[CHM 112] C effect lab Question

[Andrew J. Pounds]

On 7/3/20 10:27 PM,  wrote:
> Hi Dr. Pounds,
>
> I was doing the lab report and had a question. How do you calculate
> the initial [H2O2] for trial one? Is the initial concentration for
> [I-] in trial two the same as the [H2O2] concentration in trial 1? 
>
> Thanks!
>

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

$2H^+ + H_2O_2 + 2I^- -> I_2 + 2 H_2O$

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

$I_2 + S_2O_3^{2-} \longrightarrow S_4O_6^{2-} + 2I^-$

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.


OR -- you could just use the spreadsheet that I made for you on the
class web page in the LAB section....



-- 
Andrew J. Pounds, Ph.D.  (pounds_aj at mercer.edu)
Professor of Chemistry and Computer Science
Director of the Computational Science Program
Mercer University,  Macon, GA 31207   (478) 301-5627

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