## 15.21

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

Shannon Wasley 2J
Posts: 71
Joined: Fri Sep 29, 2017 7:06 am
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### 15.21

The problem is:
Beta blockers are drugs that are used to manage
hypertension. It is important for doctors to know how rapidly a beta blocker is eliminated from the body. A certain beta blocker is eliminated in a rst-order process with a rate constant of
7.6 10 3 min 1 at normal body temperature (37 C). A patient is given 20. mg of the drug. What mass of the drug remains in the body 5.0 h after administration?

For this problem, I used the formula [A] = [A](initial) x e^(-kt). Why can we simply use .02 grams as our initial [A]? Isn't it supposed to be a concentration?

Jasmine Botello 2F
Posts: 67
Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.21

I think my TA told me that it does not necessarily have to be concentration. As long as you keep it the same throughout the problem.

Chem_Mod
Posts: 18400
Joined: Thu Aug 04, 2011 1:53 pm
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### Re: 15.21

Hey!!
For both the [A] you would have to divide the grams by the same molar mass and volume amount to convert it into concentration. As a result, mathematically, this would give you the same result, whether you used grams for [A] or moles/L for [A]. The solution manual tends to simplify problem solutions which is why they chose to use grams but both ways are correct.
Hope this helps!

Cam Bear 2F
Posts: 60
Joined: Thu Jul 27, 2017 3:01 am

### Re: 15.21

For this problem, I was just wondering does it matter if the formula is written ln[A]=-kt+ln[A]o, ln[A]/ln[A]o=-kt, or [A]=[A]oe^-kt to solve? In discussion we derived the third one from the first to solve, but won't it give you the same answer? Why take the time to manipulate the equation?