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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
- 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.

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. abc

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)

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.

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.

(No Transcript)

2 moleculesProbability of finding 2 molecules on

the same side is 1/4

Definitions of EntropyS

- 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 O Joules/Kelvin Kb Boltzmanns

constant, the gas constant per molecule (R/NA) O

the number of available microstates of a given

state ?S q/T J / mol-K

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 Plancks Nobel Prize speech in

1918, it was pointed out that Boltzmann never

introduced the constant k, Planck did.

One He in the gas phase expands from volume V1

to 2V1

O2 2O1 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

The isothermal expansion of an ideal gas.

- Isothermal system and surroundings maintain

constant temperature. - ?E 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. ?E0

(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.

Brick A (warm)

Brick B (warm)

w(A) q(A) ?E(A) ?H(A) ?S(A)

W(B) Q(B) ?E(B) ?H(B) ?S(B)

?

Brick A (cold)

Brick B (hot)

?S(A) lt 0 (cools) ?S(B) gt 0 (heats)

?S(A) gt ?S(B) ?S(uni) ?S(A) ?S(B)

?S(uni) lt 0 This is not

a spontaneous process.

Entropy and Physical Change

Temperature Dependence of Entropy

Cp and Cv are is the heat capacities of the

system. ?S(T1 to T2) here should be written

?Ssys(T1 to T2)

Example Calculate the change in entropy that

occurs when a sample containing 1.00 mol of water

(ice) is heated from 20 C to 20C 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 0C.

- Solution
- ?S from 253K to 273K n Cp ln(T2/T1)

(1.00)(38.1)ln (273/253) 2.90 J/K - ?S phase change from liq to gas qrev/T

?Hfus/T (6010/273) 22.0 J/K - ?from 273K to 293K n Cp ln(T2/T1)

(1.00)(75.3)ln (293/273) 5.3 J/K - Total ?S ?S1 ?S2 ?S3

- 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 gt 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 ?Eforward ?Ereverse
- (Energy is conserved in both directions)
- 2nd Law of Thermodynamics
- Processes that increase ?Suniverse are

spontaneous.

?Suniv gt 0 Spontaneous Forward ?Suniv 0 At

Equilibrium ?Suniv lt 0 Spontaneous Reverse

?Suniverse ?Ssystem ?Ssurroundings

- The sign of ?Ssur depends on the direction of the

heat flow. - The magnitude of ?Ssur depends on the temperature

This is ?H of the system.

If the reaction is exothermic, ?H has a negative

sign and ?Ssurr is positive

If the reaction is endothermic, ?H has a positive

sign and ?Ssurr is negative

?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 lt S liquid lt Sgas
- ?Ssys is the Entropy Change of the system
- ?Ssur is the Entropy Change of the surroundings
- ?Suni 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

- ?G ?Hsys - T?Ssys
- Allows us to focus on the system only, without

considering the surroundings. - ?G - T?Ssur
- G is called
- Gibbs Function, or
- Gibbs Free Energy, or
- Free Energy.

Free Energy ?G lt 0 Spontaneous ?G

0 Equilibrium ?G gt 0 Spontaneous Reverse

Entropy ?Suniv gt 0 Spontaneous Forward ?Suniv

0 Equilibrium ?Suniv lt 0 Spontaneous Reverse

Gibbs Free energy

- Benzene, C6H6, boils at 80.1C (at 1 atm) and

?Hvap 30.8 kJ - a) Calculate ?Svap for 1 mole of benzene at 60C

and pressure 1 atm.

- Benzene, C6H6, boils at 80C at 1 atm.
- ?Hovap 30.8 kJ
- a) Calculate ?Svap for 1 mole of

benzene. Start with ?Gvap?Hvap-T?Svap at

the boiling point, ?Gvap 0 so ?Hvap Tb?Svap

Gibbs Free energy

- Benzene, C6H6, boils at 80.1C (at 1 atm) and

?Hvap 30.8 kJ - b) Does benzene spontaneously boil at 60C?

Effects of Temperature on ?G

- ?G ?H - T?S
- Typically ?H and ?S are almost constant over a

broad range - For the reaction above, as Temperature increases

?G becomes more positive, i.e., less negative.

