20-1 chapter 20 thermodynamics entropy, free energy and the direction of chemical reactions

20-1

Chapter 20

Thermodynamics

Entropy, Free Energy and the Direction of Chemical Reactions

20-2

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 (we willconsider later in the semester)

20-3

The Larger Question

A B

What factors must be identified in order to predictwhether the reaction will proceed spontaneously

in the direction written (as opposed to in the oppositedirection)?

Enthalpy and entropy changes are combined to yield a new term, the Gibbs free energy. The sign of the latter allows predictions of reaction spontaneity.

These are thermodynamic, not kinetic, considerations!

20-4

Limitations of the First Law of Thermodynamics

E = q + w

Euniverse = Esystem + Esurroundings

Esystem = -Esurroundings

The total energy-mass of the universe is constant.

However, these energy changes do not explain the direction of change in the universe!

Esystem + Esurroundings = 0 = Euniverse

20-5Figure 20.1

A spontaneous endothermic chemical reaction

water

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

Horxn = + 62.3 kJ

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The sign of the enthalpy change is insufficient to predictreaction spontaneity!

20-6

The Missing Factor: Entropy (S)

Entropy refers to the state of order/disorder of a system.

A change in order is a change in the number of ways of arranging the particles. It is a key factor in determining the

direction of a spontaneous process.

solid liquid gasmore order less order

crystal + liquid ions in solutionmore order less order

more order less order

crystal + crystal gases + ions in solution

20-7

Figure 20.2

The number of ways to arrange a deck of playing cards

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20-8

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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This process occursspontaneously

without a change in the total internal

energy of the system!

This is an entropy-driven process!

20-9

Ludwig Boltzmann, 1877 S = k ln W

S = entropy, W = the number of ways of arranging the components of a system having equivalent energy, and k = the Boltzmann constant =

R/NA (R = universal gas constant, NA = Avogadro’s number) = 1.38 x

10-23 J/K.

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.

Suniv = Ssys + Ssurr > 0

The Second Law of Thermodynamics

Entropy is a state function.

20-10

All processes occur spontaneously in the direction that increases

the entropy of the universe (system + surroundings).

20-11

Random motion in a crystal

The Third Law of Thermodynamics

A perfect crystal has zero entropy at a temperature of absolute zero.

Ssystem = 0 at 0 K

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Figure 20.4

(increased arrangements, greater S)

20-12

Standard Molar Entropies (So)

Standard states: 1 atm for gases; 1 M for solutions; pure substancein its most stable form for solids and liquids.

units: J/mol/K at 25 oC

The Third Law leads to absolute values for the entropies of substances!

This value equals the entropy increase in a substanceupon raising its temperature from 0 K to the specified temperature.

20-13

Predicting relative So 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

So increases as temperature rises.

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

So of a dissolved solid or liquid is usually greater than So of the pure solute. However, the extent depends on 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, increases in complexity (e.g., bond flexibility) relate directly to entropy.

20-14

Figure 20.5

The increase in entropy from solid to liquid to gas

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Large changes in S occurat phase transitions!

20-15

Figure 20.6

The entropy change accompanying the dissolution of a salt

pure solid

pure liquidsolution

MIX

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salt becomes more disordered (dissociation): S increaseswater becomes more ordered (ion-dipole interactions): S decreases

20-16

Figure 20.7

ethanol water solution of ethanol and

water

The small increase in entropy when ethanol dissolves in water

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

Freedom of movement remains essentially unchanged;S increase is due solely to random mixing effects.

