Chapter 16Chapter 16

Thermodynamics: Entropy, Free Energy, and Equilibrium

16.1 Spontaneous Processes16.1 Spontaneous ProcessesSpontaneous Process: A process that, once started, proceeds on its own without a continuous external influence.

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

catalyst

Kp = 4.4 x 105

@ 300KWill the reverse reaction spontaneous?

16.1 Spontaneous process16.1 Spontaneous process Spontaneity reaction always moves a system toward equilibrium

◦ Both forward and reverse reaction depends on Temperature Pressure Composition of reaction mixture

Q < K; reaction proceeds in the forward direction Q>K; reaction proceeds in the reverse direction Spontaneity of a reaction does not identify the speed of reaction

Spontaneous ProcessesSpontaneous Processes

16.2 Enthalpy, Entropy, and 16.2 Enthalpy, Entropy, and Spontaneous ProcessesSpontaneous Processes

S = Sfinal Sinitial

16.2Enthalpy, Entropy, and 16.2Enthalpy, Entropy, and Spontaneous ProcessesSpontaneous Processes

16.2 Enthalpy, Entropy, and 16.2 Enthalpy, Entropy, and Spontaneous ProcessesSpontaneous Processes

16.3 Entropy and Probability16.3 Entropy and Probability

S = k ln W

k = Boltzmann’s constant

= 1.38 x 1023 J/K

W =The number of ways that the state can be achieved.

16.4 Entropy and Temperature16.4 Entropy and TemperatureThird Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero.

Higher temperature• Greater molecular motion• Broader distribution of individual molecular energies• More randomness• Higher entropy

Lower temperature• Less molecular motion• Narrower distribution of individual molecular energies• Less randomness• Lower entropy

16.4 Entropy and Temperature16.4 Entropy and Temperature ΔS increases

when increasing the average kinetic energy of molecules

Total energy is distributed among the individual molecules in a number of ways

Botzman- the more way (W) that the energy can be distributed the greater the randomness of the state and higher the entropy

16.5 Standard Molar Entropies and 16.5 Standard Molar Entropies and Standard Entropies of ReactionStandard Entropies of Reaction

Standard Molar Entropy (So): The entropy of 1 mole of a pure substance at 1 atm pressure and a specified temperature.

Standard Molar Entropies and Standard Molar Entropies and Standard Entropies of ReactionStandard Entropies of Reaction

cC + dDaA + bB

So = So(products) - So(reactants)

So = [cSo(C) + dSo (D)] [aSo (A) + bSo (B)]

ReactantsProducts

16.5 Standard Molar Entropies and 16.5 Standard Molar Entropies and Standard Entropies of ReactionStandard Entropies of Reaction

Using standard entropies, calculate the standard entropy change for the decomposition of N2O4.

2NO2(g)N2O4(g)

ExampleExampleCalculate the standard entropy of reaction at 25oC for

the decomposition of calcium carbonate:

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

16.6 Entropy and the Second Law of 16.6 Entropy and the Second Law of ThermodynamicsThermodynamics

First Law of Thermodynamics: In any process, spontaneous or nonspontaneous, the total energy of a system and its surroundings is constant.

• Helps keeping track of energy flow between system and the surrounding• Does not indicate the spontaneity of the process

Second Law of Thermodynamics: In any spontaneous process, the total entropy of a system and its surroundings always increases.

• Provide a clear cut criterion of spontaneity• Direction of spontaneous change is always determined by the sign of the total entropy change

Entropy and the Second Law of Entropy and the Second Law of ThermodynamicsThermodynamics

Stotal > 0 The reaction is spontaneous.

Stotal < 0 The reaction is nonspontaneous.

Stotal = 0 The reaction mixture is at equilibrium.

Stotal = Ssys + Ssurr

or

Stotal = Ssystem + Ssurroundings

Entropy and the Second Law of Entropy and the Second Law of ThermodynamicsThermodynamics

Ssurr H

Ssurr T

1

Ssurr =

T

H

ExampleExample Consider the oxidation of iron metal

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)

Determine whether the reaction is spontaneous at 25oC

So (J/K mol) ΔHof (kJ/mol)

Fe(s) 27.3 0

O2(g) 205.0 0

Fe2O3(s) 87.4 -824.2

ExampleExample Consider the combustion of propane gas:

C3H8(g) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

a. Calculate the entropy change in the surrounding associated with this reaction occurring at 25.0oC

b. Determine the entropy change for the system

c. Determine the entropy change for the universe. Will the reaction be spontaneous?

