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2-1 Thermodynamics and kinetics Thermodynamic laws Half-cell reactions Kinetics Acid-Base Equilibrium calculations §Speciation calculation from complexation.

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Presentation on theme: "2-1 Thermodynamics and kinetics Thermodynamic laws Half-cell reactions Kinetics Acid-Base Equilibrium calculations §Speciation calculation from complexation."— Presentation transcript:

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2 2-1 Thermodynamics and kinetics Thermodynamic laws Half-cell reactions Kinetics Acid-Base Equilibrium calculations §Speciation calculation from complexation constants Provide review of concepts for applications to radiochemistry

3 2-2 Thermodynamic terms Heat Capacity (C p ) §Heat required to raise one gram of substance 1 °C Enthalpy (∆H) Energy of a system (heat content) §Internal energy, volume, pressure §Accounts for energy transferred to environment by expansion or heating ∆H = ∆H products - ∆H reactants Exothermic reactions, negative ∆H §Negative ∆H tend to be spontaneous §∆H products - ∆H reactants à2 H + + CO 3 2- CO 2 + H 2 O à-393.5 + (-285.8)-(-677.1+2(0)) = -2.2 kj/mol

4 2-3 Thermodynamic terms Endothermic §Reaction requires energy (at 25 °C) 2 HgO + 181.70 kj 2 Hg + O 2 Enthalpy (∆H) Energy of a system (heat content) §Internal energy, volume, pressure §Accounts for energy transferred to environment by expansion or heating ∆H = ∆H products - ∆H reactants Exothermic reactions, negative ∆H §Negative ∆H tend to be spontaneous

5 2-4 Entropy (∆S) and Gibbs Free Energy (∆G) 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 àK is equilibrium constant àActivity at unity Compound∆G° (kJ/mol) at 298.15 K H 2 O-237.129 OH - (aq) -157.244 H + (aq) 0 H 2 O  H + +OH - §What is the constant for the reaction? àProducts-reactants 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, K w = [H + ][OH - ]

6 2-5 Thermodynamic Laws 1st law of thermodynamics §Energy is conserved in a system àCan be changed or transferred §Heat and work are energy transfer à∆E = q (heat absorbed) + w (work) 2nd law of thermodynamics §Reactions tend towards equilibrium àIncrease in entropy of a system §Spontaneous reaction for -∆G à∆G = 0, system at equilibrium 3rd law of thermodynamics §Entropies of pure crystalline solids are zero at 0 K §Defines absolute zero

7 2-6 Half-cell potentials Cell reaction for §Zn and Fe 3+/2+ at 1.0 M §Write as reduction potentials  Fe 3+ + e - Fe 2+  °=0.77 V  Zn 2+ + 2e - Zn  °=-0.76 V Reduction potential for Fe 3+ is larger §Fe 3+ is reduced, Zn is oxidized in reaction Overall balanced equation  2Fe 3+ +Zn 2Fe 2+ + Zn 2+  °=0.77+0.76=1.53 V Use standard reduction potential Application of Gibbs Free Energy If work is done by a system  ∆G = -  °nF (n= e - ) Find ∆G for Zn/Cu cell at 1.0 M  Cu 2+ + Zn Cu + Zn 2+   °=1.10 V §2 moles of electrons (n=2) à∆G =-2(96487C/mole e - )(1.10V) à∆G = -212 kJ/mol

8 2-7 Kinetics and Equilibrium Kinetics and equilibrium important concepts in examining and describing chemistry §Identify factors which determine rates of reactions àTemperature, pressure, reactants, mixing §Describe how to control reactions §Explain why reactions fail to go to completion §Identify conditions which prevail at equilibrium General factors effecting kinetics §Nature of reactants §Effective concentrations §Temperature §Presence of catalysts §Number of steps

9 2-8 Rate Law Concentration of reactant or product per unit time Effect of initial concentration on rate can be examined §rate = k[A] x [B] y §rate order = x + y §knowledge of order can help control reaction §rate must be experimentally determined Rate=k[A] n ; A=conc. at time t, A o =initial conc., X=product conc. Orderrate equationk 0[A 0 ]-[A]=kt, [X]=ktmole/L sec 1 ln[A 0 ]-ln[A]=kt, ln[A 0 ]-ln([A o ]-[X])=kt 1/sec 2L/mole sec 3 L 2 /mole 2 sec

