HW #14.35b


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Shannon Han 2B
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Joined: Fri Sep 26, 2014 2:02 pm

HW #14.35b

Postby Shannon Han 2B » Mon Feb 16, 2015 2:21 pm

I'm confused on 35b, where the question asks to find the time it takes for the concentration of SO2Cl2 to decrease to 10% of its initial concentration. Why does the solutions manual just do ln10, and when I solved ln10/k like they did, I got 0.000819 minutes, not 819 minutes. Could someone help explain this?

Neil DSilva 1L
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Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #14.35b

Postby Neil DSilva 1L » Mon Feb 16, 2015 5:39 pm

The ln10 comes from a modified version of the first-order half-life equation. Instead of using ln2 (which equals 0.693, the number we see in the half-life equation) which comes from the following analysis:
,
the solutions manual is finding the time it takes the concentration to decrease to 10% (aka finding the "tenth-life") using the following steps:
.

Hopefully that helps you understand that portion. As for your calculations, you might have inputted the numbers incorrectly. When I do the calculations, I get and when you divide that by k (), you should get 819 minutes. How did you input your data?

Sonia Kumar 2A
Posts: 6
Joined: Fri Sep 26, 2014 2:02 pm

Re: HW #14.35b

Postby Sonia Kumar 2A » Thu Feb 19, 2015 12:06 pm

If you were also confused on the ln 10, I was originally confused on why it was ln 10 and not ln(.1) and then I noticed that by dividing the negative it flipped the fraction.

Chem_Mod
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Re: HW #14.35b

Postby Chem_Mod » Thu Feb 19, 2015 1:24 pm

Neil is correct. From the 1st order integrated rate law, if we want to find the time to drop from [A] to [A]/n, eg. the "n-th life"

ln([A]/n) = -kt + ln[A]
ln(1/n) = -kt
ln(n) / k = t

Then the n-th life is just ln(n)/k. So the half life is ln2/k, the tenth life is ln10/k, etc.


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