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Topic: Cryoscopy and Raoults law  (Read 3879 times)

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Offline kriggy

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Cryoscopy and Raoults law
« on: June 01, 2016, 05:03:47 AM »
Raoult´s law says, that adding non volatile compound into solvent reduces the partial vapor pressure of the solvent, therefore it increases its boiling point. But shouldnt it also increase the melting point? Im not realy sure why it works this way?
Can anyone comment briefly on this?

Thank you

Offline mjc123

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Re: Cryoscopy and Raoults law
« Reply #1 on: June 01, 2016, 05:22:51 AM »
Adding solute lowers the chemical potential of the solvent due to entropy of mixing, according to μ = μ° + RTlnx. Put simply, the liquid becomes more stable relative to the vapour and the solid, so it is easier to melt the solid and harder to vaporise the liquid.

Offline Enthalpy

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Re: Cryoscopy and Raoults law
« Reply #2 on: June 02, 2016, 07:55:24 AM »
The solute favours the liquid because this is where it dissolves. Which is just paraphrasing "entropy of mixing".

Offline kriggy

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Re: Cryoscopy and Raoults law
« Reply #3 on: June 05, 2016, 11:54:28 AM »
Im wondering why but that Im not sure I would be able to understand the answer :-D
Thank you guys for help

Offline mjc123

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Re: Cryoscopy and Raoults law
« Reply #4 on: June 06, 2016, 06:20:26 AM »
Perhaps you can start by saying why you think the melting point should increase?

Offline kriggy

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Re: Cryoscopy and Raoults law
« Reply #5 on: June 06, 2016, 04:06:19 PM »
Well Im not realy sure, im definitely not into physical chem but I was thinking that if the boiling point increases then the melting point increses too. I guess that it came into mind due to lack of knowledge in this area. I think the idea was that if the boiling point increases its due to stronger forces between the molecules making it harder to evaporate, then those same (stronger) intermolecular forces would make the solid more difficult to melt therefore increasing the melting point

Offline mjc123

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Re: Cryoscopy and Raoults law
« Reply #6 on: June 07, 2016, 04:56:50 AM »
First, you wouldn't have "the same (stronger) intermolecular forces" in the solid because the solid is pure solvent. The solute is soluble only in the liquid. (At least, that's the case we're considering for melting point depression. If the solute is soluble in solid solvent, you have a solid solution, which is a completely different kettle of fish thermodynamically.)
Second, it's nothing to do with stronger intermolecular forces making it harder to evaporate. It's because the liquid is not pure solvent, but a mixture, therefore its free energy is lower. Consider the process
A(s) + B(s)  :rarrow: A(l) + B(solv)
We can break this down into
A(s) + B(s)  :rarrow: A(l) + B(s)  :rarrow: A(l) + B(solv)
ΔG for the first step is 0 at Tm. ΔG for the second step is negative if B is soluble. So the overall ΔG is negative - liquid is favoured over solid at Tm in the presence of B. So you must lower T to get ΔG = 0 (the new equilibrium melting point).
See the attached diagram. If the vp of the liquid is lowered (as a result of its lower free energy) it intersects the vapour at a higher temp, and the solid at a lower temp.

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