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# Chapter 20 Thermodynamics

## 1. Chapter 20 – Thermodynamics

• 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

1

## 2. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.2Spontaneous expansion of a gas

stopcock

closed

1 atm

evacuated

stopcock

opened

0.5 atm

0.5 atm

2

## 3. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.3Expansion of a gas and the increase in number of microstates.

3

## 4. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

1877 Ludwig BoltzmanS = k ln W

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

The second law of thermodynamics.

4

## 5. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.4Random 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

5

## 6. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.5 The increase in entropy from solid to liquid to gas.6

## 7. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.6The entropy change accompanying the dissolution of a salt.

pure solid

MIX

pure liquid

solution

7

## 8. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.7The small increase in entropy when ethanol dissolves in water.

Ethanol

Water

Solution of

ethanol and

water

8

## 9. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.8The large decrease in entropy when a gas dissolves in a liquid.

O2 gas

O2 gas in H2O

9

## 10. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Figure 20.9Entropy and vibrational motion.

NO

NO2

N2O4

10

## 11. 20.1 – The Second Law of Thermodynamics: Predicting Spontaneous Change

Sample Problem 20.1:PROBLEM:

Predicting Relative Entropy Values

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.

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 mass

11

## 12. 20.2 – Calculating the Change in Entropy of a Reaction

Sample Problem 20.2:PROBLEM:

Calculating the Standard Entropy of Reaction, DS0rxn

Calculate DS0rxn for the combustion of 1mol of propane at 250C.

C3H8(g) + 5O2(g)

PLAN:

3CO2(g) + 4H2O(l)

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.

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)]

DS = - 374 J/K

12

## 13. 20.2 – Calculating the Change in Entropy of a Reaction

Sample Problem 20.3:PROBLEM:

Determining Reaction Spontaneity

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

N2(g) + 3H2(g)

2NH3(g)

DS0sys = -197 J/K

Calculate DS0rxn, and state whether the reaction occurs

spontaneously at this temperature.

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.

SOLUTION:

DH0rx = [(2 mol)(DH0fNH3)] - [(1 mol)(DH0fN2) + (3 mol)(DH0fH2)]

DH0rx = -91.8 kJ

DS0surr = -DH0sys/T =

-(-91.8x103J/298K)

DS0universe = DS0surr + DS0sys

= 308 J/K

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

DS0universe > 0 so the reaction is spontaneous.

13

## 14. 20.3 – Entropy, Free Energy, and Work

Sample Problem 20.4:PROBLEM:

Calculating DG0 from Enthalpy and Entropy Values

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).

+5

4KClO3(s)

D

+7

-1

3KClO4(s) + KCl(s)

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

PLAN:

Use Appendix B values for thermodynamic entities; place them into the

Gibbs Free Energy equation and solve.

SOLUTION:

DH0rxn = S mDH0products - S nDH0reactants

DH0rxn = (3 mol)(-432.8 kJ/mol) + (1 mol)(-436.7 kJ/mol) (4 mol)(-397.7 kJ/mol)

DH0rxn = -144 kJ

14

## 15. 20.3 – Entropy, Free Energy, and Work

Sample Problem 20.4:Calculating DG0 from Enthalpy and Entropy Values

continued

DS0rxn = S mDS0products - S nDS0reactants

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

15

## 16. 20.3 – Entropy, Free Energy, and Work

Sample Problem 20.5:PROBLEM:

PLAN:

Calculating DG0rxn from DG0f Values

Use DG0f values to calculate DGrxn for the reaction in Sample

Problem 20.4:

D

4KClO3(s)

3KClO4(s) + KCl(s)

Use the DG summation equation.

SOLUTION:

DG0rxn = S mDG0products - S nDG0reactants

DG0rxn = (3mol)(-303.2kJ/mol) + (1mol)(-409.2kJ/mol) (4mol)(-296.3kJ/mol)

DG0rxn = -134kJ

16

## 17. 20.3 – Entropy, Free Energy, and Work

Sample Problem 20.6:PROBLEM:

Determining the Effect of Temperature on DG0

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, 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:

SOLUTION:

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).

(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.

17

## 18. 20.3 – Entropy, Free Energy, and Work

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

18

## 19. 20.3 – Entropy, Free Energy, and Work

Figure B20.3The coupling of a nonspontaneous reaction to the hydrolysis of ATP.

19

## 20. 20.3 – Entropy, Free Energy, and Work

Figure B20.4 The cycling of metabolic free enery through ATP20

## 21. 20.3 – Entropy, Free Energy, and Work

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

21

## 22.

20.4 – Free Energy, Equilibrium, and Reaction DirectionFree Energy, Equilibrium and Reaction Direction

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

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

•If Q/K = 1, 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

## 23.

20.4 – Free Energy, Equilibrium, and Reaction DirectionTable 20.2 The Relationship Between DG0 and K at 250C

DG0(kJ)

K

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

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

REVERSE REACTION

9x10-36

FORWARD REACTION

200

Significance

## 24.

20.4 – Free Energy, Equilibrium, and Reaction DirectionSample Problem 20.7:

PROBLEM:

Calculating DG 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. (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.

## 25.

20.4 – Free Energy, Equilibrium, and Reaction DirectionSample Problem 20.7:

Calculating DG at Nonstandard Conditions

continued (2 of 3)

SOLUTION:

DG0

(a) Calculating K at the two temperatures:

= -RTlnK so

(DG / RT)

K e

At 298, the exponent is -DG0/RT

(DG / RT)

K e

0

= e57.2

0

=-

K e

0

= e1.50

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

= 57.2

= 7x1024

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

At 973, the exponent is -DG0/RT

(DG / RT)

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

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

= 4.5

= 1.50

## 26.

20.4 – Free Energy, Equilibrium, and Reaction DirectionSample Problem 20.7:

Calculating DG at Nonstandard Conditions

continued (3 of 3)

(b) The value of Q =

pSO32

(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 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)

DG298 = -0.9kJ/mol