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

…concentrations are ________ constant …forward and reverse rates are ________ equal Dynamic Equilibrium:

The Equilibrium Constant Consider the reaction K eq = [C] c [D] d [A] a [B] b aA + bBcC + dD At equilibrium… Rate f = Rate r k f [A] a [B] b = k r [C] c [D] d k f [C] c [D] d k r [A] a [B] b = [products] [reactants]

K c = [C] c [D] d [A] a [B] b [ ] is conc. in M NO: pure solids (s) or pure liquids (l) aA + bBcC + dD The Equilibrium Constant (K eq ) K p = [P C ] c [P D ] d [P A ] a [P B ] b P is pressure in atm K = [products] [reactants]

What Does the Value of K Mean? If K > 1, the reaction is product-favored; more product at equilibrium. If K < 1, the reaction is is reactant-favored; more reactant at equilibrium. K = [products] [reactants] Reactants  Products

K of reverse rxn = 1/K K c = = [NO 2 ] 2 [N 2 O 4 ] N2O4N2O4 2 NO 2 K c = = __1__ (0.212) [N 2 O 4 ] [NO 2 ] 2 N2O4N2O4 2 NO 2 ↔ ↔ K of multiplied reaction = K^ # (raised to power) K c = = (0.212) 2 [NO 2 ] 4 [N 2 O 4 ] 2 4 NO 2 2 N 2 O 4 ↔ Manipulating K

K of combined reaction = K 1 x K 2 … A + 3 D  3 B + 4 E K ovr = ? A  3 B + 2 CK 1 = C + 3 D  4 EK 2 = 70 Manipulating K K ovr = (1.5)(70)

Reaction Initial Change Equilibrium H 2 I 2 2 HI 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. RICE Tables

Reaction Initial Change Equilibrium0.187 H 2 I 2 2 HI M H 2 and M I 2 at 448  C is allowed to reach equilibrium. At equilibrium, the concentration of HI is M. [H 2 ] in = M[I 2 ] in = M What Do We Know? [HI] eq = M [HI] in = 0 M

Initial Change Equilibrium0.187 Reaction Initial Change Equilibrium [HI] Increases by M H 2 I 2 2 HI 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.

Reaction Initial Change Equilibrium Initial0 Change– Equilibrium0.187 Stoichiometry shows [H 2 ] and [I 2 ] decrease by half as much H 2 I 2 2 HI M HI x 1 mol H 2 = M H 2 2 mol HI

Reaction Initial0 Change Equilibrium Initial Change– Equilibrium We can now calculate the equilibrium concentrations of all three compounds… 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

Reaction Initial Change Equilibrium 2 NO N 2 O 2 + At 2000 o C the equilibrium constant for the rxn 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? Reaction Initial0.200 M0 M Change Equilibrium 2 NO N 2 O 2 + K c = 2.4 x 10 3 the initial concentration of NO is M What do we NOT know? ???

Initial Change– 2x+ x Equilibrium 2 NO N 2 O 2 + Stoichiometry shows [NO] decreases by twice as much as [N 2 ] and [O 2 ] increases. Reaction Initial00 Change Equilibrium

Reaction Initial0 Change Equilibrium Initial Change– 2x+ x Equilibrium0.200 – 2xxx We now have the equilibrium concentrations of all three compounds… (in terms of x) 2 NO N 2 O 2 + Now, what was the question again? what are the equilibrium concentrations of NO, N 2, and O 2 ?

Kc =Kc = [N 2 ] [O 2 ] [NO] 2 49 = x (0.200 – 2x) x = x 10 3 = (x) 2 (0.200 – 2x) – 98x = x 9.8 = 99x [N 2 ] eq = M [O 2 ] eq = M [NO] eq = M Equilibrium0.200 – 2xxx √ √

The Reaction Quotient (Q) A Q expression is same ratio as equilibrium (K) expression, but … …may NOT be at equilibrium. Kc = Kc = [C] c [D] d [A] a [B] b aA + bBcC + dD Calculate Q with INITIAL (or given) conc’s, then compare to K. Q = [C] c [D] d [A] a [B] b NOT given on exam

K Q rate f = rate r R P KQ Q = [P] [R] K Q If Q = K, system is at equilibrium (K). Q = KQ = K = K

