“A” students work (without solutions manual) ~ 10 problems/night

“A” students work(without solutions manual)~ 10 problems/night.

Dr. Alanah FitchFlanner Hall [email protected]

Office Hours Th&F 2-3:30 pm

Module #20Spontaneity

Effect of number ofPossible configurations(randomness) on reactions

mailto:[email protected]

What we’ve Learned So Far

A + B C + heat

exothermic

endothermicEa

Does not tell us if spontaneous or not

Energy

A + B + heat C

ra te k A B

ra te A A BE

RTa

ex p

Ea

A messy room is more probable than an organized one.

spontaneous?or probable?

Can think of this as stored energy.

heat

-heat

+randomness

reactants products

Spontaneous Reactions:

-heat -H+ randomness +S

orrandomness

OJO

Randomness = more possibilities = entropy (S)

4 configurations

1

2

1

2 configurations =21

1 coin =2 sides

2 coins

=22

What is the most probable configuration for n tossed quarters?

Most probable configuration is least organized

3 coins8 configurations = 23

4 coins? configurations = 2?

What will the pattern be for three coins?

# con figura tions n 2

1 3 3 11. Most probable configuration

is least organized2. Number of possible

configuration increases exponentially

3. No. configurations contains information about compound (sides)

Probability of finding An electron – everywhere!

Very random

How does this affect the trends in entropy of various elements?

1s electron

2p electron probabilityIs more confined – lessRandom, less entropy

Probability ofFinding an electronFor a d orbital – More spatially constrained,Less entropy

Entropy of Pure Elements

0

10

20

30

40

50

60

70

80

90

0 10 20 30 40 50 60 70 80 90

Atomic Number

En

tro

py

(J

/mo

l-K

)

row 2 3 4

Entropy and Atomic Mass

1. Entropy contraction across row – related to organization of electrons

2. Entropy increase down group – related to volume occupied

Half filled d

Entropy and Group

Entropy in a Single Group

0

10

20

30

40

50

60

70

80

90

0 1 2 3 4 5 6 7

Row of Periodic Table

So (

J/m

ol-

K)

Group 1 (Li to Cs)Group 2 (Be to Ba)Group 4A/14 (C to Pb)

Group 1 S (J/mol-K) Group 2 S (J/mol-K) Group 4 S (J/mol-K)Li 29.09 Be 9.44 C 2.43Na 51.45 Mg 32.51 Si 18.7K 64.67 Ca 41.4 Ge 42.42576Rb 76.78 Sr 54.392 Sn 44.7688Cs 85.5 Ba 63.2 Pb 68.85

What do you observe?

Same observations!





Reaction Entropy


S n S n Srx ip roducts

i ireac ts

i tan

The change in entropy in a reaction is the difference between the summed entropy of the products minus the summed entropy of the reactants scaled by the number of moles

Example Calculation 1: Calculate the change in reaction entropy that occurs for each of the following two phase changes given the data below:

1

2

. C s C s

C s C s

s

g

So (J/mol-K)Cs(s) 85.15Cs(l) 92.07Cs(g) 175.6


i ireac ts

i tan

S m oleJ

m ol Kmole

J

m ol Krx C sC s

C ssC s s

1 9 2 0 7 1 8 5 1 5

. .

SJ

K

J

Krx

9 2 0 7 8 5 1 5. .S

J

K

J

K

J

Krx

9 2 0 7 8 5 1 5 6 9 2. . .S

J

K

J

K

J

Krx

9 2 0 7 8 5 1 5 6 9 2. . .

Example Calculation 1: Calculate the change in reaction entropy that occurs for each of the following two phase changes given the data below:

1

2

. C s C s

C s C s

s

g

So (J/mol-K)Cs(s) 85.15Cs(l) 92.07Cs(g) 175.6


i ireac ts

i tan

S m oleJ

m ol Kmole

J

m ol Krx C sC s

C sC s

g

g

1 1 7 5 6 1 9 2 0 7. .

SJ

K

J

Krx

1 7 5 6 9 2 0 7. .S

J

K

J

K

J

Krx

1 7 5 6 9 2 0 7 8 3 5 3. . .

heat

-heat

+randomness

reactants products

Spontaneous Reactions:

-heat -H+ randomness +S

orrandomness

ReactionsIncrease the SpontaneityBy an increaseIn randomness

So (J/mol-K)H2O(l) 69.91H2O(g) 188.83

Hg(l) 77.40Hg(g) 174.89

Cs(s) 85.15Cs(l) 92.07Cs(g) 175.6

Entropy Changes of Cs With Phase

0

20

40

60

80

100

120

140

160

180

200

solid liquid gas

So (

J/m

ol-

K)

What do you observe?

