1. Equilibrium
AP Chem
Mr. Nelson

2. The Concept of Equilibrium
The term reversible reaction is used to describe reactions that can go in either the forward or the reverse direction
Equilibrium is used to specify that a reversible reaction has reached an equal rate for both directions

3. The Concept of Equilibrium
Chemical equilibrium occurs when opposing reactions are proceeding at equal rates
Concentrations of the reactants/products remains constant

4. The Concept of Equilibrium
Its important to note that forward and reverse reactions still occur, although outwardly no apparent change occurs

5. The Concept of Equilibrium
Note that the equilibrium concentrations are independent of initial concentrations
A mathematical relationship exists between concentration of the reactants and products once equilibrium has been reached

6. The Concept of Equilibrium
Fish Tank Example

7. Connection between Equilibrium & Kinetics
For generalized reaction:
A⇋B
Forward reaction:
AB
Forward Rate Law: Ratef = kf[A]
Reverse reaction:
BA
Reverse Rate Law: Rater = kr[B]

8. Connection between Equilibrium & Kinetics
By definition, at equilibrium forward and reverse rates must be equal, so:
Ratef = Rater
kf[A] = kr[B]
Also, concentrations remain constant,so:

9. The Equilibrium Constant
The equilibrium constant (Keq) can be determined at constant temperature for any reaction at equilibrium:
aA+ bB  cC + dD
Keq =
[C]c [D]d [A]a [B]b
Keq is unitless

10. The Equilibrium Constant
Note that only equilibrium concentrations can be placed into an equilibrium constant expression!
Also, no solids or liquids are included in an equilibrium constant expression
The reason for this is their concentrations remain fairly constant in equilibrium
Note that they are still required for the reaction to proceed however

11. The Equilibrium Constant
The Many Types of Equilibrium Constants, Keq Kc Kp
Ksp
Ka/Kb Kw

12. Concentration vs. Pressure Equilibria
Equilibrium expressions can be written in terms of concentration (Kc) measured in molarity or pressure (Kp) measured in atm
Example: or

13. Concentration vs. Pressure Equilibria
In general, Kc is not equal to Kp
The following equation converts between the two:
n = moles of gas products – moles of gas reactants
Note: When no change in moles, Kp=Kc

14. Homogeneous vs. Heterogeneous Equilibria
The term homogeneous equilibrium is used to describe reactions where all the species are in the same phase
Example:
The term heterogeneous equilibrium is used to describe reactions where species are in different phases
H2CO3 (aq) + H2O (l)  HCO3- (aq) + H3O+ (aq)

15. Multiple Equilibria A + B  C + D C + D  E + F K1 A + B  E + F K2 Kc
When the products of one reaction are the reactants of another
Example:
A + B  C + D
C + D  E + F K1 K2
A + B  E + F Kc

16. Multiple Equilibria
Using their equilibrium expressions, we can derive a mathematical relationship:

17. Manipulating Equilibria Expressions
If an equilibrium is written in the reverse, the following relationship exists:
PCl3 + Cl2  PCl5
PCl5  PCl3 + Cl2

18. Manipulating Equilibria Expressions
If an equilibrium is multiplied by a factor, the following relationship exists:
PCl3 + Cl2  PCl5
2PCl3 + 2Cl2  2PCl5

19. The Magnititude of Equilibrium Constants
What does Keq tell us?
If Keq is very large, products will be favored
This implies the numerator is large in the Keq expression compared to denominator
If Keq is very small, reactants will be favored
This implies the denominator is large in the Keq expression compared to numerator
If Keq is close to 1, then roughly equal amounts of reactants and products are present at equilibrium

20. The Reaction Quotient (Q)
Recall: In electrochemistry, we utilized “Q” which is similar in all respects to “Keq” except in definition
Q is utilized when given initial concentrations, not concentrations at equilibrium

21. Predicting the Direction of Reversible Reactions
If given an initial set of conditions for a reversible reaction, utilizing Q and Keq, a prediction of the direction of the reaction (towards reactants/towards products) can be made

22. Comparing Kc and Qc Ratio of prds to rcts is too small
Qc<Kc
Ratio of prds to rcts is too small
To reach equilibrium, rcts must go towards prds
Shifts to the right
Qc=Kc
Initial concentrations are at equilibrium concentrations
System is at equilibrium
Qc>Kc
Ratio of prds to rcts is too large
To reach equilibrium, prds must go towards rcts
Shifts to the left

23. Comparison of Kc and Qc Qc Kc Qc Kc Kc Qc

24. Calculating Equilibrium Concentrations
Tabulate all known initial and equilibrium concentrations
When both are known, calculate the change
Using coefficients, calculate changes in concentration for all participants in a rxn
Calculate equilibrium concentrations

25. Initial molarity Change in molarity Equilibrium molarity
The ICE Method
This is the process used for determining equilibrium concentrations
Initial molarity
Change in molarity
Equilibrium molarity

26. Example #1: Calculating Equilibrium Concentrations
A mixture of mol H2 and mol I2 was placed in a 1.00 L stainless-steel flask at 430 °C. The equilibrium constant Kc for the reaction: H2 (g) + I2 (g)  2 HI (g) is 54.3 at this temp. Calculate the concentrations of H2, I2, and HI at equilibrium.

