1. Kinetics
Chemical kinetics is the study of the time dependence of chemical reactions.

2. Objectives
Understand rates of reaction and the conditions affecting rates
Derive the rate equation, rate constant and reaction order from experimental data
Use integrated rate laws
Understand the collision theory of reaction rates and the role of activation energy
Relate reaction mechanisms and rate laws

3. Kinetics Reaction rate
The change in concentration of a reactant or product per unit time

4. Rate of Reaction
Rate of reaction may be described based on either the increase in concentration of a product or the decrease in concentration of a reactant per unit time
Rates decrease with time
Minus sign is required because the concentration of A decreases with time, and the rate is always expressed as a positive quantity.
Instantaneous rate can be determined by finding the slope of a line tangent to a point representing a particular time.

5. Rate of Reaction Decomposition of N2O5 2N2O5  4 NO2 + O2
Rate can be expressed as

6. Rate of Reaction
Rate could also be expressed in terms of the formation of NO2 or the rate of formation of O2. Rates expressed in these terms will have positive signs.
Formation of NO2 is 2x rate of decomposition of N2O5
Formation of O2 is one half of the rate of decomposition of N2O5

7. Rate of Reaction Example
Assume the disappearance of N2O5 between 40 min and 55 min is

8. Rate of Reaction
Rate in terms of the appearance of NO2

9. Rate of Reaction
Rate in terms of the rate at which O2 is formed

10. Rate of Reaction This procedure gives the average rate
Graphing concentration vs. time does not give a straight line because the rate of the reaction changes during the course of the reaction.
The instantaneous rate is given at a single point by drawing a line tangent to the concentration-time curve at a particular time and obtaining the slope of this line.

11. Rate of Reaction

12. Rate of Reaction
For this reaction, if the rate is measured at a range of instantaneous concentration during the experiment, a Rate vs. Concentration graph can be plotted. The resulting graph is given below:

13. Rate of Reaction

14. Rate of Reaction
This graph shows that the rate of reaction is directly proportional to the concentration of [N2O5]:
Rate = constant x [N2O5]
Rate = k[N2O5]
This expression is called a rate equation, where k is called the rate constant

15. Rate of Reaction Summarizing the rate expressions
To equate rates of disappearance or appearance, divide [reagent]/t by the stoichiometric coefficient in the balanced chemical equation

16. Rate of Reaction PROBLEM
Give the relative rates for the disappearance of reactants and formation of products for the following reaction:
4 PH3(g)  P4(g) H2(g)

17. Rate of Reaction PROBLEM
Data collected on the concentration of dye as a function of time are given in the graph below. What is the average rate of change of the dye concentration over the first 2 minutes? What is the average rate of change during the fifth minute (from t = 4 to t = 5)? Estimate the instantaneous rate at 4 minutes.

18. Rate of Reaction

19. Rate Laws Reversibility of Reactions
For the reaction 2N2O5  4 NO2 + O2 the reverse reaction may also take place
The reverse reaction effects the rate of change in concentration
[N2O5] depends on the difference in the rates of the forward, kf, and the reverse, kb, reactions

20. Reaction Conditions and Rate
Molecular collisions required for reaction to take place. Atoms and molecules are mobile (mixture of gases or using solutions of reactants)
Under these conditions several factors affect the rate of reaction

21. Reaction Conditions and Rate
Concentration
Temperature
Catalyst
Surface area (if reactant is a solid)

22. Effect of Concentration
Rate Equation
Relationship between reactant concentration and reaction rate is expressed by the rate equation or rate law. For our example, N2O5
Rate of Reaction = k[N2O5]

23. Effect of Concentration
k = proportionality constant = rate constant at a given temperature
Rate equation indicates reaction rate is proportional to the concentration of the reactant
[N2O5] doubles as the rate doubles

24. Effect of Concentration
In general,
aA + bB  xX
Rate equation form
Rate = k[A]m[B]n
m and n are not necessarily the stoichiometric coeffcients (a and b). The exponents must be determined by experiment. m and n may be positive, negative, fractions or zero

25. Order of Reaction
Order of a reaction is with respect to a particular reactant is the exponent of its concentration in the rate expression
Total reaction order is the sum of the exponents

26. Order of Reaction Example
Decomposition of H2O2 in the presence of iodide ion (clock reaction)
Reaction rate = k[H2O2][I-]
Reaction is first order with respect to H2O2 and I-; second order overall.

27. Order of Reaction So, what does it mean?
Rate doubles if either H2O2 or I- is doubled and rate increases by a factor of 4 if both concentrations are doubled.

