Enthelpy of vapourisation vs boiling point

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SUMMARY

The discussion centers on the comparison of the enthalpy of vaporization (ΔHvap) of ethanol and water, with ethanol having a ΔHvap of 43.5 kJ/mol compared to water's 41.3 kJ/mol. Despite ethanol's higher ΔHvap, water has a higher boiling point of 373 K versus ethanol's 352 K. The boiling point is determined by the relationship ΔH = TΔS, indicating that the boiling point is influenced more by the ratio of enthalpy to entropy rather than enthalpy alone. The conditions for boiling include equal pressure and temperature between the liquid and gas phases, as well as equal chemical potential.

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Miffymycat
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Ethanol has a higher ΔHvap than water (43.5 vs 41.3 kJ/mol). But water has a higher boiling point (373 vs 352K). How do we explain this?!
 
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Miffymycat said:
Ethanol has a higher ΔHvap than water (43.5 vs 41.3 kJ/mol). But water has a higher boiling point (373 vs 352K). How do we explain this?!

As the boiling point is determined by Delta H=T Delta S, the vapourisation entropy is probably even higher for alcohol than for water as compared to enthalpy.
 
Enthalpy of vapourisation of a substance is the amount of energy needed for the substance to change its state from liquid to gas.

Boiling point is the temperature at which the particles has enough kinetic energy to overcome the molecular forces of attraction between them and thus can escape into the air ( surroundings ).

I hope that's clear enough.
 
Thanks guys

Right. But even under a vacuum, liquids can still have a highish boiling point. I still find it hard to visualise the difference between the kinetic energy to overcome the IMF's by heating to the boiling point - at which point the forces between them are sufficiently low to enable the substance to overcome any external pressure and leave the liquid and the extra energy required to change the state ie enthalpy of vapourisation! What is this extra for? There no forces to overcome now!

Thank for your patience!
 
Miffymycat said:
Thanks guys

Right. But even under a vacuum, liquids can still have a highish boiling point. I still find it hard to visualise the difference between the kinetic energy to overcome the IMF's by heating to the boiling point - at which point the forces between them are sufficiently low to enable the substance to overcome any external pressure and leave the liquid and the extra energy required to change the state ie enthalpy of vapourisation! What is this extra for? There no forces to overcome now!

Thank for your patience!

You are completely right. At any temperature and pressure a certain fraction of the molecules has the sufficient amount of energy to leave the liquid and go to the gas phase.
The boiling point (of a pure substance) is characterized by three conditions:
1. The pressure of the liquid and of the gas phase are the same.
2. The temperature of the two phases is the same.
2. The chemical potential of the liquid and the gas are the same.

The last condition can be written as
\mu_g-\mu_l=\Delta G_\mathrm{vap}=\Delta H_\mathrm{vap}-T_\mathrm{vap}\Delta S_\mathrm{vap}=0
or T_\mathrm{vap}=\Delta H_\mathrm{vap}/\Delta S_\mathrm{vap}.
Hence the boiling temperature is not determined by the vaporisation enthalpy but rather by the quotient of the enthalpy and the entropy.
 

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