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Concept Development Studies in Chemistry
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Title: Concept Development Studies in Chemistry
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Subjects / Keywords: Atomic Molecular Theory, Relative Atomic Masses and Empirical Formulae, The Structure of an Atom, Quantum Energy Levels In Atoms, Covalent Bonding and Electron Pair Sharing, Molecular Geometry and Electron Domain Theory, Molecular Structure and Physical Properties, Chemical Bonding and Molecular Energy Levels, Energetics of Chemical Reactions, The Ideal …
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Abstract: This is an on-line textbook for an Introductory General Chemistry course. Each module develops a central concept in Chemistry from experimental observations and inductive reasoning. This approach complements an interactive or active learning teaching approach. Additional multimedia resources can be found at: http://cnx.org/content/col10264/1.5/
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General Note: CHM 015 - GENERAL CHEMISTRY CONCEPTS, CHM 022 - GENERAL CHEMISTRY, CHM 025 - INTRODUCTION TO GENERAL CHEMISTRY
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ConceptDevelopmentStudiesinChemistry By: JohnS.Hutchinson

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ConceptDevelopmentStudiesinChemistry By: JohnS.Hutchinson Online: < http://cnx.org/content/col10264/1.5/ > CONNEXIONS RiceUniversity,Houston,Texas

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2008JohnS.Hutchinson ThisselectionandarrangementofcontentislicensedundertheCreativeCommonsAttributionLicense: http://creativecommons.org/licenses/by/2.0/

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TableofContents 1PrefacetoConceptDevelopmentStudiesinChemistry .......................................1 2TheAtomicMolecularTheory ...................................................................3 3RelativeAtomicMassesandEmpiricalFormulae ..............................................9 4TheStructureofanAtom .......................................................................15 5QuantumEnergyLevelsInAtoms ..............................................................25 6CovalentBondingandElectronPairSharing ..................................................37 7MolecularGeometryandElectronDomainTheory ...........................................51 8MolecularStructureandPhysicalProperties .................................................63 9ChemicalBondingandMolecularEnergyLevels ..............................................71 10EnergeticsofChemicalReactions .............................................................89 11TheIdealGasLaw ..............................................................................99 12TheKineticMolecularTheory ...............................................................111 13PhaseEquilibriumandIntermolecularInteractions ........................................123 14ReactionEquilibriumintheGasPhase .....................................................135 15Acid-BaseEquilibrium ........................................................................149 16ReactionRates .................................................................................161 17EquilibriumandtheSecondLawofThermodynamics .....................................181 Index ...............................................................................................193 Attributions ........................................................................................196

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iv

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Chapter1 PrefacetoConceptDevelopmentStudies inChemistry 1 1.1WhyConceptDevelopmentStudies? ThebodyofknowledgecalledScienceconsistsprimarilyofmodelsandconcepts,basedonobservationsand deducedfromcarefulreasoning.Viewedinthisway,Scienceisacreativehumanendeavor.Themodels, concepts,andtheoriesweusetodescribenatureareaccomplishmentsequalincreativitytoanyartistic, musical,orliterarywork. Unfortunately,textbooksinChemistrytraditionallypresentthesemodelsandconceptsessentiallyas establishedfacts,strippedofthecleverexperimentsandlogicalanalyseswhichgivethemtheirhuman essence.Asaconsequence,studentsaretypicallytrainedtomemorizeandapplythesemodels,ratherthan toanalyzeandunderstandthem.Asaresult,creative,analyticalstudentsareinclinedtofeelthattheycannot "do"Chemistry,thattheycannotunderstandtheconcepts,orthatChemistryisdullanduninteresting. Thiscollectionof ConceptDevelopmentStudiesinChemistry ispresentedtoredirectthefocusof learning.Ineachconceptdevelopmentstudy,amajorchemicalconceptisdevelopedandrenedbyanalysis ofexperimentalobservationsandcarefulreasoning.Eachstudybeginswiththedenitionofaninitial Foundation ofassumedknowledge,followedbyastatementofquestionswhicharisefromtheFoundation. Analysisofthesequestionsispresentedasaseriesof observations andlogicaldeductions,followedby furtherquestions.Thisdetailedprocessisfolloweduntiltheconceptualdevelopmentofamodelprovidesa reasonableanswertothestatedquestions. ConceptDevelopmentStudiesinChemistryiswrittenwithtwobenetstothereaderinmind.First,by analyzingeachconceptdevelopmentthroughcriticalreasoning,youwillgainamuchdeeperunderstanding ofasignicantconcept.Inadditiontoknowinghowtoworkwithamodel,youwillhavebothanunderstandingofwhythemodelisbelievableandanappreciationoftheessentialbeautyofthemodel.Second, thereasoningrequiredtounderstandtheseconceptdevelopmentstudieswillenhanceyourdevelopmentof critical,analyticalthinking,askillwhichismostimportanttosuccessinScience.Asanote,thesestudies arenotintendedtobehistoricaldevelopments,althoughtheexperimentspresentedaretheoneswhichled totheconceptsdiscussed.Onlyasmallamountofhistoricalinformationhasbeenincludedforperspective. 1.2HowtoStudytheConceptDevelopmentStudies Youshouldstudyeachconceptdevelopmentstudy,notbymemorization,butbycarefullythinkingabout theexperimentsandthelogicaldevelopmentoftheconceptsandmodels.Eachstudyisshort,andismeant tobereadslowlyandmeticulously.Eachsentencecontainssubstancetobestudiedandunderstood.You 1 Thiscontentisavailableonlineat. 1

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2 CHAPTER1.PREFACETOCONCEPTDEVELOPMENTSTUDIESIN CHEMISTRY should,ateachstepintheanalysis,challengeyourselfastowhetheryoucanreproducethereasoningleading tothenextconclusion.Onegoodwaytodothisistooutlinetheconceptdevelopmentstudy,makingsure youunderstandhoweachpieceoftheargumentcontributestothedevelopmentofaconceptormodel. Itisveryimportanttounderstandthatscienticmodelsandtheoriesarealmostnever"proven,"unlike mathematicaltheorems.Rather,theyarelogicallydevelopedanddeducedtoprovidesimpleexplanationsof observedphenomenon.Assuch,youwilldiscovermanytimesintheseconceptdevelopmentstudieswhena conclusionisnotlogicallyrequiredbyanobservationandalineofreasoning.Instead,wemayarriveata modelwhichisthesimplestexplanationofasetofobservations,evenifitisnottheonlyone.Scientistsmost commonlyabidebytheprincipleof Occam'srazor ,onestatementofwhichmightbethattheexplanation whichrequirestheleastassumptionsisthebestone. Oneveryimportantwaytochallengeyourunderstandingistostudyinagroupinwhichyoutaketurns explainingthedevelopmentofthemodel.Theabilitytoexplainaconceptisamuchstrongerindicatorof yourunderstandingthantheabilitytosolveaproblemusingtheconcept.Usethequestionsattheendofthe conceptdevelopmentstudiestopracticeyourskillatexplainingtechnicalargumentsclearlyandconcisely. 1.3Acknowledgments MyownthinkinginwritingConceptDevelopmentStudiesinChemistryhasbeenstronglyinuencedbythree books: TheHistoricalDevelopmentofChemicalConcepts ,byRomanMierzecki; TheHistoryofChemistry byJohnHudson; ChemicalPrinciples ,byRichardDickerson,HarryGray,andGilbertHaight. IamdeeplyappreciativeofthecontributionsofJoannaFair,KarenAianiStevens,KevinAusman,and KarinWrightinreviewingandcriticizingearlydraftsofthemanuscriptforthistext.Iamalsoindebtedto SusanWiediger,notjustforhertechnicalexpertiseandherknowledgeoftheeducationalliterature,butalso forhercommitmenttotheconceptbehindthisbookandthisapproachtoteaching.Iappreciatethehard workofJereySilvermantoconvertthesedocumentsforuseintheConnexionsProject 2 atRiceUniversity 3 ConceptDevelopmentStudiesinChemistrywouldnothavebeenwrittenwereitnotfortheencouragementofmywifePaula,whoremindedmeatthemostdicultoftimesthatwritingitwastherightthing todo.Iwillbeforevergrateful. JSH VisionImpairedAccess: ThankstothetranslationeortsofRiceUniversity'sDisabilitySupportServices 4 ,thiscollectionisnowavailableinaBraille-printableversion.Pleaseclickhere 5 to downloada.ziplecontainingallthenecessary.dxbandimageles. 2 http://cnx.org 3 http://www.rice.edu 4 http://www.dss.rice.edu/ 5 http://cnx.org/content/m12616/latest/ConceptDevStudiesBraille.zip

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Chapter2 TheAtomicMolecularTheory 1 2.1Foundation Thereareover18millionknownsubstancesinourworld.Wewillbeginbyassumingthatallmaterials aremadefrom elements ,materialswhichcannotbedecomposedintosimplersubstances.Wewillassume thatwehaveidentiedalloftheseelements,andthatthereaverysmallnumberofthem.Allotherpure substances,whichwecall compounds ,aremadeupfromtheseelementsandcanbedecomposedintothese elements.Forexample,metallicironandgaseousoxygenarebothelementsandcannotbereducedinto simplersubstances,butironrust,orferrousoxide,isacompoundwhichcanbereducedtoelementaliron andoxygen.Theelementsarenottransmutable:oneelementcannotbeconvertedintoanother.Finally,we willassumethatwehavedemonstratedthe LawofConservationofMass Law2.1: LawofConservationofMass Thetotalmassofallproductsofachemicalreactionisequaltothetotalmassofallreactantsof thatreaction. Thesestatementsaresummariesofmanyobservations,whichrequiredatremendousamountofexperimentationtoachieveandevenmorecreativethinkingtosystematizeaswehavewrittenthemhere.By makingtheseassumptions,wecanproceeddirectlywiththeexperimentswhichledtothedevelopmentof theatomic-moleculartheory. 2.2Goals Thestatementsabove,thoughcorrect,areactuallymorevaguethantheymightrstappear.Forexample, exactlywhatdowemeanwhenwesaythatallmaterialsaremadefromelements?Whyisitthatthe elementscannotbedecomposed?Whatdoesitmeantocombineelementsintoacompound?Wewantto understandmoreaboutthenatureofelementsandcompoundssowecandescribetheprocessesbywhich elementscombinetoformcompounds,bywhichcompoundsaredecomposedintoelements,andbywhich compoundsareconvertedfromonetoanotherduringchemicalreactions. Onepossibilityforansweringthesequestionsistoassumethatacompoundisformedwhenindestructible elementsaresimplymixedtogether,asforexample,ifweimaginestirringtogetheramixtureofsugarand sand.Neitherthesandnorthesugarisdecomposedintheprocess.Andthemixturecanbedecomposed backintotheoriginalcomponents.Inthiscase,though,theresultantmixtureexhibitsthepropertiesof both components:forexample,themixturewouldtastesweet,owingtothesugarcomponent,butgritty, characteristicofthesandcomponent. 1 Thiscontentisavailableonlineat. 3

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4 CHAPTER2.THEATOMICMOLECULARTHEORY Incontrast,thecompoundwecallironrustbearslittleresemblancetoelementaliron:ironrustdoesnot exhibitelementaliron'scolor,density,hardness,magnetism,etc.Sincethepropertiesoftheelementsare notmaintainedbythecompound,thenthecompoundmustnotbeasimplemixtureoftheelements. Wecould,ofcourse,jumpdirectlytotheanswerstothesequestionsbystatingthattheelementsthemselvesarecomprisedofatoms:indivisible,identicalparticlesdistinctiveofthatelement.Thenacompound isformedbycombiningtheatomsofthecompositeelements.Certainly,theLawofConservationofMass wouldbeeasilyexplainedbytheexistenceofimmutableatomsofxedmass. However,ifwedodecidetojumptoconclusionsandassumetheexistenceofatomswithoutfurther evidenceasdidtheleadingchemistsoftheseventeenthandeighteenthcenturies,itdoesnotleadus anywhere.Whathappenstoironwhen,afterprolongedheatinginair,itconvertstoironrust?Whyisit thattheresultantcombinationofironandairdoesnotmaintainthepropertiesofeither,aswewouldexpect iftheatomsofeacharemixedtogether?Anatomicviewofnaturewouldnotyetprovideanyunderstanding ofhowtheairandtheironhaveinteractedorcombinedtoformthenewcompound,andwecan'tmake anypredictionsabouthowmuchironwillproducehowmuchironrust.Thereisnobasisformakingany statementsaboutthepropertiesoftheseatoms.Weneedfurtherobservations. 2.3Observation1:Massrelationshipsduringchemicalreactions TheLawofConservationofMass,byitselfalone,doesnotrequireanatomicviewoftheelements.Mass couldbeconservedevenifmatterwerenotatomic.TheimportanceoftheLawofConservationofMass isthatitrevealsthatwecanusefullymeasurethemassesoftheelementswhicharecontainedinaxed massofacompound.Asanexample,wecandecomposecoppercarbonateintoitsconstituentelements, copper,oxygen,andcarbon,weighingeachandtakingtheratiosofthesemasses.Theresultisthatevery sampleofcoppercarbonateis51.5%copper,38.8%oxygen,and9.7%carbon.Stateddierently,themasses ofcopper,oxygen,andcarbonareintheratioof5.3:4:1,foreverymeasurementofeverysampleofcopper carbonate.Similarly,leadsuldeis86.7%leadand13.3%sulfur,sothatthemassratioforleadtosulfur inleadsuldeisalways6.5:1.Everysampleofcoppercarbonateandeverysampleofleadsuldewill producetheseelementalproportions,regardlessofhowmuchmaterialwedecomposeorwherethematerial camefrom.Theseresultsareexamplesofageneralprincipleknownasthe LawofDeniteProportions Law2.2: LawofDeniteProportions Whentwoormoreelementscombinetoformacompound,theirmassesinthatcompoundarein axedanddeniteratio. Thesedatahelpjustifyanatomicviewofmatter.Wecansimplyarguethat,forexample,leadsuldeis formedbytakingoneleadatomandcombiningitwithonesulfuratom.Ifthisweretrue,thenwealsomust concludethattheratioofthemassofalead atom tothatofasulfur atom isthesameasthe6.5:1leadto sulfurmassratiowefoundforthebulkleadsulde.Thisatomicexplanationlookslikethedenitiveanswer tothequestionofwhatitmeanstocombinetwoelementstomakeacompound,anditshouldevenpermit predictionofwhatquantityofleadsuldewillbeproducedbyagivenamountoflead.Forexample,6.5gof leadwillproduceexactly7.5gofleadsulde,50gofleadwillproduce57.7gofleadsulde,etc. Thereisaproblem,however.Wecanillustratewiththreecompoundsformedfromhydrogen,oxygen, andnitrogen.ThethreemassproportionmeasurementsaregiveninthefollowingtableMassRelationships forHydrogen,Nitrogen,OxygenCompounds,p.5.Firstweexaminenitricoxide,tondthatthemass proportionis8:7oxygentonitrogen.Ifthisisonenitrogenatomcombinedwithoneoxygenatom,we wouldexpectthatthemassofanoxygenatomis8/7=1.14timesthatofanitrogenatom.Secondweexamine ammonia,whichisacombinationofnitrogenandhydrogenwiththemassproportionof7:1.5nitrogento hydrogen.Ifthisisonenitrogencombinedwithonehydrogen,wewouldexpectthatanitrogenatommass is4.67timesthatofahydrogenatommass.Thesetwoexpectationspredictarelationshipbetweenthemass ofanoxygenatomandthemassofahydrogenatom.Ifthemassofanoxygenatomis1.14timesthemass

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5 ofanitrogenatomandifthemassofanitrogenatomis4.67timesthemassofahydrogenatom,thenwe mustconcludethatanoxygenatomhasamasswhichis1.14 4.67=5.34timesthatofahydrogenatom. Butthereisaproblemwiththiscalculation.ThethirdlineofthefollowingtableMassRelationships forHydrogen,Nitrogen,OxygenCompounds,p.5showsthatthecompoundformedfromhydrogenand oxygeniswater,whichisfoundtohavemassproportion8:1oxygentohydrogen.Ourexpectationshould thenbethatanoxygenatommassis8.0timesahydrogenatommass.Thusthethreemeasurementsinthe followingtableMassRelationshipsforHydrogen,Nitrogen,OxygenCompounds,p.5appeartoleadto contradictoryexpectationsofatomicmassratios.Howarewetoreconciletheseresults? MassRelationshipsforHydrogen,Nitrogen,OxygenCompounds Compound Total Mass Massof Hydrogen Massof Nitrogen Massof Oxygen "Expected" Relative Atomic Massof Hydrogen "Expected" Relative Atomic Massof Nitrogen "Expected" Relative Atomic Massof Oxygen Nitric Oxide 15.0g 7.0g 8.0g 7.0 8.0 Ammonia 8.5g 1.5g 7.0g 1.5 7.0 Water 9.0g 1.0g 8.0g 1.0 8.0 Onepossibilityisthatweweremistakeninassumingthatthereareatomsoftheelementswhichcombine toformthedierentcompounds.Ifso,thenwewouldnotbesurprisedtoseevariationsinrelativemasses ofmaterialswhichcombine. Anotherpossibilityisthatwehaveerredinourreasoning.Lookingback,weseethatwehavetoassume howmanyatomsofeachtypearecontainedineachcompoundtondtherelativemassesoftheatoms.In eachoftheaboveexamples,weassumedtheratioofatomstobe1:1ineachcompound.Ifthereareatoms oftheelements,thenthisassumptionmustbewrong,sinceitgivesrelativeatomicmasseswhichdierfrom compoundtocompound.Howcouldwendthecorrectatomicratios?Itwouldhelpifweknewtheratio oftheatomicmasses:forexample,ifweknewthattheoxygentohydrogenmassratiowere8:1,thenwe couldconcludethattheatomicratioinwaterwouldbe1oxygenand1hydrogen.Ourreasoningseemsto circular:toknowtheatomicmasses,wemustknowthe formula ofthecompoundthenumbersofatoms ofeachtype,buttoknowtheformulawemustknowthemasses. Whichofthesepossibilitiesiscorrect?Withoutfurtherobservations,wecannotsayforcertainwhether matteriscomposedofatomsornot. 2.4Observation2:MultipleMassRatios Signicantinsightintotheaboveproblemisfoundbystudyingdierentcompoundsformedfromthesame elements.Forexample,thereareactuallythreeoxidesofnitrogen,thatis,compoundscomposedonlyof nitrogenandoxygen.Fornow,wewillcallthemoxideA,oxideB,andoxideC.OxideAhasoxygento nitrogenmassratio2.28:1.OxideBhasoxygentonitrogenmassratio1.14:1,andoxideChasoxygen tonitrogenmassratio0.57:1. ThefactthattherearethreemassratiosmightseemtocontradicttheLawofDeniteProportions,which onthesurfaceseemstosaythatthereshouldbejustoneratio.However,eachmasscombinationgivesrise toacompletelyuniquechemicalcompoundwithverydierentchemicalproperties.Forexample,oxideA isverytoxic,whereasoxideCisusedasananesthesia.Itisalsotruethatthemassratioisnotarbitrary orcontinuouslyvariable:wecannotpickjustanycombinationofmassesincombiningoxygenandnitrogen, ratherwemustobeyoneofonlythree.Sothereisnocontradiction:wesimplyneedtobecarefulwith

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6 CHAPTER2.THEATOMICMOLECULARTHEORY theLawofDeniteProportionstosay thateachuniquecompound hasadenitemassratioofcombining elements. Thesenewmassrationumbersarehighlysuggestiveinthefollowingway.Noticethat,ineachcase, wetooktheratioofoxygenmasstoanitrogenmassof1,andthattheresultantratioshaveaverysimple relationship: 2 : 28:1 : 14:0 : 57=2:1:0 : 5 =4:2:1 .1 Themassesofoxygenappearinginthesecompoundsareinsimplewholenumberratioswhenwetakeaxed amountofnitrogen.Theappearanceofthesesimplewholenumbersisverysignicant.Theseintegersimply thatthecompoundscontainamultipleofaxedunitofmassofoxygen.Thesimplestexplanationforthis xedunitofmassisthatoxygenis particulate .Wecallthexedunitofmassan atom .Wenowassume thatthecompoundshavebeenformedfromcombinationsofatomswithxedmasses,andthatdierent compoundshavedieringnumbersofatoms.ThemassratiosmakeitclearthatoxideBcontainstwiceas manyoxygenatomspernitrogenatomasdoesoxideCandhalfasmanyoxygenatomspernitrogenatom asdoesoxideA.Thesimplemassratiosmustbetheresultofthesimpleratiosinwhichatomscombine intomolecules.If,forexample,oxideChasthemolecularformula NO ,thenoxideBhastheformula NO 2 andoxideAhastheformula NO 4 .Thereareotherpossibilities:ifoxideBhasmolecularformula NO ,then oxideAhasformula NO 2 ,andoxideChasformula N 2 O .OrifoxideAhasformula NO ,thenoxideB hasformula N 2 O andoxideChasformula N 4 O .Thesethreepossibilitiesarelistedinthefollowingtable PossibleMolecularFormulaeforNitrogenOxides,p.6. PossibleMolecularFormulaeforNitrogenOxides Assumingthat: OxideCis NO OxideBis NO OxideAis NO OxideAis NO 4 NO 2 NO OxideBis NO 2 NO N 2 O OxideCis NO N 2 O N 4 O Wedon'thaveawayfromthesedatatoknowwhichofthesesetsofmolecularformulaeareright.But wecanassertthateitheroneofthemoroneanalogoustothemisright. Similardataarefoundforanysetofcompoundsformedfromcommonelements.Forexample,thereare twooxidesofcarbon,onewithoxygentocarbonmassratio1.33:1andtheotherwithmassratio2.66:1. Thesecondoxidemusthavetwiceasmanyoxygenatoms,percarbonatom,asdoestherst.Thegeneral statementofthisobservationisthe LawofMultipleProportions Law2.3: LawofMultipleProportions Whentwoelementscombinetoformmorethanonecompound,themassofelementAwhich combinesintherstcompoundwithagivenamountofelementBhasasimplewholenumberratio withthemassofelementAwhichcombinesinthesecondcompoundwiththesamegivenmassof elementB. Thissoundsconfusing,butanexampleclariesthisstatement.Considerthecarbonoxides,andlet carbonbeelementBandoxygenbeelementA.TakeaxedgivenmassofcarbonelementB,say1gram. Themassofoxygenwhichcombineswith1gramofcarbontoformtherstoxideis1.33grams.Themass ofoxygenwhichcombineswith1gramofcarbontoformthesecondoxideis2.66.Thesemassesareinratio 2 : 66:1 : 33=2:1 ,asimplewholenumberratio. InexplainingourobservationsoftheLawofMultipleProportionsforthecarbonoxidesandthenitrogen oxides,wehaveconcludedthatthesimplemassratioarisesfromthesimpleratioofatomscontainedin theindividualmolecules.Thus,wehaveestablishedthefollowingpostulatesofthe AtomicMolecular Theory Theory2.1: AtomicMolecularTheory

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7 theelementsarecomprisedofidenticalatoms allatomsofasingleelementhavethesamecharacteristicmass thesenumberandmassesoftheseatomsdonotchangeduringachemicaltransformation compoundsconsistofidenticalmoleculesformedofatomscombinedinsimplewholenumber ratios 2.5ReviewandDiscussionQuestions Exercise2.1 Assumethatmatterdoesnotconsistofatoms.Showbyexamplehowthisassumptionleadsto hypotheticalpredictionswhichcontradicttheLawofMultipleProportions.Dothesehypothetical examplescontradicttheLawofDeniteProportions?Arebothobservationsrequiredforconrmationoftheatomictheory? Exercise2.2 Twocompounds,AandB,areformedentirelyfromhydrogenandcarbon.CompoundAis80.0% carbonbymass,and20.0%hydrogen,whereasCompoundBis83.3%carbonbymassand16.7% hydrogen.DemonstratethatthesetwocompoundsobeytheLawofMultipleProportions.Explain whytheseresultsstronglyindicatethattheelementscarbonandhydrogenarecomposedofatoms. Exercise2.3 Inmanychemicalreactions,massdoesnotappeartobeaconservedquantity.Forexample,when atincanrusts,theresultantrustytincanhasagreatermassthanbeforerusting.Whena candleburns,theremainingcandlehasinvariablylessmassthanbeforeitwasburned.Providean explanationoftheseobservations,anddescribeanexperimentwhichwoulddemonstratethatmass isactuallyconservedinthesechemicalreactions. Exercise2.4 Thefollowingquestionwasposedonanexam: Anunknownnon-metalelementQformstwogaseousuoridesofunknownmolecularformula. A3.2gsampleofQreactswithuorinetoform10.8goftheunknownuorideA.A6.4g sampleofQreactswithuorinetoform29.2gofunknownuorideB.Usingthesedataonly, demonstratebycalculationandexplanationthattheseunknowncompoundsobeytheLawof MultipleProportions. Astudentrespondedwiththefollowinganswer: TheLawofMultipleProportionsstatesthatwhentwoelementsformtwoormorecompounds, theratiosofthemassesoftheelementsbetweenthetwocompoundsareinasimplewholenumber ratio.So,lookingatthedataabove,weseethattheratioofthemassofelementQincompound AtothemassofelementQincompoundBis 3 : 2:6 : 4=1:2 ,whichisasimplewholenumber ratio.ThisdemonstratesthatthesecompoundsobeytheLawofMultipleProportions. Assesstheaccuracyofthestudentsanswer.Inyourassessment,youmustdeterminewhatinformationiscorrectorincorrect,providethecorrectinformationwhereneeded,explainwhetherthe reasoningislogicalornot,andprovidelogicalreasoningwhereneeded.

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8 CHAPTER2.THEATOMICMOLECULARTHEORY

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Chapter3 RelativeAtomicMassesandEmpirical Formulae 1 3.1Foundation Webeginbyassumingthecentralpostulatesofthe Atomic-MolecularTheory .Theseare:theelements arecomprisedofidenticalatoms;allatomsofasingleelementhavethesamecharacteristicmass;thenumber andmassesoftheseatomsdonotchangeduringachemicaltransformation;compoundsconsistofidentical moleculesformedofatomscombinedinsimplewholenumberratios.Wealsoassumeaknowledgeofthe observednaturallawsonwhichthistheoryisbased:the LawofConservationofMass ,the Lawof DeniteProportions ,andthe LawofMultipleProportions 3.2Goals Wehaveconcludedthatatomscombineinsimpleratiostoformmolecules.However,wedon'tknowwhat thoseratiosare.Inotherwords,wehavenotyetdeterminedanymolecularformulae.Inthesecondtableof ConceptDevelopmentStudy#1PossibleMolecularFormulaeforNitrogenOxides,p.6,wefoundthatthe massratiosfornitrogenoxidecompoundswereconsistentwithmanydierentmolecularformulae.Aglance backatthenitrogenoxidedatashowsthattheoxideBcouldbe NO NO 2 N 2 O ,oranyothersimpleratio. Eachoftheseformulaecorrespondtodierentpossiblerelativeatomicweightsfornitrogenandoxygen. SinceoxideBhasoxygentonitrogenratio1.14:1,thentherelativemassesofoxygentonitrogencouldbe 1.14:1or2.28:1or0.57:1ormanyothersimplepossibilities.Ifweknewtherelativemassesofoxygenand nitrogenatoms,wecoulddeterminethemolecularformulaofoxideB.Ontheotherhand,ifweknewthe molecularformulaofoxideB,wecoulddeterminetherelativemassesofoxygenandnitrogenatoms.Ifwe solveoneproblem,wesolveboth.Ourproblemthenisthatweneedasimplewayto"count"atoms,atleast inrelativenumbers. 3.3Observation1:VolumeRelationshipsinChemicalReactions Althoughmassisconserved,mostchemicalandphysicalpropertiesarenotconservedduringareaction. Volumeisoneofthosepropertieswhichisnotconserved,particularlywhenthereactioninvolvesgasesas reactantsorproducts.Forexample,hydrogenandoxygenreactexplosivelytoformwatervapor.Ifwetake 1literofoxygengasand2litersofhydrogengas,bycarefulanalysiswecouldndthatthereactionofthese twovolumesiscomplete,withnoleftoverhydrogenandoxygen,andthat2litersofwatervaporareformed. 1 Thiscontentisavailableonlineat. 9

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10 CHAPTER3.RELATIVEATOMICMASSESANDEMPIRICALFORMULAE Notethatthetotalvolumeisnotconserved:3litersofoxygenandhydrogenbecome2litersofwatervapor. Allofthevolumesaremeasuredatthesametemperatureandpressure. Morenotableisthefactthattheratiosofthevolumesinvolvedaresimplewholenumberratios:1liter ofoxygen:2litersofhydrogen:2litersofwater.Thisresultprovestobegeneralforreactionsinvolving gases.Forexample,1literofnitrogengasreactswith3litersofhydrogengastoform2litersofammonia gas.1literofhydrogengascombineswith1literofchlorinegastoform2litersofhydrogenchloridegas. Theseobservationscanbegeneralizedintothe LawofCombiningVolumes Law3.1: LawofCombiningVolumes Whengasescombineduringachemicalreactionataxedpressureandtemperature,theratiosof theirvolumesaresimplewholenumberratios. Thesesimpleintegerratiosarestriking,particularlywhenviewedinthelightofourconclusionsfrom theLawofMultipleProportions.Atomscombineinsimplewholenumberratios,andevidently,volumes ofgasesalsocombineinsimplewholenumberratios.Whywouldthisbe?Onesimpleexplanationofthis similaritywouldbethatthevolumeratioandtheratioofatomsandmoleculesinthereactionarethesame. Inthecaseofthehydrogenandoxygen,thiswouldsaythattheratioofvolumesliterofoxygen:2liters ofhydrogen:2litersofwateristhesameastheratioofatomsandmoleculesatomofoxygen:2atoms ofhydrogen:2moleculesofwater.Forthistobetrue,equalvolumesofgaswouldhavetocontainequal numbersofgasparticlesatomsormolecules,independentofthetypeofgas.Iftrue,thismeansthatthe volumeofagasmustbeadirectmeasureofthenumberofparticlesatomsormoleculesinthegas.This wouldallowusto"count"thenumberofgasparticlesanddeterminemolecularformulae. Thereseemtobebigproblemswiththisconclusion,however.Lookbackatthedataforforminghydrogen chloride:1literofhydrogenplus1literofchlorineyields2litersofhydrogenchloride.Ifourthinkingis true,thenthisisequivalenttosayingthat1hydrogenatomplus1chlorineatommakes2hydrogenchloride molecules.Buthowcouldthatbepossible?Howcouldwemake2identicalmoleculesfromasinglechlorine atomandasinglehydrogenatom?Thiswouldrequireustodivideeachhydrogenandchlorineatom, violatingthepostulatesoftheatomic-moleculartheory. Anotherproblemappearswhenweweighthegases:1literofoxygengasweighsmorethan1literof watervapor.Ifweassumethatthesevolumescontainequalnumbersofparticles,thenwemustconclude that1oxygenparticleweighsmorethan1waterparticle.Buthowcouldthatbepossible?Itwouldseem thatawatermolecule,whichcontainsatleastoneoxygenatom,shouldweighmorethanasingleoxygen particle. Theseareseriousobjectionstotheideathatequalvolumesofgascontainequalnumbersofparticles.Our postulateappearstohavecontradictedcommonsenseandexperimentalobservation.However,thesimple ratiosoftheLawofCombiningVolumesarealsoequallycompelling.Whyshouldvolumesreactinsimple wholenumberratiosiftheydonotrepresentequalnumbersofparticles?Considertheoppositeviewpoint: ifequalvolumesofgasdonotcontainequalnumbersofparticles,thenequalnumbersofparticlesmust becontainedinunequalvolumesnotrelatedbyintegers.Nowwhenwecombineparticlesinsimplewhole numberratiostoformmolecules,thevolumesofgasesrequiredwouldproducedecidedlynon-wholenumber ratios.TheLawofCombiningVolumesshouldnotbecontradictedlightly. Thereisonlyonelogicalwayout.WewillacceptourdeductionfromtheLawofCombiningVolumesthat equalvolumesofgascontainequalnumbersofparticles ,aconclusionknownas Avogadro'sHypothesis Howdoweaccountforthefactthat1literofhydrogenplus1literofchlorineyields2litersofhydrogen chloride?Thereisonlyonewayforasinglehydrogenparticletoproduce2identicalhydrogenchloride molecules:eachhydrogenparticlemustcontainmorethanoneatom.Infact,eachhydrogenparticleor moleculemustcontainanevennumberofhydrogenatoms.Similarly,achlorinemoleculemustcontainan evennumberofchlorineatoms. Moreexplicitly,weobservethat 1literofhydrogen + 1literofchlorine 2litersofhydrogenchloride.1 Assumingthateachlitervolumecontainsanequalnumberofparticles,thenwecaninterpretthisobservation

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11 as 1 H 2 molecule + 1 Cl 2 molecule 2 HCl molecules3.2 Alternatively,therecouldbeanyxedevennumberofatomsineachhydrogenmoleculeandineachchlorine molecule.Wewillassumethesimplestpossibilityandseeifthatproducesanycontradictions. Thisisawonderfulresult,foritcorrectlyaccountsfortheLawofCombiningVolumesandeliminates ourconcernsaboutcreatingnewatoms.Mostimportantly,wenowknowthemolecularformulaofhydrogen chloride.Wehave,ineect,foundawayof"counting"theatomsinthereactionbymeasuringthevolume ofgaseswhichreact. Thismethodworkstotellusthemolecularformulaofmanycompounds.Forexample, 2litersofhydrogen + 1literofoxygen 2litersofwater.3 Thisrequiresthatoxygenparticlescontainanevennumberofoxygenatoms.Nowwecaninterpretthis equationassayingthat 2 H 2 molecules + 1 O 2 molecule 2 H 2 O molecules.4 Nowthatweknowthemolecularformulaofwater,wecandrawadeniteconclusionabouttherelative massesofthehydrogenandoxygenatoms.RecallfromtheTableMassRelationshipsforHydrogen,Nitrogen,OxygenCompounds,p.5thatthemassratioinwateris8:1oxygentohydrogen.Sincethereare twohydrogenatomsforeveryoxygenatominwater,thenthemassratiorequiresthatasingleoxygenatom weigh16timesthemassofahydrogenatom. Todetermineamassscaleforatoms,wesimplyneedtochooseastandard.Forexample,forourpurposes here,wewillsaythatahydrogenatomhasamassof1ontheatomicmassscale.Thenanoxygenatomhas amassof16onthisscale. Ourconclusionsaccountfortheapparentproblemswiththemassesofreactinggases,specically,that oxygengasweighsmorethanwatervapor.Thisseemedtobenonsensical:giventhatwatercontainsoxygen, itwouldseemthatwatershouldweighmorethanoxygen.However,thisisnowsimplyunderstood:awater molecule,containingonlyasingleoxygenatom,hasamassof18,whereasanoxygenmolecule,containing twooxygenatoms,hasamassof32. 3.4DeterminationofAtomicWeightsforGaseousElements Nowthatwecancountatomsandmoleculestodeterminemolecularformulae,weneedtodeterminerelative atomicweightsforallatoms.Wecanthenusethesetodeterminemolecularformulaeforanycompound fromthemassratiosoftheelementsinthecompound. WebeginbyexaminingdataonreactionsinvolvingtheLawofCombiningVolumes.Goingbacktothe nitrogenoxidedatagivenherePossibleMolecularFormulaeforNitrogenOxides,p.6,werecallthatthere arethreecompoundsformedfromnitrogenandoxygen.Nowwemeasurethevolumeswhichcombinein formingeach.Wendthat2litersofoxideBcanbedecomposedinto1literofnitrogenand1literofoxygen. Fromthereasoningabove,thenanitrogenparticlemustcontainanevennumberofnitrogenatoms.We assumefornowthatnitrogenis N 2 .Wehavealreadyconcludedthatoxygenis O 2 .Therefore,themolecular formulaforoxideBis NO ,andwecallitnitricoxide.Sincewehavealreadydeterminedthattheoxygento nitrogenmassratiois1.14:1,then,ifweassignoxygenamassof16,asabove,nitrogenhasamassof14. Thatis 16 1 : 14 =14 .2litersofoxideAisformedfrom2litersofoxygenand1literofnitrogen.Therefore, oxideAis NO 2 ,whichwecallnitrogendioxide.Notethatwepredictanoxygentonitrogenmassratioof 32 14 =2 : 28:1 ,inagreementwiththedata.OxideCis N 2 O ,callednitrousoxide,andpredictedtohavea massratioof 16 28 =0 : 57:1 ,againinagreementwiththedata.Wehavenowresolvedtheambiguityinthe molecularformulae. Whatifnitrogenwereactually N 4 ?Thentherstoxidewouldbe N 2 O ,thesecondwouldbe N 2 O 2 ,and thethirdwouldbe N 4 O .Furthermore,themassofanitrogenatomwouldbe7.Whydon'tweassumethis?

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12 CHAPTER3.RELATIVEATOMICMASSESANDEMPIRICALFORMULAE Simplybecauseindoingso,wewillalwaysndthattheminimumrelativemassofnitrogeninanymolecule is14.Althoughthismightbetwonitrogenatoms,thereisnoreasontobelievethatitis.Therefore,asingle nitrogenatomweighs14,andnitrogengasparticlesare N 2 3.5DeterminationofAtomicWeightsforNon-GaseousElements Wecanproceedwiththistypeofmeasurement,deduction,andpredictionforanycompoundwhichisa gasandwhichismadeupofelementswhicharegases.Butthiswillnothelpuswiththeatomicmasses ofnon-gaseouselements,norwillitpermitustodeterminethemolecularformulaeforcompoundswhich containtheseelements. Considercarbon,animportantexample.Therearetwooxidesofcarbon.OxideAhasoxygentocarbon massratio1.33:1andoxideBhasmassratio2.66:1.Measurementofreactingvolumesshowsthatwe ndthat1literofoxideAisproducedfrom0.5litersofoxygen.Hence,eachmoleculeofoxideAcontains onlyhalfasmanyoxygenatomsasdoesanoxygenmolecule.OxideAthuscontainsoneoxygenatom.But howmanycarbonatomsdoesitcontain?Wecan'tdeterminethisyetbecausetheelementalcarbonissolid, notgas.Thismeansthatwealsocannotdeterminewhatthemassofacarbonatomis. Butwecantryadierentapproach:weweigh1literofoxideAand1literofoxygengas.Theresultwe ndisthatoxideAweighs0.875timesperliterasmuchasoxygengas.Sincewehaveassumedthataxed volumeofgascontainsaxednumberofparticles,then1literofoxideAcontainsjustasmanyparticlesas 1literofoxygengas.Therefore,each particle ofoxideAweighs0.875timesasmuchasaparticleofoxygen gasthatis,an O 2 molecule.Sincean O 2 moleculeweighs32onouratomicmassscale,thenaparticleof oxideAweighs 0 : 875 32=28 .NowweknowthemolecularweightofoxideA. Furthermore,wehavealreadydeterminedfromthecombiningvolumesthatoxideAcontainsasingle oxygenatom,ofmass16.Therefore,themassofcarboninoxideAis12.However,atthispoint,wedonot knowwhetherthisisonecarbonatomofmass12,twoatomsofmass6,eightatomsofmass1.5,oroneof manyotherpossibilities. Tomakefurtherprogress,wemakeadditionalmeasurementsonothercarboncontaininggascompounds. 1literofoxideBofcarbonisformedfrom1literofoxygen.Therefore,eachoxideBmoleculecontainstwo oxygenatoms.1literofoxideBweighs1.375timesasmuchas1literofoxygen.Therefore,oneoxideB moleculehasmass 1 : 375 32=44 .SincetherearetwooxygenatomsinamoleculeofoxideB,themassof oxygeninoxideBis32.Therefore,themassofcarboninoxideBis12,thesameasinoxideA. Wecanrepeatthisprocessformanysuchgaseouscompoundscontainingcarbonatoms.Ineachcase,we ndthatthemassofcarbonineachmoleculeiseither12oramultipleof12.Wenevernd,forexamples, 6or18,whichwouldbepossibleifeachcarbonatomhadmass6.Thesimplestconclusionisthatacarbon atomhasmass12.Onceweknowtheatomicmassofcarbon,wecanconcludethatthemolecularformula ofoxideAis CO ,andthatofoxideBis CO 2 Therefore,theatomicmassesofnon-gaseouselementscanbedeterminedbymassandvolumemeasurementsongaseouscompoundscontainingtheseelements.Thisprocedureisfairlygeneral,andmostatomic massescanbedeterminedinthisway. 3.6Moles,MolecularFormulaeandStoichiometricCalculations Webeganwithacirculardilemma:wecoulddeterminemolecularformulaeprovidedthatweknewatomic masses,butthatwecouldonlydetermineatomicmassesfromaknowledgeofmolecularformulae.Sincewe nowhaveamethodfordeterminingallatomicmasses,wehaveresolvedthisdilemmaandwecandetermine themolecularformulaforanycompoundforwhichwehavepercentcompositionbymass. Asasimpleexample,weconsideracompoundwhichisfoundtobe40.0%carbon,53.3%oxygen,and 6.7%hydrogenbymass.RecallfromtheLawofDeniteProportionsthatthesemassratiosareindependent ofthesample,sowecantakeanyconvenientsampletodoouranalysis.Assumingthatwehave100.0gof thecompound,wemusthave40.0gofcarbon,53.3gofoxygen,and6.7gofhydrogen.Ifwecouldcountor

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13 otherwisedeterminethenumberofatomsofeachelementrepresentedbythesemasses,wewouldhavethe molecularformula.However,thiswouldnotonlybeextremelydiculttodobutalsounnecessary. Fromourdeterminationofatomicmasses,wecannotethat1atomofcarbonhasamasswhichis12.0 timesthemassofahydrogenatom.Therefore,themassof N atomsofcarbonisalso12.0timesthemass of N atomsofhydrogenatoms,nomatterwhat N is.Ifweconsiderthiscarefully,wediscoverthat12.0gof carboncontainsexactlythesamenumberofatomsasdoes1.0gofhydrogen.Similarly,wenotethat1atom ofoxygenhasamasswhichis 16 : 0 12 : 0 timesthemassofacarbonatom.Therefore,themassof N atomsof oxygenis 16 : 0 12 : 0 timesthemassof N atomsofcarbon.Again,wecanconcludethat16.0gofoxygencontains exactlythesamenumberofatomsas12.0gofcarbon,whichinturnisthesamenumberofatomsas1.0gof hydrogen.Withoutknowingornecessarilyevencaringwhatthenumberis,wecansaythatitisthesame numberforallthreeelements. Forconvenience,then,we dene thenumberofatomsin12.0gofcarbontobe1 mole ofatoms.Note that1moleisaspecicnumberofparticles,justlike1dozenisaspecicnumber,independentofwhat objectswearecounting.Theadvantagetodeningthemoleinthiswayisthatitiseasytodeterminethe numberofmolesofasubstancewehave,andknowingthenumberofmolesisequivalenttocountingthe numberofatomsormoleculesinasample.Forexample,24.0gofcarboncontains2.0molesofatoms,30.0g ofcarboncontains2.5molesofatoms,andingeneral, x gramsofcarboncontains x 12 : 0 molesofatoms.Also, werecallthat16.0gofoxygencontainsexactlyasmanyatomsasdoes12.0gofcarbon,andtherefore16.0g ofoxygencontainsexactly1.0moleofoxygenatoms.Thus,32.0gofoxygencontains2.0molesofoxygen atoms,40.0gofoxygencontains2.5moles,and x gramsofoxygencontains x 16 : 0 molesofoxygenatoms.Even moregenerally,then,ifwehave m gramsofanelementwhoseatomicmassis M ,thenumberofmolesof atoms, n ,is n = m M .5 Nowwecandeterminetherelativenumbersofatomsofcarbon,oxygen,andhydrogeninourunknown compoundabove.Ina100.0gsample,wehave40.0gofcarbon,53.3gofoxygen,and6.7gofhydrogen.The numberofmolesofatomsineachelementisthus n C = 40 : 0 g 12 : 0 g mol =3 : 33 moles .6 n O = 53 : 3 g 16 : 0 g mol =3 : 33 moles .7 n H = 6 : 7 g 1 : 0 g mol =6 : 67 moles .8 Wenotethatthenumbersofmolesofatomsoftheelementsareinthesimpleratio n C : n O : n H =1:1:2 Sincethenumberofparticlesin1moleisthesameforallelements,thenitmustalsobetruethatthe numberofatomsoftheelementsareinthesimpleratio1:1:2.Therefore,themolecularformulaofthe compoundmustbe COH 2 Orisit?Onfurtherreection,wemustrealizethatthesimpleratio1:1:2neednotrepresentthe exactnumbersofatomsofeachtypeinamoleculeofthecompound,sinceitisindeedonlyaratio.Thusthe molecularformulacouldjustaseasilybe C 2 O 2 H 4 or C 3 O 3 H 6 .Sincetheformula COH 2 isbasedonempirical massratiodata,werefertothisasthe empiricalformula ofthecompound.Todeterminethe molecular formula ,weneedtodeterminetherelativemassofamoleculeofthecompound,i.e.themolecularmass. OnewaytodosoisbasedontheLawofCombiningVolumes,Avogadro'sHypothesis,andthe IdealGas Law .Toillustrate,however,ifweweretondthattherelativemassofonemoleculeofthecompoundis 60.0,wecouldconcludethatthemolecularformulais C 2 O 2 H 4 .

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14 CHAPTER3.RELATIVEATOMICMASSESANDEMPIRICALFORMULAE 3.7ReviewandDiscussionQuestions Exercise3.1 StatetheLawofCombiningVolumesandprovideanexampleofyourownconstructionwhich demonstratesthislaw. Exercise3.2 ExplainhowtheLawofCombiningVolumes,combinedwiththeAtomic-MolecularTheory,leads directlytoAvogadro'sHypothesisthatequalvolumesofgasatequaltemperaturesandpressure containequalnumbersofparticles. Exercise3.3 UseAvogadro'sHypothesistodemonstratethatoxygengasmoleculescannotbemonatomic. Exercise3.4 Thedensityofwatervaporatroomtemperatureandatmosphericpressureis 0 : 737 g L .Compound Ais80.0%carbonbymass,and20.0%hydrogen.CompoundBis83.3%carbonbymassand 16.7%hydrogen.ThedensityofgaseousCompoundAis 1 : 227 g L ,andthedensityofCompoundB is 2 : 948 g L .ShowhowthesedatacanbeusedtodeterminethemolarmassesofCompoundsAand B,assumingthatwaterhasmolecularmass18. Exercise3.5 FromtheresultsaboveExercise3.4,determinethemassofcarboninamoleculeofCompound AandinamoleculeofCompoundB.Explainhowtheseresultsindicatethatacarbonatomhas atomicmass12. Exercise3.6 Explaintheutilityofcalculatingthenumberofmolesinasampleofasubstance. Exercise3.7 Explainhowwecanconcludethat28gofnitrogengas N 2 containsexactlyasmanymoleculesas 32gofoxygengas O 2 ,eventhoughwecannotpossiblycountthisnumber.

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Chapter4 TheStructureofanAtom 1 4.1Foundation Webeginasastartingpointwiththeatomicmoleculartheory.Wethusassumethatmostofthecommon elementshavebeenidentied,andthateachelementischaracterizedasconsistingofidentical,indestructible atoms.Wealsoassumethattheatomicweightsoftheelementsareallknown,andthat,asaconsequence, itispossibleviamasscompositionmeasurementstodeterminethemolecularformulaforanycompoundof interest.Inaddition,wewillassumethatithasbeenshownbyelectrochemicalexperimentsthatatomscontainequalnumbersofpositivelyandnegativelychargedparticles,calledprotonsandelectronsrespectively. Finally,weassumeanunderstandingofthePeriodicTable.Inparticular,weassumethattheelements canbegroupedaccordingtotheircommonchemicalandphysicalproperties,andthatthesechemicaland physicalpropertiesareperiodicfunctionsoftheatomicnumber. 4.2Goals Theatomicmoleculartheoryisextremelyusefulinexplainingwhatitmeanstoformacompoundfromits componentelements.Thatis,acompoundconsistsofidenticalmolecules,eachcomprisedoftheatomsof thecomponentelementsinasimplewholenumberratio.However,ourknowledgeoftheseatomsisvery limited.Theonlypropertyweknowatthispointistherelativemassofeachatom.Consequently,wecannot answerawiderangeofnewquestions.Weneedamodelwhichaccountsfortheperiodicityofchemicaland physicalpropertiesasexpressedinthePeriodicTable.Whyareelementswhichareverydissimilarinatomic massneverthelessverysimilarinproperties?Whydothesecommonpropertiesrecurperiodically? Wewouldliketounderstandwhatdeterminesthenumberofatomsofeachtypewhichcombinetoform stablecompounds.Whyaresomecombinationsfoundandothercombinationsnotobserved?Whydosome elementswithverydissimilaratomicmassesforexample,iodineandchlorineformverysimilarchemical compounds?Whydootherelementswithverysimilaratomicmassesforexample,oxygenandnitrogen formverydissimilarcompounds?Ingeneral,whatforcesholdatomstogetherinformingamolecule? Answeringthesequestionsrequiresknowledgeofthestructureoftheatom,includinghowthestructures ofatomsofdierentelementsaredierent.Ourmodelshouldtellushowthesestructuraldierencesresult inthedierentbondingpropertiesofthedierentatoms. 4.3Observation1:Scatteringof particlesbyatoms Wehaveassumedthatatomscontainpositiveandnegativechargesandthenumberofthesechargesisequal inanygivenatom.However,wedonotknowwhatthatnumberis,nordoweknowhowthosechargesare 1 Thiscontentisavailableonlineat. 15

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16 CHAPTER4.THESTRUCTUREOFANATOM arrangedinsidetheatom.Todeterminethelocationofthechargesintheatom,weperforma"scattering" experiment.Theideaisstraightforward:sincewecannot"see"theatomicstructure,thenweinstead"throw" thingsattheatomandwatchthewayinwhichtheseobjectsaredeectedbytheatom.Workingbackwards, wecanthendeducewhatthestructureoftheatommustbe. Theatomswechoosetoshootataregold,intheformofaverythingoldfoilofthicknessabout 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 cm Theobjectswe"throw"areactually particles,whicharepositivelychargedandfairlymassive,emittedby radioactivepoloniumnuclei.The particlesaredirectedinaveryprecisenarrowlineperpendiculartoand inthedirectionofthegoldfoil.Wethenlookfor particlesatvariousanglesaboutthegoldfoil,looking bothforparticleswhichhavebeendeectedastheypassthroughthefoilorwhichhavebeenreectedas theybounceoofthefoil.ThescatteringexperimentisillustratedhereFigure4.1: particleScattering fromGoldFoil. particleScatteringfromGoldFoil Figure4.1 Theresultoftheexperimentisinitiallycounter-intuitive.Mostofthe particlespassthroughthegold foilundeected,asiftherehadbeennothingintheirpath!Asmallernumberoftheparticlesaredeected sharplyastheypassthroughthefoil,andaverysmallfractionofthe particlesarereectedbackwardso ofthegoldfoil.Howcanwesimultaneouslyaccountforthelackofanydeectionformostofthe particles andforthedeectionthroughlargeanglesofaverysmallnumberofparticles? First,sincethemajorityofthepositivelycharged particlespassthroughthegoldfoilundeected,we

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17 canconcludethatmostofthevolumeofeachgoldatomisemptyspace,containingnothingwhichmight deectan particle.Second,sinceafewofthepositivelycharged particlesaredeectedverysharply, thentheymustencounterapositivelychargedmassiveparticleinsidetheatom.Wethereforeconcludethat allofthepositivechargeandmostofthemassofanatomiscontainedina nucleus .Thenucleusmust beverysmall,verymassive,andpositivelychargedifitistoaccountforthesharpdeections.Adetailed calculationbasedassumingthismodelrevealsthatthenucleusmustbeabout100,000timessmallerthan thesizeoftheatomitself.Theelectrons,alreadyknowntobecontainedintheatom,mustbeoutsideofthe nucleus,sincethenucleusispositivelycharged.Theymustmoveintheremainingspaceofthemuchlarger volumeoftheatom.Moreover,intotal,theelectronscompriselessthan0.05%ofthetotalmassofanatom. Thismodelaccountsforobservationofbothundeectedpassagemostof particlesandsharpdeection ofafew.Most particlespassthroughthevastemptyspaceoftheatom,whichisoccupiedonlybyelectron. Eventheoccasionalencounterwithoneoftheelectronshasnoeectonan particle'spath,sinceeach particleismuchmoremassivethananelectron.However,thenucleusisbothmassiveandpositivelycharged, butitisalsosmall.Therareencounterofan particlewiththenucleuswillresultinverylargedeections; ahead-oncollisionwithagoldatomnucleuswillsendan particledirectlybacktoitssource. 4.4Observation2:X-rayemission Althoughwecannowconcludethatanatomhasanuclearstructure,withpositivechargeconcentratedin averysmallnucleusandanumberofelectronsmovingaboutthenucleusinamuchlargervolume,wedo nothaveanyinformationonhowmanyelectronsthereareinanatomofanygivenelementorwhetherthis numberdependsonthetypeofatom.Doesagoldatomhavethesamenumberofelectronsasasilveratom? Allwecanconcludefromthedatagivenisthatthenumberofpositivechargesinthenucleusmustexactly equalthenumberofelectronsmovingoutsidethenucleus,sinceeachatomisneutral.Ournextdicultyis thatwedonotknowwhatthesenumbersare. Therelevantobservationseemsunrelatedtothepreviousobservations.Inthiscase,weexaminethe frequencyofx-raysemittedbyatomswhichhavebeenenergizedinanelectricalarc.Eachtypeofatom eachelementemitsafewcharacteristicfrequenciesofx-rays,whichdierfromoneatomtothenext.The lowestx-rayfrequencyemittedbyeachelementisfoundtoincreasewithincreasingpositionintheperiodic table. Mostamazingly,thereisanunexpectedrelationshipbetweenthefrequencyandtherelativemassofeach atom.Let'srankordertheelementsbyatomicmass,andassignanintegertoeachaccordingtoitsranking inorderbymass.InthePeriodicTable,thisrankordernumberalsocorrespondstotheelement'sposition inthePeriodicTable.Forexample,Hydrogenisassigned1,Heliumisassigned2,etc.Ifwenowplot thelowestfrequencyversusthepositionnumberintheperiodictable,wendthatthefrequencyincreases directlyasasimplefunctionoftherankingnumber.ThisisshownhereFigure4.2:X-rayFrequencies VersusAtomicNumber,wherewehaveplottedthesquarerootofthex-rayfrequencyasafunctionofthe rankingnumber.Afterasinglecorrection,thereisasimplestraight-linerelationshipbetweenthesenumbers. ThesinglecorrectionisthattherankingsofArgonandPotassiummustbereversed.Theseelementshave verysimilaratomicmasses.AlthoughArgonatomsareslightlymoremassivethanPotassiumatoms,the PeriodicLawrequiresthatweplaceArgonbeforePotassium,sinceArgonisamemberoftheinertgasgroup andPotassiumisamemberofthealkalimetalgroup.ByswitchingtheirordertocorrespondtothePeriodic Table,wecanmaintainthebeautifulrelationshipshownhereFigure4.2:X-rayFrequenciesVersusAtomic Number.

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18 CHAPTER4.THESTRUCTUREOFANATOM X-rayFrequenciesVersusAtomicNumber Figure4.2 Whyisthissimplerelationshipasurprise?Theintegerrankingofanelementbymasswouldnotseemto beaphysicalproperty.Wesimplyassignedthesenumbersinalistingoftheelementswhichweconstructed. However,wehavediscoveredthatthereisasimplequantitativerelationshipbetweenarealphysicalquantity thex-rayfrequencyandtherankingnumberweassigned.Moreover,thereareno"breaks"inthestraight lineshownhereFigure4.2:X-rayFrequenciesVersusAtomicNumber,meaningthatalloftheelementsin ourmasslistmustbeaccountedfor.Bothobservationsrevealthattherankingnumberofeachatommust alsobearealphysicalquantityitself,directlyrelatedtoastructuralpropertyofeachatom.Wenowcall therankingnumberthe atomicnumber ,sinceitisanumberwhichuniquelycharacterizeseachatom. Furthermore,weknowthateachatommustpossessanintegernumberofpositivecharges.Sincethex-ray datademonstratesaphysicalproperty,theatomicnumber,whichisalsoaninteger,thesimplestconclusion isthattheatomicnumberfromthex-raydataisthenumberofpositivechargesinthenucleus.Sinceeach atomisneutral,theatomicnumbermustalsoequalthenumberofelectronsinaneutralatom. Wenowknowagreatdealaboutthestructureofanatom.Weknowthattheatomhasanuclear structure,weknowthatthepositivechargesandmassoftheatomareconcentratedinthenucleus,andwe knowhowmanyprotonsandelectronseachatomhas.However,wedonotyetknowanythingaboutthe positioningandmovementoftheelectronsinthevastspacesurroundingthenucleus.