“Medicine is the Art of Observation” (Al B. Benson)

1. Large change liquid to gas2. S depends on Temp!!!

Element form S kJ/K-mol Element form S kJ/K-mol Element form S kJ/K-mol Element form S kJ/K-molAg s 0.0426 Ca s 0.0414 F- aq -0.0138 I- aq 0.1113

Ag+ aq 0.0727 CaCl2 s 0.1046 Fe s 0.0273 K s 0.0642AgBr s 0.1071 CaCO3 s 0.0929 Fe2+ aq -0.1377 K+ aq 0.0642AgCl s 0.0962 CaO s 0.0398 Fe3+ aq -0.3159 KBr s 0.0959AgI s 0.1155 Ca(OH)2 s 0.0834 Fe(OH) s 0.1067 KCl s 0.0826

AgNO3 s 0.1409 CaSO4 s 0.1067 Fe2O3 s 0.0874 KClO3 s 0.1431Ag2O s 0.1213 Cd s 0.0518 Fe3O4 s 0.14645 KClO4 s 0.151

Al s 0.0283 Cd2+ aq -0.0732 H2 g 0.1306 KNO3 s 0.133Ba s 0.0628 CdCl2 s 0.1153 H+ aq 0 Mg s 0.0327

Ba2+ aq 0.0096 CdO s 0.0548 HBr g 0.1986 Mg2+ aq -0.1381BaCl2 s 0.1237 Cl2 g 0.223 HCl g 0.1868

BaCO3 s 0.1121 Cl- aq 0.0565 HCO3- aq 0.0912BaO s 0.0704 ClO3- aq 0.1623 HF g 0.1737Br2 l 0.1522 ClO4- aq 0.182 HI g 0.2065Br- aq 0.0824 Cr s 0.0238 HNO3 l 0.1556C s 0.0057 CrO42- aq 0.0502 H2O g 0.1887

CCl4 l 0.2164 Cr2O3 s 0.0812 H2O l 0.0699ChCl3 l 0.2017 Cr2O72- aq 0.2619 H2O2 l 0.196CH4 g 0.1862 Cu s 0.0332 H2PO4- aq 0.0904C2H2 g 0.2008 Cu+ aq 0.0406 HPO42- aq -0.0335C2H4 g 0.2195 Cu2+ aq -0.0996 H2S g 0.2057C3H8 g 0.2699 CuO s 0.0426 H2SO4 l 0.1569

CH3OH l 0.1268 Cu2O s 0.0931 HSO4- aq 0.1318C2H5OH l 0.1607 CuS s 0.0665 Hg l 0.075

CO g 0.1976 Cu2S s 0.1209 Hg2+ aq -0.0322CO2 g 0.2136 CuSO4 s 0.1076 HgO s 0.0703

CO32- aq -0.0569 F2 g 0.2027 I2 s 0.1161

Some typical Standard S values

“Medicine is the Art of Observation”Let’s rearrange to make this a little easier

Element form S kJ/K-mol Element form S kJ/K-molC s 0.0057 Cr2O3 s 0.0812Cr s 0.0238 KCl s 0.0826Fe s 0.0273 Ca(OH)2 s 0.0834Al s 0.0283 Fe2O3 s 0.0874Mg s 0.0327 CaCO3 s 0.0929Cu s 0.0332 Cu2O s 0.0931CaO s 0.0398 KBr s 0.0959Ca s 0.0414 AgCl s 0.0962Ag s 0.0426 CaCl2 s 0.1046CuO s 0.0426 CaSO4 s 0.1067Cd s 0.0518 Fe(OH) s 0.1067CdO s 0.0548 AgBr s 0.1071Ba s 0.0628 CuSO4 s 0.1076K s 0.0642 BaCO3 s 0.1121CuS s 0.0665 CdCl2 s 0.1153HgO s 0.0703 AgI s 0.1155BaO s 0.0704 I2 s 0.1161

Cu2S s 0.1209average 0.0446 Ag2O s 0.1213

BaCl2 s 0.1237KNO3 s 0.133AgNO3 s 0.1409KClO3 s 0.1431Fe3O4 s 0.14645KClO4 s 0.151

average 0.11139

Solids

Pure Liquid

Element form S kJ/K-mol Element form S kJ/K-molCCl4 l 0.2164 Cr2O72- aq 0.2619CHCl3 l 0.2017 ClO4- aq 0.182H2O2 l 0.196 ClO3- aq 0.1623C2H5OH l 0.1607 HSO4- aq 0.1318H2SO4 l 0.1569 I- aq 0.1113HNO3 l 0.1556 HCO3- aq 0.0912Br2 l 0.1522 H2PO4- aq 0.0904CH3OH l 0.1268 Br- aq 0.0824Hg l 0.075 Ag+ aq 0.0727H2O l 0.0699 K+ aq 0.0642

Cl- aq 0.0565average 0.15112 CrO42- aq 0.0502

Cu+ aq 0.0406Ba2+ aq 0.0096H+ aq 0F- aq -0.0138Hg2+ aq -0.0322HPO42- aq -0.0335CO32- aq -0.0569Cd2+ aq -0.0732Cu2+ aq -0.0996Fe2+ aq -0.1377Mg2+ aq -0.1381Fe3+ aq -0.3159

average 0.021092

AqueousDo we notice anything?1. More atoms = more S2. Pure liquids higher S than s3. Aqueous species can have neg S! Why?

Element form S kJ/K-molC3H8 g 0.2699Cl2 g 0.223C2H4 g 0.2195CO2 g 0.2136HI g 0.2065H2S g 0.2057F2 g 0.2027C2H2 g 0.2008HBr g 0.1986CO g 0.1976H2O g 0.1887HCl g 0.1868CH4 g 0.1862HF g 0.1737H2 g 0.1306

average 0.20026

Gas

4. Gas highest S

Prediction reactions producing moreGas phase species will increase inentropy

solid < glass, plastic, liquid < gasSsolid < Ssolid, plastic,liquid < Sgas

S

T (K)

AbsoluteZero, no motion

Note: thisRefers to a singleMolecule or elementWater ice, liquid, gas

Possible configurations?

