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

H

OHH

OH

O

H

O

H

HO

H

H

OHH

OH

O

H

O

H

HO

H

H

OHH

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: