Spontaneity, Entropy, & Free Energy Chapter 16. 1st Law of Thermodynamics The first law of thermodynamics is a statement of the law of conservation of.

Spontaneity, Entropy, & Free Energy

Chapter 16

1st Law of Thermodynamics

The first law of thermodynamics is a The first law of thermodynamics is a statement of the law of conservation of statement of the law of conservation of energy: energy: energy can neither be created nor energy can neither be created nor destroyeddestroyed. The energy of the universe is . The energy of the universe is constant, but the various forms of energy constant, but the various forms of energy can be interchanged in physical and can be interchanged in physical and chemical processes.chemical processes.

Spontaneous Processes and Entropy

ThermodynamicsThermodynamics lets us predict lets us predict whether a whether a process will occur process will occur but gives no information but gives no information about the amount of time required for the about the amount of time required for the process.process.

A A spontaneousspontaneous process is one that process is one that occurs occurs without outside interventionwithout outside intervention..

Kinetics & ThermodynamicsChemical kinetics focuses on the pathway Chemical kinetics focuses on the pathway

between reactants and products--the between reactants and products--the kinetics of a reaction depends upon kinetics of a reaction depends upon activation energy, temperature, activation energy, temperature, concentration, and catalysts.concentration, and catalysts.

Thermodynamics only considers the initial Thermodynamics only considers the initial and final states.and final states.

To describe a reaction fully, both To describe a reaction fully, both kinetics and thermodynamics are kinetics and thermodynamics are necessary.necessary.

16_343

Ene

rgy

Reaction progress

Reactants

Products

Domain of kinetics(the reaction pathway)

Domain ofthermodynamics

(the initial andfinal states)

The rate of a reaction depends on the pathway from reactants to products. Thermodynamics tells whether the reaction is spontaneous and depends upon initial & final states only.

Entropy

The driving force for a spontaneous process The driving force for a spontaneous process is an is an increase in the entropy of the universeincrease in the entropy of the universe..

Entropy, Entropy, SS, can be viewed as a measure of , can be viewed as a measure of randomness, or disorder.randomness, or disorder.

Nature spontaneously proceeds toward the Nature spontaneously proceeds toward the states that have the highest probabilities of states that have the highest probabilities of existing.existing.

16_349

The expansion of an ideal gas into an evacuated bulb.

Positional Entropy

A gas expands into a vacuum because the A gas expands into a vacuum because the expanded state has the highest expanded state has the highest positional positional probability probability of states available to the of states available to the system.system.

Therefore, Therefore,

SSsolidsolid < < SSliquidliquid << << SSgasgas

Positional Entropy

Which of the following has higher positional Which of the following has higher positional entropy?entropy?

a) Solid COa) Solid CO22 or gaseous CO or gaseous CO22??

b) Nb) N22 gas at 1 atm or N gas at 1 atm or N22 gas at 1.0 x 10 gas at 1.0 x 10-2 -2 atm?atm?

EntropyWhat is the sign of the entropy change for the What is the sign of the entropy change for the

following?following?

a) Solid sugar is added to water to form a a) Solid sugar is added to water to form a solution? solution?

S is positiveS is positive

b) Iodine vapor condenses on a cold surface b) Iodine vapor condenses on a cold surface to form crystals? to form crystals?

S is negativeS is negative

The Second Law of Thermodynamics

. . .. . . in any spontaneous process there is in any spontaneous process there is always an always an increase in the entropy of the increase in the entropy of the universeuniverse..

SSunivuniv > 0 > 0

for a spontaneous process.for a spontaneous process.

SUniverse

SSuniverse universe is positive -- reaction is spontaneous.is positive -- reaction is spontaneous.

SSuniverseuniverse is negative -- reaction is spontaneous in the is negative -- reaction is spontaneous in the reverse direction.reverse direction.

SSuniverse universe = 0 -- reaction is at equilibrium.= 0 -- reaction is at equilibrium.

G -- Free Energy

Two tendencies exist in nature:Two tendencies exist in nature:

• tendency toward higher entropy -- tendency toward higher entropy -- SS

• tendency toward lower energy -- tendency toward lower energy -- HH

If the two processes oppose each other (e.g. If the two processes oppose each other (e.g. melting ice cube), then the direction is decided melting ice cube), then the direction is decided by the Free Energy, by the Free Energy, G, G, and depends upon the and depends upon the temperature.temperature.