A closely related idea to change in entropyIn phase changes is the change in entropy in Going from individual ligands to chelates

From Module 18 we considered the electrostaticattraction between the electron pairs on ligand functional groups and the positive nucleus of a metal ion.

Module 18 review

z

x

E kq q

r rel

1 2

1 2

y

Cu NH( )3 4

2

Module 18 review

Lead ComplexationConstants

Ligand logK1

F- 1.4Cl- 1.55Br- 1.8I- 1.9

OH- 6.3Acetate 2.7Oxalate 4.9Citrate 5.7EDTA 17.9

K K K K K K K Krx f f f f f f fn 1 2 3 4 5 6 . . . . . . .

We observed that multidentate ligands had very highKf values

Why so large?Which are polydentate?What do you observe?

logK2 LogK3 logK4 logKf

1.1 2.50.6 -0.4 -0.7 1.050.8 -0.1 -0.3 2.21.3 0.7 0.6 4.54.6 2 12.91.4 4.11.9 6.8

5.717.9

Module 18 review

Example 2: Predict the entropy change in the following reaction by considering volume occupied and number of possible configurations between the reactants and products

M NH CH X en M en X NH CH2 3 4 2 2 2 2 32 4

Note that the electrostatic attraction which shows up in the enthalpy is similar for both compounds

NH CH2 3

Example 2: Predict the entropy change in the following reaction by considering volume occupied and number of possible configurations between the reactants and products

M NH CH X en M en X NH CH2 3 4 2 2 2 2 32 4

S S

Cd NH CH en Cd en NH CHJ

mol K

J

m ol K

oca lco

ex p

. .3 3 4

2

22

3 32 4 7 9 5 5 8 5

J. Chem. Ed. 61,12, 1984, Entropy Effects in Chelation Reactions, Chung-Sun Chung

3 5

Example Calculation 3: Calculate the reaction entropy changes for the reaction shown below given:

So (J/mol-K)

S rhombic 32S orthoclinic 33Cu(s) 85.15CuS(s) 92.07Cu2S(s) 175.6

CuS Cu Cu Ss s s 2

First let’s think about this a bit1. Internal Entropy

a.Why should CuS have more entropy than Cu?

b. Why should Cu2S have much more entropy than CuS?

http://images.google.com/imgres?imgurl=http://www.unisa.edu.au/synchrotron/res/projects/chalcociteCu2S.jpg&imgrefurl=http://www.unisa.edu.au/synchrotron/res/projects/default.asp&h=242&w=180&sz=20&hl=en&start=7&um=1&tbnid=UYBN7p1lBciHwM:&tbnh=110&tbnw=82&prev=/images%3Fq%3DCu2S%2B%26svnum%3D10%26um%3D1%26hl%3Den%26client%3Dfirefox-a%26channel%3Ds%26rls%3Dorg.mozilla:en-US:official%26sa%3DG

CuS Cu Cu Ss s s 2

Which one might haveMore configurations?

http://www.haraldthielenredlich.onlinehome.de/kkch/cu1+cu2s.jpg

A single, individual, H atom can occupy 4Different locations – thus the compound will be

More random than one with three locations forThe hydrogen

So (J/mol-K)CH4 186.C2H4 219.4

This reflects internal entropy which often scales with # of Atoms in the molecule.


So (J/mol-K)


CuS Cu Cu Ss s s 2

First let’s think about this a bit2. Spatial Volume of rx

entropy Which will be more imp in rx

entropy? Internal entropy or spatial volume entropy?

2 1

We take two separate chemical units and makeThem into 1 chemical unit – implies decrease inentropy


So (J/mol-K)


CuS Cu Cu Ss s s 2


i ireac ts

i tan

S m oleJ

m ol Krx Cu SCu S

s s

s s

1 1 7 5 6.

1 8 5 1 5 1 9 2 0 7moleJ

m ol Kmole

J

m ol KCuCu

CuSCuS

s

s

s

s

. .

SJ

K

J

K

J

Krx

1 7 5 6 8 5 1 5 9 2 0 7. . .


So (J/mol-K)


CuS Cu Cu Ss s s 2

SJ

K

J

Krx

1 7 5 6 1 7 7 2 2. .

SJ

K

J

K

J

Krx

1 7 5 6 8 5 1 5 9 2 0 7. . .

SJ

K

J

K

J

Krx

1 7 5 6 1 7 7 2 2 1 6 2. . .

2 1

Example Calculation 4: Calculate the change in entropy for the allotropic forms of elemental S

So (J/mol-K)


S Ss r b ic s orthoclin ic, h o m ,

Gr: allos = others





Entropy of the surroundings


Spontaneous Process: Universal entropy increases

(The universe is winding down.)

S S Sto ta l un iverse system chem ica l rx surround ings( ) ( )

S spon eousun iverse 0 tan

Change in Sreaction?

Change in Ssurroundings?

