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Classical Mechanics (energy, position, momentum ( mv ), time?)

Classical Mechanics (energy, position, momentum ( mv ), time?). Macroscopic — E = K + V + U. U = U tr + U vib + U rot + U el + U int + U nuc. D U = q + w. Microscopic — F = ma – Fails for small particles moving at high speeds. Quantum Mechanics —

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Classical Mechanics (energy, position, momentum ( mv ), time?)

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  1. Classical Mechanics (energy, position, momentum (mv), time?) Macroscopic — E = K + V + U U = Utr + Uvib + Urot + Uel + Uint + Unuc DU = q + w Microscopic — F = ma – Fails for small particles moving at high speeds Quantum Mechanics — A theory based on wave/particle duality used to treat the energy, and motion of particles. Effective for all particles, but required for the small/quick. More difficult computationally (which is why CM is still taught).

  2. U = Utr + Uvib + Urot + Uel + Uint+ Unuc Microscopic — Quantum Mechanics The energy of a single particle can still be thought of as the sum of its kinetic and potential energy, E = K + V. The treatment of kinetic energy is mathematically distinct from CM, whereas the treatment of potential energy uses the same equations as CM. Chapter 8 – Electrochemistry and ionic solutions Coulomb’s Law – governs forces between charges Debye-Hückel Theory – predicts nonideal behavior of ions in solution

  3. Chapter 8 – Electrochemistry and ionic solutions Coulomb’s Law – defining the force between two charges, and the potential energy of a single charged particle. Force between two charges in vacuum – F = q1 • q2/(4peo • r2) N (J/m) eo = permittivity of a vacuum (C2 J-1 m-1) Charges in solution F = q1 • q2/(4peo • er • r2) er = dielectric constant (unitless) how medium shields charge forces • Electric field E = F/q1 = q2/(4peor2) |E| ≡ -df/dr Electric potential of a charge q1f = q2/(4peor) (V or J/C) 8.1 – 8.3 (assign 8.2 and 8.5)

  4. 8.1 Charge on sphere#1 attracted to sphere #2 with q2 = 1.00 C if r = 100.0 m and F = 0.0225 N? Coulomb’s Law – Force between two charges in vacuum F = q1 • q2/(4peo • r2) N q1 = (4peo • r2) • F/q2 = (4p • 8.85 x 10-12• 1002) • 0.025/1 = q1 = 2.50 x 10-8C

  5. 8.3 Two charged metal spheres in water (er = 78) with r = 6.075 cm and F = 1.55 x 10-6 N. q(-) = 2q(+) a) What are q(-) and q(+)? b) What are the electric fields of the two bodies? Charges in solution F = q+ • q-/(4peo • er • r2) 8.4 q+ • q- =F (4peo • er • r2) = 2q+2 • Electric field E+ = F/q+ = q-/(4peor2) |E| ≡ -df/dr V/m (assign 8.2 and 8.5) Electric potential of a charge q18.6 f = q2/(4peor) V (J/C)

  6. E˚rx = E˚red - E˚ox Standard reaction potentials can be determined from a table of standard reduction potentials of half reactions. E = Eº - RT/nFlnQ F = 96,485 C mol-1. The Nernst equation defines reaction potentials at non-standard concentrations. DGº = -nFEº Standard reaction potentials can be determined from standard thermodynamic tables (assuming redox reaction). DG = -nFE

  7. Ions in solution tend to deviate from ideality at low concentrations Debye-Hückel Law: lng± = A z+ z- I1/2A = 1.171 m-½ (H2O) g± = The averaged activity coefficient of ions in solution I = Ionic strength: I = ½ • Sici • zi2 The ionic strength measures the impact that ions have in a solution for properties (e.g. Conductivity – salting out proteins in biochemistry – etc.) Problem 33 Problem 41

  8. Use activities rather than [ ]s in Q when applying Nernst Equation however, each ion has to have a separate g. Extended Debye-Hückellng = - (A·z2·I1/2)/(1+B · å · I1/2) B = 2.32 x 109 m-3/2 Å represents the ionic diameter

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