Chemical Kinetics
CHAPTER 14
Chemistry: The Molecular Nature of Matter, 6th edition
By Jesperson, Brady, & Hyslop
CHAPTER 14 Chemical Kinetics
Learning Objectives:
 Factors Affecting Reaction Rate:
o Concentration
o State
o Surface Area
o Temperature
o Catalyst
 Collision Theory of Reactions and Effective Collisions
 Determining Reaction Order and Rate Law from Data
 Integrated Rate Laws
 Rate Law  Concentration vs Rate
 Integrated Rate Law  Concentration vs Time
 Units of Rate Constant and Overall Reaction Order
 Half Life vs Rate Constant (1st Order)
 Arrhenius Equation
 Mechanisms and Rate Laws
 Catalysts
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Lecture Road Map:
① Factors that affect reaction rates
② Measuring rates of reactions
③ Rate Laws
④ Collision Theory
⑤ Transition State Theory & Activation Energies
⑥ Mechanisms
⑦ Catalysts
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Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Factors that
Affect Reaction
Rates
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
The Speed at Which Reactions Occur
Kinetics: Study of factors that govern
o How rapidly reactions occur and
o How reactants change into products
Rate of Reaction:
o Speed with which reaction occurs
o How quickly reactants disappear and
products form
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Molecular Nature of Matter, 6E
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Kinetics
A
The Speed at Which Reactions Occur
B
Reaction rate is measured by
the amount of product
produced or reactants
consumed per unit time.
o [B] concentration of products
will increase over time
o [A] concentration of reactants
will decrease over time
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
Factors Affecting Reaction Rates
1. Chemical nature of reactants
o What elements, compounds, salts
are involved?
o What bonds must be formed,
broken?
o What are fundamental differences in
chemical reactivity?
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
Factors Affecting Reaction Rates
2. Ability of reactants to come in contact
(Reactants must meet in order to react)
The gas or solution phase facilitates this
o Reactants mix and collide with each other easily
o Homogeneous reaction
o All reactants in same phase
o Occurs rapidly
o Heterogeneous reaction
o Reactants in different phases
o Reactants meet only at interface between phases
o Surface area determines reaction rate
o Increase area, increase rate; decrease area,
decrease rate
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
Factors Affecting Reaction Rates
3. Concentrations of reactants
o Rates of both homogeneous and
heterogeneous reactions affected by [X ]
o Collision rate between A and B increase if we
increase [A] or increase [B ].
o Often (but not always) reaction rate
increases as [X ] increases
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
Factors Affecting Reaction Rates
4. Temperature
o Rates are often very sensitive to temperature
o Raising temperature usually makes reaction
faster for two reasons:
o Faster molecules collide more often and
collisions have more energy
o Most reactions, even exothermic reactions,
require energy to occur
o Rule of thumb: Rate doubles if temperature
increases by 10 °C (10 K)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Kinetics
Factors Affecting Reaction Rates
5. Presence of Catalysts
o Substances that increase rates of chemical
reactions without being used up
o Rate-accelerating agents
o Speed up rate dramatically
o Rate enhancements of 106 not uncommon
o Chemicals that participate in mechanism but
are regenerated at the end
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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CHAPTER 14 Chemical Kinetics
Measuring
Reaction Rates
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Measuring Rate of Reaction
o Rate = ratio with time unit in denominator
o Rate of Chemical Reaction
o Change in concentration per unit time.
D[reactant]
reaction rate =
Dtime
o Always with respect to a given reactant or product
o [reactants] decrease with time
o [products] increase with time
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Measuring Rate of Reaction
Rate X =
[X ]t - [X ]t
2
1
t2 - t1
D[X ]
=
Dt
o Concentration in M units
o Time in s units
o Units on rate:
mol/L mol M
=
=
s
Ls
s
o [product] increases by 0.50 mol/L per second
 rate = 0.50 M/s
o [reactant] decreases by 0.20 mol/L per second
 rate = 0.20 M/s
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Rate of Reaction
Always positive whether something is increasing
or decreasing in [X ]
o Reactants
D[reactant]
Rate = o Reactant consumed
Dt
o So [X ] is negative
o Need minus sign to make rate positive
o Products
o Produced as reaction goes along
o So [X ] is positive
D[product]
Rate =
o Thus rate already positive
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Dt
15
Rates
Measuring Rate of Reaction
Coefficients indicate the relative rates at which reactants are
consumed and products are formed
o Related by coefficients in balanced chemical equation
o Know rate with respect to one product or reactant
o Can use equation to determine rates with respect to all
other products and reactants.
A + B  C + D
1 DA
1 DB 1 DC 1 DD
Rate = ==
=
a Dt
b Dt g Dt d Dt
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Rate of Reaction: Example
C3H8(g) + 5O2(g)  3CO2(g) + 4H2O(g)
• O2 reacts 5 times as fast as C3H8
Rate = -
D[O2 ]
Dt
= -5
D[C3H8 ]
Dt
• CO2 forms 3 times faster than C3H8 consumed
Rate =
D[CO2 ]
Dt
= -3
D[C3H8 ]
Dt
• H2O forms 4/5 as fast as O2 consumed
D[H2O]
Dt
4 D[O2 ]
=5 Dt
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
Clorox bleach is sodium hypochlorite. It should never
be mixed with acids, (like vinegar) because it forms
chlorine gas:
NaClO + 2 HCl → Cl2 + H2O + NaCl
If Chlorine gas (Cl2) is formed at a rate of
5.0 x 10-4 mol/Ls what rate is HCl consumed?
HCl: Cl2
2:1
Therefore HCl will disappear twice as fast as Cl2 is formed.
