1-1

Thermodynamics and kinetics

• Heats of reaction• Thermodynamic laws• Electrochemistry• Kinetics• Acid-Base• Equilibrium calculations

1-2

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

• ExothermicReaction produces heat (at 25 °C)

C(s) + O2(g) <--> CO2(g) + 393.76 kj

1-3

Heats of Reaction

• EndothermicReaction 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 ∆HNegative ∆H tend to be spontaneous

1-4

Enthalpy (∆H)

• Bond energiesCan be used to estimate ∆HN2 + 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) CO3

2- H+

1-5

Bond Energies (kJ/mol) at 298 K

HH 436.0HC 418.4HN 351.5HO 464.4HF 564.8HS 338.9HCl 431.8HBr 368.2HI 297.1CC 343.1CN 267.8CO 347.3CF 426.8CS 255.2CCl 330.5CBr 276.1

Single BondsCI 217.6NN 133.9NF 234.3NCl 154.8OO 138.1OF 188.3OSi 443.5OCl 209.2OS 518.8FF 158.2PCl 326.4SS 205.0ClCl 242.7ClI 209.2BrBr 192.5II 150.6

Single Bonds

Double Bond Triple BondCC 610.9 836.8CN 615.0 891.2CO 740.6 1071.1

NN 418.4 945.6

Bond in O2 498.7

1-6

Heat Capacities

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 0.113 U 0.120

Np 0.120 Pu 0.130

Am 0.110

Elements (Jg-1K-1)

H2 28.87O2 29.50N2 29.04CO 29.16CO2 37.49CH4 35.98C2H6 53.18NH3 36.11H2O(g) 24.77HBr 27.53SnO2 56.61Al2O3 79.33Fe2O3 104.35

Compounds (Jmol-1K-1-)

1-7

Pure Substance Standard ∆H (kJ/mol) Ag(g) 284.55Ag+(g) 1021.73AgCl(c,cerargyrite) -127.068AgNO3(c) -124.39Al3+(g) 5483.17Al(OH)3 -1276BaCO3(c,witherite) -1216.3BaC2O4(c) -1368.6BaCrO4(c) -1446BaF2(c) -1207.1BaSO4(c) -1473.2BeO(c) -609.6Bi2S3(c) -143.1Br2(g) 30.907Br(g) 111.88Br-(g) -219.07C(c,diamond) 1.895C(g) 716.682CO(g) -110.525CO2(g) -393.509

COCl2(g,phosgene) -218.8CH4(g,methane) -74.81C2H2(g,ethyne) 226.73C2H2(g,ethene) 52.25C2H6(g,ethane) -84.68CH3OCH3(g) -184.05CH3OH(g,methanol) -200.66CH3OH(l,methanol) -238.66C2H5OH(g,ethanol) -235.1C2H5OH(l,ethanol) -277.69CH3COOH(l,acetic acid) -484.51(CH3)2O(g) -184.05CH3CHO(l) -192.3CH3Cl(g) -80.83CHCl3(g) -103.14CCl4(l) -135.44CaO(c) -635.09Ca(OH)2(c) -986.09CaCO3(c, calcite) -1206.92CaCO3(c, aragonite) -1207.13

1-8

Pure Substance Standard ∆H (kJ/mol) CaC2O4(c) -1360.6CaF2(c) -1219.6Ca3(PO4)2(c) -4109.9CaSO4(c,anhydrite) -1434.1Cd(g) 2623.54Cd2+(g) 112.01Cd(OH)2(c) -560.7CdS(c) -161.9Cl(g) 121.679Cl-(g) -233.13ClO2(g) 102.5Cu(g) 338.32Cu2O(c,cuprite) -168.6CuO(c,tenorite) -157.3Cu(OH)2(c) -449.8Cu2S(c,chalcocite) -79.5CuS(c,covellite) -53.1F(g) 78.99F-(g) -255.39

Fe(g) 416.3

Fe3+(g) 5712.8Fe2O3(c,hematite) -824.2H+(g) 1536.202H2O(g) -241.818H2O(l) -285.83H2O2(g) -136.31H2O2(l) -187.78H2SO4(l) -813.989HF(g) -271.1HCl(g) -92.307HBr(g) -36.4HI(g) 26.48HCN(g) 135.1PbO(c,red) -218.99PbO2(c) -277.4Pb3O4(c) -718.4PbS(c,galena) -100.4PbSO4(c) -919.94ThO2(c) -1226.4

