1

CHEM 212

Chapter 3

By Dr. A. Al-Saadi

2

Introduction to The Second Law of Thermodynamics

The two different pressures will be equalized upon removing the partition.

This is a natural (usually irreversible) process or called “spontaneous”.

3

Introduction to The Second Law of Thermodynamics

The irreversible natural “spontaneous” process takes place by the heat passing from the hot object to the cold one until we have T1 = T2.

The process that is another way around is called “nonspontaneous” and has to be done reversibly.

Hot solid

Cold solid

T1 T2

Where T1 > T2

4

Introduction to The Second Law of Thermodynamics

2H2(g) + O2(g) 2H2O(l)

Is reaction spontaneous?

It is spontaneous in going from left to right.

For the reaction to go from right to left, work has to be done on H2O(l) to get the gases, and it is nonspontaneous.

5

Entropy (S)

A measurement of spontaneity is the entropy.

If a process occurs reversible from state A to state B with infinitesimal amounts of heat (dqrev) absorbed at

each stage, the “entropy change” is defined as:

B

A

revBA T

dqS

6

Defining Entropy on the Basis of Steam Engine

It is impossible for an engine to perform work by cooling a portion of matter to a temperature below

that of the coldest part of the surroundings.

B

A

revBA T

dqS

7

Historical Development of Steam Engines

Newcomen steam engine:

• One-cylinder steam engine.

•Efficiency of converting heat into work is only 1%

8

Historical Development of Steam Engines

Watt’s steam engine:

• Two-cylinder steam engine, mainly a cylinder (kept at Th) and a condenser (kept at Tc).

•Efficiency of converting heat into work is only 10%

9

Historical Development of Steam Engines

Watt’s steam engine:

• Two-cylinder steam engine, mainly a cylinder (kept at Th) and a condenser (kept at Tc).

•Efficiency of converting heat into work is only 10%

10

The Carnot Engine

It is a theoretical engine explaining the process in terms of Th and Tc.

It assumes:-An ideal engine.-An ideal gas, and-A reversible process.

Carnot showed that such an engine would have the maximum possible efficiency for any engine working between Th and Tc.

11

The Carnot Engine

Four processes being done reversibly:

1) Isothermal expansion at Th.

2) Adiabatic expansion.3) Isothermal compression at Tc.

4) Adiabatic compression.

12

The Carnot Cycle

Pressure-volume diagram for the Carnot cycle.

AB and CD are isotherms.

BC and DA are adiabatics

13

The Carnot Cycle

The Carnot cycle for 1 mole of an ideal gas that starts at 10.0 bar and 298.15 K with the value of CP/CV = γ = 1.40 (used for N2 gas)

14

The Carnot Cycle

15

The Net Work from Carnot Cycle

16

Example 3.1

17

The Concept of Entropy

q = 0

Tc=Ta+dT

qb = qc+ dq

Tc is unchanged

18

The Concept of Entropy

Entropy is a measure of the disorder of the system.

The change in the entropy (ΔS) indicates whether a change in the disorder of the system takes place or not. The sign of ΔS also indicated whether the disorder of the system increases or decreases.

19

The Concept of Entropy

0irr

T

dq

Inequality of Clausius:

The change in entropy is generally given by:

0S

Any change that occurs irreversibly in nature is spontaneous and therefore is accompanied with a net increase in entropy.

The energy of the universe is constant. The entropy of the universe tends always towards maximum.

20

Molecular Interpretation of Entropy

When a spontaneous change takes place, the disorder of the system increases and its entropy, as a result, increases.

Entropy is a measure of disorder.

Example 1: Mixing solute and solvent

Mixing process is a spontaneous process that increases the disorder of the system.

Thus, entropy is expected to increase and ΔS is +ve.

21

Molecular Interpretation of Entropy

Example 2: Phase change

-Melting of solid. (ΔS is +ve)

-Freezing of a liquid. (ΔS is -ve)

-Evaporation of liquid. (ΔS is +ve)

Example 3: Chemical reactions

-H2(g) 2H(g) (ΔS is +ve)

-2NH2(g) 3H2(g) + N2(g) (ΔS is +ve)

Generally, in gaseous reactions, as the number of molecules increases (less pairing), entropy would increase. That is because less pairing involves more disorder and less restrictions.

22

Entropy of Mixing

The mixing of ideal gases at equal pressures and temperatures

23

Calculation of Entropy Change

24

Calculation of Entropy Change for Solids and Liquids

25

Calculation of Entropy Change

26

Calculation of Entropy Change

27

The Third Law of Thermodynamics

Nernst’s heat theorem states that: ΔS = 0 at the absolute zero provided that the states of the system are in thermodynamic equilibrium.

Equilibrium is slowly and barely achieved at extremely low temperatures.

The 3rd law of thermodynamics:The entropies (Sa , Sb , Sc , …) of all perfectly crystalline substances (thermodynamic equilibrium) must be the same at the absolute zero (Sa = Sb = Sc = …).

28

Achieving Very Low Temperatures1- Expansion of the molecules to reduce the intermolecular forces (cooling to 1K).

