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Equilibrium The Constant Chapter 6 Chemical Equilibrium Dr. E. Binamira-Soriaga Department of Chemistry Texas A&M University For the general reversible reaction: aA+bB cC+dD reactants products The equilibrium constant K at a given temperature is written as: Cc Dd K= Aa Bb equilibrium concentrations Homogeneous vs Heterogeneous Equilibria Properties of K Dimensionless quantity Reaction-specific Constant for a given temperature Independent of initial concentrations Measured experimentally Homogeneous: single-phase reaction 2H2S(g) + 3O2(g) 2H2O(g) + 2SO2(g) H 2O 2 S O 2 2 K= H 2S 2 O 2 3 2HgO(s) 2Hg(l) + O2(g) K = O2 Heterogeneous: multi-phase reaction The Magnitude of K Variations in K A measure of the extent of reaction Dependent on nature of reaction and temperature Consider: N2(g) + 3H2(g) 2NH3(g) K= NH3 N2 2 H2 3 = 3.6 x 10 8 The value of K d epends on the form of the b alanced equation for t he reaction If the reaction is reversed, the new K is the reciprocal of the original K: 2HBr(g) + Cl2(g) 2HCl(g) + Br2(g) HCl 2 B r2 K= = 4 x 10 4 HBr 2 Cl 2 2HCl(g) + Br2(g) 2HBr(g) + Cl2(g) K = HBr 2 C l 2 1 = = 2.5 x 1 0 -5 K HCl 2 Br 2 K >> 1: [Products] >> [Reactants] K << 1: [Products] << [Reactants] K = 1: [Products] ~ [Reactants] Variations in K Variations in K If the reaction i s m ultiplied by a f actor n, t he new K is obtained by raising the original K to the power n: N2(g) + 3H2(g) 2NH3(g) K = 3.5 x 108 1/2N2(g) + 3/2H2(g) NH3(g) K = (K)1/2 = (3.5 x 108)1/2 = 1.9 x 104 If reactions are summed up, the overall K is the product of individual Ks: N2 + 2O2 2NO2 2NO N2 + O2 2NO + O2 2NO2 Kp1= 6.9 x 1019 Kp2 = 2.3 x 1030 Kp3 = Kp1 x Kp2 Kp3 = 1.6 x 1012 Example The Reaction Quotient Given: H2O H+ + O HKw = [H+][OH-] = 1.0 x 10-14 at 25o C For the general reaction: aA+bBcC+dD A mass action expression or reaction quotient Q can be set up at any time during a reaction: Cc Dd Q= AaB b not necessarily equilibrium concentrations NH3(aq) + H2O NH4+ + OHKNH3 = 1.8 x 105 Find K for the reaction: NH4+ NH3(aq) + H+ Progress of a Reaction: Q vs K K= Factors That Affect Chemical Equilibrium Cc Dd Aa Bb Q = K: Q < K: Q > K: System at equilibrium; no further net reaction Forward reaction proceeds until equilibrium is established Reverse reaction proceeds until equilibrium is established A system at equilibrium will r emain in equilibrium until it is d isturbed Disturbances (stresses applied) include changes in concentration, pressure, volume, temperature, and the addition of catalysts The system will strive to reattain a new equilibrium state Le Chateliers Principle Le Chateliers Principle Suppose you have the following system: 2Cr3+ + BrO3 + 4H2O Br + Cr2O7 2 + 8H+ green orange for which the equilibrium constant is: K= If a stress is applied to a system at equilibrium (Q K), it will proceed in the direction that will reduce the stress until a new equilibrium state is reestablished (Q = K) Reaction direction is opposite of the stress applied Br C r 2 O7 2B rO3 H+ 8 2 Cr3+ = 1 x 10 11 What effect will the addition