UO2(c) -1084.9

1-9

Solution Standard ∆H (kJ/mol) Ag+ 105.579AgCl2

- -245.2Ag(NH3)2+ -111.29Ag(S2O3)2

- -1285.7Al3+ -531Br- -121.55BrO3- -67.07Ca2+ -542.83Cd2+ -75.9Cd(CN)4

2- 428Cd(NH3)4

2+ -450.2Ce3+ -696.2Ce4+ -537.2CH3COO- -486.01CH3COOH -485.76CN- 150.6CNS- 76.44Cl- -167.15ClO4- -129.33CO2 -413.8

CO2 -413.8CO3

2- -677.14H+ 0H2O2 -191.17I- -55.19I3- -51.5IO3- -221.3K+ -252.38NH3 -80.29NH4+ -132.51NO3- -205Na+ -240.12OH- -229.99O2 -11.7SO4

2- -909.27Sn2+ -8.8Sr2+ -545.8Tl3+ 196.6U4+ -591.2UO2

2+ -1019.6

1-10

∆H determination from other data

Reactants at 298 K

∆Hreactants=(Cp)(T2-298)

Reactants at T2

∆HT2Products at T2

∆Hproducts=(Cp)(298-T2)

Products at 298 K∆H298

Cp is the sum of the heat capacities

1-11

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 constantActivity at unity

1-12

CalculationsCompound ∆G° (kJ/mol) at 298.15 K

H2O -237.129

OH-(aq) -157.244

H+(aq) 0

H2OH++OH-

What is the constant for the reaction?• At 298.15 K

∆G(rxn) = 0 + -157.244 - (-273.129) = 79.9 kJ/mollnK= (79.9E3/(-8.314*298.15))=-32.2K=1E-14

1-13

Thermodynamic Laws

• 1st law of thermodynamicsIf 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 energyEnergy can be transferred, but no

direction is given

1-14

Thermodynamic Laws

• 2nd law of thermodynamicsReactions tend towards equilibrium

Increase in entropy of a systemSpontaneous reaction for -∆G∆G = 0, system at equilibrium

• 3rd law of thermodynamicsEntropies of pure crystalline solids are zero at

0K

1-15

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)

1-16

Faraday Laws

• Cu(II) is electrolyzed by a current of 10A for 1 hr between Cu electrodeanode: Cu <--> Cu2+ + 2e-

cathode: Cu2+ + 2e- <--> CuNumber of electrons

(10A)(3600 sec)/(96487 C/mol) = 0.373 F0.373 mole e- (1 mole Cu/2 mole e-) =

0.186 mole Cu

1-17

Half-cell potentials• Standard potential

Defined as °=0.00VH2(atm) <--> 2 H+ (1.000M) + 2e-

• Cell reaction for Zn and Fe3+/2+ at 1.0 MWrite as reduction potentials

Fe3+ + e- <--> Fe2+ °=0.77 VZn2+ + 2e- <-->Zn °=-0.76 V

Fe3+ is reduced, Zn is oxidized

1-18

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

1-19

Reduction PotentialsElectrode Couple "E0, V"Na+ + e- --> Na -2.7144Mg2+ + 2e- --> Mg -2.3568Al3+ + 3e- --> Al -1.676Zn2+ + 2e- --> Zn -0.7621Fe2+ + 2e- --> Fe -0.4089Cd2+ + 2e- --> Cd -0.4022Tl+ + e- --> Tl -0.3358Sn2+ + 2e- --> Sn -0.141Pb2+ + 2e- --> Pb -0.12662H+ + 2e- --> H2(SHE) 0S4O62- + 2e- --> 2S2O32- 0.0238Sn4+ + 2e- --> Sn2+ 0.1539SO42- + 4H+ + 2e- --> H2O + H2SO3(aq) 0.1576Cu2+ + e- --> Cu+ 0.1607S + 2H+ + 2e- --> H2S 0.1739AgCl + e- --> Ag + Cl- 0.2221Saturated Calomel (SCE) 0.2412UO22+ + 4H+ + 2e- --> U4+ + 4H2O 0.2682

