11

Entropy & Gibbs Free Entropy & Gibbs Free EnergyEnergy

Chapter 19Chapter 19

22

The heat taxThe heat tax

• No matter what the process, heat No matter what the process, heat always lost to surroundingsalways lost to surroundings

• More energy required to recharge More energy required to recharge batteries than energy available for batteries than energy available for workwork

33

Spontaneity Spontaneity

• Spontaneous process does NOT require Spontaneous process does NOT require outside energy outside energy – Such as a car rustingSuch as a car rusting

• Non-spontaneous process requires Non-spontaneous process requires external energyexternal energy– Imagine trying to “un-rust” the jalopy aboveImagine trying to “un-rust” the jalopy above

• Not all spontaneous processes Not all spontaneous processes exothermicexothermic– Ice melting above melting pointIce melting above melting point

44

Entropy Entropy

• Disorder/randomness in a systemDisorder/randomness in a system

• Entropy increases in universe (2Entropy increases in universe (2ndnd Law)Law)– We’re all going to dissolutionWe’re all going to dissolution

55

Entropy: the “book” Entropy: the “book” definitiondefinition

• Entropy (S) = a thermodynamic Entropy (S) = a thermodynamic function that increases with number function that increases with number of energetically equivalent ways to of energetically equivalent ways to arrange components of a system to arrange components of a system to achieve a particular stateachieve a particular state

• HUH?!HUH?!

66

Towards greater entropyTowards greater entropy

• Systems process in direction that has Systems process in direction that has largest energetically equivalent ways to largest energetically equivalent ways to arrange its partsarrange its parts

• Which configuration has greater entropy; Which configuration has greater entropy; i.e., largest energetically equivalent ways i.e., largest energetically equivalent ways to arrange itself?to arrange itself?

77

Ludwig BoltzmannLudwig Boltzmann

• Distribution of energy over different Distribution of energy over different energy states used as a way to calculate energy states used as a way to calculate entropyentropy

S = k S = k log W log W

• k = R/Nk = R/N = 1.38 x 10= 1.38 x 10-23-23 J/K = Boltzmann J/K = Boltzmann constant (R = 8.314 Jconstant (R = 8.314 J/mol●K & N = /mol●K & N = Avogadro’s # = 6.022 x 10Avogadro’s # = 6.022 x 102323 atoms/mole) atoms/mole)

• W = # of energetically equivalent ways to W = # of energetically equivalent ways to arrange components of system (unitless)arrange components of system (unitless)

88

99

Standard molar entropy, SStandard molar entropy, S°°

• For standard state elements (metals, For standard state elements (metals, gases, etc.) molar enthalpy of formation, gases, etc.) molar enthalpy of formation, HH°°ff = 0 J/mol = 0 J/mol

• For entropy, it’s different:For entropy, it’s different:– 33rdrd law of thermodynamics: law of thermodynamics:

• The entropy of a perfect crystal at absolute The entropy of a perfect crystal at absolute zero (0K) is zero.zero (0K) is zero.

• Since only one way to arrange (W = 1) Since only one way to arrange (W = 1) S = k S = k log W = k log W = k log 1 = 0 log 1 = 0

1010

More on entropyMore on entropy

• Gas entropies are much larger than Gas entropies are much larger than those for liquidsthose for liquids

• Liquid entropies are larger than solidsLiquid entropies are larger than solids

• For ex: For ex: – II2(s)2(s) vs. Br vs. Br2(l)2(l) vs. Cl vs. Cl2(g)2(g) standard entropies: standard entropies:

116.1, 152.2, 223.1 J/116.1, 152.2, 223.1 J/mol●K, respectivelymol●K, respectively– Solid C vs. gaseous C: 5.6 vs. 158.1 Solid C vs. gaseous C: 5.6 vs. 158.1

J/mol●KJ/mol●K

1111

Even more on entropy!Even more on entropy!

