1. Ch. 18 – Reversible Reactions & Equilibrium
pp. 549 – 559

2. A. Reversible Reactions
Reactions can be reversible
Conversion of reactants to products and products to reactants happens simultaneously
Forward reaction: 2SO2(g) + O2(g) → 2SO3(g)
Reverse reaction: 2SO2(g) + O2(g) ← 2SO3(g)

3. A. Reversible Reactions
Chemical equilibrium – when the rates of the forward and reverse reaction are equal
Physical Equilibrium: vapor-liquid phase or solution equilibrium
No net change occurs in the actual amounts of the components of the system (reaction will never complete)
dynamic state
Concentration at equilibrium is temperature dependent

4. A. Reversible Reactions
Equilibrium position – made up of relative concentration of reactants and products at equilibrium
A B
Different sized arrows indicate favored direction of a reaction
Catalyst speeds up forward and reverse reactions equally
decreases time it takes to reach equilibrium

5. B. Equilibrium Constant
Law of Mass Action – expression of equilibrium condition using equilibrium constant, Keq
jA + kB ↔ lC + mD
Brackets indicate concentrations at equilibrium (*pure solids and liquids do not affect equilibrium values!*)

6. B. Equilibrium Constant
Example
4NH3(aq) + 7O2(g) ↔ 4NO2(aq) + 6H2O(g)
Arrows indicate coefficients

7. B. Equilibrium Constant
Write the balanced equation and the equilibrium expression for the production of ammonia
N2 (g) + 3H2 (g)↔ 2NH3 (g)

8. C. Evaluating Equilibrium Constant
N2 (g) + 3H2 (g)↔ 2NH3 (g)
Keq > 1
Products favored at equilibrium
Keq < 1
Reactants favored at equilibrium
Keq = 1
Large amount of both products and reactants at equilibrium

9. Keq= Example [2.0]2 Keq= .10 [.625] [4.0]3
Solve for Keq in N2 (g) + 3H2 (g)↔ 2NH3 (g) if [N2] = .625, [H2] = 4.0 and [NH3] = 2.0
Plug into Keq expression
[2.0]2
Keq=
Keq= .10
[.625] [4.0]3
Are the reactants or products favored in this reaction?
Reactants

10. B. Equilibrium Constant
Write the balanced equation and the equilibrium expression for:
Zn (s) + 2 Ag+ (aq)↔ Zn2+ (aq) + 2 Ag (s)

11. Example
Solve for Keq in Zn (s) + 2 Ag+ (aq)↔ Zn2+ (aq) + 2 Ag (s)
if [Zn] = 109 M, [Ag+] = 2.0 and [Zn2+] = 24M
Plug into Keq expression
[24]
Keq=
Keq= 6.0
[2.0]2
Are the reactants or products favored in this reaction?
Products

12. Reaction Equilibrium
Concentrations in a problem, may not always be at equilibrium!
If you know initial molarities of reactants you can calculate the ending molarites of all substances if we know the Keq value.
Can also calculate the Keq if you know the initial molarities and one end molarity.
To help us with this we make a RICE chart!
R = Reaction
I = Initial
C = Change
E = Equilibrium

13. Calculating Equilibrium Constant
Hydrogen & Iodine react to form hydrogen iodide. There are initially 2.00 moles of both hydrogen and iodine present in a 1.0 L flask. After the reaction has reached equilibrium, there are 1.20 moles of hydrogen remaining. Calculate the Keq for the reaction.
1. First write out the reaction:
H2 + I2  2 HI
Reaction
[H2]
[I2]
[2 HI]
Initial
Change
Equilibrium

14. Calculating Equilibrium Constant
2. Then fill in the amounts given in molarity.
2 moles H2 / 1 L = 2.0 M H2
2 moles I2 / 1 L = 2.0 M I2
1.2 moles H2 / 1 L = 1.2 M H2
For now… the initial amount of products is always zero
Reaction
[H2]
[I2]
[2 HI]
Initial
2.00
Change
Equilibrium
1.20

15. Calculating Equilibrium constant
3. Figure out the change (subtraction)
You are given 1 ending amount,1.20 moles of H2
Change: Initial – Equilibrium = .80 moles of H2
4. Use the coefficients of the equation to figure out the rest of the ‘change’ row. (mole ratio)
Reaction
[H2]
[I2]
[2 HI]
Initial
2.00
Change
0.80
1.60
Equilibrium
1.20