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19 4.5Observation3:Ionizationenergiesoftheatoms Eachelectronmustmoveaboutthenucleusinanelectricaleldgeneratedbythepositivechargeofthe nucleusandthenegativechargesoftheotherelectrons.Coulomb'slawdeterminesthepotentialenergyof attractionofeachelectrontothenucleus: V r = + Z e )]TJ/F11 9.9626 Tf 7.748 0 Td [(e r .1 where + Z e isthechargeonthenucleuswithatomicnumber Z and )]TJ/F11 9.9626 Tf 7.748 0 Td [(e isthechargeontheelectron, andristhedistancefromtheelectrontothenucleus.Thepotentialenergyofanelectroninanatomis negative.Thisisbecausewetakethepotentialenergyoftheelectronwhenremovedtogreatdistancefrom theatomverylarge r tobezero,sincetheelectronandthenucleusdonotinteractatlargedistance.In ordertoremoveanelectronfromanatom,wehavetoraisethepotentialenergyfromitsnegativevalueto zero.AccordingtoCoulomb'slaw,weexpectelectronsclosertothenucleustohavealowerpotentialenergy andthustorequiremoreenergytoremovefromtheatom. Wecandirectlymeasurehowmuchenergyisrequiredtoremoveanelectronfromanatom.Without concerningourselveswithhowthismeasurementismade,wesimplymeasuretheminimumamountofenergy requiredtocarryoutthefollowing"ionizationreaction": A g A + g + e )]TJ/F8 9.9626 Tf 8.385 -4.114 Td [( g .2 Here, A isanatominthegasphase,and A + isthesameatomwithoneelectron e )]TJ/F15 9.9626 Tf 6.725 -3.615 Td [(removedandisthusan ion.Theminimumenergyrequiredtoperformtheionizationiscalledthe ionizationenergy .Thevalues oftheionizationenergyforeachatominGroupsIthroughVIIIoftheperiodictableareshownasafunction oftheatomicnumberhereFigure4.3.

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20 CHAPTER4.THESTRUCTUREOFANATOM Figure4.3 ThisgureisveryreminiscentofthePeriodicLaw,whichstatesthatchemicalandphysicalproperties oftheelementsareperiodicfunctionsoftheatomicnumber.Noticethattheelementswiththelargest ionizationenergiesinotherwords,themosttightlyboundelectronsaretheinertgases.Bycontrast,the alkalimetalsaretheelementswiththesmallestionizationenergies.Inasingleperiodoftheperiodictable, betweeneachalkalimetalatomandthenextinertgasatom,theionizationenergyrisesfairlysteadily,falling dramaticallyfromtheinertgastothefollowingalkalimetalatthestartofthenextperiod. Weneedamodelwhichaccountsforthesevariationsintheionizationenergy.Areasonableassumption fromCoulomb'slawisthatthesevariationsareduetovariationsinthenuclearchargeatomicnumberand inthedistanceoftheelectronsfromthenucleus.Tobegin,wecanmakeaverycrudeapproximationthat theionizationenergyisjustthenegativeofthisattractivepotentialenergygivenbyCoulomb'slaw.Thisis crudebecausewehaveignoredthekineticenergyandbecauseeachelectronmaynothavexedvalueof r Nevertheless,thisapproximationgivesawaytoanalyzethisgureFigure4.3.Forexample,from Coulomb'slawitseemstomakesensethattheionizationenergyshouldincreasewithincreasingatomic number.ItiseasiertoremoveanelectronfromLithiumthanfromNeonbecausethenuclearchargein LithiumismuchsmallerthaninNeon.Butthiscannotbethewholepicture,becausethisargumentwould implythatSodiumatomsshouldhavegreaterionizationenergythanNeonatoms,wheninfactSodium atomshaveaverymuchlowerionizationenergy.Similarly,althoughtheionizationenergyrisesaswego fromSodiumtoArgon,theionizationenergyofArgonisstilllessthanthatofNeon,eventhoughthenuclear chargeinanArgonatomismuchgreaterthanthenuclearchargeinaNeonatom.Whathaveweomitted

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21 fromouranalysis? Theansweristhatwemustconsideralsothedistanceoftheelectronsfromthenucleus.Sinceitrequires muchlessenergytoionizeaSodiumatomthantoionizeaNeonatomeventhoughSodium'snuclearcharge isgreater,itmustbethattheelectronwhichweremovefromaSodiumatomismuchfartherfromthe nucleusthantheelectronintheNeonatom.Wecanmakethesamecomparisonoftheelectronsremoved duringionizationofNeonandArgonatoms:theArgonelectronmustbefartherfromthenucleusthanthe Neonelectron. Ontheotherhand,sincetheionizationenergyfairlysmoothlyincreasesaswemovefromLithiumto Neoninthesecondperiodofelements,thisrevealsthattheelectronsareincreasinglyattractedtothenucleus forgreaternuclearchargeandsuggeststhattheelectrons'distancefromthenucleusmightnotbevarying toogreatlyoverthecourseofasingleperiodofthetable. Ifwefollowthisreasoning,wecanevenestimatehowfaranelectronmighttypicallybefromthenucleus byusingourcrudeapproximationthattheionizationenergyisequaltothenegativeoftheCoulombpotential andsolvingfor r foreachatom.Thisgivesanestimateofdistanceoftheelectronfromthenucleus: r shell = )]TJ/F1 9.9626 Tf 9.409 14.047 Td [( + Z e )]TJ/F11 9.9626 Tf 7.748 0 Td [(e ionizationenergy .3 Valuesof r shell calculatedinthiswayareshownfortherst20elementshereFigure4.4.Alsoshownfor comparisonistheionizationenergyfortheseelements.Noticethattheapproximatedistanceoftheelectrons fromthenucleusincreasesinstepsexactlycoincidingwiththeincreasesanddipsintheionizationenergy.

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22 CHAPTER4.THESTRUCTUREOFANATOM Figure4.4 Althoughthesedistanceswehavecalculateddonothaveaprecisephysicalmeaning,thisgureFigure4.4suggestsasignicantconclusion.Theelectronsintheelementsarearrangedinto"shells"ofincreasinglygreaterdistancefromthenucleus.HydrogenandHelium,withoneandtwoelectrons,haveionization energiesconsistentwithelectronsatsimilarandclosedistancefromthenucleus.Thenthesecondrowelementslithiumthroughneonhavevirtuallyidenticalsizes,thoughlargerthanthatforthersttwoelements. Thethirdrowelements,sodiumargon,haveanapproximateelectron-nucleardistancewhichuctuatesabit butisconsistentlylargerthanthesecondrowelements. Becausethesizesoftheatomsappeartogrowinstepswhichcorrespondexactlytotheperiodsofthe PeriodicTable,itseemsthattheelectronsintheatomsaregroupedintosetswhicharedieringdistances awayfromthenucleus.Thersttwoelectrons,asinHelium,areclosetothenucleus,whereasadditional electrons,asinLithiumtoNeon,arefartherfromthenucleusthanthersttwo.Thesuggeststhat,for atomsLithiumtoNeon,thersttwoelectronsareinaninner"shell",andtheremainingelectronsareinan outer"shell." Wecanrenethisshellmodelfortheelectronsinanatomwithfurtheranalysisofionizationenergies.We canremoveanynumberofelectronsinsequence,formingionswithgreatercharge.Wehavebeenexamining therstionizationenergy, IE 1 ,buteachsuccessivelyremovedelectronhassuccessivelygreaterionization energy:

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23 Firstionizationenergy IE 1 : A g A + g + e )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( g .4 Secondionizationenergy IE 2 : A + g A 2+ g + e )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( g .5 Thirdionizationenergy IE 3 : A 2+ A 3+ g + e )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( g .6 Thesequentialionizationenergiesfortheelementsinthesecondrowoftheperiodictableareshownhere SuccessiveIonizationEnergieskJ/mol,p.23. SuccessiveIonizationEnergieskJ/mol Na Mg Al Si P S Cl Ar IE 1 496 738 578 787 1012 1000 1251 1520 IE 2 4562 1451 1817 1577 1903 2251 2297 2665 IE 3 6912 7733 2745 3231 2912 3361 3822 3931 IE 4 9543 10540 11575 4356 4956 4564 5158 5770 IE 5 13353 13630 14830 16091 6273 7013 6542 7238 IE 6 16610 17995 18376 19784 22233 8495 9458 8781 IE 7 20114 21703 23293 23783 25397 27106 11020 11995 Notethatthesecondionizationenergyisalwaysgreaterthantherst,andthethirdisalwaysgreater thanthesecond,etc.Thismakessense,sinceanelectronshouldbemorestronglyattractedtoapositively chargedatomthantoaneutralatom. However,thedatainthetableSuccessiveIonizationEnergieskJ/mol,p.23showasurprisingfeature. Inmostcases,theionizationenergyincreasesafairlylargeamountforsuccessiveionizations.Butforeach atom,thereisonemuchlargerincreaseinionizationinthesequence.InNaforexample, IE 2 isnearly10 timesgreaterthan IE 1 .Similarly, IE 3 isvetimesgreaterthan IE 2 forMg,although IE 2 islessthan twice IE 1 .ThedataforNathroughSallshowasinglelargestepinadditiontothesmallerincreasesinIE. Lookingcloselyandcountingelectrons,weseethatthisunusuallylargeincreasealwaysoccursforthe ionizationwherewehavealreadyremovedalloftheoutershellelectronsandarenowremovinganelectron fromtheinnershell.Thisoccursuniformlyacrossthesecondrowelements,indicatingthatourshellmodel isinfactaveryaccuratepredictorofthehigherionizationenergies.Wecannowtellhowmanyelectrons thereareintheoutershellofeachatom:itisequaltothenumberofelectronssincethelastinertgas. Wecanconcludethataninnershellis"lled"oncewehavethenumberofelectronsequaltothenumber inaninertgasatom.Thesubsequentelectronsareaddedtoanewoutershell.Thisiscommonlyreferred toasthe valenceshell oftheatom. However,wedonotknowwhyonlyalimitednumberofelectronscanresideineachshell.Thereisno obviousreasonatthispointwhyalltheelectronsinanatomdonotresideintheshellclosesttothenucleus. Similarly,thereisnoreasongivenforwhythenumberofelectronsinaninertgasatomexactlyllstheouter shell,withoutroomforevenasingleadditionalelectron.Thesequestionsmustbeaddressedfurther.

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24 CHAPTER4.THESTRUCTUREOFANATOM 4.6ReviewandDiscussionQuestions Exercise4.1 Explainhowthescatteringof particlesfromgoldfoilrevealsthatanatomcontainsamassive, positivelychargednucleuswhosesizeismuchsmallerthanthatoftheatom. Exercise4.2 Explainthesignicanceoftherelationshipbetweenthefrequencyofx-rayemissionfromeachatom andtheatomicrankingofthatatomintheperiodictable. Exercise4.3 Provideexperimentalevidencewhichrevealsthattheelectronsinanatomaregroupedintoa valenceshellandinnershellelectrons. Exercise4.4 Stateandexplaintheevidencewhichrevealsthattheoutershellofeachinertgasatomisfull. Exercise4.5 Whydoestheionizationenergyforeachsuccessiveionizationincreaseforeveryatom?Whyisthe increasefrom IE 4 to IE 5 inSimuchlargerthananyoftheotherincreasesforSi?

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Chapter5 QuantumEnergyLevelsInAtoms 1 5.1Foundation Theatomicmoleculartheoryprovidesusaparticulateunderstandingofmatter.Eachelementischaracterizedasconsistingofidentical,indestructibleatomswithatomicweightswhichhavebeendetermined. Compoundsconsistofidenticalmolecules,eachmadeupfromaspecicnumberofatomsofeachofthe componentelements.Wealsoknowthatatomshaveanuclearstructure,meaningthatallofthepositive chargeandvirtuallyallofthemassoftheatomareconcentratedinanucleuswhichisaverysmallfraction ofthevolumeoftheatom.Finally,weknowthattheelectronsintheatomarearrangedin"shells"about thenucleus,witheachshellfartherfromthenucleusthattheprevious.Theelectronsinoutershellsaremore weaklyattachedtotheatomthantheelectronsintheinnershells,andonlyalimitednumberofelectrons cantineachshell. 5.2Goals Theshellmodeloftheatomisagoodstartinunderstandingthedierencesinthechemicalpropertiesof theatomsofdierentelements.Forexample,wecanunderstandtheperiodicityofchemicalandphysical propertiesfromourmodel,sinceelementsinthesamegrouphavethesamenumberofelectronsinthevalence shell. However,therearemanydetailsmissingfromourdescription.Otherthanaverycrudecalculationof "distance"oftheshellsfromthenucleus,wehavenodescriptionofwhatthedierencesarebetweenthe electronsindierentshells.Whatpreciselyisa"shell?" Mostimportantly,thearrangementofelementsintogroupsandtheperiodicityofchemicalproperties bothdependontheconceptthatashellis"lled"byacertainnumberofelectrons.Lookingatthenumber ofelementsineachperiod,thenumberofelectronswhichllsashelldependsonwhichshellisbeinglled. Insomecases,ashellislledbyeightelectrons,inothers,itappearstobe18electrons.Whatdetermines howmanyelectronscan"t"inashell?Whyistherealimitatall? Finally,acloserlookattheionizationenergieshereFigure4.3revealsthatourshellmodelmustbe incomplete.OurmodelimpliesthattheelementsofthesecondperiodfromLithiumtoNeonhavetheir valenceelectronsinthesecondshell.Withincreasingnuclearcharge,theionizationenergyoftheseatoms shouldincreasefromLithiumtoNeon.Asageneraltrend,thisistrue,buttherearevariations.Notethat theionizationenergyofOxygenatomsislessthanthatofNitrogenatoms.Weneedtopursueadditional detailinourmodelofthestructureoftheatom. 1 Thiscontentisavailableonlineat. 25

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26 CHAPTER5.QUANTUMENERGYLEVELSINATOMS 5.3Observation1:TheSpectrumofHydrogen Tobegin,weneedtoknowalittleaboutlight.Allformsofelectromagneticradiationtravelasanoscillating wave,withanelectriceldcomponentperpendiculartoamagneticeldcomponent.Asawave,theradiation canbecharacterizedbyits"wavelength",symbolizedas ,whichisthedistancebetweenadjacentpeaksin thewave.Dierentwavelengthscorrespondtodierentformsofelectromagneticradiation.Forexample, microwaveradiationhaswavelengthintherangeof 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 to 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 meters,whereasx-rayradiationhas wavelengthintherange 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(9 to 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(10 meters.Radiationwhichisvisibletothehumaneyehaswavelength intheverynarrowrangefrom 3 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(7 to 7 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(7 meters. Radiationcanalsobecharacterizedbythefrequencyoftheelectromagneticwave,whichisthenumber ofpeaksinthewavewhichpassapointinspacepersecond.Frequencyissymbolizedby .Thespeedwhich lighttravelsinavacuuminthesameforallformsofelectromagneticradiation, c =2 : 997 10 8 m s .Assuch, wecanrelatethefrequencyoflighttothewavelengthoflightbytheequation m )]TJ/F11 9.9626 Tf 4.567 -8.069 Td [(s )]TJ/F7 6.9738 Tf 6.226 0 Td [(1 = c m s .1 Thelongerthewavelength ,thelowerthefrequency .Thismakessensewhenwerememberthatlight travelsataxedspeed.Whenthewavelengthislonger,fewerpeakswillpassapointinspaceinasecond. Fromthisequation,thereisaspecicrelationshipbetweenfrequencyandwavelength,andeitherorboth canbeusedtocharacterizethepropertiesofradiation. Withthisbackgroundinhand,wecanuseourunderstandingoflighttopursuemoredataaboutthe energiesofelectronsinatoms.Ionizationenergiestellushowmuchenergyisrequiredtoremoveanelectron fromanatom,butdonottellwhathappensifanelectronchangesitsenergyinanatom.Toanalyzethis, weneedameanstomeasuretheenergiesgainedorlostbyanatom.Onewaytodosoistoanalyzethe "spectrum"ofanatom,whichisthesetoffrequenciesoflightemittedbytheatom.Sincehydrogenisthe simplestatom,weanalyzethehydrogenspectrumrst.Wendthat,ifwepassacurrentofelectricity throughasampleofhydrogengas,lightisemitted.Carefulanalysisshowsthat,althoughsomeofthislight isemittedby H 2 molecules,someofthelightisalsoemittedbyHatoms.Sincelightisaformofenergy, thentheseHatomsmustreleaseenergysuppliedtothembytheelectronsinthecurrent. Mostimportantly,ifwepassthelightemittedbythehydrogengassamplethroughaprism,wecan separatethecolorsasinarainbow,eachwithacharacteristicfrequency.Theresultantimageofseparated colorsiscalledthe spectrum ofhydrogen.Wendinthisexperimentthatthereareonlyfourfrequencies fourcolorsoflightintheemissionthatarevisible.Themostintenseofthelinesinthespectrumisbright red,butthereareblueandvioletlines.Itturnsoutthattherearealsomanyotherfrequenciesoflight emittedwhichareinvisibletothehumaneye. Carefulobservationandanalysisrevealsthateveryfrequencyinthehydrogenatomspectrumcanbe predictedbyaverysimpleformula,calledtheRydbergequation: = R 1 n 2 )]TJ/F8 9.9626 Tf 15.276 6.74 Td [(1 m 2 .2 where R istheRydbergconstant 3 : 29 10 15 s )]TJ/F7 6.9738 Tf 6.227 0 Td [(1 n and m areintegers,2,3,....Eachchoiceof n and m predictsasingleobservedfrequencyinthehydrogenatomspectrum. Theatomsofallelementsemitradiationwhenenergizedinanelectriccurrent,andasdoallmolecules ofallcompounds.However,wendthatthespecicfrequenciesoflightemittedarecharacteristicofeach atomormolecule.Inotherwords,thespectrumofeachelementisuniquetoeachelementorcompound. Asaresult,thespectrumofeachsubstancecanbeusedtoidentifythatsubstance.NotethattheRydberg equationtellsusonlythespectrumofhydrogen. Ourinterestisinthefactthattheradiationemittedbyanatomtellsusabouttheamountsofenergywhich canbereleasedbyanatom.Forahydrogenatom,forexample,thesechangesinenergymustcorrespondto theamountsofenergywhichtheelectronsinsidetheatomcangainorlose. Atthispoint,weneedtorelatethefrequencyofradiationemittedbyanatomtotheamountofenergy lostbytheelectronintheatom.Wethusexaminesomeobservationsabouttheenergyofradiation.

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27 5.4Observation2:ThePhotoelectricEect Whenalightsourceisdirectedatametalsurface,itisfoundundermanycircumstancesthatelectrons areejectedfromthesurface.Thisphenomenoniscalledthe"photoelectriceect."Theseelectronscanbe collectedtoproduceausableelectriccurrent.Thiseecthasavarietyofcommonpracticalapplications, forexample,in"electriceye"devices.Itisreasonabletoexpectthatacertainamountofenergyisrequired toliberateanelectronfromametalsurface,sincetheelectronisattractedtothepositivelychargednucleiin themetal.Thus,inorderfortheelectrontoescape,thelightmustsupplysucientenergytotheelectron toovercomethisattraction. Thefollowingexperimentalobservationsarefoundwhenstudyingthephotoelectriceect.First,inorder fortheeecttobeobserved,thelightmustbeofatleastaminimumfrequencywhichwecallthe threshold frequency 0 .Thisfrequencyisacharacteristicforagivenmetal.Thatis,itisthesamevalueforeach sampleofthatmetal,butitvariesfromonemetaltothenext.Forlowfrequencylight,photoelectronsare notobservedinanynumber,nomatterhowintensethelightsourceis.Forlightwithfrequencyabove 0 thenumberofphotoelectronsemittedbythemetalmeasuredbythephotoelectriccurrent, increases directlywiththeintensityofthelight.TheseresultsareshowninFigure5.1ThePhotoelectricEect. ThePhotoelectricEect a b Figure5.1: isthephotoelectriccurrent, isthefrequencyofincidentlight,and I istheintensityof incidentlight.aForphotoelectronstobeemitted,thelightfrequencymustbegreaterthanathreshold value.bIfthefrequencyishighenough,thenumberofphotoelectronsincreasesdirectlywiththelight intensity. Second,wecanmeasuretheenergiesoftheelectronsemittedbythemetal.Foragivenmetal,all photoelectronshavethesamekineticenergyforaxedfrequencyoflightabove 0 .Thisxedkineticenergy isindependentoftheintensityofthelightsource.Asthefrequencyofthelightisincreased,thekineticenergy oftheemittedelectronsincreasesproportionally.TheseresultsareshowninFigure5.2MorePhotoelectric Eect.

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28 CHAPTER5.QUANTUMENERGYLEVELSINATOMS MorePhotoelectricEect a b Figure5.2: KE isthephotoelectronkineticenergy, isthefrequencyofincidentlight,and I isthe intensityofincidentlight.aIfthefrequencyishighenough,theenergyoftheelectronsincreases directlywiththefrequency.bHowever,theenergyofthephotolectronsdoesnotdependonthelight intensity. Aretheseresultssurprising?Tothephysicistsattheendofthenineteenthcentury,theanswerwas yes,verysurprisingindeed.Theyexpectedthattheenergyofthelightsourceshouldbedeterminedbyits intensity.Hence,theenergyrequiredtoejectaphotoelectronshouldbesuppliedbylightofhighintensity, nomatterhowlowthefrequencyoftheradiation.Thus,thereshouldbenothresholdfrequency,below whichnoelectronsareemitted.Moreover,thekineticenergyoftheelectronsshouldincreasewithintensity, notwithlightfrequency.Thesepredictionsarenotobserved,sotheresultsarecountertophysicalintuition. Wecanaccountfortheseresultsinastraightforwardbutperhapsnon-obviousmanner.Einsteinprovided theexplanationin1905.Sincethekineticenergyoftheemittedphotoelectronsincreasesproportionallywith increasesinthefrequencyofthelightabovethethresholdfrequency,wecanconcludefromconservationof totalenergythattheenergysuppliedbythelighttotheejectedelectronmustbeproportionaltoitsfrequency: E / .Thisdoesnotimmediatelyaccountfortheexistenceofthethresholdfrequency,though,sinceit wouldstillseemtobethecasethatevenlowfrequencylightwouldpossesshighenergyiftheintensity weresucient.Bythisreasoning,highintensity,lowfrequencylightshouldthereforeproduceasmany photoelectronsasareproducedbylowintensity,highfrequencylight.Butthisisnotobserved. Thisisaverychallengingpuzzle,andananalogyhelpstorevealthesubtleanswer.Imaginetryingto knockpiecesoutofawallbythrowingobjectsatit.Wediscoverthat,nomatterhowmanypingpongballs wethrow,wecannotknockoutapieceofthewall.Ontheotherhand,onlyasinglebowlingballisrequired toaccomplishthetask.Theresultsofthis"experiment"aresimilartotheobservationsofthephotoelectric eect:verylittlehighfrequencylightcanaccomplishwhatanenormousamountoflowfrequencylight cannot.Thekeytounderstandingourimaginaryexperimentisknowingthat,althoughtherearemanymore ping-pongballsthanbowlingballs,itisonlytheimpactofeachindividualparticlewiththewallwhich determineswhathappens. Reasoningfromthisanalogy,wemustconcludethattheenergyofthelightissuppliedin"bundles"or "packets"ofconstantenergy,whichwewillcall photons .Wehavealreadyconcludedthatthelightsupplies energytotheelectronwhichisproportionaltothelightfrequency.Nowwecansaythattheenergyofeach

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29 photonisproportionaltothefrequencyofthelight.Theintensityofthelightisproportionaltothenumber ofthesepackets.Thisnowaccountsforthethresholdfrequencyinastraightforwardway.Foraphotonto dislodgeaphotoelectron,itmusthavesucientenergy,byitself,tosupplytotheelectrontoovercomeits attractiontothemetal.Althoughincreasingtheintensityofthelightdoesincreasethetotalenergyofthe light,itdoesnotincreasetheenergyofanindividualphoton.Therefore,ifthefrequencyofthelightistoo low,thephotonenergyistoolowtoejectanelectron.Referringbacktotheanalogy,wecansaythata singlebowlingbowlcanaccomplishwhatmanyping-pongballscannot,andasinglehighfrequencyphoton canaccomplishwhatmanylowfrequencyphotonscannot. Theimportantconclusionforourpurposesisthat lightenergyisquantizedintopacketsofenergy .The amountofenergyineachphotonisgivenbyEinstein'sequation, E = h .3 where h isaconstantcalledPlanck'sconstant. 5.5QuantizedEnergyLevelsinHydrogenAtoms Wecancombinetheobservationofthehydrogenatomspectrumwithourdeductionthatlightenergyis quantizedintopacketstoreachanimportantconclusion.Eachfrequencyoflightinthespectrumcorresponds toaparticularenergyoflightand,therefore,toaparticularenergy loss byahydrogenatom,sincethislight energyisquantizedintopackets.Furthermore,sinceonlycertainfrequenciesareobserved,thenonlycertain energylossesarepossible.Thisisonlyreasonableiftheenergyofeachhydrogenatomisrestrictedtocertain specicvalues.Ifthehydrogenatomcouldpossessanyenergy,thenitcouldloseanyamountofenergyand emitaphotonofanyenergyandfrequency.Butthisisnotobserved.Therefore,theenergyoftheelectron inahydrogenatommustberestrictedtocertain energylevels TheHydrogenatomspectrumalsotellsuswhattheseenergylevelsare.Recallthatthefrequenciesof radiationemittedbyHydrogenatomsaregivenbytheRydbergequation.2.Eachchoiceofthepositive integers n and m predictsasingleobservedfrequencyinthehydrogenatomspectrum. Eachemittedfrequencymustcorrespondtoanenergy h byEinstein'sequation.3.Thisphoton energymustbethe dierence betweentwoenergylevelsforahydrogenelectron,sincethatistheamountof energyreleasedbytheelectronmovingfromoneleveltotheother.Iftheenergiesofthetwolevelsare E m and E n ,thenwecanwritethat h = E m )]TJ/F11 9.9626 Tf 9.963 0 Td [(E n .4 BycomparingthistotheRydbergequation,eachenergylevelmustbegivenbytheformula E n = )]TJ/F11 9.9626 Tf 7.749 0 Td [(h R 1 n 2 .5 Wecandrawtwoconclusions.First,theelectroninahydrogenatomcanexistonlywithcertainenergies, correspondingtomotioninwhatwenowcallastateoran orbital .Second,theenergyofastatecanbe characterizedbyaninteger quantumnumber ,n=1,2,3,...whichdeterminesitsenergy. Theseconclusionsarereinforcedbysimilarobservationsofspectraproducedbypassingacurrentthrough otherelements.Onlyspecicfrequenciesareobservedforeachatom,althoughonlythehydrogenfrequencies obeytheRydbergformula. Weconcludethattheenergiesofelectronsinatomsare"quantized,"thatis,restrictedtocertainvalues. Wenowneedtorelatethisquantizationofenergytotheexistenceofshells,asdevelopedinapreviousstudy Chapter4.

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30 CHAPTER5.QUANTUMENERGYLEVELSINATOMS 5.6Observation3:PhotoelectronSpectroscopyofMulti-Electron Atoms Theionizationenergyofanatomtellsustheenergyoftheelectronorelectronswhichareathighestenergy intheatomandarethuseasiesttoremovefromtheatom.Tofurtheranalyzetheenergiesoftheelectrons moretightlyboundtothenucleus,weintroduceanewexperiment.Thephotoelectriceectcanbeapplied toionizeatomsinagas,inaprocessoftencalled photoionization .Weshinelightonanatomandmeasure theminimumfrequencyoflight,correspondingtoaminimumenergy,whichwillionizeanelectronfroman atom.Whenthefrequencyoflightistoolow,thephotonsinthatlightdonothaveenoughenergytoionize electronsfromanatom.Asweincreasethefrequencyofthelight,wendathresholdatwhichelectrons begintoionize.Abovethisthreshold,theenergy h ofthelightoffrequency isgreaterthantheenergy requiredtoionizetheatom,andtheexcessenergyisretainedbytheionizedelectronaskineticenergy. Inphotoelectronspectroscopy,wemeasurethekineticenergyoftheelectronswhichareionizedbylight. Thisprovidesameansofmeasuringtheionizationenergyoftheelectrons.Byconservationofenergy,the energyofthelightisequaltotheionizationenergy IE plusthekineticenergy KE oftheionizedelectron: h = IE + KE .6 Thus,ifweuseaknownfrequency andmeasure KE ,wecandetermine IE .Themoretightlyboundan electronistotheatom,thehighertheionizationenergyandthesmallerthekineticenergyoftheionized electron.Ifanatomhasmorethanoneelectronandtheseelectronshavedierentenergies,thenforagiven frequencyoflight,wecanexpectelectronstobeejectedwithdierentkineticenergies.Thehigherkinetic energiescorrespondtotheweaklyboundouterelectrons,andthelowerkineticenergiescorrespondtothe tightlyboundinnerelectrons. Theionizationenergiesforthersttwentyelementsaregiveninp.30.Wenotethatthereisasingle ionizationenergyforhydrogenandhelium.Thisisconsistentwiththeshellmodeloftheseatomssince,in bothoftheseatoms,theelectronorelectronsareintheinnermostshell.Theenergiesoftheseelectrons correspondtothe n =1 energylevelofthehydrogenatom.Inlithiumandberyllium,therearetwoionization energies.Again,thisisconsistentwiththeshellmodel,sincenowthereareelectronsinbothofthersttwo shells.Notealsothattheionizationenergyoftheinnershellelectronsincreasesaswegofromhydrogento lithiumtoberyllium,becauseoftheincreaseinnuclearcharge.Thelowerenergyelectronscorrespondto the n =1 energylevelofhydrogenandthehigherenergyelectronscorrespondtothe n =2 energylevel.

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31 Element IonizationEnergyMJ/mol H 1.31 He 2.37 Li 6.26 0.52 Be 11.5 0.90 B 19.3 1.36 0.80 C 28.6 1.72 1.09 N 39.6 2.45 1.40 O 52.6 3.12 1.31 F 67.2 3.88 1.68 Ne 84.0 4.68 2.08 Na 104 6.84 3.67 0.50 Mg 126 9.07 5.31 0.74 Al 151 12.1 7.79 1.09 0.58 Si 178 15.1 10.3 1.46 0.79 P 208 18.7 13.5 1.95 1.01 S 239 22.7 16.5 2.05 1.00 Cl 273 26.8 20.2 2.44 1.25 Ar 309 31.5 24.1 2.82 1.52 K 347 37.1 29.1 3.93 2.38 0.42 Ca 390 42.7 34.0 4.65 2.9 0.59 Surprisingly,though,boronhasthreeionizationenergies,whichdoesnotseemconsistentwiththeshell model.Fromthehydrogenatomenergylevels,wewouldhaveexpectedthatall n =2 electronswouldhave thesameenergy.Wecannotethatthetwosmallerionizationenergiesinboronarecomparableinmagnitude andsmallerbymorethanafactoroftenthantheionizationenergyoftheelectronsintheinnershell.Thus, theelectronsintheouter n =2 shellapparentlyhavecomparableenergies,buttheyarenotidentical.The separationofthesecondshellintotwogroupsofelectronswithtwocomparablebutdierentenergiesis apparentforelementsborontoneon. Assuch,weconcludefromtheexperimentaldatathatthesecondshellofelectronsshouldbedescribed astwo subshells withslightlydierentenergies.Forhistoricalreasons,thesesubshellsarereferredtoasthe asthe"2s"and"2p"subshells,with2selectronsslightlylowerinenergythan2pelectrons.Theenergiesof the2sand2pelectronsdecreasefromborontoneon,consistentwiththeincreaseinthenuclearcharge. Beginningwithsodium,weobservefourdistinctionizationenergies,andbeginningwithaluminumthere areve.Notefortheseelementsthatthefourthandfthionizationenergiesareagainroughlyafactoroften smallerthanthesecondandthirdionizationenergies,whichareinturnatleastafactoroftenlessthanthe rstionizationenergy.Thus,itappearsthattherearethreeshellsofelectronsfortheseatoms,consistent withourpreviousshellmodel.Aswith n =2 ,the n =3 shellisagaindividedintotwosubshells,nowcalled the3sand3psubshells. Thesedataalsorevealhowmanyelectronscanresideineachsubshell.Ineachnlevel,therearetwo elementswhichhaveonlytheionizationenergyforthessubshell.Hence,ssubshellscanholdtwoelectrons. Bycontrast,thereare6elementswhichhaveboththesandpsubshellionizationenergies,sothepsubshell canhold6electrons.

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32 CHAPTER5.QUANTUMENERGYLEVELSINATOMS Theshellandsubshellorganizationofelectronenergiescanalsobeobservedbymeasuringthe"electron anity"oftheatoms.Electronanityistheenergyreleasedwhenanelectronisaddedtoanatom: A g + e )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( g A )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( g .7 IfthereisastrongattractionbetweentheatomAandtheaddedelectron,thenalargeamountofenergyis releasedduringthisreaction,andtheelectronanityisalargepositivenumber.Asanote,thisconvention istheoppositeoftheoneusuallyappliedforenergychangesinreactions:exothermicreactions,whichgive oenergy,conventionallyhavenegativeenergychanges. Theelectronanitiesofthehalogensarelargepositivevalues:theelectronanitiesofF,Cl,andBrare 328.0kJ/mol,348.8kJ/mol,and324.6kJ/mol.Thus,theattachedelectronsarestronglyattractedtothe nucleusineachoftheseatoms.Thisisbecausethereisroominthecurrentsubshelltoaddanadditional electron,sinceeachatomhas5pelectrons,andthecorechargefeltbytheelectroninthatsubshellislarge. Bycontrast,theelectronanitiesoftheinertgasesare negative :theadditionofanelectrontoaninert gasatomactuallyrequiresthe input ofenergy,ineect,toforcetheelectronintoplace.Thisisbecausethe addedelectroncannottinthecurrentsubshellandmustbeaddedtoanewshell,fartherfromthenucleus. Assuch,thecorechargefeltbytheaddedelectronisveryclosetozero. Similarly,theelectronanitiesoftheelementsBe,Mg,andCaareallnegative.Thisisagainbecause thessubshellintheseatomsalreadyhastwoelectrons,sotheaddedelectronmustgointoahigherenergy subshellwithamuchsmallercorecharge. 5.7ElectronWaves,theUncertaintyPrinciple,andElectronEnergies Wenowhaveafairlydetaileddescriptionoftheenergiesoftheelectronsinatoms.Whatwedonothave isamodelwhichtellsuswhatfactorsdeterminetheenergyofanelectroninashellorsubshell.Nordowe haveamodeltoexplainwhytheseenergiesaresimilarbutdierentforelectronsindierentsubshells. Acompleteanswertothesequestionsrequiresadevelopmentofthequantumtheoryofelectronmotion inatoms.Becausethepostulatesofthisquantumtheorycannotbereadilydevelopedfromexperimental observations,wewillconcernourselveswithafewimportantconclusionsonly. Therstimportantconclusionisthatthemotionofanelectroninanatomisdescribedbyawave function.Interpretationofthewavemotionofelectronsisaverycomplicatedproposition,andwewillonly dealatpresentwithasingleimportantconsequence,namelythe uncertaintyprinciple .Acharacteristic ofwavemotionisthat,unlikeaparticle,thewavedoesnothaveadenitepositionatasinglepointinspace. Bycontrast,thelocationofaparticleisprecise.Therefore,sinceanelectrontravelsasawave,wemust concludethatwecannotdeterminethepreciselocationoftheelectroninanatom.Thisis,forourpurposes, theuncertaintyprincipleofquantummechanics.We can makemeasurementsofthelocationoftheelectron, butwendthateachmeasurementresultsinadierentvalue.Wearethenforcedtoacceptthatwecannot determinethepreciselocation.Weareallowed,however,todeterminea probabilitydistribution for wheretheelectronisobserved. Thisprobabilitydistributionisdeterminedbyquantummechanics.Themotionoftheelectronina hydrogenatomisdescribedbyafunction,oftencalledthe wavefunction orthe electronorbital and typicallydesignatedbythesymbol isafunctionofthepositionoftheelectron r ,andquantum mechanicstellsusthat j j 2 isthe probability ofobservingtheelectronatthelocation r Eachelectronorbitalhasanassociatedconstantvalueoftheelectronicenergy, E n ,inagreementwithour earlierconclusions.Infact,quantummechanicsexactlypredictstheenergyshellsandthehydrogenatom spectrumweobserve.Theenergyofanelectroninanorbitalisdeterminedprimarilybytwocharacteristics oftheorbital.Therst,ratherintuitive,propertydeterminestheaveragepotentialenergyoftheelectron: anorbitalwhichhassubstantialprobabilityinregionsoflowpotentialenergywillhavealowtotalenergy. ByCoulomb'slaw,thepotentialenergyarisingfromnucleus-electronattractionislowerwhentheelectron isnearerthenucleus.Inatomswithmorethanoneelectron,electron-electronrepulsionalsocontributesto thepotentialenergy,asCoulomb'slawpredictsanincreaseinpotentialenergyarisingfromtherepulsionof likecharges.

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33 Asecondorbitalcharacteristicdeterminesthecontributionofkineticenergy,viaamoresubtleeect arisingoutofquantummechanics.Asaconsequenceoftheuncertaintyprinciple,quantummechanics predictsthat,themoreconnedanelectronistoasmallerregionofspace,thehighermustbeitsaverage kineticenergy.Sincewecannotmeasurethepositionofelectronprecisely,wedenetheuncertaintyinthe measurementas x .Quantummechanicsalsotellsusthatwecannotmeasurethemomentumofanelectron preciselyeither,sothereisanuncertainty p inthemomentum.Inmathematicaldetail,theuncertainty principlestatesthattheseuncertaintiesarerelatedbyaninequality: x p h 4 .8 where h isPlanck'sconstant, 6 : 62 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(34 Js previouslyseeninEinstein'sequation.3fortheenergyof aphoton.Thisinequalityrevealsthat,whenanelectronmovesinasmallareawithacorrespondinglysmall uncertainty x ,theuncertaintyinthemomentum p mustbelarge.For p tobelarge,themomentum mustalsobelarge,andsomustbethekineticenergy. Therefore,themorecompactanorbitalis,thehigherwillbetheaveragekineticenergyofanelectronin thatorbital.Thisextrakineticenergy,whichcanberegardedasthe connementenergy ,iscomparablein magnitudetotheaveragepotentialenergyofelectron-nuclearattraction.Therefore,ingeneral,anelectron orbitalprovidesacompromise,somewhatlocalizingtheelectroninregionsoflowpotentialenergybut somewhatdelocalizingittoloweritsconnementenergy. 5.8ElectronOrbitalsandSubshellEnergies Weneedtoaccountforthedierencesinenergiesoftheelectronsindierentsubshells,sinceweknowthat, inaHydrogenatom,theorbitalenergydependsonlyonthenquantumnumber.Werecallthat,inthe Hydrogenatom,thereisa single electron.Theenergyofthatelectronisthusentirelyduetoitskinetic energyanditsattractiontothenucleus.Thesituationisdierentinallatomscontainingmorethanone electron,becausetheenergyoftheelectronsisaectedbytheirmutualrepulsion.Thisrepulsionisvery diculttoquantify,butourmodelmusttakeitintoaccount. Asimplewaytodealwiththeeectofelectron-electronrepulsionistoexaminetheshellstructureof theatom.Thetwo n =1 electronsinberylliumareinashellwithacomparativelyshortaveragedistance fromthenucleus.Therefore,thetwo n =2 electronsareinashellwhichis,onaverage,"outside"ofthe n =1 shell.The n =1 electronsarethusthe"core"andthe n =2 electronsareinthevalenceshell.This structureallowsustoseeinasimplewaytheeectofelectron-electronrepulsionontheenergiesofthe n =2 electrons.Each n =2 electronisattractedbythe+4chargeonthetinyberylliumnucleus,butis repelledbythetwo-1chargesfromtheinnershellformedbythetwo n =1 electrons.Net,then,an n =2 electroneectively"sees"roughlya+2nuclearcharge.Werefertothis+2asthe"corecharge"sinceitis thenetchargeonthecoreresultingfromthebalanceofattractiontothenucleusandrepulsionfromthe coreelectrons.Thenucleusispartially"shielded"fromthevalenceelectronsbythecoreelectrons. Thisshieldingeectdoesnotseemtoaccountforthedierenceinionizationenergiesbetween2sand2p orforthelowerionizationenergyofboroncomparedtoberyllium,since,ineachatom,thevalenceelectrons areinthe n =2 shell.However,theshieldingeectisnotperfect.Recallthatweonlyknowthe probabilities forobservingthepositionsoftheelectrons.Therefore,wecannotdenitelystatethatthe n =2 electrons areoutsideofthe n =1 core.Infact,thereissomeprobabilitythatan n =2 electronmightbefoundinside the n =1 core,aneectcalled"corepenetration."Whenan n =2 electrondoespenetratethecore,itisno longershieldedfromthenucleus.Inthiscase,the n =2 electronisverystronglyattractedtothenucleus anditsenergyisthuslowered.Whatistheextentofthispenetration?Wemustconsultquantumtheory. TheanswerisinFigure5.3ProbabilityforanElectronataDistancerfromaHydrogenNucleus,which showstheprobabilityofndinganelectronadistance r awayfromthenucleusforeachofthe1s,2s,and 2porbitals.Wecanseethatthereisagreaterprobabilitythoughsmallforthe2selectrontopenetrate thecorethanforthe2pelectrontodoso.

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34 CHAPTER5.QUANTUMENERGYLEVELSINATOMS ProbabilityforanElectronataDistancerfromaHydrogenNucleus Figure5.3 Asaresultofthecorepenetration,anelectronina2sorbitalfeelsagreater"eectivenuclearcharge" thanjustthecorecharge,whichwasapproximatedbyassumingperfectshielding.Thustheeectivenuclear chargefora2selectronisgreaterthantheeectivenuclearchargefora2pelectron.Therefore,theenergy ofanelectroninthe2sorbitalinberylliumislowerthanitwouldbeinthe2porbital. Adetailedanalysisfromquantummechanicsgivesthefollowingorderingoforbitalsinorderofincreasing energy: 1 s< 2 s< 2 p< 3 s< 3 p< 4 s< 3 d< 4 p< 5 s<::: .9 Thisorderingcanberationalizedonthebasisofeectivenuclearcharge,shielding,andcorepenetration. 5.9ReviewandDiscussionQuestions Exercise5.1 Thephotoelectriceectdemonstratesthatradiationenergyisquantizedinto"packets"orphotons. Explainhowandwhythisobservationisofsignicanceinunderstandingthestructureofatoms. Exercise5.2 Explainhowwecanknowthathigherfrequencylightcontainshigherenergyphotons. Exercise5.3 Electronanityistheenergyreleasedwhenanelectronisattachedtoanatom.Ifanatomhasa positiveelectronanity,theaddedelectronisattractedtothenucleustoformastablenegativeion. Whydoesn'taBerylliumatomhaveapositiveelectronanity?Explainhowthisdemonstrates thattheenergyofa2sorbitalislessthantheenergyofa2porbital. Exercise5.4 Whydoesaninertgasatomhaveahighionizationenergybutalowelectronanity?Whydo thesepropertiescombinetomaketheatomsofinertgasesunreactive?

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35 Exercise5.5 Considerelectronsfromtwodierentsubshellsinthesameatom.Inphotoelectronspectroscopy, thelowerenergyelectronhasahigherionizationenergybutisobservedtohavelowerkineticenergy afterionization.Reconcilethelowerenergywiththehigherionizationenergywiththelowerkinetic energy. Exercise5.6 Chlorineatomshave5distinctionizationenergies.Explainwhy.Predictthenumberofionization energiesforBromineatoms,andexplainyouranswer.Hint:examinethestructureoftheperiodic table. Exercise5.7 WhydoesaBromineatomhaveamuchsmallerradiusthanaPotassiumatom,eventhoughaBr atomhas16moreelectronsthandoesaKatom? Exercise5.8 Explainwhyelectronsconnedtosmallerorbitalsareexpectedtohavehigherkineticenergies. Exercise5.9 Dene"shielding"inthecontextofelectron-electronrepulsion.Whatisthesignicanceofshielding indeterminingtheenergyofanelectron?Howistheaectedbycorepenetration?

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36 CHAPTER5.QUANTUMENERGYLEVELSINATOMS

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Chapter6 CovalentBondingandElectronPair Sharing 1 6.1Foundation Webeginwithourunderstandingoftherelationshipbetweenchemicalbehaviorandatomicstructure.That is,weassumethePeriodicLawthatthechemicalandphysicalpropertiesoftheelementsareperiodic functionsofatomicnumber.Wefurtherassumethestructureoftheatomasamassive,positivelycharged nucleus,whosesizeismuchsmallerthanthatoftheatomasawhole,surroundedbyavastopenspacein whichmovenegativelychargedelectrons.Theseelectronscanbeeectivelypartitionedintoacoreanda valenceshell,anditisonlytheelectronsinthevalenceshellwhicharesignicanttothechemicalproperties oftheatom.Thenumberofvalenceelectronsineachatomisequaltothegroupnumberofthatelementin thePeriodicTable. 6.2Goals Theatomicmoleculartheoryisextremelyusefulinexplainingwhatitmeanstoformacompoundits componentelements.Thatis,acompoundconsistsofidenticalmolecules,eachcomprisedoftheatomsof thecomponentelementsinasimplewholenumberratio.However,theatomicmoleculartheoryalsoopens upawiderangeofnewquestions.Wewouldliketoknowwhatatomicpropertiesdeterminethenumberof atomsofeachtypewhichcombinetoformstablecompounds.Whyaresomecombinationsobservedand othercombinationsnotobserved?Someelementswithverydissimilaratomicmassesforexample,iodine andchlorineformverysimilarchemicalcompounds,butotherelementswithverysimilaratomicmasses forexample,oxygenandnitrogenformverydissimilarcompounds.Whatfactorsareresponsibleforthe bondingpropertiesoftheelementsinasimilargroup?Ingeneral,weneedtoknowwhatforcesholdatoms togetherinformingamolecule. Wehavedevelopedadetailunderstandingofthestructureoftheatom.Ourtasknowistoapplythis understandingtodevelopasimilarlevelofdetailabouthowatomsbondtogethertoformmolecules. 6.3Observation1:ValenceandthePeriodicTable Tobeginouranalysisofchemicalbonding,wedenethe valence ofanatombyitstendenciestoform molecules.Theinertgasesdonottendtocombinewithanyotheratoms.Wethusassigntheirvalenceas0, meaningthattheseatomstendtoform0bonds.Eachhalogenpreferstoformmoleculesbycombiningwith 1 Thiscontentisavailableonlineat. 37

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38 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING asinglehydrogenatom e.g. HF HCl .Wethusassigntheirvalenceas1,alsotakinghydrogentoalsohave avalenceof1.Whatwemeanbyavalenceof1isthattheseatomsprefertobindtoonlyoneotheratom. Thevalenceofoxygen,sulfur,etc.isassignedas2,sincetwohydrogensarerequiredtosatisfybonding needsoftheseatoms.Nitrogen,phosphorus,etc.haveavalenceof3,andcarbonandsiliconhaveavalence of4.Thisconceptalsoappliestoelementsjustfollowingtheinertgases.Lithium,sodium,potassium,and rubidiumbindwithasinglehalogenatom.Therefore,theyalsohaveavalenceof1.Correspondingly,itis notsurprisingtondthat,forexample,thecombinationoftwopotassiumatomswithasingleoxygenatom formsastablemolecule,sinceoxygen'svalenceof2isbesatisedbythetwoalkaliatoms,eachwithvalence 1.Wecanproceedinthismannertoassignavalencetoeachelement,bysimplydeterminingthenumberof atomstowhichthiselement'satomsprefertobind. Indoingso,wediscoverthattheperiodictableisarepresentationofthevalencesoftheelements: elementsinthesamegroupallshareacommonvalence.Theinertgaseswithavalenceof0sittooneside ofthetable.Eachinertgasisimmediatelyprecededinthetablebyoneofthehalogens:uorineprecedes neon,chlorineprecedesargon,bromineprecedeskrypton,andiodineprecedesxenon.Andeachhalogenhas avalenceofone.This"onestepaway,valenceofone"patterncanbeextended.Theelementsjustprior tothehalogensoxygen,sulfur,selenium,telluriumareeachtwostepsawayfromtheinertgasesinthe table,andeachoftheseelementshasavalenceoftwo e.g. H 2 O H 2 S .Theelementsjustprecedingthese nitrogen,phosphorus,antimony,arsenichavevalencesofthree e.g. NH 3 PH 3 ,andtheelementsbefore thatcarbonandsiliconmostnotablyhavevalencesoffour CH 4 SiH 4 .Thetwogroupsofelements immediatelyaftertheinertgases,thealkalimetalsandthealkalineearths,havevalencesofoneandtwo, respectively.Hence,formanyelementsintheperiodictable,thevalenceofitsatomscanbepredictedfrom thenumberofstepstheelementisawayfromthenearestinertgasinthetable.Thissystemizationisquite remarkableandisveryusefulforrememberingwhatmoleculesmaybeeasilyformedbyaparticularelement. Nextwediscoverthatthereisapatterntothevalences:forelementsingroups4through8 e.g. carbon throughneon,thevalenceofeachatom plus thenumberofelectronsinthevalenceshellinthatatomalways equals eight .Forexamples,carbonhasavalenceof4andhas4valenceelectrons,nitrogenhasavalence of3andhas5valenceelectrons,andoxygenhasavalenceof2andhas6valenceelectrons.Hydrogenis animportantspecialcasewithasinglevalenceelectronandavalenceof1.Interestingly,foreachofthese atoms,thenumberofbondstheatomformsisequaltothenumberofvacanciesinitsvalenceshell. Toaccountforthispattern,wedevelopamodelassumingthateachatomattemptstobondtoother atomssoastocompletelyllitsvalenceshellwithelectrons.Forelementsingroups4through8,thismeans thateachatomattemptstocompletean"octet"ofvalenceshellelectrons.Whyatomsshouldbehavethis wayisaquestionunansweredbythismodel.Consider,forexample,thecombinationofhydrogenand chlorinetoformhydrogenchloride, HCl .Thechlorineatomhassevenvalenceelectronsandseekstoadda singleelectrontocompleteanoctet.Hence,chlorinehasavalenceof1.Eitherhydrogenorchlorinecould satisfyitsvalenceby"taking"anelectronfromtheotheratom,butthiswouldleavethesecondatomnow needingtwoelectronstocompleteitsvalenceshell.Theonlywayforbothatomstocompletetheirvalence shellssimultaneouslyisto share twoelectrons.Eachatomdonatesasingleelectrontotheelectronpair whichisshared.Itisthissharingofelectronsthatwerefertoasachemicalbond,ormorespecically,asa covalentbond ,sonamedbecausethebondactstosatisfythevalenceofbothatoms.Thetwoatomsare thusheldtogetherbytheneedtosharetheelectronpair. 6.4Observation2:CompoundsofCarbonandHydrogen Manyofthemostimportantchemicalfuelsarecompoundscomposedentirelyofcarbonandhydrogen, i.e. hydrocarbons.Thesmallestoftheseismethane CH 4 ,aprimarycomponentofhouseholdnaturalgas.Other simplecommonfuelsincludeethane C 2 H 6 ,propane C 3 H 8 ,butane C 4 H 10 ,pentane C 5 H 12 ,hexane C 6 H 14 heptane C 7 H 16 ,andoctane C 8 H 18 .Itisinterestingtonotethatthereisaconsistencyinthesemolecular formulae:ineachcase,thenumberofhydrogenatomsistwomorethantwicethenumberofcarbonatoms,so thateachcompoundhasamolecularformulalike C n H 2 n +2 .Thissuggeststhattherearestrongsimilarities inthevalencesoftheatomsinvolvedwhichshouldbeunderstandableintermsofourvalenceshellelectron

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39 pairsharingmodel.Ineachmolecule,thecarbonatomsmustbedirectlybondedtogether,sincetheycannot bejoinedtogetherwithahydrogenatom.Intheeasiestexampleofethane,thetwocarbonatomsarebonded together,andeachcarbonatomisinturnbondedtothreehydrogenatoms.Thus,inthiscase,itisrelatively apparentthatthevalenceofeachcarbonatomis4,justasinmethane,sinceeachisbondedtofourother atoms.Therefore,bysharinganelectronpairwitheachofthefouratomstowhichitisbonded,eachcarbon atomhasavalenceshellofeightelectrons. Inmostothercases,itisnotsotrivialtodeterminewhichatomsarebondedtowhich,astheremaybe multiplepossibilitieswhichsatisfyallatomicvalences.Norisittrivial,asthenumberofatomsandelectrons increases,todeterminewhethereachatomhasanoctetofelectronsinitsvalenceshell.Weneedasystem ofelectronaccountingwhichpermitsustoseethesefeaturesmoreclearly.Tothisend,weadoptastandard notationforeachatomwhichdisplaysthenumberofvalenceelectronsintheunbondedatomexplicitly.In thisnotation,carbonandhydrogenlooklikeFigure6.1,representingthesinglevalenceelectroninhydrogen andthefourvalenceelectronsincarbon. Figure6.1 Usingthisnotation,itisnowrelativelyeasytorepresentthesharedelectronpairsandthecarbonatom valenceshelloctetsinmethaneandethane.Linkingbondedatomstogetherandpairingthevalenceshell electronsfromeachgivesFigure6.2. Figure6.2 Recallthateachsharedpairofelectronsrepresentsachemicalbond.Theseareexamplesofwhatare called Lewisstructures ,afterG.N.Lewiswhorstinventedthisnotation.Thesestructuresreveal,ata glance,whichatomsarebondedtowhich, i.e. thestructuralformulaofthemolecule.Wecanalsoeasily countthenumberofvalenceshellelectronsaroundeachatominthebondedmolecule.Consistentwithour modeloftheoctetrule,eachcarbonatomhaseightvalenceelectronsandeachhydrogenhastwointhe molecule. Inalargerhydrocarbon,thestructuralformulaofthemoleculeisgenerallynotpredictablefromthe numberofcarbonatomsandthenumberofhydrogenatoms,sothemolecularstructuremustbegivento deducetheLewisstructureandthusthearrangementoftheelectronsinthemolecule.However,oncegiven

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40 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING thisinformation,itisstraightforwardtocreateaLewisstructureformoleculeswiththegeneralmolecular formula C n H 2 n +2 suchaspropane,butane,etc.Forexample,theLewisstructurefor"normal"butanewith allcarbonslinkedoneafteranotherisfoundhereFigure6.3. Figure6.3 Itisimportanttonotethatthereexistnohydrocarbonswherethenumberofhydrogensexceedstwo morethantwicethenumberofcarbons.Forexample, CH 5 doesnotexist,nordoes C 2 H 8 .WecorrespondinglyndthatallattemptstodrawLewisstructures whichareconsistentwiththeoctetrulewillfailforthesemolecules.Similarly, CH 3 and C 2 H 5 areobserved tobesoextremelyreactivethatitisimpossibletopreparestablequantitiesofeithercompound.Againwe ndthatitisnotpossibletodrawLewisstructuresforthesemoleculeswhichobeytheoctetrule. Weconcludefromtheseexamplesthat,whenitispossibletodrawaLewisstructureinwhicheach carbonhasacompleteoctetofelectronsinitsvalenceshell,thecorrespondingmoleculewillbestableand thehydrocarboncompoundwillexistunderordinaryconditions.Afterworkingafewexamples,itisapparent thatthisalwaysholdsforcompoundswithmolecularformula C n H 2 n +2 Ontheotherhand,therearemanystablehydrocarboncompoundswithmolecularformulaewhichdonot ttheform C n H 2 n +2 ,particularlywherethenumberofhydrogensislessthan 2 n +2 .Inthesecompounds, thevalencesofthecarbonatomsarenotquitesoobviouslysatisedbyelectronpairsharing.Forexample, inethene C 2 H 4 andacetylene C 2 H 2 therearenotenoughhydrogenatomstopermiteachcarbonatomto bebondedtofouratomseach.Ineachmolecule,thetwocarbonatomsmustbebondedtooneanother.By simplyarrangingtheelectronssothatthecarbonatomsshareasinglepairofelectrons,wewindupwith ratherunsatisfyingLewisstructuresforetheneandacetylene,shownhereFigure6.4. Figure6.4 Notethat,inthesestructures,neithercarbonatomhasacompleteoctetofvalenceshellelectrons. Moreover,thesestructuresindicatethatthecarbon-carbonbondsinethane,ethene,andacetyleneshould

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41 beverysimilar,sinceineachcaseasinglepairofelectronsissharedbythetwocarbons.However,these bondsareobservedtobechemicallyandphysicallyverydierent.First,wecancomparetheenergyrequired tobreakeachbondthe bondenergy or bondstrength .Wendthatthecarbon-carbonbondenergy is347kJin C 2 H 6 ,589kJin C 2 H 4 ,and962kJin C 2 H 2 .Second,itispossibletoobservethedistance betweenthetwocarbonatoms,whichisreferredtoasthe bondlength .Itisfoundthatcarbon-carbon bondlengthis154pmin C 2 H 6 ,134pmin C 2 H 4 ,and120pmin C 2 H 2 1 picometer =1 pm =10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(12 m Theseobservationsrevealclearlythatthebondingbetweenthecarbonatomsinthesethreemoleculesmust beverydierent. Notethatthebondinetheneisaboutoneandahalftimesasstrongasthebondinethane;thissuggests thatthetwounpairedandunsharedelectronsintheethenestructureabovearealsopairedandsharedasa secondbondbetweenthetwocarbonatoms.Similarly,sincethebondinacetyleneisabouttwoandahalf timesstrongerthanthebondinethane,wecanimaginethatthisresultsfromthesharingofthreepairsof electronsbetweenthetwocarbonatoms.TheseassumptionsproducetheLewisstructureshereFigure6.5. Figure6.5 Thesestructuresappearsensiblefromtworegards.First,thetrendincarbon-carbonbondstrengthscan beunderstoodasarisingfromtheincreasingnumberofsharedpairsofelectrons.Second,eachcarbonatom hasacompleteoctetofelectrons.Werefertothetwopairsofsharedelectronsinetheneasa doublebond andthethreesharedpairsinacetyleneasa triplebond Wethusextendourmodelofvalenceshellelectronpairsharingtoconcludethatcarbonatomscanbond bysharingone,two,orthreepairsofelectronsasneededtocompleteanoctetofelectrons,andthatthe strengthofthebondisgreaterwhenmorepairsofelectronsareshared.Moreover,thedataabovetellus thatthecarbon-carbonbondinacetyleneisshorterthanthatinethene,whichisshorterthanthatinethane. Weconcludethattriplebondsareshorterthandoublebondswhichareshorterthansinglebonds. 6.5Observation3:CompoundsofNitrogen,Oxygen,andtheHalogens Manycompoundscomposedprimarilyofcarbonandhydrogenalsocontainsomeoxygenornitrogen,orone ormoreofthehalogens.Wethusseektoextendourunderstandingofbondingandstabilitybydeveloping Lewisstructuresinvolvingtheseatoms.Recallthatanitrogenatomhasavalenceof3andhasvevalence electrons.Inournotation,wecoulddrawastructureinwhicheachoftheveelectronsappearsseparately inaring,similartowhatwedrewforC.However,thiswouldimplythatanitrogenatomwouldgenerally formvebondstopairitsvevalenceelectrons.Sincethevalenceisactually3,ournotationshouldreect this.OnepossibilitylookslikethisFigure6.6.