2- Magnetization (entropy decrease and thus flow of heat from the sample to the

surrounding) and then demagnetization (done after isolating the system “adiabatic

step” causing cooling) . (cooling to about 0.005K)

29

Absolute Entropies

The absolute entropy of a substance can be calculated at another temperature by:

• cooling down that substance to nearly the absolute zero, and then

• increasing its temperature through reversible steps until the desired temperature is attained.

30

Absolute Entropies

31

Entropy Change of a Chemical Reaction

)(reactants(products) SSS

Absolute entropies for substances are tabulated at standard conditions.

http://bilbo.chm.uri.edu/CHM112/tables/thermtable.htm

32

Conditions for Equilibrium

The position of equilibrium must correspond to a state of maximum total entropy.

33

Molecular Interpretation of Gibbs Energy (G)

In spontaneous processes at constant T and P, system moves towards a state of minimum Gibbs energy. (dG < 0)

In the condition of equilibrium at constant T and P, there is no change in Gibbs energy (dG = 0)

34

Molecular Interpretation of Gibbs Energy (G)

Consider the reaction H2 ↔ 2H at T=300K

H2 2H (nonspontaneous)

2H H2 (spontaneous)

For the spontaneous 2H H2 reaction:ΔG is –ve (The system approaches equilibrium as G goes to minimum).

To show this in terms of enthalpy and entropy changes:ΔG = ΔH – TΔS

ΔH is –ve (The reaction loses heat).ΔS is –ve, (More ordered arrangement as H2 molecules are produced).

ΔH – TΔS< 0 < 0

Thus, the affecting parameter (the weighting factor) is T:T is small, ΔH > TΔS, and ΔG will be –ve (spontaneous).T is too large, ΔH < TΔS, and ΔG will be +ve (nonspontaneous dissociation of H2 molecules).

35

The Contribution to Spontaneous Change

Enthalpy Change Entropy Change Spontaneous?

ΔG = ΔH – TΔS

ΔH < 0 (Exothermic) ΔS > 0

ΔH < 0 (Exothermic) ΔS < 0

ΔH < 0 (Exothermic) T = 0

ΔH > 0 (Endothermic) ΔS > 0

ΔH > 0 (Endothermic) ΔS < 0

ΔH > 0 (Endothermic) T = 0

Yes (ΔG < 0)

Yes, if |TΔS| < |ΔH|

Yes (ΔG < 0)

Yes, if |TΔS| > |ΔH|

No (ΔG > 0)

No (ΔG > 0)

36

The vaporization of water at 100°C.

(a) Liquid water at 100°C is in equilibrium with water vapor at 1 atm pressure.

(b) Liquid water at 100°C is in contact with water vapor at 0.9 atm pressure, and there is spontaneous vaporization

37

ΔrGº for Chemical Reactions

Determine the Delta G under standard conditions using Gibbs Free Energies of Formation found in a suitable thermodynamic table for the following reaction:

4HCN(l) + 5O2(g) ---> 2H2O(l) + 4CO2(g) + 2N2(g)

http://members.aol.com/profchm/t_table.html

2 moles ( -237.2 kj/mole) = -474.4 kj = Standard Free Energy of Formation for two moles H2O(l)

4 moles ( -394.4 kj/mole) = -1577.6 kJ = Standard Free Energy of Formation for four moles CO2 (g)

2 moles(0.00 kj/mole) = 0.00 kJ = Standard Free Energy of Formation for 2 moles of N2(g)

Standard Free Energy for Products = (-474.4 kJ) + (-1577.6 kJ) + 0.00 = -2052 kJ

4 moles ( 121 kj/mole) = 484 kJ = Standard Free Energy for 4 moles HCN(l) 5 moles (0.00 kJ/mole) = 0.00 kJ = Standard Free Energy of 5 moles of O2(g) Standard Free Energy for Reactants = (484) + (0.00) = 484 kJ

Standard Free Energy Change for the Reaction =Sum of Free Energy of Products - Sum of Free Energy of Reactants

= (-2052 kJ) - (484 kJ) = -1568 kJ

38

Maxwell Relations

Important Relationships

;PV

U

S

TS

U

V

VP

H

S

; TS

H

P

PV

A

T

;

; ST

A

V

VP

G

T

ST

G

P

39

Maxwell Relations

40

Maxwell Relations

41

Applications of Maxwell Relations

Thermal expansion coefficient (α):During heat transfer, the energy that is stored in the intermolecular bonds between atoms changes. When the stored energy increases, so does the length of the molecular bond. As a result, solids typically expand in response to heating and contract on cooling. This response to temperature change is expressed as its coefficient of thermal expansion.

coefficient of linear thermal expansion α

materialα in 10-6/K at 20 °C

Mercury60BCB42Lead29Aluminum23Brass19Stainless steel17.3Copper17Gold14Nickel13Concrete12Iron or Steel11.1Carbon steel10.8Platinum9Glass8.5GaAs5.8Indium Phosphide4.6Tungsten4.5Glass, Pyrex3.3Silicon3Invar1.2

Diamond1

Quartz, fused0.59

42

Applications of Maxwell Relations

Prove that for 1 mole of a van der Walls gas, the internal pressure (∂U/∂V)T is equal to a/V2

m .

43

Fugacity

44

Fugacity