1. Chapter 15 - Chemical Equilibrium
Homework:
9, 10, 13, 14, 15, 16, 20, 21, 27, 28, 30, 31, 35, 36, 38, 39, 40, 43, 47, 48, 51, 52, 53, 54, 57, 63, 67, 69

2. 15. 1 - The Concept of Equilibrium
Imagine the following reaction
N2O4(g)  2 NO2(g)
N2O4 is colorless
NO2 is brown
We can analyze this reaction using kinetics (last chapter)
Call decomposition of N2O4 the forward reaction
Call decomposition of NO2 the reverse reaction
In both cases, these reactions are elementary

3. N2O4(g)  2 NO2(g) Forward Reverse
Rate =kf[N2O4]
f for forward
Reverse
2NO2(g)  N2O4(g)
Rate = kr[NO2]2
r for reverse
At equilibrium, the rate at which products are formed from reactants = the rate at which reactants are formed from the products
Dynamic equilibrium

4. Therefore... kf[N2O4] = kr[NO2]2 Rearrange this
So the ratio of the concentrations equals a constant
This constant will be called the equilibrium constant, which we will deal with next section
Does not matter if we start with N2O4 or with NO2, or even some mixture of the two.
At equilibrium, the ratio will equal a specific value

5. Once equilibrium has been established
Concentrations of substances no longer changes
Have created a dynamic equilibrium
Still reacting, just at the same rate

6. 3 Things to Remember From this
The fact that a mixture of reactants and products is formed in which concentrations no longer change with time indicates the reaction has reached a state of equilibrium.
For equilibrium to occur, neither reactants nor products can escape from a system
At equilibrium, the particular ratio of concentration terms equals a constant

7. 15.2 - The Equilibrium Constant
Opposing reactions naturally lead to an equilibrium, regardless of how complicated the reaction might be and regardless of the kinetics of the reaction
We can approach equilibrium either by starting with the reactants or by starting with the products

8. Law of Mass Action
For any reaction, the relationship between the concentration of the reactants and the products can be expressed by a constant.
Related to the ratio of the product and reactants in equilibrium

9. Equilibrium-Constant Expression
Given the following general equilibrium equation
aA + bB  dD + eE
The equilibrium condition is expressed by
This is called the equilibrium-constant expression
Kc is the equilibrium constant
c means that the concentrations are in molarity

10. In general
The numerator (top number) is the product of the concentrations of the products raised to their coefficient.
Denominator (bottom number) is the product of the concentrations of the reactants raised to their coefficient.
So once we know the balanced equation, we can write the equilibrium-constant expression
Depends only on stoichiometry, not as the mechanism

11. The value of the equilibrium constant AT A GIVEN TEMPERATURE does NOT depend on the initial amount of reactants and products
Does not matter if other substances are present, as long as they do not react with our reactants or products.
Equilibrium constant only depends on the particular reaction and the temperature

12. Example Write the equilibrium expression for the following reactions
2 O3(g)  3 O2(g)
2 NO(g) + Cl2(g)  2 NOCl(g)
Ag+(aq) + 2 NH3(aq)  Ag(NH3)2+(aq)

13. Equilibrium Constants in Terms of Pressure, Kp
When reactants and products are gases, we can set up the equilibrium-constant expression in terms of partial pressures
Instead of the normal molar concentrations
When we use partial pressures, we call the equilibrium constant Kp (p for pressure)

14. Using aA + bB  dD + eE, but with Pressure...
Px = partial pressure of X
Generally partial pressure here is in atmospheres
But units do not matter, as long as you are consistent

15. Kp  Kc We can perform algebra magic to convert between them
magicmagicmagicmagicmagic
Kp = Kc(RT)Δn
Δn is the change in the number of moles of gas in the equation for the reaction
Is equal to the sum of the coefficients of the gaseous products – the sum of the coefficients of the gaseous reactants
Δn = (moles of gaseous product) – (moles of gaseous reactant)

16. Example For the following reaction N2(g) + 3 H2(g)  2 NH3(g)
Kc= 9.60 at 300ºC.
Find Kp for this reaction at this temperature

17. N2(g) + 3 H2(g)  2 NH3(g) Kp = Kc(RT)Δn
Given Kc, need to find the change in moles/coefficient
2 moles of gaseous product (2 NH3) and four moles of gaseous reactant (1 N2 and 3 H2)
Therefore, Δn = 2-4 = -2
Delta always means product - reactant

