1. Kinetics: Rates and Mechanisms of organic Reactions

2. Kinetics: Rates and Mechanisms of Chemical Reactions
16.1 Factors that influence reaction rates
16.2 Expressing the reaction rate
16.3 The rate law and its components
16.4 Integrated rate laws: Concentration changes over time
16.5 Reaction mechanisms: Steps in the overall reaction
16.6 Catalysis: Speeding up a chemical reaction

3. Reaction rate: the central focus of chemical kinetics
Figure 16.1
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4. The wide range of reaction rates
Figure 16.2

5. Factors Influencing Reaction Rates
Reactant Concentration: molecular collisions are required for
reactions to occur
reaction rate a collision frequency a concentration
Physical State: molecules must mix to collide
When reactants are in different phases, the more finely divided
a solid or liquid reactant, the greater the surface area per unit
volume, the more contact it makes with other reactants,
and the faster the reaction.
Temperature: molecules must collide with sufficient energy to react
Higher T translates into more collisions per unit time and
into higher-energy collisions
reaction rate a collision energy a temperature

6. The effect of surface area on reaction rate
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Figure 16.3
steel nail in O2
steel wool in O2

7. Collision energy and reaction rate
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Collision energy and reaction rate
Other factor: reaction
trajectory (not all collisions
with sufficient energy
are productive)
Figure 16.4

8. [reactant] decreases, [product] increases
Expressing Reaction Rate
reaction rate = the changes in [reactants] or [products] per unit time;
[reactant] decreases, [product] increases
A B
Measure [A1] at t1, then measure [A2] at t2
Average rate of reaction = - ([A2] - [A1]) / (t2 - t1)
= - D[A] / Dt
rate is always
expressed as a
positive value
rate units = mol/L/s
Also, reaction rate = + D[B] / Dt

9. concentration of O3 (mol/L)
Table Concentration of O3 at various times in its reaction with C2H4 at 303 K C 2 H 4 ( g
) + O 3
) C O( )
time (s)
concentration of O3 (mol/L)
0.0
3.20 x 10-5
10.0
2.42 x 10-5
20.0
1.95 x 10-5
30.0
1.63 x 10-5
40.0
1.40 x 10-5
50.0
1.23 x 10-5
60.0
1.10 x 10-5
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10. Reaction rate varies with time as the reaction proceeds!
Different Ways to Measure Reaction Rates
Reaction rate varies with time as the reaction proceeds!
C2H4(g) + O3(g) C2H4O(g) + O2(g)
average rate = - D[O3] /Dt = - (1.10 x 10-5 mol/L) - (3.20 x 10-5 mol/L)
60.0 s s
= 3.50 x 10-7 mol/L/s
Similar calculations over earlier and later time intervals (e.g., s,
s) reveals average rates of 7.80 x 10-7 mol/L/s and
1.30 x 10-7 mol/L/s, respectively.
Data show that the average rate decreases as the
reaction proceeds!

11. Concentrations of O3 vs time during its reaction with C2H4
Curved line indicates change
in reaction rate with time.
The slope of a tangent at any
point on the curve yields the
instantaneous rate at that point.
reaction rate = (instantaneous)
reaction rate
initial rates: measured to avoid
complications due to back
reactions
Figure 16.5

12. Plots of [C2H4] and [O2] vs reaction time
Other Plots to Determine
Reaction Rate
C2H4(g) + O3(g) C2H4O(g) + O2(g)
Plots of [C2H4] and [O2] vs reaction time
Figure 16.6

13. Expressing rate in terms of changes in [reactant] and [product]
H2(g) + I2(g) HI(g)
rate = - D[H2]/Dt = - D[I2]/Dt = 1/2 D[HI]/Dt or
rate = D[HI]/Dt = -2 D[H2]/Dt = -2 D[I2]/Dt
The mathematical expression for the rate, and the numerical value of
the rate, depend on which substance is chosen as the reference.
For the general reaction: aA + bB cC + dD
rate = -1/a D[A]/Dt = -1/b D[B]/Dt = 1/c D[C]/Dt = 1/d D[D]/Dt