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42 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING Figure6.6 Notethatthisstructureleavesthreeofthevalenceelectrons"unpaired"andthusreadytojoinina sharedelectronpair.Theremainingtwovalenceelectronsare"paired,"andthisnotationimpliesthatthey thereforearenotgenerallyavailableforsharinginacovalentbond.Thisnotationisconsistentwiththe availabledata, i.e. vevalenceelectronsandavalenceof3.Pairingthetwonon-bondingelectronsseems reasonableinanalogytothefactthatelectronsarepairedinformingcovalentbonds. Analogousstructurescanbedrawnforoxygen,aswellasforuorineandtheotherhalogens,asshown hereFigure6.7. Figure6.7 Withthisnotationinhand,wecannowanalyzestructuresformoleculesincludingnitrogen,oxygen,and thehalogens.Thehydridesaretheeasiest,shownhereFigure6.8. Figure6.8 Notethattheoctetruleisclearlyobeyedforoxygen,nitrogen,andthehalogens. Atthispoint,itbecomesveryhelpfultoadoptonenewconvention:apairofbondedelectronswillnow bemoreeasilyrepresentedinourLewisstructuresbyastraightline,ratherthantwodots.Doublebonds andtriplebondsarerepresentedbydoubleandtriplestraightlinesbetweenatoms.Wewillcontinuetoshow non-bondedelectronpairsexplicitly. Asbefore,whenanalyzingLewisstructuresforlargermolecules,wemustalreadyknowwhichatoms arebondedtowhich.Forexample,twoverydierentcompounds,ethanolanddimethylether,bothhave

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43 molecularformula C 2 H 6 O .Inethanol,thetwocarbonatomsarebondedtogetherandtheoxygenatomis attachedtooneofthetwocarbons;thehydrogensarearrangedtocompletethevalencesofthecarbonsand theoxygenshownhereFigure6.9. Figure6.9 ThisLewisstructurerevealsnotonlythateachcarbonandoxygenatomhasacompletedoctetofvalence shellelectronsbutalsothat,inthestablemolecule,therearefournon-bondedelectronsontheoxygenatom. Ethanolisanexampleofan alcohol .AlcoholscanbeeasilyrecognizedinLewisstructuresbytheC-O-H group.TheLewisstructuresofallalcoholsobeytheoctetrule. Indimethylether,thetwocarbonsareeachbondedtotheoxygen,inthemiddle,shownhereFigure6.10. Figure6.10 Ethers canberecognizedinLewisstructuresbytheC-O-Carrangement.Notethat,inbothethanol anddimethylether,theoctetruleisobeyedforallcarbonandoxygenatoms.Therefore,itisnotusually possibletopredictthestructuralformulaofamoleculefromLewisstructures.Wemustknowthemolecular structurepriortodeterminingtheLewisstructure. Ethanolanddimethyletherareexamplesof isomers ,moleculeswiththesamemolecularformulabut dierentstructuralformulae.Ingeneral,isomershaveratherdierentchemicalandphysicalproperties arisingfromtheirdierencesinmolecularstructures. Agroupofcompoundscalled amines containhydrogen,carbon,andnitrogen.Thesimplestamineis methylamine,whoseLewisstructureishereFigure6.11.

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44 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING Figure6.11 "Halogenated"hydrocarbonshavebeenusedextensivelyasrefrigerantsinairconditioningsystemsand refrigerators.Thesearethenotorious"chlorouorocarbons"or"CFCs"whichhavebeenimplicatedinthe destructionofstratosphericozone.TwoofthemoreimportantCFCsincludeFreon11, CFCl 3 ,andFreon 114, C 2 F 4 Cl 2 ,forwhichwecaneasilyconstructappropriateLewisstructures,shownhereFigure6.12. Figure6.12 Finally,Lewisstructuresaccountforthestabilityofthediatomicformoftheelementalhalogens, F 2 Cl 2 Br 2 ,and I 2 .Thesingleexampleof F 2 issucient,shownhereFigure6.13. Figure6.13

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45 Wecanconcludefromtheseexamplesthatmoleculescontainingoxygen,nitrogen,andthehalogens areexpectedtobestablewhentheseatomsallhaveoctetsofelectronsintheirvalenceshells.TheLewis structureofeachmoleculerevealsthischaracterexplicitly. Ontheotherhand,therearemanyexamplesofcommonmoleculeswithapparentlyunusualvalences, including:carbondioxide CO 2 ,inwhichthecarbonisbondedtoonlytwoatomsandeachoxygenisonly bondedtoone;formaldehyde H 2 CO ;andhydrogencyanide HCN .Perhapsmostconspicuously,wehaveyet tounderstandthebondingintwoveryimportantelementaldiatomicmolecules, O 2 and N 2 ,eachofwhich hasfeweratomsthanthevalenceofeitheratom. Werstanalyze CO 2 ,notingthatthebondstrengthofoneofthe CO bondsincarbondioxideis532kJ, whichissignicantlygreaterthanthebondstrengthofthe CO bondinethanol,358kJ.Byanalogytothe comparisonofbondsstrengthsinethanetoethene,wecanimaginethatthisdierenceinbondstrengths resultsfromdoublebondingin CO 2 .Indeed,aLewisstructureof CO 2 inwhichonlysingleelectronpairs aresharedFigure6.14doesnotobeytheoctetrule,butoneinwhichwepairandsharetheextraelectrons revealsthatdoublebondingpermitstheoctetruletobeobeyedFigure6.15. Figure6.14 Figure6.15 Acomparisonofbondlengthsisconsistentwithourreasoning:thesingle CO bondinethanolis148pm, whereasthedoublebondin CO 2 is116. Knowingthatoxygenatomscandouble-bond,wecaneasilyaccountforthestructureofformaldehyde. Thestrengthofthe CO bondin H 2 CO iscomparabletothatin CO 2 ,consistentwiththeLewisstructure hereFigure6.16.

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46 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING Figure6.16 Whataboutnitrogenatoms?Wecancomparethestrengthofthe CN bondin HCN ,880kJ,tothat inmethylamine,290kJ.Thisdramaticdisparityagainsuggeststhepossibilityofmultiplebonding,andan appropriateLewisstructurefor HCN isshownhereFigure6.17. Figure6.17 Wecanconcludethatoxygenandnitrogenatoms,likecarbonatoms,arecapableofmultiplebonding. Furthermore,ourobservationsofoxygenandnitrogenreinforceourearlierdeductionthatmultiplebonds arestrongerthansinglebonds,andtheirbondlengthsareshorter. Asournalexamplesinthissection,weconsidermoleculesinwhichoxygenatomsarebondedto oxygenatoms.Oxygen-oxygenbondsappearprimarilyintwotypesofmolecules.Therstissimplythe oxygendiatomicmolecule, O 2 ,andthesecondaretheperoxides,typiedbyhydrogenperoxide, H 2 O 2 .Ina comparisonofbondenergies,wendthatthestrengthoftheOObondin O 2 is499kJwhereasthestrength oftheOObondin H 2 O 2 is142kJ.ThisiseasilyunderstoodinacomparisonoftheLewisstructuresofthese molecules,showingthattheperoxidebondisasinglebond,whereasthe O 2 bondisadoublebond,shown hereFigure6.18. Figure6.18 Weconcludethatanoxygenatomcansatisfyitsvalenceof2byformingtwosinglebondsorbyforming onedoublebond.Inbothcases,wecanunderstandthestabilityoftheresultingmoleculesbyintermsof anoctetofvalenceelectrons.

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47 6.6InterpretationofLewisStructures BeforefurtherdevelopingourmodelofchemicalbondingbasedonLewisstructures,wepausetoconsider theinterpretationandimportanceofthesestructures.Itisworthrecallingthatwehavedevelopedour modelbasedonobservationsofthenumbersofbondsformedbyindividualatomsandthenumberofvalence electronsineachatom.Ingeneral,thesestructuresareusefulforpredictingwhetheramoleculeisexpected tobestableundernormalconditions.IfwecannotdrawaLewisstructureinwhicheachcarbon,oxygen, nitrogen,orhalogenhasanoctetofvalenceelectrons,thenthecorrespondingmoleculeprobablyisnotstable. Considerationofbondstrengthsandbondlengthsenhancesthemodelbyrevealingthepresenceofdouble andtriplebondsintheLewisstructuresofsomemolecules. Atthispoint,however,wehaveobservednoinformationregardingthegeometriesofmolecules.For example,wehavenotconsideredtheanglesmeasuredbetweenbondsinmolecules.Consequently,theLewis structuremodelofchemicalbondingdoesnotatthislevelpredictorinterpretthesebondangles.Thiswill beconsideredhereChapter7.Therefore,althoughtheLewisstructureofmethaneisdrawnasshown hereFigure6.19. Figure6.19 Thisdoes not implythatmethaneisaatmolecule,orthattheanglesbetween CH bondsinmethaneis 90 .Rather,thestructuresimplyrevealsthatthecarbonatomhasacompleteoctetofvalenceelectronsina methanemolecule,thatallbondsaresinglebonds,andthattherearenonon-bondingelectrons.Similarly,one canwritetheLewisstructureforawatermoleculeintwoapparentlydierentways,shownhereFigure6.20. Figure6.20 However,itisveryimportanttorealizethatthesetwostructuresare identical intheLewismodel, becausebothshowthattheoxygenatomhasacompleteoctetofvalenceelectrons,formstwosinglebonds

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48 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING withhydrogenatoms,andhastwopairsofunsharedelectronsinitsvalenceshell.Inthesameway,thetwo structuresforFreon114shownhereFigure6.21arealso identical Figure6.21 Thesetwodrawingsdonotrepresentdierentstructuresorarrangementsoftheatomsinthebonds. Finally,wemustkeepinmindthatwehavedrawnLewisstructuresstrictlyasaconvenienttoolforour understandingofchemicalbondingandmolecularstability.Itisbasedoncommonlyobservedtrendsin valence,bonding,andbondstrengths.Thesestructuresmustnotbemistakenasobservationsthemselves, however.Asweencounteradditionalexperimentalobservations,wemustbepreparedtoadaptourLewis structuremodeltottheseobservations,butwemustneveradaptourobservationstottheLewismodel. 6.7ExtensionsoftheLewisStructureModel Withthesethoughtsinmind,weturntoasetofmoleculeswhichchallengethelimitsoftheLewismodelin describingmolecularstructures.First,wenotethatthereareavarietyofmoleculesforwhichatomsclearly mustbondinsuchawayastohavemorethaneightvalenceelectrons.Aconspicuousexampleis SF 6 ,where thesulfuratomisbondedtosixFatoms.Assuch,theSatommusthave12valenceshellelectronstoform 6covalentbonds.Similarly,thephosphorousatomin PCl 5 has10valenceelectronsin5covalentbonds, theClatomin ClF 3 has10valenceelectronsin3covalentbondsandtwolonepairs.Wealsoobservethe interestingcompoundsofthenoblegasatoms, e.g. XeO 3 ,wherenoblegasatombeginswitheightvalence electronsevenbeforeforminganybonds.Ineachofthesecases,wenotethatthevalenceoftheatomsS, P,Cl,andXearenormally2,3,1,and0,yetmorebondsthanthisareformed.Insuchcases,itisnot possibletodrawLewisstructuresinwhichS,P,Cl,andXeobeytheoctetrule.Werefertothesemolecules as"expandedvalence"molecules,meaningthatthevalenceofthecentralatomhasexpandedbeyondthe expectedoctet. Therearealsoavarietyofmoleculesforwhichtherearetoofewelectronstoprovideanoctetforevery atom.Mostnotably,BoronandAluminum,fromGroupIII,displaybondingbehaviorsomewhatdierent thanwehaveseenandthuslesspredictablefromthemodelwehavedevelopedsofar.Theseatomshave threevalenceshellelectrons,sowemightpredictavalenceof5onthebasisoftheoctetrule.However, compoundsinwhichboronoraluminumatomsformvebondsareneverobserved,sowemustconclude thatsimplepredictionsbasedontheoctetrulearenotreliableforGroupIII. Considerrstborontriuoride, BF 3 .ThebondinghereFigure6.22isrelativelysimpletomodelwith aLewisstructureifwealloweachvalenceshellelectronintheboronatomtobesharedinacovalentbond witheachuorineatom.

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49 Figure6.22 Notethat,inthisstructure,theboronatomhasonlysixvalenceshellelectrons,buttheoctetruleis obeyedbytheuorineatoms. Wemightconcludefromthisoneexamplethatboronatomsobeyasextetrule.However,boronwill formastableionwithhydrogen, BH )]TJ/F7 6.9738 Tf -0.81 -6.919 Td [(4 ,inwhichtheboronatomdoeshaveacompleteoctet.Inaddition, BF 3 willreactwithammonia NH 3 forformastablecompound, NH 3 BF 3 ,forwhichaLewisstructurecan bedrawninwhichboronhasacompleteoctet,shownhereFigure6.23. Figure6.23 Compoundsofaluminumfollowsimilartrends.Aluminumtrichloride, AlCl 3 ,aluminumhydride, AlH 3 andaluminumhydroxide, Al OH 3 ,allindicateavalenceof3foraluminum,withsixvalenceelectronsin thebondedmolecule.However,thestabilityofaluminumhydrideions, AlH )]TJ/F7 6.9738 Tf -0.81 -6.918 Td [(4 ,indicatesthatAlcanalso supportanoctetofvalenceshellelectronsaswell. Weconcludethat,althoughtheoctetrulecanstillbeofsomeutilityinunderstandingthechemistryof BoronandAluminum,thecompoundsoftheseelementsarelesspredictablefromtheoctetrule.Thisshould notbedisconcerting,however.TheoctetrulewasdevelopedinSection6.3Observation1:Valenceandthe PeriodicTableonthebasisoftheobservationthat,forelementsinGroupsIVthroughVIII,thenumberof valenceelectronsplusthemostcommonvalenceisequaltoeight.ElementsinGroupsI,II,andIIIdonot followthisobservationmostcommonly.

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50 CHAPTER6.COVALENTBONDINGANDELECTRONPAIRSHARING 6.8ResonanceStructures AnotherinterestingchallengefortheLewismodelwehavedevelopedisthesetofmoleculesforwhichitis possibletodrawmorethanonestructureinagreementwiththeoctetrule.Anotableexampleisthenitric acidmolecule, HNO 3 ,whereallthreeoxygensarebondedtothenitrogen.Twostructurescanbedrawnfor nitricacidwithnitrogenandallthreeoxygensobeyingtheoctetrule. Ineachstructure,oftheoxygensnotbondedtohydrogen,onesharesasinglebondwithnitrogenwhile theothersharesadoublebondwithnitrogen.Thesetwostructuresarenotidentical,unlikethetwofreon structuresinFigure6.12,becausetheatomsarebondeddierentlyinthetwostructures. 6.9ReviewandDiscussionQuestions Exercise6.1 Compoundswithformulaeoftheform C n H 2 n +2 areoftenreferredtoas"saturated"hydrocarbons. UsingLewisstructures,explainhowandinwhatsensethesemoleculesare"saturated." Exercise6.2 Moleculeswithformulaeoftheform C n H 2 n +1 e.g. CH 3 C 2 H 5 arecalled"radicals"andare extremelyreactive.UsingLewisstructures,explainthereactivityofthesemolecules. Exercise6.3 Stateandexplaintheexperimentalevidenceandreasoningwhichshowsthatmultiplebondsare strongerandshorterthansinglebonds. Exercise6.4 Compare N 2 to H 4 N 2 .Predictwhichbondisstrongerandexplainwhy. Exercise6.5 ExplainwhythetwoLewisstructuresforFreon114,showninFigure21Figure6.21,areidentical. DrawaLewisstructuresforanisomerofFreon114,thatis,anothermoleculewiththesame molecularformulaasFreon114butadierentstructuralformula.

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Chapter7 MolecularGeometryandElectron DomainTheory 1 7.1Foundation Webeginbyassuminga Lewisstructuremodel forchemicalbondingbasedonvalenceshellelectronpair sharingandtheoctetrule.Wethusassumethenuclearstructureoftheatom,andwefurtherassumethe existenceofavalenceshellofelectronsineachatomwhichdominatesthechemicalbehaviorofthatatom.A covalentchemicalbondisformedwhenthetwobondedatomsshareapairofvalenceshellelectronsbetween them.Ingeneral,atomsofGroupsIVthroughVIIbondsoastocompleteanoctetofvalenceshellelectrons. Anumberofatoms,includingC,N,O,P,andS,canformdoubleortriplebondsasneededtocompletean octet.Weknowthatdoublebondsaregenerallystrongerandhaveshorterlengthsthansinglebonds,and triplebondsarestrongerandshorterthandoublebonds. 7.2Goals Weshouldexpectthatthepropertiesofmolecules,andcorrespondinglythesubstanceswhichtheycomprise, shoulddependonthedetailsofthestructureandbondinginthesemolecules.Therelationshipbetween bonding,structure,andpropertiesiscomparativelysimplein diatomic molecules,whichcontaintwoatoms only, e.g. HCl or O 2 .A polyatomic moleculecontainsmorethantwoatoms.Anexampleofthecomplexitieswhicharisewithpolyatomicmoleculesismoleculargeometry:howaretheatomsinthemolecule arrangedwithrespecttooneanother?Inadiatomicmolecule,onlyasinglemoleculargeometryispossible sincethetwoatomsmustlieonaline.However,withatriatomicmoleculethreeatoms,therearetwo possiblegeometries:theatomsmaylieonaline,producingalinearmolecule,ornot,producingabent molecule.Inmoleculeswithmorethanthreeatoms,therearemanymorepossiblegeometries.Whatgeometriesareactuallyobserved?Whatdetermineswhichgeometrywillbeobservedinaparticularmolecule?We seekamodelwhichallowsustounderstandtheobservedgeometriesofmoleculesandthustopredictthese geometries. Oncewehavedevelopedanunderstandingoftherelationshipbetweenmolecularstructureandchemical bonding,wecanattemptanunderstandingoftherelationshipofhestructureandbondinginapolyatomic moleculetothephysicalandchemicalpropertiesweobserveforthosemolecules. 1 Thiscontentisavailableonlineat. 51

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52 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY 7.3Observation1:Geometriesofmolecules Thegeometryofamoleculeincludesadescriptionofthearrangementsoftheatomsinthemolecule.Ata simplelevel,themolecularstructuretellsuswhichatomsarebondedtowhich.Atamoredetailedlevel, thegeometryincludesthelengthsofallofthesebonds,thatis,thedistancesbetweentheatomswhichare bondedtogether,andtheanglesbetweenpairsofbonds.Forexample,wendthatinwater, H 2 O ,the twohydrogensarebondedtotheoxygenandeachO-Hbondlengthis95.72pmwhere 1 pm =10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(12 m Furthermore, H 2 O isabentmolecule,withtheH-O-Hangleequalto104.5 .Themeasurementofthese geometricpropertiesisdicult,involvingthemeasurementofthefrequenciesatwhichthemoleculerotates inthegasphase.Inmoleculesincrystallineform,thegeometryofthemoleculeisrevealedbyirradiating thecrystalwithx-raysandanalyzingthepatternsformedasthex-raysdiractoofthecrystal. Notalltriatomicmoleculesarebent,however.Asacommonexample, CO 2 isalinearmolecule.Larger polyatomicscanhaveavarietyofshapes,asillustratedinFigure7.1MolecularStructures.Ammonia, NH 3 ,isapyramid-shapedmolecule,withthehydrogensinanequilateraltriangle,thenitrogenabovethe planeofthistriangle,andaH-N-Hangleequalto107 .Thegeometryof CH 4 isthatofatetrahedron,with allH-C-Hanglesequalto109.5 .SeealsoFigure7.2a.Ethane, C 2 H 6 ,hasageometryrelatedtothat ofmethane.Thetwocarbonsarebondedtogether,andeachisbondedtothreehydrogens.EachH-C-H angleis109.5 andeachH-C-Cangleis109.5 .Bycontrast,inethene, C 2 H 4 ,eachH-C-Hbondangleis 116.6 andeachH-C-Cbondangleis121.7 .Allsixatomsofethenelieinthesameplane.Thus,ethene andethanehaveverydierentgeometries,despitethesimilaritiesintheirmolecularformulae.

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53 MolecularStructures Figure7.1 Webeginouranalysisofthesegeometriesbynotingthat,inthemoleculeslistedabovewhichdo not

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54 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY containdoubleortriplebonds H 2 O NH 3 CH 4 and C 2 H 6 ,thebondanglesareverysimilar,eachequalto orveryclosetothetetrahedralangle109.5 .Toaccountfortheobservedangle,webeginwithourvalence shellelectronpairsharingmodel,andwenotethat,intheLewisstructuresofthesemolecules,thecentral atomineachbondangleofthesemoleculescontainsfourpairsofvalenceshellelectrons.Formethaneand ethane,thesefourelectronpairsareallsharedwithadjacentbondedatoms,whereasin ammoniaorwater,oneortworespectivelyoftheelectronpairsarenotsharedwithanyotheratom. Theseunsharedelectronpairsarecalled lonepairs .Noticethat,inthetwomoleculeswithnolonepairs, allbondanglesare exactly equaltothetetrahedralangle,whereasthebondanglesareonlycloseinthe moleculeswithlonepairs Onewaytounderstandthisresultisbasedonthemutualrepulsionofthenegativechargesonthevalence shellelectrons.Althoughthetwoelectronsineachbondingpairmustremainrelativelyclosetogetherin ordertoformthebond,dierentpairsofelectronsshouldarrangethemselvesinsuchawaythatthedistances betweenthepairsareaslargeaspossible.Focusingforthemomentonmethane,thefourpairsofelectrons mustbeequivalenttooneanother,sincethefourC-Hbondsareequivalent,sowecanassumethatthe electronpairsareallthesamedistancefromthecentralcarbonatom.Howcanwepositionfourelectron pairsataxeddistancefromthecentralatombutasfarapartfromoneanotheraspossible?Alittle reectionrevealsthatthisquestionisequivalenttoaskinghowtoplacefourpointsonthesurfaceofasphere spreadoutfromeachotherasfarapartaspossible.Abitofexperimentationrevealsthatthesefourpoints mustsitatthecornersofatetrahedron,anequilateraltriangularpyramid,asmaybeseeninFigure7.2b. Ifthecarbonatomisatthecenterofthistetrahedronandthefourelectronpairsatplacedatthecorners, thenthehydrogenatomsalsoformatetrahedronaboutthecarbon.Thisis,asillustratedinFigure7.2a, thecorrectgeometryofamethanemolecule.Theangleformedbyanytwocornersofatetrahedronandthe centralatomis109.5 ,exactlyinagreementwiththeobservedangleinmethane.Thismodelalsoworks wellinpredictingthebondanglesinethane.

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55 TetrahedralStructureofMethane a b Figure7.2: aThedottedlinesillustratethatthehydrogensformatetrahedronaboutthecarbon atom.bThesametetrahedronisformedbyplacingfourpointsonasphereasfarapartfromone anotheraspossible. Weconcludethatmoleculargeometryisdeterminedbyminimizingthemutualrepulsionofthevalence shellelectronpairs.Assuch,thismodelofmoleculargeometryisoftenreferredtoasthe valenceshell electronpairrepulsionVSEPRtheory .Forreasonsthatwillbecomeclear,extensionofthismodel

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56 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY impliesthatabetternameisthe ElectronDomainEDTheory Thismodelalsoaccounts,atleastapproximately,forthebondanglesof H 2 O and NH 3 .Thesemolecules areclearlynottetrahedral,like CH 4 ,sinceneithercontainstherequisiteveatomstoformthetetrahedron. However,eachmoleculedoescontainacentralatomsurroundedbyfourpairsofvalenceshellelectrons.We expectfromourElectronDomainmodelthatthosefourpairsshouldbearrayedinatetrahedron,without regardtowhethertheyarebondingorlone-pairelectrons.Thenattachingthehydrogenstwoforoxygen, threefornitrogenproducesapredictionofbondanglesof109.5 ,verycloseindeedtotheobservedangles of104.5 in H 2 O and107 in NH 3 Note,however,thatwedonotdescribethegeometriesof H 2 O and NH 3 as"tetrahedral,"sincethe atoms ofthemoleculesdonotformtetrahedrons,evenifthevalenceshellelectronpairsdo.Itisworthnoting thattheseanglesarenotexactlyequalto109.5 ,asinmethane.Thesedeviationswillbediscussedlater Section7.5:Observation3:DistortionsfromExpectedGeometries. WehavedevelopedtheElectronDomainmodeltothispointonlyforgeometriesofmoleculeswithfour pairsofvalenceshellelectrons.However,thereareagreatvarietyofmoleculesinwhichatomsfromPeriod 3andbeyondcanhavemorethananoctetofvalenceelectrons.Weconsidertwosuchmoleculesillustrated inFigure7.3MoreMolecularStructures.

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57 MoreMolecularStructures Figure7.3 First, PCl 5 isastablegaseouscompoundinwhichthevechlorineatomsareeachbondedtothe phosphorousatom.Experimentsrevealthatthegeometryof PCl 5 isthatofa trigonalbipyramid :three ofthechlorineatomsformanequilateraltrianglewiththePatominthecenter,andtheothertwochlorine atomsareontopofandbelowthePatom.Thustheremustbe10valenceshellelectronsaroundthe phosphorousatom.Hence,phosphorousexhibitswhatiscalledan expandedvalence in PCl 5 .Applying ourElectronDomainmodel,weexpectthevevalenceshellelectronpairstospreadoutoptimallyto minimizetheirrepulsions.Therequiredgeometrycanagainbefoundbytryingtoplacevepointsonthe surfaceofaspherewithmaximumdistancesamongstthesepoints.Alittleexperimentationrevealsthatthis canbeachievedbyplacingthevepointstoformatrigonalbipyramid.Hence,ElectronDomaintheory accountsforthegeometryof PCl 5 .

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58 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY Second, SF 6 isafairlyunreactivegaseouscompoundinwhichallsixuorineatomsarebondedto thecentralsulfuratom.Again,itisclearthattheoctetruleisviolatedbythesulfuratom,whichmust thereforehaveanexpandedvalence.Theobservedgeometryof SF 6 ,asshowninFigure7.3MoreMolecular Structures,ishighlysymmetric:allbondlengthsareidenticalandallbondanglesare90 .TheFatoms forman octahedron aboutthecentralSatom:fouroftheFatomsformasquarewiththeSatomatthe center,andtheothertwoFatomsareaboveandbelowtheSatom.ToapplyourElectronDomainmodel tounderstandthisgeometry,wemustplacesixpoints,representingthesixelectronpairsaboutthecentral Satom,onthesurfaceofaspherewithmaximumdistancesbetweenthepoints.Therequisitegeometryis found,infact,tobethatofanoctahedron,inagreementwiththeobservedgeometry. Asanexampleofamoleculewithanatomwithlessthananoctetofvalenceshellelectrons,weconsider borontrichloride, BCl 3 .Thegeometryof BCl 3 isalsogiveninFigure7.3MoreMolecularStructures: itis trigonalplanar ,withallfouratomslyinginthesameplane,andallCl-B-Clbondanglesequalto 120 .ThethreeClatomsformanequilateraltriangle.TheBoronatomhasonlythreepairsofvalenceshell electronsin BCl 3 .InapplyingElectronDomaintheorytounderstandthisgeometry,wemustplacethree pointsonthesurfaceofaspherewithmaximumdistancebetweenpoints.Wendthatthethreepointsform anequilateraltriangleinaplanewiththecenterofthesphere,soElectronDomainisagaininaccordwith theobservedgeometry. WeconcludefromthesepredictionsandobservationsthattheElectronDomainmodelisareasonably accuratewaytounderstandmoleculargeometries,eveninmoleculeswhichviolatetheoctetrule. 7.4Observation2:MoleculeswithDoubleorTripleBonds Ineachofthemoleculesconsidereduptothispoint,theelectronpairsareeitherinsinglebondsorinlone pairs.Incurrentform,theElectronDomainmodeldoes not accountfortheobservedgeometryof C 2 H 4 inwhicheachH-C-Hbondangleis116.6 andeachH-C-Cbondangleis121.7 andallsixatomsliein thesameplane.Eachcarbonatominthismoleculeissurroundedbyfourpairsofelectrons,allofwhich areinvolvedinbonding, i.e. therearenolonepairs.However,thearrangementoftheseelectronpairs,and thusthebondedatoms,abouteachcarbonisnotevenapproximatelytetrahedral.Rather,theH-C-Hand H-C-Cbondanglesaremuchcloserto120 ,theanglewhichwouldbeexpectedif three electronpairswere separatedintheoptimalarrangement,asjustdiscussedfor BCl 3 Thisobservedgeometrycanbeunderstoodbyre-examiningtheLewisstructure.Recallthat,although therearefourelectronpairsabouteachcarbonatom,twoofthesepairsformadoublebondbetweenthe carbonatoms.Itistemptingtoassumethatthesefourelectronpairsareforcedaparttoformatetrahedron asinpreviousmolecules.However,ifthiswerethiscase,thetwopairsinvolvedinthedoublebondwouldbe separatedbyanangleof109.5 whichwouldmakeitimpossibleforbothpairstobelocalizedbetweenthe carbonatoms.Topreservethedoublebond,wemustassumethatthetwoelectronpairsinthedoublebond remaininthesamevicinity.Giventhisassumption,separatingthethree independent groupsofelectron pairsaboutacarbonatomproducesanexpectationthatallthreepairsshouldlieinthesameplaneas thecarbonatom,separatedby120 angles.Thisagreesverycloselywiththeobservedbondangles.We concludethattheourmodelcanbeextendedtounderstandingthegeometriesofmoleculeswithdoubleor triplebondsbytreatingthemultiplebondastwoelectronpairsconnedtoasingle domain .Itisforthis reasonthatwerefertothemodelasElectronDomaintheory. Appliedinthisform,ElectronDomaintheorycanhelpusunderstandthelineargeometryof CO 2 .Again, therearefourelectronpairsinthevalenceshellofthecarbonatom,butthesearegroupedintoonlytwo domainsoftwoelectronpairseach,correspondingtothetwoC=Odoublebonds.Minimizingtherepulsion betweenthesetwodomainsforcestheoxygenatomstodirectlyoppositesidesofthecarbon,producinga linearmolecule.SimilarreasoningusingElectronDomaintheoryasappliedtotriplebondscorrectlypredicts thatacetylene, HCCH ,isalinearmolecule.Iftheelectronpairsinthetriplebondaretreatedasasingle domain,theneachcarbonatomhasonlytwodomainseach.Forcingthesedomainstooppositesidesfrom oneanotheraccuratelypredicts180 H-C-Cbondangles.

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59 7.5Observation3:DistortionsfromExpectedGeometries Itisinterestingtonotethatsomemoleculargeometries CH 4 CO 2 HCCH areexactlypredictedbythe ElectronDomainmodel,whereasinothermolecules,themodelpredictionsareonlyapproximatelycorrect. Forexamples,theobservedanglesinammoniaandwatereachdierslightlyfromthetetrahedralangle. Hereagain,therearefourpairsofvalenceshellelectronsaboutthecentralatoms.Assuch,itisreasonable toconcludethatthebondanglesaredeterminedbythemutualrepulsionoftheseelectronpairs,andare thusexpectedtobe109.5 ,whichisclosebutnotexact. Oneclueastoapossiblereasonforthediscrepancyisthatthebondanglesinammoniaandwaterare both less than109.5 .Anotheristhatbothammoniaandwatermoleculeshavelonepairelectrons,whereas therearenolonepairsinamethanemolecule,forwhichtheElectronDomainpredictionisexact.Moreover, thebondangleinwater,withtwolonepairs,islessthanthebondanglesinammonia,withasinglelonepair. Wecanstraightforwardlyconcludefromtheseobservationsthatthelonepairsofelectronsmustproducea greaterrepulsiveeectthandothebondedpairs.Thus,inammonia,thethreebondedpairsofelectronsare forcedtogetherslightlycomparedtothoseinmethane,duetothegreaterrepulsiveeectofthelonepair. Likewise,inwater,thetwobondedpairsofelectronsareevenfurtherforcedtogetherbythetwolonepairs ofelectrons. Thismodelaccountsforthecomparativebondanglesobservedexperimentallyinthesemolecules.The valenceshellelectronpairsrepeloneanother,establishingthegeometryinwhichtheenergyoftheirinteractionisminimized.Lonepairelectronsapparentlygenerateagreaterrepulsion,thusslightlyreducingthe anglesbetweenthebondedpairsofelectrons.Althoughthismodelaccountsfortheobservedgeometries, whyshouldlonepairelectronsgenerateagreaterrepulsiveeect?Wemustguessataqualitativeanswerto thisquestion,sincewehavenodescriptionatthispointforwherethevalenceshellelectronpairsactually areorwhatitmeanstoshareanelectronpair.Wecanassume,however,thatapairofelectronssharedby twoatomsmustbelocatedsomewherebetweenthetwonuclei,otherwiseourconceptof"sharing"isquite meaningless.Therefore,thepowerfultendencyofthetwoelectronsinthepairtorepeloneanothermust besignicantlyosetbythelocalizationoftheseelectronsbetweenthetwonucleiwhichsharethem.By contrast,alonepairofelectronsneednotbesolocalized,sincethereisnosecondnucleustodrawtheminto thesamevicinity.Thusmorefreetomoveaboutthecentralatom,theselonepairelectronsmusthavea moresignicantrepulsiveeectontheotherpairsofelectrons. Theseideascanbeextendedbymorecloselyexaminingthegeometryofethene, C 2 H 4 .Recallthat eachH-C-Hbondangleis116.6 andeachH-C-Cbondangleis121.7 ,whereastheElectronDomain theorypredictionisforbondanglesexactlyequalto120 .WecanunderstandwhytheH-C-Hbondangle isslightlylessthan120 byassumingthatthetwopairsofelectronsintheC=Cdoublebondproducea greaterrepulsiveeectthandoeitherofthesinglepairsofelectronsintheC-Hsinglebonds.Theresultof thisgreaterrepulsionisaslight"pinching"oftheH-C-Hbondangletolessthan120 Theconceptthatlonepairelectronsproduceagreaterrepulsiveeectthandobondedpairscanbeused tounderstandotherinterestingmoleculargeometries.Sulfurtetrauoride, SF 4 ,isaparticularlyinteresting example,showninFigure7.4MolecularStructureofSF4.

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60 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY MolecularStructureofSF4 Figure7.4 Notethattwooftheuorinesformclosetoastraightlinewiththecentralsulfuratom,buttheothertwo areapproximatelyperpendiculartothersttwoandatanangleof101.5 toeachother.Viewedsideways, thisstructurelookssomethinglikeaseesaw. Toaccountforthisstructure,werstprepareaLewisstructure.Wendthateachuorineatomissingly bondedtothesulfuratom,andthatthereisalonepairofelectronsonthesulfur.Thus,withveelectron pairsaroundthecentralatom,weexpecttheelectronstoarrangethemselvesinatrigonalbipyramid,similar tothearrangementin PCl 5 inFigure7.3MoreMolecularStructures.Inthiscase,however,theuorine atomsandthelonepaircouldbearrangedintwodierentwayswithtwodierentresultantmolecular structures.Thelonepaircaneithergoontheaxisofthetrigonalbipyramid i.e. abovethesulfuroron theequatorofthebipyramid i.e. besidethesulfur. TheactualmolecularstructureinFigure7.4MolecularStructureofSF4showsclearlythatthelone pairgoesontheequatorialposition.Thiscanbeunderstoodifweassumethatthelonepairproducesa greaterrepulsiveeectthandothebondedpairs.Withthisassumption,wecandeducethatthelonepair shouldbeplacedinthetrigonalbipyramidalarrangementasfaraspossiblefromthebondedpairs.The equatorialpositiondoesabetterjobofthis,sinceonlytwobondingpairsofelectronsareatapproximately 90 anglesfromthelonepairinthisposition.Bycontrast,alonepairintheaxialpositionisapproximately 90 awayfromthreebondingpairs.Therefore,ourElectronDomainmodelassumptionsareconsistentwith theobservedgeometryof SF 4 .Notethattheseassumptionsalsocorrectlypredicttheobserveddistortions awayfromthe180 and120 angleswhichwouldbepredictedbyatrigonalbipyramidalarrangementof

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61 theveelectronpairs. 7.6ReviewandDiscussionQuestions Exercise7.1 Usingastyrofoamorrubberball,provetoyourselfthatatetrahedralarrangementprovidesthe maximumseparationoffourpointsonthesurfaceoftheball.Repeatthisargumenttondthe expectedarrangementsfortwo,three,ve,andsixpointsonthesurfaceoftheball. Exercise7.2 Explainwhyarrangingpointsonthesurfaceofaspherecanbeconsideredequivalenttoarranging electronpairsaboutacentralatom. Exercise7.3 Thevalenceshellelectronpairsaboutthecentralatomineachofthemolecules H 2 O NH 3 ,and CH 4 arearrangedapproximatelyinatetrahedron.However,only CH 4 isconsideredatetrahedral molecule.Explainwhythesestatementsarenotinconsistent. Exercise7.4 Explainhowacomparisonofthegeometriesof H 2 O and CH 4 leadstoaconclusionthatlone pairelectronsproduceagreaterrepulsiveeectthandobondedpairsofelectrons.Giveaphysical reasonwhythismightbeexpected. Exercise7.5 Explainwhytheoctetofelectronsabouteachcarbonatominethene, C 2 H 4 ,arenotarranged evenapproximatelyinatetrahedron. Exercise7.6 Assesstheaccuracyofthefollowingreasoningandconclusions: Atrigonalbipyramidformswhenthereareveelectrondomains.IfoneEDisalonepair,then thelonepairtakesanequatorialpositionandthemoleculehasaseesawgeometry.IftwoEDs arelonepairs,wehavetodecideamongthefollowingoptions:bothaxial,bothequatorial,or oneaxialandoneequatorial.Byplacingbothlonepairsintheaxialpositions,thelonepairsare asfarapartaspossible,sothetrigonalplanarstructureisfavored. Exercise7.7 Assesstheaccuracyofthefollowingreasoningandconclusions: TheCl-X-ClbondanglesinthetwomoleculesFigure7.5areidentical,becausethebondangle isdeterminedbytherepulsionofthetwoClatoms,whichisidenticalinthetwomolecules. Figure7.5

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62 CHAPTER7.MOLECULARGEOMETRYANDELECTRONDOMAIN THEORY

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Chapter8 MolecularStructureandPhysical Properties 1 8.1Foundation Webeginwithourknowledgeofthestructureandpropertiesofatoms.Weknowthatatomshaveanuclear structure,meaningthatallofthepositivechargeandvirtuallyallofthemassoftheatomareconcentrated inanucleuswhichisaverysmallfractionofthevolumeoftheatom.Inaddition,weknowthatmanyofthe propertiesofatomscanbeunderstoodbyamodelinwhichtheelectronsintheatomarearrangedinshells aboutthenucleus,witheachshellfartherfromthenucleusthattheprevious.Theelectronsinoutershells aremoreweaklyattachedtotheatomthantheelectronsintheinnershells,andonlyalimitednumberof electronscantineachshell.Withineachshellaresubshells,eachofwhichcanalsoholdalimitednumber ofelectrons.Theelectronsindierentsubshellshavedierentenergiesanddierentlocationsformotion aboutthenucleus.Wealsoassumeaknowledgeofthea Lewisstructure modelforchemicalbonding basedonvalenceshellelectronpairsharingandtheoctetrule.Acovalentchemicalbondisformedwhen thetwobondedatomsshareapairofvalenceshellelectronsbetweenthem.Ingeneral,atomsofGroups IVthroughVIIbondsoastocompleteanoctetofvalenceshellelectrons.Wenallyassumethe Electron DomainModel forunderstandingandpredictingmoleculargeometries.Thepairsofvalenceshellelectrons arearrangedinbondingandnon-bondingdomains,andthesedomainsareseparatedinspacetominimize electron-electronrepulsions.Thiselectrondomainarrangementdeterminesthemoleculargeometry. 8.2Goals Weshouldexpectthatthepropertiesofmolecules,andcorrespondinglythesubstanceswhichtheycomprise, shoulddependonthedetailsofthestructureandbondinginthesemolecules.Nowthatwehavedeveloped anunderstandingoftherelationshipbetweenmolecularstructureandchemicalbonding,weanalyzephysical propertiesofthemoleculesandcompoundsofthesemoleculestorelatetothisbondingandstructure.Simple examplesofphysicalpropertieswhichcanberelatedtomolecularpropertiesarethemeltingandboiling temperatures.Thesevarydramaticallyfromsubstancetosubstance,evenforsubstanceswhichappearsimilar inmolecularformulae,withsomemeltingtemperaturesinthehundredsorthousandsofdegreesCelsiusand otherswellbelow0 C.Weseektounderstandthesevariationsbyanalyzingmolecularstructures. Todevelopthisunderstanding,wewillhavetoapplymoredetailsofourunderstandingofatomicstructure andelectroniccongurations.Inourcovalentbondingmodel,wehaveassumedthatatomsshareelectrons toformabond.However,ourknowledgeofthepropertiesofatomsrevealsthatdierentatomsattract 1 Thiscontentisavailableonlineat. 63

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64 CHAPTER8.MOLECULARSTRUCTUREANDPHYSICALPROPERTIES electronswithdierentstrengths,resultinginverystrongvariationsinionizationenergies,atomicradii,and electronanities.Weseektoincorporatethisinformationintoourunderstandingofchemicalbonding 8.3Observation1:CompoundsofGroupsIandII WebeginbyanalyzingcompoundsformedfromelementsfromGroupsIandII e.g. sodiumandmagnesium. ThesecompoundsarenotcurrentlypartofourLewisstructuremodel.Forexample,Sodium,withasingle valenceelectron,isunlikelytogainsevenadditionalelectronstocompleteanoctet.Indeed,thecommon valenceofthealkalimetalsinGroupIis1,not7,andthecommonvalenceofthealkalineearthmetalsis2, not6.Thus,ourcurrentmodelofbondingdoesnotapplytoelementsinthesegroups. Todevelopanunderstandingofbondinginthesecompounds,wefocusonthehalidesoftheseelements. InMeltingPointsandBoilingPointsofChlorideCompounds,p.64,wecomparephysicalpropertiesofthe chloridesofelementsinGroupsIandIItothechloridesoftheelementsofGroupsIV,V,andVI,andwe seeenormousdierences.Allofthealkalihalidesandalkalineearthhalidesaresolidsatroomtemperature andhavemeltingpointsinthehundredsofdegreescentigrade.Themeltingpointof NaCl is808 C,for example.Bycontrast,themeltingpointsofthenon-metalhalidesfromPeriods2and3,suchas CCl 4 PCl 3 ,and SCl 2 ,arebelow0 C,sothatthesematerialsareliquidsatroomtemperature.Furthermore,all ofthesecompoundshavelowboilingpoints,typicallyintherangeof50 Cto80 C. MeltingPointsandBoilingPointsofChlorideCompounds MeltingPoint C BoilingPoint C LiCl 610 1382 BeCl 2 405 488 CCl 4 -23 77 NCl 3 -40 71 OCl 2 -20 4 FCl -154 -101 NaCl 808 1465 MgCl 2 714 1418 SiCl 4 -68 57 PCl 3 -91 74 SCl 2 -122 59 Cl 2 -102 -35 KCl 772 1407 CaCl 2 772 > 1600 Second,thenon-metalhalideliquidsareelectricalinsulators,thatis,theydonotconductanelectrical current.Bycontrast,whenwemeltanalkalihalideoralkalineearthhalide,theresultingliquidisanexcellent electricalconductor.Thisindicatesthatthesemoltencompoundsconsistofions,whereasthenon-metal halidesdonot. Wemustconcludethatthebondingofatomsinalkalihalidesandalkalineearthhalidesdierssignicantly frombondinginnon-metalhalides.Weneedtoextendourvalenceshellelectronmodeltoaccountforthis bonding,andinparticular,wemustaccountforthepresenceofionsinthemoltenmetalhalides.Consider theprototypicalexampleof NaCl .WehavealreadydeducedthatClatomsreactsoastoformacomplete

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65 octetofvalenceshellelectrons.Suchanoctetcouldbeachievedbycovalentlysharingthesinglevalence shellelectronfromasodiumatom.However,suchacovalentsharingisclearlyinconsistentwiththepresence ofionsinmoltensodiumchloride.Furthermore,thistypeofbondwouldpredictthat NaCl shouldhave similarpropertiestoothercovalentchloridecompounds,mostofwhichareliquidsatroomtemperature.By contrast,wemightimaginethatthechlorineatomcompletesitsoctetbytakingthevalenceshellelectron fromasodiumatom,withoutcovalentsharing.Thiswouldaccountforthepresenceof Na + and Cl )]TJ/F15 9.9626 Tf 10.293 -4.456 Td [(ions inmoltensodiumchloride. Intheabsenceofacovalentsharingofanelectronpair,though,whataccountsforthestabilityofsodium chlorideasacompound?Itisrelativelyobviousthatanegativelychargedchlorideionwillbeattracted electrostaticallytoapositivelychargedsodiumion.Wemustalsoaddtothismodel,however,thefact thatindividualmoleculesof NaCl arenotgenerallyobservedattemperatureslessthan1465 C,theboiling pointofsodiumchloride.Notethat,ifsolidsodiumchlorideconsistsofindividualsodiumionsinproximity toindividualchlorideions,theneachpositiveionisnotsimplyattractedtoasinglespecicnegativeion butrathertoallofthenegativeionsinitsnearvicinity.Hence,solidsodiumchloridecannotbeviewed asindividual NaCl molecules,butmustbeviewedratherasalatticeofpositivesodiumionsinteracting withnegativechlorideions.Thistypeofionicbonding,whichderivesfromtheelectrostaticattractionof interlockinglatticesofpositiveandnegativeions,accountsfortheveryhighmeltingandboilingpointsof thealkalihalides. WecannowdrawmodiedLewisstructurestoaccountforionicbonding,buttheseareverydierent fromourpreviousdrawings.SodiumchloridecanberepresentedasshowninFigure8.1. Figure8.1 Thisindicatesexplicitlythatthebondingisduetopositive-negativeionattraction,andnotdueto sharingofanelectronpair.Theonlysenseinwhichthe Na + ionhasobeyedanoctetruleisperhapsthat, inhavingemptieditsvalenceshellofelectrons,theremainingoutershellofelectronsintheionhasthesame octetasdoesaneonatom.Wemustkeepinmind,however,thatthepositivesodiumionisattractedto manynegativechlorideions,andnotjustthesinglechlorideiondepictedintheLewisstructure. 8.4Observation2:MolecularDipoleMoments OurLewismodelofbonding,ascurrentlydeveloped,incorporatestwoextremeviewsofthedistributionof electronsinabond.Inacovalentbond,wehaveassumeduptothispointthattheelectronpairisshared perfectly.Incompletecontrast,inionicbondingwehaveassumedthattheelectronsarenotsharedatall. Rather,oneoftheatomsisassumedtoentirelyextractoneormoreelectronsfromtheother.Wemight expectthatamoreaccuratedescriptionoftherealityofchemicalbondsfallsingeneralsomewherebetween thesetwoextremes.Toobservethisintermediatebehavior,wecanexaminemoleculardipolemoments. Anelectricdipoleisaspatialseparationofpositiveandnegativecharges.Inthesimplestcase,apositive charge + Q andanegativecharge )]TJ/F11 9.9626 Tf 7.749 0 Td [(Q separatedbyadistance R produceameasurable dipolemoment equalto Q R .Anelectriceldcaninteractwithanelectricdipoleandcanevenorientthedipoleinthe directionoftheeld.

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66 CHAPTER8.MOLECULARSTRUCTUREANDPHYSICALPROPERTIES Wemightinitiallyexpectthatmoleculesdonotingeneralhavedipolemoments.Eachatomenteringinto achemicalbondiselectricallyneutral,withequalnumbersofpositiveandnegativecharges.Consequently, amoleculeformedfromneutralatomsmustalsobeelectricallyneutral.Althoughelectronpairsareshared betweenbondednuclei,thisdoesnotaectthetotalnumberofnegativecharges.Wemightfromthese simplestatementsthatmoleculeswouldbeunaectedbyelectricormagneticelds,eachmoleculebehaving asasingleunchargedparticle. Thispredictionisincorrect,however.Toillustrate,astreamofwatercanbedeectedbyanelectrically chargedobjectnearthestream,indicatingthatindividualwatermoleculesexhibitadipolemoment.A watermoleculeisrathermorecomplicatedthanasimpleseparationofapositiveandnegativecharges, however.Recallthoughthatawatermoleculehasequaltotalnumbersofpositiveandnegativecharges, consistingofthreepositivelychargednucleisurroundedbytenelectrons.Nevertheless,measurementsreveal thatwaterhasadipolemomentof 6 : 17 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(30 Cm =1 : 85 debye .Thedebyeisaunitusedtomeasure dipolemoments: 1 debye =3 : 33 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(30 Cm .Waterisnotunique:themoleculesofmostsubstanceshave dipolemoments.AsamplingofmoleculesandtheirdipolemomentsisgiveninDipoleMomentsofSpecic Molecules,p.66. DipoleMomentsofSpecicMolecules debye H 2 O 1.85 HF 1.91 HCl 1.08 HBr 0.80 HI 0.42 CO 0.12 CO 2 0 NH 3 1.47 PH 3 0.58 AsH 3 0.20 CH 4 0 NaCl 9.00 Focusingagainonthewatermolecule,howcanweaccountfortheexistenceofadipolemomentina neutralmolecule?Theexistenceofthedipolemomentrevealsthatawatermoleculemusthaveaninternal separationofpositivepartialcharge + andnegativepartialcharge )]TJ/F11 9.9626 Tf 7.749 0 Td [( .Thus,itmustbetruethatthe electronsinthecovalentbondbetweenhydrogenandoxygenarenot equally shared.Rather,theshared electronsmustspendmoretimeinthevicinityofonenucleusthantheother.Themoleculethushasone regionwhere,onaverage,thereisanetsurplusofnegativechargeandoneregionwhere,onaverage,there isacompensatingsurplusofpositivecharge,thusproducingamoleculardipole.Additionalobservations revealthattheoxygen"end"ofthemoleculeholdsthepartialnegativecharge.Hence,thecovalentlyshared electronsspendmoretimeneartheoxygenatomthannearthehydrogenatoms.Weconcludethatoxygen atomshaveagreaterabilitytoattractthesharedelectronsinthebondthandohydrogenatoms. Weshouldnotbesurprisedbythefactthatindividualatomsofdierentelementshavedieringabilitiesto attractelectronstothemselves.Wehavepreviouslyseenthatdierentatomshavegreatlyvaryingionization energies,representinggreatvariationintheextenttowhichatomsclingtotheirelectrons.Wehavealsoseen greatvariationintheelectronanitiesofatoms,representingvariationintheextenttowhichatomsattract anaddedelectron.Wenowdenethe electronegativity ofanatomastheabilityoftheatomtoattract

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67 electronsinachemicalbond.Thisisdierentthaneitherionizationenergyorelectronanity,because electronegativityistheattractionofelectrons inachemicalbond ,whereasionizationenergyandelectron anityrefertoremovalandattachmentofelectronsinfreeatoms.However,wecanexpectelectronegativity tobecorrelatedwithelectronanityandionizationenergy.Inparticular,theelectronegativityofanatom arisesfromacombinationofpropertiesoftheatom,includingthesizeoftheatom,thechargeonthenucleus, thenumberofelectronsaboutthenuclei,andthenumberofelectronsinthevalenceshell. Becauseelectronegativityisanabstractlydenedproperty,itcannotbedirectlymeasured.Infact,there aremanydenitionsofelectronegativity,resultinginmanydierentscalesofelectronegativities.However, relativeelectronegativitiescanbeobservedindirectlybymeasuringmoleculardipolemoments:ingeneral, thegreaterthedipolemoment,thegreatertheseparationofchargesmustbe,andtherefore,thelessequal thesharingofthebondingelectronsmustbe. Withthisinmind,wereferbacktothedipolesgiveninDipoleMomentsofSpecicMolecules,p.66. Thereareseveralimportanttrendsinthesedata.Notethateachhydrogenhalide HF HCl HBr ,and HI hasasignicantdipolemoment.Moreover,thedipolemomentsincreaseaswemove up theperiodic tableinthehalogengroup.Wecanconcludethatuorineatomshaveagreaterelectronegativitythando chlorineatoms, etc. Notealsothat HF hasagreaterdipolemomentthan H 2 O ,whichisinturngreater thanthatof NH 3 .Wecanconcludethatelectronegativityincreasesaswemove across theperiodictable fromlefttorightinasingleperiod.Thesetrendsholdgenerallyincomparisonsoftheelectronegativitiesof theindividualelements.Onesetofrelativeelectronegativitiesofatomsintherstthreerowsoftheperiodic tableisgiveninElectronegativitiesofSelectedAtoms,p.67.

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68 CHAPTER8.MOLECULARSTRUCTUREANDPHYSICALPROPERTIES ElectronegativitiesofSelectedAtoms H 2.1 He Li 1.0 Be 1.5 B 2.0 C 2.5 N 3.0 O 3.5 F 4.0 Ne Na 0.9 Mg 1.2 Al 1.5 Si 1.8 P 2.1 S 2.5 Cl 3.0 Ar K 0.8 Ca 1.0 8.5Observation3:DipoleMomentsinPolyatomicMolecules Wemightreasonablyexpectfromouranalysistoobserveadipolemomentinanymoleculeformedfrom atomswithdierentelectronegativities.Althoughthismustbethecaseforadiatomicmolecule,thisis notnecessarilytrueforapolyatomicmolecule, i.e. onewithmorethantwoatoms.Forexample,carbon ismoreelectronegativethanhydrogen.However,thesimplestmoleculeformedfromcarbonandhydrogen e.g. CH 4 does not possessadipolemoment,asweseeinDipoleMomentsofSpecicMolecules,p.66. Similarly,oxygenissignicantlymoreelectronegativethancarbon,yet CO 2 isanon-polarmolecule.An analysisofmoleculardipolemomentsinpolyatomicmoleculesrequiresustoapplyourunderstandingof moleculargeometry. Notethateach CO bondisexpectedtobepolar,duetotheunequalsharingoftheelectronpairsbetween thecarbonandtheoxygen.Thus,thecarbonatomshouldhaveaslightpositivechargeandtheoxygen atomaslightnegativechargeineach CO bond.However,sinceeachoxygenatomshouldhavethesamenet negativecharge,neitherendofthemoleculewoulddisplayagreateranityforanelectriceld.Moreover, because CO 2 islinear,thedipoleinone CO bondisexactlyosetbythedipoleintheoppositedirectiondue totheother CO bond.Asmeasuredbyanelectriceldfromadistance,the CO 2 moleculedoesnotappear tohaveseparatedpositiveandnegativechargesandthereforedoesnotdisplaypolarity.Thus,inpredicting moleculardipoleswemusttakeintoaccountbothdierencesinelectronegativity,whichaectbondpolarity, andoverallmoleculargeometry,whichcanproducecancellationofbondpolarities.

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69 Usingthissameargument,wecanrationalizethezeromoleculardipolemomentsobservedforother molecules,suchasmethane,etheneandacetylene.Ineachofthesemolecules,theindividual CH bondsare polar.However,thesymmetryofthemoleculeproducesacancellationofthesebonddipolesoverall,and noneofthesemoleculeshaveamoleculardipolemoment. Asanexampleofhowamolecularpropertylikethedipolemomentcanaectthemacroscopicproperty ofasubstance,wecanexaminetheboilingpointsofvariouscompounds.Theboilingpointofacompound isdeterminedbythestrengthoftheforcesbetweenmoleculesofthecompound:thestrongertheforce, themoreenergyisrequiredtoseparatethemolecules,thehigherthetemperaturerequiredtoprovidethis energy.Therefore,moleculeswithstrongintermolecularforceshavehighboilingpoints. Webeginbycomparingmoleculeswhicharesimilarinsize,suchasthehydrides SiH 4 PH 3 ,and SH 2 from thethirdperiod.Theboilingpointsatstandardpressureforthesemoleculesare,respectively,-111.8 C, -87.7 C,and-60.7 C.Allthreecompoundsarethusgasesatroomtemperatureandwellbelow.These moleculeshaveverysimilarmassesandhaveexactlythesamenumberofelectrons.However,thedipole momentsofthesemoleculesareverydierent.Thedipolemomentof SiH 4 ,is0.0D,thedipolemomentof PH 3 is0.58D,andthedipolemomentof SH 2 is0.97D.Notethat,forthesesimilarmolecules,thehigher thedipolemoment,thehighertheboilingpoint.Thus,moleculeswithlargerdipolemomentsgenerally havestrongerintermolecularforcesthansimilarmoleculeswithsmallerdipolemoments.Thisisbecausethe positiveendofthedipoleinonemoleculecaninteractelectrostaticallywiththenegativeendofthedipole inanothermolecules,andviceversa. Wenote,however,thatonecannotgenerallypredictfromdipolemomentinformationonlytherelative boilingpointsofcompoundsofverydissimilarmolecules 8.6ReviewandDiscussionQuestions Exercise8.1 Compareandcontrastthechemicalandphysicalpropertiesof KCl and CCl 4 ,andcompareand contrasthowthechemicalbondingmodelcanbeusedtoaccountfortheseproperties. Exercise8.2 Whyisthedipolemomentof NaCl extremelylarge? Exercise8.3 Explainwhy CO hasadipolemomentbut CO 2 doesnot. Exercise8.4 Explainwhyanatomwithahighionizationenergyisexpectedtohaveahighelectronegativity. Explainwhyanatomwithahighelectronanityisexpectedtohaveahighelectronegativity. Exercise8.5 WouldyoupredictthataKratomhashighelectronegativityorlowelectronegativity?Predictthe relativeelectronegativityofKrandF. Exercise8.6 ExplainwhyShasagreaterelectronegativitythanPbutasmallerelectronegativitythanO. Exercise8.7 Natomshaveahighelectronegativity.However,Natomshavenoelectronanity,meaningthat Natomsdonotattractelectrons.Explainhowandwhythesefactsarenotinconsistent. Exercise8.8 Explainwhycompoundsformedfromelementswithlargedierencesinelectronegativitiesare ionic. Exercise8.9 Explainwhyioniccompoundshavemuchhighermeltingpointsthancovalentcompounds.

