MOLECULAR PARTITION FUNCTIONS

Transcription

1 MOLECULR PRTITIO FUCTIOS Introducton In the last chapter, we have been ntroduced to the three man ensembles used n statstcal mechancs and some examples of calculatons of partton functons were also gven. In chemstry, we are concerned wth a collecton of molecules. If the molecules are reasonably far apart as n the case of a dlute gas, we can approxmately treat the system as an deal gas system and gnore the ntermolecular forces. The present chapter deals wth systems n whch ntermolecular nteractons are gnored. The next chapters wll nclude detaled consderaton of ntermolecular forces. In ensemble theory, we are concerned wth the ensemble probablty densty,.e., the fracton of members of the ensemble possessng certan characterstcs such as a total energy E, volume V, number of partcles or a gven chemcal potental μ and so on. The molecular partton functon enables us to calculate the probablty of fndng a collecton of molecules wth a gven energy n a system. The equvalence of the ensemble approach and a molecular approach may be easly realzed f we treat part of the molecular system to be n equlbrum wth the rest of t and consder the probablty dstrbuton of molecules n ths subsystem (whch s actually qute large compared to systems contanng a small number of molecules of the order of tens or hundreds). Snce we are dealng wth number of partcles of the order of vogadro number, the ensemble descrpton and the molecular descrptons are equvalent. The energes of atoms and molecules are quantzed. Whle atoms have only electronc energy levels, molecules have quantzed energy levels arsng from electronc, vbratonal and rotatonal moton. schematc energy level dagram s shown n Fg We have already seen that n the canoncal ensemble, the probablty of a system havng energy E s proportonal to the Boltzmann factor and s gven n terms of the canoncal partton functon q by Where q s defned by P(E ) = E / k T E / k T E / k T e / e = e / q (3.1) q = e β E (3.) Here, β = 1/ kt and e β E s called the Boltzmann factor. Often the Boltzmann constant s wrtten as k B. But when there s no ambguty, we wll smply wrte k. Once we know the probablty dstrbuton for energy, we can calculate thermodynamc propertes lke the energy, entropy, free energes and heat capactes, whch are all average quanttes. To calculate P(E )s we need the energy levels of a system. The energy ("levels") of a system can be bult up from the molecular energy levels. We wll consder the smpler problem of molecular energy levels whch are pctorally shown n the Fg 3.1.

2 64 Fgure 3.1 schematc dagram showng electronc (bold lnes), vbratonal and rotatonal energy levels. The electronc quantum umbers are shown to the extreme rght. Vbratonal quantum numbers are to shown n the extreme left. The rotatonal quantum numbers are shown between the vbratonal levels. 3.1 Electronc, Vbratonal, Rotatonal and Translatonal Partton Functons The electronc energy levels are generally very wdely separated n energy compared to the thermal energy kt at room temperature. In each electronc level, there are several vbratonal levels and for each vbratonal level, there are several rotatonal states. Ths s a smplfed and useful model to start wth. The total energy s a sum of all these energes and s gven by E total = E electronc + E vbratonal + E rotatonal + E translatonal + E others (3.3) The term E others ncludes nuclear spn energy levels and may also be used later to nclude the nteractons between the frst four. ssumng the frst three to be ndependent and neglectng the last term, the molecular partton functon (e, a sum over the molecular energy states) s gven by

3 65 + β E ν rot trans q = e = e e νb e e - (E el + E b + E E ) / k T β E el β E rot β Etrans el νb rot trans (3.4) Here, the summaton s over the electronc, vbratonal and rotatonal states can be done separately snce they are assumed to be ndependent. Therefore, q = q el q vb q rot q trans (3.5) The molecular partton q functon s wrtten as the product of electronc, vbratonal, rotatonal and partton functons. The partton functon s a sum over states (of course wth the Boltzmann factor β multplyng the energy n the exponent) and s a number. Larger the value of q, larger the number of states whch are avalable for the molecular system to occupy. Snce E el > E vb > E rot > E trans, there are far too many translatonal states avalable compared to the rotatonal, vbratonal and electronc states. q el s very nearly unty, q vb and q rot are n the range of 1 to 100 whle q trans can be much n excess of We shall calculate the values of these qs and ndcate how these qs are useful n calculatng the equlbrum constants and also n certan cases, the rate constants. Usng the standard formulae for the translatonal, rotatonal and vbratonal energy levels, we wll now calculate the molecular translatonal, vbratonal and rotatonal partton functons for datomc molecules frst The Translatonal Partton Functon, q tr. Consder a molecule confned to a cubc box. molecule nsde a cubc box of length L has the translatonal energy levels gven by E tr = h (n x + n y + n z ) / 8 ml where n x, n y and n z are the quantum numbers n the three drectons (See Chapter 1 for detals). The translatonal partton functon s gven by β E tr β β E n n n E x y β Ez e = e e e. (3.6) tr x y z q tr = q x q y q z, (3.7) Whch s the product of translatonal partton functons n the three drectons. Snce the levels are very closely spaced, we can replace the sum by an ntegral - n h / (8 m L x k T) - a n B x h q x = e dn 1 x e dn 0 x, a = (3.8) 8mL k T usng a x = 1 0 e dx π / a we get, q x = 1/ ( π /a) 1/ = 1 8πm k T B L h where, Λ s the de Brogle thermal wavelength gven by = B Multpyng q x, q y and q z, and usng V = volume of the box = L 3, we have, B πm k T L = L/Λ (3.9) h h. πm k T q tr = [ π m k B T / h ] 3/ V = V / Λ 3 (3.10)) B

4 66 Ths s usually a very large number (10 0 ) for volumes of 1 cm 3 for a typcal small molecular mass. Ths means that such a large number of translatonal states are accessble or avalable for occupaton by the molecules of a gas. Example 3.1 Calculate the translatonal partton functon of an I molecule at 300K. ssume V to be 1 lter. Soluton: Mass of I s X 17 X X 10-7 kg πmk B T = X X ( X 17 X X 10-7 kg) X X 10-3 J/K X 300 K = X J kg Λ = h / ( π m k B T) 1/ = 6.66 X J s / ( X J kg) 1/ = 6.36 X 10-1 m q tr = V / Λ 3 = 1000 X 10-6 m 3 / (6.36 X 10-1 m ) 3 = 3.95 X Ths means that 3.95 X quantum states are thermally accessble to the molecular system 3.1. The Rotatonal Partton Functon of a Datomc The rotatonal energy levels of a datomc molecule are gven by E rot = B J (J + 1) where B = h / 8 π I c (3.11) Here, B s the rotatonal constant expresses n cm -1. The rotatonal energy levels are gven by ε J = J ( J + 1) h / 8π I, where I s the moment of nerta of the molecule gven by μr where, μ s the reduced mass and r, the bond length. Often, the energes are also expressed n terms of the rotatonal temperature, Θ r or Θ, whch s defned as rot Θ = h r 8π I k (3.1) In the summaton for the expresson for q rot, Eq. (3.13), we can do an explct summaton q ( J + 1) e - E, j / k T rot = rot (3.13) j f only a few terms contrbute. The factor (J+1) for each term accounts for the degeneracy of a rotatonal state J. If energy E J s degenerate wth ( J + 1) states - E rot, / k T correspondng to t, then, the Boltzmann factor j B e has to be multpled by (J+ 1) to account for all these states. If the rotatonal energy levels are lyng very close to one another, we can ntegrate smlar to what we dd for q trans above to get q rot = 0 - B J (J+1) / k T (J+1) e dj (3.14) the ntegraton can be easly be done by substtutng x = J ( J+1) and dx = (J + 1) dj q rot = kt / B (3.15) For a homonuclear datomc molecule, rotatng the molecule by 180 o brngs the molecule nto a confguraton whch s ndstngushable from the orgnal confguraton. Ths leads to an overcountng of the accessble states. To correct for ths, we dvde the partton functon by σ, whch s called the symmetry number, whch s equal to the dstnct number of ways by whch a molecule can be brought nto dentcal confguratons by rotatons. The rotatonal partton functon becomes, q kt rot = σ B (3.16) Example 3. What s the rotatonal partton functon of H at 300K?

