Download H 2

Thermodynamics and kinetics
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•
•
•
•
•
Heats of reaction
Thermodynamic laws
Electrochemistry
Kinetics
Acid-Base
Equilibrium calculations
1-1
Heats of Reaction
• 1 cal = 4.184 j
• Heat Capacity (Cp)
 Heat required to raise one gram of substance
1 °C
 Al; Cp = 0.895 j/gK
 What is the heat needed to 40 g Al 10 K
 (0.895 j/gK)(40g)(10K)= 358 j
• Exothermic
 Reaction produces heat (at 25 °C)
C(s) + O2(g) <--> CO2(g) + 393.76 kj
1-2
Heats of Reaction
• Endothermic
 Reaction requires energy (at 25 °C)
2 HgO + 181.70 kj <--> 2 Hg + O2
Enthalpy (∆H)
• Energy of a system (heat content)
 ∆H = ∆Hproducts - ∆Hreactants
 Exothermic reactions have negative ∆H
 Negative ∆H tend to be spontaneous
1-3
Enthalpy (∆H)
• Bond energies
 Can be used to estimate ∆H
 N2 + 3 H2 <--> 2 NH3
 6(351.5)-945.6-3(436.0) = -144.6 kj/2 mole
=-72.3 kj/mole (actual -46.1 kj/mol)
• Aqueous Ions (use ∆H values)
 ∆Hproducts-∆Hreactants
2 H+ + CO32- <--> CO2 + H2O
-393.5 + (-285.8)-(-677.1+2(0)) = -2.2 kj/mol
CO2(g)
H2O(l) CO32-
H+
1-4
Bond Energies (kJ/mol) at 298 K
Single Bonds
HH
HC
HN
HO
HF
HS
HCl
HBr
HI
CC
CN
CO
CF
CS
CCl
CBr
436.0
418.4
351.5
464.4
564.8
338.9
431.8
368.2
297.1
343.1
267.8
347.3
426.8
255.2
330.5
276.1
Single Bonds
CI
NN
NF
NCl
OO
OF
OSi
OCl
OS
FF
PCl
SS
ClCl
ClI
BrBr
II
217.6
133.9
234.3
154.8
138.1
188.3
443.5
209.2
518.8
158.2
326.4
205.0
242.7
209.2
192.5
150.6
CC
CN
CO
NN
Double Bond
610.9
615.0
740.6
418.4
Triple Bond
836.8
891.2
1071.1
945.6
Bond in O2 498.7
1-5
Heat Capacities
Elements (Jg-1K-1)
Compounds (Jmol-1K-1-)
Li
3.582
B
1.026
C
0.709
N
1.042
O
0.920
Al
0.900
Cr
0.449
Cu
0.385
Zr
0.278
Hf
0.140
W
0.130
Au
0.128
Hg 0.140
Pb
0.129
Th
U
0.120
Pu
0.130
H2
O2
N2
CO
CO2
CH4
C2H6
NH3
H2O(g)
HBr
SnO2
Al2O3
Fe2O3
0.113
Np 0.120
Am 0.110
28.87
29.50
29.04
29.16
37.49
35.98
53.18
36.11
24.77
27.53
56.61
79.33
104.35
1-6
Pure Substance Standard ∆H (kJ/mol)
Ag(g)
Ag+(g)
AgCl(c,cerargyrite)
AgNO3(c)
Al3+(g)
Al(OH)3
BaCO3(c,witherite)
BaC2O4(c)
BaCrO4(c)
BaF2(c)
BaSO4(c)
BeO(c)
Bi2S3(c)
Br2(g)
Br(g)
Br-(g)
C(c,diamond)
C(g)
CO(g)
CO2(g)
284.55
1021.73
-127.068
-124.39
5483.17
-1276
-1216.3
-1368.6
-1446
-1207.1
-1473.2
-609.6
-143.1
30.907
111.88
-219.07
1.895
716.682
-110.525
-393.509
COCl2(g,phosgene)
-218.8
CH4(g,methane)
-74.81
C2H2(g,ethyne)
226.73
C2H2(g,ethene)
52.25
C2H6(g,ethane)
-84.68
CH3OCH3(g)
-184.05
