Question

Calculate the half-life of benzoyl peroxide in benzene at 66 degrees C given that the rate coefficient Kd for dissociation of this initiator under these conditions is 8 x 10^-6 s^-1. What will be the change in initiator concentration after 1h at 66 degrees C

Answer 1

Solution :-

Half life t1/2 = 0.693 / Kd

                        = 0.693 / 8*10^-6 s-1

                         = 8.67*10^4 s

now lets find the change in the concentration after h

1 h = 3600 s

ln([A]t/[A]o) = - K*t

ln([A]t/1) = - 8*10^-6 s-1 * 3600 s

ln([A]t/1) = -0.0288

([A]t/1) = antiln [-0.0288]

([A]t/1) = 0.972

[A]t = 0.972* 1

[A]t = 0.972

so the change in the concnetration = 1-0.972 = 0.028

0.028 * 100 % = 2.8 %

so the change in the concnetration after 1 hr is 2.8 %