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This 36 page Class Notes was uploaded by Erika Kuvalis on Thursday October 22, 2015. The Class Notes belongs to MCDB 1B at University of California Santa Barbara taught by Staff in Fall. Since its upload, it has received 46 views. For similar materials see /class/226908/mcdb-1b-university-of-california-santa-barbara in Molecular, Cellular And Developmental Biology at University of California Santa Barbara.
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Date Created: 10/22/15
CH EM1BSagerm ann Electrochemistry Redox Reactions Galvanic cells are also called voltaic cells or batteries They use an oxidation reduction redox reaction to produce a ow of electrons AG lt0 Copper plates on magnesium and magnesium dissolves as Cu2 solution Mgs Cuzru Mg2aq Cus Feb 7 2007 Ag Plates onto Copper and Copper dissolves as Cu2 Colorless solution of Ag and N03 Blue solution of Cu2 and N03 Electrochemistry A redox reaction Cu2 aq Zn 5 9 Zn2 aq Cu S Aern0 The half reactions gain e39 Cu2 aq 2e 9 Cu 3 Zn 5 9 Zn2 aq 2e Zinc dissolves and copper is precipitated from solution if the reactants are brought into direct contact with each other AE q It is possible to separate the two half reactions linking them by a wire and a salt bridge or porous plate They do not come into direct contact with each other but the same reaction occurs AE q w Batteries Convert Internal Chemical Energy into Electrical Work Pathway 2 Fan driven w plus heating of circuit q CW2 E z Pathway 1 Only heating of coilAEq Charged battery Electrical Work H 3 w QEItE E Q charge nergy Heat AE lost by E VOItage W rk battery I current t time Distharged battery A Daniel Cell Zns a Zn2 aq 2e Oxidation Anode 0X An Cu2aq 2e39 9 Cu 5 Reduction Cathode Red Cat Zinc Copper metal 1 r y metal Zn304 cuso4 solution spl n Charges Flow in External Circuit At standard and Internal Salt concentrations Bridge to a voltage of Maintain quot 1103 V s Electrical Electrons flow measure Neutrality in the If external Daniel Cell IrCUIt Ions flow In W ysalt bridge SJquot hrulgr 15501144 r r504 Nu V 392 2 t u 504 Cu Cuso lnul 2111 n misc u 1qu CHEM1 BSagermann Feb 7 2007 A Hydrogen Electrode Ce potentials can be measured Sign Conventions for Electrodes Anode and Cathode H2057 1 lm I Hafce potential cannot be measured Porous barrier Ca mde a Mugquot 7 An arbitrary choice sets 39 the half cell potential of a standard hydrogen electrode 0 5012 HH2 o v Half Reaction is always Haq written ion aq 2 6394 H2 9 Cathode compartment Reduction occurs a Anode compartment Oxidation occurs ZnH2 Galvanic Cell Spontanteous Cell Reaction Cell Potential 0763 V The Hydrogen electrode 2Hraqn2nrspHzrgnzmaq understandardcondmons Ox Anquot 2Haq 2 e 9 E0 0 zn5quotzn2aQ2er H 1 M Voltmctei39 er Red Catquot P 1 atm H o l 2Haq2erquotH2g O l E illi l gt T 25 C Z Questions H P b m Z t u it D it Ma Cell EMF t as Standard Reduction Potentials 3mm The largerthe difference between E red values the enuncamode larger E ce In a voltaic galvanic cell spontaneous E redcathode is more positive than E redanode Recall E cell E redcathode E redan0de E ad anode more n egattve CH EM1 BSagerm ann Dependence of free energy on pressure In the book see page 433 Feb 5 2007 Spontaneous Reaction manlt0 Spontaneous Reaction At equilibrium any change G e uilibrium Greactants q Gproducts posmon Pure Pure reactants products extent of reaction N onSpontaneous Reaction IXII gt 0 equilibrium Greactants position Gl md39mts Pure Pure reactants products extent of reaction G d ts reactants requlres an uphlll pm quotc climb in energy Pu re Pu re reactants products gt extent of reactlon Fig 188 0 quotquot 5 G reactants P I F 3 G Pajamas E n y a 81 G reactants lt o gt Vth G products Reaction as E W 39 39ir r39 r System P V g reactants and products ers U S i th E efc t iis mis ts e Extent it ieactlon 3 Of E quotf G prnducts React I o n 3 e a summers GT G raactantj gt a 3 G reactants 9 3 5 Eg ttt tm g Equilibrium mixture or g reactants and products Pure Pure reactants products Extent oi reaction 1 GIBBS ENERGY SECTION 41 using AG to predict Equilibria of chem ca reactions the DIRECUON of a reaction is simply the ways it travels to find the minimum of the Z Af the equil br um is def ned as the Spontaneous composition at which ARG O Gibbs energy 639 Spontaneous 3 Equilibrium r s to the LEFT of the m nimum the forward reaction is spontaneous to the RIGHT the reverse reaction is spontaneous Pure Pure reactants products CHEM1BSagermann Feb 5 2007 The Law of Mass Action 2 5029 029 2 503g Balanced chemical equations are chara ilerl by areaction quntient Q P 2 so 2 P sozPoz Q is away to express the position or the reaction with regards to concentrations of reactants and products Q 39The Value of AG depends on the Value of Q Q can have values from 0 only reactants to 00 only products Reaction Quotient Chemical Equilibrium and AG There is a value on which leads to the minimum 39ee energy G t at the system can a ain At the minimum 39ee energy the value on is equal to the equilibrium constant KEq When Q Kw AG0 When a reaction reaches the equilibrium condition the process becomes thermodynamically reversible A dynamic equilibrium is a reversible process reactants products Relationship of AG and Q at standard conditions AG AG RT an 2 5029 029P 2 s03g standard conditions Relationship of AG and Q at Equilibrium AG A60 RT MG 2 S0291 029P 2 S03g At Chemical Equilibrium 2 2 P Psog i Q271 P2 P 24 Ps02P02 S02 02 anln10 AG0 AG AG AG RTaneq Calculating AG for a reaction AG may be calculated in several ways AGW AH TAS but usually don t know AH AS AGW AHW TASM AGw EAG products EAG reactants Calculate Aern from tables of data for AGf A3rm AG RTInQ Calculate AG from AG and Q AGW RTnKeq Calculate Keq from Aern Example Problem Nzlgl 3H2i9 a 2 NHaiQ Question a mixture of the three gases contains NZ and H2 at partial pressures of 01 atm each and NHS at partial pressure 20 atm