Chapter 12:
Chemical Kinetics


Spontaneity – inherent tendency for a
reaction to occur; does not mean speed
Reactions have different rates:



2 H2 + O2  2 H2O SLOW
Explosions FAST
To be useful, reactions must occur at a
reasonable rate.
1
Chemical Kinetics
The area of chemistry concerned
with the speed (rate) of a reaction
and reaction mechanisms.

2
Main Goal of Kinetics

Understand the steps by which a
reaction takes place




What factors determine how fast food
spoils?
Design of fast setting material for dental
fillings
What affects the rate that steel rusts?
What affects the rate the fuel burns in
cars?
3
Reaction Rate Factors

Physical state of reactants

Concentrations of reactants

Temperature

Presence of catalyst

Force of collisions
4
Reaction Rate
Change in concentration (conc) of
a reactant or product per unit time.


A  at t 2  A  at t 1
Rate 
A 
Rate 
t
t 2  t1
(where A is a reactant or product)
(note that rate is always a + value)
Unit typically M/s
5
Progress of a hypothetical rxn
6
Sample Exercise 1

Using the data given in slide 6, calculate the
average rate at which A disappears over the time
interval 20 s to 40 s.
7
2 NO2(g)  2 NO(g) + O2(g)
Time(s)
0
50
100
150
200
250
300
350
400
[NO2]
0.0100
0.0079
0.0065
0.0055
0.0048
0.0043
0.0038
0.0034
0.0031
[NO]
0
0.0021
0.0035
0.0045
0.0052
0.0057
0.0062
0.0066
0.0069
[O2]
0
0.0011
0.0018
0.0023
0.0026
0.0029
0.0031
0.0033
0.0035
8
Find the avg rate the first 50 s
of NO2 change.
We want to work with positive numbers so the equation will be:
9
Concentration of nitrogen dioxide, nitric oxide, and
oxygen versus time.
10
What do we discover about the
rate of this reaction by looking at
the graph on slide 10?


Instantaneous rate – value of the rate at
particular time; found by calculating the
slope of a line tangent to the curve at that
point
At 100 s:
NO2 
0.0026 M
5
Rate  
 2.4 10 M/s
t
110 s
11
Reaction Rates
C4H9Cl(aq) + H2O(l)  C4H9OH(aq) + HCl(aq)


All reactions slow down
over time.
Therefore, the best
indicator of the rate of a
reaction is the
instantaneous rate near
the beginning of the
reaction. (called initial
rate)
12
Using the graph on slide 12, calculate the
instantaneous rate of disappearance of
reactant at t = 0.
13
Rate Laws
Rate = k[A]m [B]n
k = rate constant, dependent
on temperature
m & n = reaction orders
14
Rate Laws
The rate law for any reaction must
be determined experimentally, it
cannot be predicted by merely
looking at the chemical equation.
*usually
involves only the concentration
of reactants
15
Types of Rate Laws
Differential Rate Law: expresses how rate
depends on concentration. (Often just called
“rate law”)

Integrated Rate Law: expresses how
concentration depends on time.

They are inter-related, so once you
experimentally determine one, the other can
be found.

16
WHY???!!!!

We can work backward from the rate
law to infer the steps by which the
reaction occurs which helps us to find
the slowest step – then we can figure
out how to speed it up!
17
Method of Initial Rates
Initial Rate: the “instantaneous
rate” just after the reaction begins
(just after t = 0)

The initial rate is determined in
several experiments using different
initial concentrations.

18
Method of Initial Rates: Sample 1
NH4+(aq) + NO2−(aq)
N2(g) + 2 H2O(l)
Rate = k [NH4+]m[NO2−]n
19
Method of Initial Rates: Sample 1
Determine the values of n and m.
Rate 2 10.8 x107 mols.L.s k(0.0200 mol/L) m (0.200 mol/L) n


Rate 1 5.4 x 10 -7 mols/L.s k(0.0100 mol/L) m (0.200 mol/L) n
(0.0200 mol/L.s)

(0.01o0 mol/L.s)
Rate 2
Rate 1
m
m
 (2.0) m
 2.00  ( 2.0)
The value of m is 1
m
20
Method of Initial Rates: Sample 1
A similar method is used to find n.
Rate 5 21.6 x10 7 mol/L.s
k(0.200M)( 0.0404 M) n


-7
n
Rate 4 10.8 x10 mol/L.s k (0.200M)(0.0202 M)
 2.00  (
)  (2.00)
.200 n
.100
n
The value of n is also 1.
21
Method of Initial Rates: Sample 1
The values of n and m are both 1 and the rate law is:

