1. Forward and reverse reactions taking place at equal rates It is a dynamic state - reactions are constantly occurring

2. (a)Start: 10 goldfish in the left tank and 10 guppies in the right. (b)Equilibrium state. with 5 of each kind of fish in each tank. The equilibrium is dynamic; an averaged state and not a static condition. The fish do not stop swimming when they have become evenly mixed. (c)If we were to observe one single fish (here a guppy among goldfish). we would find that it spends half its time in each tank.

3. Brightstorm videos Chemical Equilibrium Definition 5:01 http://www.youtube.com/watch?v=FYc_SoW2M40&list=PL06C3C4E3F 84C6A24&index=42 Crash course chemistry http://www.youtube.com/watch?v=g5wNg_dKsYYhttp://www.youtube.com/watch?v=g5wNg_dKsYY 10:56 Equilibrium http://www.youtube.com/watch?v=DP-vWN1yXrYhttp://www.youtube.com/watch?v=DP-vWN1yXrY 9:28 Equilibrium equations You don’t need to know how to do the RICE table starting at 4:40 Isaacs Teach http://www.youtube.com/watch?v=g4TKRInLdPA 10:09 Equilibriumhttp://www.youtube.com/watch?v=g4TKRInLdPA Good basic explanation! http://www.youtube.com/watch?v=4z4_rc6nsKUhttp://www.youtube.com/watch?v=4z4_rc6nsKU 12:46 What is the equilibrium constant, Keq? Also very good explanation

4. Equilibrium constant expressions aA + bB  cC + dD K eq = [C] c [D] d [A] a [B] b

5. General information about the K eq expression Equilibrium [ ] of products are placed in the numerator. Equilibrium [ ] of reactants are placed in the denominator. Each [ ] term is raised to an exponent equal to its coefficient in the balanced equation. If there is more than 1 product or reactant, the terms are multiplied. Solids and liquids (pure substances) are not included in the K eq expression. This is because their [ ] are their densities. The density of a substance does not change with changing temperatures.

6. K eq is constant for a given reaction at a given temperature. There are no units associated with the value of K eq. The value of K eq is independent of the: –individual [ ] of reactants and products – original [ ] of reactants and products –volume of the container. The value of K eq is dependent on temperature. What does the value of K eq tell you about a reaction? K eq >1: more products than reactants at equilibrium K eq < 1: more reactants than products at equilibrium

7. Using equilibrium constants Calculating equilibrium concentrations: Example: At 1405 K, hydrogen sulfide, also called rotten egg gas (because of its bad odor), decomposes to form hydrogen and a diatomic sulfur molecule,S 2. K eq = 2.27 x 10 -3. (a) Write the balanced equation for the reaction described above. Write out the K eq expression. (b) Calculate the concentration of hydrogen gas if [S 2 ] = 0.0540 M and [H 2 S] = 0.184 M.

8. Solving the problem – part (a) 2H 2 S (g)  2H 2 (g) + S 2 (g) K eq = [H 2 ] 2 [S 2 ] [H 2 S] 2

9. Solution – part (b) [H 2 ] 2 = K eq [H 2 S] 2 = [S 2 ] (2.27 x 10 -3 )(0.184 M) 2 = 0.0540 M [H 2 ] 2 = 1.42 x 10 -3 M  [H 2 ] = 3.77 x 10 -2 M

10. Le Châtelier’s principle : 1884 - Henri Le Châtelier When a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress.

11. Δ in concentration Adding more of a reactant or product: the reaction will shift in the direction to consume a portion of what was added. –more reactant  shifts right –more product  shifts left Removing some of a reactant or product: the reaction will shift in a direction to restore part of what was removed. –reactants removed  reaction shifts left (i.e. the reverse reaction) –products removed  reaction shifts right (i.e. the forward reaction).

