Question

The decomposition of hydrogen peroxide (H2O2) is a first order reaction with a rate constant 1.8×10-5 s -1 at 20°C.

(a) What is the half life (in hours) for the reaction at 20°C?

(b) What is the molarity of H2O2 after four half lives if the initial concentration is 0.30 M?

(c) How many hours will it take for the concentration to drop to 25% of its initial value?

*Help please!!!*

Answer 1

a)

Given:

k = 1.8*10^-5 s-1

use relation between rate constant and half life of 1st order reaction

t1/2 = (ln 2) / k

= 0.693/(k)

= 0.693/(1.8*10^-5)

= 3.85*10^4 s

= 3.85*10^4 / 3600 hr

= 10.7 hr

Answer: 10.7 hr

b)

after each half life concentration becomes half

after n half life,

concentration = initial concentration * (1/2)^n

= 0.30*(1/2)^4

= 0.30/16

= 0.01875 M

Answer: 0.01875 M

c)

let initial concentration = 100

final concentration = 25 % of initial concentration = 25 % of 100 = 25

use:

concentration = initial concentration * (1/2)^n

25 = 100 * (1/2)^n

1/4 = (1/2)^n

n = 2

It will take 2 half lifes

time =2*10.7 hour = 21.4 hour

Answer: 21.4 hour