Free EnergyGG = = HH TTSS (from the standpoint of the system)(from the standpoint of the system)

A process (at constant A process (at constant TT, , PP) is spontaneous in ) is spontaneous in the direction in which free energy decreases:the direction in which free energy decreases:

GGsyssys meansmeans ++SSunivuniv

Entropy changes in the surroundings are primarily determined by the heat flow. An exothermic process in the system increases the entropy of the surroundings.

G, H, & S

Spontaneous reactions are indicated by the Spontaneous reactions are indicated by the following signs:following signs:

G = negativeG = negative

H = negativeH = negative

S = positiveS = positive

Temperature Dependence

HHoo & & SSoo are are not temperature dependent.

GGoo is temperature dependent. is temperature dependent.

GG = = HH TTSS

Ssurroundings

SSsurr surr is positive -- heat flows into the is positive -- heat flows into the surroundings out of the system.surroundings out of the system.

SSsurr surr is negative -- heat flows out of the is negative -- heat flows out of the surroundings and into the system.surroundings and into the system.

Ssurr = - Hsystem

T

SbSb22SS3(s)3(s) + 3Fe + 3Fe(s)(s) ---> 2Sb ---> 2Sb(s)(s) + 3FeS + 3FeS(s) (s) H = -125 kJH = -125 kJ

SbSb44OO6(s)6(s) + 6C + 6C(s)(s) ---> 4Sb ---> 4Sb(s)(s) + 6CO + 6CO(g)(g) H = 778 kJH = 778 kJ

What is What is SSsurr surr for these reactions at 25for these reactions at 2500C & 1 atm.C & 1 atm.

SSsurr surr = - = - HHsystemsystem

TT

SSsurr surr = -(-125kJ/298K)= -(-125kJ/298K)

SSsurrsurr = 419 J/K = 419 J/K

Ssurroundings Calculations

Ssurr = - Hsystem

TSsurr = -(778kJ/298K)

Ssurr = -2.61 x 103 J/K

Effect of H and S on Spontaneity

H S Result

+ spontaneous at all temps

+ + spontaneous at high temps

spontaneous at low temps

+ not spontaneous at any temp

16_03T

Table 16.3 Interplay of Ssys and Ssurr in Determining the Sign of Suniv

Signs of Entropy Changes

Process Spontaneous?

Yes No (reaction will occur

in opposite direction)

Yes, if Ssys has a larger magnitude than Ssurr

Yes, if Ssurr has a larger magnitude than Ssys

Ssys Ssurr Suniv

Calculations showing that the melting of ice is temperature dependent. The process is spontaneousabove 0oC.

16_05T

Table 16.5 Various Possible Combinations of H and S for a Process and the Resulting Dependence of Spontaneity on Temperature

Case Result

S positive, H negative Spontaneous at all temperaturesS positive, H positive Spontaneous at high temperatures

(where exothermicity is relatively unimportant)S negative, H negative Spontaneous at low temperatures

(where exothermicity is dominant)S negative, H positive Process not spontaneous at any temperatures

(reverse process is spontaneous at all temperatures)

Free Energy G

G = H TS

GG = negative -- spontaneous = negative -- spontaneous

GG = positive -- spontaneous in = positive -- spontaneous in opposite directionopposite direction

GG = 0 -- at equilibrium = 0 -- at equilibrium

Boiling Point CalculationsWhat is the normal boiling point for liquid BrWhat is the normal boiling point for liquid Br22??

BrBr2(l)2(l) ---> Br ---> Br2(g)2(g)

HHoo = 31.0 kJ/mol & = 31.0 kJ/mol & SSo o = 93.0 J/Kmol= 93.0 J/KmolAt equilibrium, GGoo = 0 = 0

Go = Ho TS0 = 0Ho = TS0

T = Ho/S0

T = 3.10 x 104 J/mol/(93.0J/Kmol)T = 333K

The Third Law of Thermodynamics

. . . the entropy of a perfect crystal at 0 K is zero.. . . the entropy of a perfect crystal at 0 K is zero.

Because Because SS is explicitly known (= 0) at 0 K, is explicitly known (= 0) at 0 K, SS values at other temps can be calculated. values at other temps can be calculated.

See Appendix 4 for values of SSee Appendix 4 for values of S00..

