Freezing Point Depression and Vapor Pressure Lowering
Date: 2001 - 2002
Freezing Point Depression and Vapor Pressure Lowering
will freezing pt. be lowered if the solution shows positive deviation to
Is this due to the unfavorable entropy change required as the solution
separates as it freezes?
AP Chem 1994 Multiple Choice Question #14 asks
Which of the following is lower for a 1.0 molar aqueous solution of ANY
solute than it is for pure water?
The correct answer is supposedly Freezing Point
Vapor Pressure is an incorrect choice.
(Follow up Question below)
The question(s) confuse me. First, deviations from Raoult's law affect the
depression of the freezing point quantitatively, but not qualitatively.
Second, I am not sure what the terms "unfavorable entropy changes ... "
means. Third, vapor pressure IS a correct answer to the question as the
question is stated in this inquiry. Now let's address the effect of solutes
on the physical properties of a solvent. Only the case of liquid-phase
soluble/ solid and gas-phase insoluble, non-volatile, totally dissociated
solutes is considered. These restrictions affect the magnitude of the
effects, and involve more complicated computations, but they do not affect
the qualitative direction of the effects. Consider the following general
principles of thermodynamics:
*** The addition of a liquid-soluble substance to a liquid phase ALWAYS
decreases the vapor pressure of liquid solution (P) relative to that of the
pure liquid (Po). Raoult's Law:
P = Po * X (where X is the mole fraction of the solvent) is the quantitative
statement of that thermodynamic principle.
*** At temperatures less than the melting point of a pure solid, the vapor
pressure of the pure super-cooled liquid is ALWAYS greater than that of the
pure solid at the same temperature. The vapor pressure of the pure solid and
liquid phases are equal AT the melting point.
Both these principles are direct consequences of the Second Law of
Thermodynamics. To my knowledge there are no exceptions, given the
conditions imposed above.
Since the addition of a liquid-soluble, solid-insoluble solute lowers the
vapor pressure of the liquid solution, the temperature of the solid and the
solution must decrease until the vapor pressure of the pure solid is equal
to the vapor pressure of the solution in thermal contact with that solid.
This is the freezing point depression. The freezing point of the solid +
solution is less than the freezing point of the pure solid.
Now consider the vaporization process. At the normal boiling point of the
pure liquid, the vapor pressure of the liquid, Po = 1 atm. (One could elect
some other fixed pressure and the temperature corresponding to that
pressure, but that only affects the numbers, not the direction of the
effect.). The addition of a solute to the liquid again ALWAYS LOWERS the
vapor pressure of the solution relative to the vapor pressure of the pure
liquid at the same temperature. In order to bring the vapor pressure of the
solution back up to the applied pressure Po = 1 atm. (or some other fixed
pressure), the temperature of the solution must be increased by an
appropriate amount so that the vapor pressure is equal to the applied
pressure. This is the ELEVATION of the boiling point. In the idealized case,
Raoult's Law ( P = Po * X ) applies. Positive or negative deviations from
Raoult's Law only affect the quantitative "details", but any such deviations
do not alter the basic principles of the lowering of the vapor pressure of
the solution relative to the pure liquid and the concomitant increase in the
boiling point of the solution compared to the pure liquid. This is the
boiling point ELEVATION.
So, I contend that BOTH the equilibrium freezing point and vapor pressure of
a liquid are decreased upon addition of a non-volatile,
liquid-soluble/solid-insoluble solute to the liquid phase.
This is pretty strange. The addition of ANY solute to pure water will
definitely depress the freezing point. However, it will also increase the
boiling point (lower vapor pressure).
Now please review the table towards the bottom of the URL at the web site
that I have attached:
** If you will scroll down to the table that shows the relative Kb's and
Kf's of various solvents you will notice that water's Kf is greater than its
Kb. For a 1 molal solution the melting point temperature DEPRESSION for the
solution will be 1.86 C (lower than freezing point of pure water). The
boiling point ELEVATION for the solution will be 0.52 C (higher than the
boiling point of pure water.) A fair set of multiple choice answers would
include T (boiling point) and exclude vapor pressure. Of course, Tfp would
be correct, and Tbp would be incorrect to the question of "Which one is
With this being said, the AP Chem tests two answers do not make sense since
BOTH the vapor pressure and the freezing point will be decreased. Also it
is a silly comparison since the units of vapor pressure are in [ pressure
units, mm Hg, psi, atm, etc... ] and the units of freezing point are in
temperature units C, K, F, R ]
My response does not speak to your first question dealing with the effects
of the solution as it freezes. Sorry.
I hope you have made some sense of all this information.
Thanks for all of the great questions.
If a solute is volatile and has a weak attraction for water, wouldn't that
show an increased vapor pressure (a positive deviation to Raoult's Law?)
When a solute is volatile, its vapor pressure must be taken into account
since the total vapor pressure of the solution will be the sum of the
partial pressures Ptotal = Pa + Pb = Poa*Xa + Pob*Xb = Poa*Xa + Pob*(1 - Xa)
for a two component solution of 'a' and 'b' that obeys Raoult's Law. Such
solutions are referred to as ideal.
Predicting deviations from ideality on the basis of "simple" forces is not
reliable, because many factors enter into the picture in addition to like
polarities etc. Differences in molecular size (even if all intermolecular
forces are equal) result in deviations from Raoult's Law. In addition,
intermolecular forces in the liquid state are very complex and sometimes
very specific interactions occur that are not at all obvious from just
looking at the polarity or the dipole moment of the components etc.
especially when water is involved. If you have access to a chemistry library
you can look up "theory of real solutions" in a thermodynamics book as see
just how messy things can be. The website:
also will give you some insight into the complexity of the affairs.
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Update: June 2012