Rate HCl consumed = 10. x 10-4 mol/Ls
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Change of Reaction Rate with Time
Generally reaction rate changes during reaction, it
isn’t constant
o Often initially fast when lots of reactant
present
o Slower and slower as reactants are depleted
Why?
o Rate depends on the concentration of the
reactants
o Reactants being used up, so the concentration
of the reactants are decreasing and therefore
the rate decreases
Measured in 3 ways:
o instantaneous rate, average rate, initial rate
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Instantaneous & Initial Reaction Rate
Rates
Instantaneous rate
o Slope of tangent to curve at some specific time
Initial rate
o Determined at time = 0
N O 2 a p p e a ra n c e
0 .0 3 5
0 .0 3
[N O 2 ]
0 .0 2 5
0 .0 2
0 .0 1 5
0 .0 1
0 .0 0 5
0
0
200
400
600
800
T im e (s )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Average Rate of Reaction
Average Rate: Slope of line connecting starting
and ending coordinates for specified time frame
NO2 appe arance
0.035
Δ[Product]
 rate
Δtime
0.03
[NO2]
0.025
0.02
0.015
0.01
0.005
0
0
200
400
600
800
Time (s)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rates
Example Reporting Different Types of Rates
Concentration vs. Time Curve for 0.005M phenolphthalein
reacting with 0.61 M NaOH at room temperature
Rate at any time t = negative slope
(or tangent line) of curve at that point
Dy rise
Rate=–slope =
=
Dx run
http://chemed.chem.purdue.edu
22
Rates
Example Reporting Different Types of Rates
[P] (mol/L)
Time (s)
0.005
0.0045
0.004
0.0035
0.003
0.0025
0.002
0.0015
0
10.5
22.3
35.7
51.1
69.3
91.6
120.4
Initial rate = Average rate
between first two data points
(0.0045 - 0.005) M
rate = (10.5 – 0) s
–(–0.0005 M )
=
10.5 s
-5
= 4.76 ´10 M / s
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Example Reporting Different Types of Rates
Rates
Instantaneous Rate at 120.4 s
Dy rise
=
Dx run
(0.0018 - 0.0028)M
=(160 - 90) s
-0.001M
== 1.4 ´10-5 M / s
70 s
Rate = -slope =
(90,0.0028)
(160,0.0018)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Example Reporting Different Types of Rates
Rates
Average Rate between 0 and 120.4 s
Dy rise
=
Dx run
(0.0015 - 0.0045)M
=(120.4 -10.5) s
-0.003M
== 2.7 ´10-5 M / s
109.9 s
Rate = -slope =
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
A reaction was of NO2 decomposition was studied.
The concentration of NO2 was found to be 0.0258 M
at 5 minutes and at 10 minutes the concentration was
0.0097 M. What is the average rate of the reaction
between 5 min and 10 min?
A. 310 M/min
B. 3.2 × 10–3 M/min
C. 2.7 × 10–3 M/min
D. 7.1 × 10–3 M/min
(0.0258 M - 0.0097 M ) = 3.2 ´ 10
10 min - 5 min
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
-3
M / min
26
CHAPTER 14 Chemical Kinetics
Rate Laws
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Rates Based on All Reactants
A + B  C + D
D[ A]
= k[A]m[B]n
Rate = Dt
o Rate Law or Rate expression
o k is the rate constant
o Dependent on Temperature & Solvent
o m and n = exponents found experimentally
o No necessary connection between stoichiometric
coefficients (, ) and rate exponents (m, n)
o Usually small integers
o Sometimes simple fractions (½, ¾) or zero
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Rates Based on All Reactants
Below is the rate law for the reaction
2A +B → 3C
rate= 0.045 M–1s–1 [A][B]
If the concentration of A is 0.2 M and that of B is 0.3
M, and the reaction is 1st order (m & n = 1) what will
be the reaction rate?
rate=0.045 M–1 s–1 [0.2][0.3]
rate=0.0027 M/s  0.003 M/s
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Order of Reactions
Rate = k[A]m[B]n
Exponents specify the order of reaction with
respect to each reactant
Order of Reaction
o m = 1 [A]1 1st order in [A]
o m = 2 [A]2 2nd order in [A]
om=3
[A]3 3rd order in [A]
om=0
[A]0 0th order in [A]
[A]0 = 1  means A doesn't affect rate
Overall order of reaction = sum of orders (m
and n) of each reactant in rate law
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Order of Reactions: Example
Rate Laws
5Br– + BrO3– + 6H+  3Br2 + 3H2O
-
D[BrO3- ]
Dt
= k [BrO-3 ]x [Br - ]y [H+ ]z
x=1 y=1 z=2
o 1st order in [BrO3–]
o 1st order in [Br –]
o 2nd order in [H+]
o Overall order = 1 + 1 + 2 = 4
rate = k [BrO ] [Br ] [H ]
- 1
3
- 1
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Order of Reaction & Units for k
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Group
Problem
The following rate law has been observed:
