1. Introductory Chemistry, 3rd Edition Nivaldo Tro
Chapter 15
Chemical
Equilibrium
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2009, Prentice Hall

2. Equilibrium vs. Disequilibrium
When systems are at equilibrium with their surroundings, their conditions are the same as the surroundings and they stay that way.
When systems are in disequilibrium with their surroundings, their conditions are not the same as the surroundings.
Systems that are in disequilibrium tend to change until they reach equilibrium with their surroundings.
Living things are in controlled disequilibrium with their environment—they are not at the same conditions as the environment and do not tend to change toward those conditions.
Tro's Introductory Chemistry, Chapter 15

3. Tro's Introductory Chemistry, Chapter 15
Reaction Rates
Some chemical reactions proceed rapidly.
Like the precipitation reactions studied in Chapter 7 where the products form practically the instant the two solutions are mixed.
Other reactions proceed slowly.
Like the decomposition of dye molecules of a sofa placed in front of a window.
The rate of a reaction is measured in the amount of reactant that changes into product in a given period of time.
Generally moles of reactant used per second.
Like miles per hour.
Chemists study ways of controlling reaction rates.
Tro's Introductory Chemistry, Chapter 15

4. Reaction Rates, Continued
Initially, only reactants are present
After 15 seconds, the left reaction is 60% complete, but the right reaction is only 20% complete
After 30 seconds, the left reaction is complete,
whereas the right reaction is only 40% done.
After 45 seconds, the right reaction is
still not complete
Tro's Introductory Chemistry, Chapter 15

5. Tro's Introductory Chemistry, Chapter 15
2 N2O5 (g) → 4 NO2(g) + O2(g)
Over time, the
concentrations
of reactants decrease
as products increase.
Tro's Introductory Chemistry, Chapter 15

6. 2 N2O5 (g) → 4 NO2(g) + O2(g): Rate vs. Time
Because reactant concentrations
decrease, the rates of reactions
slow down over time.
Tro's Introductory Chemistry, Chapter 15

7. Tro's Introductory Chemistry, Chapter 15
Collision Theory
In order for a reaction to take place, the reacting molecules must collide with each other.
Once molecules collide they may react together or they may not, depending on two factors:
Whether the collision has enough energy to “start to break the bonds holding reactant molecules together."
Whether the reacting molecules collide in the proper orientation for new bonds to form.
Tro's Introductory Chemistry, Chapter 15

8. Tro's Introductory Chemistry, Chapter 15
Effective Collisions
Collisions in which these two conditions are met (and therefore the reaction occurs) are called effective collisions.
The higher the frequency of effective collisions, the faster the reaction rate.
There is a minimum energy needed for a collision to be effective. We call this the activation energy.
The lower the activation energy, the faster the reaction will be.
Tro's Introductory Chemistry, Chapter 15

9. Effective Collisions: Kinetic Energy Factor
For a collision to lead to overcoming the energy barrier, the reacting molecules must have sufficient kinetic energy so that when they collide, it can form the activated complex.
Tro's Introductory Chemistry, Chapter 15

10. Effective Collisions: Orientation Effect
Tro's Introductory Chemistry, Chapter 15

11. Reaction Energy Diagram
Tro's Introductory Chemistry, Chapter 15

12. Factors Effecting Reaction Rate: Reactant Concentration
The higher the concentration of reactant molecules, this increases the frequency of reactant molecule collisions the faster the reaction will generally go.
Since reactants are consumed as the reaction proceeds, the speed of a reaction generally slows over time.
Tro's Introductory Chemistry, Chapter 15

13. Effect of Concentration on Rate
Low concentrations of reactant molecules leads to fewer effective collisions, therefore a slower reaction rate.
High concentrations of reactant molecules lead to more effective collisions, therefore a faster reaction rate.
Tro's Introductory Chemistry, Chapter 15

14. Factors Effecting Reaction Rate: Temperature
Increasing the temperature increases the energy so that their collisions can overcome the activation energy.
And, increasing the temperature also increases the frequency of collisions.
Both these mean that increasing temperature increases the reaction rate.
Tro's Introductory Chemistry, Chapter 15

15. Effect of Temperature on Rate
Low temperatures lead to fewer molecules with enough energy to overcome the activation energy, and less frequent reactant collisions, therefore a slower reaction rate
High temperatures lead to more molecules with enough energy to overcome the activation energy, and more frequent reactant collisions, therefore, a faster reaction rate.