20-17

Figure 20.8

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

O2 gas

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O2 dissolved

20-18

Figure 20.9

NO

NO2

N2O4

Entropy and vibrational motion

Internal molecular motions influence entropy!

more motions, greater S

20-19

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) 1 mol of SO2(g) or 1 mol of SO3(g)(b) 1 mol of CO2(s) or 1 mol of CO2(g)(c) 3 mol of oxygen gas (O2) or 2 mol of ozone gas (O3)(d) 1 mol of KBr(s) or 1 mol of KBr(aq)

(e) seawater in mid-winter at 2 oC or in mid-summer at 23 oC(f) 1 mol of CF4(g) or 1 mol of CCl4(g)

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

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

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

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

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

(e) 23 oC - higher temperature

(f) CCl4 - larger mass

20-20

Standard Entropies of Reaction, Sorxn

By analogy to calculating standard heats of reaction: Horxn

Horxn = mHo

f (products) - nHof (reactants)

Sorxn = mSo

(products) - nSo (reactants)

From Chapter 6

As before, m and n are the appropriate coefficients in the balanced chemical equation.

20-21

Sample Problem 20.2 Calculating the standard entropy of reaction, Sorxn

PROBLEM: Calculate Sorxn for the combustion of 1 mol of propane at 25 oC.

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

PLAN: Use summation equations. Entropy is expected to decrease because the reaction goes from 6 moles of gas to 3 moles of gas.

SOLUTION: Find standard entropy values in an appropriate data table.

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

Sorxn = [(3 mol)(213.7 J/mol.K) + (4 mol)(69.9 J/mol.K)] - [(1 mol)(269.9

J/mol.K) + (5 mol)(205.0 J/mol.K)]

Sorxn = - 374 J/K

20-22

We can’t forget entropy changes in the surroundings!

For spontaneous reactions in which a decrease in the entropy of the system occurs: must be outweighed by a concomitant increase

in the entropy of the surroundings (Suniv > 0)

For an exothermic process: qsys < 0, qsurr > 0, Ssurr > 0

For an endothermic process: qsys > 0, qsurr < 0, Ssurr < 0

The initial temperature of the surroundings affects the

magnitude of Ssurr.

Ssurr -qsys Ssurr 1/Tand

20-23

Ssurr =-qsys/T

For a process of constant pressure, where qp = H:

Ssurr =-Hsys/T

Implication: Ssurr can be calculated by measuring Hsys and the temperature at which the change occurs.

Combining these ideas......

20-24

Sample Problem 20.3

SOLUTION:

Determining reaction spontaneity

PROBLEM: At 298 K, the formation of ammonia has a negative Sosys.

Calculate Souniv and state whether the reaction occurs

spontaneously at this temperature.

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

PLAN: Souniv must be > 0 in order for the reaction to be spontaneous, so

Sosurr must be > +197 J/K. To find So

surr, first find Hsys. Hosys =

Horxn which can be calculated using Ho

f values from data tables.

Then apply Souniv = So

surr + Sosys.

Horxn = [(2 mol)(Ho

f NH3)] - [(1 mol)(Hof N2) + (3 mol)(Ho

f H2)]

Horxn = -91.8 kJ

Sosurr = -Ho

sys/T = -(-91.8 x 103J/298 K) = +308 J/K (>+197 J/K)

Souniv = So

surr + Sosys

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

Souniv > 0: the reaction is spontaneous!

20-25

Figure B20.2

A whole-body calorimeter

Living systems do not violate the Second Law

Suniv = Ssys + Ssurr > 0

20-26

What happens when equilibrium is reached?

Souniv = So

surr + Sosys = 0

Examples: phase changes (fusion, vaporization)

When a system reaches equilibrium, neither the forwardnor the reverse reaction is spontaneous; neither proceeds further

because there is no driving force.

20-27

Figure 20.10

Components of Souniv for spontaneous reactions

exothermic

system becomes more disordered

exothermic

system becomes more ordered

endothermic

system becomes more disordered

20-28

But measuring Ssys and Ssurr is inconvenient to determine reaction spontaneity

More convenient to use parameters that apply only to the system!

The Gibbs free energy (G) accomplishes this goal, whereG = H - TS or G = H - TS

G combines the system’s enthalpy and entropy; thus G is a state function.

Where does the equation come from?