Rate = k[H2SeO][I–]3[H+]2. The rate with
respect to I– and the overall reaction rate is:
A. 6, 2
B. 2, 3
C. 1, 6
D. 3, 6
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Calculating k
If we know rate and concentrations, can use
rate law to calculate k
From Text Example of decomposition of HI at
508 °C
• Rate= 2.5 × 10–4 M/s
• [HI] = 0.0558 M
D[HI]
rate = = k [HI]2
Dt
rate 2.5 ´ 10-4 M / s
-1 -1
k=
=
=
0.08029
M
s
2
2
[HI]
(0.0558 M )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Exponents in Rate Law
Experimental Determination of Exponents
o Method of initial rates
o If reaction is sufficiently slow
o or have very fast technique
o Can measure [A] vs. time at very beginning of
reaction
o Before it slows very much, then
æ [ A] - [ A] ö
0÷
initial rate = - çç 1
÷
t
t
è 1 0 ø
o Set up series of experiments, where initial
concentrations vary
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Rate Law Exponents: Example
3A + 2B  products
Rate = k[A]m[B]n
Expt. # [A]0, M
1
0.10
2
0.20
3
0.20
[B]0, M
0.10
0.10
0.20
Initial Rate, M/s
1.2  10–4
4.8  10–4
4.8  10–4
 Convenient to set up experiments so
 The concentration of one species is doubled or tripled
 And the concentration of all other species are held constant
 Tells us effect of [varied species] on initial rate
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Rate Law Exponents
• If reaction is 1st order in [X],
– Doubling [X]1  21
– Doubles the rate
• If reaction is 2nd order in [X],
– Doubling [X]2  22
– Quadruples the rate
• If reaction is 0th order in [X],
– Doubling [X]0  20
– Rate doesn't change
• If reaction is nth order in [X]
– Doubling [X]n  2n times the initial rate
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Determining Rate Law Exponents: Example
Rate Laws
Expt. #
[A]0, M
[B]0, M
1
2
3
0.10
0.20
0.20
0.10
0.10
0.20
Comparing Expt. 1 and 2
o Doubling [A]
o Quadruples rate
o Reaction 2nd order in A = [A]2
m
4=
n
k éë A2 ùû éëB 2 ùû
Initial Rate, M/s
1.2  104
4.8  104
4.8  104
Rate 2 4.8  10 4

4

4
Rate 1 1.2  10
m
n
k éë0.20ùû éë0.10ùû
m
é0.20ù
ë
û
Rate 2
m
=
=
=
=
2
Rate 1 k é A ùm éB ùn k é0.10ùm é0.10ùn é0.10ùm
ë 1û ë 1û
ë
û ë
û
ë
û
2m = 4 or
m=2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Rate Law Exponents: Example
Expt. #
1
[A]0, M
0.10
[B]0, M
0.10
Initial Rate, M/s
1.2  104
2
3
0.20
0.20
0.10
0.20
4.8  104
4.8  104
Comparing Expt. 2 and 3
o Doubling [B]
o Rate does not change
o Reaction 0th order in B = [B]0 = 1
m
n
k éë A3 ùû éëB 3 ùû
Rate 3 4.8  10 4

1

4
Rate 2 4.8  10
m
n
k éë0.20ùû éë0.20ùû
n
é0.20ù
ë
û
Rate 3
n
1=
=
=
=
=
2
Rate 2 k é A ùm éB ùn k é0.20ùm é0.10ùn é0.10ùn
ë 2û ë 2û
ë
û ë
û
ë
û
2n = 1 or
n=0
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
39
Determining Rate Law Exponents: Example
Rate Laws
Expt. #
1
[A]0, M
0.10
[B]0, M
0.10
Initial Rate, M/s
1.2  10–4
2
0.20
0.10
4.8  10–4
3
0.20
0.20
4.8  10–4
o Conclusion: rate = k[A]2
o Can use data from any experiment to determine k
o Let’s choose first experiment
k=
rate
éAù
ë û
2
=
1.2 ´ 10 –4 M /s
(0.10 M )
2
= 1.2 ´ 10-2 M -1s-1
Jesperson, Brady, Hyslop. Chemistry:
The Molecular Nature of Matter, 6E
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Determining Rate Law Exponents: Ex 2
Rate Laws
2 SO2 + O2  2 SO3
Rate = k[SO2]m[O2]n
Expt
#
[SO2] M
[O2]
M
Initial Rate of SO3
Formation, M s–1
1
0.25
0.30
2.5  103
2
0.50
0.30
1.0  102
3
0.75
0.60
4.5  102
4
0.50
0.90
3.0  102
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Rate Law Exponents: Ex 2
m
n
k éëSO 2 ùû éëO 2 ùû
2
2
Rate 2
=
m
n
Rate 1
é
ù
é
ù
k ëSO 2 û ëO 2 û
1
1
4
m
n
m
n
=
k éë0.50ùû éë0.30ùû
=
é0.50ù
ë
û
k éë0.25ùû éë0.30ùû
m
4
4 = 2m
m
é0.25ù
ë
û
or
= 2m
m=2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
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Rate Laws
Determining Rate Law Exponents: Ex 2
m
n
k éëSO 2 ùû éëO 2 ùû
4
4
Rate 4
=
m
n
Rate 2
é
ù
é
ù
k ëSO 2 û ëO 2 û
2
2
m
n
m
n
3 =
k éë0.50ùû éë0.90ùû
3 =
é0.90ù
ë
û
k éë0.50ùû éë0.30ùû
n
3 = 3n
n
é0.30ù
ë
û
or
= 3n
n=1
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
43
Rate Laws
Determining Rate Law Exponents: Ex 2
Rate = k[SO2]2[O2]1
• 1st order in [O2]
• 2nd order in [SO2]
• 3rd order overall
• Can use any experiment to find k
k=
rate
[SO2 ]2 [O2 ]1
=
3.0 ´ 10 M / s
-2
(0.50 M )2 (0.90 M )
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
= 0.13 M –2s-1
44
Group
Problem
Using the following experimental data, determine the order with
respect to NO and O2 .
Expt [NO]
#
M
A.
B.
C.
D.
