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# Physical Chemistry I CH 431

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This 315 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 431 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 15 views. For similar materials see /class/223996/ch-431-north-carolina-state-university in Chemistry at North Carolina State University.

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Date Created: 10/15/15

Chemistry 431 Lecture 3 Imperfect Gases NC State University The Compression Factor One way to represent the relationship between ideal and real gases is to plot the deviation from ideality as the gas is compressed ie as the pressure is increased The compression factor is defined as Molar volume of gas Molar volume of perfect gas Compression Factor Written in symbols this becomes Vm PVm Z 39 39 W Note that perfect gases are also called ideal gases Imperfect gases are sometimes called real gases The Compression Factor A plot of the compression factor reveals that many gases exhibit Z lt 1 for low pressure This indicates that attractive forces dominate under these conditions As the pressure increases Z crosses 1 and eventually becomes positive for all gases This indicates that the finite moleCuiai volume leads to repulsions between closely packed gas molecules These repulsions are not including the ideal gas model 20 16 N I Attractive 12 Repulswe Region WKI I I 1 Region 0 200 400 600 800 1000 Patm The Virial Expansion One way to represent the deviation of a gas from ideal or perfect behavior is to expand the compress0n lactor m powers of the inverse molar volume Such an expansion is known as a virial expansion 2 1 i 2 Vm vm The coefficients B C etc are known as virial coefficients For example B is the second virial coefficient vma coefficients depend on temperature From the preUUUIlly considerations we see the B lt O for ammonia ethene methane and B gt O for hydrogen The Virial Equation of State We write Z in complete form as Wm B C 139I39Vm39l3973739l39 An then solve for the pressure RT B C P V1V2 m m m This expression is known as the virial cqu3tl0n on bldte Note that if B C etc are all equal to zero this is just the ideal gas law However if these are not zero then this equation contains corrections to ideal behavior Relating the microscopic to the macroscopic Real gases differ from ideal gases in two ways First they have a real size extent The excluded volume results in a repulsion comm partwws aw mger ressure than the corresponding ideal gas positive contribution to compressibility Secondly they have attractive forces between molecules These are dispersive forces that arise from a potential energy due to induceddipole induceddipole interactions We can relate the potential energy of a particle to the terms in the virial expansion or other equation of state While we will not do this using math in this course we wl codc c graphical form of the potential energy functions Hard Sphere Potential A hard sphere potential is the easiest potential to parameterize The hard sphere diameter corresponds to the interatomic spacmg In a Closest paCKed geometry sucn as that snown for the noble gas argon The diameter can be estimated from the density of argon in the solid state The hard sphere potential is widely used because of its simplicity ur uroo rlto ur 0 rgto The Hard Sphere Equation of State As a first correction to the ideal gas law we can consider tne TaCI tnat a gas nas finite extent Thus as we begin to decrease the volume available to the gas the pressure increases more than we would expect due to the repulsions between tne spheres of finite molar volume b of the spheres o nRT P V nb Gas m lecule of volume B The Hard Sphere Model Low density These are ideal gas conditions o o 39 0 39 o g 0 3900 0 o 0 oo o 0 o The Hard Sphere Model Increasing density the volume is V b is the molar volume of the o spheres Increasing density The Hard Sphere Model The Hard Sphere Model Increasing density The Hard Sphere Model High density At sufficiently high density the gas becomes a high densnty nwd or a liquid The Hard Sphere Model Limiting density at this density the hard spheres have condensed into an oroereo lattice They are a solid The gas cannot be compressed further If we think about the density in each of these cases we can see that it increases to a maximum value LennardJones Potential Function The LennardJones potential is a most commonly used potential function for nonbonding interactions in atomistic computer Simulations VLJR 4 12 o R R The potential has a longrange attractive tail 1r5 and negative well depth a and a steeply rising repulsive wall at R 6 Typically the parameter 6 is related to the hard sphere diameter of the molecule For a monoatomic condensed phase 6 is determined either from the solid state or from an estimate of the packing in dense liq The well depth e is related to the heat of vaporization of a monatomic fluid For example liquid argon boils at 120K at 1 atm Thus 8 kT or 138x1023 JK120 K 165x1021 J This also corresponds to 103 kJmol Graphical Representation LJ Potential The LJ potential function has a steep rise when r lt o This is the repulsive term in the potential that arises from Close contacts between mOIecuIes Ine minimum IS round for Rmin 216 o The well depth is a in units of enerov l l l l 6 4 Reduced LennardJones Potential gt 2 0 l l 10 15 20 25 R R min The van der Waal s Equation of State The microscopic terms a and o in the LJ potential can be related to the a and b parameters in the van der Waal s equation of state below P nRT n2a V nb V2 The significance of b is the same as for the hard sphere potential The parameter a Is related to the attractive force between molecules It tends to reduce the pressure compared to an ideal gas The van der Waal s Equation of State In terms of molar volume Recall that Vm Vn so that the vdW equation of state becomes RT a P vm b 39 V v m can plot this function for a variety c different temperatures As we saw for the ideal gas these are isotherms At sufficiently high temperature the isotherms of the vdW equation of state resemble thew c e Wal gas Pressure x106 Nm2 x 01 x O The argon phase diagram For argon TC 1508 K PC 487 bar 49345 Pa VC 749 cm3mol Critical Point I 100 150 200 Volume x106 m3mol Significance of the critical point Note that the vdW isotherms look very different from those of the ideal gas below the critical point Below the critical point there are two possible iq w n Vs T iq w phase is found at small molar volumes The gas phase is obseve a mgr oar volumes The shape of the isotherms is not physically th gio between the phases Note that the implication is that there is a sudden change n m for the phase transition from liquid to gas Pressure x106 Nm2 View of the liquid region OT the argon phase diagram quot 39 39 39 39 39 39 39 39 39 39 quotVQ39W39H39I39H39I H39I39H I I Volume X106 m3mol Critical Parameters The critical parameters can be derived in terms of the vdW a and b parameters as well as the gas constant R The derivation can use calculus since PC 2 the derivative of the vdW equation of 27b state is zero at the critical point T 8 a C 27Rb Given that this is also an inflection point V 3b the second derivative is also zero C Chemistry 431 Lecture 4 The Particle in a Box The Uncertainty Principle The translational partition function Tunneling NC State University The particle in a box problem Imagine that a particle is confined to a region of space The only motion possible is translation The particle has only kinetic energy While this problem seems artificial at first glance it works very well to describe translational motion in quantum mechanics 0000 0 Allowed Region 39 The solution to the Schrodinger equation with boundary conditions Suppose a particle is confined to a space of length L On either side there is a potential that is infinitely large The particle has zero probability of being found at the boundary or outside the boundary 0000 0 Allowed Region 39 The boundary conditions determine the values for the constants A and B A Bi 11sinkx 21 sin will vanish at 0 since x O and sin 0 0 sin will vanish at a if kL rm Therefore k nnL Normalized The solution to the Schrodinger equation with boundary conditions The boundary condition is that the wave function will be zero atx Oand atx L L110 AsinkO Bcosk0 O From this condition we see that B must be zero This condition does not specify A or k The second condition is L11L AsinkL O or kL arcsinO From this condition we see that kL In The conditions so far do not say anything about A Thus the solution f h 39 ort e bound state IS LPnoo AsmmnXL Note that n is a quantum number The Schrodinger equation for a free particle 2 2 h awqu 2m qz The solutions are 11 Aeikq Be ikq eikq 4 The particle in a box has boundary conditions L110 0 Voo LPL O Voo The solutions to the particle in a box 3922 2 k w E A 2m 2mt2 n L 2m L hzn2 8mL2 The uncertainty principle When we measure the properties of very small particles we cannot help but affect them The very act of measuring causes a change in the particle s properties Therefore the description of the the measurement is a probability rather than a fixed value We have seen the Born interpretation of the square of the wavefunction as a probability density The consequence of this is that certain variables are linked By the uncertainty that is inherent in the measurement Position and momentum are two such conjugate variables Note that the units of position is the reciprocal of the momentum if we factor out Planck s constant X has units of meter k has units of meter1 Momentum is p hk t has units of time v has units of time 391 Energy is E hv Where is the particle in the box Since we are using a probability function we do not really know exactly where the particle is We know that the highest probability occurs for the position L2 We can guess that this is the average position in the box However the more precisely we specify the location of the particle the less information we have about how fast the particle is moving This is a statement of the famous Uncertainty Principle AXAp 2132 Let s look at the Uncertainty Principle using the particlein abox example If we know that the particle is in the lowest level then Uncertainty in its position is approximately equal to the width of the probability distribution The location of a particle in free Space is not defined ii in iiiii position x Consider a superposition of a wave with moment hk and h11k 4 I I AmpMude po onx The sum has a characteristic envelope frequency at 032 0312 Amplitude 10 05 00 05 Mimi l l l Envelope position x The sum has a characteristic beat frequency at 032 m12 10 05 00 Amplitude O5 10 B Wm position x As we add more frequencies we can speak of a bandwidth Ak 10 05 l Amplitude O5 3 added cosines H l 40 2o 0 20 40 position x 10 As the bandwidth increases the position in Xspace becomes more defined I m H Ak02 E DonnMA AA A AAAAA E W W W W 05 Sadded cosines J 10 I I I I 40 20 O 20 40 position x The superposition of waves in space leads to the description of a location 10 05 00 Amplitude 05 15 added cosines V I I I I I 40 20 O 20 40 position x Relevance of the example Although the function used in the example is periodic it is relevant Since in a given region of space ie where a measurement can be made the probability of observing the particle in a given region of space is dependent upon the number of contributing waves If more waves contribute then the momentum of the particle is less certain Thus the we can know that moment precisely if we are totally uncertain of the position As we begin to specify the position more precisely we find that the momentum is less well known Since p hk we can also express this condition as AxAk 12 Fourier transform