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70 CHAPTER8.MOLECULARSTRUCTUREANDPHYSICALPROPERTIES

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Chapter9 ChemicalBondingandMolecularEnergy Levels 1 9.1Foundation Ourbasisforunderstandingchemicalbondingandthestructuresofmoleculesistheelectronorbitaldescriptionofthestructureandvalenceofatoms,asprovidedbyquantummechanics.Weassumeanunderstanding oftheperiodicityoftheelementsbasedonthenuclearstructureoftheatomandourdeductionsconcerning valencebasedonelectronorbitals. 9.2Goals Ourmodelofvalencedescribesachemicalbondasresultingfromthesharingofapairofelectronsinthe valenceshellofthebondedatoms.Thissharingallowseachatomtocompleteanoctetofelectronsinits valenceshell,atleastinthesensethatwecountthesharedelectronsasbelongingtobothatoms.However, itisnotclearthatthiselectroncountingpicturehasanybasisinphysicalreality.Whatismeant,more precisely,bythesharingoftheelectronpairinabond,andwhydoesthisresultinthebondingoftwoatoms together?Indeed,whatdoesitmeantosaythattwoatomsareboundtogether?Furthermore,whatis thesignicanceofsharingapairofelectrons?Whyaren'tchemicalbondsformedbysharingoneorthree electrons,forexample? Weseektounderstandhowthedetailsofchemicalbondingarerelatedtothepropertiesofthemolecules formed,particularlyintermsofthestrengthsofthebondsformed. 9.3Observation1:BondingwithaSingleElectron Webeganouranalysisoftheenergiesandmotionsoftheelectronsinatomsbyobservingtheproperties ofthesimplestatom,hydrogen,withasingleelectron.Similarly,tounderstandtheenergiesandmotions ofelectronswhichleadtochemicalbonding,webeginourobservationswiththesimplestparticlewitha chemicalbond,whichisthe H + 2 molecularion.Eachhydrogennucleushasachargeof+1.An H + 2 molecular ionthereforehasasingleelectron.Itseemsinconsistentwithournotionsofvalencethatasingleelectron, ratherthananelectronpair,cangenerateachemicalbond.However,theseconceptshavebeenbasedon observationsonmolecules,notmolecularionslike H + 2 .Anditisindeedfoundthat H + 2 isastablebound molecularion. 1 Thiscontentisavailableonlineat. 71

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72 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Whatforcesandmotionsholdthetwohydrogennucleiclosetogetherinthe H + 2 ion?Itisworthkeeping inmindthatthetwonucleimustrepeloneanother,sincetheyarebothpositivelycharged.Intheabsence oftheelectron,thetwonucleiwouldaccelerateawayfromoneanother,ratherthanremaininginclose proximity.Whatistheroleoftheelectron?Clearly,theelectronisattractedtobothnucleiatthesame time,and,inturn,eachnucleusisattractedtotheelectron.TheeectofthisisillustratedinFig.1.In Fig.1a,theelectronisoutsideofthetwonuclei.Inthisposition,theelectronisprimarilyattractedtothe nucleusontheleft,towhichitiscloser.Moreimportantly,thenucleusontherightfeelsagreaterrepulsion fromtheothernucleusthanattractiontotheelectron,whichisfartheraway.Asaresult,thenucleuson therightexperiencesastrongforcedrivingitawayfromthehydrogenatomontheleft.Thisarrangement doesnotgeneratechemicalbonding,therefore.Bycontrast,inFig.1b,theelectronisbetweenthetwo nuclei.Inthisposition,theelectronisroughlyequallyattractedtothetwonuclei,andveryimportantly, eachnucleusfeelsanattractiveforcetotheelectronwhichisgreaterthantherepulsiveforcegeneratedby theothernucleus.Focusingontheelectron'senergy,theproximityofthetwonucleiprovidesitadoubly attractiveenvironmentwithaverylowpotentialenergy.Ifwetriedtopulloneofthenucleiaway,thiswould raisethepotentialenergyoftheelectron,sinceitwouldloseattractiontothatnucleus.Hence,topullone nucleusawayrequiresustoaddenergytothemolecularion.Thisiswhatismeantbyachemicalbond:the energyoftheelectronsislowerwhentheatomsareincloseproximitythanwhentheatomsarefarpart. Thisholdsthenucleiclosetogether,sincewemustdoworkaddenergytotakethenucleiapart.

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73 Figure9.1 NotethatthechemicalbondinFig.1bresultsfromtheelectron'spositionbetweenthenuclei.On rstthought,thisappearstoanswerourquestionofwhatwemeanbysharinganelectronpairtoforma chemicalbond.Anelectronpositionedbetweentwonucleiissharedtotheextentthatitspotentialenergy isloweredduetoattractiontobothnucleisimultaneously.

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74 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Onsecondthought,though,thisdescriptionmustbeinaccurate.WehavelearnedourstudyofEnergy LevelsinAtomsthatanelectronmustobeytheuncertaintyprincipleandthat,asaconsequence,theelectron doesnothaveadeniteposition,betweenthenucleiorotherwise.Wecanonlyhopetospecifyaprobability forobservinganelectroninaparticularlocation.Thisprobabilityis,fromquantummechanics,providedby thewavefunction.Whatdoesthisprobabilitydistributionlooklikeforthe H + 2 molecularion? Toanswerthisquestion,webeginbyexperimentingwithadistributionthatweknow:the1selectron orbitalinahydrogenatom.Thiswerecallhasthesymmetryofasphere,withequalprobabilityinall directionsawayfromthenucleus.Tocreatean H + 2 molecularionfromahydrogenatom,wemustadd abaresecondhydrogennucleusan H + ion.Imaginebringingthisnucleusclosertothehydrogenatom fromaverygreatdistanceseeFig.2a.Asthe H + ionapproachestheneutralatom,boththehydrogen atom'snucleusandelectronrespondtotheelectricpotentialgeneratedbythepositivecharge.Theelectron isattractedandthehydrogenatomnucleusisrepelled.Asaresult,thedistributionofprobabilityforthe electronaboutthenucleusmustbecomedistorted,sothattheelectronhasagreaterprobabilityofbeing nearthe H + ionandthenucleushasagreaterprobabilityofbeingfartherfromtheion.Thisdistortion, illustratedinFig.2b,iscalledpolarization:thehydrogenatomhasbecomelikeadipole,withgreater negativechargetoonesideandgreaterpositivechargetotheother.

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75 Figure9.2

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76 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Thispolarizationmustincreaseasthe H + ionapproachesthehydrogenatomuntil,eventually,the electronorbitalmustbesucientlydistortedthatthereisequalprobabilityforobservingtheelectronin proximitytoeitherhydrogennucleusseeFig.2c.TheelectronprobabilitydistributioninFig.2cnow describesthemotionoftheelectron,notinahydrogenatom,butinan H + 2 molecularion.Assuch,werefer tothisdistributionasamolecularorbital. WenotethatthemolecularorbitalinFig.2cismoredelocalizedthantheatomicorbitalinFig.2a, andthisisalsoimportantinproducingthechemicalbond.WerecallfromthediscussionofAtomicEnergy Levelsthattheenergyofanelectroninanorbitalisdetermined,inpart,bythecompactnessoftheorbital. Themoretheorbitalconnesthemotionoftheelectron,thehigheristhekineticenergyoftheelectron, aneectwereferredtoastheconnementenergy.ApplyingthisconcepttotheorbitalsinFig.2,we canconcludethattheconnementenergyisloweredwhentheelectronisdelocalizedovertwonucleiina molecularorbital.Thiseectcontributessignicantlytotheloweringoftheenergyofanelectronresulting fromsharingbytwonuclei. Recallthattheelectronorbitalsinthehydrogenatomaredescribedbyasetofquantumnumbers.One ofthesequantumnumbersisrelatedtothesymmetryorshapeoftheatomicorbitalandisgenerallydepicted byaletter.Recallthatansorbitalissphericalinshape,andaporbitalhastwolobesalignedalongone axis.Similarly,themolecularorbitalsforthe H + 2 molecularionaredescribedbyasetofnumberswhich givethesymmetryorshapeoftheorbital.Forourpurposes,weneedonlyoneofthesedescriptors,based onthesymmetryoftheorbitalalongthebond:ifthemolecularorbitalhasthesymmetryofacylinder,we refertoitasa orbital.TheorbitalinFig.2csatisesthiscondition. Weconcludethatchemicalbondingresultsfromanelectroninamolecularorbitalwhichhassubstantial probabilityfortheelectrontobebetweentwonuclei.However,thisexampleillustrateschemicalbonding withasingleelectron.Ourrulesofvalenceindicatethatbondingtypicallyoccurswithapairofelectrons, ratherthanasingleelectron.Furthermore,thismodelofbondingdoesnottellushowtohandlemolecules withmanyelectronssay, F 2 wheremostoftheelectronsdonotparticipateinthebondingatall. 9.4Observation2:BondingandNon-BondinginDiatomicMolecules Wenowconsidermoleculeswithmorethanoneelectron.Theseareillustratedmosteasilybydiatomic moleculesmoleculeswithonlytwoatomsformedbylikeatoms,beginningwiththehydrogenmolecule, H 2 Themostdirectexperimentalobservationofachemicalbondistheamountofenergyrequiredtobreakit. Thisiscalledthebondenergy,orsomewhatlessprecisely,thebondstrength.Experimentally,itisobserved thatthebondenergyofthehydrogenmolecule H 2 is458kJ/mol.Bycontrast,thebondenergyofthe H + 2 molecularionis269kJ/mol.Therefore,thebondin H 2 isstrongerthanthebondin H + 2 .Thus,thepairof sharedelectronsin H 2 generatesastrongerattractiveforcethandoesthesingleelectronin H + 2 Beforededucinganexplanationofthisintermsofelectronorbitals,werstrecallthevalenceshell electronpairdescriptionofthebondingin H 2 .Eachhydrogenatomhasasingleelectron.Bysharingthese twoelectrons,eachhydrogenatomcanllitsvalenceshell,attainingtheelectroncongurationofhelium. Howdoesthistranslateintotheelectronorbitalpictureofelectronsharingthatwehavejustdescribedfor the H + 2 molecularion?Therearetwowaystodeducetheanswertothisquestion,and,sincetheyareboth useful,wewillworkthroughthemboth.Therstwayistoimaginethatweforman H 2 moleculebystarting withan H + 2 molecularionandaddinganelectrontoit.Asasimpleapproximation,wemightimaginethat therstelectron'sprobabilitydistributionitsorbitalisnotaectedbytheadditionofthesecondelectron. Thesecondelectronmusthaveaprobabilitydistributiondescribingitslocationinthemoleculeaswell.We recallthat,inatoms,itispossibletoputtwoelectronsintoasingleelectronorbital,providedthatthetwo electronshaveoppositevaluesofthespinquantumnumber,ms.Therefore,weexpectthistobetruefor moleculesaswell,andweplacetheaddedsecondelectronin H 2 intothesame orbitalastherst.This resultsintwoelectronsintheregionbetweenthetwonuclei,thusaddingtotheforceofattractionofthe twonucleiintothebond.Thisexplainsourobservationthatthebondenergyof H 2 isalmostalthoughnot quitetwicethebondenergyof H + 2 Thesecondwaytounderstandtheelectronorbitalpictureof H 2 istoimaginethatweformthemolecule

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77 bystartingwithtwoseparatedhydrogenatoms.Eachoftheseatomshasasingleelectronina1sorbital. Asthetwoatomsapproachoneanother,eachelectronorbitalispolarizedinthedirectionoftheotheratom. Oncetheatomsarecloseenoughtogether,thesetwoorbitalsbecomesuperimposed.Nowwemustrecall thattheseorbitalsdescribethewave-likemotionoftheelectron,sothat,whenthesetwowavefunctions overlap,theymustinterfere,eitherconstructivelyordestructively.InFig.3,weseetheconsequencesof constructiveanddestructiveinterference.Wecandeducethat,in H 2 theelectronorbitalsfromtheatoms mustconstructivelyinterfere,becausethatwouldincreasetheelectronprobabilityintheregionbetweenthe nuclei,resultinginbondingasbefore.Therefore,the molecularorbitaldescribingthetwoelectronsin H 2 canbeunderstoodasresultingfromtheconstructiveoverlapoftwoatomic1selectronorbitals. Wenowaddtoourobservationsofdiatomicmoleculesbynotingthat,ofthediatomicmoleculesformed fromlikeatomsofthersttenelements, H 2 Li 2 B 2 C 2 N 2 O 2 ,and F 2 arestablemoleculeswithchemical bonds,whereas He 2 Be 2 ,and Ne 2 arenotbound.Inexaminingtheelectroncongurationsoftheatoms oftheseelements,wediscoveracorrespondencewithwhichdiatomicmoleculesareboundandwhichones arenot. H;Li;B;N;andF allhaveoddnumbersofelectrons,sothatatleastoneelectronineachatomis unpaired.Bycontrast,He,Be,andNeallhaveevennumbersofelectrons,noneofwhichareunpaired.The otheratoms,CandObothhaveanevennumberofelectrons.However,asdeducedinourunderstanding oftheelectroncongurationsinatoms,electronswill,whenpossible,distributethemselvesintodierent orbitalsofthesameenergysoastoreducetheeectoftheirmutualrepulsion.Thus,inCandO,thereare three2porbitalsintowhich2and4electronsareplaced,respectively.Therefore,inbothatoms,thereare twounpairedelectrons.Weconcludethatbondswillformbetweenatomsifandonlyifthereareunpaired electronsintheseatoms. In H 2 ,theunpairedelectronsfromtheseparatedatomsbecomepairedinamolecularorbitalformed fromtheoverlapofthe1satomicelectronorbitals.Inthecaseofahydrogenatom,then,thereareofcourse nopairedelectronsintheatomtoworryabout.Inallotheratoms,therecertainlyarepairedelectrons, regardlessofwhetherthereareorarenotunpairedelectrons.Forexample,inalithiumatom,thereare twopairedelectronsina1sorbitalandanunpairedelectroninthe2sorbital.Toform Li 2 ,theunpaired electronfromeachatomcanbeplacedintoamolecularorbitalformedfromtheoverlapofthe2satomic electronorbitals.However,whatbecomesofthetwoelectronspairedinthe1sorbitalinaLiatomduring thebondingof Li 2 ? Toanswerthisquestion,weexamine He 2 ,inwhicheachatombeginswithonlythetwo1selectrons. AswebringthetwoHeatomstogetherfromalargedistance,these1sorbitalsshouldbecomepolarized, asinthehydrogenatom.Whenthepolarized1sorbitalsoverlap,constructiveinterferencewillagainresult ina molecularorbital,justasin H 2 .Yet,weobservethat He 2 isnotastableboundmolecule.The problemwhichpreventsbondingfor He 2 arisesfromthePauliExclusionPrinciple:onlytwoofthefour electronsinHe 2 canbeplacedintothis bondingmolecularorbital.Theothertwomustgointoadierent orbitalwithadierentprobabilitydistribution.Todeducetheformofthisneworbital,werecallthatthe bondingorbitaldiscussedsofararisesfromtheconstructiveinterferenceoftheatomicorbitals,asshownin Fig.3.Wecould,instead,haveassumeddestructiveinterferenceoftheseorbitals.Destructiveinterference oftwowaveseliminatesamplitudeintheregionofoverlapofthewaves,alsoshowninFig.3.Inthecase oftheatomicorbitals,thismeansthatthemolecularorbitalformedfromdestructiveinterferencedecreases probabilityfortheelectrontobebetweeninthenuclei.Therefore,itincreasesprobabilityfortheelectron tobeoutsidethenuclei,asinFig.1a.Asdiscussedthere,thisarrangementfortheelectrondoesnotresult inbonding;instead,thenucleirepeleachotherandtheatomsareforcedapart.Thisorbitalisthuscalled ananti-bondingorbital.Thisorbitalalsohasthesymmetryofacylinderalongthebondaxis,soitisalsoa orbital;toindicatethatitisananti-bondingorbital,wedesignateitwithanasterisk,

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78 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Figure9.3 In He 2 ,boththebondingandtheanti-bondingorbitalsmustbeusedinordertoaccommodatefour electrons.Thetwoelectronsinthebondingorbitallowertheenergyofthemolecule,butthetwoelectrons intheanti-bondingorbitalraiseit.SincetwoHeatomswillnotbindtogether,thentheneteectmustbe thattheanti-bondingorbitalmorethanosetsthebondingorbital. Wehavenowdeducedanexplanationforwhythepairedelectronsinanatomdonotcontributeto bonding.Bothbondingandanti-bondingorbitalsarealwaysformedwhentwoatomicorbitalsoverlap.

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79 Whentheelectronsarealreadypairedintheatomicorbitals,thentherearetoomanyelectronsforthe bondingmolecularorbital.Theextraelectronsmustgointotheanti-bondingorbital,whichraisesthe energyofthemolecule,preventingthebondfromforming. Returningtothe Li 2 examplediscussedabove,wecandevelopasimplepictureofthebonding.The two1selectronsfromeachatomdonotparticipateinthebonding,sincetheanti-bondingmorethanosets thebonding.Thus,thepairedcoreelectronsremainintheiratomicorbitals,unshared,andwecanignore themindescribingthebond.Thebondisformedduetooverlapofthe2sorbitalsandsharingofthese electronsonly.Thisisalsoconsistentwithourearlierviewthatthecoreelectronsareclosertothenucleus, andthusunlikelytobesharedbytwoatoms. Themodelwehaveconstructedseemstodescribefairlywellthebondinginthebounddiatomicmolecules listedabove.Forexample,inauorineatom,theonlyunpairedelectronisina2porbital.Recallthata2p orbitalhastwolobes,directedalongoneaxis.Iftheselobesareassumedtoliealongtheaxisbetweenthe twonucleiin F 2 ,thenwecanoverlapthemtoformabondingorbital.Placingthetwounpairedelectrons intothisorbitalthenresultsinasinglesharedpairofelectronsandastablemolecularbond. 9.5Observation3:Ionizationenergiesofdiatomicmolecule Theenergiesofelectronsinmolecularorbitalscanbeobserveddirectlybymeasuringtheionizationenergy. Thisistheenergyrequiredtoremoveanelectron,inthiscase,fromamolecule: H 2 g H + 2 g + e )]TJ/F8 9.9626 Tf 8.386 -3.615 Td [( g Themeasuredionizationenergyof H 2 is1488kJ/mol.Thisnumberisprimarilyimportantincomparison totheionizationenergyofahydrogenatom,whichis1312kJ/mol.Therefore,itrequiresmoreenergyto removeanelectronfromthehydrogenmoleculethanfromthehydrogenatom,sowecanconcludethatthe electronhasalowerenergyinthemolecule.Ifweattempttopulltheatomsapart,wemustraisetheenergy oftheelectron.Hence,energyisrequiredtobreakthebond,sothemoleculeisbound. Weconcludethatabondisformedwhentheenergyoftheelectronsinthemoleculeislowerthanthe energyoftheelectronsintheseparatedatoms.Thisconclusionseemsconsistentwithourpreviousviewof sharedelectronsinbondingmolecularorbitals. Asasecondexample,weconsiderthenitrogenmolecule, N 2 .Wendthattheionizationenergyof molecularnitrogenis1503kJ/mol,andthatofatomicnitrogenis1402kJ/mol.Onceagain,weconclude thattheenergyoftheelectronsinmolecularnitrogenislowerthanthatoftheelectronsintheseparated atoms,sothemoleculeisbound. Asathirdexample,weconsideruorine, F 2 .Inthiscase,wendthattheionizationenergyofmolecular uorineis1515kJ/mol,whichissmallerthantheionizationenergyofauorineatom,1681kJ/mol.This seemsinconsistentwiththebondingorbitalconceptwehavedevelopedabove,whichstatesthattheelectrons inthebondhavealowerenergythanintheseparatedatoms.Iftheelectronbeingionizedhasahigher energyin F 2 thaninF,whyis F 2 astablemolecule?Apparently,weneedamorecompletedescriptionof themolecularorbitalconceptofchemicalbonding. Toproceedfurther,wecomparebondenergiesinseveralmolecules.Recallthatthebondenergyorbond strengthistheenergyrequiredtoseparatethebondedatoms.Weobservethatthebondenergyof N 2 is956 kJ/mol.Thisisverymuchlargerthanthebondenergyof H 2 ,458kJ/mol,andof F 2 ,whichis160kJ/mol. Wecanaccountfortheunusuallystrongbondinnitrogenusingbothourvalenceshellelectronpairsharing modelandourelectronorbitaldescriptions.Anitrogenatomhasthreeunpairedelectronsinitsvalence shell,becausethethree2pelectronsdistributethemselvesoverthethree2porbitals,eachorientedalonga dierentaxis.Eachoftheseunpairedelectronsisavailableforsharingwithasecondnitrogenatom.The result,fromvalenceshellelectronpairsharingconcepts,isthatthreepairsofelectronsaresharedbetween twonitrogenatoms,andwecallthebondin N 2 atriplebond.Itissomewhatintuitivethatthetriple bondin N 2 shouldbemuchstrongerthanthesinglebondin H 2 orin F 2 Nowconsiderthemolecularorbitaldescriptionofbondingin N 2 .Eachofthethree2patomicorbitals ineachnitrogenatommustoverlaptoformabondingmolecularorbital,ifwearetoaccommodatethree electronpairs.Each2porbitalisorientedalongasingleaxis.One2porbitalfromeachatomisorientedin

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80 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS thedirectionoftheotheratom,thatis,alongthebondaxis.Whenthesetwoatomicorbitalsoverlap,they formamolecularorbitalwhichhasthesymmetryofacylinderandwhichisthereforea orbital.Ofcourse, theyalsoforma orbital.Thetwoelectronsarethenpairedinthebondingorbital.

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81 Figure9.4

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82 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Theothertwo2porbitalsoneachnitrogenatomareperpendiculartothebondaxis.Theconstructive overlapbetweentheseorbitalsfromdierentatomsmustthereforeresultinamolecularorbitalsomewhat dierentthatwhatwehavediscussedbefore.AsshowninFig.4,themolecularorbitalwhichresultsnowdoes nothavethesymmetryofacylinder,andinfact,lookssomethingmorelikeacylindercutintotwopieces. Thiswecalla orbital.Therearetwosuch orbitalssincetherearetwosetsofporbitalsperpendicular tothebondaxis.Figure4alsoshowsthatananti-bondingorbitalisformedfromthedestructiveoverlapof 2porbitals,andthisiscalleda orbital.Therearealsotwo orbitalsformedfromdestructiveoverlapof 2porbitals.In N 2 ,thethreesharedelectronpairsarethusinasingle orbitalandintwo orbitals.Each oftheseorbitalsisabondingorbital,thereforeallsixelectronshavetheirenergyloweredincomparisonto theseparatedatoms. ThisisdepictedinFig.5inwhatiscalledamolecularorbitalenergydiagram.Eachpairofatomic orbitals,onefromeachatom,isoverlappedtoformabondingandananti-bondingorbital.Thethree2p orbitalsfromeachatomformone and pairandtwo and pairs.Theloweringoftheenergiesofthe electronsinthe and orbitalsisapparent.Thetenn=2electronsfromthenitrogenatomsarethenplaced pairwise,inorderofincreasingenergy,intothesemolecularorbitals.Notethat,inagreementwiththePauli ExclusionPrinciple,eachpairinasingleorbitalconsistsofonespinupandonespindownelectron.

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83 Figure9.5 Recallnowthatwebeganthediscussionofbondingin N 2 becauseofthecuriousresultthattheionization energyofanelectronin F 2 islessthanthatofanelectroninanFatom.Bycomparingthemolecularorbital energyleveldiagramsfor N 2 and F 2 wearenowpreparedtoanswerthispuzzle.Therearevepelectrons ineachuorineatom.Thesetenelectronsmustbedistributedoverthemolecularorbitalswhoseenergies areshowninFig.6.Notethattheorderingofthebonding2porbitalsdierbetween N 2 and F 2 .Weplace twoelectronsinthe orbital,fourmoreinthetwo orbitals,andfourmoreinthetwo orbitals.Overall, therearesixelectronsinbondingorbitalsandfourinanti-bondingorbitals.Since F 2 isastablemolecule, wemustconcludethattheloweringofenergyfortheelectronsinthebondingorbitalsisgreaterthanthe raisingofenergyfortheelectronsintheantibondingorbitals.Overall,thisdistributionofelectronsis,net,

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84 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS equivalenttohavingtwoelectronspairedinasinglebondingorbital. Figure9.6 Thisalsoexplainswhytheionizationenergyof F 2 islessthanthatofanFatom.Theelectronwith thehighestenergyrequirestheleastenergytoremovefromthemoleculeoratom.Themolecularorbital energydiagraminFig.6clearlyshowsthatthehighestenergyelectronsin F 2 areinanti-bondingorbitals. Therefore,oneoftheseelectronsiseasiertoremovethananelectroninanatomic2porbital,becausethe energyofananti-bondingorbitalishigherthanthatoftheatomicorbitals.Recallthatthisiswhyanantibondingorbitalis,indeed,anti-bonding.Therefore,theionizationenergyofmolecularuorineislessthan

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85 thatofatomicuorine.Thisclearlydemonstratesthephysicalrealityandimportanceoftheanti-bonding orbitals. Aparticularlyinterestingcaseistheoxygenmolecule, O 2 .Incompletingthemolecularorbitalenergy leveldiagramforoxygen,wediscoverthatwemustdecidewhethertopairthelasttwoelectronsinthesame 2 p orbital,orwhethertheyshouldbeseparatedintodierent 2 p orbitals.Todeterminewhich,wenote thatoxygenmoleculesareparamagnetic,meaningthattheyarestronglyattractedtoamagneticeld.To accountforthisparamagnetism,werecallthatelectronspinisamagneticproperty.Inmostmolecules,all electronsarepaired,soforeachspinupelectronthereisaspindownelectronandtheirmagneticelds cancelout.Whenallelectronsarepaired,themoleculeisdiamagneticmeaningthatitrespondsonlyweakly toamagneticeld. Iftheelectronsarenotpaired,theycanadoptthesamespininthepresenceofamagneticeld.This accountsfortheattractionoftheparamagneticmoleculetothemagneticeld.Therefore,foramoleculeto beparamagnetic,itmusthaveunpairedelectrons.Thecorrectmolecularorbitalenergyleveldiagramfor an O 2 moleculeisshowninFig.7.

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86 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS Figure9.7 Incomparingthesethreediatomicmolecules,werecallthat N 2 hasthestrongestbond,followedby O 2 and F 2 .WehavepreviouslyaccountedforthiscomparisonwithLewisstructures,showingthat N 2 isatriple bond, O 2 isadoublebond,and F 2 isasinglebond.ThemolecularorbitalenergyleveldiagramsinFigs. 5to7castanewlightonthisanalysis.Notethat,ineachcase,thenumberofbondingelectronsinthese moleculesiseight.Thedierenceinbondingisentirelyduetothenumberofantibondingelectrons:2for N 2 ,4for O 2 ,andsixfor F 2 .Thus,thestrengthofabondmustberelatedtotherelativenumbersof bondingandantibondingelectronsinthemolecule.Therefore,wenowdenethebondorderas BondOrder = 1 2 #bondingelectrons )]TJ/F15 9.9626 Tf 9.963 0 Td [(#antibondingelectrons Notethat,denedthisway,thebondorderfor N 2 is3,for O 2 is2,andfor F 2 is1,whichagreeswith ourconclusionsfromLewisstructures.Weconcludethatwecanpredicttherelativestrengthsofbondsby

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87 comparingbondorders. 9.6ReviewandDiscussionQuestions 1.Whydoesanelectronsharedbytwonucleihavealowerpotentialenergythananelectrononasingle atom?Whydoesanelectronsharedbytwonucleihavealowerkineticenergythananelectronon asingleatom?Howdoesthissharingresultinastablemolecule?Howcanthisaectbemeasured experimentally? 2.Explainwhythebondinan H 2 moleculeisalmosttwiceasstrongasthebondinthe H + 2 ion.Explain whythe H 2 bondislessthantwiceasstrongasthe H + 2 bond. 3. Be 2 isnotastablemolecule.Whatinformationcanwedeterminefromthisobservationaboutthe energiesofmolecularorbitals? 4.Lessenergyisrequiredtoremoveanelectronfroman F 2 moleculethantoremoveanelectronfrom anFatom.Therefore,theenergyofthatelectronishigherinthemoleculethanintheatom.Explain why,nevertheless, F 2 isastablemolecule,i.e.,theenergyofan F 2 moleculeislessthantheenergyof twoFatoms. 5.Whydotheorbitalsofanatom"hybridize"whenformingabond? 6.Calculatethebondordersofthefollowingmoleculesandpredictwhichmoleculeineachpairhasthe strongerbond: a C 2 or C + 2 b B 2 or B + 2 c F 2 or F )]TJ/F7 6.9738 Tf -1.384 -6.918 Td [(2 d O 2 or O + 2 7.Whichofthefollowingdiatomicmoleculesareparamagnetic:CO, Cl 2 ,NO, N 2 ? 8. B 2 isobservedtobeparamagnetic.Usingthisinformation,drawanappropriatemolecularorbital energyleveldiagramfor B 2 .

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88 CHAPTER9.CHEMICALBONDINGANDMOLECULARENERGYLEVELS

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Chapter10 EnergeticsofChemicalReactions 1 10.1TheFoundation Webeginourstudyoftheenergeticsofchemicalreactionswithourunderstandingofmassrelationships, determinedbythestoichiometryofbalancedreactionsandtherelativeatomicmassesoftheelements.We willassumeaconceptualunderstandingofenergybasedonthephysicsofmechanics,andinparticular, wewillassumethelawofconservationofenergy.Indevelopingamolecularunderstandingofthereaction energetics,wewillfurtherassumeourunderstandingofchemicalbondingviavalenceshellelectronpair sharingandmolecularorbitaltheory. 10.2Goals Theheatreleasedorconsumedinachemicalreactionistypicallyamongstthemosteasilyobservedand mostreadilyappreciatedconsequencesofthereaction.Manychemicalreactionsareperformedroutinely specicallyforthepurposeofutilizingtheheatreleasedbythereaction. Weareinterestedhereinanunderstandingoftheenergeticsofchemicalreactions.Specically,wewish toknowwhatfactorsdeterminewhetherheatisabsorbedorreleasedduringachemicalreaction.With thatknowledge,weseektoquantifyandpredicttheamountofheatanticipatedinachemicalreaction.We expecttondthatthequantityofheatabsorbedorreleasedduringareactionisrelatedtothebondingof themoleculesinvolvedinthereaction. Priortoansweringthesequestions,wemustrstanswerafewquestionsregardingthenatureofheat. DespiteourcommonfamiliaritywithheatparticularlyinHouston,theconceptofheatissomewhatelusive todene.Werecognizeheatas"whateveritisthatmakesthingshot,"butthisdenitionistooimprecise topermitmeasurementoranyotherconceptualprogress.Exactlyhowdowedeneandmeasureheat? 10.3Observation1:MeasurementofHeatbyTemperature Wecandeneinavarietyofwaysatemperaturescalewhichpermitsquantitativemeasurementof"how hot"anobjectis.Suchscalesaretypicallybasedontheexpansionandcontractionofmaterials,particularly ofliquidmercury,oronvariationofresistanceinwiresorthermocouples.Usingsuchscales,wecaneasily showthatheatinganobjectcausesitstemperaturetorise. Itisimportant,however,todistinguishbetweenheatandtemperature.Thesetwoconceptsarenotone andthesame.Toillustratethedierence,webeginbymeasuringthetemperatureriseproducedbyagiven amountofheat,focusingonthetemperaturerisein1000gofwaterproducedbyburning1.0gofmethane gas.Wediscoverbyperformingthisexperimentrepeatedlythatthetemperatureofthisquantityofwater 1 Thiscontentisavailableonlineat. 89

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90 CHAPTER10.ENERGETICSOFCHEMICALREACTIONS alwaysrisesbyexactly13.3 C.Therefore,thesamequantityofheatmustalwaysbeproducedbyreaction ofthisquantityofmethane. Ifweburn1.0gofmethanetoheat500gofwaterinstead,weobserveatemperatureriseof26.6 C. Ifweburn1.0gofmethanetoheat1000gofiron,weobserveatemperatureriseof123 C.Therefore,the temperatureriseobservedisafunctionofthequantityofmaterialheatedaswellasthenatureofthematerial heated.Consequently,13.3 Cisnotanappropriatemeasureofthisquantityofheat,sincewecannotsay thattheburningof1.0gofmethane"produces13.3 Cofheat."Suchastatementisclearlyrevealedtobe nonsense,sotheconceptsoftemperatureandheatmustbekeptdistinct. Ourobservationsdorevealthatwecanrelatethetemperatureriseproducedinasubstancetoaxed quantityofheat,providedthatwespecifythetypeandamountofthesubstance.Therefore,wedenea propertyforeachsubstance,calledthe heatcapacity ,whichrelatesthetemperaturerisetothequantity ofheatabsorbed.Wedene q tobethequantityofheat,and T tobethetemperatureriseproducedby thisheat.Theheatcapacity C isdenedby q = C T .1 Thisequation,however,isonlyadenitionanddoesnothelpuscalculateeither q or C ,sinceweknow neitherone. Next,however,weobservethatwecanalsoelevatethetemperatureofasubstance mechanically ,thatis, bydoingworkonit.Assimpleexamples,wecanwarmwaterbystirringit,orwarmmetalbyrubbingor scrapingit.Asanhistoricalnote,theseobservationswerecrucialinestablishingthatheatisequivalentto workinitseectonmatter,demonstratingthatheatisthereforeaformofenergy.Althoughitisdicultto do,wecanmeasuretheamountofworkrequiredtoelevatethetemperatureof1gofwaterby1 C.Wend thattheamountofworkrequiredisinvariablyequalto4.184J.Consequently,adding4.184Jofenergyto1g ofwatermustelevatetheenergyofthewatermoleculesbyanamountmeasuredby1 C.Byconservationof energy,theenergyofthewatermoleculesdoesnotdependonhowthatenergywasacquired.Therefore,the increaseinenergymeasuredbya1 Ctemperatureincreaseisthesameregardlessofwhetherthewaterwas heatedorstirred.Assuch,4.184Jmustalsobetheamountofenergyaddedtothewatermoleculeswhen theyare heated by1 Cratherthanstirred.Wehavethereforeeectivelymeasuredtheheat q requiredto elevatethetemperatureof1gofwaterby1 C.Referringbackto.1,wenowcancalculatethattheheat capacityof1gofwatermustbe 4 : 184 J C .Theheatcapacity pergram ofasubstanceisreferredtoasthe specicheat ofthesubstance,usuallyindicatedbythesymbol c s .Thespecicheatofwateris 4 : 184 J C Determiningtheheatcapacityorspecicheatofwaterisanextremelyimportantmeasurementfortwo reasons.First,fromtheheatcapacityofwaterwecandeterminetheheatcapacityofanyothersubstance verysimply.Imaginetakingahot5.0gironweightat100 Candplacingitin10.0gofwaterat25 C.We knowfromexperiencethattheironbarwillbecooledandthewaterwillbeheateduntilbothhaveachieved thesametemperature.Thisisaneasyexperimenttoperform,andwendthatthenaltemperatureof theironandwateris28.8 C.Clearly,thetemperatureofthewaterhasbeenraisedby3.8 C.From.1 andthespecicheatofwater,wecancalculatethatthewatermusthaveabsorbedanamountofheat q = : 0 g 4 : 184 J g C : 8 C =159 J .Byconservationofenergy,thismustbetheamountofheat lost by the1gironweight,whosetemperaturewasloweredby71.2 C.Againreferringto.1,wecancalculate thespecicheatoftheironbartobe c s = )]TJ/F7 6.9738 Tf 6.227 0 Td [(159 J )]TJ/F7 6.9738 Tf 6.227 0 Td [(71 : 2 C : 0 g =0 : 45 J g C .Followingthisprocedure,wecan easilyproduceextensivetablesofheatcapacitiesformanysubstances. Second,andperhapsmoreimportantlyforourpurposes,wecanusetheknownspecicheatofwaterto measuretheheatreleasedinanychemicalreaction.Toanalyzeapreviousexample,weobservedthatthe combustionof1.0gofmethanegasreleasedsucientheattoincreasethetemperatureof1000gofwaterby 13.3 C.Theheatcapacityof1000gofwatermustbe g 4 : 184 J g C =4184 J C .Therefore,by.1, elevatingthetemperatureof1000gofwaterby13.3 Cmustrequire 55 ; 650 J =55 : 65 kJ ofheat.Therefore, burning1.0gofmethanegasproducesexactly55.65kJofheat. Themethodofmeasuringreactionenergiesbycapturingtheheatevolvedinawaterbathandmeasuring thetemperatureriseproducedinthatwaterbathiscalled calorimetry .Thismethodisdependentonthe

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91 equivalenceofheatandworkastransfersofenergy,andonthelawofconservationofenergy.Followingthis procedure,wecanstraightforwardlymeasuretheheatreleasedorabsorbedinanyeasilyperformedchemical reaction.Forreactionswhicharediculttoinitiateorwhichoccuronlyunderrestrictedconditionsorwhich areexceedinglyslow,wewillrequirealternativemethods. 10.4Observation2:Hess'LawofReactionEnergies Hydrogengas,whichisofpotentialinterestnationallyasacleanfuel,canbegeneratedbythereactionof carboncoalandwater: C s +2 H 2 O g CO 2 g +2 H 2 g .2 Calorimetryrevealsthatthisreactionrequiresthe input of90.1kJofheatforeverymoleof C s consumed. Byconvention,whenheatisabsorbedduringareaction,weconsiderthequantityofheattobeapositive number:inchemicalterms, q> 0 foran endothermic reaction.Whenheatisevolved,thereactionis exothermic and q< 0 byconvention. Itisinterestingtoaskwherethisinputenergygoeswhenthereactionoccurs.Onewaytoanswerthis questionistoconsiderthefactthatthereaction.2convertsonefuel, C s ,intoanother, H 2 g .To comparetheenergyavailableineachfuel,wecanmeasuretheheatevolvedinthecombustionofeachfuel withonemoleofoxygengas.Weobservethat C s + O 2 g CO 2 g .3 produces393.5kJforonemoleofcarbonburned;hence q = )]TJ/F8 9.9626 Tf 7.749 0 Td [(393 : 5 kJ .Thereaction 2 H 2 g + O 2 g 2 H 2 O g .4 produces483.6kJfortwomolesofhydrogengasburned,so q = )]TJ/F8 9.9626 Tf 7.749 0 Td [(483 : 6 kJ .Itisevidentthatmoreenergy isavailablefromcombustionofthehydrogenfuelthanfromcombustionofthecarbonfuel,soitisnot surprisingthatconversionofthecarbonfueltohydrogenfuelrequirestheinputofenergy. Ofconsiderableimportanceistheobservationthattheheatinputin.2,90.1kJ,isexactlyequalto the dierence betweentheheatevolved,-393.5kJ,inthecombustionofcarbon.3andtheheatevolved, -483.6kJ,inthecombustionofhydrogen.4.Thisisnotacoincidence:ifwetakethecombustionof carbon.3andaddtoitthe reverse ofthecombustionofhydrogen.4,weget C s + O 2 g CO 2 g 2 H 2 O g 2 H 2 g + O 2 g C s + O 2 g +2 H 2 O g CO 2 g +2 H 2 g + O 2 g .5 Cancelingthe O 2 g frombothsides,sinceitisnetneitherareactantnorproduct,10.5isequivalentto .2.Thus,takingthecombustionofcarbon.3and"subtracting"thecombustionofhydrogen.4 ormoreaccurately,addingthereverseofthecombustionofhydrogen.4yields.2.And,theheat ofthecombustionofcarbon.3 minus theheatofthecombustionofhydrogen.4equalstheheatof .2. Bystudyingmanychemicalreactionsinthisway,wediscoverthatthisresult,knownas Hess'Law ,is general. Law10.1: Hess'Law Theheatofanyreactionisequaltothesumoftheheatsofreactionforanysetofreactionswhich insumareequivalenttotheoverallreaction.

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92 CHAPTER10.ENERGETICSOFCHEMICALREACTIONS Althoughwehavenotconsideredtherestriction,applicabilityofthislawrequiresthatallreactions consideredproceedundersimilarconditions:wewillconsiderallreactionstooccuratconstantpressure. ApictorialviewofHess'Lawasappliedtotheheatof.2isillustrative.InFigure10.1APictorial ViewofHess'Law,thereactants C s +2 H 2 O g areplacedtogetherinabox,representingthe state ofthematerialsinvolvedinthereactionpriortothereaction.Theproducts CO 2 g +2 H 2 g areplaced togetherinasecondboxrepresentingthestateofthematerialsinvolvedafterthereaction.Thereaction arrowconnectingtheseboxesislabeledwiththeheatofthisreaction.Nowwetakethesesamematerials andplacetheminathirdboxcontaining C s O 2 g ,and 2 H 2 g .Thisboxisconnectedtothereactant andproductboxeswithreactionarrows,labeledbytheheatsofreactionin.3and.4. APictorialViewofHess'Law Figure10.1 ThispictureofHess'Lawrevealsthattheheatofreactionalongthe"path"directlyconnectingthe reactantstatetotheproductstateisexactlyequaltothetotalheatofreactionalongthealternative "path"connectingreactantstoproductsviatheintermediatestatecontaining C s O 2 g ,and 2 H 2 g AconsequenceofourobservationofHess'Lawisthereforethatthenetheatevolvedorabsorbedduringa reactionisindependentofthepathconnectingthereactanttoproduct.Thisstatementisagainsubjectto ourrestrictionthatallreactionsinthealternativepathmustoccurunderconstantpressureconditions. AslightlydierentviewofFigure10.1APictorialViewofHess'Lawresultsfrombeginningatthe reactantboxandfollowingacompletecircuitthroughtheotherboxesleadingbacktothereactantbox, summingthenetheatsofreactionaswego.Wediscoverthatthenetheattransferredagainprovidedthat allreactionsoccurunderconstantpressureisexactlyzero.Thisisastatementoftheconservationofenergy: theenergyinthereactantstatedoesnotdependupontheprocesseswhichproducedthatstate.Therefore, wecannotextractanyenergyfromthereactantsbyaprocesswhichsimplyrecreatesthereactants.Were thisnotthecase,wecouldendlesslyproduceunlimitedquantitiesofenergybyfollowingthecircuitouspath whichcontinuallyreproducestheinitialreactants. Bythisreasoning,wecandeneanenergyfunctionwhosevalueforthereactantsisindependentof howthereactantstatewasprepared.Likewise,thevalueofthisenergyfunctionintheproductstateis independentofhowtheproductsareprepared.Wechoosethisfunction, H ,sothatthechangeinthe function, H = H products )]TJ/F11 9.9626 Tf 9.255 0 Td [(H reactants ,isequaltotheheatofreaction q underconstantpressureconditions. H ,whichwecallthe enthalpy ,isa statefunction ,sinceitsvaluedependsonlyonthestateofthematerials underconsideration,thatis,thetemperature,pressureandcompositionofthesematerials. Theconceptofastatefunctionissomewhatanalogoustotheideaofelevation.Considerthedierence

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93 inelevationbetweentherstoorandthethirdoorofabuilding.Thisdierenceisindependentofthe pathwechoosetogetfromtherstoortothethirdoor.Wecansimplyclimbuptwoightsofstairs,or wecanclimboneightofstairs,walkthelengthofthebuilding,thenwalkasecondightofstairs.Orwe canridetheelevator.Wecouldevenwalkoutsideandhaveacraneliftustotheroofofthebuilding,from whichweclimbdowntothethirdoor.Eachpathproducesexactlythesameelevationgain,eventhough thedistancetraveledissignicantlydierentfromonepathtothenext.Thisissimplybecausetheelevation isa"statefunction."Ourelevation,standingonthethirdoor,isindependentofhowwegottothethird oor,andthesameistrueoftherstoor.Sincetheelevationthusastatefunction,theelevationgainis independentofthepath. Now,theexistenceofanenergystatefunction H isofconsiderableimportanceincalculatingheatsof reaction.ConsidertheprototypicalreactioninFigure10.2a,withreactants R beingconvertedtoproducts P .Wewishtocalculatetheheatabsorbedorreleasedinthisreaction,whichis H .Since H isastate function,wecanfollowanypathfrom R to P andcalculate H alongthatpath.InFigure10.2b,we consideronesuchpossiblepath,consistingoftworeactionspassingthroughanintermediatestatecontaining alltheatomsinvolvedinthereaction,eachinelementalform.Thisisausefulintermediatestatesinceit canbeusedforanypossiblechemicalreaction.Forexample,inFigure10.1APictorialViewofHess'Law, theatomsinvolvedinthereactionareC,H,andO,eachofwhicharerepresentedintheintermediatestate inelementalform.WecanseeinFigure10.2bthatthe H fortheoverallreactionisnowthedierence betweenthe H intheformationoftheproducts P fromtheelementsandthe H intheformationofthe reactants R fromtheelements. Calculationof H a b Figure10.2 The H valuesforformationofeachmaterialfromtheelementsarethusofgeneralutilityincalculating

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94 CHAPTER10.ENERGETICSOFCHEMICALREACTIONS H foranyreactionofinterest.Wethereforedenethe standardformationreaction forreactant R ,as elementsinstandardstate R andtheheatinvolvedinthisreactionisthe standardenthalpyofformation ,designatedby H f Thesubscript f ,standingfor"formation,"indicatesthatthe H isforthereactioncreatingthematerial fromtheelementsinstandardstate.Thesuperscript indicatesthatthereactionsoccurunderconstant standardpressureconditionsof1atm.FromFigure10.2b,weseethattheheatofanyreactioncanbe calculatedfrom H f = H f products )]TJ/F8 9.9626 Tf 9.962 0 Td [( H f reactants .6 Extensivetablesof H f havebeencompiledandpublished.Thisallowsustocalculatewithcomplete condencetheheatofreactionforanyreactionofinterest,evenincludinghypotheticalreactionswhichmay bediculttoperformorimpossiblyslowtoreact. 10.5Observation3:BondEnergiesinPolyatomicMolecules The bondenergy foramoleculeistheenergyrequiredtoseparatethetwobondedatomstogreatdistance. Werecallthatthetotalenergyofthebondingelectronsislowerwhenthetwoatomsareseparatedbythe bonddistancethanwhentheyareseparatedbyagreatdistance.Assuch,theenergyinputrequiredto separatetheatomselevatestheenergyoftheelectronswhenthebondisbroken. Wecanusediatomicbondenergiestocalculatetheheatofreaction H foranyreactioninvolvingonly diatomicmolecules.Weconsidertwosimpleexamples.First,thereaction H 2 g + Br g H g + HBr g .7 isobservedtobeendothermicwithheatofreaction 70 kJ mol .Notethatthisreactioncanbeviewedasconsisting entirelyofthebreakingofthe H 2 bondfollowedbytheformationofthe HBr bond.Consequently,wemust inputenergyequaltothebondenergyof H 2 436 kJ mol ,butinformingthe HBr bondwerecoveroutput energyequaltothebondenergyof HBr 366 kJ mol .Thereforetheheatof.7atconstantpressuremust beequaltodierenceinthesebondenergies, 70 kJ mol Nowwecananswerthequestion,atleastforthisreaction,ofwheretheenergy"goes"duringthereaction. Thereasonthisreactionabsorbsenergyisthatthebondwhichmustbebroken, H 2 ,isstrongerthanthe bondwhichisformed, HBr .Notethatenergyisreleasedwhenthe HBr bondisformed,buttheamount ofenergyreleasedislessthantheamountofenergyrequiredtobreakthe H 2 bondintherstplace. Thesecondexampleissimilar: H 2 g + Br 2 g 2 HBr g .8 Thisreactionisexothermicwith H = )]TJ/F8 9.9626 Tf 7.749 0 Td [(103 kJ mol .Inthiscase,wemustbreakan H 2 bond,withenergy 436 kJ mol ,anda Br 2 bond,withenergy 193 kJ mol .Sincetwo HBr moleculesareformed,wemustformtwo HBr bonds,eachwithbondenergy 366 kJ mol .Intotal,then,breakingthebondsinthereactantsrequires 629 kJ mol andformingthenewbondsreleases 732 kJ mol ,foranetreleaseof 103 kJ mol .Thiscalculationrevealsthatthe reactionisexothermicbecause,althoughwemustbreakoneverystrongbondandoneweakerbond,weform twostrongbonds. Therearetwoitemsworthreectionintheseexamples.First,energyisreleasedinachemicalreaction duetothe formation ofstrongbonds.Breakingabond,ontheotherhand,alwaysrequiresthe input of energy.Second,.8doesnotactuallyproceedbythetwo-stepprocessofbreakingbothreactantbonds, thusformingfourfreeatoms,followedbymakingtwonewbonds.Theactualprocessofthereactionis signicantlymorecomplicated.Thedetailsofthisprocessareirrelevanttotheenergeticsofthereaction,

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95 however,since,aswehaveshown,theheatofreaction H doesnotdependonthepathofthereaction. ThisisanotherexampleoftheutilityofHess'law. Wenowproceedtoapplythisbondenergyanalysistotheenergeticsofreactionsinvolvingpolyatomic molecules.Asimpleexampleisthecombustionofhydrogengasdiscussedpreviouslyhere.4.Thisis anexplosivereaction,producing483.6kJpermoleofoxygen.Calculatingtheheatofreactionfrombond energiesrequiresustoknowthebondenergiesin H 2 O .Inthiscase,wemustbreaknotonebuttwobonds: H 2 O g 2 H g + O g .9 Theenergyrequiredtoperformthisreactionismeasuredtobe 926 : 9 kJ mol ..4canproceedbyapathin whichwerstbreaktwo H 2 bondsandone O 2 bond,thenwefollowthereverseof.9twice: 2 H 2 g + O 2 g 4 H g +2 O g .10 4 H g +2 O g 2 H 2 O g 2 H 2 g + O 2 g 2 H 2 O g Therefore,theenergyof.4mustbetheenergyrequiredtobreaktwo H 2 bondsandone O 2 bondminus twicetheenergyof.9.Wecalculatethat H =2 )]TJ/F8 9.9626 Tf 4.567 -8.07 Td [(436 kJ mol +498 : 3 kJ mol )]TJ/F8 9.9626 Tf 9.185 0 Td [(2 )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [(926 : 9 kJ mol = )]TJ/F8 9.9626 Tf 7.749 0 Td [(483 : 5 kJ mol .It isclearfromthiscalculationthat.4isstronglyexothermicbecauseoftheverylargeamountofenergy releasedwhentwohydrogenatomsandoneoxygenatomformawatermolecule. Itistemptingtousetheheatof.9tocalculatetheenergyofanO-Hbond.Sincebreakingthe twoO-Hbondsinwaterrequires 926 : 9 kJ mol ,thenwemightinferthatbreakingasingleO-Hbondrequires 926 : 9 kJ mol 2 =463 : 5 kJ mol .However,thereaction H 2 O g OH g + H g .11 has H =492 kJ mol .Therefore,theenergyrequiredtobreakanO-Hbondin H 2 O isnotthesameasthe energyrequiredtobreaktheO-Hbondinthe OH diatomicmolecule.Stateddierently,itrequiresmore energytobreaktherstO-HbondinwaterthanisrequiredtobreakthesecondO-Hbond. Ingeneral,wendthattheenergyrequiredtobreakabondbetweenanytwoparticularatomsdepends uponthemoleculethosetwoatomsarein.Consideringyetagainoxygenandhydrogen,wendthatthe energyrequiredtobreaktheO-Hbondinmethanol CH 3 OH is 437 kJ mol ,whichdierssubstantiallyfrom theenergyof.11.Similarly,theenergyrequiredtobreakasingleC-Hbondinmethane CH 4 is 435 kJ mol buttheenergyrequiredtobreakallfourC-Hbondsinmethaneis 1663 kJ mol ,whichisnotequaltofourtimes theenergyofonebond.Asanothersuchcomparison,theenergyrequiredtobreakaC-Hbondis 400 kJ mol in trichloromethane HCCl 3 414 kJ mol indichloromethane H 2 CCl 2 ,and 422 kJ mol inchloromethane H 3 CCl Theseobservationsaresomewhatdiscouraging,sincetheyrevealthat,tousebondenergiestocalculate theheatofareaction,wemustrstmeasurethebondenergiesforallbondsforallmoleculesinvolvedinthat reaction.Thisisalmostcertainlymoredicultthanitisdesirable.Ontheotherhand,wecannotethat thebondenergiesforsimilarbondsinsimilarmoleculesareclosetooneanother.TheC-Hbondenergies inthethreechloromethanesaboveillustratethisquitewell.WecanestimatetheC-Hbondenergyinany oneofthesechloromethanesbytheaverageC-Hbondenergyinthethreechloromethanesmolecule,which is 412 kJ mol .Likewise,theaverageoftheC-Hbondenergiesinmethaneis 1663 kJ mol 4 =416 kJ mol andisthusa reasonableapproximationtotheenergyrequiredtobreakasingleC-Hbondinmethane. Byanalyzingmanybondenergiesinmanymolecules,wendthat,ingeneral,wecanapproximatethe bondenergyinanyparticularmoleculebytheaverageoftheenergiesofsimilarbonds.Theseaveragebond energiescanthenbeusedtoestimatetheheatofareactionwithoutmeasuringalloftherequiredbond energies.

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96 CHAPTER10.ENERGETICSOFCHEMICALREACTIONS Considerforexamplethecombustionofmethanetoformwaterandcarbondioxide: CH 4 g +2 O 2 g CO 2 g +2 H 2 O g .12 Wecanestimatetheheatofthisreactionbyusingaveragebondenergies.WemustbreakfourC-Hbonds atanenergycostofapproximately )]TJ/F8 9.9626 Tf 4.567 -8.07 Td [(4 412 kJ mol andtwo O 2 bondsatanenergycostofapproximately )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [(2 496 kJ mol .Formingthebondsintheproductsreleasesapproximately )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [(2 743 kJ mol forthetwoC=O doublebondsand )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [(4 463 kJ mol fortheO-Hbonds.Net,theheatofreactionisthusapproximately H = 1648+992 )]TJ/F8 9.9626 Tf 9.079 0 Td [(1486 )]TJ/F8 9.9626 Tf 9.079 0 Td [(1852= )]TJ/F8 9.9626 Tf 7.749 0 Td [(698 kJ mol .Thisisaratherroughapproximationtotheactualheatofcombustion ofmethane, )]TJ/F8 9.9626 Tf 7.749 0 Td [(890 kJ mol .Therefore,wecannotuseaveragebondenergiestopredictaccuratelytheheatofa reaction.Wecangetanestimate,whichmaybesucientlyuseful.Moreover,wecanusethesecalculations togaininsightintotheenergeticsofthereaction.Forexample,.12isstronglyexothermic,whichis whymethanegastheprimarycomponentinnaturalgasisanexcellentfuel.Fromourcalculation,we canseethatthereactioninvolvedbreakingsixbondsandformingsixnewbonds.Thebondsformedare substantiallystrongerthanthosebroken,thusaccountingforthenetreleaseofenergyduringthereaction. 10.6ReviewandDiscussionQuestions Exercise10.1 Assumeyouhavetwosamplesoftwodierentmetals,XandZ.Thesamplesareexactlythesame mass. Bothsamplesareheatedtothesametemperature.Theneachsampleisplacedintoseparate glassescontainingidenticalquantitiesofcoldwater,initiallyatidenticaltemperaturesbelowthat ofthemetals.ThenaltemperatureofthewatercontainingmetalXisgreaterthanthenal temperatureofthewatercontainingmetalZ.Whichofthetwometalshasthelargerheatcapacity? Explainyourconclusion. Ifeachsample,initiallyatthesametemperature,isheatedwithexactly100Jofenergy,which samplehasthehighernaltemperature? Exercise10.2 ExplainhowHess'Lawisaconsequenceofconservationofenergy. Exercise10.3 Considerthereaction N 2 O 4 g 2 NO 2 g DrawLewisstructuresforeachof N 2 O 4 and NO 2 .Onthebasisofthesestructures,predictwhether thereactionisendothermicorexothermic,andexplainyourreasoning. Exercise10.4 Whyisthebondenergyof H 2 notequalto H f of H 2 ?Forwhatspeciesistheenthalpyof formationrelatedtothebondenergyof H 2 ? Exercise10.5 Suggestareasonwhy H forthereaction CO 2 g CO g + O g isnotequalto H forthereaction CO g C g + O g Exercise10.6 Determinewhetherthereactionisexothermicorendothermicforeachofthefollowingcircumstances:

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97 Theheatofcombustionoftheproductsisgreaterthantheheatofcombustionofthereactants. Theenthalpyofformationoftheproductsisgreaterthantheenthalpyofformationofthe reactants. Thetotalofthebondenergiesoftheproductsisgreaterthanthetotalofthebondenergiesfor thereactants.

