GCh18-1

Chapter 18. Entropy, Free Energy, and Equilibrium

What we will learn:

• Three laws of theormodynamic

• Spontaneous processes

• Entropy

• Second law of thermodynamics

• Gibbs free energy

• Free energy and chemical reactions

GCh18-2

Thermodynamics

The scientific discipline that deals with the interconversion of heat and other forms of energy

First law of thermodynamics

Energy may be converted from one form to another, but cannot be created or destroyed. (one way to measure: DH)

Second law of thermodynamics

The law explains why chemical processes tend to favor one direction

Third law of thermodynamics

The law is an extension of the second law and only briefly examined in this course

GCh18-3

Spontaneous Processes and Entropy

Spontaneous processes

• Waterrunsdownhill

• Saltdissolvesinwater

• Heatmovestocolderobject

• Ironrusts

• Methaneburns(exothermic,DH=-value)

• Acidsneutralizebases(exothermic,DH=-value)

• Icemeltsatroomtemperature(endothermic,DH=+value)

• Ammoniumnitrate(NH4NO3)dissolvesandsolutiongetscold(endothermic, DH=+value)

GCh18-4

Two factors related to spontaneous processes

1. Enthalpy(H)

Amount of heat given or absorbed by a system Exothermicprocesses(negativeDH)tendtobespontaneous,butnot always

2. Entropy(S)

Degree of disorder or randomness

Anorderedstatehasalowprobabilityofoccurringandasmallentropy, while a disordered state has a high probability of occurring and a high entropy

GCh18-5

Macrostate (distribution) and microstates

Example

Four molecules distributed between two containers

Macrostate(I)Sissmall(fourmoleculesinleftcontainer)

Macrostate(II)Sisbig(threemoleculesinleftcontainer)

Macrostate(III)Sisverybig(twomoleculesinleftcontainer)

GCh18-6

Entropy

• Analogies:deckofcards,socksindrawer,molecularmotions

• Changeinentropy: DS=Sfinal - Sinitial

• Forareaction: DS=Sproducts - Sreactants

• PositiveDS means:

• Anincreaseindisorderasreactionproceeds

• Productsaremoredisordered(random)thanreactants

Sgas >> Sliquid > Ssolid

Ssolution >> Ssolventorsolute

• Temperatureincrease–moremolecularmotions

GCh18-7

S = k ln W

S - entropyk - constant(Boltzmanconstant=1.38x10-23J/K)W - disorder(howmanyindividualmicrostatesisinvolvedinamacrostate)

DS = Sf - Si

DS = k ln(Wf) - k ln(Wi)

DS = k ln (Wf / Wi)

GCh18-8

Standard entropy

The absolute entropy of a mole of a substance at 1 atm and 25C

Theabsoluteentropyisdifficulttodetermine,becauseitisdifficulttodeterminea number of microstates corresponding to a particular macrostate. The standard entropy of elements and compounds are already determined, and therefore we usuallyrefertheactualentropyofasubstanttoitsstandardvalue

Substance S0

(J/Kmol)H2O(l) 69.9H2O(g) 188.7Br2(l) 152.3Br2(g) 245.3

GCh18-9

Second law of thermodynamics

• Foranyspontaneousprocess,theoverallentropyoftheuniverseincreases

DSuniverse=DSsystem+DSsurroundings • Theentropyoftheuniverseincreasesinaspontaneousprocessand remains unchanged in an equilibrium process

DSuniverse=DSsystem+DSsurroundings>0(spontaneous)

DSuniverse=DSsystem+DSsurroundings=0(equilibrium)

AspontaneousprocesscanhaveanegativeDS for the system only if the surroundingshavealargerpositiveDS

GCh18-10

A. Entropychange(DS)inthesystem

aA + bB g cC + dD

DS0rxn = [ cS0(C) + dS0(D) ] - [ aS0(A) + bS0(B) ]

DS0rxn =S nDS0

products - S mDS0reactants

DS0rxn - standard entropy of reaction

n - mole number of products m - mole number of reactantsS0

products - standard entropy of prodcutsS0

reactants - standard entropy of reactants

GCh18-11

Problem

Calculate the standard entropy change for the reaction

N2 (g) + 3 H2 (g) g 2 NH3 (g)

DS0(N2) = 191.5J/KmolDS0(H2) = 131.0J/KmolDS0(NH3) = 193.0J/Kmol

Solution

DS0rxn = 2S0(NH3) - [S0(N2)+3S

0(H2)]

= -199J/Kmol

GCh18-12

Problem

Calculate the standard entropy change for the reaction

CaCO3 (s) g CaO (s) + CO2 (g)

