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Chapter 15 - Spontaneity, Entropy, and Free Energy. 1 Spontaneous Processes 2 The Isothermal Expansion and Compression of an Ideal Gas 3 The Definition of Entropy 4 Entropy and Physical Changes 5 Entropy and the Second Law of Thermodynamics 6 The Effect of Temperature on Spontaneity

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chapter 15 spontaneity entropy and free energy
Chapter 15 - Spontaneity, Entropy, and Free Energy

1 Spontaneous Processes

2 The Isothermal Expansion and Compression of an Ideal Gas

3 The Definition of Entropy

4 Entropy and Physical Changes

5 Entropy and the Second Law of Thermodynamics

6 The Effect of Temperature on Spontaneity

7 Free Energy

8 Entropy Changes in Chemical Reactions

9 Free Energy and Chemical Reactions

10 The Dependence of Free Energy on Pressure

11 Free Energy and Equilibrium

12 Free Energy and Work (skip)

13 Reversible and Irreversible Processes: A Summary (skip)

14 Adiabatic Processes (skip)

slide2
Kinetics vs Thermo

Thermodynamics predicts direction and “driving force”.

Kinetics predicts speed (rate).

Spontaneous Processes

Occur on some timescale (maybe slowly) without outside intervention (examples: a battery will discharge, a hot cup of coffee will cool to ambient temperature).

All spontaneous processes proceed toward “states” (macrostates) with the greatest number of accessible microstates.

slide3

Microstates and Macrostates:

An available microstate describes a specific detailed microscopic configuration (molecular rotations, translations, vibrations, electronic configuration) that a system can visit in the course of its fluctuations.

A macrostate describes macroscopic properties such as temperature and pressure.

For a gas at constant T: the number of available microstates increases with volume. For gas, liquid or solid, the number of available microstates increases with T (the number of available vibrational microstates, electronic microstates, etc. increases with T). When you heat anything, you increase the number of available microstates.

When a liquid vaporizes, the number of available microstates increases.

When a liquid freezes, the number of available microstates decreases.

slide4

What is a spontaneous process?

probable

not probable

This is not a spontaneous process.

The reverse process (going from right to left) is spontaneous.

a) A gas will spontaneously expand to fill the available space.

b) There is a ‘driving’ force that causes a gas to spontaneously expand to fill a vacuum.

c) The entropy of the universe increases with a gas expands to fill a vacuum.

a=b=c

slide5

Why is this not a spontaneous process?

probable

not probable

There are more available microstates on the right hand side than on the left hand side.

If a system gains degrees of freedom (more constituents, more room to move, more available quantum states, more available rotational, vibrational, translational or electronic states), then it gains entropy.

A spontaneous process increases entropy (but you must consider both the system and the surroundings)

slide6

probable

not probable

The probability of finding both molecules on the left side is ¼

(this is one available microstate out of four possible microstates that will give this arrangement)

The probability of finding one molecule on the each side is ½.

(this are two possible microstates out of four possible microstates that will give this arrangement)

The probability of finding both molecules on the right side is ¼

(this is one microstate out of four possible microstates)

The probability of occurrence of a particular arrangement (state) depends on the number of ways (microstates) in which that arrangement can be achieved. All microstates are equally probable.

slide7

There are three possible arrangements of four molecules in two chambers.

The arrangement with the greatest number of microstates is most probable. Label the molecules a,b,c,d and count the microstates.

You will see that arrangement III is most probable.

definitions of entropy s
Definitions of Entropy“S”
  • Entropy is related to probability
  • If a system has several available macrostates, it will spontaneously proceed to the one with the largest number of available microstates.
  • The macrostate with the greatest probability (largest number of available microstates) has the highest entropy.
  • When you heat something you increase its entropy.

S = kB ln ΩJoules/Kelvin

Kb = Boltzmann’s constant, the gas constant per molecule (R/NA)

Ω = the number of available microstates of a given state

∆S = q/T J / mol-K

slide11

Ludwig Boltzmann (1844-1906)

Highlights

  • Established the logarithmic connection between entropy and probability in his kinetic theory of gases.
  • The Boltzmann constant (k or kB) is the physical constant relating temperature to energy.

Moments in a Life

  • Suffered from bipolar disorder and depression
  • Ironically, in Max Planck’s Nobel Prize speech in 1918, it was pointed out that Boltzmann never introduced the constant k, Planck did.

