[CHM 112] Help with Concenctration Kinetics Prelab

[Andrew J. Pounds]

The concentration kinetics lab requires you to do numerous calculation. 
For all of the prelab calculations dealing with questions like "What is 
the concentration of ??? in flask ???" you just have to use the standard 
dilution formula $M_1V_1=M_2V_2$ where $M_1$ is the initial molarity of 
the substance and $V_1$ is the initial volume of that substance and 
$V_2$ is the volume of everything mixed together (which, by the way, is 
always 150 ml in the prelab because it was written before we made the 
change this semester to use 2 ml of starch). The hard part is 
determining the rate of disappearance of $H_2O_2$. You need to 
understand something -- ALL of the rates you
measure are for the disappearance of $H_2O_2$ -- so what I am saying 
here will apply to ALL of the rates you will fill out on the report
form. Here is what makes this difficult. The reaction you are studying is

\documentclass{article} \usepackage[utf8x]{inputenc} 
\usepackage[version=3]{mhchem} \pagestyle{empty} \pagestyle{empty} 
\begin{document} \ce{2H^+ + H2O2 + 2I^- -> I2 + 2 H2O } \end{document}

and we are using the clock reaction mentioned in the lab report to 
"time" the reaction.

\documentclass{article} \usepackage[utf8x]{inputenc} 
\usepackage[version=3]{mhchem} \pagestyle{empty} \pagestyle{empty} 
\begin{document} \ce{I_2 + 2S2O3^{2-} -> S4O6^{2-} + 2I^- } \end{document}

when the $S_2O_3^{2-}$ gets used up, the clock reaction can no longer 
regenerate $I^-$ and the solution will turn blue due to the presence of 
starch. So, what you are measuring is the amount of time for all of the 
$S_2O_3^{2-}$ to be consumed -- and then you have to relate that back, 
through stoichiometry, to the amount of $H_2O_2$ that was consumed. Here 
is the good news, its a 1:2 ratio between the $H_2O_2$ and the 
$S_2O_3^{2-}$, and the amount of $S_2O_3^{2-}$ in every reaction vessel 
is the same (you add it as $Na_2S_2O_3$, but it immediately ionizes).

So, based on what we find on the chart in the procedure manual, you had 
5.0 ml or 0.02 M $S_2O_3^{2-}$ that was then diluted to 150 ml. This 
means that the concentration of $S_2O_3^{2-}$ just after mixing but 
before reaction has to be

(5.0) (0.02) / (150) = 0.0006623 M

 From the 1:2 stoichiometry, the rate of disappaerance of $H_2O_2$ is 
thus...

( (0.0006623) / 2 ) / time (in seconds).

ALL of your RATES will be calculated this way -- just plug in the 
appropriate seconds. As always, let me know if you have any questions.

If you still are having problems with the pre-labs, email or text me 
and  I'll try to help.



-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Mercer University,  Macon, GA 31207   (478) 301-5627
http://faculty.mercer.edu/pounds_aj

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