18. Kp = Kc(RT)Δn We just plug everything in! Kp = 9.60(8.314)(573)-2
Kp= 4.34x10-3

19. Equilibrium Constants and Units
Generally
No units

20. 15.3 - Interpreting and Working with Equilibrium Constants
We are going to look at how to analyze a reaction based upon the value of its equilibrium constant

21. The Magnitude of Equilibrium Constants
Equilibrium constants can vary from very large to very small
To analyze what the constant means, we look at where it comes from
Given aA + bB  cC + dD
If we have a very large equilibrium constant, the numerator must be much larger than the denominator
When this happens, we say the equilibrium lies to the right, or, favors the product

22. If K >> 1: Equilibrium lies to the right; mostly products
If, however, the equilibrium constant is very small, then we have a large denominator compared to the numerator.
Reaction is said to lie to the left, or favor the reactants
If K >> 1: Equilibrium lies to the right; mostly products
If K << 1: Equilibrium lies to the left; mostly reactants

23. Example
The reaction of N2 with O2 to form NO is considered a means of fixing nitrogen
N2(g) + O2(g)  2 NO(g)
Kc at 25ºC is 1x10-30
Is it feasible to fix nitrogen by forming NO at 25ºC? No
Since constant is so small, reaction favors the reactants, and very little NO will be produced at equilibrium

24. The Direction of the Chemical Equation and K
Because we can approach equilibrium from either direction in a reaction (from reactants or from products), the way we write the chemical equation at equilibrium is arbitrary

25. N2O4(g)  2 NO2(g)
Could also write
2 NO2(g) N2O4(g)

26. We can see from the previous example that the equilibrium-constant expression for a reaction written in one direction is the reciprocal of the one for the reaction written in the reverse direction.
Therefore, if we know the equilibrium constant for one reaction, we can take its reciprocal to find the equilibrium constant for its reverse reaction
Note: Assuming temperature remains constant

27. Example
Given
2 NO(g)  N2(g) + O2(g)
Kc for the reverse reaction is 1x10-30
Write the equilibrium-constant expression for Kc for this reaction
Find the equilibrium-constant for this reaction

28. 2 NO(g)  N2(g) + O2(g) Kc for the reverse reaction is 1x10-30

29. Relating Chemical Equations and Equilibrium Constants
Just like the equilibrium constants of forward and reverse reactions are reciprocals of each other, the equilibrium constants of related reactions are also related.

30. Example Original equation was New equation is N2O4(g)  2 NO2(g)
Equilibrium-constant expression would be
Simply the square of the equilibrium-constant expression of the original
Because the new expression equals the original squared, the new constant equals the original squared.

31. Hess Law type problem
In problems that have multiple steps (like Hess Law), we must use equations made up of two or more steps in the whole process
We find the net equation by adding the individual equations and canceling identical terms

32. Example Consider 2 NOBr(g)  2 NO(g) + Br(g)
Br2(g) + Cl2(g)  2 BrCl(g)
Overall Equation
2 NOBr(g) + Cl2(g)  2 NO(g) + 2 BrCl(g)

33. The total equilibrium-constant expression is the product of the expressions for the individual steps
Similarly, the equilibrium-constant for the total is the product of the two individual equilibrium constants

34. Summary
The equilibrium constant of a reaction in the reverse direction is the inverse of the equilibrium constant of the reaction in the forward direction
The equilibrium constant of a reaction that has been multiplied by a number is the equilibrium constant raised to a power equal to that number
The equilibrium constant for a net reaction made up of two or more steps is the product of the equilibrium constants for the individual steps

35. 15.4 - Heterogeneous Equilibria
If, during equilibria, all of the substances are in the same phase, it is called a homogeneous equilibria.
Likewise, if the substances are in different phases, we get a heterogeneous equilibria.
Question:
What is the concentration of pure water?

36. Why is this a problem? Consider the following reaction
PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
When we try writing out the equilibrium-constant expression, we run into a problem
We cannot express the molarity of a solid!
Because there is no volume of solution or gas!
How do we solve this problem?
Ignore the solid!

