Q = reaction quotient Q = K eq only at equilibrium If Q<K eq then the forward reaction is favored If Q>K eq then the reverse reaction is favored If K eq >1, the reaction is considered spontaneous

KpKpKpKp Equilibrium constant when reactants and products are gases. Equilibrium constant when reactants and products are gases. Treated same as K eq, but using atm or kPa rather than molarities. Treated same as K eq, but using atm or kPa rather than molarities.

2HI(g)  H 2 (g) + I 2 (g) Originally, a system contains only HI at a pressure of 1.00 atm at 520°C. The equilibrium partial pressure of H 2 is found to be 0.10atm. Calculate the equilibrium K p for iodine, hydrogen, and the system. Originally, a system contains only HI at a pressure of 1.00 atm at 520°C. The equilibrium partial pressure of H 2 is found to be 0.10atm. Calculate the equilibrium K p for iodine, hydrogen, and the system. 2HI H2H2H2H2 I2I2I2I2 Initial change equil

K p =(.10)(.10) =.0156 K p =(.10)(.10) =.0156 (.80) 2 (.80) 2

Converting K c to K p PV=nRT or P=nRT P = [A]RT V K p = K c (RT) Δn Δn = (mol gaseous product) – (mol gas reactant) If Δn = 0, then K p =K c R is.0821 L·atm/mol·K

2NO(g) + O 2 (g) 2NO 2 (g) The value for K c for the above reaction is 5.6 x at 290K. What is the value of K p ? ∆n= )(290)] -1 K p = 5.6 x [(.0821)(290)] -1 = 5.6 x = 2.4 x = 5.6 x = 2.4 x )(290) (.0821)(290)

K sp -Solubility product constant- Like K eq but don’t include the solid

NaCl (s)  Na + (aq) + Cl - (aq) K sp = [Na + ][Cl - ] For CaCl 2 (s)  Ca 2+ (aq) + 2Cl - (aq) K sp = [Ca 2+ ][Cl - ] 2

SubstanceFormulaK sp Aluminum hydroxideAl(OH) x Barium carbonateBaCO x Barium chromateBaCrO x Barium fluorideBaF x Barium sulfateBaSO x Bismuth sulfideBi 2 S x Cadmium carbonateCdCO x Cadmium hydroxideCd(OH) x Cadmium iodateCd(IO 3 ) x Cadmium oxalateCdC 2 O x 10 -8

Solubility Product Constant AgBr(s)  Ag + (aq) + Br - (aq) K eq = [Ag + ][Br - ] [AgBr] [AgBr] Solids are omitted from equilibrium expressions, so... K sp = [Ag + ][Br - ] For AgBr (at 25°C), K sp = 5.01 x Find [Br - ] Hint: [Br - ] = [Ag + ][Br - ] = 7.08 x M

What is the solubility expression for a solution of Cu 3 (PO 4 ) 2 ? Cu 3 (PO 4 ) 2  3Cu 2+ (aq) + 2PO 4 3- K sp = [Cu 2+ ] 3 [PO 4 3- ] 2 What is the [Be 2+ ] in a saturated solution of Be(OH) 2 ? K sp = 1.58 x For every Be 2+ ion, there are 2OH - ions x = x(2x) 2 = 4x 3 X = 3.41 x M

Calculate the solubility product constant for lead(II) chloride, if 50.0 mL of a saturated solution of lead(II) chloride was found to contain g of lead(II) chloride dissolved in it. PbCl 2 (s)  Pb 2+ (aq) + 2 Cl - (aq) K sp = [Pb 2+ ][Cl - ] 2 Convert the amount of dissolved lead(II) chloride into moles per liter g PbCl 2 x (1 mol PbCl 2 /278.1 g PbCl 2 )= 7.94 x mol 7.94 x mol = M PbCl L

create an "ICE" table PbCl 2 (s) Pb 2+ (aq)Cl - (aq) InitialAll solid00 Change M M M equilibriumLess solid M M Substitute the equilibrium concentrations into the equilibrium expression and solve for K sp. K sp = [0.0159][0.0318] 2 = 1.61 x 10 -5

Find [IO 3 - ] in a saturated solution of copper (II) iodate. K sp of Cu(IO 3 ) 2 is 7.41 x K sp of Cu(IO 3 ) 2 is 7.41 x Cu(IO 3 ) 2  Cu IO 3 - Cu(IO 3 ) 2  Cu IO 3 - If [Cu 2+ ]=x then [IO 3 - ]= If [Cu 2+ ]=x then [IO 3 - ]= K sp =[Cu 2+ ][IO 3 - ] 2 K sp =[Cu 2+ ][IO 3 - ] 2 = x(2x) 2 = x(2x) 2 = 4x 3 = 4x x =4x 3 X= 2.65 x M and [IO 3 - ] = 5.30 x M 2x

Common Ion Effect K sp is unchanged by the addition of a solute. K sp is unchanged by the addition of a solute. The solubility of a slightly soluble salt is reduced by the presence of a second solute that produces a common ion. The solubility of a slightly soluble salt is reduced by the presence of a second solute that produces a common ion. Addition of a common ion will decrease the concentration of the ion it bonds with, creating a precipitate. Addition of a common ion will decrease the concentration of the ion it bonds with, creating a precipitate. If the ions’ concentrations > K sp, ppt will form If the ions’ concentrations > K sp, ppt will form

What is the [Tl + ] when.050mol of NaBr is added to 500.0ml of a saturated solution of TlBr? What should happen according to LeChatelier’s Principle? What should happen according to LeChatelier’s Principle? TlBr  Tl + Br - TlBr  Tl + Br - K sp = 3.39 x K sp = 3.39 x Initial Change +x +x Equilibrium x.10 + x.10 + x

K sp = [Tl + ][Br - ] = 3.39 x x = x (.10 +x) 3.39 x = x (.10 +x) X ͌ so =.10 X ͌ so = x = x (.10) 3.39 x = x (.10) 3.39 x = x 3.39 x = x