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Section 06

Section 06. General Concepts of Chemical Equilibrium. General Concepts: Chemical Equilibrium. Chemical Reactions: The Rate Concept aA + bB  cC + dD Rate f = k f [A] a [B] b Rate r = k r [C] c [D] d Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d Molar Equilibrium Constant K

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Section 06

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  1. Section 06 General Concepts of Chemical Equilibrium

  2. General Concepts: Chemical Equilibrium • Chemical Reactions: The Rate Concept • aA + bB  cC + dD • Ratef = kf[A]a[B]b • Rater = kr[C]c[D]d • Ratef = Rater • kf[A]a[B]b = kr[C]c[D]d • Molar Equilibrium Constant K • K = kf/ kr =([C]c[D]d)/([A]a[B]b) • Not Generally Valid, because reaction rates depend on mechanisms

  3. The rate of the forward reaction diminishes with time, while that of the backward reaction increases, until they are equal. A large K means the reaction lies far to the right at equilibrium. Fig. 6.1. Progress of a chemical reaction. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

  4. Equilibrium constants may be written for dissociations, associations, reactions, or distributions. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

  5. General Concepts: Chemical Equilibrium • Gibbs Free Energy & Equilibrium Constant • G = H – TS but H = E + PV • G = Gibbs Free Energy H = Enthalpy • T = Temperature S = Entropy • E = Internal Energy P = Pressure V = Volume • G = E + PV – TS but E = q – w • G = q – w + PV - TS • Derivative • dG = dq - dw + PdV + VdP – TdS – SdT

  6. General Concepts: Chemical Equilibrium • dG = dq - dw + PdV + VdP – TdS – SdT • Let’s Simplify by imposing some conditions on the reaction. • Constant Temperature: dT = 0  SdT = 0 • Reversible Reaction: dq = TdS • Expansion work only: dw = PdV • Then all terms except one cancel • dG = VdP 1mole of an ideal gas V = RT/P • dG = RTdP/P

  7. General Concepts: Chemical Equilibrium • Now lets integrate: dG = RTdP/P • Result: • G2-G1 = RTln(P2/P1) • make state 1 = standard state • G – Go = RTln(P/Po) but Po = 1 atm • Activity is defined: a = P/Po • G = Go + RTln(a)

  8. General Concepts: Chemical Equilibrium • General Expressions: • rR + sS  tT + uU • DG = tGT + uGU – rGR + sGS • Each Free Energy Term Expressed in Terms of Activity • tGT = tGTo + t RT ln aT • uGU= uGUo + u RT ln aU • rGR = rGRo + r RT ln aR • sGS = sGSo + s RT ln aS • DG = DGo + RT ln (aTt aUu/aRr aSs)

  9. General Concepts: Chemical Equilibrium • DG = DGo + RT ln (aTt aUu/aRr aSs) • At Equilibrium: DG = 0 • Reaction quotient Q = (aTt aUu/aRr aSs) = Ko • Where Ko is the thermodynamic equilibrium constant • 0 = DGo + RT ln Ko • ln Ko = - DGo/RT • Ko=e(- DGo/RT)

  10. Chemical Equilibrium • Review of Principles • Chemical reactions are never “complete” • Chemical reactions proceed to a state where ratio of products to reactants is constant • NH3 + HOH  NH4+ + OH- • [NH4+][OH-]/[NH3][HOH] = Kbo • If Kb<< 1 (little ionization) • H2SO4 + HOH  H3O+ + HSO4- • [H3O+][HSO4-] / [H2SO4][HOH] = Ka • If Ka>> 1 (mostly ionized)

  11. Chemical Equilibrium • Equilibrium • is not reached instantaneously • can be approached from either direction • is a dynamic state • amounts of reactants/products can be changed by “mass action” • (adding/ deleting products/reactants) • HCO3- + H+ CO2(g)+ HOH • Ke = [CO2][HOH]/[HCO3-][H+]

