## Homework, 6th edition, 15.27

$\frac{d[R]}{dt}=-k[R]; \ln [R]=-kt + \ln [R]_{0}; t_{\frac{1}{2}}=\frac{0.693}{k}$

Bianca Barcelo 4I
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Joined: Fri Sep 28, 2018 12:17 am
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### Homework, 6th edition, 15.27

Hi, can someone help me figure out how to do part c and d?
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monikac4k
Posts: 56
Joined: Fri Sep 28, 2018 12:25 am

### Re: Homework, 6th edition, 15.27

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!

Jennifer Su 2L
Posts: 47
Joined: Wed Nov 21, 2018 12:20 am

### Re: Homework, 6th edition, 15.27

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