SUMMARY of Unit 8: Equilibrium At Equilibrium… …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 organize info: (Number, Unit, Substance,…Time) Initial 0.100 M 0.200 M 0 M Change –0.050 +0.100 Equilibrium 0.050 M 0.150 M Reaction Initial Change Equilibrium H2 I2 2 HI + If K is NOT known…use x’s. [HI]2 [H2][I2] Kc =

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

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 absorb heat in exo dir. to produce heat no shift

Solubility Product Constant (Ksp) XaYb(s)  aX+(aq) + bY−(aq) Ksp = [X+]a [Y–]b (Always solid reactant) grams of solid (s) dissolved in 1 L (g/L) mol of solid (s) dissolved in 1 L (M) product (mult.) of M’s of ions(aq) at equilibrium solubility: molar solubility: Ksp: 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 (on solubility) adding common ion (add product) shifts left (less soluble) XY(s)  X+(aq) + Y–(aq) Basic anions, more soluble in acid (H+). H+ (remove product) shifts right (OH–, F–, etc.) H+ NO Effect on: Cl– , Br–, I–, NO3–, SO42–, ClO4– Forming complex ions… …increases solubility by (removing product) shifts right Ag(NH3)2+ NH3 AgCl(s)  Ag+(aq) + Cl−(aq)

Will a Precipitate Form? (is Q > K ?) HW 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? HW 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…