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# Introductory Physical Chemistry CH 331

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This 200 page Class Notes was uploaded by Sienna Shields on Thursday October 15, 2015. The Class Notes belongs to CH 331 at North Carolina State University taught by Stefan Franzen in Fall. Since its upload, it has received 13 views. For similar materials see /class/223995/ch-331-north-carolina-state-university in Chemistry at North Carolina State University.

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Date Created: 10/15/15

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state Universiw w Refolding kinetics of procaspase3 C1633 memos om ml 56 mm The thermodynamic intermediates are shown obtained from the equilibrium data The protein is rapidly diluted and refolding is observed on three different exponential time scal s D C Kinetic schemes In a two state model there are no intermediates L U unfolded U ltquot F F folded Kinetic schemes one intermediate For one intermediate Li F l9 unfolded intermediate U I F F folded k1 k2 In a this case the equilibrium constants are 391 39F39 39F39 KK1Kz e U I U However the time course for reaching equilibrium is biexponential The kinetics depend on whether U or F is being observed However as a general rule the approach to equilibrium for N intermediates involves N1 exponential rate constants GFP is formed by post translational modification Then several chemical transformations occur the glycine forms a chemical bond with the serine formin a new closed ring which then spontaneously dehydrates Finally over the course of an hour or so oxygen 39om the surrounding environment attacks a bond in the tyrosine forming a new double bond and creating the uorescent chromophore Since GFP makes its own chromophore i is perfect r gene ic engineering You don t have to worry about manipulating any strange chromophores you simply engineer the cell with the genetic instructions for building the GFP protein and GFP folds up by itselfand starts to glow Green uorescent Protem GFP er structure we m Ween uamscemvmmm 5 mm m ayeuv mmmhms m m mm wmmmm mm Yheyeuv sh camms mm quote v m m haemnsmethYh Wm e v m rshvmmvreenhvmwmch 5 m we Hume mmreum ths v Sa uuans m mmerpr wunde mm uvm b m mm a r M mavv awwnhabnvmvmencmar m Wequot absamsmnma uhvm mm smth 1 menermsn 5r mm Enh EFFrsusemhnsmemmcreseamhbeau H mm rm wrathmum quotNew mus mam 5 VM mmrmmmnmm Samemckrsm m BFPmamabyeummvnu m mum m wmmnv The cychzauon reaeuon We amme rm B vquotrsrn m enauvh mmme cmquot m Ser m veneme rm camuvzled resanam vs em mm m uvm 315am ech arershe dmv nebv New acmswmchhm uthv mven mm mm mm same M wamrmakcmes msunces hfvass b e hv mven bandsvwlnmA r mmavh p svhzmqvevanev GFP as a reporter system Green Marescem pm 5 vamcma vusem as 3 mm m hwvvce sand Dwamsvs mm msmnsvrwde gummy Me humquot w W nmmmechamsnsmmfevumemenumb0V mama me 2m m mm rm us humquot we m 2kmer mm m mm zrumm Wu Mame Svhzmq 5 As m BFPrsmade umumnamrhnmwrs we a39veneexvmssmn MW 5mm uorescence quenchmg m mm mm mm mmnsc quotman Mme ma ecu es mam bvaemmnvhvm mam memmrescence mm mm by Eamsmn e rmakcmarca hsmnsmsahman 2 mers emcm reanvusmnramsmvmmmvm 3 Eenmnxmns re mew o mummy 7mm sxmnsenedm an mm mm m be a mm 5an Spectra ove ap perrmts 2222 qu encm n 9 Map W D mm mm mmquot mm EDWWWW mm M mm WWW mm 3 E s W m g Spectra ove ap perrmts quenchmg 5M mum ene Vvl nskrm Spectra ove ap perrmts quenchmg 5 I g a 4 42 E L e Em oropnore 5 a mmmwmm r We m 5mm gar ms uorescent resonant energy rams t fer FR ET nummm meme gnawwinger rm 5 arganceaepemm newquot hem um We exmed 5mg mm W mm m wmch mm 51mm m danarmakcmem an 3mm make mu m a D r mg was wmymmamr vhenamnam mam mm m ma ecmaf WWW Energy transfer mechamsm namequot M j Energy transg er mechamsm v v Danar kcemar Energy transfer mechamsm numseem P mary Condmons forFRET Danm and mm naremres mus he n dase memHW Wmmw 1cm an A rne abswmmn me mm a he aeeepmrnm ayenap maresuenue emssmn Spectrum a he danm Danny and aeeemmransmn dumb anemmmns must he appmxxmmew parauer Forster and Dexter mechanisms of energy transfer Tne Dexter mecnanisrn invuives uyeriap Elf Wavefunctiuns and is essentiaiiy identicai tEI tne transitiun rnurnent fur absurptiun Witn tne difference tnat tne initiai and finai states invuive WEI rnuiecuies D A gt DA Tne Furster mecnanism is a dipuierdip ie mecnanism tnat can bberate byeribng gistances up tb inn Ai Tnis is tne mecnanism tnat is cbmmbniy used in bibibgy tEI determine tne distance between WEI rnuiecuies S1NEYLi Haugiand m ems sai rims been The Forster Radius Tne rate unstantrur energy transrer is Kim ReR Tne distance at wnicn energy transfer is SEWa is RD as x IOZXKZIU39M39AKPJOLHW K r brientatibn ractbr 23 rbr an isbtrbbic sambie n 7 index Elf refractiun 7 guantum yieig brtne gbnbr sbectrai byeriab integrai is 11 ElFnl7c d2 m Mquot Dunur Accebtbr RD A Fiuurescein Tetrametnyirnbgamine 55 iAEDANS Fiuurescein as Wu vi Brand L Ariai Biaeneni mi 143 i 990 Fluorescent resonant energy transfer probes of lipid mixing Membranes iabeieg Witn a cbmbinatibn br riubrescence b NE E anE r tiyei e i unia i mem anes Fiuurescence rescinance energy transfer detecte as rnbgamine emi t 5 nm resuiti NED exci i n at 47b n be reases wnen a e age s ti isebaratibn cirtne brcibes isi easeg ubcin rusicin cir iabeieg membranes with unia i b Tn yerse getecti n ncn ri r n eres n gy i r ubci nance e er transr r n e n rusicin cir membranes sebarateiy iabeieg Witn gbnbr ang accebtcir prubes nas aisci brbyen tci be a userui iibig mixing assay Energy transfer strategy for the observation of membrane fusion 60 gt i x H T is net bbserveg because Energy transrerET is cibserveg E because icise D and A are tun rar abart D angA are c Example of a donoracceptor pair Nitrubenzuxadiazuiyi n ED biibsbnatigyi etnanciiamine Dunur Rnudamine bnbsbnatigyi etnanciiamine Accebtbr Excimer probes of membrane fusion Pyreneriabeied ratty acids can be bibsyntneticaiiy inccirbcirateg s tci brbguce tne n tn pyrene rexcimerriuurescence Tnis excimerriubrescence is giminisneg n at abeieg membrane tn 9 s w uniabeieg membranes Fusiun CHEW C39OH can be mcinitcireg by rbiibWing O ne increase in tne ratici cir munumer4EIEI nm t excimer 47b rim EmiSSiDn Witn excitaticin at abbut 34b nm Excwmer ermss on 02 quotMWRn muquot m quotMWRn N ms quotMWRn man 0 2 w PWquot man m ononhmawvon39an w 2mm Excwmer ermsswon as a probe of membrane mmon mm mm demises m msmn Chemistry 331 Lecture 22 Colligative Properties Freezing Point Depression Boiling Point Elevation smosis NC State University Colligative properties There are a number ofproperties ofa dilute solution that depend only on the number ofparticles and not on their kind Colligative properties include the lowering oft e vap r pressure ofa solvent and elevation ofthe boiling temperature by the addition ofa nonvolatile solute the depression ofthe freezing point of a solution by a solute and osmotic pressure hey all have a common treatment using the chemical potential ofthe pure substance compared to the chemical potential ofthe mixture In this case the mixture consists of a nonvolatile solid solute Freezmg pomt depresswn At the freezing point ofa solution solid solvent is in equilibrium with the solvent in solution As we h ve seen this equilibrium implies that the chemical potential of each phase is equal to the other HISD Tms HISDWTms where subscript 1 denotes solvent and TMS is the 39eezing point ofthe solution neural n RT In a W RT In a So that we can equate uforthe liquid and solid 15min 1m RT In a Freezing point depression Solving for a we et In a1p15 Id tumRT AmeRT AfusG AfusH 39 TAfus In a1 HimldILI WRT AmsHIRT AmSSIR For pure solvent a1 1 and In 1 AmSHRT AMSSR For an activity a2 ofsolute In a1ln1 a2 n a2 a2 AmSHRT AMSSR In the above we present the fusion temperature ofpure solvent as T and the Jsion temperature ofthe solution as T Noting that ln1 0 we can eliminate AmsSR 39om the two equations AMSSR AmSHRT Freezing point depression a2 AmSHRT AmsHRT a2 AmSHRa rr 1rr The above expression can be further approximated using 1rr 1T T Tl39l39lquot s ATT 2 a2 AmsHRT ZAT The above formula can be compared with the formula for freezing point depression ATm5 fm m is molality For a dilute solution a2 a x2 n Mym10009 kg391 for small values of m Therefore KT RT 2M110009 kg391 USH is called the freezing point depression constant Justification ln1x x m dquot 7 00 x n zl I m The derivative is dn 1 x 1 dx 1 x and we are expanding about 0 so that we will need to substitute 0 in forx in the derivative The first term in the Taylor expansion is 110x x Belling p0int elevation We can determine navaqu of beurvvater Tne pnencirnencin cit beiiing puinteievatiun can be derived in a Bumpieteiyanaiuguusfasniun in both cases of cnernicai petentiai as a tunctien cit tne temperature sciia ii is We mi Tamperaxum ixi Question i n rnatnernaticai terrn expiains Wny tne cnernicai petentiai eta sciiuticin iewereg reiative tci pure iiguig7 ZAT gt i lgt in a1 E c RT D RT in a1 mini ramperaiure mi Question i n rnatnernaticai terrn expiains Wny tne cnernicai petentiai at a seiuticin iewereg reiative tci pure iiguig7 A 2M 7 AT c RT in a1 MsaiiwRTin a1 mu Sulutimi mi i Tampuratme K Question Wnicn is an accurate EXprESSiDn cittne beiiing puinteievatiun uivent c RT39ZAWHW1HDDD9 kg 1 D um u RT in a1 suiuiimr 39 7r Tampelatum Ki Question Wnicn is an accurate EXprESSiDn cit tne beiiing peint EiEvatiDn in terrns cit tne activity at tne suivent A in a1T apHR Te tT E apHR tTe tT Activity uftne suiute c RT39ZAWHM1HEIEIEIQ kg 1 Muiaiity cit tne sciiute D usequot u RT in a1 onernicai petentiai nut beiiing puint Eievatiun setiiiian 7 Tempelatule ixi Boiling point elevation in tnis piet netice tnattne siepe increases as tne cnernicai petentiai as a tuncticin cit perature is snitteg down by tne aggiticin cit sciiute Mathematicaiiy tnis is due tci wgain th RT in a1 Because tne cnernicai petentiai cittne sciiig and vapor are net snitteg bytne aggiticin cit sciiute tne intersecticin puint i e temperature cit pnase transiticin gcies duvvn furfusiun but gees up for Vapurizatiun 5 Question What is the slope ofthe chemical potential with temperature A tree energy B enthalpy C entropy D none ofthe above Question What is the slope ofthe chemical potential with temperature A free energy 0 entropy D none ofthe above Question What is Kf for water Kf RT ZIAMSH Mi1000 g kg39i A 346 K4 mol39i kg 039 067 K4 mol39i kg D 0095 K4 mol39i kg Question What is Kfforwatei Kf RT ZIAMSH Mi1000 g kg39i A 346 K4 mol39i kg 039 067 K4 mol39i kg D 0095 K4 mol39i kg Question What concentration ofsalt must be achieved to cause a1 ElC decrease in the melting temperature ofthe ice on a road A 16 mol kg 391 B 016 mol kg 391 C 54 mol kg 391 D 054 mol kg 391 Question What concentration ofsalt must be achieved to cause a 1 ElC decrease in the melting temperature ofthe ice on a road A 16 mol kg 391 B 016 mol kg 391 C 54 mol kg 391 D 054 mol kg 391 Question Which is larger A Kf RT ZIAMSH Mi1000 g kg39l B Kb RT ZIAWpH Mi1000 g kg39l Question Which is larger A Kf RT ZIAMSH Mi1000 g kg39l Kb RT ZIAWpH Mi1000 g kg39l Osmotic pressure Osmotic pressure arises from requirement that the chemical potential of a pure liquid and its solution must be the same if they are in contact through a semipermeable membrane Osmotic pressure is particularly applied to aqueous solutions where a semipermeable membrane allows water to pass back and forth from pure waterto the solution but the solute cannot diffuse into the pure water The point here is that the solute lowers the chemical potential on the solution side ofthe membrane and therefore there will be a tendency forwaterto solution side ofthe mem rane is pressure can arise Us to an increase in the hydrostatic pressure due to a rise in a column ofsolution or due to pressure inside a closed membrane The easiest to visualize is a column of water Osmotic pressure arises from an imbalance in chemical potential when solutle is added u gtu5 I I I usuln p I I I I Pure H20 The ow of solvent leads to an increase in hydrostatic pressure Pure HZO Osmotic pressure Recall that the pressure at the bottom ofa column ofa uid is given by P p g h lfwater ows into the solution the height ofthe column ofsolution increases and the hydrostatic pressure also increases At some point the chemical potential due to the concentration difference is exactly opposed to the chemical potential due to the pressure difference We express this as MU iiiSD TyPHyai The chemical potential ofthe solution is u15D TP ufTP RT In a1 u1 TP u TP I39La1 u1 TPH RT In a1 Recall that apJaT Vm subscript for molar volume so uTP n u TP deP Osmotic pressure Thus m I V1PRTln a 0 P assuming Vm does not vary with applied pressure we can write l39IVm RT In a Since a a x for a dilute solution and In x ln1x2 w x2 we have that HVm RTx which be expressed as nv nZRT The above expression bears a surprising similarity to the ideal gas law Keep in mind however thatl39I is the osmotic pressure and n2 is the number ofmoles of solute Question What is the height of a column of water that will result 39om addition ofenough NaCl to make a 01 M solution 0025 m D 0025 m Use of osmotic pressure to determine molar mass The van t Hoffequation can be modi ed to form used for the determination ofmolar mass by osmometry H cRT 11 Here we related to the concentration c in molesliterto the concentration w in gramsliter and the molar mas M in gramsmole The experimental con guration uses the measurement of height as an estimate of the osmotic pressure The equation 1391 pgh is used h Hng Osmotic pressure Thus we can compute the osmotic pressure from n anTv or cRT where c is the molarity nzN ofthe solution This equation is called the van39t Hoffequation for osmotic pressure The osmotic pressure can be used to determine the molecular masses ofsolutes particularly solutes with large molecular masses such as polymers and proteins Question What is the height of a column of water that will result from addition ofenough NaCl to make a 01 M solution 0025m 00025 m 17 RT pgh 7 7 100 mom 8 31 JmorK298 x h 39 CRTpg 39 1000 kgm 9 8 ms2 25m Use of osmotic pressure to determine molar mass Pure H ZO Pur39eIHZO Use of osmotic pressure to determine molar mass lo unkn own 1 Pu re HZQ Use of osmotic pressure to determine molar mass A sample of15 mg ofa protein ofunknown m lar mass is added to an osmometer The solution volume is 1 mL The solution height increases by 1 cm The measurement temperature is 298 K What is the molar mass of the protein A 37900 B 39700 C 79300 D 97300 Use of osmotic pressure to determine molar mass A sample of 1 5 mg ofa protein of unknown molar mass is added to an osmometer The solution volume is 1 mL The solution height increases by1 cm The measurement temperature is 298 K What is the molar mass ofthe protein A 37900 MmmW n pgh i000 kgm 98 ms 001m B39700 379kgm0 379009 mul C 79300 D 97300 Chemistry 331 Lecture 21 Properties of Mixtures NC State University Measures of concentration There are three measures of concentration molar concentration per unit volume molarity c nIiter of solution molar concentration per unit mass molality m nkg ofsolution mole fraction 7 Z n i I X Partial molar volume e have seen molar values In a mixture or a solution individual components can have partial molar values The easiest to see physically is the partial molar volume VW 6V n J The total volume is then VVi mni V2mn2 For example when 1propanol and water are mixed the nal volume V ofthe solution is not equal to the volumes of pure 1propanol and water The mixture oftwo components that can interact in a nonideal fashion leads to a solution volume that is greater or less than that ofthe pure components The partial molar volumes allow this to be quanti ed Ideal solutions Raoult39s law states Pl xJPf where Pf is the vapor pressure of pure component j The vapor pressure of component in an ideal solution is given by the product of its mole fraction and Pf The chemical potential can be expressed as J I RTnPPJ where Pf is vapor the pressure of the pure component in the standard state Ideal solutions The signi cance of this expression is that we can now consider the equilibrium between vapor and solution to write jsulri Hvap 1n RTnpJpln but for the vapor Pf39 1 bar and so jsulri Hvap 1n RTnpJ In the limit that the vapor becomes the pure vapor we have jsulri p RTnPJPf keeping in mind the notation means the pure component Ideal solutions The equation below is central equation of binary solution p1snin p RTnPPf Using Raoult39s law xJ PJwae see that the chemical potential can be expressed as p1snin I RTnXJ This equation de nes an ideal solution Question Which statement describes ideai suiutiuns A The vapurpressure uf eprhpphehti Th ah ideai spiutTph Ts given by the prpuuetpt Tts rnuiE traetTph ahu Pf ET Pf represents the vapor pressure at pureT c The eherhTeai pptehtTai uf pureT Ts u D Aii er the apuve Question Which statement describes ideai suiutiuns A The vapor pressure at eprhpphehti Th ah ideai spiutTph Ts given by the prpuuet uf Tts rnuiE traetTph ahu Pf ET Pf represents the vapor pressure at pureT c The eherhTeai pptehtTai uf pureT Ts u D Aii urthe apuve The free energy of miXing The tree ehergyterrerrhatTerT eta seiutTeh trerh individuai eurhpurTeths Ts given by AG 6 quot G e 6 Si 2 65 quot WT thz T GT WW and 32 New WE have that AG WT 772 7 NM 7 new perm xi Wh e terah idEai seiutTph there Ts he Enthaipy prrhTXThg The vuiume u the mixture aisu dues not change for an ideai seiutTeh The Entropy ehahge eah pe DbtairiEd trerh TrT agreemehtWTth a prevmus derivatiuh Which is true a AGW VTASW p AGW TASW e AGW e ASWT d AGW Asm Which is true 3 AG eTASW p AG W TAS W e AGW e ASWJT d AGW Asm Two component phase diagrams The tutai vapor pressure over an idEai seiutTph Ts given by H P TPT39 X2F39239 m up thF39z39 P w T A pint er the tutai pressure has the farm at a straight iThe Liquid composition specific example Cunsider the Exampie in the bunk pt irprupanui and Zeprupanui whieh have P1 u a turrand P2 7 5 2 turr respectiveiy 5p inthis exarnpieithe phase diagram hasthe earanee PM i laquot The vapor mole fraction ufthe muiefractun nthe Va u eguai tei that at the iiguig hum iavvvv ri h ee ssariiy in the vapor phase the reiative ers pt rneiies is given by Daitun s iaw Appiying Daitun s e ting vi F39iF39um XiF39fF39um ur V2 PzPtmi 2P2 F39uii Because ptthe this the Vapur curve is nut the sarne as the iiguig iine Deriving the vapor curve Usingthe eguatiein at the iiguig iine we ting Pm x2 P e Fquot When substituted intei yz MpgPW we have Suivihg for PM we ting Pgeyzva epi me ng y P Pam The vapor curve The is sheiwn in the Figure peieiw The purpie iine was caicuiated using the Daitun s iaw EXprESSiDh Whatiies petweenthe piue and purpie iihEs7 This is the Wu phase The two phase region it we pick a eprnpeisitiein and pressure that is inside this Boiling in a two component system it we reduce the pressure above a Wu e ppnent mixture regiein the n use a tie iine tei ingieate the eprnpeisitipn pt irprupanui and Zaprupahuiwith a muiE tractiein at u a pit eaeh phase irprupanui what is the eprnpeisitiein ptthe vapur7 A5 45 k 0 LiQuid 4 Liquid 35 a an a 25 Vapor Vapul 20 aozdon l uezwooei now sum What is the composition of the vapor What is the composition of the vapor a 2 u ED p yz u 48 e yz u 78 d El 98 5 no name E 35 e 5 25 Vapor 20 V u 02 4 A76 D E I u 02 04 DE 03 I Boiling in a two component system What i we the system muves mm s the composition BEI ifvve cuntinue tn reduce the press e We a X s a 4n yz t etweppase regiun in th t phase regiun the p X u an yz eumpusitiun ufthe iiquid and the vapor is nutthe same e x2 d xz Boiling in a two component system if WE EDNUHUE to reduce the pressure the system reaches the buundary With pure vapur 933335 Pm i an 04 a wry What is the composition iv iv iv v What is the composition The tie line The tie iine shuvvri in Figure red line is at a trial pressure ufSDturr Yuu ean readthe arid 2valuesrreirnine pint ur calculate them using the equatiuns abuve used in generate the blue arid purple curves in the eumpusitiunrpressure pint The lever rule he tlE line can be used in definethe quantity er eaen phase presentiri thetvvu phase regiun The Dial Bumpusltlun x ean be used tugethervvith x2 and v2 41 1 i qunld X7 The lever rule The leverrule statesthatthe a h e present is inversely prupurtiurial m the length ufdistanee aieng the tie liriefrurnthe phase buuridarytuthe tutal Bumpusltlun za 4 i Liqmu 9 The lever rule The leverrule states M yrzu quotW 2 g Nonideal solutions M SD utiuns are not ideal Fur ideal sulutiunstne rule at tnterrnpteeutar tnteraettpns can be ignured Thts rnay be a S In re W F the sarne tnteraettun wtth eaeh uther thatthey h v thernseiyes in uther Words strntiar suiyents wtii turrn ideal eases tnterrnetieeui suiuttuns Huvveven tn rnany tnteraet e u e deviatiunstrum Rauult sl n idertne like tnteraettpns petw n rnpieeuies ut sarne speetes and unlike tnte tt p tw rnpieeuies ptutn ts t it the unit p are t t rnpieeuie interactions theyapprp r p y a solution wtii be smaller han we uuld eaieuiate using Rauult l w it h unlikermuleeule tnteraettpns are rnpre repulsive then yapprpressure ts greaterthantur the ideal solution Henry s law The staternent ut Henry s taw ts Pt thw it looks like Rauult s taw exeepttnatyuu haye thts tunny epnstant kwins tead at F39 T the ya or pressure at eprnppnent t Thts taw ts pnty valid tpr uttute sulutiuns t e when eprnppnent t ts the solute Underthese Bundltluns the yappr pressure at eprnppnentt reatty dues not matter because eprnppnentt ts rnpstty surrpunueu py eprnppnent 2 and set tts prpperttes really quite ditterenttrum the prppertt ut pure t H nry s law can be applied to mixtures at sptyents Anu atsp tn sututtpns utgases tn liquids Furexample see the prppterns on the epneentrattpn ut oz and NZ tn waterat the end utthe teeture Henry s law Attrattive interactions between unlike molecules leads to negative deviatluns trprn Rauult