related pairs Position and momentum are related by a Fourier transform x 4 p Time and energy are related by a Fourier transform t lt gt E There is an uncertainty relationship for both of these related pairs Thus for time and energy we have At AE 3132 as well These pairs can be related by a probability function that gives the width of the distribution in each space Gaussian functions are particularly useful since the Fourier transform of a Gaussian is also a Gaussian Gaussian Functions A Gaussian Function has the Form eXp ocX XO2 The Gaussian indicated is centered about the point X0 The Fourier transform of a Gaussian in X space is Gaussian in K space Since p 39 F 1Ilt we also call this momentum space The Figure shows the inverse relationship H Question Which of the following represents the hamiltonian A k 4528 B m 6 39 23977 6X2 D m i 6X Question Which of the following represents the hamiltonian A k 4528 B m 6 39 23977 6X2 D m i 6X Question What is the hamiltonian A it is the energy B it is the momentum C it is the energy operator D it is the momentum operator Question What is the hamiltonian A it is the energy B it is the momentum C it is the energy operator D it is the momentum operator Statistical averaging over translational energy levels Quantum mechanics must agree with classical physics mechanics at high temperature or when the average quantum number becomes very large This is the case for translational energy levels since the spacing of those levels is very small compared to thermal energy kT Here we consider how to average over the energy levels given by the particleinabox solutions The translational partition function The translational partition function consists of a sum over a very large number of states 00 2 00 2 q 21607 1Bs e n 1B8dn n 1 This sum can be expressed as an integral over the states n The integral can be expressed as a Gaussian Energies from particle in the box can be used to calculate energy level spacing The difference in energy levels n in the particle inthebox solutions has the general form gn8ng2n21 where e n gives the energy of a large number of translational energy levels derived from the particle in the box solutions We can write a h28mX2 to use a factor in the Boltzmann distribution so that an en2 1 The translational partition function is a Gaussian The Gaussian in n integrates to 7122 oo n28 1 121 00 X2 1 Tc 12 e Bmin e d to Be 0 238 Substitute for e to obtain an 12 h B The translational partition function in three dimensions XY and Z The volume is V XYZ 32 32 277m 27Tm V h2 h2l3 This expression for the translational partition function derived from the particle in the box is the same as that derived classically from the integral over all velocities q The translational partition function can be expressed in terms of a thermal wavelength A 2 ka 3 2 V VT where the thermal wavelength is defined as A 271ng Tunneling Tunneling of electrons protons or other small particles is not possible according to classical mechanics However in quantum mechanics a particle can penetrate a barrier even if its kinetic energyE is less than the potential energy barrier height V V E e Tunneling Tunneling If the walls of the box are not infinitely high the The wavefunction of the particle does not decay To zero Instead it decreases exponentially Ultimately it has some probability for tunneling Through a barrier even when E lt V 4 2 2 quot 5 W 1311 ZmaXZ V E e Tunneling The evanescent wave Since V is greater than E the solutions in the Barrier are not oscillatory functions but rather are Exponentially decaying functions L11 Ce De KX K 2mlty Egt n Note that the order of EV is reversed and the i is also absorbed so the the wavefunction is no longer oscillatory The function CekX increases without bound and is not a practical solution Boundary conditions The oscillatory part and the evanescent part of the wavefunction must match up at the boundary If we think about it the wave as a propagating wave and the exponential decay as a transmission across a barrier then there will also be a reflected wave LP Ae kx Bequotquotx oscillatory part left Side LP Ce De KX exponential part LP A39e kx B39e oscillatory part right side We can use the property that the wave function and its first derivative must be continuous to find conditions at the boundary Boundary conditions The boundaries are at x O and x L Thus the conditions are ABCDatX 0 A39e B39e CeKL De KL atX L The first derivatives are ikA ikB K C K D atX O ikA39e kL ikB39e KCGKL KD9KL at X L There are four equations and six unknowns If we assume that the particle is coming from the left then we can set B 0 If we calculate the transmission coefficient then we need the ratio A IA Transmission coefficient With these constraints we can solve for the transmission coefficient T T 1 eKL e KL 1 1681 8 where a EN The tunneling phenomenon is important in electron transfer theory Electron transfer is a key aspect of energy transduction in biology eg photosynthesis and respiration among others Electron tunneling is also the effect used in the scanning tunneling microscope STM Chemistry 431 Lecture 7 Enthalpy NC State University Motivation The enthalpy change AH is the change in energy at constant pressure When a change takes place in a system that us open to the atmosphere the volume of the system changes but the pressure remains constant In any chemical reactions that involve the creation or consumption of molecules in the vapor or gas phase there is a work term associated with the creation or consum tion of the was Molar Enthalpy Enthalpy can be expressed as a molar quantity H Hm W We can also express the relationship between enthalpy and Internal energy In terms or mOIar quantities Hm Um PVm For an ideal or perfect gas this becomes Hm Um RT Usually when we write AH for a chemical or physical change we refer to a molar quantity for which the units are kJmol Enthalpy for reactions involving gases If equivalents of gas are produced or consumed in a chemical reaction the result is a change in pressurevolume work This is reflected in the enthalpy as follows AH AU PAV at const Tand P which can be rewritten for an ideal gas AH AU AnRT at const T and P The number of moles n is the number of moles cremw o absorbed during the chemical reaction For example CH2CH2Q H2g CH30H3Q A 391 We arrive at thlS value rrom the formula An n n 1 2 1 products reactants The temperature dependence of the enthalpy change Based on the discussion the heat ca acit from the last lecture we can write the temperature dependence of the enthalpy change as AHC T Note that we can use tabulated values of enthalpy at 298 K and calculate the value of the enthalpy at any temperature of interest We will see how to use this when we consider the enthalpy change of chemical reactions the standard enthalpy change The basic physics of all temperature dependence is contained in the above equation or more frequently in the equation umow as molar quamy AWQT Another view of the heat capacity At this I oint it is worth noting that the ex ressions forthe heat capacity at constant volume and constant pressure can be related to the temperature dependence of U and H respectively AH CPAT AU CVAT C AH C AU E P AT N P V AT N V The heat capacity is the rate of change of the energy with temperature The partial derivative is formal way of saying this The heat capacity is also a function of temperature We have treated the heat ca acit as a constant u to this point That is a valid approximation under many circumstances but only over a limited range of temperature In the general case the temperature dependence of the enthalpy can be described as T AHIZCPTdT cpma bT T1 The I arameters a b and c are viven in Tables Actuallv this expression is readily integrated in the general case to give AHalT2T T Tf0l Enthalpy of physical change A physical change is when one state of matter changes into another state of matter of the same substance The difference between physical and chemical changes is not always clear however hase transitions are obviousl h sical chan39es Fusion AHmS AHvap Vaporization Solid ltgt Liquid ltgt Gas Freezing AHrreeze AHcond Condensation Sublimation AHsub Solid ltgt Gas Vapor Deposition AHvapdep Properties of Enthalpy as a State Function The fact that enthalpy is a state function is useful for the additivity of enthalpies Clearly the enthalpy of fonNard and reverse rocesses must be related b AforwardH Areversell so that the phase changes are related by AfusH AfreezeH AvapH AcondH AsubH Avap depH Moreover it should not matter how the system is transformed from the solid hase to the was hase The two rocesses of fusion melting and vaporization have the same net enthalpy as sublimation Question Which is statement is false A AsubH gt O B AcondH lt U C AfusH gt 0 DA HltO vap Addivit of Enthal ies Because the enthalpy is a state function the same magnitude must be obtained for direct conversion ro sod to a as for the indirect conversion solid to liquid and then liquid to gas AsubH AfusH Avapll Of course these enthalpies must be measured at the same temperature OthenNise an appropriate correction would need to be applied as described in the section on the temperature dependence of the enthalpy Question Which statement is true A Asubl l AfusH 39 A H vap B AvapH AsubH AfusH C AfusH AsubH A H vap D AvapH AsubH AfusH Chemical Change In a chemical change the identity of substances is altered during the course of a reaction One exam le is the hydrogenation of ethene CH2CH2g H2g gt CHgCH3g A 137 kJ The negative value of AH signifies that the enthalpy of the system decreases by 137 kJ and if the reaction takes place at constant pressure 137 kJ of heat is released into the surroundin39s when 1 mol of CH2CH2 combines with 1 mol of H2 at 25 C Standard Enthalpy Changes The reaction enthalpy depends on conditions eg T and P It iS convenient LO repUIL and tabulate InIUIIIIauUII unuel a standard set of conditions Corrections can be made using heat capacity for variations in the temperature Corrections can also be made for variations in the pressure When we write AHe in a thermochemical equation we always mean the Change in enthalpy that occur when we lUdULdllLb change into the products in their respective standard states Standard Reaction Enthalpy The standard reaction enthalpy ArHe is the difference between the standard molar enthalpies of the reactants and products with each term weighted by the stoichiometric coefficient AH Z ngproducts Z ngUeactants The standard state is for reactants and products at 1 bar of pressure The unit of energy used is kJmol The temperature is not part of the standard state and it is possible to speak of the standard state of oxygen gas at 100 K 200 K etc It is conventional to report values at 298 K and unless othenNise specified all data will be reported at that temperature Enthalpies of Ionization The molar enthalpy of ionization is the enthalpy that accompanies the removal of an electron from a gas phase atom or ion Hg gtHg eg AH 1312 kJ For ions that are in higher charge states we must consider successive ionizations to reach that charge state For exam le for My we have Mgg Mgg e39g AH 738 M Mgg gtM92g e39g AH 1451 M We shall show that these are additive so that the overall enthlalpy change is 2189 kJ for the reaction Mgg M92g 269 Electron Gain Enthalpy The reverse of ionization is electron gain The corresponding enthalpy is called the electron gain enthalpy For example Cg e39g gtCIg AHe 349 kJ The sign can vary for electron gain Sometimes electron gain is endothermic The combination of ionization and electron gain enthalpy can be used to determine the enthalpy of formation of salts Other types of processes that are related include molecular dissociation reactions Enthalpies of Combustion Standard enthalpies