1. Kinetics

2. Chemical kinetics
The study of reaction mechanisms and rates (how reactions occur, and how long they take)
Mechanism - the "pathway" the reaction takes - a series of steps that leads from product to reactant.
Particles of reactant must collide in order to react
These collisions must occur with enough energy (activation energy) to react
The molecules must be oriented properly to react.
Collisions that satisfy both requirements that lead to the commencement of the reaction are called effective collisions

3. N C N C Collision Theory CO + NO2  CO2 + NO WRONG ORIENTATION
CORRECT ORIENTATION

4. A + B  C + D Reaction Mechanism STEP 3 C C N N C C N STEP 2 AB STEP 1
The mechanism can occur in a series of steps, each involving electron shifts as old bonds are broken and new bonds are formed.
STEP 3 C C O N N C O
A + B  C + D
Energy O O
CO + NO2
REACTANTS C O N
STEP 2 AB O
STEP 1
A + B O
PRODUCTS
Reaction Coordinate

5. Rate-Determining Step
Rate-determining step: the slowest step dictates the overall speed of the reaction.
If there are eleven steps, ten of which are very fast, the other step, the slow one, determines the rate of the reaction.

6. Click Below for the Video Lectures
Rate Limiting Step Video Lecture
Reaction Intermediates

7. Potential Energy Diagram

8. Potential Energy Diagram
Energy Changes in Chemical Reactions is shown in Potential Energy Diagrams
CHEMICAL EQUATIONS
A + B ---> C + D
Reactant Products
In chemical reactions, reactants react to form products with an associated release or absorption of energy.

9. DH = H (products) – H (reactants)
Enthalpy (H) is used to quantify the heat flow into or out of a system in a process that occurs at constant pressure.
DH = H (products) – H (reactants)
DH = heat given off or absorbed during a reaction at constant pressure
Hproducts < Hreactants
Hproducts > Hreactants
DH < 0
DH > 0

10. Potential Energy Diagram
Potential Energy in KJ C O N N C O
Energy O O
CO + NO2
REACTANTS C O N O O
No unit, fancy of saying “start to finish”
PRODUCTS
Reaction Coordinate

11. Potential Energy Diagram
Heat of Activated Complex C O N N C O
Energy O O
CO + NO2
REACTANTS
Heat of Reactants C O N O O
PRODUCTS
Heat of Products
Reaction Coordinate

12. Potential Energy DiagramC O N
Activation Energy
Amount of energy needed to start the reaction N C O
Energy O O
CO + NO2
REACTANTS C O N O O
ΔH: ENTHALPY
Net energy of the reaction
PRODUCTS
Reaction Coordinate

13. Click Below for the Video Lectures
Activation Energy Reaction Coordinate Catalyst

14. Exothermic and Endothermic

15. Exothermic Reaction A + B  C + D + ENERGY
Heat is released as a product.
This heat is released into the environment surrounding the reaction, causing the temperature of the surroundings to increase.
The products have less energy than the reactants after release.
Since the products have less energy stored in their chemical bonds, they are more stable than the reactants were. Burning paper is exothermic. The ash formed by the burning is not flammable and will not burn.

16. General Exothermic Reaction
A B  C
+ 70 kJ
ΔHA = 60 kJ
ΔHB = 40 kJ
ΔHc = 30 kJ
Energy Released as product Exothermic
100 KJ
 Law of conservation of energy states that the total energy of an isolated system cannot change.

