### Homework, 6th edition, 15.27

Posted:

**Mon Mar 11, 2019 9:54 pm**Hi, can someone help me figure out how to do part c and d?

Created by Dr. Laurence Lavelle

https://lavelle.chem.ucla.edu/forum/

https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=148&t=44205

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Posted: **Mon Mar 11, 2019 9:54 pm**

Hi, can someone help me figure out how to do part c and d?

Posted: **Mon Mar 11, 2019 9:59 pm**

What I did was use the half-life in order to find the k value (t1/2= ln2/k). Then I would just use the integrated rate law to solve for the time required for the other scenarios. For example, when it asks for 15% of the initial concentration, you would plug in 0.15A0 into [A] and then solve. Although you don't have a value for A, when you rearrange the integrated rate law, the A should cancel. Hope this helps!

Posted: **Mon Mar 11, 2019 10:03 pm**

For parts c and d, you need to first find the value of k by using the half life formula.

t1/2 = 0.693/k =355s, so k= 1.95x10^-3 s^-1

You can then plug this into the first order rate equation to find the time (t).

ln[A]= -kt + ln[A]not

c) ln(0.15)= -(1.95x10^-3 s^-1)*t + ln(1), so t=9.7x10^2 s

d) ln(1/9)= -(1.95x10^-3 s^-1)*t + ln(1), so t=1.1x10^3 s

t1/2 = 0.693/k =355s, so k= 1.95x10^-3 s^-1

You can then plug this into the first order rate equation to find the time (t).

ln[A]= -kt + ln[A]not

c) ln(0.15)= -(1.95x10^-3 s^-1)*t + ln(1), so t=9.7x10^2 s

d) ln(1/9)= -(1.95x10^-3 s^-1)*t + ln(1), so t=1.1x10^3 s