14. Expressing rate in terms of changes in concentration with time
Sample Problem 16.1
Expressing rate in terms of changes in concentration with time
PROBLEM:
Because it generates a nonpolluting product (water vapor), hydrogen gas is used for fuel aboard the space shuttle and may be used by automobile engines in the near future.
2H2(g) + O2(g) H2O(g)
(a) Express the rate in terms of changes in [H2], [O2], and [H2O] with time.
(b) If [O2] decreases at 0.23 mol/L/s, at what rate is [H2O] increasing?
PLAN:
Choose [O2] as the reference since its coefficient is 1. For every molecule of O2 that disappears, two molecules of H2 disappear and two molecules of H2O appear, so [O2] decreases at one-half the rates of change in [H2] and [H2O].
SOLUTION: - 1 2
D[H2] Dt
= -
D[O2] Dt
= +
D[H2O] Dt 1 2
rate =
(a)
D[O2] Dt - =
(b)
= +
D[H2O] Dt 1 2
; = mol/L.s
D[H2O] Dt
0.23 mol/L.s

15. The Rate Law rate law: rate = k[A]m[B]n........
The rate law expresses the reaction rate as a function of reactant
concentrations, product concentrations, and temperature. It is
experimentally determined.
The derived rate law for a reaction must be consistent
with the postulated chemical mechanism of the reaction!
For a general reaction: aA + bB cC dD
rate law: rate = k[A]m[B]n
k = rate constant
m and n = reaction orders (not related to a, b,...)
(for a unidirectional (one-way) reaction)!

16. How is the rate law determined experimentally?
The Procedure
1. Measure initial rates (from concentration measurements)
2. Use initial rates from several experiments to find the reaction orders
3. Calculate the rate constant

17. Reaction Order Terminology
rate = k[A] first order overall (rate a [A])
rate = k[A] second order overall (rate a [A]2)
rate = k[A]0 = k(1) = k zero order (rate independent of [A])
Examples
NO(g) + O3(g) NO2(g) + O2(g)
rate = k[NO][O3] second order overall
2NO(g) H2(g) N2(g) H2O(g)
rate = k[NO]2[H2]1 third order overall

18. Properties of Reaction Orders
Reaction orders cannot be deduced from the
balanced chemical equation.
Reaction orders are usually positive integers or zero.
Reaction orders can be fractional or negative.

19. Determining Reaction Orders
O2(g) + 2NO(g) NO2
rate = k[O2]m[NO]n
To find m and n: A series of experiments is conducted starting with different sets of reactant concentrations. The initial reaction rate is measured in each experiment.

20. Integrated Rate Laws: Change in Concentration
with Reaction Time
Consider a general first order reaction: A B
rate = - D[A]/Dt = k[A]
Upon integration over time, we obtain:
ln ([A]0/[A]t) = kt
ln = natural logarithm
[A]0 = concentration of A at t = 0
[A]t = concentration of A at any time t
ln [A]0 - ln [A]t = kt

21. For a simple second order reaction (one reactant only):
rate = - D[A]/Dt = k[A]2
Upon integration over time, we obtain:
1/[A]t /[A]0 = kt
For a zero order reaction:
rate = - D[A]/Dt = k[A]0
Upon integration over time, we obtain:
[A]t - [A]0 = -kt

22. Graphical method to determine reaction order
[A]t = -kt + [A]0
ln[A]t = -kt + ln[A]0
1/[A]t = kt + 1/[A]0
Figure 16.7

23. A plot of [N2O5] vs time for three half-lives
A first order
reaction
Figure 16.9

24. Why is t1/2 independent of reactant concentration
for a first order reaction?
ln ([A]0/[A]t) = kt
After one half-life, t = t1/2 and [A]t = 0.5[A]0. Substituting.....
ln ([A]0/0.5([A]0) = kt1/2 or
ln 2 = kt1/2
t1/2 = (ln 2)/k = /k
first order reaction: A B

25. Half-Life for Zero and Second Order Reactions
t1/2 = 1/k [A]0
simple second order (rate = k[A]2)
As a second order reaction proceeds, the half-life increases.
t1/2 = [A]0/2k
zero order (rate = k)
For zero order reactions, higher initial reactant concentrations
translate into longer half-lives.

