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<div class="moz-cite-prefix">On 07/19/2015 04:02 PM, wrote:<br>
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<p>Hi Dr. Pounds, </p>
<p>I got stuck on one of the textbook problems. On number 76 in
Burdge, I saw the solution and I don't understand why you
added 2x and x and equaled it to .318, and I also don't
understand what equation you used to find kp? Because kp =
kc(RT)^change in moles, but I don't see the temperature or R
constant used in the equation? </p>
<p>Thanks so much! </p>
<p>Srishti </p>
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<br>
<font face="serif">Yeah -- this is a good one. So in the reaction
you have a solid decomposing to become two different gases.
Therefore the ONLY thing that is responsible for the pressure is
the pressure of the independent gases. In other words,<br>
<br>
<img style="vertical-align: middle"
src="cid:part1.07020909.09070205@mercer.edu"
alt="$P_{\mathrm{tot}} = P_{\mathrm{NH_3}} + P_{\mathrm{CO_2}}$"><br>
<br>
From the stoichiometry of the reaction, for each molecule of the
solid that decomposes you have to make 2 molecules of ammonia and
one molecule of carbon dioxide. Since these are the ONLY places
where the pressure can come from, and I am told that the pressure
inside the vessel is 0.318 atm, then it is possible to write the
following:<br>
<br>
</font><br>
<font face="serif"><img style="vertical-align: middle"
src="cid:part1.07020909.09070205@mercer.edu"
alt="$P_{\mathrm{tot}} = P_{\mathrm{NH_3}} + P_{\mathrm{CO_2}}$"><br>
</font><br>
<font face="serif"><img style="vertical-align: middle"
src="cid:part3.07070209.01040804@mercer.edu"
alt="$P_{\mathrm{tot}} = 2x + x$"><br>
</font><br>
<font face="serif"><img style="vertical-align: middle"
src="cid:part4.03040208.09050901@mercer.edu"
alt="$P_{\mathrm{tot}} = 3x$"><br>
<br>
<img style="vertical-align: middle"
src="cid:part5.08010709.01080706@mercer.edu" alt="$0.318 {\
\mathrm{atm}} = 3x$"><br>
<br>
which I can then use to solve for <img style="vertical-align:
middle" src="cid:part6.05030207.03000002@mercer.edu" alt="$x$">
and then plug back in to solve for the partial pressure of each
compound.<br>
<br>
In the solution I posted I then plugged in the values for each of
the partial pressures into the law of mass action for the chemical
reaction. I think the book misappropriately uses the term <img
style="vertical-align: middle"
src="cid:part7.00070002.03030205@mercer.edu" alt="$K_p$">
because it is in fact a heterogeneous equilibrium involving solids
and gases. As such it should be written as...<br>
<br>
<img style="vertical-align: middle"
src="cid:part8.07050707.07080206@mercer.edu" alt="$K =
\frac{(P_{\mathrm{NH_3}})^2(P_{\mathrm{CO_2}})}{1}$"><br>
<br>
where the 1 in the denominator is due to the fact that the
reactant is a solid.<br>
<br>
<br>
</font>
<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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