化学原理Chemistry课件-post+2+thermochemistry

ThermochemistryChapter6Energychangeofchemicalreaction.ThermochemistryChapter6Energy1Thermalenergytherandommotion.ChemicalenergythebondsofsubstancesNuclearenergy:neutronsandprotonsintheatomElectricalenergy:theflowofelectronsPotentialenergy:virtueofanobject’spositionItisthecapacitytodowork,WhatistheEnergy?andexistsinavarietyoftheforms.Nooneknowshowmuchtheexactenergyis!Thethermodynamicsdealsonlywiththeenergychangesthataccompanychemicalreactions.Thermalenergytherandomm2Systemand

SurroundingsSystemisthespecificpartoftheuniversethatisofinterestinthestudy.Opensystemmass&energyexchangedclosedsystemonlyenergyexchangedIsolatedsystemNothingexchangedSystemandSurroundingsSystem3(1)Heat

istheenergytransferbetween

a

systemanditssurroundingsthatareatdifferenttemperatures.Whatwecanmeasureisonlytheenergychange!Temperatureisameasureofthethermalenergy.Temperature=ThermalEnergy900C400Cgreaterthermalenergy?(1)Heatistheenergytransf4(2)Work,likeheat,

istheenergytransferbetween

a

systemanditssurroundings.Work(w)=Force(F)distance(h)=PAhhAA=PDVPressure-volumework(2)Work,likeheat,istheen5ThermodynamicsStatefunctions

determinedonlybythesystemstate,regardlessofhowitwasachieved(i.e.history).Thepotentialenergyofhiker1isthesameasthatofhiker2.Energychange:DE=Efinal-EinitialOtherchanges:DP=Pfinal-PinitialDV=Vfinal-VinitialDT=Tfinal-TinitialSignificance

itmakesthereasoningandcalculationsimple.ThermodynamicsStatefunctions6DE=q+wq:theheatexchangewiththesurroundings.Attention:

Eisastatefunction,butqandwarenotastatefunction,sincethelattertwoareonlyrelatedtoaprocess.TheFirstLawofThermodynamics:

Theenergyoftheuniversecanbeconvertedfromoneformtoanother,butitcannotbecreatedordestroyed.w:theworkdoneon(orby)thesystem.“+”gainbythesystem“–”lostbythesystemDEsystem+DEsurroundings=0DE=Efinal-EinitialThesystemenergychangeDE=q+w7Exothermicprocess,givesofftheheattothesurroundings.Endothermicprocess,getstheheatfromthesurroundings.2H2

(g)+O2

(g)2H2O(l)+energyH2O(g)H2O(l)+energyenergy+2HgO(s)2Hg(l)+O2

(g)Exothermicprocess,givesoff8Asampleofnitrogengasexpandsinvolumefrom1.6Lto5.4Latconstanttemperature.Whatistheworkdoneinjoulesifthegasexpands(a)againstavacuumand(b)againstaconstantpressureof3.7atm?w=-PDV(a)DV=5.4L–1.6L=3.8LP=0atmW=-0atmx3.8L=0L•atm=0joules(b)DV=5.4L–1.6L=3.8LP=3.7atmw=-3.7atmx3.8L=-14L•atmw=-14L•atmx101.3J1L•atm=-1400JAsampleofnitrogengasexpan9DE=q+wDV=0,w=0DE=qvAtconstantvolume,theheatadsorbedorreleasedtoitssurroundingistotallyusedtoincreaseordecreasethesysteminternalenergy.AtconstantvolumeDE=q+wDV=0,w=0DE=qv10(2)AtconstantpressureAtconstantP,

DH=DE+PDVDefineenthalpy:H=E+PV

Hisastatefunction,butitisnotthesystem

energy.Instead,Hisonlyameasureofthesystemenergy!w=-PDV,DE

=qp–PDVE2-E1=qp–P(V2–V1)qp

=(E2+PV2)–(E1+PV1)qp

=H2–H1=DHatconstantpressure,theheatadsorbedorreleasedistotallyusedtoincreaseordecreasethesystementhalpy.DE=q+w(2)AtconstantpressureAtcon11AComparisonofDHandDE2Na(s)+2H2O(l)2NaOH(aq)+H2(g)DH=-367.5kJ/molDE=DH-PDVAt250Cand1atm,1moleH2=24.5LPDV=1atmx24.5L=2.5kJDE=-367.5kJ/mol–2.5kJ/mol=-370.0kJ/mol6.4AComparisonofDHandDE2Na(12DH(Enthalpychange):H(products)–H(reactants)Hproducts<HreactantsDH<0Hproducts>HreactantsDH>0H(Enthalpy):describetheheatflowinaprocessthatoccursatconstantpressure.DH(Enthalpychange):H(prod13ThermochemicalEquationsH2O(s)H2O(l)DH=6.01kJSystemabsorbsheatEndothermicDH>0Itmeansthat:

6.01kJareabsorbedbyonemoleoficethatmeltsat00Cand1atm.ThermochemicalEquationsH2O(s14H2O(s)H2O(l)DH=6.01kJThermochemicalEquationsIftheequationisreverse,thesignofDHmustchangeH2O(l)H2O(s)DH=-6.01kJIfbothsidesaremutipliedbyn,DHmustchangebyn.2H2O(s)2H2O(l)DH=2x6.01

=12.0kJThephysicalstatesmustbeallspecifiedH2O(l)H2O(g)DH=44.0kJH2O(s)H2O(l)DH=615Howmuchheatisevolvedwhen266gofwhitephosphorus(P4)burninair?P4

(s)+5O2

(g)P4O10

(s)

DH=-3013kJ266gP41molP4123.9gP4x3013kJ1molP4x=6470kJHowmuchheatisevolvedwhen16化学原理Chemistry课件-post+2+thermochemistry17theabsolutevalueofenthalpycannotbemeasured.i.e.

DHfo=0foranyelementinitsmoststableform.DH0(O2,g)=0fDH0(O3,g)=142kJ/molfDH0(C,graphite)=0fDH0(C,diamond)=1.90kJ/molfHowtomakeatheoreticalcalculation?Canweestablishareferencepoint?

DHfo

Standardenthalpyofformationtheheatchangeresultswhenonemoleofacompoundisformedat1atmfromitselementsinthemoststableform.C(s,graphite)+O2(g)=CO2(g)DHrxn0=-393.5kJThen,DHf0(CO2,g)=-393.5kJ/moltheabsolutevalueofenthalpy18DHorxn=(H2+H3)–H1

Ca(s)O2(g)C(s,graphite)32H4

Ca(s)O2(g)12H5DHoa

=H4–H1

DHoc

=H2–H5

DHod

=H3–H6

CaCO3(s)CaO(s)+CO2(g)H1H2H3O2(g)C(s,graphite)H6+DHob

=(H5+H6)–H4DHorxn=DHoa+DHob+

DHoc+

DHodDHrxn=[DH

(CaO,s)

+DH

(CO2,g)]–

DH(CaCO3,s)oofofofof=-DH(CaCO3,s)of=DH(CaO,s)of=DH(CO2,g)1atmDHorxn=(H2+H3)–H119化学原理Chemistry课件-post+2+thermochemistry20Thestandardenthalpyofreactioncarriedoutat1atm.aA+bBcC+dDDH0rxndDH0(D)fcDH0(C)f=[+]-bDH0(B)faDH0(A)f[+]Benzene(C6H6)burnsinairtoproducecarbondioxideandliquidwater.Howmuchheatisreleasedpermoleofbenzenecombusted?Thestandardenthalpyofformationofbenzeneis49.04kJ/mol.2C6H6

(l)+15O2

(g)12CO2

(g)+6H2O(l)=[12(–393.5)+6(–187.6)]–[249.04+0]=-5946kJ-5946kJ2mol=-2973kJ/molC6H6=[12(CO2,g)+6(H2O,l)]–[2(C6H6,l)+15(O2,g)]DH0rxnDH0fDH0

fDH0

fDH0

fThestandardenthalpyofreact21Hess’sLaw:Whenreactantsareconvertedtoproducts,thechangeinenthalpyisthesameregardlessofthereactiontakingplaceinonesteporinaseriesofsteps.HowmuchisDHofforCS2(l)giventhat:C(graphite)+O2

(g)CO2

(g)

DH0=-393.5kJrxnS(rhombic)+O2

(g)SO2

(g)

DH0=-296.1kJrxnCS2(l)+3O2

(g)CO2

(g)+2SO2

(g)