What is the value of AGM 1 Use AG AGW RTan 2 First calculate AGWD 3 AGMD EAGf products EAGf reactants 4 AGW 2mol17 kJmol 3 mol0 1moio 2 NH3 3 H2 N2 5 AGWD 34 kJ 6 IS AGHm more or less negative LhanAGmn lLint Consider the partial pressures of reactants and products CHEM1 B Sagermann Feb 5 2007 Example Problem cont d 2 PHZ39PN239 PM 1 atm Question a mixture of the three gases contains N2 and H2 at partial pressures of 01 atm each and NH3 Summary of AG and Reaction Conditions PH2 PN2 01 atm PM 20 atm Example C d39t39 5 at partial pressure 20 atm What is the value of Aern Kumaneuus ASantgfs kJ 0 4 7 Calculate Q G P 2 M13 3 L4x106 Equilibrium PNZPHZ 01X01Y Conditions Standard AGrXquot 0 5pm nus 8 Use Aern and Q to calculate Aern Conditions Aern Aern RTan Aernquot 34 H Aern 34 kJ 831 JK298Kln4x106 quot Mixture Aern 34 kJ 376 kJ 36 kJ 29 Hzlw has 9 Equilibrium Constant from AG N2g 3H2g gt 2 NH3Q Aomo RTnKeq 34 kJ f CHAPTER ELEVEN a aneq 34kJX1000Jk1137 C H E M I C A L l831JK i298Kl P RI N c 1 p L E s Electrochemis Keq e13 7 242 x106 try In the example We found Q 4 X 106 S l r EN 5 ZLlM lnllL Q gt Keq indicates that AG must be positive 36 kJ K llllll Illlil rl gcw sownmw o c ELECTROCHEM39STRY method to separate the oxidizing Two Broad Areas and reducing agents of a redox l l Wil c Gavanic Rechar eable Batteries Electrolysys Cells Cells AG lt 0 AG gt o MnO4 aq spontaneouS voltagecurrent source chemical Reaction drives chemical reaction Generates in nonspontaneous Voltagecurrent direCtion CHEM1 BSagermann Feb 14 2007 CHCla HO gtCHOHa HCa Rate Data for Reaction of C4H9C with Water 4 9 q 2 4 9 q C Time t s C4H9C1 M Average Rate Ms m mstamaneoug 0090 rate ail 00 01000 19 X 104 0080 Initial rate 39 16 X 10 4 3 0 000 1500 00741 4 s 1394 X 10 5 0050 2000 00671 122 gt1074 E instantaneous 3000 00549 101 X 104 g 0040 6 rateatt6005 0560 x 10 4 0020 4 A 8000 00200 mm 10000 0 100 200 300 400 500 600 700 800 9 0 T39 2 N02g gt 2 NOQ 029 Reaction Rates and Stoichiometry Rate of Reaction V m How do we write it Rate 12 A NodAt Rate 12 AlNOVAL mm I Mwill Rate 24 x 10395 Ms The molar ratios between reactants and products correspond to the rates of reaction Relative rates relationship between rates of reactant disappearance and product Rate NodAt 51er appearance at a given time 85 X10 5 W8 2 HIltggt a H2g 09 14W quot Rate43x10395Ms H A H A I r m i 2 2 At At At Practice Problem Types of Rate Laws At a given time the rate of C2H4 reaction is 023 Ms What are the rates of the other I erentlal rate law or rate law reaction components I Shows how the reaction rate changes with concentration C2H4g 3 029 gt 2 C02g 2 H20g I Must be specific in how defined htegrated rate law 023 NS 7 7 7 I Shows how concentration changes with time I Graphical determination of the order CHEM1BSagermann Feb 14 2007 Concentration amp Rate Differential Rate Law aAbBacCdD General form of rate law A B conc in M or P rate kA39quotBquot k rate constant units vary m n reaction orders Reaction orders and thus rate laws must be determined EXPERIMENTALLY Note in not necessarily a and n not necessarily b Overall order sum ofindividual orders Reaction Orders Forthe reaction A aB the rate law is rate Am a 2 NOQ 2 H2Q a N29 2 H209 rate kNO2H2 What is the order with respect to NO hat is the order with respect to H2 hat is the overall order NO is doubled What is the effect on the reaction rate Eq f H2 is halved What is the effect on the reaction rate gamma are the units of K see following page Units for the rate constant rate kNO2H2 3rd order Rate must have units of Ms molLs k mol1L1l nol l TnelLs k L2molzs Check kNOZHZ WWsXmoVI Rate molLs PtCI2NH32 H20 a PtCIHZONH32 Cl rate kPtCI2NH32 k 00090 h l malculate the rate of reaction when the concentration of PtCl2NH32 is 0020M Rate 00090 h3910020M 000018 molILhr hat is the rate of Cl production under these conditions rate APtC12NH3ZAT AC139AT 2 NOQ 2 H2Q a N2Q 2 H20Q rate kNO 2H2 k 60 x104 Mis l 1000K Calculate rate when N0 0025 M and H2 0015 M Rate 60 x 104 M 2sl0025 M20015 M Rate 0563 Ms Practice Calculate the rate when N0 0015 M and H2 0025 M CHEM1BSagermann Feb 14 2007 Determining Rate Laws Initial Rates Method Mind two experiments in which all but one reactant s concentration is constant bserve the relationship between concentration change and rate change of that reactant Repeat for other reactants NH4aq NO2aq a N2g 2 H2Ol Determine the rate law Rate kNH4 N021 Practice Problem Determine the value of the rate constant 2 NOQ C2g a 2 NOClg Exp l NOD l emu l Initial Rate M min39l 1 l 01 l l 2 l 010 l 020 l 036 3 l 020 l 020 l 145 Practice Determine rate law and rate constant Bros alt 5 BF aQ 6 HaQ 9 BE 2 H200 Practice Determine rate law and rate constant The Change of Concentration with Time aA gt products 39 A A rate kA At integrate lt 1nA kt1nA0 y mx b Plot nA vst slope k 111 AL A J kt furl1 lt AL The Change of Concentration with Time aA gt products 2 d order MA rate t A kAZ integrate lt L kt L At Ala y mx b Plot 1A vs t slope k i kt f0rAt lt AL At AL CHEM1BSagermann Feb 24 2007 Reaction Mechanisms with Elementary Steps to the balanced equation Since the rate will be determined by the slowest e ementary step this step must be consistent with the rate law No reaction intermediates can appear in the rate law Fast equilibrium steps can be used to substitute an alternate form for an intermediate The sum of the elementary steps must add up Note reaction mechanisms are theoretical and can not be proven only disprovenl Working with Reaction Mechanisms Consider a possible mechanism for the reaction 39 2 Now 2 Hue quotgt We 2 2 Rate Law 39om Experiment Rate kWNO2H2 Suggested Mechanism Step One 2NO Z NZOHE fast equil step Step Two NZOZW HM gt NZOW HZOW slow Step Three NZOW HM gt NM HZOW fast a Is this a valid mechanism for this reaction b What are the reaction intermediates c What is the molecularity of each step d Is this mechanism consistent with the rate law Is this a valid mechanism