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Delaware - CHEM - 444
Elementary reactionsReactions of interest are often complexSome reactions do occur in a single step- elementary reactionsPhysical ChemistryLecture 6Reaction mechanisms andreaction-velocity predictionsMathematics of simplereaction sequencesSimple
Delaware - CHEM - 444
Physical ChemistryLecture 8Reactions in solution andrelaxation methods in fast kineticsThe cage effectIn solution, solventis a major factor inkineticsLimited proximity ofreactantsMolecules mustdiffuse into reactionzoneDiffusion controlLimiti
Delaware - CHEM - 444
The energy problemEarly experimentsPhysical ChemistryLecture 9Introduction to QuantumMechanicsObserved only certainemission frequenciesSpectra of atoms H and other atomsColors of mineralsolutionsScientists involvedhave very familiarnamesOth
Delaware - CHEM - 444
WavesPhysical ChemistryLecture 11Waves, Matrices, Operators, andEigenvalue Equations0n(0) -Wave moving through spacewavelength in vacuumrefractive index of the medium,which varies with wavelength (dispersion)Waves in complex notationWave equat
Delaware - CHEM - 444
Postulates of quantummechanicsAny state of a dynamical system of N particles isdescribed as fully as is possible by a function, ,such that the quantity *d3r is proportional to theprobability of finding r between r and r + d3r.For every observable pr
Delaware - CHEM - 444
Angular momentumVector property thatdescribes circularmotion of a particle or asystem of particlesRigid rotor model: Aparticle of mass m fixedto a massless rodExamples Swinging a bucket ofwater Movement of theEarth around the SunPhysical Chem
Delaware - CHEM - 444
Atomic configurationsMeans of describing the energy situation of a multi-electronsystemIndependent-electron energy is a sum of hydrogenic energies ofthe electronsPhysical ChemistryLecture 20Russell-Saunders Terms, Hunds Rules,Spin-orbit Coupling,
Delaware - CHEM - 444
Geometry of dihydrogenDihydrogenPhysical ChemistryFocus on translation andCoulombic interactionsLecture 22Diatomic MoleculesHTnme = mp/1836Good approximationCenter of massdetermined by thepositions of the nucleiCenter-of-mass problemis agai
Delaware - CHEM - 444
Diatomic Molecules12th May 20091Hydrogen Molecule: Born-Oppenheimer ApproximationIn this discussion, we consider the formulation of the Schrodinger equation for diatomic molecules; this can be extended to larger molecules. First we will consider the s
Delaware - CHEM - 444
Quantum Mechanics: The Hydrogen Atom13th April 2011I. The Hydrogen AtomIn this next section, we will tie together the elements of the last severalsections to arrive at a complete description of the hydrogen atom. This will culminate in the denition of
Delaware - CHEM - 444
Helium Atom, Many-Electron Atoms, VariationalPrinciple, Approximate Methods, Spin21st April 2011I. The Helium Atom and Variational Principle: ApproximationMethods for Complex Atomic SystemsThe hydrogen atom wavefunctions and energies, we have seen, a
Delaware - CHEM - 444
Waves and the Schroedinger Equation5 april 20101The Wave EquationWe have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form based represenation. The most natural consideration a
Delaware - CHEM - 444
Waves and the Schroedinger Equation5 april 20101The Wave EquationWe have seen from previous discussions that the wave-particle duality of matter requires we describe entities through some wave-form based represenation. The most natural consideration a
Delaware - CHEM - 444
CHEMISTRY 444.11QUIZ 3Spring, 2008 (08S)March 13, 2008NAME:Score _/151. (10 points)Consider the following reaction mechanism:k2A + A A *+Ak2A *+A A + A'k2A+ M A * + M'k2A * + M A+ Mk1A *B + Ca. Write an expression for the rate of react
Delaware - CHEM - 444
1Problem 1Do the one-dimensional kinetic energy and momentum operators commute?If not, what operator does their commutator represent?h2 d2KE =2m dx21.1dP = ihdxSolutionThis question requires calculating the commutator of the operators given.K