2, 0
3,1
2, 1
1, 1
1
2
3
0.12
0.24
0.24
[O2]
M
0.25
0.25
0.50
Initial Rate
M s–1
1.5  10–3
6.0  10–3
1.2  10–2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
45
Group
Problem
R2
R1
6.0 ´ 10 M s
-3
=
-1
1.5 ´ 10-3 M s -1
=
x
y
x
y
2
y
2
y
é0.24 M ù é0.25 M ù
ë
û ë
û
é0.12 M ù é0.25 M ù
ë
û ë
û
x =2
R3
R2
=
1.2 ´ 10 M s
-1
6.0 ´ 10 M s
-1
-2
-3
=
é0.24 M ù é0.50 M ù
ë
û ë
û
é0.24 M ù é0.25 M ù
ë
û ë
û
y =1
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
46
Example : Method of Initial Rates
BrO3– + 5Br– + 6H+  3Br2 + 3H2O
Rate = -
3
D[BrO ]
Dt
= k [BrO3- ]m [Br - ]n [H+ ]p
Expt
#
1
[BrO3–]
mol/L
0.10
[Br–] [H+]
mol/L mol/L
0.10
0.10
Initial Rate
mol/(L s)
8.0  10–4
2
0.20
0.10
0.10
1.6  10–3
3
0.20
0.20
0.10
3.2  10–3
4
0.10
0.10
0.20
3.2  10–3
47
Ex.: Method of Initial Rates
Compare 1 and 2
m
n
p
Rate 2 1.6 ´ 10 M /s k (0.20 M ) (0.10 M ) (0.10 M )
=
=
-4
Rate 1 8.0 ´ 10 M /s k (0.10 M )m (0.10 M )n (0.10 M )p
-3
m
æ 0.20 M ö
÷÷ = (2.0)m
2.0 = çç
è 0.10 M ø
\m = 1
Compare 2 and 3
Rate 3 3.2 ´ 10-3 M /s k (0.20 M )m (0.20 M )n (0.10 M )p
=
=
Rate 2 1.6 ´ 10-3 M /s k (0.20 M )m (0.10 M )n (0.10 M )p
n
æ 0.20 M ö
÷÷ = (2.0)n \ n = 1
2.0 = çç
è 0.10 M ø
48
Ex.: Method of Initial Rates
Compare 1 and 4
Rate 4 3.2 ´ 10-3 M /s k (0.10 M )m (0.10 M )n (0.20 M )p
=
=
-4
Rate 1 8.0 ´ 10 M /s k (0.10 M )m (0.10 M )n (0.10 M )p
p
æ 0.20 M ö
÷÷ = (2.0)p
4.0 = çç
è 0.10 M ø
\p = 2
First order in [BrO3–] and [Br–]
Second order in [H+]
Overall order = m + n + p = 1 + 1 + 2 = 4
Rate Law is: Rate = k[BrO3–][Br–][H+]2
•
•
•
•
49
CHAPTER 14 Chemical Kinetics
Integrated Rate
Laws
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
50
Integrated
Rate Laws
Concentration & Time
Rate law tells us how speed of reaction varies with
concentrations.
Sometimes want to know
o Concentrations of reactants and products at
given time during reaction
o How long for the concentration of reactants to
drop below some minimum optimal value
 Need dependence of rate on time
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
51
Integrated
Rate Laws
First Order Integrated Rate Law
-D[A]
Rate =
= k [A]
Dt
• Corresponding to reactions
– A  products
• Integrating we get
ln
[ A ]0
[ A]t
= kt
• Rearranging gives
ln[A]t = -kt + ln[ A]0
• Equation of line
y = mx + b
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
52
Integrated
Rate Laws
First Order Integrated Rate Law
ln[ A]t   kt  ln[ A]0
Slope = –k
Yields straight line
o Indicative of first order
kinetics
o Slope = –k
o Intercept = ln [A]0
o If we don't know
already
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
53
Integrated
Rate Laws
2nd Order Integrated Rate Law
D[B]
Rate = k[B] = Dt
2
• Corresponding to special second order reaction
– 2B  products
• Integrating we get 1
[B ]t
-
1
[B ] 0
• Rearranging gives
1
[B ]t
• Equation of line
= kt +
= kt
1
[B ] 0
y = mx + b
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
54
Integrated
Rate Laws
2nd Order Integrated Rate Law
1
1
 kt 
[B ]t
[B ]0
Yields straight line
o Indicative of 2nd order
kinetics
o Slope = +k
o Intercept = 1/[B]0
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
Slope
= +k
55
Integrated
Rate Laws
Graphically determining Order
Make two plots:
1. ln [A] vs. time
2. 1/[A] vs. time
o If ln [A] is linear and 1/[A] is curved, then
reaction is 1st order in [A]
o If 1/[A] plot is linear and ln [A] is curved, then
reaction is 2nd order in [A]
o If both plots give horizontal lines, then 0th order
in [A]
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
56
Integrated
Rate Laws
Graphically determining Order
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
57
Integrated
Rate Laws
Graphically determining Order
Add examples of graphs
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
58
Integrated
Rate Laws
Example: SO2Cl2  SO2 + Cl2
Time, min
[SO2Cl2], M
ln[SO2Cl2]
1/[SO2Cl2] (L/mol)
0
0.1000
-2.3026
10.000
100
0.0876
-2.4350
11.416
200
0.0768
-2.5666
13.021
300
0.0673
-2.6986
14.859
400
0.0590
-2.8302
16.949
500
0.0517
-2.9623
19.342
600
0.0453
-3.0944
22.075
700
0.0397
-3.2264
25.189
800
0.0348
-3.3581
28.736
900
0.0305
-3.4900
32.787
1000
0.0267
-3.6231
37.453
1100
0.0234
-3.7550
Jesperson, Brady, Hyslop.