for 2 NOW 2 Hm gt N 2 H20 2NOQ Jyyegm fast equil step 7979 Him quotgtMm H20ltgJSI W gt NM HZOm fast 2Nog 2H2 9N2 2H20m I If it sums to the balanced chemical reaction it is a possible mechanism We conclude it is possible but it must lead to rate law if it is valid 9 What are the reaction intermediates 2 09 2 29 quotgt N2ltggt 2 209 2NOQ Me39m fast equil step We Him quotgtMJ Hzotg SI W a Him gt Nita Hzotg fas What appears in the mechanism that is not in the balanced equation for the reaction 39 Nzozm 39 Nzotg What is the molecularity of each step Step One 2NO 2 Nzom fast equil step Bimolecular Step Two N20 HM gt NZOQ HZOQ slow Bimolecular Step Three NZOQ Hm gt NM HZOW fast Bimolecular Is this mechanis I with the rate law I I at WNO2H r 9292 mm 11 i i 3395 30 m 2 1 Go to the rate determining step Write the rate law rate k5N202H2 Check for intermediates 8 3 N202 Express intermediates as concentrations of reactants andor products of stoichiometric reaction N202 KeqlNOIZ Express rate with concentrations of reactants and products rate j K NO2H2 i 9 CHEM1BSagermann Feb 24 2007 Catalysis Catalysis Provides a different way to perform the same reaction Catalysis Proyides a dlffererltvyay to perform tne same reaction Catalyste increasestne rate of a ieaction Witriout belrlg consumed or cnanging cnemically Catalysis Proyides a dirterent Way to perform tne same reaction Catalyst 7 increases tne rate of a reaction Witnout being consumed or cnanging cnemically Accornpll shes tnis by lowerlrlg tne acriyation energy balrler by cnangingtrie ieaction mecnanism Catalysis Proyides a dlffererltvyay to perform tne same reaction atalyste increases tne rate of a reaction Witnout being consumed or cnanging cnemically Accomplisnestnis by loweringtne actiyation energy by cnanging tne reaction mecnanism Heterogeiieotisys liomogeneousys lgiocatalysis Catalysis Proyides a dirterent Way to perform tne same reaction atalystr increases tne rate of a reaction Witnout being consumed or cnanging cnemically Accomplisnes tnis by lowerlrlg tne actiyation energy by cnanging tne reaction mecnanism Heterogeneous ys nomogeneous ys blorcatalysls Examples 7 Catalvtlccunvertel e Enzymesin llie Dudy e oznne depletion e Hetemgeneouscatalysis CHEM1 B Sagermann Feb 24 2007 Fig 1323 TwoStage Catalytic Converter Heterogeneous catalysts for an AutomObile Ceryx39s visionisto designproduce and commercialize advanced systems that balance Cost Pen ormance Emissions Reduction EXhaus man39m39d and Fuel Penalty to make the 39 economics of pollution control viable 039quot 5 i Exhaust pipe I 39 Tail Pipe We explore new ways to look at the Air compressor 7 l air quality challenges faced by Catalyzed reaCt39onS source of secondary air Catalytic converters industry and search for potential 00 02 02 solutions by combining proven 2 No N2 02 technologies with stateoftheait science Fig 1326 Substrate Binding to Enzyme Bio catalysts Substrate Binding specificity Models For Enzymatic Action Geometric Complementarity quot Silhslr alev s SUESTRATES quotW quotquot ii6 i u quot l gllitr s Mquot ranuucr Eectronic amp electrostatic Complementarity A a Induced fit vs y Lock amp Key Stereospecific enzymes and substrates are chiral CHEM1BSagermann Feb 24 2007 Chymotrypsin A Digestive Enzyme Homogeneous catalysts Ozone Layer Depleting Chemicals 0001 mb 7 Mesopause 7 001 mb chlorofluorocarbons CFCs carbon tetrachloride CCI4 methyl chloroform CH3CCI3 Mesosahere D I run I Altitude km Almude mx Suampause 1 hydrochloric acid HCI Summers 39 Chloride Tronopause methyl bromideCHsBr 400420780 80 sz 040 2080 no Rosa W O 39 39 R h CFC Reactions rlglna esearc CFC gt CI Deplete ozone 1974 Rowland and Molina 03 03 Layer In A c1 W Stratosphere c1 is F U m w F Cl39 is radiation lt CFC c1 7 F free radical CHEM1 BSagermann Feb 24 2007 Catalysis of Ozone Depletion in Stratosphere 039 09 gt 2 02 g Catalysis by Cl2 Cl2g gt 2 Clg Ultraviolet light Cg 039 gt CIOg 029 009 0g gt 029 Cg Cl39 Free Radicals free radical C10 039 09 gt 2 029 free radical Antarctic Ozone Hole Area of Antarctic Ozone Hole 210 r Antarctic ozone minimum 60 904 si Average area orozone hole 1 DUl Catalysts Lower the Activation Energy for a Chemical Reaction Movie of the Ozone Hole Uncatalyzcd pathway Antarctic season J F M A M J J A S O N D W Arcticseason JFMAMJJASOND Catalyzed pathway Enc rg Products Very low ozone Luwuzune Reaclants Rcuctiou png rcss CHEM1 B Sagermann Feb 24 2007 Heterogeneous catalytic ethylene hydrogenation CZH4 H2 CZH6 Hydrogen J W Carbmx Anive silo Heterogeneous catalysts d Q 3 Catalytic Electronic Structure of Action in Ammonia Atoms Synthesis N2 inz gt 2NH3 Chapter 12 l O 439 Atomic Structure amp Chemistry OverVieW Light Waves Photons and the Bohr Theory Wave Nature of Light Anyone Who 395 nOt ShOCked abOUt Particle Nature of Light Quantum Effects and Photons quantum theory has not understood It BohrTheory Hydrogen and Hydrogenlike atoms Nils Bohr Quantum Mechanics and Quantum Numbers Uncertainty Principle Wave Equation Quantum Numbers and Atomic Orbitals CH EM1 BSagerm ann Electromagnetic Radiation A Transverse Wave Amplitude Electric vector Magnetir mum Direction Dr propagation The Wave Nature of Light U 39 in 10 Ill 10quot m7 10 l l l l l l IU l l l l l l l i l EMkmwm u l l l l l l l l l Ki l l l ir mlquot in in in In in l i ET x mys Radio lrLquuncy l i mim l min i l l mquot 10 lt7 Frequency is 39 mi 500 60 7equot 750 um Visible light is a small portion of the electromagnetic spectrum Feb 24 2007 Wavelengthnd mpercydo 15230 00 l Wavelength MW and Frequency A 53 23 a C of Electromagnetic gtlllt V C gtNlt l cummith twin ve anmcyscydosporm A wavelength v frequency c 2997 X 108 ms The Electromagnetic Spectrum THE ELECTROMAGNETIC SPECTRUM W in mi I m m inquot w inquot H inquot mquot w W wquot m 3 Cd ryeIme M i w mm A i wavuner r 1 i mmml mquot m39 m 1ch mi l0quot ll39l39 inquot inquot in in39 i0 l0quot ioquot 2 5 4 2 2 2 2 x x x 3 393 o Nature of Sound