5 67 Soluton: The value of B for H s cm -1. The value of k B T n cm -1 can be obtaned by dvdng t by hc,.e., (k B T/hc) = 09.7 cm -1 at 300K. σ = for a homonuclear molecule. Therefore, qrot = kt / σ B = 09.7/(X60.864) = Snce the rotatonal frequency of H s qute large, only the frst few rotatonal states are accessble to at at 300K The Vbratonal Partton Functon of a Datomc The vbratonal energy levels of a datomc are gven by E n = (n +1/ ) hν (3.17) where s ν the vbratonal frequency and n s the vbratonal quantum number. In ths case, t s easy to sum the geometrc seres shown below - ( n + 1 ) h ν / k T q e ν b n = 0 - h ν / ( k BT) - h ν / k BT - h ν / k BT = e (1 + e + e +...) = (3.18) h / kt = e ν (1 + x + x + x 3...) (3.19) where x h / kbt e ν h / k T = e ν [1 / (1- x)] (3.0) = whch s less than 1. Threfore, h ν / k T e q = vb h ν/ k T 1 e 1 h / k T (3.1) q = vb 1 e ν (3.) f the zero of energy scale s at h ν /kt. nalogous to Θ r, a vbratonal temperature Θ v or Θ may be defned as hcν /k, where, ν s the vbratonal frequency n cm -1. vb Example 3.3 Calculate the vbratonal partton functon of I at 300K. Soluton: The vbratonal frequency of of I s cm -1. hν/kt = 14.57/09.7 = 1.03 e -hν/kt = q = vb 1/( ) = Ths mples, as before, that very few vbratonal states are accessble The Electronc Partton Functon Wrtng the electronc energy as E 1, E, E 3, wth degeneraces g 1, g, g 3, the electronc partton functon s gven by E1 E E3 q el = g1 e β + g e β + g3 e β (3.3) Usually, E 1 << E or E 3. Treatng E 1 to be the reference value of zero of energy, we get, q el = g 1 (3.4) whch s the ground state degeneracy of the system. Example 3.4 Fnd the electronc partton of H at 300 K. Soluton The lowest electronc energy level of H s near - 3 ev and the next level s about 5 ev hgher. Takng - 3 ev as the zero (or reference value of energy), q el = e 0 + e -5 ev / kt +...

6 68 t 300 K, T = 0.0eV and q el = 1 + e Where all terms other than the frst are nearly 0. Ths mples that q el = 1. The physcal meanng of ths s that only the ground electronc state s generally accessble at room temperature. Representatve data useful for calculatng the electronc, vbratonal, and rotatonal partton functons s gven n Table 3.1. The electronc degeneracy s represented as g. The vbratonal frequency s wrtten as ω. The dssocaton energy s wrtten as D 0. It s qute straght forward to calculate the above partton functons for these molecules. Table 3.1 Representatve molecular data for a few datomcs. Molecule g Bond ω, Θ Length cm -1 vb, B, Θ K cm -1 rot,k Force D 0 consta (kcal/ (Å) nt k mol) (dynes /cm) H D O Cl CO O HCl HI a K Example 3.5 Express the partton functon of a collecton of molecules Q n terms of the molecular partton functon q. Soluton: ssumng the molecules to be ndependent, the total energy E of molecules s a sum of ndvdual molecular energes E s and (1) () () β E β ε β ε β ε all possble Es Q = e = e e... e (3.5) Here (1) ε, () ε. come from summng over all = q.q...q = q (3.6) () ε are energes of ndvdual molecules and a sum of all E s can only ε s. Gbbs postulated that Q = q /! (3.7)

7 69 The! n the denomnator s due to the ndstngushablty of the tny molecules (or other quantum partcles n a collecton). Example 3.6 Derve the Maxwell Boltzmann dstrbuton of molecular speeds. Soluton: If we represent a molecular velocty by v r, t has three ndependent components v x, v y and v z n the three drectons x, y and z. Let us consder monatomc gas of mass m. The probablty F ( x, y, z) that gven molecule wll have velocty components lyng between x and x + dx, y and y + d y and z and z + dz can be wrtten as F (x, y, z) dz dy dz = f (x) f (y) f(z) dx dy dz. F s wrtten as a product of three functons f because x, y and z are ndependent and snce nature does not dstngush between x, y and z (unless drectonal felds lke gravtatonal or electromagnetc are present), the form of f s the same n the three drectons. gan, snce there s no dstncton between postve and negatve x, f depends on x or x. We can rephrase the above equaton as F( ) = f ( ) f ( ) f( ) (3.8) The only functon that satsfes the above equaton s an exponental functon snce x + y + z x y z e = e e e and so we conclude that f ( ) may be wrtten as b v x b v x f( ) = C e ± C e (3.9) We take only the negatve exponent (C and b are postve) because a postve exponent mples that very large veloctes have very hgh probabltes whch s hghly unlkely. To evaluate C, We nvoke the physcal argument that the velocty has to le somewhere between - to + and that the total probablty s 1.e., - b v x x x x f (v ) d v = C e d v = 1 The above ntegral s a standard ntegral Thus, C b v - a x e d x = π 1/ / 0 e x d x = C ( π / b) 1/ (3.30) - But snce we want the rght sde to be unty, C ( π / b) 1/ = 1 or C = ( b / π ) 1/ and f ( x ) = (b / π) 1/ - b v x e From a probablty dstrbuton such as f ( x), average quanttes can be determned. The averages of x and x are gven by < x > = < x > = (b / π)1/ x x (b / π) 1/ - b v x e d x = 0 (3.31) - b v x e d x =1/b (3.3) The averages have been denoted by < >. We have also used another standard ntegral, a x x e dx = 1/4 (π/a 3 ) 1/ (3.33) 0 The ntegral for < x > s zero because the value of the ntegrand for postve x s equal and opposte to ts value at x. Thus the area on the left of x = 0 s equal and opposte n sgn to the area on the rght. Ths s a specal case of a general result that the ntegral of

