1. Chapt. 17 – Chemical Equilibrium
17.1 A State of Dynamic Balance
17.2 Factors Affecting Chemical Equilibrium
17.3 Using Equilibrium Constants

2. Section 17.1 A State of Dynamic Balance
Chemical equilibrium is described by an equilibrium constant expression that relates the concentrations of reactants and products.
List the characteristics of chemical equilibrium.
Write equilibrium expressions for systems that are at
equilibrium.
Calculate equilibrium constants from concentration data.

3. Section 17.1 A State of Dynamic Balance
Key Concepts
A reaction is at equilibrium when the rate of the forward reaction equals the rate of the reverse reaction.
The equilibrium constant expression is a ratio of the molar concentrations of the products to the molar concentrations of the reactants with each concentration raised to a power equal to its coefficient in the balanced chemical equation.
The value of the equilibrium constant expression, Keq, is constant for a given temperature.

4. What is Equilibrium?
N2(g) + 3H2(g)  2NH3(g) DG0 = kJ
Spontaneous, slow reaction under standard conditions
At higher temperature (723 K) and pressure reaction proceeds at a practical rate
Next slide shows reaction progress if start with 1 mol N2 and 3 mol H2

5. N2(g) + 3H2(g)  2NH3(g) Reactant concentrations initially decrease
All concentrations stop changing – reactant concentrations are not zero
Product concentration initially increases
Concentration H2
NH3 N2
Time

6. Reversible Reactions
Reaction going to completion: almost complete conversion of reactants to products
Most reactions do not go to completion –known as reversible reactions
Forward: N2(g) + 3H2(g)  2NH3(g)
Reverse: NH3(g)  N2(g) + 3H2(g)
Reversible: N2(g) + 3H2(g) ↔ 2NH3(g)
Double arrow

7. Time prior to equilibrium
Reversible Reactions
N2(g) + 3H2(g) ↔ 2NH3(g)
Time zero
Reactants at max concentration, forward rate at maximum, reverse rate at zero
Time prior to equilibrium
Reactant concentration lower, forward rate slower, some reverse reaction
At equilibrium
Forward and reverse rates equal – no further concentration changes

8. Rateforward = Ratereverse
Chemical Equilibrium
N2(g) + 3H2(g) ↔ 2NH3(g)
Forward and reverse reactions balance each other because
Rateforward = Ratereverse
Does not mean concentrations of reactants and products are equal
Can be equal in some special cases

9. Chemical Equilibrium Equilibrium Reaction Rates Forward = Reverse Rate
Forward Rate
Reaction Rates
Equilibrium
Reverse Rate
Forward = Reverse Rate
Time

10. Dynamic Equilibrium State of equilibrium not a static one
Some reactant molecules are always changing into product molecule
Some product molecules are always changing into reactant molecules
Only net concentrations of reactants and products don’t change

11. Equilibrium Constant
Law of Chemical Equilibrium: at a given temperature, a chemical system may reach a state in which a particular ratio of reactant and product concentrations reaches a constant value
aA +bB  cC +dD
Equilibrium constant

12. Equilibrium Constant aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant [A]a[B]b
[X] = molar concentration of quantity under equilibrium conditions
Exponents come from coefficients in balanced chemical equation
Keq is temperature dependent
Units of Keq vary

13. Equilibrium Constant aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant [A]a[B]b
Keq >> 1 numerator >> denominator
Reaction as written above favors production of products
Keq <<1 denominator >> numerator
Reaction as written above favors production of reactants

14. Keq = [C]c[D]d = equilibrium constant [A]a[B]b
aA +bB ↔ cC +dD
Keq = [C]c[D]d = equilibrium constant [A]a[B]b
Rules for writing an expression for Keq differ for homogeneous and heterogeneous equilibrium
Homogeneous – all reactants and products in same physical state
Heterogeneous – not homogeneous

15. Keq – Homogeneous Equilibrium
Homogeneous when all reactants and products are in same physical state
H2(g) + I2(g) ↔ 2HI(g)
Keq = _[HI]2 [H2] [I2]
Keq = at 731 K
Note: no units for this particular reaction