4
Rate  k[NH ][ NO ]
2
This rate law is first order in both NO2- and NH4+ .
The overall reaction order is the sum of n and m.
The reaction is second order overall.
22
• For the reaction
NH4+(aq) + NO2-(aq)  N2(g) + 2H2O(l)
we can observe that
– as [NH4+] doubles with [NO2-] constant the rate doubles,
– as [NO2-] doubles with [NH4+] constant, the rate doubles,
– We conclude rate  [NH4+][NO2-].
• Rate law:

4

2
Rate  k[ NH ][ NO ]
• The constant k is the rate constant.
23
Calculate the rate constant k

4
Rate  k[NH ][ NO ]
5.4 x 10
-7
2
M/s  k(0.0100M)(0.200M)
Then
k
-7
5.4 x 10 M/s
(0.0100 M)(0.200 M)
4
 2.7 x10 l/Ms
24
Using Initial Rates to
Determine Rate Laws




A reaction is zero order in a reactant if the
change in concentration of that reactant
produces no effect.
A reaction is first order if doubling the
concentration causes the rate to double.
A reaction is nth order if doubling the
concentration causes an 2n increase in rate.
Note that the rate constant does not depend
on concentration.
25
Overall Reaction Order
Sum of the order of each component in the
rate law.



rate = k[H2SeO3][H+]2[I]3
The overall reaction order is 1 + 2 + 3 = 6.
26
Kinetic Sample Problem 1
2 H2 + 2 NO  2H2O + N2 at 800 K
Experiment
[H2]
[NO]
Initial rate in M/min
1
.001
.006
.025
2
.002
.006
.050
3
.003
.006
.075
4
.009
.001
.0063
5
.009
.002
.025
6
.009
.003
.056
27
Complete Sample Problems 24
2. Rate  kNO H 2 
2
k  2.2 1/atm min
2
3. Rate  kA  C
2
k  3.0 10 1/minM
-4
2
4. Rate  kA  B
3
k  6 10 1/minM
5
3
28
Integrated Rate Laws
A 
Rate  
 k A 
t
Using calculus to integrate the rate law for a
first-order process gives us
Where
[A]t
ln
= −kt
[A]0
[A]0 is the initial concentration of A, and
[A]t is the concentration of A at some time, t,
during the course of the reaction.
29
Integrated Rate Laws
Manipulating this equation produces…
[A]t
ln
= −kt
[A]0
ln [A]t − ln [A]0 =
ln [A]t =
− kt
− kt
On purple equation sheet
+ ln [A]0
…which is in the form
y
= mx + b
30
First-Order Processes
ln [A]t = -kt + ln [A]0
Therefore, if a reaction is first-order, a
plot of ln [A] vs. t will yield a straight
line, and the slope of the line will be -k.
31
First-Order Processes
Consider the process in
which methyl isonitrile is
converted to acetonitrile.
CH3NC
CH3CN
32
First-Order Processes
CH3NC
CH3CN
This data was
collected for this
reaction at 198.9 °C.
33
First-Order Processes


When ln P is plotted as a function of time, a
straight line results.
Therefore,


The process is first-order.
k is the negative of the slope: 5.1  10-5 s−1.
34
If you know the initial concentration and
k, the concentration at any time can be
calculated.

Example: The decomposition of a certain
insecticide in water follows first-order kinetics
with a rate constant of 1.45 yr-1. A quantity
of this insecticide is washed into a lake on
June 1, leading to a concentration of 5.0 x 107 g/cm3.
(a) What is the concentration of the
insecticide on June 1 of the following year?
(b) How long will it take for the concentration
of the insecticide to decrease to 3.0 x 10-7
g/cm3?
35
Complete sample problems on
wkst.
36
Second-Order Processes
Similarly, integrating the rate law for a
process that is second-order in reactant A,
we get
1
1

 kt
At A0
Rearranged to give
also in the form
On purple equation
sheet
1
1
= kt +
[A]t
[A]0
y = mx + b
37
Second-Order Processes
1
1
= kt +
[A]t
[A]0
So if a process is second-order in A, a
1
plot of [A] vs. t will yield a straight line,
and the slope of that line is k.
38
Second-Order Processes
The decomposition of NO2 at 300°C is described by
the equation
NO2 (g)
NO (g) +
1
2
O2 (g)
and yields data comparable to this:
Time (s)
[NO2], M
0.0
0.01000
50.0
0.00787
100.0
0.00649
200.0
0.00481
300.0
0.00380
39
Second-Order Processes
• Plotting ln [NO2] vs. t yields the graph below.
• The plot is not a straight line, so the process is not
first-order in [A].
Time (s)
[NO2], M
ln [NO2]
0.0
0.01000
−4.610
50.0
0.00787
−4.845
100.0
0.00649
−5.038
200.0
0.00481
−5.337
300.0
0.00380
−5.573
40
Second-Order Processes
1
• Graphing [NO
vs. t, however, gives this
]
2
plot.
• Because this is a straight line, the process is
second-order in [A].
Time (s)
[NO2], M
1/[NO2]
0.0
0.01000
100
50.0
0.00787
127
100.0
0.00649
154
200.0
0.00481
208
300.0
0.00380
263
41
Half-Life


