Chemical Kinetics
Spontaneity: the inherent tendency for the process to occur,
does not mean fast.
2 H2 + O2  2 H2O H2 + Cl2  2 HCl
N2 + 3 H2  2 NH3
Highly to occur from a thermodynamic, but so slow to detectable.
Chemical kinetics concerns reaction rates.
Reaction mechanism: the series of steps (a reaction takes place)
To facilitate the reaction
15.1 Reaction Rates
2 NO2(g)  2 NO(g) + O2(g)
The reaction rate: the change in the concentration of a reactant or
product per unit time
rate of reaction =
[ A] [ NO2 ] 0.0079  0.01
 4.2  10 5


50  0
t
t
2 NO2(g)  2 NO(g) + O2(g)
8.6  10-6
4.3  10-6
1
The instantaneous rate: the slope of a line tangent to the curve
2 NO2(g)  2 NO(g) + O2(g)
rate of reaction = 
1 [ NO2 ] 1 [ NO ] [O2 ]


t
2 t
2 t
15.2 Rate Laws : An Introduction
[NO2] depends on the different in the rates of the forward
and reverse reactions
A reaction at a point soon after the reactions are mixed, and the
reverse reaction makes a negligible contribution.
Rate law:
rate = k[NO2]n
k: the rate constant
n: the order of reactant
1. The concentration of the products does not appear.
2. The value of n must be determined by experiment.
Rate = 
Rate =
[ NO2 ]
d [ NO2 ]
= k[NO2]n

t
dt
dd[[O
]
O 2] = k[NO ]n
2
dt
dt
2
Rate = 2  Rate 
2 NO2(g)  2 NO(g) + O2(g)
k[NO2]n = 2 k[NO2]n
k = 2  k
The types of rate laws:
1. The differential rate law: how the rate depends on concentrations
The integrated rate law: how the concentrations depend on time
2. The reverse reaction is unimportant, concentration of reactants
3. The differential rate law  The integrated rate law
Rate/concentration  concentration/time
4. Which type of rate law is determined experimentally.
5. The individual steps involved in the reaction from the rate law.
2
15.3 Determining the Form of the Rate Laws
2 N2O5(g)  4 NO2(g) + O2(g)
[N2O5]
rate
0.90
5.4 10-4
0.45
2.7 10-4
When [N2O5]
is halved, the
rate is halved.
 First order
Rate =  d[N 2O5 ] = k[N2O5]1
dt
Method of Initial Rates:
the instantaneous rate vs the initial concentration of reactants
NH4 + NO2  N 2 + 2 H2O
+
-

rate  
d [ NH 4 ]


 k [ NH 4 ]n [ NO2 ]m
dt
rate2 2.70  107
k (0.1) n (0.01) m
2



 ( 2.0) m
rate1 1.35  107
k (0.1) n (0.005) m
m=1
rate3 5.40  107
k (0.2) n (0.01) m
2



 ( 2.0) n
rate2 2.70  107
k (0.1) n (0.01) m
n=1
2.70  107  k (0.1)( 0.01)
The overall reaction order
k  2.70  104 L  mol 1  s 1
=n+m=2
3
Example 15.1
BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O
Rate =k[BrO3-]n[Br-]m[H+]p
rate2 1.6  103
(0.2) n
2



 ( 2.0) n
rate1 8.0  104
(0.1) n
n=1
rate3 3.2  103
(0.2) m
2



 ( 2.0) m
rate2 1.6  103
(0.1) m
m=1
rate4 3.2  103
(0.2) p
Rate =k[BrO3-][Br-][H+]2
4



 ( 2.0) p p = 2
4
p
rate1 8.0  10
(0.1)
8.0  104  k (0.10)(0.10)(0.10) 2  k (1.0  104 )
k  8.00 L3  mol 3  s 1
15.4 The Integrated Rate Law
First-Order rate laws:
2 N2O5(g)  4 NO2(g) + O2(g)
O55]]
dd[N
[ N 22O
Rate = 
= k[N2O5]
dt
dt
d [ N 2O5 ]
  k [ N 2O5 ]
dt
d [ N 2O5 ]

 k  dt
[ N 2O5 ]

