Chapter 15 - Spontaneity, Entropy, and Free Energy 1 Spontaneous Processes 2 The Isothermal Expansion and Compression of an Ideal Gas 3 The Definition.

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)

• 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.

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.

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

What is a spontaneous process?


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)

Why is this not a spontaneous process?

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.

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 both molecules on the right side is ¼(this is one microstate out of four possible microstates)

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)

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.

04/19/23

21 1 1 1x = =

2 2 2 4

n1

2

2 moleculesProbability of

finding 2 molecules on the same side is 1/4

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 Ω Joules/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

04/19/23 Zumdahl Chapter 10

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.

04/19/23 Zumdahl Chapter 10 12

2 1 2 1

2 1

1 1

ln ln

2ln ln ln 2 ln 2

B B

nB B B B

S S S k k

k k k k

Ω2 = 2Ω1

Twice the number of

microstates

23

23

1 2

6 102

1

6 10 23

2

1

2

ln 2 (6 10 )( ln 2) ln 2 ln 2

ln

x

xB B A B

V V

S k x k N k R

VS nR

V

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

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


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

The isothermal expansion of an ideal gas.

∆E=0 (energy of a perfect gas depends only on T)∆E= w + q w = -q

2

1

lnV

S nRV

2

1

2

1

ln

ln

rev rev

rev

Vw nRT q

V

Vq nRT

V

revqS

T

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

Brick A (warm) Brick B (warm)

Brick A (cold) Brick B (hot)

↓w(A)q(A)ΔE(A)ΔH(A)ΔS(A)

W(B)Q(B)ΔE(B)ΔH(B)ΔS(B)

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

1 2

1 2

2

1

2

1

ln at constant Pressure

ln at constant Volume

T T p

T T V

TS nC

T

TS nC

T

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)

17

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

1. ΔS from 253K to 273K = n Cp ln(T2/T1) = (1.00)(38.1)ln (273/253) = 2.90 J/K

2. ΔS phase change from liq to gas = qrev/T = ΔHfus/T = (6010/273) = 22.0 J/K

3. Δfrom 273K to 293K = n Cp ln(T2/T1) = (1.00)(75.3)ln (293/273) = 5.3 J/K

4. Total ΔS = ΔS1 + ΔS2 + ΔS3

1 2

2T → T p

1

TΔS =nC ln at const. P

T

fusrevHq

ST T

Entropy Change from -20C to +20Cice to water

05

101520253035

250 260 270 280 290 300

Temperature (K)

En

tro

py

Ch

an

ge

(J

/K)


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


• 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

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

• The magnitude of ΔSsur depends on the temperature

surr

HS

T

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

Suniverse = Ssystem + Ssurroundings

This is ΔH of the system.

Ssystem + Ssurroundings = Suniverse

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


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.

Gibbs Free Energy

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 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.

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-TΔSvap

at the boiling point, ΔGvap = 0so ΔHvap = TbΔSvap

13

b

vapvap 87.2JK

80.1) (273.15J10 x30.8

TΔH

ΔS

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?

€

Gvap

= ΔHvap

- TΔSvap

ΔGvap

= 30,800J − (273K + 60oC)(87.2 J K-1)

ΔGvap

= +1749 J = +1.7 kJ

Since ΔGvap

> 0, benzene does not boil at 60o C, 1 atm.

Effects of Temperature on ΔG°

3NO (g) → N2O (g) + NO2 (g)

Effects of Temperature on ΔG

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

AB

CD


AB

CD

Case B

ΔH° < 0

ΔS° > 0

ΔG = ΔH - T·ΔS

ΔG = (-) - T·(+) = negative

ΔG < 0 or spontaneous at all temp.

Case C

ΔH° > 0

ΔS° < 0

ΔG = ΔH - T·ΔS

ΔG = (+) - T·(-) = positive

ΔG > 0 or non-spontaneous at all Temp.


AB

CD

Case D

ΔH° < 0

ΔS° < 0

ΔG = ΔH - T·ΔS at a low Temp

ΔG = (-) - T·(-) = negative

ΔG < 0 or spontaneous at low Temp.