16. Calculating Equilibrium constant
5. Find the Equilibrium concentration by subtracting on the reactants side
But adding on the products side
6. Substitute values into the Keq expression
Reaction
[H2]
[I2]
[2 HI]
Initial
2.00
Change
0.80
1.60
Equilibrium
1.20

17. Example 2: Reaction Initial Change Equilibrium
When 2.50 moles of nitrogen and 3.40 moles of hydrogen are mixed in a .500 liter container, they react to form ammonia. If the concentration of nitrogen is found to be 3.50 M when the reaction reaches equilibrium, what is the value of the equilibrium constant?
1. First write out the reaction:
N2 + 3 H2  2 NH3
Reaction
[N2]
[3 H2]
[2 NH3]
Initial
Change
Equilibrium

18. Calculating Equilibrium Constant
2. Then fill in the amounts given in molarity.
2.5 moles N2 / L = 5.00 M N2
3.40 moles H2 / L = 6.80 M H2
Initial amount of products is zero
3. Calculate the Change
4. Use mole ratio for rest of the Change row
Reaction
[N2]
[3H2]
[2 NH3]
Initial
5.00
6.80
Change
1.50
4.50
3.00
Equilibrium
3.50

19. Calculating Equilibrium Constant
5. Find the Equilibrium concentration by subtracting on the reactants side
But adding on the products side
6. Substitue values in Keq expression
Reaction
[N2]
[3H2]
[2 NH3]
Initial
5.00
6.80
Change
1.50
4.50
3.00
Equilibrium
3.50
2.30

20. D. Reaction Quotient
When reactants and products are mixed, it is useful to know if the reaction is at equilibrium
Use Law of Mass Action with initial concentrations instead of equilibrium (0’s mean initial)
N2 (g) + 3H2 (g)↔ 2NH3 (g)

21. D. Reaction Quotient N2 (g) + 3H2 (g)↔ 2NH3 (g)
Q = K, at equilibrium; no shift
Q > K, initial concentrations of products too high; shift to left
Q < K, initial concentrations of reactants too high; shift to right

22. D. Reaction Quotient Problem
N2 (g) + 3H2 (g)↔ 2NH3 (g)
At 500oC, Keq for the reaction above is 6.0 x If [NH3]0 = .0010M; [N2]0 = 1.0 x 10-5M; [H2]0 = M, is it at equilibrium or in which direction will it shift?
Q > K, so shift to left

23. Solubility Equilibrium
Special constant for the dissociation of solids that dissolve in water.
Ionic solids dissociate into their cations & anions
NaCl(s) Na+1 (aq) + Cl-1 (aq)
Ions formed carry an electrical current (electrolytes)
Solubility
The ratio of the maximum amount of solute to the volume of solvent in which this solute can dissolve at a specific Temperature
H2O

24. Solubility
All ionic solids are soluble up to a point… so they may not be completely soluble
Soluble: if concentration of solution is at least 0.1 room temperature
Slightly Soluble: if concentration of solution is between M and 0.1 room temp.
Insoluble: if concentration of solution is less than room temperature

25. Solubility Product Constant (Ksp)
This is the product of the equilibrium concentrations of ions in a saturated solution of a salt
*There is no denominator in solubility product equilibrium constant.
Ex: AgCl (s) Ag+ (aq) + Cl- (aq)
Ksp = [Ag+][Cl-]
H2O

26. Solubility Product Equilibrium
Example: The solubility of AgCl in pure water is 1.3 x10-5 M . Calculate the Ksp using the expression above.
The mole ratio of AgCl to both Ag+ and Cl-1 is 1:1, the solubility of each ions is equal to the solubility of AgCl.
Ksp = (1.3x10-5 ) (1.3x10-5 )
= 1.69 x10-10 ( )
Is AgCl considered soluble, insoluble or slightly soluble?
insoluble

27. Example
Calculate the solubility of CaF2 in grams/L if Ksp = 4.0 x
1. Write Balanced Reaction:
CaF2 (s) Ca2+ (aq) + 2 F-1 (aq)
2. Write the solubility expression:
Ksp = [Ca2+][F-1]2
4.0x10-8 = [Ca2+][F-1]2
There are 2 unknowns so we have to write in terms using the mole ratios. For every one mole of Ca2+ there are 2 moles of F-1
Assign a variable x for solubility of Ca2+
H2O

