1. Equilibrium

2. Equilibrium
The state where the concentrations of all reactants and products remain constant with time.

3. Reactions are reversible
A + B C + D ( forward)
C + D A + B (reverse)
Forward and Reverse Rxns can be shown by double arrow
A + B C + D

4. A + B C + D
Initially there is only A and B so only the forward reaction is possible
As C and D build up, the reverse reaction speeds up while the forward reaction slows down.
Eventually the rates are equal
So concentrations of the reactants and products no longer change with time

5. Forward Reaction
Reaction Rate
Equilibrium
Reverse reaction
Time

6. Static or Dynamic?
At equilibrium, forward and reverse reaction rates are equal
May seem like no changes are occurring but there are changes
No NET changes
On the molecular level, there is frantic activity. Equilibrium is not static, but is a highly dynamic situation.
chemical reactions take place ,but concentrations of reactants and products remain unchanged

7. Analogies and Metaphors to think about
In a football game, the number of players on the field is constant although exchange of players (substitution) changes actual persons.
Connected fish bowl analogy . Two fish tanks are connected by a tube large enough to allow passage of fish. A number of fish are placed in one of the tanks. At equilibrium, the number of fish in each tank will eventually become unchanged.
Two jugglers analogy.
Drinking fountain line:
Ten students waiting in line to get a drink of water on a hot day. As each gets a drink, the same student reenters the line (equilibrium in a closed system).
(b) Same situation as "a," except as each student gets a drink and leaves, a new student enters the line (steady-state in an open system).
Picture a number of horses and wranglers in a corral. As each wrangler mounts a horse, the wrangler is bucked off. The equilibrium is:
Horse + Wrangler Mounted wrangler

8. Molecular Simulation
In this simulation two gaseous reactants collide to produce a more dense solid.
A B C
gaseous R dense P

9. Homo vs Hetero
Homogeneous Equilibria all reacting species are in the same phase
gas phase
equilibrium constant can be expressed in terms of pressure or concentration, Kp or Kc
Solution (aqueous) phase
concentration term for the pure liquid does not appear in the expression for the equilibrium constant but aqueous substance concentrations do appear
Heterogeneous Equilibria  all reacting species are not in the same phase
concentration term for solid or liquid does not appear in the expression for the equilibrium constant

10. Equilibrium Summarized
Forward and Reverse rates are equal
Concentrations are not.
Rates are determined by concentrations and activation energy.
Molecular Motion is frantic and constantly changing
Macroscopically no net change is occurring (We can’t observe any changes)

12. Can you identify when the system reached equilbrium?

13. Distinguishing between Physical and Chemical Equilibrium
As with physical and chemical changes, physical and chemical equilibrium follow the same rules:
Physical  no changes to the chemical properties of the substances involved
Ex. equilibrium of water vapor with liquid water in a partly filled sealed bottle
Chemical  involve changes in the chemical composition of substances. Bond breaking and bond formation is involved.
Ex. dissociation of acetic acid water into acetate and hydronium ion

14. Activity Model Dynamic Equilibrium with Coins
Now lets plot the data using excel

15. Law of Mass Action For a reaction: aA + bB ⇄ cC + dD
equilibrium constant: K
Pure liquids and pure solids have concentrations of 1. c

16. Playing with K If we write the reaction in reverse. cC + dD ⇄ aA + bB
Then the new equilibrium constant is c

17. They are simply the inverse of one another.
Forward Reaction
aA + bB ⇄ cC + dD
So we call this K1
And K1= 1 = K2-1 K2
Reverse Reaction
cC + dD ⇄ aA + bB
So we call this K2
And K2= 1 = K1-1 K1

18. The units for K
Are determined by the various powers and units of concentrations.
They depend on the reaction.
will always have the same value at a certain
temperature (Why will the T effect it?)
no matter what amounts are initially added
ratio at equilibrium will always be same

19. K Has no units Is constant at any given temperature.
Is affected by temperature.
Equilibrium constants are reaction, phase, temperature and pressure dependent
There is a K for each temperature.
Equilibrium constant values are thus established for a specific reaction in a specific system and will be unchanging (constant) in that system, providing the temperatures does not change.

