Chemical & Solubility Equilibrium (K eq, K c, K p, K sp )

Transcription

1 Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 8 (Chp 15,17): Chemical & Solubility (K eq, K c, K p, K sp ) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall Dynamic (forward & reverse rxns occur simultaneously) double arrow : Dynamic : Initially, forward & reverse rxns occur at different rates. (based on collisions, one slows down; other speeds up) At equilibrium: Rate forward = Rate reverse 1

2 A System at At equilibrium, concentrations (M, mol, P, etc.) of reactant and product remain constant. At forward and reverse rates are equal concentrations are constant HW p. 660 #2 The Constant 2

3 The Constant Consider the reaction aa + bb At equilibrium cc + dd Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d k f k r [C] c [D] = d [A] a [B] b K eq = [C]c [D] d [A] a [B] b [products] [reactants] The Constant Consider the reaction aa + bb cc + dd The equilibrium constant expression (K eq ) is K = [products] [reactants] K c = [C]c [D] d [A] a [B] b [ ] is conc. in M K expressions do not include: pure solids(s) or pure liquids(l) (b/c concentrations are constant) The Constant In three experiments at the same temperature, equilibrium was achieved & data were collected. Exp. Butanoic acid (moles) Ethanol (moles) Ethyl butanoate (moles) Water (moles) Calculate K c for each experiment at this temperature and compare the values. Products Reactants (Kc) same ratio (constant) 3

4 What Does the Value of K Mean? Reactants Products K = [products] [reactants] If K > 1, the reaction is product-favored; more product at equilibrium. If K < 1, the reaction is reactant-favored; more reactant at equilibrium. The Constant Because pressure is proportional to concentration for gases in a closed system, the equilibrium expression can also be written Example: K p = (P C) c (P D ) d (P A ) a (P B ) b 3 Fe(s) + 4 H 2 O(g) Fe 3 O 4 (s) + 4 H 2 (g) Heterogeneous Equilibria The concentrations of solids and liquids do not appear in the equilibrium expression. PbCl 2 (s) Pb 2+ (aq) + 2 Cl (aq) K c =?[Pb 2+ ] [Cl ] 2 4

5 CaCO 3 (s) CO 2 (g) + CaO (s) As long as some CaCO 3 or CaO remains, the amount of CO 2 above the solid be constant. K c =? K c = [CO 2 ] K p =? K p = P CO2 Manipulating K K of reverse rxn = 1/K N 2 O 4 2 NO 2 [NO K c = 2 ] 2 = 4.0 [N 2 O 4 ] 2 NO 2 N 2 O 4 K c = [N 2 O 4 ] = 1. [NO 2 ] 2 (4.0) 2 N K of multiplied reaction 2 O 4 4 NO 2 = K^# (raised to power) K c = [NO 2 ] 4 = (4.0) 2 [N 2 O 4 ] 2 Manipulating K A 3 B + 2 C K 1 = C + 3 D 4 E K 2 = 60 A + 3 D 3 B + 4 E K ovr =? K of combined reaction = K 1 x K 2 HW p. 661 #14, 16, 20 K ovr = (2.5)(60) 5

6 Calculations Calculations A closed system initially containing M H 2 and M I 2 at 448 C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. Calculate K c at 448 C for the reaction taking place, which is H 2 (g) + I 2 (g) 2 HI (g) 1) find limiting reactant 2) mol reactant completely to mol product, but NOW we still have some reactant left and some product formed at equilibrium. RICE Tables Initial H 2 + I 2 2 HI Change M H 2 and M I 2 at 448 C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. Calculate K c at 448 C. 6

7 What Do We Know? [H 2 ] in = M [I 2 ] in = M [HI] in = 0 M Initial H 2 + I 2 2 HI M M 0 M Change M [HI] eq = M M H 2 and M I 2 at 448 C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. [HI] Increases by M H 2 + I 2 2 HI Initial M M 0 M Change M M H 2 and M I 2 at 448 C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. Calculate K c at 448 C. Stoichiometry shows [H 2 ] and [I 2 ] decrease by half as much H 2 + I 2 2 HI Initial M M 0 M Change M M HI x 1 mol H 2 = M H 2 2 mol HI 7