8 70 the product of an even functon and an odd functon of x s zero over a symmetrc nterval around zero. To evaluate b, we take the help of the knetc theory of gases. Do look up the detals. The pressure of a gas s gven n terms of the mean square velocty (speed) as p = 1/3 (/ V) m <v > Where / V = (number of molecules of the gas / volume) = the densty of the gas. But = n where n = number of moles and, the vogadro number; and snce pv = nrt, we have pv = 1/3 m < v > = 1/3 n m < v > = nrt (3.34) < v > = 3 RT / m = 3 k B T / m (3.35) Where k B = R / s the Boltzmann constant, X 10-3 J / K. Snce <v > = 3 k B T / m, we have <v x > = k B T/ m and substtutng, we get b = m / k B T. Equatons for f and F now become f(v x ) = (m / k B T ) 1/ - m v x / k B T e (3.36) F(v x, v y, v z ) = (m / k B T) 3 / m (v x + v y + v z )/kbt e (3.37) Ths s the Maxwell - Boltzmann dstrbuton of molecular speeds. F(v x, v y, v z ) dv x dv y dv z gves the probablty of fndng an arbtrary molecule wth a velocty (v x, v y, v z ) n the volume element dv x dv y dv z. more appealng nterpretaton of the same s that t s the fracton (of the total molecules) of molecules havng veloctes (v x, v y, v z ). nalogous to the radal probablty dstrbuton n coordnate space, we can now estmate the probablty of fndng a partcle n a sphercal shell of volume 4π v dv. Ths probablty n such a sphercal shell of radus v and thckness dv s gven by 3. Thermodynamc Functons 4π v dv (m / k B T) 3 / m (v x + v y + v z )/kbt e (3.38) We wll be restrctng ourselves to the canoncal ensemble here. Consder a collecton of molecules. The probablty of fndng a molecule wth energy E s equal to the fracton of the molecules wth energy E. In the collecton of molecules, how many molecules (n ) have the energy E?.Ths has to be exp -β E / Q. Ths s because the fracton of molecules n / havng the energy E s e -β E / Q whch s the same as the probablty of fndng a molecule wth energy E n the collecton. The average energy s obtanng by multplyng E wth ts probablty and summng over all. e, 1 1 < E > = E P = E / Q = - e = - e e - β E - β E - β E Q β Q β = 1 Q ln Q = Q β β (3.39)

9 71 Detaled connecton between partton functons and thermodynamc functons has been derved n Chapter, Secton.4., The relatons between Q and pressure and entropy are gven by Eqns. (.168) and (.171) respectvely. The pressure p can also be obtaned as the ensemble average of (- E/ V) T = (- d w/dv) T gvng, β E E E e 1 Q 1 ln Q p = P = = = (3.40) V V Q βq V β V The entropy s gven by S = k ln ( Q + β E ) (3.41) The notaton change s that Z s wrtten as Q here and E s wrtten as E. Let us recall the four fundamental thermodynamc relatons. de = TdS pdv (3.4) dh = TdS + Vdp (3.43) d = SdT pdv (3.44) dg = SdT + Vdp (3.45) Here, E s the nternal energy, H = E + pv, the enthalpy, = E TS, the Helmholtz free energy and G = H TS, the Gbbs free energy. Let us now replace <E> by smply E or more correctly by E E(0), where E(0) s the reference value of E at T = 0. Ths s necessary because, the absolute value of E s not measurable and only dfferences can be measured. The same dea holds for H, as well as G. Therefore, we rewrte the equaton for energy as E E(0) = - ( ln Q / β) V (3.46) From the equaton for enthalpy, H = E + pv, we get, H H (0) = ( ln Q β ) V + ktv ( ln Q V ) β (3.47) Snce p = - ( / V) T, we get from (3. ), (0) = kt ln Q (3.48) nd fnally we have for G G G(0) = kt ln Q + ktv ( ln Q V ) β (3.49) For an deal gas, pv = nrt where, n = number of moles of the gas and R = k = the gas constant and s the vogadro number. Substtutng ths and Q = q /! n the above equaton, we get, G G(0) = kt ln q + kt ln! + n RT otng that = n and usng Strlng s approxmaton, G G(0) = nrt ln q + kt ( ln ) + n RT q q G G(0) n RT ln q n RT ln n ln ln m n RT n RT (3.50) Where, q m = q / n, the molar partton functon n unts of mol -1. The functon [G(T) G(0)] / T s referred to as the Gauque functon. Example 3.6 The equpartton prncple states that each quadratc degree of freedom contrbutes ½ kt to the energy at hgh temperature. Verfy ths asserton for the rotatonal, translatonal and vbratonal motons of a datomc molecule.

10 7 Soluton Consder the molecular partton functons. The rotatonal energy s gven by rot 1 qrot ε = = σ β B = = kt q β σ B β β β rot The classcal expresson for the rotatonal energy s ( ) 1 I ω ω x y (3.51) +, where I s the moment of nerta and ω x and ω y are the angular veloctes n the x and y drectons. The rotaton along the molecular axs (the z axs here) has no meanng n quantum mechancs because the rotatons along the molecular axs lead to confguratons whch are ndstngushable from the orgnal confguraton. The two degrees of freedom have thus gven a value of kt. The translatonal contrbuton gves, 3 tr 1 qtr Λ V 3 Λ 3 3 ε = = = = = kt 3 (3.5) qtr β V β Λ Λ β β Thus, the three translatonal degrees of freedom n three dmensons satsfy the equpartton theorem. Turnng to the vbratonal contrbuton, we get, hcν β hcν β 1 qvb 1 hcν e e εvb = = = hcν kt f hcνβ << 1 q β q e vb vb hcν β hcν β ( 1 e ) ( 1 ) The classcal expresson for the vbratonal energy s ½ kx + ½ μ v x. t hgh temperature the equpartton theorem s vald but at low temperature, only a few vbratonal states are occuped and the equpartton prncple s not vald. (3.53) Example 3.7 Calculate the vbratonal heat capacty for datomc and sketch ts temperature dependence. Soluton The vbratonal energy s gven by the above expresson and the heat capacty at constant volume, C V, s gven by E/ T. We have, / T = ( β/ T) ( / β) = (-1/kT ) ( / β) = (-k β ) ( / β) (3.54) Therefore, C V = (-k β ) ( ε vb / β) = ΘV Θ hcν β hcν β hcν β V T { ( 1 e ) ( hcν ) e ( + hcν ) } e e kβ hcν k T = Θ (3.55) hcν β V ( 1 e ) T 1 e For large T, the molar C V becomes k = R and for small T, C V goes to zero as shown n the sketch below. The vbratonal temperature ΘV s defned as Θ V = hcν / k. For H, t has a value of 633 K and for I t s 309 K.

