week 3 thermodynamics

ENVE 501 – ENVE 501 – Week 3Week 3

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Quiz 1Quiz 1

Imagine you can follow a molecule of copper from the source to the mouth of the Mississippi River. Assume that the average concentration of dissolved copper remains constant at 5 g/L. The total dissolved solids, however, increases significantly from source to mouth.

Would you expect the activity to increase, decrease, or remain the same? Why?

From the perspective of a fish, what is more important, the activity or the concentration of copper? Briefly explain your answer.

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Quiz 1 - SolutionQuiz 1 - Solution

• Activity will decreaseActivity will decrease• Ionic strength will increase as the Ionic strength will increase as the

total dissolved solids (TDS) total dissolved solids (TDS) concentration increasesconcentration increases

• Activity coefficient will decrease as Activity coefficient will decrease as the ionic strength increasesthe ionic strength increases

aaii = = ii [i] [i]

• Activity is more importantActivity is more important• Toxicity involves biochemical Toxicity involves biochemical

reactionsreactions• Reactions depend on chemical activityReactions depend on chemical activity

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Teams - ITeams - I

Gobi

Jennings DianeJin WeidongKnudsen WestenMu BinbinTemino Boes ReginaXu Limeimei

Kalahari

Dai WeiDouce JordanGlick NicoleHuang JinjinJiang Yuanzan

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Teams - IITeams - II

Atacama

Mirza MuhammadSharpe JessicaShort BradleyWang Chen

Wang Zehua

Sahara

Gong QijianIzadmehr MahsaMeyer KristineMocny WilliamShe CongweiWang Xiaolang

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Teams - IIITeams - III

Sonoran

Azu LaudHernandez AlcidesHuo YechenPujari AniketRamos Tiffanie

Zhang Yue

Mojave

Cao JicongLin KehsunTadeusz BobakWen JiahongYoung Meghan

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Team assignment for Team assignment for next weeknext week

• Use a 9” by 9” slide in Use a 9” by 9” slide in LandscapeLandscape formatformat

• Introduce your consulting firmIntroduce your consulting firm• NamesNames• Photographs(? )Photographs(? )• HistoryHistory• SpecializationSpecialization• Company logo….Company logo….

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Details posted on Details posted on BlackboardBlackboard

• No more than four slidesNo more than four slides

• Email to me by 8 PM on September 14Email to me by 8 PM on September 14

• Use Blackboard tools to collaborateUse Blackboard tools to collaborate

• Select one team member to present Select one team member to present next weeknext week

• Students at remote sites should Students at remote sites should consider an electronic presentationconsider an electronic presentation

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Chapter 3 – Chapter 3 – ThermodynamicsThermodynamics

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Key ConceptsKey Concepts

Entropy Entropy pp 62-64; 70-74pp 62-64; 70-74

Gibbs (free) energy Gibbs (free) energy pp 80-93; 98-104pp 80-93; 98-104

Thermodynamics & chemical systems Thermodynamics & chemical systems pp 104-125pp 104-125

• Page numbers are from Chapter 2 of Page numbers are from Chapter 2 of our text. our text.

• The handout provides a more The handout provides a more comprehensive description. comprehensive description.

Laws of Laws of thermodynamics - Ithermodynamics - I

Zeroth law of thermodynamicsZeroth law of thermodynamics• Systems in thermal equilibrium have Systems in thermal equilibrium have

the same temperature. the same temperature.

First law of thermodynamicsFirst law of thermodynamics• Energy is conserved. Energy is conserved.

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Laws of Laws of thermodynamics - IIthermodynamics - II

Second law of thermodynamicsSecond law of thermodynamics• Nature is not symmetric; the quality Nature is not symmetric; the quality

of energy decreases. of energy decreases. • A process will occur spontaneously if A process will occur spontaneously if

it increases the total entropy of the it increases the total entropy of the universeuniverse

Third law of thermodynamicsThird law of thermodynamics• Entropy = 0 for a perfect crystal at T Entropy = 0 for a perfect crystal at T

= 0. = 0.

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An alternative look at An alternative look at the lawsthe laws

We can't winWe can't win

We are sure to loseWe are sure to lose

We can't get out of the gameWe can't get out of the game

(Garrett Hardin from an (Garrett Hardin from an unknown source)unknown source)

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Entropy and life Entropy and life

Life is an open or continuous system Life is an open or continuous system that is able to decrease its internal that is able to decrease its internal entropy at the expense of free entropy at the expense of free energy taken in from the energy taken in from the environment and subsequently environment and subsequently rejected in a degraded form. rejected in a degraded form.