Sound A Longitudinal Wave K Curmhiai Fault magma p39s ilcl Hearing 7 what is sound Loudness and pitCh The imam imcnsiq or a sound corresponds to in nmplnuLlc 7 High amplitude loud O Sound is a plasma wave usually in air Souudwmcsdikca a vary inmnpliludc 1 um nr lrmp The pitch nl39a mum con rsponds lo M frequency or uvelcnglh lll39rcquchM 7 High Frequency high pm ppm i Polarizations of Electromagnetic Radiation Plane Polarized X and Y Circular Polarized Right and Left CHEM1BSagerm ann Feb 24 2007 LI ht Passm Throu hCrossed Polanzers Polarization of Light by Surfaces 30m 9 him e ca Horizontal Reflectlon and Refraction M quot a 39i ifilii39l rifea i39 d I l Reflected Polarizers Angle Light 33332324 ngthave Figure 1 Action of Polarized Sunglasses Light Waves Vibrating i rl ee t 39tlz39y Ll mWaves Refracted y Light i o e m Figure 39 Wave properties of Light Refraction Wave Properties of Light Interference Young s Double Slit Experiment Red Orange Yellow Green Blue Violet Constructive and Destructive Interference of Waves Digram of Light Waves Interfering in the Double Slit Experiment incoming n quot1 2 Wave If Dark A l A A 1 x M V i 1 39 gt V i lt r tram UDDP em I i t 039 9 o oo Primary maxtmum m0 trom lower sm 7 e 0 l M l I mi 39 Dark 5 A 1 I A 1 r I A 1 CHEM1BSagermann Jan 192007 ThermOChem IStry Exothermic process is any process that gives off heat Energy Changes associated with Chemical transfersthermal energy from the system tothe surroundings or nuclear reactions and phase changes 2ii2 g 02 g gt 2H20 I energy Internal energy stored by chemical or nuclear bonds andor intermolecular interactions in solids liquids or gases Often reported under constant Endothermic process is any process in which heat has to be temperature conditions supplied to the system from the surroundings 39 q AE and AH are not zero for constant energy 2HgO s gt 2Hg I 02 9 temperature very different than physical changes in gases H20 9 gt H20 I energy energy H20 s gtH20 I Enthalpy H is used to quantify the heat flow into or out of a Thermochemical Equations system in a process that occurs at constant pressure DH H products H reactants Hi0 i ls DH negative or posi ive DH heat given off or absorbed during a reaction at constant pressure 2lul0tu 2HgtZlmIquot l AulltL l llllC lfiululhciliiir i m DH gt 0 H30 l Huzil iilwiilvcd m llk lutll System absorbs heat iium tllL39 tll39l39inllttliltgs Illhltlp Endothermic L url39g Iznc x mm the tn miintliiigx 601 kJ are absorbed for every 1 mole of ice that melts at 0 C and 1 atm 2H30 ill ZHgO I ll Hproducts lt Hreactants Hproducts gt Hreactants H20 S H20 I DH lt 0 DH gt 0 3 63 Thermochemical Equations Thermochemical Equations The stoichiometric coefficients always refer to the number Is DH negative or positive of moles of a substance H20 5 gt H20 1 DH 601 kJ CtliluH 303w 1 System gives off heat If you reverse a reaction the sign of DH changes H2C gtH2C DH 601 kJ DH lt 0 CL39l2lgt10ll If you multiply both sides ofthe equation by a factor n hen DH must change by the same factor n Exothermic 8904 kJ are released for every 1 mole of methane that is combusted at 25 C and 1 atm ZHZO S ZHZO I DH 2 X 63901 12390 kJ CH4 9 202 9 gt002 g 2H20 I 39 8904 kJ 63 63 CHEM1 BSagermann Jan 192007 Thermochemlcal Equatlons The speci c heat 5 of a substance is the amount of heat q required to raise the temperature of one gram of he The physical states of all reactants and products must be SUbStance by one degree CEISiUS Specme j m thermOChemlcal equatlons39 The heat capacity C of a substance is the amount of heat H20 gt H20 DH 601 kJ q required to raise the temperature of a given quantity m fth bt b onede reeCI39 H2 H20 DH44IO N o esu sance y g esius How much heat is evolved when 266 g of white Specific Heats of Some C ms Common Substances Phosphorus P4 bum in air speecamc Heat absorbed or released Substanze Jg 0 P4 5 502 ED 5 P4010 5 DH quot3013 kJ 1 3514 q msDt 2ny 1 m 4 x 3013 M 6470 kJ 2322 q CD1 4 139 1 E 4 Di t nal 39 tinitial Hg 0139 H10 4134 63 cszoH ethanol 245 64 How much heat is iven off when an 869 b l9 f 94cc t 50c Constant Pressure and Iron ar coo 5 ram 0 9 Constant Volume Calorimeters S of Fe 0444 Jg 0C Di t nal tinitial 5 C 94 C 89 C q msDt segyx 0444 141ny 890t 34000 J mermtmwior mum mm Thermometer A 739 Insulated vcsscl Wm Reaction mixture 397 Constant Pressure Constant VOIUme ConstantPressure Calorimetry l39llcrmomclcr A ConstantVolume Calorimetry qrxn 39 qwater qbomb qrx qwater qcal 39l39llcnnulnctcl Slirmr lgmllunuirr qwater mSDT Smmmnwps qwater msDt qbomb CbomeT qcal CoalDt 39 I i Constant V i 7 39alnnmticrltuckcl Constant P DE qrxn l1gtulacdiurkci m q i quu m DE DODV R Lll0n rm CZ 39 Mum mixlum DE m DnRT 39 DPV DnRT llmub m DE DnRT smnpit cup No heat enters or leaves 39 No heat enters or leaves CHEM1B Sagermann Jan 192007 Relationship of AH and AE Change in enthalpy AH or heat of reaction is Relationship of AH and AE C5H12m 8 0203 gt 5 cow 6 H2O mount of heat absorbed or released when a reaction occurs at constant pressure DHo 73523 kJmol The change in energy AE is the amount of heat absorbed or released when a reaction occurs at constant volume How much do the AH and AE for a reaction differ Problem Calculate the work and the value of DE for this reaction at 298 K and 1 atm W PAV AnRT The difference depends on the amount of 390nly count gases for DH solids or liquids occupy very work performed by the system or the small vomme39 DV 0 surroundings 39Moles product gases 5 mol 002g 39Moles reactant gases 8 mol 02 g Dn5mo8mo3mo Relationship of AH and AE Canm 8 o Kg gt 5 COZg 6 H20 g Table 62 Heats of Some Typical Reactions Measured at Constant Pres re M Type of Reaction Example AH k w 3 83141 298X IkJ 7433kJ Heat of neutralization HClaq NaOHaqgt NaClaq H20U 7562 My 1000 Heat of ionization moms Haq OH taq 562 Heat of fuslon DE DH DnRT 301 Heal ofvaporization 201 H20g 44m Heat of reaction Mgclzts ZNal A 2NaCs Mgs 71802 AE 3523kJ74kJ 3516kJ Heat given