17. C (g) + O2 (g)  CO2 (g) ΔH = -393.5 kJ
Exothermic Reaction
C (g) + O2 (g)  CO2 (g) ΔH = kJ
The minus sign in front of the enthalpy is kJ indicates that this reaction is exothermic, that energy was released.
This energy can be placed on the products side, as follows:
C (g) + O2 (g)  CO2 (g) kJ

18. If ΔHsynthesis = - 393.5 kJ/mole,
C (g) + O2 (g)  CO2 (g) ΔH = kJ
If we were to synthesize 2.3 moles of CO2, how many kJ would be released?
Decomposition
1 mol of CO2
2.3 mol CO2
393.5 kJ
= kJ
1 mol CO2
If ΔHsynthesis = kJ/mole,
then Δ Hdecomposition = kJ/mole

19. Potential Energy Diagram - Exothermic
A + B  C + D + 40 kJ
150
Activation Energy = 150 – 70
80 KJ
Energy
ΔH = 70 – 30
- 40 kJ, released
A + B 70
REACTANTS
C + D 30
PRODUCTS
Reaction Coordinate

20. Endothermic Reaction A + B + ENERGY  C + D
Heat is absorbed by the reactants.
This heat is absorbed from the environment surrounding the reaction, and the temperature of the surroundings decreases.
The products have more energy than the reactants after absorption.
This energy is stored in the chemical bonds of the products

21. General Endothermic Reaction
A B  C
+ 50 kJ
ΔHA = 40 kJ
ΔHB = 20 kJ
ΔHc = 110 kJ
Energy absorbed as product Endothermic
60 KJ
 Law of conservation of energy states that the total energy of an isolated system cannot change.

22. Potential Energy Diagram - Endothermic
A B  C + D
+ 40 kJ
150
Energy
C + D 70
PRODUCTS
ΔH = 70 – 30
+ 40 kJ, Absorbed
Activation Energy = 150 – 30
80 KJ
A + B 30
REACTANTS
Reaction Coordinate

23. Thermochemical Equations
The stoichiometric coefficients always refer to the number of moles of a substance
H2O (s) H2O (l)
DH = 6.01 kJ
If you reverse a reaction, the sign of DH changes
H2O (l) H2O (s)
DH = kJ
If you multiply both sides of the equation by a factor n, then DH must change by the same factor n.
2H2O (s) H2O (l)
DH = 2 x 6.01 = 12.0 kJ

24. Thermochemical Equations
The physical states of all reactants and products must be specified in thermochemical equations.
H2O (s) H2O (l)
DH = 6.01 kJ
H2O (l) H2O (g)
DH = 44.0 kJ
How much heat is evolved when 266 g of white phosphorus (P4) burn in air?
P4 (s) + 5O2 (g) P4O10 (s) DH = kJ
1 mol P4
123.9 g P4 x
3013 kJ
1 mol P4 x
266 g P4
= 6470 kJ

25. Click Below for the Video Lectures
Exothermic and Endothermic

26. Enthalpy

27. Standard Enthalpy of Formation
Standard enthalpy of formation (DHf⁰ ) is the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm.
The standard enthalpy of formation of any element in its most stable form is zero.
DH⁰f (C, graphite) = 0
DH⁰f(O2) = 0
DH⁰f (O3) = 142 kJ/mol
DH⁰f (C, diamond) = 1.90 kJ/mol

28. 6.5

29. The standard enthalpy of reaction (DHrxn⁰ ) is the enthalpy of a reaction carried out at 1 atm.
aA + bB cC + dD
DH0
rxn
dDH0 (D) f
cDH0 (C) = [ + ] -
bDH0 (B)
aDH0 (A)
DH0
rxn
nDH0 (products) f = S
mDH0 (reactants) -
Hess’s Law: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps.
(Enthalpy is a state function. It doesn’t matter how you get there, only where you start and end.)

