Docsity
Docsity

Prepare for your exams
Prepare for your exams

Study with the several resources on Docsity


Earn points to download
Earn points to download

Earn points by helping other students or get them with a premium plan


Guidelines and tips
Guidelines and tips

Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields, Study notes of Physical Chemistry

An overview of coulomb's law, electrostatics, electric fields, and electrostatic potential. The lecture discusses various intermolecular interactions, including charge-charge, charge-dipole, dipole-dipole, and dipole-quadruple interactions. Coulomb's law is presented, and the calculation of electrostatic interaction between two charged particles is explained. The document also covers the concept of force, electric field, and potential, and their relationships.

Typology: Study notes

Pre 2010

Uploaded on 08/07/2009

koofers-user-xj6-1
koofers-user-xj6-1 🇺🇸

10 documents

1 / 4

Toggle sidebar

Related documents


Partial preview of the text

Download Coulomb's Law and Electrostatics: Intermolecular Interactions and Electric Fields and more Study notes Physical Chemistry in PDF only on Docsity! 1 Lecture 35 Chapt 20 Coulomb’s Law (electrostatics) Announce: • Give exam distribution Outline: Coulomb’s Law Electric Field Electrostatic Potential Intermolecular Interactions Okay, for this week we are changing our focus to talk about various kinds of intermolecular interactions. Let’s look at the strongest interactions between molecules – electrostatics. (nuclear forces are stronger, but that’s about it.) Electrostatics includes charge-charge interactions, charge-dipole, dipole-dipole, dipole-quadruple, etc. (often described as monopole/multipole interactions) Coulomb’s law Let’s start with Coulomb’s law which everyone remembers from physics ( ) r qqru BA 04 1 πε = u is the interaction energy (units J = V⋅C) (note W = J/s = V⋅C/s = V⋅A) ε0 is the permittivity of vacuum (units F/m = C/V⋅m = C2/J⋅m) So, we can calculate the electrostatic interaction between two charged particles. In practice this is a difficult calculation. Why? Because the interaction falls off very slowly with distance ~ r−1 Text does an example of NaCl crystal. If you just look at one line of charges that surround a central Na+:       +−+−+−= L 6 2 5 2 4 2 3 2 2 2 1 2 4 1 2 0 a eu πε 2 we get something that converges VERY slowly. In this case there is a regular arrangement of charges, so we can find a series approximation that helps us out [defines the Madelung constant]. e.g. a eNU A 2 04 1747.1 πε −= This is great for cyrstalline solids, but for more general charge distributions (like dissolved salts, amorphous materials, or proteins you have to try to just add them all up. Fortunately, in most media charges are shielded by the surroundings. This is especially important in water. We can write Coulombs law the way it is really defined ( ) r qqru BA πε4 1= where ε = ε0D where D is the dielectric constant D ≡ 1 for vacuum D = 1.00059 ≈ 1 for air D ≈ 2 for hydrocarbons D ≈ 1.4 often used for proteins D = 33 MeOH D = 78.54 for water Book mentions Bjerrum length, which is a handy tool. Remember our golden tool for connecting microscopic properties to macroscopic behavior? Partition function. Energy always appears as E/kT or E/RT depending on units. The Bjerrum length is just the distance two charged bodies are apart that corresponds to U = RT. Book points out that Bjerrum length is 80 times shorter (i.e. a factor of D, 560 Å versus 7.13 Å) for NaCl dissolved in water than for two bare Na+ Cl− particles. Their interaction is severely weakened by the presence of the water. If we had many charges, we would just add up the u for each of them, until we had accounted for everybody. However, there are easier ways, which we will work up to now.
Docsity logo



Copyright © 2024 Ladybird Srl - Via Leonardo da Vinci 16, 10126, Torino, Italy - VAT 10816460017 - All rights reserved