December 27, 2024, 07:04:29 AM
Forum Rules: Read This Before Posting


Topic: Energetic "Cost" of molecular dipole moment?  (Read 2812 times)

0 Members and 1 Guest are viewing this topic.

Offline spirochete

  • Chemist
  • Full Member
  • *
  • Posts: 547
  • Mole Snacks: +51/-9
  • Gender: Male
Energetic "Cost" of molecular dipole moment?
« on: October 06, 2011, 06:51:19 PM »
As more of a qualitative organic chemist, it's my understanding that molecules prefer to have conformations that minimize dipole, especially in non polar solvents and the gas phase.  However I have no idea how to actually put numbers on this.  Is there an equation that relate dipole moment to some kind of free energy or enthalpy difference?  An equation that also includes the influence of solvent polarity and/or other relevant factors would be even more helpful. 

This question is not for any class, just for my own interest.

Offline Enthalpy

  • Chemist
  • Sr. Member
  • *
  • Posts: 4036
  • Mole Snacks: +304/-59
Re: Energetic "Cost" of molecular dipole moment?
« Reply #1 on: October 10, 2011, 04:23:18 PM »
The electrostatic energy can be computed, at least the difference between two conformations:
the charge at one place creates an electrostatic potential everywhere. Where the other(s) charge is, multiply by this charge (normally less than q) to get an energy. Move the other charge between both conformations, get an energy difference.

Then, convert the electrostatic energy into what thermodynamic value you want.

I don't believe the electrostatic energy is defined by the sole electric dipole, because moving 1*q over 2nm to create the dipole takes less energy than moving 2*q over 1nm, and because the same charge spread over a bigger group of atoms needs less energy - telling why a phenyl can more easily accept or give some charge.

Sponsored Links