• Generally, larger molecules have larger Generally, larger molecules have larger entropy than smaller moleculesentropy than smaller molecules– Due to greater molar massDue to greater molar mass

• Molecules with more complex structures Molecules with more complex structures have larger entropies than simpler have larger entropies than simpler moleculesmolecules– Since there are more ways for molecule to Since there are more ways for molecule to

twist, rotate, vibrate in spacetwist, rotate, vibrate in space• For ex: For ex:

– CHCH44, C, C22HH66, , CC33HH88: 186.3, 229.2, 270.3 J/mol: 186.3, 229.2, 270.3 J/molKK– Ar, COAr, CO22, C, C33HH88: 154.9, 213.7, 270.3 J/: 154.9, 213.7, 270.3 J/molmolKK

1212

SS°°rxnrxn

• Calculating standard changes in Calculating standard changes in entropyentropy

• Similar to calculating Similar to calculating HH°°rxnrxn

– Don’t forget molar coefficients!Don’t forget molar coefficients!

SS°°rxn rxn = = SS°°

products products - - SS°°reactantsreactants

• All reactants & products in standard All reactants & products in standard statesstates– J/(molJ/(molK)K)

1313

Appendix L, pages A-27-32Appendix L, pages A-27-32

• Thermodynamic values: Thermodynamic values: ∆∆H, H, ∆∆S, & S, & ∆∆GG

1414

Problem Problem

• Compute Compute SS°°rxn rxn for the following equation:for the following equation:

• 4NH4NH3(g)3(g) + 5O + 5O2(g)2(g) 4NO 4NO(g)(g) + 6H + 6H22OO(g)(g)

• Given:Given:

• 192.8 (J/mol192.8 (J/molK) = ammoniaK) = ammonia

• 205.2 = oxygen gas205.2 = oxygen gas

• 210.8 = nitrogen monoxide210.8 = nitrogen monoxide

• 188.8 = water188.8 = water

1515

3(g) 2(g) (g) 2 (g)

reactants

4NH + 5O 4NO + 6H O

S S S

S [(4 210.8 ) (6 188.8 )] [(4 192.8 ) (5 205.2 )]

178.8

rxn products

rxnJ J J Jmol mol mol molmol K mol K mol K mol K

JK

1616

Entropy changes and Entropy changes and spontaneityspontaneity• According to the 2According to the 2ndnd Law of Law of

Thermodynamics:Thermodynamics:– ““A spontaneous process is one that results in an A spontaneous process is one that results in an

increase of entropy in the universe.”increase of entropy in the universe.”

SS°°univuniv = = SS°°systemsystem + + SS°°surroundingsurrounding

• For a spontaneous process:For a spontaneous process:– SS°°univ univ > 0> 0

• For a system at equilibrium:For a system at equilibrium:– SS°°univ univ = 0= 0

• For a nonspontaneous process:For a nonspontaneous process:– SS°°univ univ < 0< 0

1717

Two questionsTwo questions

• Work on these individually and turn Work on these individually and turn them in:them in:– Does the entropy of surrounding Does the entropy of surrounding

increase or decrease in an exothermic increase or decrease in an exothermic process?process?

– Does the entropy of surrounding Does the entropy of surrounding increase or decrease in an endothermic increase or decrease in an endothermic process?process?

1818

Temperature dependence of Temperature dependence of SSsurrsurr

SSunivuniv = = SSsyssys + + SSsurrsurr

• Therefore, for water freezing:Therefore, for water freezing:– SSsys sys is negative (exothermic)is negative (exothermic)– SSsurr surr is positive (endothermic)is positive (endothermic)

S is large at low temperatures (closer to freezing S is large at low temperatures (closer to freezing point): has very little E, so big difference madepoint): has very little E, so big difference made

S is small at high temps (further away from fp): S is small at high temps (further away from fp): already has more E due to higher temp already has more E due to higher temp little little differencedifference– IOW, magnitude of IOW, magnitude of S depends on temp (at above 0S depends on temp (at above 0°C)°C)

1919

Quantifying entropy changes in Quantifying entropy changes in surroundingsurrounding

