1. Pages 588-625, Sections 18.1, 18.2, and 18.4 (excluding 605-612, Section 18.3)

2. Chemical Equilibrium In what ways have we studied equilibrium thus far in class? Examples: When dissolving sugar in water, equilibrium is reached when molecules of sugar are dissolving at the same rate that other molecules of sugar are crystallizing A liquid in a closed container reaches an equilibrium vapor pressure when the rate of evaporation from the surface of the liquid equals the rate of condensation

3. Reversible Reactions Any reaction can theoretically be reversible 2 HgO (s)  2 Hg (l) + O 2 (g) 2 Hg (l) + O 2 (g)  2HgO (s) Consider a closed container in which mercury (II) oxide is heated. The mercury and oxygen cannot escape, so what will they eventually do?

4. Reversible Reactions The two reactions will eventually reach an equilibrium. The rate of mercury (II) oxide decomposition will be equal to the rate of mercury (II) oxide formation. The concentrations of products and reactants will remain unchanged, but not necessarily equal. 2 HgO (s) 2 Hg (l) + O 2 (g) Notice the double arrows used to denote a reversible reaction.

5. Chemical Equilibrium While all reactions can theoretically be reversible, not all reaction systems are in a state of equilibrium Characteristics of an equilibrium system: Takes place in closed container Is a dynamic equilibrium Rate forward = Rate reverse Concentrations of substances remain constant; however, concentrations of different substances are not necessarily equal Initial concentrations of products and reactants do not matter – the equilibrium concentrations will remain the same under the same conditions (temp, pressure, etc.)

6. Chemical Equilibrium Most reactions will only reach an equilibrium when in a closed container: This rusting reaction will go to completion – meaning that once products are formed, the reaction ends (the reaction is never reversed). This happens because in an open system, there is virtually a limitless supply of oxygen gas.

7. Dynamic Equilibrium A chemical equilibrium is said to be dynamic This means that both the forward reaction AND the reverse reaction are occurring at the same time (one does not occur in response to the other)

8. The Equilibrium Expression Consider the following reaction: nA + mB xC + yD At the beginning of the reaction: -The concentration of products C and D is at zero. - The concentration of reactants A and B is at the maximum.

9. The Equilibrium Expression Consider the following reaction: nA + mB xC + yD Over time, as the reaction proceeds: -The rate of the forward reaction slows down as A and B are used up - The rate of the reverse reaction increases as products C and D are formed.

10. The Equilibrium Expression Consider the following reaction: nA + mB xC + yD Eventually, the two rates become equal, and an equilibrium is established.

11. The Equilibrium Expression Notice the difference in these two graphs!! Both show equilibrium being established, but one shows reaction rate and the other concentration. The concentrations of the reactants and products remain constant at equilibrium, but are not necessarily equal. The reaction rates of the forward and reverse reactions ARE equal at equilibrium.

12. The Equilibrium Expression Consider the following reaction: nA + mB xC + yD The relationship between the concentration of products and reactants at equilibrium can be expressed by the equilibrium constant, K

13. The Equilibrium Expression K – equilibrium constant for a specified temperature [ ] – concentration in molarity (mol/L)

14. The Equilibrium Expression K is independent of the initial concentrations However, it is dependent on the temperature. Only concentrations of substances that can actually change are included in K. Pure solids and liquids are omitted because their concentrations cannot change! A small amount of a solid and a large amount of a solid are still both pure.

15. Writing the K eq Expression Write the equilibrium expression for the following chemical reaction:

16. Writing the K eq Expression Write the equilibrium expression for the following chemical reaction:

17. Meaning of K If K ≈ 1, then the [reactants] is roughly equal to the [products] at equilibrium. If K is large (>>1), then the [products] is greater than the [reactants] at equilibrium. If K is small (<< 1), then the [reactants] is greater than the [products] at equilibrium.

18. Reaction Quotient Say you want to determine if a particular reaction system is at equilibrium The reaction quotient (Q) is used to determine if a reaction is at equilibrium by comparing its value to K eq Q is calculated in exactly the same way as K eq, except that the concentrations are not necessarily at equilibrium

19. Meaning of Q If Q = K, then the reaction is at equilibrium If Q < K, then the reaction is to the left of equilibrium – there are more reactants than products (compared to equilibrium) We would say that the reaction will then move to the right – producing more products until equilibrium is reached If Q > K, then the reaction is to the right of equilibrium – there are more products than reactants (compared to equilibrium) We would say that the reaction will then move to the left– producing more reactants until equilibrium is reached

21. Le Châtelier’s Principle French chemist Henri Louis Le Châtelier studied chemical equilibrium, and came up with the following principle “When a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that acts to relieve that stress” In other words, when a change is made to a system at equilibrium, the system will respond in such a way so as to restore equilibrium

22. Le Châtelier’s Principle Le Châtelier’s Principle is most easily observed when the following changes to a system occur: Changes in pressure (when gases are involved) Changes in concentration Changes in temperature** ** Remember that K eq is temperature dependent! Changes in pressure or concentration will not affect K eq, but changes in temperature will

