## HW #14.35b

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

Shannon Han 2B
Posts: 48
Joined: Fri Sep 26, 2014 2:02 pm

### HW #14.35b

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
Posts: 70
Joined: Fri Sep 26, 2014 2:02 pm

### Re: HW #14.35b

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:
$ln\frac{[SO_{2}Cl_{2}]_{0}}{0.5\times [SO_{2}Cl_{2}]_{0}} = ln\frac{1}{0.5} = ln\frac{10}{5} = ln2$,
the solutions manual is finding the time it takes the concentration to decrease to 10% (aka finding the "tenth-life") using the following steps:
$ln\frac{[SO_{2}Cl_{2}]_{0}}{0.10\times [SO_{2}Cl_{2}]_{0}} = ln\frac{1}{0.1} = ln\frac{10}{1} = ln10$.

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 $ln10 \approx 2.3025$ and when you divide that by k ($2.81 \times 10^{-3}$), 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

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
Posts: 19566
Joined: Thu Aug 04, 2011 1:53 pm
Has upvoted: 885 times

### Re: HW #14.35b

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.