8 We can now calculate the equilibrium concentrations of all three compounds Initial M M 0 M Change M M M Calculate K c at 448 C. K c = [HI]2 [H 2 ] [I 2 ] H 2 + I 2 2 HI = HW p. 662 #27, 30, 40 (0.187) 2 (0.0065)(0.1065) = 51 ICE Practice given K, solve for x HW p. 663 #43 At 2000 o C the equilibrium constant for the reaction 2 NO(g) N 2 (g) + O 2 (g) is K c = 2.4 x If the initial concentration of NO is M, what are the equilibrium concentrations of NO, N 2, and O 2? What Do We Know? Initial 2 NO N 2 + O 2 Change K c = 2.4 x 10 3 the initial concentration of NO is M 8

9 What Do We Know? What do we NOT know? 2 NO N 2 + O 2 Initial M 0 M 0 M Change??? K c = 2.4 x 10 3 the initial concentration of NO is M Stoichiometry shows [NO] decreases by twice as much as [N 2 ] and [O 2 ] increases. 2 NO N 2 + O 2 Initial Change 2x + x + x We now have the equilibrium concentrations of all three compounds (in terms of x) 2 NO N 2 + O 2 Initial Change 2x + x + x x x x Now, what was the question again? what are the equilibrium concentrations of NO, N 2, and O 2? 9

10 x x x K c = [N 2] [O 2 ] [NO] x 10 3 (x) = 2 ( x) 2 x 49 = ( x) x = x 9.8 = 99x x = HW p. 663 # 44 (next slide) [N 2 ] eq = M [O 2 ] eq = M [NO] eq = M ICE Practice given K, solve for x For the equilibrium HW p. 663 #44 Br 2 (g) + Cl 2 (g) 2 BrCl(g) at 400 K, K c = 7.0. If 0.30 mol of Br 2 and 0.30 mol of Cl 2 are introduced into a 1.0 L container at 400 K, what will be the equilibrium concentrations of Br 2, Cl 2, and BrCl? What Do We Know? Initial Br 2 + Cl 2 2 BrCl Change K c = 7.0 [Br 2 ] in = 0.30 M [Cl 2 ] in = 0.30 M the initial concentrations of Br 2 and Cl 2 are 0.30 M 10

11 What Do We Know? What do we NOT know? Br 2 + Cl 2 2 BrCl Initial 0.30 M 0.30 M 0 M Change K c = 7.0 [Br 2 ] in = 0.30 M [Cl 2 ] in = 0.30 M??? the initial concentrations of Br 2 and Cl 2 are 0.30 M Stoichiometry shows [BrCl] increases by twice as much as [Br 2 ] and [Cl 2 ] decrease. Initial 0.30 M 0.30 M 0 M Change x x + 2x Br 2 + Cl 2 2 BrCl We now have the equilibrium concentrations of all three compounds (in terms of x) Br 2 + Cl 2 2 BrCl Initial 0.30 M 0.30 M 0 M Change x x + 2x 0.30 x 0.30 x 2x Now, what was the question again? what are the equilibrium concentrations of Br 2, Cl 2, and BrCl? 11

12 0.30 x 0.30 x 2x K c = [BrCl]2 [Br 2 ][Cl 2 ] (2x) 7.0 = 2 (0.30 x) 2 2x 2.6 = (0.30 x) x = 2x 0.78 = 4.6x x = 0.17 HW p. 664 #64 (next slide) [Br 2 ] eq = 0.13 M [Cl 2 ] eq = 0.13 M [BrCl] eq = 0.34 M ICE Practice given K, solve for x For the equilibrium HW p. 664 #64 2 IBr(g) I 2 (g) + Br 2 (g) K p = 8.5 x 10 3 at 150 o C. If atm of IBr is placed in a 2.0 L container, what is the partial pressure of this substance after equilibrium is reached? What Do We Know? Initial 2 IBr I 2 + Br 2 Change K p = 8.5 x 10 3 the initial pressure of IBr is atm 12

