Chapter 5 Thermochemistry
The energy of chemical reactions How do you keep track of it? Where does it come from?
Energy
• Theabilityto:• dowork• transferheat.
Ø Work:Energyusedtocauseanobjectthathasmasstomove.
Ø Heat:Energyusedtocausethetemperatureofanobjecttorise.
Units of Energy
• TheSIunitofenergyisthejoule(J).• Anolder,non-SIunitissHllinwidespreaduse:Thecalorie(cal). 1cal=4.184J
Energyhasunitsof(mass)(velocity)2RememberkineHcenergywas1/2mv2
1 J = 1 ⎯⎯ kg m2
s2
Work • Energyusedtomoveanobjectoversomedistance.
• w=F� d,w=work,F=forced=distanceoverwhichtheforceisexerted.
Noteunits:F=ma,mass(distance/s2)W=F(d)=mass(distance2/s2)=mv2
Heat
• Energycanalsobetransferredasheat.
• Heatflowsfromwarmerobjectstocoolerobjects.
Kinetic Energy
EnergyanobjectpossessesbyvirtueofitsmoHon.
1 2
KE = ⎯ mv2
PotenHalEnergy
EnergyanobjectpossessesbyvirtueofitsposiHonorchemicalcomposiHon.
More potential E
Less P.E. as bike goes down.
TransferalofEnergy
a) AddP.E.toaballbyliUingittothetopofthewall
TransferalofEnergy
a) AddP.E.toaballbyliUingittothetopofthewall
b) Astheballfalls,P.E------>K.E.(1/2mv2)
TransferalofEnergy
a) AddP.E.toaballbyliUingittothetopofthewall
b) Astheballfalls,P.E------>K.E.(1/2mv2)Ballhitsground,K.E.=0,butEhastogo
somewhere.So1. Ballgetssquashed2. Heatcomesout.
Energy accounting
• WemustidenHfywheredifferenttypesofenergygo.
• Therefore,wemustidenHfytheplaces.
System and Surroundings
• Thesystemincludesthemoleculeswewanttostudy(here,thehydrogenandoxygenmolecules).
• Thesurroundingsareeverythingelse(here,thecylinderandpiston).
First Law of Thermodynamics • Energyisconserved.• Inotherwords,thetotalenergyoftheuniverseisaconstant;ΔESystem=-ΛEsurroundings
InternalEnergyTheinternalenergyofasystemisthesumofallkineHcandpotenHalenergiesofallcomponentsofthesystem;wecallitE.
Einternal,total= EKE + EPE + Eelectrons + Enuclei +…… Almost impossible to calculate total internal energy Instead we always look at the change in energy (ΔE).
InternalEnergyBydefiniHon,thechangeininternalenergy,ΔE,isthefinalenergyofthesystemminustheiniHalenergyofthesystem: ΔE=Efinal−EiniHal
ChangesinInternalEnergy
• IfΔE>0,Efinal>EiniHalØ Therefore,thesystemabsorbedenergyfromthesurroundings.
Ø Thisenergychangeiscalledendergonic.
ChangesinInternalEnergy
• IfΔE<0,Efinal<EiniHalØ Therefore,thesystemreleasedenergytothesurroundings.
Ø Thisenergychangeiscalledexergonic.
ChangesinInternalEnergy
• Whenenergyisexchangedbetweenthesystemandthesurroundings,itisexchangedaseitherheat(q)orwork(w).
• Thatis,ΔE=q+w.
ΔE,q,w,andTheirSigns
+q -q
hot plate adds heat to water
Surroundings suck heat out of water.
Signofwork
Truck pushes block up. Does work on block wtruck- wblock+
block pushes truck down does work on truck wblock- wtruck+
Exchange of Heat between System and Surroundings
• Whenheatisabsorbedbythesystemfromthesurroundings,theprocessisendothermic.
Exchange of Heat between System and Surroundings
• Heatabsorbedbysystemfromsurroundings,isendothermic.
• Heatreleasedbysystemtosurroundings,theisexothermic.
State Functions
Totalinternalenergyofasystem:K.E.+Eelectrons+Enucleus+P.E.totalvirtuallyimpossibletomeasure/calculate
StateFuncHons• However,wedoknowthattheinternalenergyofasystemisindependentofthepathbywhichthesystemachievedthatstate.Ø Inthesystembelow,thewatercouldhavereachedroomtemperaturefromeitherdirecHon.
StateFuncHons• Therefore,internalenergyisastatefuncHon.• becauseit’sPATHINDEPENDENT• Andso,ΔEdependsonlyonEiniHalandEfinal.
StateFuncHons
• However,qandwarenotstatefuncHons.
• Whetherthebaceryisshortedoutorisdischargedbyrunningthefan,itsΔEisthesame.Ø Butqandwaredifferentinthetwocases.
Work
processinanopencontainer(chemicalreacHoninabeaker)w?(cantherebeanywork)?Yes, evolving gases could push on the surroundings.
Catch the work, do the same process in a cylinder
Process evolves gas, pushes on piston, work done on piston
Catch the work, do the same process in a cylinder
w = F*d, F = P*A, d=Δh
w = -P*AΔh= -PΔV
Negative because an increase in Volume means that the system is doing work on the surroundings.
ΔE=q+w=q-PΔV qP = ΔE + PΔV
Example• Gasinsidecylinderwith
electricheater.• Add100jheatwith
heater.• 1.Pistoncangoupand
down• 2.Pistonstuck.• a.WhathappenstoTin
eachcase?• b.Whataboutqandw
foreachcase?• c.WhataboutΔEineach
case?
Example• Gasinsidecyclinderwithelectricheater.• Add100jheatwithheater.• 1.Pistoncangoupanddown• 2.Pistonstuck.• a.WhathappenstoTineachcase?• b.Whataboutqandwforeachcase?• c.WhataboutΔEineachcase? a.1. Piston goes up, some E
goes to expand gas, do work. T goes up less
a.2 T goes up more, all E goes to q.
b.1. both q and w positive
b.2. w 0, q larger c. ΔE the same & + in each case
WorkNowwecanmeasurethework: w=−PΔV
Zn + 2HCl ---------> H2(g) + ZnCl2
Work
Zn+2HCl--------->H2(g)+ZnCl2ImoleofZnreacts.Howmuchworkisdone(P=1atm,
densityofH2=0.0823g/L)?1moleofH2isproduced.
Work
ImoleofZnreacts.Howmuchworkisdone(P=1atm,densityofH2=
0.0823g/L)?1moleofH2isproduced.Zn+2HCl--------->H2(g)+ZnCl21mol 1mol2.014g/mol2.014gd=m/VV=m/dV=2.014g/0.0823g/L=24.47LW=-PΔV=1atm(24.47L)=-24.47L(atm)
Enthalpy(H)
H = E + PV
This is the definition of Enthalpy for any process Buy why do we care?
Enthalpy
• atconstantpressure,ΔH,is(Δ = change in thermodynamics) ΔH=Δ(E+PV)
• Thiscanbewricen(ifPconstant) ΔH=ΔE+PΔV
H = E + PV
Enthalpy
• SinceΔE=q+wandw=−PΔV(Pconst.)subsHtutetheseintotheenthalpyexpression: ΔH=ΔE+PΔV ΔH=(q+w)−w ΔH=q
• Note:trueatconstantpressure• qisastatefuncHonatconstP&onlyPVwork.
H=E+PV• Because:• Ifpressureisconstant(likeopentoatmosphere,i.e.mostthings)and
w=ΔPV.heatflow(q)=H(enthalpy)ofsystem.And:HisastatefuncHon,soqisalso.butonlyintherightcondi6ons
Endothermicvs.Exothermic
• AprocessisendothermicwhenΔHisposiHve.