up by system Meaxured a 25 C A 100 C the value is 4079 H to surroundings Work done on system by surroundings Shortcuts to Enthalpies for Chemical Reactions Standard States Mosl stable foml of an element found in nature ecause there is no Way t 0 measure the absolute value of Temperature 298 K thhe enthfalpy M a SUbsttancef thSt 39tngezsure he egTa39py IPressure 100 atm most recent IuPAc standard is 1 bar c ange or every reac Ion o In eres nswer 1o1295 atquot 105 Pa The Short Cut Establish an arbitrary scale formation DHD a expressions Standard thermodynamic conditions vs STP STP 100 atm 273 with the standard enthalpy of Standard Thermodynamic Conditions 100 bar 298 K s a reference point for all enthalpy The standard en halpy of formation of any element in its most stable form is ze Standard enthalgz offormation DH is the heat change that results Whe compound is formed from 0 it ements in their standard states at a pressure of 1 atm DHf 02 0 DH 03 142 kJmol DH f C graphite 0 DH f C diamond 190 kJmol CHEM1BSagermann Jan 192007 Standard States of Elements AHOtFOPeS 0f SUlfUF AC 2k mol Element Standard State at 298 15 K and 8 rhombic 8 monoclinic LIV 3 hydrogen Hz 5 carbon graphite nitrogen NZ g oxygen 02 g uorine FZ g phosphorous White s sulfur s8 rhombic 24 chlorine C1Z g bromine Br2 D i g d xenon Xe g 32 g 5 i Table 53 Standard Enthalpies of Formation of some c 0 Substance AH klMOI Substance AH kJmol a 39 A95 0 H1020 4575 AgCllX l17 D4 Ham 0 Aim e m 0 EXAMPLE Al103l5 4669 s Hlig 2594 ErzU n Mam a HBV 36 Moors 76013 IJlfquot for NH g kJmol clgiaphite u Mgcogu r l lll CEij L mfg 423 Write the equation for formation of 100 mol of NH3 g 2 59 39 W 9 from elements in their standard states 53 0 N01 33 35 306 75355 N1OAlg 966 03 quot2 59 20 3 55 Standard states hydrogen H2g nitrogen N2 g 3le 0 09 2494 Hlg 923 02th 0 Cuts n Ogtgl l421 12 Nz 32 Hz gt NH3 3 3105 65 5rhombiC 0 r g a 5monoclinlcl nan HZFQ 725351 Sally ZQEA Dlf 463 kJmol Hgl 2E2 SOng 395 2 H19 0 H s l 4015 Him 2 1503 734738 Use fractional balancing coef cients as needed H20l 2358 Standard Enthalpy of Reaction 5 The standard enthalpy of reaction DiLOXquot is the enthalpy of Hess 5 Law a reaction carried out at 1 atm aA bB gt cC dD Hess s law of heat summation states 0H9 cDH0fC lelofD aDHOf A bDHOfB that for a chemical equation that can be written as the sum oftwo or more steps DHEXF SnDHOr Pmductsl39 SmDH reactants the enthalpy change for the overall equation is the sum of the enthalpy Hess 5 Law When reactants are converted to products Changes for the individual steps the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps Enthalpy is a state function It doesn t matter how you get there only where you start and end CHEM1BSagermann Jan 192007 Hess s Law Both trails result in the Enthalpy and Internal Energy same altitude change Changes for a Cyclic Process AH2 cannot be mpl Conclusion measured A111 5060 kJmnl IJlZ J1 directly with A151 5071 kJmul calorimetry 8C02g 9I120 I CKHM maze 298K 298 K 1 00 atm state 2 J 00 atm state I AH Lyric 0 0 Lyric step 2 AH 5060 kJmul AEZ 50 71 kJmul Calculate the standard en halpy of formation of 032 I Enthalpy Diagram gEVZnto co mg 3935 kJ Hess s law 9 p 2 g 2 g n I Srhombic 02 g gt302 g Dl39rln 392951 kJ 0321 302 g gt 002 g 2302 g DHEn 4072 N 1 Write the enthalpy of formation reaction for 032 2 Add the given rxns so that the result is the desired rxn We gm poem man 3935 kJ gtQ2 g gt 2 o2 g Dun 2961x2 kJ CQ29 2692 sh 6 9892 9 out 1072 kJ Cgraphite 23rhombic gt C32 I DHPxn 3935 2x2961 1072 863 kJ 2 mo C graphite 2 mol 02539 AH3 221l0kl H cannot be PmdllClSI 3 2 mol COg 1 mol 02g measured directly AH 7 7870 kl 39 AH 3960kl AHland AH2 can be 2 3 measured directly 2 mol C01g Enthalpy H id AH 5660 kJ Benzene CeHe burns in air to produce carbon dioxide and liquid water How much heat is released per mole of benzene combusted The standard enthalpy of formation Of of benzene is 4904 kJmol DllI PXn S nDll products S lelf0 reactants I 502g gt12002g 6HZO I DHPXn 12DH COZ6Dll fJ H20 2DllJ CeHe Question Why don t we include 15 AHTquot 02 2 DHEXn 12 3935 6 1876 24904 5946 kJ M 2973 kJmol CBHB 2 mo Graphite Diamond Buckminsterfullerene CHEM1 BSagermann March14 2007 WAVE FUNCTIONS Y i Quantum Numbers QN Application of wave mechanics or the Schrodinger Y IS a funCtlon 0f dlStance and tWO equation yielded energies for the electrons that agreed well with the experimental data angles For1 electron Y corresponds to an The Schrodinger equation yields three quantum numbers QN which define electron energies better ORBITAL the region of space than M the Bohr theory Wlthm Wthh an eleCtron IS found39 Quantum mechanics does not allow us to describe Y does NOT describe the exact the e39 in an atom as moving in an orbit but it does allow us to make statistical statements about e39 location of the electron dens39ty 2 39 Y 395 propomonal to the prObablllty We need to know these QN and how they define the orbital of nding an e at a given p0int Energy Levels of the Hydrogen Atom Orbital energies of the hydrogen atom Negative A A5 energies are Infrared the change 4 gin Infrared in the internal zeiu n im miy energy when 3 7 T an unbonded 2 4x 4p 4d 4 Warn E E bonded to a 3 396 mm X 1049 H 4 proton This i is an a E VEIZAZl x in n 3 E exothermic 39 395 5 gt m 2J 27 process 9 in 6 3918 3 swam quot2 Lu 9 r7 Note Scale Break Here C 77 lines 2I 4179 x in n i in h EnElWlEVEl diagram at the Hvdmaen atom Schrodinger and Bohr Results for O m I Q t N b r a ua U U ers Hydrogen Atom n m m 1913 1926 An atomic orbital is defined by 3 quantum numbers n m Electrons are arranged in shells and subshells of ORBITALS n gt shell Bohr Theory Wave Mechanics a subshell 1 CircularOrbits Probability Distributions 2 Nodes don t occur Various Nodes do Occur ml gt deSIQnates an orbltal Wlthln a SUbShe 3 Quantized Energy n Quantized Energy n 4 No other QN Add land In 5 Only 1 electron Any number of electrons CHEM1BSagermann