21. What does the size of my K mean?
Large K > 1
products are "favored“
Ex. 1 x 1034
K = 1
neither reactants nor products are favored
Small K < 1
reactants are "favored“
Ex. 4 x 10-41

22. Now Let’s Calculate Your K
Using your data from the simulation calculate K

23. Different Equilibrium Constants (All of these are known as Keq)
Kc is the most used general form with molar concentrations.
Kp can be used with partial pressures when working with a gas phase reaction.
Ka is used for the dissociation of weak acids in water.
Kb is used for the dissociation of weak bases in water.
Kw is the equilibrium expression for the dissociation of water into its ions.
Ksp is used for the dissociation into ions of sparingly soluble solids in water.

24. Practice Writing the Equilibrium Expression
4NH3(g) + 7O2(g) NO2(g) + 6H2O(g)
First write the equilibrium expression using no concentration values.
What is the value for K if the concentrations are as follows
NH3 1.0 M
O M
NO M
H2O 1.8 M

25. Equilibrium with Gases
Equilibria involving only gases can be described using pressures or concentrations
If using pressures,
use pA not [A]
KP not KC
be sure all pressure are in the same units
N2(g) + 3H2(g) NH3(g)

26. Calculating Kc from Kp where
Δn is the difference in moles of gas on either side of the equation (np – nr)
R is the gas law constant:
T is Kelvin temperature
For: N2(g) + 3H2(g) NH3(g)
Δ n = (2) – (1+3) = -2 h

27. Practice Setup the expression for KP in terms of KC, R and T
2NO(g) + Cl2(g) NOCl(g)

28. What the equilibrium constant tells us…
if we know the value of K, we can predict:
tendency of a reaction to occur
if a set of concentrations could be at equilibrium
equilibrium position, given initial concentrations
If you start a reaction with only reactants:
concentration of reactants will decrease by a certain amount
concentration of products will increase by a same amount

29. Using this we can make ICE charts
The following reaction has a K of 16. You are starting reaction with 9 O3 molecules and 12 CO molecules.
Find the amount of each species at equilibrium.

33. Practice 1 ICE Charts Consider the following reaction at 600ºC
2SO2(g) + O2(g) SO3(g)
In a certain experiment 2.00 mol of SO2, 1.50 mol of O2 and 3.00 mol of SO3 were placed in a 1.00 L flask. At equilibrium 3.50 mol were found to be present. Calculate
The equilibrium concentrations of O2 and SO2, K and KP

34. Practice Problem 2 ICE Charts
Consider the same reaction at 600ºC
In a different experiment .500 mol SO2 and .350 mol SO3 were placed in a L container. When the system reaches equilibrium mol of O2 are present.
Calculate the final concentrations of SO2 and SO3 and K

35. The Reaction Quotient (Q)
Tells you the directing the reaction will go to reach equilibrium
Calculated the same as the equilibrium constant, but for a system not at equilibrium
Q = [Products]coefficient [Reactants] coefficient
Compare value to equilibrium constant

36. What Q tells us IF THEN Q = K reaction is at equilibrium
Q > K too much products,
left shift
Q < K too much reactants,
right shift

37. Example 1 Reaction Quotient
For the synthesis of ammonia at 500°C, the equilibrium constant is 6.0 x Predict the direction the system will shift to reach equilibrium in the following case:

38. Ex 1 Cont.

39. Example 2 In the gas phase, dinitrogen tetroxide
decomposes to gaseous nitrogen dioxide:
Consider an experiment in which gaseous N2O4 was placed in a flask and allowed to reach equilibrium at a T where KP = At equilibrium, the pressure of N2O4 was found to
be 2.71 atm.
Calculate the
equilibrium
pressure of NO2.

40. Example 3
At a certain temperature a 1.00 L flask initially contained mol PCl3(g) and 8.70x10-3 mol PCl5(g). After the system had reached equilibrium, 2.00x10-3 mol Cl2(g) was found in the flask.
PCl5(g) PCl3(g) + Cl2(g)
Calculate the equilibrium concentrations of all the species and the value of K.