CH3OH(g,methanol)
-200.66
CH3OH(l,methanol)
-238.66
C2H5OH(g,ethanol)
-235.1
C2H5OH(l,ethanol)
-277.69
CH3COOH(l,acetic acid) -484.51
(CH3)2O(g)
-184.05
CH3CHO(l)
-192.3
CH3Cl(g)
-80.83
CHCl3(g)
-103.14
CCl4(l)
-135.44
CaO(c)
-635.09
Ca(OH)2(c)
-986.09
CaCO3(c, calcite)
-1206.92
CaCO3(c, aragonite)
-1207.13
1-7
Pure Substance Standard ∆H (kJ/mol)
CaC2O4(c)
CaF2(c)
Ca3(PO4)2(c)
CaSO4(c,anhydrite)
Cd(g)
Cd2+(g)
Cd(OH)2(c)
CdS(c)
Cl(g)
Cl-(g)
ClO2(g)
Cu(g)
Cu2O(c,cuprite)
CuO(c,tenorite)
Cu(OH)2(c)
Cu2S(c,chalcocite)
CuS(c,covellite)
F(g)
F-(g)
-1360.6
-1219.6
-4109.9
-1434.1
2623.54
112.01
-560.7
-161.9
121.679
-233.13
102.5
338.32
-168.6
-157.3
-449.8
-79.5
-53.1
78.99
-255.39
Fe3+(g)
Fe2O3(c,hematite)
H+(g)
H2O(g)
H2O(l)
H2O2(g)
H2O2(l)
H2SO4(l)
HF(g)
HCl(g)
HBr(g)
HI(g)
HCN(g)
PbO(c,red)
PbO2(c)
Pb3O4(c)
PbS(c,galena)
PbSO4(c)
ThO2(c)
5712.8
-824.2
1536.202
-241.818
-285.83
-136.31
-187.78
-813.989
-271.1
-92.307
-36.4
26.48
135.1
-218.99
-277.4
-718.4
-100.4
-919.94
-1226.4
Fe(g)
416.3
UO2(c)
-1084.9
1-8
Solution Standard ∆H (kJ/mol)
Ag+
AgCl2Ag(NH3)2+
Ag(S2O3)2Al3+
BrBrO3Ca2+
Cd2+
Cd(CN)42Cd(NH3)42+
Ce3+
Ce4+
CH3COOCH3COOH
CNCNSClClO4CO2
105.579
-245.2
-111.29
-1285.7
-531
-121.55
-67.07
-542.83
-75.9
428
-450.2
-696.2
-537.2
-486.01
-485.76
150.6
76.44
-167.15
-129.33
-413.8
CO2
CO32H+
H2O2
II3IO3K+
NH3
NH4+
NO3Na+
OHO2
SO42Sn2+
Sr2+
Tl3+
U4+
UO22+
-413.8
-677.14
0
-191.17
-55.19
-51.5
-221.3
-252.38
-80.29
-132.51
-205
-240.12
-229.99
-11.7
-909.27
-8.8
-545.8
196.6
-591.2
-1019.6
1-9
∆H determination from other data
∆HT2
Reactants at T2
∆Hreactants=(SCp)(T2-298)
Reactants at 298 K
Products at T2
∆Hproducts=(SCp)(298-T2)
∆H298
Products at 298 K
SCp is the sum of the heat capacities
1-10
Entropy (∆S)
• Randomness of a system
 increase in ∆S tends to be spontaneous
• Enthalpy and Entropy can be used for evaluating the free
energy of a system
• Gibbs Free Energy
 ∆G = ∆H -T∆S
 ∆G=-RTlnK
K is equilibrium constant
Activity at unity
1-11
Calculations
Compound
H2O
OH-(aq)
H+(aq)
∆G° (kJ/mol) at 298.15 K
-237.129
-157.244
0
H2OH++OH What is the constant for the reaction?
• At 298.15 K
∆G(rxn) = 0 + -157.244 - (-273.129) = 79.9 kJ/mol
lnK= (79.9E3/(-8.314*298.15))=-32.2
K=1E-14
1-12
Thermodynamic Laws
• 1st law of thermodynamics
 If the state of a system is changed by
applying work or heat or both, then the
change in the energy of the system must
equal the energy applied.
