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

For first order reaction :

   T1/2 =  0.693/K

 

Here

  T1 = 300K

So,

 K1  =  0.693 / T1/2

   =  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

 

 

  • 49

my ans 

  • -2

I need an answer

  • -11

 Please answer my question 

  • -2
What are you looking for?