41. Ex 3 Cont

42. Approximations
If K is very small, we can assume that the change (x) is going to be negligible compared to the initial concentration of the substances
can be used to cancel out when adding or subtracting from a “normal” sized number to simplify algebra

43. Example 4
At 35°C, K=1.6x10-5 for the reaction 2NOCl(g) ⇄ 2NO(g) + Cl2(g)
Calculate the concentration of all species at equilibrium for the following mixtures
2.0 mol NOCl in 2.0 L flask
1.0 mol NOCl and 1.0 mol NO in 1.0 L flask
2.0 mol NO and 1.0 mol Cl2 in 1.0 L flask

44. Ex 4 Cont

45. Ex 4 Cont

46. Ex 4 Cont

47. Le Chatelier’s Principle
can predict how certain changes or stresses put on a reaction will affect the position of equilibrium
helps us determine which direction the reaction will progress in to achieve equilibrium again
system will shift away from the added component or towards a removed component

48. Change Concentration equilibrium position can change but not K
system will shift away from the added component or towards a removed component
Ex: N2 + 3H NH3
if more N2 is added, then equilibrium position shifts to right (creates more products)
if some NH3 is removed, then equilibrium position shifts to right (creates more products)

49. Adding Gas
adding or removing gaseous reactant or product is same as changing concentration
adding inert or uninvolved gas
increase the total pressure
doesn’t effect the equilibrium position

50. Change the Pressure by changing the Volume
only important in gaseous reactions
decrease V
requires a decrease in # gas molecules
shifts towards the side of the reaction with less gas molecules
increase V
requires an increase in # of gas molecules
shifts towards the side of the reaction with more gas molecules

52. Change in Temperature
all other changes alter the concentrations at equilibrium but don’t actually change value of K
value of K does change with temperature
if energy is added, the reaction will shift in direction that consumes energy
treat energy as a
reactant: for endothermic reactions
product: for exothermic reactions

54. Catalyst
The use of a catalyst may speed up a reaction, but it speeds it up in both the forward and reverse direction, therefore a catalyst has no effect on the equilibrium state of the system.

55. energy + N2(g) + O2(g) ⇄ 2NO(g)
endo or exo?
endothermic
increase temp
to right
remove O2
to left
increase volume
no shift
add N2

56. Le Chatlier’s Simulation

57. Solution Equilibria

58. Another Dynamic Equilibrium
Equilibrium occurs when the solution is saturated

59. Ksp The solubility product constant (Ksp) is similar to the Keq.
When a solid is added to water, some dissolves (and splits into ions) while some remains a solid.
Ex: NaCl(s) Na+(aq) + Cl-(aq)
The mass action expression for this is:
Ksp = [Na+][Cl-]
(remember… solids are 1)

60. Common Ion Effect
The solubility of a solid is lowered if the solution already contains ions common to the solid
Dissolving silver chloride in a solution containing silver ions
Dissolving silver chloride in a solution containing chloride ions

61. Precipitation Opposite of dissolution
Can predict whether precipitation or dissolution will occur
Use Q: ion product
Equals Ksp expression but doesn’t have to be at equilibrium
Q > K: more reactant will form, precipitation until equilibrium reached
Q < K: more product will form, dissolution until equilibrium reached

62. Example 1

63. Example 1 Cont.

64. Example 2

65. Example 2 Cont

66. Example 2 Cont

67. Example 2 Cont

68. Qualitative Analysis
Process used to separate a solution containing different ions using solubilities
Example Problem
A solution of 1.0x10-4 M Cu+ and 2.0x10-3 M Pb2+. If I- is gradually added, which will precipitate out first, CuI or PbI2?

