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# Adv Physical Chemistry CHEM 6320

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This 29 page Class Notes was uploaded by Mr. Clementine Gottlieb on Wednesday October 21, 2015. The Class Notes belongs to CHEM 6320 at Tennessee Tech University taught by Staff in Fall. Since its upload, it has received 24 views. For similar materials see /class/225694/chem-6320-tennessee-tech-university in Chemistry at Tennessee Tech University.

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CHEM 3520 Spring 2004 Basic Concepts of Thermodynamics Thermodynamics 7 literally means heat movement 7 deals mainly with the energy but it also deals with other measurables too an important aspect of thermodynamics is understanding of how these measurables relate to each other 7 is the study of various properties and particularly the relations between the various properties of systems in equilibrium 7 is primarily an experimental science that was developed in the 19th century The development of thermodynamics along these lines is called classical thermodynamics It is governed by few laws that are followed by all systems Its development did not involve knowledge about the molecular structure but also could not give insight about the system at the molecular level The statistical thermodynamics does that however 7 We will cover the material as a mixture of classical and statistical thermodynamics Terms and de nitions 7 System the part of the world Universe that it is investigated 7 Surrounding everything else 7 A system can be described by indicating 7 its volume 7 its pressure 7 its temperature 7 its chemical composition the number of atoms 7 its chemical reactivity 7 Equations of state are mathematical formulations connecting system s variables 7 The quantities variables or properties describing macroscopic systems can be divided in 7 Extensive properties 7 are directly proportional to the size of the system 7 Examples V m E or U 7 Intensive properties 7 are independent of the size of the system 7 Examples P T p V CHEM 3510 Fall 2003 Applications onuantum Theory The Particle in a Box Translational Motion 7 The timeindependent threedimensional Schrodinger equation 2 2 2 2 h a 8 2Vxyz 7 2 2 wxyzEvxyz 239quot 8x 8y 82 7 Consider only a onedimensional motion 5 drop y andz in the equation above 7 Consider that the particle experiences no potential energy between the position 0 and a 5 Vx0 forO gxga The particle is restricted to the region 0 g x g a Potenhal energy a a x Wall Wall 2 2 l E V 2m dx2 7 The mathematical solution of the equation is Vx Acoskx Bsinkx 12 2 2 Where k ZmE 27M2mE E h k h h 87r2m 7 Apply the boundary conditions and the normalization condition to determineA and B 40 0 s A 0 Va 0 5 Bsinka 0 5 ka mt Wheren12 7The allowed energy levels are The energy is quantized hzn2 gt En 8ma2 Where n 1 2quot quot n is called quantum number CHEM 3510 Fall 2003 Classically allowed h 1 1 th b 9 energies 7T e energy eve s1ncrease as e quantum num er increases 7 The energy separation between the allowed energy levels increases as the quantum number increases 7 The energy levels and the energy separation between the energy levels increases as the size of the box decreases 7 Quantum numbers appear naturally when the boundary conditions are put in the Schrodinger equation like in the string problem and are not introduced ad hoc like in Plank model for blackbody radiation or Bohr model of hydrogen atom 7 FindB by setting the condition that the function Vx is normalized 7 jwltxwxdx 71 0 a B2 sin2 2 1 1 0 gt 321 gt B J2 B is callednormalization constant 1 7 The normalized eigenfunctions 11 x are given by 12 Lxnx sin wherenggaandnl2 1 L1 543 2 I 5 Solving the Schrodinger equation for the particle in a box problem gives a set of allowed energies eigenvalues and a set of wave functions eigenfunctions CHEM 3510 Fall 2003 Represent and investigate the energies the wave functions a and the probability densities b for the first few levels for the particle in a box N n E 1M 1 11 V MUN quotaquot n We 712 3 V n71 1 E39 917 33949 J N39s quot1 H wEA39J mun 1 1 31m 2 T A 1F 1 El nfr fulltx m C quot7 II J II x a x 1391 EM 7 The particleinabox model