Zumdahl Chapter 10

slide12

One He in the gas phase expands from volume V1 to 2V1

Ω2 = 2Ω1

Twice the number of microstates

If 1 mole of He (instead of 2 He)

and the gas expands from V1 to V2

The change in entropy of a gas is dependent on the change in volume of the gas

Zumdahl Chapter 10

the isothermal expansion of an ideal gas
The isothermal expansion of an ideal gas.
  • Isothermal – system and surroundings maintain constant temperature.

ΔE = 0 = q + w

then q = – w

  • Consider only reversible and irreversible processes
    • For a reversible, cyclic process both the system and the surroundings are returned exactly to their original positions.
  • Cyclic expansion-compression process “work is converted to heat”

Work → Heat

Zumdahl Chapter 10

slide14

The isothermal expansion of an ideal gas.

∆E=0 (energy of a perfect gas depends only on T)

∆E= w + q

w = -q

This important relationship entropy (determined by number of available microscopic states) is related to a macroscopic properties of heat and temperature.

slide15

Brick A (warm)

Brick B (warm)

w(A)

q(A)

ΔE(A)

ΔH(A)

ΔS(A)

W(B)

Q(B)

ΔE(B)

ΔH(B)

ΔS(B)

Brick A (cold)

Brick B (hot)

ΔS(A) < 0 (cools) ΔS(B) > 0 (heats)

|ΔS(A)| > |ΔS(B)|

ΔS(uni) = ΔS(A) + ΔS(B)

ΔS(uni) < 0

This is not a spontaneous process.

entropy and physical change
Entropy and Physical Change

Temperature Dependence of Entropy:

Cp and Cv are is the heat capacities of the system.

ΔS(T1 to T2) here should be written ΔSsys(T1 to T2)

slide17

Example

Calculate the change in entropy that occurs when a sample containing 1.00 mol of water (ice) is heated from – 20 °C to +20°C at 1 atm pressure. The molar heat capacities of H2O (s) and H2O (l) are 38.1 J K-1mol-1 and 75.3 J K-1mol-1 respectively and the enthalpy of fusion (melting) is 6.01 kJ mol-1 at 0°C.

  • Solution
  • ΔS from 253K to 273K = n Cp ln(T2/T1) = (1.00)(38.1)ln (273/253) = 2.90 J/K
  • ΔS phase change from liq to gas = qrev/T = ΔHfus/T = (6010/273) = 22.0 J/K
  • Δfrom 273K to 293K = n Cp ln(T2/T1) = (1.00)(75.3)ln (293/273) = 5.3 J/K
  • Total ΔS = ΔS1 + ΔS2 + ΔS3
slide18
First Law of Thermodynamics

The change in the internal energy of a system is equal to the work done on it plus the heat transferred to it. The Law of Conservation of Energy

E = q + w

Second Law of Thermodynamics

For a spontaneous process the Entropy of the universe (meaning the system plus its surroundings) increases.

Suniverse > 0

Third Law of Thermodynamics

In any thermodynamic process involving only pure phases at equilibrium, the entropy change, S, approaches zero at absolute zero temperature; also the entropy of a crystalline substance approaches zero at 0K.

S = 0 at 0 K

Zumdahl Chapter 10

slide19
1st Law of Thermodynamics
    • In any process, the total energy of the universe remains unchanged: energy is conserved
    • A process and its reverse are equally allowed by the first law

0 =ΔEforward + ΔEreverse

(Energy is conserved in both directions)

  • 2nd Law of Thermodynamics
    • Processes that increase ΔSuniverse are spontaneous.

ΔSuniv > 0 Spontaneous Forward

ΔSuniv = 0 At Equilibrium

ΔSuniv < 0 Spontaneous Reverse

Zumdahl Chapter 10

slide20

Suniverse =Ssystem + Ssurroundings

  • The sign of ΔSsur depends on the direction of the heat flow.
  • The magnitude of ΔSsur depends on the temperature

This is ΔH of the system.

If the reaction is exothermic, ΔH has a negative sign and ΔSsurr is positive

If the reaction is endothermic, ΔH has a positive sign and ΔSsurr is negative

summary of entropy
Summary of Entropy
  • Entropy is a quantitative measure of the number of microstates available to the molecules in a system. It is a measure of the number of ways in which energy or molecules can be arranged.
  • Entropy is the degree of randomness or disorder in a system
  • The Entropy of all substances is positive

Ssolid < S liquid < Sgas

  • ΔSsys is the Entropy Change of the system
  • ΔSsur is the Entropy Change of the surroundings
  • ΔSuni is the Entropy Change of the universe
  • S has the units J K-1mol-1
slide23

Josiah Willard Gibbs (1839-1903)

Highlights

  • Devised much of the theoretical foundation for chemical thermodynamics.
  • Established the concept free energy

Moments in a Life

  • 1863 Yale awarded him the first American Ph.D. in engineering
  • Book: Equilibrium of Heterogeneous Substances, deemed one of the greatest scientific achievements of the 19th century.
  • Will never be famous like Michael Jackson.

