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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
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