can be applied to electrons moving freely in a molecule free electron model 7 Example butadiene has an adsorption band at 461 X 104 cm 1 As a simple approximation consider butadiene as being a onedimensional box of length 578 A 578 pm and consider the four 11 electrons to occupy the levels 0 a 3 calculated using the particle in a box model 2 h 2 32 22 mea 7 The electronic excitation is given by AE 7 The calculated excitation energy 17 454 X 104 cm 1 compared very c 1 well with the experimental value 7 This simple freeelectron model can be quite successful CHEM 3510 Fall 2003 X2 7 The probability of nding the particle between x1 and x2 is given by Ill xzn xdx 1 71f x1 0 and x2 aZ then Prob 0x a2l2 forall n 7For nl Prob0 x a4 lt Proba4 x a2 7 As n increases for example n 20 these 2 probabilities become the same 7 More general The probability density becomes uniform as n increases This is an illustration of the Correspondence Principle that says that quantum mechanics results and classical mechanics results tend to agree in the limit of large quantum numbers The large quantumnumber limit is called the classical limit a 7 The eigenfunctions are orthogonal to each other Iw xz xdx 0 Where m i n 0 9 7 Example look at z1x and 13 x a 2a 71x 371x Ill1xl3xdx Jsin sin dx 0 0 a 0 a a 7 Look at the average values and the variances for the position and the momentum of the particle a 2 a nizx a 7 Average value of position x Illn x x yn xdx Jxsin2 0 a 0 a 2 a 2 2 2 a a 7 Average value of pos1tion square x gt Jyn x x 41quot xdx 7 0 3 2n 2722 2 2 2 7Variance in position a x2gt7ltxgt2 a j n 72 2n72 3 2 2 12 7 Standard deviation in position ox ltx2gt 7 x2 a n 7 2 2n72 3 Znna nizx nizx 7 Average value of momentum p 71h Ism cos dx 0 a2 0 a a 2 2 2 2 2 2 2 2 2 h h h 7Also 72 2mltEgt 2m 2 a ma a a 12 2 2 7 Finally oxop n 3 7 2 gt Heisenberg Uncertainty Principle V n 0 Solutions 11quot quotyquot x yz FEFsin nxm sin y y 47 2 x Z a c CHEM 3510 Fall 2003 h2 62 62 7 Cons1der a twod1mens1onal motion W W E 2m 6x23y2 Way my XxgtYygt sin x nyny Fcienual energy sin a b 2 2 2 E 11 nxny x quotmy 8m a2 b2 Panicle con ned to sudace 7 Consider a generalization to athreedimensional motion Particle in a 3dimensional box 2 2 2 2 h6 11 a 11 a 11 6y2 az 2 ELxyz 6x2 nznz sin a b c 2 2 n2 2 En n n 1 nx y quot 2 x y 2 SM 612 b2 62 7 Look at the average values of the position and the momentum of the particle a b c i A a b c ltrgtjdxjdyjdw xyzRuxyz315J k 0 0 0 where liXifjZAk a 17 c A ltpgt jdxjdyjdax xyzP wxyz 0 0 0 0 where 13 7203 ji 6 i k K 6x By 62 CHEM 3510 Fall 2003 7Lookatthecaseofabc a Dcnmmt 2 6 7Three sets of quantum numbers give same energyE2n E121 E112 Sma2 7 g 512 7 The energy level E is degenerate murmur J 8ma2 cmztzum 7 The energy level Em 2 is nondegenerate 41x SM 0 7 D generacy is due to the symmetry ofthe system Once the symmetIy is Energy level Ofpm de a be destroyed then degeneracy is lifted 7 Look at the Hamiltonian for the particle in a 3dimensional box 177 V Fy i HxHy Hz A 2 2 2 2 It can be wrinen as a sum ofterms where H 7h a h a Zrnax2 2m ayZ A 12 62 Hz 72 2 The operator is said to beseparable quot1 67 7 The eigenfunctions ynxnynl are written as a product of eigenfunctions of each operator I y and 32 and the eigenvalues Enxnynl are wrinen as a sum of the eigenvalues of each ofthe operator IQ39X y and IQ39Z 7 This is a general property in quantum mechanics If the Hamiltonian or an operator in general can be written as a sum of terms involving different coordinates ie the Hamiltonian is separable then the eigenfunctions of I is a product ofthe eigenfunctions of each operator constituting the sum and the eigenvalues of ii is a sum of eigenvalues of each operator constituting the sum r 7 I93lts1 392w 7 wow 7 Kmw 2 Em EK Em 31W x 7 EM 0 32 WM 7 EmuM where CHEM3510 Fall 2003 The Harm uni Oscillztnr Vibratinnal Mntinn The harrnonre oseruatorrn e1assrea1 rneehanres eForee f ekaezn kx Hooke39s Law rde where krs the fame constant i n 2 eHooke39s Law eornbrned wrth Newton39s equation mszwq 0 dt eThe solution of the equatron x0 Aeom where QJ7 eThe potenaal energy othe oseruator wr 2 2 The tota1 energy rs conserved rt rs transferred between Kand V The harrnonre motion tn a dratornre rno1eeu1e