EndothermicityandExothermicity
• AprocessisendothermicwhenΔHisposiHve.
• AprocessisexothermicwhenΔHisnegaHve.
Enthalpies of Reaction
Thechangeinenthalpy,ΔH,istheenthalpyoftheproductsminustheenthalpyofthereactants:
ΔH=Hproducts−Hreactants
EnthalpiesofReacHonThisquanHty,ΔH,iscalledtheenthalpyofreacHon,ortheheatofreacHon.
Reaction Enthalpy summary
1. Enthalpyisanextensiveproperty.2. ΔHforareacHonintheforwarddirecHon
isequalinsize,butoppositeinsign,toΔHforthereversereacHon.
3. ΔHforareacHondependsonthestateoftheproductsandthestateofthereactants.
Enthalpy of reaction example
ConsiderthereacHon:2KClO3------->2KCl+3O2ΔH=-89.4kJ/mola.WhatistheenthalpychangeforformaHonof0.855molesofO2?
EnthalpyofreacHonexampleConsiderthereacHon:2KClO3------->2KCl+3O2ΔH=-89.4kJ/mola.WhatistheenthalpychangeforformaHonof0.855molesofO2?
2KClO3 -------> 2KCl + 3O2 ΔH = -89.4 kJ/mol 0.855 mol ΔH = -89.4 kJ/3 mol O2(.855 mol O2) = -25.5 kJ
Calorimetry
Sincewecannotknowtheexactenthalpyofthereactantsandproducts,
wemeasureΔHthroughcalorimetry,themeasurementofheatflow.
HeatCapacityandSpecificHeat
• heatcapacity:amountofErequiredtoraisethetemperatureofsomethingby1K
• specificheat:amountofErequiredtoraisethetemperatureof1gofasubstanceby1K.
Heat Capacity and Specific Heat
Specificheatis:
Specific heat = heat transferred
mass × temperature change
s = q
m ΔT
smΔT = q
Constant Pressure Calorimetry
indirectlymeasuretheheatchangeforthesystembymeasuringtheheatchangeforthewaterinthecalorimeter.
Constant Pressure Calorimetry
Becausethespecificheatforwateriswellknown(4.184J/g-K),wecanmeasureΔHforthereacHonwiththisequaHon:q=m×s×ΔT
m=masss=specificheat
ExampleWhena3.88gsampleofsolid
ammoniumnitratedisolvesin60.0gofwaterinacoffeecupcalorimeter,thetemperaturedropsfrom23.0°Cto18.4°C.(a)CalculateΔH(inkJ/molammoniumnitrate)forthesoluHonprocess.Assumethatthespecificheatisconstantand=1.0cal/gC.(b)Isthisprocessendothermicorexothermic?
ExampleWhena3.88gsampleofsolidammoniumnitratedisolvesin60.0gofwaterina
coffeecupcalorimeter,thetemperaturedropsfrom23.0°Cto18.4°C.(a)CalculateΔH(inkJ/molammoniumnitrate)forthesoluHonprocess.Assumethatthespecificheatisconstantand=4.184J/g°C.(b)Isthisprocessendothermicorexothermic?
Reaction: NH4NO3(s) ------> NH4
+(aq) + NO3-(aq)
gr 3.88 g MW 80.04 g/mol #Mol 3.88 g/80.04 g/mol = 0.0484 mol Mass of solution = 3.88 g + 60 g = 63.88 g.
System: Solid AmNO3 Surroundings: Solution
q = s(specific heat)m(mass)ΔT q = s(J/g°C)m(grams)(Tfinal - Tinitial) qsolution = 4.184(J/g°C)(63.88 g)(18.4°C - 23.0°C) = -1229 J qwater=-qammonium nitrate = +1229 J ΔH(per mol NH4NO3) = 1.229 kJ/.0484 mol = 25.39 kJ/mol (b) Endothermic
Bomb Calorimetry
ReacHonscanbecarriedoutseparatedfromthewaterina“bomb,”suchasthisone,
AndsHllmeasuretheheatabsorbedbythewater.
