1. Equilibrium
Chapter 15

2. A little Bit of History….

3. The Need for NH3 Plants need nitrogen fixation to grow
the enzyme nitrogenase provides plants with fixed nitrogen through reaction N2(g) + 3H2(g) 2NH3(g)
As demand for food became more, a way of producing NH3 was needed, but difficult, because to break the triple bond in N2, 550oC and 200atm pressure was needed

4. The Good and Bad of the Haber Process
WWI – Germans also needed NH3 for explosives
German Chemist Fritz Haber (1912) developed a way to synthesize NH3 from hydrogen and nitrogen gas.
It prolonged WWI, but led to the development of nitrogen fixing fertilizers which helped increased world food supply

5. Haber process
N2(g) + 3H2(g)↔2NH3(g)

6. Reversible Reactions Some reactions Most reactions aA + bB → cC + dD
and
cC dD → aA + bB
This can be written
aA + bB ↔ cC + dD

7. Reversible reactions and equilibrium
aA + bB ↔ cC + dD
This reaction will continue in both directions until a CHEMICAL EQUILIBRIUM is reached
this means that each of the opposite reactions
aA + bB → cC + dD (forward rxn)
cC dD → aA + bB (reverse rxn)
are occurring at the same time, at THE SAME RATE

8. Equilibrium
When a system (chemical reaction) reaches equilibrium, the reaction does not stop – but because opposite reactions are happening at the same rate,
the amount (concentration) of reactants and products at any given time is CONSTANT but NOT NECESSARILY EQUAL
Neither reactants nor products can escape the system or equilibrium is disturbed
Sometimes referred to as dynamic equilibrium
If the concentrations of reactants and products are constant, the ratio of their concentrations also equals a constant

9. Equilibrium
N2O4 (g) ↔ 2 NO2 (g)

11. Physical Equilibrium -occurs at phase changes
Solid ↔ Liquid

12. Solution Equilibrium
Occurs at saturation point

13. There are no units for Keq, or Kc.
Types of Equilibrium Constants- relates to the composition of the mixture at equilibrium
Keq =
Kc =
Kp =
Equilibrium (basic equation)
Concentrations (if molarity given)
Pressures ( if pressures given)
There are no units for Keq, or Kc.
Kp is in atm

14. equilibrium aA + bB ↔ cC + dD
For the reaction above, at any point in the reaction we can express the ratio of products to reactants (products over reactants) as a reactant quotient (Q)
𝑄= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏

15. equilibrium expressions
AT EQUILIBRIUM, the reaction quotient (Q) becomes the EQUILIBRIUM CONSTANT (Kc if concentration or Kp if pressure)
- no units
Concentrations of solids or pure liquids (not in solution) that are in the equilibrium expression (below) are assumed to be 1 since their concentrations do not change
Dependent only on the stoichiometry of the reaction, not the mechanism
𝐾= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏

16. Equilibrium expressions
Example : Haber process
N2(g) H2(g) ↔ 2 NH3(g)
Equilibrium expression if concentration is molarity:
Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] 3
Equilibrium expression if concentration is in partial pressure
Kp = (𝑃 𝑁𝐻 3 ) 2 𝑃 𝑁 ∙ (𝑃 𝐻 2 ) 3

17. Example 1:
Write the equilibrium expression, Kc, for the following reactions
a. 2 O3 (g) ↔ 3 O2 (g)
b. 2 NO (g) + Cl2 ↔ 2 NOCl (g)
c. Ag + (aq) NH3 (aq) ↔ Ag(NH3)2 + (aq)

18. Evaluating (solving for) Kp
There is a relationship between Kc and Kp
remember the ideal gas law ?
PV = nRT
P = nRT → P = n (RT) → P = [A](RT)
V V
derivation….
𝐾 𝑝 = 𝐾 𝑐 (𝑅𝑇) Δ𝑛

19. Example 2:
N2(g) + 3 H2(g) ↔ 2 NH3(g) Kc = 9.60 at 300°C. Calculate Kp for this reaction at this temperature Answer: 4.34 x 10-3

20. Equilibrium Constants
Magnitude (size) of equilibrium constants :
If K >> 1 then equilibrium is product favored (lies to the right)
If K << 1 then equilibrium is reactant favored (lies to the left)
Ex: CO (g) Cl2(g) → COCl2 (g)
What is the equilibrium expression (Kc ) for this reaction?
Kc = [COCl2]
[CO] [Cl2]
Kc = 4.56 x 109
Is the reaction product or reactant favored ?