Chemistry: The
Molecular Nature of Matter, 6E
42.735
59
Integrated
Rate Laws
Example: SO2Cl2  SO2 + Cl2
Second order plot for SO2Cl2
Decomposition
First Order Plot for SO2Cl2
Decomposition
45
-2.4
40
1/[SO2Cl2] (L/mol)
-2.2
ln[SO2Cl2]
-2.6
-2.8
-3.0
-3.2
-3.4
35
30
25
20
-3.6
15
-3.8
10
0
200
400
600
800
1000
1200
0
200
time (min)
400
600
800
1000
1200
time (min)
Reaction is 1st order in SO2Cl2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
60
Integrated
Rate Laws
Example: HI(g)  H2(g) + I2(g)
Time
(s)
[HI]
(mol/L)
ln[HI]
1/[HI]
(L/mol)
0
0.1000
-2.3026
10.000
50
0.0716
-2.6367
13.9665
100
0.0558
-2.8860
17.9211
150
0.0457
-3.0857
21.8818
200
0.0387
-3.2519
25.840
250
0.0336
-3.3932
29.7619
300
0.0296
-3.5200
33.7838
350
0.0265
-3.6306
37.7358
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
61
Integrated
Rate Laws
Example: HI(g)  H2(g) + I2(g)
Second order plot for HI
o
Decomposition at 508 C
First Order Plot for HI Decomposition
at 508 oC
-2.2
40
-2.4
35
1/[HI] (L/mol)
ln[HI]
-2.6
-2.8
-3.0
-3.2
30
25
20
-3.4
15
-3.6
10
-3.8
0
50
100 150 200 250 300 350
time (s)
0
50 100 150 200 250 300 350
time (s)
Reaction is second order in HI.
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
62
Group
Problem
A plot for a zeroth order reaction is shown. What is
the proper label for the y-axis in the plot ?
Zeroth Order Plot
A. Concentration
B. ln of Concentration
C. 1/Concentration
D. 1/ ln Concentration
0
200
400
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
600
800
time (min)
1000
1200
63
Integrated
Rate Laws
Half Life (t1/2) for first order reactions
Half-life = t½ We often use the half life to describe how fast a
reaction takes place
First Order Reactions
o Set [ A] = 1 [ A ]
t
0
2
o Substituting into
o Gives ln
[ A ]0
1
[ A ]0
2
ln
[ A ]0
[ A]t
= kt 1
= kt
2
o Canceling gives ln 2 = kt½
o Rearranging gives
t1 =
2
ln2
k1
=
0.693
k1
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
64
Integrated
Rate Laws
Half Life (t1/2) for First Order Reactions
Observe:
1. t½ is independent of [A]o
o For given reaction (and T)
o Takes same time for concentration to fall from
o 2 M to 1 M as from
o 5.0  10–3 M to 2.5  10–3 M
2. k1 has units (time)–1, so t½ has units (time)
o t½ called half-life
o Time for ½ of sample to decay
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
65
Integrated
Rate Laws
Half Life (t1/2)
Does this mean that all of sample is gone in
two half-lives (2 × t½)?
No!
o In 1st t½, it goes to ½[A]o
o In 2nd t½, it goes to ½(½[A]o) = ¼[A]o
o In 3rd t½, it goes to ½(¼[A]o) = ⅛[A]o
o In nth t½, it goes to [A]o/2n
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
66
Integrated
Rate Laws
Half Life (t1/2)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
67
Integrated
Rate Laws
Half Life (t1/2): First Order Example
131I
is used as a metabolic tracer in
hospitals. It has a half-life, t½ = 8.07 days.
How long before the activity falls to 1% of
the initial value?
N = N oe -kt
ln
N
-t ln2
= -kt =
No
t
1
t =-
N
t 1 ln
2
No
ln 2
2
æ 1 ö
(8.07 days)ln çç
÷÷
è 100 ø
== 53.6 days
ln 2
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
68
Group
Problem
The radioactive decay of a new atom occurs so that after 21
days, the original amount is reduced to 33%. What is the
rate constant for the reaction in s–1?
æA ö
ln çç 0 ÷÷ = kt
èAø
k = 0.0528 day–1
k = 6.11 × 10–7 s–1
100
ln(
) = k (21 day)
33
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
69
Group
Problem
The half-life of I-132 is 2.295 h. What
percentage remains after 24 hours?
ln 2
k
= t1/2
ln 2
k=
2.295 h
0.302 h–1 = k
 Ao 
ln
  kt
 A 
æA ö
-1
o ÷
ç
ln ç ÷ = 0.302 h ´ 24 h = 7.248
èAø
A = 0.0711% Ao
A A o e kt  Ao e 7.248
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
70
Integrated
Rate Laws
Half Life (t1/2): Carbon-14 Dating
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
71
Integrated
Rate Laws
Half Life (t1/2): Second Order Reactions
How long before [A] = ½[A]o?
t 1/2
1
=
k [ A]0
o t½, depends on [A]o
o t½, not useful quantity for a second order reaction
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
72
Group
Problem
The rate constant for the second order reaction 2A → B is 5.3
× 10–5 M–1 s–1. What is the original amount present if, after 2
hours, there is 0.35 M available?