∆E = q (heat absorbed) + w (work)
 Conservation of energy
Energy can be transferred, but no
direction is given
1-13
Thermodynamic Laws
• 2nd law of thermodynamics
 Reactions tend towards equilibrium
Increase in entropy of a system
 Spontaneous reaction for -∆G
 ∆G = 0, system at equilibrium
• 3rd law of thermodynamics
 Entropies of pure crystalline solids are zero at
0K
1-14
Faraday Laws
• In 1834 Faraday demonstrated that the quantities of
chemicals which react at electrodes are directly
proportional to the quantity of charge passed through
the cell
• 96487 C is the charge on 1 mole of electrons = 1F
(faraday)
1-15
Faraday Laws
• Cu(II) is electrolyzed by a current of 10A for 1
hr between Cu electrode
 anode: Cu <--> Cu2+ + 2e cathode: Cu2+ + 2e- <--> Cu
 Number of electrons
(10A)(3600 sec)/(96487 C/mol) = 0.373 F
0.373 mole e- (1 mole Cu/2 mole e-) =
0.186 mole Cu
1-16
Half-cell potentials
• Standard potential
 Defined as °=0.00V
H2(atm) <--> 2 H+ (1.000M) + 2e• Cell reaction for
 Zn and Fe3+/2+ at 1.0 M
 Write as reduction potentials
Fe3+ + e- <--> Fe2+
°=0.77 V
Zn2+ + 2e- <-->Zn
°=-0.76 V
 Fe3+ is reduced, Zn is oxidized
1-17
Half-Cell Potentials
• Overall
 2Fe3+ +Zn <--> 2Fe2+ + Zn2+ °=0.77+0.76=1.53 V
• Half cell potential values are not multiplied
Application of Gibbs
• If work is done by a system
 ∆G = -°nF (n= e-)
• Find ∆G for Zn/Cu cell at 1.0 M
 Cu2+ + Zn <--> Cu + Zn2+ °=1.10 V
 2 moles of electrons (n=2)
∆G =-2(96487C/mole e-)(1.10V)
∆G = -212 kJ/mol
1-18
Reduction Potentials
Electrode Couple
Na+ + e- --> Na
Mg2+ + 2e- --> Mg
Al3+ + 3e- --> Al
Zn2+ + 2e- --> Zn
Fe2+ + 2e- --> Fe
Cd2+ + 2e- --> Cd
Tl+ + e- --> Tl
Sn2+ + 2e- --> Sn
Pb2+ + 2e- --> Pb
2H+ + 2e- --> H2(SHE)
S4O62- + 2e- --> 2S2O32Sn4+ + 2e- --> Sn2+
SO42- + 4H+ + 2e- --> H2O + H2SO3(aq)
Cu2+ + e- --> Cu+
S + 2H+ + 2e- --> H2S
AgCl + e- --> Ag + ClSaturated Calomel (SCE)
UO22+ + 4H+ + 2e- --> U4+ + 4H2O
"E0, V"
-2.7144
-2.3568
-1.676
-0.7621
-0.4089
-0.4022
-0.3358
-0.141
-0.1266
0
0.0238
0.1539
0.1576
0.1607
0.1739
0.2221
0.2412
0.2682
1-19
Reduction Potentials
Hg2Cl2 + 2e- --> 2Cl- + 2Hg
Bi3+ + 3e- --> Bi
Cu2+ + 2e- --> Cu
Fe(CN)63- + e- --> Fe(CN)64Cu+ + e- --> Cu
I2 + 2e- --> 2II3- + 2e- --> 3IH3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O
2HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2ClHg2SO4 + 2e- --> 2Hg + SO42I2(aq) + 2e- --> 2IO2 + 2H+ + 2e- --> H2O2(l)
O2 + 2H+ + 2e- --> H2O2(aq)
Fe3+ + e- --> Fe2+
Hg22+ + 2e- --> Hg
Ag+ + e- --> Ag
Hg2+ + 2e- --> Hg
2Hg2+ + 2e- --> Hg22+
NO3- + 3H+ + 2e- -->HNO2(aq) + H2O
0.268
0.286
0.3394
0.3557
0.518
0.5345
0.5354
0.5748
0.6011
0.6152
0.6195
0.6237
0.6945
0.769
0.7955
0.7991
0.8519
0.9083
0.9275
1-20
Reduction Potentials
VO2+ + 2H+ + e- --> VO2+ + H2O
HNO2(aq) + H+ + e- --> NO + H2O
Br2(l) + 2e- --> 2BrBr2(aq) + 2e- --> 2Br2IO3- + 12H+ + 10e- -->6H2O + I2
O2 + 4H+ + 4e- --> 2H2O
MnO2 + 4H+ + 2e- -->Mn2+ + 2H2O
