5.62 Physical Chemistry II Spring 2008

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1 MIT OpenCourseWare Physical Chemistry II Spring 2008 For information about citing these materials or our Terms of Use, visit:

2 5.62 Lecture #12: Rotational Partition Function. Equipartition Readings: Hill, pp ; Maczek, pp Metiu, pp DEGREES OF FREEDOM A molecule with n atoms has 3n "degrees of freedom" or 3n coordinates to describe its position and therefore has 3n ways of incorporating energy due to nuclear motion where n is the number of atoms in the molecule. For a diatomic or a linear polyatomic molecule: 3 TRANSLATIONAL degrees of freedom 2 ROTATIONAL degrees of freedom 3n 5 VIBRATIONAL degrees of freedom 3n TOTAL degrees of freedom For a diatomic molecule 3n 5 = 1 vibrational degree of freedom MOLECULAR ROTATIONAL PARTITION FUNCTION q DIATOMIC ε (J) = J(J + 1) hcb e for J = 0,1,2, g J = 2J + 1 q = g(ε) e ε/kt = (2J +1) exp[ hcbe J(J+1)/kT] J=0 allowed ational energies Question: How do you do the summation? Two cases (Low-T limit case [next Lecture]) Case 1:ε/kT 1 or hcbej(j + 1)/kT 1 [More precisely, we want (E(J + 1) E(J)) kt at E(J) kt.) ational states are closely spaced in energy compared to kt since energy spacings are so close together, can consider

3 5.62 Spring 2008 Lecture 12, Page 2 so: ε as continuous and use Euler-MacLaurin Summation Formula (draw a picture!) this case is the classical or high-temperature limit. n 1 J=m n f ( J) = m f(j)dj + 2 [f(m) + f(n) ] + residue q = (2J +1) 0 exp[ hcb 1 e J(J+1)/kT] dj + [1 + 0] + 2 J = 0 J = substitute ω = J(J+1) thus dω = (2J+1)dJ q = 0 exp [ hcb e ω / kt ]dω kt e hcb e ω kt 1 = hcbe 2 = 0 kt 1 hcbe = kt 1 kt usually can ignore the 1 q hcb e + 2 hcbe 2 What happens for a 1 state where J min = 2 rather than 0? Hold on One correction needed to q SYMMETRY NUMBER σ # of equivalent orientations in space which leave appearance of molecule unchanged # of indistinguishable orientations in which molecules can be found as a result of ation. We divide by σ because otherwise we would be overcounting by counting indistinguishable orientations. A homonuclear molecule, O 2, has σ = 2 because an end over end (half) ation by π does not alter the appearance of O 2. The symmetry number is rigorously based on the nuclear spins. We ll see the details later. kt for hcb e kt So q = σhcb or ε kt e

4 5.62 Spring 2008 Lecture 12, Page 3 where σ symmetry # = 1 for heteronuclear diatomics = 2 for homonuclear diatomics Really this is q -nuc, but we ll refer to it as q. Define θ = hcb e ational temperature, θ k (has units of K) So kt T q = = for θ σhcb e σθ T also written as Molecular Rotational Partition Function for Diatomics q = 8π 2 IkcT h 2 σh 2 because Be = 8π 2 I = µre Ic Let's go back and check whether dropping extra terms in Euler-MacLaurin series was a good approximation q = kt σhcb e 2 at T = 300K MOLECULE B e (cm 1 ) σ θ (K) q = T/σθ + 1/2 % error (neglect of 1/2) HCl CO I As B e becomes smaller or equivalently as θ becomes smaller compared to T, dropping extra terms becomes better approximation; also, discrete to continuous approximation becomes better.

5 5.62 Spring 2008 Lecture 12, Page 4 Contributions of Rotation to Thermodynamic Functions for ε kt kt q = Q = (q ) N = kt σhcb e σhcb e N A = ktlnq = NkTln q = NkTln kt σhcb e A kt p = = NkTln V N,T V σhcb e N,T = 0 because ε does not depend on V E = kt 2 lnq = NkT 2 ln q T N,V T = NkT 2 ln T + nkt 2 ln(k / σhcb e ) = NkT T T T average ational energy of a diatomic molecule E = NkT for θ T or ε kt (not 1 2 NkT; why?) a "quantum" result (but based on the approximation of replacing a sum by an integral) CLASSICAL EQUIPARTITION RESULT FOR ROTATIONAL ENERGY NKT = NkT Each degree of translational and ational energy contributes (1/2)kT to total energy. For a diatomic molecule, there are 2 ational degrees of freedom [Why 2?]. Therefore, 2(1/2 NkT) = NkT. This is why C V for monatomic gases is ~(3/2)R and for most diatomic gases at moderate T is ~(5/2)R! Quantum and classical approach lead to same result for ation at 300 K. Why? Because ational energy levels are very closely spaced compared to kt. We calculated

6 5.62 Spring 2008 Lecture 12, Page 5 q by approximating a sum over energy levels as an integral over energy levels. Rotational energy levels are so closely spaced that they "look" continuous compared to kt at room temperature for most molecules. E does not depend on the properties of the molecule in the classical limit! Quantum result for Cv E T C V = = Nk = Nk = R if N = Na (or C T N,V V = nr) T for ε kt or θ T (ε needs to be better defined, see below) 1. More about high temperature limit, which is the requirement that permits the sum, q = g ( ε ( J ))e hcbj (J +1) kt, to be replaced by an integral, J min q = J min ( dj (2J + 1)e hcbj J +1) kt (2J min + 1) + 0. It is necessary that ε kt at ε kt. The ational energy level spacing must be small relative to kt. ε (J) = hcb [( J +1 )(J + 2 ) J ( J +1 )] = hcb2(j +1). This spacing must be small relative to kt when ε = kt ( ε = hcbj J +1) = kt Thus hcbj(j+1) hcb2(j + 1), which requires that This means that we want J 2. kt ε (J = 2) = 6hcB. kt 6hcB is the requirement that specifies when it is OK to replace sum by integral.

7 5.62 Spring 2008 Lecture 12, Page 6 2. Some useful stuff concerning fractional populations in ational levels. A. fraction of population in J-th level f J = ( 2J + 1)e T θ J ( J +1) T θ = hcb k σθ B. Most populated J Thus df J = 0 = dj 2e θ J (J +1) T ( 2J +1) 2 ( θ T )e θ J (J +1) T q 2 = (2J +1) 2 θ T For T / θ = 100 J max = 6.5 J max 2T 1/2 1 θ =. 2 C. Fractional population in most populated J level 1/2 1/2 1/2 θ 2T 2T 1 e 4T θ θ 2T 1/2 1/2 θ 2θ θ f e 1/2 Jmax = = T θ T T max For T θ = 100 f J = D. Fractional population in J = T. This is a simple-minded way of asking θ about the population of the last thermally accessible level.

8 5.62 Spring 2008 Lecture 12, Page 7 T T θ T 2 f J =T θ = + 1 e θ +1 θ T T θ = 2e +1 θ. T θ For T / θ 100 f J =T θ = 2e 101 = This is a very small fractional population. It would be more appropriate to ask for the fractional population of the J value for which J ( 2J +1) = T θ, J min because this sets the total number of significantly populated J, M J levels equal to q. Using J ( 2J +1) = (J +1) 2 J =0 and T/θ = 100, we get J = 9 and f = 19e 9 10)/100 = J =9 100 Low-T limit results for E and C V next time. But what do you know without any equations about E and C V in the limit T 0K? (