26. An overview of zero-order, first-order and
An overview of zero-order, first-order and simple second-order reactions
zero order
first order
second order
rate law
rate = k
rate = k[A]
rate = k[A]2
units for k
mol/L.s
1/s
L/mol.s
[A]t =
kt + [A]0
integrated rate law in straight-line form
ln[A]t =
-kt + ln[A]0
1/[A]t =
kt + 1/[A]0
plot for straight line
[A]t vs t
ln[A]t vs t
1/[A]t = t
slope, y-intercept
-k, [A]0
k, 1/[A]0
-k, ln[A]0
half-life
[A]0/2k
(ln 2)/k
1/k[A]0
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27. Temperature and Reaction Rate
Consider the rate law for a first order reaction:
rate = k[A]
Where is the temperature dependence?
Answer: It is embodied in the rate constant, k, that is,
k depends on the temperature at which the
reaction is conducted.
What does the T-dependence of k look like?
R-COOR’ + H2O R-COOH R’OH
test
reaction
ester
acid
alcohol
organic ester hydrolysis

28. Dependence of k on temperature for the hydrolysis of an organic ester
Note that both
reactant
concentrations are
held constant
k increases
exponentially!
Figure 16.10

29. Graphical determination of the activation energy, Ea
ln k = -Ea/R (1/T) + ln A
An Arrhenius plot
Figure 16.11
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30. Information Sequence to Determine the Kinetic Parameters of a Reaction
series of plots of [ ] vs time
initial rates
Determine slope of tangent at t0 of each plot
reaction orders
rate constant (k) and actual rate law
Compare initial rates when [A] changes and [B] is held constant and vice versa
Substitute initial rates, orders, and concentrations into general rate law: rate = k [A]m[B]n
Find k at varied T
A + B C + D
activation energy, Ea
rate constant and reaction order
Find k at varied T
integrated rate law (half-life, t1/2)
Rearrange to linear form and graph
Figure 16.12

31. Using Collision Theory to Explain the Effects
of [ ] and T on Reaction Rate
Model: A B products
Why are [A] and [B] multiplied in the rate law?
Consider several cases where there are a finite number
of particles each of A and B and determine the
number of possible A-B collisions.

32. The dependence of possible collisions on the
product of reactant concentrations A B
4 collisions
2 x 2 = 4 A B
Add another molecule of A A
6 collisions B
3 x 2 = 6 A B A A B
9 collisions
Add another molecule of B A B
3 x 3 = 9 A B
Figure 16.13

33. f = e-Ea/RT Temperature and Reaction Rate
increased T increased average speed of particles
increased collision frequency increased reaction rate
But,
most collisions fail to yield products!
Significance of activation energy: only those collisions with
energy equal to, or greater than, Ea can yield products.
Increasing T enhances the fraction of productive collisions, f.
f = e-Ea/RT
From this equation, we can see that both Ea and T
affect f, which in turn influences reaction rate.

34. Table 16. 5 Effect of Ea and T on the fraction (f) of collisions
Table Effect of Ea and T on the fraction (f) of collisions with sufficient energy to allow reaction
Ea (kJ/mol)
f (at 298 K) 50
1.70 x 10-9 75
7.03 x 10-14
100
2.90 x 10-18 T
f (Ea = 50 kJ/mol)
25 oC (298 K)
1.70 x 10-9
35 oC (308 K)
3.29 x 10-9
45 oC (318 K)
6.12 x 10-9
DT = 10o; f
~ doubles; reaction
rate ~ doubles!
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35. The effect of temperature on the distribution of collision energies
Figure 16.14

36. Energy-Level Diagram for an Equilibrium Reaction
A + B C + D
Collision Energy
Collision Energy
ACTIVATED STATE
Ea (forward)
Ea (reverse)
reactants
A + B
products
C + D
The forward reaction is exothermic because the reactants have more energy than the products.
Figure 16.15
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37. An energy-level diagram of the fraction of collisions exceeding Ea
A + B C + D
Figure 16.16

38. k = Ae -Ea/RT Effective Collisions
Not all collisions that occur with energy equal to, or exceeding,
the activation energy lead to products.
Molecular orientation is critical!
k = Ae -Ea/RT
the frequency factor = product of collision
frequency Z and an orientation probability
factor p (A = Zp)