1-20

Reduction PotentialsHg2Cl2 + 2e- --> 2Cl- + 2Hg 0.268Bi3+ + 3e- --> Bi 0.286Cu2+ + 2e- --> Cu 0.3394Fe(CN)63- + e- --> Fe(CN)64- 0.3557Cu+ + e- --> Cu 0.518I2 + 2e- --> 2I- 0.5345I3- + 2e- --> 3I- 0.5354H3AsO4(aq) + 2H+ + 2e- -->H3AsO3(aq) + H2O 0.57482HgCl2 + 4H+ + 2e- -->Hg2Cl2 + 2Cl- 0.6011Hg2SO4 + 2e- --> 2Hg + SO42- 0.6152I2(aq) + 2e- --> 2I- 0.6195O2 + 2H+ + 2e- --> H2O2(l) 0.6237O2 + 2H+ + 2e- --> H2O2(aq) 0.6945Fe3+ + e- --> Fe2+ 0.769Hg22+ + 2e- --> Hg 0.7955Ag+ + e- --> Ag 0.7991Hg2+ + 2e- --> Hg 0.85192Hg2+ + 2e- --> Hg22+ 0.9083NO3- + 3H+ + 2e- -->HNO2(aq) + H2O 0.9275

1-21

Reduction Potentials

VO2+ + 2H+ + e- --> VO2+ + H2O 1.0004

HNO2(aq) + H+ + e- --> NO + H2O 1.0362

Br2(l) + 2e- --> 2Br- 1.0775

Br2(aq) + 2e- --> 2Br- 1.0978

2IO3- + 12H+ + 10e- -->6H2O + I2 1.2093

O2 + 4H+ + 4e- --> 2H2O 1.2288

MnO2 + 4H+ + 2e- -->Mn2+ + 2H2O 1.1406

Cl2 + 2e- --> 2Cl- 1.3601

MnO4- + 8H+ + 5e- -->4H2O + Mn2+ 1.5119

2BrO3- + 12H+ + 10e- -->6H2O + Br2 1.5131

1-22

Nernst Equation• Compensated for non unit activity (not 1 M)• Relationship between cell potential and activities• aA + bB +ne- <--> cC + dD

• 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

2.30RT

nFlog

[C]c[D]d

[A]a[B]b

0.763 0.0592

2log

10.01

0.822V

1-23

Kinetics and Equilibrium• Kinetics and equilibrium are important concepts in

examining and describing chemistryIdentify factors which determine rates of reactions

Temperature, pressure, reactants, mixingDescribe how to control reactionsExplain why reactions fail to go to completionIdentify conditions which prevail at equilibrium

1-24

Kinetics• Rate of reaction

Can depend upon conditionsPaper 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/molThe reaction with Fe is much faster

1-25

Kinetics• Rate can depend upon reaction path

C6H12O6(s) + 6 O2(g) <--> 6 CO2(g) + 6 H2O(l)∆G = -791 kcal/molGlucose 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

1-26

Kinetics

• Kinetics are very difficult to describe from first principlesstructure, elements, behavior

• General factors effecting kineticsNature of reactantsEffective concentrationsTemperaturePresence of catalystsNumber of steps

1-27

Nature of Reactants

• Ions react rapidlyAg+ + Cl- <--> AgCl(s) Very fast

• Reactions which involve bond breaking are slowerNH4

+ + OCN- <-->OC(NH2)2

• Redox reactions in solutions are slowTransfer of electrons are faster than those of

atomic transfer• Reactions between covalently bonded molecules

are slow2 HI(g) <--> H2(g) + I2(g)

1-28

Nature of Reactants

• StructureP4

Same formula

ConcentrationReaction can occur when molecules contact

P

PP

P

P

PP

P

P

PP

PWhite Phosphorus Red Phosphorus

1-29

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

1-30

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

mixingdetector

Flow meter

1-31

Rates

Rate=k[A]n; A=conc. at time t, Ao=initial conc., X=product conc.Order rate equation k

0 [A0]-[A]=kt, [X]=kt mole/L sec

1 ln[A0]-ln[A]=kt, ln[A0]-ln([Ao]-[X])=kt 1/sec

2 L/mole sec

3 L2/mole2 sec

1[A]

1

[Ao]kt

1[Ao ] [X]