Bomb Calorimetry
• Becausethevolumeinthebombcalorimeterisconstant,whatismeasuredisreallytheΔE,notΔH.
• FormostreacHons,• ΔE≈ΔH• Why?
BombCalorimetry
H = E + PV ΔH = ΔE + ΔPV In a bomb calorimeter, ΔV = 0 For a process that doesn’t evolve gas: ΔP ≈ 0 as well. ΔH = ΔE + ΔPV = ΔE
Example • A50gsampleofgasolinewasburnedbycombusHon
(withexcessoxygen)inacalorimeterwithaheatcapacityof10kJ/°C.Thetemperatureincreasedby100°C.CalculatethechangeinEpergofgasoline.
• qsurroundings=CΔΤ =10kJ/°C(100°C)=1000kJ
• qsurroundings=-qsystem• qsystem=-1000• -1000kJ/50g=-20kJ/g
• DoesΔΕ =ΔΗ inthiscase?
Example • A50gsampleofgasolinewasburnedbycombusHon
(withexcessoxygen)inacalorimeterwithaheatcapacityof10kJ/°C.Thetemperatureincreased100°C.CalculatethechangeinEpergofgasoline.
• qsurroundings=CΔΤ =10kJ/°C(100°C)=1000kJ
• qsurroundings=-qsystem• qsystem=-1000• -1000kJ/50g=-20kJ/g
• DoesΔΕ =ΔΗ inthiscase?• NO!Pressurecan’tstayconstantinthiscase.
Hess’s Law
• ΔHisknownformanyreacHons.• measuringΔHcanbeapain• CanweesHmateΔHusingΔHvaluesforotherreacHons?
Hess’sLaw
Hess’slaw:statesthat:
ΔHfortheoverallreacHonwillbeequaltothesumoftheenthalpychangesfortheindividualsteps.
Yes!
Hess’s Law
Why? BecauseΔHisastatefuncHon,
andispathwayindependent.OnlydependsoniniHalstateof
thereactantsandthefinalstateoftheproducts.
Hess’slaw,example:• Given:• N2(g)+O2(g)---->2NO(g) ΔH= 180.7kJ• 2NO(g)+O2(g)---->2NO2(g)ΔH=-113.1kJ• 2N2O(g)---->2N2(g)+O2(g)ΔH= -163.2kJ• useHess’slawtocalculateΔHforthereacHon:• N2O(g)+NO2(g)---->3NO(g)
Hess’slaw,example:• Given:• N2(g)+O2(g)---->2NO(g)ΔH= 180.7kJ• 2NO(g)+O2(g)---->2NO2(g) ΔH= -113.1kJ• 2N2O(g)---->2N2(g)+O2(g) ΔH= -163.2kJ• useHess’slawtocalculateΔHforthereacHon:• N2O(g)+NO2(g)---->3NO(g)
• N2O(g) ----> N2(g) + 1/2O2(g) ΔH = -163.2/2 =-81.6kJ • NO2(g) ----> NO(g) + 1/2O2(g) ΔH = 113.1 kJ/2=56.6kJ • N2(g) + O2(g) ----> 2NO(g) ΔH = 180.7
• N2O(g)+NO2(g)---->3NO(g) ΔH = 155.7 kJ
EnthalpiesofFormaHonAnenthalpyofformaHon,ΔHf,isdefinedastheΔHforthereacHoninwhichacompoundismadefromitsconsHtuentelementsintheirmoststableelementalforms.