Soreaction

Calculate Calculate SS at 25 at 25 ooC for the reactionC for the reaction

2NiS(s) + 3O2(g) ---> 2SO2(g) +2NiO(s)

SS = = nnppSS(products)(products) nnrrSS(reactants)(reactants)

SS = [(2 mol SO = [(2 mol SO22)(248 J/Kmol) + (2 mol NiO)(38 )(248 J/Kmol) + (2 mol NiO)(38 J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol OO22)(205 J/Kmol)])(205 J/Kmol)]

SS = 496 J/K + 76 J/K - 106 J/K - 615 J/K = 496 J/K + 76 J/K - 106 J/K - 615 J/K

SS = -149 J/K = -149 J/K # gaseous molecules decreases!# gaseous molecules decreases!

So

SSo o increases with:increases with:

• solid ---> liquid ---> gassolid ---> liquid ---> gas

• greater complexity of molecules (have a greater complexity of molecules (have a greater number of rotations and greater number of rotations and vibrations)vibrations)

• greater temperature (if volume greater temperature (if volume increases)increases)

• lower pressure (if volume increases)lower pressure (if volume increases)

Free Energy Change and Chemical Reactions

GG = = standard free energy change standard free energy change that that occurs if reactants in their standard occurs if reactants in their standard state are converted to products in state are converted to products in their standard state.their standard state.

GG = = nnppGGff(products)(products) nnrrGGff(reactants)(reactants)

The more negative the value of Gthe further a reaction will go to the right to reach equilibrium.

GCalculationsCalculate Calculate , , SSGGfor the reactionfor the reaction

SOSO2(g)2(g) + O + O2(g)2(g) ----> 2 SO ----> 2 SO3(g)3(g)

= = nnppHHff(products)(products) nnrrHHff(reactants)(reactants)

= [(2 mol SO= [(2 mol SO33)(-396 kJ/mol)]-[(2 mol )(-396 kJ/mol)]-[(2 mol SOSO22)(-297 kJ/mol) + (0 kJ/mol)])(-297 kJ/mol) + (0 kJ/mol)]

HH = - 792 kJ + 594 kJ = - 792 kJ + 594 kJ

HH = -198 kJ = -198 kJ

GCalculationsContinued

SS = = nnppSS(products)(products) nnrrSS(reactants)(reactants)

SS = [(2 mol SO = [(2 mol SO33)(257 J/Kmol)]-[(2 mol SO)(257 J/Kmol)]-[(2 mol SO22))(248 J/Kmol) + (1 mol O(248 J/Kmol) + (1 mol O22)(205 J/Kmol)])(205 J/Kmol)]

SS = 514 J/K - 496 J/K - 205 J/K = 514 J/K - 496 J/K - 205 J/K

SS = -187 J/K = -187 J/K

GCalculationsContinued

Go = Ho TSo

Go = - 198 kJ - (298 K)(-187 J/K)(1kJ/1000J)

Go = - 198 kJ + 55.7 kJ

Go = - 142 kJ

The reaction is spontaneous at 25 oC and 1 atm.

Hess’s Law & Go CCdiamond(s)diamond(s) + O + O2(g)2(g) ---> CO ---> CO2(g)2(g) Go = -397 kJ

CCgraphite(s)graphite(s) + O + O2(g)2(g) ---> CO ---> CO2(g)2(g) Go = -394 kJ

Calculate Go for the reaction

Cdiamond(s) ---> Cgraphite(s)

CCdiamond(s)diamond(s) + O + O2(g)2(g) ---> CO ---> CO2(g)2(g) Go = -397 kJ

COCO2(g) 2(g) ---> C---> Cgraphite(s)graphite(s) + O + O2(g)2(g) Go = +394 kJ

Cdiamond(s) ---> Cgraphite(s) Go = -3 kJDiamond is kinetically stable, but thermodynamically unstable.

Go & TemperatureGo depends upon temperature. If a

reaction must be carried out at temperatures higher than 25 oC, then Go must be recalculated from the Ho & So values for the reaction.

Free Energy & Pressure

The equilibrium position represents the lowest The equilibrium position represents the lowest free energy value available to a particular free energy value available to a particular system (reaction).system (reaction).

G is pressure dependent

S is pressure dependent

is not pressure dependent

Free Energy and Pressure

GG = = GG + + RTRT ln( ln(QQ))

QQ = reaction quotient from the law = reaction quotient from the law of mass action.of mass action.

Free Energy Calculations. COCO(g)(g) + 2H + 2H2(g)2(g) ---> CH ---> CH33OHOH(l)(l)

Calculate Calculate Go for this reaction where CO(g) is 5.0 atm and H2(g) is 3.0 atm are converted to liquid methanol.

