CEE 370 Environmental Engineering Principles Lecture #6 - - PowerPoint PPT Presentation
Print version Updated: 18 September 2019 CEE 370 Environmental Engineering Principles Lecture #6 Environmental Chemistry IV: Thermodynamics, Equilibria, Acids-bases I Reading: Mihelcic & Zimmerman, Chapter 3 Davis & Masten, Chapter
David Reckhow CEE 370 L#6 1
Lecture #6 Environmental Chemistry IV: Thermodynamics, Equilibria, Acids-bases I
Reading: Mihelcic & Zimmerman, Chapter 3
Davis & Masten, Chapter 2 Mihelcic, Chapt 3
Updated: 18 September 2019
Print version
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Base Hydrolysis of dichloromethane (DCM)
Forms chloromethanol (CM) and chloride Classic second order reaction (molecularity of 2)
First order in each reactant, second order overall
dt Cl d dt CM d dt OH d dt DCM d OH DCM k Rate ] [ ] [ ] [ ] [ ] ][ [
− − −
= = − = − = =
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where, CA = concentration of reactant species A, [moles/liter] CB = concentration of reactant species B, [moles/liter] a = stoichiometric coefficient of species A b = stoichiometric coefficient of species B k = rate constant, [units are dependent on a and b]
Law of Mass Action
For elementary reactions
k
b B a AC
10 20 30 40 50 60 70 80 90 20 40 60 80 Time (min) Concentration
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Reactions of order “n” in reactant “c” When n=0, we have a simple zero-order reaction
Example: biodegradation of 2,4-D
dc dt kcn = −
k mg l = 10 / / min
Slope
10 20 30 40 50 60 70 80 90 20 40 60 80 Time (min) Concentration
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When n=1, we have a simple first-
This results in an “exponential decay”
Example: decay of
137Cs from
Chernobyl accident
dc dt kc = −
1
−
k =
−
0 032
1
. min
A => products
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dc dt kc = −
1
10 20 30 40 50 60 70 80 90 20 40 60 80 Time (min) Concentration
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This is a reaction between two identical molecules More common to have different molecules reacting (see next slide)
dc dt kc = −
2
k L mg = 0 0015 . / / min
When n=2, we have a simple second-order
reaction
A + A => products
When you have two different species reacting via 2nd
If one reactant (cB) is present in great excess versus the
constant and folded into the rate constant to get a pseudo-first order reaction
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B A B A
c kc dt dc dt dc − = =
A
A
c k dt dc − =
B
kc k =
where
From here you can use the first order equations
A + B => products
10 20 30 40 50 60 70 80 90 20 40 60 80 Time (min) Concentration
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Curvature: 2nd>1st>zero
Zero Order First Order Second Order
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For a zero order reaction: For a first order reaction:
2 1
2 1 =
−
2 1
kt
−
2 1
Try example 3.5 & 3.6 in Mihelcic
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Temperature Dependence
d k dT E RT
a a a
(ln ) =
2
T K E T RT
a
a a
− 293 293 293 ( )/
T T T
−
Typical values: θ=1.02 to 1.15
a a a
RT E T
/ −
Activ ivat ation en ener ergy Pre re-exponen ential F Fac actor
T C T C
− 20 20
Or more generally where T
“baseline” temperature R = universal gas constant = 1.987 cal/oK/mole T
a = absolute temp (oK)
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Activity is the “effective or apparent” concentration, which may be slightly different from the true “analytical” concentration
These two differ substantially in waters with high TDS, such as sea water.
