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<div class="moz-cite-prefix">On 04/23/14 15:05, Ashley Elizabeth Ray
wrote:<br>
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<p>Hi Dr. Pounds,</p>
<p> </p>
<p>I have a few more questions about this week's lab report.
After calculating the Ksp values, I found that all of the
In(Ksp) values are negative. When I made the graph of
1/Temperature and In(Ksp) the slope is negative instead of
positive. I am not sure where I made a mistake if any. Thanks
for your help.</p>
<p> </p>
<p>Respectfully,</p>
<p> </p>
<p>Ashley Ray</p>
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<img style="vertical-align: middle"
src="cid:part1.04010106.05080609@mercer.edu" alt="$\Delta G = -RT
\ln K = \Delta H - T \Delta S$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part2.03040509.03090001@mercer.edu" alt="$-RT \ln K =
\Delta H - T \Delta S$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part3.06040906.03030509@mercer.edu" alt="$ \ln K =
-\frac{\Delta H}{RT} + \frac{T \Delta S}{RT}$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part4.08030801.00070006@mercer.edu" alt="$\ln K =
-\frac{1}{T} \frac{\Delta H}{R} + \frac{\Delta S}{R}$"><br>
<br>
<br>
So the line should have a negative slope (m) which is equal to
negative Delta H over R. <br>
<br>
Now ask yourself -- is this an endothermic or exothermic process?
The sign of Delta H should reflect this.<br>
<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>
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