of combustion refer to the complete combination wi mm o caon i a ae For example for methane we have CH4g 2029 gtCO2g 2H2OI Acne 890 kJ Enthalpies of combustion are commonly measured in a bomb calorimeter a constant volume device Thus AUm is measured To convert from AUm to AHm we need to use the relationship AHm AUm AvgasRT The quantity Avgas is the change in the stoichiometric coefficients of the gas phase species we see in the above express that Avgas 2 Note that H20 is a liquid Question Fill in the missing stoichiometric coefficients for the combustion reaction CsH12g X02g YCOzg ZH20 A X4 Y8 212 B X8 Y5 Z6 C X4 Y10 Z6 D Question Fill in the missing stoichiometric coefficients for the combustion reaction CsH12g 8029 50029 6 200 A X4 Y8 212 B X8 Y5 Z6 C X4 Y10 Z6 D Question Determine Avgas for the reaction as written CsH12g 8029 50029 6 200 A Avgas 3 B Avgas 8 C Avgas 4 D Avgas 3 Question Determine Avgas for the reaction as written CsH12g 8029 50029 6 200 A Avgas 3 B Avgas 8 C Avgas 4 D Avgas 3 AHm AUm AvgasRT The quantity Avgas is the change in the stoichiometric coefficients of the gas phase species we see in the above express that Avgas 2 Note that H20 is a liquid Question What is the work term for expansion against the atmosphere CsH12g 8029 50029 6 200 A AvgasRT B AUm AvgasRT C AUm AvgasRT D Avgas Question What is the work term for expansion against the atmosphere CsH12g 8029 50029 6 200 A AvgasRT B AUm AvgasRT C AUm AvgasRT D Avgas Hess s Law We often need a value of AH that isenot in the thermochemical tables We can use the fact that AH is a state function to advantage by using sums and differences of known quantities to obtain the unknown We have already seen a simple example of this using the sum of AH of fusion and A u vaporization to obtain AH of sublimation Hess s law is a formal statement of this property The standard enthalpy of a reaction is the sum of the standard enthalpies of the reactions into which the overall reaction may be divided Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg e39g gt CIg AH 349 kJ Which statement is true about the charge transfer from H to CI to form H and CI39 A AH 963 kJ B AH 1661 kJ C AH 1312 kJ D A 349 kJ Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg e39g gt CIg AH 349 kJ Cg Hg gt C39g Hg AH 963 kJ Which statement is true about the charge transfer from H to CI to form H and CI39 A AH 963 kJ B AH 1661 kJ C AH 1312 kJ D A 349 kJ Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg e39g gt CIg AH 349 kJ What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Haq 20laq A AH of ionization and AH of electron capture B AH of formation AH of dissociation and AH of solvation C AH of ionization and AH of solvation D AH of dissociation and AH of solvation Question Consider the reactions Hg gt Hg e39g AH 1312 kJ Cg e39g gt CIg AH 349 kJ What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Haq 20laq A AH of ionization and AH of electron capture B AH of formation AH of dissociation and AH of solvation C AH of ionization and AH of solvation D AH of dissociation and AH of solvation Application of Hess s Law We can use the property known as Hess s law to obtain a standard enthalpy of combustion for propene from the two reactions CsHag H2g 03H8g AH 124 kJ 03H8g 5029 gtBCOZg 4H20I A 2220 kJ If we add these two reactions we get 9 CsHeg H2g 502g gt3002g 4H20 AH 2344 M and now we can subtract H2g1202g H20 AH9 286 kJ to obtain 9 03H6g 9202g gt 30029 3H2OI AH 2058 kJ Variation in Reaction Enthalpy with Temperature Since standard enthalpies are tabuaeu a AW K we new to determine the value of the entropy at the temperature of the reaction using heat capacity data Although we have seen this procedure in the general case the calculation for chemical reactions is easier if you start by calculating the heat ca acity difference between reactants and roducts ACP Z vCPproducts Z vCPreactants and then substitute this into the expression ArH T2 ArH T1 T2 ACPdT If the heat capacities are all constant of the temperature range then Arkraw Arkram ArCPAT Standard Enthalpies of Formation The standard enthalpy of formation AfHeis the enthalpy for formation of a substance from its elements in their standard states The reference state of an element is its most stable form at the temperature of interest The enthalpy of formation of the elements zero For example let s examine the formation ofewater H2g 12 029 gt H20l An 286 kJ Therefore we say that AfH H20 l 286 kJmol Although AfH for elements in their reference states is zero AfH is not zero for formation of an element in a different phase 9 Cs graphite Cs diamond AfH 1895 kJmol Question Consider the formation of carbon dioxide at 298 K 08 029 0029 How would you find the Heat of formation u OAyycll A Look up AfH for 03 and subtract it from that of COZ B Look it up the standard thermodynamic tables C The heat of formation of O2 is zero by definition D It is equal to the standard bond energy of two oxygen atoms Question Consider the formation of carbon dioxide at 298 K 08 029 0029 How would you find the Heat of formation u OAyycll A Look up AfH for 03 and subtract it from that of COZ B Look it up the standard thermodynamic tables C The heat of formation of O2 is zero by definition D It is equal to the standard bond energy of two oxygen atoms Question Consider the formation of carbon dioxide at 150 K 08 029 0029 How would you find the Heat of formation u Cuz Apply a correction to the enthalpy from A Hess s law B the van der Waal s equation of state C ideal gas law D none of the above Question Consider the formation of carbon dioxide at 150 K 08 029 0029 How would you find the Heat of formation u Cuz Apply a correction to the enthalpy from A Hess s law B the van der Waal s equation of state C ideal gas law D none of the above Chemistry 431 Lecture 1 Ideal Gas Behavior NC State University Macroscopic variables P T Pressure is a force per unit area P FA The force arises from the change in momentum as particles hit an object and change direction Temperature derives from molecular motion 32RT 12MltU2gt V IS m0Iar mass Greater avera39e velocit results in a higher gt temperature u IS the veIOCIty Mass and molar mass We can multiply the equation gRT M ltu2gt by the number of moles n to obtain 3 1 2 nRT nM ltu gt 2 2 If m is the mass and M is the molar of a particle then we can also write nM Nm N is the number of particles Mass and molar mass In other words nNA N where NA is Avagadro s number gnRT Nm ltu2gt Average properties ltu2gt represents the average speed Kinetic Model of Gases Assumptions 1A gas consists of molecules that move randomly 2 me snze or the m0IecuIes IS negligible 3 There are no interactions between the gas molecules Because there are such large numbers of gas molecules in any system we will interested in average quantities We have written average with an angle bracket For example the average speed is 2 2 2 2 SSSS We use s for speed 31 32 33 SN and ctor mean speed C N Velocity and Speed When we considered the derivation of pressure using a kinetic model we used the fact that the gas exchanges momentum with the wall of the container Therefore the vector directional quantity velocity was appropriate However in the energy expression the velocity enters as the square and so the sign of the velocity does not matter In essence it is the average speed that is relevant for the energy Another way to say this is the energy is a scalar 1 2 1 2 1 2 lt gt E 2m u va ch All of these notations p mu 2 mv mean the same thing The rootmeans uare s eed The ideal gas equation of state is consistent with an inter retation of tem erature as I ro ortional to the kinetic energy of a gas Mltu2gt RT If we solve for ltu2gt we have the meansquare speed 2 352T If we take the square root of both sides we have the rms speed 2 12 3RT M 17 The mean speed The mean value is more commonly used than the rootmeansquare of a value The rootmeansquare speed ls equal to the rootmeansquare velocity 02 uz The mean speed is 2 L2 Q who The rms speed of oxygen at 25 C 298 K is 482 ms Note M is converted to kgmol 2 m 3831Jm0 K298 K W 0032 kgmol 4818 m S The Maxwell Distribution Not all molecules have the same speed Maxwell assumed that the distributi0n on speeds was Gaussian M 32 2 MS2 Fs 471527tRT 5 exp RT As temperature increases the rms speed increases and the width of the distribution increases Moreover the functions is a normalized distribution This just means that the integral OVer the distribution function us equal LU I See the MAPLE F 1 0 Sds worksheets for examples Molecular Collisions Cross section c775d2 gt Center location of target molecule ltugtt distance traveled mean free path estmate volume of interaction number density nN moles er unit volume molar densit NN molecules per unit volume number density ltugtt mean free path estmate W Refinement of mean free path The analysis of molecular collisions assumed that the target atom was stationary If we include the fact that the target atom is moving we find that the relative velocity is lt u gtre E lt u gt Therefore k ltugtt 1 1 RT cltugttNv cNv EGNAnv EGNAP As the pressure increases the number density increases and the distance between collision mean free path becomes shorter As the temperature increases at constant pressure the number density must decrease and the mean free path ill increase Mean free path Collision frequency The mean free path A is the average distance that a molecule travels between collisions The collision frequency 2 is the average rate of collisions made by one molecule The collision cross section 6 is target area presented by one molecule to another When interpreted in the kinetic model it can be shown that 2 kzi zz NAGNULDacsz NAGP RT The product of the mean free path and collision frequency is equal to the room mean square speed U2 7vz Units of Pressure Force has units of Newtons F ma kg ms2 Pressure has units of Newtonsmeter2 P FA kg ms2m2 kgs2m These units are also called Pascals Pa 1 bar 105 Pa 105 Nm2 1atm 101325 x 105 Pa Units of Energy Energy has units of Joules 1 J 1 Nm Work and energy have the same units Work is defined as the result of a force acting through a distance We can also define chemical energy in terms of the energy per mole 1 kJmol 1 kcalmol 4184 kJmol Thermal Energy Thermal energy can be defined as RT Its magnitude depends on temperature R 831 JmoIK or 198 calmoIK At 298 K RT 2476 Jmol 2476 kJmol Thermal energy can also be expressw 6 a per molecule basis The molecular equivalent of R is the Boltzmann venom R NAk NA 6022 x 1023 moleculesmoi Extensive and Intensive Variables Extensive variables are proportional to the size of the system Extensive variables volume mass energy Intensive variables do not depend on the size of the system Intensnve variables pressure temperature dens y Equation of state relates P V and T The ideal gas equation of state is PV nRT An equation of state relates macroscopic properties which result from the average behavior of a large number of particles e Macroscopic Microscopic Microsopic view of momentum C 11X 0 b area bc a A particle with velocity uX strikes a wall The momentum of the particle changes to muX The momentum change is Ap 2muX X Transit time C uX e b area bc a The time between collisions with one wall is At Zaux This is also the round trip time Transit time C Round trip distance is 2a area bc a The time between collision is At 2aux velocity distancetime time distancevelocity The pressure on the wall force rate of change of momentum The pressure is the force per unit area The area is A be and the volume of the box is V abc Avera39e ro erties Pressure does not result from a single particle striking the wall but from many particles Thus the velocity is the average velocity times the number of particles Nmltu C gt P T PV Nm 132C gt