1. Ch. 18 Chemical Equilibrium

2. 18.1 The Nature of Chemical Equilibrium
Chapter 18
18.1 The Nature of Chemical Equilibrium
List two everyday processes that can easily be reversed and two that cannot.
The freezing of water and the melting of ice can be reversed
The cooking of an egg or the lighting of a match cannot be reversed.
Reactions that “go to completion” refer to a reaction in which all the reactants are converted to products

3. Chapter 18 Reversible Reactions
But theoretically, every reaction can proceed in two directions, forward and reverse.
Essentially all chemical reactions are considered to be reversible under suitable conditions.
A chemical reaction in which the products can react to re-form the reactants is called a reversible reaction.

4. Chapter 18
Chemical equilibrium is established when the rate of its forward reaction equals the rate of its reverse reaction
When this happens the amounts of products (concentrations) and reactants remain constant.
Both reactions continue, but there is no net
change in their concentration

6. Chapter 18 products of the forward reaction favored, lies to the right
products of the reverse reaction favored, lies to the left
Neither reaction is favored

7. Chapter 18
After equilibrium is reached, the individual concentrations of A, B, C, and D undergo no further change if conditions remain the same.
i.e. : the ratio of their concentrations remains constant.
The equilibrium constant is designated by the letter K.
The expression for K =

8. Chapter 18
The Equilibrium Constant Expression, K, is the ratio of the product of the concentrations of substances formed at equilibrium to the 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.
K is dependent on temperature and always determined experimentally
Only the concentrations of substances that can actually change are included in K.
So pure solids and liquids are omitted because their concentrations cannot change.

9. Chapter 18 What would be true if K = 1?
Concentration of products = concentration of reactants
What would be true if K is greater than 1?
Large K values indicate that the products are favored (forward reaction)
What would be true if K is less than 1?
Small K values indicate the reactants are favored (reverse reaction)

10. What does knowing the value of Keq help us predict?
Whether the forward of the reverse reaction is favored
Therefore, tell us if the reactants or products will be in higher concentrations when equilibrium is established
*Does not tell us how quickly the reaction reaches the equilibrium

11. Examples of Equilibrium Constant Expressions
Make sure you always balance an equation before writing an equilibrium constant expression

12. Examples of Equilibrium Constant Expressions

13. Chapter 18 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?

14. The Equilibrium Expression, continued
Section 1 The Nature of Chemical Equilibrium
Chapter 18
The Equilibrium Expression, continued
Sample Problem A Solution
Given: [N2] = 6.4  10–3 mol/L
[O2] = 1.7  10–3 mol/L
[NO] = 1.1  10–5 mol/L
Unknown: K
Solution:
The balanced chemical equation is
The chemical equilibrium expression is

15. Section 18.2 Le Châtelier’s principle
Chapter 18
Section Le Châtelier’s principle
Imagine children playing on a seesaw.
Five boys are sitting on one side and five girls on the other, and the seesaw is just balanced.
Then, one girl gets off, and the system is no longer at equilibrium.
One way to get the seesaw in balance again is for one of the boys to move toward the girls’ side.

16. Chapter 18
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.
This principle is true for all dynamic equilibria, chemical as well as physical.
Changes in pressure, concentration, and temperature illustrate Le Châtelier’s principle because they alter the equilibrium position and thereby change the relative amounts of reactants and products

17. Predicting the Direction of Shift: Pressure
Chapter 18
Predicting the Direction of Shift: Pressure
Change in pressure affects only equilibrium systems in which gases are involved.
An increase in pressure is an applied stress.
The system can reduce the total pressure by reducing the number of molecules.
Therefore, increase in pressure favors the reaction that produces fewer gas molecules

18. Chapter 18 Ex: Haber process for the synthesis of ammonia
When pressure is applied, the equilibrium will shift to the right, and produce more NH3.
4 molecules of gas molecules of gas
A reduction in the total number of molecules leads to a decrease in pressure.
Note: changes in pressure do not affect the value of the equilibrium constant

19. Chapter 18
Changes in Concentration
An increase in the concentration of a reactant is a stress on the equilibrium system.
To relieve the stress, some of the added A reacts with B to form products C and D.
Increase in concentration of one substance shifts the equilibrium to the opposite side (i.e. opposite side favored)

20. Chapter 18
Note: changes in concentration have no effect on the value of the equilibrium constant.
Such changes have an equal effect on the numerator and the denominator of the chemical equilibrium expression.

