1. more equilibrium concepts

2. LeChâtelier’s Principle states that when a stress is applied to a system at equilibrium, the system will respond in a manner that attempts to bring the system back to equilibrium…stresses are: Concentration Pressure Temperature

3. Concentration If you add more reactant, then… the rxn will shift to the right to use the excess reactant. If you add more product, then… the rxn will shift to the left to use the excess product. If you remove reactant, then… If you remove product, then…

4. Pressure 3 means of affecting pressure of gaseous systems Concentration Volume of container Addition of an inert gas

5. Pressure—concentration’s effect Same concept as concentration change

6. Pressure—volume’s effect If you decrease the volume of the container, then the pressure will increase and the system will adjust by… shifting to the side with the fewer number of moles in an effort to bring the pressure back down to what it was at equilibrium

7. Pressure—volume’s effect If you increase the volume of the container, then the pressure will decrease and the system will adjust by… shifting to the side with the greater number of moles in an effort to bring the pressure back up to what it was at equilibrium

8. Pressure—volume’s effect If you increase or decrease the volume of the container for a reaction who has equal number of moles on reactant and product side, then… the system cannot make the appropriate changes and the rxn will not be able to reach equlibrium

9. Pressure—inert gas’s effect If you add an inert gas to a system at equilibrium, then… there will be no shift because the inert gas only increases the total pressure of the system (think Dalton’s Law) and has no effect on the partial pressures of the reactants or products

10. Temperature effect depends on whether the rxn is endothermic or exothermic endothermic—absorbs heat energy, thus the energy appears as a reactant or the change in enthalpy is positive exothermic—releases heat energy, thus the energy appears as a product or the change in enthalpy is negative

11. Temperature—Endothermic Rxn If you add more heat, then… the rxn will shift to the right to use the excess heat. (think of the heat as a reactant) If you remove heat, then… the rxn will shift to the left to replenish the missing heat.

12. Temperature—Exothermic Rxn If you add more heat, then… the rxn will shift to the left to use the excess heat. (think of the heat as a product) If you remove heat, then… the rxn will shift to the right to replenish the missing heat.

13. Le Châtelier Practice 1.Consider the reaction: N 2 (g) + 3H 2 (g)  2NH 3 (g) + 33.3kJ Explain the direction of shift it will experience when the following stresses are applied: a.The system is heated. b.Its container is squeezed. c.Ammonia is added to the system. d.Neon is added to the system.

14. Le Châtelier Practice 2.Consider the reaction: 2SO 3 (g) + 33.3kJ  2SO 2 (g) + O 2 (g) Explain the direction of shift it will experience when the following stresses are applied: a.The system is cooled. b.Its container expands. c.Sulfur trioxide is added to the system. d.Argon is added to the system.

15. applying equilibrium stuff to salts

16. K sp Unlike NaCl, not all salts dissolve completely in aqueous solutions. The level to which a salt dissolves is expressed as its solubility product constant and is represented by K sp (similar to K) Remember that only gaseous and aqueous thingies can be represented in K expressions.

17. K sp So, in a salt dissociation rxn, only the products are aqueous. The reactant (or salt) is solid. Consider the salt, silver hydroxide. Write its formula. AgOH

18. K sp Now, write its dissociation into its ions… AgOH (s)  Ag 1+ (aq) + OH 1- (aq) Make sure you pay attention to the phases. Now write the K sp expression.

19. K sp K sp = [Ag 1+ ][OH 1- ] Note that the AgOH(s) is absent because it is a solid. Also note that the power of the Ag 1+ and OH 1- are both one because the dissociation only yielded one of each ion.

20. K sp This expression allows you to work three types of problems: determine the concentrations of ions that have dissolved determine the solubility of the salt (it will always equal x in your equilibrium chart) determine the K sp of a salt

21. K sp Practice Problem #1 Determine the concentration of each ion if the K sp of calcium fluoride is 4.0 x 10 -11.

22. K sp Practice Problem #1 Step 1—write the dissociation of salt. CaF 2 (s)  Ca 2+ (aq) + 2F 1- (aq) Recognize that the mol:mol for the ions is 1:2. This is very important info that you will use in the problem, so make certain that the formula for the salt is correct!

23. K sp Practice Problem #1 Step 2—make a dissociation chart like you did for the other equilibrium rxns. Remember…only (g) & (aq)!! i -- 0 0 [CaF 2 ] [Ca 2+ ] [F 1- ] Δ -- +x +2x eq -- x 2x

24. K sp Practice Problem #1 Step 3—write your K sp expression and plug in your stuff K sp = [Ca 2+ ][F 1- ] 2 Note that the mol:mol is reflected in the power as well as in your eq line of your chart.

25. K sp Practice Problem #1 Step 3—continued K sp = [Ca 2+ ][F 1- ] 2 4.0 x 10 -11 = [x][2x] 2 4.0 x 10 -11 = 4x 3 1.0 x 10 -11 = x 3 2.15 x 10 -4 = x

26. K sp Practice Problem #1 Step 4—use your value for x and the eq line of your chart to figure out the concentrations of the ions. [Ca 2+ ]= 2.15 x 10 -4 M [F 1- ]= 2(2.15 x 10 -4 ) = 4.30 x 10 -4 M

27. K sp Practice Problem #2 Determine the solubility of calcium fluoride from your work in Problem #1.

28. K sp Practice Problem #2 Step 1—You’ve already done all of the work for this problem. Since the mol:mol:mol for the salt to its ions is 1:1:2, the solubility of the salt in this case is equal to the concentration of the calcium ion. Just remember that the solubility of the salt will always equal x from your eq line of your chart

29. K sp Practice Problem #2 Step 1—Thus, the answer to this problem is 2.15 x 10 -4 M for the solubility of the calcium fluoride.

30. K sp Practice Problem #3 Determine the value of the solubility product constant of bismuth sulfide which has a solubility of 1.0 x 10 -15 M.

31. K sp Practice Problem #3 Step 1—write the dissociation of salt and the K sp expression. Bi 2 S 3 (s)  2Bi 3+ (aq) + 3S 2- (aq) K sp = [Bi 3+ ] 2 [S 2- ] 3 K sp = [2x] 2 [3x] 3 K sp = 108x 5 consolidated

32. K sp Practice Problem #3 Step 2—Since your know the solubility (or x) of the salt, you can deduce the concentrations of the ions from the mol:mol of the ions. Therefore, [Bi 3+ ] will be two times the solubility, and [S 2- ] will be three times the solubility.

33. K sp Practice Problem #3 Step 3—Plug your solubility into the expression as x and solve for K sp. K sp =[2(1.0 x 10 -15 )] 2 [3(1.0 x 10 -15 )] 3 Or since,K sp = 108x 5 … 108(1.0 x 10 -15 ) 5

34. K sp Practice Problem #3 Thus, the K sp should be 1.08 x 10 -73 M