20-1

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CHEMISTRYThe Molecular Nature of Matter and Change

Third Edition

Chapter 20

Lecture Outlines*

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Chapter 20

Thermodynamics:

Entropy, Free Energy and the Direction of Chemical Reactions

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Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions

20.1 The Second Law of Thermodynamics: Predicting Spontaneous Change

20.2 Calculating the Change in Entropy of a Reaction

20.3 Entropy, Free Energy, and Work

20.4 Free Energy, Equilibrium, and Reaction Direction

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Limitations of the First Law of Thermodynamics

Esystem = Ek + Ep

Esystem = heat + work

Esystem = q + w

Euniverse = Esystem + Esurroundings

Esystem = -Esurroundings

The total energy-mass of the universe is constant.

But, this does not tell us anything about the direction of change in the

universe.

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Figure 20.1 A spontaneous endothermic chemical reaction

water

Ba(OH)2 8H2O(s) + 2NH4NO3(s) Ba2+(aq) + 2NO3-(aq) + 2NH3(aq) + 10H2O(l).

H0rxn = + 62.3 kJ

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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The Concept of Entropy (S)

Entropy refers to the state of order.

A change in order is a change in the number of ways of arranging the particles, and it is a key factor in determining the direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solution

more order less order

more order less order

crystal + crystal gases + ions in solution

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Figure 20.2 The number of ways to arrange a deck of playing cards

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Figure 20.3

1 atm evacuated

Spontaneous expansion of a gas

stopcock closed

stopcock opened

0.5 atm 0.5 atm

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1877 Ludwig Boltzman S = k ln WS = entropy W = number of ways of arranging the components of a systemk = Boltzman constant

•A system with relatively few equivalent ways to arrange its components (smaller W) has relatively less disorder and low entropy.

•A system with many equivalent ways to arrange its components (larger W) has relatively more disorder and high entropy.

2nd law of thermodynamics

The universe is going to a higher state of entropy

Suniverse = Ssystem + Ssurroundings > 0

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Figure 20.4 Random motion in a crystal

3rd Law of thermodynamicsA perfect crystal has zero entropy at a temperature of absolute zero.

Ssystem = 0 at 0 K

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Predicting Relative S0 Values of a System

1. Temperature changes

2. Physical states and phase changes

3. Dissolution of a solid or liquid

5. Atomic size or molecular complexity

4. Dissolution of a gas

S0 increases as the temperature rises.

S0 increases as a more ordered phase changes to a less ordered phase.

S0 of a dissolved solid or liquid is usually greater than the S0 of the pure solute. However, the extent depends upon the nature of the solute and solvent.

A gas becomes more ordered when it dissolves in a liquid or solid.

In similar substances, increases in mass relate directly to entropy.

In allotropic substances (e.g. O2 & O3), increases in complexity (e.g. bond flexibility) relate directly to entropy.

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Figure 20.5The increase in entropy from solid to liquid to gas

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Figure 20.6

The entropy change accompanying the dissolution of a salt

pure solid

pure liquid

solution

MIX

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Figure 20.7

Ethanol Water Solution of ethanol and water

The small increase in entropy when ethanol dissolves in water

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Figure 20.8

The large decrease in entropy when a gas dissolves in a liquid

O2 gas

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Figure 20.9

NO

NO2

N2O4

Entropy and vibrational motion

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Sample Problem 20.1:

SOLUTION:

Predicting Relative Entropy Values

PROBLEM: Choose the member with the higher entropy in each of the following pairs, and justify your choice [assume constant temperature, except in part (e)]:

(a) 1mol of SO2(g) or 1mol of SO3(g)(b) 1mol of CO2(s) or 1mol of CO2(g)(c) 3mol of oxygen gas (O2) or 2mol of ozone gas (O3)(d) 1mol of KBr(s) or 1mol of KBr(aq)(e) Seawater in midwinter at 20C or in midsummer at 230C(f) 1mol of CF4(g) or 1mol of CCl4(g)

PLAN: In general less ordered systems have higher entropy than ordered systems and entropy increases with an increase in temperature.

(a) 1mol of SO3(g) - more atoms

(b) 1mol of CO2(g) - gas > solid

(c) 3mol of O2(g) - larger #mols

(d) 1mol of KBr(aq) - solution > solid

(e) 230C - higher temperature

(f) CCl4 - larger mass

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Sample Problem 20.4:

SOLUTION:

Calculating G0 from Enthalpy and Entropy Values ( 1of 2)

PROBLEM: Potassium chlorate, one of the common oxidizing agents in explosives, fireworks, and matchheads, undergoes a solid-state redox reaction when heated. In this reaction, note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation):

4KClO3(s) 3KClO4(s) + KCl(s)

+7 -1+5

Use H0f and S0 values to calculate G0

sys (G0rxn) at 250C for this reaction.

PLAN: Use Appendix B values for thermodynamic entities; place them into the Gibbs Free Energy equation and solve.