16. Keq – Homogeneous Equilibrium
N2(g) + 3H2(g) ↔ 2NH3(g)
Keq = [NH3] [N2] [H2]3
Units = L2/mol2

17. Practice Homogeneous Equilibrium Constants Problems 1(a-e), 2 page 601
Problem 44 (a-b) page 626
Problems 1- 5 page 988

18. Heterogeneous Equilibrium
Occurs when reactants & products present in more than one physical state
C2H5OH(l)
C2H5OH(g)
Ethanol in closed flask
C2H5OH(l) ↔ C2H5OH(g)

19. Keq – Heterogeneous Equilibrium
Rule: Pure liquids and solids don’t appear in the equilibrium expression because at constant temperature their concentrations don’t change when the amount present changes
C2H5OH(l)  C2H5OH(g)
Keq = [C2H5OH(g)]
I2 (s)  I2(g)
Keq = [I2 (g)]

20. Keq – Heterogeneous Equilibrium
2NaHCO3(s) ↔
Na2CO3(s) + CO2(g) + H2O(g)
Keq = [CO2(g)] [H2O(g)]

21. Practice Heterogeneous Equilibrium Constants
Problems 3 (a-e), 4 page 603
Problems 45 – 47, 49 page 626
Problems 6 – 8 page 988

22. Numerical Value of Keq
For a given reaction, final values of reactant and product concentrations will satisfy Keq expression regardless of initial concentrations used
Equilibrium position: set of final concentrations of reactants and products
One value of Keq, lots of possible equilibrium positions

23. Notation for Concentrations
[X]0 = molar concentration of X at time zero = initial concentration
[X]eq = molar concentration of X at equilibrium = final concentration

24. Numerical Value of Keq Position 1 Position 2 Position 3
H2(g) + I2(g) ↔ 2HI(g)
Keq = _[HI]2 [H2] [I2]
[H2]0 [I2]0 [HI]0 [H2]eq [I2]eq [HI]eq Keq
Three different equilibrium positions yield the same value of Keq
Position 1
Position 2
Position 3

25. Value of Keq – Problem 17.3 N2(g) + 3H2(g) ↔ 2NH3(g)
Keq = [NH3] [N2] [H2]3
[NH3] = mol/L
[N2] = mol/L [H2] = mol/L
Value of Keq?
Keq = [0.933] = (L2/mol2) [0.533] [1.600]3

26. Keq of Some Common Reactions
2H2(g) + O2(g) ↔ 2H2O(l)
1.4x K
CaCO3(s) ↔ CaO(s) + CO2(g)
1.9x K K
2SO2(g) + O2(g) ↔ 2SO3(g) K
C(s) + H2O(g) ↔ CO(g) + H2(g)
1.6x K K
Which ones favor production of products?

27. Practice Calculating value of Keq Problems 5 – 7, 11 page 605
Problem 48 page 626
Problems 9 – 10 page 988

28. Chapt. 17 – Chemical Equilibrium
17.1 A State of Dynamic Balance
17.2 Factors Affecting Chemical Equilibrium
17.3 Using Equilibrium Constants

29. Section 17.2 Factors Affecting Chemical Equilibrium
When changes are made to a system at equilibrium, the system shifts to a new equilibrium position.
Describe how various factors affect chemical equilibrium.
Explain how Le Châtelier’s principle applies to equilibrium systems.

30. Section 17.2 Factors Affecting Chemical Equilibrium
Key Concepts
Le Châtelier’s principle describes how an equilibrium system shifts in response to a stress or a disturbance.
When an equilibrium shifts in response to a change in concentration or volume, the equilibrium position changes but Keq remains constant. A change in temperature, however, alters both the equilibrium position and the value of Keq.

31. Le Châtelier’s Principle
Principle: If a stress is applied to a system at equilibrium, the system shifts in the direction that partially relieves the stress.
Stresses include
Change in concentration
Change in volume (pressure)
Change in temperature

32. Le Châtelier’s P. - Concentration
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
_________________________________________________________________________________________________
[CO]eq [H2]eq [CH4]eq [H2O]eq Keq
Instantaneously (by injection) raise CO concentration to M
Equilibrium becomes unbalanced
Rate of forward rxn increases, get
New equilibrium position has reduced value of CO from instantaneous value of
Stress of additional reactant relieved (partially) by producing more product

33. Le Châtelier’s P. - Concentration
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
_______________________________________________________________________________________________________________________________________
What will happen to equilibrium if:
A desiccant is used to remove H2O?
Equilibrium shifts to the right
CO is removed from reaction vessel?
Equilibrium shifts to the left
H2O is injected into the reaction vessel?

34. Le Châtelier’s P. - Concentration
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
Reactant addition
Product Removal
Reactant Removal
Product Addition

35. Le Châtelier’s P. - Volume
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
4 moles of reactant, 2 moles of product
Decrease V (increase P) at constant T

36. Le Châtelier’s P. - Volume
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
Start
Compress
Initial
Temporarily have non-equilibrium concentrations
More product forms
Compress
Final

37. Le Châtelier’s P. - Volume
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
4 moles of reactant, 2 moles of product
Decrease V (increase P) at constant T
Stress (increased pressure) relieved by formation of more product
P = n  R at constant V, T n = # moles
If n decreases, P decreases
P is lowered but not to its original value

38. Le Châtelier’s P. - Volume
When volume of equilibrium mixture of gases reduced, net change occurs in direction that produces fewer moles of gas
When volume increased, net change occurs in direction that produces more moles of gas
H2(g) + I2(g) ↔ 2HI(g) - no effect of V

39. Le Châtelier’s P. - Temperature
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
DH0 = kJ
Have exothermic reaction – can think of heat as a “product”
Effect of raising T is like adding more product – equilibrium shifts to the left
This approach explains why Keq has a temperature dependence

40. Le Châtelier’s P. - Temperature
Exothermic Reaction +
Raise T: Product addition
Lower T: Product removal
Endothermic Reaction
heat + Co(H2O) Cl CoCl H2O
heat
Raise T: Reactant addition
Lower T: Reactant removal

41. Summary: Le Châtelier’s Principle and Temperature
Raising temperature of equilibrium mixture shifts equilibrium condition in direction of endothermic reaction
Lowering temperature causes a shift in direction of exothermic reaction

42. Practice Le Châtelier’s Principle Problems 14 - 15 page 611
Problems , pages

43. Chapt. 17 – Chemical Equilibrium
17.1 A State of Dynamic Balance
17.2 Factors Affecting Chemical Equilibrium
17.3 Using Equilibrium Constants

44. Section 17.3 Using Equilibrium Constants
Equilibrium constant expressions can be used to calculate concentrations and solubilities.
Determine equilibrium concentrations of reactants and products.
Calculate the solubility of a compound from its solubility product constant.
Explain the common ion effect.

45. Section 17.3 Using Equilibrium Constants
Key Concepts
Equilibrium concentrations and solubilities can be calculated using equilibrium constant expressions.
Ksp describes the equilibrium between a sparingly soluble ionic compound and its ions in solution.
If the ion product, Qsp, exceeds the Ksp when two solutions are mixed, a precipitate will form.
The presence of a common ion in a solution lowers the solubility of a dissolved substance.

46. Calculating Equilibrium Concentrations
If know Keq and concentrations of all but one of the reactants and products, can solve for unknown concentration
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
0.850M M ?M M
Keq = [CH4(g)] [H2O(g)] [CO(g)] [H2(g)]3
[CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)]

47. Calculating Equilibrium Concentrations
CO(g) + 3H2(g) ↔ CH4(g) + H2O(g)
0.850M M ?M M
[CH4(g)] = Keq [CO(g)] [H2(g)]3/ [H2O(g)]
Keq = 3.933
[CH4(g)] =  (0.850)(1.333)3/ (0.286)
[CH4(g)] = 27.7 mol/L = [CH4(g)]eq

48. Calculating Equilibrium Concentrations
Example Problem 17.4
2H2S(g) ↔ 2H2(g) + S2(g)
1.84x10-1M ?M x10-2M
Keq = [H2(g)]2 [S2(g)] = 2.27x [H2S(g)]2
[H2(g)]2 = Keq [H2S(g)]2 / [S2(g)] = 1.42x10-3
[H2] = 3.77x10-2 mol/L

49. Practice Calculating equilibrium concentrations
Problems 16 (a-c), 17 page 613
Problem 71 page 627
Problems 11(a-b) page 988

50. Solubility Equilibria
Solubility equilibria applies to all compounds with finite solubility, but is most commonly applied to dissolution of those with low solubility
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Heterogeneous equilibrium
Keq = [Ba2+(aq)] [SO42-(aq)]
Has special symbol Ksp

51. Solubility Equilibria
BaSO4(s) ↔ Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+] [SO42-] = 1.1x @298 K
Ksp = solubility product constant
Ksp = equilibrium constant for the dissolution of a sparingly soluble ionic compound in water
Only ions appear in Ksp, but some amount of solid (however small) must be present to achieve equilibrium

52. Solubility Equilibria
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
Ksp = [Mg2+] [OH-]2 = 5.6x @298 K
Table 17.3, page 615 lists solubility product constants sorted by type of anion (carbonate, phosphate, etc.)
Ksp can be used to calculate solubility of salt or concentration of one of ions involved in equilibrium

53. Calculating Solubility from Ksp
AgI(s) ↔ Ag+(aq) + I-(aq)
Ksp = [Ag+(aq)] [I-(aq)] = 8.5x @298 K
s = solubility of AgI(s) in mol/L
1 mol Ag+(aq) forms per mol AgI dissolved
1 mol I-(aq) forms per mol AgI dissolved
Ksp = [Ag+(aq)] [I-(aq)] = s2 = 8.5x10-17
s = 9.2x10-9 K
Note: actual units of Ksp are “ignored”

54. Practice Calculating solubility of 1:1 salts
Problems 20 (a-c), 21 page 616
Problem 22(a) page 617
Problem 31, page 622
Problem 70 page 627
Problem 12 page 988

55. Calculating Ion Concentration from Ksp
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
Ksp = [Mg2+] [OH-]2 = 5.6x @298 K
[OH-] in a saturated solution?
This is a non 1:1 electrolyte – trickier
Have to pay attention to stoichiometry
s = solubility of Mg(OH)2 in mol/L
1 mol Mg2+ per mol Mg(OH)2 dissolved
2 mol OH- per mol Mg(OH)2 dissolved

56. Calculating Ion Concentration from Ksp
Mg(OH)2(s) ↔ Mg2+(aq) + 2OH-(aq)
Ksp = [Mg2+] [OH-]2 = 5.6x @298 K
[OH-] in a saturated solution?
s = solubility of Mg(OH)2 in mol/L
1 mol Mg2+ per mol Mg(OH)2 dissolved
2 mol OH- per mol Mg(OH)2 dissolved
Ksp = (s) (2 s)2 = 4s3 = 5.6x10-12
s = 1.1 x10-4 mol/L [OH-] = 2 s

57. Practice
Calculating solubility & ion concentrations for arbitrary salts
Problems 22 (b-c), 23, 24 page 617
Problems 78, 83 page 628
Problems page 988

58. Predicting Precipitates
Mix two soluble ionic compounds – sometimes a precipitate forms
Can use Ksp to predict: Use quantity called the ion product
Symbol Qsp
Same functional form as Ksp
Concentrations used may or may not be equilibrium concentration
AB(s) ↔ A+(aq) + B-(aq) Qsp=[A+][B-]
Current but not necessarily equilibrium values

59. Predicting Precipitates
AB(s) ↔ A+(aq) + B-(aq)
Qsp=[A+][B-] current concentrations
Ksp= [A+]eq[B-]eq equilibrium concentrations
Compare Ksp to Qsp
1. Qsp < Ksp unsaturated no precipitate
2. Qsp = Ksp saturated no precipitate
3. Qsp > Ksp supersaturated precipitate
For #3, solid will form until Qsp = Ksp

60. Predicting Precipitates
Mix equal volumes of (soluble) 0.1M iron(III) chloride and potassium hexacyanoferrate(II)
4FeCl3(aq)+ 3K4Fe(CN)6(aq)  12KCl(aq) + Fe4(Fe(CN)6)3(s)
Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-(aq)
Ksp = [Fe3+]4[Fe(CN)64-]3 = 3.3x10-41
Question: does a precipitate form?

61. Predicting Precipitates
Fe4(Fe(CN)6)3(s) ↔ 4Fe3+(aq) + 3Fe(CN)64-(aq)
Ksp = [Fe3+]eq4[Fe(CN)64-]eq3
Equal volumes of 0.10M FeCl3 & K4Fe(CN)6
solutions used
[Fe3+] = M [Fe(CN)64-] = M
Qsp = [Fe3+]4[Fe(CN)64-]3 = [0.050]4[0.050]3
Qsp = 7.8x Ksp = 3.3x10-41
Qsp > Ksp so precipitate will form
Possible to compute moles of solid formed

62. Predicting Precipitates
Example Problem 17.7
Mix 100 mL M NaCl with 100 mL M Pb(NO3)2
Does PbCl2 precipitate?
PbCl2(s) ↔ [Pb2+][Cl-]2 Ksp = 1.7x10-5
Qsp = [0.0100][0.0050]2 = 2.5x10-7
Qsp < Ksp
No precipitate

63. Practice Determining if precipitate will form
Problems 25(a-b), 26 page 619
Problems 72, 73 page 627
Problems 16, 17 page 988

64. Common Ion Effect
Common ion is an ion common to two or more ionic compounds
KI, AgI I- is the common ion
CaCl2, Ca(OH)2 Ca2+ is the common ion
Common ion effect is the lowering of the solubility of an ionic substance by the presence of a common ion

65. Common Ion Effect Quantities produced just from dissolution of solid
Lead chromate solubility at 298 K
PbCrO4(s)  Pb2+(aq) + CrO42-(aq)
Pure water, s = 4.8x10-7 mol/L
Much lower in 0.01M K2CrO4
Ksp = [Pb2+][CrO42-] = 2.3x10-13
Pure water: s = [CrO42-] = [Pb2+] = 4.8x10-7
K2CrO4 solution: [Pb2+][CrO ] = Ksp
Quantities produced just from dissolution of solid
Total CrO42-
[Pb2+]= 2.3x10-11
Ksp= 2.3x10-11x1.0x10-2 = 2.3x10-13
Common CrO42- has reduced Pb2+ solubility

66. Common Ion Effect & L-C’s Principle
PbCrO4(s) ↔ Pb2+(aq) + CrO42-(aq)
Pure water: [Pb2+] = [CrO42-] = s = 4.8x10-7
K2CrO4 solution: [Pb2+][CrO ] = Ksp
[Pb2+]= 2.3x10-11
CrO42- is common ion
Higher common ion concentration shifts equilibrium to left
Same effect if add Pb(NO3)2 to solution – common ion now Pb2+

67. Using Ksp and Common Ion
CuCO3(s) ↔ Cu2+(aq) + CO32-(aq)
Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10
s = solubility of CuCO3 (s) in solution of 0.10 M K2CO3 (Prob & strategy on p 621)
(s)( s) = 2.5x10-10 (use quad. eqn.)
(s)(0.10) = 2.5x10-10 (approximation)

68. Using Ksp and Common Ion
CuCO3(s) ↔ Cu2+(aq) + CO32-(aq)
Ksp = [Cu2+(aq)] [CO32-(aq)] = 2.5x10-10
(s)( s) = 2.5x10-10 (using quad. eqn.)
(s)(0.10) = 2.5x10-10 (using approximation)
For this problem, exact method (using quadratic equation) and approximate method yield same answer; s = 2.5x10-9
Commonly used “trick” to quickly solve variety of equilibrium problems