e Consrderthe movement of atorns of masses m and m 15 m m er 7 1 d1 2 I n 1er m 7km 7 Kt 7 In 122 eBy surnrnrng the two equataon above M eo Where Mmm2 and grew 122 M where er the center the W55 wardmate e Subtracting equanon 2 dwrded by m from equanon 1 dwrded m d 1q 0 where r x er etn and n W all m m where 1 rs the reduced W55 of the systern and x rs the relatxve coardmate systern wrth amass equal to the redueedrnass of the tworbody systern CHEM 3510 Fall 2003 The quantummechanical harmonic oscillator 7 The Schrodinger equation for quantummechanical harmonic oscillator if d2 7 2W mom 7 EW 21 dx But Vx kx2 Potential V dzvx 21 1 2 sh Z E7310 yx70 7 The solutions of this equation eigenvalues are Evh vijhwvljhyvij 2 2 2 Where v012 w E VL E 27 y 7 The eigenvalue representation in textbook is Wrong 7 The wave functions eigenfunctions are 7 z wvltx7NvHvltal2xe quotx Displacemem Where a k hz 14 7 The normalization constant is given by NV Z 1 2 U v 7 7 The wave functions form an orthogonal set 7 The Hva12x are polynomial functions called Hermite polynomials HV 5 in a vth degree polynomial in 5 Here are the rst few Hermite polynomials Ho 1 H1E2 11125 42 7 2 11135 7 83 712 Hug 164 7 4852 12 H3E 325 716053 120 even polynomials fx f7x odd polynomials fx 7f7x 7 Some properties of odd functions 71fthe function is continuous f0 0 A 7 Also food 0 7A CHEM3510 Fall 2003 eTlne normallzed wave funenons andthe probablllty denslty L M mull ln ml 239quot Mu l llrmr l e The exlstence of zerorpolnt energy 7 The rnrnrrnurn energy the energy ls calledthe zerapamt Eng 1 ZPE hv 2 20H 5 l lllll llllwlul groundrstate energy ls not zero even for v 0 Thls z example p 0 atthe same tlme example dlatomlc rnoleeule Ills amodel for vlbranons m dlatomlc rnoleeules 5212mm rule Av 1 frequency m the spectrum ofa dlatomlc the frequency called mdamenml yrmanmzz frequency v abs AE J vllrv 2 l Vabs ii 72 Fall 2003 CHEM 3510 7 q anuty xfrom above wru be m thrs ease the dAfference between the rnteratomre dutance dunng the mbrauon and the equrhbnum dutance x 1 71 eonstant k 272 Vom o E ltV t tt meehames regrons forbxdden by e1assrea1 meehames eThe average value of posrtron and momentum for the harmoan oseruator ltxgt Wm x WW 2ltxgt0 ltp 1w ltxgtr m jwm 2 ltp The Quantum Machanical Rigid Rntatnr Rntztinnal Mntinn eIt rs amodel for arotatrng 64amme mo1eeu1e eFor the eenter ofmass mm my I Mquot eVeloerues v n 27mm qm v2 5 27mm 54 Mm 7mm rs the angular speed where a where r 2 and a m1m2 e Srmuar to moment ofmema for a smgle rotaung pamcle 1 mt system where the mass rs replaced wrth the redueed mass 7 Angular momentum L CHEM 3510 Fall 2003 7 Solving Schrodinger equation for quantummechanical rigid rotator model 19V01I1 iv2 2 62 62 62 6x2 6y2 622 1 a 2 a 1 a a 1 62 2 I 2 I S111 6 2 r 6 quot 6 quot w r Sin 9 69 69 w r Sin 9 6 r 8 7 A special case is obtained when r is constant the rotator is rigid 19mg EYW V2 where Y 19 are the rigid rotator wave functions 2 2 h 1 sinei L 6 Y19 EY19 21 s1nt9 619 619 mm my 7 Further rearrange 2 2 sinei sin a Y 6 Y sin2 19Y 0 619 619 392 where g h 7 The rigid rotator wave functions Y 12 will be given later 7 By solving the equation above one can determine that must satisfy J J 1 7 The discrete set of allowed energy levels is given by 2 3 EJ JJ l where J 012 7 Each level has a degeneracy given by g J 2 1 7 The selection rule for the rigidrotator model A i1 only transitions between adjacent states are allowed 7 The energy and frequency for transition between levels R2 AEEJ1EJ 2 47239 I J1hv h 4721 3v Jl CHEM 3510 Fall 2003 7 The frequency of these transitions is about 1010 1011 Hz which is in the microwave range of electromagnetic radiation 3 microwave spectroscopy 7 Usually write the frequency in terms of the rotational constamB V ZBJ 1 where B 2 87239 I 7 In terms of wavenumbers 17 2 J 1 where E 2 87239 c 7 From experimental E one can determine the moment of inertia I then the bond distance r in a diatomic molecule 1 Llr2 J 4 l m 5531 3 J 39139 F 3 x 9 3565quot N 3 1 S E 1 41 I 1525 o q um 1 quot15 tammrum I n 35 39539 55 35 Fall 2003 CHEM 3510 Spherical Coordinates r 0 95 and Caltesian Coordinates x y z The position ofa point can be speci ed by using Cartesian coordinates