20-29

Deriving the Gibbs free energy equation

Suniv = Ssys + Ssurr

Ssurr =-Hsys/T

Suniv = Ssys - Hsys/T

-TSuniv = Hsys - TSsys

Gsys = Hsys - TSsys

Gsys = -TSuniv

Suniv > 0 or Gsys < 0 for a spontaneous process

Suniv < 0 or Gsys > 0 for a nonspontaneous process

Suniv = 0 or Gsys = 0 for a process at equilibrium

20-30

If a process is nonspontaneous in one direction (G > 0), then it isspontaneous in the opposite direction (G < 0).

The magnitude and sign of G are unrelated to the rate (speed) of the reaction.

Some Key Concepts

20-31

Gosys = Ho

sys - TSosys

Standard Free Energy Changes

All components are in their standards states.

A reference state; similar to Ho and So

20-32

Sample Problem 20.4

SOLUTION:

Calculating Go from enthalpy and entropy values

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

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

+7 -1+5

Use Hof and So values to calculate Go

sys (Gorxn) at 25 oC for this reaction.

PLAN: Obtain appropriate thermodynamic data from a data table; insert them into the Gibbs free energy equation and solve.

Horxn

= mHof (roducts) - nHo

f (reactants)

Horxn

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

[(4 mol)(-397.7 kJ/mol)]

Horxn

= -144 kJ

20-33

Sample Problem 20.4 (continued)

Sorxn

= mSoproducts - nSo

reactants

Sorxn

= [(3 mol)(151 J/mol.K) + (1 mol)(82.6 J/mol.K)] -

[(4 mol)(143.1 J/mol.K)]Sorxn

= -36.8 J/K

Gorxn

= Horxn - TSo

rxn

Gorxn

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

G0rxn

= -133 kJ

The reaction is spontaneous.

20-34

Gorxn = mGo

f (products) - nGof (reactants)

Another way to calculate Gorxn

From standard free energies of formation, Gof

Gof = the free energy change that occurs when 1 mol of compound

is made from its elements, with all components in their standard states.

Thus,

Gof values have properties similar to Ho

f values.

20-35

Sample Problem 20.5

PROBLEM:

Calculating Gorxn from Go

f values

Use Gof values to calculate Go

rxn for the following reaction:

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

PLAN: Use the Gorxn summation equation.

SOLUTION: Gorxn

= mGof (products) - nGo

f (reactants)

Gorxn

= [(3 mol)(-303.2 kJ/mol) + (1 mol)(-409.2 kJ/mol)] -

[(4 mol)(-296.3 kJ/mol)]

Gorxn

= -134 kJ

20-36

G and Work

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 minimum work that must be done to the system to make the process take place.

An example of a gas doing work

20-37

A reversible process: one that can be changed in either direction by aninfinitesimal change in a variable. The maximum work from a spontaneous

process is obtained only if the work is carried out reversibly.

In most real processes, work is performed irreversibly; thus, maximum work is not obtained; some free energy is lost to

the surroundings as heat and is thus unavailable to do work.

A reaction at equilibrium cannot do work.

More Key Concepts

20-38

Effect of Temperature on Reaction Spontaneity

Gosys = Ho

sys - TSosys

The sign of Gosys is T-independent when Ho

sys and So

sys have opposite signs.

The sign of Gosys is T-dependent when Ho

sys and So

sys have the same signs.

20-39

Table 20.1 Reaction Spontaneity and the Signs of

Ho, So and Go

Ho So -TSo Go 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

20-40

Sample Problem 20.6

PROBLEM:

PLAN:

SOLUTION:

Determining the effect of temperature on Go

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 298 K, Go = -141.6 kJ, Ho = -198.4 kJ, and So = -187.9 J/K.

(a) Use these data to decide if this reaction is spontaneous at 25 oC, and predict how Go will change with increasing T.

(b) Assuming Ho and So are constant with increasing T, is the reaction spontaneous at 900. oC?

