Why does depression of freezing point takes place if a non-volatile solute is added to the solution?

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Depression of freezing point

  • $\begingroup$ Better not to mention the link as for it is totally unclear and closed because that. $\endgroup$
    – Alchimista
    Jan 9, 2019 at 14:48
  • $\begingroup$ Why is the entropy of mixing not a suitable enough explanation? $\endgroup$
    – Jon Custer
    Jan 9, 2019 at 15:55
  • $\begingroup$ @JonCuster I think it is, as I just argued in my answer, but entropy is not a very intuitive concept for beginners. $\endgroup$
    – Zhe
    Jan 9, 2019 at 16:01
  • $\begingroup$ @Zhe - or, better put perhaps, intuition is often not very useful and highly variable from person to person. I just hate questions asking for 'intuition' since they are really unanswerable. $\endgroup$
    – Jon Custer
    Jan 9, 2019 at 16:03
  • $\begingroup$ Yes, for some folks saying that the Gibbs energy of reaction is concentration dependent and that the concentration of liquid water changes is quite intuitive. Or you could say K does not change, but the concentration of water does, so it is no longer at equilibrium. Short and to the point if you are familiar with the concepts. $\endgroup$
    – Karsten
    Jan 9, 2019 at 18:52

2 Answers 2


Imagine a pure liquid. If part of it freezes, what happens to the rest of the solution? Well, it looks exactly the same since it's also pure, so nothing.

However, this is not true for the system you've described.

What happens if there's a solute mixed in liquid? When, now you're concentrating the solute in the remainder of the liquid, so you're creating an entropic barrier to freezing the liquid.

Freezing is a process with $\Delta H < 0$ and $\Delta S < 0$. This means that with $\Delta G = \Delta H - T\Delta S$, the process is spontaneous when $T$ balances the favorable enthalpy change against an unfavorable entropy change. This $T$ is the freezing point. However, notice that due to the above argument, $\Delta S$ is now more negative. This means that the same balance will be achieved with a smaller value of $T$, hence, a freezing point depression.


Here is a - hopefully intuitive - explanation based on kinetics:

I start with a glass of water with ice cubes in it, sitting outside in winter on a day where the temperature happens to be equal to the freezing point of pure water. The system is at equilibrium, meaning that the rate of (solid) water molecules melting is equal to the rate of (liquid) water molecules freezing. Basically, some water molecules wiggle free from the surface of the ice, while some water molecules from the liquid gently bump into the surface of the ice and attach.

Now, I add a solute to the liquid water. The melting of the ice will not be affected because the ice is still made of pure water. However, some of the molecules of the liquid bumping into the surface of the ice can no longer attach because they are solute molecules and they don't fit. So the melting rate stays the same, while the freezing rate decreases (you could say because the concentration of liquid water got smaller by adding the solute).

As a net effect, the ice cubes melt, getting smaller. This means the freezing point of the solution is lower than that of pure water.

  • $\begingroup$ So your telling due to interaction between solute and solvent molecule depression for ex. (masterconceptsinchemistry.com/wp-content/uploads/2017/10/…) but depression in freezing point is a colligative property it does not depend on interaction between solute and solvent molecule could you clarify $\endgroup$
    – Chemist
    Jan 13, 2019 at 6:45

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