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 :
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