5_60_lecture29

5_60_lecture29 - MIT OpenCourseWare http/ocw.mit.edu 5.60...

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MIT OpenCourseWare http://ocw.mit.edu 5.60 Thermodynamics & Kinetics Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms .
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5.60 Spring 2008 Lecture #29 page 1 APPLICATIONS: CHEMICAL AND PHASE EQUILIBRIA Apply statistical mechanics to develop microscopic models for problems you’ve treated so far with macroscopic thermodynamics 0 Products Reactants Separated atoms Product & reactant energy levels E ε r = -D 0,r ε p = -D 0,p Chemical equilibria Gas phase: Calculate Kp from microscopic properties aA + bB F cC + dD Δ G 0 = − RT ln K p = cG C 0 + dG D 0 aG A 0 + bG B 0 Need G 0 for each species G = A + pV = − kT ln Q + pV = − kT ln Q + NkT Q = q N N ! q N ( trans ) int ln Q = N ln ( q trans q int ) ln N ! = N ln ( q trans q int ) N ln N + N G = − kT ln Q + NkT = − NkT ln ( q trans q int N ) = − NkT ln ( = N ln ( q trans q int N ) + N q N ) For molecules in the gas phase, internal degrees of freedom are rotations and vibrations and electronic states.
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5.60 Spring 2008 Lecture #29 page 2 q = q q q int rot vib elec D kT Dissociation energy D 0 from ground electronic level q elec = e 0 Usually no other electronic levels matter q = e n ε 0 kT = 1 + e ε 0 kT 2 ε 0 kT 3 ε 0 kT 1 vib n + e + e + " = 1 e ε 0 kT Note we’ve set the zero of vibrational energy as the lowest vibrational level. The zero-point vibrational energy has been included in q elec by using the dissociation energy rather than the bottom of the electronic potential energy. kT (300 K) 200 cm -1 Most molecular vibrational frequencies > 500 cm -1 q vib 1 - Need to calculate it, but it’s not large We have not treated q rot . Levels are NOT evenly spaced: ε rot = J(J + 1) ε 0, rot where J = 0, 1, 2,…, and ε 0, rot 1 cm -1 .
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