s iaw ipweryappr pressure than ideal and repuistye tnteraettpns leadtu pustttye deviatluns htgheryapprpressure than ideal As any solution apprpaehes a mule traettpn at one l apprpaehes a pure solution at one component tt peeprnes n ideal solution in uthervvurds P1 9 xny asx1 9t preyer as t the p Du up un llte molecules and the solution has the maximum deviation from n y s law 1 9X1lltwasgtlt1 9 u in thts expresstpn kw tsthe Henry s iaw epnstant Aithpugh we haye focused on eprnppnentt the sarne hpius true tpr eprnppnent Nonideal solutions A general type utexpressiun turnunrideal pehaytpr ts shpwn b l am A X P euaw Nute thatthe mule traettpn ut eprnppnentz appears tn the expo ent A lututtnistunetiunturat and p t piue compared tn Rauult s lavvblaellt ts shpwn tn the Figure beluvv Nonideal solutions Furthe example shown tn the pint appye we haye assumed that Pf tun turr Nu e at as x1 91 the sippe apprpaehes s taw preyer as x19 u and theretpre x2 9 t the sippe ts quite different trprn ideal pehaytpr We 5 n seethatasxz 91 the slup eeurnes F39fe l Note that thts ts yatue ut the Henry s taw epnstantkN1 Thts ts uepteteu tnthe Figure beluvv purple line The Henry s tawyatue can be quite ditterenttrurntne idealvalue tm Activtty The aettytty tn nunrideal spiuttpns eurrespundstu mule traction in ideal solution F a t The aettytty reptaees mule traettpn tn the expression turthe ehernteat pptenttat Wquot h t W l at h n Eunslderlng a nunrideal solution the appye expressions hate and thusthe mule traettpn xts nu lungerequaltu PF Huvveven asthe muletraetiun apprpaehes untty a pure substance the solution peeprnes tueat Thus asxy t ay xy Activity coefficient The ratio ofthe activity to the mole fraction is called the activity coef cient y v 1 1 x So as a solution becomes ideal y 9 1 as well If we examine the expressions above for the Henry39s law behavior ofa solvent we see that the exponential terms that were used are related to the activity coef cient Starting with the example from above a xexpux 0x3 we see that P m expiux m Activity coefficient The activity coef cient can be de ned in terms of the mole 39action as follows i expw 0 Although this expressions is generally true we will be m re o en concerned with the limiting cases which are implicit in these expressions These are summarized below Raoult39s law Asx91y 9 1 anda 9 X Henry39s law As X 9 0 y 9 kH P and a 9 xkH P P ro b e m Use the data 39om table below at 35 ElC determine the activity coef cient for acetone A and chloroform C The Henry39s law constants are k 175 ton39 and kH C 165 torr What is activity coef cient for chloroform C at X0 08 A 0747 B 0847 C 0934 D 1070 P ro bl e m Use the data 39om table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are k 175 torr and kH C 165 torr What is activity coef cient for chloroform C at X0 08 A 0747 B 0847 xc mp rm I rm F c219 c 0934 SC39FC39 203 07 79070747 01070 Mix cimww Problem Use the data 39om table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are k A 175ton39 and kH 165 torr What is activity coef cient for acetone A at X0 5 04 A 0001 B 0747 C 1934 l D Hes um um Problem Use the data 39om table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are k A 175 torr and kH 165 torr What is activity coef cient for acetone A at X0 5 04 P A 0681 a 0409 0409 B 0747 V x 0600 W Note XA 1 XC x mic tun I tun C 0934 347 D 1466 I I Standard states The de nitions of activity or chemical potential are only meaningful relative to a standard state There are two possible standard states These are the Raoult39s law standard state and Henry39s law standard state The Raoult39s law standard state applies when two solvents are completely miscible in all proportions The Henry39s law standard state applies when one component is sparingly soluble in the other The Raoult39s law standard state is the same as the standard state in an ideal solution as the mole 39action approaches one uj mujRTln F2 F5 The previous two problems were solved using the Raoult39s law standard state Normally we would expect this standard state to apply as X gt1 Henry s law standard state However if component is sparingly soluble Henry39s law states that PJ 9 XJkHJ as XJ 9 0 Thus the chemical potential is xk w u Win k u RTln4 Minx P The reference state for Henry39s law is k 07 ujRTln Problem Use the data from table below at 35 ElC determine the activity coef cient for acetone A and chloroform C The Henry39s law constants are kH A 175 ton39 and kH C 165 torr What is activity coef cient for acetone A at XA 02 using the Henry39s law standard state A 0892 B 0951 C 1055 D 1210 Problem Use the data from table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are kH A 175 torr and kH C 165 torr What is activity coef cient for acetone A at XA 02 using the Henry39s law standard state A 0892 B 0951 C 1055 D 1210 Problem Use the data from table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are kH A 175ton39 and kH C 165 torr What is activity coef cient for chloroform C at X 02 using the Henry39s law standard state A 0892 B 1060 C 1110 um mm D 1320 Problem Use the data from table below at 35 ElC determine the activity and activity coef cient for acetone A and chloroform C The Henry39s law constants are kH A 175 torr and kH C 165 torr What is activity coef cient for chloroform C at X 02 using the Henry39s law standard state a i 0212 A 0392 km 165 y g gg 1060 B 1060 X 39 C 1110 m m D 1320 Henry s law constant and the solubility of gases Hem aW COTS anggoa H20 Problem Divers get the bends if He 5 bubbles oszform in their blood N2 86 because they rise too rapidly co 57 Calculate the molarity of N in 02 43 water ie blood at sea level and Ar 40 100 m below sea level 02 1 6 At sea level aN2 PNzlkH m 08 atm86 X 103 atm 93 X 106 556 aN2 5 X 10 molL At 100 m aN2 PNzlkH m 98 atm86 X 103 atm 11X104 556 a cw N26X10393mo L A note on conversion from mole fraction to molarity The conversion from mole fraction to molarity can be solved analytically n1 X1 n1 n2 1 02 since n1 01 V and V cancels X40 02 Cl x102 011 x1 Note that for water as solvent component 1 X1 1 and the Concentration of water is c1 556 so that the conversion For a dilute solute such as a gas is c2 556 X2 Question Henry law 0005mm in H20 A species of sh requires a atm X 103 concentration of 02 gt 100 M 9 3 marine biologist is trying 0 of 57 determine the depth pro le for O2 o2 43 in sea water h Ar 40 calculate the concentration of 02 C02 l 5 in sea water at sea level As an assistant you perform the calculation and nd that the 02 concentration is 50 M B 430 M c 760 M D 4300 M Question Henry s law constants ll i H20 Most sh require a concentration atm X 103 of 02 that is greater than 100 M He 131 bsttt N 86 mai e ioogi is rying 0 20 57 determine the depth pro le for O2 o2 43 in seawater he ir Ar 40 calculate the concentration of 02 C02 l 5 in sea water at sea leve As an assistant you perform the calculation and nd that the 02 concentration is 50 M B 430 1LM a02 POZIKH Of 02 atm43 X 103 atm 47 X 106 0 760mm c02556 aO 25X104moL 2 D 4300 M Chemistry 331 Lecture 13 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 is open to the atmosphere the volume 0 t e 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 consumption of the gas Molar Enthalpy Enthalpy can be expressed as a molar quantity H n can also express the relationship between enthalpy and internal energy in terms of molar quanti ies H Uquot PVquot For an ideal or perfect gas this becomes H Uquot RT Usually when we write AH for a chemical or physical change we referto a molar quantity for which the units are kJmol Enthalpy for reactions involving gases If equivalents ofgas are produced or consumed in a chemical reaction the result is a change in pressurevolume work This is re ected in the enthalpy as follows AH AU PAV at const Tand P which can be rewritten for an ideal gas AHAUAnRT atconst TandP The number of moles n is the number of moles created or absorbed during the chemical reaction For example CH2CH2lt9 Hag gt CHECHag An 1 We arrive at this value from the formula An npruducts 39 reactants 1 39 2 391 The temperature dependence of the enthalpy change Based on the discussion the heat capacity from the last lecture we can write the temperature dependence ofthe enthalpy change as AH CFAT 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 dependence is contained in the above equation or r frequently in the equation below as molar quantity AH CHAT Another view of the heat capacity quot r 39 39 w noting 39 fur me heat capacity at constant volume and constant pressure can be related to the temperature dependence of U and H respectively AHCPAT AUCVAT AH AU 0 AT BTL CV AT 67 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 capacity as a constant up to this point That is a valid approximation under many circumstances but only over a limited range oftemperature In the general case the temperature dependence ofthe enthalpy can be described as 72 AHL candr our bT The parameters a b and c are given in Tables Actually this expression is readily integrated in the general case to give39 AHaT2 T T T Enthalpy of physical change A physical change is when one state ofmatter changes into another state of matter ofthe same substance The difference between physica and chemical changes is not always clear however phase transitions are obviously physical changes Fusion AHM AHW Vaporization Solid ltgt Liquid ltgt Gas Freezing AHfreeze AHcond Condensation Sublimation AHsub 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 forward and reverse processes mus e related by forward ArevelseH so that the phase changes are related by AfusH AfreezeH AvapH AcondH AsubH Avap asp Moreover it should not matter how the system is transformed from the solid phase to the gas phase The two processes of fusion melting and vaporization have the same net enthalpy as sublimat39on Addivity of Enthalpies Because the enthalpy is a state function the same magnitude must be obtained for direct conversion from solid to gas as for the indirect conversion solid to liquid and then liquid to gas AsubH AmH AvapH Of course these enthalpies must be measured at the same temperature Otherwise an ap ropriate correction would need to be applied as described in the section on the temperature dependence ofthe enthalpy So Gas Vapor Deposition AHVEWW Question Which is statement is false A AWH gt o B AEWH lt o c AMSH gt o D AWH lt 0 Question Which statement is true A AWH AMSH AvapH 3 AvapH AsubH 39 AfusH C AMSH AsubH AvapH D AvapH AsubH AMSH Chemical Change In a chemical change the identity of substances is altered during the course of a reaction One example is the hydrogenation of ethene CH2CH2g H2g CHSCH3g AH 137 kJ The negative value of AH signi es that the enthalpy ofthe system decreases by 137 kJ and if the reaction takes place at constant pressure 137 kJ ofheat is released into the surroundings when 1 mol of CH2CH2 combines with 1 mol 0sz at 25 Dc Standard Enthalpy Changes The reaction enthalpy depends on conditions eg T and P It is convenient to report and tabulate information under a standard set of conditions Corrections can be made using heat capacity for variations in the temperature Corrections can also be made or variations in the pressure When we write AH in a thermochemical equation we always mean the change in enthalpythat occurs when the reactants change into the products in their respective standard states Standard Reaction Enthalpy The standard reaction enthalpy AH isthe difference between the standard molar enthalpies ofthe reactants and products with each term weighted bythe stoichiometric coef cient A H 2 vH qmducts 2 vH oeactams The standard state is for reactants and products at 1 bar of pressure The unit of energy used is kJmo The temperature is not part ofthe standard state and it is possible to speak ofthe standard state of oxygen gas at 100 K 200 K etc It is conventional to report va ues a 298 K and unless otherwise speci ed 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 59 AH 1312kJ For ions that are in higher charge states we must consider successive ionizations to reach that charge state For example for Mg we have Mgg gt mm 69 AH 733 kJ MgggtM92g 59 AH 1451 kJ We shall show that these are additive so that the overall enthlalpy change is 2189 kJ for the reaction Mgggt M9219 259 Electron Gain Enthalpy The reverse ofionization is electron gain The corresponding enthalpy is called the electron gain enthalpy For examp e Cg e39g gtcrg Alr 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 ofprocesses that are related include molecular dissociation reactions Enthalpies of Combustion Standard enthalpies of combustion refer to the complete combination with oxygen to carbon dioxide and water or example for methane we have CH4g 2029 90029 2HZOI AEH0 890 kJ Enthalpies of combustion are commonly measured in a bomb calorimeter a constant volume device Thus AU is measured To convert from AU to AHm we need to use the relationship AHm AU AvgasRT The quantity Avgas is the change in the stoichiometric coef cients ofthe gas phase species We see in the above express that Avgas 2 Note that H20 is a liquid Ques on Fill in the missing stoichiometric coef cients for the combustion reaction OSH12Q X029 gt YCOz9 ZH20 xx New 90w Ques on Fill in the missing stoichiometric coef cients for the combustion reaction 05H129 3029 gt5Coz9 5H20 gtltgtltgtltgtlt maps 90w Ques on Determine Avgas for the reaction as written 05H12g 3029 5Coz9 5H20 A Avgas 3 B Avgas 8 Avgas 4 D Av as 3 Queshon Determine Avgas for the reaction as written 05H12g 3029 50029 5H20 A m Aum AvgasRT The quantity Avgas is the change in the stoichiometric coef cients ofthe gas phase species We see in the above express that Avgas 2 Note that H20 is a liquid Ques on What is the work term for expansion against the atmosphere 05H129 3029 gt5Coz9 5H20 A AvgasRT B Aum AvgasRT c AUm AvgasRT D Avgas Queshon What is the work term for expansion against the atmosphere 05H129 3029 50029 5H20 A AvgasRT B AU AvgasRT c AUm AvgasRT D Avgas Hess s Law c o o advantage by using sums and differences of known quantities to obtain the unknown We have already seen a simple example ofthis using the sum ofAH offusion and AH of vaporization to obtain AH of sublimation Hess s law is a formal statement ofthis 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 4 Hg e39g AH 1312 kJ 019 e39g 4 0179 AH 349 kJ Clg Hg Cl39g Hg AH 963 kJ Which statement is true about the charge transfer from H to CI to form H and Cl39 A AH 963 kJ B AH 1661 kJ C AH 1312 kJ D AH 349 kJ Question Consider the reactions Hg Hg 69 AH 1312 kJ Clg e39g 4 Cl39g AH 349 kJ What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Haq 2Cl39aq A AH ofionization and AH of electron capture B AH offormation AH of dissociation and AH of solvation C AH ofionization and AH of solvation D AH ofdissociation and AH of solvation Question Consider the reactions Hg 4v Hg 59 AH 1312 kJ 019 e39g 4 Cl39g AH 349 kJ Which statement is true about the charge transfer from H to CI to form H and Cl397 Question Consider the reactions Hg 4 Hg e39g AH 1312 kJ 019 e39g 4 0179 AH 349 kJ What further information do you need to calculate the enthalpy for the reaction H2 Cl2 2Haq 2Cl39aq A AH ofionization and AH of electron capture B AH offormation AH ofdissociation and AH of solvation C AH ofionization and AH of solvation D AH ofdissociation 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 39om the two reactions C3H6g H2g gtcaHEg AH39 124 kJ C3H8g 502ggt3C02g 4H20I AH 2220 kJ If we add these two reactions we get C3H6g H2g 502g gt3COZg 4H20I AH 2344 kJ and n wwe can subtract H2g 1202ggt H200 AH39 286 kJ to o tain39 C3H6g1 9202g gt3c02g 3H20I AH39 2o5a kJ Variation in Reaction Enthalpy with Temperature Since standard enthalpies are tabulated at 298 K we need 39 e enuu r 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 capacity difference between reactants and products c am App Z vcgproducfs Z vcgreacfanfs and then substitute this into the expression AMT AWE TAch lfthe heat capacities are all constant ofthe temperature range then A HaaZ ABE A CPAT Standard Enthalpies of Formation The standard enthalpy of formation AfH39is 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 ofinterest The enthalpy of formation ofthe elements is zero For example let s examine the formation of water H g 12 029 gt H20I AH39 236 kJ Therefore we say that AfH39HZO l 286 kJmol Although AfH for elements in their reference states is zero AfH39is not zero for formation of an element in a different a p Cs graphite gt Cs diamond Aflfl 1895 kJmol Qu estio n Consider the formation of carbon dioxide at 298 K 05 029 439 C029 How would you nd the heat of formation of oxygen A Look up AfH for Cs and subtract it from that of CO2 B Look it up the standard thermodynamic tables C The heat offormation of O2 is zero by de nition D It is equal to the standard bond energy oftwo oxygen atoms Q uestio n Consider the formation of carbon dioxide at 298 K 05 029 439 C029 How would you nd the heat offormation of oxygen A Look up AfH for Cs and subtract it from that of CO2 B Look it up the standard thermodynamic tables C The heat offormation of O2 is zero by de nition D It is equal to the standard bond energy oftwo oxygen atoms Qu estio n Consider the formation of carbon dioxide at 150 K 05 029 gt C029 How would you nd the heat of formation of 002 Apply a correction to the enthalpy 39om A Hess s law B the van der Waal s equation of state C ideal gas law D none of the above Q uestio n Consider the formation of carbon dioxide at 150 K 05 029 gt C029 How would you nd the heat offormation of 002 Apply a correction to the enthalpy from A Hess s law B the van der Waal s equation of state w C ideal gas la D none of the above Chemistry 331 Lecture 32 The Particle in a Box 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 0 AllowedRegion 39 The solution to the Schrddinger equation with boundary conditions Suppose a particle is confined to a space of length L On either sidet ere is a potential that is infinitely large The particle has zero probability of being found at the boundary or o tside the boundary OO 0 Allowed Region 39 The solution to the Schrodinger equation with boundary conditions The boundary condition is that the wave function will be zero at x kP0 Asink0 Bcosk0 0 From this condition we see that B must be zero This condition does not specify A or k The second condition is TL AsinkL 0 or kL arcsin0 From this condition we see that kL rm The conditions so far do not say anything about A Thus the solution for the bound state Is TAX AsinnnXL Note that n is a quantum number The probability interpretation The wave function is related to the probability for finding a particle in a given region of space The relationship is given by PIwwv If we integrate the square of the wave function over a given volume we find the probability that the particle is in that vo ume In order or this to be true the integral over all space must be one 1 Zdv mm If this equation holds then we say that the wave function is normalize The normalized bound state wave function For the wave function we have been considerin 9 all space is from 0 to L So the normalization constant A can be determined from the integral L 2 L 2 2 2 L 2 1 dx A sin X dx A si X dx n n L n L The solution to the integral is available on the downloadable MAPLE worksheet The solution is just L2 Thus we have 1 As you can see the socalled normalization constant has been determined The appearance ofthe Wave runcnme The appearance ofthe Wave runcnme Nuke ml We waxl rWeWW Me Me a e We WWW Mere they Wm IUD We Wquerarme a W lmem W a We WWW rumba rerWewwerWeWW Yhezvvezvm e r made a 2 geneu future af WWW with weveequWW W We ule Bvbmmd mm M WeW mtg ml We W 2 Menu qu 2 We szde szpea W 2 m wwW WWWe pWeWW wer We Wu eenaeee W We WWW We WWWWW wam WWW W We WWW W We hydraqu WWW Nu e lhzl We wam WWW 1e WWWWW W We WWWer VM WeWe ml We mleqnled wade WWW aflhexe rWeWW a eem i The propapmtv of ndrng the pamde n a gWen regron of space We We were we x1 1 Wane We wchWWew We We wchWpe 2 we me We We a We wwWWWW We eeweerezeweew WMWWW KM eWW uxmq We warned me puff 39EE Sf z wWWeeW I W 394 zinew a eze The appearance of the probab ty W p s k we Ha I e we vveuxe vvl peeWWe WWW with Dzmde We rec Wrer WWW w a 2 WmeW a meme Lel39 WW W We WeeszmWeWe W ever Wequot uWeenerWW W Va wnere re the pamde m the box we e m ewwaW we ea eW quotawemWWee We Were a We WawWW We WWW WWme WWe m Wequot Wee WW e2 wewWWwWeW we we We W erx We We a rWermWemW me em r me We We w WeWeWwe wwwmweWemr WeweWWWewWWWeWmWW buxexzw e We a De ne the ocatpn of the Dartwde kaeMeelhzlthewdlh0FMpmbzb hlvd xmbulmnm We WweW mere n m u mum weW imam arWe d xmbulmn x mx n sWeeWeWquch k We Me v e hmLGd WererWe FM 7 m Wzmmszm em WeteprWmWMeWWW de ne We WWW with Mere wWW We WWWWW even by We WWW W We szWrger quWW Ware vveuxe v WW 5 W r e We hm Wequot bvzhe Lhzenzmlvpnndv e The superpoemm of Waves m Space eads to the descnpnon of a ocanm 1 1 1 5 mm mm Re evance of the examp e Nlhauthhemndmnuxed nlheexi p uw w M m wuuymmm We mm k M mm In veufvzhe mm mm pymwm rm IE2 m mammlum x gxw wnawn gnaw mm my mm m andman 1 Wm Power tranaform re ated paws M mm m quotW by m mm n n m M mm e zw Fm mm M K n mm mm mm m m mg We mm man 2 Nu m m be m by MW mmWmmammmm W Gum mem m mumm K m 2 Gauss an Funct ms AGauxmniundmnhlehefovm 7x m Gaumzn mama H mm mm thevmm x m n mmn mm mm a mp 2 Hr WWW me We hm39mm m mm Chemistry 331 Lecture 25 Transport Properties NC State UniverSIty Kinetic theory of gases The study of transport properttes begtns the ilttnettctheory of gases e saw at the outset oftne presentatton of thermodynamics for an tdeai gas 3 nRT nM lt u2 gt 2 2 e dtscussed eariter ltu2gt ts the ayerage yeioctty squared The dtstrtbutton oryeioctttes ts gtyen b the MaxWeHrBoitzrnan reiattonshtp We assurne that the probabtitty of occupatton ofany quantum ieyei ts en by 7E KT P 0c are Where g ts the degeneracy and E ts the energy oftnat ieyei ilt ts the Boitzrnann constant