or removal of the dichromate ion have on the system? Conversion to A Complex Ion Copyright 1995 by Saunders College Publishing Le Chateliers Principle: Effect of Temperature Change Copyrght 1995 by Saunders College Publishing K of an endothermic reaction increases if T is raised K of an exothermic reaction decreases if T is raised AgCl(s) Ag+(aq) + Cl(aq) Ag+(aq) + 2NH3(aq) [Ag(NH3)2]+(aq) N2O4(g) + heat 2NO2(g) Solubility Equilibrium This dynamic equilibrium occurs in a saturated solution of Hg2Cl2: HO Hg2Cl2(s) 2 Hg22+(aq) + 2 Cl (aq) The equilibrium constant for this system is called the solubility product constant, Ksp, and is given by: Ksp = [Hg22+][Cl]2 = 1.2 x 1018 2+] and [Cl] are equilibrium [Hg2 concentrations expressed in mol/L; these concentrations vary depending on the conditions Solubility Product Constants In general, the equilibrium established upon dissolution of a slightly soluble compound is: MyXz(s) yMz+(aq) + zXy(aq) Ksp = [M z+]y [Xy]z equilibrium concentrations The solubility product constant Ksp is constant at constant temperature for a saturated solution of the compound Solubility Defined Some Terms to Remember Solubility: the maximum amount of solid that will dissolve in a given volume of solvent Weight solubility = g solute L saturated solution Molar solubility = moles solute L saturated solution Soluble in a particular solvent: solid dissolves to the extent of at least 0.02 mol/L; solids with lower solubilities are slightly soluble or insoluble unsaturated solution: amount dissolved < solubility saturated solution: amount dissolved = solubility supersaturated solution: amount dissolved > solubility Ksp and Molar Solubility Substance BaSO4 AgCl Ag2CrO4 Ksp* 1.1 x 10 10 1.8 x 10 10 9.0 x 10 12 Solubility(mol/L) 1.1 x 10 5 1.34 x 10 5 1.3 x 10 4 Solubility and the Common-Ion Effect MgF2 in pure water: MgF2(s) Mg2+(aq) + 2F(aq) solubility = X mol/L MgF2 in NaF: MgF2(s) Mg2+(aq) + 2F(aq) NaF(s) Na+(aq) + F(aq) solubility < X mol/L Solubility of a compound is lower in solution than in pure water when the solution contains an ion common to the compound *For aqueous solutions at 25oC For compounds with same stoichiometry, one with larger Ksp has higher solubility Complex Ion Formation Complex Ion Formation Complex ion formation results from the reaction of a Lewis acid (electron-pair acceptor) and a Lewis base (electron-pair donor), for example: Pb2+(aq) + 4I(aq) [PbI4]2(aq) Lewis Lewis complex ion acid base The bond between a Lewis acid and a Lewis base is called a dative or coordinate covalent bond Complex ion formation can be represented as stepwise reactions with stepwise formation constants K1 to K4 in the case of PbI42-: Pb2+ + I- PbI+ K1 = [PbI+]/[Pb2+][I-] PbI+ + I- PbI2(aq) K2 = [PbI2]/[PbI+][I -] PbI2(aq) + I- PbI3-(aq) K3 = [PbI3-]/[PbI2][I-] PbI3-(aq) + I- PbI42-(aq) K4 = [PbI42-]/[PbI3-][I-] Cumulative or overall formation constants are given by: Pb2+ + I- PbI+ K1 = [PbI+]/[Pb2+][I-]= 1.0 x 102 Pb2+ + 2I- PbI2 (aq) 2 = [PbI2]/[Pb2+][I-]2 = 1.4 x 103 Pb2+ + 3I- PbI3-(aq) 3 = [PbI3-]/[Pb2+][I -]3 = 8.3 x 103 Pb2+ + 4I- PbI42-(aq) 4 = [PbI42-]/[Pb2+][I -]4 = 3.0 x 104 where 4 = K1K2K3K4 All equilibria are satisfied simultaneously Complex Ion Formation Effect of Complex Ion Formation on Solubility Consider the equilibrium: PbI2(s) Pb2+ + 2IKsp = [Pb2+][I-]2 = 7.9 x 10-9 At high concentrations of I- solid PbI2 dissolves: Pb2+ + I- PbI+ K1 = [PbI+ ]/[Pb2+][I-]2 = 1.0 x 102 Pb2+ + 2I- PbI2(aq) 2 = [PbI2]/[Pb2+][I-]2 = 1.4 x 103 Pb2+ + 3I- PbI33 = [PbI3]/[Pb2+][I-]3 = 8.3 x 103 Pb2+ + 4I- PbI4 24 = [PbI42-]/[Pb2+][I-]4 = 3.0 x 104 In general, complex ion formation can be represented as stepwise reactions with stepwise formation constants K1 to Kn M + X MX K1 = [MX]/[M][X] MX + X MX2 K2 = [MX2]/[MX][X] MXn-1 + X MXn Kn = [MXn]/[MXn-1][X] Cumulative or overall formation constants are given by: M + 2X MX2 2 = [MX2]/[M][X]2 M + nX MXn n = [MXn]/[M][X]n where n = K1K2K3.Kn Autoprotolysis of Water Acids and Bases According to Brnsted-Lowry Pure water ionizes slightly: H2O(l) + H2O(l) H3O+(aq) + OH(aq) The eq uilib rium constant at 25oC for this ioniz ation (au top rotolysis) is: Kw = [H3O+][OH] = 1.0 x 1014 [H3O+] =[OH] = 1.0 x 107 M For pure water and dilute aqueous solutions at any temperature: Kw = [H3O+][OH] Acid: proton-donor HCl(aq) + H2O(l) H3O+(aq) + Cl(aq) Base: proton-acceptor NH3(aq) + H2O(l) NH4+(aq) + OH(aq) acid-to-base proton transfer NH3(aq) + HCl(aq) NH4+(aq) + Cl(aq) Neutralization: Common Strong Acids and Bases Strong Weak Monoprotic Acids acids and strong bases dissociate completely in water (if soluble, of course) Acid: proton-donor + HCl(aq) H2O(l) H3O+(aq) + Cl(aq) HCl, HBr, HI, H2SO4, HNO3, HClO4, HClO3 By definition, a weak monoprotic acid HA is only partially dissociated in water: HX(aq) + H2O(l) H3O+(aq) + X(aq) The acid ionization constant Ka is given by: Ka = [H3O+][X] [HX] For example, for acetic acid: CH3COOH + H2O H3O+ + CH3COO Base: proton-acceptor NH3(aq) + H2O(l) NH4+(aq) + OH(aq) LiOH, NaOH, KOH, RbOH, CsOH, R4NOH Ka = [H3O+][CH3COO] = 1.75 x 10-5 [CH3COOH] Most carboxylic acids (RCOOH) are weak acids Some Weak Acids acetylsalicylic acid A Look at a Polyprotic Acid Phosphoric acid is a weak acid: H3PO4 + H2O H3O+ + H2PO4 Ka1 = 7.5 x 103 H2PO4 + H2O H3O+ + HPO42 Ka2 = 6.2 x 108 HPO42 + H2O H3O+ + PO43 Ka3 = 3.6 x 1013 Carbonic acid, H2CO3 and oxalic acid, H2C2O4 are examples of polyprotic acids Copyright 1995 by Saunders College Publishing benzoic acid Things to Remember about Ka Weak Bases The equilibriu m involving a weak base dissolved in water is given by: B(aq) + H2O(l) BH+(aq) + OH(aq) T he The larger the value of Ka The greater the extent of ionization The stronger the acid The smaller the pKa Acid Ka HF 7.2 x 104 HNO2 4.5 x 104 acetic 1.8 x 105 HOCl 3.5 x 108 HCN 4.0 x 1010 *Ka at 25oC pKa 3.14 3.35 4.74 7.46 9.40 base ionization constant Kb is: Kb= [BH+][OH] [B] For examp le, for ammonia, NH3: NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Kb = [NH4+][OH] = 1.75 x 10-5 [NH3] Some Weak Bases Other Weak Bases Amines are weak bases; methylamine is a typical organic weak base: CH3NH2(aq) + H2O(l) CH3NH3+(aq) + OH(aq) RNH2 (a primary amine); R2NH (a secondary amine and R3N (a tertiary amine) The parent of all amines is ammonia, NH3 Carboxylate anions (RCO2-) are also weak bases: RCO2-(aq) + H2O(l) RCO2H(aq) + OH(aq) Pyridine Ammonia Aniline Copyright 1995 by Saunders College Publishing Things to Remember about Kb A Look at a Tribasic Species Phosphate ion is tribasic: PO43 + H2O OH + HPO42 Kb1 = 1.4 x 102 HPO42 + H2O OH + H2PO4 Kb2 = 1.59 x 107 H2PO4 + H2O OH + H3PO4 Kb3 = 1.42 x 1012 Kb1 is for the basic species with the least number of protons while Ka1 is for the acidic species with the most protons The larger the value of Kb The greater the extent of protonation The stronger the base The smaller the pKb Base Kb CH3NH2 5.0 x 104 (CH3)3NH 7.4 x 105 NH3 1.8 x 105 C5H5N 1.5 x 109 *Kb at 25oC pKb 3.30 4.13 4.74 8.82 Autoprotolysis of Water Conjugate Acid-Base Pair Water can act as an acid or a base depending on what other species are present Water is amphiprotic: it can accept or donate protons Autoionization also called autoprotolysis Species that differ by a proton H + + Cl H Cl +H O H H O H Acid2 H H H O H +H O H H O H + + O H Acid1 Base2 Base1 + + F HF +H O H H O H Conjugate Acid-Base Pair Acid-Conjugate Base Pairs Species that differ by a proton H + + Cl The weaker a base, the stronger its conjugate acid The weaker an acid, the stronger its conjugate base H Cl + H O H H O H Acid HClO4 HI HCl H3O+ HF CH3COOH HCN NH4+ H2O NH3 Greater Acid Strength Base ClO4 I Cl H2O F CH3COO CN NH3 OH NH2 Greater Base Strength Acid1 H O H Base2 H + H N H H Acid2 H N H H + + Base1 H+ + H+ O H Relationship between Conjugate Acid-Base Pairs Relationship between Conjugate Acid-Base Pairs HCl(aq) + H2O(l) H3O+(aq) + Cl(aq) acid base conjugate conjugate acid base CH3COOH + H2O CH3COO + H3O+ acid base conjugate conjugate base acid The weaker the acid, the stronger its conjugate base NH3(aq) + H2O(l) NH4+(aq) + OH(aq) conjugate conjugate acid base C5H5N + H2O C5H5NH+ + OH base acid conjugate conjugate acid base The weaker the base, the stronger its conjugate acid base acid The pH and pOH Scales Define: pX = log [X] Therefore: pH = log[H+] or 10pH = [H+] Acidic, Neutral and Basic pH H2O(l) + H2O(l) H3O+(aq) + OH(aq) Kw = [H+][OH]= 1.0 x 1014 pKw = 14 pH + pOH = 14 (25oC) +] = [OH] = 1.0 x 107 M Neutral: [H Neutral: pH = pOH = 7 Acidic: [H+] > 1.0 x 107 M > [OH] Acidic: pH < 7 < pOH Basic: [OH] > 1.0 x 107 M > [H+] Basic: pH > 7 > pOH pOH = log[OH] or pK = log[K] or 10pOH = [OH] 10pK = [K] Getting Acquainted with the pH and pOH Scales NEUTRAL [H+] = [OH] 4 oj vinegar gastric juice (1.0-2.0) 5 6 7 8 9 Relationships Among [H3O+], [OH], pH and pOH ACIDIC 0 1 2 3 BASIC 10 11 12 13 14 urine (4.8-7.5) blood (7.4) household ammonia pure water (11.5) oven cleaner (13.5) Calculating [H+] and [OH-] Determining pH and pOH What are the concentrations of H+ and OH- in pure water at 25oC? What is [OH-] if [H+] = 1.0 x 10-3 at 25oC? Is there such a thing as pure water? What is the pH and pOH of a 0.050 M HNO3 solution? Recall that HNO3 is a strong acid; construct a reaction summary: HN O3 + H2 O H 