1

[Ao]kt

1

[A]2 1

[Ao]2 kt2

1

([Ao] [X])2 1

[Ao]2 kt2

1-32

Rate Law

• Half-lifeA =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

t1/2 ln 2(30min)

ln(25.936.2

)62.1 min

1-33

Rate Law• Temperature

Reactions tend to double for every 10 °C• Catalysts

Accelerate reaction but are not usedPt surface

Thermodynamically drive, catalysts drive kineticsIf 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

1-34

Complexation Kinetics

NN

N

N

N

COOH

COOH

HOOC

NN N

N

HOOC COOH

NN N

N

HOOC

COOH

COOH

COOH

Uranium and cobalt with pyridine based ligands

111Py12 111Py14 222Py14

Examine complexation by UV-Visible spectroscopy

1-35

Complexation of U with 111Py12

0.00

0.10

0.20

0.30

0.40

0.50

0.60

250 300 350 400 450

1148

3956

8538

14153

21593

32767

34185

Ab

sorb

ance

Wavelength (nm)

Time (minutes)

1-36

Absorbance Kinetics

0.0

2.0

4.0

6.0

8.0

10.0

12.0

14.0

0 5000 10000 15000 20000 25000 30000 35000

Ab

sorb

ance

Su

m 2

50 n

m-3

25 n

m

Time (minutes)

Absorbance sum from 250 nm to 325 nm for 111Py12 and uranium at pH 4

1-37

Kinetic Data Evaluation

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 k (min-1) 95% EquilibriumTime (min)

111Py12 7.86±0.82 5.66±1.28 4.65±0.47x10-5 6.44±0.65x104

111Py14 4.82±1.70 7.06±5.76 4.24±0.80x10-5 7.07±1.33x104

Abst1

2

Abso1

2

Abseq1

2

(1 e kt )

Evaluation of change in absorbance

1-38

Acid-Base EquilibriaWater can act as solvent and reactant• Brønsted Theory of Acids and Bases

AcidSubstance which donates a proton

BaseAccepts proton from another substance

NH3 + HCl <--> NH4+ + Cl-

H2O + HCl <--> H3O+ + Cl-

NH3 + H2O <--> NH4+

+ OH-

• Remainder of acid is base• Complete reaction is proton exchange between sets• Extent of exchange based on strength

1-39

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

1-40

Relative Strengths of Acids and BasesConjugate Acid Conjugate BaseHClO4 ClO4

-

H2SO4 SO42-

HCl Cl-

H3O+ H2O

H2SO3 HSO3-

HF F-

HC2H3O2 C2H3O2-

HSO3- SO3

2-

H2S HS-

NH4+ NH3

HCO3 CO32-

H2O OH-

HS- S2-

OH O2-

H2 H-

Acid Strength

Base Strength

1-41

Dissociation Constants• Equilibrium expression for the behavior of acid

HA + H2O <--> A- + H3O+

Water concentration is constant

pKa=-logKa

• Can also be measured for base

Constants are characteristic of the particular acid or base

K [A ][H3O][HA][H2O]

Ka K[H2O] [A ][H3O]

[HA]

Kb [HB][OH ]

[B]

1-42

Dissociation Constants for Acids at 25°C

Acid Formula Ka

Acetic HC2H3O2 1.8E-5Hydrocyanic HCN 7.2E-10

Carbonic H2CO3 3.5E-7

HCO3- 5E-11

Nitrous HNO2 4.5E-4

Hydrosulfuric H2S 1E-7HS- 1E-14

Phosphoric H3PO4 7.5E-3

H2PO4- 6.2E-8

HPO42- 4.8E-13

Oxalic H2C2O4 5.9E-2

HC2O4- 6.4E-5

1-43

Dissociation Constants for Bases at 25°C

Base Formula Kb

Ammonia NH3 1.8E-5

Pyridine C5H5N 2.3E-9

Methylamine CH3NH2 4.4E-4

Protonation of amine group

1-44

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

Ka 10 (2*2.87)

0.1 10 2.87 1.84x10 5

1-45

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

• [OH-] = 1.33E-3 M, pOH = 2.87, pH ≈ 14-pOH ≈11.12

x2

0.1 x1.8x10 5

x 1.8E 5 (1.8E 5)2 4 *1.8E 6

2

1-46

Acid-Base Reactions

For Water

2 H2O <--> H3O+ + OH-

Water concentration remains constant, so for water:

Kw = [H3O+][OH-]= 1E-14 at 25°C

1-47

Common ion effect• Same products from different acids (hydronium ion)