March14 2007 Quantum Numbers Values of Quantum Numbers QN The Values of the Quantum Numbis are Related to Each Other Not Independent Are Connected to Each Other Svmbol Values Description a Principal QN n Defines the Principal Shells n 12 34 56 7 it gives the size and approximate energy of the orbital n principle 1 2 3 Orbital size and energy Rln2 b Angular Momentum QN I De nes the Subshells lt 0 s subshell 1 p subshell 2 d subshell 3 fsubshell I determines shape of an orbital l angular 0 l 2 Orbital shape or type subshell c Magneti n De nes number of orbitals for a subshell Inl magnetic l0l Orbital orientation in space m determines orlen a ton in space ofthe orbitals of a subshell s 1 orbital p 3 orbitals d 5 orbitals f 7 orbitals Total of orbitals in It subshell 2 l 1 sShells and Subshells For n 1 I 0 and m 0 LPM There is only one subshell and that subshell has a single orbital m has a single value gt 1 orbital This subshell is labeled 5 ess and we call this orbital 15 Each shell has 1 orbital labeled s Dot pattern of an electron with a 15 It is SPHERIC AL in Shape atomic orbital Each dot represents the position of the electron at a different instant in time Probability Density in Orbitals The symbol 15 is the label of a wave function W15 It means n 1 0m 0 Radial Distribution in Orbitals Shape of 1s Orbital Radial Distribution Functions Take Into Account Area of Thin Spherical Shell as You Move Outward from Nucleus 90 l 2 I A plot of the probability density as a function of distance m 0 so 100 m 0 for a oneelectron atom with i pm V a 15 electron A B CHEM1BSagermann Jan 12 2007 Nobel Prize in Physics 1965 What 5 energy quotIt is important to realize that in physics today we have no knowledge of what energy IS We do not have a picture that energ comes in little blobs of a de nite amount It is nott at way However there are formulas for calculating some numerical quantity and when we add it all together it gives always the same number quotThe Feynman Lectures on Physicsquot 41 There are two types of genius Ordinary geniuses do great things but they leave you room to believe that you could do the same if only you worked hard enough Then there are magicians and you can have no idea how they do it Feynman was a magician Hans Bethe Richard Feynman 1918 1988 Work Transfer Gas Heated at Constant Volume Closed System 1 e m V 2 wrm a my 5 gri 12m g a a No mass 5 w 39transfer I gt my is V V q U kw 1 Work negative r4 positive r1 7 r 551 s ij m l T increases Tm Sywm gt sV u cunswma l only work MW MWEWWW quot k 5 Only Heat No work 0 d How are heat and work related Work one Joule one Newton of Force moves a mass over a distance of one meter Or one The First Law in History ampere passes through a resistance of one Industrial revolution begins 1760 HOT for an se nd39 dt th Watt steam engine 1770 nSQZt f 2329an Lilli iii 3915 25 So Dalton Atomic Theory of Matter 1808 to 155 C Statements of first law 1850 Question How do we convert calories to joules Gas Kinetic Theory 1860 Elfw Beginnings of quantum theory 1900 quot the conversion of heat or caloric into mechanical effect work is probably impossible certainly undiscoveredquot llilliam Thomson Lord Kelvin 1848 Theory of special relativity 1905 CH EM1BSagerm ann Jan 122007 History of the First Law Julius Mayer Blood drawn from vein of sailors much brighter red in East lndiesthan Europe German Physician Dutch Ship Why Surgeon East Indies 1840 Color of blood from veins Heme Group Found Bonded to Proteins Hem c 39 c a c4 N39 2N w C l l C W gr H C C H J Frumnglubm C C C I Effect of oxygen on heme iron Deoxyhemoblobin H 020 Oxygen not bonded Fe not in porphyrin N plane Distal Q HiSE7 Plane of heme gt Conformational Changes between the T and R States of Hemoglobin Helix F Porphyrin r Fe 3 Parphyrln i 9o 2 Heme Effect of oxygen on heme iron Oxyhemoglobin ft 020 0 Oxygen bonds to Iron Plane of Heme gt Fe pulled into plane of porphyrin Distal Q HiSE7 Law of Conservation of Energy2 James Joule Heating occurs in electric circuit between battery that supplies energy and motor that does work Why English Brewer Cost of electric power from battery to drive electric motor 1841 CHEM1BSagermann Jan 12 2007 Joule Apparatus The Joule Apparatus 1845 Work done by stirring converted Temerature J Work w measured by distance 2 mass q CAT m drops when it rotates stirrer w mgz Figure 94 The piston moving a distance F calorie energy required to heat 1g of water 1 C Calorie unit of food energy 1 Cal 1kcal 1000cal Joule lkg tmzlsec2 The Joule Answer Conversion Factor 4184 J1000 cal a Initial b Final 39tatc 39 S stemSurroundin s and Universe The First Law of y g First Law The Whole Picture Thermodynamics Relating AE to Heat and Work UNIVERSE AEUNIVERSE AESYSTEM AESURROUNDINGS Energy cannot be created or destroyed SURROUN D IN GS AE surrounding 39 system Energy of system surroundings is constant Any energy transferred from a system must be transferred to the surroundings and vice versa SYSTEM system From the rst law of thermodynamics heat a he39d 1 when a system undergoes a physical or chemical change the change in internal energy is given by the heat added to or absorbed by the system plus the work done on or by the system AE q w AEUNIVERSE 0 CHEM1BSagermann Jan 12 2007 AE IS A STATE FUNCTION INDEPENDENT OF PATHWAY First Law of Thermodynamics State Functions State function a process that is determined by its initial and nal conditions ONLY A process that is not path dependant Work w and heat q are not state functions Energy change AE is a state function State Functions Changes in Internal Energy AE Pathway 1 Pathway 2 Only heating of coil q Fan drlven W plus heating of circuit q Example If 1200joues of heat are added to a g system in energy state E1 and the system does I 800 joules of work on the surroundings what is the 1 energy change forthe system AE Stirldfmiii t AB E2 E1 q v35 AB is astate H at They differ function Itis TEHET depending on J the