Chemistry, The Central Science, 10th edition Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten Unit 5 (Chp 15,17): Chemical & Solubility Equilibrium (Kc , Kp , Ksp , Q) John D. Bookstaver St. Charles Community College St. Peters, MO  2006, Prentice Hall

SUMMARY Equilibrium At Equilibrium… N2O4(g) 2 NO2(g) …forward and reverse rates are ________ …concentrations are ________ equal constant N2O4(g) 2 NO2(g)

The Equilibrium Constant For: aA + bB cC + dD …the equilibrium constant expression (Keq) is Kc = [C]c[D]d [A]a[B]b K = [products] [reactants] [ ] is conc. in M K expressions do not include: solids(s) or pure liquids(l) K > 1, the reaction is product-favored; more product at equilibrium. K < 1, the reaction is reactant-favored; more reactant at equilibrium. [P] [R] [P] [R]

↔ ↔ ↔ Manipulating K K of reverse rxn = 1/K N2O4 2 NO2 N2O4 2 NO2 Kc = = 4.0 [NO2]2 [N2O4] N2O4 2 NO2 ↔ Kc = = 1 . (4.0) [N2O4] [NO2]2 N2O4 2 NO2 K of multiplied reaction = K^# (raised to power) K of combined reactions = K1 x K2 … ↔ Kc = = (4.0)2 [NO2]4 [N2O4]2 4 NO2 2 N2O4 A  B K1 = 2.5 C  2 B K2 = 60 A + C  3 B Kovr = (2.5)(60)

RICE Tables Reaction Initial Change Equilibrium H2 I2 2 HI + 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C.

What Do We Know? H2 I2 2 HI + [H2]in = 0.100 M [I2]in = 0.200 M [HI]in = 0 M Reaction Initial 0.100 0.200 Change Equilibrium 0.187 H2 I2 2 HI + [HI]eq = 0.187 M 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M.

[HI] Increases by 0.187 M H2 I2 2 HI + Initial 0.100 Change +0.187 Equilibrium 0.187 Reaction Initial 0.200 Change Equilibrium H2 I2 2 HI + 0.100 M H2 and 0.200 M I2 at 448C is allowed to reach equilibrium. At equilibrium, the concentration of HI is 0.187 M. Calculate Kc at 448C.

Stoichiometry shows [H2] and [I2] decrease by half as much Reaction Initial 0.100 0.200 Change Equilibrium Initial Change –0.0935 +0.187 Equilibrium 0.187 H2 I2 2 HI + 0.187 M HI x 1 mol H2 = 0.0935 M H2 2 mol HI

Kc = [HI]2 [H2] [I2] = (0.187)2 (0.0065)(0.1065) = 51 We can now calculate the equilibrium concentrations of all three compounds… Initial 0.100 0.200 Change –0.0935 +0.187 Equilibrium 0.0065 0.1065 0.187 Reaction Initial Change Equilibrium H2 I2 2 HI + Calculate Kc at 448C. Kc = [HI]2 [H2] [I2] = (0.187)2 (0.0065)(0.1065) = 51

2 NO N2 O2 + At 2000oC the equilibrium constant for the rxn 2 NO(g) ↔ N2(g) + O2(g) is Kc = 2.4 x 103. Reaction Initial Change Equilibrium 2 NO N2 O2 + If the initial concentration of NO is 0.200 M , what are the equilibrium concentrations of NO , N2 , and O2 ?

? ? ? What Do We Know? What do we NOT know? 2 NO N2 O2 + Reaction Initial 0.200 M 0 M Change Equilibrium 2 NO N2 O2 + ? ? ? Kc = 2.4 x 103 the initial concentration of NO is 0.200 M

Stoichiometry shows [NO] decreases by twice as much as [N2] and [O2] increases. Initial 0.200 Change – 2x + x Equilibrium Reaction Initial Change Equilibrium 2 NO N2 O2 +

Now, what was the question again? We now have the equilibrium concentrations of all three compounds… (in terms of x) Initial 0.200 Change – 2x + x Equilibrium 0.200 – 2x x Reaction Initial Change Equilibrium 2 NO N2 O2 + Now, what was the question again? what are the equilibrium concentrations of NO , N2 , and O2 ?

√ √ Kc = [N2] [O2] [NO]2 2.4 x 103 = (x)2 (0.200 – 2x)2 x (0.200 – 2x) Equilibrium 0.200 – 2x x Kc = [N2] [O2] [NO]2 √ 2.4 x 103 = (x)2 (0.200 – 2x)2 √ x (0.200 – 2x) 49 = 9.8 – 98x = x [N2]eq = 0.099 M [O2]eq = 0.099 M [NO]eq = 0.0020 M 9.8 = 99x x = 0.099

Q = Reaction Quotient, Q R P [P] [R] R P Q = [P] [R] Q = K K Q K K Q ratef < rater ratef = rater ratef > rater Q = [P] [R] Q = [P] [R] Q Q = [P] [R] = K K Q K K Q Q < K Q > K Q = K too much R, shift faster at equilibrium too much P, shift  faster

[NH3]2 [NH3]2 K = [N2][H2]3 [NH3]2 [N2][H2]3 K = Add reactant: N2(g) + 3 H2(g)  2 NH3(g) Q = K K = [NH3]2 [N2][H2]3 Q < K Q = [NH3]2 [N2][H2]3 Q = K K = [NH3]2 [N2][H2]3 (same K)

Le Châtelier’s Principle System at equilibrium disturbed by change (affecting collisions) will shift ( or ) to counteract the change. add R or P: remove R or P: volume: ↓V shifts to ↑V shifts to temp. ↑T shifts ↓T shifts catalyst: shift away faster (consume) shift toward faster (replace) fewer mol of gas (↓ngas) (Ptotal) more mol of gas (↑ngas) (changes K) (H + R  P) (R  P + H) in endo dir. to use up heat in exo dir. to make more heat no shift