69. Solution to Example Problem

70. Example
The molar solubility for MgCl2 is M. Calculate Ksp

71. Salt Simulator
In groups of two use the simulator to discover the Ksp and Le Chatlier’s Principle

72. Acid-base Equilibria

73. Solutions of Acids or Bases Containing a Common Ion
Ion provided in solution by an aqueous acid (or base) as well as a salt
a. HF(aq) and NaF (F- in common)
b. HF(aq) H+(aq) + F-(aq) Excess F- added by NaF
Equilibrium shifts away from added component. Fewer H+ ions present.
pH is higher than expected.
a. NH4OH and NH4Cl (NH4+ in common)
b. NH3(aq) + H2O(l) NH4 +(aq) + OH-(aq)
Equilibrium shifts to the left. pH of the solution
decreases due to a decrease in OH- concentration

74. Equilibrium Calculations
Consider initial concentration of ion from salt when calculating values for H+ and OH-

75. Acid + Base  Salt + Water
Acid + Base  Conjugate Base + Conjugate Acid
Some solvents are amphiprotic
Water can act as an acid and a base!
Methanol can act as an acid and a base!
Autoprotolysis
Some solvents can react with themselves to produce an acid and a base
Water is a classical example
Weak acids dissociate partially, weak bases undergo partial hydrolysis. Strong acids and bases are strong electrolytes.

77. Kw (Dissociation of Water)
Water is amphiprotic it also undergoes autoprotolysis
Kw = 1.0E-14 at about 25 ˚C
This is where the pH scale we commonly use originates from!
What is the concentration of hydronium and hydroxide ions in neutral solution? What is the pH? What is the pOH?

78. Weak Acid & Weak Base Equilibria
Weak acids produce weak conjugate bases, and weak bases produce weak conjugate acids
Ka is a “special” equilibrium constant for the dissociation of a weak acid (found in standard tables)
Kb is a “special” equilibrium constant for the hydrolysis (or dissociation” of a weak base.

82. Calculations……..
What is the pH of a 1.0 M solution of acetic acid (HAc)?
What assumption can you make?
If [acid] is about 1000 times the Ka value, it’s concentration in solution won’t change much!
Use an “I-C-E” table to look at this.
The text goes into a more elaborate discussion of approximations. I will allow approximations if the concentrations or pH values do not change in the hundred’s decimal place.

83. What is the pH of a 4.0 M solution of phosphate ion?
Write reaction
Calculate Kb
Setup “I-C-E” table
Make assumptions
Solve algebraically.

85. Buffers
Buffers resist the change in pH because they have acid to neutralize bases and bases to neutralize acids.
Made from a weak acid (HA) and the salt of its conjugate base (A-, where the counter ion is gone for example), or a weak base and the salt of its conjugate acid.

86. Features of Buffers
Buffers work best at maintaining pH near the Ka of the acid component, usually about +/- 1 pH unit. This is their buffer capacity .
Buffers resist pH changes due to dilution.
All seen when we use the “Buffer Equation”

87. Henderson-Hasselbalch (Buffer) Equation
A modification of the equation for the dissociation of a weak acid.
The pH is the pH of the buffered solution, pKa is the pKa of the weak acid.
What is the pH of a buffer solution made from 1.0 M acetic acid and 0.9 M sodium acetate?
You add .10 moles of sodium hydroxide to the above solution? What is the new pH?

88. H-H Equation & Buffers….
If [A-] = [HA] pH = pKa!
This is what we see at half-way to the equivalence point in the titration of a weak acid with a strong base!
Dilution does not change the ratio of A- to HA, and thus the pH does not change significantly in most cases

89. How do you prepare buffers?
Select a buffer ‘system’ based on the pH you want to maintain.
Use the H-H equation to calculate how much acid and conjugate base you need
Take one of three approaches:
Mix acid and base forms, measure pH and adjust with strong acid or strong base
Start with a solution of weak acid, and add strong base until you reach the desired pH
Start with a solution weak base and add strong acid until you reach the desired pH
REMEMBER, STRONG Acids or Bases react completely!
Dilute as necessary, adjust pH further if needed with strong acid or base.

90. You want 1L of a buffer system that has a pH of 3.90?
What acid/conjugate base pair would you use?
How would you go about figuring out how much of each reagent you might need?
How would you prepare and adjust the pH of this solution?