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<div class="moz-cite-prefix">On 07/10/13 16:05, wrote:<br>
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<p>Hey Dr. Pounds,</p>
<p>I was working on problem #46 from Chapter 18 (part2). I had a
few questions.</p>
<p>1. Why do we use (delta G=deltaG^o +RTlnK) rather than just
finding the delta H and delta S and using the other formula to
find the values?</p>
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In this part of the problem they have given you the K value. Since
K is at the equilibrium condition, $\Delta G$ has to be zero.
Because you know that you can compute the value of $\Delta
G^{\circ}$.<br>
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<p>2. When do we know when to find just delta G vs. delta G^o?</p>
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If you are looking for an equilibrium constant, you will need
$\Delta G^{\circ}$.<br>
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<p>3. Also, on the second part of that question, we used this
formula again(delta G=deltaG^o +RTlnK), BUT rather than the K,
there is a Q? Why is that? When do we know it is a K or a Q?</p>
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<br>
So here they want to know the actual value of the Gibbs Free Energy
function when it is NOT at equilibrium, but rather at its initial
state. Using the value of $\Delta G^{\circ}$ you just calculated
and the values of the non-equilibrium condition (Q), you can
determine the value for $\Delta G$.<br>
<br>
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<p>4. Last, why do we leave delta G as joules when I thought it
is supposed to be in kilojoules?
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<br>
It's purely a matter of convenience. If you are looking for the
value of K then you will be dividing by the gas constant R in the
units of J/K mol.<br>
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<p> </p>
<p> </p>
<p> </p>
<p>Thank you so much!</p>
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<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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