• Lovelock, J. (2000) Lovelock, J. (2000) GAIA. A new look GAIA. A new look at life on Earthat life on Earth. Oxford University . Oxford University Press, New York. Press, New York.

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Entropy and deathEntropy and death

Entropy is a measure of a system's Entropy is a measure of a system's thermal disorder. It implies the thermal disorder. It implies the predestined and inevitable run-predestined and inevitable run-down and death of the Universe. down and death of the Universe.

• Lovelock, J. (2000) Lovelock, J. (2000) GAIA. A new look GAIA. A new look at life on Earthat life on Earth. Oxford University . Oxford University Press, New York. Press, New York.

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Fundamental Fundamental thermodynamic thermodynamic

propertiespropertiesGibbs (free) energy = Gibbs (free) energy = GG

• Energy available to do useful workEnergy available to do useful work

Enthalpy = Enthalpy = HH• Total energy (molecular + Total energy (molecular +

mechanical work)mechanical work)

Entropy = Entropy = SS• A measure of disorderA measure of disorder

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Changes in Changes in GG, , HH, and , and SS are relatedare related

i i iG H T S

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What do the symbols What do the symbols mean?mean?

Subscript identifies the compoundSubscript identifies the compound Gibbs energy associated with total Gibbs energy associated with total ii

The over-bar means per molThe over-bar means per mol energy per mol of energy per mol of ii

The superscript ‘The superscript ‘oo’ refers to the ’ refers to the standard statestandard state standard molar Gibbs energy of standard molar Gibbs energy of ii, ,

also the standard Gibbs energy of also the standard Gibbs energy of formationformation

iG

iG

oiG

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Thermodynamic Thermodynamic properties properties are relativeare relative

By definition, the Gibbs energy of a pure By definition, the Gibbs energy of a pure element under standard conditions is 0element under standard conditions is 0• For example (T = 25For example (T = 25C, P = 1 bar)C, P = 1 bar)

GGHH22 = 0; pure hydrogen gas = 0; pure hydrogen gas

GGCuCu = 0; pure solid copper = 0; pure solid copper

For dissolved ions (T = 25For dissolved ions (T = 25C, P = 1 bar)C, P = 1 bar)GGHH++ = 0; dissolved hydrogen ion = 0; dissolved hydrogen ion

• Other ions defined relative to this Other ions defined relative to this referencereference

Corrections for different Corrections for different activityactivity

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Effect on Gibbs energyEffect on Gibbs energy

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ln lnoi i i iG G RT c RT

Standard Gibbs energy of formation Concentration

Activity coefficient

What is the Gibbs What is the Gibbs energy per mole of energy per mole of

atmospheric oxygen? atmospheric oxygen?

Actual Gibbs energy = Actual Gibbs energy =

Standard Gibbs energy + Standard Gibbs energy +

Corrections for activityCorrections for activity

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Example


What is the Gibbs What is the Gibbs energy per mole of energy per mole of

atmospheric oxygen? atmospheric oxygen?

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Example

SubstitutingSubstituting

25

Example

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So what?So what?

Change in Gibbs energy tells us if a Change in Gibbs energy tells us if a reaction can occur. reaction can occur.

Reaction is possible:Reaction is possible:• If the change in the Gibbs energy of If the change in the Gibbs energy of

reaction is negativereaction is negative

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Find molar Gibbs energy of Find molar Gibbs energy of reaction for:reaction for:

1.0 mole hydrogen gas 1.0 mole hydrogen gas + 0.5 mole oxygen gas + 0.5 mole oxygen gas

= 1.0 mole water as a = 1.0 mole water as a gasgas

Assume 1.0 bar and T = 25Assume 1.0 bar and T = 25CC

Example

HH22(g) + ½ O(g) + ½ O22(g) (g) H H22O(g)O(g)

SubstituteSubstitute

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Example

-228.57 kJ per mol of -228.57 kJ per mol of what?what?

Reaction involves: Reaction involves:

1 mol H1 mol H22 + 0.5 mol O + 0.5 mol O22 + 1 mol H + 1 mol H22OO

Per mole of stoichiometric reaction. Per mole of stoichiometric reaction.