DH0=-1072kJrxnC(graphite)+2S(rhombic)CS2(l)rxnC(graphite)+O2

(g)CO2

(g)

DH0

=-393.5kJ2S(rhombic)+2O2

(g)2SO2

(g)

DH0

=-296.1x2kJrxnCO2(g)+2SO2

(g)CS2

(l)+3O2

(g)

DH0

=+1072kJrxn+C(graphite)+2S(rhombic)CS2(l)DH0

=-393.5+(2x-296.1)+1072=86.3kJrxnHess’sLaw:Whenreactantsar22DHsoln=Step1+Step2=788–784=4kJ/molStep1LatticeenergyStep2hydrationDissolutionProcessofNaClDHsoln=Step1+Step2Step23Whichsubstance(s)couldbeusedformeltingice?Whichsubstance(s)couldbeusedforacoldpack?Whichsubstance(s)couldbeus24Exercises:6.196.546.806.1046.113

Exercises:25ThermochemistryChapter6Energychangeofchemicalreaction.ThermochemistryChapter6Energy26Thermalenergytherandommotion.ChemicalenergythebondsofsubstancesNuclearenergy:neutronsandprotonsintheatomElectricalenergy:theflowofelectronsPotentialenergy:virtueofanobject’spositionItisthecapacitytodowork,WhatistheEnergy?andexistsinavarietyoftheforms.Nooneknowshowmuchtheexactenergyis!Thethermodynamicsdealsonlywiththeenergychangesthataccompanychemicalreactions.Thermalenergytherandomm27Systemand

SurroundingsSystemisthespecificpartoftheuniversethatisofinterestinthestudy.Opensystemmass&energyexchangedclosedsystemonlyenergyexchangedIsolatedsystemNothingexchangedSystemandSurroundingsSystem28(1)Heat

istheenergytransferbetween

a

systemanditssurroundingsthatareatdifferenttemperatures.Whatwecanmeasureisonlytheenergychange!Temperatureisameasureofthethermalenergy.Temperature=ThermalEnergy900C400Cgreaterthermalenergy?(1)Heatistheenergytransf29(2)Work,likeheat,

istheenergytransferbetween

a

systemanditssurroundings.Work(w)=Force(F)distance(h)=PAhhAA=PDVPressure-volumework(2)Work,likeheat,istheen30ThermodynamicsStatefunctions

determinedonlybythesystemstate,regardlessofhowitwasachieved(i.e.history).Thepotentialenergyofhiker1isthesameasthatofhiker2.Energychange:DE=Efinal-EinitialOtherchanges:DP=Pfinal-PinitialDV=Vfinal-VinitialDT=Tfinal-TinitialSignificance

itmakesthereasoningandcalculationsimple.ThermodynamicsStatefunctions31DE=q+wq:theheatexchangewiththesurroundings.Attention:

Eisastatefunction,butqandwarenotastatefunction,sincethelattertwoareonlyrelatedtoaprocess.TheFirstLawofThermodynamics:

Theenergyoftheuniversecanbeconvertedfromoneformtoanother,butitcannotbecreatedordestroyed.w:theworkdoneon(orby)thesystem.“+”gainbythesystem“–”lostbythesystemDEsystem+DEsurroundings=0DE=Efinal-EinitialThesystemenergychangeDE=q+w32Exothermicprocess,givesofftheheattothesurroundings.Endothermicprocess,getstheheatfromthesurroundings.2H2

(g)+O2

(g)2H2O(l)+energyH2O(g)H2O(l)+energyenergy+2HgO(s)2Hg(l)+O2

(g)Exothermicprocess,givesoff33Asampleofnitrogengasexpandsinvolumefrom1.6Lto5.4Latconstanttemperature.Whatistheworkdoneinjoulesifthegasexpands(a)againstavacuumand(b)againstaconstantpressureof3.7atm?w=-PDV(a)DV=5.4L–1.6L=3.8LP=0atmW=-0atmx3.8L=0L•atm=0joules(b)DV=5.4L–1.6L=3.8LP=3.7atmw=-3.7atmx3.8L=-14L•atmw=-14L•atmx101.3J1L•atm=-1400JAsampleofnitrogengasexpan34DE=q+wDV=0,w=0DE=qvAtconstantvolume,theheatadsorbedorreleasedtoitssurroundingistotallyusedtoincreaseordecreasethesysteminternalenergy.AtconstantvolumeDE=q+wDV=0,w=0DE=qv35(2)AtconstantpressureAtconstantP,