• --HHsys sys SSsurrsurr

SSsurr surr 1/T 1/T

SSsurr surr - -HHsyssys/T/T

exothermic processes have tendency to be exothermic processes have tendency to be more spontaneous at low temps than highmore spontaneous at low temps than high– They increase entropy of surrounding more if T is They increase entropy of surrounding more if T is

smallsmall• As temp increases (greater E in surrounding), As temp increases (greater E in surrounding), H H

decreases, producing a smaller decreases, producing a smaller SSsurrsurr

2020

ProblemProblem

CC33HH8(g)8(g) + 5O + 5O2(g)2(g) 3CO 3CO2(g)2(g) + 4H + 4H22OO(g)(g)

• Calculate entropy change in surrounding at Calculate entropy change in surrounding at 2525°C given °C given H°H°rxnrxn = -2044 kJ = -2044 kJ

• Determine Determine SSsyssys, given:, given:– HH22OO(g) (g) =188.84 J/mol=188.84 J/molKK– COCO2(g) 2(g) = 213.74 J/mol= 213.74 J/molKK– CC33HH8(g) 8(g) = 270.30 J/mol= 270.30 J/molKK– OO2(g)2(g)= 205.07 J/mol= 205.07 J/molKK

• What is What is SSunivuniv??

2121

syssurr

rxn

univ surr sys

- H ( 2044 )S = 6.86

T 298

S [(4 188.84 ) (3 213.74 ) [1 270.30 ) (5 205.07 )] 100.93

S S S 6860 100.93 6961

kJ kJKK

J J J J Jmol mol mol molmol K mol K mol K mol K KJ J JK K K

2222

Spontaneous or not?Spontaneous or not?

11 ExothermicExothermic Less orderLess order SpontaneousSpontaneous

<0<0 >0>0 Suniv>0Suniv>0

22 ExothermicExothermic More orderMore order Depends on H Depends on H & S& S

<0<0 <0<0 More More favorable at favorable at lower tempslower temps

33 EndothermicEndothermic Less orderLess order Depends on H Depends on H & S& S

>0>0 >0>0 More More favorable at favorable at higher tempshigher temps

44 EndothermicEndothermic More orderMore order NonspontaneoNonspontaneous us

>0>0 <0<0 Suniver<0Suniver<0

2323

Examples of each typeExamples of each type

• Type IType I– Exothermic (Exothermic (HHsyssys<0<0) and less order () and less order (SSsyssys>0)>0)

•Combustion reactionsCombustion reactionsCC33HH8(g)8(g) + 5O + 5O2(g)2(g) 3CO 3CO2(g)2(g) + 4H + 4H22OO(g)(g)

• Type IVType IV– Endothermic Endothermic ((HHsyssys>0>0) and more order ) and more order

((SSsyssys<0)<0)

NN2(g)2(g) + 2H + 2H2(g)2(g) N N22HH4(l)4(l)

2424

Examples of each type Examples of each type (cont.)(cont.)

• Type IIType II– Exothermic (Exothermic (HHsyssys<0<0) and more order ) and more order

((SSsyssys<0)<0)

• Temperature dependentTemperature dependent– More favorable at lower temperaturesMore favorable at lower temperatures

NN2(g)2(g) + 3H + 3H2(g)2(g) 2NH 2NH3(g)3(g)

2525

Examples of each type Examples of each type (cont.)(cont.)• Type IIIType III

– Endothermic Endothermic ((HHsyssys>0>0) and less order ) and less order

((SSsyssys>0)>0)

•Temperature dependentTemperature dependent– More favorable at higher temperaturesMore favorable at higher temperatures

NHNH44ClCl(s)(s) NH NH3(g)3(g) + HCl + HCl(g)(g)

2626

Gibbs Free EnergyGibbs Free Energy

• Measures spontaneity of Measures spontaneity of a process with a process with evaluation of only the evaluation of only the systemsystem– No longer needs No longer needs

evaluation of surroundingevaluation of surrounding

• ““Free energy” Free energy” =“available energy” to =“available energy” to do workdo work– Sum of energies available Sum of energies available

from dispersal of energy from dispersal of energy and matter and matter

univ surr sys

sysuniv sys

univ sys sys

sys univ

sys sys sys

S = S S

- HS = ( ) + S

T-T S = H - T S

G -T S

G H - T S

2727

GG

G (G (G<0) = spontaneous process G<0) = spontaneous process ((S>0)S>0)