30. REVERSE 2S(rhombic) + 2O2 (g)  2SO2 (g) DH0rxn = -296.1x2 kJ
Calculate the standard enthalpy of formation of CS2 (l) given that:
C(graphite) + O2 (g)  CO2 (g) DH0rxn = kJ
S(rhombic) + O2 (g)  SO2 (g) DH0rxn = kJ
CS2(l) + 3O2 (g)  CO2 (g) + 2SO2 (g) DH0rxn = kJ
REVERSE
1. Write the enthalpy of formation reaction for CS2
C(graphite) + 2S(rhombic) CS2 (l)
2. Add the given rxns so that the result is the desired rxn.
C(graphite) + O2 (g)  CO2 (g) DH0rxn = kJ
2S(rhombic) + 2O2 (g)  2SO2 (g) DH0rxn = x2 kJ +
CO2(g) + 2SO2 (g)  CS2 (l) + 3O2 (g) DH0rxn = kJ
C(graphite) + 2S(rhombic)  CS2 (l)
DH0rxn = (2x-296.1) = 86.3 kJ

31. 2C6H6 (l) + 15O2 (g)  12CO2 (g) + 6H2O (l)
Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. How much heat is released per mole of benzene combusted? The standard enthalpy of formation of benzene is kJ/mol.
2C6H6 (l) + 15O2 (g)  CO2 (g) + 6H2O (l)
DH0rxn
nDH0f (products) = S
mDH0f (reactants) -
= [12DH⁰ f (CO2) + 6 DH⁰ f (H2O)] – [ 2DH ⁰f (C6H6) + 15DH ⁰f (O2)]
= [12(-393.5kJ/mol)) + 6(-187.6kJ/mol)] – [ 2(49.04kJ/mol) + 15(0kJ/mol)]
-5946 kJ
2 mol
= kJ/mol C6H6
6.5

32. Click Below for the Video Lectures
Enthalpy of Reaction Reaction Intermediates

33. Spontaneous Reaction

34. Spontaneous Reaction 2. Entropy (S)
SPONTANEOUS CHEMICAL CHANGES - reactions that proceed of their own accord once initiated.
In order for a chemical reaction to be spontaneous, there has to be a balance between two factors:
1. Enthalpy (H)
2. Entropy (S)

35. Enthalpy (H)
ENTHALPY (H) - the heat content of the system. Nature favors reactions that undergo a decrease in enthalpy.
When you let go of a ball that you hold in your hand, it falls. This is a spontaneous decrease in enthalpy.
Nature favors reactions that have a net decrease in PE, so EXOTHERMIC reactions are favored.
Most exothermic reactions are spontaneous.
Most endothermic reactions are nonspontaneous, and they require a constant input of energy to keep them going.

36. Spontaneous Physical and Chemical Processes
A waterfall runs downhill
A lump of sugar dissolves in a cup of coffee
At 1 atm, water freezes below 0 0C and ice melts above 0 0C
Heat flows from a hotter object to a colder object
A gas expands in an evacuated bulb
Iron exposed to oxygen and water forms rust
spontaneous
nonspontaneous
18.2

37. spontaneous
nonspontaneous
18.2

38. YES! SPONTANEOUS NO, NONSPONTANEOUS
Does a decrease in enthalpy mean a reaction proceeds spontaneously?
YES! SPONTANEOUS
CH4 (g) + 2O2 (g) CO2 (g) + 2H2O (l) DH0 = kJ
H+ (aq) + OH- (aq) H2O (l) DH0 = kJ
H2O (s) H2O (l) DH0 = 6.01 kJ
NO, NONSPONTANEOUS
NH4NO3 (s) NH4+(aq) + NO3- (aq) DH0 = 25 kJ
H2O

39. Entropy (S)
2) ENTROPY (S) - the randomness (disorder) of the system. Nature favors reactions that undergo an INCREASE in entropy.
Substances undergo an increase in entropy as the temperature increases, as evidenced by the organization of particles in the various phases of matter at the different temperatures.
Since entropy increases as temperature increases, the gas phase has the most entropy and the solid phase has the least entropy.
Gas molecules move randomly, and solid molecules are locked into regular-shaped crystal lattices.
Nature favors reactions that INCREASE IN PHASE.

40. Processes that lead to an increase in entropy (DS > 0)
18.2

41. Spontaneous Reaction
Spontaneous Reactions: in order for a reaction to be spontaneous, nature has to favor the factors.
If both factors are unfavored, the reaction will be nonspontaneous.