10 2-9 Acid-Base Equilibria Brønsted Theory of Acids and Bases §Acid àSubstance which donates a proton §Base àAccepts proton from another substance NH 3 + HCl NH 4 + + Cl - H 2 O + HCl H 3 O + + Cl - NH 3 + H 2 O NH 4 + + OH - Remainder of acid is base Complete reaction is proton exchange between sets Extent of exchange based on strength Water can act as solvent and reactant

11 2-10 Dissociation Constants Equilibrium expression for the behavior of acid HA + H 2 O A - + H 3 O + Water concentration is constant pK a =-logK a Can also be measured for base Constants are characteristic of the particular acid or base AcidFormulaK a AceticHC 2 H 3 O 2 1.8E-5 CarbonicH 2 CO 3 3.5E-7 HCO 3 - 5E-11 PhosphoricH 3 PO 4 7.5E-3 H 2 PO 4 - 6.2E-8 HPO 4 2- 4.8E-13 OxalicH 2 C 2 O 4 5.9E-2 HC 2 O 4 - 6.4E-5

12 2-11 Buffer Solutions Buffers can be made over a large pH range Can be useful in controlling reactions and separations §Buffer range àEffective range of buffer àDetermined by pK a of acid or pK b of base HA + H 2 O A - + H 3 O - Write as pH The best buffer is when [HA]=[A - ] §largest buffer range for the conditions §pH = pK a - log1 For a buffer the range is determined by [HA]/[A - ] §[HA]/[A - ] from 0.1 to 10 §Buffer pH range = pK a ± 1 §Higher buffer concentration increase durability

13 2-12 Hydrolysis Constants Reaction of water with metal ion §Common reaction §Environmentally important §Strength dependent upon metal ion oxidation state 2 H 2 O H 3 O + + OH - §Water concentration remains constant, so for water: §K w = [H 3 O + ][OH - ]= 1E-14 at 25°C Metal ions can form hydroxide complexes with water M z+ + H 2 O MOH z-1+ + H + Constants are listed for many metal ion with different hydroxide amounts §Database at: http://www.escholarship.org/uc/item/9427347g http://www.escholarship.org/uc/item/9427347g

14 2-13 Equilibrium 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 àconcentration, pressure, temperature N 2 + 3 H 2 2 NH 3 + 22 kcal § What is the shift due to àIncreased temperature? àIncreased N 2 ? àReduction of reactor vessel volume?

15 2-14 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 Conditions under which the constant is measured should be listed §Temperature, ionic strength

16 2-15 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 Debye-Hückel (Physik Z., 24, 185 (1923)) Z A = charge of species A µ = molal ionic strength R A = hydrated ionic radius in Å (from 3 to 11) First estimation of activity Activities

17 2-16 Debye-Hückel term can be written as: Specific ion interaction theory §Uses and extends Debye-Hückel àlong range Debye-Hückel àShort range ion interaction term  ij = specific ion interaction term Pitzer §Binary (3) and Ternary (2) interaction parameters Activities

18 2-17 Ca 2+ K+K+ Al 3+ Fe(CN) 6 4- Experimental Data Cm-Humate shows change in stability constant with ionic strength Ion Specific Interaction Theory used

19 2-18 Constants Constants can be listed by different names §Equilibrium constants (K) àReactions involving bond breaking *2 HY H 2 + 2Y §Stability constants (ß), Formation constants (K) àMetal-ligand complexation *Pu 4+ + CO 3 2- PuCO 3 2+ *Ligand is written in deprotonated form §Conditional Constants àAn experimental condition is written into equation *Pu 4+ + H 2 CO 3 PuCO 3 2+ +2H + ËConstant can vary with concentration, pH Must look at equation!