Cl2 + 2e- --> 2ClMnO4- + 8H+ + 5e- -->4H2O + Mn2+
1.0004
1.0362
1.0775
1.0978
1.2093
1.2288
1.1406
1.3601
1.5119
2BrO3- + 12H+ + 10e- -->6H2O + Br2
1.5131
1-21
Nernst Equation
• Compensated for non unit activity (not 1 M)
• Relationship between cell potential and activities
• aA + bB +ne- <--> cC + dD
2.30RT
[C]c [D]d
   
log
nF
[A]a [B]b
• At 298K 2.3RT/F = 0.0592
• What is potential of an electrode of Zn(s) and 0.01 M
Zn2+
• Zn2+ +2e- <--> Zn °= -0.763 V
• activity of metal is 1
0.0592
1
  0.763 
log
 0.822V
2
0.01
1-22
Kinetics and Equilibrium
• Kinetics and equilibrium are important concepts in
examining and describing chemistry
 Identify factors which determine rates of reactions
Temperature, pressure, reactants, mixing
 Describe how to control reactions
 Explain why reactions fail to go to completion
 Identify conditions which prevail at equilibrium
1-23
Kinetics
• Rate of reaction
 Can depend upon conditions
Paper reacts slowly with oxygen, but increases with
temperature
• Free energy does not dictate kinetics
16 H+ + 2 MnO4- + 5 Sn2+ <--> 5 Sn4+ + 2 Mn2+ + 8 H2O
°= 1.39 V, ∆G = -1390 kcal/mol
8 H+ + MnO4- + 5 Fe2+ <--> 5 Fe3+ + Mn2+ + 4 H2O
°= 0.78 V, ∆G = -376 kcal/mol
 The reaction with Fe is much faster
1-24
Kinetics
• Rate can depend upon reaction path
 C6H12O6(s) + 6 O2(g) <--> 6 CO2(g) + 6
H2O(l)
 ∆G = -791 kcal/mol
 Glucose can be stored indefinitely at room
temperature, but is quickly metabolized in a
cell
• Thermodynamics is only concerned with
difference between initial and final state
• Kinetics account for reaction rates and describe
the conditions and mechanisms of reactions
1-25
Kinetics
• Kinetics are very difficult to describe from first
principles
 structure, elements, behavior
• General factors effecting kinetics
 Nature of reactants
 Effective concentrations
 Temperature
 Presence of catalysts
 Number of steps
1-26
Nature of Reactants
• Ions react rapidly
 Ag+ + Cl- <--> AgCl(s) Very fast
• Reactions which involve bond breaking are slower
 NH4+ + OCN- <-->OC(NH2)2
• Redox reactions in solutions are slow
 Transfer of electrons are faster than those of
atomic transfer
• Reactions between covalently bonded molecules
are slow
 2 HI(g) <--> H2(g) + I2(g)
1-27
Nature of Reactants
• Structure
 P4
Same formula
P
P
P
P
P
White Phosphorus
P
P
P
P
P
Red PhosphorusP
Concentration
 Reaction can occur when molecules contact
P
1-28
Concentration
• Surface area
 larger surface area increases reaction
• Mixing increases interaction
• Need to minimized precipitation or colloid formation
Rate Law
• Concentration of reactant or product per unit time
• Effect of initial concentration on rate can be examined
1-29
Rate Law
•
•
•
•
rate = k[A]x[B]y
rate order = x + y
knowledge of order can help control reaction
rate must be experimentally determined
Injection
Flow meter
mixing
detector
1-30
Rates
Rate=k[A]n; A=conc. at time t, Ao=initial conc.,
X=product conc.