42. 1. If BOTH factors are FAVORED, the reaction is SPONTANEOUS at ALL temperatures.
2C (s) + 3H2(g)  C2H6 (g) ΔH=-84.0kJ
ENTHALPY: ΔH=-84.0kJ, exothermic = FAVORED!
ENTROPY: s+g  g, increase in S= FAVORED!
BOTH FAVORED, SPONTANEOUS AT ALL TEMPERATURE!

43. BOTH UNFAVORED, NONSPONTANEOUS AT ALL TEMPERATURE
2) If BOTH factors are UNFAVORED, the reaction is NONSPONTANEOUS at ALL temperatures
N2(g) + 2O2 (g)  2NO2 (g) ΔH=66.4 kJ
ENTHALPY: ΔH=66.4 kJ, endothermic= UNFAVORED
ENTROPY: g+gg, no increase in S = UNFAVORED
BOTH UNFAVORED, NONSPONTANEOUS AT ALL TEMPERATURE

44. SPONTANEOUS at LOW TEMPERATURES
3. If ENTHALPY is FAVORED, but ENTROPY is UNFAVORED, then the reaction will only be spontaneous at LOW temperatures
2CO(g) + O2(g)  2CO2(g) ΔH= kJ
ENTHALPY: ΔH=-566.0kJ, exothermic = FAVORED
ENTROPY: g+gg, no increase in S = UNFAVORED
SPONTANEOUS at LOW TEMPERATURES

45. 4) If ENTHALPY is UNFAVORED, but ENTROPY is FAVORED, then the reaction will only be spontaneous at HIGH temperatures
NH4NO3(s)  NH4+(aq) + NO3-(aq) ΔH=+25.69kJ
ENTHALPY: ΔH=+25.69kJ, Endothermic = UNFAVORED
ENTROPY: saq , increase in S = FAVORED
SPONTANEOUS at HIGH TEMPERATURE

46. Click Below for the Video Lectures
Spontaneous Reaction Entropy NonSpontaneous Process

47. Gibbs Free Energy

48. Gibbs Free Energy Equation
The two thermodynamic quantities of enthalpy change (H) and entropy change (S) are related by the Gibbs Free Energy Equation
ΔG = ΔH - TΔ S

49. -ΔG = Spontaneous +ΔG = Nonspontaneous
ΔG = ΔH - TΔ S
-ΔG = Spontaneous +ΔG = Nonspontaneous

50. Click Below for the Video Lectures
Gibb’s Free Energy Free Energy and the Equilibrium Cosntant

51. Reaction Rate

52. Factors that affect the rate of reaction by changing the MECHANISM
Catalyst - speeds up a reaction by changing one or more steps in the mechanism to shorten the mechanism.
Generally lowers the activation energy, allowing the reaction to proceed at a faster rate.
Catalysts are not consumed by the reaction.
This is like finding a shortcut when driving!

53. 2) Inhibitor - slows down a reaction by adding additional steps to the mechanism - acts as a "roadblock" or detour.
Generally increases the activation energy.
After all, if you have to take a detour, it uses more gas to get to you destination!
Inhibitors are not consumed by the reaction.

54. Factors that affect the rate of reaction by changing the number of COLLISION
1) Nature of Reactants
Depending on what the reactants are, it will take different amounts of energy to elicit an effective collision.
2) Temperature
This directly affects the speed of the colliding particles.
The higher the temperature, the faster the particles are moving, and therefore they collide with more energy.
This gives more effective collisions in the same unit time, increasing the rate of reaction.

55. 3) Concentration
This affects the orientation of the colliding particles.
The more particles there are in a given volume, the greater the chance for collision, the greater the chance for proper orientation, the greater the chance for effective collisions, and the higher the rate.
An increase in pressure for gases is the same as an increase in concentration, since the gas molecules are being pushed closer together, resulting in more collisions and a faster rate.