k R 831JmoK 13BX10723JlK NA 6023x1023mol 39 Analogy n cornpare the equtitbnurn constant a rnacroscoptc quantity wtth The reiattye probabtitty AG AE i t pi 2 eAGORT E amp AEKT K m e R gwe Macroscopic Microscopic Boltzmann Distribution irthere are a iarge nurnberotenergy ieyeis en we can say that the probabtitty or any one oftnern betng occupted ts gtyen by g erEtKT N Z are tt EtKT The denorntnatorgtyes the totai ayerage nurnber of occupted states Thus tt ts a norrnaiization nstant Thererore the Pi yaiues wtii be between 0 and t and they are truiy probabtitttes tor a gtyen state to be occupted We can appiythts rnodei to rotattons ytbrattons and transiattons as weii as eiectrontc energy Normalization Nonnaiization means tnat the surn ofaii oftne probabiiities is t N Z P 1 it This is cieai Wnen one examines the definition oftne probabtitty rEKT gie N E KT gte To appiythts approach we can use the foiioWing energies EV Viim 72 JJ 1 ZHRz E Application to kinetic energy ses can be treated two wa s We can use the uanturn approach to deterrntne the energy ieyeis by appiytng the particierinrarbox exarnine the yeioctty or speed dtstrtbutton use the kinetic energy Since the energy ieyeis are so cioseiy spaced that they are essenttaiiy continuous we Write the surn as an integrai in thts approach the speed dtstnbutton ts dPu Fudu If Md 1 a U2 eemuzmrdu where dPu ts the probabtitty of tndtng the speed between u and u du and Fudu ts the fraction of rnoiecuies wtth speed between u and u du Gaussian Integrals Nute thattne speed distributiun is a Gaussian tunci e tartar um s present tn apprupriately aeeuunum degeneracy in three dimensiuns We have seen that a Gaussian integral can be sulyed as Ho eddx ltwe change ine limnstu zeru tn in nitythe integral is haltas large idol 39 ea 71 7t 2 a 9 WTE ltwe take ine derivative ultne integralwitn respecttu a We can sule rm even puwered pulynumials 2w ad and Ie39 dx 7 The MaxwellBoltzmann Distribution Puttingtnistugetnerwe have L uzeendnmdu govme fairy Thereturetne nurrnalized MaxwellrEultzrnann distributiun l5 uzeendznmdu dF u Fudu 271 Tne average speed ltugt and average sduaved speed ltu7gt are lt u gt f uFudu lt dz gt u Fudu WaS r t The MaxwellBoltzmann Distri 39 7 7 m M 2 rmuz l dF u e Fudu e ZMT u 9 HM Distribution Fu i Speed d Diffusion in a gradient 9 Arruvv represents the tendency furthls particle to move rEgiDn at lower concentration The flux is proportional to the concentration gradient We de ne the ux as the number of particles crossing a urf ce 39n a given time The ux 39s J umber density is N The diffusion coef cient is D Related transport properties Matter gradient diffusion J 7 DE N is number density Energy gradient thermal conductivity J 7 T is temperature Momentum gfidient viscosity J v is velocity V 77 Fick s first law Instead of the number density we can express the concentration in molarity c The number density N NAc where NA is Avagadros number Expressed as concentration Fick s law of diffusion is do J7DE The diffusion equation JzA is tne number at panicies tnat enters tne vuiume Ad per unit time Thus the time rate Elf change Elf the cuncentratiun is E JA l d1 Ad d Vuiume Ad Ha mm Ju z zg The diffusion equation JzdA is tne number at particies tnat ieave tne vuiume Ad per unit time frum tne rignt Thus veiume Ad a mm M z zg The diffusion equation Putting tegetner tne Wu eguatiens WE have The diffusion equation Tne generai ginusien eguatien is E 0 2 5 6 The seiutientetnis eguatien is a Gaussian 02 1 1 e42 The signineanee ertnis seiutien is tnattne Wiutn ufthe Distributiun gruvvs Witn time Tnis is a spreading Gaussian Units fur ginusien cuefficient are emZs distance sguaretime Diffusive spreading in generai Even ifthere is nu eeneentratien gradient partieies muve This is uften caiied Eruvvriiari mutiun The randum Metien ean aisu be reiatedtu spreading er a cuncentrated 2 3 A 5 AK AA s Ak39 a a o z Relationship between diffusion coefficient and molecular parameters in diiidsne mciticin tne diiying iciice istne ayeiage ciitne iandcim iciicestnat ccimesiicimtne ineimai kinetic energy ciitne sdncidnding mulecules The di usiun cue icient D is a measuie uftne aveiage yeicicity cit mciticin DkT wneieiistnecciemcientciiiiicticin Fciiaspneiicaimciiecdieciiiadids itne Sinkesiiicticin ccieiiicient is f 6 rmr wneie n istne yisccisity Units are pciise Neyncimseccindsmeteiz Diffusion coef cients of random coils Y ElelE maciumuleculestneie is a tendency fui cnain diffusiun rci tciaisciiciiiciwacadssiandistiipdticin Surcalled Gaussian cnain dynamics iead tci ineiciiiciWing ieiaticinsnip iciitne i trmEaanqauiE ain lt iadidsdciitnecn 2 d gt VNI N istne numbei ciimcincimei units and i istne iengtn ciieacn unit Introduction to Molecular Dynamics We ccinsidermciiecdiar dynamics as an apprciacn tci ciptain i Tnermcidynamic prciperties energy neat capacity etc 2 Dynamic intermaticin ditmsicin EDEWlElEHL dieiectric furiEtlEIriS currelated muticiri etc in cirder tci prcipagate a mulecular system using a tcirce field there are three typicai stages i Minimizatiun 2 Egdiiipraticin 3 Dynamics Force Fieldslnteraction Terms Maw Stretching Angie Bending Tcirsicin Ccidicimp Lennardrdunes The Force Field A Potential Energy Function U amingo uCoqumbr utemwd4m Stretching Angie Bending Tcirsicins 2Heie 2T1cosml Bending terms Le Cuulumb H arerEIrlES t gm it l 46 Ncineaciiiding terms The Potential Energy Surface crcimciiecdies peptides DilgDHuElEDIldES and ciiigcisaccnarides the ccintcirmaticins can inyciiye many different pcissipie states cit neariytne same energy The interccinyersicin inyciiyes tnermai mciticin Nuclear Coordinate The Potential Energy Surface Crassan frurn ene unfurrnatmn m anutner nvu ves transnevera barner fkas reatertnantne barner newgnttnen tnws Wm uncurthn n gn prubabmty n w s Esstnan the banner newgntt e systemvvm be un ned m smaHEr regmn uf unfurrnatmna spaee Nu wear nnrdmam Sedimentation Semmematmn e used m sepavate punly and ana yze pvme ne uhgunuc eumes ceHu avstvuduves andwmses We cunswdev 0151012 sedwmemauun m a gvawtauunamem we mg 7 mpg mg 1 e vie Hwe assume thaHheve s alumnna lame and sedwmemauun vebcmym we have mgm evzp e m m Whentm pamde nas veaenee the tevmmaw um y n 13ng as 351 as N can thenme accemauun dudt EI Theveluve mgm rvzp 7 fu 0 and WU V29 9 Centrifugatlon n a cemnmge the ane evauun sm x wheve u mne angu av equency And x sthe mstance mm cemev uhutauun x U mm jammy Tneeuanmyuuzxmneveweenypeyunnaeeewevauen HscaHedthe sedwmematmn cueMmem s The de nmun u s 5 WM V29 s f kwe have seen bemve memenen eeemeem m kTD Usmg the Fadtha M ynn We can Sewevenne mu av mass m myms e s D andvz M 7 RTs DU V29 ElectrophoreSIs Chavged mu ecu eswm muve m an eweemenem Tne vemm yws gNen by ZeE Wham E sme emnc em l mne pama chavge hsmemdmn cuemmem Theve umty p21 um Me ecmche d s we exemepnmene mummy U u Tne pvuceduve s named em a 32 ma evwa 1 Agavuse a pu ysacchande ubtamed 1an age 2 Pu yacvy amwde acvusshnked any andE pexyynemcn cnrcornng ne n em H cHx an an n m Chemistry 331 Lecture 10 Imperfect Gases NC State Un vers ty The molarvolume We can rewrite the ideal gas law in terms the molarvolume Vquot W The ideal gas law has the form R The molarvolume at standard T and P V 7 8 31 JmpAAXm K me F 00244m 244L icisxio Nmz The Compression Factor ohe Waytu represehtthe reiatieihship between igeai ahg reai gases is tei pieit the dEViatiDH treirh igeaiity as the gas s eprhpresseg i e ast e pressureisihereaseg The eprh ressieih taeteir is getiheg as 7 Mpiarupiume Ufgas cpmpressiph Facipr 7 WW Written in symbuis this peeprhes vm Pim View a p Nute that perteet gases are aisei eaiieg igeai gases irhperteet gases are sprhetirhes eaiieg reai gases The Compression Factor A pieit at the eprhpressieih taeteir reveais that many gases Exhibit 2 lt i teir iEIW pressure This ihgieates that attractive ditiuns tprees gprhiha e uhgerthese eeih As the pressure ihereases z ereissesi ahg eventuaiiy peeeirhes pusitivefuraii gases Thisihgieatesthatthetihite rheiiee r veiiurhe ieagstp repuisieihs between iuseiy packed gas rheiieeuies These repuisieihs are hptiheiugihgthe ideai gas rheigei Attractive Repuisive Regiph Regiph The Virial Expansion ohe Waytu represehtthe dEViatiDH pita gas treirh ideai ur perteet bEHaViDr is tei Expand the eprhpressieihtaeteirih pDWErS at the inverse m iarVDiUmE Sueh ah EXpanSiDn is WVDWH as a Viriai EXpanSiDn A C Z1VMV The euettieiehts E c ete are kriEIWri as Viriai euettieiehts Fur exampie E is the seeeihg Viriai euettieieht viriai euettieiehts gepehg uh temperature Frumthe preeegihg ephsigeratiphs We see the a lt u teir amm ia ethene methane and E gt u furhydrugen The Virial Equation of State We write 2 iri eprhpiete farm as PV rh B C RT 1 Vm V Ah theh suive furthe pressure B L P V 1 Vm V This EXprESSiDn is WVDWH as the Viriai eguatieih at state Nutet at ifE c ete are aii eguai tei zerei this isiustthe igeai gas iavv Huvveven it these are hptzerei then this eguatieih ephtaihs eprreetiphs tei ideai bEHaViDr Relating the microscopic to the macroscopic Real gases differ from ideal gases in two wa 5 First they have a real size extent The excluded volume results in a repulsion between particles and larger pressure 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 interaction We can relate the potential energy ofa 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 will consider the graphical form ofthe potential energyfunctions Hard Sphere Potential A hard sphere potential is the easiest potential to parameterize The hard sphere diameter corresponds to the interatomic spacing in a closest packed geometry such as that shown 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 ur lt cs ur 0 r gt o The Hard Sphere Equation of State As a first correction to the ideal gas law we can consider the fact that a gas has 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 the spheres of finite molar volume b ofthe spheres nRT P V nb Gas molecule volume B The Hard Sphere Model Low density These are ideal gas conditions The Hard Sphere Model Increasing density the volume is V b is the molarvolume of the spheres 10 00 O 0 O 00 0 The Hard Sphere Model Increasing density The Hard Sphere Model Increasing density The Hard Sphere Model High density At suf ciently high density the gas becomes a high density uid or a liquid The Hard Sphere Model Limiting density at this density the hard spheres have condensed into an ordered lattice They are a solid The gas cannot be compressed further l We think aboutthe densityin each ofthese CaSeSWe can see that it increases to a maximum value LennardJones Potential Function narerones otential l5 a MUST commonl used potentiai function for nonrbonding interactions in atomistic computer simuiations U 7 a u a a V 0 i EU The potentiai nas a iongrange attractive taii sine and negative Well oeptn a and a steepiv rising repuisivevvaii at R a Typicallythe parameter a is reiateo to the hard sphere diameter of the molecule For a monoatomic condensed phase UlS determined either fromthe SOild state orrrorn an estimate ortne packing in dense iiguios TheWell oeptn e is reiateo to the heat or vaporization or a rnonatornicriuio For exampie iiguio argon ooiis ati2nilt at 1 atm THUS E KTOM SEXiU JKUZU K155gtlt1E21J ThlS 6 50 corresponds to 1 US kJmol Graphical Representation LJ Potential he LJ potential function has a Stee rise When rlt for rnin 2 U The Well depth is ein units ofeneruv i 6 i i 7 A 4 g Reduced LennardJones Potential gt 2 V W i r 101 15 2 o 2 5 PM R The van derWaal 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 nRT nza P V nb 7 The signi cance ofb is the same as forthe hard sphere potential The parametera is related to the attractive force between molecules It tends to reduce the pressure compared to an ideal gas The van derWaal s Equation of State in terms of molar volume Recall that Vm Vn so that the vdW equation of state becomes RT a P Vquot b Wm We can plot this function for a variety of w for the are isotherms At suf ciently high temperature the isotherms of the vdW equation of state resemble those ofthe ideal gas Pressure xioa Nim The argon phase diagram For argon To 1503 K PE 39 87 bar 49345 Pa VE 749 cm3mol Critical Point Vaiume xio39 mJmnl Significance of the critical point Note that the vdW isotherms look very different vol mes physically reasonable in the transition region between the phases Note that the implication 39 olume is that there is a sud en c ange in v for the phase transition from liquid to gas Pressure to0 NM in L View of the liquid region of the argon phase diagram Phase Equilihr39 Region Volume xm39s in lniaii Critical Parameters The critical parameters can be derived in terms ofthe vdW a and b parameters as well as the gas constant R The derivation can use eaieuius sinee Pc L2 tne derivative ettne vdW Equatiun cit 27b state is zero at tne entieai imint 83 27Rb Given that tnis is also an intieetiun puirit V 3b e second derivative is alSEI zeru Chemistry 331 Lecture 2 Math Review The Waveparticle duality The Wave Equation Photoelectric effect NC State University Properties of exponentials 1OA1OB 10AB 10A10B 10AB Inverse function is logarithm Iog101OA A 1039091oB B Exponential to the base e eAeB eA39B Inverse function is natural logarithm lneA A eInB B Converting from one base to another 1OA ex What is x n1OA neX Take natural In of both sides Aln10 x Solve for x X 23025A Derivative rate of change a 5Y Ax 8x Slope Infinitesimal Rate of change Derivative rate of change 2500 Example function parabola Derivative rate of change 2500 20007 Slope 80 i x 40 x 15007 7 ll 3 r 1000 500 Derivative rate of change 2500 20007 Slope 40 i x 20 x 15007 7 ll 3 r 1000 5007 Derivative rate of change 2500 2000 Slope 0 r x 0 x 15007 7 II 3 t 1000 500 Derivative rate of change 2500 20007 Slope 40 x 20 x 15007 7 ll 3 r 1000 5007 Derivative rate of change Slope 80 i x 40 What is the pattern Slope 2x Derivative rate of change 2500 2000 e r Nx 1500 II 1000 r 500 e r C I I I I t t I 40 720 o 20 4o Plot the slope as the green points Derivative rate of change 100 50 2x dfxdx It is indeed a line with slope 2x Derivatives and Integrals of the logarithm Derivative dlnXdx lX Integral definition of natural logarithm 1nxf eXpX is inverse of nX The integral is the inverse of the derivative nx 150 Plot of the natural logarithm N Slope of the tangent line at 14 l l The derivative is the slope fx14 slope 4 I I I I l 10 O 2 4 6 8 lnx Lr39oo Slope of the tangent line at 12 nx Lr39oowA nx Slope of the tangent line at 1 l l 4 2 0 I 2 x1 slope 1 4 I I I I O 2 4 6 8 10 nx Slope of the tangent line at 2 Lr39oowA Plot of the slopes of the tangents l l l l 39 u Plot of slope vs x nx 4k N O N 4 nx leomA Derivative dnxdx 1x kg gx nx 39 O 2 4 6 8 10 nx Integral of daa the area under the curve l 1n21262 a 9X nX If fa 1a I 0 1 2 3 4 X Ibixo xmoo nx gLoANw Integral of daa the negative of the area under the curve 1 I I 12 1 1n12J 2 21 1 12 9X nX If fa 1a o 1 2 3 4 x 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 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 e 2 m v2 f 4 7c 8 0 r 2 r Coulomb Centrifugal Balance of forces Assume electron travels in a radius r There must be an integral number of wavelengths in the circumference 2nr rm n 123 The electron must not interfere with itself The condition for a stable orbit is 2m rm n123 The Bohr orbital shown has n 1 The DeBroine wavelength A hp or A hmv gives mvr nh27c n123 This is a condition for quantizatio of angular momentum t W 15 F a MM 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 of the radius r 47ceon2h2 o395 393 0537 me2 415 I 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 7 hp where the wavelength of light is L 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 e e E T V ZmV 4780r 8730r Substituting in for r gives 4 En quot782 88 7 The WaveParticle Duality The fact that the DeBroine 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 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 Metal Surface T Kinetic Energy T f g a E i V 5 hy q i Insuf cient energy Photoej ection occurs for photoej ection 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 Derivation of the Schrodinger Equation The Schrodinger equation is a wave equation Just as you might imagine the solution of such an equation in free space is a wave Mathematically we can express a wave as a sine or cosine function These functions are oscillating functions We will derive the wave equation in free space starting with one of its solutions sinx Before we begin it is important to realize that bound states may provide different solutions of the wave equation than those we find for free space Bound states include rotational and vibrational states as well as atomic wave functions These are important cases that will be treated once we have fundamental understanding of the origin of the wave equation or Schrodinger equation The derivative The derivative of a function is the instantaneous rate of change The derivative of a function is the slope We can demonstrate the derivative graphically We consider the function fX sinX shown below The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinO the slope is 1 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinn4 the slope is 12 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinnZ the slope is 0 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin3n4 the slope is 12 as shown by the blue line The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin3n4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sinn the slope is 1 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin5n4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope At sin5n4 the slope is 12 as shown by the blue line The slopes of all lines thus far are plotted as black squares The derivative of sinx The derivative of a function is the instantaneous rate of change The derivative of a function is the slope We see from of the black squares slopes that the derivative of sinx is cosX The derivative of sinx l sinx cosx dx 33 I 3 W 5 4 2 0 2 4 6 The derivative of cosx d d X cosx smx 10 39 05 1 0 The second derivative of sinx sinx sinx The second derivative of sinx sinx sinx 10 39 05 1 0 w w Sinx is an eigenfunction 2 If we define E as an operator G then we have X d2 39 39 sm X sm X dX2 which can be written as G Sinx Sinx This is a simple example of an operator equation that is closely related to the Schrodinger equation Sinkx is also an eigenfunction We can make the problem more general by including a constant k This constant is called a wavevector It determines the period of the sin function Now we must take the derivative of the sin function and also the function kx inside the parentheses chain rule i Sinkx kcoskX dX 2 i Sinkx kzsinkx dX2 Here we call the value k2 the eigenvalue Sinkx is an eigenfunction of the Schrodinger equation The example we are using here can easily be expressed as the Schrodinger equation for wave in space We only have to add a constant 2 2 2 2 M Sinkx Sinkx 2m dX2 2m In this equation h is Planck s constant divided by 27 and m is the mass of the particle that is traveling through space The eigenfunction is still Sinkx but the eigenvalue in this equation is actually the energy The Schrodinger equation Based on these considerations we can write a compact form for the Schrodinger equation HLP ELF hZ d2 Energy operator Hamiltonian 2m dX2 hzk2 E Energy eigenvalue Energy 2m LP sin Wavefunction The momentum The momentum is related to the kinetic energy Classically The kinetic energy is 1 E mv2 The momentum is p mv So the classical relationship is p2 E 2m If we compare this to the quantum mechanical energy hzk2 we see that p hk 2m The general solution to the Schrodinger equation in free space The preceding considerations are true in free space Since a cosine function has the same form as a sine function but is shifted in phase the general solution is a linear combination of cosine and sine functions LP ASinkX BCOSkX Wavefunction The coefficients A and B are arbitrary in free space However if the wave equation is solved in the presence of a potential then there will be boundary conditions Kinetics Experimental techniques 7 Definition of Rate First Order Processes Exponential Decay Experimental Techniqf 0 Monitor reaction progress using pressure conductivity spectrophotometry etc in real time o Quench reaction after a given time by rapid cooling or solvent trapping o Initiate process by flow stoppedflow rapid mixing or flash photolysis 77me scales for eXperImenta measurement of reaction kIz 39 Spectroscopy is a key tool for monitorin reaction progress on rapid time scales 0 Flash photolysis is a convenient method for initiating a timedependent process with real time acquisition as rapid as ten femtoseconds 0 Rapid migtlting flowed reactions can be monitored at xed distances with 100 us time resolution 0 Stopped ow has a 1 ms mixing time o Quenching method suitable for slow reactions De nition of r For any given reaction e g A B Rate of consumption of A is vA dAdt Rate of formation of C is vc dCdt Sign convention reactants are consumed and therefore dReactantdt lt 0 o In general for any species vj1njdJdt where n is the stoichiometric coefficient Rates ofappeara For a chemical reaction dJdt HJVJWh J is the stoichiometric coefficient 0 Example A ZB 3C gt D 2E o If dAdt v then the rate of disappearance is dBdt 2v dCdt 3v and the rate of appearance is dDdt v and dEdt 2v Rates of appeara o For a chemical reaction dJdt nJVJWhe J is the stoichiometric coef cient o EgtltampeAZB3CaD2E o va dAdt 2v dBdt 3v dCdt then v dDdt and 2v dEdt using the correct sign convention The rate is proportional to concentratio 5 Example The rate law is v kAB where each reactant is raised to the first power The coefficient k is called the rate constant The rate law can be determined by the isolation method The reaction is run in an excess of all but one reactant to determine the dependence on concentration Units of the rate const The units of rate constant are always s c that they