Order rate equation
0
[A0]-[A]=kt, [X]=kt
1
ln[A0]-ln[A]=kt, ln[A0]-ln([Ao]-[X])=kt
k
mole/L sec
1/sec
2
L/mole sec
3
1
1

 kt
[A] [Ao ]
1
1

 kt
[Ao ]  [X] [Ao ]
1
1
kt
1
1
kt
2 
2  2
2 
2  2
[A] [Ao ]
([Ao ]  [X]) [Ao ]
L2/mole2 sec
1-31
Rate Law
• Half-life
 A =Aoe-t
  = ln2/t1/2
 If a rate half life is known, fraction reacted or
remaining can be calculated
(CH3)2N2(g) <--> N2(g) + C2H6(g)
Time
Pressure (torr)
0
36.2
30
46.5
 ln 2(30 min)
t1/2 
 62.1 min
25.9
ln(
)
36.2
1-32
Rate Law
• Temperature
 Reactions tend to double for every 10 °C
• Catalysts
 Accelerate reaction but are not used
Pt surface
 Thermodynamically drive, catalysts drive kinetics
 If not thermodynamically favored, catalysts will not
drive reaction
• Autocatalytic reactions form products which act as
catalysts
• Catalytic amount is moles of catalysts needed to cause
reaction
1-33
Complexation Kinetics
Uranium and cobalt with pyridine based ligands
N
HOOC
N
COOH
N
N
N
HOOC
N
N
N
COOH
N
COOH
HOOC
N
N
COOH
N
COOH
COOH
111Py12
111Py14
222Py14
Examine complexation by UV-Visible
spectroscopy
1-34
Complexation of U with 111Py12
0.60
Absorbance
0.50
Time (minutes)
1148
3956
0.40
8538
14153
21593
32767
34185
0.30
0.20
0.10
0.00
250
300
350
400
Wavelength (nm)
450
1-35
Absorbance Kinetics
Absorbance sum from 250 nm to 325 nm for 111Py12 and uranium at pH 4
14.0
12.0
10.0
8.0
6.0
4.0
2.0
0.0
0
5000
10000 15000 20000 25000 30000 35000
Time (minutes)
1-36
Kinetic Data Evaluation
Evaluation of change in absorbance
2
 Abst 
1
2
 Abso 
1
2
 Abseq (1  e
kt
)
1
Evaluation of absorbance and kinetic data for 111Py12 and 111Py14 with uranium at pH 4.
The concentration of ligand and uranium is 50x10-6 mol/L.
Ligand
Abso
∆Abseq
111Py12
111Py14
7.86±0.82 5.66±1.28
4.82±1.70 7.06±5.76
k (min-1)
95% Equilibrium
Time (min)
4.65±0.47x10-5 6.44±0.65x104
4.24±0.80x10-5 7.07±1.33x104
1-37
Acid-Base Equilibria
Water can act as solvent and reactant
• Brønsted Theory of Acids and Bases
 Acid
Substance which donates a proton
 Base
Accepts proton from another substance
NH3 + HCl <--> NH4+ + ClH2O + HCl <--> H3O+ + ClNH3 + H2O <--> NH4+ + OH• Remainder of acid is base
• Complete reaction is proton exchange between sets
• Extent of exchange based on strength
1-38
Acid Strengths
• Strong acids tend towards completion
HCl + H2O <--> H3O+ + Cl Strong acids tend to have weak conjugate bases
• Weak acid forms only slightly ionized species
CH3COOH + H2O <--> CH3COO- + H3O+
 Stronger conjugate base
• Relative strengths of acids can be compared
 Can be rated relative to water
• Some salts can have acid-base properties
 NH4Cl
1-39
Relative Strengths of Acids and Bases
Acid
Strength
Conjugate Acid
Conjugate Base
HClO4
H2SO4
HCl
H3 O+
H2SO3
HF
HC2H3O2
HSO3H2S
NH4+
HCO3
H2 O
HSOH
H2
ClO4SO42ClH2 O
HSO3FC2H3O2SO32HSNH3
CO32OHS2O2H-
Base
Strength
1-40
Dissociation Constants
• Equilibrium expression for the behavior of acid
HA + H2O <--> A- + H3O+
[A  ][H3O  ]
K
Water concentration is constant
[HA][H2O]
[A ][H3O ]
Ka  K[H2 O] 
pKa=-logKa
[HA]
• Can also be measured for base
Constants are characteristic of the particular acid or base
[HB ][OH  ]
Kb 
[B]
1-41
Dissociation Constants for Acids at 25°C
Acid
Formula
Acetic
Hydrocyanic
Carbonic
HC2H3O2
HCN
H2CO3
HCO3HNO2
H2 S
HSH3PO4
H2PO4 HPO4 2H 2 C 2 O4
HC2O4-
Nitrous
Hydrosulfuric
Phosphoric
Oxalic
Ka
1.8E-5
7.2E-10
3.5E-7
5E-11
4.5E-4
1E-7
1E-14
7.5E-3
6.2E-8
4.8E-13
5.9E-2
6.4E-5
1-42
Dissociation Constants for Bases at 25°C
Base
Ammonia
Pyridine
Methylamine
Formula
NH3
C5H5N
CH3NH2
Kb
1.8E-5
2.3E-9
4.4E-4
Protonation of amine group
1-43
Calculations
