September 19, 2019, 12:50:14 AM
Forum Rules: Read This Before Posting


Topic: Boiling point of cyclic esters  (Read 892 times)

0 Members and 1 Guest are viewing this topic.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 3044
  • Mole Snacks: +269/-57
Re: Boiling point of cyclic esters
« Reply #15 on: April 17, 2019, 06:26:29 AM »
While the vaporization energy is certainly important for the boiling point, Trouton's rule is only a general approximation known to fail even on banal compounds.

It neglects one important point, that the molecules can deform and rotate. This makes it more difficult to find an orientation and conformation that give big intermolecular forces to keep a molecule in the liquid.

This concept, nice for reasoning, is numerically sterile and must be approached by the entropy to put figures. The effect is also much more important on the melting point, where order matters more than on boiling.

The explanation with the fewer degrees of freedom in a cyclic compound seduces me, as indeed it explains the higher boiling point for THF versus diethyl ether. and for cyclopentane versus pentane.

And, sure, it can't be the only cause.

Offline rolnor

  • Chemist
  • Full Member
  • *
  • Posts: 668
  • Mole Snacks: +51/-3
Re: Boiling point of cyclic esters
« Reply #16 on: April 17, 2019, 07:46:06 AM »
The dipole moment for diethylether is 1,15 for THF its 1,63 this must have a large effect on both boilingpoint and meltingpoint?

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 772
  • Mole Snacks: +86/-23
Re: Boiling point of cyclic esters
« Reply #17 on: April 17, 2019, 09:52:28 AM »
This is an answer to a previous question about the existence of hydrogen bonding in your products, etc. and notably in THF.
But moreover, significant heat amounts are absorbed for the activation energy of THF autoxidation during heating.  Macroscopically, this is observed as higher vaporization enthalpy and consequently, as a higher boiling point than ought to be.

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 772
  • Mole Snacks: +86/-23
Re: Boiling point of cyclic esters
« Reply #18 on: April 17, 2019, 10:21:39 AM »
1). The interference of dipole moment was previously explained, as increasing the permanent Van der Waals forces.
2). The exemptions from the Troutons’ rule were clearly mentioned above, too.
3). Other parameters that interfere in the vaporization enthalpy are also mentined above.
4). Degrees of freedom, etc. reflect on the vaporization entropy Sv, which numerically coincides with the specific heat (or better, the molar heat capacity Cn).
So and up to the phase transition, an amount of heat is consumed in order to reach the boiling temperature Tb and which for 1 mol, is Q = CnTb or else, Q = SvTb.
And then, an additional amount of energy is consumed for the phase transition, which is equal with the vaporization enthalpy and not with the vaporization free energy otherwise, vaporization entropy is counted twice in regards to the overall energy that is needed for evaporation.
This is the same as counting the same cost twice, in accounting.

« Last Edit: April 17, 2019, 11:45:39 AM by pgk »

Offline pgk

  • Chemist
  • Full Member
  • *
  • Posts: 772
  • Mole Snacks: +86/-23
Re: Boiling point of cyclic esters
« Reply #19 on: April 18, 2019, 09:22:52 AM »
CORRECTION:
“ΔQ = CnΔTb or else, ΔQ = SvΔTb” is more correct because evaporation starts from room temperature and not from 0oK.

Sponsored Links