12. Δ in volume Relevant when discussing gaseous equilibria and when the number of moles of gaseous reactants differ from the number of moles of gaseous products. The change in volume is a result of a change in pressure of the gaseous system. When P↓, the reaction will shift in a direction to↑ number of moles of gas. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) 1 mol  2 mol 2NH 3 (g)  N 2 (g) + 3H 2 (g) 2 mol  4 mol When P↑, the reaction will shift in a direction to ↓ number of moles gas. PCl 5 (g)  PCl 3 (g) + Cl 2 (g) 1 mol  2 mol 2NH 3 (g)  N 2 (g) + 3H 2 (g) 2 mol  4 mol

13. Δ in temperature View changes in temperature as reactants or products. When the temperature of an equilibrium system is ↑ the reaction that is endothermic (ΔH>0) will take place. * forward rxn is endothermic  more product (shifts to the right). * reverse rxn is endothermic  less product (shifts to the left) When the temperature of an equilibrium system is ↓, the rxn which is exothermic (ΔH<0) will take place. * forward rxn is exothermic – more product (shifts to the right). * reverse rxn is exothermic – less product (shifts to the left) General rule: if the forward rxn is endothermic,↑K. If the forward rxn is exothermic ↓K.

14. Animation demonstration http://www.learnerstv.com/animation/animation.php?ani= 120&cat=chemistry http://www.youtube.com/watch?v=PciV_Wuh9V8 7:00 Le Chatelier’s Principle; good explanations with visuals and excellent discussion on how to increase yield of a reaction http://www.youtube.com/watch?v=dd5p0VZ-MZg 5:36 Equilibrium disturbances This one will help you with the lab we’re doing. He also discusses the effect of disturbances (changes) in an equilibrium system and how they affect the value of K (the equilibrium constant) http://www.learnerstv.com/animation/animation.php?ani= 120&cat=chemistry http://www.youtube.com/watch?v=PciV_Wuh9V8 http://www.youtube.com/watch?v=dd5p0VZ-MZg

15. Reactions that go to completion  Formation of a gas H 2 CO 3 (aq)  H 2 O (l) + CO 2 (g)  Formation of precipitate (Double displacement reactions)  Formation of a slightly ionized product; often times H 2 O (i.e. in a neutralization reaction) H 3 O + + OH -  2H 2 O (l)

16. Solubility equilibria http://www.youtube.com/watch?v=YJ- dyEtB66A&feature=topics

17. The Solubility Product Constant, K sp Many important ionic compounds are only slightly soluble in water and equations are written to represent the equilibrium between the compound and the ions present in a saturated aqueous solution. The solubility product constant, K sp, is the product of the concentrations of the ions involved in a solubility equilibrium, each raised to a power equal to the stoichiometric coefficient of that ion in the chemical equation for the equilibrium.

18. The Solubility Equilibrium Equation And K sp

20. K sp And Molar Solubility The solubility product constant is related to the solubility of an ionic solute, but K sp and molar solubility - the molarity of a solute in a saturated aqueous solution - are not the same thing. Calculating solubility equilibria fall into two categories: –determining a value of K sp from experimental data –calculating equilibrium concentrations when K sp is known.

21. Calculating K sp From Molar Solubility

22. Calculating Molar Solubility From K sp

23. The Common Ion Effect In Solubility Equilibria The common ion effect also affects solubility equilibria. Le Châtelier’s principle is followed for the shift in concentration of products and reactants upon addition of either more products or more reactants to a solution. The solubility of a slightly soluble ionic compound is lowered when a second solute that furnishes a common ion is added to the solution.

24. Solubility Equilibrium Calculation -The Common Ion Effect

25. Determining Whether Precipitation Occurs

26. Determining Whether Precipitation Occurs – An Example The concentration of calcium ion in blood plasma is 0.0025 M. If the concentration of oxalate ion is 1.0x10 -7 M, do you expect calcium oxalate to precipitate? K sp = 2.3x10 -9. Three steps: (1)Determine the initial concentrations of ions. (2)Evaluate the reaction quotient Q. (3)Compare Q with K sp.

27. Solution

28. Summary The solubility product constant, K sp, represents equilibrium between a slightly soluble ionic compound and its ions in a saturated aqueous solution. The common ion effect is responsible for the reduction in solubility of a slightly soluble ionic compound. The solubilities of some slightly soluble compounds depends strongly on pH.

29. Equilibrium lab

30. Qualitative Inorganic Analysis Acid-base chemistry, precipitation reactions, oxidation-reduction, and complex-ion formation all come into sharp focus in an area of analytical chemistry called classical qualitative inorganic analysis. “Qualitative” signifies that the interest is in determining what is present, not how much is present. Although classical qualitative analysis is not as widely used today as instrumental methods, it is still a good vehicle for applying all the basic concepts of equilibria in aqueous solutions.