Avera39e ro erties There are three dimensions so the velocity along the xdirection is 13 the total 3 gt2 3 2 gt Nmltu2gt 3 From the kinetic theory of gases PV l 2 1 2Nmltugt 2nRT Puttin39 the results to39ether When we combine of microscopic view of pressure with the kinetic theory of gases result we find the ideal gas law PVnRT This approach aSSqucS urcu lllC IIIUICUUIUS have no size take up no space and that they have no interactions Chemistry 431 Lecture 2 Properties of Gases NC State University Puttin39 the results to39ether When we combine of microscopic view of pressure with the kinetic theory of gases result we find the ideal gas law PVnRT This approach applies to a monatomic gas like neon or argon What about internal motions of molecules RT is a natural energy scale We can rewrite the ideal gas law in terms of the molar volume I7 Vn The ideal gas law has the form P17 RT The molar volume at standard T and P V RT 831 Jmoz K298 K P 00244 m3 244 L 1013 gtlt 105Nm2 This is the volume of one mole of gas We could also write the unit as Lmol Microscopic variables Monatomic gases translation Pressure and temperature can be deSCrioeo solely in terms of the ballistic motion of the gas Diatomic gases translation vibration rotation ozo 060 Center of mass Quantized energy levels The constant h known as Planck s constant gives the scale tor quantizeo energy levels h 6626 x 103934 Js Energy hv Translation particle in a box Vibration harmonic oscillator Rotation rigid rotator The energy levels for each of these is obtained by solution of the Schodinger equa on The energy level spacing The constant h known as Planck s constant gives the scale tor quantizeo energy levels h 6626 X 103934 J Motion Formula kJmol Vibration V v 12hv 1 20 Rotation J h28W2JJ1 10393 1 Translation n h28ma2n2 1011 We will see how to obtain these in the second half of the course Levels are thermallv A 0A ulated Vibration Rotation Translation Key points regarding the microscopic view Translational energy levels are so densely spaced that these can be treated using classical methods We can treat particles as ideal even though they have vibrations and rotations The dynamics of the gas are not affected We will see that the heat capacity of the gas is affected by the internal degrees of freedom Key points regarding the microscopic view The kinetic energy of a large number of individual particles is proportional to the temperature of the system As the system heats up we can picture the molecules moving more rapidly Pressure results from the net momentum transfer between the particles and wall of the container Pressure of a dense fluid For a dense fluid or a liquid such as water we can think of the pressure arising from the weight of the column of fluid above the point where the measurement is made The force is due to the mass of water m kg accelerated by gravity g 98 ms2 F mg mghmgh h P A A Ah v pg where p is the density p mV The dependence of atomspheric pressure on altitude We can tth or the atmosphere IS a TlUlCl but it is not dense Moreover unlike water the density of the atmosphere decreases with altitude Thus at high elevations both the pressure and the density are decreased To obtain the dependence of pressure on height h above the earth s surface we use the ideal gas law to define the density of an ideal gas The dependence of atomspheric pressure on altitude The density of an ideal gas is p mV nMV MPRT The dependence of pressure on elevation Is dP pgdh dh We need to collect variables of integration on the same side of the equation nw P RTdh The barometric pressure formula Then we integrate assuming P01 at hO Isotherms We can plot the pressure as a function of the volume as shown below Each of the curves on the plot has a constant temperature Pressure P Volume V Partial pressure For any gas in a mixture of gases the partial pressure is defined as P where X is the mole fraction of component 1 and P is the total pressure The mole fraction is defined as Chemistry 431 Lecture 5 The harmonic oscillator Classical and quantum mechanical NC State University Classical Vibration of a Diatomic As was the case for rotation we can consider a simple model of a mass on a spring attached to a wall of infinite mass and a diatomic molecule as two simple examples Mass on a spring Diatomic k k M m1 m2 m k is the force constant Reduced mass Harmonic approximation 62 V 8Q2 VQ VQo 3 ng Q g Q QOY 0 At equnlibrium 5Q Assume terms higher than quadratic are zero By definition 2 k the force constant Classical approach to vibration Solution is oscillatory Any energy is possible M 8Q 2 k 2 E 59 0 k force constant lnternuclear Distance A 393 Q Q Q Energy cm 391 Classical vibrational motion A particle undergoes harmonic motion if it experiences a restoring force that is proportional to its displacement x F kQ k is a force constant F dVdQ and V 12kQZ The classical harmonic oscillator can also be written as 2 makQ wag kQO Solutions have the form of sincot or coscot depending on the initial conditions These solutions imply that 0 A u Classical potential function The potential is V 12kQZ which is a parabolic function This potential is called a harmonic potential Energy 4 4 2 0 2 Nuclear Dis lacement The force constant K has uhrts ot Newtonsmeter Nm or Joulesmeter2 Jmz The angular frequency on 27w v is the frequency in Hz Quantum approach to vibration Solution is Gaussian Energy is quantized T12 92li k 2 f 2H aQZ jQ LP ELF Energy cm 391 Evhv v is the quantum number Internuclear Distance A Allowed transitions Q v gtv1v gt v1 Vibrational wavefunctions Energy levels are given by EV v 1239h0 Wavefunctions are A XV NVHVe39V2 2 where HV is the Hermite polynomial Typical energies are of v the order of 0 3000 cm1 Solutions to harmonic oscillator The Hermite polynomials are derivatives of a Gaussian y cal20 M 0 T mm 1Vey2dVe y2 The normalization constant is NV 1 1 v Hm NV lZVV 7T 0 1 an1 4 1 2y 1 ocTt1l4 2 4y2 2 1 ocTc1l4 3 8y3 12y mmn Energy The bonding electronic state gives rise to a potential energy surface for the nuclear motion X2 BAAmoi JAAAK Nuclear Displacement Harmonic approximation W W xoQ 4e 92 x1Q 42ochaQZ H m Jgt EV Energy There is a potential energy surface that corresponds to each electronic state of the molecule Nuclear Displacement The shift in the nuclear displacement arises from the fact that the bond length increases in the 0 state compared to the c state We will show that the overlap of the vibra tional wave functions is key to understanding the shape of absorption bands The zero point energy The lowest level is E0 12hco The lowest vibrational level is not zero in energy This is consistent with the uncertainty principle If atoms were completely still at absolute zero then we would know both their position and moment to arbitrary accuracy The width of the wavefunction is related to positional uncertainty of an atom We call E0 the zero point energy Polyatomic Molecules There are 3N total degrees of freedom in a molecule that contains N atoms There are three translational degrees of freedom These correspond to motion of the center of mass of the molecule In a linear molecule there are two rotational degrees of freedom In a nonlinear molecule there are 3 rotational degrees of freedom The remaining degrees of freedom are vibrational There are 3N6 vibrational degrees of freedom in a molecule with N atoms Three degrees of freedom are required for translation Three degrees of freedom are required for rotation For example in H20 there are 9 total degrees of freedom and 3 vibrational degrees of freedom In CGH6 there are 36 degrees of freedom and 30 vibrational degrees of freedom Exception In linear molecules there are only 2 rotational degrees of freedom and therefore the number of vibrations BBN5 The vibrational degrees of freedom can be expressed as normal modes All normal modes have the same form for the harmonic oscillator wavefunction and differ only in the force constant k and mass m The total wavefunction is a product of normal modes The total nuclear wavefunction for water is X1X2x3 The normal mode wavefunctions of water correspond to the symmetric stretch bend and asymmetric stretch These are linear combinations of the stretching and bending internal coordinates of H20 Normal modes water yO sxQw Symmetric Stretch Asymmetric Stretch Bend v1 3825 cm391 v3 3935 cm391 v2 1654 cm391 There are 3 normal modes 3N 6 All of them are infrared active since all show a dipole moment change in their motion The harmonic approximation can be applied to each normal mode Normal modes 002 0 0 0 Symmetric stretch Asymmetric stretch V1 2289 0m391 v3 2349 cm391 Bends Raman active R active v2 667 cm1 IR active There are 4 normal modes 3N 5 Three of them are infrared active since they show a dipole moment change in their motion Vibrational Transitions Vibrational transitions Vibrational transitions arise because of the oscillation of the molecule about its equilibrium bond configuration As the molecule oscillates infrared radiation can interact to alter the quantum state C2 glam 8 Mvib XV1QXVdQ Transition dipoles In order for infrared light to be absorbed the polarization must be aligned with the direction of the transition moment For a vibrational mode this is determined by the directional change in the dipole moment This is shown below for the bending mode of H20 Transition dipoles The change in ground state dipole moment during vibration interacts with light 8M H Mg 8 5Q The first term is static and does not contribute to the transition Calling the vibrational wave functions xi the transition moment is 8 H10 283IX1QX0an Dipole derivatives The vibrational wavefunctions xi are Gaussians thus the transition moment for transition from vibrational state 0 to vibrational state 1 is aug 00 0L 22 0L 22 1 aug IO 6QOoe Q Q6 Q The transition dipole moment is proportional to the dipole derivative This is true for any normal mode of vibration ie harmonic Vibrational transitions As an example we can calculate the transition moment between the state v O and v 1 XO 14eaQ22 X1 14 Qeszl2 MVib Sg 2mw e aQ22Q29 aQ22dQ 8H oc12 m 8H 1 an amam Vibrational transitions Note that this result is a statement of the vibrational selection rule Within the harmonic approximation transitions can only occur between states separated by one quantum number Av 1 or Av 1 This general rule can be seen by considering integrals of the type shown in the previous slide Vibrational Transition Vibrational Transition a Vibrational Selection Rule W 1 1 Q Comparison of harmonic and anharmonic potentials Anharmonic A Harmonic 39 39 E a E De C UJ R e Internuclear Distance A Overtones of water 1 Even in water vapor v1 z v3 but symmetries are different F17 F3 However the third overtone of mode 1 has the same symmetry as the combination band r1r1r1r1r3r3 Strong anharmonic coupling leads to strong overtones at 11032 and 10613 cm1 These intense bands give water and ice their blue color v1 symmetric stretch 3825 cm391 v2 bend 1654 cm391 x v3 asymmetric stretch 3935 cm391 Frequency shift due to molecular interactions Hydrogen bonding lowers OH force constant and HOH bending force constant vapor gt liquid v V13825 gt 3657 H V21654 gt 1595 v3 3935 gt 3756 Question Which expression is correct Akag 1 law Mm Ckuo Dkuo2 N81 quotx Q m quot8 Binxd Somtoo m 23qu coEgtgt cozmmsd Question How many normal modes of vibration does methane have 5 2 00wgt GOA Question How many normal modes of vibration does methane have 5 2 00wgt GOA Analysis of isotope effects Vibrational spectra are analyzed within the harmonic approximation 0 Reduced mass 0 k u M m1 m2 m1 m2 Classical harmonic oscillator equation ya ZerkaO of xACOS wt a The vibrational partition The vibrational eneifg n ti evenly spaced with a separation of hv or ho If we take the zero point level as our