5.62 Lecture #11: Internal Degrees of Freedom for Atoms and Diatomic Molecules

5.62 Lecture #11: INTERNAL DEGREES OF FREEDOM FOR ATOMS AND DIATOMIC MOLECULES Readings: ATOMS Hill, pp. 147-159; Maczek pp. 42-53 Pages 11-1, 11-2 and 11-3 are a review of Lecture #10. — have one internal degree of freedom ELECTRONIC degree of freedom MOLECULES — have other degrees of freedom ELECTRONIC, VIBRATION, AND ROTATION which each contribute 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 N,M,L internal quantum #'s where εint = energy from internal degrees of freedom q = ∑ e −ε i kT i all molecular states − ε +ε = ∑ g(ε)e ( trans int ) kT ε all molecular energies We do not have to start from the beginning. qtrans and qint appear as separate multiplicative factors. q = ∑ e −ε trans kT translational states internal states q= Q= qtrans • ∑ e −ε int kT qint ⎞ (q trans q int )N ⎛ q N = ⎜ trans ⎟ q N int N! ⎝ N! ⎠ ← INTERNAL MOLECULAR PARTITION FUNCTION 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 them indistinguishable), requiring the factor of N! The internal motions do not interchange particles. CANONICAL PARTITION FUNCTION Q = Qtrans Qint Qtrans = CANONICAL TRANSLATIONAL PARTITION FUNCTION qN trans N! CANONICAL INTERNAL PARTITION FUNCTION Qint = q N int Classically Qcl = Qtrans,cl Qint,cl Q trans,cl qN = trans,cl = N! ⎡⎣ ∫ e − ε trans /kT dp 3dq 3 ⎤⎦ 3N N N!h 3N 3N −ε int /kT Q int,cl = q N int = ∫ dp dq e CONTRIBUTION OF INTERNAL DEGREES OF FREEDOM TO MACROSCOPIC PROPERTIES ⎛ ∂ ln Q ⎞ E = kT2 ⎜ ⎝ ∂T ⎟⎠ N,V = ⎛ ∂ ln Q transQ int ⎞ kT2 ⎜ ⎟⎠ ⎝ ∂T N,V ⎛ ∂ ln Q trans ⎞ E = kT2 ⎜ ⎟ ⎝ ∂T ⎠ N,V + ⎛ ∂ ln Q int ⎞ kT2 ⎜ ⎝ ∂T ⎟⎠ N,V E= + E int E trans A = –kT ln Q = –kT ln Qtrans Qint = –kT ln Qtrans + –kT ln Qint = Likewise: Atrans + Aint S = Strans + Sint because But: A = E − TS S = E / T − A / T = E / T − kTln Q / T ⎛ ∂ln Q ⎞ p = kT ⎜ = p trans + p int = p trans ⎟ ⎝ ∂V ⎠N,T no V dependence for internal coordinates INTERNAL DEGREE OF FREEDOM OF AN ATOM Electronic Excitation: promotion of an electron to a higher energy orbital. 1s2 He → He 1s,2s first electronically excited configuration (1S and 3S states). ground state Goal: Derive electronic molecular partition function q elec = ∑ e −ε i /kT = ∑ g(ε j )e −ε j /kT εj i electronic states allowed energies εj ≡ energy of j electronic energy level th g(εj) ≡ degeneracy — number of isolated-atom states with electronic energy εj Electronic energies and degeneracies are determined by atomic spectroscopy experiments. Tabulated in tables such as NBS circular #467 by Charlotte Moore. no simple energy level q elec = g (ε0 ) e−ε0 kT + g (ε1 ) e−ε1 kT + g (ε2 ) e−ε2 electronic molecular partition function kT +… formula except for 1eatoms H, He + , Li2+ , etc. ε0 ≡ ground electronic state — zero of energy (arbitrarily) set at ε0 = 0 ε1  kT usually. Contribution from first excited state and higher energy states usually very, very small (except when L ≠ 0 and S ≠ 0, get spin-orbit splittings). qint = qelec = g(ε0) FOR ATOMS Example: Electronic Excitation of a Hydrogen Atom – Rydberg States ⎛ 1⎞ εn = R y ⎜1 − 2 ⎟ ⎝ n ⎠ n ≡ principal quantum number (ε1 = 0, ε∞ = Ry) 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 [L-S terms: g(L,S) = (2L+1)(2S+1)] e.g. n = 4: 4s, 4p, 4d, 4f give 2L states, each with degeneracy 2× (2L + 1) Calculate qelec for H atom … At T = 1000K, kT = 1.987 kcal mol–1 = 8.315 kJ mol–1 g ( ε n ) e − εn n εn (kcal mol–1) g(εn) 1 0 2 2 2 235.3 8 2.98 × 10–51 3 278.9 18 1.98 × 10–60 kT  n 313.8 2n2 2n2(2.58 × 10–69) qelec = 2 + 2.98 × 10–51 + 1.98 × 10–60 + … = 2 Ratio of population of molecules with energy ε2 to energy ε1. nε 2 nε1 = g(ε 2 ) − ( ε 2 −ε1 ) kT 2.98 × 10 −51 e = = 1.5 × 10 −51 g(ε1 ) 2 Only ground state contributes significantly to qelec and only ground state is populated significantly. But sum includes ∞ # of nonzero terms! How do we justify neglect of infinite 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 case there 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 by nuclear 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 + 3 J = 1/2 → F = 1,0 g = 3 + 1 total 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 not