Rate = 
d [ A]
= k[A]
dt
ln[ A]   kt  ln[ A]0
 [ A] 
   kt
ln[ A]  ln[ A]0  ln 

[
]
A
 0
 [ A] 
ln  0   kt
 [ A] 
For a first-order reaction, a plot of ln[A] versus t is a straight line.
ln[ N 2O5 ]  kt  ln[ N 2O5 ]0
y = mx + b; y = ln[A], x = t, slope m = -k, intercept b = ln[A]0
4
2 N2O5(g)  4 NO2(g) + O2(g)
Example 15.2
[N2O5]
Time
ln[N2O5]
0.1000
0
-2.303
0.0707
50
-2.649
0.0500
100
-2.996
150
?
0.0250
200
-3.689
0.0125
300
-4.382
0.00625
400
-5.075
?
ln[N2O5 ]  kt  ln[N2O5 ]0
 2.996 k(100)  (2.303)
3
k  6.9310
-3.3425
Example 15.3
ln[N2O5 ]  kt  ln[N2O5 ]0
 (6.93103 )(150)  (2.303)  3.3425
ln[N2O5 ]  3.3425
[ N2O5 ]  e3.3425  0.0353
Half-Life of a First-Order Reaction
Half-life, t1/2: The time required for a reactant to reach half of its
original concentration
[N2O5]
Time
0.1000
0
0.0707
50
0.0500
100
0.0250
200
0.0125
300
0.00625
400
ln(
0.693
[ A]0
[ A]0
)  kt  ln(
)  kt1 / 2  ln 2  0.693 t1/ 2 
[ A]
[ A]0 / 2
k
5
Example 15.4
A certain first-order reaction has a half-life of 20.0 min.
a. Calculate the rate constant for this reaction.
b. How much time is required for this reaction to be 75% complete?
0.693 0.693
k

 3.47  10 2 min 1
t1 / 2
20
[ A]0
[ A]
)  ln 4  3.47  10 2 t ; t  40
ln( 0 )  kt  ln(
0.25[ A]0
[ A]
Second-order Rate Laws:
a A  products
rate 
The integrated second-order rate law:
d[A]
 k[A]2
dt
1
1
 kt 
[ A]
[ A]0
A plot of 1/[A] versus t is a straight line, with the slope equal to k.
1
1
1
1
 kt1 / 2
 kt1 / 2 
t1/ 2 
[ A]0 2
[ A]0
[ A]0
k[ A]0
Example 15.5
2 C4H6  C8H12
[C4H6]
time 1/[C4H6] ln[C4H6]
0.01000
0
100
-4.605
0.00625
1000
160
-5.075
0.00500
1630
0.00476
1800
210
-5.348
0.00370
2800
270
-5.599
0.00313
3600
319
-5.767
0.00270
4400
370
-5.915
0.00250
4890
0.00241
5200
415
-6.028
0.00208
6200
481
-6.175
ln[C4H6] vs. time is not a straight line  not a first-order reaction
1/[C4H6] vs. time is a straight line  a second-order reaction
k = slope = 6.14  10-2 t1 / 2 
1
1

 1.63  103
2
k[ A]0 (6.14  10 )(0.01)
6
Zero-order Rate Laws: the rate is constant. Rate = k
The integrated rate law for zero-order reaction: [A] = -kt + [A]0
2 N2O  2 N2 + O2
Zero-order
The rate is a constant.
First-order
The platinum surface is
completely covered by
N2O molecules.
[ A]0
 kt1 / 2  [ A]0
2
[ A]0
kt1 / 2 
2
[ A]0
t1 / 2 
2k
Integrated rate law for reactions with more than one reactant
Ex. BrO3- + 5 Br- + 6 H+  3 Br2 + 3 H2O
Rate = k [BrO3-][Br-][H+]2
A pseudo-first-order rate law:
[BrO3-]0 = 1.010-3 M << [Br-]0 = 1.0 M, [H+]0 = 1.0 M
[Br-] and [H+] remain approximately constant.
[Br-] = [Br-]0 ; [H+] = [H+]0
Rate = k [BrO3-][Br-][H+]2 = k [BrO3-][Br-]0[H+]02
= k [BrO3-]
k
k
2
[ Br  ]0 [ H  ]0
pseudo-first-order
k = k [Br-]0[H+]02
7
15.5 Rate Law: A Summary
1. Only the forward reaction is important.
2. There are two types of rate laws.
t1 / 2 
0.693
k
3. Which type of data can be collected conveniently and accurately.
4. The method of initial rates determine the differential rate laws.
A comparison of the initial rates and the initial concentration
5. The integrated rate law: plot of concentration versus time
Straight line : [A] vs time  zero-order
slope = - k
ln[A] vs time  first-order
slope = - k
1/[A] vs time  second-order
slope = k
6. Integrated rate law for reactions with more than one reactant
A pseudo rate law: [A]0 << [B]0, [C]0
Rate = k [A]n[B]m[C]p
Rate = k [A]n
n determined by straight line plot
k = k [B]0m[C]0p
8
15.6 Reaction Mechanisms
Most chemical reactions occur by a series of steps called the
reaction mechanism.
NO2 + CO  NO + CO2
rate = k [NO2]2
Elementary step: whose rate law can be written from its molecularity
NO2 + NO2 NO3+ NO
rate = k1 [NO2]2 rate-determining step
NO3 + CO  NO2 + CO2
rate = k2 [NO3][CO]
bimolecular
9
1. The sum of the elementary steps give the overall balanced reaction.
2. The mechanism agree with the experimentally determined rate law.
NO3 is an intermediate.
NO2 + CO  NO + CO2
rate = k [NO2]2
Example 15.6
NO2 + F2 NO2F+ F
rate = k1 [NO2][F2] rate-determining step
F + NO2  NO2F
rate = k2 [NO2][F]
1. The sum of the elementary steps give the overall balanced reaction.
2. The mechanism agree with the experimentally determined rate law.
2 NO2 + F2  2 NO2F
rate = k [NO2][F2]
The mechanism is acceptable.
Mechanisms with Fast Forward and Reverse First Steps
2 O3  3 O2
rate  k
[O3 ]2
[O2 ]
k1
O3  O2 + O rate  k1[O3 ]  k1[O2 ][O] Fast equilibrium
k-1
k [O ]
[O]  1 3
k1[O2 ]
k2
rate  k 2 [O][O3 ] Rate determining step
O + O3  2 O2
Assume that the rates of the forward and reverse reaction in the first
step are equal.
[O3 ]2
k1[O3 ] k1k 2 [O3 ]2