1
[A]
-
1
[A 0]
= kt
1
1
5.3 ´ 10 - 5
=
´ 7200 s
[0.35] [A 0]
M s
A0=0.40 M
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
73
Group
Problem
Add better rate law problem
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
74
CHAPTER 14 Chemical Kinetics
Collision Theory
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
75
Collision
Theory
Reaction Rates
Collision Theory
As the concentration of reactants increase
o The number of collisions increases
o Reaction rate increases
As temperature increases
o Molecular speed increases
o Higher proportion of collisions with
enough force (energy)
o There are more collisions per second
o Reaction rate increases
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
76
Collision
Theory
Reaction Rates
Rate of reaction proportional to number of effective
collisions/sec among reactant molecules
Effective collision
o One that gives rise to product
e.g. At room temperature and pressure
o H2 and I2 molecules undergoing 1010 collisions/sec
o Yet reaction takes a long time
o Not all collisions lead to reaction
Only very small percentage of all collisions lead to
net change
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
77
Collision
Theory
Molecular Orientation
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
78
Collision
Theory
Temperature
As T increases
o More molecules have Ea
o So more molecules undergo reaction
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
79
Collision
Theory
Activation Energy (Ea)
Molecules must possess certain amount of kinetic energy
(KE) in order to react
Activation Energy, Ea = Minimum KE needed for
reaction to occur
o Get energy from collision with other molecules
o If molecules move too slowly, too little KE, they just
bounce off each other
o Without this minimum amount, reaction will not occur
even when correctly oriented
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
80
Group
Problem
Summerize with a pretty picture
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
81
CHAPTER 14 Chemical Kinetics
Transition State
Theory
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
82
Transition
State
Molecular Basis of Transition State Theory
KE
KE
KE decreasing as PE increases
Is the combined KE of
both molecules
enough to overcome
Activation Energy
PE
KE
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
KE
83
Potential Energy
Transition
State
Molecular Basis of Transition State Theory
Activation energy (Ea)
= hill or barrier
between reactants
and products
Heat of reaction (H)
= difference in PE between
products and reactants
Hreaction = Hproducts – Hreactants
Products
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
84
Potential Energy
Transition
State
Exothermic Reactions
Exothermic reaction
• Products lower PE
than reactants
Exothermic
Reaction
H = –
Products
Reaction Coordinate (progress of reaction)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
85
Transition
State
Exothermic Reactions
o Hreaction < 0 (negative)
o Decrease in PE of system
o Appears as increase in KE
o So the temperature of the system increases
o Reaction gives off heat
o Can’t say anything about Ea from size of H
o Ea could be high and reaction slow even if
Hrxn large and negative
o Ea could be low and reaction rapid
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
86
Transition
State
Endothermic Reactions
Endothermic
Reaction
H = +
Hreaction = Hproducts – Hreactants
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
87
Transition
State
Endothermic Reactions
o Hreaction > 0 (positive)
o Increase in PE
o Appears as decrease in KE
o So temperature of the system decreases
o Have to add E to get reaction to go
o Ea  Hrxn as Ea includes Hrxn
o If Hrxn large and positive
o Ea must be high
o Reaction very slow
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
88
Transition
State
Activated Complex
o Arrangement of atoms at top of activation barrier
o Brief moment during successful collision when
o bond to be broken is partially broken and
o bond to be formed is partially formed
Example
H3C
N
C
H3C
N
H3C
C
N
C
Transition State
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
89
Transition
State
Example
Develop example that is not NO2Cl + Cl
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
90
Group
Problem
Draw the transition state complex, or the activated
complex for the following reaction:
CH3CH2O-
+ H3O+

CH3CH2OH + H2O
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
91
CHAPTER 14 Chemical Kinetics
Activation
Energies
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
92
Ea
Arrhenius Equation
The rate constant is dependent on Temperature, which
allows us to calculate Activation Energy, Ea
Arrhenius Equation:
Equation expressing temperature-dependence of k
k = Ae
-E a /RT
o A = Frequency factor has same units as k
o R = gas constant in energy units
= 8.314 J mol–1 K–1
o Ea = Activation Energy—has units of J/mol
o T = Temperature in K
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
93
Ea
Calculating Activation Energy
• Method 1. Graphically
• Take natural logarithm of both sides
• Rearranging
æE öæ 1 ö
ln k = ln A - çç a ÷÷ çç ÷÷
è R ø èT ø
• Equation for a line
• y = b + mx
Arrhenius Plot
• Plot ln k (y axis) vs. 1/T (x axis)  yield a
straight line
• Slope = -Ea/R
• Intercept = A
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
94
Arrhenius Equation: Graphing Example
Ea
k (M/s)
Given the following data,
predict k at 75 ˚C using
the graphical approach
Ea 1
ln k = - ´ + ln A
R T
ln (k) = –36.025/T – 6.908
T, ˚C
T, K
0.000886 25
0.000894 50
0.000908 100
298
348
398
0.000918 150
448
?
75
348
ln (k) = –36.025/(348) – 6.908 = – 7.011
k =e
-7.012
-4
= 9.01 ´ 10
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
95
Ea
Arrhenius Equation: Graphing Example
-6.99
ln k
-7.00
-7.01
-7.02
y = –36.025x – 6.908
-7.03
0.0022 0.0024 0.0026 0.0028 0.003 0.0032 0.0034
1/T (K–1)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
96
Ea
Arrhenius Equation
Sometimes a graph is not needed
o Only have two k s at two Ts
Here use van't Hoff Equation derived from
Arrhenius equation:
æ k ö -E
ln çç 2 ÷÷ = a
R
è k1 ø
æ1 1ö
ç - ÷
çT T ÷
è 2
1ø
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
97
Ea
Arrhenius Equation: Ex Vant Hoff Equation
CH4 + 2 S2  CS2 + 2 H2S
k (L/mol s)
T (˚C)
T (K)
1.1 = k1
550
823 = T1
6.4 = k2
625
898 = T2
æ 6.4 ö
æ 1
–E a
1 ö
çç
÷÷
ln çç ÷÷ =
–
è 1.1 ø 8.3145 J/K mol è 898 K 823 K ø
æ 6.4 ö
- 8.314 J/K mol ln çç ÷÷
è 1.1 ø
Ea =
= 1.4 ´ 105 J/mol
æ 1
1 ö
–
çç
÷÷
è 898 K 823 K ø
(
)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
98
Group
Problem
Given that k at 25 ˚C is 4.61 × 10–1 M/s and that at 50 ˚C it is
4.64 × 10–1 M/s, what is the activation energy for the reaction?
k2
-Ea æ 1
1ö
ln ( ) =
çç - ÷÷
k1
R èT 2 T 1 ø
æ 4.64 ´ 10-1 M/s ö
æ 1
-Ea
1 ö
÷
ln çç
÷ = 8.314J/(mol K) çç 323 K - 298 K ÷÷
-1
è
ø
è 4.61 ´ 10 M/s ø
Ea = 208 J/mol
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
99
Group
Problem
A reaction has an activation energy of 40 kJ/mol. What happens
to the rate if you increase the temperature from 70˚C to 80 ˚C?
A. Rate increases approximately 1.5 times
B. Rate increases approximately 5000 times
C. Rate does not increase
D.