20 2-19 Realistic Case Consider uranium in an aquifer §Species to consider include àfree metal ion: UO 2 2+ àhydroxides: (UO 2 ) x (OH) y àcarbonates: UO 2 CO 3 àhumates: UO 2 HA(II), UO 2 OHHA(I) §Need to get stability constants for all species àUO 2 2+ + CO 3 2- UO 2 CO 3 §Know or find conditions àTotal uranium, total carbonate, pH, total humic concentration

21 2-20 Stability constants for selected uranium species at 0.1 M ionic strength Specieslogß UO 2 OH + 8.5 UO 2 (OH) 2 17.3 UO 2 (OH) 3 - 22.6 UO 2 (OH) 4 2- 23.1 (UO 2 ) 2 OH 3+ 11.0 (UO 2 ) 2 (OH) 2+ 22.0 UO 2 CO 3 8.87 UO 2 (CO 3 ) 2 2- 16.07 UO 2 (CO 3 ) 3 4- 21.60 UO 2 HA(II)6.16 UO 2 (OH)HA(I)14.7±0.5 Other species may need to be considered. If total uranium concentration is low enough, binary or tertiary species can be excluded.

22 2-21 Equations Write concentrations in terms of species §[UO 2 ] tot = UO 2free +U-carb+U-hydroxide+U-humate §[CO 3 2- ] free =f(pH) §[OH - ] = f(pH) §[HA] tot = UO 2 HA + UO 2 OHHA+ HA free Write the species in terms of metal, ligands, and constants §[(UO 2 ) x A a B b ] = 10 -(xpUO2+apA+bpB-log(UO2)xAaBb) §pX = -log[X] free à[(UO 2 ) 2 (OH) 2 2+ ]=10 -(2pUO2+2pOH-22.0) Set up equations and solve for known terms

23 2-22 U speciation with different CO 2 partial pressure 0% CO 2 1% CO 2 10% CO 2

24 2-23 Comparison of measured and calculated uranyl organic colloid

25 2-24 Excel spreadsheets CHESS Program

26 2-25 Energy terms Constants can be used to evaluate energetic of reaction §From Nernst equation à∆G=-RTlnK §∆G=∆H-T∆S à-RTlnK = ∆H-T∆S àRlnK= - ∆H/T + ∆S *Plot RlnK vs 1/T

27 2-26 Solubility Products Equilibrium involving a solid phase §AgCl(s) Ag + + Cl - §AgCl concentration is constant àSolid activity and concentration is treated as constant àBy convention, reaction goes from solid to ionic phase in solution §Can use K sp for calculating concentrations in solution

28 2-27 Solubility calculations K sp of UO 2 = 10 -52. What is the expected U 4+ concentration at pH 6. Generalize equation for any pH §Solubility reaction: àUO 2 + 2 H 2 O  U(OH) 4  U 4+ + 4 OH - §K sp = [U 4+ ][OH - ] 4 §[U 4+ ]= K sp /[OH - ] 4 àpOH + pH =14 àAt pH 6, pOH = 8, [OH - ]=10 -8 §[U 4+ ]= 10 -52 /[10 -8 ] 4 = 10 -52 /10 -32 = 10 -20 M §For any pH à[U 4+ ]= 10 -52 /[10 -(14-pH)*4 ] àLog [U 4+ ]= -52+((14-pH)*4)

29 2-28 Limitations of K sp Solid phase formation limited by concentration §below ≈1E-5/mL no visible precipitate forms àcolloids formation of supersaturated solutions §slow kinetics Competitive reactions may lower free ion concentration Large excess of ligand may form soluble species §AgCl(s) + Cl - AgCl 2 - (aq) K sp really best for slightly soluble salts

30 2-29 Overview Understand heats of reactions §Enthalpy, entropy, Gibbs free energy §Reaction data from constituents Understand half-cell reactions §Nernst Equation Kinetics §Influence of reaction conditions Equilibrium and constants §Use to develop a speciation spreadsheet

31 2-30 Questions What is the difference between 1 st and 2 nd order kinetics? What can impact reaction rates? How can a compound act as a base and acid? Provide an example. What does the dissociation constant of an acid provide? Provide the speciation of acetic acid at pH 3.5, 4.5, and 5.5. What are the species from carbonic acid at pH 4.0, 6.0, and 8.0? Set up the equations to describe the speciation of uranyl, the uranyl monocarbonate, and the uranyl dicarbonate.

32 2-31 Pop Quiz Show the relationship between Gibbs free energy, enthalpy, and entropy


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