13 What Do We Know? What do we NOT know? 2 IBr I 2 + Br 2 Initial atm 0 atm 0 atm Change??? K p = 8.5 x 10 3 the initial pressure of IBr is atm Stoichiometry shows [NO] decreases by twice as much as [N 2 ] and [O 2 ] increases. 2 IBr I 2 + Br 2 Initial atm 0 atm 0 atm Change 2x + x + x We now have the equilibrium concentrations of all three compounds (in terms of x) 2 IBr I 2 + Br 2 Initial atm 0 atm 0 atm Change 2x + x + x x x x Now, what was the question again? What is the equilibrium partial pressure of IBr? 13

14 x x x K p = (P )(P ) I2 Br2 (P IBr ) x 10 3 (x) = 2 ( x) 2 x = ( x) x = x HW p. 664 #66 (next slide) = 1.184x (PIBr) eq = x = atm ICE Practice HW p. 664 #66 Solid NH 4 HS is introduced into an evacuated flask at 24 o C. The following reaction takes place: NH 4 HS(s) NH 3 (g) + H 2 S(g) At equilibrium the total pressure in the container is atm. What is K p for this equilibrium at 24 o C? Initial 0 atm 0 atm Change What Do We Know? Solid NH 4 HS is introduced into an evacuated flask at 24 o C. NH 4 HS(s) NH 3 + H 2 S At equilibrium, atm the total pressure (P T ) eq is atm. K p =? 14

15 What Do We Know? Initial 0 atm 0 atm Change What do we NOT know? NH 4 HS(s) NH 3 + H 2 S????? At equilibrium, atm the total pressure (P T ) eq is atm. K p =? Stoichiometry shows [NH 3 ] increases by the same amount as [H 2 S] increases. Initial 0 atm 0 atm Change x + x + x NH 4 HS(s) NH 3 + H 2 S atm We now have the equilibrium concentrations of all three compounds (in terms of x) NH 4 HS(s) NH 3 + H 2 S Initial 0 atm 0 atm Change x + x + x x x atm Now, what was the question again? What is K p? 15

16 x x K p = (P NH3 )(P H2S) P T = atm K p = (x) 2 K p = (0.307) 2 K p = P T = P NH3 + P H2S = x + x = 2x x = WS Equil Calc s III The Quotient (Q) aa + bb K c = [C]c [D] d [A] a [B] b cc + dd A Q expression gives the same ratio as the equilibrium (K) expression, but for a system that is NOT at equilibrium. Q = [C]c [D] d [A] a [B] b (given on exam) NOT (given on exam) Calculate Q by substituting INITIAL (current) concentrations into the Q expression. Q = [P] = K [R] K R rate f = P rate r Q K Q K Q Q = K If Q = K, system is at equilibrium (K). 16

17 R P rate f > rate r Q = [P] [R] Q K Q = [P] [R] R rate f = P rate r Q K Q K Q < K If Q < K, too much reactant, system will shift right faster to reach equilibrium (K). HW p. 662 #36, 38 K Q = [P] [R] R rate f = P rate r Q K R rate f < P rate r Q Q [P] = [R] K Q Q > K If Q > K, too much product, system will shift left faster to reach equilibrium (K). Le Châtelier s Principle 17