quotzero of energy then 8V vhco 00 vci3900 Ci39V 1 qVVE1eB ZleB 1 This is the partition function that we found for an infinite ladder of energy levels For high frequency modes 8 gtgt kT and qV z 1 For very low frequency modes 8 ltlt kT and Chemistry 431 Lecture 2 Breakdown of classical physics Heat capacity Photoelectric effect Waveparticle duality Atomic spectra Semiclassical hydrogen atom NC State University Breakdown of classical physics Aside from the ultraviolet catastrophe there were a number of experiment observations that did not agree with classical physics 1 The heat capacity approaches zero as the temperature approaches zero 2 The photoelectric effect Ionization of a metal depends on the frequency rather the intensity of radiation 3 Atomic and molecular spectra had discrete lines 4 The wavelike properties of electrons and other particles Heat capacity The heat capacity is the energy required to raise the temperature of substance The definition is C 5Um6ltEgt W 6T 0T Solids liquids and gases all have heat capacities Um is the molar energy of the substance This is also known as the molar internal energy and is the same as the average energy lt E gt The heat capacity is given at constant volume Internal energy of a solid Einstein first calculated the internal energy of a metal by treating it as a collection of oscillators which represent the bonds between the atoms U 3NAhv ehvkT 1 This expression assumes that the frequency of the oscillators is hv The expression has more than superficial similarity to the Planck Law The different is that the Planck Law refers to radiation modes and the Einstein formula refers to vibrational frequencies Internal energy of a solid We can define the Einstein temperature as 9 m E k Using the definition of the Einstein temperature we can rewrite the internal energy as Um Limits of the function of internal energy As the temperature approaches 0 the value of ehvkT gtgt 1 so the expression becomes Um z 3 Ahv 3NAhve hV kT hvkT As the temperature becomes large or approaches infinity we can use the expansion hvkT M e 1 kT to show that Um 3NAkT 3RT Comparison of heat capacities The classical heat capacity is cvm 3R The classical heat capacity agrees with experiment at room temperature However the classical heat capacity fails at low temperature The Einstein heat capacity is 9E2 eeEzT 2 Own T 3R Comparison of heat capacities One can also write this as follows Cvm 3Rf where the function 2 2 f E eeEzr T eGET 1 At high temperature f1 see page 248 However at low temperature f z if2e 95 7 This agrees with experiment As the temperature goes to zero the heat capacity goes to zero Photoelectric Effect Electrons are ejected from a metal surface by absorption of a photon Depends on frequency not on intensity Threshold frequency corresponds to hv 0 CD CD is the work function It is essentially equal to the ionization potential of the metal h 6 Metal Surface T Kinetic Energy T i h at f P Insuf cient energy for photoej ection Photoej ection occurs Photoelectric Effect The kinetic energy of the ejected particle is given by 12 mv2 hv CD The threshold energy is D the work function This demonstrates the particlelike behavior of photons A wavelike behavior would be indicated if the intensity produced the effect The WaveParticle Duality The fact that the DeBroglie wavelength explains the quantization of the hydrogen atom is a phenomenal success Other wavelike behavior of particles includes electron diffraction Particlelike behavior of waves is shown in the photoelectric effect De Broglie Relation The wavelike properties of particles can be described very simply in the relationship of wavelength and momentum A h p The practical importance of this expression is realized in electron microscopy By tuning the accelerating voltage in an electron microscope we can alter the momentum and therefore the wavelength of the electron The definition of a photon The waveparticle duality goes both ways If a particle can act like a wave then a wave can act like a particle Light particles are called photons The absorption of photons can explain how atoms and molecules can absorb discrete amounts of energy The energy of a photon is E hv Experimental observation of hydrogen atom Hydrogen atom emission is quantized It occurs at discrete wavelengths and therefore at discrete energies The Balmer series results from four visible lines at 410 nm 434 nm 496 nm and 656 nm The relationship between these lines was shown to follow the Rydberg relation Atomic spectra Atomic spectra consist of series of narrow lines Empirically it has been shown that the wavenumber of the spectral lines can be fit by 1 2 2 quot1 quot2 n2gtn1 where R is the Rydberg constant and n1 and n2 are integers The hydrogen atom semiclassical approach Why should the hydrogen atom care about integers What determines the value of the Rydberg constant R109677 cm1 Bohr model for the hydrogen atom 6 2 m 12 f 47 8 0r2 r Coulomb Centrifugal Balance of forces Assume electron travels in a radius r There must be an integral number of wavelengths in the circumference 27cr rm n 123 The electron must not interfere with itself The condition for a stable orbit is 27cr rm n123 The Bohr orbital shown has n 16 The DeBroglie wavelength A hp or A hmv it H jU tI gives mvr nh27 n123 I This is a condition for quantization of angular momentum Wat Example of selfinterference According to the Bohr picture the condition shown will lead to cancellation of the wave and is not a stable orbit The quantization of angular momentum implies quantization Lr of the radius a git 7 r 615 0 as me Mr fly05w H is v gf tj The significance of quantized orbits The Bohr model is consistent with quantized orbits of the electron around the nucleus This implies a relationship between quantized angular momentum and the wavelength Einstein argued based on relativity that X hp where the wavelength of light is 7 and the momentum of a photon is p DeBroglie argued that the same should hold for all particles The Bohr Model Predicts Quantized Energies The radii of the orbits are quantized and therefore the energies are quantized According to classical electrostatics 2 2 L 2 639 639 E T V ZmV 4780r 8780r Substituting in for r gives 4 En me2 12 Sean n Chemistry 431 Lecture 8 The variational theorem NC State University The variational theorem The variation method allows us to obtain an approximation to the ground state energy of the system without solving the Schrodinger equation The variation method is based on the following theorem Given a system with hamiltonian operator H then if I is any normalized wellbehaved function that satisfies the boundary conditions it is true that fpHpd C 2 EO where E0 is the true value of the lowest energy eigenvalue of H This important theorem allows us to calculate an upper bound for the ground state energy Practical significance The variation method serves as the basis for all methods that use combinations of hydrogenlike orbitals to solve for the eigenfunctions Wm funi and gwm ege of atoms and molecules The radial part of the hydrogenlike wave functions is modified by a variational parameter which is minimized The theorem allows us to set the derivative with respect to any parameter or equal to zero to find the value of that parameter that minimizes the energy a gtxlt a p Hpd t 0 we can be sure that the energy calculated in this way Will be greater than the true energy an upper bound The hamiltonian for H The electronic hamiltonian for the hydrogen atom consists of a kinetic energy term for the electron and the Coulomb attraction of the electron and proton nucleus W 2 7 Z 2 V1 Of course the nuclear charge of hydrogen is Z Z is included for completeness We know that the solutions Of the Schrodinger equation HLP ELF gives energy levels 2 n is the principal quantum number En 267 a0 is the Bohr radius 0 The hamiltonian for He For helium the same kinetic energy and Coulomb attraction terms are present but there is also a Coulomb repulsion between the two electrons that must be included V2 Because of the Coulomb repulsion there is no exact solution for He To solve the problem we use two 1s orbitals from the solution for hdroven and then a I I the variational method The4eNavefunc on The hydrogen 1s wave functions for electrons 1 and 2 are 32 32 ra Z ra frag ems am The aufbau approach for atoms assumes that the total wave function for a manyelectron atom is just a product of one electron wave functions In the present case m Note that the hydrogen wave functions are normalize o l mmf mm1 Variational approach for the He atom The He wave function used for the variation method is a product of two hydrogen 1s orbitals However instead of the nuclear charge Z we use a variational parameter Q C 3 p e eraOe Qrzao 5 0 Q has a physical interpretation Since one electron tends to screen the other from the nucleus each electron is subject to a nuclear charge that is les than Z The hamiltonian is H Vf E Vi EQ ZQ Zi 71 2m Evaluation of the integrals If we consider only the part of the hamiltonian in parentheses We have the solution to a hydrogen atom with two electrons in the 1s orbital 2 2 12 2 2 Ce 2 C6 262 W T E r2t Cao l where the right hand side is twice the energy of a 13 electron Using this result we have 2e2 2 9pdr 2 9pdr 2 9pdr IpHpdT Ca O ppd tC ZefTl C ZkITte T The integrals have the following values ltpltpdr ltpltpdr C 2 9de 5C 62 gtllt Immdr 1 71 f 72 a0 6 712 8610 Evaluation of the variational parameter Q We have pHpdt Q2 22g gag 2 now vary Q to minimize the variatio ral 39Cl f pHpdt Q2 22g QZ 0 2g 2z0 i C Z 16 The variational energy is t 2 25e2 ize2 IQer ZSZ 256 Z 16 The variational energy comparison With experiment The experimental ionizawn eergy of He is 245 eV Our first guess would be to calculate the energy of the 1s Electron in He using the hydrogen energy level with a nuclear charge Z 2 E Ze2a0 This gives 2136 eV 272 eV Using the value obtained by the variational method we have E 2716e2a0 2716136 eV 2295 eV The value is much closer to the true value In accord with the variational theorem the true ground state energy is less than that given by variational method Summary The h droen atom is the onl atom with an exact solution Hydrogen wave functions are used as the approximation for atomic wave functions in multielectrc moms The variational principle states that any wave function we choose that satisfies the Schrodinger equation will give an energy greater than the true energy of the system The variation method provides a general prescription for improving on any wave function with a parameter by minimizing LlIdL lunuuun VVILII IUDIJUUL tu Lllc pdldlnctcr Chemistry 431 Lecture 4 The Partition Function Statistical Thermodynamics NC State University Molecular Partition Functions In general gj is the degeneracy 8 is the energy gig5amp1 We assume that the energy of the lowest energy level the ground state is so 0 Recall that B 1kT Examples ATwo level system Blnfinite energy ladder Two Level System Assume that devenerac g0g11ie single state is found at each level q 1e BS Note that as T gt 0 q gt 1 and asT gtoo q gt 2 The ratio of the population in the two is states e 38 where e is the energy difference between the two states 00 Ensemble Partition Function We distinguish here between the partition function of the ensemble Q and that of an individual molecule q Since Q represents a sum over all states accessible to the system it can written as QNVT e B8i8jgk jk where the indices ijk represent energy levels of different particles The molecular partition function q represents the energy levels of one individual molecule We can rewrite the above sum as 0 Q qiqjqk or Q qN for N particles Note that qi means a sum over states or energy levels accessible to molecule i and qj means the same for molecule