27. H2 (g) + I2 (g)  2 HI (g) 0.500 -x +2x 0.500 – x 2x Initial (M)
Change (M) -x
+2x
Equilibrium (M)
0.500 – x 2x

28. Taking the square root of both sides:
Substituting, we get:
Taking the square root of both sides:
x = M

29. So, the concentrations at equilibrium would be the following:
[H2] = (0.500 – 0.393) M = M
[I2] = (0.500 – 0.393) M = M
[HI] = 2 x M = M

30. Example #2: Calculating Equilibrium Concentrations
When 3.0 mol of I2 and 4.0 mol of Br2 are placed in a 2.0 L reactor at 150oC, the following reaction occurs until equilibrium is reached:
I2(g) + Br2(g)  2IBr(g)
Chemical analysis then shows that the reactor contains 3.2 mol of IBr. What is the value of the equilibrium constant Kc for the reaction?

31. I2 (g) + Br2 (g)  2 IBr (g) 1.5 2 -x +2x 1.5 – x 2 – x 1.6
Initial (M)
1.5 2
Change (M) -x
+2x
Equilibrium (M)
1.5 – x
2 – x
1.6

32. When initial and final concentrations are known, calculate the change:
0 + 2x = 1.6
x = 0.8
Then determine all equilibrium concentrations:
[I2] = 1.5 – 0.8 = 0.7 M
[Br2] = 2 – 0.8 = 1.2 M
[IBr] = 1.6 M

33. Plug equilibrium concentrations into equation and solve:

34. Calculating Equilibrium Concentrations when Keq is small
When solving equilibrium concentrations and Keq is small, assume the effect of the change (“x”) is negligible (for rcts only)

35. Example: Calculating Equilibrium Concentrations when Keq is small
At 100 °C the equilibrium constant (Kc) for the reaction:
COCl2 (g)  2 CO (g) + Cl2 (g)
is 2.19 x If the initial concentration of [COCl2] = 1.5 M, calculate the concentrations of all species at equilibrium.

36. COCl2 (g)  2 CO (g) + Cl2 (g) 1.5 -x +2x +x 1.5 – x Initial (M)
Change (M) -x
+2x +x
Equilibrium (M)
1.5 – x

38. Le Chatelier’s Principle
If a stress is applied to a system in equilibrium, the equilibrium shifts to relieve that stress
Forms of “stress”:
Concentration
Pressure
Temperature

39. Change in Reactant or Product Concentration
Adding a substance shifts a system at equilibrium away from the added substance
System must consume added item
Removing a substance from a system at equilibrium shifts the reaction toward the removed substance
System must replace removed item

40. Changes in Pressure
Increase in pressure forces an equilibrium to shift in the direction that reduces the number of moles of gas in the system
Example:
An increase in pressure causes a shift to the left
N2O4 (g)  2 NO2 (g)

41. Changes in Pressure
Recall: Changes in volume can also affect changes in pressure
Increase volume = decrease pressure
Decrease volume = increase pressure
Also the addition of an inert gas (one not in the reaction) can cause an increase in pressure as well

42. Changes in Temperature
Enthalpy of the reaction plays a major roll in how heat effects the equilibrium
Recall: Endothermic (+∆H) means heat is a reactant, while exothermic (- ∆H) means it is a product
Treat changes in heat in the same manner that you treat changes in concentration

43. Effects of Catalysts on Equilibrium
Recall: Catalysts lower energy barrier between reactants and products
Ea for both forward and reverse are lowered to the same extent
Catalysts increase the rate at which equilibrium is reached
However, it does NOT change the composition of the equilibrium mixture.

