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MIT 5 62 - Lecture #11: INTERNAL DEGREES OF FREEDOM FOR ATOMS AND DIATOMIC MOLECULES

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MIT OpenCourseWare http://ocw.mit.edu 5.62 Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit: http://ocw.mit.edu/terms.5.62 Lecture #11: INTERNAL DEGREES OF FREEDOM FOR ATOMS AND DIATOMIC MOLECULES Readings: Hill, pp. 147-159;Maczek pp. 42-53Pages 11-1, 11-2 and 11-3 are a review of Lecture #10. ATOMS — have one internal degree of freedomELECTRONIC degree of freedom MOLECULES — have other degrees of freedom ELECTRONIC, VIBRATION, AND ROTATION which eachcontribute to total energy and to other macroscopic properties. Nuclear hyperfine? [Nuclear spin degeneracy factors. LATER.] Internal energy adds to translational energy to get total energy ε = εtrans + εint quantum #'s internal quantum #'sN,M,L where εint = energy from internal degrees of freedom q = ∑ e−εi kT = ∑g(ε)e− ε( trans +εint ) kT i ε all molecular all molecular states energies We do not have to start from the beginning. qtrans and qint appear as separate multiplicativefactors. q = ∑ e−εtrans kT∑ e−εint kT translational internal states states q = qtrans • qint ← INTERNAL MOLECULAR PARTITION FUNCTION (qtransqint )N N! qNtrans N! Q = = ⎛ ⎞ ⎜⎝ ⎟⎠ qintN5.62 Spring 2008 Lecture 11, Page 2 NOTE: N! is included with qtrans. This is because it's the translational motion that causes the positions of identical particles to be interchanged (thus rendering themindistinguishable), requiring the factor of N! The internal motions do not interchangeparticles. Q = Qtrans Qint Qtrans = qtrans N N! Qint = qintN CANONICAL PARTITION FUNCTION CANONICAL TRANSLATIONAL PARTITION FUNCTION CANONICAL INTERNAL PARTITION FUNCTION Classically Qcl = Qtrans,cl Qint,cl Qtrans,cl = qtrans,clN N! = e−εtrans /kT ∫ dp3dq⎡⎣ N!h3N 3 ⎤⎦ N Qint,cl = qintN = ∫ dp3Ndq3Ne−εint /kT CONTRIBUTION OF INTERNAL DEGREES OF FREEDOM TO MACROSCOPIC PROPERTIES E = kT2 ⎛∂lnQ ⎞ = kT2 ⎛∂lnQtransQint ⎞ ⎝⎜ ∂T ⎠⎟ N ,V ⎝⎜ ∂T ⎠⎟ N ,V ⎛∂lnQtrans ⎞ ⎛∂lnQint ⎞E = kT2 ⎝⎜ ∂T ⎠⎟ N ,V + kT2 ⎝⎜ ∂T N ,V ⎠⎟ E = Etrans + Eint revised 1/10/08 12:44 PM5.62 Spring 2008 Lecture 11, Page 3 A = –kT ln Q = –kT ln Qtrans Qint = –kT ln Qtrans + –kT ln Qint = Atrans + Aint Likewise: S = Strans + Sint because A = E − TS S = E / T − A / T = E / T − kTln Q / T But: p = kT ∂lnQ ∂V ⎛⎜⎝ ⎞⎟⎠ N ,T for internal coordinates INTERNAL DEGREE OF FREEDOM OF AN ATOM Electronic Excitation: promotion of an electron to a higher energy orbital. He 1s2 → He 1s,2sground state first electronically excited configuration (1S and 3S states). Goal: Derive electronic molecular partition function qelec = ∑ e−εi/kT = ∑g(ε j)e−ε j/kT i ε j electronic states allowed energies εj ≡ energy of jth electronic energy level g(εj) ≡ degeneracy — number of isolated-atom states with electronic energy εj Electronic energies and degeneracies are determined by atomic spectroscopyexperiments. Tabulated in tables such as NBS circular #467 by Charlotte Moore. no simple energy level = ptrans + pint = ptrans no V dependence qelec = g ε0( )e−ε0 kT + g ε1( )e−ε1 kT + g ε2( )e−ε2 kT +… formula except for 1e-atoms H, He+, Li2+, etc. electronic