DS0(CaCO3) = 92.9J/KmolDS0(CaO) = 39.8J/KmolDS0(CO2) = 213.6J/Kmol

Solution

DS0rxn = S0(CaO)+S0(CO2)-S

0(CaCO3)

= 160.5J/Kmol

GCh18-13

Problem

Calculate the standard entropy change for the reaction

CaCO3 (s) g CaO (s) + CO2 (g)

DS0(CaCO3) = 191.5J/KmolDS0(CaO) = 131.0J/KmolDS0(CO2) = 193.0J/Kmol

Solution

DS0rxn = S0(CaO)+S0(CO2)-S

0(CaCO3)

= 160.5J/Kmol

GCh18-14

General Rules

• Inthegasphasetherearemoremicrostatesthanincorrespondingliquid and solid phasses

• Ifareactionproducesmoregasmoleculesintheproductsthaninthe reactants, DS0

rxn>0

• Ifareactionproduceslessgasmoleculesintheproductsthaninthe reactants, DS0

rxn<0

• Ifthereisnonetchangeinthenumberofgasmolecules, DS0rxn~0

GCh18-15

Problem

Predicttheentropychangeofthesystemineachofthefollowingreactionsispositiveornegative

2 H2 (g) + O2 (g) g 2 H2O (l) DS<0

Two molecules combine to form a product. There is a net decrease of one moleculeandgasesareconvertedintoliquid.Thenumberofmicrostatesofproductthatwillbesmaller,andentropywillbenegative

NH4Cl (s) g NH3 (g) + HCl (g) DS>0

Asolidisconvertedintotwogaseousproducts H2 (g) + Br2 (g) g 2HBr (g) DS=0

Twodiatomicmoleculesareinvolvedinreactantsandproducts.Allmoleculeashavesimilarcomplexity,andthereforeweexpectsimilarentropyofreactantsandproducts

GCh18-16

Third law of thermodynamics and absolute entropy

• The entropy of a pure crystalline substance equals zero at absolute zero

S = 0 at T = 0 K

any substance above 0 K has entropy > 0 any impure crystal has entropy > 0

• Standard Entropy (at 25 C) = S0 entropy at 25 C which is relative to 0 (zero) entropy at absolute zero

Atabsolutezeroallmolecularmovesarefrozen,andthereforethereisonlyonemicrostate which forms a particular macrostate.

S = k ln (W) W = 1 ln(1) = 0

S = 0

GCh18-17

Entropy changes in surroundings

Forexothermicprocessestheheatistranferedfromasystemtosurroundingsincreasing the number of microstates in the surroundings, therefore the entropy of surroundings increases

Ifthetemperatureofthesurroundingsishigh,thetransferedheatdoesnotchangethenumberofmicrostatesmuch,howeverifthetemperatureislow,thetransfered heat changes the number of microstates more strongly

DSsurr = -DHsys / T

Example

N2 (g) + 3 H2 (g) g 2 NH3 (g) -DH0rxn=-92.6kJ/mol

DSsys = -199J/Kmol

DSsurr = -(-92.6x1000J/mol)/298K = 311J/Kmol

DSuniv = DSsys + DSsurr

GCh18-18

DSuniv =112J/Kmol

BecausetheDSunivispositive,weexpectthatthereactionisspontaneousat 25 C.

GCh18-19

Gibbs free energy (G)

AnotherthermodynamicfunctionthathelpsdeterminewhetherareactionisspontaneousisGibbsfreeenergy,alsoknownasfreeenergy.Twodrivingforcesinthenature(onerelatedtotheenergychange,andanotherrelatedtothedissorderchange)arecombinedinoneequation

The term “free” meanes that Gibbs free energy is an amount of energy of a molecularsystem,whichcanbeusedforwork(exchangedfromaheattomechanicalwork,orinanoppositedirection).Therestoftheenergy,whichcannotbeconveredintowork,isrelatedtoamoleculardissorder(entropy)

G = H - TS

Ifareactionisspontaneous

DSuniv =DSsys+DSsurr > 0

DSsurr = -DHsys / T

DSuniv =DSsys+-DHsys/T > 0

GCh18-20

-TDSuniv= DHsys - TDSsys < 0

-TDSuniv= DG

DG = DH-TDS < 0

• Foranyspontaneouschange,DGisnegative (thefreeenergydecreases)

• Gisastatefunction(likeHandS)-independentofpathway

Summary

• DG<0 reactionisspontaneousinforwarddirection

• DG>0 reactionisnonspontaneousoritis spontaneous in the opposite direction

• DG=0 atequilibrium

GCh18-21

Standard Free Energy Changes DG0

• Freeenergychangeforareactionwhenitoccursunderstandard-state conditions(reactantsandproductsareintheirstandardstates)

