1. 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

2. 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.

3. 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 Initial1.0000 change equil -0.20+0.10+0.10 0.800.10 0.10

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

5. 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

6. 2NO(g) + O 2 (g) 2NO 2 (g) The value for K c for the above reaction is 5.6 x 10 12 at 290K. What is the value of K p ? ∆n=-1.0821)(290)] -1 K p = 5.6 x 10 12 [(.0821)(290)] -1 = 5.6 x 10 12 = 2.4 x 10 11 = 5.6 x 10 12 = 2.4 x 10 11.0821)(290) (.0821)(290)

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

8. 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

9. SubstanceFormulaK sp Aluminum hydroxideAl(OH) 3 4.6 x 10 -33 Barium carbonateBaCO 3 2.6 x 10 -9 Barium chromateBaCrO 4 1.2 x 10 -10 Barium fluorideBaF 2 1.0 x 10 -6 Barium sulfateBaSO 4 1.1 x 10 -10 Bismuth sulfideBi 2 S 3 1.8 x 10 -99 Cadmium carbonateCdCO 3 6.2 x 10 -12 Cadmium hydroxideCd(OH) 2 5.3 x 10 -15 Cadmium iodateCd(IO 3 ) 2 2.4 x 10 -8 Cadmium oxalateCdC 2 O 4 1.5 x 10 -8

10. 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 10 -13 Find [Br - ] Hint: [Br - ] = [Ag + ][Br - ] = 7.08 x 10 -7 M

11. 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 10 -22 For every Be 2+ ion, there are 2OH - ions. 1.58 x 10 -22 = x(2x) 2 = 4x 3 X = 3.41 x 10 -8 M

12. 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 0.2207 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. 0.2207 g PbCl 2 x (1 mol PbCl 2 /278.1 g PbCl 2 )= 7.94 x 10 -4 mol 7.94 x 10 -4 mol = 0.0159 M PbCl 2.0500L

13. create an "ICE" table PbCl 2 (s) Pb 2+ (aq)Cl - (aq) InitialAll solid00 Change- 0.0159 M+ 0.0159 M+ 0.0318 M equilibriumLess solid0.0159 M0.0318 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

14. Find [IO 3 - ] in a saturated solution of copper (II) iodate. K sp of Cu(IO 3 ) 2 is 7.41 x 10 -8 K sp of Cu(IO 3 ) 2 is 7.41 x 10 -8 Cu(IO 3 ) 2  Cu 2+ + 2IO 3 - Cu(IO 3 ) 2  Cu 2+ + 2IO 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 3 7.41 x 10 -8 =4x 3 X= 2.65 x 10 -3 M and [IO 3 - ] = 5.30 x 10 -3 M 2x

15. 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

16. 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 10 -6 K sp = 3.39 x 10 -6 Initial 0 0.10 0.10 Change +x +x Equilibrium x.10 + x.10 + x

17. K sp = [Tl + ][Br - ] = 3.39 x 10 -6 3.39 x 10 -6 = x (.10 +x) 3.39 x 10 -6 = x (.10 +x) X ͌ 0.00184 so.10 + 0.00184 =.10 X ͌ 0.00184 so.10 + 0.00184 =.10 3.39 x 10 -6 = x (.10) 3.39 x 10 -6 = x (.10) 3.39 x 10 -5 = x 3.39 x 10 -5 = x