• 1 L of 0.1 M acetic acid has pH = 2.87
What is the pKa for acetic acid
CH3COOH + H2O <--> CH3COO- + H3O+
[CH3COO-] = [H3O+] =10-2.87
pKa=4.73
10 (2*2.87)
5
Ka 

1.84x10
0.1 10 2.87
1-44
Calculations
• What is pH of 0.1 M NH3
Kb=1.8E-5
NH3 + H2O <-->NH4+ + OH[NH4+] = [OH -] = x
[NH3] = 0.1 - x
x2 +
1.8E-5x -1.8E-6 = 0
x2
 1.8x10 5
0.1  x
• [OH-] = 1.33E-3 M, pOH = 2.87, pH ≈ 14-pOH ≈11.12
1.8E  5  (1.8E  5)2  4 * 1.8E  6
x
2
1-45
Acid-Base Reactions
For Water
2 H2O <--> H3O+ + OHWater concentration remains constant, so
for water:
Kw = [H3O+][OH-]= 1E-14 at 25°C
1-46
Common ion effect
• Same products from different acids (hydronium ion)
0.2 M CH3COOH in 0.1 M HCl
What is the free acetic acid concentration
HCl is totally dissociated
x=[CH3COO-], [CH3COOH]=0.2-x, [H3O+] = 0.1 + x
• Ka = 1.8E-5
• Small Ka, x is much smaller than 0.1
x=3.6E-5 M
0.1x
Ka 
0.2
1-47
Hydrolysis Constants
• Reaction of water with metal ion
 Common reaction
 Environmentally important
 Dependent upon metal ion oxidation state
• Mz+ + H2O <--> MOHz-1+ + H+
• Constants are listed for many metal ion with
different hydroxide amounts
1-48
Buffers
• Weak acid or weak base with conjugate salt
• Acetate as example
 Acetic acid, CH3COONa
 CH3COOH + H2O <--> CH3COO- + H3O+
large quantity
 If acid
is addedhuge quantity large quantity small quantity
hydronium reacts with acetate ion, forming
undissociated acetic acid
 If base is added
Hydroxide reacts with hydronium, acetic acid
dissociates to removed hydronium ion
1-49
Buffer Solutions
• Buffers can be made over a large pH range
• Can be useful in controlling reactions and separations
 Buffer range
Effective range of buffer
Determined by pKa of acid or pKb of base
HA + H2O <--> A- + H3O-
Write as pH
[A  ][H3O  ]
Ka 
[HA]
Ka [HA]
[H3O ] 
[A ]

1-50
Buffer Solutions
[HA]
pH  pKa  log 
[A ]
• The best buffer is when [HA]=[A-]
 largest buffer range for the conditions
 pH = pKa - log1
• For a buffer the range is determined by [HA]/[A-]
 [HA]/[A-] from 0.1 to 10
 Buffer pH range = pKa ± 1
 Higher buffer concentration increase durability
1-51
Buffers
• What is a good buffer system for maintaining a solution
at pH 4? Want 0.1 M total in 1L total volume
• For acetic acid, pKa=4.75
[CH 3COOH]
4.00  4.75  log
[CH 3COO  ]
We have 0.1 M each
Volume CH3COO- =1/6.62 L = 0.15 L
Acid =1-0.15 =0.85 L
[CH 3COOH]
log
  0.75
[CH 3 COO ]
[CH 3COOH]
  5.62
[CH 3COO ]
1-52
Buffers
• You have a buffer with 0.1 M acetic acid and 0.1 M
sodium acetate in 1L. 0.01 moles of NaOH(s) is added.
What is the pH?
NaOH reacts with 0.01 moles of proton from the acid.
• At equilibrium, we have 0.11 M sodium acetate and
0.09 M acetic acid
• [H3O+]=1.5E-5, pH =4.8
 ][0.11]
[H
O
 (pKa=4.75)K  1.8E  5  3
a
[0.09]
1-53
Equilibrium
•
Reactions proceed in the forward and reverse direction
simultaneously
 N2 + 3 H2 <--> 2 NH3
 Initially contains nitrogen and hydrogen
Forward rate decreases as concentration
(pressure) decreases
Ammonia production increase reverse rate
Eventually, forward rate is equal to reverse rate
No net change in concentration
Reaction still occurring
1-54
Equilibrium
• Some reactions have a negligible reverse rate
 Proceeds in forward direction
 Reaction is said to go to completion
Le Châtelier’s Principle
• At equilibrium, no further change as long as external
conditions are constant
• Change in external conditions can change equilibrium
 A stressed system at equilibrium will shift to reduce
stress
concentration, pressure, temperature
1-55
Equilibrium
• N2 + 3 H2 <--> 2 NH3 + 22 kcal
 What is the shift due to
Increased temperature?