56. 4) Surface area
The more surface that is exposed, the more surface that can undergo effective collisions.
Crushing a salt cube to dissolve it exposes fresh surface to react.
Smaller particles of the same mass of a large cube of substance have a higher surface area.
Affects only solid reactants, since liquids and gases have maximum surface area (no chunks to break up).

57. Click Below for the Video Lectures
Rate of Reaction

58. Rate Law

59. Reaction Rate
Reaction rate: A positive quantity that expresses how the concentration (Molarity) of a reactant or product changes with time.

60. Rate Law Rate of reaction = k [A]x[B]y
Rate Law: Determined experimentally (There is no way to calculate this theoretically).
A + B --> C + D
Rate of reaction = k [A]x[B]y
k = rate constant
x = order of reaction with respect to A
y = order of reaction with respect to B
x + y = overall order of reaction

61. The initial rate of decomposition of acetaldehyde, CH3CHO,
The initial rate of decomposition of acetaldehyde, CH3CHO, CH3CHO --> CH4 + CO at 600C
was measured at a series of concentrations with the following results:
(a) Determine the reaction order.
(b) Write the rate law.
Experiments 1 2 3 4
[CH3CHO] (M)
0.10
0.20
0.30
0.40
rate (mol/L s)
0.085
0.34
0.76
1.4
(The power of the [ ] of a reactant)
Choose any two experiments
4 = 2x
x = 2
Rate of reaction = k [A]x[B]y
Rate = k [CH3CHO]2

62. Determine the rate constant, k.
(d) Calculate the rate of the above reaction if the concentration of CH3CHO is 0.50 M
Units of k:
Choose data from any experiment
Rate = k [CH3CHO]2
1.4 = k [0.40]2 =
k = 8.75 = 8.8 L/mol • s
Units of k depend on the rate law.
Rate = k [CH3CHO]2
Units of rate:
Rate = 8.8 (0.50)2 =
Rate = 2.2 mol/L • s

63. 5. Consider the following reaction: A + B  C + D
A series of reactions were carried out with the following results:
a. Find the order of the reaction with respect to A and B.
Experiment 1 2 3
[A] (M)
0.100
0.400
[B] (M)
0.200
0.600
Rate (mol/L s)
1.1 x 10-6
9.9 x 10-6    
Find data where the other reactant remains constant.
A  Use expt. 2 & 3
1 = 4x
x = 0
B  Use expt. 1 & 2
9 = 3y
y = 2

64. Write the rate law of the reaction.
Determine the rate constant, k.
d. Determine the rate of the reaction if [A] is M and [B] is M.
Rate of reaction = k [A]x[B]y
Rate = k [A]0 [B]2
Rate = k [B]2
Choose data from any experiment
Rate = k [B]2
Units of k:
1.1 x 10-6 = k (0.200)2 =
k = 2.8 x 10-5 L/mol•s
Rate = k [B]2
Rate = 2.8 x 10-5 (0.800)2
Rate = 1.8 x 10-5 mol/L•s

65. Click Below for the Video Lectures
Rate Law

66. Equilibrium

67. Equilibrium A + B ↔ C + D + energy A + B  C + D + energy forward
A state of rate balance between two opposing changes. The rate of the forward change is equal to the rate of the reverse change.
Most reactions are reversible:
A + B  C + D + energy forward
C + D + energy  A + B reverse
A + B ↔ C + D + energy
when the rate of the forward reaction equals the rate of the reverse reaction, a state of equilibrium is reached.

68. Properties of Equilibrium
Equilibrium is dynamic (in motion), with molecules of reactant reacting to form molecules of product at the same rate at which molecules of product react to form molecules of reactant.
The system is in constant motion, but the equal rates give the impression that the system is at a standstill.
2) Equilibrium can only occur in a closed system, that is, the system is set apart and sealed from the environment, so that nothing gets in or out.