  12. Chemical Equilibrium • Equilibrium Constants • 2 A + 3 B  C + 4 D • Ke = [C][D]4/[A]2[B]3 • Concentrations [ ] : • molar for solutes • partial pressures (atm) for gases • [1.0] for pure liquid, solid, or solvent

  13. Chemical Equilibrium • Important Equilibria in Analytical Chemistry • Solubility: • AgCl(s)  Ag+ + Cl- • Ag3AsO4(s)  3 Ag+ + AsO43- • BaSO4(s)  Ba2+ + SO42- • Ksp(AgCl) = [Ag+][Cl-] = 1.0 x 10-10 • Ksp(Ag3AsO4) = [Ag+]3[AsO43-] = 1.0 x 10-22 • Ksp(BaSO4) = [Ba2+][SO42-] = 1.0 x 10-10

  14. Chemical Equilibrium • Important Equilibria in Analytical Chemistry • Autoprotolysis: • HOH + HOH  H3O+ + OH- • Ke = [H3O+][OH-]/[HOH]2 • Ke[HOH]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 @ 25oC • In pure water @ 25oC [H3O+] = [OH-] = 10-7 • Acid Dissociation: • H2CO3 + HOH  H3O+ + HCO3- • Ka = [H3O+][HCO3-]/[H2CO3] = 4.3 x 10-7

  15. Chemical Equilibrium • Important Equilibria in Analytical Chemistry • H2CO3 + HOH  H3O+ + HCO3- • acid1 base1 • Dissociation of Conjugate Base1: • HCO3- + HOH  H3O+ + CO32- • Ka(HCO3-) = [H3O+][CO32-]/[HCO3-] = 4.8 x 10-11 • Hydrolysis of Conjugate Base1: • HCO3- + HOH  H2CO3 + OH- • Kb(HCO3-) = Kw/Ka(H2CO3) = 10-14/4.3x 10-7 • Kb(HCO3-) = 2.3 x 10-8

  16. Chemical Equilibrium • Important Equilibria in Analytical Chemistry • Base Dissociation: • NH3 + HOH  NH4+ + OH- • Kb(NH3) = [NH4+][OH-]/[NH3] = 1.75 x 10-5 • Hydrolysis of Salts: • NH4Cl(s)  NH4+ + Cl- • NH4+ + HOH  NH3 + H3O+ • Ka(NH4+) = Kw/Kb(NH3) = 10-14/1.75 x 10-5 • Ka(NH4+) = 5.7 x 10-10

  17. Chemical Equilibrium • Some Useful Calculations • Common Ion Effects on Solubility: • What is the solubility of BaF2 in pure water? • What is the solubility of BaF2 in 0.1 M NaF?

  18. Chemical Equilibrium • Some Useful Calculations • pH of Weak Acid or Base Solutions:

  19. Chemical Equilibrium Electrolyte Effects • Electrolytes: • Substances producing ions in solutions • Can electrolytes affect chemical equilibria? • (A) “Common Ion Effect”  Yes • Decreases solubility of BaF2 with NaF • F- is the “common ion” • (B) No common ion: “inert electrolyte effect”or “diverse ion effect” • Add Na2SO4 to saturated solution of AgCl • Increases solubility of AgCl Why???

  20. Activity and Activity Coefficients • Activity of an ion, ai = Ciƒi • Ci = concentration of the ion • ƒi= activity coefficient( @ Ci < 10-4M )= 1 • Ionic Strength, m = ½SCiZi2 • Zi = charge on each individual ion.

