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Page 1: Ch. 17: Spontaneity, Entropy, and Free Energyflemingapchem.weebly.com/uploads/2/4/6/5/24658308/... · Free Energy and Spontaneity • At what temperatures is the following process

Ch.17:Spontaneity,Entropy,andFreeEnergy

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Recall

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Equations

9

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Spontaneous Processes •  Spontaneity is determined by ΔH (enthalpy)

and ΔS (entropy)

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The First Law of Thermodynamics •  Law of conservation of energy

– The 1st Law of Thermodynamics: – Application of the law of conservation of

energy to heat and thermodynamic processes.

– Energy cannot be created or destroyed, but it can be changed (the energy of the universe is constant)

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The Second Law of Thermodynamics •  Second Law of Thermodynamics =

– The entropy of the universe always increases for a spontaneous process

– Recall: The First Law of Thermodynamics. Energy is conserved, but entropy is not

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The Second Law of Thermodynamics •  Changes in Entropy of the Universe •  ΔSunivispositive

– Entropyoftheuniverseincreases– Processisspontaneousinthedirectionwritten

•  ΔSunivisnegative– Processisspontaneousintheoppositedirection

•  ΔSuniviszero– Processhasnotendencytooccur– Systemisatequilibrium

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The Second Law of Thermodynamics •  ΔS is usually positive (ΔS > 0) when:

1.  Solid à Liquid à Gas 2.  The total number of gas molecules

increases (look at your products vs. reactants)

3.  A larger molecule is broken into 2 or more smaller molecules

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GibbsFreeEnergy

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Gibbs Free Energy •  ΔG = Gibbs Free Energy

– “Available” energy à Energy that we can actually use to do work

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Gibbs Free Energy •  ΔG = Gibbs Free Energy

– This is contrast to enthalpy (H), which represents the total energy of the system

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Gibbs Free Energy •  ΔG = Gibbs Free Energy

– The energy that is actually available has to factor in the entropy of the system

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Gibbs Free Energy • ΔG = ΔH – TΔS •  ΔG = Δ Gibbs free energy •  ΔH = Δ Enthalpy •  T = Temperature in K •  ΔS = Δ Entropy

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ΔG = ΔH – TΔS •  Relationship between ΔG and Spontaneity:

– ΔG < 0 , – Spontaneous process in the forward

direction; releases energy

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ΔG = ΔH – TΔS •  Relationship between ΔG and Spontaneity:

– ΔG > 0 , – Nonspontaneous process in the forward

direction

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ΔG = ΔH – TΔS •  Relationship between ΔG and Spontaneity:

– ΔG = 0 , – System is at equilibrium; no net change

occurs

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ΔG = ΔH – TΔS

ΔH ΔS ΔG ReactionOutcome

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Rxnisspontaneousat_________________________________

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Rxnis__________________________________

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Rxnisspontaneousat_________________________________

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Rxnisspontaneousat_________________________________

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Pair-Share-Respond1.  Whatdoesthe2ndlawofthermodynamicstellus?

2.  WhatarethreewaysthatΔSwillbecomepositive?

3.  DefineGibbsFreeEnergy4.  WhatistheequationforGibbsFreeEnergy?

5.  UnderwhatΔHandΔSconditionswillΔGalwaysbepositive?

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MoreonFreeEnergy

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RelationshipbetweenFreeEnergy(G)andSpontaneity

univΔΔ = at constant andGS T PT

•  ProcessesthatoccuratconstantTandParespontaneousinthedirectioninwhichthefreeenergydecreases– NegativeΔGmeanspositiveΔSuniv

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FreeEnergyandSpontaneity

•  Atwhattemperaturesisthefollowingprocessspontaneousat1atm?

Br2(l)àBr2(g)ΔH°=31.0kJ/molandΔS°=93.0kJ/mol

•  WhatisthenormalboilingpointofliquidBr2?•  Hint:

– Thereactionhasapositive(favored)entropy,butapositive(notfavored)enthalpy

– Howshouldthesevaluescompareattheboilingpoint?

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Solution

•  Attemperaturesabove333K,TΔS°hasalargermagnitudethanΔH°,andΔG°isnegative– Above333K,thevaporizationprocessisspontaneous– Theoppositeprocessoccursspontaneouslybelowthistemperature

•  At333K,liquidandgaseousBr2coexistinequilibrium

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Solution•  Summaryofobservations(thepressureis1atmineachcase)– T>333K

•  ThetermΔS°controls,andtheincreaseinentropywhenliquidBr2isvaporizedisdominant

– T<333K•  Theprocessisspontaneousinthedirectioninwhichitisexothermic,andthetermΔH°controls

– T=333K•  Theopposingdrivingforcesarejustbalanced(ΔH°=0),andtheliquidandgaseousphasesofbrominecoexist

•  Thisisthenormalboilingpoint

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•  Amixtureofhydrogenandchlorineremainsunreacteduntilitisexposedtoultravioletlightfromaburningmagnesiumstrip–  Thenthefollowingreactionoccursveryrapidly:

ΔG=–45.54kJ H2(g)+Cl2(g)à2HCl(g) ΔH=–44.12kJ ΔS=–4.76J/K

–  Selectthestatementthatbestexplainsthisbehavior a.  Reactantsarethermodynamicallymorestablethanthe

productsb.  Reactionhasasmallequilibriumconstantc.  Ultravioletlightraisesthetemperatureofthesystemand

makesthereactionmorefavorabled.  NegativevalueforΔSslowsdownthereactione.  Reactionisspontaneous,butthereactantsarekinetically

stable

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SampleQuestion•  Forthevaporizationofaliquidatagivenpressure,

a.  ΔGispositiveatalltemperaturesb.  ΔGisnegativeatalltemperaturesc.  ΔGispositiveatlowtemperatures,butnegativeat

hightemperaturesd.  ΔGisnegativeatlowtemperatures,butpositiveat

hightemperatures

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SampleQuestion•  Inwhichcasemustareactionbespontaneousatalltemperatures?a.  ΔHispositive,andΔSispositiveb.  ΔHisnegative,andΔSisnegativec.  ΔHispositive,andΔSisnegatived.  ΔHisnegative,andΔSispositivee.  ΔH=0andΔS=0

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EntropyandAqueousSolutions

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Entropy and Aqueous Solutions •  What about solutions? •  Solutions will have a greater entropy than

pure lquids as the particles in a solution are more separated and solvent molecules separate the solute particles

•  So in order of increasing entropy: Solids < Liquids < Aqueous Solutions < Gases

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FormationofSolutions•  FollowingvaluesofΔSforKCl(s),LiF(s),andCaS(s)formingaqueoussolutionsillustratetheunusualnatureofwaterasasolvent

osolnSΔ

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EntropyChangesinAqueousSolutions•  Recall:Hydration

osolnSΔ

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EntropyChangesinAqueousSolutions•  Polarwatermoleculesareattractedtotheionstoformhydratedspecies– Assemblingofagroupofwatermoleculesaroundtheionsisanorder-producingphenomenonandwouldbeexpectedtomakeanegativecontributiontoΔS

– Hydrationeffectwillbegreaterwhenanionpossessesmorechargedensity

•  Enthalpychangesandentropiceffectsmustbeconsideredtopredictsolubility

osolnSΔ

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SampleQuestion•  ForthedissociationreactionoftheacidHF,ΔSisnegative

– WhichstatementbestexplainswhyΔSisnegative?a.  EachHFmoleculeproducestwoionswhenitdissociatesb.  Ionsarehydratedc.  Reactionisexpectedtobeexothermic,soΔSshouldbe

negatived.  Reactionisexpectedtobeendothermic,soΔSshouldbe

negative

–aq aq a( ) ( ) )q(à àÜá àà +HF    H + F

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EntropyChanges•  Thirdlawofthermodynamics

– Entropyofaperfectcrystalat0Kiszero•  ThereisonlyonemicrostateavailableatzeroK•  Anotherwaytolookatthislaw:

– Theentropyofaperfectcrystalapproacheszeroastheabsolutetemperatureapproacheszero

•  Entropyofasubstanceincreaseswithtemperature

40

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EntropyValues•  Standardentropyvalues(S°)representincreaseinentropythatoccurswhenasubstanceisheatedfrom0Kto298Kat1atm– Morecomplexthemolecule,thehigherthestandardentropyvalue

41

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EntropyChangeforaGivenChemicalReaction

•  Entropyisastatefunctionofachemicalsystem•  Entropychangescanbecalculatedasfollows:

reaction p products r reactantsΔ =S n S n S° ° − °∑ ∑

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FreeEnergyandPressure•  SystemunderconstantPandTproceedsspontaneouslyinthedirectionthatlowersitsfreeenergy– Freeenergyofareactionsystemchangesasthereactionproceeds

•  Dependentonthepressureofagasorontheconcentrationofspeciesinsolution

•  Equilibrium-Pointwherefreeenergyvalueisatitslowest

43

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FreeEnergyandPressure•  Foridealgases:

– Enthalpyisnotpressure-dependent– Entropydependsonpressureduetoitsdependenceonvolume

•  Atagiventemperaturefor1moleofidealgas:– Slargevolume>Ssmallvolume

Or– Slowpressure>Shighpressure

44

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Equations

45

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TheMeaningofΔGforaChemicalReaction

•  EvenwhenvalueofΔGprovidesinformationregardingwhetherthesystemisfavoredunderagivensetofconditions:– Systemmaynotproceedtopureproducts(ifΔGisnegative)

– Systemmaynotremainatpurereactants(ifΔGispositive)

•  Asystemwillspontaneouslyseekequilibrium

46

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TheMeaningofΔGforaChemicalReaction

•  Systemcanachievethelowestpossiblefreeenergybygoingtoequilibrium,notbygoingtocompletion

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EquilibriumPoint(continued2)•  Quantitativerelationshipbetweenfreeenergyandthevalueoftheequilibriumconstantisgivenby:

( )Δ = 0 = Δ ° + RT lnG G K

( )Δ ° = RT ln−G K