21. Or, stated one more way…
If Keq has a positive exponent then its mostly products at equilibrium
If a negative exponent, then it’s mostly reactants

22. Example 3:
H2 (g) + I2 (g) → 2 HI(g) At 298 K , Kp = 794 At 700 K , Kp = 55 Is the formation of HI favored more at the higher or lower temperature? Answer : lower

23. Direction of a reaction and k
The value of the equilibrium constant is dependent on how you write the chemical reaction
For example, The equilibrium-constant expression for a reaction written in one direction is the reciprocal of the expression for the reaction written in the reverse direction

24. Example 4: For the given reaction N2O4 (g) ↔ 2 NO2 (g)
Kc = [NO2]2 = at 100°C
[N2O4 ]
a. What is the equilibrium expression for the reverse reaction?
Kc = [N2O4 ]
[NO2]2
b. What is the value of the equilibrium constant for the reverse reaction?
Kc = 4.72

25. Example 5: Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] 3 = 9.60
For the following reaction at 300 K, N2(g) H2(g) ↔ 2 NH3(g)
Kc = [𝑁𝐻 3 ] 2 [ 𝑁 2 ] [𝐻 2 ] = 9.60
What would the value of the equilibrium constant be for the following reaction?
2 N2(g) H2(g) ↔ 4 NH3(g)

26. Equilibrium constant
We can calculate equilibrium constant for a reaction if we know the equilibrium constants for the other reactions that add up to give us the target equation
For example :

27. Equilibrium constant cont.
Reaction 1 : 2 NOBr(g) ↔ 2 NO(g) + Br2(g) Kc (1) = [NO]2 [Br2] = [NOBr] Reaction 2 : Br2(g) + Cl2(g) ↔ 2 BrCl (g) Kc (2) = [BrCl]2 = 7.2 [Br2] [Cl2] The sum of these equations Reaction 3: 2 NOBr(g) + Cl2(g) ↔ 2 NO(g) + 2 BrCl (g) If you write the equilibrium expression for this reaction, it is equal to the two individual equilibrium expressions multiplied by each other, so we can multiply the individual equilibrium constants Kc (3) = [NO]2 [BrCl]2 = Kc (1) x Kc (2) = Kc (3) = 0.10 [NOBr] [Cl2]

28. Example 6: Given the following reactions,
HF (aq) ↔ H + (aq) + F - (aq) Kc = 6.8 x 10 -4
H2C2O4 (aq) ↔ H + (aq) + C2O4 -2 (aq) Kc = 3.8 x 10 -6
determine the value of Kc for the reaction
2 HF (aq) + C2O4 -2 (aq) ↔ 2 F - (aq) H2C2O4 (aq)
Answer = 0.12

29. Equilibrium constant manipulations summary
1. the equilibrium constant of a reaction in the revers direction is the inverse (reciprocal) of the equilibrium constant of the reaction in the forward direction
2. if the coefficients of a reaction have been multiplied by a number, the new equilibrium constant is equal to the original equilibrium constant raised to a power equal to that number
3. The equilibrium constant for a net reaction made up of two or more reactions is the product of the individual equilibrium constants

30. Heterogeneous equilibria
If substances in a chemical reaction are all in the same phase at equilibrium (usually a gas or liquid) – homogeneous equilibria
If substances in different phases – heterogeneous equilibria
Substances that have constant composition are not included in equilibrium expression - Solids and pure liquids are not included
Only substances that can be expressed in terms of concentration (molarity) or pressure can be included in the equilibrium expression
If a solvent is a reactant or product it is not written in in the equilibrium expression
As long as the concentration of the reactants and products are low, the solvent is considered a pure liquid

31. example
PbCl2 (s) ↔ Pb 2+ (aq) + Cl – (aq)
Kc = [Pb 2+] [Cl – ]

32. Example 7:
Write the equilibrium expression, Kc, for the following reactions
a. CO2 (g) + H2 (g) ↔ CO (g) H2O (l)
b. SnO2 (s) CO (g) ↔ Sn (s) CO2 (g)

33. Calculating equilibrium constants when all concentrations are known, Example 8:
An aqueous solution of acetic acid is found to have the following equilibrium concentrations at 25°C: [CH3COOH] = 1.65 x 10-2 M [H+] = 5.44 x 10-4 M [CH3COO -] = 5.44 x 10-4 M Calculate the equilibrium constant, Kc, for the ionization of acetic acid according to the following reaction CH3COOH (aq) ↔ H+ (aq) + CH3COO – (aq) Answer : 1.79 x 10-5

