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Chapter 19. Thermodynamics:. Entropy, Free Energy and the Direction of Chemical Reactions. Thermodynamics: Entropy, Free Energy, and the Direction of Chemical Reactions. The Second Law of Thermodynamics: Predicting Spontaneous Change. Calculating the Change in Entropy of a Reaction.

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slide1

Chapter 19

Thermodynamics:

Entropy, Free Energy and the Direction of Chemical Reactions

slide2

Thermodynamics:

Entropy, Free Energy, and the Direction of Chemical Reactions

The Second Law of Thermodynamics:

Predicting Spontaneous Change

Calculating the Change in Entropy of a Reaction

Entropy, Free Energy, and Work

Free Energy, Equilibrium, and Reaction Direction

slide3

Limitations of the First Law of Thermodynamics

Euniverse= Esystem + Esurroundings

DEsystem = -DEsurroundings

DEsystem + DEsurroundings = 0 = DEuniverse

The total energy-mass of the universe is constant.

However, this does not tell us anything about the direction of change in the universe.

slide4

water

.

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

Figure20.1

A spontaneous endothermic chemical reaction.

DH0rxn = +62.3 kJ

slide5

more order

less order

solid liquid gas

more order

less order

crystal + liquid ions in solution

more order

less order

crystal + crystal gases + ions in solution

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.

slide6

S = k ln W

1877 Ludwig Boltzman

where S is entropy, W is the number of ways of arranging the components of a system, and k is a constant (the Boltzman constant), R/NA (R = universal gas constant, NA = Avogadro’s number.

  • 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.

DSuniverse = DSsystem + DSsurroundings > 0

This is the second law of thermodynamics.

slide7

Random motion in a crystal

Figure 20.4

The third law of thermodynamics.

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

Ssystem = 0 at 0 K

slide8

Predicting Relative S0 Values of a System

1. Temperature changes

S0 increases as the temperature rises.

2. Physical states and phase changes

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

phase.

3. Dissolution of a solid or liquid

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.

4. Dissolution of a gas

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

5. Atomic size or molecular complexity

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

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

slide10

MIX

solution

Figure 20.6

The entropy change accompanying the dissolution of a salt.

pure solid

pure liquid

slide11

Solution of ethanol and water

Ethanol

Water

Figure 20.7

The small increase in entropy when ethanol dissolves in water.

slide12

O2 gas

O2 gas in H2O

Figure 20.8

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

slide13

N2O4

NO2

Entropy and vibrational motion.

Figure 20.9

NO

slide14

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)]:

PLAN:

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

Sample Problem 20.1:

Predicting Relative Entropy Values

(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)

SOLUTION:

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

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

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

(e) 230C - higher temperature

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

(f) CCl4 - larger with more orbitals/e-

slide15

Hsystem

Ssurroundings = -

T

Entropy Changes in the System

S0rxn - the entropy change that occurs when all reactants

and products are in their standard states.

S0rxn = S0products - S0reactants

The change in entropy of the surroundings is directly

related to an opposite change in the heat of the system

and inversely related to the temperature at which the

heat is transferred.

slide16

PROBLEM:

Calculate DS0rxn 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 6 mols of gas to 3 mols of gas.

SOLUTION:

Find standard entropy values in the Appendix or other table.

Sample Problem 20.2:

Calculating the Standard Entropy of Reaction, DS0rxn

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

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

Unrealistic – entropy for gaseous water = 188.7

DS = - 374 J/K

slide17

PROBLEM:

At 298K, the formation of ammonia has a negative DS0sys;

N2(g) + 3H2(g) 2NH3(g) DS0sys = -197 J/K

PLAN:

DS0universe must be > 0 in order for this reaction to be spontaneous, so DS0surroundings must be > 197 J/K. To find DS0surr, first find DHsys; DHsys = DHrxn which can be calculated using DH0f values from tables. DS0universe = DS0surr + DS0sys.

Sample Problem 20.3:

Determining Reaction Spontaneity

Calculate DH0rxn, and state whether the reaction occurs spontaneously at this temperature.

SOLUTION:

DH0rxn = [(2 mol)(DH0fNH3)] - [(1 mol)(DH0fN2) + (3 mol)(DH0fH2)]

DH0rxn = -91.8 kJ

DS0surr = -DH0sys/T =

-(-91.8x103J/298K)

= 308 J/K

DS0universe = DS0surr + DS0sys

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

DS0universe > 0 so the reaction is spontaneous.

slide18

Gibbs Free Energy (G)

G, the change in the free energy of a system, is a measure of the spontaneity of the process and of the useful energy available from it.

DG0system = DH0system - TDS0system

G < 0 for a spontaneous process

G > 0 for a nonspontaneous process

G = 0 for a process at equilibrium

DG0rxn = S mDG0products - SnDG0reactants

slide19

Components of DS0universe for spontaneous reactions

Figure 20.10

exothermic

endothermic

system becomes more disordered

exothermic

system becomes more disordered

system becomes more ordered

slide20

-

+

-

-

+

-

+

+

+

+

-

+ or -

-

-

+

+ or -

Table 20.1 Reaction Spontaneity and the Signs of DH0, DS0, and DG0

DH0

DS0

-TDS0

DG0

Description

Spontaneous at all T

Nonspontaneous at all T

Spontaneous at higher T;

nonspontaneous at lower T

Spontaneous at lower T;

nonspontaneous at higher T

slide21

PROBLEM:

Potassium chlorate, a common oxidizing agent in fireworks and matchheads, undergoes a solid-state disproportionation reaction when heated.

