<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=ISO-8859-1">
</head>
<body bgcolor="#FFFFFF" text="#000000">
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 <img
style="vertical-align: middle"
src="cid:part1.07080106.06080407@mercer.edu" alt="$M_1V_1=M_2V_2$">
where <img style="vertical-align: middle"
src="cid:part2.02000205.09030804@mercer.edu" alt="$M_1$"> is the
initial molarity of the substance and <img style="vertical-align:
middle" src="cid:part3.04070705.09090004@mercer.edu" alt="$V_1$">
is the initial volume of that substance and <img
style="vertical-align: middle"
src="cid:part4.00030903.03000908@mercer.edu" alt="$V_2$"> is the
volume of everything mixed together (which, by the way, is always
150 ml).<br>
<br>
The hard part is determining the rate of disappearance of <img
style="vertical-align: middle"
src="cid:part5.07000801.01010009@mercer.edu" alt="$H_2O_2$">. You
need to understand something -- ALL of the rates you measure are for
the disappearance of <img style="vertical-align: middle"
src="cid:part5.07000801.01010009@mercer.edu" alt="$H_2O_2$"> -- 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.<br>
<br>
The reaction you are studying is<br>
<br>
<pre><img style="vertical-align: middle" src="cid:part7.05060502.01030204@mercer.edu" title="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document} \ce{2
H^+ + H2O2 + 2 I^- -> I2 + 2 H2O }
\end{document}" alt="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document} \ce{2
H^+ + H2O2 + 2 I^- -> I2 + 2 H2O }
\end{document}"></pre>
<br>
and we are using the clock reaction mentioned in the lab report to
"time" the reaction.<br>
<br>
<img style="vertical-align: middle"
src="cid:part8.02060403.05070106@mercer.edu"
title="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\ce{I2 + 2S2O3^{2-} -> S4O6^{2-} + 2I^- } \end{document}"
alt="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\ce{I2 + 2S2O3^{2-} -> S4O6^{2-} + 2I^- } \end{document}"><br>
<br>
when the <img style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">
gets used up, the clock reaction can no longer regenerate <img
style="vertical-align: middle"
src="cid:part10.00010805.01080804@mercer.edu" alt="$I^-$"> 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 <img
style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">
to be consumed -- and then you have to relate that back, through
stoichiometry, to the amount of <img style="vertical-align: middle"
src="cid:part5.07000801.01010009@mercer.edu" alt="$H_2O_2$"> that
was consumed. Here is the good news, its a 1:2 ratio between the <img
style="vertical-align: middle"
src="cid:part5.07000801.01010009@mercer.edu" alt="$H_2O_2$"> and
the <img style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">,
and the amount of <img style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">
in every reaction vessel is the same (you add it as <img
style="vertical-align: middle"
src="cid:part16.07080003.02080003@mercer.edu" alt="$Na_2S_2O_3$">,
but it immediately ionizes).<br>
<br>
So, based on what we find on the chart in the procedure manual, you
had 5.0 ml or 0.02 M <img style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">
that was then diluted to 150 ml. This means that the concentration
of <img style="vertical-align: middle"
src="cid:part9.09060709.04020006@mercer.edu" alt="$S_2O_3^{2-}$">
just after mixing but before reaction has to be<br>
<br>
(5.0) (0.02) / (150) = 0.000667 M<br>
<br>
From the 1:2 stoichiometry, the rate of disappaerance of <img
style="vertical-align: middle"
src="cid:part5.07000801.01010009@mercer.edu" alt="$H_2O_2$"> is
thus...<br>
<br>
( (0.000667) / 2 ) / time (in seconds).<br>
<br>
ALL of your RATES will be calculated this way -- just plug in the
appropriate seconds.<br>
<br>
As always, let me know if you have any questions.<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
<a class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
</pre>
</body>
</html>