Half-life is defined as the
time required for one-half
of a reactant to react.
Because [A] at t1/2 is onehalf of the original [A],
[A]t = 0.5 [A]0.
42
Half-Life
For a first-order process, this becomes
0.5 [A]0
ln
=
−kt
1/2
[A]0
ln 0.5 = −kt1/2
−0.693 = −kt1/2
NOTE: For a first-order
process, then, the half-life
does not depend on [A]0.
0.693
= t1/2
k
43
Half-Life
For a second-order process,
1
1
= kt1/2 +
0.5 [A]0
[A]0
2
1
= kt1/2 +
[A]0
[A]0
2 − 1 = 1 = kt
1/2
[A]
[A]0
0
1
= t1/2
k[A]0
NOTE: For a second
order reaction, The
half-life is dependent
upon the initial
concentration.
44
Testing for a Rate Law
Plot [A] vs t.
Plot ln[A] vs t.
Plot 1/[A] vs t.
45
Summary of the Kinetics for Reactions
that are Zero, First, and Second Order in
[A]
Order
Zero
First
Second
Rate law:
Rate = k
Rate = k[A]
Rate = k[A]2
Integrated rate
law:
[A] = -kt +
[A]0
ln[A] = -kt +
ln[A]0
1/[A] = -kt +
1/[A]0
Plot needed to
give a straight
line:
[A] vs. t
ln[A] vs. t
1/[A] vs. t
Relationship of
rate constant to
the slope of
straight line:
Slope = -k
Slope = -k
Slope = k
Half-life:
t1/2 =
[A]0/2k
t1/2 = 0.693/k
t1/2 = 1/k[A]0
46
Temperature and Rate


Most reactions speed up as temperature
increases. (E.g. food spoils when not
refrigerated.)
Two light sticks are placed in water, one at room
temp. and one in ice




the one at room temp. is brighter than the one in ice.
The chemical reaction, responsible for
chemiluminescence is dependent on temp.
The higher the temp., the faster the reaction and
the brighter the light.
As temperature increases, the rate increases.
47
Collision Model

Key Idea: Molecules must collide to react.
However, only a small fraction of collisions
produces a reaction. Why?

Arrhenius: An activation energy must be
overcome.

48
The Collision Model


The more molecules present, the greater
the probability of collision and the faster
the rate.
The higher the temperature, the more
energy avail. to the molecules and the
faster the rate.
49
The Collision Model


Complication: not all collisions lead to
products. In fact, only a small fraction
of collisions lead to product.
In order for reaction to occur the
reactant molec. must collide in the
correct orientation and with enough
energy to form products.
50
Arrhenius Equation

Arrhenius: molecules must possess a
minimum amount of energy to react.
Why?



In order to form products, bonds must
be broken in the reactants.
Bond breakage requires energy.
Activation energy, Ea, is the minimum
energy required to initiate a chemical
reaction.
51
Arrhenius Equation
52
Arrhenius Equation
Consider the rearrangement of acetonitrile:



In H3C-NC, the C-NC bond bends until the C-N bond
breaks and the NC portion is perpendicular to the H3C
portion. This structure is called the activated complex
or transition state.
The energy required for the above twist and break is
the activation energy, Ea.
Once the C-N bond is broken, the NC portion can
continue to rotate forming a C-CN bond.
53
Arrhenius Equation
54
Arrhenius Equation
(continued)
k = Ae-Ea/RT
ln k =
-Ea 1
R





k = rate constant
A = frequency factor
favorable collision)
+ ln A
T
(a measure of the probability of a
Ea = activation energy
T = temperature
R = gas constant (8.31 J/molK)
Both A and Ea are specific to a given reaction
55
Catalysis
Catalyst: A substance that speeds up a
reaction without being consumed

Enzyme: A large molecule (usually a
protein) that catalyzes biological
reactions.

Homogeneous catalyst: Present in the
same phase as the reacting molecules.

Heterogeneous catalyst: Present in a
different phase than the reacting
molecules.

56
Catalysis
57
Enzymes
•Enzymes are biological catalysts.
•Most enzymes are protein molecules with
large molecular masses (10,000 to 106 amu).
•Enzymes have very specific shapes.
•Most enzymes catalyze very specific
reactions.
•Substrates undergo reaction at the active site
of an enzyme.
58