k
rate  k 2 [O][O3 ]  k 2 [O3 ]
[O2 ]
k 1[O2 ] k 1 [O2 ]
k
k1k 2
k 1
It is an acceptable mechanism for the decomposition
of ozone to oxygen
10
Example 15.7
rate  k[Cl2 ]1/ 2 [CHCl3 ]
Cl2 + CHCl3  HCl + CCl4
k1
Mechanism:
Cl2  2 Cl
k-1
fast equilibrium
1/ 2
 k [Cl ] 
rate  k1[Cl2 ]  k 1[Cl ]  [Cl ]   1 2 
 k 1 
k2
slow
Cl + CHCl3  HCl + CCl3
2
k3
CCl3 + Cl  CCl4
fast
 k [Cl ] 
rate  k 2 [Cl ][CHCl3 ]  k 2 [CHCl3 ] 1 2 
 k 1 
1/ 2
1/ 2
k 
 k 2  1  [Cl2 ]1/ 2 [CHCl3 ]  k[Cl2 ]1/ 2 [CHCl3 ]
 k 1 
k 
k  k2  1 
 k1 
1/ 2
15.7 The Steady-State Approximation
The concentration of any intermediate remains constant.
2 NO + H2  N2O + H2O
k1
2 NO  N2O2
k-1
k2
N2O2 + H2 
N2O + H2O
Rate of production of N2O2
rate  k1[NO]2
Rate of consumption of N2O2
rate  k1[ N 2O2 ]  k2 [ N 2O2 ][ H 2 ]
The steady-state condition:
k1[ NO ]2  k1[ N 2O2 ]  k2 [ N 2O2 ][ H 2 ]
k1[ NO ]2  (k1  k2 [ H 2 ])[ N 2O2 ]
k1[ NO ]2
[ N 2O2 ] 
k1  k2 [ H 2 ]
11
2 NO + H2  N2O + H2O
d[H 2 ]
 k2 [ N 2O2 ][ H 2 ]
dt
k1[ NO ]2
 k2 [ H 2 ]
k1  k2 [ H 2 ]
Rate of reaction = 