Increased N2?
Reduction of reactor vessel volume?
1-56
Equilibrium Constants
• For a reaction
 aA + bB <--> cC + dD
• At equilibrium the ratio of the product to reactants is a
constant
 The constant can change with conditions
 By convention, constants are expressed as products
over reactants
[C]c [D]d
K
[A]a [B]b
• Conditions under which the constant is measured
should be listed
 Temperature, ionic strength
1-57
Activities
• Strictly speaking, activities, not concentrations should be
used
 C[C]c  D [D]d
K
 A [A]a  B[B]b
• At low concentration, activities are assumed to be 1
• The constant can be evaluated at a number of ionic
strengths and the overall activities fit to equations
1-58
Activities
• Debye-Hückel (Physik Z., 24, 185 (1923))
0.5085Z 2a 
 log A 
1  0.3281R A 
ZA = charge of species A
µ = molal ionic strength
RA = hydrated ionic radius in Å (from 3 to 11)
First estimation of activity
1-59
Activities
• Debye-Hückel term can be written as:
0.5107 
D
1  1.5 
• Specific ion interaction theory
 Uses and extends Debye-Hückel
long range Debye-Hückel
Short range ion interaction term
ij = specific ion interaction term
log  i  Z 2D  ij
log ß()  logß(0)  Z2i D   ij
1-60
Activities
• Pitzer
 Binary (3) and Ternary (2) interaction parameters
K+
Ca2+
Al3+
Fe(CN)641-61
Cm-Humate at pH 6
6.6
Experimental Data shows change in stability
constant with ionic strength
6.5
Ion Specific Interaction Theory used
logß
6.4
6.3
6.2
6.1
6.0
0.0
0.5
1.0
1.5
I mIm
sqrt
2.0
2.5
3.0
1-62
Constants
• Constants can be listed by different names
 Equilibrium constants (K)
Reactions involving bond breaking
* 2 HY <--> H2 + 2Y
 Stability constants (ß), Formation constants (K)
Metal-ligand complexation
* Pu4+ + CO32- <--> PuCO32+
* Ligand is written in deprotonated form
1-63
Constants
 Conditional Constants
An experimental condition is written into
equation
* Pu4+ + H2CO3 <--> PuCO32+ +2H+
Constant can vary with
concentration, pH
Must look at equation!
Often the equation is not explicitly written
1-64
Using Equilibrium Constants
• Constants and balanced equation can be used to evaluate
concentrations at equilibrium
[H2 ][Y]2
 2 HY <--> H2 + 2Y,
K
 K=4E-15
[HY]2
 If you have one mole of HY initially, what are the
concentration of the species at equilibrium?
 Try to write species in terms of one unknown
Start will species of lowest concentration
[H2] =x, [Y]=2x, [HY]=1-2x
 Since K is small, x must be small, 1-2x ≈ 1
 K=4x3, x =1E-5, 2x=2E-5
[x][2x]2
K
[1  2x]2
1-65
Realistic Case
• Consider uranium in an aquifer
 Species to consider include
free metal ion: UO22+
hydroxides: (UO2)x(OH)y
carbonates: UO2CO3
humates: UO2HA(II), UO2OHHA(I)
 Need to get stability constants for all species
UO22+ + CO32- <--> UO2CO3
 Know or find conditions
Total uranium, total carbonate, pH, total humic
concentration
1-66
Stability constants for selected uranium species at 0.1 M
ionic strength
Species
logß
UO2 OH+
8.5
UO2(OH)2
17.3
UO2(OH)322.6
UO2(OH)4223.1
Other species may need to be
(UO2)2OH3+
11.0
considered. If the total
(UO2)2(OH)2+
22.0
uranium concentration is low
UO2CO3
8.87
enough, binary or tertiary
UO2(CO3)2216.07
species can be excluded.