generally included in qnuc because all hyperfine levels are equally populated except at extremely 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. INTERNAL DEGREES OF FREEDOM — DIATOMIC MOLECULES INTRAMOLECULAR POTENTIAL OF A DIATOMIC AB MOLECULE POTENTIAL ENERGY U(R) separated atoms in their electronic ground states [Spectroscopists use X,A,B… a,b,c… instead of 0, 1, 2…] INTERNAL ENERGY LEVELS OF DIATOMIC MOLECULES ELECTRONIC ENERGIES — same situation as for atoms — no analytical expression — consult tables in which these energies, as measured in a spectroscopy experiment, are tabulated — denoted as ε0, ε1, ε2 … Degeneracy: (2S + 1)(2-δ0,Λ). Notation: ΛΩ 2S+1 VIBRATIONAL ENERGIES — harmonic oscillator model used to approximate intramolecular potential in solution of Schrödinger equation. vibrational energy levels given by … 1⎞ 1⎞ ⎛ ⎛ ε vib (v) = ⎜ v + ⎟ hν = ⎜ v + ⎟ hcω e ⎝ ⎝ 2⎠ 2⎠ (ωe in cm–1 units) where v = 0,1,2 … vibrational quantum number ν (sec–1) = c/λ = frequency [3 × 1013 s–1 typical] (ωe = 1000 cm–1) ν = cωe ωe = 1/λ (cm–1) ≡ wavenumber [all spectroscopists use cm–1 units synonymously with Energy and Frequency!] • need ωe or ν to calculate εvib(v) — get from IR spectroscopy expt. • vibrational frequencies have different values in each electronic state because force constants are different (e.g. consider bonding↔antibonding orbital excitation in N2) v0 ≡ cω 0e ≡ vibrational frequency in ground electronic state v1 ≡ cω1e ≡ vibrational frequency in 1st electronically excited state • degeneracy of each vibrational level = 1. gvib = 1 • note that lowest energy is 1 hν or “zero point energy”. 2 [There are two common choices of the zero of vibrational energy, at v = 0 or at v = –1/2. We will return to this when we consider isotope effects on reaction equilibrium constants.] ROTATIONAL ENERGIES — use rigid rotor approximation; bond length does not change with rotational excitation. rotational energy levels given by … h2 ε rot (J) = J(J + 1)hcBe = J(J + 1) 2 2 8π µR e (Βe in cm–1 units, B = 1 cm–1 typical) [to get from cm–1 to E units, multiply by hc] where J = 0,1,2 … rotational quantum number [J can be 1/2 integer or integer depending on whether there is an odd or even # of e–.] Be ≡ rotational constant — property of molecule [always given in cm–1 units.] Be = h h = 2 ⎡⎣ cm −1 ⎤⎦ 2 8π µR e c 8π Ic ↑ I = moment of inertia = µR 2e for diatomics m Am B µ= reduced mass m A + mB 2 • degeneracy of each rotational level = 2J + 1 gJ = 2J + 1 • rotational constants Be have different values in each electronic state B0e rotational constant in ground electronic state B1e rotational constant in first electronically excited state INTERNAL MOLECULAR PARTITION FUNCTION — qint — DIATOMICS GOAL: Obtain an expression for qint including all degrees of freedom q int = ∑ ε(n,vn ,J n ) ( ) g n, vn , J n e ( −ε n,vn ,J n ) kT (n here refers to electronic state, not principal quantum number) sum over all internal energy levels Assume internal energies are separable and additive (not quite true): εint(n,vn,Jn) = εel(n) + εvib(vn) + εrot(Jn) This approximation (Born-Oppenheimer) is okay for separation of electronic from internal degrees of freedom, but ωe and Be both depend on electronic state. But it is OK here because only the lowest electronic state usually contributes (except rare earths). Separability of vibration from rotation okay for low vibrational energies. Because energies are additive, qint factors into terms for each electronic state q int = g 0 e−ε0 /kT ∑ e−ε0,vib /kT ∑ g (ε J 0 ) e−ε0,rot /kT v0 +g1e−ε1 /kT ∑ e J0 −ε1,vib /kT ∑ g (ε 1) e J v1 −ε1,rot /kT +… J1 qint = q0,elq0,vibq0,rot + q1,elq1,vibq1,rot + q2,elq2,vibq2,rot + … q int = g 0 e−ε0 kT q 0,vib q 0,rot + g1e−ε1 kTq1,vib q1,rot + g 2 e−ε2 kT q 2,vib q 2,rot +… usually only 1st term contributes significantly because g1e−ε1 kT  g 0 e−ε0 kT Only the ground electronic state contributes significantly except at extremely high temperature or when electronic ground state has  S ≠ 0. (“Multiplet states” have low-lying Ω-components. Ω is projection of J on body axis.) Need to find qrot and qvib!!! But usually only for the lowest electronic state.