molecular partition function ε0 ≡ ground electronic state — zero of energy (arbitrarily) set at ε0 = 0 revised 1/10/08 12:44 PM5.62 Spring 2008 Lecture 11, Page 4 ε1  kT usually. Contribution from first excited state and higher energystates usually very, very small (except when L ≠ 0 and S ≠ 0, get spin-orbit splittings). FOR ATOMS Example: Electronic Excitation of a Hydrogen Atom – Rydberg States qint = qelec = g(ε0) 1− Ry ≡ Rydberg constant = 2.180 × 10–18 J/molecule = 313.76 kcal/mol = hc(109737 cm–1) When n = ∞ ε∞ = 313.76 kcal/mol = 13.60 eV ionization potential of H atom Units: 350 cm–1 = 1kcal/mol, k = 0.695 cm–1/K Degeneracy of a Rydberg energy level is g(εn) = 2n2 e.g. n = 4: 4s, 4p, 4d, 4f[L-S terms: g(L,S) = (2L+1)(2S+1)] give 2L states, each withdegeneracy 2× (2L + 1) ⎛⎜⎝ Calculate qelec for H atom … At T = 1000K, kT = 1.987 kcal mol–1 = 8.315 kJ mol–1 kT( )e− εnn εn (kcal mol–1) g(εn) g εn1 ⎞ = Ry n ≡ principal quantum number (ε1 = 0, ε∞ = Ry)εn⎟⎠ n2 1 0 2 2 2 235.3 8 2.98 × 10–51 3 278.9 18 1.98 × 10–60  n 313.8 2n2 2n2(2.58 × 10–69) qelec = 2 + 2.98 × 10–51 + 1.98 × 10–60 + … = 2 revised 1/10/08 12:44 PM5.62 Spring 2008 Lecture 11, Page 5 Ratio of population of molecules with energy ε2 to energy ε1. nε2 g(ε2) ( 2.98 × 10−51 = 1.5 × 10−51 g(ε1) e− ε2 −ε1 )kT= = 2nε1 Only ground state contributes significantly to qelec and only ground state is populatedsignificantly. But sum includes ∞ # of nonzero terms! How do we justify neglect ofinfinite number of positive, non-zero terms in qelect? Hint: 〈r〉n = a0n2. What about the nuclear partition function qnuc? 2I + 1. Nuclear spin degeneracy?Excited states of nucleus? Changing the nuclear state generally requires huge energies, so as for the electronic casethere is only one nuclear energy level that must to be considered at normal temperatures. However, the nuclear ground state has an associated spin angular momentum denoted bynuclear spin quantum number I. There is a degeneracy g(I) = 2I + 1 = qnuc. e.g. for H(I = 1/2) 2P J = 3/2 → F = 2,1 g = 5 + 3J = 1/2 → F = 1,0 g = 3 + 1total g = 12 = 2 × 2 × 3 ↑ ↑ ↑ S I L This should affect the entropy for I ≥ 1/2 and L ≠ 0 or S ≠ 0 since there are several available nuclear states at the lowest energy level. However, this contribution is notgenerally included in qnuc because all hyperfine levels are equally populated except atextremely low T. Thus there is no nuclear spin contribution to S except at extremely low T. In the case of thermonuclear reactions, I may change and qnuc must be accounted for. revised 1/10/08 12:44 PM5.62 Spring 2008 Lecture 11, Page 6 INTERNAL DEGREES OF FREEDOM — DIATOMIC MOLECULES INTRAMOLECULAR POTENTIAL OF A DIATOMIC AB MOLECULE POTENTIAL ENERGY U(R) separated atoms intheir electronic ground states [Spectroscopists use X,A,B… a,b,c… instead of 0, 1, 2…] revised 1/10/08 12:44 PM5.62 Spring 2008 Lecture 11, Page 7 INTERNAL ENERGY LEVELS OF DIATOMIC MOLECULES ELECTRONIC ENERGIES — same situation as for atoms — no analytical expression—


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MIT 5 62 - Lecture #11: INTERNAL DEGREES OF FREEDOM FOR ATOMS AND DIATOMIC MOLECULES

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