Rate increases approximately 3 times
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
100
Group
Problem
A reaction has an activation energy of 40 kJ/mol. What happens
to the rate if you increase the temperature from 70˚C to 80 ˚C?
Rate is proportional to the rate constant
k2
k1
=
e
e
æ
ö
ç
÷
40000
J
÷
-ç
ç
÷
J
´ (80+273) K ÷
ç 8.314
è
ø
mol K
æ
ö
ç
÷
40000
J
÷
-ç
ç
÷
J
´ (70+273) K ÷
ç 8.314
è
ø
mol K
= 1.49
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
101
CHAPTER 14 Chemical Kinetics
Mechanisms of
Reactions
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
102
Mechanisms
Overall vs Individual Steps
Sometimes rate law has simple form
– N2O5  NO2 + NO3
Rate = -
d [N2O5 ]
dt
= k 1 [N2O5 ]
– NO2 + NO3  N2O5
d [NO2 ]
Rate = = k 2 [NO2 ][NO3 ]
dt
But others are complex
– H2 + Br2  2 HBr
Rate = -
d [H2 ]
dt
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
=
k [H2 ][Br2 ]1/2
1+
k ¢[HBr]
[Br ]103
Mechanisms
Overall vs Individual Steps
Some reactions occur in a single
step, as written
Others involve a sequence of steps
o Reaction Mechanism
o Entire sequence of steps
o Elementary Process
o Each individual step in mechanism
o Single step that occurs as written
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
104
Mechanisms
Overall vs Individual Steps
o Exponents in rate law for elementary process are equal to
coefficients of reactants in balanced chemical equation for that
elementary process
o Rate laws for elementary processes are directly related to
stoichiometry
o Number of molecules that participate in elementary process
defines molecularity of step
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
105
Mechanisms
Unimolecular Process
o Only one molecule as reactant
o H3C—NC  H3C—CN
o Rate = k[CH3NC]
o 1st order overall
o As number of molecules increases, number that rearrange
in given time interval increases proportionally
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
106
Mechanisms
Bimolecular Process
o Elementary step with two reactants
o NO(g) + O3(g)  NO2(g) + O2(g)
o Rate = k[NO][O3]
o 2nd order overall
o From collision theory:
o If [A] doubles, number of collisions between A and B will
double
o If [B] doubles, number of collisions between A and B will
double
o Thus, process is 1st order in A, 1st order in B, and 2nd order
overall
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
107
Mechanisms
Termolecular Process
o Elementary reaction with three molecules
o Extremely rare
o Why?
o Very low probability that three molecules
will collide simultaneously
o 3rd order overall
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
108
Mechanisms
Elementary Processes
Molecularity Elementary Step
Rate Law
Unimolecular
A  products
Rate = k[A]
Bimolecular
Bimolecular
A + A  products
A + B  products
Rate = k[A]2
Rate = k[A][B]
Significance of elementary steps:
o If we know that reaction is elementary step
o Then we know its rate law
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
109
Mechanisms
Multi-step Mechanisms
o Contains two or more steps to yield net reaction
o Elementary processes in multi-step mechanism must always
add up to give chemical equation of overall process
o Any mechanism we propose must be consistent with
experimentally observed rate law
o Intermediate = species which are formed in one step and
used up in subsequent steps
o Species which are neither reactant nor product in overall
reaction
o Mechanisms may involve one or more intermediates
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
110
Mechanisms
Example
The net reaction is:
NO2(g) + CO(g)  NO(g) + CO2(g)
The proposed mechanism is:
NO2(g) + NO2(g)  NO3(g) + NO(g)
NO3(g) + CO(g)  NO2(g) + CO2(g)
1
2NO2(g) + NO3(g) + CO(g)  NO2(g) + NO3(g) + NO(g) + CO2(g)
or
NO2(g) + CO(g)  NO(g) + CO2(g)
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
111
Mechanisms
Rate Determining Step
o If process follows sequence of steps, slow step determines
rate = rate determining step.
o Think of an assembly line
o Fast earlier steps may cause intermediates to pile up
o Fast later steps may have to wait for slower initial steps
o Rate-determining step governs rate law for overall
reaction
o Can only measure rate up to rate determining step
Jesperson, Brady, Hyslop. Chemistry: The
Molecular Nature of Matter, 6E
112
Mechanisms
Example: Rate Determining Step
(CH3)3CCl(aq) + OH–(aq)  (CH3)3COH(aq) + Cl–(aq)
chlorotrimethylmethane
trimethylmethanol
o Observed rate = k[(CH3)3CCl]
o If reaction was elementary
o Rate would depend on both reactants
o Frequency of collisions depends on both concentrations
o Mechanism is more complex than single step
o What is mechanism?
o Evidence that it is a two step process
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Mechanisms
Rate Determining Step as Initial Step
Step 1: (CH3)3CCl(aq)  (CH3)3C+(aq) + Cl–(aq) (slow)
Step 2: (CH3)3C+(aq) + OH–(aq)  (CH3)3COH(aq) (fast)
o Two steps each at different rates
o Each step in multiple step mechanism is elementary process,
so
o Has its own rate constant and its own rate law
o Hence only for each step can we write rate law directly
o Observed rate law says that step 1 is very slow compared to
step 2
o In this case step 1 is rate determining
o Overall rate = k1[(CH3)3CCl]
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Mechanisms
Mechanisms with Fast Initial Step
1st step involves fast, reversible reaction
Ex. Decomposition of ozone (No catalysts)
Net reaction: 2O3(g)  3O2(g)
Observed Rate 
k [O 3 ]2
[O2 ]
Proposed mechanism:
O3(g)  O2(g) + O(g)
(fast)
O(g) + O3(g)  2O2(g) (slow)
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Mechanisms
Is the Mechanism Rate Law Consistent?
o Rate of formation of O2 = Rate of reaction 2
= k2[O][O3]
o But O is intermediate
o Need rate law in terms of reactants and products
o and possibly catalysts
o Rate (forward) = kf[O3]
o Rate (reverse) = kr[O2][O]
o When step 1 comes to equilibrium
o Rate (forward) = Rate (reverse)
o kf[O3] = kr[O2][O]
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Is the Mechanism Rate Law Consistent?
