C H E M I S T R Y

Chapter 16Chapter 16Thermodynamics: Entropy, Free Energy, Thermodynamics: Entropy, Free Energy, and Equilibriumand Equilibrium

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

Spontaneous processSpontaneous processSpontaneity 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 directionSpontaneity of a reaction does not identify the

speed of reaction

Spontaneous ProcessesSpontaneous Processes

Enthalpy, Entropy, and Spontaneous Enthalpy, Entropy, and Spontaneous ProcessesProcesses

S = Sfinal Sinitial

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

Enthalpy, Entropy, and Spontaneous Enthalpy, Entropy, and Spontaneous ProcessesProcesses

Entropy and ProbabilityEntropy 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.

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

Entropy and TemperatureEntropy and TemperatureΔS 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

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

Standard Molar Entropies and Standard Molar Entropies and Standard Entropies of ReactionStandard Entropies of ReactionUsing 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 sign of the entropy change for the system

c. Determine the sign of 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 - TS

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 - TS = -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)

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

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

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?

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 16/28

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)