More organized (fewer molecules)implies less entropy, less randomSreaction

<0

heat willincreasekinetic energy of gases =Ssurround >0

Example 1: Ssurroundings

C O CO hea ts g g 2 2, ,

+ heat

Although this process requiresheat, it is spontaneous, drivenby entropy of chemical reaction

Reaction less random rx more random

From surroundings, withdraw heat,Less kinetic energy, less motion, less entropy

Example 2: Ssurroundings

H O hea t H O g2 2

H O2

H O g2

The two reactions (the system)

Interact with the surroundings by exchange of heat

Heat of reaction must be related to entropy of surroundings

C O CO heats g g 2 2, ,

H O hea t H O g2 2





Randomness of the“surroundings” affectedBy enthalpy


related to enthalpyor heat of reaction


S Hsurround ings reaction

proportional

Where will impact on Ssurroundings be greatest?a. 1 J at 600oCb. 1 J at 25oC

Predict entropy change is largest at low temperatures

sign change accounts for thefact that entropy increases withexothermic reactions

S as Tsurround ings

SH

Tsurround ingsreaction

SH


Context Slide for a calculation on entropy of the surroundingsHistorically Ag was mined as Ag2S found in the presence of PbS, galena. Part of the process of releasing the silver required oxidizing the galena. The lead oxide recovered was used in glass making. Thefumes often killed animals near by and have left a permanent record in the artic ice. Large regions near silver mines were deforested.One reason that this process was discovered so early in history wasThe low temperature at which it could be carried out.

Yorkshire, near old lead mines

Lead in Artic Ice

Who has the “honor” of most contaminatingThe artic ice? Medicine is the art of observation

Compare the change in entropy of the surroundings for this reaction at room temperature and at the temperature of a campfire (~600 oC).

Know: Don’t know red herrings?

Calculating Ssurrounding Example 1

reaction noneentropy

2 3 2 22 2PbS O PbO SOs g s g , ,

SH


H n H n HOf productso

f reac tso , , tan

T Co 2 5T Co 6 0 0

Substance Hf0 (kJ/mole)

O2(gas) 0PbS -100PbO -219SO2(gas) -297 H n H n HO

f productso

f reac tso , , tan

2 3 2 22 2PbS O PbO SOs g s g , ,

H molekJ

m oleO

PbOPbO

s

s

2

2 1 9

H kJ kJ kJO 4 3 8 5 9 4 2 0 0

H kJO 1 0 3 2 2 0 0 8 3 2

SH


S

kJ

Tsurround ings 8 3 2

H molekJ

m olem ole

kJ

m oleO

PbOPbO

SOSO

s

s

g

g

2

2 1 92

2 9 72

2

,

,

H molekJ

m olem ole

kJ

m olem ole

kJ

m oleO

PbOPbO

SOSO

PbSPbS

s

s

g

g

s

s

2

2 1 92

2 9 72

1 0 02

2

,

,

H molekJ

m olem ole

kJ

m olem ole

kJ

m olem ole

kJ

m oleO

PbOPbO

SOSO

PbSPbS

OO

s

s

g

g

s

s

g

g

2

2 1 92

2 9 72

1 0 03

02

2

2

2

,

,

,

,

Compare the change in entropy of the surroundings for this reaction at room temperature and at the temperature of a campfire (~600 oC).

Calculating Ssurrounding Example 1

2 3 2 22 2PbS O PbO SOs g s g , ,

T Co 2 5 T Co 6 0 0

S

kJ


S

kJsurround ings

8 3 2

2 9 8

T K 2 9 8 T K 8 7 3

S

kJsurround ings

8 3 2

8 7 3

S

kJkJsurround ings

8 3 2

2 9 82 7 9 2.

SkJ

kJsurround ings

8 3 2

8 7 30 9 5 3.

Our prediction was right! S surround ings Larger at low T





Total Entropy changeWith reaction enthalpy



SH


S SH

Tto ta l un iverse system chem ica l rx

chem ica l rx

( ) ( )

Substance S0 (J/K-mole)O2(gas) 205PbS(solid) 91PbO(solid) 66.5SO2(gas) 248

Reaction Entropy Example Calculation 3 Compare the total entropy change for the following reaction at 25oC and 600oC

2 3 2 22 2PbS O PbO SOs g s g , ,

S m oleJ

K molm ole

J

K mol

m oleJ

K moleM ole

J

K mole

rx PbOPbO

SOSO

PbSPbS

OO

2 6 6 5 2 2 4 8

2 9 1 3 2 0 5

2

2

2

2

.

S n S n Srx ioi p roducts i

oi reac ts , , tan

Reaction Entropy Example Calculation 3 Compare the total entropy change for the following reaction at 25oC and 600oC

2 3 2 22 2PbS O PbO SOs g s g , ,

S m oleJ

K molm ole

J

K mol

m oleJ

K moleM ole

J

K mole

rx PbOPbO

SOSO

PbSPbS

OO

2 6 6 5 2 2 4 8

2 9 1 3 2 0 5

2

2

2

2

.

