Question

18. A certain first order reaction has a half-life of 0.35 hours. How long will it take for 12.5 % of the reactant to react?

19. For a certain first order reaction, it was found that after 35 seconds; only 15 % of the reactants had reacted. What is the numerical value of the rate constant?

20. A certain first-order reaction has a half-life of 25 seconds. What percentage of the reactant will remain after 125 seconds?

21. A certain first-order reaction has a half-life of 12 minutes. What is the numerical value of the rate constant for the reaction?

Answer 1

1)
Given:
Half life = 0.35 hr
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(0.35)
= 1.98 hr-1

we have:
[A]o = 100 (Initial concentration be 100)
12 % has reacted.
So, remaining is,
[A] = 87.5
k = 1.98 hr-1

use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(87.5) = ln(1*10^2) - 1.98*t
4.472 = 4.605 - 1.98*t
1.98*t = 0.1335
t = 6.744*10^-2 hr
Answer: 6.74*10^-2 hr

2)
we have:
[A]o = 100
[A] = 85
t = 35.0 s


use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln(85) = ln(1*10^2) - k*35
4.443 = 4.605 - k*35
k*35 = 0.1625
k = 4.643*10^-3 s-1
Answer: 4.64*10^-3 s-1

3)
Given:
Half life = 25 s
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(25)
= 2.772*10^-2 s-1

Given:
[A]o = 100
t = 125.0 s
k = 2.772*10^-2 s-1

use integrated rate law for 1st order reaction
ln[A] = ln[A]o - k*t
ln[A] = ln(1*10^2) - 2.772*10^-2*1.25*10^2
ln[A] = 4.605 - 2.772*10^-2*1.25*10^2
ln[A] = 1.14
[A] = e^(1.14)
[A] = 3.13 M

3.13 of 100 remains which is 3.13 %
Answer: 3.13 %

4)
Given:
Half life = 12 min
use relation between rate constant and half life of 1st order reaction

k = (ln 2) / k
= 0.693/(half life)
= 0.693/(12)
= 5.775*10^-2 min-1

Answer: 5.78*10^-2 min-1