39. Molecular orientation and effective collisions
NO + NO NO2
Figure 16.17
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40. Transition State Theory
Addresses the limitations of collision theory in explaining
chemical reactivity
Key principle: During the transformation of reactant into
product, one or more very short-lived chemical species
form that resemble (but are different from) reactant or product;
these transitional species contain partial bonds; they are called
transition states (TS) or activated complexes.
The activation energy is used to stretch/deform specific bonds
in the reactant(s) in order to reach the transition state.
Example reaction:
CH3Br OH CH3OH Br-
What might the TS look like for this substitution reaction?

41. The proposed transition state in the reaction between CH3Br and OH-
The TS is trigonal bipyramidal; note the elongated C-Br
and C-O bonds
Figure 16.18

42. Reaction Energy Diagrams
Shows the potential energy of the system during a
chemical reaction
(a plot of potential energy vs reaction progress)

43. Reaction energy diagram for the reaction between CH3Br and OH-
Figure 16.19
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44. General Principles of TS Theory
Every reaction (and each step in an overall reaction) goes
through its own TS.
All reactions are reversible.
Transition states are identical for the individual forward and reverse
reactions in an equilibrium reaction.

45. Drawing reaction energy diagrams and transition states
Sample Problem 16.7
PROBLEM:
A key reaction in the upper atmosphere is
O3(g) + O(g) O2(g)
The Ea(fwd) is 19 kJ, and DHrxn for the reaction is -392 kJ. Draw a reaction energy diagram for this reaction, postulate a transition state, and calculate Ea(rev).
PLAN:
Consider relationships between reactants, products and transition state. The reactants are at a higher energy level than the products, and the transition state is slightly higher in energy than the reactants.
Ea= 19 kJ
transition state
Ea(rev) = ( ) kJ =
411 kJ
SOLUTION:
DHrxn = -392 kJ

46. Reaction energy diagrams and possible
transition states for three reactions
endothermic
exothermic
exothermic
Figure 16.20
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47. Reaction Mechanisms The Specific Steps in an Overall Reaction
How a reaction works at the molecular level
2A + B E + F
A + B C
possible
steps
C + A D
D E + F
C and D are reaction intermediates.
All postulated steps must sum to yield the chemical equation for the overall reaction.
Mechanisms of reactions are proposed and then tested.

48. Elementary Reactions and Molecularity
individual steps = elementary reactions (or steps)
Elementary steps are characterized by their molecularity.
molecularity = number of reactant particles involved in the step
2O3(g) O2(g)
O3(g) O2(g) + O(g)
unimolecular
O3(g) + O(g) O2(g)
bimolecular
Termolecular elementary steps are rare!
The rate law for an elementary reaction can be deduced
from the reaction stoichiometry; equation coefficients are
used as the reaction orders.

49. Table 16.6 Rate Laws for General Elementary Steps
molecularity
rate law
A product
unimolecular
rate = k [A]
2A product
bimolecular
rate = k [A]2
A + B product
bimolecular
rate = k [A][B]
2A + B product
termolecular
rate = k [A]2[B]
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50. The Rate-Determining Step of a Reaction Mechanism
All of the elementary steps of a mechanism do
not occur at the same rate. Usually one step is
much slower than the others. This step is called the
rate-determining (or rate-limiting) step.
The overall rate of a reaction is related to the rate of the rate-determining step. That is, the rate law for the rate-determining step represents the rate law for the overall reaction!

51. Correlating the Mechanism with the Rate Law
Chemical mechanisms can never be proven
unequivocally! But potential chemical mechanisms
can be eliminated based on experimental data.
Three Key Criteria for Elementary Steps
1. The elementary steps must add up to the overall equation.
2. The elementary steps must be physically reasonable.
3. The mechanism must correlate with the rate law.