69. Properties of Equilibrium
3) As long as the system is closed, a system at equilibrium will remain that way forever.
Changing any conditions of the equilibrium will change the equilibrium.
4) Equilibrium occurs at different concentrations of product and reactant.
Depending on the nature of the species involved, assuming that one starts with the forward reaction, the rate of the reverse reaction will increase as product is formed by the forward reaction.
When the rates are equal, equilibrium occurs. This may occur at different concentrations of product and reactant.

70. Click Below for the Video Lectures
Equilibrium Reaction Quotient Reversible Reactions

71. Systems at Equilibrium
Equilibrium Constants (Keq) is the ratio of product concentration to reactant concentration
aA + bB  cC + dD
Keq = [C]c [D]d Keq = [product]
[A]a [B]b [reactant]
Keq < 1, reactants favored at equal.
Keq > 1, products favored at equal.
Does not include pure solids and liquids

72. Equilibrium Expressions
Write the equilibrium expressions, Keq, for the following reaction:
2 NO2(g) ↔ 2 NO(g) + O2(g)
Keq =
[Product]
[Reactant]
Keq =
[NO]2 [O2]
[NO2]2

73. = 0.67 M Write a equilibrium expression for PCl5(g)  PCl3(g) + Cl2(g)
Keq =
[PCl3] [Cl2]
[PCl5]
For the above reaction, the equilibrium constant is 35. If the concentrations of PCl5 and PCl3 are M and 0.78 M respectively. What is the concentration of Cl2?
Keq [PCl5] =
[PCl3] [Cl2]
[Cl2] =
Keq [PCl5]
[PCl3] =
35 [0.015]
[0.78]
= M

74. Click Below for the Video Lectures
Equilibrium Constant

75. Changing Equilibrium

76. Dynamic Equilibrium
Equilibrium systems are dynamic, meaning they are constantly in motion.
Equilibrium must be in a closed system, or the system cannot remain at equilibrium.
But sometimes you don’t want your system to be at equilibrium!
How irritating it would be if you went to all the trouble to make a product, just to have it decompose back into the reactants again and reach equilibrium!

77. Le Chatelier's Principle
If a system at equilibrium is subjected to a stress, the equilibrium will shift in a way that relieves the stress, in the direction of whichever reaction was made faster by the stress.
This will cause a change in concentration of both the reactants and products until the equilibrium is re-established
Reverse
Forward
Reactant ↔ Product
Reactant
Product
Product

78. LeChatelier Example #1 Ice + Energy (ΔH)  Water
A closed container of ice and water at equilibrium. The temperature is raised.
Ice + Energy (ΔH)  Water
right
The equilibrium of the system shifts to the _______ to use up the added energy.
Forward Reaction is favored,
more water will be produced.

79. LeChatelier Example #2 N2O4 (g) + Energy (ΔH)  2 NO2 (g)
A closed container of N2O4 and NO2 at equilibrium. NO2 is added to the container.
N2O4 (g) + Energy (ΔH)  2 NO2 (g)
The equilibrium of the system shifts to the _______ to use up the added NO2.
left
Reverse Reaction is favored,
more N2O4 (g), Energy (ΔH) will be produced.

80. LeChatelier Example #3 water + Energy (ΔH)  vapor
A closed container of water and its vapor at equilibrium. Vapor is removed from the system.
water + Energy (ΔH)  vapor
The equilibrium of the system shifts to the _______ to replace the vapor.
right
Forward Reaction is favored,
more vapor will be produced.

81. LeChatelier Example #4 N2O4 (g) + Energy (ΔH)  2 NO2 (g)
A closed container of N2O4 and NO2 at equilibrium. The pressure is increased.
N2O4 (g) + Energy (ΔH)  2 NO2 (g)
1 mol
2 mol
left
The equilibrium of the system shifts to the _______ to lower the pressure, because there are fewer moles of gas on that side of the equation.
Reverse Reaction is favored,
more N2O4 (g), Energy (ΔH) will be produced.

82. Click Below for the Video Lectures
Equilibrium Disturbance Le Chatelier’s Principle