The decomposition of hydrogen peroxide in dilute sodium hydroxide solution is de
ID: 506983 • Letter: T
Question
The decomposition of hydrogen peroxide in dilute sodium hydroxide solution is described by the equation: 2H2O2 (aq) 2H2O(l) + O2 (g) The reaction is first order with respect to H2O2 , and the rate constant for the consumption of H2O2 at 20 degree C is 1.8x10-5 s -1 .
a) What is the half life (in hours) of the reaction at 20 degree C?
b) What is the molarity of H2O2 after 4 half-lives if the initial concentration of H2O2 is 0.30 M?
c) How many hours will it take for the concentration to drop to 25% of its initial value?
Explanation / Answer
a)
k = 1.8*10^-5 s-1
use formula for half life for 1st order reaction,
half life = 0.693 / k
= 0.693 / (1.8*10^-5)
= 38500 s
= 38500/3600 hr
= 10.7 hour
Answer: 10.7 hours
b)
after n half life,
[R] = [Ro]*(1/2)^n
here n=4
[R] = [Ro]*(1/2)^4
[R] = [Ro] / 16
= 0.30/16
= 0.019 M
Answer: 0.019 M
c)
25 % means
[R] = [Ro]/4
which is 2 half life
so,
time taken = 2* half life
= 2*10.7 hours
= 21.4 hours
Answer: 21.4 hours
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