Question

Consider the reaction A → products at 312 K. For this reaction it was observed that the first three half-lives were 11.3 h, 22.6 h, and 45.2 h when [A]o 1.343 M. How long will it take for [A] to decrease by 63 %? Time for [A] to decrease by 63 % (in hours)=

second order

Answer 1

half life for zero order reaction = [R]o/2k

half life for 1st order reaction = 0.693/k

half life for 2nd order reaction = 1/2k[R]o

where [R]o is initial concentration

For 0 order reaction, subsequent half life will keep decreasing. Half life will become half each time

For 1st order reaction, half life will be constant

For 2nd order reaction, subsequent half life will keep increasing. Also half life will double each time

Here half life is getting doubled each time.

So, it is 2nd order reaction.

Given:

Half life = 11.3 h

use relation between rate constant and half life of 2nd order reaction

k = 1/([A]o*half life)

= 1/(100.0*11.3)

= 8.85*10^-4 M-1.h-1

use integrated rate law for 2nd order reaction

1/[A] = 1/[A]o + k*t

1/(63) = 1/(100) + 8.85*10^-4*t

0.016 = 0.01 +8.85*10^-4*t

8.85*10^-4*t = 0.006

t = 6.64 h

Answer: 6.64 h