Page 393 - Physical Chemistry

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Chapter 12 The two freezing-point curves intersect at point E. For a solution with x to the
B
Multicomponent Phase Equilibrium left of E, solid C will freeze out as T is lowered. For x to the right of E, solid B will
l
B
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freeze out. At the values of T and x corresponding to point E, the chemical potentials
B
of B and C in the solution equal the chemical potentials of pure solid B and C, re-
spectively, and both B and C freeze out when a solution with the eutectic composition
x is cooled. Point E is the eutectic point (Greek eutektos, “easily melted”).
B
The portion of line DE with x very close to 1 can be calculated from the ideally
B
dilute–solution equation (12.14) with A and B replaced by B and C, respectively.
Likewise, the portion of line AFGE with x very close to 0 (and hence x very close
B C
to 1) can be calculated from (12.14) with A and B replaced by C and B, respectively.
Away from the ends of these lines, Eq. (12.7) with A replaced by B or C applies. This
exact equation is hard to use to find the freezing point T as a function of x or x . To
f B C
get a rough idea of the shape of the curves DE and AE, we neglect the temperature de-
pendences of H and H ; and we approximate g and g as 1 over the full
fus m,B fus m,C B C
range of solution composition (this ideal-solution approximation is usually quite
poor). Thus the very approximate equation for DE is
1 1
R ln x ¢ H m,B a b for DE (12.46)
B
fus
T* T
B
The approximate equation for AE is obtained by replacing B with C in (12.46).
Suppose we start at point R in Fig. 12.19 and isobarically cool a liquid solution of
B and C with composition x
. The overall composition of the closed system remains
B
constant at x
, and we proceed vertically down from R. When T reaches T , solid C
1
B
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starts to freeze out. As C freezes out, x increases and (since B is the solute here) the
B
freezing point is lowered further. To freeze out more of the solvent (C), we therefore
must lower the temperature further. At a typical temperature T , there is an equilibrium
2
between a solution whose composition is given by point G as x and solid C, whose
B
composition is given by point I as x 0. As usual, the points at the ends of the tie
B
line (GHI) give the compositions of the two phases in equilibrium. The lever rule gives
s
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s
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n HI (n n )HG , where n is the number of moles of solid C in equilibrium with
C B C C
a solution of n moles of B plus n moles of C. At point F, the lever rule gives n
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s
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B C C
0. As T drops along the line FHK, the horizontal distance to the line AFGE increases,
s
indicating an increase in n .
C
As T is lowered further, we finally reach the eutectic temperature T at point K.
3
Here, the solution has composition x (point E), and now both solid C and solid B
B
freeze out, since both solids freeze out when a solution with the eutectic composition
is cooled. The relative amounts of B and C that freeze out at E correspond to the eu-
tectic composition x , and the entire remaining solution freezes at T with no further
B 3
change in composition. At K, three phases are in equilibrium (solution, solid B, and
solid C), so the lever rule (12.41) does not apply. With three phases, we have f 2
3 2 1 degree of freedom; this degree of freedom has been eliminated by the spec-
ification that P is fixed at 1 atm. Hence there are no degrees of freedom for the three-
phase system, and the temperature must remain constant at T until all the solution has
3
frozen and the number of phases has dropped to 2. Below T , we are simply cooling a
3
mixture of solid B plus solid C. The endpoints of a horizontal tie line drawn through
point S lie at x 0 and x 1, and these are the compositions of the phases [pure
B B
C(s) and pure B(s)] present at S.
If we reverse the process and start at point S with solid B plus solid C, the first
liquid formed will have the eutectic composition x . The system will remain at point K
B
until all the B has melted, along with enough C to give a solution with eutectic composi-
tion. Then the remaining solid C will melt over the temperature range T to T . (Sharpness
3 1
of melting point is one test organic chemists use for the purity of a compound.) A solid