# Freezing point depression

Freezing point depression  ( GPE ) (also melting point depression  ( SPE ) or melting point depression  ( SPD )) describes the phenomenon that the melting point of solutions is lower than the melting point of pure liquid solvents .

## definition

For dilute solutions, the depression of the freezing point is proportional to the molality b of the particles of the dissolved substance (i.e. the concentration of all dissolved particles in the substance in moles per  kilogram of solvent and not the initial concentration of  the substance): ${\ displaystyle \ Delta T}$ ${\ displaystyle \ Delta T = E_ {n} \ cdot b}$ The freezing point per mole of dissolved substance per kilogram of solvent is reduced by a solvent-specific value.

When calculating the molality of the dissolved substances, it should be noted that e.g. B. Dissociate salts in aqueous solution . Table salt  (NaCl) breaks down e.g. B. into the ions Na + and Cl - . 1 mol of sodium chloride produces 2 mol of particles, and this value must be taken into account in the above equation. This is done by multiplying by the Van 't Hoff factor , which indicates how many particles the dissolved substance dissociates into on average: ${\ displaystyle i}$ ${\ displaystyle \ Delta T = E_ {n} \ cdot b \ cdot i}$ ${\ displaystyle E_ {n}}$ is the cryoscopic constant , which only depends on the solvent and not on the dissolved substance (with water as the solvent, this value is 1.86 ( K kg) / mol). It can be from the Raoult's law and the Clausius-Clapeyron equation derived to

${\ displaystyle E_ {n} = R {\ frac {T_ {g} ^ {2}} {L_ {S}}}}$ ,

With

• the gas constant = 8.314472 J / (mol K)${\ displaystyle R}$ • the freezing point of the solvent in K${\ displaystyle T_ {g}}$ • the specific enthalpy of fusion of the solvent in J / kg.${\ displaystyle L_ {S}}$ This relationship only applies to very dilute solutions (concentrations <0.1 mol / L ); the activity of the ions and the water must be taken into account for more concentrated solutions . Very highly concentrated solutions also have a triple point at which the salt solution freezes, before only water freezes out of the solution, the solution is increasingly concentrated.

Since the freezing point drops by exactly 1.86 K every time you dissolve one mole of particle in one kilogram of water, the associated temperature difference is also called the molar lowering of the freezing point . This effect is independent of the type of solute, it is a colligative property .

The boiling point of solutions also depends on the molality or the concentration of the dissolved substances; it increases . One speaks here of a molar boiling point increase . The cause of these effects is a lowering of the chemical potential of the solution compared to the pure solvent due to the entropy of mixing .

## Examples

solvent Freezing point in ° C Freezing point depression
in K kg / mol
water 0 −1.86
naphthalene 80.2 −6.80
chloroform −63.5 −4.68
benzene 5.5 −5.12
Camphor 179 −39.7
Ethanol −114.6 −1.99
Cyclohexane 6.4 −20.2
Carbon tetrachloride −23 −30

## background

In addition to the increase in the boiling point, the lowering of the freezing point is another consequence that is related to the reduced vapor pressure of solutions.

If a liquid mixture of the solid substance A and the solvent B is in equilibrium with the solid substance A , this results in the approach (with as chemical potential ). It still applies to the differentials . ${\ displaystyle \ mu _ {A, l} = \ mu _ {A, s}}$ ${\ displaystyle \ mu}$ ${\ displaystyle \ mathrm {d} \ mu _ {A, l} = \ mathrm {d} \ mu _ {A, s}}$ The index denotes the liquid phase, while the solid phase denotes. The total differentials can be set up from the above equation : ${\ displaystyle l}$ ${\ displaystyle s}$ ${\ displaystyle V_ {A, l} \ cdot \ mathrm {d} p-S_ {A, l} \ cdot \ mathrm {d} T + R \ cdot T \ cdot \ mathrm {d} \ ln x_ {A} = V_ {A, s} \ cdot \ mathrm {d} p-S_ {A, s} \ cdot \ mathrm {d} T}$ where is the mole fraction of the solute in the solvent. If you work at constant pressure , the equation takes the simplified form: ${\ displaystyle x}$ ${\ displaystyle (S_ {A, l} -S_ {A, s}) \ mathrm {d} T = R \ cdot T \ cdot \ mathrm {d} \ ln x_ {A}}$ The difference between the entropy of the solid and the liquid state ( ) corresponds to the molar entropy of fusion of the substance A . This size can be described as. ${\ displaystyle S_ {A, l} -S_ {A, s}}$ ${\ displaystyle \ Delta H _ {\ text {Schm.}} / T _ {\ text {Schm.}}}$ As is the melting point temperature refers to the pure solid phase. If you insert this relationship into the above equation and integrate between the limits of the temperatures and T or 1 and the mole fraction , the result is: ${\ displaystyle T _ {\ text {Schm.}}}$ ${\ displaystyle T _ {\ text {Schm.}}}$ ${\ displaystyle x_ {A}}$ ${\ displaystyle \ int _ {T} ^ {T _ {\ text {Schm.}}} {\ frac {\ Delta H _ {\ text {Schm.}}} {R \ cdot T ^ {2}}} \ mathrm {d} T = \ int _ {1} ^ {x_ {A}} \ mathrm {d} \ ln x_ {A}}$ ${\ displaystyle {\ frac {\ Delta H _ {\ text {Schm.}}} {R}} \ cdot \ left [{\ frac {1} {T}} - {\ frac {1} {T _ {\ text {Schm.}}}} \ Right] = \ ln x_ {A}}$ with and and you get the equation ${\ displaystyle x_ {A} = 1-x_ {B}}$ ${\ displaystyle \ lim _ {x_ {B} \ rightarrow 0} ^ {} \ ln (1-x_ {B}) = - x_ {B}}$ ${\ displaystyle T _ {\ text {Schm.}} - T = \ Delta T}$ ${\ displaystyle {\ frac {\ Delta H _ {\ text {Schm.}}} {R}} \ cdot {\ frac {\ Delta T} {T \ cdot T _ {\ text {Schm.}}}} = x_ {B}}$ Replacing by , where and or is, leads to the following equations when introducing molalities (with T · TT ): ${\ displaystyle x_ {B}}$ ${\ displaystyle {\ frac {n_ {B}} {n}}}$ ${\ displaystyle n \ approx n_ {A}}$ ${\ displaystyle n_ {B} = {\ frac {m_ {B}} {M_ {B}}}}$ ${\ displaystyle n_ {A} = {\ frac {m_ {A}} {M_ {A}}}}$ ${\ displaystyle _ {s}}$ ${\ displaystyle _ {s} ^ {2}}$ ${\ displaystyle \ Delta T = K_ {kry} \ cdot {\ frac {m_ {B} \ cdot 1000g} {M_ {B} \ cdot m_ {A}}}}$ With

${\ displaystyle K_ {kry} = {\ frac {M_ {A} \ cdot R \ cdot T _ {\ text {Schm.}} ^ {2}} {\ Delta H _ {\ text {Schm.}} \ cdot 1000 \ mathrm {g}}}}$ By transforming the equation, one can determine the molar mass of the dissolved substance from the observed lowering of the freezing point. The following applies:

${\ displaystyle M_ {B} = K _ {\ text {kry.}} \ cdot {\ frac {m_ {B} \ cdot 1000g} {m_ {A} \ cdot \ Delta T}} = {\ frac {M_ { A} \ cdot R \ cdot T _ {\ text {Schm.}} ^ {2}} {\ Delta H _ {\ text {Schm.}}}} \ Cdot {\ frac {m_ {B}} {m_ {A } \ cdot \ Delta T}}}$ 