11 73 Fgure 3.3 Vbratonal heat capacty of a datomc as a functon of Θ V /T. 3.3 Rotatonal and Vbratonal Partton Functons of Polyatomc Molecules For a polyatomc molecule contanng n atoms, the total number of coordnate degrees of freedom s 3n. Out of these, three degrees of freedom are taken up for the translatonal moton of the molecule as a whole. The translatonal partton functon s gven by the same formula as Eq. (3.0) wth the mass now gven by the total mass of the molecule, m where m s the mass of a consttuent atom. We have to consder the three rotatonal degrees of freedom and the 3n 6 vbratonal degrees. For a lnear molecule, the rotatonal motonal moton along the molecular axs s quantum mechancally not meanngful as the rotated confguraton s ndstngushable from the orgnal confguraton. Therefore, for a lnear molecule, there are two rotatonal degrees of freedom and 3n 5 vbratonal degrees of freedom. To nvestgate the rotatonal moton, we need to fx the center of mass of the molecule and calculate the three prncpal moments of nerta I, I B and I C of the ellpsod of nerta. The center of mass s defned as the pont for whch the followng denttes hold. m x = m y = m z = 0 (3.56) The nerta products are defned by ( I xx = m y + z ) ; I xy = m x y (3.57) The other components I yy, I xz,.. are defned analogously. To fnd the drecton cosnes ( α β γ ) of the three prncple moments of nerta, we need to solve the followng matrx equatons. α ( I xx η ) β I xy γ I xz = 0 (3.58) α I xy + β ( I yy η ) γ I yz = 0 (3.59) α I xz β I yz + γ ( I zz η ) = 0 (3.60) If the off dagonal terms I xy are zero n the above equatons then the x, y, z axs wll be the prncpal axs. The energy of a rotor wth the three moments of nerta I, I B and I C s gven by

12 L LB LC ε = I ω + I BωB + ICωC = + + = (3.61) I I I B C Here, ω, ωb and ω C are the three angular speeds and L, L B and L C are the three angular momenta. For a symmetrc top molecule such as ammona, or chloromethane, two components of the moments of nerta are equal,.e., I B = I C. The rotatonal energy levels of such a molecule are specfed by two quantum numbers J and K. The total angular momentum s determned by J and the component of ths angular momentum along the unque molecular axs s determned by K. The energy levels are gven by h h ε = J, K B J ( J + 1) + ( B) K ; B =, 8π ci = B 8π ci n cm-1 (3.6) Where, J takes on values 0,1,,.and K = - J, - J + 1, - J +, 0, 1,, J. The rotatonal partton functon s gven by J 1 - BJ(J+1)/ k T (- B)K / k T q ( J + 1) e e rot = σ (3.63) J = 0 K= J Ths can be converted to an ntegral as n (3.14) and the result s 1/ π 8π I BkT 8π I kt qrot = (3.64) σ h h For asymmetrc tops, the expressons for rotatonal energes are more complex and the conversons to ntegratons are not easy. One can actually calculate the sum of terms usng a computer. n ntutve answer can be obtaned by ntegratng over the angular momenta L, L B and L C as H ( p, q) / k T 1/ 1/ 1/ e dl dlc dlc = ( π I k T ) ( π I B k T ) ( π IC k T ) (3.65) nd then multplyng by a factor of 8 π /σ h 3 we get the rotatonal partton functon. The factor of 8π accounts for the angular ntegraton. For any axs chosen n a molecule, a complete rotaton contrbutes a factor of π. Integraton over all possble orentatons of ths axs contrbute another factor of 4π. The symmetry number σ corrects for overcountng of rotatonal confguratons and the factor of h 3 s for the converson from the classcal phase space to the quantum mechancal phase space. The fnal result s 1/ 1/ I kt π IBkT 8π 1/ π 8π 8 ICkT qrot = (3.66) σ h h h We can explctly obtan the classcal rotatonal partton functon of an asymmetrc top. by wrtng the classcal expresson for the rotatonal energy n terms of the Euler angles. The orentaton of a rgd rotor can be specfed by there Euler angles θ, φ and ψ wth the ranges of angles 0 to π, 0 to π and 0 to π respectvely. The rotatonal Hamltonan for the knetc energy can be wrtten n terms of the angles and ther conjugate momenta p θ, pϕ and p ψ as sn ψ cosψ H = pθ ( pϕ cos θ pψ ) I snθ snψ 1/ 1/

13 75 cos ψ snψ 1 pθ ( pϕ cos θ pψ ) pψ I B cosθ cosψ IC The rotatonal partton functon s gven by π π π 1 H ( p, q) / k T 3 h qrot e dpθ dpϕ dpψ dθ dϕ dψ (3.67) = (3.68) The ntegratons can be smplfed by rewrtng H(p,q) / kt as 1 sn ψ cos ψ 1 1 snψ cosψ H / kt = + pθ + ( p cos p ) ϕ θ ψ I kt I IB I B I sn ψ cos ψ snθ + I I B ( p cos θ p ) + p k T I I sn + I I B ϕ ψ ψ B θ sn ψ cos ψ KTIC (3.69) Usng the followng ntegral, a ( x + b) a x Integraton over p θ gves usng the above expresson,, 1/ sn ψ cos ψ ( π kt ) + I I B Integraton over p φ gves the factor, e dx = e dx = π / a (3.70) 1/ 1/ 1/ sn ψ cos ψ π k T I IB θ + I I B ( ) sn (3.71) (3.7) Ths cancels partly the second square root n the earler expresson. Integraton over p ψ gves the factor 1/ ( π k T I C ) (3.73) Integraton over θ, φ and ψ gves a factor of 8 π. π π π snθdθ = ; dϕ = π; dψ = π (3.74) Combnng all the ntegrals, we fnally get, 1/ 1/ I kt π IBkT 8π 1/ π 8π 8 ICkT qrot = σ h h h Here, we have rentroduced the symmetry number σ as before. nd wth 1/ (3.75)

