1. CHEMISTRY CHAPTER 18. CHEMICAL EQUILIBRIUM
SECTION 1. THE NATURE OF CHEMICAL EQUILIBRIUM
Reversible Reactions
Theoretically, every reaction can proceed in two directions, forward and reverse.
A chemical reaction in which the products can react to re-form the reactants is called a reversible reaction.

2. Ex. (p. 589): Decomposition of mercury (II) oxide: 2HgO(s) → 2Hg(l) + O2(g)
If this is carried out in a closed container, some of the Hg and O2 can react to produce HgO:
2Hg(l) + O2(g) → 2HgO(s)
When both are happening, we use two arrows to show that the reaction is reversible:

3. With time, the system reaches a point where forward and reverse reactions are happening at equal rates. Mercury (II) oxide is being broken down at the same rate that it is being formed. This is called equilibrium. At equilibrium, the concentrations of reactants and products do not change with time.

4. At equilibrium, the reaction does not stop
At equilibrium, the reaction does not stop. It continues in both directions, but there is no net change in composition.

5. Reaction Rate Over Time for an Equilibrium System

6. The Equilibrium Constant
For a reaction
After equilibrium is reached, the individual concentrations of A, B, C, and D undergo no further change if conditions remain the same.

7. A ratio of their concentrations should also remain constant.
The equilibrium constant is designated by the letter K.

8. The equilibrium constant, K, is the ratio of the mathematical product of the concentrations of substances formed at equilibrium to the mathematical product of the concentrations of reacting substances. Each concentration is raised to a power equal to the coefficient of that substance in the chemical equation.

9. The constant K is independent of the initial concentrations.
K is dependent on the temperature of the system.
The numerical value of K for a particular equilibrium system is obtained experimentally.

10. If the value of K is small, the reactants are favored.
A large value of K indicates that the products are favored.
Only the concentrations of substances that can actually change are included in K.
Pure solids and liquids are omitted because their concentrations cannot change.

11. We have already seen one equilibrium constant, the ionization of water.
The self-ionization of water is an equilibrium reaction.

12. Equilibrium is established with a very low concentration of H3O+ and OH− ions.
Kw=[H3O+][OH–] = 1.0  10-14
The concentration of water is not included because it is a liquid and its concentration does not change.

13. Equilibrium Constants

14. Equilibrium Constants

15. The equilibrium constant is
Example: the H2, I2, HI Equilibrium System
For the reaction
The equilibrium constant is
This is constant for any system of H2, I2, and HI at equilibrium at a given temperature.

16. At 425°C, the equilibrium constant for this equilibrium reaction system has the average value of

17. The balanced chemical equation for an equilibrium system is necessary to write the expression for the equilibrium constant.
Once the value of the equilibrium constant is known, the equilibrium constant expression can be used to calculate concentrations of reactants or products at equilibrium.

18. What is the equilibrium constant for the system at this temperature?
Sample Problem A
An equilibrium mixture of N2, O2 , and NO gases at 1500 K is determined to consist of 6.4  10–3 mol/L of N2, 1.7  10–3 mol/L of O2, and 1.1  10–5 mol/L of NO.
What is the equilibrium constant for the system at this temperature?

19. Solution:
The balanced chemical equation is
The chemical equilibrium expression is

20. SECTION 2. SHIFTING EQUILIBRIUM
Le Châtelier’s principle states that if a system at equilibrium is subjected to a stress, the equilibrium is shifted in the direction that tends to relieve the stress.

21. Henri Louis Le Châtelier
wikipedia

22. a. Changes in pressure
This applies only to reactions involving gases, and only when the number of moles on the left side is different from the number of moles on the right side.

23. Ex.: the Haber process for the synthesis of ammonia
What happens if the system is at equilibrium, and then the pressure is increased?

24. The stress (increased pressure) can be relieved by decreasing the number of particles.
How many particles are on the left? On the right? Stress will be relieved by shifting the reaction

25. The stress (increased pressure) can be relieved by decreasing the number of particles.
How many particles are on the left? On the right? Stress will be relieved by shifting the reaction to the right.

26. So when equilibrium is reached again, the ratio of NH3 to N2 and H2 will be higher than before.
b. Changes in concentration
For a reaction A + B ↔ C + D
An increase in [A] creates a stress. It can be relieved by reacting some of the added A with B.

27. So: [A] will increase (compared with the original).
[B] will decrease.
[C] and [D] will increase.
Predict the effects of each of these (to the left or to the right):
Increase in [B] decrease in [A]
Decrease in [D] increase in [C]

28. So: [A] will increase (compared with the original).
[B] will decrease.
[C] and [D] will increase.
Predict the effects of each of these (to the left or to the right):
Increase in [B] R decrease in [A] L
Decrease in [D] R increase in [C] L

29. c. Changes in temperature
c. Changes in temperature. The effect depends on whether the reaction is exothermic or endothermic.
Adding energy in the form of heat by raising the temperature shifts the reaction in the direction that will relieve the stress by absorbing the heat.

30. To help predict the effect, remember that we can include heat as if it were a reactant or product of a reaction.
Exothermic reaction
Ex.: N2 + 3H2 ↔ 2NH kJ
An increase in temperature shifts the reaction to the left, in order to take up the extra heat. A decrease shifts it to the right.

31. Endothermic reaction
Ex.:
CaCO kJ ↔ CaO + CO2
An increase in temperature shifts the equilibrium to the right.

32. Temperature Changes Affect an Equilibrium System

33. Le Chatelier's Principal 75348

34. Reactions That Go to Completion
Some reactions essentially go completely in one direction because a product is removed.

35. Formation of a gas Ex.:
H2CO3(aq) → H2O(l) + CO2(g)
If the container is open, the CO2 escapes. The reaction will continue to go to the right until H2CO3 is nearly all gone.

36. Formation of a precipitate: the product is essentially insoluble, so it is removed.
Ex.:
Na+(aq) + Cl-(aq) + Ag+(aq) + NO3-(aq) →
Na+(aq) + NO3-(aq) + AgCl(s)

37. SECTION 3. EQUILIBRIA OF ACIDS, BASES, AND SALTS
Ionization Constant of a Weak Acid
The term Ka is called the acid ionization constant.
Ex. - Acetic acid, CH3COOH, is a weak acid and most of the CH3COOH molecules remain unionized.
An acetic solution contains CH3COOH molecules, H3O+ ions, and acetate ions, CH3COO.

38. The equilibrium equation for the ionization of acetic acid is
The equation for Ka is
Because water is the solvent, one can assume that the molar concentration of H2O molecules remains constant. The concentration of water is not included in the equilibrium expression.

39. Ionization data and constants for some dilute acetic acid solutions at 25°C are given below.
The numerical value of Ka is almost identical for each solution molarity shown.

40. An increase in the concentration of CH3COO− ions through the addition of sodium acetate, NaCH3COO, disturbs the equilibrium.
This disturbance causes a decrease in [H3O+] and an increase in [CH3COOH].

41. The equilibrium is reestablished with the same value of Ka
The equilibrium is reestablished with the same value of Ka. But there is a higher concentration of nonionized acetic acid molecules and a lower concentration of H3O+ ions.
So the pH will change.

42. example: CH3COOH and NaCH3COO
Buffers
A buffered solution resists changes in pH. It contains both a weak acid and a salt of the weak acid.
example: CH3COOH and NaCH3COO
Buffered solution can react with either an acid or a base. When small amounts of acids or bases are added, the pH of the solution remains nearly constant.

43. Buffered Vs. Nonbuffered Solutions

44. If a small amount of acid is added to the acetic acid–sodium acetate solution, acetate ions react with most of the added hydronium ions to form nonionized acetic acid molecules.
The hydronium ion concentration and the pH of the solution remain practically unchanged.

45. If a small amount of a base is added, the OH− ions of the base react with and remove hydronium ions to form nonionized water molecules. Acetic acid molecules then ionize and mostly replace the hydronium ions neutralized by the added OH− ions.
The hydronium ion concentration and the pH of the solution remain practically unchanged.

46. A solution of a weak base containing a salt of the base also behaves as a buffered solution.
Buffer action has many important applications in chemistry and physiology.
Human blood is naturally buffered to maintain a pH of between 7.3 and 7.5.

47. SECTION 4. SOLUBILITY EQUILIBRIA
Solubility Product
A saturated solution contains the maximum amount of solute possible at a given temperature in equilibrium with an undissolved excess of the substance.
A saturated solution is not necessarily a concentrated solution.

48. The equilibrium principles developed in this chapter apply to all saturated solutions of sparingly soluble salts.
Ex.: the heterogeneous equilibrium system in a saturated solution of silver chloride containing an excess of the solid salt:

49. The solubility product constant, Ksp, of a substance is the product of the molar concentrations of its ions in a saturated solution, each raised to the power that is the coefficient of that ion in the balanced chemical equation.
The equation for the solubility equilibrium expression for the dissolution reaction of AgCl is

50. The equilibrium expression is written without including the solid species.
For a saturated solution of CaF2, the equilibrium equation is
The expression for the solubility product constant is

51. Using Solubility Product Constants 75356

52. Determining Ksp for Reactions at Chemical Equilibrium

53. CaF2 dissociates to yield twice as many F− ions as Ca2+ ions.
Calculating Solubility Product Constant from Solubility
The solubility of CaF2 at 25˚C is 8.6 g/100 g water, which is equal to 1.1 x 10-3 M.
CaF2 dissociates to yield twice as many F− ions as Ca2+ ions.
So: [Ca2+] = 1.1  10−3 mol/L
and [F− ] = 2.2  10−3 mol/L

54. Ksp = 5.3  10-9

55. Solubility Product Constants at 25°C

56. Solubility Product Constant 75358

57. The solubility product constant is an equilibrium constant; it has only one value for a given solid at a given temperature.
The solubility of a solid is the amount of the solid required to form a saturated solution with a specific amount of solvent. Its value at a given temperature is dependent on other conditions, such as the presence of a common ion.

58. Calculating Solubility from Solubility Product Constant
The solubility product constant can be used to determine the solubility of a sparingly soluble salt.
Ex.: How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?

59. The molar solubility of BaCO3 is
7.1  10−5 mol/L.

60. Precipitation Calculations
The equilibrium condition does not require that the two ion concentrations be equal. Equilibrium will still be established so that the ion product does not exceed the value of Ksp for the system.

61. If the ion product is less than the value of Ksp at a particular temperature, the solution is unsaturated.
If the ion product is greater than the value for Ksp, solid precipitates.
The solubility product can be used to predict whether a precipitate forms when two solutions are mixed.

62. Sample Problem: Will a precipitate form if 20. 0 mL of 0
Sample Problem: Will a precipitate form if 20.0 mL of M BaCl2 is mixed with 20.0 mL of M Na2SO4?
The two possible new pairings of ions are NaCl and BaSO4. BaSO4 is a sparingly soluble salt.

63. mol Ba2+ ion:
mol SO42- ion:

64. total volume of solution:
0.020 L L = L
concentration Ba2+ ion in combined solution:
concentration SO42- ion in combined solution:

65. The ion product:
Precipitation occurs.