For example:For example:

a a A + A + b b BB c c C + C + d d DD

1.0 mole of stoichiometric reaction: 1.0 mole of stoichiometric reaction: aa mol A + mol A + bb mol B mol B cc mol C + mol C + dd mol D mol D

Stoichiometry is importantStoichiometry is important

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For the reaction: For the reaction:

,reactants , ,

, ,ln ln

r r A r B

o or A r BA B

G G G

G RT a G RT a

a a A + A + b b B B c c C + C + d d DD

, , ,

, ,ln ln

r products r C r D

o or C r DC D

G G G

G RT a G RT a

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ln

ln

o o o o C Dr C D A B

A B

o C Dr

A B

a aG G G G G RT

a a

a aG RT

a a

lnc dc d

C Dor r a ba b

A B

C DG G RT

A B

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For dilute solutionsFor dilute solutions

As As ii 1.0, 1.0,

ln

ln

c do

r r a b

or

C DG G RT

A B

G RT Q

Q = reaction Q = reaction

quotientquotient

NOTE: From here on we often assume a dilute solution

HH22(g) + ½ O(g) + ½ O22(g) (g) H H22O O (g (g

• Average atmospheric partial Average atmospheric partial pressurespressures

ppHH22 = 6 = 61010-7-7 bar bar

ppOO22 = 0.21 bar = 0.21 bar

ppHH22OO = 0.02 bar = 0.02 bar

• Gibbs energy under conditions Gibbs energy under conditions different from the standard state different from the standard state is : is :

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lnoi i iG G RT a

Example

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Under standard conditions: Under standard conditions:

RT = 8.314 (J/mol.K) RT = 8.314 (J/mol.K) 298 (K) 298 (K)

RT = 2.48 (kJ/mol)RT = 2.48 (kJ/mol)

aaii = = ii p pii

Example

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Example

Substituting for each Substituting for each termterm

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Quiz 2Quiz 2

• Match the reaction described in the Match the reaction described in the first column with the dimensions of first column with the dimensions of the rate constant in the final the rate constant in the final columncolumn

• You might not need all the choices You might not need all the choices for for kk

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Interpreting values of Interpreting values of

reaction can proceed in forward reaction can proceed in forward direction direction

reaction can proceed in reverse directionreaction can proceed in reverse direction

forward and reverse reaction rates are forward and reverse reaction rates are the same; reaction is at equilibriumthe same; reaction is at equilibrium

rG0rG

0rG

0rG

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A special reaction quotient A special reaction quotient

whenwhen

KKeqeq = equilibrium constant = equilibrium constant

0rG

lnor eqG RT K

expc do

req a b

C DGK

RT A B

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How much oxygen should be How much oxygen should be dissolved in Lake Michigan when T dissolved in Lake Michigan when T = 25= 25C?C?

Reaction is: Reaction is:

OO2(g)2(g) O O2(aq)2(aq)

Gibbs energy for the reaction as Gibbs energy for the reaction as written is:written is:

Example

2

2lnor r

O

OG G RT

p

What are the dimensions of the What are the dimensions of the term in brackets? term in brackets?

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Recall the “hidden” terms: Recall the “hidden” terms:

Numerator is:Numerator is:

Denominator is:Denominator is:

Change in standard state Gibbs Change in standard state Gibbs energy for this reaction is: energy for this reaction is:

= 16.32 - 0 = 16.32 = 16.32 - 0 = 16.32 (kJ/mol)(kJ/mol)

Example

orG

LmolLmol

cO

0.1

2

bar

barpO0.12

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Solving for equilibrium dissolved Solving for equilibrium dissolved oxygen: oxygen:

[O[O22] = 1.38] = 1.381010-3-3 0.21 0.21

[O[O22] = 2.9] = 2.91010-4-4 (mol/L) (mol/L) 9.3 mg/L 9.3 mg/L

Dimensions follow from “hidden” Dimensions follow from “hidden” terms. terms.

Example

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exp

16.32 10exp 1.38 10

8.314 298

or

eqG

KRT

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Estimate Lake Michigan dissolved Estimate Lake Michigan dissolved oxygen concentration for T = oxygen concentration for T = 11C. C.

It looks as if :It looks as if :

Example

expor

eqG

KRT

orGKnowing Knowing

Can we substitute T = 1Can we substitute T = 1C?C?

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No! No!

Superscript ‘o’ refers to the Superscript ‘o’ refers to the reference state. reference state.

These values apply only for T = These values apply only for T = 2525CC

Example

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Equilibrium constant at T = 1Equilibrium constant at T = 1C C estimated from the van’t Hoff estimated from the van’t Hoff equation. equation.