37. PbCl2(s)  Pb2+(aq) + 2 Cl-(aq)
So the equilibrium-constant expression for the above reaction would just be
Kc = [Pb2+][Cl-]2
Even though PbCl2 is not shown in the expression, it must still be present for there to be equilibrium.
Likewise, we can also ignore pure liquids
Such as water

38. Why can we ignore these?
Concentration of a solid or pure liquid has a constant value
If we double the mass of a solid, we also double its volume
So its concentration (relates mass to volume) stays the same
Since equilibrium-constant expressions only have terms for things whose concentrations can change, these are left out

39. Example
Write the equilibrium-constant expressions for Kc for each of the following
CO2(g) + H2(g)  CO(g) + H2O(l)
SnO2(s) + 2 CO(g)  Sn(s) + 2 CO2(g)

40. Gas-Style Reaction CaCO3(s)  CaO(s) + CO2(g) Kc = [CO2] and Kp = PCO2
So at a given temperature, an equilibrium between CaCO3, CaO and CO2 will ALWAYS lead to the same partial pressure of CO2
As long as all three components are present

41. Example Given this reaction
CaCO3(s)  CaO(s) + CO2(g)
If each of the following mixtures was placed in a closed container and allowed to stand, which could attain the above equilibrium?
pure CaCO3
CaO and CO2 pressure greater than the value of Kp
Some CaCO3 and CO2 pressure greater than the value of Kp
CaCO3 and CaO

42. 15.5 - Calculating Equilibrium Constants
The simplest way to calculate the equilibrium constant is when the concentrations of all substances is known

43. Example 7.38 atm H2, 2.46 atm N2, 0.166 atm NH3
N2(g) + 3 H2(g)  2 NH3(g)
Create the equilibrium-constant expression, solve for Kp
And that is it!

44. Unfortunately
We oftentimes dont know the equilibrium concentrations of everything
If we know the equilibrium concentration of at least ONE thing, we can generally use stoichiometry of the reaction to find the concentrations of the others
Then once we have found the equilibrium concentrations of the others, we can solve for the equilibrium constant.

45. Steps to Solve for Unknown Equilibrium Concentrations
Tabulate all the known initial and equilibrium concentrations
For those that we have both the initial and equilibrium concentrations, find the change in concentration
Use stoichiometry to find the changes for the rest

46. Example
A closed system initially contains 1.000x10-3 M H2, and 2.000x10-3M I2 at 448ºC. It is allowed to reach equilibrium. We find that at equilibrium, the concentration of HI is 1.87x10-3M.
Calculate Kc at this temperature for this reaction
H2(g) + I2(g)  2 HI(g)

47. H2(g) + I2(g)  2 HI(g) ICE = Initial, Change, Equilibrium H2(g) I2(g)
All of these are concentration values
H2(g)
I2(g)
HI(g)
Initial
1.000x10-3M
2.000x10-3 M
0 M
Change
Equilibrium
1.87x10-3 M

48. If we know the initial and final of ANY of the substances, we can find the change in concentration
Change in [HI]
1.87x10-3 M - 0 = 1.87x10-3 M
Once we know the change, we use stoichiometry to find the change in the others

49. H2(g) + I2(g)  2 HI(g) Change in HI = 1.87x10-3 M
So in this case, both reactants had a change of 0.935x10-3 M

50. H2(g) I2(g) HI(g) Initial 1.000x10-3M 2.000x10-3 M 0 M Change
Equilibrium
0.065x10-3 M
1.065x10-3 M
1.87x10-3 M

51. So we now have our equilibrium concentrations
Can now find the equilibrium constant
Does this reaction favor the products or reactants?

52. Example 2
Sulfur trioxide decomposes at high temperature in a sealed container
2 SO3(g)  2 SO2(g) + O2(g)
Initially, the container is charged at 1000 K with SO3(g) at a partial pressure of atm.
At equilibrium, the SO3 partial pressure is atm.
Calculate the value of Kp at 1000 K.