Mechanisms
o Solving this for intermediate O gives:
[O] =
k f [O3 ]
k r [O2 ]
o Substitution into rate law for step 2 gives:
Observed Rate =
o Rate of reaction 2 = k2[O][O3] =
o where
k obs =
k 2k f
k [O3 ]2
[O2 ]
k 2k f [O3 ]2
k r [O2 ]
kr
o This is observed rate law
o Yes, mechanism consistent
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117
Group
Problem
The reaction mechanism that has been proposed for the
decomposition of H2O2 is
1. H2O2 + I– → H2O + IO– (slow)
2. H2O2 + IO– → H2O + O2 + I– (fast)
What is the expected rate law?
First step is slow so the rate determining step defines
the rate law
rate=k [H2O2][I–]
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Molecular Nature of Matter, 6E
118
Group
Problem
The reaction: A + 3B → D + F was studied and the
following mechanism was finally determined:
1. A + B  C
(fast)
2. C + B → D + E
(slow)
3. E + B → F
(very fast)
What is the expected rate law?
Rate Step 2=k2[C][B]
Rate = kobs[A][B]2
Rate forward = kf[A][B]
Rate reverse = kr[C]
kf[A][B] = kr[C]
[C]= kf[A][B]/kr
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119
Terminal Reaction for Superoxide Radical
Catalyst
H
N
O
OH2
OH
NHis
OAsp FeIII
O2 -
NHis
OH
Asp 156 O
O2-
Fe
N
His 73
NH
H+
NH
OH2
NHis
OAsp FeIII
His 160
His 26 N
O2
NHis
N
NHis
OAsp FeII
NHis
NHis
NHis
NHis
OH2
HOOH
H+
2H+
OAsp FeII
NHis
NHis
O2-
O2NHis
* O2- hydrogen bonds to residues in secondary
coordination sphere, positioning it near Fe(II),
Miller, Anne-Frances. “Fe Superoxide Dismutase” Handbook of Metalloproteins. John Wiley & Sons, Ltd, Chinchester, 2001
Rodrigues, J. V; Abreu, I. A.; Cabelli, D; Teixeira, M. Biochemistry 2006, 45, 9266-9278.
120
CHAPTER 14 Chemical Kinetics
Catalysts
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Molecular Nature of Matter, 6E
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Catalyst
Definition
o Substance that changes rate of chemical reaction
without itself being used up
o Speeds up reaction, but not consumed by reaction
o Appears in mechanism, but not in overall reaction
o Does not undergo permanent chemical change
o Regenerated at end of reaction mechanism
o May appear in rate law
o May be heterogeneous or homogeneous
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Molecular Nature of Matter, 6E
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Catalyst
Activation Energy
o By providing alternate
mechanism
o One with lower Ea
o Because Ea lower, more
reactants and collisions
have minimum KE, so
reaction proceeds faster
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Catalyst
Activation Energy
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Catalyst
Homogeneous Catalyst
• Same phase as reactants
Consider : S(g) + O2(g) + H2O(g)  H2SO4(g)
S(g) + O2(g)  SO2(g)
NO2(g) + SO2(g)  NO(g) + SO3(g) Catalytic pathway
SO3(g) + H2O(g)  H2SO4(g)
NO(g) + ½O2(g)  NO2(g)
Regeneration of catalyst

Net: S(g) + O2(g) + H2O(g)  H2SO4(g)
• What is Catalyst?
– Reactant (used up) in early step
– Product (regenerated) in later step
• Which are Intermediates?
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Catalyst
o
o
o
o
Heterogeneous Catalyst
Exists in separate phase from reactants
Usually a solid
Many industrial catalysts are heterogeneous
Reaction takes place on solid catalyst
Ex. 3H2(g) + N2(g)  2NH3(g)
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Molecular Nature of Matter, 6E
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Catalyst
H2 and N2
approach
Fe catalyst
H2 and N2
bind to Fe
& bonds
break
Heterogeneous Catalyst
N—H
bonds
forming
N—H
bonds
forming
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Molecular Nature of Matter, 6E
NH3
formation
complete
NH3
dissociates
127
Catalyst
Methane Oxidation to Methanol
CH4  CH3OH
o C-H has a high bond strength (~ 410
kJ/mol)
o Thermodynamically favorable to oxidize
CH4 to CO2
o Selective catalysts can stop the
oxidation at methanol
GOAL: Develop a selective and robust catalyst
that produces high yields of methanol at low
temperatures and pressures
o Syngas synthesis of methanol (CO + H2):
o High Pressure with ZnO/Cr2O3
catalyst
o Cu-Zeolite catalyst
o Shilov cycle with a platinum catalyst
Catalyst
Methane Monooxygenase
CH4  CH3OH
o Oxidoreductase Enzyme found in
methanotrophic bacteria that help cycle
carbon in anarobic sediments.
o Bioinorganic chemistry tries to replicate
the highly specialized chemistry found
in nature using smaller molecules that
can be synthesized in a lab.
Catalyst
Splitting Water
Requirments: Very Endothermic
o Need a minimum of 1.23 V to split water
o Kinetically infrared light could do this,
but the reaction is very slow
o The potential really needs to be at
least 3.0 V to utilize the full spectrum
of light
Catalyst
Photosystem II
PQ + H2O --> PQH2 + O2 (g)
The overall reaction of Photosystem II is
the oxidation of water and the reduction
of plastoquinone.
Catalyst
Splitting Water
Increase efficiency and decrease activation energy of electrolysis of water with a
catalyst that will work at room temperature: Co3+ HPO4-