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**Unformatted text preview: **< P*A.
This lowering of vapor pressure is responsible for both lowering the freezing point
as well as raising the boiling point of the solvent.
Freezing Point Depression:
For the equilibrium between the liquid and solid phases, we require µlA =µsA, or
l ,l + RT ln x A = l s, from which we get
A
A
l s − l ,l −D fus G A,m
, at constant T and P.
ln x A = A RT A =
RT
Differentiating both sides with respect to T, we get
éæ
öù
D fus H m
d ln x A −1 ê ¹ è D fus G A,m /T ø ú
= Rê
ú = RT 2 , at the equilibrium temperature.
dT
¹T
ê
ú
ë
ûP
Now we rearrange and integrate both sides, the LHS from a mole fraction of xA =
1 (pure solvent) to the solvent mole fraction in solution, xA < 1, and
correspondingly, the RHS from a liquid-solid equilibrium temperature of T*f
(freezing point of the pure solvent) to T, the freezing point of the solution. This
yields
DH
1
1
ln x A = fus m æ T − T ö .
÷
(5.115)
R çf
è
ø
We may re-write this as
D fus H m æ T − T ö
ln(1 − x B ) = R ç T T f ÷ .
èf
ø (5.116) We now make two simplifications: for small xB, ln(1–xB) ≈ –xB (see Eq. 5.117) and
T*f Tf ≈ T*f2. T...

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