k1k2 [ H 2 ][ NO]2
k1  k2 [ H 2 ]
[H2] is high
k2[H2] >> k-1
[H2] is low
k-1 >> k2[H2]
rate 
k1k2 [ H 2 ][ NO]2
 k1[ NO]2
k2 [ H 2 ]
k1k2
[ H 2 ][ NO ]2  k[ H 2 ][ NO ]2
k1
Summary:
1. Write the proposed mechanism (the elementary steps)
2. Construct a steady-state expression for each intermediate.
3. From the steady-state approximation for each intermediate,
solve for [I1]
4. Construct the rate law for the overall reaction in terms of one of
the reacts or products
5. Use the expression form Step 3 for [I1] to substitute for the
concentration of intermediated found in the rate law for Step 4.
An overall rate law contains only the reactant and/or product
concentration.
12
15.8 A Model for Chemical Kinetics
a A + b B  products
Rate = k [A]n[B]m
the order of each reactant depends on the detailed reaction mechanism
All rate constants show an exponential
increase with absolute temperature.
The collision model:
Molecules must collide to react.
[reactant] increase
 increase the collision of reactant
 increase the reaction rate
Temperature increase
 increase molecular velocities
 increase the frequency of collisions
25
1880s Savante Arrhenius: the activation energy
a threshold energy
2 BrNO   2 NO + Br2
Two Br-N broken, Br-Br formed the activated complex, or
transition state
13
The rate depends on Ea.
Neff = N e-Ea/RT
Rate  Neff  e-Ea/RT
The observed reaction rate is considerably smaller than the rate of
collisions with enough energy to surmount the barrier.
The molecular orientations
O
enough energy
O
right orientations
X
k  zpe  Ea / RT  Ae Ea / RT
z: the collision frequency
E 1
ln( k )  ln A  a ( )
R T
p: the steric factor
The Arrhenius equation:
A: pre-exponential factor
frequency factor
A plot of lnk vs. 1/T gives a straight line; slope = - Ea / R.
14
2 N2O5(g)  4 NO2(g) + O2(g)
T(oC) T(K)
1/T
(10-3)
k
(10-5)
ln(k)
20
293
3.41
2.0
-10.82
30
303
3.30
7.3
-9.53
40
313
3.19
27
-8.22
50
323
3.10
91
-7.00
60
333
3.00
290
-5.84
Ea
R
Ea   R ( slope)
slope  
 (8.3145)  (1.2 10 4 )
 1.0 105
ln( k1 )  ln A 
Ea 1
( )
R T1
ln( k2 )  ln A 
Ea 1
( )
R T2
k2
E 1 1
) a(  )
k1
R T1 T2
15.9 Catalysis: speeds up a reaction without being consumed itself.
ln(
Enzymes increase the rates of
reactions at body temperature.
Carbonic anhydrase: 600000 CO2/sec
CO2 + H2O  HCO3- + H+
Provide a new pathway
 A lower activation energy
15
A pathway with lower activation energy, a much larger fraction of
collisions is effective at a given temperature, and the reaction rate is
increased.
A homogeneous catalyst is present in the same phase as the reacting
molecules.
A heterogeneous catalyst exists in different phase, usually as a solid.
Adsorption: the collection of on substance on the surface
Absorption: the penetration of one substance into another
Heterogeneous: Haber process, Fe, K and Ca
N-N and H-H bonds are weaken when the H2
and N2 are bound to iron atoms on the surface of
the metal.
Hydrogenation of unsaturated hydrocarbons
Pt, Pd, Ni
H
H
C
H
C
H
+ H2
H
H
H
C
C
H
H
H
1. Adsorption and activation of the reactants
2. Migration of the adsorbed reactants.
3. Reaction
4. Escape, or desorption, of the products
16
The oxidation of SO2 to SO3.
negative side: acid rain, SO2 + O2  SO3 + H2O  H2SO4
positive side: the manufacture of H2SO4, Pt and VO2
TiO2 is a photocatalyst, it required light for its catalyst activity.
It also required the present of oxygen and water.
The mechanism for the catalytic behavior of TiO2 results from it s
behavior as a semiconductor.
TiO2
hole + electron
OH + O2-
Free radicals, with unpaired electrons, reactive radicals can react
with and destroy virtually all organic materials.
Catalyze nitrogen oxides to form nitric acid.
Radicals are lethal to microorganisms, bactericide.
Antiknocking: tetraethyl lead
NOx  N2
CH  CO2 + H2O
CO  CO2
SO2  SO3
34
17
The atmosphere closest to earth: troposphere
NO catalyzes ozone production
N2 + O2  2 NO
2 NO + O2  2 NO2
NO2  NO + O
O + O2  O3
3 O2  2 O3 : a powerful oxidizing agent
The upper atmosphere : NO the depletion ozone
NO + O3  NO2 + O2
O + NO2  NO + O2
O + O3  2 O2
Chlorofluorocarbons (CFC), refrigeration and air conditioning
Freons (1996, banned)
nontoxic and chemically inert
Ultraviolet light cause them to decompose, producing chlorine atoms
that catalyze the decomposition of the ozone in the stratosphere.
Freon-11 CCl3F1
Freon-12 CCl2F2  CClF2 + Cl
Cl + O3  ClO + O2
O + ClO  Cl + O2
O + O3  2 O2
18
Enzymes: Nature’s catalysts
Homogeneous
Protein  amino acid
Carboxypeptidase-A
A zinc-containing protein
Carboxypeptidase-A: A zinc-containing protein
38
19