j qV77 Z 9 3 The molecular partition function counts the energy levels accessible to molecule i only Q counts not only the states of all of the molecules but all of the possible combinations of occupations of those states However if the particles are not distinguishable then we will have counted N states too many N NN1N2 This factor is exactly how many times we can swap the indices in QNVT and get the same value again provided that the particles are not distinguishable If we consider 3 particles we have ijkjik kij kji jki ikj or 6 3 Thus we write the partition function as Q qN for distinguishable pariicies Q X7 for indistinguishable pariicies Translational Partition Function The translational partition function is the most important one for statistical thermodynamics Pressure is caused by translational motion ie momentum exchange with the walls of a container For this reason it is important to understand the origin ofthe translational partition function Translational energy levels are so closely spaced as that they are essentially a continuous distribution The quantum mechanical description ofthe energy levels is obtained from the quantum mechanical particle in a box Particleinabox energy levels The energy levels are enxnynz n n z nXnynz 1 2 The box is a cube of length a The average quantum numbers will be very large for a typical molecule This is very different than what we find for vibration and electronic levels where the quantum numbers are small ie only one or a few levels are populated Many translational levels are populated thermally The translational partition function is qfrans Z 67 anxnynZkT nxnynz 1 oo oo 00 If 73 E13 Wkw Mimi The three summations are identical and so they can be written as the cube of one summation 39 39 3 w 2 2 trans ex h n 7 E1 p 8ma2kT The fact that the energy levels are essentially continuous and that the average quantum number is very large allows us to rewrite the sum as an integral mil ml 3 The translational partition function is proportional to volume The sum started at 1 and the integral at 0 This difference is not important if the average value of n is ca 109 If we have the substitution a h28masz we can rewrite the integral as 3 3 trans 00 ocn2 i12 q lie dnl 4a This is a Gaussian integral The solution of Gaussian integrals is discussed the math section of the Website If we now plug in for a and recognize that the volume of the box is V a3 we have 32 gm 2 V Probability in the ensemble The ensemble partition function is ElkT Q 6 Where the ensemble energy is E The population of a particular state J with energy EJ is given by e EJkT p f e EJkT Q J0 This known as the Boltzmann distribution The normalization constant of the above probability s nu The sum of all of the probabilities must equal 1 e EJkT Calculation of average properties The importance of the canonical ensemble is evident once we begin to calculate average thermodynamic quantities The basic approach is to sum over the probability of a state being occupied times the value of the property in a given state In general for an average property M we can write ltMgt Pij I M could be energy or pressure etc PJ is the Boltzmann probability given by P e BSJQ Average energy If we denote the average energy ltEgt then 00 ltEgt 10 PJEJ J J EJe BE 0 6 BE 0 M8 This can be written corn actl as 3an Equot 68 Consistency check with kinetic theory of gases The average energ pv olecule is given by trans trans Strains alnq sz V V as aT sz lnT terms independentof T 2 1 a kT7a TT kT which agrees with the kinetic heory of gases The second step follows from the fact that nabc na nb nc We can rewrite the logarithm as a sum Ine terms that do not depend on temperature will vanish Chemistry 431 Lecture 9 Spinorbit coupling NC State University Spinorbit Hamiltonian The magnitude of spinorbit coupling is measured spectroscopically as a splitting of spectral lines The Hamiltonian is Hso h0Ajj 1 1 ss 1 Where I is the orbital angular momentum quantum number and s is the spin quantum number The total angular moment is s and A is the magnitude of the spinorbit coupling in wavenumbers HsohcAlsl s1 l1 ss1 ghcAl2s2lsslls l2 l s2 shcAIs Spinorbit Hamiltonian The magnitude of the spin orbit coupling can be calculated in terms of molecular parameters t substitution hcA LS 2amp2 lt13 gtLS 2 r where or is the fine structure constant or e2hc47ceo which is a dimensionless constant or 1137037 L and S are operators Z is an effective atomic number The spin orbit coupling splitting can be calculated from ES 3 Such 1350 VHSOsUdF Z 50 21372 r Spinorbit Hamiltonian This can be evaluated using the above identity that can be recast to give an spinorbit coupling energy in terms of molecular parameters Z l Eso lt1gt IltI1gt slts1gt21372 r3 where ltgtl T1713 WT Spinorbit Hamiltonian We can evaluate this integral explicity for a given atomic orbital For example for T210 we have 32 T210 1 l Ae z 2quotquotocos 9 4m lt90 30 so that the integral is 5 271 TI 00 dd coszesinede rzle 0r2dr which integrates to l Ll 5 2 a02 i 3 ltr3gt 32nao2 3z2 24ao or 2324 in atomic units Spinorbit Hamiltonian Therefore we have ltl gt Z3 r3 n363It121l Therefore In general the spinorbit sp IttIng IS given by E Z4 jj1 1 SS1 so 21372agn3l 2121 Note that the spinorbit coupling increases as the fourth power of the effective nuclear charge Z but only as the third power of the principal quantum number n This indicates that spin orbitcoupling interactions are significantI larger for atoms that are further down a particular column of the periodic table The Sodium D Line One notable atomic spectral line of sodium vapor is the socalled Dline which may be observed directly as the sodium flametest line and also the major light output of lowpressure sodium lamps these produce an unnatural yellow The Dline is one of the classified Fraunhofer lines Sodium vapor in the upper layers of the sun creates a dark line in the emitted spectrum of electromagnetic radiation b absorbin39 visible li39ht in a band of wavelengths around 5895 nm This wave length corresponds to transitions in atomic sodium in which the valenceelectron transitions from a 3s to 3p electronic state The Splitting of the D Line Closer examination of the visible spectrum of atomic sodium reveals that the Dline actually consists of two lines called the D1 and D2 lines at 5896 nm and 5890 nm respectively The splitting between These lines arises because of spinorbit cou lin The constantA is usually given in cm391 For Na it is 115 cm391 Na has one unpaired electron s 12 If we consider the s gt p transition then for the excited state p we have I 1 Thus j 32 or 12 Practical calculations using the Spinorbit Hamiltonian The two energy levels can be calculated in terms or the constant A 1 2A 1532 A3232 1 11 1 1212 1 E12 N A1212 1 11 1 1212 1 A The energy difference between the lines is 32A Thus the energy splitting for Na is 173 cm391 Chemistry 431 Lecture 1 Ideal Gas Behavior NC State University Macroscopic variables P T Pressure is a force per unit area P FA The force arises from the change in momentum as particles hit an object and change direction Temperature derives from molecular motion 32RT 12Mltu2gt M is molar mass Greater average velocity results in a 39 temperature u is the velocity Mass and molar mass We can multiply the equation g l lt 2gt 2RT 2M u by the number of moles n to obtain 3 2 nRT nM ltu gt 2 2 If m is the mass and M is the molar of a particle then we can also write nM Nm N is the number of particles Mass and molar mass In otherwords nNA N where NA is Avagadro s number 3 2 EnRT Nm ltu gt Average properties ltu2gt represents the average speed Kinetic Model of Gases Assumptions 1A gas consists of molecules that move randomly 2 The size of the molecules is negligible 3 There are no interactions between the gas molecules Because there are such large numbers of gas molecules in any system we will interested in average quantities We have written average with an angle bracket For example the average speed is 2 2 2 2 ltllzgt5251szsasN We use s for speed 512 53 5N and cfor mean speed f Velocity and Speed When we considered the derivation of pressure using a kinetic model we used the fact that the gas exchanges e tum with the wall of the container Therefore the vector directional quantity velocity was appropriate er in the energy expression the velocity enters as the square and so the sign ofthe velocity does not matter In essence it is the average speed that is relevant forthe energy Another way to say this is the energy is a scalar A 2gt l 2 i z E 2m 2m 2quot All ofthese notations p mu quot1 mean the same thing The rootmeansquare speed The ideal gas equation of state is consisten 39 t With an interpretation of temperature as proportional to the kinetic energy of a Mu2 RT Ifwe solve for ltu2gt we have the meansquare speed 2 7 3RT u 7 M Ifwe take the square root ofboth sides we have the rms spee 2 12 7 3RT M 17 The mean speed The mean value is more commonly used than the rootmeansquare ofa value The rootmeansquare speed ls equal to the rootmeansquare velocity 02 12 The mean speed is 7 8 z 7 c a 37 The rms speed of oxygen at 25 ElC 298 K is 482 ms Note M is converted to kgmol Um 31 Jmo Kz98 K 0 032 legmoi 4818ms The Maxwell Distribution Not all molecules have the same speed Maxwell assumed that the distribution of speeds was Gaussian M 32 2 7 M52 175 4n2 RT 5 exp RT As temperature increases the rms speed increases and the width of the distribution increases Moreover the functions is a normalized distribution Thisjust means that the integral over the distribution function is equal to 1 m See the MAPLE L Foods 1 worksheets for examples Molecular Collisions Cross section d2 0 a Center location of target molecule ltugtt distance traveled mean ee path animate volume of interaction number density nN moles per unit volume molar density NN molecules per unit volume number density lt u gt t mean free path estimate W Refinement of mean free path The analysis of molecular collisions assumed that the target atom was stationary Ifwe include the fact that the target atom is moving we nd that the relative velocity is lt u gtm if lt u gt Therefore 9 As the pressure increases the number density increases and the distance between collision mean free path becomes shorter As the temperature increases at constant pressure the number density must decrease and the mean free path ill increase ltugtt 1 1 RT cltugttlvlv cNV oNAnV EGNAP Mean free path Collision frequency The mean free path A is the average distance that a molecule travels between collisions The collision frequency 2 is the average rate of collisions made by one mo ec The collision cross section 0 is target area presented by one molecule to another When interpreted in the kinetic model it can be shown that 7 RT 7ENAUV UP 7 2 17 27 67nd ENAGP RT The product of the mean free path and collision frequency is equal to the room mean square speed Wm E m 8222008 Units of Pressure Units of Energy Energy has units of Joules 1 J 1 Nm Work and energy have the same units Work is defined as the result of a force Force has units of Newtons F ma kg ms2 Pressure has units of Newtonsmeter2 2 2 2 P HA Kg mS m KgS m1 acting through a distance These units are also called Pascals Pa 5 P 105 N 2 We can also define chemical energy in 1 bar T 10 a T m 39 terms ofthe energy per mole 1atm101325x105 Pa 1kJmol 1 kcalmol 4184 kJmol Thermal Energy Extensive and Intensive Variables Thermal energy can be de ned as RT Extensive variables are proportional to the Its magnitude depends on temperature size of the system R 831 JmOIK or 198 GalmOIK Extensive variables volume mass energy At 298 K RT 2476 Jmol 2 476 kJmol Thermal energy can also be expressed on 8 Intensive variables do not depend on the per molecule basis The molecular size ofthe