18 Le Châtelier s Principle: Systems at equilibrium disturbed by a change (that affects collision frequency) like: amount (M, PP) of reactants or products volume (V) of container temperature (T) (changes K value) will shift ( or ) to counteract the change. Because R rate f > P rate r or the rxn goes faster in one direction (Q K) until rate f = rate r (Q = K again, R P ). R P rate f < rate r The Haber Process The conversion of nitrogen (N 2 ) & hydrogen (H 2 ) into ammonia (NH 3 ) is tremendously significant in agriculture, where ammonia-based fertilizers are of utmost importance. N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Add (changes Q) reactant: N 2 (g) + 3 H 2 (g) 2 NH 3 (g) Q = K K = [NH 3] 2 [N 2 ][H 2 ] 3 Q < K Q = [NH 3] 2 Q = K (same K) [N 2 ][H 2 ] 3 K = [NH 3] 2 [N 2 ][H 2 ] 3 18

19 Remove product: (changes Q) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) This apparatus pushes the equilibrium to the right by removing the product ammonia (NH 3 ) from the system as a liquid. Change Volume (moles of gas) (changes Q) N 2 (g) + 3 H 2 (g) 2 NH 3 (g) V ( P T ) shifts tofewer mol of gas ( n gas ) V ( P T ) shifts tomore mol of gas ( n gas ) V has no shift if equal mol of gas R & P P T by adding noble gas? no shift b/c same R & P pressures 1. 2 NO(g) N 2 (g) + O 2 (g), V will shift. NOT 2. N 2 O 4 (g) 2 NO 2 (g), V will shift. 3. CaCO 3 (s) CO 2 (g) + CaO(s), shift by _V. Change Temperature Changing temp. is the ONLY way to change the value of. K Why? T (add heat) shifts in the thermic direction to heat (smaller) K 2 = [P] [R] (original) K 1 = [P] [R] T (add heat) shifts in the thermic direction to heat (larger) K [P] 2 = [R] 19

20 Change Temperature heat + CoCl H 2 O(l) 4 Cl + Co(H 2 O) heat H = Add product: Remove product: ice Add heat Remove heat Catalysts increase rate f and rate r. R (same) [P] K = [R] occurs faster, but P at no shift (composition [P]/[R] is same). Le Châtelier s Principle N 2 (g) + 3 H 2 (g) 2 NH 3 (g) + heat H = 92 Change in external factor Increase pressure (decrease volume) H = Increase temp. Increase [N 2 ] Increase [NH 3 ] Add a catalyst Shift to restore equilibrium Right Left Right Left No Shift (practice) Reason P ( V) shifts to side of fewer moles of gas H =, adding heat shifts in the endo- dir. Adding reactant shifts right to consume it Adding product shifts left to consume it Catalysts change how fast, but not how far. 20

21 Le Châtelier s Principle heat + 2 H 2 O(g) + O 2 (g) 2 H 2 O 2 (l) H = +108 Change in external factor Decrease pressure ( Increase volume) H = + Decrease temp. Increase PO 2 Decrease PH 2 O Decrease H 2 O 2 Shift to restore equilibrium Left Left Right Left No Shift (practice) Reason P ( V) shifts to side of more moles of gas H = +, remove heat shifts in the exo- dir. Adding reactant shifts right to consume it Removing reactant shifts left to produce it (s) & (l) do not affect Q & K (usually) Le Châtelier s Principle (summary) R P (M, PP R, PP P ) - add R or P: shift away faster (consume) - remove R or P: shift toward faster (replace ) - volume: V shifts to fewer mol of gas ( n gas ) (P total ) V shifts to more mol of gas ( n gas ) - temp. (changes K) (H + R P) (R P + H) T shifts in endo dir. to use up heat T shifts in exo dir. to make more heat - catalyst: no shift WS Eq Pract. 2 #4 Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 8 (Chp 15, 17): Chemical & Solubility Equilibria (K eq, K c, K p, K sp ) John D. Bookstaver St. Charles Community College St. Peters, MO 2006, Prentice Hall 21