  21. Activity and Activity Coefficients • Calculation of Activity Coefficients • Debye-Huckel Equation: • -log ƒi = 0.51Zi2(m)½/(1+0.33ai (m)½) • ai = ion size parameter in angstrom (Å) • 1 Å = 100 picometers (pm, 10-10 meters) • Limitations: singly charged ions = 3 Å -log ƒi = 0.51Zi2(m)½/(1+(m)½)

  22. Chemical Equilibria Electrolyte Effects • Diverse ion (Inert) electrolyte effect • For m < 0.1 M, electrolyte effect depends on m only, NOT on the type of electrolyte • Solute activities: • ax = activity of solute X • ax = [X]x • x= activity coefficient for X • Asm  0, x  1, ax  [X]

  23. Chemical Equilibria Electrolyte Effects • Diverse Ion (Inert Electrolyte) Effect: • Add Na2SO4 to saturated solution of AgCl • Kspo = aAg+ . aCl- = 1.75 x 10-10 • At high concentration of diverse (inert) electrolyte: higher ionic strength, m • aAg+< [Ag+] ; aCl-< [Cl-] • aAg+ . aCl- < [Ag+] [Cl-] • Kspo < [Ag+] [Cl-] ;  Kspo < [Ag+] = solubility • Solubility = [Ag+] >  Kspo

  24. Chemical Equilibria Electrolyte Effects • “Diverse ion (Inert) electrolyte effect” • Is dependent on parameter called “ionic strength (m)” • m = (1/2) {[A]ZA2 + [B]ZB2 + … + [Y]Zy2} • 0.1 M Na2SO4 ; [Na+] = 0.2M [SO4] = 0.1M • m = (1/2) {[A]ZA2 + [B]ZB2} • m = (1/2) {[0.2](1+)2 + [0.1](2-)2}= 0.3M

  25. Chemical Equilibria Electrolyte Effects • Solute activities: • When m is notzero, ax= [X]x • Equilibrium effects: • mM + xX  zZ • Ko=(az)z/(am)m(ax)x • Ko=([Z]Z)z/([M]M)m([X]x)x • Ko={([Z])z/([M])m([X])x }{Zz/Mm xx} • Ko = K {Zz/Mm xx} • K = Ko {Mm xx / Zz}

  26. The Diverse Ion Effect • The Thermodynamic Equilibrium Constant and Activity Coefficients • thermodynamic equilibrium constant, Ko • case extrapolated to infinite dilution • At infinite dilution, activity coefficient, ƒ= 1 • Dissociation AB  A+ + B- • Ko = aA aB/aAB = [A+] ƒA. [B-] ƒB / [AB] ƒAB • Ko = K (ƒA. ƒB / ƒAB) • K = Ko (ƒAB/ ƒA. ƒB )

  27. Calculation of Activity Coefficients Debye-Huckel Equation: -log ƒx= 0.51Zi2(m)½/(1+0.33ai (m)½) Where ax = effective diameter of hydrated ion, X (in angstrom units, 10-8cm), Å Chemical Equilibria Electrolyte Effects

  28. Chemical Equilibrium Electrolyte Effects • Equilibrium calculations using activities: • Solubility of PbI2 in 0.1M KNO3 • m = 0.1 = {0.1(1+)2 + 0.1(1-)2}/2 (ignore Pb2+,I-) • ƒPb = 0.35 ƒI = 0.76 • Kspo= (aPb)1(aI)2 = ([Pb2+]Pb )1([I-]I )2 • Kspo = ([Pb2+] [I-]2)(Pb I2) = Ksp (Pb I2) • Ksp = Kspo / (Pb I ) • Ksp = 7.1 x 10-9 /((0.35)(0.76)2) = 3.5 x 10-8 • (s)(2s)2 = Ksp s = (Ksp/4)1/3 s =2.1 x 10-3 M • Note: If s = (Kspo/4)1/3 then s =1.2 x 10-3M • Solubility calculation difference approx. –43%

  29. Multiple Chemical EquilibriaCompositional Calculations • Setting up the problem: • Write balanced equations for all equilibria • Write Ke expressions and find values • Write mass and charge balance equations • Write expression for sought for substance • Determine in No. independent equations (N) at least equals No. of unknowns (U) • (if N < U can approximations reduce U?) • Make approximations to simplify math • Solve set of equations for all unknowns • Check validity of assumptions • (re-solve with second approximation if needed)

