Physical Chemistry week 6 Monday February 11, 2013 page 1
ΔP = -χBP*A A is solvent.
B is solute.
Conclusion: ΔP is independent of the nature of A and depends only on its mole fraction (i.e. the
number of solute particles).
II freezing point depression and boiling point elevation
Consider a graph with ΔP on the vertical axis and χB on the horizontal axis. A dotted line with a negative
slow represents an ideally dilute solution. A solid line that starts at the same left side point and follows
it for a while then curves down represents a real solution. This means at low concentration a real
solution behaves like an ideally dilute solution.
All colligative properties depend on quantity not the nature of the solute particles.
∂Gi,m
∂μi
(
) = ( ) = -Si,m
∂T P
∂T P
dG = -SdT + VdP
At constant P:
dG = -SdT
∂G
( ) = -S
∂T P
Consider a graph with µ*H2O on the vertical axis and T on the horizontal axis. From the left side near the
top is a straight line with a slightly negative slope. This is the line for the solid state. To the right is a line
with a more negative slope. This is the line for the liquid state. Where they intersect is Tf*. Tf* is the
freezing point or melting point. To the right is a line with an even more negative slope. This is the line
for the gaseous state. Where this line intersects with the line for the liquid state is Tb*. Tb* is the boiling
point. As seen above, the slope of each of these lines is –S. At Tf*: µ*solid=µ*liquid. At Tb*: µ*liquid=µ*gas. In
the solid region: µ*solid<µ*liquid. In the liquid region: µ*liquid<µ*solid. In the gas region: µ*gas<µ*liquid. As T↑,
S↑ so the slope gets more negative (steeper).
Now consider a graph just like the previous one, but with a dotted line below and parallel to the line for
the liquid state. The original line was for the pure H2O. The dotted line is for the H2O after adding some
solute. Adding solute doesn’t change the slope of the line, just shifts it. The original freezing point was
T*f. The dotted line intersects the line for the solid state farther to the left, so the freezing point for the
solvent in solution Tf is left of the original T*f. This is called freezing point depression. The original
boiling point was T*b. The dotted line intersects the line for the gaseous state farther to right, so the
boiling point of the solvent in solution is right of the original T*b. This is called boiling point elevation.
Boiling point elevation is not caused by energy. It happens even when ΔHmix=0. ΔHmix=0 for ideal
solutions. Boiling point elevation is caused by entropy. Adding solute increases the entropy of the
solution, which decreases the chemical potential in the solution, so there’s less reason for molecules to
escape into the gas phase. ΔTf is usually larger than ΔTb since the solid curve has the lowest slope.
µA = µ*A + RTlnχA
χA=1 for pure solvent
χA<1 after adding solute
T>0
lnχA<0
so µA<µ*A
µ*A is constant at given T
For ideally dilute solution:
H̄ A = H*m,A
S̄A = S*m,A+RTlnχA
S̄A>0 µA = H̄ A – TS̄A and S̄A >0 so µA decreases
Derivation of formula:
Assumption: Only pure A (the solvent) freezes out when the solution freezes. That’s why ice is pure
water, even if it freezes out of a salt water ocean.
µA(soln)=µ*A(s)
µA(soln)=µ°A(liquid)+RTlnaA=µ*A,l+RTlnaA
ai = γiχi
For ideal or ideally dilute solution: γi=1
µ*A,l+RTlnaA=µ*A,s
µ*A,l(Tf,P)+RTflnaA = µ*A,s(Tf,P)
P = 1 atm
lnaA = -
*
*
GA,m,s
-GA,m,l
(Tf )
RTf
For s⇌l: ΔGfus=G*m,A,l-G*m,A,s
From Gibbs equations:
(
∂G*
) =-S*
∂T P
∂∆Gm
∴(
) = -∆Sm
∂T P
∂lnaA
∂
1
∂ 1 ∆Gm,fus,A (Tf )
(
) = - (∆Gm,fus,A (Tf )) +
( )
∂Tf
∂T
Tf ∂Tf Tf
RTf2
Common denominator and split ΔG into ΔH and ΔS:
∂lnaA
Tf ∆Sm,fus,A (Tf )+∆Hm,fus,A -Tf ∆Sm,fus,A
(
) =
∂Tf P
RTf2
∂lnaA
∆Hm,fus,A
(
) =
∂Tf P
RTf2
Notice the similarity between the previous equation and the Vant Hoff equation in chapter 6.
dlnk °P ∆H °
=
dT
RT 2
Multiply by dT to integrate:
2
2
∫ dlnaA = ∫
1
State 1: Pure A:
State 2: aA,2=aA
Tf,1=T*f
1
∆Hm,fus,A
dTf
RT 2
aA,1=1
Tf,2=Tf
ln (
2
aA,2
∆Hm,fus,a (Tf )
dTf
)= ∫
aA,1
RTf2
1
Tf is a dummy integration variable. Call it T.
aA = γAχA
χA = 1-χB for only 1 solute, assuming no dissociation of B, no association of B, B is not electrolyte.
lnχAγA = lnχA + lnγA = ln(1-χB)+ lnγ
Chapter 8 has Taylor series.
lny about y=1:
lny= (y-1)-
(y-1)2 (y-1)3 (y-1)4
+
…
2!
3!
4!
ln(1-χB )= -χB lnγA = B2 χ2B +B3 χ3B +…
χ2B χ3B
- …
2! 3!
B=B(T,P) B ' s are coefficients
From statistical mechanical theory of solution:
1
lnγA χA = -χB + (B2 - ) χ2B +….
2
Since χB is small, χB2 and so on can be ignored.
lnγAχA ≈-χB
continued tomorrow