Page 176 - Physical Chemistry

lev38627_ch05.qxd 3/3/08 9:33 AM Page 157

157
molecules obey quantum mechanics, not classical mechanics. Provided T is not close to Section 5.7
absolute zero, it is an adequate approximation to use classical mechanics to treat the mole- Conventional Entropies and
the Third Law
cular motions. When T is close to absolute zero, one must use quantum mechanics. It is
found that quantum ideal gases do obey the third law.
Although the third law does apply to mixtures, it is hard to achieve the required con-
dition of internal equilibrium in solid mixtures at very low T, so to avoid error, it is safest
to apply the third law only to pure substances.

Determination of Conventional Entropies
To see how (5.25) is used to find conventional entropies of compounds, consider the
process
H 1s2 1 2 O 1s2 S H O1s2 (5.26)
2
2
2
where the pure, separated elements at 1 bar and T are converted to the compound H O
2
at 1 bar and T. For this process,
¢S S° 1H O2 S° 1H 2 1 2 S° 1O 2 (5.27)
2
m
m
2
m
2
Our arbitrary choice of the entropy of each element as zero at 0 K and 1 bar
[Eq. (5.22)] gives lim S° (H ) 0 and lim S° (O ) 0. The third law, Eq.
T→0 m 2 T→0 m 2
(5.25), gives for the process (5.26): lim S 0. In the limit T → 0, Eq. (5.27) thus
T→0
becomes lim S° (H O) 0, which we write more concisely as S° (H O) 0.
T→0 m 2 m,0 2
Exactly the same argument applies for any compound. Hence S° 0 for any ele-
m,0
ment or compound in internal equilibrium. The third law (5.25) shows that an isother-
mal pressure change of a substance in internal equilibrium in the limit of absolute zero
has S 0. Hence we can drop the superscript degree (which indicates P 1 bar).
Also, if S 0, then S 0 for any amount of the substance. Our conclusion is that
m,0 0
the conventional entropy of any element or compound in internal equilibrium is zero
in the limit T → 0:
S 0 element or compound in int. equilib. (5.28)*
0
Now that we have the conventional standard-state entropies of substances at T 0,
their conventional standard-state entropies at any other T are readily found by using
the constant-P equation S S S T 2 (C /T) dT [Eq. (3.28)], with inclusion
0
T 2 0 T 2 P
also of the S of any phase changes between absolute zero and T . For example, for
¢
2
a substance that is a liquid at T and 1 bar, to get S° we add the entropy changes for
2 m,T 2
(a) warming the solid from 0 K to the melting point T , (b) melting the solid at T
fus fus
[Eq. (3.25)], and (c) warming the liquid from T to T :
fus 2
m,T 2 T fus C° 1s2 ¢ H° m T 2 C° 1l2
P,m
fus
P,m
S° dT dT (5.29)
T T T
0 fus T fus
where H is the molar enthalpy change on melting (fusion) and C (s) and
fus m P,m
C (l) are the molar heat capacities of the solid and liquid forms of the substance.
P,m
Since the standard pressure is 1 bar, each term in (5.29) is for a pressure of 1 bar.
Thermodynamic properties of solids and liquids change very slowly with pressure
(Sec. 4.4), and the difference between 1-bar and 1-atm properties of solids and liquids
is experimentally undetectable, so it doesn’t matter whether P is 1 bar or 1 atm in
(5.29). At 1 atm, T is the normal melting point of the solid (Sec. 7.2).
fus
Frequently a solid undergoes one or more phase transitions from one crystalline
form to another before the melting point is reached. For example, the stable low-T
form of sulfur is orthorhombic sulfur; at 95°C, solid orthorhombic sulfur is trans-
formed to solid monoclinic sulfur (whose melting point is 119°C); see Fig. 7.9a. The