convert into a rate expressed as concentration divided by time Rate k A B mol L391 s391 L mol391 s391mol L391mol L391 Rate k A2 B mol L391 s391 L2 mol392 s391mol2 L392mol L391 Question What are the units of k in vA kA quot A mol L391 s391 B L mol391 S391 C mol L391 D s391 Question 7 What are the units of k in vA kA 39 A mol L391 s391 B L mol391 S391 C mol L391 D s391 Historder Kine 39 o A first order rate process dAdt kA dA di A MM 2 HM k d A A x M At ln 40 kl Firstorder Kine 39 o A first order rate process dAdt kA Inc110 A10 A 1 1410 o The rst order rate constant is give Historder Kine 39 dAdt kA o The solution is A A loe em 0 Known as single exponential kinetics Halflife kcm nAD2A D n12 is the de nition ofa halflife 112 ln2k Halflife and rate cons t The halflife for 23EU is 45 x 109 years What is the rate constant for radioactive decay of 23EU A 15 x 10m year B 75 x 10 3 ymr 1 C 225 x 109 year D 15 x 10399 year1 Hal The halflife What is the rate constant for radioactive fe and rate cons for 23EU is 45 x 109 years decay of 23EU A 15 x 10m year B 75 x 10399 year C 225 x 109 year D 15 x 10399 year 4 5 Time Ari exponential process rs typical of rst order kineucs Decay of radioacuve nuclei rs one texlbook example Exponential knetji o one yraxls These are two W 3 IO S a eXpHZ 3 5 expat E 04 exprm e g 02 e o r r r r U 1 4 G E 1 ne We can plot popu anon or concenuauon i av of saying the same mrng Exponential knetii Q 1 n e s 05 eXpHZ k 12 E 05 eXpH 047 aner k 2 e e 5 027 e U 2 4 6 B Time Ari exponenual process nas a le urne mat corresponds to z 1k Where k is line rate constant Exponential kineti pager Concentration At 4 e Eemnds We focus on dne exponenua process WNW a rate constant dk1w Exponential kineti Q m t t t t 7 08 s d 7d 7 g as Slope7 e 7 ke a 04 7 E 02 8 e O t t t 1 U 2 4 6 B 1 39 me Tne mma rate 1 tne s ope at Dme Zero TH 15 gwen by dne bme hne m dne gure It s obtamed from dne denyauye Exponential kineti 9 L0 7 2 on d n E 05 HERE stape71 o E 0V4 rune lmalUx g 027 7 E o t t t 1 U z 4 5 a 1 me Tne mma rate 1 dne s ope at Dme zero Ms S gwen by dne bme hne m dne gure It s obtamed notn dne denyauye Exponential kineti Q 0 dl7e 397c E 08 5101767 ke e 7 E as HERE stope71eez7182 o E 047 rareo3o7motLc b 5 m7 7 E o t t 1 U 2 s S i w At t 7 1 second dne sobe s agam gwen by dne denyauye at 1 second Now dne rate 1 SmaHer m Q o 2 1 a m 1 9 1o 7 E as 7 g 5 p 71ee27182 E up rate 0367malLx 5 027 7 E o t t 1 U z a 1 4 Tune At t 7 1 second dne s ope s agam g ven by dne denyauye at 1 second Now dne rate 1 SmaHer k 1 per second 9 10 E oz 7 7 g 05 HERE Slope 1ee27182 E M rare 0367malLx 5 n27 7 E o t t 1 U z a A s Tyne At t 7 1 second dne sobe s agam g ven by dne denyauye at 1 second Now dne rate 1 SmaHer Exponential kneti E g 75 HERE stape31e1ez7182 54 rate 0135malLx 2 5 n2 e E o r r r U 2 s a 1 4 Time At t 2 seconds the siope is again given by me derwauve at 2 seconds The conunues to decrease as A is used up Exponential kneti Q quot0 419323 goa Slap k9 a2 e g 05 HERE stape31e2ez71823 Lu VUIGZOBSMOIUx g n2 RATECONSTANT15quot e E o r U 2 6 i 4 s Time At t 2 seconds the siope is again given by me dewauve at 2 seconds The conunues to decrease as A is used up Quesnon A first order process occurs with a rate constant f 01 39 e in39 what is the concentration after 30 seconds 91 n e 3 n m 3 3 a 5 3 a II B 005 M C 0005 M D 00005 M Question A first order process occurs with a rate constant of 039 m i 91 n o x n m x q a 5 x 2 what is the concentration after 30 seconds B 005 M C 0005 M D 00005 M K Equiiibnum isapproached as the sumo the forward and reverse rate constans US Approach to equiIllgr m Let X be the deviation from equiiibrium X A Aq 5 then A Aeq X and B 5 X Rate equauons are 39dWdt MA 39 MES kAq X 39 NOEL X d Therefore rdXdt k kgtlt X x expi 39 09 MN 39 Mwdt MAL 39 Kim 0 reverse rate constants L kil 0 From before B km 161 k4 K Question A protein folds with a rate constant of 100 s391 Alsl melt temperaturequot the free energy of the folding transition is AGO 0 at is the obsened rate constant for folding at this temperature B 200 s391 C 50 s391 D 0 s391 Question A protein folds with a rate constant of 100 s391 At39t melt temperatu requot the free energy of the folding transition is AGO 0 What is the obsened rate constant for folding at this temperature A 100 s391 B 200 s391 C 50 s391 D 0 s391 Summay of rstorder proc ses 1 Firstorder procsses have an exponential time course 2 The rate constant k can be related in a 1e time I ora halftime 112 k 11 k ln2 112 3 The units of the rate constant are s391 4 The approach to equilibrium is the sum of forward and reverse rate constants 5 The rate v is the instantaneous change slope 5 The units for the rate are moleslitersec Ms39l Chemistry 331 Lecture 4 Introduction to Spectroscopy Quantum Mechanics Background Electronic states and energies Transitions between states ection Absorption and Emission NC State University Postulates of quantum mechanics are assumptions found to be consistent with observation The first postulate states that the state of a system can be represented by awavefunction KPqr q2 3H t The q are coordinates of the particles in the system and t is time The wavefunction can also be time independent or stationary wqr q2 qgn Postulate 2 The probability offinding a particle in a region of space is given by Pa r V Pdr Postulate 2 Assumptions 1 KP is real 6 is Hermitian 2 The wavefunction is normalized 3 We integrate over all relevant space Normalization is needed so that probabilities are meaningful Normalization means that the integral of the square of thewavefunction probability density over all space is equal to one I V Pdt 1 all We The significance of this equation is that the probability of finding the particle somewhere in the universe is one Postulate 3 Every physical observable is associated with a linear Hermitian operator Observables are energy momentum position dipole moment etc operator P gt observable P The fact that the operator is Hermitian ensures that the observable will be real Postulate 4 The average value of a physical property can be calculated by Normalization Postulate 4 The calculation of a physical observable can be written as an eigenvalue equation Postulate 5 The form of the operators Position 6 q Momentum 13 17322 P P aq This is an operator equation that returns the I I same wavefunction multiplied by the constant P A a P is an eigenvalue An eigenvalue is a number Ener H Stationary State Wave Equation Quantum Mechanical Description Hamiltonian Eigenvalue Energy Operator Energy value I I HO P EO P Wavefunction The Hamiltonian and wavefunction are timeindependent The wavefunction is composed of electronic and nuclear parts LI l l electronicx nuclear Al Al Al Total Electronic Nuclear The wavefunction represents the probability amplitude of electrons and nuclei The wave equation can be separated into electronic and nuclear parts Hamiltonian Eigenvalue Energy perat0r Eriergy value I V Helecw Eelecw A H mafia E nuchE Wavefu nctions The interaction of electromagnetic radiation with a transition moment The electromagnetic wave has an angular momentum of 1 Therefore an atom or molecule must have a change of 1 in its orbital angular momentum to conserve this quantity This can be seen for hydrogen atom 4 Electric vector of radi ion 1 1 V 10 Spectroscopy Electromagnetic radiation Dipole moment Transition moment Selection rules Experimental techniques Intensities The ongln of spectral llnes ll l molecular spectroscopy ls the ernlsslon oraosorptlon of a photon when the energy of a molecule changes Characteristics of electromagnetic radiation Electromagnetic radiation can be described as a wave with an oscillating electric field Light can be linearly polarized along the x y orz axes The electric eld vector is L to the direction of propagation F electric eld i E magnetic eld i E EOCOS21wt B Basic Phenomena Emission spectroscopy a molecule undergoes a transition from a state ofhigh ergy E to a state of lower energy E2 and emits the excess energy as a p oton Absorption spectroscopy the energy of an incident photon drives a polarization 39om the molecular ground state to the excited state An absorbed photon has a 39equency given by the Bohr relation hv E1 E2 Definition of the Dipole Moment The dipole moment operator is H 62 21 where ze is the electronic charge at a nucleus and r is avector from an arbitrary origin Along the xdirection u elm I x l dx n Examples of ground state dipole moments The ground state dipole moment of hydrogen halides can be calculated from the fractional charges 5 8 1 Debye 333x1t3D Cm 0 639 Rpm MX 103 Cm HF 042 042 917 637 Rs 19 D HCI 016 016 1275 344m10 D HBr 011 011 1414 264m08D HI 005 005 1609 140m04 D Definition of the transition dipole moment The transition dipole moment results from the interaction of electromagnetic radiation with the molecule where E E cos2nvt and the hamiltonian for interaction is H a uxEmcosant where pxis the transition dipole moment My cf 39Px391 edx Selection Rules A transition will be allowed only if the transition dipole moment integral Sources of Radiation I Nerst lament infrared is nonzero The general rules are Arc lamp Xe UVVis I Tungstenhalogen visible nearIR Electronic Al r 1 Am 0 I Lasers Rotational A r 1 AM 0 Excimer Vibrational Av r l Ion YAG For mathematical description see the T Lsapphire workshop on selection rules Dis ersion is essential p Detectors A dispersing element separates different frequencies into different spatial directions A prism separates different frequencies because of the optical beam using the variation of the index of refraction such that high frequency radiation undergoes a greater de ection than low frequency radiation A diffraction grating consists of grooves cut ca m apart Interference from re ected waves gives rise to specific angles of propagation Thermistor bolometer far ir Mercury Cadmium Telluride MCT ir Germanium near ir Silicon visible Photomultiplier tube Ampli cation by dynodes genera es current for each photon hit CCD detector Anay of detectors on a chip UV coatings ef ciency ease of use Absorption Spectrometer Dispersing element is in the spectrograph Sample l J x Comparison of sample and reference is essential Light Source Intensities of Spectral Lines I Intensity of absorption for a sample that has thickness d is given by logL0 scd I sis the molarabsorption coef cient I c is the concentration I d is the pathlength I I is the transmitted intensity I0 is the incident intensity Beeriznherl Law mm r rsrmiai c in mi iiaimmi rmiewie is r it Theory Absorption and Emission siwimrwiii in ii iii M by aminoquot n vhman Stimuizled mm mm iaw am nitmisiai n vhman ai summed mm 7 A spammismsoi spamming WSW is uorescence swimsWiggimisustmis 5 sponta ous 5 Wm WM N282 Elnsheln39s derivation of absorption and emission coemcients yi mmmisim riimhevirime mason mi demandsthm a W rigiiiai equiiioii WW ie Ni ii Azi 2 we aiminicnimmsnioigi ip warming smiianinipiaicmisiinin a in 1 N l N aiz Mi blackbody distribution rigminimiiimiimimisiii ism sis iii am rim disibmiaipimeo i aw Hzi i c in rim disiibmiai is J Interpretation of Einstein Coef cients The intrinsic coefficient for absorption B12 is related to k1Z NBnp wherep is the energy density I of pvdv Einstein showed that the rate of absorption and stimulated emission are equal The spontaneous emission rate has a de nite relation to the stimulated emission ra e 321 BIZ 3 xpanmrieaus Sithv A21 7 C3 321 The transition probability is proportional to the square ofthe transition momen The absorption of radiation involves time dependence because the eld is timevarying Easin21rvt Ifwe solve for the trans39tion probability we nd that it is proportional to the square of transition moment 3 WM Connection with experiment Beer s law states that v cd 01 0 4 l 2v is the molar extinction coef cient c is the concentrat39on d is the pathlength ln differential form this is written dl 2303evcl dx A comparable expression in terms of the individual transition rates is given by dl thk12 dX Experimental determination of the transition moment by absorption spectroscopy Nh 23038VC thkm willch Lv ax quotNIH2 2 6909mm 1 n A W FC Lvdv The transition moment is related to the integrated extinction coef cient Chemistry 331 Lecture 19 Applications NC State UnIverSIty Thermodynamics of glycolysls ReactTorT kJrnot Dngtuc TP eDegTucoseesephosphate ADP Ac 46 7 DrgTucoser rphosphate sDermctosesephosphate Ac T 7 Dntructosen sedTphosphate ATP gt DTructoseT BndTphosphate ADP AGquot 44 2 DafructoserT BdTphosphate gt gTyceraTdehydeearphosphate y roxyacetohe phospha A dThydroxyacetorTe phosphate gt gTyceraTdehydesephosphate Ac gTyceraTdehydeearphosphate phospha e NAD gt T 9 I 3dTp os ogTycerate Ac T SVdTphosphogTycerate ADP gt Z rphosphogTycerate ATP Ac Z rphosphogTycerate gt erhosphogTycerate Ac erhosphogTycerate e2ephosph enoprmyate H20 Ac 27 phosphoenoprmyate ADP gt pyruvate ATP Ac pyruvate NADH h gtTactate N D pyruvate gt acetaTdehyde co2 acetaTdehyde NADH h gt ethanoT NAD Phosphorylation of glucose Dngtucose ATP sDegTdcosesephosphate ADP AGquot 716 7 The reactTon can be decomposed Ttho two reactTorTs DrgTucose phosphate gt DrgTucoser rphosphate H20 Ac T4 3 ATP H20 aADP phosphate AGquot 731 0 change under standard conthTons Tn thTs mannerthe strongTy TT39T the ceH Note that the vaTues forAGquot assume concentratToT39Ts 0M M Ctearty the change The role ofenzymes ATT of the reactTons Tn the gTycoTytTc pathway are cataTyzed by enzymes For exampTe the reactTon consTdered on the preyTods sTTde Ts cataTyzed oy hexokTrTase The roTe ofthe enzyme Ts to speed up the reactTon but the enzyme does not change thennodynamTcs ofthe process The roTe of enz mes Tst hat of any cataTyst CataTysts affectthe hnetTcs ofthe reactTon but notthe thennodynamTcs We WTH consTder the roTe ofcataTysts Tnthe second hanofthe course NotTce that Ac for certaTn steps Ts posTtTye PorexampTe DagTucoser rphosphate gt Drfructoser rphosphate Ts cataTyzed oy phosphogTucose Tsomerase AGquot 1 7 The engTTondm constant forthTs process Ts Tlt expfAGDRT epo7008 3T3T0T 0 5 Under standard conthTons there WTH be onTy a smaTT concentratTon of Drfmctoser rphosphate Question cTyen that Tlt 0 5 forthe reactTon DagTucoser rphosphate eDtructosesephosphate Catcutate the concentratTon of anructosen nphosphate at equTTTbrTurn under standard coerTtTorTs ATM 805M C0333M DOT5M Question cTyen that Tlt 0 5 forthe reactTon DagTucoser rphosphate gt Drfructoser rphosphate Catcutate the concentratTon of anmctosen nphosphate at equTTTbrTurn under standard coerTtTorTs ATM K 7 KTrxx KTKx 805M 0333 comm DOT5M Intracellular conditions are not equilibrium conditions the subsequent step in a series or reactions is highly spontaneous this Will tend to deplete the product forthe previous reaction Thus ore orthe product Will tend to be formed by t cnateliers principle can observe this quantitatively by considering the value ofO the reaction quotient since AG Tin o the value of AG rnay not be zero in otherwords the coupled series e cell are not at equilibrium but ratherthey proceed under s eady state conditions Where the concentrations are not at i M but are poised so that the overall or effect on a series or reactions is to produce a net spontaneous change 3 a m g a 3 3 339 Sample Problem in Metabolism e zyrne aldolase catalyzes the conversion orrructose the concentrations orthese species in red blood cells erythrocytes are FDP 35 uM DHAP 30 pm and GAP i5 pm VWi the conversion occur spontaneously underthese conditions Solution The standard tree energy change forthe reaction is FDP gtDHAP GAP AGquot 238 w nd DHAPGAPFDP 30 x loel5 X may 35 x i0 6 5 57 x i0 5 AG AGquot RT in O 23800 Jrnol 8 Si JrnOMXSiO Kln5 57X105 4434 Jrnol oral 43 kJrnol The reaction Will occur Spontaneously undertne conditions ortne cell Thermodynamics of D NA hybridization A combination of spectroscopy and calorimetry was used to determine the ea energies of melting of short oligo nucleotides Based on these measurements the ea energy ofa helix can be determined based on10 sets of nearestneighbor pairs shown on the next slide In addition to these values we need to know the free energy ofthe initiation ie the rst base pair The overall free energy Is then calculated 39om AG AG initiation 2 AG nearest neighbors Sample problem Determine the melt temperature for the oligonucleotide 5 ATAGCA 3 5 ATAGCA 3 3 TATCGT5 3 TATCGT5 SOlutiOh Ac AG initiation 2 AG neare5t neighbors A6160 init A61 D Ae AefT g AGigg Aeig e 3 8 6 7 209 763 7130 769 Note that We use CC initiation irtnei39e is a Single CC base pair OnlyuSeAT initiation ortne Strands are all AandT Sample problem Determine the melt temperature for the oligonucleotide 5 ATAGCA 3 5 ATAGCA 3 3 TATCGT5 3 TATCGT5 Solution contd AG 46 8 VJ Notice that the the tree energyotinitiation is positive initiation is unfavorable because orthe en ropy tha rnust be overcorne to bring the chains together TO calculate the melt temperature We heed the enthalpy or reaction as Well A TA AG so CA AH AH tTZ A AT AH1TC AH CG AH GT 7360 7251 7326 7464 7243 hi 64 4 w Sample problem Given that AG 168 kJmol and AH 1644 kJmol for the hexamer determine the melt temperature A 42 C B 48 C C 52 C D 58 C Sample problem Given that A D 168 kJrnol and D 1644 kJmol for the hexamer determine the melt temperature 42 ElC A B48 ElC C 52 DC As Anus AleT 464 Mb 8298 x1000 D39 58 DC 4 rnoilt 95 3 J The melt temperature occurs When AGquot 0 T NlaASquot 464AOO495 3 331K 58 C Henry s law constant and the solubility of gases Henry s law constants ll l H20 102 Problem Divers get the bends if 9 1 bubbles of N2 form in their blood 2 85 because they rise too rapidly O 57 Calculate the molarity of N2 in 2 43 water ie blood at sea level and 0 m below sea level At sea level aN2 PNZkH NZ 08 atm86 x103 atm 93 x105 5 aN2 5x10 moVL At 100 m aN2 PNZkH NZ 98 atm86x103 atm11x10 556 aN2 6x103 moVL O uestIo n Henry s law constants ll i HZO A species of sh requires a atm X 103 concentration of O2 gt100 uM He 131 A b I N 86 marIne Io oglst Is tryIng to of 57 determine the depth pro le for O 2 02 43 in sea water The rst step is to N 40 calculate the concentration of O2 4 in sea water at sea level As an assistant you perform the calculation and nd that the 02 concentration is A 25 M D 4300 Mill Question Henry s law constants ll l H20 trnx 102 131 Most sh require a concentration a ofO2 that is greater than 100 M e 86 A marine biologist is trying to f 57 determine the depth pro le for O2 2 43 in sea water The rst step is to 40 ca lculate the concentration of 02 In sea water at sea level As an assistant you perform the calculation and nd that the n is 02 concentratio A 250 5430 FM a02 Pozlku oz 02 atm43 x 103 atm 47 x105 c 760 NM c02 556 a02 25x10 molL D 4300 M Use of osmotic pressure to determine molar mass The van t Hoffequation can be modi ed to form used for the determination of molar mass by osmometry ML H CRT H MRT Here we related to the concentration c in molesliter to the concentration w in gramsliter and the molar ma M in gramsmole The experimental con guration uses the measurement of height as an estimate ofthe osmotic pressure The equation H pgh is used h 139Ipg Use of osmotic pressure to determine molar mass Pur 3911 o Patele o Use of osmotic pressure to determine molar mass Use of osmotic pressure to determine molar mass A sample of 15 mg ofa protein ofunknown molar mass is added to an osmometer The solution volume is 1 mL The solution height increases by 10 cm The measurement temperature is 298 K What is the molar mass ofthe protein A 37900 B 39700 C 79300 D 97300 Use of osmotic pressure to determine molar mass A sample of 1 5 mg ofa protein ofunknown molar mass is added to an osmometer The solution volume is1 mL The solution height increases by 10 c Th measurement temperature is 298 K What is the molar mass ofthe protein A 37900 M 7M7 M7 1 5 kgm3E 31jmoleKZBE K 39 n 39 pgh 39 mun kgmlxa EmEE1 m B 39700 37auu g mol C 79300 D 97300 Chemistry 331 Lecture 4 Vibrational Spectroscopy NC State University 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 5H o H an where p is the dipole operator 000 00 0 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 MQ Mu bowOf The total wave function The total wave function can be factored into an electronic a vibrational and a rotational wave function P We va Mm 1 1 if mm mm 1 mvdoj I nmmvmsinededw 1 uvmulxvvmsinedeol Interaction with radiation An oscillating electromagnetic field enters as Eocosmt such that the angular frequency hm is equal to a vibrational energy level difference and the transition moment is Mm Bf Yj Mcos9YyMsin9ded bow9i Interaction with radiation The choice of cose means that we consider zpolarized microwave light In general we could considerx orypolarized as well X S39neoos um W MN 216 y S39nesln u p sinecostbz sinesin jcosek EEK H3 3 V brat ona trans t ons V brat ona trans t ons usmuanun mum mu ecme abum Ms ethbnumbundmnhguvauun Asm mam usmHales myan vamauun can mlevac u ahenhe quamum 51am 40 gig o M 1 wm o mumem bemeen m smev n and v 1 1 1 quot J oe a MM Big rmf e Q equot quot dQ V brat ona trans t ons Nme mm m 125ml 5 a s atemem mum wbvatmna se ecuun 1M2 VWhmthe havmum appmwmatmnhansnmnscannn yuccuv bemeenstatessepavale byunequamum numbevtAv MFA Tm genem We can be seen by cunsmymg Megva s uHhetype Shawn mm pvewuus 5th C ass ca V brahon of a D atormc Axmaxthzcaselnnmmxmmmcancnn mum mad mamam axpmg mammna ma mm quotmm a dmnmcmnbcu easmmwh mm NBS an 3 mg Mum mm m mniam mm ms Harmomc approwmahon www ggjwigw jguzigw AI ethbnum U ksume tevms mghenhan quadvauc ave 22ch By de nmun 67 V 7 C asswca approach to wbrauon smmmn s asmHatmv Anv enevgv s passme Enevgylcm k 3351 Q Q C ass ca Wbrat ona mot on a name unamemmmmmmn mmmm reame m m s pmpnmnnm m n d w amme x lmmmm vmqanavvmx We chmcamamnmcnw mvmn mm mm 3 mt a 1 7 kn Sumnan mam mi lmmlm mm mm m mma mam u Waxexnmmxwv what C ass ca potenha funchon We pmer m xxv 1rwmnma mmmmnn m pmemaw mm a hanmmcpmema mmcummmaa mnmmmw r News my We