DH=DE+PDVDefineenthalpy:H=E+PV

Hisastatefunction,butitisnotthesystem

energy.Instead,Hisonlyameasureofthesystemenergy!w=-PDV,DE

=qp–PDVE2-E1=qp–P(V2–V1)qp

=(E2+PV2)–(E1+PV1)qp

=H2–H1=DHatconstantpressure,theheatadsorbedorreleasedistotallyusedtoincreaseordecreasethesystementhalpy.DE=q+w(2)AtconstantpressureAtcon36AComparisonofDHandDE2Na(s)+2H2O(l)2NaOH(aq)+H2(g)DH=-367.5kJ/molDE=DH-PDVAt250Cand1atm,1moleH2=24.5LPDV=1atmx24.5L=2.5kJDE=-367.5kJ/mol–2.5kJ/mol=-370.0kJ/mol6.4AComparisonofDHandDE2Na(37DH(Enthalpychange):H(products)–H(reactants)Hproducts<HreactantsDH<0Hproducts>HreactantsDH>0H(Enthalpy):describetheheatflowinaprocessthatoccursatconstantpressure.DH(Enthalpychange):H(prod38ThermochemicalEquationsH2O(s)H2O(l)DH=6.01kJSystemabsorbsheatEndothermicDH>0Itmeansthat:

6.01kJareabsorbedbyonemoleoficethatmeltsat00Cand1atm.ThermochemicalEquationsH2O(s39H2O(s)H2O(l)DH=6.01kJThermochemicalEquationsIftheequationisreverse,thesignofDHmustchangeH2O(l)H2O(s)DH=-6.01kJIfbothsidesaremutipliedbyn,DHmustchangebyn.2H2O(s)2H2O(l)DH=2x6.01

=12.0kJThephysicalstatesmustbeallspecifiedH2O(l)H2O(g)DH=44.0kJH2O(s)H2O(l)DH=640Howmuchheatisevolvedwhen266gofwhitephosphorus(P4)burninair?P4

(s)+5O2

(g)P4O10

(s)

DH=-3013kJ266gP41molP4123.9gP4x3013kJ1molP4x=6470kJHowmuchheatisevolvedwhen41化学原理Chemistry课件-post+2+thermochemistry42theabsolutevalueofenthalpycannotbemeasured.i.e.

DHfo=0foranyelementinitsmoststableform.DH0(O2,g)=0fDH0(O3,g)=142kJ/molfDH0(C,graphite)=0fDH0(C,diamond)=1.90kJ/molfHowtomakeatheoreticalcalculation?Canweestablishareferencepoint?

DHfo

Standardenthalpyofformationtheheatchangeresultswhenonemoleofacompoundisformedat1atmfromitselementsinthemoststableform.C(s,graphite)+O2(g)=CO2(g)DHrxn0=-393.5kJThen,DHf0(CO2,g)=-393.5kJ/moltheabsolutevalueofenthalpy43DHorxn=(H2+H3)–H1

Ca(s)O2(g)C(s,graphite)32H4

Ca(s)O2(g)12H5DHoa

=H4–H1

DHoc

=H2–H5

DHod

=H3–H6

CaCO3(s)CaO(s)+CO2(g)H1H2H3O2(g)C(s,graphite)H6+DHob

=(H5+H6)–H4DHorxn=DHoa+DHob+

DHoc+

DHodDHrxn=[DH

(CaO,s)

+DH

(CO2,g)]–

DH(CaCO3,s)oofofofof=-DH(CaCO3,s)of=DH(CaO,s)of=DH(CO2,g)1atmDHorxn=(H2+H3)–H144化学原理Chemistry课件-post+2+thermochemistry45Thestandardenthalpyofreactioncarriedoutat1atm.aA+bBcC+dDDH0rxndDH0(D)fcDH0(C)f=[+]-bDH0(B)faDH0(A)f[+]Benzene(C6H6)burnsinairtoproducecarbondioxideandliquidwater.Howmuchheatisreleasedpermoleofbenzenecombusted?Thesta