1. Reaction Rates and Equilibrium
CHAPTER 12
Reaction Rates and Equilibrium
12.2 Chemical Equilibrium

2. CITY
traffic in
traffic out
There is a balance or equilibrium between cars that go into the city
and cars that leave the city

3. Concept of equilibrium
Physical equilibrium
There is a balance or equilibrium between cars that go into the city
and cars that leave the city
Physical equilibrium

4. Concept of equilibrium
Physical equilibrium
Chemical equilibrium
There is also a “balance” in a chemical system
Reactants Products
N2(g) + 3H2(g) NH3(g)
Chemical equilibrium
Physical equilibrium

5. Concept of equilibrium
Physical equilibrium
Chemical equilibrium
Le Châtelier’s principle
Effect of temperature
Effect of concentration
Effect of pressure / volume
Reactants Products
N2(g) + 3H2(g) NH3(g)
Chemical equilibrium
Products
Reactants

6. Concept of equilibrium
Physical equilibrium
Chemical equilibrium
Le Châtelier’s principle
Effect of temperature
Effect of concentration
Effect of pressure / volume
The equilibrium expression
Calculating the equilibrium constant, K
Determining where the equilibrium lies
Reactants Products
N2(g) + 3H2(g) NH3(g)
Chemical equilibrium

7. Physical equilibrium H2O(l) H2O(g)
An equilibrium between two phases of water:
H2O(l) H2O(g)

8. Physical equilibrium H2O(l) H2O(g)
An equilibrium between two phases of water:
H2O(l) H2O(g)

9. H2O(l) H2O(g) An equilibrium between two phases of water:
physical equilibrium: a “balance” in a physical system where there is an equilibrium between two or more phases at the same time.

10. Physical equilibrium
Pouring 50 mL from A into B
Pouring 50 mL from B into A
It looks like nothing is happening, but there is in fact a dynamic equilibrium
50 mL
50 mL
50 mL
50 mL

11. and the rate of transfer are the same from both sides
Physical equilibrium
Equilibrium does not mean that there must be equal amounts on each side
It is an equilibrium when the level of neither tank changes over time. This occurs when the amount transferred
and the rate of transfer are the same from both sides
50 mL
50 mL
50 mL
50 mL

12. Physical equilibrium The illustration below shows an example of .
non-equilibrium
25 mL
50 mL
50 mL
25 mL

13. Physical equilibrium The illustration below shows an example of .
non-equilibrium
25 mL
50 mL
50 mL
25 mL

14. Chemical equilibrium N2O4(g) 2NO2(g) a closed system:
clear
red-brown
a closed system:
no exchange of matter with the surroundings;
both reactant and product are present as a mixture
Depending on the color of the gas, we can tell whether there is more reactant (clear) or more product (red-brown)

15. Chemical equilibrium N2O4(g) 2NO2(g) clear red-brown more reactant
more product

16. Chemical equilibrium
In each case, an equilibrium has been established.
(Remember: amounts of reactants and products don’t have to be equal.)
N2O4(g) NO2(g)
clear
red-brown
more reactant
more product

17. Chemical equilibrium N2O4(g) 2NO2(g)
Nitrogen oxide (NO2) is a poisonous gas responsible for respiratory problems!
N2O4(g) NO2(g)
clear
red-brown
The smoggy skyline of Los Angeles

18. an equilibrium is reached
Chemical equilibrium
N2O4(g) NO2(g)
clear
red-brown
Start: Only N2O4 is present
Over time,
an equilibrium is reached
End: The concentrations of N2O4 and NO2 are constant

19. an equilibrium is reached
Chemical equilibrium
N2O4(g) NO2(g)
clear
red-brown
Start: Only N2O4 is present
Over time,
an equilibrium is reached
End: The concentrations of N2O4 and NO2 are constant

20. Notice that the equilibrium concentrations of N2O4 and NO2 are the same in cases A and B
N2O4(g) NO2(g)
N2O4
NO2
N2O4
NO2

21. The equilibrium favors the reverse reaction
At equilibrium, there is more N2O4 (reactant) than NO2 (product)
The equilibrium favors the reverse reaction
N2O4(g) NO2(g)
N2O4
NO2
equilibrium position: the favored direction of a reversible reaction, determined by each set of concentrations for the reactant(s) and product(s) at equilibrium.

