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<th align="RIGHT" nowrap="nowrap" valign="BASELINE">Subject:
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<td>Re: [CHM 112] Kinetics Lab</td>
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<th align="RIGHT" nowrap="nowrap" valign="BASELINE">Date: </th>
<td>Sun, 01 Mar 2015 21:46:24 -0500</td>
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<th align="RIGHT" nowrap="nowrap" valign="BASELINE">From: </th>
<td>Andrew J. Pounds <a class="moz-txt-link-rfc2396E" href="mailto:pounds_aj@mercer.edu"><pounds_aj@mercer.edu></a></td>
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<td><a class="moz-txt-link-abbreviated" href="mailto:pounds_aj@mercer.edu">pounds_aj@mercer.edu</a></td>
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<td>undisclosed-recipients:;</td>
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On 03/01/2015 09:23 PM, wrote:<br>
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<blockquote
cite="mid:C40B2F181831EF44A88CD73525827803130E1990BC@MERCERMAIL.MercerU.local"
type="cite">
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<p>âHi, Dr. Pounds!</p>
<p><br>
</p>
<p>I had a few questions on the lab. I was wondering what
the purpose of finding the slope of the line was and if we
would use that for any other calculations. Last lab we
used it to find k, but it seems that the values of k are the
same for each trial and therefore would cancel out to be 1
when put in the equation. Is there a use for the slope or
am I missing something in the calculations? </p>
<p><br>
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<p>Thank you!</p>
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Whoa.. hold on there..<br>
<br>
<br>
<br>
So you the lab description states "However, we won't actually
measure the rate constant k. Insteas we assume that since the
concentration of reactants is the same in each trial, the measured
rate is equal to k times some constant."<br>
<br>
So you have the Arrhenius equation<br>
<br>
<img style="vertical-align: middle"
src="cid:part1.09070807.07040908@mercer.edu" alt="$k =
Ae^{-E_a/RT}$"><br>
<br>
the log of which is<br>
<br>
<img style="vertical-align: middle"
src="cid:part2.05030108.06080900@mercer.edu" alt="$\ln k = \ln A
- \frac{E_a}{R} \frac{1}{T}$"><br>
<br>
based on the statement from the lab, we are going to replace k
with rate<br>
<br>
<img style="vertical-align: middle"
src="cid:part3.01060709.05030305@mercer.edu" alt="$\ln
\mathrm{rate} = \ln A - \frac{E_a}{R} \frac{1}{T}$"><br>
<br>
If we then plot the <img style="vertical-align: middle"
src="cid:part4.08060206.08060807@mercer.edu" alt="$\ln
\mathrm{rate}$"> on the y-axis and <img style="vertical-align:
middle" src="cid:part5.07040908.01060100@mercer.edu"
alt="$\frac{1}{T}$"> on the x-axis we should get a linear
relationship with the slope equal to <img style="vertical-align:
middle" src="cid:part6.09040508.05060100@mercer.edu"
alt="$-\frac{E_a}{R}$">. So from the slope you can determine
the activation energy.<br>
<br>
Got it? <br>
<br>
<pre class="moz-signature" cols="72">\mathrm{rate}--
Andrew J. Pounds, Ph.D. (<a moz-do-not-send="true" 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 moz-do-not-send="true" class="moz-txt-link-freetext" href="http://faculty.mercer.edu/pounds_aj">http://faculty.mercer.edu/pounds_aj</a>
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