Zumdahl Chapter 10

gibbs free energy
Gibbs Free Energy

ΔG = ΔHsys - TΔSsys

  • Allows us to focus on the system only, without considering the surroundings.
  • ΔG = - TΔSsur
  • G is called
      • Gibbs Function, or
      • Gibbs Free Energy, or
      • Free Energy.
slide25

Free Energy

ΔG < 0 Spontaneous

ΔG = 0 Equilibrium

ΔG > 0 Spontaneous Reverse

Entropy

ΔSuniv > 0 Spontaneous Forward

ΔSuniv = 0 Equilibrium

ΔSuniv < 0 Spontaneous Reverse

gibbs free energy1
Gibbs Free energy

Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ

  • a) Calculate ΔSvap for 1 mole of benzene at 60°C and pressure = 1 atm.
slide27
Benzene, C6H6, boils at 80°C at 1 atm.

ΔHovap = 30.8 kJ

  • a) Calculate ΔSvap for 1 mole of benzene.Start with ΔGvap=ΔHvap-TΔSvapat the boiling point, ΔGvap = 0 so ΔHvap = TbΔSvap
gibbs free energy2
Gibbs Free energy

Benzene, C6H6, boils at 80.1°C (at 1 atm) and ΔHvap = 30.8 kJ

  • b) Does benzene spontaneously boil at 60°C?
effects of temperature on g
Effects of Temperature on ΔG°

ΔG° = ΔH° - TΔS°

  • Typically ΔH and ΔS are almost constant over a broad range
  • For the reaction above, as Temperature increases ΔG becomes more positive, i.e., less negative.

3NO (g) → N2O (g) + NO2 (g)

slide32

For temperatures other than 298K or 25CΔG = ΔH - T·ΔS

Case C

ΔH° > 0

ΔS° < 0

ΔG = ΔH - T·ΔSΔG = (+) - T·(-) = positive

 ΔG > 0 or non-spontaneous at all Temp.

B

A

Case B

ΔH° < 0

ΔS° > 0

ΔG = ΔH - T·ΔSΔG = (-) - T·(+) = negative

 ΔG < 0 or spontaneous at all temp.

C

D

slide33

For temperatures other than 298K or 25CΔG = ΔH - T·ΔS

Case D

ΔH° < 0

ΔS° < 0

ΔG = ΔH - T·ΔS at a low Temp

ΔG = (-) - T·(-) = negative

 ΔG < 0 or spontaneous at low Temp.

B

A

Case A

ΔH° > 0

ΔS° > 0

ΔG = ΔH - T·ΔSΔG = (+) - T·(+) at a High Temp

 ΔG < 0 or spontaneous at high Temp.

C

D

entropies of reaction
Entropies of Reaction

ΔSrxn° = ΣS°products – ΣS°reactants

ΔSrxn° is the sum of products minus the sum of the reactants, for one mole of reaction (that is what ° means)

For a general reaction

a A + b B → c C + d D

Appendix 4 tabulates standard molar entropy values, S° in units JK-1mol-1

example
Example
  • Calculate ΔSr° at 298.15 K for the reaction

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

(b) Calculate ΔS° of the system when 26.71 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at 298.15K

slide37
Calculate ΔSr° at 298.15 K for the reaction

2H2S(g) + 3O2(g) → 2SO2(g) + 2H2O(g)

Solution

(a) Look up each S° of formation [Note this is for “one mole of the reaction”as written: i.e. 2 moles of H2S, 3 moles of O2, etc]

ΔSrxn°= 2S°(SO2(g) ) + 2S°(H2O(g)) -2S°(H2S(g) ) - 3S°(O2(g))

ΔSrxn°= 2(248) + 2(189) -2(206) - 3(205)

= – 153 JK-1mol-1

slide38
2H2S(g) + 3O2(g) → 2SO2(g) +2H2O(g)

(a) ΔSrxn= -153 JK-1mol-1

(b) Calculate ΔS° when 26.7 g of H2S(g) reacts with excess O2(g) to give SO2(g) and H2O(g) and no other products at 298.15K

Solution:

free energy and chemical reactions
Free Energy and Chemical Reactions

ΔG = ΔH - T·ΔS

  • ΔGf° is the standard molar Gibbs function of formation
  • Because G is a State Property, for a general reaction

a A + b B → c C + d D

slide40
Calculate ΔG° for the following reaction at 298.15K. Use Appendix 4 for additional information needed.