0.2 M CH3COOH in 0.1 M HClWhat is the free acetic acid concentrationHCl 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 Ka 0.1x0.2

1-48

Hydrolysis Constants

• Reaction of water with metal ionCommon reactionEnvironmentally importantDependent upon metal ion oxidation state

• Mz+ + H2O <--> MOHz-1+ + H+

• Constants are listed for many metal ion with different hydroxide amounts

1-49

Buffers• Weak acid or weak base with conjugate salt• Acetate as example

Acetic acid, CH3COONaCH3COOH + H2O <--> CH3COO- + H3O+

If acid is addedhydronium reacts with acetate ion, forming

undissociated acetic acidIf base is added

Hydroxide reacts with hydronium, acetic acid dissociates to removed hydronium ion

large quantity huge quantity large quantity small quantity

1-50

Buffer Solutions

• Buffers can be made over a large pH range• Can be useful in controlling reactions and separations

Buffer rangeEffective range of bufferDetermined by pKa of acid or pKb of base

HA + H2O <--> A- + H3O-

Write as pHKa [A ][H3O]

[HA][H3O]

Ka[HA]

[A ]

1-51

Buffer Solutions

• The best buffer is when [HA]=[A-] largest buffer range for the conditionspH = pKa - log1

• For a buffer the range is determined by [HA]/[A-] [HA]/[A-] from 0.1 to 10Buffer pH range = pKa ± 1

Higher buffer concentration increase durability

pH pKa log[HA]

[A ]

1-52

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

We have 0.1 M each

Volume CH3COO- =1/6.62 L = 0.15 LAcid =1-0.15 =0.85 L

4.00 4.75 log[CH3COOH]

[CH3COO ]

log[CH3COOH]

[CH3COO ]0.75 [CH3COOH]

[CH3COO ]5.62

1-53

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 (pKa=4.75)Ka 1.8E 5

[H3O][0.11]

[0.09]

1-54

Equilibrium• Reactions proceed in the forward and reverse

direction simultaneously N2 + 3 H2 <--> 2 NH3

Initially contains nitrogen and hydrogenForward rate decreases as concentration

(pressure) decreasesAmmonia production increase reverse rateEventually, forward rate is equal to reverse rateNo net change in concentrationReaction still occurring

1-55

Equilibrium• Some reactions have a negligible reverse rate

Proceeds in forward directionReaction 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 stressconcentration, pressure, temperature

1-56

Equilibrium

• N2 + 3 H2 <--> 2 NH3 + 22 kcal

What is the shift due toIncreased temperature?Increased N2?

Reduction of reactor vessel volume?

1-57

Equilibrium Constants• For a reaction

aA + bB <--> cC + dD• At equilibrium the ratio of the product to reactants is a

constantThe constant can change with conditionsBy convention, constants are expressed as products over

reactants

• Conditions under which the constant is measured should be listedTemperature, ionic strength

K [C]c[D]d

[A]a[B]b

1-58

K C[C]cD[D]d

A[A]aB[B]b

Activities

• Strictly speaking, activities, not concentrations should be used

• 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

1-59

Activities

• Debye-Hückel (Physik Z., 24, 185 (1923))

ZA = charge of species Aµ = molal ionic strength

RA = hydrated ionic radius in Å (from 3 to 11)

First estimation of activity

logA 0.5085Za

2 10.3281RA

1-60

• Debye-Hückel term can be written as:

• Specific ion interaction theory Uses and extends Debye-Hückel

long range Debye-HückelShort range ion interaction term

ij = specific ion interaction term

Activities

D 0.5107 11.5

logi Z2D ij

log ß() logß(0) Zi2D ij

1-61

Activities• Pitzer

Binary (3) and Ternary (2) interaction parameters

Ca2+

K+

Al3+

Fe(CN)64-

1-62

Cm-Humate at pH 6

6.0

6.1

6.2

6.3

6.4

6.5

6.6

0.0 0.5 1.0 1.5 2.0 2.5 3.0

log

ß

sqrt ImIm

Experimental Data shows change in stability constant with ionic strength

Ion Specific Interaction Theory used

1-63

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

1-64

Constants

Conditional ConstantsAn experimental condition is written into

equation* Pu4+ + H2CO3 <--> PuCO3

2+ +2H+

Constant can vary with concentration, pHMust look at equation!

