## Book Problem 15.21

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

704578485
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### Book Problem 15.21

Problem 15.21 of the book involves a scenario in which you administer to a patient 20.mg of a beta blocking drug. The reaction that eliminates the drug from the body is a first order reaction with a k of 7.6x10^-3 min^-1. The question ask what mass of the beta drug will still be in the body after 5 hours. The answer the book gives is 2.0mg and I cannot see how they are getting that answer.

Erin Murashige 1L
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Joined: Fri Sep 25, 2015 3:00 am

### Re: Book Problem 15.21

I think you use the formula ln[A]t/[A]0=-kt. Then k will be the 7.6 x 10^-3 and t will be 300 minutes since it is 5 hours. Then after simplifying the equation you'll get that 10% of the initial drug concentration remains in the body. Then 10% of 20 mg will be 2.0 mg.

Gurkriti Ahluwalia 1K
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### Re: Book Problem 15.21

note that in solving for this the ln's DO NOT cancel, instead you use the inverse operation (e^) to get rid of the lns, and then divide both sides by A0 to get At=10% of A initial.

Beza Ayalew 1I
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Joined: Fri Sep 29, 2017 7:07 am

### Re: Book Problem 15.21

the solution doesn't plug in 20mg as the initial concentration, but if you do that the same values come out, and i found doing that easier to grasp because you can just directly solve for the final concentration

torieoishi1A
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Joined: Sat Jul 22, 2017 3:00 am

### Re: Book Problem 15.21

Then plug in k(7.6x10^-3/min) and t (5 hours but use 300 min)
Second you will now have: ln(A/A(initial)=-2.28
Cancel out on by using e: (A/A(initial)=e^-2.28
A(initial)= 20mg
Solve for A which is 2.0 mg

Gwyneth Huynh 1J
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Joined: Fri Sep 29, 2017 7:05 am

### Re: Book Problem 15.21

So to clarify, can we substitute mass for concentration? Or are we supposed to find the ratio of the initial and final concentration and use that to find the final mass?