1. University of Louisiana at Lafayette
Chapter 6
Lecture
Outline
Prepared by
Andrea D. Leonard
University of Louisiana at Lafayette
Modified by
Peggy Sleevi
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Energy Energy -- the ability to do work.
Law of conservation of energy:
The total energy in a system does not change. Energy can be transformed from one form to another, but cannot be created or destroyed.

3. Types of Energy Kinetic – Energy of motion Potential – Stored energy
Chemical – Energy stored in chemical bonds
Thermal – Kinetic energy of molecules/atoms – Transferred as heat

4. Types of Energy Electrical – Energy due to the flow of electric charge
Nuclear – Energy released during nuclear fission or nuclear fusion
Solar – Radiant energy from the sun

5. Chemical Energy Chemical bonds store potential energy.
How is this potential energy released?

6. Chemical Energy
A compound with lower potential energy is more stable than a compound with higher potential energy.
Reactions that form products having lower potential energy than the reactants are favored.

7. Energy Units
Calorie (cal) -- amount of energy needed to raise the temperature of 1 g of water by 1 oC.
Joule (J) is another unit of energy.
1 cal = J
1,000 J = 1 kJ
1,000 cal = 1 kcal
1 kcal = kJ

8. Example
Lecture Problem #1

9. Energy Changes in Reactions
Chemical Reaction:
bonds are broken in the reactants,
new bonds are formed in the products.
Bond breaking requires an input of energy.
Bond formation releases energy.

10. Energy Changes in Reactions
Enthalpy:
Heat content of a system at
constant pressure
ΔH = change in enthalpy
= heat of reaction

11. Energy Changes in Reactions
H = heat of reaction or enthalpy change
H > 0
Energy is absorbed
Reaction is endothermic
H < 0
Energy is released
Reaction is exothermic

12. Bond Dissociation Energy
H for breaking a covalent bond by
equally dividing e− ’s between the two
atoms.

13. Bond Dissociation Energy
Bond dissociation energies are positive values,
because bond breaking is endothermic (H > 0). H H
H H
H = +104 kcal/mol
Bond formation always has negative values,
because bond formation is exothermic (H < 0).
H H H H
H = −104 kcal/mol

14. Bond Dissociation Energy
To cleave this bond,
H = +58 kcal/mol. Cl Cl
To form this bond,
H = −58 kcal/mol.

15. Bond Dissociation Energy
The stronger the bond, the higher its bond
dissociation energy.
Bond dissociation energies generally decrease going down within a group.

16. Bond Dissociation Energies for Some Common Bonds
ΔH (kcal/mol)
H – H
+104
Br – Br
+38
F – F
+58
Cl – Cl
+46
I – I
+36
H – OH
+119
H – F
+136
H – Cl
+103
H – Br
+88
H – I
+71

17. Example
Lecture Problem #2-3

18. Energy Changes in Reactions Calculations Involving H Values
H indicates the relative strength of the bonds broken and formed in a reaction.
When H is negative:
Exothermic
More energy is released forming bonds than
needed to break the bonds.
The bonds formed in the products are stronger
than the bonds broken in the reactants.
CH4(g) O2(g)
CO2(g) H2O(l)
H = −213 kcal/mol
Heat is released.

19. Energy Changes in Reactions Calculations Involving H Values
When H is positive:
Endothermic
More energy is needed to break bonds than is
released in the formation of new bonds.
The bonds broken in the reactants are stronger
than the bonds formed in the products.
6 CO2(g) H2O(l)
C6H12O6(aq) O2(g)
ΔH = +678 kcal/mol
Heat is absorbed.

20. Energy Changes in Reactions Calculations Involving H Values

21. Example
Lecture Problem #4-5

22. Energy Diagrams
For a reaction to occur, two molecules must collide with enough kinetic energy to break bonds.
This bond breaks
This bond forms

23. Energy Diagrams
The orientation of the two molecules must be correct as well.

24. Energy Diagrams Ea, the energy of activation, is the difference in
energy between the reactants and the transition
state.

25. Energy Diagrams
The Ea is the minimum amount of energy that the reactants must possess for a reaction to occur.
Ea is called the energy barrier and the height of the barrier determines the reaction rate.

26. Energy Diagrams
When the Ea is high, few molecules have enough energy to cross the energy barrier, and the reaction is slow.
When the Ea is low, many molecules have enough energy to cross the energy barrier, and the reaction is fast.

27. Energy Diagrams The difference in energy between the reactants
and the products is the H.
If H is negative, the reaction is exothermic:

28. Energy Diagrams
If H is positive, the reaction is endothermic:

29. Example
Lecture Problem #6

30. Reaction Rates Increasing the concentration of the reactants:
increases the number of collisions
increases the reaction rate
Increasing the temperature:
increases the kinetic energy of the molecules
increases the reaction rate

31. Reaction Rates – Catalysts
A catalyst is a substance that speeds up the rate of a reaction.
A catalyst is recovered unchanged in a reaction; does not appear in the product.

32. Reaction Rates – Catalysts
In the following reaction, Pd acts as a catalyst:
Catalysts accelerate a reaction by lowering Ea
without affecting H.

33. Reaction Rates – Catalysts
The uncatalyzed reaction (higher Ea) is slower.
The catalyzed reaction (lower Ea) is faster.
H is the same for both reactions.