• 2Al + Fe2O3 -------> Al2O3 + 2Fe
• What is the heat of reaction given: • 2Fe + 3/2O2 -----> Fe2O3 ΔH = -825.5 KJ (heat of formation) • 2Al + 3/2O2 -----> Al2O3 ΔH = -1675.7 KJ (heat of formation)
CalculaHonofΔH
• Imaginethisasoccurringin3steps:
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) 4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
CalculaHonofΔH
• Imaginethisasoccurringin3steps:
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) 4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
CalculaHonofΔH
• Imaginethisasoccurringin3steps:
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) 4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
C3H8 (g) ⎯→ 3 C(graphite) + 4 H2 (g)
3 C(graphite) + 3 O2 (g) ⎯→ 3 CO2 (g) 4 H2 (g) + 2 O2 (g) ⎯→ 4 H2O (l)
C3H8 (g) + 5 O2 (g) ⎯→ 3 CO2 (g) + 4 H2O (l)
CalculaHonofΔH
• ThesumoftheseequaHonsis:
Make each reactant or product from its elements This is called the heat of formation of a compound
CalculaHonofΔH
WecanuseHess’slawinthisway:
ΔH=Σ n ΔHf(products)-Σ m ΔHf(reactants)wherenandmarethestoichiometriccoefficients.
° °
StandardEnthalpiesofFormaHon
StandardenthalpiesofformaHon,ΔHf,aremeasuredunderstandardcondiHons(25°Cand1.00atmpressure).
°
CalculaHonofΔH• CalculateΔHusingthetable:• C3H8(g)+5O2(g)----->3CO2(g)+4H2O(l)
CalculaHonofΔH• C3H8+5O2----->3CO2+4H2OΔH = [3(ΔHfCO2) + 4(ΔHfH2O)] - [(ΔHf C3H8) + (5ΔHf O2)]
= [3(-393.5 kJ) + 4(-285.8 kJ)] - [(-103.85 kJ) + 5(0)
= [-1180.5 kJ + (-1143.2 kJ)] - [(-103.85 kJ)+ 0 kJ
= [-2323.7 kJ] - [-103.85 kJ)
= -2219.9 kJ
Phase Changes • Conversionfromonestateofmacertoanotherisaphasechange.
• Energyiseitheraddedorreleasedinaphasechange.
• .
Why is E (Q,H) transferred in phase change? • Atoms/molecules
sHcktoeachotherinliquid/solid,alwayssHckmoreinsolid.
• Atoms/moleculesdon’ttouchingas.
• Liquidtogas,Eisneededtopulltheatoms/moleculesfromeachother.
Heating Curves
• AplotofTvs.q• Withinaphase:• q=ms• Thetemperatureofthesubstancedoesnotriseduringaphasechange.
• For the phase changes, the product of mass and the heat of fusion of vaporization is heat.
Heating Curves
• AplotofTvs.q• Withinaphase:• q=ms• Thetemperatureofthesubstancedoesnotriseduringaphasechange.
• For the phase changes, the product of mass and the heat of fusion of vaporization is heat.
Heating Curves
• AplotofTvs.q• Withinaphase:• q=msΔT• DuringmelHng:• q=ΔHFusm• Duringboiling:• q=ΔHvapm
• For the phase changes, the product of mass and the heat of fusion of vaporization is heat.
Example:
• For the phase changes, the product of mass and the heat of fusion of vaporization is heat.
• Calculateqfortaking10gicefromAtoF:
Example:
ΔHFus water: 334 J/g Δhvap water: 2444 J/g s water: 4.184 J/g°K s ice: 2.05 J/gK s vapor: 2.00 J/gK
Example:
• Calculateqfortaking10gicefromAtoF:• A->B:(10g)2.05J/gK(25°C)=512.5J• B->C:10g(334J/g)=3340J• C->D:10g(4.184J/gK)(100K)=4184J• D->E:10g(2444J/g)=24440J• E->F:10g(2.00J/gK)(25K)=500J• TOTAL =32977J
ΔHFus water: 334 J/g Δhvap water: 2444 J/g s water: 4.184 J/g°K s ice: 2.05 J/gK s vapor: 2.00 J/gK
EnergyinFoodsMostofthefuelinthefoodweeatcomesfromcarbohydratesandfats.