H0rxn

= mHf0products - nHf

0reactants

H0rxn

= (3mol)(-432.8kJ/mol) + (1mol)(-436.7kJ/mol) -

(4mol)(-397.7kJ/mol)

H0rxn

= -144kJ

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Sample Problem 20.4: Calculating G0 from Enthalpy and Entropy Values

continued

S0rxn

= mS0products - nS0

reactants

S0rxn

= (3mol)(151J/mol*K) + (1mol)(82.6J/mol*K) -

(4mol)(143.1J/mol*K)

S0rxn

= -36.8J/K

G0rxn

= H0rxn - T S0

rxn

G0rxn

= -144kJ - (298K)(-36.8J/K)(kJ/103J)

G0rxn

= -133kJ

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Sample Problem 20.2: Calculating the Standard Entropy of Reaction, S0rxn

PROBLEM: Calculate S0rxn for the combustion of 1mol of propane at 250C.

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

PLAN: Use summation equations. It is obvious that entropy is being lost because the reaction goes from 6mols of gas to 3mols of gas.

SOLUTION: Find standard entropy values in the Appendix or other table.

S = [(3mol)(S0 CO2) + (4mol)(S0 H2O)] - [(1mol)(S0 C3H8) + (5mol)(S0 O2)]

S = [(3mol)(213.7J/mol*K) + (4mol)(69.9J/mol*K)] - [(1mol)(269.9J/mol*K) + (5mol)(205.0J/mol*K)]

S = - 374 J/K

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Figure B20.1

Lavoisier studying human respiration as a form of combustion

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Figure B20.2 A whole-body calorimeter

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Figure 20.10 Components of S0universe for spontaneous reactions

exothermic

system becomes more disordered

exothermic

system becomes more ordered

endothermic

system becomes more disordered

Suniverse = Ssystem + Ssurroundings

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Sample Problem 20.3:

SOLUTION:

Determining Reaction Spontaneity

PROBLEM: At 298K, the formation of ammonia has a negative S0sys;

Calculate S0rxn, and state whether the reaction occurs

spontaneously at this temperature.

N2(g) + 3H2(g) 2NH3(g) S0sys = -197J/K

PLAN: S0universe must be > 0 in order for this reaction to be spontaneous, so

S0surroundings must be > 197J/K. To find S0

surr, first find Hsys; Hsys = Hrxn which can be calculated using H0

f values from tables. S0universe =

S0surr + S0

sys.

H0rx = [(2mol)(H0

fNH3)] - [(1mol)(H0fN2) + (3mol)(H0

fH2)]

H0rx = -91.8kJ

S0surr = -H0

sys/T = -(-91.8x103J/298K) = 308J/K

S0universe = S0

surr + S0sys = 308J/K + (-197J/K) = 111J/K

S0universe > 0 so the reaction is spontaneous.

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Different Ways to Calculate G0rxn

0 = Std conds: 25oC, 1 atm, 1 M solutions

(Eqn 1) G0

rxn = H0rxn - TS0

rxn

(Eqn 2) G0rxn = mGf

0products - nGf

0reactants

(Eqn 3) G0 = - RT lnKeq R = 8.314 J/mol.K

When not at standard conditions

(Eqn 4) Grxn = G0rxn + RT ln Q (Q = Reaction Quotient)

Use these equations with equation #1

S0rxn

= mS0products - nS0

reactants

H0rxn

= mHf0

products - nHf0

reactants

Useful Thermodynamic Relationships

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Sample Problem 20.6:

PROBLEM:

PLAN:

SOLUTION:

Determining the Effect of Temperature on G0 (1 of 2)

An important reaction in the production of sulfuric acid is the oxidation of SO2(g) to SO3(g):

2SO2(g) + O2(g) 2SO3(g)

At 298K, G0 = -141.6kJ; H0 = -198.4kJ; and S0 = -187.9J/K

(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how G0 will change with increasing T.

(b) Assuming H0 and S0 are constant with increasing T, is the reaction spontaneous at 900.0C?

The sign of G0 tells us whether the reaction is spontaneous and the signs of H0 and S0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b).

(a) The reaction is spontaneous at 250C because G0 is (-). Since H0 is (-) but S0 is also (-), G0 will become less spontaneous as the temperature increases.

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Sample Problem 20.6:

continued

(b) G0rxn

= H0rxn - T S0

rxn

G0rxn

= -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J)

G0rxn

= 22.0 kJ; the reaction will be nonspontaneous at 900.0C

Determining the Effect of Temperature on G0 (2 of 2)

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G and the Work a System Can Do

For a spontaneous process, G is the maximum work obtainable from the system as the process takes place: G = workmax

For a nonspontaneous process, G is the maximum work that must be done to the system as the process takes place: G = workmax

An example

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Table 20.1 Reaction Spontaneity and the Signs of H0, S0, and G0

H0 S0 -TS0 G0 Description

- + - -

+ - + +

+ + - + or -

- - + + or -

Spontaneous at all T

Nonspontaneous at all T

Spontaneous at higher T;nonspontaneous at lower T

Spontaneous at lower T;nonspontaneous at higher T

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Sample Problem 20.5:

PROBLEM:

Calculating G0rxn from G0

f Values

Use G0f values to calculate Grxn for the reaction in

Sample Problem 20.4:

4KClO3(s) 3KClO4(s) + KCl(s)

PLAN: Use the G summation equation.