angmavmquencvm 2m wwnveqmm Hz Quantum approach to wbrauon Sumuun sGausswan Enevgy s quarmzed L n aw zueg f7395 v an giwm wsthe quantum numhev memudcavomancHN A avuedtmnsmans Q vqwmqw V brahona Wavefunct ons Energy eva szrggwm by rwrzm Typmz wages are a memevmo 73000 cm So uhons to harmomc oscmator m mm pn wnmakm ammmm Gauanvmquot 0 a HM 1 e vequot V W m quotmum mm x N 2 ow 3 aywzy The Zero pom energy Mmmamhi mm We mum mum M m1 22m Margy mecnnaaemmnhmunmnamvvnmme u m we mmmwam at 3mm zemthen m m anhmmhewpnxmnnand rmmemn aanTYEN acmraEV We wmhmthemmummn memed39n mum WW1an an 3mm We E a m 22m pmm anew Po yatormc Mo ecu es mm a 3mm degnexwmdnmm amnbcme mm mm N stuns mm mm mmnamegmmmanm m mmwnnd m mm mm mam1 quotEx mm mm a hmanmhm emm awm mannnamwmxm ednm m a Norma modes 7 Water A symme mstmch Asymmmsm am y axzsm wavasm Mum ya man hrmvmdem emm 3723 WWW aegmmmanm m mm dammmdnmm WW Yhere ave 3 nama Nudes m is An a hem ave mm saws snug 3H shawa mpme mamem mangemthewmmmn m hamm appm mmmn an be appuem saw mm made am oO oo Norma modes 7 co2 Svmmemcslmch Asmmmc V zzxm wzuvm Ben 07mm mm v m mm Yhereave nwmam avemw mass m is was a hem Bred saws snue hEV mange m mew mmmn mm mm mamem V brahona Transmon V b rat ona Transmon V brat ona Se echon Ru e Overtones of water Even in water vapor v1 n4 v3 but symmetries are different 1quot1 F3 However the third overtone of mode 1 has the same symmetry as the combination vl symmetric stretch 3825 cmquot REFEREE vzbend1654cm39l Strong anharmonic coupling leads to strong overtones at 11032 and 10613 cm39l I These intense bands give water and ice their blue color 71 v3 asymmemc stretch 3935 cm Frequency shift due to molecular interactions Hydrogen bonding lowers OH force constant and HOH bending force constant 0 O c vapor liquid v1 3825 gt3657 v2 1654 gt 1595 v3 3935 gt 3756 Question Which expression is correct AkJg Question Which expression is correct AkJg Bk Fla Bk plat k c M c kttw 03 E D IltJum2 D IltJum2 m2 A u k umz Question Question How many normal modes of vibration does methane have How many normal modes of vibration does methane have Chermstry 331 Lecture 1 mtmuuetmn Eteetmmagnette Spectrum E ack new Ramanun NC state t t OveNteW Quantum Mechamcs Faullmlmczguwlvxu ne ms Rmm mm mun B ack body Radtahon An mea emmev mmmmmn scaHed a mack haw Obsether mm Peak a he enevev m emssmn shmsm shanev mve enmhs Esme tempevmme ts meeseu vuenuteuteeemem aWMJnyS Yhesecandmmatmncansamcz1 Mm K 2 ea x m nmeK M The wmetengm and the frequency We mve englh Ms he mstance between the Peaksmatvavehngvmve nc assm thsushgmxsavmvemm mve s themc v eu Yhemvenumhevvvlc Yhev avenumhev hasunnsmcm The etectromagnettc Spectrum A maeasu g gt v deeeasmg We mve englh and weeuenw are mvevse v etueu DHemma for C asstca Phystcs Yhemaxmum neHEVEVVUHhehadlt may spe mm snm Em amed byeteseeet thsus ruemvnymauesmmemwmeme Pvemded m be EnkgT F 1 Wm Mme mdmmenUVVdenSW we umquot meeesesumu baund es 1 en mm tsknawn esme mew Jeans mu The Sun 5 a B ack Body rne em and stavsave mack hames rne Peak mne emssanspe mm depends amempevmme asmdmated hevwend m acemem aw The P anck D st but on Law mm assAmed quamuslmn m mm madesE nu m2 rne mnstam n detevmnesthat anmnase madesvwlh an enevevspeenee bvthe muse amaums wen een e exmed Yn e upmatmna heeve vaHsquav ev enevev euenmn numbev Nudes avev neney enevees P anck Assumptmn nnpnes ma average Energy 5 temperamre dependent enegv evehsltz r Ouarmzed eve s npmnenne avevage enevev n eeen asmHatm e e e nvewe 1 Ismael Mv ecanmsa netmsas eeee cWeMMJJ Mamthe Wan2K mynme Smp vvemace w hvthelt ewe1W1 emesemn thed bv euemzemn Mathemat ca Form of the P anck Law 2th Meenevgv eve sw hepapu med amammgm athema v ewmmg rne neney eve s Wu he esspa at dthamhemv ev eve s mne mam meawthe enereveensnv became e mm 1 n W 9 y 1 P anck D st but on Law Conswstency Wnn Expenment Yhe empevmuve behavmv mne Reneen Jeans avws recavered bemuse ewe mka as r e e ne Magva auneme enevev epmpamane mY Wmmgwesme emnr anmann aW weenwmnen xavene ayes Yhevwend m auemem avwsremveved nan dmevermatmn MP 521mg am e Ewesthe maximum mne msmbmmn avv Consequences of Planck Law Classical physics fails to describe black body radiation A model that includes quantized modes of electromagnetic radiation succeeds The constant h 6626 x 1034 Js is a fundamental constant the determines the scale of energy quantization What is the radiant power ofthe sun Use the StefanBoltzman law w a T39 w is the flux or power per unit area c 56704 x 10 kg s3 Kquot WattsImZIK Assuming that the temperature at the surface is 5500 K and the diameter is 14 x 106 km The way that the sun39s drameter Is measured Is hytakmu ahuuiar drameter measurements and then trahsiatmu them In hhear drameter measure he h I m u ereury w eh Is m trahsrt m trum utthe sun The rst series at measurements were taken m the early 17 39s by Jean Freud m pans rrahee What is the radiant power ofthe sun First calculate the area and then the ux power The area is A 47rR2 A 4314167 x1032 A 616x1Ole2 The ux is W o T W 56704 X 108 5500 W 519 x 107 Wattsm2 The total power is P WA 519 x107 Wattsm7616 x 1018 m2 P 32 X1026 Natts What is the radiant power at the surface of the earth We use the distance 39om the earth to the sun to obtain the ux at the earth The earth is R2 15 x 108 km 39om the sun The area irradiated is 7ng A2 283 x1023 m2 What is the radiant flux at the surface of the earth The ux in space above the earth is called the insolation The insolation is the power coming 39om the sun divided by t otal area at the radius of the earth wE PAE 32 x 1026 Watts283 x 1023 2 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 7rRearth2 Ac 3141613x 107 m I x 10 m2 Pap 39 eAc Pab 113x103Wattsm253x10 m7 Pabs 6 x 10 7Watts 7 What is the temperature at the surface of the earth Pabs Pemit 6 X 1017Watts Pemit G Tearth4 Aearth Aearth 4 Rearthz 21 X 1015012 Tearth PemitG Aearth 14 Tearth 6 X rot56704 x 10821 X 1015v4 Tearth 266 K Why is this I L 1 What is the temperature at the surface of the earth Pabs Pemit 6 X 1017Watts Pemit G Tearth4 Aearth Aearth 4 Rearthz 21 X 1015012 Tearth PemitG Aearth 14 Teall 6 x rot56704 x 10021 x 1015 Tearth 266 K This is close but it is a little frosty l Why is this We ignored the fact that 1 the earth has an atmosphere The atmosphere does two things 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 limxT 288 X 106 an Thus for the sun with T 5500 K lbw 523 nm For the earth wi h T 266 K low 10800 nm 108 pm 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 BLACK BODY lol g CURVES 5 5500 K 266 K 1 Electronic Vibrational Rotational Chemistry 331 Lecture 25 Collision Theory NC State University Kinetics Gas Phase Kinetics Elementary Reactions Collision Theory Experimental Rate Law Application to Formation of NO2 Elementary Reactions Reactions usually occur in a series of steps involving one or two molecules The molecularity of an elementary reaction is equal to the number of molecules that come together in an elementary step The reaction order is an empirical quantity that does not have to equal the molecularity Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier E H AA Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier f Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier wx Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H Al Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier E H m AA Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier l Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier E H Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier i Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier 99 Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier W Hi Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H m gt No reaction occurs because the kinetic energy is lower than F7 H In order to react molecules must collide with enough energy to surmount the energy barrier Collision theory Hi Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H E A H In order to react molecules must collide with enough energy to surmount the energy barrier Collision theory Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H M Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H quotAl Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H M E A Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H w l Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier HM E A Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier HM Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier H99 Collision theory In order to react molecules must collide with enough energy to surmount the energy barrier HQ gt The reaction H BrZwHBr Br occurs because the kinetic energy is larger than Ea Relationship to Arrhenius theory For the preceeding reaction the molecularity is two It is a bimolecular reaction Thus rate of collision oc HBr But as we saw the collisions must occur with an energy of Ea or greater for a reaction to occur The kinetic theory of gases tells us that there is a connection between the kinetic energy and the temperature The rate law and the probability The probability that a given reaction will occur at given temperature is f eraRT Thus the rate of reaction oc HBr eEaW If we compare to the second order rate law rate of reaction kHBr2 If follows that k oc eEaRT The preexponential factor The value of A can be calculated from the kinetic theory of gases 8k T 2 T5 p mlmZm1 ml is the reduced mass 6 is the collision cross section There is also an orientational factor P If not all orientations lead to producB then P lt 1 Ni The formulation of rate laws The empirical rate law for the men 2 N0g 029 39 2 N02g is Rate of formation of NO2 kNO2O Does this rate law mean that N02 is formed in a termolecular process Such collisions have very low probability A combination of bimolecular processes is much more likely The following reaction mechanism has been propose Elementary steps in the formation of NO2 Step 1 Two NO molecular collide to form a dimer NO NO gtN202 Step 2 The O2 molecule collides the dimerto form N02 02 N202 gt N02 No2 For step 2 the rate of consumption of N202 is kbN20202 and the rate of formation of N02 is ZlqjN20202 Applying the steady state approximation to intermediate N202 The rate of formation of NO2 2kbNZOZOZ is not acceptable as a rate law since it contains the concentration of the intermediate N202 J2d Naggl kNo2 Iciwzoz N20202 t Now we use the steady state approximation 0 kaNO2 ka N202 kbN20202 How to deal with the intermediate NZO2 Solving the steady state equation we find 2 N 0 2 2 k kitozi so that the rate of formation of NO2 can be written 2k k NO2O V 2k N 0 0 b 2 2 2 kg 02 Note that this agrees with the empirical rate law provided ka gtgt kb02 Thus Rate of formation of N02 ZkakbkaNO2OZ Rate limiting step If the concentration of O2 is increased to a very high level then the second step is very rapid compared to the first The first step becomes the rate limiting step Under these conditions the rate law becomes k 2k3N02 Such conditions are used experimentally to determine the order of NO in the reaction Chermstry 331 Lemme E Ahsmpheh speeha e1 amms The SehTeeThgeTeeuaheh TeT hydmgen The eTemTehTe s1mcluve e1 amms NC State T T The SoTaT Spectrum quotwere an vapSThme shah emssmh eeuee Theeehhhhehhhes The have me hm speehhe ms h he eh rm mm mamaquot ExpeTTmemaT obsewauon of hydrogen atom means 31 msmete v avebnmhs aha thevemve m msmete eheTeTes The aeheh senesvesuhs m TewshTe Thesmmhmhmhmhasshmhessshh e The TeTehahsth hetweenmese hnesv as shavmm VuHaW he Mhehe TeTehah mhehuhheh auhe medva hnescan he mbv AtomTcspectTa Emphheem hesheeh ehwmhmhe emgth e h h Wneve R Tsthe Mhehe mheemh and h and e hueeehe unr equalinn inr lnnlrnunn The SchmdngYEmzhumm hyumgm m anzmes zSemrahun RaeTszh argmzmzns a eha veTuhnTms mantahunvdues E e Spmmseepy a ammo hyumgm The sewnheehehmeh Thhhhee ahheheehm 2 L 7 7 P W E The upemnmeT Squared T a a v 7 7 sh sf The vmmdum ma pheheannTaTmnmhmexwehh nmad dxwandzthe mnn amexan e mahah 11 lm nillm nnlnnlill 2 me 0mm mlemz bawaemhe emvm 2m H mm WNW mmmwm WE Pm quot 5 ZWAMX mnh au mm mama munanenwy 5 me mmunm my mlmhe pmnn mu gecan when a mom Munm a hymn m vanah e m s n w H T vkiv o v N r m any are Semrahunm nm mvand dmmmc an 1 5mm m mew mle resuns m anewenmmc equahummhe 2mm 2 was mm was hymns mass cmmmzbzs H 7 am 7 220472 3 WWW aaeshvzw W mlm quotMl n We whencamanmmchnleyml mun t me wavmmmm summs mmezng zv 2mm are 5 mm mvmumcs vquot 6 mm we gemvayngmnamahsw 2ng argubvmmerlum mm mm s m 1h may rm mmuav We shape mm nfnamsemhbxme m Whema hanmn v am mmmmm mmmv 2 Wm mmmm been mm mamaquot m hneav mmmnm mm anvmmmexnumhm quotMl quotMl m v Wm hanmmchaxthe nrmman nfn m v v gnaw mhenmmammmmm mpnr nak39 39 mm are m mgmavmmmm mum mm x mm anguanmnn nu n mm 4 w A mman mm 39 a u t mm mammguammnwn quotIE niinngivn lllll lllill sllll ni tammmlmndemrg EmzHu v 7 mm ms ermcunznsmecunnbmumtmm magmm mg szms m 21mm mumM quotIE linhrrlllills m manna dmmanm unnlanavsxandww ma 1quot ha um a mm mm nllnn nillmralii mumIns Em rm radxa eqmnn WWW mm quotmm anewnm m Wennmalzann n amed mum mam R MRMEm 1 m mm mm m mmnm m v m mm mm mum mm mm smemeam lqAPllwu lslllzl a MIIPLE vmmmt mmm m My mum m mmumnmmgmm rad ahmmm m vmwmt mm mm mm Wm WM mmnnga pmkaepm mm mat Wumn rim g a a 5 g x 3 5 E i We mayy aE smm quotMam am 22 smcmed We pnnmpd manum numbs n E 3h 225 n 7 m 512125 Wm m2 same mm rumba n mvathe same mergy m 512125 m neganva may 22 mum 512125 51312 mpcsmvaenagyze mbmm mdzre ma 1 m2 cumnmm The Rydberg constant The energy levels calculated using the Schrodinger equation permit calculation of the Rydberg constant w One major issue is units Spectroscopists often use units of wavenumber or cml At rst this seems odd but hv hcA hcv where i isthe value of the transition in wavenumbers 64 R I1632n2802 2 in cml The simnle form b Using the Rydberg constant the energy of the hydrogen atom can be written as where R 107636 cm1 Shells and subshells All of the orbitals of a given value of n for a shell n 1 2 3 4 correspond to shells K L M N Orbitals with the same value of n and different values of Iform subshells 3 0 1 2 correspond to subshells s p d Using the quantum numbers that emerge from solution of the Schrodinger equation the subshells can be described as orbitals Hvdrouenic orbitals s orbitals are spherically symmetrical The 1s wavefunction decays exponentially from a maximum value of 11903 2 at the nucleus p orbitals have zero amplitude at r0 and the electron possesses an angular momentum of hwl 1 The orbital with m 0 has zero angular momentum about the z axis The angular variation is cose which can be written as r leading to the name pZ orbital l39lllllf g 1 S radial wavefunction e The 15 orbital has no nodes and decays exponentially D R15 21aUYZe39 z in n 1 and I 0 are the quantum numbers for this orbital 5 n The Radial Ilistrihution in Hvdrouen 2s and 2H orbitals sneetroseonv ot atomie htnlrooen Spectra reported inwavenumbers Rydberg t all of the series of hydrogen spectra with a single equation Absorption or emission of a photon of frequency v occurs in resonance with an energy change AE hv Bohr frequency condition Solutions of Schrodinger equation result in further selection rules SIIBGIIIJSGIJIHG transitions A transition requires a transfer from one state with its quantum numbers ml m to another state n2 mz Not all transitions are possible there are selection rules Al r 1 m 0 r 1 These rules demand conservation of angular momentum Since a photon carries an intrinsic angular momentum of 1 Illlantreleotron atoms The orbital annrottimation The orbital approximation to the total many electron wavefunction I r r2 is to rewrite it as a product of oneelectron wave functions ulrmlrz sothat I r r2 ulrtpr2 The configuration of an atom is the list of occupied orbitals The Pauli exclusion principle states that the spins must be paired if two electrons are to occupy one orbital and no more than two electrons may occupy an orbital Penetration antl shieltlinti ln manyelectron atoms the s p d etc orbitals of each shell are not degenerate An electron distance r from the nucleus experiences the nuclear charge Z shielded by all of the other electrons The effective charge is Ze ZEff Z owhere o is the shielding parameter An s electron has a greater penetration through inner shells than a p electron of the same shell because the p electron has a node at the nucleus The Authau nrinoinle u Aufbau means buildingup in German The configurations of atoms are built up by population of the hydrogenic orbitals Imagine a bare nucleus with charge Z and then add Z electrons to the orbitals in the following order 1s 2s 2p 3s 3p 4s 3d 4p 5s 4d 5p 6s etc Hund s rule an atom in its ground state adopts a configuration with the greatest number of unpaired spins The nrohlem ot multinle electrons The central difficulty with application of the Schrodinger equation is the presence of electronelectron interaction terms in the potential energy I 2 U 2 7 Ze e V7714T8ur1l 472an run 7 nuclear electrcn r electrcn elect w No analytical solutions HartreeFock procedure Find solutions that optimize the electron in each orbital in the presence of the field of all of the other orbitals HartreeFock nrocedure As an example for He we can write the two electron wave function as a product of orbitals r1 r2 r1 r2 The probability distr bution at r2ntgtr2drZ for electron 2 corresponds to a charge density in classical hysics Therefore we can say that the effective potential felt by electron 1 is Vim I ltr2gt ltr2gtdr2 Tl39le selfconsistent field method The effective or average potential can be used in a one electron hamiltonian operator for electron 1 A 4 1 2 2 ff H1 r1T 2V1 7Jr V1201 The Schrodinger equation is solved for electron 1 WW 21w Start with a trial function r2 and solve for Mn Using r1 calculate an effective potential for 2 and solve for r2 Continue until convergence is eached Atomic term SVIIIIIIJIS The letter indicates the total orbital angular momentum quantum number of all electrons in an atom Ll1lz 4421 will lzl The left superscript gives the multiplicity 281whereS s1s s The right subscn39pt gives the total angular momentum J LS LS 1 L S The term symbol is EHL J I39llIIIII39S rules determine the term SVIIIIIIJI attire ground state Each state is designated by a term symbol corresponds to a wave function that is an eigenfunction of L2and 9 with unique energy r The state with the larges value ofS is the most stable v For states with the same value of S the state with the largest value ofL is the most stable lfthe states have the same value of L and S subshell less than half lled smallestJ is most stable subshell more than half lled the state with the largest value OH is the most stable First Law Summary Statements of the first law AU w q work plus heat The internal energy is conserved The internal energy is a state function The internal energy change is not pathdependent The internal energy change of a closed cycle is zero The internal energy change of an isolated system is zero Work is path dependent w PdV Definitions pressure always refers to external pressure Sign convention work done by system work done on system Path 1 constant pressure path w PextAV Path 2 constant volume path w 0 because AV 0 Path 3 isothermal path AU 0 q w w nRT ln VJVi Path 4 adiabatic path q 0 AU w The internal energy of a system increases as its temperature is raised In a constant volume process the constant of proportionality is called heat capacity at constant volume CV extensive units JK39I Molar heat capacity is intensive CV intensive units JK39lmol39l Specific heat capacity is heat capacity per unit mass JK391 g39l CV dUdT V dU CV dT AU CV AT CV 32nR for a monatomic ideal gas CV 52nR for a diatomic ideal gas CV 3nR for a polyatomic ideal gas The constant pressure energy is called the enthalpy AH AU PAV PAV is the work term from moles of gas created or consumed We can also write this as AH AU AnRT where An is the change in moles of gas For an ideal gas Cp CV nR Chemistry 331 Lecture 14 Entropy and the Second Law NC State University Spontaneity of Chemical Reactions One might be tempted based on the results ofthermo chemistry to predict that all exothermic reactions would be spontaneous The corollary this would be the statement endothermic reactions that are spontaneous Of course heat must be taken up from the surroundings in orderfor such processes to occur Nonetheless the enthalpy ofthe reaction does not determine whether or not the reaction will occur only how much heat will be required or generated by the reaction The observation that gases n e s spontaneously mix when introduced into the same ve sel are further examples that require quantitative explanation Spontaneity of Chemical Reactions As you might guess by now we are going to de ne a new state function that will explain all ofthese observations and de ne the direction of spontaneous processes This state function is the entropy Entropy is related to heat and heat ow and yet heat is not a state Jnction Recall that q is a path function It turns out that the state function needed to describe spontaneous change is the heat divided by the temperature Here we simply state this result Engines Historically people were interested in understanding the ef ciency with which heat is converted into work This was a very important question at the dawn ofthe industrial revolution since it was easy to conceive of an engine powered by steam but it turned out to be quite dif cult to build one that was ef cient enough to get anything done In an engine there is a cycle in which fuel is burned to heat gas inside the piston The ex ansion ofthe piston leads to cooling and work Compression readies the piston for the next cycle A state function should have zero net change forthe cycle It is onlyt e state that matterst Jnction not the path required to get there Heat is a path function As we all know in an internal combustion engine or a steam engine there is a net release of heat Therefore we all understand that Sq 2 0 forthe cycle A cyclic heat