28. Order of Reaction Let’s consider the following rate law
Rate = k[NO]2[Cl2]
Reaction rate is second order in NO
Reaction rate is first order in Cl2
Overall third order

29. Order of Reaction Let’s look at some data Exp [NO] M [Cl2] M
Rate M∙s-1 1
0.250
1.43 x 10-6 2
0.500
5.72 x 10-6 3
2.86 x 10-6 4
11.4 x 10-6

30. Order of Reaction
Exp 1 & 2 – [Cl2] held constant; [NO] is 2x; reaction rate increases by factor of 4
Exp 1, 3, 4 – Compare 1 & 3; [NO] held constant, [Cl2] 2x; rate doubles compare 1& 4; [NO] and [Cl2] 2x; rate 8x original value

31. Rate Constant k
Reaction rates ([A]/t) have units of mol/L∙time when [ ] are given as moles/L
Rate constants must have units consistent with the units for the other terms in the rate equation

32. Rate Constant k 1st order reactions: units of k are time-1
2nd order reactions: units of k are L/mol∙time
zero order reactions: units of k are mol/L∙time

33. Determining a Rate Equation
“Method of Initial Rates”
Initial rate (instantaneous reaction rate at start of reaction (rate at t = 0)
Mix reactants then determine [products]/t or -[reactants]/t after 1% or 2% of limiting reactant has been used up.

34. Determining a Rate Equation
Example
CH3CO2CH3(aq) + OH-(aq)  CH3CO2-(aq) + CH3OH(aq)
Initial Concentrations
Initial Rxn Rate
Exp
[CH3CO2CH3 ] M
[OH-] M
M∙s-1 1
0.050
 no change
 x 2 2
0.10 4

35. Determining a Rate Equation
When initial [reactant] is 2x while other [reactant] held constant, initial rxn rate 2x
Reaction is directly proportional to the concentrations of both reactants
Reaction is first order in each reactant
Rate = k[CH3CO2CH3][OH-]

36. Determining a Rate Equation
If rate equation is known, the value of k can be found by substituting values for the rate and concentration into rate equation

37. Determining a Rate Equation
Using our example
Substitute data from one of the experiments into the rate equation

38. Integrated Rate Laws What are they?
Equations that describe concentration-time curves
Why use them?
Calculate a concentration at any given time
Find length of time needed for a given amount of reactant to react

39. First Order Reactions
If reaction ‘A  products’ is first order, the reaction rate is directly proportional to [A]1
A little calculus…

40. First Order Reactions Integrated rate equation
Where [A]0 and [A]t are concentrations of reactants at time t = 0 and at a later time, t
This ratio is the fraction of reactant that remains after a given time has elapsed.

41. First Order Reactions
Negative sign (-) means ratio is less than 1 because [A]t is always less than [A]0.
Logarithm of ratio is negative therefore other side of equation must be negative

42. First Order Reactions Equation is useful in three ways:
If [A]t/[A]0 is measured after some amount of time has elapsed then k can be calculated
2. If [A]0 and k are known than [A]t­ can be calculated (amount remaining after a given amount of time
3. If k is known, then time elapsed until a specific fraction ([A]t/[A]0) remains can be calculated

43. First Order Reactions Note
k for 1st order reactions is independent of concentration; k has units of time-1
Therefore any unit for [A]t and [A]0 can be chosen (M, mol, g, atoms, molecules, pressure)

44. First Order Reactions PROBLEM
When heated cyclopropane rearranges to propene in a first order process
Rate = k[cyclopropane] k = 5.4 x 10-2 h-1
If the initial concentration of cyclopropane is mol/L, how much time (in hours) must elapse for its concentration to drop to mol/L

45. First Order Reactions
SOLUTION

46. Second Order Reactions
If reaction ‘A  products’ is second order, the reaction rate equation is
A little calculus again…

47. Second Order Reactions
Integrated rate equation
In this case k is the second-order rate constant (L/mol ∙ time)

48. Second Order Reactions
PROBLEM
The gas-phase decomposition of HI
HI(g)  ½ H2(g) + ½ I2(g)
has the rate equation
Where k = 30. L/mol ∙ min at 443C. How much time does it take for the concentration of HI to drop from mol/L to mol/L at 443C?

