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.}}}$

Freezing point depression Δ T in a p , T diagram

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:

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