Case A

ΔH° > 0

ΔS° > 0

ΔG = ΔH - T·ΔS

ΔG = (+) - T·(+) at a High Temp

ΔG < 0 or spontaneous at high Temp.

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

(B)S(A)S(D)S(C)SΔS ooooo badc

Example

(a) 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

(a) 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

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:

€

S° = 26.7 g H2S •1

34 g H2S

mol

•1 mol reaction

2 mol H2S•

(−153 J K−1)

1 mol reaction

ΔS° = −60.0 J K−1

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

(B)ΔG(A)ΔG(D)ΔG(C)ΔGΔG of

of

of

of

or badc


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


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

conditions anybB

aA

dD

cC

PP

PPQ

mequilibriubB

aA

dD

cC

PP

PPK

The Dependence of Free Energy on Pressure


At Equilibrium conditions, ΔG = 0

ΔG° = -RT ln KNOTE: we can now calculate equilibrium constants

(K) for reactions from standard ΔGf functions of formation

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

ba

dc

[B][A]

[D][C]K

(B)ΔG(A)ΔG

(D)ΔG(C)ΔGΔGof

of

of

of

or

ba

dc

Calculate the equilibrium constant for this reaction at 25C.

3NO(g) ↔ N2O(g) + NO2(g)

• StrategyUse - ΔG ° = RT ln K

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

3NO(g) ↔ N2O(g) + NO2(g)

• Solution

Use – ΔG ° = RT ln K

Rearrange

ΔGrxn°= – 105 kJ mol-1

RTK

G

ln

€

ln K =−(−105,000 J mol−1)

(8.3145 J K−1 mol−1)(298.15 K)= 42

K = e42 = 2 x 1018

183

11

1.8x10][

][][22

NO

NOON

P

PPK


ΔG = ΔG° + RT ln QWhere Q is the reaction quotient

a A + b B ↔ c C + d D

Spontaneous Processes

Equilibrium Processes

Non-spontaneous

Processes

Conditions

ΔSuniv > 0 ΔSuniv = 0 ΔSuniv < 0 All conditions

ΔGf < 0 ΔGf = 0 ΔGf > 0 Constant P and T

Q < K Q = K Q > K Constant P and T

Criteria for Spontaneity in a Chemical Reaction


The Temperature Dependence of Equilibrium Constants

• Where does this come from?• Recall ΔG = ΔH - T·ΔS• Divide by RT, then multiply by -1

lnKR

ΔSRT

ΔHRT

ΔG

R

ΔSRT

ΔHRT

ΔG

RTK

G

ln

RRTRTK

S H G ln


• 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

R ΔS

RT ΔH

lnK

RΔS

intercept-Y

R ΔH

slope


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

Equation Hofft van'

T

1

T

1

R

ΔH-

K

Kln

121

2

• 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

Equation Hofft van'

T

1

T

1

R

ΔH

K

Kln

211

2

or


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

dissolved solute of moles

solutionin particles of moles i

van’t Hoff Factor (i)

ΔT = − i m K

Equation Hofft van'

T

1

T

1

R

ΔH-

K

Kln

121

2


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.

5600

2.83

600

3600

1-1-

1-

443

600

121

2

1.5x10K

16e8.8x10

K

2.8)8.8x10

Kln(

443K

1

600K

1

mol8.315JK

Jmol 39,800-

K

Kln

Eq. Hofft van'T

1

T

1

R

ΔH-

K

Kln


VP = mg/A

2 step expansion 6 step expansion Infinite-step expansion

2

1

2

1

ln

ext

V

extV

rev rev

w P V

Work P dV

Vw nRT q

V

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

P = mg/A

2

1

2

1

ln

ext

V

extV

rev rev

w P V

Work P dV

Vw nRT q

V

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

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

V

Chapter 15 - Spontaneity, Entropy, and Free Energy 1 Spontaneous Processes 2 The Isothermal Expansion and Compression of an Ideal Gas 3 The Definition.

Documents

possible microstates

electronic microstates

number of ways microstates

available space

available quantum states

entropy changes

definition of entropy

spontaneous processesoccur