GG = = nnppGGff(products)(products) nnrrGGff(reactants)(reactants)

GG = = mol CH mol CH33OH)(- 166 kJ/mol)]-[(1 OH)(- 166 kJ/mol)]-[(1 mol CO)(-137 kJ/mol) + (0 kJ)]mol CO)(-137 kJ/mol) + (0 kJ)]

GG = = kJ + 137 kJkJ + 137 kJ

GG = = x 10x 1044 J J

Free Energy CalculationsContinued

)P)((P

1 QH2CO

2(5.0)(3.0)1 Q

Q = 2.2 x 10-2

Free Energy CalculationsContinued

GG = = GG + + RTRT ln( ln(QQ) )

GG = (-2.9 x 10 = (-2.9 x 104 J4 J/mol rxn) + (8.3145 /mol rxn) + (8.3145 J/Kmol)(298 K) ln(2.2 x 10J/Kmol)(298 K) ln(2.2 x 10-2-2))

GG = = x 10x 104 4 J/mol rxn) - (9.4 x 10J/mol rxn) - (9.4 x 1033 J/mol rxn)J/mol rxn)

GG = - 38 kJ/ mol rxn = - 38 kJ/ mol rxn

Note: Note: GG is significantly more negative is significantly more negative than than GGimplying that the reaction is implying that the reaction is more spontaneous at reactant pressures more spontaneous at reactant pressures greater than 1 atm. Why?greater than 1 atm. Why?

16_352

A

B

A

B

(a) (b)

C

A system can achieve the lowest possible free energyby going to equilibrium, not by going to completion.

16_353

GA

GB

(a)

GA

GB

(b)

(PA decreasing)

(PB increasing)

GA GB

(c)

G

G

G

As A is changed into B, the pressure and free energy of A decreases, while the pressure and free energy of B increasesuntil they become equal at equilibrium.

16_354

0Fraction of A reacted

0.5 1.0

G

Equilibriumoccurs here

0Fraction of A reacted

0.5 1.0

G

Equilibriumoccurs here

0Fraction of A reacted

0.5 1.0

Equilibriumoccurs here

(a) (b) (c)

G

Graph a) represents equilibrium starting from only reactants, while Graph b) starts from products only. Graph c) represents the graph for the total system.

Free Energy and Equilibrium

GG = = RTRT ln( ln(KK))

KK = equilibrium constant = equilibrium constant

This is so because This is so because GG = 0 and = 0 and QQ = = KK at at equilibrium.equilibrium.

Go & K GGoo = 0 = 0 K = 1K = 1

GG < 0 < 0 K > 1 (favored)K > 1 (favored)

GGnot favored)not favored)

Equilibrium Calculations4Fe4Fe(s)(s) + 3O + 3O2(g)2(g) <---> 2Fe <---> 2Fe22OO3(s)3(s)

Calculate K for this reaction at 25 Calculate K for this reaction at 25 ooC.C.Go = - 1.490 x 106 J Ho = - 1.652 x 106 JSo = -543 J/KGG = = RTRT ln( ln(KK))

K = e K = e - - GGRR

K = e K = e 601 601 or 10or 10261261

K is very large because K is very large because GG is very negative. is very negative.

Temperature Dependence of K

yy = = mxmx + + bb

((HH and and SS independent of temperature over a independent of temperature over a small temperature range) small temperature range)

If the temperature increases, K decreases for If the temperature increases, K decreases for exothermic reactions, but increases for exothermic reactions, but increases for endothermic reactions.endothermic reactions.

ln ( ) SR

o

K HR

T

( / )1

Free Energy & Work

The maximum possible useful work The maximum possible useful work obtainable from a process at constant obtainable from a process at constant temperature and pressure is equal to the temperature and pressure is equal to the change in free energy:change in free energy:

wmax = G

Reversible vs. Irreversible Processes

ReversibleReversible: The universe is : The universe is exactly the same exactly the same as it was before the cyclic process.as it was before the cyclic process.

IrreversibleIrreversible: The universe is : The universe is differentdifferent after after the cyclic process.the cyclic process.

All real processes are irreversible All real processes are irreversible -- (some -- (some work is changed to heat). work is changed to heat). w < w < G

Work is changed to heat in the surroundings and the entropy of the universe increases.

Laws of Thermodynamics

First Law: You can’t win, you can only First Law: You can’t win, you can only break even.break even.

Second Law: You can’t break even.Second Law: You can’t break even.