SOLUTION: G0rxn

= mGf0products - nGf

0reactants

G0rxn

= (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) - (4mol)(-296.3kJ/mol)

G0rxn

= -134kJ

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Figure 20.11 The effect of temperature on reaction spontaneity

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The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

1st Rxn of Glycolysis catalyzed by Hexose Kinase:

Glucose + HPO42- Glucose-6-phosphate + H2O G0 = +13.8 kJ

ATP4- + H2O ADP3- + HPO42- G0 = -30.5 kJ

Glucose + HPO42- Glucose-6-phosphate + H2O G0 = +13.8 kJ

Glucose + ATP Glucose-6-phosphate + ADP3- G0 = -16.7 kJ

Enzyme

Glucose

Glucose-6-phosphateEnzyme: hexose kinase

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Figure B20.4 The cycling of metabolic free energy through ATP

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Figure B20.5 Why is ATP a high-energy molecule?

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Free Energy, Equilibrium and Reaction Direction

•If Q/K < 1, then ln Q/K < 0; the reaction proceeds to the right (G < 0)

•If Q/K > 1, then ln Q/K > 0; the reaction proceeds to the left (G > 0)

•If Q/K = 1, then ln Q/K = 0; the reaction is at equilibrium (G = 0)

G = RT ln Q/K = RT lnQ - RT lnK

Under standard conditions (1M concentrations, 1atm for gases), Q = 1 and lnQ = 0 so...

G0 = - RT lnK

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FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

Table 20.2 The Relationship Between G0 and K at 250C

G0(kJ) K Significance

200

100

50

10

1

0

-1

-10

-50

-100

-200

9x10-36

3x10-18

2x10-9

2x10-2

7x10-1

1

1.5

5x101

6x108

3x1017

1x1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to same extent

Forward reaction goes to completion; essentially no reverse reaction

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Sample Problem 20.7:

PROBLEM:

Calculating G at Nonstandard Conditions

The oxidation of SO2, which we considered in Sample Problem 20.6

2SO2(g) + O2(g) 2SO3(g)

is too slow at 298K to be useful in the manufacture of sulfuric acid. To overcome this low rate, the process is conducted at an elevated temperature.

(a) Calculate K at 298K and at 973K. (G0298 = -141.6kJ/mol of reaction as

written using H0 and S0 values at 973K. G0973 = -12.12kJ/mol of reaction

as written.)(b) In experiments to determine the effect of temperature on reaction spontaneity, two sealed containers are filled with 0.500atm of SO2, 0.0100atm of O2, and 0.100atm of SO3 and kept at 250C and at 700.0C. In which direction, if any, will the reaction proceed to reach equilibrium at each temperature?(c) Calculate G for the system in part (b) at each temperature.

PLAN: Use the equations and conditions found on slide .

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SOLUTION:

Sample Problem 20.7: Calculating G at Nonstandard Conditions

continued (2 of 3)

(a) Calculating K at the two temperatures:

G0 = -RTlnK so

K e (G 0 /RT)

At 298, the exponent is -G0/RT = -(-141.6kJ/mol)(103J/kJ)

(8.314J/mol*K)(298K)= 57.2

K e (G 0 /RT) = e57.2 = 7x1024

At 973, the exponent is -G0/RT(-12.12kJ/mol)(103J/kJ)

(8.314J/mol*K)(973K)= 1.50

K e (G 0 /RT) = e1.50 = 4.5

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Sample Problem 20.7: Calculating G at Nonstandard Conditions

continued (3 of 3)

(b) The value of Q =pSO3

2

(pSO2)2(pO2)=

(0.100)2

(0.500)2(0.0100)= 4.00

Since Q is < K at both temperatures the reaction will shift right; for 298K there will be a dramatic shift while at 973K the shift will be slight.

(c) The nonstandard G is calculated using G = G0 + RTlnQ

G298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)

G298 = -138.2kJ/mol

G973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)

G298 = -0.9kJ/mol

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Figure 20.12

The relation between free energy and the extent of reaction

G0 < 0 K >1 G0 > 0

K <1

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Different Ways to Calculate Grxn

(Eqn 1) G0rxn = H0

rxn - TS0rxn

(Eqn 2) G0rxn = mGf

0products - nGf

0reactants

(Eqn 3) G0 = - RT lnKeq R = 8.314 J/mol.K

(Eqn 4) G = G0 + RT lnQ Q = Rxn QuotientUse these with equation #1

S0rxn

= mS0products - nS0

reactants

H0rxn

= mHf0

products - nHf0

reactants

Useful Thermodynamic Relationships