system eqU39Va39ent 0f R 395 the Bonzmann congantv k39 Intensive variables pressure temperature R NAk density NA 6022 x 1023 moleculesmoi Equation of state relates P V and T Microsopic view of momentum The ideal gas equation of state is c V nRT An equation of state relates macroscopic ux b properties which result from the average f O 39 behavior of a large number of particles or w a lt as 3 0 V0 A particle with velocity uX strikes a wall Macroscopic Microscopic The momentum of the particle changes from muX to muX The momentum change is Ap 2mux Transit time C ux a The time between collisions with one wall is At 2aux This is also the round trip time The pressure on the wall force rate of change of momentum Ap Zmux mug E Zaux 1 The pressure is the force per unit area The area is A bc and the volume of the box is V abc F Transit time C Round trip ux distance Is 21 2 b a The time between collision is At 2aux velocity distancetime time distancevelocity Average properties Pressure does not result from a single particle striking the wall but from many particles Thus the velocity is the average velocity times the number of particles Nmltux2 gt P 7 T PV Nm u gt Average properties There are three dimensions so the velocity along he xdirection is 13 the total ltui lt 2gt Nmltu2 me T From the kinetic theory of gases l 2 1 2Nmltugt 2riRT Putting the results together When we combine of microscopic view of pressure with the kinetic theory of gases result we find the ideal gas law PVnRT This approach assumes that the molecules have no size take up no space and that they have no interactions Chemistry 431 Lecture 1 Introduction Statistical Averaging Electromagnetic Spectrum Black body Radiation NC State University Overview Quantum Mechanics Failure of classical physics Wave equation Rotational vibrational and translational motion Electronic structure Spectroscopy Statistical Mechanics Partition function Average energy Entropy Macroscopic view of equilibrium For a chemical reaction in which X is the reactant and Y is the product X Y with free energy change AGO the equilibrium constant is AG RT K e m Microscopic view of equilibrium For a system where there are two energy states which we can call X and Y Y AE X with energy difference AE the ratio of population is WE NY 6 Nx where NY and NX are the numbers of particles in each state Boltzmann s constant Note that we k instead of R in the exponent K is called Boltzmann s constant and it is related to R by k R NA where NA is Avagadro s number kT has units of Joules and is a measure of the thermal energy of a molecule RT has units of Joulesmole and is a measure of the thermal energy of a mole Microscopic probability When we look at the world in terms of energy levels we can see that the probability of being in a given level depends on temperature We define the probability of being in levte as number of molecules in state j or P total number of molecules N Pl l Microscopic probability We can calculate the probability of levels A and B P Nx 1 1 X NX NY 1 NyNX 1 eAEW Py N y N yNX e AEkT NX NY 39 1 NyNX 39 1 e AEW Of course PX PY 1 Average energy for a two level system We defined the relative energy of the two states as AE EY EX We can arbitrarily set the energy of state X to zero EX 0 Then the energy of state Y is EY AE The average energy of the system is lt Egt PIE PXEX PyEy AEkT AEkT e e 1 e AEkT Y 1 e AEkT Extension to many levels If we have a system with an infinite number of equally spaced energy levels we can calculate the probability and energy in the same way 6 jAEkT AE j 1 e AEkT e ZAEkT e 3AEkT e AEkTAE 26 2AEkTAE 26 3AEkTAE 1 e AEkT e ZAEkT e 3AEkT Partition function for ladder of energy levels The denominator gives the average number of levels that are accessible at a given temperature We can simplify it as follows Q 1 e AEkT e ZAEkT e 3AEkT Letx e39AE kT then Q 1 x x2 x3 Q 1XX2X3X3 and XQXX2X3X3 since Q 1 XQ Q11 1X 1e AEkT Calculating the energy It is not obvious how to obtain the energy for this system We can consider the general formula and make the substitution 3 1kT e BAEAE 2e 2BAEAE 2e 3BAEAE 1e BAE e ZBAE e 3BAE aQaB 39 T For the energy ladder we have 1 G BAEX G BAEXAE AE 1 eBAE2 e BAE 1 ltEgt High temperature limit As the temperature approaches infinity all of the levels become equally populated The average energy can be calculated assuming 3 1kT ltlt 1 ltEgt AE AE 1 e BAE1 1 BAE 1 B H Thus kT is a classical energy for an averaged system at high temperature In units ofjoulesmole ltEgtRT The energy of a gas is related to its PVm product Vm is the molar volume thus PVm RT The wavelength and the frequency of electromagnetic radiation i9 wavelength si The wavelength k is the distance between the peaks in a traveling wave In classical physics light is a wave that travels with velocity 0 The frequency is v CI The wavenumber V v c The wavenumber has units of cm39l The electromagnetic spectrum 7 increasing gt llll lllllm v decreasing The wavelength and frequency are inversely related Black body Radiation An ideal emitter of radiation is called a black body Observation that peak of the energy of emission shifts to shorter wavelengths as the temperature is increased Wien displacement law kmaXT 288 x 106 nmK Dilemma for Classical Physics The maximum in energy for the black body spectrum is not explained by classical physics The cavity modes of the black body are predicted to be p 2 8n k B T k 4 Where p is the radiant energy density This function increases without bound as A gt 0 This law is known as the RayleighJeans law p x103 Wm393 40 30 20 UV Catastrophe RayleighJean 8nk5T 9L4 f I 1000 2000 3000 Wavelength nm 4000 5000 The Planck Distribution Law Planck assumed quantization of cavity modes E nhv n012 The constant h determines that only those modes with an energy specified by the precise amounts given can be excited The population of the levels will favor lower energy quantum number modes over higher energies Planck Assumption implies that average energy is temperature dependent In classical physics the average energy in an energy level is lt E gt kT Quantized levels imply that the average energy in each oscillator is lt E gt hvthkT 1 Since c w we can also write this as lt E gt hcveh0 kT 1 To obtain the Planck formula simply replace kT by the lt E gt hc Memka 1 expression implied by quantization Planck s Innovation Classical radiation Quantized radiation All frequencies Only frequencies nhv are possible are allowed Mathematical Form of the Planck Law The energy levels will be populated according to a thermal weighting The higher levels will be less populated than the lower levels In the Planck theory the energy density becomes 87thC 1 9 k5 ehcMBT 1 p x103 Wm393 40 30 20 UV Catastrophe RayleighJean 8nk5T 9L4 f I 1000 2000 3000 Wavelength nm 4000 5000 Resolution of UV Catastrophe 8757C 1 9 k5 ehcxk5T 1 I I I I 40 RayleighJean Planck E 30 E S 20 5 O 10 O I I I I O 1000 2000 3000 4000 5000 Wavelength nm q 11 39x10 Wmquot 40 30 20 10 Planck Distribution Law T2500K T2000K T1500K T1000K 2000 3000 4000 Wavelength nm 1000 5000 Consistency with Experiment The temperature behavior of the Rayleigh Jeans law is recovered because ehClkT 1 z hckkT as T gt oo The integral of the total energy is proportional to T4 which gives the StefanBoltzmann law W GT4 W is the flux or energyarea The Wien displacement law is recovered from differentiation of p Setting dpdk 0 gives the maximum in the distribution law p x103 Wm39s Wien displacement law 0 6 87T7C 1 ax ax k5 ehckk5T 1 40 30 20 10 0 I I I I I 1000 2000 3000 4000 Wavelength nm 5000 StefanBoltzmann law 0 008757C 1 4 k5 de CT 40 30 20 p x103 Wm39s 10 0 0 1000 2000 3000 4000 5000 Wavelength nm Consequences of Planck Law Classical physics fails to describe black body radiation A model that includes quantized modes of electromagnetic radiation succeeds Planck s constant h 6626 x 1034 Js is a fundamental constant the determines the scale of energy quantization The Sun is a Black Body 5 ica 235w x I ll 777 Lil l lll V 2Ill Mahmunwm The sun and stars are black bodies The peak of the emission spectrum depends on temperature as indicated by the Wien displacement law What is the radiant power of the sun Use the StefanBoltzman law W o T4 W is the flux or power per unit area 6 56704 x 10398 kg s3 K4 Wattsm2K4 Assuming that the temperature at the surface is 5500 K and the diameter is 14 x 106 km The way that the Sun39s diameter is measured is by taking angular diameter measurements and then translating them to linear diameter measurements The angular diameter of the Sun can be measured using a telescope during a total solar eclipse or by timing Mercury when it is in transit in front of the Sun The first series of measurements were taken in the early 170039s by Jean Picard in Paris France What is the radiant power of the sun First calculate the area and then the flux power The area is A 47cR2 A 4314167 x1082 A 616 x1018 m2 The flux is W G T4 W 56704 X 10398 55004 W 519 x 107 Wattsm2 The total power is P WA 519 x 107 Wattsm2616 x1018 m P 32 x1026 Watts What is the radiant power at the surface of the earth We use the distance from the earth to the sun to obtain the flux at the earth The earth is Re 15x 108 km from the sun The area irradiated is Ae 475Re2 Ae 283 x1023 m2 What is the radiant flux at the surface of the earth The flux in space above the earth is called the insolation The insolation is the power coming from the sun divided by the total area at the radius of the earth We PAe 32 x 1026 Watts283 x 1023 m2 We 113 x103 Wattsm2 This is very close to the measured value for radiation in space above the earth How much energy does the earth absorb The earth has a crosssectional area of Ac TERearthz AC 3141613 x 107 m 2 AC 53 x1014 m2 Pabs WeAc Pabs 113 x103 Wattsm2 53 x1014 m2 Pabs 6 x 1017Watts What is the temperature at the surface of the earth Pabs Pemit 6 X 1017 Watts Pemit G Tearth4 Aearth Aearth 47 Rearch 21 x 1015 m2 Tearth PemitG Aearth 14 Tearth 6 X 101756704 X 1039821 X 1015 4 T earth 266 K This is close but it is a little frosty Why is this PS What is 266 K in 0C What is 266 K in 0F What is the temperature at the surface of the earth Pemit 6 x 1017Watts emit G Tearth4 Aearth Aearth 43975 Rearth2 2391 X 1015 m2 P Tearth PemitG Aearth 14 Tearth 6 x101756704 X 10821 x101514 T This is close but it is a little frosty Why is this We ignored the fact that the earth has an atmosphere The atmosphere reflects back some of the radiated heat What is the role of the atmosphere 1 Some molecules in the atmosphere absorb incident light Ozone absorbs UV light and prevents harmful radiation from reaching the surface of the earth 2 Molecules can also absorb emitted or radiated light What is the wavelength of such light It can be obtained from the Wien displacement law 7 T 288 X 106 nmK max Thus for the sun with T 5500 K max 523 nm For the earth with T 266 K Amax 10800 nm 108 um The sun s emission is peaked in the visible region of the Electromagnetic spectrum and the earth emits in the infrared Absorption by gases in the atmosphere ll normalized ABSORBTION l ABSORFHVITY quotIll BLACK BODY CURVES 5500 K a H20 cacao on J C I I l 0 15 02 03 Electronic 1 5 2 3 5 WAVELENGTH u Vibrational Rotational Chemistry 431 Lecture 6 Rotational motion NC State University Classical Rotation In a circular trajectory JZ pr and E JZZZI is the moment of inertia Mass in a circle I mr2 Diatomic Mr2 m FR 1 m2 m1m2 Reduced mass Ll m m 1 2 Rotation in two dimensions The angular momentum ist pr J Z p m Using the deBroglie relation p hk we also have a condition for quantization of angular motion JZ hrk The 2D rotational hamiltonian The wavelength must be a whole number fraction of the circumference for the ends to match after each circuit The condition 27rrm 9 combined with the deBroine