23. Changes in Pressure Changes in pressure only affect equilibria involving gases 1) Change in pressure: When the pressure on a system is increased, the system acts to reduce pressure by reducing the number of gas particles present When the pressure is decreased, the number of gas particles will increase For equilibria where the number of moles of gaseous reactants is equal to the number of moles of gaseous products, changes in pressure cannot affect the system

24. Changes in Pressure Example 1: Predict whether each of the following pressure changes would favor the forward or reverse reaction: (a)increased pressure (b)decreased pressure a) Favors forward rxn b) Favors reverse rxn

25. Changes in Pressure Example 2: Predict whether each of the following pressure changes would favor the forward or reverse reaction: (a)increased pressure (b)decreased pressure a) no effect b) no effect

26. Changes in Pressure Adding an inert gas into the reaction vessel does not affect the equilibrium Although this would increase the total pressure of the system, it does not change the partial pressures of the reactants/products “Increasing pressure by adding a gas that is not a reactant or product cannot affect the equilibrium position of the system.”

27. Changes in Pressure Example 3: What happens to the solution when the bottle is opened? Equilibrium shifts to the left, and bubbles form in the solution

28. Changes in Concentration 2) Change in concentration If the concentration of a substance is lowered, the reaction will shift to increase it If the concentration of a substance is increased, the reaction will shift to lower it Remember that the concentrations of solids and liquids do not change! Changing the amount of a solid or liquid does not affect the equilibrium expression.

29. Changes in Concentration Example 4: Predict the direction of the shift in the equilibrium position of this reaction when (a) CO is added (b) As 4 is removed (c) How would an increase in P affect this system?

30. Changes in Concentration Example 4: Predict the direction of the shift in the equilibrium position of this reaction when (a) Shifts to left (b) Shifts to right (c) Shifts to left

31. Changes in Temperature 3) Change in temperature If heat is added to the system, the system reacts to absorb this heat – favoring the endothermic reaction If heat is added to the system, then the observed temperature will increase. So, temp increase favors endothermic reaction If heat is removed from the system, the system reacts to release more heat – favoring the exothermic reaction If heat is removed from the system, then the observed temperature will decrease. So, temp decrease favors exothermic reaction

32. Changes in Temperature Example 5: Predict the direction of the shift in the equilibrium position of this reaction when (a) the temperature is increased (b) the temperature is decreased

33. Changes in Temperature Example 5: Predict the direction of the shift in the equilibrium position of this reaction when (a) Shifts to left (b) Shifts to right

34. Changes in Temperature Example 6: What kind of reaction will shift to the right with the addition of heat, exothermic or endothermic? Endothermic

35. Addition of a Catalyst 4) Catalyst Adding a catalyst increases both the forward and reverse reaction rates Therefore, addition of a catalyst does not affect the equilibrium position

37. Solubility Equilibrium Ionic solids dissolve in water until they are in equilibrium with their ions We can write equilibrium expressions to describe a substance’s dissociation This solubility product (K sp ) can be used to determine if precipitation will occur when two solutions are mixed

38. Solubility Product K sp is calculate the same way as other equilibrium expressions AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ] [Cl - ]

39. Solubility Product The more the ionic compound dissolves, the larger the K sp value. The more it dissolves, the more product (ions) there is, so the larger the K sp value. Solubility product is only useful for describing solutions of sparingly soluble salts (those substances not classified as “soluble” on your solubility rules)

40. Solubility Equilibrium Write the expression for the solubility product for Fe(OH) 3.

41. Solubility Equilibrium Write the expression for the solubility product for PbI 2.

42. Solubility Equilibrium At 25 o C, the concentration of Pb +2 ions in a saturated solution of PbF 2 is 1.9 x 10 -3 M. What is the value of K sp for PbF 2 ?

43. Solubility Equilibrium At 25 o C, the concentration of Cd +2 ions in a saturated solution of Cd(OH) 2 is 1.7 x 10 -5 M. What is the value of K sp for Cd(OH) 2 ?

44. Solubility Equilibrium Looking at the previous two examples, which solute dissolves in water to a greater extent, PbF 2 or Cd(OH) 2 ? Whichever one has the larger K sp value dissolves better: PbF 2

45. Solubility Equilibrium What will be the equilibrium concentrations in a saturated solution of Ni(OH) 2 if the K sp = 1.6 x 10 -16 ?

46. Solubility Equilibrium What does this mean?

47. Ion Product We can use the ion product to determine if a precipitate will form if two solutions are mixed AgCl (s) Ag + (aq) + Cl - (aq) K sp = [Ag + ] [Cl - ] = 1.8 x 10 -10 If [Ag + ] [Cl - ] < K sp, the solution is unsaturated (no precipitate forms) If [Ag + ] [Cl - ] > K sp, a precipitate forms

48. Ion Product Will a precipitate form if 20.0 mL of 0.010 M BaCl 2 solution is mixed with 20.0 mL of 0.0050 M Na 2 SO 4 (aq)? Try additional practice problems on page 620