14 76 Γ = h h h,, I kt Γ = I kt Γ = I kt (3.76) B C 8π 8π B 8π C The partton functon smplfes to ln q rot 1 π = ln Γ Γ Γ B C σ (3.77) nd the molar thermodynamc functon can be readly calculated as RT Γ ΓBΓCσ rot 3 1 π rot = RT ln qrot = ln ; Srot = R ln π T = + Γ Γ Γ σ (3.78) V B C nd 3 3 Erot = RT; CV, rot = R (3.79) Improvements over the classcal approxmaton for the rotatonal partton functon have been obtaned and one of the mproved verson s 0 h IC I I B = qrot π kt I I B IC I IB I BIC I IC (3.80) 0 qrot q rot where s the classcal approxmaton. Let us now consder the vbratnal partton functon. We have already consdered the Lagrangan L for the oscllators n Eq. (1.43). We saw there that there were cross terms nvolvng x, y and z components of dfferent atoms. Ths makes the soluton of the problem dffcult. However, t s possble to take approprate lnear combnatons of the coordnates so that the cross terms are elmnated and the classcal Hamltonan as well as the operator correspondng to t contans no cross terms and n terms of the new coordnates, the Hamltonan can be wrtten as, f f h k H = + ξ µ ξ = 1 = 1 Here, the degrees of freedom f s 3n - 5 for a lnear molecule and 3n - 6 for a nonlnear molecule. Here, k s the force constant and μ s the reduced mass for that partcular vbratonal mode whch s referred to as a normal mode. The Eq. (3.81) represents f lnearly ndependent harmonc oscllators and the total energy for such a system s (3.81) The vbratonal frequences are gven by ε f 1 vb = ( n + ) hν = 1 (3.8) 1 k ν = (3.83) π µ The vbratonal partton functon s gven by the product of f vbratonal functons for each frequency.

15 77 q vb Θvb, / T f e = Θvb, / T, Θ vb, j = hν / k (3.84) = 1 (1 e ) Where, the Θ vb, j s called the characterstc vbratonal temperature. The molar energes and the heat capactes are gven by f Θvb, / T Θvb, Θvb, e Evb = k + Θvb, / T (3.85) = 1 1 e C f Θvb, / T Θvb, e V, vb = k Θvb, / T = 1 T (1 e ) (3.86) Fgure 3.3 The three vbratonal modes of water, the symmetrc stretch (3656.7, 5160), the asymmetrc stretch (3755.8, 5360) and the bendng mode (1594.8,90) are shown. The vbratonal frequences n cm -1 and the characterstc temperature n K for each mode are shown n parenthess. Example 3.6 The three characterstc rotatonal temperatures for O are 11.5 K, 0.64 K and K and the three vbratonal temperatures n Kelvn are 1900, 1980 and 330. calculate the rotatonal and vbratonal partton functons at 300K. Soluton: The rotatonal temperature s gven by Θ = h 8π I k The rotatonal partton functon becomes, q rot 1/ 1/ 1/ π T T T = σ Θ Θ Θ B C 1/ π q rot = σ = (1.77 / ) X X 1.97 X.55 = 4.4 ( 300 /11.5) ( 300 / 0.64) ( 300 / 0.590) 1/ 1/ 1/ The vbratonal partton s calculated by takng the zero pont energes as the reference ponts wth respect to whch the other energes are measured. f Θvb, / T f e qvb = Θvb, / T = Θvb, / T 1900/ / / 300 = 1 (1 e ) = 1 (1 e ) 1 e 1 e 1 e = X X = The mplcaton s that whle several rotatonal state sare accessble at room temperature, very few vbratonal states (other than the ground vbratonal state) are accessble 1/

16 Ortho and Para Hydrogens The molecules of hydrogen can exst n two forms dependng on the spns on the two hydrogen nucle. If both the nuclear spns are parallel, the molecule s called ortho and f the spns are antparallel, t s referred to as para (In dsubsttuted benzene, para refers to the two groups at two opposte ends, whle n ortho, they are adjacent or parallel to each other). The spn on the hydrogen nucleus has a magntude of ½ ħ. The presence of nuclear spns leads to very nterestng consequences for the populatons of the rotatonal states and on a macroscopc scale, has consequences on measured entropes and heat capactes as well. The total partton functon of H can be wrtten as q tot = q elec q vb q rot q trans q nucl (3.87) where, the subscrpts refer to the respectve motons. fter half a rotaton, the nucle are supermposed on each other. Snce a proton s a spn half nucleus, the total wavefuncton must be antsymmetc wth respect to the exchange of the partcles..e., Ψ (1,) = Ψ (,1) (3.88) The translatonal moton refers to the moton of the molecular center of mass and has no nfluence on the symmetry of the nuclear wavefuncton. Vbratonal moton depends on the magntude of the nternuclear dstance and has no effect on the partcle exchange. The electronc moton also has no effect on the symmetry propertes of the nuclear wavefuncton. Therefore, the product of the nuclear spn and rotatonal wavefunctons must be antsymmetrc wth respect to the partcle exchange. For the nuclear spn functons, there are four combnatons. One combnaton s a snglet ψ nu, s = α(1) β () α() β (1) (3.89) nd the other three combnatons are the three states of a trplet. ψ nu, t = α(1) α () α(1) β () + α() β (1) (3.90) β (1) β () The rotatonal wavefunctons ( ψ r ) are gven n terms of the assocated Legendre polynomals P m ( x) where x = cos θ. l m m ψ = e ϕ P (cos θ ) (3.91) r l m l l m m / d Pl ( x) 1 d ( x 1) Pl ( x) = ( 1 x ) ; P ( ) m l x = (3.9) l l dx l! dx When the nucle are nterchanged, θ becomes π - θ and φ s changed to φ + π. The polynomals change as P x = P x P x = P x (3.93) l m l m m l ( ) ( 1) l ( ); l ( ) ( 1) l ( )

17 79 The exponental functon changes as m ( ϕ + π ) m π m ϕ m m ϕ e = e e = ( 1) e (3.94) Therefore, the rotatonal wavefuncton changes as l m m l ψ ( π θ, ϕ + π ) = ( 1) ( 1) ψ ( θ, ϕ) = ( 1) ψ ( θ, ϕ) (3.95) r r r ψ r = symmetrcal for even j, and antsymmetrcal for odd j (3.96) Combnng the nuclear spn and the rotatonal parts, we see that, the product ψ r ψ nu must be antsymmetrcal (wth respect to the exchange of nucle)for half ntegral nuclear spns and symmetrcal for ntegral spns. To accomplsh ths, the snglet nuclear states (para) must be combned wth the even rotatonal functons and the trplet nuclear states must be combned wth the odd rotatonal states. The rotatonal partton functons for ortho and para hydrogens are, thus, j ( j 1) R / T qortho = qnu, t qr, odd = 3 ( j + 1) e + Θ (3.97) nd para nu, s r, even j= 1,3,5.. j ( j 1) R / T (3.98) j= 0,,4.. q = q q = 1 ( j + 1) e + Θ Where Θ R s the rotatonal temperature defned by Eq. (3.1). The total partton functon consstng both ortho and para hydrogens s gven by rot, nu j ( j 1) R / T j ( j 1) R / T + Θ (3.99) j= j= 1,3,5.. q = 1 ( j + 1) e + 3 ( j + 1) e The rato of ortho to para hydrogens at thermal equlbrum s gven by 3 ( j + 1) e o j= 1,,3.. = ( j + 1) e p j= j ( j+ 1) ΘR / T j ( j+ 1) ΘR / T t hgh temperature, the two summatons become equal and therefore, the hgh temperature lmt of o p s 3. t low temperature, the rato becomes, (3.100) o p ΘR / T 3(3 e +...) 5 ΘR / T 1(1+ 5 e +...) = 0, as T 0 (3.101) good expermental verfcaton of the above analyss s a comparson between the calculated rotatonal heat capactes at constant volume (C V ) rot, nu (calculated from < E > T, where <E> = ln q rot, nu β ). The heat capactes are shown as a functon of temperature n Fg. 3.4.