28. Example [Ca2+] = x and [F-] = 2x
Substitute the values into the equilibrium expression & solve for x
4.0x10-8 = [x] [2x]2
4.0x10-8 = [x] [4x2]
4.0x10-8 = 4x3
1.0x10-8 = x3
√ 1.0x10-8 = √ x3
X = 2.2 x M 4 4 3 3

29. Example
X = 2.2 x M
We have assigned x as the solubility of the Ca2+ ion which is equal to the solubility of the salt, CaF2.
Now convert from mol/L to g/L
78.1 g CaF2
1 mole CaF2
2.2x10-3 mol CaF L
= g/L of CaF2

30. E. Factors Affecting Equilibrium
Lechâtelier’s Principle
If a stress is applied to a system in dynamic equilibrium, the system will change in a way to relieve the stress
Reaction shifts towards products or reactants in order to relieve stress
Stresses that upset equilibrium include changes in:
Concentration
Temperature
Pressure

31. E. Factors Affecting Equilibrium
Concentration
 in reactants causes forward rate to increase, then increase in reverse rate and equilibrium is re-established, but is shifted in direction of products
H2CO3 (aq) CO2 (aq) + H2O (l)
<1% add H2CO >99%
 in products has reverse effect
 in concentration shifts equilibrium toward that substance

32. E. Factors Affecting Equilibrium
Temperature
 shifts away from side that contains the heat
Think of heat as a reactant or product, same as before with addition or removal
Remove heat (cool)
2SO2 (g) + O2 (g) SO3 (l) + heat
‘ Add heat

33. Temperature Exothermic Reactions: give off heat
Heat in kJ or Joules is a product
Endothermic Reactions: take in heat
Heat in kJ or Joules is a reactant
Equilibrium is temperature dependent, Keq can only be affect overall by this change T!

34. E. Factors Affecting Equilibrium
Pressure
Only affects gaseous solution with unequal number of moles of reactants and products
Increase shifts towards side w/ least amount of gas
Decrease shifts towards side w/ most amount of gas
Decrease pressure
N2 (g) + 3H2 (g) NH3 (g)
Increase pressure

35. Example N2 (g) + 3 H2 (g)  2NH3 (g) + 92 kJ left right
What direction is the shift if you reduce the pressure?
How would cooling the system shift the reaction?
How would adding N2(g) shift the system?
What direction is the shift if you remove NH3 ?
left
right

36. Reaction Energy and Reaction Kinetics
I. Rates of Reaction
Reaction Energy and Reaction Kinetics

37. Collision Theory
Reaction rate depends on the collisions between reacting particles.
The particles collide and make new substances.
Successful collisions occur if the particles...
collide with each other
have the correct orientation
have enough kinetic energy to break bonds

38. Unsuccessful Collisions
Collision Theory
Particle Orientation
Required Orientation
Unsuccessful Collisions
Successful Collision

39. Activation Energy Activation Energy (Ea)
minimum energy required for a reaction to occur
Activated Complex: the transitional structure in a collision that exists while old bonds are breaking and new bonds are being formed.
Activation
Energy
A. Collision Theory
Activation Energy (Ea)
minimum energy required for a reaction to occur

40. Activation Energy Cont…
- depends on reactants
- is always positive
- low Ea = fast reaction rate
- takes less energy for the
reaction to start.
Reaction Rate: the change in concentration of reactants per unit time as a reaction proceeds. Ea

41. Factors Affecting Rxn Rate
1. Nature of Reactants
- substances vary greatly in their tendencies to react.
- bonds are broken and other bonds are formed in reactions.
- the rate of reaction depends on the particular reactants and the bonds involved.
2. Surface Area
high SA = fast rxn rate
more opportunities for collisions
Increase surface area by…
using smaller particles – if we make the pieces of the reactants smaller, we increase the number of particles on the surface which can react.
dissolving in water – gases & dissolved particles can mix & collide freely. Reactions happen rapidly.