Change in external factor Shift to restore equilibrium Le Châtelier’s Principle (practice) N2(g) + 3 H2(g)  2 NH3(g) ∆H = –92 WS + heat Change in external factor Shift to restore equilibrium Reason Increase pressure (decrease volume) Increase temp. Increase [N2] Increase [NH3] Add a catalyst ↑P (↓V) shifts to side of fewer moles of gas Right  ∆H = – –∆H, heat as product, adding prod. shifts left  Left Adding reactant shifts right faster to consume Right  Adding product shifts left faster to consume  Left Catalysts inc. both rates, but not how far. No Shift

Solubility Product Constant (Ksp) XaYb(s)  aX+(aq) + bY−(aq) Ksp = [X+]a [Y–]b (Always solid reactant) solubility: molar solubility: Ksp: grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product of conc.’s (M) of ions(aq) at equilibrium R I C E XaYb  aX+ + bY– # 0 M 0 M –# +a# +b# 0 M a# b# Ksp = [X+]a[Y–]b

1 PbBr2 dissociates into… Ksp Calculations HW p. 763 #48a If solubility (or molar solubility) is known, solve for Ksp . [PbBr2] is 0.010 M at 25oC . (maximum that can dissolve) R I C E PbBr2(s)  Pb2+ + 2 Br– 0.010 M 0 M 0 M –0.010 +0.010 +0.020 0 M 0.010 M 0.020 M Ksp = [Pb2+][Br–]2 (all dissolved = saturated) (any excess solid is irrelevant) Ksp = (0.010)(0.020)2 1 PbBr2 dissociates into… 1 Pb2+ ion & 2 Br– ions Ksp = 4.0 x 10–6

Ksp Calculations If only Ksp is known, solve for x (M). Ksp for PbCl2 is 1.6 x 10–5 . R I C E PbCl2(s)  Pb2+ + 2 Cl– x 0 M 0 M –x +x +2x 0 M x 2x Ksp = [Pb2+][Cl–]2 Ksp = (x)(2x)2 Ksp = 4x3 (molar solubility) 1.6 x 10–5 = 4x3 3√4.0 x 10–6 = x 0.016 = x [PbCl2] = 0.016 M [Pb2+] = 0.016 M [Cl–] = 0.032 M

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) BaSO4(s) Ba2+(aq) + SO42−(aq) BaSO4 would be least soluble in which of these 1.0 M aqueous solutions? Na2SO4 BaCl2 Al2(SO4)3 NaNO3 most soluble?

Common Ion  less soluble (in pure H2O) (in 0.010 M KF) LaF3(s)  La3+ + 3 F– x 0 0 –x +x +3x 0 x 3x LaF3(s)  La3+ + 3 F– x 0 –x +x +3x 0 x 0.010 + 3x 0.010 ≈ 0.010 b/c K <<<1 Solubility is lower in _________________ sol’n w/ common ion Ksp = [La3+][F–]3 Ksp = (x)(3x)3 2 x 10–19 = 27x4 x = 9 x 10–6 M LaF3 Ksp = [La3+][F–]3 Ksp = (x)(0.010 + 3x)3 2 x 10–19 = (x)(0.010)3 x = 2 x 10–13 M LaF3

Basic anions, more soluble in acidic solution. H+ NO Effect on: Cl– , Br– , I– NO3–, SO42–, ClO4– Adding H+ would cause… shift  , more soluble. H+ Mg(OH)2(s)  Mg2+(aq) + 2 OH−(aq)

more soluble by forming complex ions Adding :NH3 causes… shift  , more soluble. Ag(NH3)2+ NH3 AgCl(s)  Ag+(aq) + Cl−(aq)

Will a Precipitate Form? (OR…is Q > K ?) XaYb(s)  aX+(aq) + bY−(aq) Ksp = [X+]a [Y–]b Q = [X+]a [Y−]b In a solution, If Q = Ksp, at equilibrium (saturated). If Q < Ksp, more solid will dissolve (unsaturated) until Q = Ksp . (products too small, shift right→) If Q > Ksp, solid will precipitate out (saturated) until Q = Ksp . (products too big, shift left←)

Will a Precipitate Form? (is Q > K ?) p. 764 #62b AgIO3(s)  Ag+ + IO3– Ksp = [Ag+][IO3–] Ksp = 3.1 x 10–8 100 mL of 0.010 M AgNO3 10 mL of 0.015 M NaIO3 Q = [Ag+][IO3–] (mixing changes M and V) M1V1 = M2V2 Q = (0.0091)(0.0014) Q = [Ag+] = ________ (0.010 M)(100 mL) = M2(110 mL) 0.0091 M Q = 1.3 x 10–5 Q > K , so… rxn shifts left prec. will form [IO3–] = ________ (0.015 M)(10 mL) = M2(110 mL) 0.0014 M

When Will a Precipitate Form? p. 764 #66a (BaSO4) (SrSO4) BaSO4(s)  Ba2+ + SO42– SrSO4(s)  Sr2+ + SO42– Ksp = [Ba2+][SO42–] Ksp = [Sr2+][SO42–] 1.1 x 10–10 = (0.010)(x) 3.2 x 10–7 = (0.010)(x) x = 1.1 x 10–8 x = 3.2 x 10–5 [SO42–] = 1.1 x 10–8 M [SO42–] = 3.2 x 10–5 M #66b less SO42– is needed to reach equilibrium (Ksp). Ba2+ will precipitate first b/c…