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At the freezing point of a substance, the solid phase is in dynamic equilibrium with the liquid phase. My textbook defines the freezing point of a substance as: "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase". Why is this so? How does the vapor phase even come into the picture during freezing? Shouldn't it be all about the solid and the liquid?

My book also states that when we add a non volatile solute to a liquid solvent, "the solution will freeze at the temperature when its vapor pressure is equal to the vapor pressure of the pure solid solvent". This doesn't make sense to me as well. Assuming the first statement is true, shouldn't the second statement read, "...is equal to the vapor pressure of the solid solution"? Are the two statements somehow related to each other? And why are they true?

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2 Answers 2

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When two phases are in equilibrium the chemical potential of the atoms/molecules in the two phases are the same. If you're not familiar with the concept of chemical potential it basically just means that the molar Gibbs free energies of the two phases are equal so the $\Delta G$ for the phase change is zero.

The argument is:

  1. If the liquid and vapour are in equilibrium then the chemical potential in the liquid $\mu_l$ and vapour $\mu_v$ must be the same: $\mu_l = \mu_v$.

  2. If the solid is also in equilibrium with the vapour then the chemical potential of the solid $\mu_s$ and vapour $\mu_v$ are also the same: $\mu_s = \mu_v$.

  3. And that means that the chemical potential of the liquid and solid must be the same, because both are the same as the vapour: $\mu_s = \mu_l = \mu_v$.

  4. And finally if the chemical potential of the liquid and solid are the same then it means the liquid and solid are in equilibrium i.e. we are at the freezing/melting point.

Re the second paragraph: when the solvent freezes it will freeze to form pure solid solvent i.e. it excludes the solute. So we have pure solid solvent in equilibrium with the vapour and "solvent + solute" in equilbrium with the vapour. So it's the same argument as above.

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  • $\begingroup$ How is 'chemical potential' related to vapor pressure? $\endgroup$
    – Newton
    Commented Dec 29, 2016 at 11:27
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    $\begingroup$ @kalyan: the chemical potential of the vapour is a function of temperature and partial pressure. And at equilibrium the partial pressure of the vapour is equal to the vapour pressure of the solid/liquid. Since it's the same vapour in contact with the liquid and solid the partial pressure is the same for both, all everything is at the same temperature, so that implies the vapour pressure of the liquid and solid have to be the same. $\endgroup$ Commented Dec 29, 2016 at 11:56
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    $\begingroup$ Thank you so much @John, you are so helpful!!!! Cleared my doubt at once! God will always bless you! $\endgroup$
    – Selena
    Commented Feb 22, 2018 at 16:55
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What you have to remember to understand this is that everything has an interface, either be solid or liquid. It's at the interface (typically, but not necessarily, at the outermost interface) that it makes sense to talk about vapor pressure of a solid or a liquid.

At an interface, the atoms or molecules at a certain temperature will have a higher probability of bonding/breaking from the rest of the solid or liquid. That probability depends on the temperature and can be regarded as the vapour pressure. There is no problem in calling it vapour pressure because in fact you are talking about single atoms or molecules.

The freezing point is defined as "the temperature at which the vapor pressure of the substance in its liquid phase is equal to its vapor pressure in the solid phase" which means that at that temperature the same amount of atoms/molecules are bonding/breaking between solid and liquid at the same rate. Above that temperature, there will be more atoms/molecules breaking bonds with the solid interface (leading to a liquid state), and at lower temperatures the opposite.

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  • $\begingroup$ You appear to be making a kinetics argument for a thermodynamic problem. $\endgroup$
    – Jon Custer
    Commented Dec 29, 2016 at 23:22
  • $\begingroup$ @JonCuster My interpretation of the question was that the misunderstanding was related to the word "vapour" and why is it called vapour pressure if we are talking about solids and liquids. From that point of view, I believed that this was the best way to explain it. If one was trying to understand what the freezing point was from a thermodynamic point of view, I agree that the answer from JohnRennie is more appropriate. $\endgroup$
    – cinico
    Commented Dec 30, 2016 at 10:17

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