A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction.

(Given: log 2 = 0.3010, log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

Rate constant for first order reaction is given by:

Where k is the rate constant, t is the time, R_o is the initial concentration and R is the final condition.

At 300K,

=0.058 × log 2

=0.058 × 0.301

=0.017

At 320K,

=0.11× log 2

=0.11 × 0.3010=0.034

Now,

log = ()

Where E_a is the activation energy

K₁, T₁ is the temperature of 300 K

K₂, T₂ is the temperature of 320 K

Putting the value in the above equation,

K

E_a=28,805.7 j mol^-1