34. Calculating equilibrium constant when not all concentrations are known
Process
Make a table of all known initial and equilibrium concentrations
Calculate the change in the known concentrations as the system reaches equilibrium using the coefficients of the balanced equation
Use the initial concentrations and changes in concentrations to determine the equilibrium concentrations
Plug in concentrations into equilibrium expressions and solve for the equilibrium constant
This is also known as making an ICE chart

35. Change in [HI] = 1.87  103 M  0 = 1.87  103 M
A closed system initially containing  103 M H2 and  103 M I2 at 448 C is allowed to reach equilibrium, and at equilibrium the HI concentration is 1.87  103 M. Calculate Kc at 448 C for the reaction taking place, which is
the initial and
equilibrium concentrations of as many
species as we can. We also provide space
in our table for listing the changes in concentrations.
use the chemical equation as the heading
for the table.
calculate the change in HI
concentration, which is the difference
Between the equilibrium
and initial values:
Change in [HI] = 1.87  103 M  0 = 1.87  103 M 35

36. Notice that the entries for the changes are negative when a reactant
Third, we use the coefficients in the balanced equation to relate the change in [HI] to the changes in [H2] and [I2]:
Fourth, we calculate the equilibrium concentrations of H2 and I2, using initial concentrations and changes in concentration. The equilibrium concentration equals the initial concentration minus that consumed:
[H2] =  103 M   103 M =  103 M
[I2] =  103 M   103 M =  103 M
Notice that the entries for the changes are negative when a reactant
is consumed and positive when a product is formed. 36

37. Finally, we use the equilibrium-constant
expression to calculate the equilibrium
constant: 37

38. Example 10: Calculating concentrations from initial concentrations
H2 + I2 ↔ 2HI
Given:
Initial concentration H2 = .2 M
Initial concentation I2 = .2 M
Keq = 64
What are the equilibrium constants for each substance? (in other words, what is their molarity at equilibrium)

39. Step 1: Write Keq expression for reaction
Keq = [HI]2
[H2] [ I2]

40. Chapter 15, Unnumbered Table, Page 629

41. Step 2: Ice Chart/Box I initial conc.
2HI
I initial conc.
C Change in conc. (use balanced eq.)
E equilibrium conc.

42. Step 2: Ice Box H2 + I2 2HI I C -x 2x E (.2M- x) (.2M - x) (0 + 2x)
Initial conc.
.2M C
Change in conc. -x 2x E
Equilibrium conc.
(.2M- x)
(.2M - x)
(0 + 2x)
Fill in value for change in conc. Based on stoich. shown in equation (+ toward substance being produced)

43. Chapter 15, Unnumbered Table 1, Page 630

44. Plug in values into Keq equation
64 = ____[2x]2____
[.2-x] [.2-x]

45. How to put Ice Box Problem into calculator………

46. 64 = (2x)2
(.2-x) (.2-x)
64 = x2______
(.2-x ) ( .2 – x)
Multiply top first
64 = x 2_______
x + x2
“Foil” bottom
4x2 = 64x2 – 25.6x
Cross multiply
0 = 60 x2 – 25.6x
Subtract 4x2 from both sides
60 x2 – 25.6x = 0
“a” “b” “c”
Use quadratic equation in calculator to solve for x
*Program ---quad---fill in values for “a”,”b” and “c”
x = or x = .16
You get 2 possible answers. Choose value that makes sense . If you started with 0.2 M and the molarity of the reactants goes down as it approaches equilibrium, then the correct value for X = .166
Plug in the value of x for the equilibrium concentration in the ICE Box

48. Step 3.
Ans. X=.16
After solving for x, use it to determine the equilibrium concentrations
So H2 = = .04M at equilibrium
I2 = = .04M at equilibrium
HI = 2X = 2x .16= .32M at equilibrium

49. How does q fit into all of this?
Remember that Q is the reaction quotient, the concentrations that are used in the expression are not equilibrium concentrations
𝑄= [𝐶] 𝑐 [𝐷] 𝑑 [𝐴] 𝑎 [𝐵] 𝑏
The reaction quotient can vary throughout the reaction

50. How does q fit into all of this?
We can use the value of Q to determine if a reaction is at equilibrium or not
You can experimentally determine concentrations of reactants and products throughout a reaction and calculate Q
If you know what Kc is, then you can compare it to determine if you are at equilibrium or not

