[CHM 112] Help with Concentration Kinetics Prelab

[Andrew J. Pounds]

X <#>LatexIt! run report...

*** Found expression $H_2O_2$
*** Found expression $S_2O_3^{2-}$
*** Found expression $S_2O_3^{2-}$
Image was already generated
*** Found expression $Na_2S_2O_3$
*** Found expression $S_2O_3^{2-}$
Image was already generated
*** Found expression $S_2O_3^{2-}$
Image was already generated
*** Found expression $H_2O_2$
Image was already generated
*** Found expression $H_2O_2$
Image was already generated
*** Found expression $S_2O_3^{2-}$
Image was already generated
*** Found expression $I^-$
*** Found expression $S_2O_3^{2-}$
Image was already generated
*** Found expression $H_2O_2$
Image was already generated
*** Found expression $H_2O_2$
Image was already generated
*** Found expression $V_2$
*** Found expression $V_1$
*** Found expression $M_1$
*** Found expression $M_1V_1=M_2V_2$

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).

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} \begin{document} \ce{2 
H^+ + H2O2 + 2 I^- -> I2 + 2 H2O } %this is where your LaTeX expression 
goes \end{document}

and we are using the clock reaction that I mentioned in lab yesterday...

\documentclass{article} \usepackage[utf8x]{inputenc} 
\usepackage[version=3]{mhchem} \pagestyle{empty} \begin{document} \ce{I2 
+ 2S2O3^{2-} -> S4O6^{2-} + 2I^- } %this is where your LaTeX expression 
goes \end{document}

Remember, when the $S_2O_3^{2-}$ gets used up, the clock reaction can no 
longer regenerate $I^-$ and the solution will turn blue.  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 ration 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) = 0.000667 M

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

( (0.000667) / 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.

-- 
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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