What’sthedealwithfat?• Carbohydrates:• CnH2nOn+nO2-->-->-->nCO2+nH2O+Energy
• Fats:• CnH2nO2+mO2-->-->-->-->-->-->nCO2+nH2O
Fat storage.
more steps
It also clogs your arteries.
Energy and oxidation states
• OxidaHonstateofCinafacyacid:
• OxidaHonstateofCinglucose:
• C6H8O6
CO2-
H3C
CO2-
HH
HH
HH
HH
HHHH
HHHH
Energy and oxidation states
• OxidaHonstateofCinafacyacid:
• OxidaHonstateofCinglucose:
• C6H12O6
• O
• BothgotoCO2C:+4
CO2-
H3C
CO2-
HH
HH
HH
HH
HHHH
HHHH
-3 -2 -2 -2 -2 -2 -2 -2 +3
FuelsThevastmajorityoftheenergyconsumedinthiscountrycomesfromfossilfuels.
Majorissues• Portablefuel(liquid,relaHvelylight),transportaHon• Non-portablefuel(makeselectricity).
transportation
TheEnergycycle:• Us(andalmosteverythingelsealiveontheearth):• C6H12O6+6O2---->6CO2+6H2O
• FossilfuelproducHon:• CH4+2O2---->CO2+2H2O(formethane)
• Plants:• 6CO2+6H2O+light---->C6H12O6+6O2
Net CO2 production could therefore be 0.
Energy research on Campus ADREC
Anaerobic digestor
Energy research on Campus ADREC
Anaerobic digestor • AnaerobicDigesHon:
• C6H12O6---->3CO2+3CH4
Basically:Poop---àCO2+CH4
bacteria
https://www.egr.msu.edu/bae/adrec/feature/south-campus-anaerobic-digester
Hydrogen,theperfectfuel?2H2 + O2 -----> 2H2O ΔH = -285 kJ/mol H2(1mol/2g)= -142 kJ/g
This is literally what fuel cells do. You get nothing but water!
Theproblemwithoil
• Not“renewable”(willrunout)• PolluHon(combusHonnotperfect).• Globalwarming
CO2absorbsheat.CnH2n+2+(3n+1/2)O2----->nCO2+(n+1)H2O
Efficiency/conservaHon
• U.S.coulddecreaseenergyneedsby20-50%bybeinglesswasteful.
• Highmileagecars• moreenergyefficientbuilding/homes.
Hybridcar
• Gasenginepluselectricmotor• Why?• AlltheenergyissHllcomingfromburninggasoline.
Hybrids• Electricmotorsareway
moreefficientthangasengines.(94%)
• Note,yourengineisveryhot,
• Itmustbecooled• FlushallthatEdowndrain.
Nowork,onlyheat.gas engines are 24-30% efficient
Problem: batteries suck! Heavy, expensive, limited recharging cycles, limited current etc.
Liionbacery
x e- +xLi+ + Li1-xCo(IV)O2 -----> LiCo(III)O2
LixC6 ------> xLi+ + xe- + C6
Lithium is really light. Dissolves in organic solvents which are also light. Li is at the top of the activity series. Means a higher potential (more voltage per battery cell)
Hybrids• Electricmotorsworkatlowspeeds• gasengineshutsoffwhennotneeded• atlowspeeds,stoplights,etc.• (infinitetorque,reallygofrom0-15)• Gasenginechargesbaceryandisusedathigherspeeds
• HybridsgetBETTERgasmilageintownversushighway
OthersourcesHowmuchbangforyourbuck?
TheproblemwithHydrogenStorage
gas, less dense, hard to get enough in the car and have trunk space
Kaboom (Hindenburg)
Where do you get the hydrogen?