18 80 Fgure 3.4 The heat capactes of o- and p- hydrogens as a functon of temperature. The curve marked Exp depcts the expermental data, the curve eq represents the data for an equlbrated mxture of o- and p- at a gven temperature and the curves o- and p- represent the heat capactes of pure o- and pure p- hydrogens. The curve marked Exp gves the expermental data, the curve eq represents the data for an equlbrated mxture of o- and p- at a gven temperature. The curves o- and p- represent the heat capactes of pure o- and pure p- hydrogens calculated from the o- and p- partton functons gven by Eqs. (3.97) and (3.98) respectvely. Intally t was a puzzle as to why the expermental data dffers from the calculated values. In fact, the expermental data seemed to agree very well wth the followng equaton. 3 1 C = C ( ortho) + C ( para) (3.10) ( ) ( ) ( ) V rot, nu V rot, nu V rot, nu 4 4 The reason for ths s that, when H s cooled down from a hgher temperature, the ortho para rato contnues to reman 0.75 / 0.5 (the hgh temperature value) because the ortho para nterconverson rate s very very small and we do not reach the equlbrum composton unless a catalyst such as actvated charcoal s added to the gas mxture. Eq. (3.10) corresponds to a frozen hgh temperature mxture of ortho para hydrogens. In the presence of the catalyst, the experments also gve the curve labeled as eq n the graph. Ths n ndeed a very nce case where the experments support not only the methods of statstcal thermodynamcs but also of the antsymmetry prncple for bosons 16 and fermons. If we consder the case of O, where the nuclear spns are zero, the rotatonal wavefuncton has to be symmetrc as only symmetrc wavefunctons are permtted for bosons. Thus, only even rotatonal states contrbute to the partton functon.

19 81 16 rot, nu j ( j 1) R / T (3.103) j= 0,,4.. q ( O ) = ( j + 1) e + Θ Chemcal Equlbra Consder a reacton whch may be represented as a + b B c C + d D + e E (3.104) The equlbrum constant K eq for the reacton s gven by o G = RT ln K eq (3.105) Where the superscrpt o represents the standard state,.e., the state at 1 bar pressure at each temperature. The free energy of a speces s gven by eq. (3.39) as q G = G(0) n RT ln m (3.106) nd the molar free energy n the standard state for a speces J s gven by o o o qj, m GJ, m = GJ, m(0) n RT ln (3.107) The standard free energy change for ths reacton s gven by o o o o o o o G = G = cg (0) + d G (0) + eg (0) ag (0) bg (0) reacton r C, m D, m E, m, m B, m o o o o o qc, m qd, m qe, m q, m qb, m c RT ln d RT ln e RT ln + a RT ln + b RT ln (3.108) But snce G(0) s the same as E(0), o o o o o o o G = E = cg (0) + d G (0) + eg (0) ag (0) bg (0) (3.109) ( ) ( ) 0 r 0 r C, m D, m E, m, m B, m nd the equlbrum constant s gven by o ( 0 ) o c o o ( qc, m ) ( qd, m ) d ( qe, m ) a b ( q, m ) ( qb, m ) RT ln K = E RT ln (3.110) r o o e c o o o ( qc, m ) ( qd, m ) d ( qe, m ) o a o b ( q, m ) ( qb, m ) e o ( E ) K = e 0 / RT r (3.111) It s nstructve to analyse K of Eq. (33.111) n terms of molecular energy levels and ther populatons. Consder the energy levels for the followng equlbrum, B (3.11) The energy levels of systems contanng only, only B and both and B are schematcally shown n Fg. 3.5.

20 8 Fgure 3.5 (a) Energy levels of ndvdual molecules, B and the ndvdual populatons of these levels, (b) Energy levels of the combned system and B and the relatve populatons of the combned levels at equlbrum (schematc). If the molecules of and B are kept n separate contaners, each contaner wll be n a state of equlbrum wth the populatons of and B gven by ther separate Boltzmann dstrbutons. Ths s shown n Fg 3.5(a). The partton functons of the systems and B are gven by q and q B. Fg 3.5(b) shows the combned system and the populaton of the levels of the combned system. The populaton of the th level of the combned system s gven by β E e n = P = (3.113) q Where q s the partton functon of the combned system and E s the energy of the th level of the combned system. In the combned system, the total populaton of molecules at equlbrum s gven by summng n for molecules. Smlarly, for B. We thus have, β ( E ) = n = e = q q q (3.114) ll levels of levels of = n = e = e = q e (3.115) β ( E ) B β [( E ) B + E0 ] E0 / RT B B ll levels of B q levels of B q levels of B q

21 83 The prme n eq (33.16) ndcates that the levels (E )' B are now measured from the lowest level of the combned states of and B, whch s (E 0 ). Expressng E' n terms of E, we have, E = E + E E = E + E (3.116) ( ) ( ) ( ) ( ) ( ) B B 0 B 0 B 0 The equlbrum constant s now gven by K = B/ = (q B / q ) 0 e E / RT (3.117) For a more general reacton + B I C + D, the equlbrum constant s gven by K = [ q C q D / q q B ] e (3.118) Where, ΔE 0 = (E 0 ) C + (E 0 ) D - (E 0 ) - (E 0 ) B (3.119) 3.5 ctvated Complex theory In many reactons, mxng the reactants does not lead to products nstantaneously. short lved hgh energy confguraton of molecules, referred to as the actvated complex s formed whch may lead to products or the reactants dependng on the specfc motons n the actvated complex. lthough the terms transton state and actvated complex are often used synonymously, the transton state does not have a chemcally sgnfcant lfe tme. Consder the reacton + B C P (3.10) The potental energy dagram for ths reacton s gven n Fg 3.6. In Eq. (3.10), C s the actvated complex whch forms the products P wth a unmolecular rate constant k. Fgure 3.6. Potental energy of reacton as a functon of the reacton coordnate. The energy levels of the shallow actvated complex C # are shown n the nset to the rght. The ordnate s the potental energy (PE) and the absssa s the reacton coordnate. t a specfc confguraton of the reactants, the potental energy s maxmum, the slope s zero and the PE falls to lower values n both the forward and the reverse drecton. ll the structures n the vcnty of ths transton state or confguraton may be consdered as the "actvated complex", whch s very reactve. moton along the "forward" drecton wll