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98 CHAPTER10.ENERGETICSOFCHEMICALREACTIONS

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Chapter11 TheIdealGasLaw 1 11.1Foundation Weassumeasourstartingpointtheatomicmoleculartheory.Thatis,weassumethatallmatteriscomposed ofdiscreteparticles.Theelementsconsistofidenticalatoms,andcompoundsconsistofidenticalmolecules, whichareparticlescontainingsmallwholenumberratiosofatoms.Wealsoassumethatwehavedetermined acompletesetofrelativeatomicweights,allowingustodeterminethemolecularformulaforanycompound. 11.2Goals Theindividualmoleculesofdierentcompoundshavecharacteristicproperties,suchasmass,structure, geometry,bondlengths,bondangles,polarity,diamagnetismorparamagnetism.Wehavenotyetconsidered thepropertiesofmassquantitiesofmatter,suchasdensity,phasesolid,liquidorgasatroomtemperature, boilingandmeltingpoints,reactivity,andsoforth.Thesearepropertieswhicharenotexhibitedbyindividual molecules.Itmakesnosensetoaskwhattheboilingpointofonemoleculeis,nordoesanindividualmolecule existasagas,solid,orliquid.However,wedoexpectthatthesematerialorbulkpropertiesarerelatedto thepropertiesoftheindividualmolecules.Ourultimategoalistorelatethepropertiesoftheatomsand moleculestothepropertiesofthematerialswhichtheycomprise. Achievingthisgoalwillrequireconsiderableanalysis.InthisConceptDevelopmentStudy,webeginat asomewhatmorefundamentallevel,withourgoaltoknowmoreaboutthenatureofgases,liquidsand solids.Weneedtostudytherelationshipsbetweenthephysicalpropertiesofmaterials,suchasdensityand temperature.Webeginourstudybyexaminingthesepropertiesingases. 11.3Observation1:Pressure-VolumeMeasurementsonGases Itisanelementaryobservationthatairhasa"spring"toit:ifyousqueezeaballoon,theballoonrebounds toitsoriginalshape.Asyoupumpairintoabicycletire,theairpushesbackagainstthepistonofthepump. Furthermore,thisresistanceoftheairagainstthepistonclearlyincreasesasthepistonispushedfartherin. The"spring"oftheairismeasuredasapressure,wherethepressure P isdened P = F A .1 F istheforceexertedbytheaironthesurfaceofthepistonheadand A isthesurfaceareaofthepiston head. 1 Thiscontentisavailableonlineat. 99

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100 CHAPTER11.THEIDEALGASLAW Forourpurposes,asimplepressuregaugeissucient.Wetrapasmallquantityofairinasyringea pistoninsideacylinderconnectedtothepressuregauge,andmeasureboththevolumeofairtrappedinside thesyringeandthepressurereadingonthegauge.Inonesuchsamplemeasurement,wemightndthat,at atmosphericpressuretorr,thevolumeofgastrappedinsidethesyringeis29.0ml.Wethencompress thesyringeslightly,sothatthevolumeisnow23.0ml.Wefeeltheincreasedspringoftheair,andthisis registeredonthegaugeasanincreaseinpressureto960torr.Itissimpletomakemanymeasurementsin thismanner.AsamplesetofdataappearsinSampleDatafromPressure-VolumeMeasurement,p.100. Wenotethat,inagreementwithourexperiencewithgases,thepressureincreasesasthevolumedecreases. ThesedataareplottedhereFigure11.1:MeasurementsonSpringoftheAir. SampleDatafromPressure-VolumeMeasurement Pressuretorr Volumeml 760 29.0 960 23.0 1160 19.0 1360 16.2 1500 14.7 1650 13.3 MeasurementsonSpringoftheAir Figure11.1 Aninitialquestioniswhetherthereisaquantitativerelationshipbetweenthepressuremeasurements andthevolumemeasurements.Toexplorethispossibility,wetrytoplotthedatainsuchawaythatboth

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101 quantitiesincreasetogether.Thiscanbeaccomplishedbyplottingthepressureversustheinverseofthe volume,ratherthanversusthevolume.ThedataaregiveninAnalysisofSampleData,p.101andplotted hereFigure11.2:AnalysisofMeasurementsonSpringoftheAir. AnalysisofSampleData Pressuretorr Volumeml 1/Volume/ml Pressure Volume 760 29.0 0.0345 22040 960 23.0 0.0435 22080 1160 19.0 0.0526 22040 1360 16.2 0.0617 22032 1500 14.7 0.0680 22050 1650 13.3 0.0752 21945 AnalysisofMeasurementsonSpringoftheAir Figure11.2 Noticealsothat,withelegantsimplicity,thedatapointsformastraightline.Furthermore,thestraight lineseemstoconnecttotheorigin f 0 ; 0 g .Thismeansthatthepressuremustsimplybeaconstantmultiplied by 1 V : P = k 1 V .2

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102 CHAPTER11.THEIDEALGASLAW Ifwemultiplybothsidesofthisequationby V ,thenwenoticethat PV = k .3 Inotherwords,ifwegobackandmultiplythepressureandthevolumetogetherforeachexperiment,we shouldgetthesamenumbereachtime.TheseresultsareshowninthelastcolumnofAnalysisofSample Data,p.101,andweseethat,withintheerrorofourdata,allofthedatapointsgivethesamevalueofthe productofpressureandvolume.Thevolumemeasurementsaregiventothreedecimalplacesandhence areaccuratetoalittlebetterthan1%.Thevaluesof Pressure Volume areallwithin1%ofeachother, sotheuctuationsarenotmeaningful. Weshouldwonderwhatsignicance,ifany,canbeassignedtothenumber 22040 torrml wehave observed.Itiseasytodemonstratethatthis"constant"isnotsoconstant.Wecaneasilytrapanyamount ofairinthesyringeatatmosphericpressure.Thiswillgiveusanyvolumeofairwewishat760torrpressure. Hence,thevalue 22040 torrml isonlyobservedfortheparticularamountofairwehappenedtochoosein oursamplemeasurement.Furthermore,ifweheatthesyringewithaxedamountofair,weobservethat thevolumeincreases,thuschangingthevalueofthe 22040 torrml .Thus,weshouldbecarefultonotethat the productofpressureandvolumeisaconstantforagivenamountofairataxedtemperature .This observationisreferredtoas Boyle'sLaw ,datingto1662. ThedatagiveninSampleDatafromPressure-VolumeMeasurement,p.100assumedthatweusedairfor thegassample.That,ofcourse,wastheonlygaswithwhichBoylewasfamiliar.Wenowexperimentwith varyingthecompositionofthegassample.Forexample,wecanputoxygen,hydrogen,nitrogen,helium, argon,carbondioxide,watervapor,nitrogendioxide,ormethaneintothecylinder.Ineachcasewestart with29.0mlofgasat760torrand25 C.WethenvarythevolumesasinSampleDatafromPressureVolumeMeasurement,p.100andmeasurethepressures.Remarkably,wendthatthepressureofeach gasisexactlythesameaseveryothergasateachvolumegiven.Forexample,ifwepressthesyringetoa volumeof16.2ml,weobserveapressureof1360torr,nomatterwhichgasisinthecylinder.Thisresult alsoappliesequallywelltomixturesofdierentgases,themostfamiliarexamplebeingair,ofcourse. Weconcludethatthepressureofagassampledependsonthevolumeofthegasandthetemperature, butnotonthecompositionofthegassample.Wenowaddtothisresultaconclusionfromaprevious studyChapter3.Specically,werecalltheLawofCombiningVolumesLaw3.1,LawofCombining Volumes,p.10,whichstatesthat,whengasescombineduringachemicalreactionataxedpressureand temperature,theratiosoftheirvolumesaresimplewholenumberratios.Wefurtherrecallthatthisresult canbeexplainedinthecontextoftheatomicmoleculartheorybyhypothesizingthatequalvolumesofgas containequalnumbersofgasparticles,independentofthetypeofgas,aconclusionwecall Avogadro's Hypothesis .CombiningthisresultwithBoyle'slawrevealsthatthe pressure ofagasdependsonthe number ofgasparticles,the volume inwhichtheyarecontained,andthe temperature ofthesample.The pressuredoes not dependonthetypeofgasparticlesinthesampleorwhethertheyareevenallthesame. WecanexpressthisresultintermsofBoyle'slawbynotingthat,intheequation PV = k ,the"constant" k isactuallyafunctionwhichvarieswithbothnumberofgasparticlesinthesampleandthetemperature ofthesample.Thus,wecanmoreaccuratelywrite PV = k N;t .4 explicitlyshowingthattheproductofpressureandvolumedependson N ,thenumberofparticlesinthe gassample,and t ,thetemperature. Itisinterestingtonotethat,in1738,Bernoullishowedthattheinverserelationshipbetweenpressure andvolumecouldbeprovenbyassumingthatagasconsistsofindividualparticlescollidingwiththewalls ofthecontainer.However,thisearlyevidencefortheexistenceofatomswasignoredforroughly120years, andtheatomicmoleculartheorywasnottobedevelopedforanother70years,basedonmassmeasurements ratherthanpressuremeasurements.

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103 11.4Observation2:Volume-TemperatureMeasurementsonGases WehavealreadynotedthedependenceofBoyle'sLawontemperature.Toobserveaconstantproduct ofpressureandvolume,thetemperaturemustbeheldxed.Wenextanalyzewhathappenstothegas whenthetemperatureisallowedtovary.Aninterestingrstproblemthatmightnothavebeenexpected isthequestionofhowtomeasuretemperature.Infact,formostpurposes,wethinkoftemperatureonlyin therathernon-quantitativemannerof"howhotorcold"somethingis,butthenwemeasuretemperature byexaminingthelengthofmercuryinatube,orbytheelectricalpotentialacrossathermocoupleinan electronicthermometer.Wethenbrieyconsiderthecomplicatedquestionofjustwhatwearemeasuring whenwemeasurethetemperature. Imaginethatyouaregivenacupofwaterandaskedtodescribeitas"hot"or"cold."Evenwithouta calibratedthermometer,theexperimentissimple:youputyourngerinit.Onlyaqualitativequestionwas asked,sothereisnoneedforaquantitativemeasurementof"howhot"or"howcold."Theexperimentis onlyslightlymoreinvolvedifyouaregiventwocupsofwaterandaskedwhichoneishotterorcolder.A simplesolutionistoputonengerineachcupandtodirectlycomparethesensation.Youstilldon'tneed acalibratedthermometerorevenatemperaturescaleatall. Finally,imaginethatyouaregivenacupofwatereachdayforaweekatthesametimeandareasked todeterminewhichday'scupcontainedthehottestorcoldestwater.Sinceyoucannolongertrustyour sensorymemoryfromdaytoday,youhavenochoicebuttodeneatemperaturescale.Todothis,we makeaphysicalmeasurementonthewaterbybringingitintocontactwithsomethingelsewhoseproperties dependonthe"hotness"ofthewaterinsomeunspeciedway.Forexample,thevolumeofmercuryina glasstubeexpandswhenplacedinhotwater;certainstripsofmetalexpandorcontractwhenheated;some liquidcrystalschangecolorwhenheated; etc. Weassumethatthispropertywillhavethesamevaluewhen itisplacedincontactwithtwoobjectswhichhavethesame"hotness"ortemperature.Somewhatobliquely, thisdenesthetemperaturemeasurement. Forsimplicity,weillustratewithamercury-lledglasstubethermometer.Weobservequiteeasilythat whenthetubeisinsertedinwaterweconsider"hot,"thevolumeofmercuryislargerthanwhenweinsertthe tubeinwaterthatweconsider"cold."Therefore,thevolumeofmercuryisameasureofhowhotsomething is.Furthermore,weobservethat,whentwoverydierentobjectsappeartohavethesame"hotness,"they alsogivethesamevolumeofmercuryintheglasstube.Thisallowsustomakequantitativecomparisonsof "hotness"ortemperaturebasedonthevolumeofmercuryinatube. Allthatremainsistomakeupsomenumbersthatdenethescaleforthetemperature,andwecan literallydothisinanywaythatweplease.Thisarbitrarinessiswhatallowsustohavetwodierent,but perfectlyacceptable,temperaturescales,suchasFahrenheitandCentigrade.Thelatterscalesimplyassigns zerotobethetemperatureatwhichwaterfreezesatatmosphericpressure.Wetheninsertourmercury thermometerintofreezingwater,andmarkthelevelofthemercuryas"0".Anotherpointonourscale assigns100tobetheboilingpointofwateratatmosphericpressure.Weinsertourmercurythermometer intoboilingwaterandmarkthelevelofmercuryas"100."Finally,wejustmarkoinincrementsof 1 100 ofthe distancebetweenthe"0"andthe"100"marks,andwehaveaworkingthermometer.Giventhearbitrariness ofthiswayofmeasuringtemperature,itwouldberemarkabletondaquantitativerelationshipbetween temperatureandanyotherphysicalproperty. Yetthatiswhatwenowobserve.Wetakethesamesyringeusedintheprevioussectionandtrapinita smallsampleofairatroomtemperatureandatmosphericpressure.Fromourobservationsabove,itshould beclearthatthetypeofgasweuseisirrelevant.Theexperimentconsistsofmeasuringthevolumeofthe gassampleinthesyringeaswevarythetemperatureofthegassample.Ineachmeasurement,thepressure ofthegasisheldxedbyallowingthepistoninthesyringetomovefreelyagainstatmosphericpressure.A samplesetofdataisshowninSampleDatafromVolume-TemperatureMeasurement,p.103andplotted hereFigure11.3:Volumevs.TemperatureofaGas.

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104 CHAPTER11.THEIDEALGASLAW SampleDatafromVolume-TemperatureMeasurement Temperature C Volumeml 11 95.3 25 100.0 47 107.4 73 116.1 159 145.0 233 169.8 258 178.1 Volumevs.TemperatureofaGas Figure11.3 Wendthatthereisasimplelinearstraightlinerelationshipbetweenthevolumeofagasandits temperatureasmeasuredbyamercurythermometer.Wecanexpressthisintheformofanequationfora line: V = t + .5 where V isthevolumeand t isthetemperaturein C. and aretheslopeandinterceptoftheline,and

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105 inthiscase, =0 : 335 and, =91 : 7 .Wecanrewritethisequationinaslightlydierentform: V = t + .6 Thisisthesameequation,exceptthatitrevealsthatthequantity mustbeatemperature,sincewecan addittoatemperature.Thisisaparticularlyimportantquantity:ifweweretosetthetemperatureofthe gasequalto )]TJ/F1 9.9626 Tf 9.409 11.058 Td [( = )]TJ/F8 9.9626 Tf 7.749 0 Td [(273 C ,wewouldndthatthevolumeofthegaswouldbeexactly0!Thisassumes thatthisequationcanbeextrapolatedtothattemperature.Thisisquiteanoptimisticextrapolation,since wehaven'tmadeanymeasurementsnearto )]TJ/F8 9.9626 Tf 7.748 0 Td [(273 C .Infact,ourgassamplewouldcondensetoaliquidor solidbeforeweeverreachedthatlowtemperature. SincethevolumedependsonthepressureandtheamountofgasBoyle'sLaw,thenthevaluesof and alsodependonthepressureandamountofgasandcarrynoparticularsignicance.However, whenwerepeatourobservationsformanyvaluesoftheamountofgasandthexedpressure,wendthat the ratio )]TJ/F1 9.9626 Tf 9.409 11.059 Td [( = )]TJ/F8 9.9626 Tf 7.749 0 Td [(273 C doesnotvaryfromonesampletothenext.Althoughwedonotknowthe physicalsignicanceofthistemperatureatthispoint,wecanassertthatitisatrueconstant,independent ofanychoiceoftheconditionsoftheexperiment.Werefertothistemperatureas absolutezero ,since atemperaturebelowthisvaluewouldbepredictedtoproduceanegativegasvolume.Evidently,then,we cannotexpecttolowerthetemperatureofanygasbelowthistemperature. Thisprovidesusan"absolutetemperaturescale"withazerowhichisnotarbitrarilydened.Thiswe denebyadding273thevalueof totemperaturesmeasuredin C,andwedenethisscaletobein unitsofdegreesKelvinK.ThedatainSampleDatafromVolume-TemperatureMeasurement,p.103are nowrecalibratedtotheabsolutetemperaturescaleinAnalysisofVolume-TemperatureData,p.105and plottedhereFigure11.4:Volumevs.AbsoluteTemperatureofaGas. AnalysisofVolume-TemperatureData Temperature C TemperatureK Volumeml 11 284 95.3 25 298 100.0 47 320 107.4 73 350 116.1 159 432 145.0 233 506 169.8 258 531 178.1

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106 CHAPTER11.THEIDEALGASLAW Volumevs.AbsoluteTemperatureofaGas Figure11.4 Notethat thevolumeisproportionaltotheabsolutetemperature indegreesKelvin, V = kT .7 providedthatthepressureandamountofgasareheldconstant.Thisresultisknownas Charles'Law datingto1787. AswithBoyle'sLaw,wemustnownotethatthe"constant" k isnotreallyconstant,sincethevolume alsodependsonthepressureandquantityofgas.AlsoaswithBoyle'sLaw,wenotethatCharles'Lawdoes notdependonthe type ofgasonwhichwemakethemeasurements,butratherdependsonlythenumberof particlesofgas.Therefore,weslightlyrewriteCharles'Lawtoexplicitindicatethedependenceofkonthe pressureandnumberofparticlesofgas V = k N;P T .8 11.5TheIdealGasLaw Wehavebeenmeasuringfourpropertiesofgases:pressure,volume,temperature,and"amount",which wehaveassumedabovetobethenumberofparticles.Theresultsofthreeobservationsrelatethesefour propertiespairwise.Boyle'sLawrelatesthepressureandvolumeatconstanttemperatureandamountof gas: P V = k 1 N;T .9

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107 Charles'Lawrelatesthevolumeandtemperatureatconstantpressureandamountofgas: V = k 2 N;P T .10 TheLawofCombiningVolumesleadstoAvogadro'sHypothesisthatthevolumeofagasisproportionalto thenumberofparticles N providedthatthetemperatureandpressureareheldconstant.Wecanexpress thisas V = k 3 P;T N .11 Wewilldemonstratebelowthatthesethreerelationshipscanbecombinedintoasingleequationrelating P V T ,and N .Jumpingtotheconclusion,however,wecanmoreeasilyshowthatthesethreerelationships canbeconsideredasspecialcasesofthemoregeneralequationknownasthe IdealGasLaw : PV = nRT .12 where R isaconstant, n isthenumberofmolesofgas,relatedtothenumberofparticles N byAvogadro's number, N A n = N N A .13 InBoyle'sLaw,weexaminetherelationshipof P and V when n or N and T arexed.IntheIdealGas Law,when n and T areconstant, nRT isconstant,sotheproduct PV isalsoconstant.Therefore,Boyle's LawisaspecialcaseoftheIdealGasLaw.If n and P arexedintheIdealGasLaw,then V = nR P T and nR P isaconstant.Therefore,Charles'LawisalsoaspecialcaseoftheIdealGasLaw.Finally,if P and T areconstant,thenintheIdealGasLaw, V = RT P n andthevolumeisproportionalthenumberofmolesor particles.Hence,Avogadro'shypothesisisaspecialcaseoftheIdealGasLaw. WehavenowshownthattheeachofourexperimentalobservationsisconsistentwiththeIdealGasLaw. Wemightask,though,howdidwegettheIdealGasLaw?WewouldliketoderivetheIdealGasLawfrom thethreeexperiementalobservations.Todoso,weneedtolearnaboutthefunctions k 1 N;T k 2 N;P k 3 P;T WebeginbyexaminingBoyle'sLawinmoredetail:ifwehold N and P xedinBoyle'sLawandallow T tovary,thevolumemustincreasewiththetemperatureinagreementwithCharles'Law.Inotherwords,with N and P xed,thevolumemustbeproportionalto T .Therefore, k 1 inBoyle'sLawmustbeproportional to T : k 1 N;T = k 4 N T .14 where k 4 isanewfunctionwhichdependsonlyon N ..9thenbecomes P V = k 4 N T .15 Avogadro'sHypothesistellsusthat,atconstantpressureandtemperature,thevolumeisproportionalto thenumberofparticles.Therefore k 4 mustalsoincreaseproportionallywiththenumberofparticles: k 4 N = k N .16 where k isyetanothernewconstant.Inthiscase,however,therearenovariablesleft,and k istrulya constant.Combining.15and.16gives P V = k N T .17 ThisisveryclosetotheIdealGasLaw,exceptthatwehavethenumberofparticles, N ,insteadofthe numberofthenumberofmoles, n .Weputthisresultinthemorefamiliarformbyexpressingthenumber

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108 CHAPTER11.THEIDEALGASLAW ofparticlesintermsofthenumberofmoles, n ,bydividingthenumberofparticlesbyAvogadro'snumber, N A ,from11.13.Then,from.17, P V = k N A n T .18 Thetwoconstants, k and N A ,canbecombinedintoasingleconstant,whichiscommonlycalled R ,thegas constant.Thisproducesthefamiliarconclusionof11.12. 11.6Observation3:PartialPressures Wereferredbrieyabovetothepressureofmixturesofgases,notinginourmeasurementsleadingtoBoyle's Lawthatthetotalpressureofthemixturedependsonlyonthenumberofmolesofgas,regardlessofthe typesandamountsofgasesinthemixture.TheIdealGasLawrevealsthatthepressureexertedbyamole ofmoleculesdoesnotdependonwhatthosemoleculesare,andourearlierobservationaboutgasmixtures isconsistentwiththatconclusion. Wenowexaminetheactualprocessofmixingtwogasestogetherandmeasuringthetotalpressure. Consideracontainerofxedvolume25.0L.Weinjectintothatcontainer0.78molesof N 2 gasat298K. FromtheIdealGasLaw,wecaneasilycalculatethemeasuredpressureofthenitrogengastobe0.763atm. Wenowtakeanidenticalcontainerofxedvolume25.0L,andweinjectintothatcontainer0.22molesof O 2 gasat298K.Themeasuredpressureoftheoxygengasis0.215atm.Asathirdmeasurement,weinject 0.22molesof O 2 gasat298Kintothe rst containerwhichalreadyhas0.78molesof N 2 .Notethatthe mixtureofgaseswehavepreparedisverysimilartothatofair.Themeasuredpressureinthiscontaineris nowfoundtobe0.975atm. Wenotenowthatthe total pressureofthemixtureof N 2 and O 2 inthecontainerisequaltothesumof thepressuresofthe N 2 and O 2 samplestakenseparately.Wenowdenethe partialpressure ofeachgas inthemixturetobethepressureofeachgasasifitweretheonlygaspresent.Ourmeasurementstellus thatthepartialpressureof N 2 P N 2 ,is0.763atm,andthepartialpressureof O 2 P O 2 ,is0.215atm. Withthisdenition,wecannowsummarizeourobservationbysayingthatthetotalpressureofthe mixtureofoxygenandnitrogenisequaltothesumofthepartialpressuresofthetwogases.Thisisa generalresult: Dalton'sLawofPartialPressures Law11.1: Dalton'sLawofPartialPressures Thetotalpressureofamixtureofgasesisthesumofthepartialpressuresofthecomponentgases inthemixture 11.7ReviewandDiscussionQuestions Exercise11.1 SketchagraphwithtwocurvesshowingPressurevs.Volumefortwodierentvaluesofthenumber ofmolesofgas,with n 2 >n 1 ,bothatthesametemperature.Explainthecomparisonofthetwo curves. Exercise11.2 SketchagraphwithtwocurvesshowingPressurevs.1/Volumefortwodierentvaluesofthe numberofmolesofgas,with n 2 >n 1 ,bothatthesametemperature.Explainthecomparisonof thetwocurves. Exercise11.3 SketchagraphwithtwocurvesshowingVolumevs.Temperaturefortwodierentvaluesofthe numberofmolesofgas,with n 2 >n 1 ,bothatthesamepressure.Explainthecomparisonofthe twocurves.

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109 Exercise11.4 SketchagraphwithtwocurvesshowingVolumevsTemperaturefortwodierentvaluesofthe pressureofthegas,with P 2 >P 1 ,bothforthesamenumberofmoles.Explainthecomparisonof thetwocurves. Exercise11.5 Explainthesignicanceofthefactthat,inthevolume-temperatureexperiments, isobservedto havethesamevalue,independentofthequantityofgasstudiedandthetypeofgasstudied.What isthesignicanceofthequantity ?Whyisitmoresignicantthaneither or ? Exercise11.6 Amonton'sLaw saysthatthepressureofagasisproportionaltotheabsolutetemperaturefora xedquantityofgasinaxedvolume.Thus, P = k N;V T .DemonstratethatAmonton'sLaw canbederivedbycombiningBoyle'sLawandCharles'Law. Exercise11.7 UsingBoyle'sLawinyourreasoning,demonstratethatthe"constant"inCharles'Law, i.e. k 2 N;P ,isinverselyproportionalto P Exercise11.8 ExplainhowBoyle'sLawandCharles'Lawmaybecombinedtothegeneralresultthat,forconstant quantityofgas, P V = kT Exercise11.9 UsingDalton'sLawandtheIdealGasLaw,showthatthepartialpressureofacomponentofagas mixturecanbecalculatedfrom P i = PX i .19 Where P isthetotalpressureofthegasmixtureand X i isthe molefraction ofcomponent i denedby X i = n i n total .20 Exercise11.10 Dryairis78.084%nitrogen,20.946%oxygen,0.934%argon,and0.033%carbondioxide.Determine themolefractionsandpartialpressuresofthecomponentsofdryairatstandardpressure. Exercise11.11 Assesstheaccuracyofthefollowingstatement: Boyle'sLawstatesthat PV = k 1 ,where k 1 isaconstant.Charles'Lawstatesthat V = k 2 T where k 2 isaconstant.Inserting V fromCharles'LawintoBoyle'sLawresultsin Pk 2 T = k 1 Wecanrearrangethistoread PT = k 1 k 2 = aconstant.Therefore,thepressureofagasisinversely proportionaltothetemperatureofthegas. Inyourassessment,youmustdeterminewhatinformationiscorrectorincorrect,providethe correctinformationwhereneeded,explainwhetherthereasoningislogicalornot,andprovide logicalreasoningwhereneeded.

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110 CHAPTER11.THEIDEALGASLAW

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Chapter12 TheKineticMolecularTheory 1 12.1Foundation Weassumeanunderstandingoftheatomicmoleculartheorypostulates,includingthatallmatteriscomposed ofdiscreteparticles.Theelementsconsistofidenticalatoms,andcompoundsconsistofidenticalmolecules, whichareparticlescontainingsmallwholenumberratiosofatoms.Wealsoassumethatwehavedetermined acompletesetofrelativeatomicweights,allowingustodeterminethemolecularformulaforanycompound. Finally,weassumeaknowledgeofthe IdealGasLaw ,andtheobservationsfromwhichitisderived. 12.2Goals Ourcontinuinggoalistorelatethepropertiesoftheatomsandmoleculestothepropertiesofthematerials whichtheycomprise.Assimpleexamples,wecomparethesubstanceswater,carbondioxide,andnitrogen. Eachoftheseiscomposedofmoleculeswithfewtwoorthreeatomsandlowmolecularweight.However, thephysicalpropertiesofthesesubstancesareverydierent.Carbondioxideandnitrogenaregasesatroom temperature,butitiswellknownthatwaterisaliquidupto100 C.Toliquefynitrogen,wemustcoolit to-196 C,sotheboilingtemperaturesofwaterandnitrogendierbyabout300 C.Waterisaliquidover aratherlargetemperaturerange,freezingat0 C.Incontrast,nitrogenisaliquidforaverynarrowrange oftemperatures,freezingat-210 C.Carbondioxideposesyetanotherverydierentsetofproperties.At atmosphericpressure,carbondioxidegascannotbeliqueedatall:coolingthegasto-60 Cconvertsit directlytosolid"dryice."Asiscommonlyobserved,warmingdryicedoesnotproduceanyliquid,asthe solidsublimesdirectlytogas. Whyshouldthesematerials,whosemoleculesdonotseemallthatdierent,behavesodierently?What aretheimportantcharacteristicsofthesemoleculeswhichproducethesephysicalproperties?Itisimportant tokeepinmindthatthesearepropertiesofthebulkmaterials.Atthispoint,itisnotevenclearthatthe conceptofamoleculeisusefulinansweringthesequestionsaboutmeltingorboiling. Thereareatleasttwoprincipalquestionsthatariseaboutthe IdealGasLaw .First,itisinterestingto askwhetherthislawalwaysholdstrue,orwhetherthereareconditionsunderwhichthepressureofthegas cannotbecalculatedfrom nRT V .Wethusbeginbyconsideringthelimitationsofthevalidityofthe Ideal GasLaw .Weshallndthattheidealgaslawisonlyapproximatelyaccurateandthattherearevariations whichdodependuponthenatureofthegas.Second,then,itisinterestingtoaskwhytheidealgaslaw shouldeverholdtrue.Inotherwords,whyarethevariationsnottheruleratherthantheexception? Toanswerthesequestions,weneedamodelwhichwillallowustorelatethepropertiesofbulkmaterials tothecharacteristicsofindividualmolecules.Weseektoknowwhathappenstoagaswhenitiscompressed intoasmallervolume,andwhyitgeneratesagreaterresistingpressurewhencompressed.Perhapsmost 1 Thiscontentisavailableonlineat. 111

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112 CHAPTER12.THEKINETICMOLECULARTHEORY fundamentallyofall,weseektoknowwhathappenstoasubstancewhenitisheated.Whatpropertyofa gasismeasuredbythetemperature? 12.3Observation1:LimitationsoftheValidityoftheIdealGasLaw Todesignasystematictestforthevalidityofthe IdealGasLaw ,wenotethatthevalueof PV nRT ,calculated fromtheobservedvaluesof P V n ,and T ,shouldalwaysbeequalto1,exactly.Deviationof PV nRT from1 indicatesaviolationofthe IdealGasLaw .Wethusmeasurethepressureforseveralgasesunderavariety ofconditionsbyvarying n V ,and T ,andwecalculatetheratio PV nRT fortheseconditions. HereFigure12.1:ValidityoftheIdealGasLaw,thevalueofthisratioisplottedforseveralgases asafunctionofthe"particledensity"ofthegasinmoles, n V .Tomaketheanalysisofthisplotmore convenient,theparticledensityisgivenintermsoftheparticledensityofanidealgasatroomtemperature andatmosphericpressure i.e. thedensityofair,whichis 0 : 04087 mol L .InthisgureFigure12.1:Validity oftheIdealGasLaw,aparticledensityof10meansthattheparticledensityofthegasis10timesthe particledensityofairatroomtemperature.Thex-axisinthegureFigure12.1:ValidityoftheIdealGas Lawisthusunitless. ValidityoftheIdealGasLaw Figure12.1 Notethat PV nRT onthey-axisisalsounitlessandhasvalueexactly1foranidealgas.Weobservein thedatainthisgureFigure12.1:ValidityoftheIdealGasLawthat PV nRT isextremelycloseto1for particledensitieswhichareclosetothatofnormalair.Therefore,deviationsfromthe IdealGasLaw are notexpectedunder"normal"conditions.Thisisnotsurprising,since Boyle'sLaw Charles'Law ,and the LawofCombiningVolumes wereallobservedundernormalconditions.ThisgureFigure12.1: ValidityoftheIdealGasLawalsoshowsthat,astheparticledensityincreasesabovethenormalrange,the valueof PV nRT startstovaryfrom1,andthevariationdependsonthetypeofgasweareanalyzing.However, evenforparticledensities10timesgreaterthanthatofairatatmosphericpressure,the IdealGasLaw is accuratetoafewpercent.

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113 Thus,toobserveanysignicantdeviationsfrom PV = nRT ,weneedtopushthegasconditionsto somewhatmoreextremevalues.TheresultsforsuchextremeconditionsareshownhereFigure12.2: DeviationsfromtheIdealGasLaw.Notethatthedensitiesconsideredarelargenumberscorrespondingto veryhighpressures.Undertheseconditions,wendsubstantialdeviationsfromthe IdealGasLaw .In addition,weseethatthepressureofthegasandthus PV nRT doesdependstronglyonwhichtypeofgaswe areexamining.Finally,thisgureFigure12.2:DeviationsfromtheIdealGasLawshowsthatdeviations fromthe IdealGasLaw cangeneratepressureseithergreaterthanorlessthanthatpredictedbythe Ideal GasLaw DeviationsfromtheIdealGasLaw Figure12.2 12.4Observation2:DensityandCompressibilityofGas Forlowdensitiesforwhichthe IdealGasLaw isvalid,thepressureofagasisindependentofthenature ofthegas,andisthereforeindependentofthecharacteristicsoftheparticlesofthatgas.Wecanbuildon thisobservationbyconsideringthesignicanceofalowparticledensity.Evenatthehighparticledensities consideredinthisgureFigure12.2:DeviationsfromtheIdealGasLaw,allgaseshavelowdensityin comparisontothedensitiesofliquids.Toillustrate,wenotethat1gramofliquidwateratitsboilingpoint hasavolumeverycloseto1milliliter.Incomparison,thissame1gramofwater,onceevaporatedintosteam, hasavolumeofover1700milliliters.Howdoesthisexpansionbyafactorof1700occur?Itisnotcredible thattheindividualwatermoleculessuddenlyincreaseinsizebythisfactor.Theonlyplausibleconclusion isthatthedistancebetweengasmoleculeshasincreaseddramatically. Therefore,itisacharacteristicofagasthatthemoleculesarefarapartfromoneanother.Inaddition, thelowerthedensityofthegasthefartherapartthemoleculesmustbe,sincethesamenumberofmolecules occupiesalargervolumeatlowerdensity. Wereinforcethisconclusionbynotingthatliquidsandsolidsarevirtuallyincompressible,whereasgases areeasilycompressed.Thisiseasilyunderstoodifthemoleculesinagasareveryfarapartfromoneanother, incontrasttotheliquidandsolidwherethemoleculesaresocloseastobeincontactwithoneanother.

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114 CHAPTER12.THEKINETICMOLECULARTHEORY WeaddthisconclusiontotheobservationsinFigure12.1ValidityoftheIdealGasLawandFigure12.2 DeviationsfromtheIdealGasLawthatthepressureexertedbyagasdependsonlyonthenumberof particlesinthegasandisindependentofthetypeofparticlesinthegas,providedthatthedensityislow enough.Thisrequiresthatthegasparticlesbefarenoughapart.Weconcludethatthe IdealGasLaw holdstruebecausethereissucientdistancebetweenthegasparticlesthattheidentityofthegasparticles becomesirrelevant. Whyshouldthislargedistanceberequired?IfgasparticleAwerefarenoughawayfromgasparticleB thattheyexperiencenoelectricalormagneticinteraction,thenitwouldnotmatterwhattypesofparticles AandBwere.NorwoulditmatterwhatthesizesofparticlesAandBwere.Finally,then,weconcludefrom thisreasoningthatthevalidityoftheidealgaslawrestsofthepresumptionthattherearenointeractions ofanytypebetweengasparticles. 12.5PostulatesoftheKineticMolecularTheory Werecallatthispointourpurposeintheseobservations.Ourprimaryconcerninthisstudyisattempting torelatethepropertiesofindividualatomsormoleculestothepropertiesofmassquantitiesofthematerials composedoftheseatomsormolecules.Wenowhaveextensivequantitativeobservationsonsomespecic propertiesofgases,andweproceedwiththetaskofrelatingthesetotheparticlesofthesegases. Bytakinganatomicmolecularviewofagas,wecanpostulatethatthepressureobservedisaconsequence ofthecollisionsoftheindividualparticlesofthegaswiththewallsofthecontainer.Thispresumesthatthe gasparticlesareinconstantmotion.Thepressureis,bydenition,theforceappliedperarea,andtherecan benootheroriginforaforceonthewallsofthecontainerthanthatprovidedbytheparticlesthemselves. Furthermore,weobserveeasilythatthepressureexertedbythegasisthesameinalldirections.Therefore, thegasparticlesmustbemovingequallyinalldirections,implyingquiteplausiblythatthemotionsofthe particlesarerandom. Tocalculatetheforcegeneratedbythesecollisions,wemustknowsomethingaboutthemotionsofthe gasparticlessothatweknow,forexample,eachparticle'svelocityuponimpactwiththewall.Thisistoo muchtoask:thereareperhaps 10 20 particlesormore,andfollowingthepathofeachparticleisoutofthe question.Therefore,weseekamodelwhichpermitscalculationofthepressurewithoutthisinformation. Basedonourobservationsanddeductions,wetakeasthepostulatesofourmodel: Agasconsistsofindividualparticlesinconstantandrandommotion. Theindividualparticleshavenegligiblevolume. Theindividualparticlesdonotattractorrepeloneanotherinanyway. Thepressureofthegasisdueentirelytotheforceofthecollisionsofthegasparticleswiththewalls ofthecontainer. Thismodelisthe KineticMolecularTheoryofGases .Wenowlooktoseewherethismodelleads. 12.6DerivationofBoyle'sLawfromtheKineticMolecularTheory Tocalculatethepressuregeneratedbyagasof N particlescontainedinavolume V ,wemustcalculatethe force F generatedperarea A bycollisionsagainstthewalls.Todoso,webeginbydeterminingthenumber ofcollisionsofparticleswiththewalls.Thenumberofcollisionsweobservedependsonhowlongwewait. Let'smeasurethepressureforaperiodoftime t andcalculatehowmanycollisionsoccurinthattime period.Foraparticletocollidewiththewallwithinthetime t ,itmuststartcloseenoughtothewallto impactitinthatperiodoftime.Iftheparticleistravellingwithspeed v ,thentheparticlemustbewithin adistance v t ofthewalltohitit.Also,ifwearemeasuringtheforceexertedonthearea A ,theparticle musthitthatareatocontributetoourpressuremeasurement. Forsimplicity,wecanviewthesituationpictoriallyhereFigure12.3:CollisionofaParticlewithaWall withintime t.Weassumethattheparticlesaremovingperpendicularlytothewalls.Thisisclearlynot

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115 true.However,veryimportantly,thisassumptionisonlymadetosimplifythemathematicsofourderivation. Itisnotnecessarytomakethisassumption,andtheresultisnotaectedbytheassumption.Inorderfor aparticletohitthearea A markedonthewall,itmustliewithinthecylindershown,whichisoflength v t andcross-sectionalarea A .Thevolumeofthiscylinderis Av t ,sothenumberofparticlescontained inthecylinderis )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [( Av t N V CollisionofaParticlewithaWallwithintime t Figure12.3 Notalloftheseparticlescollidewiththewallduring t ,though,sincemostofthemarenottravelingin thecorrectdirection.Therearesixdirectionsforaparticletogo,correspondingtoplusorminusdirection inx,y,orz.Therefore,onaverage,thefractionofparticlesmovinginthecorrectdirectionshouldbe 1 6 assumingaswehavethatthemotionsareallrandom.Therefore,thenumberofparticleswhichimpactthe wallintime t is )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [( Av t N 6 V TheforcegeneratedbythesecollisionsiscalculatedfromNewton'sequation, F = ma ,where a isthe accelerationduetothecollisions.Considerrstasingleparticlemovingdirectlyperpendiculartoawall withvelocity v asinFigure12.3CollisionofaParticlewithaWallwithintime t.Wenotethat,when theparticlecollideswiththewall,thewalldoesnotmove,sothecollisionmustgenerallyconservetheenergy oftheparticle.Thentheparticle'svelocityafterthecollisionmustbe )]TJ/F11 9.9626 Tf 7.749 0 Td [(v ,sinceitisnowtravellinginthe oppositedirection.Thus,thechangeinvelocityoftheparticleinthisonecollisionis 2 v .Multiplyingby thenumberofcollisionsin t anddividingbythetime t ,wendthatthetotalaccelerationchangein velocitypertimeis 2 ANv 2 6 V ,andtheforceimpartedonthewallduecollisionsisfoundbymultiplyingbythe massoftheparticles: F = 2 ANmv 2 6 V .1

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116 CHAPTER12.THEKINETICMOLECULARTHEORY Tocalculatethepressure,wedividebythearea A ,tondthat P = Nmv 2 3 V .2 or,rearrangedforcomparisonto Boyle'sLaw PV = Nmv 2 3 .3 Sincewehaveassumedthattheparticlestravelwithconstantspeed v ,thentherightsideofthisequationis aconstant.Thereforetheproductofpressuretimesvolume, PV ,isaconstant,inagreementwith Boyle's Law .Furthermore,theproduct PV isproportionaltothenumberofparticles,alsoinagreementwith the LawofCombiningVolumes .Therefore,themodelwehavedevelopedtodescribeanidealgasis consistentwithourexperimentalobservations. Wecandrawtwoveryimportantconclusionsfromthisderivation.First,theinverserelationshipobserved betweenpressureandvolumeandtheindependenceofthisrelationshiponthetypeofgasanalyzedareboth duetothelackofinteractionsbetweengasparticles.Second,thelackofinteractionsisinturnduetothe greatdistancesbetweengasparticles,afactwhichwillbetrueprovidedthatthedensityofthegasislow. 12.7InterpretationofTemperature Theabsenceoftemperatureintheabovederivationisnotable.Theothergaspropertieshaveallbeen incorporated,yetwehavederivedanequationwhichomitstemperaturealltogether.Theproblemisthat, aswediscussedatlengthabove,thetemperaturewassomewhatarbitrarilydened.Infact,itisnotprecisely clearwhathasbeenmeasuredbythetemperature.Wedenedthetemperatureofagasintermsofthe volumeofmercuryinaglasstubeincontactwiththegas.Itisperhapsthennowonderthatsuchaquantity doesnotshowupinamechanicalderivationofthegasproperties. Ontheotherhand,thetemperaturedoesappearprominentlyinthe IdealGasLaw .Therefore,there mustbeagreatersignicanceandlessarbitrarinesstothetemperaturethanmighthavebeenexpected. Todiscernthissignicance,werewritethelastequationaboveintheform: PV = 2 3 N 1 2 mv 2 .4 Thelastquantityinparenthesiscanberecognizedasthekineticenergyofanindividualgasparticle,and N )]TJ/F7 6.9738 Tf 5.762 -4.147 Td [(1 2 mv 2 mustbethetotalkineticenergy KE ofthegas.Therefore PV = 2 3 KE .5 Nowweinsertthe IdealGasLaw for PV tondthat KE = 3 2 nRT .6 Thisisanextremelyimportantconclusion,foritrevealstheanswertothequestionofwhatpropertyis measuredbythetemperature.Weseenowthatthetemperatureisameasureofthetotalkineticenergyof thegas.Thus,whenweheatagas,elevatingitstemperature,weareincreasingtheaveragekineticenergy ofthegasparticles,causingthentomove,onaverage,morerapidly. 12.8AnalysisofDeviationsfromtheIdealGasLaw WeareatlastinapositiontounderstandtheobservationsaboveSection12.3:Observation1:Limitationsof theValidityoftheIdealGasLawofdeviationsfromthe IdealGasLaw .Themostimportantassumption

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117 ofourmodelofthebehaviorofanidealgasisthatthegasmoleculesdonotinteract.Thisallowedusto calculatetheforceimpartedonthewallofthecontainerduetoasingleparticlecollisionwithoutworrying aboutwheretheotherparticleswere.Inorderforagastodisobeythe IdealGasLaw ,theconditionsmust besuchthatthisassumptionisviolated. Whatdothedeviationsfromidealitytellusaboutthegasparticles?Startingwithverylowdensityand increasingthedensityasinFigure12.1ValidityoftheIdealGasLaw,wendthat,formanygases,the valueof PV nRT fallsbelow1.Onewaytostatethisresultisthat,foragivenvalueof V n ,and T ,thepressure ofthegasislessthanitwouldhavebeenforanidealgas.Thismustbetheresultoftheinteractionsof thegasparticles.Inorderforthepressuretobereduced,theforceofthecollisionsoftheparticleswiththe wallsmustbelessthanispredictedbyourmodelofanidealgas.Therefore,theeectoftheinteractions istoslowtheparticlesastheyapproachthewallsofthecontainer.Thismeansthatanindividualparticle approachingawallmustexperienceaforceactingtopullitbackintothebodyofthegas.Hence,thegas particlesmustattractoneanother.Therefore,theeectofincreasingthedensityofthegasisthatthegas particlesareconnedincloserproximitytooneanother.Atthiscloserrange,theattractionsofindividual particlesbecomesignicant.Itshouldnotbesurprisingthattheseattractiveforcesdependonwhatthe particlesare.WenoteinFigure12.1ValidityoftheIdealGasLawthatdeviationfromthe IdealGas Law isgreaterforammoniathanfornitrogen,andgreaterfornitrogenthanforhelium.Therefore,the attractiveinteractionsofammoniamoleculesaregreaterthanthoseofnitrogenmolecules,whichareinturn greaterthanthoseofheliumatoms.Weanalyzethisconclusionismoredetailbelow. Continuingtoincreasethedensityofthegas,wendinFigure12.2DeviationsfromtheIdealGas Lawthatthevalueof PV nRT beginstorise,eventuallyexceeding1andcontinuingtoincrease.Underthese conditions,therefore,thepressureofthegasisgreaterthanwewouldhaveexpectedfromourmodelof non-interactingparticles.Whatdoesthistellus?Thegasparticlesareinteractinginsuchawayasto increasetheforceofthecollisionsoftheparticleswiththewalls.Thisrequiresthatthegasparticlesrepel oneanother.Aswemovetohigherdensity,theparticlesareforcedintocloserandcloserproximity.Wecan concludethatgasparticlesatverycloserangeexperiencestrongrepulsiveforcesawayfromoneanother. Ourmodelofthebehaviorofgasescanbesummarizedasfollows:atlowdensity,thegasparticlesare sucientlyfarapartthattherearenointeractionsbetweenthem.Inthiscase,thepressureofthegasis independentofthenatureofthegasandagreeswiththe IdealGasLaw .Atsomewhathigherdensities, theparticlesareclosertogetherandtheinteractionforcesbetweentheparticlesareattractive.Thepressure ofthegasnowdependsonthestrengthoftheseinteractionsandislowerthanthevaluepredictedbythe IdealGasLaw .Atstillhigherdensities,theparticlesareexcessivelyclosetogether,resultinginrepulsive interactionforces.Thepressureofthegasundertheseconditionsishigherthanthevaluepredictedbythe IdealGasLaw 12.9Observation3:BoilingPointsofsimplehydrides Thepostulatesofthe KineticMolecularTheory provideusawaytounderstandtherelationshipbetween molecularpropertiesandthephysicalpropertiesofbulkamountsofsubstance.Asadistinctexampleofsuch anapplication,wenowexaminetheboilingpointsofvariouscompounds,focusingonhydridesofsixteen elementsinthemaingroupGroupsIVthroughVII.ThesearegivenhereBoilingPointsofHydridesof GroupsIVtoVII,p.117.

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118 CHAPTER12.THEKINETICMOLECULARTHEORY BoilingPointsofHydridesofGroupsIVtoVII BoilingPoint C CH 4 -164 NH 3 -33 H 2 O 100 HF 20 SiH 4 -111.8 PH 3 -87.7 H 2 S -60.7 HCl -85 GeH 4 -88.5 AsH 3 -55 H 2 Se -41.5 HBr -67 SnH 4 -52 SbH 3 -17.1 H 2 Te -2.2 HI -35 Intabularform,therearenoobvioustrendshere,andthereforenoobviousconnectiontothestructure orbondinginthemolecules.ThedatainthetableBoilingPointsofHydridesofGroupsIVtoVII,p.117 aredisplayedinasuggestiveform,however,inFigure12.4BoilingPointsofMainGroupHydrides,the boilingpointofeachhydrideisplottedaccordingtowhichperiodrowoftheperiodictablethemaingroup elementbelongs.Forexample,thePeriod2hydrides CH 4 NH 3 H 2 O ,and HF aregroupedinacolumn totheleftofthegure,followedbyacolumnforthePeriod3hydrides SiH 4 PH 3 H 2 S HCl ,etc. Nowafewtrendsaremoreapparent.First,thelowestboilingpointsineachperiodareassociatedwith theGroupIVhydrides CH 4 SiH 4 GeH 4 SnH 4 ,andthehighestboilingpointsineachperiodbelongto theGroupVIhydrides H 2 O H 2 S H 2 Se H 2 Te .Forthisreason,thehydridesbelongingtoasinglegroup havebeenconnectedinFigure12.4BoilingPointsofMainGroupHydrides.

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119 BoilingPointsofMainGroupHydrides Figure12.4 Second,wenoticethat,withtheexceptionsof NH 3 H 2 O ,and HF ,theboilingpointsofthehydrides alwaysincreaseinasinglegroupaswegodowntheperiodictable:forexample,inGroupIV,theboiling pointsincreaseintheorder CH 4
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120 CHAPTER12.THEKINETICMOLECULARTHEORY First,themostdominanttrendintheboilingpointsisthat,withinasinglegroup,theboilingpoints ofthehydridesincreaseaswemove down theperiodictable.ThisistrueinallfourgroupsinFigure12.4 BoilingPointsofMainGroupHydrides;theonlyexceptionstothistrendare NH 3 H 2 O ,and HF .Wecan concludethat,withnotableexceptions,intermolecularinteractionsincreasewithincreasingatomicnumber ofthecentralatominthemolecule.Thisistruewhetherthemoleculesofthegroupconsideredhavedipole momentsasinGroupsV,VI,andVIIornotasinGroupIV.Wecaninferthatthelargeintermolecular attractionsformoleculeswithlargecentralatomsarisesfromthelargenumberofchargedparticlesinthese molecules. Thistypeofinteractionarisesfromforcesreferredtoas Londonforces or dispersionforces .These forcesarebelievedtoarisefromtheinstantaneousinteractionsofthechargedparticlesfromonemolecule withthechargedparticlesinanadjacentmolecule.Althoughthesemoleculesmaynotbepolarindividually, thenucleiinonemoleculemayattracttheelectronsinasecondmolecule,thusinducinganinstantaneous dipoleinthesecondmolecule.Inturn,thesecondmoleculeinducesadipoleintherst.Thus,twonon-polar moleculescaninteractasifthereweredipole-dipoleattractionsbetweenthem,withpositiveandnegative chargesinteractingandattracting.Thetendencyofamoleculetohaveaninduceddipoleiscalledthe polarizability ofthemolecule.Themorechargedparticlesthereareinamolecule,themore polarizable amoleculeisandthegreatertheattractionsarisingfromdispersionforceswillbe. Second,wenotethat,withoutexception,theGroupIVhydridesmusthavetheweakestintermolecular interactionsineachperiod.Asnotedabove,thesearetheonlyhydridesthathavenodipolemoment. Consequently,ingeneral,moleculeswithoutdipolemomentshaveweakerinteractionsthanmoleculeswhich arepolar.Wemustqualifythiscarefully,however,bynotingthatthenonpolar SnH 4 hasahigherboiling pointthanthepolar PH 3 and HCl .Wecanconcludefromthesecomparisonsthattheincreasedpolarizability ofmoleculeswithheavieratomscanosetthelackofamoleculardipole. Third,andmostimportantly,wenotethattheintermolecularattractionsinvolving NH 3 H 2 O ,and HF mustbeuniquelyandunexpectedlylarge,sincetheirboilingpointsaremarkedlyoutoflinewith thoseoftherestoftheirgroups.Thecommonfeatureofthesemoleculesisthattheycontainsmallatomic numberatomswhicharestronglyelectronegative,whichhavelonepairs,andwhicharebondedtohydrogen atoms.Moleculeswithoutthesefeaturesdonothaveunexpectedlyhighboilingpoints.Wecandeduce fromtheseobservationsthatthehydrogenatomsineachmoleculeareunusuallystronglyattractedtothe lonepairelectronsonthestronglyelectronegativeatomswiththesamepropertiesinothermolecules.This intermolecularattractionofahydrogenatomtoanelectronegativeatomisreferredtoas hydrogenbonding Itisclearfromourboilingpointdatathathydrogenbondinginteractionsaremuchstrongerthaneither dispersionforcesordipole-dipoleattractions. 12.10ReviewandDiscussionQuestions Exercise12.1 Explainthesignicancetothedevelopmentofthekineticmolecularmodeloftheobservationthat theidealgaslawworkswellonlyatlowpressure. Exercise12.2 Explainthesignicancetothedevelopmentofthekineticmolecularmodeloftheobservationthat thepressurepredictedbytheidealgaslawisindependentofthetypeofgas. Exercise12.3 Sketchthevalueof PV nRT asafunctionofdensityfortwogases,onewithstrongintermolecular attractionsandonewithweakintermolecularattractionsbutstrongrepulsions. Exercise12.4 Giveabriefmolecularexplanationfortheobservationthatthepressureofagasatxedtemperature increasesproportionallywiththedensityofthegas.

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121 Exercise12.5 Giveabriefmolecularexplanationfortheobservationthatthepressureofagasconnedtoaxed volumeincreasesproportionallywiththetemperatureofthegas. Exercise12.6 Giveabriefmolecularexplanationfortheobservationthatthevolumeofaballoonincreasesroughly proportionallywiththetemperatureofthegasinsidetheballoon. Exercise12.7 Explainwhythereisacorrelationbetweenhighboilingpointandstrongdeviationfromthe Ideal GasLaw Exercise12.8 ReferringtoFigure12.4BoilingPointsofMainGroupHydrides,explainwhythehydrideofthe Group4elementalwayshasthelowestboilingpointineachperiod. Exercise12.9 ExplainwhythePeriod2hydridesexcept CH 4 allhavehighboilingpoints,andexplainwhy CH 4 isanexception.

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122 CHAPTER12.THEKINETICMOLECULARTHEORY

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Chapter13 PhaseEquilibriumandIntermolecular Interactions 1 13.1Foundation The"phase"ofasubstanceistheparticularphysicalstateitisin.Themostcommonphasesaresolid,liquid, andgas,eacheasilydistinguishablebytheirsignicantlydierentphysicalproperties.Agivensubstancecan existindierentphasesunderdierentconditions:watercanexistassolidice,liquid,orsteam,butwater moleculesare H 2 O regardlessofthephase.Furthermore,asubstancechangesphasewithoutundergoing anychemicaltransformation:theevaporationofwaterorthemeltingoficeoccurwithoutdecompositionor modicationofthewatermolecules.Indescribingthedieringstatesofmatterchangesbetweenthem,we willalsoassumeanunderstandingoftheprinciplesofthe AtomicMolecularTheory andthe Kinetic MolecularTheory .Wewillalsoassumeanunderstandingofthebonding,structure,andpropertiesof individualmolecules. 13.2Goals Wehavedevelopedaveryclearmolecularpictureofthegasphase,viatheKineticMolecularTheory. Thegasparticlesatomsormoleculesareverydistantfromoneanother,sucientlysothatthereareno interactionsbetweentheparticles.Thepathofeachparticleisindependentofthepathsofallotherparticles. Wecandeterminemanyofthepropertiesofthegasfromthisdescription;forexample,thepressurecanbe determinedbycalculatingtheaverageforceexertedbycollisionsofthegasparticleswiththewallsofthe container. Todiscussliquidsandsolids,though,wewillbeforcedtoabandonthemostfundamentalpiecesofthe KineticMolecularTheoryofGases.First,itisclearthattheparticlesintheliquidorsolidphasesarevery muchclosertogetherthantheyareinthegasphase,becausethedensitiesofthese"condensed"phasesare oftheorderofathousandtimesgreaterthanthetypicaldensityofagas.Infact,weshouldexpectthatthe particlesintheliquidorsolidphasesareessentiallyincontactwitheachotherconstantly.Second,sincethe particlesinliquidorsolidareinclosecontact,itisnotreasonabletoimaginethattheparticlesdonointeract withoneanother.Ourassumptionthatthegasparticlesdonotinteractisbased,inpart,ontheconcept thattheparticlesaretoofaraparttointeract.Moreover,particlesinaliquidorsolidmustinteract,for withoutattractionsbetweentheseparticles,randommotionwouldrequirethatthesolidorliquiddissipate orfallapart. Inthisstudy,wewillpursueamodeltodescribethedierencesbetweencondensedphasesandgases andtodescribethetransitionswhichoccurbetweenthesolid,liquid,andgasphases.Wewillndthat 1 Thiscontentisavailableonlineat. 123

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124 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS intermolecularinteractionsplaythemostimportantroleingoverningphasetransitions,andwewillpursue anunderstandingofthevariationsoftheseintermolecularinteractionsfordierentsubstances. 13.3Observation1:Gas-LiquidPhaseTransitions Webeginbyreturningtoourobservationsof Charles'Law .Recallthatwetrapanamountofgasin acylinderttedwithapiston,andweapplyaxedpressuretothepiston.Wevarythetemperatureof thegas,andsincethepressureappliedtothepistonisconstant,thepistonmovestomaintainaconstant pressureofthetrappedgas.Ateachtemperature,wethenmeasurethevolumeofthegas.Fromourprevious observations,weknowthatthevolumeofthegasisproportionaltotheabsolutetemperatureindegrees Kelvin.ThusagraphofvolumeversusabsolutetemperatureFigure13.1:Vapor-LiquidPhaseTransition isastraightline,whichcanbeextrapolatedtozerovolumeat0K. Vapor-LiquidPhaseTransition Figure13.1 Consider,then,tryingtomeasurethevolumeforlowerandlowertemperaturestofollowthegraph Figure13.1:Vapor-LiquidPhaseTransition.Tobespecic,wetakeexactly1.00molofbutane C 4 H 10 at1atmpressure.Aswelowerthetemperaturefrom400Kto300K,weobservetheexpectedproportional decreaseinthevolumefrom32.8Lto24.6Landthisproportionalityworksverywellfortemperaturesjust slightlyabove272.6K,wherethevolumeis22.4L.However,whenwereach272.6K,thevolumeofthebutane dropsveryabruptly,fallingtoabout0.097Lattemperaturesjustslightlybelow272.6K.Thisislessthanonehalfofonepercentofthepreviousvolume!ThestrikingchangeinvolumeisshowninthegraphFigure13.1: Vapor-LiquidPhaseTransitionasaverticallineat272.6K. Thisdramaticchangeinphysicalpropertiesatonetemperatureisreferredtoasa phasetransition Whencoolingbutanethroughthetemperature272.6K,thebutaneisabruptlyconvertedatthattemperature fromonephase,gas,toanotherphase,liquid,withverydierentphysicalproperties.Ifwereversethe

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125 process,startingwithliquidbutaneat1atmpressureandtemperaturebelow272.6Kandthenheating,we ndthatthebutaneremainsentirelyliquidfortemperaturesbelow272.6Kandthenbecomesentirelygas fortemperaturesabove272.6K.Werefertothetemperatureofthephasetransitionasthe boilingpoint temperature.Wewilldiscussthephasespresent at theboilingpoint,ratherthanaboveandbelowthat temperature,inanothersectionSection13.4:Observation2:Vaporpressureofaliquid. Wenowconsiderhowthephasetransitiondependsonavarietyoffactors.First,weconsidercapturing 2.00molofbutaneinthecylinderinitially,stillat1atmpressure.Thevolumeof2.00molistwicethatof 1.00mol,by Avogadro'shypothesis .Theproportionaldecreaseinthevolumeof2.00molofgasisshown inFigure13.2VariationofPhaseTransitionwithPressurealongwiththepreviousresultfor1.00mol. Notethatthephasetransitionisobservedtooccuratexactlythesametemperature,272.6K,eventhough thereisdoublethemassofbutane. VariationofPhaseTransitionwithPressure Figure13.2 Considerinsteadthenvaryingtheappliedpressure.Theresultforcooling1.00molofbutaneata constant2.00atmpressureisalsoshowninFigure13.2VariationofPhaseTransitionwithPressure.We observethenowfamiliarphasetransitionwithasimilardramaticdropinvolume.However,inthiscase, wendthatthephasetransitionoccursat293.2K,over20Khigherthanatthelowerpressure.Therefore, thetemperatureofthephasetransitiondependsonthepressureapplied.Wecanmeasuretheboilingpoint temperatureofbutaneasafunctionoftheappliedpressure,andthisresultisplottedhereFigure13.3: BoilingPointversusPressure.