51. How does Q fit into all of this?
Q = K then the reaction is at equilibrium
Q > K then the concentration of products is too large and that of reactants too small, the reaction needs to proceed to from right to left in order to reach equilibrium – reverse reaction takes over
Q < K then the concentration of products is too small and the reaction needs to proceed from right to left in order to reach equilibrium – forward reaction takes over

52. Figure 15.8 Predicting the direction of a reaction by comparing Q and K at a given temperature.

53. Example 11: Predicting direction of equilibrium
At 448 °C the equilibrium constant, Kc, for the reaction
H2 + I2 ↔ 2HI
Is Predict which direction the reaction proceeds to reach equilibrium if we start with
2.0 x 10-2 mol of HI, 1.0 x 10-2 mol of H2 , and 3.0 mol of I2 in a 2.0 L container.
Find the initial concentrations of reactants and products, then plug into Q expression

54. Because Qc < Kc, the concentration of HI must increase
and the concentrations of H2 and I2 must decrease to
Reach equilibrium; the reaction as written proceeds left
to right to attain equilibrium.

55. Le Chatlier’s principle
How can we increase the yield (amount of products) of a reaction?
If we change the conditions of the reaction, can we change the yield?

56. Figure 15.9 Effect of temperature and pressure on NH3 yield in the Haber process.

57. Le Chatlier’s principle
If a system at equilibrium is disturbed by a change in temperature, pressure, or a component concentration, the system will shift its equilibrium position in order to counteract the effect of the disturbance
We can use this principle to predict how a system at equilibrium will respond when we add or remove a reactant or product, change the pressure by changing the volume, or changing the temperature

58. Chapter 15, Unnumbered Figure, Page 631

59. Le chatlier’s principle: concentration
Shift means that the reactant or product concentration will change to accommodate the new situation THERE IS NO CHANGE IN EQUILIBRIUM CONSTANT
If a chemical system in equilibrium and the concentration of any substance in the mixture is increased the system reacts to consume that substance OR if the concentration of a substance is decreased, the system will react to produce more of that substance

60. Le chatlier’s principle: concentration
For example N2(g) H2(g) ↔ 2 NH3(g)
if H2(g) is added, the reaction will shift to the right to use up the excess hydrogen
If we remove NH3(g) what will happen?
If we add NH3(g) what will happen?
So in order to increase the yield of this reaction, we could remove NH3(g) as it is produced

61. Le chatlier’s principle: volume and pressure changes
At constant temperature, reducing the volume of a gaseous equilibrium mixture causes the system to shift in the direction that reduces the number of moles of gas
Increasing the volume causes a shift in the direction that produces more gas molecules

62. Le chatlier’s principle: volume and pressure changes
For example N2(g) H2(g) ↔ 2 NH3(g)
4 molecules of reactants react to produce 2 moles of product
If we increase the pressure, by reducing the volume, this will shift in the direction that produces fewer molecules of gases , which in this reaction is a shift to the right, formation of product
How about this one? Which way will the reaction shift?
H2(g) + I2(g) ↔ 2 HI (g)

63. Le chatlier’s principle: temperature
When temperature changes, the equilibrium constant changes
Treat temperature as a reactant or product
Endothermic : reactants + heat ↔ products , increasing temp increases K
Exothermic : reactants ↔ products + heat , decreasing temp decreases K
When the temperature of a system at equilibrium is increased, the system reacts as if we added a reactant to an endothermic reaction or a product to an exothermic reaction. The equilibrium shifts in the direction that consumes the excess reactant (or product)

64. Le chatlier’s principle: temperature
For example, if you cool an endothermic reaction the equilibrium shifts equilibrium to the left
Endothermic : reactants + heat ↔ products
If you cool an exothermic reaction the equilibrium will shift to the right
Exothermic : reactants ↔ products + heat

65. Example 12: N2O4 (g) ↔ 2 NO2 (g) ΔH = 58.0 kJ
In which direction will the equilibrium shift when
N2O4 is added
NO2 is removed
Pressure is increased by addition of N2 (g)
Volume is increased
The temperature is decreased

66. A. Shift to the right to the products to decrease concentration of reactant
B. Shift to the right to compensate for removal of NO2
C. Adding nitrogen gas will increase the pressure, but nitrogen is not involved in the reaction so no shift in equilibrium
D. If the volume is increased, reaction shifts to the right to create more gas molecules
E. The reaction is endothermic so a decrease in temperature will shift to the direction that produces heat, which will shift to the left

67. Catalyst effect on equilibrium
Catalysts change the rate of a reaction, but not the value of K
So catalysts increases the rate at which equilibrium is achieved but not the composition of the equilibrium mixture