Textbook Problem 7B.13

[Francesca_Borchardt_2D]

Hello, I am having trouble with the textbook problem 7B.13.

"The half-life of A in a second-order reaction is 50.5 s when
[A]0=0.84 mol/L. Calculate the time needed for the concentration of A to decrease to (a) one-sixteenth; (b) one-fourth; (c) onefifth of its original value."

I understand that the half-life equation for second-order reactions should be used to find k and the integrated rate law to find the time needed for each concentration. However, I am still having trouble figuring out this problem. Could someone help?

Thank you!

[derickngo3d]

For a and b, you can simply use the fact that they are exponentials of the half life, which is 1/2 of the value of the initial concentration. So 1/16 would be (1/2)^4. So multiply the half-life in seconds by 4.
Hope this helps!

[Christine Ma 3L]

You're pretty much on the right track!
We have the half-life equation for a second-order reaction:
and its integrated rate law: which can be rearranged into since we're only interested in time.


First we use the half-life equation to solve for k:
k=

For part a, [A] should be equal to 1/16 of [A]₀, and after substituting we should get:


.


You solve parts b and c the same way, but for part b [A] should equal 1/4 of [A]₀ and for part c [A] should equal 1/5 of [A]₀.

Hope this helps!

[Zoe Kaiser]

I get how it's solved but why do we have to solve it this way when in #7 we could just multiply by the half life?