S n S n Srx ioi p roducts i

oi reac ts , , tan

SJ

Krx 1 6 8

SJ

K

J

K

J

K

J

Krx

1 4 3 4 9 6 1 8 2 6 1 5

SJ

Kreaction 1 6 8

S SH


chem ica l x

( ) ( )

T Co 2 5 T Co 6 0 0

S

kJ


S J

kJ

Tto ta l un iverse( )

1 6 88 3 2

SJ

K

kJ

Kto ta l ) . 1 6 8 2 7 9 1

SJ

K

J

K 1 6 8 2 7 9 1,

SJ

K 2 6 2 3

S

J

K

kJ

Kto ta l un iverse( )

1 6 88 3 2

2 9 8

Spontaneous

S

J

K

kJ

Kto ta l un iverse( )

1 6 88 3 2

8 7 3

SJ

K

kJ

Kto ta l ) . 1 6 8 0 9 5 3

SJ

K

J

Kto ta l 1 6 8 9 5 3

SJ

Kto ta l 7 8 5

Less spontaneous





“Free energy” is a Way of accountingFor contribution of randomness


S SH


chem ica l x

( ) ( )

T S Gto ta l un iverse free energy rx ( )

T S T S Hto ta l un iverse system chem ica l rx chem ica l rx ( ) ( )

G H T Sfree energy r x rx

Gibb’s free energya) enthalpy of bondsb) organization of atomsc) randomness of surroundings

G free energy 0

Spontaneous reaction

S SH

TTto ta l un iverse system chem ica l rx

chem ica l r x

( ) ( )

T S H T Sto ta l un iverse chem ica l rx system chem ica l rx ( ) ( )

Define

Marie the Jewess, 300 Jabir ibn Hawan, 721-815

Galen, 170 Jean Picard1620-1682

Galileo Galili1564-1642

Daniel Fahrenheit1686-1737

Evangelista Torricelli1608-1647

Isaac Newton1643-1727

Robert Boyle, 1627-1691

Blaise Pascal1623-1662

Anders Celsius1701-1744

Charles Augustin Coulomb 1735-1806

John Dalton1766-1844

B. P. Emile Clapeyron1799-1864

Jacques Charles 1778-1850

Germain Henri Hess1802-1850

Fitch Rule G3: Science is Referential

William ThompsonLord Kelvin, 1824-1907

James Maxwell1831-1879

Johannes D.Van der Waals1837-1923

Justus von Liebig (1803-1873

Johann Balmer1825-1898

James Joule (1818-1889)

Johannes Rydberg1854-1919

Rudolph Clausius1822-1888

Thomas Graham1805-1869

Heinrich R. Hertz, 1857-1894

Max Planck1858-1947

J. J. Thomson1856-1940

Linus Pauling1901-1994

Werner Karl Heisenberg1901-1976

Wolfgang Pauli1900-1958

Count Alessandro GA A Volta, 1747-1827

Georg Simon Ohm 1789-1854

Henri Louis LeChatlier1850-1936

Svante Arrehenius1859-1927

Francois-Marie Raoult1830-1901

William Henry1775-1836

Gilbert N Lewis1875-1946

Fritz Haber1868-1934

Michael Faraday1791-1867

Luigi Galvani1737-1798

Walther Nernst1864-1941

Lawrence Henderson1878-1942

Amedeo Avogadro1756-1856

J. Willard Gibbs1839-1903

Niels Bohr1885-1962

Erwin Schodinger1887-1961

Louis de Broglie (1892-1987)

Friedrich H. Hund1896-1997

Fritz London1900-1954

An alchemist

Ludwig Boltzman1844-1906

Richard AC E Erlenmeyer1825-1909

Johannes Bronsted1879-1947

Thomas M Lowry1874-1936

James Watt1736-1819

Dmitri Mendeleev1834-1907

Marie Curie1867-1934

Henri Bequerel1852-1908

Rolf Sievert, 1896-1966

Louis Harold Gray1905-1965

Jacobus van’t Hoff1852-1911

Conceptually:

reaction Sreaction Spontaneous? - + + + - - + -

alwaysat high T, 2nd term lg.at lowT, 2nd term sm

never

G H T Sfree energy r x rx G free energy 0

Gibbs Free Energy Example 1When will this reaction be spontaneous, hi or lo T?

At LowT!!!

2 3 2 22 2PbS O PbO SOs g s g , ,


SJ

rx

1 6 8

HkJ

rxO

8 3 2

reaction Sreaction Spontaneous? - + + + - - + -

alwaysat high T, 2nd term lg.at lowT, 2nd term sm

never

-1000

-800

-600

-400

-200

0

200

0 1000 2000 3000 4000 5000 6000 7000

o C

Free

Ene

rgy

(kJ)

Spontaneous ReactionsFree Energy <0

Non Spontaneous ReactionFree Energy >0

2 3 2 22 2PbS O PbO SOs g s g , ,G kJ T

J

Kfree energy

8 3 2 1 6 8

Can we figureOut exactly at what Tthis reaction becomesSpontaneous?

K

More spontaneous

To find when a reaction will just go Spontaneous (or not)1. Use the equation:

2. Set Go to zero (equilibrium)

3. Solve for T.

4. Depending upon sign of enthalpy entropy

determine if temperature decrease/increasecauses Go to go negative


0 H T Sr x rx

TH

Sbecom es spon eous

r x

rxtan

T S Hrx r x

G H T Sfree energy r x rx 2 3 2 22 2PbS O PbO SOs g s g , ,

G kJ TJ

Kfree energy

8 3 2 1 6 8

Gibbs Free Energy Example 2 At what T will this reaction become change betweenSpontaneous and non-spontaneous?

0 8 3 2 1 6 8

kJ T

J

K

8 3 2 1 6 8kJ TJ

K

8 3 2

0 1 6 8

kJkJ

K

T.

