Chapt. 17 – Chemical Equilibrium 17.1 A State of Dynamic Balance 17.2 Factors Affecting Chemical Equilibrium 17.3 Using Equilibrium Constants

Section 17.1 A State of Dynamic Balance Chemical equilibrium is described by an equilibrium constant expression that relates the concentrations of reactants and products. List the characteristics of chemical equilibrium. Write equilibrium expressions for systems that are at equilibrium. Calculate equilibrium constants from concentration data.

Section 17.1 A State of Dynamic Balance Key Concepts A reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction. The equilibrium constant expression is a ratio of the molar concentrations of the products to the molar concentrations of the reactants with each concentration raised to a power equal to its coefficient in the balanced chemical equation. The value of the equilibrium constant expression, Keq, is constant for a given temperature.

What is Equilibrium? N2(g) + 3H2(g)  2NH3(g) DG0 = -33.1 kJ Spontaneous, slow reaction under standard conditions At higher temperature (723 K) and pressure reaction proceeds at a practical rate Next slide shows reaction progress if start with 1 mol N2 and 3 mol H2

N2(g) + 3H2(g)  2NH3(g) Reactant concentrations initially decrease All concentrations stop changing – reactant concentrations are not zero Product concentration initially increases Concentration H2 NH3 N2 Time

Reversible Reactions Reaction going to completion: almost complete conversion of reactants to products Most reactions do not go to completion –known as reversible reactions Forward: N2(g) + 3H2(g)  2NH3(g) Reverse: 2NH3(g)  N2(g) + 3H2(g) Reversible: N2(g) + 3H2(g) ↔ 2NH3(g) Double arrow

Time prior to equilibrium Reversible Reactions N2(g) + 3H2(g) ↔ 2NH3(g) Time zero Reactants at max concentration, forward rate at maximum, reverse rate at zero Time prior to equilibrium Reactant concentration lower, forward rate slower, some reverse reaction At equilibrium Forward and reverse rates equal – no further concentration changes

Rateforward = Ratereverse Chemical Equilibrium N2(g) + 3H2(g) ↔ 2NH3(g) Forward and reverse reactions balance each other because Rateforward = Ratereverse Does not mean concentrations of reactants and products are equal Can be equal in some special cases

Chemical Equilibrium Equilibrium Reaction Rates Forward = Reverse Rate Forward Rate Reaction Rates Equilibrium Reverse Rate Forward = Reverse Rate Time

Dynamic Equilibrium State of equilibrium not a static one Some reactant molecules are always changing into product molecule Some product molecules are always changing into reactant molecules Only net concentrations of reactants and products don’t change

Equilibrium Constant Law of Chemical Equilibrium: at a given temperature, a chemical system may reach a state in which a particular ratio of reactant and product concentrations reaches a constant value aA +bB  cC +dD Equilibrium constant

Equilibrium Constant aA +bB ↔ cC +dD Keq = [C]c[D]d = equilibrium constant [A]a[B]b [X] = molar concentration of quantity under equilibrium conditions Exponents come from coefficients in balanced chemical equation Keq is temperature dependent Units of Keq vary

Equilibrium Constant aA +bB ↔ cC +dD Keq = [C]c[D]d = equilibrium constant [A]a[B]b Keq >> 1 numerator >> denominator Reaction as written above favors production of products Keq <<1 denominator >> numerator Reaction as written above favors production of reactants

Keq = [C]c[D]d = equilibrium constant [A]a[B]b aA +bB ↔ cC +dD Keq = [C]c[D]d = equilibrium constant [A]a[B]b Rules for writing an expression for Keq differ for homogeneous and heterogeneous equilibrium Homogeneous – all reactants and products in same physical state Heterogeneous – not homogeneous

Keq – Homogeneous Equilibrium Homogeneous when all reactants and products are in same physical state H2(g) + I2(g) ↔ 2HI(g) Keq = _[HI]2 [H2] [I2] Keq = 49.7 at 731 K Note: no units for this particular reaction

Keq – Homogeneous Equilibrium N2(g) + 3H2(g) ↔ 2NH3(g) Keq = [NH3]2 [N2] [H2]3 Units = L2/mol2

Practice Homogeneous Equilibrium Constants Problems 1(a-e), 2 page 601 Problem 44 (a-b) page 626 Problems 1- 5 page 988

Heterogeneous Equilibrium Occurs when reactants & products present in more than one physical state C2H5OH(l) C2H5OH(g) Ethanol in closed flask C2H5OH(l) ↔ C2H5OH(g)

Keq – Heterogeneous Equilibrium Rule: Pure liquids and solids don’t appear in the equilibrium expression because at constant temperature their concentrations don’t change when the amount present changes C2H5OH(l)  C2H5OH(g) Keq = [C2H5OH(g)] I2 (s)  I2(g) Keq = [I2 (g)]

Keq – Heterogeneous Equilibrium 2NaHCO3(s) ↔ Na2CO3(s) + CO2(g) + H2O(g) Keq = [CO2(g)] [H2O(g)]

Practice Heterogeneous Equilibrium Constants Problems 3 (a-e), 4 page 603 Problems 45 – 47, 49 page 626 Problems 6 – 8 page 988

Numerical Value of Keq For a given reaction, final values of reactant and product concentrations will satisfy Keq expression regardless of initial concentrations used Equilibrium position: set of final concentrations of reactants and products One value of Keq, lots of possible equilibrium positions

Notation for Concentrations [X]0 = molar concentration of X at time zero = initial concentration [X]eq = molar concentration of X at equilibrium = final concentration

Numerical Value of Keq Position 1 Position 2 Position 3 H2(g) + I2(g) ↔ 2HI(g) Keq = _[HI]2 [H2] [I2] [H2]0 [I2]0 [HI]0 [H2]eq [I2]eq [HI]eq Keq 1.00 2.00 0.00 0.066 1.07 1.86 49.70 0.0 0.0 5.00 0.55 0.55 3.90 49.70 1.00 1.00 1.00 0.332 0.332 2.34 49.70 Three different equilibrium positions yield the same value of Keq Position 1 Position 2 Position 3

Value of Keq – Problem 17.3 N2(g) + 3H2(g) ↔ 2NH3(g) Keq = [NH3]2 [N2] [H2]3 [NH3] = 0.933 mol/L [N2] = 0.533 mol/L [H2] = 1.600 mol/L Value of Keq? Keq = . [0.933]2 .= 0.399 (L2/mol2) [0.533] [1.600]3

Keq of Some Common Reactions 2H2(g) + O2(g) ↔ 2H2O(l) 1.4x1083 298 K CaCO3(s) ↔ CaO(s) + CO2(g) 1.9x10-23 298 K 1.0 1200 K 2SO2(g) + O2(g) ↔ 2SO3(g) 3.4 1000 K C(s) + H2O(g) ↔ CO(g) + H2(g) 1.6x10-21 298 K 10.0 1100 K Which ones favor production of products?

Practice Calculating value of Keq Problems 5 – 7, 11 page 605 Problem 48 page 626 Problems 9 – 10 page 988

Chapt. 17 – Chemical Equilibrium 17.1 A State of Dynamic Balance 17.2 Factors Affecting Chemical Equilibrium 17.3 Using Equilibrium Constants

Section 17.2 Factors Affecting Chemical Equilibrium When changes are made to a system at equilibrium, the system shifts to a new equilibrium position. Describe how various factors affect chemical equilibrium. Explain how Le Châtelier’s principle applies to equilibrium systems.

Section 17.2 Factors Affecting Chemical Equilibrium Key Concepts Le Châtelier’s principle describes how an equilibrium system shifts in response to a stress or a disturbance. When an equilibrium shifts in response to a change in concentration or volume, the equilibrium position changes but Keq remains constant. A change in temperature, however, alters both the equilibrium position and the value of Keq.

Le Châtelier’s Principle Principle: If a stress is applied to a system at equilibrium, the system shifts in the direction that partially relieves the stress. Stresses include Change in concentration Change in volume (pressure) Change in temperature

Le Châtelier’s P. - Concentration CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) _________________________________________________________________________________________________ [CO]eq [H2]eq [CH4]eq [H2O]eq Keq 0.30000 0.10000 0.05900 0.02000 3.933 1.0000 0.10000 0.05900 0.02000 1.178 0.99254 0.07762 0.06648 0.02746 3.933 Instantaneously (by injection) raise CO concentration to 1.0000 M Equilibrium becomes unbalanced Rate of forward rxn increases, get New equilibrium position has reduced value of CO from instantaneous value of 1.0000 Stress of additional reactant relieved (partially) by producing more product

Le Châtelier’s P. - Concentration CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) _______________________________________________________________________________________________________________________________________ What will happen to equilibrium if: A desiccant is used to remove H2O? Equilibrium shifts to the right CO is removed from reaction vessel? Equilibrium shifts to the left H2O is injected into the reaction vessel?

Le Châtelier’s P. - Concentration CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) Reactant addition Product Removal Reactant Removal Product Addition

Le Châtelier’s P. - Volume CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) 4 moles of reactant, 2 moles of product Decrease V (increase P) at constant T

Le Châtelier’s P. - Volume CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) Start Compress Initial Temporarily have non-equilibrium concentrations More product forms Compress Final

Le Châtelier’s P. - Volume CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) 4 moles of reactant, 2 moles of product Decrease V (increase P) at constant T Stress (increased pressure) relieved by formation of more product P = n  R at constant V, T n = # moles If n decreases, P decreases P is lowered but not to its original value

Le Châtelier’s P. - Volume When volume of equilibrium mixture of gases reduced, net change occurs in direction that produces fewer moles of gas When volume increased, net change occurs in direction that produces more moles of gas H2(g) + I2(g) ↔ 2HI(g) - no effect of V

Le Châtelier’s P. - Temperature CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) DH0 = - 206.5 kJ Have exothermic reaction – can think of heat as a “product” Effect of raising T is like adding more product – equilibrium shifts to the left This approach explains why Keq has a temperature dependence

Le Châtelier’s P. - Temperature Exothermic Reaction + Raise T: Product addition Lower T: Product removal Endothermic Reaction heat + Co(H2O)62+ + 4Cl- CoCl42- + 6H2O heat Raise T: Reactant addition Lower T: Reactant removal

Summary: Le Châtelier’s Principle and Temperature Raising temperature of equilibrium mixture shifts equilibrium condition in direction of endothermic reaction Lowering temperature causes a shift in direction of exothermic reaction

Practice Le Châtelier’s Principle Problems 14 - 15 page 611 Problems 54 - 63, pages 626-27

Chapt. 17 – Chemical Equilibrium 17.1 A State of Dynamic Balance 17.2 Factors Affecting Chemical Equilibrium 17.3 Using Equilibrium Constants

Section 17.3 Using Equilibrium Constants Equilibrium constant expressions can be used to calculate concentrations and solubilities. Determine equilibrium concentrations of reactants and products. Calculate the solubility of a compound from its solubility product constant. Explain the common ion effect.

Section 17.3 Using Equilibrium Constants Key Concepts Equilibrium concentrations and solubilities can be calculated using equilibrium constant expressions. Ksp describes the equilibrium between a sparingly soluble ionic compound and its ions in solution. If the ion product, Qsp, exceeds the Ksp when two solutions are mixed, a precipitate will form. The presence of a common ion in a solution lowers the solubility of a dissolved substance.

Calculating Equilibrium Concentrations If know Keq and concentrations of all but one of the reactants and products, can solve for unknown concentration CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) 0.850M 1.333M ?M 0.286M Keq = [CH4(g)] [H2O(g)] [CO(g)] [H2(g)]3 [CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)]

Calculating Equilibrium Concentrations CO(g) + 3H2(g) ↔ CH4(g) + H2O(g) 0.850M 1.333M ?M 0.286M [CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)] Keq = 3.933 [CH4(g)] = 3.933  (0.850)(1.333)3/ (0.286) [CH4(g)] = 27.7 mol/L = [CH4(g)]eq

Calculating Equilibrium Concentrations Example Problem 17.4 2H2S(g) ↔ 2H2(g) + S2(g) 1.84x10-1M ?M 5.40x10-2M Keq = [H2(g)]2 [S2(g)] = 2.27x10-3 [H2S(g)]2 [H2(g)]2 = Keq [H2S(g)]2 / [S2(g)] = 1.42x10-3 [H2] = 3.77x10-2 mol/L

Practice Calculating equilibrium concentrations Problems 16 (a-c), 17 page 613 Problem 71 page 627 Problems 11(a-b) page 988

Solubility Equilibria Solubility equilibria applies to all compounds with finite solubility, but is most commonly applied to dissolution of those with low solubility BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) Heterogeneous equilibrium Keq = [Ba2+(aq)] [SO42-(aq)] Has special symbol Ksp

Solubility Equilibria BaSO4(s) ↔ Ba2+(aq) + SO42-(aq) Ksp = [Ba2+] [SO42-] = 1.1x10-10 @298 K Ksp = solubility product constant Ksp = equilibrium constant for the dissolution of a sparingly soluble ionic compound in water Only ions appear in Ksp, but some amount of solid (however small) must be present to achieve equilibrium

Solubility Equilibria Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K Table 17.3, page 615 lists solubility product constants sorted by type of anion (carbonate, phosphate, etc.) Ksp can be used to calculate solubility of salt or concentration of one of ions involved in equilibrium

Calculating Solubility from Ksp AgI(s) ↔ Ag+(aq) + I-(aq) Ksp = [Ag+(aq)] [I-(aq)] = 8.5x10-17 @298 K s = solubility of AgI(s) in mol/L 1 mol Ag+(aq) forms per mol AgI dissolved 1 mol I-(aq) forms per mol AgI dissolved Ksp = [Ag+(aq)] [I-(aq)] = s2 = 8.5x10-17 s = 9.2x10-9 mol/L @298 K Note: actual units of Ksp are “ignored”

Practice Calculating solubility of 1:1 salts Problems 20 (a-c), 21 page 616 Problem 22(a) page 617 Problem 31, page 622 Problem 70 page 627 Problem 12 page 988

Calculating Ion Concentration from Ksp Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K [OH-] in a saturated solution? This is a non 1:1 electrolyte – trickier Have to pay attention to stoichiometry s = solubility of Mg(OH)2 in mol/L 1 mol Mg2+ per mol Mg(OH)2 dissolved 2 mol OH- per mol Mg(OH)2 dissolved

Calculating Ion Concentration from Ksp Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq) Ksp = [Mg2+] [OH-]2 = 5.6x10-12 @298 K [OH-] in a saturated solution? s = solubility of Mg(OH)2 in mol/L 1 mol Mg2+ per mol Mg(OH)2 dissolved 2 mol OH- per mol Mg(OH)2 dissolved Ksp = (s) (2 s)2 = 4s3 = 5.6x10-12 s = 1.1 x10-4 mol/L [OH-] = 2 s

Practice Calculating solubility & ion concentrations for arbitrary salts Problems 22 (b-c), 23, 24 page 617 Problems 78, 83 page 628 Problems 13 -15 page 988

Predicting Precipitates Mix two soluble ionic compounds – sometimes a precipitate forms Can use Ksp to predict: Use quantity called the ion product Symbol Qsp Same functional form as Ksp Concentrations used may or may not be equilibrium concentration AB(s) ↔ A+(aq) + B-(aq) Qsp=[A+][B-] Current but not necessarily equilibrium values

Predicting Precipitates AB(s) ↔ A+(aq) + B-(aq) Qsp=[A+][B-] current concentrations Ksp= [A+]eq[B-]eq equilibrium concentrations Compare Ksp to Qsp 1. Qsp < Ksp unsaturated no precipitate 2. Qsp = Ksp saturated no precipitate 3. Qsp > Ksp supersaturated precipitate For #3, solid will form until Qsp = Ksp

Predicting Precipitates Mix equal volumes of (soluble) 0.1M iron(III) chloride and potassium hexacyanoferrate(II) 4FeCl3(aq)+ 3K4Fe(CN)6(aq)  12KCl(aq) + Fe4(Fe(CN)6)3(s) Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-(aq) Ksp = [Fe3+]4[Fe(CN)64-]3 = 3.3x10-41 Question: does a precipitate form?

Predicting Precipitates Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-(aq) Ksp = [Fe3+]eq4[Fe(CN)64-]eq3 Equal volumes of 0.10M FeCl3 & K4Fe(CN)6 solutions used [Fe3+] = 0.050 M [Fe(CN)64-] = 0.050 M Qsp = [Fe3+]4[Fe(CN)64-]3 = [0.050]4[0.050]3 Qsp = 7.8x10-10 Ksp = 3.3x10-41 Qsp > Ksp so precipitate will form Possible to compute moles of solid formed

Predicting Precipitates Example Problem 17.7 Mix 100 mL 0.0100M NaCl with 100 mL 0.0200M Pb(NO3)2 Does PbCl2 precipitate? PbCl2(s) ↔ [Pb2+][Cl-]2 Ksp = 1.7x10-5 Qsp = [0.0100][0.0050]2 = 2.5x10-7 Qsp < Ksp No precipitate

Practice Determining if precipitate will form Problems 25(a-b), 26 page 619 Problems 72, 73 page 627 Problems 16, 17 page 988

Common Ion Effect Common ion is an ion common to two or more ionic compounds KI, AgI I- is the common ion CaCl2, Ca(OH)2 Ca2+ is the common ion Common ion effect is the lowering of the solubility of an ionic substance by the presence of a common ion

Common Ion Effect Quantities produced just from dissolution of solid Lead chromate solubility at 298 K PbCrO4(s)  Pb2+(aq) + CrO42-(aq) Pure water, s = 4.8x10-7 mol/L Much lower in 0.01M K2CrO4 Ksp = [Pb2+][CrO42-] = 2.3x10-13 Pure water: s = [CrO42-] = [Pb2+] = 4.8x10-7 K2CrO4 solution: [Pb2+][CrO42- + 0.01] = Ksp Quantities produced just from dissolution of solid Total CrO42- [Pb2+]= 2.3x10-11 Ksp= 2.3x10-11x1.0x10-2 = 2.3x10-13 Common CrO42- has reduced Pb2+ solubility

Common Ion Effect & L-C’s Principle PbCrO4(s) ↔ Pb2+(aq) + CrO42-(aq) Pure water: [Pb2+] = [CrO42-] = s = 4.8x10-7 K2CrO4 solution: [Pb2+][CrO42- + 0.01] = Ksp [Pb2+]= 2.3x10-11 CrO42- is common ion Higher common ion concentration shifts equilibrium to left Same effect if add Pb(NO3)2 to solution – common ion now Pb2+

Using Ksp and Common Ion CuCO3(s) ↔ Cu2+(aq) + CO32-(aq) Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10 s = solubility of CuCO3 (s) in solution of 0.10 M K2CO3 (Prob. 17.5 & strategy on p 621) (s)(0.10 + s) = 2.5x10-10 (use quad. eqn.) (s)(0.10) = 2.5x10-10 (approximation)

Using Ksp and Common Ion CuCO3(s) ↔ Cu2+(aq) + CO32-(aq) Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10 (s)(0.10 + s) = 2.5x10-10 (using quad. eqn.) (s)(0.10) = 2.5x10-10 (using approximation) For this problem, exact method (using quadratic equation) and approximate method yield same answer; s = 2.5x10-9 Commonly used “trick” to quickly solve variety of equilibrium problems