22 Solubility Product Constant (K sp ) Write the K expression for a saturated solution of PbI 2 in water: PbI 2 (s) Pb 2+ (aq) + 2 I (aq) K sp= [Pb 2+ ][I ] 2 X a Y b (s) ax + (aq) + by (aq) K sp = [X + ] a [Y ] b For insoluble solids, the equilibrium constant, K sp, is the solubility product constant, or the product of M s of dissolved ions at equilibrium. K sp is NOT the Solubility solubility: grams of solid (s) dissolved in 1 L (g/l) molar solubility: mol of solid (s) dissolved in 1 L (M) K sp : product of conc. s (M) of ions(aq) at equilibrium X a Y b ax + + by [X a Y b ] [X + ] [Y ] molar solubility molar conc. s of ions K sp = [X + ] a [Y ] b K sp is NOT the Solubility solubility: grams of solid (s) dissolved in 1 L (g/l) molar solubility: mol of solid (s) dissolved in 1 L (M) K sp : product of conc. s (M) of ions (aq) at equilibrium 4.2 g/l M CaCrO M Ca 2+ molar mass: g/mol M CrO 2 4 CaCrO 4 (s) Ca 2+ + CrO 2 4 [CaCrO 4 ] [Ca 2+ ] [CrO 2 4 ] HW p. 763 #46 K sp = [Ca 2+ ][CrO 4 2 ] 22

23 HW p. 763 #48a R I C E K sp Calculations If solubility (or molar solubility) is known, solve for K sp. [PbBr 2 ] is M at 25 o C. (maximum that can dissolve) PbBr 2 (s) Pb Br M 0 M 0 M M M M K sp = [Pb 2+ ][Br ] 2 (all dissolved = saturated)(any excess solid is irrelevant) K sp = (0.010)(0.020) 2 1 PbBr 2 dissociates into 1 Pb 2+ ion & 2 Br ions K sp = 4.0 x 10 6 HW p. 763 #50 R I C E K sp Calculations Solubility (or molar solubility) is known, solve for K sp. [PbI 2 ] = M K sp = (0.0012)(0.0024) 2 PbI 2 (s) Pb I M 0 M 0 M M M M solubility: 0.54 g/l Molar solubility: M K sp = [Pb 2+ ][I ] 2 K sp = 6.9 x g x 1 mol = mol 461 g R I C E If only K sp is known, solve for x (M). K sp for AgCl is 1.8 x AgCl(s) Ag + + Cl x 0 M 0 M x +x +x 0 M x x (molar solubility) [AgCl] = 1.3 x 10 5 M [Ag + ] = 1.3 x 10 5 M [Cl ] = 1.3 x 10 5 M K sp Calculations K sp = [Ag + ][Cl ] K sp = x x = x x = x 1.3 x 10 5 = x 23

24 R I C E HW p. 663 #47 If only K sp is known, solve for x (M). K sp for CaSO 4 is 2.4 x CaSO 4 (s) Ca 2+ + SO 2 4 x 0 M 0 M x +x +x 0 M x x (molar solubility) [CaSO 4 ] = M [Ca 2+ ] = M [SO 2 4 ] = M K sp Calculations K sp = [Ca 2+ ][SO 4 2 ] K sp = x x 10 5 = x x 10 5 = x = x R I C E If only K sp is known, solve for x (M). K sp for PbCl 2 is 1.6 x PbCl 2 (s) Pb Cl x 0 M 0 M x +x +2x 0 M x 2x (molar solubility) [PbCl 2 ] = M [Pb 2+ ] = M [Cl ] = M K sp Calculations K sp = [Pb 2+ ][Cl ] 2 K sp = (x)(2x) 2 K sp = 4x x 10 5 = 4x x 10 6 = x = x K sp Calculations If only K sp is known, solve for x (M). K sp for Cr(OH) 3 is 1.6 x R Cr(OH) 3 (s) Cr OH K sp = [Cr 3+ ][OH ] 3 I x 0 M 0 M K sp = (x)(3x) 3 C x +x +3x K sp = 27x 4 E 0 M x 3x (molar solubility) [Cr(OH) 3 ] = 1.6 x 10 8 M 1.6 x = 27x x = x [Cr 3+ ] = 1.6 x 10 8 M [OH ] = 4.8 x 10 8 M 1.6 x 10 8 = x 24