4 9 5 2 K TRx spontaneous at T<4952K

Gibbs Free Energy Example 2: The only good substitute for PbCO3 for white paint is TiO2. To manufacture this paint need to be able to process titanium ore TiO2. (Different allotrope). At what temperature does the following reaction become spontaneous?

T iO C T i COs s s g 2 2

Substance Hf0 (kJ/mole) S0(J/K-mole)

Tisolid 485 179.45CO(gas) -110.5 198TiO2(solid) -945 50O2(gas) 0 205SO2(gas) -297 248Csolid 0 0 G H T Sfree energy r x rx

G free energy


G kJ

TJ

K

free energy

4 8 5 2 2 1 9 4 5

5 7 5 5 0

G kJ TJ

Kfree energy 1 6 5 2 5 2 5


T moleJ

K mole1

1 7 9 4 5.

14 8 5

211 0 5

19 4 5

molekJ

m olem ole

kJ

m olem ole

kJ

m ole

.1

4 8 5mole

kJ

m ole

14 8 5

211 0 5

molekJ

m olem ole

kJ

m ole

.14 8 5

211 0 5

19 4 5

20

molekJ

m olem ole

kJ

m olem ole

kJ

m olem ole

kJ

m ole

.

T moleJ

K molem ole

J

K mole1

1 7 9 4 52

1 9 8.

T moleJ

K molem ole

J

K molem ole

J

K mole1

1 7 9 4 52

1 9 81

5 0.

T moleJ

K molem ole

J

K molem ole

J

K molem ole

J

K mole1

1 7 9 4 52

1 9 81

5 02

0.


Tisolid 485 179.45CO(gas) -110.5 198TiO2(solid) -945 50O2(gas) 0 205SO2(gas) -297 248Csolid 0 0


G kJ TJ

Kfree energy 1 6 5 2 5 2 5

0 1 6 5 2 0 5 2 5 T .

T 0 5 2 5 1 6 5 2.

T K 3 1 4 4


When is this reaction spontaneous:at high or low temp?

Rx spontaneous > 3144K


TiO2solid -945 50Ti 485 179.45O2(gas) 0 205SO2(gas) -297 248Csolid 0 0CO(gas) -110.5 198

WWII

Context Slide 1

titanium was not routinely processed until after WWII (jet engine technology). So TiO2 purifiednot available cheaply for paint until after WWII





Reference states forFree Energy


f means formation at standard state 25 oC!!!!!

As for enthalpy and entropy, there are tablesOf values obtained via Hess’s Law

G n G n Grxo

iof i p roducts i

ofi reac ts, , , , tan

Properties and MeasurementsProperty Unit Reference StateSize m size of earthVolume cm3 mWeight gram mass of 1 cm3 water at specified Temp

(and Pressure)Temperature oC, K boiling, freezing of water (specified

Pressure)1.66053873x10-24g amu (mass of 1C-12 atom)/12quantity mole atomic mass of an element in gramsPressure atm, mm Hg earth’s atmosphere at sea levelEnergy, General

Animal hp horse on tread millheat BTU 1 lb water 1 oF

calorie 1 g water 1 oCKinetic J m, kg, sElectrostatic 1 electrical charge against 1 Velectronic states in atom Energy of electron in vacuum

Electronegativity F

f means formation at standard state 25 oC!!!!!

As for enthalpy and entropy, there are tablesOf values obtained via Hess’s Law

G n G n Grxo

iof i p roducts i

ofi reac ts, , , , tan

State of Matter Standard (Reference) State

Solid Pure solidLiquid Pure liquidGas 1 atm pressureSolution 1 M concentrationElements Gf

o0

Substance Gf 0 (kJ/mole)PbO -188.9SO2(gas) -300PbS -99O2(gas) 0

Gibbs Standard Free Energy Example Calc. 1: What Is the standard free energy change of the following Reaction? 2 3 2 22 2PbS O PbO SOs g s g , ,

G n G n Grxo

iof i p roducts i

ofi reac ts , , , tan

G rxo 2 1 8 8 9 2 3 0 0

2

2

moleskJ

m olesm oles

kJ

m olePbOPbO

SOSO

g

.

,

2 9 9 3 0

2

2

moleskJ

m olem ole

kJ

m olePbSPbS

OO

s

s

g

g

,

,


G rxo

G kJ kJ

kJrxo

3 7 7 8 6 0 0

1 9 8

.

G kJrxo 7 7 9 8.

2 1 8 8 9 2 3 0 02

2

moleskJ

m olesm oles

kJ

m olePbOPbO

SOSO

g

.

,

2 9 9 3 0

2

2

moleskJ

m olem ole

kJ

m olePbSPbS

OO

s

s

g

g

,

,


G n G n Grxo

iof i p roducts i

ofi reac ts , , , tan

G kJrxo 7 7 9 8.

G kJ TJ

Kfree energy

8 3 2 1 6 8

For comparison, we calculated from before:

G kJ KJ

Kfree energy

8 3 2 2 9 8 1 6 8

G kJfree energy 7 8 2SJ

Krx 1 6 8H kJrx 8 3 2

Not too bad ofAgreement!





Summing Reactions


Rx# Greaction

1 A + B nC Greaction 1

2 nC + D E Greaction 2

3 A + B + D E Greaction 1 + Greaction 2

Summing Free Energy Example Calculation Why was lead one of the first elements first processed by man? A. Calculate the standard free energy of the Combined reactions. B. Calculate the free energy of the reaction at 600 oC (campfire temp).

