## 15.37(B) homework

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

Jack Papciak 2F
Posts: 31
Joined: Fri Sep 29, 2017 7:06 am

### 15.37(B) homework

I couldn't figure out this question so I checked the solutions manual and it says eventually you get ln10/2.8x10^-3. I don't understand where the ln10 is coming from. Any help would be appreciated

Joshua Xian 1D
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### Re: 15.37(B) homework

Since final concentration is 10% of the initial, dividing initial by final would result in 10. Therefore, you get ln(10)/k.

Shanmitha Arun 1L
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### Re: 15.37(B) homework

For problems like this that give % of the initial like 10%, it would be ln(1/0.1) which ends up being ln(10). For 15%, it would be ln(1/0.15) since its ln of [A]/[Ao]

Sophie 1I
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Joined: Fri Sep 29, 2017 7:04 am

### Re: 15.37(B) homework

So then it wouldn't matter whether you had the actual initial concentration?