21. Chapter 18 Changes in Temperature
Reversible reactions are exothermic in one direction and endothermic in the other.
The addition of energy in the form of heat shifts the equilibrium so that energy is absorbed.
An increase in temperature favors the endothermic reaction
A decreased in temperature (removal of energy) favors the exothermic reaction

22. Chapter 18 A rise in temperature increases the rate of any reaction.
In an equilibrium system, the rates of the opposing reactions are raised unequally.
**The value of the equilibrium constant for a given system is affected by the temperature.

23. Chapter 18 Haber Process: exothermic
Section 2 Shifting Equilibrium
Chapter 18
Haber Process: exothermic
A rise in temperature increases the rate of any reaction, but the rates of the opposing reactions are raised unequally.
The value of the equilibrium constant for a given system is affected by the temperature.
A high temperature favors the decomposition of ammonia, the endothermic reaction.
At low temperatures, the forward reaction is too slow to be commercially useful.

24. Temperature Changes Affect an Equilibrium System

25. Reactions That Go to Completion
Chapter 18
Reactions That Go to Completion
Some reactions involving compounds formed by the chemical interaction of ions in solutions appear to go to completion in the sense that the ions are almost completely removed from solution.

26. Chapter 18 3 Examples of Reactions that go to Completion
1. Formation of a Gas
H2CO3(aq) H2O(l) + CO2(g)
Reaction goes practically to completion because one of the products, CO2, escapes as a gas if the container is open to the air.

27. Chapter 18 2. Formation of a Precipitate
If chemically equivalent amounts of the two solutes are mixed, almost all of the Ag+ ions and Cl− ions combine and separate from the solution as a precipitate of AgCl.
AgCl is only very sparingly soluble in water.
The reaction thus effectively goes to completion because an essentially insoluble product is formed.

28. Chapter 18 3. Formation of a Slightly Ionized Product
Neutralization reactions between H3O+ ions from aqueous acids and OH− ions from aqueous bases result in the formation of water molecules, which are only slightly ionized.
Hydronium ions and hydroxide ions are almost entirely removed from the solution.
The reaction effectively runs to completion because the product is only slightly ionized.

29. Chapter 18 Common-Ion Effect
The phenomenon in which the addition of an ion common to two solutes brings about precipitation or reduced ionization of a reactant
Ex: hydrogen chloride gas is bubbled into a saturated solution of sodium chloride.

30. Common-Ion Effect, continued
Chapter 18
Common-Ion Effect, continued
As the hydrogen chloride dissolves in sufficient quantity, it increases the concentration of Cl− ions in the solution, which is a stress on the equilibrium system.
The system can compensate by forming some solid NaCl. The NaCl precipitates out, relieving the stress of added chloride (reactions shifts to the left)
The new equilibrium has a greater concentration of Cl− ions but a decreased concentration of Na+ ions.

31. Particle Model for the Common-Ion Effect
Section 2 Shifting Equilibrium
Chapter 18
Particle Model for the Common-Ion Effect

32. Chapter 18 Solubility Product
Soluble ionic compounds dissolve in water until they are at equilibrium with their ions
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.

33. Solubility Product, continued
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
For dissolution of the ionic compound AaBb (A cation, B anion), the solubility-production expression is
AaBb (s) → aA + bB
The equilibrium expression is written without including the solid species (or water)
The numerical value of Ksp helps us determine how soluble a compound is in water
Ksp = [A]a [B]b

34. Solubility Product, continued
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
Compounds with large Ksp values are more soluble
Compounds with smaller Ksp values are less soluble
The solubility of a solid is an equilibrium position that represents the amount of the solid required to form a saturated solution with a specific amount of solvent.
It has an infinite number of possible values at a given temperature and is dependent on other conditions,
such as the presence of a common ion.

35. Solubility Product, continued
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
For a saturated solution of CaF2, the equilibrium equation is
The expression for the solubility product constant is
Ksp = [Ca] [F]2
The solubility of CaF2 is 8.6  10−3 g/100 g of water at 25°C. Expressed in moles per liter this
concentration becomes 1.1  10−3 mol/L.

36. Ionization Constant of Acids & Bases
Chapter 18
Section 3 Equilibria of Acids, Bases, and Salts
Ionization Constant of Acids & Bases
When we write the equiblium constant expression for the ionization of an acid/base we call them
Ka :acid ionization constant.
Kb: base ionization constant.
The acid ionization constant, Ka , is constant for a specified temperature
Weak acids, like CH3COOH, remainly largely unionized as will have small Ka values.
Strong acids, like HNO3, ionize completely and will have larger Ka values.

37. Ionization Constant of Water, Kw
Section 3 Equilibria of Acids, Bases, and Salts
Chapter 18
Ionization Constant of Water, Kw
The self-ionization of water is an equilibrium reaction.
Equilibrium is established with a very low concentration of H3O+ and OH− ions.
Kw=[H3O+][OH–] = 1.0  10-14

38. Determining Ksp for Reactions at Chemical Equilibrium
Section 4 Solubility Equilibrium
Chapter 18
Determining Ksp for Reactions at Chemical Equilibrium

39. [Ca2+] = 1.1  10−3 mol/L [F− ] = 2.2  10−3 mol/L
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
CaF2 dissociates to yield twice as many F− ions as Ca2+ ions.
[Ca2+] = 1.1  10−3 mol/L [F− ] = 2.2  10−3 mol/L
Ksp = 5.3  10-9
Calculations of Ksp ordinarily should be limited to
two significant figures.

40. Solubility Product Constants at 25°C
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product Constants at 25°C

41. Solubility Product, continued
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
Sample Problem B
Calculate the solubility product constant, Ksp ,for
copper(I) chloride, CuCl, given that the solubility
of this compound at 25°C is  10–2 g/100. g
H2O.

42. [Cu+] = [Cl–] = solubility in mol/L
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
Sample Problem B Solution
Given: solubility of CuCl = 1.08  10−2 g CuCl/100. g H2O
Unknown: Ksp
Solution:
Ksp=[Cu+][Cl–]
[Cu+] = [Cl–] = solubility in mol/L

43. Solubility Product, continued
Section 4 Solubility Equilibrium
Chapter 18
Solubility Product, continued
Sample Problem B Solution, continued
1.09  10-3 mol/L CuCl
[Cu+] = [Cl–]=1.09  10-3 mol/L
Ksp = (1.09  10-3)(1.09  10-3) =
1.19  10-6

44. Calculating Solubilities
Section 4 Solubility Equilibrium
Chapter 18
Calculating Solubilities
The solubility product constant can be used to determine the solubility of a sparingly soluble salt.
How many moles of barium carbonate, BaCO3, can be dissolved in 1 L of water at 25°C?
The molar solubility of BaCO3 is 7.1  10−5 mol/L.

45. Calculating Solubilities, continued
Section 4 Solubility Equilibrium
Chapter 18
Calculating Solubilities, continued
Sample Problem C
Calculate the solubility of silver bromide, AgBr,
in mol/L, using the Ksp value for this compound.

46. [Ag+] = [Br−], so let [Ag+] = x and [Br−] = x
Section 4 Solubility Equilibrium
Chapter 18
Calculating Solubilities, continued
Sample Problem C Solution
Given: Ksp = 5.0 10−13
Unknown: solubility of AgBr
Solution:
[Ag+] = [Br−], so let [Ag+] = x and [Br−] = x

47. Limitations on the Use of Ksp
Section 4 Solubility Equilibrium
Chapter 18
Limitations on the Use of Ksp
The solubility product principle can be very useful when applied to solutions of sparingly soluble substances.
It cannot be applied very successfully to solutions of moderately soluble or very soluble substances.
The positive and negative ions attract each other, and this attraction becomes appreciable when the ions are close together.
Sometimes it is necessary to consider two equilibria simultaneously.

48. Equilibrium Calculations
Section 4 Solubility Equilibrium
Chapter 18
Equilibrium Calculations

49. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
1. A chemical reaction is in equilibrium when
A. forward and reverse reactions have ceased.
B. the equilibrium constant equals 1.
C. forward and reverse reaction rates are equal.
D. No reactants remain.

50. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
1. A chemical reaction is in equilibrium when
A. forward and reverse reactions have ceased.
B. the equilibrium constant equals 1.
C. forward and reverse reaction rates are equal.
D. No reactants remain.

51. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
2. Which change can cause the value of the equilibrium
constant to change?
A. temperature
B. concentration of a reactant
C. concentration of a product
D. None of the above

52. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
2. Which change can cause the value of the equilibrium
constant to change?
A. temperature
B. concentration of a reactant
C. concentration of a product
D. None of the above

53. Chapter 18 Multiple Choice 3. Consider the following reaction:
Standardized Test Preparation
Chapter 18
Multiple Choice
3. Consider the following reaction:
The equilibrium constant expression for this reaction is
A C.
B D.

54. Chapter 18 Multiple Choice 3. Consider the following reaction:
Standardized Test Preparation
Chapter 18
Multiple Choice
3. Consider the following reaction:
The equilibrium constant expression for this reaction is
A C.
B D.

55. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
4. The solubility product of cadmium carbonate, CdCO3, is 1.0  10−12. In a saturated solution of this salt, the concentration of Cd2+(aq) ions is
A −13 mol/L.
B −12 mol/L.
C −6 mol/L.
D −7 mol/L.

56. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
4. The solubility product of cadmium carbonate, CdCO3, is 1.0  10−12. In a saturated solution of this salt, the concentration of Cd2+(aq) ions is
A −13 mol/L.
B −12 mol/L.
C −6 mol/L.
D −7 mol/L.

57. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
5. Consider the following equation for an equilibrium system:
Which concentration(s) would be included in the denominator of the equilibrium constant expression?
A. Pb(s), CO2(g), and SO2(g)
B. PbS(s), O2(g), and C(s)
C. O2(g), Pb(s), CO2(g), and SO2(g)
D. O2(g)

58. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
5. Consider the following equation for an equilibrium system:
Which concentration(s) would be included in the denominator of the equilibrium constant expression?
A. Pb(s), CO2(g), and SO2(g)
B. PbS(s), O2(g), and C(s)
C. O2(g), Pb(s), CO2(g), and SO2(g)
D. O2(g)

59. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
6. If an exothermic reaction has reached equilibrium,
then increasing the temperature will
A. favor the forward reaction.
B. favor the reverse reaction.
C. favor both the forward and reverse reactions.
D. have no effect on the equilibrium.

60. Chapter 18 Multiple Choice
Standardized Test Preparation
Chapter 18
Multiple Choice
6. If an exothermic reaction has reached equilibrium,
then increasing the temperature will
A. favor the forward reaction.
B. favor the reverse reaction.
C. favor both the forward and reverse reactions.
D. have no effect on the equilibrium.

61. Chapter 18 Multiple Choice 7. Le Châtelier’s principle states that
Standardized Test Preparation
Chapter 18
Multiple Choice
7. Le Châtelier’s principle states that
A. at equilibrium, the forward and reverse reaction
rates are equal.
B. stresses include changes in concentrations, pressure, and temperature.
C. to relieve stress, solids and solvents are omitted from equilibrium constant expressions.
D. chemical equilibria respond to reduce applied stress.

62. Chapter 18 Multiple Choice 7. Le Châtelier’s principle states that
Standardized Test Preparation
Chapter 18
Multiple Choice
7. Le Châtelier’s principle states that
A. at equilibrium, the forward and reverse reaction
rates are equal.
B. stresses include changes in concentrations, pressure, and temperature.
C. to relieve stress, solids and solvents are omitted from equilibrium constant expressions.
D. chemical equilibria respond to reduce applied stress.