2 3 2 22 2PbS O PbO SOs g s g , ,

2 2 2 2PbO C Pb COs s s g

2 3 2 2 2 22 2PbS O C Pb SO COs g s s g g , ,

Summing Free Energy Example Calculation Why was lead one of the first elements first processed by man? A. Calculate the standard free energy of the Combined reactions. B. Calculate the free energy of the reaction at 600 oC (campfire temp).

2 3 2 22 2PbS O PbO SOs g s g , ,

SJ

Krx0 1 6 8 H kJrx

O 8 3 2 G kJrxo 7 7 9 8.


Need standard free energy to solve ABut! Will also need standard enthalpy and S

To solve B – so solve for those


Pb 0 0CO(gas) -110.5 198PbS -100 91PbO -219 66.5O2(gas) 0 205SO2(gas) -297 248Csolid 0 0

ΔH= [{(2(0)+2(-110.5)}-{2(-219)+2(0)}]=+217kJ

ΔS=[{2(0)+2(198)}-{2(66.5)+2(0)}]=263J/K

2PbOsolid + 2Csolid 2Pbsolid + 2CO(gas) ?


SJ

Krx0 2 6 3H kJrx

O 2 1 7


G kJ TJ

K

kJ

Jfree energy

2 1 7 2 6 3

1 0 3

At standard conditions

G kJ TkJ

Ko

2 1 7 0 2 6 3.

G kJ KkJ

Ko

2 1 7 2 5 2 7 3 0 2 6 3 1 3 8 6( ) . .

At campfire conditions

G kJ KkJ

Ko

2 1 7 8 7 3 0 2 6 3 1 2 5 6( ) . .

G kJrxo 1 3 8 6.

2PbS(solid) + 3O2(gas) PbO(solid) + 2SO2(gas) -779.8 kJ

2PbOsoloid + 2Csolid 2Pbsolid + 2CO(gas) +138.6kJ

Grx

2PbS + 3O2(gas) + 2Csolid 2Pbsolid + 2SO2gas + 2COgas

sum = -641kJ

Net reaction at 25 oC

The net standard free energy for the coupled two reactions is -641 kJ, spontaneous

2PbS(solid) + 3O2(gas) PbO(solid) + 2SO2(gas)

2PbOsoloid + 2Csolid 2Pbsolid + 2CO(gas)

Grx

2PbS + 3O2(gas) + 2Csolid 2Pbsolid + 2SO2gas + 2COgas

sum = -697kJ

The reduction of Pb in PbS to metal and oxidation of S in PbS to sulfur dioxide gas is spontaneous at campfire temperatures of 600oC

Net reaction at 600 oC

-685 kJ

-12.6kJ

Context point: can manufacture pure lead in a campfire

Context Slide

Decade Estimate lbs white lead/housing unit1914-23 1101920-29 871930-39 421940-49 221950-59 71960-69 31970-1979 1

Where did all the lead go?

TiO2 makes inroadsparticularly in Europe

3 0 0

1 0 6

gPb

gso il

White lead restricted

300ppm = “background” level of Chicago soil leadDepth: doesNot move downBecause of Oh Card me PleaSe

Context Slide

3 0 0

1 0

1

4 5 3 5 96

gPb

gso ilx

lbPb

gPb.

3 0 0

1 0

1

4 5 3 5 9

1 26 3

gPb

gso ilx

lbPb

gPbx

gso il

cm.

.3 0 0

1 0

1

4 5 3 5 9

1 2 2 2 8 4 46 3

2gPb

gso ilx

lbPb

gPbx

gso il

cmx

m i

Chicago.

. .3 0 0

1 0

1

4 5 3 5 9

1 2 2 2 8 4 4 1 6 0 96 3

2 2gPb

gso ilx

lbPb

gPbx

gso il

cmx

m i

Chicagox

km

m i.

. . .

3 0 0

1 0

1

4 5 3 5 9

1 2 2 2 8 4 4 1 6 0 9 1 06 3

2 2 5 2gPb

gso ilx

lbPb

gPbx

gso il

cmx

m i

Chicagox

km

m ix

cm

km.

. . .

3 0 0

1 0

1

4 5 3 5 9

1 2 2 2 8 4 4 1 6 0 9 1 036 3

2 2 5 2gPb

gso ilx

lbPb

gPb

gso il

cmx

m i

Chicagox

km

m ix

cm

kmx cm

.

. . .

1 4 7 9 0 2 3 9, , lbPb

ugPb/gsoil<250250-500500-1000>1000

5 0 5 Miles

N

EW

S

Percent of Children with Elevated Blood Lead Levels

Chicago, 1999

Percent of Childrenwith Elevated BloodLead Levels

< 5%5% - 15%16% - 25%> 25%

ugPb/gsoil<250250-500500-1000>1000

5 0 5 Miles

N

EW

S

Percent of Children with Elevated Blood Lead Levels

Chicago, 1999

Percent of Childrenwith Elevated BloodLead Levels

< 5%5% - 15%16% - 25%> 25%

Relates to a) Age of Housingb) “Gentrification”

Relevant Chem 102 Concepts:1. Temperature dependence

of spontaneous reactions2. Stability of soil lead form

(Oh Card me PleaSe)