22. From this graph of concentrations versus time, which side of the reaction is favored?
2SO2(g) + O2(g) SO3(g)

23. The forward reaction is favored
From this graph of concentrations versus time, which side of the reaction is favored?
2SO2(g) + O2(g) SO3(g)
The forward reaction is favored
Higher concentration

24. Temperature has an effect on the equilibrium
Concept of equilibrium
Physical equilibrium
Chemical equilibrium
Le Châtelier’s principle
Effect of temperature
Effect of concentration
Effect of pressure/ volume
N2O4(g) NO2(g)
more product
more reactant
Temperature has an effect on the equilibrium

25. Le Châtelier’s principle
If an equilibrium has been established…
N2(g) + 3H2(g) NH3(g)
You no longer have an equilibrium!
… what happens when two NH3 molecules are removed?

26. Le Châtelier’s principle
If an equilibrium has been established…
N2(g) + 3H2(g) NH3(g)
… what happens when two NH3 molecules are removed?
The forward reaction is favored so there is more product
You no longer have an equilibrium!
The equilibrium is re-established

27. Favored reaction
If removed
N2(g) + 3H2(g) NH3(g)
Le Châtelier’s principle: principle that states that when a “change” is made to a system at equilibrium, the system will shift in a direction that partially offsets the “change.”

28. Temperature and equilibrium
In an endothermic reaction, energy is considered a reactant
N2O4(g) kJ/mole NO2(g)
clear
red-brown
Raising the temperature is equivalent to adding more reactants
The equilibrium shifts to the right
More NO2 (red-brown gas) is produced

29. Temperature and equilibrium
In an endothermic reaction, energy is considered a reactant
N2O4(g) kJ/mole NO2(g)
clear
red-brown
Lowering the temperature is equivalent to having
less reactants
The equilibrium shifts to the left
More N2O4 (clear gas) is produced

30. Concentration and equilibrium
less of it
2SO2(g) + O2(g) SO3(g)
reactants
product
In a reversible equilibrium reaction, forming product means “using up” or consuming reactants
More product means less of the reactants.

31. Concentration and equilibrium
2SO2(g) + O2(g) SO3(g)
reactants
product
shifts left
In a reversible equilibrium reaction, forming product means “using up” or consuming reactants

32. Forming reactants means consuming some of the products
Concentration and equilibrium
2SO2(g) + O2(g) SO3(g)
reactants
product
In a reversible equilibrium reaction, forming product means “using up” or consuming reactants
shifts left
less of it
More of the reactants means less of the product.
2SO2(g) + O2(g) SO3(g)
reactants
product
Forming reactants means consuming some of the products

33. Forming reactants means consuming some of the products
Concentration and equilibrium
2SO2(g) + O2(g) SO3(g)
reactants
product
In a reversible equilibrium reaction, forming product means “using up” or consuming reactants
shifts left
2SO2(g) + O2(g) SO3(g)
reactants
product
shifts right
Forming reactants means consuming some of the products

34. Concentration and equilibrium
Using the equilibrium system
predict the direction in which the system will shift as a result of an increase in the concentration of Cl2.
PCl5(g) PCl3(g) + Cl2(g)

35. Concentration and equilibrium
Using the equilibrium system
predict the direction in which the system will shift as a result of an increase in the concentration of Cl2.
PCl5(g) PCl3(g) + Cl2(g)
Asked: Direction that the system will shift
Given: Increase in Cl2
Relationships: Note that Cl2 is a product on the right side of the equation.

36. Concentration and equilibrium
Using the equilibrium system
predict the direction in which the system will shift as a result of an increase in the concentration of Cl2.
PCl5(g) PCl3(g) + Cl2(g)
Asked: Direction that the system will shift
Given: Increase in Cl2
Relationships: Note that Cl2 is a product on the right side of the equation.
Solve: The system will shift toward the reactant to consume some of the added Cl2.

