Page 66 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch06 Elementary Physical Chemistry

Phase and Chemical Equilibria 51

Comment: The way the chemical potential behaves is similar to the
behavior of the free energy. Thus, we have ∆µ ≤ 0, where the equal
sign pertains to equilibrium and the unequal sign to non-equilibrium
situations. If a system can exist in two different states, the most
favored state is the one with the lowest µ. When liquid and vapor
∗
are at equilibrium (coexist), say at the normal boiling point, T ,
b
the chemical potentials of the two phases must be equal. Above the
normal boiling temperature, the vapor phase is the more stable state,
and the chemical potential of the vapor will be lower than that of the
pure liquid. Below the normal boiling temperature, the liquid phase
is the stable phase, and the chemical potential of the liquid will be
the lowest. Similar considerations apply to the chemical potentials
responsible for the lowering of the freezing point (see Fig. 6.4).

Let us denote the normal boiling point and normal freezing point of the
∗
∗
pure liquid respectively as T and T and the boiling and freezing points
b
f
of the actual solutions as T b and T f .We can write
∗
∆T b = T b − T = m bK b (6.14a)
b
∗
∆T f = T − T f = m f K f (6.14b)
f
where m b is the molality of solute and K b and K f are respectively the
boiling point elevation constant and freezing point depression constant.
(These constants can readily be related to thermodynamic functions, but
will not be developed in this course.)
3
Example 6.3. Calculate the boiling point of a solution of 250 cm of
H 2 O(l) containing 7.5 g of a solute. The normal boiling point of H 2 O(l)
is 373.25 K. The boiling-point elevation constant, K b is 0.51K kg mol −1 .
The molecular weight of the solute is M = 342 g mol −1 .

Solution
3
7.5 g in 250 cm of water is equivalent to 30.0g of solute in 1 L of water. The
mole fraction of 30.0 g of solute in 1 L of H 2 O(l) is m =30.0 g/342g mol −1 =
8.77 × 10 −2 mol and thus the molarity of the solute is 8.77 × 10 −2 mol/L.
Assuming that 1 L of water at the normal boiling point weighs essentially
1 kg, we can equate the molarity m to the molality, m b ,and write m b =