Answer 1

The activation energy of the reaction is 12.103 KJ-1 mol-1

Explanation:

The half-lives of the 2 reactions are t1/2 and t’1/2 mins, i.e., 50 and 25 mins, respectively, at Temperatures, T1 and T2, i.e., 300 and 350K

For the first order reaction, the rate of reaction is:

K = 0.693 / half life

Hence;

K1 = 0.693 / 50  

K2 = 0.693 / 25  

Now;

Log k1/ k2 = Ea / 2.303 x R x (1/T1 – 1/T2)

Value of R = 8.314 JK-1 mol -1

Log (0.693/25) / (0.693/50) = Ea / 2.303 x 8.314 x (1 /300 – 1/ 350)

Log 2 x 2.303 x 8.314 x 300 x 350/50 = Ea

12103 J mol-1

Thus, Ea = 12.103 KJ-1 mol -1

The activation energy of the reaction is 12.103 KJ-1 mol-1