relation leads to a quantized expressionJZ rrTh The hamiltonian is 125ng E 2 549 The 2D rotational hamiltonian Solutions of the 2D rotational hamiltonian are sine and cosine functions just like the particle in a box Here the boundary condition is imposed by the circle and the fact that the wavefunction must not interfere with itself The 2D model is similar to condition in the Bohr model of the atom The 2D rotational hamiltonian The wavelength must be a whole number fraction of the circumference for the ends to match after each circuit The condition 27rrm 9 combined with the deBroine relation leads to a quantized expressionJZ m 39 r2 2 The hamiltonian IS n D Eq5 Quantization of rotational motion solution of the 4 equation The corresponding wavefunctions are 1 12 CD e m m 2 with the constraint that 12 m L H Since the energy is constrained to values JZZZI we find that m 0 i1 i2 The wavefunctions of a rigid rotor are called spherical harmonics The solutions to the 9 and 1 equation angular part are the spherical harmonics Y9 9CD Separation of variables using the functions 99 and 134 allows solution of the rotational wave equation 2 I 2 7 1 ay 1 asin6 EY Z sin26 5P2 Sine 59 59 We can obtain a 9 and 1 equation from the above equation Spherical Polar Coordinates The volume element in spherical polar coordinates To solve the Schrodinger equation we need to integrate of all space This is the same thing as performing a volume integral The volume element is dV rzdr sinede dd This integrates to 47 which is the normalization constant 47 stearadians also gives the solid angle of a sphere Separation of variables The spherical harmonics arise from the product of 9CD after substituting Y 91 acho 6 m2 sin9sineage sinZGECDG For sake of simplicity we have factored the term 72 W into the energy At the end of the calculation we Will multiply by this term to obtain the energy Separation of variables The operators in variables 9 and 1 operate on function 9 and CD respectively so we can write 2 G8 I 8p2 sinGa sinG g sinZGECDG When we divide by Y 91 we obtain 162 1 a a 2 paw sm989sm9 89 Sln GE 0 Now these equations can be separated Separation of variables We use a separation constant equal to m2 to write two equations The 9 and 1 equations are sinega82 2 8 69sm989 sm 9E m 1 3ng 2 5Wquot m We have already seen the solution to the 1 equation from the example of rotation in two dimensions The solution of 9 equation gives Legendre polynomials Substitute x c059 and the equation becomes 2 2 1 x2 2x E L P 0 8X2 5X 1 X2 The solution requires that Efh22lJJ1 with JO12 WhereJ is the rotational quantum number The azimuthal quantum number m is also called J2 The magnitude of JZltJ The solutions are Legendre polynomials P0Xl P2X12 3X2 1 P1XX P3X12 5X3 3X The spherical harmonics as solutions to the rotational hamiltonian The spherical harmonics are the product of the solutions to the 9 and 1 equations With norm aization these solutions are YJM9ltIgt NJMPJ39M39COS 99 M The M quantum number corresponds to JZ the 2 component of angular momentum The norm aization constant is N 2J1J Ml1l2 JM 2 J M The form of the spherical harmonics Including normalization the spherical harmonics are Y Y20t300329 1 Y10 t cose Y2i1 33 Sinecoseei l 1 A 39 ii l 2 ii Y 8Tc 322 Sln 9e 24 The form commonly used to represent p and d orbitals are linear combinations of these functions Euler relation Linear combinations are formed using the Euler relation em 0039 i isine 94 e i SInG 2 id il 0039 Z Projection along the z axis is usually taken using 2 rcosG Projection in the xy plane is taken using x rsinGcoscl and y rsinGsind The 3D rotational hamiltonian There are two quantum numbers J is the total angular momentum quantum number M is the zcomponent of the angular momentum The spherical harmonics called Y M are functions whose probability YJM2 has the well known shape of the s p and d orbitals etc JO iss MO J1isp M101 J2isd M21012 Space quantization in 3D Solutions of the rotational Schrodinger equation have energies E h2JJ 121 Specification of the azimuthal quantum number mZ implies that the angular momentum about the zaxis is JZ hmz This implies a fixed orientation between the total angular momentum and the 2 component The x and y components cannot be known due to the Uncertainty principle Spherical harmonic for I15 P0cose Plot in polar coordinates represents Y002 where 415 Y0014n 2 t Solution corresponds to rotational quantum numbers j J O f M orJ O 2 Z Polynomial is valid for n21 quantum numbers of hydrogen wavefunctions Spherical harmonic for P1cose Plot in polar coordinates represents Y112 where Y1112327 2 sineeilh Solution corresponds to rotational quantum numbers J 1 JZ i1 Polynomial is valid for n22 quantum numbers of hydrogen wavefunctions Spherical harmonic for P1cose Plot in polar coordinates ff Hmmm represents Y102 where I Y101237 2 cose 0539 x with normalization I Rah Eff Solution corresponds to M w D Ema 394 rotational quantum K numbersJ 1 JZ O 05 K Polynomial is valid for a 54xr n22 quantum numbers of hydrogen wavefunctions Spherical harmonic for P2cose Plot in polar coordinates of Y222 where l 2 2 2i 4 Y2 141527 2COS 6e 4 a 05V Solution corresponds to a oa r I14 I16 rOtational quantum Km numbers J 2 JZ i 2 39 t Polynomial is valid for L M J n23 quantum numbers of hydrogen wavefunctions Spherical harmonic for P2cose Plot in polar coordinates m m of Y212 where t quot n1 I Y21158n 2sinecoseel Hanvi J o Solution corresponds to in 431 um aky 015 rotational quantum aft K numbers J 2 JZ i 1 39 XX Polynomial is valid for ix 39 5 Us n23 quantum numbers of hydrogen wavefunctions Spherical harmonic for P2cose Plot in polar coordinates quotEms of Y202 where U l Y201457c 23cos261 x a I o Solution corresponds to rotational quantum quotf x numbers J 1 JZ O 39239 3 Polynomial is valid for warfj n23 quantum numbers of hydrogen wavefunctions Rotational Wavefunctions These are the spherical harmonics YJM Which are solutions of the angular Schrodinger equation The degeneracy of the solutions The solutions form a set of 2J 1 functions at each energy the energies are E 1 12JJ 121 A set of levels that are equal in energy is called a degenerate set l2 l1 Rotational Transitions Electromagnetic radiation can interact with a molecule to change the rotational state Typical rotational transitions occur in the microwave region of the electromagnetic spectrum There is a selection rule that states that the quantum number can change only by or 1 for an allowed rotational transition AJ i1 l2 A l1 lo Orthogonality of wavefunctions The rotational wavefunctions can be represented as the product of sines and cosines Ignoring normalization we have s 1 p 0039 sinecosd sinesind d 12300329 1 0032900324 coszesinZd cosesinecosd cosesinesind The differential angular element is sin9d9d 4rc over the limits 9 O to 7 and l O to 27 The angular wavefunctions are orthogonal Orthogonality of wavefunctions For the theta integrals we can use the substitution x cose and dx sinede For example for s and ptype rotational wave functions we have n 1 2 1 1 ltSpgtoclcosesinedel deX 0 1 Question Which of the following statements is true A The number of zprojection of the quantum numbers is 2J1 B The spacing between rotational energy levels increases as 2J1 C Rotational energy levels have a degeneracy of 2J1 D All of the above Question Which of the following statements is true A The number of zprojection of the quantum numbers is 2J1 B The spacing between rotational energy levels increases as 2J1 C Rotational energy levels have a degeneracy of 2J1 D All of the above AEJ2J 1 JJ 1 2J 1 Question The fact that rotational wave functions are orthogonal means that A They have no overlap B They are normalized C They are linear functions D None of the above Question The fact that rotational wave functions are orthogonal means that A They have no overlap B They are normalized C They are linear functions D None of the above The moment of inertia The kinetic energy of a rotating body is 12Im2 The moment of inertia is given by 00 I 2 min2 i1 The rigid rotor approximation assumes that molecules do not distort under rotation The types or rotor are with moments la lb lc Spherical Three equal moments CH4 SFG Note No dipole moment Symmetric Two equal moments NH3 CH3CN Linear One moment COz HCI HCN Note Dipole moment depends on asymmetry Asymmetric Three unequal moments H20 Pure rotational spectra A pure rotational spectrum is obtained by microwave absorption The range in wavenumbers is from 0200 cml Rotational selection rules dictate that the change in quantum number must beAJi1andAMJO A molecule must possess a ground state dipole moment in order to have a pure rotational spectrum The Dipole Moment Expansion The permanent dipole moment of a molecule oscillates about an equilibrium value as the molecule vibrates Thus the dipole moment depends on the nuclear coordinate Q 8H uQ no wk where u is the dipole operator Rotational Transitions Rotational transitions arise from the rotation of the permanent dipole moment that can interact with an electromagnetic field in the microwave region of the spectrum The total wave function The total wave function can be factored into an electronic a vibrational and a rotational wave function 5 WeXVYJM Mrot XLY M f WigH WedTeXvYMdrnuc xLchQf f xLIYWquVYJMsinecedl I I XTYI M HOXVYIMSin9d9dd Interaction with radiation An oscillating electromagnetic field enters as EOCOSoot such that the angular frequency hm is equal to a vibrational energy level difference and the transition moment is 27 Tc Mm of I 1McoseYJMsinaded 0 O Interaction with radiation The choice of cos9 means that we consider zpolarized microwave light In general we could consider x or ypolarized as well X Sln9cos Mo M W M Z 0 0sinecoscln39 sinGsincbj cosek Energy level spacing Energy levels 2 EJJJ1 Energy Differences 0fAJi1 l2 2AM EJH39EJ The rotational constant The spacing of rotational levels in spectra is given by AE EJ1 EJ according to the selection rule AE J 1J 2 JJ 1 2U 1 The line spacing is thus proportional to the rotational constant 2 If n h h th B 4750 8720 8n20pR2 A pure rotational spectrum m i i i d co i TM 40 E X105709 a 3 H E 5 39 m a E n E a m E Q 39 o c i i 200 400 600 800 Rest Freqv GHz A pure rotational spectrum is observed in the microwave range of electromagnetic spectrum Question Which molecule found in the atmosphere has a pure rotational spectrum A Diatomic oxygen B Diatomic nitrogen C Water D Carbon Dioxide Question Which molecule found in the atmosphere has a pure rotational spectrum A Diatomic oxygen B Diatomic nitrogen C Water D Carbon Dioxide 14 Rbranoh Rm Qbranch quot3931 Pbranch R321 quot113 pm quot4 334 line WEI 1 53 quot5 W26 Fm m 3 13 pm mm mm quotth ll L 391 me me me 295 m 2350 23m 2751 27m 2166 m 39W39auenumbers Note that the rotational spectrum is centered a vibrational frequency The rotational partition function The degeneracy of rotational levels is JJ1 The energies are given by 8 thJJ1 where J is the rotational quantum number and B is the rotational constant qr i 1e thBJJ1 JO A large number of rotational levels populated since the rotational constant is of the order of 10 cm 391 for many molecules and kT gt 10 cm391 for temperatures higher than about 15 K The high temperature form of the rotational partition function The partition function can be expressed as an integral at high temperature qr 00 1e thBJJ1dJ 0 It turns out that the integral can be solved analytically by making the substitution u JJ1 u JJ1du 2J1dJ The rotational partition function is an exponential integral Making the substitution u JJ1 the integral reduces to an easily soluble exponential integral 00 3hCBu 1 qr L9 d thB The rotational constant B is B i Where I is the moment of inertia 47tCI

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