Page 371 - Physical Chemistry

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Chapter 12 As noted in Sec. 10.3, measurement of solution vapor pressures enables determination
Multicomponent Phase Equilibrium of g . Use of the Gibbs–Duhem equation then gives g of the solute.
A
If the solution is very dilute, then g 1 and
A
¢P 1x 12P* ideally dil. soln., nonvol. solute (12.3)
A
A
For a single nondissociating solute, 1 x equals the solute mole fraction x and
A
B
P x P*. Under these conditions, P is independent of the nature of B and de-
B A
pends only on its mole fraction in solution. Figure 12.1 plots P versus x for su-
B
crose(aq) at 25°C. The dotted line is for an ideally dilute solution.

12.3 FREEZING-POINT DEPRESSION AND
BOILING-POINT ELEVATION
The normal boiling point (Chapter 7) of a pure liquid or solution is the temperature at
which its vapor pressure equals 1 atm. A nonvolatile solute lowers the vapor pressure
(Sec. 12.2). Hence it requires a higher temperature for the solution’s vapor pressure to
Figure 12.1 reach 1 atm, and the normal boiling point of the solution is elevated above that of the
pure solvent.
Vapor-pressure lowering P
versus sucrose mole fraction for Addition of a solute to A usually lowers the freezing point. Figure 12.2 plots m A
aqueous sucrose solutions at 25°C for pure solid A, pure liquid A, and A in solution (sln) versus temperature at a fixed
(solid line). The dotted line is for pressure of 1 atm. At the normal freezing point T*of pure A, the phases A(s) and A(l)
f
an ideally dilute solution.
are in equilibrium and their chemical potentials are equal: m* m* . Below T*,
A(l)
A(s)
f
pure solid A is more stable than pure liquid A, and m* m* , since the most stable
A(l)
A(s)
pure phase is the one with the lowest m (Sec. 7.2). Above T* , A(l) is more stable than
f
A(s), and m* m* . Addition of solute to A(l) at constant T and P always lowers
A(s)
A(l)
m (Sec. 12.1), so m A(sln) m* at any given T, as shown in the figure. This makes the
A
A(l)
intersection of the A(sln) and A(s) curves occur at a lower T than the intersection of
the A(l) and A(s) curves. The solution’s freezing point T (which occurs when m A(sln)
f
m* , provided pure A freezes out of the solution) is thus less than the freezing point
A(s)
T*of pure A(l). The lowering of m stabilizes the solution and decreases the tendency
f
A
of A to escape from the solution by freezing out.
We now calculate the freezing-point depression due to solute B in solvent A. We
shall assume that only pure solid A freezes out of the solution when it is cooled to
its freezing point (Fig. 12.3). This is the most common situation. For other cases, see
Sec. 12.8. The equilibrium condition at the normal (that is, 1-atm) freezing point is
that the chemical potentials of pure solid A and of A in the solution must be equal. m A
in the solution is m A(sln) m° RT ln a m* RT ln a [Eqs. (10.4) and (10.9)],
A(l)
A
A
A(l)
Figure 12.2
where m* is the chemical potential of pure liquid A and a is the activity of A in the
A(l)
A
Chemical potential of A as a solution. Equating m* and m A(sln) at the solution’s normal freezing point T , we have
f
A(s)
function of T (at fixed P) for pure
solid A, pure liquid A, and A in m* 1T , P2 m A1sln2 1T , P2
f
A1s2
f
solution (the dashed line). The
lowering of m by addition of m* 1T , P2 m* 1T , P2 RT ln a
f
A
f
f
A1s2
A1l2
A
solute to A(l) lowers the freezing
point from T* to T . f where P is 1 atm. The chemical potential m* of a pure substance equals its molar
f
Gibbs energy G* [Eq. (4.86)], so
m
G* 1T 2 G* 1T 2 ¢ G m,A 1T 2
fus
f
m,A1s2
f
f
m,A1l2
ln a (12.4)
A
RT f RT f
where G m,A G* G* is G for fusion of A. Since P is fixed at 1 atm,
m
m,A(s)
m,A(l)
fus
the pressure dependence of G* is omitted.
m