TheproblemwithHydrogenWhere do you get the hydrogen? (petroleum)
CH4(g) + H2O(g) ---à CO(g) + H2(g)
CO(g) +H2O(g) -à CO2(g) + H2(g)
Ethanol,wheredoesitcomefrom
• AlcoholicfermentaHon:• C6H12O6---->2CO2+2C2H5OH(ethanol)ΔH=-76 kJ/mol • -1270 2(-393) 2(-280) • (anaerobic, bacteria & yeast can do this, we can’t)
Exactly the same place it comes from in your beer.
Ethanol• AlcoholicfermentaHon:• C6H12O6---->2CO2+2C2H6O(ethanol)ΔH=-76 kJ/mol • -1270 2(-393) 2(-280) • (anaerobic, yeast can do this, we can’t) only to 10%. • Distillation (requires energy) to purify.
bug
Alcohol combustion: C2H6O + O2 ---> 2CO2 + 3H2O ΔH = -1367 kJ/mol(1mol/46g)=-29.7kJ/g
But why would this be better for global warming?
Ethanol,problems• Lotsoflandtogrow(yield2-4tons/acre)• AllpresentagriculturallandinU.S.wouldnotbeenoughforall
transportaHonneeds.• requiresferHlizer,tractors,etc.forgrowing(energy)• DisHllaHonrequiresenergy• Forevery1.4kJneed1.0kJ,muchmorethanoil• Brazil,however,isapproaching50%ethanolfortransportaHon• Why?Sugarcane,largeststarchorsugaryield/acre.• But,youcan’tgrowsugarcaneonthegreatplains.
Ethanol
• However,presentlyweonlyuseStarch,
O
H
HO
H
HO
H
OHH
HO
O
H
O
H
HOH
OH
H
H
OH
O
H
O
H
HOH
HO
OH
H
H
OH
O
H
HO
H
HO
H
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OHH
OH
O
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O
H
HO
H
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OHH
OH
O
H
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OH
OH
not cellulose Most stuff in plants is cellulose
Two major types of carbohydrates in plants
Cellulosicethanol
• 10+tons/acre(asopposedto2-4tons/acre)• Canuseanycrop,notjustfoodcropswithhighstarch(“switchgrass”).
• Problem:Breakingitdowntosmallsugarsthatyeastcanferment.
• Needcellulase,theenzymethatbreaksthisup.• ThisisacomparaHvelyeasyproblemtosolve• (comparedtohydrogen.)
Ethanol can work.
Thingstoconsider
• Energyyield(howmuchEoutversusEin)?• Breakevenprice(howmuch/gallonofgasequivalents(presentcornethanolis2.25/gallonjusttomake).
• WhereisthetechnologyNOW?• Isstoragerequired,&ifso,howyougonnadoit• (solarwhenthesundoesn’tshine)• Remember,atpresentBaceriessuck!
TheChemistryNobelPrize
• DanielShechtman,• Technion,Israel• For:• Thediscoveryof“quasi-crystals”in1984
TheChemistryNobelPrize• AnHo-Mg-Znquasi-crystal
Note, the five-fold symmetry of the faces! This was thought to be impossible! Is this a solid?
TheThermitereacHon
• 2Al+Fe2O3------->Al2O3+2Fe
• WhatkindofreacHonisthis?• Whydoesithappen?• Usedforweldingrailroadtracks• WhatistheheatofreacHongiven:
• 2Fe+3/2O2----->Fe2O3ΔH=-825.5KJ• 2Al+3/2O2----->Al2O3 ΔH=-1675.7KJ
TheThermiteReacHon • 2Al+Fe2O3------->Al2O3+2Fe
• WhatistheheatofreacHongiven:• 2Fe+3/2O2----->Fe2O3ΔH=-825.5KJ• 2Al+3/2O2----->Al2O3 ΔH=-1675.7KJ
• 2Al+3/2O2----->Al2O3 ΔH=-1675.7KJ• Fe2O3----->2Fe+3/2O2ΔH=825.5KJ
• 2Al+Fe2O3------->Al2O3+2FeΔH=-850.2KJ
http://www.youtube.com/watch?v=BnHR4cMXiyM
A thermite mystery: