Chapt 17 Chemical Equilibrium 17 1 A State

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? N 2(g) + 3 H 2(g) 2 NH 3(g) DG 0 = -33. 1 k. J 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 N 2 and 3 mol H 2

Concentration N 2(g) + 3 H 2(g) 2 NH 3(g) Reactant All concentrations stop changing – concentrations reactant concentrations are not zero initially decrease Product concentration initially increases H 2 NH 3 N 2 Time

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

Reversible Reactions N 2(g) + 3 H 2(g) ↔ 2 NH 3(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

Chemical Equilibrium N 2(g) + 3 H 2(g) ↔ 2 NH 3(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

Reaction Rates Chemical Equilibrium Forward Rate Reverse Rate Equilibrium 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 a. A +b. B c. C +d. D Equilibrium constant

Equilibrium Constant a. A +b. B ↔ c. C +d. D 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 a. A +b. B ↔ c. C +d. D 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

Equilibrium Constant a. A +b. B ↔ c. C +d. D 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 H 2(g) + I 2(g) ↔ 2 HI(g) Keq = _[HI]2 [H 2] [I 2] Keq = 49. 7 at 731 K • Note: no units for this particular reaction

Keq – Homogeneous Equilibrium N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) Keq = [NH 3]2 [N 2] [H 2]3 Units = L 2/mol 2

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 C 2 H 5 OH(g) C 2 H 5 OH(l) Ethanol in closed flask C 2 H 5 OH(l) ↔ C 2 H 5 OH(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 C 2 H 5 OH(l) C 2 H 5 OH(g) Keq = [C 2 H 5 OH(g)] I 2 (s) I 2(g) Keq = [I 2 (g)]

Keq – Heterogeneous Equilibrium 2 Na. HCO 3(s) ↔ Na 2 CO 3(s) + CO 2(g) + H 2 O(g) Keq = [CO 2(g)] [H 2 O(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 H 2(g) + I 2(g) ↔ 2 HI(g) Keq = _[HI]2 [H 2] [I 2] [H 2]0 [I 2]0 [HI]0 [H 2]eq [I 2]eq Position 1 Position 2 Position 3 [HI] K eq eq 1. 00 2. 00 0. 066 1. 07 1. 86 49. 70 0. 0 5. 00 0. 55 3. 90 49. 70 1. 00 0. 332 2. 34 49. 70 Three different equilibrium positions yield the same value of Keq

Value of Keq – Problem 17. 3 N 2(g) + 3 H 2(g) ↔ 2 NH 3(g) Keq = [NH 3]2 [N 2] [H 2]3 [NH 3] = 0. 933 mol/L [N 2] = 0. 533 mol/L [H 2] = 1. 600 mol/L Value of Keq? Keq =. [0. 933]2. = 0. 399 [0. 533] [1. 600]3 (L 2/mol 2)

Keq of Some Common Reactions Reaction Keq 2 H 2(g) + O 2(g) ↔ 2 H 2 O(l) 1. 4 x 1083 298 K Ca. CO 3(s) ↔ Ca. O(s) + CO 2(g) 2 SO 2(g) + O 2(g) ↔ 2 SO 3(g) 1. 9 x 10 -23 298 K 1. 0 1200 K 3. 4 1000 K -21 298 K 1. 6 x 10 C(s) + H 2 O(g) ↔ CO(g) + H 2(g) 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: 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) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) ________________________________________________ __ [CO]eq [H 2]eq [CH 4]eq [H 2 O]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 New equilibrium (by position injection) has reduced raise COvalue concentration of CO from instantaneous to 1. 0000 M value of 1. 0000 Equilibrium Stress of additional becomes reactant unbalanced relieved (partially) by producing more product Rate of forward rxn increases, get

Le Châtelier’s P. - Concentration CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) ____________________________________________________________________ What will happen to equilibrium if: A desiccant is used to remove H 2 O? Equilibrium shifts to the right CO is removed from reaction vessel? Equilibrium shifts to the left H 2 O is injected into the reaction vessel? Equilibrium shifts to the left

Le Châtelier’s P. - Concentration CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) Product Removal Reactant addition Reactant Removal Product Addition