engine ismnermai Amman IV T i 11 q 3 Wl 39 Wm Forthe adiabatic steps I Ch Cliv Forthe isothermal steps AU 0 III Work and Heat for the Cycle Neitherthe work northe heat is a state function Neither one is zero for the cycle as should be the case for a state function The work is Wiv VTuldThutn RTculdln VANa CvTnm Train nRT D lnV N3 since wH ww 1N2 hutlnVzN nRTmdInV2V property oflogarithms at is cI Ch Clm Since 5Wu My 0 w wm since dU 0 for isothermal steps 4 3 q nRThDIn2N nRTmdlnVN2 since VANa VN2 q nRThDIn2N nRTmdln2N property of logarithms A new state function Entropy The heat is not a state Jnction The sum q qm is not zero From this point on we will make the following de nitions Ch qhm Clm qculd V V q mm a nRTmln VZ nervn VZ c o r r w M thd quotR1quot nRn 0 T m Tm V1 V1 However the heat divided by temperature is a state function This reasoning leads to the idea ofa state function called the entropy We can write qrev AS T Thermodynamics of an Engine The cycle just described could be the cycle for a piston in a steam engine orin an internal combustion engine The hot gas that expands following combustion ofa small quantity offossil fuel drives the cycle If you think about the fact that the piston is connected to the cranksha you will realize that the external pressure on the piston is 39 m t39 f 39 d39 hel ingto realize o m 1 9 1 Lo m m m 1 o 6 1 o 3 m m 1 1 above reasoning to our car The formalism above for the entropy can be used to tell us the thermodynamic ef ciency ofthe engine Thermodynamic Efficiency We de ne the ef ciency as the work extracted divided by the total heat input work done ef ciency W Tl thotaI IHRTcold 720939 V2 VOI Tmz Tm qnor quotRTvan V2 V1 That The ef ciency de ned here is the ideal best case It assumes a reversible process with no losses due to friction e 39 t gine he temperature Tmd is the temperature of the exhaust Tmd cannot be less than the temperature of the surroundings Question Your car has an operating temperature of 400 K lfthe ambient temperature is 300 K what is the thermodynamic ef ciency of the the engine Question Your car has an operating temperature of400 K lfthe ambient temperature is 300 K what is the thermodynamic ef ciency of the the engine A 75 B 50 c 25 D 5 IWmal Tmz Tmd Tm 300K T39 qmz T 1 Tm 400K 03925 Question The thermodynamic cycle was derived for reversible expansions What are the consequences ifthe cycle is not perfectly reversible A The work of expansion will decrease B The work of compression will decrease C There will be no adiabatic expansion D There can be no cycle Question The thermodynamic cycle was derived for reversible expansions What are he consequences ifthe cycle is not perfectly reversible A The work of expansion will decrease B The work of compression will decrease C There will be no adiabatic expansion D There can be no cycle The Thermodynamic Temperature Scale The de nition of entropy is qml39l39hut qm dTm d 0 W can write this as elmTm qculdTmld Since qm d is negative we can combine the minus sign with qm d and write the expression as Iqmlfrm Iqwmlmum and nally Iqhotlllqcoldl ThatTrain The ratio ofthe heats is equal to the ratio oftemperatures for two steps in a thermodynamic cycle This de nes a t t u t as well ie this scheme represents a thermometer Both this expression and the thermodynamic ef ciency further imply that there is an absolute zero oftemperature Heat Transfer To examine the function that we havejust de ned let us imagine that we place to identical metal bricks in contact with one another If one of the bricks is at equilibrium at 300 K and the other at 500 K what will the new equilibrium temperature be lntuitively you would say 400 K and you would imagine that heat ows spontaneouslyfrom the warmer brick to the colder brick The entropy function makes these ideas quantitative at G i AS 72 E T500K Usingthisd efnition of entropy change as the heat ow divided by the temperature Heat Transfer Let s assume that heat ows from the hot bodyto the cold body Then q1 is negative the ow from the hot body and q2 is positive the ow into the cold body Moreover q2 39Cll Cl This means that we can substitute in q to obtain 11 ll 4 As T2 E T2 E 300 500 000130 q T1 500 K Using this de nition 39 s AS gt 0 that the process is spontaneous Calculating reversible and irreversible paths It is important to reiterate that the calculation ofthe entropy ofthe system always follows a reversible a h You might ask well what happens ifthe process is not reversible To consider this let us the example of expansion of gas in a cylinder The process can occur along different paths a constant pressure expansion w PEXAV PEX1If b reversible isothermal expansion the work w nRTlnVV For both a irreversible and b reversible we will calculate the entropy ofthe system along a reversible path A w IT wT annV For the reversible path we can use the fact that ASSW A8 to obtain ASSW annVV sys System and surroundings The heat transfer example shows us that we always must consider the surroundings Any time the system releases u t u t Au 4 4 t overall entropy change Thus the total entropy is total Assurr Assys For a reversible process the total entropy is zero If a process is irreversible and spontaneous the entropy change is positive This implies that we must treat the system and the surroundings differently when we calculate entropy The rule is always calculate the entropy ofthe system along a reversible path If the process is truly reversible then total 0 and Assurr 39 Assys Understanding the irreversible path For the irreversible path we use the actual work ofthe constant pr sure expa io PMAV PEXKVf V to calculate ASSW qlT wlT Pextlf V T where T is the same temperature we used for the isothermal expansion Note that the sign is opposite since the heat is owing into the surround ings and out of the system We see that in this case the entropy change ASSW is smaller in magnitude than ASS We know this 39om the rst law where we saw that the irreversible work of expansion is always less than the reversible work of expansion Thus Astotal gt 0 for the irreversible process The dependence of the entropy on volume For a constant temperature isothermalexpansion we have as a wE The logic behind this statement is that the internal energy change is zero for a constant temperature process and 5me To calculate the reversible work we simply plug in 6wEV PdV According to the ideal gas law P Nso dSanVN S2 V2 av as R L V v The result ofthis equation is that AS annV2N1 at constant temperature The dependence of the entropy on temperature The entropy change as a function ofthe temperature is derived at constant volume using the fact that dU Sqv nCVdT The reversible heat in this case qw is a constant volume he tan soi can be rep aced by d8 aqmvT nCVdTlT at constant volume To obtain AS we need to integrate both sides fusqcvfig We obtain s s2 s1 nCVnTZIT1 Exactlythe same reasoning applies at constant pressure so that AS nCpInTZIT1 Chermstry 331 Propemes of exponenha s LedumZ WEB mm mmua 1W MathR awew The Waxerpamde duamy H Werse functwon 5 oga thm TheWave Equa mn PhD uE EdHE EWEE WWWquot A was Exponenna to the base e owemg W ehe base to another We EM WUAZ 2quot What s 7 ENE eAB n1EIAnEquotTaKE natura h nve 5e fLH39VCUOH Elf bum SMES 5 hamrax oga thm SD VE fDrX We Ema a Denvatwe rate of chahge Denvatwe rate of chahge Ax ax S upe H mtES ma Rateufchange EXamp emnmun payahma Denvatwe rate of change Denvatwe rate ofchange M R mama A Denvatwe rate of change S ave e w x 2n Whahsthepa evr Smile 2x Denvatwe rate of change Pmnhe gape Esme ween Pmms Denvatwe rate of enange u smdeed a hne Mn SW2 2x Derivatives and Integrals 0 th t39 Dawatwe dnXdx e lx Inteqm amen afnatuvd annthm 110 A ewgtlt smvevseafWX Yhemtetzvd t5 mewase D the dawatwe P oto he natura ogarnhm S ope of the tangent hne at 14 mm Yhedevwatwe t5 the S DDe ZKFIH smve 4 Mix S ope of the tangent hne at 12 S ope ottne tangent We at 1 mm 12 smve e 2 mm ntastm nzt nm S ope ofthe tangent We at 2 mm wot of the pres of the tangenm B mtafs avevs x Derwatwe dnxdx lx Intega ofdaa the area under the curve 9 I mm Intega ofdaa Expe menta obsewauon of ydrog om Hwyagenm mama n aamwms me my M WNW nesamunm mnm oasnmandsasnm mg ve atmnsmp hammse Wm mm mm RWhevg yam Atormc spectra We eeeeue We m Empmeem eeeeeenemmeme wemeey We eeeeue nee be m s m navmvvhnes Wneve R sthe RWhevg mnsame ene m ene n ave Meyers e hydrogen atom sem ecxass cax approach meme dusr Mme en new numberm wam enmhs m me emmeem 2 2 23 The e ectron must no mterfere thh tse f neemmewmmmme 2 m ne 2 Me Warm Shawn has we DeBtheuam enmh w MW We W Mwm Yms sa candman Ourquammmn manvmarmamemum m3 5 EXamp e of Se frmterference mng a We saw mauve We enmwmeeem atmnaHhe ve nms e We euemzemn mangu av mamemum mphes euemzemn mmevadws Meww e7 The Swgm cance of quarmzed arms We saw made e mnsmem m euemzee mhnsa he e edmn amund he nuc eus De mghe avguedtha he we enema ham my en Pam es The BohrMode Predwcts Ouarmzed Energwes Meeemmemeneeeeuemzeeene hevemve We energxes are euemzee A mmmgmc ca e edmsmms eey veemre e1 Suhstmmng W Van ewes 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 Photoelectric Effect Electrons are ejected from a metal surface by hv 5 absorption of a photon Depends on frequency sity not on inten Metal Surface TT T Threshold 39equency Em gy corresponds to hv n I I is the work function It is essentially equal to the ionization potential of the metal Insuffmmt energy Fhufuejecnm u ccurs fm39phmuqecum Photoelectric Effect The kinetic energy of the ejected particle is given b 12 mvZ hv CD The threshold energy is D the work function This demonstrates the particleI ke behavior of photons A wavelike behavior would be indicated if the intensity produced the effect Myoglobin Kinetics of ligand recombination Photosynthesis Electron transfer kinetics Protein Folding Polymer chain kinetics Photolysis in heme pro Deoxy Heme Excitation by light is a method for initiating ligand dynamics or enzyme intermediates for kinetic studies Teng Srajer Moffat Nature Shuct Bol 1994 1701 is shown as a ribbon that follows the ochelical structure of myoglobin The structure shown is at equilibrium Conformational substates are called A states The photopro Iron moves out of the heme plane when CO is photolyzed CO moves to a docking site and is parallel to the heme plane Conformational substates are called has no CO ligand The protein backbone has shifted to permit a water to enter the distal pocket This form is often referred to an S state Ligand recombination is a sum of single exponential processes at room tem ture S CDgee new CDb His FeP co HisFeP CO lr mume l1 quot i Difference spectrum from nanosecond transient absorption spectroscopy HisFepCO emu Wm quotquot pm mmm pm mema awty mum Abe co recombmaam kmeacs aspena m my me End but 65ch w 2 am m Ham oommqu 7th mm w m hm mm m 7m bur Wm WM WW m mm mm WW I m mmmk 23 31522 Photusmb msis the ImpMg of 5mm ofme Phomsynmem photnlndured eler tron a Reacuon Denmr 0be Sph WWWquot Wammmmuwm A it Pegm b r Mumquot qu evzmmn mg m 2 mm mquumu w M m w J Mmwwmwaqumww aH LWJ an id mmm nle m mm mm m Monq mmquot mm m OTDmLpWDFES m the Battera m KW v2 m mm mm mummy a p x 5mm me Email maresieme quanmm yrsd om F39 arm om me mmpernm by elem f9 m 7 k N d i H mf nkmm i mmmewm memw 5 m m a m m m Wmmmm m mm the awe om me mmpenmn by elem WWW marge 5 WWW mmkm Mam mm P mm km NW m 4mm k a 7 mam Fm k km k l m 394 m K m W mm m 2 mm mama 1mmquot WWW mm m m A we d mnymm L m n m in 3n m m mam Folding Ufa cyc 0m hexzmev 39 mam tho ded mmwm m w W W3quot W W W W Wquot Wu mmmwu m m mummy 0F 7 m vevnda w u 1 mm mm affo qu 3 Absorbancex10 reference sample 107 1 1oquot 1039 r2 Time s 3 DeltaAbsorbanoex10 sample reference Infrared kinetic data Generic r ihairpin r r39 39 t ellx melting a J a I E r Infrared thermal melt data E N G iiHeatIng 7 o a V K i 39 TWPtOPhan J E 6 These experiments T T fluorescence if g em on were performed on I Y r 1 qr 2 i the H3 peptide 39 Time us D G G R R 72 7 Wavenumbercm39 Kinetics The Arrhmius rate tenstant Transitian state ra tenstant cataiysis The Arrhenius equation I ihe enpiiitai ahsavahan is that in k in A e E favmaneathans ihis memsthataviataf nkv lY qivesastvaiqhthne Aisthe vesewanenhai air a is the athvatian enei re uenEY rattai w Aisa AeesJAr Experimenmi determination of Arrhenius paramemr We tan piat Wk vs Mr te uetennin e annatian enng Aphtuf inkvs Jryieiusa v Activated complex meora intermediate g neeaienie eiaauers MiniIr teammate ihe inteineeiate is the athvatsd Dmviex The transition state ihe athvatsd tanpiex is a eistaitee sun the is inteineeiate between the stiutruie is that the transitian un With the ieattants and piaeutts A a i The assumpuon Ofequihbrium between me reactanm and me transition 5 the raimanan ar the activated tamp i t in equiiitiiun With the ieattants wetan emiess the equiiitiiun mnslant as i e 93 7 A15 and the iate mnstant is given by the viaduct ar a fve uenw fact ai kJh rai the rainatan ar the tamviex tines the equiiitiiun anslant Arrhenius paramemrs Yhe martian iae m aisa he Mina K kJne new wheve AG is meactwatian Gibbs enmqv Yhis Dmvides an inrepieraran in the rimming Davame is n kYhe 5 i39 Yhe hetiuentv rami devmds an the emanentiai D the activatian enlmw a M whe ie a is the activatian mthaipy Camiysis involves lowering of me energy barrier Rea M Fradums AH39 may mini Mamie n atdvsl Wavides m ailemahve ieacrian pairian Win a imi activatian enaqv a Kristian enmaipv Types of camiysis phase as Wigwam Em a We a mi h e stews Hetemqmm 5 aa YSisr the dam is in a aimimtpmg naii the veactants Empie maai mmpiexes suvfiesi zeahhes Enzvmah ata Vssr the skews is a Dvatein ma has a Substvahe pinning Size ma antvaiied veathan Path Zeohtes an important ciass of cataivs Dalanxe 0F miiie mine in iiWiu mam avgdzubzxexl Enmvie mu pi ms mii ii imam zn ninA r n n be zip wim m is a 7m 3 uz A w g2mmqnn Basis for heterogeneous cataivsia in zeohtes zeaiiia m quotmin mid We up afso ruining were leluhedn iii iaii mama Farm Mizi airmen i imam mumimeiiiiirMeme ii iiurriiiiriiai WWW mind with in high geiuiimiiziiiiig fwm hzve geiaiimii 2nd i2ng iiimi Mm m mzke in miiie 2i iaui i duxlmi aim Zeohtes shape seiecrive cataiv MWquotriieimiewiipmweiw WW paTiWTi pm ie W rne newquot Dave sue zemee Rm as qnee v veactMlY rm umene rennemn than may Dave sue aa Vgs Cdm aed enEHZY Suvface fm D usan af umene m zeahhe as Zie IereNama catalyst for polymerization of emylen new annex e n nnnm en en enenenen ennwnee neneneee fovmeuemhxvg eKezllhED mnumnmennxwm In n eenmnnne e daub eufnan uvban mm In new 0F nnen ma ezuhvvmqm ene NEWamend lereavequ x x eeeu neenenneneen lhelisn Vlheav enenu v Zmz ev quothe x r m n 2 Inc We ne Z eq eve v avldz uu w eenmnne 0F We afmzuum leuxMaHde dev aved invth mummy afth Zmz ev uu w taxed an m nqu an the nezhzmxmaflheva wiunlmn Zie IereNatna camlyst for polymerization of emylen no nee e nenbennnm ne n2 nee neeeee n n awed end I weekeyneenne blm wen e V n e we sen genenee e m enwenne zlamxn Mn azuhedn eeenew ZieglerrNatba camlyst for polyme39 39 n of emylen m the ma a afth mm 2 nlzuum m n uvmunded an ene Knee by va enwenne zlamxn km the alhevmdebvemvlvwzze m ezvu numum enwenne Khan mnnen e ene afth lnnxman meuk m we mum e ezlvan neH Ad m h E u emvlvavb u enewn 2 en n 775 ZieglerrNatba camlyst for polymerization of ethylen39 n nen m In nu H mm a len MCMLG mm the meme 1 dame ene em emu Emuvn the nnwenene nunmm but kkk we ene mne enwenneennnnepmeen Wexl hzmznemvlvafmu a 2 mean gtan nm nen el ZieglerNatta catalyst for ZieglerNatta catalyst for polymerization of ethylen polymerization of ethylene This process forms the active polymerization ca The the aluminum is coordinated though not cova bonded to the CH2 carbon atom of the ethy group it l which happens to be insoluble unlike the 2 components just donated to the titanium and to one of the chlorine that make up he complex so we h ve what is commonly atoms adjacent to the titanium termed a heterogeneous catalyst also known as a solid Cl There is still a Elmoquot vaca nt site where polymerization ca n occur ZieglerNatta catalyst for ZieglerNatta catalyst for polymerization of ethylen polymerization of ethylen Upon binding ethylene forms bonds with the Ti at The growing polymer chain and the carbon of the ethylene ligand t is im iated CH 3 Hzc Cl Cl Cquot3 I H cu A CH c Al HSCHzc l V H3CH2C c c Cl H2 bTi u gtTi H Cl Cl l c c1 c1 ZieglerNatta catalyst for Enzymatic catalysis polymerization of ethylen Alcohol dehydrogenase Thevacaquott s39te 395 The enzyme alcohol dehydrogenase EC 111115 ava39lable f r the also known as aldehyde reductase This enzyme quotat Ethylene belongs to theoxidoreductase class of enzymes Alcohol molecule to dehydrogenase catalyzes the oxidation of ethane t bmd39 acetaldehyde with the reduction of NAD to NADH The alcohol can be in the arm of rimary secon ary cyclic secondary or a hemiacetal to produce these products aldehyde ketone and N H This reaction occurs during the glycolysis pathway in mitochondria of animal cells Cofactors include zinc or iron that act on p imary and secondary alcohols or hemiacetals chHzc A Enzymatic camlysi Binding site 0 N Alcohol dehydrogenase Obligatory amino f AD acid residu 2m Wm 22 2 Lew 22m 2 WWW e 2mm 2222222222quot my 2 mm Mm 242 WWW 2 mmm thuqh mam a mm 2 en the WW mmdwgmmmmy e Subshate binding Site Key poims he pm We ezxxumvlmn aflunxman 222m Marv m 2222 2mm 2 m Numbnum m Mum z mmm mam 2m 222 MW 2222 m o Auu vxx Wm M bzmevfov 2 mmquot n mm 2quot 2m mmquot mm but due M 2m mm s mama v Pelevaqeneau 22w 5 52m m m 2 222m hzve mm 222 b qu evNauz va wemzlmnuu vil Nwha dehv mqenzxe znenlwe Chemistry 331 Lecture 17 Free Energy Functions NC State University System and surroundings both play in role in the entropy In an isolated system the criterion dS gt 0 indicates that a process is spontaneous In general we must consider dSSys for the system and dSSW for surroundings Since we can think of the entire universe as an isolated system dSmta gt 0 The entropy tends to increase forthe universe as a whole lfwe decompose dSmta into the entropy change for the system and that forthe surroundings we have a criterion for spontaneity forthe system that also requires consideration oft e entropy change in the surroun ings The 39ee ener functions will allow us to eliminate consideration ofthe surroundings and to express a criterion for spontaneity solely in terms of parameters that depend on the system Free Energy at Constant T and V Starting with the First Law At constant temperature and volume we have SW 0 and U 6 cl Recall that dS 2 6qlT so we have dU gTdS which leads to dU TdS g 0 Since T and V are constant we can write this as d U TS lt0 7 The quantity in parentheses is a measure ofthe spontaneity ofthe system that depends on known state functions Definition of Helmholtz Free Energy We de ne a new state function A U TS such that dA g 0 We call A the Helmholtz 39ee energ At constant T and V the Helmholtz free energy will decrease until all possible spontaneous processes have occurred At that point the system will be in equilibrium The condition for equilibrium is dA 0 time Question The statement dA g 0 means A The condition for equilibrium is dA 0 B Processes are not spontaneous if dA lt 0 C dA cannot be greater than 0 D All ofthe above time Question The statement dA g 0 means A The condition for equilibrium is dA 0 B Processes are not spontaneous if dA lt 0 C dA cannot be greater than 0 D All ofthe above time Definition of Helmholtz Free Energy Expressing the change in the Helmholtz 39ee energy we have AA AU TAS for an isothermal change from one state to another The condition for spontaneous change is that AA is less than zero and the condition for equilibrium is that AA 0 We write AA AU TAS g 0 at constant T and V lfAA is greaterthan zero a process is not spontaneous It can occur ifwork is done on the system however The Helmholtz free energy has an important physical interpretation Noting the qEV TAS we have AA A U qmv According to the rst law AU qEV wEV so AA M reversible isothermal A represents the maximum amount of reversible work that can be extracted from the system Question AA wrgv means that the Helmholtz 39ee energy is equal to the maximum amount of reversible work that can be extracted from the system This follows from the fact that A The reversible heat is equal to TAS B A is a state function 0 The reversible work is the maximum work D All ofthe above Question AA wrgv means that the Helmholtz 39ee energy is equal to the maximum amount of reversible work that can be extracted from the system This follows from the fact that A The reversible heat is equal to TAS B A is a state function 0 The reversible work is the maximum work D All ofthe above De nition of Gibbs Free Energy Most reactions occur at constant pressure rather than constantvolume Using the facts that SqEV g TdS and 6wEv PdV we have dU g TdS PdV which can be written dU TdS V g 0 The sign applies to an equilibrium condition and the lt sign means that the process is spontaneous Therefore dU TS PV g 0 at constantT and P We de ne a state function V TS H TS Thus dG g 0 at constant T and P The quantity G is called the Gibb39s free energy In a system at constant T and P the Gibb39s energy will decrease as the result of spontaneous processes until the system reaches equilibrium where dG 0 Comparing Gibbs and Helmholtz The quantity G is called the Gibb39s free energy In a system at constant T and P the Gibb39s energy will decrease as the result ofspontaneous processes until the system reaches equilibrium where dG 0 Comparing the Helmholtz and Gibb39s free energies we see that AVT and GPT are completely analogous except that A is valid at constant V and G is valid at constant P We can see that G A PV which is exactly analogous to U PV the relationship between enthalpy and internal energy For chemical processes we see AG AH TAS g 0 at constant T and P AA AU TAS g 0 at constant T and V Conditions for Spontaneity We will not use the Helmholtz free energy to describe chemical