37. Concentration and equilibrium
Using the equilibrium system
predict the direction in which the system will shift as a result of an increase in the concentration of Cl2.
PCl5(g) PCl3(g) + Cl2(g)
Asked: Direction that the system will shift
Given: Increase in Cl2
Relationships: Note that Cl2 is a product on the right side of the equation.
Solve: The system will shift toward the reactant to consume some of the added Cl2.
Answer: Shifts left to produce some PCl5
Discussion: The shift in direction partially offsets the increase in Cl2.

38. The number of particles in the system is unchanged
Pressure and equilibrium
Pressure or volume only affect gaseous equilibrium systems
Higher pressure
Smaller volume
Lower pressure
Larger volume
The number of particles in the system is unchanged

39. Pressure and equilibrium
2SO2(g) O2(g) SO3(g)
2 moles
1 mole
2 moles
3 moles
2 moles
Reactants take up more space (larger volume) than the product
If we increase the pressure, the volume will decrease
The equilibrium will shift to the right

40. Pressure and equilibrium
2SO2(g) O2(g) SO3(g)
2 moles
1 mole
2 moles
3 moles
2 moles
Reactants take up more space (larger volume) than the product
If we decrease the pressure, the volume will increase
The equilibrium will shift to the left

41. Le Châtelier’s principle helps to determine where the equilibrium lies when the system undergoes a change in:
Temperature
Concentration
Pressure or volume (for gaseous systems)
Le Châtelier’s principle: principle that states that when a “change” is made to a system at equilibrium, the system will shift in a direction that partially offsets the “change.”

42. The equilibrium position depends on a ratio between products and reactants, called the equilibrium expression:
aA + bB cC + dD
raise to the power of coefficients
Equilibrium constant
[A] means “molarity of A”

43. aA + bB cC + dD
Write the equilibrium expression for the following reaction:
CH4(g) + 2H2S(g) CS2(g) + 4H2(g)
Answer:
Molarities of products
Molarities of reactants

44. Experimental results for:
N2(g) + 3H2(g) NH3(g) at 500oC

45. Experimental results for:
N2(g) + 3H2(g) NH3(g) at 500oC

46. Experimental results for: N2(g) + 3H2(g) 2NH3(g) at 500oC
K is constant at the same temperature

47. 2H2(g) + O2(g) 2H2O(l) K = 1.4 x 1083 at 25oC
Large K favors product
CaCO3(s) CaO(s) + CO2(g) K = 1.9 x 10–23 at 25oC
Small K favors reactant

48. Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC
[H2] = 0.22 M, and [I2] = 0.22 M
H2(g) + I2(g) HI(g)

49. Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC
[H2] = 0.22 M, and [I2] = 0.22 M
H2(g) + I2(g) HI(g)
Answer: Calculate the concentration of HI at equilibrium
Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M
Relationships:

50. Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC
[H2] = 0.22 M, and [I2] = 0.22 M
H2(g) + I2(g) HI(g)
Answer: Calculate the concentration of HI at equilibrium
Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M
Relationships:
Solve:

51. Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC
[H2] = 0.22 M, and [I2] = 0.22 M
H2(g) + I2(g) HI(g)
Answer: Calculate the concentration of HI at equilibrium
Given: K = 50 at 450oC, [H2] = [I2] = 0.22 M
Relationships:
Solve:
Answer: The equilibrium concentration of [HI] is 1.56 M

52. Calculate the equilibrium concentration of hydrogen iodide [HI], given the following information: K = 50 at 450oC
[H2] = 0.22 M, and [I2] = 0.22 M
H2(g) + I2(g) HI(g)
Answer: Calculate the concentration of HI at equilibrium
Relationships:
Solve:
Answer: The equilibrium concentration of [HI] is 1.56 M
Discussion: Here we used a known equilibrium constant (K) to solve for the unknown equilibrium concentration of the product.

53. Le Châtelier’s principle helps to determine where the equilibrium lies when the system undergoes a change in:
Temperature
Concentration
Pressure or volume (for gaseous systems)
The equilibrium constant helps to determine where the equilibrium lies:
Large K favors products
Small K favors reactants
aA + bB cC + dD
raise to the power of coefficients
Equilibrium constant
[A] means “molarity of A”