xyz or by using spherical coordinates r t9 1 Z Z 2 E xrsint9c0s r x y 2 y rsint9sin and c0st9 Z 1 zrc0st9 x2y2zz7 tan X z oo oo oo 007r27r dVdxdydzr2sin6drdad j j j dxdydz H Irzsin drd6d 700700700 00 0 oo oo oo 00 It 27 j j j Fxyzdxdydz jrzdrjsinada jd Fr 9 wiqu 0 0 0 1 i a 2 16 262 6x2 6y2 622 6 quot 62 62 1 a 2 a 1 a a 2 r 2 s1nt9 r 6 quot 94 r s1nt96 9 6 9 r r s1n t9 CHEM 3520 Spring 2004 The Second Law of Thermodynamics The second law of thermodynamics shows that isolated systems not in equilibrium evolve such to increase their disorder It introduces another state function called entropy that gives a quantitative measure of the disorder of the system 7 What is the force that determines if a process will proceed spontaneously or not 7 A criterion for a reaction or process to proceed spontaneously is to be exothermic or to evolve energy 7 This idea is coming from the classical idea of rolling down hill The variational principle of quantum mechanics also said that a system seek its lowest energy state 7 There is another condition or force that drives spontaneous processes The following processes obey first law of thermodynamics but cannot be explained by it An analysis in a microscopic or molecular point of view shows that in these processes there is an increase in disorder or randomness of the system 7 Expansion against vacuum two bulbs one with gas and one empty separated by a stop clock By opening the stop clock the gas occupies both bulbs although AU or AH is essentially zero 7 Mixing of gases two bulbs with different gases By opening the stop clock the gases mix and both occupy both bulbs although AU or AH is essentially zero 7 Melting of ice at temperature higher than 0 C AquHO 60 kJmol endothermic process but still spontaneous 7 Reaction of BaOH2 with NH 4NO3 BaOH2 s 2NH4NO3 s gt BaNO32 s 2H20l 2NH3 g reaction is endothermic and the system can cool down to 20 C 3 The systems evolve spontaneously in a direction that lowers their energy and that increases their disorder Entropy a quantitative form of disorder 7 It is a state function 7 Consider the heat transfer associated with a reversible process in which both T and Vof an ideal gas are changed 5qu dU 5me CV TdT PdV CVTdT nRT VdV CHEM 3520 Spring 2004 62 f 62 f the cross derivative are not equal I Eqrev 1s an 1nexact d1fferent1al axay an quotRT depends on T CV T does not depend on Vbut V Eqrev CVTdT dV T T V Egrev T is an exact differential so it is the derivative of a state function 7 De ne this function as entropy and denote it by S dS agrev T 7 The entropy S is a function of T and V 7 At lower Tthe increase in S is higher than higher T for the same amount of heat 7 For a cyclic process AS 0 or dS 0 where means cyclic process 3 fgquev 0 7 The formulas are general applicable to both ideal gas and other systems 7 Examples of entropy calculations for various reversible transformations of ideal gas P 7 Isothermal transformation path A 2 V2 nRT V ASA j Eqm A ji 1dV mm 2 1 T V T1 V V1 7 Adiabatic transformation path B 2 EquVB ASE l T 0 7 Isochoric transformation path C and path E 2 T1 T2 T2 C T C T ASczj39gqrewCzquot V dTJ39 V dej39 1 T T T T T 2 1 2 T1 C T ASE J396qrevE J V dT T T 1 T3 7 Isobaric transformation path D 2 T3 C T V ASD j 5qrew j V dTann 2 1 T T T V1 1 CHEM 3520 Spring 2004 172 1 171 1 7Fora P Vie Tb gas A A Rlnbb A BC 0Rln 7 Entropy change for processes in an isolated system 7 By adding energy as heat to a system then its entropy increases because its thermal disorder increases didn t show explicitly that the entropy is related to the disorder of the system 7 The entropy of an isolated system increases as a result of a spontaneous process 7 Consider an isolated system and two compartments of this system separated by a divider that allows heat ow Also no work can be done UA UB constant S SA SB dUA Egrev wrev TAdSA dUB Egrev wrev TBdSB But wrev 0 because VA ct amp VB ct dU dU T T 13 Mgg UAUBct3dUA dUB TB TA 3dSdSA dSB If TB gtTA 3dUB lt0 3 dSgt0 If TB ltTA3dUB gt0 3 dSgt0 If T B T A 3 dS 0 7 For an in nitesimal spontaneous change in an isolated system energy remains constant so driving force is the increase in entropy But because system is isolated the increase in entropy is created within