42. Factors Affecting Rxn Rate Cont.
3. Concentration
- high conc = fast rxn rate
- more opportunities for collisions because there are more particles in the same volume that can react.
There are less red particles in the same volume so there is less chance of a collision
There are more red particles in the same volume so there is more chance of a collision so the reaction goes faster

43. Factors Affecting Rxn Rate Cont.
4. Temperature
- high temp. = fast rxn. rate
- high KE
- when we increase the temperature, we give the
particles energy
- this makes the particles move faster
- so there are more opportunities for collision
- it is easier to reach activation energy

44. Factors Affecting Rxn Rate Cont.
5. Catalyst
substance that increases rxn rate without being consumed in the rxn
lowers the activation energy

45. Ch. 17 – Reaction Energy and Reaction Kinetics
II. Energy Diagrams
Ch. 17 – Reaction Energy and Reaction Kinetics

46. Terms and Symbols
Reactants: the chemicals you start with in a reaction.
Products: the chemicals formed during the reaction.
Heat of reaction (DE or DHrxn): the difference in energy between reagents and products (units are Joules).
Transition State: the highest point on the energy diagram, representing the point at which the reaction is half-completed.
Activation Energy (Ea): the amount of energy required for the reaction to take place. The higher the Ea, the slower the reaction (units are Joules).

47. Energy Diagrams
Show relationship between time and energy during the course of a chemical reaction.
Activated Complex (transition state) Ea
Ea’
Energy (E) in kJ/mol
Reactants
- - DE
Products
Forward Rxn
(exothermic)
Reverse Rxn
(endothermic)
Course of Reaction (time)

48. Endothermic Reaction A reaction in which heat is absorbed
Products have higher potential energy than the reactants.
The pink curve shows the uncatalyzed reaction. The blue curve shows what happens when a catalyst is present.
The energies and amounts of the products and reactants stays the same, and the DE stays the same. The catalyst just allows the reaction to reach equilibrium faster.
Energy
Course of Reaction

49. Exothermic Reactions Reactions in which heat is released.
Products have lower potential energy than the reactants.
The blue curve shows the uncatalyzed reaction. The red curve shows what happens when a catalyst is present.
Again, nothing changes but the amount of time it takes for the reaction to reach equilibrium.
Exothermic rxns are referred to as “spontaneous” because they can proceed to products without outside intervention.
Energy
Course of Reaction

50. Formulas * DEforward = Eproducts – Ereactants
* DEreverse = Ereactants – E products
* Ea = energy of activated complex – energy of reactants
* Ea’ = energy of activated complex – energy of products

51. Practice #1
For the energy diagram provided, label the reactants, products, DE, Ea, and Ea’. Also, determine the values of DE for the forward and reverse reactions, and the values of Ea and Ea’. 80 60 40 20
-20
Ea’
Energy (kJ/mol)
products Ea DE
reactants
Forward Reverse

52. Practice #1 Continued
* DEforward = Eproducts – Ereactants = 55 kJ/mol – (-20 kJ/mol)
= 75 kJ/mol
* DEreverse = Ereactants – Eproducts = -20 kJ/mol – 55 kJ/mol
= -75 kJ/mol 80 60 40 20
-20
Ea’
Energy (kJ/mol)
products Ea DE
reactants
Forward Reverse

53. Practice #1 Continued
* Ea = energy of activated complex – energy of reactants
= 80 kJ/mol – (-20 kJ/mol) = 100 kJ/mol
* Ea’ = energy of activated complex – energy of products
= 80 kJ/mol – 55 kJ/mol = 25 kJ/mol 80 60 40 20
-20
Ea’
Energy (kJ/mol)
products Ea DE
reactants
Forward Reverse

54. Practice #2
Draw and label an energy diagram that depicts the following reaction. Place the reactants at energy level of 0 kJ/mol. DEforward = -10 kJ/mol and Ea’ = 40 kj/mol
DEforward = Eproducts – Ereactants
-10 kJ/mol = Eproducts – 0 kJ/mol = -10 kJ/mol
Ea’ = activated complex – Eproducts
40 kJ/mol = a.c – (-10 kJ/mol) = 20 kJ/mol 40 30 20 10
-10 40 30 20 10
-10 40 30 20 10
-10 40 30 20 10
-10 40 30 20 10
-10
Energy (kJ/mol)
Energy (kJ/mol) Ea
Ea’
reactants DE
products
Course of Reaction