3NO (g) ? N2O (g) NO2 (g)

Effects of Temperature on ?G

For temperatures other than 298K or 25C?G ?H -

T?S

A

B

C

D

For temperatures other than 298K or 25C?G ?H -

T?S

Case C ?H gt 0 ?S lt 0 ?G ?H - T?S?G ()

- T(-) positive ? ?G gt 0 or non-spontaneous at

all Temp.

A

B

Case B ?H lt 0 ?S gt 0 ?G ?H - T?S?G (-)

- T() negative ? ?G lt 0 or spontaneous at all

temp.

C

D

For temperatures other than 298K or 25C?G ?H -

T?S

Case D ?H lt 0 ?S lt 0 ?G ?H - T?S at a

low Temp ?G (-) - T(-) negative ? ?G lt 0 or

spontaneous at low Temp.

A

B

Case A ?H gt 0 ?S gt 0 ?G ?H - T?S?G

() - T() at a High Temp ? ?G lt 0 or

spontaneous at high Temp.

C

D

(No Transcript)

Entropies of Reaction

- ?Srxn SSproducts SSreactants
- ?Srxn 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

- Calculate ?Sr at 298.15 K for the reaction
- 2H2S(g) 3O2(g) ? 2SO2(g) 2H2O(g)
- (b) Calculate ?S 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

- Calculate ?Sr 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 reactionas written i.e. 2

moles of H2S, 3 moles of O2, etc - ?Srxn 2S(SO2(g) ) 2S(H2O(g)) -2S(H2S(g) )

- 3S(O2(g)) - ?Srxn 2(248) 2(189) -2(206) - 3(205)
- 153 JK-1mol-1

- 2H2S(g) 3O2(g) ? 2SO2(g) 2H2O(g)
- (a) ?Srxn -153 JK-1mol-1
- (b) Calculate ?S 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

- ?G ?H - T?S
- ?Gf 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

- Calculate ?G for the following reaction at

298.15K. Use Appendix 4 for additional

information needed. - 3NO(g) ? N2O(g) NO2(g)

Solution From Appendix 4 ?Gf(N2O) 104 kJ

mol-1 ?Gf(NO2) 52 ?Gf (NO) 87 ?G 1(104)

1(52) 3(87) ?G - 105 kJ therefore, spontaneous

The Dependence of Free Energy on Pressure

- ?G ?G RT ln Q
- Where Q is the reaction quotient

- a A b B ? c C d D
- If Q gt 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 lt 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

- ?G ?G RT ln Q
- Where Q is the reaction quotient
- a A b B ? c C d D
- If Q lt K the rxn shifts towards the product side
- If Q K equilibrium
- If Q gt K the rxn shifts toward the reactant side

At Equilibrium conditions, ?G 0 ?G -RT ln

K NOTE we can now calculate equilibrium

constants (K) for reactions from standard ?Gf

functions of formation

- Calculate the equilibrium constant for this

reaction at 25C. - 3NO(g) ? N2O(g) NO2(g)
- Strategy
- Use - ?G RT ln K

Use ?G - 105 kJ mol -1 (from previous)

- 3NO(g) ? N2O(g) NO2(g)
- Solution
- Use ?G RT ln K
- Rearrange

?Grxn 105 kJ mol-1

?G ?G RT ln Q Where Q is the reaction

quotient a A b B ? c C d D

The Temperature Dependence of Equilibrium

Constants

- Where does this come from?
- Recall ?G ?H - T?S
- Divide by RT, then multiply by -1

- 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

- If we have two different Temperatures and Ks

(equilibrium constants)

or

- Now given ?H and T at one temperature, we can

calculate K at another temperature, assuming that

?H and ?S are constant over the temperature range

The Person Behind the Science

J.H. vant 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

vant Hoff Factor (i)

- ?T - i m K

- The reaction
- 2 Al3Cl9 (g) ? 3 Al2Cl6 (g)
- Has an equilibrium constant of 8.8X103 at 443K

and a ?Hr 39.8 kJmol-1 at 443K. Estimate the

equilibrium constant at a temperature of 600K.

P mg/A

V

Infinite-step expansion

2 step expansion

6 step expansion

Expansion (V2 gt V1) Work flows out of the system

and the Work sign is negative

(No Transcript)

P mg/A

V

Compression (V2 lt V1) Work is put into system,

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

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