the system itself so the entropy is not necessarily conserved it actually increases until the system reaches equilibrium dS 0 at equilibrium 7 Not only dS 0 at equilibrium but also dS 0 for a reversible process that remains essentially in equilibrium dS gt 0 spontaneous process 7 In an 1solated system dS 0 revers1ble process 7 Entropy change for processes in a closed system dS dSprod dSexch dSprod dSexch is due to exchange of energy as heat with the surroundings dSprod is created by irreversible process CHEM 3520 Spring 2004 7 But the in nitesimally change in heat 5g can be 7 for a reversible process Eqmv 7 for an irreversible process qm 7 For reversible processes dSProd 0 and g Eqrev 3 dS quiev 7 For irreversible processes dSprod gt 0 and Eq Eqirr 3 dS gt 5 7 Combining the two expressions dS Z T or AS 2J T Inequallty of Claus1us where equal for reversible and greater for irreversible process 7 This is one way of expressing the Second Law of Thermodynamics 7 Formal Statement of the Second Law of Thermodynamics 7 There is a thermodynamic state function of a system called the entropy S such that for any change in the thermodynamic state of the system dS Z where the equality sign applies if the change is carried out reversibly and the inequality sign applies if the change is carried out irreversibly at any stage 7 Example of entropy increase during a spontaneous irreversible process 7 Assume two transformations between state 1 and state 2 one done irreversibly from 1 to 2 and one done reversibly from 2 to l 2 641 1 641 IA 0 because the system is isolated J rev S1 S2 lt 0 T T 1 2 3 AS S2 S1 gt 0 S increases going irreversibly from state 1 to state 2 7 Considering Universe as an isolated system and all naturally occurring processes as irreversible the second law states that the entropy of Universe constantly increasing 7 Clausius summarizes rst two laws of thermodynamics The energy of the Universe is constant the entropy is tending to a maximum CHEM 3520 Spring 2004 7 Molecular interpretation of entropy 7 The entropy is related to the disorder of a system but how to measure the disorder 7 Consider an ensemble of A isolated systems each with energy E volume V and number ofparticles Nwhere E EN V 7 Systems have same energy but they may be in different quantum states because of degeneracy Consider the degeneracy associated with energyE to be QE and label the degenerate levels as j l2QE 7 Consider that the A systems are distinguishable and let 611 be the number of systems in the ensemble that are in state j the number of ways of having all systems in state 1 a2 in state 2 is given by I I A39 A39 where ZajA Wa1502 a1a2a3 Haj J f can be taken as a quantitative measure of ensemble disorder 7AllA systems in one state 611 A ell 0 3 W 1 7 All aj are equal highest disorder 3 highest value for W 7 Boltzmann proposed the relationship between S and W as S kB an 7 the most famous equation of statistical thermodynamics 7 correlates a thermodynamics quantity S with a statistical quantity W 7 When W l 3 S 0 3 completely ordered system 7 Completely disordered system 3 W maximum 7 The logarithm dependence has advantage that for a system with 2 parts X and Y Stotal SX SY because WXY WxVVY 7 The relationship between S and degeneracy Q 7 Consider Q degenerate quantum states occurring in an ensemble with equal probability An ensemble of isolated systems should contain equal number of systems in each degenerate quantum state Sensemble AS nQS Ssystem S kB In Q system system CHEM 3520 Spring 2004 7 Example A system of N distinguishable spins or dipoles that can be oriented in one of two possible directions with equal probability 7 Each spin has degeneracy of 2 7 Degeneracy of N spins 2N 3 Entropy S NkB ln2 7 Example Ideal gas QE CN f E VN Determine the entropy change during an isothermal expansion of 1 mole of ideal gas Q AS kB 11192 7kB 1an kB 111 2 Q1 N E VN V AS k 1nM 22N R1n2 CN f E1 V1 V1 7 Example Change in S per mole for mixing of 2 gases AmiR y11ny1 y2 lnyz Amix gt 0 because ln yl lt 0 7 Calculations of entropy for various processes 7 To determine the entropy changes assume a reversible process for which use equality 7 Example Isothermal expansion of ideal gas against vacuum V1 