Physical Chemistry , 1st ed.

(Darren Dugan) #1
Solution
a.The specific chemical process that is the freezing of mercury is
Hg () →Hg (s)
which occurs at 38.9°C or 234.3 K. When the liquid phase goes to the solid
phase, heat must be lost, so the process is inherently exothermic. Therefore
transHis actually 2.33 kJ/mol, or 2330 J/mol (not the positive 2.33 kJ/mol
given for fusHfor Hg). To determine the entropy change, we have

S


 2

2

3

3

3

4

0

.3

J/

K

mol
9.94
mo

J

lK
The entropy change is negative, meaning the entropy decreases. This is what’s
expected for a liquid-to-solid phase transition.
b.The vaporization of carbon tetrachloride is represented by the reaction
CCl 4 () →CCl 4 (g)
which at normal atmospheric pressure occurs at 77.0°C, or 350.2 K. In order
to go from the liquid phase to the gas phase, energy must be put into the
system, which means that this change is inherently endothermic. Therefore
we can use vapHdirectly. For the entropy change, we have

S ^29
3

,

5

8

0

9

0

2

J

K

/mol85.35
mo

J

lK

It was noted as early as 1884 that many compounds have a vapSof around
85 J/molK. This phenomenon is called Trouton’s rule.Deviations from
Tr o u t o n’s rule are marked for substances that have strong intermolecular in-
teractions, like hydrogen bonding. Table 6.2 gives a list ofvapHand vapS
values for some compounds. Hydrogen and helium have very small entropies
of vaporization. Compounds that have strong hydrogen bonding, like water
(H 2 O) and ethanol (C 2 H 5 OH), have higher entropies of vaporization than ex-
pected. Table 6.2 also lists fusHand fusSvalues for these compounds.

6.4 The Clapeyron Equation


The previous discussion detailed general trends in the behavior of equilibria.
In order to get more quantitative, we need to derive some new expressions.
Equation 6.3, when generalized, states that the chemical potential of two
phases of the same component are equal at equilibrium:
phase1phase2
By analogy to the natural variable expression for G, at a constant total amount
of substance the infinitesimal change in ,d, as pressure and temperature
change infinitesimally is given by the equation
dSdTVdp (6.8)
(Compare this to equation 4.17.) If the multiphase equilibrium experienced an
infinitesimal change in Tor p, the equilibrium would shift infinitesimally but
would still be at equilibrium. This means that the changein phase1would
equal the change in phase2. That is,
dphase1dphase2

148 CHAPTER 6 Equilibria in Single-Component Systems

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