Delaware - CHEM - 444
1Problem 1The d orbitals have the nomenclature d z 2 , dxy , dxz ,dyz , and dx2 y2 . Showhow the d orbital given below can be written in the form y z F (r ).3dyz1.12=81 1ao3/2r 2 r/3aoesin( ) cos( ) sin()a2oSolutionz = r cos ( )x = r si
Delaware - CHEM - 444
1Problem 1In previous discussion, we have considered approximate methods for obtaining energies, wavefunctions, or both for quantum systems more complicatedthan the Hydrogen atom. In this exercise, consider the quantum harmonicoscillator and use the v
Delaware - CHEM - 444
1Problem 1For the atomic term 3 P , what are the energy dierences between the levelsarising due to spin-orbit coupling.1.1SolutionFor the atomic term 3 P , show the energy dierences between the levelsarising due to spin-orbit coupling.Answer For t
Delaware - CHEM - 444
CHEMISTRY 444.10/80QUIZ 1NAME:Spring, 2011February 18, 2011EMIL JANNINGSScore _/20[Numbers without decimal points are considered infinitely precise. Show reasonable significant figuresand proper units. Answers should be in reasonable units for the
Delaware - CHEM - 444
1Problem 1The rotational constant, Be , for IBr determined from microwave spectroscopyis 0.2241619 cm1 . Approximate the bond length of this molecule.1.1SolutionFrom the Handbook:Be =h28 2 ReRearranging gives the required form for solving for t
Delaware - CHEM - 444
Quiz Solutions: S10 (Section 011, CHEM444)27th February 20101Quiz 11.1Problem 1What is the ratio of the following values for a particle of Argon and Kryptonat 298K?AraveKraveArave=KraveArmostprobableArmostprobableKrmost83.798= 1.
Delaware - CHEM - 444
CHEMISTRY 444.11QUIZ 1Spring, 2011 (11S)Feb. 17, 2011NAME:Score _/20Problem 1 (8 points)Starting from the continuous probability distribution for one-dimensional diffusion, show how one woulddetermine the average squared displacement in a model de
Delaware - CHEM - 444
Quiz Solutions: S10 (Section 011, CHEM444)27th February 201011.1Quiz 1Problem 1What is the ratio of the following values for a particle of Argon and Krypton at 298K? Ar ave Kr aveAr ave = Kr avemKr = mAr83.798 = 1.45 39.948Ar most Kr mostproba
Delaware - CHEM - 444
Transport Properties: Momentum Transport,Viscosity13th February 20111IntroductionMuch as mass(material) is transported within uids (gases and liquids),linear momentum is also associated with transport, in this case, due togradients in velocity.See
Delaware - CHEM - 419
CORRECTIONS TO PHYSICAL CHEMISTRY FOR THE LIFE SCIENCES BYENGEL, DROBNY, AND REIDPage 337: The third sentence from the top of the page should read For E< V(x) (insidethe box), the solutions have.Page 622: In Example Problem 24.1, the equation in the s
Delaware - CHEM - 419
CorrectionsEssential Data and Equations for a Course in PhysicalChemistryCecil Dybowski and Andrew V. TeplyakovPage 3-2, Table 3.41 joule = 9.47810-4 BTUPage 3-2, Table 3.61 Pa = 7.5006210-3 torr1 torr = 1.3157910-3 atm1 psi = 6.804610-2 atmPage
Delaware - CHEM - 419
NAME:KEY LARGOCHEMISTRY 419, SPRING, 2010 (2103)Examination 1, March 11,2010Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILYunderstand your work; only work on the exam
Delaware - CHEM - 419
NAME:MONKEYCHEMISTRY 419, SPRING, 2010 (2103)Midterm Examination 2, April 22, 2010Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILYunderstand your work; only work on the
Delaware - CHEM - 419
NAME:KEY, FRANCIS SCOTTCHEMISTRY 419, SPRING, 2010 (2103)Final Examination, May 22, 2010Answer each question in the space provided; use back of page if extra space is needed. Answer questions so the grader can READILYunderstand your work; only work o
Delaware - CHEM - 419
5/23/2010Student ID Midterm 1 Midterm 2 Quizzes Final Total5695153.55135.060.553700066944364040.538.238700101354927877.583.584700106045765850.563.764700127845554455.581.462700152593806975.574.775700170036535047.85
Delaware - CHEM - 419
