C H E M I S T R Y

Chapter 16Thermodynamics: Entropy, Free Energy, and Equilibrium

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

Spontaneous processSpontaneity reaction always moves a system

toward equilibriumBoth 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

State Function: A function or property whose value depends only on the present state, or condition, of the system, not on the path used to arrive at that state.

Enthalpy Change (H): The heat change in a reaction or process at constant pressure;H = E + PV

Entropy (S): The amount of molecular randomness in a system

= +40.7 kJHvap

CO2(g) + 2H2O(l)CH4(g) + 2O2(g)

H2O(l)H2O(s)

H2O(g)H2O(l)

2NO2(g)N2O4(g)

= +3.88 kJHo

= +6.01 kJHfusion

= 890.3 kJHo

= +55.3 kJHo

Endothermic:

Exothermic:

Na+(aq) + Cl(aq)NaCl(s)H2O

S = Sfinal Sinitial

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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.

Third Law of Thermodynamics: The entropy of a perfectly ordered crystalline substance at 0 K is zero.

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

cC + dDaA + bB

So = So(products) - So(reactants)

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

ReactantsProducts

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

2NO2(g)N2O4(g)

ExampleCalculate the standard entropy of reaction at

25oC for the decomposition of calcium carbonate:

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

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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

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

Ssurr H

Ssurr T

1

Ssurr =

T

H

ExampleConsider 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) Hof (kJ/mol)

Fe(s) 27.3

O2(g) 205.0

Fe2O3(s) 87.4 -824..2

ExampleConsider 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 25.0oCb. Determine the sign of the entropy change for the systemc. Determine the sign of the entropy change for the universe. Will the reaction be spontaneous?

G = H - TS = -TStotal

Free Energy: G = H - TS

G = H - TS

S = Ssys

Using:

ssurr = T

- H

Stotal = Ssys + Ssurr

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

∆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)

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°

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

ExampleOne of the possible initial steps in the

formation of acid rain is the oxidation of the pollutant of SO2 to SO3 by the following reaction

SO2(g) + ½ O2(g) SO3(g)ΔHo = -98.9 kJ ΔSo = -94.0 J/K

Calculate the ΔGo for this reaction at 25oCIs the reaction spontaneous at standard-state condition?Does the reaction become spontaneous at higher temperature?

Go = Gof (products) Go

f (reactants)

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

ReactantsProducts

cC + dDaA + bB

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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)

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

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?

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:

At equilibrium, G = 0 and Q = K.

Go = RT ln K

G = Go + RT ln Q

Calculate Kp at 25 oC for the following reaction:

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

ExampleThe value of ΔGo

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