3O + + N O 3 ini 0.050 M ~0 M 0M chn 0.050 M +0.050 M +0.050 M final 0M +0.050 M +0.050 M + = +0.050 M H 3O pH = log 0.050 = ( 1.30) = 1.30 pOH = 14 1.30 = 12.7 Hydrolysis Hydrolysis The reaction of a substance with water. For example, when NaCH3COO is dissolved in water: NaCH3COO(s) Na+(aq) + CH3COO(aq) The reaction of a substance with water. For example, when NH4Cl is dissolved in water: NH4Cl(s) NH4+(aq) + Cl(aq) NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) CH3COO(aq) + H2O(l) CH3COOH(aq) + OH(aq) hydrolysis } Excess OH produced; solution is basic Hydrolysis Constants Ka and Kb The equilibrium constant for a hydrolysis reaction is called an acid (Ka) or base hydrolysis constant (Kb): CH3COO(aq) + H2O(l) CH3COOH(aq) + OH(aq) base Kb = [ C H 3COOH ][ OH ] [ C H 3COO ] NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) acid Ka = [N H 3][ H 3O +] [N H 4+] } hydrolysis Excess H3O+ produced; solution is acidic Evaluating Ka or Kb For the hydrolysis reaction with the equilibrium (hydrolysis) constant Kb: CH3COO(aq) + H2O(l) CH3COOH(aq) + OH(aq) base conjugate acid [C H 3COOH ][ OH ] Kb = [C H 3COO ] It can be shown that Kb for CH3COO is related to Ka for CH3COOH by: K bK a = K w (valid for any conjugate acid-base pair) } Evaluating Ka or Kb Salts Kb for methylamine is 4.47 x Find Ka for methylammonium ion Derive the Ka and Kbs for diprotic and triprotic acids and their conjugate bases 10-4. Ion ic compou nds formed from the reaction of an acid with a base If soluble, are strong electrolytes that dissociate 100% in water May hydrolyze to alter solution pH Salts of Strong Bases and Strong Acids Salts of Strong Bases and Weak Acids Do not hydrolyze Yield solutions that are neutral For example: NaCl(s) Na+(aq) + Cl(aq) H2O(l) + H2O(l) OH(aq) + H3O+(aq) React with water (hydrolyze) Yield solutions that are basic For example: NaCH3COO(s) Na+(aq) + CH3COO(aq) CH3COO(aq) + H2O(l) CH3COOH(aq) + OH(aq) Salts of Weak Bases and Strong Acids Salts of Weak Bases and Weak Acids React with water (hydrolyze) Yield solutions that are acidic For example: NH4Cl(s) NH4+(aq) + Cl(aq) NH4+(aq) + H2O(l) NH3(aq) + H3O+(aq) React with water (hydrolyze) Yield solutions that are acidic, basic or neutral depending on the relative s trength of acid a nd base: When Kb = Ka neutral When Kb > Ka basic When Kb < Ka acidic Salts for Which Kb = Ka Salts for Which Kb > Ka When NH4CH3COO dissolves in water, both ions hydrolyze to same extent so solution is neutral When NH4CN dissolves in water, CN hydrolyzes to a greater extent t han N H4+ so the solution is basic N H 4+ + H 2 O N H 3 + H 3O + K a = 5.6 x 10 10 K = 1.8 x 10 5 NH NH ++ H O H + H O+ N 4 2 3 3 K a = 5.6 x 10 10 C H 3COO + H 2O C H 3COOH + OH K b = 5.6 x 10 10 C N + H 2O HCN + OH K b = 2.5 x 10 5 K = 4.0 x 10 10 HCN 3 Salts for Which Kb < Ka When NH4F dissolves in water, NH4+ hydrolyzes to a greater extent than F so solution is acidic N H 4+ + H 2O N H 3 + H 3O + K a = 5.6 x 10 10 K = 1.8 x 10 5 F + H 2O HF + OH 11 K b = 1.4 x 10 K = 7.2 x 10 4 HF N H3 ... 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