3NO(g) → N2O(g) + NO2(g)

  • Solution From Appendix 4
      • ΔGf°(N2O) = 104 kJ mol-1
      • ΔGf°(NO2) = 52
      • ΔGf° (NO) = 87
  • ΔG°= 1(104) + 1(52) – 3(87)
  • ΔG°= − 105 kJ therefore, spontaneous

Zumdahl Chapter 10

slide41

The Dependence of Free Energy on Pressure

ΔG = ΔG° + RT ln Q

Where Q is the reaction quotient

  • a A + b B ↔ c C + d D
      • If Q > K the rxn shifts towards the reactant side
        • The amount of products are too high relative to the amounts of reactants present, and the reaction shifts in reverse (to the left) to achieve equilibrium
      • If Q = K equilibrium
      • If Q < K the rxn shifts toward the product side
        • The amounts of reactants are too high relative to the amounts of products present, and the reaction proceeds in the forward direction (to the right) toward equilibrium

compare

Zumdahl Chapter 10

slide42

ΔG = ΔG° + RT ln Q

    • Where Q is the reaction quotient
  • a A + b B ↔ c C + d D
      • If Q < K the rxn shifts towards the product side
      • If Q = K equilibrium
      • If Q > K the rxn shifts toward the reactant side
  • At Equilibrium conditions, ΔG = 0
    • ΔG° = -RT ln K
  • NOTE: we can now calculate equilibrium constants (K) for reactions from standard ΔGf functions of formation

Zumdahl Chapter 10

slide43
Calculate the equilibrium constant for this reaction at 25C.

3NO(g) ↔ N2O(g) + NO2(g)

  • Strategy

Use -ΔG ° = RT ln K

Use ΔG°= - 105 kJ mol -1 (from previous)

slide44
3NO(g) ↔ N2O(g) + NO2(g)
  • Solution

Use–ΔG ° = RT ln K

Rearrange

ΔGrxn°= – 105 kJ mol-1

slide45

ΔG = ΔG° + RT ln Q

    • Where Q is the reaction quotient
  • a A + b B ↔ c C + d D

Zumdahl Chapter 10

the temperature dependence of equilibrium constants
The Temperature Dependence of Equilibrium Constants
  • Where does this come from?
  • Recall ΔG = ΔH - T·ΔS
  • Divide by RT, then multiply by -1

Zumdahl Chapter 10

slide47
Notice that this is y = mx + b the equation for a straight line
  • A plot of y = mx + b or
  • ln K vs. 1/T

Zumdahl Chapter 10

slide48
If we have two different Temperatures and K’s (equilibrium constants)

or

  • Now given ΔH and T at one temperature, we can calculate K at another temperature, assuming that ΔH and ΔS are constant over the temperature range

Zumdahl Chapter 10

slide49

The Person Behind the Science

J.H. van’t Hoff (1852-1901)

Highlights

  • Discovery of the laws of chemical dynamics and osmotic pressure in solutions
  • his work led to Arrhenius's theory of electrolytic dissociation or ionization
  • The Van't Hoff equation in chemical thermodynamics relates the change in temperature to the change in the equilibrium constant given the enthalpy change.

Moments in a Life

  • 1901 awarded first Noble Prize in Chemistry

van’t Hoff Factor (i)

  • ΔT= − i m K

Zumdahl Chapter 10

slide50
The reaction

2 Al3Cl9 (g) → 3 Al2Cl6 (g)

Has an equilibrium constant of 8.8X103 at 443K and a ΔHr°= 39.8 kJmol-1 at 443K. Estimate the equilibrium constant at a temperature of 600K.

Zumdahl Chapter 10

slide51

P = mg/A

V

Infinite-step expansion

2 step expansion

6 step expansion

Expansion (V2 > V1): Work flows out of the system and the Work sign is negative

Zumdahl Chapter 10

slide53

P = mg/A

V

Compression (V2 < V1): Work is put into system,

ln(V2/V1) is negative and the Work is positive