Example

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,

3

11.71 10 1 11.38 10 exp

8.314 298 274

2.09 10

eq TK

[[OO22] = 2.09] = 2.091010-3-3 0.21 0.21

[O[O22] = 4.39] = 4.391010-4-4 (mol/L) (mol/L) 14 14 mg/Lmg/L

2 1, ,1 2

1 1exp

or

eq T eq TH

K KR T T

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Compare available energy Compare available energy

Aerobic versus anaerobic process Aerobic versus anaerobic process

Degradation of acetic acid Degradation of acetic acid

Aerobic processAerobic process

CC22HH44OO22(aq) + 2 O(aq) + 2 O22(aq) = 2 H(aq) = 2 H22COCO33 (aq) (aq)

= 2= 2(-623.2) – (-396.6) - (-623.2) – (-396.6) - 22(16.32) (16.32)

= -882.44 kJ/mol= -882.44 kJ/mol

Example

rG

rG

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Anaerobic process Anaerobic process

CC22HH44OO22(aq) + H(aq) + H22O(l) = CHO(l) = CH44(aq) + H(aq) + H22COCO33(aq)(aq)

= (-34.39)+(-623.2)-(-396.6)-(-237.18)= (-34.39)+(-623.2)-(-396.6)-(-237.18)

= -23.81 kJ/mol= -23.81 kJ/mol

Relative to anaerobic processes, aerobic Relative to anaerobic processes, aerobic process release more energyprocess release more energy

Example

rG

rG

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Temperature affects equilibrium:Temperature affects equilibrium:

Dissolution of ODissolution of O22

= -11.71 kJ/mol= -11.71 kJ/mol

Precipitation of CaCOPrecipitation of CaCO33(s)(s)

CaCa2+2+(aq) + CO(aq) + CO332-2-(aq) = CaCO(aq) = CaCO33(s)(s)

= (-1207.4) - (-542.83) – (-= (-1207.4) - (-542.83) – (-677.1) 677.1)

= 12.53 kJ/mol= 12.53 kJ/mol

Example

rH

rH

rH

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Sign of Sign of HH is important is important

OO22 solubility increases with solubility increases with decreasing temperaturedecreasing temperature

Implications for fish? Implications for fish?

CaCOCaCO33(s) solubility decreases with (s) solubility decreases with decreasing temperature decreasing temperature

Implications for water Implications for water softening? softening?

Example

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Precipitation of FeCOPrecipitation of FeCO33(s)(s)

Find changes in:Find changes in:

Gibbs energyGibbs energy

EnthalpyEnthalpy

EntropyEntropy

FeFe2+2+ + CO + CO332-2- = FeCO = FeCO33(s)(s)

Example

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= -666.7 – (-78.87 – 527.9) = -666.7 – (-78.87 – 527.9)

= -59.93 kJ/mol= -59.93 kJ/mol

rH

Example

rG

= -737.0 – (-89.10 – = -737.0 – (-89.10 – 677.1) 677.1)

= 29.2 kJ/mol= 29.2 kJ/mol

= 105 – (-138 – 56.9) = 105 – (-138 – 56.9)

= 299.90 J/mol.K= 299.90 J/mol.K

Entropy increases for formation of Entropy increases for formation of this solid. this solid.

Why? Why?

rS

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Summary ISummary I


products reactantsrG G G

i i iG H T S

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Summary IISummary II

expc do

req a b

C DGK

RT A B

1 2, ,1 2

1 1exp

or

eq T eq TH

K KR T T

Some time for Some time for teamsteams

ororMaxwell’s Demon

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0.5 N0.5 N22 + O + O22 NO NO22 + + heatheat

The expression above shows an The expression above shows an exothermic reaction (heat is exothermic reaction (heat is generated) wherein nitrogen gas generated) wherein nitrogen gas is oxidized to NOis oxidized to NO22. .

What effect would each of the What effect would each of the following have on the reaction? following have on the reaction? • Decrease the temperature. Decrease the temperature. • Increase the volume. Increase the volume. • Decrease the ODecrease the O22 concentration. concentration.

• Add a catalyst. Add a catalyst.

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Solutions…Solutions…

Decrease the temperature. Decrease the temperature. Equilibrium shifts to the right to Equilibrium shifts to the right to

increase the temperature.increase the temperature.

Increase the volume. Increase the volume. Equilibrium shifts to the left to increase Equilibrium shifts to the left to increase

the pressure. the pressure.

Decrease the ODecrease the O22 concentration. concentration.

Equilibrium shifts to the left to increase Equilibrium shifts to the left to increase OO22. .

Add a catalyst. Add a catalyst. A catalyst will affect the rate of the A catalyst will affect the rate of the

reaction, but not the equilibrium. reaction, but not the equilibrium.

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week 3 thermodynamics

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