Problem 24.2 part a 23k 1.3806505 10nm 101 J K23N 6.02211415 105M 0.131293 kgD 0.5 102 1m s3T 273 K1 3 8k T m a2mVaver 209.82sD 7.149 10Vaverk T2Problem 24.2 part bmN 2 1 N N 8k T m N2 P N0.0140067 kgN2 Vav
Delaware - CHEM - 419
Macroscopic characterizationwith distribution functionsPhysical ChemistryLecture 1Distributions and TransportProcessesParticles in a BoxExample distributionfunction for particles in abox, showing two regionsLeft side has more particlesthan the
Delaware - CHEM - 419
Macroscopic versus microscopicFicks laws describe timeevolution macroscopicallyPhysical ChemistryLecture 2Random walks; microscopictheory of diffusionDiffusion can be understoodin terms of a microscopicprocess the random walkProbability of event
Delaware - CHEM - 419
Origins of viscosityFluids try to achieve uniformflow across any region (i.e. aconstant speed independent ofposition)Fluids resist a gradient ofspeedsPhysical ChemistryLecture 3Viscosity and sedimentationDescribed by a drag force thatslows fast
Delaware - CHEM - 419
Electrophoresis in practiceGel electrophoresisPhysical ChemistryUses an acrylamide or agarose gelApply electric field for a given timeAfter electrophoresis, the gel isvisualized with a stainLecture 4Electrophoresis and lightscatteringFrom J. Vin
Delaware - CHEM - 419
RatesDescription of time dependence requires astatement of the time derivatives of theconcentrations of reactants and productsPhysical ChemistryLecture 5Introduction to chemical kineticsH2 ( gas) 1O ( gas) 22H2 O ( liquid )RatesRate of change
Delaware - CHEM - 419
Example Arrhenius plotPhysical ChemistryLecture 6Mechanisms of chemical reactionsUseful linear formSlope of plot givesthe activationenergy directlyMany rate constantsseem to obey thisrelationshipChemical kineticsUnderstand the nature of reacti
Delaware - CHEM - 419
Finding reaction velocityPhysical ChemistryLecture 7Multi-step reaction mechanismsOne knows the form of the reaction rate for everyelementary reactionExpress disappearance of reactant or appearance ofproduct in terms of the rates of these elementar
Delaware - CHEM - 419
Rate constants for bimolecularreactions in solutionReactionk (298 K)/ (dm3 mol-1 s-1)-H + HS H2S7.5 1010H+ + CH3OH CH3OH2+1 108OH- + HCO3- CO32- + H2O6 109Physical ChemistryLecture 8Reactions in solution andrelaxation methods in fast kinetic
Delaware - CHEM - 419
Pre-equilibrium approximationProduct formed only bysecond stepPhysical ChemistryElementary ratedetermined byintermediatesconcentrationABIIAssumption of rapidequilibriumLecture 9Kinetics of enzymatic reactionsSubstitution gives anexpression
Delaware - CHEM - 419
Absorption of energyEnergy is conserved in thetransfer of energy from theelectromagnetic field to amoleculePhysical ChemistryLecture 10Photochemistry and photophysicsin biological systemsAbsorption and emission onlyindicate energy differencesbe
Delaware - CHEM - 419
Photoelectric effectEmission of particles (electrons)by a metal due to interactionwith lightEinsteins explanation (for whichhe got the Nobel Prize) involvedassuming that the energy of aphoton was proportional to itsfrequencyPhysical ChemistryLec
Delaware - CHEM - 419
Conserved quantitiesQuantities that do not change in time are said to beconserveddp xExamples 0 p x is a conserved quantityPhysical ChemistrydtdEdtdp zdtLecture 12Mathematics of QuantumMechanics 0 E is a conserved quantity0p z is not cons
Delaware - CHEM - 419
Free particle in one dimensionFor a value of E there aretwo solutionsPhysical ChemistryLecture 13Solving Schroedingers Equationfor Simple SystemsGeneral solution for theenergy problem is a linearcombination of particularsolutionsNo limits on va
Delaware - CHEM - 419
Particle in three-dimensionalspaceThe world has threespatial dimensionsGeneralize simpleproblem to reflect thethree- dimensionalaspects of problemExample: a free particlein 3-D spaceAppropriate model for agas molecule in anunconfined spaceThe
Delaware - CHEM - 419