49. Second Order Reactions
SOLUTION

50. Zero Order Reactions
If reaction ‘A  products’ is zero order, the reaction rate equation is
A little calculus again…

51. Zero Order Reactions Integrated rate equation
Where k has the units mol/L ∙ s

52. Graphical Methods for Determining Reaction Order and the Rate Constant
First Order
[A]t = - kt + [A]0
  
y mx b

53. Graphical Methods for Determining Reaction Order and the Rate Constant
Second Order
1/[A]t = + kt + 1/[A]0
  
y mx b

54. Half-Life and First Order Reactions
Half-life t1/2
Time required for the concentration of a reaction to decrease to one-half its initial value; reactant R remaining is ½
Used primarily when dealing with 1st order reactions
Indicates the rate at which a reactant is consumed – is reaction fast or slow?
Longer the half-life – slower the reaction

55. Half-Life and First Order Reactions
Where:
[R]0 = initial concentration
[R]t = concentration after the reaction is half complete

56. Half-Life and First Order Reactions
Let’s evaluate t1/2 for 1st order reaction
Substitute [R]t/[R]0 = ½ and t = t1/2 in integrated rate equation

57. Half-Life and First Order Reactions
Rearrange equation (ln 2 = 0.693)
Equations relates half-life to 1st order rate constant
t1/2 is independent of concentration

58. Half-Life, Zero Order & Second Order Reactions
Zero order reaction
Second order reaction

59. Half-Life
PROBLEM
Sucrose, C12H22O11, decomposes to fructose and glucose in acid solution with the rate law
Rate =k[sucrose]
k = h-1 at 25C
What amount of time is required for 87.5% of the initial concentration of sucrose to decompose?

60. Half-Life
SOLUTION
After 87.5% of sucrose has decomposed, 12.5% remains (fraction remaining = 0.125). This will require 2 half-lives to reach this point.
Time elapsed = 3 x 3.33 h = 9.99 h

61. NH2NO2(aq)  N2O(g) + H2O(l)
Half-Life
YOUR TURN
The following reaction is first order with respect to [NH2NO2]. The rate constant, k, is 9.3 x 10-5 s-1. What is the half-life of this reaction?
NH2NO2(aq)  N2O(g) + H2O(l)
The following reaction at 400K is second order with respect to [CF3] and the value of the rate constant, k, is 2.51 x 1010 M-1s-1. If the initial [CF3] = 2.0M, what is the half-life of the reaction?
2 CF3(g)  C2F6(g)

62. Collision Theory of Reaction Rates
Theory states that for any reaction to occur 3 conditions must be met
The reacting molecules must collide with one another
The reacting molecules must collide with sufficient energy to break bonds
The molecules must collide in an orientation that can lead to rearrangement of the atoms

63. Collision Theory of Reaction Rates
So,
molecules must collide with one another. The rate of their reaction is primarily related to the number of collisions, which is in turn related to their concentration
The number of collisions between the two reactant molecules is directly proportional to the concentration of each reactant, and the rate of the reaction shows a first order dependence on each reactant

64. Collision Theory of Reaction Rates

65. Temperature, Reaction Rate, & Activation Energy
How and why does temperature affect reaction rates?
In any sample of gas or liquid some molecules have very low energies, others have high energies, most have intermediate energies.
As temperature increases average energy of the molecules in a sample increases as does the fraction having higher energies

66. Activation Energy
Molecules require some minimum energy to react; ‘energy barrier’
Energy required to surmount the barrier is called the activation energy, Ea
Barrier low, energy requirement (Ea) is low, a high proportion of molecules in a sample have sufficient energy to react – reaction is fast
Barrier high, energy requirement (Ea) is high and only a few reactant molecules in a sample have sufficient energy to react – reaction is slow

67. Activation Energy Reaction Coordinate Diagram Consider HI  H2 + I2
Reaction can not occur without input of energy
Energy reaches maximum at transition state; sufficient energy has been concentrated in bonds, bonds can break, reaction can move forward.

68. Activation Energy

69. Effects of Temperature Increase
Raising the temperature always increases the reaction rate by increasing the fraction of molecules with enough energy to surmount the activation energy barrier

70. Effects of Molecular Orientation on Reaction Rate
The lower the probability of achieving alignment, the smaller the value of k, the slower the reaction

71. The Arrhenius Equation
Reaction rates depend on energy, frequency of collision between reactants, temperature, and correct geometry. Arrhenius equation summarizes these factors:
R = gas constant x 10-3 kJ/K∙mol
A = frequency factor L/mol∙s (related to number of collision and fraction of collisions w/correct geometry. A is specific to each reaction and temperature dependent
Fraction of molecules having the minimum energy required for reaction. Value is always <1