So (J/K mol) ΔHof (kJ/mol)

C3H8(g) 270.2 -103.8

O2(g) 205.0 0

CO2(g) 213.6 -393.5

H2O(g) 188.7 -241.8

16.7 Free Energy16.7 Free Energy

G = -TStotal

Free Energy: a thermodynamic quantity defined by G = H - TS

G = H - TSsys

Stotal = Ssys -

Using:

ssurr = T

- H

Stotal = Ssys + Ssurr

H

T

Constant pressure and temperature

Free EnergyFree Energy

G < 0 The reaction is spontaneous.

G > 0 The reaction is nonspontaneous.

G = 0 The reaction mixture is at equilibrium.

Using the second law and G = H - TSsys = -TStotal

16.8 Standard Free-Energy Changes for 16.8 Standard Free-Energy Changes for ReactionsReactions

Thermodynamic Standard State: Most stable form of a substance at 1 atm pressure and at a specified temperature, usually 25 °C; 1 M concentration for all substances in solution.

G° = H° - TS°

The effect of H, S and T on SpontaneityThe effect of H, S and T on Spontaneity

∆H ∆S Low Temperature High Temperature

- + Spontaneous (G<0) Spontaneous (G<0)

+ - Nonspontaneous (G > 0) Nonspontaneous (G > 0)

- - Spontaneous (G< 0) nonSpontaneous (G>0)

+ + Nonspontaneous (G>0) Spontaneous (G<0)

ExampleExample What is the standard free-energy change, ΔoG, for the following

reaction at 25oC?

2KClO3(g) 2KCl(s) + 3O2(g)

ΔHo = -78.0kJ ΔSo =493.9J/K

Is the reaction spontaneous at standard-state condition?

Does the reaction become nonspontaneous at higher temperature?

16.916.9 Standard Free-Energy Changes Standard Free-Energy Changes for Reactionsfor Reactions

Calculate the standard free-energy change at 25 oC for the Haber synthesis of ammonia using the given values for the standard enthalpy and standard entropy changes:

So = 198.7 J/K

2NH3(g)N2(g) + 3H2(g) Ho = 92.2 kJ

Will the reaction become nonspontanous at higher temperature? If so, then determine at what temperature will it become nonspontanous?

Standard Free Energies of FormationStandard Free Energies of Formation

Go = Gof (products) Go

f (reactants)

Go = [cGof (C) + dGof (D)] [aGof (A) + bGof (B)]

ReactantsProducts

cC + dDaA + bB

Standard Free Energies of Standard Free Energies of FormationFormation

Chaptr 16/27

Standard Free Energies of FormationStandard Free Energies of Formation

Using table values, calculate the standard free-energy change at 25 °C for the reduction of iron(III) oxide with carbon monoxide:

2Fe(s) + 3CO2(g)Fe2O3(s) + 3CO(g)

Fe2O3(s)

3CO(g) 2Fe(s) 3CO2(g)

Gof (kJ/mol) -743.5 -137.2 0 -394.4

16.10 Free Energy Changes and the 16.10 Free Energy Changes and the Reaction MixtureReaction Mixture

G = G° + RT ln Q

G = Free-energy change under nonstandard conditions.

For the Haber synthesis of ammonia:

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

P2

H2PN2

P3

Free Energy Changes and the Reaction Free Energy Changes and the Reaction MixtureMixture

C2H4(g)2C(s) + 2H2(g)

Calculate G for the formation of ethylene (C2H4) from carbon and hydrogen at 25 °C when the partial pressures are 100 atm H2 and 0.10 atm C2H4.

Is the reaction spontaneous in the forward or the reverse direction?

C(s) H2(g) C2H4(g)

∆ Gof (kJ/mol) 0 0 61.8

16.11 Free Energy and Chemical 16.11 Free Energy and Chemical EquilibriumEquilibrium

G = G° + RT ln Q

Q >> 1 RT ln Q >> 0 G > 0

The total free energy decreases as the reaction proceeds spontaneously in the reverse direction.

• When the reaction mixture is mostly products:

Q << 1 RT ln Q << 0 G < 0

The total free energy decreases as the reaction proceeds spontaneously in the forward direction.

• When the reaction mixture is mostly reactants:

Free Energy and Chemical Free Energy and Chemical EquilibriumEquilibrium

At equilibrium, G = 0 and Q = K.

Go = RT ln K

G = Go + RT ln Q

Free Energy and Chemical EquilibriumFree Energy and Chemical EquilibriumCalculate Kp at 25 oC for the following reaction:

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

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

Gof (kJ/mol) -1128.8 -603.5 -394.4

ExampleExample The value of ΔGo

f at 25oC for gaseous mercury is 31.85 kJ/mol. What is the vapor pressure of mercury at 25oC?

Hg(l) Hg(g)