StandardTemperatureandPressure(STP) 1atmosphere,solutionsat1M,DGf=0fortheelements

• DG0(atSTP)isgenerallyusedtodecideifareactionisspontaneous. ItisspontaneousifDG0

rxnisnegative

• TwowaystoobtainDG0rxn for a reaction:

(1) FromDH0f and DS0 requireshavingDH0

f and DS0 data for all reactants and products

DG0=DH0 - TDS0

DH0=S DH0f(products)-S DH0

f(reactants)

DS0=S DS0(products)-S DS0(reactants)

GCh18-22

(2) Fromstandard“FreeEnergiesofFormation”DG0f

DG0rxn =S nDG0

f(products)-S mDG0f(reactants)

where DG0f is the free energy change for the formation of one mole of

the compound from its elements

Standard free energy of formation

The free energy change that occurs when 1 mole of the compound is synthesized from its elements in their standard states

GCh18-23

Example

DG0fforAl2(SO3)3 equals DG0

f for the following reaction

2 Al (s) + 3 S (s) + 9/2 O2 (g) D Al2(SO3)3 (s)

Asinthecaseofthestandardenthalpyofformation,wedefinethestandardfreeenergy of formation of any element in its allotropic form at 1 atm and 25 C

Example

DG0f(O2) = 0

DG0f(C,graphite) = 0

GCh18-24

Problem

Calculate the standard free energy changes for the following reactions at 25 C

a) CH4 (g) + 2 O2(g)g CO2 (g) + 2 H2O (l)

DG0rxn =DG0

f(CO2)+2DG0f(H2O)-DG0

f(CH4)-DG0f(O2)

DG0f(O2)=0becauseitisastableallotropicelementat1atmand25C

DG0rxn = -818.0kJ/mol

b) 2 MgO (s) g 2 Mg (s) + O2 (g)

DG0rxn =2DG0

f(Mg)+DG0f(O2)-2DG0

f(MgO)

DG0rxn = 1139kJ/mol

GCh18-25

Temperature and chemical reactions

“Ballpark”estimatesoftemperatureatwhichareactionbecomesspontaneous

Procedure:

• FindDH°andDS°at25C

• SetDH-TDS=0

• SolveforT,estimatedtemperatureatwhichreactionbecomes spontaneous

GCh18-26

Problem

Calciumoxide(CaO)isproducedaccordingtothereaction

CaCO3 (s) D CaO (s) + CO2 (g)

Whatisthetemperatureatwhichthereactionfavorstheproducts

First we calculate DH0 and DS0 using the standard data of the reactants and products

DH0= DH0f(CaO)+DH

0f(CO2)-DH

0f(CaCO3)

DH0= 177.8kJ/mol

DS0= S0(CaO)+S0(CO2)-S0(CaCO3)

DS0= 160.5J/Kmol

GCh18-27

DG0= DH0 - TDS0

DG0= 130.0kJ/mol at298K

BecauseDG0isalargepositivevalue,weconcludethatthereactionisnotfavoredforproductsat25C(298K).InordertomakeDGnegative,firstwefindthe temperature at which DGiszero

0 = DH0 - TDS0

T = DH0/ DS0

T = 1108K = 835C

Attemperaturehigherthat835CDG0becomesnegativeindicatingthatthereactionnowfavorstheproductformation

GCh18-28

Phase Transitions

Ataphasetransitionthesystemisatequilibrium

• DG=0 atthetemperatureofaphasetransition

• DG=DH-TDS

• Thus DH-TDS=0

• So DS=DH/T

GCh18-29

Problem

Themolarheatoffusionofwaterare6010J/mol.Calculatetheentropychanges for the ice g water transitions.

Theentropychangeformelting1moleofwaterat0Cis

DSice g water = DHfus / Tf

= (6010J/mol)/(273)K

= 22J/Kmol

Thuswhen1moleoficemeltsat0Cthereisanincreaseinentropyof22J/Kmol. The increase in entropy is consistent with the increase in microstates from solid to liquid. Consequently for the liquid g solid transition the decrease of entropy is

DSwater g ice = -DHfus / Tf

DSwater g ice = -22J/Kmol

GCh18-30

Problem

Themolarheatoffusionandvaporizationofbenzeneare10.9kJ/moland31kJ/mol.Calculatetheentropychangesforthesolidg liquid and liquid gvaportransitions.At1atm,benzenemeltsat5.5Candboilsat80.1C.

