## Problem 15.37

$aR \to bP, Rate = -\frac{1}{a} \frac{d[R]}{dt} = \frac{1}{b}\frac{d[P]}{dt}$

allyz1F
Posts: 58
Joined: Sat Jul 22, 2017 3:00 am

### Problem 15.37

This problem asks for the time needed for SO2Cl2 concentration to decrease to 10% of the initial concentration. In the answer it says t= ln(SO2Cl2 initial/SO2Cl2 ) / k is equal to ln10/k. Why is this?

Joshua Xian 1D
Posts: 51
Joined: Fri Sep 29, 2017 7:04 am
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### Re: Problem 15.37

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

Lauren Seidl 1D
Posts: 51
Joined: Fri Sep 29, 2017 7:06 am

### Re: Problem 15.37

Because they divide initial concentration by final concentration, you get a value of 10 in the ln and would therefore set ln(10) = k*t. If you divided final concentration by initial concentration instead, you would get a value of 0.1 in the ln, but you would have to set ln(0.1) = - k*t according to the equation ln[A] = -k*t + ln[A]0. They just flipped it around so that positive k*t = ln([A]0/[A]).

Shanmitha Arun 1L
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Joined: Fri Sep 29, 2017 7:04 am
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### Re: Problem 15.37

You would do ln(1/0.1) which is the same as ln(10).