Spontaneity, Entropy, & Free Energy

Spontaneity, Entropy, & Free Energy

Chapter 16

1st Law of Thermodynamics1st 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

Spontaneous Processes and Entropy

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

A A spontaneous spontaneous process is one that process is one that occurs occurs without outside intervention. without outside intervention. This is also This is also considered considered thermodynamically favored.thermodynamically favored.

EntropyEntropy

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

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.

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The expansion of an ideal gas into an evacuated bulb.

Positional EntropyPositional 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 EntropyPositional 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?

EntropyEntropy

What is the sign of the entropy change for What is the sign of the entropy change for the following?the 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

The Second Law of Thermodynamics

. . .. . . in any spontaneous(thermodynamically in any spontaneous(thermodynamically favored) process there is always an favored) process there is always an increase in the entropy of the universeincrease in the entropy of the universe..

SSunivuniv > 0 > 0

for a spontaneous process.for a spontaneous process.

SUniverse 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.

S = npS(products) nrS(reactants)

S = npS(products) nrS(reactants)

HH = = nnppHH(products) (products) nnrrHH(reactants)(reactants)

Soreaction

Soreaction

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

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

SO2 (248 J/Kmol) NiO (38 J/Kmol)

O2 (205 J/Kmol)

NiS (53 J/Kmol)

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

SS = [(2 mol SO = [(2 mol SO22)(248 J/Kmol) + (2 mol )(248 J/Kmol) + (2 mol NiO)(38 J/Kmol)] - [(2 mol NiS)(53 NiO)(38 J/Kmol)] - [(2 mol NiS)(53 J/Kmol) + (3 mol OJ/Kmol) + (3 mol O22)(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 # gaseous molecules decreases!decreases!

Effect of H and S on Spontaneity

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

G -- Free EnergyG -- 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 melting ice cube), then the direction is decided by the Free Energy, decided by the Free Energy, G, G, and depends and depends upon the temperature.upon the temperature.

Free EnergyFree Energy

GG = = 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.

Free Energy GFree 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

G, H, & S 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

Free Energy Change and Chemical Reactions

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)

Temperature DependenceTemperature Dependence

HHoo & & SSoo are are not temperature dependent.

GGoo is temperature dependent. is temperature dependent.

GG = = HH TTSS

GCalculationsGCalculationsCalculate 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

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

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.

The Third Law of ThermodynamicsThe Third Law of Thermodynamics

. . .. . . The third law of thermodynamics: The third law of thermodynamics: the entropy of a perfect the entropy of a perfect crystal at 0K is zero.crystal at 0K is zero.

[not a lot of perfect crystals out there so, entropy values are RARELY [not a lot of perfect crystals out there so, entropy values are RARELY ever zero—evenelements]ever zero—evenelements]

So what? This means the absolute entropy of a substance can then be So what? This means the absolute entropy of a substance can then be determined at any temp. higherdetermined at any temp. higher

than 0 K. (Handy to know if you ever need to defend than 0 K. (Handy to know if you ever need to defend why why G & H for G & H for elements = 0. . . . BUT S doeselements = 0. . . . BUT S does

not!).not!).

Hess’s Law & Go 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.

SoSo

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)

Go & TemperatureGo & Temperature

Go 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 & PressureFree Energy & Pressure

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

G is pressure dependent

S is pressure dependent

is not pressure dependent

Free Energy and PressureFree 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 CalculationsFree 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

Free Energy CalculationsContinued

)P)((P

1 Q

H2CO

2(5.0)(3.0)

1 Q

Q = 2.2 x 10-2

Free Energy CalculationsContinued

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?

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A

B

A

B

(a) (b)

C

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

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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.

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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

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 Go & K

GGoo = 0 = 0 K = 1K = 1

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

GGnot favored)not favored)

Equilibrium CalculationsEquilibrium 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 J

So = -543 J/K

GG = = 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 KTemperature Dependence of K

yy = = mxmx + + bb

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

R

o

KH

RT

( / )1

Free Energy & WorkFree 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

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 ThermodynamicsLaws 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.