If Q < K, too much reactant, system will proceed right to reach equilibrium (K). rate f > rate r R P rate f = rate r R P Q =Q = [P] [R] K Q KQK Q Q < KQ < K Q =Q = [P] [R]

If Q > K, too much product, system will proceed left to reach equilibrium (K). rate f < rate r R P rate f = rate r R P Q = [P] [R] K Q KQK Q Q > KQ > K Q =Q = [P] [R]

N 2 (g) + 3 H 2 (g)  2 NH 3 (g) K =K = [NH 3 ] 2 [N 2 ][H 2 ] 3 Q =Q = [NH 3 ] 2 [N 2 ] [H 2 ] 3 K =K = [NH 3 ] 2 [N 2 ] [H 2 ] 3 Q < KQ < K Q = K (same K) Q = K Add reactant:

increase rate f and rate r. R P Equilibrium occurs faster, but… at no shift (composition [P]/[R] is same). [P][R][P][R] K =K = (same) Catalysts

shift away faster (consume) shift toward faster (replace ) fewer mol of gas (↓n gas ) Le Châtelier’s Principle more mol of gas (↑n gas ) R  P (summary) no shift (H + R  P) (R  P + H) (changes K) (P total ) (M, PP R, PP P ) in endo dir. to use up heat in exo dir. to make more heat

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

K sp = [Ag + ] 2 [SO 4 2− ] Saturated solutions are in equilibrium: Ag 2 SO 4 (s) 2 Ag + (aq) + SO 4 2− (aq) Solubility Product Constant (K sp ) K sp is solubility-product constant solubility is g/L molar solubility is mol/L (M) X a Y b (s)  a X + (aq) + b Y – (aq) [X a Y b ] [X + ] [Y – ] molar solubility molar concentrations of solid of ions K sp = [X + ] a [Y – ] b

[PbBr 2 ] is M at 25 o C. K sp Calculations PbBr 2 (s)  Pb Br – – K sp = [Pb 2+ ][Br – ] 2 If solubility (or molar solubility) is known, solve for K sp. K sp = (0.010)(0.020) 2 K sp = 4.0 x 10 –6 1 PbBr 2 dissociates into… 1 Pb 2+ ion and 2 Br – ions ICEICE

If only K sp is known, solve for x (M). AgCl (s)  Ag + + Cl – x 0 0 –x +x +x 0 x x K sp = [Ag + ][Cl – ] K sp = x x 10 –10 = x 2 √1.8 x 10 –10 = x 1.3 x 10 –5 = x K sp for AgCl is 1.8 x 10 –10. [AgCl] = 1.3 x 10 –5 M [Ag + ] = 1.3 x 10 –5 M [Cl – ] = 1.3 x 10 –5 M (molar solubility) K sp Calculations ICEICE

Cr(OH) 3 (s)  Cr OH – x 0 0 –x +x +3x 0 x 3x K sp = [Cr 3+ ][OH – ] 3 K sp = (x)(3x) 3 K sp = 27x 4 [Cr(OH) 3 ] = 1.6 x 10 –8 M [Cr 3+ ] = 1.6 x 10 –8 M [OH – ] = 4.8 x 10 –8 M 1.6 x 10 –30 = 27x 4 4 √5.9 x 10 –32 = x 1.6 x 10 –8 = x If only K sp is known, solve for x (M). K sp for Cr(OH) 3 is 1.6 x 10 –30. (molar solubility) K sp Calculations ICEICE

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? adding common ion shifts left (less soluble) OR 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.

LaF 3 (s)  La F – x 0 0 –x +x +3x 0 x 3x K sp = [La 3+ ][F – ] 3 K sp = (x)(3x) 3 2 x 10 –19 = 27x 4 x = 9 x 10 –6 M LaF 3 Common Ion  less soluble (in pure H 2 O)(in M KF) LaF 3 (s)  La F – x 0 –x +x +3x 0 x x K sp = [La 3+ ][F – ] 3 K sp = (x)( x) 3 2 x 10 –19 = (x)(0.010) 3 x = 2 x 10 –13 M LaF ≈ b/c K <<<1 Solubility is lower in _________________ sol’n w/ common ion

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.

more soluble by forming complex ions AgCl (s)  Ag + (aq) + Cl − (aq) NH 3 Ag(NH 3 ) 2 + Adding :NH 3 causes… shift , more soluble.

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