Freezing-point depression

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Freezing-point depression is the decrease in temperature of the freezing point of a solution compared to that of the pure solvent. It is a colligative property: for dilute solutions of non-electrolytes, the difference in freezing point is proportional to the amount of solute.[1]

Cryoscopy is the experimental technique that uses freezing-point depression to measure the molecular weight of compounds: it is of historical interest only.[1]

Description

Solvent Kf
K kg mol−1
Acetic acid 3.90
Acetone 2.40
Aniline 5.87
Benzene 5.12
Camphor 40   
Carbon disulfide 3.8  
Chloroform 4.90
Cyclohexane 20.1  
Diethyl ether 1.79
Naphthalene 6.94
Nitrobenzene 6.90
Phenol 7.27
Pyridine 4.75
Tetrachloromethane 30   
Water 1.86
Data: Kaye & Laby Tables of Physical
& Chemical Constants[2]

The freezing-point depression ΔTf is proportional to the molality of the solution m: the proportionality constant is called the cryoscopic constant Kf.

ΔTf = Kfm

The molality of the solution is the amount of solute divided by the mass of solvent. For dilute solutions, where the mass of the solution can be approximated by the mass of solvent, the molality is equal to the mass fraction w of solute divided by its molar mass M.[Note 1] Hence, the molar mass of a solute can be calculated from the freezing-point depression by

M = KfwTf

This description only applies for dilute solutions. It also breaks down if the solute dissociates (e.g. electrolytes) or associates (e.g. acetic acid in non-polar solvents) in solution.

Derivation

We assume ideal solution behaviour, so that the chemical potential of the solvent in the solution μ(ℓ) is given by

μ(ℓ) = μ0(ℓ) + RTln(1−x)

where μ0(ℓ) is the chemical potential of the pure solvent and x is the amount fraction of solute (so that 1−x is the amount fraction of solvent). The solution freezes when the solvent in the solution is in equilibrium with pure solid solvent, so μ(ℓ) must be equal to μ0(s), the chemical potential of the pure solid solvent.

μ0(s) = μ(ℓ) = μ0(ℓ) + RTln(1−x)
ln(1−x) = [μ0(s) − μ0(ℓ)]/RT = ΔfusG/RT

From the definition of the Gibbs energy change of fusion ΔfusG = ΔfusH − TΔfusS, we can write two equations, the first for the solution and the second for the pure solvent (where T0 is the freezing point of the pure solvent):

ln(1−x) = ΔfusH/RT − ΔfusS/R
ln(1) = ΔfusH/RT0 − ΔfusS/R

The difference of the two equations, remembering that ln(1) = 0, is

ln(1−x) = (ΔfusH/R)[(1/T) − (1/T0)]

Approximating ln(1−x) ≈ −x (for x ≪ 1, i.e. dilute solution) gives

x = (ΔfusH/R)[(1/T0) − (1/T)]

As T ≈ T0 (small depression of the freezing point), we can approximate (1/T0) − (1/T) ≈ ΔT/T20, so

ΔT = (RT20fusH)x

The amount fraction of solute is equal to the molality of the solution m multiplied by the molar mass of the sovent M0, so

ΔT = (RT20M0fusH)m
Kf = RT20M0fusH

This derivation shows why freezing-point depression is proportional to the amount fraction of solute and hence, for dilute solutions, to the molality. Nevertheless, cryoscopic constants Kf are usually treated as empirical constants to be determined experimentally rather than being calculated from enthalpies of fusion.[1]

See also

Notes and references

Notes

  1. In formal terms, w = m2/(m1+m2) ≈ m2/m1 if m2 ≪ m1 (here, m2 is the mass of solute and m1 is the mass of solvent).

References

  1. 1.0 1.1 1.2 Atkins, P. W. Physical Chemistry, 4th ed.; University Press: Oxford, 1990; pp 169–70. ISBN 0-19-855283-1.
  2. Crysocopic and ebullioscopic constants and enthalpies of fusion and of evaporation of some common solvents. In Kaye & Laby Tables of Physical & Chemical Constants, 16th ed., 1995; Chapter 3.10.4, <http://www.kayelaby.npl.co.uk/chemistry/3_10/3_10_4.html>. (accessed 27 March 2011).

Further reading

External links

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