processes It is an important concept in the derivation of the Gibbs energy However from this point we will consider the implications ofthe Gibbs energy for physical and chemical processes There are four possible combinations of the sign of AH and AS in the Gibbs free energy change Description of process gt Endothermic sontaneous for T gt AHAS Exothermic sontaneous for T lt AHAS lt Exothermic sontaneous for all T Question Fora given reaction we have AH gt 0 and AS lt 0 When will the reaction will be spontaneous A never B when T gt AHAS 0 always D when T lt AHAS Question For a given reaction we have AH gt 0 and AS lt 0 When will the reaction will be spontaneous A never B when T gt AHAS C always D when T lt AHAS E Description of process Endothermic sontaneous for T gt AHAS Exothermic sontaneous for T lt AHAs Exothermic sontaneous for all T Gibbs energy for a phase change For a phase transition the two phases are in equilibrium Therefore AG 0 for a phase transition For example for water liquid and vapor are in equilibrium at 37315 K at 1 atm of pressure We can write Ame GHzoggt GHm where we have ex ressed G as a molar 39ee energy From the de nition offree energy we have Ame AvapHm TAvapsn The magnitude ofthe molar enthalpy ofvaporization is 407 kJmol and that ofthe entropy is 1089 JmolK Thus Avian 651d malquotr37315Kl89 Kquot matquotu Question Which statement is true for a phase transition AAG0andAS0 DAS0andAH 0 Question Which statement is true for a phase transition AAG0andAS0 DAS0andAH 0 Gibbs energy for a phase change However ifwe were to calculate the tree energ of vaporization at 36315 K we would nd that it is 11 kJmol so vaporization is not spontaneous at that temperature lfwe consider the free energy ofvaporization at 38315 it is 108 kJmol and so the process is spontaneous AG lt 0 State Function Summary At this point we summarize the state functions that we have developed U internal energy H U PV enthalpy S entropy A U TS Helmholtz free energy G U PV TS H TS Gibbs free energy Please note that we can express each ofthese in a differential f This simply refers to the possible changes in eac function expressed in terms of its dependent variables 2 3 dH dU PdVVdP dAdUTdSSdT dG dH TdS SdT Question We have shown the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables of the enthalpy H TdS VdP entropy and pressure d d d d T VdP temperature and pressure SdT PdV temperature and volume Question We have shown the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables of the enthalpy A dH TdS VdP entropy and pressure B d TdS PdV entropy and volume 0 dH SdT VdP temperature and pressure D dH SdT PdV temperature and volume Question We have shown the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables of the Gibbs free energy A dG TdS VdP entropy and pressure TdS PdV entropy and volu e C dG SdT VdP temperature and pressure D dG SdT PdV temperature and volume Question We have shown the dU TdS PdV This means that the natural variables of internal energy are entropy and volume What are the natural variables ofthe Gibbs 39ee energy A dG TdS VdP entropy and pressure B d TdS PdV entropy and vol e C dG SdT VdP temperature and pressure D dG SdT PdV temperature and volume Chemistry 331 Lecture 15 Second Law Applications NC State University Summary of entropy calculations In the last lecture we derived formula for the calcu the entropy change as a function oftemperature a lation of nd volume changes These are summarized in the table below Sometimes we only have pressure information and the entropy change can be rewritten as fo ows Pitsave 2 i AS annM annP2 Entropy of mixing When two substances can mix there is a spontaneous tendency for this occur We quantify this using the entropy state function lfwe consider two containers separated by a stopcock lt N2 gas is in one and Br2 gas is in the other we know 39om experience that the gases will mix once the stopcock is opened Entropy of mixing For each gas we can describe the mixing as a volume change The N2 gas is originally on the right side contained in volume V at pressure P After opening the stopcock the available volume is 2V and the partial pressure of P x The same is true for Brz lts initial pressure is P2 and nal pressure is P2 xZP We treat the entropy as the sum of two expansions ie an expansion for each gas Amixs Aewsi Aexnsz P P AWS nR ln3 an ln32 AWS nR lnx an lnx2 AWS nRxlnx lenxz Note that the moie traction appiies to the hhai composition Entropy of phase transition The entropy ofphase transition can be calculated using the enthalpy and temperature ofthe transition Tm Tm A muss qF Jus meH Example using the data in the Table calculate the entropy ofvaporization forthe following compounds Solution Use the following relation A a A s W van Rep Entropy of phase transition The entropy of phase transition can be calculated using the enthalpy and temperature ofthe transition AMS qum AMH we we Example using the data in the Table calculate the entropy ofvaporization for the following compound KJmul K JmuHK Note the similarityin ass 2 av 2 the values for the 4 79 6 entropy of ya etizetien arm n as 7 m 5 knownpas 73 2 BE 5 Trouton s rule 208 E77 Conformational entropy The entropy of a polymer or a protein depends on the number ofpossible conformations This concept was realized rst more than 100 years ago by Boltzmann The entropy is proportional to the natural logarithm ofthe number of possible conformations W S R In W For a polymerW MN where M is the number ofpossible conformations per monomer and N is the number ofmonomers For a typical polypeptide chain in the unfolded state M could be a number like 6 where the conformations include different any angles and side chain angles On the other hand when the protein is folded the conformational entropy is reduced to W 1 in the theoretical limit ofa uniquel folded structure Thus we can use statistical considerations to estimate the entropy barrier to protein folding Problem solving We can identify the following main types of problems that involve entropy change Isothermal expansioncom pression reversiblyirreversible n e Phase Transition Statistical or Conformational Entropy Adiabatic trick question ifq 0 then AS 0 never you are solving an entropy problem remember to consider both and surroundings The system is always calculated along a reversible path Isothermal compression Calculate the entropy for an irreversible compression of oxygen gas The initial pressure of the gas is 1 bar in a volume of100 L The nal pressure of the of the gas is 10 bar and the temperature is 400 K Solution Note that you need to obtain the number of mo es The problem does not ask you for a molar entropy Write down the expression for the entropy V2 Pl A8 nR Inv nR In We need either the ratio ofvolumes or the ratio ofpressures h e We are given the pres ures so we can uset os PZIP 10 bar1bar10 Isothermal com pression We obtain the number ofmoles using the ideal gas law 105 Pa01 m3 831 JImoI K400 K 1 bar100 L 00831 L barmol 400 K 300 moles n Now we can substitute into the entropy expression AS nR In 30 mola31 Jmol KIn 10 249 JK Equilibration We have seen a simple example where there are two metal blocks both made of the same material However this need not be the case For any arbitrary materials 1 and 2 in contact we need to know the initial temperatures and heat capacities to calculate the nal temperature q1 72 cm Ta 09272 Ta Now we solve for the equilibrium temperature Tm 91 T1 091 Tea GMT szEa CM 09sz Cpl 092Ta T ea CR1 CD 2 Equilibration Calculating the entropy Once you have obtained the equilibrium temperature the entropy is easily calculated rom T T A8 now In AS2 nCp 2 In In such problems you are assuming that the two objects in thermal contact are a closed system The overall entropy for heat ow should be positive since heat ow from a hotter to a colder body is a spontaneous process Example The coffee cup problem A hlkeruses an alumlnum coffee cup Ifthe mass of the cu IS 4 grams and the amblerlt temperature I e the cups temperature IS 40 ac and the hlkerpours 50 ml of coffee B Calculate the entropy change Gas exchange in a green leaf The cells orthe spohgy layer are irregular h shape aho iooseiypacheo Their rharh ruhctroh seems to pe the temporary storage of sugars aho arhrho acros syhthesrzeo intne pahsaoe layer They also are rh petweeh the ear am the ehyrrohrheht During the day these cells grye offoxygen aho wateryaporto the air spacesthat surrouho them They also prch up carpoh dioxidefrorntne air spaces The air spaces are rhtercohhecteo am open to the outsroe through pores caiieo storhata singular stoma Entropy O2 mixing in a Duhhg photosynthesis 02 pro u eorh tnyl rherhprah orgreehiea es gas e tstheatrhosphere r the sto r Assuming thattherohow mole rractrohs exist calculate the molar ehtropy or rhrwhg ihsroe xuz m 5 am M m n 5 Outsi e 32M n a o 2 us Assume thatthe storhata closeswrth ah eouaiyoiurhe or outsroe air aho rhsroe gas reSentln ah ehcioseo space Solution Asstrrhe thatthe rhrtrai state is premixed gas hath rhsroe ahtt outsme all oxygen cohrpouhtt l and hrtrogeh corhpouh 2 The hhaicorhposruoh is x1x r wz tn 5 n M n 35 Xe x2 s x hz u 5 n M n 55 Entropy O2 mixing in a the stomata of a leaf outsroe Thereis ah ehtropy or rhrwhg oreac o t etwo reactahts that must he suptracteo from the ehtropy of the that mix Notethatthere rs hairas much or each at the ieactant gases as the product Asntal Asher ASMM Ashram awn x xnln xquot L2RX JH x Xazlrl x Tl xnlh x X ln xquot shce the propierh asks rorthe rhoiar ehtropywe reaiiyheett Asmwgh so we can whte the rorrhtria as Entropy O2 mixing in a Whte oowh the total ehtropy chahge as prooucts final mixed h e W6 r eactahts that must he subti actett from the ehtropy of the that mix Notetnaunere is hairas rhuch oreach or the r actarrtgases as the product ASH a a ah gar grquot X 8 st umorwo 05 042T JmUPK This is a small ehtropy Part orthe reasoh rsthat the gases were aireaoy mixed it they hao been completely unmixed the ehtropy wouio haye peeh 5 4 lrnoHK Phase Transition Compare the ehtropy or suphrhatroh or water to the ehtropy oryapohzatroh at n to Which is larger Why Solution The ehtropy ora phase transition rs gryeh py om ASV m From the throrrhatroh h Athrhs page Hi we me that quotDH 45 M KJrnol and WH 51 no kJrnol Using these yaiues ahtt the temperature or 273 x we hhtt AS W 165 t fWUFK 51080 Jmnl Awa e W 71871 Jmnl x Conformational entropy of a protein Estimate the conformational entropy of myoglobin Myoglobin has 150 residues with 6 possible Solution The oenrhtroh s R anis xhowh as the Statistical ehtropy R is the gas constant aho W is the hurhher of possrhie cohror rhatrohs fol a structure and W N M For the uhroioeo pioteln W ate and s lSEIP ln For the roioett plotehi W i one s I There cohrormatrohai ehtropy is Aws a AWWS aho rs thererore 2233 Jrnol or 2 kJrnol The Levinthal Paradox Th L 39 th I d 3939 e evm a para ox A assumes that all of the possible conformations will be sampled with equal probability until the proper one N native is found Thus the funnel N Conformational states that if a protein samples Entropy all 6M conformations it will take a time longer than the age of the universe to nd the native fold N for a polypeptide where M 100 and if it takes 1tii seconds to sample each possible conformation The Pathway Model Imagine that the a unique pathway winds through T the surface to the hole gt The path starts at A and E the folding goes through E a unique and wellde ned set of conformational changes Here the entropy must decrease rapidly since N the number of degrees of Conformat39onal gt freedom in the folding pathway is quite Entropy small compared to GM On this diagram the configurational entropy is given by the width of the funnel and the relative by the height relative to the bottom folded state The vertical axis is energy NOT free energy Evidence for folding pathways One piece of evidence for folding pathways comes from trapping disul de intermediates This method was pioneered by Creighton using BPTI a as only been used on other p 39oteins 30515714 30751333 H 4 Ej E ll msmma I M Reduced 7 W333 ss esswaa 5 Native Creighton etal Prog Biophys Mol Biol 33 231 1978 Beyond pathways Kim et al showed that some ofthe previous data and interpretations M n r were wrong The major 2DS species contains the two native DS39s 3051 and 555 The third disul de is formed quite slowly because it is quite buried It is possible to isolate a stable species with only the first two disul des formed and the third remaining in the reduced form Studies with a mutant in which the third DS was replaced by 2 Ala and which folded at a similar rate to the wild type support the idea that the trapped disul de species have partial nativelike structure These observations and others like it can be used to the idea ofa pathway into question Kim 1993 Nature 365 185 The Folding Funnel The folding funnel shown here represents the change gt in energy for a large number of folding paths that lead to the native configuration There are n nergy barriers This implies that all paths have N an equal probability leading COHfOFmatlonal gt to the folded state The funnel py shown here has no energy barriers and all paths lead directly to the native state Thus this funnel is consistent with two state folding behavior Energ D111 and chan Nature Strucl Biol 1997 4 10719 Barriers and misfolding The energy surface does not have to be a smooth trajectory There can be barriers that will trap intermediate states These an intermediate is observed C f t I there will be a question as to on orma Iona whether this is part of a pathway Entropy gt or whether a funnel description is more applicable Note that the funnel provides the possibility for misfolding This will typically result in multiple minima in the energy landscape D111 and Chan Nature Strucl B101 19974 10719 Chemistry 331 Lecture 39 Vibrational Spectroscopy NC State University 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 5H ca W my where p 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 0 Mu 3 me The total wave function The total wave function can be factored into an electronic a vibrational and a rotational wave function P WeXVYJM M I x if mm mm I amdoI I nmmymsinededw I I uvmulxyvmsinedeol Interaction with radiation An oscillating electromagnetic field enters as Eocosmt such that the angular frequency hm is equal to a vibrational energy level difference and the transition moment is 2n 7 MM DID In l l McosIQIlyMsinIQIdedaj gxwff Interaction with radiation The choice of cose means that we consider zpolarized microwave light In general we could consider x or ypolarized as well X Sime 00 um W w 1126 ysinesin 4 e e 9k 2 008 um sm costbz sm smtbj cos ismxi 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 ire leb Vibrational transitions As an example we can calculate the transition moment between the state v 0 and v 1 KB lAeim221x1lAeriuQ22 Mm Ba 5er e mzleze WZZdQ all a Ii all 1 my m2am m 21 Vibrational transitions Note that this result is a statement of the vibrational selection rule V thin 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 Chemistry 331 Lecture 9 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 n change directio Temperature derives from molecular motion 32RT 12Mltu2gt M is molar mass Greater average velocity results in a higher temperature u is the velocity Mass and molar mass We can multiply the equation g lt 2gt 2 RT 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 gnRT Nm ltu2gt Kinetic Model of Gases Assumptions 1 A gas consists ofmolecules that move randomly 2 The size ofthe molecules is negligible 3 There are no interactions between the gas molecules Because there are such large numbers ofgas 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 ltu2gtc ssz53sN We use 5 for speed 7 s szs3sN andcformean speed N Velocity and Speed When we considered the derivation ofpressure 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 energy Anotherway to say this is the energy is a scalar 1 2 7 1 21 2 lt gt E 2 quot M 2m 2quot All ofthese notations pmum mean the same thing The rootmeansquare speed The ideal gas equation of state is consistent with an interpretation of temperature as proportional to the kinetic energy of a gas 1 2 3Mu RT Ifwe solve for ltu2gt we have the meansquare speed 2 7 3RT u 7 r M Ifwe take the square root of both sides we have the rms spee 2 12 7 3RT u a v 7 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 Hz The mean speed is i 2 2 376 The rms speed of oxygen at 25 ElC 298 K is 482 ms Note M is converted to kgmol Um 38 31 J 1 K298 K 0 032 legmoi 4818ms The Maxwell Distribution Not all molecules have the same speed Maxwell assumed that the distribution ofspeeds was Gaussian M 32 2 isz Ft 47r2nRT sexp RT As temperature increases the rms speed increases and the width ofthe distribution increases Moreover the fucntions is a normalized distribution This just means that the integral overthe distribution function is equal to 1 I 5 1 See the MAPLE u worksheets for examples Diffusion and Effusion Diffusion process by which substances mix with one another Ef Jsion escape ofa gas through a small hole Graham s law of ef Jsion 1 Ra te of effusron oc W Can be related to the rms speed ofa gas given by the kinetic theory of gases Molecular Collisions The mean 39ee path A is the average distance that a molecule travels between collisions The collision 39equency z is the average rate ofcollisions made by one mo 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 RT JEN46v WP RT 1 o 1 7w 2 6 rrdZ ENAUP 1 In other words Units of Pressure Force has units of Newtons F ma Kg msz Pressure has units of Nevvtonsmeter2 P FA kg ms2m2 KgsZm These units are also called Pascals Pa 1 bar 105 Pa 105 Nmz 1atm 101325x 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 lts magnitude depends on temperature R 831 JmolK or 198 calmolK At 298 K RT 2476 Jmol 2476 kJmol Thermal energy can also be expressed on a per molecule basis The molecular equivalent of R is the Boltzmann constant k R NAk NA 6022 x 1023 moleculesmoi Extensive and Intensive Variables Extensive variables are proportional to the size of the system Intensive variables do not depend on the size of the system Extensive variables volume mass energy Intensive variables pressure temperature density Microsopic view of momentum C X f 0 b area bc a A particle with velocity uX strikes a wall The momentum of the particle changes from muX to muX The momentum change is Ap 2mux 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 cf 9 9 EIH Macroscopic Microscopic Transit time C The time between collision is At 2aux Round trip ux distance is 2a b Transit time C area bcl a The time between collision is At 2aux velocity distancetime time distancevelocity The pressure on the wall force rate of change of momentum F 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 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 Nm u gt Pe jr PV Nm u gt Average properties There are three dimensions so the velocity along the xdirection is 13 the total 1 3ltuzgt PV Nm uz From the kinetic theory of gases l 2 1 2Nmltugt 2nRT 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 applies to a monatomic gas ike neon or argon What about internal motions of molecules RT is a natural energy scale We can rewrite the ideal gas law in terms ofthe molar volume n The ideal gas law has the form PV RT The molar volume at standard T and P Microscopic variables Monatomic gases translation Pressure and temperature can be described solely in terms of the ballistic motion of the gas Diatomic gases translation vibration rotation 54 0 Center ofmass Quantized energy levels The constant h known as Planck s constant gives the scale for quantized energy levels h 6626 x1034 J Translation particle in a box Vibration harmonic oscillator Rotation rigid rotator The energy levels for each ofthese is obtained by solution ofthe Schrodinger equation The energy level spacing The constant h known as Planck s constant gives the scale for quantized energy levels h 6626 x1034 J We will see how to obtain these in the second half of the course Levels are thermally populated 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 uid or a liquid such as water we can think ofthe pressure arising from the weight of the column of uid above the point where the measurement is made The force is due to the mass ofwater m kg accelerated by gravity g 98 msz Lmm9hms P A A Ah v pgh where p is the densityp mN The dependence of atomspheric pressure on altitude We can think of the atmosphere is a uid but it is not dense Moreover unlike water the density ofthe atmosphere decreases with altitu e 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 de ne the density ofan ideal gas The dependence of atomspheric pressure on altitude The density of an ideal gas is p mN nMN MPRT The dependence of pressure on elevation is dP pg dh dh We need to collect variables of integration on the same side ofthe equation dP Mg quotWquot The barometric pressure formula Then we integrate assuming PU1 at h0 P Puexp or P exp atm lsotherms We can plot the pressure as a function of the volume as shown below Each ofthe curves on the plot has a constant temperature Partial pressure For any gas in a mixture of gases the partial pressure is de ned as P XJP where X is the mole 39action of component and P is the total pressure The mole fraction is de ned as Chemistry 331 Lecture 40 Electronic Spectroscopy NC State University Biomolecular Spectroscopy uantum mechanics Electronic states and energies Transitions between states Absorption and emission Biopolymers Postulates of quantum mechanics are assumptions found to be consistent with observation The first postulate states that the state of a system can be represented by a wavefunction KPql q2 qgn The q are coordinates of the particles in the system and t is time The wavefunction can also be time independent or stationary Vq1 q2 qgn Postulate 2 The probability of nding a palticle in a region