53. Workspace SO3(g) SO2(g) O2(g) Initial .500 atm 0 atm Change
Equilibrium
0.200 atm

54. Workspace - Find change

55. SO3(g)
SO2(g)
O2(g)
Initial
.500 atm
0 atm
Change
Equilibrium
0.200 atm

56. Workspace - Stoichiometry

57. Workspace - Calculation

58. 15.6 - Applications of Equilibrium Constants
We know that the magnitude of K shows how a reaction will proceed
If K is large, reaction tends to proceed toward the product
If K is small, reaction tends to proceed towards the reactants
We can also use equilibrium constant to
determine the direction a reaction will go (based on a reaction mixture)
find the concentrations of reactants and products once equilibrium has been achieved.

59. Predicting the Direction of Reaction
We know which direction of a reaction is favored (based upon value of K)
But what if we have a mixture of products and reactants, and we want to see which way the reaction goes to achieve equilibrium?
Meaning do we produce more reactants or more products?

60. Reaction Quotient
The reaction quotient, Q, lets us find which direction a reaction goes
To find Q, we substitute initial reactant and product concentrations (or partial pressures) into our normal equilibrium-constant expression
So we find Qc (for concentration) or Qp (for pressure)

61. Determining Reaction Direction
We compare the value of Q with the K value for the reaction
If Q = K: We are already at equilibrium.
If Q > K: Concentration of products too large. Reaction moves from right to left (products to reactants)
If Q < K: Concentration of products too small. Reaction moves by forming products (left to right)

62. Example
At 448ºC the equilibrium constant Kc for the following reaction is 50.5.
H2(g) + I2(g)  2 HI(g)
Predict in which direction the reaction will proceed to reach equilibrium at this temperature if we start with 2.0x10-2 mol of HI, 1.0x10-2 mol of H2, and 3.0x10-2 mol of I2 in a 2.00 L container

63. Step 1: Find actual concentrations
[HI]
[H2]
[I2]

64. Step 2: Find Q and compare

65. Calculating Equilibrium Concentrations
Oftentimes we must find the amount of reactants and products at equilibrium.
We do this similar to how we found the equilibrium constant
We use ICE method or just the equilibrium values, depending on information given

66. Example Kp for the following process is 1.45x10-5 at 500ºC
N2(g) + 3 H2(g)  2 NH3(g)
In an equilibrium mixture of the three gases at this temperature, the partial pressure of H2 is atm, and that of N2 is atm.
What is the partial pressure of NH3 in this equilibrium mixture?

67. Start with equilibrium-constant expression
Rearrange to solve for x
x2 = (1.45x10-5)(0.432)(0.928)3
x2 = 5.01x10-6
x = 2.24x10-3 atm = PNH3

68. We then solve for the equilibrium amounts.
Usually we know the value of the equilibrium constant and the initial amount of everything.
We then solve for the equilibrium amounts.
We use the change in concentration as a variable
So NOW we use ICE

69. Example
A L flask is filled with mol of H2and mol of I2 at 448ºC. Kc for this reaction is 50.5
H2(g) + I2(g)  2 HI(g)
What are the equilibrium concentrations of H2, I2and HI in molarity?

70. Start the ICE tabulation
H2(g)
I2(g)
HI(g)
Initial
1.000 M
2.000 M
0 M
Change
Equilibrium

71. Use stoichiometry
We use the stoichiometric relationships to determine the change in concentration
Since H2 and I2 have a 1:1 mole ratio, they will both lose the same concentration
H2(g)
I2(g)
HI(g)
Initial
1.000 M
2.000 M
0 M
Change -x
+2x
Equilibrium

72. Use change to find equilibrium
Use the same data to find out the equilibrium amount
H2(g)
I2(g)
HI(g)
Initial
1.000 M
2.000 M
0 M
Change -x
+2x
Equilibrium
(1.000-x)M
( x)M
2x M

73. Substitute equilibrium amounts into equation
Now substitute the amounts of the equilibrium into the equilibrium-constant expression
Using algebra, solve for x
4x2 = 50.5(x x )
46.5x x = 0
x = or (quadratic equation gives two answers)

74. x = or 0.935?
Substitute each value into the equilibrium amounts, and see if it makes sense.
Start with = x
[H2] = x = = a negative concentration
We cannot have a negative concentration, therefore the correct value of x is hopefully 0.935
[H2] = = M
[I2] = = M
[HI] = 2x = M
Whenever we do this, we will always have to choose between two answers. Only one of them will be chemically meaningful

