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<p class="MsoNormal"><span style="font-family:"Arial",sans-serif">19.34(b) asks you how the enropy changes for the processes in
<br>
19.12. The mistake is for 19.12(e). <br>
<br>
19.12(e) is HCl(g) dissolving to give HCl(aq). The solutions' manual <br>
states the entropy for this process increases due to more motional <br>
freedom (the HCl ionizes and you get ions). This is incorrect. <br>
<br>
It is true the HCl ionizes to give ions. That process itself would have <br>
a positive entropy change. However, the HCl(g) dissolving in water <br>
has a negative entropy change. I've pointed out several times how <br>
the entropy change for a gas dissolving in a liquid is negative. This <br>
is a pretty large negative entropy change. Negative enough to <br>
outweigh the positive entropy change of the HCl ionizing to form <br>
ions. You can calculate the entropy change using the standard <br>
entropy values in the back of the book (at 25 C). <br>
<br>
HCl (g) ----> H+ (aq) + Cl- (aq) <br>
<br>
S 186.69 0 56.5 <br>
<br>
del(S) = [ 1 (0) + 1 (56.5)] - 1 (186.69) = -130.2 J/K <br>
<br>
This reaction is very spontaneous (at 25 C) even with this rather large <br>
negative entropy change because it is very exothermic, del(H) = -74.9 kJ <br>
and del(G) = -35.9 kJ. <br>
<br>
Dr. Zellmer <o:p></o:p></span></p>
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