4UO2(CO3)3
21.60
UO2HA(II)
6.16
UO2(OH)HA(I)
14.7±0.5
1-67
Equations
• Write concentrations in terms of species
 [UO2]tot= UO2free+U-carb+U-hydroxide+U-humate
 [CO32-]free=f(pH)
 [OH-] = f(pH)
 [HA]tot = UO2HA + UO2OHHA+ HAfree
• Write the species in terms of metal, ligands, and
constants
 [(UO2)xAaBb] = 10-(xpUO2+apA+bpB-log(UO2)xAaBb)
 pX = -log[X]free
[(UO2)2(OH)22+]=10-(2pUO2+2pOH-22.0)
• Set up equations and solve for known terms
1-68
U speciation with different CO2 partial
pressure
0% CO2
1.0
UO 2(OH)
1.0
3
UO 2
UO 2OHHA(I)
UO 2(OH)
Mole Fraction of U(VI) Species
0.8
UO 2HA(II)
2
0.6
0.4
0.2
0.0
2.0
4.0
6.0
8.0
UO 22+
0.8
UO 2HA(II)
UO 2OHHA(I)
UO 2(CO 3) 34-
0.6
0.4
UO 2(CO 3) 22-
0.2
0.0
10.0
2.0
pH
4.0
6.0
8.0
10.0
pH
1.0
Mole Fraction of U(VI) Species
Mole Fraction of U(VI) Species
2+
1% CO2
0.8
UO 22+
UO 2HA(II)
UO 2OHHA(I)
UO 2(CO 3) 34-
0.6
10% CO2
0.4
UO 2(CO 3) 22-
0.2
0.0
2.0
4.0
6.0
pH
8.0
10.0
1-69
Comparison of measured and
calculated uranyl organic colliod
1.0
0.8
10%
1%
total
[U(VI)]
[U-colloid]
100%
0.6
0.4
0%
0.035%
0.2
0.0
2.0
4.0
6.0
pH
8.0
10.0
1-70
Energy terms
• Constants can be used to evaluate energetic of
reaction
 From Nernst equation
∆G=-RTlnK
 ∆G=∆H-T∆S
-RTlnK = ∆H-T∆S
RlnK= - ∆H/T + ∆S
* Plot RlnK vs 1/T
1-71
Temperature effect on Np-Humate stability
T emp (°C)
56
48
40
32
24
16
76
74
Rlnß
72
70
²H = -22.2 ± 2.8 kJ/mol
²G 298=-21.7 kJ/mol
²S=1.2±1.4 J/molK
68
66
64
0.003
0.0031
0.0032
0.0033
1/T (K)
0.0034
0.0035
1-72
Solubility Products
• Equilibrium involving a solid phase
 AgCl(s) <--> Ag+ + Cl[Cl  ][Ag  ]
K
[AgCl]
 AgCl concentration is constant
Solid activity and concentration is treated
as constant
By convention, reaction goes from solid
to ionic phase in solution
 Can use Ksp for calculating concentrations
Ksp  K[AgCl]  [Cl  ][Ag ]
1-73
Solubility calculations
• AgCl(s) at equilibrium with water at 25°C gives
1E-5 M silver ion in solution. What is the Ksp??
 AgCl(s) <--> Ag+ + Cl-: [Ag+] = [Cl-]
 Ksp = 1E-52 = 1E-10
• What is the [Mg2+] from Mg(OH)2 at pH 10?
 Ksp = 1.2E-11= [Mg2+] [OH]2
 [OH] = 10-(14-10)
[Mg
2
1.2E  11
]
 1.2E  3
1E  8
1-74
Solubility Calculations
• 50 mL of 0.02 M Ag+ added to 50 mL 0.1 M
HCl, what is the silver ion concentration in
solution??
 So assume all Ag+ forms AgCl
 0.001 moles Ag+, 0.005 moles Cl [Cl-] = 0.004 moles/0.1 L =0.04 M
 Ksp= 1E-10
 [Ag+] =1E-10/0.04 =2.5E-9 M
1-75
Solubility Calculations
• What is the maximum pH of a 0.1 M Mg2+ solution in
which Mg(OH)2 will not precipitate??
 Ksp = 1.2E-11= [Mg2+][OH-]2
 [OH-]=(1.2E-11/0.1)0.5 =1.09E-5
 pOH = 4.96
 pH = 9.04
• How can this be used to separate metal ions?
 Separate Mg2+ from Fe2+, 0.1 M each metal ion.
Want minimum 500:1 Mg2+:Fe2+
 Ksp(Fe(OH)2)=1.6E-14
1-76
Limitations of Ksp
• Solid phase formation limited by concentration
 below ≈1E-5/mL no visible precipitate forms
colloids
• formation of supersaturated solutions
 slow kinetics
• Competitive reactions may lower free ion concentration
• Large excess of ligand may form soluble species
 AgCl(s) + Cl- <--> AgCl2-(aq)
Ksp really best for slightly soluble salts
1-77