75. 15.7 - Le Châteliers Principle
Le Châteliers principle is stated as:
If a system at equilibrium is disturbed by a change in temperature, pressure, or the concentration of one of the components, the system will shift its equilibrium position so as to counteract the effect of the disturbance.
This means if we mess with a reaction at equilibrium, the system will change to accommodate our change

76. Change in Reactant or Product
This is the easiest shift
If a reaction is in equilibrium, and we add a substance (either reactant or product), the reaction will shift to reestablish equilibrium by consuming part of the added substance.
Similarly if we remove a substance, the reaction moves in the direction that forms more of that substance

77. What This Means Consider this familiar equation What would happen if I
N2(g) + 3 H2(g)  2 NH3(g)
What would happen if I
Added more nitrogen?
Reaction shifts to the right, forming more ammonia (but using more hydrogen)
Added more hydrogen?
Same as above. Shifts to the right, forms more ammonia, using more nitrogen
Added more ammonia?
Shifts to the left, forming more nitrogen and hydrogen
Removed ammonia?
Shifts to the right, forming more ammonia, using up nitrogen and hydrogen

78. Effect of Volume and Pressure Change
If volume of system is decreased (increasing pressure), system will shift equilibrium to reduce the pressure
We reduce pressure by reducing the total number of gas molecules
More gas molecules = more pressure
So lowering the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas

79. Example: N2O4(g) 2 NO2(g)
If I decrease the volume of the container this reaction occurred in
Reaction shifts to the side of fewer moles
So the reaction would shift towards the reactants
So increasing pressure (decreasing volume) causes reaction to produce more N2O4(g)

80. How about N2(g) + 3 H2(g)  2 NH3(g)
There are 4 reactant molecules to every 2 NH3 molecules
That means that in increase in pressure (decrease in volume) shifts reaction towards side with fewer molecules
So reaction moves to the right
What if we decreased the pressure in the reaction?
Reaction would move to the left

81. Remember Pressure-volume changes DO NOT change the K value
as long as T remains constant
All we do is change the partial pressures of the substances

82. Example N2(g) + 3 H2(g)  2 NH3(g)
At 472ºC, Kp = 2.79x10-5
7.38 atm H2, 2.46 atm N2 and atm NH3 at equilibrium
Consider what will happen when we suddenly reduce the volume by ½.
If the equilibrium did not shift, then the partial pressure of each would double
Giving us atm H2, 4.92 atm N2 and atm for NH3

83. These new pressures would mean the reaction quotient would no longer equal the equilibrium constant
Qp  Kp
Because Qp < Kp
Reaction no longer at equilibrium
Equilibrium reestablished by increasing PNH3 and decreasing the reactants
Therefore, reaction shifts to the right

84. Pressure Change w/o Volume Change?
We can change pressure of system without changing the volume
If additional amounts of any of the reacting components is added/removed
Adding a gas not involved in reaction
Since this would not change any of the partial pressures, no shift in equilibrium

85. Effect of Temperature Change
Almost every equilibrium constant changes with temperature changes
Look at temperature dependence through Le Châteliers principle.
Treat heat as if it were a chemical reagent
In an endothermic reaction, treat heat as a reactant
In an exothermic reaction, treat heat is a product

86. When the temperature of a system at equilibrium is increased, it is like we added a reactant to an endothermic reaction or a product to an exothermic reaction
The equilibrium will shift in the direction that consumes the excess reactant (or product), meaning heat

87. What This Means Endothermic: Exothermic:
Increasing T results in an increase in K
Exothermic:
Increasing T results in a decrease in K
Cooling is just the opposite of heating

88. Example 1 Consider the equilibrium
N2O4(g)  2 NO2(g) ΔHº = 58.0 kJ
In which direction will the equilibrium shift when
N2O4is added?
To the right
NO2 is removed

89. N2O4(g)  2 NO2(g) ΔHº = 58.0 kJ
The total pressure is increased by addition of N2?
No shift at all
N2 does not change the partial pressures of either of these
The volume is increased?
Shifts towards the right
The temperature is decreased?
To the right
Endothermic reaction, heat is treated as a reactant

90. The Effect of Catalysts
Remember, when we add a catalyst, we lower the activation energy between reactants and products
Also lowers the activation energy for the reverse reaction
Catalysts DO NOT change the composition of an equilibrium
It just changes the rate at which equilibrium is achieved