We identify these two as follows:
Curved brackets ({X}) indicate activity Square brackets ([X]) indicate concentration
Usually this is molar concentration This may also be used when we’re not very concerned about the differences between activity and concentration
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Mostly long-range interactions between uninterested bystanders (chemical species that are not involved in the reaction) and the two dancers of interest (those species that are reacting)
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Equilibrium quotients are really written for activities, not concentrations in most natural waters activities are nearly equal to the molar concentrations In saline waters, we must account for differences between the two
activity coefficients (f or γ) are used for this Ionic Strength (I or µ) is used to determine the extent of correction
2 2 1
i iz
A
{ } { } { } { }
b a d c
B A D C K =
A A ≈
A
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Mihelcic & Zimmerman
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µ vs. specific conductance: Russell Approximation
µ = 1.6 x 10-5 x K (in µmho/cm)
µ vs. TDS: Langlier approximation
µ ~ 2.5 x 10-5 x TDS (in mg/L)
Equ 3.2 Equ 3.3
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Approximation Equation Applicable Range for I
Debye-Hückel
I z f
2
5 . log − =
<10-2.3 Extended Debye-Hückel
I a I z f 33 . 1 5 . log
2
+ − =
<10-1 Güntelberg
I I z f + − = 1 5 . log
2
<10-1, solutions
electrolytes Davies
− + − = I I I z f 2 . 1 5 . log
2
<0.5
From Stumm & Morgan, Table 3.3 (pg.103) 0.3, based on Mihelcic
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Consider a reaction as follows: Since all reactions are reversible, we have two possibilities The rates are: And at equilibrium the two are equal, rf=rb We then define an equilibrium constant (Keq)
D C B A D C B A
b f
k k
+ ← + + → +
} }{ { B A k r
f f =
A + B = C + D
} }{ { D C k r
b b =
} }{ { } }{ { B A D C k k K
b f eq
= ≡
} }{ { } }{ { D C k B A k
b f
=
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In terms of molar concentrations, the rates are: And at equilibrium the two are equal, rf=rb And solving for the equilibrium constant (Keq)
[ ] [ ] B
A f f
B A k r γ γ =
C b b
D C k r γ γ =
[ ] [ ] [ ] [ ] [ ][ ] [ ][ ]
= = ≡
B A D C B A D C b f eq
B A D C B A D C k k K γ γ γ γ γ γ γ γ
[ ] [ ] [ ] [ ] D
C b B A f
D C k B A k γ γ γ γ =
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Tells us what direction the reaction is headed in Doesn’t tell us how fast the reaction is going (kinetics) Solving equilibrium problems identify reactants and products formulate equations
equilibrium equations mass balance equations electroneutrality equation
solve equations
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Need ∆H (enthalpy change)
∆H < 0, exothermic (heat evolved) ∆H > 0, endothermic (heat absorbed)
The Van’t Hoff Equation:
recall that:
( )
1 2 1 2 1 2
303 . 2
T RT T T H K K
∆
Log K 1/T
∆ ∆
=
i
H ν
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pH of most mineral-bearing waters is 6 to 9. (fairly constant) pH and composition of natural waters is regulated by reactions of acids & bases
chemical reactions; mostly with minerals
carbonate rocks: react with CO2 (an acid)
CaCO3 + CO2 = Ca+2 + 2HCO3
acids from volcanic activity: HCl, SO2
Biological reactions: photosynthesis & resp. Sillen: Ocean is result of global acid/base titration
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Equilibrium is rapidly established
proton transfer is very fast
we call [H+] the Master Variable
because Protons react with so many chemical species, affect equilibria and rates
Strength of acids & bases
strong acids have a substantial tendency to donate a
well as the base accepting the proton (often water).
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Alkalinity: a capacity factor
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Mathematical Expression of Acid/Base Strength
acids: HA = H+ + A-
HCl + H2O = H3O+ + Cl- HCl = H+ + Cl-
Bases: B + H2O = BH+ + OH-
NH3 + H2O = NH4
+ + OH-
3
10 ≈ =
− +
HCl Cl H Ka
[ ]
76 . 4 3 4
10−
− +
= = NH OH NH Kb
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Classical Thermodynamics Ideal Gas Law Equilibrium Chemistry
Working with equations Phase Transfer Acids/Bases
The Carbonate System
Precipitation/Dissolution Adsorption
Review material
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∆ ∆ ∆ ∆ ∆ H = (p H + r H ) + (a H + b H )
rxn P R A B
° ° ° ° °
For a generic reaction:
where, A,B = reactant species P,R = product species a,b = stoichiometric coefficients of the reactants p,r = stoichiometric coefficients of the products ∆H°A,B = enthalpy of reactants A and B ∆H°P,Q = enthalpy of products P and Q
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∆
H = a a
rxn i=1 n i prod j=1 m j rxtnts
° ° °
where, ai = stoichiometric coefficient of product species i aj = stoichiometric coefficient of reactant species j ∆H°i = enthalpy of product species i, [kcal/mol] ∆H°j = enthalpy of reactant species j, [kcal/mol] Or more generally:
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Species ∆G°, kcal/mol ∆H°, kcal/mol Ca
2+(aq)CaCO3(s), calcite