34. Example
Lecture Problem #7

35. Reaction Rates – Biological Catalysts
Enzymes (usually protein molecules) are biological catalysts held together in a very specific three-dimensional shape.
The active site binds a reactant, which then under- goes a very specific reaction with an enhanced rate.
The enzyme lactase converts the carbohydrate lactose into the two sugars glucose and galactose.
People who lack adequate amounts of lactase suffer from abdominal cramping and diarrhea because they cannot digest lactose when it is ingested.

36. Equilibrium A reversible reaction can occur in either direction,
from reactants to products or from products to
reactants.
The forward reaction
proceeds to the right.
CO(g) H2O(g)
CO2(g) H2(g)
The reverse reaction
proceeds to the left.
The system is at equilibrium when the rate of the
forward reaction equals the rate of the reverse reaction.
The net concentrations of reactants and products
do not change at equilibrium.

37. Equilibrium forward reaction CO(g) + H2O(g) CO2(g) + H2(g)
reverse reaction
The system is at equilibrium when the rate of the
forward reaction equals the rate of the reverse reaction.
The net concentrations of reactants and products
do not change at equilibrium.

38. Equilibrium Constant The relationship between the concentration of the
products and the concentration of the reactants
[moles per Liter] is the equilibrium constant, K.
For the reaction:
a A b B
c C d D
equilibrium
constant
[products]
[reactants]
[C]c [D]d = K = =
[A]a [B]b

39. Equilibrium Constant For the following balanced chemical equation:
N2(g) O2(g)
2 NO(g)
[NO]2
equilibrium
constant = K =
[N2] [O2]
The coefficient becomes the exponent.

40. Example
Lecture Problem #10

41. Magnitude of the Equilibrium Constant
When K is much greater than 1 (K > 1):
[products]
The numerator is
larger.
[reactants]
Equilibrium favors the products and lies to the right.
When K is much less than 1 (K < 1):
[products]
The denominator is
larger.
[reactants]
Equilibrium favors the reactants and lies to the left.

42. Magnitude of the Equilibrium Constant
When K is around 1 (0.01 < K < 100):
[products]
Both are similar in magnitude.
[reactants]
Both reactants and products are present.
For the reaction:
2 H2(g) O2(g)
2 H2O(g)
K = 2.9 x 1082
The product is favored because K > 1.
The equilibrium lies to the right.

43. Equilibrium
Equilibrium favors the products when they are lower in energy than the reactants. K > 1 and ΔH < 0

44. Equilibrium
The value of K does not impact the rate of the reaction

45. Examples
Lecture Problems #11-12

46. Calculating the Equilibrium Constant
HOW TO Calculate the Equilibrium Constant for
a Reaction
Calculate K for the reaction between the
general reactants A2 and B2. The
equilibrium concentrations are as follows:
Example
[A2] = 0.25 M
[B2] = 0.25 M
[AB] = 0.50 M
A B2
2 AB
Write the expression for the equilibrium
constant from the balanced equation.
Step [1]
[AB]2
K =
[A2][B2]

47. Calculating the Equilibrium Constant
HOW TO Calculate the Equilibrium Constant for
a Reaction
Substitute the given concentrations in
the equilibrium expression and calculate K.
Step [2]
[AB]2
[0.50]2
K = =
[A2][B2]
[0.25][0.25]
0.25 = =
4.0
0.0625
Since the concentration is always reported in
mol/L, these units are omitted during the
calculation.

48. Example
Lecture Problem #13

49. Le Châtelier’s Principle
If a chemical system at equilibrium is disturbed or
stressed, the system will react in a direction that
counteracts the disturbance or relieves the stress.
Some of the possible disturbances:
concentration changes
temperature changes
pressure changes

50. Le Châtelier’s Principle Concentration Changes
2 CO(g) O2(g)
2 CO2(g)
What happens if [CO(g)] is increased?
The concentration of O2(g) will decrease.
The concentration of CO2(g) will increase.

51. Le Châtelier’s Principle Concentration Changes
2 CO(g) O2(g)
2 CO2(g)
What happens if [CO2(g)] is increased?
The concentration of CO(g) will increase.
The concentration of O2(g) will increase.

52. Le Châtelier’s Principle Concentration Changes
What happens if a product is removed?
The concentration of ethanol will decrease.
The concentration of the other product (C2H4) will increase.

53. Le Châtelier’s Principle Temperature Changes
When the temperature is increased, the reaction
that absorbs heat is favored.
An endothermic reaction absorbs heat, so increasing the temperature favors the forward reaction.

54. Le Châtelier’s Principle Temperature Changes
An exothermic reaction releases heat, so increasing the temperature favors the reverse reaction.
Conversely, when the temperature is decreased,
the reaction that adds heat is favored.

55. Le Châtelier’s Principle Pressure Changes
When pressure increases, equilibrium shifts in
the direction that decreases the number of moles
in order to decrease pressure.

56. Le Châtelier’s Principle Pressure Changes
When pressure decreases, equilibrium shifts in the
direction that increases the number of moles in
order to increase pressure.

57. Le Châtelier’s Principle Summary

58. Le Chatlier Principle video in Sway

59. Example
Lecture Problem #14
Lecture Problems #8-9