1. …concentrations are ________ constant …forward and reverse rates are ________ equal At Equilibrium…

2. The Equilibrium Constant …the equilibrium constant expression (K eq ) is K c = [C] c [D] d [A] a [B] b [ ] is conc. in M K expressions do not include: solids(s) or pure liquids( l ) K = [products] [reactants] For: aA + bBcC + dD 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]

3. K of reverse rxn = 1/K K c = = 4.0 [NO 2 ] 2 [N 2 O 4 ] N2O4N2O4 2 NO 2 K c = = 1. (4.0) [N 2 O 4 ] [NO 2 ] 2 N2O4N2O4 2 NO 2 ↔ ↔ K of multiplied reaction = K^ # (raised to power) K c = = (4.0) 2 [NO 2 ] 4 [N 2 O 4 ] 2 4 NO 2 2 N 2 O 4 ↔ Manipulating K K of combined reactions = K 1 x K 2 … A + C  3 B K ovr = (2.5)(60) A  BK 1 = 2.5 C  2 BK 2 = 60

4. Reaction Initial Change Equilibrium Initial0.100 M0.200 M0 M Change–0.0935 +0.187 Equilibrium0.0065 M0.1065 M0.187 M RICE Tables organize info: (Number, Unit, Substance,…Time) H 2 I 2 2 HI + Calculate K c at 448  C. Kc =Kc = [HI] 2 [H 2 ] [I 2 ] = (0.187) 2 (0.0065)(0.1065) = 51

5. Reaction Initial0 Change Equilibrium Initial0.20000 Change– 2x+ x Equilibrium0.200 – 2xxx RICE Tables: Use x if K is known (in terms of x) 2 NO N 2 O 2 + what are the equilibrium concentrations of NO, N 2, and O 2 ?

6. Kc =Kc = [N 2 ] [O 2 ] [NO] 2 49 = x (0.200 – 2x) x = 0.099 2.4 x 10 3 = (x) 2 (0.200 – 2x) 2 9.8 – 98x = x 9.8 = 99x [N 2 ] eq = 0.099 M [O 2 ] eq = 0.099 M [NO] eq = 0.0020 M Equilibrium0.200 – 2xxx Avoid polynomials by rooting

7. K Q rate f = rate r R P KQ Q =Q = [P] [R] K Q at equilibrium Q = KQ = K = K Q =Q = [P] [R] Q =Q = [P] [R] Q < KQ < KQ > KQ > K too much R, shift  faster too much P, shift  faster rate f > rate r R P rate f < rate r R P Reaction Quotient, Q

8. shift away faster (consume) shift toward faster (replace) fewer mol of gas (↓n gas ) Le Châtelier’s Principle more mol of gas (↑n gas ) no shift (H + R  P) (R  P + H) (changes K) (P total ) in endo dir. to use up heat in exo dir. to make more heat System at equilibrium disturbed by change (affecting collisions) will shift (  or  ) to counteract the change.

9. K sp = [X + ] a [Y – ] b Solubility Product Constant (K sp ) X a Y b (s)  a X + (aq) + b Y − (aq) 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 solubility: molar solubility: K sp : X a Y b  a X + + b Y – [X a Y b ] [X + ] [Y – ] molar solubility molar conc.’s of ions K sp = [X + ] a [Y – ] b (Always solid reactant)

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

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

12. Common-Ion Effect (more Le Châtelier) adding common ion (product) shifts left (less soluble) BaSO 4 (s)  Ba 2+ (aq) + SO 4 2− (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?

13. Basic anions, more soluble in acidic solution. Mg(OH) 2 (s)  Mg 2+ (aq) + 2 OH − (aq) H+H+ H + NO Effect on: Cl –, Br –, I –, NO 3 –, SO 4 2–, ClO 4 – Adding H + would cause… shift , more soluble.

14. AgCl (s)  Ag + (aq) + Cl − (aq) NH 3 Ag(NH 3 ) 2 + Forming complex ions… …increases solubility

15. AgIO 3 (s)  Ag + + IO 3 – K sp = [Ag + ][IO 3 – ] K sp = 3.1 x 10 –8 HW p. 764 #62b (is Q > K ?) Q = [Ag + ][IO 3 – ] Q = 100 mL of 0.010 M AgNO 3 10 mL of 0.015 M NaIO 3 [Ag + ] = ________ (0.010 M)(100 mL) = M 2 (110 mL) (mixing changes M and V) M 1 V 1 = M 2 V 2 0.0091 M [IO 3 – ] = ________ (0.015 M)(10 mL) = M 2 (110 mL) 0.0014 M Will a Precipitate Form? Q = (0.0091)(0.0014) Q = 1.3 x 10 –5 Q > K, so… rxn shifts left prec. will form

16. BaSO 4 (s)  Ba 2+ + SO 4 2– HW p. 764 #66a (BaSO 4 )(SrSO 4 ) SrSO 4 (s)  Sr 2+ + SO 4 2– K sp = [Sr 2+ ][SO 4 2– ] When Will a Precipitate Form? K sp = [Ba 2+ ][SO 4 2– ] 1.1 x 10 –10 = (0.010)(x) 3.2 x 10 –7 = (0.010)(x) x = 1.1 x 10 –8 [SO 4 2– ] = 1.1 x 10 –8 M x = 3.2 x 10 –5 [SO 4 2– ] = 3.2 x 10 –5 M #66b Ba 2+ will precipitate first b/c… less SO 4 2– is needed to reach equilibrium (K sp ).