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126 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS BoilingPointversusPressure Figure13.3 Finally,weconsidervaryingthesubstancewhichwetrapinthecylinder.Ineachcase,wediscoverthat theboilingpointtemperaturedependsonbothwhatthesubstanceisandontheappliedpressure,butdoes notdependontheamountofthesubstancewetrap.InFigure13.3BoilingPointversusPressure,wehave alsoplottedtheboilingpointasafunctionofthepressureforseveralsubstances.Itisveryclearthatthe boilingpointsfordierentsubstancescanbeverydierentfromoneanother,althoughthevariationofthe boilingpointwithpressurelookssimilarfromonesubstancetothenext. 13.4Observation2:Vaporpressureofaliquid Ourpreviousobservationsindicatethat,foragivenpressure,thereisaphasetransitiontemperaturefor liquidandgas:belowtheboilingpoint,theliquidistheonly stable phasewhichexists,andanygaswhich mightexistatthatpointwillspontaneouslycondenseintoliquid.Abovetheboilingpoint,thegasisthe onlystablephase. However,wecanalsocommonlyobservethatanyliquidleftinanopencontainerwill,undermost conditions,eventuallyevaporate,evenifthetemperatureoftheliquidiswellbelowthenormalboilingpoint. Forexample,weoftenobservethatliquidwaterevaporatesattemperatureswellbelowtheboilingpoint.This observationonlyseemssurprisinginlightofthediscussionofabove.Whywouldliquidwaterspontaneously evaporateifliquidisthemorestablephasebelowtheboilingpoint?Weclearlyneedtofurtherdevelopour understandingofphasetransitions. Thetendencyofaliquidtoevaporateisreferredtoasits volatility :amorevolatileliquidevaporates

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127 morereadily.Tomakeaquantitativemeasureofliquidvolatility,weslightlymodifyourpreviouscylinderpistonapparatusbyaddingagaugetomeasurethepressureofgasinsidethecylinder.HereFigure13.4: MeasuringVaporPressureisanillustration.Webeginwithliquidwateronlyinthecylinderwithan appliedpressureof1atmatatemperatureof25 C.Wenowpullbackthepistonbyanarbitraryamount, andthenwelockthepistoninplace,xingthevolumetrappedinsidethecylinder.Wemightexpectto havecreatedavacuuminthecavityabovetheliquidwater,andassuchwemightexpectthatthepressure insidethecylinderissmallorzero. MeasuringVaporPressure Figure13.4 Althoughtherewasinitiallynogasinthecontainer,weobservethatthepressureinsidethecontainer risestoaxedvalueof23.8torr.Clearly,theobservationofpressureindicatesthepresenceofgaseouswater insidethecontainer,arisingfromevaporationofsome,butnotall,oftheliquidwater.Therefore,someof theliquidwatermusthaveevaporated.Ontheotherhand,alookinsidethecontainerrevealsthatthereis stillliquidwaterpresent.Sincebothaliquidphaseandagasphasearepresentatthesametime,wesay thattheliquidwaterandthewatervapormustbein phaseequilibrium .Theterm equilibrium inthis caseindicatesthatneitherthevapornortheliquidspontaneouslyconvertsintotheotherphase.Rather, bothphasesarestableatequilibrium. Veryinterestingly,wecanrepeatthismeasurementbypullingthepistonbacktoanyotherarbitrary positionbeforelockingitdown,and,providedthatthereisstillsomeliquidwaterpresent,thepressurein thecontainerineverycaserisestothesamexedvalueof23.8torr.Itdoesnotmatterwhatvolumewe havetrappedinsidethecylinder,nordoesitmatterhowmuchliquidwaterwestartedwith.Aslongas thereisstillsomeliquidwaterpresentinthecylinderatequilibrium,thepressureofthevaporabovethat liquidis23.8torrat25 C. Notethat,invaryingeithertheamountofliquidinitiallyorthexedvolumeofthecontainer,the amount ofliquidwaterthatevaporatesmustbedierentineachcase.Thiscanbeseenfromthefactthatthevolume availableforvapormustbedierentinvaryingeitherthevolumeofthecontainerortheinitialvolumeof theliquid.Sinceweobservethatthepressureofthevaporisthesameataxedtemperature,thediering volumesrevealdieringnumbersofmolesofwatervapor.Clearlyitisthe pressure ofthevapor,notthe amount,whichisthemostimportantpropertyinestablishingtheequilibriumbetweentheliquidandthe vapor.Wecanconcludethat,atagivenxedtemperature,thereisasinglespecicpressureatwhicha givenliquidanditsvaporwillbeinphaseequilibrium.Wecallthisthe vaporpressure oftheliquid. Wecanimmediatelyobservesomeimportantfeaturesofthevaporpressure.First,foragivensubstance, thevaporpressurevarieswiththetemperature.Thiscanbefoundbysimplyincreasingthetemperature

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128 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS ontheclosedcontainerintheprecedingexperiment.Ineverycase,weobservethattheequilibriumvapor pressureincreaseswithincreasesinthetemperature. ThevaporpressuresofseveralliquidsatseveraltemperaturesareshownhereFigure13.5:VaporPressuresofVariousLiquids.Thevaporpressureforeachliquidincreasessmoothlywiththetemperature, althoughtherelationshipbetweenvaporpressureandtemperatureisdenitelynotproportional. VaporPressuresofVariousLiquids Figure13.5 Second,Figure13.5VaporPressuresofVariousLiquidsclearlyillustratesthatthevaporpressure dependsstronglyonwhattheliquidsubstanceis.Thesevariationsreectthediering volatilities ofthe liquids:thosewithhighervaporpressuresaremorevolatile.Inaddition,thereisaveryinterestingcorrelation betweenthevolatilityofaliquidandtheboilingpointoftheliquid.Withoutexception,thesubstanceswith highboilingpointshavelowvaporpressuresandviceversa. Lookingmorecloselyattheconnectionbetweenboilingpointandvaporpressure,wecanndanimportantrelationship.LookingatFigure13.5VaporPressuresofVariousLiquids,wediscoverthatthevapor pressureofeachliquidisequalto760torrwhichisequalto1atmattheboilingpointforthatliquid. Howshouldweinterpretthis?Atanappliedpressureof1atm,thetemperatureofthephasetransitionfrom liquidtogasisthetemperatureatwhichthevaporpressureoftheliquidisequalto1atm.Thisstatement isactuallytrueregardlessofwhichpressureweconsider:ifweapplyapressureof0.9atm,theboilingpoint temperatureisthetemperatureatwhichtheliquidasavaporpressureof0.9atm.Statedgenerally,the liquidundergoesphasetransitionatthetemperaturewherethevaporpressureequalstheappliedpressure.

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129 13.5Observation3:PhaseDiagrams Sincetheboilingpointisthetemperatureatwhichtheappliedpressureequalsthevaporpressure,thenwe canviewFigure13.5VaporPressuresofVariousLiquidsinadierentway.Considerthespeciccaseof water,withvaporpressuregivenhereFigure13.6:VaporPressureofLiquidWater.Tondtheboiling pointtemperatureat1atmpressure,weneedtondthetemperatureatwhichthevaporpressureis1atm. Todoso,wendthepointonthegraphwherethevaporpressureis1atmandreadothecorresponding temperature,whichmustbetheboilingpoint.Thiswillworkatanygivenpressure.Viewedthisway,for waterFigure13.6VaporPressureofLiquidWatergivesus both thevaporpressureasafunctionofthe temperature and theboilingpointtemperatureasafunctionofthepressure.Theyarethesamegraph. VaporPressureofLiquidWater Figure13.6 Recallthat,attheboilingpoint,weobservethatbothliquidandgasareatequilibriumwithoneanother. Thisistrueateverycombinationofappliedpressureandboilingpointtemperature.Therefore,forevery combinationoftemperatureandpressureonthegraphinFigure13.6VaporPressureofLiquidWater,we observeliquid-gasequilibrium. Whathappensattemperature/pressurecombinationswhicharenotonthelineinFigure13.6Vapor PressureofLiquidWater?Tondout,werststartatatemperature-pressurecombinationonthegraph andelevatethetemperature.Thevaporpressureoftheliquidrises,andiftheappliedpressuredoesnot alsoincrease,thenthevaporpressurewillbegreaterthantheappliedpressure.Wemustthereforenotbe atequilibriumanymore.Alloftheliquidvaporizes,andthereisonlygasinthecontainer.Conversely,ifwe startatapointonthegraphandlowerthetemperature,thevaporpressureisbelowtheappliedpressure, andweobservethatallofthegascondensesintotheliquid. Now,whatifwestartatatemperature-pressurecombinationonthegraphandelevatetheapplied

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130 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS pressurewithoutraisingthetemperature?Theappliedpressurewillbegreaterthanthevaporpressure, andallofthegaswillcondenseintotheliquid.Figure13.6VaporPressureofLiquidWaterthusactually revealstouswhatphaseorphasesarepresentateachcombinationoftemperatureandpressure:alongthe line,liquidandgasareinequilibrium;abovetheline,onlyliquidispresent;belowtheline,onlygasis present.WhenwelabelthegraphwiththephaseorphasespresentineachregionasinFigure13.6Vapor PressureofLiquidWater,werefertothegraphasa phasediagram Ofcourse,Figure13.6VaporPressureofLiquidWateronlyincludesliquid,gas,andliquid-gasequilibrium.Weknowthat,ifthetemperatureislowenough,weexpectthatthewaterwillfreezeintosolid.To completethephasediagram,weneedadditionalobservations. WegobacktoourapparatusinFigure13.4MeasuringVaporPressureandweestablishliquid-gas waterphaseequilibriumatatemperatureof25 Cand23.8torr.Ifweslowlylowerthetemperature,the vaporpressuredecreasesslowlyaswell,asshowninFigure13.6VaporPressureofLiquidWater.If wecontinuetolowerthetemperature,though,weobserveaninterestingtransition,asshowninthemore detailedFigure13.7WaterPhaseTransitions.Theverysmoothvariationinthevaporpressureshowsa slight,almostunnoticeablebreakverynearto0 C.Belowthistemperature,thepressurecontinuestovary smoothly,butalongaslightlydierentcurve. WaterPhaseTransitions Figure13.7 Tounderstandwhatwehaveobserved,weexaminethecontentsofthecontainer.Wendthat,at temperaturesbelow0 C,thewaterinthecontainerisnowanequilibriummixtureofwatervaporandsolid waterice,andthereisnoliquidpresent.Thedirecttransitionfromsolidtogas,withoutliquid,iscalled sublimation .Forpressure-temperaturecombinationsalongthisnewcurvebelow0 C,then,thecurve showsthesolid-gasequilibriumconditions.Asbefore,wecaninterpretthistwoways.Thesolid-gascurve givesthevaporpressureofthesolidwaterasafunctionoftemperature,andalsogivesthesublimation temperatureasafunctionofappliedpressure.

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131 Figure13.7WaterPhaseTransitionsisstillnotacompletephasediagram,becausewehavenotincluded thecombinationsoftemperatureandpressureatwhichsolidandliquidareatequilibrium.Asastarting pointfortheseobservations,welookmorecarefullyattheconditionsnear0 C.Verycarefulmeasurements revealthatthesolid-gaslineandtheliquid-gaslineintersectinFigure13.7WaterPhaseTransitionswhere thetemperatureis0.01 C.Undertheseconditions,weobserveinsidethecontainerthatsolid,liquid,and gasareallthreeatequilibriuminsidethecontainer.Assuch,thisuniquetemperature-pressurecombination iscalledthe triplepoint .Atthispoint,theliquidandthesolidhavethesamevaporpressure,soallthree phasescanbeatequilibrium.Ifweraisetheappliedpressureslightlyabovethetriplepoint,thevapormust disappear.Wecanobservethat,byveryslightlyvaryingthetemperature,thesolidandliquidremainin equilibrium.Wecanfurtherobservethatthetemperatureatwhichthesolidandliquidareinequilibrium variesalmostimperceptiblyasweincreasethepressure.Ifweincludethesolid-liquidequilibriumconditions onthepreviousphasediagram,wegetthisFigure13.8:PhaseDiagramofWater,wherethesolid-liquid lineisverynearlyvertical. PhaseDiagramofWater Figure13.8 Eachsubstancehasitsownuniquephasediagram,correspondingtothediagraminFigure13.8Phase DiagramofWaterforwater. 13.6Observation4:DynamicEquilibrium Thereareseveralquestionsraisedbyourobservationsofphaseequilibriumandvaporpressure.Therst wewillconsideriswhythepressureofavaporinequilibriumwithitsliquiddoesnotdependonthevolume ofthecontainerintowhichtheliquidevaporates,orontheamountofliquidinthecontainer,oronthe

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132 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS amountofvaporinthecontainer.Whydowegetthesamepressureforthesametemperature,regardless ofotherconditions?Toaddressthisquestion,weneedtounderstandthecoexistenceofvaporandliquidin equilibrium.Howisthisequilibriumachieved? Toapproachthesequestions,letuslookagainatthesituationinFigure13.4MeasuringVaporPressure. Webeginwithacontainerwithaxedvolumecontainingsomeliquid,andequilibriumisachievedatthe vaporpressureoftheliquidatthexedtemperaturegiven.Whenweadjustthevolumetoalargerxed volume,thepressureadjuststoequilibriumatexactlythesamevaporpressure. Clearly,therearemoremoleculesinthevaporafterthevolumeisincreasedandequilibriumisreestablished,becausethevaporexertsthesamepressureinalargercontaineratthesametemperature.Also clearly,moreliquidmusthaveevaporatedtoachievethisequilibrium.Averyinterestingquestiontopose hereishowtheliquidrespondedtotheincreaseinvolume,whichpresumablyonlyaectedthespacein whichthegasmoleculesmove.Howdidtheliquid"know"toevaporatewhenthevolumewasincreased? Themoleculesintheliquidcouldnotdetecttheincreaseinvolumeforthegas,andthuscouldnotpossibly berespondingtothatincrease. Theonlyreasonableconclusionisthatthemoleculesintheliquidwerealwaysevaporating,evenbefore thevolumeofthecontainerwasincreased.Theremustbeaconstantmovementofmoleculesfromtheliquid phaseintothegasphase.Sincethepressureofthegasabovetheliquidremainsconstantwhenthevolume isconstant,thentheremustbeaconstantnumberofmoleculesinthegas.Ifevaporationisconstantly occurring,thencondensationmustalsobeoccurringconstantly,andmoleculesinthegasmustconstantly beenteringtheliquidphase.Sincethepressureremainsconstantinaxedvolume,thenthenumberof moleculesenteringthegasfromtheliquidmustbeexactlyosetbythenumberofmoleculesenteringthe liquidfromthegas. Atequilibrium,therefore,thepressureandtemperatureinsidethecontainerareunchanging,butthereis constantmovementofmoleculesbetweenthephases.Thisiscalled dynamicequilibrium .Thesituation is"equilibrium"inthattheobservablepropertiesoftheliquidandgasinthecontainerarenotchanging,but thesituationis"dynamic"inthatthereisconstantmovementofmoleculesbetweenphases.Thedynamic processesthattakeplaceoseteachotherexactly,sothatthepropertiesoftheliquidandgasdonotchange. Whathappenswhenweincreasethevolumeofthecontainertoalargerxedvolume?Weknowthat thepressureequilibratesatthesamevaporpressure,andthatthereforetherearemoremoleculesinthe vaporphase.Howdidtheygetthere?Itmustbethecasethatwhenthevolumeisincreased,evaporation initiallyoccursmorerapidlythancondensationuntilequilibriumisachieved.Therateofevaporationmust bedeterminedbythenumberofmoleculesintheliquidwhichhavesucientkineticenergytoescapethe intermolecularforcesintheliquid,andaccordingtothekineticmoleculartheory,thisnumberdependsonly onthetemperature,notonthevolume.However,therateofcondensationmustdependonthefrequencyof moleculesstrikingthesurfaceoftheliquid.AccordingtotheKineticMolecularTheory,thisfrequencymust decreasewhenthevolumeisincreased,becausethedensityofmoleculesinthegasdecreases.Therefore, therateofcondensationbecomessmallerthantherateofevaporationwhenthevolumeisincreased,and thereforethereisanetowofmoleculesfromliquidtogas.Thiscontinuesuntilthedensityofmolecules inthegasisrestoredtoitsoriginalvalue,atwhichpointtherateofevaporationismatchedbytherateof condensation.Atthispoint,thispressurestopsincreasingandisthesameasitwasbeforethevolumewas increased. 13.7ReviewandDiscussionQuestions Exercise13.1 InthephasediagramforwaterinFigure13.6VaporPressureofLiquidWater,startatthepoint wherethetemperatureis60 Candthepressureis400torr.Slowlyincreasethetemperature withconstantpressureuntilthetemperatureis100 C.Statewhathappensphysicallytothewater duringthisheatingprocess. Exercise13.2 InthephasediagramforwaterinFigure13.6VaporPressureofLiquidWater,startatthepoint

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133 wherethetemperatureis60 Candthepressureis400torr.Slowlylowerthepressureatconstant temperatureuntilthepressureis80torr.Statewhathappensphysicallytothewaterduringthis process. Exercise13.3 ExplainwhyFigure13.6VaporPressureofLiquidWaterisbothagraphoftheboilingpointof liquidwaterasafunctionofappliedpressureandagraphofthevaporpressureofliquidwateras afunctionoftemperature. Exercise13.4 Weobservethat,whentheappliedpressureislessthanthevaporpressureofaliquid,allofthe liquidwillspontaneouslyevaporate.Intermsofdynamicequilibrium,explainwhynoliquidcan bepresentundertheseconditions. Exercise13.5 UsingargumentsfromtheKineticMolecularTheoryandtheconceptofdynamicequilibrium, explainwhy,atagivenappliedpressure,therecanbeoneandonlyonetemperature,theboiling point,atwhichaspecicliquidanditsvaporcanbeinequilibrium. Exercise13.6 Usingdynamicequilibriumarguments,explainwhythevaporpressureofaliquidisindependent oftheamountofliquidpresent. Exercise13.7 Usingdynamicequilibriumarguments,explainwhythevaporpressureofaliquidisindependent ofthevolumeavailableforthevaporabovetheliquid. Exercise13.8 Usingdynamicequilibriumarguments,explainwhyasubstancewithweakerintermolecularforces hasagreatervaporpressurethanonewithstrongerintermolecularforces. Exercise13.9 AccordingtoFigure13.5VaporPressuresofVariousLiquidsthevaporpressureofphenolis muchlessthanthevaporpressureofdimethylether.Whichofthesesubstanceshasthegreater intermolecularattractions?Whichsubstancehasthehigherboilingpoint?Explainthedierence intheintermolecularattractionsintermsofmolecularstructure. Exercise13.10 Thetextdescribesdynamicequilibriumbetweenaliquidanditsvaporattheboilingpoint.Describe thedynamicequilibriumbetweenaliquidanditssolidatthemeltingpoint.Usingthisdescription, explainwhythemeltingpointofasolidvariesverylittleasthepressureincreases.

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134 CHAPTER13.PHASEEQUILIBRIUMANDINTERMOLECULAR INTERACTIONS

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Chapter14 ReactionEquilibriumintheGasPhase 1 14.1Foundation Inbeginningourstudyofthereactionsofgases,wewillassumeaknowledgeofthephysicalpropertiesof gasesasdescribedbythe IdealGasLaw andanunderstandingofthesepropertiesasgivenbythepostulates andconclusionsofthe KineticMolecularTheory .Weassumethatwehavedevelopedadynamicmodel ofphaseequilibriumintermsofcompetingrates.Wewillalsoassumeanunderstandingofthebonding, structure,andpropertiesofindividualmolecules. 14.2Goals Inperformingstoichiometriccalculations,weassumethatwecancalculatetheamountofproductofa reactionfromtheamountofthereactantswestartwith.Forexample,ifweburnmethanegas, CH 4 g ,in excessoxygen,thereaction CH 4 g +2 O 2 g CO 2 g +2 H 2 O g .1 occurs,andthenumberofmolesof CO 2 g producedisassumedtoequalthenumberofmolesof CH 4 g westartwith. Fromourstudyofphasetransitionswehavelearnedtheconceptofequilibrium.Weobservedthat,in thetransitionfromonephasetoanotherforasubstance,undercertainconditionsbothphasesarefound tocoexist,andwerefertothisasphaseequilibrium.Itshouldnotsurpriseusthatthesesameconcepts ofequilibriumapplytochemicalreactionsaswell.Inthereaction.1,therefore,weshouldexamine whetherthereactionactuallyproducesexactlyonemoleof CO 2 foreverymoleof CH 4 westartwithor whetherwewindupwithanequilibriummixturecontainingboth CO 2 and CH 4 .Wewillndthatdierent reactionsprovideuswithvaryinganswers.Inmanycases,virtuallyallreactantsareconsumed,producing thestoichiometricamountofproduct.However,inmanyothercases,substantialamountsofreactantare stillpresentwhenthereactionachievesequilibrium,andinothercases,almostnoproductisproducedat equilibrium.Ourgoalwillbetounderstand,describeandpredictthereactionequilibrium. Animportantcorollarytothisgoalistoattempttocontroltheequilibrium.Wewillndthatvarying theconditionsunderwhichthereactionoccurscanvarytheamountsofreactantsandproductspresentat equilibrium.Wewilldevelopageneralprincipleforpredictinghowthereactionconditionsaecttheamount ofproductproducedatequilibrium. 1 Thiscontentisavailableonlineat. 135

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136 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE 14.3Observation1:Reactionequilibrium Webeginbyanalyzingasignicantindustrialchemicalprocess,thesynthesisofammoniagas, NH 3 ,from nitrogenandhydrogen: N 2 g +3 H 2 g 2 NH 3 g .2 Ifwestartwith1moleof N 2 and3molesof H 2 ,thebalancedequationpredictsthatwewillproduce2moles of NH 3 .Infact,ifwecarryoutthisreactionstartingwiththesequantitiesofnitrogenandhydrogenat298K ina100.0Lreactionvessel,weobservethatthenumberofmolesof NH 3 producedis1.91mol.This"yield" islessthanpredictedbythebalancedequation,butthedierenceisnotduetoalimitingreagentfactor. Recallthat,instoichiometry,thelimitingreagentistheonethatispresentinlessthantheratioofmoles givenbythebalancedequation.Inthiscase,neither N 2 nor H 2 islimitingbecausetheyarepresentinitially ina1:3ratio,exactlymatchingthestoichiometry.Notealsothatthisseemingdecitintheyieldisnotdue toanyexperimentalerrororimperfection,norisitduetopoormeasurementsorpreparation.Rather,the observationthat,at298K,1.91molesratherthan2molesareproducediscompletelyreproducible:every measurementofthisreactionatthistemperatureinthisvolumestartingwith1moleof N 2 and3molesof H 2 givesthisresult.Weconcludethatthereaction.2achieves reactionequilibrium inwhichallthree gasesarepresentinthegasmixture.Wecandeterminetheamountsofeachgasatequilibriumfromthe stoichiometryofthereaction.When n NH 3 =1 : 91 mol arecreated,thenumberofmolesof N 2 remainingat equilibriumis n N 2 =0 : 045 mol and n H 2 =0 : 135 mol Itisimportanttonotethatwecanvarytherelativeamountof NH 3 producedbyvaryingthetemperature ofthereaction,thevolumeofthevesselinwhichthereactionoccurs,ortherelativestartingamountsof N 2 and H 2 .Weshallstudyandanalyzethisobservationindetailinlatersections.Fornow,though,we demonstratethattheconceptofreactionequilibriumisgeneraltoallreactions. Considerthereaction H 2 g + I 2 g 2 HI g .3 Ifwebeginwith1.00moleof H 2 and1.00moleof I 2 at500Kinareactionvesselofxedvolume,weobserve that,atequilibrium, n HI =1 : 72 mol ,leavingintheequilibriummixture n H 2 =0 : 14 mol and n I 2 =0 : 14 mol Similarly,considerthedecompositionreaction N 2 O 4 g 2 NO 2 g .4 At298Kina100.0Lreactionask,1.00molof N 2 O 4 partiallydecomposestoproduce,atequilibrium, n NO 2 =0 : 64 mol and n N 2 O 4 =0 : 68 mol Somechemicalreactionsachieveanequilibriumthatappearstobeverynearlycompletereaction.For example, H 2 g + Cl 2 g 2 HCl g .5 Ifwebeginwith1.00moleof H 2 and1.00moleof Cl 2 at298Kinareactionvesselofxedvolume,we observethat,atequilibrium, n HCl isalmostexactly2.00mol,leavingvirtuallyno H 2 or Cl 2 .Thisdoes notmeanthatthereactionhasnotcometoequilibrium.Itmeansinsteadthat,atequilibrium,thereare essentiallynoreactantsremaining. Ineachofthesecases,theamountsofreactantsandproductspresentatequilibriumvaryastheconditions arevariedbutarecompletelyreproducibleforxedconditions.Beforemakingfurtherobservationsthat willleadtoaquantitativedescriptionofthereactionequilibrium,weconsideraqualitativedescriptionof equilibrium. Webeginwithadynamicequilibriumdescription.Weknowfromourstudiesofphasetransitionsthat equilibriumoccurswhentherateoftheforwardprocess e.g. evaporationismatchedbytherateofreverse process e.g. condensation.Sincewehavenowobservedthatgasreactionsalsocometoequilibrium,we postulatethatatequilibriumtheforwardreactionrateisequaltothereversereactionrate.Forexample,

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137 inthereactionhere.4,therateofdecompositionof N 2 O 4 moleculesatequilibriummustbeexactly matchedbytherateofrecombinationor dimerization of NO 2 molecules. Toshowthattheforwardandreversereactionscontinuetohappenatequilibrium,westartwiththe NO 2 and N 2 O 4 mixtureatequilibriumandwevarythevolumeoftheaskcontainingthemixture.We observethat,ifweincreasethevolumeandthereactionisallowedtocometoequilibrium,theamountof NO 2 atequilibriumislargerattheexpenseofasmalleramountof N 2 O 4 .Wecancertainlyconcludethatthe amountsofthegasesatequilibriumdependonthereactionconditions.However,iftheforwardandreverse reactionsstoponcetheequilibriumamountsofmaterialareachieved,themoleculeswouldnot"know"that thevolumeofthecontainerhadincreased.Sincethereactionequilibriumcananddoesrespondtoachange involume,itmustbethatthechangeinvolumeaectstheratesofboththeforwardandreverseprocesses. Thismeansthatbothreactionsmustbeoccurringatequilibrium,andthattheirratesmustexactlymatch atequilibrium. Thisreasoningrevealsthattheamountsofreactantandproductpresentatequilibriumaredetermined bytheratesoftheforwardandreversereactions.Iftherateoftheforwardreaction e.g. decomposition of N 2 O 4 isfasterthantherateofthereversereaction,thenatequilibriumwehavemoreproductthan reactant.Ifthatdierenceinratesisverylarge,atequilibriumtherewillbemuchmoreproductthan reactant.Ofcourse,theconverseoftheseconclusionsisalsotrue.Itmustalsobethecasethattherates oftheseprocessesdependson,amongstotherfactors,thevolumeofthereactionask,sincetheamountsof eachgaspresentatequilibriumchangewhenthevolumeischanged. 14.4Observation2:Equilibriumconstants Itwasnotedabovethattheequilibriumpartialpressuresofthegasesinareactionvarydependingupon avarietyofconditions.Theseincludechangesintheinitialnumbersofmolesofreactantsandproducts, changesinthevolumeofthereactionask,andchangesinthetemperature.Wenowstudythesevariations quantitatively. Considerrstthereactionhere.4.Followingonourpreviousstudyofthisreaction,weinjectan initialamountof N 2 O 4 g intoa100Lreactionaskat298K.Now,however,wevarytheinitialnumberof molesof N 2 O 4 g intheaskandmeasuretheequilibriumpressuresofboththereactantandproductgases. TheresultsofanumberofsuchstudiesaregivenhereEquilibriumPartialPressuresinDecomposition Reaction,p.137. EquilibriumPartialPressuresinDecompositionReaction Initial n N 2 O 4 P N 2 O 4 atm P NO 2 atm 0.1 0.00764 0.033627 0.5 0.071011 0.102517 1 0.166136 0.156806 1.5 0.26735 0.198917 2 0.371791 0.234574 2.5 0.478315 0.266065 3 0.586327 0.294578 3.5 0.695472 0.320827 4 0.805517 0.345277 4.5 0.916297 0.368255 5 1.027695 0.389998

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138 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE Wemighthaveexpectedthattheamountof NO 2 producedatequilibriumwouldincreaseindirectproportiontoincreasesintheamountof N 2 O 4 webeginwith.EquilibriumPartialPressuresinDecomposition Reaction,p.137showsthatthisisnotthecase.Notethatwhenweincreasetheinitialamountof N 2 O 4 by afactorof10from0.5molesto5.0moles,thepressureof NO 2 atequilibriumincreasesbyafactorofless than4. Therelationshipbetweenthepressuresatequilibriumandtheinitialamountof N 2 O 4 isperhapsmore easilyseeninagraphofthedatainEquilibriumPartialPressuresinDecompositionReaction,p.137,as showninFigure14.1EquilibriumPartialPressuresinDecompositionReaction.Therearesomeinteresting featureshere.Notethat,whentheinitialamountof N 2 O 4 islessthan1mol,theequilibriumpressureof NO 2 isgreaterthanthatof N 2 O 4 .Theserelativepressuresreverseastheinitialamountincreases,asthe N 2 O 4 equilibriumpressurekeepstrackwiththeinitialamountbutthe NO 2 pressurefallsshort.Clearly,the equilibriumpressureof NO 2 doesnotincreaseproportionallywiththeinitialamountof N 2 O 4 .Infact,the increaseisslowerthanproportionality,suggestingperhapsasquarerootrelationshipbetweenthepressure of NO 2 andtheinitialamountof N 2 O 4 EquilibriumPartialPressuresinDecompositionReaction Figure14.1 WetestthisinFigure14.2RelationshipofPressureofProducttoInitialAmountofReactantbyplotting P NO 2 atequilibriumversusthesquarerootoftheinitialnumberofmolesof N 2 O 4 .Figure14.2Relationship

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139 ofPressureofProducttoInitialAmountofReactantmakesitclearthatthisisnotasimpleproportional relationship,butitiscloser.NoteinFigure14.1EquilibriumPartialPressuresinDecompositionReaction thattheequilibriumpressure P N 2 O 4 increasesclosetoproportionallywiththeinitialamountof N 2 O 4 .This suggestsplotting P NO 2 versusthesquarerootof P N 2 O 4 .ThisisdoneinFigure14.3EquilibriumPartial Pressures,wherewediscoverthatthereisaverysimpleproportionalrelationshipbetweenthevariables plottedinthisway.Wehavethusobservedthat P NO 2 = c p P N 2 O 4 .6 where c istheslopeofthegraph..6canberewritteninastandardform K p = P NO 2 2 P N 2 O 4 .7 Totesttheaccuracyofthisequationandtondthevalueof K p ,wereturntoEquilibriumPartialPressures inDecompositionReaction,p.137andaddanothercolumninwhichwecalculatethevalueof K p for eachofthedatapoints.EquilibriumPartialPressuresinDecompositionReaction,p.140makesitclear thatthe"constant"in.7trulyisindependentofboththeinitialconditionsandtheequilibriumpartial pressureofeitheroneofthereactantorproduct.Wethusrefertotheconstant K p in.7asthe reaction equilibriumconstant RelationshipofPressureofProducttoInitialAmountofReactant Figure14.2

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140 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE EquilibriumPartialPressures Figure14.3 EquilibriumPartialPressuresinDecompositionReaction Initial n N 2 O 4 P N 2 O 4 atm P NO 2 atm K p 0.1 0.00764 0.0336 0.148 0.5 0.0710 0.102 0.148 1 0.166 0.156 0.148 1.5 0.267 0.198 0.148 2 0.371 0.234 0.148 2.5 0.478 0.266 0.148 3 0.586 0.294 0.148 3.5 0.695 0.320 0.148 4 0.805 0.345 0.148 4.5 0.916 0.368 0.148 5 1.027 0.389 0.148 Itisveryinterestingtonotethefunctionalformoftheequilibriumconstant.Theproduct NO 2 pressure appearsinthenumerator,andtheexponent2onthepressureisthestoichiometriccoecienton NO 2 in thebalancedchemicalequation.Thereactant N 2 O 4 pressureappearsinthedenominator,andtheexponent 1onthepressureisthestoichiometriccoecienton N 2 O 4 inthechemicalequation. Wenowinvestigatewhetherotherreactionshaveequilibriumconstantsandwhethertheformofthis equilibriumconstantisahappycoincidenceorageneralobservation.Wereturntothereactionforthe synthesisofammonia.2.

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141 InaprevioussectionSection14.2:Goals,weconsideredonlytheequilibriumproducedwhen1mole of N 2 isreactedwith3molesof H 2 .Wenowconsiderarangeofpossibleinitialvaluesoftheseamounts, withtheresultantequilibriumpartialpressuresgiveninEquilibriumPartialPressuresoftheSynthesisof Ammonia,p.141.Inaddition,anticipatingthepossibilityofanequilibriumconstant,wehavecalculated theratioofpartialpressuresgivenby: K p = P NH 3 2 P N 2 P H 2 3 .8 InEquilibriumPartialPressuresoftheSynthesisofAmmonia,p.141,theequilibriumpartialpressures ofthegasesareinaverywidevariety,includingwhetherthenalpressuresaregreaterforreactantsor products.However,fromthedatainEquilibriumPartialPressuresoftheSynthesisofAmmonia,p.141,it isclearthat,despitethesevariations, K p in.8isessentiallyaconstantforalloftheinitialconditions examinedandisthusthe reactionequilibriumconstant forthisreaction.2. EquilibriumPartialPressuresoftheSynthesisofAmmonia VL n N 2 n H 2 P N 2 P H 2 P NH 3 K p 10 1 3 0.0342 0.1027 4.82 6 : 2 10 5 10 0.1 0.3 0.0107 0.0322 0.467 6 : 0 10 5 100 0.1 0.3 0.00323 0.00968 0.0425 6 : 1 10 5 100 3 3 0.492 0.00880 0.483 6 : 1 10 5 100 1 3 0.0107 0.0322 0.467 6 : 0 10 5 1000 1.5 1.5 0.0255 0.00315 0.0223 6 : 2 10 5 Studiesofmanychemicalreactionsofgasesresultinthesameobservations.Eachreactionequilibriumcan bedescribedbyanequilibriumconstantinwhichthepartialpressuresoftheproducts,eachraisedtotheir correspondingstoichiometriccoecient,aremultipliedtogetherinthenumerator,andthepartialpressures ofthereactants,eachraisedtotheircorrespondingstoichiometriccoecient,aremultipliedtogetherinthe denominator.Forhistoricalreasons,thisgeneralobservationissometimesreferredtoasthe LawofMass Action 14.5Observation3:TemperatureDependenceoftheReactionEquilibrium Wehavepreviouslyobservedthatphaseequilibrium,andinparticularvaporpressure,dependonthetemperature,butwehavenotyetstudiedthevariationofreactionequilibriumwithtemperature.Wefocus ourinitialstudyonthisreaction.3andwemeasuretheequilibriumpartialpressuresatavarietyof temperatures.Fromthesemeasurements,wecancompilethedatashowingthetemperaturedependenceof theequilibriumconstant K p forthisreactioninEquilibriumConstantfortheSynthesisofHI,p.141.

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142 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE EquilibriumConstantfortheSynthesisofHI TK K p 500 6 : 25 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 550 8 : 81 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 650 1 : 49 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 700 1 : 84 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 720 1 : 98 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 Notethattheequilibriumconstantincreasesdramaticallywithtemperature.Asaresult,atequilibrium, thepressureof HI mustalsoincreasedramaticallyasthetemperatureisincreased. Thesedatadonotseemtohaveasimplerelationshipbetween K p andtemperature.Wemustappeal toargumentsbasedonThermodynamics,fromwhichitispossibletoshowthattheequilibriumconstant shouldvarywithtemperatureaccordingtothefollowingequation: lnK p = )]TJ/F1 9.9626 Tf 9.409 17.037 Td [( H RT + S R .9 If H and S donotdependstronglyonthetemperature,thenthisequationwouldpredictasimple straightlinerelationshipbetween lnK p and 1 T .Inaddition,theslopeofthislineshouldbe )]TJ/F1 9.9626 Tf 9.409 11.059 Td [( H R .We testthispossibilitywiththegraphinFigure14.4InverseofTemperaturevs.NaturalLogofEquilibrium Constant.

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143 InverseofTemperaturevs.NaturalLogofEquilibriumConstant Figure14.4 Infact,wedoobserveastraightlinethroughthedata.Inthiscase,thelinehasanegativeslope.Note carefullythatthismeansthat K p is increasing withtemperature.Thenegativeslopevia.9meansthat )]TJ/F1 9.9626 Tf 9.409 11.059 Td [( H R mustbenegative,andindeedforthisreaction.3inthistemperaturerange, H =15 : 6 kJ mol ThisvaluematcheswellwiththeslopeofthelineinFigure14.4InverseofTemperaturevs.NaturalLogof EquilibriumConstant. Giventhevalidityof.9indescribingthetemperaturedependenceoftheequilibriumconstant,we canalsopredictthatanexothermicreactionwith H < 0 shouldhaveapositiveslopeinthegraph of lnK p versus 1 T ,andthustheequilibriumconstantshould decrease withincreasingtemperature.A goodexampleofanexothermicreactionisthesynthesisofammonia.2forwhich H = )]TJ/F8 9.9626 Tf 7.749 0 Td [(99 : 2 kJ mol EquilibriumconstantdataaregiveninEquilibriumConstantfortheSynthesisofAmmonia,p.143.Note that,aspredicted,theequilibriumconstantforthisexothermicreactiondecreasesrapidlywithincreasing temperature.ThedatafromEquilibriumConstantfortheSynthesisofAmmonia,p.143isshownin Figure14.5InverseofTemperaturevs.NaturalLogofEquilibriumConstant,clearlyshowingthecontrast betweentheendothermicreactionandtheexothermicreaction.Theslopeofthegraphispositiveforthe exothermicreactionandnegativefortheendothermicreaction.From.9,thisisageneralresultforall reactions.

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144 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE EquilibriumConstantfortheSynthesisofAmmonia TK K p 250 7 10 8 298 6 10 5 350 2 10 3 400 36 InverseofTemperaturevs.NaturalLogofEquilibriumConstant Figure14.5 14.6Observation4:ChangesinEquilibriumandLeChtelier'sPrinciple Oneofourgoalsattheoutsetwastodeterminewhetheritispossibletocontroltheequilibriumwhichoccurs duringagasreaction.Wemightwanttoforceareactiontoproduceasmuchoftheproductsaspossible.In thealternative,ifthereareunwantedby-productsofareaction,wemightwantconditionswhichminimize

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145 theproduct.Wehaveobservedthattheamountofproductvarieswiththequantitiesofinitialmaterials andwithchangesinthetemperature.Ourgoalisasystematicunderstandingofthesevariations. AlookbackatEquilibriumPartialPressuresinDecompositionReaction,p.137andEquilibriumPartial PressuresinDecompositionReaction,p.140showsthattheequilibriumpressureoftheproductofthe reactionincreaseswithincreasingtheinitialquantityofreaction.Thisseemsquiteintuitive.Lessintuitive isthevariationoftheequilibriumpressureoftheproductofthisreaction.2withvariationinthevolume ofthecontainer,asshowninEquilibriumPartialPressuresoftheSynthesisofAmmonia,p.141.Notethat thepressureof NH 3 decreasesbymorethanafactoroftenwhenthevolumeisincreasedbyafactorof ten.Thismeansthat,atequilibrium,therearefewermolesof NH 3 producedwhenthereactionoccursina largervolume. Tounderstandthiseect,werewritetheequilibriumconstantin.8toexplicitshowthevolumeof thecontainer.Thisisdonebyapplying Dalton'sLawofPartialPressures ,sothateachpartialpressure isgivenbytheIdealGasLaw: K p = n NH 3 2 RT V 2 n N 2 RT V n H 2 3 RT V 3 = n NH 3 2 n N 2 n H 2 3 RT V 2 .10 Therefore, K p RT V 2 = n NH 3 2 n N 2 n H 2 3 .11 Thisformoftheequationmakesitclearthat,whenthevolumeincreases,theleftsideoftheequation decreases.Thismeansthattherightsideoftheequationmustdecreasealso,andinturn, n NH 3 must decreasewhile n N 2 and n H 2 mustincrease.Theequilibriumisthusshiftedfromproductstoreactantswhen thevolumeincreasesforthisreaction.2. Theeectofchangingthevolumemustbeconsideredforeachspecicreaction,becausetheeectdepends onthestoichiometryofthereaction.Onewaytodeterminetheconsequenceofachangeinvolumeisto rewritetheequilibriumconstantaswehavedonein.11. Finally,weconsiderchangesintemperature.Wenotethat K p increaseswith T forendothermicreactions anddecreaseswith T forexothermicreactions.Assuch,theproductsareincreasinglyfavoredwithincreasing temperaturewhenthereactionisendothermic,andthereactantsareincreasinglyfavoredwithincreasing temperaturewhenthereactionisexothermic.Onreection,wenotethatwhenthereactionisexothermic,the reversereactionisendothermic.Puttingthesestatementstogether,wecansaythatthereactionequilibrium alwaysshiftsinthedirectionoftheendothermicreactionwhenthetemperatureisincreased. Alloftheseobservationscanbecollectedintoasingleunifyingconceptknownas LeChtelier's Principle Principle14.1: LeChtelier'sPrinciple Whenareactionatequilibriumisstressedbyachangeinconditions,theequilibriumwillbe reestablishedinsuchawayastocounterthestress. Thisstatementisbestunderstoodbyreectiononthetypesof"stresses"wehaveconsideredinthis section.Whenareactantisaddedtoasystematequilibrium,thereactionrespondsbyconsumingsome ofthataddedreactantasitestablishesanewequilibrium.Thisosetssomeofthestressoftheincrease inreactant.Whenthetemperatureisraisedforareactionatequilibrium,thisaddsthermalenergy.The systemshiftstheequilibriumintheendothermicdirection,thusabsorbingsomeoftheaddedthermalenergy, counteringthestress. Themostchallengingofthethreetypesofstressconsideredinthissectionisthechangeinvolume.By increasingthevolumecontainingagasphasereactionatequilibrium,wereducethepartialpressuresofall gasespresentandthusreducethetotalpressure.Recallthattheresponseofthisreaction.2tothe volumeincreasewastocreatemoreofthereactantsattheexpenseoftheproducts.Oneconsequenceofthis shiftisthatmoregasmoleculesarecreated,andthisincreasesthetotalpressureinthereactionask.Thus,

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146 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE thereactionrespondstothestressofthevolumeincreasebypartiallyosettingthepressuredecreasewith anincreaseinthenumberofmolesofgasatequilibrium. LeChtelier'sprincipleisausefulmnemonicforpredictinghowwemightincreaseordecreasetheamount ofproductatequilibriumbychangingtheconditionsofthereaction.Fromthisprinciple,wecanpredict whetherthereactionshouldoccurathightemperatureorlowtemperature,andwhetheritshouldoccurat highpressureorlowpressure. 14.7ReviewandDiscussionQuestions Exercise14.1 Inthedatagivenforequilibriumofthisreaction.3,thereisnovolumegiven.Showthat changingthevolumeforthereactiondoesnotchangethenumberofmolesofreactantsandproducts presentatequilibrium, i.e. changingthevolumedoesnotshifttheequilibrium. Exercise14.2 Forthisreaction.4thenumberofmolesof NO 2 atequilibriumincreasesifweincreasethe volumeinwhichthereactioniscontained.Explainwhythismustbetrueintermsofdynamic equilibrium,giveareasonwhytheratesoftheforwardandreversereactionsmightbeaected dierentlybychangesinthevolume. Exercise14.3 Wecouldbalance.2bywriting 2 N 2 g +6 H 2 g 4 NH 3 g .12 Writetheformoftheequilibriumconstantforthereactionbalancedasin.12.Whatisthe valueoftheequilibriumconstant?RefertoEquilibriumPartialPressuresoftheSynthesisof Ammonia,p.141.Ofcourse,thepressuresatequilibriumdonotdependonwhetherthereaction isbalancedasin.2orasin.12.Explainwhythisistrue,eventhoughtheequilibrium constantcanbewrittendierentlyandhaveadierentvalue. Exercise14.4 Showthattheequilibriumconstant K p in.8forthisreaction.2canbewrittenintermsof theconcentrationsorparticledensities, e.g. [ N 2 ]= n N 2 V ,insteadofthepartialpressures.Inthis form,wecalltheequilibriumconstant K c .Findtherelationshipbetween K p and K c ,andcalculate thevalueof K c Exercise14.5 Foreachofthesereactions,predictwhetherincreasesintemperaturewillshiftthereactionequilibriummoretowardsproductsormoretowardsreactants. 2 CO g + O 2 g 2 CO 2 g O 3 g + NO g NO 2 g + O 2 g 2 O 3 g 3 O 2 g Exercise14.6 PlotthedatainEquilibriumConstantfortheSynthesisofHI,p.141onagraphshowing K p on they-axisand T onthex-axis.Theshapeofthisgraphisreminiscentofthegraphofanother physicalpropertyasafunctionofincreasingtemperature.Identifythatproperty,andsuggesta reasonwhytheshapesofthegraphsmightbesimilar. Exercise14.7 UsingLeChtelier'sprinciple,predictwhetherthespecied"stress"willproduceanincreaseora decreaseintheamountofproductobservedatequilibriumforthereaction: 2 H 2 g + CO g CH 3 OH g .13

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147 H = )]TJ/F8 9.9626 Tf 7.749 0 Td [(91 kJ mol Volumeofcontainerisincreased. Heliumisaddedtocontainer. Temperatureofcontainerisraised. Hydrogenisaddedtocontainer. CH 3 OH isextractedfromcontainerasitisformed.

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148 CHAPTER14.REACTIONEQUILIBRIUMINTHEGASPHASE

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Chapter15 Acid-BaseEquilibrium 1 15.1Foundation Wehavedevelopedanunderstandingof equilibrium involvingphasetransitionsandinvolvingreactions entirelyinthegasphase.Wewillassumeanunderstandingoftheprinciplesofdynamicequilibrium,reaction equilibriumconstants,and LeChtelier'sPrinciple .Tounderstandapplicationoftheseprinciplesto reactionsinsolution,wewillnowassumeadenitionofcertainclassesofsubstancesasbeingeitheracidsor bases.Anacidisasubstancewhosemoleculesdonatepositivehydrogenionsprotonstoothermolecules orions.Whendissolvedinpurewater,acidmoleculeswilltransferahydrogeniontoawatermolecule ortoaclusterofseveralwatermolecules.Thisincreasestheconcentrationof H + ionsinthesolution. Abaseisasubstancewhosemoleculesaccepthydrogenionsfromothermolecules.Whendissolvedin purewater,basemoleculeswillacceptahydrogenionfromawatermolecule,leavingbehindanincreased concentrationof OH )]TJ/F15 9.9626 Tf 11.126 -3.615 Td [(ionsinthesolution.Tounderstandwhatdeterminesacid-basebehavior,wewill assumeanunderstandingofthebonding,structure,andpropertiesofindividualmolecules. 15.2Goals Acidsandbasesareverycommonsubstanceswhosepropertiesvarygreatly.Manyacidsareknowntobe quitecorrosive,withtheabilitytodissolvesolidmetalsorburnesh.Manyotheracids,however,arenotonly benignbutvitaltotheprocessesoflife.Farfromdestroyingbiologicalmolecules,theycarryoutreactions criticalfororganisms.Similarly,manybasesarecausticcleanserswhilemanyothersaremedicationstocalm indigestionpains. Inthisconceptstudy,wewilldevelopanunderstandingofthecharacteristicsofmoleculeswhichmake themeitheracidsorbases.Wewillexaminemeasurementsabouttherelativestrengthsofacidsandbases, andwewillusethesetodevelopaquantitativeunderstandingoftherelativestrengthsofacidsandbases. Fromthis,wecandevelopaqualitativeunderstandingofthepropertiesofmoleculeswhichdeterminewhether amoleculeisastrongacidoraweakacid,astrongbaseoraweakbase.Thisunderstandingisvaluablein predictingtheoutcomesofreactions,basedontherelativequantitativestrengthsofacidsandbases.These reactionsarecommonlyreferredtoasneutralizationreactions.Asurprisinglylargenumberofreactions, particularlyinorganicchemistry,canbeunderstoodastransferofhydrogenionsfromacidmoleculestobase molecules. 1 Thiscontentisavailableonlineat. 149

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150 CHAPTER15.ACID-BASEEQUILIBRIUM 15.3Observation1:StrongAcidsandWeakAcids FromthedenitionofanacidgivenintheFoundation,atypicalacidcanbewrittenas HA ,representing thehydrogenionwhichwillbedonatedandtherestofthemoleculewhichwillremainasanegativeion afterthedonation.Thetypicalreactionofanacidinaqueoussolutionreactingwithwatercanbewrittenas HA aq + H 2 O l H 3 O + aq + A )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( aq .1 Inthisreaction, HA aq representsanacidmoleculedissolvedinaqueoussolution. H 3 O + aq isanotation toindicatethatthedonatedprotonhasbeendissolvedinsolution.Observationsindicatethattheprotonis associatedwithseveralwatermoleculesinacluster,ratherthanattachedtoasinglemolecule. H 3 O + isa simpliednotationtorepresentthisresult.Similarly,the A )]TJ/F8 9.9626 Tf 8.385 -3.615 Td [( aq ionissolvatedbyseveralwatermolecules. .1isreferredtoas acidionization .1impliesthata0.1Msolutionoftheacid HA inwatershouldproduce H 3 O + ionsinsolutionwith aconcentrationof0.1M.Infact,theconcentrationof H 3 O + ions, [ H 3 O + ] ,canbemeasuredbyavarietyof techniques.Chemistscommonlyuseameasureofthe H 3 O + ionconcentrationcalledthe pH ,denedby: pH = )]TJ/F1 9.9626 Tf 9.409 8.07 Td [()]TJ/F11 9.9626 Tf 4.566 -8.07 Td [(log H 3 O + Wenowobservetheconcentration [ H 3 O + ] producedbydissolvingavarietyofacidsinsolutionata concentrationof0.1M,andtheresultsaretabulatedinH3O+pHfor0.1MAcidSolutions,p.150. H3O+pHfor0.1MAcidSolutions Acid [ H 3 O + ] M pH H 2 SO 4 0.1 1 HNO 3 0.1 1 HCl 0.1 1 HBr 0.1 1 HI 0.1 1 HClO 4 0.1 1 HClO 3 0.1 1 HNO 2 6 : 2 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.2 HCN 7 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 5.1 HIO 1 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 5.8 HF 5 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.3 HOCN 5 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.3 HClO 2 2 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 1.6 CH 3 COOH aceticacid 1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.9 CH 3 CH 2 COOH propionicacid 1 : 1 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.9 Notethatthereareseveralacidslistedforwhich [ H 3 O + ]=0 : 1 M ,and pH =1 .Thisshowsthat,forthese acids,theacidionizationiscomplete:essentiallyeveryacidmoleculeisionizedinthesolutionaccording to.1.However,thereareotheracidslistedforwhich [ H 3 O + ] isconsiderablylessthan0.1Mandthe pHisconsiderablygreaterthan1.Foreachoftheseacids,therefore,notalloftheacidmoleculesionize accordingto.1.Infact,itisclearinH3O+pHfor0.1MAcidSolutions,p.150thatintheseacidsthe vastmajorityoftheacidmoleculesdonotionize,andonlyasmallpercentagedoesionize.

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151 Fromtheseobservations,wedistinguishtwoclassesofacids: strongacids and weakacids .Strong acidsarethoseforwhichnearly100%oftheacidmoleculesionize,whereasweakacidsarethoseforwhich onlyasmallpercentageofmoleculesionize.TherearesevenstrongacidslistedinH3O+pHfor0.1M AcidSolutions,p.150.Frommanyobservations,itispossibletodeterminethatthesesevenacidsarethe onlycommonlyobservedstrongacids.Thevastmajorityofallsubstanceswithacidicpropertiesareweak acids.Weseektocharacterizeweakacidionizationquantitativelyandtodeterminewhatthedierencesin molecularpropertiesarebetweenstrongacidsandweakacids. 15.4Observation2:PercentIonizationinWeakAcids H3O+pHfor0.1MAcidSolutions,p.150showsthatthepHof0.1Macidsolutionsvariesfromoneweak acidtoanother.Ifwedissolve0.1molesofacidina1.0Lsolution,thefractionofthoseacidmolecules whichwillionizevariesfromweakacidtoweakacid.Forafewweakacids,usingthedatainH3O+pHfor 0.1MAcidSolutions,p.150wecalculatethepercentageofionizedacidmoleculesin0.1Macidsolutionsin PercentIonizationof0.1MAcidSolutions,p.151. PercentIonizationof0.1MAcidSolutions Acid [ H 3 O + ] M %ionization HNO 2 6 : 2 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 6.2% HCN 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.007% HIO 1 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.001% HF 5 : 5 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 5.5% HOCN 5 : 5 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 5.5% HClO 2 2 : 8 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 28.2% CH 3 COOH aceticacid 1 : 3 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 1.3% CH 3 CH 2 COOH propionicacid 1 : 1 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 1.1% WemightbetemptedtoconcludefromPercentIonizationof0.1MAcidSolutions,p.151thatwecan characterizethestrengthofeachacidbythepercentionizationofacidmoleculesinsolution.However,before doingso,weobservethepHofasingleacid,nitrousacid,insolutionasafunctionoftheconcentrationof theacid. HNO 2 aq + H 2 O l H 3 O + aq + NO )]TJ/F7 6.9738 Tf -0.276 -6.918 Td [(2 aq .2 Inthiscase,"concentrationoftheacid"referstothenumberofmolesofacidthatwedissolvedperliter ofwater.Ourobservationsarelistedin%IonizationofNitrousAcid,p.151,whichgives [ H 3 O + ] ,pH,and percentionizationasafunctionofnitrousacidconcentration.