25 HW p. 763 #52a K sp Calculations (in pure H 2 O) LaF 3 (s) La F x 0 M 0 M x +x +3x 0 M x 3x K sp = [La 3+ ][F ] 3 K sp = (x)(3x) 3 2 x = 27x 4 x = 9 x 10 6 M LaF 3 (in M KF) K sp = [La 3+ ][F ] 3 #52b LaF 3 (s) La F x 0 M M x +x +3x 0 M x x b/c K <<<1 Solubility is lower in sol n w/ common ion K sp = (x)( x) 3 2 x = (x)(0.010) 3 x = 2 x M LaF 3 Factors Affecting Solubility Common-Ion Effect (more Le Châtelier) If a common ion is added to an equilibrium solution, the equilibrium will shift left and the solubility of the salt will decrease. OR adding common ion shifts left (less soluble) BaSO 4 (s) Ba 2+ (aq) + SO 2 4 (aq) BaSO 4 would be least soluble in which of these 1.0 M aqueous solutions? Na 2 SO 4 BaCl 2 Al 2 (SO 4 ) 3 NaNO 3 most soluble? WS Ksp #1-2 Factors Affecting Solubility Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH (aq) Addition of HCl (acid) to this solution would cause Greater solubility because Added acid (H + ) reacts with OH thereby removing product causing a shift to the right to dissolve more solid Mg(OH) 2. 25

26 HW p. 763 #55 Basic anions, more soluble in acidic solution. H + NO Effect on: Cl, Br, I, NO 3, SO 4 2, ClO 4 Adding H + would cause shift, more soluble. H + Mg(OH) 2 (s) Mg 2+ (aq) + 2 OH (aq) Factors Affecting Solubility Complex Ions Metal ions can and form complex ions with lone e pairs in the solvent. Ag(NH 3 ) 2 + forming complex ions increases solubility p. 765 #59 NH 3 AgCl(s) Ag + (aq) + Cl (aq) 26

27 Will a Precipitate Form? WS K sp #4 X a Y b (s) ax + (aq) + by (aq) K sp = [X + ] a [Y ] b In a solution, If Q = K sp, at equilibrium (saturated). Q = [X + ] a [Y ] b HW p. 764 #62b, 66 If Q < K sp, more solid will dissolve (unsaturated) until Q = K sp. (products too small, proceed right ) If Q > K sp, solid will precipitate out (saturated) until Q = K sp. (products too big, proceed left ) Will a Precipitate Form? HW p. 764 #62b (OR is Q > K?) AgIO 3 (s) Ag + + IO 3 K sp = [Ag + ][IO 3 ] 100 ml of M AgNO 3 K sp = 3.1 x ml of M NaIO 3 (mixing changes M and V) Q = [Ag + ][IO 3 ] M 1 V 1 = M 2 V 2 Q = (0.0091)(0.0014) Q = [Ag + ] = M Q = 1.3 x 10 5 (0.010 M)(100 ml) = M 2 (110 ml) Q > K, so [IO 3 ] = M rxn shifts left (0.015 M)(10 ml) = M 2 (110 ml) prec. will form HW p. 764 #65 (AgI) (PbI 2 ) AgI(s) Ag + + I K sp = [Ag + ][I ] Which Will Precipitate First? = 8.3 x PbI 2 (s) Pb I K sp = [Pb 2+ ][I ] 2 = 7.9 x 10 9 AgI will precipitate first b/c less I is needed to reach equilibrium at a smaller K sp. K sp = [Ag + ][I ] 8.3 x = (2.0 x 10 4 )(x) 7.9 x 10 9 = (1.5 x 10 3 )(x) 2 x = 4.2 x [I ] = 4.2 x M K sp = [Pb 2+ ][I ] 2 x = 2.3 x 10 3 [I ] = 2.3 x 10 3 M 27