Ca(OH)2(s), lime
CO2(g)
CO2(aq)
CO
Cl
Cl2(aq)
OCl
H
+(aq)HNO3(aq)
NO(aq)
HOCl(aq)
OH
O2(aq) 3.9
H2O(l)
H2(g) O2(g)
Table 4.2, pg. 57
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Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3-1 from Snoeyink & Jenkins) Part I
Species
H
∆
kcal/mole
G
∆
kcal/mole
Species
H
∆
kcal/mole
G
∆
kcal/mole Ca+2(aq)
CO3
CaC03(s), calcite
CH3COO-, acetate
CaO (s)
H+ (aq) C(s), graphite H2 (g) CO2(g)
Fe+2 (aq)
CO2(aq)
Fe+3 (aq)
CH4 (g)
Fe(OH)3 (s)
H2CO3 (aq)
Mn+2 (aq)
HCO3
MnO2 (s)
Conversion: 1kcal = 4.184 kJ
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Thermodynamic Constants for Species of Importance in Water Chemistry (Table 3-1 from Snoeyink & Jenkins) Part II
Species
H
∆
kcal/mole
G
∆
kcal/mole
Species
H
∆
kcal/mole
G
∆
kcal/mole
Mg+2 (aq)
O2 (g)
Mg(OH)2 (s)
OH- (aq)
NO3
H2O (g)
NH3 (g)
H2O (l)
NH3 (aq)
SO4
NH4
+ (aq)
HS (aq)
3.01 HNO3 (aq)
H2S(g)
O2 (aq)
3.93
H2S(aq)
Conversion: 1kcal = 4.184 kJ
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∆ ∆ ∆ G = H - T S where, ∆G = Gibbs free energy, [kcal/mol] ∆H = enthalpy, [kcal/mol] ∆S = entropy, [kcal/K-mol] T = temperature, [K]
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Combines enthalpy and entropy
1st and 2nd laws of thermodynamics
Determines whether a reaction is favorable or spontaneous Practical form is based on an arbitrary datum
the pure and most stable form of each element at standard state
∆ ∆ ∆
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∆ ∆ ∆ ∆ ∆ G = (p G + r G ) - (a G + b G )
rxn P R A B
° ° ° ° °
where, A,B = reactant species P,R = product species a,b = stoichiometric coefficients of the reactants p,r = stoichiometric coefficients of the reactants ∆G°A,B = Gibbs free energy of reactants A and B ∆G°P,Q = Gibbs free energy of reactants P and Q
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∆
G = a
rxn prod i=1 n i i rxtnts j=1 m j j
° ° °
where, ai = stoichiometric coefficient of product species i aj = stoichiometric coefficient of reactant species j ∆G°i = standard Gibbs free energy of product species i, [kcal/mol] ∆G°j = standard Gibbs free energy of reactant species j, [kcal/mol]
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∆ ∆ G = G + RT P R A B
p r a b
° ln
Q = P R A B
p r a b
And we define:
∆G = - RT Keq ° ln
At equilibrium, ∆G=0, and Q=Keq
End equilibria
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Parameter Standard State or Condition Temperature 25°C Gas 1 atm Solid Pure solid Liquid Pure liquid Solution 1 M Element
*
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∆Go
react
A reaction with a negative ∆G will proceed from reactants to products. A reaction with a positive ∆G will proceed from products to reactants (backwards). A reaction in which ∆G is zero is at equilibrium
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Lime, or calcium hydroxide, is commonly used to improve sedimentation processes or to precipitate toxic metals. A 500 L reaction vessel contains an excess amount of solid calcium hydroxide. The solution contains 10-8 M of Ca2+ and 10-8 M of OH─. Is the lime still dissolving, is the solution at equilibrium, or is lime precipitating?
Ca(OH ) (s) Ca + 2OH
2 2+
Example 4.6 from Ray
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° ∆ ° ∆ + ° ∆ ° ∆
) Ca(OH OH Ca rxn
2 + 2
G
G 2 G ( = G
∆ G = [(-132.3 kcal/ mol) + 2 x (-37.6 kcal/ mol) - (-214.7 kcal/ mol)]
rxn
°
∆ G = + 7.22 kcal/ mol
rxn
°
First calculate the Gibbs Free Energy under standard state conditions
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∆ ∆
rxn 2+ 2
G = G + RT ln [Ca ][OH ] °
Now we need to adapt to actual conditions:
( )
2 8
) 10 )( 10 ( ln x K) (298 x ) cal 1000 kcal ( x ) mol K- cal (1.987 + ) mol K- kcal (7.22 = G ∆
∆ ∆ G = G + RT P R A B
p r a b
° ln
rxn
Ca(OH ) (s) Ca + 2OH
2 2+
Since ∆G is negative, the reaction will proceed as written
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Kinetics
Rates of reactions Dynamic, complex Best approach for slow reactions
Oxidation reactions, biochemical transformations, photochemical reactions, radioactive decay
Equilibrium: thermodynamics
Final stopping place Static, simple Best approach for fast reactions
Acid/base reaction, complexation, some phase transfer
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A polluted stream with a temperature of 25°C has a dissolved oxygen concentration of 4 mg/L. Use Gibbs free energy to determine if oxygen from the atmosphere is dissolving into the water, the oxygen is at equilibrium, or oxygen from the stream is going into the atmosphere.
2 2
Example 4.7 from Ray
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∆ ∆ ∆ G = G
= (0 kcal mol ) - (-3.9 kcal mol )
rxn O (g) O (aq)
2 2
° ° °
∆ G = 3.9 kcal mol
rxn
°
∆ ∆ G = G + RT ln p [O (aq)]
2
O 2
°
[O (aq)] = 4 mg O L x g O 1000 mg O x mol O 32 g O = 1.25 x 10 M
2 2 2 2 2 2
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∆ G = 3.9 kcal mol x 1000 cal kcal + . cal K- mol (298K) ln 0.209 atm 1.25 x 10 M
− 1987
∆ G = cal mol > 0 491
Since ∆G is positive, the reaction will proceed in the reverse direction as written. From the atmosphere to the water.
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