72. The Arrhenius Equation
Use to calculate:
Value of Ea
Rate constant for a given temperature if Ea and A are known

73. The Arrhenius Equation
A little rearrangement…
y mx b

74. The Arrhenius Equation
PROBLEM
Calculate Ea for the reaction
2N2O (a)  2N2 (g) + O2 (g)
Exp
T (K)
K (L/mols) 1
1125
11.59 2
1053
1.67 3
1001
0.380 4
838
0.0011

75. The Arrhenius Equation
Kinetics problems that deal with changing rates, or rate constants (k), and temperature changes require use of the Arrhenius equation

76. The Arrhenius Equation
PROBLEM
What is the activation energy of a reaction that has a rate constant of 2.50 x 102 kJ/mol at 325K and a rate constant of 5.00 x 102 kJ/mol at 375K?

77. Catalysis
Catalyst increases the rate of a reaction without being consumed by it.
Catalyst lowers the activation energy required thus speeding up the reaction

78. Heterogeneous & Homogeneous Catalysis
Heterogeneous – catalyst is in a different phase from the reaction mixture; generally a solid that increases the rate of a gas phase or liquid phase reaction Au
N2O(g)  N2(g) + ½ O2(g)
N2O is chemically absorbed on the surface of the solid catalyst.
A bond is formed between the N2O molecule and Au (covalent bonds) thus weakening the bond between nitrogen and oxygen making it easier for N2O to break apart.

79. Heterogeneous & Homogeneous Catalysis
N  N – O(g) + Au(s)  N  N --- O --- Au(s)  N  N(g) + O2(g) + Au(s)
Heterogeneous catalyst generally used in industrial processes:
Preparation of
ammonia
sulfuric acid
methanol
Pt catalyst used to reduce automobile emissions

80. Heterogeneous & Homogeneous Catalysis
Homogeneous – catalyst that is present in the same phase as the reactants.
Speeds up reaction by forming a reactive intermediate that decomposes to give products thus providing an alternative path of lower activation energy for the reaction

81. Heterogeneous & Homogeneous Catalysis
Clock reaction (decomposition of hydrogen peroxide)
Uncatalyzed
2H2O2 (aq)  2H2O + O2 (g)
Catalyzed
Step 1:H2O2 (aq) + I- (aq)  H2O + IO- (aq)
Step 2:H2O2 (aq) + IO- (aq)  H2O + O2 (g) + I- (aq)
2H2O2 (aq)  2H2O + O2(g)

82. Reaction Mechanisms
Sequence of bond-making and bond-breaking steps during a reaction
A ‘guess’ to help better understand the chemistry
Nano level (atoms and molecules)
Use rate equation to understand mechanisms

83. Reaction Mechanisms Most reactions occur in a sequence of steps
Br2(g) + 2 NO(g)  2 BrNO(g)
Written as a single step, this reaction would require the 3 reactants to collide simultaneously with the right velocity and orientation to react.
Low probability of this happening

84. Reaction Mechanisms
Using a sequence of steps involving only one or two molecules increases the probability of a collision (production of an intermediate)
Step 1 Br2(g) + NO(g)  Br2NO(g)
Step 2 Br2NO(g) + NO(g)  2 BrNO(g)
Br2(g) + 2 NO(g)  2 BrNO(g)

85. Reaction Mechanisms
Each step of mechanism is referred to as an elementary step
Describes a single molecular event
Each step has its own Ea and k
Steps must add up to give overall balance chemical equation

86. Molecularity
Classification of elementary steps based on # of reactant molecules colliding
Unimolecular – one molecule in an elementary step
O2(g)  O2(g) + O(g)
Bimolecular – two molecules (same or different) 
O2(g) + O(g)  2 O2(g)

87. Molecularity Termolecular – three molecules (same or different)
O(g) + O2(g) + N2(g)  O3(g) + energetic N2(g)

88. Molecularity Elementary Step Molecularity Rate Equation A  product
Unimolecular
Rate = k[A]
A + A  product
Bimolecular
Rate = k[A]2
A + B  product
Rate = k[A][B]
2 A + B  product
Termolecular
Rate = k[A]2[B]

89. Molecularity
The rate equation of an elementary step is defined by the reaction stoichiometry. The rate equation of an elementary step is given by the product of the rate constant and the concentrations of the reactants in that step. 
The molecularity of an elementary step and its order are the same. This is not true for the overall reaction.

90. Rate Equations & Mechanisms
Products of a reaction can never be produced at a rate faster than the rate of the slowest step
Slowest step = Rate-determining step

91. Rate Equations & Mechanisms
For the slowest step Rate = k1[A][B]
Note: the rate law must be written with respect to the reactants only