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

Le Châtelier’s P. - Volume CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) Start Compress Initial More Compress product Final forms Temporarily have nonequilibrium concentrations

Le Châtelier’s P. - Volume CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(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 H 2(g) + I 2(g) ↔ 2 HI(g) - no effect of V

Le Châtelier’s P. - Temperature CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) DH 0 = - 206. 5 k. J 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 + Lower T: Product removal + + + Raise T: Product addition Endothermic Reaction heat + Co(H 2 O)62+ + 4 Cl. Co. Cl 42 - + 6 H 2 O Raise T: Reactant addition heat + Co(H 2 O)62+ + 4 Cl. Co. Cl 42 - + 6 H 2 O 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 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) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) 0. 850 M 1. 333 M ? M 0. 286 M Keq = [CH 4(g)] [H 2 O(g)] [CO(g)] [H 2(g)]3 [CH 4(g)] = Keq [CO(g)] [H 2(g)]3/ [H 2 O(g)]

Calculating Equilibrium Concentrations CO(g) + 3 H 2(g) ↔ CH 4(g) + H 2 O(g) 0. 850 M 1. 333 M ? M 0. 286 M [CH 4(g)] = Keq [CO(g)] [H 2(g)]3/ [H 2 O(g)] Keq = 3. 933 [CH 4(g)] = 3. 933 (0. 850)(1. 333)3/ (0. 286) [CH 4(g)] = 27. 7 mol/L = [CH 4(g)]eq

Calculating Equilibrium Concentrations Example Problem 17. 4 2 H 2 S(g) ↔ 2 H 2(g) + S 2(g) 1. 84 x 10 -1 M ? M 5. 40 x 10 -2 M Keq = [H 2(g)]2 [S 2(g)] = 2. 27 x 10 -3 [H 2 S(g)]2 [H 2(g)]2 = Keq [H 2 S(g)]2 / [S 2(g)] = 1. 42 x 10 -3 [H 2] = 3. 77 x 10 -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 Ba. SO 4(s) ↔ Ba 2+(aq) + SO 42 -(aq) Heterogeneous equilibrium Keq = [Ba 2+(aq)] [SO 42 -(aq)] Has special symbol Ksp

Solubility Equilibria Ba. SO 4(s) ↔ Ba 2+(aq) + SO 42 -(aq) Ksp = [Ba 2+] [SO 42 -] = 1. 1 x 10 -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) ↔ Mg 2+(aq) + 2 OH-(aq) Ksp = [Mg 2+] [OH-]2 = 5. 6 x 10 -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 Ag. I(s) ↔ Ag+(aq) + I-(aq) Ksp = [Ag+(aq)] [I-(aq)] = 8. 5 x 10 -17 @298 K s = solubility of Ag. I(s) in mol/L 1 mol Ag+(aq) forms per mol Ag. I dissolved 1 mol I-(aq) forms per mol Ag. I dissolved Ksp = [Ag+(aq)] [I-(aq)] = s 2 = 8. 5 x 10 -17 s = 9. 2 x 10 -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) ↔ Mg 2+(aq) + 2 OH-(aq) Ksp = [Mg 2+] [OH-]2 = 5. 6 x 10 -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 Mg 2+ per mol Mg(OH)2 dissolved 2 mol OH- per mol Mg(OH)2 dissolved

Calculating Ion Concentration from Ksp Mg(OH)2(s) ↔ Mg 2+(aq) + 2 OH-(aq) Ksp = [Mg 2+] [OH-]2 = 5. 6 x 10 -12 @298 K [OH-] in a saturated solution? s = solubility of Mg(OH)2 in mol/L 1 mol Mg 2+ per mol Mg(OH)2 dissolved 2 mol OH- per mol Mg(OH)2 dissolved Ksp = (s) (2 s)2 = 4 s 3 = 5. 6 x 10 -12 s = 1. 1 x 10 -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 Current but not called the ion product necessarily equilibrium values • 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-]

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. 1 M iron(III) chloride and potassium hexacyanoferrate(II) 4 Fe. Cl 3(aq)+ 3 K 4 Fe(CN)6(aq) 12 KCl(aq) + Fe 4(Fe(CN)6)3(s) ↔ 4 Fe 3+(aq) + 3 Fe(CN)64 -(aq) Ksp = [Fe 3+]4[Fe(CN)64 -]3 = 3. 3 x 10 -41 Question: does a precipitate form?

Predicting Precipitates Fe 4(Fe(CN)6)3(s) ↔ 4 Fe 3+(aq) + 3 Fe(CN)64 -(aq) Ksp = [Fe 3+]eq 4[Fe(CN)64 -]eq 3 Equal volumes of 0. 10 M Fe. Cl 3 & K 4 Fe(CN)6 solutions used [Fe 3+] = 0. 050 M [Fe(CN)64 -] = 0. 050 M Qsp = [Fe 3+]4[Fe(CN)64 -]3 = [0. 050]4[0. 050]3 Qsp = 7. 8 x 10 -10 Ksp = 3. 3 x 10 -41 Qsp > Ksp so precipitate will form Possible to compute moles of solid formed

Predicting Precipitates Example Problem 17. 7 Mix 100 m. L 0. 0100 M Na. Cl with 100 m. L 0. 0200 M Pb(NO 3)2 Does Pb. Cl 2 precipitate? Pb. Cl 2(s) ↔ [Pb 2+][Cl-]2 Ksp = 1. 7 x 10 -5 Qsp = [0. 0100][0. 0050]2 = 2. 5 x 10 -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 is an ion common to two or more ionic compounds • KI, Ag. I I- is the common ion • Ca. Cl 2, Ca(OH)2 Ca 2+ 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 Lead chromate solubility at 298 K Pb. Cr. O 4(s) Pb 2+(aq) + Cr. O 42 -(aq) Quantities -7 mol/L Pure water, s = 4. 8 x 10 produced just from 2 Total Cr. O 4 Much lower in 0. 01 Mof K 2 solid Cr. O 4 dissolution Ksp = [Pb 2+][Cr. O 42 -] = 2. 3 x 10 -13 Pure water: s = [Cr. O 42 -] = [Pb 2+] = 4. 8 x 10 -7 K 2 Cr. O 4 solution: [Pb 2+][Cr. O 42 - + 0. 01] = Ksp [Pb 2+]= 2. 3 x 10 -11 -2 = 2. 3 x 10 -13 2 -x 1. 0 x 10 2+ solubility Common Cr. O-11 has reduced Pb Ksp= 2. 3 x 10 4

Common Ion Effect & L-C’s Principle Pb. Cr. O 4(s) ↔ Pb 2+(aq) + Cr. O 42 -(aq) Pure water: [Pb 2+] = [Cr. O 42 -] = s = 4. 8 x 10 -7 K 2 Cr. O 4 solution: [Pb 2+][Cr. O 42 - + 0. 01] = Ksp [Pb 2+]= 2. 3 x 10 -11 Cr. O 42 - is common ion Higher common ion concentration shifts equilibrium to left Same effect if add Pb(NO 3)2 to solution – common ion now Pb 2+

Using Ksp and Common Ion Cu. CO 3(s) ↔ Cu 2+(aq) + CO 32 -(aq) Ksp = [Cu 2+(aq)] [CO 32 -(aq)] = 2. 5 x 10 -10 s = solubility of Cu. CO 3 (s) in solution of 0. 10 M K 2 CO 3 (Prob. 17. 5 & strategy on p 621) (s)(0. 10 + s) = 2. 5 x 10 -10 (use quad. eqn. ) (s)(0. 10) = 2. 5 x 10 -10 (approximation)

Using Ksp and Common Ion Cu. CO 3(s) ↔ Cu 2+(aq) + CO 32 -(aq) Ksp = [Cu 2+(aq)] [CO 32 -(aq)] = 2. 5 x 10 -10 (s)(0. 10 + s) = 2. 5 x 10 -10 (using quad. eqn. ) (s)(0. 10) = 2. 5 x 10 -10 (using approximation) For this problem, exact method (using quadratic equation) and approximate method yield same answer; s = 2. 5 x 10 -9 Commonly used “trick” to quickly solve variety of equilibrium problems