  30. Multiple Chemical Equilibria • Dissolve NaHCO3 in water: • NaHCO3 Na+ + HCO3- • HCO3- + HOH  H2CO3 + OH- Ke = Kb = Kw/Ka1 • HCO3- + HOH  H3O+ + CO32-Ke = Ka2 • HOH + HOH  H3O+ + OH- Ke = Kw • 5 chemical species affected by 3 equilibria • Equilibrium constants do NOT change with chemical additions/ deletions • Add Ba2+: Ba2+ + CO32-  BaCO3(s) Ke = Ksp • (Now there are 6 species affected by 4 equilibria) • Note: For Polyprotic Acids (HNA): K(step) = Ka1, Ka2,-- KaN • H2A + HOH  H3O+ + HA- Ke = Ka1 • HA- + HOH  H3O+ + A2- Ke = Ka2

  31. Multiple Chemical EquilibriaCompositional Calculations • What are concentrations of individual species? • For N species, M equilibria • Need, N independent algebraic expressions: • Equilibrium expressions (M < N) • Mass balance statements • Charge Balance statements

  32. Multiple Chemical EquilibriaCompositional Calculations • Mass Balance Equations: • Relate equilibrium concentrations of species • Stoichiometric relationships • How the solution was prepared • What kinds of equilibria exist • E.g. 0.1 M HNO2(CHA = 0.1 M) • HNO2 + HOH  H3O+ + NO2- Ke = Ka • CHA (x) (x) • HOH + HOH  H3O+ + OH- Ke = Kw • (w) (w) • CHA = [HNO2] + [NO2-] all forms of “HNO2” • [H3O+] = [OH-] + [NO2-] = w + x [H3O+] from 2 sources

  33. Multiple Chemical EquilibriaCompositional Calculations • Charge Balance Equations: • In any electrolyte solution, • amt. of positive charge = amt. of negative charge • solution charge for each species = [conc.][charge/ion] • E.g. for solution MgCO3 • MgCO3 _ Mg2+ + CO32- leads to equilibria: • CO32- + HOH  HCO3- + OH- • HCO3- + HOH  H2CO3 + OH- • Charge Balance: • 2[Mg2+] + [H3O+] = 2[CO32-] + [HCO3-] + [OH-]

  34. Systematic Approach to Equilibrium Calculations • How to Solve Any Equilibrium Problem • 1. Write balanced chemical reactions • 2. Write equilibrium constant expressions • 3. Write all mass balance expressions • 4. Write the charge balance expression • 5. Equations >= Chemical Species sol possible • 6. Make assumptions where possible • 7. Calculate answer • 8. Check validity of assumptions

  35. Multiple Chemical EquilibriaExample Problem #1 • What is pH of Mg(OH)2 solution ? (assume ax= [X]) • Equilibria: • Mg(OH)2(s) Mg2+ + 2 OH- Ksp= [Mg2+][OH-] = 1.8 x 10-11 • HOH + HOH  H3O+ + OH- Ke = Kw = 1.0 x 10-14 • Mass Balance: [OH-] = [H3O+] + 2 [Mg2+] • Charge Balance: [OH-] = [H3O+] + 2 [Mg2+] • Expression for unknown : [H3O+] = Kw/ [OH-]

  36. Multiple Chemical EquilibriaExample Problem #1 • What is pH of Mg(OH)2 solution ? (assume ax= [X]) • Expression for unknown : [H3O+] = Kw/ [OH-] • Solution: Assume [H3O+] << 2 [Mg2+] ; [OH-] = 2 [Mg2+] • Substitute into Ksp, Ksp = ([OH-] /2)([OH-])2 • [OH-]3 = 2 Ksp ; [OH-] = (2 Ksp)1/3 = 3.3 x 10-4 M • [H3O+] = Kw/(3.3 x 10-4) = 3.0 x 10-11 M • pH = -log [H3O+] = 10.52 (2 sig figs) • Note: original approximation was OK! • i.e. [H3O+] << [OH-] (3.0 x 10-11 << 3.3 x 10-4) • Assumed [OH-] = 2 [Mg2+] ; [H3O+] << 2 [Mg2+] • Assumed [H3O+] << [OH-]  OK!

