Download - 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
ln lnoi i i iG G RT c RT
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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
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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)
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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
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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
ln lnoi i i iG G RT c RT
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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