gt V2 gqgev Eqrev dU gwrev gwrev PdV dV 2 Asj 1 V2 3ASann7 V2gtV1ASgt0 1 7 A similar result is obtained for real gas 7 This result is same for both reversible and irreversible processes What is the difference Look at ASsm 7 Reversible process 2 qrev V2 ASsm T ann V1 2 AStotal 0 7 Irreversible process V2 ASsm 0 3 AStOtal quotR1117 gt 0 1 9m 0 because AU 0 5W 0 CHEM 3520 Spring 2004 7 Example Mixing of two ideal gases 7 Each gas act independently of the other so each gas expands from Vnitial t0 V nal V V V V ASA nARlnM ASE nBRlnM V VB ASASAASB 7 Transform volume to number of moles and this to mole fraction A S 3 JAIHJ A yB lnyB 7 More general N AmiXS RZyj lnyj AmiXS gt0 j1 7 Example Two equal pieces of metal at different temperatures 7Let Tc cold Th hot T T CVTh TCVT TcT d3 6gquot although process is irreversible Eqrev dU CVdT because no work done C dT T AS IV CV ln 2 assuming CV ct between T1 and T2 T T1 T1 T T T T ASh CV1n ASC CV 1mg 2Th 2T T T 2 ASAShASc CV1n h gt0 4Tth because Th Tc2 gt 4TthcgtTh Tc2 gt 0 7 Example Entropy changed during isobar heating 7 Given CP fT T2 C T AS 63 T T1 CHEM 3520 Spring 2004 Heat steam engines and the second law of thermodynamics 7 The second law of thermodynamics was determined by Hm source investigation of the ef ciency of the steam engines Tn 7 The steam engine withdraws energy as heat from a high temperature thermal reservoir uses some energy to do work and discharges rest of energy as heat to a lower temperature thermal reservoir 7 The amount of work is in practice even smaller due to 7 the fact that reversible processes are idealized 7 the iction and other mechanical factors 7 The steam engine is based on a cyclic process AUengine W 4th qrevc 0 Cold Slnk 4th 4m c AS v 0 engine h To 5 7 w qmvyh qmvyc 7 wis positive Adlabat 7 The ef ciency of a steam engine i lsothen n 7 The maximum ef ciency is obtained in a heat engine following an ideal circle called Carnot circle consisting on two isothermal and two Pressure 9 adiabatic transformations W 7 qrevh qrevc 7 Tc 7 Th 7Tc 77mm 7 717 4mm 4mm T h T h Volume v 7 Example heat engine working between 273 K and 573 K 5737 273 17m T 534 7 The ef ciency is higher as Tc is lower and T11 is higher 7 If T c T h 17 0 ltgt no net work can be obtained in isothermal cyclic process 7 Kelvin statement of second law 7 A closed system operating in an isothermal cyclic manner cannot convert heat into work without some accompanying change in the surroundings CHEM 3520 Spring 2004 Introduction to SolidState and Surface Chemistry The solid materials can be 7 crystalline substances that have a periodic structure 7 amorphous substances that don t have a periodic structure The structure of crystals 7 The smallest collection of atoms or molecules in the crystals such that its replication in the three directions will generate the crystal is called the unit cell The unit cell cannot have any arbitrary shape it should be a geometric structure that will ll all space when replicated The unit cell is not unique for a crystal 7 The most general unit cell is three dimensional parallelepipedic Consider the origin of the coordinate system in the lower left corner and the 5quot positive a b and c axes pointing along b the sides ofthe unit cell a a 7 The geometry of the unit cell can be described by specifying the length along the a b and c axes that is a b and c respectively and the angles between pairs of axes that is a and y respectively 7 The atoms or molecules will always occupy the corners of the unit cell and in addition the center of the unit cell and centers of faces In the case of a crystal containing more than one type of particles like ionic crystals other positions can be occupied 7 The number of atoms in a unit cell is given by the formula atomsunit cell atoms in the corners atoms on edges atoms on the faces atoms inside the unit cell 7 August Bravais showed that only 14 distinct unit cells are necessary to generate all possible crystal lattices and they are called Bravais lattices The lattices are organized in seven classes triclinic monoclinic orthorhombic tetragonal hexagonal trigonal and cubic based on the general geometric features of the unit cell CHEM 3520 Spring 2004 P C F R i I Trlcilnjg