At 300K a certain reaction is 50% complete is 20 minutes . At 350 K , the same reaction is 50% complete in 5 minutes . Calculate the activation energy

Solution:

For first order reaction :

   T_1/2 =  0.693/K

Here

  T₁ = 300K

So,

 K₁  =  0.693 / T_1/2

   =  0.693/ (20 × 60)

   = 0.693 /1200

   =5.775 × 10^-4  s^-1

Similarly at T₂  = 350K

 K₂ = 0.693/(5×60)

  = 0.693/300

   = 2.31 × 10^-3 s^-1

Now we know that . 

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

so

  ln[(2.31 × 10^-3)/( 5.775 × 10^-4) ]×R =  Ea (1/300 – 1/350)

   Ea  =  1.3962 × 210 × 8.314

   Ea  =  24.21 KJ/mol