Context Slide





Relating Free EnergyTo Concentrations


G G RT Qo ln

The free energy of the reaction related to a) standard free energy changeb) and the ratio of concentrations of

products to reactants, Q

In this equation you can use (simultaneously)PressuresConcentrations

The ln(Q) is treated as unitless

G kJJ

KK

Pb P P

PbS P Crx

s SO CO

s O s

6 4 1 8 3 1 4 2 9 8

2 2 2

2 3 22

2

. ln

Free Energy and Conc. Example Calc. Calculate the free energy of the reaction if the partial pressures of the gases are each 0.1 atm, 298 K. Remember, we calculated ΔGrx

to be -641 kJ at 298K (25 oC)

2 3 2 2 2 22 2PbS O C Pb SO COs s s s g g( ) ( ) ( ) ( ) ( )

G kJ JP P

Prx

SO CO

o

6 4 1 2 4 7 7 5 7

1

1 1

2 2 2

2 3 22

2

. ln

G kJ kJP P

Prx

SO CO

O

6 4 1 2 4 7 7 5 7 2

2

2 2

3. ln

G kJ kJrx

SO CO

O

6 4 1 2 4 7 7 5 7

0 1 0 1

0 12

2

2 2

3. ln. .

.

G kJ kJrx 6 4 1 2 4 7 7 5 7 2 3 0 2. . )

G kJ kJrx 6 4 1 5 7 0 3.

G G RT Qo ln

GkJ

m olrx 6 4 7

G kJ kJrx 6 4 1 2 4 7 7 5 7 0 1. ln .

When Q = K (equilibrium):

K = 1 -RT ln (1) = 0

K >1 -RT ln (>1) = -(+) < 0

K < 1 -RT ln (<1) = -(-) > 0

G G RT Qo ln

0 KAt equilbrium noEnergy to driveRx one way or other

0 G RT Ko ln

RT K G oln

G RT Ko ln

Example Problem 2 Free Energy and Equilibrium:What is the equilibrium constant for the reactionat a campfire temperature?

Go = -697kJ/mol rx

K e

kJ

m ol

xkJ

m ol KK

6 9 7

8 3 1 4 1 0 2 9 83.

K e 2 8 1 1 0 01 0

2 3 2 2 2 22 2PbS O C Pb SO COs g s s g g , ,

G RT Ko ln

GRT

Ko

ln

e KG

RT

o

Example 3 Free Energy and Equilibrium:The corrosion of Fe at 298 K is K = 10261

.What is the equilibrium constant for corrosionof lead?

2Pbsolid + O2gas 2PbOsolid

We don’t have any K values so we need To go to appendix for various enthalpy andEntropies to come at K from the backside


PbS -100 91PbO -219 66.5Pb 0 0O2(gas) 0 205SO2(gas) -297 248Csolid 0 0CO(gas) -110.5 198

Go = Ho - TSo

Ho = 2(-219) - {2(0) + 2(0)} = -438kJSo = 2(.0665) - {2(0) + 2(.205)} = -0.277kJ/KGo = -438 - T(-.277) = -438 - (298)(-0.277) = -355kJ

2 22Pb O PbOs g ( )

RT ln K - Go

so: even though the reaction is favorableit is less so than for iron.Lead rusts less than iron = used for plumbing

K for rusting of Fe = 10261

K for rusting of Pb = 1.27x1062

K e eG

RT

kJ

m ol

xkH

mol KK

0

3

3 5 5

8 3 1 4 1 0 2 9 8.

K e x 1 4 3 6 21 2 7 1 0

G kJ 3 5 5

“A” students work(without solutions manual)~7 problems/night.

What you need toknow


1. Be able to rank the entropy of various phases of materials, including allotropes

2. Be able to rank the entropy of various compounds3. Explain entropy concepts as related to chemical

geometry4. Calc. standard entropy change for a reaction5. Relate surrounding entropy to reaction enthalpy6. Calc. temperature at which a reaction becomes

spontaneous7. Explain why TiO2 was relatively late in replacing

PbCO3 as a white pigment; why lead was one of first pure metals obtained by humanity

8. Convert standard free energy to equilibrium constant

“A” students work(without solutions manual)~7 problems/night.

END


Entropy and Molecular Structure:

# con figura tions z nS,J/mol-K

O(g) 161O2(g) 205O3(g) 237.6

C (g) 158.OCO (g) 197.9CO 2(g) 213.6

Cl (g) 165.2Cl2(g) 2232.96

Pb 68.85PbO 68.70PbO2 76.98Pb3O4 209.2

CH4 186.4C2H4 219.4

C2H2 200.8C2H4 219.4C2H6 229.5

z

z

z

z

“z” = 2

“z” = 4

“z” =6

“z” = 8

Have weConvincedOurselves Yet?

Internal Entropy and Molecular Structure:

# con figura tions z n

Entropy of various Solids

0

2

4

6

8

10

12

14

0 50 100 150 200 250 300 350

Entropy (J/mol-K)

Nu

mb

er

ob

serv

ed

1, average=38.26

2, average=62.15

3, average = 95.42

5, average 103

4, average=137.9

6, average- 124

Reliability of averagedecreases with numberof averaged data points

Internal Entropy is generally increasingWith number of atoms in the moleculeBecause the number of locations within the moleculeWhere an atom could be found is increasing andBecause the possible orientations of the molecule increases

Documents

“A” students work (without solutions manual) ~ 10 problems/night