same for Heat AE 1051 pathway for bah Pathways L w rk 1 battery 322 f AB 400 J AEsyS 400 J Discharged battery Changes In Internal Energy AE Changes in Internal Energy AE Summary 2 energy change ofthe surroundings AESW AEuniv 0 AEsys AEsulr AEsulr AEsys AE 400 J q 1200 J System w800J AE 400J Slll l Surroundings ABM 400 J CH EM1 BSagerm ann Jan 122007 Application of the First Law to Ideal Gases PV nRT P V and T are variables Gases can absorb heat or transfer heat to surroundings T tells the story Gases can do work on surroundings or have work done on them by the surroundings P and V tell the story Heat Absorption by Diatomic Molecules 0 m H 070 GHQ 00 M g Rotational Energy Translational Energy Vibr am 1 Energy Gas Heated at Constant Volume T AE T nal initial g swam gt System Pf nal Pinitial I Ideal rnonatornic gas 3 v Gas Heated at Constant Volume No work can occur because no mass is moved over a distance w 0 T and P must increase When heat is absorbed the internal energy ofthe gas must increase AE q w q Intemal energy is distributed between translations vibrations and rotation ofgas particles For a rnonatornic gas only translations are possible T and q are related by a heat capacity at constant volume CV J mol R 1247 2 K Heat Changes q from Heat Capacities and Temperature General q CA T C is heat capacity of object Molar q nCnm T 07 is heat capacity per mole Gram q mgCgm T 0g is heat capacity per gram Ideal Monatomic Gas at constant volume qVn RArL47nAT Example Gas heated at constant volume 100 liters of helium gas at STP heated to 375 K at constant V Calculate AE System surroundings 10 0 liters He Surroundings 10 0 liters He 1375 K 375 K increases 1V is positive Initial State Final State 3 J AE w C 3R1247K I W W 0 CHEM1BSagermann Mar 52006 Polarizations of Electromagnetic Radiation Polarization of Light by Surfaces Reflection and Refraction 2 Th quotfir cl rr itiF ntiarr m parallel and per 01m War in me plane inf Incidenu L Reflected Light Brevaler a 7 Kg Plane Polarized XandY Circular Polarized Right and Left 39 H L Refracted Light Ligm Fasslng WW1 Palm Wave Properties of Light Refraction Vemcal Honxonlall Incident Beam Unpalar lxe Polarizers Ve y Polarized Ligm Wave and Action of Polarized Sunglasses Sun Glasses Perpendicular to ma Highway Liam Waves Ihratin Parana to me Highway Figure 4 39 Wave Properties of Light Interference Young s Double Slit Experiment a quot andr quot quot ofWaves Constructive Interference Destructive Interference CHEM1B Sagermann Mar 5 2006 Digram of Light Waves Interfering History In the Double Sllt Experiment Blackbody Radiation m2quot ll llioml I Wave g I Dark introduced gt 1 quantum number n r In 1900 Max Planck Photoelectric Effect 1FQETALUEquot Elli In 1905 Albert pooogt Primary maxxmum mU Einstein s work b led to particle Wenquot lowe39 9039 nature of EMR and Ehv Dark In 1913 Niels Bohr gt intro uced electronic energy levels Hydrogen Atom Spectrum Blackbody Radiation All material objects emit electromagnetic radiation the distribution of photon energies and fluxes emitted depend primarily on the object39s temperature This phenomenon is known as blackbody radiation Because the amount of radiation and its spectrum depends on the temperature it is sometimes called thermal radiation or heat radiation Blackbody Radiation Wavelength and Intensity of Blackbody Wavelength Depends on Temperature Radiation Depend 0 Temperature HEIDI Shorter Wavelength Higher Intensity Intensity 7 5 all 2000 K 4000 K 6 Wavelength Wien s Law we 2898 x 10 quotm K 39quottens39ty 1000 3000 5000 A max T 1 Wavelength nm CHEM1BSagermann Mar 5 2006 HRclassification of stars The HR Diagram MassLuminosityRadii Relations The HeitzsprungRussell HR diagram is the fundamental quottoolquot for understanding stellar astrophysics The fundamental quantities plotted are log L vs log T Variants ofthe HR diagram include the colormagnitude diagram plotting absolute visual magnitude MV vs color BV Below is an example of an HR diagram The surface temperature of the sun is around 5780K The FIRclassi cation pennits the identi cation of chemical composition and size The color of a star indicates its temperature from the very cool 5 by stellar standards that is red M stars that radiate heavily in Z the infrared to the very blue 0 stars that radiate largely in the ultraviolet Color of Stars Blackbody Radiation and is Used to Determine the UltraVIolet Catastrophe their ExPeriment zThe umam ef Electromagnetic TeBIQQIQtEWE 50W Ca as wphe Theory 20000 K g 391 2898 x 10 an E max T I Rayleighr Jeans Law em 145nm Planck Radiation Formula What color is 145 nm Why do we see blue Hub HBFIEARE i 000 2 000 3000 Wavelength of radiation in rim Planck s Blackbody Radiation Law Fits Planck s Blackbody Radiation Law Experimental Data H 7 Toward the l39 iirig Planck s constant h Requires 1 V determined from best Rayleighdeans Law fit to the law to Quantlzatlon 4 m l gt experimental curve Of Energy E h6626x1034J s Levels Of 3hv 39 739 E Blackbody o u 2 h 1 g OscHlators V t g hv I Curves agree 3 ve low re uenmes A5 hV W t q l I ll Frequency CHEM1BSagermann Mar 5 2006 Summary of Quantization in Blackbodies Energy of Radiation We can also say that light energy is quantized This is used to explain the light given offby hot objects PROBLEM Calculate the energy Of 100 mol of photons of red light Max Plank theorized that energy released or absorbed by an I 700 nm n 429 x1014 sec1 atom is in the form of chunks oflight quanta E h v E h n h plank s constant 6626 x 10 34J0s X 10 34 J39S429 X 1014 sec l 285 x 103919 J per photon Energy must be in packets ofhv 2hv 3hv etc E per mol 285 x 10 19 Jph602 x 1023 phmol Planck determined h by nding the value that gave the best t 1716 kJmol ofhis blackbody radiation law to experimental result the range of energies that can break bonds Transmon Spectroscopy Physrcal baSIS Of absorption 4 Biological molecules may also 1 i absorb light and get excited to 335 mmw