16. Tro's Introductory Chemistry, Chapter 15
Reaction Dynamics
If the products of a reaction are removed from the system as they are made, then a chemical reaction will proceed until the limiting reactants are used up.
However, if the products are allowed to accumulate; they will start reacting together to form the original reactants. This is called the reverse reaction.
Reactions that can proceed in both the forward and reverse directions are called reversible reactions.
Tro's Introductory Chemistry, Chapter 15

17. Reaction Dynamics, Continued
The forward reaction slows down as the amounts of reactants decreases.
At the same time, the reverse reaction speeds up as the concentration of the products increases.
Eventually, the forward reaction is using reactants and making products as fast as the reverse reaction is using products and making reactants. This is called chemical equilibrium.
Dynamic equilibrium is reached when the rates of two opposite processes are the same.
Tro's Introductory Chemistry, Chapter 15

18. Tro's Introductory Chemistry, Chapter 15
Chemical Equilibrium
When a reaction reaches equilibrium, the production of both reactants and products is constant.
The forward and reverse reactions still continue.
Because they go at the same rate, the amounts of materials does not appear to change.
Tro's Introductory Chemistry, Chapter 15

19. Equilibrium
As the reaction proceeds, the forward reaction slows down as the reactants get used up. At the same time, the reverse reaction speeds up as product concentration increases.
Initially, we only have reactant molecules in the mixture. The reaction can only proceed in the forward direction, making products.
Once equilibrium is established, the concentrations of the reactants and products in the final mixture do not change, (unless conditions are changed).
Eventually, the forward and reverse rates are equal. At this time equilibrium is established.

20. Equilibrium, Continued
As the forward reaction proceeds
it makes products and uses reactants.
Because the reactant concentration
decreases, the forward reaction slows.
As the products accumulate, the
reverse reaction speeds up.
Once equilibrium is established,
the forward and reverse reactions
proceed at the same rate, so the
concentrations of all materials
stay constant.
Initially, only the forward
reaction takes place.
Eventually, the reaction proceeds
in the reverse direction as fast as
it proceeds in the forward direction.
At this time equilibrium is established.
Rate
Rate forward
Rate reverse
Time
Tro's Introductory Chemistry, Chapter 15

21. Hypothetical Reaction 2 Red  Blue
Time
[Red]
[Blue]
0.400
0.000 10
0.208
0.096 20
0.190
0.105 30
0.180
0.110 40
0.174
0.113 50
0.170
0.115 60
0.168
0.116 70
0.167
0.117 80
0.166 90
0.165
0.118
100
110
0.164
120
130
140
150
The reaction slows over time,
but the red molecules never run out!
At some time between 100 and 110 sec,
the concentrations of both the red and
the blue molecules no longer change—
equilibrium has been established.
Notice that equilibrium does not mean
that the concentrations are equal!
Once equilibrium is established, the rate
of red molecules turning into blue is the
same as the rate of blue molecules turning
into red.

22. Hypothetical Reaction 2 Red  Blue, Continued
Tro, Chemistry: A Molecular Approach

23. Tro's Introductory Chemistry, Chapter 15
Equilibrium  Equal
The rates of the forward and reverse reactions are equal at equilibrium.
But that does not mean the concentrations of reactants and products are equal.
Tro's Introductory Chemistry, Chapter 15

24. Equilibrium, Continued
As the forward reaction proceeds
it makes products and uses reactants.
Because the reactant concentration
decreases, the forward reaction slows.
As the products accumulate, the
reverse reaction speeds up.
Once equilibrium is established,
the forward and reverse reactions
proceed at the same rate, so the
concentrations of all materials
stay constant.
Initially, only the forward
reaction takes place.
Eventually, the reaction proceeds
in the reverse direction as fast as
it proceeds in the forward direction.
At this time equilibrium is established.
Rate
Rate forward
Rate reverse
Time
Tro's Introductory Chemistry, Chapter 15

25. What occurs in a reaction at equilibrium.
When the number of people moving up is the same as the number of people moving down, the number of people on each floor remains constant, and the two populations are in equilibrium.
Equilibrium occurs when the forward and reverse reactions are the same.
Chapter Seven

26. An Analogy: Population Changes
When Narnians feel overcrowded, some will emigrate to Middle Earth.
However, as time passes, emigration will occur in both directions at the same rate, leading to populations in Narnia and Middle Earth that are constant, though not necessarily equal.
Tro's Introductory Chemistry, Chapter 15

27. Tro's Introductory Chemistry, Chapter 15
Equilibrium Constant
Even though the concentrations of reactants and products are not equal at equilibrium, there is a relationship between them.
For the reaction H2(g) + I2(g)  2HI(g) at equilibrium, the ratio of the concentrations raised to the power of their coefficients is constant.
Tro's Introductory Chemistry, Chapter 15

28. Equilibrium Constant, Continued
For the general equation aA + bB  cC + dD, the relationship is given below:
The lowercase letters represent the coefficients of the balanced chemical equation.
Always products over reactants.
The constant is called the equilibrium constant, Keq.
Tro's Introductory Chemistry, Chapter 15

29. Writing Equilibrium Constant Expressions
For aA + bB  cC + dD, the equilibrium constant expression is:
So for the reaction
2 N2O5  4 NO2 + O2, the
equilibrium constant
expression is:
Tro's Introductory Chemistry, Chapter 15

30. Equilibrium Constants for Heterogeneous Equilibria
Pure substances in the solid and liquid state have constant concentrations.
Adding or removing some does not change the concentration because they do not expand to fill the container or spread throughout a solution.
Therefore, these substances are not included in the equilibrium constant expression.
For the reaction CaCO3(s) + 2 HCl(aq)  CaCl2(aq) + CO2(g) + H2O(l):
Tro's Introductory Chemistry, Chapter 15

31. Tro's Introductory Chemistry, Chapter 15
Write the Equilibrium Constant Expressions, Keq, for Each of the Following:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba+2(aq) + SO4-2(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 15

32. Write the Equilibrium Constant Expressions, Keq, for Each of the Following, Continued:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba+2(aq) + SO4-2(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Keq =
[CO]2•[O2]
[CO2]2
Keq =
[Ba+2]•[SO4-2]
Keq =
[CO2]
[CH4]•[O2]2

33. What Does the Value of Keq Imply?
When the value of Keq > > 1, we know that when the reaction reaches equilibrium, there will be many more product molecules present than reactant molecules.
The position of equilibrium favors products.
When the value of Keq < < 1, we know that when the reaction reaches equilibrium, there will be many more reactant molecules present than product molecules.
The position of equilibrium favors reactants.
Tro's Introductory Chemistry, Chapter 15

34. A Large Equilibrium Constant
Tro's Introductory Chemistry, Chapter 15

35. A Small Equilibrium Constant
Tro's Introductory Chemistry, Chapter 15

36. Tro's Introductory Chemistry, Chapter 15
Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following:
2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95
2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17
Tro's Introductory Chemistry, Chapter 15

37. Write the Equilibrium Constant Expressions, Keq, and Predict the Position of Equilibrium for Each of the Following, Continued:
2 HF(g) Û H2(g) + F2(g) Keq = 1 x 10-95
2 SO2(g) + O2(g) Û 2 SO3(g) Keq = 8 x 1025
N2(g) + 2 O2(g) Û 2 NO2(g) Keq = 3 x 10-17
Favors reactants.
Favors products.
Favors reactants.

38. Tro's Introductory Chemistry, Chapter 15
Calculating Keq
The value of the equilibrium constant may be determined by measuring the concentrations of all the reactants and products in the mixture after the reaction reaches equilibrium, then substituting in the expression for Keq.
Although you may have different amounts of reactants and products in the equilibrium mixture, the value of Keq will always be the same.
The value of Keq depends only on the temperature.
The value of Keq does not depend on the amounts of reactants or products with which you start.
Tro's Introductory Chemistry, Chapter 15

39. Initial and Equilibrium Concentrations for H2(g) + I2(g)  2HI(g)
Constant
[H2]
[I2]
[HI]
0.50
0.0
0.11
0.78
0.055
0.39
0.165
1.17
1.0
0.5
0.53
0.033
0.934
Tro's Introductory Chemistry, Chapter 15

40. Tro's Introductory Chemistry, Chapter 15
Example 15.3—Find the Value of Keq for the Reaction from the Given Concentrations: 2 CH4(g)  C2H2(g) + 3 H2(g).
Given:
Find:
[CH4] = M, [C2H2] = M, [H2] = M
Keq
Solution Map:
Relationships:
[CH4], [C2H2], [H2]
Keq
Solve:
Check:
Keq is unitless.
Tro's Introductory Chemistry, Chapter 15

41. Tro's Introductory Chemistry, Chapter 15
Example 15.3:
A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature?
2 CH4(g)  C2H2(g) + 3 H2(g)
Tro's Introductory Chemistry, Chapter 15

42. Tro's Introductory Chemistry, Chapter 15
Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g)
Write down the given quantity and its units.
Given: [CH4] = M
[C2H2] = M
[H2] = M
Tro's Introductory Chemistry, Chapter 15

43. Tro's Introductory Chemistry, Chapter 15
Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g)
Information:
Given: [CH4] = M
[C2H2] = M
[H2] = 0.112M
Write down the quantity to find and/or its units.
Find: Keq
Tro's Introductory Chemistry, Chapter 15

44. Tro's Introductory Chemistry, Chapter 15
Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g)
Information:
Given: [CH4] = M
[C2H2] = M
[H2] = 0.112M
Find: Keq
Collect needed equation:
Tro's Introductory Chemistry, Chapter 15

45. Tro's Introductory Chemistry, Chapter 15
Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g)
Information:
Given: [CH4] = M
[C2H2] = M
[H2] = 0.112M
Find: Keq
Write a solution map:
[CH4],
[C2H2],
[H2]
Keq
Tro's Introductory Chemistry, Chapter 15

46. Tro's Introductory Chemistry, Chapter 15
Example: A mixture of CH4, C2H2, and H2 is allowed to come to equilibrium at 1700 °C. The measured equilibrium concentrations are [CH4] = M, [C2H2] = M, and [H2] = M. What is the value of the equilibrium constant at this temperature? 2CH4(g)  C2H2(g) + 3 H2(g)
Information:
Given: [CH4] = M
[C2H2] = M
[H2] = 0.112M
Find: Keq
Solution Map: [ ]s → Keq
Apply the solution map:
Tro's Introductory Chemistry, Chapter 15

47. Tro's Introductory Chemistry, Chapter 15
Practice—Calculate Keq for the Reaction 2 NO2(g)  N2O4(g) at 100 C if the Equilibrium Concentrations Are [NO2] = M and [N2O4] = M.
Tro's Introductory Chemistry, Chapter 15

48. Practice—Find the Value of Keq for the Reaction from the Given Concentrations: 2 NO2(g)  N2O4(g).
[NO2] = M, [N2O4] = M
Keq
Solution Map:
Relationships:
[NO2], [N2O4]
Keq
Solve:
Check:
Keq is unitless.

49. Tro's Introductory Chemistry, Chapter 15
Example 15.4—Find the Value of [HI] for the Reaction at Equilibrium from the Given Concentrations and Keq: H2(g) + I2(g)  2HI(g)
Given:
Find:
[I2] = M, [H2] = M, Keq = 69
[HI]
Solution Map:
Relationships:
[I2], [H2], Keq
[HI]
Solve:
Tro's Introductory Chemistry, Chapter 15

50. H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Example 15.4:
In an equilibrium mixture, the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI?
H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Tro's Introductory Chemistry, Chapter 15

51. Tro's Introductory Chemistry, Chapter 15
Example: In an equilibrium mixture the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Write down the given quantity and its units.
Given: [H2] = M
[I2] = M
Keq = 69
Tro's Introductory Chemistry, Chapter 15

52. Tro's Introductory Chemistry, Chapter 15
Example: In an equilibrium mixture the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Information:
Given: [H2] = 0.020M
[I2] = 0.020M; Keq = 69
Write down the quantity to find and/or its units.
Find: concentration of HI, M
Tro's Introductory Chemistry, Chapter 15

53. Tro's Introductory Chemistry, Chapter 15
Example: In an equilibrium mixture the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Information:
Given: [H2] = 0.020M
[I2] = 0.020M; Keq = 69
Find: [HI]
Collect needed equation:
Tro's Introductory Chemistry, Chapter 15

54. Tro's Introductory Chemistry, Chapter 15
Example: In an equilibrium mixture the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Information:
Given: [H2] = 0.020M
[I2] = 0.020M; Keq = 69
Find: [HI]
Write a solution map:
[H2], [I2]
Keq
[HI]
Tro's Introductory Chemistry, Chapter 15

55. Tro's Introductory Chemistry, Chapter 15
Example: In an equilibrium mixture the concentrations of H2 and I2 are both M. What is the value of the equilibrium concentration of HI? H2(g) + I2(g)  2 HI(g) Keq = 69 at 340 °C
Information:
Given: [H2] = 0.020M
[I2] = 0.020M; Keq = 69
Find: [HI]
SM: [ ] & Keq → [ ]
Apply the solution map:
Tro's Introductory Chemistry, Chapter 15

56. Tro's Introductory Chemistry, Chapter 15
A Sample of PCl5(g) Is Placed in a L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq =
Tro's Introductory Chemistry, Chapter 15

57. A Sample of PCl5(g) Is Placed in a 0
A Sample of PCl5(g) Is Placed in a L Container and Heated to 160 °C. The PCl5 Is Decomposed into PCl3(g) and Cl2(g). At Equilibrium, Moles of PCl3 and Cl2 Are Formed. Determine the Equilibrium Concentration of PCl5 if Keq = , Continued.
PCl5 Û PCl Cl2
Equilibrium
Concentration, M ?
if in addition you calculate Kp from Kc you find that it is 2.26, slightly greater than 1 – showing that the equilibrium constant may be unreliable for predicting the position of equilibrium when it is close to 1

58. Disturbing and Re-Establishing Equilibrium
Once a reaction is at equilibrium, the concentrations of all the reactants and products remain the same.
However, if the conditions are changed, the concentrations of all the chemicals will change until equilibrium is re-established.
The new concentrations will be different, but the equilibrium constant will be the same.
Unless you change the temperature.
Tro's Introductory Chemistry, Chapter 15

59. Le Châtelier’s Principle
Le Châtelier’s principle guides us in predicting the effect on the position of equilibrium when conditions change.
“When a chemical system at equilibrium is disturbed, the system shifts in a direction that will minimize the disturbance.”
Tro's Introductory Chemistry, Chapter 15

60. An Analogy: Population Changes, due to a Gold Rush!
When the populations of Narnia and Middle Earth are in equilibrium, the emigration rates between the two states are equal so the populations stay constant.
When an influx of population enters Middle Earth from somewhere outside Narnia, it disturbs the equilibrium established between Narnia and Middle Earth.
The result will be people moving from Middle Earth into Narnia faster than people moving from Narnia into Middle Earth. This will continue until a new equilibrium between the populations is established, However, the new populations will have different numbers of people than the old ones.

61. The Effect of Concentration Changes on Equilibrium
Adding a reactant will decrease the amounts of the other reactants and increase the amount of the products until a new position of equilibrium is found.
That has the same Keq.
Removing a product will increase the amounts of the other products and decrease the amounts of the reactants.
You can use to this to drive a reaction to completion!
Remember: Adding more of a solid or liquid does not change its concentration and, therefore, has no effect on the equilibrium.
Tro's Introductory Chemistry, Chapter 15

62. The Effect of Concentration Changes on Equilibrium, Continued
When NO2
is added,
some of it
combines
to make
more N2O4.
Tro's Introductory Chemistry, Chapter 15

63. The Effect of Concentration Changes on Equilibrium, Continued
When N2O4
is added,
some of it
decomposes
to make
more NO2.
Tro's Introductory Chemistry, Chapter 15

64. Tro's Introductory Chemistry, Chapter 15
Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 15

65. Practice—Predict the Effect on the Equilibrium When the Underlined Substance Is Added to the Following Systems, Continued:
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Shift right, removing some of the added CO2 and
increasing the concentrations of CO and O2.
Shift left, removing some of the added Ba2+ and
reducing the concentration of SO42-.
Shift right, removing some of the added CO2 and decreasing the O2,
while increasing the concentration of CO2.

