1. Chapter 16 Principles of Chemical Reactivity: Equilibria

2. Chemical Equilibrium: A Review
All chemical reactions are reversible, at least in principle.
The concept of equilibrium is fundamental to chemistry.
The general concept of equilibrium was introduced in Chapter 3 to explain the limited dissociation of weak acids.
The goals of this and the following chapter will be to consider chemical equilibria in quantitative terms.
The extent to which that equilibrium lies (product favored verses reactant favored) will be discussed.
See Chapter 16 Video Presentation Slide 4

3. Equilibrium
At some point in time during the progress of a reaction, if the concentration of the reactants and products remains constant, equilibrium is said to be achieved.
The concentrations are NOT equal.

4. Equilibrium
Moving towards equilibrium
Equilibrium established

5. Properties of Chemical Equilibria
Equilibrium systems are said to be:
Dynamic (in constant motion)
Reversible
Equilibrium can be approached from either direction.
Pink to blue
Co(H2O)6Cl Co(H2O)4Cl H2O
See Chapter 16 Video Presentation Slide 5 
Blue to pink
Co(H2O)4Cl H2O Co(H2O)6Cl2 

6. Chemical Equilibrium Fe3+(aq) + SCN (aq)  Fe(SCN)2+ (aq)
After a period of time, the concentrations of reactants and products are constant.
The forward and reverse reactions continue after equilibrium is attained.
They are equal and opposite in rate.
See Chapter 16 Video Presentation Slides 6 & 7

7. Examples of Chemical Equilibria
Phase changes such as H2O(s) vs. H2O(liq)
See Chapter 16 Video Presentation Slide 8

8. The Equilibrium Constant
When a general chemical reaction
is at equilibrium, the equilibrium constant is given by:
If K > 0 then the reaction is said to be product favored.
If K < 0 then the reaction is said to be reactant favored.

9. The Equilibrium Constant
Product-favored
K > 1
Reactant-favored
K < 1

10. The Equilibrium Constant
For the formation of HI(g) the equilibrium constant is given by:

11. N2O4 (g) 2NO2 (g) equilibrium equilibrium equilibrium Start with NO2
Start with NO2 & N2O4

12. Disturbing a Chemical Equilibrium
The equilibrium between reactants and products may be disturbed in three ways:
(1) by changing the temperature
(2) by changing the concentration of a reactant
(3) by changing the volume (for systems involving gases)
A change in any of these factors will cause a system at equilibrium to shift back towards a state of equilibrium.
This statement is often referred to as Le Chatelier’s principle.

13. Disturbing a Chemical Equilibrium
Effect of the Addition or Removal of a Reactant or Product
If the concentration of a reactant or product is changed from its equilibrium value at a given temperature, equilibrium will be reestablished eventually.
The new equilibrium concentrations of reactants and products will be different, but the value of the equilibrium constant expression will still equal K

14. Le Châtelier’s Principle
If an external stress is applied to a system at equilibrium, the system adjusts in such a way that the stress is partially offset as the system reaches a new equilibrium position.
Changes in Concentration
N2 (g) + 3H2 (g) NH3 (g)
Add
NH3
Equilibrium shifts left to offset stress

15. Le Châtelier’s Principle
Changes in Concentration continued
Remove
Add
Add
Remove
aA + bB cC + dD
Change
Shifts the Equilibrium
Increase concentration of product(s)
left
Decrease concentration of product(s)
right
Increase concentration of reactant(s)
right
Decrease concentration of reactant(s)
left

16. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
For a reaction that involves gases, what happens to equilibrium concentrations or pressures if the size of the container is changed? (Such a change occurs, for example, when fuel and air are compressed in an automobile engine.)
To answer this question, recall that concentrations are in moles per liter. If the volume of a gas changes, its concentration therefore must also change, and the equilibrium composition can change.

17. Le Châtelier’s Principle
Changes in Volume and Pressure
A (g) + B (g) C (g)
Change
Shifts the Equilibrium
Increase pressure
Side with fewest moles of gas
Decrease pressure
Side with most moles of gas
Increase volume
Side with most moles of gas
Decrease volume
Side with fewest moles of gas
When the number of gas moles on either side is the same, there is no effect.

18. Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase Equilibria
Consider the reaction of nitrogen and oxygen to form nitric oxide:
As the temperature of the reaction is increased, the equilibrium constant increases. K
Temperature (K)
4.5  1031
298
6.7  1010
900
1.7  103
2300
Why?

19. Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase Equilibria
Lets write the reaction in this manner:
Notice that energy is included as a reactant!

20. Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase Equilibria
Lets write the reaction in this manner:
As temperature (Energy) is increased, equilibrium shifts to the right, favoring products.

21. Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase Equilibria
Lets write the reaction in this manner:
As the concentration of products increases so does the value of the equilibrium constant.

22. Disturbing a Chemical Equilibrium
Effect of Temperatue Changes on Gas-Phase Equilibria
This explains the increase in the equilibrium constant with increasing temperature. K
Temperature (K)
4.5  1031
298
6.7  1010
900
1.7  103
2300

23. Disturbing a Chemical Equilibrium
Effect of Temperature Changes on Gas-Phase Equilibria
Conclusion:
Increasing the temperature of an endothermic reaction favors the products, equilibrium shifts to the right.
Increasing the temperature of an exothermic reaction favors the reactants, equilibrium shifts to the left.
Lowering temperature results in the reverse effects.

24. Temperature Effects on Equilibrium
See Chapter 16 Video Presentation Slide 11
Kc (273 K) =
Kc (298 K) =

25. Le Châtelier’s Principle
Adding a Catalyst
does not change K
does not shift the position of an equilibrium system
system will reach equilibrium sooner
Catalyst lowers Ea for both forward and reverse reactions.
Catalyst does not change equilibrium constant or shift equilibrium.

26. Le Chatelier’s Principle Practice
Change
Reaction Shift
Adding more N2(g)
Right
Removing H2(g)
Left
Decreasing the container volume
Increasing the container temperature
Increasing the container volume
Decreasing the container temperature
adding a catalyst
no effect
adding argon to the container
See Chapter 16 Video Presentation Slide 16

27. Le Châtelier’s Principle - Summary
Change
Shift Equilibrium
Change Equilibrium
Constant
Concentration
yes no
Pressure
yes* no
Volume
yes* no
Temperature
yes
yes
Catalyst no no
*Dependent on relative moles of gaseous reactants and products

28. Writing Equilibrium Constant Expressions
In an equilibrium constant expression, all concentrations are reported as equilibrium values.
Product concentrations appear in the numerator, and reactant concentrations appear in the denominator.
Each concentration is raised to the power of its stoichiometric balancing coefficient.
Values of K are dimensionless.
The value of the constant K is particular to the given reaction at a specific temperature.

29. Writing Equilibrium Constant Expressions
Reactions Involving Solids
So long as a solid is present in the course of a reaction, its concentration is not included in the equilibrium constant expression.
Equilibrium constant:

30. Writing Equilibrium Constant Expressions
Reactions in Solution
If water is a participant in the chemical reaction, its concentration based on magnitude is considered to remain constant throughout.
Equilibrium constant:

31. Writing Equilibrium Constant Expressions
Reactions Involving Gases: Kc and Kp
Concentration data can be used to calculate equilibrium constants for both aqueous and gaseous systems.
In these cases, the symbol K is sometimes given the subscript “c” for “concentration,” as in Kc.
For gases, however, equilibrium constant expressions can be written in another way: in terms of partial pressures of reactants and products.

32. Writing Equilibrium Constant Expressions
Reactions Involving Gases: Kp
Notice that the basic form of the equilibrium constant expression is the same as for Kc. In some cases, the numerical values of Kc and Kp are the same. They are different when the numbers of moles of gaseous reactants and products are different.

33. Writing Equilibrium Constant Expressions
Reactions Involving Gases: Kp & Kc
The general relationship between Kp and Kc is derived in chapter 26, pa726.
When the number of gas mole is equivalent on either side of the chemical equation, the two equilibrium constants are the same value.
R = the gas constant
T = the absolute temperature
n = (mols gas product)  mols gas reactant)

34. The Reaction Quotient, Q
In general, ALL reacting chemical systems are characterized by their REACTION QUOTIENT, Q.
If Q = K, then system is at equilibrium.

35. The Reaction Quotient, Q
If Q < K then the system is heading towards equilibrium: There are more reactants than products as expected at equilibrium. The reaction is said to be headed to the “right”.
If Q > K the system has gone past equilibrium. There are more products than reactants as expected at equilibrium. The reaction is said to be headed to the left.
If Q = K then the system is at equilibrium.

36. Equilibrium Will K = [C]c[D]d [A]a[B]b aA + bB cC + dD K >> 1
Lie to the right
Favor products
K << 1
Lie to the left
Favor reactants

37. The equilibrium concentrations for the reaction between carbon monoxide and molecular chlorine to form COCl2 (g) at 740C are [CO] = M, [Cl2] = M, and [COCl2] = 0.14 M. Calculate the equilibrium constants Kc and Kp.
CO (g) + Cl2 (g) COCl2 (g)
[COCl2]
[CO][Cl2] =
0.14
0.012 x 0.054
Kc =
= 220
Kp = Kc(RT)Dn
Dn = 1 – 2 = -1
R =
T = = 347 K
Kp = 220 x ( x 347)-1 = 7.7 37

38. Heterogenous equilibrium applies to reactions in which reactants and products are in different phases.
CaCO3 (s) CaO (s) + CO2 (g)
[CaO][CO2]
[CaCO3]
Kc = ′
[CaCO3] = constant
[CaO] = constant
[CaCO3]
[CaO]
Kc x ′
Kc = [CO2] =
Kp = PCO 2
The concentration of solids and pure liquids are not included in the expression for the equilibrium constant.

39. Determining the Equilibrium Constant
The value of a reaction’s equilibrium constant is determined by measuring the concentrations of the reactants and products when a system is at equilibrium.
The equilibrium constant can also determined by looking at the changes in concentrations as a system achieves equilibrium.
This is know as an “ICE” table.

40. Determining the Equilibrium Constant
ICE tables:
I = Initial concentration
C = change in concentration
E = concentrations at equilibrium
Reactants
Products I C E

41. Determining K
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?

42. Determining K
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?
[NOCl]
[NO]
[Cl2]
Initial
Change
Equilibrium

43. Determining K
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?
[NOCl]
[NO]
[Cl2]
Initial
2.00
Change
Equilibrium

44. note the reaction stoichiometry
Determining K
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?
[NOCl]
[NO]
[Cl2]
Initial
2.00
Change
-0.66
+0.66
0.33
Equilibrium
note the reaction stoichiometry

45. Determining K
2.00 mol of NOCl is added to 1.00 L flask. At equilibrium the concentration of NO(g) is found to be 0.66 mol/L. What is the value of K?
[NOCl]
[NO]
[Cl2]
Initial
2.00
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33

46. Determining K [NOCl] [NO] [Cl2] Initial 2.00 Change -0.66 +0.66 +0.33
Change
-0.66
+0.66
+0.33
Equilibrium
1.34
0.66
0.33

47. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
See Chapter 16 Video Presentation Slides 9 & 10

48. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
[H2]
[I2]
[HI]
Initial
1.00
Change
Equilibrium

49. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
[H2]
[I2]
[HI]
Initial
1.00
Change
- x
+ 2x
Equilibrium

50. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
[H2]
[I2]
[HI]
Initial
1.00
Change
- x
+ 2x
Equilibrium x
+ 2 x
Where x is defined as amount of H2 and I2 consumed on approaching equilibrium in moles.

51. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
[H2]
[I2]
[HI]
Initial
1.00
Change
- x
+ 2x
Equilibrium x
+ 2 x

52. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?
[H2]
[I2]
[HI]
Initial
1.00
Change
- x
+ 2x
Equilibrium x
+ 2 x

53. Equilibrium Concentrations from K
1.00 mol each of H2 and I2 in a 1.00 L flask. What are the equilibrium concentrations of all species. Kc = 55.3?

54. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

55. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
[N2O4]
[NO2]
Initial
0.50
Change
Equilibrium

56. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
[N2O4]
[NO2]
Initial
0.50
Change -x
+2x
Equilibrium

57. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
[N2O4]
[NO2]
Initial
0.50
Change -x
+2x
Equilibrium x 2x

58. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
[N2O4]
[NO2]
Initial
0.50
Change -x
+2x
Equilibrium x 2x

59. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

60. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
Solving this requires the Quadratic Equation:

61. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

62. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?

63. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
x = or 0.028
The negative value is not reasonable which gives the equilibrium values of concentration to be:

64. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
x = or 0.028
The negative value is not reasonable which gives the equilibrium values of concentration to be:
x = M
[N2O4] = M  x = M
[NO2] = 2x = M

65. Equilibrium Concentrations from K
Consider the following reaction:
If initial concentration of N2O4 is 0.50 M, what are the equilibrium concentrations?
x = or 0.028
The negative value is not reasonable which gives the equilibrium values of concentration to be:
x = M
[N2O4] = M  x = M
[NO2] = 2x = M
The results are in agreement with the magnitude of K: The
reactant concentration is favored due to the small value of K

66. Approximations in Equilibrium Concentrations
What happens when the change in concentration for a reaction is much less than the initial concentration of the reactants?
It turns out that when K  100 < [A]0, then the quadratic equation is not required.
since [A]0  x  [A]0

67. Approximations in Equilibrium Concentrations
Let’s look at what happens to the previous problem if the initial concentration of N2O4(g) is doubled to 1.00 M.

68. Approximations in Equilibrium Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?

69. Approximations in Equilibrium Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?
Solving for the equilibrium concentration using the quadratic equation yields:

70. Approximations in Equilibrium Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?
Solving for the equilibrium concentration using the quadratic equation yields:
x = M, [N2O4] = & [N2O] = 0.076

71. Approximations in Equilibrium Concentrations
Consider the following reaction:
If initial concentration of N2O4 is 1.00 M, what are the equilibrium concentrations?
Solving for the equilibrium concentration using the approximation, ([N2O4]0  x) = [N2O4]0 also yields:
x = M, [N2O4] = & [N2O] = 0.076
The two results are in agreement.

72. More About Balanced Equations & Equilibrium Constants
Multiplying the coefficients of a reaction:
The second reaction is 2 the first.
The general relationship is:
When “n” is the multiplication factor.

73. More About Balanced Equations & Equilibrium Constants
Reversing a reaction
The second reaction is the reverse the first.
The general relationship is:
The equilibrium constant is the inverse of the original.

74. More About Balanced Equations & Equilibrium Constants
Adding reactions at equilibrium:
Net.
The new equilibrium constant is the product of the individual.

75. Butane-Isobutane Equilibrium

76. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = M. K = 2.50
If 1.50 M butane is added to the system, what is the mew concentration of each when the system returns to equilibrium?

77. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium.

78. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium.

79. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s first calculate the value of Q to see which direction the reaction will take to reestablish equilibrium.
Q < K: There are more reactants than products as expected at equilibrium. The reaction shift to the “right”. Reactants must react to form products.

80. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up an “ICE” table.
[butane]
[isobutane]
Initial
Change
Equilibrium

81. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up an “ICE” table.
[butane]
[isobutane]
Initial
1.25
Change
Equilibrium

82. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up an “ICE” table.
[butane]
[isobutane]
Initial
1.25
Change
 x +x
Equilibrium

83. Butane-Isobutane Equilibrium
At equilibrium [isobutane] = 1.25 M and [butane] = 0.50 M.
If 1.50 M butane is added to the system, what is the new concentration of each when the system returns to equilibrium?
Let’s now calculate the changes in concentration by setting up an “ICE” table.
[butane]
[isobutane]
Initial
1.25
Change
 x +x
Equilibrium
2.00  x x

84. Butane-Isobutane Equilibrium
Initial
1.25
Change
 x +x
Equilibrium
2.00  x x
Using the value of K, determine the change in concentration, “x”.

85. Butane-Isobutane Equilibrium
Initial
1.25
Change
 x +x
Equilibrium
2.00  x x
Using the value of K, determine the change in concentration, “x”.

86. Butane-Isobutane Equilibrium
Initial
1.25
Change
 x +x
Equilibrium
2.00  x x
Using the value of K, determine the change in concentration, “x”.
X = 1.07

87. Butane-Isobutane Equilibrium
Initial
1.25
Change
 x +x
Equilibrium
2.00  x x
The new concentrations for butane and isobutane are:
[butane] = 0.93 M [isobutane] = 2.32 M
Equilibrium is shifted to the right, forming more products.

88. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
Consider the following reaction:
At Equilibrium, the concentrations are:
[N2O4] = M & [NO2] = M
What would happen if the total volume of the system was suddenly doubled?

89. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
What would happen if the total volume of the system was suddenly doubled?
½ [N2O4] = ½  M = M
½ [NO2] = ½  M = M

90. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
What would happen if the total volume of the system was suddenly doubled?
½ [N2O4] = ½  M = M
½ [NO2] = ½  M = M
Calculate Q:

91. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
What would happen if the total volume of the system was suddenly doubled?
Q = 342 > K: Therefore some products must shift to reactants (left) to reestablish equilibrium.

92. Disturbing a Chemical Equilibrium
Effect of Volume Changes on Gas-Phase Equilibria
Conclusion:
Increasing the volume of a container favors the side of equilibrium with the greatest number of gas moles.
Decreasing the volume favors the side with the least number of moles.
When the number of gas moles on either side is the same, there is no effect.

94. Chemical Kinetics and Chemical Equilibrium
ratef = kf [A][B]2
A + 2B AB2 kf kr
rater = kr [AB2]
Equilibrium
ratef = rater
kf [A][B]2 = kr [AB2] kf kr
[AB2]
[A][B]2 =
Kc =

95. Modifying the Chemical Equation (cont’d)
Consider the reaction: 2 NO(g) + O2(g) NO2(g)
[NO2]2
Kc = ––––––––– = 4.67 x 1013 (at 298 K)
[NO]2 [O2]
Now consider the reaction: NO2(g) NO(g) + ½ O2(g)
What will be the equilibrium constant K"c for the new reaction?
[NO] [O2]1/ /2
K"c = ––––––––– = –––
[NO2] Kc
= x 10–14 = x 10–7

96. Calculate K of reversed reaction, ½ of a reaction or doubled
Example: 2SO2(g) + O2(g)  2SO3(g)
4SO2(g) + 2O2(g)  4SO3(g)
SO2(g) + ½ O2(g)  SO3(g)
2SO3(g)  O2(g) + 2SO2(g)
Example: 2CO(g) + O2(g)  2CO2(g) K = 2.75 x @1000K
CO2(g)  CO(g) + ½ O2(g) K = 6.03 x @1000K
Summarize:
If the coef in the reaction is:
Then K is:
Doubled
Squared
Halved
Square root
Reversed in sign
Inverted
Multiplied by a constant n
Raised to the nth power