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Chapter 5 From Fig. 5.9, the integral of C d ln T for the gas from 263.1 to 298.15 K
P
Standard Thermodynamic is the product of 10 cal/(mol K) and ln 298.15 ln 263.1 0.125. This integral
Functions of Reaction
equals 1.2 cal/(mol K). [Accurate evaluation gives 1.22 cal/(mol K).]
5
So far, we have gone from the solid at 0 K and 1 atm to the real gas at 298.15 K
and 1 atm. We next add in the given value S m,id S m,re 0.07 cal/(mol K) to
reach the ideal gas at 298.15 K and 1 atm. The final step is to include S for
m
changing the ideal gas from 1 atm to 1 bar at 298.15 K. For an isothermal ideal-
gas process, Eq. (3.30) and Boyle’s law give S R ln (V /V ) R ln (P /P ).
1
2
m
1
2
The S for going from 1 atm to 1 bar ( 750 torr) is thus R ln (760/750)
m
0.03 cal/(mol K).
Sº m,298
Adding everything, we get
S° m,298
10.28 20# 8.95 6.0 22.65 1.2 0.07 0.032 cal>1mol K2
5
6
S° m,298 59# cal>1mol K2
8
[The accurate values give S° 59.28 cal/(mol K) 248.0 J/(mol K).]
m,298
Exercise
1
Use Fig. 5.9 to estimate S° S° for SO (s). (Answer: 11 cal mol 1 K .)
m,148 m,55 2
Figure 5.10 plots some conventional S° values. The Appendix tabulates S°
m,298 m,298
for various substances. Diamond has the lowest S° of any substance. The
m,298
Appendix S° values show that (a) molar entropies of gases tend to be higher than
m,298
those of liquids; (b) molar entropies of liquids tend to be higher than those of solids;
(c) molar entropies tend to increase with increasing number of atoms in a molecule.
Conventional entropies are often called absolute entropies. However, this name is
inappropriate in that these entropies are not absolute entropies but relative (conven-
tional) entropies. Since full consideration of this question requires statistical mechan-
ics, we postpone its discussion until Sec. 21.9.
Figure 5.10 Since C ( H / T) , integration of C° from 0 K to T with the addition of
P,m m P P,m
H° for all phase transitions that occur between 0 and T gives H° H° , where
S° m,298 values. The scale is m m,T m,0
logarithmic. H° m,T and H° are the standard-state molar enthalpies of the substance at T and of the
m,0
corresponding solid at 0 K. For solids and liquids, H° H° is essentially the
m,T m,0
same as U° U° . Figure 5.11 plots H° H° versus T and plots S° versus
m,T m,0 m,T m,0 m,T
T for SO . Both H and S increase as T increases. Note the large increases in S and
2 m m
H that occur on melting and vaporization.
S° m,T (J mol K)
(H° m,T H° ) (kJ mol)
m,0
SO 2
SO 2
Figure 5.11
S° and H° H° versus T for
m,0
m,T
m,T
SO , where H° is for solid SO .
m,0
2
2