I I II I a bpc I I I up 395 I 39 I l Manacllnlt b Evbuc LA quotI LIIn 90 onh 39rhombic 1 c a U 3 anti r1 mr I Tumstuna a I Triganal F Ff H033 anal alb L Equot EV a m Lquot In F 1 fr f 39J 39 a 39 a tl II I I 5quot T fl avidly a 9 If L F Iii 90 iiidf T flit Cubic a b c u ijg w CHEM 3520 Spring 2004 7 These lattices are also labeled 7 P refers to a primitive unit cell atoms only in the comers of the unit cell one atom or lattice point per unit cell 7 I refers to a bodycentered unit cell atoms in the comers and in the center of the unit cell two atoms per unit cell 7 C refers to a endcentered unit cell atoms in the comers and on two opposite faces two atoms per unit cell 7 F refers to a facedcentered unit cell atoms in the comers and on all faces four atoms per unit cell 7 R refers to a rhombohedral unit cell atoms only in the comers of the unit cell one atom per unit cell 7 Examples of cubic Bravais lattices 7 Primitive cubic Example polonium 7 Bodycentered cubic Example potassium 7 Facecentered cubic also known as cubic closedpacked Example copper 7 Calculation of the crystallographic radius 7 Example copper has a density of 8930 g cm 3 at 20 C Assuming that copper atoms are spheres that touch along a diagonal face calculate the radius 4 X 6355g mol 1 722 23 71 422lgtlt10 g 6022 X 10 mol mass of unit cell 4221 X 10 22 g 3 4727 gtlt10723 cm3 8930gcm volume of unit cell length of an edge a Vunit 081018 3616 gtlt1078 cm 3616pm 12 radius of Cu 2 a l278pm 7 Calculation of the fraction of the volume of the unit cell occupied by Cu atoms 7 Volume of Cu atom V i7W3 7113 because 4r2 2a 12 212 7 Fraction occupied 4Va3 0740 2 CHEM 3520 Spring 2004 The Miller indices 7 They describe the orientation of the lattice planes 7 The coordinates of atoms in the unit cell are expressed in units of a b 0 lengths of the edges 7 For example the atom on the origin has the coordinates 0a 0b 00 which is written as 000 The coordinates of the other atoms in the primitive cubic unit cell are 100 010 001 110 101 011 and 111 7 The coordinates ofthe atom in the center ofa cubic unit cell are 121212 7 The lattice can be seen as a set of equally spaced parallel planes containing lattice points These planes are important in understanding Xray diffraction patterns 7 These planes are described in terms ofthe lengths ofthe three sides ofthe unit cells 1 1 1 7 Consider that the plane intersects the a b and c axes of the unit cells at points a39 b39 and 039 One ortwo ofthe a39 b39 and c39 can be in nity 7 Designed the plane by three indices h k and I called Miller indices where 1 k1b1 I 1 000 100 b39b bquot c39c c39 m a 7 The Miller indices should be integer numbers so if the formulas above give fractional values those values should be multiply with the smallest number that give all three Miller indices as integers 7 The indices hkl are associated with a family of parallel planes separated by a distance ah along the a axis bk along the b axis and 01 along the c axis 7 If either one of the Miller indices h k or Z is negative the respective negative number is designated by a bar over it 7 Try to visualize the following planes 100 110 111 220 and 11T for a cubic unit cell The 110 and 220 planes are parallel but are not the same 7 The perpendicular distance between adjacent hkl planes for an orthorhombic unit cell is given by L2 Q g and for a cubic unit cell is given by L2 M d a b c d a 7 Use notation hkl to denote an individual plane and hkl for a set of parallel planes 7 How to denote the planes going through the origin 7 It would be impossible to define without using 00 7 It actually belongs to a set of planes that can be defined CHEM 252 Spring 2mm Th2 crystal surface 7 Th 9mm mndzl Ufa cryshl surface s prfecdy m wnh ms dmmce between mums bung ms same as m m bulk Even m thls case m s hm afthz m use 1s mt us and E j a E 5 E s f i E S a 73 E 5 a E i E mm m mm mm 7 Um y ms surface mums accnpysms um are sh zd M m mum pssms m m bulk Mssmrsmmw man s exhbna agm cm camncmnk 40srmm1syuaums bethen ms rm and secand layus afamms onsn mm are campnsmng ex 11 bethen ms secand and um layquot 7 and smile expumanbethen ms um and ms ilngmznl mthnntspcuhxemmzm ms ammo smwmre afacxyslalsnrface myshnw numzm Irregulannes