  37. Multiple Chemical EquilibriaExample Problem #2 • What is pH of 0.1 M Na3PO4 solution? (Cs = 0.1 M) • Na3PO4(s)  3Na+ + PO43- • Equilibria: • (1) PO43- + HOH  HPO42- + OH- Kw/Ka3 = 2.38 x 10-2 • (2) HPO42- + HOH  H2PO4- + OH- Kw/Ka2 = 1.6 x 10-9 • (3) H2PO4- + HOH  H3PO4 + OH- Kw/Ka1 = 1.4 x 10-12 • (4) HOH + HOH  H3O+ + OH- Kw = 1.0 x 10-14 • Mass Balance Equations: • (5) Cs= [PO43-] + [HPO42-] + [H2PO4-] + [H3PO4] • (6) [OH-] = [H3O+] + [HPO42-] + [H2PO4-] + [H3PO4] • (*) Ka1 = 7.1 x 10-3 Ka2 = 6.3 x 10-8 Ka3 = 4.2 x 10-13

  38. Multiple Chemical EquilibriaExample Problem #2 • What is pH of 0.1 M Na3PO4 solution? (Cs = 0.1 M) • Charge Balance: • (7) [H3O+] + [Na+] = [OH-] + [H2PO4-] + 2 [HPO42-] + 3 [PO43-] • Note: [Na+] = 3 Cs • Is problem solvable? • 6 unknowns [H3O+] ,[OH-],[H2PO4-],[HPO42-],[PO43-], [H3PO4] • 7 equations (see 1-7) • Thus, the problem should be solvable.

  39. Multiple Chemical EquilibriaExample Problem #2 (solution) • What is pH of 0.1 M Na3PO4 solution? (Cs = 0.1 M) • Assume [H3O+] + [H2PO4-] + [H3PO4] << [HPO42-] • Then from equation (6): [OH-] = [HPO42-] = x • Also Assume [HPO42-] + [H2PO4-] + [H3PO4] << [PO43-] • Then from equation (5): [PO43-] = Cs = 0.1 M • equation (1): [HPO42-] [OH-]/ [PO43-]= Kw/Ka3 = 2.38 x 10-2 • Thus x2/Cs = 2.38 x 10-2 x = [OH-] = 4.9 x 10-2 • [H3O+] = Kw/[OH-] = 2.0 x 10-13M • pH = 12.69 (2 sig fig) • Note: Check assumptions

  40. Multiple Chemical EquilibriaExample Problem #2 • Check Assumptions:

  41. Solve 0.70x2 + 0.21x – 0.30 = 0. First prepare a spreadsheet containing: • Cells containing the constants a, b, and c to be used in the formula, 0.70, 0.21, -0.30: (B3, B4, B5). • Cell for the variable, x, to be solved for: ($C$7). • Cell containing the formula 0.70x2 +0.21x – 0.30 (ES) (do not enter = 0): • =B3*C7^2+B4*C7+B5 (continued next slide) Excel Solver to solve the quadratic formula for Example 6.1. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

  42. Click on Solver to open the parameters dialogue box. Need to enter 3 parameters: • Set Target Cell: enter the cell containing the formula (E5). • Equal To: enter the value the equation is set to (0). • By Changing Cells: enter the cell containing the variable, x (C7). • Then click Solve. The variable x will be changed by iteration until the equation equals zero. (continued next slide) Excel Solver to solve the quadratic formula for Example 6.1. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

  43. Click on Solve, and you receive a message that “Solver found a solution.” The answer is x = 0.10565. The formula after iteration is equal to –8E-08, essentially equal to zero. The solved quadratic formula. ©Gary Christian, Analytical Chemistry, 6th Ed. (Wiley)

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