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152 CHAPTER15.ACID-BASEEQUILIBRIUM %IonizationofNitrousAcid c 0 M [ H 3 O + ] pH %Ionization 0.50 1 : 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 1.8 3.3% 0.20 1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 2.0 5.1% 0.10 7 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.2 7.0% 0.050 4 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.3 9.7% 0.020 2 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.5 14.7% 0.010 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.7 20.0% 0.005 1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2.9 26.7% 0.001 4 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 3.3 49.1% 0.0005 3 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 3.5 60.8% Surprisingly,perhaps,thepercentionizationvariesconsiderablyasafunctionoftheconcentrationof thenitrousacid.Werecallthatthismeansthatthefractionofmoleculeswhichionize,accordingto.2, dependsonhowmanyacidmoleculesthereareperliterofsolution.Sincesomebutnotalloftheacid moleculesareionized,thismeansthatnitrousacidmoleculesarepresentinsolutionatthesametimeasthe negativenitriteionsandthepositivehydrogenions.Recallingourobservationofequilibriumingasphase reactions,wecanconcludethat.2achievesequilibriumforeachconcentrationofthenitrousacid. Sinceweknowthatgasphasereactionscometoequilibriumunderconditionsdeterminedbytheequilibriumconstant,wemightspeculatethatthesameistrueofreactionsinaqueoussolution,includingacid ionization.Wethereforedeneananalogytothegasphasereactionequilibriumconstant.Inthiscase, wewouldnotbeinterestedinthepressuresofthecomponents,sincethereactantsandproductsareallin solution.Instead,wetryafunctioncomposedoftheequilibriumconcentrations: K = [ H 3 O + ] NO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(2 [ HNO 2 ][ H 2 O ] .3 Theconcentrationsatequilibriumcanbecalculatedfromthedatain%IonizationofNitrousAcid,p.151for nitrousacid. [ H 3 O + ] islistedand NO )]TJ/F7 6.9738 Tf -0.276 -6.918 Td [(2 =[ H 3 O + ] .Furthermore,if c 0 istheinitialconcentrationoftheacid denedbythenumberofmolesofaciddissolvedinsolutionperliterofsolution,then [ HA ]= c 0 )]TJ/F8 9.9626 Tf 9.826 0 Td [([ H 3 O + ] Notethatthecontributionof [ H 2 O l ] tothevalueofthefunction K issimplyaconstant.Thisisbecause the"concentration"ofwaterinthesolutionissimplythemolardensityofwater, n H 2 O V =55 : 5 M ,whichis notaectedbythepresenceorabsenceofsolute.Alloftherelevantconcentrations,alongwiththefunction in.3arecalculatedandtabulatedinEquilibriumConcentrationsandKforNitrousAcid,p.152.

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153 EquilibriumConcentrationsandKforNitrousAcid c 0 M [ H 3 O + ] NO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(2 [ HNO 2 ] K 0.50 1 : 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 1 : 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 0.48 1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 0.20 1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 0.19 9 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.10 7 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 7 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 9 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 9 : 6 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.050 4 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 4 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 4 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 9 : 4 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.020 2 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 4 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 9 : 4 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.010 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 8 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 8 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.005 1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 3 : 6 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 8 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.001 4 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 4 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 5 : 1 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 8 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 0.0005 3 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 3 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 8 : 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(6 Wenotethatthefunction K in.3isapproximately,thoughonlyapproximately,thesameforall conditionsanalyzedin%IonizationofNitrousAcid,p.151.Variationoftheconcentrationbyafactor of1000producesachangein K ofonly10%to15%.Hence,wecanregardthefunction K asaconstant whichapproximatelydescribestheacidionizationequilibriumfornitrousacid.Byconvention,chemists omittheconstantconcentrationofwaterfromtheequilibriumexpression,resultinginthe acidionization equilibriumconstant K a ,denedas: K a = [ H 3 O + ] NO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(2 [ HNO 2 ] .4 FromanaverageofthedatainEquilibriumConcentrationsandKforNitrousAcid,p.152,wecancalculate that,at25 Cfornitrousacid, K a =5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 .AcidionizationconstantsfortheotherweakacidsinPercent Ionizationof0.1MAcidSolutions,p.151arelistedinWeakAcidIonizationConstants,KaandpKa,p.153. WeakAcidIonizationConstants,KaandpKa Acid K a p K a HNO 2 5 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 3.3 HCN 4 : 9 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(10 9.3 HIO 2 : 3 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(11 10.6 HF 3 : 5 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 3.4 HOCN 3 : 5 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 3.4 HClO 2 1 : 1 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 2.0 CH 3 COOH aceticacid 1 : 7 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 4.8 CH 3 CH 2 COOH propionicacid 1 : 4 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 4.9 WemaketwonalnotesabouttheresultsinWeakAcidIonizationConstants,KaandpKa,p.153. First,itisclearthelargerthevalueof K a ,thestrongertheacid.Thatis,when K a isalargernumber, thepercentionizationoftheacidislarger,andviceversa.Second,thevaluesof K a veryovermanyorders ofmagnitude.Assuch,itisoftenconvenienttodenethequanity pK a ,analogoustopH,forpurposesof comparingacidstrengths: pK a = )]TJ/F8 9.9626 Tf 9.41 0 Td [( logK a .5

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154 CHAPTER15.ACID-BASEEQUILIBRIUM Thevalueof pK a foreachacidisalsolistedinWeakAcidIonizationConstants,KaandpKa,p.153.Note thatasmallvalueof pK a impliesalargevalueof K a andthusastrongeracid.Weakeracidshavelarger valuesof pK a K a and pK a thusgiveasimplequantitativecomparisonofthestrengthofweakacids. 15.5Observation3:AutoionizationofWater SincewehavetheabilitytomeasurepHforacidsolutions,wecanmeasurepHforpurewateraswell. Itmightseemthatthiswouldmakenosense,aswewouldexpect [ H 3 O + ] toequalzeroexactlyinpure water.Surprisingly,thisisincorrect:ameasurementonpurewaterat25 Cyields pH =7 ,sothat [ H 3 O + ]=1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(7 M .Therecanbeonlyonepossiblesourcefortheseions:watermolecules.Theprocess H 2 O l + H 2 O l H 3 O + aq + OH )]TJ/F8 9.9626 Tf 8.385 -4.114 Td [( aq .6 isreferredtoasthe autoionization ofwater.Notethat,inthisreaction,somewatermoleculesbehaveas acid,donatingprotons,whileotheracidmoleculesbehaveasbase,acceptingprotons. Sinceatequilibrium [ H 3 O + ]=1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(7 M ,itmustalsobetruethat [ OH )]TJ/F8 9.9626 Tf 6.725 -3.616 Td [(]=1 : 0 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(7 M .Wecan writetheequilibriumconstantfor.6,followingourpreviousconventionofomittingthepurewaterfrom theexpression,andwendthat,at25 C, K w =[ H 3 O + ][ OH )]TJ/F8 9.9626 Tf 6.725 -3.616 Td [(] =1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(14 M .7 Inthiscase,thesubscript"w"refersto"water". .6occursinpurewaterbutmustalsooccurwhenionsaredissolvedinaqueoussolutions.This includesthepresenceofacidsionizedinsolution.Forexample,weconsiderasolutionof0.1Maceticacid. Measurementsshowthat,inthissolution [ H 3 O + ]=1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 M and [ OH )]TJ/F8 9.9626 Tf 6.725 -3.615 Td [(]=7 : 7 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(12 M .Wenotetwo thingsfromthisobservation:rst,thevalueof [ OH )]TJ/F8 9.9626 Tf 6.725 -3.616 Td [(] isconsiderablylessthaninpurewater;second,the autoionizationequilibriumconstantremainsthesameat 1 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(14 .Fromthesenotes,wecanconcludethat theautoionizationequilibriumofwateroccursinacidsolution,buttheextentofautoionizationissuppressed bythepresenceoftheacidinsolution. Weconsideranalnoteontheautoionizationofwater.ThepHofpurewateris7at25 C.Addingany acidtopurewater,nomatterhowweaktheacid,mustincrease [ H 3 O + ] ,thusproducingapHbelow7.As such,wecanconcludethat,forallacidsolutions,pHislessthan7,orontheotherhand,anysolutionwith pHlessthan7isacidic. 15.6Observation4:BaseIonization,NeutralizationandHydrolysis ofSalts Wehavenotyetexaminedthebehaviorofbasemoleculesinsolution,norhavewecomparedtherelative strengthsofbases.Wehavedenedabasemoleculeasonewhichacceptsapositivehydrogenionfrom anothermolecule.Oneofthemostcommonexamplesisammonia, NH 3 .Whenammoniaisdissolvedin aqueoussolution,thefollowingreactionoccurs: NH 3 aq + H 2 O l NH + 4 aq + OH )]TJ/F8 9.9626 Tf 8.386 -4.113 Td [( aq .8 DuetothelonepairofelectronsonthehighlyelectronegativeNatom, NH 3 moleculeswillreadilyattach afreehydrogenionformingtheammoniumion NH + 4 .Whenwemeasuretheconcentrationof OH )]TJ/F15 9.9626 Tf 10.82 -3.615 Td [(for variousinitialconcentrationof NH 3 inwater,weobservetheresultsinEquilibriumConcentrationsand KbforAmmonia,p.155.Weshouldanticipatethatabaseionizationequilibriumconstantmightexist

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155 comparabletotheacidionizationequilibriumconstant,andinEquilibriumConcentrationsandKbfor Ammonia,p.155,wehavealsocalculatedthevalueofthefunction K b denedas: K b = NH + 4 [ OH )]TJ/F8 9.9626 Tf 6.724 -3.615 Td [(] [ NH 3 ] .9 EquilibriumConcentrationsandKbforAmmonia c 0 M [ OH )]TJ/F8 9.9626 Tf 6.725 -3.616 Td [(] K b pH 0.50 3 : 2 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 11.5 0.20 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 11.3 0.10 1 : 4 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 11.1 0.050 9 : 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 1 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 11.0 0.020 6 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 1 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 10.8 0.010 4 : 2 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 1 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 10.6 0.005 3 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 1 : 9 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 10.5 0.001 1 : 3 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 1 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 10.1 0.0005 8 : 7 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 1 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(5 9.9 Giventhatwehavedissolvedabaseinpurewater,wemightbesurprisedtodiscoverthepresenceof positivehydrogenions, H 3 O + ,insolution,butameasurementofthepHforeachofthesolutionsreveals smallamounts.ThepHforeachsolutionisalsolistedinEquilibriumConcentrationsandKbforAmmonia, p.155.Thesourceofthese H 3 O + ionsmustbetheautoionizationofwater.Note,however,thatineach caseinbasicsolution,theconcentrationof H 3 O + ionsislessthanthatinpurewater.Hence,thepresence ofthebaseinsolutionhassuppressedtheautoionization.Becauseofthis,ineachcasethepHofabasic solutionisgreaterthan7. Baseionizationisthereforequiteanalogoustoacidionizationobservedearlier.Wenowconsidera comparisonofthestrengthofanacidtothestrengthofabase.Todoso,weconsideraclassofreactions called"neutralizationreactions"whichoccurwhenwemixanacidsolutionwithabasesolution.Sincethe aciddonatesprotonsandthebaseacceptsprotons,wemightexpect,whenmixingacidandbase,toachievea solutionwhichisnolongeracidicorbasic.Forexample,ifwemixtogetherequalvolumesof0.1M HCl aq and0.1M NaOH aq ,thefollowingreactionoccurs: HCl aq + NaOH aq Na + aq + Cl )]TJ/F8 9.9626 Tf 8.385 -4.456 Td [( aq + H 2 O l .10 Theresultantsolutionissimplyasaltsolutionwith NaCl dissolvedinwater.Thissolutionhasneither acidicnorbasicproperties,andthepHis7;hencetheacidandbasehaveneutralizedeachother.Inthis case,wehavemixedtogetherastrongacidwithastrongbase.Sincebotharestrongandsincewemixed equalmolarquantitiesofeach,theneutralizationreactionisessentiallycomplete. Wenextconsidermixingtogetheraweakacidsolutionwithastrongbasesolution,againwithequal molarquantitiesofacidandbase.Asanexample,wemix100mlof0.1Maceticacid HA solutionwith 100mlof0.1Msodiumhydroxide.Inthisdiscussion,wewillabbreviatetheaceticacidmolecularformula CH 3 COOH as HA andtheacetateion CH 3 COO )]TJ/F15 9.9626 Tf 10.045 -3.616 Td [(as A )]TJ/F15 9.9626 Tf 6.725 -3.616 Td [(.Thereactionof HA and NaOH is: HA aq + NaOH aq Na + aq + A )]TJ/F8 9.9626 Tf 8.385 -4.114 Td [( aq + H 2 O l .11 A )]TJ/F8 9.9626 Tf 8.386 -3.616 Td [( aq istheacetateioninsolution,formedwhenanaceticacidmoleculedonatesthepositivehydrogen ion.Wehavethuscreatedasaltsolutionagain,inthiscaseofsodiumacetateinwater.Notethatthevolume ofthecombinedsolutionis200ml,sotheconcentrationofsodiumacetate NaA insolutionis0.050M.

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156 CHAPTER15.ACID-BASEEQUILIBRIUM Unlikeourprevious NaCl saltsolution,ameasurementinthiscaserevealsthatthepHoftheproduct saltsolutionis9.4,sothesolutionisbasic.Thus,mixingequalmolarquantitiesofstrongbasewithweak acidproducesabasicsolution.Inessence,theweakaciddoesnotfullyneutralizethestrongbase.To understandthis,weexaminethebehaviorofsodiumacetateinsolution.SincethepHisgreaterthan7, thenthereisanexcessof OH )]TJ/F15 9.9626 Tf 9.871 -3.615 Td [(ionsinsolutionrelativetopurewater.Theseionsmusthavecomefromthe reactionofsodiumacetatewiththewater.Therefore,thenegativeacetateionsinsolutionmustbehaveas abase,acceptingpositivehydrogenions: A )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( aq + H 2 O aq HA aq + OH )]TJ/F8 9.9626 Tf 8.385 -4.113 Td [( l .12 Thereactionofanionwithwatertoformeitheranacidorabasesolutionisreferredtoas hydrolysis Fromthisexample,thesaltofaweakacidbehavesasabaseinwater,resultinginapHgreaterthan7. Tounderstandtheextenttowhichthehydrolysisofthenegativeionoccurs,weneedtoknowthe equilibriumconstantforthisreaction.Thisturnsouttobedeterminedbytheacidionizationconstantfor HA .Toseethis,wewritetheequilibriumconstantforthehydrolysisof A )]TJ/F15 9.9626 Tf 10.045 -3.616 Td [(as K h = [ HA ][ OH )]TJ/F8 9.9626 Tf 6.724 -3.616 Td [(] [ A )]TJ/F8 9.9626 Tf 6.725 -2.878 Td [(] .13 Multiplyingnumeratoranddenominatorby [ H 3 O + ] ,wendthat K h = [ HA ] [ OH )]TJ/F8 9.9626 Tf 5.894 -3.754 Td [(] [ A )]TJ/F7 6.9738 Tf 5.894 -1.992 Td [(] [ H 3 O + ] [ H 3 O + ] = K w K a .14 Therefore,forthehydrolysisofacetateionsinsolution, K h =5 : 8 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(10 .Thisisfairlysmall,sotheacetate ionisaveryweakbase. 15.7Observation5:Acidstrengthandmolecularproperties Wenowhaveafairlycompletequantitativedescriptionofacid-baseequilibrium.Tocompleteourunderstandingofacid-baseequilibrium,weneedapredictivemodelwhichrelatesacidstrengthorbasestrength tomolecularproperties.Ingeneral,weexpectthatthestrengthofanacidisrelatedeithertotherelative easebywhichitcandonateahydrogenionorbytherelativestabilityoftheremainingnegativeionformed afterthedepartureofthehydrogenion. Tobegin,wenotethattherearethreebasiccategoriesofacidswhichwehaveexaminedinthisstudy. First,therearesimple binaryacids : HF ; HCl ; HBr ; HI .Second,thereareacidsformedfrommaingroup elementscombinedwithoneormoreoxygenatoms,such H 2 SO 4 or HNO 3 .Thesearecalled oxyacids Third,therearethe carboxylicacids ,organicmoleculeswhichcontainthecarboxylicfunctionalgroupin Figure15.1CarboxylicFunctionalGroup.

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157 CarboxylicFunctionalGroup Figure15.1 Weconsiderrstthesimplebinaryacids. HCl HBr ,and HI areallstrongacids,whereas HF isaweak acid.Incomparingtheexperimentalvaluesof pK a valuesinH-XBondStrengthsandpKa,p.157,wenote thattheacidstrengthincreasesintheorder HF
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158 CHAPTER15.ACID-BASEEQUILIBRIUM Whywouldelectronegativityplayaroleinacidstrength?Therearetwoconclusionswemightdraw.First, agreaterelectronegativityoftheatomoratomsattachedtotheH-Ointheoxyacidapparentlyresultsina weakerH-Obond,whichisthusmorereadilyionized.Weknowthatanelectronegativeatompolarizesbonds bydrawingtheelectronsinthemoleculetowardsit.Inthiscase,theClin HOCl andtheBrin HOBr must polarizetheH-Obond,weakeningitandfacilitatingtheionizationofthehydrogen.Incomparing HOCl to HOClO ,theaddedoxygenatommustincreasethepolarizationoftheH-Obond,thusweakeningthebond furtherandincreasingtheextentofionization. Asecondconclusionhastodowiththeioncreatedbytheacidionization.Thenegativeionproduced hasasurpluselectron,andtherelativeenergyofthisionwilldependonhowreadilythatextraelectron isattractedtotheatomsofion.Themoreelectronegativethoseatomsare,thestrongeristheattraction. Therefore,the OCl )]TJ/F15 9.9626 Tf 10.172 -4.456 Td [(ioncanmorereadilyaccommodatethenegativechargethancanthe OBr )]TJ/F15 9.9626 Tf 10.171 -4.345 Td [(ion.And the OClO )]TJ/F15 9.9626 Tf 10.044 -3.616 Td [(ioncanmorereadilyaccommodatethenegativechargethancanthe OCl )]TJ/F15 9.9626 Tf 10.045 -4.456 Td [(ion. Weconcludethatthepresenceofstronglyelectronegativeatomsinanacidincreasesthepolarizationof theH-Obond,thusfacilitatingionizationoftheacid,andincreasestheattractionoftheextraelectronto thenegativeion,thusstabilizingthenegativeion.Bothofthesefactorsincreasetheacidstrength.Chemists commonlyusebothoftheseconclusionsinunderstandingandpredictingrelativeacidstrength. Therelativeacidityofcarboncompoundsisamajorsubjectoforganicchemistry,whichwecanonlyvisit brieyhere.Ineachofthecarboxylicacids,theH-OgroupisattachedtoacarbonylC=Ogroup,whichisin turnbondedtootheratoms.Thecomparisonweobservehereisbetweencarboxylicacidmolecules,denoted as RCOOH ,andotherorganicmoleculescontainingtheH-Ogroup,suchasalcoholsdenotedas ROH .Ris simplyanatomorgroupofatomsattachedtothefunctionalgroup.Theformerareobviouslyacidswhereas thelattergroupcontainsmoleculeswhicharegenerallyextremelyweakacids.Oneinterestingcomparison isfortheacidandalcoholwhenRisthebenzenering, C 6 H 5 .Benzoicacid, C 6 H 5 COOH ,has pK a =4 : 2 whereasphenol, C 6 H 5 OH ,has pK a =9 : 9 .Thus,thepresenceofthedoublybondedoxygenatomonthe carbonatomadjacenttotheO-Hclearlyincreasestheacidityofthemolecule,andthusincreasesionization oftheO-Hbond. Thisobservationisquitereasonableinthecontextofourpreviousconclusion.Addinganelectronegative oxygenatominnearproximitytotheO-HbondbothincreasesthepolarizationoftheO-Hbondand stabilizesthenegativeionproducedbytheacidionization.Inadditiontotheelectronegativityeect, carboxylateanions, RCOO )]TJ/F15 9.9626 Tf 6.725 -3.615 Td [(,exhibitresonancestabilization,asseeninFigure15.2. Figure15.2 Theresonanceresultsinasharingofthenegativechargeoverseveralatoms,thusstabilizingthenegative ion.Thisisamajorcontributingfactorintheacidityofcarboxylicacidsversusalcohols. 15.8ReviewandDiscussionQuestions Exercise15.1 Strongacidshaveahigherpercentionizationthandoweakacids.Whydon'tweusepercent ionizationasameasureofacidstrength,ratherthan K a ?

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159 Exercise15.2 UsingthedatainEquilibriumConcentrationsandKforNitrousAcid,p.152fornitrousacid, plot [ H 3 O + ] versus c 0 ,theinitialconcentrationoftheacid,andversus [ HNO 2 ] theequilibrium concentrationoftheacid.Onasecondgraph,plot [ H 3 O + ] 2 versus c 0 ,theinitialconcentration oftheacid,andversus [ HNO 2 ] theequilibriumconcentrationoftheacid.Whichoftheseresults givesastraightline?Usingtheequilibriumconstantexpression,explainyouranswer. Exercise15.3 UsingLeChtelier'sprinciple,explainwhytheconcentrationof [ OH )]TJ/F8 9.9626 Tf 6.724 -3.615 Td [(] ismuchlowerinacidic solutionthanitisinneutralsolution. Exercise15.4 Weconsideredmixingastrongbasewithaweakacid,butwedidnotconsidermixingastrongacid withaweakacid.Considermixing0.1M HNO 3 and0.1M HNO 2 .PredictthepHofthesolution andthepercentionizationofthenitrousacid.RationalizeyourpredictionusingLeChtelier's principle. Exercise15.5 Imaginetakinga0.5Msolutionofnitrousacidandslowingaddingwatertoit.Lookingat% IonizationofNitrousAcid,p.151,weseethat,astheconcentrationofnitrousaciddecreases,the percentionizationincreases.Bycontrast, [ H 3 O + ] decreases.RationalizetheseresultsusingLe Chtelier'sprinciple. Exercise15.6 Weobservedthatmixingastrongacidandastrongbase,inequalamountsandconcentrations, producesaneutralsolution,andthatmixingastrongbasewithaweakacid,inequalamounts andconcentrations,producesabasicsolution.Imaginemixingaweakacidandaweakbase,in equalamountsandconcentrations.Predictwhethertheresultingsolutionwillbeacidic,basic,or neutral,andexplainyourprediction. Exercise15.7 UsingtheelectronegativityargumentspresentedaboveSection15.7:Observation5:Acidstrength andmolecularproperties,explainwhy,ingeneral,compoundslikeM-O-Harebasesratherthan acids,whenMisametalatom.Predicttherelationshipbetweenthepropertiesofthemetalatom Mandthestrengthofthebase MOH Exercise15.8 Ionizationofsulfuricacid H 2 SO 4 produces HSO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(4 ,whichisalsoanacid.However, HSO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(4 isa muchweakeracidthan H 2 SO 4 .UsingtheconclusionsfromaboveSection15.7:Observation5: Acidstrengthandmolecularproperties,explainwhy HSO )]TJ/F7 6.9738 Tf -0.277 -6.918 Td [(4 isamuchweakeracid. Exercise15.9 Predictandexplaintherelativeacidstrengthsof H 2 S and HCl .Predictandexplaintherelative acidstrengthsof H 3 PO 4 and H 3 AsO 4 Exercise15.10 UsingargumentsfromaboveSection15.7:Observation5:Acidstrengthandmolecularproperties,predictandexplaintherelativeacidityofphenolFigure15.3a:Phenolandmethanol Figure15.3b:Methanol.

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160 CHAPTER15.ACID-BASEEQUILIBRIUM aPhenol bMethanol Figure15.3

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Chapter16 ReactionRates 1 16.1Foundation Wewillassumeanunderstandingofthepostulatesofthe KineticMolecularTheory andoftheenergetics ofchemicalreactions.Wewillalsoassumeanunderstandingofphaseequilibriumandreactionequilibrium, includingthetemperaturedependenceofequilibriumconstants. 16.2Goals Wehavecarefullyexaminedtheobservationthatchemicalreactionscometoequilibrium.Dependingon thereaction,theequilibriumconditionscanbesuchthatthereisamixtureofreactantsandproducts,or virtuallyallproducts,orvirtuallyallreactants.Wehavenotconsideredthetimescaleforthereactionto achievetheseconditions,however.Inmanycases,thespeedofthereactionmightbeofmoreinterestthan thenalequilibriumconditionsofthereaction.Somereactionsproceedsoslowlytowardsequilibriumasto appearnottooccuratall.Forexample,metallicironwilleventuallyoxidizeinthepresenceofaqueoussalt solutions,butthetimeissucientlylongforthisprocessthatwecanreasonablyexpecttobuildaboatout ofiron.Ontheotherhand,somereactionsmaybesorapidastoposeahazard.Forexample,hydrogengas willreactwithoxygengassorapidlyastocauseanexplosion.Inaddition,thetimescaleforareactioncan dependverystronglyontheamountsofreactantsandtheirtemperature. Inthisconceptdevelopmentstudy,weseekanunderstandingoftheratesofchemicalreactions.Wewill deneandmeasurereactionratesanddevelopaquantitativeanalysisofthedependenceofthereactionrates ontheconditionsofthereaction,includingconcentrationofreactantsandtemperature.Thisquantitative analysiswillprovideusinsightintotheprocessofachemicalreactionandthusleadustodevelopamodel toprovideanunderstandingofthesignicanceofreactantconcentrationandtemperature. Wewillndthatmanyreactionsproceedquitesimply,withreactantmoleculescollidingandexchanging atoms.Inothercases,wewillndthattheprocessofreactioncanbequitecomplicated,involvingmany molecularcollisionsandrearrangementsleadingfromreactantmoleculestoproductmolecules.Therateof thechemicalreactionisdeterminedbythesesteps. 16.3Observation1:ReactionRates Webeginbyconsideringafairlysimplereactiononaratherelegantmolecule.Oneoxidizedformofbuckminsterfullerene C 60 is C 60 O 3 ,withathreeoxygenbridgeasshowninFigure16.1OxidizedBuckminsterfullerene. 1 Thiscontentisavailableonlineat. 161

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162 CHAPTER16.REACTIONRATES OxidizedBuckminsterfullerene Figure16.1 C 60 O 3 ispreparedfrom C 60 dissolvedintoluenesolutionattemperaturesof 0 C orbelow.Whenthe solutioniswarmed, C 60 O 3 decomposes,releasing O 2 andcreating C 60 O inareactionwhichgoesessentially tocompletion.Wecanactuallywatchthisprocesshappenintimebymeasuringtheamountoflightofa specicfrequencyabsorbedbythe C 60 O 3 molecules,calledthe absorbance .Theabsorbanceisproportional

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163 totheconcentrationofthe C 60 O 3 inthetoluenesolution,soobservingtheabsorbanceasafunctionoftime isessentiallythesameasobservingtheconcentrationasafunctionoftime.Onesuchsetofdataisgivenin OxidizedBuckminsterfullereneAbsorbanceduringThermalDecompositionat23 C,p.163,andisshown inthegraphinFigure16.2OxidizedBuckminsterfullereneAbsorbance. OxidizedBuckminsterfullereneAbsorbanceduringThermalDecompositionat23 C timeminutes C 60 O 3 absorbance 3 0.04241 9 0.03634 15 0.03121 21 0.02680 27 0.02311 33 0.01992 39 0.01721 45 0.01484 51 0.01286 57 0.01106 63 0.00955 69 0.00827 75 0.00710 81 0.00616 87 0.00534 93 0.00461 99 0.00395

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164 CHAPTER16.REACTIONRATES OxidizedBuckminsterfullereneAbsorbance Figure16.2 Therateatwhichthedecompositionreactionisoccurringisclearlyrelatedtotherateofchangeofthe concentration [ C 60 O 3 ] ,whichisproportionaltotheslopeofthegraphinFigure16.2OxidizedBuckminsterfullereneAbsorbance.Therefore,wedenetherateofthisreactionas Rate = )]TJ/F1 9.9626 Tf 9.409 14.047 Td [( d dt [ C 60 O 3 ] )]TJ/F1 9.9626 Tf 19.925 14.047 Td [( [ C 60 O 3 ] t .1 Wewanttherateofreactiontobepositive,sincethereactionisproceedingforward.However,because wearemeasuringtherateofdisappearanceofthereactantinthiscase,thatrateisnegative.Weincludea negativesigninthisdenitionofratesothattheratein.1isapositivenumber.Notealsothattheslope ofthegraphinFigure16.2OxidizedBuckminsterfullereneAbsorbanceshouldbetakenasthederivative ofthegraph,sincethegraphisnotastraightline.Wewillapproximatethatderivativebyestimatingthe slopeateachtimeinthedata,takingthechangeintheabsorbanceofthe C 60 O 3 dividedbythechangein timeateachtimestep.Therate,calculatedinthisway,isplottedasafunctionoftimeinFigure16.3Rate

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165 ofDecomposition. RateofDecomposition Figure16.3 ItisclearthattheslopeofthegraphinFigure16.2OxidizedBuckminsterfullereneAbsorbancechanges overthecourseoftime.Correspondingly,Figure16.3RateofDecompositionshowsthattherateofthe reactiondecreasesasthereactionproceeds.Thereactionisatrstveryfastbutthenslowsconsiderablyas thereactant C 60 O 3 isdepleted. TheshapeofthegraphforrateversustimeFigure16.3RateofDecompositionisverysimilartothe shapeofthegraphforconcentrationversustimeFigure16.2OxidizedBuckminsterfullereneAbsorbance. Thissuggeststhattherateofthereactionisrelatedtotheconcentrationof C 60 O 3 ateachtime.Therefore, inFigure16.4RateversusConcentration,weplottherateofthereaction,denedin.1andshownin Figure16.3RateofDecomposition,versustheabsorbanceofthe C 60 O 3 .

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166 CHAPTER16.REACTIONRATES RateversusConcentration Figure16.4 Wendthatthereisaverysimpleproportionalrelationshipbetweentherateofthereactionandthe concentrationofthereactant.Therefore,wecanwrite Rate = )]TJ/F1 9.9626 Tf 9.41 8.07 Td [()]TJ/F10 6.9738 Tf 7.266 -4.147 Td [(d dt [ C 60 O 3 ] = k [ C 60 O 3 ] .2 where k isaproportionalityconstant.Thisequationshowsthat,earlyinthereactionwhen [ C 60 O 3 ] is large,thereactionproceedsrapidly,andthatas C 60 O 3 isconsumed,thereactionslowsdown..2isan exampleofa ratelaw ,expressingtherelationshipbetweentherateofareactionandtheconcentrations ofthereactantorreactants.Ratelawsareexpressionsoftherelationshipbetweenexperimentallyobserved ratesandconcentrations. Asasecondexampleofareactionrate,weconsiderthedimerizationreactionofbutadienegas, CH 2 = CH CH = CH 2 .Twobutadienemoleculescancombinetoformvinylcyclohexene,showninFigure16.5DimerizationofButadienetoVinylcyclohexene.

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167 DimerizationofButadienetoVinylcyclohexene Figure16.5 DimerizationofButadieneat250 C,p.167providesexperimentaldataonthegasphaseconcentration ofbutadiene [ C 4 H 6 ] asafunctionoftimeat T =250 C DimerizationofButadieneat250 C Times [ C 4 H 6 ] M RateM/s Rate [ C 4 H 6 ] Rate [ C 4 H 6 ] 2 0 0.0917 9 : 48 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 1 : 03 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 1 : 13 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 500 0.0870 8 : 55 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 9 : 84 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 13 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 1000 0.0827 7 : 75 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 9 : 37 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 13 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 1500 0.0788 7 : 05 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 8 : 95 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 14 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2000 0.0753 6 : 45 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 8 : 57 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 14 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 2500 0.0720 5 : 92 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 8 : 22 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 14 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 3000 0.0691 5 : 45 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 7 : 90 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 14 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 3500 0.0664 5 : 04 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 7 : 60 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 14 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 4000 0.0638 4 : 67 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(6 7 : 32 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(5 1 : 15 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 Wecanestimatetherateofreactionateachtimestepasin.1,andthesedataarepresentedin DimerizationofButadieneat250 C,p.167aswell.Againweseethattherateofreactiondecreasesasthe concentrationofbutadienedecreases.Thissuggeststhattherateisgivenbyanexpressionlike.2.To testthis,wecalculate Rate [ C 4 H 6 ] inDimerizationofButadieneat250 C,p.167foreachtimestep.Wenote thatthisis not aconstant,so.2doesnotdescribetherelationshipbetweentherateofreactionandthe concentrationofbutadiene.Insteadwecalculate Rate [ C 4 H 6 ] 2 inDimerizationofButadieneat250 C,p.167. Wediscoverthatthisratioisaconstantthroughoutthereaction.Therefore,therelationshipbetweenthe rateofthereactionandtheconcentrationofthereactantinthiscaseisgivenby Rate = )]TJ/F1 9.9626 Tf 9.409 8.069 Td [()]TJ/F10 6.9738 Tf 7.267 -4.147 Td [(d dt [ C 4 H 6 ] = k [ C 4 H 6 ] 2 .3 whichistheratelawforthereactioninFigure16.5DimerizationofButadienetoVinylcyclohexene.This isaveryinterestingresultwhencomparedto.2.Inbothcases,theresultsdemonstratethattherateof

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168 CHAPTER16.REACTIONRATES reactiondependsontheconcentrationofthereactant.However,wenowalsoknowthatthewayinwhich theratevarieswiththeconcentrationdependsonwhatthereactionis.Eachreactionhasitsownratelaw, observedexperimentally. 16.4Observation2:RateLawsandtheOrderofReaction Wewouldliketounderstandwhatdeterminesthespecicdependenceofthereactionrateonconcentration ineachreaction.Intherstcaseconsideredabove,theratedependsontheconcentrationofthereactantto therstpower.Werefertothisasa rstorderreaction .Inthesecondcaseabove,theratedependson theconcentrationofthereactanttothesecondpower,sothisiscalleda secondorderreaction .Thereare also thirdorderreactions ,andeven zerothorderreactions whoseratesdonotdependontheamount ofthereactant.Weneedmoreobservationsofratelawsfordierentreactions. TheapproachusedintheprevioussectionSection16.3:Observation1:ReactionRatestodeterminea reaction'sratelawisfairlyclumsyandatthispointdiculttoapply.Weconsiderhereamoresystematic approach.First,considerthedecompositionof N 2 O 5 g 2 N 2 O 5 g 4 NO 2 g + O 2 g Wecancreateaninitialconcentrationof N 2 O 5 inaaskandmeasuretherateatwhichthe N 2 O 5 rst decomposes.Wecanthencreateadierentinitialconcentrationof N 2 O 5 andmeasurethenewrateatwhich the N 2 O 5 decomposes.Bycomparingtheserates,wecanndtheorderofthedecompositionreaction.The ratelawfordecompositionof N 2 O 5 g isofthegeneralform: Rate = k [ N 2 O 5 ] m .4 soweneedtodeterminetheexponent m .Forexample,at 25 C weobservethattherateofdecomposition is 1 : 4 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(3 M s whentheconcentrationof N 2 O 5 is 0 : 020 M .Ifinsteadwebeginwe [ N 2 O 5 ]=0 : 010 M ,we observethattherateofdecompositionis 7 : 0 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(4 M s .Wecancomparetheratefromtherstmeasurement Rate totheratefromthesecondmeasurement Rate .From.4,wecanwritethat Rate Rate = k [ N 2 O 5 ] 1 m k [ N 2 O 5 ] 2 m = 1 : 4 10 )]TJ/F6 4.9813 Tf 5.396 0 Td [(3 M s 7 : 0 10 )]TJ/F6 4.9813 Tf 5.397 0 Td [(4 M s = k : 020 M m k : 010 M m .5 Thiscanbesimpliedonbothsidesoftheequationtogive 2 : 0=2 : 0 m Clearly,then m =1 ,andthedecompositionisarstorderreaction.Wecanalsothenndtherstorder rateconstant k forthisreactionbysimplyplugginginoneoftheinitialratemeasurementsto.4.We ndthat k =0 : 070 s )]TJ/F7 6.9738 Tf 6.227 0 Td [(1 Thisapproachtondingreactionorderiscalledthemethodofinitialrates,sinceitreliesonxingthe concentrationatspecicinitialvaluesandmeasuringtheinitialrateassociatedwitheachconcentration. Sofarwehaveconsideredonlyreactionswhichhaveasinglereactant.Considerasecondexampleofthe methodofinitialratesinvolvingthereactionofhydrogengasandiodinegas: H 2 g + I 2 g 2 HI g .6 Inthiscase,weexpecttondthattherateofthereactiondependsontheconcentrationsforbothreactants. Assuch,weneedmoreinitialrateobservationstodeterminetheratelaw.InHydrogenGasandIodineGas InitialRateDataat700K,p.169,observationsarereportedfortheinitialrateforthreesetsofinitial concentrationsof H 2 and I 2 .

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169 HydrogenGasandIodineGasInitialRateDataat700K Experiment [ H 2 ] 0 M [ I 2 ] 0 M RateM/sec 1 0.10 0.10 3 : 00 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 2 0.20 0.10 6 : 00 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(4 3 0.20 0.20 1 : 19 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 Followingthesameprocessweusedinthe N 2 O 5 example,wewritethegeneralratelawforthereaction as Rate = k [ H 2 ] n [ I 2 ] m .7 Bycomparingexperiment1toexperiment2,wecanwrite Rate Rate = k [ H 2 ] 1 n [ I 2 ] 1 m k [ H 2 ] 2 n [ I 2 ] 2 m = 3 : 00 10 )]TJ/F6 4.9813 Tf 5.396 0 Td [(4 M s 6 : 00 10 )]TJ/F6 4.9813 Tf 5.396 0 Td [(4 M s = k : 10 M m : 10 M n k : 20 M m : 10 M n .8 Thissimpliesto 0 : 50=0 : 50 m 1 : 00 n fromwhichitisclearthat m =1 .Similarly,wecanndthat n =1 .Thereactionisthereforerstorderin eachreactantandissecondorderoverall. Rate = k [ H 2 ][ I 2 ] .9 Onceweknowtheratelaw,wecanuseanyofthedatafromHydrogenGasandIodineGasInitialRate Dataat700K,p.169todeterminetherateconstant,simplybyplugginginconcentrationsandrateinto .9.Wendthat k =3 : 00 10 )]TJ/F7 6.9738 Tf 6.227 0 Td [(2 1 Ms Thisprocedurecanbeappliedtoanynumberofreactions.Thechallengeispreparingtheinitialconditions andmeasuringtheinitialchangeinconcentrationpreciselyversustime.RateLawsforVariousReactions, p.169providesanoverviewoftheratelawsforseveralreactions.Avarietyofreactionordersareobserved, andtheycannotbeeasilycorrelatedwiththestoichiometryofthereaction. RateLawsforVariousReactions Reaction RateLaw 2 NO g + O 2 g 2 NO 2 g Rate = k [ NO ] 2 [ O 2 ] 2 NO g +2 H 2 g 2 N 2 g +2 H 2 O g Rate = k [ NO ] 2 [ H 2 ] 2 ICl g + H 2 g 2 HCl g + I 2 g Rate = k [ ICl ][ H 2 ] 2 N 2 O 5 g 4 NO 2 g + O 2 g Rate = k [ N 2 O 5 ] 2 NO 2 g + F 2 g 2 NO 2 F g Rate = k [ NO 2 ][ F 2 ] 2 H 2 O 2 aq 2 H 2 O l + O 2 g Rate = k [ H 2 O 2 ] H 2 g + Br 2 g 2 HBr g Rate = k [ H 2 ][ Br 2 ] 1 2 O 3 g + Cl g O 2 g + ClO g Rate = k [ O 3 ][ Cl ]

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170 CHAPTER16.REACTIONRATES 16.5ConcentrationsasaFunctionofTimeandtheReactionHalf-life Onceweknowtheratelawforareaction,weshouldbeabletopredicthowfastareactionwillproceed. Fromthis,weshouldalsobeabletopredicthowmuchreactantremainsorhowmuchproducthasbeen producedatanygiventimeinthereaction.Wewillfocusonthereactionswithasinglereactanttoillustrate theseideas. Considerarstorderreactionlike A products ,forwhichtheratelawmustbe Rate = )]TJ/F1 9.9626 Tf 9.409 8.07 Td [()]TJ/F10 6.9738 Tf 7.267 -4.147 Td [(d dt [ A ] = k [ A ] .10 FromCalculus,itispossibletouse.10tondthefunction [ A ] t whichtellsustheconcentration [ A ] asafunctionoftime.Theresultis [ A ]=[ A ] 0 e )]TJ/F7 6.9738 Tf 6.227 0 Td [( kt .11 orequivalently ln [ A ]= ln [ A ] 0 )]TJ/F11 9.9626 Tf 9.962 0 Td [(kt .12 .12revealsthat,ifareactionisrstorder,wecanplot ln [ A ] versustimeandgetastraightlinewith slopeequalto )]TJ/F11 9.9626 Tf 7.749 0 Td [(k .Moreover,ifweknowtherateconstantandtheinitialconcentration,wecanpredictthe concentrationatanytimeduringthereaction. Aninterestingpointinthereactionisthetimeatwhichexactlyhalfoftheoriginalconcentrationof A hasbeenconsumed.Wecallthistimethe halflife ofthereactionanddenoteitas t 1 2 .Atthattime, [ A ]= 1 2 [ A ] 0 .From.12andusingthepropertiesoflogarithms,wendthat,forarstorderreaction t 1 2 = ln 2 k .13 Thisequationtellsusthatthehalf-lifeofarstorderreactiondoesnotdependonhowmuchmaterialwe startwith.Ittakesexactlythesameamountoftimeforthereactiontoproceedfromallofthestarting materialtohalfofthestartingmaterialasitdoestoproceedfromhalfofthestartingmaterialtoone-fourth ofthestartingmaterial.Ineachcase,wehalvetheremainingmaterialinatimeequaltotheconstant half-lifein.13. Theseconclusionsareonlyvalidforrstorderreactions.Considerthenasecondorderreaction,suchas thebutadienedimerizationdiscussedabovep.166.Thegeneralsecondorderreaction A products has theratelaw Rate = )]TJ/F1 9.9626 Tf 9.409 8.07 Td [()]TJ/F10 6.9738 Tf 7.267 -4.148 Td [(d dt [ A ] = k [ A ] 2 .14 Again,wecanuseCalculustondthefunction [ A ] t from.14.Theresultismosteasilywrittenas 1 [ A ] = 1 [ A ] 0 + k t .15 Notethat,as t increases, 1 [ A ] increases,so [ A ] decreases..15revealsthat,forareactionwhichissecond orderinthereactant A ,wecanplot 1 [ A ] asafunctionoftimetogetastraightlinewithslopeequalto k Again,ifweknowtherateconstantandtheinitialconcentration,wecanndtheconcentration [ A ] atany timeofinterestduringthereaction. Thehalf-lifeofasecondorderreactiondiersfromthehalf-lifeofarstorderreaction.From.15,if wetake [ A ]= 1 2 [ A ] 0 ,weget t 1 2 = 1 k [ A ] 0 .16 Thisshowsthat,unlikearstorderreaction,thehalf-lifeforasecondorderreactiondependsonhowmuch materialwestartwith.From.16,themoreconcentratedthereactantis,theshorterthehalf-life.

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171 16.6Observation3:TemperatureDependenceofReactionRates Itisacommonobservationthatreactionstendtoproceedmorerapidlywithincreasingtemperature.Similarly,coolingreactantscanhavetheeectofslowingareactiontoanearhalt.Howisthischangein ratereectedintheratelawequation, e.g. .9?Onepossibilityisthatthereisaslightdependenceon temperatureoftheconcentrations,sincevolumesdovarywithtemperature.However,thisisinsucient toaccountforthedramaticchangesinratetypicallyobserved.Therefore,thetemperaturedependenceof reactionrateisprimarilyfoundintherateconstant, k Considerforexamplethereactionofhydrogengaswithiodinegasathightemperatures,asgivenin .6.Therateconstantofthisreactionateachtemperaturecanbefoundusingthemethodofinitialrates, asdiscussedabove,andwendinRateConstantforHydrogenGasandIodineGas,p.171thattherate constantincreasesdramaticallyasthetemperatureincreases. RateConstantforHydrogenGasandIodineGas TK k 1 Ms 667 6 : 80 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 675 9 : 87 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(3 700 3 : 00 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 725 8 : 43 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(2 750 2 : 21 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(1 775 5 : 46 10 )]TJ/F7 6.9738 Tf 6.226 0 Td [(1 800 1.27 AsshowninFigure16.6RateConstant,therateconstantappearstoincreaseexponentiallywith temperature.Afteralittleexperimentationwiththedata,wendinFigure16.7RateConstantthatthere isasimplelinearrelationshipbetween lnk and 1 T .

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172 CHAPTER16.REACTIONRATES RateConstant Figure16.6

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173 RateConstant Figure16.7 FromFigure16.7RateConstant,wecanseethatthedatainRateLawsforVariousReactions,p.169 ttheequation lnk = a 1 T + b .17 where a and b areconstantforthisreaction.Itturnsoutthat,forourpurposes,allreactionshaverate constantswhicht.17,butwithdierentconstants a and b foreachreaction.Figure16.7RateConstant isreferredtoasan Arrheniusplot ,afterSvanteArrhenius. Itisveryimportanttonotethattheformof16.17andtheappearanceofFigure16.7RateConstant areboththesameastheequationsandgraphsforthetemperaturedependenceoftheequilibriumconstant foranendothermicreaction.Thissuggestsamodeltoaccountforthetemperaturedependenceoftherate constant,basedontheenergeticsofthereaction.Inparticular,itappearsthatthereactionrateisrelated totheamountofenergyrequiredforthereactiontooccur.Wewilldevelopthisfurtherinthenextsection.

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174 CHAPTER16.REACTIONRATES 16.7CollisionModelforReactionRates Atthispoint,wehaveonlyobservedthedependenceofreactionratesonconcentrationofreactantsandon temperature,andwehavetthesedatatoequationscalledratelaws.Althoughthisisveryconvenient,it doesnotprovideusinsightintowhyaparticularreactionhasaspecicratelaworwhythetemperature dependenceshouldobey.17.Nordoesitprovideanyphysicalinsightsintotheorderofthereactionor themeaningoftheconstants a and b in.17. Webeginbyaskingwhythereactionrateshoulddependontheconcentrationofthereactants.To answerthis,weconsiderasimplereactionbetweentwomoleculesinwhichatomsaretransferredbetween themoleculesduringthereaction.Forexample,areactionimportantinthedecompositionofozone O 3 by aerosolsis O 3 g + Cl g O 2 g + ClO g Whatmusthappenforareactiontooccurbetweenan O 3 moleculeanda Cl atom?Obviously,forthesetwo particlestoreact,theymustcomeintocloseproximitytooneanothersothatan O atomcanbetransferred fromonetotheother.Ingeneral,twomoleculescannottradeatomstoproducenewproductmolecules unlesstheyarecloseenoughtogetherfortheatomsofthetwomoleculestointeract.Thisrequiresacollision betweenmolecules. Therateofcollisionsdependsontheconcentrationsofthereactants,sincethemoremoleculesthere areinaconnedspace,themorelikelytheyaretorunintoeachother.Towritethisrelationshipinan equation,wecanthinkintermsofprobability,andweconsiderthereactionabove.Theprobabilityforan O 3 moleculetobenearaspecicpointincreaseswiththenumberof O 3 molecules,andthereforeincreases withtheconcentrationof O 3 molecules.Theprobabilityfora Cl atomtobenearthatspecicpointisalso proportionaltotheconcentrationof Cl atoms.Therefore,theprobabilityforan O 3 moleculeanda Cl atom tobeincloseproximitytothesamespecicpointatthesametimeisproportionaltothe [ O 3 ] times [ Cl ] Itisimportanttorememberthatnotallcollisionsbetween O 3 moleculesand Cl atomswillresultina reaction.Thereareotherfactorstoconsiderincludinghowthemoleculesapproachoneanother.Theatoms maynotbepositionedproperlytoexchangebetweenmolecules,inwhichcasethemoleculeswillsimply bounceoofoneanotherwithoutreacting.Forexample,ifthe Cl atomapproachesthecenter O atom ofthe O 3 molecule,that O atomwillnottransfer.Anotherfactorisenergyassociatedwiththereaction. Clearly,though,acollisionmustoccurforthereactiontooccur,andthereforethererateofthereactioncan benofasterthantherateofcollisionsbetweenthereactantmolecules. Therefore,wecansaythat,ina bimolecularreaction ,wheretwomoleculescollideandreact,therateof thereactionwillbeproportionaltotheproductoftheconcentrationsofthereactants.Forthereactionof O 3 with Cl ,theratemustthereforebeproportionalto [ O 3 ][ Cl ] ,andweobservethisintheexperimental ratelawinRateLawsforVariousReactions,p.169.Thus,itappearsthatwecanunderstandtheratelaw byunderstandingthecollisionswhichmustoccurforthereactiontotakeplace. Wealsoneedourmodeltoaccountforthetemperaturedependenceoftherateconstant.Asnotedatthe endofthelastsectionp.173,thetemperaturedependenceoftherateconstantin.17isthesameas thetemperaturedependenceoftheequilibriumconstantforanendothermicreaction.Thissuggeststhatthe temperaturedependenceisduetoanenergeticfactorrequiredforthereactiontooccur.However,wend experimentallythat.17describestherateconstanttemperaturedependenceregardlessofwhetherthe reactionisendothermicorexothermic.Therefore,whatevertheenergeticfactoristhatisrequiredforthe reactiontooccur,itisnotjusttheendothermicityofthereaction.Itmustbethatallreactions,regardless oftheoverallchangeinenergy,requireenergytooccur. Amodeltoaccountforthisistheconceptof activationenergy .Forareactiontooccur,atleast somebondsinthereactantmoleculemustbebroken,sothatatomscanrearrangeandnewbondscanbe created.Atthetimeofcollision,bondsarestretchedandbrokenasnewbondsaremade.Breakingthese bondsandrearrangingtheatomsduringthecollisionrequirestheinputofenergy.Theminimumamountof energyrequiredforthereactiontooccuriscalledtheactivationenergy, E a .ThisisillustratedinFigure16.8 ReactionEnergy,showingconceptuallyhowtheenergyofthereactantsvariesasthereactionproceeds. InFigure16.8aEndothermicReaction,theenergyislowearlyinthereaction,whenthemoleculesare

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175 stillarrangedasreactants.Asthemoleculesapproachandbegintorearrange,theenergyrisessharply, risingtoamaximuminthemiddleofthereaction.Thissharpriseinenergyistheactivationenergy,as illustrated.Afterthemiddleofthereactionhaspassedandthemoleculesarearrangedmoreasproducts thanreactants,theenergybeginstofallagain.However,theenergydoesnotfalltoitsoriginalvalue,so thisisanendothermicreaction. Figure16.8bExothermicReactionshowstheanalogoussituationforanexothermicreaction.Again, asthereactantsapproachoneanother,theenergyrisesastheatomsbegintorearrange.Atthemiddleof thecollision,theenergymaximizesandthenfallsastheproductmoleculesform.Inanexothermicreaction, theproductenergyislowerthanthereactantenergy. Figure16.8ReactionEnergythusshowsthatanenergybarriermustbesurmountedforthereaction tooccur,regardlessofwhethertheenergyoftheproductsisgreaterthanFigure16.8aEndothermic ReactionorlessthanFigure16.8bExothermicReactiontheenergyofthereactants.Thisbarrier accountsforthetemperaturedependenceofthereactionrate.Weknowfromthekineticmoleculartheory thatastemperatureincreasestheaverageenergyofthemoleculesinasampleincreases.Therefore,as temperatureincreases,thefractionofmoleculeswithsucientenergytosurmountthereactionactivation barrierincreases.

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176 CHAPTER16.REACTIONRATES ReactionEnergy aEndothermicReaction bExothermicReaction Figure16.8

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177 Althoughwewillnotshowithere,kineticmoleculartheoryshowsthatthefractionofmoleculeswith energygreaterthan E a attemperature T isproportionalto e )]TJ/F8 9.9626 Tf 6.227 -0.747 Td [( E a RT .Thismeansthatthereactionrateand thereforealsotherateconstantmustbeproportionalto e )]TJ/F8 9.9626 Tf 6.226 -0.747 Td [( E a RT .Thereforewecanwrite k T = Ae )]TJ/F8 9.9626 Tf 6.226 -0.747 Td [( E a RT .18 where A isaproportionalityconstant.Ifwetakethelogarithmofbothsidesof.18,wendthat ln k T = )]TJ/F1 9.9626 Tf 9.409 14.047 Td [( E a RT + lnA .19 Thisequationmatchestheexperimentallyobserved.17.Werecallthatagraphof lnk versus 1 T is observedtobelinear.Nowwecanseethattheslopeofthatgraphisequalto )]TJ/F1 9.9626 Tf 9.409 8.07 Td [()]TJ/F10 6.9738 Tf 5.762 -3.985 Td [(E a R Asanalnoteon.19,theconstant A musthavesomephysicalsignicant.Wehaveaccounted fortheprobabilityofcollisionbetweentwomoleculesandwehaveaccountedfortheenergeticrequirement forasuccessfulreactivecollision.Wehavenotaccountedfortheprobabilitythatacollisionwillhavethe appropriateorientationofreactantmoleculesduringthecollision.Moreover,noteverycollisionwhichoccurs withproperorientationandsucientenergywillactuallyresultinareaction.Thereareotherrandomfactors relatingtotheinternalstructureofeachmoleculeattheinstantofcollision.Thefactor A takesaccountfor allofthesefactors,andisessentiallytheprobabilitythatacollisionwithsucientenergyforreactionwill indeedleadtoreaction. A iscommonlycalledthe frequencyfactor 16.8Observation4:RateLawsforMoreComplicatedReactionProcesses Ourcollisionmodelintheprevioussectionaccountsfortheconcentrationandtemperaturedependenceof thereactionrate,asexpressedbytheratelaw.Theconcentrationdependencearisesfromcalculatingthe probabilityofthereactantmoleculesbeinginthesamevicinityatthesameinstant.Therefore,weshould beabletopredicttheratelawforanyreactionbysimplymultiplyingtogethertheconcentrationsofall reactantmoleculesinthebalancedstoichiometricequation.Theorderofthereactionshouldthereforebe simplyrelatedtothestoichiometriccoecientsinthereaction.However,RateLawsforVariousReactions, p.169showsthatthisisincorrectformanyreactions. Considerforexampletheapparentlysimplereaction 2 ICl g + H 2 g 2 HCl g + I 2 g .20 Basedonthecollisionmodel,wewouldassumethatthereactionoccursby 2 ICl moleculescollidingwith asingle H 2 molecule.Theprobabilityforsuchacollisionshouldbeproportionalto [ ICl ] 2 [ H 2 ] .However, experimentallyweobserveseeRateLawsforVariousReactions,p.169thattheratelawforthisreaction is Rate = k [ ICl ][ H 2 ] .21 Asasecondexample,considerthereaction NO 2 g + CO g NO g + CO 2 g .22 Itwouldseemreasonabletoassumethatthisreactionoccursasasinglecollisioninwhichanoxygenatom isexchangedbetweenthetwomolecules.However,theexperimentallyobservedratelawforthisreactionis Rate = k [ NO 2 ] 2 .23 Inthiscase,the [ CO ] concentrationdoesnotaecttherateofthereactionatall,andthe [ NO 2 ] concentration issquared.Theseexamplesdemonstratethattheratelawforareactioncannotbepredictedfromthe

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178 CHAPTER16.REACTIONRATES stoichiometriccoecientsandthereforethatthecollisionmodeldoesnotaccountfortherateofthereaction. Theremustbesomethingseriouslyincompletewiththecollisionmodel. Thekeyassumptionofthecollisionmodelisthatthereactionoccursbyasinglecollision.Sincethis assumptionleadstoincorrectpredictionsofratelawsinsomecases,theassumptionmustbeinvalidin atleastthosecases.Itmaywellbethatreactionsrequiremorethanasinglecollisiontooccur,evenin reactionsinvolvingjusttwotypesofmoleculesasin.22.Moreover,ifmorethantwomoleculesare involvedasin.20,thechanceofasinglecollisioninvolvingallofthereactivemoleculesbecomesvery small.Weconcludethatmanyreactions,includingthosein.20and.22,mustoccurasaresultof severalcollisionsoccurringinsequence,ratherthanasinglecollision.Therateofthechemicalreactionmust bedeterminedbytheratesoftheindividualstepsinthereaction. Eachstepinacomplexreactionisasinglecollision,oftenreferredtoasan elementaryprocess .In singlecollisionprocessstep,ourcollisionmodelshouldcorrectlypredicttherateofthatstep.Thesequence ofsuchelementaryprocessesleadingtotheoverallreactionisreferredtoasthe reactionmechanism Determiningthemechanismforareactioncanrequiregainingsubstantiallymoreinformationthansimply theratedatawehaveconsideredhere.However,wecangainsomeprogressjustfromtheratelaw. Considerforexamplethereactionin.22describedbytheratelawin.23.Sincetheratelaw involved [ NO 2 ] 2 ,onestepinthereactionmechanismmustinvolvethecollisionoftwo NO 2 molecules. Furthermore,thisstepmustdeterminetherateoftheoverallreaction.Whywouldthatbe?Inanymultistepprocess,ifonestepisconsiderablyslowerthanalloftheothersteps,therateofthemulti-stepprocess isdeterminedentirelybythatsloweststep,becausetheoverallprocesscannotgoanyfasterthantheslowest step.Itdoesnotmatterhowrapidlytherapidstepsoccur.Therefore,thesloweststepinamulti-step processisthuscalledthe ratedetermining or ratelimiting step. Thisargumentsuggeststhatthereactionin.22proceedsviaaslowstepinwhichtwo NO 2 molecules collide,followedbyatleastoneotherrapidstepleadingtotheproducts.Apossiblemechanismistherefore Step1 NO 2 + NO 2 NO 3 + NO .24 Step2 NO 3 + CO NO 2 + CO 2 .25 IfStep1.24:Step1ismuchslowerthanStep2.25:Step2,therateofthereactionisentirely determinedbytherateofStep1.24:Step1.Fromourcollisionmodel,theratelawforStep1.24: Step1mustbe Rate = k [ NO 2 ] 2 ,whichisconsistentwiththeexperimentallyobservedratelawforthe overallreaction.Thissuggeststhatthemechanismin.24Step1and.25Step2isthecorrect descriptionofthereactionprocessfor.22,withtherststepastheratedeterminingstep. Thereareafewimportantnotesaboutthemechanism.First,oneproductofthereactionisproduced intherststep,andtheotherisproducedinthesecondstep.Therefore,themechanismdoesleadtothe overallreaction,consumingthecorrectamountofreactantandproducingthecorrectamountofreactant. Second,therstreactionproducesanewmolecule, NO 3 ,whichisneitherareactantnoraproduct.The secondstepthenconsumesthatmolecule,and NO 3 thereforedoesnotappearintheoverallreaction,.22. Assuch, NO 3 iscalleda reactionintermediate .Intermediatesplayimportantrolesintheratesofmany reactions. Iftherststepinamechanismisratedeterminingasinthiscase,itiseasytondtheratelawforthe overallexpressionfromthemechanism.Ifthesecondsteporlaterstepsareratedetermining,determining theratelawisslightlymoreinvolved.Theprocessforndingtheratelawinsuchacaseisillustratedin Exercise16.11. 16.9ReviewandDiscussionQuestions Exercise16.1 When C 60 O 3 intoluenesolutiondecomposes, O 2 isreleasedleaving C 60 O 3 insolution.