G (G (G>0) = non-spontaneous G>0) = non-spontaneous process (process (S<0)S<0)

2828

The effect of The effect of H, H, S, & T on S, & T on ∆G∆G

• Type IType I– Exothermic (-Exothermic (-H) H) – Entropy > 0 (+Entropy > 0 (+S) S) – --G G

•Spontaneous at all tempsSpontaneous at all temps

2N2N22OO(g)(g) 2N 2N2(g)2(g) + O + O2(g)2(g)

2929

The effect of The effect of H, H, S, & T on S, & T on ∆G∆G

• Type IVType IV– Endothermic (+Endothermic (+H) H) – Entropy < 0 (-Entropy < 0 (-S)S)– ++G G

•Nonspontaneous at all tempsNonspontaneous at all temps

3O3O2(g)2(g) 2O 2O3(g)3(g)

3030

The effect of The effect of H, H, S, & T on S, & T on ∆G∆G• Type IIType II

– Exothermic (-Exothermic (-H) H) – Entropy < 0 (-Entropy < 0 (-S) S) – G is temp-dependentG is temp-dependent

HH22OO(l)(l) H H22OO(s)(s) --HH

Water freezing is temp-dependentWater freezing is temp-dependentSpontaneous @ low tempsSpontaneous @ low temps

Non-spontaneous @ high tempsNon-spontaneous @ high temps

3131

The effect of The effect of H, H, S, & T on S, & T on ∆G∆G• Type IIIType III

– Endothermic (+Endothermic (+H)H)– Spontaneous (+Spontaneous (+S)S)– G is temp-dependentG is temp-dependent

HH22OO(l)(l) H H22OO(g)(g) ++HH

Water boiling is temp-dependentWater boiling is temp-dependentSpontaneous @ high tempsSpontaneous @ high tempsNonspontaneous @ low tempsNonspontaneous @ low temps

3232

Calculating Calculating GG°°rxnrxn

• Calculate Calculate GG°°rxn rxn (at 25°C)(at 25°C) forfor

SOSO2(g)2(g) + ½ O + ½ O2(g) 2(g) SO SO3(g)3(g)

• SOSO22; ; HH°°f f = -296.8 kJ/mol & = -296.8 kJ/mol & SS° ° = 248.2 = 248.2

J/(molJ/(molK)K)

• OO22; ; HH°°f f = ?, = ?, SS° ° = = 205.2J/(mol205.2J/(molK)K)

• SOSO33; ; HH°°ff = -395.7 kJ/mol, = -395.7 kJ/mol, SS° ° = 256.8 = 256.8

J/(molJ/(molK)K)

• Is it spontaneous under standard conditions?Is it spontaneous under standard conditions?

3333

rxn

rxn

3rxn

kJ kJH° [(1 -395.7 )] [(1 -296.8 )] 98.9mol molJ J J J° =[(1mole 256.8 )]-[(1mol 248.2 )+(1/2mole 205.2 )]=-94.0mol K mol K mol K K

J° = 98.9 10 (298 -94.0 ) 70888K

The rxn is spontaneous a

mol mol kJ

S

G J K J

t 25 C.

3434

Calculating Calculating GG°°rxn rxn from free from free

energies of formationenergies of formation

• Free E of formation of pure Free E of formation of pure elements in standard states = 0 elements in standard states = 0 kJ/molkJ/mol

GG°°rxnrxn = = GG°°

products products - - GG°°reactantsreactants

• CHCH4(g)4(g) + 8O + 8O2(g)2(g) CO CO2(g)2(g) + 2H + 2H22OO(g)(g)

• -50.5 ? -394.4 -228.6-50.5 ? -394.4 -228.6

• Calculate Calculate GG°°rxnrxn

3535

rxn products reactants

rxn

G° = G° - G°

G° =[(1mol 394.4 ) (2 228.6 )]-[(1mol -50.5 )+(8mol 0.0 )]=-801.1kJ kJ kJ kJmol kJmol mol mol mol

3636

Determine Determine GG°°rxn rxn for for

3C3C(s)(s) + 4H + 4H2(g)2(g) C C33HH8(g)8(g) • Given: Given:

CC33HH8(g)8(g) + 5O + 5O2(g)2(g) 3CO 3CO2(g)2(g) + 4H + 4H22OO(g)(g) GG°°

rxn rxn = -2074 kJ= -2074 kJ

CC(s)(s) + O + O2(g)2(g) CO CO2(g)2(g)

GG°°rxnrxn = -394.4 kJ = -394.4 kJ

2H2H2(g)2(g) + O + O2(g)2(g) 2H 2H22OO(g)(g)

GG°°rxnrxn = -457.1 kJ = -457.1 kJ

3737

2(g) 2 (g) 3 8(g) 2(g) rxn

(s) 2(g) 2(g) rxn

2(g) 2(g) 2 (g) rxn

(s) 2(g)

3CO + 4H O C H + 5O ; G° = +2074 kJ

3C + 3O 3CO ; G° = -394.4 kJ 3

4H + 2O 4H O ; G° = -457.1 kJ 2

_______________________________________________

3C + 4H C

3 8(g)H ; 23kJ

3838

Non-standard conditionsNon-standard conditions

GG°°rxn rxn is at standard conditionsis at standard conditions

– 25 °C, 273 K, 1 atm, 1M, etc.25 °C, 273 K, 1 atm, 1M, etc.

GGrxn rxn = non-standard conditions= non-standard conditions– Explains why water evaporates off floor Explains why water evaporates off floor

even though even though GG°°rxnrxn = 8.59 kJ/mol & = 8.59 kJ/mol & is a is a

non-spont process.non-spont process.

• Partial pressure of water well below 1 Partial pressure of water well below 1 atm atm Non-standard condition! Non-standard condition!

3939

Free E change under non-Free E change under non-standardstandard conditions conditions

GGrxn rxn = = GG°°rxn rxn + RT(lnQ)+ RT(lnQ)

• Q = rxn quotient, R = 8.314 J/(molQ = rxn quotient, R = 8.314 J/(molK)K)

4040

Problem Problem

• Compute Compute GGrxn rxn under the following under the following conditions for: conditions for:

2NO2NO(g)(g) + O + O2(g)2(g) 2NO 2NO2(g)2(g)

– PPNONO = 0.100 atm; P = 0.100 atm; POO22 = 0.100 atm; P = 0.100 atm; PNONO22

= =

2.00 atm 2.00 atm

– GG°°rxn rxn = -71.2 kJ= -71.2 kJ

– At 25°CAt 25°C

4141

(g) 2(g) 2(g)

rxn rxn

2 232

2 22

3

rxn

2NO + O 2NO

G = G° + RT(lnQ)

(NO ) (2.00)Q= 4.00 10

(NO) (O ) (0.100) (0.100)

ln(4.00 10 ) 8.29

JG = (-71.2 kJ) + (8.314 )(298K)(8.29)=20467.9kJmol K

4242

Relating Relating GG°°rxn rxn to Kto K

GGrxn rxn = = GG°°rxn rxn + RT(lnQ)+ RT(lnQ)

At equilibrium, Q = K & At equilibrium, Q = K & GGrxn rxn = 0= 000 = = GG°°

rxn rxn + RT(lnK)+ RT(lnK)

GG°°rxn rxn = -RT(lnK)= -RT(lnK)

• When K<1 (reactant-favored)When K<1 (reactant-favored)– lnK = “-” & lnK = “-” & GG°°

rxn rxn = “+”= “+” spontaneous in reverse directionspontaneous in reverse direction

• When K>1 (product-favored)When K>1 (product-favored)– lnK = “+” & lnK = “+” & GG°°

rxn rxn = “-’= “-’ spontaneous in fwd directionspontaneous in fwd direction

• When K = 1, lnK = 0 & When K = 1, lnK = 0 & GG°°rxn rxn = 0= 0

– Rxn @ equilibrium under standard conditionsRxn @ equilibrium under standard conditions

4343

Temp dependence of KTemp dependence of K

• How does one obtain the equation on the How does one obtain the equation on the left?left?

• Useful for obtaining Useful for obtaining HH°°rxn rxn & & SS°°

rxnrxn

• Does the equation look familiar?Does the equation look familiar?

• How would we plot this? How would we plot this?