of space is given by Pa I Td E Postulate 2 Assumptions 1 KP is real T is Hermitian 2 The wavefunction is normalized 3 We integrate over all relevant space Normalization is needed so that probabilities are meaningful Normalization means that the integral of the square of the wavefunction probability density over all space is equal to one I V Pdt 1 all We The significance of this equation is that the probability of finding the particle somewhere in the universe is one Postulate 3 Every physical observable is associated with a linear Hermitian operator Observables are energy momentum position dipole moment etc operatorp gt observable P The fact that the operator is Hermitian ensures that the observable will be real Postulate 4 The average value of a physical property can be calculated by I PT Td E P Normalization Postulate 4 The calculation of a physical observable can be written as an eigenvalue equa on Ps1qu This is an operator equation that returns the same wavefunction multiplied by the constant P P is an eigenvalue An eigenvalueis a number Postulate 5 The form of the operators is Position 51 q Momentum 13 iii1 661 Time f I 5 Energy H 1725 The Hamiltonian and wavefunction are timeindependent Stationary State Wave Equation Quantum Mechanical Description Hamiltonian Eigenvalue EnergyOperator Energyvalue V V HO P E0 Wavefunction The interaction of electromagnetic radiation with a transition moment The electromagnetic wave has an angular momentum of 1 Therefore an atom or molecule must have a change of 1 in its orbital angular momentum to conserve this quantity This can be seen for hydrogen atom Electric vector of radiation gt 1 o 1 1 V Electronic States and Energies for Molecules The wavefunction is The wave equation can be separated composed of electronic and into electronic and nuclear parts nuclear parts Hamiltonian Eigenvalue LI Energy Operator Energy value l l electronicx nuclear I V Helec ll Eelecw Total Electronic NUCICZI HmiX E nuch The wavefunction represents the probability amplitude of electrons and nuclei Wavefu notions The molecular orbitals in a diatomic molecule are formed from linear combinations of atomic orbitals 00 00090 The electronic part AntiBondingm Constructive overlap between two Destructive overlap between two atomic orbitals gives rise to a atomic orbitals gives rise to an bonding state antibonding state No NODES UNKNODE o o o O 0 00 k1 k1 gt 1N2szsz 1 1 S S gt Maggie Fur mammm hydmgen We answer Absorpuon ofvwswme or u trawo et runean Energy magrarn transmons lt M mun 5 s O n 5t occupwed mo ecu arorbwta of ethy ene e owest unoccupwed mo ecu arorbwta of ethy ene 1 n uni orbwta s of benzene usmg the w e ectrons m p orbwta s L seems m A sen pm Benzene Stm uve aemane EHEVWLEve s corresponds to eacn energy eve quotmiss 2 1 ms of the transmon mo The Uansmun mumemceltwdqh zgt can he Separauon of eTearomc and nudear pa mem ha depends Em e amhe nudeavwaver mc mn 1mm dues hm CWT WWZFWHX 2v These emenhe Tate EXpYESSmn as We squave 92ltmeW2zltXthvz Much The eTectromc transmon m oment Me re vJHWf LTgM M be ahmvhed menthe eTemmve m e ehehee thhehansman hehem LTgM Wu hm he ahsmhed Wnen the eTeehe ve m e Pemenmm avmthetvansman hehem Transmon moment of TOWeSt 1k1t transmon of ethy ene Thehahsmeh memem Tspeypehmemanmhe change h heeaT shuduve ETememaghehe pu anzada ungWSdHEE mn 4 p Wmehe 0T2 mahmumhahsueh pmhahhw The eTectromc transmon moment determmes the Tmensm of an absorpuve transmon The 7 and 7x quot sums ave w ew we and we ew foe The uahsmeh NDNEM TS MM I Pft l wdr The magmtude of the transmon moment for H can be caTcuTated from a STmpTe modeT quotI h Wm my 1 wwer max Wnevex rszsthecdi enhmehe h DH 35 A One marge mm acedthmugm A has a d pa e hehemmeo M1326DmszHe The nuclear part uu tuu wtth exctted state v 0 1 etc mm thhmmunl wtth exctted state v 70 1 etc A A mm thhmmml MW mum may rat v mm Dtivhwmm l uu e v0 etc t to the overtap at ground stat e V wtth exctted 5 at 7 construct a suck spectrum m mum wavemmev ym cm 39t Ahsowlton Estimated assmtmg Emu anuu cm and wbvmmnat madeaMDDDcm t 1 2VEDBSBEM t Nudwir Dtinhwmml The Franchcmuuh racmruetehmhes a er VTbratTonaT reTaXann i 2 e Numwmm The FrahewCOthh phheTpTe The FrahewCOthh factons Transmons are vemeah the same for absorbahee and absorphon and ermsmon u 1 Seva 1 sumha 2 WWW 2 WWW g WW 5 WW 5 FTuavesuence S 3 FTuavesuence mew DTivTawmml mew DTsszwmml The Teads to a mhor Tmage reTannsmp between absorphon and uorescence bands Absmpnan Flumescznce lt mxy Dr T T T r T T 3r a wviuam r BTopoTymere The carbony group has n 7 1H and 7H transwtwons m mm mm Determmat on of the dwrectwon ofthe transmon moment m the Peptwde absorbence mteracuuns betweenme nagnnunng amme gmups absurptmn spectrum Exmtun mteracnun rsheet whehx 7 4 Excwton mteract ons Fuvexamp e we cunsmym newghbuvmg awe gmups o 1 12 and o 191239 we emmmcuansmuns Excuon energwes and Wavefuncuons mg exman v avemndmns Buss m hneav mmhmatmns a he exmed S1512 v avemnmans 0 w my V E er V 701 and me armde Uansmuns ave mdependem mm was use m Mu emmmcbanus Excwton tranawtwon momenta We aMmm hneav cumbmauuns mum uansmun mumems w m w Wm T Smue these add as vectmsthe dwedmn m u sneav va hgana tamm aw Yheexam es mm have mm mamems an be extended mamem bmthe hasquot punmme sthe same Aromat c ammo adds The ammu amds pheny a amne tyvusme andtvypmphan have urn tvansnmns Nmethe Panen mv eak bands m 2m 7 sun nm by whmmc muphng L wvaum ow Tryptophan mm was ahsmm m hamsmwmphan asmmm W and m We We mg Apamd n a a de ma an my spews mmm we WVLM 2 u s Equot Yhev eak hand hasa avge mm anyway mamemumchange Yhestmng in mm DNA base absorptwon spectra mg manam e edmmctmnsmns MA 9 c tvanshansthm are mm mum M Smg e mam pmanzed ahsamtmn speamsmpv s usedm determmemetvansman mamem dwedmn mm gea hesexsxmpanam snce he e e m ve m a he hm must he augneuwmmeuansmanmamemmmamze an absmmmn s gna rw Mum mum 39 gt Applications and mechanism of DNA hypochromism DNA hypochromism is useful for determining the hybridizationmelting of DNA While the origin of the effect is at least partly excitonic base stacking also contributes by creating a hydrophobic environment for the bases Since water is excluded the dielectric environment is quite different in DNA and this may have an effect on the absorption spectrum as well Chermstry 331 Lemme A WMaHuna Mum NC Stale Mnletlllarlnnllnll mm pmmpvam m mmm an m enequ em mm We mm mamquot Delenmulmn ammugwemk u m m abummq ma ezuhvvizmaer ma meme am mhm Mm am quotam mne mamaan c m ne a mmWmmmWth f W 6V WW E n m m ksume mvms hwghenhan quadvauc ave 22m Eyde nmun gagzkmfommmm memudelr New My Q 1 v 91 n g VQ mmunaemnmamnmmnmm remnrmginmethatSpmpnmnnaunnsmxmac m ahmecnnaan we max12 We damca hamnmc a inrm m mm m mm m an m m mmmm a 7 F ma m 5mm m m a m m mum an am he mm a 7 d m n 27mm x mm Yhepmema w mmwmsa parabnhcmncnnn mxpmzmahxcaueda hamnmcpmzrmm memo mnaamkhawnn mmnwmmx m Jnu eme lym l Weangu avmquencvm 2mwmmnuencwm smuan s Gausswan Enevgy s quarmzed n a w k 1 5 m ag rw W as v a Evhv 7 v s he mm numbev memudmersLAMHN Mamuansmm Q v avot a w vmm nmuamummechamc GWWIW uweschmdmgevequannm d v x 13w 7mm 55v M w 9 wnmemwmmnmmmmmm a mm V M Wennrmahmlmnmmzmx amnmmmm n 1 5mmmmmaeammummm 1 2y Nr Hemnepn anawvand annmahmmncnnaamm 2 M 2 3 81 121 lillralinnal wamunninn I Inquot unimgneruv a Energy eva s mm by Am i Wmmmms ave w WM where w 5 m Hermne pdynurmz a mom ensges are m We umev m 0 r 3000 cm me umes1 max 5 En mm mmmnzxwmsmzemmmgy c a me mm mm wavmmmm s M2131 m pusmunz uncenzny m an aim is We 22 am 121 pmm mergy Pnlvalnmil MIIIEEIIIES a mm are swam dagvaes mwemum m2 mmecme mat curlzms N gum is mm are myeemnsmmz degaes m quot22mm mesa cumspumtummun mm mmquot mass mm mdacm thee r23m a me remzmmgdegrezs m39reaium are mbvamnz a Manual llllllllIWallf A symme mstmch Amman am mm mam masocm Yheve ave 3 nmma Nudes m4 7 5 AM mm ave nvmved saws smce au mm mm mamem changemthewmmmn Yhehamamcappm mmmn can he apphed a each nmma made NIIrIIIaIInIIIIEIIHIz 40 0 u O oo i aO u Wmemc sum Asymemcs mch q a mum WW m WWW Wm WM mm m Yheve mum madeS Nr mum ave nvmved saws smce au mm mm mamem mange m mew mmmn Villrallnllal39lrallsillnll l39llra 39nnalsele 39 ll llle Allllarlnnllil llnlenlial 7 mmnm Enemy cm Memuc ear DMEHCE w Ernalmuwn nillm harmnnil allllnl unalinn Expenmema cunsequences 1 Fvequencywsmmpevamve dependem Vuvasmg emude 2 Ovenunes ave auuweu 3 Mbmwuna mum ave cuup ed A Fevmnesunance a nghhequencymude equency dependsunma ecu anmevactmns Fermi remnant Smmmg mm mm anhamamcwuphng Sandman Fvequenwmn r u 1 Symmewn 1 r1 Immune niwam 3R3 m m mm m w m mums mm Msszsm mm m r Ham mm mm mm quotmum mmmsm mmquot m Si mr vv nha m mmcauvhn 7 mm mam WV avmanes mum an l m wens mu m maker and mmm ca ar n mm mm m Frequent Illlllll mnlnnularimnraninn Hydvugen bundmg uwevs 07H me cumam and HVOVH bendmgmvce cumam vapure Mama 2 v 3825 3557 v2 1554 1595 5393593756 Chemistry 331 Lecture 7 Absorption in the atmosphere The electronic structure of molecules NC State University Absorption by gases in the atmosphere BLACK BODY l0 CURVES 5500 K 266 K Electronic Vibrational Rotational The electronic absorption of oxygen wk 118nm The major electronic absorption of ozone 26 F39I39E 201 nm 20 non Molecular Orbital Theory ln MO theory electrons are treated as including the entire molecule Each MO is built up from a linear combination of atomic orbitals LCAO I 39PMO cpl where p are atomrcorbrtas The coefficients are optimized by the self consistent field SCF method The variational principle justifies minimization of the energy by adjustment of the coefficientsci The molecular orbitals in a diatomic molecule are formed from linear combinations of atomic orbitals Bonding o S S AntiBondingor alarm urmaxs gwes nse m a bundmg state quotmm 00 gt 42mm alarm urmaxs gwes nsEtu an ammunan state 00 15 gt mm 7 w Fur mammm hydrugen We cunsmer me sand 1 mmemar umaxs m me ruuuwmg Energy magram 0 9 0 MO rea mem 01 H2 2 2 Wewme mmnxmmhaw mum wmxuand u Wm We am 5mmquot mm mm mm ymwnwz7 mw mmu WM smmwmm Nmogen Mo ecu ar Orb ms Exarrwe u 2 Hurmnuc ezv Dwztumc in susmaces vepvesentw u uvbna shuwmg am uHuta pmbabuny The spam Wavemm mn s an LOAD Cave 2 mans ave mm mm Dmmogen 16 M0 2d Them ave N2 ammunva each N am m x adnuh vdegenem arm omym m m m hnm m x adnuh vdegenem arm omym m m m hnm Energy eve dwagram or N2 Negatwe enemas On vva ence mama s ave shavm Molecular Properties I Bond length structure I Vibrational frequency Calculated at stationary point Depends on accuracy of second derivative matrix with respect to nuclear displacement I Dipole moment clearly zero for N2 I Absorption spectrum Electronic transitions I lsosurfaces representKI of orbital showing 90 of total probability I The spatial wavefunction is an LCAO I Core electrons are not included I There are five electrons for each N atom The wavefunction is composed of electronic and nuclear parts LI l l electronicx nuclear 4 4 4 Total Electronic Nuclear The wavefunction represents the probability amplitude of electrons and nuclei The wave equation can be separated into electronic and nuclear parts Hamiltonian Eigenvalue Energy OPerat0r Enfrgy value I V Helec ll E elec ll A H mch E rurch Wavefu nctions The electronic part nm For diatomic hydrogen we consider the O39 and 6 molecular orbitals in the following energy diagram Absorpth ofvtstble orultxavlolet radlatlon leads to eleetxonte txanstttons munll the ehshte h nadal Wm atsa thettes e chanve h amnal anvular mamemum The hlghest occupled molecular orbltzl of ethylene t dquot The lowest unoccupled molecular orbltzl of ethylene l ttnntt Separatlon of electrontc and nuclear parts of the transttton moment Thettahstttnh NDNEM eeltwtlnlw2gt can he sepavaled thtethe eleettnhte Wavetuncllun tn e e n she the 2a waver e huel t mm wthlwzattht These ehtetthe tate Expvesslun as e squave 62wthlw2 atttatzt2 Mtz2FC The electronlc tmnsttton moment the eteenmetnhsttm hehett ts M 76mm tteht WH e absmbed When the eteetteveeet ts sttehee vwlh the ttshsttm hehett tteht WH nm he absmbed When the eteette Vaclav ts hemehetmtstte the tnhsttm hehett Transltlon moment of lowest 171 transltlon of ethylene Thettahstttnh moment ts petpeheteulattnthe ehahnethhnealsttuetute Etedmmagnetlc pnlattzee alnhn thts etteettnh wtll gNE the maxlmumtvansmun pmbablhty for and transttton ustng a strnpte rnodet The 7 and 7K quot gates ave w t Q and wr m as We transrtren rnernent rs wrer rnornenttor cm can be catcutated from a stmpte modet 7 mt thwr Wde Umrxwrsj mdx x Wnevex wstne cc hand tenglh mt 35 A one unaree ursptaeeutnmuent Anas a urpute mamemaNED M32 DMCIH The nuclear part lm ancKCondon ractor ts due to tne overtap or ground state v0 wttn exctted statev 0 etc Emma state 070 Gmund state Nudair thrammu l r due to tne overtap or ground state v0 thh exctted Stae V ZU 1 etc Exctted state 071 Ensng Ground state Nudsir ornament r due to tne overtap or ground state v0 wttn exctted statev 0 1 etc Exctted state 072 Ground state Nudsir statement Tne FranckeCondon factor rs due to tne overtao of ground state v0 WW1 exctted Stae V ZU 1 etc Emma state 073 Gmund state utueer Dtivhmmml Tne Franckecondon factor rs due to tne overtao or ground state v0 WW1 exctted Stae V ZU 1 etc Emma state 04 Gmund state Nuahir D iv ammml Based on tne FC factors We can construct a suck pectrurn can Absorvuon wavenumhe x m ceroueteu eseunrne Ewen e Euuu W W Mbrslmna mademm m t 12Ve DBSBEM tne envetop ortne absorptan Hneshape uorescence usuaHy occurs artervroratronat retaxatron 1 Ahsmmmn 2 Abrslmna rerexetran 3 rruareseenee Nudeir statement Tne FranckeConoon onncrote Transrtrons are vertrcat m ootn aosorotron and ernrssron 1 Ahsmmmn 2 Abrslmna rerexetran 3 rruareseenee Energy Nudeir statement Tne FranckyCondon factorrs tne Sarne for a Sorbance and uorescence 1 Ahsmmmn 2 Abrslmna ve amhan 3 ames22mm Nudsir umpmwm Tms eadsto a rmrror wage re auonamp between absorpuon and uorescence bands Ahsmpnun numescence an M m LEI zen Biopolymers l The carbony group has n 7 1H andnenwansmons m m mam Deterrn nat on of ne drrect on of ne transm n rnornent n tne o ecL e reference frarne Smg e mam Pa anzed ahswmmn spemasmpv rs used m determmethetvanslmn mamem dwedmn m s Wpanam s ncet a he hem mus he MK 3 gm m an ahsarmmn sqna Pepude absorbence mteracuu new Excwton mteract ons Fuvexamp e WW2 cunswdenwu newghbuvmg armde gmups The sma y waned W m and Hsthesewavemnmmnsmahmevac u N2 exmtumcuansmuns Excuon emerges andwavefuncuons We exumnv avemndmns ansewam hneav camhmatmns a he exmed stmev avemnmans my w m a vandum emu pepudetheve e We uvdev andthe armde quotansmens ave mdependem amwdesgwes use mmu emmmc bands Excxton transmon moments We can mm hneav enmbmauuns uHhe uansnmn mumems w w 2 Smue these see as vedmsthe dweman m e neav vanhgana m mm m We exemnee mm here mm mamemscan be extended Aromat c ammo adds The ammu amds phenyxaxamne ymsme andtvy tuphanhae H uansmens me the Pattern m weak N bandswam 2m 7 sun nm We weak hands ave euwee bv whmmc mupune L Trypto pH an Mew Wee eeemn We meme me a Me warm and em We me my A Famee an e e ee me e mamemum change We Vang hand mm 1 DNA base absorptwon spectra mg mpanam e e mmctvanstmns m A e c snm hee e n ma hehgmmusthe shyned m hetvanstmn mamemm mamas an absarmmn sqna DNA base absorptwon spectra L 2 Elm M39 DNA hypocmormsm quotA Apphcatwona and mechamsm of DNA hypocmormsm WWW s useM my delevmnmg mnmemng mDNA Wh ethe an enact anthe ahsmmmn spectrum asva Chemistry 331 Lecture 24 Applications to Myoglobin and Hemoglobin NC State Universrty Ligand binding in myoglobin Myuglubin isine pieiein in museieinai is iespensibieiei uxygen binding and sieiage it is eeinbesed era whEllEES in be glubular siideidie shuwn beluw The iron in heme is the binding site for oxygen and peroxide Heme is irol i protoporphyrih IX 0 Fuhcbohal aspecB in Nb 1 Discnrninauon against co binding 2 oZ is the physiologically relevant ligand but it can oxidize iron autooxidatioh Ligand binding in myoglobin Ligand binding lei belween a rid myugiubin is desenbed as a ebeinieai egdiiibiidin bed siai nd a sulvenl slaie Mb e MbCO a disseeiaied s1aie Mb 00 a co MbCO 4 MbCO aMbco iiwe igneieine iniennediaie slaieine binding can be desenbed by An uverall egdiiibiidin pvucess Mb 00 4v MbCO Wiin eguiiibiiunn eensiani K We have K 7 MbCO 7 MbHCO Thelractiun buund is f Mbcoi llb MDCO Ligand binding curve Tneiiaeiien beund can be ieiaied lei be binding cuns1anl K E g KllbCO 7 Kim W 7 W 7 W Thistype eibinding curve is platted beluw quotI an as d4 02 d iaiszddsm Ligand binding in terms of partial pressure We aie mien inieies1ed in measuiing bindingie myugiubin in be piesenee era given paiiiai piessuie era gas Tu ieiaieine eeneeniiaiien urine gas in sulutiun lei ine piessuie we use Heniy s law Poo NCO Tne binding in seidiien is guvemed byine inieiaeiien brine iigandWiin iien im an MbCO The peptide backbone is shown as a ribbon that follows the ahelical structure of myoglobin The structure shown is at equilibrium Conformational substates are called A states m s Muml Mas am mm mm CO The photoproduct Iron moves out of the heme p ne when C0 is photolyzed C0 moves to a docking site and is parallel to me plane Conformational substates are called B states The deoxy structure has no C0 ligand The protein backbone has shifted to permit water to enter the distal pocket This form is often referred to an S state Deoxy Mb Deoxy Mb MbCO MbCO here We will compare UVvis and infrared Tm ier Momf sznSlncl m we 17m The Perimeter Model The aromatic hhg has 18 electrons The 7 system approximates circular elec tron Dam Q Q The four orbital model is used to represent the highest occupied and lowest unoccupied MOs of porphyrins The two highest occupied A orbitals aWaZU are nearly equal in energy The eg orbitals are equal in energy Ml Transitions occur from an eg an a2h gt eg a t a The transitions from ground state 7 orbitals am and 32 to excited state nquot orbitals eg can mix by con guration interaction Two electronic transitions 5 n l s are observed One Is very strong B or Soret and the other is wea 1 The transition moments are MEM1M2 39 MQM139M2 aha 4 1 41 a Absorption spectra for MbCO and deoxy Mb Surat Band 1 Band n 150 B MbCO 39 3 too Deoxy Mb 7 50 Q s 480 I 5so GED Wavelength rim The spectrum ufthe herhe has Band is alluvved and th it is ubserved because u bands The E band urSuret erefure intense The a band isf rbidden vaibruni eeupiihg With the Suret band The ligation of CO changes the spin state of the heme iron 2 lt9 Low spin FeII High spin FeII The heme iron center moves out ofthe heme plane and the porphyrin macrocycle domes upon deligation ofCO C0 is PhOtOIYZEd Fe displacement Planar Dom Heme Heme Low spin FeII High spin FeII Metal bonding to CO lowers the IR frequency and increases the intensity O V Ziziaern1 Weak Nigm Em strung Free CD Eluurid CO Effect of metal bonding on C0 increases the dipole moment 0 DO pnN015 tp gt015 do 90 nrbackbundlng 3 Fe E ufid CO Free CD can in ahdoti 2th respectively calculate the g urid state dl ule rnurnerit a N122 er co The co band length isi 27A uDZEZd um eE 2mm 27 A 254 eA Given thatthe partiai charges eh c and o are iZZD Dipole calculation Fe E ufid CO Dipole calculation 000 68 Free CO 015 Given that the partial charges on C and O are C 00255 and O 00255 respectively calculate the ground state dipole moment of CO The CO bond length Is 1 23 A Dipole calculation 0 CDC c6 0015 Free CO Given that the partial charges on C and O are C 00255 and O 00255 respectively calculate the ground state dipole moment of CO The CO bond length ls 123 A p0 ezd p0 e00255123 A 0031 eA 0150 D Difference dipole gives rise to infrared intensity CD00 653 Free CO 015 The difference dipole moment is 0H 6Q As the the bond length changes the dipole moment also changes This occurs at the frequency of vibration From physics we knowthat the vibrational frequency is L mim2 V 211 H p39 m1m2 Interaction of the metal has two effects The iron atom injects electron density into an antibonding orbital of CO This causes the CO bond length to increase The infrared intensity increases because the charge density changes more as the C0 bond length changes It weakens the bond so k decreases and the frequency decreases i V 211 H p39 m1m2 mimZ 0 I 0 85 Bound CO bmubanx i xia Comparison of A states and B states in myoglobin at low temperature i i i 1900i 194 1950 1960 1970 Wavenunmer cum 393 A state CO bands Ahsarlmme mo i ii ii 2120 2 30 2140 i Wavenmnber cm y B state CO bands The origin of the A states is the hydrogen bonding conformations to CO 1915 is 1932 194 1951 In 1914 1900 1920 um I960 mm mm Frequency cm l sw vem sw 12 sw wildWW 5w HMQ Pig V li l39 sw usv sw HMV Pig H ovVm Infrared spectra of myoglobin Wild type Mt IR SpeCtra Show multiple bands Distal gt Y m0 Distal sw V6839N sw vom SW 12 SW 12 sw mid1w sw mid1w sw HMQ 5w no Pig vwr Pig vam sw F46V sw F46V sw H64V sw H64V P13 ewVsquot 91 ewVsquot 5 7 i 1 Proximal Proximal 1900 1920 1940 19m 1930 2000 1900 1920 1940 19m 1930 2000 Quillin Phillips et al WW 1m Protein Data Bank 2MGK Frequcmy cm1 3MB 19931341140155 Hydrogen bonding affects the Mutants at V68 position show multiple frequency of CO bands depending on hydrogen bonding L29 mo m0 Distal 11 gtkV68 O sw wad1w m g sw wad1w swnm C swnm PigV B l39 m lt PigV B l39 1900 1920 1940 19m 1930 2000 Quillin Phillips etal Fe 1900 1920 1940 19m 1930 2000 Proxnnal mm M1 JMB 1993 234 140155 mm M1 The H64V mutant shows a single Cooperativity hemoglobin IR lfthere are multiple binding sites in a macromolecule there can be L29 two possible cases First they can be completely noninteracting m0 Distal and independent In that case they behave exactly as a collection a g g y of monomers On the other hand the sites can interact In that case 3 lt we can speak of cooperativity or anticooperativity ofthe binding sites We can use hemoglobin as an example 39 SWV 39N 1 SWL29F quot O 1H64 hr sw wild1w 39 SWHMQ lt PigV B l39 gt 1 WK swmv syn164V J P1 newvsquot 1 quot Proximal 1900 1920 1m 1950 I930 2000 Frequency cm l

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