Note that the oxidation number of Cl in the reactant is higher in one of the products and lower in the other (disproportionation).

D

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

PLAN:

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

Sample Problem 20.4:

Calculating DG0 from Enthalpy and Entropy Values

+5

+7

-1

Use DH0f and S0 values to calculate DG0sys (DG0rxn) at 250C for this reaction.

SOLUTION:

DH0rxn= SmDH0products - SnDH0reactants

DH0rxn= (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) -

(4 mol)(-397.7 kJ/mol)

DH0rxn= -144 kJ

slide22

Sample Problem 20.4:

Calculating DG0 from Enthalpy and Entropy Values

continued

DS0rxn= SmDS0products - SnDS0reactants

DS0rxn= (3 mol)(151 J/mol*K) + (1 mol)(82.6 J/mol*K) -

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

DS0rxn= -36.8 J/K

DG0rxn= DH0rxn - T DS0rxn

DG0rxn= -144 kJ - (298K)(-36.8 J/K)(kJ/103 J)

DG0rxn= -133 kJ

slide23

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

Sample Problem 20.6:

Determining the Effect of Temperature on DG0

PROBLEM:

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

At 298K, DG0 = -141.6kJ; DH0 = -198.4kJ; and DS0 = -187.9J/K

(a) Use the data to decide if this reaction is spontaneous at 250C, and predict how DG0 will change with increasing T.

(b) Assuming DH0 and DS0 are constant with increasing T, is the reaction spontaneous at 900.0C?

PLAN:

The sign of DG0 tells us whether the reaction is spontaneous and the signs of DH0 and DS0 will be indicative of the T effect. Use the Gibbs free energy equation for part (b).

SOLUTION:

(a) The reaction is spontaneous at 250C because DG0 is (-). Since DH0 is (-) but DS0 is also (-), DG0 will become less spontaneous as the temperature increases.

slide24

Sample Problem 20.6:

Determining the Effect of Temperature on DG0

continued

(b) DG0rxn= DH0rxn - T DS0rxn

DG0rxn= -198.4kJ - (1173K)(-187.9J/mol*K)(kJ/103J)

DG0rxn= 22.0 kJ; the reaction will be nonspontaneous at 900.0C

slide25

DG and the Work a System Can Do

For a spontaneous process, DG is the maximum work obtainablefrom the system as the process takes place: DG = workmax

For a nonspontaneous process, DG is the minimum work that must be done to the system as the process takes place: DG = workmin

An example

work = pΔv

slide26

D

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

PLAN:

Use the DG summation equation.

SOLUTION:

DG0rxn= SmDG0products - SnDG0reactants

Sample Problem 20.5:

Calculating DG0rxn from DG0f Values

PROBLEM:

Use DG0f values to calculate DGrxn for the reaction in Sample Problem 20.4:

DG0rxn= (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) -

(4mol)(-296.3kJ/mol)

DG0rxn= -134kJ

slide27

Figure B20.3

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

slide30

Free Energy, Equilibrium and Reaction Direction

  • If Q<K, then ln Q/K is neg; the reaction proceeds to the right (DG is neg)
  • If Q>K, then ln Q/K is pos; the reaction proceeds to the left (DG is pos)
  • If Q=K, then ln Q/K = 0; the reaction is at equilibrium (DG = 0)

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

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

DG0 = - RT lnK

slide31

Essentially no forward reaction; reverse reaction goes to completion

Forward and reverse reactions proceed to same extent

FORWARD REACTION

REVERSE REACTION

Forward reaction goes to completion; essentially no reverse reaction

Table 20.2 The Relationship Between DG0 and K at 250C

DG0(kJ)

K

Significance

200

9x10-36

100

3x10-18

50

2x10-9

10

2x10-2

1

7x10-1

0

1

-1

1.5

-10

5x101

-50

6x108

-100

3x1017

-200

1x1035

slide32

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

Sample Problem 20.7:

Calculating DG at Nonstandard Conditions

PROBLEM:

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

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. (DG0298 = -141.6kJ/mol of reaction as written using DH0 and DS0 values at 973K. DG0973 = -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 DG for the system in part (b) at each temperature.

PLAN:

Use the equations and conditions found on slide .

slide33

(-12.12kJ/mol)(103J/kJ)

(-141.6kJ/mol)(103J/kJ)

At 298, the exponent is -DG0/RT

= -

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

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

Sample Problem 20.7:

Calculating DG at Nonstandard Conditions

continued (2 of 3)

SOLUTION:

(a) Calculating K at the two temperatures:

DG0 = -RTlnK so

= 57.2

= e57.2

= 7x1024

At 973, the exponent is -DG0/RT

= 1.50

= e1.50

= 4.5

slide34

pSO32

(0.100)2

=

(pSO2)2(pO2)

(0.500)2(0.0100)

Sample Problem 20.7:

Calculating DG at Nonstandard Conditions

continued (3 of 3)

(b) The value of Q =

= 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 DG is calculated using DG = DG0 + RTlnQ

DG298 = -141.6kJ/mol + (8.314J/mol*K)(kJ/103J)(298K)(ln4.00)

DG298 = -138.2kJ/mol

DG973 = -12.12kJ/mol + (8.314J/mol*K)(kJ/103J)(973K)(ln4.00)

DG973 = -0.9kJ/mol