Particle in three-dimensionalspaceThe world has threespatial dimensionsGeneralize simpleproblem to reflect thethree- dimensionalaspects of problemExample: a free particlein 3-D spaceAppropriate model for agas molecule in anunconfined spaceThe
Delaware - CHEM - 419
Quantum angular-momentumoperatorsVector definitionsPhysical ChemistryLecture 15Angular momentum and the rigidrotorL Lx i L y j Lz kL2 L L L2 L2y L2xzExpression by correspondenceLxL2 i y z y z L2 L2 L2xyz i z x z xLyLz i x y x
Delaware - CHEM - 419
Particle in a finite 1-D boxFor E > V0Wave functions are oscillatoryfunctions of xSums of exponential functions withimaginary argumentsAmplitudes are related by boundaryconditionsPhysical ChemistryFor E , it approaches thebehavior of the free pa
Delaware - CHEM - 419
Simplification of the hydrogenatom problemBy substitution, the hydrogen-atomHamiltonian becomesPhysical ChemistryHLecture 17The Hydrogen Atom, a CentralForce Problem12PCM2M H CM12p2 H relDecompose into two problemsH CM CM ECM CMH rel rel
Delaware - CHEM - 419
The helium atomConsists of a nucleus andtwo electronsHamiltonian has four kindsof terms Nuclear kinetic energy Electronic kineticenergy Nuclear-electronCoulombic potentialenergy Electron-electronCoulombic potentialenergyPhysical ChemistryLe
Delaware - CHEM - 419
IndistinguishabilityIn practice, one cannot distinguish electrons by labelsPhysical ChemistryLecture 19Excited States of Helium; TermsTheory must reflect the ambiguity this fact requiresInterchange of labels cannot have a noticeable effect onproper
Delaware - CHEM - 419
5/18/2010Student ID5695170006694470010135470010604570012784570015259370017003670026641870026691470030093170030759870032981170032983970033117670034850070058264570061672770070102970070628870071418370071424070071592170071991470072028
Delaware - CHEM - 419
CHEMISTRY 419QUIZ 1NAME:Spring, 2010 (2103)February 18, 2010KEY ONEScore _/10[Numbers without decimal points are considered infinitely precise. Show reasonable significant figuresand proper units. Answers should be in reasonable units for the quan
Cal Poly - BUS - 438
Chapter 14Capital Structure in a Perfect Market Solutions to Chapter 14 Data CaseNumerical solutions to this case are available in Excel spreadsheet form, available on the Instructors Resource Center: http:/www.aw-bc.com, with the files for the Solutio
Cal Poly - BUS - 438
Chapter 14Capital Structure in a Perfect Market Solutions to Chapter 14 Data CaseNumerical so lutions to this case are av ailab le in Excel spreadsheet form, av ailab le on th e Instru ctors ResourceCen ter : http:/www.aw-bc.com, w ith the files for t
Cal Poly - BUS - 438
Chapter 14 Data Case Unlevered rU 1 a Stock Price Shares Outst c Current YTM 2 Debt Value Equity Value Total D/E 3 rE 4 WACC 12%38.61 2,040,871,000 4.94% $4,085,000,000 $78,798,029,310 $82,883,029,310 0.05 12.37% 12.00%5 Swap $1 billion in Debt for Equi
Cal Poly - BUS - 438
Chapter 14Capital Structure in a Perfect Market14-1. E C (1) = = 1 2(130, 000 + 180, 000) = 155, 000,1.20 100, 000 = 129, 167 100, 000 = 29, 167 155, 000 1.20a.a.155, 000b. b. c. d. e. 14-2. f. g. a. c.Equity value = (1) = = 129, 167Debt paymen
Cal Poly - BUS - 438
Chapter 14Capital Structure in a Perfect Market14-1.a.()E C1=NPV =12(130,000 + 180,000) = 155,000,155,0001.20100,000 = 129,167 100,000 = $29,167( ( )155,000b.Equity value = PV C 1 =c.Debt payments = 100,000, equity receives 20,000 or 70
Cal Poly - BUS - 438
ProblemsProblem 14-4Problem 14-7Problem 14-15Problem 14-4Suppose there are no taxes. Firm ABC has no debt, and firm XYZ has debt of $5000 on which it pays interest of 10% each year. Both companies haveidentical projects that generate free cash flows
Cal Poly - BUS - 438
Chapter14CapitalStructureinaPerfectMarket14.1 EquityVersusDebtFinancing1) Whichofthefollowingstatementsisfalse?A) Therelativeproportionsofdebt,equity,andothersecuritiesthatafirmhasoutstandingconstituteitscapitalstructure.B) Themostcommonchoicesarefi