um gm mughmss m m surface U m h ammlc m m m ndz andmgt smwmrea surfacevm pnndzxshndmchzmwalnammnsanthzsnr cesmkz 75am a Impx39feconns afthz surfaces are mmsss steps masasmms rm nachvnyaf nzse xmpn39ec mns 1s susnymrmm mm um htafaprfect surface WWW m mm m mm 7 Turn CHEM 3520 Spring 2004 Adsorption on crystal surfaces 7Adsorption is the process of trapping molecules or atoms on a surface It was con rmed to be the first step in a surfacecatalyzed reaction The adsorbed molecule atom is called adsorbate and the surface is called the substrate 7 Adsorption is always an exothermic process cgt AadsH lt 0 VZ 7Physisorption or physical adsorption full line is the process in which the attractive forces between substrate and adsorbate are weak ATB lt 20 kJmol and are mainly van der Waals in nature 7 Chemisorption or chemical adsorption dotted line is the process in which the adsorbate is bound to the substrate by covalent or ionic forces The interaction is strong 7 250500 kJmol Chemisorption involves a I I I I I I I I I I I I I II bond broken 1n the molecule and new bond made between the molecular I I I I I I I I I I II III fragments and the surface dissociative chemisorption Only a single J layer of molecules monolayer can chemisorb on the surface 7 LennardJones modeled the adsorption in terms of onedimensional potential energy curves The potential energy depends on the distance between the substrate and adsorbate z Vz 0 at infinite separation between the substrate and nondissociated adsorbate 7 The equilibrium substrateadsorbate bond length for the physisorbed molecule zph is longer than the substrateadsorbate bond length for the chemisorbed molecule 2011 7 Molecules that chemisorb can be initially trapped in a physisorbed state The physisorbed molecule is a precursor to the chemisorbed molecule The physisorption and chemisorption potential energy curves cross at a distance 20 The transformation from the physisorbed state to the chemisorbed state can be seen as a chemical reaction characterized by activation energy Ea In some cases the energy at 2C is bigger than 0 Monitoring the adsorption 7 An adsorption isotherm is a plot of surface coverage as a function of the gas pressure at constant temperature The adsorption isotherms can be used to determine the equilibrium constant for the adsorptiondesorption reaction the concentration of surface sites active sites for the adsorption and the enthalpy of adsorption 7 A simple model was proposed by Langmuir in 1918 Assumptions adsorbed molecules do not interact to each other the enthalpy of adsorption is independent of surface coverage there are a finite number of surface sites where the molecule can adsorb CHEM 3520 Spring 2004 9 7 The processes of adsorption and desorption are reversible elementary processes k Ag ss A ss Kc d d where ka and kd are the rate constants for adsorption and desorption 7 Consider 0390 to be the concentration of surface sites in n f2 9 to be the 1 6 fraction of active sites occupied by adsorbates 3 concentration of empty surface sites is 0390 600 l 900 7 Rate of adsorption va ka 1 t9O39OA Rate of desorption vd kdt90390 l 1 1 1 7Ateuilibrium l or l A P k TbK k T q KCA 6 bPA ll A B c B 113A 7 Denote by Vm the volume adsorbed onto surface at 9 l 3 t9 V Vm i L A linear representation of i versus i has a slope of V Pme Vm V P l and an 1ntercept of V 7 Vm can be used to determine the concentration of surface sites 0390 by knowing the are of the surface and transforming Vm to the number of molecules number of active sites 7 The Langmuir adsorption isotherm for a diatomic molecule k a ka A S2 A g2ss 32A Ss Kc 4 2 362d kd Azns2 7 Rate of adsorption va kaA2l 9203 Rate of desorption vd kdt92039g l l 7 At equ111br1um va vd 3 E l W A2 A2 l l 7 A representatlon of E versus 12 Wlll be a 11near representatlon P A2 7 The rate constant of desorption from a surface obey Arrheniuslike expression kd raleiEadSRT where Eads E AadsH and 1390 m 10 12 s is a constant with units oftime The reciprocal of kd is called residence time ofa molecule on the surface 139 reradSRT 7 There are more advanced models that Langmuir model that can account also multilayer adsorption

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