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<div class="moz-cite-prefix">On 07/13/2015 08:23 PM, wrote:<br>
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<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130FFFABCB@MERCERMAIL.MercerU.local"
type="cite">
<pre wrap="">
Dear Dr. Pounds.
I have a few questions on the labs that are due tomorrow. On the “Kinetics: Temperature Effect” lab, I was wondering if the change in concentration of the MnO4 would just be the stock solution molarity, which is 0.0050 M, since the MnO4 is used up in the overall reaction. Therefore to find the rate would I just multiply the 0.0050 by one half and divide by the overall reaction time to find the rate?</pre>
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<br>
Yes. Think about it you took a stock solution and mixed it with
the same volume of another compound. This effectively halved the
concentration. The rate was then this concentration divided by the
time. The lab report form is a little confusing -- because the
column for the rate includes a header that includes the one half.
Just make sure you (1) divide the stock concentration by two and
then (2) divide that by the time to get the rate. <br>
<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130FFFABCB@MERCERMAIL.MercerU.local"
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<pre wrap=""> On the “Kinetics: Concentration Effect” lab I am having problems trying to find the initial concentrations. I know I have to use M1V1=M2V2, would the V1 be the amount we use in the reaction or 1 liter since the molarity is defined as moles of solute over liters of solvent. </pre>
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Its the amount you use for each reaction. For example, in the first
reaction the conentration of I- would be (0.30 M)(2 ml)/(150
ml)=0.004 M. This assumes I only used 1 ml of starch, so my final
volume is 150 ml.<br>
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cite="mid:C40B2F181831EF44A88CD73525827803130FFFABCB@MERCERMAIL.MercerU.local"
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<pre wrap=""> After finding the rates and the order of the reactions would I then make two separate rate laws, RATE=k[I-]^m and RATE=k[H2O2]^n to solve for the bottom portion of the lab report where it asks for the “Effect Trials”? Thank you for your time.
</pre>
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Nope -- it is one rate law <img style="vertical-align: middle"
src="cid:part1.05060008.05030808@sandbox.mercer.edu" alt="$RATE =
k [I^-]^m [H_2O_2]^n$">. The two plots you create let you
determine the orders. Once you have the orders, you can use your
eight rates and eight different initial concentrations to solve for
<img style="vertical-align: middle"
src="cid:part2.01090003.02030906@sandbox.mercer.edu" alt="$k$">.
You will get eight different values of <img style="vertical-align:
middle" src="cid:part2.01090003.02030906@sandbox.mercer.edu"
alt="$k$"> -- you take the average of all these and put that in
the last blank.<br>
<br>
<pre class="moz-signature" cols="72">--
Andrew J. Pounds, Ph.D. (<a class="moz-txt-link-abbreviated" href="mailto:pounds@theochem.mercer.edu">pounds@theochem.mercer.edu</a>)
Professor of Chemistry and Computer Science
Mercer University, Macon, GA 31207 (478) 301-5627
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