Theentropychangeformelting1moleofbenzeneat5.5Cis

DSfus = DHfus / Tf

= (10.9kJ/mol)(1000J/kJ)/(5.5+273)K

= 39.1J/Kmol

Theentropychangeforboiling1moleofbenzeneat80.1Cis

DSvap = DHvap / Tbp

= (31kJ/mol)(1000J/kJ)/(80.1+273)K

= 87.8J/Kmol

GCh18-31

Free energy and chemical equilibrium

During a chemical reactions not all reactants and products will be at their standard states. Under this condition, there is a relationship between DG and DG0

• Foranychemicalsystem:

DG=DG0 + (RT) ln Q

Q - concentration quotient

• IfDGisnotzero,thenthesystemisnotatequilibriumanditwill spontaneously shift toward the equilibrium state

• But,atequilibrium:DG=0andQ=K

DG0 = - (RT) ln K

GCh18-32

• RelatesequilibriumconstanttostandardfreeenergyDG0

• AllowsthecalculationofKifDG0isknownandviceversa,e.g.,whenKis verysmallorlarge,maybedifficulttomeasuretheconcentrationofa reactantorproduct,soitiseasiertofindDG0 and calculate K

• Forgaseousreactions: K=Kp

• Forsolutionreactions: K=Kc

• UnitsofDGmustmatchthoseofRTvalue

GCh18-33

(a) G0prod < G0

reac so DG0<0(negative)

(b) G0reac < G0

prod so DG0>0(positive)

(c) Theactualfreeenergychange(DG)ofthesystemvariesasthereaction progressesandbecomeszeroatequilibrium.Howeverthestandardfree energy(DG0)changeisaconstantforaparticularreactionataaprticular temperature

GCh18-34

• Gproducts=Greactants and DG=0

• Since DG=DH-TDS=0

• DH=TDS or T=DH/DS

Problem

Giventhefollowing,determinethetemperatureofthenormalboilingpointofmercury(Hg)

Hgliq D Hggas

Data DHvaporization =60.7kJ/moleS0((liquid) =76.1J/moleKS0(vapor) =175J/moleK

GCh18-35

Atequilibrium DG=0

DH-TDS=0

T=DH/DS

T =(60.7x103J/mole)/[(175-76.1)J/moleK]

T =(60.7x103J/mole)/(273K) =614K

GCh18-36

Problem

Calculate the equilibrium constant, KP, for the reaction shown below at 25 C

2 H2O (l) D 2 H2 (g) + O2 (g)

Solution

First we calculate DG0

DG0 = 2DG0f(H2)+DG0

f(O2)-2DG0f(H2O)

= 474.4kJ/mol

GCh18-37

DG0 = - (RT) ln Kp

474.4kJ/mol(1000J/kJ) = -(8.314J/Kmol)(298K)ln(Kp)

ln(Kp) = -191.5

Kp = e-191.5

= 7x10-84

GCh18-38

Problem

Calculate the standard free energy change, DG0, of the reaction shown below at 25 C

AgCl (s) D Ag+ (aq) + Cl- (aq)

The solubility product Kspofthereactionis1.6x10-10.

Solution

DG0 = - (RT) ln Ksp

Ksp = [Ag+] [Cl-] = 1.6x10-10

DG0 = -(8.314J/Kmol)(298K)ln(1.6x10-10)

= 56kJ/mol

GCh18-39

Problem

Theequilibriumconstant(Kp)forthereaction

N2O4 (g) D 2 NO2 (g)

is0.113at298K,whichcorrespondstoastandardfreeenergychangeof5.4kJ/mol.TheinitialpressuresarePNO2=0.122atmandPN2O4=0.453atm.CalculateDG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium

Solution

DG = DG0 + (RT) ln Qp

Qp = P2NO2 / PN2O4

DG = DG0 + (RT) ln (P2NO2 / PN2O4)

= 5.6x103J/mol+(8.314J/Kmol)(298K)ln[(0.122)2/(0.453)]

GCh18-40

DG = 5.4x103J/mol-8.46x103 J/mol

= -3.06x103 J/mol

BecauseDG < 0,thenetreactionproceedsfromlefttoright(fromreactantstoproducts)toreachequilibrium

GCh18-41

Problem

The DG 0 for the reaction

H2 (g) + I2 (g) D 2 HI(g)

is2.6kJ/molat25K.TheinitialpressuresarePH2=4.26atmandPI2=0.024atmandPHI=0.23.CalculateDG for the reaction at these pressures and predict the direction of the net reaction toward equilibrium

Solution

DG = DG0 + (RT) ln Qp

Qp = P2HI / (PH2 PI2 )

DG = DG0 + (RT) ln [P2HI / (PH2 PI2)]

= 2.6x103J/mol+(8.314J/Kmol)(298K)ln[(0.23)2/(4.26x0.024)]

= 0.97x103J/mol = 0.97kJ/mol