<html>
<head>
<meta http-equiv="content-type" content="text/html; charset=utf-8">
</head>
<body bgcolor="#FFFFFF" text="#000000">
The concentration kinetics lab requires you to do numerous
calculation. <br>
For all of the prelab calculations dealing with questions like "What
is <br>
the concentration of ??? in flask ???" you just have to use the
standard <br>
dilution formula <img style="vertical-align: middle"
src="cid:part1.03090703.09090208@mercer.edu" alt="$M_1V_1=M_2V_2$">
where <img style="vertical-align: middle"
src="cid:part2.07050700.07090803@mercer.edu" alt="$M_1$"> is the
initial molarity of <br>
the substance and <img style="vertical-align: middle"
src="cid:part3.05070805.02030400@mercer.edu" alt="$V_1$"> is the
initial volume of that substance and <br>
<img style="vertical-align: middle"
src="cid:part4.06050005.08040002@mercer.edu" alt="$V_2$"> is the
volume of everything mixed together (which, by the way, is <br>
always 150 ml). The hard part is determining the rate of
disappearance <br>
of <img style="vertical-align: middle"
src="cid:part5.07090006.08050106@mercer.edu" alt="$H_2O_2$">. You
need to understand something -- ALL of the rates you <br>
measure are for the disappearance of <img style="vertical-align:
middle" src="cid:part5.07090006.08050106@mercer.edu"
alt="$H_2O_2$"> -- so what I am saying <br>
here will apply to ALL of the rates you will fill out on the report
<br>
form. Here is what makes this difficult. The reaction you are
studying is<br>
<br>
<img style="vertical-align: middle"
src="cid:part7.04040804.01060407@mercer.edu"
title="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\pagestyle{empty} \begin{document}
\begin{document} \ce{2H^+ + H2O2 + 2 I^- -> I2 + 2 H2O } %this
is where your LaTeX expression goes
\end{document}" alt="\documentclass{article}
\usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\pagestyle{empty} \begin{document}
\begin{document} \ce{2H^+ + H2O2 + 2 I^- -> I2 + 2 H2O } %this
is where your LaTeX expression goes
\end{document}"><br>
<br>
and we are using the clock reaction mentioned in the lab report to <br>
"time" the reaction.<br>
<br>
<img style="vertical-align: middle"
src="cid:part8.06040308.06080606@mercer.edu"
title="\documentclass{article} \usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\pagestyle{empty} \begin{document} \ce{I_2 + 2S2O3^{2-} ->
S4O6^{2-} + 2I^- } \end{document}" alt="\documentclass{article}
\usepackage[utf8x]{inputenc}
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document}
\pagestyle{empty} \begin{document} \ce{I_2 + 2S2O3^{2-} ->
S4O6^{2-} + 2I^- } \end{document}"><br>
<br>
when the <img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">
gets used up, the clock reaction can no longer <br>
regenerate <img style="vertical-align: middle"
src="cid:part10.00090301.05040109@mercer.edu" alt="$I^-$"> and the
solution will turn blue due to the presence of <br>
starch. So, what you are measuring is the amount of time for all of
the <br>
<img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">
to be consumed -- and then you have to relate that back, <br>
through stoichiometry, to the amount of <img style="vertical-align:
middle" src="cid:part5.07090006.08050106@mercer.edu"
alt="$H_2O_2$"> that was consumed. Here <br>
is the good news, its a 1:2 ratio between the <img
style="vertical-align: middle"
src="cid:part5.07090006.08050106@mercer.edu" alt="$H_2O_2$"> and
the <br>
<img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">,
and the amount of <img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">
in every reaction vessel <br>
is the same (you add it as <img style="vertical-align: middle"
src="cid:part16.09070805.01010607@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 <br>
5.0 ml or 0.02 M <img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">
that was then diluted to 150 ml. This <br>
means that the concentration of <img style="vertical-align: middle"
src="cid:part9.01050201.04090905@mercer.edu" alt="$S_2O_3^{2-}$">
just after mixing but <br>
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:part5.07090006.08050106@mercer.edu" alt="$H_2O_2$"> is <br>
thus...<br>
<br>
( (0.000667) / 2 ) / time (in seconds).<br>
<br>
ALL of your RATES will be calculated this way -- just plug in the <br>
appropriate seconds. As always, let me know if you have any
questions.<br>
<br>
<br>
If you still are having problems with the pre-labs, email or text me
and <br>
I'll try to help.<br>
<br>
<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>