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Unformatted text preview: Electrostatics and the thermodynamics of solutions Phillip Geissler March 19, 2010 We come together Cuz opposites attract. Paula Abdul Opposite charges attract one another; like charges repel. This simple and familiar fact plays a central role in determining the structure of molecules and the forces that act between them. Here, we will explore some of the most important effects of electrostatic forces on the thermodynamics of dilute solutions, including the very large solvation energies of ions in polar solvents and the screening of charge by counterions in salt solutions. 1 T h e b a s i c l a w o f e l e c t r o s t a t i c s The force between two charges acts along the line connecting them, as shown in Fig. 1. Mathematically, it can be written F = zz 4 π( r 2 ˆ r . (1) (Throughout these notes vector quantities such as force and position will be written in bold face.) Here, z and z specify the magnitude and sign of the charges. As you may recall, the strength of this “Coulomb force” is inversely proportional to the square of the distance r between charges. Its direction is either parallel (for opposite charges) or antiparallel (for like charges) to the separation vector r . (Whether r should point from z to z or instead from z to z depends on whether we are calculating the force on z or instead the force on z . Just keep in mind that opposite charges attract and like charges repel, and you shouldn’t get confused.) This direction is indicated in Eq. 1 by the unit vector ˆ r = r /r , which is parallel to r but has magnitude  ˆ r  =  r  /r = 1. The “permittivity of free space” ( ≈ 8 . 85 × 10 12 C 2 / J m is a fundamental physical constant that sets the scale of electrostatic forces and energies. To make our equations easier on the eyes, we will bundle the factor 4 π( into the charges, q = z √ 4 π( , q = z √ 4 π( With this new “unit of charge”, we have F = qq r 2 ˆ r . (2) When you plug numbers into the formulas in these notes, don’t forget to account for the factor √ 4 π( wherever charge appears. 2 P o t e n t i a l e n e r g y Force tells us how much work ¯ dw = F · d r is required to “push” something over a small displacement d r . For the situation sketched in Fig. 1, moving one charge z from an initial position r 1 , along the direction of 1 r z z F Figure 1: The Coulomb force F exerted by one charge z onto another charge z of the same sign is parallel to the vector r pointing from z to z . separation ( d r = ˆ r dr ), to a new position r 2 , while keeping the other charge z fixed at the origin, requires an amount of work w = F · d r = r 2 r 1 dr qq r 2 = qq 1 r 2 1 r 1 . Since there is no possibility of heat flow in our twoparticle system, we can immediately identify the change in electrostatic energy as Δ E = w . Furthermore, since we have made no changes in velocity, only potential energy u is involved, Δ E = u ( r 2 ) u ( r 1 ). The electrostatic potential energy for a pair of charges is evidently)....
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 Spring '09
 JAMESAMES
 Physical chemistry, pH, Electrostatics, Energy, Electric charge, Ion

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