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

(a) The reaction is spontaneous at 25 oC because Go is (-). Since Ho is (-) and So is (-), Go will become less negative and the reaction less spontaneous as temperature increases.

20-41

Sample Problem 20.6 (continued)

(b) Gorxn

= Horxn - TSo

rxn

Gorxn

= -198.4 kJ - [(1173 K)(-187.9 J/mol.K)(kJ/103J)]

Gorxn

= +22.0 kJ; the reaction will be nonspontaneous at 900.oC

20-42

Determining the Temperature at which a Reaction Becomes Spontaneous

(only when H and S have the same sign)

Go = Ho - TSo = 0 (solve for this condition)

Ho = TSo

T = Ho/So

20-43

Figure 20.11

The effect of temperature on reaction spontaneity

Ho: relatively insensitive to T

So: increases

with increasing T

Cu2O(s) + C(s)2Cu(s) + CO(g)

Ho = +58.1 kJSo = +165 J/K

20-44

Figure B20.4

The cycling of metabolic free energy through ATP

20-45

Figure B20.3

The coupling of a nonspontaneous reaction to the hydrolysis of ATP

D-glucose + ATP D-glucose 6P + ADP

(catalyzed by the enzyme, hexokinase)

20-46

Figure B20.5

Why is ATP a high-energy molecule?

The Go of ATP hydrolysis is -31 kJ/mol; considerably more (-) under in vivo conditions!

20-47

End of Assigned Material

20-48

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 ln Q - RT ln K

Under standard conditions (1 M concentrations, 1 atm for gases), Q = 1 and ln Q = 0. Therefore:

Go = - RT ln K

20-49

FO

RW

AR

D R

EA

CT

ION

RE

VE

RS

E R

EA

CT

ION

Table 20.2 The relationship between Go and K at 25 oC

Go (kJ) K significance

200

100

50

10

1

0

-1

-10

-50

-100

-200

9 x 10-36

3 x 10-18

2 x 10-9

2 x 10-2

7 x 10-1

1

1.5

5 x 101

6 x 108

3 x 1017

1 x 1035

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to the same extent

Forward reaction goes to completion; essentially no reverse reaction

20-50

Sample Problem 20.7

PROBLEM:

Calculating G at non-standard conditions

The oxidation of SO2,

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

is too slow at 298 K 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 298 K and at 973 K. (Go298 = -141.6 kJ/mol for the

reaction as written using Ho and So values. At 973 K, Go973 =

12.12 kJ/mol for the reaction as written.)(b) In experiments to determine the effect of temperature on reaction

spontaneity, two sealed containers are filled with 0.500 atm of SO2,

0.0100 atm of O2, and 0.100 atm of SO3 and kept at 25 oC and at

700. oC. 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 previous slides.

20-51

SOLUTION:

Sample Problem 20.7 (continued)

(a) Calculating K at the two temperatures:

Go = -RTln K so

€

K =e−(ΔG 0 /RT)

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

(8.314 J/mol.K)(298 K)

= 57.2

€

K =e−(ΔG 0 /RT) = e57.2 = 7 x 1024

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

(8.314 J/mol.K)(973 K)

= 1.50

€

K =e−(ΔG 0 /RT) = e1.50 = 4.5

20-52

Sample Problem 20.7 (continued)

(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 298 K there will be a dramatic shift while at 973 K the shift will be slight.

(c) The non-standard G is calculated using G = Go + RTlnQ

G298 = -141.6 kJ/mol + (8.314 J/mol.K)(kJ/103J)(298 K)(ln 4.00)G298 = -138.2 kJ/mol

G973 = -12.12 kJ/mol + (8.314 J/mol.K)(kJ/103J)(973 K)(ln 4.00)

G298 = -0.9 kJ/mol

20-53

Figure 20.12

The relation between free energy and the extent of reaction

Go < 0 K >1

Go > 0 K <1

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