66. Effect of Volume Change on Equilibrium
For solids, liquids, or solutions, changing the size of the container has no effect on the concentration.
Changing the volume of a container changes the concentration of a gas.
Same number of moles, but different number of liters, resulting in a different molarity.
Tro's Introductory Chemistry, Chapter 15

67. Effect of Volume Change on Equilibrium, Continued
Decreasing the size of the container increases the concentration of all the gases in the container.
This increases their partial pressures.
If their partial pressures increase, then the total pressure in the container will increase.
According to Le Châtelier’s principle, the equilibrium should shift to remove that pressure.
The way to reduce the pressure is to reduce the number of molecules in the container.
When the volume decreases, the equilibrium shifts to the side with fewer molecules.
Tro's Introductory Chemistry, Chapter 15

68. The Effect of Volume Change on Equilibrium, Continued
Since there are more gas molecules on the reactants side of the reaction, when the pressure is increased the position of equilibrium shifts toward the products.
When the pressure is decreased by increasing the volume, the position of equilibrium shifts toward the side with the greater number of molecules—the reactant side.
Tro's Introductory Chemistry, Chapter 15

69. Tro's Introductory Chemistry, Chapter 15
Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced.
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Tro's Introductory Chemistry, Chapter 15

70. Practice—Predict the Effect on the Equilibrium When the Volume Is Reduced, Continued.
2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l)
Shift left because there are fewer gas molecules
on the reactant side than on the product side.
No effect because none of the substances are gases.
Shift right because there are fewer gas molecules
on the product side than on the reactant side.

71. The Effect of Temperature Changes on Equilibrium
Exothermic reactions release energy and endothermic reactions absorb energy.
If we write heat as a product in an exothermic reaction or as a reactant in an endothermic reaction, it will help us use Le Châtelier’s principle to predict the effect of temperature changes.
However, heat is not matter and not written in a proper equation.
Tro's Introductory Chemistry, Chapter 15

72. Tro's Introductory Chemistry, Chapter 15
The Effect of Temperature Changes on Equilibrium for Exothermic Reactions
For an exothermic reaction, heat is a product.
Increasing the temperature is like adding heat.
According to Le Châtelier’s principle, the equilibrium will shift away from the added heat.
The concentrations of C and D will decrease and the concentrations of A and B will increase.
The value of Keq will decrease.
How will decreasing the temperature effect the system?
aA + bB  cC + dD + heat
Tro's Introductory Chemistry, Chapter 15

73. Tro's Introductory Chemistry, Chapter 15
The Effect of Temperature Changes on Equilibrium for Endothermic Reactions
For an endothermic reaction, heat is a reactant.
Increasing the temperature is like adding heat.
According to Le Châtelier’s principle, the equilibrium will shift away from the added heat.
The concentrations of C and D will increase and the concentrations of A and B will decrease.
The value of Keq will increase.
How will decreasing the temperature effect the system?
Heat + aA + bB  cC + dD
Tro's Introductory Chemistry, Chapter 15

74. The Effect of Temperature Changes on Equilibrium
Tro's Introductory Chemistry, Chapter 15

75. Tro's Introductory Chemistry, Chapter 15
Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced.
Heat + 2 CO2(g) Û 2 CO(g) + O2(g)
BaSO4(s) Û Ba2+(aq) + SO42-(aq) (endothermic)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) (exothermic)
Tro's Introductory Chemistry, Chapter 15

76. Practice—Predict the Effect on the Equilibrium When the Temperature Is Reduced, Continued.
Heat + 2 CO2(g) Û 2 CO(g) + O2(g)
Heat + BaSO4(s) Û Ba2+(aq) + SO42-(aq)
CH4(g) + 2 O2(g) Û CO2(g) + 2 H2O(l) + Heat
Shift left, reducing the value of Keq.
Shift left, reducing the value of Keq.
Shift right, increasing the value of Keq.

77. Solubility and Solubility Product
Even “insoluble” salts dissolve somewhat in water.
Insoluble = less than 0.1 g per 100 g H2O.
The solubility of insoluble salts is described in terms of equilibrium between undissolved solid and aqueous ions produced.
AnYm(s) Û n A+(aq) + m Y-(aq)
Equilibrium constant for this process is called the solubility product constant, Ksp.
Ksp = [A+]n[Y-]m
If there is undissolved solid in equilibrium with the solution, the solution is saturated.
Larger Ksp = more soluble.
For salts that produce the same number of ions.
Tro's Introductory Chemistry, Chapter 15

78. Example—Determine the Ksp of PbBr2 if its Solubility Is 1.44 x 10-2 M.
PbBr2(s) Û Pb2+(aq) + 2 Br–(aq)
init
equil
Ksp = [Pb2+][Br–]2 = (0.0144)(0.0288)2 = 1.19 x 10-5
Tro's Introductory Chemistry, Chapter 15