Often the equation is not explicitly written

1-65

Using Equilibrium Constants• Constants and balanced equation can be used to evaluate

concentrations at equilibrium2 HY <--> H2 + 2Y,K=4E-15If 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 ≈ 1K=4x3, x =1E-5, 2x=2E-5

K [H2 ][Y]2

[HY]2

K [x][2x]2

[1 2x]2

1-66

Realistic Case• Consider uranium in an aquifer

Species to consider includefree metal ion: UO2

2+

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 conditionsTotal uranium, total carbonate, pH, total humic

concentration

1-67

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)3- 22.6

UO2(OH)42- 23.1

(UO2)2OH3+ 11.0

(UO2)2(OH)2+ 22.0

UO2CO3 8.87

UO2(CO3)22- 16.07

UO2(CO3)34- 21.60

UO2HA(II) 6.16

UO2(OH)HA(I) 14.7±0.5

Other species may need to be considered. If the total uranium concentration is low enough, binary or tertiary species can be excluded.

1-68

Equations• Write concentrations in terms of species

[UO2]tot= UO2free+U-carb+U-hydroxide+U-humate [CO3

2-]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

1-69

U speciation with different CO2 partial pressure

0.0

0.2

0.4

0.6

0.8

1.0

2.0 4.0 6.0 8.0 10.0

Mo

le F

rac

tio

n o

f U

(VI)

Sp

ec

ies

pH

UO2

2+

UO2HA(II)

UO2OHHA(I) UO

2(OH)

2

UO2(OH)

3

-

0.0

0.2

0.4

0.6

0.8

1.0

2.0 4.0 6.0 8.0 10.0

Mo

le F

rac

tio

n o

f U

(VI)

Sp

ec

ies

pH

UO2

2+

UO2HA(II) UO

2OHHA(I)

UO2(CO

3)

2

2-

UO2(CO

3)

3

4-

0.0

0.2

0.4

0.6

0.8

1.0

2.0 4.0 6.0 8.0 10.0

Mo

le F

rac

tio

n o

f U

(VI)

Sp

ec

ies

pH

UO2

2+UO

2HA(II) UO

2OHHA(I)

UO2(CO

3)

2

2-

UO2(CO

3)

3

4-

0% CO2 1% CO2

10% CO2

1-70

Comparison of measured and calculated uranyl organic colliod

0.0

0.2

0.4

0.6

0.8

1.0

2.0 4.0 6.0 8.0 10.0pH

0%

0.035%

1%10%

100%

[U-c

ollo

id]

[U(V

I)]

tota

l

1-71

Energy terms

• Constants can be used to evaluate energetic of reactionFrom Nernst equation

∆G=-RTlnK∆G=∆H-T∆S

-RTlnK = ∆H-T∆SRlnK= - ∆H/T + ∆S

* Plot RlnK vs 1/T

1-72

64

66

68

70

72

74

76

0.003 0.0031 0.0032 0.0033 0.0034 0.0035

Temperature effect on Np-Humate stability

162432404856

Rln

ß

1/T (K)

Temp (°C)

²H = -22.2 ± 2.8 kJ/mol²G

298=-21.7 kJ/mol

²S=1.2±1.4 J/molK

1-73

Solubility Products

• Equilibrium involving a solid phaseAgCl(s) <--> Ag+ + Cl-

AgCl concentration is constantSolid activity and concentration is treated

as constantBy convention, reaction goes from solid to

ionic phase in solutionCan use Ksp for calculating concentrations

K [Cl ][Ag]

[AgCl]

Ksp K[AgCl] [Cl ][Ag]

1-74

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)

[Mg2] 1.2E 11

1E 81.2E 3

1-75

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 AgCl0.001 moles Ag+, 0.005 moles Cl-

[Cl-] = 0.004 moles/0.1 L =0.04 MKsp= 1E-10

[Ag+] =1E-10/0.04 =2.5E-9 M

1-76

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-5pOH = 4.96pH = 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

1-77

Limitations of Ksp

• Solid phase formation limited by concentrationbelow ≈1E-5/mL no visible precipitate forms

colloids

• formation of supersaturated solutionsslow kinetics

• Competitive reactions may lower free ion concentration

• Large excess of ligand may form soluble speciesAgCl(s) + Cl- <--> AgCl2

-(aq)

Ksp really best for slightly soluble salts