52. Reaction Energy Diagram for the two-step NO2-F2 reaction
2NO2(g) + F2(g) NO2F(g)
rate = k[NO2][F2]
Two transition states and
one intermediate are involved.
The first step is
rate-determining.
The reaction is exothermic
(thermodynamics).
Figure 16.21
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53. What happens when a fast reversible step(s) occurs prior to the
rate-determining step?
2NO(g) + O2(g) NO2 (g)
Rate law: rate = k[NO]2[O2]
Proposed Mechanism
Step 1: NO(g) + O2(g) NO3(g) fast and reversible
Step 2: NO3(g) + NO(g) NO2(g) slow; rate-determining
Rate laws for the two elementary steps:
Step 1: rate1 (fwd) = k1[NO][O2]; rate1 (rev) = k-1[NO3]
Step 2: rate2 = k2[NO3][NO]
Step 2 (rate-determining) contains [NO3] which does
not appear in the experimental rate law!

54. Conclusion: The proposed mechanism is consistent
If Step 1 is at equilibrium, then....
rate1 (fwd) = rate1 (rev) or
k1[NO][O2] = k-1[NO3]
Solving for [NO3]:
[NO3] = (k1/k-1)[NO][O2]
Substituting into the rate law for rate-limiting Step 2:
rate2 = k2[NO3][NO] = k2 (k1/k-1)[NO][O2][NO] = [(k1k2)/k-1][NO]2[O2]
Conclusion: The proposed mechanism is consistent
with the kinetic data.

55. Catalysis: Enhancing Reaction Rates
Catalyst: increases reaction rate but is not consumed in the reaction
A catalyst increases reaction rate (via increasing k) by lowering the
activation energy (barrier) of the reaction.
Both forward and reverse reactions are catalyzed; reaction
thermodynamics is unaffected!
The catalyzed reaction proceeds via a different mechanism
than the uncatalyzed reaction.

56. Note that both reactions
Energy diagram of an uncatalyzed and catalyzed reaction
Note that both reactions
exhibit the same
thermodynamics!
The catalyzed and
uncatalyzed
reactions occur
via different
pathways.
Figure 16.22
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57. Uncatalyzed reaction:
A + B product
Catalyzed reaction:
A + catalyst C
C + B product + catalyst
Types of Catalysts
Homogeneous: exists in solution with the reaction mixture
Heterogeneous: catalyst and reaction mixture are in different phases

59. The Mechanism of Acid-catalyzed Organic Ester Hydrolysis
Homogeneous Catalysis
The Mechanism of Acid-catalyzed
Organic Ester Hydrolysis
R-COOR’ + H2O + H R-COOH R’OH H+
ester
acid
alcohol
The reaction rate is slow at neutral pH (pH 7.0), but
increases significantly at acidic pH (pH 2).
H+ ion catalyzes the reaction at low pH; what is the
molecular basis for the catalysis by H+?

60. Proposed reaction mechanism for the H+-catalyzed
hydrolysis of an organic ester +
Figure 16.23
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61. Heterogeneous Catalysis: Metal-catalyzed hydrogenation of ethylene2 C C H 2 ( g ) + H 3 C C H 3 ( g )
Ni, Pd or Pt
Figure 16.24

62. Kinetic isotope effect
Kinetic isotope effect: the decrease in the rate of a chemical reaction upon replacement of one atom in a reactant by a heavier isotope.
Primary kinetic isotope effect: the kinetic isotope effect observed when the rate-determining step requires the scission of a bond involving the isotope:
with
Secondary kinetic isotope effect: the variation in reaction rate even though the bond involving the isotope is not broken to form product:

63. Kinetic isotope effect (primary)

64. Kinetics isotope effect (secondary)

65. Chain reactions
Chain reactions: a reaction intermediate produced in one step generates an intermediate in a subsequent step, then that intermediate generates another intermediate, and so on.
Chain carriers: the intermediates in a chain reaction. It could be radicals (species with unpaired electrons), ions, etc.
Initiation step:
Propagation steps:
Termination steps:

66. Example: The hydrogen-bromine reaction has a complicated rate law rather than the second order reaction as anticipated.
H2(g) + Br2(g) → 2HBr(g)
Yield
The following mechanism has been proposed to account for the above rate law.
1. Initiation: Br M → Br Br. + M ki
2. Propagation: Br H2 → HBr + H kp1
H Br2 → HBr Br kp2
3. Retardation: H HBr → H Br kr
4. Termination: Br Br. + M → Br M* kt
derive the rate law based on the above mechanism.