79. Example 15.9— Calculating Molar Solubility from Ksp
Tro's Introductory Chemistry, Chapter 15

80. Tro's Introductory Chemistry, Chapter 15
Example 15.9:
Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Tro's Introductory Chemistry, Chapter 15

81. Tro's Introductory Chemistry, Chapter 15
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Write down the given quantity and its units.
Given: Ksp = 1.07 x 10-10
Tro's Introductory Chemistry, Chapter 15

82. Tro's Introductory Chemistry, Chapter 15
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Information:
Given: Ksp = 1.07 x 10-10
Write down the quantity to find and/or its units.
Find: molar solubility, M
Tro's Introductory Chemistry, Chapter 15

83. Collect needed equations: BaSO4(s)  Ba2+(aq) + SO42-(aq)
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Information:
Given: Ksp = 1.07 x 10-10
Find: [BaSO4], M
Collect needed equations:
BaSO4(s)  Ba2+(aq) + SO42-(aq)
Ksp = [Ba2+][SO42-]
because for each BaSO4 dissolved you produce one Ba2+:
[BaSO4] = [Ba2+] = [SO42-]
Tro's Introductory Chemistry, Chapter 15

84. Tro's Introductory Chemistry, Chapter 15
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Information:
Given: Ksp = 1.07 x 10-10
Find: [BaSO4], M = [Ba2+] = [SO42-]
Equation: Ksp = [Ba2+][SO42-]
Write a solution map:
Ksp
[Ba2+]
Ksp = [Ba2+][SO42-]
Tro's Introductory Chemistry, Chapter 15

85. Tro's Introductory Chemistry, Chapter 15
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Information:
Given: Ksp = 1.07 x 10-10
Find: [BaSO4], M = [Ba2+] = [SO42-]
Equation: Ksp = [Ba2+][SO42-]
Solution Map: Ksp → [Ba2+]
Apply the solution map:
Tro's Introductory Chemistry, Chapter 15

86. Example: Calculate the molar solubility of BaSO4. Ksp = 1
Example: Calculate the molar solubility of BaSO4. Ksp = 1.07 x at 25 °C
Information:
Given: Ksp = 1.07 x 10-10
Find: [BaSO4], M = [Ba2+] = [SO42-]
Equation: Ksp = [Ba2+][SO42-]
Solution Map: Ksp → [Ba2+]
Check the answer:
The molar solubility of BaSO4 is 1.03 x 10-5 M.
The units, M, are correct.
The magnitude makes sense as BaSO4 is classified insoluble,
so its solubility should be very small.
Tro's Introductory Chemistry, Chapter 15

87. Tro's Introductory Chemistry, Chapter 15
Activation Energy
The energy barrier that prevents any collision between molecules from being an effective collision is called the activation energy.
The larger the activation energy of a reaction, the slower it will be.
At a given temperature.
Tro's Introductory Chemistry, Chapter 15

88. Tro's Introductory Chemistry, Chapter 15
Exothermic Reaction
Activation
energy,
large
Activation
energy,
small
Reactants
Relative potential energy
DHreaction
Products
Progress of reaction
Tro's Introductory Chemistry, Chapter 15

89. Tro's Introductory Chemistry, Chapter 15
Endothermic Reaction
Activation
energy
Products
Relative potential energy
DHreaction
Reactants
Progress of reaction
Tro's Introductory Chemistry, Chapter 15

90. Tro's Introductory Chemistry, Chapter 15
Catalysts
A catalyst is a substance that increases the rate of a reaction, but is not consumed in the reaction.
Catalysts lower the activation energy of a reaction.
Catalysts work by providing an easier pathway for the reaction.
Tro's Introductory Chemistry, Chapter 15

91. Catalyst Effect on Activation Energy
Tro's Introductory Chemistry, Chapter 15

92. Catalyst Effect on Activation Energy, Continued
Tro's Introductory Chemistry, Chapter 15

93. Tro's Introductory Chemistry, Chapter 15
Enzymes
Enzymes are protein molecules produced by living organisms that catalyze chemical reactions.
The enzyme molecules have an active site to which organic molecules bind.
When the organic molecule is bound to the active site, certain bonds are weakened.
This allows a particular chemical change to occur with greater ease and speed.
i.e., the activation energy is lowered.
Tro's Introductory Chemistry, Chapter 15

94. Tro's Introductory Chemistry, Chapter 15
Sucrase
Tro's Introductory Chemistry, Chapter 15