m higher energy levels T This energy can also be emitted g G d 4 p H l in quanta m E H Etzf ng i k quotmam Vibrational energy A Rotational energy Wan 1 H7 dKVLmtt h 1 h 1 ul Absorbing electrons are promoted 39 AE The glow Icomp exdt he m0 CC es to a higher energy level from Which a I v T e tense y Space t C energy they can sequentially release the energy H quan a to return to the ground state UltravioletVisible if S AMP Example absorption spectra Chromophore absorb speci c wavelengths Hr strde J J i Chromophore max i V l l v y mMr cmr a a 4 M 39 39 Phen alanine f I L l L y MW a H W M Tryptophan 280nm 56 g 219 470 mm m Tyrosine 274 14 to h C The absorpt10n phenomenon can be expressed 1n the wky c mme Phenylalanine 257 02 BeerLambert laW quotm m w m log Io e 1 m w Adenosine 260 149 I DNA 260 66 RNA 260 74 e Molar extinction coef cient 1 pathlength Lysozyme Bovine Serum Albumin 280 409 CH EM1 BSagermann Mar 5 2006 Planck s Blackbody Radiation Law Fits Blackbody Radiation and Experimental Data the Ultraviolet Catastrophe EQerimem Electromagnetic Planck s constant h Theory V determined from best Rayleighdeans Law m to the aw o xperimental curve h 6626 x10 Js anielloii Rammed iniensdy Frequency loan Planck s Blackbody Radiation Law summary quugn zg an in Bluckbm es inn We can also say that light energy is quantized This is used to explain the light givereoit by hot objects 1 Hugr Requires I H Max lflanh theorized that energy relea ed or absorbed by an atorn rs in the form or chunks of hght quanta Quantization W I Ehv Jo Levels of W h plank s consmnt66l6x10 J s gla frl 393de 2 W J Energy nnrst be in packets of hv 2hv 3hv ete SCI a ors l Planck determined h by iinding the value that gave the best lit 1 of his blackbody radiation law to experimental res t Energy of Radiation PROBLEM Calculate the energy of 100 mol 39 h of photons of red ig 700nm n 429X10M secl The Photelectric Effect E hon 663 x 103 Jos429 x 10 sec39l 235 x 10719 J per photon Demonstration of the particle nature of Ii ht E per mol 285 x 1039 Jph602 x 1013 phmol 171 6 Jmol the range of energies that can break bonds CHEM1BSagermann Mar 5 2006 Where Would Albert Einstein Go to Ride a Bicycle Emstem Photoelectnc Effect Sa nta Ba rba ra California 1933 A Photocell is Used to Study the Photoelectric Effect Incnming blue light collector 1 late Einstein Photoelectric Effect emitter plate Ehv Ebinding quot39 Ekinetic turr ent flows Frequency 0 quotgm Influence of Light Intensity incident on the Photoelectric Effect on phototubes u gt Highintensity light I 7 Low intensity light Electron Photo cathode Anode Intensity has no effect below Lnu urIntmmn39 light threshold frequency Energy threshold Highereintensity light 39 ejected by cathode 4 C G x a 3 no of e Number of electrons ejected r Frequency a Photoelectric Effect CH EM1BSagerm ann Mar 5 2006 Photoelectric effect Excitation of an electrode With light above a certain frequency can cause the release of electrons Photons can be viewed as particles m Evucuaicu mm Lighl 39 b 3 O P r 63 m c I v Sn I tar slimquot ill 1 V W Release of electrons 1s dependent on the light energy not intensity Wtquot Where Does Quantum Theory Come From VFquot 522xi0 mls 30533 WNW Light can eject electrons 400m from a metal but only if 0 EV 0 its frequency is above a a threshold frequency characteristic for each Photoelectric effect metal Classically for light as a wave its energy is proportional to the square of its amplitude For particles energy is proportional to frequency Einstein 1905 proposed that light has particle nature as we as ane nature light is quantized photons Millikan Determination of Planck s Constant Ekinetic hV 39 Ebinding Ermgw pielvww1ptm rrurnmdiunl metal hamp4i x 10quot5evS A AEI2SeV AU 3 x 10 Hz The im um naveitlr rmmluu F39lancl39s l l I H l39 ID A J 1 alai rnrnltillilan t ltn Quantized Energy and Photons The Photoelectric Effect The photoelectric effect provides evidence for the particle nature oflight It also provides evidence for quantization If light shines on the surface ofa metal there is a point at which electrons are ejected from the metal The electrons will only be ejected once the threshold frequency is reached Below the threshold frequency no electrons are ejected Above the threshold frequency the number ofelectrons ejected depend on the intensity ofthe light Quantized Energy and Photons The Photoelectric Effect Einstein assumed that light traveled in energy packets he called energy quanta Later G N Lewis suggested the name phOtOH and it stuck The energy ofone photon E hn This equation means that the energy ofthe photon is proportional to its frequency Light Waves or Particles CHEM1 BSagerm ann Mar 5 2006 Light Photons and Waves Epzoton h v Particle Pronen v Wave Pronertvl Light Waves lromlhe Photoelectric Effect Que ion Potassium metal must absorb radiation with a minimum frequency of 557 x 1014 Hz before it can emit electrons from its surface via the photoelectric effect If Ks is irradiated with light of wavelength 510 nm what is the maximum possible velocity of an emitted electron 1r momquot VM 7 622x10 m5 7 5V ssonm 1M 1 gain39 Hrs 4oonm g e 6 r v mums my Elavnn Photoelectric effect Electrons 39 ngm photons Blamed a f surlace quot Bright i f and Dark Areas t 4 l Sawmrnetali E Photoelectric Effect Interference Answer Step 1 Convert threshold frequency to binding energy Eb h n 663 x 1034 Js 557 x 1014 sl 369 X 10 19 J Ev h Em V2ltmv2gt Energy of photon required to eject electron from potassium surface a binding energy of electron Step 2 Determine the photon energy of 510 nm light we 663 1110 34 Js 300 x108 ms 510 110 7 m 390 X 103919 J Step 3 Determine the kinetic energy of the emitted electrons 390 x 1019 J 369 x 1019J 210 X 103920 J Step 4 Calculate the velocity of the emitted electrons EK l mv2 210 10 20J 2 2210 x 1020 J V 911 x 1031 kg E ho Ema 12 va P f 215 X 105 ms Eb potassium
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