22 84 lead to the products. The actvated complex theory or the transton state theory provdes a way to calculate the rate constant for the reacton. The assumptons nvolved n the transton state theory are: 1) The electronc moton (whch can only be descrbed only quantum mechancally) may be separated from the moton of the nucle and a classcal descrpton of the nuclear moton s used to evaluate the rate constant. ) The energy dstrbuton of the reactants s descrbed by the equlbrum Boltzmann dstrbuton throughout the reacton. 3) The actvated complex whch has crossed the transton state can not return to the reactant confguraton. In the transton state, moton along the reacton coordnate may be separated from all other motons. 4) The actvated complex s also dstrbuted accordng to the Boltzmann dstrbuton. The rate of the reacton can thus be expressed as d [C ] / dt = k [C ] (3.11) The concentraton of C s expressed n terms of the concentratons of and B through the proportonalty constant K (whch s not an equlbrum constant). [C ] = K [] [B] (3.1) substtutng Eq. (3.1) nto Eq. (3.11), we have, d [C ] dt = k [] [B] where k = k K. (3.13) We now wll determne the values of the constants k and K. It should be obvous that the process represented n Eq. (3.1) s not a regular equlbrum process because, as the products are formed, [] and [B] are decreasng wth tme and [C ] s not ncreasng wth tme snce the collsons responsble for the formaton of C are themselves decreasng n frequency as the reacton evolves. The actvated complex can exhbt several motons such as translatons, rotatons and vbratons, snce t has a fnte lfe tme. The moton leadng to the product s assumed to be a specfc vbraton wth a frequency ν. Ths vbratonal moton along the reacton coordnate s equated to the constant k. The transton state s n a shallow well wth a vbratonal frequency much smaller than that of the reactants and the products. The energy levels of the shallow actvated complex C are shown to the rght n Fg k ν (3.14) Our next task s the determnaton of the concentraton of C. Let us express the equlbrum constant for the reacton + B C n terms of the partal pressures of, B and C (p, p B and p C, wrtten smply as p C ). K = [C ] / [ ] [ B] = [ p C / p ] / [ ( p / p ) ( p B / p ) ] (3.15) Where p s the standard pressure of 1 bar. These partal pressure can be expressed n terms of molar concentratons of, B and C as follows. [] = n / V ; V = n RT / p; p = x p = n p / n (3.16)

23 85 [] = n p / nrt = x p / R T = p / RT or p = RT [] (3.17) In Eq. (3.16), V = total volume, p = total pressure, n = total number of moles, n = number of moles of, x = n / n, the mole fracton, R s the gas constant and T absolute temperature Substtutng each partal pressure n terms of the mole fractons, we have K = RT [C ] / p / [(RT [] / p ) (RT [B] / p )] (3.18) = [C ] p / [] [B] RT (3.19) or C = [] [B] RT / p K (3.130) Comparng ths equaton wth Eq. (3.1), we have or K = K RT / p (3.131) In Eq. (3.1), K was merely a proportonalty constant and now t has been related to the equlbrum constant for the formaton of the actvated complex Specfc vbraton of the actvated complex along the reacton coordnate leads to product formaton. When we have a reacton such as + B C, the equlbrum constant gven n terms of partton functons by (see Eq ). K = ( q C / q q B) E 0 (3.13) Where ΔE 0 = (E 0 ) C - (E 0 ) - (E 0 ) B. Here s the vogadro number. In the above equaton, we have suppressed the standard state and molarty labels o and m of Eq. (3.111). Returnng to Eq. (3.13), k = k K, the unmolecular rate constant k s taken as proportonal to the vbratonal frequency along the reacton coordnate k = ν. The partton functons for ths vbratonal moton of the transton state s e β h / kbt ( q V ) C = 1/ (1- ) (3.133) Usually ths frequency s qute small because the transton state falls apart nto products wth ths frequency. Very hgh frequences of the transton state are mprobable and they would lead, not to the usual products but to rather hghly dsntegrated products. Expandng the denomnator, e -hν/k B T 1- hν/k B T, ( q V ) C = k BT / h ν (3.134) Wrtng q C n terms of the vbratonal modes (q v ) C [ whch s approxmated as k B T / h ν] and all the other remanng modes [ qc ], we have for q C, q C = (k B T / h ν) qc ( 3.135) e ν Combnng Eqs. (3.13), (3.131) and (3.13), we have, k = k K = ν ( RT / p ) K = (ν k BT / h ν)[( q / q q B) E C e β 0 RT / p ) (3.136)

24 86 = k B T / h K (3.137) Where K s the second, square bracketed expresson n Eq. (3.136) whch s akn to an equlbrum constant wth the vbratonal mode of C removed. There are often factors not ncluded n Eq. (3.137) and they are ncluded through a transmsson coeffcent κ and the rate constant n the transton state theory becomes, k = κ ( k B T / h ) K (3.138) We want to express K n terms of molecular partton functons. Let us obtan a formula for smple speces of and B where and B are atoms. The partton functon of atoms s smply the translatonal partton functon (as rotatons and vbratons are absent). The translatonal partton functon of, usng the molar volume, becomes q 0 3 tr, Vm / ( ) 3/ 0 π q m k T V tr, = B m = Λ Λ = h /, where ( π m k T ) 1/ B (3.139) (3.140) Where V 0 m s the standard molar volume, gven by RT / p. For the actvated complex, the partton functon s the product of translatonal, vbratonal and rotatonal partton functons, because n our present model, B s a datomc. However, we have already consdered the vbratonal partton functon n Eq. (3.135) and we need to consder q C now whch ncludes only rotaton and translaton. Ths s gven by q = ( I k T / h ) V Λ (3.141) C B 0 3 m C The frst term n the brackets s the rotatonal partton functon of the actvated complex. The moment of nerta I s gven by μ r where r s the "bond length" of the actvated complex and μ= (m m B ) / ( m + m B ) s the reduced mass. Substtutng the values of q, q B and q C n equatons (3.136) and (3.138), we get k = κ (k B T / h)(rt / p ) Λ Λ / Λ ( I k B T / h E ) 0 (3.14) V B C m Cancelng RT / p and V 0 m ( whch are equal ) and substtutng the values of Λ C, we get e β Λ, Λ B and k = (8 k BT /πμ) 1/ (κ π r E ) 0 (3.143) e β If we dentfy the reacton cross secton as κ π r, ths equaton s dentcal to the equaton derved usng a smple collson theory. It s ndeed remarkable that two very dfferent approaches gve the same result! Does t mean that ths result s more "correct" than the result that s usually obtaned from a sngle theory? Whle t would be temptng to say yes, what ths means s that we have captured some of the essental features relevant n the dynamcs of chemcal reactons n both approaches. Further mprovements wll come when we consder the cross sectons of each of the states of the reactng speces and also when we remove the requrement that C s not n equlbrum wth and B. In the next secton, we relate k to the actvaton parameters for the reacton

