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General Chemistry IIChemical Equilibria and ThermodynamicsModule 2Equilibrium Reaction2

At equilibrium rate of the forward reaction = rate of the reverse reaction

Rate of the forward reaction = k1[A]a[B]bRate of the reverse reaction = k2[C]c[D]d3The Equilibrium ConstantThe equilibrium constant for a chemical reaction can be written in terms of a ratio of the concentrations of the products over the reactants each raised to their stoichiometric coefficients.e.g. aA + bB cC + dDK = [C]c[D]d/[A]a[B]b For any of the reactants or products that are gases, the concentrations are replaced by partial pressures PA, PB, PC, or PD.K = PCcPDd/PAaPBb* A reaction is favored whenever K > 1.4The Equilibrium ConstantEquilibrium constants are dimensionless. Each quantity in the equilibrium constant is actually a ratio of that species divided by its concentration in its standard state Concentration of solutes are expressed in mol/L Concentrations of gases are expressed in bar(1bar = 105Pa) Concentrations of pure solids, pure liquids, and solvents are omitted because they are unity

Equilibrium Constant5Write an equilibrium constant for the reaction

BrO3- + 2Cr3+ + 4H2O Br- + Cr2O72- + 8H+Water is the solvent as well as a reactant and is in its standard state

Always write the equilibrium constant with products as the numerator and reactants as the denominator

K = [Br-][Cr2O72- ][H+]8[BrO3- ][Cr3+]2Equilibrium Constant6Write the equilibrium constant expressions for the following reactions. C2H4 (g) + 3 O2 (g) 2 CO2 (g) + H2O(g)K = P2CO2PH2O / PC2H4 P3O2HF (aq) + H2O (l) F- (aq) + H3O+ (aq)K = [F-][H3O+]/[HF] (H2O is in its standard state and has an activity of 1)4 NH3 (g) + 5 O2 (g) 4 NO (g) + 6 H2O (g)K = PNO4PH2O6 / PNH34PO257Example:Write equilibrium expressions for the following heterogeneous equilibria.

CaCO3(s) CaO(s) + CO2(g)(at 500C)K = PCO2SO2(g) + H2O(g) H2SO3(aq)(at 25C)K = [H2SO3] / PH2OPSO2CaF2(s) Ca2+(aq) + 2 F-(aq)(at 25C)K = [Ca2+][F-]23 Fe(s) + 4 H2O(g) Fe3O4(s) + 4 H2(g)(at 500C)K = PH24 / PH2O48The Direction of Change

The Reaction Quotient, QQ PCcPDd/PAaPBb

In the reaction quotient, the concentrations or partial pressures do not necessarily have their equilibrium values.The equilibrium constant K is determined by the concentrations or partial pressures of reactants and products at equilibrium and is a constant dependent only upon temperature.9The Direction of Change

Q varies with time or the extent of the reaction. As equilibrium is approached, Q K.Qo PCocPDod/PAoaPBob the initial reaction quotient at t = 0If Qo < K, the reaction proceeds from left to right and leads to the formation of more product.If Qo > K, the reaction proceeds from right to left and leads to the formation of more reactant.If Qo = K, the reaction has reached equilibrium, no net reaction occurs.10Example:The equilibrium constant for the following reaction is 49 at 450C. H2 (g) + I2 (g) 2 HI (g)K = 49If 0.22 mole of I2, 0.22 mole of H2 and 0.66 mole of HI were put into an evacuated 1.00-L container, would the system be at equilibrium? Qo = [HI]o2/[H2]o[I2]o = (0.66)2/(0.22)(0.22) = 9.0Qo < K The system is not at equilibrium.If not, what must occur to establish equilibrium?To establish equilibrium the reaction must proceed to the right, that is the reaction must produce more HI product. This will lead to an increase in Q. The reaction will proceed until Q increases to K or becomes 49.11External Effects and Le Chateliers PrincipleIn 1884, Henri LeChatelier stated what has become known as LeChateliers PrincipleA system in equilibrium that is subjected to a stress will react in a way that tends to counteract the stress.Most industrial processes are designed in such a way that products can be removed continuously (pushing the reaction towards the right, leading to the generation of more product) to achieve high overall yields, even for reactions with small equilibrium constants.12External Effects and Le Chateliers Principle

An endothermic reaction absorbs heat from the surroundings. K for an endothermic reaction with T.An exothermic reaction gives off heat to the surroundings. K for an exothermic reaction with T.13Example:Given the reaction below at equilibrium in a closed container at 500C. N2(g) + 3 H2(g) 2 NH3(g) + 92kJHow would the following influence the equilibrium? Twant to absorb heat, pushed to the left Twant to produce heat, pushed to the right Vwant to decrease ngas, pushed to rightintroducing a platinum catalystno effectforcing more H2 into the systemwant to react H2 away, pushed to the rightremoving NH3want to produce NH3, pushed to the right 14Equilibrium and Thermodynamics The equilibrium constant is directly related to the thermodynamics of a chemical reaction. The heat absorbed or released (enthalpy) and the degree of disorder of reactants and products (entropy) independently contribute to the degree to which the reaction is favored or disfavored.15Enthalpy The enthalpy change, DH, for a reaction is the heat absorbed or released when the reaction takes place under constant applied pressure. The standard enthalpy change, DHo, refers to the heat absorbed when all reactants and products are in their standard states. when DH is (+), heat is absorbed, the reaction is endothermic when DH is (-), heat is released, the reaction is exothermic 16Entropy The entropy, S, of a substance is a measure of its disorder. The greater the disorder, the greater the entropy.In general S (gas) > S (liquid) > S (solid)S (ions in soln.) > S (salt)DSo is the change in entropy when all species are in their standard states. when DS is (+), disorder increases, order decreases when DS is (-),disorder decreases, order increases 17Free Energy The change in Gibbs free energy, DG, balances the opposing tendencies of enthalpy (to lower values) and entropy (to higher values)DG = DH - TDS* A reaction is favored if DG is negative.The equilibrium constant depends on DGoK = e-DG/RTThe more (-) the value of DGo, the larger is K A chemical reaction is favored by the liberation of heat (DH negative) and/or an increase in disorder (DS positive). 18DG vs K The standard free energy change for a reaction is DG and corresponds to the free energy change that accompanies complete conversion of reactants, initially present in their standard state, to products in their standard state. DG is the free energy change for other concentrations and pressures. The relationship between DG and DG isDG = DG + RTlnQWhen a system is at equilibrium, DG = 0 and Q = K0 = DG + RTlnKTherefore, the standard free energy change and the equilibrium constant are related byDG = -RTlnK 19DG vs K When DG < 0, K > 1, reaction is spontaneous, products are favored over the reactants at equilibrium

DG > 0, K < 1, reaction is not spontaneous, reactants are favored over the products at equilibrium

DG = 0, K = 1, system is at equilibrium under standard conditions 20Example:Calculate the equilibrium constant, Kp, for the following reaction at 25C from the given thermodynamic data.N2O4(g) 2 NO2(g)DGf(NO2) = 51.30 kJ/molDGf(N2O4) = 97.82 kJ/mol

DGrxn = 2DGf(NO2) - DGf(N2O4) = {2(51.30) - 97.82}kJ/mol = 4.78kJ/mol

Kp = e-DG/RT = e{-(4.78kJ/mol)(1000J/kJ)/(8.314JK-1mol-1)(298.15K)} = 0.19321Equilibrium vs Kinetics The equilibrium constant tells us about the thermodynamics of the reaction, the relative ratios of reactants and products at equilibrium. It tells us NOTHING of the kinetics of the reaction, i.e., how rapidly equilibrium is reached. 22Solubility Product

Dissolution and precipitation reactions by which solids pass into and out of solution are of great practical importancepermit isolation of single products from reaction mixturespurification of impure solidscontrol formation of mineral depositsaffect ecologies of rivers, lakes and oceans 23The Nature of Solubility Equilibria

Recrystallization one of the most powerful methods for purification of solids, relies on differences between the solubilities of the desired substance and its contaminants

An impure product is dissolved and re-precipitated, repeatedly if necessary, with careful control of the factors that influence solubility.In recrystallization, a solution begins to deposit a compound when it is brought to the point of saturation with respect to that compound. 24The Nature of Solubility Equilibria

Saturated solution one in which a dissolutionprecipitation equilibrium exists between the solid substance and its dissolved form.Unsaturated solution one in which solubility equilibrium ceases to exist and all of the solid is dissolvedSupersaturated solutions condition in which the concentration of dissolved solids exceed their equilibrium value(s)Dissolution-precipitation reactions frequently come to equilibrium very slowly taking days, weeks, months or even years to establish equilibrium! 25The Solubility of Salts

Although all ionic compounds dissolve to some extent in water, those having solubilities (at 25C) of < 0.1 g/L are referred to as insoluble. > 10 g/L are soluble 0.1 g/L < S < 10 g/L are slightly soluble 26Solubility Product, Ksp

AgCl(s) Ag+(aq ) + Cl-(aq)Ksp = [Ag+][Cl-] = 1.6 x 10-10

If no solid is present, then the product of the two ion concentrations is no longer constrained by the solubility product expression, and[Ag+][Cl-] Kspmolar solubility the # of moles of a solid that will dissolve at a given temperature in one liter of solution Solubility of a Sparingly Soluble ElectrolyteCalculate the solubility of silver oxalate (Ag2C2O4) in a saturated solution Ksp = 1.3 x 10-14Ag2C2O4 2 Ag+ + C2O42-Let the solubility of Ag2C2O4 be x moles/literFrom the stoichiometry of the reactionx 2x + x[Ag+] = 2x and [C2O42-] = x Ksp = [Ag+]2[C2O42-] = 1.3 x 10-14 Ksp = [2x]2[x] = 1.3 x 10-144x3 = 1.3 x 10-14[Ag] = 2x = 2.96 x 10-4x = 1.48 x 10-4 moles/liter [C2O42-] = x = 1.48 x 10-4 2728Example:At 25 C, in 1.00 L of solution, 0.5 mol of A2B salt is soluble. What is the solubility product constant of A2B?A2B(s) 2A+(aq ) + B2-(aq)Initial excess 0 0Change -0.5 +2(0.5) 0.5Final 1.0 0.5 Ksp = [A+]2[B2-]Ksp = [1]2[0.5]Ksp = 0.529Precipitation and the Solubility Products

Computing Ksp from solubility, and solubility from Ksp is valid if the solution is ideal, and if there are no side reactions that reduce the concentrations of the ions after they enter solution.

A solubility product is more general than dissolving a single solid in solution.

Two solutions can be mixed to give a precipitate, or when another salt is present that contains an ion common to the salt under consideration. 30Precipitation from Solution

Use the reaction quotient to determine whether or not a precipitate will formQo = [Ag+]o[Cl-]oif Qo < Ksp no precipitateif Qo > Ksp a precipitate forms until Q = KspThe equilibrium concentrations of ions after mixing two solutions to give a precipitate are most easily calculated by assuming that the reaction first goes to completion (consuming one type of ion) and subsequent dissolution of the solid restores some of that ionic species to solution. 31The Common Ion Effect

*If a solution and the solid salt to be dissolved in it have an ion in common, then the solubility of the salt is depressed.

Example:Calculate the solubility, S, of BaSO4BaSO4(s) Ba2+(aq ) + SO42-(aq)If S = the molar solubility of BaSO4, then S = [Ba2+] = [SO42-]Ksp = [Ba2+][SO42-] = [S][S] = S2 = 1.1 x 10-10 S = 1.0 x 10-5 M 32Example:Calculate the solubility, S, of BaSO4 in 1.0 x 10-2 M BaCl2.BaSO4(s) Ba2+(aq ) + SO42-(aq)mass balances:Ba2+[Ba2+] = S + 1.0 x 10-2 (because of common Ba Ion)SO42- [SO42-] = S

Ksp = [Ba2+][SO42-] = 1.1 x 10-10 = S(S + 1.0 x 10-2)S2 + 1.0 x 10-2S 1.1 x 10-10 = 0 (quadratic equation)S = -1.0 x 10-2 [(1.0 x 10-2)2 + 4(1.1 x 10-10)]/2 = 1.1 x 10-8M Solubility is decreased by almost a factor of 103 compared to pure water.PrecipitationA precipitate will form when the ion product concentration exceeds the KspHow many mL of 0.10 M K2CrO4 solution would have to be added to 50 mL of 1.00 mM Cu(NO3)2 to begin to precipitate CuCrO4. Ksp = 3.6 x 10-6 for CuCrO4

Cu(NO3)2 + K2CrO4 CuCrO4 + 2 KNO3After addition of x mL of K2CrO4 solution[Cu2+] = 1 x 10-3(50/ 50 + x)[CrO42-] = 0.1(x / 50 + x)Ksp = [Cu2+] [CrO42-] = 3.6 x 10-6 33PrecipitationAfter substitution of concentration terms into Ksp and rearrangement3.6 x 10-6 x2 4.64 x 10-3 x + 9 x 10-3 = 0a = 3.6 x 10-6, b = 4.64 x 10-3, and c = 9 x 10-3 x = {-b (b2 4ac)}/2ax = 1.94 mL1.94 ml of K2CrO4 solution is the minimum volumerequired to just form a permanent precipitate 34SeparationsPrecipitation reactions can sometimes be used to separate ions from each otherA solution contains 0.01 M Ca2+ and 0.01 M Ce3+. Is it possible to selectively precipitate 99.9% of Ce3+ from this solution by addingoxalate without precipitating any Ca2+ ?Ca(C2O4) Ca2+ + C2O42-Ksp = 1.3 x 10-8 Ce2(C2O4)3 2 Ce3+ + 3 C2O42-Ksp = 3.0 x 10-29 Original [Ce3+] = 0.01 MIf 99.9% is to be precipitated then [Ce3+] = 0.01 (0.01/100) = 1 x 10-5 MLet x = [C2O42-] in equilibrium with Ca2+ with the required [Ce3+] Ksp = [Ce3+] 2 [C2O42-] 3 = 3.0 x 10-29[Ce3+] = 1 x 10-5 and [C2O42-] = xx3 = (3.0 x 10-29 / 1 x 10-10) and x = 6.69 x 10-735Separations- Part 2Will the 6.69 x 10-7 M of C2O42- cause precipitation with a 0.01 M solution of Ca2+?

[Ca2+] = 0.01 M[C2O42-] = 6.69 x 10-7 M

Q = [Ca2+][C2O42-] = [0.01][6.69 x 10-7] = 6.69 x 10-9

Since Q (=6.69 x 10-9) is less than Ksp (= 1.3 x 10-8) it is possible to precipitate 99.9% cerium from this solution3637Complex Ions and Solubility

Coordination Complexes metal ion surrounded by a group of anions or neutral molecules called ligands.

Bonding interaction involves the sharing of a lone pair of electrons on each ligand with the metal providing a partially covalent bond.Overall reaction M(H2O)n + nL MLn + n H2O38Coordination Complexes

Complexes formed in a series ofSequential stepwise equilibrium reactionsCu2+(aq) + NH3(aq) Cu(NH3)2+(aq)K1 = [Cu(NH3)2+] / [Cu2+][NH3] = 1.0 x 104

Cu(NH3)2+(aq) + NH3(aq) Cu(NH3)22+(aq)K2 = [Cu(NH3)22+] / [Cu(NH3)2+][NH3] = 2.0 x 103

Cu(NH3)22+(aq) + NH3(aq) Cu(NH3)32+(aq)K3 = [Cu(NH3)32+] / [Cu(NH3)22+][NH3] = 5.0 x 102

39Coordination Complexes

Cu(NH3)32+(aq) + NH3(aq) Cu(NH3)42+(aq)K4 = [Cu(NH3)42+] / [Cu(NH3)32+][NH3] = 90.

Overall reaction:Cu2+(aq) + 4 NH3(aq) Cu(NH3)42+(aq)Kf = [Cu(NH3)42+] / [Cu2+][NH3]4 = K1K2K3K4 = 1.1 x 1012

Kf is the formation constant of the fully complexed ion.The larger Kf is, the more stable the corresponding complex ion is (for ions with the same # of ligands). Common Ion Effect versus Formation of Coordination Complexes

Complex ion formation affects solubility in the opposite direction from the common ion effect. Common ion effect reduces solubility by increasing the concentration of one of the ions in solution. Complex ion formation increases solubility by decreasing the concentration of one of the ions in solution.Protic Acids and Bases

In aqueous solutions, an acid is a substance that increases the concentration of H+ (or more accurately H3O+, the hydronium ion) when added to the solution.A base decreases the concentration of H+ (or more accurately, H3O+) or increases the concentration of OH-The word protic refers to chemistry involving the transfer of protons between species.42Brnsted-Lowry Acids and Bases

A Brnsted-Lowry acid is a proton donor

HCl (aq) + H2O () H3O+ (aq) + Cl- (aq)

A Bronsted-Lowry base is a proton acceptor.The Bronsted-Lowry definition of acids and bases does not require an aqueous solution, and applies to nonaqueous solutions and to the gas phase as well.

43Salts

HCl(g) + NH3(g) NH4+Cl-(s) acid base saltAn ionic solid is called a salt. It can be thought of as a product of an acid base reaction.When an acid and a base react, they are said to neutralize each other and produce a salt.Most salts composed of singly charged cations and anions such has Na+Cl- and NH4+Cl- are strong electrolytes and dissociate completely in dilute aqueous solution.44Conjugate Acids and Bases

The products of a reaction between an acid and a base are also classified as acids and bases

HA + B A- + HB+acid + base conjugate base + conjugate acid

46The Nature of H+ (H3O+) and OH-

The proton H+ does not exist on its own, it is tightly associated with at least one (H3O+) or maybe even multiple H2O molecules in solution (H5O2+, H7O3+, H9O4+, etc.) and with other molecules in crystals.

Most often H+ in solution is designated as H3O+ rather than H+ because the first H2O molecule is very strongly bound, (while the strength of binding of additional H2O molecules falls of rapidly with the extent of solvation).

47Autoprotolysis

Water undergoes self-ionization, a reacton called autoprotolysis, in which it acts as both an acid and a base

2 H2O () H3O+ + OH-Protic solvents (water, methanol, ethanol, acetic acid, etc.) have a reactive H+ and all protic solvents undergo autoprotolysis to some extent.

2 CH3CO2H () CH3CO2H2+ (aq) + CH3CO2- (aq)48pH

The autoprotolysis constant for H2O has a special designation Kw2 H2O () H3O+ (aq) + OH- (aq)

Kw = [H3O+][OH-] = 1.0 x 10-14 at 25C

In general equilibrium constants depend upon temperature, Kw increases with T because this reaction is endothermic.49pH

An approximate definition of pH

pH = -log [H3O+].

The exact definition depends upon activities (Chapter 7). pH generally lies between 0 and 14 but can extend beyond these ranges in very strongly acidic and basic solutions.

pOH = -log [OH-]pH + pOH = 14.00A solution is acidic if [H3O+] > [OH-], and pH < 7.00A solution is basic if [H3O+] < [OH-] and pH > 7.0050Strengths of Acids and Bases

The strength of an acid or base is a measure of how completely it dissociates to produce H3O+ or OH-.

Strong acids dissociate completely or nearly completely in waterHA(aq) + H2O() H3O+(aq) + A-(aq)

Weak acids dissociate only partially and thus their reactivity is described by an equilibrium constant.HA(aq) + H2O() H3O+(aq) + A-(aq)51Common Strong Acids and Bases

Strong AcidsHCl, HBr, HI, H2SO4, HNO3, HClO4

For H2SO4, only the first proton ionization is complete. Dissociation of a second proton has an equilibrium constant of 1.0x10-2

Strong BasesLiOH, NaOH, KOH, RbOH, CsOH, R4NOH (where R = H, CH3, C2H5, or any organic group)52pH of Various Substances

53Weak Acids and Bases

Dissociation of a weak acid HA (aq) + H2O () H3O+ (aq) + A- (aq) Acid dissociation constantKa = [H3O+][A-]/[HA]Base HydrolysisB (aq) + H2O () BH+ (aq) + OH- (aq)Base hydrolysis or dissociation constantKb = [BH+][OH-]/[B]

Dissociation of Acetic Acid

Dissociation of Weak Acid/Base56Common Classes of Weak Acids and Bases

Carboxylic acids RCO2H where R = an organic substituent. Most carboxylic acids are weak acids and carboxylate anions are weak bases.Amines are (RNH2, R2NH, R3N) weak bases.

Ammonium ions (RNH3+, R2NH2+, R3NH+) are weak acids.

57Polyprotic Acids and Bases

Polyprotic acids and bases are compounds that can donate or accept more than one proton.H3A(aq) + H2O() H2A-(aq) + H3O+(aq)K1 = [H2A-][H3O+]/[H3A]H2A-(aq) + H2O() HA2-(aq) + H3O+(aq)K2 = [HA2-][H3O+]/[H2A-]HA2-(aq) + H2O() A3-(aq) + H3O+(aq)K3 = [A3-][H3O+]/[HA2-]58Relation Between Ka and KbRelationship between Ka and Kb of a conjugate acid-base pairHA(aq) + H2O() H3O+(aq) + A-(aq)Ka = [H3O+][A-]/[HA]A-(aq) + H2O() HA(aq) + OH-(aq)Kb = [HA][OH-]/[A-]2 H2O() H3O+(aq) + OH-(aq)Kw = KaKb = [H3O+][OH-]pKw = pKa + pKbpKa + pKb = 14Similar relationships hold for polyprotic acids/bases.