44. Solubility Equilibria
Saturated solutions of salts are another type of chemical equilibria
Slightly soluble salts establish a dynamic equilibrium with the hydrated cations and anions in solution

45. The Solubility Product, Ksp
The equilibrium constant, the Ksp, is no more than the product of the ions in solution (Recall: solids do not appear)
For a saturated solution of AgCl, the equation would be:
The solubility product expression would be:

46. Sequence of Steps for Solubility Equilibria
Solubility of compound
Molar solubility of compound
Concentrations of cations and anions
Ksp of compound
Ksp of compound
Concentrations of cations and anions
Molar solubility of compound
Solubility of compound

47. Solubility Solubility can be expressed two ways:
Molar solubility is the moles of solute per 1L of saturated solution (mol/L)
Solubility is the number of grams of solute per 1 L of saturated solution (g/L)
Ksp expressions require molar solubility

48. Determining Ksp by Experimental Measurements
Use ICE table, but with “s” for change (instead of “x”)
“s” is not only the change, but the molar solubility

49. Determining Ksp: Example #1
The solubility of CaSO4 is found to be 0.67 g/L. Calculate the value of Ksp for calcium sulfate.
Solubility of CaSO4 in g/L
Molar solubility of CaSO4
[Ca2+] and [SO42-]
Ksp of CaSO4

50. Determining Ksp: Example #1
Initial (M)
Change (M) -s +s
Equilibrium (M) s

51. Determining Ksp: Example #1
As CaSO4 dissolves, we get 1 mole Ca2+, and 1 mole SO42-, so at equilibrium:

52. Determining Ksp: Example #2
Lead (II) chloride dissolves to a slight extent in water according to the equation:
Calculate the Ksp if the solubility of lead (II) chloride has been found to be x 10-2 M

53. Determining Ksp: Example #2
Molar solubility of PbCl2
[Pb2+] and [Cl-]
Ksp of PbCl2
Initial (M)
Change (M) -s +s
+2s
Equilibrium (M) s 2s

54. Determining Ksp: Example #2

55. Calculating Solubility from Ksp: Example #1
The Ksp for CaCO3 is 3.8 x 10-9 at 25 °C. Calculate the solubility (in g/L) of calcium carbonate in pure water.
Ksp of CaCO3
Concentrations of [Ca2+] and [CO32-]
Molar solubility of CaCO3
Solubility of CaCO3

56. Calculating Solubility from Ksp: Example #1
Initial (M)
Change (M) -s +s
Equilibrium (M) s

57. Calculating Solubility from Ksp: Example #1

58. Predicting Precipitation Reactions using Q
Using the reaction quotient, Q, we can
Decide whether a ppt will form
Calculate ion concentration required to begin the ppt. of an insoluble salt
Q < Ksp
System has too much reactant
No ppt
Q = Ksp
System at equilibrium
Q > Ksp
System has too much product
Forms ppt!

59. Predicting Precipitation Reactions
Steps
Determine insoluble product using solubility rules
Calculate cation and anion concentration of insoluble product (if mixed)
Calculate Q
Compare to Ksp

60. Example: Predicting Precipitation
Exactly 200 mL of M BaCl2 are mixed with exactly 600 mL of M K2SO4. Will a precipitate form? (Ksp of insoluble salt is 1.1 x )

61. The Common Ion Effect
Le Chatelier’s Principle can describe the effect of an added common ion
For example, if NaCl were added to a saturated solution of PbCl2, what effect would occur?
Increase product concentration shifts equilibrium towards reactants, forming more ppt.

62. Solubility and the Common Ion Effect
The presence of a common ion will cause the equilibrium to shift so that even less of the substance (with smaller Ksp value) will dissolve

63. The Common Ion Effect Be aware: pH can affect solubility
The addition of an acid or base could have a side reaction, decreasing concentration of an ion
If an acid were added to the following, what effect would be observed?
H+ combines with OH-, so [OH-] is decreased, forcing equilibrium to the right

64. Example: Common Ion Effect
Calculate the solubility of silver chloride (in g/L) in a 6.5 x 10-3 M silver nitrate solution. (Ksp = 1.6 x 10-10)
Since silver nitrate is a soluble, strong electrolyte:
6.5 x 10-3 M
6.5 x 10-3 M

65. Example: Common Ion Effect
Initial (M)
6.5 x 10-3
0.00
Change (M) -s +s
Equilibrium (M)
(6.5 x s) s
Because Ksp is so small, s must be very small in comparison to 6.5 x 10-3, so we approximate that 6.5 x s  6.5 x 10-3

66. Selective Precipitation
In certain problems, you will have to determine which of two possible precipitates will form
The salt with the lower Ksp will precipitate first
To determine the amount of an ion needed to precipitate, recall that Ksp concentrations are saturated, so anything greater than this will form a ppt.

67. Example: Selective Precipitation
A solution contains 1.0 x 10-4 M Cu+ and 2.0 x 10-3 M Pb2+. If a source of I- is added gradually to this solution, will PbI2 (Ksp = 1.4 x 10-8) or CuI (Ksp = 5.3 x 10-12) precipitate first? Specify the concentration of I- necessary to begin precipitation of each salt