25 87 ctvaton Parameters In the rrhenus theory, the only actvaton parameter that was ntroduced was the actvaton energy. In the transton state theory developed n ths chapter, we have related the concentraton of the actvated complex to the reactant concentratons through an equlbrum constant K. Treatng ( p / RT) K of Eqs. (3.19) and (3.136) as an equlbrum constant (although one vbratonal mode s removed from C ), we can defne the Gbbs free energy of actvaton ΔG as ΔG = - RT ln ( K p / RT) (3.144) the rate constant k becomes k = κ k B T /h (RT / p ) e G / RT e (3.145) The free energy of actvaton can be dvded nto enthalpy and entropy terms (analogous to G = ΔH TS ) through ΔG = ΔH - T ΔS (3.146) For the tme beng let us ether take κ = 1 or nclude t n the entropy term. The rate constant then becomes S / R H / RT = wth B = (k B T / h)( R T / p ) (3.147) k Be e The actvaton energy of the rrhenus equaton k = e- Ea/RT, E a s defned through E a = R T ( d ln k / dt). (3.148) Substtutng k nto ths, we get Ea = H + RT (3.149) Substtutng ths n Eq. (3.147), we get k = e - B S / R e - E a / RT (3.150) and the rrhenus factor becomes = e - S / R e (3.151) Usually, when and B form a complex (B), more orderng s created and entropy s reduced. ΔS s thus negatve. Ths s based on a collson complex model for the reacton. In addton to ths decrease, there s a further orentatonal or sterc decrease due to the preference of only certan confguratons for the actvated complex (as n the case of a reacton between molecules such as C H 4 and H ). The sterc factor P of collson S / theory may be assocated wth sterc R e,.e., / P = S sterc R e, (3.15) where, S = Scollsonal + Ssterc (3.153) Smlar to actvaton enthalpes and entropy, actvaton volume can also be defned through ( G ) p = V (3.154) For reactons n soluton, these ΔV values are small, but for gas phase reactons, they may be approxmated by the deal gas values e

26 88 P V = n RT (3.155) Example 3.10 For the K + Br KBr + Br reacton at 300 K, fnd the thermodynamc actvaton parameters ΔG, ΔH and ΔS of the transton state theory. The rrhenus parameters for the above reacton are gven to be = 1.0 * 10 1 M -1 s -1 and E a = 0 kj/mol. Soluton: ΔS = R ln { h p / e k BT } Here h = 6.65 * Js, p = 1 bar = 10 5 J/m 3, = 6.0 * 10 3 mol -1 k B = 1.38 * 10-3 J/ K, = 10 1 M -1 s -1 = 10 1 mol -1 dm +3 s -1 Substtute all these values and show that h p / e k BT = 7.81 * (/M -1 s -1 ) / (T/K) Substtutng the above value n the equaton for ΔS, we have ΔS = R ln { 7.81 X X 10 1 X 300 } = - 58 JK -1 mol -1 ΔH = E a - RT = 0 X * 300 J / mol = -.5 kj / mol ΔG = ΔH - TΔS = X (-58) / 10-3 kj / mol = 15 kj / mol Summary The populatons or the occupaton numbers of dfferent levels that are avalable for molecules are expressed n terms of partton functons whch measure the number of avalable states wth each state weghted by the Boltzmann factor, e - E / kt. Snce the total molecular energy can be wrtten as a sum over electronc vbratonal, rotatonal and translaton energes, the total partton functon was expressed as the product of electronc, rotatonal, vbratonal and translatonal partton functons. Separate expressons were obtaned for these partton functons ther estmates were gven. The partton functon for a collecton of molecules and the average energy of a system was obtaned n terms of partton functons. The relaton of partton functons to thermodynamc functons and equlbrum constants was outlned. The rato of ortho and para hydrogens n a H mxture, ther heat capactes was derved n terms of ther partton functons and the nterestng connecton wth experment was shown. The use of partton functons n chemcal knetcs and ther use n estmatng actvaton parameters of chemcal reactons has been llustrated. References tkns, P. W. Physcal Chemstry, 7 th ed, Oxford Unversty Press, 006. McQuarre, D.. and Smon, Physcal Chemstry, Molecular pproach, 003 Hll, T. L. n Introducton to Statstcal Thermodynamcs, Dover Publcatons, Mayer, J. E. and Mayer, M. G. Statstcal Mechancs, nd ed, John Wley and Sons, Davdson,. Statstcal Mechancs, McGraw-Hll Book company, 196. McQuarre, D.. Statstcal Mechancs, Vva Books Pvt. Ltd., 003 Problems

27 89 1) If the nucleus has a spn of s n, then ts spn degeneracy g n, = s n + 1. The datomc molecule formed from such a nucleus wll have g n spn functons whch have to be combned to form symmetrc and antsymmetrc functons. Carry out an analyss smlar to that of H for D where the deuterum nucleus has a spn of 1. ) Derve the thermodynamc functons from the polyatomc rotatonal partton functon. 3) Carry out the ntegraton for the rotatonal partton functon of the symmetrc top. 4) Calculate the total partton functon and the thermodynamc functons of water at 1000K. The three moments of nerta of water are 1.0, 1.91 and.9 n kg m. The symmetry number s. The vbratonal data n gven n Fg ssume a nondegenerate electronc ground state. 5) Verfy that the symmetry numbers for methane, benzene and SF 6 are 1, 1 and 4 respectvely. 6) The ground state of a s a doublet (two states wth the same energy). ssumng ths to be the zero of energy and assumng that the next energy level to be ev hgher than the ground state, calculate q el. 7) The bond length r e of O s 1. Å. The moment of nerta I s m r e / where m of O s 16 X 1.66 X10-7 kg. Calculate B and the rotatonal partton functon of O at 300 K. 8) The vbratonal frequency ν of ICl s 384 cm -1. What s ts vbratonal partton functon at 300 K? What s the fracton of molecules n the ground state (n = 0) and the frst excted state n = 1? 9) Calulate the translatonal partton functon of at 300 K. For volume, use the molar volume at 300 K. 10) n sotope exchange reacton between sotopes of bromne s Br Br + Br Br Br Br The fundamental vbratonal frequency of Br Br cm -1. ll the molecules can be assumed to have the same bond length and have a snglet ground electronc state. Calculate the equlbrum constant at 300K and 1000K. 11) For the reacton I I, calculate the equlbrum constant at 1000K. The relevant data are as follows. The ground electronc state of I s P 3/ whose degeneracy s 4. The rotatonal and vbratonal frequences of I are cm -1 and cm -1 respectvely. The dssocaton energy of I s 1.54 ev. 1) The representatve molecular data for a few molecules s gven n table 3.1. Usng the relevant data, calculate the equlbrum constant for the reacton H + Cl I HCl at 1000K. What s the value of the equlbrum constant as T?