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<div class="moz-cite-prefix">On 07/12/13 18:22, wrote:<br>
</div>
<blockquote
cite="mid:C40B2F181831EF44A88CD735258278030261C41C73@MERCERMAIL.MercerU.local"
type="cite">
<pre wrap="">Dr Pounds,
I have a question with regards to the Delta G circle = -RT ln K equation,
In number 82 of Burdge, you used Delta G= -RT ln K to solve, how do we know when to use either equation for Delta G circle or delta G.
Thank you
</pre>
</blockquote>
<br>
I apparently left off the superscript zero when I wrote that
solution. Here is the rule of thumb you need to remember -- if you
are looking for an equilibrium constant then you are looking for
where the Gibbs Free Energy function has a slope of zero (<img
style="vertical-align: middle"
src="cid:part1.02080604.02000109@mercer.edu" alt="$\Delta G$">=
0}. <br>
<br>
<img style="vertical-align: middle"
src="cid:part2.08030709.03080503@mercer.edu" alt="$\Delta G =
\Delta G^{\circ} + RT lnQ$"><br>
<br>
at equilibrium<br>
<br>
<img style="vertical-align: middle"
src="cid:part3.05010204.02090600@mercer.edu" alt="$0 = \Delta
G^{\circ} + RT lnK$"><br>
<br>
or <br>
<br>
<img style="vertical-align: middle"
src="cid:part4.01040009.04010106@mercer.edu" alt="$\Delta
G^{\circ} = -RT lnK$"><br>
<br>
So basically -- if you are looking for an eqauilibrium constant you
will want to calculate <img style="vertical-align: middle"
src="cid:part5.08060506.08010504@mercer.edu" alt="$\Delta
G^{\circ}$"> using the thermo tables and then find K.<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>
</pre>
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