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179 BasedonthedatainFigure16.2OxidizedBuckminsterfullereneAbsorbanceandFigure16.3 RateofDecomposition,plottheconcentrationof C 60 O asafunctionoftime. Howwouldyoudenetherateofthereactionintermsoftheslopeofthegraphfromabove Section16.9.1?Howistherateofappearanceof C 60 O relatedtotherateofdisappearanceof C 60 O 3 ?Basedonthis,plottherateofappearanceof C 60 O asafunctionoftime. Exercise16.2 Thereaction 2 N 2 O 5 g 4 NO 2 g + O 2 g wasfoundinthisstudytohaveratelawgivenby Rate = k [ N 2 O 5 ] with k =0 : 070 s )]TJ/F7 6.9738 Tf 6.227 0 Td [(1 Howistherateofappearanceof NO 2 relatedtotherateofdisappearanceof N 2 O 5 ?Which rateislarger? Basedontheratelawandrateconstant,sketchaplotof [ N 2 O 5 ] [ NO 2 ] ,and [ O 2 ] versustime allonthesamegraph. Exercise16.3 ForwhichofthereactionslistedinRateLawsforVariousReactions,p.169canyoubecertain thatthereactiondoesnotoccurasasinglestepcollision?Explainyourreasoning. Exercise16.4 Considertwodecompositionreactionsfortwohypotheticalmaterials,AandB.Thedecomposition ofAisfoundtoberstorder,andthedecompositionofBisfoundtobesecondorder. Assumingthatthetworeactionshavethesamerateconstantatthesametemperature,sketch [ A ] and [ B ] versustimeonthesamegraphforthesameinitialconditions, i.e. [ A ] 0 =[ B ] 0 Comparethehalf-livesofthetworeactions.Underwhatconditionswillthehalf-lifeofBbeless thanthehalf-lifeofA?Underwhatconditionswillthehalf-lifeofBbegreaterthanthehalf-lifeof A? Exercise16.5 Agraphofthelogarithmoftheequilibriumconstantforareactionversus 1 T islinearbutcan haveeitheranegativeslopeorapositiveslope,dependingonthereaction,aswasobservedhere Chapter14.However,thegraphofthelogarithmoftherateconstantforareactionversus 1 T hasanegativeslopeforessentiallyeveryreaction.Usingequilibriumarguments,explainwhythe graphfortherateconstantmusthaveanegativeslope. Exercise16.6 Using.18andthedatainRateConstantforHydrogenGasandIodineGas,p.171,determine theactivationenergyforthereaction H 2 g + I 2 g 2 HI g Exercise16.7 Wefoundthattheratelawforthereaction H 2 g + I 2 g 2 HI g is Rate = k [ H 2 ][ I 2 ] Therefore,thereactionis secondorder overallbut rstorder in H 2 .Imaginethatwestartwith [ H 2 ] 0 =[ I 2 ] 0 andwemeasure [ H 2 ] versustime.Willagraphof ln [ H 2 ] versustimebelinearorwill agraphof 1 [ H 2 ] versustimebelinear?Explainyourreasoning. Exercise16.8 Asaroughestimate,chemistsoftenassumea ruleofthumb thattherateofanyreactionwilldouble whenthetemperatureisincreasedby 10 C Whatdoesthissuggestabouttheactivationenergiesofreactions? Using.18,calculatetheactivationenergyofareactionwhoseratedoubleswhenthetemperatureisraisedfrom 25 C to 35 C Doesthisruleofthumbestimatedependonthetemperaturerange?Tondout,calculatethe factorbywhichtherateconstantincreaseswhenthetemperatureisraisedfrom 100 C to 110 C assumingthesameactivationenergyyoufoundaboveSection16.9.2.Doestheratedoublein thiscase? Exercise16.9 Consideraverysimplehypotheticalreaction A + B $ 2 C whichcomestoequilibrium.

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180 CHAPTER16.REACTIONRATES Atequilibrium,whatmustbetherelationshipbetweentherateoftheforwardreaction, A + B 2 C andthereversereaction 2 C A + B ? Assumethatboththeforwardandreversereactionsareelementaryprocessesoccurringbya singlecollision.Whatistheratelawfortheforwardreaction?Whatistheratelawforthereverse reaction? UsingthepreviousresultsfromhereSection16.9.1andhereSection16.9.2,showthatthe equilibriumconstantforthisreactioncanbecalculatedfrom K c = k f k r ,where k f istherateconstant fortheforwardreactionand k r istherateconstantforthereversereaction. Exercise16.10 Consideraverysimplehypotheticalreaction A + B $ C + D .ByexaminingFigure16.8Reaction Energy,provideandexplaintherelationshipbetweentheactivationenergyintheforwarddirection, E a;f ,andinthereversedirection, E a;r .Doesthisrelationshipdependonwhetherthereactionis endothermicFigure16.8aEndothermicReactionorexothermicFigure16.8bExothermic Reaction?Explain. Exercise16.11 Forthereaction H 2 g + I 2 g 2 HI g ,theratelawis Rate = k [ H 2 ][ I 2 ] .Althoughthissuggests thatthereactionisaone-stepelementaryprocess,thereisevidencethatthereactionoccursintwo steps,andthesecondstepistheratedeterminingstep: Step1 I 2 $ 2 I .26 Step2 H 2 +2 I 2 HI .27 WhereStep1.26:Step1isfastandStep2.27:Step2isslow. IftheboththeforwardandreversereactionsinStep1.26:Step1aremuchfasterthan Step2.27:Step2,explainwhyStep1.26:Step1canbeconsideredtobeatequilibrium. Whatistheratelawfortheratedeterminingstep? SincetheratelawaboveSection16.9.2dependsontheconcentrationofanintermediate I weneedtondthatintermediate.Calculate [ I ] fromStep1.26:Step1,assumingthatStep1 .26:Step1isatequilibrium. Substitute [ I ] fromaboveSection16.9.3intotheratelawfoundpreviouslySection16.9.2to ndtheoverallratelawforthereaction.Isthisresultconsistentwiththeexperimentalobservation?

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Chapter17 EquilibriumandtheSecondLawof Thermodynamics 1 17.1Foundation Wehaveobservedanddenedphasetransitionsandphaseequilibrium.Wehavealsoobservedequilibriumin avarietyofreactionsystems.Wewillassumeanunderstandingofthepostulatesofthe KineticMolecular Theory andoftheenergeticsofchemicalreactions. 17.2Goals Wehavedevelopedanunderstandingoftheconceptofequilibrium,bothforphaseequilibriumandreaction equilibrium.Asanillustration,atnormalatmosphericpressure,weexpecttond H 2 O insolidformbelow 0 C,inliquidformbelow100 C,andingaseousformabove100 C.Whatchangesaswemovefromlow temperaturetohightemperaturecausethesetransitionsinwhichphaseisobserved?Vieweddierently,ifa sampleofgaseouswaterat120 Ciscooledtobelow100 C,virtuallyallofthewatervaporspontaneously condensestoformtheliquid: H 2 O g H 2 O l spontaneousbelow100 C Bycontrast,verylittleofliquidwaterat80 Cspontaneouslyconvertstogaseouswater: H 2 O l H 2 O g notspontaneousbelow100 C Wecanthusrephraseourquestionas,whatdetermineswhichprocessesarespontaneousandwhicharenot? Whatfactorsdeterminewhatphaseis"stable"? Asweknow,atcertaintemperaturesandpressures,morethanonephasecanbestable.Forexample,at 1atmpressureand0 C, H 2 O s $ H 2 O l equilibriumat0 C Smallvariationsintheamountofheatappliedorextractedtotheliquid-solidequilibriumcauseshiftstowards liquidorsolidwithoutchangingthetemperatureofthetwophasesatequilibrium.Therefore,whenthetwo phasesareatequilibrium,neitherdirectionofthephasetransitionisspontaneousat0 C.Wethereforeneed tounderstandwhatfactorsdeterminewhentwoormorephasescanco-existatequilibrium. Thisanalysisleavesunansweredaseriesofquestionsregardingthedierencesbetweenliquidsandgases. Theconceptofagasphaseoraliquidphaseisnotacharacteristicofanindividualmolecule.Infact,itdoes 1 Thiscontentisavailableonlineat. 181

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182 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS notmakeanysensetorefertothe"phase"ofanindividualmolecule.Thephaseisacollectiveproperty oflargenumbersofmolecules.Althoughwecandiscusstheimportanceofmolecularpropertiesregarding liquidandgasphases,wehavenotdiscussedthefactorswhichdeterminewhetherthegasphaseortheliquid phaseismoststableatagiventemperatureandpressure. Thesesamequestionscanbeappliedtoreactionequilibrium.Whenamixtureofreactantsandproducts isnotatequilibrium,thereactionwilloccurspontaneouslyinonedirectionortheotheruntilthereaction achievesequilibrium.Whatdeterminesthedirectionofspontaneity?Whatisthedrivingforcetowards equilibrium?Howdoesthesystem know thatequilibriumhasbeenachieved?Ourgoalwillbetounderstand thedrivingforcesbehindspontaneousprocessesandthedeterminationoftheequilibriumpoint,bothfor phaseequilibriumandreactionequilibrium. 17.3Observation1:SpontaneousMixing Webeginbyexaminingcommoncharacteristicsofspontaneousprocesses,andforsimplicity,wefocuson processesnotinvolvingphasetransitionsorchemicalreactions.Averyclearexampleofsuchaprocessis mixing.Imagineputtingadropofblueinkinaglassofwater.Atrst,thebluedyeintheinkishighly concentrated.Therefore,themoleculesofthedyearecloselycongregated.Slowlybutsteadily,thedye beginstodiusethroughouttheentireglassofwater,sothateventuallythewaterappearsasauniformblue color.Thisoccursmorereadilywithagitationorstirringbutoccursspontaneouslyevenwithoutsucheort. Carefulmeasurementsshowthatthisprocessoccurswithoutachangeintemperature,sothereisnoenergy inputorreleasedduringthemixing. Weconcludethat,althoughthereisnoenergeticadvantagetothedyemoleculesdispersingthemselves, theydosospontaneously.Furthermore,thisprocessis irreversible inthesensethat,withoutconsiderable eortonourpart,thedyemoleculeswillneverreturntoformasinglelocalizeddrop.Wenowseekan understandingofhowandwhythismixingoccurs. Considerthefollowingratherabstractmodelforthedyemoleculesinthewater.Fortheglass,wetake arowoftensmallboxes,eachoneofwhichrepresentsapossiblelocationforamolecule,eitherofwateror ofdye.Forthemolecules,wetakemarbles,clearforwaterandblueforink.Eachboxwillaccommodate onlyasinglemarble,sincetwomoleculescannotbeinthesameplaceasthesametime.Sinceweseeadrop ofdyewhenthemoleculesarecongregated,wemodela"drop"asthreebluemarblesinconsecutiveboxes. Noticethatthereareonlyeightwaystohavea"drop"ofdye,assumingthatthethreedye"molecules"are indistinguishablefromoneanother.TwopossibilitiesareshowninFigure17.1aandFigure17.1b.Itis notdiculttondtheothersix.

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183 ArrangementofThreeInkMolecules a b c d Figure17.1: aAnunmixedstate.bAnotherunmixedstate.cAmixedstate.dAnother mixedstate. Bycontrast,therearemanymorewaystoarrangethedyemoleculessothattheydonotformadrop, i.e. sothatthethreemoleculesarenottogether.TwopossibilitiesareshowninFigure17.1candFigure17.1d. Thetotalnumberofsuchpossibilitiesis112.Thetotalnumberofallpossiblearrangementscanbecalculated asfollows:thereare10possiblelocationsfortherstbluemarble,9forthesecond,and8forthethird.This gives720possiblearrangements,butmanyoftheseareidentical,sincethemarblesareindistinguishable. Thenumberofduplicatesforeacharrangementis6,calculatedfromthreechoicesfortherstmarble,two forthesecond,andoneforthethird.Thetotalnumberofnon-identicalarrangementsofthemoleculesis 720 6 =120 .Weconcludethat,ifwerandomlyplacethe3marblesinthetrayof10boxes,thechancesare only8outof120or1outof15ofobservingadropofink. Now,inarealexperiment,therearemany,manytimesmoreinkmoleculesandmany,manytimesmore possiblepositionsforeachmolecule.Toseehowthiscomesintoplay,considerarowof500boxesand5blue marbles.The molefraction ofinkisthus0.01.Thetotalnumberofdistinctcongurationsoftheblue marblesintheseboxesisapproximately 2 10 11 .Thenumberofthesecongurationswhichhaveallve inkmarblestogetherinadropis496.Ifthearrangementsaresampledrandomly,thechancesofobserving adropofinkwithallvemoleculestogetherarethusaboutonein500million.Thepossibilitiesareremote evenforobservingapartial"droplet"consistingoffewerthanallvedyemolecules.Thechanceforfour ofthemoleculestobefoundtogetherisaboutonein800,000.Evenifwedeneadroplettobeonlythree moleculestogether,thechancesofobservingonearelessthanonein1600. Wecould,withsomediculty,calculatetheprobabilityforobservingadropofinkwhenthereare 10 23 molecules.However,itisreasonablydeducedfromoursmallcalculationsthattheprobabilityisessentially zerofortheinkmolecules,randomlydistributedintothewatermolecules,tobefoundtogether.Weconclude

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184 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS fromthisthatthereasonwhyweobserveinktodisperseinwateristhattheprobabilityisinnitesimally smallforrandomlydistributeddyemoleculestobecongregatedinadrop. Interestingly,however,whenwesetuptherealinkandwaterexperiment,wedidnotrandomlydistribute theinkmolecules.Rather,webeganinitiallywithadropofinkinwhichthedyemoleculeswerealready congregated.Weknowthat,accordingtoourkinetictheory,themoleculesareinconstantrandommotion. Therefore,theymustbeconstantlyrearrangingthemselves.Sincetheserandommotionsdonotenergetically favoranyonearrangementoveranyotheronearrangement,wecanassumethatallpossiblearrangements areequallyprobable.Sincemostofthearrangementsdonotcorrespondtoadropofink,then mostofthe time wewillnotobserveadrop.Inthecaseabovewithvebluemarblesin500boxes,weexpecttoseea droponlyonceinevery500milliontimeswelookatthe"glass".Inarealglassofwaterwitharealdropof ink,thechancesareverymuchsmallerthanthis. Wedrawtwoveryimportantconclusionsfromourmodel.First,therandommotionsofmoleculesmake everypossiblearrangementofthesemoleculesequallyprobable.Second,mixingoccursspontaneouslysimply becausetherearevastlymanymorearrangementswhicharemixedthanwhicharenot.Therstconclusion tellsus"how"mixingoccurs,andthesecondtellsus"why."Onthebasisoftheseobservations,wededucethe followingpreliminarygeneralization:aspontaneousprocessoccursbecauseitproducesthemostprobable nalstate. 17.4ProbabilityandEntropy Thereisasubtletyinourconclusiontobeconsideredinmoredetail.Wehaveconcludedthatallpossible arrangementsofmoleculesareequallyprobable.Wehavefurtherconcludedthatmixingoccursbecause thenalmixedstateisoverwhelminglyprobable.Placedtogether,thesestatementsappeartobeopenly contradictory.Toseewhytheyarenot,wemustanalyzethestatementscarefully.Byan"arrangement"of themolecules,wemeanaspecicationofthelocationofeachandeverymolecule.Wehaveassumedthat, duetorandommolecularmotion,eachsucharrangementisequallyprobable.Inwhatsense,then,isthe nalstate"overwhelminglyprobable"? RecallthesystemillustratedinFigure17.1ArrangementofThreeInkMolecules,whereweplaced threeidenticalbluemarblesintotenspaces.Wecalculatedbeforethatthereare120uniquewaystodo this.IfweaskfortheprobabilityofthearrangementinFigure17.1a,theansweristhus 1 120 .Thisisalso theprobabilityforeachoftheotherpossiblearrangements,accordingtoourmodel.However,ifwenow askinsteadfortheprobabilityofobservinga"mixed"statewithnodrop,theansweris 112 120 ,whereasthe probabilityofobservingan"unmixed"statewithadropisonly 8 120 .Clearly,theprobabilitiesarenotthe samewhenconsideringthelessspeciccharacteristics"mixed"and"unmixed". Inchemistrywearevirtuallyneverconcernedwith microscopic details,suchasthelocationsofspecic individualmolecules.Rather,weareinterestedinmoregeneralcharacteristics,suchaswhetherasystemis mixedornot,orwhatthetemperatureorpressureis.Thesepropertiesofinteresttousare macroscopic .As such,werefertoaspecicarrangementofthemoleculesasa microstate ,andeachgeneralstatemixedor unmixed,forexampleasa macrostate .Allmicrostateshavethesameprobabilityofoccurring,according toourmodel.Assuch,themacrostateshavewidelydieringprobabilities. Wecometoanimportantresult:theprobabilityofobservingaparticularmacrostate e.g. ,amixed stateisproportionaltothenumberofmicrostateswiththatmacroscopicproperty.Forexample,from Figure17.1ArrangementofThreeInkMolecules,thereare112arrangementsmicrostateswiththe "mixed"macroscopicproperty.Aswehavediscussed,theprobabilityofobservingamixedstateis 112 120 whichisobviouslyproportionalto112.Thus,onewaytomeasuretherelativeprobabilityofaparticular macrostateisbythenumberofmicrostates W correspondingtothatmacrostate. W standsfor"ways", i.e. thereare112"ways"togetamixedstateinFigure17.1ArrangementofThreeInkMolecules. Nowwerecallourconclusionthataspontaneousprocessalwaysproducestheoutcomewithgreatest probability.Since W measuresthisprobabilityforanysubstanceorsystemofinterest,wecouldpredict, using W ,whethertheprocessleadingfromagiveninitialstatetoagivennalstatewasspontaneousby simplycomparingprobabilitiesfortheinitialandnalstates.Forreasonsdescribedbelow,weinsteaddene

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185 afunctionof W S W = klnW .1 calledthe entropy ,whichcanbeusedtomakesuchpredictionsaboutspontaneity.The k isaproportionalityconstantwhichgives S appropriateunitsforourcalculations.Noticethatthemoremicrostatesthere are,thegreatertheentropyis.Therefore,amacrostatewithahighprobability e.g. amixedstatehasa largeentropy.Wenowmodifyourpreviousdeductiontosaythataspontaneousprocessproducesthenal stateofgreatestentropy.Followingmodicationsaddedbelow,thisstatementformsthe SecondLawof Thermodynamics Itwouldseemthatwecoulduse W forourcalculationsandthatthedenitionofthenewfunction S is unnecessary.However,thefollowingreasoningshowsthat W isnotaconvenientfunctionforcalculations. Weconsidertwoidenticalglassesofwateratthesametemperature.Weexpectthatthevalueofanyphysical propertyforthewaterintwoglassesistwicethevalueofthatpropertyforasingleglass.Forexample,if theenthalpyofthewaterineachglassis H 1 ,thenitfollowsthatthetotalenthalpyofthewaterinthetwo glassestogetheris H total =2 H 1 .Thus,theenthalpyofasystemisproportionaltothequantityofmaterial inthesystem:ifwedoubletheamountofwater,wedoubletheenthalpy.Indirectcontrast,weconsider thecalculationinvolving W forthesetwoglassesofwater.Thenumberofmicrostatesofthemacroscopic stateofoneglassofwateris W 1 ,andlikewisethenumberofmicrostatesinthesecondglassofwateris W 1 However,ifwecombinethetwoglassesofwater,thenumberofmicrostatesofthetotalsystemisfound fromtheproduct W total = W 1 W 1 ,whichdoesnotequal 2 W 1 .Inotherwords, W isnotproportionalto thequantityofmaterialinthesystem.Thisisinconvenient,sincethevalueof W thusdependsonwhether thetwosystemsarecombinedornot.Ifitisnotclearthatweshouldmultiplythe W values,considerthe simpleexampleofrollingdice.Thenumberofstatesforasingledieis6,butfortwodicethenumberis 6=36 ,not 6+6=12 Wethereforeneedanewfunction S W ,sothat,whenwecombinethetwoglassesofwater, S total = S 1 + S 1 .Since S total = S W total S 1 = S W 1 ,and W total = W 1 W 1 ,thenournewfunction S must satisfytheequation S W 1 W 1 = S W 1 + S W 1 Theonlyfunction S whichwillsatisfythisequationisthelogarithmfunction,whichhasthepropertythat ln x y = lnx + lny .Weconcludethatanappropriatestatefunctionwhichmeasuresthenumberof microstatesinaparticularmacrostateis.1. 17.5Observation2:AbsoluteEntropies Itispossible,thoughexceedinglydicult,tocalculatetheentropyofanysystemunderanyconditionsof interestfromtheequation S = klnW .Itisalsopossible,usingmoreadvancedtheoreticalthermodynamics, todetermine S experimentallybymeasuringheatcapacitiesandenthalpiesofphasetransitions.Valuesof S determinedexperimentally,oftenreferredtoas"absolute"entropies,havebeentabulatedformanymaterials atmanytemperatures,andafewexamplesaregiveninAbsoluteEntropiesofSpecicSubstances,p.185. Wetreatthesevaluesasobservationsandattempttounderstandtheseinthecontextof.1.

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186 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS AbsoluteEntropiesofSpecicSubstances T C S J mol C H 2 O g 25 188.8 H 2 O l 25 69.9 H 2 O l 0 63.3 H 2 O s 0 41.3 NH 3 g 25 192.4 HN 3 l 25 140.6 HN 3 g 25 239.0 O 2 g 25 205.1 O 2 g 50 207.4 O 2 g 100 211.7 CO g 25 197.7 CO g 50 200.0 CO 2 g 24 213.7 CO 2 g 50 216.9 Br 2 l 25 152.2 Br 2 g 25 245.5 I 2 s 25 116.1 I 2 g 25 260.7 CaF 2 s 25 68.9 CaCl 2 s 25 104.6 CaBr 2 s 25 130 C 8 H 18 s 25 361.1 ThereareseveralinterestinggeneralitiesobservedinAbsoluteEntropiesofSpecicSubstances,p.185. First,incomparingtheentropyofthegaseousformofasubstancetoeitheritsliquidorsolidformatthe sametemperature,wendthatthegasalwayshasasubstantiallygreaterentropy.Thisiseasytounderstand from.1:themoleculesinthegasphaseoccupyaverymuchlargervolume.Thereareverymanymore possiblelocationsforeachgasmoleculeandthusverymanymorearrangementsofthemoleculesinthegas. Itisintuitivelyclearthat W shouldbelargerforagas,andthereforetheentropyofagasisgreaterthan thatofthecorrespondingliquidorsolid. Second,weobservethattheentropyofaliquidisalwaysgreaterthanthatofthecorrespondingsolid. Thisisunderstandablefromourkineticmolecularviewofliquidsandsolids.Althoughthemoleculesinthe liquidoccupyacomparablevolumetothatofthemoleculesinthesolid,eachmoleculeintheliquidisfreeto movethroughoutthisentirevolume.Themoleculesinthesolidarerelativelyxedinlocation.Therefore, thenumberofarrangementsofmoleculesintheliquidissignicantlygreaterthanthatinthesolid,sothe liquidhasgreaterentropyby.1. Third,theentropyofasubstanceincreaseswithincreasingtemperature.Thetemperatureis,ofcourse,a measureoftheaveragekineticenergyofthemolecules.Inasolidorliquid,then,increasingthetemperature increasesthetotalkineticenergyavailabletothemolecules.Thegreatertheenergy,themorewaysthere aretodistributethisenergyamongstthemolecules.Althoughwehavepreviouslyonlyreferredtotherange

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187 ofpositionsforamoleculeasaecting W ,therangeofenergiesavailableforeachmoleculesimilarlyaects W .Asaresult,asweincreasethetotalenergyofasubstance,weincrease W andthustheentropy. Fourth,theentropyofasubstancewhosemoleculescontainmanyatomsisgreaterthanthatofasubstance composedofsmallermolecules.Themoreatomsthereareinamolecule,themorewaystherearetoarrange thoseatoms.Withgreaterinternalexibility, W islargerwhentherearemoreatoms,sotheentropyis greater. Fifth,theentropyofasubstancewithahighmolecularweightisgreaterthanthatofsubstancewithalow molecularweight.Thisresultisahardertounderstand,asitarisesfromthedistributionofthemomentaof themoleculesratherthanthepositionsandenergiesofthemolecules.Itisintuitivelyclearthatthenumber ofarrangementsofthemoleculesis not aectedbythemassofthemolecules.However,evenatthesame temperature,therangeofmomentaavailableforaheaviermoleculeisgreaterthanforalighterone.To seewhy,recallthatthemomentumofamoleculeis p = mv andthekineticenergyis KE = mv 2 2 = p 2 2 m Therefore,themaximummomentumavailableataxedtotalkineticenergy KE is p = p 2 KE .Sincethis islargerforlargermassmolecules,therangeofmomentaisgreaterforheavierparticles,thusincreasing W andtheentropy. 17.6Observation3:CondensationandFreezing Wehaveconcludedfromourobservationsofspontaneousmixingthataspontaneousprocessalwaysproduces thenalstateofgreatestprobability.Afewsimpleobservationsrevealthatourdeductionneedssome thoughtfulrenement.Forexample,wehaveobservedthattheentropyofliquidwaterisgreaterthanthat ofsolidwater.Thismakessenseinthecontextof.1,sincethekinetictheoryindicatesthatliquidwater hasagreatervalueof W .Nevertheless,weobservethatliquidwaterspontaneouslyfreezesattemperatures below0 C.Thisprocessclearlydisplaysadecreaseinentropyandthereforeevidentlyashiftfromamore probablestatetoalessprobablestate.Thisappearstocontradictdirectlyourconclusion. Similarly,weexpecttondcondensationofwaterdropletsfromsteamwhensteamiscooled.Ondays ofhighhumidity,waterspontaneouslyliqueesfromtheaironcoldsurfacessuchastheoutsideofaglass oficewaterorthewindowofanairconditionedbuilding.Inthesecases,thetransitionfromgastoliquidis clearlyfromahigherentropyphasetoalowerentropyphase,whichdoesnotseemtofollowourreasoning thusfar. Ourpreviousconclusionsconcerningentropyandprobabilityincreaseswerecompelling,however,andwe shouldbereluctanttoabandonthem.Whatwehavefailedtotakeintoconsiderationisthatthesephase transitionsinvolvechangesofenergyandthusheatow.Condensationofgastoliquidandfreezingofliquid tosolidbothinvolveevolutionofheat.Thisheatowisofconsequencebecauseourobservationsalsorevealed thattheentropyofasubstancecanbeincreasedsignicantlybyheating.Onewaytopreserveourconclusions aboutspontaneityandentropyistoplaceaconditionontheirvalidity:aspontaneousprocessproducesthe nalstateofgreatestprobabilityandentropy providedthat theprocessdoesnotinvolveevolutionofheat. Thisisanunsatisfyingresult,however,sincemostphysicalandchemicalprocessesinvolveheattransfer.As analternative,wecanforcetheprocessnottoevolveheatby isolating thesystemundergoingtheprocess: noheatcanbereleasedifthereisnosinktoreceivetheheat,andnoheatcanbeabsorbedifthereisno sourceofheat.Therefore,weconcludefromourobservationsthataspontaneousprocess inanisolated system producesthenalstateofgreatestprobabilityandentropy.Thisisonestatementofthe Second LawofThermodynamics 17.7FreeEnergy HowcantheSecondLawbeappliedtoaprocessinasystemthatisnotisolated?Onewaytoviewthe lessonsofthepreviousobservationsisasfollows:inanalyzingaprocesstounderstandwhyitisorisnot spontaneous,wemustconsiderboththechangeinentropyofthesystemundergoingtheprocess and the eectoftheheatreleasedorabsorbedduringtheprocessontheentropyofthesurroundings.Althoughwe

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188 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS cannotproveithere,theentropyincreaseofasubstanceduetoheat q attemperature T isgivenby S = q T FromanotherstudyChapter10,wecancalculatetheheattransferforaprocessoccurringunderconstant pressurefromtheenthalpychange, H .Byconservationofenergy,theheatowintothesurroundings mustbe )]TJ/F8 9.9626 Tf 9.409 0 Td [( H .Therefore,theincreaseintheentropyofthesurroundingsduetoheattransfermustbe S surr = )]TJ/F1 9.9626 Tf 9.409 8.069 Td [()]TJ/F7 6.9738 Tf 5.761 -4.147 Td [( H T .Noticethat,ifthereactionisexothermic, H< 0 so S surr > 0 AccordingtoourstatementoftheSecondLaw,aspontaneousprocessinanisolatedsystemisalways accompaniedbyanincreaseintheentropyofthesystem.Ifwewanttoapplythisstatementtoanonisolatedsystem,wemustincludethesurroundingsinourentropycalculation.Wecansaythenthat,fora spontaneousprocess, S total = S sys + S surr > 0 Since S surr = )]TJ/F1 9.9626 Tf 9.409 8.07 Td [()]TJ/F7 6.9738 Tf 5.762 -4.148 Td [( H T ,thenwecanwritethat S )]TJ/F7 6.9738 Tf 11.183 3.923 Td [( H T > 0 .Thisiseasilyrewrittentostatethat,fora spontaneousprocess: H )]TJ/F11 9.9626 Tf 9.963 0 Td [(T S< 0 .2 .2isreallyjustadierentformoftheSecondLawofThermodynamics.However,thisformhas theadvantagethatittakesintoaccounttheeectsonboththesystemundergoingtheprocessandthe surroundings.Thus,thisnewformcanbeappliedtonon-isolatedsystems. .2revealswhythetemperatureaectsthespontaneityofprocesses.Recallthatthecondensationof watervaporoccursspontaneouslyattemperaturebelow100 Cbutnotabove.Condensationisanexothermic process;toseethis,considerthatthereverseprocess,evaporation,obviouslyrequiresheatinput.Therefore H< 0 forcondensation.However,condensationclearlyresultsinadecreaseinentropy,therefore S< 0 also.Examining.2,wecanconcludethat H )]TJ/F11 9.9626 Tf 9.527 0 Td [(T S< 0 willbelessthanzeroforcondensationonlyif thetemperatureisnottoohigh.Athightemperature,theterm )]TJ/F8 9.9626 Tf 9.41 0 Td [( S ,whichispositive,becomeslarger than H ,so H )]TJ/F11 9.9626 Tf 9.517 0 Td [(T S> 0 forcondensationathightemperature.Therefore,condensationonlyoccursat lowertemperatures. Becauseoftheconsiderablepracticalutilityof.2inpredictingthespontaneityofphysicalandchemicalprocesses,itisdesirabletosimplifythecalculationofthequantityontheleftsideoftheinequality.One waytodothisistodeneanewquantity G = H )]TJ/F11 9.9626 Tf 10.117 0 Td [(TS ,calledthe freeenergy .Ifwecalculatefromthis denitionthechangeinthefreeenergywhichoccursduringaprocessatconstanttemperature,weget G = G final )]TJ/F11 9.9626 Tf 9.963 0 Td [(G initial = H final )]TJ/F11 9.9626 Tf 9.963 0 Td [(TS final )]TJ/F8 9.9626 Tf 9.963 0 Td [( H initial )]TJ/F11 9.9626 Tf 9.963 0 Td [(TS initial = H )]TJ/F11 9.9626 Tf 9.963 0 Td [(T S andthereforeasimpliedstatementoftheSecondLawofThermodynamicsin.2isthat G< 0 .3 foranyspontaneousprocess.Thus,inanyspontaneousprocess,thefreeenergyofthesystemdecreases. Notethat G isastatefunction,sinceitisdenedintermsof H T ,and S ,allofwhicharestatefunctions. Since G isastatefunction,then G canbecalculatedalonganyconvenientpath.Assuch,themethods usedtocalculate H inanotherstudyChapter10canbeusedjustaswelltocalculate G 17.8ThermodynamicDescriptionofPhaseEquilibrium Aswerecall,theentropyofvaporismuchgreaterthantheentropyofthecorrespondingamountofliquid. AlookbackatAbsoluteEntropiesofSpecicSubstances,p.185showsthat,at25 C,theentropyofone moleofliquidwateris 69 : 9 J K ,whereastheentropyofonemoleofwatervaporis 188 : 8 J K .Ourrstthought, basedonourunderstandingofspontaneousprocessesandentropy,mightwellbethatamoleofliquidwater at25 Cshouldspontaneouslyconvertintoamoleofwatervapor,sincethisprocesswouldgreatlyincrease theentropyofthewater.Weknow,however,thatthisdoesnothappen.Liquidwaterwillexistinaclosed containerat25 Cwithoutspontaneouslyconvertingentirelytovapor.Whathaveweleftout?

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189 Theanswer,basedonourdiscussionoffreeenergy,istheenergyassociatedwithevaporation.The conversionofonemoleofliquidwaterintoonemoleofwatervaporresultsinabsorptionof 44 : 0 kJ ofenergy fromthesurroundings.Recallthatthislossofenergyfromthesurroundingsresultsinasignicantdecrease inentropyofthesurroundings.Wecancalculatetheamountofentropydecreaseinthesurroundingsfrom S surr = )]TJ/F1 9.9626 Tf 9.409 8.069 Td [()]TJ/F7 6.9738 Tf 5.762 -4.147 Td [( H T .At25 C,thisgives S surr = )]TJ/F7 6.9738 Tf 6.226 0 Td [(44 : 0 kJ 298 : 15 K = )]TJ/F8 9.9626 Tf 7.749 0 Td [(147 : 57 J K forasinglemole.Thisentropy decreaseisgreaterthantheentropyincreaseofthewater, 188 : 8 J K )]TJ/F8 9.9626 Tf 10.85 0 Td [(69 : 9 J K =118 : 9 J K .Therefore,the entropyoftheuniverse decreases whenonemoleofliquidwaterconvertstoonemoleofwatervaporat 25 C. Wecanrepeatthiscalculationintermsofthefreeenergychange: G = H )]TJ/F11 9.9626 Tf 9.963 0 Td [(T S G =44000 J mol )]TJ/F8 9.9626 Tf 9.962 0 Td [( : 15 K 118 : 9 J Kmol G =8 : 55 kJ mol > 0 Sincethefreeenergyincreasesinthetransformationofonemoleofliquidwatertoonemoleofwatervapor, wepredictthatthetransformationwillnotoccurspontaneously.Thisissomethingofarelief,becausewe havecorrectlypredictedthatthemoleofliquidwaterisstableat25 Crelativetothemoleofwatervapor. Wearestillfacedwithourperplexingquestion,however.Whydoesanywaterevaporateat25 C?How canthisbeaspontaneousprocess? Theansweristhatwehavetobecarefulaboutinterpretingourprediction.Theentropyofonemoleof watervaporat25 C and1.00atmpressure is 188 : 8 J K .Weshouldclarifyourpredictiontosaythatone moleofliquidwaterwillnotspontaneouslyevaporatetoformonemoleofwatervaporat25 Cand1.00 atmpressure.Thispredictionisinagreementwithourobservation,becausewehavefoundthatthewater vaporformedspontaneouslyaboveliquidwaterat25 Chaspressure23.8torr,wellbelow1.00atm. Assumingthatourreasoningiscorrect,thenthespontaneousevaporationofwaterat25 Cwhen no watervaporispresentinitiallymusthave G< 0 .And,indeed,aswatervaporformsandthepressureof thewatervaporincreases,evaporationmustcontinueaslongas G< 0 .Eventually,evaporationstopsina closedsystemwhenwereachthevaporpressure,sowemustreachapointwhere G isnolongerlessthan zero,thatis,evaporationstopswhen G =0 .Thisisthepointwherewehaveequilibriumbetweenliquid andvapor. Wecanactuallydeterminetheconditionsunderwhichthisistrue.Since G = H )]TJ/F11 9.9626 Tf 10.789 0 Td [(T S ,then when G =0 H = T S .Wealreadyknowthat H =44 : 0 kJ fortheevaporationofonemoleofwater. Therefore,thepressureofwatervaporatwhich G =0 at25 Cisthepressureatwhich S = H T =147 : 6 J K forasinglemoleofwaterevaporating.Thisislargerthanthevalueof S foronemoleand1.00atmpressure ofwatervapor,whichaswecalculatedwas 118 : 9 J K .Evidently, S forevaporationchangesasthepressure ofthewatervaporchanges.Wethereforeneedtounderstandwhytheentropyofthewatervapordepends onthepressureofthewatervapor. Recallthat1moleofwatervaporoccupiesamuchsmallervolumeat1.00atmofpressurethanitdoesat theconsiderablylowervaporpressureof23.8torr.Inthelargervolumeatlowerpressure,thewatermolecules haveamuchlargerspacetomovein,andthereforethenumberofmicrostatesforthewatermoleculesmust belargerinalargervolume.Therefore,theentropyofonemoleofwatervaporislargerinalargervolumeat lowerpressure.Theentropychangeforevaporationofonemoleofwateristhusgreaterwhentheevaporation occurstoalowerpressure.Withagreaterentropychangetoosettheentropylossofthesurroundings,it ispossiblefortheevaporationtobespontaneousatlowerpressure.Andthisisexactlywhatweobserve. Tondouthowmuchtheentropyofagaschangesaswedecreasethepressure,weassumethatthe numberofmicrostates W forthegasmoleculeisproportionaltothevolume V .Thiswouldmakesense,

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190 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS becausethelargerthevolume,themoreplacesthereareforthemoleculestobe.Sincetheentropyisgiven by S = klnW ,then S mustalsobeproportionalto lnV .Therefore,wecansaythat S V 2 )]TJ/F11 9.9626 Tf 9.963 0 Td [(S V 1 = RlnV 2 )]TJ/F11 9.9626 Tf 9.963 0 Td [(RlnV 1 = Rln V 2 V 1 .4 Weareinterestedinthevariationof S withpressure,andwerememberfromBoyle'slawthat,foraxed temperature,volumeisinverselyrelatedtopressure.Thus,wendthat S P 2 )]TJ/F11 9.9626 Tf 9.963 0 Td [(S P 1 = Rln P 1 P 2 = )]TJ/F1 9.9626 Tf 9.409 11.059 Td [( Rln P 2 P 1 .5 Forwatervapor,weknowthattheentropyat1.00atmpressureis 188 : 8 J K foronemole.Wecanuse thisandtheequationabovetodeterminetheentropyatanyotherpressure.Forapressureof 23 : 8 torr = 0 : 0313 atm ,thisequationgivesthat S : 8 torr is 217 : 6 J K foronemoleofwatervapor.Therefore,atthis pressure,the S forevaporationofonemoleofwatervaporis 217 : 6 J K )]TJ/F8 9.9626 Tf 10.502 0 Td [(69 : 9 J K =147 : 6 J K .Wecanuse thistocalculatethatforevaporationofonemoleofwaterat25 Candwatervaporpressureof23.8torr is G = H )]TJ/F11 9.9626 Tf 10.613 0 Td [(T S =44 : 0 kJ )]TJ/F8 9.9626 Tf 10.613 0 Td [( : 15 K )]TJ/F8 9.9626 Tf 4.566 -8.07 Td [(147 : 6 J K =0 : 00 kJ .Thisistheconditionweexpectedfor equilibrium. Wecanconcludethattheevaporationofwaterwhennovaporispresentinitiallyisaspontaneousprocess with G< 0 ,andtheevaporationcontinuesuntilthewatervaporhasreacheditstheequilibriumvapor pressure,atwhichpoint G =0 17.9Thermodynamicdescriptionofreactionequilibrium Havingdevelopedathermodynamicunderstandingofphaseequilibrium,itprovestobeevenmoreuseful toexaminethethermodynamicdescriptionofreactionequilibriumtounderstandwhythereactantsand productscometoequilibriumatthespecicvaluesthatareobserved. Recallthat G = H )]TJ/F11 9.9626 Tf 9.343 0 Td [(T S< 0 foraspontaneousprocess,and G = H )]TJ/F11 9.9626 Tf 9.343 0 Td [(T S =0 atequilibrium. Fromtheserelations,wewouldpredictthatmostbutnotallexothermicprocesseswith H< 0 are spontaneous,becauseallsuchprocessesincreasetheentropyofthesurroundingswhentheyoccur.Similarly, wewouldpredictthatmostbutnotallprocesseswith S> 0 arespontaneous. Wetryapplyingtheseconclusionstosynthesisofammonia N 2 g +3 H 2 g 2 NH 3 g .6 at298K,forwhichwendthat S = )]TJ/F8 9.9626 Tf 7.748 0 Td [(198 J molK .Notethat S < 0 becausethereactionreduces thetotalnumberofgasmoleculesduringammoniasynthesis,thusreducing W ,thenumberofwaysof arrangingtheatomsinthesemolecules. S < 0 suggeststhat.6shouldnotoccuratall.However, H = )]TJ/F8 9.9626 Tf 7.749 0 Td [(92 : 2 kJ mol .Overall,wendthat G = )]TJ/F8 9.9626 Tf 7.749 0 Td [(33 : 0 kJ mol at298K,whichaccordingto.3suggeststhat .6isspontaneous. Giventhisanalysis,wearenowpressedtoask,if.6ispredictedtobespontaneous,whydoesthe reactioncometoequilibriumwithoutfullyconsumingallofthereactants?Theanswerliesinamorecareful examinationofthevaluesgiven: S H ,and G arethevaluesforthisreaction atstandardconditions whichmeansthatallofthegasesinthereactantsandproductsaretakentobeat1atmpressure.Thus, thefactthat G < 0 for.6atstandardconditionsmeansthat,ifallthreegasesarepresentat1atm pressure,thereactionwillspontaneouslyproduceanincreaseintheamountof NH 3 .Notethatthiswill reducethepressureofthe N 2 and H 2 andincreasethepressureofthe NH 3 .Thischangesthevalueof S andthusof G ,becauseaswealreadyknowtheentropiesofallthreegasesdependontheirpressures.As thepressureof NH 3 increases,itsentropydecreases,andasthepressuresofthereactantsgasesdecrease,

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191 theirentropiesincrease.Theresultisthat S becomesincreasinglynegative.Thereactioncreatesmore NH 3 untilthevalueof S issucientlynegativethat G = H )]TJ/F11 9.9626 Tf 9.962 0 Td [(T S =0 Fromthisanalysis,wecansaybylookingat S H ,and G that,since G < 0 for.6, reactionequilibriumresultsinproductionofmoreproductandlessreactantthanatstandardconditions. Moreover,themorenegative G is,themorestronglyfavoredaretheproductsoverthereactantsat equilibrium.Bycontrast,themorepositive G is,themorestronglyfavoredarethereactantsoverthe productsatequilibrium. 17.10ThermodynamicDescriptionoftheEquilibriumConstant Thermodynamicscanalsoprovideaquantitativeunderstandingoftheequilibriumconstant.Recallthatthe conditionforequilibriumisthat G =0 .Asnotedbefore, G dependsonthepressuresofthegasesin thereactionmixture,because S dependsonthesepressures.Thoughwewillnotproveithere,itcanbe shownbyapplicationof.5toareactionthattherelationshipbetween G andthepressuresofthegases isgivenbythefollowingequation: G = G + RTlnQ .7 Recallagainthatthesuperscript referstostandardpressureof1atm. G isthedierencebetweenthe freeenergiesoftheproductsandreactantswhenallgasesareat1atmpressure.Inthisequation, Q isa quotientofpartialpressuresofthegasesinthereactionmixture.Inthisquotient,eachproductgasappears inthenumeratorwithanexponentequaltoitsstoichiometiccoecient,andeachreactantgasappearsin thedenominatoralsowithitscorrespondingexponent.Forexample,forthereaction H 2 g + I 2 g 2 HI g .8 Q = P HI 2 P H 2 P I 2 .9 Itisimportanttonotethatthepartialpressuresin Q neednotbetheequilibriumpartialpressures.However, ifthepressuresin Q aretheequilibriumpartialpressures,then Q hasthesamevalueas K p ,theequilibrium constant,bydenition.Moreover,ifthepressuresareatequilibrium,weknowthat G =0 .Ifwelookback at.7,wecanconcludethat G = )]TJ/F8 9.9626 Tf 9.409 0 Td [( RTlnK p .10 Thisisanexceptionallyimportantrelationship,becauseitrelatestwoverydierentobservations.To understandthissignicance,considerrstthecasewhere G < 0 .Wehavepreviouslyreasonedthat,in thiscase,thereactionequilibriumwillfavortheproducts.From.10wecannotethat,if G < 0 ,it mustbethat K p > 1 .Furthermore,if G isalargenegativenumber, K p isaverylargenumber.By contrast,if G isalargepositivenumber, K p willbeaverysmallthoughpositivenumbermuchless than1.Inthiscase,thereactantswillbestronglyfavoredatequilibrium. Notethatthethermodynamicdescriptionofequilibriumandthedynamicdescriptionofequilibrium arecomplementary.Bothpredictthesameequilibrium.Ingeneral,thethermodynamicargumentsgiveus anunderstandingoftheconditionsunderwhichequilibriumoccurs,andthedynamicargumentshelpus understandhowtheequilibriumconditionsareachieved. 17.11ReviewandDiscussionQuestions Exercise17.1 Eachpossiblesequenceofthe52cardsinadeckisequallyprobable.However,whenyoushuea deckandthenexaminethesequence,thedeckisneverordered.Explainwhyintermsofmicrostates, macrostates,andentropy.

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192 CHAPTER17.EQUILIBRIUMANDTHESECONDLAWOF THERMODYNAMICS Exercise17.2 Assessthevalidityofthestatement,"Inallspontaneousprocesses,thesystemmovestowarda stateoflowestenergy."Correctanyerrorsyouidentify. Exercise17.3 Ineachcase,determinewhetherspontaneityisexpectedatlowtemperature,hightemperature, anytemperature,ornotemperature: H > 0 S > 0 H < 0 S > 0 H > 0 S < 0 H < 0 S < 0 Exercise17.4 Usingthermodynamicequilibriumarguments,explainwhyasubstancewithweakerintermolecular forceshasagreatervaporpressurethanonewithstrongerintermolecularforces. Exercise17.5 Whydoestheentropyofagasincreaseasthevolumeofthegasincreases?Whydoestheentropy decreaseasthepressureincreases? Exercise17.6 Foreachofthefollowingreactions,calculatethevaluesof S H ,and G at T =298 K andusethesetopredictwhetherequilibriumwillfavorproductsorreactantsat T =298 K .Also calculate K p 2 CO g + O 2 g 2 CO 2 g O 3 g + NO g NO 2 g + O 2 g 2 O 3 g 3 O 2 g Exercise17.7 Predictthesignoftheentropyforthereaction 2 H 2 g + O 2 g 2 H 2 O g Giveanexplanation,basedonentropyandtheSecondLaw,ofwhythisreactionoccursspontaneously. Exercise17.8 Forthereaction H 2 g 2 H g ,predictthesignofboth H and S .Shouldthisreaction bespontaneousathightemperatureoratlowtemperature?Explain. Exercise17.9 ForeachofthereactionsinExercise17.6,predictwhetherincreasesintemperaturewillshiftthe reactionequilibriummoretowardsproductsormoretowardsreactants. Exercise17.10 Using.7and.9,showthatforagivensetofinitialpartialpressureswhere Q islargerthan K p ,thereactionwillspontaneouslycreatemorereactants.Alsoshowthatif Q issmallerthan K p thereactionwillspontaneouslycreatemoreproducts.

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INDEX 193 IndexofKeywordsandTerms Keywords arelistedbythesectionwiththatkeywordpagenumbersareinparentheses.Keywords donotnecessarilyappearinthetextofthepage.Theyaremerelyassociatedwiththatsection. Ex. apples,1.1 Terms arereferencedbythepagetheyappearon. Ex. apples,1 A absolutezero,11,105 absorbance,162 acidionization,150 acidionizationequilibriumconstant, 15,153 Activationenergy,16,174 alcohol,6,43 amine,6 amines,43 Amonton'sLaw,11,109 anti-bondingorbital,9 Arrhenius,16 Arrheniusplot,173 atom,2,6 AtomicMolecularTheory,2,6,13, 123 atomicnumber,4,18 Atomic-MolecularTheory,3,9 autoionization,15,154 Avogadro'sHypothesis,3,10,11, 102,13,125 B binaryacids,15,156 boilingpoint,13,125 bondenergy,6,41,10,94 bondlength,6,41 bondstrength,6,41 bondingorbital,9 Boyle'sLaw,11,102,12,112, 116,116 C calorimetry,10,90 carboxylicacids,15,156 Charles'Law,11,106,12,112, 13,124 ChemicalConcepts,1 ChemicalKinetics,16 Chemicalreactionhalf-life,16 compounds,2,3 ConceptDevelopmentStudiesinChemistry,1 connementenergy,5,33 covalentbond,6,38 D Dalton'sLawofPartialPressures,11, 108,14,145 diatomic,7,51 dimerization,14,137 dipolemoment,8,65 dispersionforces,12,120 domain,58 doublebond,6,41 dynamicequilibrium,13,132 E ED,7 ElectronDomainEDTheory,56 electrondomainmodel,8,63 ElectronDomainTheory,7 electronorbital,5,32 electronegativity,8,66 elementaryprocess,178 elements,2,3 empiricalformula,3,13 endothermic,10,91 energylevels,5,29 enthalpy,10,92 entropy,17,185 equilibrium,127,15,149 ether,6 Ethers,43 exothermic,10,91 expandedvalence,7,57 F rstorderreaction,168 formula,2,5 Foundation,1 freeenergy,17,188 frequencyfactor,177 G GeneralChemistry,1 H halflife,170 heatcapacity,10,90 Hess'Law,10,91 hybridorbital,9 hydrogenbonding,12,120

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194 INDEX hydrolysis,15,156 I IdeaGasLaw,3 IdealGasLaw,13,107,12,111,111, 111,112,112,112,112,113,113,113,113, 114,116,116,116,117,117,117,117,117, 121,14,135 IntroductoryChemistry,1 ionizationenergy,4,19 irreversible,17,182 isomer,6 isomers,43 K KineticMolecularTheory,117,13,123, 14,135,161,17,181 KineticMolecularTheoryofGases,12, 114 L LawofCombiningVolumes,3,10, 12,112,116 LawofConservationofMass,2,3,3, 9 LawofDeniteProportions,2,4,3,9 LawofMassAction,14,141 LawofMultipleProportions,2,6,3, 9 LeChatelier'sPrinciple,14,15 LeChtelier'sPrinciple,14,145, 15,149 Lewisstructure,8,63 Lewisstructuremodel,7,51 Lewisstructures,6,39 Londonforces,12,120 lonepairs,751,54 M macroscopic,17,184 macrostate,17,184 microscopic,17,184 microstate,17,184 mole,3,13 molefraction,11,109,17,183 molecularformula,3,13 molecularorbitaltheory,9 N nucleus,415,17 O observations,1 Occam'srazor,2 octahedron,7,58 orbital,5,29 oxyacids,15,156 P partialpressure,11,108 particulate,2,6 phasediagram,133,130 phaseequilibrium,13,127 phasetransition,13,124 photoionization,5,30 photons,5,28 polarizability,12,120 polarizable,12,120 polyatomic,7,51 probabilitydistribution,5,32 Q quantumnumber,5,29 R ratedetermining,178 ratelaw,166 ratelimiting,178 reactionequilibrium,14,136 reactionequilibriumconstant,14,139, 141 reactionintermediate,178 Reactionmechanism,16,178 Reactionrate,16 S SecondLawofThermodynamics,17, 185,187 secondorderreaction,168 specicheat,10,90 spectrum,5,26 standardenthalpyofformation,10,94 standardformationreaction,10,94 state,92 statefunction,10,92 strongacids,151 sublimation,13,130 subshells,5,31 T thirdorderreactions,168 thresholdfrequency,5,27 trigonalbipyramid,7,57 trigonalplanar,7,58 triplebond,6,41 triplepoint,133,131 U uncertaintyprinciple,5,32 V valence,6,37 valencebondtheory,9 valenceshell,23 valenceshellelectronpairrepulsionVSEPR theory,55 valenceshellelectronpairrepulsiontheory, 7 vaporpressure,13,127

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INDEX 195 volatilities,128 volatility,13,126 VSEPR,7 W wavefunction,5,32 weakacids,151 Z zerothorderreactions,168

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196 ATTRIBUTIONS Attributions Collection: ConceptDevelopmentStudiesinChemistry Editedby:JohnS.Hutchinson URL:http://cnx.org/content/col10264/1.5/ License:http://creativecommons.org/licenses/by/2.0/ Module:"PrefacetoConceptDevelopmentStudiesinChemistry" By:JohnS.Hutchinson URL:http://cnx.org/content/m12616/1.4/ Pages:1-2 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"TheAtomicMolecularTheory" By:JohnS.Hutchinson URL:http://cnx.org/content/m12432/1.6/ Pages:3-7 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0 Module:"RelativeAtomicMassesandEmpiricalFormulae" By:JohnS.Hutchinson URL:http://cnx.org/content/m12431/1.7/ Pages:9-14 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0 Module:"TheStructureofanAtom" By:JohnS.Hutchinson URL:http://cnx.org/content/m12433/1.2/ Pages:15-24 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0 Module:"QuantumEnergyLevelsInAtoms" By:JohnS.Hutchinson URL:http://cnx.org/content/m12451/1.2/ Pages:25-35 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0 Module:"CovalentBondingandElectronPairSharing" By:JohnS.Hutchinson URL:http://cnx.org/content/m12584/1.5/ Pages:37-50 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0

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ATTRIBUTIONS 197 Module:"MolecularGeometryandElectronDomainTheory" By:JohnS.Hutchinson URL:http://cnx.org/content/m12594/1.1/ Pages:51-61 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"MolecularStructureandPhysicalProperties" By:JohnS.Hutchinson URL:http://cnx.org/content/m12595/1.1/ Pages:63-69 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"ChemicalBondingandMolecularEnergyLevels" By:JohnS.Hutchinson URL:http://cnx.org/content/m14777/1.3/ Pages:71-87 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"EnergeticsofChemicalReactions" By:JohnS.Hutchinson URL:http://cnx.org/content/m12592/1.1/ Pages:89-97 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"TheIdealGasLaw" By:JohnS.Hutchinson URL:http://cnx.org/content/m12598/1.2/ Pages:99-109 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"TheKineticMolecularTheory" By:JohnS.Hutchinson URL:http://cnx.org/content/m12450/1.2/ Pages:111-121 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/1.0 Module:"PhaseEquilibriumandIntermolecularInteractions" By:JohnS.Hutchinson URL:http://cnx.org/content/m12596/1.1/ Pages:123-133 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"ReactionEquilibriumintheGasPhase" By:JohnS.Hutchinson URL:http://cnx.org/content/m12597/1.2/ Pages:135-147 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/

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198 ATTRIBUTIONS Module:"Acid-BaseEquilibrium" By:JohnS.Hutchinson URL:http://cnx.org/content/m12591/1.2/ Pages:149-160 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"ReactionRates" By:JohnS.Hutchinson URL:http://cnx.org/content/m12728/1.2/ Pages:161-180 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/ Module:"EquilibriumandtheSecondLawofThermodynamics" By:JohnS.Hutchinson URL:http://cnx.org/content/m12593/1.3/ Pages:181-192 Copyright:JohnS.Hutchinson License:http://creativecommons.org/licenses/by/2.0/

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ConceptDevelopmentStudiesinChemistry "ConceptDevelopmentStudiesinChemistry"isanon-linetextbookforanIntroductoryGeneralChemistry course.EachmoduledevelopsacentralconceptinChemistryfromexperimentalobservationsandinductive reasoning.Thisapproachcomplementsaninteractiveoractivelearningteachingapproach. AboutConnexions Since1999,Connexionshasbeenpioneeringaglobalsystemwhereanyonecancreatecoursematerialsand makethemfullyaccessibleandeasilyreusablefreeofcharge.WeareaWeb-basedauthoring,teachingand learningenvironmentopentoanyoneinterestedineducation,includingstudents,teachers,professorsand lifelonglearners.Weconnectideasandfacilitateeducationalcommunities. Connexions'smodular,interactivecoursesareinuseworldwidebyuniversities,communitycolleges,K-12 schools,distancelearners,andlifelonglearners.Connexionsmaterialsareinmanylanguages,including English,Spanish,Chinese,Japanese,Italian,Vietnamese,French,Portuguese,andThai.Connexionsispart ofanexcitingnewinformationdistributionsystemthatallowsfor PrintonDemandBooks .Connexions haspartneredwithinnovativeon-demandpublisherQOOPtoacceleratethedeliveryofprintedcourse materialsandtextbooksintoclassroomsworldwideatlowerpricesthantraditionalacademicpublishers.